, with the same "Reed" voice. This book is different from the four arithmetics: it is theorems and proofs, not drill. So each lesson is a short read-along of the book itself, with its figures, followed by a quiz and, where the section measures something, number problems. This post covers what is new for Book 5 and how to rebuild it. The app, the server and the voice recorder are explained in the Book One post, the text-recognition script in the Book 2 post, and the family accounts in the family accounts post. Every block has a Copy button.

What is in Book 5

Each section of the book's table of contents became a lesson: 64 lessons, 863 parts of the book to study, 261 figures, 810 quiz questions and 273 number answers.

Lessons 1-10, Introductory: logical terms, axioms, ratio and proportion, definitions, the postulates, lines and surfaces, problems with straight lines, broken lines, angles.
Lessons 11-35, Plane Geometry: perpendicular, oblique and parallel lines; circumferences (arcs, tangents, secants, chords, angles at the center, intercepted arcs, two circumferences); triangles (equality, similarity, the Pythagorean theorem); quadrilaterals and measure of area; polygons, regular polygons and isoperimetry; circles, the value of pi and the area of the circle.
Lessons 36-53, Geometry of Space: lines and planes in space, diedrals, parallel planes, triedrals, polyedrals; tetraedrons, pyramids, prisms, measure of volume, similar and regular polyedrons; cones, cylinders, spheres, spherical angles and triangles, spherical areas and volumes, mensuration. (The book's long "Spheres" section is split in two lessons.)
Lessons 54-64, Trigonometry: measure of angles, the six functions, signs and limits, the relations between the functions, the tables, right angled triangles, solving plane triangles and measuring heights and distances, spherical trigonometry (Napier's analogies and rule of circular parts), and common logarithms.

New in the app for Book 5

Study step. Reed reads each numbered article of the book (theorem, corollary, definition) while the book's own figure is shown beside it (above it on a phone held upright). Tap any article's number to hear it again, or Read to me to hear the rest of the lesson. Figure letters are read one by one ("A O E"), and the book's cross-references like "(103)" are skipped when reading.
Quiz step. Pick-the-answer and true-or-false questions written from that section. A wrong pick shakes and greys out; two misses (one for true or false) show and say the answer, and the question comes back at the end of the quiz.
Number problems reuse the problems step of Books 1-4: areas, volumes, angles of polygons, chords and tangents, trigonometric values, logarithms.

The book has no answer key, so the quiz questions and number problems were written for the app from the book's own articles, and every answer was checked by hand.

How the book was typed up

The free PDF (raysarithmetic.wordpress.com, "9-10 Ray's New Geometry and Trigonometry", 420 pages, about 19 MB) has no text layer. All 420 pages were read with Windows' text recognition (ocr-pdf.ps1 from the Book 2 post), which reads the words well but loses most formulas and many letters on figures. Book page = PDF page. The text recognition takes a while; start it and do something else.

The book source, geometry.txt, has four new kinds of lines:

art: 99 | Corollary | Vertical angles are equal. Thus, a and i are each supplements of e.
fig: 37 0.15 0.25 0.56 0.76
mc: Two angles whose sum is one right angle are... | *complements | supplements | vertical angles
tf: All right angles are equal. {T}

art: number | heading | text is one article of the book, read in the Study step.
fig: page top bottom [left right] cuts the figure for the article above it out of the scan; the numbers are fractions of the page (0 is the top or left edge, 1 the bottom or right). An article can have several.
mc: is a quiz question; the right answer has a star, and the choices are shuffled in the app.
tf: is a true-or-false question.
q: and sol: are number problems, as in Books 1-4.

The figures are not kept as files in this post: make-figs.ps1 cuts all 261 of them again from the PDF using the fig: lines, trims the white edges, and saves them in public\fig\geometry\. It needs pdftoppm (Poppler), the same tool ocr-pdf.ps1 uses.

Placing a figure box takes two helper scripts. fig-sheet.ps1 makes a sheet of page pictures with red marks every twentieth of the page, so the box can be read off. fig-montage.ps1 shows all the cut figures with their names, to see at a glance which ones caught a line of text or cut off a letter.

Things found on the way:

Text creeps into figures. The book wraps text beside most figures, so the left edge of a box has to stop where the text ends, and a box's top or bottom often catches half a line. Every figure was checked on a montage and many were nudged two or three times.
No minus sign on the number pad. "The cosine of 180 degrees is minus 1" had to be a pick-the-answer question.
Bracketed numbers are not spoken. Because "(103)" is dropped when reading, a question that said "inside (1), on (2), or outside (3)" would have lost its numbers; it now says "Answer 1 for inside, 2 for on, 3 for outside".
Answers that differ only in capitals. The voice list has keys like "any two" and "Any two". The browser handles them fine, but PowerShell's ConvertFrom-Json refuses the file, so check it in the browser, not with PowerShell.
The offline cache serves the old app. After changing app.js, bump rays-shell-v16 in sw.js, or the old version keeps showing.

The voice

The same recorder as Books 1-4: 3,653 lines, 3,350 of them new at each speed. It took about 3 minutes per speed on the PC.

Files

Set up the Book One post first, and download the Book 5 PDF from raysarithmetic.wordpress.com (Ray's Free Arithmetic page, "Ray's New Geometry and Trigonometry (9-10)") as C:\RaysArithmetic\source\pdf\rays-geometry-and-trigonometry-9-10.pdf. Book 5 adds or replaces the files below. Each block writes one file into /root/rays-rebuild/ when pasted into a Proxmox host shell (Proxmox web UI, pve, Shell). Then copy that folder to the PC and put:

geometry.txt in C:\RaysArithmetic\source\
make-book.ps1, make-audio.ps1, make-figs.ps1, fig-sheet.ps1, fig-montage.ps1 in C:\RaysArithmetic\tools\
app.js, index.html, prompts.json, sw.js in C:\RaysArithmetic\public\ (they replace the earlier versions and run all five books)

Proxmox host shell:

mkdir -p /root/rays-rebuild

geometry.txt: the whole of Book 5: articles, figure boxes, quizzes and problems.

cat > /root/rays-rebuild/geometry.txt <<'EOF'
@book: geometry | Ray's Geometry and Trigonometry | Elements of Geometry - Book Five | #7a6a1f
# Ray's New Geometry and Trigonometry (Eli T. Tappan), typed from source/pdf/rays-geometry-and-trigonometry-9-10.pdf
# (book page N = PDF page N; no text layer: OCR in rays-geometry-and-trigonometry-9-10.ocr.txt, checked against the
# page images). Each lesson is one section of the book's Table of Contents.
# Format (Book Five):
#   art: <article no.> | <Theorem/Corollary/... or empty> | <the book's text>   -> Study step (read in the Reed voice)
#   fig: <pdf page> <top> <bottom> [<left> <right>]   figure for the art above, fractions of the page (tools/make-figs.ps1)
#   mc: <question> | *right answer | wrong | wrong    -> Quiz step (choices shuffled)
#   tf: <statement> {T|F}                             -> Quiz step, True / False
#   q: / sol: as in the other books (number answers) -> Problems step
# The quiz questions are written for this app from the book's own articles; the book has no answer key.

== 1 | Preliminary | Logical terms
art: 1 |  | Before the student begins the study of geometry, he should know certain principles and definitions, which are of use, though they are not peculiar to this science. They are very briefly presented in this chapter.
art: 2 |  | Every statement of a principle is called a proposition. Every proposition contains the subject of which the assertion is made, and the property or circumstance asserted. When the subject has some condition attached to it, the proposition is said to be conditional. The subject, with its condition, if it have any, is the hypothesis of the proposition, and the thing asserted is the conclusion. Each of two propositions is the converse of the other, when the two are such that the hypothesis of either is the conclusion of the other.
art: 3 |  | A proposition is either theoretical, that is, it declares that a certain property belongs to a certain thing; or it is practical, that is, it declares that something can be done. Propositions are either demonstrable, that is, they may be established by the aid of reason; or they are indemonstrable, that is, so simple and evident that they can not be made more so by any course of reasoning. A theorem is a demonstrable, theoretical proposition. A problem is a demonstrable, practical proposition. An axiom is an indemonstrable, theoretical proposition. A postulate is an indemonstrable, practical proposition. A proposition which flows, without additional reasoning, from previous principles, is called a corollary.
art: 4 |  | The reasoning by which a proposition is proved is called the demonstration. The explanation how a thing is done constitutes the solution of a problem. A direct demonstration proceeds from the premises by a regular deduction. An indirect demonstration attains its object by showing that any other hypothesis than the one advanced would involve a contradiction, or lead to an impossible conclusion. Such a conclusion may be called absurd, and hence the Latin name of this method of reasoning, reductio ad absurdum.
art: 5 | Remark | The student should learn each proposition, so as to state separately the hypothesis and the conclusion, also the condition, if any. He should also learn, at each demonstration, whether it is direct or indirect. It is a good exercise to state the converse of a proposition.
mc: Every statement of a principle is called a... | *proposition | demonstration | figure
mc: The subject of a proposition, with its condition, is called the... | *hypothesis | conclusion | corollary
mc: The thing asserted in a proposition is called the... | *conclusion | hypothesis | axiom
mc: When the hypothesis of each of two propositions is the conclusion of the other, each is the ___ of the other. | *converse | corollary | postulate
mc: A demonstrable, theoretical proposition is a... | *theorem | problem | axiom | postulate
mc: A demonstrable, practical proposition is a... | *problem | theorem | axiom | postulate
mc: An indemonstrable, theoretical proposition is an... | *axiom | theorem | problem | postulate
mc: An indemonstrable, practical proposition is a... | *postulate | axiom | theorem | problem
mc: A proposition which flows, without more reasoning, from earlier principles is a... | *corollary | converse | postulate
mc: The reasoning by which a proposition is proved is called the... | *demonstration | solution | hypothesis
mc: Showing that any other supposition leads to something impossible is called... | *an indirect demonstration | a direct demonstration | a postulate
tf: An axiom is so simple and evident that reasoning can not make it more so. {T}
tf: A practical proposition declares that something can be done. {T}
tf: A theorem is a practical proposition. {F}

== 2 | Preliminary | General axioms
art: 6 | Axiom | Quantities which are each equal to the same quantity, are equal to each other.
art: 7 | Axiom | If the same operation be performed upon equal quantities, the results will be equal. For example, if the same quantity be separately added to two equal quantities, the sums will be equal.
art: 8 | Axiom | If the same operation be performed upon unequal quantities, the results will be unequal. Thus, if the same quantity be subtracted from two unequal quantities, the remainder of the greater will exceed the remainder of the less.
art: 9 | Axiom | The whole is equal to the sum of all the parts.
art: 10 | Axiom | The whole is greater than a part.
art: 11 | Exercise | What is the hypothesis of the first axiom? What is the subject? What is the condition? What is the conclusion? Give an example of this axiom.
mc: If a equals c and b equals c, then a and b are... | *equal to each other | unequal | not known
mc: Add 5 to two equal quantities. The two sums are... | *equal | unequal | not known
mc: Take 3 from two unequal quantities. Then the remainder of the greater is... | *greater than the remainder of the less | equal to the remainder of the less | less than the remainder of the less
mc: The whole is equal to... | *the sum of all the parts | any one of its parts | the greatest part
mc: In the first axiom, "quantities which are each equal to the same quantity" is the... | *hypothesis | conclusion | converse
mc: In the first axiom, "are equal to each other" is the... | *conclusion | hypothesis | condition
tf: The whole is greater than a part. {T}
tf: If the same quantity is doubled two times, the results can be unequal. {F}

== 3 | Preliminary | Ratio and proportion
art: 12 |  | All mathematical investigations are conducted by comparing quantities, for we can form no conception of any quantity except by comparison.
art: 13 |  | In the comparison of one quantity with another, the relation may be noted in two ways: either, first, how much one exceeds the other; or, second, how many times one contains the other. The result of the first method is the difference between the two quantities; the result of the second is the ratio of one to the other. Every ratio, as it expresses how many times one quantity contains another, is a number.
art: 14 |  | A ratio is usually written a : b, and is read, a is to b. When the ratio is expressed in the form of a fraction, the first term, or antecedent, becomes the numerator, and the second, or consequent, the denominator.
art: 15 |  | A proportion is the equality of two ratios, and is generally written a : b :: c : d, and is read, a is to b as c is to d. The first and last terms are the extremes, and the second and third are the means of a proportion. The fourth term is called the fourth proportional of the other three. When a series of three quantities has the same ratio from each to the next, the second is the mean proportional of the other two.
art: 16 | Proposition | The product of the extremes of any proportion is equal to the product of the means. For any proportion is the equation of two fractions; multiplying these equals by the product of the denominators, we have a times d equal to b times c.
art: 17 | Corollary | The square of a mean proportional is equal to the product of the extremes. A mean proportional of two quantities is the square root of their product.
art: 18 | Proposition | When the product of two quantities is equal to the product of two others, either two may be the extremes and the other two the means of a proportion.
art: 19 | Proposition | The order of the terms may be changed without destroying the proportion, so long as the extremes remain extremes, or both become means.
art: 20 |  | When we say the first term is to the third as the second is to the fourth, the proportion is taken by alternation. When we say the second term is to the first as the fourth is to the third, the proportion is taken inversely.
art: 21 | Proposition | Ratios which are equal to the same ratio are equal to each other.
art: 22 | Proposition | If two quantities have the same multiplier, the multiples will have the same ratio as the given quantities.
art: 23 | Proposition | In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. This is called proportion by composition.
art: 24 | Proposition | The difference between the first and second terms of a proportion is to the second, as the difference between the third and fourth is to the fourth. This is called proportion by division.
art: 25 | Proposition | If four quantities are in proportion, their same powers are in proportion, also their same roots.
mc: How many times one quantity contains another is their... | *ratio | difference | proportion
mc: A proportion is... | *the equality of two ratios | the difference of two ratios | the product of two numbers
mc: In a : b :: c : d, the extremes are... | *a and d | b and c | a and b
mc: In a : b :: c : d, the means are... | *b and c | a and d | c and d
mc: In a ratio a : b, the first term a is the... | *antecedent | consequent | mean
mc: In any proportion, the product of the extremes equals... | *the product of the means | the sum of the means | the ratio of the means
mc: The square of a mean proportional is equal to... | *the product of the extremes | the sum of the extremes | the difference of the extremes
mc: Saying "the second term is to the first as the fourth is to the third" takes the proportion... | *inversely | by alternation | by composition
mc: Saying "the first term is to the third as the second is to the fourth" takes the proportion... | *by alternation | inversely | by division
tf: Every ratio is a number. {T}
tf: If 2 : 3 :: 4 : 6, then 2 : 4 :: 3 : 6. {T}
tf: If two quantities are both multiplied by 5, their ratio changes. {F}
q: In the proportion 2 : 3 :: 4 : x, find the fourth proportional x. {6}
sol: The product of the extremes equals the product of the means: 2 times x = 3 times 4 = 12, so x = 6.
q: In the proportion 3 : 5 :: 9 : x, find x. {15}
sol: 3 times x = 5 times 9 = 45, so x = 15.
q: Find the mean proportional of 4 and 9. {6}
sol: The mean proportional is the square root of the product: 4 times 9 = 36, and the square root of 36 is 6.
q: Find the mean proportional of 2 and 8. {4}
sol: 2 times 8 = 16, and the square root of 16 is 4.
q: In the proportion x : 4 :: 6 : 8, find x. {3}
sol: x times 8 = 4 times 6 = 24, so x = 3.

== 4 | The subject stated | Definitions
art: 26 |  | We know that every material object occupies a portion of space, and has extent and form. For example, this book occupies a certain space; it has a definite extent, and an exact form.
art: 27 |  | Such a limited portion of space is called a solid. Be careful to distinguish the geometrical solid, which is a portion of space, from the solid body which occupies space.
art: 28 |  | The limit or boundary which separates a solid from the surrounding space is a surface. A surface is like a solid in having only these two properties, extent and form; but a surface differs from a solid in having no thickness or depth.
art: 29 |  | The limits or boundaries of a surface are lines. The intersection of two surfaces is a line. A line has neither thickness nor breadth.
art: 30 |  | The ends or limits of a line are points. The intersections of lines are also points. A point has neither extent nor form.
art: 32 | Definitions | A point has only position, without extent. A line has length, without breadth or thickness. A surface has length and breadth, without thickness. A solid has length, breadth, and thickness.
art: 33 |  | A line has only one dimension. A surface has two, and a solid has three dimensions. The extent of a line is called its length; of a surface, its area; and of a solid, its volume.
art: 34 |  | Whatever has only extent and form is called a magnitude. Geometry is the science of magnitude. It establishes the principles upon which all measurements are made. It is the basis of Surveying, Navigation, and Astronomy.
mc: A limited portion of space is called a... | *solid | surface | line
mc: The boundary which separates a solid from the space around it is a... | *surface | line | point
mc: The boundaries of a surface are... | *lines | points | solids
mc: The ends of a line are... | *points | surfaces | lines
mc: What has only position, without extent? | *A point | A line | A surface
mc: What has length, without breadth or thickness? | *A line | A point | A surface
mc: What has length and breadth, without thickness? | *A surface | A line | A solid
mc: How many dimensions has a solid? | *Three | Two | One
mc: How many dimensions has a surface? | *Two | One | Three
mc: The extent of a surface is called its... | *area | length | volume
mc: The extent of a solid is called its... | *volume | area | length
mc: Geometry is the science of... | *magnitude | number | motion
tf: The place where two surfaces cut each other is a line. {T}
tf: A point has extent but no form. {F}

== 5 | The subject stated | Postulates of extent and form
art: 35 | Postulate of Extent | Magnitudes may have any extent. We may conceive lines, surfaces, or solids, which do not extend beyond the smallest spot, or of such extent as to reach across the universe. A magnitude may be made to have any extent whatever.
art: 36 | Postulate of Form | Every geometrical conception is composed of only two kinds of elementary thoughts: directions and distances. The directions determine its form, and the distances its extent. The points of a magnitude may be made to have from each other any directions whatever, thus giving the magnitude any conceivable form.
art: 37 |  | Magnitudes which have the same form while they differ in extent, are called similar. Any point, line, or surface in a figure, and the similarly situated point, line, or surface in a similar figure, are called homologous. Magnitudes which have the same extent, while they differ in form, are called equivalent.
art: 38 | Motion | Since the parts of a magnitude may have any position, they may change position. The path of a point is a line, the path of a line may be a surface, and the path of a surface may be a solid.
art: 39 | Superposition | By the idea of motion, one magnitude may be mentally applied to another, and their form and extent compared. This is called the method of superposition, and is the most simple and useful of all the methods of demonstration used in geometry.
art: 40 | Equality | When two equal magnitudes are compared, it is found that they may coincide; that is, each contains the other. Conversely, if two magnitudes are composed of parts respectively equal and similarly arranged, one may be applied to the other till the wholes coincide, showing the two magnitudes to be equal.
mc: How many postulates of geometry are there? | *Two | Four | Ten
mc: The two kinds of thoughts in every geometrical figure are... | *directions and distances | numbers and letters | lines and colors
mc: The directions of a figure determine its... | *form | extent | color
mc: The distances of a figure determine its... | *extent | form | name
mc: Magnitudes with the same form but different extent are... | *similar | equivalent | equal
mc: Magnitudes with the same extent but different form are... | *equivalent | similar | homologous
mc: Similarly situated parts of two similar figures are called... | *homologous | equivalent | vertical
mc: The path of a moving point is a... | *line | surface | solid
mc: Laying one figure on another to compare them is the method of... | *superposition | division | composition
tf: Two equal magnitudes can be made to coincide. {T}
tf: A magnitude can be made as large as we please. {T}

== 6 | The subject stated | Classification of lines
art: 41 |  | Any magnitude or combination of magnitudes which can be accurately described, is called a geometrical figure.
art: 42 |  | A small spot is commonly called a point, and a long mark a line. But these have color, weight, and other properties; and, therefore, they are not geometrical points and lines. The figure which is the subject of a proposition is said to be given. The additions to the figure made for the purpose of demonstration constitute the construction.
art: 43 |  | In the diagrams in this work, points are designated by capital letters. Thus, the points A and B are at the extremities of the line. Figures are usually designated by naming some of their points, as the line AB, and the figure CDEF. When it is more convenient to designate a figure by a single letter, the small letters are used. Thus, the line a, the figure b.
fig: 22 0.52 0.78 0.585 0.82
art: 44 |  | A straight line is one which has the same direction throughout its whole extent. It may be regarded as the path of a point moving in one direction, turning neither up nor down, to the right or left.
art: 45 |  | A curved line is one which constantly changes its direction. The word curve is used for a curved line.
art: 46 |  | A line composed of straight lines, is called broken. A line may be composed of curves, or of both curved and straight parts.
fig: 23 0.24 0.35 0.55 0.85
art: 47 | Problem | A straight line may be made to pass through any two points.
art: 48 | Problem | There may be a straight line from any point, in any direction, and of any extent.
art: 49 | Axiom of Direction | In one direction from a point, there can be only one straight line.
art: 50 | Corollary | From one point to another, there can be only one straight line.
art: 51 | Theorem | If a straight line have two of its points common with another straight line, the two lines must coincide throughout their mutual extent. For, if they could separate, there would be from the point of separation two straight lines having the same direction, which is impossible.
art: 52 | Corollary | Two fixed points, or one point and a certain direction, determine the position of a straight line.
art: 53 |  | If a straight line were turned upon two of its points as fixed pivots, no part of the line would change place. If the curve BC were to revolve upon the points B and C, the straight line connecting these points would remain at rest. A straight line about which any thing revolves, is called its axis.
fig: 24 0.297 0.355 0.47 0.78
art: 54 | Axiom of Distance | The straight line is the shortest which can join two points. Therefore, the distance from one point to another is reckoned along a straight line.
art: 55 |  | There have now been given two postulates and two axioms. The science of geometry rests upon these four simple truths.
mc: A line which has the same direction throughout is a... | *straight line | curved line | broken line
mc: A line which constantly changes its direction is a... | *curve | straight line | broken line
mc: A line made of straight lines is called... | *broken | curved | similar
mc: In one direction from a point, how many straight lines can there be? | *Only one | Two | Any number
mc: What determines the position of a straight line? | *Two fixed points | One point | Its length
mc: The shortest line which can join two points is... | *the straight line | a curve | a broken line
mc: A straight line about which a figure revolves is its... | *axis | vertex | secant
mc: In this book, points in the diagrams are marked by... | *capital letters | small letters | numbers
mc: The science of geometry rests upon two postulates and two... | *axioms | theorems | corollaries
tf: Two straight lines which have two points in common must coincide. {T}
tf: There can be two different straight lines from one point to another. {F}
tf: The chalk line on a blackboard is a true geometrical line. {F}

== 7 | The subject stated | Classification of surfaces
art: 56 |  | Surfaces, like lines, are classified according to their uniformity or change of direction. A plane is a surface which never varies in direction. A curved surface is one in which there is a change of direction at every point.
art: 57 |  | The plane surface and the straight line have the same essential character, sameness of direction. The plane is straight in every direction that it has. A straight line and a plane, unless the extent be specified, are always understood to be of indefinite extent.
art: 58 | Theorem | A straight line which has two points in a plane, lies wholly in it, so far as they both extend.
art: 59 | Theorem | Two planes having three points common, and not in the same straight line, coincide so far as they both extend. Let A, B, and C be three points common to two planes m and p, and let straight lines join them. Each of these lines lies wholly in each plane. Any point D in the plane m may be joined to a point E of the line AC; the line DE, produced, meets AB or BC at F. Then F and E are both in the plane p, so the whole line FD, with the point D, is in the plane p. Therefore the two planes coincide.
fig: 25 0.48 0.62 0.52 0.8
art: 60 | Corollary | Three points not in a straight line, or a straight line and a point out of it, fix the position of a plane.
art: 61 |  | A part of a plane on one side of any straight line in it, may revolve about the line till it meets the other part, when the two will coincide.
art: 62 | Exercises | To try if a surface is plane, the mechanic applies a straight rule to it in many directions, observing whether the two touch throughout. What geometrical principle is used? In a diagram two letters suffice to mark a straight line. Why? But it may require three letters to designate a curve. Why?
art: 63 | Division of the subject | Elementary Geometry is divided into Plane Geometry, which treats of plane figures, and Geometry in Space, which treats of figures whose points are not all in one plane.
mc: A surface which never varies in direction is a... | *plane | curved surface | solid
mc: A straight line with two points in a plane lies... | *wholly in the plane | partly out of the plane | across the plane
mc: How many points, not in one straight line, fix the position of a plane? | *Three | Two | Four
mc: Plane geometry treats of... | *plane figures | figures not all in one plane | numbers only
mc: Why do two letters suffice to mark a straight line? | *Two points determine a straight line | Every line has two ends | Lines are short
tf: A straight line and a point out of it fix the position of a plane. {T}
tf: The plane and the straight line both have sameness of direction. {T}
tf: Two planes with two points in common must coincide. {F}

== 8 | Straight lines | Problems
art: 64 | Problem | Straight lines may be added together, and one straight line may be subtracted from another. For a straight line may be produced to any extent, so its length may be increased by the length of another line. Again, one line may be subtracted from another, by applying the less to the greater and noting the difference.
art: 65 | Problem | A straight line may be multiplied by any number. For several equal lines may be added together.
art: 66 | Problem | A straight line may be divided by another. By repeating the process of subtraction.
art: 67 | Problem | A straight line may be decreased in any ratio, or it may be divided into several equal parts.
art: 68 | Problems in drawing | Exercises in linear drawing afford the best applications of the principles of geometry. Except the paper and pencil, the only instruments used are the ruler and compasses.
art: 70 |  | With the compasses, the length of a given line may be taken by opening the legs till the fine points are one on each end of the line. Then this length may be measured on the greater line as often as it will contain the less.
art: 73 | Theorem | Any two straight lines are similar figures. For each has one invariable direction. Hence, two straight lines have the same form, and can differ only in their extent.
art: 74 |  | Any straight line may be divided in any ratio. The points in two lines which divide them in the same ratio are homologous points. Thus, if the lines AB and ED are divided at the points C and F, so that AC : CB :: EF : FD, then C and F are homologous points, and AC and EF are homologous parts.
fig: 30 0.37 0.45 0.47 0.86
art: 75 | Corollary | Two homologous parts of two straight lines have the same ratio as the two whole lines.
art: 76 | Problem in drawing | To find the ratio of two given straight lines, apply each of them a number of times in succession along an indefinite straight line from a point A. If the ends coincide for the first time at E, then AE is the least common multiple of the two lines, and the numbers of applications express the ratio of the two lines.
fig: 31 0.15 0.2
mc: The only drawing instruments used are the ruler and the... | *compasses | protractor | square
mc: Any two straight lines are... | *similar | equal | equivalent
mc: Why are any two straight lines similar? | *Each has one invariable direction | They have the same length | They have two ends
mc: Points which divide two lines in the same ratio are... | *homologous | vertical | adjacent
mc: Two homologous parts of two straight lines have the same ratio as... | *the two whole lines | their difference | one right angle
tf: A straight line may be divided into several equal parts. {T}
tf: One straight line may be subtracted from another. {T}
q: A line 12 inches long is divided at C in the ratio 1 to 2. How long is the shorter part, in inches? {4}
sol: The parts are as 1 to 2, so the whole is 3 parts; one part is 12 / 3 = 4 inches.
q: A line AB of 10 inches is divided at C, and a line ED of 15 inches at F, so that the points are homologous. If AC is 4 inches, how long is EF, in inches? {6}
sol: Homologous parts have the ratio of the whole lines: AC : EF :: 10 : 15, so EF = 4 times 15 / 10 = 6.
q: A line b goes 5 times into a line AE, and a line c goes 3 times into the same line AE. What is the ratio of b to c? Give the first number, b : c :: ? : 5 {3}
sol: 5 times b equals 3 times c, so b : c :: 3 : 5.

== 9 | Straight lines | Broken lines
art: 78 |  | A curve or a broken line is said to be concave on the side toward the straight line which joins two of its points, and convex to the other side.
art: 79 | Theorem | A broken line which is convex toward another line that unites its extreme points, is shorter than that line. The line ABCD is shorter than the line AEGD, toward which it is convex. Produce AB and BC till they meet the outer line in F and H. Since CD is shorter than CHD, the line ABCD is shorter than ABHD. For a similar reason, ABHD is shorter than AFGD, and AFGD is shorter than AEGD. Therefore, ABCD is shorter than AEGD.
fig: 32 0.15 0.24 0.5 0.8
art: 80 | Exercise | Vary the above demonstration by producing the lines DC and CB to the left, instead of AB and BC to the right; also, by substituting a curve for the outer line.
art: 81 |  | A fine thread being tightly stretched, and thus forced to take the shortest path between its ends, is a good representation of a straight line. Hence, a stretched cord is used for marking straight lines. The word straight is derived from stretch.
mc: A curve is ___ on the side toward the straight line joining two of its points. | *concave | convex | vertical
mc: A curve is ___ on the side away from that straight line. | *convex | concave | oblique
mc: A broken line convex toward an outer line with the same ends is... | *shorter than the outer line | longer than the outer line | equal to the outer line
mc: Why is a stretched cord used for marking straight lines? | *It takes the shortest path between its ends | It is easy to see | It never breaks
mc: The word straight comes from the word... | *stretch | strong | street
tf: In the figure, ABCD is longer than AEGD. {F}
tf: The demonstration works the same if the outer line is curved. {T}

== 10 | Straight lines | Angles
art: 82 |  | An angle is the difference in direction of two lines which have a common point.
art: 83 | Theorem | The two lines which form an angle lie in one plane, and determine its position. For the plane may pass through the common point and another point in each line, making three in all.
art: 84 |  | Let the line AB be fixed, and the line AC revolve in a plane about the point A. The angle will increase from zero, when AC coincides with AB, till AC takes the direction exactly opposite that of AB. After a complete revolution, AC will again coincide with AB. The lines which form an angle are called the sides, and the common point is called the vertex. The angle depends upon the directions only, and not upon the length of the sides.
fig: 33 0.2 0.29 0.56 0.8
fig: 33 0.315 0.43 0.43 0.8
art: 85 |  | Three letters may be used to mark an angle, the one at the vertex being in the middle, as the angle BAC. When there can be no doubt, one letter may answer, as the angle C. Angles may also be marked with letters between the sides, as the angles a and b. Two angles are adjacent when they have the same vertex and one common side between them.
fig: 33 0.56 0.655 0.55 0.82
fig: 33 0.677 0.735 0.66 0.82
art: 86 |  | Every straight line has two directions, which are the opposite of each other. One line meeting another at some point other than the end makes two angles with it, as BDF and BDC. When two lines cut each other, four angles are formed. The opposite angles formed by two lines cutting each other are called vertical angles. A line which cuts another, or which cuts a figure, is called a secant.
fig: 34 0.256 0.335 0.5 0.8
art: 87 | Problem | One angle may be added to another. Let the angles ADB and BDC be adjacent and in the same plane. The angle ADC is plainly equal to the sum of the other two. An angle may be subtracted from a greater one, for the angle ADB is the difference between ADC and BDC.
fig: 34 0.64 0.755 0.55 0.82
art: 88 | Theorem | The sum of all the successive angles formed in a plane upon one side of a straight line, is an invariable quantity; that is, all such sums are equal to each other. For the line AE may be placed on CF, the point E on the point F. Then EB will fall on FD, and the sum of all the angles upon AB coincides with the sum of all those upon CD.
fig: 35 0.43 0.615 0.52 0.8
art: 89 |  | When one line meets another, making the adjacent angles equal, the angles are called right angles. One line is perpendicular to the other when the angle which they make is a right angle. Two lines are oblique to each other when they make an angle which is greater or less than a right angle.
fig: 35 0.71 0.808 0.585 0.8
art: 90 | Corollary | All right angles are equal. For each is half of the sum of the angles upon one side of a straight line.
art: 91 | Corollary | The sum of all the successive angles formed in a plane and upon one side of a straight line, is equal to two right angles.
art: 92 | Corollary | The sum of all the successive angles formed in a plane about a point, is equal to four right angles.
fig: 36 0.28 0.435 0.53 0.72
art: 93 | Corollary | When two lines cut each other, if one of the angles thus formed is a right angle, the other three must be right angles.
art: 94 |  | An angle less than a right angle is called acute. An angle greater than one right angle and less than two, is called obtuse. When the sum of two angles is equal to a right angle, each is the complement of the other. When the sum of two angles is equal to two right angles, each is the supplement of the other.
art: 95 | Corollary | Angles which are the complements of the same or of equal angles are equal.
art: 96 | Corollary | Angles which are the supplements of the same or of equal angles are equal.
art: 97 | Corollary | The supplement of an obtuse angle is acute.
art: 98 | Corollary | The greater an angle, the less is its supplement.
art: 99 | Corollary | Vertical angles are equal. Thus, a and i are each supplements of e.
fig: 37 0.15 0.25 0.56 0.76
art: 100 | Theorem | When the sum of several angles in a plane having their vertices at one point is equal to two right angles, the extreme sides form one straight line. If the sum of AGB, BGC, and so on, be equal to two right angles, then AGF is one straight line.
fig: 37 0.32 0.445 0.5 0.78
art: 101 | Exercises | Which is the greater angle, a or b, and why? What is the greatest number of points in which two straight lines may cut each other? In which three may cut each other? Four?
fig: 37 0.61 0.68 0.5 0.78
mc: An angle is the difference in ___ of two lines which have a common point. | *direction | length | position
mc: The common point of the two sides of an angle is the... | *vertex | axis | secant
mc: An angle depends upon... | *the directions of its sides only | the length of its sides | both length and direction
mc: In the angle BAC, which letter is at the vertex? | *A | B | C
mc: Two angles with the same vertex and one common side between them are... | *adjacent | vertical | right
mc: The opposite angles formed by two lines cutting each other are... | *vertical angles | adjacent angles | right angles
mc: A line which cuts another line is called a... | *secant | vertex | side
mc: When a line meets another making the adjacent angles equal, they are... | *right angles | acute angles | obtuse angles
mc: The sum of the angles on one side of a straight line is... | *two right angles | one right angle | four right angles
mc: The sum of all the angles about a point is... | *four right angles | two right angles | three right angles
mc: An angle less than a right angle is... | *acute | obtuse | straight
mc: An angle between one and two right angles is... | *obtuse | acute | right
mc: Two angles whose sum is one right angle are... | *complements | supplements | vertical angles
mc: Two angles whose sum is two right angles are... | *supplements | complements | adjacent
mc: The supplement of an obtuse angle is... | *acute | obtuse | a right angle
tf: All right angles are equal. {T}
tf: Vertical angles are equal. {T}
tf: The longer the sides, the greater the angle. {F}
tf: If one of the four angles made by two cutting lines is a right angle, all four are right angles. {T}
tf: The greater an angle, the greater its supplement. {F}
q: Two straight lines cut each other. One of the angles is 3/5 of a right angle. How many of the four angles are equal to it? {2}
sol: Its vertical angle is equal to it; the other two are its supplements. So 2 of the 4 angles are equal to it.
q: What is the greatest number of points in which two straight lines may cut each other? {1}
sol: Two straight lines can cut each other in only one point, for if they had two points in common they would coincide.
q: What is the greatest number of points in which three straight lines may cut each other? {3}
sol: Each pair of lines may cut in one point, and there are 3 pairs.
q: What is the greatest number of points in which four straight lines may cut each other? {6}
sol: Each pair of lines may cut in one point, and there are 6 pairs: 4 times 3 / 2 = 6.

== 11 | Straight lines | Perpendicular and oblique lines
art: 103 | Theorem | There can be only one line through a given point perpendicular to a given straight line. For, since all right angles are equal, all lines lying in one plane and perpendicular to a given line, must have the same direction. Now, through a given point in one direction there can be only one straight line. Therefore, since the perpendiculars have the same direction, there can be through a given point only one perpendicular to a given straight line. When the point is in the given line, this theorem must be limited to one plane.
art: 104 | Theorem | If a perpendicular and oblique lines fall from the same point upon a given straight line, the perpendicular is shorter than any oblique line. If AD is perpendicular and AC oblique to BE, then AD is shorter than AC. Let the figure revolve upon BE as upon an axis, the point A falling upon F, and the lines AD and AC upon FD and FC. Now, the angle CDF is equal to the angle CDA, and both are right angles. Therefore, the sum of those two angles being equal to two right angles, ADF is a straight line, and is shorter than ACF. Therefore, AD, the half of ADF, is shorter than AC, the half of ACF.
fig: 38 0.515 0.66 0.537 0.73
art: 105 | Corollary | The distance from a point to a straight line is the perpendicular let fall from the point to the line.
art: 106 | Theorem | If a perpendicular and several oblique lines fall from the same point upon a given straight line, and if two oblique lines meet the given line at equal distances from the foot of the perpendicular, the two are equal. Let AD be the perpendicular and AC and AE the oblique lines, making CD equal to DE. Then AC and AE are equal. Let that portion of the figure on the left of AD turn upon AD. Since the angles ADB and ADF are equal, DB will take the direction DF; and since DC and DE are equal, the point C will fall on E. Therefore, AC and AE will coincide, and are equal.
fig: 39 0.235 0.34 0.495 0.74
art: 107 | Corollary | When the oblique lines are equal, the angles which they make with the perpendicular are equal. For CAD may coincide with DAE.
art: 108 | Theorem | If a line be perpendicular to another at its center, then every point of the perpendicular is equally distant from the two ends of the other line. For straight lines extending from any point of the perpendicular to the two ends of the other line must be equal.
art: 109 | Corollary | Since two points fix the position of a line, if a line have two points each equidistant from the ends of another line, the two lines are perpendicular to each other, and the second line is bisected. The two points may be on the same side, or on opposite sides of the second line.
art: 110 | Theorem | If a perpendicular and several oblique lines fall from the same point on a given straight line, of two oblique lines, that which meets the given line at a greater distance from the perpendicular is the longer. If AD be perpendicular to BG, and DF is greater than DC, then AF is greater than AC. On the line DF take a part DE equal to DC, and join AE. Then let the figure revolve upon BG, the point A falling upon H, and the lines AD, AE, and AF upon HD, HE, and HF. Now, AEH is shorter than AFH; therefore, AE, the half of AEH, is shorter than AF, the half of AFH. But AC is equal to AE. Hence, AF is longer than AC, or AE, or any line from A meeting the given line at a less distance from D than DF.
fig: 40 0.24 0.45 0.39 0.72
art: 111 | Corollary | A point may be at the same distance from two points of a straight line, one on each side of the perpendicular; but it can not be at the same distance from more than two points.
art: 112 | Theorem | If a line be perpendicular to another at its center, every point out of the perpendicular is nearer to that end of the line which is on the same side of the perpendicular. If BF is perpendicular to AC at its center B, then D, a point not in BF, is nearer to C than to A. Join DA and DC, and let the perpendicular DE fall from D upon the line AC. This perpendicular must fall on the same side of BF as the point D, for if it crossed the line BF, there would be from the point of intersection two perpendiculars on AC, which is impossible. Now, since AB is equal to BC, AE must be greater than EC. Hence, AD is greater than CD.
fig: 40 0.74 0.85 0.495 0.7
art: 113 | Theorem | Every point of the line which bisects an angle is equidistant from the sides of the angle. Let BCD be the given angle, and AC the bisecting line. Then the distance of the two sides from any point A of that line is measured by perpendiculars to the sides, as AF and AE. Since the angles BCA and DCA are equal, that part of the figure upon the one side of AC may revolve upon AC, and the line BC will take the direction of CD, and coincide with it. Then the perpendiculars AF and AE must coincide, and the point F fall upon E. Therefore, AF and AE are equal, and the point A is equally distant from the sides of the given angle.
fig: 41 0.417 0.535 0.51 0.735
art: 114 | Application | Perpendicular lines are constantly used in architecture, carpentry, stone-cutting, machinery, and so on. The carpenter's square is used for drawing perpendicular lines, and for testing the correctness of right angles. The square itself should be tested in the following manner: on any plane surface draw an angle, as BAC, with the square. Extend BA in the same straight line to D. Then turn the square so that the edges by which the angle BAC was described, may be applied to the angle DAC. If the coincidence is exact, the square is correct as to these edges.
fig: 42 0.265 0.37 0.48 0.73
art: 115 | Theorem | Of any two lines which may extend from two given points outside of a straight line to any point in it, those are together least which make equal angles with that line. Let CD be the line and A and B the points, and AEB the shortest line that can be made from A to B through any point of CD. Then it is to be proved that AEC and BED are equal angles. Make AH perpendicular to CD, and produce it to F, making HF equal to AH. Now every point of the line CD is equally distant from A and F. Therefore, every line joining B to F through some point of CD, is equal to a line joining B to A through the same point. But BEA is, by hypothesis, the shortest line from B to A through any point of CD. Therefore, BEF is the shortest line from B to F, and is a straight line. Since BEF is one straight line, the angles FEH and BED are vertical and equal. But the angles FEH and AEH are equal. Therefore, AEH and BED are equal.
fig: 42 0.645 0.79 0.43 0.72
art: 116 |  | When several magnitudes are of the same kind but vary in extent, the least is called a minimum, and the greatest a maximum. When a ray of light is reflected from a polished surface, the incident and reflected parts of the ray make equal angles with the surface. We learn from this geometrical principle that light, when reflected, still adheres to that law of its nature which requires it to take the shortest path.
mc: Through a given point, how many lines can be perpendicular to a given straight line in one plane? | *Only one | Two | Any number
mc: From a point to a straight line, the shortest line is... | *the perpendicular | any oblique line | the longest oblique line
mc: The distance from a point to a straight line is measured along... | *the perpendicular let fall to the line | any line to the line | a curve
mc: Two oblique lines from one point meet a line at equal distances from the foot of the perpendicular. Then the two oblique lines are... | *equal | unequal | perpendicular
mc: Of two oblique lines from one point, the one that meets the line farther from the perpendicular is... | *the longer | the shorter | equal to the other
mc: Every point of the perpendicular at the center of a line is... | *equally distant from the two ends of that line | nearer to one end | on the line
mc: A point not on that perpendicular is nearer to... | *the end on the same side of the perpendicular | the end on the other side | neither end
mc: From how many points of a straight line can one point be equally distant? | *At most two | Three | Any number
mc: Every point of the line which bisects an angle is... | *equidistant from the sides of the angle | nearer to one side | on one side
mc: In the theorem of the minimum line, the two lines which are together least make ___ with the given line. | *equal angles | right angles | no angles
mc: The least of several magnitudes of the same kind is called a... | *minimum | maximum | mean
mc: The greatest of several magnitudes of the same kind is called a... | *maximum | minimum | mean
mc: A reflected ray of light takes... | *the shortest path | the longest path | a curved path
mc: If a line has two points each equidistant from the ends of another line, then the two lines are... | *perpendicular, and the second is bisected | parallel | oblique
tf: The perpendicular from a point to a line is shorter than any oblique line from that point. {T}
tf: Equal oblique lines make equal angles with the perpendicular. {T}
tf: A point can be equally distant from three points of one straight line. {F}
tf: The carpenter's square is tested by drawing an angle, turning the square over, and seeing if the edges fit the angle next to it. {T}
q: From a point A, the perpendicular AD to a line is 3 inches, and an oblique line AC meets the line 4 inches from D. Another oblique line AE meets the line 4 inches from D on the other side. If AC is 5 inches, how long is AE, in inches? {5}
sol: Oblique lines that meet the line at equal distances from the foot of the perpendicular are equal, so AE = AC = 5 inches.
q: A point on the perpendicular at the center of a line AB is 7 inches from A. How many inches is it from B? {7}
sol: Every point of the perpendicular at the center is equally distant from the two ends, so it is 7 inches from B.
q: A point on the line which bisects an angle is 2 inches from one side of the angle. How many inches is it from the other side? {2}
sol: Every point of the bisecting line is equidistant from the two sides, so 2 inches.

== 12 | Straight lines | Parallel lines
art: 117 |  | Parallel lines are straight lines which have the same directions.
art: 118 | Corollary | Two lines which are each parallel to a third are parallel to each other.
art: 119 | Corollary | From the above definition, and the Axiom of Direction, it follows that there can be only one line through a given point parallel to a given line.
art: 120 | Corollary | From the same premises, it follows that two parallel lines can never meet, or have a common point.
art: 121 | Theorem | Two parallel lines both lie in one plane and determine its position. The position of a plane is determined by either line and one point of the other line. Now the plane has the direction of the first line and can not vary from it, and the second line has also the same direction and can not vary from it. Therefore, the second line must also lie wholly in the plane.
art: 122 | Names of angles | When two straight lines are cut by a secant, the eight angles thus formed are named as follows: The four angles between the two lines are interior; as, f, g, h, and k. The other four are exterior; as, b, c, l, and m. Two angles on the same side of the secant, and on the same side of the two lines cut by it, are called corresponding angles. The angles h and b are corresponding. Two angles on opposite sides of the secant, and on opposite sides of the two lines cut by it, are called alternate angles. The angles f and k are alternate; also, b and m.
fig: 44 0.41 0.51 0.47 0.71
art: 123 | Corollary | The corresponding and the alternate angles of any given angle are vertical to each other, and therefore equal.
art: 124 | Theorem | When two parallel lines are cut by a secant, each of the eight angles is equal to its corresponding angle. If the straight lines AB and CD have the same directions, then the angles FHB and FGD are equal. For, since the directions GD and HB are the same, the direction GF differs equally from them. Therefore, the angles are equal. In the same manner, it may be shown that any two corresponding angles are equal.
fig: 45 0.28 0.42 0.475 0.73
art: 125 | Corollary | When two parallel lines are cut by a secant, each of the eight angles is equal to its alternate.
art: 126 | Corollary | Two interior angles on the same side of the secant are supplements of each other. For, since GHB is the supplement of FHB, it is also the supplement of its equal HGD. Two exterior angles on the same side of the secant are supplementary, for a similar reason.
art: 127 | Corollary | When a secant is perpendicular to one of two parallels, it is also perpendicular to the other, and all the angles are right.
art: 128 | Corollary | When the secant is oblique to the parallels, four of the angles formed are obtuse and are equal to each other; the other four are acute, and equal; and any acute angle is the supplement of any obtuse.
art: 129 | Theorem | When two straight lines, being in the same plane, are cut by a third, making the corresponding angles equal, the two lines so cut are parallel. If AB and CD lie in the same plane, and if the angles AHF and CGF are equal, then AB and CD are parallel. For, suppose a straight line to pass through the point H, parallel to DC. Such a line makes a corresponding angle equal to CGF, and therefore equal to AHF. This supposed parallel line lies in the same plane as CD and H; that is, by hypothesis, in the same plane as AB. But if it lies in the same plane with AB and makes the same angle with the same line EF, at the same point H, then it must coincide with AB. Therefore, AB is parallel to CD.
fig: 46 0.215 0.36 0.47 0.74
art: 130 | Corollary | If the alternate angles are equal, the lines are parallel.
art: 131 | Corollary | The same conclusion must follow when the interior angles on the same side of the secant are supplementary.
art: 132 | Theorem | Two parallel lines are everywhere equally distant. The distance between two parallel lines is measured by a line perpendicular to them, since it is the shortest from one to the other. Let AB and CD be two parallels. Then any perpendiculars to them, as EF and GH, are equal. From M, the center of FH, erect the perpendicular ML. Let that part of the figure to the left of ML revolve upon ML. All the angles of the figure being right angles, MC will fall upon MD. Since MF is equal to MH, the point F will fall on H, and FE will take the direction HG, and the point E will be on the line HG. But since the angles at L are equal, the point E will also fall on LB, and being on both LB and HG, it must be on G. Therefore, FE and HG coincide and are equal.
fig: 47 0.18 0.258 0.46 0.71
art: 133 | Corollary | The parts of parallel lines included between perpendiculars to them, must be equal. For the perpendiculars are parallel.
art: 134 | Theorem | If several equally distant parallel lines be cut by a secant, the secant will be divided into equal parts. If the parallels BC, DF, GH, and KL are at equal distances, then the parts EI, IO, and OU of the secant AY are equal. For that part of the figure included between BC and DF may be placed upon and will coincide with that part between DF and GH; for the parallels are everywhere equally distant. Let them be so placed that the point E may fall upon I. Then, since the angles BEI and DIO are equal, the line EI will take the direction IO. And since DF and GH coincide, the point I will fall on O. Therefore, EI and IO coincide and are equal.
fig: 47 0.645 0.825 0.41 0.71
art: 135 | Corollary | Conversely, if several parallel lines intercept equal segments of a secant, then the several distances between the parallels are equal.
art: 136 | Corollary | When the distances between the parallels are unequal, the segments of the secant are unequal. And conversely, when the segments of the secant are unequal, the distances are unequal.
art: 137 | Theorem | If two straight lines are in the same plane and are not parallel, they will meet if sufficiently produced. Let AB and CD be two lines. Let the line EF, parallel to CD, pass through any point of AB, as H. From H let the perpendicular HG fall upon CD. Since AB and EF have different directions, they cut each other at the point H. Take any point, as I, in that part of AB which lies between EF and CD, and extend a line IK parallel to CD through the point I. Now divide HG into parts equal to HK until one of the points of division falls beyond G. Then along HB, take parts equal to HI, as often as HK was taken along HG. Lastly, from each point of division of HB, extend a line perpendicular to HG. These perpendiculars are parallel to each other and to CD. These parallels by construction intercept equal parts of HB. Therefore, they are equally distant from each other. Hence, HG is divided by them into equal segments; that is, each one passes through one of the previously ascertained points of the line HG. But the last of these points was beyond the line CD, and as the parallel can not cross CD, the corresponding point of HB is beyond CD. Therefore, HB and CD must cross each other.
fig: 48 0.58 0.705 0.39 0.73
art: 138 | Theorem | When the sides of one angle are parallel to the sides of another, and have respectively the same directions from their vertices, the two angles are equal. If the directions BA and DC are the same, and the directions DE and BF are the same, then the angles ABF and CDE are equal. For each of these angles is equal to the angle CGF.
fig: 49 0.57 0.775 0.49 0.73
art: 139 | Exercise | Let the student demonstrate that when two of the parallel sides have opposite directions, and the other two have the same direction, then the angles are supplementary. Let him also demonstrate that if both sides of one angle have directions respectively opposite to those of the other, then again the angles are equal.
art: 140 | Theorem | Two angles which have their sides respectively perpendicular are equal or supplementary. If AB is perpendicular to DG, and BC is perpendicular to EF, then the angle ABC is equal to one, and supplementary to the other of the angles formed by DG and EF. Through B extend BI parallel to GD, and BH parallel to EF. Now, ABI and CBH are right angles, and therefore equal. Subtracting the angle HBA from each, the remainders HBI and ABC are equal. But HBI is equal to FGD, and is the supplement of EGD. Therefore, the angle ABC is equal or supplementary to any angle formed by the lines DG and EF.
fig: 50 0.24 0.445 0.395 0.73
art: 141 | Application | The instrument called the T square consists of two straight and flat rulers fixed at right angles to each other, as in the figure. It is used to draw parallel lines. Draw a straight line in a direction perpendicular to that in which it is required to draw parallel lines. Lay the cross-piece of the T ruler along this line. The other piece of the ruler gives the direction of one of the parallels. The ruler being moved along the paper, keeping the cross-piece coincident with the line first described, any number of parallel lines may be drawn.
fig: 50 0.675 0.805 0.555 0.74
art: 142 | Application | The uniform distance of parallel lines is the principle upon which numerous instruments and processes in the arts are founded. If two systems, each consisting of several parallel lines, cross each other at right angles, all the parts of one system included between any two lines of the other system will be equal. The ordinary framing of a window consists of two systems of lines of this kind; the shelves and upright standards of book-cases and the paneling of doors also afford similar examples.
art: 143 | Application | The joiner's gauge is a tool with which a line may be drawn on a board parallel to its edge.
art: 144 | Application | It is frequently important in machinery that a body shall have what is called a parallel motion; that is, such that all its parts shall move in parallel lines, preserving the same relative position to each other. The piston of a steam-engine, and the rod which it drives, receive such a motion.
mc: Parallel lines are straight lines which have... | *the same directions | the same length | a common point
mc: Two lines which are each parallel to a third are... | *parallel to each other | perpendicular to each other | oblique to each other
mc: Through a given point, how many lines can be parallel to a given line? | *Only one | Two | Any number
mc: Two parallel lines can... | *never meet | meet once | meet twice
mc: Two lines cut by a secant form how many angles? | *Eight | Four | Six
mc: The four angles between the two lines cut by a secant are called... | *interior | exterior | vertical
mc: Two angles on the same side of the secant, and on the same side of the two lines, are... | *corresponding angles | alternate angles | vertical angles
mc: Two angles on opposite sides of the secant, and on opposite sides of the two lines, are... | *alternate angles | corresponding angles | adjacent angles
mc: When two parallels are cut by a secant, two interior angles on the same side of the secant are... | *supplements of each other | equal | complements of each other
mc: When two parallels are cut by a secant, each angle is equal to... | *its corresponding angle and its alternate angle | every other angle | a right angle
mc: When a secant is oblique to two parallels, how many of the eight angles are obtuse? | *Four | Two | Eight
mc: The distance between two parallel lines is measured by... | *a line perpendicular to them | any secant | the longest line between them
mc: Two lines in one plane which are not parallel will... | *meet if far enough produced | never meet | always be perpendicular
mc: Two angles whose sides are parallel and have the same directions are... | *equal | supplementary | complementary
mc: Two angles whose sides are respectively perpendicular are... | *equal or supplementary | always equal | always complementary
mc: The T square is used to draw... | *parallel lines | circles | angles of any size
tf: Two parallel lines lie in one plane. {T}
tf: If the alternate angles made by a secant are equal, the lines are parallel. {T}
tf: Two parallel lines get nearer together the farther they are produced. {F}
tf: If equally distant parallels cut a secant, the parts of the secant between them are equal. {T}
tf: When a secant is perpendicular to one of two parallels, it is oblique to the other. {F}
q: Two parallel lines are cut by a secant. One of the eight angles is 1/3 of a right angle. How many of the eight angles are equal to it? {4}
sol: When the secant is oblique, the four acute angles are equal to each other, so 4 angles are equal to it.
q: Two parallels are cut by a secant, and one interior angle is 3/4 of a right angle. The other interior angle on the same side of the secant is how many fourths of a right angle? {5}
sol: Two interior angles on the same side are supplements, so together they make two right angles = 8 fourths; 8 - 3 = 5 fourths.
q: Four equally distant parallel lines cut a secant. The part of the secant between the first two is 3 inches. How long is the part between the first and the fourth, in inches? {9}
sol: Equally distant parallels divide the secant into equal parts. There are 3 parts from the first to the fourth line: 3 times 3 = 9 inches.

== 13 | Circumferences | General properties of circumferences
art: 145 |  | Let the line AB revolve in a plane about the end A, which is fixed. Then the point B will describe a line which returns upon itself, called a circumference of a circle. Hence, the following definitions: A circle is a portion of a plane bounded by a line called a circumference, every point of which is equally distant from a point within called the center.
fig: 52 0.34 0.505 0.515 0.73
art: 146 | Theorem | A circumference is curved throughout. For a straight line can not have more than two points equally distant from a given point.
art: 147 | Corollary | A straight line can not cut a circumference in more than two points.
art: 148 |  | The circumference is the only curve considered in elementary geometry. Let us examine the properties of this line, and of the straight lines which may be combined with it.
art: 149 | Theorem | Three points not in the same straight line fix a circumference both as to position and extent. The three given points, as A, B, and C, determine the position of a plane. Let the given points be joined by straight lines AB and BC. At D and E, the middle points of these lines, let perpendiculars be erected in the plane of the three points. By the hypothesis, AB and BC make an angle at B. Therefore, DG and EH are not parallel, and they will meet if produced. Let L be their point of intersection. Since every point of DG is equidistant from A and B, and since every point of EH is equidistant from B and C, their common point L is equidistant from A, B, and C. Therefore, with this point as a center, a circumference may be described through A, B, and C. There can be no other circumference through these three points, for there is no other point besides L equally distant from all three. Therefore, these three points fix the position and the extent of the circumference which passes through them.
fig: 53 0.15 0.36 0.41 0.72
mc: A portion of a plane bounded by a line every point of which is equally distant from a point within is a... | *circle | circumference | radius
mc: The line which bounds a circle is called the... | *circumference | diameter | center
mc: The point within a circle, equally distant from every point of the circumference, is the... | *center | vertex | axis
mc: The path of the end B, when the line AB revolves in a plane about the fixed end A, is a... | *circumference | straight line | broken line
mc: In how many points, at most, can a straight line cut a circumference? | *Two | One | Three
mc: How many points, not in one straight line, fix a circumference? | *Three | Two | Four
mc: The only curve considered in elementary geometry is the... | *circumference | broken line | spiral
mc: The center of the circle through A, B, and C is where the ___ of AB and BC meet. | *perpendiculars at the middle points | parallels | radii
tf: A circumference is curved throughout. {T}
tf: Three points in one straight line always lie on one circumference. {F}
tf: Through three points not in a straight line, there can be two different circumferences. {F}

== 14 | Circumferences | Arcs and radii
art: 150 |  | An arc is a portion of a circumference. A radius is a straight line from the center to the circumference. A diameter is a straight line passing through the center, and limited at both ends by the circumference. A chord is a straight line joining the ends of an arc.
art: 151 | Corollary | All radii of the same circumference are equal.
art: 152 | Corollary | In the same circumference, a diameter is double the radius, and all diameters are equal.
art: 153 | Corollary | Every point of the plane at greater distance from the center than the length of the radius, is outside of the circumference. Every point at a less distance from the center, is within the circumference. Every point whose distance from the center is equal to the radius, is on the circumference.
art: 154 | Theorem | Circumferences which have equal radii are equal. Let the center of one be placed on that of the other. Then the circumferences will coincide. For if it were otherwise, then some points would be unequally distant from the common center, which is impossible when the radii are equal. Therefore, the circumferences are equal.
art: 155 | Corollary | A circumference may revolve upon, or slide along its equal.
art: 156 | Corollary | Two arcs of the same or of equal circles may coincide so far as both extend.
art: 157 | Theorem | Every diameter bisects the circumference and the circle. For that part upon one side of the diameter may be turned upon that line as its axis. When the two parts thus meet, they will coincide; for if they did not, some points of the circumference would be unequally distant from the center.
art: 158 |  | A line which divides any figure in this manner, is said to divide it symmetrically; and a figure which can be so divided is symmetrical.
art: 159 | Theorem | A diameter is greater than any other chord of the same circumference. To be demonstrated by the student.
art: 160 | Problem | Arcs of equal radii may be added together, or one may be subtracted from another. For an arc may be produced till it becomes an entire circumference, or it may be diminished at will. Therefore, the length of an arc may be increased or decreased by the length of another arc of the same radius; and the result, that is, the sum or difference, will be an arc of the same radius.
art: 161 | Corollary | Arcs of equal radii may be multiplied or divided in the same manner as straight lines.
art: 162 |  | The sum of several arcs may be greater than a circumference.
art: 163 |  | Two arcs not having the same radius may be joined together, and the result may be called their sum; but it is not one arc, for it is not a part of one circumference.
art: 164 | Application | The circumference is the only line which can move along itself, around a center, without suffering any change. It is in virtue of this property that the axles of wheels, shafts, and other solid bodies which are required to revolve within a hollow mold or casing of their own form, must be circular.
art: 165 | Application | Wheels which are intended to maintain a carriage always at the same hight above the road on which they roll, must be circular, with the axle in the center.
art: 166 | Application | The art of turning consists in the production of the circular form by mechanical means. The substance to be turned is placed in a machine called a lathe, which gives it a rotary motion. The edge of a cutting tool is placed at a distance from the axis of revolution equal to the radius of the intended circle.
art: 167 | Problems in drawing | The compasses enable us to draw a circumference, or an arc of a given radius and given center. Open the instrument till the points are on the two ends of the given radius. Then fix one point on the given center, and the other point may be made to revolve around in contact with the surface, thus tracing out the circumference.
art: 168 |  | It is evident that with the ruler and compasses, 1. A straight line can be drawn through two given points. 2. A given straight line can be produced any length. 3. A circumference can be described from any center, with any radius.
art: 169 |  | The foregoing are the three postulates of Euclid. Since the straight line and the circumference are the only lines treated of in elementary geometry, these Euclidian postulates are a sufficient basis for all problems. Hence, the rule that no instruments shall be used except the ruler and the compasses.
art: 171 | Problem | To bisect a given straight line. With A and B as centers, and with a radius greater than the half of AB, describe arcs which intersect in the two points D and E. The straight line joining these two points will bisect AB at C.
fig: 57 0.155 0.331 0.465 0.74
art: 172 | Problem | To erect a perpendicular on a given straight line at a given point. Take two points in the line, one on each side of the given point, at equal distances from it. Describe arcs as in the last problem, and their intersection gives one point of the perpendicular.
art: 173 | Problem | To let fall a perpendicular from a given point on a given straight line. With the given point as a center, and a radius long enough, describe an arc cutting the given line BC in the points D and E. With D and E as centers, and with a radius greater than the half of DE, describe arcs cutting each other in F. The straight line joining A and F is perpendicular to DE.
fig: 57 0.43 0.612 0.46 0.74
art: 174 | Problem | To draw a line through a given point parallel to a given line. Let a perpendicular fall from the point on the line. Then, at the given point, erect a perpendicular to this last. It will be parallel to the given line.
art: 175 | Problem | To describe a circumference through three given points. The solution of this problem is evident, from Article 149.
art: 176 | Problem | To find the center of a given arc or circumference. Take any three points of the arc, and proceed as in the last problem.
art: 177 |  | The student is advised to make a drawing of every problem. First draw the parts given, then the construction requisite for solution. Afterward demonstrate its correctness.
mc: A portion of a circumference is an... | *arc | chord | radius
mc: A straight line from the center to the circumference is a... | *radius | diameter | chord
mc: A straight line through the center, ended at both ends by the circumference, is a... | *diameter | radius | tangent
mc: A straight line joining the ends of an arc is a... | *chord | radius | secant
mc: In the same circumference, a diameter is... | *double the radius | half the radius | equal to the radius
mc: A point whose distance from the center is greater than the radius is... | *outside of the circumference | within the circumference | on the circumference
mc: A point whose distance from the center equals the radius is... | *on the circumference | within it | outside of it
mc: Circumferences which have equal radii are... | *equal | similar but unequal | always concentric
mc: Every diameter bisects... | *the circumference and the circle | only the circle | only the radius
mc: The greatest chord of a circumference is... | *a diameter | a radius | any chord
mc: A figure which a line divides into two parts that fold onto each other is... | *symmetrical | equivalent | concave
mc: Why must wheels be circular, with the axle in the center? | *To keep the carriage always at the same hight | To be lighter | To turn faster
mc: The three postulates of Euclid need only the... | *ruler and compasses | protractor and square | ruler alone
mc: To find the center of a given arc, take ___ points of the arc. | *any three | any two | the two end
tf: All radii of the same circumference are equal. {T}
tf: The sum of several arcs may be greater than a circumference. {T}
tf: Two arcs with different radii, joined together, make one arc. {F}
tf: To bisect a line AB, describe arcs from A and B with a radius less than half of AB. {F}
q: The radius of a circle is 6 inches. How long is its diameter, in inches? {12}
sol: A diameter is double the radius: 2 times 6 = 12 inches.
q: The diameter of a circle is 18 inches. How long is its radius, in inches? {9}
sol: The radius is half the diameter: 18 / 2 = 9 inches.
q: A circle has a radius of 5 inches. A point is 7 inches from the center. Is it inside, on, or outside the circumference? Answer 1 for inside, 2 for on, 3 for outside. {3}
sol: 7 is greater than the radius 5, so the point is outside the circumference.
q: A circle has a radius of 10 inches. How long, at most, can a chord of it be, in inches? {20}
sol: A diameter is the greatest chord, and the diameter is 2 times 10 = 20 inches.

== 15 | Circumferences | Tangents
art: 178 | Theorem | A straight line which is perpendicular to a radius at its extremity, touches the circumference in only one point. Let AD be perpendicular to the radius BC at its extremity B. Then it is to be proved that AD touches the circumference at B, and at no other point. If the center C be joined by straight lines with any points of AD, the perpendicular BC will be shorter than any such oblique line. Therefore, every point of the line AD, except B, is outside of the circumference.
fig: 58 0.545 0.75 0.465 0.73
art: 179 |  | A tangent is a line touching a circumference in only one point. The circumference is also said to be tangent to the straight line. The common point is called the point of contact.
art: 180 | Application | Tangent lines are frequently used in the arts. A common example is when a strap is carried round a part of the circumference of a wheel, and extending to a distance, sufficient tension is given to it to produce such a degree of friction between it and the wheel, that one can not move without the other.
art: 181 | Problem in drawing | To draw a tangent at a given point of an arc. Draw a radius to the given point, and erect a perpendicular to the radius at that point. It will be necessary to produce the radius beyond the arc, as the student has not yet learned to erect a perpendicular at the extremity of a line without producing it.
mc: A line touching a circumference in only one point is a... | *tangent | secant | chord
mc: The common point of a tangent and the circumference is the... | *point of contact | center | vertex
mc: A straight line perpendicular to a radius at its end... | *touches the circumference in only one point | cuts it in two points | passes through the center
mc: Every point of the tangent, except the point of contact, is... | *outside of the circumference | inside the circumference | on the circumference
mc: To draw a tangent at a given point of an arc, first draw... | *a radius to the given point | a chord | a diameter at right angles
mc: A strap carried round a wheel is an example of a... | *tangent | diameter | radius
tf: The radius to the point of contact is shorter than any other line from the center to the tangent. {T}
tf: A tangent may touch a circumference in two points. {F}
tf: The circumference is said to be tangent to the straight line. {T}
q: A circle has a radius of 4 inches. A tangent touches it at B. What is the shortest distance, in inches, from the center to the tangent? {4}
sol: The radius to the point of contact is perpendicular to the tangent, and the perpendicular is the shortest line, so the distance is the radius, 4 inches.

== 16 | Circumferences | Secants
art: 182 | Theorem | A straight line which is oblique to a radius at its extremity, cuts the circumference in two points. Let AD be oblique to the radius CB at its extremity B. Then it will cut the circumference at B, and at some other point. From the center C, let CE fall perpendicularly on AD. On ED, take EF equal to EB. Then the distance from C to any point of the line AD between B and F is less than the length of the radius CB, and to any point of the line beyond B and F, it is greater than the length of CB. Therefore, that portion of the line AD between B and F is within, and the parts beyond B and F are without the circumference. Hence, the oblique line cuts the circumference in two points.
fig: 59 0.528 0.757 0.493 0.72
art: 183 | Corollary | A tangent to the circumference is perpendicular to the radius which extends to the point of contact. For, if it were not perpendicular, it would be secant.
art: 184 | Corollary | At one point of a circumference, there can be only one tangent.
mc: A straight line oblique to a radius at its end... | *cuts the circumference in two points | touches it in one point | misses it
mc: A tangent is ___ to the radius at the point of contact. | *perpendicular | oblique | parallel
mc: At one point of a circumference, how many tangents can there be? | *Only one | Two | Any number
mc: The part of the secant AD between B and F lies... | *within the circumference | outside the circumference | on the circumference
mc: A line which cuts a circumference is called a... | *secant | tangent | radius
tf: If a line through the end of a radius is not perpendicular to it, the line is a secant. {T}
tf: A secant cuts a circumference in three points. {F}
tf: The parts of the secant beyond B and F are without the circumference. {T}

== 17 | Circumferences | Chords
art: 185 | Theorem | The radii being equal, if two arcs are equal their chords are also equal. If the arcs AOE and BCD are equal, and their radii are equal, then AE and BD are equal. For, since the radii are equal, the circumferences are equal; and the arcs may be placed one upon the other, and will coincide, so that A will be upon B, and E upon D. Then the two chords, being straight lines, must coincide, and are equal.
fig: 60 0.415 0.595 0.15 0.75
art: 186 |  | Every chord subtends two arcs, which together form the whole circumference. Thus the chord AE subtends the arcs AOE and AIE. The arc of a chord always means the smaller of the two, unless otherwise expressed.
art: 187 | Theorem | The radius which is perpendicular to a chord bisects the chord and its arc. Let CD be perpendicular at E to the chord AB, then will AE be equal to EB, and the arc AD to the arc DB. Produce DC to the circumference at F, and let that part of the figure on one side of DF be turned upon DF as upon an axis. Then the semi-circumference DAF will coincide with DBF. Since the angles at E are right, the line EA will take the direction of EB, and the point A will fall on the point B. Therefore, EA and EB will coincide, and are equal; and the same is true of DA and DB, and of FA and FB.
fig: 61 0.165 0.355 0.495 0.74
art: 188 | Corollary | Since two conditions determine the position of a straight line, if it has any two of the four conditions mentioned in the theorem, it must have the other two. These four conditions are, 1. The line passes through the center of the circle, that is, it is a radius. 2. It passes through the center of the chord. 3. It passes through the center of the arc. 4. It is perpendicular to the chord.
art: 189 | Theorem | The radii being equal, when two arcs are each less than a semi-circumference, the greater arc has the greater chord. If the arc AMB is greater than CND, and the radii of the circles are equal, then AB is greater than CD. Take AME equal to CND. Join AE, OE, and OB. Then AE is equal to CD. Since the arc AMB is less than a semi-circumference, the chord AB will pass between the arc and the center O. Hence, it cuts the radius OE at some point I. Now, the broken line OIB is greater than OB, or its equal OE. Subtracting OI from each, the remainder IB is greater than the remainder IE. Adding AI to each of these, we have AB greater than AIE. But AIE is greater than AE. Therefore, AB, the chord of the greater arc, is greater than AE, or its equal CD, the chord of the less.
fig: 62 0.168 0.343 0.15 0.75
art: 190 | Corollary | When the arcs are both greater than a semi-circumference, the greater arc has the less chord.
art: 191 | Theorem | When the radii are equal, equal chords are equally distant from the center. Let the chords AB and CD be equal, and in the equal circles ABG and CDF; then the distances of these chords from the centers E and H will also be equal. Let fall the perpendiculars EK and HL from the centers upon the chords. Now, since the chords AB and CD are equal, the arcs AB and CD are also equal; and we may apply the circle ABG to its equal CDF, so that they will coincide, and the arc AB coincide with its equal CD. Therefore, the chords will coincide. Since K and L are the middle points of these coinciding chords, K will fall upon L. Therefore, the lines EK and HL coincide and are equal. But these equal perpendiculars measure the distance of the chords from the centers.
fig: 62 0.625 0.81 0.15 0.75
art: 192 | Theorem | When the radii are equal, the less of two unequal chords is the farther from the center. Let AB be the greater of two chords, and FG the less, in the same or an equal circle. Then FG is farther from the center than AB. Take the arc AC equal to FG. Join AC, and from the center D let fall the perpendiculars DE and DN upon AB and AC. Since the arc AC is less than AB, the chord AB will be between AC and the center D, and will cut the perpendicular DN. Then DN, the whole, is greater than DH, the part cut off; and DH is greater than DE. So much the more is DN greater than DE. Therefore, AC and its equal FG are farther from the center than AB.
fig: 63 0.535 0.765 0.43 0.74
art: 193 | Corollary | Conversely of these two theorems, when the radii are equal, chords which are equally distant from the center are equal; and of two chords which are unequally distant from the center, the one nearer to the center is longer than the other.
art: 194 | Problem in drawing | To bisect a given arc. Draw the chord of the arc, and erect a perpendicular at its center.
art: 195 |  | The most simple case of the division of an arc, after its bisection, is its trisection, or its division into three equal parts. Its object was to determine means of dividing any given arc into three equal parts, without any other instruments than the rule and compasses permitted by the postulates prefixed to Euclid's Elements. Simple as the problem appears to be, it never has been solved, and probably never will be, under the above conditions. (Lardner's Treatise.)
mc: The radii being equal, if two arcs are equal, their chords are... | *equal | unequal | parallel
mc: Every chord subtends how many arcs? | *Two | One | Four
mc: Unless otherwise said, the arc of a chord means... | *the smaller of the two arcs | the greater of the two arcs | the whole circumference
mc: The radius which is perpendicular to a chord... | *bisects the chord and its arc | is parallel to it | is equal to it
mc: How many of the four conditions (through the center, through the center of the chord, through the center of the arc, perpendicular to the chord) determine the line? | *Any two | All four | Only one
mc: For two arcs each less than a semi-circumference, the greater arc has... | *the greater chord | the less chord | an equal chord
mc: For two arcs each greater than a semi-circumference, the greater arc has... | *the less chord | the greater chord | an equal chord
mc: In equal circles, equal chords are... | *equally distant from the center | at different distances from the center | always diameters
mc: Of two unequal chords in the same circle, the less is... | *farther from the center | nearer to the center | at the same distance
mc: To bisect a given arc, draw its chord and... | *erect a perpendicular at the center of the chord | draw a tangent | draw a diameter parallel to it
mc: Dividing any arc into three equal parts with ruler and compasses alone is... | *the trisection, never solved | easy | the bisection
tf: The distance of a chord from the center is measured by the perpendicular to it from the center. {T}
tf: Of two chords in one circle, the one nearer the center is the longer. {T}
tf: A line through the center of a circle and the center of a chord is perpendicular to the chord. {T}
q: A radius perpendicular to a chord of 16 inches cuts the chord at E. How long is the part AE, in inches? {8}
sol: The radius perpendicular to a chord bisects it: 16 / 2 = 8 inches.
q: In one circle, a chord of 10 inches is 3 inches from the center. Another chord of the same circle is also 3 inches from the center. How long is it, in inches? {10}
sol: Chords equally distant from the center are equal, so it is 10 inches.

== 18 | Circumferences | Angles at the center
art: 196 |  | Angles which have their vertex at the center of a circle are called, for this reason, angles at the center. The arc between the sides of an angle is called the intercepted arc of the angle.
art: 197 | Theorem | The radii being equal, any two angles at the center have the same ratio as their intercepted arcs. This theorem presents the three following cases: 1st. If the arcs are equal, the angles are equal. For the arcs may be placed one upon the other, and will coincide. Then BC will coincide with AO, and DC with EO. Thus the angles may coincide, and are equal. The converse is proved in the same manner.
fig: 65 0.21 0.37 0.15 0.75
art: 197 | 2d case | If the arcs have the ratio of two whole numbers, the angles have the same ratio. Suppose, for example, the arc BD : arc AE :: 13 : 5. Then, if the arc BD be divided into thirteen equal parts, and the arc AE into five equal parts, these small arcs will all be equal. Let radii join to their respective centers all the points of division. The small angles at the center thus formed are all equal, because their intercepted arcs are equal. But BCD is the sum of thirteen, and AOE of five of these equal angles. Therefore, angle BCD : angle AOE :: 13 : 5; that is, the angles have the same ratio as the arcs.
fig: 65 0.45 0.63 0.15 0.75
art: 197 | 3d case | It remains to be proved, that, if the ratio of the arcs can not be expressed by two whole numbers, the angles have still the same ratio as the arcs. If this proportion is not true, then the fourth term is either too large or too small. If it were too large, then some smaller angle, as AOI, would verify the proportion. Let the arc BD be divided into equal parts, so small that each of them shall be less than EI. Let one of these parts be applied to the arc AE, beginning at A. One of those points must necessarily fall between I and E, say at the point U. Now the arcs BD and AU have the ratio of two whole numbers, and it follows that arc AE : arc AU :: angle AOI : angle AOU. But this last proportion is impossible, for the first antecedent is greater than its consequent, while the second antecedent is less than its consequent. It may be shown, in the same way, that the fourth term is not too small. Therefore, the arc BD is to the arc AE as the angle BCD is to the angle AOE.
fig: 66 0.265 0.43 0.15 0.75
art: 198 | Demonstration by limits | We have this definition of a limit: Let a magnitude vary according to a certain law which causes it to approximate some determinate magnitude. Suppose the first magnitude can, by this law, approach the second indefinitely, but can never quite reach it. Then the second, or invariable magnitude, is said to be the limit of the first, or variable one.
art: 199 |  | Any curve may be treated as a limit. The straight parts of a broken line, having all its vertices in the curve, may be diminished at will, and the broken line made to approximate the curve indefinitely. Hence, a curve is the limit of those broken lines which have all their vertices in the curve.
art: 200 |  | The arc BC, which is cut off by the secant AD, may be diminished by successive bisections, keeping the remainders next to B. Thus AD, revolving on the point B, may approach indefinitely the tangent EF. Hence, the tangent at any point of a curve is the limit of the secants which may cut the curve at that point.
fig: 68 0.15 0.33 0.49 0.73
art: 201 |  | The principle upon which all reasoning by the method of limits is governed, is that, whatever is true up to the limit is true at the limit. We admit this as an axiom of reasoning, because we can not conceive it to be otherwise.
art: 202 |  | Having proved that, when two arcs have the ratio of two whole numbers, the angles at the center have the same ratio, we may then suppose that the ratio of BD to BF can not be expressed by whole numbers. We may conceive the arc BF to be divided into any number of equal parts, and by increasing this number, the point O, the point of division nearest to D, may be made to approach within any conceivable distance of D. Although the arc BD is itself incommensurable with BF, yet it is the limit of the arcs BO, and the angle BCD is the limit of the angles BCO. Therefore, since whatever is true up to the limit is true at the limit, arc BD : arc BF :: angle BCD : angle BCF.
fig: 68 0.685 0.833 0.5 0.73
art: 203 | Method of infinites | Modern geometers have made much use of a kind of reasoning which may be called the method of infinites. It consists in supposing that any line of definite extent and form is composed of an infinite number of infinitely small straight lines. A surface is supposed to consist of an infinite number of infinitely narrow surfaces, and a solid of an infinite number of infinitely thin solids. These thin solids, narrow surfaces, and small lines, are called infinitesimals.
art: 204 |  | The method of infinites is a much abbreviated form of the method of limits. The student must be careful how he adopts it. For when the infinite is brought into an argument by the unskillful, the conclusion is very apt to be absurd.
art: 205 | Corollary | If two diameters are perpendicular to each other, they divide the whole circumference into four equal parts.
art: 206 |  | A quadrant is the fourth part of a circumference.
fig: 70 0.425 0.575 0.53 0.74
art: 207 |  | Since the angle at the center varies as the intercepted arc, mathematicians have adopted the same method of measuring both angles and arcs. As a right angle is the unit of angles, so a quadrant of a certain radius may be taken as the standard for the measurement of arcs that have the same radius. For the same reason, we usually say that the intercepted arc measures the angle at the center. Thus, the right angle is said to be measured by the quadrant; half a right angle, by one-eighth of a circumference; and so on.
art: 208 | Application | In the applications of geometry to practical purposes, the quadrant and the right angle are divided into ninety equal parts, each of which is called a degree. Each degree is marked thus °, and is divided into sixty minutes, marked thus ′; and each minute is divided into sixty seconds, marked thus ″. Hence, it appears that there are in an entire circumference, or in the sum of all the successive angles about a point, 360°, or 21600′, or 1296000″.
art: 209 | Application | Instruments for measuring angles are founded upon the principle that arcs are proportional to angles.
art: 210 | Application | An instrument called a protractor is used, in drawing, for measuring angles, and for laying down, on paper, angles of any required size. It consists of a semicircle of brass or mica, the circumference of which is divided into degrees and parts of a degree.
art: 211 | Problem | To draw an angle equal to a given angle. Let it be required to draw a line making, with the given line BC, an angle at B equal to the given angle A. With A as a center, and any assumed radius AD, draw the arc DE cutting the sides of the angle A. With B as a center, and the same radius as before, draw an arc FG. Join DE. With F as a center, and a radius equal to DE, draw an arc cutting FG at the point G. Join BG. Then GBF is the required angle. For, joining FG, the arcs DE and FG have equal radii and equal chords, and therefore are equal. Hence, they subtend equal angles.
fig: 71 0.59 0.745 0.48 0.74
art: 213 | Problem | To draw an angle equal to the sum of two given angles. Let A and B be the given angles. First, make the angle DCE equal to A, and then at C, on the line CE, draw the angle ECF equal to B. The angle FCD is equal to the sum of A and B.
fig: 72 0.16 0.297 0.455 0.74
art: 216 | Problem | To erect a perpendicular to a given line at its extreme point, without producing the given line. A right angle may be made separately, and then, at the end of the given line, an angle be made equal to the given angle.
art: 217 | Problem | To draw a line through a given point parallel to a given line. This has been done by means of perpendiculars. It may be done with an oblique secant, by making the alternate or the corresponding angles equal.
mc: An angle whose vertex is at the center of a circle is an... | *angle at the center | inscribed angle | exterior angle
mc: The radii being equal, two angles at the center have the same ratio as... | *their intercepted arcs | their chords | their radii
mc: If the arc BD : arc AE :: 13 : 5, then angle BCD : angle AOE is... | *13 : 5 | 5 : 13 | 1 : 1
mc: A magnitude which a variable approaches indefinitely, but never quite reaches, is its... | *limit | maximum | quadrant
mc: A curve is the limit of the ___ which have all their vertices in the curve. | *broken lines | tangents | radii
mc: The tangent at a point of a curve is the limit of the... | *secants which cut the curve at that point | chords | radii
mc: Whatever is true up to the limit is... | *true at the limit | false at the limit | not known at the limit
mc: The fourth part of a circumference is a... | *quadrant | degree | semicircle
mc: A right angle is said to be measured by... | *the quadrant | the whole circumference | one degree
mc: How many degrees are in a right angle? | *90 | 60 | 360
mc: One degree is divided into sixty... | *minutes | seconds | parts called grades
mc: An instrument for measuring and laying down angles on paper is the... | *protractor | T square | joiner's gauge
mc: To draw an angle equal to a given angle, the arcs DE and FG are made with equal radii and equal... | *chords | tangents | diameters
tf: Two perpendicular diameters divide the circumference into four equal parts. {T}
tf: Half a right angle is measured by one-eighth of a circumference. {T}
tf: The method of infinites is a shortened form of the method of limits. {T}
tf: In an entire circumference there are 180 degrees. {F}
q: How many degrees are in an entire circumference? {360}
sol: A right angle, or quadrant, has 90 degrees, and the circumference has four quadrants: 4 times 90 = 360.
q: How many minutes are in an entire circumference? {21600}
sol: 360 degrees times 60 minutes = 21600 minutes.
q: How many seconds are in one degree? {3600}
sol: One degree is 60 minutes, and each minute is 60 seconds: 60 times 60 = 3600.
q: How many degrees are in half a right angle? {45}
sol: A right angle is 90 degrees; half of it is 45 degrees.
q: An angle at the center intercepts one-eighth of the circumference. How many degrees is the angle? {45}
sol: The intercepted arc measures the angle: 360 / 8 = 45 degrees.
q: Two angles at the center of the same circle intercept arcs of 3 inches and 9 inches. The first angle is 20 degrees. How many degrees is the second? {60}
sol: The angles have the same ratio as their arcs, 3 : 9, so the second is 3 times 20 = 60 degrees.

== 19 | Circumferences | Intercepted arcs
art: 218 |  | An arc which is included between two parallel lines, or between the sides of an angle, is called intercepted.
art: 219 | Theorem | Two parallel lines intercept equal arcs of a circumference. The two lines may be both secants, or both tangents, or one a secant and one a tangent. 1st. When both are secants. The arcs AC and BD intercepted by the parallels AB and CD are equal. For, let fall from the center O a perpendicular upon CD, and produce it to the circumference at E. Then OE is also perpendicular to AB. Therefore, the arcs EA and EB are equal; and the arcs EC and ED are equal. Subtracting the first from the second, there remains the arc AC equal to the arc BD.
fig: 73 0.17 0.325 0.5 0.735
art: 219 | 2d case | When one is a tangent. Extend the radius OE to the point of contact. This radius is perpendicular to the tangent AB. Hence, it is perpendicular to the secant CD, and therefore it bisects the arc CED at the point E. That is, the intercepted arcs EC and ED are equal.
fig: 73 0.425 0.588 0.5 0.735
art: 219 | 3d case | When both are tangents. Extend the radii OE and OI to the points of contact. These radii being perpendicular to the parallels, must form one straight line. Therefore, EI is a diameter, and divides the circumference into equal parts. But these equal parts are the arcs intercepted by the parallel tangents. Therefore, in every case, the arcs intercepted by two parallels are equal.
fig: 73 0.652 0.825 0.5 0.74
art: 220 |  | An inscribed angle is one whose sides are chords or secants, and whose vertex is on the circumference. An angle is said to be inscribed in an arc, when its vertex is on the arc and its sides extend to or through the ends of the arc. In such a case the arc is said to contain the angle. Thus, the angle AEI is inscribed in the arc AEI, and the arc AEI contains the angle AEI. An angle is said to stand upon the arc intercepted between its sides. Thus, the angle AEI stands upon the arc AOI.
fig: 74 0.235 0.405 0.5 0.735
art: 221 | Corollary | The arc in which an angle is inscribed, and the arc intercepted between its sides, compose the whole circumference.
art: 222 | Theorem | An inscribed angle is measured by half of the intercepted arc. This demonstration also presents three cases. The center of the circle may be on one of the sides of the angle, or it may be inside, or it may be outside of the angle. 1st. One side of the angle, as AB, may be a diameter. Make the diameter DE, parallel to BC, the other side of the angle. Then the angle B is equal to its alternate angle BOD, which is measured by the arc BD. This arc is equal to CE, and also to EA. Therefore, the arc BD is equal to the half of AC, and the inscribed angle B is measured by half of its intercepted arc.
fig: 74 0.65 0.838 0.51 0.73
art: 222 | 2d case | The center of the circle may be within the angle. From the vertex B extend a diameter to the opposite side of the circumference at D. As just proved, the angle ABD is measured by half of the arc AD, and the angle DBC by half of the arc DC. Therefore, the sum of the two angles, or ABC, is measured by half of the sum of the two arcs, or half of the arc ADC.
fig: 75 0.228 0.37 0.505 0.74
art: 222 | 3d case | The center of the circle may be outside of the angle. Extend a diameter from the vertex as before. The angle ABC is equal to ABD diminished by DBC, and is, therefore, measured by half of the arc DA diminished by half of DC; that is, by the half of AC.
fig: 75 0.419 0.569 0.508 0.735
art: 223 | Corollary | When an inscribed angle and an angle at the center have the same intercepted arc, the inscribed angle is half of the angle at the center.
art: 224 | Corollary | All angles inscribed in the same arc are equal, for they have the same measure.
fig: 75 0.637 0.779 0.528 0.74
art: 225 | Corollary | Every angle inscribed in a semi-circumference is a right angle. If the arc is less than a semi-circumference, the angle is obtuse. If the arc is greater, the angle is acute.
art: 226 | Theorem | The angle formed by a tangent and a chord is measured by half the intercepted arc. The angle CEI, formed by the tangent AC and the chord EI, is measured by half the intercepted arc IDE. Through I, make the chord IO parallel to the tangent AC. The angle CEI is equal to its alternate EIO, which is measured by half the arc OME, which is equal to the arc IDE. Therefore, the angle CEI is measured by half the arc IDE. The sum of the angles AEI and CEI is two right angles, and is therefore measured by half the whole circumference. Hence, the angle AEI is measured by half the whole circumference diminished by half the arc IDE; that is, by half the arc IOME.
fig: 76 0.203 0.368 0.48 0.73
art: 228 | Theorem | Every angle whose vertex is within the circumference, is measured by half the sum of the arcs intercepted between its sides and its sides produced. Thus, the angle OAE is measured by half the sum of the arcs OE and IU. To be demonstrated by the student, using the previous theorems.
fig: 76 0.677 0.85 0.495 0.73
art: 229 | Theorem | Every angle whose vertex is outside of a circumference, and whose sides are either tangent or secant, is measured by half the difference of the intercepted arcs. Thus, the angle ACF is measured by half the difference of the arcs AF and AB; the angle FCG, by half the difference of the arcs FG and BI; and the angle ACE, by half the difference of the arcs AFGE and ADIE. This, also, may be demonstrated by the student, by the aid of the previous theorems on intercepted arcs.
fig: 77 0.24 0.46 0.465 0.74
art: 230 | Problem | Through a given point out of a circumference, to draw a tangent to the circumference. Let A be the given point, and C the center of the given circle. Join AC. Bisect AC at the point B. With B as a center and BC as a radius, describe a circumference. It will pass through C and A, and will cut the circumference in two points, D and E. Draw straight lines from A through D and E. AD and AE are both tangent to the given circumference. Join CD and CE. The angle CDA is inscribed in a semi-circumference, and is therefore a right angle. Since AD is perpendicular to the radius CD, it is tangent to the circumference. AE is tangent for the same reasons.
fig: 77 0.615 0.785 0.418 0.745
art: 231 | Problem | Upon a given chord to describe an arc which shall contain a given angle. Let AB be the chord, and C the angle. Make the angle DAB equal to C. At A erect a perpendicular to AD, and erect a perpendicular to AB at its center. Produce these till they meet at the point F. With F as a center, and FA as a radius, describe a circumference. Any angle inscribed in the arc BGHA will be equal to the given angle C. For AD, being perpendicular to the radius FA, is a tangent. Therefore, the angle BAD is measured by half of the arc AIB. But any angle contained in the arc AHGB is also measured by half of the same arc, and is therefore equal to BAD, which was made equal to C.
fig: 78 0.2 0.41 0.365 0.74
mc: Two parallel lines intercept ___ arcs of a circumference. | *equal | unequal | opposite
mc: An angle whose vertex is on the circumference and whose sides are chords is an... | *inscribed angle | angle at the center | exterior angle
mc: An inscribed angle is measured by... | *half of the intercepted arc | the whole intercepted arc | twice the intercepted arc
mc: An inscribed angle and an angle at the center stand on the same arc. The inscribed angle is... | *half of the angle at the center | double the angle at the center | equal to it
mc: All angles inscribed in the same arc are... | *equal | supplementary | right angles
mc: Every angle inscribed in a semi-circumference is... | *a right angle | acute | obtuse
mc: An angle inscribed in an arc less than a semi-circumference is... | *obtuse | acute | a right angle
mc: The angle formed by a tangent and a chord is measured by... | *half the intercepted arc | the whole intercepted arc | half the circumference
mc: An angle whose vertex is within the circumference is measured by... | *half the sum of the arcs between its sides and its sides produced | half their difference | the greater arc
mc: An angle whose vertex is outside the circumference is measured by... | *half the difference of the intercepted arcs | half their sum | the less arc
mc: The arc in which an angle is inscribed and the arc between its sides together make... | *the whole circumference | a semi-circumference | a quadrant
mc: To draw a tangent from a point A outside a circle with center C, first... | *bisect AC and describe a circumference on it | draw a diameter through A | draw a chord from A
tf: An angle inscribed in an arc greater than a semi-circumference is acute. {T}
tf: Two parallel tangents touch the circumference at the ends of a diameter. {T}
tf: An inscribed angle is equal to the angle at the center on the same arc. {F}
q: An inscribed angle stands on an arc of 80 degrees. How many degrees is the angle? {40}
sol: An inscribed angle is measured by half its intercepted arc: 80 / 2 = 40 degrees.
q: An angle at the center is 70 degrees. How many degrees is an inscribed angle standing on the same arc? {35}
sol: The inscribed angle is half of the angle at the center: 70 / 2 = 35 degrees.
q: How many degrees is an angle inscribed in a semi-circumference? {90}
sol: It stands on a semi-circumference of 180 degrees, and half of 180 is 90; it is a right angle.
q: A tangent and a chord make an angle which intercepts an arc of 100 degrees. How many degrees is the angle? {50}
sol: The angle formed by a tangent and a chord is measured by half the intercepted arc: 100 / 2 = 50.
q: Two chords cut each other inside a circle. The arcs between the sides of one angle and its sides produced are 70 degrees and 30 degrees. How many degrees is the angle? {50}
sol: Half the sum of the arcs: (70 + 30) / 2 = 50 degrees.
q: Two secants meet outside a circle. The intercepted arcs are 100 degrees and 40 degrees. How many degrees is the angle? {30}
sol: Half the difference of the arcs: (100 - 40) / 2 = 30 degrees.

== 20 | Circumferences | Positions of two circumferences
art: 232 | Theorem | Two circumferences can not cut each other in more than two points. For three points determine the position and extent of a circumference. Therefore, if two circumferences have three points common, they must coincide throughout.
art: 233 |  | Let us investigate the various positions which two circumferences may have with reference to each other. Let A and B be the centers of two circles, and let these points be joined by a straight line, which therefore measures the distance between the centers. First, suppose the sum of the radii to be less than AB. Then AC and BD, the radii, can not reach each other. At C and D, where the curves cut the line AB, let perpendiculars to that line be erected. These perpendiculars are parallel to each other. They are also tangent respectively to the two circumferences. It follows, therefore, that CD, the distance between these parallels, is also the least distance between the two curves.
fig: 79 0.15 0.274 0.475 0.735
art: 234 |  | Next, let the sum of the radii AC and BC be equal to AB, the distance between the centers. Then both curves will pass through the point C. At this point let a perpendicular be erected as before. This perpendicular CG is tangent to both the curves; that is, it is cut by neither of them. Therefore, the curves have only one common point C.
fig: 79 0.362 0.512 0.455 0.73
art: 235 |  | Next, let AB be less than the sum, but greater than the difference, of the radii AC and BD. Then the point C will fall within the circumference DF. For the same reasons, the point D will fall within the circumference CG. Therefore, these circumferences cut each other, and have two points common.
fig: 79 0.605 0.755 0.462 0.735
art: 236 |  | Next, let the difference between the two radii AC and BC be equal to the distance AB. A perpendicular to this line at the point C will be a tangent to both curves, and they have a common point at C. They have no other common point, for the two curves are both symmetrical about the line AC, and, therefore, if they had a common point on one side of that line, they would have a corresponding common point on the other side; but this can not be, for they would then have three points common.
fig: 80 0.19 0.345 0.445 0.76
art: 237 |  | Lastly, suppose the distance AB less than the difference of the radii AC and BD, by the line CD. That is, AB + BD + DC = AC. Join A, the center of the larger circle, with F, any point of the smaller circumference, and join BF. Then AB and BD are together equal to AB and BF, which are together greater than AF. Therefore, AD is greater than AF. Hence, the point D is farther from A than any other point of the circumference DF. It follows that CD is the least distance between the two curves.
fig: 80 0.482 0.632 0.42 0.735
art: 238 | Theorem | Two circumferences may have, with reference to each other, five positions: 1st. Each may be entirely exterior to the other, when the distance between their centers is greater than the sum of their radii. 2d. They may touch each other exteriorly, having one point common, when the distance between the centers is equal to the sum of the radii. 3d. They may cut each other, having two points common, when the distance between the centers is less than the sum and greater than the difference of the radii. 4th. One may be within the other and tangent, having one point common, when the distance between the centers is equal to the difference of the radii. 5th. One may be entirely within the other, when the distance between the centers is less than the difference of the radii.
art: 239 | Corollary | Two circumferences can not have more than one chord common to both.
art: 240 | Corollary | The common chord of two circumferences is perpendicular to the straight line which joins their centers and is bisected by it. For the ends of the chords are equidistant from each of the centers, the ends of the other line.
art: 241 | Corollary | When two circumferences are tangent to each other, the two centers and the point of contact are in one straight line.
art: 242 | Corollary | When two circumferences have no common point, the least distance between the curves is measured along the line which joins the centers.
art: 243 | Corollary | When the distance between the centers is zero, that is, when they coincide, a straight line through this point may have any direction in the plane; and the two curves are equidistant at all points. Such circles are called concentric.
art: 244 |  | A locus is a line or a surface all the points of which have some common property, which does not belong to any other points. Thus, the circumference of a circle is the locus of all those points in the plane, which are at a given distance from a given point. A straight line perpendicular to another at its center is the locus of all those points in the plane, which are at the same distance from both ends of the second line.
art: 245 | Suggestions | In his efforts to discover the solutions of the more difficult problems in drawing, the student will be much assisted by the following suggestions: 1. Suppose the problem solved, and the figure completed. 2. Find the geometrical relations of the different parts of the figure thus formed, drawing auxiliary lines, if necessary. 3. From the principles thus developed, make a rule for the solution of a problem. This is the analytic method of solving problems.
art: 245 | Exercises | From a given point, to draw the shortest line possible to a given straight line. With a given length of radius, to draw a circumference through two given points. In each of the five cases of the last theorem, how many straight lines can be tangent to both circumferences? The number is different for each case. What is the locus of the centers of those circles which have a given radius, and are tangent to a given circle?
mc: In how many points, at most, can two circumferences cut each other? | *Two | Three | Four
mc: How many positions can two circumferences have with reference to each other? | *Five | Three | Two
mc: When the distance between the centers is greater than the sum of the radii, the circumferences... | *are entirely exterior to each other | touch exteriorly | cut each other
mc: When the distance between the centers equals the sum of the radii, the circumferences... | *touch exteriorly, with one common point | cut in two points | have no common point
mc: When the distance between the centers is less than the sum and greater than the difference of the radii, the circumferences... | *cut each other in two points | touch | are one within the other
mc: When the distance between the centers equals the difference of the radii, one circumference is... | *within the other and tangent | entirely exterior | concentric
mc: Circles whose centers coincide are called... | *concentric | tangent | equal
mc: The common chord of two circumferences is ___ to the line joining their centers. | *perpendicular | parallel | equal
mc: When two circumferences are tangent, the two centers and the point of contact are... | *in one straight line | at the corners of a square | on a circle
mc: How many chords, at most, can two circumferences have in common? | *One | Two | Three
mc: A line or surface all of whose points have some common property, which no other points have, is a... | *locus | quadrant | limit
mc: The circumference of a circle is the locus of all points in the plane... | *at a given distance from a given point | on a given line | at equal distances from two points
mc: Supposing the problem solved, and studying the finished figure, is the ___ method of solving problems. | *analytic | indirect | infinite
tf: When two circumferences have no common point, their least distance is measured along the line joining the centers. {T}
tf: Two circumferences with three points in common must coincide. {T}
tf: Two concentric circles of different radii have one point in common. {F}
q: Two circles have radii of 5 and 3 inches, and their centers are 10 inches apart. How many points have they in common? {0}
sol: 10 is greater than the sum of the radii, 8, so each is entirely exterior to the other: no common point.
q: Two circles have radii of 5 and 3 inches, and their centers are 8 inches apart. How many points have they in common? {1}
sol: 8 equals the sum of the radii, so they touch exteriorly at one point.
q: Two circles have radii of 5 and 3 inches, and their centers are 6 inches apart. How many points have they in common? {2}
sol: 6 is less than the sum 8 and greater than the difference 2, so they cut each other in two points.
q: Two circles have radii of 5 and 3 inches, and their centers are 2 inches apart. How many points have they in common? {1}
sol: 2 equals the difference of the radii, so one is within the other and tangent, with one common point.
q: Two circles have radii of 5 and 3 inches, and their centers are 1 inch apart. How many points have they in common? {0}
sol: 1 is less than the difference of the radii, 2, so one is entirely within the other.
q: Two circles have radii of 5 and 3 inches, and their centers are 10 inches apart. What is the least distance between the two curves, in inches? {2}
sol: The least distance is along the line joining the centers: 10 - 5 - 3 = 2 inches.
q: How many straight lines can be tangent to both of two circumferences that are entirely exterior to each other? {4}
sol: Two lines touch both on the same side of each (outer tangents) and two cross between them (inner tangents): 4 in all.
q: How many straight lines can be tangent to both of two circumferences that cut each other? {2}
sol: Only the two outer tangents; no line can pass between the circles, since they overlap.

== 21 | Triangles | General properties of triangles
art: 246 |  | Next in regular order is the consideration of those plane figures which inclose an area; and, first, of those whose boundaries are straight lines. A polygon is a portion of a plane bounded by straight lines. The straight lines are the sides of the polygon. The perimeter of a polygon is its boundary, or the sum of all the sides.
art: 247 |  | A triangle is a polygon of three sides. Less than three straight lines can not inclose a surface, for two straight lines can have only one common point. Therefore, the triangle is the simplest polygon. From a consideration of its properties, those of all other polygons may be derived.
art: 248 | Problem | Any three points not in the same straight line may be made the vertices of the three angles of a triangle. For these points determine the plane, and straight lines may join them two and two, thus forming the required figure.
art: 249 | Corollary | Any three points of a circumference may be made the vertices of a triangle. A circumference may pass through the vertices of any triangle, for it may pass through any three points not in the same straight line.
art: 250 | Theorem | Within every triangle there is a point equally distant from the three sides. In the triangle ABC, let lines bisecting the angles A and B be produced until they meet. The point D, where the two bisecting lines meet, is equally distant from the two sides AB and BC, since it is a point of the line which bisects the angle B. For a similar reason, the point D is equally distant from the two sides AB and AC. Therefore, it is equally distant from the three sides of the triangle.
fig: 86 0.27 0.39 0.495 0.74
art: 251 | Corollary | The three lines which bisect the several angles of a triangle meet at one point. For the point D must be in the line which bisects the angle C.
art: 252 | Corollary | With D as a center, and a radius equal to the distance of D from either side, a circumference may be described, to which every side of the triangle will be a tangent.
art: 253 |  | When a circumference passes through the vertices of all the angles of a polygon, the circle is said to be circumscribed about the polygon, and the polygon to be inscribed in the circle. When every side of a polygon is tangent to a circumference, the circle is inscribed and the polygon circumscribed.
art: 254 |  | The angles at the ends of one side of a triangle are said to be adjacent to that side. Thus, the angles A and B are adjacent to the side AB. The angle formed by the other two sides is opposite. Thus, the angle A and the side BC are opposite to each other.
fig: 86 0.79 0.865 0.535 0.74
art: 255 | Theorem | The sum of the angles of a triangle is equal to two right angles. Let the line DE pass through the vertex of one angle, B, parallel to the opposite side, AC. Then the angle A is equal to its alternate angle DBA. For the same reason, the angle C is equal to the angle EBC. Hence, the three angles of the triangle are equal to the three consecutive angles at the point B, which are equal to two right angles. Therefore, the sum of the three angles of the triangle is equal to two right angles.
fig: 87 0.34 0.44 0.455 0.73
art: 256 | Corollary | Each angle of a triangle is the supplement of the sum of the other two.
art: 257 | Corollary | At least two of the angles of a triangle are acute.
art: 258 | Corollary | If two angles of a triangle are equal, they are both acute. If the three are equal, they are all acute, and each is two-thirds of a right angle.
art: 259 |  | An acute angled triangle is one which has all its angles acute, as a. A right angled triangle has one of the angles right, as b. An obtuse angled triangle has one of the angles obtuse, as c.
fig: 87 0.72 0.795 0.56 0.75
fig: 87 0.805 0.86 0.555 0.75
fig: 88 0.13 0.188 0.562 0.76
art: 260 | Corollary | In a right angled triangle, the two acute angles are complementary.
art: 261 | Corollary | If one side of a triangle be produced, the exterior angle thus formed, as BCD, is equal to the sum of the two interior angles not adjacent to it, as A and B. So much the more, the exterior angle is greater than either one of the interior angles not adjacent to it.
fig: 88 0.266 0.348 0.497 0.765
art: 262 | Corollary | If two angles of a triangle are respectively equal to two angles of another, then the third angles are also equal.
art: 263 |  | Either side of a triangle may be taken as the base. Then the vertex of the angle opposite the base is the vertex of the triangle. The altitude of the triangle is the distance from the vertex to the base, which is measured by a perpendicular let fall on the base produced, if necessary.
art: 264 | Corollary | The altitude of a triangle is equal to the distance between the base and a line through the vertex parallel to the base.
art: 265 |  | When one of the angles at the base is obtuse, the perpendicular falls outside of the triangle. When one of the angles at the base is right, the altitude coincides with the perpendicular side. When both the angles at the base are acute, the altitude falls within the triangle.
art: 266 | Theorem | Each side of a triangle is smaller than the sum of the other two, and greater than their difference. The first part of this theorem is an immediate consequence of the Axiom of Distance; that is, AC is less than AB + BC. Subtract AB from both members of this inequality, and AC - AB is less than BC. That is, BC is greater than the difference of the other sides.
fig: 89 0.24 0.318 0.495 0.73
art: 267 |  | An equilateral triangle is one which has three sides equal. An isosceles triangle is one which has only two sides equal. A scalene triangle is one which has no two sides equal.
art: 268 | Theorem | When two sides of a triangle are equal, the angles opposite to them are equal. If the triangle BCD is isosceles, the angles B and D, which are opposite the equal sides, are equal. Let the angle C be divided into two equal parts, and let the dividing line extend to the opposite side of the triangle at F. Then, that portion of the figure upon one side of this line may be turned upon it as upon an axis. Since the angle C was bisected, the line BC will fall upon DC; and, since these two lines are equal, the point B will fall upon D. But F, being a point of the axis, remains fixed; hence, BF and DF will coincide. Therefore, the angles B and D coincide, and are equal.
fig: 89 0.68 0.828 0.51 0.73
art: 269 | Corollary | The three angles of an equilateral triangle are equal.
art: 270 |  | In an isosceles triangle, the angle included by the equal sides is usually called the vertex of the triangle, and the side opposite to it the base.
art: 271 | Corollary | If a line pass through the vertex of an isosceles triangle, and also through the middle of the base, it will bisect the angle at the vertex, and be perpendicular to the base. The straight line which has any two of these four conditions must have the other two.
art: 272 | Theorem | When two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite to the less side. If in the triangle BCD the side BC is greater than DC, then the angle D is greater than the angle B. Let the line CF bisect the angle C, and be produced to the side BD. Then let the triangle CDF turn upon CF. CD will take the direction CB; but, since CD is less than CB, the point D will fall between C and B, at G. Join GF. Now, the angle FGC is equal to the angle D, because they coincide; and it is greater than the angle B, because it is exterior to the triangle BGF. Therefore, the angle D is greater than B.
fig: 90 0.645 0.787 0.517 0.745
art: 273 | Corollary | When one side of a triangle is not the largest, the angle which is opposite to that side is acute.
art: 274 | Corollary | In a scalene triangle, no two angles are equal.
art: 275 | Theorem | If two angles of a triangle are equal, the sides opposite them are equal. For if these sides were unequal, the angles opposite to them would be unequal, which is contrary to the hypothesis.
art: 276 | Corollary | If a triangle is equiangular, that is, has all its angles equal, then it is equilateral.
art: 277 | Theorem | If two angles of a triangle are unequal, the side opposite to the greater angle is greater than the side opposite to the less. If, in the triangle ABC, the angle C is greater than the angle A, then AB is greater than BC. For, if AB were not greater than BC, it would be either equal to it or less. If AB were equal to BC, the opposite angles A and C would be equal; and if AB were less than BC, then the angle C would be less than A; but both of these conclusions are contrary to the hypothesis. Therefore, AB being neither less than nor equal to BC, must be greater.
fig: 91 0.645 0.728 0.448 0.74
art: 278 | Corollary | In an obtuse angled triangle, the longest side is opposite the obtuse angle; and in a right angled triangle, the longest side is opposite the right angle.
art: 279 |  | The hypotenuse of a right angled triangle is the side opposite the right angle. The other two sides are called the legs. The student will notice that some of the above propositions are but different statements of the principles of perpendicular and oblique lines.
art: 280 | Exercises | 1. How many degrees are there in an angle of an equilateral triangle? 2. If one of the angles at the base of an isosceles triangle be double the angle at the vertex, how many degrees in each? 3. If the angle at the vertex of an isosceles triangle be double one of the angles at the base, what is the angle at the vertex? 4. To circumscribe a circle about a given triangle. 5. To inscribe a circle in a given triangle. 8. Can a triangular field have one side 436 yards, the second 547 yards, and the third 984 yards long?
mc: A portion of a plane bounded by straight lines is a... | *polygon | circle | triangle only
mc: The perimeter of a polygon is... | *the sum of all the sides | its longest side | its area
mc: The simplest polygon is the... | *triangle | square | circle
mc: The three lines which bisect the angles of a triangle meet at... | *one point | three points | no point
mc: The point where the bisectors of the angles meet is equally distant from the three... | *sides | vertices | angles
mc: When every side of a polygon is tangent to a circumference, the circle is... | *inscribed | circumscribed | concentric
mc: The sum of the angles of a triangle is equal to... | *two right angles | one right angle | four right angles
mc: At least how many angles of a triangle are acute? | *Two | One | Three
mc: A triangle which has one right angle is... | *right angled | acute angled | obtuse angled
mc: In a right angled triangle, the two acute angles are... | *complementary | supplementary | equal
mc: An exterior angle of a triangle is equal to... | *the sum of the two interior angles not adjacent to it | the adjacent interior angle | two right angles
mc: The distance from the vertex of a triangle to the base is its... | *altitude | perimeter | hypotenuse
mc: When one angle at the base is obtuse, the altitude falls... | *outside of the triangle | within the triangle | on a side
mc: Each side of a triangle is smaller than... | *the sum of the other two | the difference of the other two | either other side
mc: A triangle with only two sides equal is... | *isosceles | equilateral | scalene
mc: A triangle with no two sides equal is... | *scalene | isosceles | equilateral
mc: When two sides of a triangle are equal, the angles opposite them are... | *equal | unequal | right angles
mc: The angle opposite the greater of two sides is... | *the greater angle | the less angle | equal to the other
mc: In a right angled triangle, the longest side is opposite... | *the right angle | the smallest angle | either acute angle
mc: The side opposite the right angle is the... | *hypotenuse | leg | base always
tf: An equiangular triangle is also equilateral. {T}
tf: A triangle can have two obtuse angles. {F}
tf: In a scalene triangle, two angles may be equal. {F}
tf: A circumference may pass through the vertices of any triangle. {T}
q: How many degrees are there in an angle of an equilateral triangle? {60}
sol: The three angles are equal and together make two right angles, 180 degrees: 180 / 3 = 60.
q: Two angles of a triangle are 50 and 70 degrees. How many degrees is the third angle? {60}
sol: The sum is 180 degrees: 180 - 50 - 70 = 60.
q: In a right angled triangle, one acute angle is 35 degrees. How many degrees is the other? {55}
sol: The two acute angles are complementary: 90 - 35 = 55.
q: The two interior angles A and B of a triangle are 40 and 65 degrees. How many degrees is the exterior angle at C? {105}
sol: The exterior angle equals the sum of the two interior angles not adjacent to it: 40 + 65 = 105.
q: One angle at the base of an isosceles triangle is double the angle at the vertex. How many degrees is the angle at the vertex? {36}
sol: Call the vertex angle 1 part; each base angle is 2 parts; 5 parts = 180 degrees, so 1 part = 36 degrees.
q: The angle at the vertex of an isosceles triangle is double one of the angles at the base. How many degrees is the vertex angle? {90}
sol: Each base angle is 1 part, the vertex 2 parts; 4 parts = 180, so a part is 45 and the vertex is 90 degrees.
q: Can a triangular field have sides of 436, 547, and 984 yards? Answer 1 for yes, 0 for no. {0}
sol: 436 + 547 = 983, which is less than 984; each side must be smaller than the sum of the other two, so no.
q: Two sides of a triangle are 7 and 3 inches. The third side is a whole number of inches. What is the greatest it can be? {9}
sol: It must be less than 7 + 3 = 10, so the greatest whole number is 9 (and it must be greater than 7 - 3 = 4).

== 22 | Triangles | Equality of triangles
art: 281 |  | The three sides and three angles of a triangle may be called its six elements. It may be shown that three of these are always necessary, and they are generally enough, to determine the triangle.
art: 282 | Theorem | Two triangles are equal when the three sides of the one are respectively equal to the three sides of the other. Let the side BD be equal to AI, the side BC equal to AE, and CD to EI; then the two triangles are equal. Apply the line AI to its equal BD, so that the point A will fall upon B. Then I will fall upon D, since the lines are equal. Next, turn one of the triangles, if necessary, so that both shall fall on the same side of this common line. Now, the point A being on B, the points E and C are at the same distance from B, and therefore they are both in the circumference, which has B for its center, and BC or AE for its radius. For a similar reason, the points E and C are both in the circumference, which has D for its center and DC or IE for its radius. These two circumferences have only one point common on one side of the line BD, which joins their centers. Hence, E and C are both at this point. Therefore, AE coincides with BC, and EI with CD; that is, the two triangles coincide throughout, and are equal.
fig: 93 0.408 0.665 0.432 0.72
art: 283 |  | Every plane figure may be supposed to have two faces, which may be termed the upward and the downward faces. In order to place the triangle m upon l, we may conceive it to slide along the plane without turning over; but, in order to place n upon l, it must be turned over, so that its upward face will be upon the upward face of l. There are, then, two methods of superposition; the first, called direct, when the downward face of one figure is applied to the upward face of the other; and the second, called inverse, when the upward faces of the two are applied to each other.
art: 284 | Theorem | Two triangles are equal when they have two sides and the included angle of the one, respectively equal to two sides and the included angle of the other. If the angle A is equal to D, and the side AB to the side DF, and AC to DE, then the two triangles are equal. Apply the side AC to its equal DE, turning one triangle, if necessary, so that both shall fall upon the same side of that common line. Then, since the angles A and D are equal, AB must take the direction DF, and these lines being equal, B will fall upon F. Therefore, BC and FE, having two points common, coincide; and the two triangles coincide throughout, and are equal.
fig: 94 0.675 0.787 0.15 0.78
art: 285 | Theorem | Two triangles are equal when they have one side and two adjacent angles of the one, respectively equal to a side and the two adjacent angles of the other. If the triangles ABC and DEF have the side AC equal to DE, and the angle A equal to D, and C equal to E, then the triangles are equal. Apply the side AC to its equal DE, so that the vertices of the equal angles shall come together, A upon D, and C upon E. Then, since the angles A and D are equal, AB will take the direction DF, and the point B will fall somewhere in the line DF. Since the angles C and E are equal, CB will take the direction EF, and B will also be in the line EF. Therefore, B falls upon F, the only point common to the two lines DF and EF. Hence, the sides of the one triangle coincide with those of the other, and the two triangles are equal.
fig: 95 0.448 0.573 0.15 0.75
art: 286 | Theorem | Two triangles are equal when they have one side and any two angles of the one, respectively equal to the corresponding parts of the other. For the third angle of the first triangle must be equal to the third angle of the other. Then this becomes a case of the preceding theorem.
art: 287 | Theorem | Two triangles are equal when one of them has two sides, and the angle opposite to the side which is equal to or greater than the other, respectively equal to the corresponding parts of the other triangle. Let the sides AE and EI, EI being equal to or greater than AE, and the angle A, be respectively equal to the sides BC, CD, and the angle B. Then the triangles are equal. For the side AE may be placed on its equal BC. Then, since the angles A and B are equal, AI will take the direction BD, and the points I and D will both be in the common line BD. Since EI and CD are equal, the points I and D are both in the circumference whose center is at C, and whose radius is equal to CD. Now, this circumference cuts a straight line extending from B toward D in only one point; for B is either within or on the circumference, since BC is equal to or less than CD. Therefore, I and D are both at that point. Hence, AI and BD are equal, and the triangles are equal.
fig: 96 0.47 0.575 0.15 0.75
art: 288 | Corollary | Two triangles are equal when they have an obtuse or a right angle in the one, together with the side opposite to it, and one other side, respectively equal to those parts in the other triangle.
art: 289 | Corollary | In equal triangles each part of one is equal to the corresponding part of the other.
art: 290 | Corollary | In equal triangles the equal parts are similarly arranged, so that equal angles are opposite to equal sides.
art: 291 | Exceptions to the general rule | There are two exceptions. 1. When the three angles are given. For two very unequal triangles may have the angles of one equal to those of the other. 2. When two unequal sides and the angle opposite to the less are given. For with the sides AB and BC and the angle A given, there are two triangles, ABC and ABD.
fig: 97 0.535 0.638 0.375 0.72
fig: 97 0.725 0.818 0.474 0.73
art: 292 |  | The student may show that two parts alone are never enough to determine a triangle.
art: 293 | Theorem | When two triangles have two sides of the one respectively equal to two sides of the other, and the included angles unequal, the third side in that triangle which has the greater angle, is greater than in the other. Let BCD and AEI be two triangles, having BC equal to AE, and BD equal to AI, and the angle A less than B. Then, it is to be proved that CD is greater than EI. Apply the triangle AEI to BCD, so that AE will coincide with its equal BC. Since the angle A is less than B, the side AI will fall within the angle CBD. Let BG be its position, and EI will fall upon CG. Then let a line BF bisect the angle GBD. Join FG. The triangles GBF and BDF have the side BF common, the side GB equal to the side DB, since each is equal to AI, and the included angles GBF and DBF equal by construction. Therefore, the triangles are equal, and FG is equal to FD. Hence, CD, the sum of CF and FD, is equal to the sum of CF and FG, which is greater than CG. Therefore, CD is greater than CG, or its equal EI.
fig: 98 0.272 0.442 0.395 0.735
art: 294 | Theorem | Conversely, if two triangles have two sides of the one equal to two sides of the other, and the third sides unequal, then the angles opposite the third sides are unequal, and that is greater which is opposite the greater side. For if it were less, then the opposite side would be less, and if it were equal, then the opposite sides would be equal; both of which are contrary to the hypothesis.
art: 295 | Problem | To draw a triangle when the three sides are given. Let a, b, and c be the given lines. Draw the line IE equal to c. With I as a center, and with the line b as a radius, describe an arc, and with E as a center and the line a as a radius, describe a second arc, so that the two may cut each other. Join O, the point of intersection of these arcs, with I and with E. IOE is the required triangle. If c were greater than the sum of a and b, what would have been the result?
fig: 99 0.31 0.488 0.485 0.74
art: 297 | Problem | To draw a triangle, two sides and the included angle being given. Let a and b be the given lines, and E the angle. Draw FC equal to b. At C make an angle equal to E. Take DC equal to a, and join FD. Then FDC is a triangle having the required conditions.
fig: 99 0.662 0.825 0.49 0.745
art: 298 | Problem | To draw a triangle when one side and the adjacent angles are given. Let a be the given line, and D and E the angles. Draw BC equal to a. At B make an angle equal to D, and at C an angle equal to E. Produce the sides till they meet at the point F. FBC is a triangle having the given side and angles.
fig: 100 0.195 0.375 0.485 0.75
art: 299 | Problem | To draw a triangle when one side and two angles are given. If one of the angles is opposite the given side, find the supplement of the sum of the given angles. This will be the other adjacent angle. Then proceed as in Article 298.
art: 300 | Problem | To draw a triangle when two sides and an angle opposite to one of them are given. Construct an angle equal to the given angle. Lay off on one side of the angle the length of the given adjacent side. With the extremity of this adjacent side as a center, and with a radius equal to the side opposite the given angle, draw an arc. This arc may cut the opposite side of the angle. Join the point of intersection with the end of the adjacent side which was taken as a center. A triangle thus formed has the required conditions.
mc: The three sides and three angles of a triangle are called its... | *six elements | six vertices | six bases
mc: How many elements are always necessary to determine a triangle? | *Three | Two | Six
mc: Two triangles are equal when the three sides of one are equal to... | *the three sides of the other | the three angles of the other | two sides of the other
mc: Two triangles are equal when they have two sides and ___ of the one equal to those of the other. | *the included angle | any angle | the opposite angle to the less side
mc: Two triangles are equal when they have one side and ___ equal. | *the two adjacent angles | one angle | no angles
mc: Sliding one figure onto another without turning it over is superposition called... | *direct | inverse | indirect
mc: Turning one figure over so the upward faces meet is superposition called... | *inverse | direct | similar
mc: In equal triangles, equal angles are opposite to... | *equal sides | the longest side | the shortest side
mc: Which three parts do NOT always fix a triangle? | *The three angles | The three sides | Two sides and the included angle
mc: Two triangles have two sides equal, but the included angle is greater in the first. Then the third side of the first is... | *greater | less | equal
mc: How many parts alone are never enough to determine a triangle? | *Two | Three | Four
mc: To draw a triangle with three given sides, use... | *two arcs from the ends of one side | a protractor | parallel lines
tf: Two triangles with equal angles must be equal. {F}
tf: Two parts alone are never enough to determine a triangle. {T}
tf: Two unequal sides with the angle opposite the less may give two different triangles. {T}
tf: In equal triangles each part of one is equal to the corresponding part of the other. {T}
q: Two triangles are equal. In the first, the angle opposite a side of 5 inches is 40 degrees. How many degrees is the angle opposite the equal side of 5 inches in the second? {40}
sol: In equal triangles equal angles are opposite equal sides, so it is 40 degrees.
q: A triangle is to be drawn with sides of 3, 4, and 8 inches. How many triangles can be drawn? {0}
sol: 3 + 4 = 7 is less than 8; each side must be smaller than the sum of the other two, so the arcs do not meet and no triangle can be drawn.
q: To draw a triangle, one side and two angles of 50 and 60 degrees are given, the 60 degree angle being opposite the given side. How many degrees is the other angle adjacent to the given side? {70}
sol: It is the supplement of the sum of the given angles: 180 - (50 + 60) = 70 degrees.

== 23 | Triangles | Similar triangles
art: 302 |  | Similar magnitudes have been defined to be those which have the same form while they differ in extent.
art: 303 |  | Let the student bear in mind that the form of a figure depends upon the relative directions of its points, and that angles are differences in direction. Therefore, the definition may be stated thus: Two figures are similar when every angle that can be formed by lines joining points of one, has its corresponding equal and similarly situated angle in the other.
art: 304 | Theorem | Two triangles are similar, when the three angles of the one are respectively equal to the three angles of the other. This may appear to be only a case of the definition of similar figures; but it may be shown that every angle that can be made by any lines whatever in the one, may have its corresponding equal and similarly situated angle in the other. Let the angles A, B, and C be respectively equal to the angles D, E, and F. Suppose GH and IR to be any two lines in the triangle ABC. Join IC and GR. From F, the point homologous to C, extend FL, making the angle LFE equal to ICB. Now, the triangles LFE and ICB have the angles B and E equal, by hypothesis, and the angles at C and F equal, by construction. Therefore, the third angles, ELF and BIC, are equal. Then show, by reasoning in the same manner, that the angles at N are equal to the corresponding angles at G; and so on, throughout the two figures. Therefore, the relative directions of all their points are the same in both triangles; that is, they have the same form. Therefore, they are similar figures.
fig: 102 0.165 0.325 0.15 0.75
art: 305 | Corollary | Two similar triangles may be divided into the same number of triangles respectively similar, and similarly arranged.
art: 306 | Corollary | Two triangles are similar, when two angles of the one are respectively equal to two angles of the other. For the third angles must be equal also.
art: 307 | Corollary | If two sides of a triangle be cut by a line parallel to the third side, the triangle cut off is similar to the original triangle.
fig: 103 0.29 0.378 0.552 0.74
art: 308 | Theorem | Two triangles are similar, when the sides of one are parallel to those of the other; or, when the sides of one are perpendicular to those of the other. We know that the angles formed by lines which are parallel are either equal or supplementary; and that the same is true of angles whose sides are perpendicular. We will show that the angles can not be supplementary in two triangles. If even two angles of one triangle could be respectively supplementary to two angles of another, the sum of these four angles would be four right angles; and then the sum of all the angles of the two triangles would be more than four right angles, which is impossible. Hence, when two triangles have their sides respectively parallel or perpendicular, at least two of the angles of one triangle must be equal to two of the other. Therefore, the triangles are similar.
fig: 103 0.58 0.688 0.15 0.75
art: 309 | Theorem | One side of a triangle is to the homologous side of a similar triangle as any side of the first is to the homologous side of the second. If AE and BC are homologous sides of similar triangles, also EI and CD, then, AE : BC :: EI : CD. Take CF equal to EA, and CG equal to EI, and join FG. Then the triangles AEI and FCG are equal, and the angles CFG and CGF are respectively equal to the angles A and I, and therefore equal to the angles B and D. Hence, FG is parallel to BD. Suppose BC divided at the point F into parts which have the ratio of two whole numbers, for example, four and three. Then let the line CF be divided into four, and BF into three equal parts. Let lines parallel to BD extend from the several points of division till they meet CD. Since BC is divided into equal parts, the distances between these parallels are all equal. Therefore, CD is also divided into seven equal parts, of which CG has four. That is, CF : CB :: CG : CD :: 4 : 7. If the lines BC and CF have not the ratio of two whole numbers, CF and CG are the limits of lines which are commensurable with BC and CD. Hence, whatever be the ratio of CF to CB, it is the same as that of CG to CD. By substituting for CF and CG the equal lines AE and EI, we have, AE : BC :: EI : CD.
fig: 104 0.25 0.385 0.15 0.75
art: 310 | Corollary | The ratio is the same between any two homologous lines of two similar triangles.
art: 311 |  | This ratio of any side of a triangle to the homologous side of a similar triangle, is called the linear ratio of the two figures.
art: 312 | Corollary | The perimeters of similar triangles have the linear ratio of the two figures. For, AE : BC :: EI : CD :: IA : DB. Therefore, AE + EI + IA : BC + CD + DB :: AE : BC.
art: 313 | Corollary | If two sides of a triangle are cut by one or more lines parallel to the third side, the two sides are cut proportionally. For the triangles so formed are similar.
art: 314 | Corollary | When several parallel lines are cut by two secants, the secants are divided proportionally. For the secants being produced till they meet, form several similar triangles.
fig: 106 0.18 0.327 0.51 0.74
art: 315 | Theorem | If two sides of a triangle be cut proportionally by a straight line, the secant line is parallel to the third side. Let BCD be the triangle, and FG the secant. A line parallel to CD may pass through F, and such a line must divide BD in the same ratio as BC. But, by hypothesis, BD is so divided at the point G. Therefore, a line through F parallel to CD, must pass through G, and coincide with FG. Hence, FG is parallel to CD.
fig: 106 0.425 0.56 0.535 0.74
art: 316 | Theorem | Two triangles are similar when the ratios between each side of the one and a corresponding side of the other are the same. Suppose AE : BC :: EI : CD :: AI : BD. Take CF equal to EA and CG equal to EI, and join FG. Then, CF : CB :: CG : CD. Therefore, FG is parallel to BD, the triangles CFG and CBD are similar, and CF : CB :: FG : BD. But, by hypothesis, EA : CB :: AI : BD. Hence, since CF is equal to EA, FG is equal to AI, and the triangles AEI and FCG are equal. Therefore, the triangles AEI and BCD have their angles equal, and are similar.
fig: 106 0.685 0.804 0.393 0.75
art: 317 | Theorem | Two triangles are similar when two sides of the one have respectively to two sides of the other the same ratio, and the included angles are equal. Suppose AE : BC :: AI : BD; and let the angle A be equal to B. Take BF equal to AE, and BG equal to AI, and join FG. Then the triangles AEI and BFG are equal, and the angle BFG is equal to E, and BGF is equal to I. Since the sides of the triangle BCD are cut proportionally by FG, the angle BFG is equal to C, and BGF is equal to D. Therefore, the triangles AEI and BCD are mutually equiangular and similar.
fig: 107 0.36 0.483 0.33 0.72
art: 318 |  | If two similar triangles have two homologous lines equal, since all other homologous lines have the same ratio, they must also be equal, and consequently the two figures are equal. Thus, the equality of figures may be considered as a case of similarity.
art: 319 | Problem | To find a fourth proportional to three given straight lines. Let a be the given extreme, and b and c the given means. Take DG equal to a, the given extreme. Produce it, making DH equal to c, one of the means. From G draw GF equal to b. Then, from D draw a line through F, and from H a line parallel to GF. Produce these two lines till they meet at the point K. HK is the required fourth proportional. For the triangles DGF and DHK are similar. Hence, DG : GF :: DH : HK. That is, a : b :: c : HK.
art: 320 | Problem | To divide a given line into parts having a certain ratio. Let LD be the line to be divided into parts proportional to the lines a, b, and c. From L draw the line LE equal to the sum of a, b, and c, making LF equal to a, FG equal to b, and GE equal to c. Join DE, and draw GI and FH parallel to DE. LH, HI, and ID are the parts required.
fig: 108 0.425 0.607 0.445 0.75
art: 321 | Problem | To divide a given line into any number of equal parts. To divide the line AB into ten equal parts; draw AC indefinitely, and take on it ten equal parts. Join BC, and from the several points of division of AC, draw lines parallel to AB, and produce them to BC. The parallel nearest to AB is nine-tenths of AB, the next is eight-tenths, and so on. This also depends upon similarity of triangles.
fig: 108 0.712 0.8 0.47 0.75
art: 323 | Right angled triangles | Every triangle may be divided into two right angled triangles, by a perpendicular let fall from one of its vertices upon the opposite side. Thus the investigation of the properties of right angled triangles leads to many of the properties of triangles in general.
art: 324 | Theorem | If in a right angled triangle, a perpendicular be let fall from the vertex of the right angle upon the hypotenuse, then, 1. Each of the triangles thus formed is similar to the original triangle; 2. Either leg of the original triangle is a mean proportional between the hypotenuse and the adjacent segment of the hypotenuse; and, 3. The perpendicular is a mean proportional between the two segments of the hypotenuse. The triangles AEO and AEI have the angle A common, and the angles AEI and AOE are equal, being right angles. Therefore, these two triangles are similar. That the triangles EOI and EIA are similar, is proved by the same reasoning. Since the triangles are similar, the homologous sides are proportional, and we have AI : AE :: AE : AO; that is, the leg AE is a mean proportional between the whole hypotenuse and the segment AO which is adjacent to that leg. Lastly, the triangles AEO and EIO are similar, and therefore, AO : OE :: OE : OI. That is, the perpendicular is a mean proportional between the two segments of the hypotenuse.
fig: 109 0.632 0.737 0.47 0.72
art: 325 | Corollary | A perpendicular let fall from any point of a circumference upon a diameter, is a mean proportional between the two segments which it makes of the diameter.
fig: 110 0.355 0.443 0.535 0.775
art: 326 |  | In the several proportions just demonstrated, in place of the lines we may substitute those numbers which constitute the ratios. Indeed, it is only upon this supposition that the proportions have a meaning.
art: 327 | Theorem | The second power of the length of the hypotenuse is equal to the sum of the second powers of the lengths of the two legs of a right angled triangle. Let h be the hypotenuse, a the perpendicular let fall upon it, b and c the legs, and d and e the corresponding segments of the hypotenuse made by the perpendicular. By the second conclusion of the last theorem, we have h : b :: b : d, and h : c :: c : e. Hence, h d = b squared, and h e = c squared. By adding these two, h times (d + e) = b squared + c squared. But d + e = h. Therefore, h squared = b squared + c squared.
fig: 110 0.69 0.79 0.51 0.75
art: 328 | Theorem | If, in any triangle, a perpendicular be let fall from one of the vertices upon the opposite side as a base, then the whole base is to the sum of the other two sides, as the difference of those sides is to the difference of the segments of the base. Let a be the perpendicular, b the base, c and d the sides, and e and i the segments of the base. Then, two right angled triangles are formed, in one of which we have a squared + i squared = d squared; and in the other, a squared + e squared = c squared. Subtracting, i squared - e squared = d squared - c squared. Factoring, (i + e) (i - e) = (d + c) (d - c). Whence, i + e : d + c :: d - c : i - e.
fig: 111 0.355 0.418 0.365 0.71
art: 329 | Theorem | If a line bisects an angle of a triangle, it divides the opposite side in the ratio of the adjacent sides. If BF bisects the angle CBD, then CF : FD :: CB : BD. From C and from D, let perpendiculars DG and CH fall upon BF, and BF produced. Then, the triangles BDG and BCH are similar, for they have equal angles at B, by hypothesis, and at G and H, by construction. Hence, CB : BD :: CH : DG. But the triangles DGF and CHF are also mutually equiangular and similar. Hence, CF : FD :: CH : DG. Therefore, CF : FD :: CB : BD.
fig: 111 0.708 0.833 0.42 0.72
art: 330 | Problem in drawing | To find a mean proportional to two given straight lines. Make a straight line equal to the sum of the two. Upon this as a diameter, describe a semi-circumference. Upon this diameter, erect a perpendicular at the point of meeting of the two given lines. Produce this to the circumference. The line last drawn is the required line.
art: 331 | Theorem | If two chords of a circle cut each other, the parts of one may be the extremes, and the parts of the other the means, of a proportion. Join AD and CB. Then the two triangles AED and CEB have the angle A equal to the angle C, since they are inscribed in the same arc. For the same reason, the angles D and B are equal. Therefore, the triangles are similar; and we have AE : EC :: DE : EB.
fig: 112 0.62 0.795 0.515 0.745
art: 332 | Theorem | If from the same point, without a circle, two lines cutting the circumference extend to the farther side, then the whole of one secant and its exterior part may be the extremes, and the whole of the other secant and its exterior part may be the means, of a proportion. Joining BC and AD, the triangles AED and CEB are similar; for they have the angle E common, and the angles at B and D equal. Therefore, AE : EC :: DE : EB.
fig: 113 0.18 0.338 0.415 0.73
art: 333 | Corollary | If from the same point there be a tangent and a secant, the tangent is a mean proportional between the secant and its exterior part. For the tangent is the limit of all the secants which pass through the point of meeting.
art: 334 | Problem in drawing | To divide a given straight line into two parts so that one of them is a mean proportional between the whole line and the other part. This is called dividing a line in extreme and mean ratio. Let AC be the given line. At C erect a perpendicular, CI, equal to half of AC. Join AI. Take ID equal to CI, and then AB equal to AD. The line AC is divided at the point B in extreme and mean ratio. That is, AC : AB :: AB : BC.
fig: 113 0.57 0.722 0.368 0.72
art: 335 | Analysis and synthesis | Geometrical analysis is a process employed both for the discovery of the solution of problems and for the investigation of the truth of theorems. Analysis is the reverse of synthesis. Synthesis commences with certain principles, and proceeds by undeniable and successive inferences. The whole theory of geometry is an example of this method. In a word, analysis is used in geometry in order to discover truths, and synthesis to demonstrate the truths discovered.
art: 336 |  | A problem is said to be determinate when it admits of one definite solution. A problem is indeterminate when it admits of more than one definite solution. The solution of an indeterminate problem frequently amounts to finding a geometrical locus; as, to find a point equidistant from two given points; or, to find a point at a given distance from a given line.
art: 337 | Exercises | 2. The semi-perimeter of a triangle is greater than any one of the sides, and less than the sum of any two. 8. The middle point of a hypotenuse is equally distant from the three vertices of a right angled triangle. 20. If in a right angled triangle one of the acute angles is equal to twice the other, then the hypotenuse is equal to twice the shorter leg. 28. The perpendiculars erected upon the several sides of a triangle at their centers, meet in one point. 30. The altitudes of a triangle meet in one point.
mc: Two figures are similar when every angle formed by lines joining their points has... | *a corresponding equal angle in the other | a supplement in the other | a right angle in the other
mc: Two triangles are similar when the three angles of the one are... | *equal to the three angles of the other | supplements of the other | twice those of the other
mc: How many pairs of equal angles are enough to show two triangles similar? | *Two | One | Three are always needed
mc: A line parallel to one side of a triangle cuts off a triangle which is... | *similar to the original | equal to the original | a right angled triangle
mc: Two triangles whose sides are respectively parallel are... | *similar | always equal | never similar
mc: In similar triangles, the homologous sides are... | *proportional | equal | perpendicular
mc: The ratio of a side of a triangle to the homologous side of a similar triangle is the... | *linear ratio | perimeter | mean proportional
mc: The perimeters of similar triangles have the ratio of... | *their homologous sides | their areas | their angles
mc: If a line cuts two sides of a triangle proportionally, it is ___ to the third side. | *parallel | perpendicular | equal
mc: Two triangles with two sides proportional and the included angles equal are... | *similar | always equal | never similar
mc: In a right angled triangle, the perpendicular from the right angle to the hypotenuse is a mean proportional between... | *the two segments of the hypotenuse | the two legs | the hypotenuse and a leg
mc: Each leg of a right angled triangle is a mean proportional between the hypotenuse and... | *the adjacent segment of the hypotenuse | the other leg | the perpendicular
mc: The square of the hypotenuse equals... | *the sum of the squares of the two legs | the square of one leg | twice the square of a leg
mc: A line which bisects an angle of a triangle divides the opposite side in the ratio of... | *the adjacent sides | the angles | the altitudes
mc: When two chords cut each other, the parts of one are the extremes and the parts of the other are... | *the means of a proportion | equal to them | parallel to them
mc: A tangent from a point is a mean proportional between a secant from that point and... | *its exterior part | the radius | the diameter
mc: Dividing a line so one part is a mean proportional between the whole and the other part is dividing it in... | *extreme and mean ratio | equal parts | the linear ratio
mc: Reasoning from principles forward by successive inferences is... | *synthesis | analysis | superposition
tf: Equal triangles may be thought of as a case of similar triangles. {T}
tf: Two triangles can have two angles of one supplementary to two angles of the other. {F}
tf: When parallel lines are cut by two secants, the secants are divided proportionally. {T}
tf: A problem with more than one definite solution is called determinate. {F}
q: Two triangles are similar, with linear ratio 2 : 3. A side of the first is 8 inches. How long is the homologous side of the second, in inches? {12}
sol: 2 : 3 :: 8 : x, so 2 times x = 3 times 8 = 24, and x = 12.
q: Two similar triangles have homologous sides of 4 and 6 inches. The perimeter of the first is 18 inches. What is the perimeter of the second, in inches? {27}
sol: Perimeters have the linear ratio: 4 : 6 :: 18 : x, so x = 6 times 18 / 4 = 27.
q: The legs of a right angled triangle are 3 and 4 inches. How long is the hypotenuse, in inches? {5}
sol: The square of the hypotenuse is 9 + 16 = 25, and the square root of 25 is 5.
q: The legs of a right angled triangle are 6 and 8 feet. How long is the hypotenuse, in feet? {10}
sol: 36 + 64 = 100, and the square root of 100 is 10.
q: The hypotenuse of a right angled triangle is 13 inches and one leg is 5 inches. How long is the other leg, in inches? {12}
sol: 13 squared - 5 squared = 169 - 25 = 144, and the square root of 144 is 12.
q: The perpendicular from the right angle divides the hypotenuse into segments of 4 and 9 inches. How long is the perpendicular, in inches? {6}
sol: The perpendicular is a mean proportional between the segments: the square root of 4 times 9 = 36 is 6.
q: The chords AB and CD cut each other at E. AE is 4, EC is 6, and DE is 2 inches. In the proportion AE : EC :: DE : EB, how long is EB, in inches? {3}
sol: AE and EB, the parts of one chord, are the extremes; EC and DE are the means. 4 times EB = 6 times 2 = 12, so EB = 3.
q: From a point outside a circle, a secant of 9 inches has an exterior part of 4 inches. How long is a tangent from that point, in inches? {6}
sol: The tangent is a mean proportional between the secant and its exterior part: the square root of 9 times 4 = 36 is 6.
q: In triangle CBD, BF bisects the angle B. CB is 6 inches, BD is 9 inches, and CD is 10 inches. How long is CF, in inches? {4}
sol: CF : FD :: CB : BD = 6 : 9, so CD is 15 parts of which CF has 6: CF = 10 times 6 / 15 = 4.

== 24 | Quadrilaterals | General properties of quadrilaterals
art: 338 |  | In a polygon, two angles which immediately succeed each other in going round the figure, are called adjacent angles. A diagonal of a polygon is a straight line joining the vertices of any two angles which are not adjacent. Sometimes a diagonal is exterior, as the diagonal BD of the figure ABCD. A convex polygon has all its diagonals interior. A concave polygon has at least one diagonal exterior, as in the above diagram. Angles, such as BCD, are called reentrant.
fig: 119 0.43 0.545 0.53 0.72
art: 339 |  | A quadrilateral is a polygon of four sides.
art: 340 | Corollary | Every quadrilateral has two diagonals.
art: 341 | Corollary | An interior diagonal of a quadrilateral divides the figure into two triangles.
art: 342 | Theorem | Two quadrilaterals are equal when they are each composed of two triangles, which are respectively equal, and similarly arranged. For, since the parts are equal and similarly arranged, the wholes may be made to coincide.
art: 343 | Corollary | Conversely, two equal quadrilaterals may be divided into equal triangles similarly arranged. In every convex quadrilateral this division may be made in either of two ways.
art: 344 | Theorem | Two quadrilaterals are equal when the four sides and a diagonal of one are respectively equal to the four sides and the same diagonal of the other. By the same diagonal is meant the diagonal that has the same position with reference to the equal sides. For, since all their sides are equal, the triangles AEI and BCD are equal, also the triangles AIO and BDF. Therefore, the quadrilaterals are equal.
fig: 120 0.396 0.524 0.15 0.75
art: 345 | Theorem | Two quadrilaterals are equal when the four sides and an angle of the one are respectively equal to the four sides and the similarly situated angle of the other. By the similarly situated angle is meant the angle included by equal sides. For, if the sides AE, IE, and the included angle E are respectively equal to the sides BC, DC, and the included angle C, then the triangles AEI and BCD are equal; and AI is equal to BD. But since the three sides of the triangles AIO and BDF are respectively equal, the triangles are equal. Hence, the quadrilaterals are equal.
fig: 120 0.69 0.832 0.435 0.74
art: 346 | Theorem | The sum of the angles of a quadrilateral is equal to four right angles. For the angles of the two triangles into which every quadrilateral may be divided, are together coincident with the angles of the quadrilateral. Therefore, the sum of the angles of a quadrilateral is twice the sum of the angles of a triangle. In applying this theorem to a concave figure, the value of the reentrant angle must be taken on the side toward the polygon, and therefore as amounting to more than two right angles.
art: 347 | Problem | Any four points of a circumference may be joined by chords, thus making an inscribed quadrilateral.
art: 348 | Theorem | The opposite angles of an inscribed quadrilateral are supplementary. For the angle A is measured by half of the arc EIO, and the angle I by half of the arc EAO. Therefore, the two together are measured by half of the whole circumference, and their sum is equal to two right angles.
fig: 121 0.715 0.855 0.505 0.71
mc: Two angles which come one after the other in going round a polygon are... | *adjacent angles | vertical angles | opposite angles
mc: A straight line joining the vertices of two angles of a polygon which are not adjacent is a... | *diagonal | side | base
mc: A polygon with all its diagonals interior is... | *convex | concave | reentrant
mc: A polygon with at least one diagonal exterior is... | *concave | convex | regular
mc: A polygon of four sides is a... | *quadrilateral | triangle | pentagon
mc: How many diagonals has every quadrilateral? | *Two | One | Four
mc: An interior diagonal divides a quadrilateral into... | *two triangles | two quadrilaterals | four triangles
mc: The sum of the angles of a quadrilateral is... | *four right angles | two right angles | three right angles
mc: The opposite angles of an inscribed quadrilateral are... | *supplementary | complementary | equal
tf: Two quadrilaterals made of two equal triangles, similarly arranged, are equal. {T}
tf: A reentrant angle is greater than two right angles. {T}
tf: Four sides alone are enough to fix a quadrilateral. {F}
q: Three angles of a quadrilateral are 80, 90, and 100 degrees. How many degrees is the fourth? {90}
sol: The four angles make four right angles, 360 degrees: 360 - 80 - 90 - 100 = 90.
q: One angle of a quadrilateral inscribed in a circle is 70 degrees. How many degrees is the opposite angle? {110}
sol: Opposite angles of an inscribed quadrilateral are supplementary: 180 - 70 = 110.
q: How many degrees are in the sum of the angles of a quadrilateral? {360}
sol: Four right angles: 4 times 90 = 360.

== 25 | Quadrilaterals | Trapezoids
art: 349 |  | If two adjacent angles of a quadrilateral are supplemental, the remaining angles are also supplemental. Then, one pair of opposite sides must be parallel. A trapezoid is a quadrilateral which has two sides parallel. The parallel sides are called its bases.
fig: 122 0.19 0.275 0.44 0.75
art: 350 | Corollary | If the angles adjacent to one base of a trapezoid be equal, those adjacent to the other base must also be equal. For if A and D are equal, their supplements, B and C, must be equal.
art: 351 | Application | The figure described in the last corollary is symmetrical. For it can be divided into equal parts by a line joining the middle points of the bases. The symmetrical trapezoid is used in architecture, sometimes for ornament, and sometimes as the form of the stones of an arch.
fig: 122 0.52 0.608 0.545 0.75
art: 352 | Exercises | 1. To construct a quadrilateral when the four sides and one diagonal are given. For example, the side AB, 2 inches; the side BC, 5; CD, 3; DA, 4; and the diagonal AC, 6 inches. 4. The sum of two opposite sides of any quadrilateral which is circumscribed about a circle, is equal to the sum of the other two sides.
mc: A quadrilateral which has two sides parallel is a... | *trapezoid | rectangle | rhombus
mc: The parallel sides of a trapezoid are called its... | *bases | legs | diagonals
mc: If two adjacent angles of a quadrilateral are supplemental, then one pair of opposite sides is... | *parallel | equal | perpendicular
mc: If the angles adjacent to one base of a trapezoid are equal, those adjacent to the other base are... | *equal | supplementary to them | right angles
mc: A trapezoid whose angles at each base are equal is... | *symmetrical | concave | a parallelogram
mc: The symmetrical trapezoid is used as the form of the stones of... | *an arch | a wall | a floor
tf: In a trapezoid, the two angles next to one of the slanting sides are supplementary. {T}
tf: A symmetrical trapezoid is divided into equal parts by the line joining the middle points of its bases. {T}
q: In a trapezoid ABCD with bases AD and BC, the angle A is 65 degrees. How many degrees is the angle B? {115}
sol: A and B lie between the parallels on one side of the secant AB, so they are supplements: 180 - 65 = 115.
q: A quadrilateral circumscribed about a circle has sides 5, 7, and 6 inches, taken in order. How long is the fourth side, in inches? {4}
sol: The sums of opposite sides are equal: 5 + 6 = 7 + x, so x = 4.

== 26 | Quadrilaterals | Parallelograms
art: 353 |  | A parallelogram is a quadrilateral which has its opposite sides parallel.
art: 354 | Corollary | Two adjacent angles of a parallelogram are supplementary. The angles A and B, being between the parallels AD and BC, and on one side of the secant AB, are supplementary.
fig: 123 0.393 0.472 0.47 0.72
art: 355 | Corollary | The opposite angles of a parallelogram are equal. For both D and B are supplements of the angle C.
art: 356 | Theorem | The opposite sides of a parallelogram are equal. For, joining AC by a diagonal, the triangles thus formed have the side AC common; the angles ACB and DAC equal, for they are alternate; and ACD and BAC equal, for the same reason. Therefore, the triangles are equal, and the side AD is equal to BC, and AB to CD.
fig: 123 0.64 0.761 0.5 0.72
art: 357 | Corollary | When two systems of parallels cross each other, the parts of one system included between two lines of the other are equal.
fig: 123 0.78 0.86 0.55 0.73
art: 358 | Corollary | A diagonal divides a parallelogram into two equal triangles. But the diagonal does not divide the figure symmetrically, because the position of the sides of the triangles is reversed.
art: 359 | Theorem | If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. Join AC. Then, the triangles ABC and CDA are equal. For the side AD is equal to BC, and DC is equal to AB, by hypothesis; and they have the side AC common. Therefore, the angles DAC and BCA are equal. But these angles are alternate with reference to the lines AD and BC, and the secant AC. Hence, AD and BC are parallel, and, for a similar reason, AB and DC are parallel. Therefore, the figure is a parallelogram.
fig: 124 0.297 0.375 0.46 0.75
art: 360 | Theorem | If, in a quadrilateral, two opposite sides are equal and parallel, the figure is a parallelogram. If AD and BC are both equal and parallel, then AB is parallel to DC. For, joining BD, the triangles thus formed are equal, since they have the side BD common, the side AD equal to BC, and the angle ADB equal to its alternate DBC. Hence, the angle ABD is equal to BDC. But these are alternate with reference to the lines AB and DC, and the secant BD. Therefore, AB and DC are parallel, and the figure is a parallelogram.
fig: 124 0.6 0.69 0.5 0.755
art: 361 | Theorem | The diagonals of a parallelogram bisect each other. The diagonals AC and BD are each divided into equal parts at H, the point of intersection. For the triangles ABH and CDH have the sides AB and CD equal, the angles ABH and CDH equal, and the angles BAH and DCH equal. Therefore, the triangles are equal, and AH is equal to CH, and BH to DH.
fig: 125 0.153 0.243 0.385 0.73
art: 362 | Theorem | If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. To be demonstrated by the student.
art: 363 | Rectangle | If one angle of a parallelogram is right, the others must be right also. A rectangle is a right angled parallelogram. The rectangle has all the properties of other parallelograms, and the following peculiar to itself, which the student may demonstrate.
fig: 125 0.503 0.565 0.53 0.73
art: 364 | Theorem | The diagonals of a rectangle are equal.
art: 365 | Rhombus | When two adjacent sides of a parallelogram are equal, all its sides must be equal. A rhombus, or, as sometimes called, a lozenge, is a parallelogram having all its sides equal.
fig: 125 0.738 0.823 0.505 0.74
art: 366 | Theorem | The diagonals of a rhombus are perpendicular to each other.
art: 367 | Theorem | The diagonals of a rhombus bisect its angles.
art: 368 | Square | A square is a quadrilateral having its sides equal, and its angles right angles. The square may be shown to have all the properties of the parallelogram, of the rectangle, and of the rhombus.
art: 369 | Corollary | The rectangle and the square are the only parallelograms which can be inscribed in a circle.
art: 370 | Theorem | Two parallelograms are equal when two adjacent sides and the included angle in the one, are respectively equal to those parts in the other. For the remaining sides must be equal, and this becomes a case of Article 345.
art: 371 | Corollary | Two rectangles are equal when two adjacent sides of the one, are respectively equal to those parts of the other.
art: 372 | Corollary | Two squares are equal when a side of the one is equal to a side of the other.
art: 373 | Application | The rectangle is the most frequently used of all quadrilaterals. The walls and floors of our apartments, doors and windows, books, paper, and many other articles, have this form. Carpenters make an ingenious use of a geometrical principle in order to make their door and window-frames exactly rectangular. Having made the frame, with its sides equal and its ends equal, they measure the two diagonals, and make the frame take such a shape that these also will be equal. In this operation, what principle is applied?
art: 374 | Application | A rhombus inscribed in a rectangle is the basis of many ornaments used in architecture and other work.
fig: 127 0.183 0.25 0.5 0.72
art: 375 | Application | An instrument called parallel rulers, used in drawing parallel lines, consists of two rulers, connected by cross pieces with pins in their ends. The rulers may turn upon the pins, varying their distance. The distances between the pins along the rulers, that is, AB and CD, must be equal; also, along the cross pieces, that is, AC and BD. Then the rulers will always be parallel to each other.
fig: 127 0.29 0.405 0.43 0.73
art: 376 | Exercises | 3. What parts need be given for the construction of a rectangle? 4. What must be given for the construction of a square? 5. If the four middle points of the sides of any quadrilateral be joined by straight lines, those lines form a parallelogram.
mc: A quadrilateral which has its opposite sides parallel is a... | *parallelogram | trapezoid | rhombus only
mc: Two adjacent angles of a parallelogram are... | *supplementary | equal | complementary
mc: The opposite angles of a parallelogram are... | *equal | supplementary | right angles
mc: The opposite sides of a parallelogram are... | *equal | unequal | perpendicular
mc: A diagonal divides a parallelogram into... | *two equal triangles | two unequal triangles | two trapezoids
mc: If the opposite sides of a quadrilateral are equal, the figure is a... | *parallelogram | trapezoid | concave figure
mc: The diagonals of a parallelogram... | *bisect each other | are always equal | are always perpendicular
mc: A right angled parallelogram is a... | *rectangle | rhombus | trapezoid
mc: The diagonals of a rectangle are... | *equal | perpendicular | unequal
mc: A parallelogram having all its sides equal is a... | *rhombus | rectangle | trapezoid
mc: The diagonals of a rhombus are... | *perpendicular to each other | always equal | parallel
mc: A quadrilateral with its sides equal and its angles right angles is a... | *square | rhombus | trapezoid
mc: Which parallelograms can be inscribed in a circle? | *Only the rectangle and the square | Every parallelogram | Only the rhombus
mc: Carpenters make a frame exactly rectangular by making ___ equal. | *the two diagonals | the four angles by eye | two sides
mc: Two squares are equal when... | *a side of one equals a side of the other | they have one equal angle | they are both inscribed
tf: A rhombus may be called a lozenge. {T}
tf: The diagonals of a rhombus bisect its angles. {T}
tf: If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. {T}
tf: A diagonal divides a parallelogram symmetrically. {F}
q: One angle of a parallelogram is 70 degrees. How many degrees is an adjacent angle? {110}
sol: Two adjacent angles of a parallelogram are supplementary: 180 - 70 = 110.
q: One angle of a parallelogram is 70 degrees. How many degrees is the opposite angle? {70}
sol: The opposite angles of a parallelogram are equal.
q: Two adjacent sides of a parallelogram are 5 and 8 inches. What is its perimeter, in inches? {26}
sol: The opposite sides are equal: 5 + 8 + 5 + 8 = 26 inches.
q: A diagonal of a parallelogram is 14 inches. The other diagonal cuts it at H. How long is each part, in inches? {7}
sol: The diagonals bisect each other: 14 / 2 = 7 inches.
q: A rectangle is 6 inches by 8 inches. How long is each diagonal, in inches? {10}
sol: Each diagonal is the hypotenuse of a right angled triangle with legs 6 and 8: 36 + 64 = 100, and the root is 10.

== 27 | Quadrilaterals | Measure of area
art: 377 |  | The standard figure for the measure of surfaces is a square. That is, the unit of area is a square, the side of which is the unit of length, whatever be the extent of the latter. The square has been almost universally adopted, because 1. Its form is regular and simple; 2. The two dimensions of the square, its length and breadth, are the same; and, 3. A plane surface can be entirely covered with equal squares.
art: 378 |  | Any side of a polygon may be taken as the base. The altitude of a parallelogram is the distance between the base and the opposite side. Hence, the altitude of a parallelogram may be taken in either of two ways.
art: 379 | Theorem | The area of a rectangle is measured by the product of its base by its altitude. That is, if we multiply the number of units of length contained in the base, by the number of those units contained in the altitude, the product is the number of units of area contained in the surface. Suppose that the base AB and the altitude AD are multiples of the same unit of length, for example, four and three. Divide AB into four equal parts, and through all the points of division extend lines parallel to AD. Divide AD into three equal parts, and through the points of division extend lines parallel to AB. All the intercepted parts of these two sets of parallels must be equal; and all the angles, right angles. Thus, the whole rectangle is divided into equal squares. The number of these squares is equal to the number in one row multiplied by the number of rows; that is, the number of units of length in the base multiplied by the number in the altitude. In the example taken, this is three times four, or twelve.
fig: 129 0.188 0.312 0.515 0.735
art: 380 | Corollary | The area of a square is expressed by the second power of the length of its side. Anciently the principles of arithmetic were taught and illustrated by geometry, and we still find the word square in common use for the second power of a number.
art: 381 |  | If a straight line move in a direction perpendicular to itself, it describes a rectangle, one of whose dimensions is the given line, and the other is the distance which it has moved. A rectangle is said to be contained by its base and altitude. Thus, also, the area of any figure is called its superficial contents.
art: 382 | Application | All enlightened nations attach great importance to exact and uniform standard measures. The standards generally used for the measure of surface, are the square described upon a yard, a foot, a mile, or some other certain length; but the acre, one of the most common measures of surface, is an exception. The public lands belonging to the United States are divided into square townships, each containing thirty-six square miles, called sections.
art: 383 | Theorem | The area of a parallelogram is measured by the product of its base by its altitude. At the ends of the base AB erect perpendiculars, and produce them till they meet the opposite side, in the points E and F. Now the right angled triangles AED and BFC are equal, having the side BF equal to AE, since they are perpendiculars between parallels; and the side BC equal to AD, by hypothesis. If each of these equal triangles be subtracted from the entire figure, ABCE, the remainders ABFE and ABCD must be equivalent. But ABFE is a rectangle having the same base and altitude as the parallelogram ABCD. Hence, the area of the parallelogram is measured by the same product as that which measures the area of the rectangle.
fig: 131 0.205 0.315 0.435 0.72
art: 384 | Corollary | Any two parallelograms have their areas in the same ratio as the products of their bases by their altitudes. Parallelograms of equal altitudes have the same ratio as their bases, and parallelograms of equal bases have the same ratio as their altitudes.
art: 385 | Corollary | Two parallelograms are equivalent when they have equal bases and altitudes; or, when the two dimensions of the one are the extremes, and the two dimensions of the other are the means, of a proportion.
art: 386 | Theorem | The area of a triangle is measured by half the product of its base by its altitude. For any triangle is one-half of a parallelogram having the same base and altitude.
fig: 131 0.783 0.865 0.475 0.72
art: 387 | Corollary | The areas of triangles are in the ratio of the products of their bases by their altitudes.
art: 388 | Corollary | Two triangles are equivalent when they have equal bases and altitudes.
art: 389 | Corollary | If a parallelogram and a triangle have equal bases and altitudes, the area of the parallelogram is double that of the triangle.
art: 390 | Theorem | If from half the sum of the three sides of a triangle each side be subtracted, and if these remainders and the half sum be multiplied together, then the square root of the product will be the area of the triangle. Let DEF be any triangle, DF being the base and EG the altitude. Let DF = a, EF = b, DE = c, and DE + EF + FD = s. The area of the triangle is half the product of the base a by the altitude h. In this expression, we have the area of the triangle in terms of the three sides. For greater facility of calculation it is reduced to the following: area = one fourth of the square root of (a + b + c) (a + b - c) (a - b + c) (- a + b + c). But, by hypothesis, a + b + c = s. Substituting in the equation of the area, it becomes, area = the square root of s/2 times (s/2 - a) times (s/2 - b) times (s/2 - c).
fig: 132 0.408 0.49 0.4 0.78
art: 391 | Theorem | The areas of similar triangles are to each other as the squares of their homologous lines. Let AEI and BCD be similar triangles, and IO and DH homologous altitudes. Then, IO : DH :: AE : BC. Multiply by AE : BC :: AE : BC. Then, AE times IO : BC times DH :: AE squared : BC squared. But AE times IO : BC times DH :: area AEI : area BCD. Therefore, area AEI : area BCD :: AE squared : BC squared.
fig: 133 0.725 0.818 0.15 0.75
art: 392 | Theorem | The area of a trapezoid is equal to half the product of its altitude by the sum of its parallel sides. The trapezoid may be divided by a diagonal into two triangles, having for their bases the parallel sides. The altitude of each of these triangles is equal to that of the trapezoid. The area of each triangle being half the product of the common altitude by its base, the area of their sum, or of the whole trapezoid, is half the product of the altitude by the sum of the bases.
art: 393 | Exercises | 1. Measure the length and breadth, and find the area of the blackboard; of the floor. 2. To divide a given triangle into any number of equivalent triangles. 5. The area of a triangle is equal to half the product of the perimeter by the radius of the inscribed circle.
mc: The unit of area is a... | *square whose side is the unit of length | circle | triangle
mc: The area of a rectangle is measured by... | *the product of its base by its altitude | the sum of its sides | twice its base
mc: The area of a square is expressed by... | *the second power of its side | four times its side | twice its side
mc: The area of a parallelogram is measured by... | *the product of its base by its altitude | the product of two adjacent sides | half its base
mc: The area of a triangle is measured by... | *half the product of its base by its altitude | the product of its base by its altitude | the sum of its sides
mc: Two triangles with equal bases and equal altitudes are... | *equivalent | always equal | similar
mc: A parallelogram and a triangle have equal bases and altitudes. The parallelogram's area is... | *double the triangle's | equal to the triangle's | half the triangle's
mc: The areas of similar triangles are to each other as... | *the squares of their homologous lines | their homologous lines | their perimeters
mc: The area of a trapezoid is half the product of its altitude by... | *the sum of its parallel sides | one base | its two slanting sides
mc: Parallelograms of equal altitudes have the same ratio as their... | *bases | angles | diagonals
mc: A straight line moving perpendicular to itself describes a... | *rectangle | circle | triangle
mc: The area of any figure is also called its... | *superficial contents | perimeter | altitude
tf: A plane surface can be entirely covered with equal squares. {T}
tf: Two parallelograms with equal bases and altitudes are equivalent. {T}
tf: The acre is the square on some whole number of feet. {F}
q: A rectangle has a base of 4 feet and an altitude of 3 feet. How many square feet are in its area? {12}
sol: Base times altitude: 4 times 3 = 12 square feet.
q: How many square inches are in a square whose side is 9 inches? {81}
sol: The area of a square is the second power of its side: 9 times 9 = 81.
q: A parallelogram has a base of 10 inches and an altitude of 6 inches. What is its area, in square inches? {60}
sol: Base times altitude: 10 times 6 = 60.
q: A triangle has a base of 10 inches and an altitude of 6 inches. What is its area, in square inches? {30}
sol: Half the product of base and altitude: 10 times 6 / 2 = 30.
q: A trapezoid has parallel sides of 8 and 12 feet and an altitude of 5 feet. What is its area, in square feet? {50}
sol: Half the altitude times the sum of the parallel sides: 5 times (8 + 12) / 2 = 50.
q: The sides of a triangle are 13, 14, and 15 inches. What is its area, in square inches? {84}
sol: Half the sum of the sides is 21. The remainders are 21 - 13 = 8, 21 - 14 = 7, 21 - 15 = 6. 21 times 8 times 7 times 6 = 7056, and its square root is 84.
q: Two similar triangles have homologous sides of 2 and 3 inches. The area of the first is 8 square inches. What is the area of the second, in square inches? {18}
sol: The areas are as the squares of the sides, 4 : 9, so 4 : 9 :: 8 : x, and x = 9 times 8 / 4 = 18.
q: A township contains 36 sections, each a square mile. How many miles long is a side of the square township? {6}
sol: The area of a square is the square of its side, and 6 times 6 = 36.

== 28 | Quadrilaterals | Equivalent surfaces
art: 394 |  | Isoperimetrical figures are those whose perimeters have the same extent.
art: 395 | Theorem | Of all equivalent triangles of a given base, the one having the least perimeter is isosceles. The equivalent triangles having the same base, AE, have also the same altitude. Hence, their vertices are in the same line parallel to the base, that is, in DB. Now, the shortest line that can be made from A to E through some point of DB, will constitute the other two sides of the triangle of least perimeter. This shortest line is the one making equal angles with DB, as ACE, that is, making ACD and ECB equal. The angle ACD is equal to its alternate A, and the angle ECB to its alternate E. Therefore, the angles at the base are equal, and the triangle is isosceles.
fig: 135 0.44 0.546 0.46 0.72
art: 396 | Corollary | Of all isoperimetrical triangles of a given base, the one having the greatest area is isosceles.
art: 397 |  | To draw a square equivalent to a given figure, is called the squaring, or quadrature of the figure.
art: 398 | Problem | To draw a rectangle with a given base, equivalent to a given parallelogram. With the given base as a first term, and the base and altitude of the given figure as the second and third terms, find a fourth proportional. This is the required altitude.
art: 399 | Problem | To draw a square equivalent to a given parallelogram. Find a mean proportional between the base and altitude of the given figure. This is the side of the square.
art: 400 | Problem | To draw a triangle equivalent to a given polygon. Let ABCDE be the given polygon. Join DA. Produce BA, and through E draw EF parallel to DA. Join DF. Now, the triangles DAF and DAE are equivalent, for they have the same base DA, and equal altitudes, since their vertices are in the line EF parallel to the base. To each of these equals, add the figure ABCD, and we have the quadrilateral FBCD equivalent to the polygon ABCDE. In this manner, the number of sides may be diminished till a triangle is formed equivalent to the given polygon. In this diagram it is the triangle FDG.
fig: 136 0.437 0.595 0.43 0.76
art: 401 | Problem | To draw a square equivalent to a given triangle. Find a mean proportional between the altitude and half the base of the triangle. This will be the side of the required square.
art: 402 |  | Having shown how an area is expressed by the product of two lengths, it follows that an equation will represent equivalent surfaces, if each of its terms is composed of two factors which represent lengths. For example, let a and b represent the lengths of two straight lines. Now we know, from algebra, that whatever be the value of a and b, (a + b) squared = a squared + 2ab + b squared.
art: 403 | Theorem | The square described upon the sum of two lines is equivalent to the sum of the squares described on the two lines, increased by twice the rectangle contained by these two lines. Since the truths of algebra are universal in their application, this theorem is demonstrated by the truth of the above equation. Such a proof is called algebraic.
fig: 137 0.465 0.605 0.53 0.73
art: 404 | Theorem | The square described on the difference of two straight lines is equivalent to the sum of the squares described on the two lines, diminished by twice the rectangle contained by those lines. This is a consequence of the truth of the equation, (a - b) squared = a squared - 2ab + b squared.
art: 405 | Theorem | The rectangle contained by the sum and the difference of two straight lines is equivalent to the difference of the squares of those lines. This, again, is proved by the principle expressed in the equation, (a + b) (a - b) = a squared - b squared.
art: 406 |  | These two theorems may also be demonstrated by purely geometrical reasoning. The algebraic method is sometimes called the modern, while the other is called the ancient geometry. The algebraic method was invented by Descartes, in the seventeenth century, while the other is twenty centuries older.
art: 407 | The Pythagorean theorem | This is called the Pythagorean Theorem, because it was discovered by Pythagoras. It is also known as the Forty-seventh Proposition, that being its number in the First Book of Euclid's Elements. The fame of this theorem makes it proper to give here the demonstration from Euclid.
art: 408 | Theorem | The square described on the hypotenuse of a right angled triangle is equivalent to the sum of the squares described on the two legs. Let ABC be a right angled triangle, having the right angle BAC. Through A make AL parallel to BD, and join AD and FC. Then, because each of the angles BAC and BAG is a right angle, the line GAC is one straight line. For the same reason, BAH is one straight line. The angles FBC and DBA are equal, since each is the sum of a right angle and the angle ABC. The two triangles FBC and DBA are equal, for the side FB in the one is equal to BA in the other, and the side BC in the one is equal to BD in the other, and the included angles are equal. Now, the area of the parallelogram BL is double that of the triangle DBA, because they have the same base BD, and the same altitude DL. And the area of the square BG is double that of the triangle FBC, because these also have the same base BF, and the same altitude FG. But doubles of equals are equal. Therefore, the parallelogram BL and the square BG are equivalent. In the same manner, by joining AE and BK, it is demonstrated that the parallelogram CL and the square CH are equivalent. Therefore, the whole square BE, described on the hypotenuse, is equivalent to the two squares BG and CH, described on the legs of the right angled triangle.
fig: 139 0.145 0.418 0.375 0.72
art: 409 | Corollary | The square described on one leg is equivalent to the difference of the squares on the hypotenuse and the other leg.
art: 410 |  | If from the extremities of one line perpendiculars be let fall upon another, then the part of the second line between the perpendiculars is called the projection of the first line on the second. If one end of the first line is in the second, then only one perpendicular is necessary.
art: 411 | Theorem | The square described on the side opposite to an acute angle of a triangle, is equivalent to the sum of the squares described on the other two sides, diminished by twice the rectangle contained by one of these sides and the projection of the other on that side. Let A be the acute angle, and from B let a perpendicular fall upon AC, produced if necessary. Then, AD is the projection of AB upon AC. For, BD squared = AB squared - AD squared; and CD squared = AC squared + AD squared - 2 AC times AD. By addition, BD squared + CD squared = AB squared + AC squared - 2 AC times AD. But the square on BC is equivalent to BD squared + CD squared. Therefore, it is also equivalent to AB squared + AC squared - 2 AC times AD.
fig: 140 0.422 0.562 0.15 0.75
art: 412 | Theorem | The square described on the side opposite an obtuse angle of a triangle, is equivalent to the sum of the squares described on the other two sides, increased by twice the rectangle of one of those sides and the projection of the other on that side. In the triangle ABC, the square on BC which is opposite the obtuse angle at A, is equivalent to the sum of the squares on AB and on AC, and twice the rectangle contained by CA and AD. For, BD squared = AB squared - AD squared; and CD squared = AC squared + AD squared + 2 AC times AD. By addition, BD squared + CD squared = AB squared + AC squared + 2 AC times AD. But, BC squared = BD squared + CD squared. Therefore, BC squared is equivalent to AB squared + AC squared + 2 AC times AD.
fig: 141 0.263 0.395 0.44 0.675
art: 413 | Corollary | If the square described on one side of a triangle is equivalent to the sum of the squares described on the other two sides, then the opposite angle is a right angle. For the last two theorems show that it can be neither acute nor obtuse.
art: 414 | Exercises | 1. When a quadrilateral has its opposite angles supplementary, a circle can be circumscribed about it. 7. What is the area of a lot, which has the shape of a right angled triangle, the longest side being 100 yards, and one of the other sides 36 yards? 12. Of all equivalent parallelograms having equal bases, what one has the minimum perimeter?
mc: Figures whose perimeters have the same extent are... | *isoperimetrical | equivalent | similar
mc: Of all equivalent triangles on a given base, the one with the least perimeter is... | *isosceles | scalene | right angled
mc: Of all isoperimetrical triangles on a given base, the one with the greatest area is... | *isosceles | scalene | obtuse angled
mc: Drawing a square equivalent to a given figure is called its... | *quadrature | projection | rectification
mc: The side of a square equivalent to a parallelogram is a mean proportional between... | *the base and the altitude | the two diagonals | two adjacent sides
mc: The square on the sum of two lines equals the sum of their squares increased by... | *twice their rectangle | their rectangle | four times their rectangle
mc: The rectangle of the sum and difference of two lines equals... | *the difference of their squares | the sum of their squares | the square of their sum
mc: The algebraic method of geometry was invented by... | *Descartes | Euclid | Pythagoras
mc: The square on the hypotenuse is equivalent to... | *the sum of the squares on the two legs | twice the square on one leg | the rectangle of the legs
mc: The Pythagorean theorem is the ___ proposition of Euclid's First Book. | *forty-seventh | first | tenth
mc: The part of a second line between perpendiculars let fall from the ends of a first line is the first line's... | *projection | altitude | diagonal
mc: The square on the side opposite an obtuse angle is ___ the sum of the squares on the other two sides. | *greater than | less than | equal to
mc: If the square on one side equals the sum of the squares on the other two, the opposite angle is... | *a right angle | acute | obtuse
tf: A proof made by the truth of an algebraic equation is called algebraic. {T}
tf: The square on the side opposite an acute angle is greater than the sum of the squares on the other two sides. {F}
tf: The square on one leg equals the difference of the squares on the hypotenuse and the other leg. {T}
q: Two lines are 3 and 2 inches. How many square inches is the square on their sum? {25}
sol: (3 + 2) squared = 9 + 4 + 2 times 3 times 2 = 25.
q: Two lines are 7 and 3 inches. How many square inches is the rectangle of their sum and difference? {40}
sol: It equals the difference of their squares: 49 - 9 = 40; and (7 + 3) times (7 - 3) = 40.
q: The legs of a right angled triangle are 5 and 12 inches. How long is the hypotenuse, in inches? {13}
sol: 25 + 144 = 169, and the square root of 169 is 13.
q: In triangle ABC the angle A is acute. AB is 13, AC is 14, and the projection AD of AB on AC is 5. How long is BC? {15}
sol: BC squared = AB squared + AC squared - 2 AC times AD = 169 + 196 - 140 = 225, and the root is 15.
q: In triangle ABC the angle A is obtuse. AB is 10, AC is 9, and the projection AD of AB on CA produced is 6. How long is BC? {17}
sol: BC squared = AB squared + AC squared + 2 AC times AD = 100 + 81 + 108 = 289, and the root is 17.
q: The sides of a triangle are 6, 8, and 10 inches. Is the angle opposite the 10 inch side a right angle? Answer 1 for yes, 0 for no. {1}
sol: 36 + 64 = 100, the square of 10; so the opposite angle is a right angle.
q: A parallelogram has a base of 9 inches and an altitude of 4 inches. How long is the side of an equivalent square, in inches? {6}
sol: The side is a mean proportional between base and altitude: the square root of 9 times 4 = 36 is 6.

== 29 | Polygons | General properties of polygons
art: 415 |  | Hitherto the student's attention has been given to polygons of three and of four sides only. In the present chapter, some principles of polygons of any number of sides will be established. A pentagon is a polygon of five sides; a hexagon has six sides; an octagon, eight; a decagon, ten; a dodecagon, twelve; and a pentedecagon, fifteen.
art: 416 | Theorem | The number of diagonals from any vertex of a polygon, is three less than the number of sides. For, from each vertex a diagonal may extend to every other vertex except itself, and the one adjacent on each side. Thus, the number is three less than the number of vertices, or of sides.
art: 417 | Corollary | The diagonals from one vertex divide a polygon into as many triangles as the polygon has sides, less two. Polygons may be divided into this number, or into a greater number of triangles, in various ways; but a polygon can not be divided into a less number of triangles than here stated.
art: 418 | Corollary | The whole number of diagonals possible in a polygon of n sides, is one half of n times (n - 3). For, if we count the diagonals at all the n vertices, we have n times (n - 3), but this is counting each diagonal at both ends. This last product must therefore be divided by two.
art: 419 | Theorem | Two polygons are equal when they are composed of the same number of triangles respectively equal and similarly arranged. This is an immediate consequence of the definition of equality.
art: 420 | Corollary | Conversely, two equal polygons may be divided into the same number of triangles respectively equal and similarly arranged.
art: 421 | Theorem | Two polygons are equal when all the sides and all the diagonals from one vertex of the one, are respectively equal to the same lines in the other, and are similarly arranged. For each triangle in the one would have its three sides equal to the similarly situated triangle in the other, and would be equal to it. Therefore, the polygons would be equal.
art: 422 | Theorem | Two polygons are equal when all the sides and the angles of the one are respectively equal to the same parts of the other, and are similarly arranged. For each triangle in the one is equal to its homologous triangle in the other, since they have two sides and the included angle equal. It is enough for the hypothesis of this theorem, that all the angles except three be among the equal parts.
art: 423 | Theorem | The sum of all the angles of a polygon is equal to twice as many right angles as the polygon has sides, less two. For the polygon may be divided into as many triangles as it has sides, less two; and the angles of these triangles coincide altogether with those of the polygon. The sum of the angles of each triangle is two right angles. Therefore, the sum of the angles of the polygon is equal to twice as many right angles as it has sides, less two.
art: 424 |  | Let R represent a right angle; then the sum of the angles of a polygon of n sides is 2 (n - 2) R; or, it may be written thus, (2n - 4) R.
art: 425 | Theorem | If each side of a convex polygon be produced, the sum of all the exterior angles is equal to four right angles. Let the sides be produced all in one way; that is, all to the right or all to the left. Then, from any point in the plane, extend lines parallel to the sides thus produced, and in the same directions. The angles thus formed are equal in number to the exterior angles of the polygon, and are respectively equal to them. But the sum of those formed about the point is equal to four right angles. Therefore, the sum of the exterior angles of the polygon is equal to four right angles.
fig: 146 0.152 0.298 0.405 0.76
art: 426 |  | This theorem will also be true of concave polygons, if the angle formed by producing one side of the reentrant angle is considered as a negative quantity. Thus, the remainder, after subtracting the angle formed at A by producing GA, from the sum of the angles formed at B, C, D, E, F, and G, is four right angles.
fig: 146 0.425 0.555 0.49 0.75
art: 427 | Exercises | 1. What is the number of diagonals that can be in a pentagon? In a decagon? 2. What is the sum of the angles of a hexagon? Of a dodecagon? 3. What is the greatest number of acute angles which a convex polygon can have?
art: 428 | Problem | To draw a polygon equal to a given polygon. By diagonals divide the given polygon into triangles. The problem then consists in drawing triangles equal to given triangles.
art: 429 | Problem | To draw a polygon when all its sides and all the diagonals from one vertex, are given in their proper order. This consists in drawing triangles with sides equal to three given lines.
art: 430 | Problem | To draw a polygon when the sides and angles are given in their order. It is enough for this problem if all the angles except three be given. For, suppose first that the angles not given are consecutive, as at D, B, and C. Then, draw the triangles a, e, i, and o. Then, having DC, complete the polygon by drawing the triangle DBC from its three known sides.
fig: 147 0.425 0.56 0.515 0.73
mc: A polygon of five sides is a... | *pentagon | hexagon | octagon
mc: A polygon of six sides is a... | *hexagon | pentagon | decagon
mc: A polygon of eight sides is an... | *octagon | hexagon | dodecagon
mc: A dodecagon has how many sides? | *Twelve | Ten | Fifteen
mc: The number of diagonals from one vertex of a polygon is... | *three less than the number of sides | two less than the number of sides | equal to the number of sides
mc: The diagonals from one vertex divide a polygon into as many triangles as it has sides, less... | *two | three | one
mc: The sum of the angles of a polygon is twice as many right angles as it has sides, less... | *two | three | four
mc: The sum of the exterior angles of a convex polygon is... | *four right angles | two right angles | as many right angles as sides
mc: To draw a polygon equal to a given one, divide the given polygon by diagonals into... | *triangles | squares | parallelograms
tf: Two polygons made of the same number of equal triangles, similarly arranged, are equal. {T}
tf: A polygon can be divided into fewer triangles than its sides less two. {F}
tf: To fix a polygon by its sides and angles, all the angles except three are enough. {T}
q: How many diagonals can be in a pentagon? {5}
sol: One half of n times (n - 3): 5 times 2 / 2 = 5.
q: How many diagonals can be in a decagon? {35}
sol: One half of 10 times 7 = 35.
q: How many diagonals can be drawn from one vertex of an octagon? {5}
sol: Three less than the number of sides: 8 - 3 = 5.
q: Into how many triangles do the diagonals from one vertex divide an octagon? {6}
sol: As many as the sides, less two: 8 - 2 = 6.
q: What is the sum of the angles of a hexagon, in degrees? {720}
sol: 2 (n - 2) right angles = 2 times 4 = 8 right angles = 8 times 90 = 720 degrees.
q: What is the sum of the angles of a dodecagon, in degrees? {1800}
sol: 2 (12 - 2) = 20 right angles = 20 times 90 = 1800 degrees.
q: What is the greatest number of acute angles which a convex polygon can have? {3}
sol: Each acute angle has an obtuse exterior angle, greater than one right angle. The exterior angles together make only four right angles, so there can be at most 3 acute angles.

== 30 | Polygons | Similar polygons
art: 431 | Theorem | Similar polygons are composed of the same number of triangles, respectively similar and similarly arranged. Since the figures are similar, every angle in one has its corresponding equal angle in the other. If, then, diagonals be made to divide one of the polygons into triangles, every angle thus formed may have its corresponding equal angle in the other. Therefore, the triangles of one polygon are respectively similar to those of the other, and are similarly arranged.
fig: 147 0.733 0.85 0.15 0.75
art: 432 | Theorem | If two polygons are composed of the same number of triangles which are respectively similar and are similarly arranged, the polygons are similar. By the hypothesis, all the angles formed by the given lines in one polygon have their corresponding equal angles in the other. It remains to be proved that angles formed by any other lines in the one have their corresponding equal angles in the other polygon. This may be shown by reasoning, in the same manner as in the case of triangles.
art: 433 | Theorem | Two polygons are similar when the angles formed by the sides are respectively equal, and there is the same ratio between each side of the one and its homologous side of the other. Let all the diagonals possible extend from a vertex A of one polygon, and the same from the homologous vertex B of the other polygon. Now the triangles AEI and BCD are similar, because they have two sides proportional, and the included angles equal. Therefore, EI : CD :: AI : BD. But, by hypothesis, EI : CD :: IO : DF. Then, AI : BD :: IO : DF. Also, if we subtract the equal angles EIA and CDB from the equal angles EIO and CDF, the remainders AIO and BDF are equal. Hence, the triangles AIO and BDF are similar. In the same manner, prove that each of the triangles of the first polygon is similar to its corresponding triangle in the other. Therefore, the figures are similar.
fig: 148 0.63 0.755 0.15 0.75
art: 434 | Theorem | In similar polygons the ratio of two homologous lines is the same as of any other two homologous lines. For, since the polygons are similar, the triangles which compose them are also similar, and AE : BC :: EI : CD :: AI : BD :: IO : DF, and so on. This common ratio is the linear ratio of the two figures.
fig: 149 0.548 0.697 0.15 0.75
art: 435 | Theorem | The perimeters of similar polygons are to each other as any two homologous lines.
art: 436 | Theorem | The area of any polygon is to the area of a similar polygon, as the square on any line of the first is to the square on the homologous line of the second. Let the polygons BCD, and so on, and AEI, and so on, be divided into triangles by homologous diagonals. The triangles thus formed in the one are similar to those formed in the other. Therefore, the areas of the homologous triangles have the same ratio, the squares of homologous lines. Therefore, the whole polygons have the same ratio as the areas of any two homologous parts; or, as the squares of any homologous lines.
fig: 150 0.34 0.518 0.325 0.73
art: 437 | Corollary | The superficial ratio of two similar polygons is always the second power of their linear ratio.
art: 438 | Exercises | 2. To draw a triangle similar to a given triangle, but with double the area. 3. What is the relation between the areas of the equilateral triangles described on the three sides of a right angled triangle?
mc: Similar polygons are made of the same number of triangles which are... | *similar and similarly arranged | equal | equivalent
mc: Two polygons are similar when their angles are equal and their homologous sides are... | *proportional | equal | parallel
mc: In similar polygons, the common ratio of homologous lines is the... | *linear ratio | superficial ratio | perimeter
mc: The perimeters of similar polygons are to each other as... | *any two homologous lines | the squares of homologous lines | their areas
mc: The areas of similar polygons are to each other as... | *the squares of homologous lines | homologous lines | their perimeters
mc: The superficial ratio of two similar polygons is ___ of their linear ratio. | *the second power | the square root | one half
mc: For two polygons to be similar, how many angles may be left out of the equal parts? | *Three | One | None
tf: Two similar polygons may be divided into similar triangles similarly arranged. {T}
tf: If the linear ratio of two similar polygons is 1 : 2, their areas are as 1 : 2. {F}
tf: The equilateral triangle on the hypotenuse of a right angled triangle equals the sum of those on the two legs. {T}
q: Two similar polygons have homologous sides of 3 and 5 inches. The perimeter of the first is 24 inches. What is the perimeter of the second, in inches? {40}
sol: Perimeters are as homologous lines: 3 : 5 :: 24 : x, so x = 5 times 24 / 3 = 40.
q: Two similar polygons have homologous sides of 2 and 5 feet. The area of the first is 12 square feet. What is the area of the second, in square feet? {75}
sol: Areas are as the squares of homologous lines, 4 : 25, so x = 25 times 12 / 4 = 75.
q: The linear ratio of two similar polygons is 1 : 3. The area of the larger is how many times the area of the smaller? {9}
sol: The superficial ratio is the second power of the linear ratio: 3 times 3 = 9.

== 31 | Polygons | Regular polygons
art: 439 |  | A regular polygon is one which has all its sides equal, and all its angles equal. The square and the equilateral triangle are regular polygons.
art: 440 | Theorem | Within a regular polygon there is a point equally distant from the vertices of all the angles. Let ABCD, and so on, be a regular polygon, and let lines bisecting the angles A and B extend till they meet at O. In the triangle ABO, the angles at A and B are equal, being halves of the equal angles of the polygon. Therefore, the opposite sides AO and BO are equal. Join OC. Now, the triangles ABO and BCO are equal, for they have the side AO of the first equal to BO of the second, the side AB equal to BC, because the polygon is regular, and the included angles OAB and OBC equal, since they are halves of angles of the polygon. Hence, BO is equal to OC. In the same manner, it is proved that OC is equal to OD, and so on. Therefore, the point O is equally distant from all the vertices.
fig: 151 0.505 0.648 0.48 0.75
art: 441 | Corollary | Every regular polygon may have a circle circumscribed about it. For, with O as a center and OA as a radius, a circumference may be described passing through all the vertices of the polygon.
art: 442 | Theorem | The point which is equally distant from the vertices is also equally distant from the sides of a regular polygon. The triangles OAB, OBC, and so on, are all isosceles. If perpendiculars be let fall from O upon the several sides, these sides will be bisected. Then, the perpendiculars will be equal, for they will be sides of equal triangles. But they measure the distances from O to the several sides of the polygon. Therefore, the point O is equally distant from all the sides of the polygon.
fig: 152 0.445 0.572 0.495 0.74
art: 443 | Corollary | Every regular polygon may have a circle inscribed in it.
art: 444 | Corollary | A regular polygon is a symmetrical figure.
art: 445 |  | The center of the circumscribed or inscribed circle is also called the center of a regular polygon. The radius of the circumscribed circle is also called the radius of a regular polygon. The apothem of a regular polygon is the radius of the inscribed circle.
art: 446 | Theorem | If the circumference of a circle be divided into equal arcs, the chords of those equal arcs will be the sides of a regular polygon. For the sides are all equal, being the chords of equal arcs; and the angles are all equal, being inscribed in equal arcs.
art: 447 | Corollary | An angle formed at the center of a regular polygon by lines from adjacent vertices, is an aliquot part of four right angles, being the quotient of four right angles divided by the number of the sides of the polygon.
art: 448 | Theorem | If a circumference be divided into equal arcs, and lines tangent at the several points of division be produced until they meet, these tangents are the sides of a regular polygon. Let A, B, C, and so on, be points of division, and F, D, and E points where the tangents meet. Join GA, AB, and BC. Now, the triangles GAF, ABD, and BCE have the sides GA, AB, and BC equal, as they are chords of equal arcs; and the angles at G, A, B, and C equal, for each is formed by a tangent and chord which intercept equal arcs. Therefore, these triangles are all isosceles, and all equal; and the angles F, D, and E are equal. Also, FD and DE, being doubles of equals, are equal. Therefore, it is a regular polygon.
fig: 153 0.6 0.8 0.455 0.71
art: 449 | Theorem | Regular polygons of the same number of sides are similar. Since the polygons have the same number of sides, the sum of all the angles of the one is equal to the sum of all the angles of the other. But all the angles of a regular polygon are equal. Dividing the equal sums by the number of angles, it follows that an angle of the one polygon is equal to an angle of the other. Again: all the sides of a regular polygon are equal. Hence, there is the same ratio between a side of the first and a side of the second, as between any other side of the first and a corresponding side of the second. Therefore, the polygons are similar.
art: 450 | Corollary | The areas of two regular polygons of the same number of sides are to each other as the squares of their homologous lines.
art: 451 | Corollary | The ratio of the radius to the side of a regular polygon of a given number of sides, is a constant quantity.
art: 452 | Corollary | The same is true of the apothem and side, or of the apothem and radius.
art: 453 | Problem | To inscribe a square in a given circle. Draw two diameters perpendicular to each other. Join their extremities by chords. These chords form an inscribed square.
art: 454 | Problem | To inscribe a regular hexagon in a circle. Suppose the problem solved and the figure completed. Join two adjacent angles with the center, making the triangle ABC. Now, the angle C, being measured by one-sixth of the circumference, is equal to one-sixth of four right angles, or one-third of two right angles. Hence, the sum of the two angles, CAB and CBA, is two-thirds of two right angles. But CA and CB are equal, being radii; therefore, the angles CAB and CBA are equal, and each of them must be one-third of two right angles. Then, the triangle ABC being equiangular, is equilateral. Therefore, the side of an inscribed regular hexagon is equal to the radius of the circle. The solution of the problem is now evident: apply the radius to the circumference six times as a chord.
fig: 155 0.378 0.553 0.475 0.74
art: 455 | Corollary | Joining the alternate vertices makes an inscribed equilateral triangle.
art: 456 | Problem | To inscribe a regular decagon in a given circle. Divide the radius CA in extreme and mean ratio, at the point B. Then BC is equal to the side of a regular inscribed decagon. That is, if we apply BC as a chord, its arc will be one-tenth of the whole circumference.
fig: 156 0.19 0.329 0.535 0.775
art: 457 | Corollary | By joining the alternate vertices of a decagon, we may inscribe a regular pentagon.
art: 458 | Corollary | A regular pentedecagon, or polygon of fifteen sides, may be inscribed, by subtracting the arc subtended by the side of a regular decagon from the arc subtended by the side of a regular hexagon. The remainder is one-fifteenth of the circumference, for one sixth - one tenth = one fifteenth.
art: 459 | Problem | Given a regular polygon inscribed in a circle, to inscribe a regular polygon of double the number of sides. Divide each arc subtended by a given side into two equal parts. Join the successive points into which the circumference is divided.
art: 460 |  | We have now learned how to inscribe regular polygons of 3, 4, 5, and 15 sides, and of any number that may arise from doubling either of these four. The problem, to inscribe a regular polygon in a circle by means of straight lines and arcs of circles, can be solved in only a limited number of cases.
art: 461 |  | Notice that the regular decagon was drawn by the aid of two isosceles triangles composing a third, one of the two being similar to the whole. Now, if we could combine three isosceles triangles in this manner, we could draw a regular polygon of fourteen, and then one of seven sides. However, this can not be done by means only of straight lines and arcs of circles. The regular polygon of seventeen sides has been drawn in more than one way, using only straight lines and arcs of circles.
fig: 157 0.15 0.27 0.448 0.73
art: 462 | Problem | Given a regular polygon inscribed in a circle, to circumscribe a similar polygon. The vertices of the given polygon divide the circumference into equal parts. Through these points draw tangents. These tangents produced till they meet, form the required polygon.
art: 464 | A plane of regular polygons | In order that any plane surface may be entirely covered by equal polygons, it is necessary that the figures be such, and such only, that the sum of three or more of their angles is equal to four right angles. Each angle of an equilateral triangle is equal to one-third of two right angles. Therefore, six such angles exactly make up four right angles; and the equilateral triangle is such a figure as is required.
fig: 158 0.33 0.43 0.6 0.765
art: 465 |  | Each angle of the square is a right angle, four of which make four right angles. So that a plane can be covered by equal squares. One angle of a regular pentagon is the fifth part of six right angles. Three of these are less than, and four exceed four right angles; so that the regular pentagon is not such a figure as is required.
fig: 158 0.49 0.582 0.605 0.75
art: 466 |  | Each angle of a regular hexagon is one-sixth of eight right angles. Three such make up four right angles. Hence, a plane may be covered with equal regular hexagons. This combination is remarkable as being the one adopted by bees in forming the honeycomb.
fig: 158 0.68 0.782 0.617 0.765
art: 467 |  | Since each angle of a regular polygon evidently increases when the number of sides increases, and since three angles of a regular hexagon are equal to four right angles, therefore, three angles of any regular polygon of more than six sides, must exceed four right angles. Hence, no other regular figures exist for the purpose here required, except the equilateral triangle, the square, and the regular hexagon.
mc: A polygon with all its sides equal and all its angles equal is... | *regular | similar | symmetrical only
mc: Every regular polygon may have a circle... | *both circumscribed about it and inscribed in it | only circumscribed | only inscribed
mc: The radius of the circle inscribed in a regular polygon is its... | *apothem | radius | diagonal
mc: The chords of equal arcs of a circumference are the sides of a... | *regular polygon | trapezoid | concave polygon
mc: Regular polygons of the same number of sides are... | *similar | always equal | never similar
mc: The side of a regular hexagon inscribed in a circle is equal to... | *the radius | the diameter | half the radius
mc: Joining the alternate vertices of an inscribed regular hexagon makes an inscribed... | *equilateral triangle | square | pentagon
mc: To inscribe a regular decagon, divide the radius in... | *extreme and mean ratio | two equal parts | three equal parts
mc: Joining the alternate vertices of a regular decagon makes a regular... | *pentagon | hexagon | octagon
mc: Which regular polygons can cover a plane? | *The equilateral triangle, the square, and the hexagon | Only the square | All regular polygons
mc: Bees use which regular polygon in the honeycomb? | *The hexagon | The pentagon | The octagon
mc: An angle at the center of a regular polygon of n sides is four right angles divided by... | *n | 2 | n - 2
tf: A regular polygon of seven sides can be drawn with only straight lines and arcs of circles. {F}
tf: A regular polygon is a symmetrical figure. {T}
tf: Four angles of a regular pentagon exceed four right angles. {T}
q: How many degrees is each angle of a regular hexagon? {120}
sol: The sum is 720 degrees, and 720 / 6 = 120.
q: How many degrees is each angle of a regular pentagon? {108}
sol: The sum is 2 (5 - 2) = 6 right angles = 540 degrees, and 540 / 5 = 108.
q: How many degrees is each angle of a regular octagon? {135}
sol: The sum is 2 (8 - 2) = 12 right angles = 1080 degrees, and 1080 / 8 = 135.
q: How many degrees is the angle at the center of a regular decagon, subtended by one side? {36}
sol: Four right angles divided by the number of sides: 360 / 10 = 36.
q: A regular hexagon is inscribed in a circle of radius 7 inches. How long is each side, in inches? {7}
sol: The side of an inscribed regular hexagon equals the radius.
q: A regular hexagon is inscribed in a circle of radius 5 inches. What is its perimeter, in inches? {30}
sol: Each side equals the radius, 5 inches; 6 times 5 = 30.
q: How many equilateral triangles fit together around one point to cover the plane? {6}
sol: Each angle is 60 degrees, and 6 times 60 = 360 degrees, four right angles.

== 32 | Polygons | Isoperimetry
art: 468 | Theorem | Of all equivalent polygons of the same number of sides, the one having the least perimeter is regular. Of several equivalent polygons, suppose AB and BC to be two adjacent sides of the one having the least perimeter. It is to be proved, first, that these sides are equal. Join AC. Now, if AB and BC were not equal, there could be constructed on the base AC an isosceles triangle equivalent to ABC, whose sides would have less extent. Then, this new triangle, with the rest of the polygon, would be equivalent to the given polygon, and have a less perimeter, which is contrary to the hypothesis. It follows that AB and BC must be equal. So of every two adjacent sides. Therefore, the polygon is equilateral.
fig: 159 0.42 0.558 0.41 0.72
art: 468 | Part second | It remains to be proved that the polygon will have all its angles equal. Suppose AB, BC, and CD to be adjacent sides. Produce AB and CD till they meet at E. Now the triangle BCE is isosceles. For if EC, for example, were longer than EB, we could then take EI equal to EB, and EF equal to EC, and we could join FI, making the two triangles EBC and EIF equal. Then, the new polygon, having AFID for part of its perimeter, would be equivalent and isoperimetrical to the given polygon having ABCD as part of its perimeter. Its sides AF, FI, and ID must also be equal. But this is absurd, for AF is less than AB, and ID is greater than CD. Hence, EB and EC must be equal. Therefore, the angles EBC and ECB are equal, and their supplements ABC and BCD are equal. It being proved that the polygon has its sides equal and its angles equal, it is regular.
fig: 160 0.155 0.29 0.445 0.77
art: 469 | Corollary | Of all isoperimetrical polygons of the same number of sides, that which is regular has the greatest area.
art: 470 | Theorem | Of all regular equivalent polygons, that which has the greatest number of sides has the least perimeter. It will be sufficient to demonstrate the principle, when one of the equivalent polygons has one side more than the other. In the polygon having the less number of sides, join the vertex C to any point, as H, of the side BG. Then, on CH construct an isosceles triangle, CKH, equivalent to CBH. Then HK and KC are less than HB and BC; therefore, the perimeter GHKCDF is less than the perimeter of its equivalent polygon GBCDF. But the perimeter of the regular polygon AO is less than the perimeter of its equivalent irregular polygon of the same number of sides, GHKCDF. So much more is it less than the perimeter of GBCDF.
fig: 161 0.17 0.37 0.15 0.75
art: 471 | Corollary | Of two regular isoperimetrical polygons, the greater is that which has the greater number of sides.
art: 472 | Exercises | 6. Three houses are built with walls of the same aggregate length; the first in the shape of a square, the second of a rectangle, and the third of a regular octagon. Which has the greatest amount of room, and which the least? 7. Of all triangles having two sides respectively equal to two given lines, the greatest is that where the angle included between the given sides is a right angle.
mc: Of all equivalent polygons of the same number of sides, the one with the least perimeter is... | *regular | concave | a triangle
mc: Of all isoperimetrical polygons of the same number of sides, the one with the greatest area is... | *regular | the longest | concave
mc: Of regular equivalent polygons, the one with the least perimeter has... | *the greatest number of sides | the least number of sides | four sides
mc: Of two regular polygons with equal perimeters, the greater is the one with... | *more sides | fewer sides | longer sides
mc: Houses with walls of equal total length as a square, a rectangle, and a regular octagon. Which has the most room? | *The octagon | The square | The rectangle
mc: Of those three houses, which has the least room? | *The rectangle | The square | The octagon
mc: Of all triangles with two given sides, the greatest has the angle between them... | *a right angle | acute | obtuse
tf: Isoperimetrical figures have equal perimeters. {T}
tf: Of equivalent triangles on one base, the isosceles has the greatest perimeter. {F}
q: A square has a perimeter of 24 feet. What is its area, in square feet? {36}
sol: Each side is 24 / 4 = 6 feet, and 6 times 6 = 36.
q: A rectangle 8 feet by 4 feet has a perimeter of 24 feet. What is its area, in square feet? {32}
sol: 8 times 4 = 32, less than the 36 square feet of the square with the same perimeter.
q: Two sides of a triangle are 6 and 10 inches. What is the greatest area it can have, in square inches? {30}
sol: The greatest triangle has a right angle between the given sides; then one side is the base and the other the altitude: 6 times 10 / 2 = 30.

== 33 | Circles | Limit of inscribed polygons
art: 473 |  | The properties of the curve which bounds a circle, and of some straight lines connected with it, were discussed in a former chapter. Having now learned the properties of polygons, or rectilinear figures inclosing a plane surface, the student is prepared for the study of the circle as a figure inclosing a surface. The circle is the only curvilinear figure treated of in Elementary Geometry.
art: 474 |  | A segment of a circle is that portion cut off by a secant or a chord. Thus, ABC and CDE are segments. A sector of a circle is that portion included between two radii and the arc intercepted by them. Thus, GHI is a sector.
fig: 163 0.627 0.787 0.15 0.8
art: 475 | Theorem | A circle is the limit of the polygons which can be inscribed in it, also of those which can be circumscribed about it. Having a polygon inscribed in a circle, a second polygon may be inscribed of double the number of sides. Then, a polygon of double the number of sides of the second may be inscribed, and the process repeated at will. We may conceive the number of sides to go on increasing to any number whatever. At each step the inscribed polygon grows larger and the circumscribed grows smaller, both becoming more nearly identical with the circle. Now, it is evident that by the process described, the polygons can be made to approach as nearly as we please to equality with the circle, but can never entirely reach it. The circle is therefore the limit of the polygons.
art: 476 | Corollary | A circle is the limit of all regular polygons whose radii are equal to its radius. It is also the limit of all regular polygons whose apothems are equal to its radius. The circumference is the limit of the perimeters of those polygons.
art: 477 |  | By the method of infinites, the circle is considered as a regular polygon of an infinite number of sides, each side being an infinitesimal straight line. But the method of limits is preferred in this place, because, strictly speaking, the circle is not a polygon, and the circumference is not a broken line.
art: 478 | Theorem | A curve is shorter than any other line which joins its ends, and toward which it is convex. For the curve BDC is the limit of those broken lines which have their vertices in it. Then, the curve BDC is less than the line BFC.
fig: 165 0.36 0.465 0.45 0.72
art: 479 | Corollary | The circumference of a circle is shorter than the perimeter of a circumscribed polygon.
art: 480 | Corollary | The circumference of a circle is longer than the perimeter of an inscribed polygon. This is a corollary of the Axiom of Distance.
art: 481 | Theorem | A circle has a less perimeter than any equivalent polygon. For, of equivalent polygons, that has the least perimeter which is regular, and has the greatest number of sides.
art: 482 | Corollary | A circle has a greater area than any isoperimetrical figure.
art: 483 | Theorem | Circles are similar figures. For angles which intercept like parts of a circumference are equal. Hence, whatever lines be made in one circle, homologous lines, making equal angles, may be made in another. This theorem may be otherwise demonstrated, thus: Inscribed regular polygons of the same number of sides are similar. The number of sides may be increased indefinitely, and the polygons will still be similar at each successive step. The circles being the limits of the polygons, must also be similar.
art: 484 | Theorem | Two sectors are similar when the angles made by their radii are equal.
art: 485 | Theorem | Two segments are similar when the angles which are formed by radii from the ends of their respective arcs are equal. These two theorems are demonstrated by completing the circles of which the given figures form parts.
art: 486 | Exercise | When the Tyrian Princess stretched the thongs cut from the hide of a bull around the site of Carthage, what course should she have pursued in order to include the greatest extent of territory?
mc: The portion of a circle cut off by a chord is a... | *segment | sector | quadrant
mc: The portion of a circle between two radii and their arc is a... | *sector | segment | chord
mc: A circle is the limit of the polygons... | *inscribed in it and circumscribed about it | only inscribed in it | of three sides
mc: As the number of sides is doubled again and again, an inscribed polygon... | *grows larger | grows smaller | stays the same
mc: The circumference of a circle is ___ the perimeter of a circumscribed polygon. | *shorter than | longer than | equal to
mc: The circumference of a circle is ___ the perimeter of an inscribed polygon. | *longer than | shorter than | equal to
mc: Of all figures with the same perimeter, the greatest area belongs to the... | *circle | square | equilateral triangle
mc: Circles are... | *similar figures | never similar | equal figures
mc: Two sectors are similar when... | *the angles made by their radii are equal | their radii are equal | their arcs are equal in length
mc: To include the most land within a thong of given length, the Tyrian Princess should lay it in a... | *circle | square | long rectangle
tf: Strictly speaking, the circle is a polygon. {F}
tf: A circle has a less perimeter than any equivalent polygon. {T}
tf: The circumference is the limit of the perimeters of inscribed regular polygons. {T}

== 34 | Circles | Rectification of the circumference
art: 487 | Theorem | The ratio of the circumference to its diameter is a constant quantity. Two circumferences are to each other in the ratio of their diameters. For the perimeters of similar regular polygons are in the ratio of homologous lines; and the circumference is the limit of the perimeters of regular polygons. Then, designating any two circumferences by C and C', and their diameters by D and D', C : C' :: D : D'. Hence, by alternation, C : D :: C' : D'. That is, the ratio of a circumference to its diameter is the same as that of any other circumference to its diameter.
art: 488 |  | The ratio of the circumference to the diameter is usually designated by the Greek letter pi, the initial of perimeter. If we can determine this numerical ratio, multiplying any diameter by it will give the circumference, or a straight line of the same extent as the circumference. This is called the rectification of that curve.
art: 489 |  | The number pi is less than 4 and greater than 3. For, if the diameter is 1, the perimeter of the circumscribed square is 4; but this is greater than the circumference. And the perimeter of the inscribed regular hexagon is 3, but this is less than the circumference.
art: 490 | Theorem | Given the apothem, radius, and side of a regular polygon; the apothem of a regular polygon of the same length of perimeter, but double the number of sides, is half the sum of the given apothem and radius; and the radius of the polygon of double the number of sides, is a mean proportional between its own apothem and the given radius. Let CD be the apothem, CB the radius, and BE the side of a regular polygon. Produce DC to F, making CF equal to CB. Join BF and EF. The angle BFE is equal to the angle at the center of a regular polygon of double the number of sides of the given polygon. GH is equal to the side of a regular polygon, with the same length of perimeter as the given polygon, and double the number of sides. We have GH the side, IF the apothem, and GF the radius of the polygon of double the number of sides. FI is one-half of FD. But FD is, by construction, equal to the sum of CD and CB. Therefore, the apothem of the second polygon is equal to half the sum of the given apothem and radius. Again, in the right angled triangle GCF, FC : FG :: FG : FI. Therefore, FG, the radius of the second polygon, is a mean proportional between the given radius and the apothem of the second.
fig: 168 0.15 0.415 0.555 0.77
art: 491 |  | Let us represent the given apothem by a, the radius by r, and the side by s, the apothem of the polygon of double the number of sides by x, and its radius by y. Then, x = (a + r) / 2, and x : y :: y : r. Hence, y squared = x r, and y = the square root of x r.
art: 492 |  | Again, since, in any regular polygon, the apothem, radius, and half the side form a right angled triangle, we always have, r squared = a squared + (s/2) squared. Hence, a = one half of the square root of (4 r squared - s squared).
art: 493 | Problem | To find the approximate value of the ratio of the circumference to the diameter of a circle. Suppose a regular hexagon whose perimeter is unity. Then its side is one sixth, or .166667, and its radius is the same. By the formula, the apothem is .144338. Then the apothem of the regular polygon of twelve sides, the perimeter being unity, is .155502. The radius of the same is .160988. Proceeding in the same way, a table may be constructed for 24, 48, 96, 192, 384, 768, and 1536 sides. These ratios gradually approach each other, till they agree for six places of decimals. Hence, we conclude, that when the circumference is 1, the radius is .159155, with a near approximation to exactness. The diameter, being double the radius, is .31831. Therefore, pi = 1 / .31831 = 3.14159.
art: 494 |  | It was shown by Archimedes, by methods resembling the above, that the value of pi is less than 3 1/7, and greater than 3 10/71. This number, 3 1/7, is in very common use for mechanical purposes. It is too great by about one eight-hundredth of the diameter. About the year 1640, Adrian Metius found the nearer approximation 355/113, which is true for six places of decimals. It is easily retained in the memory, as it is composed of the first three odd numbers, in pairs, 113355, taking the first three digits for the denominator, and the other three for the numerator.
art: 495 | Exercises | 1. Two wheels, whose diameters are twelve and eighteen inches, are connected by a belt, so that the rotation of one causes that of the other. The smaller makes twenty-four rotations in a minute; what is the velocity of the larger wheel? 2. Two wheels, whose diameters are twelve and eighteen inches, are fixed on the same axle, so that they turn together. A point on the rim of the smaller moves at the rate of six feet per second; what is the velocity of a point on the rim of the larger wheel?
mc: The ratio of the circumference to its diameter is... | *a constant quantity | different for every circle | always 4
mc: Two circumferences are to each other as... | *their diameters | the squares of their diameters | their areas
mc: The ratio of the circumference to the diameter is written with the Greek letter... | *pi | alpha | delta
mc: Finding a straight line equal to a circumference is called... | *rectification | quadrature | bisection
mc: The number pi is... | *less than 4 and greater than 3 | less than 3 | greater than 4
mc: The perimeter of a regular hexagon inscribed in a circle of diameter 1 is... | *3 | 4 | 2
mc: The value of pi to five places is... | *3.14159 | 3.14285 | 3.16667
mc: Archimedes showed that pi is less than... | *3 1/7 | 3 | 3 1/10
mc: Adrian Metius's approximation for pi is... | *355/113 | 22/7 | 333/106
mc: The common value 3 1/7 for pi is... | *a little too great | a little too small | exact
tf: The circumscribed square of a circle of diameter 1 has a perimeter of 4. {T}
tf: The value of pi has been found exactly as a fraction. {F}
tf: Doubling the sides of polygons with perimeter 1 brings their apothems and radii nearer together. {T}
q: Using 3 1/7 for pi, what is the circumference of a circle whose diameter is 14 inches, in inches? {44}
sol: Circumference = pi times diameter = 22/7 times 14 = 44 inches.
q: Using 3 1/7 for pi, what is the circumference of a circle whose radius is 21 feet, in feet? {132}
sol: The diameter is 42 feet; 22/7 times 42 = 132 feet.
q: Using 3 1/7 for pi, a circumference is 88 inches. What is the diameter, in inches? {28}
sol: Diameter = circumference / pi = 88 times 7 / 22 = 28 inches.
q: Two wheels of 12 and 18 inch diameters are joined by a belt. The smaller makes 24 rotations a minute. How many rotations a minute does the larger make? {16}
sol: The belt moves the same length on each rim, so the rotations are inversely as the diameters: 24 times 12 / 18 = 16.
q: Two wheels of 12 and 18 inch diameters are on the same axle. A point on the rim of the smaller moves 6 feet a second. How many feet a second does a point on the rim of the larger move? {9}
sol: They turn together, so the rim speeds are as the circumferences, that is, as the diameters: 6 times 18 / 12 = 9.
q: Two circumferences have diameters of 3 and 5 feet. The first is 9.42 feet long. How long is the second, in feet? {15.7}
sol: Circumferences are as diameters: 9.42 times 5 / 3 = 15.7 feet.

== 35 | Circles | Quadrature of the circle
art: 496 |  | The quadrature or squaring of the circle, that is, the finding an equivalent rectilinear figure, is a problem which excited the attention of mathematicians during many ages, until it was demonstrated that it could only be solved approximately. The solution depends, indeed, on the rectification of the circumference, and upon the following
art: 497 | Theorem | The area of any polygon in which a circle can be inscribed, is measured by half the product of its perimeter by the radius of the inscribed circle. From the center C of the circle, let straight lines extend to all the vertices of the polygon ABDEF, also to all the points of tangency, G, H, I, K, and L. The lines extending to the points of tangency are radii of the circle, and are therefore perpendicular to the sides of the polygon, which are tangents of the circle. The polygon is divided by the lines extending to the vertices into as many triangles as it has sides. Regarding the sides of the polygon as the bases of these several triangles, they all have equal altitudes, for the radii are perpendicular to the sides of the polygon. Now, the area of each triangle is measured by half the product of its base by the common altitude. But the area of the polygon is the sum of the areas of the triangles, and the perimeter of the polygon is the sum of their bases. It follows that the area of the polygon is measured by half the product of the perimeter by the common altitude, which is the radius.
fig: 172 0.42 0.737 0.49 0.75
art: 498 | Corollary | The area of a regular polygon is measured by half the product of its perimeter by its apothem.
art: 499 | Theorem | The area of a circle is measured by half the product of its circumference by its radius. For the circle is the limit of all the polygons that may be circumscribed about it, and its circumference is the limit of their perimeters.
art: 500 | Theorem | The area of a circle is equal to the square of its radius, multiplied by the ratio of the circumference to the diameter. For, let r represent the radius. Then, the diameter is 2 r, and the circumference is pi times 2 r, and the area is one half of pi times 2 r times r, or pi r squared; that is, the square of the radius multiplied by the ratio of the circumference to the diameter.
art: 501 | Corollary | The areas of two circles are to each other as the squares of their radii; or, as the squares of their diameters.
art: 502 | Corollary | When the radius is unity, the area is expressed by pi.
art: 503 | Theorem | The area of a sector is measured by half the product of its arc by its radius. For, the sector is to the circle as its arc is to the circumference.
art: 504 |  | Since that which is true of every polygon may be shown, by the method of limits, to be true also of plane figures bounded by curves, it follows that in any two similar plane surfaces the ratio of the areas is the second power of the linear ratio.
art: 506 | Exercises | 2. What is the value, in degrees, of the angle at the center, whose arc has the same length as the radius? 6. Two irregular garden-plats, of the same shape, contain, respectively, 18 and 32 square yards; required their linear ratio. 7. To describe a circle equivalent to two given circles.
art: 507 | Exercises in review | 11. To construct a triangle with angles in the ratio 1, 2, 3. 18. In the triangle ABC, the side AB = 13, BC = 15, the altitude = 12; required the base AC. 27. A chord is 8 inches, and the altitude of its segment 3 inches; required the area of the circle. 32. In a quadrilateral inscribed in a circle, the sum of the two rectangles contained by opposite sides, is equivalent to the rectangle contained by the diagonals. This is known as the Ptolemaic Theorem.
mc: Finding a rectilinear figure equivalent to a circle is its... | *quadrature | rectification | projection
mc: The area of a polygon in which a circle can be inscribed is half the product of its perimeter by... | *the radius of the inscribed circle | its longest side | the diameter
mc: The area of a regular polygon is half the product of its perimeter by its... | *apothem | radius | side
mc: The area of a circle is half the product of its circumference by its... | *radius | diameter | area
mc: The area of a circle equals the square of its radius multiplied by... | *pi | 2 | 4
mc: The areas of two circles are to each other as... | *the squares of their radii | their radii | their circumferences
mc: When the radius is unity, the area of the circle is... | *pi | 1 | 2
mc: The area of a sector is half the product of... | *its arc by its radius | its chord by its radius | its angle by its radius
mc: In any two similar plane figures, the ratio of the areas is ___ of the linear ratio. | *the second power | the square root | the same
mc: The Ptolemaic Theorem is about a quadrilateral... | *inscribed in a circle | circumscribed about a circle | with equal diagonals
tf: The quadrature of the circle can be solved exactly with ruler and compasses. {F}
tf: A circle with double the radius has four times the area. {T}
tf: The area of a sector is to the area of its circle as its arc is to the circumference. {T}
q: Using 3 1/7 for pi, what is the area of a circle whose radius is 7 inches, in square inches? {154}
sol: Area = pi r squared = 22/7 times 49 = 154 square inches.
q: Using 3 1/7 for pi, what is the area of a circle whose diameter is 28 feet, in square feet? {616}
sol: The radius is 14; 22/7 times 196 = 616 square feet.
q: A circle has a circumference of 44 inches and a radius of 7 inches. What is its area, in square inches? {154}
sol: Half the product of the circumference by the radius: 44 times 7 / 2 = 154.
q: Two circles have radii of 2 and 3 inches. The area of the larger is how many times the smaller? Give the answer as a fraction. {9/4}
sol: Areas are as the squares of the radii: 9 : 4, so 9/4 times.
q: A sector has an arc of 10 inches and a radius of 6 inches. What is its area, in square inches? {30}
sol: Half the product of its arc by its radius: 10 times 6 / 2 = 30.
q: A regular hexagon has a perimeter of 36 inches and an apothem of 5.2 inches. What is its area, in square inches? {93.6}
sol: Half the product of the perimeter by the apothem: 36 times 5.2 / 2 = 93.6.
q: Two garden-plats of the same shape contain 18 and 32 square yards. Their linear ratio is 3 : ? {4}
sol: The areas are as the squares of the linear ratio. 18 : 32 = 9 : 16, and the square roots are 3 : 4.
q: In the triangle ABC, AB = 13, BC = 15, and the altitude from B is 12. How long is the base AC? {14}
sol: The altitude cuts AC into parts: the square root of 169 - 144 = 5, and the square root of 225 - 144 = 9; AC = 5 + 9 = 14.
q: A chord is 8 inches, and the altitude of its segment is 3 inches. How many inches is the radius of the circle? Give a fraction. {25/6}
sol: The perpendicular from the middle of the chord is a mean proportional between the segments of the diameter: 4 times 4 = 3 times the other part, so the other part is 16/3; the diameter is 3 + 16/3 = 25/3, and the radius is 25/6.

== 36 | Lines and planes in space | Lines and planes
art: 508 |  | The elementary principles of those geometrical figures which lie in one plane, furnish a basis for the investigation of the properties of those figures which do not lie altogether in one plane. The student should bear in mind that when straight lines and planes are given by position merely, without mentioning their extent, it is understood that the extent is unlimited.
art: 509 | Theorem | Through a given point in space there can be only one line parallel to a given straight line.
art: 510 | Theorem | Two straight lines in space parallel to a third, are parallel to each other.
art: 511 | Problem | There may be in space any number of straight lines, each perpendicular to a given straight line at one point of it. For we may suppose that while one of two perpendicular lines remains fixed as an axis, the other revolves around it, remaining all the while perpendicular.
art: 512 | Exercises | 1. Designate two lines which are everywhere equally distant, but which are not parallel. 2. Designate two straight lines which are not parallel, and yet can not meet. 3. Designate four points which do not lie all in one plane.
art: 513 | Theorem | The position of a plane is determined by any plane figure except a straight line. Hence, we say, the plane of an angle, of a circumference, etc.
art: 514 | Theorem | A straight line and a plane can have only one common point, unless the line lies wholly in the plane.
art: 515 |  | When a line and a plane have only one common point, the line is said to pierce the plane, and the plane to cut the line. The common point is called the foot of the line in the plane. When a line lies wholly in a plane, the plane is said to pass through the line.
art: 516 | Theorem | The intersection of two planes is a straight line. For two planes can not have three points common, unless those points are all in one straight line.
art: 517 | Theorem | A straight line which is perpendicular to each of two straight lines at their point of intersection, is perpendicular to every other straight line which lies in the plane of the two, and passes through their point of intersection. Suppose D, B, and C to be on the plane of the paper, the point A being above, and I below that plane. If AB is perpendicular to BC and to BD, it is also perpendicular to BE. Produce AB, making BI equal to BA, and let any line, as FH, cut BC, BE, and BD in F, G, and H. The triangles AFH and IFH are equal, so the angle AHF is equal to IHF. Then the triangles AHG and IHG are equal, and AG is equal to IG. Therefore BG, having two points each equally distant from A and I, is perpendicular to AI at its center B.
fig: 179 0.45 0.6 0.52 0.72
art: 518 | Theorem | Conversely, if several straight lines are each perpendicular to a given line at the same point, then these several lines all lie in one plane. Thus, if BA is perpendicular to BC, to BD, and to BE, then these three all lie in one plane.
fig: 180 0.31 0.5 0.57 0.78
art: 519 |  | A straight line is said to be perpendicular to a plane when it is perpendicular to every straight line which passes through its foot in that plane, and the plane is said to be perpendicular to the line. Every line not perpendicular to a plane which cuts it, is called oblique.
art: 520 | Corollary | If a plane cuts a line perpendicularly at the middle point of the line, then every point of the plane is equally distant from the two ends of the line.
art: 521 | Corollary | If one of two perpendicular lines revolves about the other, the revolving line describes a plane which is perpendicular to the axis.
art: 522 | Corollary | Through one point of a straight line there can be only one plane perpendicular to that line.
art: 523 | Theorem | Through a point out of a plane there can be only one straight line perpendicular to the plane. For, if there could be two perpendiculars, then each would be perpendicular to the line in the plane which joins their feet. But this is impossible.
art: 524 | Theorem | Through a point in a plane there can be only one straight line perpendicular to the plane. Let BA be perpendicular to the plane MN at the point B. Then any other line, BC for example, will be oblique to the plane MN. For the intersection of the plane of ABC with MN is a straight line DE, and AB is perpendicular to DE. Hence BC is not perpendicular to DE, and therefore not perpendicular to the plane MN.
fig: 181 0.33 0.51 0.385 0.75
art: 525 | Corollary | The direction of a straight line in space is fixed by the fact that it is perpendicular to a given plane. The directions of a plane are fixed by the fact that it is perpendicular to a given line.
art: 526 | Corollary | All straight lines which are perpendicular to the same plane, have the same direction; that is, they are parallel to each other.
art: 527 | Corollary | If one of two parallel lines is perpendicular to a plane, the other is also.
art: 528 |  | The axis of a circle is the straight line perpendicular to the plane of the circle at its center.
art: 529 | Theorem | If from a point without a plane, a perpendicular and oblique lines be extended to the plane, then two oblique lines which meet the plane at equal distances from the foot of the perpendicular, are equal. Let AB be perpendicular, and AC and AD oblique to the plane MN, and the distances BC and BD equal. Then the triangles ABC and ABD are equal, and AC is equal to AD.
fig: 182 0.265 0.41 0.5 0.78
art: 530 | Corollary | A perpendicular is the shortest line from a point to a plane. Hence, the distance from a point to a plane is measured by a perpendicular line.
art: 531 | Corollary | All points of the circumference of a circle are equidistant from any point of its axis.
art: 532 |  | If from all points of a line perpendiculars be let fall upon a plane, the line thus described upon the plane is the projection of the given line upon the given plane.
art: 533 | Theorem | The projection of a straight line upon a plane is a straight line. Let AB be the given line, and MN the given plane. From A and B let the perpendiculars AC and BD fall upon the plane MN. AC and BD, being perpendicular to the same plane, are parallel, and lie in one plane. Every perpendicular to MN from a point of AB lies in the plane AD, and meets MN in some point of CD. Hence CD is the projection of AB. When the given line is perpendicular to the plane, its projection is a point.
fig: 182 0.685 0.83 0.44 0.78
art: 534 | Corollary | A straight line and its projection on a plane, both lie in one plane.
art: 535 | Theorem | The angle which a straight line makes with its projection on a plane, is smaller than the angle it makes with any other line in the plane. Let AC be the given line, and BC its projection on the plane MN. With C as a center and BC as a radius, describe a circumference in the plane MN, cutting CD at D. The triangles ACD and ACB have two sides of the one equal to two sides of the other, but AD is longer than AB. Therefore, the angle ACD is greater than the angle ACB.
fig: 183 0.465 0.665 0.425 0.74
art: 537 |  | The angle which a line makes with its projection in a plane, is called the angle of inclination of the line and the plane.
art: 538 | Theorem | If a straight line in a plane is parallel to a straight line not in the plane, then the second line and the plane can not have a common point.
art: 539 |  | Such a line and plane, having the same direction, are called parallel.
art: 540 | Corollary | If one of two parallel lines is parallel to a plane, the other is also.
art: 541 | Corollary | A line which is parallel to a plane is parallel to its projection on that plane.
art: 542 | Corollary | A line parallel to a plane is everywhere equally distant from it.
art: 543 | Applications | Three points, however placed, must always be in the same plane. It is on this principle that stability is more readily obtained by three supports than by a greater number. A three-legged stool must be steady, but if there be four legs, their ends should be in one plane, and the floor should be level.
art: 544 |  | A ready way of constructing a line perpendicular to a plane is by the use of two squares. Place the angle of each at the foot of the desired perpendicular, one side of each square resting on the plane surface. Bring their perpendicular sides together. Their position must then be that of a perpendicular to the plane, for it is perpendicular to two lines in the plane.
art: 545 |  | Two circular millstones are placed so as to have the same axis, to which their faces are perpendicular, being, therefore, parallel to each other. The lower stone is fixed, while the upper one is made to revolve. Their distance being properly regulated, all the grain which passes between them will be ground with the same degree of fineness.
mc: The intersection of two planes is... | *a straight line | a point | a curve
mc: A line which has only one point in common with a plane is said to ___ the plane. | *pierce | pass through | be parallel to
mc: The common point of a line and a plane which it pierces is called the line's... | *foot | vertex | axis
mc: A line perpendicular to two lines at their point of meeting is perpendicular to... | *every line through that point in their plane | only those two lines | no other line
mc: Through a point out of a plane, how many lines can be perpendicular to the plane? | *Only one | Two | Any number
mc: All straight lines perpendicular to the same plane are... | *parallel to each other | perpendicular to each other | equal
mc: The shortest line from a point to a plane is... | *the perpendicular | an oblique line | the projection
mc: The straight line perpendicular to the plane of a circle at its center is the circle's... | *axis | radius | diameter
mc: The projection of a straight line upon a plane is a straight line, except when the line is perpendicular to the plane; then it is... | *a point | a circle | a plane
mc: The angle a line makes with its projection on a plane is called the angle of... | *inclination | projection | elevation
mc: A line parallel to a plane is everywhere ___ from it. | *equally distant | farther | nearer
mc: Why is a three-legged stool always steady? | *Any three points lie in one plane | Three is an odd number | The legs are equal
tf: Two straight lines which are not parallel must meet. {F}
tf: Two oblique lines meeting a plane at equal distances from the foot of the perpendicular are equal. {T}
tf: If one of two parallel lines is perpendicular to a plane, the other is also. {T}
tf: A line makes its smallest angle with its own projection on the plane. {T}
q: From a point 12 inches above a plane, an oblique line meets the plane 5 inches from the foot of the perpendicular. How long is the oblique line, in inches? {13}
sol: The perpendicular, the 5 inches on the plane, and the oblique line form a right triangle: the square root of 144 + 25 = 169 is 13.
q: A point is 8 feet from a plane. A circle on the plane has its center at the foot of the perpendicular and a radius of 6 feet. How far is the point from every point of the circumference, in feet? {10}
sol: All points of the circumference are equidistant from a point of its axis: the square root of 64 + 36 = 100 is 10.
q: A rod 10 feet long leans against a plane so that its projection on the plane is 6 feet long. How high is its upper end above the plane, in feet? {8}
sol: The rod, its projection, and the perpendicular form a right triangle: the square root of 100 - 36 = 64 is 8.

== 37 | Lines and planes in space | Diedral angles
art: 547 |  | A diedral angle is formed by two planes meeting at a common line. This figure is also called simply a diedral. The planes are its faces, and the intersection is its edge. In naming a diedral, four letters are used, one in each face, and two on the edge, the letters on the edge being between the other two.
art: 548 |  | The quantity of a diedral, as is the case with a linear angle, depends on the difference in the directions of the faces from the edge, without regard to the extent of the planes. Hence, two diedrals are equal when they can be so placed that their planes will coincide.
art: 549 | Problem | One diedral may be added to another. AB, AC, and AD represent three planes having the common intersection AE. Evidently the sum of BEAC and CEAD is equal to BEAD.
fig: 186 0.23 0.444 0.575 0.78
art: 550 | Corollary | Diedrals may be subtracted one from another. A diedral may be bisected or divided in any required ratio by a plane passing through its edge.
art: 552 | Theorem | One diedral is to another as the plane angle, formed in the first by a line in each face perpendicular to the edge, is to the similarly formed angle in the other. Thus, if FO, GO, and HO are each perpendicular to AE, then the diedral CEAD is to the diedral BEAD as the angle GOH is to the angle FOH.
fig: 186 0.65 0.86 0.46 0.78
art: 553 | Corollary | A diedral is said to be measured by the plane angle formed by a line in each of its faces perpendicular to the edge.
art: 554 | Corollary | Accordingly, a diedral angle may be acute, obtuse, or right. In the last case, the planes are perpendicular to each other.
art: 555 |  | Many of the principles of plane angles may be applied to diedrals. All right diedral angles are equal. When the sum of several diedrals is measured by two right angles, the outer faces form one plane. When two planes cut each other, the opposite or vertical diedrals are equal.
art: 556 | Theorem | If a line is perpendicular to a plane, then any plane passing through this line is perpendicular to the other plane. If AB in the plane PQ is perpendicular to the plane MN, then AB is perpendicular to RQ, the intersection of the two planes, and to BC, which is made perpendicular to RQ. Then the angle ABC measures the inclination of the two planes, and is a right angle. Therefore, the planes are perpendicular.
fig: 187 0.6 0.765 0.495 0.75
art: 557 | Corollary | Conversely, if a plane is perpendicular to another, a straight line, which is perpendicular to one of them, at some point of their intersection, must lie wholly in the other plane.
art: 558 | Corollary | If two planes are perpendicular to a third, then the intersection of the first two is a line perpendicular to the third plane.
art: 559 | Theorem | If from a point within a diedral, perpendicular lines be made to the two faces, the angle of these lines is supplementary to the angle which measures the diedral. Let M and N be two planes whose intersection is AB, and CF and CE perpendiculars upon them from the point C. The angle FDE measures the diedral. But in the quadrilateral FDEC, the two angles F and E are right angles. Therefore, the other two angles at C and D are supplementary.
fig: 188 0.43 0.575 0.5 0.75
art: 560 | Theorem | Every point of a plane which bisects a diedral is equally distant from its two faces. The angles HOA and IOA measure the two halves of the diedral, which are equal; so AO bisects the angle IOH, and the point A is equally distant from the lines OH and OI, and therefore from the two faces.
fig: 189 0.225 0.365 0.36 0.66
art: 561 | Applications | Diedrals are illustrated by a door turning on its hinges. In every position it is perpendicular to the floor and ceiling. As it turns, it changes its inclination to the wall, the angle of inclination being that which is formed by the upper edge of the door and the lintel.
art: 562 |  | The theory of diedrals is important in the science of crystallography, which teaches us how to classify mineral substances according to their geometrical forms. Crystals of one kind have edges of which the diedral angles measure a certain number of degrees, and crystals of another kind have edges of a different number of degrees.
art: 563 |  | The plane of the surface of a liquid at rest is called horizontal, or the plane of the horizon. The direction of a plumb-line when the weight is at rest, is a vertical line. The vertical line is perpendicular to the horizon. Every plane passing through a vertical line is called a vertical plane, and every vertical plane is perpendicular to the horizon. Floors and ceilings are examples of horizontal planes, and walls of vertical planes.
art: 564 | Exercises | 1. If, from a point without a plane, several equal oblique lines extend to it, they make equal angles with the plane. 3. What is the locus of those points in space, each of which is equally distant from two given points?
mc: A figure formed by two planes meeting at a common line is a... | *diedral angle | triedral | plane angle
mc: The common line of the two faces of a diedral is its... | *edge | vertex | axis
mc: In naming a diedral with four letters, the two letters on the edge are written... | *between the other two | first | last
mc: A diedral is measured by the plane angle formed by a line in each face... | *perpendicular to the edge | parallel to the edge | through any point
mc: When the diedral is a right angle, the two planes are... | *perpendicular to each other | parallel | equal
mc: Any plane passing through a line perpendicular to another plane is ___ to that plane. | *perpendicular | parallel | oblique
mc: If two planes are perpendicular to a third, their intersection is... | *perpendicular to the third plane | parallel to the third plane | a point
mc: From a point within a diedral, perpendiculars to the two faces make an angle that is ___ to the diedral's angle. | *supplementary | equal | complementary
mc: Every point of a plane which bisects a diedral is... | *equally distant from its two faces | on its edge | nearer one face
mc: The direction of a plumb-line at rest is... | *vertical | horizontal | oblique
mc: The locus of points in space equally distant from two given points is... | *a plane perpendicular to the line joining them at its middle | a circle | a straight line
tf: Two diedrals are equal when their planes can be made to coincide. {T}
tf: The size of a diedral depends on how far its planes extend. {F}
tf: Every vertical plane is perpendicular to the horizon. {T}
tf: Vertical diedrals, made by two planes cutting each other, are equal. {T}
q: Two planes cut each other. One of the four diedrals is measured by 70 degrees. How many degrees measure the diedral next to it? {110}
sol: Adjacent diedrals on one side of a plane are together two right angles, 180 degrees: 180 - 70 = 110.
q: A diedral is measured by 50 degrees. From a point within it, perpendiculars are let fall on the two faces. How many degrees is the angle of these perpendiculars? {130}
sol: The angle of the perpendiculars is the supplement of the angle which measures the diedral: 180 - 50 = 130.
q: A door is opened so that its upper edge makes an angle of 35 degrees with the lintel. Through how many more degrees must it turn to be perpendicular to the wall? {55}
sol: Perpendicular means a right angle, 90 degrees: 90 - 35 = 55.

== 38 | Lines and planes in space | Parallel planes
art: 565 |  | Two planes which are perpendicular to the same straight line, at different points of it, are both fixed in position, and they have the same directions. If the parallel lines AB and CD revolve about the line EF, to which they are both perpendicular, then each of the revolving lines describes a plane. Two planes which have the same directions are called parallel planes. Parallelism consists in having the same direction, whether it be of two lines, of two planes, or of a line and a plane.
fig: 190 0.68 0.805 0.55 0.76
art: 566 | Corollary | Two planes parallel to a third are parallel to each other.
art: 567 | Corollary | Two planes perpendicular to the same straight line are parallel to each other.
art: 568 | Corollary | A straight line perpendicular to one of two parallel planes is perpendicular to the other.
art: 569 | Corollary | Every straight line in one of two parallel planes has its parallel line in the other plane. Therefore, every straight line in one of the planes is parallel to the other plane.
art: 571 | Theorem | Two parallel planes can not meet. For, if they had a common point, being parallel, they would have the same directions from that point, and therefore would coincide throughout, and be only one plane.
art: 572 | Theorem | The intersections of two parallel planes by a third plane are parallel lines. Let AB and CD be the intersections of the two parallel planes M and N, with the plane P. A line through C parallel to AB must lie in the plane P, and also in the plane N. Therefore, it is the intersection CD, and the two intersections are parallel lines.
fig: 192 0.175 0.3 0.515 0.76
art: 573 | Theorem | The parts of two parallel lines intercepted between parallel planes are equal. For, if the lines AB and CD are parallel, they lie in one plane. Then AC and BD are the intersections of this plane with the two parallel planes M and P. Hence, AC is parallel to BD, and ABDC is a parallelogram. Therefore, AB is equal to the opposite side CD.
fig: 192 0.405 0.515 0.51 0.72
art: 574 | Theorem | Two parallel planes are everywhere equally distant. For the shortest distance from any point of one plane to the other, is measured by a perpendicular. But these perpendiculars are all parallel, and therefore equal to each other.
art: 575 | Theorem | If the two sides of an angle are each parallel to a given plane, then the plane of that angle is parallel to the given plane. From A let the perpendicular AD fall upon the plane M, and let DE and DF be the projections of AB and AC. Then DA is perpendicular to AB and to AC, and therefore to the plane of BAC; and the two planes, being perpendicular to the same line, are parallel.
fig: 193 0.155 0.336 0.46 0.7
art: 576 | Theorem | If two straight lines which cut each other are respectively parallel to two other straight lines which cut each other, then the plane of the first two is parallel to the plane of the second two.
fig: 193 0.47 0.645 0.405 0.71
art: 577 | Corollary | The angles made by the first two lines are respectively the same as those made by the second two. For they are the differences between the same directions.
art: 578 | Theorem | Straight lines cut by three parallel planes are divided proportionally. If the line AB is cut at the points A, E, and B, and the line CD at the points C, F, and D, by the parallel planes M, N, and P, then AE : EB :: CF : FD. Join AD, which pierces N at G. Then EG is parallel to BD, so AE : EB :: AG : GD. For a like reason, AG : GD :: CF : FD. Therefore, AE : EB :: CF : FD.
fig: 194 0.165 0.365 0.485 0.76
art: 579 | Application | The general problem of perspective in drawing consists in representing upon a plane surface the apparent form of objects in sight. All the visual rays from one straight object lie in one plane, and their intersection with the plane of the picture must be a straight line. Therefore, all straight objects must be drawn as straight lines. Two parallel straight objects which are also parallel to the plane of the picture remain parallel in the perspective; two parallel lines which are not parallel to the plane of the picture will meet in the perspective.
art: 580 | Exercises | 1. A straight line makes equal angles with two parallel planes. 2. Two parallel lines make the same angle of inclination with a given plane. 3. The projections of two parallel lines on a plane are parallel. 5. If two straight lines be not in the same plane, one straight line, and only one, may be perpendicular to both of them.
mc: Two planes which have the same directions are... | *parallel | perpendicular | oblique
mc: Two planes perpendicular to the same straight line are... | *parallel to each other | perpendicular to each other | the same plane
mc: The intersections of two parallel planes by a third plane are... | *parallel lines | perpendicular lines | one point
mc: Parts of two parallel lines between two parallel planes are... | *equal | unequal | perpendicular
mc: Two parallel planes are everywhere... | *equally distant | meeting | perpendicular
mc: If both sides of an angle are parallel to a plane, the plane of the angle is ___ to it. | *parallel | perpendicular | oblique
mc: Straight lines cut by three parallel planes are divided... | *proportionally | equally | in halves
mc: Parallelism consists in having the same... | *direction | length | position
mc: In a picture, two parallel lines that are not parallel to the plane of the picture... | *meet in the perspective | stay parallel | disappear
tf: Two parallel planes can meet if they are made large enough. {F}
tf: A line perpendicular to one of two parallel planes is perpendicular to the other. {T}
tf: All straight objects must be drawn in perspective as straight lines. {T}
q: Two parallel planes are 9 inches apart. A line cuts both planes, making them intercept 15 inches of it. How far, in inches, is the foot of the perpendicular from the point where the line pierces the lower plane, if the perpendicular is let fall from where it pierces the upper plane? {12}
sol: The perpendicular is 9, the line between the planes is 15; the square root of 225 - 81 = 144 is 12.
q: Three parallel planes cut a line into parts of 4 and 6 inches. Another line is cut by them, and its first part is 6 inches. How long is its second part, in inches? {9}
sol: Lines cut by three parallel planes are divided proportionally: 4 : 6 :: 6 : x, so x = 36 / 4 = 9.
q: Three parallel planes cut a line 20 inches long into two parts in the ratio 2 : 3. How long is the longer part, in inches? {12}
sol: 2 + 3 = 5 parts; one part is 20 / 5 = 4; the longer is 3 times 4 = 12.
== 39 | Lines and planes in space | Triedrals
art: 581 |  | When three planes cut each other, three cases are possible. 1st. The intersections may coincide. Then six diedrals are formed, having for their common edge the intersection of the three planes. 2d. The three intersections may be parallel lines. Then one plane is parallel to the intersection of the other two.
fig: 195 0.545 0.68 0.46 0.7
fig: 195 0.725 0.865 0.46 0.7
art: 581 | Third case | 3d. The three intersections may meet at one point. Then the space about the point is divided by the three planes into eight parts. For two intersecting planes make four diedrals; and if a third plane cut through the intersection of the first two, it will divide each of the diedrals into two parts, making eight in all. Each of these parts is called a triedral, because it has three faces. A fourth case is impossible.
fig: 196 0.165 0.375 0.385 0.7
art: 582 |  | A triedral is the figure formed by three planes meeting at one point. The point where the planes and intersections all meet, is called the vertex of the triedral. The intersections are its edges, and the planes are its faces. The corners of a room, or of a chest, are illustrations of triedrals with rectangular faces. The point of a triangular file has the form of a triedral with acute faces.
art: 582 | Elements | The triedral has many things analogous to the plane triangle. The three faces, combined two and two, make three diedrals, and the three intersections, combined two and two, make three plane angles. These six are the six elements or principal parts of a triedral. Two triedrals are equal when their several edges respectively coincide.
art: 583 |  | A triedral which has one rectangular diedral, that is, whose measure is a right angle, is called a rectangular triedral. If it has two, it is birectangular; if it has three, it is trirectangular. A triedral which has two of its faces equal, is called isosceles; if all three are equal, it is equilateral.
art: 584 | Symmetrical triedrals | If the edges of a triedral be produced beyond the vertex, they form the edges of a new triedral. The faces of these two triedrals are respectively equal, for the angles are vertical; and the diedrals are equal, being vertically opposite. But the two triedrals are not equal, for they can not be made to coincide, although composed of parts which are respectively equal. The difference consists in the opposite order in which the parts are arranged. This may be illustrated by two gloves: the right hand glove will not fit the left hand.
fig: 197 0.65 0.83 0.5 0.72
art: 585 |  | When two magnitudes are composed of parts respectively equal, but arranged in reverse order, they are said to be symmetrical magnitudes. Two symmetrical plane figures are also equal, for one can be turned over to coincide with the other. But this is not possible with figures that are not in one plane.
art: 586 | Theorem | Each plane angle of a triedral is less than the sum of the other two. Let ASB be the largest of the three angles of the triedral S. From ASB take the part ASD equal to ASC. Join SA and SB by any straight line AB, take SC equal to SD, and join AC and BC. Then AD is equal to AC, and BD is less than BC. Therefore the angle BSD is less than BSC; and adding the equal angles ASD and ASC, the angle ASB is less than the sum of ASC and BSC.
fig: 199 0.455 0.58 0.42 0.74
art: 587 | Theorem | The sum of the plane angles which form a triedral is always less than four right angles. Through three points, one in each edge, let the plane ABC pass. The sum of the angles of the triangle ABC, two right angles, is less than the sum of the six angles at the bases of the triangles on the faces. The sum of all the angles of these three triangles is six right angles. Therefore the sum of those at the vertex S must be less than four right angles.
fig: 200 0.265 0.35 0.46 0.74
art: 588 |  | Take any three points on the paper for A, B, and C, and S at some distance from the surface. Then let S approach the surface of the triangle ABC. The angles at S become larger and larger, until the point S touches the surface of the triangle, when the sum of the angles becomes four right angles, and the triedral becomes one plane.
art: 589 | Theorem | If, from a point within a triedral, perpendicular lines fall on the several faces, these lines will be the edges of a second triedral, whose faces will be supplements respectively of the diedrals of the first; and the faces of the first will be respectively supplements of the diedrals of the second triedral.
fig: 201 0.335 0.525 0.48 0.73
art: 590 |  | Two triedrals, in which the faces and diedral angles of the one are respectively the supplements of the diedral angles and faces of the other, are called supplementary triedrals.
art: 591 | Theorem | In every triedral the sum of the three diedral angles is greater than two right angles, and less than six. For the three diedrals of the given triedral and the three faces of its supplementary triedral make six right angles, and the faces of the supplementary triedral are together less than four right angles and more than zero.
art: 592 | Theorem | When two triedrals have two faces, and the included diedral of the one respectively equal to the corresponding parts of the other, then the remaining face and diedrals of the first are respectively equal to the corresponding parts of the other. When the parts are arranged in the same order, the two triedrals are equal; when in reverse order, they are symmetrical.
fig: 203 0.325 0.49 0.31 0.72
fig: 204 0.275 0.572 0.34 0.75
art: 593 | Theorem | When two triedrals have one face and the two adjacent diedrals of the one respectively equal to the corresponding parts of the other, then the remaining faces and diedral of the first are respectively equal to the corresponding parts of the other.
fig: 205 0.235 0.395 0.315 0.72
art: 594 | Theorem | An isosceles triedral and its symmetrical are equal. Let ABCD be an isosceles triedral, having the faces BAC and DAC equal, and let AEFG be its symmetrical triedral. The faces BAC, DAC, FAG, and FAE are all equal, and the diedrals whose edges are AC and AF, being vertical, are equal. Hence equal parts may be taken in the same order in both triedrals, and the two triedrals are equal.
fig: 205 0.63 0.825 0.535 0.72
art: 595 | Corollary | In an isosceles triedral, the diedrals opposite the equal faces are equal.
art: 596 | Corollary | Conversely, if in any triedral two of the diedral angles are equal, then the faces opposite these diedrals are equal, and the triedral is isosceles.
art: 597 | Theorem | When two triedrals have two faces of the one respectively equal to two faces of the other, and the included diedrals unequal, then the third faces are unequal, and that face is greater which is opposite the greater diedral.
fig: 206 0.485 0.68
art: 599 | Theorem | When two triedrals have their three faces respectively equal, their diedrals will be respectively equal; and the two triedrals are either equal, or they are symmetrical.
art: 600 | Theorem | Two triedrals which have their diedrals respectively equal, have also their faces respectively equal; and the two triedrals are either equal, or they are symmetrical. This theorem has nothing analogous in plane geometry.
art: 602 | Corollary | All trirectangular triedrals are equal.
art: 604 | Exercises | 1. In any triedral, the greater of two faces is opposite to the greater diedral angle; and conversely.
mc: Three planes whose intersections meet at one point divide the space about it into how many parts? | *Eight | Six | Four
mc: A figure formed by three planes meeting at one point is a... | *triedral | diedral | triangle
mc: How many elements or principal parts has a triedral? | *Six | Three | Eight
mc: A triedral with three right diedrals is called... | *trirectangular | birectangular | equilateral
mc: A triedral with two of its faces equal is called... | *isosceles | equilateral | rectangular
mc: Magnitudes made of equal parts arranged in reverse order are... | *symmetrical | equal | similar
mc: Which pair is an example of symmetrical magnitudes? | *A right glove and a left glove | Two right gloves | Two equal squares
mc: Each plane angle of a triedral is ___ the sum of the other two. | *less than | equal to | greater than
mc: The sum of the plane angles of a triedral is always less than... | *four right angles | two right angles | six right angles
mc: The sum of the three diedrals of a triedral is greater than two right angles and less than... | *six right angles | four right angles | three right angles
mc: In an isosceles triedral, the diedrals opposite the equal faces are... | *equal | supplementary | unequal
mc: Two triedrals with their three faces respectively equal are... | *either equal or symmetrical | always equal | never equal
mc: Which theorem about triedrals has nothing like it for triangles? | *Equal diedrals give equal faces | Two faces and the included diedral | One face and two diedrals
tf: The corner of a room is a triedral with rectangular faces. {T}
tf: Two symmetrical figures in space can always be turned to coincide. {F}
tf: An isosceles triedral and its symmetrical are equal. {T}
tf: All trirectangular triedrals are equal. {T}
q: Two faces of a triedral are 50 and 70 degrees. The third face must be less than how many degrees? {120}
sol: Each plane angle of a triedral is less than the sum of the other two: 50 + 70 = 120.
q: Two faces of a triedral are 100 and 140 degrees. The third face must be less than how many degrees? {120}
sol: The three faces together are less than four right angles, 360 degrees: 360 - 100 - 140 = 120. (It must also be less than 100 + 140, which is more.)
q: The faces of a triedral are 60, 80, and 90 degrees. By how many degrees is their sum less than four right angles? {130}
sol: 60 + 80 + 90 = 230, and 360 - 230 = 130.
q: The three diedrals of a triedral are 80, 90, and 100 degrees. What is the sum of the faces of its supplementary triedral, in degrees? {270}
sol: Each face of the supplementary triedral is the supplement of a diedral: 100 + 90 + 80 = 270.

== 40 | Lines and planes in space | Polyedrals
art: 605 |  | A polyedral is the figure formed by several planes which meet at one point. Thus, a polyedral is composed of several angles having their vertices at a common point, every edge being a side of two of the angular faces. The triedral is a polyedral of three faces.
art: 606 | Problem | Any polyedral of more than three faces may be divided into triedrals. For a plane may pass through any two edges which are not adjacent. Thus, a polyedral of four faces may be divided into two triedrals; one of five faces, into three; and so on.
art: 607 |  | This is like the division of a polygon into triangles. The plane passing through two edges not adjacent is called a diagonal plane. A polyedral is called convex, when every possible diagonal plane lies within the figure; otherwise it is called concave.
fig: 209 0.545 0.665 0.335 0.73
art: 608 | Corollary | If the plane of one face of a convex polyedral be produced, it can not cut the polyedral.
art: 609 | Corollary | A plane may pass through the vertex of a convex polyedral, without cutting any face of the polyedral.
art: 610 | Corollary | A plane may cut all the edges of a convex polyedral. The section is a convex polygon.
art: 611 |  | When any figure is cut by a plane, the figure that is defined on the plane by the limits of the figure so cut, is called a plane section.
art: 612 | Theorem | The sum of all the angles of a convex polyedral is less than four right angles. Suppose the polyedral to be cut by a plane; the section is a polygon of n sides, and the plane cuts off n triangles on the faces. The angles of the polygon are 2n - 4 right angles, and the angles of all the triangles are 2n right angles. If v right angles is the sum at the vertex, the angles at the bases are 2n - v. Since two faces of a triedral are greater than the third, 2n - v is greater than 2n - 4. Therefore 4 is greater than v.
art: 613 | Theorem | In any convex polyedral, the sum of the diedrals is greater than the sum of the angles of a polygon having the same number of sides that the polyedral has faces.
art: 614 | Descriptive geometry | Descriptive Geometry is that branch of mathematics which teaches how to represent and determine, by means of drawings on a plane surface, the absolute or relative position of points or magnitudes in space. Its elements were first reduced to a system by Monge, one of the founders of the Polytechnic School at Paris.
art: 615 | Exercises | 1. What is the locus of those points in space, each of which is equally distant from three given points? 2. What is the locus of those points in space, each of which is equally distant from two given planes? 7. If each diedral of a triedral be bisected, the three planes have one common intersection.
mc: A figure formed by several planes meeting at one point is a... | *polyedral | polygon | polyedron
mc: A plane through two edges of a polyedral which are not adjacent is a... | *diagonal plane | plane section | face
mc: A polyedral is convex when every diagonal plane lies... | *within the figure | outside the figure | on a face
mc: Into how many triedrals may a polyedral of five faces be divided? | *Three | Two | Five
mc: The sum of all the angles of a convex polyedral is less than... | *four right angles | two right angles | eight right angles
mc: A plane cutting all the edges of a convex polyedral makes a section which is a... | *convex polygon | triangle | circle
mc: The branch of mathematics which shows figures in space by drawings on a plane is... | *descriptive geometry | trigonometry | perspective only
mc: The figure a plane cuts out of a solid figure is called a... | *plane section | diagonal plane | projection
mc: The locus of points equally distant from two given parallel planes is... | *a plane midway between them, parallel to both | a straight line | a point
tf: The plane of one face of a convex polyedral, produced, can not cut the polyedral. {T}
tf: A triedral is a polyedral of three faces. {T}
tf: Six angles of 70 degrees each can form a convex polyedral. {F}
q: Into how many triedrals may a polyedral of seven faces be divided? {5}
sol: Like a polygon divided into triangles: 7 - 2 = 5.
q: A convex polyedral has five equal face angles. Each must be less than how many degrees? {72}
sol: The sum is less than 360 degrees, so each of the five is less than 360 / 5 = 72.
q: A convex polyedral of six faces is cut by a plane. How many right angles is the sum of the angles of the section? {8}
sol: The section is a polygon of six sides: 2 times 6 - 4 = 8 right angles.
== 41 | Polyedrons | Tetraedrons
art: 616 |  | A polyedron is a solid, or portion of space, bounded by plane surfaces. Each of these surfaces is a face, their several intersections are edges, and the points of meeting of the edges are vertices of the polyedron.
art: 617 | Corollary | The edges being intersections of planes, must be straight lines. It follows that the faces of a polyedron are polygons.
art: 618 |  | A diagonal of a polyedron is a straight line joining two vertices which are not in the same face. A diagonal plane is a plane passing through three vertices which are not in the same face.
art: 619 |  | Three planes can not inclose a space. But if any point be taken on each edge of a triedral, a plane passing through these three points would, with the three faces of the triedral, cut off a portion of space inclosed by four triangular faces. A tetraedron is a polyedron having four faces.
fig: 213 0.695 0.833 0.45 0.72
art: 620 | Problem | Any four points whatever, which do not all lie in one plane, may be taken as the four vertices of a tetraedron. For they may be joined two and two, by straight lines, thus forming the six edges; and these bound the four triangular faces of the figure.
art: 621 |  | Either face of the tetraedron may be taken as the base. Then the other faces are called the sides, the vertex opposite the base is called the vertex of the tetraedron, and the altitude is the perpendicular distance from the vertex to the plane of the base.
art: 623 | Theorem | There is a point equally distant from the four vertices of any tetraedron. In the plane of the face BCF, suppose a circle whose circumference passes through B, C, and F, and at its center erect a line perpendicular to that plane; every point of it is equally distant from B, C, and F. In the same manner erect a perpendicular for the face BDF. These two perpendiculars lie in one plane and meet at a point equally distant from the four vertices.
fig: 214 0.575 0.725 0.48 0.78
art: 624 | Corollary | The six planes which bisect perpendicularly the several edges of a tetraedron all meet in one point. But this point is not necessarily within the tetraedron.
art: 625 | Theorem | There is a point within every tetraedron which is equally distant from the several faces. Every point of the line OB, where the planes bisecting the diedrals at AO and EO meet, is equally distant from three faces. The plane bisecting the diedral at EI cuts OB at C, which is equally distant from all four faces.
fig: 215 0.375 0.505 0.435 0.72
art: 626 | Corollary | The six planes which bisect the several diedral angles of a tetraedron all meet at one point.
art: 627 | Theorem | Two tetraedrons are equal when three faces of the one are respectively equal to three faces of the other, and they are similarly arranged.
art: 628 | Corollary | Two tetraedrons are equal when the six edges of the one are respectively equal to those of the other, and they are similarly arranged.
art: 631 |  | When tetraedrons are composed of equal parts in reverse order, they are symmetrical.
art: 632 | Model tetraedron | The student may easily construct a model of a tetraedron when the six edges are given. Draw one face, as ABC; on each side of it draw a triangle equal to the corresponding face. Then cut out the whole figure and carefully fold it at the lines AB, BC, and CA, uniting the points F, D, and E to form a vertex. But he must never forget that the geometrical figure exists only as an intellectual conception.
fig: 217 0.16 0.265 0.44 0.8
art: 633 | Similar tetraedrons | When two polyedrons are similar, their homologous faces are similar polygons, their homologous edges are of equal diedral angles, and their homologous vertices are of equal polyedrals.
art: 634 | Theorem | When two tetraedrons are similar, any edge or other line in the one is to the homologous line in the second, as any other line in the first is to its homologous line in the second. Thus the altitudes AK and BH have the same ratio as two homologous edges.
fig: 218 0.195 0.379
art: 635 | Theorem | Two tetraedrons are similar when their faces are respectively similar triangles, and are similarly arranged.
art: 639 | Corollary | Two tetraedrons are similar when the edges of one are respectively proportional to those of the other, and they are similarly arranged.
art: 640 | Theorem | The areas of homologous faces of similar tetraedrons are to each other as the squares of their edges.
art: 643 | Corollary | The area of the entire surface of one tetraedron is to that of a similar tetraedron as the squares of homologous lines.
art: 644 | Theorem | If a plane cut a tetraedron parallel to the base, the tetraedron cut off is similar to the whole.
art: 645 | Theorem | If two tetraedrons, having the same altitude and their bases on the same plane, are cut by a plane parallel to their bases, the areas of the sections will have the same ratio as the areas of the bases.
fig: 221 0.156 0.335
art: 646 | Corollary | When the bases are equivalent the sections are equivalent.
art: 648 | Regular tetraedron | There is one form of the tetraedron which has all its faces equilateral. This is called a regular tetraedron. Its faces are equal triangles, its vertices are of equal triedrals, and its edges are of equal diedral angles.
art: 651 | Exercises | 2. In any tetraedron, the lines which join the centers of the opposite edges bisect each other. 3. If one of the vertices of a tetraedron is a trirectangular triedral, the square of the area of the opposite face is equal to the sum of the squares of the areas of the other three faces.
mc: A solid bounded by plane surfaces is a... | *polyedron | polyedral | polygon
mc: The faces of a polyedron are... | *polygons | circles | triedrals
mc: A straight line joining two vertices of a polyedron not in the same face is a... | *diagonal | edge | altitude
mc: A polyedron having four faces is a... | *tetraedron | triedral | prism
mc: How many edges has a tetraedron? | *Six | Four | Eight
mc: Any four points not all in one plane may be the vertices of a... | *tetraedron | square | triedral
mc: There is a point equally distant from the four ___ of any tetraedron. | *vertices | edges | diagonals
mc: The point within a tetraedron equally distant from its faces is where the planes ___ meet. | *bisecting its diedrals | bisecting its edges | through its vertices
mc: A plane parallel to the base of a tetraedron cuts off a tetraedron which is... | *similar to the whole | equal to the whole | symmetrical to the whole
mc: The areas of homologous faces of similar tetraedrons are as... | *the squares of their edges | their edges | the cubes of their edges
mc: A tetraedron with all its faces equilateral is called... | *regular | isosceles | symmetrical
tf: Three planes can inclose a space. {F}
tf: The point equally distant from the four vertices is always inside the tetraedron. {F}
tf: Two tetraedrons with their six edges respectively equal and similarly arranged are equal. {T}
q: Two similar tetraedrons have homologous edges of 2 and 6 inches. The area of a face of the larger is how many times that of the homologous face of the smaller? {9}
sol: The areas are as the squares of the edges: 36 : 4 = 9.
q: The altitudes of two similar tetraedrons are 3 and 5 inches. An edge of the first is 9 inches. How long is the homologous edge of the second, in inches? {15}
sol: Homologous lines are proportional: 3 : 5 :: 9 : x, so x = 45 / 3 = 15.
q: A tetraedron 10 inches high is cut by a plane parallel to its base, 4 inches from the vertex. Its base has an area of 50 square inches. What is the area of the section, in square inches? {8}
sol: The tetraedron cut off is similar; areas are as the squares of the altitudes: 50 times 16 / 100 = 8.
q: The entire surface of a tetraedron is 20 square inches. A similar tetraedron has edges 3 times as long. What is its entire surface, in square inches? {180}
sol: Surfaces are as the squares of homologous lines: 20 times 9 = 180.

== 42 | Polyedrons | Pyramids
art: 652 |  | If a polyedral is cut by a plane which cuts its several edges, the section is a polygon, and a portion of space is cut off, which is called a pyramid. A pyramid is a polyedron having for one face any polygon, and for its other faces, triangles whose vertices meet at one point. The polygon is the base of the pyramid, the triangles are its sides, and their intersections are the lateral edges. The perpendicular distance from the vertex to the plane of the base is its altitude. Pyramids are called triangular, quadrangular, pentagonal, etc., according to the polygon which forms the base. The tetraedron is a triangular pyramid.
art: 653 | Problem | Every pyramid can be divided into the same number of tetraedrons as its base can be into triangles. Let a diagonal plane pass through the vertex of the pyramid and each diagonal of the base, and the solution is evident.
art: 654 | Theorem | Two pyramids are equal when the base and two adjacent sides of the one are respectively equal to the corresponding parts of the other, and they are similarly arranged.
art: 655 | Theorem | Two similar pyramids are composed of tetraedrons respectively similar, and similarly arranged; and, conversely, two pyramids are similar when composed of similar tetraedrons, similarly arranged.
art: 656 | Theorem | When a pyramid is cut by a plane parallel to the base, the pyramid cut off is similar to the whole.
art: 657 |  | A regular pyramid is one whose base is a regular polygon, and whose vertex is in the line perpendicular to the base at its center.
art: 658 | Corollary | The lateral edges of a regular pyramid are all equal, and the sides are equal isosceles triangles.
art: 659 |  | The slant hight of a regular pyramid is the perpendicular let fall from the vertex upon one side of the base. It is therefore the altitude of one of the sides of the pyramid.
art: 660 | Theorem | The area of the lateral surface of a regular pyramid is equal to half the product of the perimeter of the base by the slant hight. For the area of each side is half the product of its base by its altitude, which is the slant hight; and the sum of the bases of the sides is the perimeter of the base.
art: 661 |  | When a pyramid is cut by a plane parallel to the base, that part of the figure between this plane and the base is called a frustum of a pyramid, or a truncated pyramid.
art: 662 | Corollary | The sides of a frustum of a pyramid are trapezoids; and the sides of the frustum of a regular pyramid are equal trapezoids.
art: 663 |  | The section made by the cutting plane is called the upper base of the frustum. The slant hight of the frustum of a regular pyramid is that part of the slant hight of the original pyramid which lies between the bases of the frustum.
art: 664 | Theorem | The area of the lateral surface of the frustum of a regular pyramid is equal to half the product of the sum of the perimeters of the bases by the slant hight.
art: 665 | Corollary | The area of the lateral surface of a frustum of a regular pyramid is equal to the product of the perimeter of a section midway between the two bases, multiplied by the slant hight.
art: 667 | Model pyramids | The student may construct a model of a regular pyramid. First, draw a regular polygon of any number of sides. Upon these sides, as bases, draw equal isosceles triangles, taking care that their altitude be greater than the apothem of the base. The figure may then be cut out and folded.
art: 668 | Exercises | 1. Find the area of the surface of a regular octagonal pyramid whose slant hight is 5 inches, and a side of whose base is 2 inches. 2. What is the area in square inches of the entire surface of a regular tetraedron, the edge being one inch? Answer, the square root of 3.
mc: A polyedron with one face any polygon and the other faces triangles meeting at one point is a... | *pyramid | prism | tetraedral
mc: A pyramid whose base is a triangle is a... | *tetraedron | prism | frustum
mc: A pyramid whose base is a regular polygon and whose vertex is over its center is... | *a regular pyramid | a frustum | a right prism
mc: The sides of a regular pyramid are equal... | *isosceles triangles | rectangles | trapezoids
mc: The perpendicular from the vertex of a regular pyramid upon one side of the base is the... | *slant hight | altitude | lateral edge
mc: The lateral surface of a regular pyramid is half the product of the perimeter of the base by the... | *slant hight | altitude | lateral edge
mc: The part of a pyramid between the base and a parallel cutting plane is a... | *frustum | prism | tetraedron
mc: The sides of a frustum of a pyramid are... | *trapezoids | triangles | rectangles
mc: A pentagonal pyramid can be divided into how many tetraedrons? | *Three | Five | Two
tf: The lateral edges of a regular pyramid are all equal. {T}
tf: The slant hight of a regular pyramid is the same as its altitude. {F}
tf: A pyramid cut by a plane parallel to the base leaves a pyramid similar to the whole. {T}
q: A regular octagonal pyramid has a slant hight of 5 inches, and each side of its base is 2 inches. What is the area of its lateral surface, in square inches? {40}
sol: The perimeter is 8 times 2 = 16; half of 16 times 5 is 40.
q: A regular square pyramid has a base 6 feet on a side and a slant hight of 10 feet. What is the area of its lateral surface, in square feet? {120}
sol: Perimeter 24; half of 24 times 10 = 120.
q: The same square pyramid, 6 feet on a side with slant hight 10 feet: what is the area of its entire surface, base included, in square feet? {156}
sol: Lateral surface 120, plus the base 6 times 6 = 36: 156.
q: A frustum of a regular square pyramid has bases 4 and 6 inches on a side and a slant hight of 5 inches. What is its lateral surface, in square inches? {100}
sol: Perimeters 16 and 24; half their sum, 20, times the slant hight 5 = 100.
q: A regular hexagonal pyramid has base sides of 3 inches and a slant hight of 8 inches. What is its lateral surface, in square inches? {72}
sol: Perimeter 18; half of 18 times 8 = 72.
q: How many tetraedrons can a pyramid with an octagonal base be divided into? {6}
sol: As many as the base can be divided into triangles: 8 - 2 = 6.

== 43 | Polyedrons | Prisms
art: 669 |  | A prism is a polyedron which has two of its faces equal polygons lying in parallel planes, and the other faces parallelograms. The parallel polygons are called the bases, the parallelograms the sides of the prism, and the intersections of the sides are its lateral edges. The altitude of a prism is the perpendicular distance between the planes of its bases.
fig: 226 0.65 0.845 0.575 0.8
art: 670 | Corollary | The lateral edges of a prism are all parallel to each other, and therefore equal to each other.
art: 671 |  | A right prism is one whose lateral edges are perpendicular to the bases. A regular prism is a right prism whose base is a regular polygon.
art: 672 | Corollary | The altitude of a right prism is equal to one of its lateral edges; and the sides of a right prism are rectangles. The sides of a regular prism are equal.
art: 673 | Theorem | If two parallel planes pass through a prism, so that each plane cuts every lateral edge, the sections made by the two planes are equal polygons.
art: 674 | Corollary | The section of a prism made by a plane parallel to the base is equal to the base, and the given prism is divided into two prisms.
art: 675 | Problem | Every prism can be divided into the same number of triangular prisms as its base can be into triangles. Homologous diagonals of the two bases, as EO and CF, lie in one plane; through each pair of them a diagonal plane may pass, dividing the prism into triangular prisms.
fig: 228 0.285 0.445 0.5 0.78
art: 676 | Problem | A triangular prism may be divided into three tetraedrons, which, taken two and two, have equal bases and equal altitudes. A diagonal plane through B, C, and H cuts off the tetraedron BCDH. The remainder is a quadrangular pyramid with base BCGF and vertex H, which the plane through H, G, and B divides into the tetraedrons HBCG and HBFG.
fig: 228 0.58 0.765 0.53 0.78
art: 678 | Remark | On account of the importance of the above problem in future demonstrations, the student is advised to make a model triangular prism, and divide it into tetraedrons. A potato may be used for this purpose.
art: 679 | Theorem | Two prisms are equal, when a base and two adjacent sides of the one are respectively equal to the corresponding parts of the other, and they are similarly arranged.
art: 680 | Corollary | Two right prisms are equal when they have equal bases and the same altitude.
art: 682 | Theorem | The area of the lateral surface of a prism is equal to the product of one of the lateral edges by the perimeter of a section, made by a plane perpendicular to those edges. For each side is a parallelogram whose base is an edge of the prism and whose altitude is a side of the section HN; and the edges are equal.
fig: 231 0.18 0.48 0.515 0.76
art: 683 | Corollary | The area of the lateral surface of a right prism is equal to the product of the altitude by the perimeter of the base.
art: 684 | Corollary | The area of the entire surface of a regular prism is equal to the product of the perimeter of the base by the sum of the altitude of the prism and the apothem of the base.
art: 685 | Exercises | 1. A right prism has less surface than any other prism of equal base and equal altitude; and a regular prism has less surface than any other right prism of equivalent base and equal altitude.
mc: A polyedron with two equal polygons in parallel planes and parallelograms for its other faces is a... | *prism | pyramid | tetraedron
mc: The lateral edges of a prism are all... | *parallel and equal | perpendicular to each other | meeting at one point
mc: A prism whose lateral edges are perpendicular to the bases is a... | *right prism | regular pyramid | frustum
mc: A right prism whose base is a regular polygon is a... | *regular prism | regular pyramid | cube
mc: The sides of a right prism are... | *rectangles | triangles | trapezoids
mc: A section of a prism parallel to the base is ___ the base. | *equal to | similar but smaller than | larger than
mc: A triangular prism may be divided into how many tetraedrons? | *Three | Two | Four
mc: The lateral surface of a right prism equals the altitude times... | *the perimeter of the base | the area of the base | the slant hight
tf: The altitude of a right prism is equal to one of its lateral edges. {T}
tf: Two right prisms with equal bases and the same altitude are equal. {T}
tf: A hexagonal prism can be divided into four triangular prisms. {T}
q: A right prism is 10 inches high, and its base is a triangle with sides 3, 4, and 5 inches. What is its lateral surface, in square inches? {120}
sol: The perimeter of the base is 12; 12 times 10 = 120.
q: A regular hexagonal prism has base sides of 2 inches and an altitude of 7 inches. What is its lateral surface, in square inches? {84}
sol: Perimeter 12, times 7 = 84.
q: A cube has edges of 3 inches. What is the area of its entire surface, in square inches? {54}
sol: Six square faces of 9 square inches each: 54.
q: An oblique prism has lateral edges 8 inches long. A section perpendicular to the edges has a perimeter of 15 inches. What is its lateral surface, in square inches? {120}
sol: One lateral edge times the perimeter of the perpendicular section: 8 times 15 = 120.
q: A regular square prism has base sides of 4 inches and an altitude of 9 inches. What is its entire surface, in square inches? {176}
sol: Lateral surface 16 times 9 = 144, plus two bases of 16 each: 144 + 32 = 176.
q: Into how many triangular prisms can a prism with a pentagonal base be divided? {3}
sol: As many as the base can be divided into triangles: 5 - 2 = 3.
== 44 | Polyedrons | Measure of volume
art: 686 |  | A parallelopiped is a prism whose bases are parallelograms. Hence, a parallelopiped is a solid inclosed by six parallelograms.
art: 687 | Theorem | The opposite sides of a parallelopiped are equal. For example, the faces AI and BD are equal. For IO and DF are equal, being opposite sides of the parallelogram IF; EI is equal to CD; and since these equal sides are also parallel, the included angles EIO and CDF are equal. Hence, the parallelograms are equal.
fig: 232 0.345 0.485 0.475 0.78
art: 689 |  | A parallelopiped is called right in the same case as any other prism. When the bases also are rectangles, it is called rectangular. Then all the faces are rectangles.
art: 690 |  | A cube is a rectangular parallelopiped whose length, breadth, and altitude are equal. Then a cube is a solid, bounded by six equal squares. The cube has the simplest form of all geometrical solids. The cube is taken, therefore, as the unit of measure of volume. Whatever straight line is taken as the unit of length, the cube whose edge is of that length is the unit of volume.
art: 691 | Theorem | The volume of a rectangular parallelopiped is equal to the product of its length, breadth, and altitude. If the altitude, the length, and the breadth have a common measure, let planes parallel to the faces pass through all the points of division. The prism is divided into equal cubes. The number at the base is the number of rows multiplied by the number in each row, and there are as many layers as there are units of altitude. In the diagram, the dimensions being four, three, and two, the volume is twenty-four.
fig: 233 0.4 0.575 0.43 0.73
art: 692 | Theorem | The volume of any parallelopiped is equal to the product of its length, breadth, and altitude. For any parallelopiped is equivalent to a rectangular one having the same linear dimensions. Let the altitude be divided into an infinite number of equal parts, and through each point of division pass a plane parallel to the base. Every section in either prism is equal to its base, and the bases are equivalent; therefore the two parallelopipeds are equivalent.
fig: 235 0.22 0.365
fig: 235 0.675 0.84 0.45 0.78
art: 693 | Corollary | The volume of a cube is equal to the third power of its edge. Thence comes the name of cube, to designate the third power of a number.
art: 694 | Model cubes | Draw six equal squares, as in the diagram. Cut out the figure, fold at the dividing lines, and glue the edges. It is well to have at least eight of one size.
fig: 236 0.39 0.48 0.505 0.76
art: 695 | Corollary | The volume of any parallelopiped is equal to the product of its base by its altitude.
art: 696 | Corollary | The volumes of any two parallelopipeds are to each other as the products of their three dimensions.
art: 697 | Theorem | The volume of any triangular prism is equal to the product of its base by its altitude. The base of a right triangular prism may be considered as one-half of the base of a right parallelopiped; the whole parallelopiped is double the given prism.
art: 698 | Corollary | The volume of any prism is equal to the product of its base by its altitude. For any prism is composed of triangular prisms, having the common altitude of the given prism, and the sum of their bases forming the given base.
art: 700 | Theorem | Two tetraedrons of equivalent bases and of the same altitude are equivalent. Suppose the bases in the same plane; the vertices then lie in a plane parallel to the bases. Planes parallel to the base cut both into infinitesimal frustums, and each section of one tetraedron is equivalent to the section made by the same plane in the other.
fig: 237 0.632 0.805 0.05 0.72
art: 701 | Theorem | The volume of a tetraedron is equal to one-third of the product of the base by the altitude. Upon the base of the tetraedron a triangular prism may be erected, of the same altitude. This prism may be divided into three tetraedrons, the given one and two others, which, taken two and two, have equal bases and altitudes. These three are equivalent, and the given tetraedron is one-third of the prism.
art: 702 | Corollary | The volume of any pyramid is equal to one-third of the product of its base by its altitude.
art: 703 | Corollary | The volumes of two prisms of equivalent bases are to each other as their altitudes, and the volumes of two prisms of equal altitudes are to each other as their bases. The same is true of pyramids.
art: 704 | Corollary | Symmetrical prisms are equivalent. The same is true of symmetrical pyramids.
art: 705 |  | The volume of a frustum of a pyramid is found by subtracting the volume of the pyramid cut off from the volume of the whole. When the altitude of the whole is not given, it may be found by this proportion: the area of the lower base of the frustum is to the area of its upper base, as the square of the whole altitude is to the square of the altitude of the part cut off.
art: 706 | Exercises | 1. What is the ratio of the volumes of a pyramid and prism having the same base and altitude?
mc: A prism whose bases are parallelograms is a... | *parallelopiped | pyramid | cube
mc: A parallelopiped whose faces are all rectangles is... | *rectangular | oblique | regular
mc: A rectangular parallelopiped whose length, breadth, and altitude are equal is a... | *cube | prism | square
mc: The unit of measure of volume is... | *a cube | a square | a straight line
mc: The volume of a rectangular parallelopiped is the product of its... | *length, breadth, and altitude | length and breadth | base and slant hight
mc: The volume of a cube is the ___ power of its edge. | *third | second | fourth
mc: The volume of any prism is the product of its base by its... | *altitude | lateral edge | perimeter
mc: The volume of a pyramid is ___ of the product of its base by its altitude. | *one-third | one-half | two-thirds
mc: A pyramid and a prism have the same base and altitude. The pyramid's volume is what part of the prism's? | *One-third | One-half | The same
mc: Two prisms of equal altitudes have volumes in the ratio of their... | *bases | lateral edges | perimeters
tf: Opposite faces of a parallelopiped are equal. {T}
tf: Two tetraedrons with equivalent bases and the same altitude are equivalent. {T}
tf: Symmetrical pyramids are equivalent. {T}
q: A rectangular box is 4 feet long, 3 feet broad, and 2 feet high. What is its volume, in cubic feet? {24}
sol: 4 times 3 times 2 = 24 cubic feet.
q: What is the volume of a cube whose edge is 5 inches, in cubic inches? {125}
sol: The third power of the edge: 5 times 5 times 5 = 125.
q: A prism has a base of 12 square inches and an altitude of 7 inches. What is its volume, in cubic inches? {84}
sol: Base times altitude: 12 times 7 = 84.
q: A pyramid has a base of 12 square inches and an altitude of 7 inches. What is its volume, in cubic inches? {28}
sol: One-third of base times altitude: 84 / 3 = 28.
q: A right triangular prism has a base which is a right triangle with legs 3 and 4 inches, and an altitude of 10 inches. What is its volume, in cubic inches? {60}
sol: The base is half of 3 times 4 = 6 square inches; 6 times 10 = 60.
q: A square pyramid has a base 6 feet on a side and an altitude of 9 feet. What is its volume, in cubic feet? {108}
sol: The base is 36; one-third of 36 times 9 = 108.
q: A pyramid 12 feet high is cut parallel to its base 6 feet from the vertex. The base is 48 square feet. What is the volume of the frustum, in cubic feet? {168}
sol: The whole pyramid is 48 times 12 / 3 = 192. The part cut off has half the altitude, so its base is 48 / 4 = 12 and its volume 12 times 6 / 3 = 24. The frustum is 192 - 24 = 168.
q: Two parallelopipeds measure 2 by 3 by 4 and 4 by 6 by 8 inches. The second is how many times the first? {8}
sol: Volumes are as the products of the three dimensions: 192 : 24 = 8.

== 45 | Polyedrons | Similar polyedrons
art: 707 |  | The propositions upon the ratios of the areas of the surfaces of similar tetraedrons may be applied to any similar polyedrons. These propositions are equally applicable to polyedrons that are symmetrically similar.
art: 708 | Problem | Any two similar polyedrons may be divided into the same number of similar tetraedrons, which shall be respectively similar, and similarly arranged.
art: 709 | Theorem | The volumes of similar polyedrons are proportional to the cubes of homologous lines. First, for two tetraedrons with altitudes AH and BG: their bases are as the squares of homologous lines, and their altitudes as the lines themselves. Multiplying, the volumes are as the cubes of homologous lines. Any two similar polyedrons may then be divided into the same number of similar tetraedrons.
fig: 240 0.3 0.52
art: 710 |  | Notice that in the measure of every area there are two linear dimensions; and in the measure of every volume, three linear, or one linear and one superficial.
art: 711 | Exercise | What is the ratio between the edges of two cubes, one of which has twice the volume of the other? This problem of the duplication of the cube was one of the celebrated problems of ancient times. It is said that the oracle of Apollo demanded of the Athenians a new altar, of the same shape, but of twice the volume of the old one. The Greek geometers tried to draw the edge using no instruments but the rule and compasses. In this they failed. The student will find no difficulty in making an arithmetical solution, within any desired degree of approximation.
mc: The volumes of similar polyedrons are proportional to the ___ of homologous lines. | *cubes | squares | first powers
mc: The surfaces of similar polyedrons are proportional to the ___ of homologous lines. | *squares | cubes | first powers
mc: How many linear dimensions are in the measure of every volume? | *Three | Two | One
mc: The ancient problem of making a cube of twice the volume of another is the... | *duplication of the cube | quadrature of the circle | trisection of the angle
mc: The Greek geometers failed to solve the duplication of the cube with... | *the rule and compasses only | arithmetic | any method at all
tf: Two similar polyedrons can be divided into the same number of similar tetraedrons. {T}
tf: If the edges of a cube are doubled, its volume is doubled. {F}
q: Two similar polyedrons have homologous edges of 1 and 3 inches. The volume of the larger is how many times the smaller? {27}
sol: Volumes are as the cubes of homologous lines: 27 : 1.
q: A model of a building is made with every line 1/10 of the real one. The model holds 2 cubic feet. How many cubic feet does the building hold? {2000}
sol: Volumes are as the cubes: 10 times 10 times 10 = 1000, and 2 times 1000 = 2000.
q: Two similar pyramids have altitudes of 2 and 4 feet. The smaller has a volume of 5 cubic feet. What is the volume of the larger, in cubic feet? {40}
sol: The ratio of altitudes is 2, so the volumes are as 8 : 1; 5 times 8 = 40.
q: Two similar prisms have volumes of 8 and 64 cubic inches. An edge of the smaller is 3 inches. How long is the homologous edge of the larger, in inches? {6}
sol: The volumes are as 1 : 8, so the lines are as the cube roots, 1 : 2; 3 times 2 = 6.
q: A cube has twice the volume of a cube whose edge is 1 inch. Its edge, to the nearest hundredth of an inch, is how many inches? {1.26}
sol: The edge is the cube root of 2, which is 1.2599..., about 1.26 inches.

== 46 | Polyedrons | Regular polyedrons
art: 712 |  | A regular polyedron is one whose faces are equal and regular polygons, and whose vertices are equal polyedrals. The regular tetraedron and the cube, or regular hexaedron, have been described. The regular octaedron has eight, the dodecaedron twelve, and the icosaedron twenty faces. This class of figures must not be confounded with regular pyramids or prisms.
art: 713 | Problem | It is not possible to make more than five regular polyedrons. Each angle of a regular triangle is one-third of two right angles. Three, four, or five of these may form a vertex, the sum being less than four right angles; but six are not less than four right angles. So there are three with triangular faces: the tetraedron, octaedron, and icosaedron. Only one is possible with square faces, the cube; and only one with pentagonal faces, the dodecaedron. Three angles of a regular hexagon are not less than four right angles; therefore hexagons, and polygons of more sides, can not form a regular polyedron.
art: 714 | Model regular polyedrons | For the octaedron, draw eight equal regular triangles, as in the diagram. For the dodecaedron, draw twelve equal regular pentagons. For the icosaedron, draw twenty equal regular triangles. There are many crystals, which, though not regular in the geometrical rigor of the word, yet present a certain regularity of shape.
fig: 243 0.12 0.21 0.4 0.8
fig: 243 0.24 0.365 0.38 0.8
fig: 243 0.385 0.48 0.38 0.8
art: 715 | Exercises | 1. How many edges and how many vertices has each of the regular polyedrons? 6. The diagonals of a rectangular parallelopiped are equal. 7. The square of the diagonal of a rectangular parallelopiped is equivalent to the sum of the squares of its length, breadth, and altitude. 8. A cube is the largest parallelopiped of the same extent of surface. 14. The area of the lower base of a frustum of a pyramid is five square feet, of the upper base one and four-fifths square feet, and the altitude is two feet; required the volume.
mc: A polyedron whose faces are equal regular polygons and whose vertices are equal polyedrals is... | *regular | similar | symmetrical
mc: How many regular polyedrons are possible? | *Five | Six | Any number
mc: How many faces has a regular octaedron? | *Eight | Six | Twelve
mc: How many faces has a regular dodecaedron? | *Twelve | Ten | Twenty
mc: How many faces has a regular icosaedron? | *Twenty | Twelve | Eight
mc: The faces of a regular dodecaedron are... | *pentagons | triangles | squares
mc: The cube is also called the regular... | *hexaedron | octaedron | tetraedron
mc: Which polygons can not be the faces of a regular polyedron? | *Hexagons | Squares | Pentagons
mc: How many regular polyedrons have triangles for faces? | *Three | One | Two
tf: Five regular triangles can meet at a vertex of a regular polyedron. {T}
tf: Six regular triangles can meet at a vertex of a regular polyedron. {F}
tf: The diagonals of a rectangular parallelopiped are equal. {T}
q: How many edges has a cube? {12}
sol: Six square faces have 24 sides, and each edge belongs to two faces: 24 / 2 = 12.
q: How many vertices has a regular octaedron? {6}
sol: Eight triangles have 24 corners, and four meet at each vertex: 24 / 4 = 6.
q: How many edges has a regular dodecaedron? {30}
sol: Twelve pentagons have 60 sides, and each edge belongs to two faces: 60 / 2 = 30.
q: How many vertices has a regular icosaedron? {12}
sol: Twenty triangles have 60 corners, and five meet at each vertex: 60 / 5 = 12.
q: A rectangular box is 3 feet long, 4 feet broad, and 12 feet high. How long is its diagonal, in feet? {13}
sol: The square of the diagonal is 9 + 16 + 144 = 169, and the square root is 13.
q: A frustum of a pyramid has a lower base of 5 square feet, an upper base of 1 4/5 square feet, and an altitude of 2 feet. What is its volume, in cubic feet? Give a fraction. {98/15}
sol: The bases are as 25 : 9, so the altitudes of the whole pyramid and of the part cut off are as 5 : 3; they differ by 2, so they are 5 and 3 feet. The whole is 5 times 5 / 3 = 25/3; the part cut off is 9/5 times 3 / 3 = 9/5. The frustum is 25/3 - 9/5 = 98/15, or 6 8/15 cubic feet.
== 47 | Solids of revolution | Cones
art: 716 |  | Of the infinite variety of forms there remain but three to be considered in this elementary work. These are formed or generated by the revolution of a plane figure about one of its lines as an axis. Figures formed in this way are called solids of revolution.
art: 717 |  | A cone is a solid formed by the revolution of a right angled triangle about one of its legs as an axis. The other leg revolving describes a plane surface, which is a circle. The hypotenuse describes a curved surface. The plane surface of a cone is called its base. The opposite extremity of the axis is the vertex. The altitude is the distance from the vertex to the base, and the slant hight is the distance from the vertex to the circumference of the base.
fig: 245 0.455 0.585 0.555 0.74
art: 719 |  | As the curved surfaces of the cone and of the cylinder are generated by the motion of a straight line, each of these surfaces is straight in one direction. A straight line from the vertex of the cone to the circumference of the base must lie wholly in the surface. One surface is tangent to another when it meets, but being produced does not cut it.
art: 720 | Conic sections | If a cone or a cylinder be cut by a plane parallel to the base, the section is a circle. If a cone is cut by a plane in other positions, the curves formed are the ellipse, parabola, and hyperbola. These conic sections are not usually considered in Elementary Geometry, as their properties can be better investigated by the application of algebra.
art: 721 |  | A cone is said to be inscribed in a pyramid, when their bases lie in one plane, and the sides of the pyramid are tangent to the curved surface of the cone. A cone is said to be circumscribed about a pyramid, when their bases lie in one plane, and the lateral edges of the pyramid lie in the curved surface of the cone.
art: 722 | Theorem | A cone is the limit of the pyramids which can be circumscribed about it; also of the pyramids which can be inscribed in it. By increasing the number of sides of the pyramid, it can be made to approximate to the cone within less than any appreciable difference.
fig: 247 0.595 0.755 0.455 0.74
art: 723 | Corollary | The area of the curved surface of a cone is equal to one-half the product of the slant hight by the circumference of the base.
art: 724 | Corollary | The area of the entire surface of a cone is equal to half of the product of the circumference of the base by the sum of the slant hight and the radius of the base.
art: 725 | Corollary | The volume of a cone is equal to one-third of the product of the base by the altitude.
art: 727 | Corollary | The area of the curved surface of the frustum of a cone is equal to half the product of its slant hight by the sum of the circumferences of its bases.
art: 728 | Corollary | If a cone be cut by a plane parallel to the base, the cone cut off is similar to the whole.
art: 729 | Exercises | 1. Two cones are similar when they are generated by similar triangles, homologous sides being used for the axes. 2. A section of a cone by a plane passing through the vertex, is an isosceles triangle.
mc: A cone is formed by the revolution of... | *a right angled triangle about one of its legs | a rectangle about one side | a semicircle about its diameter
mc: The plane surface of a cone is its... | *base | vertex | slant hight
mc: The distance from the vertex of a cone to the circumference of the base is its... | *slant hight | altitude | radius
mc: A cone cut by a plane parallel to its base gives a section which is a... | *circle | triangle | parabola
mc: The ellipse, parabola, and hyperbola are... | *conic sections | spheres | polyedrons
mc: A cone is the limit of the ___ which can be inscribed in it or circumscribed about it. | *pyramids | prisms | cylinders
mc: The curved surface of a cone is half the product of the slant hight by... | *the circumference of the base | the radius of the base | the altitude
mc: The volume of a cone is ___ of the product of its base by its altitude. | *one-third | one-half | the whole
mc: A section of a cone by a plane through the vertex is... | *an isosceles triangle | a circle | a rectangle
tf: The curved surface of a cone is straight in one direction. {T}
tf: A cone cut by a plane parallel to its base leaves a cone similar to the whole. {T}
q: Using 3 1/7 for pi: a cone has a base of radius 7 inches and a slant hight of 10 inches. What is its curved surface, in square inches? {220}
sol: The circumference is 22/7 times 14 = 44; half of 44 times 10 = 220.
q: Using 3 1/7 for pi: the same cone, radius 7 and slant hight 10 inches. What is its entire surface, in square inches? {374}
sol: Half the circumference times the sum of slant hight and radius: 22 times 17 = 374.
q: Using 3 1/7 for pi: a cone has a base of radius 7 inches and an altitude of 6 inches. What is its volume, in cubic inches? {308}
sol: The base is 22/7 times 49 = 154; one-third of 154 times 6 = 308.
q: A cone has an altitude of 8 inches and a base of radius 6 inches. How long is its slant hight, in inches? {10}
sol: The altitude, radius, and slant hight form a right triangle: the square root of 64 + 36 = 100 is 10.
q: A cone with a base of 30 square inches is 12 inches high. What is its volume, in cubic inches? {120}
sol: One-third of 30 times 12 = 120.

== 48 | Solids of revolution | Cylinders
art: 718 |  | A cylinder is a solid described by the revolution of a rectangle about one of its sides as an axis. The sides adjacent to the axis describe circles, while the opposite side describes a curved surface. The plane surfaces of a cylinder are called its bases, and the perpendicular distance between them is its altitude.
fig: 245 0.705 0.835 0.56 0.74
art: 730 |  | A cylinder is said to be inscribed in a prism, when their bases lie in the same planes, and the sides of the prism are tangent to the curved surface of the cylinder. A cylinder is said to be circumscribed about a prism, when their bases lie in the same planes, and the lateral edges of the prism lie in the curved surface of the cylinder.
fig: 249 0.4 0.56 0.545 0.74
fig: 249 0.565 0.725 0.545 0.74
art: 731 | Theorem | A cylinder is the limit of the prisms which can be circumscribed about it; also of those which can be inscribed in it.
art: 732 | Corollary | The area of the curved surface of a cylinder is equal to the product of the altitude by the circumference of the base.
art: 733 | Corollary | The area of the entire surface of a cylinder is equal to the product of the circumference of the base by the sum of the altitude and the radius of the base.
art: 734 | Corollary | The volume of a cylinder is equal to the product of the base by the altitude.
art: 735 | Model cones and cylinders | Models of cones and cylinders may be made from paper, taking a sector of a circle for the curved surface of a cone, and a rectangle for the curved surface of a cylinder. Make the bases separately.
art: 736 | Exercises | 2. A section of a cylinder made by a plane perpendicular to the base is a rectangle. 3. The axis of a cone or of a cylinder is equal to its altitude.
mc: A cylinder is formed by the revolution of... | *a rectangle about one of its sides | a right triangle about a leg | a semicircle about its diameter
mc: The plane surfaces of a cylinder are its... | *bases | sides | axes
mc: A cylinder is the limit of the ___ inscribed in it or circumscribed about it. | *prisms | pyramids | cones
mc: The curved surface of a cylinder is the altitude times... | *the circumference of the base | the area of the base | the radius
mc: The volume of a cylinder is the product of... | *the base by the altitude | the circumference by the altitude | the radius by the altitude
mc: A section of a cylinder by a plane perpendicular to the base is a... | *rectangle | circle | triangle
mc: A paper model of the curved surface of a cylinder is cut as a... | *rectangle | sector of a circle | triangle
tf: The axis of a cylinder is equal to its altitude. {T}
tf: A cylinder has the same volume as a cone of equal base and altitude. {F}
q: Using 3 1/7 for pi: a cylinder has a radius of 7 inches and an altitude of 10 inches. What is its curved surface, in square inches? {440}
sol: The circumference is 44; 44 times 10 = 440.
q: Using 3 1/7 for pi: the same cylinder, radius 7 and altitude 10 inches. What is its entire surface, in square inches? {748}
sol: The circumference times the sum of altitude and radius: 44 times 17 = 748.
q: Using 3 1/7 for pi: a cylinder has a radius of 7 inches and an altitude of 10 inches. What is its volume, in cubic inches? {1540}
sol: The base is 22/7 times 49 = 154; 154 times 10 = 1540.
q: A cylinder and a cone have equal bases and equal altitudes. The cylinder holds 90 cubic inches. How many cubic inches does the cone hold? {30}
sol: The cone is one-third of the product of base and altitude: 90 / 3 = 30.
q: Using 3 1/7 for pi: a round cistern is 14 feet across and 5 feet deep. How many cubic feet does it hold? {770}
sol: The radius is 7; the base is 154 square feet; 154 times 5 = 770.

== 49 | Solids of revolution | Spheres
art: 737 |  | A sphere is a solid described by the revolution of a semicircle about its diameter as an axis. The center, radius, and diameter of the sphere are the same as those of the generating circle. The spherical surface is described by the circumference.
fig: 250 0.685 0.815 0.545 0.74
art: 738 | Corollary | Every point on the surface of the sphere is equally distant from the center. This property of the sphere is frequently given as its definition.
art: 739 | Corollary | All radii of the same sphere are equal. The same is true of the diameters.
art: 741 | Corollary | A plane passing through the center of a sphere divides it into equal parts. The halves of a sphere are called hemispheres.
art: 742 | Theorem | A plane which is perpendicular to a radius of a sphere at its extremity is tangent to the sphere. For if straight lines extend from the center to any other point of the plane, they are oblique and longer than the radius, which is perpendicular. Therefore, every point of the plane except one is beyond the surface of the sphere.
fig: 251 0.43 0.545 0.51 0.74
art: 743 |  | The spherical surface is curved in every direction. Unlike those surfaces which are generated by the motion of a straight line, every possible section of it is a curve.
art: 744 | Theorem | Every section of a sphere made by a plane is a circle. If the section DGF does not pass through the center, let CI fall perpendicularly on the cutting plane from the center C. The oblique lines CD and CG being equal radii, the points D and G must be equally distant from I. Therefore, DGF is the circumference of a circle of which I is the center.
fig: 252 0.2 0.35 0.48 0.74
art: 745 | Corollary | The circle formed by the section through the center is larger than one formed by any plane not through the center.
art: 746 |  | When the plane passes through the center of a sphere, the section is called a great circle; otherwise it is called a small circle.
art: 747 | Corollary | All great circles of the same sphere are equal.
art: 748 | Corollary | Two great circles bisect each other, and their intersection is a diameter of the sphere.
art: 749 | Corollary | If a perpendicular be let fall from the center of a sphere on the plane of a small circle, the foot of the perpendicular is the center of the circle; and conversely, the axis of any circle is a diameter of the sphere. The two points where the axis of a circle pierces the spherical surface, are the poles of the circle.
art: 750 | Corollary | Circles whose planes are parallel to each other have the same axis and the same poles.
art: 751 | Theorem | The shortest line which can extend from one point to another along the surface of a sphere, is the arc of a great circle, passing through the two points. Let ABCDEFG be any curve on the sphere from G to A, and AKG the arc of a great circle. The arcs AD and DG are greater than AKG, for any one of the angles of the triedral at the center is less than the sum of the other two. Taking more and more intermediate points, the lines grow greater and approach the curve, which is their limit.
fig: 253 0.465 0.585 0.42 0.74
art: 752 | Theorem | Every plane passing through the axis of a circle is perpendicular to the plane of that circle, and its section is a great circle.
art: 754 | Corollary | The distance of the pole of a great circle from any point of the circumference is a quadrant.
art: 755 | Applications | The equator is a great circle of the earth, which is nearly a sphere. The parallels of latitude are small circles, all having the same poles as the equator. The meridians are great circles perpendicular to the equator. A vessel crossing the ocean from a port in a certain latitude to a port in the same latitude, should not sail along a parallel of latitude, for that is the arc of a small circle.
art: 756 |  | The curvature of the sphere in every direction renders it impossible to construct an exact model with plane paper. But the student is advised to procure or make a globe, upon which he can draw the diagrams of all the figures.
mc: A sphere is formed by the revolution of... | *a semicircle about its diameter | a circle about a tangent | a rectangle about a side
mc: Every point on the surface of a sphere is equally distant from the... | *center | poles | equator
mc: The halves of a sphere are called... | *hemispheres | lunes | wedges
mc: A plane perpendicular to a radius at its extremity is ___ to the sphere. | *tangent | secant | parallel
mc: Every section of a sphere made by a plane is a... | *circle | ellipse | point
mc: A section through the center of a sphere is a... | *great circle | small circle | pole
mc: The points where the axis of a circle pierces the sphere are its... | *poles | centers | vertices
mc: The shortest path between two points on a sphere is along... | *the arc of a great circle | a parallel of latitude | any small circle
mc: On the earth, the equator is... | *a great circle | a small circle | a pole
mc: The parallels of latitude are... | *small circles | great circles | meridians
mc: The distance from the pole of a great circle to its circumference is... | *a quadrant | a semicircle | a diameter
tf: All great circles of the same sphere are equal. {T}
tf: Two great circles bisect each other. {T}
tf: A ship sailing between two ports at the same latitude takes the shortest path by following the parallel. {F}
tf: An exact model of a sphere can be folded from flat paper. {F}
q: A sphere has a radius of 10 inches. A plane cuts it 6 inches from the center. What is the radius of the section, in inches? {8}
sol: The radius of the sphere, the distance from the center, and the radius of the section make a right triangle: the square root of 100 - 36 = 64 is 8.
q: A sphere has a radius of 13 inches. A small circle on it has a radius of 5 inches. How far is its plane from the center, in inches? {12}
sol: The square root of 169 - 25 = 144 is 12.
q: How many degrees of a great circle make a quadrant? {90}
sol: A quadrant is a quarter of the circumference: 360 / 4 = 90.

== 50 | Solids of revolution | Spherical angles and triangles
art: 757 |  | A spherical angle is the difference in the directions of two arcs of great circles at their point of meeting. The direction of an arc at a given point is the same as the direction of a straight line tangent to the arc at that point. Thus, the direction of the arc BDF at the point B is the same as the direction of the tangent BH.
fig: 255 0.43 0.62 0.535 0.74
art: 758 | Corollary | A spherical angle is the same as the plane angle formed by lines tangent to the given arcs at their point of meeting.
art: 759 | Corollary | A spherical angle is the same as the diedral angle formed by the planes of the two arcs.
art: 760 | Corollary | A spherical angle is measured by the arc of a circle included between the sides of the angle, the pole of the arc being at the vertex. Thus, if DG is an arc of a great circle whose pole is at B, then the spherical angle DBG is measured by the arc DG.
fig: 256 0.13 0.31 0.565 0.78
art: 761 |  | A lune is that portion of the surface of a sphere included between two halves of great circles. That portion of the sphere included between the two planes is called a spherical wedge. Hence, two great circles divide the surface into four lunes, and the sphere into four wedges.
art: 762 |  | A spherical polygon is that portion of the surface of a sphere included between three or more arcs of great circles. If C, the center of a sphere, is the vertex of a convex polyedral, the planes of its faces cut the surface in arcs of great circles, which form the polygon BDFGH. The angles of the polygon are the same as the diedral angles of the polyedral.
fig: 256 0.63 0.76 0.525 0.8
art: 763 | Theorem | The sum of all the sides of a spherical polygon is less than a circumference of a great circle. For the sides measure the faces of the polyedral at the center, and their sum is less than four right angles.
art: 764 | Theorem | A spherical polygon is always within the surface of a hemisphere. For a plane may pass through the vertex of the polyedral, having all of the polyedral on one side of it. Its section is a great circle, as KLM.
fig: 257 0.425 0.575 0.455 0.7
art: 765 |  | That portion of a sphere which is included between a spherical polygon and its corresponding polyedral is called a spherical pyramid, the polygon being its base.
art: 766 | Spherical triangles | If the three planes which form a triedral at the center of a sphere be produced, they divide the sphere into eight parts, each having its triedral at the center, and its spherical triangle at the surface. The triangle vertically opposite the given triangle, as GKH to FDB, has its sides respectively equal to those of the given triangle, but arranged in reverse order. Such spherical triangles are called symmetrical.
fig: 258 0.165 0.33 0.51 0.78
art: 769 | Corollary | The three sides and the three angles of a spherical triangle are respectively the measures of the three faces and the three diedrals of the triedral at the center.
art: 771 | Corollary | The sum of the angles of a spherical triangle is greater than two, and less than six right angles.
art: 772 | Corollary | An isosceles spherical triangle is equal to its symmetrical, and has equal angles opposite the equal sides.
art: 773 | Corollary | The radius being the same, two spherical triangles are equal: 1st, when they have two sides and the included angle respectively equal; 2d, when they have one side and the adjacent angles equal; 3d, when the three sides are equal; 4th, when the three angles are equal; the parts being similarly arranged. When the arrangement is reversed, the triangles are symmetrical.
art: 775 | Polar triangles | If the vertex of a triedral is at the center of a sphere, the triangles of the triedral and of its supplementary triedral are such that the vertex of each angle of one triangle is the pole of the opposite side of the other. Hence, such triangles are called polar triangles.
art: 777 | Theorem | Each angle of a spherical triangle is the supplement of the opposite side of its polar triangle. Let ABC be the given triangle, and EF, DF, and DE arcs of great circles whose poles are A, B, and C. Then ABC and DEF are polar triangles.
fig: 260 0.225 0.395 0.47 0.77
art: 779 | Inscribed and circumscribed | A sphere is inscribed in a polyedron when the faces are tangent to the curved surface. A sphere is circumscribed about a polyedron when the vertices all lie in the curved surface.
art: 780 | Problem | Any tetraedron may have a sphere inscribed in it; also, one circumscribed about it. For within any tetraedron there is a point equally distant from all the faces, and a point equally distant from all the vertices.
art: 781 | Corollary | A spherical surface may be made to pass through any four points not in the same plane.
art: 782 | Exercises | 1. In a spherical triangle, the greater side is opposite the greater angle; and conversely. 3. When two spherical surfaces intersect each other, the line of intersection is a circumference of a circle; and the straight line which joins the centers of the spheres is the axis of that circle.
mc: A spherical angle is the difference in direction of two arcs of... | *great circles | small circles | any curves
mc: A spherical angle is the same as the ___ formed by the planes of its two arcs. | *diedral angle | triedral | lune
mc: The surface between two halves of great circles is a... | *lune | wedge | spherical triangle
mc: Two great circles divide the surface of a sphere into how many lunes? | *Four | Two | Eight
mc: A part of the spherical surface bounded by three or more arcs of great circles is a... | *spherical polygon | lune | zone
mc: The sum of the sides of a spherical polygon is less than... | *a circumference of a great circle | a semicircle | a quadrant
mc: A spherical polygon always lies within the surface of a... | *hemisphere | lune | quadrant
mc: The sum of the angles of a spherical triangle is greater than two and less than ___ right angles. | *six | four | three
mc: Two spherical triangles with sides equal but in reverse order are... | *symmetrical | polar | similar
mc: In polar triangles, each angle of one is the ___ of the opposite side of the other. | *supplement | complement | equal
mc: How many points, not in one plane, fix a spherical surface? | *Four | Three | Two
tf: Two spherical triangles with their three angles equal and similarly arranged are equal, on the same sphere. {T}
tf: The angles of a spherical triangle can add up to exactly two right angles. {F}
tf: Any tetraedron may have a sphere circumscribed about it. {T}
q: A spherical triangle has angles of 90, 90, and 90 degrees. By how many degrees does their sum exceed two right angles? {90}
sol: 90 + 90 + 90 = 270, and 270 - 180 = 90.
q: An angle of a spherical triangle is 70 degrees. How many degrees is the opposite side of its polar triangle? {110}
sol: Each angle is the supplement of the opposite side of the polar triangle: 180 - 70 = 110.
q: Two sides of a spherical polygon are 100 and 120 degrees. The other sides together must be less than how many degrees? {140}
sol: The sides together are less than a circumference, 360 degrees: 360 - 100 - 120 = 140.
== 51 | Solids of revolution | Spherical areas
art: 783 |  | Let AHF be a right angled triangle and BFD a semicircle, the hypotenuse AF being a secant, and the vertex F in the circumference. From E, where AF cuts the arc, let a perpendicular EG fall upon AD. Suppose the whole figure to revolve about AD as an axis. The triangle AFH describes a cone, the trapezoid EGHF describes the frustum of a cone, and the semicircle describes a sphere. A frustum of a cone is said to be inscribed in a sphere, when the circumferences of its bases lie in the surface of the sphere.
fig: 261 0.615 0.83 0.555 0.75
art: 784 | Theorem | The area of the curved surface of an inscribed frustum of a cone, is equal to the product of the altitude of the frustum by the circumference of a circle whose radius is the perpendicular let fall from the center of the sphere upon the slant hight of the frustum.
fig: 262 0.385 0.57 0.575 0.78
art: 786 | Theorem | The area of the surface of a sphere is equal to the product of the diameter by the circumference of a great circle. Inscribe in the generating semicircle half of a regular polygon, and let it revolve with it about the diameter AB. The surface described by each side is the circumference whose radius is CI times the part of the axis under that side. Adding, the whole surface is that circumference times the diameter AB. At the limit, the polygon becomes the circle and CI becomes the radius of the sphere.
fig: 263 0.63 0.81 0.555 0.75
art: 787 | Corollary | The area of the surface of a sphere is four times the area of a great circle. For the area of a circle is equal to the product of its circumference by one-fourth of the diameter.
art: 788 | Corollary | The area of the surface of a sphere is equal to the area of the curved surface of a circumscribing cylinder; that is, a cylinder whose bases are tangent to the surface of the sphere.
fig: 264 0.465 0.605 0.545 0.76
art: 789 | Areas of zones | A zone is a part of the surface of a sphere included between two parallel planes. That portion of the sphere itself, so inclosed, is called a segment. The circular sections are the bases of the segment, and the distance between the parallel planes is the altitude of the zone or segment.
art: 790 | Theorem | The area of a zone is equal to the product of its altitude by the circumference of a great circle.
art: 791 | Theorem | The area of a lune is to the area of the whole spherical surface as the angle of the lune to four right angles. The angle of the lune is measured by the arc of a great circle whose pole is at the vertex, and the lune varies exactly with that angle.
fig: 265 0.45 0.6 0.53 0.72
art: 792 | Trirectangular triangle | If the planes of two great circles are perpendicular to each other, they divide the surface into four equal lunes. If a third circle be perpendicular to these two, each lune is divided into two equal triangles, which have their angles all right angles and their sides all quadrants. This triangle is the eighth part of the whole surface; its area is one-half that of a great circle. The area of the trirectangular triangle is frequently assumed as the unit of spherical areas.
fig: 266 0.195 0.325 0.565 0.78
art: 793 | Theorem | Two symmetrical spherical triangles are equivalent. Through the vertices of each pass a small circle; these circles are equal. Joining the pole of each circle to the vertices divides each triangle into isosceles triangles which are equal, two and two. Therefore, the two triangles, being composed of equal parts, are equivalent.
fig: 266 0.555 0.7 0.42 0.8
art: 794 |  | The sum of the angles of any spherical polygon is greater than the sum of the angles of a plane polygon of the same number of sides. The difference is called the spherical excess.
art: 795 | Theorem | The area of a spherical triangle is equal to the area of a trirectangular triangle, multiplied by the ratio of the spherical excess of the given triangle to one right angle. Produce the sides of the triangle AEI around the sphere. The three pairs of triangles so formed have areas 2a, 2e, and 2i times the trirectangular triangle. Adding, they cover the hemisphere once and the given triangle three times, so the area of AEI is a + e + i - 2 trirectangular triangles.
fig: 268 0.298 0.505 0.46 0.78
art: 797 | Theorem | The area of any spherical polygon is equal to the area of the trirectangular triangle multiplied by the ratio of the spherical excess of the polygon to one right angle.
fig: 269 0.65 0.765 0.51 0.74
art: 798 | Exercises | 1. What is the area of the earth's surface, supposing it to be in the shape of a sphere, with a diameter of 7912 miles? 3. Upon the same hypothesis, what portion of the whole surface is between two meridians which are ten degrees apart?
mc: The surface of a sphere is the product of the diameter by... | *the circumference of a great circle | the area of a great circle | the radius
mc: The surface of a sphere is how many times the area of a great circle? | *Four | Two | Three
mc: The surface of a sphere equals the curved surface of... | *a circumscribing cylinder | an inscribed cone | a great circle
mc: The part of a spherical surface between two parallel planes is a... | *zone | lune | segment
mc: The part of the sphere itself between two parallel planes is a... | *segment | zone | sector
mc: The area of a zone is its altitude times... | *the circumference of a great circle | the area of a great circle | its radius
mc: A lune is to the whole spherical surface as its angle is to... | *four right angles | two right angles | one right angle
mc: The trirectangular triangle is what part of the whole surface of the sphere? | *One-eighth | One-fourth | One-half
mc: The amount by which the angles of a spherical triangle exceed two right angles is the... | *spherical excess | lune | polar angle
mc: Two symmetrical spherical triangles are... | *equivalent | unequal | similar only
tf: All zones of the same altitude on one sphere have equal areas. {T}
tf: The angles of a spherical triangle add up to exactly two right angles. {F}
q: Using 3 1/7 for pi: what is the surface of a sphere whose radius is 7 inches, in square inches? {616}
sol: Four times the area of a great circle: 4 times 22/7 times 49 = 616.
q: Using 3 1/7 for pi: a sphere has a diameter of 14 inches. What is the area of a zone 3 inches high, in square inches? {132}
sol: The circumference of a great circle is 44; 44 times 3 = 132.
q: A lune has an angle of 45 degrees. What part of the whole spherical surface is it? Give a fraction. {1/8}
sol: The lune is to the whole as its angle to four right angles: 45 / 360 = 1/8.
q: A spherical triangle has angles of 100, 60, and 50 degrees. What is its spherical excess, in degrees? {30}
sol: 100 + 60 + 50 = 210, and 210 - 180 = 30.
q: A sphere has a surface of 80 square inches. A spherical triangle on it has a spherical excess of one right angle. What is its area, in square inches? {10}
sol: The area is the trirectangular triangle, 80 / 8 = 10, times the excess in right angles, 1: 10.
q: Using 3.1416 for pi, what is the area of the earth's surface, if it is a sphere 7912 miles in diameter, in millions of square miles, to the nearest million? {197}
sol: Diameter times circumference: 3.1416 times 7912 times 7912 = 196,663,000 nearly, about 197 million square miles.

== 52 | Solids of revolution | Spherical volumes
art: 799 | Theorem | The volume of any polyedron in which a sphere can be inscribed, is equal to one-third of the product of the entire surface of the polyedron by the radius of the inscribed sphere. For planes through each edge to the center of the sphere divide the polyedron into pyramids whose bases are the faces and whose altitude is the radius.
art: 800 | Theorem | The volume of a sphere is equal to one-third of the product of the surface by the radius. For the sphere is the limit of the polyedrons in which it is inscribed.
art: 801 | Corollary | The volume of a spherical pyramid, or of a spherical wedge, is equal to one-third of the product of its spherical surface by the radius.
art: 802 |  | A spherical sector is that portion of a sphere which is described by the revolution of a circular sector about a diameter of the circle. If AB is the axis and the generating sector is AEC, the sector has one spherical and one conical surface; but if the generating sector is FCG, it has one spherical and two conical surfaces.
fig: 271 0.205 0.405 0.49 0.76
art: 803 | Corollary | The volume of a spherical sector is equal to one-third of the product of its spherical surface by the radius.
art: 804 |  | The volume of a spherical segment of one base is found by subtracting the volume of a cone from that of a sector. For the sector ABCD is composed of the segment ABC and the cone ACD. The volume of a spherical segment of two bases is the difference of the volumes of two segments each of one base.
fig: 271 0.545 0.665 0.43 0.73
art: 805 |  | All spheres are similar, since they are generated by circles which are similar figures.
art: 806 | Theorem | The areas of the surfaces of two spheres are to each other as the squares of their diameters; and their volumes are as the cubes of their diameters. For the surface of any sphere is pi times the square of the diameter, and the volume is one-sixth of pi times the cube of the diameter.
art: 807 | Useful formulas | Let R be the radius of a circle or a sphere, or of the base of a cone or cylinder; D the diameter; A the altitude; and H the slant hight. Circumference of a circle, 2 pi R. Area of a circle, pi R squared. Curved surface of a cone, pi R H. Volume of a cone, one-third pi R squared A. Curved surface of a cylinder, 2 pi R A. Volume of a cylinder, pi R squared A. Surface of a sphere, 4 pi R squared. Volume of a sphere, four-thirds pi R cubed. Pi is 3.1415926535.
mc: The volume of a sphere is one-third of the product of its surface by its... | *radius | diameter | circumference
mc: The volume of a polyedron in which a sphere is inscribed is one-third of its surface times... | *the radius of the inscribed sphere | its longest edge | its altitude
mc: A solid formed by revolving a circular sector about a diameter is a spherical... | *sector | segment | wedge
mc: A spherical segment of one base is a sector less a... | *cone | cylinder | pyramid
mc: The volumes of two spheres are as the ___ of their diameters. | *cubes | squares | first powers
mc: The surfaces of two spheres are as the ___ of their diameters. | *squares | cubes | first powers
mc: The volume of a sphere in terms of its radius is... | *four-thirds pi R cubed | 4 pi R squared | pi R squared A
mc: The surface of a sphere in terms of its radius is... | *4 pi R squared | pi R squared | 2 pi R
tf: All spheres are similar. {T}
tf: A sphere of twice the diameter has twice the volume. {F}
q: Using 3 1/7 for pi: what is the volume of a sphere whose radius is 21 inches, in cubic inches? {38808}
sol: Four-thirds pi R cubed: 4/3 times 22/7 times 9261 = 38808.
q: A polyedron with an entire surface of 600 square inches has a sphere of radius 6 inches inscribed in it. What is its volume, in cubic inches? {1200}
sol: One-third of the surface times the radius: 600 times 6 / 3 = 1200.
q: Two spheres have diameters of 2 and 6 inches. The volume of the larger is how many times the smaller? {27}
sol: Volumes are as the cubes of the diameters: 216 : 8 = 27.
q: Two spheres have diameters of 3 and 12 inches. The surface of the larger is how many times the smaller? {16}
sol: Surfaces are as the squares of the diameters: 144 : 9 = 16.
q: A spherical sector has a spherical surface of 88 square inches, in a sphere of radius 6 inches. What is its volume, in cubic inches? {176}
sol: One-third of its spherical surface times the radius: 88 times 6 / 3 = 176.

== 53 | Solids of revolution | Mensuration
art: 810 |  | Mensuration, or the art of measuring, consists in rules for the measurement of lines, surfaces, and solids. Exercises have been given in the previous pages requiring the application of the principles of Geometry to various kinds of measurement. The useful formulas also afford many rules of Mensuration, each of which should be applied by the student to particular examples. A few trigonometrical principles will, with the exercises given hereafter, complete all that is usually included in this branch of applied mathematics.
art: 808 | Exercises | 1. What is the locus of those points in space which are at the same distance from a given point? 2. What is the locus of those points in space which are at the same distance from a given straight line?
art: 809 | Exercises for general review | 21. What is the area of the sector whose arc is 50 degrees and whose radius is 10 inches? 36. In any polyedron, the sum of the number of vertices and the number of faces exceeds by two the number of edges. 38. A frustum of a pyramid is equivalent to the sum of three pyramids having the same altitude as the frustum, and having for their bases the lower base of the frustum, the upper base, and a mean proportional between them. 41. If about a sphere a cylinder be circumscribed, also a cone whose slant height is equal to the diameter of its base, then the area and volume of the sphere are two-thirds of the area and volume of the cylinder; and the area and volume of the cylinder are two-thirds of the area and volume of the cone.
mc: The locus of points in space at the same distance from a given point is... | *a spherical surface | a circle | a plane
mc: The locus of points in space at the same distance from a given straight line is... | *a cylindrical surface | a sphere | a plane
mc: Mensuration is the art of... | *measuring | drawing | proving
mc: The volume of a sphere is what part of the volume of its circumscribing cylinder? | *Two-thirds | One-half | One-third
mc: In any polyedron, vertices plus faces equals edges plus... | *two | one | four
tf: The surface of a sphere is two-thirds of the entire surface of its circumscribing cylinder. {T}
tf: A frustum of a pyramid equals three pyramids on the lower base, the upper base, and their mean proportional, with the same altitude. {T}
q: A cube has 8 vertices and 6 faces. How many edges has it, by the rule that vertices and faces exceed the edges by two? {12}
sol: 8 + 6 = 14, and 14 - 2 = 12.
q: Using 3.1416 for pi: what is the area of a sector whose arc is 50 degrees and whose radius is 10 inches, to the nearest hundredth of a square inch? {43.63}
sol: The circle is 3.1416 times 100 = 314.16; the sector is 50/360 of it: 314.16 times 5 / 36 = 43.633, about 43.63.
q: A frustum of a pyramid has bases of 16 and 4 square feet and an altitude of 3 feet. What is its volume, in cubic feet? {28}
sol: Three pyramids of altitude 3 on bases 16, 4, and their mean proportional 8: one-third of 3 times (16 + 4 + 8) = 28.
q: Using 3 1/7 for pi: a cylinder is circumscribed about a sphere of radius 7 inches. What is the volume of the cylinder, in cubic inches? {2156}
sol: Its base is 154 and its altitude the diameter 14: 154 times 14 = 2156.
q: Using 3 1/7 for pi: the sphere in that cylinder, of radius 7 inches. What is its volume, in cubic inches? Give a fraction. {4312/3}
sol: Two-thirds of the cylinder: 2156 times 2 / 3 = 4312/3, or 1437 1/3 cubic inches.
q: A cistern in the form of a rectangular box measures 2 feet by 3 feet by 4 feet. A cubic foot holds about 7 1/2 gallons. How many gallons does it hold? {180}
sol: 2 times 3 times 4 = 24 cubic feet; 24 times 7 1/2 = 180 gallons.
== 54 | Plane trigonometry | Measure of angles
art: 811 |  | Trigonometry is the science in which the relations subsisting between the angles, sides, and area of any triangle are investigated. The science was originally called Plane Trigonometry or Spherical Trigonometry, according as the triangle was plane or spherical. Plane Trigonometry has now a wider meaning, comprising algebraic investigations concerning angles and their functions, and the methods of calculating these functions.
art: 812 |  | In Elementary Geometry, the unit for the measure of angles is usually the right angle. The frequent fractions which the use of this unit gives rise to, render it inconvenient for calculation. It has been divided into degrees, minutes, and seconds. This sexagesimal division of angles has been in use since the second century. Efforts have been made to substitute the centesimal division, making the right angle contain one hundred grades; but this plan has never been generally in use.
art: 813 |  | There is another unit which has been called the circular measure of an angle, also called the analytical unit. It is that angle at the center of a circle whose intercepted arc has the same linear extent as the radius. Thus, if the arc AB has the same linear extent as the radius AC, then the angle C is the unit of circular measure. Hence, this unit is equal to 180 degrees divided by pi, or 57.29578 degrees, which is 57 degrees, 17 minutes, 44.8 seconds. Also, one degree is .017453 times the circular measure.
fig: 278 0.19 0.325 0.585 0.79
art: 814 |  | Various instruments are used for the measure of angles. A protractor is used to measure the angle of two lines in a drawing. It is usually shaped like a semicircumference with its diameter, the arc being marked with the degrees from 0 to 180. To measure the angle ABC, place the center of the straight edge at the vertex B; let the edge lie along one side, as BC; then read the degree marked where the other side BA passes the arc. The same instrument is used for drawing angles of a known size.
fig: 278 0.615 0.695 0.465 0.79
art: 815 | Exercises | 2. Draw a triangle having one side two inches, another three inches, and the included angle 100 degrees. Find the other angles and side by measurement. 3. Draw a triangle with the sides three, four, and five inches in length. Find the angles by measurement.
mc: Trigonometry investigates the relations between the angles, sides, and ___ of a triangle. | *area | color | position
mc: Dividing the right angle into degrees, minutes, and seconds is the ___ division. | *sexagesimal | centesimal | decimal
mc: The plan of dividing the right angle into one hundred grades is the ___ division. | *centesimal | sexagesimal | circular
mc: The unit of circular measure is the angle whose arc is as long as the... | *radius | diameter | circumference
mc: The unit of circular measure is about how many degrees? | *57.3 | 45 | 90
mc: An instrument for measuring the angle of two lines in a drawing is a... | *protractor | compass | square
mc: A protractor's arc is usually marked with degrees from 0 to... | *180 | 90 | 360
tf: The sexagesimal division of angles has been in use since the second century. {T}
tf: The centesimal division is the one in general use. {F}
q: How many degrees are in a right angle? {90}
sol: The right angle is divided into 90 degrees.
q: How many minutes are in one degree? {60}
sol: Each degree is divided into 60 minutes, and each minute into 60 seconds.
q: How many seconds are in one degree? {3600}
sol: 60 minutes times 60 seconds = 3600.
q: In the centesimal division, how many grades are in two right angles? {200}
sol: Each right angle contains 100 grades: 2 times 100 = 200.
q: Using 3.1416 for pi, how many degrees is the unit of circular measure, to the nearest hundredth? {57.3}
sol: 180 / 3.1416 = 57.2958, about 57.30 degrees.
q: What is the circular measure of an angle of 180 degrees, to the nearest ten-thousandth? {3.1416}
sol: 180 degrees is pi times the unit of circular measure: about 3.1416.

== 55 | Plane trigonometry | Functions of angles
art: 816 |  | When two quantities are so related that any variation in one causes a variation in the other, each is a function of the other. Thus, the area of a circle is a function of its radius. A quantity may be a function of several others; the area of a triangle is a function of its base and altitude. The angles of a triangle are functions of the ratios of the sides, and the ratios of the sides are functions of the angles. For another example, if the triangle ACD has its angles 30, 60, and 90 degrees, then AC is twice AD.
fig: 280 0.105 0.245 0.565 0.78
art: 817 | Theorem | If from any point in one side of an angle, a perpendicular fall upon the other side, a right angled triangle is formed, and the ratios of the sides of this triangle are functions of the given angle. For, if any number of triangles were thus formed with a given angle, all of them would be similar, and their sides would have the same ratios. When the given angle is greater than a right angle, one side may be produced to meet the perpendicular.
fig: 280 0.4 0.495 0.485 0.76
fig: 280 0.505 0.615 0.5 0.72
art: 818 | Definitions | If from any point on one side of a given angle a perpendicular fall on the other side as a base, then: the sine of the angle is the ratio of the perpendicular to the hypotenuse; the tangent is the ratio of the perpendicular to the base; the secant is the ratio of the hypotenuse to the base; the cosine is the ratio of the base to the hypotenuse; the cotangent is the ratio of the base to the perpendicular; and the cosecant is the ratio of the hypotenuse to the perpendicular. These six are called the simple functions of an angle. The versed sine is the ratio of the excess of the hypotenuse over the base, to the hypotenuse.
art: 819 |  | A table of sines of every degree from 0 to 90 may be made by drawing and measurement. Draw a right angled triangle with an angle at the base equal to the given angle, making the hypotenuse the length of a certain unit. Divide the length of the perpendicular by that of the hypotenuse. The quotient is the sine. By calculation, the functions may all be found to any required degree of accuracy.
art: 820 |  | Formerly, with any radius an arc was described about the vertex C, and from B, one extremity of the arc, a perpendicular BD fell upon CA; then BD was called the sine of the arc BA, or of the angle C. A perpendicular to AC at A meeting CB at E gave AE, the tangent, and CE, the secant. The curve BAF is like the bow of an archer, the chord BF joins its ends, and BD touches the breast or sinus of the archer. DA was called the sagitta or arrow; now it is called the versed sine. The oldest work on Trigonometry now extant is the Almagest of Ptolemy, written in the second century. The use of sines was introduced by the Arabian mathematicians about the eighth or ninth century. Napier, a Scotch Baron of the early seventeenth century, has done more for the science than any other one man.
fig: 282 0.105 0.34 0.565 0.79
art: 821 | Exercises | 1. Demonstrate that the tangent of 45 degrees is 1; also, that the sine of 60 degrees is one-half the square root of 3.
mc: The sine of an angle is the ratio of the perpendicular to the... | *hypotenuse | base | other angle
mc: The cosine of an angle is the ratio of the base to the... | *hypotenuse | perpendicular | radius only
mc: The tangent of an angle is the ratio of the perpendicular to the... | *base | hypotenuse | arc
mc: The secant is the ratio of the hypotenuse to the... | *base | perpendicular | radius
mc: The cotangent is the ratio of the base to the... | *perpendicular | hypotenuse | secant
mc: The cosecant is the ratio of the hypotenuse to the... | *perpendicular | base | tangent
mc: How many simple functions has an angle? | *Six | Three | Four
mc: The word sine comes from the Latin for the ___ of the archer. | *breast | bow | arrow
mc: The oldest work on Trigonometry now extant is the... | *Almagest of Ptolemy | Elements of Euclid | Logarithms of Napier
mc: The man who has done more for Trigonometry than any other is said to be... | *Napier | Ptolemy | Monge
tf: The functions of an angle do not depend on how far from the vertex the perpendicular is drawn. {T}
tf: The sine of an angle depends on the length of its sides. {F}
q: In a right triangle with sides 3, 4, and 5, what is the sine of the angle opposite the side 3? Give a decimal. {0.6}
sol: The sine is the perpendicular over the hypotenuse: 3 / 5 = 0.6.
q: In the same 3, 4, 5 triangle, what is the cosine of that angle? {0.8}
sol: The cosine is the base over the hypotenuse: 4 / 5 = 0.8.
q: In the same triangle, what is the tangent of that angle? {0.75}
sol: The tangent is the perpendicular over the base: 3 / 4 = 0.75.
q: What is the tangent of 45 degrees? {1}
sol: In a right triangle with a 45 degree angle, the two legs are equal, so the perpendicular over the base is 1.
q: What is the sine of 30 degrees? Give a decimal. {0.5}
sol: In the triangle with angles 30, 60, 90, the side opposite 30 degrees is half the hypotenuse.
q: In a right triangle with sides 5, 12, and 13, what is the secant of the angle between the side 12 and the hypotenuse? Give a fraction. {13/12}
sol: The secant is the hypotenuse over the base: 13 / 12.

== 56 | Plane trigonometry | Signs and limits of functions
art: 822 |  | An angle may be conceived to be generated by the revolution of a line about a point. Thus, the line AB beginning at AX may take the positions AB, AB', AB'', AB''', AX, and so on indefinitely. In Trigonometry, the amount of this revolution is considered as an angle, so that an angle may be greater than two or four right angles. Quantities in a certain direction are called positive, marked plus; those in the opposite direction are negative, marked minus. Angles generated by the motion of AB up from AX are positive. Directions to the right and upwards are positive; to the left and downward, negative.
fig: 283 0.335 0.515 0.32 0.74
art: 823 | Theorem | The functions of any acute angle are positive. For when the revolving line is in the first quarter of its revolution, all the sides of the triangle are positive.
art: 824 | Theorem | The tangent, secant, cosine, and cotangent of obtuse angles are negative, while the sine and cosecant of obtuse angles are positive. For in the second quarter, the base AC' is the only negative term.
art: 825 |  | In this manner the signs of all the functions may be found. In the first quarter, all are positive. In the second quarter, the sine and cosecant are positive, the others negative. In the third quarter, the tangent and cotangent are positive, the others negative. In the fourth quarter, the cosine and secant are positive, the others negative.
art: 826 | Corollary | The functions of two angles are the same, when one of the angles is greater than the other by four right angles.
art: 828 | Theorem | Any given simple function, taken irrespective of its sign, belongs to four different angles within each revolution. For the revolving line makes, in each quarter, an acute angle with XX' equal to the angle BAC.
fig: 285 0.67 0.805 0.375 0.74
art: 829 | Corollary | Any simple function of an angle is numerically equal to the same function of: the supplement of the angle; the angle increased by two right angles; and the angle taken negatively. Thus the sine of A equals the sine of 180 degrees less A.
art: 832 | Theorem | The sine of any angle can not be greater than 1, nor less than minus 1; and the cosine has the same limits. For the leg of a right angled triangle can not be greater than the hypotenuse.
art: 833 | Theorem | The secant and cosecant can not have any values between 1 and minus 1; and the tangent and cotangent have no limits.
art: 834 |  | As the revolving line passes through the first quarter, the sine increases from 0 to 1. The sine of a right angle is unity. Then the sine decreases till the angle is two right angles, when the sine becomes 0; it continues to decrease to minus 1 at three right angles, and increases again to 0 at the end of the revolution. The cosine of 0 degrees is 1, of 90 degrees is 0, and of 180 degrees is minus 1. The tangent of 0 degrees is 0; it increases without limit, and the tangent of a right angle is infinite.
mc: In trigonometry, an angle generated by the line moving up from AX is... | *positive | negative | zero
mc: The functions of any acute angle are... | *positive | negative | zero
mc: Which functions of an obtuse angle are positive? | *The sine and cosecant | The tangent and cotangent | The cosine and secant
mc: In the third quarter, which functions are positive? | *The tangent and cotangent | The sine and cosecant | The cosine and secant
mc: In the fourth quarter, which functions are positive? | *The cosine and secant | The sine and cosecant | The tangent and cotangent
mc: The sine of an angle can never be greater than... | *1 | 0 | 2
mc: Which functions have no limits? | *The tangent and cotangent | The sine and cosine | The secant and cosecant
mc: The sine of an angle equals the sine of... | *its supplement | its complement | twice the angle
mc: The tangent of a right angle is... | *infinite | 0 | 1
tf: The functions of 30 degrees and of 390 degrees are the same. {T}
tf: The secant of an angle can be one-half. {F}
tf: The cosine of an obtuse angle is negative. {T}
mc: What is the cosine of 180 degrees? | *Minus 1 | 0 | 1
q: What is the sine of 90 degrees? {1}
sol: The sine of a right angle is unity, for the perpendicular coincides with the hypotenuse.
q: What is the cosine of 90 degrees? {0}
sol: The base vanishes when the angle is a right angle.
q: What is the sine of 810 degrees? {1}
sol: 810 - 720 = 90, and the functions repeat every four right angles: the sine of 90 degrees is 1.
q: The sine of 30 degrees is 0.5. What is the sine of 150 degrees? {0.5}
sol: An angle and its supplement have the same sine: 180 - 150 = 30.
q: In how many different angles within one revolution does a given sine, taken without its sign, belong? {4}
sol: One in each quarter of the revolution.

== 57 | Plane trigonometry | Relations between the functions
art: 836 |  | A simple function of an angle, being a ratio, may be expressed as a fraction. Let a be the perpendicular, b the base, and c the hypotenuse. Then the sine of A is a over c; the cosine is b over c; the tangent is a over b; the cotangent is b over a; the secant is c over b; and the cosecant is c over a.
art: 837 | Corollary | The sine and cosecant of an angle are reciprocals; also, the tangent and cotangent are reciprocals; and the cosine and secant are reciprocals. Hence the cosecant, secant, and cotangent are less used than the other simple functions.
art: 838 |  | By means of the Pythagorean Theorem, any function may be expressed in terms of any other function of the same angle. Since a squared plus b squared equals c squared, dividing by c squared: the sum of the squares of the sine and cosine of any angle is equal to unity.
art: 839 |  | The sine of an angle is equal to the product of the tangent by the cosine. Hence the tangent is the sine divided by the cosine, and the cotangent is the cosine divided by the sine.
art: 841 | Theorem | The cosine of an angle is the sine of its complement. For, in the right angled triangle of the definitions, the acute angles are complementary, and the cosine of A, b over c, is the sine of B.
art: 842 | Corollary | The cotangent and cosecant are respectively the tangent and secant of the complementary angle. It is from this property that these functions derive their names.
art: 843 | Theorem | The sine of 90 degrees plus A is the cosine of A, and the cosine of 90 degrees plus A is minus the sine of A.
art: 845 | Functions of two angles | Let the angle DCF be A and FCG be B; then DCG is A plus B. From G let fall GH and GF perpendicular to CD and CF; from F let fall FD and FK perpendicular to CD and GH. Then the angle FGK equals A. It follows that the sine of A plus B is the sine of A times the cosine of B, plus the cosine of A times the sine of B. Also, the cosine of A plus B is the cosine of A times the cosine of B, less the sine of A times the sine of B. Putting minus B for B: the sine of A minus B is sine A cosine B less cosine A sine B; and the cosine of A minus B is cosine A cosine B plus sine A sine B.
fig: 292 0.64 0.805 0.49 0.79
art: 847 | Multiples and parts | Putting B equal to A: the sine of 2A is twice the sine of A times the cosine of A; the cosine of 2A is the square of the cosine of A less the square of the sine of A. Also, twice the square of the sine of half A is 1 less the cosine of A; and twice the square of the cosine of half A is 1 plus the cosine of A. By these formulas, from the cosine of an angle may be calculated the sine and cosine of its half, fourth, eighth, and so on.
mc: The sine and the ___ of an angle are reciprocals. | *cosecant | cosine | secant
mc: The tangent and the ___ of an angle are reciprocals. | *cotangent | sine | secant
mc: The cosine and the ___ of an angle are reciprocals. | *secant | sine | cotangent
mc: The sum of the squares of the sine and cosine of any angle is... | *1 | 0 | 2
mc: The tangent of an angle is the sine divided by the... | *cosine | secant | cotangent
mc: The cosine of an angle is the sine of its... | *complement | supplement | double
mc: The sine of A plus B is sine A cosine B plus... | *cosine A sine B | sine A sine B | cosine A cosine B
mc: The sine of 2A equals... | *twice sine A cosine A | twice sine A | sine A squared
mc: The cosine of A plus B is cosine A cosine B... | *less sine A sine B | plus sine A sine B | times sine A sine B
tf: The cotangent is so named because it is the tangent of the complement. {T}
tf: The sine of 90 degrees plus A is the cosine of A. {T}
q: The sine of an angle is 0.6. What is its cosine, if the angle is acute? {0.8}
sol: The square of the cosine is 1 - 0.36 = 0.64, and the square root is 0.8.
q: The sine of an angle is 0.6 and its cosine 0.8. What is its tangent? {0.75}
sol: The tangent is the sine divided by the cosine: 0.6 / 0.8 = 0.75.
q: The tangent of an angle is 4. What is its cotangent? Give a fraction. {1/4}
sol: The tangent and cotangent are reciprocals.
q: The cosine of an angle is 0.8. What is its secant? {1.25}
sol: The cosine and secant are reciprocals: 1 / 0.8 = 1.25.
q: The sine of A is 0.6 and the cosine of A is 0.8. What is the sine of 2A? {0.96}
sol: Twice sine A cosine A: 2 times 0.6 times 0.8 = 0.96.
q: The same angle: what is the cosine of 2A? {0.28}
sol: The square of the cosine less the square of the sine: 0.64 - 0.36 = 0.28.
q: The sine of 60 degrees is the cosine of how many degrees? {30}
sol: The cosine of an angle is the sine of its complement: 90 - 60 = 30.
== 58 | Plane trigonometry | Construction and use of tables
art: 849 | Formulas for logarithmic use | In order to render a formula fit for logarithmic calculation, products and quotients must be substituted for sums and differences. From the formulas for the sine and cosine of A plus and minus B it follows that the sum of the sines of two angles is to their difference, as the tangent of half their sum is to the tangent of half their difference.
art: 851 | Trigonometrical tables | The student has found that the sine of 30 degrees is one-half. From such values, and the formulas for multiples and parts of angles, the sines of smaller and smaller angles may be calculated, and then the functions of their multiples. This method is tedious and is not used in practice; it serves to show the possibility of calculating these functions by elementary algebra and geometry. These numerical functions are called the natural sines, tangents, and so on, to distinguish them from the logarithmic functions.
art: 852 |  | The table of natural sines and tangents gives these functions to six places of figures for every 10 minutes from 0 to 90 degrees. It also serves as a table of cosines and cotangents. The sine of an intermediate angle may be found by taking a proportional part of the difference. For example, the sine of 34 degrees 20 minutes is .564007; since 3 minutes 30 seconds is .35 of 10 minutes, multiply the difference to the next sine, 2399 millionths, by .35, and add the product: the sum .564847 is the natural sine of 34 degrees 23 minutes 30 seconds. The tangent of a large angle is found more accurately by taking the reciprocal of its cotangent.
art: 853 | Logarithmic functions | A logarithmic sine, tangent, and so on, means the logarithm of the sine, of the tangent, and so on. In the tables, the characteristic of every logarithmic trigonometric function is increased by 10. For example, the logarithm of the sine of 30 degrees, which is the log of one-half, is minus 1 plus .698970; but the tabular logarithmic sine of 30 degrees is 9.698970. The object is simply to avoid negative characteristics. Whenever a tabular logarithmic function is added, 10 must be subtracted from the result; and whenever one is subtracted, 10 must be added.
art: 854 |  | The table gives the logarithmic sine, tangent, cosine, and cotangent for every minute from 0 to 90 degrees. The columns marked P. P. 1 second contain the proportional part for one second. Remember that the cosine and cotangent decrease when the angle increases.
art: 855 | To find the function of an angle | If the angle has seconds, take the logarithmic function for the degrees and minutes; multiply the proportional part for 1 second by the number of seconds; and add the product for the sine and tangent, but subtract it for the cosine and cotangent. For example, the tabular log sine of 40 degrees 13 minutes is 9.810017, the P. P. for 1 second is 2.5; 2.5 times 14 is 35; so the tabular log sine of 40 degrees 13 minutes 14 seconds is 9.810052.
art: 856 | To find the angle | If the given function is not in the table, take that which is next less; subtract it from the given function; divide the remainder by the proportional part for 1 second; the quotient is the number of seconds, to be added for a sine or tangent, and subtracted for a cosine or cotangent. For example, the angle whose tabular log tangent is 10.456789: the table gives 10.456501 for 70 degrees 44 minutes, P. P. 6.75; 288 divided by 6.75 is 43; the angle is 70 degrees 44 minutes 43 seconds.
mc: Sines and tangents themselves, not their logarithms, are called... | *natural functions | logarithmic functions | tabular functions
mc: The table of natural sines and tangents gives them for every ___ from 0 to 90 degrees. | *10 minutes | degree | second
mc: In the tables, the characteristic of every logarithmic function is increased by... | *10 | 1 | 100
mc: Why is 10 added to the characteristics? | *To avoid negative characteristics | To make them larger | To use fewer figures
mc: When a tabular logarithmic function is added in a calculation, you must then... | *subtract 10 | add 10 | multiply by 10
mc: As the angle increases, the cosine and cotangent... | *decrease | increase | stay the same
mc: The tangent of a large angle is found most accurately from... | *the reciprocal of its cotangent | its sine | its secant
tf: The table of natural sines also serves as a table of cosines. {T}
tf: The tabular logarithmic sine of 30 degrees is 9.698970. {T}
q: The tabular log sine of 40 degrees 13 minutes is 9.810017, and the P. P. for 1 second is 2.5 (in the last figures). What are the last two figures of the tabular log sine of 40 degrees 13 minutes 14 seconds? It is 9.8100?? {52}
sol: 2.5 times 14 = 35; 9.810017 + .000035 = 9.810052.
q: The natural sine of 34 degrees 20 minutes is .564007 and the difference to the next 10 minutes is .002399. What fraction of 10 minutes is 3 minutes 30 seconds? Give a decimal. {0.35}
sol: 3 1/2 minutes divided by 10 minutes = .35.
q: A tabular log tangent is 10.456789. The table gives 10.456501 for 70 degrees 44 minutes, with P. P. 6.75 for 1 second. How many seconds must be added? {43}
sol: 10.456789 - 10.456501 = .000288; 288 / 6.75 = 42.7, about 43 seconds.
q: The true logarithm of a sine is 1.698970 with a characteristic of minus 1. What is the tabular logarithmic sine? {9.69897}
sol: Add 10 to the characteristic: minus 1 + 10 = 9, giving 9.698970.

== 59 | Plane trigonometry | Right angled triangles
art: 857 |  | The principles have now been established, by which, whenever certain parts of a triangle are known, the remaining parts can be calculated. Since the trigonometrical functions are the ratios between the sides of a right angled triangle, the problems concerning such triangles need no other demonstration than is contained in the definitions. The sum of the acute angles being 90 degrees, when one is known, the other is found by subtraction.
art: 858 | Problem | Given the hypotenuse and one angle, to find the other parts. The product of the hypotenuse by the sine of either acute angle, is the side opposite that angle. The product of the hypotenuse by the cosine of either acute angle, is the side adjacent to that angle.
art: 859 | Problem | Given one leg and one angle, to find the other parts. The quotient of one leg divided by the sine of the opposite angle is the hypotenuse. The product of one leg by the tangent of the adjacent angle is the other leg.
art: 860 | Problem | Given one leg and the hypotenuse, to find the other parts. The quotient of one leg divided by the hypotenuse is the sine of the angle opposite that leg, and the cosine of the adjacent angle.
art: 861 | Problem | Given the two legs, to find the other parts. The quotient of one leg divided by the other is the tangent of the angle opposite the dividend. When two sides are given, the third may also be found by the Pythagorean Theorem.
art: 862 |  | Since none of the above problems requires addition or subtraction, the operations may all be performed by logarithms. For example: a railroad track, 463 feet 3 inches long, has a uniform grade of 3 degrees. How high is one end above the other? Here the hypotenuse and one acute angle are given, to find the opposite side. The log of 463.25 is 2.665815, the tabular log sine of 3 degrees is 8.718800; omitting the tabular 10, the sum 1.384615 is the logarithm of 24.2446. Hence, the ascent is nearly 24 feet 3 inches.
art: 863 | Exercises | 3. Two posts on the bank of a river are one hundred feet apart; the line joining them is perpendicular to the line from the first post to a certain point on the opposite bank; and the same line makes an angle of 78 degrees 52 minutes with the line from the second post to the same point. How wide is the river? 12. A tower 103 feet high throws a shadow 51.5 feet long upon the level plane; what is the angle of elevation of the sun?
mc: The hypotenuse times the sine of an acute angle gives... | *the side opposite that angle | the side adjacent to it | the other angle
mc: The hypotenuse times the cosine of an acute angle gives... | *the side adjacent to that angle | the side opposite it | the area
mc: One leg times the tangent of the adjacent angle gives... | *the other leg | the hypotenuse | the area
mc: One leg divided by the sine of the opposite angle gives... | *the hypotenuse | the other leg | the angle
mc: One leg divided by the other leg is the tangent of the angle opposite... | *the dividend | the divisor | the hypotenuse
mc: In a right triangle, if one acute angle is known, the other is found by... | *subtracting it from 90 degrees | doubling it | the sine table
tf: Problems on right triangles may all be worked by logarithms. {T}
tf: In a right triangle, the two acute angles together make 180 degrees. {F}
q: A right triangle has a hypotenuse of 20 feet and an angle of 30 degrees. The sine of 30 degrees is 0.5. How long is the side opposite that angle, in feet? {10}
sol: The hypotenuse times the sine of the angle: 20 times 0.5 = 10.
q: One acute angle of a right triangle is 37 degrees. How many degrees is the other? {53}
sol: 90 - 37 = 53.
q: A ladder 25 feet long reaches a window, making an angle with the ground whose cosine is 0.28. How far is its foot from the wall, in feet? {7}
sol: The hypotenuse times the cosine of the angle: 25 times 0.28 = 7.
q: A tree casts a shadow 40 feet long when the tangent of the sun's elevation is 0.75. How tall is the tree, in feet? {30}
sol: One leg times the tangent of the adjacent angle: 40 times 0.75 = 30.
q: A tower 103 feet high throws a shadow 51.5 feet long. What is the tangent of the angle of elevation of the sun? {2}
sol: The tower over the shadow: 103 / 51.5 = 2. (The angle is about 63 degrees 26 minutes.)
q: A railroad track 463.25 feet long rises at a grade whose sine is .052336 (3 degrees). How many feet does it rise, to the nearest tenth? {24.2}
sol: 463.25 times .052336 = 24.24, about 24.2 feet, nearly 24 feet 3 inches.
q: A leg of a right triangle is 6 inches and the hypotenuse is 10 inches. What is the sine of the angle opposite that leg? Give a decimal. {0.6}
sol: The leg divided by the hypotenuse: 6 / 10 = 0.6.

== 60 | Plane trigonometry | Solution of plane triangles
art: 864 |  | One angle of a triangle being the supplement of the sum of the other two, when two are known the third may be found by subtraction. The letters a, b, and c represent the sides of a triangle respectively opposite the angles A, B, and C.
art: 865 | Theorem | The square of one side of a triangle is equal to the sum of the squares of the other two sides, less twice the product of those sides by the cosine of their included angle. That is, a squared equals b squared plus c squared, less 2 b c times the cosine of A. These three equations suffice for the solution of all problems on plane triangles, but they are not suitable for logarithmic calculations.
fig: 305 0.205 0.335 0.47 0.74
fig: 305 0.345 0.475 0.47 0.74
art: 867 | Theorem | The sides of any triangle are proportional to the sines of the opposite angles. That is, a is to b as the sine of A is to the sine of B. For, whether A is acute or obtuse, the perpendicular BD is AB times the sine of A, and also BC times the sine of C.
fig: 307 0.175 0.31 0.48 0.72
fig: 307 0.32 0.445 0.48 0.72
art: 868 | Theorem | One side of a triangle is equal to the sum of the products found by multiplying each of the other sides by the cosine of the angle which it forms with the first side.
art: 869 | Theorem | The sum of any two sides of a triangle is to their difference as the tangent of half the sum of the two opposite angles is to the tangent of half their difference.
art: 870 | Problem | Given the sides of a triangle, to find the angles. From half the sum of the sides, subtract each of the sides adjacent to the required angle; multiply together these remainders; divide this product by the product of the two adjacent sides, and extract the square root of the quotient. This root is the sine of half the angle sought.
art: 871 | Problem | Given two angles and a side, to find the other angle and sides. Find the third angle by subtracting the sum of the given two from 180 degrees. Then find the remaining sides by the proportion: the sine of A is to the sine of B as a is to b.
art: 872 | Problem | Given two sides and an angle opposite one of them, to find the other parts. Find the angle opposite the other given side by the same proportion. When the side opposite the given angle is less than the other given side, there may be two solutions. This is called the ambiguous case, for a given sine belongs to two angles, one acute and one obtuse. If the sine is found greater than unity, the triangle is impossible.
art: 873 | Problem | Given two sides and the included angle, to find the other parts. Find the sum of the other angles by subtraction, and their difference by the tangent proportion. Half the sum plus half the difference is the greater angle; half the sum less half the difference is the less.
art: 874 | Theorem | The area of a triangle is equal to half the product of any two sides multiplied by the sine of the included angle.
art: 875 | Applications | 1. To measure the distance between two points A and B when the line between them can not be passed over, take a point C from which both are visible; measure AC, BC, and the angle C; then AB can be calculated. 2. To find the height and distance of an inaccessible object P, observe at A the angle of elevation PAC; measure AB directly toward the object, and at B observe the angle PBC. In the triangle APB, AB and the angles are known; then the height PC is AP times the sine of A.
fig: 311 0.265 0.35 0.435 0.75
fig: 311 0.575 0.7 0.48 0.75
art: 875 | More applications | 3. To find the distance between two visible but inaccessible objects P and N, measure a base CB and observe the angles at C and B; find CP and CN, and then PN. 4. To find the width of a river without an instrument for observing angles, take points B, C, and D on this side, in line with P and A, and measure the lines; the triangles give the angles, and then AP may be calculated.
fig: 312 0.416 0.575 0.445 0.75
fig: 312 0.74 0.86 0.46 0.75
art: 876 | Exercises | 1. The sides of a triangle being 70, 80, and 100, what are the angles? 7. A person observing the elevation of a spire to be 35 degrees, advances 80 yards nearer to it, and then finds the elevation is 70 degrees; required the height of the spire.
mc: In a triangle, the sides are proportional to the ___ of the opposite angles. | *sines | cosines | tangents
mc: The square of a side equals the sum of the squares of the other two, less twice their product by the cosine of... | *their included angle | the opposite angle | a right angle
mc: The area of a triangle is half the product of two sides by the ___ of the included angle. | *sine | cosine | tangent
mc: Given two sides and an angle opposite one of them, there may be two solutions. This is the... | *ambiguous case | impossible case | right case
mc: If the sine found for an angle is greater than unity, the triangle is... | *impossible | right angled | isosceles
mc: The sum of two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of... | *half their difference | their sum | the third angle
mc: To find the height of an object you can not reach, you observe its... | *angle of elevation from two points | shadow only | weight
tf: Knowing two angles of a triangle, the third is found by subtracting their sum from 180 degrees. {T}
tf: The law of sines works only for acute triangles. {F}
q: Two angles of a triangle are 76 and 54 degrees. How many degrees is the third? {50}
sol: 180 - 76 - 54 = 50.
q: A triangle has sides b = 5 and c = 8, with the included angle A whose cosine is 0.5 (60 degrees). How long is the side a? {7}
sol: a squared = 25 + 64 - 2 times 5 times 8 times 0.5 = 89 - 40 = 49, and a = 7.
q: Two sides of a triangle are 10 and 12 feet, and the included angle is 30 degrees, whose sine is 0.5. What is the area, in square feet? {30}
sol: Half the product of the sides by the sine: 10 times 12 times 0.5 / 2 = 30.
q: In a triangle, the sine of A is 0.6 and the sine of B is 0.8, and the side a is 9. How long is the side b? {12}
sol: a : b :: sine A : sine B, so b = 9 times 0.8 / 0.6 = 12.
q: A triangle has sides 3, 5, and 7. The cosine of the angle opposite the side 7 is minus one-half. How many degrees is that angle? {120}
sol: Check: 9 + 25 - 2 times 3 times 5 times (minus 1/2) = 34 + 15 = 49. The angle whose cosine is minus one-half is 120 degrees.
q: From a point 100 feet from the foot of a spire, its elevation has a tangent of 1.5. How tall is the spire, in feet? {150}
sol: The horizontal distance times the tangent of the elevation: 100 times 1.5 = 150.
== 61 | Spherical trigonometry | Spherical arcs and angles
art: 877 |  | Spherical Trigonometry is the investigation of the relations which exist between the sides and angles of spherical triangles. Each side of a spherical triangle being an arc, is the measure of an angle. It may be measured by degrees, minutes, and seconds, as an angle is measured, and it has its sine, tangent, and other functions; the sine of an arc being the sine of the angle at the center which that arc subtends. These relations apply equally well to the faces and diedral angles of a triedral.
art: 878 | Theorem | The cosine of any side of a spherical triangle is equal to the product of the cosines of the other two sides, increased by the product of the sines of those sides and the cosine of their included angle. Let ABC be a spherical triangle, O the center, AD and AE tangents to the arcs AB and AC. Then the angle EAD is the angle A, and EOD is measured by the side a. From the triangles EOD and EAD, the cosine of a equals the cosine of b times the cosine of c, plus the sine of b times the sine of c times the cosine of A. These are called the fundamental equations of Spherical Trigonometry; by their aid, when any three of the elements of a spherical triangle are known, the others may be calculated.
fig: 315 0.16 0.37 0.335 0.75
fig: 316 0.135 0.25 0.405 0.76
fig: 316 0.575 0.745 0.44 0.76
art: 879 | A side and the three angles | Since these formulas are true of all spherical triangles, they apply to the polar triangle. Hence, the cosine of A equals minus the cosine of B times the cosine of C, plus the sine of B times the sine of C times the cosine of a.
art: 880 | Formulas for logarithmic use | Let p represent the perimeter. Then the sine of half of A is the square root of: the sine of half p less b, times the sine of half p less c, divided by the sine of b times the sine of c. Similar formulas give the cosine and tangent of half an angle.
art: 881 |  | Let E represent the spherical excess, that is, A plus B plus C less 180 degrees. By reasoning upon the polar triangle, similar formulas give the sides in terms of the angles. Any angle of a spherical triangle is greater than half the spherical excess.
art: 882 | Theorem | The sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. From any point P in OA, let PD fall perpendicular to the plane BOC, and make DE, DF perpendicular to BO, OC. Then PD is OP times the sine of c times the sine of B, and also OP times the sine of b times the sine of C. Therefore, the sine of B is to the sine of C as the sine of b is to the sine of c.
fig: 320 0.44 0.615 0.445 0.77
art: 883 | Four contiguous parts | The product of the cotangent of one side by the sine of another, is equal to the product of the cosine of the included angle by the cosine of the second side, plus the product of the sine of the included angle by the cotangent of the angle opposite the first side.
art: 884 | Formulas of Delambre | Four formulas give the sine and cosine of half the sum and half the difference of two angles in terms of the sides and the third angle. These four formulas of Delambre were published by him in 1807.
art: 885 | Napier's analogies | Dividing the formulas of Delambre, one by another, gives four proportions, which are called Napier's Analogies, from their inventor, analogy being formerly used as synonymous with proportion. The first: the tangent of half the sum of A and B, is to the cotangent of half C, as the cosine of half the difference of a and b, is to the cosine of half their sum.
art: 886 |  | It follows from the analogies that half the sum of two angles and half the sum of the opposite sides are either both less or both greater than ninety degrees. Therefore, in any spherical triangle, the greater angle is opposite the greater side, and conversely.
mc: Each side of a spherical triangle, being an arc, is the measure of... | *an angle at the center | a straight line | an area
mc: The relations of spherical trigonometry apply equally well to the faces and diedrals of a... | *triedral | prism | polygon
mc: The cosine of a side equals the product of the cosines of the other two sides, increased by the product of their sines and the cosine of... | *their included angle | the opposite angle | the third side
mc: The formulas for the cosine of a side are called the ___ equations of spherical trigonometry. | *fundamental | logarithmic | polar
mc: In a spherical triangle, the sines of the angles are proportional to the sines of the... | *opposite sides | adjacent sides | polar angles
mc: Napier's Analogies are so named because analogy formerly meant... | *proportion | similarity | logarithm
mc: The four formulas of Delambre were published in... | *1807 | 1614 | 1700
mc: In any spherical triangle, the greater angle is opposite... | *the greater side | the less side | a quadrant
tf: Any angle of a spherical triangle is greater than half the spherical excess. {T}
tf: The fundamental formulas are well suited to logarithmic calculation. {F}
q: In a spherical triangle, the sides b and c are each 90 degrees. The cosine of 90 degrees is 0 and its sine is 1. If the angle A between them has a cosine of 0.5, what is the cosine of the side a? {0.5}
sol: cos a = cos b cos c + sin b sin c cos A = 0 + 1 times 1 times 0.5 = 0.5.
q: An angle of a spherical triangle is 110 degrees. How many degrees is the opposite side of its polar triangle? {70}
sol: Each angle is the supplement of the opposite side of the polar triangle: 180 - 110 = 70.
q: The angles of a spherical triangle are 80, 70, and 60 degrees. What is the spherical excess, in degrees? {30}
sol: 80 + 70 + 60 - 180 = 30.
q: In that triangle (excess 30 degrees), every angle must be greater than how many degrees? {15}
sol: Any angle is greater than half the spherical excess: 30 / 2 = 15.

== 62 | Spherical trigonometry | Right angled spherical triangles
art: 888 |  | The foregoing formulas may be applied to right angled triangles by supposing one of the angles, as A, to be right. Then: the cosine of the hypotenuse equals the product of the cosines of the sides; the cosine of the hypotenuse equals the product of the cotangents of the angles; the sine of a side equals the sine of the opposite angle times the sine of the hypotenuse; the tangent of a side equals the tangent of the hypotenuse times the cosine of the included angle; the tangent of a side equals the tangent of the opposite angle times the sine of the other side; and the cosine of an angle equals the cosine of the opposite side times the sine of the other angle.
art: 889 |  | Since the cosine of the hypotenuse has the same sign as the product of the cosines of the other two sides, in a right angled spherical triangle either all the sides are less than quadrants, or two are greater and one is less. Also, an oblique angle and its opposite side are either both less or both greater than 90 degrees. This is expressed by saying they are of the same species.
art: 890 | Napier's rule of circular parts | A rule for remembering these formulas was invented by Napier, and published with his description of logarithms in 1614. The right angle being omitted, the two sides which include it, the complements of the other angles, and the complement of the hypotenuse are called the circular parts. Any one may be called the middle part; the two next to it are the adjacent parts, and the remaining two are the opposite parts. Napier's rule is: The sine of the middle part is equal to the product of the tangents of the adjacent parts, also to the product of the cosines of the opposite parts. The words sine and middle have their first vowel the same, also tangent and adjacent, also cosine and opposite.
art: 891 | Problem | Given the hypotenuse and an oblique angle, to find the other parts. For example, given the hypotenuse 64 degrees 17 minutes 35 seconds and an angle of 70 degrees, the side opposite is found by its sine, the sine of the angle times the sine of the hypotenuse: the side is 57 degrees 51 minutes 11 seconds. It is known to be acute because its opposite angle is acute.
art: 895 |  | In the cases of right angled triangles there is no ambiguous case, except when a side and its opposite angle are given; for whenever a part is found by its sine, its species is determined by the rule of the same species.
art: 896 | Problem | Given a side and its opposite angle, to find the other parts. Here the triangle is ambiguous, as all the parts are found by their sines. If BAC is right angled at A, produce CB and CA to meet in C'. Then the triangle C'AB has a right angle at A, the given side BA, and the angle C' equal to C.
fig: 328 0.16 0.26 0.43 0.78
art: 897 |  | The solution of an oblique triangle may sometimes depend upon that of a right angled triangle. If a triangle has one of its sides a quadrant, its polar triangle has a right angle. If a triangle is isosceles, an arc from the vertex to the middle of the base divides it into two equal right angled triangles. If a triangle has two sides supplementary, the sides may be produced to make an isosceles triangle.
fig: 328 0.625 0.735 0.405 0.78
art: 898 | Exercises | 5. Show that if the hypotenuse is 90 degrees, one of the sides must be 90 degrees, and conversely. 7. A lateral edge of a pyramid whose base is a square, makes angles of 60 and 65 degrees with the two conterminous sides of the base; find the diedral angle of that edge.
mc: In a right angled spherical triangle, the cosine of the hypotenuse equals the product of the ___ of the sides. | *cosines | sines | tangents
mc: The sine of a side equals the sine of the opposite angle times the sine of the... | *hypotenuse | other side | right angle
mc: An oblique angle and its opposite side, both less or both greater than 90 degrees, are said to be of the same... | *species | kind of circle | quadrant
mc: Napier's rule: the sine of the middle part equals the product of the ___ of the adjacent parts. | *tangents | cosines | sines
mc: Napier's rule: the sine of the middle part equals the product of the ___ of the opposite parts. | *cosines | tangents | secants
mc: Which is NOT one of Napier's circular parts? | *The right angle | A side including the right angle | The complement of the hypotenuse
mc: Napier published his rule of circular parts in... | *1614 | 1807 | 1700
mc: In a right angled spherical triangle, which case is ambiguous? | *A side and its opposite angle | The two sides | The hypotenuse and an angle
tf: In a right spherical triangle, all three sides may be less than quadrants. {T}
tf: In a right spherical triangle, exactly one side may be greater than a quadrant while the other two are less. {F}
tf: If a triangle has one side a quadrant, its polar triangle has a right angle. {T}
q: A right angled spherical triangle has sides of 60 and 60 degrees about the right angle. The cosine of 60 degrees is 0.5. What is the cosine of the hypotenuse? {0.25}
sol: The cosine of the hypotenuse is the product of the cosines of the sides: 0.5 times 0.5 = 0.25.
q: A right angled spherical triangle has a side whose cosine is 0.8 and another whose cosine is 0.5. What is the cosine of the hypotenuse? {0.4}
sol: 0.8 times 0.5 = 0.4.
q: The hypotenuse of a right spherical triangle is 90 degrees, and one side is 40 degrees. How many degrees is the other side? {90}
sol: The cosine of the hypotenuse is 0, so the product of the cosines of the sides is 0; one side, not being 90, makes the other 90 degrees.
q: In a right spherical triangle, the sine of the hypotenuse is 0.8 and the sine of an oblique angle is 0.5. What is the sine of the side opposite that angle? {0.4}
sol: The sine of a side is the sine of the opposite angle times the sine of the hypotenuse: 0.5 times 0.8 = 0.4.

== 63 | Spherical trigonometry | Solution of spherical triangles
art: 899 | Problem | Given the sides, to find the angles. Either of the angles may be found by the formulas for half an angle. When all the angles are required, the formula for the tangent is to be preferred.
art: 900 | Problem | Given the angles, to find the sides. Either of the sides may be found by the formulas in terms of the spherical excess.
art: 901 | Problem | Given two sides and the included angle, to find the other angles and side. The half sum of the other angles may be found by the first of Napier's Analogies, and the half difference by the second; and hence, the angles themselves. Then the third side may be found by the proportion of the sines. For example, given a = 76 degrees 35 minutes 36 seconds, b = 50 degrees 10 minutes 30 seconds, and C = 34 degrees 15 minutes 3 seconds, the analogies give half the sum of A and B, 81 degrees 55 minutes 47 seconds, and half their difference, 39 degrees 40 minutes 33 seconds. Hence A is 121 degrees 36 minutes 20 seconds, and B is 42 degrees 15 minutes 14 seconds.
art: 902 | Problem | Given one side and the adjacent angles, to find the other sides and angle. The half sum of the other sides may be found by the third analogy, and the half difference by the fourth; and hence, the sides themselves. Then the third angle may be found by the proportion of the sines.
art: 903 | Problem | Given two sides and an angle opposite one of them, to find the other parts. The angle opposite the other given side may be found by the proportion of the sines. Since the sine is used, there may be two solutions. The ambiguity is sometimes removed by observing that the greater angle is opposite the greater side. When both values are consistent with this principle, there are two distinct spherical triangles, as ABC and AB'C, which have the given elements. When neither value is consistent, the triangle is impossible.
fig: 332 0.28 0.435 0.425 0.76
art: 904 | Problem | Given two angles and a side opposite one of them, to find the other parts. This case is precisely analogous to the last; it presents the same ambiguity, and the ambiguity is resolved in the same manner.
art: 905 | Exercises | 2. Find the diedral angle of a regular tetraedron. 7. Given the latitude of Paris 48 degrees 50 minutes 12 seconds, the latitude of New York 40 degrees 17 minutes 17 seconds, and the longitude of New York west of Paris 76 degrees 20 minutes 27 seconds, to find the distance between these points, along an arc of a great circle; the earth being considered a sphere of a radius of 3956 miles.
mc: Given two sides and the included angle, the half sum of the other angles is found by... | *the first of Napier's Analogies | the formula for half an angle | Delambre's formulas only
mc: Given one side and the adjacent angles, the half sum of the other sides is found by... | *the third analogy | the first analogy | the law of cosines
mc: Which given parts may have two solutions? | *Two sides and an angle opposite one of them | The three sides | Two sides and the included angle
mc: The ambiguity is sometimes removed by noticing that the greater angle is opposite... | *the greater side | the smaller side | a quadrant
mc: When all three angles are wanted from the three sides, which formula is preferred? | *The tangent of half the angle | The sine of half the angle | The cosine of the side
mc: The shortest distance between two cities on the earth is found along... | *an arc of a great circle | a parallel of latitude | a meridian always
tf: When given two angles and a side opposite one, the case is analogous to two sides and an angle opposite one. {T}
tf: Given the three angles of a spherical triangle, the sides can be found. {T}
q: In the example, half the sum of A and B is 81 degrees 55 minutes 47 seconds, and half their difference is 39 degrees 40 minutes 33 seconds. The angle A is 121 degrees 36 minutes and how many seconds? {20}
sol: A is half the sum plus half the difference: 81 55 47 + 39 40 33 = 121 degrees 36 minutes 20 seconds.
q: In the same example, B is 42 degrees, how many minutes, and 14 seconds? Give the minutes. {15}
sol: B is half the sum less half the difference: 81 55 47 - 39 40 33 = 42 degrees 15 minutes 14 seconds.
q: The earth is a sphere of radius 3956 miles. Using 3.1416 for pi, how many miles long is one degree of a great circle, to the nearest mile? {69}
sol: The circumference is 2 times 3.1416 times 3956 = 24856 miles; divided by 360 degrees, about 69 miles.
q: Two places on the equator differ in longitude by 10 degrees. About how many miles apart are they along the equator, at 69 miles to a degree? {690}
sol: The equator is a great circle: 10 times 69 = 690 miles.
== 64 | Logarithms | Use of common logarithms
art: 906 |  | Nearly all trigonometrical calculations are made by means of logarithms. To understand this chapter, the student must be acquainted with the algebraic theory of positive and negative exponents.
art: 907 | Common logarithms | The common logarithm of a number is the exponent of that power of 10 which is equal to the number. Hence, the logarithm of 1 is 0, of 10 is 1, of 100 is 2, of 1000 is 3; and the logarithm of one-tenth is minus 1, of one-hundredth is minus 2. Numbers greater than unity have positive logarithms; numbers less than unity have negative logarithms. No other numbers than the powers of 10 and their reciprocals have integral logarithms.
art: 908 | Characteristic | The characteristic of a logarithm is its integral part. The mantissa is the decimal part. The mantissa is always kept positive, and only the characteristic of a negative logarithm is negative; the negative sign is written over the characteristic. The characteristic of the logarithm of a number greater than 1 is one less than the number of integral places in the number. The characteristic denotes how many places the first significant figure of the number is to the left of the unit's place.
art: 909 | Table of logarithms | Multiplying a number by 10 does not change the mantissa of its logarithm, and it is the same when the number is multiplied or divided by any power of 10. In other words: if two numbers have the same significant figures, their logarithms have the same mantissas. The characteristics are not given in the tables, but must be found by the rule. The large tables of Adrien Vlacq give the logarithms to ten places of decimals of all numbers from 1 to 100000.
art: 910 | To find the logarithm of a number | Find the mantissa in the table and prefix the proper characteristic. For example, the logarithm of 1245 is 3.095169; of 124500, 5.095169. If the number has more figures than the table, find the mantissa for the first four or five figures, and add the proportional part of the tabular difference. Thus the logarithm of 3.1416 is .497068, the mantissa for 3141, plus six-tenths of the difference 138, or 83: .497151.
art: 911 | To find the number from its logarithm | If the mantissa is not in the table, take the next less, and annex to its number two figures of the quotient found by dividing the excess by the tabular difference. Fix the decimal point by the rule of the characteristic.
art: 912 | Multiplication and division | To multiply, add the logarithms of the factors. The sum is the logarithm of the product. To divide, subtract the logarithm of the divisor from that of the dividend. The remainder is the logarithm of the quotient.
art: 913 |  | It saves labor, instead of subtracting a logarithm, to add its arithmetical complement, which is the excess of 10 over the logarithm. Each time that an arithmetical complement is added, 10 must be subtracted from the result. The complement may be written at once, beginning at the left, and subtracting each figure from 9, except the last significant figure, which is subtracted from 10.
art: 914 | Involution and evolution | To raise a number to a required power, multiply its logarithm by the exponent of the power. The product is the logarithm of the power. To extract any root of a number, divide its logarithm by the index of the required root. The quotient is the logarithm of the root.
art: 915 |  | In order that an arithmetical problem may be solved by logarithms, it should not contain any additions or subtractions. Mathematicians have therefore given much attention to formulas which require only multiplication, division, involution, and evolution.
art: 916 | Exercises | 2. Find the area of a circle, the radius being 3 feet. 3. What is the diameter of a circle whose circumference is 314 feet 3 inches? 5. The diameter of the earth at the equator being 41850000 feet, what is the length in miles of one degree of longitude on the equator, there being 5280 feet in one mile?
mc: The common logarithm of a number is the exponent of the power of ___ equal to the number. | *10 | 2 | e
mc: The integral part of a logarithm is its... | *characteristic | mantissa | complement
mc: The decimal part of a logarithm is its... | *mantissa | characteristic | difference
mc: Numbers less than unity have ___ logarithms. | *negative | positive | zero
mc: Two numbers with the same significant figures have logarithms with the same... | *mantissa | characteristic | value
mc: To multiply numbers by logarithms, you ___ their logarithms. | *add | multiply | subtract
mc: To divide by logarithms, you ___ the logarithm of the divisor. | *subtract | add | divide by
mc: To raise a number to a power, ___ its logarithm by the exponent. | *multiply | divide | add
mc: To extract a root, ___ the logarithm by the index of the root. | *divide | multiply | subtract
mc: The arithmetical complement of a logarithm is its excess under... | *10 | 1 | 100
mc: A problem worked by logarithms should not contain... | *additions or subtractions | multiplications | roots
tf: The logarithm of 1 is 0. {T}
tf: The characteristics of logarithms are printed in the tables. {F}
tf: Multiplying a number by 10 changes the mantissa of its logarithm. {F}
q: What is the logarithm of 1000? {3}
sol: 1000 is 10 to the third power.
q: What is the logarithm of 100000? {5}
sol: 100000 is 10 to the fifth power.
q: What is the characteristic of the logarithm of 5000? {3}
sol: 5000 has four integral places; the characteristic is one less: 3.
q: What is the characteristic of the logarithm of 365.242? {2}
sol: 365.242 has three integral places; the characteristic is 2.
q: The logarithm of a number is 4. What is the number? {10000}
sol: 10 to the fourth power is 10000.
q: The logarithm of 2 is .301030. What is the logarithm of 200? {2.30103}
sol: Same significant figures, same mantissa; 200 has three integral places, so the characteristic is 2: 2.301030.
q: The logarithm of 2 is .301030 and of 3 is .477121. What is the logarithm of 6? {0.778151}
sol: To multiply, add the logarithms: .301030 + .477121 = .778151.
q: The logarithm of 2 is .301030. What is the logarithm of 8? {0.90309}
sol: 8 is the cube of 2: multiply the logarithm by 3: .903090.
q: The logarithm of 64 is 1.806180. What is the logarithm of its square root? {0.90309}
sol: Divide the logarithm by 2: .903090, the logarithm of 8.
q: What is the arithmetical complement of the logarithm 3.538574? {6.461426}
sol: 10 - 3.538574 = 6.461426: each figure from 9, the last from 10.
q: The diameter of the earth at the equator is 41850000 feet. Using 3.1416 for pi, how many miles long is one degree of longitude on the equator, to the nearest mile? {69}
sol: The circumference is 3.1416 times 41850000 = 131,476,000 feet; divided by 360 and by 5280 feet in a mile: about 69.2 miles.
EOF

make-book.ps1: builds any book: -Book geometry writes public\books\geometry.json.

cat > /root/rays-rebuild/make-book.ps1 <<'EOF'
# Builds public/books/<id>.json from a hand-corrected book in source/<Book>.txt (format at the top of source/primary.txt).
# Generates every table exactly, works out {?} answers, and checks typed answers wherever the question is plain
# arithmetic ("How many are 3 and 5?", "One from 9?", "Nine and how many make 17?", "How many twos in 8?").
# Usage: powershell -ExecutionPolicy Bypass -File RaysArithmetic\tools\make-book.ps1 [-Book primary|intellectual|...]
param([string]$Book = 'primary')
$root = Split-Path -Parent (Split-Path -Parent $MyInvocation.MyCommand.Path)
$src = Get-Content "$root\source\$Book.txt" -Encoding UTF8
# the book's id, title, shelf line and color: "@book: primary1877 | Ray's New Primary Arithmetic | 1877 - Book One | #2f6f8f"
$head = $src | Where-Object { $_ -match '^@book: ' } | Select-Object -First 1
if (-not $head) { throw "source\$Book.txt has no @book: line" }
$bk = ($head -replace '^@book: ', '') -split '\s*\|\s*'
$script:errors = New-Object System.Collections.ArrayList

# ---------- numbers ----------
$ones = @{ naught = 0; zero = 0; one = 1; two = 2; three = 3; four = 4; five = 5; six = 6; seven = 7; eight = 8; nine = 9; ten = 10; eleven = 11; twelve = 12; thirteen = 13; fourteen = 14; fifteen = 15; sixteen = 16; seventeen = 17; eighteen = 18; nineteen = 19 }
$tens = @{ twenty = 20; thirty = 30; forty = 40; fifty = 50; sixty = 60; seventy = 70; eighty = 80; ninety = 90 }
$plural = @{ ones = 1; twos = 2; threes = 3; fours = 4; fives = 5; sixes = 6; sevens = 7; eights = 8; nines = 9; tens = 10 }
function WordNum([string]$w) {
  $w = $w.ToLower()
  if ($w -match '^\d+$') { return [long]$w }
  if ($w -eq 'one hundred' -or $w -eq 'hundred') { return 100 }
  if ($ones.ContainsKey($w)) { return $ones[$w] }
  if ($tens.ContainsKey($w)) { return $tens[$w] }
  if ($w -match '^(\w+)-(\w+)$' -and $tens.ContainsKey($Matches[1]) -and $ones.ContainsKey($Matches[2])) { return $tens[$Matches[1]] + $ones[$Matches[2]] }
  return $null
}
$numRx = '\b(?:\d+|one hundred|(?:twenty|thirty|forty|fifty|sixty|seventy|eighty|ninety)(?:-(?:one|two|three|four|five|six|seven|eight|nine))?|naught|zero|eleven|twelve|thirteen|fourteen|fifteen|sixteen|seventeen|eighteen|nineteen|one|two|three|four|five|six|seven|eight|nine|ten)\b'

# Plain arithmetic in a question part -> its value, or $null when the part isn't plain arithmetic.
# Reads left to right, the way the book works them ("2 and 5 are 7; 7 less 3 are 4"); every x comes first in the book.
function Calc([string]$t) {
  $s = ' ' + $t.ToLower() + ' '
  $s = $s -replace '[?:]', ' ' -replace '\s*=\s*', ' ' -replace ',', ' '
  # "Twelve are how many times 3?" -> 12 / 3
  # "What are 2/3 of 9?" -> 6
  if ($s -match '^\s*(?:what (?:are|is)\s+)?(\d+)/(\d+) of (\d+)\s*$') { $v = [int]$Matches[3] * [int]$Matches[1]; if ($v % [int]$Matches[2]) { return $null }; return [int]($v / [int]$Matches[2]) }
  if ($s -match "^\s*($numRx)\s+(?:are|is) how many times\s+($numRx)\s*$") { $a = WordNum $Matches[1]; $b = WordNum $Matches[2]; if ($a % $b) { return $null }; return [int]($a / $b) }
  $s = $s -replace '\bare how many\b', ' ' -replace '\bhow many (are|is)\b', ' ' -replace '\bhow many\b', ' '
  # "N from M leaves how many" / "N from M" / "N taken away from M"
  if ($s -match "^\s*($numRx)\s+(?:taken\s+)?(?:away\s+)?from\s+($numRx)\s*(?:leaves)?\s*$") { return (WordNum $Matches[2]) - (WordNum $Matches[1]) }
  # "N and how many make M"
  if ($s -match "^\s*($numRx)\s+and\s+make\s+($numRx)\s*$") { return (WordNum $Matches[2]) - (WordNum $Matches[1]) }
  # "how many twos in 8"
  if ($s -match "^\s*(ones|twos|threes|fours|fives|sixes|sevens|eights|nines|tens)\s+in\s+($numRx)\s*$") { return [int]((WordNum $Matches[2]) / $plural[$Matches[1]]) }
  # "times is 2 contained in 4" / "2 in 4"
  if ($s -match "^\s*(?:times is\s+)?($numRx)\s+(?:contained\s+)?in\s+($numRx)\s*$") { return [int]((WordNum $Matches[2]) / (WordNum $Matches[1])) }
  $s = $s -replace '\bmultiplied by\b', ' x ' -replace '\bdivided by\b', ' / ' -replace '\btimes\b', ' x ' -replace '\btime\b', ' x ' -replace '\bplus\b', ' + ' -replace '\bminus\b', ' - ' -replace '\bless\b', ' - ' -replace '\band\b', ' + '
  $tok = @([regex]::Matches($s, "$numRx|[-+x/]") | ForEach-Object { $_.Value })
  $rest = ([regex]::Replace($s, "$numRx|[-+x/]", ' ')).Trim()
  if ($rest -or $tok.Count -lt 3 -or $tok.Count % 2 -eq 0) { return $null }
  $v = WordNum $tok[0]; if ($null -eq $v) { return $null }
  for ($i = 1; $i -lt $tok.Count; $i += 2) {
    $b = WordNum $tok[$i + 1]; if ($null -eq $b) { return $null }
    switch ($tok[$i]) { '+' { $v += $b } '-' { $v -= $b } 'x' { $v *= $b } '/' { if ($v % $b) { return $null }; $v = $v / $b } default { return $null } }
  }
  return [long]$v
}

# ---------- exact fractions ----------
# {~ 1/2 + 1/4} works the answer out exactly: whole numbers, a/b, mixed numbers written 1_3/4, + - * / and ( ).
# A fraction is written without spaces (3/4); division between numbers needs spaces (12 / 3).
# The answer is written the book's way: 3/4, 2 1/3, 5.
function Gcd([long]$a, [long]$b) { $a = [Math]::Abs($a); $b = [Math]::Abs($b); while ($b) { $t = $a % $b; $a = $b; $b = $t }; [long][Math]::Max($a, 1) }
function Rat([long]$n, [long]$d) { if ($d -lt 0) { $n = -$n; $d = -$d }; $g = Gcd $n $d; ,@([long]($n / $g), [long]($d / $g)) }
function RatStr($r) {
  $n = $r[0]; $d = $r[1]
  if ($d -eq 1) { return "$n" }
  $w = [long][Math]::Truncate($n / $d); $f = $n - $w * $d
  if ($w) { "$w $([Math]::Abs($f))/$d" } else { "$n/$d" }
}
function EvalFrac([string]$e) {
  $script:tk = @([regex]::Matches(($e -replace '(?<=\d),(?=\d{3})', ''), '\d*\.\d*\(\d+\)|\d+_\d+/\d+|\d+/\d+|\d*\.\d+|\d+|[-+*/()]') | ForEach-Object { $_.Value }); $script:ti = 0
  function Atom {
    $x = $script:tk[$script:ti]; $script:ti++
    if ($x -eq '(') { $v = Sum; $script:ti++; return ,$v }
    # a circulate: .45(3) = .4533..., 104.(14) = 104.1414...; the figures in ( ) repeat
    if ($x -match '^(\d*)\.(\d*)\((\d+)\)$') {
      $w = [long]('0' + $Matches[1]); $a = $Matches[2]; $rp = $Matches[3]
      $num = [long]('0' + $a + $rp) - [long]('0' + $a); $den = [long](('9' * $rp.Length) + ('0' * $a.Length))
      return ,(Rat ($w * $den + $num) $den)
    }
    if ($x -match '^(\d+)_(\d+)/(\d+)$') { return ,(Rat ([long]$Matches[1] * [long]$Matches[3] + [long]$Matches[2]) ([long]$Matches[3])) }
    if ($x -match '^(\d+)/(\d+)$') { return ,(Rat ([long]$Matches[1]) ([long]$Matches[2])) }
    if ($x -match '^\d+$') { return ,(Rat ([long]$x) 1) }
    if ($x -match '^(\d*)\.(\d+)$') { return ,(Rat ([long]($Matches[1] + $Matches[2])) ([long][Math]::Pow(10, $Matches[2].Length))) }
    throw "bad token '$x' in $e"
  }
  function Prod {
    $v = Atom
    while ($script:ti -lt $script:tk.Count -and $script:tk[$script:ti] -match '^[*/]$') {
      $op = $script:tk[$script:ti]; $script:ti++; $b = Atom
      if ($op -eq '*') { $v = Rat ($v[0] * $b[0]) ($v[1] * $b[1]) } else { $v = Rat ($v[0] * $b[1]) ($v[1] * $b[0]) }
    }
    ,$v
  }
  function Sum {
    $v = Prod
    while ($script:ti -lt $script:tk.Count -and $script:tk[$script:ti] -match '^[-+]$') {
      $op = $script:tk[$script:ti]; $script:ti++; $b = Prod
      $n = $(if ($op -eq '+') { $v[0] * $b[1] + $b[0] * $v[1] } else { $v[0] * $b[1] - $b[0] * $v[1] })
      $v = Rat $n ($v[1] * $b[1])
    }
    ,$v
  }
  $r = Sum
  if ($script:ti -ne $script:tk.Count) { throw "couldn't read all of: $e" }
  $script:lastRat = $r
  RatStr $r
}
# a value as a circulate, the repeating figures in ( ): 77/111 -> ".(693)", 5/36 -> ".13(8)"
function AsRep($r) {
  $n = [long]$r[0]; $d = [long]$r[1]; $w = [long][Math]::Floor($n / $d); $rem = $n - $w * $d
  $digits = New-Object Text.StringBuilder; $seen = @{}
  while ($rem -and -not $seen.ContainsKey($rem)) { $seen[$rem] = $digits.Length; $rem *= 10; [void]$digits.Append([long][Math]::Floor($rem / $d)); $rem = $rem % $d }
  $s = $digits.ToString(); $lead = $(if ($w) { "$w" } else { '' })
  if (-not $rem) { if ($s) { return "$lead.$s" } else { return "$w" } }
  $k = $seen[$rem]; "$lead.$($s.Substring(0, $k))($($s.Substring($k)))"
}
# a value as the book's complex decimal, carried to $p places: 41 1391/1875 -> "41.7422 2/3"; exact decimals stay pure
function AsCplx($r, [int]$p) {
  $n = [long]$r[0]; $d = [long]$r[1]; $scale = [long][Math]::Pow(10, $p)
  $N = $n * $scale; $q = [long][Math]::Floor($N / $d); $rem = $N - $q * $d
  $ip = [long][Math]::Floor($q / $scale); $dec = ($q % $scale).ToString().PadLeft($p, '0')
  if (-not $rem) { $dec = $dec.TrimEnd('0'); if (-not $dec) { return "$ip" } }
  $s = $(if ($ip) { "$ip" } else { '' }) + '.' + $dec
  if ($rem) { $g = Gcd $rem $d; $s += " $($rem / $g)/$($d / $g)" }
  $s
}
# the same value written as a decimal when it ends (denominator only 2s and 5s): "2 39/50" -> "2.78"; else unchanged
function AsDec([string]$v) {
  if ($v -notmatch '^(-?\d+)( (\d+)/(\d+)|/(\d+))?$') { return $v }
  if (-not $Matches[2]) { return $v }
  if ($Matches[3]) { $w = [long]$Matches[1]; $n = [long]$Matches[3]; $d = [long]$Matches[4] } else { $w = 0; $n = [long]$Matches[1]; $d = [long]$Matches[5] }
  $t = $d; while ($t % 2 -eq 0) { $t /= 2 }; while ($t % 5 -eq 0) { $t /= 5 }; if ($t -ne 1) { return $v }
  $s = ([decimal]$w + [decimal]$n / [decimal]$d).ToString('0.##########################')
  if ($s.StartsWith('0.')) { $s.Substring(1) } else { $s }
}

# ---------- the book's tables ----------
function Table([string]$op, [int]$n) {
  $f = New-Object System.Collections.ArrayList
  # t = the book's line, q = the same fact asked as a question (the app's drill and the voice use both)
  function F($t, $a, $b, $r) {
    $q = switch ($op) { 'add' { "How many are $a and $b?" } 'sub' { "$a from $b leaves how many?" } 'mul' { if ($a -eq 1) { "How many is 1 time $b?" } else { "How many are $a times $b?" } } 'div' { "How many times is $a contained in $b?" } }
    [void]$f.Add([ordered]@{ t = $t; q = $q; a = $a; b = $b; r = $r })
  }
  switch ($op) {
    'add' { foreach ($k in 1..10) { F "$n and $k are $($n + $k)" $n $k ($n + $k); if ($k -ne $n) { F "$k and $n are $($n + $k)" $k $n ($n + $k) } } }
    'sub' { F "$n from $n leaves 0" $n $n 0
            foreach ($k in 1..9) { $m = $n + $k; F "$n from $m leaves $k" $n $m $k; if ($k -ne $n) { F "$k from $m leaves $n" $k $m $n } } }
    'mul' { if ($n -eq 1) { foreach ($k in 2..10) { F "1 time $k is $k" 1 $k $k; F "$k times 1 are $k" $k 1 $k } }
            else { F "$n times 1 are $n" $n 1 $n; foreach ($k in 2..10) { F "$k times $n are $($k * $n)" $k $n ($k * $n); if ($k -ne $n) { F "$n times $k are $($k * $n)" $n $k ($k * $n) } } } }
    'div' { F "$n in $n, 1 time" $n $n 1
            foreach ($k in 2..10) { $m = $n * $k; F "$n in $m, $k times" $n $m $k; if ($k -ne $n) { F "$k in $m, $n times" $k $m $n } } }
  }
  return ,$f.ToArray()
}

# ---------- algebra (Book 6) ----------
# Math in a text is written between backticks in plain ASCII: `3a^2b - 4(x+y)/2`. It is shown the book's way
# (3a²b − 4(x+y)/2) and said for the voice ("3 A squared B minus 4 times the quantity X plus Y, over 2").
# An answer {$ expr} is typed on the algebra pad; {$ expr @ same} checks it has the value of `same`, {$~ (a+b)(c+d)} multiplies it out itself,
# {$ 12 @x: 6x+2x = 3x+60} checks it solves the equation for x, {15 @= 5^2-2*5} checks a plain number.
# chk: x=5, y=4 | x+y=9 | x-y=1   checks a pair of answers against the equations.
Add-Type -Path "$root\tools\alg.cs"
function MathSplit([string]$t, [scriptblock]$plain, [scriptblock]$math) {
  $bits = $t -split '`'; $o = ''
  for ($i = 0; $i -lt $bits.Count; $i++) { $o += $(if ($i % 2) { & $math $bits[$i] } elseif ($bits[$i]) { & $plain $bits[$i] } else { '' }) }
  $o
}
$SUPS = @{ '0' = [char]0x2070; '1' = [char]0x00B9; '2' = [char]0x00B2; '3' = [char]0x00B3; '4' = [char]0x2074; '5' = [char]0x2075; '6' = [char]0x2076; '7' = [char]0x2077; '8' = [char]0x2078; '9' = [char]0x2079; '-' = [char]0x207B }
function AlgPretty([string]$m) {
  $m = [regex]::Replace($m, '\^(-?\d+)', { param($x) -join ($x.Groups[1].Value.ToCharArray() | ForEach-Object { $SUPS[[string]$_] }) })
  $m.Replace('-', [string][char]0x2212).Replace('*', [string][char]0x00D7)
}
function AlgSay([string]$m) {
  $tk = @([regex]::Matches(($m -replace [char]0x2212, '-'), '\d+\.?\d*|\.\d+|[A-Za-z]|\^-?\d+|[-+*/()=<>,;:.]|\S') | ForEach-Object { $_.Value })
  $w = New-Object System.Collections.ArrayList; $prev = ''
  for ($i = 0; $i -lt $tk.Count; $i++) {
    $t = $tk[$i]; $nx = $(if ($i + 1 -lt $tk.Count) { $tk[$i + 1] } else { '' })
    if ($t -match '^\^(-?\d+)$') { $n = [int]$Matches[1]; [void]$w.Add($(switch ($n) { 0 { 'to the zero power' } 2 { 'squared' } 3 { 'cubed' } default { "to the $(if ($n -lt 0) { 'minus ' })$(OrdWord ([Math]::Abs($n))) power" } })) }
    elseif ($t -match '^[a-z]$') { [void]$w.Add($t.ToUpper()) }
    elseif ($t -match '^[\d.]+$') {
      # a plain number fraction 3/4 (not 3/4x) is said "3 fourths"
      if ($nx -eq '/' -and $i + 2 -lt $tk.Count -and $tk[$i + 2] -match '^\d+$' -and [int]$tk[$i + 2] -gt 1 -and ($i + 3 -ge $tk.Count -or $tk[$i + 3] -notmatch '^([a-z(]|\^)') -and $prev -notmatch '^[a-z)]$') { [void]$w.Add((FracWords $t $tk[$i + 2])); $i += 2; $t = 'frac' }
      else { if ($prev -match '^[a-z)]$') { [void]$w.Add('times') }; [void]$w.Add($t) }
    }
    elseif ($t -eq '(') { [void]$w.Add($(if ($prev -match '^([\d.]+|[a-z)]|\^-?\d+)$') { 'times the quantity' } else { 'the quantity' })) }
    elseif ($t -eq ')') { [void]$w.Add(',') }
    else { [void]$w.Add($(switch ($t) { '+' { 'plus' } '-' { 'minus' } '*' { 'times' } '/' { 'over' } '=' { 'equals' } '^' { 'to the power' } '<' { 'is less than' } '>' { 'is greater than' } ([string][char]0x221A) { $(if ($prev -match '^([\d.]+|[a-z)]|\^-?\d+)$') { 'times the square root of' } else { 'the square root of' }) } ([string][char]0x00F7) { 'divided by' } default { $t } })) }
    $prev = $t
  }
  (($w -join ' ') -replace ' ,', ',' -replace ',(\s*,)+', ',' -replace ',\s*$', '').Trim()
}
# the pad's extra keys: the letters used, and a root sign when there is one
function AlgLetters([string]$s) { (-join (@([regex]::Matches(($s -replace '\^-?\d+', ''), '[a-z]') | ForEach-Object { $_.Value }) | Sort-Object -Unique)) + $(if ($s.Contains([string][char]0x221A)) { [string][char]0x221A } else { '' }) }

# ---------- questions ----------
# display: the book's x and / become real signs; spoken: signs become words for the voice
$TIMES = " $([char]0x00D7) "; $DIV = " $([char]0x00F7) "; $MINUS = " $([char]0x2212) "   # plain-ASCII script: no BOM needed
function Display([string]$t) { if ($t.Contains('`')) { return MathSplit $t { param($x) Display $x } { param($x) AlgPretty $x } }; ($t -replace '(?<=[\d)]) x (?=[\d.(])', $TIMES -replace '(?<=[\d)]) / (?=[\d.(])', $DIV -replace '(?<=[\d)]) - (?=[\d.(])', $MINUS) }
# fractions said the book's way: "3/4" -> "3 fourths", "2 1/2" -> "2 and 1 half"
$FRN = @{ 2 = 'half'; 3 = 'third'; 4 = 'fourth'; 5 = 'fifth'; 6 = 'sixth'; 7 = 'seventh'; 8 = 'eighth'; 9 = 'ninth'; 10 = 'tenth'; 11 = 'eleventh'; 12 = 'twelfth'; 13 = 'thirteenth'; 14 = 'fourteenth'; 15 = 'fifteenth'; 16 = 'sixteenth'; 17 = 'seventeenth'; 18 = 'eighteenth'; 19 = 'nineteenth'; 20 = 'twentieth'; 100 = 'hundredth' }
# any other denominator as an ordinal word: 23 -> twenty-third, 115 -> one hundred and fifteenth
function OrdWord([int]$d) {
  $o = 'zero one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen' -split ' '
  $t = ',,twenty,thirty,forty,fifty,sixty,seventy,eighty,ninety' -split ','
  $oo = @{ one = 'first'; two = 'second'; three = 'third'; five = 'fifth'; eight = 'eighth'; nine = 'ninth'; twelve = 'twelfth' }
  function Card([int]$n) { if ($n -lt 20) { return $o[$n] }; $w = $t[[Math]::Floor($n / 10)]; if ($n % 10) { $w += '-' + $o[$n % 10] }; $w }
  $words = @()
  if ($d -ge 1000) { $words += (Card ([Math]::Floor($d / 1000))) + ' thousand'; $d = $d % 1000 }
  if ($d -ge 100) { $words += (Card ([Math]::Floor($d / 100))) + ' hundred'; $d = $d % 100 }
  if ($d) { if ($words.Count) { $words += 'and' }; $words += (Card $d) }
  $s = $words -join ' '
  if ($s -match '^(.*?)(\w+)$') { $head = $Matches[1]; $last = $Matches[2] }
  $last = $(if ($oo[$last]) { $oo[$last] } elseif ($last -match 'y$') { $last -replace 'y$', 'ieth' } else { $last + 'th' })
  $head + $last
}
function FracWords([int]$n, [int]$d) {
  $name = $FRN[$d]; if (-not $name) { $name = OrdWord $d }
  if ($n -gt 1) { $name = $(if ($name -eq 'half') { 'halves' } else { $name + 's' }) }
  "$n $name"
}
function SayFractions([string]$t) {
  $t = [regex]::Replace($t, '(?<![\d/])(\d+) (\d+)/(\d+)(?![\d/])', { param($m) "$($m.Groups[1].Value) and $(FracWords $m.Groups[2].Value $m.Groups[3].Value)" })
  [regex]::Replace($t, '(?<![\d/])(\d+)/(\d+)(?![\d/])', { param($m) FracWords $m.Groups[1].Value $m.Groups[2].Value })
}
function SayRep([string]$t) { [regex]::Replace($t, '(\d*)\.(\d*)\((\d+)\)', { param($m) $w = $m.Groups[1].Value; $a = ($m.Groups[2].Value.ToCharArray() -join ' '); $r = ($m.Groups[3].Value.ToCharArray() -join ' '); (($(if ($w) { "$w " } else { '' }) + 'point ' + $(if ($a) { "$a, " } else { '' }) + "$r repeating")) }) }
function Spoken([string]$t) { if ($t.Contains('`')) { return MathSplit $t { param($x) Spoken $x } { param($x) AlgSay $x } }; $t =($t -replace '(?<=[\d)]) x (?=[\d.])', ' times ' -replace '(?<=[\d)]) / (?=[\d.])', ' divided by ' -replace '(?<=[\d)]) - (?=[\d.])', ' minus ' -replace '(?<=[\d)]) \+ (?=[\d.])', ' plus ' -replace '= \?', 'equals what?'); SayFractions (SayRep $t) }
function Question([string]$q, [string]$where) {
  $parts = New-Object System.Collections.ArrayList; $script:reps = New-Object System.Collections.ArrayList
  $pos = 0
  foreach ($m in [regex]::Matches($q, '\{([^}]*)\}')) {
    $seg = $q.Substring($pos, $m.Index - $pos).Trim(); $pos = $m.Index + $m.Length
    $ans = $m.Groups[1].Value.Trim(); if ($ans -notmatch '^~p\s') { $ans = $ans -replace '(?<=\d),(?=\d{3})', '' }   # 1,234 -> 1234 (not in a compound chain: 9,144)
    $algE = $null; if ($ans -match '^\$\s*(.+)$') { $algE = $Matches[1].Trim(); $ans = '$' }   # algebra answer, below
    # {15 @= 5^2-2*5}: a plain number checked with the algebra reader (powers, letters' values written in)
    # {12 @x: 6x+2x = 3x+60}: a plain number that must solve the equation(s) for x (several split by ;)
    if ($ans -match '^([^@]+?)\s*@([a-z]):\s*(.+)$') { $ans = $Matches[1].Trim(); $r = [RaysAlgCheck]::Holds("$($Matches[2])=" + ($ans -replace '^(\d+) (\d+)/(\d+)$', '($1+$2/$3)'), @($Matches[3] -split '\s*;\s*')); if ($r) { [void]$script:errors.Add("$where : $r") } }
    if ($ans -match '^([^@]+?)\s*@=\s*(.+)$') { $ans = $Matches[1].Trim(); $r = [RaysAlgCheck]::Same(($ans -replace '^(\d+) (\d+)/(\d+)$', '($1+$2/$3)'), $Matches[2]); if ($r) { [void]$script:errors.Add("$where : $r") } }
    # {1234 @ 567 + 667}: the book's printed answer, checked against the sum worked out from the question (catches misread digits)
    if ($ans -match '^([^@]+?)\s*@\s*(.+)$') {
      $ans = $Matches[1].Trim(); $chk = $Matches[2]
      try { $want = EvalFrac $chk; $got = EvalFrac (($ans -replace '^(\d+) (\d+/\d+)$', '$1_$2') -replace '\|.*$', '')
            if ($want -ne $got) { [void]$script:errors.Add("$where : printed answer $ans, but $chk = $want") } } catch { [void]$script:errors.Add("$where : $_") }
    }
    # [[+ 184 216 135]] / [[- 5000 1234]] / [[x 345 27]]: a sum written in a column, the book's way; its result checks the answer
    $cv = $null; $colSay = $null
    if ($seg -match '\[\[([-+x])\s+([\d,. ]+)\]\]') {
      $op = $Matches[1]; $ns = @($Matches[2].Trim() -split '\s+'); $seg = $seg.Replace($Matches[0], '').Trim()
      $expr = ($ns | ForEach-Object { $_ -replace ',', '' }) -join $(switch ($op) { '+' { ' + ' } '-' { ' - ' } 'x' { ' * ' } })
      try { $cv = EvalFrac $expr } catch { [void]$script:errors.Add("$where : $_") }
      $colSay = switch ($op) { '+' { 'Add ' + (($ns | Select-Object -SkipLast 1) -join ', ') + ', and ' + $ns[-1] + '.' } '-' { "From $($ns[0]) take $($ns[1])." } 'x' { "Multiply $($ns[0]) by $($ns[1])." } }
      if ($ans -eq '?') { $ans = "$cv" } elseif ($ans -match '^[\d.]+( \d+/\d+)?$' -and $ans -ne $cv -and (EvalFrac ($ans -replace ' ', '_')) -ne $cv) { [void]$script:errors.Add("$where : the column comes to $cv, not $ans") }
    }
    $p = [ordered]@{ t = (Display $seg) }
    if ($null -ne $cv) { $p.col = [ordered]@{ op = $op; n = $ns } }
    if ($ans -match '^(-?\d+)\.\.(-?\d+) by (-?\d+)$') {
      $a = [int]$Matches[1]; $b = [int]$Matches[2]; $st = [int]$Matches[3]
      $seq = @(); for ($v = $a; ($st -gt 0 -and $v -le $b) -or ($st -lt 0 -and $v -ge $b); $v += $st) { $seq += $v }
      if ($seq[-1] -ne $b) { [void]$script:errors.Add("$where : count by $st from $a never lands on $b") }
      $p.seq = $seq
    } elseif ($ans -match '^\d+(\|\d+)+$') {   # more than one right answer, e.g. {100|1} = 100 cents or 1 dollar
      $all = @($ans -split '\|' | ForEach-Object { [int]$_ }); $p.a = $all[0]; $p.alt = @($all | Select-Object -Skip 1)
    } elseif ($ans -match '^[\d./ ]+(\|[\d./ ]+)+$') {   # either form, e.g. money to mills or exact: {23.062|23.0625}
      $all = @($ans -split '\|' | ForEach-Object { $_.Trim() }); $p.f = $all[0]; $p.falt = @($all | Select-Object -Skip 1)
    } elseif ($ans -match '^~\s*(.+)$') {   # worked out exactly: {~ 1/2 + 1/4} -> 3/4
      # {~d expr} or an expression with a decimal in it: show a decimal answer when it ends (2.78, not 2 39/50)
      # {~c4 expr}: the answer as a complex decimal carried to 4 places (.0154 1/4), the book's way with complex decimals
      # {~r3 expr}: rounded to 3 decimal places ("preserving three places"); the cut-off value is taken too
      $ex = $Matches[1]
      # {~m expr}: money to the cent; the book works interest by aliquot parts carried to mills, so a cent either way is right too
      if ($ex -match '^m\s+(.+)$') {
        try { [void](EvalFrac $Matches[1]); $r = $script:lastRat; $x = [decimal]$r[0] / [decimal]$r[1]
              $c = [Math]::Round($x * 100, [MidpointRounding]::AwayFromZero)
              $p.f = ($c / 100).ToString('0.00'); $p.falt = @((($c - 1) / 100).ToString('0.00'), (($c + 1) / 100).ToString('0.00')) } catch { [void]$script:errors.Add("$where : $_") }
        $ex = $null
      }
      elseif ($ex -match '^r(\d+)\s+(.+)$') {
        $rp = [int]$Matches[1]; $rex = $Matches[2]
        try { [void](EvalFrac $rex); $r = $script:lastRat; $sc = [decimal][Math]::Pow(10, $rp); $x = [decimal]$r[0] / [decimal]$r[1]
              $rnd = [Math]::Round($x * $sc, [MidpointRounding]::AwayFromZero) / $sc; $cut = [Math]::Truncate($x * $sc) / $sc
              $fmt = { param($v) $s = $v.ToString('0.' + ('0' * $rp)); if ($rp -eq 0) { $s = $v.ToString('0') }; if ($s.StartsWith('0.')) { $s.Substring(1) } else { $s } }
              $p.f = & $fmt $rnd; if ($cut -ne $rnd) { $p.falt = @(& $fmt $cut) } } catch { [void]$script:errors.Add("$where : $_") }
        $ex = $null
      }
      # {~p 320,16_1/2,12 # 2 # expr}: a compound number. expr is the whole quantity in the lowest unit (inches);
      # the chain gives how many of each unit make one of the unit above (mi.>rd.>ft.>in.); # 2 asks for the 3rd unit (ft.).
      # Every unit but the last is a whole number; the last takes the remainder (decimal when it ends, else a fraction).
      if ($null -ne $ex -and $ex -match '^p\s+([^#]*)#\s*(\d+)\s*#\s*(.+)$') {
        $chain = @($Matches[1].Trim() -split '\s*,\s*'); $idx = [int]$Matches[2]; $pex = $Matches[3].Trim()
        if ($pex -eq '$') { $pex = $script:lastPex; $chain = $script:lastChain } else { $script:lastPex = $pex; $script:lastChain = $chain }   # {~p # 1 # $}: same quantity, next unit
        try {
          [void](EvalFrac $pex); $rem = $script:lastRat
          for ($k = 0; $k -le $chain.Count; $k++) {
            $sz = $(if ($k -lt $chain.Count) { [void](EvalFrac (($chain[$k..($chain.Count - 1)]) -join ' * ')); $script:lastRat } else { ,@([long]1, [long]1) })
            if ($k -eq $chain.Count) { $part = $rem } else {
              $qq = [long][Math]::Floor(([decimal]$rem[0] * $sz[1]) / ([decimal]$rem[1] * $sz[0]))
              $part = Rat $qq 1; $rem = Rat ($rem[0] * $sz[1] - $qq * $sz[0] * $rem[1]) ($rem[1] * $sz[1]) }
            if ($k -eq $idx) { $ps = RatStr $part; if ($ps -match '^\d+$') { $p.a = [long]$ps } else { $p.f = $(if ($pex -match '\d\.\d') { AsDec $ps } else { $ps })
                # the book often gives the last unit "nearly": accept it rounded or cut to 0-4 places too
                $x = ([decimal]($part[0])) / ([decimal]($part[1])); $alts = @()
                foreach ($dp in 0..4) { $sc = [decimal][Math]::Pow(10, $dp); foreach ($v in @(([Math]::Round($x * $sc, [MidpointRounding]::AwayFromZero) / $sc), ([Math]::Truncate($x * $sc) / $sc))) { $vs = $v.ToString('0.####'); if ($vs.StartsWith('0.')) { $vs = $vs.Substring(1) }; if ($alts -notcontains $vs) { $alts += $vs } } }
                if ($pex -match '\d\.\d' -or $part[1] -gt 100) { $p.falt = $alts } }; break }
          }
        } catch { [void]$script:errors.Add("$where : $_") }
        $ex = $null
      }
      if ($null -ne $ex) {
      # {~q expr}: as a common fraction, with the circulate shown after the question (.45(3))
      $wantRep = $ex -match '^q\s'; $ex = $ex -replace '^q\s+', ''
      $cp = $(if ($ex -match '^c(\d+)\s') { [int]$Matches[1] } else { -1 }); $ex = $ex -replace '^c\d+\s+', ''
      $wantDec = $ex -match '^d\s' -or $ex -match '\d*\.\d'; $ex = $ex -replace '^d\s+', ''
      try { $v = EvalFrac $ex; if ($wantRep) { [void]$script:reps.Add((AsRep $script:lastRat)) }
            if ($cp -ge 0) { $c = AsCplx $script:lastRat $cp; if ($c -match '^\d+$') { $p.a = [long]$c } else { $p.f = $c } }
            elseif ($v -match '^\d+$') { $p.a = [long]$v } else { $p.f = $(if ($wantDec) { AsDec $v } else { $v }) } } catch { [void]$script:errors.Add("$where : $_") }
      }
    } elseif ($ans -match '^=(.+)$') {   # exact written form, e.g. {=1/2} for "reduce to lowest terms"; {=3/6|1/2} more than one
      $ex = @($Matches[1] -split '\|' | ForEach-Object { $_.Trim() }); $p.f = $ex[0]; $p.ex = $ex
    } elseif ($ans -eq '$') {         # algebra: typed on the algebra pad, checked by alg.js (same value, written as simply)
      $e = $algE; $chk = $null
      # {$~ (a+b)(c+d)}: multiplied out exactly, written the book's way
      $expand = { param($x) try { [RaysAlgCheck]::Expand($x) } catch { [void]$script:errors.Add("$where : $($_.Exception.InnerException.Message) in $x"); '0' } }
      if ($e -match '^~\s*(.+)$') { $e = & $expand $Matches[1] }
      else {
        if ($e -match '^(.+?)\s+@\s*(.+)$') { $e = $Matches[1].Trim(); $chk = $Matches[2].Trim() }
        # {$ x^2(x+a) | ~x^2(x+a)}: other ways that count too (~ = multiplied out)
        $alts = @($e -split '\s*\|\s*'); $e = $alts[0]
        if ($alts.Count -gt 1) { $p.ealt = @($alts[1..($alts.Count - 1)] | ForEach-Object { if ($_ -match '^~\s*(.+)$') { & $expand $Matches[1] } else { $_ } }) }
      }
      # @gcd: A; B  = the answer divides each evenly;  @lcm: A; B  = each divides the answer evenly
      if ($chk -match '^(gcd|lcm):\s*(.+)$') {
        $mode = $Matches[1]
        foreach ($qq in @($Matches[2] -split '\s*;\s*')) {
          try { [void][RaysAlgCheck]::Expand($(if ($mode -eq 'gcd') { "($qq)/($e)" } else { "($e)/($qq)" })) } catch { [void]$script:errors.Add("$where : $e is not a $(if ($mode -eq 'gcd') { 'divisor' } else { 'multiple' }) of $qq") }
        }
        $chk = $null
      }
      foreach ($x in @($p.ealt)) { if ($x) { $r = [RaysAlgCheck]::Same($x, $e); if ($r) { [void]$script:errors.Add("$where : alternative $r") } } }
      $p.e = $e; $p.es = AlgSay $e
      # pad letters: the answer's and the question's (so they don't give the answer away); none for a plain number
      $p.v = $(if ($e -match "[a-z$([char]0x221A)]") { AlgLetters ($e + ' ' + (@([regex]::Matches($seg, '`([^`]*)`') | ForEach-Object { $_.Groups[1].Value }) -join ' ')) } else { '' })
      if ($chk) {
        # @sq: radicand = the answer squared gives it (a square root)
        $r = $(if ($chk -match '^sq:\s*(.+)$') { [RaysAlgCheck]::Same("($e)^2", $Matches[1]) } elseif ($chk -match '^([a-z]):\s*(.+)$') { [RaysAlgCheck]::Holds("$($Matches[1])=$e", @($Matches[2] -split '\s*;\s*')) } else { [RaysAlgCheck]::Same($e, $chk) })
        if ($r) { [void]$script:errors.Add("$where : $r") }
      } else { $r = [RaysAlgCheck]::Same($e, $e); if ($r) { [void]$script:errors.Add("$where : can't read answer: $r") } }
    } elseif ($ans -eq 'open') {      # answered out loud to a grown-up (e.g. "Give two numbers which make 10")
      $p.open = $true
    } elseif ($ans -match '^(\d+ )?\d+/\d+$' -or $ans -match '^\d*\.\d+$' -or $ans -match '^\d*\.\d+ \d+/\d+$') {   # fraction, mixed number, decimal, or complex decimal (.0154 1/4)
      $p.f = $ans
    } elseif ($ans -eq '?') {
      $v = Calc $seg
      if ($null -eq $v) { [void]$script:errors.Add("$where : can't work out {?} for '$seg'") } else { $p.a = $v }
    } elseif ($ans -match '^\d+$') {
      $p.a = [long]$ans
      $v = Calc $seg
      if ($null -ne $v -and $v -ne $p.a) { [void]$script:errors.Add("$where : '$seg' is $v, not $ans") }
    } else { [void]$script:errors.Add("$where : bad answer {$ans}") }
    if ($colSay) { $p.s = ((Spoken $seg) + ' ' + $colSay).Trim() } elseif ((Spoken $seg) -ne $seg) { $p.s = Spoken $seg }
    [void]$parts.Add($p)
  }
  # "x = {A} or {B}": either order (the two roots of a quadratic); app.js swaps them when B comes first
  for ($k = 1; $k -lt $parts.Count; $k++) { if ($parts[$k].t -match '^or\b') { $parts[$k - 1].pr = 1 } }
  $tail = $q.Substring($pos).Trim()
  if ($tail) { if ($parts.Count) { $parts[$parts.Count - 1].t += ' ' + $tail } else { [void]$script:errors.Add("$where : question has no {answer}") } }
  $res = [ordered]@{ parts = $parts.ToArray() }
  if ($script:reps.Count) { $res.sol = 'As a circulate: ' + ($script:reps -join '; ') + '.' }
  return $res
}

# ---------- lessons ----------
$lessons = New-Object System.Collections.ArrayList
$L = $null; $lastQ = $null; $qn = 0
foreach ($line in $src) {
  if ($line -match '^\s*(#|$)' -or $line -match '^@book: ') { continue }
  if ($line -match '^== (\d+) \| ([^|]+) \| (.+)$') {
    $L = [ordered]@{ n = [int]$Matches[1]; part = $Matches[2].Trim(); title = $Matches[3].Trim(); kind = 'problems' }
    $L.review = $L.part -eq 'Review'
    [void]$lessons.Add($L); $lastQ = $null; $lastArt = $null; $qn = 0; continue
  }
  if (-not $L) { continue }
  $where = "Lesson $($L.n)"
  if ($line -match '^kind: (count|numbers) (\d+) (\d+)$') { $L.kind = $Matches[1]; $L.from = [int]$Matches[2]; $L.to = [int]$Matches[3] }
  elseif ($line -match '^kind: (read|write)$') { $L.kind = $Matches[1] }
  elseif ($line -match '^list: (.+)$') { $L.list = @($Matches[1] -split '\s+' | ForEach-Object { [int]$_ }) }
  elseif ($line -match '^note: (.+)$') { $L.note = $Matches[1] }
  elseif ($line -match '^table: (add|sub|mul|div) (\d+)$') { $L.op = $Matches[1]; $L.table = Table $Matches[1] ([int]$Matches[2]) }
  elseif ($line -match '^table: \+(add|sub|mul|div) (\d+) (\d+)( from 0)?$') {   # Book Two style "2 + 1 = 3" ... "2 + 12 = 14"; several add up
    $L.op = $Matches[1]; $op = $Matches[1]; $n = [int]$Matches[2]; $max = [int]$Matches[3]; $k0 = $(if ($Matches[4]) { 0 } else { 1 })   # Book Three: "2 + 0 = 2"
    $rows = New-Object System.Collections.ArrayList; if ($L.table) { [void]$rows.AddRange(@($L.table)) }
    foreach ($k in $k0..$max) {
      switch ($op) {
        'add' { $r = $n + $k; [void]$rows.Add([ordered]@{ t = "$n + $k = $r"; q = "How many are $n and $k?"; a = $n; b = $k; r = $r }) }
        'sub' { $m = $n + $k - 1; $r = $k - 1; [void]$rows.Add([ordered]@{ t = "$m - $n = $r"; q = "$n from $m leaves how many?"; a = $n; b = $m; r = $r }) }
        'mul' { $r = $n * $k; [void]$rows.Add([ordered]@{ t = "$k x $n = $r"; q = "How many are $k times $n?"; a = $k; b = $n; r = $r }) }   # the book's columns: 1 x 4, 2 x 4 ... 12 x 4
        'div' { $m = $n * $k; [void]$rows.Add([ordered]@{ t = "$m / $n = $k"; q = "How many times is $n contained in $m?"; a = $n; b = $m; r = $k }) }
      }
    }
    foreach ($f in $rows) { $f.t = Display $f.t }
    $L.table = $rows.ToArray()
  }
  elseif ($line -match '^rule: (\d+) (.+) make 1 (.+)$') {
    if (-not $L.rules) { $L.rules = New-Object System.Collections.ArrayList }
    $r = [ordered]@{ t = "$($Matches[1]) $($Matches[2]) make 1 $($Matches[3])"; q = "How many $($Matches[2]) make 1 $($Matches[3])?"; n = [int]$Matches[1]; a = $Matches[2]; b = $Matches[3] }
    if ($r.a -match '\d|^and ') { $r.quiz = $false }   # "365 days and 6 hours": show it, don't quiz it
    [void]$L.rules.Add($r)
  }
  elseif ($line -match '^q: (.+)$') {
    $qn++; if (-not $L.qs) { $L.qs = New-Object System.Collections.ArrayList }
    $lastQ = Question $Matches[1] "$where q$qn"; [void]$L.qs.Add($lastQ)
  }
  elseif ($line -match '^sol: (.+)$') { if ($lastQ) { $lastQ.sol = $Matches[1] } else { [void]$script:errors.Add("$where : sol without a question") } }
  # Book Five (Geometry): the book's articles to study, each with its figure cut from the scan, then a quiz
  elseif ($line -match '^art: (\d+) \| ([^|]*) \| (.+)$') {
    if (-not $L.arts) { $L.arts = New-Object System.Collections.ArrayList }
    $lastArt = [ordered]@{ n = [int]$Matches[1]; h = $Matches[2].Trim(); t = $Matches[3].Trim() }; [void]$L.arts.Add($lastArt)
    if ("$($lastArt.h)$($lastArt.t)".Contains('`')) { $lastArt.s = "$($lastArt.n). $(if ($lastArt.h) { "$(Spoken $lastArt.h). " })$(Spoken $lastArt.t)"; $lastArt.h = Display $lastArt.h; $lastArt.t = Display $lastArt.t }
  }
  elseif ($line -match '^chk: ([^|]+)\|(.+)$') {   # chk: x=5, y=4 | x+y=9 | x-y=1
    $r = [RaysAlgCheck]::Holds($Matches[1].Trim(), @($Matches[2] -split '\s*\|\s*' | ForEach-Object { $_.Trim() }))
    if ($r) { [void]$script:errors.Add("$where : $r") }
  }
  elseif ($line -match '^fig: (\d+) ([\d.]+) ([\d.]+)( [\d.]+ [\d.]+)?$') {   # page top bottom [left right], fractions of the page
    if (-not $lastArt) { [void]$script:errors.Add("$where : fig without an art"); continue }
    $fk = "p$($Matches[1])-" + [int]([double]$Matches[2] * 1000)
    $lastArt.figs = @($lastArt.figs | Where-Object { $_ }) + "fig/$($bk[0])/$fk.png"   # an article may have several
  }
  elseif ($line -match '^mc: (.+)$') {
    $parts = @($Matches[1] -split '\s*\|\s*'); $ch = @($parts[1..($parts.Count - 1)])
    $right = @(for ($i = 0; $i -lt $ch.Count; $i++) { if ($ch[$i] -like '[*]*') { $i } })
    if ($ch.Count -lt 2 -or $right.Count -ne 1) { [void]$script:errors.Add("$where : mc needs 2+ choices and one *right one: '$line'"); continue }
    if (-not $L.quiz) { $L.quiz = New-Object System.Collections.ArrayList }
    $cs = @($ch | ForEach-Object { $_ -replace '^\*', '' }); $item = [ordered]@{ q = $parts[0]; c = $cs; a = $right[0] }
    if ($line.Contains('`')) { $item.q = Display $parts[0]; $item.c = @($cs | ForEach-Object { Display $_ }); $item.qs = Spoken $parts[0]; $item.cs = @($cs | ForEach-Object { Spoken $_ }) }   # algebra: shown the book's way, and what the voice says
    [void]$L.quiz.Add($item)
  }
  elseif ($line -match '^tf: (.+) \{([TF])\}$') {
    if (-not $L.quiz) { $L.quiz = New-Object System.Collections.ArrayList }
    $tq = $Matches[1]; $item = [ordered]@{ q = $tq; c = @('True', 'False'); a = $(if ($Matches[2] -eq 'T') { 0 } else { 1 }); tf = $true }
    if ($tq.Contains('`')) { $item.q = Display $tq; $item.qs = Spoken $tq }
    [void]$L.quiz.Add($item)
  }
  else { [void]$script:errors.Add("$where : can't read line '$line'") }
}
foreach ($L in $lessons) {
  if ($L.rules) { $L.rules = $L.rules.ToArray() }
  if ($L.qs) { $L.qs = $L.qs.ToArray() }
  if ($L.arts) { $L.arts = $L.arts.ToArray() }
  if ($L.quiz) { $L.quiz = $L.quiz.ToArray() }
  if ($L.kind -eq 'problems' -and -not $L.qs -and -not $L.rules -and -not $L.table -and -not $L.arts) { [void]$script:errors.Add("Lesson $($L.n) : no questions") }
}
$nums = @($lessons | ForEach-Object { $_.n })
for ($i = 0; $i -lt $nums.Count; $i++) { if ($nums[$i] -ne $i + 1) { [void]$script:errors.Add("lesson numbers out of order at $($nums[$i])"); break } }

if ($script:errors.Count) { "PROBLEMS:"; $script:errors; exit 1 }
$out = [ordered]@{ id = $bk[0]; title = $bk[1]; sub = $bk[2]; color = $bk[3]; lessons = $lessons.ToArray() }
New-Item -ItemType Directory -Force "$root\public\books" | Out-Null
[IO.File]::WriteAllText("$root\public\books\$($bk[0]).json", ($out | ConvertTo-Json -Depth 10 -Compress), (New-Object Text.UTF8Encoding $false))
$q = ($lessons | ForEach-Object { @($_.qs).Count } | Measure-Object -Sum).Sum
$a = ($lessons | ForEach-Object { foreach ($x in @($_.qs)) { if ($x) { $x.parts.Count } } } | Measure-Object -Sum).Sum
"$($lessons.Count) lessons, $q questions, $a answers -> public\books\$($bk[0]).json"
EOF

make-audio.ps1: records every line of a book with Eloquence: -Id geometry.

cat > /root/rays-rebuild/make-audio.ps1 <<'EOF'
# Records everything Ray's Arithmetic says with the PC's ETI-Eloquence (Reed, the same voice as ABC Books) at each
# speed, then writes public/audio/<id>.json (what the app asks for -> clip id). WAVs land in tools/out/s<speed>/,
# shared by every book (a line two books both say is recorded once).
# Usage: powershell -ExecutionPolicy Bypass -File RaysArithmetic\tools\make-audio.ps1 [-Id primary1877|intellectual|...]
# The keys must match app.js: prompts, "lesson:N", "n:N", "find:N", "ans:N", and the book's own texts.
param([string]$Id = 'primary1877', [int[]]$Speeds = @(60, 75))
$root = Split-Path -Parent (Split-Path -Parent $MyInvocation.MyCommand.Path)
$book = Get-Content "$root\public\books\$Id.json" -Raw -Encoding UTF8 | ConvertFrom-Json
$P = Get-Content "$root\public\prompts.json" -Raw -Encoding UTF8 | ConvertFrom-Json
$kids = 'Layla', 'Lavon', 'Lilly'

$say = New-Object System.Collections.Specialized.OrderedDictionary ([StringComparer]::Ordinal)
function Add($key, $spoken) { if ($key -and -not $say.Contains([string]$key)) { $say[[string]$key] = $(if ($spoken) { $spoken } else { [string]$key }) } }
foreach ($pr in $P.PSObject.Properties) { if ($pr.Value -like '*{kid}*') { foreach ($k in $kids) { Add ($pr.Value -replace '\{kid\}', $k) } } else { Add $pr.Value } }
foreach ($k in $kids) { Add "Great job, $k!" }
foreach ($n in 0..100) { Add "n:$n" "$n."; Add "find:$n" "Find $n." }
# "2 1/2" -> "2 and 1 half", "3/4" -> "3 fourths": how the book says fractions
# any denominator as an ordinal word: 23 -> twenty-third, 115 -> one hundred and fifteenth
function OrdWord([int]$d) {
  $o = 'zero one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen' -split ' '
  $t = ',,twenty,thirty,forty,fifty,sixty,seventy,eighty,ninety' -split ','
  $oo = @{ one = 'first'; two = 'second'; three = 'third'; five = 'fifth'; eight = 'eighth'; nine = 'ninth'; twelve = 'twelfth' }
  function Card([int]$n) { if ($n -lt 20) { return $o[$n] }; $w = $t[[Math]::Floor($n / 10)]; if ($n % 10) { $w += '-' + $o[$n % 10] }; $w }
  $words = @()
  if ($d -ge 1000) { $words += (Card ([Math]::Floor($d / 1000))) + ' thousand'; $d = $d % 1000 }
  if ($d -ge 100) { $words += (Card ([Math]::Floor($d / 100))) + ' hundred'; $d = $d % 100 }
  if ($d) { if ($words.Count) { $words += 'and' }; $words += (Card $d) }
  $s = $words -join ' '
  if ($s -match '^(.*?)(\w+)$') { $head = $Matches[1]; $last = $Matches[2] }
  $last = $(if ($oo[$last]) { $oo[$last] } elseif ($last -match 'y$') { $last -replace 'y$', 'ieth' } else { $last + 'th' })
  $head + $last
}
function Frac([string]$f) {
  $den = @{ 2 = 'half'; 3 = 'third'; 4 = 'fourth'; 5 = 'fifth'; 6 = 'sixth'; 7 = 'seventh'; 8 = 'eighth'; 9 = 'ninth'; 10 = 'tenth'; 11 = 'eleventh'; 12 = 'twelfth'; 20 = 'twentieth'; 100 = 'hundredth' }
  if ($f -match '^\d*\.\d+$' -or $f -match '^\d+$') { return $f }   # decimals, and whole numbers too long for a number ({=digits})
  if ($f -match '^(\d*\.\d+) (\d+/\d+)$') { return "$($Matches[1]) and $(Frac $Matches[2])" }   # complex decimal: .0154 1/4
  $w = ''; if ($f -match '^(\d+) (.+)$') { $w = "$($Matches[1]) and "; $f = $Matches[2] }
  $n, $d = $f -split '/'; $name = $den[[int]$d]; if (-not $name) { $name = OrdWord ([int]$d) }
  if ($name -eq 'half' -and [int]$n -gt 1) { $name = 'halve' }
  return "$w$n $name$(if ([int]$n -gt 1) { 's' })"
}
function GeoSay([string]$s) {
  $s = $s -replace '\s*\(\d+(, ?\d+)*\)', ''
  [regex]::Replace($s, '\b[A-Z]{2,6}\b', { param($m) ($m.Value.ToCharArray() -join ' ') })
}
# must match artKey() in app.js
function ArtKey($a) { "$($a.n). $(if ($a.h) { "$($a.h). " })$($a.t)" }
$answers = New-Object 'System.Collections.Generic.SortedSet[long]'
foreach ($L in $book.lessons) {
  Add "lesson:$($L.n)" "Lesson $($L.n). $($L.title)."
  foreach ($v in @($L.list)) { if ($null -ne $v) { [void]$answers.Add([long]$v); if ([long]$v -gt 100) { Add "n:$v" "$v." } } }
  foreach ($f in @($L.table)) { if ($f) { Add $f.t; Add $f.q; [void]$answers.Add([long]$f.r) } }
  foreach ($r in @($L.rules)) { if ($r) { Add $r.t; Add $r.q } }
  foreach ($q in @($L.qs)) {
    if (-not $q) { continue }
    foreach ($p in $q.parts) {
      $s = if ($p.s) { $p.s } else { $p.t }; Add $s
      if ($null -ne $p.a) { [void]$answers.Add([long]$p.a) }
      if ($p.f) { Add "ans:$($p.f)" "The answer is $(Frac $p.f)." }
      if ($null -ne $p.e) { Add "ans:$($p.e)" "The answer is $($p.es)." }   # algebra: es = the answer in words
      foreach ($v in @($p.seq)) { if ($null -ne $v) { [void]$answers.Add([long]$v); if ($v -gt 100) { Add "n:$v" "$v." } } }
    }
    if ($q.sol) { Add $q.sol }
  }
  # Geometry: figure letters are said one by one ("AOE" -> "A O E"), article references "(103)" are not read
  # Algebra: arts and quiz with math carry what to say (s, qs, cs) since "3a²" can't be read off the screen text
  foreach ($a in @($L.arts)) { if ($a) { $k = ArtKey $a; Add $k $(if ($a.s) { $a.s } else { GeoSay $k }) } }
  foreach ($x in @($L.quiz)) {
    if (-not $x) { continue }
    $qs = $(if ($x.qs) { $x.qs } else { GeoSay $x.q })
    if ($x.tf) { Add $x.q "True or false? $qs" } else { Add $x.q $qs }
    for ($i = 0; $i -lt @($x.c).Count; $i++) { $c = @($x.c)[$i]; Add "ansc:$c" $(if ($x.cs) { "The answer is: $(@($x.cs)[$i])." } else { GeoSay "The answer is: $c." }) }
  }
}
foreach ($a in $answers) { Add "ans:$a" "The answer is $a." }
Add "Answer out loud, then tap Next."
"$($say.Count) clips per speed"

$sha = [Security.Cryptography.SHA1]::Create()
function Id($s) { (($sha.ComputeHash([Text.Encoding]::UTF8.GetBytes($s)) | Select-Object -First 6 | ForEach-Object { $_.ToString('x2') }) -join '') }
$items = New-Object System.Collections.Specialized.OrderedDictionary ([StringComparer]::Ordinal)
foreach ($k in $say.Keys) { $items[$k] = Id $say[$k] }
$ps32 = 'C:\Windows\SysWOW64\WindowsPowerShell\v1.0\powershell.exe'
foreach ($sp in $Speeds) {
  $dir = "$root\tools\out\s$sp"; New-Item -ItemType Directory -Force $dir | Out-Null
  $jobs = @(); $seen = @{}
  foreach ($k in $say.Keys) { $cid = $items[$k]; if (-not $seen[$cid] -and -not (Test-Path "$dir\$cid.wav")) { $seen[$cid] = 1; $jobs += [ordered]@{ text = $say[$k]; out = "$dir\$cid.wav" } } }
  "speed $sp : $($jobs.Count) to record"
  for ($i = 0; $i -lt $jobs.Count; $i += 400) {   # batches, so a long run shows progress
    $jf = "$dir\jobs.json"; ConvertTo-Json @($jobs[$i..([Math]::Min($i + 399, $jobs.Count - 1))]) | Set-Content -Encoding utf8 $jf
    & $ps32 -NoProfile -ExecutionPolicy Bypass -File "$root\tools\eloq-render.ps1" -Jobs $jf -Speed $sp | Out-Null
    "  $([Math]::Min($i + 400, $jobs.Count)) / $($jobs.Count)"
  }
  Remove-Item "$dir\*.wav.json" -ErrorAction SilentlyContinue   # word times aren't used here
}
$out = [ordered]@{ speeds = $Speeds; items = $items }
New-Item -ItemType Directory -Force "$root\public\audio" | Out-Null
[IO.File]::WriteAllText("$root\public\audio\$Id.json", ($out | ConvertTo-Json -Depth 4 -Compress), (New-Object Text.UTF8Encoding $false))
"manifest: $root\public\audio\$Id.json"
EOF

make-figs.ps1: cuts the figures out of the scan.

cat > /root/rays-rebuild/make-figs.ps1 <<'EOF'
# Cuts every figure a book's source names ("fig: <pdf page> <top> <bottom> [<left> <right>]", fractions of the page)
# out of the scan, trims the white edges, and saves public/fig/<id>/p<page>-<top*1000>.png (the name make-book.ps1 uses).
# Usage: powershell -ExecutionPolicy Bypass -File RaysArithmetic\tools\make-figs.ps1 [-Book geometry] [-Dpi 200] [-Force]
param([string]$Book = 'geometry', [int]$Dpi = 200, [switch]$Force)
$root = Split-Path -Parent (Split-Path -Parent $MyInvocation.MyCommand.Path)
$src = Get-Content "$root\source\$Book.txt" -Encoding UTF8
$pdf = @{ geometry = "$root\source\pdf\rays-geometry-and-trigonometry-9-10.pdf" }[$Book]
$out = "$root\public\fig\$Book"; New-Item -ItemType Directory -Force $out | Out-Null
$tmp = Join-Path $env:TEMP "figs-$Book"; New-Item -ItemType Directory -Force $tmp | Out-Null
Add-Type -AssemblyName System.Drawing
Add-Type -ReferencedAssemblies System.Drawing -TypeDefinition @'
using System; using System.Drawing; using System.Drawing.Imaging; using System.Runtime.InteropServices;
public static class FigTrim {
  public static int[] Ink(Bitmap b, int x0, int y0, int x1, int y1) {
    var d = b.LockBits(new Rectangle(0, 0, b.Width, b.Height), ImageLockMode.ReadOnly, PixelFormat.Format24bppRgb);
    var px = new byte[d.Stride * b.Height]; Marshal.Copy(d.Scan0, px, 0, px.Length); b.UnlockBits(d);
    int minX = x1, minY = y1, maxX = x0, maxY = y0;
    for (int y = y0; y < y1; y++) for (int x = x0; x < x1; x++)
      if (px[y * d.Stride + x * 3] < 170) { if (x < minX) minX = x; if (x > maxX) maxX = x; if (y < minY) minY = y; if (y > maxY) maxY = y; }
    return new[] { minX, minY, maxX, maxY };
  }
}
'@
$n = 0
foreach ($line in $src) {
  if ($line -notmatch '^fig: (\d+) ([\d.]+) ([\d.]+)(?: ([\d.]+) ([\d.]+))?$') { continue }
  $pg = [int]$Matches[1]; $t = [double]$Matches[2]; $b = [double]$Matches[3]
  $l = $(if ($Matches[4]) { [double]$Matches[4] } else { 0.05 }); $r = $(if ($Matches[5]) { [double]$Matches[5] } else { 0.95 })
  $file = "$out\p$pg-$([int]($t * 1000)).png"
  if ((Test-Path $file) -and -not $Force) { continue }
  $png = "$tmp\pg-$pg.png"
  if (-not (Test-Path $png)) { & pdftoppm -r $Dpi -gray -png -f $pg -l $pg -singlefile $pdf "$tmp\pg-$pg" }
  $img = [Drawing.Bitmap]::FromFile($png)
  $x0 = [int]($img.Width * $l); $x1 = [int]($img.Width * $r); $y0 = [int]($img.Height * $t); $y1 = [int]($img.Height * $b)
  # trim white edges (a pixel darker than 170 counts as ink)
  $bb = [FigTrim]::Ink($img, $x0, $y0, $x1, $y1); $minX = $bb[0]; $minY = $bb[1]; $maxX = $bb[2]; $maxY = $bb[3]
  if ($maxX -le $minX) { "EMPTY: $line"; $img.Dispose(); continue }
  $pad = 8; $minX = [Math]::Max(0, $minX - $pad); $minY = [Math]::Max(0, $minY - $pad); $maxX = [Math]::Min($img.Width - 1, $maxX + $pad); $maxY = [Math]::Min($img.Height - 1, $maxY + $pad)
  $w = $maxX - $minX; $h = $maxY - $minY
  $crop = New-Object Drawing.Bitmap $w, $h; $g = [Drawing.Graphics]::FromImage($crop); $g.Clear([Drawing.Color]::White)
  $g.DrawImage($img, (New-Object Drawing.Rectangle 0, 0, $w, $h), (New-Object Drawing.Rectangle $minX, $minY, $w, $h), [Drawing.GraphicsUnit]::Pixel)
  $crop.Save($file, [Drawing.Imaging.ImageFormat]::Png); $g.Dispose(); $crop.Dispose(); $img.Dispose(); $n++
}
"$n figures -> $out"
EOF

fig-sheet.ps1: page pictures with marks, for placing figure boxes.

cat > /root/rays-rebuild/fig-sheet.ps1 <<'EOF'
# Contact sheet for placing "fig:" boxes: PDF pages side by side, red ticks every 0.05 down the left edge and
# along the top (labelled in tenths), so a figure's top/bottom/left/right fractions can be read off.
# Usage: powershell -File fig-sheet.ps1 -Pages 177,178,179 [-Out sheet.png] [-Dpi 60] [-Cols 4]
param([Parameter(Mandatory)] [string[]]$Pages, [string]$Out = "$env:TEMP\fig-sheet.png", [int]$Dpi = 60, [int]$Cols = 4, [string]$Pdf)
$Pages = @($Pages -join ',' -split '[,\s]+' | Where-Object { $_ } | ForEach-Object { [int]$_ })   # "-File" passes "80,103" as one string
if (-not $Pdf) { $Pdf = "$(Split-Path -Parent $PSScriptRoot)\source\pdf\rays-geometry-and-trigonometry-9-10.pdf" }
$d = Join-Path $env:TEMP 'fig-sheet'; New-Item -ItemType Directory -Force $d | Out-Null
Add-Type -AssemblyName System.Drawing
$imgs = foreach ($p in $Pages) { $f = "$d\s$p-$Dpi"; if (-not (Test-Path "$f.png")) { & pdftoppm -r $Dpi -gray -png -f $p -l $p -singlefile $Pdf $f }; [Drawing.Image]::FromFile("$f.png") }
$w = $imgs[0].Width; $h = $imgs[0].Height; $rows = [Math]::Ceiling($imgs.Count / $Cols)
$s = New-Object Drawing.Bitmap ($w * $Cols), ($h * $rows); $g = [Drawing.Graphics]::FromImage($s); $g.Clear([Drawing.Color]::White)
$pen = New-Object Drawing.Pen ([Drawing.Color]::Red), 1; $font = New-Object Drawing.Font 'Arial', 7; $big = New-Object Drawing.Font 'Arial', 11, ([Drawing.FontStyle]::Bold)
for ($i = 0; $i -lt $imgs.Count; $i++) {
  $x = ($i % $Cols) * $w; $y = [Math]::Floor($i / $Cols) * $h; $g.DrawImage($imgs[$i], $x, $y, $w, $h)
  for ($k = 1; $k -lt 20; $k++) {
    $yy = $y + $h * $k / 20; $xx = $x + $w * $k / 20; $len = $(if ($k % 2) { 5 } else { 10 })
    $g.DrawLine($pen, $x, $yy, $x + $len, $yy); $g.DrawLine($pen, $xx, $y, $xx, $y + $len)
    if ($k % 2 -eq 0) { $g.DrawString("$($k/2)", $font, [Drawing.Brushes]::Red, $x + 10, $yy - 6); $g.DrawString("$($k/2)", $font, [Drawing.Brushes]::Red, $xx - 3, $y + 10) }
  }
  $g.DrawString("p$($Pages[$i])", $big, [Drawing.Brushes]::Blue, $x + $w - 45, $y + $h - 20)
  $g.DrawRectangle([Drawing.Pens]::LightGray, $x, $y, $w - 1, $h - 1)
}
$s.Save($Out); $imgs | ForEach-Object { $_.Dispose() }; $Out
EOF

fig-montage.ps1: all cut figures on one picture, for checking them.

cat > /root/rays-rebuild/fig-montage.ps1 <<'EOF'
# Shows cut figures in a grid with their names, to check that no text bled in and no letter was cut off.
# Usage: powershell -File fig-montage.ps1 [-FromPage 177] [-ToPage 212] [-Out montage.png]
param([int]$FromPage = 0, [int]$ToPage = 9999, [string]$Out = "$env:TEMP\fig-montage.png", [string]$Book = 'geometry', [int]$Cols = 5)
$root = Split-Path -Parent (Split-Path -Parent $MyInvocation.MyCommand.Path)
Add-Type -AssemblyName System.Drawing
$fs = Get-ChildItem "$root\public\fig\$Book\*.png" | Where-Object { $p = [int]($_.BaseName -replace '^p(\d+)-.*', '$1'); $p -ge $FromPage -and $p -le $ToPage } |
  Sort-Object { [int]($_.BaseName -replace '^p(\d+)-.*', '$1') }, { [int]($_.BaseName -replace '^p\d+-', '') }
if (-not $fs) { "no figures"; return }
$cw = 300; $ch = 200; $rows = [Math]::Ceiling($fs.Count / $Cols)
$s = New-Object Drawing.Bitmap ($cw * $Cols), ($ch * $rows); $g = [Drawing.Graphics]::FromImage($s); $g.Clear([Drawing.Color]::LightGray)
$font = New-Object Drawing.Font 'Arial', 10
for ($i = 0; $i -lt $fs.Count; $i++) {
  $im = [Drawing.Image]::FromFile($fs[$i].FullName); $sc = [Math]::Min(($cw - 10) / $im.Width, ($ch - 26) / $im.Height)
  $x = ($i % $Cols) * $cw + 5; $y = [Math]::Floor($i / $Cols) * $ch + 20
  $g.DrawImage($im, $x, $y, [int]($im.Width * $sc), [int]($im.Height * $sc)); $g.DrawString($fs[$i].BaseName, $font, [Drawing.Brushes]::Red, $x, $y - 18); $im.Dispose()
}
$s.Save($Out); "$($fs.Count) figures -> $Out"
EOF

app.js: the game, for the whole shelf (new: the Study and Quiz steps).

cat > /root/rays-rebuild/app.js <<'EOF'
// Ray's Arithmetic: gamified Ray's New Primary Arithmetic (1877). Progress is saved per kid in localStorage ('rays').
// Same look and voice as ABC Books; the kid picked in ABC Books (same site) is picked here too.
const Rays = (() => {
  const $ = id => document.getElementById(id);
  // Readers come from ABC Books on this site ('abcbooks'.readers: names a grown-up added there). Same rule as ABC
  // Books: no 'abcbooks' (or no kid and no progress) = no readers yet; an old device without .readers = the three
  // original kids plus any names in its progress.
  const NAME_OK = n => typeof n === 'string' && /^\p{L}[\p{L}' .-]{0,15}$/u.test(n.trim());
  function readers() {
    let c = null; try { c = JSON.parse(localStorage.getItem('abcbooks')); } catch (e) {}
    let r = [];
    if (c && Array.isArray(c.readers)) r = c.readers.filter(NAME_OK);
    else if (c && (c.kid || (c.progress && Object.keys(c.progress).length))) r = ['Layla', 'Lavon', 'Lilly'].concat(Object.keys(c.progress || {}).filter(n => NAME_OK(n) && !['Layla', 'Lavon', 'Lilly'].includes(n)));
    if (S && S.kid && r.length && !r.includes(S.kid)) r = r.concat(S.kid);
    return r;
  }
  // Ray's series in order (raysarithmetic.wordpress.com/rays-free-arithmetic). A book is playable once books/<id>.json exists.
  const SHELF = [
    { id: 'primary1877', title: "Ray's New Primary Arithmetic", sub: 'Book 1 · Grades 1-2', color: '#2f6f8f' },
    { id: 'intellectual', title: "Ray's New Intellectual Arithmetic", sub: 'Book 2 · Grades 3-4', color: '#6a4c93' },
    { id: 'practical', title: "Ray's New Practical Arithmetic", sub: 'Book 3 · Grades 5-6', color: '#2e7d4f' },
    { id: 'higher', title: "Ray's New Higher Arithmetic", sub: 'Book 4 · Grades 7-8', color: '#a0522d' },
    { id: 'geometry', title: "Ray's Geometry and Trigonometry", sub: 'Book 5 · Grades 9-10', color: '#7a6a1f' },
    { id: 'algebra1', title: "Ray's New Elementary Algebra", sub: 'Book 6 · Grades 9-10', color: '#8f2f4f' },
    { id: 'algebra2', title: "Ray's New Higher Algebra", sub: 'Book 7 · Grade 11', color: '#3b4a8f', soon: true }
  ];
  const BOOKS = {};
  let P = {};
  const STICKERS = ['🦊', '🐢', '🦉', '🐝', '🦋', '🐞', '🐸', '🐧', '🦄', '🐬', '🦁', '🐼', '🐨', '🐙', '🦀', '🐳', '🦒', '🦓', '🐘', '🦔', '🐿️', '🦜', '🐠', '🐤', '🐰', '🐻', '🐯', '🐮', '🐷', '🐵', '🌻', '🌈', '⭐', '🚂', '🚀', '⛵', '🎈', '🍎', '🍓', '🍩'];
  const STEPS = {
    count: { name: 'Count', icon: '●' }, numbers: { name: 'Numbers', icon: '#' }, find: { name: 'Find the number', icon: '👂' },
    say: { name: 'Read the numbers', icon: '💬' }, write: { name: 'Write the numbers', icon: '✎' }, trace: { name: 'Trace the numbers', icon: '1' },
    learn: { name: 'Learn the table', icon: '📖' }, rules: { name: 'Learn the table', icon: '📖' }, drill: { name: 'Practice', icon: '⚡' },
    problems: { name: 'Story problems', icon: '?' }, study: { name: 'Study', icon: '📐' }, quiz: { name: 'Quiz', icon: '✔' }
  };

  // ---------- storage ----------
  let S = load();
  function load() {
    let s = null; try { s = JSON.parse(localStorage.getItem('rays')); } catch (e) {}
    s = s || {};
    s.progress = s.progress || {}; s.pos = s.pos || {}; s.best = s.best || {}; s.stickers = s.stickers || {};
    if (!s.speed) s.speed = 60;
    if (!s.kid) {   // the kid picked in ABC Books or BlockCraft on this site
      for (const k of ['abcbooks', 'blockcraft_cfg']) { try { const c = JSON.parse(localStorage.getItem(k)); const n = c && (k === 'abcbooks' ? c.kid : c.name); if (NAME_OK(n)) { s.kid = n.trim(); break; } } catch (e) {} }
    }
    return s;
  }
  function save() { try { localStorage.setItem('rays', JSON.stringify(S)); } catch (e) {} }
  const kidMap = obj => (obj[S.kid] = obj[S.kid] || {});
  const prog = b => (kidMap(S.progress)[b] = kidMap(S.progress)[b] || {});

  // ---------- sound + voice (recorded Eloquence clips, else the device voice) ----------
  let AC;
  function beep(good) {
    try {
      AC = AC || new (window.AudioContext || window.webkitAudioContext)();
      const o = AC.createOscillator(), g = AC.createGain(), t = AC.currentTime;
      o.type = 'triangle'; o.connect(g); g.connect(AC.destination);
      const notes = good ? [523, 659, 784] : [220, 180];
      notes.forEach((f, i) => o.frequency.setValueAtTime(f, t + i * 0.1));
      g.gain.setValueAtTime(0.25, t); g.gain.exponentialRampToValueAtTime(0.001, t + notes.length * 0.1 + 0.2);
      o.start(t); o.stop(t + notes.length * 0.1 + 0.25);
    } catch (e) {}
  }
  const hasTTS = 'speechSynthesis' in window;
  let voice = null;
  function pickVoice() { if (!hasTTS) return; const v = speechSynthesis.getVoices().filter(v => /^en/i.test(v.lang)); voice = v.find(v => /female|samantha|zira|aria|jenny|google us/i.test(v.name)) || v[0] || null; }
  if (hasTTS) { pickVoice(); speechSynthesis.onvoiceschanged = pickVoice; }
  let AUDIO = null;   // audio/primary1877.json: spoken text -> clip id (tools/make-audio.ps1)
  const clip = new Audio(); clip.preload = 'auto';
  let clipDone = null, tok = 0, unlocked = false;
  const blobs = new Map();
  function clipURL(url) {   // fetched whole and played from memory: works offline and on Safari
    if (!blobs.has(url)) blobs.set(url, fetch(url).then(r => { if (!r.ok) throw new Error(r.status); return r.blob(); }).then(b => URL.createObjectURL(b)));
    const p = blobs.get(url); p.catch(() => blobs.delete(url)); return p;
  }
  const speedOf = () => AUDIO.speeds.includes(S.speed) ? S.speed : AUDIO.speeds[0];
  const urlFor = key => { const c = AUDIO && AUDIO.items[key]; return c ? `audio/s${speedOf()}/${c}.mp3` : null; };
  function stopAll() { tok++; clip.pause(); if (clipDone) { const d = clipDone; clipDone = null; d(); } if (hasTTS) speechSynthesis.cancel(); }
  function playURL(u, my) {
    return new Promise(res => {
      clipURL(u).then(b => {
        if (my !== tok) return res();
        clipDone = res; clip.muted = false; clip.src = b;
        clip.onended = clip.onerror = () => { if (clipDone) { const d = clipDone; clipDone = null; d(); } };
        clip.play().catch(() => clip.onended());
      }).catch(() => res());
    });
  }
  // speak(...keys): one clip per key, played one after the other; a key with no clip uses the device voice
  async function speak(...keys) {
    stopAll(); const my = tok;
    for (const k of keys) {
      if (my !== tok || k == null || k === '') continue;
      const u = urlFor(String(k));
      if (u) await playURL(u, my); else await tts(String(k).replace(/^(n|find|ans|lesson):/, ''));
    }
  }
  function tts(text) {
    return new Promise(res => {
      if (!hasTTS) return setTimeout(res, 300);
      const u = new SpeechSynthesisUtterance(text); if (voice) u.voice = voice; u.rate = 0.85;
      u.onend = u.onerror = res; speechSynthesis.speak(u);
    });
  }
  const sayNum = n => speak('n:' + n);
  window.addEventListener('pointerdown', () => {   // iOS: unlock audio inside the first tap
    if (AC && AC.state === 'suspended') AC.resume();
    const u = urlFor(P.yes); if (unlocked || !u) return; unlocked = true;
    clipURL(u).then(b => { if (!clip.src) { clip.muted = true; clip.src = b; clip.play().then(() => { clip.pause(); clip.muted = false; }).catch(() => { clip.muted = false; }); } }).catch(() => {});
  }, { passive: true });

  // ---------- helpers ----------
  function toast(t, ms) { const e = $('toast'); e.textContent = t; e.style.display = 'block'; clearTimeout(toast.t); toast.t = setTimeout(() => e.style.display = 'none', ms || 1100); }
  function el(tag, cls, html) { const e = document.createElement(tag); if (cls) e.className = cls; if (html != null) e.innerHTML = html; return e; }
  function btn(label, fn, cls) { const b = el('button', cls, label); b.onclick = fn; return b; }
  const shuffle = a => { a = a.slice(); for (let i = a.length - 1; i > 0; i--) { const j = Math.floor(Math.random() * (i + 1)); [a[i], a[j]] = [a[j], a[i]]; } return a; };
  const range = (a, b) => Array.from({ length: b - a + 1 }, (_, i) => a + i);
  const starStr = n => '★'.repeat(n) + '☆'.repeat(3 - n);
  const starsFor = (miss, n) => miss === 0 ? 3 : miss <= Math.max(2, Math.round(n * 0.15)) ? 2 : 1;
  const wait = ms => new Promise(r => setTimeout(r, ms));
  function go(id) { document.querySelectorAll('.screen').forEach(s => s.classList.toggle('on', s.id === id)); if (id === 'home') home(); if (id === 'stickers') stickers(); }
  const ONES = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen'];
  const TENS = ['', '', 'twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety'];
  const words = n => n < 20 ? ONES[n] : n === 100 ? 'one hundred' : TENS[Math.floor(n / 10)] + (n % 10 ? '-' + ONES[n % 10] : '');

  // pictures: dots for a fact (a and b in two colors, crossed out when taken away, rows for times, groups for division)
  function dotsFor(op, a, b) {
    const box = el('div', 'dots');
    const grp = (n, cls, boxed) => { const g = el('div', 'grp' + (boxed ? ' box' : '')); for (let i = 0; i < n; i++) g.appendChild(el('i', 'dot ' + (cls || ''))); return g; };
    if (op === 'add') { box.append(grp(a), el('span', 'op', '+'), grp(b, 'b')); }
    else if (op === 'sub') { const g = el('div', 'grp'); for (let i = 0; i < b; i++) g.appendChild(el('i', 'dot ' + (i >= b - a ? 'x' : ''))); box.appendChild(g); }
    else if (op === 'mul') { if (a * b > 100) return box; for (let i = 0; i < a; i++) box.appendChild(grp(b, i % 2 ? 'b' : '', true)); }
    else if (op === 'div') { if (b > 100) return box; for (let i = 0; i < b / a; i++) box.appendChild(grp(a, i % 2 ? 'b' : '', true)); }
    return box;
  }
  const SIGN = { add: '+', sub: '−', mul: '×', div: '÷' };
  const expr = (op, f) => op === 'sub' ? `${f.b} − ${f.a}` : op === 'div' ? `${f.b} ÷ ${f.a}` : `${f.a} ${SIGN[op]} ${f.b}`;
  // a picture for a story problem, from the first thing it counts
  const PICS = [['cent|dime|money', '💰'], ['dollar', '💵'], ['apple', '🍎'], ['pear', '🍐'], ['peach', '🍑'], ['orange', '🍊'], ['lemon', '🍋'], ['cherr', '🍒'], ['plum|raisin|fig', '🍇'], ['melon', '🍈'], ['pine-apple', '🍍'], ['strawberr|berr', '🍓'], ['nut|chestnut|pecan|walnut', '🌰'],
    ['marble', '🔵'], ['top', '🎯'], ['kite', '🎏'], ['ball', '⚽'], ['doll', '🧸'], ['bird|swallow|crow|pigeon', '🐦'], ['owl', '🦉'], ['bat', '🦇'], ['duck', '🦆'], ['geese|goose', '🦢'], ['chicken|hen', '🐔'], ['fish', '🐟'], ['horse', '🐴'], ['cow|calf|calves', '🐄'], ['sheep', '🐑'], ['pig|hog', '🐖'], ['rabbit', '🐇'], ['cat|kitten', '🐈'], ['mice|mouse', '🐭'],
    ['egg', '🥚'], ['cake|candy', '🍬'], ['milk', '🥛'], ['book|primer|reader|geograph', '📚'], ['pen|pencil|slate', '✏️'], ['ribbon|tape|cloth|calico|muslin|silk', '🎀'], ['tree|orchard', '🌳'], ['flower|rose', '🌹'], ['leaves|leaf', '🍃'], ['ship|boat', '⛵'], ['car', '🚃'], ['mile|travel', '🛤️'], ['hour|minute', '⏰'], ['day|week|month|year', '📅'], ['wood|cord', '🌲'], ['flour|wheat|corn|oats|seed', '🌾'],
    ['boy|girl|pupil|scholar|men|man|persons', '🧒'], ['window|pane', '🏠'], ['chair|bench', '💺'], ['foot|feet|inch|yard|rod', '📏'], ['pound|ounce|ton', '⚖️'], ['quart|pint|gallon|gill|peck|bushel', '🍶']];
  function picFor(text) { const t = text.toLowerCase(); let best = null, at = 1e9; for (const [rx, e] of PICS) { const m = t.match(new RegExp('\\b(' + rx + ')')); if (m && m.index < at) { at = m.index; best = e; } } return best || '🧮'; }

  // ---------- spoken answers ----------
  // The kid can say the answer. First choice: our own Vosk server (CT 307, /api/recognize, shared with ABC Books),
  // told to listen only for number words. Fallback: the browser's own recognizer. No mic at all: the button hides.
  const NUM_WORDS = 'zero oh one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty thirty forty fifty sixty seventy eighty ninety hundred thousand and a';
  const SR = window.SpeechRecognition || window.webkitSpeechRecognition;
  const GUM = !!(window.isSecureContext && navigator.mediaDevices && navigator.mediaDevices.getUserMedia);
  let micOK = GUM || !!SR, useServer = GUM;
  const W1 = { zero: 0, oh: 0, one: 1, two: 2, three: 3, four: 4, for: 4, five: 5, six: 6, seven: 7, eight: 8, ate: 8, nine: 9, ten: 10, eleven: 11, twelve: 12, thirteen: 13, fourteen: 14, fifteen: 15, sixteen: 16, seventeen: 17, eighteen: 18, nineteen: 19 };
  const W10 = { twenty: 20, thirty: 30, forty: 40, fifty: 50, sixty: 60, seventy: 70, eighty: 80, ninety: 90 };
  // "nine thousand six hundred" -> 9600; "four two" (digits one at a time) -> 42; "7" -> 7
  function wordsToNumber(words) {
    // "a hundred" = one hundred; a stray "a" (Vosk often hears it for "one"/"and" in fast speech) alone is one
    const w = words.join(' ').toLowerCase().replace(/-/g, ' ').split(/\s+/).filter(x => x && x !== 'and').map((x, i, all) => x === 'a' ? (all.length === 1 || /^(hundred|thousand)$/.test(all[i + 1] || '') ? 'one' : '') : x).filter(Boolean);
    if (!w.length) return null;
    if (w.every(x => /^\d+$/.test(x))) return parseInt(w.join(''), 10);
    if (w.length > 1 && w.every(x => x in W1 && W1[x] < 10)) return parseInt(w.map(x => W1[x]).join(''), 10);
    let total = 0, cur = 0, any = false;
    for (const x of w) {
      if (/^\d+$/.test(x)) { cur += parseInt(x, 10); any = true; }
      else if (x in W1 || x in W10) {   // "nineteen twenty" (said like a year) -> 1920; "twenty four" -> 24
        const v = x in W1 ? W1[x] : W10[x];
        cur = v >= 10 && cur % 100 ? cur * 100 + v : cur + v; any = true;
      }
      else if (x === 'hundred') { cur = (cur || 1) * 100; any = true; }
      else if (x === 'thousand') { total += (cur || 1) * 1000; cur = 0; any = true; }
      else return null;   // a word that isn't a number: didn't understand
    }
    return any ? total + cur : null;
  }
  let stopListening = () => {};
  async function hearNumber() {
    if (useServer) {
      try { const w = await listenServer(7000); return w ? wordsToNumber(w) : null; }
      catch (e) { useServer = false; if (!SR) { micOK = false; return null; } }
    }
    const alts = await listenBrowser(7000);
    for (const a of alts) { const n = wordsToNumber(a); if (n != null) return n; }
    return null;
  }
  // records until the kid stops talking (1.2 s quiet) or the time runs out (same as ABC Books)
  async function listenServer(timeoutMs) {
    let stream;
    try { stream = await navigator.mediaDevices.getUserMedia({ audio: { echoCancellation: true, noiseSuppression: true } }); }
    catch (e) { micOK = false; return null; }
    const ctx = new (window.AudioContext || window.webkitAudioContext)();
    try { await ctx.resume(); } catch (e) {}
    const src = ctx.createMediaStreamSource(stream), proc = ctx.createScriptProcessor(4096, 1, 1);
    const chunks = []; let spoke = false, quiet = 0; const t0 = performance.now();
    await new Promise(res => {
      let done = false; const fin = () => { if (!done) { done = true; res(); } };
      stopListening = fin;
      proc.onaudioprocess = e => {
        const d = e.inputBuffer.getChannelData(0); chunks.push(new Float32Array(d));
        let s = 0; for (let i = 0; i < d.length; i++) s += d[i] * d[i];
        if (Math.sqrt(s / d.length) > 0.015) { spoke = true; quiet = 0; } else if (spoke) quiet += d.length / ctx.sampleRate * 1000;
        if ((spoke && quiet > 1200) || performance.now() - t0 > timeoutMs) fin();
      };
      src.connect(proc); proc.connect(ctx.destination);
    });
    stopListening = () => {};
    proc.disconnect(); src.disconnect(); stream.getTracks().forEach(t => t.stop());
    const rate = ctx.sampleRate; ctx.close();
    if (!spoke) return null;
    const r = await fetch('/api/recognize?near=0&words=' + encodeURIComponent(NUM_WORDS), { method: 'POST', headers: { 'Content-Type': 'application/octet-stream' }, body: to16k(chunks, rate) });
    if (!r.ok) throw new Error('speech ' + r.status);
    const j = await r.json();
    return j.words || [];
  }
  function to16k(chunks, rate) {   // Float32 at the device rate -> 16 kHz 16-bit PCM (what Vosk wants)
    const n = chunks.reduce((a, c) => a + c.length, 0), all = new Float32Array(n); let o = 0;
    for (const c of chunks) { all.set(c, o); o += c.length; }
    const ratio = rate / 16000, m = Math.floor(n / ratio), out = new Int16Array(m);
    for (let i = 0; i < m; i++) {
      const a = Math.floor(i * ratio), b = Math.min(n, Math.max(a + 1, Math.floor((i + 1) * ratio))); let s = 0;
      for (let k = a; k < b; k++) s += all[k];
      out[i] = Math.max(-1, Math.min(1, s / (b - a))) * 32767;
    }
    return out.buffer;
  }
  function listenBrowser(timeoutMs) {
    return new Promise(res => {
      if (!SR) return res([]);
      const r = new SR(); r.lang = 'en-US'; r.interimResults = false; r.maxAlternatives = 5;
      let best = [], done = false;
      const fin = () => { if (done) return; done = true; try { r.stop(); } catch (e) {} res(best); };
      stopListening = fin;
      r.onresult = e => { best = [...e.results[0]].map(a => a.transcript.split(/\s+/)); fin(); };
      r.onerror = e => { if (/not-allowed|service-not-allowed/.test(e.error || '')) micOK = false; fin(); };
      r.onend = fin;
      try { r.start(); } catch (e) { micOK = false; return fin(); }
      setTimeout(fin, timeoutMs);
    });
  }

  // number pad (big buttons; a computer keyboard works too; 🎤 to say the answer)
  function numPad(onOK) {
    const wrap = el('div', 'col'); wrap.style.height = 'auto'; wrap.style.gap = '1vmin';
    const entry = el('div', 'entry', '&nbsp;'); let val = '';
    // long whole numbers are shown with commas, the book's way (4,576,791)
    const show = () => { entry.textContent = (algMode ? RaysAlg.pretty(val) : /^\d{5,}$/.test(val) ? val.replace(/\B(?=(\d{3})+$)/g, ',') : val) || ' '; entry.classList.toggle('long', val.length > 9); entry.classList.remove('no', 'ok'); };
    const pad = el('div', 'pad');
    let fracMode = false;   // answers like 3/4 or 2 1/2: extra keys, and OK hands back the text
    let algMode = false;    // Algebra: + − ( ) / power and the question's letters; OK hands back the text (alg.js checks it)
    const press = k => {
      if (k === '⌫') val = val.slice(0, -1);
      else if (k === 'OK') { if (val.trim()) onOK(fracMode || algMode ? val.trim() : parseInt(val, 10)); return; }
      else if (algMode) { if (val.length < 200 && !(k === '^' && !/[\da-z)]$/.test(val))) val += k === '−' ? '-' : k; }
      // "and" after a decimal makes a complex decimal (.0154 1/4); the fraction after it may then have a /
      else if (k === 'and') { if (fracMode && val && !/[ /]/.test(val) && !val.endsWith('.')) val += ' '; }
      else if (k === '/') { if (fracMode && val && !val.endsWith(' ') && !val.includes('/') && (!val.includes('.') || /\.\d+ \d+$/.test(val))) val += '/'; }
      else if (k === '.') { if (fracMode && !/[ /.]/.test(val)) val += '.'; }
      else if (val.length < 40) val += k;   // the book's long numbers (up to 36 figures) are checked as written
      show();
    };
    ['1', '2', '3', '4', '5', '6', '7', '8', '9', '⌫', '0', 'OK'].forEach(k => pad.appendChild(btn(k, () => press(k), k === 'OK' ? 'okb' : '')));
    const fracRow = el('div', 'pad'); fracRow.style.display = 'none';
    fracRow.append(btn('/', () => press('/')), btn('and', () => press('and')), btn('.', () => press('.')));
    const algRow = el('div', 'pad alg'), letRow = el('div', 'pad alg'); algRow.style.display = letRow.style.display = 'none';
    ['+', '−', '(', ')', '/', '^'].forEach(k => algRow.appendChild(btn(k === '^' ? 'xⁿ' : k, () => press(k))));
    const onKey = e => { if (algMode && /^[a-z+\-()/^*]$/.test(e.key)) press(e.key); else if (/^[0-9]$/.test(e.key)) press(e.key); else if (e.key === '/' || e.key === '.') press(e.key); else if (e.key === ' ' && fracMode) { e.preventDefault(); press('and'); } else if (e.key === 'Backspace') press('⌫'); else if (e.key === 'Enter') press('OK'); };
    window.addEventListener('keydown', onKey);
    const mic = btn('🎤 Say it', async () => {
      if (mic.classList.contains('live')) return stopListening();
      stopAll(); mic.classList.add('live'); mic.textContent = '🎤 Listening…';
      const n = await hearNumber();
      mic.classList.remove('live'); mic.textContent = '🎤 Say it';
      if (!micOK) { mic.style.display = 'none'; return toast('This device cannot listen. Type the answer.', 2000); }
      if (n == null) return toast('I did not catch that. Say it again!', 1500);
      val = String(n); show(); onOK(n);
    }, 'micb');
    if (!micOK) mic.style.display = 'none';
    wrap.append(entry, pad, fracRow, algRow, letRow, mic);
    return { el: wrap, entry, frac(on) { fracMode = !!on; fracRow.style.display = on ? '' : 'none'; mic.style.display = on || !micOK ? 'none' : ''; },
      alg(keys) {   // keys = the question's letters ("abx"); null turns the algebra keys off
        algMode = keys != null; wrap.classList.toggle('algmode', algMode); algRow.style.display = letRow.style.display = algMode ? '' : 'none'; letRow.innerHTML = '';
        [...(keys || '')].forEach(c => letRow.appendChild(btn(c, () => press(c), 'let')));
        letRow.style.display = algMode && keys ? '' : 'none';
        if (algMode) { fracMode = false; fracRow.style.display = 'none'; mic.style.display = 'none'; }
      }, show(on) { wrap.style.display = on ? '' : 'none'; }, clear() { val = ''; show(); }, set(v, cls) { val = String(v); show(); if (cls) entry.classList.add(cls); }, bad() { entry.classList.remove('no'); void entry.offsetWidth; entry.classList.add('no'); }, stop() { stopListening(); window.removeEventListener('keydown', onKey); } };
  }
  // fractions compare by value: 5/2 = 2 1/2 = 2.5
  function fracVal(s) {
    s = String(s).trim(); let m;
    if ((m = s.match(/^(\d+) (\d+)\/(\d+)$/))) return +m[1] + m[2] / m[3];
    if ((m = s.match(/^(\d+)\/(\d+)$/))) return m[1] / m[2];
    if ((m = s.match(/^(\d*)\.(\d+) (\d+)\/(\d+)$/))) return parseFloat((m[1] || '0') + '.' + m[2]) + m[3] / m[4] / Math.pow(10, m[2].length);   // complex decimal
    return /^\d*\.?\d+$/.test(s) ? parseFloat(s) : NaN;
  }
  // relative check, so tiny decimals (.00000002) and large ones (2900.000029) both compare exactly
  const sameVal = (a, b) => { const x = fracVal(a), y = fracVal(b); return Math.abs(x - y) <= 1e-13 + 1e-10 * Math.abs(y); };
  // counters tray: tap beads to count with
  function initTray() {
    const f = $('frame'); f.innerHTML = '';
    for (let i = 0; i < 30; i++) f.appendChild(btn('', e => { e.currentTarget.classList.toggle('on'); count(); }));
    const count = () => { $('trayN').textContent = f.querySelectorAll('.on').length; };
    $('trayClear').onclick = () => { f.querySelectorAll('.on').forEach(b => b.classList.remove('on')); count(); };
    $('trayClose').onclick = () => $('tray').classList.remove('on');
  }

  // "Show your work": a pencil button that opens scratch paper beside the question (scratch.js). The paper comes
  // back open on the next question and the next lesson if the kid left it open.
  function workPad() {
    const w = RaysWork.Pad($('stage'), S.kid, bookId, toast);
    const b = btn('', () => { label(w.toggle()); });
    const label = on => { b.textContent = on ? '✏️ Put paper away' : '✏️ Show work'; b.classList.toggle('on', on); };
    label(w.start());
    return { btn: b, page: w.page, result: w.result, stop: w.stop };
  }
  // grown-ups: every page of work one kid has in one book, newest first, grouped by lesson
  // With a family account, the pages saved there from other devices are added (fetched after the first draw).
  let workBack = 'map';
  const gone = new Set();   // deleted in this visit: the account may not have heard yet
  function showWork(from, fromAcct) {
    workBack = from; const id = bookId || 'primary1877', B = BOOKS[id], kid = S.kid;
    const pages = RaysWork.list(kid, id), box = $('workList'); box.innerHTML = '';
    if (fromAcct) {
      const at = new Map(pages.map((pg, i) => [pg.key, i]));
      fromAcct.filter(pg => !gone.has(id + '|' + pg.key) && !gone.has(id + '|*')).forEach(pg => {
        if (!at.has(pg.key)) pages.push(pg); else if ((pg.t || 0) > (pages[at.get(pg.key)].t || 0)) pages[at.get(pg.key)] = pg;
      });
    } else RaysWork.remote(kid, id).then(r => { if (r.length && $('work').classList.contains('on') && S.kid === kid && (bookId || 'primary1877') === id) showWork(from, r); });
    // lessons with the newest work first; inside a lesson, the pages in question order
    const newest = {}; pages.forEach(pg => { newest[pg.l] = Math.max(newest[pg.l] || 0, pg.t || 0); });
    pages.sort((a, b) => newest[b.l] - newest[a.l] || a.key.localeCompare(b.key, undefined, { numeric: true }));
    $('workTitle').textContent = `${S.kid}: work in ${B ? B.title : ''} (${pages.length} page${pages.length === 1 ? '' : 's'})`;
    $('workDel').style.display = pages.length ? '' : 'none';
    $('workDel').onclick = () => { if (confirm(`Delete all of ${S.kid}'s work in this book?`)) { RaysWork.remove(S.kid, id); gone.add(id + '|*'); showWork(from); } };
    if (!pages.length) box.appendChild(el('p', 'note', 'No work saved yet. In Story problems and Practice, the ✏️ Show work button opens scratch paper; each page is kept here with its question.'));
    const tw = Math.min(300, Math.floor((window.innerWidth - 60) / 2));
    let shown = 0, lastL = null;
    function more() {
      pages.slice(shown, shown + 40).forEach(pg => {
        if (pg.l !== lastL) { lastL = pg.l; const Ls = B && B.lessons.find(x => x.n === pg.l); box.appendChild(el('h3', 'wles', '')).textContent = `Lesson ${pg.l}${Ls ? ': ' + Ls.title : ''}`; }
        const card = el('button', 'wpage'); card.appendChild(RaysWork.picture(pg, tw));
        const cap = el('div', 'wcap'); cap.textContent = pg.q;
        const res = el('div', 'wres' + (pg.r === 'miss' ? ' miss' : pg.r === 'ok' ? ' ok' : ''));
        res.textContent = (pg.a != null ? `Answer: ${pg.a} ` + (pg.r === 'ok' ? '✓' : pg.r === 'miss' ? '(missed a try)' : '') : 'Not answered') + ' · ' + new Date(pg.t).toLocaleString([], { month: 'short', day: 'numeric', hour: 'numeric', minute: '2-digit' });
        card.append(cap, res); card.onclick = () => bigPage(pg, id, from); box.appendChild(card);
      });
      shown += 40; const m = box.querySelector('.wmore'); if (m) m.remove();
      if (shown < pages.length) box.appendChild(btn('Show more', more, 'wmore'));
    }
    more(); go('work');
  }
  function bigPage(pg, id, from) {
    const v = $('workBig'); v.innerHTML = ''; v.classList.add('on');
    const q = el('div', 'wcap'); q.textContent = `Lesson ${pg.l}: ${pg.q}`;
    const w = Math.min(window.innerWidth - 40, 1000), h = window.innerHeight - 190;
    v.append(q, RaysWork.picture(pg, Math.round(Math.min(w, h / (pg.h || 0.75)))),
      el('div', 'row'));
    v.lastChild.append(btn('Close', () => v.classList.remove('on'), 'big'),
      btn('Delete this page', () => { if (confirm('Delete this page?')) { RaysWork.remove(S.kid, id, pg.key); gone.add(id + '|' + pg.key); v.classList.remove('on'); showWork(from); } }, 'ghost parent'));
  }
  $('mapWork').onclick = () => showWork('map');
  $('stkWork').onclick = () => { if (S.kid) showWork('stickers'); };
  $('workBack').onclick =() => { $('workBig').classList.remove('on'); go(workBack); };

  // ---------- home, map, stickers ----------
  let addingReader = false;
  function home() {
    const k = $('kids'); k.innerHTML = '';
    // names go in as text (never innerHTML); a name with no recorded greeting gets the line without a name
    const rs = readers();
    const signed = window.ABCAccount && window.ABCAccount.signedIn;
    if (!rs.length && S.kid) { S.kid = null; save(); }
    $('whoHint').textContent = rs.length ? '' : 'Grown-ups: add each learner (a first name or nickname).';
    rs.forEach(n => { const b = btn('', () => { S.kid = n; save(); home(); const g = P.greet.replace('{kid}', n); speak(urlFor(g) ? g : P.hello); }, S.kid === n ? 'sel' : ''); b.textContent = n; k.appendChild(b); });
    if (signed) {   // a family account: names are added and removed on the account page
      if (!rs.length) $('whoHint').innerHTML = 'Grown-ups: add your learners on the <a href="/account/">account page</a>.';
      else { const a = el('a', 'parent'); a.href = '/account/'; a.textContent = 'Add or remove learners'; k.appendChild(a); }
    } else if (addingReader || !rs.length) {
      // the names live in 'abcbooks'.readers, shared with ABC Books, so a learner added here is in both
      const inp = el('input'); inp.maxLength = 16; inp.placeholder = 'Name'; inp.autocomplete = 'off';
      const add = () => {
        const n = inp.value.trim().replace(/\s+/g, ' ');
        if (!NAME_OK(n)) return toast('Letters only, please');
        if (rs.some(x => x.toLowerCase() === n.toLowerCase())) return toast(`${n} is already here`);
        let c = null; try { c = JSON.parse(localStorage.getItem('abcbooks')); } catch (e) {}
        c = c || { kid: null, progress: {}, boards: {} };
        c.readers = rs.concat(n); if (!c.kid) c.kid = n;
        try { localStorage.setItem('abcbooks', JSON.stringify(c)); } catch (e) {}
        S.kid = n; addingReader = false; save(); home();
      };
      inp.onkeydown = e => { if (e.key === 'Enter') add(); };
      k.append(inp, btn('Add', add, 'big'));
      if (rs.length) k.appendChild(btn('Cancel', () => { addingReader = false; home(); }, 'ghost parent'));
      setTimeout(() => inp.focus(), 50);
    } else if (rs.length < 8) k.appendChild(btn('＋', () => { addingReader = true; home(); }, 'add'));
    const sb = $('speedBtn'); sb.textContent = 'Voice: ' + (S.speed === 75 ? 'faster' : 'normal');
    sb.onclick = () => { S.speed = S.speed === 75 ? 60 : 75; save(); home(); speak(P.yes); };
    const sh = $('shelf'); sh.innerHTML = '';
    // the kid's current book glows: the first open book they haven't finished
    const doneIn = id => S.kid && BOOKS[id] ? Object.values(prog(id)).filter(x => x && x.stars).length : 0;
    const cur = S.kid && SHELF.find(b => BOOKS[b.id] && doneIn(b.id) < BOOKS[b.id].lessons.length);
    SHELF.forEach(b => {
      const B = BOOKS[b.id], n = B ? B.lessons.length : 0, d = doneIn(b.id);
      const pr = !B ? 'Coming soon' : S.kid ? (d >= n ? '★ Finished!' : d ? `Lesson ${d + 1} of ${n}` : `Lessons: ${n}`) : '';
      const e = btn(`<b>${b.title}</b><span class="bp">${pr}</span><small>${b.sub}</small>`, () => {
        if (!B) return toast('Coming soon!');
        if (!S.kid) return toast('Pick your name first!');
        openBook(b.id);
      }, 'book' + (!B ? ' soon' : '') + (cur && cur.id === b.id ? ' cur' : ''));
      e.style.background = b.color; sh.appendChild(e);
    });
  }
  let book, bookId;
  let skipArm = null;   // a locked tile tapped once: a 2nd tap within 4 s skips ahead to it
  function openBook(id) {
    bookId = id; book = BOOKS[id]; $('mapTitle').textContent = `${S.kid}: ${book.title}`;
    const t = $('tiles'); t.innerHTML = ''; const p = prog(id);
    // Lessons are open up to one past the furthest one finished, so after a skip the kid carries on from there.
    let far = -1; book.lessons.forEach((L, i) => { if ((p[L.n] || {}).stars) far = i; });
    const paidAt = i => window.ABCAccount && window.ABCAccount.locked('math', i);
    book.lessons.forEach((L, i) => {
      const st = (p[L.n] || {}).stars || 0, open = i <= far + 1, isNext = i === far + 1;
      // past the free lessons without a subscription (/account/sync.js; only when the paywall is switched on)
      const paid = paidAt(i);
      const tile = btn(`${L.n}<small>${L.part}</small><span class="st">${st ? starStr(st) : paid ? '🔒' : ''}</span>`, () => {
        if (paid) return window.ABCAccount.unlockScreen('math');
        if (!open && !st && !(skipArm && skipArm.i === i && Date.now() - skipArm.at < 4000)) {
          skipArm = { i, at: Date.now() };
          return toast(`Lesson ${L.n} is ahead. Tap it again to skip to it.`, 3500);
        }
        skipArm = null; startLesson(L);
      }, 'tile' + (isNext ? ' next' : '') + (!open && !st ? ' locked' : '') + (L.review ? ' rev' : ''));
      tile.title = `Lesson ${L.n}: ${L.title}` + (st ? ', done' : !open ? ', ahead' : '');
      tile.setAttribute('aria-label', tile.title); t.appendChild(tile);
    });
    // "Go to lesson": every lesson by title, grouped by part, starting on the next one; Go starts it.
    const sel = $('jumpSel'); sel.innerHTML = ''; let grp = null;
    book.lessons.forEach((L, i) => {
      if (!grp || grp.label !== L.part) { grp = document.createElement('optgroup'); grp.label = L.part || ''; sel.appendChild(grp); }
      const o = el('option'); o.textContent = `Lesson ${L.n}: ${L.title}` + ((p[L.n] || {}).stars ? ' ★' : '');
      o.value = i; if (i === Math.min(far + 1, book.lessons.length - 1)) o.selected = true; grp.appendChild(o);
    });
    $('jumpGo').onclick = () => { const i = +sel.value; paidAt(i) ? window.ABCAccount.unlockScreen('math') : startLesson(book.lessons[i]); };
    go('map'); speak(P.map);
    const nx = t.querySelector('.next'); if (nx) nx.scrollIntoView({ block: 'center' });
  }
  const stkKey = (b, n) => b === 'primary1877' ? n : `${b}:${n}`;   // Book One keeps its first keys
  function stickers() {   // the sticker book for the book the kid is in (Book One when none is open yet)
    const g = $('stkList'); g.innerHTML = ''; const got = kidMap(S.stickers), id = bookId || 'primary1877', B = BOOKS[id];
    const n = B ? B.lessons.filter(L => got[stkKey(id, L.n)]).length : 0;
    $('stkTitle').textContent = `${S.kid || ''} sticker book: ${B ? B.title : ''}, ${n} of ${B ? B.lessons.length : 0}`;
    (B ? B.lessons : []).forEach(L => { const v = got[stkKey(id, L.n)]; const s = el('span', v ? 'got' : '', v || ''); s.title = 'Lesson ' + L.n; g.appendChild(s); });
  }
  $('toStickers').onclick = () => { if (!S.kid) return toast('Pick your name first!'); go('stickers'); };

  // ---------- lesson flow ----------
  let L, steps, stepIdx, scores, cleanup = null, sayAgain = null, stepTok = 0;
  function stepsFor(L) {
    switch (L.kind) {
      case 'count': return ['count', 'find', 'trace'];
      case 'numbers': return ['numbers', 'find', 'trace'];
      case 'read': return ['say', 'find'];
      case 'write': return ['write', 'trace'];
    }
    const s = [];
    if (L.arts) s.push('study'); if (L.quiz) s.push('quiz');
    if (L.table) s.push('learn', 'drill');
    if (L.rules) s.push('rules'); if (L.rules && L.rules.some(r => r.quiz !== false) && !L.table) s.push('drill');
    if (L.qs) s.push('problems');
    return s;
  }
  function startLesson(lesson) {
    L = lesson; steps = stepsFor(L); scores = {}; stepIdx = 0;
    $('lesTitle').textContent = `Lesson ${L.n}: ${L.title}`;
    go('lesson'); runStep(true);
  }
  function leave() { stopStep(); openBook(bookId); }
  function stopStep() { stopAll(); $('tray').classList.remove('on'); if (cleanup) { cleanup(); cleanup = null; } }
  function runStep(first) {
    stopStep();
    const id = steps[stepIdx];
    const dots = $('steps'); dots.innerHTML = '';
    steps.forEach((x, i) => dots.appendChild(el('i', i < stepIdx ? 'done' : i === stepIdx ? 'cur' : '', STEPS[x].icon)));
    if (!id) return finish();
    $('stepTitle').textContent = STEPS[id].name; $('stage').innerHTML = ''; $('controls').innerHTML = ''; $('hint').textContent = P[id] || '';
    sayAgain = null; const my = ++stepTok;
    const intro = (first ? speak('lesson:' + L.n, P[id]) : speak(P[id]));
    $('sayBtn').onclick = () => sayAgain ? sayAgain() : speak(P[id]);
    ({ count: stepCount, numbers: stepNumbers, find: stepFind, say: stepSay, write: stepWrite, trace: stepTrace, learn: stepLearn, rules: stepRules, drill: stepDrill, problems: stepProblems, study: stepStudy, quiz: stepQuiz })[id](intro.then(() => my === stepTok));
  }
  function next(stars) { scores[steps[stepIdx]] = stars; stepIdx++; runStep(); }
  const afterIntro = (intro, fn) => intro.then(ok => { if (ok) fn(); });

  // Lesson 1: count the balls, one tap per ball
  function stepCount(intro) {
    let n = L.from;
    const st = $('stage'); const col = el('div', 'col'); st.appendChild(col);
    function show() {
      col.innerHTML = ''; let c = 0;
      const balls = el('div', 'balls'); const big = el('div', 'big-n', '&nbsp;'); const w = el('div', 'word', '&nbsp;');
      for (let i = 0; i < n; i++) balls.appendChild(btn('', e => {
        const b = e.currentTarget; if (b.classList.contains('on')) return;
        b.classList.add('on'); b.textContent = ++c; sayNum(c);
        if (c === n) { big.textContent = n; w.textContent = words(n); setTimeout(() => { beep(true); speak('n:' + n).then(() => wait(700)).then(() => { if (++n > L.to) next(3); else show(); }); }, 650); }
      }, 'ball'));
      col.append(balls, big, w);
      $('hint').textContent = `Tap each ball and count. (${n - L.from + 1} of ${L.to - L.from + 1})`;
    }
    sayAgain = () => speak(P.count);
    show();
  }
  // Lessons 2-4: each number with its tens and ones
  function stepNumbers(intro) {
    let n = L.from; const st = $('stage'); const col = el('div', 'col'); st.appendChild(col);
    let auto = false;
    function show() {
      col.innerHTML = '';
      col.append(el('div', 'big-n', n), el('div', 'word', words(n)));
      const rods = el('div', 'rods');
      for (let t = 0; t < Math.floor(n / 10); t++) { const r = el('div', 'rod'); for (let i = 0; i < 10; i++) r.appendChild(el('i')); rods.appendChild(r); }
      const ones = el('div', 'grp'); for (let i = 0; i < n % 10; i++) ones.appendChild(el('i', 'dot')); rods.appendChild(ones);
      col.appendChild(rods);
      col.appendChild(el('div', 'note', `${Math.floor(n / 10)} tens and ${n % 10} ones`));
      $('hint').textContent = `${n - L.from + 1} of ${L.to - L.from + 1}`;
      return sayNum(n);
    }
    async function playAll() { auto = true; playB.disabled = true; while (auto && n <= L.to) { await show(); await wait(700); if (!auto) break; if (n < L.to) n++; else break; } auto = false; playB.disabled = false; nextB.disabled = false; }
    const prevB = btn('◀', () => { auto = false; if (n > L.from) { n--; show(); } });
    const fwdB = btn('▶', () => { auto = false; if (n < L.to) { n++; show(); } else nextB.disabled = false; });
    const playB = btn('▶ Play all', playAll, 'big'), nextB = btn('Next step ▶', () => next(3));
    nextB.disabled = true; $('controls').append(prevB, playB, fwdB, nextB);
    sayAgain = show; cleanup = () => { auto = false; };
    col.append(el('div', 'big-n', n), el('div', 'word', words(n)));
    afterIntro(intro, playAll);
  }
  // hear a number, tap it
  function stepFind(intro) {
    const pool = L.list || range(L.from, L.to);
    const rounds = shuffle(pool).slice(0, 8); let r = 0, miss = 0, busy = false, target;
    const st = $('stage'); const col = el('div', 'col'); st.appendChild(col);
    const grid = el('div', 'choices'); col.appendChild(grid);
    function round() {
      if (r >= rounds.length) { toast('Great job! ⭐'); return setTimeout(() => next(starsFor(miss, rounds.length)), 900); }
      target = rounds[r++]; grid.innerHTML = '';
      const opts = shuffle([target, ...shuffle(pool.filter(x => x !== target)).slice(0, 3)]);
      opts.forEach(v => grid.appendChild(btn(v, e => {
        if (busy) return; const b = e.currentTarget;
        if (v === target) { busy = true; beep(true); b.classList.add('ok'); setTimeout(() => { busy = false; round(); }, 800); }
        else { miss++; beep(false); b.classList.remove('no'); void b.offsetWidth; b.classList.add('no'); }
      })));
      $('hint').textContent = `Find the number you hear (${r} of ${rounds.length})`;
      sayAgain = () => speak('find:' + target); sayAgain();
    }
    afterIntro(intro, round);
  }
  // Lesson 5: numbers to be read, out loud to a grown-up
  function stepSay(intro) {
    const g = el('div', 'grid'); $('stage').appendChild(g);
    L.list.forEach(v => g.appendChild(btn(v, e => { e.currentTarget.classList.add('seen'); sayNum(v); })));
    $('controls').appendChild(btn('Done ✓', () => next(3), 'big'));
  }
  // Lesson 6: numbers to be written: hear it, type it
  function stepWrite(intro) {
    const items = shuffle(L.list).slice(0, 12); let i = 0, miss = 0, tries = 0;
    const col = el('div', 'col'); $('stage').appendChild(col);
    const pad = numPad(v => {
      if (v === items[i]) { beep(true); pad.set(v, 'ok'); tries = 0; setTimeout(() => { i++; ask(); }, 700); }
      else if (++tries < 2) { miss++; beep(false); pad.bad(); speak(P.tryAgain, 'n:' + items[i]); }
      else { miss++; pad.set(items[i], 'ok'); speak('ans:' + items[i]); tries = 0; setTimeout(() => { i++; ask(); }, 1800); }
    });
    col.appendChild(pad.el);
    function ask() {
      if (i >= items.length) return next(starsFor(miss, items.length));
      pad.clear(); $('hint').textContent = `Write the number you hear (${i + 1} of ${items.length})`;
      sayAgain = () => sayNum(items[i]); sayAgain();
    }
    cleanup = () => pad.stop();
    afterIntro(intro, ask);
  }
  // trace the figures (same engine as ABC Books)
  function stepTrace(intro) {
    const nums = L.kind === 'count' ? ['1 2 3', '4 5 6', '7 8 9', '10'] : L.kind === 'write' ? ['0 1 2 3 4', '5 6 7 8 9', '10 20 30'] : (() => { const r = range(L.from, L.to), out = []; for (let i = 0; i < r.length && out.length < 4; i += 4) out.push(r.slice(i, i + 4).join(' ')); return out; })();
    let i = 0, misses = 0, tracer;
    const box = el('div', 'grow paper'); box.style.height = '100%'; $('stage').appendChild(box);
    const cv = el('canvas'); box.appendChild(cv);
    ABCTrace.Tracer.onMiss = why => { misses++; beep(false); $('hint').textContent = why === 'start' ? 'Start at the green dot!' : 'Follow the arrow all the way.'; };
    function page() {
      if (tracer) tracer.stop();
      $('hint').textContent = `Page ${i + 1} of ${nums.length}: start at the green dot and follow the arrow.`;
      tracer = ABCTrace.Tracer(cv, nums[i], 'print', () => { beep(true); toast('⭐', 700); setTimeout(() => { if (++i < nums.length) page(); else next(misses < 3 ? 3 : misses < 8 ? 2 : 1); }, 800); });
    }
    $('controls').appendChild(btn('Skip page', () => { if (++i < nums.length) page(); else next(1); }, 'ghost parent'));
    setTimeout(page, 50);
    const onRes = () => page(); window.addEventListener('resize', onRes);
    cleanup = () => { tracer && tracer.stop(); window.removeEventListener('resize', onRes); };
  }

  // the book's table, one line at a time with its picture
  function stepLearn(intro) { learnList(L.table.map(f => ({ t: f.t, pic: () => dotsFor(L.op, f.a, f.b), ex: `${expr(L.op, f)} = ${f.r}` })), intro); }
  function stepRules(intro) { learnList(L.rules.map(r => ({ t: r.t, pic: () => el('div', 'pic', picFor(r.t)) })), intro); }
  function learnList(items, intro) {
    const col = el('div', 'col'); $('stage').appendChild(col);
    const face = el('div', 'col'); face.style.height = 'auto'; const chips = el('div', 'chips'); col.append(face, chips);
    let i = 0, auto = false; const seen = new Set();
    const chipEls = items.map((it, k) => { const c = btn(it.t, () => { auto = false; i = k; show(); }, 'chip'); chips.appendChild(c); return c; });
    function show() {
      face.innerHTML = ''; const it = items[i];
      const f = el('div', 'fact', it.t); if (it.ex) f.appendChild(el('span', 'sm', it.ex));
      face.append(f, it.pic());
      chipEls.forEach((c, k) => c.classList.toggle('cur', k === i)); chipEls[i].classList.add('seen'); seen.add(i);
      chipEls[i].scrollIntoView({ block: 'nearest' });
      if (seen.size === items.length) nextB.disabled = false;
      return speak(it.t);
    }
    async function playAll() { auto = true; playB.disabled = true; for (i = 0; i < items.length && auto; i++) { await show(); await wait(600); } i = Math.min(i, items.length - 1); auto = false; playB.disabled = false; nextB.disabled = false; }
    const playB = btn('▶ Read to me', playAll, 'big'), nextB = btn('Next ▶', () => next(3));
    nextB.disabled = true; $('controls').append(playB, nextB);
    sayAgain = show; cleanup = () => { auto = false; };
    face.appendChild(el('div', 'fact', items[0].t));
    $('hint').textContent = 'Tap ▶ Read to me, or tap each line.';
    afterIntro(intro, playAll);
  }

  // practice: every fact in the table (or every rule), timed; a missed one comes back later
  function stepDrill(intro) {
    const items = L.table ? L.table.map(f => ({ q: f.q, ex: expr(L.op, f) + ' = ?', a: f.r, say: f.t, pic: () => dotsFor(L.op, f.a, f.b) }))
      : L.rules.filter(r => r.quiz !== false).map(r => ({ q: r.q, ex: `? ${r.a} = 1 ${r.b}`, a: r.n, say: r.t, pic: () => el('div', 'pic', picFor(r.t)) }));
    let queue = shuffle(items), done = 0, miss = 0, tries = 0, streak = 0, t0 = 0, cur;
    const key = `${bookId}:${L.n}`, best = kidMap(S.best);
    const col = el('div', 'qa'); $('stage').appendChild(col);
    const top = el('div', 'row'); const timer = el('div', 'timer'); const fire = el('div', 'timer'); top.append(timer, fire);
    const ex = el('div', 'expr'); const qd = el('div', 'note'); const picBox = el('div', 'picbox'); picBox.style.minHeight = '0';
    const pad = numPad(check);
    col.append(top, ex, qd, picBox, pad.el);
    const showB = btn('👀 Show me', () => { picBox.innerHTML = ''; picBox.appendChild(cur.pic()); }, 'ghost');
    const work = workPad();
    $('controls').append(showB, work.btn);
    const tick = setInterval(() => { if (t0) timer.textContent = `⏱ ${Math.round((Date.now() - t0) / 1000)}s` + (best[key] ? `  (best ${best[key]}s)` : ''); }, 500);
    function ask() {
      if (!queue.length) {
        const secs = Math.round((Date.now() - t0) / 1000);
        let msg = `${secs} seconds!`;
        if (!best[key] || secs < best[key]) { if (best[key]) msg += ' New best time! 🏆'; best[key] = secs; save(); }
        toast(msg, 1800); return setTimeout(() => next(starsFor(miss, items.length)), 1800);
      }
      cur = queue[0]; tries = 0; pad.clear(); picBox.innerHTML = '';
      ex.textContent = cur.ex; ex.className = 'expr' + (cur.ex.length > 12 ? ' long' : ''); qd.textContent = cur.q;
      work.page(`${L.n}:d:${cur.ex}`, { q: `${cur.ex}   ${cur.q}`, l: L.n });
      $('hint').textContent = `${done + 1} of ${items.length}`;
      sayAgain = () => speak(cur.q); sayAgain();
      if (!t0) t0 = Date.now();
    }
    function check(v) {
      work.result(v, v === cur.a);
      if (v === cur.a) {
        beep(true); pad.set(v, 'ok'); streak++; fire.textContent = streak >= 3 ? `🔥 ${streak}` : '';
        queue.shift(); done++; setTimeout(ask, 600);
      } else if (++tries < 2) { beep(false); pad.bad(); miss++; streak = 0; fire.textContent = ''; speak(P.tryAgain); }
      else {   // show it, say the book's line, and ask it again later
        pad.set(cur.a, 'ok'); picBox.innerHTML = ''; picBox.appendChild(cur.pic());
        const it = queue.shift(); queue.splice(Math.min(3, queue.length), 0, it);
        speak(cur.say).then(() => wait(600)).then(ask);
      }
    }
    cleanup = () => { clearInterval(tick); pad.stop(); work.stop(); };
    afterIntro(intro, ask);
  }

  // story problems: the book's questions, one answer at a time; two misses shows the answer and the book's solution
  function stepProblems(intro) {
    const qs = L.qs, posKey = `${bookId}:${L.n}`, pos = kidMap(S.pos);
    let qi = pos[posKey] || 0, pi = 0, tries = 0, miss = 0, si = 0, busy = false;
    if (qi >= qs.length) qi = 0;
    const col = el('div', 'qa'); $('stage').appendChild(col);
    const pic = el('div', 'pic'); const qd = el('div', 'q'); const seq = el('div', 'seq'); const sol = el('div', 'sol');
    const pad = numPad(check);
    col.append(pic, qd, seq, sol, pad.el);
    initTray();
    const work = workPad();
    if (!/^algebra/.test(bookId)) $('controls').append(btn('🧮 Counters', () => $('tray').classList.toggle('on')));   // no counting beads in Algebra
    $('controls').append(work.btn, btn('Finish later', () => { pos[posKey] = qi; save(); leave(); }, 'ghost parent'));
    if (L.note) $('hint').textContent = L.note;
    const Q = () => qs[qi], Pt = () => Q().parts[pi];
    function render() {
      const q = Q(); qd.innerHTML = ''; qd.classList.toggle('has-stack', q.parts.some(p => p.col));
      q.parts.forEach((p, k) => {
        qd.appendChild(el('span', 'pt' + (k === pi ? ' cur' : k < pi ? '' : ''), p.t + ' '));
        if (p.col) {   // a sum written in a column, the book's way: numbers right-aligned, the sign on the last line, a rule under
          const st = el('div', 'stack'), sign = { '+': '+', '-': '−', x: '×' }[p.col.op];
          p.col.n.forEach((v, j) => st.appendChild(el('div', '', (j === p.col.n.length - 1 ? `<b>${sign}</b>` : '') + v)));
          if (k < pi) st.appendChild(el('div', 'sum', String(p.f || p.a))); qd.appendChild(st);
        }
        if (k < pi) qd.appendChild(el('span', 'got' + (p.missed ? ' miss' : ''), p.seq || p.open ? '✓' : p.e != null ? RaysAlg.pretty(p.e) : (p.f || p.a)));
        if (k < q.parts.length - 1) qd.append(' ');
      });
      pic.textContent = picFor(q.parts[0].t);
      seq.innerHTML = '';
      const p = Pt(); if (p && p.seq) p.seq.forEach((v, k) => seq.appendChild(el('span', k < si ? 'got' : k === si ? 'cur' : '', k < si ? v : '?')));
      const cur = seq.querySelector('.cur'); if (cur) cur.scrollIntoView({ block: 'nearest' });
    }
    function ask(sayIt) {
      if (qi >= qs.length) { pos[posKey] = 0; save(); return next(starsFor(miss, qs.reduce((n, q) => n + q.parts.length, 0))); }
      tries = 0; pad.clear(); sol.textContent = ''; if (pi === 0) Q().shown = false;
      const p = Pt(); if (p.seq && si === 0) si = 1;   // the first number of a count is given
      pad.frac(p.f); pad.alg(p.e != null ? p.v || '' : null); pad.show(!p.open); openB.style.display = p.open ? '' : 'none';
      render();
      work.page(`${L.n}:${qi}`, { q: Q().parts.map(x => x.t).join(' '), l: L.n });   // one page for all the parts of a question
      $('hint').textContent = `Question ${qi + 1} of ${qs.length}`;
      sayAgain = () => speak(p.s || p.t, p.open ? 'Answer out loud, then tap Next.' : null);
      if (sayIt !== false) sayAgain();
    }
    function advance() {
      busy = false; const q = Q();
      if (++pi < q.parts.length) { si = 0; return ask(); }
      if (q.sol && !q.shown) { q.shown = true; sol.textContent = q.sol; }
      qi++; pi = 0; si = 0; pos[posKey] = qi; save();
      setTimeout(() => ask(), q.sol && sol.textContent ? 600 : 0);
    }
    function check(v) {
      if (busy) return; const p = Pt();
      const norm = s => String(s).trim().replace(/\s+/g, ' ');
      const isRight = (p, v) => { const want = p.seq ? p.seq[si] : p.e != null ? p.e : p.f || p.a;
        return p.e != null ? RaysAlg.ok(v, p.e) || (p.ealt || []).some(x => RaysAlg.ok(v, x)) : p.ex ? p.ex.includes(norm(v)) : p.f ? sameVal(v, p.f) || (p.falt || []).some(x => sameVal(v, x)) : v === want || (!p.seq && p.alt && p.alt.includes(v)); };
      let right = isRight(p, v);
      // "x = {A} or {B}" (the two roots of a quadratic): B typed first counts too; the next box then wants A
      const nx = Q().parts[pi + 1];
      if (!right && p.pr && nx && isRight(nx, v)) {
        for (const k of ['e', 'es', 'ealt', 'a', 'alt', 'f', 'falt', 'v']) { const t = p[k]; if (nx[k] === undefined) delete p[k]; else p[k] = nx[k]; if (t === undefined) delete nx[k]; else nx[k] = t; }
        right = true;
      }
      const want = p.seq ? p.seq[si] : p.e != null ? p.e : p.f || p.a;
      if (!p.seq || !right) work.result(v, right);
      if (right) {
        beep(true); pad.set(v, 'ok'); tries = 0;
        if (p.seq) { if (++si >= p.seq.length) { busy = true; render(); return setTimeout(() => { si = 0; advance(); }, 700); } setTimeout(() => { pad.clear(); render(); }, 350); return; }
        busy = true; speak(P.yes); setTimeout(advance, 900); return;
      }
      miss++;
      if (++tries < 2) { beep(false); pad.bad(); speak(P.tryAgain); return; }
      // second miss: show the answer, say it, and the book's solution when there is one
      busy = true; pad.set(want, 'ok'); tries = 0;
      const q = Q(), last = !p.seq && pi === q.parts.length - 1;
      if (last && q.sol) { sol.textContent = q.sol; q.shown = true; }
      if (p.seq) { si++; if (si >= p.seq.length) { si = 0; p.missed = true; return speak('ans:' + want).then(() => advance()); } return speak('n:' + want).then(() => { busy = false; pad.clear(); render(); }); }
      p.missed = true;
      speak('ans:' + want, last && q.sol ? q.sol : null).then(() => wait(800)).then(advance);
    }
    // a question answered out loud to a grown-up: no typing, just Next. When the book gives the answer (sol:), the first
    // Next shows and says it (e.g. "57 is written LVII."), the second moves on.
    const openB = btn('Next ▶', () => {
      if (busy) return; const q = Q();
      if (pi === q.parts.length - 1 && q.sol && !q.shown) { q.shown = true; sol.textContent = q.sol; speak(q.sol); return; }
      busy = true; beep(true); setTimeout(advance, 300);
    }, 'big'); openB.style.display = 'none';
    col.appendChild(openB);
    cleanup = () => { pad.stop(); work.stop(); };
    afterIntro(intro, () => ask());
  }

  // Book Five: the book's articles, each beside its figure cut from the scan (same key as make-audio.ps1 ArtKey)
  const artKey = a => `${a.n}. ${a.h ? a.h + '. ' : ''}${a.t}`;
  function stepStudy(intro) {
    const arts = L.arts; let i = 0, run = 0; const seen = new Set();
    const stop = () => { run++; playB.disabled = false; };   // a tap cancels reading-to-me
    const col = el('div', 'col study'); $('stage').appendChild(col);
    const face = el('div', 'art'); const chips = el('div', 'chips'); col.append(face, chips);
    const chipEls = arts.map((a, k) => { const c = btn('§' + a.n, () => { stop(); i = k; show(); }, 'chip'); chips.appendChild(c); return c; });
    function draw() {
      const a = arts[i]; face.innerHTML = '';
      if (a.figs) { const fb = el('div', 'figs'); a.figs.forEach(f => { const im = el('img', 'fig'); im.src = f; im.alt = 'Figure for article ' + a.n; fb.appendChild(im); }); face.appendChild(fb); }
      face.appendChild(el('p', '', `<b>${a.n}.</b> ${a.h ? `<b>${a.h}.</b> ` : ''}${a.t}`));
      chipEls.forEach((c, k) => c.classList.toggle('cur', k === i));
    }
    function show() {
      const a = arts[i]; draw(); chipEls[i].classList.add('seen'); seen.add(i);
      chipEls[i].scrollIntoView({ block: 'nearest' }); face.scrollTop = 0;
      $('hint').textContent = `Part ${i + 1} of ${arts.length}`;
      if (seen.size === arts.length) nextB.disabled = false;
      return speak(artKey(a));
    }
    async function playAll() {   // from the part on screen to the end
      const my = ++run; playB.disabled = true;
      for (; i < arts.length; i++) { await show(); if (my !== run) return; await wait(700); if (my !== run) return; }
      i = arts.length - 1; playB.disabled = false; nextB.disabled = false;
    }
    const prevB = btn('◀', () => { stop(); if (i > 0) { i--; show(); } });
    const fwdB = btn('▶', () => { stop(); if (i < arts.length - 1) { i++; show(); } else nextB.disabled = false; });
    const playB = btn('▶ Read to me', playAll, 'big'), nextB = btn('Next step ▶', () => next(3));
    nextB.disabled = true; $('controls').append(prevB, playB, fwdB, nextB);
    sayAgain = show; cleanup = () => { run++; };
    draw(); afterIntro(intro, () => { if (!seen.size) playAll(); });
  }
  // pick the answer (or True / False); a miss comes back at the end, two misses show the answer
  function stepQuiz(intro) {
    let queue = shuffle(L.quiz.map(x => ({ ...x }))), done = 0, miss = 0, tries = 0, busy = false, cur;
    const total = queue.length;
    const col = el('div', 'qa'); $('stage').appendChild(col);
    const qd = el('div', 'q'); const grid = el('div', 'mc'); col.append(qd, grid);
    function ask() {
      if (!queue.length) { toast('Great job! ⭐'); return setTimeout(() => next(starsFor(miss, total)), 900); }
      cur = queue[0]; tries = 0; busy = false; grid.innerHTML = '';
      qd.textContent = (cur.tf ? 'True or false? ' : '') + cur.q;
      const order = cur.tf ? [0, 1] : shuffle(cur.c.map((_, k) => k));
      order.forEach(k => grid.appendChild(btn(cur.c[k], e => pick(k, e.currentTarget))));
      $('hint').textContent = `${done + 1} of ${total}`;
      sayAgain = () => speak(cur.q); sayAgain();
    }
    function pick(k, b) {
      if (busy) return;
      if (k === cur.a) { busy = true; beep(true); b.classList.add('ok'); queue.shift(); done++; speak(P.yes); return setTimeout(ask, 900); }
      miss++; beep(false); b.classList.remove('no'); void b.offsetWidth; b.classList.add('no'); b.disabled = true;
      if (++tries < 2 && !cur.tf) return speak(P.tryAgain);
      busy = true; [...grid.children].forEach(x => { if (x.textContent === cur.c[cur.a]) x.classList.add('ok'); });
      const it = queue.shift(); it.again = true; queue.push(it);   // asked again at the end
      speak('ansc:' + cur.c[cur.a]).then(() => wait(700)).then(ask);
    }
    afterIntro(intro, ask);
  }

  function finish() {
    const vals = Object.values(scores), stars = Math.max(1, Math.round(vals.reduce((a, b) => a + b, 0) / Math.max(1, vals.length)));
    const p = prog(bookId), old = (p[L.n] || {}).stars || 0;
    p[L.n] = { stars: Math.max(old, stars), at: Date.now() };
    const got = kidMap(S.stickers), sk = stkKey(bookId, L.n); const sticker = got[sk] || STICKERS[(L.n * 7 + SHELF.findIndex(b => b.id === bookId) * 3) % STICKERS.length]; got[sk] = sticker;
    save();
    $('stepTitle').textContent = `${S.kid}, you finished Lesson ${L.n}!`;
    $('stage').innerHTML = `<div class="col"><div class="stars">${starStr(stars)}</div><div class="stk">${sticker}</div><div class="note">A new sticker for your sticker book!</div></div>`;
    $('hint').textContent = ''; $('controls').innerHTML = '';
    $('controls').append(btn('Lessons', () => openBook(bookId), 'big'), btn('Sticker book', () => go('stickers')));
    const gj = `Great job, ${S.kid}!`; beep(true); speak(urlFor(gj) ? gj : P.greatJob, P.done);
  }

  // ?preview opens the "Coming soon" books that already have a book file (testing a book before it goes live)
  const ready = b => !b.soon || /[?&]preview\b/.test(location.search);
  Promise.all([
    fetch('prompts.json').then(r => r.json()).then(j => { P = j; }),
    ...SHELF.filter(ready).map(b => fetch(`books/${b.id}.json`).then(r => r.ok ? r.json() : null).then(j => { if (j) BOOKS[b.id] = j; })),
    ...SHELF.filter(ready).map(b => fetch(`audio/${b.id}.json`).then(r => r.ok ? r.json() : null).then(j => {
      if (!j) return; if (!AUDIO) AUDIO = { speeds: j.speeds, items: {} }; Object.assign(AUDIO.items, j.items);
    }).catch(() => {}))
  ]).then(home, () => toast('Could not load the book. Check the internet.', 4000));
  // the account answer can arrive after the first draw: redraw the home screen
  if (window.ABCAccount) window.ABCAccount.ready.then(() => { if ($('home').classList.contains('on') && Object.keys(BOOKS).length) home(); });
  return { go, leave, debug: () => ({ S, L, steps, stepIdx, scores }), toStep: id => { const i = steps.indexOf(id); if (i >= 0) { stepIdx = i; runStep(); } }, wordsToNumber };
})();
EOF

index.html: the page and its styles (new: the study card and quiz buttons).

cat > /root/rays-rebuild/index.html <<'EOF'
<!doctype html>
<html lang="en"><head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width,initial-scale=1,maximum-scale=1,user-scalable=no">
<meta name="theme-color" content="#121212">
<title>Ray's Arithmetic</title>
<link rel="icon" type="image/png" sizes="192x192" href="icon-192.png">
<link rel="manifest" href="manifest.webmanifest">
<link rel="apple-touch-icon" href="icon-192.png">
<meta name="apple-mobile-web-app-capable" content="yes">
<meta name="mobile-web-app-capable" content="yes">
<meta name="apple-mobile-web-app-status-bar-style" content="black">
<meta name="apple-mobile-web-app-title" content="Ray's Math">
<style>
:root{color-scheme:dark;--paper:#121212;--ink:#ececec;--soft:#a8a8a8;--accent:#2f8fbf;--warm:#d9733f;--good:#3fae6e;--bad:#ff6b5e;--board:#1d3527;--card:#1e1e1e;--line:#6a6a6a;--hi:#1f3340;--a:#f0a03c;--b:#5ab4f0}
*{box-sizing:border-box;-webkit-tap-highlight-color:transparent}
html,body{margin:0;height:100%;background:var(--paper);color:var(--ink);font-family:Georgia,"Times New Roman",serif;overflow:hidden;user-select:none;-webkit-user-select:none;-webkit-text-size-adjust:100%;text-size-adjust:100%}
.screen{position:absolute;inset:0;display:none;flex-direction:column;padding:10px 16px}
.screen.on{display:flex}
h1{margin:4px 0 10px;font-size:clamp(26px,6vmin,44px);text-align:center;letter-spacing:.04em}
h2{margin:4px 0;font-size:clamp(19px,4.2vmin,30px);text-align:center}
button{font:inherit;font-size:clamp(18px,4vmin,26px);border:3px solid var(--line);background:var(--card);color:var(--ink);border-radius:14px;padding:8px 16px;min-height:50px;cursor:pointer}
button:active{transform:scale(.96)}
button.big{background:var(--accent);color:#fff;border-color:#164a63;font-size:clamp(21px,4.6vmin,30px);padding:12px 26px}
button.ghost{border-color:transparent;background:transparent}
button:disabled{opacity:.35}
.row{display:flex;gap:10px;justify-content:center;flex-wrap:wrap;align-items:center}
.grow{flex:1;min-height:0;position:relative}
.kids button{min-width:30vmin;font-size:clamp(24px,6vmin,40px)}
.kids button.sel{background:var(--accent);border-color:var(--accent);color:#fff}
.kids button.add,.kids button.ghost,.kids button.big{min-width:0}
.kids input{font:inherit;font-size:clamp(22px,5vmin,34px);width:min(60vw,320px);padding:8px 14px;border-radius:14px;border:3px solid var(--line);background:var(--card);color:var(--ink)}
#home{overflow-y:auto}
.shelf{display:flex;gap:3vmin;justify-content:center;flex-wrap:wrap;margin-top:3vmin;padding:8px}
.book{width:min(34vmin,210px);aspect-ratio:3/4;border-radius:6px 14px 14px 6px;color:#fff;padding:14px;display:flex;flex-direction:column;justify-content:space-between;box-shadow:inset 10px 0 0 rgba(0,0,0,.18),0 4px 10px rgba(0,0,0,.25);border:none;text-align:left}
.book b{font-size:clamp(18px,4vmin,26px)}.book small{font-size:clamp(13px,2.6vmin,17px);opacity:.9}
.book .bp{font-size:clamp(14px,3vmin,19px);font-weight:bold;background:rgba(0,0,0,.35);border-radius:8px;padding:4px 8px;align-self:flex-start}
.book.soon{filter:grayscale(.6);opacity:.6}
.book.cur{outline:5px solid #ffd166;outline-offset:3px;animation:pulse 1.6s infinite}
.map{display:grid;grid-template-columns:repeat(auto-fill,minmax(min(22vmin,120px),1fr));grid-auto-rows:min(24vmin,120px);gap:12px;overflow-y:auto;padding:6px;align-content:start}
.tile{height:100%;min-height:0;border-radius:18px;display:flex;flex-direction:column;align-items:center;justify-content:center;font-size:clamp(22px,6vmin,38px);padding:4px;line-height:1.1}
.tile small{font-size:clamp(11px,2.4vmin,15px);color:var(--soft)}
.tile .st{font-size:clamp(14px,3vmin,20px);color:#e0a000;letter-spacing:2px}
.tile.next{background:var(--hi);border-color:var(--accent);animation:pulse 1.4s infinite}
.tile.locked{opacity:.35}
.jump{display:flex;align-items:center;gap:8px;flex-wrap:wrap;margin:6px 0;font-size:clamp(16px,3.4vmin,22px)}
.jump select{flex:1;min-width:0;max-width:100%;font:inherit;padding:8px;background:var(--card);color:var(--ink);border:2px solid var(--line);border-radius:12px}
.tile.rev{border-style:dashed}
@keyframes pulse{50%{transform:scale(1.05)}}
.top{display:flex;align-items:center;gap:8px}
.top .t{flex:1;text-align:center;font-size:clamp(17px,3.8vmin,26px)}
.steps{display:flex;gap:6px;justify-content:center;margin:4px 0}
.steps i{width:clamp(18px,4vmin,28px);height:clamp(18px,4vmin,28px);border-radius:50%;border:3px solid var(--soft);font-style:normal;font-size:12px;display:flex;align-items:center;justify-content:center}
.steps i.done{background:var(--good);border-color:var(--good)}.steps i.cur{border-color:var(--accent);background:var(--hi)}
.hint{font-size:clamp(15px,3.2vmin,20px);color:var(--soft);text-align:center;min-height:1.4em}
.toast{position:fixed;left:50%;top:40%;transform:translate(-50%,-50%);background:#2a2a2a;border:3px solid var(--accent);color:#fff;padding:18px 28px;border-radius:18px;font-size:clamp(24px,6vmin,40px);display:none;z-index:9;text-align:center}
.stars{font-size:clamp(40px,12vmin,90px);color:#e0a000;text-align:center;letter-spacing:6px}
.parent{font-size:15px;color:var(--soft)}
.col{display:flex;flex-direction:column;align-items:center;gap:1.2vmin;height:100%;overflow-y:auto;justify-content:center}
.big-n{font-size:clamp(70px,22vmin,170px);line-height:1;font-weight:bold}
.word{font-size:clamp(22px,5vmin,36px);color:var(--soft)}
.fact{font-size:clamp(30px,8vmin,64px);text-align:center;line-height:1.15}
.fact .sm{display:block;font-size:clamp(16px,3.4vmin,22px);color:var(--soft)}
.dots{display:flex;flex-wrap:wrap;gap:1.4vmin;justify-content:center;align-items:center;max-width:94vw}
.grp{display:grid;grid-template-columns:repeat(5,auto);gap:.8vmin;padding:1vmin;border-radius:10px;border:2px dashed transparent}
.grp.box{border-color:var(--line)}
.dot{width:clamp(14px,3.6vmin,26px);height:clamp(14px,3.6vmin,26px);border-radius:50%;background:var(--a)}
.dot.b{background:var(--b)}.dot.x{background:#444;position:relative}.dot.x::after{content:'';position:absolute;left:-2px;right:-2px;top:45%;height:3px;background:var(--bad);transform:rotate(-35deg)}
.op{font-size:clamp(26px,6vmin,44px);color:var(--soft)}
.chips{display:flex;flex-wrap:wrap;gap:6px;justify-content:center;max-height:24vh;overflow-y:auto}
.chip{font-size:clamp(14px,3vmin,19px);min-height:38px;padding:4px 10px;border-radius:10px}
.chip.cur{background:var(--hi);border-color:var(--accent)}.chip.seen{border-color:var(--good)}
.balls{display:flex;flex-wrap:wrap;gap:2vmin;justify-content:center;max-width:90vw}
.ball{width:clamp(44px,11vmin,80px);height:clamp(44px,11vmin,80px);border-radius:50%;border:none;background:radial-gradient(circle at 35% 30%,#ffd9a8,var(--warm) 60%,#7a3412);padding:0;min-height:0;font-size:clamp(18px,4vmin,28px);color:#fff;font-weight:bold}
.ball.on{background:radial-gradient(circle at 35% 30%,#b8f0c8,var(--good) 60%,#1b5a35)}
.rods{display:flex;gap:1vmin;align-items:flex-end;justify-content:center;flex-wrap:wrap}
.rod{display:grid;grid-template-rows:repeat(10,1fr);gap:2px;height:clamp(110px,26vmin,190px);width:clamp(14px,3.4vmin,22px)}
.rod i{background:var(--b);border-radius:3px}
.choices{display:grid;grid-template-columns:repeat(2,minmax(0,1fr));gap:2vmin;width:min(90vw,520px)}
.choices button{font-size:clamp(40px,12vmin,86px);min-height:clamp(80px,18vmin,140px)}
.choices button.ok{background:var(--good);border-color:var(--good);color:#fff}.choices button.no{animation:shake .4s}
@keyframes shake{25%{transform:translateX(-8px)}75%{transform:translateX(8px)}}
.grid{display:flex;flex-wrap:wrap;gap:8px;justify-content:center;overflow-y:auto;max-height:100%;align-content:flex-start;padding:4px}
.grid button{min-width:clamp(56px,14vmin,90px);font-size:clamp(24px,6vmin,40px)}
.grid button.seen{border-color:var(--good)}
/* question + number pad */
.study{gap:1vmin}.study .chips{max-height:15vh;flex-shrink:0}.art{flex:1;min-height:0;overflow-y:auto;width:min(94vw,760px);background:#fffdf6;border-radius:12px;padding:10px 14px;color:#222}
.art{display:flex;flex-direction:column;gap:10px}.art .figs{display:flex;flex-wrap:wrap;gap:8px;justify-content:center;align-items:center}
.art img.fig{display:block;max-width:100%;max-height:26vh;mix-blend-mode:multiply}
@media (min-aspect-ratio:5/4){.art{flex-direction:row;align-items:flex-start}.art .figs{flex:0 0 38%;flex-direction:column;position:sticky;top:0}.art img.fig{max-height:30vh}.art p{flex:1}}
.art p{margin:0;font-size:clamp(18px,3.8vmin,28px);line-height:1.45}
.mc{display:flex;flex-direction:column;gap:1.4vmin;width:min(92vw,640px)}
.mc button{font-size:clamp(18px,4vmin,28px);min-height:56px;padding:8px 14px;text-align:left;line-height:1.3}
.mc button.ok{background:var(--good);border-color:var(--good);color:#fff}.mc button.no{animation:shake .4s;opacity:.55}
.qa{display:flex;flex-direction:column;align-items:center;gap:1vmin;height:100%;min-height:0}
.q{font-size:clamp(20px,4.6vmin,34px);line-height:1.35;text-align:center;overflow-y:auto;max-height:42%;padding:4px 10px;border-radius:12px}
.q .pt{opacity:.55}.q .pt.cur{opacity:1;background:var(--hi);border-radius:8px;padding:0 4px}
.q .got{display:inline-block;background:var(--good);color:#fff;border-radius:8px;padding:0 8px;margin:0 4px;font-weight:bold}
.q .miss{background:var(--warm)}
.q.has-stack{max-height:60%}.q.has-stack .pt{font-size:.85em}
.stack{display:table;margin:.2em auto;font-family:'Courier New',monospace;font-weight:bold;text-align:right;line-height:1.05;font-size:1.05em;border-bottom:3px solid currentColor;padding:0 .3em}
.stack b{float:left;margin-right:.6em}.stack .sum{color:var(--good);border-top:3px solid currentColor;margin-bottom:-3px}
.pic{font-size:clamp(30px,8vmin,56px);line-height:1}
.expr{font-size:clamp(46px,13vmin,100px);font-weight:bold;line-height:1.05;text-align:center}
.micb{min-width:min(86vw,360px);font-size:clamp(20px,4.6vmin,30px)}.micb.live{background:var(--bad);border-color:var(--bad);color:#fff;animation:pulse 1s infinite}
.expr.long{font-size:clamp(28px,7vmin,52px)}
.entry{min-width:3.4em;min-height:1.3em;font-size:clamp(36px,9vmin,64px);text-align:center;border-bottom:4px solid var(--accent);font-weight:bold;letter-spacing:.05em}
.entry.long{font-size:clamp(24px,6vmin,44px)}.entry.no{animation:shake .4s;border-color:var(--bad)}.entry.ok{color:var(--good)}
.pad{display:grid;grid-template-columns:repeat(3,1fr);gap:1.2vmin;width:min(86vw,360px)}
.pad button{font-size:clamp(24px,6vmin,38px);min-height:clamp(46px,9.5vmin,70px);padding:0}
.pad button.okb{background:var(--good);border-color:var(--good);color:#fff}
.algmode .entry{max-width:92vw;overflow-wrap:anywhere;letter-spacing:0}.algmode .entry.long{font-size:clamp(18px,4.2vmin,32px)}
.pad.alg,.algmode .pad{grid-template-columns:repeat(6,1fr);width:min(92vw,480px)}.pad.alg button,.algmode .pad button{font-size:clamp(20px,5vmin,32px);min-height:clamp(40px,8vmin,60px)}.pad.alg button.let{font-style:italic;font-family:Georgia,serif}
.seq{display:flex;flex-wrap:wrap;gap:6px;justify-content:center;max-height:22vh;overflow-y:auto}
.seq span{min-width:2.2em;text-align:center;border:2px solid var(--line);border-radius:8px;padding:2px 6px;font-size:clamp(18px,4vmin,28px)}
.seq span.got{border-color:var(--good);color:var(--good)}.seq span.cur{border-color:var(--accent);background:var(--hi)}
.sol{font-size:clamp(16px,3.4vmin,22px);color:#ffd166;text-align:center;max-width:90vw}
.tray{position:fixed;left:50%;bottom:2vh;transform:translateX(-50%);background:#1a1a1a;border:3px solid var(--accent);border-radius:16px;padding:10px;z-index:20;display:none;flex-direction:column;align-items:center;gap:8px}
.tray.on{display:flex}
.frame{display:grid;grid-template-columns:repeat(10,1fr);gap:5px}
.frame button{width:clamp(26px,6.5vmin,44px);height:clamp(26px,6.5vmin,44px);min-height:0;border-radius:50%;padding:0;border-width:2px}
.frame button.on{background:var(--a);border-color:var(--a)}
.timer{font-size:clamp(16px,3.4vmin,22px);color:var(--soft)}
.stk{font-size:clamp(40px,10vmin,72px)}
.stickers{display:flex;flex-wrap:wrap;gap:10px;justify-content:center;overflow-y:auto;font-size:clamp(34px,9vmin,60px)}
.stickers span{width:1.5em;height:1.5em;display:flex;align-items:center;justify-content:center;border:2px dashed var(--line);border-radius:12px}
.stickers span.got{border:2px solid #e0a000;background:#2a2410}
canvas{position:absolute;inset:0;width:100%;height:100%;touch-action:none;border-radius:14px}
.paper{background:#1a1a1a;border:3px solid var(--line);border-radius:14px}
.note{font-size:clamp(13px,2.8vmin,17px);color:var(--soft);text-align:center;max-width:90vw}
/* show your work: scratch paper beside the question (landscape) or under the number pad (portrait), never over them */
#controls button.on{background:var(--hi);border-color:var(--accent)}
#stage.working{display:flex;flex-direction:column;gap:8px}
#stage.working .qa{flex:1 1 auto;height:auto;min-height:0;overflow-y:auto}
#stage.working .qa .pic{display:none}
#stage.working .pad button{min-height:clamp(42px,7vmin,60px)}#stage.working .micb{min-height:44px;padding:4px 16px;font-size:clamp(17px,3.6vmin,24px)}#stage.working .entry{font-size:clamp(28px,6.5vmin,50px)}#stage.working .expr{font-size:clamp(32px,7.5vmin,64px)}#stage.working .expr.long{font-size:clamp(24px,5.5vmin,44px)}#stage.working .qa{gap:.6vmin;overflow:hidden}#stage.working .qa>*{flex-shrink:0}#stage.working .qa>.q{flex:0 1 auto;min-height:2.8em;max-height:24vh}#stage.working .qa>.picbox{flex:0 1 auto;overflow-y:auto}
.scratch{flex:0 0 35%;display:flex;flex-direction:column;gap:6px;min-height:0;min-width:0}
.sarea{flex:1;position:relative;min-height:0;border:3px solid var(--line);border-radius:14px;overflow:hidden;background:#fffdf6}
.sarea canvas{border-radius:0}
.stools{display:flex;gap:6px;justify-content:center;flex-wrap:wrap}
.stools button{font-size:clamp(16px,3.2vmin,22px);min-height:clamp(46px,8vmin,60px);padding:4px 12px;flex:1 1 0;max-width:180px;white-space:nowrap}
.stools button.on{background:var(--accent);border-color:#164a63;color:#fff}
.erz{display:inline-block;width:1.1em;height:.65em;background:linear-gradient(90deg,#f4a3b4 60%,#5a8fd0 60%);border-radius:3px;transform:rotate(-20deg);vertical-align:middle}
@media (min-aspect-ratio:5/4){#stage.working{flex-direction:row}#stage.working .qa{flex:0 0 44%;height:100%}#stage.working .qa .pic{display:block}.scratch{flex:1 1 0}}
.worklist{display:flex;flex-wrap:wrap;gap:12px;justify-content:center;align-content:flex-start;overflow-y:auto;padding:6px}
.wles{flex-basis:100%;margin:10px 0 0;font-size:clamp(17px,3.6vmin,24px)}
.wpage{display:flex;flex-direction:column;gap:4px;padding:6px;text-align:left;font-size:15px;line-height:1.3;max-width:min(316px,46vw);border-width:2px}
.wpage canvas,.workbig canvas{position:static;border-radius:6px}
.wcap{font-size:clamp(14px,2.8vmin,18px);max-height:4.2em;overflow:hidden}
.workbig .wcap{max-height:none;text-align:center;max-width:94vw}
.wres{color:var(--soft);font-size:14px}.wres.ok{color:var(--good)}.wres.miss{color:var(--a)}
.workbig{position:fixed;inset:0;background:var(--paper);z-index:30;display:none;flex-direction:column;align-items:center;justify-content:center;gap:10px;padding:12px}
.workbig.on{display:flex}
</style>
</head><body>

<section id="home" class="screen on">
  <h1>Ray's Arithmetic</h1>
  <h2>Who is doing math today?</h2>
  <div class="row kids" id="kids"></div>
  <div class="hint" id="whoHint"></div>
  <div class="shelf" id="shelf"></div>
  <div class="row" style="margin-top:auto"><button class="ghost parent" id="speedBtn">Voice: normal</button><button class="ghost parent" id="toStickers">Sticker book</button><a class="parent" href="/account/">Grown-ups: account</a><a class="parent" href="/readers/">Readers</a><a class="parent" href="/">About</a></div>
</section>

<section id="map" class="screen">
  <div class="top"><button onclick="Rays.go('home')">◀</button><div class="t" id="mapTitle"></div></div>
  <div class="jump"><label for="jumpSel">Go to lesson</label><select id="jumpSel"></select><button id="jumpGo">Go</button><button id="mapWork" class="parent" title="Grown-ups: the scratch paper saved with each question">✏️ See the work</button></div>
  <div class="map grow" id="tiles"></div>
</section>

<section id="lesson" class="screen">
  <div class="top"><button onclick="Rays.leave()">◀</button><div class="t" id="lesTitle"></div><button id="sayBtn" title="Say it again">🔊</button></div>
  <div class="steps" id="steps"></div>
  <h2 id="stepTitle"></h2>
  <div class="grow" id="stage"></div>
  <div class="hint" id="hint"></div>
  <div class="row" id="controls"></div>
</section>

<section id="stickers" class="screen">
  <div class="top"><button onclick="Rays.go('home')">◀</button><div class="t" id="stkTitle">Sticker book</div></div>
  <div class="stickers grow" id="stkList"></div>
  <div class="row"><button id="stkWork" class="parent">✏️ See the work</button></div>
</section>

<section id="work" class="screen">
  <div class="top"><button id="workBack">◀</button><div class="t" id="workTitle">Work</div></div>
  <div class="worklist grow" id="workList"></div>
  <div class="row"><button id="workDel" class="ghost parent">Delete all work in this book</button></div>
  <div class="workbig" id="workBig"></div>
</section>

<div class="tray" id="tray"><div class="frame" id="frame"></div><div class="row"><b id="trayN">0</b><button id="trayClear">Clear</button><button id="trayClose">Close</button></div></div>
<div class="toast" id="toast"></div>
<script src="hershey.js"></script>
<script src="trace.js"></script>
<script src="/account/sync.js"></script>
<script src="scratch.js"></script>
<script src="alg.js"></script>
<script src="app.js"></script>
<script>if ('serviceWorker' in navigator) navigator.serviceWorker.register('sw.js').catch(() => {});</script>
</body></html>
EOF

prompts.json: the spoken instructions (new: Study and Quiz).

cat > /root/rays-rebuild/prompts.json <<'EOF'
{
  "home": "Who is doing math today? Tap your name. Then tap the book.",
  "greet": "Hi, {kid}! Tap the book to start.",
  "hello": "Hi! Tap the book to start.",
  "greatJob": "Great job!",
  "map": "Pick a lesson. The glowing one is next.",
  "count": "Count the balls. Tap each ball, and count out loud.",
  "numbers": "Learn the numbers. Listen, and say each number with me.",
  "find": "Find the number. Listen, then tap the number you hear.",
  "say": "Read the numbers. Say each number out loud to a grown-up. Tap a number to hear it.",
  "write": "Write the numbers. Listen, then type the number you hear, and tap OK.",
  "trace": "Time to trace the numbers. Start at the green dot, and follow the arrow.",
  "learn": "Learn the table. Listen, and say it with me.",
  "rules": "Learn the table. Listen, and say it with me.",
  "drill": "Practice time. Type the answer, then tap OK. Try to beat your best time!",
  "problems": "Story problems. Listen to each question. Type the answer, then tap OK. Tap the counters if you need help.",
  "study": "Study the book. Listen to each part, and look at the figure. Tap a part to hear it again.",
  "quiz": "Quiz time. Listen to each question, then tap the right answer.",
  "yes": "Yes!",
  "tryAgain": "Not quite. Try again.",
  "done": "You finished the lesson! You earned a sticker."
}
EOF

sw.js: offline support (now keeps Book 5 too).

cat > /root/rays-rebuild/sw.js <<'EOF'
// Offline support. App files: network first (so updates show), cached copy when offline.
// Voice clips (audio/*.mp3) never change for a given name: cache first, saved the first time they play.
const SHELL = 'rays-shell-v19', CLIPS = 'rays-clips-v1';
const FILES = ['./', 'index.html', 'app.js', 'prompts.json', 'trace.js', 'scratch.js', 'alg.js', 'hershey.js', 'books/primary1877.json', 'audio/primary1877.json', 'books/intellectual.json', 'audio/intellectual.json', 'books/practical.json', 'audio/practical.json', 'books/higher.json', 'audio/higher.json', 'books/geometry.json', 'audio/geometry.json', 'books/algebra1.json', 'audio/algebra1.json', 'icon-192.png', 'manifest.webmanifest'];
self.addEventListener('install', e => { e.waitUntil(caches.open(SHELL).then(c => Promise.all(FILES.map(f => c.add(f).catch(() => {})))).then(() => self.skipWaiting())); });
self.addEventListener('activate', e => {
  e.waitUntil(caches.keys().then(ks => Promise.all(ks.filter(k => k.startsWith('rays-') && k !== SHELL && k !== CLIPS).map(k => caches.delete(k)))).then(() => self.clients.claim()));
});
self.addEventListener('fetch', e => {
  const r = e.request; if (r.method !== 'GET' || new URL(r.url).origin !== location.origin) return;
  if (r.url.endsWith('.mp3')) {
    e.respondWith(caches.open(CLIPS).then(c => c.match(r).then(hit => hit || fetch(r).then(res => { if (res.ok && res.status === 200) c.put(r, res.clone()); return res; }))));
    return;
  }
  e.respondWith(fetch(r).then(res => { if (res.ok) caches.open(SHELL).then(c => c.put(r, res.clone())); return res; }).catch(() => caches.match(r)));
});
EOF

Rebuild, step by step

1. Cut the figures

PC (PowerShell). It prints "261 figures":

powershell -ExecutionPolicy Bypass -File C:\RaysArithmetic\tools\make-figs.ps1 -Book geometry

To check them, make a montage of a range of pages and open the picture it names:

powershell -ExecutionPolicy Bypass -File C:\RaysArithmetic\tools\fig-montage.ps1 -FromPage 1 -ToPage 420

2. Build the book data

PC (PowerShell). It prints "64 lessons ...", or a list of problems:

powershell -ExecutionPolicy Bypass -File C:\RaysArithmetic\tools\make-book.ps1 -Book geometry

3. Record the voice

PC (PowerShell), about 6 minutes:

powershell -ExecutionPolicy Bypass -File C:\RaysArithmetic\tools\make-audio.ps1 -Id geometry

4. Put it on the server

PC (PowerShell). -Audio sends only the recordings CT 306 (abc-web) doesn't have yet; the figures go with the app files. It ends with "RAYS-DONE":

powershell -ExecutionPolicy Bypass -File C:\RaysArithmetic\deploy.ps1 -Audio

5. Check it

Open https://abcbooks.fyi/math/ , tap a name, and tap the olive Book 5. Lesson 1 starts reading article 1 ("Before the student begins the study of geometry..."); Lesson 10 shows the figures of angles beside the text.

Troubleshooting

Book 5 says "Coming soon": app.js still has soon: true on the geometry line of the shelf, or books/geometry.json is missing on the server.
A figure is missing (empty space by the text): check the figures are on the server. Proxmox web UI, pve, Shell; it should print 261:

pct exec 306 -- ls /var/www/abcbooks/rays/fig/geometry | wc -l

Book 5 is silent: check that its voice list exists (Proxmox host shell):

pct exec 306 -- ls -l /var/www/abcbooks/rays/audio/geometry.json

and count the MP3s (all five books together are about 19,400 at each speed):

pct exec 306 -- ls /var/www/abcbooks/rays/audio/s60 | wc -l

*make-book.ps1 says "mc needs 2+ choices and one right one":** a quiz line has no star, two stars, or a missing |.
make-book.ps1 says "bad answer {-1}": the pad has no minus sign; make it an mc: question.
make-figs.ps1 prints "EMPTY": that box has no ink in it; the page number or the fractions are wrong. Make a sheet of the page and read the box again:

powershell -ExecutionPolicy Bypass -File C:\RaysArithmetic\tools\fig-sheet.ps1 -Pages 37

A re-cut figure doesn't change: make-figs.ps1 skips figures that already exist; delete that PNG first, or run it with -Force.
An old version shows after an update: change rays-shell-v16 in sw.js to the next number, deploy, and reload the page twice.

Notes

Stickers are kept per book and per reader, so Book 5 has its own sticker page.
Next on the shelf: Ray's New Elementary Algebra and Ray's New Higher Algebra, which will need new answer types.

Ray's Geometry and Trigonometry as a game (Ray's Arithmetic Book 5 rebuild guide) 15OCT2026

Book 5 of Ray's, Grades 9-10: the whole of the Elements of Geometry and the Trigonometry, every article read aloud beside the book's own figure, with a quiz and number problems for each section