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ELEMENTS
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GEOMETRY;
PHACTICAL A V^Jj I G:A%i QjNS
MENSURATION.
Jbr BENJAMIN GKEENLEAF, A. M.,
AOTHOS OF "THK NATIONAL ARIxmCBTIC," " TREATI8K ON ALQTBMJJ* «XC.
Nineteenth Electrotype Edition. r*fcrJ^ "^
fi?in^
BOSTON:
PUBLISHED BY ROBERT S. DAVIS & CO.
NEW YORK: OAKLEY AND MASON, AND W. I. POOLKY. PHILA.DELPHIA : J. B. LIPPINCOTT AND COMPANY. CINCINNATI: G. 8. BLANCHARD fc CO- ST. LOUIS: KEITH AND WOODS.
1868.
GREENLEAF'S ^7 MATHEMATICAL SERIES.
I. Greenleap's New Primary Arithmetic ; an attractive book of easy lessons for beginners. 84 pp.
II. Greenleaf's New Intellectual Arithmetic ; a late work, with models of analysis. 180 pp.
III. Greenleaf's Common School Arithmetic ; a com- plete «ystem of* Writ^eJii Arithmetic for Common Schools. 324 pp.
IV.' (jtREenleaf's NATiotXAL ARITHMETIC ; a thorough course for-,Hi^ '«ScKools,'Aca3eimee,.* Normal Schools, and Commercial Colleges.' 444 pp.
V. Greenleaf's New Elementary Algebra ; or the First Principles of Analysis for Schools and Academies. 324 pp.
VI. Greenleaf's New Higher Algebra; an advanced Analytical Course,, for High Schools and Colleges. 394 pp.
VII. Greenleaf's Elements of Geometry ; with applica- tions to Mensuration. 320 pp. L^ ~7 "?■ '■^ ■
VIII. Greenleaf's Elements of Trigonometry; with Practical Applications and Tables. 170 pp.
IX. Greenleaf's Geometry and Trigonometry; or the last two works in one volume. 490 pp.
X. Greenleaf's Surveying and Navigation, with Prac- tical Applications and Tables. \In 'pre'paration.\
S:^ Keys to the Arithmetics, Algebras, Geometry and Trigonometry. For Teachers only. 6 volumes.
Entered according to Act of Congress, in tlie year 1868, by
BENJAMIN GREENLEAF,
In the Clerk's Office of ttie District Court of tlie District of Massaohosetts.
KIVEHSIDK, CAMB'^IDGE: PBIMTSP BT B. 0. HOUQHXON AlfD OOMFANT.
PREFACE.
The preparation of this treatise has been undertaken at the earnest solicitation of many teachers, who, having used the au- thor's Arithmetics and Algebra with satisfaction, have been de- sirous of seeing his series rendered more complete by the addition of the Elements of Geometry.
That there are peculiar advantages in a graded series of text- books on the same subject, few, if any, properly qualified to judge, will doubt. The author, therefore, feels justified in intro- ducing this volume to the attention of the public.
In common with most compilers of the present day, he has followed, in the main, the simple and elegant order of arrange- ment adopted by Legendre ; but in the methods of demonstra- tion no particular authority has been closely followed, the aim having been to adapt the work fully to the latest and most ap- proved modes of instruction. In this respect, it is believed, there will be found incorporated a considerable number of important improvements.
More attention than is usual in elementary works of this kind has been given to the converse of propositions. In almost all cases where it was possible, the converse of a proposition has been demonstrated.
The demonstration of Proposition XX. of tiie first book is essentially the one given by M. da Cunha in the Principes Mathi-
IV PREFACE.
matiques, which has justly been pronounced by the highest mathematical authorities to be a very important improvement in elementary geometry. It has, however, never before been intro- duced into a text-book by an American author.
The Application of Geometry to Mensuration, given in the eleventh and twelfth books, are designed to show how the theo- retical principles of the science are connected with manifold practical results.
The Miscellaneous Geometrical Exercises, which follow, are calculated to test the thoroughness of the scholar's geometrical knowledge, besides being especially adapted to develop skill and discrimination in the demonstration of theorems and the solution of problems unaided except by principles.
Sufficient Applications of Algebra to Geometry are given to show the relation existing between these two branches of the mathematics. The problems introduced in connection therewith will be found to be, not only of a highly interesting character, but well calculated to secure valuable mental discipline.
In the preparation of this work the author has received valua- ble suggestions from many eminent teachers, to whom he would here express his sincere thanks. Especially would he acknowl- edge his great obligations to H. B. Maglathlin, A. M., who for many months has been associated with him in his labors, and to whose experience as a teacher, skill as a mathematician, and ability as a writer, the value of this treatise is largely due.
BENJAMIN GREENLEAF.
Bradford, Mass., June 25, 1858.
NOTICE.
A Key, comprising the Solutions of the Problems contained in the last four Books of this Geometry, has been published, for Teachers only ; and the same will be mailed, post-paid, to the address of any Teacher who will forward fifty cents in stamps to the Publishers.
C ONTENTS
PLANE GEOMETRY. BOOK I.
MOP BLEMENTARY PRINCIPLES . ..... 7
BOOK II. RATIO AND PROPORTION <9
BOOK III. THE CIRCLE, AND THE MEASURE OF ANGLES ... 55
BOOK IV.
PROPORTIONS, AREAS, AND SIMILARITY OP FIGURES . . 76
BOOK V. PROBLEMS RELATING TO THE PRECEDING BOOKS . . .118
BOOK VI.
BEGULAR POLYGONS, AND THE AREA OF THE CIRCLE . . 148
1*
Tl CONTENTS.
SOLID GEOMETRY. BOOK VII.
PLANES. — DIEDRAL AND POLYEDRAL ANGLES . . . 165
BOOK VIII.
POLYEDRONS 184
BOOK IX.
THE SPHERE, AND ITS PROPERTIES 214
BOOK X. THE THREE ROUND BODIES . 238
MENSURATION. BOOK XI.
j».^PLlCATIONS OF GEOMETRY TO THE MENSURATION OP
PLANE FIGURES 253
BOOK XII..
APPLICATIONS OF GEOMETRY TO THE MENSURATION OF
SOLIDS '-^8
BOOK XIII.
MISCELLANEOUS GEOMETRICAL EXERCISES . . • 801
BOOK XIV.
APPLICATION OP ALGEBRA TO GEOMETRY . . .811
ELEMENTS TTr^^lTOMETRY.
BOOK I.
ELEMENTARY PRINCIPLES. DEFINITIONS.
1. Geometry is the science of Position and Extension. The elements of position are direction and distance. The dimensions of extension are length, breadth, and
height or thickness.
2. Magnitude, in general, is that which has one or more of the three dimensions of extension.
3. A Point is that which has position, without magni- tude.
4. A Line is that which has length, without either breadth or thickness.
5. A Straight Line, or Right
Line, is one which has the same .
direction in its whole extent; as the line A B.
The word line is frequently used alone, to designate a straight line.
6. A Curved Line is one which ^^^ ..^^^ continually changes its direction ; C ^^^ ^""^ D as the line CD.
The word curve is frequently used to designate a curved line.
8
ELEMENTS OP GEOMETRY.
7. A Broken Line is one which is composed of straight lines, not lying in the same direction ; as the line E F.
8. A Mixed Line is one which is composed of straight lines' kiid of curved Inies.
9. A Surface is that which has length and breadth, witiiout height or thickness.
10. A Plane Surface, or simply a Plane, is one in which any two points being taken, the straight line that joins them will lie wholly in the surface.
11. A Curved Surface is one that is not a plane sur- face, nor made up of plane surfaces.
12. A Solid, or Volume, is that which has length, breadth, and thickness.
ANGLES AND LINES.
13. A Plane Angle, or simply an Angle, is the difference in the direc- tion of two lines, which meet at a point ; as the angle A.
The point of meeting. A, is tlie vertex of the angle, and the lines AB, AC are the sides of the angle.
An angle may be designated, not only by the letter at its vertex, as C, but by three letters, particularly when two or more angles have the same vertex ; as the angle A C D or ^
D C B, the letter at the vertex always occupying the mid- dle place.
The quantity of an angle does not depend upon the length, but entirely upon the position, of the sides ; for the angle remains the same, however the lines containing it be increased or diminished.
BOOK I.
9
D
14. Two straight lines are said to B be perpendicular to each other, when their meeting forms equal adjacent angles ; thus the lines A B and C D are perpendicular to each other. ^ ~
Two adjacent angles, as C AB and BAD, have a com mon vertex, as A ; and a common side, as A B.
15. A Right Angle is one which is formed hy a straight line and a perpendicular to it ; as the angle CAB.
B
D
16. An Acute Angle is one which is less than a right angle ; as the angle D E F.
An Obtuse Angle is one which is greater than a right angle ; as the angle EFG.
Acute and obtuse angles have their sides oblique to each other, and are sometimes called oblique angles.
17. Parallel Lines are such as, being in the same plane, cannot meet, however far either way both of them may be produced ; as the lines A B, CD.
18. When a straight line, as E F, intersects two parallel lines, as AB, CD, the angles formed by the intersecting or secant line take particular names, thus : —
Interior Angles on the same Side are those w^liich lie within the parallels, and on the same
B
10 ELEMENTS OP GEOMETRY.
side of the secant line ; as the angles BGH, GHD, and also AGH, GHC. A-
ExTERiOR Angles on the same Side are those which lie without the parallels, and on the same side of the secant line ; as the angles BGE, DHF, and also the angles AGE, CHF.
Alternate Interior Angles lie within the parallels, and on different sides of the secant line, but are not adja- cent to each other; as the angles BGH, GHC, and also AGH, GHD.
Alternate Exterior Angles lie without the parallels, and on different sides of the secant line, but not adjacent to each other ; as the angles E G B, CHF, and also the angles AGE, DHF.
Opposite Exterior and Interior Angles lie on the same side of the secant line, the one without and the other within the parallels, but not adjacent to each other ; as the angles E G B, GHD, and also EGA, G H C, are, respec- tively, the opposite exterior and interior angles.
PLANE FIGURES.
19. A Plane Figure is a plane terminated on all sides by straight lines or curves.
The boundary of any figure is called its perimeter.
20. When the boundary lines are straight, the space |hey enclose is called a Rectilineal Figure, or Polygon ; as the figure A B C D E.
A B
21. A polygon of three sides is called a triangle ; one of four sides, a quadrilateral ; one of five, a pentagon ; one of six, a hexagon; one of seven, a heptagon; one
BOOK I.
11
of eight, an octagon ; one of nine, a nonagon ; one of ten, a decagon ; one of eleven, an undecagon ; one of twelve, a dodecagon ; and so on.
22. An Equilateral Triangle is one which has its three sides equal ; as the triangle ABC.
An Isosceles Triangle is one which has two of its sides equal ; as the triangle D E F.
A Scalene Triangle is one which has no two of its sides equal ; as the triangle G H I.
^3. A Right-angled Triangle is one which lias a right angle ; as the triangle J K L.
The side opposite to the right angle is called tlie hy- pothenuse ; as the side J L.
24. An Acute-angled Triangle is one which has three acute angles ; as the triangles ABC and D E F, Art. 22.
An Obtuse-angled Triangle is one which has an ob- tuse angle ; as the triangle G H I, Art. 22.
Acute-angled and obtuse-angled triangles are also called oblique-angled triangles.
25. A Parallelogram is a quadrilateral which has its opposite sides parallel.
26. A Rectangle is any parallel- ogram whose angles are riglit angles ; as the parallelogram A B C D. ^
12
ELEMENTS OF GEOMETRY.
A Square is a rectangle whose sides are equal ; as the rectangle EFGH.
27. A Rhomboid is any parallelo- gram whose angles are not right an- gles ; as the parallelogram IJKL.
G
K
A Rhombus is a rhomboid whose sides are equal ; as the rhomboid MNOP.
M^
28. A Trapezoid is a quadrilateral which has only two of its sides par- allel ; as the quadrilateral R S T U..
W
A Trapezium is a quadrilateral which has no two of its sides paral- lel ; as the quadrilateral V WXY.
29. A Diagonal is a line joining the vertices of any two angles which are opposite to each other ; as the lines E C and E B in the polygon ABODE.
30. A Base of a polygon is the side on which the poly- gon is supposed to stand. But in the case of the isosceles triangle, it is usual to consider that side the base which is not equal to either of the other sides.
31. An equilateral polygon is one which has all its sides equal. An equiangular poly^n is one which has
BOOK I. 13
all its angles equal. A re^lar polygon is one which is equilateral and equiangular.
32. Two polygons are mutually equilateral, when all the sides of the one equal the corresponding sides of the other, each to each, and are placed in the same order.
Two polygons are mutually equiangular, when all the angles of the one equal the corresponding angles of the other, each to each, and are placed in the same order.
33. The corresponding equal sides, or equal angles, of polygons mutually equilateral, or mutually equiangular, are called homologous sides or angles.
AXIOMS.
34. An Axiom is a self-evident truth ; such as, —
1. Things which arc equal to the same thing, are equal to each other.
2. If equals be added to equals, the sums will be equal.
3. If equals be taken from equals, the remainders will be equal.
4. If equals be added to unequals, the sums will be unequal.
5. If equals be taken from unequals, the remainders will be unequal.
6. Things which are double of the same thing, or of equal things, are equal to each other.
7. Things which are halves of the same thing, or of equal things, are equal to each other.
8. The whole is greater than any of its parts.
9. The whole is equal to the sum of all its parts.
10. A straight line is the shortest line that can be drawn from one point to another.
11. From one point to another only one straight line can be drawn.
12. Through the same point only one parallel to a straight line can be drawn. •
2
14 ELEMENTS OF GEOMETRY.
13. All right angles are equal to one another.
14. Magnitudes which coincide throughout their whole extent, are equal.
POSTULATES.
35. A Postulate is a self-evident problem ; such as, —
1. That a straight line may be drawn from one point to another.
2. That a straight line may be produced to any length.
3. That a straight line may be drawn through a given point parallel to another straight line.
4. That a perpendicular to a given straight line may be drawn from a point either within or without the line.
5. That an angle may be described equal to any given angle.
PROPOSITIONS.
36. A Demonstration is a course of reasoning by which a truth becomes evident.
37. A Proposition is something proposed to be demon- strated, or to be performed.
A proposition is said to be the converse of another, when the conclusion of the first is used as the supposition in the second.
38. A Theorem is something to be demonstrated.
39. A Problem is something to be performed.
40. A Lemma is a proposition preparatory to the dem- onstration or solution of a succeeding proposition.
41. A Corollary is an obvious consequence deduced from one or more propositions.
42. A Scholium is a remark made upon one or more preceding propositions.
43. An Hypothesis is a supposition, made either in the
BOOK I.
15
enunciation of a proposition, or in the course of a demon- stration.
Proposition I. — Theorem.
44. The adjacent angles which one straight line makes by meeting another straight line, are together equal to two right angles.
Let the straight line D C meet E AB, making the adjacent angles A C D, D C B ; these angles to- gether will be equal to two right angles. A . [/ g
From the point C suppose C E ^
to be drawn perpendicular to A B ; then the angles ACE and E C B will each be a right angle (Art. 15) . But the angle A C D is composed of the right angle ACE and the angle ECD (Art. 34, Ax. 9), and the angles ECD and D C B compose the other right angle, E C B ; hence the angles AC D, D C B together equal two right angles.
45. Cor. 1. If one of the angles A C D, D C B is a right angle, the other must also be a right angle.
46. Cor. 2. All the successive angles, B A C, CAD, DAE, EAF, formed on the same side of a straight line, B F, are equal, when taken together, to two right angles ; for their sum is equal to g that of the two adjacent angles, -^ BAC, CAF.
Proposition II. — Theorem.
47. If one straight line meets two other straight lines at a common point, making adjacent angles, ivhich to- gether are equal to two right angles, the two lines form one and the same straight line.
16
ELEMENTS OF GEOMETRY.
Let the straight line D C meet the two straight lines AC, C B at the common point C, making the adjacent angles A C D, D C B to- gether equal to two right angles ;
then the lines AC and CB will ^ C z:;::;; ^
- -p
form one and the same straight line.
If C B is not the straight line A C produced, let C E be that line produced ; then the line ACE being straight, the sum of the angles A C D and D C E will be equal to two right angles (Prop. I.). But by hypothesis the angles A C D and D C B are together equal to two right angles ; therefore the sum of the angles A C D and D C E must be equal to the sum of the angles A C D and D C B (Art. 34, Ax. 2). Take away the common angle A C D from each, and there will remain the angle D C B, equal to the angle D C E, a part to the whole, which is impossible ; therefore C E is not the line A C produced. Hence A C and C B form one and the same straight line.
Proposition III. — Theorem,
48. Two straight lines, ivhich have tivo points co^nmon, coincide ivith each other throughout their ivhole extent^ and form one and the same straight line.
Let the two points which are F
common to two straight lines be A and B.
The two lines must coincide
between tlie points A and B, for a
otherwise there would be two ^ ^
straight lines between A and B, which is impossible (Art.
34, Ax. 11).
Suppose, however, that, on being produced, the lines begin to separate at the point C, the one taking the direc-
BOOK I; 17
tion C D, and the other C E. From the point C let the line C F be drawn, making, with C A, the right angle A C F. Now, since A C D is a straight line, the angle F C D will be a right angle (Prop. I. Cor. 1) ; and since ACE is a straight line, the angle F C E will also be a right angle ; therefore the angle F C E is equal to the angle F C D (Art. 34, Ax. 13), a part to the whole, which is impossible ; hence two straight lines which have two points common, A and B, cannot separate from each other when produced ; hence they must form one and the same straight line.
Proposition TV. — Theorem.
49. When ttvo straight lines intersect each other, the opposite or vertical angles which they form are equal.
Let the two straight lines A B, ^
C D intersect each other at the point E ; then will the angle AE C be equal to the angle DEB, and the angle CEB to AED.
For the angles AEC, CEB, D
which the straight line C E forms by meeting the straight line AB, are together equal to two right angles (Prop. I.) ; and the angles CEB, BED, which the straight line B E forms by meeting the straight line C D, are equal to two right angles ; hence the sum of the angles AEC, CEB is equal to the sum of the angles CEB, BED (Art. 34, Ax. 1). Take away from each of these sums the common angle CEB, and there will remain the angle AEC, equal to its opposite angle, BED (Art. 34, Ax. 3).
In the same manner it may be shown that the angle C E B is equal to its opposite angle, AED.
50. Cor. 1. The four angles formed by two straight lines intersecting each other, are together equal to four right angles.
2*
18 ELEMENTS OF GEOMETRY.
51. Cor. 2. All the successive angles, around a com- mon point, formed by any number of straight lines meet- ing at that point, are together equal to four right angles.
Proposition Y. — Theorem.
52. If two triangles have tioo sides and the included angle in the one equal to two sides and the included angle in the other, each to each, the two triangles will be equal.
In the two triangles ABC, DEF, let the side A B be equal to the side DE, the side AC to the side D F, and the angle A to the angle D ; then the triangles ABC, DEF will be equal. ^
Conceive the triangle ABC to be applied to the triangle D EF, so that the side AB shall fall upon its equal, D E, the point A upon D, and the point B upon E ; then, since the angle A is equal to the angle D, the side AC will take the direction DF. But AC is equal to D F ; therefore the point C will fall upon F, and the third side B C will co- incide with the third side E F (Art. 34, Ax. 11). Hence tlie triangle ABC coincides with the triangle DEF, and they are therefore equal (Art. 34, Ax. 14).
53. Cor. When, in two triangles, these three parts are equal, namely, the side A B equal to D E, the side A C equal to D F, and the angle A equal to D, the other three corresponding parts are also equal, namely, the side B C equal to E F, the angle B equal to E, and the angle C equal to F.
Proposition YI. — Theorem.
54. If two triangles have two angles and the included side in the one equal to two angles and the included side in the other, each to each, the two triangles will he equal.
BOOK L
19
In the two triangles ABC, DEF, let the angle B be equal to the angle E, the angle C to the angle F, and the side BC to the side EF; then the triangles ABC, DEF will be equal.
Conceive the triangle A B C to be applied to the triangle DEF, so that the side B C shall fall upon its equal, E F, the point B upon E, and the point C upon F. Then, since the angle B is equal to the angle E, the side B A will take the direction E D ; therefore the point A will be found somewhere in the line ED. In like manner, since the angle C is equal to the angle F, the line C A will take the direction F D, and the point A will be found some- where in the Ime F D. Hence the point A, falling at the same time in both of the straight lines E D and F D, must fall at their intersection, D. Hence the two triangles ABC, DEF coincide with each other, and are therefore equal (Art. 34, Ax. 14).
55. Cor, When, in two triangles, these three parts are equal, namely, the angle B equal to the angle E, the angle C equal to tlie angle F, and the side B C equal to the side EF, the other three corresponding parts are also equal; namely, the side B A equal to E D, the side C A equal to F D, and the angle A equal to the angle D.
Proposition YII, — Theorem.
56. In an isosceles triang'le, the angles opposite the equal sides are equal.
Let ABC be an isosceles triangle, in which the side AB is equal to the side A C ; then will the angle B be equal to the angle C.
Conceive the angle B A C to be bisect- ed, or divided into two equal parts, by ^
20
ELEMENTS OF GEOMETRY.
the straight line AD, making the angle
BAD equal to D A C. Then the two
triangles BAD, CAD have the two
sides A B, A D and the included angle
in the one equal to the two sides A C,
A D and the included angle in the other, B
each to each ; hence the two triangles are equal, and the
angle B is equal to the angle C (Prop. V.).
57. Cor. 1. The line bisecting the vertical angle of an isosceles triangle bisects the base at right angles.
58. Cor. 2. Conversely, the line bisecting the base of an isosceles triangle at right angles, bisects also the verti- cal angle.
59. Cor. 3. Every equilateral triangle is also equian- gular.
Proposition YIII. — Theorem.
60. ff two angles of a triangle are equal, the opposite sides are also equal, and the triangle is isosceles.
Let A B C be a triangle having the an- A
gle B equal to the angle C ; then will the side A B be equal to the side A C.
For, if the two sides are not equal, one of them must be greater than the other. Let A B be the greater ; then take D B equal to AC the less, and draw CD. -^ ^
Now, in the two triangles D B C, A B C, we have D B equal to A C by construction, the side- B C common, and the angle B equal to the angle A C B by hypothesis ; therefore, since two sides and the included angle in the one are equal to two sides and the included angle in the other, each to each, the triangle D B C is equal to the triangle ABC (Prop. Y.), a part to the whole, wliich is impossible (Art. 34, Ax. 8). Hence tlie sides AB and A C cannot be unequal ; therefore the triangle A B C is isosceles.
BOOK I.
21
61. Cor. lateral.
Therefore every equiangular triangle is equi-
B
Proposition IX. — Theorem.
62. Any side of a triangle is less than the sum of the other two.
In the triangle ABC, any one side, C
as AB, is less than the sum of the other two sides, A C and C B.
For the straight line AB is the shortest line that can be drawn from the point A to the point B (Art. 34, Ax. 10) ; hence the side A B is less than the sum of the sides A C and C B.
In like manner it may be proved that the side A C is less than the sum of A B and B C, and the side B C less than the sum of B A and A C.
63. Cor. Since the side AB is less than the sum of AC and C B, if we take away from each of tliese two unequal s the side C B, we shall have the difference between A B and C B less than A C ; that is, the difference between any tivo sides of a triangle is less than the other side.
Proposition X. — Theorem.
64. The greater side of any triangle is opposite the greater angle.
In the triangle CAB, let the angle A C be greater than B ; then will the side A B, opposite to C, be greater than A C, opposite to B.
Draw the straight line C D, making the angle BCD equal to B. Then, in the triangle B D C, we shall have the side BD equal to D C (Prop. VIII.). But the side A C is less than the sum of AD and D C (Prop. IX.), and the
22 ELEJVIENTS OF GEOMETRY.
sum of A D and D C is equal to the sum of A D and D B, which is equal to A B ; therefore the side A B is greater than AC.
65. Cor. 1. Therefore the shorter side is opposite to the less angle.
66. Cor. 2. In the right-angled triangle the hypothe- nuse is the longest side.
Proposition XI. — Theorem.
67. The greater angle of any triangle is opposite the greater side.
In the triangle CAB, suppose the a side A B to be greater than A C ; then will the angle C, opposite to A B, be greater than the angle B, opposite to AC.
For, if the angle C is not greater than B, it miist either be equal to it or less. If the angle C were equal to B, then would the side AB be equal to the side AC (Prop. VIII.), which is contrary to the hypothesis ; and if the angle C were less than B, then would the side AB be less than AC (Prop. X. Cor. 1), which is also contrary to the hypothesis. Hence, the angle C must be greater than B.
68. Cor. It follows, therefore, that the less angle is opposite to the shorter side.
Proposition XII. — Theorem,
69. If, from any point within a triangle, tioo straight lines are draivn to the extremities of either side, their sum will he less than that of the other two sides of the triangle.
BOOK I.
23
Let the two straight lines B 0, C 0 be drawn from the point O, within the triangle ABC, to the extremities of the side B C ; then will the sum of the two lines B 0 and 0 C be less than the sum of the sides B A and A C.
Let the straight line BO be pro- duced till it meets the side A C in the point D ; and be- cause one side of a triangle is less than the sum of the other two sides (Prop. IX.), the side OC in the triangle C D 0 is less than the sum of 0 D and DC. To each of these inequalities add B 0, and we have the sum of B O and 0 C less than the sum of BO, OD, and DC (Art. 34, Ax. 4) ; or the sum of B 0 and 0 C less than the sum of B D and D C. Again, because the side B D is less than the sum of B A and A D, by adding D C to each, we have the sum of B D and D C less than the sum of B A and A C. But it has been just shown that the sum of B 0 and 0 C is less than the sum of B D and D C ; much more, then, is the sum of B 0 and 0 C less than B A and A C.
Proposition XIII. — Theorem.
70. From a point ivithout a straight line, only one per- pendicular can be drawn to that line.
Let A be the point, and DE the given straight line ; then from the point A only one perpendicular can be drawn to D E:
Let it be supposed that we can draw two perpendiculars, AB and AC. Produce one of them, as AB, till BF is equal to AB, and join F C. Then, in the triangles ABC and C B F, the angles C B A and CBF are both right angles (Prop. I. Cor. 1), the side C B is common to both, and the side B F is equal to
24
ELEMENTS OF GEOMETRY.
the side A B ; hence the two triangles are equal, and the angle B C F is equal to the angle B C A (Prop. Y .) But the angle BCA is, by hypothesis, a right angle ; therefore B C F must also be a right angle ; and if the two adja- cent angles, BCA and B C F, are to- gether equal to two right angles, the two lines A C and C F must form one and the same straight line (Prop. II.). Whence it follows, that be- tween the same two points, A and F, two straight lines can be drawn, which is impossible (Art. 34, Ax. 11) ; hence no more than one perpendicular can be drawn from the same point to the same straight line.
71. Cor. At the same point C, in the line AB, it is likewise impossible to erect more than one perpendicular to that line. For, if C D and C E were each perpendicular to A B, the angles BCD, B C E would be right angles '/^ ^
hence the angle BCD would be equal to the angle B C E, a part to the whole, which is impossible.
E
D
/
/
Proposition XIY. — Theorem.
72. If^ from a point without a straight line, a perpen- dicular be let fall on that line, and oblique lines be drawn to different points in the same line ; —
1st. The perpendicular will be shorter than any oblique line.
2d. Any tiuo oblique lines, which meet the given line at equal distances from the perpendicular, will be equal.
3d. Of any two oblique lines, that vjhich meets the given line at the greater distance from the perpendicular will be the longer.
BOOK I.
25
Let A be the given point, an4. D E the given straight line. Draw A B perpendicular to D E, and the oblique lines AE, AC, AD. Produce AB till BF is equal to A B, and join C F, D F.
First. The triangle B C F is equal to the triangle B C A, for -^
they have the side C B common, the side A B equal to the side B F, and the angle ABC equal to the angle F B C, both being right angles (Prop. I. Cor. 1) ; hence the third sides, CF and AC, are equal (Prop. V. Cor.). But A B F, being a straight line, is shorter than A C F, which is a broken line (Art. 34, Ax. 10) ; therefore A B, the half of A B F, is shorter than A C, the half of A C F ; hence the perpendicular is shorter than any oblique line.
Secondly. If B E is equal to B C, then, since A B is common to the triangles, ABE, ABC, and the angles ABE, ABC are right angles, the two triangles are equal (Prop. Y.), and the side A E is equal to the side AC (Prop. V. Cor.). Hence the two oblique lines, meeting the given line at equal distances from the perpendicular, are equal.
Thirdly. The point C being in the triangle ADF, the sum of the lines A C, C F is less than the sum of the sides AD, DF (Prop. XII.) But AC has been shown to be equal to C F ; and in like manner it may be shown that A D is equal to D F. Therefore A C, the half of the line A C F, is shorter than AD, the half of the line ADF; hence the oblique line which meets the given line the greater distance from the perpendicular, is the longer.
73. Cor. 1, The perpendicular measures the shortest distance of any point from a straight line.
74. Cor. 2. From the same point to a giren straight line only two equal straight lines can be drawn.
26 ELEMENTS OF GEOMETRY.
75. Cor. 3. Of any two straight lines drawn from a point to a straight line, that whicli is not shorter than the other will be longer than any straight line that can be drawn between them, from the same point to the same line.
Proposition XY. — Theorem.
76. If from the middle point of a straight line a per- pendicular to this line be drawn, —
1st. Any point in the perpendicular will he equally dis- tant from the extremities of the line.
2d. Any point out of the perpendicular will be un- equally distant from those extremities.
Let D C be drawn perpendicular to the straight line A B, from its middle point C.
First. Let D and E be points, taken at pleasure, in the perpendicular, and join DA, DB, and also AE, EB. Then, since A C is equal to C B, the two oblique lines D A, D B meet points which are at the same distance from the perpendicular, and are therefore equal (Prop. XIY.). So, likewise, the two oblique lines E A, E B are equal ; therefore any point in the perpendic- ular is equally distant from the extremities A and B.
Secondly. Let F be any point out of the perpendicular, and join F A, F B. Then one of those lines must cut the perpendicular, in some point, as E. Join E B ; then we have EB equal to EA. But in the triangle F E B, the side F B is less than the sum of the sides E F, E B (Prop. IX.), and since the sum of FE, E B is equal to the sum of F E, E A, which is equal to F A, F B is less than F A. Hence any point out of the perpendicular is at unequal distances from the extremities A and B.
77. Cor. If a straight line have two points, of whicli each is equally distant from the extremities of another
BOOIv I.
27
straight line, it will be perpendicular to that line at its middle pohit.
Proposition XYI. — Theorem.
78. If tivo triangles have tioo sides of the one equal to two sides of the other, each to each, and the included an- gle of the one greater than the included angle of the other, the third side of that which has the greater angle will be greater than the third side of the other.
Let ABC, DEF be two triangles, having the side A B equal to D E, and AC equal to D F, and the angle A greater than D ; ^ then will the side BC be greater than EF.
Of the two sides D E, D F, let D F be the side which is not shorter than the other ; make the angle E D G equal to B A C ; and make D G equal to A C or D F, and join EG, GF.
Since D F, or its equal D G, is not shorter than D E, it is longer than D H (Prop. XIY. Cor. 3) ; therefore its extremity, F, must fall below the line E G. The two tri- angles, ABC and D E G, have the two sides AB, AC equal to the two sides D E, D G, each to each, and the '.eluded angle BAC of the one equal to the included angle E D G of the other ; hence the side B C is equal to EG (Prop. Y. Cor.).
In the triangle D F G, since D G is equal to D F, the angle D F G is equal to the angle D G F (Prop. YII.) ; but the angle D G F is greater than the angle E G F ; therefore the angle D F G is greater than E G F, and much more is the angle EFG greater than the angle
28 ELEMENTS OF GEOMETRY.
E G F. Because tlie angle E F G in the triangle E F G is greater than EGF, and because the greater side is opposite the greater angle (Prop. X.), the side E G is greater than E F ; and E G has been shown to be equal to B C ; hence B C is greater than E F.
Proposition XYII. — Theorem. .
79. If two triangles have tivo sides of the one equal to tiuo sides of the other ^ each to each, but the third side of the one greater than the third side of the other, the angle coyitained by the sides of that ivhich has the greater third side ivill be greater than the angle contained by the sides of the other.
Let ABC, DEF be two triangles, the side AB equal to D E, and AC equal to DF, and the side C B greater than EF, then will the angle b~~ ^q
A be greater than D. ^"^ F
For, if it be not greater, it must either be equal to it or less. But the angle A cannot be equal to D, for tlicn the side B C would be equal to E F (Prop. Y. Cor.), which is contrary to the hypothesis ; neither can it be loss, for then the side B C would be less than E F (Prop. XVI.), which also is contrary to the hypothesis ; there- fore the angle A is not less than the angle D, and it has been shown that is not equal to it ; hence the angle A must be greater than the angle D.
Proposition XVIII. — Theorem.
80. If tioo triangles have the three sides of the one equal to the three sides of tlie other, each to each, the triangles themselves will be equal.
BOOK
29
Let the triangles ABC, D E F have the side A B equal to D E, A C to D F, and B C to E F ; then will the angle A be equal to D, the angle B to the angle
E, and the angle C to the angle F, and tlie two triangles will also be equal.
For, if the angle A were greater than the angle D, since the sides A B, A C are equal to the sides D E, D F, each to each, the side B C would be greater than E F (Prop. XVI.) ; and if the angle A were less than D, it would follow that the side B C would bo less than E F. But by hypothesis B C is equal to E F ; hence tlie angle A can neither be greater nor less than D ; therefore it must bo equal to it. In the same manner, it may be shown that the angle B is equal to E, and the angle C to F ; hence tlie two triangles must be equal.
81. Scholium. In two triangles equal to each other, the equal angles are opposite the equal sides ; thus the equal angles A and D are opposite the equal sides B C and EP.
Proposition XIX. — Theorem.
82. If Uvo right-angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, each to each, the triangles are equal.
Let the two right-an- A D
gled triangles ABC, DEF, have the hypothenuse A C equal to D F, and the side A B equal to D E ; then will the triangle A B C be equal to the triangle DEF. B G C E F
The two triangles are evidently equal, if the sides B C and EF are equal (Prop. XVIII.) . If it be possible, let
3*
30
ELEMENTS OF GEOMETRY.
these sides be unequal, and let B C be the greater. Take B G equal to E F, the less side, and join A G. Then, in the two triangles A B G, D E F, the angles B and E are equal, both B G C E F
being right angles, the side A B is equal to D E by hy- pothesis, and the side B G to E F by construction ; hence these triangles are equal (Prop. V.) ; and therefore A G is equal to D F. But by hypothesis D F is equal to A C, and therefore A G is equal to A C. But the oblique line A C cannot be equal to A G, which meets the same straight line nearer the perpendicular A B (Prop. XIV.) ; there- fore B C and E F cannot be unequal, hence they must be equal; therefore the triangles ABC and DEF are equal.
E
Proposition XX. — Theorem.
SS. If a straight line, intersecting tivo other straight lines, makes the alternate angles equal, the two lines are parallel.
Let the straight line EF intersect the two straight lines A B, CD, making the alternate an- gles BGH, CHG equal; I then the lines A B, C D will be parallel.
For, if the lines A B, C D are not parallel, let them meet in some point K, and through 0, the middle point of GH, draw the straight line IK, making 10 equal to 0 K, and join H I. Then the opposite angles K 0 G, I 0 H, formed by the intersection of the two straight lines IK, GH, are equal (Prop. IV.) ; and the triangles K 0 G,
K
BOOK I. 31
1 0 H have the two sides K 0, 0 (x and the included angle in the one equal to the two sides 1 0, 0 H and tlie includ- ed angle in the other, each to each ; hence the angle K G 0 is equal to the angle I H 0 (Prop. V. Cor.). But, by hypothesis, the angle KGO is equal to the angle CHO, therefore the angle 1 H 0 is equal to CHO, so that H I and H C must coincide ; that is, the line C D when pro- duced meets IK in two points, I, K, and yet does not form one and the same straight line, which is impossible (Prop. III.) ; therefore the lines A B, CD cannot meet, conse- quently they are parallel (Art. 17).
Note. — The demonstration of the proposition is substantially that given by M. da Cunha in the Prmcipes Malhematiques. This demon- stration Young pronounces " superior to every other that has been given of the same proposition " ; and Professor Playfair, in the Edinburgh lie- view, Vol. XX., calls attention to it, as a most important improvement in elementary Geometry.
Peoposition XXI. — Theorem.
84. If a straight line, intersecting tivo other straight lines, makes any exterior angle equal to the interior and opposite angle, or makes the interior angles on the same side together equal to tiuo right angles, the two lines are parallel.
Let the straight line E F inter- E
sect the two straight lines A B, \
C D, making the exterior angle ^ yr,
E Gr B equal to the interior and opposite angle, G H D ; then the lines A B, CD are parallel. ^ 1^^- d
For the angle A G H is equal \
to the angle E G B (Prop. IV.) ; F
and E G B is equal to G H D, by hypothesis ; therefore the angle A G II is equal to the angle G H D ; and they are alternate angles ; hence the lines A B, CD are parallel (Prop. XX.).
32 ELEMENTS OF GEOMETRY.
Again, let the interior angles on the same side, B GH, GH D, be together equal to two right angles ; then the lines A B, C D are parallel.
For the sum of the angles BGH, GHD is equal to two right angles, by hypothesis ; and F
the sum of AG H, BGH is also equal to two right an- gles (Prop. I.) ; take away BGH, which is common to both, and there remains the angle GHD, equal to the angle A G H ; and these are alternate angles ; hence the lines AB, CD are parallel.
E
B
D
85. Cor, If two straight lines A- are perpendicular to another, they are parallel ; thus AB, CD, per- pendicular to E F, are parallel. *
Proposition XXH. — Theorem.
86. If a straight line intersects two parallel lines, it viakes the alternate angles equal; also any exterior angle equal to the interior and opposite angle ; and the tivo in- terior angles upon the same side together equal to two right angles.
Let the straight line E F inter- sect the parallel lines AB, CD ; the alternate angles AGH, GHD are equal ; the exterior angle E G B is equal to the interior and
opposite angle GHD; and the q l\ d
two interior angles BGH, GHD upon the same side are together equal to two right angles.
L
A Y-|-- r-B
K
Ji^
BOOK I. 33
For if the angle A G H is not equal to G H D, draw the straiglit line K L through the point G, making the angle K G H equal to G H D ; then, since the alternate angles GHD, KGH are equal, KL is parallel to CD (Prop. XX.) ; but by hypothesis A B is also parallel to C D, so that through the same point, G, two straight lines are drawn parallel to C D, which is impossible (Art. 34, Ax. 12). Hence the angles A G H, GHD are not unequal ; that is, they are equal.
Now, the angle E G B is equal to the angle A G H (Prop. IV.)? ^i^d AGH has been shown to be equal to GHD; hence E G B is also equal to G H D.
Again, add to each of these equals the angle B G H ; then the sum of tlie angles E G B, B G H is equal to the sum of the angles B G H, G H D. But E G B, B G H are equal to two right angles (Prop. I.) ; hence B G H, GHD are also equal to two right angles.
87. Cor. If a line is perpendicular to one of two parallel lines, it is perpendicular to the other ; thus EF (Art. 85), perpendicular to A B, is perpendicular to CD.
Proposition XXIII. — Theorem.
88. If tvw straight lines intersect a third line, and make the two interior angles on the same side together less than tivo right angles, the tivo lines ivill meet on being j)roduced.
Let the two lines KL, CD make j.
with EF the angles KGH, GHC, togetlicr less than two right angles ; then KL and CD will meet on being produced.
For if tliey do not meet, they are parallel (Art. 17). But they are not parallel ; for then the sum
34
ELEMENTS OF GEOMETRY.
of tlie interior angles K G IT, G II C would be equal to two right angles (Prop. XXII.) ; bat by hypothesis it is less ; therefore the lines K L, CD will meet on being- prod u cod.
89. Scholium. The two lines K L, CD, on being pro- duced, must meet on the side of E F, on which are the two interior angles whose sum is less than two right angles.
E
Proposition XXIY. — Theorem.
90. Slraig-M lines which are parallel to the same line are parallel to each other.
Let the straight lines A B, C D be each parallel to the line E F ; then are they parallel to each other.
Draw GHI perpendicular to EF. Then, since AB is parallel to EF, Gl will be perpendicular to AB (Prop. XXII. Cor.) ; and since C D is parallel to EF, GI will for a like reason be perpendicular to CD.
n
D
B
straight line ;
Conse- the same hence they are parallel (Prop. XXI. Cor.).
quently A B and C D are perpendicular to
H
G
Proposition XXV. — Theorem.
91. Tvw parallel straight lines are everywhere equallij distant from each other.
Let AB, CD be two parallel straight lines. Through any two points in A B, as E and F, draw the straight lines E G, F H, per- pendicular to A B. These lines will be equal to each other.
For, if GF be joined, the angles GF E, F G H, consid- ered in reference to the parallels AB, CD, will be alter-
E
D
B
BOOK I. 85
nate interior angles, and tlierefore equal to each other (Prop. XXII.). Also, since the straight lines E G, F H are perpendicular to the same straiglit line AB, and con- sequently parallel (Prop. XXI. Cor.), the angles EGF, GFli, considered in reference to the parallels EG, FH, will be alternate interior angles, and therefore equal. Hence, the two triangles E F G, F G H, have a side and the two adjacent angles of the one equal to a side and the two adjacent angles of the other, eacli to each ; therefore these triangles are equal (Prop. VI.) ; hence the side E G, which measures the distance of the parallels A B, CD, at the point E, is equal to the side F H, which measures the distance of the same parallels at the point F. Hence two parallels are everywhere equally distant.
Proposition XXVI. — Theorem.
92. If Uvo angles have their sides parallel, each to each, and lying^ in the same direction, the two angles are equal.
Let A B C, D E F be two angles, ^ j^
which have the side AB parallel / /
to HE, and BC parallel to EF; / ^/ ^
then these angles are equal. / 7
For produce HE, if necessary, ^ 7^ CJ
till it meets B C in the point G. h ^ F
Then, since E F is parallel to G C,
the angle D E F is equal to D G C (Prop. XXII.) ; and since D G is parallel to A B, the angle H G C is equal to ABC; hence the angle H E F is equal to ABC.
93. Scholiu7n. This proposition is restricted to the case where the side E F lies in the same direction with B C, since if F E were produced toward H, the angles D E H, ABC would only be equal when they are right angles.
36
ELEMENTS OF GEOMETRY.
Proposition XXYII. — Theorem.
94. If any side of a triang-Ie be produced^ the exterior ang-le is equal to the sum of the tivo interior and opposite angles.
Let ABC be a triangle, and let one of its sides, B C be pro- duced towards D ; then the ex- terior angle A C D is equal to the two interior and opposite g angles, CAB, ABC.
For, draw E C parallel to the side AB ; then, since AC meets the two parallels A B, EC, the alternate angles B AC, A C E are equal (Prop. XXII.).
Again, since BD meets the two parallels AB, EC, the exterior angle E C D is equal to the interior and opposite angle ABC. But the angle ACE is equal to BAC; therefore, the whole exterior angle A C D is equal to the two interior and opposite angles CAB, ABC (Art. 34, Ax. 2).
Proposition XXYIII. — Theorem.
95. In every triangle the sum of the three angles is equal to two right angles.
Let A B C be any triangle ; then will the sum of the angles ABC, BCA, CAB be equal to two right angles.
For, let the side B C be pro- g duced towards D, making the exterior angle A C D ; then the angle A C D is equal to CAB and ABC (Prop. XXVII.). To each of these equals add the angle A C B, and we shall have the sum of
BOOK I. 37
A C B and A C D, equal to the sum of A B C, B C A, and CAB. But the sum of A C B and A C D is equal to two right angles (Prop. I.) ; hence the sum of the three an- gles ABC, B C A, and C A B is equal to two right angles (Art. 34, Ax. 2).
96. Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.
97. Cor. 2. If two angles in one triangle be respective- ly equal to two angles in another, their third angles will also be equal.
98. Cor. 3. A triangle cannot have more than one an- gle as great as a right angle.
99. Cor. 4. And, therefore, every triangle must have at least two acute angles.
100. Cor. 5. In a right-angled triangle the right angle is equal to tlie sum of the other two angles.
101. Cor. 6. Since every equilateral triangle is also equiangular (Prop. VII. Cor. 3), each of its angles will be equal to two thirds of one right angle.
Proposition XXIX. — Theorem.
102. The sum of all the interior angles of any polygon is equal to tivice as many right angles, less four, as the figure has sides.
Let A B C D E be any polygon ; then the sum of all its interior angles. A, B, C, D, E, is equal to twice as many right angles as the figure has sides, less four right angles.
For, from any point P within the pol- A B
ygon, draw tlie straight lines PA, PB, PC, P D, P E, to the vertices of all the angles, and the polygon will be
4
38 ELEMENTS OF GEOMETRY.
divided into as many triangles as it has sides. Now, the sum of the three angles in each of these triangles is equal to two right angles (Prop. XXVIII.) ; therefore the sum of the angles of all these triangles is equal to twice as many right angles as there are triangles, or sides, to the polygon. But the sum of all the angles about the point P is equal to four right angles (Prop. TV. Cor. 2), which sum forms no part of the interior angles of the polygon ; therefore, deducting the sum of the angles about the point, there remain the angles of the polygon equal to twice as many right angles as the figure has sides, less four right angles.
103. Cor. 1. The sum of the angles in a quadrilateral is equal to four right angles ; hence, if all the angles of a quadrilateral are equal, each of them is a right angle ; alsoy if three of the angles are right angles, the fourth is likewise a right angle.
104. Cor. 2. The sum of the angles in a pentagon is equal to six right angles ; in a hexagon^ the sum is equal to eight right angles, &c.
105. Cor. 3. In every equiangular figure of more than four sides, each angle is greater than a right angle ; thus, in a regular pentagon^ each angle is equal to one and one fifth right angles ; in a regular hexagon^ to one and one third right angles, <fec.
106. Scholium. In applying this prop- d osition to polygons which have re-en- trant angles, or angles whose vertices are directed inward, as BPC, each of these angles must be considered greater than two right angles. But, in order to avoid ambiguity, we shall hereafter limit our reasoning to polygons with salient angles, or with angles directed outwards, and which may be called convex polygons. Every convex polygon is such that a
BOOK I. 39
straight line, however drawn, cannot meet the perimeter of the polygon in more than two points.
Proposition XXX. — Theorem.
107. The sutn of all the ex^rior angles of any pohjgon^ formed hy producing each side in the same direction^ is equal to four right angles.
Let each side of the polygon ABODE he produced in the same direction ; then the sum of the exterior angles A, B, C, D, E, will be equal to four right angles.
For each interior angle, together with its adjacent exterior angle, is equal to two right angles (Prop. 1.) ; hence the sum of all tlie angles, both interior and exterior, is equal to twice as many right angles as there are sides to the polygon. But the sum of the interior angles alone, less four right angles, is equal to the same sum (Prop. XXIX.) ; therefore the sum of the exterior angles is equal to four right angles.
Proposition XXXI. — Theorem.
108. The opposite sides and angles of every parallelo- gram are equal to each other.
Let A B C D be a parallelogram ; p) C
then the opposite sides and angles are y^x 7
equal to each other. / ... /
Draw the diagonal B D, then, since /_ x /
the opposite sides AB, DC are paral- A B
lei, and BD meets them, the alternate angles ABD, BDC are equal (Prop. XXII.) ; and since AD, BC are parallel, and B D meets them, the alternate angles A D B, D B C are likewise equal. Hence, the two triangles AD B, D B C liave two angles, A B D, A D B, in tlie one, equal to two angles, BDC, DBC, in the otlier, eacli to each; and since
40 ELEMENTS OF GEOMETRY.
the side BD included between these equal angles is common to the two tri- angles, they are equal (Prop. VI.) ; hence the side AB opposite the angle A D B is equal to the side D C opposite the angle DBG (Prop. Yl. Cor.) ; and, in like manner, the side A D is equal to the side B C ; hence the opposite sides of a parallelogram are equal.
Again, since the triangles are equal, the angle A is equal to the angle C (Prop. VI. Cor.) ; and since the two angles D B C, A B D are respectively equal to the two angles A D B, B D C, the angle ABC is equal to the angle ADC.
109. Cor. 1. The diagonal divides a parallelogram into two equal triangles.
110. Cor. 2. The two parallels AD, B C, included be- tween two other parallels, AB, CD, are equal.
Proposition XXXII. — Theorem.
111. If the opposite sides of a quadrilateral are equal, each to each, the equal sides are parallel, and the figure is a parallelogram.
Let A B C D be a quadrilateral D C
having its opposite sides equal ; then /\ 7
will the equal sides be parallel, and / \ / the figure be a parallelogram. / jj /
For, having drawn the diagonal ^ ^
B D, the triangles A B D, B D C have all the sides of tlie one equal to the corresponding sides of the other ; there- fore they are equal, and the angle A D B opposite the side A B is equal to D B C opposite C D (Prop. XVIII. Sch.) ; hence the side A D is parallel to B C (Prop. XX.). For a like reason, AB is parallel to C D ; therefore the quad- rilateral A B C D is a parallelogram.
BOOK I. 41
Proposition XXXIII. — Theorem.
112. If two opposite sides of a quadrilateral are equal and parallel, the other sides are also equal and parallel, and the figure is a parallelogram.
Let A B C D be a quadrilateral, D C
having the sides A B, CD equal and ^\ 7
parallel ; then will the other sides / /
also be equal and parallel. [_ '''■■■■. /
Draw the diagonal BD ; then, since ^ ^
A B is parallel to C D, and B D meets them, tlie alternate angles ABD, BD C are equal (Prop. XXII.) ; moreover, in the two triangles ABD, D B C, the side B D is com- mon ; therefore, two sides and the included angle in the one are equal to two sides and the included angle in the other, each to each ; hence these triangles are equal (Prop. Y.), and the side A D is equal to B C. Hence the angle A D.B is equal to D B C, and consequently A D is parallel to B C (Prop. XX.) ; therefore the figure A B C D is a parallelogram.
Proposition XXXIV. — Theorem.
113. The diagonals of every parallelogram bisect each other.
Let A B C D be a parallelogram, D C
and AC, DB its diao'onals, intersect- /x
ing at E ; then will A E equal EC, / .'.:.-:;.;'
and B E equal E D. /■•--••""" "'•-•
For, since A B, C D are parallel, A and B D meets them, the alternate angles C D E, ABE are equal (Prop. XXII.) ; and since A C meets the same parallels, tlie alternate angles BAE, ECD are also equal; and the sides AB, CD arc equal (Prop. XXXI.). Hence the triangles ABE, CDE have two angles and tlie in-
4*
42 ELEMENTS OF GEOMETRY.
eluded side in the one equal to two angles and the includ- ed side in the other, each to each ; hence the two triangles are equal (Prop. VI.) ; therefore the side A E opposite the angle A B E is equal to C E opposite C D E ; hence, also, the sides B E, D E opposite the other equal angles are equal.
114. Scholium. In the case of a rhom- bus, the sides A B, B C being equal, the triangles A E B, E B C have all the sides of tlie one equal to the corresponding sides of the other, and are, therefore, equal ; whence it follows that the angles A E B, BEG are equal. Therefore the diagonals of a rhombus bisect each other at right angles.
Peoposition XXXY. — Theorem.
115. If the diagonals of a quadrilateral bisect each other ^ the figure is a parallelogram.
Let A B C D be a quadrilateral, and D C
AC, D B its diagonals intersecting at E ; /\ .■■•■-""/
then will the figure be a parallelogram. / j'>C /
For, in the two triangles ABE, CDE, /•••••-•"" '''•'••.../
the two sides A E, E B and the hicluded ^ ^
angle in the one are equal to the two sides C E, E D and the included angle in the other ; hence the triangles are equal, and the side AB is equal to the side CD (Prop. Y. Cor.). For a like reason, A D is equal to C B ; therefore the quadrilateral is a parallelogram (Prop. XXXII.).
BOOK II.
RATIO AND PROPORTION. DEFINITIONS.
116. Ratio is the relation, in respect to quantity, which one magnitude bears to another of tlie same kind ; and is tlie quotient arising from dividing the first by the second.
A ratio may be written in the form of a fraction, or witli the sign : .
Thus the ratio of A to B may be expressed either by
g, or by A : B.
IIT. The two magnitudes necessary to form a ratio are called the terms of tlie ratio. The first term is called the ANTECEDENT, and tlic last, the consequent.
118. Ratios of magnitudes may be expressed by num- bers, either exactly, or approximately.
This may be illustrated by the operation of finding the numerical ratio of two straight lines, AB, CD.
From the crreater line ^
O Q, , ,jy
AB cut off a part equal ^
to the less C D, as many
times as possible ; for ex- E G
ample, twice, with the remainder B E.
From the line C D cut off a part equal to the remainder B E as many times as possible ; once, for example, with the remainder D F.
From the first remainder B E, cut off a part equal to the second D F, as many times as possible ; once, for ex- ample, with tlie remainder B G.
44 ELEMENTS OF GEOMETRY.
-• ^D
A*-
From the second re- mainder DF, cut off a part equal to B G, the
thh^d, as many times as ^' ' 15 G~
possible.
Proceed tlms till a remainder arises, which is exactly contained a certain number of times in the preceding one.
Then this last remainder will be the common measure of the proposed lines ; and, regarding it as unity, we sliall easily find the values of the preceding remainders ; and, at last, those of the two proposed lines, and hence their ratio in numbers.
Suppose, for instance, we find G B to be contained ex- actly twice in F D ; B G will be the common measure of the two proposed lines. Let B G equal 1 ; then will F D equal 2.. But EB contains FD once, plus GB; there- fore we have E B equal to 3. CD contains E B once, plus FD ; therefore we have CD equal to 5. AB con- tains C D twice, plus E B ; therefore we have A B equal to 13. Hence the ratio of the two lines is that of 13 to 5. If the line C D were taken for unity, the line A B would be V- ; if A B were taken for unity, C D would be i\.
It is possible that, however far the operation be con- tinued, no remainder may be found which shall be con- tained an exact number of times in the preceding one. In that case there can be obtained only an approximate ratio, expressed in numbers, more or less exact, according as the operation is more or less extended.
119. When the greater of two magnitudes contains the less a certain number of times without having a remain- der, it is called a multiple of the less ; and the less is then called a submultiple, or measure of the greater.
Thus, 6 is a multiple of 2 ; 2 and 3 are submultiples, or measures, of 6.
120. Equimultiples, or like multiples, are those which contain their respective submultiples the same number of
BOOK II. 45
times ; and equisubmultiples, or like submultiples, are tliose contained in their respective multiples the same number of times.
Thus 4 and 5 arc like submultiples of 8 and 10 ; 8 and 10 are like multiples of 4 and 5.
121. Commensurable magnitudes are magnitudes of the same kind, which have a common measure, and whose ratio therefore may be exactly expressed in numbers.
122. Incommensurable magnitudes are magnitudes of the same kind, which have no common measure, and whose ratio, therefore, cannot be exactly expressed m numbers.
123. A direct ratio is the quotient of the antecedent by the consequent ; an inverse ratio, or recipIiocal ratio, is the quotient of the consequent by the antecedent, or the reciprocal of the direct ratio.
Thus the direct ratio of a line 6 feet long to a line 2 feet long is I or 3 ; and the inverse ratio of a line 6 feet long to a line 2 feet long is | or ~, which is the same as the reciprocal of 3, the direct ratio of 6 to 2.
The word ratio when used alone means the direct ratio.
124. A compound ratio is the product of two or more ratios.
Thus the ratio compounded of A : B and C : D is A C A X C B -^ 5' ^'' B X D*
125. A proportion is an equality of ratios.
Four magnitudes are in proportion, when the ratio of the first to the second is the same as that of the third to the fourth.
Thus, the ratios of A : B and X : Y, being equal to
A X each other, when written A : B = X : Y, or — = y)
form a proportion.
46 ELEMENTS OF GEOMETRY.
126. Proportion is written not only with the sign =, but, more often, with the sign : : between the ratios.
Thus, A : B : : X : Y, expresses a proportion, and is read. The ratio of A to B is equal to the ratio of X to Y ; or, A is to B as X is to Y.
- 127., TliQ first and third terms of a proportion are called the antecedents; the second and fourth, the consequents. T\iQ first mid fourth are also called the extremes, and the second and third the means.
Thus, in the proportion A : B : : C : D, A and C are the antecedents ; B and D are the consequents ; A and D are the extremes ; and B and C are the means.
The antecedents are called homologous or like terms, and so also are the consequents.
128. All the terms of a proportion are called propor- tionals ; and the last term is called a fourth propor- tional to the other three taken in their order.
Thus, in the proportion A : B : : C : D, D is the fourth proportional to A, B, and C.
129. When both the means are the same magnitude, either of them is called a mean proportional between the -extremes ; and if, in a series of proportional magnitudes, each consequent is the same as the next antecedent, those magnitudes are said to be in continued proportion.
Thus, if we have A : B : : B : C : : C : D : : D : E, B is a mean proportional between A and C, C between B and D, J) between C and E ; and the magnitudes A, B, C, D, E are said to be in continued proportion.
130. When a continued proportion consists of but three terms, the middle term is said to be a mean proportional between the other two ; and the last term is said to be the third proportional to tlie first and second.
Thus, when A, B, and C are in proportion, A : B : : B : C ; in which case B is called a moan proportional between A and C ; and C is called the third proportional to A and B.
BOOK II. 47
131. Magnitudes are in proportion by inversion, or INVERSELY, when each antecedent takes the place of its consequent, and each consequent the place of its antece- dent.
Thus, let A : B : : C : D ; then, by inversion, B : A : : D : C.
132. Magnitudes are in proportion by alternation, or ALTERNATELY, whcu antecedent is compared with antece- dent, and consequent with consequent.
Thus, let A ; B : : D : C ; then, by alternation, A : D : : B : C.
133. Magnitudes are in proportion by composition, when the sum of the first antecedent and consequent is to the first antecedent, or consequent, as the sum of the second antecedent and consequent is to the second ante- cedent, or consequent.
Thus, let A : B : : C : D ; then, by composition, A + B : A : : C + D : C, or A + B : B : : C + D : D.
134. Magnitudes are in proportion by division, when the difference of the first antecedent and consequent is to the first antecedeiit, or consequent, as the difference of the second antecedent and consequent is to the second ante- cedent, or consequent.
Thus, let A : B : : C : D ; then, by division, A — B:A::C — D:C, or A — B : B : : C — D : D.
Proposition I. — Theorem.
135. If four magnitudes are in proportion^ the product of the tvjo extremes is equal to the product of the tivo means.
Let A : B : : C : D ; then will A X I) = B ^^^^^^^^ For, since the magnitudes arc in proportioj^^^ <^
B = D' /f%iV* v*:^:
48 ELEMENTS OF GEOMETRY.
and reducing the fractions of this equation to a common denominator, we have
AX D ^ B X C B X i) B X D' or, the common denominator being omitted, AX D = B X C.
Proposition II. — Theorem.
136. If the product of two magnitudes is equal to the product of two others^ these four magnitudes form a proportion.
Let A X D = B X C ; then will A : B : : C : D. For, dividing each member of the given equation by B X D, we have
AX D _ BX C B X D B X D' which, reduced to the lowest terms, gives A ^ 0 B D' Whence A : B : : 0 : D.
Proposition III. — Theorem.
137. If three magnitudes are in proportion^ the product of the two extremes is equal to the square of the mean.
Let A : B : : B : C ; then will A X C = B^ For, since the magnitudes are in proportion, A ^ B B C' and, by Prop. L,
AXC:=-BXB, or A X C = B^
BOOK II. 49
Proposition IV . — Theorem.
138. If the product of any tvw quantities is equal to the square of a third, the third is a mean proportional between the other two.
Let Ax C =: B^ ; then B is a mean proportional be- tween A and C. For, dividing each member of the given equation hj
B X C, we have
A ^ B
B ~ C'
whence A : B : : B : C.
Proposition Y. — Theorem.
139. If four mag-nitudes are in proportion, they will be in proportion when taken inversely.
Let A : B : : C : D ; then will B : A : : D : C. For, from the given proportion, by Prop. I., we have AXD = BXC, or BxC = AxD. Hence, by Prop. II.,
B : A : : D : C.
Proposition VI. — Theorem.
140. If four magnitudes are in proportion, they will be in proportion when taken alternately.
Let A : B : : C : D ; then will A : C : : B : D. For, since the magnitudes are in proportion, A C
B==D'
B
and multiplying each member of this equation by p, we
have
A X B _ B X C ~
50 ELEMENTS OF GEOMETRY.
which, reduced to the lowest terms, gives
A _ B
C ~ D' whence A : C : : B : D.
Proposition VII. — Theorem.
141. If four magnittides are in proportion^ they will be in proportion by composition.
Let A : B : : C : D ; then will A + B : A : : C + D : C. For, from the given proportion, by Prop. I., we have
BX C = AX D. Adding A X C to each side of this equation, we have
AXC + BXC = AXC + AXD, and resolving each member into its factors,
(A + B) X C = (C + D) X A. Hence, by Prop. II.,
A + B:A::C + D:C.
Proposition YIII. — Theorem.
142. If four magnitudes are in proportion^ they ivill be in proportion by division.
Let A : B : : C : D ; then will A--- B : A : : C — D : C. For, from the given proportion, by Prop. I., we have B X C = A X D.
Subtracting each side of this equation from A X C, we
have
AxC-^BxC = AxC — AxD,
and resolving each member into its factors,
(A — B) X C = (C — D) X A.
Hence, by Prop. II. ,
A — B : A : : C -^ D : C.
BOOK II. 51
Proposition IX. — Theorem.
143. Equimultiples of tivo magnitudes have the same ratio as the magnitudes themselves.
Let A and B be two magnitudes, and m X A and m X B their equimultiples, then will m X A : m X B : ; A : B. For A X B = B X A ;
Multiplying each side of tins equation by any number, m, we have
m X A X B = wi X B X A ; therefore
Qn X A) X B = (wi X B) X A.
Hence, by Prop. II.,
w X A : wi X B : : A : B.
Proposition X. — Theorem.
144. Magnitudes which are proportional to the same proportionals^ will be proportional to each other.
Let A : B : : E : F, and C : D : : E : F ; then will
A : B : : C : D.
For, by the given proportions, we have
A E C E
- = -, and - = ^.
Therefore, it is evident (Art. 34, Ax. 1),
A __ C
B ~~ D* Hence A : B : : C : D.
145. Cor. 1. If two proportions have an antecedent and its consequent the same in both, the remaining terms will be in proportion.
146. Cor. 2. Therefore, by alternation (Prop. VL), if two proportions have the two antecedents or the two con-
62 ELEMENTS OF GEOMETRY.
sequeiits the same in both, the remainmg terms will be in proportion.
Proposition XI. — Theorem.
147. If any number of magnitudes are proportional^ any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents.
Let A : B : : C : D : : E : F ; then will
A:B::A+C + E:BH-D + F. For, from the given proportion, we have
A X D = B X C, and A X F == B X E.
By adding A X B to the sum of the corresponding sides of these equations, we have
AxB+AxD+AxF=AxB+BxC+Bxfi. Therefore,
A X (B + D + F) = B X (A + C + E).
Hence, by Prop. II.,
A:B::A + C + E:B + D + F.
Proposition XII. — Theorem.
148. If four magnitudes are in proportion^ the sum of the first and second is to their difference as the sum of the third and fourth is to their difference.
Let A : B : : C : D ; then will
A + B:A — B::C + D:C — D. For, from the given proportion, by Prop. YII., we have A + B:A::C + D:C; and from the given proportion, by Prop. VIII., we have A — B:A::C — D:C.
Hence, from these two proportions, by Prop. X. Cor. 2,
we have
A + B:A — B::C + D:C — D.
BOOK II. 63
Proposition XIII. — Theorem.
149. If there be two sets of proportional magnitudes, the products of the corresponding terms will be propor- tionals.
Let A : B : : C : D, and E : F : : G : H ; then wiU AXE:BXF::CXG:DXH.
For, from the first of the given proportions, by Prop. I., we have
A X D = B X C;
and from the second of the given proportions, by Prop. I., we have
E X H = F X G.
Multiplying together the corresponding members of these equations, we have
AxDxExH = BxCxFxG.
Hence, by Prop. II.,
AxE:BxF::CxG:DxH.
Proposition XIY. — Theorem.
150. If three viagnitudes are proportionals, the first will be to the third as the square of the first is to the square of the second.
Let A : B : : B : C ; then will A : C : : A^ : B^
For, from the given proportion, by Prop. III., we have
A X C = B^
Multiplying each side of this equation by A gives A^ X C = A X B^
Hence, by Prop. II.,
A : C : : A'* : B^
5*
54 ELEMENTS OF GEOMETRY.
Proposition XV. — Theorem.
151. If four magnitudes are proportionals^ their like powers and roots will also be proportional.
Let A : B : : C : D ; then will
A" : B" : : C" : D% and A^^ : B^ : : C* : DK
For, from the given proportion, we have
A _ C
B ~ D*
Raising both members of this equation to the wtli power, we have
^ _ 51
and extracting the n\h root of each member, we have
A* _ C^
B^ ~~ D^' Hence, by Prop. II., the last two equations give
A^' : B'^ : : C' : D", and
A^ : B» : : C» : D».
BOOK III
THE CIRCLE, AND THE MEASURE OF ANGLES.
DEFINITIONS.
152. A CIRCLE is a plane fig^ure bounded by a curved line, all tlie points of wliicli are equally distant from a point witbin called tlie centre; as tbe figure A D B E.
153. Tbe CIRCUMFERENCE or PERIPHERY of a circle is its entire bounding line ; or it is a curved line, all points of wbicb are equally distant from a point witbin called tbe centre.
154. A RADIUS of a circle is any straigbt line drawn from tbe centre to tbe circumference ; as tbe line C A, CD, or CB.
155. A DIAMETER of a circlc is any straigbt line drawn tbrougb tbe centre, and terminating in botb directions in tbe circumference ; as tbe line A B.
All tbe radii of a circle are equal ; all tbe diameters are also equal, and eacb is double tbe radius.
156. An ARC of a circle is any part of tbe circumference ; as tbe part AD, AE, or EGF.
157. Tbe CHORD of an arc is tbe straigbt line joining its extremities ; tbus E F is tbe cbord of tbe arc EGF.
66
ELEMENTS OF GEOMETRY.
158. The SEGMENT of a circle is the part of a circle iiichided be- tween an arc and its chord ; as the surface included between the arc ^ EGF and the chord EF. E
159. The SECTOR of a circle is the part of a circle included between an ^
arc, and the two radii drawn to the eHtremities of the arc ; as the surface included between the arc A D, and the two radii CA, CD.
160. A SECANT to a circle is a straight line which meets the cir- cumference in two points, and lies partly within and partly without the circle ; as the line A B.
161. A TANGENT to a circle is a straight line which, how far so ever produced, meets the circumference in but one point; as the line CD. The point of meeting is called the point of contact ; as the point M.
162. Two circumferences touch each other, when they have a point /-— ^\ / ^ — -^ of contact without cutting one an- other ; thus two circumferences touch each other at the point A, and two at the point B.
163. A STRAIGHT LINE is IN- SCRIBED in a circle when its ex- tremities are in the circumference; as the line A B, or B C.
164. An INSCRIBED ANGLE is onc which has its vertex in the circumference, and is formed by two chords ; as the angle ABC.
BOOK III.
57
165. An INSCRIBED POLYGON is oiie
which has the vertices of all its angles in the circumference of the circle ; as the triangle ABC.
166. The circle is then said to be circumscribed about the polygon.
167. A POLYGON is CIRCUMSCRIBED
about a circle when all its sides are tangents to the circumference ; as the polygon ABCDEF.
168. The circle is then said to be inscribed in the polygon.
Proposition I. — Theorem.
169. Every diameter divides the circle and its circum- ference each into two equal parts.
Let A E B F be a circle, and A B F
a diameter ; then the two parts A E B, A F B are equal.
For, if the figure A E B be applied to AF B, tlieir common base A B re- taining its position, the curve line A E B must fall exactly on the curve lino AFB; otherwise there would be points in the one or the other unequally distant from the centre, which is contrary to the definition of the circle (Art. 152). Hence a diameter divides the circle and its circumference into two equal parts.
170. Cor. 1. Conversely^ a straight line dividing the circle into two equal parts is a diameter.
58 ELEMENTS OF GEOMETRY.
For, let the line AB divide the
■p
circle A E B C F into two equal parts ; ^ — ^^^
then, if the centre is not in A B, let /^ N.
A C be drawn through it, which is / \ ^
therefore a diameter, and conse- Ar"" ^'b
quently divides the circle into two \ j
equal parts ; hence the surface AFC \^ /
is equal to the surface A F C B, a part ^^~E^^^ to the whole, which is impossible.
171. Cor. 2. The arc of a circle, whose chord is a diameter, is a semi-circumference, and the included seg- ment is a semicircle.
Pkoposition II. — Theorem.
172. A straight line cannot meet the circumference of a circle in more than two points.
For, if a straight Ihie could meet the circumference ABD, in three points, A, B, D, join each of these points with the centre, C ; then, since the straight lines C A, C B, C D are radii, they are equal (Art. 155) ; hence, three equal straight lines can be drawn from the same point to the same straight line, which is impossible (Prop. XIV. Cor. 2, Bk. L).
Proposition III. — Theorem.
173. In the same circle^ or in equal circles, equal arcs are subtended by equal chords; and, conversely, equal chords subtend equal arcs.
Let A D B and E G F be two equal circles, and let the arc A D be equal to E C ; then will the chord A D be equal to the chord E G.
' BOOK III. 69
For, since the diameters A B, EF are equal,
the semicircle [ r i^ -^
A D B may be applied to the semicircle E G F ; and tlie curve line A D B will coincide with the curve line EGF (Prop. I.). But, by hypothesis, the arc AD is equal to the arc E G ; hence the point D will fall on G ; hence the chord A D is equal to the chord E G (Art. 34, Ax. 11).
Conversely^ if the chord A D is equal to the chord E G, the arcs A D, E G will be equal.
For, if the radii CD, 0 G are drawn, the' triangles A C D, E 0 G, having the three sides of the one equal to the three sides of the other, each to each, are themselves equal (Prop. XVIII. Bk. I.) ; therefore the angle A C D is equal to the angle E 0 G (Prop. XVIII. Sch., Bk. I.).
If now the semicircle A D B be applied to its equal EGF, with the radius A C on its equal E 0, since the angles A C D, E 0 G are equal, the radius C D will fall on OG, and the point D on G. Therefore the arcs AD and EG coincide with each other ; hence they must be equal (Art. 34, Ax. 14).
Proposition IV. — Theorem.
174. In the same circle^ or in equal circles, a greater arc is subtended by a greater chord; and, conversely, the greater chord subtends the greater arc.
In the circle of which C is the centre, let the arc A B 1)0 greater than the arc A D ; then will the chord A B be greater than the chord AD.
Draw the radii CA, CD, and C B. Tlie two sides AC^
GO
ELEMENTS OF GEOlvfETRY.
C B in the triangle A C B are equal to the two A C, C D in the triangle A CD, and the angle ACB is greater than the angle A C D ; therefore the third side A B is gi-eater than the third side A D (Prop. XVI. Bk. I.) ; hence the chord which subtends the greater arc is the greater.
Conversely^ if the chord A B be greater than the chord AD, the arc A B will be greater than the arc AD.
For the triangles ACB, ACD have two sides, AC, CB, in the one, equal to two sides, AC, CD, in the other, while the side A B is greater tlian the side AD ; therefore the angle AQB is greater than the angle ACD (Prop. XVII. Bk. I.) ; hence the arc A B is greater than the arc AD.
175. Scholium. Tlie arcs here treated of are each less than the senii-circumfercnce. If they were greater, the contrary would be true ; in which case, as the arcs in- creased, the chords would diminish, and conversely.
Proposition V. — Theorem.
176. In the same circle^ or in equal circles^ radii ivhich make equal ang-Ies at the centre intercept equal arcs on the circumference ; and, conversely, if the intercepted arcs are equal, the ang-les made by the radii are also equal.
Let ACB and I) C E be equal angles made by radii at the centre of equal cir- cles ; then will the \ intercepted arcs A B /, and DE be also equal.
First. Since the angles A 0 B, D C E are equal, the one may bo applied to the otlicr; and since their sides,
BOOK III.
61
being radii of equal circles, are equal, the point A will coincide with D, and the point B with E. Therefore the arc A B must also coincide with the arc D E, or there would be points in the one or tlie other unequally distant from the centre, which is impossible ; hence tiie arc A B is equal to the arc D E.
Second. If the arcs A B and D E are equal, the angles A C B and D C E will be equal.
For, if these angles are not equal, let ACB be tha greater, and let A C F be taken equal to D C E. From what has been shown, we shall have the arc AF equal to the arc D E. But, by hypothesis, A B is equal to D E ; hence A F must be equal to A B, the part to the whole, which is impossible ; hence the angle A C B is equal to
the angle D C E.
Proposition YI. — Theorem.
177. The radius which is perpendicular to a chord bi- sects the chord, and also the arc subtended by the chord.
Let the radius C E be perpendicu- lar to the chord A B ; then will C E bisect the chord at D, and the arc AB at E.
Draw the radii C A and C B. Then C A and C B, with respect to tlie perpendicular C E, are equal ol)lique lines drawn to the chord AB ; therefore their extremities are at equal distances from the perpendicular (Prop. XIV. Bk. I.) ; hence A I) and I) B are equal.
Again, since the triangle ACB has the sides A C and CB equal, it is isosceles; and the line CE bisects the base A B at right angles ; therefore C E bisects also the angle ACB (Prop. VII. Cor. 2, Bk. I.). Since the an- gles A C D, D C B are equal, the arcs A E, E B are equal 6
62 ELEMENTS OP GEOMETRY.
(Prop. Y.) ; hence the radius C E, which is perpendicular to the chord A B, bisects the arc A B subtended by the chord.
178. Cor. 1. Any straight line which joins the centre of the circle and the middle of the chord, or the middle of the arc, must be perpendicular to the chord.
For the perpendicular from the centre C passes through tlie middle, D, of the chord, and the middle, E, of the arc subtended by the chord. Now, any two of these three points in the straight line C E are sufficient to determine its position.
179. Cor. 2. A perpendicular at the middle of a chord passes through the centre of tlie circle, and through the middle of the arc subtended by the chord, bisecting at the centre the angle which the arc subtends.
Proposition VII. — Theorem.
180. Through three given points, not in the same straight line, one circumference can he made to pass, and but one.
Let A, B, and C be any three ^ .^^
points not in the same straight line ; /^ ^v
one circumference can be made to / \ / \
pass through them, and but one. \^\. /
_ Join A B and B C ; and bisect \ V J^ I these straight lines by the perpendic- \ \ / / ulars D E and F E. Join J) F ; then, ^^^^-^
the angles B D E, B F E, being each a right angle, are together equal to two right angles ; therefore the angles E D F, E F I) are togetlier loss than two right angles; hence D E, FE, produced, must meet in some point E (Prop. XXIII. Bk. I.).
Now, since the point E lies in the perpendicular D E, it is equally distant from the two points A and B (Prop. XY. Bk. I.) ; and since the same point E lies in the per-
BOOK III. 63
pendicular FE, it is also equally distant from the two points B and C ; therefore the three distances, E A, E B, E C, are equal ; hence a circumference can be described from the centre E passing through the three points A, B, C. Again, the centre, lying in the perpendicular D E bi- secting the chord AB, and at the same time in the per- pendicular F E bisecting the chord B C (Prop. VI. Cor. 2), must be at the point of their meeting, E. There- fore, since there can be but one centre, but one circum- ference can be made to pass through three given points.
181. Cor. Two circumferences can intersect in only two points ; for, if they have three points in common, they must have the same centre, and must coincide.
Proposition VIII. — Theorem.
182. Equal chords are equally distant from the centre ; and J conversely^ chords ivhich are equally distant from the centre are equal.
Let A B and D E bo equal chords, and C the centre of the circle ; and draw C F perpendicular to A B, and C G perpendicular to D E ; then these perpendiculars, which measure the distance of the chords from the centre, are equal.
Join CA and CD. Then, in the right-angled triangle C AF, C D G, the hypothenuses C A, C D are equal ; and the side A F, the half of A B, is equal to the side D G, the half of D E ; therefore the triangles are equal, and C F is equal to C G (Prop. XIX. Bk. I.) ; hence the two equal chords A B, D E are equally distant from the centre.
Conversely^ if the distances C F and C G are equal, the chords A B and D E are equal.
For, in the right-angled triangles A C F, D C G, the hypothenuses C A, C D are equal ; and the side C F is
64
ELEMENTS OF GEOMETRY.
equal to the side C G ; therefore the triangles are equal, and AF is equal to D Gr ; hence AB, the dovible of AF, is equal to D E, the double of D G (Art. 34, Ax. 6).
Proposition IX. — Theorem.
183. Of tivo unequal chords , the less is the farther from the centre.
Of the two chords D E and A II, let A H be the greater ; then will D E be the farther from the cen- tre C.
Since the chord AH is greater than the chord D E, the arc A H is greater than the arc D E (Prop. IV.). Cut off from the arc A H a part, A B, equal D E ; draw CF perpendicular to this chord, CI perpendicular to AH, and C G perpendicular to D E. C F is greater than C 0 (Art. 34, Ax. 8), and C 0 than C I (Prop. XIV. Bk. I.) ; therefore C F is greater than C I. But C F is equal to C G, since the chords A B, D E are equal (Prop. VIII.) ; therefore, C G is greater than C I ; hence, of two unequal chords, the less is the farther from the centre.
— D
Proposition X. — Theorem.
184. A straig-ht line perpendicular to a radius at its termination in the circu/mference, is a tang-ent to the circle.
Let the straight line B D be per- pendicular to the radius C A at its termination A ; then will it be ia tangent to the circle.
Draw from the centre C to BD any other straight line, as C E. Then, since C A is perpendicular to B D, it is shorter than the oblique
BOOK in. 65
line C E (Prop. XIY. Bk. I.) ; hence the point E is with- out the circle. The same may be shown of any other point in the line BD, except the point A ; tlierefore BD meets the circumference at A, and, being produced, does not cut it ; hence B D is a tangent (Art. 161).
Proposition XI. — Theorem.
185. If a line is A tangent to a circumference^ the ra- dius drawn to the point of contact with it is perpendicular to the tangent.
Let B D be a tangent to the cir- cumference, at the point A ; then will the radius C A be perpendicu- lar to BD.
For every point in BD, except A, being without the circumference (Prop. X.), any line CE drawn from the centre C to B D, at any point other than A, must terminate at E, without the cir- cumference ; therefore the radius C A is the shortest line that can be drawn from the centre to B D ; hence C A is perpendicular to the tangent B D (Prop. XIY. Cor. 1, Bk. I.).
186. Cor. Only one tangent can be drawn through the same point in a circumference ; for two lines cannot both be perpendicular to a radius at the same point.
Proposition XII. — Theorem.
187. Tivo parallel straight lines intercept equal arcs of the circumference.
First. When the two parallels are secants, as AB, DE.
Draw the radius C H perpendicular to A B ; and it will
also be perpendicular to D E (Prop. XXII. Cor., Bk. I.) ;
6*
G6
ELEMENTS OF GEOMETRY.
therefore the point H will be at the same time the middle of the arc A H B and of the arc D H E (Prop. YI.) ; therefore, the arc AH is equal to the arc H B, and the arc D H is equal to the arc H E ; hence AH diminished by D H is equal to H B diminished by H E ; that is, the in- tercepted arcs AD, BE are equal.
Second. When of the two parallels, one, as AB, is a secant, and the other, as D E, is a tangent.
Draw the radius C H to the point of contact H. This radius will be D perpendicular to the tangent D E (Prop. X.), and also to its parallel AB (Prop. XXn. Cor., Bk. I.). But, since C H is perpendicular to the chord AB, the point H is the middle of the arc A H B ; hence the arcs AH, HB, included between the parallels AB, D E, are equal.
Third. When the two parallels are tangents, as D E, IL.
Draw the secant AB parallel to either of the tangents, and it will be parallel to the other (Prop. XXI Y. Bk. I.) ; then, from what has been just shown, the arc AH is equal to the arc H B, and also the arc A G is equal to the arc G B ; hence the whole arc HAG is equal to the whole arc H B G.
It is further evident, since the two arcs HAG, H B G are equal, and together make up the whole circumference, that each of them is a semi-circumference.
188. Cor. Two parallel tangents meet the circumfer- ence at the extremities of the same diameter.
BOOK III.
67
Proposition XIII. — Theorem.
189. If tivo circumferences touch each other externally or internally^ their centres and the point of contact are in the same straight line.
Let the two circumferences, whose centres are C and D, touch each other externally in the point A ; the points C, B, and A will be all in the same straight line.
Draw from the point of con- tact A the common tangent A B. Then the radius C A of tlie one circle, and the radius D A of the other, are each perpendicular to A B (Prop. XI.) ; but there can be but one straight line drawn through the point A perpendicular to AB (Prop. XIII. Bk. I.) ; therefore the points C, D, and A are in one perpendicular ; hence they are in one and the same straight line.
Also, let the two circumferences touch each other internally in A ; tlion their centres, C and D, and the point of contact, A, will be in the same straight line.
Draw the common tangent AB. Then a straight line perpendicular to A B, at the point A, on being suf- ficiently produced, must pass through the two centres C and D (Prop. XI.) ; but from the same point there can be but one perpendicular ; therefore the points C, D, and A are in that perpendicular ; hence they are in the same straight line.
190. Cor. 1. When two circumferences touch each other externally, the distance between their centres is equal to the sum of their radii.
68
ELEMENTS OF GEOMETRY.
191. Cor. 2. And when two circumferences touch each other internally, the distance between their centres is equal to the difference of their radii.
Proposition XIY. — Theorem.
192. ^ two circumferences cut each other, the straight line passing through their centres will bisect at right an- gles the chord which joins the points of intersection.
Let two circumferences cut each other at the points A and B; then the straight line passing through the
centres C and D will bisect at right angles the chord A B common to the two circles.
For, if a perpendicular be erected at the middle of this chord, it will pass through each of the two centres C and D (Prop. VI. Cor. 1). But no more than one straight line can be drawn through two points ; hence the straight line C D, passing through the centres, must bisect at right angles the common chord A B.
193. Cor. The straight line joining the points of inter- section of two circumferences is perpendicular to the straight line which passes through their centres.
Proposition XY. — Theorem.
194. If two circumferences cut each other, the distance betiveen their centres ivill be less than the sum of their radii, and greater than their difference.
BOOK III.
69
Let two circumferences whose centres are C and D cut each other in the point A, and draw the radii C A and D A. Then, in order that the intersec- tion may take place, the triangle CAD must be possible. And in this triangle the side CD must be less tlian the sum of AC and AD (Prop. IX. Bk. I.) ; also C D must be greater than the difference between DA and C A (Prop. IX. Cor., Bk. I.).
Proposition XYI. — Theorem.
195. In the same circle^ or in equal circles^ if two an- gles at the centre are to each other as tiuo ivhole numbers, the intercepted arcs will be to each other as the same numbers.
Let us suppose, for example, that the angles ACB, D C E, at the centre of equal circles, are to each other as 7 to 4 ; or, which amounts to the same thing, that the angle M, which will serve as a common measure, is con-
tained seven times in the angle ACB, and four times in the angle D C E. The seven partial angles A C m, mCn, nCp, &G. into which ACB is divided, being each equal to any of the four partial angles into whicli D C E is divided, each of the partial arcs A m, m n, np, <fec. will
70 ELEMENTS OF GEOMETRY.
be also equal to each of the partial arcs Dx, xy^ <fec. (Prop. V.) ; therefore the whole arc A B will be to the whole arc D E as 7 to 4. But the same reasoning would apply, if in place of 7 and 4 any numbers whatever were employed ; hence, if the ratio of the angles A C B, D C E can be expressed in whole numbers, the arcs A B, D E will be to each other as the angles ACB, DCE.
196. Cor. Conversely^ if the ai'cs A B, D E are to each other as two whole numbers, the angles A C B, D C E will be to each other as the same whole numbers, and we shall have ACB:DCE::AB:DE. For, the partial arcs A?M, mw, <fec. and Da:, .^•y, &c. being equal, the partial angles A C m^ m C w, &c, and D C x, x C «/, <fec. will also be equal.
Proposition XVII. — Theorem.
197. In the same circle, or in equal circles, any two ang-les at the centre are to each other as the arcs inter- cepted hetiveen their sides.
Let ACB be the greater, and A C D the less angle ; then will the angle ACB be to the angle A C D as the arc A B is to the arc AD.
Conceive the less angle to be placed on the greater ; then, if the proposition be not true, the angle ACB will be to the angle A C D as the arc A B is to an arc greater or less than AD. Suppose this arc to be greater, and let it be represented by A 0 ; we shall have the angle ACB: angle A C D : : arc A B : arc A 0. Conceive, now, the arc A B to be divided into equal parts, each of which is less
BOOK III. 71
than D 0 ; there will be at least one point of division be- tween D and 0 ; let I be that point ; and join C I. The arcs A B, A I will be to each other as two whole numbers, and, by tlie preceding proposition, we shall have tlie an- gle A C B : angle A C I : : arc A B : arc A I. Comparing these two proportions with each other, and observing that tlie antecedents are the same, we infer that the conse- quents are proportional (Prop. X. Cor. 2, Bk. II.) ; hence the angle ACD : angle ACl : : arc AO : arc A I. But tlic arc A 0 is greater than the arc A I ; therefore, if this proportion is true, the angle ACD must be greater than the angle A C I. But it is less ; hence the angle A C B cannot be to the angle A C D as the arc A B is to an arc greater than AD.
By a process of reasoning entirely siniihir, it may be shown that the fourth term of the proportion cannot be less than A D ; therefore it must be A D ; hence we have,
Angle A C B : angle A C D : : arc A B : arc A D.
198. Scholium 1. Since the angle at the centre of a circle, and the arc intercepted by its sides, have sucli a connection, that, if the one be increased or dimhiished in any ratio, the other will be increased or diminished in the same ratio, we are authorized to take the one of tliese magnitudes as the measure of the other. Henceforth we shall assume the arc AB as the measure of the angle A C 1^ It is to be observed, in the comparison of angles with each other, that the arcs which serve to measure them must be described with equal radii.
199. Scholium 2. Sectors taken in the same circle, or in equal circles, arc to each other as their arcs ; for sec- tors are equal wlicn tlieir angles are so, and therefore are in all respects proportional to their angles.
72
ELEMENTS OF GEOMETRY.
Proposition XVIII. — Theorem.
200. An inscribed angle is measured by half the arc included between its sides.
Let B A D be an inscribed angle, whose sides include the arc B D ; then the angle BAD is measured by half of the arc B D.
First. Suppose the centre of the circle C to lie within the angle BAD. Draw the diameter AE, and the radii CB, CD.
The angle B C E, being exterior to the triangle ABC, is equal to the sum of the two interior angles CAB, ABC (Prop. XXVII. Bk. I.). But the triangle BAC being isosceles, the angle C A B is equal to A B C ; hence, the angle B C E is double BAC. Since B C E lies at the centre, it is measured by the arc B E (Prop. XVII. Sch. 1) ; hence BAC will be measured by half of B E. For a like reason, the angle CAD will be measured by the half of ED; hence BAC and CAD together, or BAD, will be measured by the half of B E and E D, or half B D.
Second. Suppose that the centre ^
C lies without the angle BAD. Then, drawing the diameter AE, the angle B A E will be measured by the half of B E ; and the angle D A E is measured by the half of D E ; hence, their difference, BAD, will be measured by the half of BE minus the half of ED, or by the half of BD.
Hence every inscribed angle is measured by the half of the arc included between its sides.
BOOK III.
73
201. Cor, 1. All the angles, BAG, BDC, inscribed in the same segment, are equal ; because they are all measured by the half of the same arc, BOG.
202. Cor. 2. Every angle, BAD, inscribed in a semicircle, is a right angle ; because it is measured by half the semi-circumference, BOD ; that is, by the fourth part of the whole circumference.
203. Cor. 3. Every angle, BAG, inscribed in a segment greater than a semicircle, is an acute angle ; for it is measured by the half of the arc BOG, less than a semi-circum- ference.
And every angle, BOG, inscribed in a segment less than a semicircle, is an obtuse angle ; for it is meas- ured by half of the arc BAG, greater than a semi-circumference.
204. Cor. 4. The opposite an- gles, A and D, of an inscribed quadrilateral, ABD G, are together equal to two right angles ; for the angle BAG is measured by half the arc BDG, and the angle BDG is measured by half the arc BAG; hence the two angles BAG, BDG, taken together, are measured by half the circumference ; hence their sum is equal to two right angles.
7
74
ELEMENTS OF GEOMETRY.
Proposition XIX. — Theorem.
205. The angle formed by the inter sectio7i of two chords is measured by half the sum of the tivo intercepted arcs.
Lot the two chords A B, C D mter- sect each other at the point E ; then will the angle DEB, or its equal, A E C, be measured by half the sum of the two arcs D B and A C.
Draw A F parallel to D C ; then will the arc FD be equal to the arc AC (Prop. XII.), and the an- gle FAB equal to the angle DEB (Prop. XXII. Bk. I.). But the angle F A B is measured by half the arc F D B (Prop. XVIII.) ; that is, by half the arc D B, plus half the arc FD. Hence, since FD is equal to A C, the angle DEB, or its equal angle A EC, is measured by half the sum of the intercepted arcs D B and A C
Proposition XX. — Theorem.
206. The angle formed by a tangent and a chord is measured by half the intercepted arc.
Let the tangent B E form, with the chord AC, the angle BAC; then BAC is measured by half the arc AMC.
From A, the point of contact, draw the diameter AD. The an- gle BAD is a right angle (Prop. X.), and is measured by half of the semi-circumference AMD (Prop. XVIII.) ; and the angle D A C is measured by half the arc D C ; hence the sum of the angles BAD, D A C, or B A C, is measured by the half of AMD, plus the half of D C ; or by half the whole arc A M D C.
In like manner, it may be shown that the angle C A E is measured by half the intercepted arc A C.
BOOK III.
75
Proposition XXI. — Theorem.
207. The ang-le formed by two secants is measured hy half the difference of the tivo intercepted arcs.
Let A B, AC be two secants A
forming the angle BAG; then will that angle be measured by half the difference of the two arcs BEG and D F.
Draw D E parallel to AG ; then will the arc E G be equal to the arc D F (Prop. XII.) ; and the angle B D E be equal to the an- gle B A G (Prop. XXII. Bk. I.). But the angle B D E is measured by half the arc B E (Prop. XVIII.) ; hence the equal angle B A G is also measured by half the arc B E ; that is, by half the difference of tlie arcs BEG and E G, or, since E G is equal to D F, by half the differeucc of the intercepted arcs BEG and D F.
Proposition XXII. — Theorem.
208. The angle formed by a secant and a tangent is meas- ured by half the difference of the tivo intercepted arcs.
Let the secant AB form, with A
the tangent A G, the angle BAG; then BAG is measured by half the difference of the two arcs BEF and FD.
Draw D E parallel to A G ; then will the arc E F be equal to the arc DF (Prop. XII.), and the angle BDE be equal to the angle BAG. But the angle B D E is measured by half of the arc B E (Prop. XVIII.) ; hence the equal angle BAG is also measured ])y lialf the arc B E ; that is, by lialf the difference of the arcs P>EF and EF, or, since EF is equal to DF, by half the difference of tlie intercepted arcs BEF and DF.
BOOK IV
PROPORTIONS, AREAS, AND SIMILARITY OF FIGURES.
DEFINITIONS.
209. The area of a figure is its quantity of surface, aud is expressed by the number of times which the surface contains some other area assumed as a unit of measure.
Figures have equal areas, when they contain the same unit of measure an equal number of times.
210. Similar figures are such as have the angles of the one eqvial to those of the other, each to each, and the sides containing the equal angles proportional.
211. Equivalent figures are such as have equal areas. Figures may be equivalent which are not similar.
Thus a circle may be equivalent to a square, and a tri- angle to a rectangle.
212. Equal figures are such as, when applied the one to the other, coincide throughout (Art. 84, Ax. 14). Thus circles having equal radii are equal ; and triangles having the three sides of the one equal to the three sides of the other, each to each, are also equal.
Equal figures are always similar ; but similar figures may be very unequal.
213. In diiferent circles, similar arcs, segments, or sectors are such as correspond to equal angles at the centres of the circles.
BOOK IV.
77
Thus, if the angles A and E are equal, the arc B C will be similar to the arc FG; the segment BDC to the segment FHG, and the sector ABC to the sector EFG.
214. The ALTITUDE OF A TRIANGLE
is the perpendicular, which measures the distance of any one of its vertices from the opposite side taken as a base ; as the perpendicular A D let fall on the base B C in the triangle ABC.
215. The ALTITUDE OF A PARALLEL- OGRAM is the perpendicular which measures the distance betw^een its opposite sides taken as bases ; as the perpendicular E F measuring the dis- tance between the opposite sides, A B allelogram A B C D.
216. The ALTITUDE OF A TRAPEZOID
is the perpendicular distance between its parallel- sides ; as the distance measured by the perpendicular EF between the parallel sides, AB, D C, of the trapezoid A B C D.
Proposition I. — Theorem.
217. Parallelograms which have equal bases and equal altitudes are equivalent.
Let ABCD, ABEF be two D C F E
parallelograms having equal bases and equal altitudes ; then these paj-allelograms are equivalent.
Let the base of the one paral-
7*
DC,
F B
of the par-
78 ELEMENTS OF GEOMETRY.
lelogram be placed on that of the other, so that A B shall be the common base. Now, since the two parallelograms are of the same altitude, their upper bases, DC, F E, will be in the same straight line, D C E F, parallel to A B. From the nature of parallelograms D C is equal to A B, and F E is equal to A B (Prop. XXXI. Bk. I.) ; therefore D C is equal to F E (Art. 34, Ax. 1) ; hence, if D C and F E be taken away from tiie same line, D E, the remain- ders C E and D F will be equal (Art. 34, Ax. 3). But A D is equal to B C and A F to B E (Prop. XXXI. Bk. I.) ; therefore the triangles D A F, C B E, are mutu- ally equilateral, and consequently equal (Prop. XVIII. Bk. L).
If from the quadrilateral ABED, we take away the tri- angle A D F, there will remain the parallelogram ABEF; and if from the same quadrilateral A B E D, we take away the triangle C B E, there will remain the parallelogram A B C D. Hence the parallelograms A BC D, ABE F, which have equal bases and equal altitude, are equivalent.
218. Cor. Any parallelogram is equivalent to a rec- tangle having the same base and altitude.
Proposition II. — Theorem.
219. If a triang-le and a parallelogram have the same base and altitude^ the triangle is equivalent to half the parallelogram.
Let A B E be a triangle, and D C F E
A B C D a parallelogram having V V'' S^^^/
the same base, A B, and the \ /0^^^\/ same altitude ; then will the v>-^""^^ \x
triangle be equivalent to half A B
the parallelogram.
Draw AF, FE so as to form the parallelogram ABEF. Then the parallelograms A B C D, ABEF, having tlie same base and altitude, are equivalent (Prop. I.). But
BOOK IV. • 79
the triangle A B E is half the parallelogram A B E P (Prop. XXXI. Cor. 1, Bk. I.) ; hence the triangle ABE is equivalent to half the parallelogi*am A B C D (Ai't. 34, Ax. T).
220. Cor. 1. Any triangle is equivalent to half a rec- tangle having the same base and altitude, or to a rectangle either having the same base and half of the same altitude, or having the same altitude and half of the same base.
221. Cor. 2. All triangles which have equal bases and altitudes are equivalent.
Proposition III. — Theorejm.
222. Two rectangles having equal altitudes are to each other as their bases.
Let ABCD, AEFD be ^ F C
two rectangles having tlie common altitude AD ; they are to each other as tlieir bases A B, A E. A E B
First. Suppose that the bases AB, AE are commensu- rable, and are to each other, for example, as the numbers 7 and 4. If A B is divided into seven equal parts, A E will contain four of those parts. At each point of division draw lines perpendicular to the base ; seven rectangles will thus be formed, all equal to each other, since they have equal bases and the same altitude (Prop. I.). The rectangle A B C D will contain seven partial rectan- gles, while A E F D will contain four ; hence the rectangle A B C D is to A E F D as 7 is to 4, or as A B is to A E. The same reasoning may be applied, whatever be the numbers expressing the ratio of the bases ; hence, what- ever be that ratio, when its terms are commensurable, we shall have
ABCD:AEFD::AB:AE.
80 • ELEMENTS OF GEOMETRY.
Second. Suppose that the bases A B, A E are incommensurable ; we shall still have
A B C D : A E F D : : A B : A E. For, if this proportion be not true, the A E I O B
first three terms remaining the same, the fourth term must be either greater or less than A E. Suppose it to be greater, and that we have
ABCD:AEFD::AB:AO. Conceive A B divided hito equal parts, each of which is less than EG. There will be at least one point of division, I, between E and 0. Through this point, I, draw the perpendicular I K ; then the bases A B, A I will be com- mensurable, and we shall have
ABCD : AIKD: : AB: AI. But, by hypothesis, we have
ABCD:AEFD::AB:AO.
In these two proportions the antecedents are equal ; hence the consequents are proportional (Prop. X. Cor. 2, Bk. 11.), and we have
AIKD:AEFD::AI:AO.
But A 0 is greater than A I ; therefore, if this proportion is correct, the rectangle A E F D must be greater than the rectangle AIKD (Art. 125) ; on the contrary, however, it is less (Art. 34, Ax. 8) ; therefore the proportion is impossible. Hence, ABCD cannot be to A E F D as A B is to a line greater than AE.
In the same manner, it may be shown that the fourtli term of the proportion cannot be less than A E ; therefore it must be equal to AE. Hoiice, any two rectangles ABCD, AEFD, having equal altitudes, are to each other as their bases A B, A E.
BOOK IV. 81
Proposition IV. — Theorem.
223. Any two rectangles are to eacn, otJier as the pro- ducts of their bases multiplied by their attitudes.
Let ABCD, AEGF be two n d C
rectangles ; then will ABCD be to A E G F as A B multiplied by AD is to A E multiplied by A F. E - Having placed the two rectangles so that the angles at A arc verti- cal, produce the sides GE, CD
G
till they meet iji H. The two rectangles ABCD, AEHD, having the same altitude, AD, are to each other as their bases, A B, AE. In like manner the two rectan- gles AEHD, AEGF, having the same altitude, A E, are to each other as their bases, AD, AF. Heuce wc have the two proportions,
A B C D : A E H D : : A B : A E, AEHD : AEGF: : AD : AF.
Multiplying the corresponding terms of these propor- tions together (Prop. XIH. Bk. H.), and omitting the factor AEHD, which is common to both the antecedent and the consequent (Prop. IX. Bk. II.), we shall have
ABCD:AEGF::ABXAD:AEXAF.
224. Scholium. Hence, we may assume as tlie measure of a rectangle, the product of its base by its altitude, pro- vided we understand by this product the product of two numbers, one of which represents the number of linear units contained in the base, the other the number of linear units contained in the altitude.
The product of two lines is often used to designate their rectangle ; but the term square is used to designate the product of a number multiplied by itself.
82
ELEMENTS OF GEOMETRY.
Proposition V. — Theorem.
225. The area of any parallelogram is equal to the pro- duct of its base by its altitude.
Let A B C D be any parallelogram, A B its base, and B E its altitude ; then will its area be equal to the pro- duct of A B by B E.
Draw BE and AF perpendicular A B
to A B, and produce CD to F. Then the parallelogram A B C D is equivalent to the rectangle A B E F, which has the same base, AB, and the same altitude, BE (Prop. I. Cor.). But the rectangle A B E F is measured by A B X BE (Prop. lY. Sch.) ; therefore AB X BE is equal to the area of the parallelogram A B C D.
226. Cor. Parallelograms having equal bases are to each other as their altitudes, and parallelograms having equal altitudes are to each other as their bases ; and, in general, parallelograms are to each other as the products of their bases by their altitudes.
E
Proposition YI. — Theorem.
227. The area of any triangle is equal to the product of its base by half its altitude
Let A B C be any triangle, B C its base, and A D its altitude ; then its area will be equal to the product of B C by half of A D.
Draw A E and C E so as to form the parallelogram A B C E ; then the triangle A B C is half the parallelogram A B C E, wliich has the same base B C, and the same altitude A D (Prop. II.) ; but the area of the parallelogram is equal to B C X A D (Prop. Y.) ; hence the area of the triangle must be ^ BC X AD, or B C X | AD.
BOOK IV. 83
228. Cor, Triangles of equal altitudes are to each other as their bases, and triangles of equal bases are to each other as their altitudes ; and, in general, triangles are to each other as the products of their bases and alti- tudes.
Proposition YII. — Theorem.
229. The area of any trapezoid is equal to the product of its altitude by half the sum of its parallel sides.
Let A B C D be a trapezoid, E F its altitude, and A B, CD its par- allel sides ; then its area will be equal to the product of E F by half the sum of A B and C D.
Through I, the middle point of A F LB
the side B C, draw K L parallel to AD ; and produce DO till it meet K L. In the triangles I B L, I C K, we have the sides I B, I C equal, by construction ; the vertical an- gles LIB, C I K are equal (Prop. IV. Bk. I.) ; and, since CK and BL are parallel, the alternate angles IBL, ICK are also equal (Prop. XXII. Bk. I.) ; therefore tlie trian- gles I B L, I C K are equal (Prop. YI. Bk. I.) ; hence the trapezoid ABCD is equivalent to the parallelogram A D K L, and is measured by the product of E F by A L (Prop. v.).
But we have AL equal DK; and since the triangles IBL and K C I are equal, the sides B L and C K are equal ; therefore the sum of A B and C D is equal to the sum of A L and D K, or twice A L. Hence A L is lialf the sum of the bases A B, CD; hence the area of the trapezoid A B, C D is equal to the product of the altitude E F by half the sum of the parallel sides AB, CD.
Cor. If through I, the middle point of B C, the line IH be drawn parallel to the base AB, the point H will also be the middle point of AD. For, since the figure AH IL
84 ELEMENTS OF GEOMETRY.
is a parallelogram, as is likewise D H I K, their opposite sides being parallel, we have A H equal to I L, and D H equal to IK. But since the triangles BIL, CIK are equal, we have I L equal to I K ; hence A H is equal to D H.
Now, the line H I is equal to A L, which has been shown to be equal to half tlie sum of A B and C D ; there- fore the area of the trapezoid is equal to the product of E F by HI. Hence, the area of a trapezoid is equal to the product of its altitude by the line connecting tlie mid- dle points of tlie sides wliich are not parallel.
Proposition VIII. — Theorem.
230. If a straight line he divided into tivo parts, the sqvare described on the v^hole line is equivalent to the sum of the squares described on the parts, together with twice the rectangle contained, by the parts.
Let AC be a straight line, divided e H D
into two parts, AB, BC, at the point B;
then the square described on AC is F _j G
equivalent to the sum of the squares described on the parts AB, B C, togetli- er with twice the rectangle contained by A B, B C ; that is,
AC^ = AB^ 4- B~C^ + 2 AB X B C.
On A C describe the square A C D E ; take AF equal to A B ; draw F G parallel to A C, and B H parallel to A E.
The square A C D E is divided into four parts; the first, A B I F, is the square described on A B, since A F was taken equal to A B. The second, I G D H, is the square described upon B C ; for, since A C is equal to A E, and A B is equal to AF, AC minus AB is equal to AE minus A F, which gives B C equal to E F. But I G is equal to B C, and D G to EF, since tlie lines are parallels ; there- fore I G D H is equal to the square described on B C.
|
I |
B C
BOOK IV.
85
These two parts being taken from the whole square, there remain two rectangles B C G I, E F I H, eacli of which is measured by A B X B C ; hence the square on the whole line AC is equivalent to the squares on the parts AB, BC, together with twice the rectangle of the parts.
E H D
281. Cor. The square described on the whole line A C is equivalent to four times the square described on the half AB.
|
1 |
G
B
232. Scholium. This proposition is equivalent to the algebraical formula,
L F
K
H
Proposition IX. — Theorem.
233. The square described on the difference of tivo straight lines is equivalent to the sum of the squares de- scribed on the tivo lines, diminished by twice the rectangle contained by the lines.
Let AB and BC be two lines, and A C their difference ; then will the square described on A C be equiva- lent to the sum of the squares de- scribed on AB, B C, diminished by twice the rectangle AB, B C ; that is, a C B
(AB — B Cy or AC' = AB' + FC^ — 2 A B X BC.
On A B describe the square A B I F ; take A E equal to A C ; draw C Gr parallel to B I, H K parallel to AB, and complete the square E F L K.
Since AF is equal to AB, and AE to AC, EF is equal to B C, and L F to G I ; therefore L G is equal to F I ; hence the two rectangles CBIG, GLKD are each
|
E |
D |
86 ELEMENTS OP GEOMETRY.
measured by A B X B C. Take these rectangles from the whole figure ABILKE, which is equivalent to A B2 -f- B C\ and there will evidently remain the square A C D E ; hence the square on A C is equivalent to the sum of the squares on A B, B C, diminished by twice the rectangle contained by AB, B C.
234. Scholium. This proposition is equivalent to the algebraical formula,
Proposition X. — Theorem.
235. The rectangle contained by the sum and difference of two straight lines is equivalent to the difference of the squares of these lines.
F G I
Let A B, B C be two lines ; then will the rectangle contained by the sum and difference of A B, B C, be equivalent to the difference of the squares of A B, B C ; that is, A C B K
(AB + BC) X (AB — BC) = AB'— BCl
On AB describe the square ABIF, and on AC the square A C D E ; produce CD to G ; and produce A B until B K is equal to B C, and complete the rectangle AKLE.
The base A K of the rectangle is the sum of the two lines A B, B C ; and its altitude A E is tlie difference of the same lines ; therefore the rectangle AKLE is that contained by the sum and the dififercnco of the lines A B, B C. But this rectangle is composed of the two parts A B H E and B H L K ; and the part B H L K is equal to the rectangle E D GrF, since B H is equal to D E, and BK to EF. Hence the rectangle AKLE is equivalent to A B H E plus EDGE, which is equivalent to the dif-
|
H |
||
|
D |
BOOK IV.
87
fereiice between the square A B I F described on A B, and D H I Gr described on B C ; hence
AB^
BC\
(AB + BC) X (AB — BC)
236. Scholium. This proposition is equivalent to the algebraical formula,
(a + b) X Ca — b) = a'' — b\
Proposition XI. — Theorem.
237. The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other tivo sides.
Let ABC be a right-angled L
triangle, having the right angle at A; then the square described H, on the hypothenuse B C will be equivalent to the sum of the squares on the sides BA, AC.
On B C describe the square BCGF, andon AB, AC the squares ABHL, ACIK; and through A draw AE parallel to BFor CG, and join AF,HC.
The angle A B F is composed of the angle A B C, to- gether with the right angle C B F ; the angle C B H is composed of the same angle ABC together with the right angle A B H ; therefore the angle A B F is equal to the angle H B C. But we have A B equal to B H, being sides of the same square ; and B F equal to B C, for the same reason ; therefore the triangles A B F, H B C liave two sides and the included angle of the one equal to two sides and the included angle of the other ; hence they are themselves equal (Prop. Y. Bk. I.).
But the triangle A B F is equivalent to half the rectan- gle B D E F, since they have the same base B F, and the
E G
88
ELEMENTS OP GEOMETRY.
same altitude BD (Prop. II. Cor. 1). The triangle HBC is, in like manner, equivalent H to half the square A B H L ; for the angles BAG, BAL being both right, AC and AL form one and the same straiglit line parallel to HB (Prop. II. Bk. I.) ; and consequently the tri- angle and the square have the same altitude A B (Prop. ^ ^ ^
XXy. Bk. I.) ; and they also have the same base B H ; hence the triangle is equivalent to half the square (Prop.
II.).
The triangle A B F has already been proved equal to the triangle HBC; hence the rectangle BDEF, which is double the triangle A B F, must be equivalent to the square ABHL, which is double the triangle HBC. In the same manner it may be proved that the rectangle C D E G is equivalent to the square A C I K. But the two rectangles BDEF, CDEG, taken together, compose the square BCGF; therefore the square B C G F, de- scribed on the hypotlienuse, is equivalent to the sum of the squares ABHL, ACIK, described on the two other sides ; that is,
FC^ is equivalent to AB^ + AC^-
238. Cor. 1. The square of either of the sides which form the right angle of a right-angled triangle is equiva- lent to the square of the hypothenuse diminished by the square of the other side ; thus,
A B^ is equivalent to B C — A C^.
239. Cor. 2. The square of the hypothenuse is to the square of either of the other sides, as the hypothenuse is to the part of the hypothenuse cut off, adjacent to that side,
BOOK IV. 89
hy the perpendicular let fall from the vertex of the rig'ht angle. For, on account of the common altitude B F, the square BCGF is to the rectangle BDEF as the base BC is to the base B D (Prop. III.) ; now, the square A B H L has been proved to be equivalent to the rectangle BDEF; therefore we have,
BC^ AB^ :BC:BD.
In like manner, we have,
W& : AC^ : : B C : C D.
240. Cor, ^. If a perpendicular he drawn from the vertex of the right angle to the hf/pothenuse, the squares of the sides about the right angle will be to each other as the adjacent segments of the hypothenuse. For the rec- tangles BDEF, DC GE, having the same altitude, are to each other as their bases, B D, C D (Prop. III.). But these rectangles are equivalent to the squares A B H L, A C I K ; therefore we have,
AB" : AC^ : : B D : D C.
241. Cor, 4. The square described on the diagonal of a square is equivalent to double the square described on a side. For let A B C D be a square, and A C its diagonal ; the triangle ABC being right-angled and isosceles, we have.
B
AC"= AB'-f BC"=2AB^=2 X ABCD.
242. Cor. 5. Since A C^ is equal to 2 A B^, we have AC^ : AB' : : 2 : 1 ; and, extracting the square root, we have AC : AB : : V2: 1;
hence, the diagonal of a square is incommensurable with
a side.
8*
90
ELEMENTS OP GEOMETRY.
M
243. Note. — The proposition may also be demonstrated as follows : —
Let ABC be a right-angled triangle, having the right angle at A ; then the square described on the hypothenuse B C will be equivalent to the sum of tlie squares on the sides BA, AC.
On B C describe the square BCGF, and on AB, ACthe squares ABHL, AC IK; pro- duce F B to N, HL £nd I K to M ; and through A draw EDA parallel to FBN, and meeting the prolongation of H L in M.
Then, since the angles H B A, N B C are both right an- gles, if the common angle N B A be taken from each of these cquxils, there will remain the equal angles H B N, ABC; and, consequently, since the triangles H B N, ABC are both right-angled, and have also the sides B H, B A equal, their hypothenuses B N, B C are equal (Prop. yi. Cor., Bk. I.). But B C is equal to B F ; therefore B N is equal to B F ; hence the parallelograms B A M N, B D E F, of which the common altitude is B D, have equal bases ; therefore the two parallelograms are equivalent (Prop. I.). But the parallelogram B AM N is equivalent to the square ABHL, since they have the same base B A, and the same altitude A L ; hence the parallelogram B D E F is also equivalent to the square ABHL. In like manner it may be shown that the rectangle D C G E is equivalent to the square A C I K ; hence the two rectan- gles together, that is, the square BCGF, are equivalent to the sum of the squares ABHL, A C I K.
BOOK IV. 91
Proposition XII. — Theorem.
244. In any triangle, the square of the side opposite an acute angle is less than the sum of the squares of the base and the other side, by tiuice the rectangle contained by the base and the distance from the vertex of the acute angle to the perpendicular let fall from the vertex of the opposite angle on the base, or on the base produced.
Let ABC be any triangle, A A
C one of its acute angles, /K jK
and AD the perpendicular / I \ l\\
let fall on the base B C, or / i \ \ \
on B C produced ; then, in / j \ \ \
cither case, will the square ^ ^ -^ il" "p p
of A B be less than the sum
of the squares of AC, B C, by twice the rectangle B C X
CD.
First. Wlien the perpendicular falls within the triangle ABC, we have BD = BC — CD; and consequently, BD' = BC^ + CD^ — 2 B C X C D (Prop. IX.). By adding A D to each of tliese equals, we have BD' + ad' = BC' + CD' + AD' _ 2 BC X C D. But the two right-angled triangles A D B, ADC give
AB' = BD' + AD', and AC' = CD' + AD' (Prop. XI.) ; therefore,
AB' = BC' + AC' — 2 B C X C D.
Secondly. When the perpendicular A D falls without the triangle ABC, we have BD = CD — BC; and conse- quently, B D' == CD' + BC' _ 2 C D X B C. By add- ing A D to each of these equals, we find, as before,
AB' = B C' + AC' — 2 B C X C D.
92 ELEMENTS OF GEOMETRY.
Proposition XIII. — Theorem.
245. In any ohtuse-angled triangle^ the square of the side opposite the obtuse angle is equivalent to the sum of the squares of the two other sides plus twice the rectangle contained by the one of those sides into the distance from the vertex of the obtuse angle to the perpendicular let fall from the vertex of the opposite angle to that side produced.
Let A C B be an obtuse-angled triangle, A having the obtuse angle at C, and let A D be perpendicular to the base B C produced ; then the square of A B is greater than the sum of the squares of B C, AC, by twice the rectangle B C X C D. Since B D is the suna of the lines B C + C D, we have
B"D^ = Blf + CD' + 2 BC X CD
(Prop. VIII.). By adding AD to each of these equals, we have
BD' + AD'=B"C'+CD' + AD' + 2BC X CD. But the two right-angled triangles A D B, A D C give
AB' = BD' + AD^ . and AC' = CD' + AD^ (Prop. XI.) ; therefore,
AB' = BC' + AC'-|-2BC X CD.
246. Scholium. The right-angled triangle is the only one in which the sum of the squares of two sides is equiv- alent to the square of the third ; for if the angle contained by the two sides is acute, the sum of their squares will be greater than the square of the opposite side ; if obtuse, it will be less.
Proposition XIY. — Theorem.
247. In any triangle, if a straight line be drawn from the vertex to the middle point of the base, the sum of the
BOOK IV.
93
squares of the other two sides is equivalent to twice the square of the bisecting line, together with twice the square of half the base.
In any triangle ABC, draw the line A
A E from the vertex A to the middle of the base B C ; then the sum of the squares of the two sides, A B, AC, is equivalent to twice the square of A E together with twice the square of B E.
On B C let fall the perpendicular AD ; ^ then, in the triangle ABE,
AB^ = AE^ + EB^ + 2 E B X E D (Prop. XIII.), and, in triangle AE C,
AC^ = AE^ + W(f — 2 E C X E D (Prop. XII.). Hence, by adding the corresponding sides together, observing that since E B and E C are equal,
EB^ is equal to E~C^ and EBxEDtoECxED, we have
AB^ + A C^ = 2 AE^ + 2 EB^*-
Proposition XV. — Theorem.
248. In any parallelogram the sum of the squares of the four sides is equivalent to the sum of the squares of the tivo diagonals.
Let A B C D be any parallelogram, d q
the diagonals of which are AC, BD ; then the sum of the squares of A B, B C, CD, DA is equivalent to the sum of the squares of A C, B D.
For the diagonals AC, B D bisect each other (Prop. XXXIV. Bk. I.) ; gle ABC, Xb' + B^' == 2 AE' XIV.) ; also, in the triangle A D C,
hence, in the trian- + 2 be' (Prop.
AD' + D C' = 2 A E' + 2 D E^
94 ELEMENTS OF GEOMETRY.
Hence, by adding the corresponding sides together, and observing that, since B E and D E are equal, B E^ and D E^ must also be equal, we shall have,
AB' + bo' + ad' + DC' = 4 AE' + 4 DE'.
But 4 A E is the square of 2 A E, or of A C, and 4 D E' is the square of 2 DE, or of BD (Prop. YIII. Cor.); hence,
BA' + BC' + C'D' + ad' = AC' + BD'.
Proposition XVI. — Theorem.
249. In any quadrilateral the sum of the squares of the sides is equivalent to the sum of the squares of the diag- onals^ plus four times the square of the straight line that joins the middle points of the diagonals.
Let A B C D be any quadrilateral, the diagonals of which are AC, D B, and E P a straight line joining their mid- dle points, E, F; then the sum of the squares of AB, B C, C D, AD is equiv- alent to AC' + BD' + 4 EP'. A
Join E B and E D ; then in the triangle ABC,
AB' + BC' = 2 AE' + 2 BE^ (Prop. XIY.), and in the triangle ADC,
AD' + CD' = 2 AE' -f 2 DE'. Hence, by adding the corresponding sides, we have
AB' + BC' + AD'+ CD'=: 4 AE' + 2 5^+ 2 DE'. But 4 AE' is equivalent to AC' (Prop. VHI. Cor.), and 2 BT]' + 2 D"e' is equivalent to 4 B~p' + 4 EP' (Prop. XIV.); hence,
Al3' + BC' -f ad' + C^' = AC' + BD^ + 4 EP'.
BOOK IV. 95
250. Cor. If the quadrilateral is a parallelogram, the points E and F will coincide ; then the proposition will be the same as Prop. XY.
251. Scholium. Proposition XY. is only a particular case of this proposition.
Proposition XYII. — Theorem.
252. If a straight line be draivn in a triangle parallel to one of the sides, it vnll divide the other tivo sides pro- portionally.
Let ABC be a triangle, and D E a straiglit line drawn within it parallel to the side B C ; then will
AD : DB:: AE:EC.
Join B E and D C ; then the two trian- gles B D E, DEC have the same base, D E ; they have also the same altitude, since the vertices B and C lie in a line parallel to the base ; therefore the triangles are equivalent (Prop. 11. Cor. 2).
The triangles A D E, B D E, having their bases in the same line A B, and having the common vertex E, have the same altitude, and therefore are to each other as their bases (Prop. YI. Cor.) ; hence
ADE:BDE::AD:DB.
The triangles A D E, D E C, whose common vertex is D, liave also tlie same altitude, and therefore are to each other as their bases ; hence
ADE:DEC::AE:EC.
But the triangles B D E, DEC have been sliown to be equivalent ; therefore, on account of the common ratio in the two proportions (Prop. X. Bk. II.),
A D : D B : : A E : E C.
253. Cor. 1. Hence, by composition (Prop. YII. Bk.
96
ELEMENTS OF GEOMETRY.
IL), AB
we have AD + D B : AD : : AE + E C : AE, or AD : : AC : AE; also, AB: BD : : AC :EC.
254. Co7\ 2. If two or more straight lines be drawn in a triangle parallel to one of the sides, they will divide the other two sides proportionally.
For, in the triangle ABC, since D E ♦ A
is parallel to B C, by the theorem, A D : D B : : A E : E C ; and, in the triangle A D E, since F G is parallel to D E, by the preceding corollary, AD : FD : : A E : G E. Hence, since the antece- dents are the same in the two propor- tions (Prop. X. Cor. 2, Bk. II.), F D : D B : : G E : E C.
Proposition XVIII. — Theorem.
255. If a straight line divides tivo sides of a triangle proportionally^ the line is parallel to the other side of the triangle.
Let ABC be a triangle, and D E a ^
straight line drawn in it dividing the sides AB, A C, so that AD : D B : : AE : E C ; then will the line D E be parallel to the side B C.
Join B E and D C ; then the triangles A D E, B D E, having their bases in the same straight line AB, and having a common vertex, E,
are to each other as Cor.) ; that is,
ADE
their bases AD, DB (Prop. VI.
BDE: : AD : DB.
Also, the triangles ADE, DEC, having the common vertex D, and their bases in the same line, are to each other as these bases, A E, EC; that is,
A D E : D E C : : A E : E C.
BOOK IV. 97
But, by hypothesis, A D : D B : : A E : E C ; hence (Prop. X. Bk. II.),
ADE:BDE::ADE:DEC;
that is, BDE, DE C have the same ratio to ADE ; there- fore the triangles BDE, DEC have the same area, and. consequently are equivalent (Art. 211). Since these tri- angles have the same base, D E, their altitudes are equal (Prop. VI. Cor.) ; hence the line B C, in which their ver- tices are, must be parallel to D E.
Proposition XIX. — Theorem.
256. The straight line bisecting' any angle of a triangle divides the opposite side into parts, which are proportional to the adjacent sides.
In any triangle, ABC, let the an- E gle B A C be bisected by the straight (■■••..... line A D ; then will
BD : DC::AB:AC.
Through the point C draw C E parallel to AD, meeting BA pro- ^ ^ ■"
duced in E. Then, since the two parallels AD, EC are met by the straight line A C, the alternate angles D A C, ACE are equal (Prop. XXII. Bk. I.) ; and the same parallels being met by the straight line B E, the op- posite exterior and interior angles BAD, A E C are also equal (Prop. XXII. Bk. I.). But, by hypothesis, tlie an- gles DAC, BAD are equal; consequently the angle ACE is equal to the angle A E C ; hence the triangle A C E is isosceles, and the side AE is equal to the side A C (Prop. VIII. Bk. I.). Again, since A D, in the triangle E B C, is parallel to E C, we have BD:DC::AB:AE (Prop. XVII.), and, substituting A C in place of its equal A E,
BD:DC::AB:AC.
98 ELEMENTS OF GEOMETRY.
Proposition XX. — Theorem.
257. If a straight line draiim from the vertex of any angle of a triangle divides the opposite side into parts tvhich are proportional to the adjacent sides, the line bi- sects the angle.
Let the straight line AD, drawn E from the vertex of the angle BAG, f •.. in the triangle ABC, divide the op- posite side B C, so that B D : D C : : A B : A C ; then will the line A D bisect the angle BAG.
Through the point G draw G E parallel to AD, meeting B A produced in E. Then, by hypothesis, B D : D G : : A B : A G ; and since AD is parallel to E G, B D : D G : : A B : A E (Prop. XVII.) ; then AB:AG::AB:AE (Prop. X. Bk. II.) ; consequently A G is equal to A E ; hence the angle A E G is equal to the angle AGE (Prop. VII. Bk. L). But, since G E and AD are parallels, the angle AE G is equal to the opposite exterior angle BAD, and the angle A G E is equal to the alternate angle DAG (Prop. XXTL Bk. I.) ; hence tlie angles BAD, DAG are equal, and consequently the straight line A D bisects the angle BAG.
Proposition XXI. — Theorem.
258. If the exterior angle formed by producing one of the sides of any triangle be bisected by a straight line which meets the base produced, the distances from the ex- tremities of the base to the point where the bisecting line meets the base produced, will be to each other as the other two sides of the triangle.
Let the exterior angle G AE, formed by producing the side B A of the tria igle A B G, be bisected 1)y the straight
BOOK IV.
99
line AD, which meets the side B C produced in D, then will BD : DC : : AB: AC. Through C draw CF parallel to A D ; then the angle A C F is equal to the alternate angle CAD, and the exterior angle D A E is equal to the interior and opposite angle C F A (Prop. XXII. Bk. I.). But, by hypothesis, the angles CAD, DAE are equal ; consequently the angle A C F is equal to the angle C F A ; hence the triangle A C F is isosceles, and the side AC is equal to the side AF (Prop. VIII. Bk. I.). Again, since A D is parallel to F C, B D : D C : : BA: AF (Prop. XVII. Cor. 1), and substituting AC in the place of its equal A F, we have
BD:DC::BA:AC.
Proposition XXII. — Theorem.
259. Equiangular triangles have their homologous sides proportional^ and are similar.
Let the two triangles A B C, D C E ^
be equiangular; the angle BAC / \
being equal to the angle CDE, the angle A B C to the angle D C E, and the angle A C B to the angle DEC, then the homologous sides will be proportional, and we shall have
BC:CE::AB:CD::AC:DE.
For, let the two triangles be placed so that two homol- ogous sides, B C, C E, may join each other, and be in the same straight line ; and produce the sides B A, E D till they meet in F.
Since B C E is a straight line, and the angle B C A is equal to the angle C E D, AC is parallel to F E (Prop. XXI. Bk. I.) ; also, since the angle A B C is equal to the
100 ELEMENTS OF GEOMETRY.
angle D C E, the line B F is parallel to the line CD. Hence the figure A C D F is a parallelogram ; and, consequently, A F is equal to CD, and AC to FD (Prop. XXXI. Bk. I.). B C E
In the triangle B E F, since the line A C is parallel to the side F E, we have BC : CE : : BA : AF (Prop. XYII.) ; or, substituting CD for its equal, AF,
BC:CE::BA:CD.
Again, C D is parallel to B F ; therefore, B C : C E : : F D : D E ; or, substituting A C for its equal F D,
B C : C E : : A C : D E.
And, since both tliese proportions contain the same ratio B C : C E, we have (Prop. X. Bk. II.)
AC : DE : : BA: CD. Hence, the equiangular triangles B A C, C D E have their homologous sides proportional ; and consequently the two triangles are similar (Art. 210).
260. Cor. Two triangles having two angles of the one equal to two angles of the other, each to each, are similar; since the third angles will also be equal, and the two tri- angles be equiangular.
261. Scholium. In similar triangles, the homologous sides are opposite to the equal angles ; thus the angle A C B being equal to DEC, the side A B is homologous to D C ; in like manner, A C and D E are homologous.
Proposition XXIII. — Theorem.
262. Triangles lohich have their homologous sides pro- portional, are equiang"ular and similar.
Let the two triangles ABC, D E F have their sides pro- portional, so that we have BC : EF : : AB : DE ; : AC : D F;
« BOOK IV. 101
tlien will the triangles have their angles equal ; namely, the angle A equal to the angle D, the angle B to the angle E, and the angle C to the angle F.
At the point E, in the B C G
straight line EF, make the angle FEG equal to the angle B, and at the point F, the angle E F G equal the angle C ; the third angle G will be equal to the third angle A (Prop. XXVIII. Cor. 2, Bk. I.) ; and the two triangles ABC, E F G will be equiangular. Therefore, by the last theorem, we have
B C : E F : : A B : E G ;
but, by hypothesis, we have
BC : EF: : AB : DEj ^ »..„.,......
hence, E G is equal to D E.
By the same theorem, we also have
BC:EF: : AC :FG;
and, by hypothesis,
BC:EF::AC:DF;
hence F G is equal to D F. Hence, the triangles E G F, D E F, having their three sides equal, each to each, are themselves equal (Prop. XVIII. Bk. I.). But, by con- struction, the triangle E G F is equiangular with the tri- angle ABC; hence the triangles D E F, ABC are also equiangular and similar.
263. Scholium. The two preceding propositions, togctlier with that relating to the square of the hypothenuse (Art. 237), are the most important and fertile in results of any in Geometry. They are almost sufficient of themselves for all applications to subsequent reasoning, and for the 9*
102
ELEMENTS OF GEQ^IETRY.
solution of all problems ; since the general properties of triangles include, by implication, those of all figures.
Proposition XXIY. — Theorem.
264. Tvw triangles, which have an angle of the one equal to an angle of the other, and the sides containing these angles proportional, are similar.
Let the two triangles ABC, I) E F have the angle A equal to the angle D, and the sides contain- ing these angles proportional, so that AB : DE : : AC : DF; then tlie triangles are similar.
Take AG equal D E, and draw GH parallel to BC. The angle AGH will b^ equal to fhe angle' A 'H (3 (Prop. XXII. Bk. I.); and the triangles AQ.H, A.BC will "be equiangular; hence we shall have <'i>"U.'^ ':^ .c AB: AG: : AC : AH. But, by hypothesis,
A B : D E : : A C : D F ;
and, by construction, AG is equal to D E ; hence AH is equal to D F. Therefore the two triangles A G H, D E F, having two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, are themselves equal (Prop. Y. Bk. I.). But the triangle A G H is similar to ABC; therefore D E F is also similar to ABC.
Proposition XXY . : — Theorem.
265. Tivo triangles, which have their sides, taken two and two, either parallel or perpendicular to each other, are similar.
Let the two triangles ABC, D E F have the side A B parallel to the side D E, B C parallel to E F, and A C
BOOK IV. 103
parallel to D F ; these triangles will then be similar.
For, since the side AB is parallel to the side D E, and B C to E F, the angle ABC is equal to the angle DEF (Prop. XXYI. Bk. I.). Also, since AC is parallel to DF, the angle ACB is equal to the angle D FE, and the angle BAC to EDF; therefore the triangles ABC, DEF are equiangular; hence they are similar (Prop. XXII.) .
Again, let the two triangles ^
ABC, DEF have the side D E perpendicular to the side AB, DF perpendicular to AC, and E F perpendicular to B C ; these triangles are similar.
Produce F D till it meets A C at Gr ; tlien the angles D G A, D E A of the quadrilateral A E D G are two right angles ; and since all the four an- gles are together equal to four right angles (Prop. XXIX. Cor. 1, Bk. I.), the remaimng two angles, E D G, E A G, are together equal to two right angles. But the two angles E D G, EDF are also together equal to two right angles (Prop. I. Bk. I.) ; hence the angle E D F is equal to E AG or BAC.
The two angles, G F C, G C F, in the right-angled trian- gle F G C, are together equal to a right angle (Prop. XXYIII. Cor. 5, Bk. I.), and the two angles GFC, GFE are together equal to the right angle E F C (Art. 34, Ax. 9) ; therefore G F E is equal to G C F, or D F E to B C A. Therefore the triangles ABC, DEF have two angles of the one equal to two angles of the other, each to each ; hence they are similar (Prop. XXII. Cor.).
266. Scholium. When the two triangles have their sides parallel, the parallel sides are homologous ; and when- they have them perpendicular, the perpendicular sides are
104 ELEMENTS OP GEOMETRY.
homologous. Thus, D E is homologous with A B, D F with A 0, aiidEF with B C.
Proposition XX YI. — Theorem.
267. In any triangle, if a line he drawn parallel to the base, all lines drawn from the vertex icill divide the par- allel and the base proportionally.
In the triangle BAG, let DE a
be drawn parallel to the base B C ; then will the lines A F, A G, A H, drawn from the vertex, divide the parallel D E, and the base B C, so that DI:BF::IK:FG::KL:GH.
For, since D I is parallel to B F, the triangles AD I and A B F are equiangular ; and we have (Prop. XXII.),
DI: BF: : AI: AF;
and since I K is parallel to F G, we have in like manner,
AI:AF::IK:FG;
and, since these two propositions contain the same ratio, A I : AF, we shall have (Prop. X. Cor. 1, Bk. II.),
DI:BF::IK:FG.
In the same manner, it may be shown that
IK:FG: :KL: GH: :LE:HC.
Therefore the line D E is divided at the points I, K, L, as the base B C is, at the points F, G, H.
268. Cor. If B C were divided into equal parts at the points F, G, H, the parallel D E would also be divided into equal parts at the points I, K, L.
Proposition XXVII. — Theorem.
269. In a right-angled triangle, if a perpendicular is drawn from the vertex of the right angle to the hypothec
BOOK IV. f^ 105
nuse, the triangle icill be divided into ttvo triangles simi- lar to the given triangle and to each other.
In the right-angled triangle ABC, from the vertex of the right angle B A C, let AD be drawn perpendicu- lar to the hypothenuse B C ; then the triangles BAD, D A C will be simi- lar to the triangle ABC, and to each other.
For the triangles BAD, B A C have the common angle B, the right angle B D A equal to the right angle B xV C, and therefore the third angle, B AD, of the one, equal to the third angle, C, of the other (Prop. XXVIII. Cor. 2, Bk. I.) ; hence these two triangles are equiangular, and consequently are similar (Prop. XXII.). In the same manner it may be shown that the triangles D A C and B A C are equiangular and similar. The triangles BAD and D A C, being each similar to the triangle B A C, are similar to each other.
270. Cor, 1. Each of the sides containing the right angle is a mean proportional between the hypothenuse and the part of it which is cut off adjacent to that side by the perpendicular from the vertex of the right angle.
For, the triangles BAD, BAC being similar, their homologous sides are proportional ; hence
BD : BA: :BA: BC;
and, the triangles D A C, B A C being also similar,
DC:AC::AC:BC;
hence each of the sides A B, A C is a mean proportional between the hypothenuse and the part cut off adjacent to that side.
271. Cor. 2. The perpendicular from the vertex of the right angle to the hypothenuse is a mean proportional be- tween the two parts into which it divides the hypothenuse.
106 ELEMENTS OF GEOMETRY.
For, since the triangles ABD, ADC are similar, by comparing their homologous sides we have
BD:AD::AD:DC;
hence, the perpendicular A D is a mean proportional be- tween the parts D B, D C into which it divides the hy- pothenuse B C.
Proposition XXYIII. — Theorem.
272. Two triangles, having an angle in each equal, are to each other as the rectangles of the sides which contain the equal angles.
A
Let the two triangles ABC, A D E have the angle A in common ; then will the triangle ABC be to the tri- angle ADEasABxAC toADX AE.
Join B E ; then the triangles A B E, B C
A D E, having the common vertex E, and their bases in the same line, AB, have the same altitude, and are to each other as their bases (Prop. VI. Cor.) ; hence
ABE:ADE::AB:AD.
In like manner, since the triangles ABC, ABE have the common vertex B, and their bases in the same line, A C, we have
ABC:ABE::AC:AE.
By multiplying together the corresponding terms of these proportions, and omitting the common term ABE, we have (Prop. XIII. Bk. II.),
ABC:ADE::ABxAC:ADxAE.
273. Cor. If the rectangles of the sides containing the equal angles were equivalent, the triangles would be equivalent.
BOOK 17.
'%
107
C E F
proportional (Art.
Proposition *XXIX. — Theorem.
274. Similar triang-les are to each other as the squares described on their homologous sides.
Let A B C, D E F be two similar A triangles, and let AC, D F be ho- mologous sides ; then the triangle ABC will be to the triangle D E F as the square on AC is to the square on D F.
For, the triangles being similar, they have their homologous sides 210); therefore
AB: DE : : AC: DF;
and multiplying the terms of this proportion by the cor- responding terms of the identical proportion,
AC:DF: : AC:DF, we have (Prop. XIII. Bk. II.),
ABXAC:DEXDF:: AC^ : DF^
But, by reason of the equal angles A and D, the triangle ABC is to the triangle DEFasAB X AC is toDExDF (Prop. XXVIII.) ; consequently (Prop. X. Bk. II.),
A B C : D E F : : AC^ : DFI
Therefore, the two similar triangles ABC, D E F are to each other as the squares described on the homologous sides AC, D F, or as the squares described on any other two homologous sides.
Proposition XXX. — Theorem.
275. Similar polygons may be divided into the same number of triangles similar each to each, and similarly situated.
108
Elements of geometry.
Let ABODE, FGHIK be two similar polygons ; they may be divid- ed into the same number of trian- gles similar each to each, and similarly situated. From the homologous an- gles A and F, draw the diagonals A C, A D and F H, F I.
The two polygons being similar, the angles B and G, which are homologous, must be equal, and the sides AB, B C must also be proportional to F G, G H (Art. 210) ; that is, A B : F G : : B C : G H. Therefore the triangles A B C, F GH have an angle of the one equal to the angle of the other, and the sides containing these angles propor- tional ; hence they are similar (Prop. XXIV.) ; conse- quently the angle B C A is equal to the angle G H F. These equal angles being taken from the equal angles BCD, G H I, the remaining angles A C D, F H I will be equal (Art. 34, Ax. 3). But, since the triangles ABC, F G H are similar, we have
AC:FH: : BC: GH;
and, since the polygons are similar (Art. 210),
BC:GH::CD:HI; hence (Prop. X. Cor. 1, Bk. II.),
AC:FH::CD:HI.
But the terms of the last proportion are the sides about the equal angles A C D, F H I ; hence the triangles A C I), FHI are similar (Prop. XXIY.). In the same mannjer, it may be shown that the corresponding triangles A D E, F I K are similar ; hence the similar polygons may be divided into the same number of triangles similar each to each, and similarly situated.
276. Cor. Conversely^ if tiuo polygons are composed
BOOK IV.
109
of the same numher of similar triangles^ and similarly situated, the tvw polygons are similar.
For tlie similarity of the corresponding triangles give the angles ABC equal to F G H, B C A equal to G H F, and A C D equal to F H I ; hence, BCD equal to G H I, likewise C D E equal to H 1 K, <fec. Moreover, we have
AB:FG: :BC: GH: : AC:FH::CD:HI,&c.;
therefore the two polygons have their angh their sides proportional ; hence they are sii
Proposition XXXI. — Theor
277. The perimeters of similar polygons are^ other as their homologous sides ; and their areas art to each other as the squares described on these sides.
Let ABCDE, FGHIK be two similar polygons ; then their perim- eters are to each other as their ho- mologous sides A B and F G, B C and GH, <S:c each other as AB^ is to F~G^ BC^ to GH^ <fec.
First, Since the two polygons are similar, we have AB:FG: :BC : GH: : CD :HI, cfcc.
Now the sum of the antecedents A B, B C, CD, <fec., which compose the perimeter of the first polygon, is to the sum of the consequents F G, GH, HI, <fec., which compose the perimeter of the second polygon, as any one antecedent is to its consequent (Prop. XI. Bk. II.) ; there- fore, as any two homologous sides are to each other, or as A B is to F G.
Secondly. From the homologous angles A and F, draw
and their areas are to
2
110 ELEMENTS OF GEOMETRY.
the diagonals A C, A D and F H, F I. Then, smce the tri- angles ABC, F GH are similar, the triangle
A B C : F G H : : AC^ : FH^
(Prop. XXIX.) ; and, since the triangles AC D, F HI are
similar, the triangle A C D : F H I : : AC^ : FhI But
the ratio AC : FH is common to both of the propor- tions; therefore (Prop. X. Bk. II.),
ABC:FGH::ACD:FHI.
By the same mode of reasoning, it may be proved that ACD :FHI: : ADE:FIK,
and so on, if there were more triangles. Therefore the sum of the antecedents ABC, ACD, ADE, which com- pose the area of the polygon A B C D E, is to the sum of the consequents FGH, FHI, FIK, which compose the area of the polygon F G H I K, as any one antecedent AB C is to its consequent FGH (Prop. XI. Bk. II.), or as A B" is to F G ; hence the areas of similar polygons are to each other as the squares described on their ho- mologous sides.
278. Cor. 1. The perimeters of similar polygons are also to each other as their corresponding diagonals.
279. Cor. 2. The areas of similar polygons are to each other as the squares described on their corresponding diagonals.
Proposition XXXII. — Theorem.
280. A chord in a circle is a mean proportional between the diameter and the part of the diameter cut off between one extremity of the chord and a perpendicular drawn from the other extremity to the diameter.
Let A B be a chord in a circle, B C a diameter drawn from one extremity of A B, and A D a perpendicular
BOOK 17. Ill
drawn from the other extremity to BC; then
BD:AB::AB:BC.
Join A C ; then the triangle ABC, described in a semicircle, is right-angled at A (Prop. XVIII. Cor. 2, Bk. III.) ; and the triangle B A D is similar to the triangle B A C (Prop. XXVIl.) ; hence, we have (Prop. XXVII. Cor. 1), BD : AB: : AB:BC; therefore the chord A B is a mean proportional between the diameter B C, and the part, B D, cut off between the extremity of the chord and the perpendicular from the other extremity.
281. Cor. If from any point. A, in the circumference of a circle, a perpendicular, AD, be drawn to the diameter BC, the perpendicular will be a mean proportional be- tween the parts BD, D C into which it divides the diam- eter.
For, joining AB and AC, we have the triangle ABC, right-angled at xV, and the triangles BAD, D A C similar to it and to each other (Prop. XXYII.) ; therefore (Prop. XXYII. Cor. 2),
BD : AD : : AD : DC,
or, what amounts to the same thing (Prop. III. Bk. II.),
B D X D C = ADI
Scholium. A part of a straight line cut off by another is called a segment of the line. Thus BD, D C are seg- ments of the diameter B C.
Proposition XXXIII. — Theorem.
282. If tivo chords in a circle intersect each other ^ the segments of the one are reciprocally proportional to the segments of the other.
112
ELEMENTS OF GEOMETRY.
Let A B, CD be two chords, which intersect each other at E ; then will AE:DE::EC: EB.
Join A C and B D. In the triangles A EC, BED, the angles at E are equal being vertical angles (Prop. IV. Bk. I.) ; the angle A is equal to the angle D, being measured by half the same arc, BC (Prop. XVIII. Cor. 1, Bk. III.) ; for the same reason, the angle C is equal to the angle B ; the triangles are there- fore similar (Prop. XXII.), and their homologous sides give the proportion,
AE:DE::EC:EB.
283. Cor. Hence, AExEB = DExEC; there- fore the rectangle of the two segments of the one chord is equal to the rectangle of the two segments of the other.
Proposition XXXIV . — Theorem .
284. If from the same point without a circle tioo secants he dravm^ terminating in the concave arc^ the whole se- cants will be reciprocally proportional to their external
sesrments.
p. Let E B, E C be two secants drawn
from the point E without a circle, and
terminating in the concave arc at the
points B and C ; then will
EB:EC::ED:EA.
For, joining AC, BD, the triangles A E C, BED have the angle E com- mon ; and the angles B and C, being ^ measured by lialf the same arc, A D, are equal (Prop. XVIII. Cor. 1, Bk. III.) ; these triangles are therefore similar (Prop. XXII. Cor.), and their homologous sides give the proportion,
EB: EC : : ED :EA.
BOOK IV.
/
113
285. Cor, Hence, EBxEA = ECxED; therefore the rectangle contained bj the whole of one secant and its external segment is equivalent to the rectangle contained by the whole of the other secant and its external segment.
Proposition XXXY . — Theorem.
286. If from a point without a circle there he draicn a tangent terminating- in the circumference^ and a secant terminating in the concave arc^ the tafigent ivill be a mean proportional between the whole secant and its external segment.
From the point E let the tangent E A, and the secant E C, be drawn ; then will EC:EA::EA:ED.
For, joining A D and A C, the triangles E A D, E A C have the angle E common ; also, the angle E A D formed by a tangent and a chord has for its measure half the arc AD (Prop. XX. Bk. III.), and the angle C has the same measure ; therefore the angle E A D is equal to the angle C ; hence the two triangles are similar (Prop. XXII. Cor.), and give the proportion,
EC:EA::EA:ED.
287. Cor. Hence, EA^ = E C X E D ; therefore the square of the tangent is equivalent to the rectangle con- tained by the whole secant and its external segment.
Proposition XXXY I. — Theorem.
288. If any angle of a triangle is bisected by a line terminating in the opposite side, the rectangle of the other tvjo sides is equivalent to the square of the bisecting line plus the rectangle of the segments of the third side.
10*
114
ELEMENTS OF GEOMETRY.
Let the triangle ABC have the angle BAG bisected by the straight line AD terminating in the oppo- site side B C ; then the rectangle B A X AC is equivalent to the square of AD plus the rectangle B D X DC. Describe a circle througli the three points A, B, C ; produce A D till A meets the circumference at E, and join CE.
The triangles BAD, EAC have, by hypothesis, the angle BAD equal to the angle EAC; also the angle B equal to the angle E, being measured by half of the same arc AC (Prop. XVIII. Cor. 1, Bk. III.) ; these triangles are therefore similar (Prop. XXII. Cor.), and their ho- mologous sides give the proportion,
B A : A E : : A D : A C ;
hence, BAxAC = AExAD.
But A E is equal to A D + D E, and multiplying each of these equals by AD, we have,
A E X A D = AD' + A D X D E ;
now, AD X D E is equivalent to B D X DC (Prop. XXXIII. Cor.) ; hence
B A X A C = ad' + B D X D C.
Proposition XXXVII. — Theorem.
289. The rectangle contained by any two sides of a triangle is equivalent to the rectangle contained by the diameter of the circumscribed circle and the perpendicular drawn to the third side from the vertex of the opposite angle.
In any triangle A B C, let A D be drawn perpendicular to B C ; and let E C be the diameter of the circle circum-
BOOK IV.
115
scribed about the triangle ; then will A B X A C be equivalent to A D X 0 E.
For, joining AE, the angle E AC is a right angle, being inscribed in a semicircle (Prop. XVIII. Cor. 2, Bk. III.) ; and the angles B and E are equal, being measured by half of the same arc, A C (Prop. XVIII. Cor. 1, Bk. III.) ; hence the two right- angled triangles are similar (Prop. XXII. Cor.), and give the proportion AB:CE::AD:AC; hence ABXAC = CEXAD.
290. Cor. If these equals be multiplied by BC, we shall have
ABXACXBC = CEXADXBC.
But AD X B C is double the area of the triangle (Prop. YI.) ; therefore the product of the three sides of a triangle is equal to its area multiplied by twice the diameter of the circumscribed circle.
Proposition XXXVIII. — Theorem.
291. Tlie rectangle contained by the diagonals of a quadrilateral inscribed in a circle is equivalent to the sum of the two rectangles of the opposite sides.
Let A B C D be any quadrilat- eral inscribed in a circle, and A C, B D its diagonals ; then the rec- tangle A C X B D is equivalent to the sum of the two rectangles AB X CD, AD X BC.
For, draw BE, making the angle ABE equal to the angle C B D ; to each of these equals add the angle E BD, and we shall have the angle A B D equal to the angle E B C ; and th©
IIG
ELEMENTS OF GEOMETRY.
angle ADB is equal to the angle BCE, being in the same segment (Prop. XVm. Cor. 1, Bk. III.) ; therefore the triangles ABD, B C E are similar ; hence the proportion,
AD : BD : : CE : BC; and, consequently,
ADXBC = BDXCE. Again, since the angle A B E is equal to the angle C B D, and the angle B A E is equal to the angle B D 0, being in the same segment (Prop. XVIII. Cor. 1, Bk. HI.)? the triangles A B E, B C D are similar ; hence,
AB:AE::BD:CD; and consequently,
ABxCD = AExBD. By adding the corresponding terms of the two equations obtained, and observing that
BDxAE + BDxCE = BD(AE + CE) = BDxAC, we have
BD X AC = AB X CD + AD X BC.
E
Proposition XXXIX. — Theorem.
292. The diagonal of a square is incommensurable with its side.
Let A B C D be any square, and A C its diagonal ; then A C is in- commensurable with the side A B.
To find a common measure, if there be one, we must apply A B, or its equal CB, to C A, as often as it can be done. In order to do this, from the point C as a centre, with a radius C B, describe the semicircle F B E, and produce AC to E. It is evident that CB is contained once in AC,
y
BOOK IV. 117
with a remainder AF, which remainder must be com- pared with B 0, or its equal, A B.
The angle ABC being a right angle, A B is a tangent to the circumference, and A E is a secant drawn from the same point, so that (Prop. XXXV.)
A F : A B : : A B : A E.
Hence, in comparing A F with A B, the equal ratio of A B to A E may be substituted ; but A B or its equal C F is contained twice in A E, with a remainder A F ; which remainder must again be compared with A B.
Thus, the operation again consists in comparing AF with A B, and may be reduced in the same manner to the comparison of A B, or its equal C F, with A E ; which will result, as before, in leaving a remainder A F ; hence, it is evident that the process will never terminate ; conse- quently the diagonal of a square is incommensurable with its side.
293. Scholium, The impossibility of finding numbers to express the exact ratio of the diagonal to the side of a square has now been proved ; but, by means of the con- tinued fraction which is equal to that ratio, an approxi- mation may be made to it, sufficiently near for every practical purpose.
BOOK V.
PROBLEMS RELATING TO THE PRECEDING BOOKS.
Problem I.
294. To bisect a given straight line, or to divide it into two equal parts.
Let A B be a straight line, which it ;k'c
is required to bisect. I
From the point A as a centre, with j^ [E ^
a radius greater than the half of A B, I
describe an arc of a circle ; and from I
the point B as a centre, with the same ^^
radius, describe another arc, cutting the former in the points C and D. Through C and D draw the straight line C D ; it will bisect A B in the point E.
For the two points C and D, being each equally distant from the extremities A and B, must both lie in the per- pendicular raised from the middle point of AB (Prop. XY. Cor., Bk. I.). Therefore the line CD must divide the line A B into two equal parts at the point E.
Problem II.
295. From a given point, without a straight line, to draiv a perpendicular to that line.
Let A B be the straight line, and let C be a g'iven point without the line.
BOOK V.
119
From the point C as a centre, and with a radius sufficiently great, describe an arc cutting the line AB in two points, A and B ; then, from the points
A and B as centres, with a radius ^'::
greater than half of A B, describe two
arcs cutting each other in D, and draw
the straight line C D ; it will be the ^
perpendicular required.
For, the two points C and D are each equally distant from the points A and B ; hence, the line CD is a perpen- dicular passing through the middle of A B (Prop. XV. Cor., Bk. L).
D
B
Problem III.
296. At a given point in a straight line to erect a per pendicular to that line.
Let A B be the straight line, and let q
D be a given point in it.
In the straight line AB, take the points A and B at equal distances from D ; then from the points A and B as £- centres, with a radius greater than AD, describe two arcs cutting each other at C ; through C and D draw the straight line C D ; it will be the perpendicular required.
For the point C, being equally distant from A and B, must be in a line perpendicular to the middle of A B (Prop. XY. Cor., Bk. I.) ; hence C D has been drawn perpendicular to A B at the point D.
297. Scholium. Tlie same construction serves for mak- ing a right angle, A D C, at a given point, D, on a given straight line, A B.
120
ELEMENTS OF GEOMETRY.
Problem IY.
298. To erect a perpendicular at the end of a g-ioen straight line.
Let A B be the straight line, and B the end of it at which a perpen- dicular is to be erected.
From any point, D, taken without the line A B, with a radius equal to the distance D B, describe an arc cutting the line AB at the points A and B ; through the point A, and the centre D, draw the diameter AC. Then through C, where the diameter meets the arc, draw the straight line C B, and it will be the perpendicular required.
For the angle ABC, being inscribed in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. 111.)-.
Problem Y.
299. At a point in a given straight line to make an angle equal to a given angle.
Let A be the given point, A B the given line, and E F G the given angle.
From the point F as a centre, with any radius, describe an arc, GE, terminating in the sides of the angle ; from the point A as a centre, with the same radius, describe the indefinite arc BD. Draw the chord G E ; then from B as a centre, with a radius equal to G E, describe an arc cutting the arc B D in C. Draw AC, and the angle CAB will be equal to the given angle E F G.
For the two arcs, BC and GE, have equal radii and equal chords ; therefore they are equal (Prop. 111. Bk.
BOOK V. 121
III.) ; hence the angles C A B, E F G, measured by these arcs, are also equal (Prop. V. Bk. III.).
Problem YI.
300. To bisect a given arc^ or a given angle. First. Let A D B be the given arc C
which it is required to bisect.
Draw the chord A B ; from the cen- /
tre C draw the line C D perpendicular /
to AB (Prob. III.) ; it will bisect the a^--- arc A D B in the point D.
For C D being a radius perpendicu- lar to a chord A B, must bisect the arc A D B which is subtended by that chord (Prop. VI. Bk. III.).
Secondly. Let A C B be the angle which it is required to bisect. From C as a centre, with any radius, describe the arc A D B ; bisect this arc, as in the first case, by drawing the line C D ; and this line will also bisect the angle ACB.
For the angles A C D, BCD are equal, being measured by the equal arcs AD, D B (Prop. Y. Bk. III.).
301. Scholium. By the sam^ construction, we may bi- sect each of the halves AD, D B ; and thus, by successive subdivisions, a given angle or a given arc may be divided into four equal parts, into eight, into sixteen, <fec.
Problem YIL,
302. Through a given point, to draw a straight line parallel to a given straight line.
Let A be the given point, and A -^^r:::; B
C D the given straight line.
From A draw a straight line, "'•■••..,
AE, to any point, E, in CD. E
Then draw A B, making the angle E A B equal to the 11
122 ELEMENTS OF GEOMETRY.
angle AEC (Prob. Y.) ; and a^ B
A B is parallel to C D. ■"-••.....
For the alternate angles E AB, ■■•••-..
AEC, made by the straight line E ^
AE meeting the two straight lines AB, CD, being equal, the lines AB and CD must be parallel (Prop. XX. Bk. I.).
Problem VIII.
303. Tivo angles of a triangle being given, to find the third angle.
Draw the indefinite straight line ABE. At any point, B, make the angle ABC equal to one of the given angles (Prob. Y.), and the angle C B D equal to the other given ABE angle ; then the angle D B E will be the third angle re- quired.
For these three angles are together equal to two right angles (Prop. I. Cor. 2, Bk. I.), as are also the three an- gles of every triangle (Prop. XXYIII. Bk. I.) ; and two of the angles at B having been made equal to two angles of tlie triangle, the remaini^ig angle D B E must be equal to the third angle.
Problem IX.
304. Tivo sides of a triangle and the included angle being given, to construct the triangle.
Draw the straight line A B equal to one of the two given sides. At the point A make an angle, CAB, equal to the given angle (Prob. Y.) ; and take AC equalto the other given side. Join A^ B C ; and the triangle ABC will be the one required (Prop. Y. Bk. I.).
BOOI^ V.
123
Problem X.
305. One side and two angles of a triangle being given, to construct the triangle.
The two given angles will either be ^
both adjacent to the given side, or one adjacent and the other opposite. In the latter case, find the third angle (Prob. VIII.) ; and the two angles ad- jacent to the given side will then be known.
In*the former case, draw the straight line AB eqnal to the given side ; at the point A, make an angle, BAG, equal to one of the adjacent angles, and at B an angle, ABC, equal to the other. Then the two sides AC, BC will meet, and form with AB the triangle required (Prop. VI. Bk. I.)
Problem XI.
306. Two sides of a triangle and an angle opposite one of them being given,) to construct the triangle.
Draw the indefinite straight line A B. At the point A make an angle BAC equal to the given angle, and make A C equal to that side which is adjacent to the given angle. Then from C, as a centre, with a radius equal to the other side, describe an arc, which must either touch the line AB in D, or cut it in the points E and F, otherwise a tri- angle could not be formed.
When the arc touches A B, a straight line drawn from C to the point of contact, D, will be perpendicular to AB (Prop. XI. Bk. III.), and the right-angled triangle CAD will be the triangle required.
When the arc cuts A B in two points, E and F, lying
124 ELEMENTS OF GEOMETRY.
on the same side of tlie point A, draw the straight lines C E, C F ; and each of the two triangles C A E, C A F will satisfy the conditions of the problem. If, however, the two points E and F should lie on different sides of the point A, only one of the triangles, as C A F, will satisfy all the conditions; hence that will be the triangle required.
307. Scholium. The problem would be impossible, if the side opposite the given angle were less than the per- pendicular let fall from the point C on the straight line AB.
Problem XII. «
308. The three sides of a triangle being' given, to con- struct the triangle.
Draw the straight line AB equal to C
one of the given sides ; from the point A as a centre, with a radius equal to either of the other two sides, describe an arc ;. from the point B, with a radius equal to the third side, describe another ^ ^
arc cutting the former in the point C ; draw the straight lines AC, BC; and the triangle ABC will be the one required (Prop. XVIII. Bk. I.).
309. Scholium. The problem would be impossible, if one of the given sides were equal to or greater than the sum of the other two.
Problem XIII.
310. Two adjacent sides of a parallelogram and the irir- eluded angle being given, to construct the parallelogram.
Draw the straight line A B equal to one of the given sides. At the point A make an angle, BAD, equal to the given angle, and take AD equal to the other given side. From
BOOK V. 125
the point D, with a radius equal to AB, describe an arc ; and from the point B as a centre, with a radius equal to A D, describe anotlier arc cutting the former in the point C. Draw the straight lines CD, C B ; and the parallel- ogram A B C D will be the one required.
For the opposite sides are equal, bj construction ; hence the figure is a parallelogram (Prop. XXXII. Bk. I.) ; and it is formed with the given sides and the given angle.
311. Cor. If the given angle is a right angle, the figure will be a rectangle ; and if the adjacent sides are albo equal, the figure will be a square.
Problem XIV.
312. A circumference^ or an arc^ being given, to find the centre of the circle.
Take any three points. A, B, C, on the given circumference, or arc. Draw the chords A B, B C, and bisect them by the perpendiculars DE and FE (Prob. I.) ; the point E, in which these perpendiculars meet, is the centre required.
For the perpendiculars DE, FE must both pass through the centre (Prop. VI. Cor. 2, Bk. III.), and E being the only point through which they both pass, E must be the centre.
313. Scholium. By the same construction, a circumfer- ence may be made to pass through three given points, A, B, C, not in the same straight line ; and also a cir- cumference described in which a given triangle, ABC, shall be inscribed.
Problem XV.
314. T/irovgh a given point to draw a tangent to a
given circle.
11*
126
ELEMENTS OP GEOMETRY.
First. Let the given point A be in the circumference.
Find the centre of the circle, C (Prob. XIY.) ; draw the radius C A ; flirough the point A draw AB perpendicular to C A (Prob. IV.) ; and A B will be the tangent required (Prop. X. Bk. III.).
Secondly. Let the given point B be witliont the circum- ference.
Join the point B and the centre C by the straight line B C ; bisect B C in D ; and from D as a centre, with a radius equal to C D or D B, describe a circumference intersecting the given circumference in the points A and E. Draw AB and E B, and each will be a tangent as required.
For, drawing C A, the angle CAB, being inscribed in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.) ; therefore A B is perpendicular to the radius C A at its ex- tremity, A, and consequently is a tangent (Prop. X. Bk. III.). In like manner it may be shown that EB is a tangent.
Phoblem XVI.
315. On a given straight line to construct a segment of a circle that shall contain an angle equal to a given angle.
Let AB be the given straight line. Through the point B draw the straight line BD, making the angle A B D equal to the given angle ; draw B E perpendicular to B D ; bisect A B, and from F erect the perpendicular F E. From the point E, where these perpendiculars meet, as a centre.
with the distance EB
BOOK V. 127
or E A, describe a circumference, and A C B will be the segment rcqnircd.
For, since B D is a perpendicular at the extremity of the radius E B, it is a tangent (Prop. X. Bk. III.) ; and the angle A B D is measured by half the arc A G B (Prop. XX. Bk. III.). Also, the angle ACB, being an inscribed angle, is measured by half the arc A G- B ; there- fore the angle ACB is equal to the angle A BD. But, by construction, the angle ABD is equal to the givea angle ; hence the segment ACB contains an angle equal to the given angle.
316. Scholium, If the given angle were acute, the centre must lie within the segment (Prop. XYIII. Cor. 3, Bk. III.) ; and if it were right, the centre must be in the middle of the line A B, and the required segment would be a semicircle.
Problem XVII.
317. To inscribe a circle in any given triangle. Bisect any two of the angles,
as A and B, by the straight lines A E and B E, meeting in the point E (Prob. YI.) . From the point E let fall the perpen- diculars ED, EF, EG (Prob. II.) on the three sides of the triangle ; these perpendiculars will all be equal.
For, by construction, we have the angle DAE equal to the angle EAG, and the right angle ADE equal to the right angle AGE; hence the third angle A E D is equal to the third angle A E G. Moreover, A E is common to the two triangles A ED, AEG; hence the triangles them- selves are equal, and E D is equal to E G. In the same manner it may be shown that the two triangles BEF,
128 ELEMENTS OP GEOMETRY.
BEG are equal ; therefore E F is equal to E G ; hence tlie three perpendiculars ED, E F, E G are all equal, and if, from the point E as a centre, with the radius ED, a circle be described, it must pass througli the points F and G.
318. Scholium. The three lines whicli bisect the angles of a triangle all meet in the centre of the inscribed circle.
Problem XYIII.
319. To inscribe a circle in a given square. Draw the diagonals AC, D B, and
from the point E, where the diagonals mutually bisect each other (Prop. XXXIV. Bk. I.), draw the straight line EF perpendicular to a side of the square. From E as a centre, with a radius equal to EF, describe a circle, and it will touch each side of the square.
For the square is divided by the diagonals into four equal isosceles triangles ; hence, the perpendicular, from the vertex E to the base, is the same in each triangle ; therefore the circumference described from the centre E, with the radius E F, passes through the extremities of each perpendicular ; consequently, the sides of the square are tangents to the circle (Prop. X. Bk. III.).
Problem XIX.
320. To find the side of a square tvhich shall be equiv- alent to the sum of two given squares.
Draw the two straight lines A B, B C perpendicular to each other, taking A B equal to a side of one of the given squares, and B C equal to a side of tlic other. Join AC; this will be the side of the square required.
BOOK V,
129
For, the triangle ABC being riglit-angled, the square that can be described upon the hypothenuse A C is equiv- alent to tlie sum of the squares that can be described upon tlie sides A B and B C (Prop. XI. Bk. lY.).
321. Scholium, A square may thus be found equivalent to the sum of any number of squares ; for the construc- tion which reduces two of them to one, will reduce three of them to two, and these two to one.
Problem XX.
322. To find the side of a square vjhich shall be equiv- alent to. the difference of tuw given squares.
Draw the two straight lines A B, A C C perpendicular to each other, making A C equal to the side of the less square. Then from C as a centre, with a radius equal to tlie side of the otlier square, describe an arc intersecting A B in the point B, and A B will be the side of the required square.
For, join B C, and the square that can be described upon AB is equivalent to the difference of the squares that can be described on B C and A C (Prop. XI. Cor. 1, Bk. IV.).
Problem XXI.
323. To construct a rectangle that shall he equivalent to a given triangle.
Let ABC be the given triangle.
Draw the indefinite straight line C D parallel to -the base AB ; bi- sect A B by the perpendicular EF, and make E G equal to F B. Then, by drawing G B, tlie rec- tangle E F B G is equal to the tri- angle ABC.
ISO ELEMENTS OF GEOMETRY.
For the rectangle EFBG has the same altitude, EF, as the triangle ABC, and half its base (Prop. II. Cor. 1, Bk. lY.).
Problem XXII.
324. To construct a triangle that shall he equivalent to a given -polygon.
Let A B C D E be the given poly- gon.
Draw the diagonal CE, cutting off the triangle C D E ; through the point D draw D F parallel to C E, and meeting A E produced in F. ^ -^ ^ ^ Draw C F ; and tlie polygon A B C D E will be equivalent to the polygon A B C F, whicli has one side less than the given polygon.
For the triangles C D E, C F E have the base C E com- mon ; they have also the same altitude, »ince their ver- tices, D, F, are situated in a line, DF, parallel to the base ; these triangles are therefore equivalent (Prop. II. Cor. 2, Bk. lY.). Add to each of them the figure AB C E, and the polygon A B C D E will be equivalent to the polygon A B C F.
In like manner, the triangle C G A may be substituted for tlie equivalent triangle ABC, and tluis the pentagon A B C D E will be changed into an equivalent triangle GCF.
The same process may be applied to every other polygon ; for, by successively diminishing the number of its sides, one at each step of the process, the equivalent triangle will at last be found. *
Problem XXIII.
325. To divide a given straight line into any number of equal parts*
BOOK V.
131
Let A B be the given straight line proposed to be divided into any number of equal parts ; for example, six.
Through the extremity A draw the indefinite straight
line A E, making any angle with A B. Take A C of any convenient length, and apply it six times upon A E. Join the last point of division, E, and the extremity B by tlie straight line E B ; and through the point C draw C D par- allel to E B ; then A D will be the sixth part of the line A B, and, being applied six times to A B, divides it into six equal parts.
For, since C D is parallel to E B, in the triangle ABE, we have the proportion (Prop. XVII. Bk. IV.))
AD : AB : : AC: AE. But A C is the sixth part of A E ; hence A D is the sixth part of A B.
Problem XXIV.
32G. To divide a given straight line into parts that shall he proportional to other given lines.
Let A B be the given straight line proposed to be divided into parts proportional to the given lines AC, CD, D E.
Through the point A draw the indefinite straight line AE, mak- ing any angle with A B. On A E lay off A C, C D, and D E. Join the points E and B by the straight line E B, and through the, points C and D draw C G and D H par- allel to E B ; and the line A B will be divided into parts proportional to the given lines.
For, since C G and D H are each parallel to E B, wo have the proportion (Prop. XVII. Cor. 2, Bk. IV.), A C : A G : : C D : G H : : D E : H B.
182 ELE^IENTS OF GEOMETRY
Problem XXV.
827. To find a fourth proportional to three given straight lines.
Draw the two indefinite straight lines AB, AE, forming any angle with each other.
On AB make AD equal to the first of the proposed lines, and A B equal to the second ; and on A E A D B
make AE equal to the third. Join BE ; and through the point D draw D C parallel to B E, and A C will be the fourth proportional required.
For, since D C is parallel to B E, we have the propor- tion (Prop. XVII. Cor. 1, Bk. IV.),
AB : AD : : AE : AC.
828. Cor. A third proportional to two given lines, A and B, may be found in the same manner, for it will be the same as a fourth proportional to the three lines, A, B, and B.
Problem XXVI.
329. To find a mean proportional between tioo given straight lines.
Draw the indefinite straight line AB. On AB take AC equal to ^
the first of the given lines, and C B equal to the second. On A B, as a diameter, describe a semicircle, and at the point C draw the perpendic- ular C D, meeting the semi-circumference in D ; CD will be the mean proportional rcfquired.
For the perpendicular C D, drawn from a point in the circumference to a point in the diameter, is a mean pro-
B
BOOK V. 133
portional between the two segments of the diameter A C, C B (Prop. XXXII. Cor., Bk. lY.) ; and these segments are equal to the given lines.
Problem XXYII.
330. To divide a given straight line into two such parts^ that the greater part shall be a mean proportional betioeen the whole line and the other part.
Let A B be the given straight ^ ^.^
line. / \j«
At the extremity, B, of the line /
AB, erect the perpendicular BC, 1. ^^
equal to the half of A B. From ^^\
the point C as a centre, with the
radius C B, describe a circle. -^ E B
Draw AC cutting the circumference in D ; and take AE equal to A D. The line A B will be divided at the point E in the manner required ; that is,
AB: AE : : AE.EB.
For A B, being perpendicular to the radius at its ex- tremity, is a tangent (Prop. X. Bk. III.) ; and if A C be produced till . it again meets the circumference, in F, we shall have (Prop. XXXV. Bk. IV.),
AF: AB: : AB: AD; hence, by division (Prop. VIII. Bk. II.),
AF — AB:AB::AB — AD:AD.
But, since the radius is the half of A B, the diameter D F is equal to A B, and consequently A F — A B is equal to AD, which is equal to A E ; also, since A E is equal to A D, we have A B — AD equal to E B ; hence,
AE : AB : : EB : AD, or AE; and, by inversion (Prop. V. Bk. II.),
AB: AE: : AE:EB.
12
134
ELEMENTS OF GEOMETRY.
331. Scholium. This sort of division of the line A B is called division in extreme and mean ratio.
Problem XXVIII.
332. Tlirough a given point in a given angle, to draio a straight line, ivhich shall have the parts included be- tiveen that point and the sides of the angle equal to each other.
Let E be the given point, and ABC the given angle.
Through the point E draw E F paral- lel to BC, make AF equal to BF. Through the points A and E draw the straight line A E C, and it will be the line required.
For, E F being parallel to B C, we have (Prop. XYII. Bk. lY.),
AF:FB: : AE:EC;
but A F is equal to F B ; therefore A E is equal to E 0.
Problem XXIX.
333. On a given straight line to construct a rectangle that shall be equivalent to a given rectangle.
Let AB be the given straight p E
line, and CDEF the given H G
rectangle.
Find a fourth proportional to the three straiglit lines AB, CD,DE (Prob.XXY.); and let B G be that fourth proportional. The rectangle constructed on A B and B G will be equivalent to the rec- tangle CDEF.
For, since AB : CD : : D E: B G, it follows (Prop. I. Bk. II.) that
AB X BG=CD X DE;
BOOK V.
135
hence, the rectangle A B G H, which is constructed on the line A B, is equivalent to the rectangle C D E F.
Problem XXX.
334. To construct a square that shall be equivalent to a given parallelogram, or to a given triangle.
First. Let A.B C D be the given D C
parallelogram, A B its base, and D E /
its altitude. /
Find a mean proportional between LA
A B and D E (Frob. XXYI.) ; and ^ ^ the square constructed on that proportional will be equiv- alent to the parallelogram A B C D.
For, denoting the mean proportional by xy, we have, by construction,
A B : a; y : : a; y : D E ; therefore,
5c]/^ = A B X D E ;
but A B X B E is the measure of the parallelogram, and X y that of the square ; hence they are equivalent.
Secondly. Let A B C be the given tri- angle, BC its base, and AD its altitude.
Find a mean proportional between /
BC and the half of AD, and let xy /
denote that proportional ; the square / constructed on xy will be equivalent B DC
to the.triangle ABC.
For since, by construction,
B C : xy : : xy : j A D, it follows that
5r^^=BC X ^ AD;
hence the square constructed on xy is equivalent to the triangle ABC.
136 ELEMENTS OP GEOMETRY.
Problem XXXI.
335. To construct a rectangle equivalent to a given square, and having the sum of its adjacent sides equal to a given line.
Let the straight line A B be equal to the sum of the adjacent sides of
/
the required rectangle.
Upon A B as a diameter describe a semicircle ; at the point A, draw ^ E B
A D perpendicular to A B, making A D equal to the side of the given square ; then draw the line D C parallel to the diameter A B. From the point C, where the parallel meets the circumference, draw C E perpendicular to the diameter ; A E and E B will be the sides of the rectangle required.
For their sum is equal to A B ; and their rectangle A E X E B is equivalent to the square of C E, or to the square of A D (Prop. XXXII. Cor., Bk. lY.) ; hence, this rectangle is equivalent to the given square.
336. Scholium. The problem is impossible, when the distance A D is greater than the half the given line A B, for then the line D C will not meet the circumference.
Problem XXXII.
337. To construct a rectangle that shall be equivalent to a given square, and the difference of whose adjacent sides shall be equal to a given line.
Let the straight line A B be equal to the difference of the adjacent sides of the required rectangle.
Upon A B as a diameter, describe a circle. At the ex- tremity of the diameter, draw the tangent AD, making it equal to the side of the given square.
BOOK V.
137
Through the point D and the centre C draw the secant D C F, intersecting the circumference in E ; then D E and D F will be the adjacent sides of the rectangle required.
For the difference of these lines is equal to the diameter E F or A B ; and the rectangle D E X D F is equal to AD^ (Prop. XXXY. Cor., Bk. IV.) ; hence it is equiv- alent to the given square.
Problem XXXIII.
338. To construct a square that shall be to a given square as one given line is to another given line.
Draw the indefinite line AB, on which take A C equal to one of the given lines, and C B equal to the other. Upon A B as a di- ameter, describe a semicircle, and at the point C draw the perpendicular C D, meeting the circumference in D. Through the points A and B draw the straight lines D E, D F, making the former equal to the side of the given square ; and through the point E draw E F parallel to A B ; D F will be the side of the square required.
For, since E F is parallel to A B,
DE : DF: : D A: DB; consequently (Prop. XY. Bk. II.),
5^^ : DF^ : : D^ : D'bI
But in the right-angled triangle A D B the square of A D is to the square of D B as the segment A C is to the seg- ment C B (Prop. XI. Cor. 3, Bk. TV.) ; hence,
DE^ DF': : AC: CB.
12*
ld» ELEMENTS OP GEOMETRY.
But, by construction, D E is equal to the side of the given square ; also, A C is equal to one of the given lines, and C B to the other ; hence, the given square is to that con- structed on D F as the one given line is to the other.
Problem XXXIY.
339. Upon a given base to construct an isosceles tri- angle^ having each of the angles at the base double the vertical angle.
Let A B be the given base.
Produce A B to some point C till the -^
rectangle A C X B C shall be equiva- A
lent to the square of A B (Prob. / \
XXXII.) ; then, with the base A B and y v \
sides each equal to A C, construct the / ^sV
isosceles triangle DAB, and the angle A B C
A will double the angle D.
For, make DE equal to AB, or make AE equal to BC, and join E B. Then, by construction,
AD : AB: : AB: AE;
for AE is equal to BC ; consequently the triangles DAB, B A E have a common angle. A, contained by proportional sides ; hence they are similar (Prop. XXIY. Bk. IV.) ; therefore these triangles are both isosceles, for D A B is isosceles by construction, so that A B is equal to E B ; but A B is equal to D E ; consequently D E is equal to E B, and therefore the angle D is equal to the angle E B D ; hence the exterior angle A E B is equal to double the an- gle D, but the angle A is equal to the angle AEB ; there- fore the angle A is double the angle D.
Problem XXXY.
340. Upon a given straight line to construct a polygon similar to a given polygon.
BOOK V. 139
Let ABC DE C
be the given poly- ^^^'^7^\ H
ffon, and F G the ^^f"^ /. ^\ .^^<^
given straight line. \ /' -/-^ \ ..-■■' \
Draw the diag- ^ liij^ /^ fV /
onals AC, AD. ^^^^^.X \^X
At the point F in ^ ^
the straight line F G, make the angle G F H equal to the angle B A C ; and at the point G make the angle F G H equal to the angle ABC. The lines FH, GPI will cut each other in H, and FGH will be a triangle similar to ABC. In the same manner, upon F H, homologous to A C, construct the triangle F I H similar to A D C ; and upon FI, homologous to AD, construct the triangle FIK similar to A D E. The polygon F G H I K will be similar to A B C D E, as required.
For these two polygons are composed of the same num- ber of triangles, similar each to each, and similarly situ- ated (Prop. XXX. Cor., Bk. lY.).
Problem XXXYI.
341. Two similar polygons being given, to construct a similar polygon, which shall be equivalent to their sum or their differetice.
Let A and B be two homologous sides of the given poly- gons.
Find a square equal to the sum or to the difference of tlie squares described up- on A and B ; let x be the side of that square ; then will x in the polygon required be the side which is homologous to the sides A and B in tlie given polygons. The polygon itself may then be constructed on x, by the last problem.
140 ELEMENTS OF GEOMETRY.
For similar figures are to each other as the squares of their homologous sides ; but the square of the side x is equal to the sum or the difference of the squares described upon the homologous sides A and B ; therefore the figure described upon the side x is equivalent to the sum or to the difference of the similar figures described upon the sides A and B.
Problem XXXYII.
342. To construct a polygon similar to a given polygon^ and which shall have to it a given ratio.
Let A be a side of the given polygon.
Find the side"B of a square, which is to the square on A in the given ratio of the polygons (Prob. XXXIIL).
Upon B construct a polygon similar to the given polygon (Prob. XXXV.), and B will be the polygon required.
For the similar polygons constructed upon A and B have the same ratio to each other as the squares con- structed upon A and B (Prop. XXXI. Bk. IV.).
Problem XXXVIII.
343. To construct a polygon similar to a given polygon, P, and which shall be equivalent to another polygon, Q.
Find M, the side of a square, equivalent to the polygon P, and N, the side of a square equivalent to the polygon Q. Let a; be a fourth propor- tional to the three given lines M, N, A B ; vipon the side x, homologous to A B, describe a polygon similar to the polygon P (Prob. XXXV.) ; it will also be equivalent to the polygon Q.
BOOK Y. Ml
For, representing the polygon described upon the side xhj y^ we have
P : y : : AB' : x" ;
but, by construction,
A B : rr : : M : N, or AB^ : x" : : M^ : N^;
hence.
But, by construction also, M^ is equivalent to P, and N^ is equivalent to Q ; therefore,
P:7/::P:Q;
consequently 1/ is equal to Q ; hence the polygon 1/ is similar to the polygon P, and equivalent to the poly- gon Q.
BOOK VI
REGULAR POLYGONS, AND THE AREA OF THE CIRCLE.
DEFINITIONS.
344. A Regular Polygon is one which is both equi- lateral and equiangular.
345. Regular polygons may have any number of sides : the equilateral triangle is one of three sides ; the square is one of four.
Proposition I. — Theorem.
346. Regular polygons of the same number of sides are similar figures.
LetABCDEF, E D
GHIKLM, be two regular poly- gons of the same number of sides ; then these poly- gons are similar. A B G H
For, since the two polygons have the same number of sides, they have the same number of angles ; and the sum of all the' angles is the same in the one as in the other (Prop. XXIX. Bk. I.). Also, since the polygons are equiangular, each of the angles A, B, C, <fec. is equal to each of the angles G, H, I, <fec. ; hence the two polygons are mutually equiangular.
BOOK VI. . 143
Again ; the polygons being regular, the sides A B, B C, CD, &G. are equal to each other; so likewise are the sides G H, H I, I K, &c. Hence,
AB:GH::BC:HI::CD:IK, &c.
Therefore the two polygons have their angles equal, and their homologous sides proportional ; hence they are simi- lar (Art. 210).
347. Cor. The perimeters of two regular polygons of the same number of sides, are to each other as their ho- mologous sides, and their areas are to each other as the squares of those sides (Prop. XXXI. Bk. IV.).
348. Scholium. The angle of a regular polygon js de- termined by the number of its sides (Prop. XXIX. Bk. I.).
Proposition II. — Theorem.
349. A circle map be circumscribed about, and another inscribed in, any regular polygon.
Let ABCDEFGH be any reg- ular polygon ; then a circle may be circumscribed about, and another inscribed in it.
Describe a circle whose circum- ference shall pass through the three points A, B, C, the centre being 0 ; let fall the perpendicular 0 P from ' 0 to the middle pohit of the side B C ; and draw the straight lines 0 A, OB, 0 C, OD.
Now, if the quadrilateral 0 P C D be placed upon the quadrilateral OP B A, they will coincide ; for the side OP is common, and the angle 0 P C is equal to the angle OPB, each being a right angle; consequently the side P C will fall upon its equal, P B, and the point C on B. Moreover, from tlie nature of the polygon, the angle PCD is equal to the angle PB A ; therefore C D will take the
144 .ELEMENTS OF GEOMETRY.
direction B A, and C D being equal to B A, tlie point D will fall upon A, and the two quadrilaterals will coincide throughout. Therefore OD is equal to AO, and the circum- ference which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it may be shown that the circle which passes through the three vertices B, C, D, will also pass through the vertex E, and so on. Hence, the circumfer- ence which passes through the three points A, B, C, passes through the vertices of all the angles of the polygon, and is circumscribed about the polygon (Art. 166).
Again, with respect to this circumference, all the sides, AB, B C, CD, <fec., of the polygon are equal chords ; con- sequently they are equally distant from the centre (Prop. YIII. Bk. III.). Hence, if from the point 0, as a centre, and with the radius 0 P, a circle be described, the circum- ference will touch the side B C, and all the other sides of the polygon, each at its middle point, and the circle will be inscribed in the polygon (Art. 168).
350. Scholium 1. The point 0, the common centre of the circumscribed and inscribed circles, may also be re- garded as the centre of the polygon. The angle formed at the centre by two radii drawn to the extremities of the same side is called the angle at the centre ; and the per- pendicular from the centre to a side is called the apothegm of the polygon.
Since all the chords AB, B C, CD, <fec. are equal, all the angles at the centre must likewise be equal ; therefore tlie value of each may be found by dividing four right an- gles by the number of sides of the polygon.
351. Scholium 2. To inscribe a regular polygon of any number of sides in a given circle, it is only necessary to
BOOK VI. 145
divide the circumference into as
many equal parts as the polygon
has sides ; for the arcs being equal,
the chords AB, BC, CD, &c. are
also equal (Prop. III. Bk. III.) ;
hence likewise the triangles A 0 B,
B 0 C, COD, &c. must be equal, since their sides are
equal each to each (Prop. XVIII. Bk. I.) ; therefore all
the angles ABC, BCD, CDE, &c. are equal ; hence
the figure ABCDEF is a regular polygon.
Proposition III. — Theorem.
352. If from a common centre a circle can he circum- scribed about, and another circle inscribed within, a poly' gon, that polygon is regular .
Suppose that from the point 0, as a centre, circles can be circum- scribed about, and inscribed in, the polygon ABCDEF; then that polygon is regular.
For, supposing it to be described, the inner one will touch all the A B
sides of the polygon ; therefore these sides are equally dis- tant from its centre ; and consequently, being chords of tlie outer circle, they are equal ; therefore tliey include equal angles (Prop. XVIII. Cor. 1, Bk. III.). Hence the polygon is at once equilateral and equiangular ; con- sequently it is regular (Art. 344).
Proposition IV. — Problem.
353. To inscribe a square in a given circle.
Draw two diameters, AC, BD, intersecting each other at riglit angles ; join their extremities, A, B, C, D, and the figure A B C D will be a square.
13
146
ELEMENTS OF GEOMETRY.'
For, the angles A OB, BOC, &c. be- ing equal, the chords AB, BC, <fec. are also equal (Prop. Ill, Bk. III.) ; and the angles ABC, BCD, <fec., being in- scribed in semicircles, are right angles (Prop. XVIII. Cor. 2, Bk. III.). Hence A B C D is a square, and it is inscribed in the circle ABCD.
354. Cor. Since the triangle A 0 B is right-angled and isosceles, we have (Prop. XI. Cor. 5, Bk. IV.),
AB: AO: : V2 : 1;
hence, the side of the inscribed square is to the radius as the square root of 2 is to unity.
Proposition V. — Theorem.
355. The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.
Let ABCDEF be a regular hexagon inscribed in a circle, the centre of which is 0 ; then any side, as B C, will be equal to the radius OA.
Join BO; and the angle at the centre, A OB, is one sixth of four right angles (Prop. II. Sch. 1), or one third of two right angles ; therefore the two other angles, 0 A B, 0 B A, of the same triangle, are together equal to two tliirds of two right angles (Prop. XXVIII. Bk. I.). But A 0 and B O being equal, the angles OAB, OB A are also equal (Prop. VII. Bk. I.) ; consequently, each is one third of two right angles. Hence the triangle A 0 B is equiangular ; there- fore A B, the side of the regular hexagon, is equal to A 0, the radius of the circle (Prop. VIII. Cor. Bk. I.).
BOOK VI.
147
356. Cor. 1. To inscribe a regular hexagon iii a given circle, apply the ra- dius, A 0, of the circle six times, as a chord to the circumference. Hence, beginning at any point A, and applying A 0 six times as a chord to the circum- ference, we are brought round to the point of beginning, AB CD EF.
figure
and the inscribed
thus formed, is a regular hex- agon.
357. Cor. 2. By joining the alternate angles of the in- scribed regular hexagon by the straight lines A C, C E, E A, the figure ACE, thus inscribed in the circle, will be an equilateral triangle, since its sides subtend equal arcs, ABC, CDE, EFA, on the circumference (Prop. III. Bk. III.).
358. Cor. 3. Join 0 A, 0 C, and the figure A B C 0 is a rhombus, for each side is equal to the radius. Hence, the sum of the squares of the diagonals A C, 0 B is equiv- alent to the sum of the squares of the sides (Prop. XV. Bk. IV.) ; or to four times the square of the radius 0 B ; that is, A C^ -f OB^ is equivalent to 4 AB^ or 4 6 B^ ; and taking away 0 B" from both, there remains A C^ equivalent to 3 0 B ; hence
AC^0B'::3:1, or AC : 0 B : : Vs : 1 ;
hence, the side of the inscribed equilateral triangle is to the radius as the square root of 3 is to unit?/.
Proposition VI. — Problem.
359. To inscribe a regular decagon in a given circle.
Divide tlie radius, 0 A, of the given circle, in extreme and mean ratio, at the point M (Prob. XXVII. Bk. V.).
148 ELEMENTS OF GEOMETRY.
Take the chord AB equal to OM, and AB will be tlie side of a regular decagon inscribed in the circle. For we have by construction,
A 0 : 0 M : : 0 M : A M ;
or, since A B is equal to 0 M, A 0 : A B : : A B : A M.
Draw MB and BO; and the triangles ABO, AMB have a common angle, A, included between proportional sides ; hence the two triangles are similar (Prop. XXIV.. Bk. IV.). Now, the triangle OAB being isosceles, AMB must also be isosceles, and A B is equal to B M ; but A B is equal to 0 M, consequently M B is equal to M 0 ; hence the triangle M B 0 is isosceles.
Again, the angle AMB, being exterior to the isosceles triangle BMO, is double the interior angle 0 (Prop. XXVII. Bk. L). But the angle AMB is equal to the angle M A B ; hence the triangle 0 A B is such, that each of the angles at the base, OAB, OB A, is double the angle 0, at its vertex. Hence the three angles of tlie triangle are together equal to five times the angle 0, which conse- quently is a fifth part of two right angles, or the tenth part of four right angles ; therefore the arc A B is the tenth part of the circumference, and the chord A B is the side of an inscribed regular decagon.
360. Cor. 1. By joining the vertices of the alternate angles A, C, &g. of the regular decagon, a regular penta- gon may be inscribed. Hence, the chord A C is the side of an inscribed regular pentagon.
361. Cor. 2. A B being the side of the inscribed regu- lar decagon, let AL be tlie side of an inscribed regular hexagon (Prop. V. Cor. 1). Join BL; then BL will be the side of an inscribed regular pentedecagon, or regular polygon of fifteen sides. For AB cuts off an arc equal to a tenth part of the circumference ; and A L subtends an
BOOK VI. 149
arc equal to a sixth of the circumference ; therefore B L, tlie difference of these arcs, is a fifteenth part of the cir- cumference ; and since equal arcs are subtended by equal chords, it follows that the chord B L may be a])|)licd ex- actly fifteen times around the circumference, thus forming a regular pentedecagon.
362. Scholium. If the arcs subtended by the sides of any inscribed regular polygon be severally bisected, the chords of those semi-arcs will form another inscribed polygon of double the number of sides. Thus, from having an inscribed square, there may be inscribed in suc- cession polygons of 8, 10, 32, 64, <fec. sides ; from the hexagon may be formed polygons of 12, 24, 48, 96, <fec. sides ; from the decagon, polygons of 20, 40, 80, <fec. sides ; and from the pentedecagon, polygons of 30, 60, 120, &c. sides.
Note. — For a long time the polygons above noticed were supposed to include all that could be inscribed in a ciix:le. In the year 1801, M. Gauss, of Gottingen, made known the curious discovert' that the circum- ference of a circle could be divided into any number of equal parts capable of being expressed by the formula 2" -[- 1, provided it be a prime number. Now, the number 3 is the simplest of this kind, it being the value of the above formula when the exponent n is 1 ; the next prime number is 5, and this is contained in the formula. But the poly- gons of 3 and of 5 sides have already been inscribed. The next prime number expressed by the fonnula is 1 7, so that it is possible to inscribe a regular polygon of 17 sides in a circle. The investigations, however, -which establish this geometrical fact involve considerations of a nature that do not enter into the elements of Geometry.
Proposition YII. — Problem.
363. A regular inscribed polygon being given, to cir- cumscribe a similar polygon about the same circle.
Let ABCDEF be a regular polygon inscribed in a circle whose centre is 0.
Through M, the middle point of the arc A B, draw the tangent, G II ; also draw tangents at the middle points of
13*
150 ELEMENTS OP GEOMETRY.
tlie arcs B C, C D, <fec. ; these tan- gents are parallel to the chords AB, BC, CD, &c. (Prop. XL Bk. IIL, and Prop. YI. Cor. 1, Bk. III.), and by their intersec- tions form the regular circum- scribed polygon GHI, &c. similar to the one inscribed.
Since M is the middle point of the arc A B, and N tlie middle point of the equal arc B C, the arcs B M, B N are halves of equal arcs, and therefore are equal ; that is, the vertex, B, of the inscribed polygon is at the middle point of the arc MN. Draw OH ; the line OH will pass through the point B. For, the right-angled triangles OMH, ONH, having the common hypothenuse 0 H, and the side 0 M equal to ON, must be equal (Prop. XIX. Bk. I.), and consequently the angle M 0 H is equal to HON, where- fore the line 0 H passes through the middle point, B, of the arc M N. In like manner, it may be shown that the line 01 passes through the middle point, C, of the arc N P ; and so with the other vertices.
Since O H is parallel to A B, and II I to B C, the angle G II I is equal to the angle ABC (Prop. XXVI. Bk. I.) ; in like manner, H I K is equal to B C D ; and so with the other angles ; hence, the angles of the circumscribed poly- gon are equal to those of the inscribed polygon. And, further, by reason of these same parallels, we have
GH:AB::OH:OB, and H I : B C : : 0 H : OB ;
therefore (Prop. X. Bk. II.),
G H : A B : : H I : B C.
But A B is equal to B C, therefore G H is equal to II I. For the same reason, H I is equal to I K, <fec. ; conse- quently, the sides of the circumscribed polygon are all equal ; hence this polygon is regular, and similar to the inscribed one.
BOOK VI. 151
364. Cor. 1. Converselj, if the circumscribed polygon G H I K, <fcc. is given, and it is required, by means of it, to construct a similar inscribed polygon, draw the straight lines 0 G, OH, <fcc. from the vertices of the angles G, H, I, <fec. of the given polygon to the centre ; tlie lines will meet the circumference in tlie points A, B, C, <fec. Join these points by the chords A B, B C, <fec., which will form the inscribed polygon. Or simply join the points of con- tact, M, N, P, <fec., by chords, MN, N P, <fcc., which like- wise would form an inscribed polygon similar to the cir- cumscribed one.
365. Cor. 2. Hence, we may circumscribe about a cir- cle any regular polygon similar to an inscribed one, and conversely.
366. Cor. 3. It has been shown that N H and H M are equal ; therefore the sum of N H and H M, which is equal to the sum of H M and M G, is equal to H G, one of the equal sides of the polygon.
367. Scholium. From having a circumscribed regular polygon, another having double the number of sides may be readily constructed, by drawing tangents to the points of bisection of tlie arcs, intercepted by the sides of the pro- posed polygon, limiting these tangents by those sides. In like manner other circumscribed polygons may be formed ; l)ut it is plain that each of the polygons so formed will be less than the preceding polygon, being entirely compre- hended in it.
Proposition YIH. — Theorem.
368. Tlie area of a regular poly ^^on is equivalent to the product of its perimeter hy half of the radius of the in- scribed circle.
Let A B C B E F be a regular polygon, and 0 tlie centre of the inscribed circle.
From 0 let the straight lines 0 A, 0 B, <fec. be drawn to
152
ELEMENTS OP GEOMETRY.
|
E |
D |
||
|
^ |
0 |
^ |
y |
|
•■';■■.'" |
■"]■"/ |
||
|
/ • '■•• |
■•-•.. |
// |
|
|
/ |
1 |
\ y |
/n |
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^^ |
^-.7 |
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A |
M |
B |
the vertices of all the angles of the polygon, and the polygon will be divided into as many equal triangles as it has sides ; and let the radii 0 M, ON, &c. of the in- scribed circle be drawn to the centres of the sides of the polygon, or to the points of tangency M, N, <fec., and these radii are perpendicular to the sides re- spectively (Prop. XI. Bk. III.) ; therefore the radius of the circle is equal to the altitude of the several triangles.
Now, the triangle A 0 B is measured by the product of AB by half of 0 M (Prop. YI. Bk. IV.) ; the triangle O B C by the product of B C by half of 0 N. But 0 M is equal to 0 N ; hence the two triangles taken together are measured by the sum of A B and B C by half of 0 M. In like manner the measure of the other triangles may be found ; hence, the sum of all the triangles, or the whole polygon, is equal to the sum of the bases AB, B C, ttc, or the perimeter of tho polygon, multiplied by half of OM, or half the radius of the inscribed circle.
Proposition IX. — Theorem.
369. The perimeters of two reg-vlar polygons^ having the same number of sides, are to each other as the radii of the circumscribed circles, and, also, as the radii of the inscribed circles ; and their areas are to each other as the squares of those radii.
Let A B be a side of one polygon, O the centre, and consequently 0 A the radius of the circumscribed cir- cle, and 0 M, perpendicular to A B, the radius of the inscribed circle. Let G H be a side of the other poly- gon, C the centre, C G and C N the A B G radii of the circumscribed and the inscribed circles.
The perimeters of tlie two polygons arc to-e*eli otIi'Sr' the sides AB and G H (Prop. XXXI. Bk. iyr)rf)til^ angles A and G are equal, being eacli luilf oi' tlio angle of the polygon ; so also are the angles B and H ; hence, drawing OB and OH, the isosceles triangles ABO, GHC are similar, as are likewise the right-angled triangles A M 0, G N C ; hence
A B : G H : : A 0 : G C : : M 0 : N C.
Hence, the perimeters of the polygons are to each other as the radii A 0, GO of tlie circumscribed circles, and, also, as tlie radii M 0, N C of the inscribed circles.
The surfaces of these polygons are to each other as the squares of the homologous sides A B, G H (Prop. XXXI. Bk. IV.) ; tliey are therefore to each other as the squares of AO, GO, the I'adii of the circumscribed circles, or as the squares of 0 M, ON, the radii of the inscribed circles.
Proposition X. — Pboblem.
370. The surface of a reg'iilar inscribed poii/g'on, and that of a similar circumscribed pol/jg-on, bei/i^ g-iven ; to find the surfaces of regular inscribed and circumscribed polygons having double the number of sides.
Let AB be a side of the given ^ -r. ,, ^ „
M J 1 T^Ti iTix EPMQF
inscribed polygon ; l!j F, parallel to
AB, a side of the circumscribed
polygon, and C the centre of tlie
circle. Draw the chord AM, and
the tangents A P, B Q ; then A M
will be a side of the inscribed poly- v
gon, having twice the number of ^
sides ; and P Q, the double of P M, will be a side of the
similar circumscribed polygon.
Let A, then, be the surface of the inscribed polygon
whose side is AB, B tliat of tlie similar circumscribed
polygon ; A' the surface of the polygon whose side is A ^M,
154 ELEMENTS OP GEOMETRY.
B' that of the similar circumscribed ^ ^ ^. ^ -,, 1 A A Ty • E P M Q F
polygon : A and B are given ; we . — -7—
have to find A' and B'. 4^^^
First. The triangles ACD,ACM, \\ f /
whose common vertex is A, are to each other as their bases CD, CM (Prop. VI. Cor., Bk. lY.) ; they are likewise as the polygons A and A' ;
hence
A : A' : : C D : C M.
Again, the triangles CAM, C M E, whose common vertex is M, are to each other as their bases C A, C E ; they are likewise to each other as the polygons A' and B ; hence
A' : B : : C A : C E.
But, since A D and M E are parallel, we have,
CD : CM: : CA: CE; hence
A : A' : : A' : B ;
hence, the polygon A! is a mean proportional between tliei two given polygons.
Secondly. The altitude C M being common, the tri- angle CPM is to the triangle C PE as PM is to PE ; but since CP bisects the angle MCE, we have (Prop. XIX. Bk. IV.),
PM:PE::CM:CE::CD:CA::A:A';
hence
CPM: CPE: : A: A';
and, consequently,
C P M : C P M + C P E or C M E : : A : A + A'. But CMPAor2 CMP and C M E are to each other as the polygons B' and B ; hence
B' : B : : 2 A: A + A';
which gives
J\i — 2 Ax B -^ — A + A' '
BOOK VI. 155
or, the polygon B' is equal to the qvotient of ttoice the product of the given polygons divided by the sum of the inscribed polygons.
Thus, by means of the polygons A and B, it is easy to find the polygons A' and B', whicli have double the num- ber of sides.
Proposition XI. — Theorem.
371. A circle being given ^ tivo similar polygons can alvmys be formed^ the one circumscribed about the circle^ the other inscribed in it^ which shall differ from each other by less than any assignable surface.
Let Q be the side of a square I
less than the given surface. H,^^^^^^^^:^^ |r
Bisect A C, a fourth part of ^^\. 3)U
the circumference, and then bi- /jsr-\.\ "^••^.. \V
sect the half of this fourth, and ^ u a "o W/^'^ so proceed until an arc is found \ o //
whose chord AB is less tliau ^^>J Z-^
Q. As this arc must be an ex- ^^^"rtp^^^"^
act part of the circumference, if we apply the chords AB, B C, <fec., each equal to AB, the last will terminate at A, and there will be inscribed in the circle a regular polygon, ABODE, &c. Next describe about the circle a similar polygon, G H I K L, &c. (Prop. VII.) ; and the difference of tliese two polygons will be less than the square of Q.
Find the centre, 0 ; from the points G and H draw the straight lines GO, HO, and they will pass through the points A and B (Prop. VII.). Draw also OM to the point of tangency, M ; and it will bisect A B hi N, and be per- pendicular to it (Prop. VI. Cor. 1, Bk. III.)- Produce A 0 to E, and draw B E.
Let P represent the circumscribed polygon, and p the inscribed polygon. Thou, since these polygons are simi- lar, they are as the squares of the homologous sides G H,
1.G6 ELEMENTS OF GEOMETRY.
A B (Prop. XXXI. Bk. lY.) ; but the triangles G 0 H, A O B are similar (Prop. XXIV. Bk. IV.) ; hence tliey are to each otlier as the squares of the homologous sides OG and 0 A (Prop. XXIX. Bk. IV.) ; therefore
P : ;? : : OG^ OA' or 0 III Again, the triangles O G M, E A B, having their sides respectively parallel, are similar ; therefore
F:p:: OW : OM^ : : A E' : Blf ; and, hy division,
P : P —p : : A E^ : AE^ — ¥B^ or AB"l But P is less than tlie square described on the diameter AE; therefore P — p is less than tlie square described on AB, that is, less than the given square Q. Hence, tlie difference between the circumscribed and inscribed polygons may always be made less than any given sur- face.
372. Cor. Since the circle is obviously greater than any inscribed polygon, and less than any circumscribed one, it follows that a po/j/g-on may be inscribed or circumscribed^ which ivill differ from the circle by less than any assign- able magnitude.
Proposition XII. — Problem.
373. To find the approximate area of a circle vjhose radius is unity.
Let the radius of the circle ])o 1, and let tlie first in- scribed and circumscribed polygons be squares ; the side of the inscribed square will be V:^ (Prop. IV. Cor.), and that of the circumscribed square will be equal to tlie di- ameter 2. Hence the surface of the inscribed square is 2, and that of the circumscribed square is 4. Let, there- fore A = 2, and B = 4. Now it has been proved, in Proposition X., that tlie surface of the inscribed octagon, or, as it has been represented, A', is a mean proportional
BOOK VI. 157
between the two squares A and B, so that A' = v' 8 =. 2.8284271; and it has also been proved, in the same prop- osition, that the circumscribed octagon, represented by B',
= ^A^^l so that B' = -i^ = 3.3137085. The
inscribed and the circumscribed octagons being thus determined, we can easily, by means of them, determine the polygons having twice the number of sides. We have only in this case to put A = 2.8284271, B = 3.3137085 ; and we shall find A' = V A X B = 3.0G14674, and
B' = 2^iX^ = 3.1825979. A-f A'
In like manner may be determined the area of polygons of sixteen sides, and thence the area of polygons of thirty- two sides, and so on till we arrive at an inscribed and a circumscribed polygon differing so little from each other, and consequently fi'om the circle, that the difference shall be less than any assignable magnitude (Prop. XI- Cor.).
The subjoined table exhibits the area, or numerical ex- pression for the surface, of these polygons, carried on till they agree as far as the seventh place of decimals.
Number of sides. Inscribed Polvsjons. Circumscribed Polygons.
4 . . . . 2.0000000 .... 4.0000000
8 . . . . 2.8284271 .... 3.3137085
IG . . . . 3.0614674 .... 3.1825979
32 ... . 3.1214451 .... 3.1517249
64 ... . 3.1365485 .... 3.1441148
128 ... . 3.1403311 .... 3.1422236
256 .... 3.1412772 .... 3.1417504
512 ... . 3.1415138 .... 3.1416321
1024 .... 3.1415729 .... 3.1416025
2048 .... 3.1415877 .... 3.1415951
4096 .... 3.1415914 .... 3.1415933
8192 ..... 3.1415923 .... 3.1415928
16384 . . ." . 3.1415925 .... 3.1415927
32768 .... 3.1415926 .... 3.1415926
14
158 ELEMENTS OF GEOMETRY.
It appears, therefore, that the inscribed and circum- scribed polygons of 32768 sides differ so little from each other that the numerical value of each, as far as seven places of decimals, is absolutely the same ; as the circle is between the two, it cannot, strictly speaking, differ from either so much as they do from each other ; so that the number 3.1415926 expresses the area of a circle whose radius is 1, correctly, as far as seven places of decimals.
Some doubt may exist, perhaps, about the last decimal figure, owing to errors proceeding from the parts omitted ; but the calculation has been carried on with an additional figure, that the final result here given might be absolutely correct even to the last decimal place.
374. Cor. Since the inscribed and circumscribed poly- gons are regular, and have the same number of sides, they' are similar (Prop. I.) ; therefore, by increasing the num- ber of the sides, the corresponding polygons formed will approach to an equality with the circle. Now if, by con- tinual bisections, the polygons formed shall have their number of sides indefinitely great, each side will become indefinitely small, and the inscribed and circumscribed polygons will ultimately coincide with each other. But when they coincide with each other, they must each co- incide with the circle, since no part of an inscribed poly- gon can be without the circle, nor can any part of a circumscribed one be within it ; hence, the perimeters of the polygons must coincide with the circumference of the circle^ and be equal to it.
375. Scholium. Every circle, therefore, may be regard- ed as a polygon of an infinite number of sides.
Note. — This new definition of the circle, if it does not appear at first view to be very strict, has at least the advantage of introducing more simplicity and precision into demonstrations. {Cows de Ge'ome- trie Elementaire, par Vincent et Bourdon.)
BOOK VI.
159
Proposition XIII. — Theorem.
376. TJ\e circumferences of circles are to each other as their radii, and their areas are to each other as the squares of their radii. Let C denote tlie circumfer- ence of one of tlie circles, R its radius OA, A its area; and let C denote tlie circumfer- ence of the otlier circle, r its radius 0 B, A' its area ; tlien will
C : C : : R : r, and
A : A' : : R^ : r\
Inscribe within the given circles two regular polygons of the same number of sides ; and, whatever be the num- ber of sides, the perimeters of the polygons will be to each other as the radii 0 A and OB (Prop. IX.). Now, con- ceive the arcs subtending the sides of the polygon to be continually bisected, forming other inscribed polygons, until polygons are formed of an indefinite number of sides, and therefore having perimeters coinciding with the cir- cumference of the circumscribed circles (Prop. XII. Cor.) ; and we shall have
C : C : : R : r.
Again, the areas of the inscribed polygons are to each other as 0 A to OB (Prop. IX.). But when the num- ber of sides of tlie polygons is indefinitely increased, the areas of the polygons become equal to the areas of the circles ; hence we shall have
A : A' : : R2 : r^
160 ELEMENTS OF GEOMETRY.
377. Cor. 1. The circumferences of circles are to each other as twice their radii, or as tlieir diameters.
For, multiplying the terms of the second ratio in the first proportion by 2, we have
C : C : : 2 R : 2 r.
378. Cor. 2. The areas of circles are to each other as the squares of their diameters.
For, multiplying the second ratio of the second propor- tion by 4, or 2 squared, we have
A: A': :4R^•4r^
Proposition XI Y. — Theorem.
379. Similar arcs are to each other as their radii; and similar sectors are to each other as the squares of their radii.
Let A B, D E be similar arcs ; A C B, DOE, similar sectors ; and denote the radii C A and OD by R and r ; then will
A B : D E : : R : r,
and A C B : D 0 E : : R2 : r^.
For, since the arcs are similar, the angle C is equal to the angle 0 (Art. 213). But the angle C is to four right angles as the arc AB is to the whole circumference de- scribed with the radius C A (Prop. XVII. Sch. 2, Bk. III.) ; and the angle 0 is to four right angles as the arc D E is to the circumference described with the radius OD. Hence, the arcs A B, D E are to each other as the circum- ferences of which they form a part. But these circumfer- ences are to each other as their radii, C A, 0 D (Prop. XIII.) ; therefore
^rc A B : ^rc D E : : R : r.
By like reasoning, the sectors A C B, DOE are to each
BOOK VI.
IGl
otlicr as tlie whole circles of which they are a part ; and these are as the squares of tlieir radii (Prop. XIII.) ; therefore
Sector A C B : Sector D 0 E : : R^ : r^.
Proposition XY. — Theorem.
380. The area of a circle is equal to the product of the tircumference by half the radius.
Let C denote the circumference of the circle, whose centre is 0, R its radius 0 A, and A its area ; then will A = C X ^ R.
For, inscribe in the circle any reg- ular polygon, and from the centre draw OP perpendicular to one of the sides. The area of the polygon, whatever be the number of sides, will be equal to its perimeter multiplied by half of OP (Prop. VIII.). Conceive the arcs subtending the sides of the polygon to be continually bisected, until a polygon is formed having an indefinite number of sides ; its perimeter will be equal to the circumference of tlie circle (Prop. XII. Cor.), and 0 P be equal to the radius 0 A ; therefore the area of the polygon is equal to that of the circle ; hence
A = C X i- R.
381. Cor. 1. The area of a sector is equal to the product of its arc by half of its radius.
For, let C denote the circumference of the circle of which the sector DOE is a part, R its radius 0 D, and A its area ; tlien we sliall have (Prop. XYII. Sch. 2, Bk. III.),
Sector D 0 E : A : : Arc D E : C ;
162 ELEMENTS OF GEOMETRY.
hence, since equimultiples of two magnitudes have the same ratio as tlie magnitudes themselves (Prop. IX. Bk. IL),
Sector D 0 E : A : : ^rc D E X i^ R : C X i R.
But A, or the area of the whole circle, is equal to C X 2" R ; hence, the area of the sector D 0 E is equal to the arc D E X ^ R.
382. Cor. 2. Let the circumference of the circle whose diameter is unity be denoted by n (which is called pi), the radius by R, and the diameter by D ; and the circum- ference of any other circle by C, and its area by A. Then, since circumferences are to each other as their diameters (Prop. XIII. Cor. 1), we shall have,
C : D : : tt: 1; therefore
C = DX^=2RX^.
Multiplying both numbers of this equation by ^ R, we have
C X i- R = R' X ^, or A = R^ X ^ ; tliat is, the area of a circle is equal to the product of the square of its radius by the constant number n.
383. Cor. 3. The circumference of every circle is equal to the product of its diameter, or twice its radius, by the constant number n.
384. Cor. 4. The constant number n denotes the ratio of the circumference of any circle to its diameter ; for
C
D = "-
385. Scholium 1. The exact numerical value of the ratio denoted by n can be only approximately expressed. The approximate value found by Proposition XII. is 3.1415926 ; but, for most practical purposes, it is suf- ficiently accurate to take n = 3.1416. The symbol n is the first letter of tlie Greek word ireplfMerpop, pcrimetron, which signifies circumference.
BOOK VI. 163
386. Scholium 2. The Quadrature of the Circle is the problem "vvhich requires tiie finding of a square equiv- alent in area to a circle having a given radius. Now, it has just been proved that a circle is equivalent to the rec- tangle contained by its circumference and half its radius ; and this rectangle may be changed into a square, by find- ing a mean proportional between its length and its breadth (Prob. XXVI. Bk. Y.). To square the circle, therefore, is to find the circumference when the radius is given ; and for effecting this, it is enough to know the ratio of the cir- cumference to its radius, or its diameter.
But this ratio has never been determined except approx- imately ; but the approximation has been carried so far, that a knowledge of the exact ratio would afford no real advantage whatever beyond that of the approximate ratio. Professor Rutherford extended the approximation to 208 places of decimals, and Dr. Clausen to 250 places. The value of 71, as developed to 208 places of decimals, is 3 . 14159265358979323846264338327950288419716939937 5105820974944592307816406286208998628034825342717 0G79821480865132823066470938446095505822317253594 0812848473781392038633830215747399600825931259129 40183280651744.
Such an approximation is evidently equivalent to per- fect correctness ; the root of an imperfect power is iu no case more accurately known.
Proposition XYI. — Problem.
387. To divide a circle into any number of equal parts by means of concentric circles.
Let it be proposed to divide the circle, whose centre is 0, into a certain number of equal parts, — three for in- stance,— by means of concentric circles.
Draw the radius AO; divide AO into three equal parts, A B, B C, C 0. Upon A 0 describe a semi-circumference,
164
ELEMENTS OF GEOMETRY.
and draw the perpendiculars, B E, C D, meeting that semi-circumfer- ence in the points E, D. Join 0 E, 0 D, and with these Hnes as radii from the centre, 0, describe circles ; these circles will divide the given circle into the required number of equal parts.
For join A E, AD; then the angle ADO, being in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.) ; hence the triangles D A 0, D C 0 are similar, and consequently are to each other as the squares of their homologous sides ; that is,"
but
hence
DAO:DCO: : 0 A^- 0 D^
DAO: DCO: : OA: 00;
OA': OD^: : OA: 00;
consequently, since circles are to each other as the squares of their radii (Prop. XIII.), it follows that tlie circle whose radius is OA, is to that whose radius is OD, as OA to 0 C ; that is to say, the latter is one third of tlie former. In the same manner, by means of the right-angled tri- angles E A 0, E B 0, it may be proved that tlie circle whose radius is 0 E, is two thirds that whose radius is 0 A. Hence, the smaller circle and the two surrounding annular spaces are all equal.
Note. — This useful problem was first solved by Dr. Hutton, the justly distinguished English mathematician.
BOOK VII
PLANES. — DIEDRAL AND POLYEDRAL ANGLES.
DEFINITIONS.
388. A STRAIGHT line is per- pendiciilar to a plane ^ wlieii it is perpendicular to every straight line which it meets in that plane.
Conversely, the plane, in the same case, is perpendicular to the line.
The foot of tlie perpendicular is the point in which it meets the plane.
Thus the straight line A B is perpendicular to the plane M N ; the plane M N is perpendicular to the straight line A B ; and B is the foot of the perpendicular A B.
389. A line is parallel to a plane when it cannot meet the plane, however far both of them may be produced.
Conversely, the plane, in the same case, is parallel to the line.
390. Two planes are parallel to each other, when they cannot meet, however far both of them may be produced.
391. A PiEDRAL Angle is an angle formed by the intersec- tion of two planes, and is meas- ured by the inclination of two straight lines drawn from any point in the line of intersection, perpendicular to tliat line, one being drawn in each plane.
166
ELEMENTS OF GEOMETRY.
The line of common section is called tlie edge^ and the two planes are called the faces, of the diedral angle.
Thus the two planes ABM, A B N, whose line of intersec- tion is AB, form a diedral angle, of which the line AB is the edge, and the planes ABM, ABN are the faces.
392. A diedral angle may be acute, right, or obtuse.
If the two faces are perpendicular to eacli other, the angle is right.
303. A POLYEDRAL AnGLE IS
an angle formed by the meeting at one point of more than two plane angles, which are not in the same plane.
The common point of meeting of the planes is called the vertex, each of the plane angles a face, and the line of common section of any two of the planes an edge of the polyedral angle.
Thus the three plane angles ASB, BSC, CSA form a polyedral angle, whose vertex is S, whose faces are the plane angles, and whose edges are the sides, AS, BS, CS, of the same angles.-
394. A polyedral angle formed by three faces is called a triedral angle ; by four faces, a tetraedral ; by five faces, a pentaedral, &q.
Proposition I. — Theorem.
395. A straight line cannot he partly in a plane, and partly out of it. For, by the definition of a plane (Art. 10), a straight
BOOK VII.
167
line wliicli has two points in common with a plane lies ivholly in that plane.
396. Scholium. To determine whether a surface is a 3lane, apply a straight line in different directions to that surface, and ascertain whether the line throughout its rvhole extent touches the surface.
Proposition II. — Theorem.
397. Tivo straight lines which intersect each other lie n the same plane and determine its position.
Let A B, A C be two straight lines vliich intersect each other in A ; then hese lines will be in the same plane.
Conceive a plane to pass through I B, and to be turned about A B, mtil it pass through the point C ; hen, the two points A and C being in this plane, the line ^C lies wholly in it (Art. 10). Hence, the position of he plane is determined by the condition of its containing he two straight lines A B, AC.
398. Cor. 1. A triangle, ABC, or three points. A, B, ^, not in a straight line, determine the position of a ►lane.
399. Cor. 2. Hence, also, two parallels, A B, CD, determine he position of a plane ; for, [rawing the secant E F, the >lane of the two straight lines LB, E F is that of the parallels LB, CD.
Proposition III. — Theorem.
400. If two planes cut each other ^ their common section s a straight line.
168
ELEMENTS OF GEOMETRY.
Let the two planes A B, CD cut each other, and let E, F be two points in their common section. Draw the straight line E F. Now, since the points E and F arc in the plane A B, and also in the plane CD, the straight line E F, joining E and F, must be wholly in each plane, or is com- mon to both of them. Therefore, the common section of the two planes AB, CD is a straight line.
B
D
Proposition TV. — Theorem.
401. If a straig-ht line is perpendicular to each of tiao straight lines, at their point of intersection, it is perpen- dicular to the plane in ivhich the tioo lines lie.
Let the straight line AB be perpendicular to each of the straight lines CD, E F, at B, the point of their intersection, and MN the plane in which the lines C D, E F lie ; then will A B be perpendicular to the plane M N.
Through the point B draw any straight line, B G, in tlie plane M N ; and through any point G draw D G F, meet- ing the lines CD, E F in such a manner that D G shall be equal to G F (Frob. XXVIII. Bk. V.). Join AD, A G, A F.
The line D F being divided into two equal parts at the point G, the triangle DBF gives (Prop. XIV. Bk. IV.)
B p^ + bT)^ = 2 B^^ + 2 GFl The triangle D AF, in like manner, gives
A F' + A D- = 2 A Cr + 2 G F'. Subtracting the first equation from the second, and ob-
BOOK VII. 1G9
serving that the triangles ABF, ABD, each being right- angled at B, give
AF' — BF' = AB', and AD' — BD^=AB^, we shall have
AB' + AB' = 2 AG' — 2 BGl Therefore, by taking the halves of both members, we have
AB' = AG'— BG', or AG' = AB' + BG' ; hence, the triangle A B G is right-angled at B, and the side A B is perpendicular to B G.
In the same manner, it may be shown that AB is per- pendicular to any other straight line in the plane M N, which it may meet at B ; therefore A B is perpendicular to the plane MN (Art. 388).
402. Scholium. Thus it is evident, not only that a straight line may be perpendicular to all the straight lines which pass tlirough its foot, in a plane, but it always must be so whenever it is perpendicular to two straight lines drawn in the plane ; which shows the accuracy of the first definition (Art. 388).
403. Cor, 1. The perpendicular AB is shorter than any oblique line A G ; therefore it measures the shortest distance from the point A to the plane M N.
404. Cor. 2. From any given point, B, in a plane, only one perpendicular to that plane can be drawn. For if there could be two, conceive a plane to pass through them, intersecting the plane MN in BG; the two perpendiculars would then be perpendicular to the straight line B G at the same point, and in the same plane, which is impossible (Prop. XIII. Cor., Bk. I.).
It is also impossible to let fall from a given point out of a plane two perpendiculars to that plane. For, suppose A B, A G to be two such perpendiculars, then the triangle A B G will have two right angles, A B G, A G B, which is impossible (Prop. XXVIII. Cor. 3, Bk. I.).
ELEMENTS OF GEOMETRY.
Proposition Y. — Theorem.
405. Oblique lines drawn from a point to a plane at equal distances from a perpendicular dravm from the same point to it, are equal; and of tiuo oblique lines unequally distant from the perpendicular, the more remote is the longer.
Let A B be perpendicular to the plane M N ; and A C, AD, AE be oblique lines, from the point A, meeting the plane at equal distances, B C, B D, B E, from the per- pendicular ; and A P a line meeting the plane more remote from the perpendicular ; then will A C, A D, A E be equal to each other, and AF be longer than A C.
For, the angles ABC, ABD, ABE being right angles, and the distances BC, BD, BE being equal to each other, the triangles ABC, ABD, ABE have in each an equal angle contained by equal sides ; consequently they are equal (Prop. V. Bk. I.) ; therefore, the hypothenuscs, or the oblique lines A C, AD, AE, are equal to each other.
In like manner, since the distance B F is greater than BC, or its equal BE, the oblique line AF must be greater than AE, or its equal A C (Prop. XIV. Bk. I.).
406. Cor. All the equal oblique lines AC, AD, AE, &c. terminate in the circumference of a circle, C D E, de- scribed from B, the foot of the perpendicular, as a centre ; therefore, a point, A, being given out of a plane, the point B, at which the perpendicular let fall from it would meet that plane, may be found by taking upon the plane three points, C, D, E, equally distant from the point A, and then finding the centre of the circle which passes through these points ; this centre will be the point B required.
BOOK VII.
171
407. Scholium. The angle A C B is called the inclina- tion of the oblique line AC to the plane M N ; which inclination is evidently equal with respect to all such lilies, AC, AD, A E, as are equally distant from the per- pendicular ; for all the triangles A C B, A D B, A E B, &c. are equal to each other.
Proposition YI. — Theorem.
408. ff from the foot of a perpendicular a straight line be draivn at right angles to any straight line of the plane, and a straight line be draivn from the point of intersection to any point of the perpendicular, this last line will be perpendicular to the line of the plane.
Let A B be perpendicular to the plane M N, and B D a straight line drawn through B, cutting at right angles the straight line C E in the plane ; draw the straight line AD from the point of intersection, D, to any point. A, in the perpendicular AB; and AD will be perpendicular to CE.
For, take DE equal to D C, and join BE, BC, AE, AC. Since D E is equal to D C, the two right-angled triangles B D E, B D C are equal, and the oblique line B E is equal to B C (Prop. V. Bk. I.) ; and shice B E is equal to B C, the oblique line AE is equal to A C (Prop. Y. Bk. I.) ; therefore the line AD has two of its points, A and D, equally distant from tlie extremities E and C ; hence, AD is a perpendicular to EC, at its middle point, D (Prop. XY. Cor., Bk. I.).
409. Cor. It is also evident that C E is perpendicular to the plane of the triangle xl B D, since C E is perpendic- ular at the same time to the two straight lines A D and BD (Prop. lY.).
172
ELEMENTS OF GEOMETRY.
Proposition VII. — Theorem.
410. If a straight line is perpendicular to a plane, every plane ivhich passes through that line is also perpendicular to the plane.
Let A B be a straight line perpendicular to the plane M N ; then will any plane, A C, passing througli A B, be perpendicular to MN.
For, let C D be the inter- section of the planes A C, MN; in the plane MN draw E F, through the point B, perpendicular to C D ; then the line A B, being perpendicular to the plane M N, is perpen- dicular to each of the two straight lines CD, EF (Art. 388). But the angle ABE, formed by the two perpen- diculars AB, EF to their common section, CD, measures the angle of the two planes A 0, MN (Art. 891) ; hence, since that angle is a right angle, the two planes are per- pendicular to each other.
411. Cor. When three straight lines, as AB, CD, EF, are perpendicular to each other, each of those lines is per- pendicular to the plane of the other two, and the three planes are perpendicular to each other.
Proposition Y III . — Theorem .
412. If two planes are perpendicular to each other, a straight line drawn in one of them, perpendicular to their common section, will he perpendicular to the other plane.
Let AC, MN be two planes perpendicular to each other, and let the straight line AB be drawn in the plane A C perpendicular to the common section C D ; then will A B be perpendicular to the plane M N.
BOOK VII.
173
For, in the plane MN, draw E F, through the point B, perpendicular to CD ; then, since the planes AC, M N are perpendicular, the angle A B E is a right angle (Art. 391) ; therefore the line A B is perpendicular to the two straight lines C D, E F, at the point of their intersection ; hence it is perpendicular to their plane, MN (Prop. IV.).
413. Cor. If the plane A C is perpendicular to the plane M N, and if at a point B of the common section we erect a perpendicular to the plane M N, that perpendicular will be in the plane AC. For, if not, there may be drawn in the plane A C a line, A B, perpendicular to the common section C D, which would be at the same time perpendicu- lar to the plane M N. Hence, at the same point B there would be two perpendiculars to the plane MN, which is impossible (Prop. IV. Cor. 2).
Proposition IX. — Theorem.
414. If two planes ivhich cut ea,ch other are perpendic- ular to a third plane, their common section is perpendicu- lar to the same plane.
Let the two planes C A, D A, which cut each other in the straight line A B, be each perpendicular to the plane M N ; then will their common section A B be per- pendicular to M N.
For, at the point B, erect
My
a perpendicular to the plane MN; that perpendicular must be at once in the plane C A and in the plane D A (Prop. VIII. Cor.) ; hence, it is their common section, A B.
15*
174 ELEMENTS OP GEOMETRY.
Proposition X. — Theorem.
415. If one of two parallel straight lines is perpendic- ular to a plane, the other is also perpendicular to the same plane.
Let AB, CD be two parallel A
straight lines, of which A B is per- pendicular to the plane MN; then will C D also be perpendicular to it.
For, pass a plane through the parallels A B, CD, cutting the plane MN in the straight line BD. In the plane M N draw the straight line E F, at right angles with BD; and join AD.
Now, E F is perpendicular to the plane A B D C (Prop. VI. Cor.) ; therefore the angle C D E is a right angle ; but the angle C D B is also a right angle, since A B is per- pendicular to B D, and C D parallel to A B (Prop. XXII. Cor., Bk. I.) ; therefore the line C D is perpendicular to the two straight lines E F, B D ; hence it is perpendicular to their plane, MN (Prop. IV,).
416. Cor. 1. Conversely, if the straight lines AB, CD are perpendicular to ihe same plane, M N, they must be parallel. For, if they be not so, draw, through the point D, a line parallel to AB ; this parallel will be perpendic- ular to the plane M N ; hence, through the same point D more than one perpendicular may be erected to the same plane, which is impossible (Prop. IV. Cor. 2).
417. Cor. 2. Two lines, A and B, parallel to a third, C, are parallel to each other ; for, conceive a plane per- pendicular to the line C ; the lines A and B, being parallel to C, will be perpendicular to the same plane ; hence, by the preceding corollary, they will be parallel to each other.
The three lines are supposed to be not in the same plane ; otherwise the proposition would be already demon- strated (Prop. XXIV. Bk. I.).
BOOK VII. 175
Proposition XI. — Theorem.
418. If a straight line without a plane is parallel to a line loithin the plane ^ it is parallel to the plane.
Let the straight line A B, with- A B
out the plane M N, be parallel to the line C D in that plane ; then will AB be parallel to the plane M N.
Conceive a plane A B C D to pass through the parallels AB, CD. Now, if the line A B, wliich lies in the plane A B C D, could meet the plane M N, it could only be in some point of the line CD, the common section of the two planes ; but the line A B cannot meet C D, since they are parallel (Art. 17) ; til ere fore it will not meet the plane M N ; hence it is par- allel to that plane (Art. 389).
Proposition XII. — Theorem.
419. If tivo planes are perpendicular to the same straight line, the?/ are parallel to each other.
Let the planes M N,
|
M/ |
1 |
I |
I |
|
■■-■"/ |
/n |
||
|
/ |
|||
|
/ |
I |
5 |
/ |
PQ,be each perpendic- ular to the straight line O.; -■ "' A B ; then will they be parallel to each other.
For, if they can meet, on being produced, let O be one of their com- Q
mon points ; and join 0 A, 0 B. The line A B, which is perpendicular to the plane M N, is perpendicular to the straight line OA, drawn through its foot in that plane (Art. 388). For the same reason, AB is perpendicular to B 0. Therefore O A and 0 B are two perpendiculars let fall from the same point, 0, upon the same straight line, AB, which is impossible (Prop. XIIL Bk. I.).
1T6
ELEMENTS OP GEOMETRY.
Therefore, the planes M N, P Q cannot meet on being produced ; hence they are parallel to each other.
Proposition XIII. — Theorem.
420. If tioo parallel planes are cut hy a third plane ^ the tvjo intersections are parallel.
Let the two parallel planes F
MN and P Q be cut by the plane E F G H, and let their intersec- tions with it be E F, G H ; then E F is parallel to G H.
For, if the lines E F, G H, ly- ing in the same plane, were not parallel, they would meet each other on being produced ; therefore the planes MN, PQ, •ju which those lines are situated, would also meet, which is impossible, since these planes are parallel.
M/
N
Proposition XIY. — Theorem.
421. A straight line which is perpendicular to one of two parallel planes, is also perpendicular to the other plane.
Let M N, P Q be two parallel planes, and A B a straight line perpendicular to the plane M N ; then A B is also perpendicular to the plane PQ.
Draw any line, B C, in the plane P Q ; and through the lines A B, B C, conceive a plane, ABC, to pass, intersecting the plane M N in A D ; the intersection A D will be parallel to B C (Prop. XIJL). But the line A B, being perpendicular to the plane M N, is perpendic- ular to the straight line A D ; consequently it will be per- pendicular to its parallel BC (Prop. XXII. Cor., Bk. I.).
BOOK VII. 177
Hence the line A B, being perpendicular to any line, B C, drawn through its foot in the plane P Q, is consequently perpendicular to the plane PQ (Art. 388).
Proposition X Y. — Theorem.
422. Parallel straight lines included between tivo par- allel planes are equal.
Let E F, G H be two parallel
straight planes, included between / ^, — ^"^ i /
two parallel planes, M N, P Q ; then E F and G H are equal.
For, through the parallels EF, G H conceive the plane E F G H to pass, intersecting the parallel / ' /
planes in EG, F H. The inter- ^
sections EG, F H are parallel to each other (Prop. XIII.) ; and E F, G H are also parallel ; consequently the figure E F G H is a parallelogram ; hence E F is equal to GH (Prop. XXXI. Bk. I.).
423. Cor. Two parallel planes are everyivhere equi- distant. For, if E F, G H are perpendicular to the two planes M N, P Q, they will be parallel to each other (Prop. X. Cor. 1) ; and consequently equal.
Proposition XVI. — Theorem.
424. If two angles not in the same plane have their sides parallel and lying in the same direction., these an- gles will be equal., and their planes ivill be parallel.
Let B A C, E D F be two tri- ^[,-
angles, lying in different planes, M N and PQ, having their sides parallel and lying in the same direction ; then the angles BAG, EDF will be equal, and their planes, MN, PQ, be parallel.
ITS
ILKIMKNTS OF () KOM KTI! V.
|
M, |
/ |
TT |
TT |
/ |
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i |
/ |
".if |
^ |
?c |
/. |
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For, tako A 15 equal to K I), and A C cuiual to J) F ; and join BC, EF, JiE, AD, OF. Sinco A B is equal and parallel to E D, tlio liguro A B F I) is a parallolograni (Prop. XXXIII. Bk. 1.) ; thoroforo A D is equal and parallel to B K. For a similar reason, CF is equal and parallel to A 1); licnco, also, BE is equal and parallel to C F ; henee the figure B C F E is a parallelogram, and the side B C is e(iual and ])arallel to E F ; therefore the triangles B A C, E I) F have their sides equal, eaeh to each ; henee the angle B A C is equal to the angle E D F.
Again, the j)lane BAG is ])arallel to the plane E D F. For, if not, sup|)ose a i)lane to pass through the j)oint A, parallel to E I) F, meeting the lines BE, C F, in points diderent from B and C, for instanee G and II. Then the three lines G E, AD, II F will bo equal (Prop. XV.). But the three lines B E, A D, C F are already known to lui ecpial ; henee B E is equal to G E, and J I F is ecpial to (/ F, whieh is absurd ; henee the plane B A C is parallel to the plane E D F.
425. Cor, If two parallel ])lanes M N, P Q, are met by two other planes, A B E D, A C F D, the angles B A C, E 1) F, formed by the interseetions of the parallel planes, are equal ; for the interseetion A 15 is ])arallel to ED, and AC to DF (Prop. Xlll.) ; therefore the angle BAG is equal to the angle E 1) F.
Proposition XVII. — Theorem.
420. If three straight lines not in the same plane are equal, and parallel^ the triangles formed bj/ joining- the ex- trcinitirs of these lines ivill be equals and their planes will be pa rail eL
Lot BE, AD, GF bo three equal and parallel straight linos, not in the same plane, and let B A C, E D F be two
BOOK VII. 179
triangles formed by joining the cxtrcniitics of tlicsc lines ; then will these triangles be equal, and their planes parallel.
For, since B E is equal and parallel to AD, the figure ABED is a parallelogi-am ; hence, the side A B is equal and parallel to D E (Prop. XXXIII. Bk. I.). For a like reason, the sides BC, EF are equal and parallel ; so also are AC, D F ; hence, the two triangles B A C, E D F, having their sides equal, are themselves equal (Prop. XVIII. Bk. I.) ; consequently, as showu in the last proposition, their planes are parallel.
Proposition XVIII. — Theorem.
427. If tioo straight lines are cnt by three parallel planes^ they will be divided proportionally.
Let the straight line A B meet tlie parallel planes, M N, P Q, R S, at the points A, E, B ; and the straight line C D meet the same planes at the points C, F, D ; then will
AE:EB::CF:FD.
Draw the line AD, meeting the plane P Q in G, and draw A C, E G, B D. Then the two parallel planes PQ, RS, being cut by the plane A B D, tlie intersections E G, B D are parallel (Prop. XIII.) ; and, in the triangle A BD, Wo have (Prop. XVII. Bk. IV.),
AETeB: : AG: GD.
In like manner, the intersections A C, G F being paral- lel, in the triangle ADC, we have
AG: GD : : CF:FD;
|
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/■m.'r |
|
/ V ^ |
F / LvO |
|
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|
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180 ELEMENTS OP GEOMETRY.
hence, since the ratio A Gr : G D is common to both pro- portions, we have
AE : EB : : OF : ED.
Proposition XIX. — Theorem.
428. TJie sum of any two of the plane angles which form a triedral angle is greater than the third.
The proposition requires dem- g
onstration only when the plane angle, which is compared to the .
sum of the other two, is greater /
than either of them. /
Let the triedral angle whose / vertex is S be formed by the three "•-..... \ /'-^^
plane angles ASB, ASC, BSC; """"--\X
and suppose the angle A S B to
be greater than either of the other two ; then the angle A S B is less than the sum of the angles ASC, BSC.
In the plane ASB make the angle BSD equal to BSC; draw the straight line A D B at pleasure ; make S C equal S D, and draw A C, B C.
The two sides B S, S D are equal to the two sides B S, S C, and the angle B S D is equal to the angle BSC; therefore the triangles BSD, BSC are equal (Prop. V. Bk. I.) ; hence the side B D is equal to the side B C. But AB is less than the sum of A C and BC; taking B D from the one side, and from the other its equal, B C, there remains A D less than AC. The two sides A S, S D of the triangle A S D, are equal to the two sides A S, S C, of the triangle ASC, and the third side A D is less than the third side A C ; hence the angle A S D is less than the angle ASC (Prop. XVII. Bk. I.). Adding B S D to one, and its equal, BSC, to the other, we shall have the sum of A S D, B S D, or A S B, less than the sum of A S C, BSC. '
BOOK VII. 181
Proposition XX. — Theorem.
429. The sum of the plane angles which form any poly- edral angle is less than four right angles.
Let the polyedral angles whose vertex is S be formed by any number of plane angles, A S B, B S C, C S D, &G. ; the sum of all these plane angles is less than four right angles.
Let the planes forming the poly- edral angle be cut by any plane, ABCDEF. From any point, 0, B C
in this plane, draw the straight lines A 0, BO, CO, DO, E 0, F 0. The sum of the angles of tlie triangles A SB, BSC, (fee. formed about the vertex S, is equal to the sum of the angles of an equal number of triangles AOB, BOC, &c. formed about the point 0. But at the point B the sum of the angles ABO, 0 B C, equal to ABC, is less than the sum of the angles A B S, SBC (Prop. XIX.) ; in the same manner, at the point C we have the sum of B C 0, 0 C D less than the sum of B C S, S C D ; and so with all the angles at the points D, E, <fec. Hence, the sum of all the angles at the bases of the triangles whose vertex is 0, is less than the sum of all the angles at the bases of the triangles whose vertex is S ; therefore, to make up the deficiency, the sum of the angles formed about the point 0 is greater tlian- the sum of the angles formed about the point S. But the sum of the angles about the point 0 is equal to four right angles (Prop. IV. Cor. 2, Bk. L) ; therefore the sum of the angles about S must be less than four right angles.
430. Scholium. This demonstration supposes that the polyedral angle is convex ; that is, that no one of the faces would, on being produced, cut the polyedral angle ; if it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude.
16
182
ELEMENTS OP GEOMETRY.
Proposition XXI. — Theorem.
431. If two triedral angles are formed by plane angles which are equal each to each, the planes of the equal an- gles will be equally inclined to each other.
Let the two triedral an- gles whose vertexes are S and T, have the angle ASC equal to D T F, the angle A S B equal to DTE, and the angle B S C equal to ETF; then will the incli- nation of the planes ASC, ASB be equal to that of the planes D T F, DTE.
For, take S B at pleasure ; draw B 0 perpendicular to the plane ASC; from the point 0, at which the perpen- dicular meets the plane, draw OA, 0 C, perpendicular to S A, S C ; and join A B, B C. Next, take TE equal S B ; draw E P perpendicular to the plane DTE; from the point P draw PD, PF, perpendicular respectively to TD, T F ; and join D E, E F.
The triangle S A B is right-angled at A, and the trian- gle T D E at D ; and since the angle A S B is equal to DTE, we have S B A equal to T E D. Also, S B is equal to T E ; therefore the triangle S AB is equal to T D E ; hence S A is equal to T D, and A B is equal to D E.
In like manner it may be shown that S C is equal to T F, and B C is equal to E F. We can now show that the quadrilateral ASCO is equal to the quadrilateral DTFP; for, place the angle ASC upon its equal D T F ; since S A is equal to T D, and S C is equal to T F, the point A will fall on D, and the point C on F ; and, at the same time, AO, which is perpendicular to S A, will fall on D P, which is perpendicular to TD, and, in like manner, CO on F P ; wherefore the point 0 will fall on the point P, and A 0 will be equal to D P.
BOOK vir. 183
But the triangles A 0 B, D P E are right-angled at 0 and P; tiie hypotenuse AB is equal to D E, and the side A 0 is equal to D P ; hence the two triangles are equal (Prop. XIX. Bk. I.) ; and, consequently, the angle OAB is equal to the angle PDE. The angle OAB is the incli- nation of the two planes A S B, A S C ; and the angle PDE is that of the two planes DTE, DTP; hence, those two inclinations are equal to each other.
432. Scholium 1. It must, however, be observed, that the angle A of the right-angled triangle 0 A B is properly the inclination of the two planes A S B, A S C only when the perpendicular B 0 falls on the same side of S A with S C ; for if it fell on the other side, the angle of the two planes would be obtuse, and joined to the angle A of the triangle 0 A B it would make two right angles. But, in the same case, the angle of the two planes DTE, D T F would also be obtuse, and joined to the angle D of the triangle D P E it would make two right angles ; and the angle A* being thus always equal to the angle D, it would follow in the same manner that the inclination of the two planes A S B, A S C must be equal to that of the two planes DTE, DTP.
433. Scholium 2. If two triedral angles are formed by three plane angles respectively equal to each other, and if at the same time the equal or homologous angles are similarly situated^ the two angles are equal. For, by the proposition, the planes which contain the equal angles of tlie triedral angles are equally inclined to each other.
434. Scholium 3. When the equal plane angles forming tlie two triedral angles are not similarly situated^ these angles are equal in all their constituent parts, but, not admitting of superposition, are said to be equal by sym- metry, and are called symmetrical angles.
BOOK VIII
POLYEDRONS.
DEFINITIONS.
435. A PoLYEDRON is a solid, or volume, bounded by planes.
The bounding planes are called the face$ of the polye- dron ; and the lines of intersection of the faces are called tlie edges of the polyedron.
436. A Prism is a polyedron having two of its faces equal and parallel pol- ygons, and the other faces parallelo- grams.
The equal and parallel polygons are called the bases of the prism, and the parallelograms its lateral faces. The lateral faces taken together constitute the lateral or convex surface of the prism.
Thus the polyedron ABCDE-K is a prism, having for its bases the equal "and parallel polygons ABODE, F G H I K, and for its lateral faces the parallelograms A B G F, B C H G, <fec.
The principal edges of a prism are those which join the corresponding angles of the bases ; as A F, B G, <fec.
437. The altitude of a prism is a perpendicular drawn from any point in one base to the plane of the other.
438. A Right Prism is one whose principal edges are perpendicular to the planes of its bases. Each of the
BOOK VIII.
185
edges is then equal to the altitude of the prism. Every other prism is oblique^ and has each edge greater than the altitude.
439. A prism is triangular^ quadrangular, pentangular, hexangular, &c., according as its base is a triangle, a quadrilateral, a pentagon, a hexagon, &c.
440. A Parallelopipedon is a prism whose bases are parallelograms ; as the prism A B C D - H.
The parallelopipedon is rectangular when all its faces are rectangles ; as the parallelopipedon A B C D - H.
441. A Cube, or Regular Hexaedron, is a rectangular parallelopipedon having all its faces equal squares ; as the paral- A lelopipedon A B C D - H.
442. A Pyramid is a polyedron of which one of the faces is any polygon, and all the others are triangles meeting at a common point.
The polygon is called the base of the pyramid, the triangles its lateral faces, and the point at which the triangles meet its vertex. The lateral faces taken to- B
gether constitute the lateral or convex surface of the pyramid.
Thus the polyedron ABCDE-Sisa pyramid, having for its base tlie polygon A B C D E, for its lateral faces tlie triangles ASB, BSC, CSD, &c., and for its vertex the pohit S.
16*
186 ELEMENTS OF GEOMETRY.
443. The Altitude of a pyramid is a perpendicular drawn from the vertex to the plane of the base.
444. A pyramid is triang-ular, quadrangular, &g., ac- cording as its base is a triangle, a quadrilateral, &c.
445. A Right Pyramid is one whose base is a regular polygon, and the perpendicular drawn from the vertex to the base passes through the centre of the base, in this case tlie perpendicular is called the axis of the pyramid.
446. The Slant Height -of a right pyramid is a line drawn from the vertex to the middle of one of the sides
of the base.
'i
447. A Frustum of a pyramid is the part of the pyra- mid included between the base and a plane cutting the pyramid parallel to the base.
448. The Altitude of the frustum of a pyramid is the perpendicular distance between its parallel bases.
449. The Slant Height of a frustum of a right pyra- mid is that part of the slant height of the pyramid which is intercepted between the bases of the frustum.
450. The Axis of the frustum of a pyramid is that part of the axis of the pyramid which is intercepted between the bases of the frustum.
451. The Diagonal of a polyedron is a line joining the vertices of any two of its angles which are not in tlie same face.
452. Similar Polyedrons are those which are bounded by the same number of similar faces, and have their poly- edral angles respectively equal.
453. A Regular Polyedron is one whose faces are all equal and regular polygons, and whoso polyedral angles are all equal to each otlicr.
B C
BOOK YIII. 187
Proposition I. — Theorem.
454. The convex surface of a rig-ht prism is equal to the perimeter of its base multiplied by its altitude.
Let ABCDE-K be a right prism; K
then will its convex surface be equal p to the perimeter of its base,
AB+.BC + CD + DE + EA,
multiplied by its altitude A F.
For, the convex surface of the prism is equal to the sum of the parallelo- grams AG, BH, CI, DK, EF (Art. 436). Now, the area of each of those parallelograms is equal to its base, AB, B C, CD, &c., multiplied by its altitude, AF, B G, CH, &c. (Prop. Y. Bk. IV.). But the altitudes AF, BG, C H, (fcc. are each equal to A F, the altitude of the prism. Hence, the area of these paral- lelograms, or the convex surface of the prism, is equal to
(AB + BC + CD + DE + EA)X AF; or the product of the perimeter of the prism by its alti- tude.
455. Cor. If two right prisms have the same altitude, their convex surfaces are to each other as the perimeters of their bases.
Proposition II. — Theorem.
456. In everij prism, the sections formed by parallel planes are equal polygons.
Let the prism ABCDE-K be intersected by the parallel planes N P, S Y ; then are the sections NOPQR, STYXY equal polygons.
For the sides ST, NO are parallel, being the intersec- tions of two parallel planes with a third plane A B G F
188
ELEMENTS OF GEOMETRY.
(Prop. XIII. Bk. YII.) ; these same sides ST, NO, are included between the parallels N S, OT, which are sides of the prism ; hence N 0 is equal to S T. For like reasons, the sides 0 P, P Q, Q R, &c. of the section N 0 P Q R, are respectively equal to the sides TV, YX, XY, &c. of the section S T Y X Y ; and since the equal sides are at the same time parallel, it fol- lows that the angles NOP, OP Q, &c. of the first section are respectively equal to the angles STY, TYX of the second (Prop. XYI. Bk. YII.). Hence, the two sections NOPQR, S T Y X Y, are equal polygons.
457. Cor, Every section made in a prism parallel to its base, is equal to that base.
Proposition III. — Theorem.
458. Two prisms are equal, when the three faces which form a triedral angle in the one are equal to those vjhich form a triedral angle in the other, each to each, and are similarly situated.
Let the two prisms A.BCDE-K and LMOPQ-Y have the faces which form the triedral angle B equal to the faces which form the tri- edral angle M ; that is,the base ABODE equal to the base L M N 0 P Q, the parallelogram A B G F equal to the parallelogram LMSR, and the parallelogram B 0 H G equal to MOTS; then the two prisms are equal.
BOOK VIII.
189
For, apply the base ABODE to the equal base LMOPQ ; then, the triedral an- gles B and M, being equal, will coincide, since the plane an- gles which form these triedral angles are B C MO
equal each to each, and similarly situated (Prop. XXI. Sch. 2, Bk. yil.) ; hence the edge B G will fall on its equal M S, and the face B H will coincide with its equal MT, and the face BF with its equal MR. But the upper bases are equal to their corresponding lower bases (Art. 436) ; therefore the bases F G H I K, R S T V Y are equal ; hence they coincide with each other. Therefore H I coin- cides with TV, IK with VY, and KF with YR; and consequently the lateral faces coincide. Hence the two prisms coincide throughout, and are equal.
459. Cor, Two right prisms, which have equal bases and equal altitudes, are equal.
For, since the side AB is equal to LM, and the altitude B G to M S, the rectangle A B G F is equal to the rectan- gle L M S R ; so, also, the rectangle B G H C is equal to M S T 0 ; and thus the three faces which form the triedral angle B, are equal to the three faces which form the trie- dral angle M. Hence the two prisms are equal.^^^X-»X.^
Proposition I Y. — Theorem. V\C> ^^>v
460. In every parallelopipedon the opposite ./^^^Os,^-. equal and parallel, ^^^>^ tj
Let ABCD-H be a parallelopipedon ; then its opp5^ site faces are equal and parallel.
The bases ABC D, EFGH are equal and parallel (Art. 436), and it remains only to be shown that the same is
190
ELEMENTS OF GEOMETRY.
H
X>
B
true of any two opposite lateral faces, as B CGF, ADHE. Now, since the base A B C D is a parallelogram, the side
A D is equal and parallel to B C. For r) ^G
a similar reason, AE is equal and par- allel to B F ; hence the angle DAE is equal to the angle C B F (Prop. XYI. Bk. VII.), and the planes DAE, CBF are parallel ; hence, also, the parallelogram B C G 1^^ is equal to the parallelogram A D H E. In the same way, it may be- shown that the opposite faces A B F E, D C G II are equal and parallel.
461. Cor. Any two opposite faces of a parallelopipe- don may be assumed as its bases, since any face and the cue opposite to it are gqual and parallel.
Proposition V. — Theorem.
G
462. The diagonals of every parallelopipedon bisect each other.
Let A B C D - H be a parallelo- pipedon ; then its diagonals, as B H, D F, will bisect each other.
For, since B F is equal and par- allel to D H, the figure B F H D is a parallelogram ; hence the diago- nals B H, D F bisect each other at the point 0 (Prop. XXXIV. Bk. I.). In the same man- ner it may be shown that the two diagonals A G and C E bisect each other at the point 0 ; hence the several diag- onals bisect each other.
463. Scholium. The point at which the diagonals mu- tually bisect each other may be regarded as the centre of the parallelopipedon.
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BOOK VIII.
191
Proposition VI. — Theorem.
464. Amj paraUelopipedon may be divided into tico equivalent triangular prisms by a plane passing through its opposite diagonal edges.
Let any paraUelopipedon, ABC D-H, q
be divided into two prisms, A B C - G, AD C-G, by a plane, A C G E, passing through opposite diagonal edges; then will the two prisms be equivalent.
Through the vertices A and E, draw the planes A K L M, E N 0 P, perpen- dicular to the edge AE, and meeting BE, CG, DH, the three other edges of the paraUelopipedon, in the points K, L, M, and in N, 0, P. The sections A K L M, E N O P are equal, since they are formed by planes perpendicular to the same straight lines, and hence parallel (Prop. II.). They are parallelograms, since the two opposite sides of the same section, AK, LM, are the intersections of two parallel planes, A B F E, D C G H, by the same plane, AKLM (Prop. XIII. Bk. VII.).
For a like reason, the figure A M P E is a parallelo- gram ; so, also, are A K N E, K L 0 N, L M P 0, the other lateral faces of the solid AKLM-P; consequently, this solid is a prism (Art. 436) ; and this prism is right, since the edge AE is perpendicular to the plane of its base. This right prism is divided by the plane ALOE into the two right prisms AKL-0, AML-0, which, liaving equal bases, A K L, A M L, and the same altitude, AE, are equal (Prop. III. Cor.).
Now, since AEHD, AEPM are parallelograms, the sides D H, MP, being each equal to AE, are equal to each other ; and taking away the common part, D P, there re- mains D M equal to H P. In the same manner it may be shown that C L is equal to G 0.
192
ELEMENTS OP GEOMETRY.
Conceive now E P 0, the base of the solid E P 0 - G, to be applied to its equal A M L, the point P falling upon M, and the point 0 upon L ; the edges G 0, H P will coincide with their equals C L, DM, since they are all perpen- dicular to the same plane, A K L M. Hence tlie two solids coincide through- out, and are therefore equal. To each of these equals add the solid ADC-P, and the right prism AML-0 is equivalent to the prism ADC-G.
In the same manner, it may be proved that the riglit prism AKL-0 is equivalent to the prism ABC-G. Tlie two riglit prisms A K L - 0, A M L - 0 being equal, it fol- lows that two triangular prisms, ABC-G, ADC-G, are equivalent to each other.
465. Cor. Every triangular prism is half of a parallelo- pipedon having the same triedral angle, with the same edges.
Proposition VII. — Theorem.
466. Two parallelopipedons, having a common lower base, and their upper bases in the same plane and between the same parallels, are equivalent to each other.
Let the two parallelo- pipedons A G, A L have tlie commonbase ABCD, and their upper bases, EFGH, IKLM, in the same plane, and be- tween the same paral- lels, EK, HL; then the parallelopipedons will be equivalent.
BOOK VIII. 193
There may be throe cases, according as E I is greater or less than, or equal to, EF; but the demonstration is the same for each.
Since AE is parallel to BF, and HE to GF, the pla]io anjrle A E I is equal to B F K, H E I to G F K, and H E A to G F B. Of these six plane angles, the three first form the polyedral angle E, the three last the polyedral angle F ; consequently, since these plane angles are equal each to each, and similarly situated, the polyedral angles, E, F, must be equal. Now conceive the prism A E I - M to be applied to the prism B F K - L ; the base A E I, being placed upon the base B F K, will coincide witli it, since they are equal ; and, since the polyedral angh^ E is equal to the polyedral angle F, the side E II will fall upon its equal, F G. But the base A E I and its edge I'] II deter- mine the prism A E I-M, as the base B F K and its edge FG determine tlje prism BFK-L (Prop. III.) ; hcuce the two prisms coincide throughout, and therefore are equal to each other.
Take away, now, from the whole solid AELC, the prism AEI-M, and tliere will remain the parallelopipcdon AL; and take away from the same solid A L the prism BFK-L, and there will remain the parallelopipcdon A G ; hence the two parallelopipedons A L, A G are equivalent.
Proposition YIII. — Theorem.
467. Tvjo parallelopipedons having- the same base and the same altitude are equivalent.
Let the two parallelopipedons AG, A L have tlie com- mon base ABCD, and the same altitude ; then will the two parallelopipedons be equivalent.
For, the upper bases EFGH, IKLM being in the same plane, produce the edges E F, HG, L K, IM, till ]>y their intersections they form the parallelogram N 0 P Q ; this parallelogram is eq\ial to eitlier of the bases I L, E G, and
17
194
ELEMENTS OF GEOMETRY.
is between the same par- allels ; hence N 0 P Q is equal to the common base A B C D, and is parallel to it.
Now, if a third paral- lelopipedon be conceived, which, with tlie same lower base A B C D, has for its upper base NOPQ,
this tliird parallelopipe- A"^ ^B
don will be equivalent to the parallelopipedon A G, since the lower base is the same, and tlie upper bases lie in the same plane and between the same parallels, GQ, FN (Prop. VII.).
For the same reason, this third parallelopipedon will also be equivalent to the parallelopipedon A L ; hence the two parallelopipedons AG, AL, which have the same base and the same altitude, are equivalent.
Proposition IX. — Theorem.
468. Any oblique 'parallelopipedon is equivalent to a rectangular parallelopipedon hauing- the same altitude and an equivalent base.
Let A G be any paral- H G
lelopipedon ; then A G will be equivalent to a rectangular parallelopip- edon having the same altitude and an equiv- alent base.
From the points A, B, C, D,draw AI, BK, C L, D M, perpendicular to the lower base, and equal in altitude to A G ; there will thus be formed the
BOOK VIII.
195
parallelopipedon AL, equivalent to A G (Prop. YIII.), and liaving its lateral faces, AK, B L, <fec., rectangular. Now, if the base ABCD is a rectangle, AL will be a rec- tangular parallelopipedon equivalent to A G.
But if ABCD is not a rectangle, j^jq ^p
draw A 0, B N, each perpendicular to CD; also 0 Q, N P, each perpendicular to the base ; then we shall have a rec- tangular parallelopipedon A B N 0 - Q. For, by construction, the bases ABNO, I K P Q are rectangles ; so, also, are the lateral faces, the edges A I, 0 Q, &c. being perpendicular to the plane of the base ; therefore the solid A P is a rectangular parallelopipedon. But the two parallelopip- edons A P, A L may be considered as having the same base, A BKI, and tlie same altitude, AO ; hence they are equivalent. Hence the parallelopipedon A G, which was shown to be equivalent to the parallelopipedon AL, is also equivalent to tlie rectangular parallelopipedon A P, having the same altitude, A I, and a base, ABNO, equivalent to the base ABCD.
Proposition X. — Theorem.
469. Two rectangular parallelopipedons, tvhich the same base, are to each other as their altitudes.
Let the two parallelopipedons AG, -^ AL have the same base, ABCD ; then they are to each other as their altitudes, AE, AL
First. Suppose tlie altitudes AE, AI are to each other as two whole numbers ; for example, as 15 is to 8. Divide A E into 15 equal parts, of which A I will contain 8. Through .^•, y, z, <fec., the points of division, conceive planes to
have
196
ELEMENTS OP GEOMETRY.
pass parallel to the common base. These planes will divide the solid A G into 15 small parallelopipedons,all equal to each other, having equal bases and equal altitudes ; equal bases, since every section, as 1 K L M, parallel to the base A B C D, is equal to that base (Prop. II.), and equal altitudes, since the alti- tudes are the equal divisions Ax, xy^ y 2r, &c. But of those 15 equal parallel- opipedons, 8 are contained in A L ; hence the parallelo- pipedon A G is to the parallelopipedon A L as 15 is to 8, or, in general, as the altitude A E is to the altitude A I.
Secondly, If the ratio of A E to A I cannot be exactly expressed by numbers, we shall still have tlie proportion,
Solid A G : Solid A L : : A E : A I. For, if this proportion is not correct, suppose we have
Solid A G : Solid A L : : A E : A 0 greater than A I.
Divide A E into equal parts, each of which shall be less than I 0 ; there will be at least one point of division, m, between I and 0. Let P represent the parallelopipedon, whose base is A B C D, and altitude Am ; since the alti- tudes AE, Am are to each other as two whole numbers, we shall have
Solid AG : P : : AE : Am.
But, by hypothesis, we have
Solid A G : Solid A L : : A E : A 0 ; hence (Prop. X. Cor. 2, Bk. II.),
Solid AL : P : : AG : Am. But AO is greater than Am ; lience, if the proportion is correct, the parallelopipedon A L must be greater than P. On the contrary, however, it is less ; consequently the solid A G cannot be to the solid A L as the line A E is to a line greater than A I.
BOOK VIII.
197
By the same mode of reasoning, it may be shown that tlie fourth term of the proportion cannot be less tlian A I ; therefore it must be equal to A I. Hence rectangular parallelopipedons, having the same base, are to each other as their altitudes.
Proposition XI. — Theorem.
470. Tivo rectangular parallelopipedons^ having' the same altitude, are to each other as their bases.
Let the two rectan-
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gular parallelopipedons AG, AK have the same altitude, A E ; then they are to each other as their bases.
Place the two solids so that their faces, BE, 0 E, may have the com- mon angle B A E ; pro- duce the plane 0 N K L till it meets the plane D C G H in P Q ; we shall thus have a third parallelopipedon, A Q, wliich may be compared with each of the parallelopipedons A G, A K. The two solids, A G, A Q, having the same base, A E H D, are to each other as their altitudes A B, A 0 (Prop. X.) ; in like manner, the two solids A Q, AK, having the same base, A 0 L E, arc to each other as their altitudes AD, AM. Hence we have the two proportions,
Solid A G : Solid A Q : : A B : A 0, Solid A Q : Solid A K : : A D : A M. Multiplying together the corresponding terms of these
B
198^ *
ELEMENTS OP GEOMETRY.
proportions, and omitting, in the result, the common factor Solid A Q, we shall have,
Solid A G : Solid AK::ABxAD:AOxAM.
But A B X A D measures the base A B C D (Prop. lY. Sell., Bk. lY.) j a^id AO X AM measures the base A M N 0 ; hence two rectangular parallelopipcdons of the same altitude are to each other as their bases.
M
Proposition XII. — Theorem.
471. Any tivo rectang-ular parallelopipedons are to each other as the product of their bases by their altitudes.
Let A G, A Z be two i v vi
rectangular parallelo- pipedons ; then they are to each other as the product of their bases, ABCD, AMNO, by their altitudes, A E, AX.
Place the two solids so that their faces, B E, 0 X, may have the com- mon angle B A E ; pro- duce the planes neces- sary for completing the third parallelopipedon, A K, having the same altitude with the parallelopipedon A G. By the last proposition, we shall have
Solid A G : Solid AK::ABCD:AMNO.
But the two parallelopipedons A K, A Z, having the same base, A M N 0, are to each other as their altitudes, A E, A X (Prop. X.) ; hence we have
Solid AK : Solid A Z : : A E : A X. Multiplying together the corresponding terms of these
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BOOK VIII. 199
proportions, and omitting, in the result, the common fac- tor Solid A K, we shall have
SolidKGc'.Solid^Z::A.BOJ)X^^: AMNOx AX. Hence, any two rectangular parallelopipedons are to each other as the products of their bases by their altitudes.
472. Scholium 1. We are consequently authorized to assume, as the measure of a rectajigular parallelopipedon, the product of its base by its altitude ; in other words, the product of its three dimensions. But by the product of two or more lines is always meant the product of the num- bers which represent them ; those numbers themselves being determined by the particular linear unit, which may be assumed as the standard. It is necessary, therefore, in comparing magnitudes, that the measuring unit be the same for each of the magnitudes compared.
473. Scholium 2. The measured magnitude of a solid, or volume, is called its volume, solidit?/, or solid contents. We assume as the unit of volume, or solidity, the cube, each of whose edges is the linear unit, and each of whose faces is the unit of surface.
Proposition XIII. — Theorem.
474. The solid contents of a parallelopipedon, and of any other prism, are equal to the product of its base by its altitude.
First. Any parallelopipedon is equivalent to a rectan- gular parallelopipedon having the same altitude and an equivalent base (Prop. IX.). But the solid contents of a rectangular parallelopipedon are equal to the product of its base by its altitude; therefore the solid contents of any parallelopipedon are equal to the product of its base by its altitude.
Second. Any triangular prism is half of a parallelopip- edon, so constructed as to have the same altitude, and a
200 ELEMENTS OF GEOMETRY.
base twice as great (Prop. YI.). But the solid contents of the parallelopipedon are equal to the product of its base by its altitude ; hence, that of the triangular prism is also equal to the product of its base, or half that of the paral- lelopipedon, by its altitude.
Third. Any prism may be divided into as many trian- gular prisms of the same altitude, as there are triangles in the polygon taken for a base. But the solid contents of each triangular prism are equal to the product of its base by its altitude; and, since the altitude is the same in each, it follows that the sum of all these prisms is equal to the sum of all the triangles taken as bases multiplied by the common altitude.
Hence the solid contents of any prism are equal to the product of its base by its altitude.
475. Cor. When any two prisms have the same altitude, the products of the bases by the altitudes will be as the bases (Prop. IX. Bk. 11.) ; hence, prisms of the same altitude are to each other as their bases. For a like reason, prisms of the same base are to each other as their altitudes.
Proposition XI Y. — Theorem.
476. Similar prisms are to each other as the cubes of their homologous edges.
Let ABC-E, FHI-M e be two similar prisms ; these prisms are to each other as the cubes of their homologous edges, A B and FII.
For, from I) and K, ho- mologous angles of the two prisms, draw the perpendiculars DN, KO, to the bases ABC, FHI. Take AK' equal to F K, and join AN.
BOOK VIIT.
201
Draw K' 0' perpendicular to A N in the plane AND, and K' 0 will be perpendicular to tlie plane ABC, and equal to K 0, the altitude of the prism F H I - M. For, con- ceive the triedral angles A and F to be applied the one to the other ; the planes containing them, and therefore the perpendiculars K' 0', K 0, will coincide.
Now, since the bases A B C, F H I are similar, we have (Prop. XXIX. Bk. IV.),
Base ABC: Base F H I : : AB^ : FH^ ;
and, because of the similar triangles DAN, K F 0, and of the similar parallelograms D B, K H, we have
DN:KO::DA:KF::AB:FH.
Hence, multiplying together the corresponding terms of these proportions, we have
Base ABC XT)^ : Base FRlXKO: AB^ : FW.
But the product of the base by the altitude is equal to the solidity of a prism (Prop. XIII.) ; hence
PmmABC-E: Pmm F H I-M : : AB' : FH'.
Proposition XV. — Theorem.
477. The convex surface of a rig-lU pyramid is to the perimeter of its base, multiplied by half the height.
Let A B C D E - S be a right pyra- mid, and S M its slant heiglit ; then the convex surface is equal to the perimeter AB + BC + CD + DE + EA mul- tiplied by J- S M.
The triangles SAB, SBC, S C D, <fcc. are all equal ; for the sides A B, BC, CD, &c. are equal (Art. 445), and the sides S A, S B, S C, &c., being ob- lique lines meeting the base at equal
equal slant
202
ELEMENTS OF GEOMETRY.
distances from a perpendicular let fall from the vertex S to the centre of the base, are also equal (Prop. V. Bk. VII.). Hence, these triangles are all equal (Prop. XVIII. Bk. I.) ; and the altitude of each is equal to the slant height S M. But the area of a triangle is equal to the product of its base mul- tiplied by half its altitude (Prop. VI. Bk. IV.). Hence, the areas of the tri- angles SAB, SBC, SCD, &c. are equal to the sum of the bases A B, B C, CD, &c. multiplied by half the common altitude, S M ; that is, the convex surface of the pyramid is equal to the perimeter of the base multiplied by half the slant height.
478. Cor. The lateral faces of a right pyramid are equal isosceles triangles, having for their bases the sides of the base of the pyramid.
Proposition XVI. — Theorem.
479. If a pyramid be cut bi/ a plane parallel to its base, —
1st. The edges and the altitude will be divided propor- tionally.
2d. The section ivill be a polygon similar to the base.
Let the pyramid A B C D E - S, whose S
altitude is SO, be cut by a plane, G H I K L, parallel to its base ; then will the edges S A, S B, S C, &c., with the altitude SO, be divided proportion- ally ; and the section G H I K L will be similar to the base ABODE.
First. Since the planes ABC, GHI are parallel, their intersections A B, GH, by the third plane SAB, are parallel (Prop. XIII. Bk. VII.); hence
BOOK vm. 20.S
tlie triangles SAB, SGH are similar (Prop. XXY. Bk. IV.), and we have
SA:SG::SB:SH.
For the same reason, we have
SB: SH: : SC: SI;
and so on. Hence all the edges, S A, S B, S C, &c., are cut proportionally in G, H, I, &c. The altitude S 0 is likewise cut in the same proportion, at the point P ; for B 0 and H P are parallel ; therefore we have
SO:SP::SB:SH.
Secondly, Since G H is parallel to AB, H I to B C, I K to C D, (fee. the angle G H 1 is equal to A B C, the angle H I K to B C D, and so on (Prop. XVI. Bk. VII.). Also, by reason of the similar triangles SAB, S G H, we have
AB:GH::SB:SH;
and by reason of the similar triangles SBC, SHI, we have
SB: SH: : BC : HI;
hence, on account of the common ratio S B : S H,
A B : G H : : B C : H I. For a like reason, we have
B C : H I : : C D : I K,
and so on. Hence the polygons ABCDE, GHIKL have their angles equal, each to each, and their homolo- gous sides proportional ; hence they are similar.
480. Cor. 1. If two pyramids have the same altitude, and their bases in the same plane, their sections made by a plane parallel to the plane of their bases are to each other as their bases.
Let ABCPE-S, MNO-S be two pyramids, having the same altitude, and their bases in the same plane ; and let G H I K L, P Q R be sections made by a plane parallel
204
ELEMENTS OF GEOMETRY.
: S A : S G.
:S^^
to the plane of their bases ; then these sections are to each other as the bases ABODE, M N 0.
For, the two polygons ABODE, GHIKL be- ing similar, their surfaces are as the squares of the homologous sides AB, GH (Prop. XXXI. Bk. IV.). But
AB: GH Hence,
ABODE: GHIKL: :SA'
For the same reason,
M N O : P Q R : : S M^ : SP^- But since GHIKL and P Q R. are in the same plane, we have also (Prop. XVIII. Bk. VII.),
SA: SG::SM: SP; hence
' ABODE: GHIKL: :MNO:PQR;
therefore the sections GHIKL, PQR are to each other as the bases ABODE, M N 0.
481. Cor. 2. If the bases A B 0 D E, M N O are equiv- alent, any sections, GHIKL, PQR, made at equal dis- tances from those bases, are likewise equivalent.
Proposition XVII. — Theorem.
482. The convex surface of a frustum of a right pi/ra- viid is equal to half the sum of the perimeters of its two bases, multiplied by its slant height.
Let ABODE-L be the frustum of a riglit pyramid, and MN its slant height; then the convex surface is equal to the sum of the perimeters of the two bases ABODE, G H I K L, multiplied by half of M N.
BOOK VIII.
205
For the upper base G H I K L is similar to the base ABODE (Prop. XVI.), and ABODE is a regular polygon (Art. 445) ; hence the sides G H, HI, IK, K L, and L G are all equal to each other. The angles GAB, A B H, HBO, &c. are equal (Prop. XV. Oor.), and the edges AG, B H, 01, &c. are also equal (Prop. XVI.) ; therefore the faces A H, B I, OK, &c. are all equal trapezoids (Art. 28), havuig a common altitude, M N, the slant height of the frustum. But the area of either trapezoid, as AH, is equal to ^(AB+GH)XMN (Prop. VII. Bk. IV.) ; hence the areas of all the trapezoids, or the convex surface of frustum, are equal to half the sum of the perimeters of the two bases multiplied by the slant height.
Proposition XVIII. — Theorem.
483. Triangular pyramids^ having equivalent bases and the same altitude, are equivalent. T S __
B B'
Let A B 0 - S, A' B' 0' - S' be two triangular pyramids, having equivalent bases, ABO, A' B' 0', situated in the same plane ; and let them have the same altitude, A T ; then these pyramids are equivalent.
For, if the two pyramids are not equivalent, let A' B' 0 - S' be the smaller, and suppose A X to be the
18
206
ELEMENTS OF GEOMETRY.
altitude of a prism, which, having ABC for its base, is equal to thoir difference.
T S
B B'
Divide the altitude A T into equal parts, each less than A X ; through each point of division pass a plane parallel to the plane of the base, thus forming corresponding sections in the two pyramids, equivalent each to each, namely, DEF to D'E'F, GHI to G'HT, &c.
Upon the triangles ABC, DEF, GHI, <fec., taken as bases, construct exterior prisms, having for edges the parts AD, D G, GK, &c. of the edge S A ; in like manner, on the bases D' E' F', G' H' T, &c. in the second pyramid, construct interior prisms, having for edges the correspond- ing parts of S' A'. It is plain that the sum of all the ex- terior prisms of the pyramid. A B C-S is greater than this pyramid ; and also that the sum of all the interior prisms of the pyramid A'B'C'-S' is less than this pyramid. Hence, the difference between the sum of all tlie exterior prisms and the sum of all the interior ones, must be greater than the difference between the two pyramids themselves.
Now, beginning with the bases ABC, A'B'C', the second exterior prism, D E F - G, is equivalent to the first interior prism, D'E'F'-A', since they have equal alti- tudes, and their bases, DEF, D'E'F', are equivalent. For a like reason, the third exterior prism, GHI-K, and tlio second interior prism, G'HT-D', are equivalent; and so
BOOK VIII.
207
on to the last in each series. Hence, all the exterior prisms of the pyramid A B C - S, excepting the first prism, ABC-D, have equivalent correspondhig ones in the in- terior prisms of the pyramid A B C'-S'. Therefore the prism ABC-D is the difference hetween the sum of all the exterior prisms of the pyramid ABC-S, and the sum of the interior prisms of the pyramid A' B' C - S'. But the difference between these two sets of prisms has been proved to be greater than that of the two pyramids, which latter difference we supposed to be equal to tlie prism A B C - X. Hence, the prism ABC-D must be greater than tlie prism A B C -X, which is impossible, since they have the same base, ABC, and the altitude of the first is less than A X, the altitude of the second. Hence, the supposed inequality between the two pyramids cannot exist ; therefore the two pyramids ABC-S, A' B' C- S', having the same altitude and equivalent bases, are them- selves equivalent.
Proposition XIX. — Theorem.
484. Ever?/ triangular pyramid is a third part of a tri- angular prism having the same base and the same altitude.
Let A B C - F be a triangu- lar pyramid, and A B C - D E F a triangular prism of the same base and the same altitude ; tlien the pyramid is one third of the prism.
C ut off the pyramid A B C - F from the prism, by the plane F A C ; there will remain the solid A C D E - F, which may be considered as a quadrangu- lar pyramid, whose vertex is F, and whose base is the parallelogram A C D E. Draw the
208
ELEMENTS OP GEOMETRY.
diagonal CE, and pass the plane F C E, which will cut the quad- rangular pyramid into two tri- angular ones, ACE-F, EDC-F. These two triangular pyramids have for their common altitude the perpendicular let fall from F on the plane A C D E ; they have equal bases, since the tri- angles A C E, C D E are halves of the same parallelogram ; hence the two pyramids ACE-F, C D E-F are equivalent (Prop. XYIII.). But the pyra- mid CDE-F and the pyramid ABC-F have equal bases, ABC, D E F ; they have also the same altitude, namely, the distance between the parallel planes ABC, D E F ; hence the tw^o pyramids are equivalent. Now, the pyramid CDE-F has been proved equivalent to ACE-F; hence the three pyramids ABC-F, CDE-F, A C E-F, which compose the whole prism A B C -D E F, are all equivalent; therefore, either pyramid, as ABC-F, is the third part of the prism, which has the same base and the same altitude.
485. Cor, 1. Every triangular prism may be divided into three equivalent triangular pyramids.
486. Cor. 2. The solidity of a triangular pyramid is equal to a third part of the product of its base by its altitude.
Proposition XX. — Theorem.
487. The solidity of every pyramid is equal to the pro- duct of its base hy one third of its altitude.
Let ABCDE-S be any pyramid, whose base is ABCDE, and altitude SO; then its solidity is equal toABCDEx^SO.
BOOK VIII. 209
Draw the diagonals AC, AD, and pass the pliuies SAC, SAD tlirough these diagonals and the A'ertex S ; the polygonal pyramid A B C D E - S will be divided into several triangular pyra- mids, all having the same altitude, S 0. But each of these pyramids is measured by the product of its base, B AC, CAD, D A E, by a third part of its altitude, S 0 (Prop. XIX. Cor. 2) ; hence, the sum of these triangular pyramids, or the polygonal pyra- mid A B C D E - S, will be measured by the sum of the triangles B A C, CAD, D A E, or the polygon A B C D E, multiplied by one third of S 0 ; hence, every ])yramid is measured by the product of its base by one third of its altitude.
488. Cor. 1. Every pyramid is the third part of the prism which has the same base and the same altitude.
489. Cor. 2. Pyramids having the same altitude are to each other as their bases.
490. Cor. 3. Pyramids having the same base, or equiv- alent bases, are to each other as their altitudes.
491. Cor. 4. Pyramids are to each other as the pro- ducts of their bases by their altitudes.
492. Scholium. The solidity of any polyedron may be found by dividing it into pyramids, by passing planes through its vertices.
Proposition XXI. — Theorem.
493. A fruslnm of a pyramid is equivalent to the sum of three pyramids^ having for their common altitude the altitude of the frustum , and whose bases are the two bases of the frustum and a mean proportional bettveen them.
First. Let A B C-D E F be the frustum of a pyramid, whose base is a triangle. Pass a plane through the points
210
ELEMENTS OP GEOMETRY.
A, E, C ; it cuts off the triangular pyramid A I> C - E, whose altitude is that of the frustum,' and whose base, ABC, is the lower base of the frustum. Pass another plane through the points D, E, C; it cuts off the triangular pyramid D E F - C, whose altitude is that of the frus- tum, and whose base, DEE, is the upper base of the frustum.
There now remains of the frus- tum the pyramid A C D - E. Draw E G parallel to AD ; join C G and D G. Then, since E G is parallel to AD, it is parallel to the plane A C D (Prop. XL Bk. VII.) ; and the pyramid A C D - E is equivalent to the pyramid AC. D-G, since they have the same base, A C D, and their vertices, E and G, lie in the same straight line par- allel to the common base. But the pyramid A C D - G is the same as the pyramid AGC-D, whose altitude is that of the frustum, and whose base, A G C, as will be proved, is a mean proportional between the bases ABC and DEE.
The two triangles A G C, DEE have the angles A and D equal to each other (Prop. XVI. Bk. VII.) ; hence we have (Prop. XXVIII. Bk. IV.),
AGC:DEF::AGXAC:DEXDF;
but since A G is equal to D E,
AGC : DEE : : AC : D F.
We have, also (Prop. VI. Cor., Bk. IV.),
A B C : A G C : : A B : A G or D E.
But the similar triangles ABC, DEE give
A B : D E : : A C : D F ;
hence (Prop. X. Bk. II.),
ABC:AGC::AGC:DEF;
BOOK VIII.
211
tliat is, the base A G C is a mean proportional between the bases A B C, D E F of the frustum.
Secondly. Let G H I K L - M N 0 P Q be the frustum of a pyramid, whose base is any polygon.
Let ABC-S be a triangular pyramid liaving the same alti- tude, and an equiva- lent base, with any polygonal pyramid, G H I K L - T ; these pyramids are equiva- lent (Prop. XX. Cor. 3.)
The bases of the two pyramids may be regarded as situ- ated in the same plane, in which case the plane MNOPQ produced will form in the triangular pyramid a section, D E F, at the same distance above the common plane of the bases ; and therefore the section D E F will be to the section MNOPQ as the base ABC is to the base G H I K L (Prop. XVI. Cor. 1) ; and since the bases are equivalent, the sections will be so likewise. Hence, the pyramids MNOPQ- T, DEF-S, having the same al- titude and equivalent bases, are equivalent. For the same reason, the entire pyramids GHIKL-T, ABC-S are equivalent; consequently, the frustums GHIKL- M N 0 P Q, A B C - D E F, are equivalent. But the frustum A B C - D E F has been shown to be equivalent to the sum of three pyramids having for their common altitude the altitude of the frustum, and whose bases are the two bases of the frustum, and a mean proportional between tlicm. Hence the proposition is true of the frustum of any pyramid.
Proposition XXII. — Theorem.
494. Similar pyramids are to each other as the cubes of their homolog-ous edg-es.
212
ELEMENTS OP GEOMETRY.
Let ABC-S and DEF-S be two sim- ilar pyramids ; these pyramids are to each otlier as the cubes of their homologous edges AB and D E, or BC and EF, &c.
For, the two pyra- mids being similar, the homologous polyedral angles at the vertices are equal (Art. 452) ; hence the smaller pyramid may be so applied to the larger, that the polyedral angle S shall be common to both.
In that case, the bases ABC, D E F will he parallel ; for, since the homologous faces are similar, the angle S D E is equal to S A B, and S E F to SBC; hence the plane A B C is parallel to the plane D E F (Prop. XVI. Bk. VII.). Then let S 0 be drawn from the vertex S perpendicular to the plane ABC, and let P be the point where this perpendicular meets the plane D E F. From what has already been shown (Prop. XVI.), we shall have
SO:SP::SA:SD::AB:DE; and consequently,
^SO: ^SP: : AB:I)E. But the bases ABC, DEF are similar; hence (Prop. XXIX. Bk. IV.),
A B C : D E F : : AB^ : We\ Multiplying together the corresponding terms of these two proportions, we have
ABCXi^SO:DEFXiSP:: AB' : DEI Now, A B C X 1 S O represents the solidity of the pyra- mid ABC-S, and D E F X ^ S P that of the pyramid DEF-S (Prop. XX.) ; hence two similar pyramids are to each other as the cubes of their homologous edges.
BOOK VIII. 213
Proposition XXIII. — Theorem.
495. There can be no more than Jive regular pohjedrons.
For, since regular polyedroiis have equal regular poly- gons for tlieir faces, and all their polyedral angles equal, there can be but few regular pulyedrons.
Firsl. If tlie faces are equilateral triangles, polyedrons may be formed of them, having each polyedral angle con- tained by tliree of these triangles, forming a solid bounded by fjur equal equilateral triangles ; or hy four, forming a solid bounded by eight equal equilateral triangles ; or by Jire, forming a solid bounded by twenty equal equilateral triangles. No others can be formed with equilateral triangles. For six of these angles are equal to four riglit angles, and cannot form a polyedral angle (Prop. XX. Bk. VII.).
Secondly/, If the faces are squares, their angles may be arranged by threes, forming a solid bounded by six equal squares. Four angles of a square are equal to four right angles, and cannot form a polyedral angle.
Thirdly/. If the faces are regular pentagons, their angles may be arranged by threes, forming a solid bounded by twelve equal and regular pentagons.
We can proceed no farther. Three angles of a regular hexagon are equal to four right angles ; three of a hepta- gon are greater. Hence, there can be formed no more than five regular polyedrons, — three with equilateral tri- angles, one with squares, and one with pentagons.
496. Scholium. The regular polyedron bounded by four equilateral triangles is called a tetraedron ; the one bounded by eight is called an octaedron ; tlie one bound- ed by twenty is called an icosaedron. The regular polj^e- dron bounded by six equal squares is called a hexaedron, or CUBE ; and tlie one boundod by twelve equal and regu- lar pentagons is called a dodecaedron.
BOOK IX.
THE SPHERE, AND ITS PROPERTffiS.
DEFINITIONS.
497. A Sphere is a solid, or Yolume, bounded by a curved surface, all points of which are equally distant from a point within, called the centre.
The sphere may be con- ceived to be formed by the revolution of a semicircle, DAE, about its diameter, D E, which remains fixed.
498. The Radius of a sphere is a straiglit line drawn from the centre to any point in surface, as the line C B.
The Diameter, or Axis, of a sphere is a line passing through the centre, and terminated both ways by the sur- face, as the line D E.
Hence, all the radii of a sphere are equal ; and all the diameters are equal, and each is double the radius.
499. A Circle, it will be shown, is a section of a sphere. A Great Circle of the sphere is a section made by a
plane passing through the centre, and having the centre of the sphere for its centre ; as the section A B, whose centre is C.
500. A Small Circle of the sphere is any section made by a plane not passing througli tlie centre.
501. The Pole of a circle of tlie sphere is a point in the
BOOK IX.
215
surface equally distant from every point in the circumfer- ence of the circle.
502. It will be shown (Prop. V.) that every circle, great or small, has two poles.
503. A Plane is tangent to a sphere, when it meets the sphere in but one point, however far it may be produced.
504. A Spherical Angle is the difference in the direc- tion of two arcs of great cir- cles of the sphere ; as A E D, formed by the arcs E A, D E.
It is the same as the angle resulting from passing two planes through those arcs ; as the angle formed on the edge EP, by the planes ExVF, EDP.
505. A Spherical Triangle is a portion of the surface of a sphere bounded by three arcs of great circles, each arc being less than a semi-circumference ; as A E D.
These arcs are named the sides of the triangle ; and the angles which their planes form with each other arc the ang-les of the triangle.
506. A spherical triangle takes the name of ri^ht-ang-Ied, isosceles^ equilateral, in the same cases as a plane triangle.
507. A Spherical Polygon is a portion of the surface of a sphere bounded by several arcs of great circles.
508. A LuNE is a portion of the surface of a sphere com- prehended between semi-cir- cumferences of two great cir- cles; as AIGBDF.
509. A Spherical Wedge, or Ungula, is that portion of a sphere comprehended between
2i(J
ELEMENTS OF GEOMETiiY.
two great semicircles having a comnioii diameter.
510. A Zone is a portion of the surface of a sphere cut ofT by a plane, or comprehended between two parallel planes ; as EIFK-A, or CGDII- EIFK.
511. A Spherical Segment is a portion of the sphere cut off by a plane, or compre- hended between two parallel ))lanes.
612. The Altitude of a Zone or of a Spherical Seg- ment is the perpendicular distance between the two parallel planes which comprehend the zone or segment.
In case the zone or segment is a portion of the sphere cut off, one of the planes is a tangent to tlie sphere.
513. A Spherical Sector is a solid described by the revolution of a circular sector, in the same manner as the semicircle of which it is a part, by revolving round its diameter, describes a sphere.
514. A Spherical Pyramid is a portion of the sphere comprehended between the planes of a polyedral angle whose vertex is the centre.
The base of the pyramid is the spherical polygon inter- cepted by the same planes.
Proposition I. — Theorem.
515. Every section of a sphere made by a plane is a circle.
Let ABE be a section made by a plane in the sphere whose centre is C. From the centre, C, draw CD per- pendicular to tlie plane ABE; and draw the lines C A, C B, C E, to different points of the curve ABE, which bounds the section.
BOOK IX.
217
The oblique lines C A, C B, C E are equal, being radii of the sphere ; therefore they are equally distant from the perpendieular, CD (Prop. y. Cor., Bk. VII.). Hence, the lines DA, D B, D E, and, in like manner, all the lines drawn from I) to the boundary of the section, are ccpial ; and therefore the section ABE is a circle whose centre is D.
516. Cor. 1. If the section passes through the centre of the sphere, its radius will be the radius of the sphere ; hence all great circles are cijual.
517. Chr. 2. Two great circles always bisect each other. For, since the two circles have the same centre, tlicir com- mon intersection, passing through the centre, must \)Q a common diameter bisecting both circles.
518. Cor. 3. Every great circle divides the splierc and its surface into two equal parts. For if the two hemi- splieres were separated, and afterwards placed on the com- mon base, with their convexities turned the same way, the two surfaces would exactly coincide.
519. Cor. 4. The centre of a small circle, and that of the sphere, are in a straight line perpendicular to the j)lane of the small circle.
520. Cor. 5. ^ Small circles are less according to their distance from the centre ; for, the greater the distance CD, the smaller the chord AB, the diameter of the small circle ABE.
521. Cor. 6. The arc of a great circle may be made to pass through any two points on the surface of a sphere ; for the two given points and the centre of the sj)here deter- mine the position of a plane. If, however, the two given points be the extremities of a diameter, these two points
19
218
ELEMENTS OF GEOMETRY,
and the centre would be in a straight line, and any num- ber of great circles may be made to pass through the two given points.
Proposition II. — Theorem.
522. Any one side of a spherical triangle is less than the sum of the other two.
Let AB C be any spherical triangle ; then any side, as A B, is less than the sum of the other two sides, A C, B C.
For, draw the radii OA, OB, OC, and the plane angles A 0 B, A 0 C, COB will form a triedral angle, 0. The angles A OB, AOC, COB will be measured by AB, AC, B C, the side of the spherical triangle. But each of the tlu^ce plane angles forming a triedral angle is less than the sum of the other two (Prop. XIX. Bk. YII.). Hence, any side of a spherical triangle is less than the sum of the other two.
Proposition III. — Theorem.
523. The shortest path from one point to another^ on the surface of a sphere^ is the arc of the great circle luhich joins the tivo given points.
Let ABD be the arc of the great circle which joins tlie points ^ A and D ; then the line A B D is the shortest path from A to D on the surface of the sphere.
For, if possible, let the shortest path on the surface from A to D pass through the point C, out of the arc of the great circle ABD. Draw A C, D C, arcs of great circles, and take D B equal to D C. Then in the spherical triangle A B D C the side A B D is less than tlie sum of the sides AC, D C^ (Prop. II.) ; and
BOOK IX. 219
subtracting the equal D B and D C, there will remain AB less tlian AC.
Now, the shortest path, on the surface, from D to C, whether it is the arc D C, or any other line, is equal to the shortest path from D to B ; for, revolving J) C about the diameter which passes through D, the point C may be brouglit into the position of the point B, and the shortest path from D to C be made to coincide with the shortest path from D to B. But, by hypothesis, the shortest path from A to D passes through C ; consequently, the shortest path on the surface from A to C cannot be greater than that from A to B.
Now, since AB has been proved to be less than AC, the shortest path from A to C must be greater tlian that from A to B ; but this has just been shown to be impos- sible. Hence, no point of the shortest path from A to D can lie out of the arc A B D ; consequently, this arc of a great circle is itself the shortest path between its extrem- ities:
524. Cor. The distance between any two points of sur- face, on the surface of a sphere, is measured by the arc of a great circle joining the two points.
Proposition TV. — Theorem.
525. The sum of all the sides of any spherical polygon is less than the circumference of a great circle.
Let A B C D E be a spherical polygon ; E
then the sum of the sides AB, B C, CD, / I^^^D
<fec. is less than the circumference of a A^ j /j
great circle. \ \B|Tc/
For, from 0, the centre of the sphere, \\\ I
draw the radii OA, OB, OC, <fec., and \\|//
the plane angles A 0 B, B 0 C, COB, M li'
&Q,. will form a polyedral angle at 0. Y
Now, the sum of the plane angles which ^
220
ELEMENTS OF GEOMETRY.
form a poljcdral angle is less than four right angles (Prop. XX. Bk. YIL). Hence, the sum of the arcs AB, B C, CD, &G., which measure these angles, and bound the spherical polygon, is less than the circumference of a great circle.
"526. Cor. The sum of the three sides of a spherical tri- angle is less than the circumferejice of a great circle, since a triangle is a polygon of three sides.
Proposition Y. — Theorem.
527. The extremities of a diameter of a sphere are the poles of all circles of the sphere whose planes are j)erpen- dicular to that diameter.
Let D E be a diameter per- pendicular to AHB, a great circle of a sphere, and also to tlie small circle FIG; then D and E, the extremities of this diameter, are the poles of these two circles.
For, since D E is perpendic- ular to the plane AHB, it is perpendicular to all the straight lines, AC, H C, B C, &c., drawn through its foot in this plane ; hence, all the arcs D A, D H, D B, <fec. are quar- ters of the circumference. So, likewise, are all the arcs E A, EH, E B, <fec. ; hence the points D and E are each equally distant from all the points of the circumference, AHB; consequently D and E are poles of that circum- ference (Art. 501).
Again, since the i-adius I) C is perpendicular to the plane AHB, it is perpendicular to the parallel plane FIG; hence it passes through 0, the centre of the circle FIG (Prop. I. Cor. 4). Hence, if the oblique lines D F, D I, D G, &c. be drawn, these lines will be equally distant from
BOOK IX. 221
the perpendicular D 0, and will themselves he equal (Prop. V. Bk. VII.). But the chords heing equal, the arcs are equal ; hence the point D is a pole of the small circle FIG; and, for like reasons, the point E is tlie other pole.
528. Cor. 1. Every arc of a great circle, D II, drawn from a point in the arc of a great circle, AH'B, to its pole, is a quarter of tl^e circumference, and is called a quad- rant. This quadrant makes a riglit angle with the arc AH. For, the line D C heing perpendicular to the plane AH C, every plane B H C passing through the line D C is perpendicular to the plane A H C (Prop. YII. Bk. YII.) ; hence the angle of those planes, or the angle AHD, is a right angle (Art. 50G).
529. Cor. 2. To find the pole of a given arc, AH, draw the indefinite arc HD perpendicular to AH, and take HD equal to a quadrant ; the point D will he one of the poles of the arc AHD ; or at each of the two points A and H, draw the arcs AD and HD perpendicular to AH; the point of their intersection, D, will be the pole required.
530. Cor. 3. Conversely, if the distance of the point D from each of tlie points A and H is equal to a quadrant, the point D will be the pole of the arc A H ; and the an- gles D A H, A H D will be right.
For, let C be the centre of the sphere, and draw the radii C A, C D, C H. Since the angles A C D, H C D are right, the line C D is perpendicular to the two straight lines C A, C H ; hence it is perpendicular to their plane (Prop. IV. Bk. VII.). Hence the point D is the pole of the arc A H ; and consequently the angles D A H, Jl II D are right angles.
531. Scholium. A circle may be described on the surface of a sphere with the same facility as on a plane surface. For instance, by turning the arc D F, or any other lino extending to the same distance, round the point D tho
222 ELEMENTS OF GEOMETRY.
extremity, F, will describe the small circle FIG; and by turning the quadrant D F A round the point D, its ex- tremity, A, will describe the great circle A H B.
Proposition VI. — Theorem.
532. A plane perpendicular to a radius^ at its terminal Hon, in the surface^ is tangent to the sphere.
Let A D B be a plane per- a * D E B pendicular to a radius, C D, at its termination, D ; tlien the plane A D B is a tangent to the sphere.
For, draw from the centre, C, any other straight line, CE, to the plane, ADB. Then, since C D is perpendicular to the plane, it is shorter than the oblique line C E ; hence the radius C F is shorter than C E ; consequently the point E is without the sphere. The same may be shown of any other point in the plane ADB, except the point D ; hence the plane can meet the sphere in but one point, and therefore is a tangent to the sphere (Art. 503).
533. Scholimu. In the same manner, it may be proved that two spheres ^re tangent to each other, wlien the dis- tance between their centres is equal to the sum or the dif- ference of their radii ; in which case the centres and the point of contact lie in the same straight line.
Proposition YII. — Theorem.
534. The angle formed by two arcs of great circles is equal to the angle formed by the tangents of those arcs at the point of their intersection, and is measured by the arc of a great circle described from its vertex as a pole, and intercepted between its sides, produced if nScessary,
BOOK IX.
223
Let B A C be an angle formed by the two arcs AB, AC ; then will it be equal to the angle E A F, formed by the tangents A E, A F, and it is measured by B C, the arc of a great circle described from the vertex A as a pole.
For the tangent AE, drawn in the plane of the arc A B, is perpendicular to the radius A 0 (Prop. X. Bk. III.) ; and the tangent A F, drawn in the plane of the arc A C, is per- pendicular to the same radius AO. Hence the angle E A F is equal to the angle of the planes A 0 B, A 0 C (Art. 391) ; which is that of the arcs AB, AC.
Also, if the arcs A B, A C are both quadrants, the lines 0 B, 0 C will be perpendicular to A 0, and the angle BOC will be equal to the angle of the planes A OB, AOC; hence the arc B C is the measure of tlie angle of tliese planes, or the measure of the angle CAB.
535. Cor. 1. The angles of spherical triangles may be compared together, by means of the arcs of great circles described from their vertices as poles, and included be- tween their sides ; hence it is easy to make an angle of this kind equal to a given angle.
536. Cor. 2. Vertical angles, such as A 0 C and BOD, are equal ; for each of them is equal to the angle formed by the two planes A 0 B, COD.
It is also evident that the two adjacent angles, AOC, COB, taken together, are equal to two right angles.
224
ELEMENTS OF GEOMETRY.
Proposition YIII. — Theorem.
637. If from the vertices of any spherical triangle, as poles, arcs of great circles are described, a second trian- gle is formed, whose vertices will he poles to the sides of the first triangle.
Let A B C be any spher- ical triangle ; and from the vertices, A, B, C, as poles, let the arcs E F, FD, DE be described, and a second triangle, D E F, is formed, whose vertices, D, E, F, will be poles to the sides of the triangle ABC.
For, the point A being the pole of the arc E F, the dis- tance AE is a quadrant; the point C being the pole of tlie arc D E, the distance C E is also a quadrant ; hence tlie point E is at the distance of a quadrant from each of the points A and C ; hence it is the pole of the arc A C (Prop. V. Cor. 3). In like manner, it may be shown that D is the pole of the arc B C, and F that of the arc A B.
538. Scholium. Hence the triangle ABC may be de- scribed by moans of D E F, as D E F may be by means of ABC. Spherical triangles tlius described are said to l)e polar to each other ^ and are called polar or svpplcmental triangles.
Proposition IX. — Theorem.
539. Each of the angles of a spherical tnangle is mcas- vred by a semi-circumference minus the side lying opposite to it in the polar triangle.
Let A B C be a spherical triangle, and D E F a triangle polar to it ; then each of the angles of A B C is measured
BOOK IX.
225
by a semi-circumference minus the side lying oppo- site to it in D E F.
For, produce the sides A B, A C, if necessary, till they meet EF in G and II. The point A being the pole of the arc GH, the angle A will be measured by that arc (Prop. VII.). But, E being the pole of A H, the arc E H is a quadrant ; and F being the pole of A G, F G is a quadrant. Hence, E H and G F together are equal to a semi-circumference. Now, the sum of E H and G F is equal to the sum of E F and G H ; hence the arc G H, which measures the angle A, is equal to a semi-circumference minus the side E F. In like manner, the angle B will be measured by a semi- circumference minus D F ; and the angle C by a semi- circumference minus D E.
540. Cor. This property must be reciprocal in the two triangles, since they are polar to each other. The angle D, for example, of the triangle D E F, is measured by the arc I K ; but the sum of 1 K and B C is equal to the sum of I C and B K, which is equal to a semi-circumference ; licnce the arc IK, the measure of D, is equal to a semi- circumference minus B C. In like manner, it may be shown tliat E is measured by a semi-circumference miuus A C, and F by a semi-circumference minus A B.
Proposition X. — Theorem.
541. Tlie sum of the ang-les in any spherical triangles is less than six right angles^ and greater than two.
First. Every angle of a spherical triangle is less than two right angles ; hence, the sum of the three is less than bix right angles.
226 ELEMENTS OF GEOMETRY.
Secondly. The measure of each angle of a spherical tri- angle is equal to the semi-circumference minus the corre- sponding side of the polar triangle (Prop. IX.) ; hence, the sum of the three is measured hy three semi-circumfer- ences minus the sum of the sides of the polar triangle. Now, this latter sum is less than a circumference (Prop. lY. Cor.) ; therefore, taking it away from three semi- circumferences, the remainder will bo greater than ono semi-circumference, which is the measure of two right angles ; hence, the sum of the three angles of a spherical triangle is greater than two right angles.
542. Cor. 1. The sum of the angles of ^ spherical tri- angle is not constant, like that of the angles of a rectilineal triangle. It varies between two right angles and six, without ever arriving at either of these limits. Two given angles, therefore, do not serve to determine tlie third.
543. Cor. 2. A spherical triangle may have two, or even three right angles, or obtuse angles.
544. Scholium. If a spherical triangle has two riglit angles, it is said to be bi-rectang-ular ; and if it has thi-ee right angles, it is said to be tri-rectang-ular, or quadrantal. The quadrantal triangle is evidently contained eight times in the surface of the sphere.
Proposition XI. — Theorem.
545. If around the vertices of any two ang-fes of a i^iven spherical triangle, as poles, the circumferences of two cir- cles be described, which shall pass through the third angle of the triangle, and then if through the other point in vjhich these circmnferences intersect, and the vertices of the first two angles of the triangles, arcs of tivo great circles be drawn, the triangle thus formed ivill liave all its parts equal to those of the given triangle, each to each.
B
BOOK IX. 227
Let A B C be the given spherical triangle, and CED, DFC arcs de- scribed about the vertices of any two of its angles, A and B, as poles; then will the triangle A D B have all its parts equal to those of ABC.
For, by construction, the side AD is equal to AC, D B is equal to BC, and A B is common ; hence the two triangles have tlieir sides equal, each to each. We are now to show that the angles opposite these equal sides are also equal.
If the centre of the sphere is supposed to be at 0, a tri- edral angle may be conceived as formed at O by the three plane angles A OB, AOC, BOC; also, another triedral angle may be conceived as formed by the three plane an- gles A OB, AOD, BOD. Now, since the sides of the triangle ABC are equal to those of the triangle A D B, the plane angles forming the one of these triedral angles are equal to the plane angles forming the other, each to each. Therefore the planes, in which the equal angles lie, are equally inclined to each other (Prop. XXI. Bk. VII.) ; hence, all the angles of the spherical triangle DAB are respectively equal to those of the triangle CAB; namely, DAB is equal to B AC, DBA to ABC, and A D B to A C B ; hence, the sides and angles of the triangle A D B are equal to the sides and the angles of the triangle A C B, each to each.
546. Scholium. The equality of these triangles is not, however, an absolute equality, or one of superposition ; for it would be impossible to apply them to each other exactly, unless they were isosceles. The equality here meant is that by symmetry ; therefore the triangles AC B, A D B are termed symmetrical triangles.
228
ELEMENTS OF GEOMETRY.
Proposition XII. — Theorem.
547. If two triangles on the same sphere^ or on equal spheres^ are mutual/// equilateral, the// are mutually equi- angular; and their equal angles are opposite to equal sides.
Let A B C, A B D be two triangles on j^
tlie same sphere, or on eqvial spheres, having the sides of the one respective- ly equal to those of the other ; then the angles opposite to the equal sides, in the two triangles, are equal.
For, with three given sides, A B, AC, B C, there can be constructed only two trianglcj?, A C B, A B D, and these triangles will be equal, each to each, in the magnitude of all their parts (Prop. XI.). Hence, these two triangles, which are mutually equilat- eral, must be either absolutely equal, or equal by symme- try; in either case tliey are mutually equiangular, and the equal angles lie opposite to equal sides.
Proposition XIII. — Theorem.
548. If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they are mutually equi- lateral.
Let A and B be the two given triangles ; P and Q, their polar triangles.
Since the angles are equal in the triangles A and B, the sides will be equal in the polar triangles P and Q (Prop. IX.). But since the triangles P and Q are mutually equi- lateral, they jnust also be mutually equiangular (Prop. XII.) ; and, tlio angles being equal in the triangles P and Q, it follows that the sides are equal in their polar trian- gles A and B. Hence, the triangles A and B, which are
BOOK IX.
229
mutually equiangular, are at the same time mutually equilateral.
Proposition XIY. — Theorem.
549. If hvo triangles on the same sphere, or on equal spheres, have tvm sides and the included angle in the one equal to tivo sides and the included angle in the other ^ each to each, the tivo triangles are equal in all their parts.
lu the two triangles A. ABC, DEF, let the side A B be equal to the side D E, the side A C to the side D F ; and the angle B A C to the angle E D F ; then the triangles will he equal in all their parts. B E
Let the triangle D E G be symmetrical with the triangle DEF (Prop. XL Sch.), having the side E G equal to E F, the side G D equal to F D, and the side E D common, and consequently the angles of the one equal to those of the other (Prop. XIL).
Now, the triangle ABC may be applied to the triangle D E F, or to D E G symmetrical with DEF, just as two rectilineal triangles are applied to each other, when they have an equal angle included between equal sides. Heiice, all the parts of the triangle ABC will be equal to all the parts of the triangle D E ¥, each to each ; that is, besides the three parts equal by hypothesis, we shall have the side B C equal to E F, the angle ABC equal to DEF, and the angle A C B equal to D F E.
550. Cor. If two triangles, ABC, DEF, on tlie same sphere, or on equal spheres, have two angles and the in- cluded side in the one equal to two angles and the included side in the other, each to each, the two triangles are equal in all their parts.
20
230 ELEMENTS OF GEOMETRY.
For one of these triangles, or the triangle symmetrical with it, may be applied to the other, as is done in the cor- responding case of rectilineal triangles.
Proposition XY. — Theorem.
551. In every isosceles spherical triangle, the angles op> posite the equal sides are equal ; and, conversely, if tivo angdes of a spherical triang-le are equal, the triangle is isosceles.
Let ABC be an isosceles spherical A
triangle, in which the side A B is equal to the side A C ; then will the angle B be equal to the angle C.
For, if the arc A D be drawn from the«\^ertex A to the middle point, D, of the base, the two triangles ABD, ACD will have all the sides of the one re- spectively equal to the corresponding sides of tlie other, namely, A D common, B D equal to D C, and A B equal to A C ; hence their angles must be equal ; consequently, the angles B and C are equal.
Conversely. Let the angles B and C be equal ; then will the side A C be equal to A B.
For, if A C and A B are not equal, let A B be the great- er of the two ; take B 0 equal to A C, and draw 0 C. The two sides BO, B C in the triangle B 0 C are equal to the two sides A B, B C in the triangle BAG; the angle 0 B C, contained by the first two, is equal to A C B, con- tained by the second two. Hence, the two triangles BOG, BAG have all their other parts equal (Prop. XIY. Gor.) ; hence the angle 0 G B is equal to A B G. But, by hypothesis, the angle A B G is equal to A G B ; hence we have 0GB equal to AG B, which is impossible ; there- fore A B cannot be unequal to A C ; consequently the sides A B, AG, opposite the equal angles B and G, are equal.
BOOK IX. 231
552. Cor. The angle B A D is equal to D A C, and the angle B D A is equal to A D C ; the last two are therefore right angles ; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to the base, and bisects the vertical angle.
Proposition XVI. — Theorem.
553. In a spherical triangle ^ the greater side is opposite the. greater angle ; and, conversely, the greater angle is opposite the greater side.
In the triangle ABC, let the angle A be greater than B ; then will the side B C, opposite to A, be greater than A C, opposite toB.
Take the angle BAD equal to the angle B ; then, in the triangle ABD, we shall have the side AD equal to DB (Prop. XV.). But the sum of AD plus D C is greater than AC; hence, putting DB in the place of AD, we shall have the sum of D B plus DC, or B C, greater than AC.
Conversely. Let the side B C be greater than A C ; then the angle B A C will be greater than ABC. For, if B A C were equal to A B C, we should have B C equal to A C ; and if B A C were less than ABC, we sliould then have, as has just been shown, B C less than A C. Both of these results are contrary to the hypothesis ; hence the angle B A C is greater than ABC.
Proposition XVII. — Theorem.
554. If two triangles on the same sphere, or on equal spheres, are mutually equilateral, they are equivalent.
232
ELEMENTS OF GEOMETRY.
the three points A, B, C ; and they will all be equal At the point F make the angle D F P make the arc V P equal to C 0 ; and
Let ABC, DEFbe two triangles, having the three sides of the one equal to the three sides of the other, each to each, namely, AB to D E, ACtoDF, andCBtoEF; then their triangles will be equivalent.
Let 0 be the pole of the small circle passing through draw the arcs 0 A, OB, 0 0 (Prop. V. Sch.). equal to A C 0 ; draw DP, E P.
The sides D F, F P are equal to the sides A C, C 0, and the angle D F P is equal to the angle A 0 0 ; hence the two triangles DFP, AGO are equal in all their parts (Prop. XIV.) ; hence the side D P is equal to AO, and the angle D P F is equal to A 0 0.
In the triangles D F E, ABC, the angles D FE, ACB, opposite to the equ^al sides D E, A B, are equal (Prop. XII.). Taking away the equal angles DFP, A C 0, there will remain the angle PFE, equal to 0 C B. The sides P F, F E are 'equal to the .sides 0 C, C B ; hence the two triangles F P E, COB are equal in all their parts (Prop. XIV.) ; hence the side PE is equal to OB, and the angle F P E is equal to C 0 B.
Now, the triangles D F P, A C 0, which have the sides equal, each to each, are at the same time isosceles, and may be applied the one to the other. For, having placed 0 A upon its equal P D, the side 0 C will fall on its equal P F, and thus the two triangles will coincide ; conse- quently they are equal, and the surface D P F is equal to A 0 C. For a like reason, the surface F P E is equal to C 0 B, and the surface D P E is equal to A 0 B ; hence we have
BOOK IX.
233
AOC + COB — AOB = DPF + FPE — DPE, or, A B C = D E F.
Hence the two triangles A B C, D E F are equivalent.
555. Cor. 1. If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they are equiva- lent. For in that case the triangles will be mutually equilateral.
556. Cor. 2. Hence, also, if two triangles on the same sphere, or on equal spheres, have two sides and the includ- ed angle, or have two angles and the included side, in the one equal to those in the other, the two triangles are equivalent.
557. Scholium. The poles 0 and P might lie within the triangles A B C, D E F ; in which case it would be requi- site to add the three triangles DPF, FPE, DPE together, to form the triangle D E F ; and in like manner to add the three triangles AOC, COB, AOB together, to form the triangle ABC; in all other respects the demonstra- tion would be the same.
Peoposition XYHI. — Theorem.
558. The area of a lune is to the surface of the sphere as the angle of the lune is to four right angles^ or as the arc which measures that angle is to the circumference.
Let A C B D be a lune upon a sphere whose diameter is A B ; then will the area of the lune be to the surface of the sphere as the angle D 0 C to four right angles, or as the arc D C to the circumference of a great circle.
For, suppose the arc C D to be to tlie circumference C D E F in the ratio of two whole numbers, as 5 to 48, for example.
20*
234
ELEMENTS OP GEOMETRY.
Then, if the circumference CDEF be divided into 48 equal parts, C D will contain 5 of them ; and if the pole A he joined with the several points of division by as many quadrants, we shall have 48 triangles on the surface of the hemisphere ACDEF, all equal, since all their parts are equal. Hence, the whole sphere must contain 96 of these trian- gles, and the lune A C B D 10 of them ; consequently, the lune is to the sphere as 10 is to 96, or as 5 to 48 ; that is, as the arc C D is to the circumference.
If the arc C D is not commensurable with the circum- ference, it may still be shown, by a mode of reasoning ex- emplified in Prop. XVI. Bk. III., that the lune is to the sphere as C D is to the circumference.
559. Cor. 1. Two lunes on the same sphere, or on equal spheres, are to each other as the angles included between their planes.
560. Cor. 2. It has been shown that the wiiole surface of the sphere is equal to eight quadrantal triangles (Prop. X. Sell.). Hence, if the area of a quadrantal triangle be represented by T, the surface of the sphere will be repre- sented by 8T. Now, if the right angle be assumed as unity, and the angle of the lune be represented by A, we
have,
Area of the lune : 8 T : : A : 4,
which gives the area of lune equal to 2 A X T.
561. Cor. 3. The spherical ungula included by tlie planes A C B, A D B, is to the whole sphere as the angle D O C is to four right angles. For, the lunes being equal, the spherical ungulas will also be equal ; hence, two spherical ungulas on the same sphere, or on equal spheres,
BOOK IX. 235
are to each other as the angles included between their planes.
Proposition XIX. — Theorem.
562. If two great circles intersect each other on the sur- face of a hemisphere^ the sum of the opposite triangles thus formed is equivalent to a lune, whose angle is equal to the angle formed by the circles.
Let the great circles BAD, C A E intersect on the surface of a hemisphere, ABODE; then will the sum of the oppo- site triangles, BAG, DAE, be equal to a lune whose angle is DAE.
For, produce the arcs AD, A E till they meet in F ; and the arcs BAD, ADF will each be a semi-circumference. Now, if we take away A D from both, we shall have D F equal to B A. For a like reason, we have E F equal to C A. D E is equal to BC. Hence, the two triangles BxlC, DEF are mutually equilateral; therefore they are equivalent (Prop. XVII.). But the sum of the triangles DEF, D A E is equivalent to the lune A D F E, whose angle is DAE.
Proposition XX. — Theorem.
563. The area of a spherical triangle is equal to the excess of the sum of its three angles above two right an- gles^ multiplied by the quadrantal triangle.
Let ABC be a spherical triangle ; its area is equal to tlie excess of the sum of its angles, A, B, C, above two right angles multiplied by the quadrantal triangle.
For produce the sides of the triangle ABC till they
236
ELEMENTS OF GEOMETRY.
meet the great circle DEFGHI, drawn without the triangle. Tlie two triangles ADE, AGH are together equivalent to the lune whose angle is A (Prop. XIX.), and whose area is expressed by 2 A X T (Prop. XVIII. Cor. 2). Hence we have
ADE + AGH = 2AX T;
and, for a like reason,
BGF+BID=2BxT, andCIH + CFE=2CxT.
But the sum of these six triangles exceeds the hemisphere by twice the triangle ABC; and the hemisphere is repre- sented by 4 T ; consequently, twice the triangle A B C is equivalent to
2AXT + 2BXT + 2CXT — 4T;
therefore, once the triangle A B C is equivalent to
(A + B + C — 2) X T.
Hence the area of a spherical triangle is equal to the excess of the sum of its three angles above two right angles mul- tiplied by the quadrantal triangle.
564. Cor, If the sum of the three angles of a spherical triangle is equal to three right angles, its area is equal to the quadrantal triangle, or to an eighth part of the suiface of the sphere ; if the sum is equal to four right angles, the area of the triangle is equal to two quadrantal triangles, or to a fourth part of the surface of the sphere, <fec.
Proposition XXI. — Theorem.
^^b. The area of a spherical polygon is equal to the excess of the sum of all its angles above tivo right angles taken as many times as the polygon has sides, less twOy multiplied by the quadrantal triangle.
BOOK IX.
237
Let ABODE be any spherical polygon. From one of the vertices, A, draw the arcs AC, AD to the opjjosite vertices ; tlie polygon will be divided into as many spherical triangles as it has sides less two. But tlie area of each of these trian- gles is equal to the excess of the sum of its three angles above two right angles multiplied by the quadrantal tri- angle (Prop. XX.) ; and the sum of the angles in all the triangles is evidently the same as that of all the angles in the polygon ; hence the area of the polygon ABODE is equal to the excess of the sum of all its angles above two right angles taken as many times as the polygon has sides, less two, multiplied by the quadrantal triangle.
566. Cor. If the sum of all the angles of a spherical polygon be denoted by S, the number of sides by w, the quadrantal triangle by T, and the right angle be regarded as unity ^ the area of the polygon will be expressed by
2 (n — 2) X T= (S-^2/j + 4) X T,
BOOK X
THE THREE ROUND BODIES.
DEFINITIONS.
567. A Cylinder is a solid, which may be described by the revolution of a rectan- gle turning about one of its sides, which remains immovable ; as the solid described by the rectangle A B C D revolving about its side A B.
The BASES of the cylinder are the circles described by the sides, AC, BD, of the revolving rectangle, which are adjacent to the immovable side, A B.
The AXIS of the cylinder is the straight line joining the centres of its two bases ; as the immovable line A B.
The CONVEX SURFACE of the cylinder is described by the side C D of the rectangle, opposite to the axis A B.
568. A Cone is a solid which may be ^ described by the revolution of a right- angled triangle turning about one of its perpendicular sides, which remains im- movable ; as the solid described by the right-angled triangle ABC revolving about its perpendicular side A B.
The BASE of the cone is the circle de- scribed by the revolution of the side B C, which is perpendicular to the im- movable side.
BOOK X. 239
The CONVEX SURFACE of a cone is described by the hy- potlienuse, A C, of the revolving triangle.
The VERTEX of tlie cone is the point A, where the hy- pothenuse meets the immovable side.
The AXIS of the cone is the straight line joining the vertex to the centre of the base ; as the line A B.
The ALTITUDE of a cone is a line drawn from the vertex perpendicular to the base ; and is the same as the axis, AB.
The SLANT HEIGHT, or SIDE, of a cone, is a straight line drawn from tlie vertex to the circumference of the base ; as the line A C.
569. The frustum of a cone is the part of a cone included between the base and a plane parallel to the base ; as the solid CD -P.
The AXIS, or altitude, of the frus- tum, is the perpendicular line A B in- cluded between the two bases; and the SLANT HEIGHT, or SIDE, is that portion of the slant height of the cone which lies between the bases ; as F C.
570. Similar Cylinders, or Cones, are tliose whose axes are to each other as the radii, or diameters, of their bases.
571. Tlie sphere, cylinder, and cone are termed the Three Round Bodies of elementary Geometry.
Proposition I. — Theorem.
572. Tlie convex surface of a cylinder is equal to the' circumference of its base multiplied by its altitude.
Let ABCDEF-Gbe a cylinder, whose circumference is the circle A B C D E F, and whose altitude is the line A G ; then its convex surface is equal to A B C D E F multiplied by AG.
240
ELEMENTS OF GEOMETRY.
In tliG base of the cylinder inscribe any regular polygon, A B C D E F, and on this polygon construct a right prism of the same altitude with the cylinder. The prism will be inscribed in the con- vex surface of the cylinder. The con- vex surface of this prism is equal to the perimeter of its base multiplied by its altitude, AG (Prop. I. Bk. VIII.). B ~ C
Conceive now the arcs subtending the sides of the poly- gon to be continually bisected, until a polygon is formed having an indefinite number of sides ; its perimeter will then be equal to the circumference of the circle ABCDEF (Prop. XII. Cor., Bk. VI.) ; and thus the convex surface of the prism will coincide with the convex surface of the cylinder. But the convex surface of the prism is always equal to the perimeter of its base multiplied by its alti- tude ; hence, the convex surface of the cylinder is equal to the circumference of its base multiplied by its altitude.
573. Cor. 1. If two cylinders have the same altitude, their convex surfaces are to each other as the circumfer- ences of their bases.
574. Cor. 2. If H represent the altitude of a cylinder, and R the radius of its base, then we shall have the cir- cumference of the base represented by 2R X n (Prop. XV. Cor. 3, Bk. VI.), and the convex surface of the cyl- inder by 2 R X TT X H.
Proposition II. — Theorem.
575. The solid contents of a cylinder are equal to the product of its base by its altitude.
Let ABCDEF-Gr be a cylinder whose base is the circle ABCDEF, and whoso altitude is the line A G ; then its solid contents are equal to the product of ABCDEF by AG.
BOOK X.
241
/F
£
In the base of the cylhider inscribe any regular polygon, A B C D E F, and on this polygon construct a right prism of the same altitude with the cylinder. The prism will be inscribed in the con- vex surface of the cylinder. The solid contents of this prism are equal to the product of its base by its altitude (Prop. XIII. Bk. VIIL). F~^C
Conceive now the number of the sides of the polygon to be indefinitely increased, until its perimeter coincides with the circumference of the circle A B C D E F (Prop. XII. Cor., Bk. VI.), and the solid contents of the prism will equal those of the cylinder. But the solid contents of the prism will still be equal to the product of its base by its altitude ; hence the solid contents of the cylinder are equal to the product of its base by its altitude.
576. Co?'. 1. Cylinders of the same altitude are to each other as their bases ; and cylinders of equal bases are to each other as their altitudes.
577. Cor. 2. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases. For the bases are as the squares of their radii (Prop. XIII. Bk. VI.), and the cylinders being simi- lar, the radii of their bases are to each other as their alti- tudes (Art. 570) ; therefore the bases are as the squares of the altitudes ; hence, the products of the bases by the altitudes, or the cylinders themselves, are as the cubes of the altitudes.
578. Cor. 3. If the altitude of a cylinder be represented by H, and the area of its base by R* X ^ (Prop. XV. Cor. 2, Bk. VI.), the solid contents of the cylinder will bo rep- resented by R- X ^ X H.
21
242 ELEMENTS OF GEOMETRY.
Proposition III. — Theorem.
579. The convex surface of a cone is equal to the cir- cumference of the base multiplied h\j half the slant height.
Let A B C D E F-S be a cone S
whose base is the circle A B C D E F, and whose slant height is the line S A ; then its convex surface is equal to A B C D E F multiplied by I SA.
In the base of the cone inscribe any regular polygon, A B C D E F, and on this polygon construct a reg- ular pyramid having tlie same ver- tex, S, with the cone. Then a right pyramid will be in- scribed in the cone.
From S draw S H perpendicular to B C, a side of the polygon. The convex surface of tlie pyramid is equal to the perimeter of its base, multiplied by half its slant height, SH (Prop. XV. Bk. YIII.). Conceive now the arcs subtending the sides of the polygon to be continually bisected, until a polygon is formed having an indefinite number of sides ; its perimeter will equal the circumfer- ence of the circle A B C D E F ; its slant height, S H, will equal that of the cone, and its convex surface coincide with the convex surface of the cone. But the convex surface of every right pyramid is equal to the perimeter of its base, multiplied by half the slant height ; hence the convex surface of the cone is equal to the circumference of its base multiplied by half its slant height.
580. Cor. If S A represent the slant height of a cone, and R the radius of the base, then, since the circumference of the base is represented by 2 R X ^r (Prop. XY. Cor. 3, Bk. YI.), the convex surface of the cone will be repre- sented by 2 R X 71 X ^ S A, equal to tt X R X S A.
BOOK X.
243
Proposition IY. — Theorem.
581. The convex surface of a frustum of a cone is eqnal to half the sum of the circumference of the two bases multiplied by its slant height.
Let A B C D E F - M be the frustum of a cone, and AG its slant height; tlieu the convex surface is equal to half the sum of the circumferences of the two bases ABCDEF, GHIKLM, multiplied by A G. .
For, inscribe in the bases of the frus- tum two regular polygons of tlie same ^ ^ number of sides, having their sides parallel, each to each. Draw the straight lines AG, B H, CI, <fcc., joining the vertices of the corresponding angles, and these lines will be the edges of the frustum of a pyramid inscribed in the frustum of the cone. The convex surface of the frustum of the pyramid is^^qual to half the sum of the perimeters of the two bases multiplied by its slant height, ON (Prop. XVII. Bk. VIII.) .
Conceive now the number of sides of the inscribed poly- gons to be indefinitely increased ; the perimeters of the polygons will then coincide with the circumferences of the circles ABCDEF, GHIKLM; and the slant height, ON, of the frustum of the pyramid, will equal the slant height, A G, of the frustum of the cone ; and the surfaces of the two frustums will coincide.
But the convex surface of every frustum of a right pyramid is equal to half the sum of the perimeters of its two bases, multiplied by its slant height ; hence, the con- vex surface of the frustum of the cone is equal to half the sum of the circumference of its two bases multiplied by half its slant height.
582. Cor. Through R, the middle point of the- side KD,
244
ELEMENTS OF GEOMETRY.
draw the diameter RST, parallel to the
diameter AQD, and the straight lines
RU, KY, parallel to the axis P Q.
Then, since D R is equal to R K, D U
is equal to UY (Prop. XYII. Cor. 2,
Bk. lY.) ; hence, the radius S R is equal
to half the sum of the radii QD, PK.
But the circumferences of circles being to each other as
their radii (Prop. XIII. Bk. YI.), the circumference of
the section of which S R is the radius is equal to half the
sum of the circumferences of which QP, PK are the
radii ; hence, the convex surface of a frustum of a cone is
equal to the slant height multiplied by the circumference
of a section at equal distances between the two bases.
Proposition Y
Theorem.
583. The solidity of a cone is equal to the product of its base by one third of its altitude.
Let ABCDEF-S be a cone, whose base is A B C D E F, and alti- tude S II ; then its solidity is equal to ABCDEF X ^SII.
In the base of the cone inscribe any regular polygon, ABCDEF, and on this polygon construct a reg- ular pyramid, having the same vertex, S, with the cone. Then a right pyra- mid will be inscribed in the cone ; and its solidity will be equal to the product of its base by one third of its altitude (Prop. XX. Bk. YIIL).
Conceive, now, the number of sides of the polygon to bo indefinitely increased, and its perimeter will become equiil to the circumference of the cone, and the pyramid will exactly coincide with the cone. But the solidity of every right pyramid is equal to the product of the base by one
BOOK X.
245
third of its altitude ; licnce, the solidity of a cone is equal to the product of its base by one third of its altitude.
584. Cor. 1. A cone is the third of a cylinder havuig the same base and the same altitude ; hence it follows, —
1. That cones of equal altitudes are to each other as their bases ;
2. That cones of equal bases are to each other as their altitudes ;
3. That similar cones are as the cubes of the diameters of their bases, or as the cubes of their altitudes.
585. Cor. 2. If the altitude of a cone be represented by H, and the radius of its base by R, the solidity of the cone will be represented by
R^ X 71 X * H, or i 71 X R' X H.
Proposition YI. — Theorem.
586. The solidity of the frustum of a cone is equivalent to the sum of three cones ^ having- for their common alti- tude the altitude of the frustum, and lohose bases are the two bases of the frustum, and a mean proportional between them.
Let A B C D E F - M be the frur- turn of a cone ; then will its solidity be equivalent to the sum of three cones having the same altitude as the frustum, and whose bases arc tlic two bases of the frustum, and a mean proportional between them.
For, inscribe in the two bases of the frustum two regular polygons having the same number of sides, ^ ^
and having their sides parallel, each to each. Lot the vertices of the corresponding angles be joined by the straight lines BII, CI, &c., and there is inscribed in the
21*
246 ELEMENTS OF GEOMETRY.
frustum of tlio cone the frustum of a regular pyramid. The solidity of the frustum of this pyramid is equivalent to the sum of three pyramids, having for their common altitude the altitude of the frustum, and whose hases are the two bases of the frustum, and a mean proportional between them.
Conceive now the number of the sides of the polygons to be indefinitely increased ; and the bases of the frustum of the pyramid will equal the bases of the frustum of the cone ; and the two frustums will coincide. Hence the frustum of a cone is equivalent to the sum of three cones, having for their common altitude the altitude of the frus- tum, and whose bases are the two bases of the frustum, and a mean proportional between them.
Proposition YII. — Theorem.
587. If any regular semi-polygon be revolved about a line passing through the centre and the vertices of oppo- site angles, the surface described will be equal to the pro- duct of its axis by the circumference of its inscribed circle.
Let the regular semi-polygon A B C D E F be revolved about AF as an axis; tlicn the surface described by the sides AB, BC, CD, &c. will equal the product of AF by the inscribed circle.
For, from the vertices B, C, D, E of the semi-polygon, draw B G, CH, DM, EN, perpendicular to the axis A F ; and from the centre, 0, draw 0 1 perpendicular to one of the sides; also draw IK perpendicular to AF, anc] B L perpendicular to C H.
Now 0 I is the radius of the inscribed circle (Prop. II. Bk. VI.) ; and the surface described by the revolution of a side, B C, of a regular polygon, is equal to B C multiplied by the circumference, IK (Prop. IV. Cor.).
BOOK X. 247
The two triangles OIK, B C L, having their sides per- pendicular to each other, are similar (Prop. XXV. Bk. IV.) ; therefore,
B C : BL or GH : : 01 : IK : : Circ.Ol : Circ. IK.
Hence (Prop. I. Bk. II.),
BC X Circ. IK= GH X Circ. 01;
that is, the surface described by B C is equal to the pro- duct of the altitude G H b j the circumference of the in- scribed circle. The same may be shown of each of the other sides ; hence, the surface described by all the sides taken together is equal to the product of the sum of the altitudes AG, GH, HM, MN, N F, by the circ. 0 I, or to the product of the axis A F by the circ. 0 1.
Proposition VIII. — Theorem.
588. The surface of a sphere is equal to the product of its diameter by the circumference of a great circle.
Let ABCDEF be a semicircle in x
which is inscribed any regular scmi-poly- b^^#^
gon ; from the centre, 0, draw 0 1 per- ]U^
pendicular to one of the sides. C r ^-j."
If now the semicircle and the semi- polygon be revolved about the axis A F, D v~
tho surface described by the semicircle \v
will be the surface of a sphere (Art. 497), ^^
and that described by the semi-polygon ^
will be equal to the product of its axis, AF, by the cir- cumference, 0 1 (Prop. VII.) ; and the same is true, whatever be the number of sides of the polygon.
Conceive the number of sides of the semi-polygon to be made, by continual bisections, indefinitely great ; tlicu its perimeter will coincide with the semi-circumferenco ABCDEF, and the perpendicular 0 1 will be equal to the radius 0 A ; hence, the surface of the sphere is equal
248 ELEMENTS OF GEOMETRY.
to the product of the diameter by the circumference of a great circle. ^^i^.
589. Cor. 1. The surface of a sphere //\ is equal to the area of four of its great I circles. I
For the area of a circle is equal to the -^V ^^
product of the circumference by half the e^^^^
radius, or one fourth of the diameter ^^^^^^F
(Prop. XY. Bk.VI.).
590. Cor. 2. The surface of a zone or seg-ment is equal to the product of its altitude hy the circumference of a great circle.
For the surface described by the sides B C, C I) of the inscribed polygon is equal to the product of the altitude G M by the circumference of the inscribed circle 0 1. If, now, the number of the sides of an inscribed polygon be indefinitely increased, its perimeter will equal the circle, and B C, C D will coincide with tlie arc BCD; conse- quently, the surface of the zone described by the revolution of B C D is equal to the product of its altitude by the cir- cumference of a great circle. In like manner, the same may bo proved true of a segment, or a zone having but one base.
591. Cor. 3. The surfaces of two' zones, or segments upon the same sphere, are to eacli other as their altitudes ; and any zone or segment is to the surface of the sphere as the altitude of that zone or segment is to the diameter.
592. Cor. 4. If the radius of a sphere is represented by R, and its diameter by D,its surface will be represented by
4 71 X B-, or 71 X I)^
593. Cor. 5. Hence, the surfaces of spheres are to each otlicr as the squares of their radii or diameters.
594. Cor. 6. If the altitude of a zone or segment is
BOOK X. 249
represented by H, the surface of a zone or segment will be represented by
2 7t X R X H, or TT X D X H.
Proposition IX. — Theorem.
595. The solidity of a sphere is equal to the product of its surface by one third of its radius.
For a sphere may be regarded as composed of an indefi- nite number of pyramids, each having for its bp-se a part of the surface of tlie sphere, and for its vertex the centre of the sphere ; consequently, all these pyramids have the radius of the sphere as their common altitude.
Now, the solidity of every pyramid is equal to tlie pro- duct of its base by one third of its altitude (Prop. XX. Bk. VIII.); hence, the sum of the solidities of these pyra- mids is equal to the product of tlie sum of their bases by one third of their common altitude. But the sum of their bases is the surface of the spliere, and their common altitude its radius ; consequently, the solidity of the sphere is equal to the product of its surface by one third of its radius.
596. Cor. 1. The solidity of a spherical pyramid or sector is equal to the product of the polygon or zone ivhich forms its base, by one third of the radius.
For the polygon or zone forming the base of the spheri- cal pyramid or sector may be regarded as composed of an indefinite number of planes, each serving as a base to a pyramid, having for its vertex the centre of the sphere.
597. Cor. 2. Spherical pyramids, or sectors of the same sphere or of equal spheres, are to eacli other as their bases.
598. Cor. 3. A spherical pyramid or sector is to tlie sphere of which it is a part, as its base is to the surface of the sphere.
599. Cor. 4. Hence, spherical sectors upon the same
250 ELEMENTS OF GEOMETRY.
sphere are to each other as the altitudes of the zones form- ing their bases (Prop. YIII. Cor. 3) ; and any spherical sector is to the sphere as the altitude of the zone forming its base is to the diameter of the sphere.
GOO. Cor. 5. If the radius of a sphere is represented by R, its diameter by D, and its surface by S, its solidity will be represented by
SxiR = 47rXR'X-^R==|^XR'ori7rXr)^
601. Cor 6. Hence, the solidities of spheres are to each other as the cubes of their radii.
602. Cor. 7. If the altitude of the zone which forms the base of a sector be represented by H, the solidity of the sector will be represented by
2 71 X R X H X ^ B = § ^ X H' X H.
603. Scholium. The solidity of the spher- ical segment less than a hemisphere, and of one base, formed by the revolution of a por- tion, AB C, of a semicircle about the radius OA, is equivalent to the solidity of the spherical sector formed by A 0 B, less the solidity of the cone formed by 0 B C.
The solidity of the spherical segment greater than a hemisphere, and of one base, formed by the revolution of ADE, is equivalent to the solidity of the splierical sector formed by A 0 D, plus the solidity of the cone formed by 0 D E.
The solidity of the spherical segment of two bases formed by the revolution of C B D E about the axis A F, is equivalent to the solidity of the segment formed by ADE, less the solidity of the segment formed by ABC.
Proposition X. — Theorem.
604. The surface of a sphere is equivalent to the coniiex surface of the circumscribed cylinder^ and is two thirds
BOOK X.
of the whole surface of the cylinder ; also, the solidity of the spite re is two thirds of that of the circumscribed cyl- inder.
Let ABFI be a great circle of the sphere ; D E G H the circum- scribed square ; then, if the semi- circle A B F and the semi-square A D E F be revolved about the di- ameter AF, the semicircle will describe a sphere, and the semi- square a cylinder circumscribing the sphere.
The convex surface of the cylinder is equal to the cir- cumference of its base multiplied by its altitude (Prop. I.). But the base of the cylinder is equal to the great circle of the sphere, its diameter E G being equal to the diameter B I, and the altitude D E is equal to the diam- eter AF ; hence, the convex surface of the cylinder is equal to the circumference of the great circle multiplied by its diameter. This measure is the same as that of the surface of the sphere (Prop. VIII.) ; hence, the surface of the sphere is equal to the convex surface of the circum- scribed cylinder.
But the surface of the sphere is equal to four great cir- cles of the sphere (Prop. VIII. Cor. 1) ; hence, the convex surface of the cylinder is also equal to four great circles ; and adding the two bases, each equal to a great circle, the whole surface of the circumscribed cylinder is equal to six great circles of the sphere ; hence, the surface of the sphere is f or f of the wliole surface of the circum- scribed sphere.
In the next place, since the base of the circumscribed cylinder is equal to a great circle of the sphere, and its altitude to the diameter, the solidity of the cylinder is equal to a great circle multiplied by its diameter (Prop. II.). But the solidity of the splierc is equal to its sur-
252 ELEMENTS OP GEOMETRY.
face, or four great circles, multiplied by one third of its radius (Prop. IX.), which is the same as one great circle multiplied by | of the radius, or by f of the diameter ; hence, the solidity of the sphere is equal to f of that of the circumscribed cylinder.
605. Cor. 1. Hence the sphere is to the circumscribed cylinder as 2 to 3 ; and their solidities are to each other as their surfaces.
606. Cor. 2. Since a cone is one third of a cylinder of the same base and altitude (Prop. V. Cor. 1), if a cone has tlie diameter of its base and its altitude each equal to the diameter of a given sphere, the solidities of the cone and sphere are to each other as 1 to 2 ; and the solidities of the cone, sphere, and circumscribing cylinder are to each other, respectively, as 1, 2, and 3.
BOOK XI.
APPLICATIONS OF GEOMETRY TO THE MENSU- RATION OF PLANE FIGURES.
DEFINITIONS.
607. Mensuration of Plane Figures is the process of determining the areas of plane surfaces.
608. The area of a figure, or its quantity of surface, is determined by the number of times the given surface con- tains some other area, assumed as the unit of measure.
609. The measuring unit assumed for a given surface is called the svperficial unit, and is usually a square, tak- ing its name from the linear unit forming its side ; as a square whose side is 1 inch, 1 foot, 1 yard, &c.
Some superficial units, however, have no corresponding linear unit ; as the rood, acre, &c.
|
610. |
Table of Line. |
iR Measures. |
|
|
12 |
Inches i: |
aakc |
1 Foot. |
|
3 |
Feet |
u |
1 Yard. |
|
bh |
Yards |
u |
1 Rod or Pole. |
|
40 |
Rods |
u |
1 Furlong. |
|
8 |
Furlongs |
(i |
1 Mile. |
|
Also, |
|||
|
7y^^^ Inches |
££ |
1 Link. |
|
|
25 |
Links |
u |
1 Rod or Pole. |
|
100 |
Luiks |
u |
1 Chain. |
|
10 |
Chains |
u |
1 Furlong. |
|
8 |
Furlongs |
(( |
1 Mile. |
|
Note. — For other linear measures, |
see National Arithmetic, Art. |
||
|
133, 134, 136. |
22 |
254
ELEMENTS OF GEOMETRY.
611. Table of Surface Measures. 144 Square Inches make 1 Square Foot.
|
9 |
Square Feet |
a |
1 Square Yard. |
|
30|. |
Square Yards |
a |
1 Square Rod or Pole |
|
40 |
Square Rods |
u |
1 Rood. |
|
4 |
Roods |
li |
1 Acre. |
|
640 |
Acres |
a |
1 Square Mile. |
|
625 |
Square Links |
u |
1 Square Rod. |
|
16 |
Square Rods |
u |
1 Square Chain. |
|
10 |
Square Chains |
a |
1 Acre. |
Also,
612. Since an acre is equal to 10 chains, or 100,000 links, square chains may be readily reduced to acres by pointingoff one decimal place from the right, and square links by pointing off five decimal places from the right.
Problem I.
613. To find the area of a parallelogram.
Multiply the base by the altitude^ and the product ivill be the area (Prop. Y. Bk. lY.).
Examples. p q
1. What is the area of a square, A B C D, whose side is 25 feet ?
25 X 25 = 625 feet, Ans.
2. What is the area of a square field whose side is 35.25 chains ? Ans. 124 A. 1 R. 1 P.
3. How many square feet of boards are required to lay a floor 21 ft. 6 in. square ?
4. Required the area of a square farm, whose side is 3,525 links.
5. What is the area of the rectangle ABCD, whose length, AB, is bQ feet, and whose width, AD, is 37 feet ?
56 X 37 = 2,072 feet, Ans.
V>
D
B
BOOK XI. 255
6. How many square feet in a plank, of a rectangular form, which is 18 feet long and 1 foot 6 inches wide ?
7. How many acres in a rectangular garden, whose sides are 326 and 153 feet ? Ans. 1 A. 23 P. 6^ yd.
8. A rectangular court 68 ft. 3 in. long, by 56 ft. 8 in. broad, is to be paved with stones of a rectangular form, each 2 ft. 3 in. by 10 in. ; how many stones will be re- quired ? Ans. 2,062| stones.
9. Required the area of the rhom- boid A B C D, of which the side A B is 354 feet, and the perpendicular dis- tance, E F, between A B and the oppo- site side C D, is 192 feet.
354 X 192 = 67,968 feet, Ans.
10. How many square feet in a flower-plat, in the form of a rhombus, whose side is 12 feet, and the perpendicular distance between two opposite sides of which is 8 feet ?
11. How many acres in a rhomboidal field, of wliich the sides are 1,234 and 762 links, and the perpendicular dis- tance between the longer sides of which is 658 links ?
Ans. 8 A. 19 P. 4 yd. 6^ ft.
Problem II.
614. The area of a square being given, to find the side. Extract the square root of the area. Scholium. This and the two following problems are the converse of Prob. I.
Examples.
1. What is the side of a square containing 625 square feet ?
a/ 625 = 25 feet, the side required.
2. The area of a square farm is 124 A. 1 R. 1 P. ; how many links in length is its side ?
3. A certain corn-field in the form of a square contains
256 ELEMENTS OF GEOMETRY.
15 A. 2 R. 20 p. If the corn is planted on the margin, 4 hills to a rod in lengtli, how many hills are there on the margin of the field ? Ans. 800 hills.
Problem III.
615. The area of a rectangle and either of its sides being given, to find the other side.
Divide the area by the given side, and the quotient will
be the other side.
Examples.
1. The area of a rectangle is 2,072 feet, and the length of one of the sides is 6Q feet ; what is the length of the other side ?
2072 H- 56 = 37 feet, the side required.
2. How long must a rectangular board be, which is 15 inches in width, to contain 11 square feet ?
3. A rectangular piece of land containing 6 acres is 120 rods long ; what is its width ? Ans. 8 rods.
4. The area of a rectangular farm is 266 A. 3 R. 8 P., and the breadth 46 chains ; what is the length ?
Ans. 58 chains.
Problem IY.
616. The area of a rhomboid or rhombus and the length of the base being given, to find the altitude ; or the area and the altitude being given, to find the base.
Divide the area by the length of the base, and the quo- tient will be the altitude ; or divide the area by the alti- tude, and the quotient will be the length of the base.
Examples.
1. The area of a rhomboid is 67,968 square feet, and the length of the side taken as its base 354 feet ; what is the altitude ?
67,968 -^ 354 = 192 feet, the altitude required.
2. The area of a piece of land in the form of a rhombus
BOOK XI. 257
is 69,452 square feet, and the perpendicular distance be- tween two of its opposite sides is 194 feet ; required the length of one of the equal sides. Ans. 358 ft.
3. On a base 12 feet in length it is required to find the altitude of a rhomboid containing 968 square feet.
4. The area of a rhomboidal-shaped park is lA. 3R. 34 P. 5^ yd. ; and the perpendicular distance between the two shorter sides is 96 yards ; required the length of each of these sides ? Ans. 18 rods.
Problem V.
617. The diagonal of a square being given, to find the area.
Divide the square of the diagonal by 2, and the quotient will he the area, (Prop. XI. Cor. 4, Bk. IV.)
Examples. D C
1. The diagonal, AC, of the square A B C D, is 30 feet ; what is the area ?
80^ = 900 ; 900 -^ 2 = 450 square feet, [the area required.
B
2. The diagonal of a square field is 45 chains ; how many acres does it contain ?
3. The distance across a public square diagonally is 27 rods ; what is the area of the square ?
Problem YI.
618. The area of a square being given, to find the diagonal.
Extract the square root of double the area. Scholium. This problem is the converse of the last.
Examples. 1. The area of a square is 450 square feet ; what is its diagonal ?
450 X 2 = 900 ; /v/"900 = 30 feet, the diagonal required.
22*
258 ELEMENTS OF GEOMETRY.
2. The area of a public square is 4 A. 2 R. 9 P. ; what is the distance across it diagonally ?
3. The area of a square farm is 57.8 acres ; what is the diagonal in chains ? Ans. 34 chains.
Problem YII.
619. The sides of a rectanCxLe being given, to cut off a given area by a line parallel to either side.
Divide the given area by the side which is to retain its length or width, and the quotient will be the length or width of the part to be cut off. (Prop. lY. Sch., Bk. IV.)
Examples.
1. If the sides of a rectangle, ABCD, are 25 and 14 feet, how wide an area, E B C F, to contain 154 square feet, can be cut off by a line parallel to the bide AD?
154 -r- 14 = 11 feet, the width required.
2. A farmer has a field 16 rods square, and wishes to cut off from one side a rectangular lot containing exactly one acre ; what must be the width of the lot ?
3. A carpenter sawed off, from the end of a rectangular plank, in a line parallel to its width, 5 square feet. From the remainder he then sawed off', in a line parallel to the length, 8 square feet. Required the dimensions of the part still remaining, provided the original dimensions of the plank were 20 feet by 15 inches.
Ans. 16 feet by 9 inches.
4. The length of a certain rectangular lot is 64 rods, and its width 50 rods ; how far from the longer side must a parallel line be drawn to cut off an area of 4 acres, and how far from the shorter side of the remaining portion to cut off 5 acres and 2 roods ? How many acres will re- main after the two portions are cut off?
BOOK XI. 259
Probleivi YIII.
620. To find the area of a triangle, the base and alti- tnde being given.
Multiply the base by half the altitude (Prop. \L Bk. lY.).
621. Scholium. The same result can be obtained by nuiltiplymg the altitude by half the base, or by multiply- ing together the base and altitude and taking half the product.
Examples.
1. Required the area of the triangle A B C, whose base, B C, is 210, and alti- tude, AD, is 190 feet.
190
210 X 2^ = 19,950 square feet, the [area required.
ij JJ L"
2. A piece of land is in tlie form of a right-angled tri- angle, having the sides about the right angle, the one 254 and the other 136 yards ; required the area in acres.
Ans. 3A. 2R. 10 P. 29j^yd.
3. Required the number of square feet in a triangular board whose base is 27 inches and altitude 27 feet.
4. What is the area of a triangle whose base is 15.75 chains, and the altitude 10.22 chains?
5. What is the area of a triangular field whose base is 97 rods, and the perpendicular distance from the base to the opposite angle 40 rods ? Ans. 12 A. 20 P.
Problem IX.
622. To find the area of a triangle, the three sides being given.
From half the sum of the three sides subtract each
2G0
ELEMENTS OF GEOMETRY,
side ; muUiply the half sum and the three remainders to- gether, and the square root of the product will be the area required.
For, let ABC be a triangle whose three c
sides, A B, B C, AC, are given, but not the altitude C D, and let the side B C be represented by <z, AC by 6, and AB by c.
Now, since A is an acute angle of the triangle ABC, we have (Prop. XII. Bk. IV.),
fl^2 = ^^2_|_ c'_2 6- X AD, or AD =
Hence, in the right-angled triangle ADC, we have (Prop. XI. Cor. 1, Bk. lY.),
C D^ = ^2
(&2 _|- c2 _ «9)2 4 &2 c2 _ (62 ^
2\2
«2)
4c2
4c9
and, by extracting the square root CD = ^^iZZ
{b-^^r
cfiyi
2c
But the area of the triangle ABC is equivalent to the product of c by half of C D (Prob. YIII.) ; hence
A B C = i a/ 4 tn-'-i — (69 -f- ca — a2)a.
The expression 4 h'^
(h^ + c2 — ay, being the
difference of two squares, can be decomposed into
(2 & c + Z>^ + c^ — «0 X (2bc — h^ — 6- + aO-
Now, the first of these factors may be transformed to (b -\- cy — a^, and consequently may be resolved into (^b -\- c -\- a') X (6 + c — a) ; and the second is tlie same thing as tt^ — (ft — cy, which is equal to (^a-\-b — c) X (^a — 6 + 6'). We have then
^b^c^—(b'' + c^ — ay= (a + b + c) X (b-\-c — a^ X (a + c — b) X (a + b — c^.
BOOK XI. 261
Let S represent half the sum of the three sides of the triangle ; tlicu
a-{.b+ c = 2S; b + c — a =^2 (^ — a) ; a+c — b = 2(8 — b); a + b — c = 2(S — c^; hence
A B C = i V 16 S (S — «) X (S^^6)"xTS^=^)» which, being reduced, gives as the area of the triangle, as given above,
V S(S — a) X (S — 6) X (S — c).
Examples. C
1. What is the area of a triangle, ABC, whose sides, AB, BC, CA, are 40, 30, and 50 feet?
30 + 40 + 50 4- 2 = 60, half the sum of the three sides.
60 — 30 = 30, first remainder.
60 — 40 = 20, second remainder.
60 — 50 = 10, third remainder. 60 X 30 X 20 X 10 = 180,000; Vl80~000 = 424.26 square feet, the area required.
2. How many square feet in a triangular floor, whose sides are 15, 16, and 21 feet ?
3. Required the area of a triangular field whose sides are 834, 658, and 423 links.
Ans. lA. IR. 20 P. 4 yd. 1.6 ft.
4. Required the area of an equilateral triangle, of which each side is 15 yards.
5. What is the area of a garden in the form of a paral- lelogram, whose sides are 432 and 263 feet, and a diagonal 342 feet ? Ans. 2 A. 10 P. 11.46 yd.
6. Required the area of an isosceles triangle, whoso base is 25 and each of its equal sides 40 rods.
7. Wliat is the area of a rhoml)oidal field, whose sides are 57 and 83 rods, and the diagonal 127 lods ?
Ans. 22 A. 3 R. 21 P. 26 yd. 5 ft.
262 ELEMENTS OF GEOMETRY.
Problem X.
623. Any two sides of a right-angled triangle being given, to find the third side.
To the square of the base add the square of the perpen- dicular ; and the square root of the sum will give the hy- pothenvse (Prop. XI. Bk. lY.).
From the square of the hypothenuse subtract the square of the given side, and the square root of the difference loiil be the side required (Prop. XI. Cor. 1, Bk. IV.).
Examples.
1. The base, AB, of the triangle ABC is 48 feet, and the perpendicular, B C, 36 feet ; what is the hypothenuse ?
48^ + 36^ = 3600 ; V 3600 = 60 feet,
[the hypothenuse required. .
2. The hypothenuse of a triangle is 53 feet, and the per- pendicular 28 feet ; what is the base ?
3. Two ships sail from tlie same port, one due west 50 miles, and the other due south 120 miles ; how far are they apart ? Ans. 130 miles.
4. A rectangular common is 25 rods long and 20 rods wide ; what is the distance across it diagonally ?
5. If a house is 40 feet long and 25 feet wide, with a pyramidal-shaped roof 10 feet in lieight, how long is a rafter which reaches from the vertex of the roof to a cor- ner of the building ?
6. There is a park in the form of a square containing 1 0 acres ; how many rods less is the distance from the centre to each corner, than the length of the side of the square ? Ans. 11.716 rods.
Problem XI. 624. The sum of the hypothenuse and perpendicular
BOOK XI. 263
and the base of a right-angled triangle being given, to find the hypothenuse and the perpendicular.
To the square of the svm add the square of the base, and divide the amount by twice the sum of the hypothe- nuse and perpendicular , and the quotient will be the hy- pothenuse.
From the sum of the hypothenuse and perpendicidar subtract the hypothenuse, and the remainder will be the perpendicular.
625. Scholium. This problem may be regarded as equiv- alent to the sum of two numbers and the difference of their squares being given, to find the numbers (National Arithmetic, Art. 553).
Note. — The learner should be required to give a geometrical demon- stration of the problem, as an exercise in the application of principles.
Examples.
1. The sum of the hypothenuse and the perpendicular of a right-angled triangle is 160 feet, and the base 80 feet ; required the hypothenuse and the perpendicular.
Ans. Hypothenuse, 100 ft. ; perpendicular, 60 ft.
1602 + 80^ = 32,000 ; 32,000 -f- (160 X 2) = 100 ; 160 — 100 = 60.
2. Two ships leave the same anchorage ; the one, sailing due north, enters a port 50 miles from the place of depart- ure, and the other, sailing due east, also enters a port, but by sailing thence in a direct course enters the port of the first ; now, allowing that the second passed over, in all, 90 miles, how far apart are the two ports ?
3. A tree 100 feet high, standing perpendicularly on a horizontal plane, was broken by the wind, so that, as it fell, while the part broken off remained in contact with the upright portion, the top reached the ground 40 feet from tlie foot of the tree; what is the length of each part?
Ans. The part broken off, 58 ft. ; the upriglit, 42 ft.
264 ELEMENTS OF GEOMETRY.
Problem XII.
626. The area and the base of a triangle being given, to find the altitude ; or the area and altitude being given, to find the base.
Divide double the area by the base, and the qnotient will be the altitude ; or divide double the area by the alti- tude^ and the quotient ivill be the base.
627. Scholium. This problem is the converse of Prob. VIII.
Examples.
1. The area of a triangle is 1300 square feet, and the base Qb feet ; what is the altitude ?
1300 X 2 = 2600 ; 2600 -^ 65 = 40 ft., altitude required.
2. The area of a right-angled triangle is 17,272 yards, of which one of the sides about the right angle is 136 yards ; required the other perpendicular side.
3. The area of a triangle is 46.25 chains, and the alti- tude 5.2 chains ; what is the base ?
4. A triangular field contains 30 A. 3 R. 27 P. ; one of its sides is 97 rods ; required the perpendicular distance from the opposite angle to that side. Ans. 102 rods.
Problem XIII.
628. To find the area of a trapezoid.
Multiply half the sum of its parallel sides by its altitude (Prop. YII. Bk. lY.).
Examples. DEC
1. What is the area of the trapezoid /
ABCD, whose parallel sides, A B, /
D C, are 32 and 24 feet, and the alti- [_
tude, E F, 20 feet ? A F B
32 + 24 = bQ) 56 ~ 2 = 28 ; 28 X ^0 = 560 ^q. ft.,
[tlie area required.
BOOK XI. 205
2. How many square feet in a board in the form of a trapezoid, whose width at one end is 2 feet 3 inches, and at tlie other 1 foot 6 inches, the length being 16 feet ?
3. Required the area of a garden in the form of a trape- zoid, wliose parallel sides are 786 and 473 links, and the perpendicular distance between them 986 links.
Ans. 6A. 38P. 3 yd.
4. How many acres in a quadrilateral field, having two parallel sides 83 and 101 rods in length, and which are distant from each other 60 rods ?
Problem XIV.
629. To find the area of a regular polygon, the pe- rimeter and apothegm being given.
Multiply the perimeter by half the apothegm^ and t/ie product will be the area (Prop. VIII. Bk. VI.).
630. Scholium. This is in effect resolving the polygon into as many equal triangles as it has sides, by drawing lines from the centre to all the angles, then finding their areas, and taking their sum.
Examples,
1. Bequired the area of a regu- lar hexagon, A B C D E F, whose sides, A B, B C, &c. are each 15 yards, and the apothegm, O M, 13 yards.
15X6 = 90; 90-^V- = 585yd., ____
[the area required. A M B
2. What is the area of a regular pentagon, whose sides are each 25 feet, and the perpendicular from the centre to a side 17.205 feet ?
3. A park is laid out in the form of a regular heptagon, whose sides are each 19.263 chains ; and the perpendicular
23
2G6
ELEMENTS OF GEOMETRY.
distance from the centre to each of the sides is 20 chains. How many acres does it contain ?
Ans. 134 A. 311. 14 P.
Problem XY.
631. To find the area of a regular polygon, its side or perimeter being given.
Multiply the square of the side of the polygon hy the area of a similar polygon whose side is unity or 1 (Prop. XXXI. Bk. IV.).
632. A Table of Regular Polygons whose Side is 1.
|
NAMES. |
AREAS. |
NAMES. |
AUEAS. |
|
Triangle, Square, Pentagon, Hexagon, Heptagon, |
0.4330127 1.0000000 1.7204774 2.5980762 3.6339124 |
Octagon, Nonagon, Decagon, Undecagon, Dodecagon, |
4.8284271 6.1818242 7.6942088 9.3656399 11.1961524 |
The apothegm of any regular polygon wliose side is 1 being ascertained, its area is computed readily, by Prob. XIY.
Examples.
1. Required the area of an equilateral triangle, whose side is 100 feet.
100= = 10,000 ; 10,000 X 0.4830127 = 4380.127 square
[feet, the area required.
2. What is the area of a regular pentagon, whose side is 87 yards ?
8. How many acres in a field in the form of a regular undecagon, whose side is 27 yards ?
Ans. 1 A. 1 R. 25 P. 21 yd. 2.7 ft.
BOOK XI. 267
4. What is the area of an octagonal floor, whose side is 15 ft. 6 in. ?
5. How many acres in a regular nonagon, whose perim- eter is 2286 feet ? Ans. 9 A. 24 P. 28 yd.
Problem XYI.
633. To find the side of any regular polygon, its area heing given.
Divide the given area by the area of a similar polygon whose side is 1, and the square root of the quotient ivill be the side required.
634. Scholium, This problem is the converse of Prob. XY.
Examples.
1. The area of an equilateral triangle is 4330.127 square' feet ; what is its side ?
4330.127 -^ .4330127 = 10,000; V 10,000 = 100 feet,
[the side required.
2. The area of a regular hexagon is 1039.23 feet ; what is its side ?
3. The area of a regular decagon is 7 P. 18 yd. 5 ft. 128.55 in. ; what is its side ? Ans. 16 ft. 5 in.
Problem XVII.
635. To find the area of an irregular polygon. Divide the polygon into triangles, or triangles and
trapezoids, and find the areas of each of them separately ; the sum of these areas will be the area required.
636. Scholium. When the irregular polygon is a quad- rilateral, the area may be found by multiplying together the diagonal and lialf the sum of the perpendiculars drawn from it to the opposite angles.
268 ELEMENTS OF GEOMETRY.
Examples.
1. Kequired the area of the irregular pentagon A B C D E, of which the diag- onal AC is 20 feet, and A D 30 feet ; and the perpendicular distance from the angle B to A C is 8 feet, from C to AD 12 feet, and from E to AD 6 feet.
20 X I = 80 ; 36 X ¥- = 216 ; 36 X | = 108 ; 80 + 216 + 108 = 504 sq. ft., the area required.
2. What is the area of a trapezium, whose diagonal is 42 feet, and the two perpendiculars from the diagonal to the opposite angles are 16 and 18 feet ?
3. In an irregular hexagon, A B C D E F, are given the sides A B 536, B C 498, CD 620, DE 580, EF 308, and AF 492 linlfs, and the diagonals AC 918, CE 1048, and A E 652 links ; required the area.
Ans. 6A. 2R. 9P. 23 yd. 8.4 ft;
4. In measuring along one side, AB, of a quadrangular field, A B C D, that side and the perpendiculars let fall on it from two opposite corners measured as follows : A B 1110, AE 110, AF 745, DE 352, CF 595 links. What is the area of the field ? Ans. 4 A. 1 R. 5 P. 24 yd.
5. In a four-sided rectilineal field, A B C D, on account of obstructions, there could be taken only the following measures : the two sides B C 265 and AD 220 yards, the diagonal A C 378, and the two distances of the perpendic- ulars from the ends of the diagonal, namely, A E 100, and C F 70 yards. Required the area in acres.
Problem XVIII.
637. To find the circumference of a circle, wlien the diameter is given, or the diameter when the circumference is given.
Multiply the diameter hy 3.1416, and the product vnll he the circumference ; or, divide the circumference by
BOOK XT. 269
3.1416, and the quotient tvill he the diameter (Prop. XY. Cor. 3, Bk. YI.).
638. Scholium, The diameter may also be found by multiplying the circumference by .31831, the reciprocal of 3.1416.
Examples.
1. The diameter, AB, of the cir- cle A E B F is 100 feet ; what is its circumference ?
100 X 3.1416 == 314.16 feet, the [circumference required.
2. Required the circumference of a circle whose diam- eter is 628 luiks. Ans. Ifur. 38 rd. 5 yd. 1.56 in.
3. If the diameter of the earth is 7912 miles, what is its circumference ?
4. Required the diameter of a circular pond whose cir- cumference is 928 rods.
Ans. 7 fur. 15 rd. 2 yd. 5.55 in.
5. The circumference of a circular garden is 1043 feet ; what is its radius ? Ans. 10 rd. 1 ft.
Problem XIX.
639. To find the length of an arc of a circle containing any number of degrees, the radius or diameter being given.
Multiply the number of degrees in the given arc by 0.01745, and the product by the radius of the circle.
For, when the diameter of a circle is 1, the circumfer- ence is 3.1416 (Prop. XY. Sch. 1, Bk. YI.) ; hence, when the radius is 1, the circumference is 6.2832 ; which, divided by 360, the number of degrees into which every circle is supposed to be divided, gives 0.01745, the length of the arc of 1 degree, wlien the radius is 1.
640. Scholium. Each of the 360 degrees of a circle,
23*
270 ELEMENTS OP GEOMETRY.
marked tlius, 360°, is divided into 60 minutes, marked thus, 60', and each minute into 60 seconds, marked thus, 60" (National Arithmetic, Art. 1-18).
Examples.
1. What is the length of an arc, A D, containing 60° 30' on the cir- cumference of a circle whose radius, AC, is 100 feet? 60° 30' = 60.5° ; 60.5 X 0.01745 =
1.055725; 1.055725 x 100 = 105.5725 ft., arc required.
2. Required the length of an arc of 31° 15', the radius being 12 yards.
3. Required tlie lengtli of an arc of 12° 10', the diam- eter being 20 feet. Ans. 2.1231 feet.
4. What is the length of an arc of 57° 17' 44i", the radius being 25 feet ? Ans. 25 feet.
Problem XX.
641. To find the area of a circle.
Multiply the circumference by half the radius (Prop. XV. Bk. VI.) ; or, multiply the square of the radius by 3.1416 (Prop. XV. Cor. 2, Bk VI.).
642. Scholium. Multiplying the circumference by half the radius is the same as multiplying the circumference and diameter together, and taking one fourth of the pro- duct. Now, denoting the circumference by c, and the diameter by d, since c = 3.1416 X d (Prob. XVIII.), we have (d X 3.1416 x d) -^ 4 = d^ X 0.7854 = the area of a circle. Again, since d=c-T- 3.1416 (Prob. XVIII.), we have c ^ 3.1416 X c -^ 4 = c^ -^ 12.5664, which is, by taking the reciprocal of 12.5664, equal to c^ X 0.07958 = the area of the circle. Hence the area of the circle may also be found by multiplying' the square of the diam-
BOOK XI. 271
eler by 0.7854 ; or by multiplying^ the square of the cir- cumference by 0.07958.
Examples.
1. The circumference of a circle is 814.16 feet, and its radius 50 feet ; what is its area ?
814.16 X ^i- = 7854 feet, the area required.
2. If the circumference of a circle is 855 feet, and its diameter 113 feet, what is the area?
8. What is the area of a circular garden, whose radius is 281 J links ? Ans. 2 A. 1 R. 88 rd. 9yd. 5 ft.
4. A horse is tethered in a meadow by a cord 89.25075 yards long ; over how much ground can he graze ?
5. Required the area of a semicircle, the diameter of the whole circle beinji 751 feet.
Ans. 5 A. 18 P. 16 yd.
•«>
Problem XXI.
643. To find the diameter or circumference, the area being given.
Divide the area by 0.7854, and the square root of the quotient will be the diameter; or^ divide the area by 0.07958, and the square root of the quotient will he the circumference .
644. Scholium, This problem is the converse of Prob. XX.
EXAMPLES-
1. The area of a circle is 814,16 feet ; what is the diameter ?
814.16 ^ 0.7854 = 400 ; V 400 = 20 feet, the diameter
[required.
2. What must be the length of a cord to be used as a radius in describing a circle which shall contain exactly 1 acre ?
8. The area of a circular pond is 6 A. 1 R. 27 P. 18.2 yd. ; what is the circumference ? Ans. 625 yd.
272 ELEMENTS OF GEOMETRY.
4. The area of a circle is T856 feet ; what is the cir- cumference ?
5. The length of a rectangular garden is 32, and its width 18 rods ; required the diameter of a circular garden having the same area. Ans. 27 rd. 1 ft. 4 in.
Problem XXII.
645. To find the area of a sector of a circle. MuUiphj the arc of the sector by half of its radius
(Prop. XV. Cor. 1, Bk. VI.) ; or,
As 3G0'' are to the degrees in the arc of the sector, so is the area of the circle to the area of the sector.
Examples.
1. Required the area of a sector, D E, whose arc is 80 feet, and its radius, 0 E, 70 feet.
80 X -V- = 2800 square feet, the area
[required.
2. Required the area of a sector, of which the arc is 90 and the radius 112 yards.
3. Required the area of a sector, of which the angle is 137" 20', and the radius 456 links.
Ans. 2 A. IR. 38 P. 21.92 yd.
Problem XXIII.
646. To find the area of a segment of a circle.
Find the area of the sector having- the same arc ivith the segment, and also the area of the triangle formed by the chord of the segment and the radii of the sector. Then, if the segment is less than a semicircle, take the difference of these areas ; but if greater, take their sum.
647. "Scholium. When the heiglit of the segment and
BOOK XI.
273
the diameter of the circle are given, the area may be readily found by means of a table of segments, by divid- ing the height by the diameter^ and looking in the table for the quotient in the column of heights, and taking ovt, in the next column on the right hand, the corresponding area; lahich, multiplied by the square of the diameter, will give the area required.
When the quotient cannot be exactly found in the table, proportions may be instituted so as to find the area be- tween the next higher and the next lower, in the same ratio that the given height varies from the next higher
and lower heights.
648. Table of Segments.
|
*s a .01 |
Seg. Area. |
i ^ ; tp 'S 03 |
Seg. Area. |
33 |
Seg. Area. |
[3) 'S !I3 .31 |
Seg. Area. |
.41 |
Seg. Area. |
|
.00133 |
.11 |
.04701 |
.21 |
.11990 |
.20738 |
.30319 |
|||
|
.02 |
.00375 |
.12 |
.05339 |
.22 |
.12811 |
.32 |
.21667 |
.42 |
.31304 |
|
.03 |
.00687 |
;.i3 |
.06000 |
.23 |
.13646 |
.33 |
.22603 |
.43 |
.32293 |
|
.04 |
.01054 |
.14 |
.06683 |
.24 |
.14494 |
.34 |
.23547 |
.44 |
.33284 |
|
.05 |
.01468 |
.15 |
.07387 |
.25 |
.15354 |
.35 |
.24498 |
.45 |
.34278 |
|
.06 |
.01924 |
.16 |
.08111 |
.26 |
.16226 |
.36 |
.25455 |
.46 |
.35274 |
|
.07 |
.02417 |
.17 |
.08853 |
.27 |
.17109 |
.37 |
.26418 |
.47 |
.36272 |
|
.08 |
.02944 |
.18 |
.09613 |
.28 |
.18002 |
.38 .27386 |
.48 |
.37270 |
|
|
.09 |
.03502 |
.19 |
.10390 |
.29 |
.18905 |
.39 .28359 |
.49 |
.38270 |
|
|
.10 |
.04088 |
.20 |
.11182 |
.30 |
.19817 |
.40 .29337 1 |
.50 |
.39270 |
^The segments in the table are those of a circle whoso diameter is 1, and the first column contains the corre- sponding heights divided by the diameter. The method of calculating the areas of segments from the elements in the table depends upon the principle that similar plane figures are to each other as the squares of their like linear dimensions.
Examples. 1. Wliat is the area of the segment ABE, its arc A E B being 73.74% its chord A B being 12 feet, and
274 ELEMENTS OF GEOMETRY.
the radius, C B, of the circle 10 feet?
0.7854 X 20' = 314.16, area of circle ; then 3G0° : 73.74° : : 314.16 : 64.3504, area of sector A E B C ; and, by Prob- lem IX., 48 is the area of the trian- gle ABC; 64.3504 — 48 = 16.3504 feet, the area required.
2. Eequired the area of a segment whose height is 18, and the diameter of the circle 50 feet.
18 -7-50 = .36 ; to which the corresponding area in the ta- ble is .25455 ; .25455 X 50^ = 636.375, area required.
3. Required the area of a segment whose arc is 100°, chord 153.208 feet, and the diameter of the circle 200 feet.
4. What is the area of a segment whose height is 4 feet, and the radius 51 feet ? Ans. 106 feet.
5. Required the area of a segment, the arc being 160°, chord 196.9616 feet, and the radius of the circle 100 feet.
Problem XXIY.
649. To find tlie area of a circular zone, or tlie space included between two parallel chords and their intercepted arcs.
From the area of tlie inliole circle subtract the areas of the segments on the sides of the zone.
Examples.
1. What is the area of a zone whose chords are each 12 feet, subtending eacli an arc of 73.74°, when the radius of the circle is 10 feet ?
Area of the whole circle by Prob. XX. = 314.16 ; area of each segment by Prob. XXIII. = 16.3504 ; 16.3504 X 2 = 32.7008 = area of both segments ; 314.16 — 32.7008 =: 281.4592, the area required.
BOOK XI. i^K)
2. What is the area of a circular zone whose longer chord is 20 yards, subtending an arc of 60°, and the shorter chord 14.66 yards, subtending an arc of 43°, the diameter of tlie circle being 40 yards ?
3. A circle whose diameter is 20 feet is divided into three parts by two parallel chords ; one of the segments cut off is 8 feet in height, and the other 6 feet ; what is the area of the circular zone ? Ans, 117.544 ft.
Problem XXV.
650. To find the area of a crescent.
Find the difference of the areas of the two segments formed hy the arcs of the crescent and its chord.
Examples.
1. The arcs A C B, A E B, of circles having the same radius, 50 rods, intersecting, form the crescent A C B E ; the height, D C, of the segment A C B is 60 rods, and the height, D E, of the segment ABE is 40 rods ; what is the area of the crescent ?
The area of the segment A C B, by Prob. XXIII., is 4920.3 rods, and that of the segment A B E is 2933.7 rods ; 4920.3 — 2933.7 = 1986.6 rods, the area of the crescent.
2. If the arc of a circle whose diameter is 24 yards in- tersects a circle whose diameter is 20 yards, forming a crescent, so that the height of the segment of the first cir- cle is 5.072 yards, an-d that of the segment of the second circle is 8 yards, what is the area of the crescent ?
Problem XXVI.
651. To find the area of a circular ring, or the space included between two concentric circles.
Find the areas of the two circles separately (Prob. XX.), and take the difference of these areas; or sub-
276 ELEMENTS OF GEOMETRY.
tract the square of the less diameter from the square of the g-reater^ and multiply their difference by 0.T851 (Prob. XX, Sell.).
Examples,
1. Required the area of the ring formed by two circles ■\yhose diameters are 30 and 50 feet.
50-^— £0^ = 1400; 1400 X 0.7854=1099.56 sq. feet,
[tlie area of the ring.
2. What is tlie area of a ring formed by two circles whose radii are 36 and 24 feet ?
3. A circular park, 256 yards in diameter, has a car- riage-way running around it 29 feet wide ; what is tlie area of the carriage-way 1
Ans. lA. 2R. 26P. 21.5 yd.
Problem XXVII.
652. The diameter or circumference of a circle being given, to find the side of an equivalent square.
Multiply the diameter hy 0.8862, or the circumference by 0.2821 ; the product in either case will be the side of an equivalent square.
For, since 0.7854 is the area of a circle whose diameter is 1 (Prob. XX. Sch.), the square root of 0.7854, which is 0.8862, is the side of a square wliich is equivalent to a circle whose diameter is 1. Now when the circumference is 1, the side of an equivalent square must have the same ratio to 0.8862 as the diameter 1 has to its circumference 3.1416 (Prop. XV. Cor. 4, Bk. VI.) ; and 0.8862-^-3.1416 gives 0.2821 as the side of the equivalent square when the circumference is 1,
ExAMPij:s. 1. The diameter of a circle is 120 feet ; what is the side of an equivalent square ?
120 X 0.8862 = 106.344 feet, the side required.
BOOK XI. 277
2. The circumference of a circle is 100 yards ; what is the side of an equivalent square ? Ans. 28.21yd.
3. There is a circular floor 30 feet in diameter ; what is the side of a square floor containing the same area ?
4. If 500 feet is the circumference of a circular island, what is the side of a square of equal area ?
Ans. 141.05 ft.
Problem XXVIII.
653. The diameter or circumference of a circle being given, to find the side of the inscribed square.
Multiplij the diameter by 0.7071, or the circumference by 0.2251 ; the product in either case will be the side of the inscribed square.
For 0.7071 is tlie side of the inscribed square when the diameter of the circumscribed circle is 1, since the side of the inscribed square is to the radius of the circle as the square root of 2 to 1 (Prop. IV. Cor., Bk. VI.) ; conse- quently, the side is to the diameter, or twice the radius, as half the square root of 2 is to 1, and half the square root of 2 is 0.7071, approximately. Now, the ratio of the diameter of a circle to the side of its inscribed square being as 1 to 0.7071, and the ratio of the circumference of a circle to its diameter as 3.1416 to 1, the ratio of the inscribed square is to the circumference of the circle as 0.7071 to 3.1416 ; and 0.7071 -^ 3.1416 gives 0.2251 as the side of the inscribed square when the circumference is 1.
Examples.
1. The diameter, AC, of a circle is 110 feet ; what is the side, A B, of the inscribed square ?
110 X 0.7071 = 77.781 feet, the side
[required.
24
278 ELEMENTS OF GEOMETRY.
2. The circumference of a circle is 300 feet ; what is the side of the inscribed square ? Aiis. 67.53 ft.
3. A log is 36 inches in diameter ; of how many inches square can a stick be hewn from it ?
4. There is a circular field 1000 rods in circuit ; what is the side of the largest square that can be described in it ? Ans. 225.10 rods.
Problem XXIX.
654. The diameter or circumference of a circle beinir given, to find the side of an inscribed equilateral tri- angle.
MuUipli/ the diameter by 0.8660, or the circvmference by 0.2757 ; the product in either case will be the side of the inscribed equilateral triangle.
For 0.8660 is the side of the inscribed equilateral trian- gle when the diameter of the circumscribed circle is 1, since the side of the inscribed equilateral triangle is to the radius of the circle as the square root of 3 is to 1 (Prop. V. Cor. 3, Bk. YI.) ; consequently, the side is to the diam- eter, or twice the radius, as half the square root of 3 is to 1, and half the square root of 3 is 0.8660, approximately. Also, since the ratio of the circumference of a circle to its diameter is as 3.1416 to 1, the side of the inscribed equilateral triangle, when the circumference is 1, equals 0.8660 4- 3.1416, or 0.2757.
Examples.
1. Required the side of an equilateral triangle that may be inscribed in a circle 101 feet in diameter.
101 X 0.8660 = 87.4660 feet, tlie side required.
2. Required the side of an equilateral triangle that may be inscribed in a circle 80 rods in circumference.
Ans. 22.05 rods,
3. Required the side of the largest equilateral triangular beam that can be hewn from a piece of round timber 36 inches in diameter.
BOOK XI. 279
4. Required the side of an equilateral triangle that can be inscribed in a circle 251.33 feet in circumference.
5. How much less is the area of an equilateral triangle that can be inscribed in a circle 100 feet in diameter, than the area of the circle itself? Ans. 4606.4 sq. ft.
The Ellipse. „
655. An Ellipse is a plane figure bounded by^curycr;" ^^ from any point of which the sum of the distances to two fixed points is equal to a straight line drawu through those two points, and terminated both ways by the curve.
Thus A D B C is an ellipse. The two fixed points G and H are called the foci. The longest diameter, AB, of the ellipse is called its major or transverse axis, and its sliortest di- ameter, CD, is called its minor ov conjugate axis.
QbQ. The AREA of an ellipse is a mean proportional between the areas of two circles whose diameters are the two axes of the ellipse.
This, however, can only be well demonstrated by means of Analytical Geometry, a branch of tlie matliematics with which the learner here is not supposed to be acquainted^
Problem XXX.
657. To find the area of an ellipse, the major and minor axes being given.
Multiply the axes together, and their product by 0.7854, and the result will be the area.
For AB^ X 0.7854 expresses the area of a circle whose diameter is AB, and C D^ X 0.7854 expresses the area of a circle whose diameter is C D ; and the product of these two areas is equal to A B-* X C D^ X 0.7854^, which is
280 ELEMENTS OP GEOMETRY.
equal to the square of A B X C D X 0.7854 ; hence, A B X C D X 0.7854 is a mean proportional between the areas of the two circles whose diameters are A B and C D (Prop. TV. Bk. II.) ; consequently it measures the area of an ellipse whose axes are AB and CD (Art. QdQ}.
Examples.
1. Required the area of an ellipse, of which the major axis is 60 feet, and the minor axis 40 feet.
60 X 40 X 0.7854 = 1884.96 sq. ft., the area required.
2. "What is the area of an ellipse whose axes are 75 and 85 feet ?
8. Required the area of an ellipse whose axes are 526 and 854 inches. Ans. 112 yd. 7 ft. 84.62 in.
4. How many acres in an elliptical pond whose semi- axes are 436 and 254 feet ?
Ans. 7A. 8R. 37 P. 27 yd. 7 ft.
BOOK XII.
APPLICATIONS OF GEOMETRY TO THE IVIENSU- RATION OF SOLIDS.
DEFINITIONS.
658. Mensuration op Solids, or Volumes, is the pro- cess of determining their contents.
The SUPERFICIAL CONTENTS of a body is its quantity of surface.
Tlie SOLID CONTENTS of a body is its measured magni- tude, volume, or solidity.
659. The unit of volume, or solidity, is a cube, whose faces are each a superficial unit of the surface of the body, and whose edges are each a linear unit of its linear di-
|
mensions. |
||||
|
660. |
Table of 1 |
Solid Measures. |
||
|
1728 Cubic |
Inches make 1 Cubic Foot |
|||
|
27 |
u |
Feet |
u |
1 " Yard. |
|
4492i |
u |
Feet |
u |
1 " Rod. |
|
32,768,000 |
u |
Rods |
a |
1 " Mile. |
|
Also, |
||||
|
231 |
u |
Inches |
a |
1 Liquid Gallon. |
|
268i |
u |
Inches |
a |
1 Dry Gallon. |
|
2150xVo |
. ii |
Inches |
a |
1 Bushel. |
|
128 |
u |
Feet |
u |
1 Cord. |
Problem I.
6(31. To find the surface of a right prism. MiiUipIy the perimeter of the base by the altitude , and the product will be the convex surface (Prop. I. Bk.
282 ELEMENTS OF GEOMETRY.
VIII.). To this add the areas of the tiuo bases, and the result ivilL be the entire surface.
Examples.
1. Required the entire surface of a pentangular prism, having each side of its base, ABODE, equal to 2 feet, and its altitude, AF, equal to 5 feet.
2 X 5 = 10 ; 10 X 5 = 50 square feet, [the surface required.
2. The altitude of a hexangular prism is 12 feet, two of its faces are each 2 feet wide, three are each 2^ feet wide, and the remaining face is 9 inches wide ; what is the convex surface of the prism ?
3. Required the entire surface of a cube, the length of each edge being 25 feet.
4. Required, in square yards, the wall surface of a rec- tangular room, whose height is 20 feet, width 30 feet, and length 50 feet. Ans. 355t sq. yd.
Problem II.
662. To find the solidity of a prism. Multiply the area of its base by its altitude, and the product will be its solidity (Prop. XIII. Bk. YIII.).
Examples.
1. Required the solidity of a pentangular prism, having each side of its base equal to 2 feet, and its altitude equal to 5 feet.
2^ X 1.72048 = 6.88192 ; 6.88192 X 5 = 34.40960 cubic
[feet, the solidity reqviired.
2. Required the solidity of a triangular prism, whose length is 10 feet, and the three sides of whose base are 3, 4, and 5 feet. Ans. 60.
3. A slab of marble is 8 feet long, 3 feet wide, and 6 inches tliick ; required its solidity.
BOOK XII. 283
4. There is a cistern in the form of a cube, whose edge is 10 feet ; what is its capacity in liquid gallons ?
Ans. 7480.519 gallons.
5. Required the solid contents of a quadrilateral prism, the length being 19 feet, the sides of the base 43, 54, 62, and 38, and the diagonal between the first and second sides, 70 inches. Ans. 306.047 cu. ft.
6. How many cords in a range of wood cut 4 feet long, the range being 4 feet 6 inches high and 160 feet long ?
Problem III.
663. To find the surface of a right pyramid. MuUiphj the perimeter of the base by half its slant
heig-ht, and the product ivill be the convex surface (Prop. XV, Bk. VIII.). To this add the area of the base, and the result ivill be the entire surface.
664. Scholium. The surface of an oblique pyramid is found by taking the sum of the areas of its several faces.
S Examples.
1. Required the convex surface of a pentangular pyramid, A B C D E - S, each side of whose base, A B C D E, is 5 feet, and whose slant height, S M, is 20 feet. 5 X 5 = 25 ; 25 X -^^- = 250 square
[feet, the surface required. g q
2. What is the entire surface of a triangular pyramid, of which the slant height is 18 feet, and each side of the base 42 inches ? Ans. 99.804 sq. ft.
3. Required the convex surface of a triangular pyramid, tlie slant height being 20 feet, and each side of the base 3 feet.
4. What is the entire surface of a quadrangular pyra- mid, the sides of the base being 40 and 30 inches, and the slant height upon the greater side 20.04, and upon the less side 20.07 feet ? Ans. 125.308 ft.
284 ELEMENTS OF GEOMETRY.
Problem IV.
665. To find the surface of a frustum of a right
PYRAMID.
Multiply half the sum of the perimeters of its tvm bases hy its slant heig-ht, and the product will be the convex surface (Prop. XVII. Bk. VIII.) ; to this add the areas of the two bases f and the result will be the entire surface.
Examples.
1. What is the entire surface of a rectangular frustum whose slant height is 12 feet, and the sides of whose bases are 5 and 2 feet ?
5 X 4 = 20 ; 2X4 = 8; 20+ 8 = 28 ; 2/ X 12= 168 ; 52 + 22 = 29 ; 168 + 29 = 197 sq. ft., area required.
2. Required the convex surface of a regular hexangular frustum, whose slant height is 16 feet, and the sides of whose bases are 2 feet 8 inches and 3 feet 4 inches.
3. What is the entire surface of a regular pentangular frustum, whose slant height is 11 feet, and the sides of whose bases are 18 and 34 inches ?
Ans. 136.849 sq. ft.
Problem V.
666. To find the solidity of a pyramid.
Multiply the area of its base by one third of its altitude (Prop. XX. Bk. VIII.).
Examples.
1. Required the solidity of a pen- tangular pyramid, A B C D E - S, each side of whose base, ABODE, is 5 feet, and whose altitude, SO, is 15 feet.
6^ X 1.7205 = 43.0125 ; 43.0125 X J^ = 215.0575 cu. ft., the solidity required.
COOK XII. 285
2. What is the solidity of a hexangular pyramid, the altitude of which is 9 feet, and each side of the base 29 inches ?
3. What is the solidity of a square pyramid, each side of whose base is 30 feet, and whose perpendicular height is 25 feet ? Ans. 7500.
4. Required the solid contents of a triangular pyramid, the perpendicular height of which is 24 feet, and the sides of the base 34, 42, and 50 inches. Ans. 39.2354 cu. ft.
Problem YI.
667. To find the solidity of a frustum op a pyramid.
Add together the areas of the tivo bases and a mean proportional beticeen them, and multipli/ that sum by one third of the altitude of the frustum (Prop. XXI. Bk. YIIL).
ExAMri-KS.
1. Required the solidity of the frustum of a quadran- gular pyramid, the sides of whose bases are 3 feet and 2 feet, and whose altitude is 15 feet.
3X3 = 9; 2x2 = 4; V"y><4=6 (Prop. lY. Bk. 11.) ; (9-1-4 + 6) X V- = 95 cu. ft., solidity required.
2. How many cubic feet in a stick of timber in the form of a quadrangular frustum, the sides of whose bases are 15 inches and 6 inches, and whose altitude is 20 feet ?
3. Required the solid contents of a pentangular frus- tum, whose altitude is 5 feet, each side of whose lower base is 18 inches, and each side of whose upper base is 6 inches. Ans. 9.319 cu. ft.
4. Required the solidity of the frustum of a triangular pyramid, the altitude of which is 14 feet, the sides of the lower base 21, 15, and 12, and those of the upper base 14, 10, and 8 feet. Ans. 868.752 cu. ft.
286
ELEMENTS OF GEOMETRY.
The Wedge.
668. A Wedge is a polyedron bounded by a rectangle, called tlie base of tlie wedge ; by two trapezoids, called the sides, which meet in an edge parallel to the base ; and l)y two triangles, called the ends of the wedge.
Thus ABCD-GH is a wedge, of which ABCD is the rectangular base ; ABHG, DC II G, the tra- pezoidal sides, which meet in the edge GH ; and ADG, B C H, the triangular ends.
The altitude of a wedge is the perpendicular distance from its edge to the plane of its base ; as G P.
Problem YII.
669. To find the solidity of a wedge.
Add the length of the edg-e to twice the length of the base ; multiply the sum by one sixth of the product of the altitude of the wedge and the breadth of the base.
For, let L equal AB, the lengtli of the base ; / equal GH, the length of the edge ; b equal B C, the breadth of the base ; and h equal PG, the height of the wedge. Tlien L — / = A B — G H = A M.
Now, if the length of tlie base and the edge be equal, the polyedron is equal to half a parallclopipcdon having the same base and altitude (Prop. VI. Bk. Vlll.), and its solidity will be equal to ^ b I h (Prop. XHI. Bk. VIII.).
If the length of the base is greater than that of the edge, let a section, M N G, be made parallel to B C 11.
BOOK XII. 287
This section will divide tlie whole wedge into the quad- rangular pyramid A M N D - G, and the triangular prism BC^II-G.
The solidity of A M N D - G is equal to ibhx (L — /) (Prob. V.) ; and the solidity of BCH-G is equal to -J- b I h ; hence the solidity of the wliole wedge is equal to
ibhl+ibh X (L — 0 = ibh^l+^bh'2L — ibh 21= ibh X (2L + 0-
But, if the length of the base is less than that of tlie edge, the solidity of the wedge will be equal to the prism less the pyramid ; or to
^bhl — ^bhx (l — L) = ^b h ^ I — ib h 2 I + hbh2L = ibhX (2L + 0.
Examples.
1. Required the solidity of a wedge, the edge of which is 10 inches, the sides of the base 12 inches and 6 inches, and the altitude 14 inches.
10 4- (12 X 2) = 34 ; 34 X ^^^ = 476 cu. in., the
[solidity required.
2. "What is the solidity of a wedge, of which the edge is 24 inches, the sides of the base 36 inches and 9 inches, and the altitude 22 inches ?
3. How many solid feet in a wedge, of which the sides of tlie base are 35 inches and 15 inches, the length of the edge 55 inches, and the altitude 17/o inches ?
Ans. 3 cu. ft. 17of cu. in.
Rectangular Prismoid.
670. A RECTANGULAR PRISMOID is a polyedron bounded by two rectangles, called the bases of the prismoid, and by four trapezoids called the lateral faces of the prismoid.
The altitude of a prismoid is the perpendicular distance between its bases.
288 elements op geometry.
Problem YIII.
671. To find the solidity of a rectangular prtsmotd.
Add the area of the tvm bases to four times the area of a parallel section at equal distances from the bases ; mul- tiply the sum by one sixth of the altitude.
Let L and B be the length and breadth J^:^ — i k
of the lower base, / and b the length and / \ l-A
/III breadth of the upper base, M and m tlie f''^"'\ [\
length and breadth of the parallel section / " t
equidistant from the bases, and h the \^/ \|
altitude of the prismoid.
If a plane be passed through tlie opposite edges L and /, the prismoid will be divided into two wedges, having for bases the bases of the prismoid, and for edges L and /.
The solidity of these wedges, which compose the pris- moid, is (Prob. Vll.),
^BAX (2L + /) + iZ;Ax (2Z+L) = iA(2BL + But M being equally distant from L and /, 2 M = L + ^, and 2m = B + ^ (Prop. YII. Cor., Bk. lY.) ; conse- quently,
4Mm=(L + /)X(B + ^^) = BL + B/ + ^^L + Z^/.
Substituting 4M7yi for its base, in the preceding equation, we have, as the expression of the solidity of a prismoid,
^h (BL-f ^Z + 4Mm).
672. Scholium. This demonstration applies to prismoids of other forms. For, whatever be the form of the two bases, there may be inscribed in each such a number of small rectangles that the sum of them in each base shall differ less from that base than any assignable quantity ; so that tlie sum of the rectangular prismoids that may be
BOOK XII. 289
constructed on these rectangles will differ from the given prismoid by less than any assignable quantity.
Examples.
1. Kcquired the solidity of a prismoid, tlie larger base of which is 80 inches by 27 inclies, the smaller base 24 inches by 18 inches, and the altitude 48 inches.
QA _j_ 24 27 -4- 18
SO X 27 == 810 ; 21 X 18 = 432 ; ~\~ X — ^ X 4 = 2430 ; (810 + 432 + 2430) X ¥ = 29,376 cu. in. = 17 cu. ft., the solidity required.
2. What is the solidity of a stick of timber, whose larger end is 24 inches by 20 inches, the smaller end 16 inches by 12 inches, and the length 18 feet ?
3. What is the solidity of a block, whose ends are re- spectively 30 by 27 inches and 24 by 18 inches, and whose length is 36 inches ?
4. What is the capacity in gallons of a cistern 47:|^ inches deep, whose inside dimensions are, at the top 81^ and 55 inches, and at the bottom 41 and 29^ inches ?
Ans. 546.929 gall.
Problem IX.
673. To find the surface of a regular polyedron.
Multiphj the area of one of the faces by the number of faces ; or multiply the square of one of the edges of the polyedron by the surface of a similar polyedron whose edges are 1.
For, since the faces of a regular polyedron are all equal, it is evident that the area of one face multiplied by the number of faces will give the area of the whole surface. Also, since the surfaces of regular polyedrons of the same name are bounded by the same number of similar poly- gons (Prop. I. Bk. VI.), their surfaces are to each other as the squares of the edges of the polyedrons (Prop. I. Cor., Bk. VI.).
25
290
ELEMENTS OF GEOMETRY.
674. Table of Surfaces and Solidities of Polyedrons WHOSE Edge is 1.
|
NAMES. |
NO. OF FACES. |
SURFACES. |
SOI.ID1TIES. |
|
Tetraedron, Hexaedron, Octaedron, Dodecaedron, Icosaedron, |
4 6 8 12 20 |
1.7320508 6.0000000 3.4641016 20.6457288 8.6602540 |
0.1178511 1.0000000 0.4714045 7.603118a 2.1816050 |
The surfaces in the table are obtained by multiplying the area of one of the faces of the polyedron, as given in Art. 632, by the number of faces.
Examples.
1. What is the surface of an octaedron whose edge is 16 inches ?
16' X 3.4641016 -- 886.81 sq. in., the area required.
2. Required the surface of an icosaedron whose edge is 20 inches.
3. Required the surface of a dodecaedron whose edge is 12 feet. Ans. 2972.985 sq. ft.
Problem X.
6T5. To find the solidity of a regular polyedron.
Multiply the surface by one third of the perpendicular distance from the centre to one of the faces ^ or multiply the cube of one of the edg-es by the solidity of a similar polyedron ivhose edge is 1.
For any regular polyedron may be divided into as many equal pyramids as it has faces, the common vertex of the pyramids being the centre of the polyedron ; lience, the solidity of the polyedron must equal the product of the areas of all its faces by one third the perpendicular dis- tance from the centre to each face of the polyedron.
BOOK XII.
201
Also, since similar pyramids arc to each other as the cubes of their homologous edges (Prop. XXII. Bk. VIIL), two polyedrons containing the same number of similar pyramids are to each other as the cubes of their edges ; hence, the solidity of a polyedron whose edge is 1 (Art. 673), may be used to measure other similar polyedrons.
Examples.
1. Required the solidity of an octaedron whose edge is 16 inches.
IG' X 0.4714045 = 1930.8728 cu.in., solidity required.
2. What is the solidity of a tetraedron whose edge is 2 feet ?
3. Required the solidity of au icosaedron whose edge is 15 inches. Aiis. 7303. 220G cii. in.
pROnLEM XI.
676. To find tlie surface of a cylinder.
MuUipIfj the circumference of its base by its altitude^ and the product will be the convex surface (Prop. I. Bk. X.). To this add the areas of its two bases, and the re- sult will be the entire surface.
Examples.
1. What is the entire surface of a cylin- der, tlie altitude of wliicli, AB, is 10 feet, and the circumference of the base 20 feet ?
10 X 20 = 200 ; 20^ x 0.07958 X 2 = 63.264; 200 + 63.264 = 263.264 sq. ft., the surface required.
2. Required the convex surface of a cylinder whose alti- tude is 16 feet, and the circumference of whose base is 21 feet.
3. Wliat is the entire surface of a cylinder whose alti- tude is 10 inches, and whose circumference is 4 feet ?
292 ELEMENTS OP GEOMETRY.
4. How many times must a cylinder 5 feet 3 inches long, and 21 inches in diameter, revolve, to roll an acre ?
Ans. 1509.18 times.
Problem XII.
677. To find the solidity of a cylinder.
Multiply the area of the base by the altitude^ and the product will be the solidity (Prop. II. Bk. X.).
Examples.
1. What is the solidity of a cylinder, whose altitude is 10 feet, and the circumference of whose base is 20 feet ?
20^ X 0.07958 X 10 == 318.32 cu. ft., solidity required.
2. Hequired the solidity of a cylindrical log, whose length is 9 feet, and the circumference of wliose base is 6 feet.
Ans. 25.7831 cu. ft.
3. The Winclicster bushel is a liollow cylinder 18j- inches in diameter, and 8 inches deep ; what is its ca- pacity in cubic inches ?
Problem XIII. ^
678. To find the surface of a cone.
Multiply the circumference of the base by half the slant heig-ht (Prob. III. Bk. X.), and the product will be the convex surface. To this add the area of the base, and the result will be the entire surface.
Examples.
1. What is the convex surface of a cone, whose slant height is 28 feet, and the circumference of whose base is 40 feet ?
40 X ^/ = 560 sq. ft., the surface required.
2. Bequired tlie entire surface of a cone, whose slant height is 14 feet, and the circumference of whose base is 92 inches.
BOOK XII. 293
3. Wliat is the surface of a cone, whose slant height is 9 feet, and the diameter of whose base is 36 inches ?
4. How many yards of canvas are required for the cov- ering of a conical tent, the slant height of which is 30 feet, and the circumference of the base 900 feet ?
Ans. 1500 sq. yd.
Problem XIY.
679. To find the surface of a frustum of a cone.
MuUiphj half the sum of the circumferences of its tivo bases hy its slant hei^-ht, and the product ivill be the con- vex surface (Prop. IV. Bk. X.). To this add the area of its bases, and the result will he the entire surface.
G80. Scholium. The convex surface of a frustum of a cone may also be found by multiplying the slant height by the circumference of a section at equal distances between the two bases (Prop. lY. Cor., Bli. X.).
Examples.
1. Required the convex surface of a frustum of a cone, whose slant height is 20 feet, and the circumferences of wliose bases are 30 feet and 40 feet.
— "-^ — X 20 = 700 sq. ft., the surface required.
2. Required tlie surface of a frustum of a cone, the di- ameters of the bases being 43 inches and 23 inches, and the slant height 9 feet.
3. What is the convex surface of a frustum of a cone, of which a section equidistant from its two bases is 24 feet in circumference, the slant height of X\\q frustum being 19 feet ?
4. From a cone the circumference of wliose base is 10 feet, and whose slant heiglit is CO feet, a cone has been cut off, whose slant height is 8 feet. What is the convex surface of the frustum ? Ans. 139^ sq. ft.
294 ELEMENTS OP GEOMETRY.
Problem XV.
681. To find the solidity of a cone.
MuUiply the area of its base by one third of its altitude, and the product will be the solidity (Prop. Y. Bk. X.).
Examples.
1. What is tlie solidity of a cone whose altitude is 42 feet, and the diameter of wliose base is 10 feet ?
102 X 0.7854 X ¥- = 1099.56 cu. ft., solidity required.
2. Required tlie solidity of a cone wliose altitude is 63 feet, and the radius of whose base is 12 feet 6 inches.
8. Plow many cubic feet in a conical stick of timber, whose length is 18 feet, the diameter at the larger end being 42 inches ? Ans. 57.7269 cu. ft.
Problem XVI.
682. To fmd the solidity of the frustum of a cone. Add together the areas of the two bases and a mean
proportional bettoeen them, and midliply that sum by one third of the altitude of the frustum ; and the result ivill be the solidity required (Prop. VI. Ek. X.).
Examples.
1. What is the solidity of a frustum of a cone, C D E F, whose altitnde, A B, is 21 feet, and the area of whose bases, F E, CD, are 80 square feet and 300 square feet ?
(80 + 300 + V80 X 300) X -V = 3732.96 cu. ft., solidity required.
2. Required the solidity of a frustum of a cone, the diameters of the bases being 38 and 27 inches, and the altitude 11 feet.
3. If a cask, which is two equal frustums of cones joined together at the larger bases, have its bung diameter 28
LOOK XII.
295
inches, the head diameter 20 inches, and length 40 inches, how many gallons of wine will it hold ? Ans. 79.06.
Problem XVII.
683, To find the surface of a sphere.
Multiply the diameter by the circumference of a great circle of the sphere (Prop. VIII. Bk. X.) ; or multiply the area of one great circle of the sphere by 4 (Proj). VIII. Cor 1, Bk. X.) ; or multiply 3.1416 by the square of the diameter (Prop. VIII. Cor. 4, Bk. X.).
D
Examples.
1. What is the surface of a sphere, whose diameter, ED, is 40 feet, and whose circum- ference, A E B D, is 125.664?
125.664 X 40 = 5026.56 sq. [ft., the surface required.
2. Required the surface of a sphere whose diameter is 30 inches.
3. What is the surface of a globe whose diameter is 7 feet and circumference 21.99 feet ? Ans. 153.93.
4. How many square miles of surface has the earth, its diameter being 7912 miles ?
Problem XVIII. 684. To find the surface of a zone or segment op a
SPHERE.
Multiply the altitude of the zone or segment by the cir- cumference of a great circle of the sphere (Prop. VIII. Cor. 2, Bk. X.) ; or mtdtiply the product of the diameter and altitude by 3.1416 (Prop. VIII. Cor. 6, Bk. X.).
296 ELEMENTS OF GEOMETRY.
EXAMPLKS.
1. What is the surface of a segment of a sphere, the altitude of the segment being 10 feet, and the diameter of the sphere 50 feet ?
50 X 10 X 3.1416 = 1570.80 sq. ft., surface required.
2. The altitude of a segment of a spliere is 88 inches, and the circumference of the sphere is 25 feet ; Avhat is the surface of the segment ?
3. Required the surface of a zone or segment, the diam- eter of the sphere being 72 feet, and the altitude of the zone 24 feet. Ans. 5428.6848 sq. ft.
4. If the earth be regarded as a perfect sphere whose axis is 7912 miles, and tlie part of the axis corresponding to each of the frigid zones is 327.192848, to each of the temperate zones 2053.468612, and to the torrid zone 8150.67708 miles ; what is the surface of each zone ? Ans. Each frigid zone 8132797.39568 ; each temperate zone
51041592.99898; torrid zone 78314115.07768 miles.
Problem XIX.
685. To find the solidity of a sphere.
Multiply the surface of the sphere by one third of its radius (Prop. IX. Bk. X.) ; or midtiply the cube of the di- ameter of the sphere by 0.5236 (Prop. IX. Cor. 5, Bk. X.).
Examples.
1. What is the solidity of a sphere whose diameter is 40 inches ?
40^ X 0.5236 = 33510.4 cu. in., the solidity required.
2. Required the solidity of a globe whose circumference is 60 inches.
3. What is the solidity of the moon in cubic miles, sup- posing it a perfect sphere with a diameter of 2160 miles ?
4. Required the solidity of the eartli, supposing it to Co a perfect sphere, whose diameter is 7912 miles.
Ans. 259332805349.80493 cu. miles.
BOOK XII.
207
Problem XX.
686. To fiiid the surface of a spherical polygon.
From the sum of all the angles subtract the product of two right angles by the number of sides less tivo ; divide the remaimler by 90°, and multiply the quotient by one eighth of the surface of the sphere ; and the result ivill be the surface of the spherical polygon (Prop. XX. Bk. IX.).
Examples.
1. Required the surface of a spherical polygon having five sides, described on a sphere whose diameter is 100 feet, tlie sum of the angles being 720 degrees.
2 X 90° X (5 — 2) = 540° ; (720° — 540°) -^- 90° = 2 ; 100^ X 3.1416 = 31416 ; 2 X ^-^F^ = 7854 sq. ft., the surface required.
2. What is the surface of a triangle on a sphere whose diameter is 20 feet, the angles being 150°, 90°, and 54° ?
Problem XXI. 687. To find the solidity of a spherical pyramid or
SECTOR.
Multiply the area of the polygon or zone which forms the base of the pyramid or sector by one third of the radius (Prop. IX. Cor. 1, Bk. X.) ; or multiply the altitude of the base by the square of the radius, and that product by 2.0944 (Prop. IX. Cor. 7, Bk. X.).
Examples. 1. Required the solidity of a spherical sector, A C B E, the al- titude, ED, of the zone forming its base being 5 feet, and the radius, C B, of the sphere being 12 feet.
5 X 24 X 3.1416 = 376.992 ; 376.992 X -V- = 1507.968 cu. ft., the solidity required.
298 ELEMENTS OF GEOMETRY.
2. What is the solidity of a spherical pyramid, the area of its base being 3G4 square feet, and the diameter of the sphere 60 feet ?
3. Required the solidity of a spherical sector, whose base is a zone 16 inches in altitude, in a sphere 3 feet in diameter.
4. What is the solidity of a spherical sector, whose base is a zone 6 feet in altitude, iu a sphere 18 feet in diam- eter ? Ans. 1017.88 cu. ft.
Problem XXII.
688. To find the solidity of a spherical segment. When the segment is less than a hemisphere^ from the
solidity of the spherical sector luhose base is the zone of the seg-ment, take the soliditij of the cone ivhose vertex is the centre of the sphere^ and ivhose base is the circular base of the segment; but ivhen the segment is greater than a hemisphere, take the sum of these solidities (Prop. IX. Sch., Bk. X.).
689. Scholium, If the segment has two plane bases, its solidity may be found by taking- the difference of the tiro segments which lie on the same side of its two bases (Prop. IX. Sch., Bk. X.).
Examples.
1. What is the solidity of a segment, ABE, whose altitude, E D, is 5 feet, cut from a sphere whose radius, C E, is 20 feet ? The altitude of the cone A B C is
equal to CE — ED, or 20 — 5,
which is equal to 15 feet ; and
the radius of its base is equal to
VC A2"-i:CD^ or V20^ — 15S
which is equal to 13.23 ; consequently the diameter
A B is equal to 26.46 feet ; 5 X 20^ X 2.0944 = 4188.8
BOOK XII. 299
cubic feet, the solidity of the sector A C D E (Prob. XXI.) ; 20.40^ X 0.7854 X -V'- = 294.6,99 cubic feet, the solidity of the cone A B - C (Prob. XV.) ; 4188.8 — 2946.99 == 1241.81 cubic feet, the solidity of the segment ABE required.
2. Required tlie solidity of a segment, whose altitude is 57 inches, the diameter of the sphere being 153 inches.
3. What is the solidity of a spherical segment, whose altitude is 13 feet, and tlie diameter of the sphere S3 feet 6 inches ?
4. Required the solidity of the segments of the earth which are bounded severally by its five zones, the earth's diameter being 7912 miles, and the part of the diameter corresponding to each of the frigid zones being 327.19, to each temperate zone 2053.47, and to the torrid zone 3150.68.
Ans. Each frigid zone 1293793463.32, each temperate zone 55013912318.45, and the torrid zone 146717393786.26 cubic miles.
The Spheroid.
690. A SPHEROID is a solid which may be described by the revolution of an ellipse about one of its axes, which remains immovable.
An oblate spheroid is one described by the revolution of the ellipse about its minor or conjugate axis.
A prolate spheroid is one described by the revolution of the ellipse about its major or transverse axis.
Problem XXIII.
691. To find the solidity of a spheroid.
Multiply the square of the axis of revolution by the fixed axis, and that product by 0.5 236.
A full demonstration of this, which is based upon the principle that a spheroid is two thirds of its circumscribing
800 ELEMENTS OF GEOMETRY.
cylinder, would require a knowledge of Conic Sections, or of the DitTerential and Integral Calculi, with neither of which is the learner here supposed to be acquainted.
The relation, however, of the spheroid to its circumscrib- ing cylinder, is that which the sphere sustains to its cir- cumscribing cylinder (Prop. X. Bk. X.).
Now the area of the base of the cylinder is fovmd by multiplying the square of the axis of revolution by 0.7854, and the solidity of the cylinder by multiplying that pro- duct by the fixed axis (Prop. II. Bk. X.). But the solid- ity of the spheroid is only two thirds of that of the cylin- der ; hence, to obtain the solidity of the former, instead of multiplying by 0.7854, we must use a factor only two thirds as large, which will be 0.5236.
Examples.
1. What is the solidity of the ob- late spheroid ACBD, whose fixed axis, C D, is 30 inches, and the axis of revokition, A B, 40 inches. 402 X 30 X 0.5236 = 25132.8 cubic
inches, the solidity required.
2. Required the solidity of a prolate spheroid, whose fixed axis is 50 feet, and the axis of revolution 36 feet.
3. What is the solidity of a prolate, and also of an oblate spheroid, the axes of each being 25 and 15 inches ?
Ans. Prolate, 2945.25 cu. in. ; oblate, 4908.75 cu. in.
4. What is the solidity of a prolate, and also of an ob- late spheroid, the axes of each being 3 feet 6 inches and 2 feet 10 inches ?
5. Required the solidity of the earth, its figure being that of an oblate spheroid whose axes are 7925.3 and 7898.9 miles. Ans. 259774584886.834 cubic miles.
BOOK XIII.
MISCELLANEOUS GEOMETRICAL EXERCISES.
1. If the opposite angles formed by four lines meeting at a point are equal, these lines form but two straight lines.
2. If the equal sides of an isosceles triangle are pro- duced, the two exterior angles formed with the base will be equal.
3. The sum of any two sides of a triangle is greater than the third side.
4. If from any point within a triangle two straight lines are drawn to the extremities of either side, they will in- clude a greater angle than that contained by the other two sides.
5. If two quadrilaterals have the four sides of the one equal to the four sides of the otlier, each to each, and the angle included by any two sides of tlie one equal to the angle contained by the corresponding sides of the other, the quadrilaterals are themselves equal.
6. The sum of the diagonals of a trapezium is less than the sum of any four lines which can be drawn to the four angles from any point within the figure, except from the intersection of the diagonals.
7. Lines joining the corresponding extremities of two equal and parallel straight lines, are themselves equal and parallel, and the figure formed is a parallelogram.
8. If, in the sides of a square, at equal distances from the four angles, points be taken, one in each side, the straight lines joining these points will form a square.
26
302 ELEMENTS OP GEOMETRY.
9. If one angle of a parallelogram is a right angle, all its angles are riglit angles.
10. Any straight line drawn throngli the middle point of a diagonal of a parallelogram to meet the sides, is bi- sected in that point, and likewise bisects the parallelogram.
11. If four magnitudes are proportionals, the first and second may be multiplied or divided by the same magni- tude, and also the third and fourth by the same magni- tude, and the resulting magnitudes will be proportionals.
12. If four magnitudes are proportionals, the first and third may be multiplied or divided by the same magni- tude, and also the second and fourth by the same magni- tude, and the resulting magnitudes will be proportionals.
13. If there be two sets of proportional magnitudes, the quotients of the corresponding terms will be proportionals.
14. If any two points be taken in the circumference of a circle, the straight line joining them will lie wholly within the circle.
15. The diameter is the longest straight line that can be inscribed in a circle.
16. If two straight lines intercept equal arcs of a circle, and do not cut each other within the circle, the lines will be parallel.
17. If a straight line be drawn to touch a circle, arfd be parallel to a chord, the point of contact Avill be the middle point of the arc cut off by that chord.
18. If two circles cut each other, and from either point of intersection diameters be drawn, the extremities of these diameters and the other point of intersection will be in the same straight line.
19. If one of the equal sides of an isosceles triangle be the diameter of a circle, the circumference of the circle will bisect the base of the triangle.
20. If the opposite angles of a quadrilateral be together equal to two right angles, a circle may be circumscribed about the quadrilateral.
BOOK XIII. 303
21. Parallelograms wliicli have two sides and the in- cluded angle equal in each, are themselves equal.
22. Equivalent triangles upon the same base, and upon the same side of it, are between the same parallels.
23. If the middle points of the sides of a trapezoid, which are not parallel, be joined by a straight line, that line will be parallel to each of the two jjarallel sides, and be equal to half their sum.
24. If, in opposite sides of a parallelogram, at equal distances from opposite angles, points be taken, one in each side, the straight line joining these points will bisect the parallelogram.
25. The perimeter of an isosceles triangle is greater than the perimeter of a rectangle, which is of the same altitude with, and equivalent to, the given triangle.
26. If the sides of the square described upon the hypoth- enuse of a right-angled triangle be produced to meet the sides (produced if necessary) of the squares described upon the other two sides of the triangle, the triangles thus formed will be similar to the given triangle, and two of them will be equal to it.
27. A square circumscribed about a given circle is double a square inscribed in the same circle.
28. If the sum of the squares of the four sides of a quadrilateral be equivalent to the sum of the squares of the two diagonals, the figure is a parallelogram.
29. Straight lines drawn from the vertices of a triangle, so as to bisect the opposite sides, bisect also the triangle.
30. The straight lines which bisect the three angles of a triangle meet in the same point.
31. The area of a triangle is ecfual to its perimeter mul- tiplied by half the radius of the inscribed circle.
32. If the points of bisection of the sides of a given tri- angle be joined, the triangle so formed will be one fourth of the given triangle.
33. To describe a square upon a given straight line.
304 ELEMENTS OF GEOMETRY.
34. To find in a given straight line a point equally dis- tant from two given points.
35. To construct a triangle, the base, one of the angles at the base, and the sum of the other two sides being given.
36. To trisect a right angle.
37. To divide a triangle into two parts by a line drawn parallel to a side, so that these parts shall be to each other as two given straight lines.
38. To divide a triangle into two parts by a line drawn perpendicular to tlie base, so that these parts shall be to each other as two given lines.
39. To divide a triangle into two parts by a line drawn from a given point in one of the sides, so that the parts shall be to each other as two given lines.
40. To divide a triangle into a square number of equal triangles, similar to eacli other and to the original triangle.
41. To trisect a given straight line.
42. To inscribe a square in a given right-angled isosceles triangle.
43. To inscribe a square in a given quadrant.
44. To describe a circle that shall pass through a given point, have a given radius, and touch a given straight line.
45. To describe a circle, the centre of wliich shall be in the perpendicular of a given right-angled triangle, and the circumference of which shall pass through the right angle and touch the hypotlienuse.
46. To describe three circles of equal diameters which shall touch each other, and to describe another circle wliicli shall touch the three circles.
47. If, on the diameter of a semicircle, two equal circles be described, and in the curvilinear space included by the three circumferences a circle be inscribed, its diameter will be to that of the equal circles in the ratio of two to three.
48. If two points be taken in tlie diameter of a circle,
BOOK XIII. 305
equidistant from the centre, the sum of tlie squares of two lines drawn from these points to any point in the circum- ference will always be the same.
49. Given the vertical angle, and the radii of the in- scribed and circumscribed circles, to construct the triangle.
50. If a diagonal cuts off three, five, or any odd number of sides from a regular polygon, the diagonal is parallel to one of the sides.
51. The area of a regular hexagon inscribed in a circle is double that of an equilateral triangle inscribed in the same circle.
52. The side of a square circumscribed about a circle is equal to the diagonal of a square inscribed in the same circle.
53. To describe a circle equal to half a given circle.
54. A regular duodecagon is equivalent to three fourths of the square constructed on the diameter of its circum- ocribed circle ; or is equal to the square constructed on the side of the equilateral triangle inscribed in the same circle.
55. If semicircles be described on the sides of a right- angled triangle as diameters, the one described on the hypothenuse will be equal to the sum of the other two.
56. If on the sides of a triangle inscribed in a semi- circle, semicircles be described, the two crescents thus formed will together equal the area of the triangle.
57. If the diameter of a semicircle be divided into any number of parts, and on them semicircles be described, their circumferences will together be equal to the circum- ference of the given semicircle.
58. To divide a circle into any number of parts, which shall all be equal in area and equal in perimeter, and not have the parts in the form of sectors.
59. To draw a straight line perpendicular to a plane, from a given point above the plane.
60. Two straight lines not in the same plane being
20*
306 ELEMENTS OF GEOMETRY.
given ill position, to draw a straight line which shall be perpendicular to them both.
61. The solidity of a triangular prism is equal to the product of the area of either of its rectangular sides as a base multiplied by half its altitude on that base.
62. All prisms of equal bases and altitudes are equal in solidity, whatever be tiie figure of tiieir bases.
63. The convex surface of a regular pyramid exceeds the area of its base in the ratio that the slant height of the pyramid exceeds the radius of the circle inscribed in its base.
64. If from any point in the circumference of the base of a cylinder, a straight line be drawn perpendicular to the plane of the base, it will be wholly in the surface of the cylinder.
65. A cylinder and a parallelopipedon of equal bases and altitudes are equivalent to each other.
6Q. If two solids have the same height, and if their sec- tions made at equal altitudes, by planes parallel to the bases, have always the same ratio which the bases have to one another, the solids have to one another the same ratio which their bases have.
67. The side of the largest cube that can be inscribed in a sphere, is equal to the square root of one tliird of the square of the diameter of the sphere.
68. To cut off just a square yard from a plank 14 feet 3 inches long, and of a uniform width, at what distance from the edge must a line be struck ? Ans. 7^^ in.
69. How much carpeting a yard wide will be required to cover the floor of an octagonal hall, whose sides arc 10 feet each ?
70. The perambulator, or surveying-wheel, is so con- structed as to turn just twice in the length of a rod ; what is its diameter ? Ans. 2.626 ft.
71. What is the excess of a floor 50 feet long by 00 broad, above two others, each of half its dimensions ?
BOOK XIII. 307
72. The four sides of a trapezium are 13, 13.4, 24, and 18 feet, and the first two contain a right angle. Required tlie area. Ans. 253.38 sq. ft.
73. If an equilateral triangle, whose area is equal to 10,000 square feet, be surrounded with a walk of uniform width, and equal to the area of the mscribed circle, what is the width of the walk ? Ans. 11.701 ft.
74. A right-angled triangle has its base 16 rods, and its perpendicular 12 rods, and a triangle is cut off from it by a line parallel to its base, of which the area is 24 rods. Re- quired the sides of tliat triangle. Ans. 8, 6, and 10 rods.
75. There is a circular pond whose area is 5028i square feet, in the middle of which stood a pole 100 feet high ; now, the pole having been broken off, it was observed that the top portion resting on the stump just reached the brink of the pond. What is the height of the piece left stand- ing ? Ans. 41.9968 ft.
76. The area of a square inscribed in a circle is 400 square feet ; required the diagonal of a square circum- scribed about the same circle.
77. The four sides of a field, whose diagonals are equal, are known to be 25, 35, 31, and 19 rods, in a successive order ; what is the area of the field ?
Ans. 4A. IR. 38^ p.
78. The wheels of a chaise, each 4 feet high, in turning within a ring, moved so that the outer wheel made two turns while the inner made one, and their distance from one another was 5 feet ; what were the circumferences of the tracks described by them ?
Ans. Outer, 62.8318 ft.; inner, 31.4159 ft.
79. The girt of a vessel round the outside of the hoop is 22 inches, and the hoop is 1 inch thick ; required the true girt of the vessel.
80. If one of the Egyptian pyramids is 490 feet high, having each slant side an equilateral triangle and the base a square, what is the area of the base ?
Ans. 11 A. 3 rd. 2234 ft.
308 ELEMENTS OF GEOMETRY.
81. An ellipse is surrounded by a wall 14 inclies thick ; its axes are 840 links and 612 links ; required the quan- tity of ground enclosed, and the quantity occupied by the wall.
Ans. 4 A. 6 rd. enclosed, and 1760.49 sq. ft., area oc- cupied by the wall.
82. There is a meadow of 1 acre in the form of a square ; what must be the length of the rope by which a horse, tied equidistant from eacli angle, can be permitted to graze over the entire meadow ?
83. A gentleman has a rectangular garden, whose length is 100 feet and breadth 80 feet ; what must be the uni- form width of a walk half-way round the same, to take up just half the garden ? Ans. 25.9688 ft.
84. Two trees, 100 feet asunder, are placed, the one at the distance of 100 feet, and the other 50 feet from a wall ; what is the distance that a person must pass over in run- ning from one tree to touch the wall, and then to the other tree, the lines of distance making equal angles with the wall ? Ans. 173.205 ft.
85. There is a rectangular park 400 feet long and 300 feet l)road, all round which, and close by the wall, is a border 10 feet broad ; close by the border there is a walk, and also two others, crossing each other and the park at right angles, m the middle of the garden. The walks are all of one breadth, and their area takes up one tenth of the whole park ; required the breadth of the walks.
Ans. 6.2375 ft.
86. A farmer borrowed a cubical pile of Avood, which measured 6 feet every way, and repaid it by two cubical piles, of which the sides were 3 feet each ; what part of the quantity borrowed lias he returned ?
87. A board is 10 feet long, 8 inches in breadth at the greater end, and 6 inches at the less ; how much must be cut off from tlie less end to make a square foot ?
Ans. 23.2493 in.
88. A piece of timber is 10 feet long, each side of the
BOOK XIII. 309
greater base 9 inches, and each side of tlie less 6 inches ; how much must be cut off from the less end to contain a solid foot? Ans. 3.39214ft.
89. What must be the inside dimensions of a cubical box to hold 200 balls, each 2^ inches in diameter ?
90. Near my house I intend making a hexagonal or six- sided seat around a tree, for which I have procured a pine plank 16|^ feet long and 11 inches broad ; what must be the inner and outer lengths of each side of the seat, that there may be the least loss in cutting up tlie plank ?
Ans. 26.64915 in. inner, and 39.35085 in. outer length.
91. Required the capacity of a tub in the form of a frustum of a cone, of which the greatest diameter is 48 inches, the inside length of the staves 30 inches, and the diagonal between the farthest extremities of the diameters 50 inches. Ans. 165.34 gals.
92. The front of a house is of such a height, that, if the foot of a ladder of a certain length be placed at the dis- tance of 12 feet from it, the top of the ladder will just reach to the top of the house ; but if the foot of the ladder be placed 20 feet from the front, its top will fall 4 feet be- low the top of the house. Required the height of the house, and the length of the ladder.
Ans. 34 feet, the height of the building ; 36.0555 feet, the length of the ladder.
93. A sugar-loaf in form of a cone is 20 inches high ; it is required to divide it equally among three persons, by sec- tions parallel to the base ; what is the height of each part ?
Ans. Upper 13.8672, next 3.6044, lowest 2.5284 in.
94. Within a rectangular court, whose length is four chains, and breadth three chains, there is a piece of water in the form of a trapezium, whose opposite angles are in a direct line with those of the court, and the respective dis- tances of tlie angles of the one from those of the other are 20, 25, 40, and 45 yards, in a successive order ; required the area of the water. Ans. 960 sq. yd.
310 ELEMENTS OF GEOMETRY.
95. What will the diameter of a sphere be, when its solidity and the area of its surface are expressed by the same numbers ? Ans. 6.
96. There is a circular fortification, which occupies a quarter of an acre of ground, surrounded by a ditch coin- ciding with the circumference, 24 feet wide at bottom, 2G at top, and 12 deep ; how much water will fill the ditch, if it slope equally on both sides ? Ans. 135483.25 cu. ft.
97. A father, dying, left a square field containing CO acres to be divided among his five sons, in such a manner that the oldest son may have 8 aci-es, the second 7, the third 6, the fourth 5, and the fifth 4 acres. Now, tlic division fences are to be so made that the oldest son's share shall be a narrow piece of equal breadth all around the field, leaving the remaining four shares in the form of a square ; and in like manner for each of the other shares, leaving always the remainders in form of squares, one within another, till tlie share of the youngest be the inner- most square of all, equal to 4 acres. Ilcquired a side of each of the enclosures.
Ans. 17.3205, 14.8324, 12.2474, 9.4868, and 6.3246 chains.
98. Required the dimensions of a cone, its solidity be- ing 282 inches, and its slant height being to its base diam- eter as 5 to 4.
Ans. 9.796 in. the base diameter ; 12.246 in. the slant height; and 11.223 in. the altitude.
99. A gentleman has a piece of ground in form of a square, the difference between whose side and diagonal is 10 rods. He would convert two thirds of the area into a garden of an octagonal form, but would have a fish-pond at the centre of tlie garden, in the form of an equilateral triangle, whose area must equal five square rods. Re- quired the length of each side of the garden, and of each side of the pond.
Ans. 8.9707 rods, each side of the garden, and 3.398 rods, each side of the pond.
BOOK XIV.
APPLICATIONS OF ALGEBRA TO GEOMETRY.
692. When it is proposed to solve a geometrical prob- lem by aid of Algebra, draw a figure wbich shall represent the several parts or conditions of the problem, both known and required.
Represent the known parts by the first letters of the alphabet, and the required parts by the last letters.
Then, observing the geometrical relations that the parts of the figure have to each other, make as many indepen- dent equations as there are unknown quantities intro- duced, and the solution of these equations will determine the unknown quantities or required parts.
To form these equations, howeyer, no definite rules can be given ; but the best aids may be derived from ex[)eri- ence, and a thorough knowledge of geometrical principles.
It should be the aim of the learner to effect the simplest solution possible of each problem.
Problem I.
693. In a right-aiigled triangle, having' given the hy- pothenuse, and the sum of the other two sides, to deter- mine these sides.
C
Let A B C be the triangle, right-an- gled at B. Put A C :=j a, the sum A B + B C = 5, AB =. .T, and B C = 2/-
312
ELEMENTS OF GEOMETRY.
Then, x -\- 1/ = s,
and (Prop. XI. Bk. lY.),
x' + 7/ = a\ From the first equation, x =^ s — p.
Substitute in second equation this vahie of .r,
Or, 2p^ — 2sij=^ a' — s%
Or, ,f^s7j = ia'—is\
By completing the square,
2/' — si/ + is'=^a^ — is',
Extracting sq. root, 1/ — ^ 5 = ± V ^ a- — ^ s^^
Or, 7/ = ^s ± V^a'—is\
If A C = 5, and the sum AB + BC = 7, 2/ = 4or3, and X = 3 or 4.
Problem II.
694. Having given the base and perpendicular of a tri- angle^ to find the side of an inscribed square.
Let ABC be the triangle, C
and H E F G the inscribed square. Put A B = 6, C D = a, and GForOH = DI = x-; then will CI = CD — DI = a — X.
Since the triangles ABC, A Gr F C are similar,
AB:CD::GF:CI, or b : a : : X : a — x.
Hence, a b — b x = a .x,
a b
or, * = « + &•
DE B
BOOK XTV. 313
that is, the side of the inscribed square is equal to the pro- duct of the base by the altitude , divided by their sum.
Problem III.
695. Having' given the lengths of two straight lines drawn from the acute angles of a right-angled trian- gle to the 7niddle of the opposite sides, to determine those sides.
Let A B C be tho given triangle, A
and A D, B E the given lines.
Put A D = «, B E = Z;, CD or ^ CB = .^, and CE or ^ CA = y; then, since C D^ + C A=^ = A D\ and / / ^\P'
CE^+CB^ = BES we have x"^ -)- 4 y^ = a^.
and y^ + 4 x^ = b^, B D C
By subtracting the second equation from four timQs the first,
lby^=4:a^ — b%
jfi — h^
J2
by subtracting the first equation from four times the second,
15 x^ = 4.b^ — a\
"I
or, x= \^^-_^',
15
which values of x and y are half the base and perpendic- ulars of the triangle.
Problem IV.
696. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a point vnthin to the three sides, to determine these sides.
27
314 ELEMENTS OF GEOMETRY.
Let A B C be the equilateral trian- gle, and D E, D F, D G the given per- pendiculars from the point D. Draw D A, D B, D C to the vertices of the three angles, and let fall the perpen- dicular, OH, on the base, AB.
Put DE = 6t, DF = 6, DG = c, ^ ^^ ^
and A H or B H, half the side of the equilateral triangle, = X. Then A C or B C = 2 .r , and C H = ^ KQ^^^KW = \f\x^ — o;^ = V 8 x,"^ = a; V 3. Now, since the area of a triangle is equal to the product of half its base by its altitude (Prop. VI. Bk. IV.), The triangle ACB = J^ A B X CH = x X x^/l^^x'^l, ABD = ^ABxBG = .^Xc =cx, BCD = ^BCxr)E=a;X^ =ax. ACD = ^ACxDF=a;X^> =hx. . But the three triangles ABD, BCD,ACD are together equaf to the triangle A C B.
Hence, x^ ^ '^ = ax -\- b x -\- c x = x (a -\- h -\- c) ,
or, .^• V 3 = « + & + c ;
a 4-& -\-c or, X = ^ .
Hence each side, or 2 a; = -^ .
697. Cor, Since the perpendicular, CH, is equal to X V 3, it is equal io a -\- b -\- c ; that is, the ivhole per- pendicular of an equilateral triang-le is equal to the sum of all the perpendiculars let fall from any point in the tri- angle to each of its sides.
Problem V.
698. To determine the radii of three equal circles de- scribed within and tangent to a given circle, and also tan- iX'ent to each other.
BOOK XIV.
315
Let A F be the radius of the given circle, and B E the radius of one of the equal circles de- scribed within it. Put A¥ = a, and BE = x; then each side of the equilateral triangle, BCD, formed by joining the centres of the required circles, will be rep- resented by 2x, and its altitude, C E, by V 47^3^^% or x Vs.
The triangles B C E, ABE are similar, since the angles B C E and ABE are equal, each being half as great as one of the angles of the equilateral triangle, and the angle B E C is common.
|
Hence, |
CE :BE::BC; AB, |
|
or |
xV~S:x::2x: AB, |
|
and |
AB_^!. V3 |
|
But |
AB + BF = AF; |
|
hence, |
2x |
|
or |
2x + x^S = aV^, |
|
or |
(2 -f VS) x = aV^, |
|
Hence, |
^- ;/;.- 2.1. -»x 0.4641. |
|
Problem YI. |
699. In a right-angled triangle, having given the base, and the sum of the perpendicular and hypothenuse, to find these two sides.
Problem YH.
YOO. In a rectangle, having given the diagonal and perimeter, to find the sides.
316 elements of geometry.
Problem YIII.
701. In a riglit-aiigled triangle, having given the base, and the difference between the hypothenuse and perpen- dicular, to find both these two sides.
Problem IX.
702. Having given the area of a rectangle inscribed in a given triangle, to determine the sides of the rectangle.
Problem X. . 703. In a triangle, having given the ratio of the two sides, together with both» the segments of the base, made by a perpendicular from the vertical angle, to determine the sides of the triangle.
Problem XI.
704. In a triangle, having given the base, the sum of the other two sides, and the length of a line drawn from the vertical angle to the middle of the base, to find the sides of the triangle.
Problem XII.
705. In a triangle, having given the two sides about the vertical angle together with the line bisecting that angle, and terminating in the base, to find the base.
Problem XIII.
706. To determine a right-angled triangle, having given the perimeter and the radius of its inscribed circle.
Problem XIY.
707. To determine a triangle, having given the base, the perpendicular, and the ratio of the two sides.
Problem XY.
708. To determine a right-angled triangle, having given the hypothenuse, and the side of the inscribed square.
BOOK XIV. 317
Problem XYI.
709. In a right-angled triangle, having given the perim- eter, or sum of all the sides, and the perpendicular let fall from the right angle on the hypothenuse, to determine the triangle, that is, its sides.
Problem XYII.
710. To determine a right-angled triangle, having given the hypothenuse, and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle.
' Problem XYIII.
711. To determine a triangle, having given the base, the perpendicular, and the difference of the two other sides.
Problem XIX.
712. To determine a triangle, having given the lengths of three lines drawn from the three angles to the middle of the opposite sides.
Problem XX.
713. In a triangle, having given all the three sides, to find the radius of the inscribed circle.
Problem XXI.
714. To determine a right-angled triangle, having given the side of the inscribed square, and the radius of the inscribed circle.
Problem XXII.
715. To determine a triangle, having given the base, tlie perpendicular, and the rectangle of the two other sideg.
27*
L
318 ELEMENTS OP GEOMETRY.
Problem XXIII.
716. To determine a right-angled triangle, having given the hypothenuse, and the radius of the inscribed circle.
Problem XXIY.
717. To determine a right-angled triangle, having given the hypothenuse and the difference between a side and the radius of the inscribed circle.
Problem XXY.
718. To determine a triangle, having given the base, the line bisecting the vertical angle, and the diameter of the circumscribing circle.
Problem XXVI.
719. There are two stone pillars in a garden, whose perpendicular heights are 20 and 30 feet, and the distance between them 60 feet. A ladder is to be placed at a cer- tain point in the line of distance, of such a length, that it may just reach the top of both the pillars. What is the length of the ladder, and how far from each pillar must it be placed ?
Ans. 39.5899 feet, length of the ladder ; 34J feet, dis- tance of the foot of the ladder from the bottom of the lower pillar ; and 25f feet, distance of the foot of the ladder from the bottom of the higher pillar.
Problem XXVII.
720. There is a cistern, the sum of the length and breadth of which is 84 inches, the diagonal of the top 60 inches, and the ratio of the breadth to the depth as 25 to 7. What are its dimensions, provided it has the form of a rectangular parallelopipedon ?
Ans. Length 48 inches ; width 36 inches ; depth 10.08 inches.
BOOK XIV. 319
Problem XXYIII.
721. The three distances from an oak, growing in an open plain, to the three visihle corners of a square field,' lyin^at some distance, are known to be 78, 59.161, and 78 poles, in successive order. What are the dimensions of the field, and its area ?
Ans. Side of the square 21 rd. ; area 3 A. 2 R. 16 rd.
Problem XXIX.
722. There is a house of three equal stories in height. Now a ladder being raised against it, at 20 feet distance from the foot of the building, reaches the top ; whilst another ladder, 12 feet shorter, raised from the same point, reaches only to the top of the second story. What is the height of the building ? Ans. 41.696 ft.
Problem XXX.
723. The solidity of a cone is 2513.28 cubic inches, and the slant side of a frustum of it, whose solidity is 2474.01, is 19.5 inches. Required the dimensions of the cone.
Ans. Altitude 24 inches ; base diameter 20 inches.
Problem XXXI.
724. Within a rectangular garden containing just an acre of ground, I have a circular fountain, whose circum- ference is 40, 28, 52, and 60 yards distant from the four angles of the garden. From these dimensions, the length and breadth of the garden, and likewise the diameter of the fountain, are required.
Ans. Length 94.996 yds. ; width 50.949 yds. ; diameter of the fountain 20 yds.
Problem XXXII.
725. There is a vessel in the form of a frustum of a cone, standing on its lesser base, whose solidity is 8.67 feet, the depth 21 inches, its greater base diameter to that
320 ELEMENTS OF GEOMETRY.
of the lesser as 7 to 5, into whicli a globe had accidentally been put, whose solidity was 2i times the measure of its surface. Required the diameters of the vessel and of the globe, and how many gallons of water would be requisite just to cover the latter within the former.
A.ns. 35 and 25 inches, top and bottom diameters of tlie
frustum ; 15 inches, diameter of the globe ; and 34.2
gallons, the water required.
Problem XXXIII.
726. Three trees, A, B, C, whose respective heights are 114, 110, and 98 feet, are standing on a horizontal plane, and the distance from A to B is 112, from B to C is 104, and from A to C is 120 feet. What is the distance from the top of each tree to a point in the plane which shall be equally distant from eacli ? Ans. 126.634 ft.
Problem XXXIV.
727. A person possessed a rectangular meadow, the fences of which had been destroyed, and the only mark left was an oak-tree in the east corner ; he however recol- lected the following particulars of the dimensions. It had once been resolved to divide the meadow into two parts by a hedge running diagonally ; and he recollected that a segment of the diagonal intercepted by a perpen- dicular from one of the corners was 16 chains, and the same perpendicular, produced 2 chains, met the other side of the meadow. Now the owner has bequeathed it to four grandchildren, whose shares are to be bounded by the diagonal and perpendicular produced. What is the area of the meadow, and what are the several shares ?
Ans. Area of the whole meadow, 16 acres ; shares, 1 B,. 24 rd.; 1 A. 2 R. 16 rd. ; 6 A. IB. 24 rd. ; 7 A. 2 B. 16 rd.
THE END
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