^iffrrf^ffl-irr IN MEMORIAM FLORIAN CAJORl 7rfk . IV This book is intended as a sequel to the " First lessons in Geometry," and, therefore, presupposes some acquaint- ance with that little treatise. I think it better, however, that some interval should elapse between the study of that book and of this, — during which time the child may be occupied in the study of Arithmetic. Geometrical facts and conceptions are easier to a child than those of Arithmetic, but arithmetical reasoning is easier than geometrical. The true scientific order in a mathematical education would therefore be, to begin with the facts of Geometry, then take both the facts and rea- soning of Arithmetic, and afterwards return to Geometry, not to its facts only, but to its proofs. The object of "First Lessons in Geometry" is to develop the child's powers of imagination ; the object of this book is to develop his powers of reasoning. That book I con- sider adapted to children from six to twelve years of age, this to children from thirteen to eighteen years old. (3) 918241 CONTENTS. PART I. CHAPTER I. PAGE Prelbiinary, 11 1, Geometry the science of form. 2, Imagination. 3, Reasoning. 4| Application, CHAPTER II. Definitions, 12 6, Geometry. 6, Point. 7, Line. 8, Surface. 9, Solid. 10, Straight line. 11, Curve. 12, Plane. 1.3-14, Angle. 15, Parallels. 16, Triangle. 17, Right triangle. 18, Parallelogram. 19, Redtangle. 20, Square. CHAPTER III. Reasoning, • 15 21, Experiment. 22, Doubt. 23, Error. 24, No error small. 25, To avoid error. 20-30, Sum of angles in a triangle. 31, Generality. 32, Analysis. 33, Synthesis. 34, Vertical angles. .35, Alternate internal angles. 36, Proof of Art. 26. 37, Experimental proof. 38, Analysis. 39, Synthesis. 40, Reasoning. CHAPTER IV. Analysis and Synthesis, 19 41, Synthesis. 42-44, Must follow analysis. 45-47, Reasoning is in- sight. 48-49, Assumptions to be avoided. 50, Axioms. 61, Inevitable inferences. 62, Illustrated by a string. 63, Aristotle's dictum. 1* (5) 6 CONTENTS. CHAPTER V. Variety of Paths, . 24 54-5C, Various proofs of same truth. 67, 58, Two proofs of Art. 20. 59, 60, Algebraic proof of Art. 20. 01, 02, Kinematic proofs of Art. 20. 63, All proofs useful. — Examples. I. External internal angles equal. II. If angles equal, tlie linos parallel. III., IV., V., Idem. VI. Paral- lels do not meet. VII. Two perpendiculars from one point impossible. CHAPTER VI. The Pythagorean Proposition, 28 64-75, Analysis of Pythagorean proposition. 70, Ratio. 77, Its nota- tion. 78, Ratio unchanged by multiplying both terms. 79, Proportion. 80, Extremes and means. 81, Product of means equals that of ex- tremes. 82, Mean proportional. 83, Equals square root of product of extremes. 84, 85, Units. 80, Area. 87, Straight line makes same angle witli parallels. 88, 89, If same angles, the lines are parallel. 90, Coinci- dence by Buperimposition. CHAPTER VII. The Pythagorean Proposition, (continued), . . 33 91, One side and adjacent angles given, gives the triangle. 92, Parallelogram has its opposite sides equal. 93, Rotation at right angles to perpendicular on axis. 94, 95, The three sides given gives the triangle. 90, Quadrangle with equal opposite sides is a parrallelogram. 97, Area of rectangle. 98, Infinitesimals neglected. 99, 100, Angles given gives ratio of sides in triangle. 101, Right triangle and its parts. 102, Same as Art. 20. 103, Sum of acute angles 90^. 104, One acute angle gives form of right triangle. 105, Leg mean propor- tional between hypothenuse and segment. 100, Pythagorean. 107, Scho- lium. 108,Newproof of Art. 100. 109, Base and altitude. 110, Rectangle equivalent to parallelogram. Ill, Triangle half rectangle. 112, Pythago- rean. 113, Criticism invited. — Examples. VIII. Two sides and included angle gives triangle. IX. Oblique lines equal. X. Perpendicular line shortest. XI. Equal sides, equal angles. XII. Equilateral, equiangular. XIII. Bisecting isosceles triangle. XIV. Equal angles, equal sides. XV. Equiangular, equilateral. XVI. External angle. XVII. Longer side opposite greater angle. XVIII. Angles in isosceles triangle. XIX. Proof of Art. 93. XX. Square on diagonal. XXI. Quadrangle with equal opposite angles is parallelogram. XXII. Sum of angles of polygon. XXIII. Perpendicular hypothenuse mean proportional. XXIV. Areas proportional to altitudes. CONTENTS. 7 CHAPTER VIII. TuE Maximum Area, 42 114-116, Circle maximum. 117, rolygon. 118, Perimeter. 119, Isoperimetrical. 120, Maximum. 121, Circle and arc. 122, Centre. 12:{, 124, Radius, diameter. 125, Chord. 12fl, 127, Tangent. 128-130, Inscribed and circumacrilKid polyg-ona, and circles. 131, Regular poly- gons. 132, Analysis of Art, 114. 133, Syntliesis. 134, Area of triangle. i:]5, Pythagorean. 136, Shorter line nearer perpendicular. 137, Perpen- dicular shortest. 138, Radius and tangent at IK)^. 139, Radius perpen- dicular to arc. 140, Either side of triangle shorter than the sum of the others. 141, Maximum with two sides given. 142, Measure of angles. 143, Equal sides prove equal angles. 144, External angle. 145-147, Angle of two chords, and corollaries. 148, Maximum polygon with one Hide undetermined. 149, Maximum polygon of given sides. 150, Maximum isoperimetrical triangle, with one side given. 151, Perpen- dicular to base of isosceles triangle bisects the base. 152, New proof of Art. 150. 153, 154, Maximum isoperimetrical polygon of given number of sides. 155, Circle regular polygon. 150, Area of regular polygon. 157, Area of circle. 158, Perimeter of circumscribed polygon. 159-101, The circle the maximum of isoperimetrical figures. 162-161, Scholia. PART II. CHAPTER I. Geometrical Construction, . . 55 105, Truth invaluable. 166, A fortiori valuable. 167, Construction defined. 168, Rule, or ruler, how to make it. 169, Compasses. 170, Their use. 171, A scale, how made. 172, Multiplication by construc- tion. 173, Use of construction. CHAPTER II. Postulates, 59 174, Postulates. 175, Plane paper. 176, Straight line from point to point. 177, Straight line may be prolonged. 178, Circles may be drawn witli any centre and any radius. 179, Scholium. CHAPTER III. Straight Lines and Angles, 61 180, To divide a line mto equal parts^ 181, To divide an angle into equal parts. 182, Equal chords, equal arcs. 183, To draw angle of given degrees. 184, Protractor. 185, Its use. 186, To draw angle equal to given angle. 187, To draw a line making given angle with given line. 188, Another mode. 1S9, Parallel ruler, and triangles. 190, 191, To raise a perpendicular at a point. 192, To let one fall from a point. 8 CONTENTS. CHAPTER IV. Triangles, 69 193, Three sides given. 194, Two sides and an angle given. 195, One side and two angles given. 196, Choice of unit! 197-199, Impossibili- ties. 200-204. Examples. 205, 200, Scholia. 207, 208, Peculiar case. CHAPTER V. Quadrangles, 72 209, 210, Four sides and one angle given. 211, Three sides and two angles given. 212, Two sides and three angles given. 213, Examples. CHAPTER VI. Circles, 74 214, To draw an arc of given radius. 215, First solution. 216, Second solution. 217, Third solution. 218, Fourth solution. 219, To draw a tangent through a given point. 220, Examples. 221, To inscribe a cir- cle in a triangle. 222, To circumscribe a circle about a triangle. 223, To find centre of an arc. 224, To find radius when the arc is very flat. 225, Examples. 226, 227, To inscribe a hexagon and equilateral triangle in a circle. 228, To inscribe a square. 229, To inscribe a pen- tagon. CHAPTER VII. Areas, 81 230, Area. 231, Unit. 232, Rectangle. 233, Parallelogram. 234, Tri- angle. 235, Polygon. 236, To find a product. 237, Circle. 238, To find n. 239, 240, arcle. 241, The length of an arc. 242, Sector. 243, Segment. 244-246, Examples. CHAPTER VIII. Double Position, . 85 247, Approximation. 248, Double position. 249, Double position by construction. 250, Examples. CHAPTER IX. Interpolation and Average, 89 251, Problem explained. 252, Application. 253, Extension of double position. 254-256, Examples. 257, Scholium. CHAPTER X. Surveying, 93 258. Boy's instruments. 259, Ten-foot pole. 260, Horizontal circle. 261, 262, Artificial horizon. 263, 264, Quadrant. 265-267, Use of these instruments. 268, Lengths and angles. 269, Lines and instruments to be level. CONTENTS. .9 CHAPTER XI. Heights and Distances, 97 270, Hcijfht of tower when accessible. 271, Examples. 272, Object on a plain seen from a height. 273, Inaccessible tower. 274, 276, When tlie ground is not level. 27C, Example. 277-279, Examples. CHAPTER XII. Miscellaneous Examples, 100 280, Diagonals of rectangle. 281, Parallels equidistant. 282, One Bide, one jungle, and the ratio of %he pthqr two given. 283, One side and the ratio of tiie three angles given. 284, Circumscribed and inscribed squares. 285-288, Angle of two cliords, and chord and tangent. 289, Chord and radius perpendicular. 290, Two radii and length of common chord given. 291, 292, If two opposite angles are supplements, the quadrangle may be circumscribed. 293, Chords systematically arranged. 294, Parallels intercepted between parallels. 295, Angle between tan- gents. 290, 297, Two sides of a triangle and one perpendicular given. 298-300, One wde of two perpendiculars given. 301-304, One angle, with perpendiculars, or a side, given. 305, To describe a segment capable of containing a given angle. 300, Base, altitude, and vertical angle given. 307, Case of equivalent triangles, 308, Diagonals and their angle, with diameter of circumscribed circle, given. 309, Given angle at the vertex and segments of base. 310, 311, Diagonal of parallelopiped. 312, Square root of sum of squares. 313, 314, Base, vertical angle, and line to the middle of base, given. 315, 310, An angle, the altitude, and radius of circumscribed circle given. iil7. Circles tangent. 318, 319, Mean proportional between two lines. 320, 321, Tangent and secant. 322, Tripod always steady. 323, Intersection of two planes. 324, Sections of a sphere by a plane. 325, 32C, Map and profile of a railroad. PART III. SOLID GEOMETRY. CHAPTER I. Ratio and Proportion, . .107 327-340, Explanation of algebraic language. 341, Doctrine of propor- tions. CHAPTER II. Planes AND Angles, . . 110 342-345, Planes and angles. 346, 347, Parallel lines in a plane. 348- 351, Perpendiculars to plane. 352, Intersection of planes. 353, 354. Perpendicular planes. 355-357, Diedral angle. 358, 359, Line and 10 CONTENTS. plane. SCO, 362, 364, Parallel planes. 363, Kight lines not in a plane. 305, 367, 308, Triedral angle. 369, When equal. 371, 372, Sum of plane angles CHAPTER III. POLYEDRONS, 115 373, 375, Tetraedons. 376, 378, Similar. 379-381, Pyramids. 382, Prism. 383, 38t, Parallelopiped. 385, 386, Frustum. 387, 389, All Bolids divisible into pyramids. CHAPTER IV. Areas, 119 390-393, Similarity. 394, Areas of triangles as squares on homolo- gous sides. 395, Of polygons the same. 396, 397, Surfaces of polye- drons. 398, Pyramids cut by a plane equidistant from vertices. CHAPTER V. Volumes, 121 399, Unit. 400-403, Parallelopipedon. 404-406, Prism. 407-410, Pyra- mid. 411, 412, Truncated prism. 413, 414, Volumes as cubes. CHAPTER VI. The Cone, 124 415-419, Definitions. 420, Volume. 422, 425, Right cone. 426, Cones of equal height as their bases. 427, 430, Cylinder. 431, Cone one third a cylinder. CHAPTER VII. The Sphere, 126 432-435, Definitions. 436, Radii equal. 437, 438, Sections by planes. 439-443, Spherical triangle. 444, Shortest patli on a sphere. 445, 448, Sum of sides and angles in triangle. 449-451, Poles and radii. 452, Tangent plane. 453-455, Polar triangles. 450, 457, Three sides given gives the angles of a spherical triangle. 458, Symmetrical triangles. 459, The three angles given gives the sides of a spherical triangle. 460, Spherical triangles given in other ways. 461, Degree of surface. 462, 463, Lune. 464, Symmetrical triangles equivalent. 465, Surface of triangle. 466, 467, Surface of conic frustum. 468-470, Sur- face of sphere. 471, Solidity. 472, Usual decimals, CHAPTER VIII. Problems and Theorems, 134 SECOND BOOK IN GEOMETRY. PART I. CHAPTER I. PRELIMINARY. 1. Geometry is the science of forai. We really begin to learn Geometry when we first begin to notice the forms of things about us. Some pei*sons observe forms much more closely than others do ; partly owing to their nat- ural taste, and partly to their peculiar education. The study of plants, animals, and minerals, the practice of drawing, and the use of building blocks and geometrical puzzles, are good modes of leading one to notice, quickly and accurately, differences of form. 2. The second step in learning Geometry is to become able to imagine perfect forms, without seeing them drawn. The Httle book called "First Lessons in Geometry" was chiefly designed to help in the attainment of this power. It is filled with descriptions of forms that cannot be ex- actly drawn. This is especially true of many of tlie curves, which cannot be drawn so exactly as straight-lined figures and circles, but which we can, with equal ease, imagine perfect. (11) 12 '. ; : :.•'. ; : definitions. : -^l .'Tiift' 't£1^4 sfep^*in Ibaming Geometry is to learn to reason aborrt. 'foriiisi and to prove the truth of the inter- esting facts which we think that we have observed. This is the only way in which we can become able to find out new truths, and to be certain that they are true. And the firet part of this second book is written to teach the scholar how to reason out, or prove, geometrical truths. 4. After learning to reason out or prove geometrical tmths, it is pleasant to know how to use them. This is not the only object of Geometry. It is worth while to know a truth, simply because it is true. But it is also pleasant to be able to apply that truth to practical use, for the benefit of our fellow-men. And the second part of this book is written to show in what way we can turn Geometry to practical use. CHAPTER II. DEFINITIONS. 5. Geometry is the science o^ form. Every form or shape is, in general, enclosed by a surface ; every surface can be imagined as bounded, or else as divided, by lines ; and in every line we can imagine an endless number of points. 6. A point is a place without any size. It has a position, but no dimensions ; neither length, breadth, nor depth. 7. A line is a place having length, without breadth or depth. As we attempt to mark the position of a point by making a dot with the point of a pen or pencil, and the position of a line by moving the pencil point along the surface of tlie paper, we find it convenient to speak of a geometrical line as if it were made by the motion of a geo- fnetrical point. As the eye runs along the pencil line, so DEFINITIONS. 13 the eye of the mind runs along the geometrical line from point to point. 8. A surface is a place having length and breadth, with- out depth, that is, without thickness. 9. A solid is a place having length, breadth, and depth, A geometrical solid is not a solid body, but is simply the space that a solid body would occupy, if it were of tliat shape and in that place. In like manner a geometrical surface is not the surface of a solid body, but simply the surface of a geometrical solid. 10. A straight line is a line that does not bend j^ in any part. A point moving in it never changes the direction of its motion, unless it reverses its di- rection. 11. A cui*ve is a line that bends imperceptibly at every point. It must not have any ^q\ straight portion, nor any corners; A ^^.JV that is to say, it must bend at every point, but the bend must be too small, at each place, to be measurable. 12. A plane is a geometrical surface, such that B a point, moving in a straight line from any one point in the surface to any other, never leaves the surfece. The common name of a plane is ^'ajlat surface." 13. An angle is the difference of two directions in ono plane. If the line C O should turn around the point O so as to make the arc D C grow longer, the difference of the directions of O C and O D would increase, and we should say that the angle DOC grew larger and larger until the point C arrived at A, so that the two lines O D and O C were opposite in direction. 14. If the point C were earned round halfway to the opposite point A, that is, to the point E, the angle DOC 2 14 DEFINITIONS. -± would be a right angle, as D O E is. A right angle is a difference of direction half as great as oppositeness of di- rection. The difference between an angle and a right an- gle is called the complement of the angle. The difference between an angle and two right angles is called the sup- plement of the angle. Thus C O E is the complement of DOC, and C O A is the supplement of D O C. 15. When two lines make no angle with each y/ other, or make two right angles, they are called C — parallel lines. That is to — say, parallel lines are those that lie in the same direction or in opposite direc- tions. When two lines in a plane are not parallel, the point where they cross, or would cross if prolonged, is called the vertex of the angle. Lines making a right angle with each other are called perpendicular to each other. 16. A triangle is a figure enclosed by three straight lines in one plane. 17. A right triangle is a triangle in which two of the sides make a right angle with each other. These sides are then called the legs of the triangle, while the third side is called the hypothenuse. 18. A parallelogram is a figure bounded by four straight lines in a plane, with its opposite sides parallel. 19. A rectangle is a parallelo- ^ gram with its angles all right angles. 20. A square is a rectangle with its sides all equal. "/ REASONING. 15 CHAPTER III. REASONING. 21. Suppose that we wished to make another person believe that the tliree angles of a triangle are, together, equal to two right angles. One way of convin- cing him would be to take a trian- gular piece of card, or of paper, cut off the corners by a waving line, and lay the three comei^s together, to show him that the outer edges will make a straight line, as two square comers put together will do. 22. Yet he might not be satisfied that the line was per- fectly straight. Or perhaps he might say that if the angles of the triangle were in a different proportion, the corners put together would not make a straight line with their outer edges. 23. A gentleman once came to me and said, " I have found out that if you draw such and such lines, you will always find these two, A B and C D, equal. At least my most careful measurement shows no difference between them ." I said to another gentleman, who knew something of Geometry, " Can you prove that these lines will be equal if the figure is drawn exactly as direct- ed ? " He said he would try, and in a few days he sent me what he called a proof. But on reading it I found it only amounted to saying that " if the lines are equal, they are equal." I then examined the matter myself, and found that the lines were, in reality, never equal, although the difference was always very small, — too small to be easily discovered by measurement. C D 16 REASONING. 24. Such errors, too small to be discovered by measure- ment, are sometimes large enough to do great mischief; and at any rate, however small, they are still errors, and it is best to get rid of errors, and to find the exact truth, whether the error is mischievous or not. In order to do this we must leani how to reason, how to prove truths. And in order to avoid such mistakes as that of my friend, who thought he had proved the false proposition of which I have been speaking, we must learn to reason correctly. 25. When we put the comers of a paper triangle to- gether to make a straight line, we may say, Perhaps there is some slight error here, too small to be detected by meas- urement. How then shall we prove that there is no such error in a perfect geometrical triangle ? 26. The first thought that occurs to us will be, that if any straight line be dra^vn through one vertex of a trian- gle, as D E is drawn through the point ^ A, without passing through the trian- A^ gle, the three angles on one side of the -pv line, about the point A, are equal to two right angles, and if the sum of the three C li angles of the triangle is equal to two right angles, it must be equal to that of the three about the point A. 27. But as the central angle at A is already an angle of the triangle, it follows that the other two angles must be equal in their sum to the sum of the angles B and C 28. Now, this w^ill be true in whatever direction the line D E is drawn, only provided it does not pass through the triangle. Let us then imagine it to pass in such a direc- tion as to make the angle B A E equal to the angle ABC, and it will only be necessary to prove that D A C is then equal to A C B. For if D A C is equal to A C B, then, since we have sup- posed B A E equal to A B C, and BAG is one of the REASONING. 17 angles of the triangle, we shall have the three angles about A equal to the three angles of the triangle ; and as the three ano-les about A are equal in their sum to two right angles, the three angles of the triangle will be equal to two right angles, which is what we wish to prove. 29. But to say that D A C is equal to A C B is equiva- lent to saying that A D and C B differ equally in their direction from the direction of A C ; and since A C is a straight line and its direction from A is opposite to its direction from C, this is equivalent to saying that AD and C B go in opposite directions. 30. All that we have now to prove is, that the line A D or E D goes in the same direction as the line B C. But this needs no proof, because we have already supposed that E A makes the same angle with A B that B C does ; and as A B is a straight line, the direction of E A and C B must be opposite. But as E A is part of the same straight line with A D, it has the same direction as A D. The proof is now complete. 31. And this mode of proof does not depend at all upon the particular shape of the triangle. We have made no supposition concerning the shape of A B C, except that it should be a triangle. We have, therefore, proved that the sum of the three angles of any triangle is equivalent to two right angles. 32. Thus we have analyzed the proposition that the sum of the three angles of a triangle is equivalent- to two right angles, and found that it resolved itself at last into saying that two lines making equal angles on opposite sides at the end of a straight line must point in opposite directions — a proposition which is easily shown to be true. 33. But this mode of analyzing is very tedious when stated in words. A geometer usually does not state it ; he passes through it very rapidly in his own mind, and then 2* 18 REASONING. restates the process carefully in an inverted order, as fol- lows in articles 34, 35, and 36. 34. When one straight line crosses another, the oppo- site or vertical angles are equal. For since each line has but one direction, the difference of direction on one side of the vertex must be the same as on the other side. 35. When one straight line crosses two parallel straight lines, the alternate inter- nal angles are equal, or j^ in the figure A G E is ~7g ^ equal to D H F. For c "Zl D DHF is equal to B GF, y^ having its sides pointing ' ^ -in the same direction as those of B G F, and A G E is equal to B G F by article 34. 36. Through the vertex of any triangle, as through A, draw a straight line D E parallel to the opposite side B C. Then E A B will be D -tJt ^ equal to its alternate internal angle y^ \ \ ABC; and for the same reason D A C c -^ ^ — -^B will be equal to A C B. So that the three angles of the triangle will be equal to the three an- gles about the point A, and their sum is plainly two right angles. 37. The mode of proving that the sum of the three an- gles of a triangle is equal to two right angles, by cutting a piece of card, is called experimental proof. It is of very little use in mathematics, but of great use in the study of physics, especially in mechanics and chemistry. 38. The mode of proof used in articles 26-31 is called, by metaphysicians and by writers on Arithmetic, analysis. But as geometers, in their writings, almost never use this method, they have no name for it; and when they speak of analysis, or of analytical methods, they usually refer to something else apparently of a very different <;liaracter. ANALYSIS AND SYNTHESIS. 19 39. The mode of proof in articles 34-36, called by metaphysicians synthesis, by geometers demonstration or deduction, is that usually employed in stating geomet- rical results. This mode is chiefly applicable to mathe- matics, and must be used with very great caution in rea- soning upon other subjects. 40. A proposition, which we wish to prove, may be compared to a mountain peak which we wish to show is accessible from the highway. The method of articles 26-31 may be compared to taking a flight by a balloon to the top of the peak, and then finding a path down to the highway; while the method of articles 34-36 maybe com- pared to the direct ascent of the mountain. In either case we show that the peak is accessible, because we actually pass over all the steps of a connected pathway between the road and the mountain top. Thus in geometrical demonstration we pass through every step connecting the simplest self-evident truths with the highest deductions of the science ; while in the process which writers on Arithmetic call analysis, we pass over ever^ step from the latter truths down to the simplest. In either case we prove that the higher truth really stands on the same basis as the simpler, and must, therefore, be true. CHAPTER IV. ANALYSIS AND SYNTHESIS. 41. What I have called demonstration or deduction, but which is better called synthesis, because it is a putting together, one by one, of the parts of a complex truth, is the only mode of proof that you will usually find in works on 20 ANALYSIS AND SYNTHESIS. geometry. And if such works are carefully read they are always intelligible to a child of good geometrical reason- ing powers. 42. But the study of such works does not always teach a child to reason for himself. The pupil says, " Yes, I un- derstand all this, and yet I could not have done it without aid ; I do not see how the writer knew where to begin ; how he knew that by starting from these particular truths, and going in that particular path, he could reach that proposition." A pupil who had never studied geometry could not, for instance, tell why in articles 34-36 we should begin with showing that vertical angles are equal. He would not see any contiection between that truth and the desired proof, and would not know that this synthesis had been preceded, in the mind of the writer, by a rapid anal- ysis, such as that of articles 26-31. 43. It is as though a mountain guide, wishing to make for a child a path up to a mountain peak, should lead him along the highway until the peak was hidden by the lower hills about its base, and then begin boldly to clear a road, through the brush-wood and trees, until he reached the top. The child might say. " How did you dare begin at once to cut down the bushes and clear the path ? How did you know that the road you were making would not lead you to the edge, or to the foot, of some precipice, or that it would not take you to a different peak from that which you wished to climb ?" And if the child received no answer to his questions, — if he was not told that the guide had already climbed to the summit and again de- scended, — he would have learned little to help him in laying out paths for himself. 44. In like manner, although the descent from difficult propositions to more simple is more tedious than the as- cent, it will be more useful to a learner, because it will show him the maimer in which, by a mental process, we ANALYSIS AND SYNTHESIS. 21 discover the points from wliich wc are to- start in our ascent. That is to say, if we follow a good analysis, we shall learn how to perform synthesis for ourselves ; but if we were simply to follow a writer's synthesis, we should not learn how to analyze, which must nevertheless always go before synthesis, 45. Among the first requisites in reasoning is a clear understanding of the object in view ; that is, of the point to be proved ; and next, a clear perception of each partic- ular part of the demonstration, and of the connection of each part wuth the adjacent parts. Thus, in laying out a path up a mountain, it is necessary to know exactly from what point yon wish to start, and to what point you wish to go. It is also necessary to exam- ine carefully each point of the road, for a single impassable place would destroy the value of the whole road. 46. Each step of the proof must be a simple step, and clearly true ; that is, it must be so simple and self-evident as to be beyond all doubt. 47. The analysis must end, or the synthesis begin, with truths that are self-evident, or else that have been already proved. Your mountain path must begin on level, or at least on accessible ground. 48. Care must be taken not to introduce any thing as true which has not been proved. This would be like starting your mountain road in two places at once. You might afterwards find impassable barriei-s between the two parts of your road, and perhaps find that one of them could not be made to the top of the mountain, nor the other to its base. For example, in Art. 36, I drew a straight line through A, parallel to B C. This was very well — for no one can possibly doubt such a line might be drawn. But if, instead of that, I had said. Let us draw a straight line through A in such a manner as to make the angles on the two sides of A equal to the angles B and C, I should 22 ANALYSIS AND SYNTHESIS. have done what I had no right to do. For that would be taking for granted a thing which I must prove ; namely, that a straight line can be thus drawn. It would be start- ing half way up my mountain, and taking for granted that the lower part of the path could be built afterwards. It would require a straight line to fulfil two conditions at once, without having shown that one condition does not exclude the other. 49. Whether we reason by synthesis or analysis, we must therefore reason very carefully, in order to connect the proposition which we wish to prove by a stairway of self-evident steps with a self-evident foundation. 50. By a self-evident truth, I mean a truth which can- not be made any plainer, and which is already perfectly plain to an intelligent person who looks steadily at it. For instance, that two straight lines can cross each other only in one place at once ; that any cui-ve can be cut by a straight line in at least two places ; that either side of a triangle is shorter than the sum of the other two ; tl\at if three strings, and no more than three, come from one point, one of them must have an end at that point, — these are self-evident truths. 51. By a self-evident step in reasoning, I mean the statement of the relation of one truth to another, or of the dependence of one truth upon another, when that depend- ence or that relation is itself a self-evident truth. Self evident steps in reasoning are simply the statement of self-evident truths of connection. For instance, when w^ have explained the meaning of " a straight line " by calling it a line that has in every part the same direction, and have explainedthe word "angle" to mean the difference of two directions in one plane, then it follows that the angle which two straight lines make with each other is the same in one part of the lines as in any other, and that tlie two different angles apparently made by two straight lines ANALYSIS AND SYNTHESIS. 23 cannot really bo made, unless one of the lines goes in two o[)posite directions at the same time. No reasoning can make the connection between these definitions and the equality of vertical angles any more plain. It is a self- evident connection. 52. Or, suppose that we say that you cannot make one rope go from a centre post to the four corners of a square, and also around the square, and have but a single rope from post to post. We should prove it in this way. Let there be a rope around the square, and going also from each post to the centre. This of course can be imagined. It is a defi- nite and allowable conception. But we will prove that this rope must be in two pieces. For each of the four corners will have three lines coming from it, one towards each adjacent corner, and one towards the centre. Thus it follows by self-evident connection, from the conception of the rope going around the square and to each corner, that there will be four points, from each of which three lines come. But it is a self-evident truth that at each of these points there must be one end of a rope. Hence, by self-evident connection, there will be four ends of rope about the square. Hence, by self-evident connection with the self-evident truth that one piece of rope can have but two, and must have two, ends, it follows that there must be two pieces of rope, and cannot be only one. Now, the whole of this proof is simply the statement of self-evident connections betvv^een the proposition that one rope cannot go around a square and also from each corner to the cen- tre without doubling, and the self-evident truths that a piece of rope must have two, and cannot have more than two, ends ; and that when only three lines of rope come from one point, one of them must end at that point. The proof is simple ; and yet intelligent men have spent hours in experimenting with a string and five posts thus arranged, or with a pencil and five dots representing posts. 24 VARIETY OP PATHS. 53. Many self-evident truths are general, and self-evi- dent steps are generally the recognition of general rela- tions; and therefore most writers on reasoning say that reasoning consists simply in showing that a particular case comes under a general class, that is, that the only self- evident connection of propositions is the actual inclusion of one proposition in another. But in the mathematics, there are many self-evident truths which it is difficult to state in a general form ; and I therefore think that the explanation which I have given of the process of reasoning will be of more use to you in your geometrical studies. CHAPTER V. VARIETY or PATHS. 54. As there are usually many paths by which we may ascend a hill, so there are usually many modes by which we may demonstrate a proposition. In the case of a sim- ple proposition, it is not usually worth while to try more than one mode. But with more difficult problems, it is sometimes worth while to spend a great deal of labor in discovering the simplest mode of demonstration. There are geometrical truths which can be demonstrated in so simple a manner as to require only twenty lines to write down the demonstration ; and yet some writers, from ig- norance of this simple mode, have written more thantwent)^ pages to prove the same truths. 55. It will therefore be useful to you, to show you, by a simple example, such as that of the equality of the sum of the angles in a triangle to two right angles, the great variety of methods by which a single proposition can be proved. VARIETY OF PATHS. 25 56. In the proofs of this proposition, which I will now give yon, I will not be careful to follow out every step. It will be enough, /br the purpose ice now have in vieWy simply to show you the general line of the paths, without taking you through every step of the way. 57. The line D E might be drawn so as to coincide in direction with one of the other sides of the triangle, as A B, which would give us the figure in the margin. And by imagining the dotted line A F par- allel to C B, we should have F A C equal to A C B, and FAD equal to C B A, which would make the three angles at A equal to the three angles of the triangle, as in the former proof. (See Chapter III.) 58. By prolonging the lines C A and B A through the ^oint A, D E being parallel to B C, we should have NAD equal to ^ \A^ ABC, JAE equal to A CB, and NAJ equal to CAB. So that the three angles of the triangle will be equal to the three angles on the upper side of the line D E, which are manifestly equal to two right angles. 59. The three methods of proving this proposition that I have now given, are strictly geometrical. Others might be given, that are something more like algebraic reasoning. 60. Let us, for instance, imagine each side of a triangle prolonged at its right hand end, as in this figure, and also a line ^ drawn from one vertex, as C, \ parallel to the opposite side B A. Now, the external angles F C B, ^ D B A, E A F, are plainly equal ^ to the three angles F C B, B C G, G C F, and these amount to four right angles. But each external angle is plainly 3 26 VAEIETY OF PATHS. the supplement of one of the angles of the triangle ; that is, it is equal to the difference between two right angles and one angle of the triangle. The sum of the three ex- ternal angles must therefore be equal to the difference between six right angles and the angles of the triangle. But as this difference is four right angles, the three angles of the triangle must be equivalent to two right angles. 61. If we introduce the idea of motion, we can devise quite a different sort of demonstration. p Suppose, for instance, that I stand at the point A, w^ith my face towards C. Let me now turn to the right until I face towards B. I have now changed the direction of my flice by an amount w^hich is equal to the angle at A. Suj)pose that I now walk to B, without turning ; I shall have my back towards A, and if standing still, I turn to the right, until I have cliangcd my direction by an amount equal to the angle ABC; I shall have my back towards C. Let me now walk back- ward without turning, until I reach C, and I shall have my face tOAvards B. I will now turn a third time to tlie right, until I face the point A. My three turnhigs, or changes of direction, have been equal to the three angles of the triangle; they have all been to the right ; thci-efore my whole change of direction is equal to tlie sum of these angles ; I am now looking in exactly tlie opposite direction to that from which I started ; I am looking from C to A, instead of from A to C; I have turned half way round; that is, through two right angles. Whence, the sum of the three angles of the triangle is equivalent to tv/o right angles. 62. Another demonstration, by means of motion, may be obtained as follows : Suj)posc an arrow, longer than either side of the triangle, to be laid upon the side A C, pointing in the direction from A to C. Taking hold of the pointed end beyond C, turn the arrow round upon the VARIETY OF PATHS. 27 point A, as a pivot, until the arrow lies upon the line A B. Taking now hold of the further end, beyond A, turn the aiTow upon B as a pivot, until the arrow lies upon the line B C. Using C as a pivot, turn it now until the point of the arrow is over A. The arrow has thus been reversed in direction, turned half way round, or through two right angles. It has been turned successively through tlie three angles of a triangle, and every time in the same direction, like the hands of a watch ; so that its total change of direction, two right angles, is equivalent to the sum of the three angles. 63. You have thus seen how a single proposition may be proved in a variety of ways. We have shown what is the value of the sum of the angles in a triangle in six different ways ; in three, by what is called rigid geometry ; in one, by a partly algebraical process ; and in two, by introducing the idea of motion. And I wish you to ob- serve, that every one of the six ways is satisfactory. They are all proofs that are certain, because they lead you from self-evident truths by self-evident steps. One is not more certain than the other, because they are all absolutely cer- tain. The only choice between them is, that some are more purely geometrical ; some are better adapted to the peculiar tastes of different students ; and some are neater, and more quickly perceived by untaught persons. Examples. By aid of the principles and methods of the five preced- ing chapters, the learner may demonstrate (sometimes in a variety of Avays) the following simple projiositions X — I. When parallel lines are crossed by a third, the /^ external ' internal angles ~7g are equal ; that is, F G B C ^/^ . ■ ■ D = GHD,4fcc. X 28 THE PYTHAGOREAN PROPOSITION. II. If two lines,' cut by a third, make the alternate- internal, the external-internal or the opposite-external angles equal, the lines are parallel. III. If two lines, crossed by a third, make the adjacent internal angles (as B G H, D H G) supplements to each other, the lines are parallel. IV. If two lines make the same angle with a third, they are parallel to each other. V. State this proposition for the cases when the angle is zero, one right angle, and two right angles. VI. Parallel lines can never meet. [Note. To prove a negative of this kind, the easiest mode is to show the absurdity of the affirmative. In the present case, grant that the lines met at a certain point, and show from the nature of the straight line, that the parallel lines must in this case be one line, which is absurd.] VII. Only one perpendicular can pass through a given point to a given straight line. [Proof by VI.] CHAPTER VI. the PYTHAGOREAN PROPOSITION. 64. YoTJ recollect that the square built on the hypoth- enuse of a right triangle is equivalent in its area to the sum of the squares built upon its legs. This is one of the most useful of all geometrical truths. Let us first an- alyze it in one or two modes, and then build it up synthetically by the same paths. We may afterwards, if we like, devise other modes of anal- ysis and synthesis ; for this proposi- tion, like all others, may be ap- proached in various ways. THE PYTHAGOREAN PROPOSITION. 29 65. The Pythagorean proposition or theorem might bo suggested in different ways. But in whatever way wo were led to suspect that the square on the liypothenuse is equivalent to the sum of the squares on the legs, we should, in reflecting upon it, probably begin by drawing a right triangle with a square built upon each side. CG. We should inquire whether the square on the hy- pothenuse could be divided into two parts that should be respectively equal to the other two squares. And we should judge that these parts should be somewhat similar to each otlier in shape, because the legs do not differ in their relations to the hypothenuse, except in size, and in the angles they make with it. 67. But we cannot readily conceive of any division of the square into two somewhat similar parts, except into two rectangles. And then it is ap- parent that two rectangles, bearing respectively the same relations to the squares on the legs, may be formed by drawing a line from the vertex of the right angle at right angles with the hypothenuse, and continuing it through the square, as C F is here drawn. 68. It will now only be necessary to show that one of these rectangles is equivalent to its corresponding square ; because the same mode of proof will obviously answer for the other rectangle and its square. 69. Now, if we know, or can prove, that the area of a rectangle is measured by the product of its sides, we shall have to prove that A E X A B, or A E X A B, is equiva- lent to A C X A e. 70. But by the doctrine of proportion it may be shown that this would be equivalent to saying that A E is to A C 35 A C is to A B. 3* 30 THE rYTIIAGOREAN niOrOSITIOX. 71. Again, it may be shown by geometry that this pro- portion between the lines A B, A C, and A E, would be true if the triangle A E C were similar to A C B, and that A E stood in one to AC as A C stood to A B in the other ; so that all that remains for us to do is to show that these triangles are similar. 72. But we can show by geometry that two triangles are similar when their angles are equal. 73. And it is easy to show that the angles of these tri- angles are equal to each other. 74. For CAB and C A E are the same angle ; A C B and A E C are both right angles ; and therefore ABC and ACE are each complements of C A E. Moreover, A C and A E are situated in the triangle A E C, in the same manner that A B and A C are situated in the triangle ABC. 75. We have thus, in articles 66-74, sufficiently ana- lyzed the Pythagorean proposition to enable us to build it up again in a deductive form. This analysis, however, has been partly algebraical, as it has introduced the idea of multiplying two lines to produce a surface. Let us now begin and build up the proposition by the same road. We shall find 31 articles necessary ; and I will number them from 76 to 106. First Proof of the Pythagorean Proposition. 76. Pefinition, The comparative size of two quantities is called their ratio ; thus, if one is twice as large as the other, they are said to be in the same ratio as that of 2 to 1 ; or to be in the ratio 2 to 1 ; or it is said, in a looser way, that their ratio equals 2. 77. Notation, Ratio is written by means of the marks :, —, and by writing one quantity over the other. Thus, A : B, A -f- B, and jg, are each used to signify the ratio THE PYTHAGOREAN PROPOSITION. 81 of A to 1^>. These marks arc the same as those used in iirithni'v lie. to .sigiiity Quotient, because the meanmg of a (juotiuiit is '-d number havmg the same ratio to 1 that the dividend has to the divisor." The ratio of A to B is not the quotient of A divided by 13, but it is the ratio of that quotient to iniity. 78. Axiom. If each of two quantities is multipUed or divided by the same number, the ratio of the products or quotients will be tlie same as that of the quantities them- selves. Thus twenty inches is in the same ratio to twenty' rods as one inch to one rod, or as the twentieth of an inch to the twentieth of a rod. 79. Dejinition. A proportion is the equality of two r:itios. Thus (if we use the sign i= to signify "is equal to ") A : 13 = C : D is the statement of a proportion. It signilies that A is in the same ratio to B that C is to D. 80. Definition, When a proportion is written as in article 79, the first and last terms, that is, A and D, are called the extremes, and the others, that is, B and C, arc c;illed the means. 81. Theorem, In every proportion the product of the means is equal to that of the extremes. Proof, In any pro]^ortion, as M : N" = P : Q, we wish to prove (using the mark X to signify « multipHed by") that M X Q = N X P. Xow, in order to do this, we must use only self-evident truths. The only truth of this character that we have given above is that of article 78. But in order, by means of the multiplications of article 78, to change the first ratio ]\[ : N^ into M X Q, we must, whatever else we do, at least m ultiply each term by Q, and this will give us M X Q : X X Q = P : Q ; and in order to change the second ratio P : Q into N X P? we must, at all events, multiply each term by N, and this will give us M X Q • ^ X Q = N X P : N X Q. Thus, from the self-evident truth of article 78, we find 32 THE PYTHAGOREAN PROPOSITION. that the product of the means bears the same ratio to the product N X Q that is borne to it by the product of the extremes. And as it is self-evident that two quantities, bearing the same ratio to a third, must be equal to each other, we have proved that the product of the means is equal to that of the extremes. 82. Definition, When both the means are the 'same quantity, that quantity is called a mean proportional be- tween the extremes. 83. Corollary. It follows from article 81, that the product of the mean proportional multiplied by itself is equal to the product of the extremes. 84. Definitions » A unit of length is a line taken as a standard of comparison for lengths. Thus an inch, a foot, a pace, a span, &c., are units. The length of any line is its ratio to the unit of length. 85. Definition, A unit of surface is a surface taken as a standard of comparison. The most common unit of surface is a square whose side is a unit of length. 86. Definition. The area of a surface is the ratio of the surface to the unit of surface. 87. Theorem, Any straight line in the same plane with two parallel lines makes the same angle with one that it does with the other. Proof, For as the straight line has but one direction, and each of the parallel lines may al- ways be considered as going in the same direction as the other, the difference of that direction from the direction of the third straight line must be the same for each of the parallel lines. 88. Corollary, If a straight hne is parallel to one of two parallel lines, it is parallel to the other ; if at right angles to one of the two, it is at right angles to the other. 89. Theorem, If a straight line makes on the same side of itself the same angle with two other straight lines in the same plane, those other straight lines must be parallel. THE PYTHAGOREAN PROPOSITION. 33 Scholium. The line must not be conceived as reversing its direction at any point. Proof, For if two directions differ equally from a third, they must be equal to each oUier. Second Scholium, If the straight line reverses its di- rection between the other lines, and makes equal angles with them, it shows that it crosses each at an equal dis- tance from their point of mutual intersection. 90. Axiom, If the boundaries of one plane surface are similar to those of another in such a way that the two surfaces would coincide in extent if laid one upon the other, the two surfaces are equivalent. CHAPTER VII. THE PYTHAGOREAN PROPOSITION CONTINUED 91. Theorem, If a triangle has one side and the adja- cent angles equal respectively to a side and the adjacent angles in another triangle, the two ^ j, triangles are equal. Proof, Let us suppose that, in the triangles ABC and D E F, we have the side A B equal to the side D E, the angle at A equal to the angle at D, and that at B equal to that at E. Let us imagine the triangle D E F to be laid upon ABC in such a manner as to place E upon B, and D upon A, which can be done, because A B is equal to D E. Now, as the angle A is equal to D, the line D F will run in the same direction as A C, and, as it starts from the same point, will coincide with it. Also, since the angle B is equal to E, the line E F will coincide wnth B C. The 34 THE PYTHAGOREAN PROPOSITION. point (F) of intersection of D F and E F must therefore coincide witli C, the point of intersection of A C and B C. Whence, by article 90, the triangles are equal. 92. Theorem, The opposite sides of a parallelogram are equal. Proof, Article 90 gives us the only test of geometrical equality. So that, in order to prove this the- orem, we must show that in a parallelogram like A B C D, A B may be made to coincide with D C, 5^— . -jj^y C and B C with A D. And this would ev- idently be done if we could show that the triangle A B C is equal to A D C. But in these tri- angles the line A C is the same,, and by article 87 the adja- cent angles A C B and CAB are equal to the adjacent angles CAD and A C D ; whence, by article 91, the two triangles are equal, and A D is equal to B C, and A B equal to D C. 93. Axiom, * If one end of a straight line stands still while the other tunis round, the end that moves will hegin to move in a direction at right angles to that of the line itself Thus if A B were to ^ begin to turn about the point A B A, B would hegin to move either towards C or towards D. [If this proposition is not acknowledged as an axiom, the proof is in Ex. XIX. at the close of the chapter.] 94. Theorem, The angles of a triangle cannot be al- tered without altering the length of the sides. Proof. If in any triangle, as ABC, the sides were unchangeable, any alteration of the angles A and B would, by article 93, make the point C move in two directions at once, (namely, at right angles to A C, and at right angles to B C,) which is impossible, and therefore the angles cannot be altered. THE PYTHAGOBEAN PROPOSITION. 35 95. Corollary, If the tlireo sides of a tri.in and C B ^ will now be composed of equal sides, and we have already proved (Arts. 91-95) that they must have equal angles ; that is, the angle at A is equal to that at C. 144. Theorem, If one side of a triangle is prolonged, the external angle is equal to the sum of the opposite in- ternal angles. This has been proved in Art. 57. 145. Two chords starting from one point in a circum- ference intercept double the arc that would be intercepted by radii making the same angle ; that is, the angle of the chords is measured by half the arc included between them. Proof, If one chord, as AB, passes through the centre D of the circle, it is plain that by drawing D C the angle C D B will be equal to the sum of the angles CAD and D C A. But since D A and D C are equal, these angles arc equal, and C D B is equal to twice CAD. If neither chord passes through the centre of the circle, we can draw a third chord, starting from A, passing through the centre of the circle, and apply this reasoning to the two angles formed with this third chord by the other two. The angle of the other two chords will simply be the sum or the difference of these two angles. 146. Corollary, If the vertex of a right angle be placed in the circumference, the sides will intercept a semicircle. 147. Corollary, If a circle be circumscribed about a triangle, and one side of the triangle passes through the centre of the circle, the opposite angle is a right angle. 48 THE MAXIMUM AREA. 148. Theorem. The maximum of polygons, having all the sides given but one, may have a circle circumscribed about it, having the unknown side for a diameter. Proof. Let A B C D E be the maxi- mum polygon, formed of given sides A B, B C, &c., and the unknown side, A E. Join B E by a straight line. Now, since the polygon is a maximum, we cannot, leaving B E unaltered, by altering A E enlarge the triangle ABE, because that would enlarge the polygon. The angle ABE is therefore a right angle, by Art. 141, and a circumference, having A E for its di- ameter, would pass througli the point B. In like manner it can be shown that a circumference having the same diameter Svould pass through each of the other points. 149. Theorem. The maximum of polygons formed with given sides can be inscribed in a circle. Proof, Let A B C D E be a polygon, formed of given sides, with a circle circumscribed about it. Draw the diameter A F, and ' join F C and F D. The polygons A B € F and A E D F are now maxi- mum polygons, and therefore ABODE must also be a maximum, since its en- largement would enlarge the sum of the other two. We have thus proved the converse of the proposition, and the proposition is true, unless there is more than one maximum form of the polygon. The converse is more easily proved than the proposition, and I therefore proved it, on the assumption that there is but one maximum form. That is, I have proved that a polygon of given sides, when inscribed in a circle, is a maximum ; but that does not strictly prove that the maxi- THE MAXIMUM AREA. 49 mum can always be inscribed in a circle ; except on the assumption, which is, however, a safe one, that a polygon formed of given sides, arranged in a given order of succes- sion, can have but one maximum form. 150. Theorem, Of isopcrimetrical triangles with one side given, the maximum has the two undetermined sides equal. Proof, In order to prove this we have only to show that the point A is at its greatest dis- tance from the base B C, w^hen opposite the middle of it. This might seem scarcely to need proof For when we use a string and stick to illustrate the problem, we can see that by sliding the finger from the middle of the string, it can be brought down into a line with the stick; and the greatest height from the stick is near the middle of the string. Further consideration shows it must be exactly at the centre of the string, because the finger and string have precisely the same rela- tion to one end of the stick as to the other ; and- a motion towards either end must afiect the height of the finger in a similar manner. This reasoning is doubtless satisfactory to every fair mind. Yet it is not a good mathematical demonstration, and I have given it to you for the purpose of illustrating the peculiar nature of mathematical reasoning. The rea- soning just given leaves no real doubt on the mind, but it is rather because we see with the eye that the finger is highest in the middle, than because we see with the mind that it must be. There is another step still lacking, to prove to us that the highest points are not on each side of the exact middle, as that would satisfy the conditions of symmetry and of declination towards each end. Let us 50 THE MAXIMUM AREA. then seek a proof which shall not force us to consider the whole motion of the finger, but which shall simply compare two forms of the triangle, one with the finger in the mid- dle of the string, and one with the finger on one side, ^y 151. Theorem, If a straight line be drawn from the vertex of two equal sides in a triangle, at right angles to the third side, it divides the third side into equal parts. Proof, Let c and a be ^ equal sides in a triangle e^.^''^'T^'*^**-^ ABC. Since the angles . ^^^^^"^ { ^^""""^^ ^^ at A and C are equal, the ^ angles SBC and ^ B A are also equal. If, therefore, the triangle B 5 C be folded over on the line B ^, the line a will take the same direction as the line c, and, being of the same length, will coincide with it. Hence, b C will also coincide with h A, and the two lines must be of equal length. •152. Mio 2'>roof of Art, 150. Let A B C and A B C be isoperimetrical, and let A B and B C be equal. Continue A B to D, making B D =: B A rzr: B C, and join D C. Then, by Art. 147, the angle D C A is a right angle. Draw B' E making it equal to B' C. Join A E. A E will be less than the sum of A B' and B' E, that is, less than A B' and B'C, that is, less than ^ ^ A B and B C, that is, less than A D. But if A E is less than A D, then C E must be less than C D, by Art. 136. Draw B H and B' I at right angles to CD; we have C I, which is half C E, less than C H which is half CD. But C I and C H are the altitudes of the triangles ABC and A B' C above their common base A C. The triangle with the undetermined sides equal has the greatest altitude, and must be the largest triangle. 153. Theorem, The maximum of isoperimetrical poly- A I> /) E ^ld coincide. That is, equal chords in the same circle subtend equal arcs, and arcs can be equally divided by a pair of dividers, in tlie same manner as straight lines and angles. STRAIGHT LINES AND ANGLES. 63 183. Frohlem, To draio an angle of a given number of degrees, — From any point A, as centre, in a straight line ; r--.,^^ ^ A B, with any radius A B, dc- j /'^n scribe an arc B C. Keeping • A^ the compasses open at the same j \ width, place one foot at B, and j J^D with the other mark the point I ^.^^^^^ \ C. The arc B C is then (Ex. j ^^ XII.) an arc of 60^ Closing \k^^^ \^ the compasses until they will pass over the same arc in four steps, you obtain (by Art. 182) arcs of 15°. Selecting either of these, according to the degrees required, close the compasses, until they divide it into arcs of 5°. By dividing one of these arcs of 5° into ^\Q equal parts, you can obtain the required degree, count- ing from B up to D. Join D A, and you manifestly have the angle required. Thus, if the given number of degrees were twenty-seven, we should take the second arc of 15°, the third arc of 5^ in that arc of 15°, and the second de- gree of those five. 184. The formation of a protractor. — Take a piece of hard, smooth card, draw a fine, straight line, as A B (see fig. above), and with a convenient radius, say three inches, draw the arc B C. Measure carefully the arc B of 60° by having the compasses, while yet unaltered from the radius with which you drew the arc, step from B to C. Divide the arc as accurately as possible into four equal arcs of 15*^ each, and set off two such arcs beyond C, so as to make the whole arc 90°. Divide each arc of 15° carefully into three equal parts, which will each be 5°. Divide each also into five parts, each of which will be 3°. By stepping over the whole arc with the compasses open for three degrees, first stepping over it lightly to make sure that twenty steps will exactly make 60°, and then with a heavier step, so as 64 STRAIGHT LINES AND ANGLES. to leave footprints ; repeating this heavier stepping from each point of division of the 5° arcs, you can divide t\ie prolonged arc into 90 equal degrees. The first divisions, starting from B, will give 3, 6, 9, 12, 15, 18, 21, &c. The second, starting from the 5° point, will give 2, 6, 8, 11, 14, 17, 20, &G, The third, starting from the 10° point, will give 1, 4, 7, 10, 13, 16, 19, 22, &c. ; and these three series evidently embrace all numbers. Mark each fifth point with a longer mark, and number them from B towards C. 185. The graduated arc and its centre, described in Art. 183, is called a protractor, and may be found for sale, en- graved on wood, ivory, or brass. It is used for measuring angles, and also for drawing angles of a given size. There are two ways in which it can be finished and used. The first way of measuring an angle, is to draw an arc between its sides, prolonged if necessary, with the same radius as that of your protractor, its centre being exactly at the vertex. Set the compasses so as to reach exactly across this arc from side to side of the angle ; then, placing one foot of the compasses at the point B of the protractor, the other will mark out on the graduated arc the size of the angle. The reverse process of drawing a given angle consists in drawing an arc of the same radius as that of the protractor, and then with the compasses taking the chord of the given number of degrees from the protractor and setting it upon the arc ; lines drawn from these two points of the arc to the centre from which it was drawn, will make the required angle. The second method of using the protractor is to cut off all the card below the line A B and all outside the gradu- ated arc B C. Placing then the point A over the vertex of the angle, and making A B coincide with one side of the angle, the other side prolonged if necessary, -will P^^s out under the graduated edge of the card, and the degree of the angle can be at once read. Or, if you wish to draw an STRAIGHT LINES AND ANGLES. 65 angle of a given size, having placed the edge A B as just directed, make a dot on the paper, at the right degree on the graduated edge, and then, removing the protractor, join the dot by a straight line with the vertex that was under A. Protractors may be purchased having graduated arcs of 180°, or of 360°. In the latter case, the central part of the plate is removed, and a piece of transparent mica in- serted, with a fine dot upon it to mark the exact centre. 186. To draw an angle equal to a given angle. — If the given angle is given in degrees, the required angle mjiy be drawn by Art. 185. But if the given angle is one sim- ply drawn on paper, as A B C, then from the vertex of the given angle as a centre, with any radius, draw an arc be- tween the sides of the angle ; and with the vertex of the required angle as a centre, with the same radius, describe an arc of equal length. (Art. 182.) Lines drawn through the extremities of this arc to the vertex will make the required angle. Thus, if it be required to draw a line from the point A, making the angle E with the line A B, draw with any radius the arc F G, and with the same radius the arc C D, making C D equal to F G. A line drawn through the points A and D will ^ ^ -^ make the required angle. 187. To draw tkrougJi a given pointy as C, a line paral- lel to a given line^ as A B. — Join C to any point in the given line by a straight line, as C A. Make the angle D C A equal to CAB, and the line DC will be manifestly par- allel to A B. To draw through C a line making any angle 6* D C E .-^ y^" \ 66 STRAIGHT LINES AND ANGLES. with A B, we need only draw, from any point in A B, a line making the required angle with A B, and then draw through C a line parallel with the line so drawn. 188. A simpler mode of doing the same thing, though not allowed by the postulates of Chaj). II., is to open the compasses until, with one foot on the point C, the other will describe an arc touching the line A B, but not cutting it. With the same radius and one foot at B, describe an- other arc at E. Draw a line through C, touching, but not cutting, the arc E, and it will be parallel to A B. The proof may be readily discovered by the learner. 189. The instrument called a parallel ruler is simply two rulers with par- i ; 1 . allel edges, joined ^A^V — ' ^^*Nr by two strips of . XL _^ \5~ brass, riveted to the " '■ rulers, but the rivets allowing motion in the plane of the paper on which it is laid. Great care must be taken to have the rivet holes in the two pieces of brass at equal distances, and also those in the rulers at equal distances. If this is done, then, while one ruler is held still and the other moved, the moving ruler must remain parallel to its first position. Additional care is usually taken, in making the instni- ment, that this position shall be parallel to that of the sta- tionary ruler, by having the holes in each ruler on a line parallel with its edges. Another kind of parallel ruler is made by simply mounting a ruler on rollers. This is less accurate. The readiest and most accurate mode of drawing paral- lel lines is, however, to use a flat triangle of wood, one side of wliich is slid against the edge of a straight ruler, held firmly stationary, while the other sides remain paral- lel to their first position. 190. The drawing of a parallel line is simply the draw- STRAIGHT LIXES AND ANGLES. 67 ing of an angle equal to zero. Another angle of pecnliar interest is the right angle, and there are better ways than that of Art. 185 for drawing a right angle. 191. 2'o raise a perpefidicular at a given point A upon a straight line A 13. — I^irst Method, From any point B, in the line A B, with Dn any radius B C, describe an arc, and from the point A, with the same radius, describe an arc cutting the first at C. Draw the line B C, prolonging C D to equal C B. Join D A by a straight line, and it will be perpendicular to A B, by Ait. 147. Second Method, Opening the compasses to any con- venient width, step off ^nq equal portions of A B, begin- ning at A. Let B be the fourth point of division. From B as a centre, with a radius equal to five of these parts, (^•aw an arc above A, and from A as a centre, with a ra- dius equal to three parts, draw a second arc intersecting the first at D. Join D to A by a straight line, and it will be at right angles to A B, by Art. 106. Third Method, Measure with the compasses equal dis- tances on each side of A, and bisect the line thus meas- ured off by the method of Art. 180, and you have a line passing through A at right angles to A B. Fourth Method, Visiting or business cards are usually cut very exactly at right angles. By applying one corner of a card at A, and making one edge coincide with A B, the other edge will be at right angles to A B. The accu- racy of this right angle may be tested by drawing perpen- diculars on opposite sides of A. Fifth Method, If you have neither card nor compasses, fold a piece of writing paper carefully, and then double the folded edge carefully on itself. This corner of four thick- 68 STRA.IGHT LINES AND ANGLES. nesses of paper will be a square corner, to be used as the card. Sixth Method. From any point outside the line, as C, with a radius equal to C A, draw an arc cutting A B, say- in B, and prolonged to a point where the radius B C, pro- longed through C, may cut it, say in D. DA will then, by Art. 147, be the perpendicular required. 192. To let fall a perpendicular from a pointy as D, upon a straight line, as A B. D^ — I^irst Method. Join any point of the line A B, as, for in- stance, the point B, to the point D, by a straight line D B. From C, the middle of D B, with a radius equal to CB or CD, draw an arc cutting A B at E, and D E will be (by Art. 147) the peipendicular required. /Second Method. From D as a p centre, describe any arc cutting the line A B in two places. The middle point between these places will be E, the foot of the perpen- dicular from D. Third Method. Make one side of the square card coin- cide with A B, and slip the card along until the end passes through D. Or if the triangle of Art. 189 be made with an- gles of 30\ 60°, and 90°, its square corner may be used. Fourth Method. If the perpendicular is to be drawn merely for the sake of measuring its length, that is, for finding the distance of D from the line A B, it need not be drawn ; but you may simply place one foot of the com- I^asses in D, and then open them wide enough to desciibe an arc touching, but not cutting, A B. Although the pos- sibility of doing this is not claimed in the postulates, yet it is practically equivalent to the postulate of Art. 178. TKIANGLES. 69 CHAPTER ly. TRIANGLES. 195. To draio a triaiigle of three given sides, — If the eides are not drawn, but are given in numbers, open the compasses to extend upon the scale to a number corre- sponding to one of the sides. If the lines are drawn, open the compasses to the length of one of them. Set the dividers on paper with sufficient pressure to mark the points where the feet touch. From these points as cen- tres, with radii equal to the other sides of the triangle, draw arcs cutting each other. This point of intersection and the points used as centres will be the vertices of the required triangle, and must be joined by straight lines. 194. To draw a triangle xolien tioo sides and one angle are given. — First Case. When the angle is included heticeen the given sides. Draw two lines making the re- quired angle ; and upon each line set off with the com- passes, from the vertex, the length of the given sides ; join their extremities by a straight line, and you evidently have the required triangle. Second Case. When the angle is opposite one of the given sides. Draw two lines, A B and A C, making the given angle. Set off from the vertex A the given adjacent side A B, and from B as a centre, with the other given side B C as a radius, draw an arc cutting A C in C or c. Ei- ther A B C or A B c will be the required triangle. If B C is greater than A B, only on crosses A D will be one corner of the quadrangle. From this point D draw D C parallel to d c, and the quadrangle is manifestly completed. 213. My garden is an iiTegular quadrangle, the sides being 150, 207, 315, and 97 feet. The sides are placed in that order, and the angle between the first and second is 96°. Draw me a plot. The south front of a lot is 37 feet, the east side 63, the west 52 feet, and the south-west corner is 92'', the south- east 62°. Draw a map. The south front of a lot of land being 31 feet, the east side 53, the west 46 feet, the south-west corner is a right angle, the north-east corner measures 78°. Draw a map. The sides of another lot, and its south front, measure \\\q same as in the last example, but the two corners in the rear are square corners. Draw a map. CHAPTER VI, CIRCLES. 214. To dram an arc of a circle^ the radius being given, — As a practical question this is one of great importance, as it concerns not only the process of geometrical cofl^- CIECLES. 75 struction, but many processes of mechanical construction also. In geometrical construction the compasses usually afford the readiest means of draAving arcs and circumfer- ences ; but in mechanical construction of machines, roads, and other things, the compasses are frequently of no value. Sometimes a thread, string, or rope is fastened by one end to the spot selected for the centre of the arc, while the other end is carried round. It is plain thataf the line is kept stretched, and equally stretched, its moving end remains always at the same distance from the station- ary end, and the curve must be the circumference of a circle. Sometimes the thing on which you wish to describe a circle is turned round, as in a lathe. Sometimes, at the blacksmith's shop, a circle is made by bending a strip of iron equally at every part. This is done by passing it between three rollers. The amount of bend- ing at each point must be inconceivably small, because any perceptible bend at any one point would make an angle there. Yet small as the bending at each point must be, it must be practically measured by a screw that raiseg or de- presses the middle roller, and thus alters the curvature of the tire as it passes between them. In laying out railroads, arcs of circles are drawn by measuring off equal angles from one point, as A, and setting off equal chords be- tween the sides of the angles, beginning at the point A. That these will be chords to a circle may be proved from Art. 145. Engineers have tables prepared, telling them what the angle must be in order to have chords of 100 76 CIRCLES. feet each from circles of 300 feet, 400 feet, or any other radius. In laying out garden paths and walks, it is convenient to have a wooden " square," such as A B D. A B should be straight, and have a -q mark in the middle ^ c E at C. BD should ' be divided into small divisions. Putting down two stakes, one at A afid one at C, a third one may be placed on D, at such a distance from B as is desirable. Taking up the square, now place the end A at the stake which was at C, and place the point C at the stake which was at D. Put down a fourth stake on the side D as far from B as the third stake was placed. The size of the circle will depend upon the distance from B at which the stake on the side D is placed. For a small circle, divide C B in the middle at E, and use the instrument as though it had been cut off at C. By using the whole for a small circle, D is carried too far from B, and thus C D becomes longer than A C, which will make the first few stakes irregular. 215. I^irst Solution of Art 214. Open the compasses to the given radius, and draw the arc as usual. 216. Second Solution, If the radius is too long for the compasses to be opened to that width, we may use a string, a strip of wood or of paper. If a pin be thrust through one end of a strip of stiff paper, and a pencil point be in- serted through a small hole at the required distance on the strip, very accurate arcs of circles may be drawn, and the radius measured beforehand with accuracy. The pin is held at the centre of the circle, and the pencil carried round. 217. Third Solution. When there is no convenient place to set the central foot of the compasses, or the pin in the paper strip, upon, other plans may be adopted. Draw two lines on a piece of stiff paper at an angle of CIRCLES. 77 165° 31'. Trim the paper off to these lines, leaving a small piece about the intersec- tion. At the point of intersec- tion make a small hole in the paper to insert a lead pencil. Placing two pins P P at a distance apart equal to half the radius, a neat arc may thus be drawn. Other angles may be used with corresponding parts of the radius. For in- stance, 170° 24', with pins at a distance apart equal to one third the radius. Of course, in drawing these angles it is easier to measure the supplement of the angle, that is, the remainder after subtracting it from 180°, and then simply , let the lines cross. 218. Fourth Solution, If we draw a tangent to a cir- cle, and measure off upon it, from the point of contact, dis- tances equal to one tenth the radius, we shall find the dis- tance of the tangent to the circumference at the first three points of division equal to .005, .020, .046 of the radius. Hence we may draw an arc of a given radius by drawing a straight line, and marking upon it half a dozen spots equally distant, at a distance equal to one tenth the radius. If over any one of these we make a dot at the distance of five thousandths of the radius, over the next a dot at the distance of twenty thousandths, &c., a curve drawn care- fully through these dots will be an arc as required. If we take twentieths of the radius instead of tenths, the distance of the curve will be .001, .005, .011, .020, .031, .046. 219. To draw through a given point a tangent to a given circle, — First Case, If the point is inside the cir- cle, the problem is insoluble. Second Case, If the point is in the circumference, draw aradiustothe circumference at that point. Draw a line at right angles to the end of the radius, and it will be a tangent. 78 CIRCLES. Third Case. When the point is outside the circle, let A be the point and C the cen- tre of the circle. Join the point A to the centre C by a straight line. From the middle point B, of that line, with a radius B C, equal to half the line, draw an arc which will cut the circumference at the points through which the tan- gents from A must pass. For it is manifest that if the angle ADC were drawn, it would be a right angle, by Art. 147. Practically^ it is only necessary to lay the ruler with its edge upon A, and touching the circumference without cutting. This is not, however, practically useful when A is very nearly touching the circumference, and not at all practical when A is actually on the circumference, as in that case the angle of the line with the radius to the point of contact might fail to be a right angle. 220. Draw a circle of two inches radius. A semicircle of three inches diameter. Draw, by Art. 217, an arc three inches long, with a radius of sixteen inches. Draw, by Art. 218, an arc five inches long, with a radius of ten inches, also of twenty inches. What angle does a tangent through A make with a straight line drawn from A to the centre of the circle, the radius being 1.5 inches, and the distance from A to the centre 3 inches ? Over how many degrees of latitude could you then look from a balloon at a height of four thousand miles above the sea, supposing it possible to rise that height? Over how many degrees of latitude could you look from a balloon at the height of one thousand miles ? 221. To inscribe a circle in a triangle, — Bisect two of the angles of the triangle; that is, divide each angle into two equal parts by a straight line. The point where these I CIRCLES. 79 two lines intersect each other will be the centre of the required circle, and the radius will be the length of a perpendicular to either side of the triangle. That these perpendiculars will be equal in length, may readily be shown from the equality of the triangles which they form. Thus the triangles A F D and A E D are equal, because they have the same line A D for hy- pothenuse, and equal angles, by the construction of the figure. Therefore D F rz: D E. In like manner, D G may may be shown to be equal to D F. 222. To circumscribe a circle about a triangle. In other words. To draio a circumference that will pass through three given points^ as the vertices of a triangle. — As the centre of the circle must be equally distant from each of the points, it must be found on a line perpendicu- lar to the middle of a line joining any two points. (Art. 136.) In other words, if we draw lines perpendicular to two sides of the triangle, at the middle of those sides, the centre will be in both these lines ; that is, will be found at their intersection. 223. To find the centre of a given arc, — It is manifest from Art. 222, that we only need draw any two chords in the arc, and erect perpendiculars at the middle of each chord. 224. To find the radius of a circumference that will pass through three given points when they lie nearly in a straight line. In other words, WTien the radius is large^ to draio the arc without finding the centre, — The arc may be drawn by setting up pins at the extreme points, and cutting a piece of paper with straight edges at the angle that is made by lines to the intermediate point, then proceeding as in Art. 217. The length of the radius may be found by measuring the supplement of this angle. 80 CIRCLES. According as it measures 1% 2°, 3^ 4°, 5°, 6°, 7^ 8% 9°, or 10°, so must you multiply the distance between the ex- treme points by the number 28.65, 14.33, 9.55, 7.17, 5.74, 4.78, 4.12, 3.59, 3.20, or 2.88, to obtain the radius. It will be noticed how nearly any of these numbers can be obtained by dividing 28.65 by the number of degrees in the supple- ment ; this will enable you to obtain it for angles not con- sisting of whole degrees. Thus for the angle 2J degr^s, you will divide 28.65 by five halves. 225. Make three dots upon paper nearly in a straight line, and discover by Art. 224 the radius of a circle that will pass through them. 226. To describe an equilateral triangle in a circle, — Step round the circumference with the radius, and you will require six steps (since by joining two of the points thus marked with each other and with the centre of the circle you form an equilateral triangle, whose angles must each equal \ of 180°, or \ of 360) : by joining the alternate points of division with straight lines you draw the triangle required. ^ 227. To describe a hexagon in a circle, — Step round the circumference with the radius, and join each point, thus marked, with the adjacent points, by straight lines : the hexagon is drawn. 228. To describe a square in a given circle. — Bisect the arcs to which two opposite sides of the hexagon are chords, join the points of bisection with the two vertices of the hexagon that are at 90° from them, and you will mani- festly have drawn a square. 229. To describe a regular pen- tagon or jive-sided figure in a cir- cle, — Draw a diameter H G, and erect a radius C F perpendicular to it. From D, at the bisection of C G, measure D E equal to AREAS. 81 D F. Join E to F by a straight line, and it will be equal in length to one side of a pentagon. The proof of this proposition is somewhat intricate, and would require more use of algebraic language than is con- sistent with the design of this little book. Supposing the radius C F to be 1 ; then C D would equal ^. From this we should find the value of D F by the Pythagorean proposition. E C would then be found by subtracting ^ from D F. Hence, by the Pythagorean proposition, we could find E F. Next, supposing a chord of 72° (the side of a pentagon) drawn in a circle of radius of unity, we could show that its length would be precisely the same as that of E F. But the simplest way to draw a pentagon is to open the compasses to, as nearly as you can estimate it, the fifth of a circumference, and after stepping round once, alter their width, as nearly as you can estimate it, the fifth part of the resulting error. This really conforms to the spirit of the postulate, that one can open the compasses to a given radius ; and the preceding method is given simply to show a way of drawing a pentagon in conformity with the let- ter of the postulate. If the compasses are opened so as to step round five times without any apparent resulting error, their real error is probably but one fifth as great as it usually is in measuring a radius. CHAPTER VII. AREAS. 230. Areas are the numbers which measure surfaces; that is, which express the ratio of the surfaces to a unit surface, or surface adopted as a unit, or standard of ref- erence. 82 AREAS. 231. The usual unit of surface is a square whose side is a linear unit ; for instance, a square inch, square mile, &c. 232. To find the area of a rectangle. — Multiply the length of a side by that of an end, and the product will be the area. (Art. 97.) 233. To find the area of a parallelogram, — Multiply the length of a side by the distance to the opposite side ; the product will be the area. (Art. 110.) 234. To find the area of a triangle, — Multiply the length of either side by the distance from the opposite vertex; the product will be twice the area. (Art. 111.) 235. To find the area of any polygon. — Divide the polygon by diagonals — that is, by lines drawn through vertices not adjacent — into triangles, and measure these triangles. 236. To multiply two numbers by geometrical construc- tion. — This problem is not commonly practically useful, and yet to one who wishes only approximative results, and dislikes numerical computation, it may be made to yield good results, especially if care be taken in using the par- allel ruler. Have prepared, on a piece of hard, smooth paper, two lines at an angle of 30° or 40°, and on one of them a unit, AB, measured from the vertex A, and permanently marked. Lay off either number from A to C, the other from A to D. Join B C, and draw D E parallel to B C. A E is the required product. That is, if A B were an inch, A C 1^ inches, and A D 2 inches, then A E would be found to be 3 inches. Proof. The triangles ABC and A D E given, by Art. 99 the proportion A E is to A D as A C is to A B, that is, as A C is to unity. But this is the definition of a product, that it is a quantity bearing the same ratio to the multipli- cand that the multiplier does to unity. cmcLEs. 83 287. To find the area of a circle, — If the circumfer- ence, instead of being a curve, consisted of many millions of short, straight sides, it is plain that the circle could be divided into many millions of little triangles, by means of many millions of radii. Now, the circumference may be thus conceived of, and the area of these triangles may be found, according to Art. 234, by multiplying the circum- ference (which is the sum of the short sides) by the radius (the distance to the opposite vertex), and dividing by two. The circumference or the radius may be divided by two before they are multiplied. (Art. 157.) 238. To find the circumference of a circle from knowing its radius. — It is manifest from Arts. 237 and 99 that the ratio of the circumference to the radius is the same in all circles, and we have only to find the circumference of the circle whose radius is unity. Suppose, then, that B E is a chord of 60°, bi- sected at D, and that A is the centre of the circle. Knowing D B is equal to i, and A B to 1, we can, by the Pythagorean proposition, calculate the length of A D. Subtracting this from 1 gives us D C. Then, knowing D C and D B, we can, by the same proposition, calculate B C. Bisecting B C at c?, we know A C and C d, and can therefore calculate A d, and thus find d c. Hence we get the chord C c, or chord of ^V ^^ ^ circle. By continuing this process of applying the Pythagorean proposition we can find the chord of the 48th, or 96th, or 192d of a circumference. Multiplying these chords by 48, or 96, or 192, gives us nearly the length of the circumference ; and the greater the number of times that we bisect the arc of 60°, the more nearly will we attain the exact length. The length of the semi-circumference, with a radius of unity, or of a circum- ference with a diameter of unity, is called n. 84 CIRCLES. Vastly more rapid ways of calculating n have been found by the Differential Calculus. Its exact value cannot be obtained, because the diameter and circumference are not in the ratio of any two numbers whatever. For ordi- nary calculations it may be taken as equal to 3.1416, and for the most accurate osculations 3.14159265. 239. As the area of a circle is, by Art. 237, equal to the ra- dius multiplied by the semi-circumference, and as the semi- circumference is, by Art. 238, equal to the radius multiplied by TT, it follows that the area is equal to the radius multiplied by itself and then by n. In other words, the area of a circle is 3.1416 times as large as the square on the radius. 240. As the square on a radius is one fourth the square on the diameter, we may find the area of a circle by mul- tiplying the square of the diameter by one fourth of 3.1416, that is, by .7854 ; or, for very accurate calculations, by .785398. 241. To find the length of an arc of any number of degreeSy and of a given radius, — The semi-circumference with a radius of 1 is 3.14159265, and if we divide this by 180 it gives the length of one degree, .017453. If we multiply this decimal (.017453) by the number of degrees in the given arc, it will give the length of that arc in a circumference whose radius is 1. Multiplying this by the given radius will give the required arc. 242. To find the area of a sector of a circle ; that is^ of a figure included between two radii and an arc, — Multi- ply the arc by the radius, and the product will be twice the area. Or find the area of the circle, divide it by 360, and multiply the quotient by the number of degrees in the arc. 243. To find the area of a segment of a circle ; that is^ pf a figure included between an arc and its chord. — Find the area of the sector, having the same arc, and also of the triangle included between the radii and the chord. If the DOUBLE POSITION. 85 arc is less than 180°, subtract the triangle from the sector; if the arc is more than 180°, add the triangle to the sector. Examples. 244. With a radius of 7.3 inches what is the length of the circumference ? Of an arc of 79° ? of 53° ? of 58^° ? What is the area of the circle? Of a sector of 51°? of 37° ? Of a segment of 63° ? of 79° ? of 176° ? of 183° ? 245. In measuring the altitude of triangles, or the dis- tance of any point from a line, the simplest mode, justified by the spirit, although not the letter, of the usual postu- lates, is to place one foot of the compasses on the point, and open them until the other foot, swinging near the line, will touch it without crossing it. 246. What is the area of a triangle whose sides are 17, 23, 31? What is the area of the four lots in Art. 213? Of the triangles in Arts. 204, 202, 200. CHAPTER VIII. DOUBLE POSITION. 247. In many practical problems there may be no direct method of solution, and nevertheless there may be direct modes of testing the accuracy of a solution. In these cases the arithmetical rule of "Double Position" is a most valuable means of obtaining the number sought. It pro- ceeds upon the simple supposition that the errors of a result are in proportion to the errors of the data from which the result is obtained. 248. To solve a question by double position, you must first discover a mode of testing an answer by subjecting it to calculations which, if the answer is correct, will yield a 86- DOUBLE POSITION-. given number. Make two " positions," or supposed answers, test them, and note the errors of the results. Then the difference of the results is to the difference of the positions as the error of either result is to the error of its position, and the solution of this problem in the " Rule of Three" will enable you to correct your "position." 249. To solve the question of double position by geometri- cal construction. — Draw a straight line A B of any length. Mark upon it, at any convenient place, a point C, to represent the smaller number of your two po- sitions. Measure C D equal to the a ^^r t b difference of your positions, taken on the scale. Above or below C and D, at right angles to A B, at a distance equal to the error of their results, mark the dots d and c measured on the same scale as C D, or on a different scale, and join them by a straight line, cutting A B in E. Measure C E on the same scale as that on which C D was measured, and subtract it from the smaller posi- tion. This will give a new position, more correct than C or D. Try now two new positions, nearly equal to this corrected one, and employ a larger scale in constructing them, and thus proceed until you find a position nearly enough exact for your purpose. If either result is too small, the corresponding point c or d will be below the line A B, and E will fall between C and D, so that C E must be added to the position C. 1^ T> d is less than C c, then E will fall beyond D, and it will be better to measure D E, and add it to the position D. If the line c d is too nearly parallel to A B, then the distance CD must be measured on a smaller scale, ox X> d and C c on a larger scale. In many cases a more exact result can be more rapidly attained by making three positions, plotting three points like d and c, and then drawing an arc instead of a straight DOUBLE POSITION. 8t line through them. The intersection of this arc with the line A B will then show most exactly the true point for a " position " which will stand the test. 250. A few examples will show more clearly the mean- ing of the above directions. (a.) What two iiumbers are they whose sum is 11 and the sum of their squares 76 ? Supposing the least number to be 3, the greater will be 8, and the sum of the squares will be 9 -j- 64 i= 73. It is therefore plain that 3 is too large, or rather that 8 is too small. Supposing, therefore, the least number to be 2.5, the greater number will be 8.5, and the sum of the squares will be 6.25 + 72.25 = 78.5. On a straight line A B, I now make two dots, C and D, at the dis- Cv tance of .5 of an inch apart, because my \^ P positions for the smallest number were \ 2.5 and 3, and 3 — 2.5 = .5. Over C, I ^, b\ c', and of the other by a\ Z>, 5', c' ; and thus the notation shows that three cor- ners in one coincide with three in the other, and that the fourth, a, is analogous in position to the fourth, a'^ in the other. RATIO AND rROPOKTION. 107 CHAPTER I. RATIO AND rilOrOUTION. 327. In algebraic notation letters are used to represent numbers, cither known or unknown, and the results of arithmetical operations on those numbers are represented by signs. 328. The sum of the two numbers a and x is written a -j- i^*5 and is called a plus x, 329. The difference between a and x is written a — a*, and is called a minus x, 330. The product of a by x is written cither a X ^t or a • a*, or simply a x^ and is called simply a^ x, 331. The quotient of a divided by x is written a -^ cc, a or a : x. or — . X 332. A power or root is written either by means of ex- ponents or of the radical sign. Thus, the ccth power of a is 1 rr . cC^ and the x\h root of a is either a^ or V a, 333. A bar over two quantities indicates that they are to be considered together, and a parenthesis is used for the same purpose. Thus, Va + x is the sum of x and of the square root of a ; but ^ a-\-x is the square root of the sum of a and x. 334. The notation thus far explained may be illustrated by an example; such as '^ i{a xr- -\- {a yY — tf -r- tS- Here the number a must be multiplied by. the second power of ic, and the product added to the second power of the product a times y. From this sum we must sub- tract the number U The cube root of this difference must 108 tlATIO AND PROPOETION. be raised to the second power, and then divided by the number % and finally the square root of the quotient must be extracted. 335. The sign z=l signifies that the sums of the quan- tities on either side of it are numerically equal. Thus, V zzmt-^m signifies that the number P is equal to the sum of m and the product n times L 336. The signs > and <^ are signs of inequality. Thus, P > Q and Q <^ P signify that P is greater than Q, and Q is less than P. 337. The first letters of the alphabet usually signify known, and the last letters unknown, quantities. 338. The signs =, >, <^, are the only verbs in alge- braic language, so that each sentence must contain one of them. Such a sentence is called an equation. Equations containing the sign >, or <;, are sometimes called ine- qualities. 339. An equation may be transposed in any form what- ever, if we are but careful to preserve the equality of the two members; that is, to add or subtract from one side of the sign precisely what we add or subtract from the other, &c. Thus, suppose we have the equation v and wish to find the value of x in terms of t and a. We may first extract the square root of each member, which simply gives a- — ta=:x^ — t. We may now add t to each member, producing the equa- tion a^ — j5a + ^ = cc^. From this we may at once infer that x'^=za'^ — ta-^-t ; and extracting the square of each member, obtain 340. The quantity under the radical sign may be put jnto other forms, thus : — RATIO AND PROPORTION. 109 x=: \/ a^-\-t — at. aj=^a^+(l — a)t. x=i yj a^ — {a — 1)^. an: ^ a{a — 0+^- x=^J t — (t — a) a. 341. The equality of two ratios is called a proportion. Thus, the proportion a is to c as A is to C is announced also by saying that the r;jtio of a to c is the same as that of A to C. Writing this as an equality between two quo- tients, we obtain, — a A (^•) c = C Multiplying both members by the quantity C c, we get, — (2.) aC=iAc, Dividing by A C will then give us a c (^•) A = C Adding A a to both members of (2.), we obtain, — (4.) Aa + Caiz:Aa + Ac. Multiplying each member of (2.) by 2, and subtracting the product, member by member, from (4.), gives,— (5.) A a — C a = A a — A c. Equations (4.) and (5.) may be divided into factors, and written as in Art. 340. (6.) «(A+C) = A(« + c). (7.) a(A — C)=:A(a — c). Dividing (6.) by ( A + C ) ( a + c ), and (7.) by (A — C ) (a — c ), will give us, — a A . (^•) "^+^~A + C a A (^•^ "^ir^~"A— c Hence, by equation (3.), we obtain, — 10 a a + c c A" a "A+C" a — c "C c 110 PLANES AND ANGLES. (10.) ("•) A-A-C-C Adding a c to both members of (2.), gives us, --- (12.) ac-|-aC=iac-f-Ac. Which, divided into factors, becomes (13.) a(c + C)z:rc(a + A). And this divided by c(c-f-C) furnishes us with the equation, — a a-X- K A great variety of results may thus be obtained from the primitive equation (1.), all of which may evidently bo of use in treating the subject of similar figures in Geometry. CHAPTER II. PLANES AND ANGLES. 342. If of two straight lines, having a point in common, and at right angles to each other, the first remain station- ary as an axis, while the second revolves, the second will generate a surface called a plane, and the first will be a line at right angles to the plane ; that is, a perpendicular to the plane. 343. It is manifest that all points in the plane will be equidistant from any two points in the perpendicular taken equidistant from its foot. A plane may be defined from the converse ; that is, a surface, every point of which is equidistant from two given points, is a plane. 344. Any two points in a plane being joined by a PLANES AND ANGLES. Ill straight line, every part of that line is in tlie plane. Proof, Let the foot of a perpendicular be 1*, the given points be B and C, and let A and A' be points in the per- pendicular equidistant from P. The triangles ABC and A' B C are equal, because their sides are equal. Let D be any point in the line B C. Then the triangles A' B D and A B D have the sides A B = A' B, and B D common, and the angles at B equal ; whence A D = A' D, and the point D is in the plane. 345. Three points fix the position of a plane. For if a plane, passing through two of the points, be swung round upon the line joining them, as an axis, it can evidently take but one position, including the third point. 346. Two parallel lines are of necessity in one plane. For if through any point in the second line we draw a line parallel to the first line, and in the same plane with it, it must coincide with the second line ; therefore, the sec- ond line is in that plane. 347. If a straight line move in such manner that any two points in it move in parallel straight lines, it generates a plane. 348. Two lines, having a point in common, lie in one plane. For a straight line from any point in the second line drawn through its intersection with the first line, and in the same plane with the first line, must coincide with the second line. 349. A perpendicular to two lines at their point of inter- section is perpendicular to their plane ; that is, by Art. 1, is perpendicular to every line in the plane drawn through its foot. Proof. Using the notation of Art. 344, let B and C be chosen in the given lines, and we have only to show that the angle D P A is a riglit angle. But this follows at once from the fact that in the triangle A A' D we have A D 1= A^ D, and the base A A' is bisected at P. 350. When a line is perpendicular to a plane, the plane is also said to be perpendicular to the line. 112 PLANES AND ANGLES. 851. It is manifest from the Pythagorean proposition, that a perpendicular measures the shortest distance from a point to a plane, and that of two lines from a point to a plane, that more nearly perpendicular is shorter. 352. The intersection of two planes is a straight line. For, by Art. 344, the straight line joining two points of the intersection, lies wholly in both planes. 353. A second plane including a perpendicular to the first is said to be perpendicular to the first. 354. If a straight line be drawn in the first plane, from the foot of the perpendicular, at right angles to the inter- section, it will be at right angles to two lines in the second plane, and be a perpendicular to it. Hence, each plane is perpendicular to the other, and they are said to be at right angles to each other. 355. The angle made by two planes may be called a diedral angle. A diedral angle is measured by the angle made by two lines, one in each plane, each perpendicular to the intersection of the planes. For it is manifest, that if the planes be brought to coincidence, these lines coin- cide, and that if the planes be then swung open the angle of these lines is generated with exactly the velocity of the motion of the planes. 356. Parallel lines, making the angle zero, may be con- ceived as meeting at an infinite distance in either direction. In like manner, when a diedral angle is zero, the planes do not of necessity coincide, but are parallel, having their intersection at an infinite distance. 357. As the intersection of parallel planes may be at an infinite distance in any direction, any two parallel lines, of which one is in either plane, may be considered as measuring their angle. 858. A line parallel to a line in a plane is said to be parallel to the plane. 859. A straight line, neither perpendicular to a plane PLANES AND ANGLES. 113 nor parallel to it, makes an angle Avitli it, said lo bo equal to the angle which the line makes with the intersec- tion of the plane by a perpendicular plane, including the given line. 360. When two parallel planes are cut by a third, the intersections are two parallel lines. They are straight lines by Art. 352, and although in the. same plane, cannot ap- proach each other in either direction, because the inter- section of the parallel planes is at an infinite distance, in any direction. 361. A straight line makes the same angle with either of two parallel planes, whether the angle be zero, a right angle, or of intermediate value. 362. Parallel lines intercepted between parallel planes are equal. For, joining the points of interception by straight lines in the planes, gives, by Art. 360, a parallelo- gram, whose sides are of course equal. Hence it is evident that parallel planes are every where equidistant. 363. When two lines neither intersect nor are parallel, it may be made evident by Art. 351 that the shortest dis- tance between them is the distance of a point in one from a plane parallel to it drawn through the other. Hence, when two lines neither intersect nor are parallel, it is evi- dent that the right line joining their points of nearest approach is- perpendicular to each. 364. If two right lines are intercepted between parallel planes, a third parallel plane will divide the intercepts in the same proportion. Proof, Let A, C, and A', C^, be tho points of intersection of the right lines with the first planes, and B, B', the points of intersection with the third plane. Draw A B'' C^' parallel to A' C, and Ave have, by Art. 362, A B" = A' B' and B'' 0" = B' C^ Completing the simi- lar triangles ABB" and A C O' gives us A B : B C = AB'' ;B' C := A'B' :B'C'. 10* 114 PLANES AND ANGLES. 365. The intersection of three planes produces triedral angles, the point common to the three planes being called their vertex. In English, the vertex of a diedral angle is usually called an edge, that of a triedral angle a solid corner. 366. If a third plane is j^erpendicular to each of two planes, it is 2:)erpendicular to their intersection. Proof, From the triedral vertex raise a perpendicular to the third plane, and as it must be in both the other planes, it will coincide with their line of intersection. 367. The vertex of a triedral angle is the vertex of three plane angles, constituting the faces {edrai) of the triedral angle. 368. The sum of any two of the angles of the faces is greater than the third. For, if they were simply equal, the two faces would be brought into the same plane with tlie third, and thus reduce the solid corner to one plane ; and if the sum of two were less than the third, the solid corner would be impossible. 369. If two triedral angles have the same angles on the fices, the diedral angles between equal plane angles are equal. Proof. Let A be the vertex in one, A^ in the other, and the plane angles B A C, B A D, D A C, be re- spectively equal to B' A' C, B' A' D', and D' A' C. Make A' B' ~ A B, and the angles A B C, A B D, A' B^ C^ A' B' D' all right angles. We have now to prove that the angle D B C equals the angle D' B' C, — which is done if we prove that the triangle D B C equals the triangle J}' B' O. But the triangles ABC and A' B' C" have the side A B and its adjacent angles equal, by construction, to the side A^ B' and its adjacent angles. Hence, B C = B' a, and A C = A^ C In like manner B D = B'D', and AD — A'D'. Then, in the triangles A C D and A' C D^, we have two sides in one, with their included angle, equal to two sides of the other with their included POLYEBRONS. Il5 angle. Ilcnco, the third sides arc equal, that is, D C = D' C. The three sides of the triangle D B C being thus proved equal to the three sides of D' B' C\ the angles are equal, and the diedral angle on the line A B is equal to that on A' B'. By proj^er changes in the figure the same may be proved of the other diedral angles. 370. The two triedral angles mayjn this case be either equal or symmetrically equivalent. Place one plane, say that of A B C, and A' B' C horizontal with the vertices from you and th6 lines A B, and A' B' on your left. If, now, the* lines A D and A' D' are both above, or both below, the horizontal plane, the triedral angles arc equal ; but if one is above and one below, they are merely equivalent, 371. A solid corner made of several planes may bo called a polyedral angle. If the polygon produced by a new ])lane, cutting off a solid piece from this corner, has no reentering angles, the corner is called a convex polye- dral angl^, 372. If the sum of the plane angles about a convex polyedral angle is zefo, the polyedral angle becomes a nee(llo-l>oint, a Kne; and if the sum of the angles is 2 ti, that is four right angles, the polyedral angle becomes a plane. The sum is always, therefore, less than 2 n. CHAPTER III. POLYEDRONS. • 873. The least number of planes that can enclose a space is four. The solid thus enclosed has four triangular faces, and is called a tetraedron. 374. If two tetraedrons have each a solid angle enclosed in three triangles, equal and similarly arranged in one and 116 POLYEDRONS. in the other, the tetraedrons are equal. For, if one solid angle be imagined laid in the other, so as to have one of the three triangles in one coincide with the corresponding triangle in the other, the other two will coincide, by Art. 369 and by hypothesis ; and the boundaries of the fourth triangle in each thus coinciding, the fourth triangles them- selves will coincide. The entire surface of one solid thus coinciding with that of -the other, the two solids are equal. 375. If two triangles in one tetraedron are equal to two in another, and similarly disposed, and enclose the same diedral angle, then the two tetraedrons are equal. For it is manifest that the two triangles of the one may be ima- gined superimposed upon the two triangles of the other, and will coincide. Two sides of each of the unknown triangles in one tetraedron will then coincide with two sides in the unknown triangles of the other, and thus the whole surfaces will coincide. 376. Polyedrons, like polygons, are called similar when their homologous angles are equal and their homologous sides are proportional. It follows, by induction from the preceding sections, that polyedrons are similar when their homologous faces are similar polygons, similarly arranged. 377. Two tetraedrons are similar if a triedral angle in one and its homologous angle in the other are composed of similar triangles, similarly arranged. For, if these two angles are superimposed, they will coincide, by Arts. 369 and 374, and the fourth planes will be parallel to each other. Hence follows, by Arts. 360 and 374, the similarity of the fourth triangles, and the equality of ratios in the homologous sides. 378. It will also be easy to show that, if two triangles in one tetraedron are similar to two in another, and simi- larly arranged, and enclose an equal diedral angle, the two tetraedrons are similar. 379. If all the planes of a polyedron except one have a POLYEDRONS. 117 common point of intersection, tlie polycdron is called a pyramid ; the common point of intersection is called the vertex of the pyramid ; the face, which does not reach the vertex, is called the base. 380. A pyramid is called triangular, quadrangular, &c., from the shape of its base. The other faces are, of course, always triangles. 381. A tetraedron is, therefore, a triangular pyramid, any face of which may be taken as its base. 382. If two faces of a polyedron are equal, and their homologous sides are parallel, and if each of the other faces is a plane joining a pair of these parallel sides, the polyedron is called a prism. The parallel faces are called the bases of the prism. The other faces are evidently parallelograms. A section parallel to the base of a prism is readily shown to be a polygon equal to the base. 383. When the bases of a prism are parallelograms, the prism is called a parallelopipedon. 384. A right parallelopipedon is a prism of which every face is a rectangle. When each face is a square, the prism is called a cube. 385. When a pyramid is intersected by a plane parallel to the base, the part intercepted between the bases is called the frustum of the pyramid. The part above the cutting plane is easily shown to be a pyramid, Avith all its angles equal to those of the given pyramid, and therefore similar to it. 386. Let a be the length of one side of the base of a pyra- mid, a' that of the homologous line on the upper end of the frustum; /ithe height of the pyramid, h' that of the similar pyramid cut off; b and h' the slant heights of the pyramids on the edge at the left end of a and a'. By similar triangles we have a: a' =Lh',h', Also h : h' = h : h'. Whence h: h' =:a: a^. Whence, by the theory of proportions, a — a' laz^h — h! \h. Thus the total 118 POLYEDKONS. dimensions of the pyramid are obtained from that of the frustum, since h — W \'$> shnply the vertical height of the frustum ; and h = r a — a Example. What is the height of a pyramid "whose base has sides of 3, 6, 4^, and 6 inches, and at the perpendicular heiglit of tAvo inches the sides of the frustum are 4, 8, 6, and 8 inches ? What is the slant height on the corner, on which the shxnt height of the frustum is 3 inches? 387. Any polyedron can be divided into pyramids by simply selecting a point within the polyedron for a com- mon vertex, and taking the faces of the polyedron as bases for the pyramids. By taking the common vertex for the pyramids in the surface of the polyedron, the number of the pyramids may be reduced. Thus a right parallelopipedon may be divided into six pyramids ; but by bringing the common vertex up to one of the faces, the pyramid of which that face was base becomes zero, and the pyramids are reduced to five ; on moving the vertex to one of the edges, a second pyramid becomes zero, reducing the number to four ; and on taking a vertex of the parallelopipedon as the common vertex, the pyramids are reduced to three. 388. Any polyedron can be divided into triangular pyramids by simply dividing each base in Art. 387 into triangles. 389. Two bodies which are composed of equal and simi- larly arranged triangular pyramids are evidently equal. ABEAS. 119 CHAPTER IV. AREAS. 390. When two polygons have all the angles of one equal to those of the other, and similarly arranged, and their homologous sides proportional, — i. e., each pair hav- ing the same ratio to each other, the polygons are called similar. 391, Similar polygons may evidently be divided into snnilar triangles by diagonals from homologous vertices. 392 Lines drawn in a similar manner, in two similar polygons, may evidently be made the sides of similar trian- gles, and shown to have the same ratio as homologous sides of the polygons. 393. Hence the altitudes of similar triangles have the same ratio as their bases. 394. Let h be the base and h the altitude of a triangle, and X be their ratio to the base and altitude of a similar triangle. The base of the second triangle will then be h X and its altitude h x. The area of the first will be J- h h^ and of the second ^ hb x^. The ratio of these areas will therefore be x"^. Calling now the bases and altitudes h and B, h and H, and the areas s and S, we have B:h = ll:h = x\ ^ : s zizx^ =:W-.h'' = ir : h\ 395. It may easily be shown, by help of equation (10.) (Art. 341), in the theory of proportions, that the areas of -similar polygons, and of any homologous areas in or about similar polyedrons, are in the same ratio ; in other words, that in similar figures homologous lines have all the same 120 AREAS. ratio, and that this ratio multiplied by itself will give the ratio of homologous surfaces ; or, in other words, that in similar figures homologous surfaces are in the ratio of squares on homologous lines, or of circles on homologous lines as diameters. 396. The similar polyedrons spoken of in Arts. 395, 385, 376, may be defined as polyedrons capable of being divided into similar tetraedrons similarly arranged. 397. By the reasonmg alluded to in Art. 395, it may be shown that the external surfaces of similar polyedrons are in the ratio of squares built upon their homologous edges, or upon any homologous lines ; also that any pair of ho- mologous faces are in the same ratio. 398. If any two pyramids be cut by a plane passing at equal distances from their summits, the areas of the sec- tions have a fixed ratio, whatever be that distance. Proof, Let the plane pass first at the distance A, and secondly at the distance h' from the summits, and the areas of the sections be in the first case A and B, in the second a and h. We have, by Art. 397, — ' K h? B A2 A B ' . Hence? — =t") and by (3.), in the theory of proportions, we have :^=-i7> which is what we wish to prove. B h VOLUMES. 121 CHAPTER V. VOLUMES. 399. The volume of a solid is the ratio which it bears to a solid uuit. The solid unit generally emjDloyed is a cube, whose faces are units of area, and wliose edges are units of length. ^ 400. It will readily be perceived that the volume of a right parallelopiped is the product of the lengths of its three edges. If each edge is commensurable with the unit of length, the right parallelopiped may be divided into cubes by three series of planes parallel to its faces, evidently equal in number to the product of the numbers into which each edge is divided. If the edges are incommensurable, we can choose a unit as small as we please, and so multi- ply our planes that the parallelopipedon shall need but an infinitesimal change to render it capable of being divided into cubes. If the right parallelopided cannot be divided into cubic inches, it may be into cubic tenths, or hun- dredths, or thousandths, &c., of an inch. 401. Any parallelopipedon is equivalent in volume to any other parallelopipedon of equivalent base and equal al- titude. Proof, Set the two parallelopipeds upon one plane, and move a second plane, parallel to the bases, from above steadily down until the two planes coincide. This mov- ing plane moves with equal velocity through each paral- lelopiped, and the sections of the two are constantly equiv- alent from first to last. The two solids therefore pass equal volumes through the moving plane in equal times, and the volumes of the two are equal when the two planes coincide. 402. Corollary. Any parallelopipedon is equivalent to a right parallelopiped on of equivalent base and altitude. 11 122 VOLUMES. 403. Let the six planes of a parallelopiped be A and A', B and B\ C and C ; similar letters denoting parallel faces. Cut the solid by a plane, A'^, perpendicular to B and to C. Transpose the parts, so that A shall coincide with A', making the surfaces on A'' new bases for the solid. Cut this new solid by a plane, C, perpendicular to A'^ and to B, and transpose so that C shall coincide with C. The solid is thus, without loss or addition, converted into a right parallelo- piped ; two bases, B and B', remaining, in altered form, of the same size as before, and their distance apart, as measured on the intersection of A'' with C, being un- changed. From Art. 402 thus proved. Art. 401 may be drawn as a corollary. But if the plane A" cuts A or A', this proof needs modification, by resections. 404. The diagonal of B being parallel to that of B\ a plane may be passed through these two diagonals, dividing every section made by a plane parallel to B into two equal triangles, and the parallelopiped into two triangular prisms, equivalent to each other. Conversely a triangular prism may be considered as one half a parallelopiped. 405. Two triangular prisms of equivalent base and equal altitude are, by Arts. 401 and 404, equivalent in volume. 406. The volume of a parallelopiped, or of a prism, is, by Arts. 400, 402, 404, 405, found by multiplying the area of its base by its altitude. 407. Pyi-amids of equivalent base and equal altitude are equal. Proof, Let the pyramids be set upon a plane, and a second plane, parallel to the first, move steadily from the vertex of the pyramids to their base. As the sections of the pyramids made by the second plane are, by Art. 398, at each instant equivalent, the pyramids must be passing with equal velocity through the second plane, and the total amounts passed through at any instant are equivalent, and the whole amounts, when the two planes coincide, will be equal. 408. Every triangular prism is divisible into tlirce VOLUMES. 123 equivalent tetraeclrons. Proof. Lot the ends of tlie prism be the equal triangles ah c and a! h' (/, a and a! being at the ends of tlie same edge, etc. Pass two planes, aV d and a h d^ through the prism, and you have manifestly di- vided it into three triangular pyramids. But of these, two have the common vertex a, and the equal bases hU d and hcd^ and are therefore equal ; while the third has a com- mon vertex h'=.^^. But Bc'znB^' + ^/c', and B 5' + C ^>' zz: B C. Hence, ^h> -\-h' d -^-Qh' = 180° = V>Q-^h' d. But y d is the measure of the angle a. Hence, the angle a and the side B C are supplements of each other ; their sum is TT HZ 180°. 456. If the three sides of a triangle are respectively equal to the three, sides of another, the two triangles are said to be equilateral with respect to each other. 457. If two triangles on the same sphere, or on equal spheres, are equilateral with respect to each other, they are also equiangular with each other ; which follows from Arts. 440 and 369. 458. Place the centres of the equal spheres together, and bring one side of one triangle, say A B, into coinci- dence with the equal homologous side A' B' on the other Iriangle. ^ If, now, the equal angles A and A' lie on the same side of AB, the triangles evidently coincide. But if the angles A and A' lie on opposite sides of the com- mon side, these triangles, equilateral and equiangular with respect to each other, are called symmetrical. 459. If two triangles are equiangular with respect to each other, they are also mutually equilateral. Proof, For their polar triangles are, by Art. 455, equilateral, and there- fore, by Art. 457, equiangular with respect to each other. Hence, by Art. 455, the triangles themselves are equilateral with respect to each other. 460. Spherical triangles, having one side and the adja- cent angles, or two sides and the included angle, in one, equal to the like parts in the other, are either equal or symmetrical. For it may readily be shown that one would OP THE SPHlERE. 131 coincide with the other, or with a triangle symmetrical with the other. 461. In measuring spherical surfaces a peculiar unit is sometimes used, namely, the surface of a triangle whose tliree sides are 90°, 90°, and 1°. This surface is called one degree of surface, and is jij^th of the surface of a sphere. 462. The surface included between two great semicir- cles is called a lune. 463. The surface of a lune is double the angle of the lune. lu other words, the surface of the lune is to that of the sphere as the angle of the lune is to 360°, or double the angle is to 720°. 464. Symmetrical spherical tnangles are equivalent in area. Proof, Place the vertex of one angle upon the vertex of the equal angle in the other triangle, giving the sides the same direction. Part of the triangles will coincide. The non-coincident parts will be new triangles, mutually equiangular, and therefore symmetiically equi- lateral. The same operation may be repeated upon them, and upon their * non-coincident parts, until, finally, the non-coincident parts are infinitesimal symmetrical trian- gles, or zeros, mutually equiangular and equilateral, whose difference will differ nothing from zero. 465. To measure the surface of a spherical triangle. — Solution. Prolong one side, say A C, into a great circle. I'rolong A B into a semicircle meeting A C prolonged in A'; and in like manner prolong CB to C Now, the triangle A' B C is symmetrical or equivalent to the triangle required to complete the triangle ABC into a lune with the angle B. And the sui-face of the hemisphere (=360°) is equal to the lune C A B C plus the triangle A' B O, plus the trian- gle A'BC. The triangle A' B C is, however, the lune .\ B C A' minus the triangle ABC. Substituting for each lune the double of its angle gives us, — 132 OF THE SPHERE. 360° = 2 C + 2 B — triangle AB C + 2 A — triangle AB C. 360° z=:2(A + B + C)— 2 triangle ABC. 180° = A + B + C — triangle ABC. Triangle ABCznA + B + C — 180°. That is, the surface of a triangle is to that of its sphere as the excess of the sum of its angles over 180° is to 720°. 466. To measure the surface of afrustutn of a right coue. — Let S be the area of the curved surface of the frustum, S" that of the cone required to fill the deficiency of the frustum, S' that of the cone thus completed ; and let 5, s"^ and s' be the slant heights of the three bodies. Also, let r and r' be the radii of the two bases of the frus- tum. We then have, by Art. 424, and by the evident relations of the bodies, — S' = ^ r s', S" zzinr' s", ^=:^' — ^" = 71 {rs' — r' s"). But, on jDassing a plane through the apex of the cone and the centre of the base, it will be evident, from similarity of triangles, that r\r'=:s'\s"\ whence, r' s' z=r s^' ; and it cannot alter the value of any quantity to add to it r' s' — r s". We may, therefore, write, remembering that s=is' — s", S =171 (r s' — r' s" -\-r' s' — r s'^ ) r= tt ( r -f- ^' ) ^' The quantity 7t(r -^r') is the circumference of a circle, whose radius is one half the sum of the radii of the bases ; and it may readily be shown, by Art. 421, and reasoning similar to that of Art. 364, that this circle is a section of the frustum, parallel to the bases, bisecting the distance between them. Hence, the convex surface of the frustum of a right cone is measured by the product of its slant height into the circumference of a section midway between the bases. 467. Pass a plane through the centres of the bases of the frustum, and calling the length of half the sections of OP TOE SPHERE. 133 the bases r and r', as in the preceding article, and the sec- tion of convex side 5, let fall a perpendicular of the length h from the extremity of r' upon r. From the mid-point of s draw a perpendicular to the axis of the cone, and its length is readily shown to be ^ ( r -f- r' ). From the same point draw a perpendicular to 5, until it touches the axis of the cone, prolonged if need be, and call the length of this line R. We have now two right triangles, of which the sides h and s in one are respectively perpendicular to two sides i (>' + ^') ^^^ K i^^ the other. They therefore enclose equal angles, and the two right triangles are simi- lar, which gives S = 2 TT R A. That is, the convex surface of a frustum of a right cone is measured by the product of its altitude into the circum- ference of a circle whose radius is a perpendicular to the slant surface reaching from its mid-point to the axis of the cone. 468. A portion of a sphere enclosed between two paral- lel planes — that is, a spherical segment with parallel bases — may, if the bases are brought infinitely near together, be considered as an infinitesimal frustum of a right cone, whose axis passes through the centre of the segment and of the sphere, and whose slant surface is perpendicular to the radii of the sphere touching the surface. 469. A hemisphere may thus be cut, by an infinite num- ber of planes parallel to its base, into an infinite number of frustums of different cones. But all these frustums agree in having their axis pass through the centre of the sphere, and in having also perpendiculars to their slant suiface pass through the centre of the sphere. Moreover, the sum of their altitudes is the radius of the sphere. Hence, by Art. 467, the surface of a hemisphere with the radius R is S = 27rR2. 12 134 PEOBLEMS A^B THEOREMS. But as the area of the base is 2 nR x i^ ^= ^ K>\ it follows that the convex surface of a hemisiDhere is double the plane surface of its base. 470. The surface of a sphere is therefore equivalent to that of four great circles. 471. As a sphere may evidently be divided into pyra- mids, with their common apex at the centre, and the sum of their bases constituting the surface of the sphere, we have, for itssoHdity, by Art. 410, — 472. Using D = 2 R as the diameter of the circle and sphere, we have for the area of a circle, — A = 7rR2 = i7rD2=: .7854 X D 2. V = |7rR3=:j7rD3 = .5236 X D^ CHAPTER VIII. PROBLEMS AND THEOREMS. 473. Given the angle which a straight line makes with a plane. Find, by geometrical construction, the altitude from the plane of a point in the line at a given distance from its intersection with the plane. For example, what is the height of one end of a yardstick, the other end rest- ing on the floor at an angle of 37° ? 474. Find, by geometrical construction, the angle which a line makes with a plane, having given its altitude above two given points in the plane, vertically under it ; and find the place of its intersection with the plane. For instance, let two posts on level ground be seven feet apart, and be, one four, the other six feet high. What is the inclination to the horizon of a line joining their summits, and where would it strike the ground ? PROBLEMS AND THEOREMS. 135 475. Given the altitude of a second plane above three given points in a first plane, to find by geometrical con- struction the line of intersection of the planes, and their diedral angle. 476. A right line drawn from one vertex of a parallelo- pipedon to the vertex, which has no plane in common with the first, is called a diagonal of the parallelopip- edon. Prove that a plane containing one diagonal may- be rotated upon it until it contains a second. Prove that all four diagonals have a common point of intersection. Prove that each diagonal is bisected at this point. 477. The convex surface of the frustum of a sphere is called a zone. Prove that the area of a zone is the prod- uct of the altitude of the frustum into the circumference of a great circle. 478. Prove that the two tangents from a given point to a circle are equal. Prove that a right line from the given point to the centre of the circle bisects the angle between the tangents. 479. Prove that when a chord is bisected by a diameter, the semicbord is a mean proportional between the segments of the diameter. Find a mean proportional between two given lines. 480. Prove — what is assumed in Art. 468 — that the line of tangency of surfaces, when a sphere is enclosed in a hollow cone, is the circumference of a circle whose plane is at right angles to the axis of the cone. 481. Prove that, if two chords are prolonged until they meet, the entire lines thus produced are inversely propor- tioned to the parts without the circle. Calling these entire lines secants, prove that if, from a given point, a tangent and a secant be drawn to a circle, the tangent will be a mean proportional between the secant and the part with- out the circle. 482. Find the centres, by construction, of the two oir^ 136 PROBLEMS AND THEOREMS. cles whose circumferences pass through two given points and are tangent to a given right line in one plane with them. When would this problem become absurd ? 483. Find the centres of the two circles whose circum- ferences pass through one given point and are tangent to two given right lines all in one plane. Remember that a right line passing through the centres is readily found, and that a perpendicular upon it from the given point is a semi-chord in both circles. 484. Find the centres of the four circles which are tan- gent to three given right lines in one plane. In what case would the problem be impossible ? 485. It is evident that circles concentric with those of Art. 483 would pass equally near the given points and the given right lines. Find, then, the centres of the four circles which are tangent to a given circle and to two given right lines. 486. Given the radius of a sphere and distance at which a plane passes from its centre. Find the radius of the sec- tion. When does this radius equal that of the sphere ? when become zero ? and when become impossible ? 487. A given sphere has its centre on the axis of a given cone, at a given distance from the vertex. Find the. radii of the two circles made by the intersection of the surfaces. In what cases will they be equal ? In what case will the two circles coincide? When will they coincide with a great circle ? 488. Given a great circle in a sphere, and the radii of two small circles parallel to it. Find the hollow cones tangent to the sphere on the small circles. Find also the cones whose intersections with the sphere would give the circumferences of both circles. Four cones are required in each case. a \ si h - ^,^ \ /^ N ^ "■\ J2\J §^466 467 THIS BOOK IS DUB ON THB LAST DATB STA3MPED BBLOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. •^CT 20 19?' -• }:| ^ •^^^ i 918241 \ THE UNIVERSITY OF CALIFORNIA LIBRARY^ ^