B B M: A 3 Fig. 321. F B If two right lines LM, FG be drawn, in these two planes, perpendicular to the line ! formed by the two planes. revolve about itself, with- an obtuse angle with AB, will describe in the revolution a convex surface MAD. scribe in the revolution a surface which will be neither concave nor convex, but plane: and the line A B will be perpendi. cular to the plane MLGFPN, because it will make right angles with the lines AM, AL, AG, &c. drawn from the point X in that plane. 6. Two planes are parallel when all perpendiculars drawn from one to the other are equal. See fig. 325., wherein AB, CD are equal between the surfaces LM, FG. Fig. 345. Because the line BA is perpendicular to the plane DF, the angle BAC is a right angle. 971. Pror. XCI. A perpendicular measures the distance of any point from a plane. The distance of one point from another is measured by a right line, because it is the shortest line which can be drawn from one point to another. So the distance from a point to a line is measured by a perpendicular, because this line is the shortest which can be drawn from the point to the line. In like manner, the distance from a point to a plane must be measured by a perpendicular drawn from that point to the plane, because this is the shortest line which can be drawn from the point to the plane. 972. Prop. XCII. The common intersection of two planes is a right line, Let the two planes ALBMA, AFBGA (fig. 327.) intersect each other; the line which is common to both is a right line. Draw a right line from the point A to the point B. Because the right line AB touches the two planes in the points A and B, it will touch them (Defin. 1.) in all other points; this line therefore, is common to the two planes. Wherefore the common intersection of the two planes is a right line. 973. Prop. XCIII. If three points, not in a right line, are com. mon to two planes, these two planes are one and the same plane. Let two planes be supposed to be placed upon one another, in such Because the right line AB (fig. 328.) touches the two planes in the points A and B, it Again, because the right line CD touches the two planes in the If a right line be perpendicular to two right lines which cut each other, it will be perpendicular to the plane of these right lines. D A Fig. 326. L F B B A G B L B Let the line AB (fig. 329.) make right angles with the lines AC, AD, it will be perpen. dicular to the plane which passes through these lines. If the line AB were not perpendicular to the FDCG, another plane might be made to pass through the point A, to which the AB would le perpendicular. But this is impossible ; for, since the angles BAC, BAD are right angles, this other plane (Defin. 2.) must pass through the points C, D; it would therefore (Prop. 93.) be the same with the plane FDCG, since these two planes would have three common points ! A, C, D. 975. Prop. XCV. From a given point in a plane to raise a perpen Fig. 329 ricular to that plane. Let it be required to raise a perpendicular from the point A (fy. 380.) in the plane LM. Form a rectangle CDFG, divide it into two rectangles, having a common section AB, and place these rectangles upon the plane LM in D such a manner that the bases of the two rect. angles AC, AG shall be in the plane LM, and form any angle with each other; the line AB shall be perpendicular to the plane LM. Thc line A B makes right angles with the two lines AC, AG, which, by supposition, are in the plane C LM; it is therefore (Prop. 94.) perpendicular to the plane LM. Fig. 330. 976, Prop. XCVI. If two planes cut each other at right angles, and a right line be drawn in one of the planes perpendicular to their common intersection, it will be perpendicular to the other plane. Let the two planes AFBG, ALBM (fig. 331.), cut each other at right angles; if the line LC be perpendicular to their common intersection, it is also perpendicular to the plane AFBG. Draw CG perpendicular to AB. Because the lines CL, CG are perpendicular to the common in. tersection AB, the angle LCG (Defin. 3.) is the angle of inclination of the two planes. Since the two planes cut each other perpendicularly, the angle of inclination LCG is therefore a right angle. And because the line LC is perpendicular to the two lines CA, CG in the plane ABFG, it is (Prop. 94.) perpendicular to the plane AFBG. 977. Prop. XCVII. If one plane meet another plane, it makes Fig. 331. angles with that other plane, which are together equal to two right angles. Let the plane ALBM (fig. 332.) meet the plane AFBG; these planes will make with each other two angles, which will together be equal to two right angles. Through any point C draw the lines FG, LM perpendicular to the line AB. The line CL makes with the line FG two angles together equal to two right angles. But these two angles are (Defin. 3.) the angles of inclination of the two planes. Therefore the two planes make angles with each other, which are together equal to two right angles. COROLLARY. It may be demonstrated in the same manner that planes which intersect each other have their vertical angles equal, that parallel planes have their alternate angles equal, &c. 978. Prop. XCVIII. If two plancs be parallel to each other, Fig. 332. a right line, which is perpendicular to one of the planes, will be also perpendicular to the other. Let the two planes LM, FG (fig. 333.) be parallel. If the line BA 1: be perpendicular to the plane FG, it will also be perpendicular to the plane LM. From any point C in the plane LM draw CD perpendicular to the plane FG, and draw BC, AD. Because the lines BA, CD are perpendicular to the plane FG, the angles A, D are right angles. Because the planes LM, FG are parallel, the perpendiculars AB, F DC (Defin. 6.) are equal; whence it follows that the lines BC, AD are parallel. Fig. 333. The line BA, being at right angles to the line AD, will also (Prop. 13.) be at right angles to the parallel line BC. The line BA is therefore perpendicular to the line BC. In the same manner it may be demonstrated that the line BA is at right angles to all other lir:es which can be drawn from the point B in the plane LM. Wherefore Defin. 2.) the line BA is perpendicular to the plane LM. M L F B M AL SOLIDS. a Fig. 331. Fig. 335. Fig. 337. Fig. 338. Fig. 339. 979. DEFINITIONS. - 1. A solid, as we have before observed, is that which has length, breadth, and thickness. 2. A polyhedron is a solid terminated by plane surfaces. 3. A prism is a solid terminated by two identical plane bases parallel to each other, and by surfaces which are parallelu. grains. (Fig. 334.) 4. A parallelopiped is a prism the bases of which are parallelo grams. (Fig. 335.) 5. A cube is a solid terminated by six square surfaces : a die, for example, is a cube. (Fig. 336.) 6. If right lines be raised from every point in the perimeter of any rectilineal figure, and meet in one common point, these lines together with the rectilineal figure inclose a solid which is called a pyramid. (Fig. 337.) planes, which are equal and parallel cir- Fig. 336. in the circumference of a circle, and meet in one common point, these lines together with the circle inclose a solid, which is called a cone. (Fig. 339.). 9. A semicircle revolving about its diame ter forms a solid, which is called a sphere. (Fig. 340.) dicular be let fall upon the opposite Fig. 310. Fig. 311. 11. Solids are said to be equal, if they inclose an equal space: thus a cone and a pyramid are equal solids if the space inclosed within the cone be equal to the space inclosed within the pyramid. 12. Similar solids are such as consist of an equal number of physical points disposed in the same manner. Thus (in the fig. Defin. 10.) the larger pyramid ACD and the smaller pyramid Acd are similar solids if every point in the larger pyramid has a point corresponding to it in the smaller pyramid. A hundred musket balls, and the same number of cannon balls, disposed in the same manner, form two similar solids. 980. Prop. XCIX. The solid content of a cube is equal to the product of one of its sides trice multiplied by itself. Let the lines AB, XD (fig. 342.) be equal. Let the line AD, drawn perpendicular to AB, be supposed to move through the whole length of AB; when it arrives at BC, and coincides with it, it will have formed the square DABC, and will have been multiplied by the line AB. Next let the line AF be drawn equal to AD, and perpendicular to the plane DABC; suppose the plane DABC to move perpendicularly through dl the whole length of the line AF; when it arrives at the plane MFGL, and coincides with it, it will have formed the cube AFLC, and will have Fig. 312. been multiplied by the line AF. Hence it appears, that to form the cube AFLC, it is necessary, first, to multiply the side AD by the side AB equal to AD; and then to multiply the product, that is, the square of AC, by the side AF equal to AD; that is, it is necessary to multiply AD by AD, and to multiply the product again by AD 981. Prop. C. Similar solids have their homologous lines proportional. Let the two solids A, a (fig. 343.) be similar ; and let their homologous lines be AB, ab, BG, bg; AB will be to BG at ab to bg. Because the solids A, a are similar, every point in the solid A has a point corresponding to it, and disposed in the same M A A с a 1 A B F 6 side bf. manner, in the solid a. Thus, if the line AB is composed of 20 physical points, and the line BG of 10, the linc ab will be composed of 20 corresponding points, and the line bg of 10. Now it is evident that 20 is to 10 as 20 is to 10: therefore AB is to BG as ab to lg. 982. Prop. CI. Similar solids are equiangular. Let the solids (see fig. to preced. Prop.) A, a be similar ; their corresponding angles are equal, Because the solids A, a are similar, the surfaces BAF, baf are composed of an equial number of points disposed in the same manner. These surfaces are therefore similar figures, and consequently (Prop. 88.) equiangular. The angles B, A, F are therefore equal to the angles b, a, f. In the same manner it may be demonstrated that the other correspondent angles are equal. 983. Pror. CII. Solids which hare their angles equal and their sides proportional arc similar. If the solids A, a (fig. 344.) have their angles equal and their sides proportional, tłey are similar, For if the solids A, a were not similar, another solid might be formed upon the line BF siinilar to the solid a. But this is impossible; for, in order to form this other solid, some angle or some side of the solid A must be increased or diminished ; and then this new solid would not have all its angles equal and all its sides proportional to those of the solid a, that is (Prop. 100, 101.), would not be similar. 984. Prop. CIII. Similar solids are to one another as the cubes Fig. 341. of their homologous sides. Let A, a (see fig. to preced. Prop.) be two similar solids, the solid A contains the solid a as many times as the cube formed upon the side BF contains the cube formed upon the Because the solid A is similar to the solid a, every point in the solid A has its corresponding point in the solid a. From whence it follows, that if the side BF is composed, for example, of 50 points, the side of will also be composed of 50 points: and consequently the cubes formed upon the sides BF, bf will be composed of an equal number of points. Let it then be supposed that the solid d is composed of 4000 points, and the cube of the side BF of 5000 points; the solid A must be composed of 4000 points, and the cube of the side bf of 5000 points. Now it is evident that 4000 is to 5000 as 4000 to 5000. Wherefore the solid A is to the cube of BF as the solid a to the cube of us; and, alternately, the solid A is to the solid a as the cube of BF to the cube of bf. Corollary. It may be demonstrated in the same manner that the spheres A, a (fig. 345.), which are similar solids, are to one another as the cubes of their radii AB, ab. 985. Prop. CIV. The solid content of a perpendicular prism is equal to the product of its base and height. The solid content of the perpendicular prism ABCD ( fig. 346.) is equal to the Fig. 315 product of its base AD, and height AB. If the lower base AD be supposed to move perpendicularly along the height A B till it coincides with the upper base BC, it will have formed the prism ABCD. Now the base AD will have been repeated as many times as there are physical points in the height AB. Therefore the solid content of the prism ABCD is equal to the product of the base multiplied by the height. COROLLARY. In the same manner it may be demonstrated that the solid content of the perpendicular cylinder ABCD is equal to the product of its base AD and height AB. 986. Pror. CV. The solid content of an inclined prism is equal to the product of its base and height. Let the inclined prism be CP (fig. 317.), it is equal to the product of its base RP and its height CD. Conceive the base NB of the perpendicular prism NA, and the base RP of the inclined prism PC, to move on in the same time parallel to themselves; when they have reached the points A and C, each of them will have been taken over again the same number of times. But the base NB will have been taken over again (Prop. 104.) as many times as there are physical points in the height CD. The base RP will therefore have been taken over again as many times as there are physical points in CD. Fig. 347. Consequently the solid content of the inclined prism CP is equal to the product of its base RP and the height CD. с B C B A A-- B a b D D 987. Prop. CVI. In a pyramid, a section parallel to the base is similar to the base. Let the section cd be parallel to the base CD (fig. 348.); this section is a figure similar to the base. Draw AB perpendicular to the base CD; draw also BC, bc, BE, be. Because the planes cd CD are parallel ; AB, being perpendicular to the plane CD, will also (Prop. 98.) be perpendicular to the plane cd: whence the triangles Abc, ABC, having the angles b, B right angles, and the angle A common, are equiangular. Therefore (Prop. 61.) A6 is to AB as bc to BC, and as Ac to AC. In like manner it may be proved that Ab is to AB as be to BE, and as Ae to AE. Consequently if Ab be one third part of AB, C be will be one third part of BC, be the same of BE, Ac of AC, and Ae of AE. Fig. 318. Again, in the two triangles cAe, CAE, there are about the angle A, common to both, two sides proportional; they are therefore. (Prop. 63.) equiangular, and consequently (Prop. 61.) have their other sides proportional. Therefore ce will be proportional to CE. The two triangles cbe, CBE, having their sides proportional, are therefore (Prop. 89.) similar. The same may be demonstrated concerning all the other triangles which form the planes cd, CD. Therefore the section cd is similar to the base CD. REMARK. If the perpendicular AB fall out of the base; by drawing lines from thic points b, B, it may be demonstrated in the same manner that the section is similar to the base. 988. Pror. CVII. In a pyramid, sections parallel to the base are to one another as the squares of their heights. Let CD cd (fig. 349.) be parallel sections. From the vertex A draw a perpendicular AB to the plane CD: the plane cd is to the plane CD as the square of the height Ab is to the square of the height AB. Draw BC, bc. The line AB, being perpendicular to the plane CD, will also (Prop. 3 98.) be perpendicular to the parallel plane cd: whence the angle Abc is a right angle, and also the angle ABC. Moreover, the angle at A is coninon to the two triangles Abc, ABC; these two triangles, therefore, are equiangular. Therefore (Prop. 61.) the side cb is to the side CB as the side Ab is to the side AB; and consequently the square of cb is to the square of CB as the square of Ab to the square of AB. The planes cd, CD, being (Prop. 106.) similar figures, are to one another (Prop. 82.) as the squares of the homologous lines cb, CB; Fig. 319. they are therefore also as the squares of the heights Ab, AB. COROLLARY. In the same manner it may be demonstrated that in a cone the sections parallel to the base are to one another as the squares of the heights or perpendicular distances from the vertex. 989. Pror. CVIII. Pyramids of the same height are to one another as their bascs. Let A, F (fig. 350.) be two pyramids. If the perpendicular AB be equal to the perpendicular FG, the pyramid A is to the pyramid F as the base CD to the base LM. Supposing, for example, the base CD to be triple of the base LM, the pyramid A will be triple of the pyramid F. Two sections cd, lm, being taken at equal heights Ab, Fg, the section cd is (Prop. 107.) to the base CD as the square of the height Ab to the square of the height AB; and the section Im is to the base LM as the square of the Fig. 330. height Fg to the square of the height FG. And because the heights are equal, AB to FG, and Ab to Fg, the section cd is to the base CD as the section Im to the base LM; and, alternately, the section cd is to the section Im as the base CD is to the base LM. But the base CD is triple of the base LM, therefore the section cd is also triple of the section Im. Because the heights AF, FG are equal, it is manifest that the two pyramids are composed of an equal number of physical surfaces placed one upon another. Now it inay be demonstrated in the same manner that every surface or section of the pyramid A is triple of the corresponding surface or section of the pyramid F. Therefore the whole pyramid A is triple of the whole pyramid F. COROLLARY. Pyramids of the same height and equal bases are equal, since they are to one another as their bases. 990. PROP. CIX. A pyramid whose base is that of a cube and whose rertex is at the centre of the cube is equal to a third part of the product of its height and base. A d in M D G E L |