17. Two polyedrons are similar when they are contained by the same number of similar planes, similarly situated, and having like inclinations with each other. PROPOSITION I. THEOREM. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude. Let ABCDE-K be a right prism: then will its convex surface be equal to (AB+BC+CD+DE+EA) x AF. For, the convex surface is equal to the sum of all the rectangles AG, BH, CI, DK, EF, which compose it. Now, the altitudes AF, BG, EH, &c. of the rectangles, are equal to the altitude of the prism. Hence, the sum of these rectangles, or the convex surface of the prism, is equal to (AB+BC+CD+DE+EA) x AF; that is, to the perimeter of the base of the prism multiplied by its altitude. Cor. If two right prisms have the same altitude, their convex surfaces will be to each other as the perimeters of their bases. PROPOSITION II. THEOREM. In every prism, the sections formed by parallel planes, are equal polygons. Let the prism AH be intersected by the parallel planes NP, SV; then are the polygons NOPQR, STVXY equal. S For, the sides ST, NO, are parallel, being the intersections of two parallel planes with a third plane ABGF; these same sides, ST, NO, are included between the parallels NS, OT, which are sides of N the prism hence NO is equal to ST. For like reasons, the sides OP, PQ, QR, &c. of the section NOPQR, are equal E to the sides TV, VX, XY, &c. of the sec- A tion STVXY, each to each. And since B D H the equal sides are at the same time parallel, it follows that the angles NOP, OPQ, &c. of the first section, are equal to the angles STV, TVX, &c. of the second, each to each (Book VI. Prop. XIII.). Hence the two sections NOPQR, STVXY, are equal polygons. Cor. Every section in a prism, if drawn parallel to the base, is also equal to the base. PROPOSITION III. THEOREM. If a pyramid be cut by a plane parallel to its base, 1st. The edges and the altitude will be divided proportionally. 2d. The section will be a polygon similar to the base. Let the pyramid S-ABCDE, of which SO is the altitude, be cut by the plane abcde; then will Sa: SA:: So: SO, and the same for the other edges: and the polygon abcde, will be similar to the base ABCDE. First. Since the planes ABC, A abc, are parallel, their intersections AB, ab, by a third plane SAB will also be parallel 2 20 E C B (Book VI. Prop. X.); hence the triangles SAB, Sab are similar, and we have SA: Sa :: SB: Sb; for a similar reason, we have SB: Sb :: SC: Sc; and so on. Hence the edges SA, SB, SC, &c. are cut proportionally in a, b, c, &c. The altitude SO is likewise cut in the same proportion, at the point o; for BO and bo are parallel, therefore we have SO: So :: SB: Sb. Secondly. Since ab is parallel to AB, bc to BC, cd to CD, &c. the angle abc is equal to ABC, the angle bcd to BCD, and so on (Book VI. Prop. XIII.). Also, by reason of the similar triangles SAB, Sab, we have AB: ab: : SB: Sb; and by reason of the similar triangles SBC, Sbc, we have SB: Sb BC: bc; hence AB ab: BC: bc; we might likewise have BC: bc: CD: cd, and so on. Hence the polygons ABCDE, abcde have their angles respectively equal and their homolo-, gous sides proportional; hence they are similar. Cor. 1. Let S-ABCDE, S S-XYZ be two pyramids, hav-b ing a common vertex and the apt to same altitude, or having their bases situated in the same plane; if these pyramids are cut by a plane parallel to the plane of their bases, giving the sections abcde, xyz, then will A the sections abcde, xyz, be to each other as the bases ABCDE, XYZ. C 60 For, the polygons ABCDE, abcde, being similar, their surfaces are as the squares of the homologous sides AB, ab; but AB: ab: SA Sa; hence ABCDE: abcde: SA2: Sa2. For the same reason, XYZ xyz:: SX2: Sz2. But since abc and xyz are in one plane, we have likewise SA: Sa : : SX: Sz (Book VI. Prop. XV.); hence ABCDE: abcde: : XYZ xyz; hence the sections abcde, xyz, are to each other as the bases ABCDE, XYZ. Cor. 2. If the bases ABCDE, XYZ, are equivalent, any sections abcde, xyz, made at equal distances from the bases, will be equivalent likewise. PROPOSITION IV. THEOREM. The convex surface of a regular pyramid is equal to the perime ter of its base multiplied by half the slant height. For, since the pyramid is regular, the point O, in which the axis meets the base, is the centre of the polygon ABCDE (Def. 14.); hence the lines OA, OB, OC, &c. drawn to the vertices of the base, are equal. In the right angled triangles SAO,SBO, the bases and perpendiculars are equal: hence the hypothenuses are equal: and it may be proved in the same way that all the sides of the right pyramid are equal. The triangles, therefore, which form the convex surface of the prism are all equal to each other. But the area of either of these triangles, as ESA, is equal to its base EA multiplied by half the perpendicular SF, which is the slant height of the pyramid: hence the area of all the triangles, or the convex surface of the pyramid, is equal to the perimeter of the base multiplied by half the slant height. Cor. The convex surface of the frustum of a regular pyramid is equal to half the perimeters of its upper and lower bases multiplied by its slant height. For, since the section abcde is similar to the base (Prop. III.), and since the base ABCDE is a regular polygon (Def. 14.), it follows that the sides ea, ab, bc, cd and de are all equal to each other. Hence the convex surface of the frustum ABCDE-d is formed by the equal trapezoids EAae, ABba, &c. and the perpendicular distance between the parallel sides of either of these trapezoids is equal to Ff, the slant height of the frustum. But the area of either of the trapezoids, as AEea, is equal to (EA+ea) x Ff (Book IV. Prop. VII.): hence the area of all of them, or the convex surface of the frustum, is equal to half the perimeters of the upper and lower bases multiplied by the slant height. PROPOSITION V. THEOREM, 944 If the three planes which form a solid angle of a prism, are equal to the three planes which form the solid angle of another prism, each to each, and are like situatedly the two prisms will be equal to each other. Let the base ABCDE be equal to the base abcde, the parallelogram ABGF equal to the parallelogram abgf, and the parallelogram BCHG equal to bchg; then will the prism ABCDE-K be equal to the prism abcde-k. K T For, lay the base ABCDE upon its equal abcde; these two bases will coincide. But the three plane angles which form the solid angle B are respectively equal to the three plane angles, which form the solid angle b, namely, ABC-abe, ABG abg, and GBC gbe; they are also similarly situated : hence the solid angles B and bare equal (Book VI. Prop. XXI. Sch.); and therefore the side BG will fall on its equal bg. It is likewise evident, that by reason of the equal parallelograms ABGF, abgf, the side GF will fall on its equal gf, and in the same manner GH on gh; hence, the plane of the upper base, FGHIK will coincide with the plane fghik (Book VI. Prop. II.). But the two upper bases being equal to their corresponding lower bases, are equal to each other: hence HI will coincide with hi, IK with ik, and KF with kf; and therefore the lateral faces of the prisms will coincide: therefore, the two prisms coinciding throughout are equal (Ax. 13.). Cor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to ab, and the altitude BG to bg, the rectangle ABGF will be equal to abgf; so also will the rectangle BGHC be equal to bghc; and thus the three planes, which form the solid angle B, will be equal to the three which form the solid angle b. Hence the two prisms are equal. PROPOSITION VI. THEOREM. In every parallelopipedon the opposite planes are equal and parallel. By the definition of this solid, the bases ABCD, EFGH, are equal parallelograms, and their sides are parallel: it remains only to show, that the same is true of any two opposite lateral faces, such as AEHD, BFGC. Now AD is equal and parallel to BC, because the figure ABCD is a pár |