AH, ah perpendicular to the bases BCD, bcd: join BH; take Ba=ba, and in the plane BHA draw ah perpendicular to BH: then ah will be perpendicular to the plane CBD, and equal to ah the altitude of the other prism; for if the solid angles B and b were applied the one to the other, the planes which contain them, and consequently the perpendiculars ah, ah would coincide. Now because of the similar triangles ABH, abh, and the similar figures P, p, we have AH: ah:: AB: ab:: BC: bc; and because of these similar bases, the base BCD base bcd:: BC2: be2. (B. III., T. XXVII.) Taking the product of the corresponding terms of these proportions, we have AH x base BCD :ah x base bed::BC:bở. But AH x base BCD expresses the volume of the prism P, and ah base bcd expresses the volume of the other prism p; therefore. prism P: prism p:: BC3: bo3. THEOREM XIV. If a pyramid be cut by a plane parallel to its base, the section thus formed will be a polygon similar to the base, and the lateral edges and the altitude will be cut into proportional parts. First, since the planes ABCDEF, abcdef are parallel, it follows that AB and ab, BC and bc, CD and cd, etc., are parallel (B. V., T. X.); consequently the angles ABC and abc, BCD and bcd are respectively equal (B. V., T. XIII.). Moreover the couples of similar triangles SAB and Sab, SBC and Sbc, etc., give the following equal ratios: SA: Sa:: AB: ab:: SB: Sb, E S D &c. &c. &c. B Hence, by equality of ratios, we have AB: ab:: BC: bc:: CD: cd:: etc. Consequently the two polygons ABCDEF, abcdef are similar (B. III., T. X.). Now, considering only the ratios between the lateral edges, we have SA: Sa::SB: Sb:: SC: Sc:: etc., from which we see that the edges are cut into proportional parts. Finally, if we pass a plane through the edge SB and the altitude SO, its intersection bo with the plane abcde will be parallel to BO (B. V., T. X.), and the two similar triangles SBO, Sbo will give SO: So:: SB: Sb:: SA: Sa:: etc., which establishes the theorem. Cor. When two pyramids S-ABCDE, T- MNP have equivalent bases situated in the same plane, and equal altitudes, the sections made by a plane parallel to their bases, are also equivalent. For, since the polygons ABCDE, abcde are similar, we have the proportion ABCDE: abcde:: AB2: ab2; we also have, by reason of the foregoing relations, ABCDE: abcde :: SO2 : So2. In the same manner the two polygons MNP, mnp give MNP: mnp:: TQ2 : Tq2, = we have, moreover, by supposition, SO TQ, So=Tq; hence, ABCDE: abcde:: MNP: mnp. Now, by hypothesis, ABCDE=MNP; consequently, abcde= mnp. THEOREM XV. The lateral surface of a right pyramid is equal to the perimeter of its base multiplied by half the slant height. Since the lateral surface is composed of equal isosceles triangles SAB, SBC, SCD, etc., each one of which is measured by its base into one half its altitude, which altitude is the same as SF the slant height of the pyramid, it follows that the lateral surface is equal to the sum of the bases of all these triangles, or the perimeter of the pyramid's base, into one half the slant height. Cor. I. If two right pyramids have the same slant heights, their lateral surfaces will be to each other as the perimeters of their bases. Cor. II. The lateral surface of a frustum of a regular pyramid is measured by half the sum of the perimeters of its two bases multiplied by its slant height. For, since the frustum is formed from a regular pyramid, the two bases are similar and regular polygons (T. XIV.). Consequently the lateral surface of a frustum of a right pyramid is composed of equal trapezoids, the altitude of each being the same as the slant height of the frustum. The area of each trapezoid is measured by half the sum of its parallel bases multiplied into its altitude (B. III., T. XXIII., C. II.). Hence the sum of all these trapezoids, or the lateral surface of the frustum, is measured by half the sum of the perimeters of its bases into its slant height. THEOREM XVI. Two triangular pyramids having equivalent bases and equal altitudes, are equivalent, or equal in volume. Let S-ABC, S-abc be those two pyramids; let their equivalent bases ABC, abc be situated in the same plane, and let AT be their common altitude. If they are not equivalent, let Sabc be the smaller; and suppose Aa to be the altitude of a prism, which, having ABC for its base, is equal to their difference. Divide the altitude AT into equal parts Ax, xy, yz, etc., each less than Aa, and let k be one of those parts; through the points of division, pass planes parallel to the plane of the bases: the corresponding sections formed by these planes in the two pyramids will be respectively equivalent (T. XIV., C.), namely, DEF to def, GHI to ghi, etc. This being granted, upon the triangles ABC, DEF, GHI, etc., taken as bases, construct exterior prisms, having for edges the parts AD, DG, GK, etc., of the edge SA. In like manner, on the bases def, ghi, klm, etc., in the second pyramid, construct interior prisms, having for edges the corresponding parts of sa. It is plain that the sum of all the exterior prisms of the pyramid S- ABC will be greater than this pyramid; and, also, that the sum of all the interior prisms of the pyramid 8 - abc will be less than this. Hence the difference between the sum of all the exterior prisms and the sum of all the interior ones, must be greater than the difference between the two pyramids themselves. Now, beginning with the bases ABC, abc, the second exterior prism DEFG is equivalent to the first interior prism defa, because they have the same altitude k, and their bases DEF, def are equivalent: for like reasons, the third exterior prism GHIK, and the second interior prism ghid, are equivalent; the fourth exterior, and the third interior; and so on, to the last in each series. Hence all the exterior prisms of the pyramid S - ABC, excepting the first prism DABC, have equivalent corresponding ones in the interior prisms of the pyramid s— abc: hence the prism DABC is the difference between the sum of all the exterior prisms of the pyramid S - ABC, and the sum of all the interior prisms of the pyramid 8- abc. But the difference between these two sets of prisms has already been proved to be greater than that of the two pyramids, which latter difference we supposed to be equal to the prism aABC: hence the prism DABC must be greater than the prism aABC; but in reality it is less, for they have the same base ABC, and the altitude Ax of the first is less than Aa the altitude of the second. Hence the supposed inequality between the two pyramids cannot exist; hence the two pyramids S - ABC, 8- abc, having equal altitudes and equivalent bases, are themselves equivalent. s THEOREM XVII. Every triangular pyramid is the third of the triangular prism having the same base and altitude. Let FABC be a triangular pyramid, ABCDEF a triangular prism of the same base and altitude; the pyramid will be equal to one third of the prism. E B D Conceive the pyramid F-ABC to be cut off from the prism by a section made along the plane FAC, and there will remain the solid FACDE, which may be considered as a quadrangular pyramid whose vertex is F, and base the parallelogram ACDE. Draw |