THEOREM V. 767. Quaders having equal altitudes are to each other as their bases. HYPOTHESIS. Let the rectangles bc and b'd be the bases of two quaders, and Q, of altitude a. PROOF. Make P a third quader of altitude a and base be. Now, considering the rectangles ab as the bases of Q and P, by 766, Q : P : : c : c'; considering the rectangles ac as bases of P and Q, by 766, P : Q :: b : b'. Therefore, compounding the ratios, Q : & :: bc : b'c. THEOREM VI. 768. Two quaders are to each other in the ratio compounded of the ratios of their bases and altitudes. HYPOTHESIS. Let Q and be two quaders, of altitude a and a', and base bc and b'c', respectively. CONCLUSION. Q: Q :: abc : a'b'c. PROOF. Make P a third quader of altitude a' and base bc. and, by 767, Therefore, compounding, Q : P : : a : d' ; P: Q :: bc : b'd. Q: Q: abc: a'b'd. THEOREM VII. 769. Any parallelopiped is equivalent to a quader of equivalent base and equal altitude. For, supposing AB an oblique parallelopiped on an oblique base, prolong the four edges parallel to AB, take sect CD = AB, and draw CEI CD, and DF || CE. Through CE and DF pass parallel planes. Now the solids DFB and CEA are congruent, having all their angles and edges respectively equal. Taking each in turn from the whole solid DFA, leaves parallelopiped AB = CD. In the line CD take GH = CD, and through G and H pass planes. perpendicular to GH. The solids DFH and CEG are congruent, therefore parallelopiped CD = GH. Now prolong the four edges not parallel to GH, and take LM = HK, and through Z and M pass planes perpendicular to LM. As before, parallelopiped GH = LM; but LM is a quader of equivalent base and equal altitude to parallelopiped AB. THEOREM VIII. 770. A plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms. If the lateral faces are all rectangles, the two prisms are congruent ; if not, the prisms are still equivalent. For draw planes perpendicular to AA' at the points A and A'. Then the prism ABC A'B'C' is equivalent to the right prism AEM A'E'M', because the pyramid AEB CM is congruent to the pyramid A'E'B'C'M'. In the same way, ADC A'D'C' is equivalent to ALM A'L'M'. But AEM A'E'M' and ALM A'L'M' are congruent, :. ABC A'B'C' and ADC A'D' C' are equivalent. 771. COROLLARY. Any triangular prism is half a parallelopiped of twice its base but equal altitude. THEOREM IX. 772. If a pyramid be cut by a plane parallel to its base, the section is to the base as the square of the perpendicular on it, from the vertex, is to the square of the altitude of the pyramid. Ꮴ A The section and base are similar, since corresponding diagonals cut them into triangles similar in pairs because having all their sides respectively proportional. For A'C': AC :: VC' : VC :: VO: VO, and B'C' : BC :: VC' : VC, AC': AC :: B'C': BC; and in the same way, B'C': BC :: A'B' : AB, Δ Α' Β' Γ' ~ ▲ ABC, etc., section base :: AC: AC2 :: VO2: Vo2. . 773. COROLLARY. In pyramids of equivalent bases and equal altitudes, two sections having equal perpendiculars from the vertices are equivalent. |
