Proposition 21. Theorem.* 637. Two tetraedrons which have a triedral angle of the one equal to a triedral angle of the other, are to each other as the products of the three edges of these triedral angles. To prove Hyp. Let S-ABC, S-DEF be the two given tetraedrons, having the common triedral ZS; and let Vand V' denote their volumes. V SA X SB X SC V SD X SE X SF Proof. Draw CO, FP I to the plane SAB. Then we may take the faces SAB, SDE as the bases, Х FP: (633) SAB SA X SB But SDE SD X SE (375) And since the rt. AS SCO, SEP are similar, CO SC (307) = V = SF Proposition 22. Theorem.* 638. A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the lower base of the prism, and whose vertices are the three vertices of the upper base. Hyp. Let ABC-DEF be a truncated triangular prism. To prove ABC-DEF equivalent to the sum of the three pyramids E-ABC, D-ABC, F-ABC. Proof. Pass the planes AEC and DEC, dividing the truncated prism into the three pyramids E-ABC, E-ACD, and E-CDF. The first pyramid E-ABC has the base ABC and the vertex E. The second pyramid E-ACD and the pyramid B-ACD, have the same base ACD, and the same altitude, since their vertices E and B are in the line EB || to the base ACD. .. E-ACD is equivalent to B-ACD. (633) But the pyramid B-ACD is the same as D-ABC. ... E-ACD is equivalent to D-ABC. The third pyramid E-CDF and the pyramid B-ACF, have equivalent bases CDF and ACF (367), and the same altitude, since their vertices E and B are in the line EB || to CF, or to the plane ACFD. ... E-CDF is equivalent to B-ACF. (633) But the pyramid B-ACF is the same as F-ABC. ..E-CDF is equivalent to F-ABC. ... ABC-DEF is equivalent to the sum of the three pyramids E-ABC, D-ABC, F-ABC. Q.E.D. 639. CoR. 1. The volume of a truncated right triangular prism is equal to the product of its base by one-third the sum* of its lateral edges. For, the lateral edges AD, BE, CF, being I to the base, are the altitudes of the three pyramids whose sum is equivalent to the truncated prism. (638) ... volume ABC-DEF = {ABC X AD +ŽABC X BE + JABC X CF (632) = ABC x }(AD + BE + CF). 640. COR. 2. The volume of any truncated triangular prism is equal to the product of its right section by onethird the sum of its lateral edges. For, the rt. section GHK divides the truncated triangular prism ABC-DEF into two truncated rt. prisms whose vol F E K umes are H GHK X }(AG+BH+CK), (639) A and GHKX }(GD+HE+KF). ... their sum is GHK X }(AD+BE+CF). B * Called the arithmetic mean of its lateral edges, also its mean height. SIMILAR POLYEDRONS. 641. Similar polyedrons are those whose corresponding polyedral angles are equal, and which have the same number of faces, similar each to each, and similarly placed. Homologous faces, edges, angles, etc., in similar polyedrons are faces, edges, angles, etc., which are similarly placed. 642. COR. 1. The homologous edges of two similar polyedrons are proportional to each other. (317) 643. CoR. 2. The homologous faces of two similar polyedrons are proportional to the squares of any two homologous edges. (379) 644. COR. 3. The entire surfaces of two similar polyedrons are proportional to the squares of any two homologous edges. (296) Proposition 23. Theorem.* F G h 645. Two similar polyedrons may be decomposed into the same number of tetruedrons, similar each to each, and similarly placed. Hyp. Let ABCDE-H and abcde-l be two similar poly H edrons, H and h being homologous vertices. A To prove that they may be decomposed into the same number of tetraedrons, similar each to each, and similarly placed. Proof. Decompose the polyedron AG into tetraedrons, by dividing all the faces not adjacent to H, into As, and drawing st. lines from H to their vertices. d B In the same manner, decompose the polyedron ag into tetraedrons having their common vertex at h homologous to H. The two polyedrons are then decomposed into the same number of tetraedrons, similarly placed. In the two homologous tetraedrons H-ABC and l-abc, since the faces AD, HA, HC are similar respectively to the faces ad, ha, hc, (641) .:. the As ABC, HAB, HBC are similar respectively to the As abc, hab, hbc. (321) Also, the corresponding triedral (s of these tetraedrons are equal. (567) ... the tetraedrons H-ABC, h-abc are similar. (641) In the same way it may be proved that any two homologous tetraedrons are similar. ... the two similar polyedrons may be decomposed into the same number of tetraedrons, similar each to each, and similarly placed. Q.E.D. 646. COR. Any two homologous lines in two similar polyedrons are proportional to any two homologous edges. |