PROPOSITION XXI. THEOREM. 657. The volume of the frustum of any pyramid is equal to the sum of the volumes of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and the mean proportional between the bases of the frustum. Let B and b denote the lower and upper bases, H the altitude, and V the volume of the frustum ABCD-EFGI. To prove that V = &H(B+ b + √ B × b). Proof. Let T-KLM be a triangular pyramid having the same altitude as S-ABCD and its base KLM equivalent to ABCD, and lying in the same plane. Then T-KLM ≈ S-ABCD. § 653 Let the plane EFGI cut T-KLM in NOP. Hence, if we take away the upper pyramids, we have left the equivalent frustums NOP-KLM and EFGI-ABCD. But the volume of the frustum NOP-KLM is equal to H(B+b+√B × b). . . V = } H (B + b + √ B × b). § 656 Q. E. D. PROPOSITION XXII. THEOREM. 658. The volumes of two triangular pyramids, having a trihedral angle of the one equal to a trihedral angle of the other, are to each other as the products of the three edges of these trihedral angles. Trim C shad Let V and V' denote the volumes of the two triangular pyramids S-ABC and S'-A'B'C', having the trihedral angles S and S' equal. V To prove that V SA× SB × SC Proof. Place the pyramid S-ABC upon S'-A'B'C' so that the trihedral S shall coincide with S'. Draw CD and C'D' L to the plane S'A'B', and let their plane intersect S'A'B' in S'DD'. The faces S'AB and S'A'B' may be taken as the bases, and CD, C'D' as the altitudes, of the triangular pyramids C-S'AB and C-S'A'B', respectively. 659. A truncated triangular prism is equivalent to the sum of three pyramids, whose common base is the base of the prism and whose vertices are the three vertices of the inclined section. Let ABC-DEF be a truncated triangular prism whose base is ABC, and inclined section DEF. Pass the planes AEC and DEC, dividing the truncated prism into the three pyramids E-ABC, E-ACD, and E-CDF. To prove ABC-DEF equivalent to the sum of the three pyramids, E-ABC, D-ABC, and F-ABC. Proof. E-ABC has the base ABC and the vertex E. The pyramid E-ACD≈ B-ACD. § 650 For they have the same base ACD, and the same altitude since their vertices E and B are in the line EB to ACD. But the pyramid B-ACD may be regarded as having the base ABC and the vertex D; that is, as D-ABC. The pyramid E-CDF ≈ B-ACF. For their bases CDF and ACF are equivalent, § 650 § 404 since the CDF and ACF have the common base CF and equal altitudes, their vertices lying in the line AD || to CF; and the pyramids have the same altitude, since their vertices E and B are in the line EB || to the plane of their bases. But the pyramid B-ACF may be regarded as having the base ABC and the vertex F; that is, as F-ABC. Therefore, the truncated triangular prism ABC-DEF is equivalent to the sum of the three pyramids E-ABC, D-ABC, and F-ABC. Q.E. D. 660. 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 DA, EB, FC (Fig. 1), being perpendicular to the base ABC, are the altitudes of the three pyramids whose sum is equivalent to the truncated prism. 661. COR. 2. The volume of any truncated triangular prism is equal to the product of its right section by one third the sum of its lateral edges. For the right section DEF (Fig. 2) divides the truncated prism into two truncated right prisms. GENERAL THEOREMS OF POLYHEDRONS. PROPOSITION XXIV. THEOREM. (EULER'S.) 662. In any polyhedron the number of edges increased by two is equal to the number of vertices increased by the number of faces. in prove whad. Let E denote the number of edges, V the number of vertices, F the number of faces, of the polyhedron AG. Proof. Beginning with one face BCGF, we have E = V. Annex a second face ABCD, by applying one of its edges to a corresponding edge of the first face, and there is formed a surface of two faces, having one edge BC and two vertices B and C common to the two faces. Therefore, for 2 faces E = V + 1. Annex a third face ABFE, adjoining each of the first two faces; this face will have two edges, AB, BF, and three vertices, A, B, F, in common with the surface already formed. Therefore, for 3 faces E = V + 2. In like manner, for 4 faces E = V3, and so on. Therefore, for (F-1) faces E = V + (F− 2). But F1 is the number of faces of the polyhedron when only one face is lacking, and the addition of this face will not increase the number of edges or vertices. Hence, for F faces E VF-2, or E+2 = V + F. = Q. E.D. |