582. A rectangular parallelopiped is one whose faces are all rectangles. The three edges of a parallelopiped which meet at a vertex are its dimensions. 583. A cube is a rectangular parallelopiped whose six faces are all squares. 584. The cube whose edge is the unit of length is taken as the unit of volume. (571) REM. In a right parallelopiped the adjacent lateral faces may make any angle with each other, while in a rectangular parallelopiped the faces are perpendicular to each other. B Proposition 1. Theorem. 585. The sections of the lateral faces of a prism made by parallel planes are equal polygons. Hyp. Let the prism MN be cut by the || planes AD, A'D'. E? To prove ABCDE = A'B'C'D'E'. D' Proof. Because the intersections of two Il planes by a third plane are parallel, (519) : . AB, BC, CD, etc., are || respec M tively to A'B', B'C', C'D', etc. ::: ABC = A'B'C', BCD = _ B'O'D'. (525) _ Z () Because || s included between || s are equal, (131) .. AB = A'B', BC = B'C', etc. ... the polygons ABCDE, A'B'C'D'E' are mutually equilateral and mutually equiangular. .:. ABCDE = A'B'C'D'E'. (146) Q. E. D. 586. COR. Any section of a prism by a plane parallel to the base is equal to the base; and all right sections are equal. = Proposition 2. Theorem. 587. If three faces including a triedral angle of a prism are equal respectively to three faces including a triedral angle of a second prism, and similarly placed, the two prisms are equal. Hyp. Let the faces AD, K · AG,.,BH, of the triedral ZB, be = respectively to the faces A'D', A'G', B'H' of the triedral B'. To prove A prism AI= prism A'I'. Proof. Since the face Ls ABC, ABG, CBG = the face Zs A'B'C', A'B'G', C'B'G', respectively, (Hyp.) .:. triedral _B= triedral _B'. (567) ... the prism AI may be applied to the prism A'l' so that the base AD shall coincide with A'D', the face AG with A’G’, and the face BH with B'H', the pts. D, E falling on D', E'. Since the pts. F, G, H coincide with F', G', H', Since the lateral edges of the prisms are equal and parallel, (574) .:. the edges DI, EK will coincide with D'I', E'K', respectively, and the pts. I, K with the pts. I', K'. .. the prisms coincide throughout, and ... are equal. Q.E.D. 588. COR. 1. Two right prisms which have equal bases and equal altitudes are equal. 589. COR. 2. Two truncated prisms are equal if three faces including a triedral angle of the one are equal respectively to three faces including a triedral angle of the other, and similarly placed. Proposition 3. Theorem. 590. An oblique prism is equivalent to a right prism having for its base a right section of the oblique prism and for its altitude a lateral edge of the oblique prism. F! To prove = B Hyp. Let ABCDE-I be an oblique prism, and A'D' art. section of it. Produce AF to F', making A'F' = AF, and through F' pass the plane F'I' I to AF', cutting all the edges BG, CH, etc., produced, in the pts. G', H', etc., and forming the rt. section F'I'll to A'D'. (516) prism AI = prism A'I'. . Proof. In the truncated prisms AD' and FI', ABCDE= FGHIK. (574) Because AF = A'F', and BG = B'G', ,. AA'=FF', and BB' = GG'. Because AG and A'G' are Os, (574) ... AB is = and || to FG, and A'B' is =and || to F'G'. .. quadl. AB' = quadl. FG', being mutually equilateral and equiangular (146). Similarly, BC = GH'. ... the truncated prisms AD' and FI' are equal. (589) Taking AD' from the whole solid, the right prism A'I' remains, and taking FI' from the same solid, the oblique prism AI remains. .: prism AI = prism A'I'. Q.E.D. = Proposition 4. Theorem. 591. The lateral area of a prism is equal to the product of the perimeter of a right section by a lateral edge. E' Hyp. Let FGHIK be a rt. section, and AA' a lateral edge of the prism AD'. To prove lateral area of AD' = AA (FG + GH + etc.). Proof. Since the rt. section FI is I to the lateral edges, (578) ::. FG, GH, .etc., are I to AA', BB', etc., (487) and .:. FG, GH, etc., are the altitudes of the OS AB', BC', etc., which form the lateral area of the prism. .. area AA'B'B = AA' X FG, (363) and .. area BB'C'C = BB' X GH, etc. Because AA' = BB' = etc., (574), and FG + GH+ etc., = the perimeter of the rt. section, .. adding the above areas, we have lateral area AD' = AA' X FG + AA' X GH + etc. AA'(FG+GH+HI+etc.). Q.E.D. 592. COR. The lateral area of a right prism is equal to the product of the perimeter of its base by its altitude. Proposition 5. Theorem. 693. The opposite faces of a parallelopiped are equal and parallel. To prove = Hyp. Let ABCD-G be a parallelopiped. face AF and || to DG. Proof. Since AC is a o, (580) .:. AB = and || to DC. Since BG is a o, (580) .. BF = and || to CG. Since BA and BF are || respectively to CD and CG, ...ZABF = _ DCG, and plane AF is || to plane DG.(525) O AF = ODG, ... face AF = and || to face DG. face AH = and || to face BG. Also, face AC = and || to face EG, by definition. (574) Q.E.D. 594. Sch. Since the opposite faces of a parallelopiped are equal and parallel parallelograms, any face may be taken for the base. 595. Cor. The twelve edges of a parallelopiped may be divided into three sets, each set having four equal and parallel lines. |