619. Two rectangular parallelopipeds having equal altitudes are to each other as their bases. Let a, b, c, and a', b', c, be the three dimensions, respectively, of the two rectangular parallelopipeds P and P ́. Proof. Let be a third rectangular parallelopiped whose dimensions are a', b, and c. Now has the two dimensions b and c in common with P, and the two dimensions a' and e in common with P'. The products of the corresponding members of these two equalities give P = P ахъ a' x b' Q. E. D. 620. COR. Two rectangular parallelopipeds which have one dimension in common are to each other as the products of their other two dimensions. PROPOSITION IX. THEOREM. 621. Two rectangular parallelopipeds are to each other as the products of their three dimensions. Let a, b, c, and a', b', c', be the three dimensions, respectively, of the two rectangular parallelopipeds P and P'. Ex. 637. Find the ratio of two rectangular parallelopipeds, if their altitudes are each 6 inches, and their bases 5 inches by 4 inches, and 10 inches by 8 inches, respectively. Ex. 638. Find the ratio of two rectangular parallelopipeds, if their dimensions are 3, 4, 5, and 9, 8, 10, respectively. PROPOSITION X. THEOREM. 622. The volume of a rectangular parallelopiped is equal to the product of its three dimensions. Let a, b, and c be the three dimensions of the rectangular parallelopiped P, and let the cube U be the unit of volume. To prove that the volume of P = a × b × c. 623. COR. 1. The volume of a cube is the cube of its edge. 624. COR. 2. The volume of a rectangular parallelopiped is equal to the product of its base by its altitude. 625. SCHOLIUM. When the three dimensions of a rectangular parallelopiped are each exactly divisible by the linear unit, this proposition is rendered evident by dividing the solid into cubes, each equal to the unit of volume. Thus, if the three edges which meet at a common vertex contain the linear unit 3, 5, and 8 times respectively, planes passed through the several points of division of the edges, perpendicular to the edges, will divide the solid into 3 × 5 × 8 cubes, each equal to the unit of volume. PROPOSITION XI. THEOREM. 626. The volume of any parallelopiped is equal to the product of its base by its altitude. Let P be an oblique parallelopiped no two of whose faces are perpendicular, whose base B is a rhomboid, and whose altitude is H. To prove that the volume of P = B × H. Proof. Prolong the edge EF and the edges || to EF, and cut them perpendicularly by two parallel planes whose distance apart GI is equal to EF. We then have the oblique parallelopiped whose base C is a rectangle. Prolong the edge IK and the edges to IK, and cut them perpendicularly by two planes whose distance apart MN is equal to IK. We then have the rectangular parallelopiped R. PROPOSITION XII. THEOREM. 627. The volume of a triangular prism is equal to the product of its base by its altitude. Let V denote the volume, B the base, and H the altitude of the triangular prism CEA-E'. Proof. Upon the edges CE, EA, EE', construct the parallelopiped CEAD-E'. Then CEA-E'C'EAD-E'. § 615 Now the volume of CEAD-E' = CEAD × H. § 626 But CEAD 2 B. § 179 .. V = {(2 B x H) = B× H. Q.E. D. Ex. 639. Two triangular prisms are equal if their lateral faces are equal, each to each, and similarly placed. Ex. 640. The square of a diagonal of a rectangular parallelopiped is equal to the sum of the squares of the three dimensions. Ex. 641. The sum of the squares of the four diagonals of a parallelopiped is equal to the sum of the squares of the twelve edges. Ex. 642. The volume of a triangular prism is equal to half the product of any lateral face by the perpendicular dropped from the opposite edge on that face. |