and ZEFG=2 ABC, ZFGH= Z BCD, etc. (468) (289) 587. CoR. 1. A section of a pyramid parallel to the base is to the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. 588. Cor. 2, If two pyramids that have equal altitudes are cut by planes parallel to the bases, and at equal distances from the vertices, the sections have the same ratio as the bases. Hint. - Use Cor. 1. 589. CoR. 3. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to the bases, and at equal distances from the vertices, are equivalent. Ex. 1104. The altitude of a regular triangular pyramid, each side of whose base is 4 ft., is 12 ft. Find the area of a section made by a plane parallel to the base and 4 ft. from the vertex. Ex. 1105. Find the area of a section of a regular quadrangular pyramid, each side of whose base is 3 ft. and whose altitude is 8 ft., made by a plane 3 ft. from the vertex and parallel to the base. Ex. 1106. The base of a regular hexagonal pyramid measures 10 in. on a side. If the altitude be 5 in., how far from the vertex must a plane be passed to form a section the area of which is 3V3? U PROPOSITION XV. THEOREM 590. The lateral area of a regular pyramid is equal to half the product of the slant height by the perimeter of the base. To prove Hyp. A is the lateral area, SI. H. the slant height, and P the perimeter of the base of the regular pyramid 0-ABCDE. A= SI. H. X P. A= { Sl. H. x its base. = į SI. H. X P. Q.E.D. Ex. 1107. Find the lateral area of a regular quadrangular prism whose altitude is 12, and whose base has sides equal to 10. Ex. 1108. Find the lateral area of a frustum of a regular pyramid of five sides whose slant height is 5, and the sides of whose bases are 7 and 3, respectively. 592. Two triangular pyramids which have equivalent bases and equal altitudes are equivalent. Hyp. Two triangular pyramids 0-ABC and 0'-A'B'C" have equal altitudes and the base ABC - base A'B'C'. Proof. If they are not equivalent, assume 0'-A'B'C">O-ABC. Place the pyramids so that their bases are in the same plane. Divide the altitudes into n equal parts, calling the length of each part h. Through the points of division pass planes parallel to the bases. The corresponding sections thus formed are equivalent. (589) On the base A'B'C', and on each parallel section of O'-A'B'C' as a lower base, construct a prism with lateral edges parallel to O'C" and altitude equal to h. On each section of 0-ABC as an upper base construct a prism with lateral edges parallel to OC and altitude equal to h. Each prism of 0-ABC is equivalent to the prism next above it in O'-A'B'C', hence the difference between the two sets of prisms is the lowest prism of the first set. Now the sum of the volumes of the circumscribed prisms is evidently greater than the volume of the pyramid O'-A'B'C', and the sum of the volumes of the inscribed prisms is less than the volume of the pyramid 0-ABC, hence the difference between the volumes of the two sets of prisms (which is the lowest prism of the first set) must be greater than the difference between the pyramids. By increasing indefinitely the number of parts into which the altitude is divided, each part h becomes smaller and finally less than any assignable quantity, hence the volume of the lowest prism (which is greater than the difference between the pyramids) may be made less than any assignable volume. ::. Pyramid O'-A'B'C' is not greater than pyramid 0-ABC. In the same manner we may prove that pyramid 0-ABC is not greater than pyramid 0'-A'B'C'. .:. 0-ABC = O'-A'B'C". PROPOSITION XVII. THEOREM 593. The volume of a triangular pyramid is equal to one third of the product of its base by its altitude. Hyp. V is the volume, B the base, and a the altitude of the triangular pyramid E-ABC. Proof. On ABC construct the prism ABC-DEF, having its lateral edges equal and parallel to EB. The prism is composed of the pyramid E-ABC and the quadrangular pyramid E-ACFD. Through ED and EC pass a plane cutting ACFD in DC, forming two triangular pyramids, E-DCF and E-ADC. E-DCF - E-ADC. (1) (Same altitude and equivalent bases.) Pyramid E-ADC is the pyramid C-ADE, since any face of a tetrahedron may be taken as base, . . C-ABE or E-ABC= į ABC-EDF. (From (1) and (2)) 594. CoR. 1. The volume of any pyramid is equal to onethird the product of its base by its altitude. For the pyramid may be divided into triangular pyramids by passing planes through any edge and the corresponding diagonals of the base. The bases of the triangular pyramids together form the base of the given pyramid, and the sum of the volumes of the triangular pyramids equals the volume of the given pyra- E mid. Therefore the volume of any pyramid equals one-third the product of its base by its altitude. A B В |