121. Corollary. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to the bases at equal distances from the vertices are equivalent. 1. The base of a pyramid is an equilateral triangle 2 in. on a side. Find the area of a section parallel to the base and halfway between the vertex and the base. 2. A section of a pyramid parallel to the base is equal to half the base. If the altitude of the pyramid is 10 in., how far is the section from the base? 3. Solve Ex. 2 when the section is one nth of the base and the altitude of the pyramid is h feet. 4. The perimeter of the base of a regular pyramid is 40 mm. and the lateral area is 320 sq. mm. Find the slant height of the pyramid. 5. The top of a certain obelisk is a regular pyramid with a square base 225 sq. in. in area and an altitude of 30 in. Find to the nearest 0.1 in. the slant height of the pyramid, and then find the lateral area. Proposition 12. Triangular Pyramids 122. Theorem. Two triangular pyramids with equivalent bases and equal altitudes are equivalent. Given the triangular pyramids P and P' with the equivalent bases ABC and A'B'C' and the equal altitudes h. Proof. Place the bases in the plane m, divide h into any number of equal parts, as a, and through the points of division of h let the planes n, p, q, pass II to m, as shown. ... Let the prisms y, z, with lateral edges || to VA have DEF, GHI,... for their upper bases, and let the prisms x', y', z', with lateral edges || to V'A' have A'B'C', D'E'F', G'H'I', for their lower bases. ... In the above figure there are two prisms in one case and three in the other, but the proof may be applied to any number. Hence x' + y' + z' − (y + z) = x'. Ax. 2 Then by substituting P', which is less than x'+y'+z', for x'+y'+z', and by substituting P, which is greater than y+z, for y+2, we have P'-P<x'. That is, the difference between the pyramids is less than x', which is the difference between the sets of prisms. Now by increasing indefinitely the number of parts into which h is divided, and consequently decreasing a indefinitely, x' can be made as small as we please. Hence whatever difference we assume to exist between the pyramids, x' can be made smaller than that difference. But this is impossible, since we have shown that x' is greater than the difference, if any exists. Hence it leads to an impossibility to suppose that .. P=P'. Exercises. Review 1. The slant height of a regular pyramid is 12 in., and the base is an equilateral triangle whose altitude is 4√3 in. Find the lateral area. 2. The slant height of a regular triangular pyramid is equal to the altitude of the base, and the area of the base is √5 sq. ft. Find the area of the total surface. 3. If one pyramid has for its base a right triangle with hypotenuse 10 and shortest side 6, and another pyramid of equal altitude has for its base an equilateral triangle which is 4√2√3 on a side, the pyramids are equivalent. 4. The base of one of two equivalent triangular pyramids 6 in. high is 9 sq. in. in area, and that of the other is an equilateral triangle. Find the side of this triangle. Proposition 13. Volume of a Triangular Pyramid 123. Theorem. The volume of a triangular pyramid is one third the product of the base and the altitude. Given P-QRS, a triangular pyramid; V, the volume; B, the area of the base; and h, the altitude. Proof. On QRS as base let there stand the prism QRS-XPY, and let the plane XPS pass through XP and S. Then the prism is composed of the three triangular pyra mids P-QRS, P-SYX, and P-QSX. Now P-SYX and P-QSX have the same altitude, § 113 Now S-XPY has the same altitude as P-QRS, $ 58 Proposition 14. Volume of any Pyramid 124. Theorem. The volume of any pyramid is one third the product of the base and the altitude. Given P-QRSTU, a pyramid; V, the volume; B, the area of the base; and h, the altitude. Proof. From any vertex of the base draw diagonals to the other vertices. In the above figure, let these diagonals be TQ and TR. Then the planes determined by PT and the diagonals TQ, TR divide the given pyramid into three triangular pyramids, each of which has the altitude h. The volume of each of these triangular pyramids is h times the area of the base. Hence the volume of P-QRSTU, the sum of the triangular pyramids, ish times the sum of the bases. But the sum of the bases is B. .' . V = }} Bh. § 123 Ax. 1 Ax. 10 Ax. 5 The proof is evidently the same whatever the number of the triangular pyramids into which the given pyramid is divided. |