2. Show that, in general, the midsection of a prismatoid has as many sides as the two bases combined. When will this not be so? 3. Find the volume of the frustum of a square pyramid, the lower base being 8 ft. square, the upper base 6 ft. square, and the altitude 12 ft. Solve both by the frustum formula and the prismatoid formula, and compare. 4. Both bases of a prismatoid of altitude h are squares, and the lateral faces isosceles triangles. The sides of the upper base are parallel to the diagonals of the lower base and half the length of these diagonals. If b is a side of the lower base, find the volume. Ans. 5b2h. 5. Find the volume of the prismatoid with dimensions as shown in the figure. The bases are right triangles having their corresponding sides parallel. Ans. 3420 cu. in. 2011 90 30 6. Use the prismatoid rule to find the volume of a frustum of a pyramid whose bases are regular hexagons 10 in. and 6 in. on a side respectively, and whose altitude is 18 in. 7. Find the volume of the frustum of the preceding exercise by § 741 and compare the result with that of the preceding. 8. The volume of a truncated triangular prism is readily determined by the prismatoid formula. To do this consider one face and the opposite edges as the bases. Find the volume of the truncated triangular prism shown in the figure. The base ABC, which has a right angle at C, is a right section. D 15' E B 9. Show that the volume of any truncated triangular prism is equal to the product of the area of a right section and D one-third the sum of the three edges. SUGGESTION. Let the edges be c, d, and e, and the altitude and base of a right section a and b respectively. Consider the prism as a prismatoid with bases ABED and CF, and midsection M. Then Valb(c+d)+0+4M]. с A 8 751 E M e B .. Vab• (c+d+e) = area of right section(c+d+e). 10. A truncated right triangular prism has for base an isosceles triangle whose sides are 10 in., 10 in., and 8 in. Find its volume if the three lateral edges are 5 in., 7 in., and 11 in. respectively. POLYEDRONS 753. Regular Polyedrons. A polyedron whose faces are congruent regular polygons and whose polyedral angles are congruent is called a regular polyedron. A polyedron of four faces is a tetraedron; one of six faces, a hexaedron; one of eight faces, an octaedron; one of twelve faces, a dodecaedron; and one of twenty faces, an icosaedron. It is instructive to construct the regular polyedrons as shown in the following figures. Draw on cardboard the diagrams shown below, cut along the full lines, and fold along the dotted lines. The edges that meet may be fastened with gummed paper. Durable models may be constructed from tin and soldered at the edges. There are many interesting facts connected with the five regular polyedrons. The so-called fourteenth, fifteenth, and sixteenth books of Euclid give many theorems and problems concerning these solids. A A XXX 754. Theorem. There can be no more than five regular polyedrons. The proof of this theorem depends upon the facts: (a) At least three planes must meet to form a polyedral angle. (§ 603). (b) The sum of the face angles of a polyedral angle is less than 360°. (§ 609). (c) The faces of a regular polyedron are congruent regular polygons. (1) Faces equilateral triangles. A polyedral angle can be formed of three, four, or five equilateral triangles, and no more. Why? Therefore no more than three regular polyedrons can be formed having equilateral triangles as faces. (2) Faces squares. A polyedral angle can be formed of three squares, and no more. Why? Therefore no more than one regular polyedron can be formed having squares for faces. (3) Faces regular pentagons. A polyedral angle can be formed of three regular pentagons, and no Why? more. Therefore no more than one regular polyedron can be formed having regular pentagons for faces. Regular polygons with a greater number of sides than five cannot be used to form polyedral angles because the sum of three or more angles of such polygons is equal to or greater than 360°. Hence there can be no more than five regular polyedrons. 755. Relation between the Number of Faces, Edges, and Vertices of a Regular Polyedron. It is interesting to note the relation between the number of faces, edges, and vertices of a regular polyedron. Complete the following table, and compare with the table on page 67. 756. Euler's Theorem. This theorem is stated because of its interest. The proof is too difficult to be given here. As an exercise it may be tested for the regular polyedrons. Use the preceding table. Theorem. In any polyedron the number of edges plus two is equal to the sum of the number of faces and the number of vertices. NOTE. Leonhard Euler was born in Switzerland in 1707 and died in Petrograd in 1783. He was one of the greatest physicists, astronomers, and mathematicians of the eighteenth century. 757. Theorem. The sum of the face angles of any polyedron is equal to 360° multiplied by two less than the number of vertices. The proof of this theorem is based upon the preceding theorem. Test it for the regular polyedrons. EXERCISES 1. Find the area of the surface of a regular tetraedron 3 in. on an edge. Of a regular octaedron 4 in. on an edge. Of a regular icosaedron 6 in. on an edge. Ans. 15.588+ sq. in.; 55.426 — sq. in.; 311.769+ sq. in. 2. Find the area of the surface of a regular dodecaedron 2 in. on an edge. Ans. 82.583- sq. in. 3. If the edge of a regular tetraedron is e, show that its volume is given by the formula V=11⁄2e3√2. SOLUTION. V=ON×area of ▲ABC. DC=e√3. Area of AABC-ABXDC=e2 √3. N See p. 166, Ex. 7 (3). § 211. B D A 4. If the edge of a regular octaedron is e, show that its volume is given by the formula Ve3√2. 5. Compare the volumes of a regular tetraedron, a cube, and a regular octaedron, each 2 in. on an edge. 6. Find the area and volume of a regular tetraedron having an altitude of 8 in. 758. Theorem. The volume of two tetraedrons that have a triedral angle of one congruent to a triedral angle of the other are to each other as the products of the three edges of these triedral angles. C B' N M A' Given the tetraedrons O-ABC and O'-A'B'C', with triedral 20= triedral ≤O'. To prove V OA OB OC = V' O'A' O'B'.O'C" volumes of the tetraedrons. where V and V' denote the Proof. Place tetraedron O-ABC so that triedral 20 will coincide with triedral ZO'. Draw CN and C'M perpendicular to O'A'B'. CN and C'M determine a plane which intersects O'A'B' in = C'M O'C' V O'A O'B.O'C OA OB OC V'.O'A'.O'B' O'C' O'A'.O'B'.O'C' Compare this theorem and the proof with § 375. 759. Theorem. Two parallelepipeds that have a triedral angle of one congruent to a triedral angle of the other are to each other as the products of the three edges of these triedral angles. = § 111 |
