Or, 2. Multiply the circumference by .31881 ( = = 2.14159). From the foregoing rules we may readily deduce the following additional rules for finding the area of a circle. 548. Multiply the square of the diameter by .7854. Multiply the square of the radius by 3.14159. Multiply the square of the circumference by .079577. PROBLEMS. Find the area of a triangle whose 1. Base is 25 ft., altitude 6. 2. Base is 100 rd., altitude 50. 3. Base is 16 yards, altitude 16. Ans. 75 sq. ft. Ans. 2500 sq. rd. Ans. 128 sq. yd. Ans. 4683 sq. ft. Ans. 29.933 sq. rd. Ans. 63.277 sq. yd. Ans. 139.77 sq. ft. Find the area of a parallelogram whose 8. Base is 13, altitude 20 ft. 9. Base is 250, altitude 10 rd. 10. Base is 123, altitude 16 yd. 11. Base is 37 in., altitude 6 ft. Ans. 260 sq. ft. Ans. 2500 sq. rd. Ans. 200 sq. yd. Ans. 2700 sq. in. whose bases Find the area of a trapezoid 12. Are 6 and 10 ft., altitude 5 ft. 13. Are 25 and 35 yd., altitude 9 ft. 14. Are 25 and 35 ft., altitude 9 yd. 15. Are 75 ft. and 17 yd., altitude 10 ft. ft. Ans. 40 sq. Find the area of a trapezuim whose 16. Is 21, perpendiculars 3 and 5 ft. 17. Is 36, perpendiculars 21 and 30 ft. 18. Is 31, perpendiculars 27 and 16 yd. 19. Is 15, perpendiculars 8 and 10 rd. Find the area of a regular figure of 20. Five sides, each 10 ft.; per. 6.88 ft. 21. Six sides, each 10 ft.; per. 8.66 ft. 22. Eight sides, each 20 ft.; per. 24.14 ft. Ans. 172 sq. ft. Ans. 259.8 sq. ft. Ans. 1931.2 sq. ft. Find the circumference of 23. A circle whose diameter is 10. 24. A circle whose diameter is 25. 25. A circle whose radius is 10. 26. A circle whose radius is 100. Find the diameter of 27. A circle whose circumference is 36. 28. A circle whose circumference is 25. Ans. 31.4159. Ans. 78.54. Ans. 628.318. Ans. 11.45916. Find the area of a circle whose 29. Circumf. is 20 ft.; radius, 3.1831. 30. Circumf. is 25 ft.; radius, 3.97887. 31. Diameter is 13 rd. 32. Diameter is 16 yd. 33. Radius is 35 ft. 34. Circumference 13.5 ft. 35. Circumference 12.5 rd. FIG. 1. B Ans. 31.831. Ans. 49.7359. Ans. 3848.46 sq. ft. SOLIDS. 549. A Solid, or Body, is that which has length, breadth and thickness, or three dimensions. A Prism is a solid, bounded by plane surfaces, two of which, the ends, or bases of the prism, are equal and similar plane figures; and the sides, or faces, parallelograms; as Figs. 1, 2, and 3, and (215, Fig. 2.) · Both plane figures and solids are often regarded as having two bases, an upper and a lower base. In Fig. 1, A B C is the lower, and DEF, the upper base. In Fig. 2, ABCD, the lower, and E F G H the upper base. Prisms are named from the form of their bases. Fig. 1 represents a triangular prism, because its bases A B C and D E F are triangles; Fig. 2, a quadrangular prism, because its bases A B C D and E F G H are quadrilaterals; Fig. 3, a pentagonal prism, because its bases ABCDE and FGHIJ are pentagons. The edges of a prism are the lines in which the bounding surfaces meet. Thus, in Fig. 1, A D, C B and B E are edges; in Fig. 2, A E, B F, &c. are edges. A Right Prism is one whose faces are perpendicular to the base. Fig. 3 is a right prism, because each of its faces, as A B G F, is perpendicular to the base ABCDE. A Pyramid is a solid having for its bases any polygon, and for the rest of its surface, plane triangles, Fig. 4. It terminates in a point called its vertex, S, Fig. 4. Pyramids are named from their bases. The one represented in the margin is a pentangular pyramid, because its base is a pentagon. The altitude of a pyramid is the perpendicular distance from its vertex to its base. If the pyramid is a right pyramid, the perpendicular will fall on the middle of the base; as, SO, Fig. 4. The slant height is the length of a line drawn from the vertex perpendicular to one side of the base; as, S M, Fig. 4. The slant height of a frustum is the perpendicular distance between the two parallel sides of one of its faces; as, L K, Fig. 5. SURFACES OF SOLIDS. 550. The Convex Surface of a prism, or pyramid, or frustum, is all of its surface except the base, or bases. 551. The Entire Surface is the convex surface and the surface of the base or bases. By inspection we can readily see that in order to obtain the surface of solids we have simply to apply the rules already learned. For example-to find the convex surface of the prism, Fig. 1, we have only to find the area of the three parallelograms A BED, BEF C, and ACF D, and then find their sum, which is just the same as multiplying the sum of A B, B C, and CA by the altitude A D. If we wish the entire surface, we add the area of the two triangles, A B C, and DEF. Hence, to find the surface of a prism we have this 552. RULE.-Multiply the perimeter of the base by the altitude. To this result, add the area of the bases for the entire surface. In like manner, to find the convex surface of a pyramid, we find the sum of the areas of the triangles that form its surface; the slant height of the pyramid being the altitude of the triangles. From which we deduce this 553. RULE.—Multiply the perimeter of the base by onehalf the slant height. To this add the area of base, for the entire surface. Again, since the surface of the frustum of a pyramid is composed of trapezoids, we see that the surface can be obtained by the following: 554. RULE.-Multiply the sum of the perimeters of the parallel bases, by one-half the slant height. Add the areas of the bases, for the entire surface. PROBLEMS. What is the convex surface of a prism whose 1. Altitude is 10 ft. and perimeter of base 16 ft. ? Ans. 160 ft. 2. Altitude is 13 yd. and perimeter of base 14 yd.? Ans. 182 sq. yd. 3. Altitude is 6 in. and perimeter of base 14 in. ? Ans. 87 in. 4. What is the entire surface of a square prism whose altitude is 25 ft. and each side of the base 7 ft.? 5. What is the entire surface of a triangular prism whose altitude is 5 yd. and each side of the base, 6 yd.? Ans. 121.176 sq. yd. 6. What is the convex surface of a pentangular pyramid whose slant height is 30 ft. and each side of its base is 3 ft.? Ans. 225 sq. ft. 7. What is the entire surface of a quadrangular pyramid whose slant height is 25 ft. and each side of the base is 6 ft.? Ans. 336 sq. ft. |