(2) The inner diameter of a ring is 7 inches, and the outer diameter is 8 inches. The difference of these diameters is twice the diameter of the circular section; therefore the radius of the circular section is of an inch; and the area of the circular section is 19635 square inches. The outer boundary of the ring is 7 x 31416 inches, and the inner boundary is 8 × 31416 inches; half the sum of these numbers is 23 562 inches, which is therefore the length of the ring. Therefore the volume of the ring in cubic inches =*19635 × 23 562=46264 very nearly. 1. Shew that the length of a ring is equal to the difference of the outer boundary and the circumference of the cross section. 2. Shew that the length of a ring is equal to the sum of the inner boundary and the circumference of the cross section. Find in cubic inches the volumes of the rings having the following dimensions: 3. Length 20 inches, radius of cross section of an inch. 4. Length 16 inches, diameter of cross section 1·1 inches. 5. Outer diameter 4.8 inches, inner 4'2 inches. 6. Inner diameter 12.3 inches, diameter of cross section 3.2 inches. 7. Outer diameter 19 inches, diameter of cross section 3 inches. 8. Outer boundary 15 inches, circumference of cross section 16 inches. 9. The volume of a ring is 800 cubic inches, the radius of the cross section is 2 inches: find the length of the ring. 10. The volume of a ring is 100 cubic inches, and the length is 20 inches: find the inner diameter. (1) The base of a pyramid is a square, each side of which is 3 feet 6 inches, and the height of the pyramid is 3 feet 9 inches. (2) The radius of the base of a cone is 10 inches, and the height of the cone is 18 inches. 1 3 10 × 10 × 3·1416=314.16, × 18 × 314 16 6 × 314·16=1884.96. Thus the volume is about 1884.96 cubic inches. 265. If we know the volume of a pyramid or a cone, and also the area of the base, we can find the height by dividing three times the number which expresses the volume by the number which expresses the area; and similarly, if we know the volume and the height, we can find the area of the base. 266. Examples. (1) The volume of a pyramid is a cubic yard, and the area of the base is 18 square feet: find the height. 3 x 27 9 A cubic yard=27 cubic feet; 42. 18 Thus the height is 4 feet. (2) The volume of a cone is half a cubic foot, and its height is 27 inches: find the area of the base. Half a cubic foot = 864 cubic inches; 3 x 864 =96. Thus the area of the base is 96 square inches. 267. We will now solve some exercises. (1) The base of a pyramid is a square, each side of which measures 10 feet; the length of each of the four edges which meet at the vertex is 18 feet: find the volume. We must determine the height of the pyramid. Let ABCD be the base, and E the vertex of the pyramid. Let EF be the height of the pyramid, that is, the perpendicular from E on the base; then F will be the middle point of the diagonal AC. Now, by Art. 55, we shall find that the number of feet in AC is 102; and thus the number of feet in AF is 5/2. In the right-angled triangle AEF, the hypotenuse AE is 18 feet; and the number of feet in D A E AF is 5 √2; therefore, by Art. 60, the number of feet in EF is the square root of 324-50, that is, the square root of 274, that is, 16 5529454. Hence the volume of the pyramid in cubic feet × 100 × 16.5529454-551.76484. (2) The base of a pyramid is a square, each side of which is 10 feet; the length of the straight line drawn from the vertex to the middle point of any side of the base is 13 feet: find the volume. We must determine the height of the pyramid. Using the same figure as in the preceding Exercise, let G be the middle point of AD. Then EG is 13 feet; and GF is 5 feet: therefore, by Art. 60, the number of feet in EF is the square root of 169–25, that is, the square root of 144, that is, 12. Hence the volume of the pyramid in cubic feet We may observe that EG is sometimes called the slant height of the pyramid. (3) A corner of a cube is cut off by a plane which meets the edges at distances 3, 4, and 5 inches respectively from their common point: find the volume of the piece cut off. The piece cut off is a triangular pyramid: we may take for the base the right-angled triangle the sides of which are 3 and 4 inches respectively, and then the height of the pyramid is 5 inches. Hence the volume of the pyramid in 3 × 4 cubic inches = 1 3 X 2 x5=10. EXAMPLES. XXV. Find in cubic feet and inches the volumes of the pyramids having the following dimensions: 1. Base 7 square feet 102 square inches; height 2 feet 5 inches. 2. Base 14 square feet 96 square inches; height 3 feet 7 inches. 3. Base 20 square feet 120 square inches; height 4 feet 8 inches. 4. Base 23 square fect 21 square inches; height 5 feet 11 inches. Find in cubic feet and decimals the volumes of the triangular pyramids having the following dimensions: 5. Sides of the base 4, 5, and 7 feet; height 6 feet. 6. Sides of the base 7, 9, and 11 feet; height 8 feet. 7. Sides of the base 15, 19, and 20 feet; height 22 feet. 8. Sides of the base 23, 27, and 30 feet; height 24 feet. Find in cubic feet and decimals the volumes of the cones having the following dimensions: 9. Radius of base 2 feet; height 4 feet. 10. Radius of base 3 feet 6 inches; height 5 feet. 12. Radius of base 10 feet; height 10 feet. Find the heights of the pyramids which have the following volumes and bases: 13. Volume 17 cubic feet 363 cubic inches; base 2 square feet 143 square inches. 14. Volume 33 cubic feet 309 cubic inches; base 4 square feet 83 square inches. 15. Volume 91 cubic feet 792 cubic inches; base 9 square feet 21 square inches. 16. Volume 114 cubic feet 1152 cubic inches; base 10 square feet 96 square inches. Find the radii of the bases of the cones which have the following volumes and heights: 17. Volume 4000 cubic inches; height 5 feet. 18. Volume 40 cubic feet; height 5'3 feet. |