and of similar Examples, we can see that the following theorem is true: a vessel of given capacity is to be made in the form of a rectangular parallelepiped on a square base; if there is to be no lid the internal surface will be least when the height is half the length. Thus, with the view of saving material, the most advantageous shape is that in which the height is half the length. In like manner from the results of Examples 42... 46, and of similar examples, we learn that if there is to be a lid the cube is the most advantageous shape. And suppose that we have to make a vessel in the form of a rectangular parallelepiped on a square base, out of a given quantity of material: then if there is to be no lid the capacity will be greatest when the height is half the length; and if there is to be a lid the capacity will be greatest when the vessel is a cube. EXAMPLES. XXXIII. Find the area of the whole surface of the cubes which have the following lengths: 1. 2 feet 6 inches. 3. 5 feet 10 inches. 2. 3 feet 8 inches. 4. 6 feet 7 inches. Find the area of the whole surfaces of rectangular parallelepipeds which have the following dimensions: 5. 2 feet 6 inches, 3 feet, 5 feet. 6. 2 feet 4 inches, 3 feet 6 inches, 4 feet. 7. 2 feet 8 inches, 3 feet 2 inches, 4 feet 10 inches. 8. 2 feet 11 inches, 3 feet 7 inches, 5 feet 2 inches. Find the area of the whole surface of right triangular prisms having the following dimensions: 9. Sides of the base, 3, 4, and 5 feet; height 8 feet. 10. Sides of the base 8, 15, and 17 feet; height 10 feet. 11. Sides of the base 1 foot 4 inches, 2 feet 1 inch, 3 feet 3 inches; height 7 feet 6 inches. 12. Sides of the base 2 feet 1 inch, 2 feet 9 inches, 4 feet 4 inches; height 8 feet. 13. Find the area of the whole surface of a pyramid on a square base; each side of the base is 2 feet 7 inches, and the length of the straight line drawn from the vertex to the middle point of any side of the base is 3 feet 5 inches. 14. Find the area of the whole surface of a pyramid on a square base; each side of the base is 3 feet 4 inches, and the length of the straight line drawn from the vertex to the middle point of any side of the base is 5 feet 8 inches. 15. Find the area of the whole surface of a pyramid on a square base; each side of the base is 3 feet 4 inches, and each of the other edges is 8 feet 5 inches. 16. Find the area of the whole surface of a pyramid on a square base; each side of the base is 28 feet, and each of the other edges is 16 feet 1 inch. 17. Find the area of the whole surface of a pyramid on a square base, having its other faces equal; each side of the base is 17 feet 6 inches, and the height of the pyramid is 17 feet 4 inches. 18. Find the area of the whole surface of a pyramid on a square base, having its other faces equal; each side of the base is 29 feet 2 inches, and the height of the pyramid is 24 feet. 19. Find the area of the whole surface of a frustum of a pyramid; the ends are squares, the sides of which are 2 feet and 3 feet respectively, and the distance between the parallel sides of each trapezoidal face is 6 inches. 20. Find the area of the whole surface of a frustum of a pyramid; the ends are squares, the sides of which are 2 feet 3 inches and 4 feet 9 inches respectively, and the distance between the parallel sides of each trapezoidal face is 18 inches. 21. Find the area of the whole surface of a frustum of a pyramid; the ends are squares, the sides of which are 3 feet 4 inches and 3 feet 10 inches respectively; and each of the remaining edges is 5 inches. 22. Find the area of the whole surface of a frustum of a pyramid; the ends are squares, the sides of which are 3 feet 2 inches and 4 feet respectively; and each of the remaining edges is 13 inches. 23. Find the area of the whole surface of a prismoid; the ends are rectangles; one measures 7 feet by 6 feet, and the corresponding dimensions of the other are 4 feet 6 inches and 4 feet 10 inches; each of the remaining edges is 25 inches. 24. The four faces of a triangular pyramid are equilateral triangles, the edge of each being 10 feet: find the area of the whole surface. 25. Find the area of the whole surface of a pyramid on a rectangular base which measures 4 feet 6 inches by 6 feet 8 inches, each of the remaining edges being 6 feet. 26. The area of the whole surface of a cube is 7 square feet 6 square inches: find the volume. 27. The dimensions of a rectangular parallelepiped are 3, 7, and 9 feet: find the edge of the cube of equivalent surface. 28. The edge of a wedge is 12 inches, the length of the base is 10 inches, and its breadth 2 inches; each of the other sides of the trapezoidal faces is 25 inches: find the area of the whole surface. 29. The edge of a wedge is 14 inches, the length of the base is 6 inches, and its breadth 2 inches; each of the other sides of the trapezoidal faces is 9 inches: find the area of the whole surface. 30. Find the area of the whole surface of a right prism, the ends of which are regular hexagons, each edge of the solid being 2 feet. 31. The base of a pyramid is a regular octagon, each side being 4 feet; each of the other edges of the pyramid is 12 feet 1 inch: find the area of the surface excluding the base. 32. The dimensions of a rectangular parallelepiped are 18 inches, 16 inches and 6 inches: find the area of the whole surface of a cube of equivalent volume. 33. The length, breadth, and height of a rectangular parallelepiped are respectively 8, 18 and 21 inches: find its surface. Also find the surface of a rectangular parallelepiped of the same height and volume on a square base. If two rectangular parallelepipeds have the same height and volume, and one of them have a square base, the whole surface of this will be less than the whole surface of the other verify this statement by comparing the surfaces of the following rectangular parallelepipeds with the surface of others having respectively the same volume, and height, and square bases: 34. Base 3 feet by 4 feet; height 5 feet. 35. Base 3 feet by 7 feet; height 9 feet. . 36. Base 8 feet by 15 feet; height 19 feet. The following examples involve the extraction of the cube root: A vessel without a lid in the form of a rectangular parallelepiped on a square base is to be made to hold 1000 cubic inches: find in square inches the area of the whole external surface in the following cases: 37. The height equal to the length. 38. The height equal to twice the length. 41. The height equal to one-third of the length. A vessel with a lid in the form of a rectangular parallelepiped on a square base is to made to hold 1000 cubic inches; find in square inches the area of the whole internal surface in the following cases: 42. The height equal to the length. 43. The height equal to twice the length. 44. The height equal to half the length. 45. The height equal to three times the length. 46. The height equal to one-third of the length. XXXIV. RIGHT CIRCULAR CYLINDER. 318. The surface of a right circular cylinder consists of two circular ends, and another portion which we shall call the curved surface. 319. Let ABCD be a rectangle. Cut it out of paper or cardboard. Then let it be bent until the edge AB just comes into contact with the edge DC. It is easy to see that by proper adjustment, we can thus obtain a thin cylindrical shell; AB becomes the height of the shell, and BC the circumference of the B base. Hence it will follow that the curved surface of a right circular cylinder is equal to a rectangle one dimension of which is the height of the cylinder, and the other dimension the circumference of the base of the cylinder: thus we obtain the Rule which we shall now give. 320. To find the area of the curved surface of a right circular cylinder. RULE. Multiply the circumference of the base by the height of the cylinder. 321. Examples. (1) The radius of the base of a right circular cylinder is 3 feet, and the height is 2 feet: find the area of the curved surface. The circumference of the base in feet 2×3×3·1416 = Thus the area of the curved surface is about 47.124 square feet. |