area of the triangular base multiplied by the altitude, must be the solid content of the prism. If the base of the prism be not a triangle, it may be divided into triangles, by drawing lines from any point in the base taken at pleasure, to the several vertices of the plane rectilineal figure forming that base; the triangular prisms erected upon the triangular bases thus furnished, and all having the same altitude as the original prism, unite to compose that prism, so that the solid content of this is the sum of the solid contents of the component triangular prisms, the solid content of one of these is found by multiplying the area of its triangular base by the altitude, and therefore the solid content of the body formed from the union of all, is found by multiplying the sum of the triangular bases by the altitude; in other words, the solid content of the original prism is found by multiplying the area of its base by its altitude, which is the rule for the prism. 2d. For the Cylinder.—Reasoning similar to that employed above, applies to the cylinder, whatever be the outline of the plane figure forming its base. By assuming a point within this base, and drawing from it lines to the boundary of it, we may conceive the base to be divided into as many component triangles as we please; as we increase their number, so of course do we diminish their individual magnitude, and the lengths of their bases, or sides opposite the common vertex. The solid being a cylinder, some at least, of these bases will be curved; if we draw chords to these curved bases, we shall have, in connexion with the curved portion of the original boundary, a polygonal line approaching nearer and nearer to coincidence with the curve line as the before-said bases are multiplied, and thus individually diminished. As there is no limit to this approach, we may, as in the case of the circle, substitute the polygonal for the curve boundary, and therefore we may treat the cylinder as a prism whose base has the same area as the base of the cylinder: and hence the truth of the rule given above. EXAMPLES. 1. Required the solid content of a quadrilateral prism of which the altitude is 19 feet, and the four sides of the base 43, 54, 62, and 38 inches; the diagonal between the first and last of these sides being 70 inches. And (75.5 X 32.5 X 37.5X5·5)=√506085-9375= 711-397173 Inches, in one triangle. (93X39 x 31 x 23)=√2586051=1608-12033 Inches, The solid content of the prism = 306·0474483 Feet 2. The length of an oblique elliptic cylinder is 18 feet, the axes of the elliptic base are 35 and 28 inches, and the axis of the cylinder is inclined to the major axis of the base at an angle of 56°: required the solid content of the cylinder. sine 56°=82904 .. ·82904 × 18=14·9227, the altitude in feet. Also 35 x 28 x 7854-769-692 Inches 5.345 area of the base. Feet, the .. 14.9227×5·34579.763 Feet, the solid content. 3. A regular hexagonal prism is 11 feet 2 inches in length. and each side of the base is 14 inches: required its solid content. Ans. 39-48835 Feet. 4. The diameter of a rolling-stone is 18.7 inches, and its length 4 feet 9 inches: required its solid content. Ans. 9.0595 Feet. 5. The diameter of a well is 3 feet 9 inches, and its depth 45 feet: what did the sinking of it cost at 7s. 3d. per cubic foot? Ans. £6: 13:51. 6. Required the solid content of an oblique circular cylinder whose axis inclines at an angle of 60°, the length of this axis being 25 feet, and the diameter of the base 30 inches. Ans. 106 F. 4791⁄21⁄2 I. 7. Required the solid content of an upright elliptic cylinder 24 feet high, the axes of the base being 32 and 24 inches. 4 ns. 100 F. 9171. 8. A regular decagonal prism is 30 feet high, and its perimeter is 4 feet 2 inches: required its solidity and surface. Ans. Solidity, 40-074 Feet. Surface, 127.672 Feet. 9. The circular base of an oblique cylinder is 15 inches in circumference, its slant height is 5 inches, and the inclination of its axis to the horizon is 64° 45': what is its solid content? Ans. 80.97 Inches. 10. The gallery of a church is supported by 16 regular octagonal prisms of wood, each of which is 4 feet 8 inches in perimeter, and 12 feet high: required the solid content of the whole, and what they cost painting at 9d. per square yard. Ans. Solid content, 315.457 Feet. Cost of painting, £3:14:8. 11. What is the surface and solid content of a circular cylinder, the diameter of whose base is 20.75 inches, and its length 4 feet 7 inches. Ans. Surface, 29.595 Feet. Solidity, 10-763 Feet. 12. The length of a hollow iron roller is 4 feet, the diameter of the exterior circumference 2 feet, and the thickness of the metal inches. Required its solid content. Ans. 1.5217 Feet. PROBLEM III.-To find the solid content of a pyramid or cone. RULE. Multiply the area of the base by the altitude, and onethird of the product will be the solid content. The truth of this rule is an immediate consequence from the 7th and 10th Propositions of Euclid's Twelfth Book, where it is proved that a pyramid is the third part of a prism of the same base and altitude, and that a cone is the third part of a cylinder of the same base and altitude. The rule for the cone may be inferred from that for the pyramid, for the pyramid merges into a cone when the faces of the former are indefinitely narrowed. To find the surface of a pyramid.—Find the areas of all the triangles which form its sides or faces, and to the sum of these add the area of the base. If the base be a regular polygon and the pyramid upright, the triangular faces will all be equal; the perpendicular distance of the vertex from the base of either of these triangles will be the slant height of the pyramid. The area of each triangle will therefore be obtained by multiplying its base by half this slant height, so that the sum of the areas will be found by multiplying the perimeter of the base of the regular pyramid by half the slant height: this product, added to the area of the base, will give the whole surface. To find the surface of a right cone, we have only to consider the circular base to be replaced by a regular polygon inscribed in it, and to conceive the sides of the polygon to be multiplied in number till the perimeter differs insensibly from the circumference of the circle, when the surface of the pyramid upon the polygon will differ insensibly from the surface of the cone upon the circle. The above rule for the regular upright pyramid, may thus be extended to the upright circular cone; that is, the circumference of the base multiplied by half the slant height, will give the convex surface; to which, if the area of the base be added, the whole surface will be obtained. In the case of the oblique cone, the method of finding the surfacerequiring, as it does, the integral calculus-is too difficult and complicated to be treated of here. NOTE. It is of importance to remember that similar surfaces are to one another as the squares of any two similar dimensions, and that similar solids are to one another as the cubes of any two similar dimensions-similar triangles, for instance, are to one another as the squares of their bases or of their altitudes; and similar pyramids or cones are to one another as the cubes of their slant heights or of their altitudes, or of any two lines similarly situated in their bases, &c. These propositions are proved in Euclid, Book VI. Prop. 20, and Book XII. Props. 8 and 12. In triangles, pyramids, and cones, the portions cut off, by parallels to the base, are always similar to the whole; and the bases of the two similar pyramids or cones, are to one another as the squares of their altitudes. EXAMPLES. 1. Each side of the base of an upright heptagonal pyramid is 15 inches, and the altitude is 13 feet: required the solid content, as also the surface of the pyramid. By the table at page 112, the area of a regular heptagon, whose side is 1, is 3.6339126, and the square of a side of the proposed heptagon is 225 inches or 11 feet. 4) 3.6339126=A 4) ⚫9084782 •2271195 9 2.0440755 A 2) 5.6779881=1A=Area of the base. 13 170339643 28389941 3) 76-6528394 25-5509465 Feet, the solid content. To find the surface, we must first determine the slant height; this will be the hypotenuse of a right-angled triangle of which the perpendicular is the altitude of the pyramid, and the base the perpendicular from the centre of the heptagon upon one of its sides. The corresponding perpendicular in the heptagon whose side is 1, by the table (page 112) is 1·0382608; and therefore the perpendicular in the proposed heptagon is 1.0382608 x 11.297826 feet. And √{(13·5)2 + (1·297826)2}=13·56224, the slant height; and if this be multiplied by 14 feet, and then half the product by 7, we shall have the areas of all the triangular faces of the pyramid. 4) 13.56224 2) 16.95280 5.6780 Area of heptagonal base. 65 0128 Area of the whole surface in Feet. 2. The diameter of the base of an upright circular cone is 26 inches, and the altitude is 16 feet: required the solid content of the surface of the cone. First, to find the area of the base (pages 74, 80). 26.5 26.5 1325 1590 530 702.25 11 2)7724-75 7) 3862-375 55,1-768 221 551.547 Inches, area of base. |