5. The area of a circle is equal to the rectangle of its radius, and a right line equal to half its circumference. 6. The area of a circle is found by squaring the diameter, and multiplying by the decimal 7854; or by multiplying the circumference by the radius, and dividing the product by 2? EXAMPLE 1. Required the area of a circle, the diameter being 30.5. 30.52 × 7854 = 730·618350, the area required. EXAMPLE 2. What is the area of a circle when the diameter is 1? In this case the circumference is 3.1416, half of which is 1.5708, and half of 1 = 5; then 1·5708 × 5 = 7854, the area. Having the area of a circle given, to find the diameter. RULE. As 355 is to 452, so is the area to the square of the dia meter. Or, multiply the square root of the area by 1'12837, and the product will be the diameter. Or, divide the area by the decimal 7854, and extract the square root. EXAMPLE. Required the diameter of that circle whose area is 122.71875. √ 122·71875 × 452 355 12.5, diameter. Or, √122-71875 = 11·077; and 11·077 × 1·12837 = 12·49895, or 12.5, diameter. PROBLEM XI. To find the area of a sector of a circle. RULE. Multiply the length of the arc by the radius of the circle, and half the product will be the area. B A. EXAMPLE Required the area of a sector of a circle whose arc A B C 26'666, and radius BO 16.9. To find the area of a segment of a circle. RULE. Multiply the versed sine by the decimal 626, to the square of the product add the square of half the chord; multiply twice the square root of the sum by of the versed sine, and the product will be the area. EXAMPLE. Required the area of a segment of a circle whose chord A B 48, and versed sine CD = 18. 18 × ·626 = 11·2682 = 126·967824; 11.2682126.967824; which add to 576, being the square of half the chord = 702967824, twice the square root of which is 53 026 × 12; being versed sine = 636·312, the area. of the B D A The following is a near approximate to the preceding rule: To the cube of the versed sine, divided by twice the length of the chord, add of the product of the chord, multiplied by the versed sine; and the sum will be the area of the segment nearly. Take the last example: Versed sine 18, and chord 48, then, = 183 48 × 2 48 x 18 x 2 576 + 60′′7 = 636'7, the area nearly. 3 60.7; and Or, the area of a segment may be found by finding the area of a sector having the same radius as the segment; then deducting the area of the triangle, leaves the area of the segment. PROBLEM XIII. To find the area of a circular ring or space included between two concentric circles. RULE. Add the inside and outside diameters together, multiply the sum by their difference, and by 7854, and the product will be the area. EXAMPLE. The diameters of two concentric circles, A B and CD, are 10 and 6; required the area of the ring or space contained between them. 10 + 6 × 4 × 7854 = 50*2656, the area. PROBLEM XIV. To find the area of an ellipsis. B D CA RULE. Multiply the transverse or longer diameter by the conjugate or shorter diameter, and by 7854, and the product will be the area. EXAMPLE. Required the area of an ellip- B sis whose longer diameter A B = 12, and shorter diameter CD 9. 13 × 9 × 7854 — 84 8232, the area. D C Note. If half the sum of the two diameters be multiplied by 3'1416, the product will be the circumference of the ellipsis. 3'1416 × 21 Thus 12 + 9 = 21, and 36 1384, the circumference. 2 Mensuration of Solids. By solids are meant all bodies, whether solid, fluid, or bounded space, that can be comprehended within length, breadth, and thickness. PROBLEM. I. To find the convex surface and solid content of a cylinder. RULE 1. Multiply the circumference of the base by the height of the cylinder, and the product is the convex surface. RULE 2. Multiply the area of the base by the height of the cylinder, and the product is the solid content. EXAMPLE 1. Required the convex surface of the cylinder A B CD, whose base AB 32 inches, and perpendicular height BC = 6 C D B feet. 3+1416 x 32 x 72 inches: 2 inches = 7238 2464 square or superficial inches, and 7238-2464 ÷ 144 = 50·2658 superficial feet. EXAMPLE 2. Required the solid content, in cubic inches and cubic feet, of the cylinder as above. 322 × 7854 × 72 = 57905 9712 cubic inches, and EXAMPLE 3. Suppose the cylinder ABCD be intended to contain a fluid, and that the sides and bottom are each one inch in thickness, how many imperial gallons would it contain? 32-230 inches diameter; and 72-1=71 inches deep; 302 × 7854 X 71 = 181 gallons. then 277.274 Or, 50187 06 X× ‘003607 = 181, as before. PROBLEM II. To determine the dimensions of any cylindrical vessel, whereby to contain the greatest cubical contents, bounded by the least superficial surface. RULE. Multiply the given cubical contents by 2:56, and the cube root of the product equal the diameter, and half the diameter equal the depth. EXAMPLE. Suppose a cylindrical vessel is to be made so as to contain 600 cubic feet, and of such dimensions as to require the least possible materials by which it is constructed, what must be its depth and diameter? 600 × 2·56 = √1536 = 11·5379 feet diameter, and 11.5379 ÷ 2 = 576895 feet in depth. Note. If the vessel is to be constructed with two ends, then the cube root of four times the solidity divided by 3'1416 equal both the length and diameter, so as to expose the least possible surface, or be composed of the least possible materials, of which to be constructed. PROBLEM III. To find the surface and solid content of a cone or pyramid. RULE 1. Multiply the circumference of the base by the slant height, and half the product will be the slant surface; to which add the area of the base, and the product will be the whole surface. RULE 2. Multiply the area of the base by the perpendicular height, and of the product will be the solid content. EXAMPLE 1. Required the convex surface of a cone whose base A B = 20 inches, and slant height. BD 29.5. 3·1416 × 20 × 29.5 2 D 926-772 square inches, and divided by 144 = 6·435 superficial feet. EXAMPLE 2. Required the solidity of the cone as above, the perpendicular CD being 28 inches. 202 × 7854 × 28 3 1.697 cubic feet. 2932.16 cubic inches, and divided by 1728 = PROBLEM IV. To find the surface of the frustum of a cone or pyramid. RULE. Multiply the sum of the perimeters of the two ends by the slant height, and half the product will be the slant surface; to which add the areas of the two ends, and the product will be the whole surface. EXAMPLE. Required the convex surface of the frustum of a cone ABCD, whose base 20 inches, the slant height BC= 19, AB and top end CD = 11. 31416 × 20 + 3·1416 × 11 × 19 2 =925-2012 square inches, and divided by 144 B. = 6'425 feet nearly. E D A PROBLEM V. To find the solid content of the frustum of a cone. RULE. To the product of the diameters of the two ends add the sum of their squares; multiply this sum by the perpendicular height and by 2618; the product is the solid content. EXAMPLE 1. Required the solid content of the frustum in Problem IV., whose perpendicular E F = 18 inches. 20 × 11 = 220, and 220 + 202 + 11a × 18 × 26183491·8884 cubic inches, and divided by 1728 20208 cubic feet nearly. EXAMPLE 2. Required the content, in imperial gallons, of the inverted frustum of a cone ABCD, whose inner dimensions are 3 feet deep, 18 inches diameter at bottom, and 22 inches diameter at top. To find the solid content of the frustum of a pyramid. RULE. To the sum of the areas of the two ends add the square root of their product; multiply this sum by the perpendicular height, and of the product is the solid content. C EXAMPLE. Required the solid content of the frustum of a pyramid ABCD, whose perpendicular height 24 inches, the area of the base 144 inches, and area of the top end = 64. 144 + 64 — 208, and √144 × 64 96; then 20896 × 24 3 2432 cubic inches, and 1728 =1.4074 cubic feet nearly. PROBLEM VII. To find the solidity of a wedgc. RULE. To the length of the wedge add twice the length of the base; multiply that sum by the height, and by the breadth of the base, and one-sixth of the product will be the solidity. EXAMPLE. Required the content in cubic inches of the wedge ABCDE, whose base ABC = 12 inches long and 4 inches broad, the length of the edge DE 10 inches, and perpendicular height r E 20 inches. To find the convex surface and solid content of a sphere or globe. RULE 1. Multiply the square of the diameter by 3·1416; the product will be the convex superficies. |