EXAMPLE 3. Given the length of side DC=7·64235; required the radius D O. 7-64235 × 1.307 = 9.98855145, or 9.99 in. nearly. PROBLEM VI. Having the diameter of a circle given, to find the circumference. ;| or the circumference given, to find the diameter. RULE 1. As 7 is to 22, so is the diameter to the circumference. Or, as 22 is to 7, so is the circumference to the diameter. 2. As 1 is to 3.1416, so is the diameter to the circumference. Or, as 31416 is to 1, so is the circumference to the diameter. EXAMPLE 1. Required the circumference of a circle when the diameter is 23.5. 23.5 × 22 EXAMPLE 2. The circumference of a circle is 739, required the diameter. 739 × 7 22 × 23.5, the diameter. EXAMPLE 3. Required the circumference of a circle whose diameter is 30. 3.1416 × 30 94.248, the circumference. diameter of a circle when the circum EXAMPLE 4. What is the ference is 94.248? To find the length of any arc of a circle. RULE. Subtract the chord of the whole are from eight times the chord of half the arc; and of the remainder is the length of the arc nearly. EXAMPLE. Required the length of the arc ABC; the chord of half the are AB 198, and chord of the whole are A C 34.4. To find the diameter of a circle, by having the chord and versed sine given. RULE. Divide the square of half the chord by the versed sine, to the quotient of which add the versed sine, and the sum will be the diameter. Or, if the sum of the squares of the semichord and versed sine be divided by the versed sine, the quotient will be the diameter of the circle to which that segment corresponds. = EXAMPLE. Given the chord AB 24, and versed sine CD8; required the diameter of the circle CE. Half the chord 12, and 1228 18 +8= 26, the diameter. D C 122+82 Or, 26, as before. E PROBLEM IX. To find the area of an ellipsis, or oval. RULE. Multiply the longest diameter by the shortest, and the product by 7854; the result is the area. An oval is 25 inches by 16:5: what are its superficial contents? 25 x 16'0 412.5 x 7854 323-9775 inches, the area. Note.-Multiply half the sum of the two diameters by 3'1416, and the product is the circumference of the oval or ellipsis. PROBLEM X. To find the area of a parabola, or its segment. RULE. Multiply the base by the perpendicular height, and twothirds of the product is the area. 12? What is the area of a parabola whose base is 20 feet and height Some of the properties of a circle. 1. It is the most capacious of all plane figures, or contains the greatest area within the same perimeter or outline. 2. The areas of circles are to each other as the squares of their diameters, or of their radii. 3. Any circle whose diameter is double that of another, contains four times the area of the other. 4. The area of a circle is equal to the area of a triangle whose base is equal to the circumference, and perpendicular equal to the radius. 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 21 EXAMPLE 1. Required the area of a circle, the diameter being 30.5. 30.52 x 7854730 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 × 57854, 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 112837, 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. 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. 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. 35 EXAMPLE. Required the area of a segment of a circle whose chord A B 48, and versed sine CD 18. 18x 626 11 2682 126 967824; which add to 576, being the square of half the chord 702967824, twice the square root of which is 53 026 × 12; being of the versed sine 636 312, the area. B 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, 48 x 18 x 2 3 576607 189 48 × 2 =60.7; and 6367, the area nearly. 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 + B) To find the area of a circular ring or space included between twoconcentric circles. RULE. Add the inside and outside diameters together, multiply the sun by their difference, and by 7854, and the product will be the area. EXAMPLE. The diameters of two concentric circles, AB and CD, are 10 and 6; required the area of the ring or space contained between them. 10+ 6 × 4 × 785450-2656, the area. PROBLEM XIV. To find the area of an ellipsis. 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 sie whose longer diameter A B12, B shorter diameter CD 9. 13 × 9 × 7854 84 8232, the area. and の Note-Ifhalf 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 cylin der, and the product is the solid content. EXAMPLE 1. Required the convex surface of the cylinder ABCD, whose base A B 32 inches, and perpendicular height B C 6 feet. 3.1416 x 32 x 72 inches 7238 2464 square or su perficial inches, and 7238′2464 ÷ 144 = 50·2658 superficial feet. EXAMPLE 2. Required the solid content, in cubic BA inches and cubic feet, of the cylinder as above. 322 x 7854 × 72 57905 9712 cubic inches, and 57905 9712 1728 83.5104 cubic feet. EXAMPLE 3. Suppose the cylinder ABCD be intended to contain a fluid, and that the sides and bottom are each one inch in thick ness, how many imperial gallons would it contain ? · 32 -230 inches diameter; and 72 1 71 inches deep; 302 × 7854 X 71 then 277.274 181 gallons. Or, 50187 06 003607181, 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 X 2.56 = √1536 11:5379 feet diameter, and 11 53792576895 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. |