Ex. 2. To find the area of a sector, whose radius is 10, and arc 20. Ans. 100. Ex. 3. Required the area of a sector, whose radius is 25, and its arc containing 147° 29'. Ans. 804-3986. PROBLEM XII. To find the Area of a Segment of a Circle. RULE I. FIND the area of the sector having the same arc with the segment, by the last problem. Find also the area of the triangle, formed by the chord of the segment and the two radii of the sector. Then add these two together for the answer, when the segment is greater than a semicircle: or subtract them when it is less than a semicircle. As is evident by inspection. Ex. 1. To find the area of the segment ACBDA, its chord AB being 12, and the radius AE or CE 10. First, As AE : sin. ▲ D 90° :: AD : sin. 36° 52'36.87 degrees, the degrees in the LAEC or arc Ac. Their double, 73.74, are the degrees in the whole arc ACB. Now 7854 × 400 = 314-16, the area of the whole circle. E Therefore 360°: 73·74 :: 314 16: 64 3504, area of the Again, sector ACBE. AE2— AD2: = ✓100-36√648 = DE. Theref. AD X DE = 6 x 848, the area of the tri angle AEB. Hence sector ACBE seg. ACBDA. RULE II. Divide the height of the segment by the diameter, and find the quotient in the column of heights in the following tablet: Take out the corresponding area in the next colunin on the right hand; and multiply it by the square of the circle's diameter, for the area of the segment * Note. * The truth of this rule depends on the principle of similar plane figures, which are to one another as the square of their like linea dimensions. The segments in the table are those of a which is the same thing, subtract the square of the less diameter from the square of the greater, and multiply their difference by 7854.—Or lastly, multiply the sum of the diameters by the difference of the same, and that product by 7854; which is still the same thing, because the product of the sum and difference of any two quantities, is equal to the difference of their squares. Ex. 1. The diameters of two concentric circles being 1 and 6, required the area of the ring contained between the circumferences. Here 10+ 6 = 16 the sum, and 10 6 = 4 the d Ex. 2. What is the area of the ring, the diameter whose bounding circles are 10 and 20 ? Ans. 23 PROBLEM XI. To find the Area of the Sector of a Circle. RULE I. MULTIPLY the radius, or half the diame half the arc of the sector, for the area. Or, multi whole diameter by the whole arc of the sector, and of the product. The reason of which is the same as first rule to problem 9, for the whole circle. RULE II. Compute the area of the whole circle: 1 as 360 is to the degrees in the arc of the sector, area of the whole circle, to the area of the sector. This is evident, because the sector is proportio: length of the arc, or to the degrees contained in it Ex. 1. To find the area of a circular sector, who. tains 18 degrees; the diameter being 3 feet? 1. By the 1st Rule. First, 3.1416 × 3 = 9.4248, the circumferenc 2. By the 2d Rule. First, 7854 × 327·0686, the area of the wi sector. PROBLEM XIII, To measure long Irregular Figures. TAKE or measure the breadth at both ends, and at se, veral places, at equal distances. Then add together all these intermediate breadths and half the two extremes, which sum multiply by the length, and divide by the number of parts, for the area * Note. If the perpendiculars or breadths be not at equal distances, compute ali the parts separately, as so many trapezoids, and add them all together for the whole area. Or else, add all the perpendicular breadths together, and divide their sum by the number of them for the mean breadth, to multiply by the length; which will give the whole area, not far from the truth. Ex. 1. The breadths of an irregular figure, at five equi distant places, being 8.2, 74, 92, 102, 8'6; and the whole length 39; required the area? 8.2 8.6 35.2 sum. 39 in order be denoted by the corre D F K H a A sponding letters a, b, c, d, e, and the whole length AB by l; then compute the areas of the parts into which the figure is divided by the perpendiculars, as so many trapezoids, by prob. 3, and add them all together. Thus, the sum of the parts is, d + e a + b b+c × GI + X IB 2 2 =(a+b+c+d+ c) x 4 l = (m + b + c + d) il, Ex. 2. The length of an irregular figure being 84, and the breadths at six equidistant places 17 4, 206, 14°2, 16·5, 20°1, 24.4; what is the area? Ans. 1550 64. PROBLEM XIV. To find the Area of an Ellipsis or Oval. MULTIPLY the longest diameter, or axis, by the shortest ; then multiply the product by the decimal 7854, for the As appears from cor. 2, theor. 3, of the Ellipse, in the Conic Sections. area. Ex. 1. Required the area of an ellipse whose two axes are 70 and 50. Ans. 2748.9. Ex. 2. To find the area of the oval whose two axes are 24 and 18. Ans. 339.2928. PROBLEM XV. To find the Area of an Elliptic Segment. FIND the area of a corresponding circular segment, having the same height and the same vertical axis or diameter. Then say, as the said vertical axis is to the other axis, parallel to the segment's base, so is the area of the circular segment before found, to the area of the elliptic segment sought. This rule also comes from cor. 2, theor. 3, of the Ellipse. Otherwise thus. Divide the height of the segment by the vertical axis of the ellipse; and find, in the table of circular segments to prob. 12, the circular segment having the above quotient for its versed sine: then multiply all together, this segment and the two axes of the ellipse, for the area. Ex. 1. To find the area of the elliptic segment, whose height is 20, the vertical axis being 70, and the parallel axis 50. which is the whole area, agreeing with the rule: m being the arithmetical mean between the extremes, or half the sum of them both, and 4 the number of the parts. And the same for any other number of parts whatever. Here Here 2070 gives 284 the quotient or versed sine; to which in the table answers the seg. 18518 Ex. 2. Required the area of an elliptic segment, cut off parallel to the shorter axis; the height being 10, and the two axes 25 and 35. Ans. 162.03. Ex. 3. To find the area of the elliptic segment, cut off parallel to the longer axis; the height being 5, and the axes 25 and 35. Ans. 97.8425. PROBLEM XVI. To find the Area of a Parabola, or its Segment. MULTIPLY the base by the perpendicular height; then take two-thirds of the product for the area. As is proved in theorem 17 of the Parabola, in the Conic Sections. Ex. 1. To find the area of a parabola; the height being 2, and the base 12. Here 2 x 12 = 24. Then 3 of 24 = 16, is the area. Ex. 2. Required the area of the parabola, whose height is 10, and its base 16. Ans. 106. MENSURATION OF SOLIDS. BY the Mensuration of Solids are determined the spaces included by contiguous surfaces; and the sum of the measures of these including surfaces, is the whole surface or superficies of the body. The measure of a solid, is called its solidity, capacity, or content. Solids are measured by cubes, whose sides are inches, or feet, or yards, &c. And hence the solidity of a body is said. to be so many cubic inches, feet, yards, &c, as will fill its capacity or space, or another of an equal magnitude. The |