Ex. 2. The length of an irregular figure being 84, and the breadths at six equidistant places 17 4, 206, 142, 165, 20*1, 24•4; what is the area? Ans. 1550.64. PROBLEM XIV. To find the Area of an Ellipsis or Oval. area. 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. Ex. 1. Required the area of an llipse 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 carithmetical mean between the extremes, or half the sum of thein ?, and 4 the number of the parts. And the same for any other er of paris whatever. Here Here 20 = 70 gives .284 the quotient or versed sine; to which in the table answers the seg. •18518 then 70 1296260 50 648.13000 the area. 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 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 Ex. 2. The length of an irregular Egare being 34, an breadths at sixes istars places 17 4,0-6, 112, 10:5, 244; what is the area? Ans. 15:. MULTIPLr the longest discreter, or asis, by the sho then multiply the product by the decimal 1854, f area. As appears from cor. 2, theor. S, of the Elli the Conic Sections. Ex. 1. Required the area of an ellipse whose tw are 10 and 50. Ans. 2 Ex. 2. To find the area of the oral whose two ax 24 and is. Ans. 339 PROBLEM XY. To find the Area of an Elliptic Segment. Find the area of a corresponding circular segment, 1 the same height and the same vertical axis or diameter. say, as the said vertical axis is to the other axis, paral. the segment's base, so is the area of the circular seg before found, to the area of the elliptic segment so This rule also comes from cor. 2, theor. 3, of the Ellip: Otheruwe hus. Divide the height of the segment by vertical axis of the ellipse; and find, in the table of circ gments to prob. 12, the circular segment having the a jotient for its versed sine: then multiply all together, of two axes of the ellipse, for the area. the area of the elliptic segment, wl. greeing with the rule: m being th the extremes, or half the sum of the Here 47 amid or Cone. ltiply that area by the the product for the 1 square pyramid, each endicular height 25. Ans. 7500. ingular pyramid, whose side of the base 3. Ans. 38.97117. triangular pyramid, its le three sides of its base Ans. 71.0352. pentagonal pyramid, its of its base 2 feet? Ans. 275276. the hexagonal pyramid, le of its base 6 inches? Ans. 1.38564 feet. ? cone, its height being s base 9 feet. Ans, 22:56093, Note. When the quotient is not found exactly in the table, proportion may be made between the next less and greater area, in the same manner as is done for logarithms, or any other table, Table of the Areas of Circular Segments. 01/.0013311.04701:21•11990:31:2073841|:30319 02.00375|.12 :05339|22:12811:32.21667|.42 31304 •03.00687.13 -06000-23:13646-33 22603:43.32293 "04.01054 -14 06083-24:14494|o34.23547||-4433284 05 01468.15.07387-25 1535435\21498||:45:34276 06.01924:16)•(18111•26•16226|1:36 •25455 -46.35275 •07.024171.171.08853|27|.171091.371.36418||-47-36272 08.029+4.18.09613-28 -18002||:38 27386.4837270 0903502:19:10390-29-18905):39 •28359-49-38270 •10.04088 l.20:11182.301.1981711401:29337||50:39270 Ex. 2. Taking the same example as before, in which are given the chord AB 12, and the radius 10, or diameter 20. And having found, as above, De = 8; then ce DE - CD = 10 8 = 2. Hence, by the rule, CD CF = 2 • 20 = 1 the tabular height. This being sought in the first column of the table, the corresponding tabular area is found = '04088. Then 04088 X 202 = .04088 x 400 = 16.352, the area, nearly the same as before. Ex. 3. What is the area of the segment, whose height is 18, and diameter of the circle 50 ? Ans. 636.375. Ex. 4. Required the area of the segment whose chord is 16, the diameter being 20 ? Ans. 44728 circle whose diameter is 1; and the first column contains the corresponding heights or versed sines divided by the diameter. Thus then, the area of the similar segment, taken from the table, and multiplied by the square of the diameter, gives the area of the segment to this diameter, PROBLEM |