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, 704, 9.2, 10-2, 8:6; and the whole length-39; required the area ? 8:2 35•2 sum. 8.6 39 2) 16.8 sum of the extremes. 3168 8.4 mean of the extremes. 1056 4 ) 1972.8 9.2 343-2 Ans. 10:2 35•2 sum. Ex. D С 6 EG, GI, IB. * This rule is made out as follows: -Let ABCD be the irregular piece ; F K H having the several breadths AD, El', d CH, IK, BC, at the equal distances a E, A Let the several breadths in order be denoted by the corresponding letters a, b, c, d, e, and the whole length A B 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, @ + b btc X IB 2 2 2 atb btc cta x 2 = (12+6+c+d+1) X*1 = (m + b + c d) 11, c+d d te 2 d te 2 2 which Ex. 2. The length of an irregular figure being 84, and the breadths at six equidistant places 17 4, 206, 14.2, 16:5, 20ʻl, 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 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 20 = 70 gives .284 the quotient or versed sine; to which in the table answers the seg. :18518 then 70 12.96260 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 Ans. 97.8425. 25 and 35. 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. 1063 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 The least solid measure is the cubic inch,, other cubes being taken from it according to the proportion in the following table, which is formed by cubing the linear proportions. Table of Cubic or Solid Measures. 1728 cubic inches make 27 cubic feet 166} cubic yards 64000 cubic poles 512 cubic furlongs I cubic foot PROBLEM IS To find the Superficies of a Prism or Cylinder. MULTIPLY the perimeter of one end of the prism by the length or height of the solid, and the product will be the surface of all its sides. To which add also the area of the two ends of the prism, when required *. Or, compute the areas of all the sides and ends separately, and add them all together. Ex. 1. To find the surface of a cube, the length of each side being 20 feet. Ans. 2400 feet. Ex. 2. To find the whole surface of a triangular prism, whose length is 20 feet, and each side of its end or base 18 inches. Ans. 91.948 feet. Ex. 3. To find the convex surface of a round prism, or cylinder, whose length is 20 feet, and the diameter of its base is 2 feet. Ans. 125.664. Ex. 4. What must be paid for lining a rectangular cistern with lead, at 3d. a pound weight, the thickness of the lead being such as to weigh 71b. for each square foot of surface; the inside dimensions of the cistern being as follow, viz. the length 3 feet 2 inches, the breadth 2 feet 8 inches, and depth 2 feet 6 inches ? Ans. 31. 55. 9 d. * The truth of this will easily appear, by considering that the sides of any prism are parallelograms, whose cominon length is the same as the length of the solid, and their breadths taken all together make up the perimeter of the ends of the same. And the rule is evidently the same for the surface of a cylinder. PROBLEM PROBLEM II. To find the Surface of a Pyramid or Cone. MULTIPLY the perimeter of the base by the slant height, or length of the side, and half the product will evidently be the surface of the sides, or the sum of the areas of all the triangles which form it. To which add the area of the end or base, if requisite. Ex. 1. What is the upright surface of a triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet? Ans. 90 feet. Ex. 2. Required the convex surface of a cone, or circular pyramid, the slant height being 50 feet, and the diameter of its başe 8 feet. Ans. 667.59. PROBLEM III. To find the Surface of the Frustum of a Pyramid or Cone, being the lower part, when the top is cut off by a plane parallel to the base. Add together the perimeters of the two ends, and multiply their sum by the slant height, taking half the product for the answer. -As is evident, because the sides of the solid are trapezoids, having the opposite sides parallel. Ex. 1. How many square feet are in the surface of the frustum of a square pyramid, whose slant height is 10 feet; also each side of the base or greater end being 3 feet 4 inches, and each side of the less end 2 feet 2 inches ? Ans. 110 feet. Ex.2. To find the convex surface of the frustum of a cone, the slant height of the frustum being 12 feet, and the circunferences of the two ends 6 and 8.4 feet. Ans. 90 feet. PROBLEM IV. To find the Solid Content of any Prism or Cylinder. Find the area of the base, or end, whatever the figure of it may be; and multiply it by the length of the prism or cylinder, for the solid content *. Note. * This rule appears from the Geom. theor. 110, cor. 2. The same is more particularly shown as follows: Let the annexed rectangular |