Find also the area of the triangle, formed by the chord of the segment and the two radii of the sector. Then take the sum of these two for the answer, when the segment is greater than a semicircle: or take their difference for the answer, 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: AD :: sin. angle D 90°: sin. 36° 52'} = 36.87 degrees, the degrees in the angle AEC or arc AC. Their double, 73-74, are the degrees in the whole arc ACB. Now, 7854 X 400 = 314·16, the area of the whole circle, Therefore, 360°: 73-74 :: 314-16:64-3504, area of the whole sector ACBE Again, AE AD = √100 - 36 = √64 = 8 = DE. Therefore, AD × DE = 6 × 8=48, the area of the triangle AEB. Hence, sector ACBE — triangle AEB = 16·3504, area of seg. ACBDA. RULE 11.-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 column on the right hand; and multiply it by the square of the circle's diameter, for the area of the segment.* 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. 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 = 9 Hence, by the rule, CD ÷ CF = 2 ÷ 20 = ·1, the tabular height. The truth of this rule depends on the principle of similar plane figures, which are to one another as the square of their like linear dimensions. The segments in the table are those of a circle whose diameter is ; 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. This being found in the first column of the table, the corresponding tabular area is -04088. Then 04088 X 20204088 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. 44.7292. PROBLEM XIII. To measure long irregular figures. Take or measure the breadth in several places at equal distances; then add all these breadths together, and divide the sum by the number of them, for the mean breadth; which multiply by the length for the area. * Note 1.-Take half the sum of the extreme breadths for one of the said breadths. Note 2.—If the perpendiculars or breadths be not at equal distances, compute all 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 equidistant places, being 8.2, 7.4, 9.2, 10.2, 8.6; and the whole length 39: required the area? First, (8.28·6) ÷ 2 = 8'4, the mean of the two extremes. Hence, 8.8 X 39 = 343-2, the answer. 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. This rule is made out as follows: Let ABCD be the irregular piece; having the several breadths AD, EF, GH, IK, BC, at the equal distances AE, EG, GI, IB. Let the several breadths in order be denoted by the corresponding letters a, b, c, d, e, and the whole length AB by ; then compute the areas of the parts into which the figure is divided by the perpendi. culars, as so many trapezoids by Problem 3, and add them all together. Thus, the sum of the parts is, a+b × AE+ = (a+b+c+ d + že) × }l = (m+b+c+ d) }l, which is the whole area, agreeing with the rule; m being the arithmetic mean between the extremes, and 4 the number of the parts. And the same for any other number of parts. 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 equal magnitude. The least solid measure is the cubic inch, other cubes being taken from it according to the proportion in the following table: To find the superficies of a prism. 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, length is 20 feet, and diameter of its base is 2 feet. or cylinder, whose Ans. 125 664. Ex. 4.—What must be paid for lining a rectangular cistern with lead, at 2d. a pound weight, the thickness of the lead being such as to weigh 7 lb. 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. £2. 3s. 101d. * Before perusing this chapter the student must make himself master of the treatise on the "Geometry of Solids," which immediately follows the "Geometry of Planes." The principle upon which the rules are founded are explained in the Differential Calculus. The truth of this will easily appear, by considering that the sides of any prism are parallelograms, whose common 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 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 base 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 circumferences of the two ends 6 and 8.4. 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. Ex. 1.—To find the solid content of a cube, whose side is 24 inches. Ans. 13824. Ex. 2.-How many cubic feet are in a block of marble, its length being 3 feet 2 inches, breadth 2 feet 8 inches, and thickness 2 feet 6 inches? Ans. 21 Ex. 3.-How many gallons of water will the cistern contain, whose dimensions are the same as in the last example, when 277-274 cubic inches are contained in one gallon? Ans. 131.566. Ex. 4.-Required the solidity of a triangular prism, whose length is 10 feet, and the three sides of its triangular end or base, are 3, 4, 5 feet. Ans. 60. Ex. 5.—Required the content of a round pillar, or cylinder, whose length is 20 feet, and circumference 5 feet 6 inches. Ans. 48-1459. PROBLEM V. To find the content of any pyramid or cone. Find the area of the base, and multiply that area by the perpendicular height; then take of the product for the content. Ex. 1.-Required the solidity of the square pyramid, each side of its base being 30, and its perpendicular height 25. Ans. 7500. Ex. 2. To find the content of a triangular pyramid, whose perpendicular height is 30, and each side of the base 3. Ans. 38.97117. Ex. 3. To find the content of a triangular pyramid, its height being 14 feet 6 inches, and the three sides of its base 5, 6, 7. Ans. 71-0352. Ex. 4. What is the content of a pentagonal pyramid, its height being 12 feet, and each side of its base 2 feet? Ans. 27.5276. Ex. 5.-What is the content of the hexagonal pyramid, whose height is 6-4, and each side of its base 6 inches ? Ans. 1.38564 feet. Ex. 6.-Required the content of a cone, its height being 10 feet, and the circumference of its base 9 feet. Ans. 22-56093. PROBLEM VI. To find the solidity of the frustum of a cone or pyramid. Add into one sum, the areas of the two ends, and the mean proportional between them, or the square root of their product; and of that sum will be a mean area; which, being multiplied by the perpendicular height or length of the frustum, will give its content. Ex. 1.-To find the number of solid feet in a piece of timber, whose bases are squares, each side of the greater en being 15 inches, and each side of the less end 6 inches; also, the length or perpendicular altitude 24 feet? Ans. 19. Ex. 2.—Required the content of a pentagonal frustum, whose height is 5 feet, each side of the base 18 inches, and each side of the top or less end 6 inches. Ans. 9-31925 feet. Ex. 3.-To find the content of a conic frustum, the altitude being 18, the greatest diameter 8, and the least diameter 4. Ans. 527-7888. Ex. 4.—What is the solidity of the frustum of a cone, the altitude being 25, also the circumference at the greater end being 20, and at the less end 10? Ans. 464.216. Ex. 5.—If a cask, which is two equal conic frustums joined together at the bases, have its bung diameter 28 inches, the head diameter 20 inches, and length 40 inches; how many gallons of wine will it hold? Ans. 79-0613. PROBLEM VIL To find the surface of a sphere, or any segment. RULE I.-Multiply the circumference of the sphere by its diameter, and the product will be the whole surface of it. RULE II.—Multiply the square of the diameter by 3·1416, and the product will be the surface. Note. For the surface of a segment or frustum, multiply the whole circumference by the height of the part required. Ex. 1.-Required the convex superficies of a sphere, whose diameter is 7, and circumference 22. Ans. 154. Ex. 2.-Required the superficies of a globe, whose diameter is 24 inches. Ans. 1809-5616. |