1. What is the circumference of a circle whose diameter is 25? Ans. 78.54. 2. If the diameter of the earth is 7921 miles, what is the circumference? Ans. 24884.6136, 3. What is the diameter of a circle whose circumference is 11652.1904? Ans. 37.09. 4. What is the diameter of a circle whose circumference is 6850? Ans. 2180.41. PROBLEM IX To find the length of an arc of a circle containing any number of degrees. RULE.-Multiply the number of degrees in the given arc by 0.0087266, and the product by the diameter of the circle. 3.1416 Since the circumference of a circle whose diameter is 1, is 3.1416, it follows, that if 3.1416 be divided by 360 degrees, the quotient will be the length of an arc of 1 degree: that is, =0.0087266= arc of one degree to the diameter 1. This being multiplied by the number of degrees in an arc, the product will be the length of that arc in the circle whose diameter is 1; and this product being then multiplied by the diameter, will give the length of the arc for any diameter whatever. 360 REMARK. When the arc contains degrees and minutes, reduce the minutes to the decimal of a degree, which is done by dividing them by 60. 1. To find the length of an arc of 30 degrees, the diameter being 18 feet. Ans. 4.712364. 2. To find the length of an arc of 12° 10′, or 121°, the diameter being 20 feet. Ans. 2.123472. 3. What is the length of an arc of 10° 15', or 101°, in a circle whose diameter is 68? Ans. 6.082396. PROBLEM X. To find the area of a circle. RULE I.-Multiply the circumference by half the radius (Book V. Prop. XII.). RULE II.-Multiply the square of the radius by 3.1416 (Book V. Prop. XII. Cor. 2). 1. To find the area of a circle whose diameter is 10 and circumference 31.416. Ans. 78.54. 2. Find the area of a circle whose diameter is 7 and circumference 21.9912. Ans. 38.4846. 3. How many square yards in a circle whose diameter is 3 feet? Ans. 1.069016. 4. What is the area of a circle whose circumference is 12 feet? Ans. 11.4595. PROBLEM XI. To find the area of the sector of a circle. RULE I.-Multiply the arc of the sector by half the radius (Book V. Prop. XII. Cor. 1). RULE II.-Compute the area of the whole circle: then say, as 360 degrees is to the degrees in the arc of the sector, so is the area of the whole circle to the area of the sector. 1. To find the area of a circular sector whose arc contains 18 degrees, the diameter of the circle being 3 feet. Ans. 0.35343. 2. To find the area of a sector whose arc is 20 feet, the radius being 10. Ans. 100. 3. Required the area of a sector whose arc is 147° 29′, and radius 25 feet. Ans. 804.3986. PROBLEM XII. To find the area of a segment of a circle. RULE.-1. Find the area of the sector having the same arc, by the last problem. 2. Find the area of the triangle formed by the chord of the segment and the two radii of the sector. 3. Then add these two together for the answer when the segment is greater than a semicircle, and subtract them when it is less. 1. To find the area of the segment ACB, its chord AB being 12, and the radius EA, 10 feet. C Then, 0.0087266 × 73.74 × 20 = 12.87=arc ACB, nearly Again, ✓ EA2-AD2=√100-36=√64=8=ED; 6x8 48 the area of the triangle EAB. Hence, sect. EACB-EAB=64,35-48-16.35=ACB. and = 2. Find the area of the segment whose height is 18, the diameter of the circle being 50. Ans. 636.4834. 3. Required the area of the segment whose chord is 16, the diameter being 20. Ans. 44.764. PROBLEM XIII. To find the area of a circular ring: that is, the area included between the circumferences of two circles which have a common centre. RULE.-Take the difference between the areas of the two circles. Or, subtract the square of the less radius from the square of the greater, and multiply the remainder by 3.1416. Their difference, or the area of the ring, is 1. The diameters of two concentric circles being 10 and 6, required the area of the ring contained between their circumferences. Ans. 50.2656. 2. What is the area of the ring when the diameters of the circles are 10 and 20? Ans. 235.62. PROBLEM XIV. To find the area of an ellipse, or oval.* RULE.-Multiply the two semi-axes together, and their product by 3.1416. # Although this rule, and the one for the following problem, cannot be de. monstrated without the aid of principles not yet considered, still it was thought best to insert them, as they complete the rules necessary for the mensuration of planes. 2. Required the area of an ellipse whose axes are 24 and 18. Ans. 339.2928. PROBLEM XV. To find the area of any portion of a parabola. RULE.-Multiply the base by the perpendicular height, and take two-thirds of the product for the required area. 1. To find the area of the parabola ACB, the base AB being 20 and the altitude CD, 18. Ans. 240. A D B 2 Required the area of a parabola, the base being 20 and the altitude 30. Ans. 400. MENSURATION OF SOLIDS. The mensuration of solids is divided into two parts. 1st. The mensuration of their surfaces; and, 2dly. The mensuration of their solidities. We have already seen, that the unit of measure for plane surfaces is a square whose side is the unit of length. A curved line which is expressed by numbers is also referred to a unit of length, and its numerical value is the number of times which the line contains its unit. If, then, we suppose the linear unit to be reduced to a right line, and a square constructed on this line, this square will be the unit of measure for curved surfaces. The unit of solidity is a cube, the face of which is equal to the superficial unit in which the surface of the solid is estimated, and the edge is equal to the linear unit in which the linear dimensions of the solid are expressed (Book VII. Prop. XIII. Sch.). The following is a table of solid measures: 1728 cubic inches 27 cubic feet 44921 cubic feet 282 cubic inches 231 cubic inches 2150.42 cubic inches OF POLYEDRONS, OR SURFACES BOUNDED BY PLANES. PROBLEM I. To find the surface of a right prism. RULE.-Multiply the perimeter of the base by the altitude, and the product will be the convex surface (Book VII. Prop. I.). To this add the area of the two bases, when the entire surface is required. 1. To find the surface of a cube, the length of each side being 20 feet. Ans. 2400 sq. ft. 2. To find the whole surface of a triangular prism, whose base is an equilateral triangle, having each of its sides equal to 18 inches, and altitude 20 feet. Ans. 91.949. 3. What must be paid for lining a rectangular cistern with lead at 2d. a pound, the thickness of the lead being such as to require 7lbs. for each square foot of surface; the inner dimensions of the cistern being as follows, viz. the length 3 feet 2 inches, the breadth 2 feet 8 inches, and the depth 2 feet 6 inches? Ans. 21. 3s. 10ğd. PROBLEM II. To find the surface of a regular pyramid. RULE.-Multiply the perimeter of the base by half_the_ slant height, and the product will be the convex surface (Book VII. Prop. IV.): to this add the area of the base, when the entire surface is required. 1. To find the convex surface of a regular triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet. Ans. 90 sq. ft. 2. What is the entire surface of a regular pyramid, whose slant height is 15 feet, and the base a pentagon, of which each side is 25 feet? Ans. 2012.798. PROBLEM III. To find the convex surface of the frustum of a regular pyramid. RULE.-Multiply the half-sum of the perimeters of the two bases by the slant height of the frustum, and the product wili be the convex surface (Book VII. Prop. IV. Cor.). |