The areas of similar polygons are to each other as the squares of their homologous sides (Book IV., Prop. XXVII.). Denoting the area of a regular polygon whose side is s by Q, and that of a similar polygon whose side is 1 by T, the tabular area, we have Q: T :: s2: 12; = Q Ts2; hence, the following RULE.-Multiply the corresponding tabular area by the square of the given side; the product will be the area required. Examples. What is the area of a regular hexagon, each of whose sides is 20? We have T = 2.5980762, and s2 400: hence, Q = 2.5980762 × 400 = 1039.23048, Ans. 2. Find the area of a pentagon, whose side is 25. Ans. 1075.298375. 3. Find the area of a decagon, whose side is 20. Ans. 3077.68352. To find the circumference of a circle, when the diameter is given. 102. From the principle demonstrated in Book V., Prop. XVI., we may write the following RULE.-Multiply the given diameter by 3.1416; the product will be the circumference required. Examples. 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 ? 24884.6136. Ans. To find the diameter of a circle, when the circumference is given. 103. From the preceding case, we may write the following RULE.-Divide the given circumference by 3.1416; the quotient will be the diameter required. Examples. 1. What is the diameter of a circle, whose circumference is 11652.1944? Ans. 3709. 2. What is the diameter of a circle, whose circumference is 6850? Ans. 2180.41. To find the length of an arc containing any number of degrees. 104. The length of an arc of 1°, in a circle whose diameter is 1, is equal to the circumference, or 3.1416, divided by 360; that is, it is equal to 0.0087266: hence, the length of an arc of n degrees will be nx 0.0087266. To find the length of an arc containing n degrees, when the diameter is d, we employ the principle demonstrated in Book V., Prop. XIII., C. 2: hence, we may write the following RULE -Multiply the number of degrees in the arc by .0087266, and the product by the diameter of the circle; the result will be the length required. Examples. 1. What is the length of an arc of 30 degrees, the diameter being 18 feet? Ans. 4.712364 ft. 2. What is the length of an arc of 12° 10′, or 124°, the diameter being 20 feet? Ans. 2.123472 ft. To find the area of a circle. 105. From the principle demonstrated in Book V., Prop. XV., we may write the following RULE.-Multiply the square of the radius by 3.1416; the product will be the area required; Examples. 1. Find the area of a circle, whose diameter is 10 and circumference 31.416. Ans. 78.54. 2. How many square yards in a circle whose diameter is 3 feet? Ans. 1.069016. 3. What is the area of a circle whose circumference is 12 feet? Ans. 11.4595. To find the area of a circular sector. 106. From the principle demonstrated in Book V., Prop. XIV., C. 1 and 2, we may write the following RULE.-I. Multiply half the length of the arc by the radius; or, II. Find the area of the whole circle, by the last rule; then write the proportion, 360 is to the number of degrees in the arc of the sector, as the area of the circle is to the area of the sector. Examples. 1. Find the area of a circular sector, whose arc contains 18°, the diameter of the circle being 3 feet. Ans. 0.35343 sq. ft. 2. 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° 291 and radius 25 feet. Ans. 804.3986 sq. ft. To find the area of a circular segment. 10%. Let AB represent the chord corresponding to the two segments ACB and AFB. Draw AE and BE. The seg ment ACB is equal to the sector EACB, minus the triangle AEB. The segment AFB is equal to the sector EAFB, plus the triangle AEB. Hence, we have the following D RULE.-Find the area of the corresponding sector, and also of the triangle formed by the chord of the segment and the two extreme radii of the sector; subtract the latter from the former when the segment is less than a semicircle, and add the latter to the former when the segment is greater than a semicircle; the result will be the area required. Examples. 1. Find the area of a segment, whose chord is 12 and whose radius is 10. Solving the triangle AEB, we find the angle AEB is equal to 73° 44', the area of the sector EACB equal to 64.35, and the area of the triangle AEB equal to 48; hence, the segment ACB is equal to 16.35. 2. Find the area of a segment, whose height is 18, the diameter of the circle being 50. Ans. 636.4834. 3. Required the area of a segment, whose chord is 16, the diameter being 20. Ans. 44.764. To find the area of a circular ring contained between the circumferences of two concentric circles. 108. Let R and r denote the radii of the two circles, R being greater than r. The area of the outer circle is R2 x 3.1416, and that of the inner circle is rx 3.1416; hence, the area of the ring is equal to (R2 — r2) × 3.1416. Hence, the following RULE. Find the difference of the squares of the radii of the two circles, and multiply it by 3.1416; the product will be the area required. Examples. 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. |