To find the area of a regular polygon. 101. Let AB, denoted by s, represent one side of a regular polygon, whose centre is C. Draw CA and CB, and from C draw CD perpendicular to AB. Then will CD be the apothem, and we shall have AD BD. D A Denote the number of sides of the polygon by C n; then 360° will the angle ACB, at the centre, be equal to กน (B. V., Page 138, D. 2), and the angle ACD, which is half In the right-angled triangle ADC, we shall have, Formula (3), Art. 37, Trig., But CAD, being the complement of ACD, we have, tan CAD = cot ACD; hence, CD is cot 180° a formula by means of which the apothem may be computed. But the area is equal to the perimeter multiplied by half the apothem (Book V., Prop. VIII.): hence the following RULE. Find the apothem, by the preceding formula; multiply the perimeter by half the apothem; the product will be the area required. EXAMPLES. 1. What is the area of a regular hexagon, each of whose sides is 20 ? We have, The perimeter is equal to 120: hence, denoting the area by Q, 2. What is the area of an octagon, one of whose sides is 20 ? Ans. 1931.36886. The areas of some of the most important of the regular polygons have been computed by the preceding method, on the supposition that each side is equal to 1, and the results are given in the following 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 T's2; Multiply the corresponding tabular area by the square of the given side; the product will be the area required. EXAMPLES. 1. What is the area of a regular hexagon, each of whose sides is 20? We have, T = 2.5980762, and s2 =400 hence, 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? Ans. 24884.6136. 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, n x 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. XII., C. 1: 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? 4.712364 ft. Ans. 2. What is the length of an arc of 12° 10′, or 121°, the diameter being 20 feet? Ans. To find the area of a circle. 2.123472 ft. 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. Ans. 11.4595. 3. What is the area of a circle whose circumference is 12 feet ? 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 arc by the radius; or, II. Find the area of the whole circle, by the last rule; then write the proportion, as 360 is to the number of degrees in the sector, so is the area of the circle to the area of the sector. EXAMPLES. 1. Find the area of a circular sector, whose are contains 18°, the diameter of the circle being 3 feet. 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° 29', and radius 25 feet. Ans. 804.3986 sq. ft. To find the area of a circular segment. 107. Let AB represent the chord corresponding to the two segments ACB and AFB. Draw AE and BE The segment ACB is equal to the sector EA CB, minus the triangle AEB. The segment AFB is equal to the sector EAFB, plus the triangle AEB. Hence, we have the following RULE. F 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 take their sum when the segment is greater than a semicircle; the result will be the area required. |