12. To find the area of a regular polygon. Multiply half the perimeter of the polygon by the apothem, or perpendicular let fall from the centre on one of the sides, and the product will be the area required (B. V., P. 8). REMARK 1.—The following is the manner of determining the perpendicular when one side and the number of sides of the regular polygon are known: First, divide 360 degrees by the number of sides of the polygon, and the quotient will be the angle at the centre; that is, the angle subtended by one of the equal sides. Divide this angle by 2, and half the angle at the centre will then be known. Now, the line drawn from the centre to an angle of the polygon, the perpendicular let fall on one of the equal sides, and half this side, form a right-angled triangle, in which there are known the base, which is half the side of the polygon, and the angle at the vertex. Hence, the perpendicular can be determined. Ex. 1. To find the area of a reg. ular hexagon, whose sides are 20 feet each. 6)360° 60°, 60° = ACB, the angle at the centre. 30o = ACD, half the angle at centre. Also, CAD = 90° – ACD= 60°; and, AD = 10. Then, sin ACD : 30°, ar. comp. . 0.301030 : sin CAD . 9.937531 :: AD . . 10, . . . 1.000000 : CD . 17.3205 1.288561 Perimeter = 120, and half the perimeter = 60. Then, 60 X 17.3205 = 1039.23, the area. 2. What is the area of an octagon whose şide is 20 ? Ans. 1931.36886. REMARK II.—The area of a regular polygon of any number of sides is easily calculated by the above rule. Let the areas of the regular polygons whose sides are unity, or 1, be calculated and arranged in the following Now, since the areas of similar polygons are to each other as the squares of their homologous sides (B. IV., P. 27), we have, 1 : any side squared :: tabular area : area. Hence, to find the area of any regular polygon, 1. Square the side of the polygon. 2. Then multiply that square by the tabular area set opposite the polygon of the same number of sides, and the product will be the required area. Er. 1. What is the area of a regular hexagon whose side is 20 ? 20° = 400, tabular area = 2.5980762. Hence, 2.5980762 X 400 = 1039.2304800, as before. 2. To find the area of a pentagon whose side is 25. Ans. 1075.298375. 3. To find the area of a decagon whose side is 20. Ans. 3077.68352. 13. To find the circumference of a circle when the diame. ter is given, or the diameter when the circumference is given. Multiply the diameter by 3.1416, ant the product will be the circumference; or, divide the circumference by 3.1416, and the quotient will be the diameter. It is shown (B. V., P. 16, s. 1), that the circumference of a circle whose diameter is 1, is 3.1415926, or 3.1416. But, since the circumferences of circles are to each other as their radii or diameters, we have, by calling the diameter of the second circle d, 1 :d :: 3.1416 : circunference, hence, d x 3.1416 = circumference. , circumference Hence, also, d= 3.1416 Ex. 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. 3709. 4. What is the diameter of a circle whose circumference is 6850 ? Ans. 2180.41. 14. To find the length of an arc of a circle containing any number of degrees. Multiply the number of degrees in the given arc by 0.0087266, and the product by the diameter of the circle. 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 : 3.1416 degree: that is, = 0.0087266 = arc of one degree 360 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, the product is the length of the arc for any diameter whatever. 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. " Ex. 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. 15. To find the area of a circle. 1. Multiply the circumference by half the radius (B. V., P. 15). Or, 2. Multiply the square of the radius by 3.1416 (B. V., P. 16). Ex. 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 34 feet? Ans. 1.069016. 4. What is the area of a circle whose circumference is 12 feet? Ans. 11.4591. 16. To find the area of a sector of a circle. 1. Multiply the arc of the sector by half the radius (B. V., P. 15, c). Or, 2. 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. Ex. 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. 17. To find the area of a segment of a circle. 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. Then add these two together for the answer when the segment is greater than a semicircle, and subtract the triangle from the sector when it is less. Ex. 1. To find the area of the segment ACB, its chord AB being 12, and the radius EA, 10 feet. EA 10 ar. comp. 9.000000 : AD 6 . : 0.778151 :: sin D 90° . . 10.000000 : sin AED 36° 52' = 36.87 9.778151 2 73.74 = the degrees in the arc ACB. Then, 0.0087266 x 73.74 × 20 = 12.87 = arc ABC nearly. 64.35 = area EACB. Again, V EAP – AD2 = V 100 – 36 = V 64 = 8 = ED. and, 6 x 8 = 48 = the area of the triangle EAB. Hence, sect. EACB – EAB= 64.35 – 48 = 16.35 = ACB. 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. 18. To find the area of a circular ring: that is, the area included between the circumferences of two circles which have a common centre. 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. For the area of the larger is . . . RRT, and of the smaller . . . . . . go? . Their difference, or the area of the ring, is (R“ — 2)#. Er. 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. |