REMARK I. The following is the manner of determining the perpendicular when only 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 equal side of the polygon, and the angle at the vertex. Hence, the perpendicular can be determined. 1. To find the area of a regular hexagon, whose sides are 20 feet each. 6)360° 60°=ACB, the angle at the centre. 30°=ACD, half the angle at the centro Also, CAD=90°—ACD=60°; and AD=10. 'Then, as sin ACD . . . 30°, ar. comp. ... : sin CAD. 60° .: AD.. · 10 : CD . . . 17.3205 Perimeter 120, and half the perimeter =60. 2. What is the area of an octagon whose side 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 calcu lated and arranged in the following or, Hence, also. Now, since the areas of similar polygons are to each other as the squares of their homologous sides (Book IV. Prop. XXVII.), we shall have 12: tabular area :: any side squared area. Or, to find the area of any regular polygon, we have RULE II.-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. 1. What is the area of a regular hexagon whose side is 20? 202-400, tabular area =2.5980762. Hence, 2.5980762 × 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. PROBLEM VIII. To find the circumference of a circle when the diameter is given, or the diameter when the circumference is given. RULE.-Multiply the diameter by 3.1416, and the product will be the circumference; or, divide the circumference by 3.1416, and the quotient will be the diameter. It is shown (Book V. Prop. XIV.), 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, 1d: 3.1416 circumference, d= circumference 3.1416 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. 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. 3.1416 360 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 diam eter, will give 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. 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° eter being 20 feet. 3. What is the length of an arc of 10° cle whose diameter is 68 ? 10′, or 121°, the diamAns. 2.123472. 15', or 104°, in a cirAns. 6.082396. PROBLEM X. To find the area of a circle. RULE_1.-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, oy 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. Then, 0.0087266 × 73.74 × 20=12.87=arc ACB, nearly 5 and Again, ✓ EA2-AD2= √100-36=√64-8=ED; 6×8=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. 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. 64.35 area EACB. 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 (R2-p2)π. 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. 1. Required the area of an ellipse whose semi-axes AE, EC, are 35 and 25. Ans. 2748.9. EG B D 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. |