PROBLEM VIII. To find the Length of any Arc of a Circle. MULTIPLY the decimal 01745 by the degrees in the given arc, and that product by the radius of the circle, for the length of the arc*. Ex. 1. To find the length of an arc of 30 degrees, the radius being 9 feet. Ans. 4.7115. Ex. 2. To find the length of an arc of 12° 10', or 12°, the radius being 10 feet. PROBLEM IX. To find the Area of a Circle+. Ans. 2.1231. RULE 1. MULTIPLY half the circumference by half the diameter. Or multiply the whole circumference by the whole diameter, and take of the product. 44 RULE the perimeter of the inner polygon, but less than that of the outer, it must consequently be greater than 6.2831788, but less than 6·2831920, and must therefore be nearly equal their sum, or 6-2831854, which in fact is true to the last figure, which should be a 3 instead of the 4. Hence, the circumference being 6.2831854 when the diameter is 2, it will be the half of that, or 3.1415927, when the diameter is 1, to which the ratio in the rule, viz. 1 to 3·1416 is very near. Also the other ratio in the rule, 7 to 22 or 1 to 3 31428 &c. is another near approximation. *It having been found, in the demonstration of the foregoing problem, that when the radius of a circle is 1, the length of the whole circumference is 6.2831854, which consists of 360 degrees; therefore as 360° : 6·2831854 :: 1° : 01745 &c. the length of the arc of 1 degree. Hence the decimal 01745 multiplied by any number of degrees, will give the length of the arc of those degrees. And because the circumferences and arcs are in proportion as the diameters, or as the radii of the circles, therefore as the radius 1 is to any other radius r, so is the length of the arc above mentioned, to 0.745 x degrees in the arc xr, which is the length of that arc, as in the rule. + The first rule is proved in the Geom. theor. 94. And the 2d and 3d rules are deduced from the first rule, in this manner. By that rule, dc4 is the area, when denotes the RULE II. Square the diameter, and multiply that square by the decimal 7854, for the area. RULE III. Square the circumference, and multiply that square by the decimal 07958. Ex. 1. To find the area of a circle whose diameter is 10, and its circumference 31·416. So that the area is 78.54 by all the three rules. Ex. 2. To find the area of a circle, whose diameter is 7, and circumference 22. Ans. 38. Ex. 3. How many square yards are in a circle whose diameter is 3 feet? Ans. 1'069. Ex. 4. To find the area of a circle whose circumference is 12 feet. Ans. 11.4595 PROBLEM X. To find the Area of a Circular Ring, or of the Space included between the Circumferences of two Circles; the one being.. contained within the other. TAKE the difference between the areas of the two circles, as found by the last problem, for the area of the ring.-Or, the diameter, and c the circumference. But, by prob. 7, c is = 3.1416d; therefore the said area de 4, becomes d x 31416d 47854d2, which gives the 2d rule.-Also, by the same prob. 7, disc31416; therefore again the same first area dc4, becomes c31416 x c ÷ 4 = c2 ÷ 12:5664, which is =c2x 07958, by taking the reciprocal of 12:5664, or changing that divisor into the multiplier 07958; which gives the 3d rule. Corol. Hence, the areas of different circles are in proportion to one another, as the square of their diameters, or as the square of their circumferences; as before proved in the Geom. theor. 93. which is the same thing, subtract the square of the less diameter from the square of the greater, and multiply their difference by 7854.-Or lastly, multiply the sum of the diameters by the difference of the same, and that product by 7854; which is still the same thing, because the product of the sum and difference of any two quantities, is equal to the difference of their squares. Ex. 1. The diameters of two concentric circles being 10 and 6, required the area of the ring contained between their circumferences. Here 10+ 6 = 16 the sum, and 10 64 the diff. the area. Ex. 2. What is the area of the ring, the diameters of whose bounding circles are 10 and 20? Ans. 235.62. PROBLEM XI.. To find the Area of the Sector of a Circle. RULE I. MULTIPLY the radius, or half the diameter, by half the arc of the sector, for the area. Or, multiply the whole diameter by the whole arc of the sector, and take of the product. The reason of which is the same as for the first rule to problem 9, for the whole circle. RULE II. Compute the area of the whole circle: then say, as 360 is to the degrees in the arc of the sector, so is the area of the whole circle, to the area of the sector. This is evident, because the sector is proportional to the length of the arc, or to the degrees contained in it. Ex. 1. To find the area of a circular sector, whose arc contains 18 degrees; the diameter being 3 feet? 1. By the 1st Rule. First, 3:1416 × 3 = 9.4248, the circumference... And 360 18: 94248: 47124, the length of the arc. : Then 47124 × 3 ÷ 4 = 1·41372 ÷4=⚫35343, the area. 2. By the 2d Rule. First, 7854 x 327·0686, the area of the whole circle. Then, as 360 18: 70686: 35343, the area of the sector. Ex. 2. To find the area of a sector, whose radius is 10, and arc 20. Ans. 100. Ex. 3. Required the area of a sector, whose radius is 25, and its arc containing 147° 29'. Ans. 804-3986. PROBLEM XII. To find the Area of a Segment of a Circle. RULE I. FIND the area of the sector having the same arc with the segment, by the last problem. Find also 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: or subtract them when it is less than a semicircle.-As is evident by inspection. Ex. 1. To find the area of the segment ACBDA, its chord AB being 12, and the radius AE or CE 10. = First, As AE sin. 4D 90°:: AD: sin. 36° 52′ 36.87 degrees, the degrees in the LAEC or arc AC. Their double, 73.74, are the degrees in the whole arc ACE. Now 7854 × 400 = 314-16, the area of the whole circle. Therefore 360°: 73-74 :: 314-16: 64-3504, area of the sector ACBE. - Again, AE2 - AD2 = √/ 100 36= √64 = 8 = DÉ. Theref. AD X DE = 6 x 8 = 48, the area of the trian RULE II. Divide the height of the segment by the diameter, and find the quotient in the column of heights in the following tablet: Take out the corresponding area in the next column on the right hand; and multiply it by the square of the circle's diameter, for the area of the segment *. Note. * The truth of this rule depends on the principle of similar plane figures, which are to one another as the square of their like linear dimensions. The segments in the table are those of a Note. When the quotient is not found exactly in the table, proportion may be made between the next less and greater area, in the same manner as is done for logarithms, or any other table. Table of the Areas of Circular Segments. 01001331104701211199031207384130319 Ex. 2. Taking the same example as before, in which are given the chord AB 12, and the radius 10, or diameter 20. And having found, as above, DE = 8; then CE - DE CD 10 8 = 2. Hence, by the rule, CDCF = 2 20 = 1 the tabular height. This being found in the first column of the table, the corresponding tabular area is 04088. Then 04088 x 20' 04088 x 400 = 16.352, the area, nearly the same as before. Ex. 3. What is the area of the segment, whose height is 18, and diameter of the circle 50? Ans. 636-375. Ex. 4. Required the area of the segment whose chord is 16, the diameter being 20? Ans. 44 728. circle whose diameter is 1; and the first column contains the corresponding heights or versed sines divided by the diameter. Thus then, the area of the similar segment, taken from the table, and multiplied by the square of the diameter, gives the area of the segment to this diameter. PROBLEM |