By the first rule, as 7:22::20: 62, the answer. Ex. 2. If the circumference of the earth be 25,000 miles, what is its diameter? By the 2d rule, as 3.1416:1::25000:79573, nearly the diameter. Corol. To find the circumference from the radius, multiply the latter by 6.2832. PROBLEM VIII. To find the length of any arc of a circle. Multiply the degrees in the given arc by the radius of the circle, and the product, again, by the decimal 01745, 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. Ans. 2.1231. PROBLEM IX. To find the area of a circle. RULE I.† Multiply half the circumference by half the diameter. Or multiply the whole circumference by the whole diameter, and take of the product. 14 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 (th. 71), 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 343.1428, &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::10: 01745, &c., the length of the arc of 1 degree. Hence the number 01745, multiplied by any number of degrees, will give the length of the arc of those degrees. And, because the circumferences and arcs are as the diameters, or 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 r 01745 × degrees in the arc, which is the length of that arc, as in the rule. This first rule is proved in the Geometry, theor. 73. And the second rule is derived from formula (3), sehol., of the same theorem, π being 3.1416. Rule 3 is derived from the same formula, observing that r2 = 4ď2, and π = ·7854. == RULE II. Square the radius, and multiply that square by 3.1416. RULE III. Square the diameter, and multiply that square by the decimal 7854, for the area. Ex. 1. To find the area of a circle whose diameter is 10, and its circumference 31.416. 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. Find the area of a circle whose radius is 10. Ans. 314 16. PROBLEM X. To find the area of a circular ring or space included between two concentric circles. Take the difference between the areas of the two circles, as found by the last problem. Or, since circles are as the squares of their diameters, 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 dif ference 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 10-6=4, the difference. * In this last method logarithms may be advantageously applied. 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 one quarter of the product. The reason for which is, that the sector bears the same proportion to the whole circle that its arc does to the whole circumference. RULE II. 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. area. 2. By the 2d Rule. First, 7854 × 32=7′0686, the area of the whole circle. Then, as 360:18::7·0686:·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'. PROBLEM XII. Ans. 804-4017. 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 take the sum of these two for the answer, when the segment is greater than a semicircle: or take their difference for the answer, when it is less than a semicircle; as is evident by inspection. A C B Ex. 1. To find the area of the segment ACBDA, its chord AB being 12, and the radius AE or CE 10. *First, AD AE= sin. angle D= sin. 36° 52'36-87 degrees, the degrees in the angle AEC or arc AC. Their double, 73-74, are the degrees in the whole arc ACB. Now 7854 × 400314·16, the area of the whole circle. Therefore 360°: 73-74:: 314 16:64 3504, area of the whole sector ACBE. = Therefore, AD × DE 6 x 8=48, the area of the triangle AEB. Hence sector ACBE-triangle AEB = 16.3504, area of seg. ACBDA. 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. When the quotient is not found exactly in the *This requires a knowledge of plane trigonometry. The truth of this rule depends on the principle of similar plane figures, which have the ratio of their like lines (as the height and radius of a segment) equal, and are to one another as the square of their like linear dimensions. The segments in the table are those of a circle whose diameter is 1; and the first column contains the quotients of corresponding heights, or versed sines, divided by the diameter, which are the same for similar segments of all diameters. 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 corresponding to this diameter. 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. 01.00133 11 04701 21 11990 31 20738.41.30319 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 CEDE CD-10-8 2. Hence, by the rule, CD÷ CF=220='1, the tabular height. This being found in the first column of the table, the corresponding tabular area is 04088. Then 04088 × 202 = 04088 × 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. 44.7292. PROBLEM XIII. To measure long irregular figures. Take or measure the breadth in several places at equal distances; then add all these breadths together, |