PROBLEM VII. To find the Diameter and Circumference of any Circle, the one from the other. THIS may be done nearly by either of the two following proportions, viz. As 7 is to 22, so is the diameter to the circumference. Or, As 1 is to 3.1416, so is the diameter to the circum-, ference *. Ex. 1. To find the circumference of the circle whose diameter is 20. By the first rule, as 7: 22 :: 20.62%, the answer. Ex. 2. * For, let ABCD be any circle, whose centre is E, and let AB, BC be any two equal arcs. Draw the several chords as in the figure, and join BE; also draw the diameter DA, which produce to F, till BF be equal to the chord BD. T Then the two isosceles triangles DEB, DBF, are equiangular, because they have the angle at D common; consequently DE: DB::DB: DF. But the two triangles AFB, DCB are identical, or equal in all respects, because they have the angle F the angle BDC, being each equal to the angle A DB, these being subtended by the equal arcs AB, BC; also the exterior angle FAB of the quadrangle ABCD, is equal to the opposite interior angle at c; and the two triangles have also the side BF the side BD; therefore the side AF is also equal to the side DC. Hence the proportion above, viz. DE: DB::DB: DFDA AF, becomes DE: DB::DB: 2DE+ DC. Then, by taking the rectangles of the extremes and means, it is DB2 2DE2 + DE. DC. Now, if the radius DE be taken = 1, this expression becomes DB2 DC, and hence the root DB =√2 + DC. That is, if the measure of the supplemental chord of any arc be increased by the number 2, the square root of the sum will be the supplemental chord of half that arc. Now, to apply this to the calculation of the circumference of the circle, let the arc AC be taken equal to of the circumference, and be successively bisected by the above theorem: thus, the chord AC of of the circumference, is the side of the inscribed regular hexagon, and is therefore equal to the radius AE or 1 hence, in the right-angled triangle ACD,.it will be DC= AD-AC Ex. 2. If the circumference of the earth be 25000 miles, what is its diameter ? By the 2d rule, as 3.1416 1: 25000 7957 nearly the diameter. PROBLEM VAD2 AC2= √22 - 12√3 1.7320508076, the supplemental chord of of the periphery. Then, by the foregoing theorem, by always bisecting the arcs, and adding 2 to the last square root, there will be found the supplemental chords of the 12th, the 24th, the 48th, the 96th, &c, parts of the periphery; thus, ✔3.7320508076 = 1·9318516525′ 3.93185165251.9828897227 √3.98288972271.9957178465 √3.9957178465 = 1·9989291743 3 9989291743 = 19997322757 √3.99973227571.9999330678 3.9999330678 = 1′9999832669 ✔3.9999832669 = for the supplemen- I 384 738 T536 of the periphery. Since then it is found that 3'9999832669 is the square of the supplemental chord of the 1536th part of the periphery, let this number be taken from 4, which is the square of the diameter, and the remainder 0.0000167331 will be the square of the chord of the said 1536th part of the periphery, and consequently the root ✔00000167331 = 0·0040906112 is the length of that chord; this number then being multiplied by 1536 gives 6.2831788 for the perimeter of a regular polygon of 1536 sides inscribed in the circle; which, as the sides of the polygon nearly coincide with the circumference of the circle, must also express the length of the circumference itself, very nearly. But now, to show how near this determination is to the truth, let AQP = 0.0040906112 represent one side of such a regular polygon of 1536 sides, and SRT a side of another similar polygon described about the circle; and from the centre E let the perpendicular EQR be drawn, bisecting AP and ST in Q and R. Then since AQ is AP = 0.0020453050, and EA = 1, therefore EQ2 =EA2 - E AQ2=9999958167, and consequently its root gives EQ 9999979084; then because of the parallels AP, ST, it is EQ: ER :: AP: ST:: as the whole inscribed perimeter to the circumscribed one, that is, as 9999979084: 1 :: 6.2831788: 6.2831920 the perimeter of the circumscribed polygon. Now, the circumference of the circle being greater than D 2 the 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. 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. RULE the perimeter of the inner polygon, but less than that of the outer, it must consequently be greater and must therefore be nearly equal to which in fact is true to the last figure, stead of the 4. than 6.2831788, but less than 6.2831920, their sum, or 6·2831854, which should be a 3, in 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 31416 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 :: 1o: 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 01745 × 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, dc÷ 4 is the area, when d 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 3'1416 d ÷4=7854d2, which gives the 2d rule.-Also by the same prob. 7, d is c÷31416; therefore again the same first area dc 4, becomes c 31416 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. Therefore 7854 x 16 x 4 = 7854 × 64 50 2656, 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 32 7·0686, the area of the whole circle. Then, as 360. 18:: 7.0686: 35343, the area of the sector. |