Note 3.-The circumferences of circles have the same ratio to each other as their radii or diameters. Thus, if the diameter of a circle is double that of another, then its circumference is also double that of the other; and if its diameter is three times as long, then its circumference is three times that of the other; and so on. Note 4.-Concentric circles are those which are drawn from a common centre. And in questions which involve the finding of the space AA' between two concentric circles, when their circumferences are given, we must find the diameters AC and A'C' of the two circles by Rule 2. Then, half the difference of these two diameters is the width AA'. A Example 1.-Find the circumference of a circle whose diameter is 11.27 ft. The circumference diameter x 22=11.27 x 22=247.94 =35.42 ft. Example 2.-The circumference of a circle is 37 ft. 5 in.; find its diameter. Now, 37 ft. 5 in.=449 in. Then the diameter 449+2=449 × 7=2143-14212in. -11 ft. 1012 in. Example 3.-The fencing of a circular plot of land, at 4s. 2d. per yd., costs £59 11s. 8d. ; find its diameter. First, £59 118. 8d.+4s. 2d. gives 286 yards of fencing, which is also the circumference of the circle. Find the diameter of a circle whose circumference is (9) Circumference 12 ft. 10 in. (10) Circumference 36 ft. 8 in. (11) Circumference 61 yds. 0 ft. 4 in. (12) Circumference 6 fur. 24 poles. (13) Circumference 3 ch. 30 lks. (14) Circumference 38.5 ft. (15) Circumference 314-16 ft. (16) The diameter of a carriage wheel is 3 ft.; how many revolutions does it make in traversing one-fourth of a mile ? (17) If a carriage wheel makes 220 revolutions in traversing half a mile, find its diameter. (18) There are two concentric circles: the circumference of the inner circle is 16 ft. 6 in., and of the outer one is 18 ft. 4 in. Find the width of the ring. (19) There are two concentric circles: the circumference of the outer circle is 440 ft., and of the inner one is 330 ft. Find the width of the ring. (20) The walling-in of a circular plot of land, at 12s. 6d. per yd., costs £123 15s.; find its diameter. (21) In raising water from the bottom of a well by means of a wheel, it is found that the wheel, whose diameter is 2 ft. 4 in., makes 30 revolutions in raising the bucket; find the depth of the well. (22**) If the diameter of a well is 3 ft. 9 in., what is mference? (23**) Find the radius of a circle whose perimeter is 100 chains. (24) The minute-hand of a clock is 5 in. long; find the length of the circle that its point will describe in an hour's time. XII. THE AREA OF A CIRCLE. Definitions.-See Problem XI. B Fig. 1. RULES.-(1) To find the area of a circle, when its radius or diameter is given. Multiply the square of the radius by 22; or, multiply the square of the diameter by 1. (2) To find the area of a circle, when its circumference is given. Multiply the square of the circumference by Note 1.-To find the diameter or radius of a circle, when its area is given.-Divide the area by 1, and the square root of the quotient is the diameter; or, divide the area by 22, and the square root of the quotient is the radius. Note 2.-To find the circumference of a circle, when its area is given.-Divide the area by g, and the square root of the quotient is the circumference. Note 3.—The above rules give the area of a circle with sufficient accuracy for all practical purposes; but whenever greater accuracy is desirable, then the following rules may be used for finding the area of a circle:-Multiply the square of the diameter by 7854, and the product is the area; or, multiply the square of the circumference by 07958, and the product is the area. Note 4.-The area of a circle is equal to that of a triangle whose base is equal to the circumference, and whose perpendicular is equal to the radius of the circle. Note 5.-Of all plane figures, the circle is that which contains the greatest area within the same perimeter; that is, if we have a square, a rectangle, a circle, or any other plane figure, all of which have the same perimeter, the circle will contain the greatest area. Note 6.-The diameter of a circle inscribed in a square, that is, of a circle which touches each side of a square ABCD, is EF. A Hence, the diameter of an inscribed circle is equal to a side of B the square. Note 7.-To find the side AB of a square inscribed in a circle ABCD.Since AOB is a right-angled triangle, having two equal sides AO, OB (each of which is half the diameter of the circle), we shall be able to obtain AB, the hypothenuse, by Rule 1, Prob. I. Or thus the area of the square= Prob. IV.)= 2 A B E D F Fig. 2. C Fig. 3. ACX BD (see Note 4, 2 Having thus found the area of the inscribed square, its side will be found by taking the square root of the area. Note 8.-To find the side of a square that is equal in area to a given circle.-In this case, we must first find the area of the given circle; and the square root of this area is the side of the required square. D F E B Note 9.-To find the area between two concentric circles ABC and DEF.-The area between the two circles is the difference between the area of the larger circle ABC and of the smaller circle DEF; and these areas can easily be found by the rules given in this chapter. Or, the area may be found by either of the following rules (a) Multiply the sum of the radii of the two circles by their difference, and the product by 22. (b) Multiply the sum of the outer and inner diameters by their difference, and the product by 14. : Fig. 4. Area of smaller circle larger circle :: 22x0A2: 22 X OD2 ::OA2: OD2. In the same manner, it may be shown that their areas are proportional to their diameters. (a) For instance, if the radius (AO) of the smaller circle is 10 ft., and it is required to find the radius (OD) of the larger circle, whose area is twice that of the smaller circle, we shall have the following proportion : OD2: 0A2:: area of larger circle: area of smaller circle. Substituting the known terms, we have OD2 102: 2:1; therefore OD2=2×102. And also the square roots of these quantities are equal; therefore OD=10 √2=10 × 1·414=14·14+ft. |