Making p equal to 2.8284271, and P equal to 3.3137085, we have, from the same equations, By a continued application of these equations, we find the areas indicated below, Now, the figures which express the areas of the two last polygons are the same for six decimal places; hence, those areas differ from each other by less than one-millionth of the measuring unit. But the circle differs from either of the polygons by lesɛ than they differ from each other. Hence, 12 taken 3.141592 times, expresses the area of a circle whose radius is 1, to less than one millionth of the measuring unit; and by increasing the number of sides of the polygons, we should obtain an area still nearer the true one. Denote the number of times which the square of the radius is taken, by, we have, π × 12 = 3.141592; that is, the area of a circle whose radius is 1, is 3.141592, in which the unit of measure is the square on the radius. Sch. For ordinary accuracy, is taken equal to 3.1416. The circumferences of circles are to each other as their radii, and the areas are to each other as the squares of their radii. Let C and O be the centres of two circles whose radii are CA and OB: then will the circumferences be to each other as their radii, and the areas will be to each other as the squares of their radii. For, let similar regular polygons MNPST and EFGKL be inscribed in the circles: then will the perimeters of these polygons be to each other as their apothems, and the areas will be to each other as the squares of their apothems, whatever may be the number of their sides (P. IX.). If the number of sides be made infinite (P. X. Sch.), the polygons will coincide with the circles, the perimeters with the circumferences, and the apothems with the radii: hence, the circumferences of the circles are to each other as their radii, and the areas are to each other as the squares of the radii; which was to be proved. Cor. 1. Diameters of circles are proportional to their radii: hence, the circumferences of circles are proportional to their diameters, and the areas are proportional to the squares of the diameters. Cor. 2. Similar arcs, as AB and DE, are like parts of the circumferences to which they belong, and similar sectors, as ACR and DOE, are like parts of the circles to which they belong : hence, similar arcs are to each other as their radii, and similar sectors are A B D to each other as the squares of their radii. Scholium. The term infinite, used in the proposition, is to be understood in its technical sense. When it is proposed to make the number of sides of the polygons infinite, by the method indicated in the scholium of Proposition X., it is simply meant to express the condition of things, when the inscribed polygons reach their limits; in which case, the difference between the area of either circle and its inscribed polygon, is less than any appreciable quantity. We have seen (P. XII.), that when the number of sides is 16384, the areas differ by less than the millionth part of the measuring unit. By increasing the number of sides, we approximate still nearer. PROPOSITION XIV. THEOREM. The area of a circle is equal to half the product of its circumference and radius. Let be the centre of a circle, OC its radius, and ACDE its circumference: then will the area of the circle be equal to half the product of the circumference and radius. For, inscribe in it a regular polygon ACDE Then will the area of this polygon be equal to half the pro duct of its perimeter and apothem, whatever may be the number of its sides (P. VIII.). If the number of sides be made infinite, the polygon will coincide with the circle, the perimeter with the circumference, and the apothem with the radius: hence, the area of the ircle is equal to half the product of its circumference and Vadius; which was to be proved. Cor. 1. The area of a sector is equal to half the product of its arc and radius. Cor. 2. The area of a sector is to the area of the circle, as the arc of the sector to the circumference. To find an expression for the area of any circle in terms of its radius. Let C be the centre of a circle, and CA its radius. Denote its area by area CA, its radius by R, and the area of a circle whose radius is 1, by π × 1a (P. XII., S.). Then, because the areas of circles are to each other as the squares of their A radii (P. XIII.), we have, area CA : TX 1' R: 1; area CA = «R2. whence, That is, the area of any circle is 3.1416 times the square of the radius PROPOSITION XVI. PROBLEM. To find an expression for the circumference of a circle, in terms of its radius, or diameter. Let C be the centre of a circle, and CA its radius. Denote its circumference by circ. CA, its radius by R, and its diameter by D. From the last Proposition, we have, circ. CA 2xR, or, circ. CA = «D. That is, the circumference of any circle is equal to 3.1416 times its diameter. Scholium 1. The abstract number, equal to 3.1416, denotes the number of times that the diameter of a circle is contained in the circumference, and also the number of times that the square constructed on the radius is contained in the area of the circle (P. XV.). Now, it has been proved by the methods of Higher Mathematics, that the value of is incommensurable with 1; hence, it is impossible to express, by means of numbers, the exact length of a circumference in terms of the radius, or the exact area in terms of the square described on the radius. We may also infer that it is impossible to square the circle; that is, to construct a square whose area shall be exactly equal to that of the circ.e. Scholium 2. Besides the approximate value of, 3.1416, usually employed, the fractions and are also used to express the ratio of the diameter to the circumference. |