until the difference between the inscribed and circumscribed polygons is less than any given surface (P. 10). Since the circle lies between the polygons, it will differ from either polygon by less than the polygons differ from each other: and hence, in so far as the figures which express the areas of the two polygons agree, they will be the true figures to express the area of the circle. We have subjoined the computation of these polygons, carried on till they agree as far as the seventh place of decimals. The approximate area of the circle, we infer, therefore, is equal to 3.1415926. Some doubt may exist perhaps about the last decimal figure, owing to errors proceeding from the parts omitted; but the calculation has been carried on with an additional figure, that the final result here given might be absolutely correct even to the last decimal place. The number generally used, for computation, is 3.1416, a number very near the true area. Scholium 1. Since the inscribed polygon has the same number of sides as the circumscribed polygon, and since the two polygons are regular, they will be similar (P. 1): and, therefore, when their areas approach to an equality with the circle, their perimeters will approach to an equality with the circumference. Scholium 2. That magnitude to which a varying magnitude approaches continually, and which it cannot pass, is called a limit. Having shown that the inscribed and circumscribed polygons may be made to differ from each other by less than any given surface (P. 10), and since each differs from the circle less than from the other polygon, it follows that the circle is the limit of all inscribed and circumscribed polygons, resulting from continued bisections, and that the circumference is the limit of their perimeters. Hence, no sensible error can arise in supposing that what is true of a polygon resulting from an indefinite number of bisections of the subtended arcs, is also true of its limit, the circle. Indeed, the circle is but a polygon of an infinite number of sides. PROPOSITION XIII. THEOREM. 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 us designate the circumference of the circle whose radius is CA by circ. CA; and its area, by area CA: it is then to be shown that circ. CA : circ. OB :: CA: OB, and that Inscribe within the circles two regular polygons of the same number of sides. Then, whatever be the number of sides, their perimeters will be to each other as the radii CA and OB (P.9). Now, if the arcs subtending the sides of the polygons be continually bisected, the perimeter of each polygon will continue to approach the circumference of the circumscribed circle, and when we pass to the limit (P. 12, s. 2), we shall have circ. CA : circ. OB :: CA : OB. Again, the areas of the inscribed polygons are to each other as CA to OB (P. 9). But when the number of sides of the polygons is increased, and we pass to the limit; we shall have area CA : area OB CA: OB2. :: Cor. 1. It is plain that the limit of any portion of the perimeter of an inscribed regular polygon lying between the vertices of two angles, is the corresponding arc of the circumscribed circle. Thus, the limit of the perimeter intercepted between G and E is the arc GFE. Cor. 2. If we multiply the antecedent and consequent of the second couplet of the first proportion by 2, and of the second by 4, we shall have that is, the circumferences of circles are to each other as their diameters, and their areas are to each other as the squares of their diameters. PROPOSITION XIV. THEOREM. Similar arcs are to each other as their radii: and similar sectors are to each other as the squares of their radii. Let AB, DE, be similar arcs, and ACB, DOE, similar sectors: then will angle C 4 right angles 4 right angles :: AB : AB circ. CA, and, angle 0 : 4 right angles :: DE : circ. OD; hence (B. II., P. 4, C.), AB DE :: circ. CA: circ. OD; : but these circumferences are as the radii AC, DO (P. 13); hence, AB : DE :: CA OD. : For a like reason, the sectors ACB, DOE, are to each other as the whole circles: which again are as the squares of their radii (P. 13); therefore, CA2: OD2. sect. ACB : sect. DOE :: CA PROPOSITION XV. THEOREM. The area of a circle is equal to the product of half the radius by the circumference. Let ACDE be a circle whose centre is 0 and radius OA: then will area OA=10AX circ. OÀ. For, inscribe in the circle any regular polygon, and draw OF perpendicular to one of its sides. The area of A F D the polygon is equal to OF, mul- the area becomes A F E area of the polygon is the area of the Passing to the limit, the expression for area OA=10A×circ. OA; consequently, the area of a circle is equal to the product of half the radius by the circumference. Cor. The area of a sector is equal to the arc of the The area of a circle is equal to the square of the radius multiplied by the ratio of the diameter to the circumference. Let the circumference of the circle whose diameter is unity be denoted by : then, since the diameters of circles are to each other as their circumferences (P. 13, c. 2), T will denote the ratio of any diameter to its circumference. We shall then have 1 : T therefore, :: 2CA Multiplying both members by CA, we have 2 1CA×circ. CA= ̃× CA3, |