than CBX circ. CB. This last quantity is, by hypothesis, the measure of the circle of which CA is the radius; consequently the polygon would be less than the inscribed circle which is absurd; it is impossible, then, that the circumference of a circle multiplied by half of the radius should be the measure of a less circle. Therefore the circumference of a circle multiplied by half of the radius is the measure of this circle. 290. Corollary 1. The surface of a sector is equal to the arc of this sector multiplied by half of the radius. For the sector ACB (fig. 168) is to the entire circle, as the Fig. 168 arc AMB is to the entire circumference ABD (125), or as AMB × AC is to ABD × AC. But the entire circle is equal to ABD × AC; therefore the sector ACB has for its measure AMB × AC. 291. Corollary 11. Since the circumferences of circles are as their radii, or as their diameters, calling the circumference of a circle whose diameter is one, we have this proportion; the diameter of a circle 1 is to its circumference, as the diameter 2 CA is to the circumference of a circle whose radius is CA, 1: :: 2 CA: circ. CA; circ. CA2′′ × CA. or hence Multiplying each member by CA, we have or ¿CA × circ. CA = « × CÅ, surf. CA=«× CÂ; therefore, the surface of a circle is equal to the product of the square of the radius by the constant number, which represents the circumference of a circle whose diameter is 1, or the ratio of the circumference to the diameter. In like manner, the surface of a circle, whose radius is OB, is equal to × OB. But TX CA: TX CÂ : « × OB :: CÅ: OB; therefore, the surfaces of circles are to each other as the squares of their radii, which agrees with the preceding theorem. 292. Scholium. Scholium. We have already said, that the problem of the quadrature of the circle consists in finding a square equal in surface to a circle whose radius is known; now we have just shown that a circle is equivalent to a rectangle contained by the circumference and half of the radius, and this rectangle is changed into a square by finding a mean proportional between its two dimensions (243). Thus the problem of the quadrature of the circle reduces itself to finding the circumference, when the radius is known; and, for this purpose, it is sufficient to know the ratio of the circumference to the radius or to the diameter. Hitherto we have not been able to obtain this ratio except by approximation; but the process has been carried so far, that a knowledge of the exact ratio would have no real advantage over the approximate ratio. Indeed, this question, which occupied much of the attention of geometers, when the methods of approximation were less known, is now ranked among those idle questions which engage the attention of such only as have scarcely attained to the first principles of geometry. Archimedes proved that the ratio of the circumference to the diameter is comprehended between 34% and 341; thus 34 or 22 is a value already approaching very near to the number, which we have represented by ; and this first approximation is much in use on account of its simplicity. Metius gave a much nearer value of this ratio in the expression 55. Other calculators have found the value of TM, developed to a certain number of decimals, to be 3,1415926535897932, &c., and some have had the patience to extend these decimals to the hundred and twenty-seventh, and even to the hundred and fortieth. Such an approximation may evidently be taken as equivalent to the truth, and the roots of imperfect powers are not better known. We shall explain, in the following problems, two elementary methods, the most simple, for obtaining these approximations. Fig. 169. PROBLEM. 293. The surface of a regular inscribed polygon and that of a similar circumscribed polygon being given, to find the surfaces of regular inscribed and circumscribed polygons of double the number of sides. Solution. Let AB (fig. 169) be the side of a given inscribed polygon, EF parallel to AB that of a similar circumscribed polygon, C the centre of the circle; if we draw the chord AM, and the tangents AP, BQ, the chord AM will be the side of an inscribed polygon of double the number of sides, and PQ double of PM will be that of a similar circumscribed polygon (277); and, as the different angles of the polygon equal to ACM will admit of the same construction, it is sufficient to consider the angle ACM only, and the triangles here contained will be to each other as the entire polygons. Let A be the surface of the inscribed polygon whose side is AB, B the surface of a similar circumscribed polygon, A' the surface of a polygon whose side is AM, B' the surface of a similar circumscribed polygon. A and B are known, and it is proposed to find A' and B'. 1. The triangles ACD, ACM, the common vertex of which is A, are to each other as their bases CD, CM; moreover, these triangles are as the polygons A and A', of which they are respectively a part; hence A: A':: CD : CM. The triangles CAM, CME, the common vertex of which is M, are to each other as their bases CA, CE; these same triangles are also as the polygons A' and B, of which they are respectively a part; hence that is, the polygon A', one of those which is sought, is a mean proportional between the two known polygons A and B ; consequently AJAX B. 2. On account of the common altitude CM, the triangle CPM is to the triangle CPE as PM is to PE; but, as the line CP bisects the angle MCE (201), PM: PE :: CM: CE :: CD: CA or CM :: A : A' ; hence and also CPM: CPM+ CPE or CME :: A: A+ A'; 2 CPM or CMPA : CME :: 2A : A + A'. But CMPA and CME are to each other as the polygons B' and B, of which they are respectively a part; we have, then, B': B:2A: A+ A'. Now A' has already been determined; and this new proportion will give the determination of B', namely, therefore, by means of the polygons A and B, it is easy to find the polygons A' and B', which have double the number of sides. PROBLEM. 294. To find the approximate ratio of the circumference of a circle to its diameter. Solution. Let the radius of the circle be = 1, the side of the inscribed square will be 2 (270), that of the circumscribed square will be equal to the diameter 2; hence the surface of the inscribed square 2, and that of the circumscribed square = 4. Now, if we make A = 2, and B = 4, we shall find, by the preceding problem, the inscribed octagon A' == 2,8284271, and the circumscribed octagon B' = =- = 3,3137085. Knowing 2+√8 16 thus the inscribed and circumscribed octagons, we can find, by means of them, the polygons of double the number of sides ; we now suppose A= 2,8284271, B 3,3137085, and we shall have = = A' = √AX B = 3,0614674, and B' = 2AX B A+A' =3,1825979. These polygons of 16 sides will serve to find those of 32 sides; and we may proceed in this manner, till there is no difference between the inscribed and circumscribed polygons, at least for the number of decimals to which the calculation is carried, which, in this example, is seven. Having arrived at this point, we conclude that the circle is equal to the last result, for the circle must always be comprehended between the inscribed and circumscribed polygons; therefore, if these do not differ from each other for a certain number of decimals, the circle will not differ from them for the same number. See the calculation of these polygons continued till they give the same result for the seven first decimals. Hence we conclude that the surface of the circle 3,1415926. There might be some doubt with respect to the last decimal, on account of the error arising from the parts neglected; but we have extended the calculation to one decimal more, in order to be assured of the correctness of the above result to the last figure. Since the surface of a circle is equal to the product of the semicircumference by the radius, the radius being 1, the semicircumference will be 3,1415926; or, the diameter being 1, the circumference will be 3,1415926; therefore the ratio of the circumference to the diameter, above denoted by, is equal to 3,1415926. LEMMA. 295. The triangle CAB (fig. 170) is equivalent to the isosceles Fig 170 triangle DCE, which has the same angle C, and of which the side CE equal to CD is a mean proportional between CA and CB. Moreover, if the angle CAB is a right angle, the perpendicular CF, let fall upon the base of the isosceles triangle, will be a mean proportional between the side CA and the half sum of the sides CA, CB. Demonstration. 1. On account of the common angle C, the triangle ABC is to the isosceles triangle DCE as AC × CB is to DC X CE or DC (216); consequently these triangles are ᎠᏟ equivalent, when DC-AC × CB, or when DC is a mean proportional between AC and CB. 2. As the perpendicular CGF bisects the angle ACB, AG: GB:: AC: CB (201), whence, by composition, AG: AG + GB or AB :: AC: AC + CB; but AG: AB :: triangle ACG triangle ACB or 2 CDF; moreover, if the angle A is a right angle, the right-angled triangles ACG, CDF, are similar; whence If we multiply the two terms of the second ratio by AC, the antecedents will become equal, and we shall consequently have 2CF=AC × (AC + CB), or CF=AC × (CCB), |
