D G B с AF: AG::P:Q ; and therefore (V. 11.) AHG: AHD::P:Q. Hence (V. E. cor.) AHG : ADG :: P:P + Q: whence by doubling the antecedents, 2 AHG: ADG::2P: P + Q. But it is evident, that whatever part the triangle ADG is of P', the same part of the polygon Q' is the triangle AHK, which is double of the triangle AHG. Hence the last analogy becomes Qʻ: P':: 2P:P + Q. Now (V. Sup. 2. cor. 1.). the product of the extremes is equal to the product of the means ; and therefore Q' will be computed by dividing twice the product of P and P' by P.+Q; and the mode of finding Q has been pointed out already. Schol. This proposition enables us to find the approximate ratio of the diameter and circumference of a circle. Thus, let the radius be 1, and it is plain that the area of the circumscribed square is 4, and that of the inscribed one 2, being (I. 41.) half of the circumscribed one. Put therefore P = 2 and P' = 4; and by the methods found above, we have the area of the inscribed octagon Q=V8, 16 and the area of the circumscribed octagon Q'= 2 +18 : or by actual computation, Q = 2.8284271, and Q'= 3:3137085. By using these numbers instead of P and P', we should find the areas of the inscribed and circumscribed regular polygons of sixteen sides. From them, in like manner, we should derive the areas of regular polygons of thirty-two sides : and thus, we may proceed as far as we please, each succeeding operation giving the areas of the inscribed and circumscribed polygons of double the number of sides of those employed in finding them. By this means the following table has been calculated, the decimals having been carried out to eight places, and the last rejected, so as to secure their accuracy for seven places. Number of Sides. Inscribed Polygon. Circumscribed Polygon. 4.......... 2.0000000... ........... 4.0000000 8........... ..2.8284271 3:3137085 16...............3.0614674...............3.1825979 .....3.1214451. ..3:1517249 ........3.1441184 3.1416321 3:1416025 ..3:1415951 ....3•1415933 ...3:1415923. ..........3•1415928 .3:1415925...............3.1415927 32768 .3.1415926 32......... 128... ..... ..... 16384..... ......... 3:1415926...... * Now, as the area of the circle is intermediate between the areas of the inscribed and circumscribed polygons, and as the areas of the last two in the table are the same in all the figures to which they are carried out, it follows that the area of the circle must also be 3.1415926: and as, by the last proposition, this is equal to the product of the radius 1, and half the circumference, the half circumference to the radius I, or the whole circumference to the diameter 1, is the same number 3.1415926. This number is usually denoted by 7. PROP. XLI. THEOR. If the diameter of a circle be divided into any two parts, AB, BC, and if semicircles ADB, BEC be described on opposite sides of these, the circle is divided by their arcs into two figures GDE, FED, the boundary of each of which is equal to the circumference of FG ; and which are such that AC: BC :: FG: FED, and AC: AB:: FG: GDE. For (App. I. 38.) the circumferences of circles, and consequently the halves of their circumferences, are to one another as their diameters: therefore AB is to AC, as the arc ADB to AFC, and BC is to AC, as the arc BEC to AFC. Hence (V. 24.) AC is to AC, as the compound arc ADEC to AFC: therefore ADEC is equal to half the circunference; and the entire boundaries of the figures GDE, FED are each equal to the circumference of FG. Again, (XII. 2.) circles, and consequently semicircles, are to one another as the squares of their diameters: therefore AC2 is to AB? as the semicircle AFC to the semicircle ADB. Hence, since (II. 4.) AC? = AB' + 2 AB.BC + BC”, we find by conversion, that AC is to 2 AB.BC + BC, as the semicircle AFC to the remaining space BDAFC; whence, by inversion, 2 AB.BC F D A * This number, when carried out to more places, is found to be 3:14159265358979, &c.; and by the method now pointed out, or by others still easier, especially by series' furnished by the modern analysis, it may be continued to any extent whatever. It cannot be determined, however, with complete precision ; and hence the celebrated problem of the quadrature of the circle can be solved only by approximation. The ratio of 1 to - is nearly that of 7 to 22, which gives by division 3-1428. This agrees with the value of -, so far as to produce an error in excess in the circumference, which is only a little more than the eight-hundredth part of the radius : and on account of the smallness of the numbers, it is much used, when great accuracy is not required. The approximate ratio of 113 to 355, discovered by Metius, is also very convenient, and is so near the truth, as to give an error in excess in the circumference, not much exceeding a four-millionth part of the radius. + BC is to AC, as BDAFC to the semicircle AFC. But BC is to AC, as the semicircle BEC to the semicircle AFC: and therefore (V. 24.) 2 AB.BC + 2BC is to AC, as the compound figure FDE to the semicircle AFC. But (II. 3.) 2 AB.BC + 2BC = 2AC.BC, and (VI. 1.) 2 AC.BC : AC :: BC: JAC. Hence the preceding analogy becomes BC to JAC, as FED to the semicircle, or, by doubling the consequents, and hy inversion, AC to BC, as FG to FED: and it would be proved in the same manner, that AC : AB::FG : GDE. Cor. Hence we can solve the curious problem, in which it is required to divide a circle into any proposed number of parts, equal in area and boundary; as it is only necessary to divide the diameter into the proposed number of equal parts, and to describe the semicircles as in the annexed diagram. Then, whatever part AB is of AD the same part is LEF of the circle KL; and AC being double of AB, LGH is two of these parts : GEFH is therefore one of them; and so on whatever is their number. Their boundaries are also equal, the boundary of each being equal to the circumference of the circle. * G K D B H L PROP. XLII. PROB. To divide a given circle ABC into any proposed number of equal parts, by means of concentric circles. Divide the radius AD into the proposed number of equal parts, suppose three, in the points E, F, and through these points draw perpendiculars to AD, meeting a semicircle described on it as diameter, in G, H; from D as centre, at the distances DG, DH, describe the circles GI, HK: their circumferences divide the circle into equal parts. Join AH, DH. Then (VI. 20. cor. 2.) AD, DH, DF being continual proportionals, AD is to DF, as a square described on AD is to one described on DH. But (XII. 2.) circles are proportional to the squares of their diameters, and consequently to the squares of their radii. Hence (V. 11.) AD is to FD, as the circle ABC to the circle HK; and therefore, since FD is a third of AD, HK is a third of ABC. It would be proved in a similar manner, that Another solution would be obtained, if the circumference were divided into the proposed number of equal parts, and radii drawn to the points of division. This division, however, can be effected only in some particular cases by means of elementary geometry. H B E L K AD is to ED, as ABC to GI. But ED is two-thirds of AD, and therefore GI is two-thirds of ABC ; wherefore the space between the circumferences of GI, HK, is one-third of ABC, as is also the remaining space between the circumferences of ABC and GI. Cor. Hence it is plain, that the area of any annulus, or ring, between the circumferences of two concentric circles, such as that between the circumferences of ABC and GI, is to the circle ABC, as the difference of the squares of the radii DM, DL to the square of DM; or (II. 5. cor. 1. and III. 35.) as the rectangle AL.LM, or the square of the perpendicular LB, to the square of DM: and it therefore follows (XII. 2.) that the ring is equal to a circle described with a radius equal to LB. PROP. XLIII. THEOR. D G If on BC, the hypotenuse of a right-angled triangle ABC, a semicircle BFAGC be described on the same side as the triangle, and if semicircles ADB, AEC be described on the legs, falling without the triangle, the lunes or crescents, ADBF, AECG, bounded by the arcs of the semicircles, are together equal to the right-angled triangle ABC. For (XII. 2. and V. 15.) the semicircle ADB is to BFGC, as the square of AB to the square of BC, and AEC to BFGC, as the square of AC to the square of BC; whence (V. 24.) the two semicircles ADB, AEC, taken together, have to BFGC, the same ratio, as the sum of the squares of AB, AC to the square of BC, that is, the ratio of equality. From these equals, take the segments AFB, AGC, and there remain the lunes DF, EG equal to the triangle ABC. Cor. If the legs AB, AC be equal, the arcs AFB, AGC are equal, and each of them an arc of a quadrant, and the radius drawn from A is perpendicular to BC: and since the halves of equals are equal, each lune is equal to half of the triangle ABC. If, therefore, ABC be a quadrant, and on its chord a semicircle be described, the lune D comprehended between the circumferences, is equal to the triangle ABC: and since (II. 14.) a square can be found equal to ABC, we can thus effect the quadrature of a space D bounded by arcs of circles; though, as we have seen already (App. I. 40. note.) we cannot effect the quadrature of the circle itself, except by approximation. А PROP. XLIV. THEOR. The side and diagonal of a square are incommensurable to one another; that is, there is no line which is a measure of both. *. Let ABCD be a square, and BD one of its diagonals : AB, BD are incommensurable. Cut off DE equal to DA, and join AE. Then, since (I. 5.) the angle DEA is equal to the acute angle DAE, AEB is obtuse, and therefore (I. 19.) in the triangle A BE, BE is less than AB, or than wherefore AD is not a measure of BD. Draw EF perpendicular to BD. Then, the angles, FAE, FEA, being the complements of the equal angles DAE, DEA, are equal, and therefore AF, FE are equal. But (1. 32. cor. 4.) ABD is half a right angle; as is also BFÉ, since BEF is a right angle: wherefore BE is equal to FE, and therefore to AF. From FB, AD; D E B F B * To illustrate this proposition, let AB, CD be two magnitudes of which it is required to find a common measure. From A B the greater cut off as many parts as possible, AE, EF, each equal to CD, and let there be the remainder FB, which must be less than CD. From CD cut off parts CG, GH, each equal to FB, and let HD remain, which is less than FB. In like manner, from FB cut off parts each equal to HD, and suppose that HD is contained A without remainder in FB: tben HD is a common measure of AB and CD. C_A_HD For, since HD measures FB, it measures also CH, which is a multiple of FB; and it must also measure the sum of CH and HD, that is, CD, which is one of the proposed lines. But, since HD is a measure of CD, it is a measure of AF, which is a multiple of CD; and being also a measure of FB, it is a measure of the sum of AF, FB, that is, of AB: it is therefore a common measure of AB, CD. Now, any measure wha.ever of AB, CD is either HD, or a submultiple of it. For let K be a measure of AB, CD; then K is also a measure of AF, which is a multiple of CD; and being a measure of A B, it is a measure of FB, the difference of AB, AF. Hence, since K is a measure of FB, it is a measure of CH, which is a multiple of FB; and being a measure of CD and CH, it must also be contained without remainder in their difference HD; therefore K is either equal to HD, or is a submultiple of it: and hence it follows, that HD is the greatest common measure of AB, CD, every other being a submultiple of HD. It is also plain, that if, by the continuation of processes such as that employed in finding HD, no magnitude can be found which will be contained without remainder in the one preceding it, the magnitudes have no common measure, or are incommensurable, as is the case with regard to the side and diagonal of a square. This subject will receive farther illustration from attempting to express numerically (I. 47. cor. 1.) the length of the diagonal of a square ; as it will be found, that if the side be expressed by a whole number, the diagonal cannot be exactly expressed by any number whatever, but may be approximated as nearly as we please, by means of a decimal or other fraction, |