BD a mean proportional between AB and BF, the excess of the radius above CF. Join AD ; and because ADB is a right angle, being an angle in a semicircle ; and because CGB is also a right angle, the triangles ABD, CBG are equiangular, and, AB : AD :: BC : CG (4. 6.), or alternately, AB : BC :: A'D : CG ; and therefore. because AB is double of BC, AD is double of CG, and the square of AD therefore equal to four times the square of CG. But, because ADB is a right angled triangle, and DF a perpendicular on AB, AD is a mean proportional between AB and AF (8. 6.), and ADP=AB.AF (17. 6.), or since AB is =4AH, AD>=4AH.AÉ. Therefore also, because 4CGP=AD, 4CGC4AH.AF, and CG?= AH. AF; wherefore CG is a mean proportional between AH and AF, that is, between half the radius and the line made up of the radius, and the perpendicular on the chord of twice the arch BD. Again, it is evident, that BD is a mean proportional between AB and BF (8. 6.), that is, between the diameter and the excess of the radius above the perpendicular, on the chord of twice the arch DB. Therefore, &c. Q. E. D. PROP. IX. THEOR.* less than ten of the parts, of which the diameter contains seventy, but Let ABD be a circle, of which the centre is C, and the diameter AB; the circumference is greater than three times AB, by a line 10 1 10 less than of AC, but greater than of AC. 70' 7 71 In the circle ABD apply the straight line BD equal to the radius 1 or In this proposition, the character + placed after a number, signifies that something is to be added to it. and the character -, on the other hand, signifies that something is to be taken away from it. BC : Draw DF perpendicular to BC, and let it meet the circumference again in E; draw also CG perpendicular to BD : produce BC to A, bisect AC in H, and join CD. It is evident, that the arches BD, 'BE are each of them one-sixth of the circumference (Cor. 15. 4.), and that therefore the arch DBE is one-third of the circumference. Wherefore, the line (8. 1. Sup.) CG is a mean proportional between AH, half the radius, and the line AF. Now because the sides BD, DC, of the triangle BDC are equal, the angles DCF, DBF are also equal ; and the angles DFC, DFB being equal, and the side DF common to the triangles DBF, DCF, the base BF is equal to the base CF, and BC is bisected in F. Therefore, if AC or BC=1000, AH=500, CF=500, AF=1500, and CG being a mean proportional between AH and AF, CG"=(17. 6.) AH.AF=500 X 1500=750000 ; wherefore CG=866.0254+, because (866.0254)2 is less than 750000. Hence also, AC+CG=1866. 0254+. Now, as CG is the perpendicular drawn from the centre C, on the chord of one-sixth of the circumference, if P = the perpendicular from C on the chord of one-twelfth of the circumference, P will be a mean proportional between AH (8. 1. Sup.) and AC+CG, and P == AH (AC+CG) = 500 X (1866.0254 +) = 933012.7 +. Therefore, P=965.9258 +, because (965.9258) is less than 933012.7. Hence also, AC+P=1965.9258+. Again, if Q=the perpendicular drawn from C on the chord of one twenty-fourth of the circumference, Q will be a mean proportional between AH and AC+P, and Q2 =AH (AC+P) =500(1965.9258+) =982962.9+; and therefore Q=991.4449+, because (991.4449) is less than 982962.9. Therefore also AC + Q=1991.4449+. In like manner, if S be the perpendicular from C on the chord of one forty-eighth of the circumference, Sa= AH (AC+Q) = 500 (1991.4449+) = 995722.45+; and S = 997.8589+, because (997. 8589) is less than 995722.45. Hence also, AC+S=1997.8589+. Lastly, if T be the perpendicular from C on the chord of one ninety-sixth of the circumference, T2 = AH (AC+S)= 500 (1997.8589 +) = 998929.45+, and T 999.46458 +. Thus T, the perpendicular on the chord of one ninety-sixth of the circumference, is greater than 999.46458 of those parts of which the radius contains 1000. But by the last proposition, the chord of one ninety-sixth part of the circumference is a mean proportional between the diameter and the excess of the radius above S, the perpendicular from the centre on the chord of one forty-eighth of the circumference. Therefore, the square of the chord of one ninety-sixth of the circumference AB (AC—S) = 2000 X (2.1411–») = 4282.2-; and therefore the chord itself = 65.4386, because (65.4386)2 is greater than 4282.2. Now, the chord of one ninety-sixth of the circumference, or the side of an equilateral polygon of ninety-six sides inscribed in the circle, being 65.4386, the perimeter of that polygon will be = (65.4386 --) 96 = 6282.1056m. Let the perimeter of the circumscribed polygon of the same num, her of sides, be M, then (2. Cor. 2. 1. Sup.) T: AC :: 6282.1056– M, that is, (since T = 999.46458+, as already shewn), 999.46458 + : 1000 :: 6282.1056: M ; if then N be such, that 999.46458 : 1000 :: 6282.1056_:N; ex æquo perturb. 999.46458 + : 999.46458 ::N:M; and, since the first is greater than the second, the third is greater than the fourth, or N is greater than M. 12 Now, if a fourth proportional be found to 999.46458, 1000 and 6282.1056, viz. 6285.461–, then, because, 999.46458 : 1000 :: 6282.1056 : 6285.461, and as before, 999.46458 : 1000 :: 8282.1056 : N ; therefore, 6282.1056 : 6282.1056 ::6285.461-:N, and as the first of these proportionals is greater than the second, the third, viz. 6285. 2 461- is greater than N, the fourth. But N was proved to be greater than M ; much more, therefore, is 6285.461 greater than M, the perimeter of a polygon of ninety-six sides circumscribed about the circle ; that is, the perimeter of that polygon is less than 6285.461 ; now, the circumference of the circle is less than the perimeter of the polygon ; much more, therefore, is it less than 6285.461 ; wherefore the circumference of a circle is less than 6285.461 of those parts of which the radius contains 1000. The circumference, therefore, has to the diameter a less ratio (8. 5.) than 6285.461 has to 2000, or than 3142.7305 has to 1000 : but the ratio of 22 to 7 is greater thab the ratio of 3142.7305 to 1000, therefore the circumference h:s a less ratio to the diameter than 22 has to 7, or the circumference is less than 22 of the parts of which the diameter contains 7. It remains to demonstrate, that the part by which the circumference 10 exceeds the diameter is greater than of the diameter. 71 It was before shewn, that CG2 = 750000 ; wherefore CG=866. 02545, because (866.02545)2 is greater than 750000; therefore AC+CG=1866.02545. Now, P being, as before, the perpendicular from the centre on the chord of one-twelfth of the circumference, P = AH (AC+CG) = 500 X (1866.02545) 933012.73~; and P=965.92585, because (965.92585)2 is greater than 933012.73. Hence also, AC+P =1965.92585-. Next, as Q=the perpendicular drawn from the centre on the chord of one twenty-fourth of the circumference, Q2 = AH (AC+P) = 500 X (1965.92585–) =982962.93—; and Q=991.44495—, because (991.44495)2 is greater than 982962.93. Hence also, AC+Q =1991.44495-. In like manner, as S is the perpendicular from C on the chord of one forty-eighth of the circumference, Sa=AH(AC+Q)=500/1991. 44495,-) = 995722.475, and S = (997.85895–) because (997. 85895)2 is greater than 995722.475. But the square of the chord of the ninety-sixth part of the circumference = AB (AC—S) = 2000 (2.14105+) = 4282.1+, and the chord itself 65.4377 + because (65.4377)2 is less than 4282.1 : Now the chord of one ninety-sixth part of the circumference being = 65.4377 +, the perimeter of a polygon of ninety-six sides inscribed in the circle (65.4377+) 96=6282,019+. But the circumference of the circle is greater than the perimeter of the inscribed polygon ; therefore the circumference is greater than 6282.019, of those parts of which the radius contains 1000 ; or than 3141.009 of the parts of which the radius contains 500, or the diameter contains 10 1000. Now, 3141.009 has to 1000 a greater ratio than 3+ 71 therefore the circumference of the circle has a greater ratio to the 10 diameter than 3+ has to 1; that is, the excess of the circumference 71 above three times the diameter is greater than ten of those parts of which the diameter contains 71 ; and it has already been shewn to be less than ten of those of which the diameter contains 70. Therefore, &c. Q. E. D. Cor. 1. Hence the diameter of a circle being given, the circumference may be found nearly, by making as 7 to 22, so the given diameter to a fourth proportional, which will be greater than the circumfer 10 And if as 1 to 3+ 71' or as 71 to 223, so the given diameter to a fourth proportional, this will be nearly equal to the circumference, hut will be less than it. Z to l; ence. an 1 10 1 Cor. 2. Because the difference between 7 is there's 71 497' 1 fore the lines found by these proportions differ by of the diame 497 ter. Therefore the difference of either of them from the circumference must be less than the 497th part of the diameter. Cor. 3. As 7 10 22, so the square of the radius to the area of the circle nearly For it has been shewn, that (1. Cor. 5. 1. Sup.) the diameter of a circle is to its circumference as the square of the radius to the area of the circle ; but the diameter is to the cirkumference nearly as 7 to 22, therefore the square of the radius is to the area of the circle nearly in that same ratio. SCHOLIUM. It is evident that the method employed in this proposition, for finding the limits of the ratio of the circumference to the diameter, may be carried to a greater degree of exactness, by finding the perimeter of an inscribed and of a circumscribed polygon of a greater number of sides than 96. The manner in which the perimeters of such polygons approach nearer to one another, as the number of their sides increases, may be seen from the following Table, which is constracted on the principles explained in the foregoing Proposition, and in which the radius is supposed = 1. The part that is wanting in the numbers of the second column, tu inake up the entire perimeter of any of the inscribed polygons, is less than unit in the sixth decimal place ; and in like manner, the part by which the numbers in the last column exceed the perimeter of any of the circumscribed polygons is less than a unit in the sixth |