« ΠροηγούμενηΣυνέχεια »
BC: Draw DF perpendicular to BC, and let it meet the eircumfer, ence 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 eaoh 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 beoause 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 ÂC or BC=1000, AH=500, CF=500, AF=1500, and CG being a mean proportional between AH and AF, CGʻ=(17. 6.) AH. AF=500 X 1500=750000; wherefore CG=866.0254+, because (866.0254) 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 (1856.0254 +) =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 Con the chord of one twenty-fourth of the circumference, Q will be a mean proportional between AH and AC+P, and Qo=AH (AC+P)=300(1965.9258+) =982962.9+; and therefore Q=991.4419+, because (991.4449)" is less than 982962.9. Therefore also AC+Q=1991.4449+.
In like manner, if S he the perpendicular from C on the chord af one forty-eighth of the circumference, S'=AH (AC+Q)=500 (1991.4449+). =995722.45+; and S=997.8589+, because (997 8589)2 is less than 999722.45. Hence also, AC+S=1997.8589+.
Lastly, if T be the perpendicular from Con the chord of one nine, ty-sixth of the circumference, To=AH(AC+S)=500 (1997.8589 +)=998929.45+, and T=999.46458+. Thus T. the perpen, dicular on the chord of one ninety sixth of the circumference, is greater thąn 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-8) = 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=(68.4386) 96=6282.1056,
Let the perimeter of the circumscribed polygon of the same number of sides, be M, then (2. Cor. 2. 1. Sup.) T: AC :; 6282.1056; M, that is, (since T=999.46458+, as already shown),
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.
Now, if a fourth proportional be found to 999.46458, 1000 and 6282.1056, viz. 6285.461-, then,
because, 999.46458 : 1000 :: 6282.1056 : 6285,4614, and as before, 999.46458 : 1000 :: 6282.1056—: N; therefore, 6282.1056 : 6282.1056—:: 6283.461-N, and as the first of these proportionals is greater than the second, the third, viz. 6285,
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 than the ratio of 3142.7305 to 1000, therefore the circumference has 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 show'u, that CGʻ=760000; wherefore CG=866. 02315_, becouse (866.02545)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 (1866.02545) 933012.73—; and P=965.92585_, be bange (965.92587) is greater than 933012.73. Hence also, AC+P *1965.92585
Next, as V=the perpendicular drawn from the centre on the chord of one twenty-feurth of the circumference, Q* = AH (AC+P)= 500 X (1965.92585—-) = 982962.93—; and Q =991.44495---, because (991.44495)® is greater than 982962.93. Hence also, AC+Q
In like manner, as S is the perpendicular from C on the chord of one forty-eighth of the circumf rence, Sø=AH(AC+Q)=500 (1991. 4149.3–) = 995722.473—, and S = (997.858954) because (997. 83895)2 is greater than 995722.475.
But the square of the chord of the ninety-sixth part of the eireumference = AB (AC–S) = 2000 (2.44105+) = 4282.1 +, and the chord itself 65.4377 + because (65.4377)a 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=165.4377+). 96=6282.019+. But the circumfers enee of the circle is greater than the perimeter of the inscribed polygon; therefore the circumference is greater than 6 282.019, 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+7to 1; 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 shown 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 circumfer. ence may be found nearly, by making as 7 to 22, so the given diameter to a fourth proportional, which will be greater than the circumference. And if as 1 to 3+29, or as 71 to 223, so the given diameter to a fourth proportional, this will be nearly equal to the cireumfer. ence, but will be less than it.
Cor. 2. Because the difference between 5 and
fore the lines found by these proportions differ by 497 of the diameter. Therefore the difference of either of them from the circumference must be less than the 497th part of the diameter.
Cor. 3. As 7 to 22, so the square of the radius to the area of the cirele nearly
For it has been shown, 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 circumference nearly as 7 to 82, therefore the square of the radius is to the area of the eirele 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. T'he 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 constructed on the principles explained in the foregoing Proposition, and in which the radius is supposed
No. of sides of
6 12 24 48
Perimeter of the Perimeter of the
192 384 768 1536 3072 6144
The part that is wanting in the numbers of the second column, to make 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 the circumscribed polygons is less than a unit is the sixth
1 decimal place, that is than 1000000
of the radius. Also, as the numbers in the second column, are less than the perimeters of the inscribed polygons, they are each of them less than the circumference, of the circle; and for the same reason, each of those in the third column is greater than the cireumference. But when the arch of h of the circumference is bisected ten times, the number of sides in the polygon is 6144, and the numbers in the Table differ from one
1 another only by part of the radius, and therefore the peri
1000000 meters of the polygons differ by less than that quantity; and conse. quently the circumference of the circle, which is greater than the least, and less than the greatest of these numbers, is determined with in less than the millioneth part of the radius.
Hence also, if R be the radius of any circle, the circumference is greater than RX6.283185, or than 2RX3.141592, but less than 2R X3.141593; and these numbers differ from one another only by a millioneth part of the radius. So also Ro+3.141592 is less, and RX3.141593 greater than the area of the circle; and these numbers differ from one another only by a millioneth part of the square of the radius.
In this way, also, the circumference and the area of the circle may be found still nearer to the truth; but neither by this, nor by any other method yet known to geometers, can they be exactly determin. ed, though the errors of both may be reduced to a less quantity than any that can be assigned.