Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/49

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THE FIRST PERIOD

35

On the diameter through $A$ take $AD=OB$ , and draw $DE$ parallel to $OC$ . Then ${\frac {AE}{AD}}={\frac {AC}{AO}}={\frac {13}{5}}$ ; therefore $AE=r{\,.\,}{\frac {13}{5}}{\sqrt {1+\left({\frac {11}{5}}\right)^{2}}}=r{\,.\,}{\frac {13}{25}}{\sqrt {146}}$ ; thus $AE=r{\,.\,}6{\cdot }2831839\ldots$ , so that $AE$ is less than the circumference of the circle by less than two millionths of the radius.

The rectangle with sides equal to $AE$ and half the radius $r$ has very approximately its area equal to that of the circle. This construction was given by Specht (Crelle's Journal, vol. 3, p. 83).

(4) Let $AOB$ be the diameter of a given circle. Let

$\textstyle OD={\frac {3}{5}}r,\qquad OF={\frac {3}{2}}r,\qquad OE={\frac {1}{2}}r$ .

Describe the semi-circles $DGE$ , $AHF$ with $DE$ and $AF$ as diameters; and let the perpendicular to $AB$ through $O$ cut them in $G$ and $H$ respectively. The square of which the side is $GH$ is approximately of area equal to that of the circle.

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Fig. 19.

We find that $GH=r{\,.\,}1{\cdot }77246\ldots$ , and since ${\sqrt {\pi }}=1{\cdot }77245$ we see that $GH$ is greater than the side of the square whose area is equal to that of the circle by less than two hundred thousandths of the radius.