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

 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

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.

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.