Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/48

 Let $$J$$ be on the tangent at $$B$$, and such that $$\ang BAJ = 30^\circ$$. Then $$JL$$ is approximately equal to the semi-circular arc $$BCD$$. Taking the radius as unity, it can easily be proved that

the correct value to four places of decimals.

(2) The value $$\textstyle \frac{355}{113} = 3{\cdot}141592\ldots$$ is correct to six decimal places.

Since $$\frac{355}{113} = 3+\frac{4^2}{7^2 + 8^2}$$, it can easily be constructed.

Let $$CD = 1$$, $$\textstyle CE = \frac{7}{8}$$, $$\textstyle AF = \frac{1}{2}$$; and let $$FG$$ be parallel to $$CD$$ and $$FH$$ to $$EG$$; then $$AH = \frac{4^2}{7^2 + 8^2}$$.

This construction was given by Jakob de Gelder (Grünert's Archiv, vol. 7, 1849).

(3) At $$A$$ make $$\textstyle AB = (2+\frac{1}{5})$$ radius on the tangent at $$A$$ and let