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

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

$JL={\sqrt {{\frac {40}{3}}-{\sqrt {12}}}}=3{\cdot }141533\ldots$,

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

$\textstyle BC={\frac {1}{5}}.{\text{radius}}$.