Let $PQR$ be a circle with centre $O$, of which a diameter is $PR$. Bisect $PO$ at $H$ and let $T$ be the point of trisection of $OR$ nearer $R$. Draw $TQ$ perpendicular to $PR$ and place the chord $RS=TQ$.
Join $PS$, and draw $OM$ and $TN$ parallel to $RS$. Place a chord $PK=PM$, and draw the tangent $PL=MN$. Join $RL$, $RK$ and $KL$. Cut off $RC=RH$. Draw $CD$ parallel to $KL$, meeting $RL$ at $D$.
Then the square on $RD$ will be equal to the circle $PQR$ approximately.
For

$RS^{2}={\frac {5}{36}}d^{2}$,


where
$d$ is the diameter of the circle.
Therefore

$PS^{2}={\frac {31}{36}}d^{2}$.


But $PL$ and $PK$ are equal to $MN$ and $PM$ respectively.
Therefore

$PK^{2}={\frac {31}{144}}d^{2}$, and $PL^{2}={\frac {31}{324}}d^{2}$.


Hence

$RK^{2}=PR^{2}PK^{2}={\frac {113}{144}}d^{2}$,


and

$RL^{2}=PR^{2}+PL^{2}={\frac {355}{324}}d^{2}$.


But

${\frac {RK}{RL}}={\frac {RC}{RD}}={\frac {3}{2}}{\sqrt {\frac {113}{355}}}$,


and

$RC={\frac {3}{4}}d$.


Therefore

$RD={\frac {d}{2}}{\sqrt {\frac {355}{113}}}=r{\sqrt {\pi }}$, very nearly.


Note.—If the area of the circle be $140,000$ square miles, then $RD$ is greater than the true length by about an inch.