Squaring the circle

Let ${\displaystyle PQR}$ be a circle with centre ${\displaystyle O}$, of which a diameter is ${\displaystyle PR}$. Bisect ${\displaystyle PO}$ at ${\displaystyle H}$ and let ${\displaystyle T}$ be the point of trisection of ${\displaystyle OR}$ nearer ${\displaystyle R}$. Draw ${\displaystyle TQ}$ perpendicular to ${\displaystyle PR}$ and place the chord ${\displaystyle RS=TQ}$.

Join ${\displaystyle PS}$, and draw ${\displaystyle OM}$ and ${\displaystyle TN}$ parallel to ${\displaystyle RS}$. Place a chord ${\displaystyle PK=PM}$, and draw the tangent ${\displaystyle PL=MN}$. Join ${\displaystyle RL}$, ${\displaystyle RK}$ and ${\displaystyle KL}$. Cut off ${\displaystyle RC=RH}$. Draw ${\displaystyle CD}$ parallel to ${\displaystyle KL}$, meeting ${\displaystyle RL}$ at ${\displaystyle D}$.

Then the square on ${\displaystyle RD}$ will be equal to the circle ${\displaystyle PQR}$ approximately.

For

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

${\displaystyle d}$

Therefore

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

But ${\displaystyle PL}$ and ${\displaystyle PK}$ are equal to ${\displaystyle MN}$ and ${\displaystyle PM}$ respectively.

Therefore

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

Hence	${\displaystyle RK^{2}=PR^{2}-PK^{2}={\frac {113}{144}}d^{2}}$,
and	${\displaystyle RL^{2}=PR^{2}+PL^{2}={\frac {355}{324}}d^{2}}$.

But	${\displaystyle {\frac {RK}{RL}}={\frac {RC}{RD}}={\frac {3}{2}}{\sqrt {\frac {113}{355}}}}$,
and	${\displaystyle RC={\frac {3}{4}}d}$.

Therefore

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

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