# Squaring the circle

Squaring the circle  (1913)
by Srinivasa Ramanujan

Squaring the circle

(Journal of the Indian Mathematical Society, v, 1913, 138)

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. 