# Page:Squaring the circle.djvu

5

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.