# Squaring the circle

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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. This work is in the public domain in the United States because it was published before January 1, 1924.

The author died in 1920, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.