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.
Approximately squaring the circle.svg
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, 1923.

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.