Page:Squaring the circle.djvu

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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.
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.