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.

where $$d$$ is the diameter of the circle.

But $$PL$$ and $$PK$$ are equal to $$MN$$ and $$PM$$ respectively.



Note.—If the area of the circle be $$140,000$$ square miles, then $$RD$$ is greater than the true length by about an inch.