Squaring the circle

[Paramjit]

Please help me with the first step of Ramanujan's method of squaring the circle:
rs2(square)=5/36 d2(square)

 

[Jamest]

Solution:Given rs²=5/36 d²
Let PQR be a circle with center O of which a diameter is PR. Bisect PO at H and let T is the point of trisection 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 RD at D.
Then the square on RD will be equal to circle PQR approximately. For
RS2 =5/36d2
where d is diameter of circle. Therefore, PS2=31/36d
But PL and PK are equal to MN and PM
respectively.
Therefore,PK= 31/144 d2,
and PL2=31/324 d2.
Hence, RK2=PR2-PK2=113/144 d2,
and RL2=PR2+PL2=355/324d²
=>RK/RL=RC/RD
=>3/2√(113/355)
=>RC=3/4d
=>RD=d/2√(355/113)


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Prime Factorization

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[Deleted member 4993]

Solution:Given rs²=5/36 d²
Let PQR be a circle with center O of which a diameter is PR. Bisect PO at H and let T is the point of trisection 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 RD at D.
Then the square on RD will be equal to circle PQR approximately. For
RS2 =5/36d2
where d is diameter of circle. Therefore, PS2=31/36d
But PL and PK are equal to MN and PM
respectively.
Therefore,PK= 31/144 d2,
and PL2=31/324 d2.
Hence, RK2=PR2-PK2=113/144 d2,
and RL2=PR2+PL2=355/324d²
=>RK/RL=RC/RD
=>3/2√(113/355)
=>RC=3/4d
=>RD=d/2√(355/113)


Learn more about
​

Prime Factorization

355/113 ≈ \(\displaystyle \displaystyle \pi\)