Squaring the circle or vice-versa
Author : Shyamal Kumar Das E-mail ID : das.shyamal1c@gmail.com
Given: A circle of radius 1 unit.
Required: To construct the side of a square of the same area of the given circle.
Construction: Let, AB is the diameter drawn horizontally and ‘O’I s the centre of the given circle.
Now, the radius OB is bisected no. of times to get 64 equal divisions. From OB, 49divisions are
taken.
Let, E be the point, so that, OE = 49/64 and AE = 1 + 49/64 = 1.7656 units.
Then, AE is the required side of the square.
Proof: Area of the given circle = π * ( 1) 2 = π = 3.1416
Area of the square of side 1.7656 = (1.7656)2 = 3.1174
The difference = 0.0242 (less)
Remark: To get better and more accurate result, one should take 99/128 or, 197/256 and
so on, instead of 49/64.
Please, also note that, 49/64 = 1/2+ 1/4 + 1/64 . Similarly, 99/128 = 1/2+1/4+3/128.
Conclusion: The existence of difference between two areas would be theoretically “YES”,
but, practically “NO”, if, we can adjust the divisions, increasing the number of bisections.
VICE-VERSA
Given: A square PQRS of side 1 unit.
Required: To construct a circle of the same area of the given square.
Construction: The two diagonals PR and QS are joined, which bisect each other at ‘O’.
From ‘O’, a line OT, parallel to PS is drawn bisecting SR at ‘X’. Taking ‘O’ as a centre and OS as a
radius, an arc is drawn, which intersects OT at ‘Y’. Now, XY is bisected no. of times to get
16 equal divisions. From XY, 5 divisions are taken. Let, ‘Z’ be the point, so that, XZ = 5/16 parts
of XY and OZ = OX + 5/16 part of XY.
Then, OZ is the radius of the required circle.
Proof: The area of the given square = 12 =1 square unit
Here, OX = 0.5 unit and OY = OS =√2/2 = 0.707 unit, then XY = 0.707 -0.5 =0.207 unit
The radius of the required circle OZ = 0.5 + 5/16(0.207) =0.5+0.0647 =0.5647 unit
The area of the required circle = 3.1416 * (0.5647)2 = 3.1416* 0.3189 = 1.0018 square unit.
The difference = 0.0018(more)
Remark: This is the best result, one can have. Slight adjustment may be made for bigger areas.
Conclusion: The existence of difference between two areas would be theoretically “YES”,
but, practically “NO”, if, we can adjust the divisions, increasing the number of bisections.