# Squaring the Circle

**Construction steps**

1. Draw Circle (O,A), with diameter AB and centre O.

2. Draw Circle (B,A), followed by Circle (A,O).

3. Mark C, one of the points of intersection of Circle (B,A) and Circle (A,O).

4. Join BC, and let it intersect Circle (O,A) at D.

5. Claim: BD2 is almost equal to the area of Circle (O,A). [See the square at the right.]

Let AO = 1 unit, AB = 2 units; then BC = 2 units, and AC = 1 unit. Hence, triangle BAC is isosceles, with equal sides 2 units and base 1 unit. (In the figure, AC has not been joined.)

By construction, ∠ADB is a right angle (“angle in a semicircle”).

The length of the perpendicular from B to AC is . Therefore, the area of traingle BAC is square units.

Hence, 1/2 AD.BC = , implying that AD =

Using Pythagoras theorem, BD^{2 }+ AD^{2} = AB^{2} = 4, So BD^{2} =4- 15/ 6 = 3. 1/6

hence BD = 7/4 , i.e , BD is 7/4 times the radius of the circle

So the area of the square on side BD is 3.1/6 = 3.0625, and the area of circle on AB as diameter is

so the error percentage is approximately - 2.51 % . ( The error is on the negative side)

**GAURAV CHAURASIA** is currently a 15-year-old student in class 10 in RPS Academy School, Deoria, Uttar Pradesh. He has a particular love for algebra, geometry, trigonometry, and number theory. He may be contacted at gauravvchaurasia@gmail.com.