Circling the Square and Approximating the Square Root of 2: Exploring the Baudhayana Shulbasutra

Posted on Jun 24, 2024 in Mathematics

Circling the Square – Baudhayana Shulbasutra

This method describes how to construct a circle with an area equal to that of a given square.

It means we have to convert or construct a circle equal to the square.

-> By Pythagoras, OA’ = √(a²+a²)

                                OA = √(2a²) = a√2

-> Here OA=OE=√2

-> Make a point F, such that OF= OG+1/3GE

-> Draw a circle with center F and radius OF. This circle’s area is equal to the square’s area.

Proof

OF= OG+1/3GE

OF=a+1/3(OE-OG)

OF=a+1/3(a√2-a)

OF=a[1+1/3(√2-1)]

OF=a[3/3+1/3(√2-1)]

OF=a[2/3+√2/3]

OF=a/3[2+√2] – required radius of the circle

For square, area = (2a)² = 4a²

Area of Circle = πr²

                          = (3.1415) * a²(1.293)

                          = a²(3.9954)

                          ~ a² (4)

Approximating the Square Root of 2 – Baudhayana Shulbasutra

This method explains how to find an approximation for the square root of 2.

Let two squares ABCD & PQRS of side 1 unit each.

Consider another square whose area is equal to the sum of the squares ABCD & PQRS.

The length of the side of the combined square is √2

Then the area of the combined square is 2*1 = 2 sq. units

But we will get a rectangle on combining, but we want a square.

Cut square PQRS into 3 equal parts & place them on square ABCD.

• The 1st and 2nd parts will be attached to the square ABCD.
• But for the third piece, divide it into a 1:2 ratio.

By dividing square PQRS into 7 pieces, we made a square.

-> {area of ABCD}+{area of PQRS}+{1/12*1/12} = Area of the combined square

-> {area of ABCD}+{area of PQRS} = Area of the combined square – {1/12*1/12}

-> If we remove the excess from the combined square, then it will no longer be a square.

-> So we will decrease very little along the length & breadth equally from any one corner of the combined square such that it amounts to a decrease of the excess area of 1/12*1/12

-> Removing these rectangles, we will maintain the square shape.

We know length = 1+1/3+1/12 = 17/12

Let B=breadth, A = area = 1/12*1/12

So A = l*b*2 (We have 2 rectangles)

1/12*1/12 = 17/12*b*2

b = 1/(17*12*2)

We have 2 rectangles so b=1/(34*12)=1/(3*4*34)

Hence √2=1+1/3+1/(3*4)-1/(3*4*34)