Please Help Me With This Geometry Problem

Remember that the area of a square of sidelength L is:

A = L^2

And the area of a circle of diameter D is:

A = pi*(D/2)^2

If we inscribe a square in a circle, we will get four segments, like the ones shaded in the image below:

Notice that the diameter of the circle will be equal to the diagonal of the square.

And the diagonal of a square of side length L is:

d = √(2)*L

knowing that the side length of our square is 6 inches, the diameter of the circle will be:

D = √2*6in

Now, the total area of the four shaded parts will be equal to the difference between the area of the circle and the area of the square.

The area of the circle is:

A = pi*(√2*6in/2)^2 = (pi/2)*36in^2

The area of the square is:

A' = (6in)^2 = 36in^2

The difference is:

A - A' =  (pi/2)*36in^2 - 36in^2 = (pi/2 - 1)*36in^2

And there are 4 of these segments, then the area of every single one is one-fourth of that:

a = (1/4)*(pi/2 - 1)*36in^2 = (pi/2 - 1)*9 in^2

The area of each segment is:

a =  (pi/2 - 1)*9 in^2

if we replace pi by 3.14, the exact area will be:

a = (3.14/2 - 1)*9in^2 = 5.13 in^2