Squares in circles

1) consider a semicircle of radius R with center O and having a sqare ABCD inscribed in it.

2) make a perpendicular from O to AB. Let it intersect at P.

3) Now PODB is a rectangle therefore OB = PD = R (Radius)

4) also as can be seen from figure 2PB = 2OD = PO = BD

5) Let PB = x hence BD = 2x

6) using Ptolemy's theorm for cyclic quadrilaterals => x^{2} + (2x)^{2} = r^{2}

7) therefore side of square 2x = 2r/sqrt{5}

8) area of square in semicircle is 4r/5

9) area of square in circle is 2r square

on dividing both we will get 2:5

For circle: (2r)^2 = a^2 + a^2 => a^2= 2(r^2) ANd for semi circle: a=(2*r/(sqrt5) => a^2 = (4 r^2)/5, Hence ratio is 2:5

The concept is the radius of the circle is equal to kroot5/2 for the semicircle square with side k and the circle radius is lroot2/2 for the complete circle contains square of side l Hence k/l = root(2/5) So the area ratio = k^2/l^2= 2/5 That simple