First let ABCD be the square inscribed in the circle.
Let diameter of circle be 2r.
Since square ABCD is inscribed in the circle , diagonal AC and BD will be equal to the diameter (2r)
therefore, BD = 2r
In triangle BCD inside square ABCD ,
BC2 + CD2 = BD2
BC2 + BC2 = (2r)2 ( BD = 2r , BC = CD)
2BC2 = 4r2
BC2 = 2r2
Ratio of Areas of circle and the square = area of circle / area of square = pie r2 / BC2 = 22 / 7 * r2 / 2r2 (BC2=2r2)
= 22 / 7 / 2
= 11 / 7