A square and a regular hexagon are drawn such that all

Solution

Here ABCD is a square and PQRSTU is Hexagon Let o is the center of circle then diagonal of square ABCD will 2r then area of square will be ½ (d)² => ½ (2r)² => 2r² Now we have to find the area of hexagon we know that O angle is 60° then and ∠ S = ∠ T, then ∆ OST will be equilateral ∆, having side r. ∆OST = √3/4 × r² In the Hexagon there will be 6 triangle and all are same then area of Hexagon will be => 6 × √3/4 × r² (3√3/2) × r² Ratio => Square : Hexagon => (2r/2) : (3√3/2) × r² => 4 : 3√3