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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
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