Quantitative Aptitude :: Area - Discussion

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Explanation:
let x be side of square
perimeter of square=4x=perimeter of triangle=3*side of triangle
so side of eq. triangle=(4/3)*x
diameter of circle circumscribing the square=sqrt(2)*x
area of circle circumscribing the square=pi*(sqrt(2)*x)^2/4=(pi/2)*x^2 ----(1)
to find radius of the circle inscribed in the triangle
area of triangle=r*s=sqrt(3)/4 * (4x/3)^2
now s=(4/3)*x+(4/3)*x+(4/3)*x/2=2x
so sqrt(3)/4 * (4x/3)^2=r*2x gives
r={2/3*(3^1/2)}*x
area of the circle inscribed in the triangle=pi*({2/3*(3^1/2)}*x)^2
=pi*(4/27)*x^2 -------(2)
so reqd ratio= eqn(1)/eqn(2)
=((pi/2)*x^2)/(pi*(4/27)*x^2)=27/8
so reqd ratio=27:8