A square and an equilateral triangle have the same perimeter. Let A be the area of the circle circumscribed about the square and-B be the area of the circle circumscribed about thetriangle. Then the ratio A: B is

Let the common perimeter be $$c$$

So, each side of square = $$\frac{c}{4}$$

=> Circum radius of square = $$r_s=\frac{s}{\sqrt2}=\frac{c}{4\sqrt2}$$

Thus, area of circle circumscribed about the square will be

=> $$A=\pi (r_s)^2=\pi\frac{c^2}{32}$$ -------------(i)

Now, each side of triangle = $$\frac{c}{3}$$

=> Height of triangle = $$\frac{\sqrt3}{2}\times\frac{c}{3}$$

=> Circum radius of triangle = $$r_t=\frac{2}{3}\times(\frac{\sqrt3}{2}\times\frac{c}{3})=\frac{c}{3\sqrt3}$$

So area of the circle circumscribed about the triangle 

=> $$B=\pi (r_t)^2=\pi\frac{c^2}{27}$$ -----------(ii)

$$\therefore$$ Required ratio $$A:B=27:32$$

=> Ans - (C)