# 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 cicumscribed about the triangle. Then A/B=?

Jul 29, 2018

Let the common perimeter be $c$
So each side of the square $\frac{c}{4}$
And the circumradius of the square $\left({r}_{s}\right) = \frac{1}{2} \sqrt{2} \times \frac{c}{4} = \frac{c}{4 \sqrt{2}}$

So the area of the circle circumscribed about the square will be

$A = \pi {r}_{s}^{2} = \pi {c}^{2} / 32$

Each side of the triangle will be $\frac{c}{3}$

Its height $h = \frac{\sqrt{3}}{2} \times \frac{c}{3}$

The circumradius of the triangle $\left({r}_{t}\right) = \frac{2}{3} \times \frac{\sqrt{3}}{2} \times \frac{c}{3} = \frac{c}{3 \sqrt{3}}$

So area of the circle cicumscribed about the triangle.

$B = \pi {r}_{t}^{2} = \pi {c}^{2} / 27$

Hence the ratio $\frac{A}{B} = \frac{27}{32}$