Question

A square and 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 the triangle then the ratio $\frac{A}{B}$ is

• A. $\frac{9}{16}$
• B. $\frac{3}{4}$
• C. $\frac{27}{32}$
• D. $\frac{3\sqrt{6}}{8}$

Answer 1

The correct option is C $\frac{27}{32}$

From given, we have,

Let the common perimeter be $c$

So each side of the square $\frac{c}{4}$. As the perimeter of square = 4*side

And the circumradius of the square $r_s=\frac{1}{2}\sqrt{2}\times \frac{c}{4}=\frac{c}{4\sqrt{2}}$

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

$A=\pi r_s^2=\pi \frac{c^2}{32}$........(1)

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

The height of triangle $h=\frac{\sqrt{3}}{2}\times \frac{c}{3}$

The circumradius of the triangle $r_t=\frac{2}{3}\times \frac{\sqrt{3}}{2}\times \frac{c}{3}=\frac{c}{3\sqrt{3}}$

So area of the circle cicumscribed about the triangle.

$B=\pi r_t^2=\pi \frac{c^2}{27}$..........(2)

Dividing (1) and (2), we get,

The ratio,

$\frac{A}{B}=\frac{27}{32}$