Answer:
Option D
Explanation:
Let the perimeter of each be $a$.
Then, side of the equilateral triangle $=\frac{a}{3}$: side of the square $=\frac{a}{4}$;
Radius of the circle $=\frac{a}{2\pi}$.
$\therefore T$ $=\frac{\sqrt{3}}{4}$ $\times\left(\frac{a}{3}\right)^{2}$
$=\frac{\sqrt{3}a^{2}}{36}$; $S=\left(\frac{a}{4}\right)^{2}$
$=\frac{a^{2}}{16}$; $C$ $=\pi\times\left(\frac{a}{2\pi}\right)^{2}$ $=\frac{a^{2}}{4\pi}$
$=\frac{7a^{2}}{88}$.
So, $C>S>T$.