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Explanation:

Let radius of the circle be r, a side of the equilateral triangle be a, and a side of the square be x.

The circumference/perimeter of the circle, triangle and square are equal. Hence,

2πr = 3a = 4x = k

$\therefore r=\frac{k}{2\pi },a=\frac{k}{3},$ and $x=\frac{k}{4}$

The areas of the circle, triangle and square are c, t, s respectively. Hence,

$\begin{array}{rl}& c=\pi r^2=\frac{\pi k^2}{4{\pi }^2}=\frac{k^2}{4\pi },\\ & t=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times \frac{k^2}{9}=\frac{k^2}{12\sqrt{3}}\\ & s=x^2=\frac{k^2}{16}\\ & \because \frac{1}{\pi }>\frac{1}{4}>\frac{1}{3\sqrt{3}}\\ & \therefore c>s>t\end{array}$

Hence, option (c).