Solution :
Let the radius of the circle be $r$ and side of the square be $a$.
Then, according to question,
$2 \pi r= 4a$
=> $a = \large\frac{2 \pi r }{4}$
$\qquad= \large\frac{\pi r}{2}$ ----(i)
Now, ratio of their areas,
$ \pi r^2 $ and $a^2$
=> $ \pi r^2$ and $ \bigg( \large\frac{\pi r}{4}\bigg)^2$ [From eq(i)]
=> $ \pi r^2 $ and $ \large\frac{\pi^2 r^2}{4}$
Therefore, Area of the circle > Area of the square Report Error