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# If the circumference of a circle and the perimeter of a square are equal, then

$\begin{array}{1 1}(A) \; \text{Area of the circle =area of the square} \\(B)\;none\;of\;these \\ (C)\;\text{Area of the circle} > \text{area of the square}\\(D)\; \text{Area of the circle} < \text{area of the square} \end{array}$

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A)
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