If the circumference of a circle and

(b) According to the given condition,

Circumference of a circle = Perimeter of square

$2 \pi r=4 a$

[where, $r$ and a are radius of circle and side of square respectively]

$\Rightarrow \quad \frac{22}{7} r=2 a \Rightarrow 11 r=7 a$

$\Rightarrow \quad a=\frac{11}{7} r \Rightarrow r=\frac{7 a}{11}$ $\ldots($ (i)

Now, $\quad$ area of circle, $A_{1}=\pi r^{2}$

$=\pi\left(\frac{7 a}{11}\right)^{2}=\frac{22}{7} \times \frac{49 a^{2}}{121}$ [from Eq. (i)]

$=\frac{14 a^{2}}{11}$....(ii)

and area of square, $A_{2}=(a)^{2}$ ....(iii)

From Eqs. (i) and (iii), $\quad A_{1}=\frac{14}{11} A_{2}$

$\therefore \quad A_{1}>A_{2}$

Hence, Area of the circle $>$ Area of the square.