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

Solution:

(b) Area of the circle > Area of the square
Let r be the radius of the circle.
We know:

Circumference of the circle $=2 \pi r$

Now,
Let a be the side of the square.
We know:
Perimeter of the square = 4a
Now,

$2 \pi r=4 a$

$\Rightarrow r=\frac{4 a}{2 \pi}$

$\therefore$ Area of the circle $=\pi r^{2}$

$=\pi \times\left(\frac{4 a}{2 \pi}\right)^{2}$

$=\pi \times \frac{16 a^{2}}{4 \pi^{2}}$

$=\frac{4 a^{2}}{\pi}$

$=\frac{4 \times 7 a^{2}}{22}$

 

$=\frac{14 a^{2}}{11}$

Also,

Area of the square $=a^{2}$

Clearly, $\frac{14 a^{2}}{11}>a^{2}$

∴ Area of the circle > Area of the square