# The area of a sector whose perimeter

Question:

The area of a sector whose perimeter is four times its radius r units, is

(a) $\frac{r^{2}}{4}$ sq. units

(b) $2 r^{2}$ sq. units

(c) $r^{2}$ sq.units

(d) $\frac{r^{2}}{2}$ sq. units

Solution:

We know that perimeter of the sector $=2 r+\frac{\theta}{360} \times 2 \pi r$.

We have given that perimeter of the sector is four times the radius.

$2 r+\frac{\theta}{360} \times 2 \pi r=4 r$

Subtracting 2r from both sides of the equation,

$\therefore \frac{\theta}{360} \times 2 \pi r=4 r-2 r$

$\therefore \frac{\theta}{360} \times 2 \pi r=2 r$

Dividing both sides of the equation 2r we get,

$\frac{\theta}{360} \times \pi=1$

$\therefore \frac{\theta \pi}{360}=1$.......(1)

Let us find the area of the sector.

$\therefore$ Area of the sector $=\frac{\theta}{360} \pi r^{2}$

Substituting $\frac{\theta \pi}{360}=1$ we get,

Area of the sector $=r^{2}$

Hence, area of the sector is $r^{2}$ sq.units

Therefore, the correct answer is (c).