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
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).