If the perimeter of a sector of a circle

Question:

If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area is

(a) $58 \mathrm{~cm}^{2}$

(b) $52 \mathrm{~cm}^{2}$

(c) $25 \mathrm{~cm}^{2}$

(d) $56 \mathrm{~cm}^{2}$

Solution:

We know that perimeter of a sector of radius $t=2 r+\frac{\theta}{360} \times 2 \pi r$

We have given perimeter of the sector and radius of the sector and we are asked to find the area of the sector. For that we have to find the sector angle.

Therefore, substituting the corresponding values of perimeter and radius in equation (1) we get,

$29=2 \times 6.5+\frac{\theta}{360} \times 2 \pi \times 6.5$.......(2)

We will simplify equation (2) as shown below,

$29=2 \times 6.5\left(1+\frac{\theta}{360} \times \pi\right)$

Dividing both sides of the equation by $2 \times 6.5$, we get,

$\frac{29}{2 \times 6.5}=\left(1+\frac{\theta}{360} \times \pi\right)$

Subtracting 1 from both sides of the equation we get,

$\frac{29}{2 \times 6.5}-1=\frac{\theta}{360} \times \pi$......(3)

We know that area of the sector $=\frac{\theta}{360} \times \pi r^{2}$

From equation (3), we get

Area of the sector $=\left(\frac{29}{2 \times 6.5}-1\right) r^{2}$

Substituting  we get,

Area of the sector $=\left(\frac{29}{2 \times 6.5}-1\right) 6.5^{2}$

$\therefore$ Area of the sector $=\left(\frac{29 \times 6.5^{2}}{2 \times 6.5}-6.5^{2}\right)$

$\therefore$ Area of the sector $=\left(\frac{29 \times 6.5}{2}-6.5^{2}\right)$

$\therefore$ Area of the sector $=\left(\frac{188.5}{2}-42.25\right)$

$\therefore$ Area of the sector $=(94.25-42.25)$

Therefore, area of the sector is $52 \mathrm{~cm}^{2}$.

Hence, the correct answer is option (b).