Question

If the adjoining figure is a sector of a circle of radius 10.5 cm, what is the perimeter of the sector? (Take $$\pi =22/7$$) 

 

Answer 1

Given figure is a quadrant of a circle. We have given radius of sector that is 10.5 cm. Arc AB subtended an angle of 60° at the centre of the circle.

Perimeter of the sector =${\displaystyle \frac{\theta }{360}\times 2\pi r+2r}$

Substituting the values we get, 

${\displaystyle \text{Perimeter of the sector}=\frac{60}{360}\times 2\pi \times 10.5+2\times 10.5}$ ..............(1) 

Now we will simplify equation (1) as shown below,

${\displaystyle \text{Perimeter of the sector}=\frac{1}{6}\times 21\pi +21}$

Now we will substitute pi=${\displaystyle \frac{22}{7}}$

${\displaystyle \text{Perimeter of the sector}=\frac{1}{6}\times 21\times \frac{22}{7}+21}$

${\displaystyle \text{Perimeter of the sector}=\frac{1}{6}\times 3\times 22+21}$

${\displaystyle \text{Perimeter of the sector}=\frac{1}{2}\times 22+21}$

${\displaystyle \text{Perimeter of the sector}11+21}$

${\displaystyle \text{Perimeter of the sector}=32}$

Therefore, perimeter of the given sector is ${\displaystyle 32cm}$