Chapter 11.3, Problem 39PPS

To determine

The area of the shaded sector.

Answer to Problem 39PPS

The required area of the sector is 30 $m{m}^2$ .

Explanation of Solution

Given: A sector of a circle with perimeter 22 mm and radius 6 mm as show below-

  

Concepts Used: The formula for the area of the sector of a circle with central angle $\theta$ and radius r is given by-

Area of sector $=\frac{\theta }{360°}\times \pi r^2$

The length of an arc s in a circle of radius r which subtends an angle $\theta$ at the centre is given by-

  $s=r\theta$ , where $\theta$ is in radian measure.

Any angle in radian measure can be converted into degree measure using the formula-

Degree measure $=\frac{180°}{\pi }\times$ Radian measure

Calculations: The perimeter of the given sector is 22 mm. Length of the radius is 6 mm.

Now,

Perimeter of sector $=$ $2\times$ (Radius) + Arc length

Let the length of the arc in the given sector be s. Thus,

  $22=2\times 6+s$

  $\Rightarrow s=22-12$

  $\Rightarrow s=10$

Let the angle subtended by this arc at the centre be $\theta$ . Thus,

  $\theta =\frac{s}{r}$

  $=\frac{10}{6}$ rad

  $=\frac{5}{3}$ rad

Converting the central angle into degree measure gives-

Degree measure $=\frac{180°}{\pi }\times \frac{5}{3}$

  $=\frac{300°}{\pi }$

Hence, area of the shaded sector is given by-

Shaded area $=\frac{\left(\frac{300°}{\pi }\right)}{360°}\times \pi {\left(6\right)}^2$ $m{m}^2$

  $=\frac{300°}{360°\times \pi }\times \pi \times 36$ $m{m}^2$

  $=\frac{5}{6}\times 36$ $m{m}^2$

  $=30$ $m{m}^2$

Thus, area of the shaded region is 30 $m{m}^2$ .