Question

If the area of the sector of a circle with angle ${45}^{\circ }$ is $11c{m}^2$ , then the area of corresponding major sector is equal to.

• A. $88c{m}^2$
• B. $66c{m}^2$
• C. $77c{m}^2$

Answer 1

The correct option is C $77c{m}^2$


Given, area of the sector of a circle with angle ${45}^{\circ }$ is $11c{m}^2$
In the above figure, green portion represents a minor sector and yellow portion represents a major sector.
So, angle of minor sector $\left(\theta \right)={45}^{\circ }$
Area of minor sector $=\frac{\theta }{360}\times \pi r^2$
$\Rightarrow$ $11=\frac{45}{360}\times \frac{22}{7}$ $\times r^2$
$\Rightarrow$ $11=\frac{1}{8}\times \frac{22}{7}$ $\times r^2$
$\Rightarrow$ $r^2=\frac{11\times 8\times 7}{22}=28$
$\Rightarrow$ $r^2=28$

Now, area of circle $=\pi r^2$ $=\frac{22}{7}$$\times 28$
$\Rightarrow$ Area of circle$=88c{m}^2$

$\because$ Area of circle = Area of major sector + Area of minor sector
$\Rightarrow$ 88 = Area of major sector + 11
$\Rightarrow$ Area of major sector = 88 - 11 $=77c{m}^2$