Question

Find the area of the sector whose arc is making an angle of ${60}^{\circ }$ at the center of a circle of radius 3.2 cm. What will be the area of the balance sector? [3 MARKS]

Answer 1

Formula for area of a sector: 1 Mark
Solution: 2 Marks

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

Area of the sector
$=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi r^2$

$=0.1666\times \frac{22}{7}\times 3.2\times 3.2$

$=5.36c{m}^2$

b) The area of the balance of the sector will be as follows



Area of the circle - Area of the sector

$\pi r^2-\frac{60}{{360}^{\circ }}\times \pi r^2$

$\Rightarrow \pi r^2-\frac{1}{6}\pi r^2$

$\Rightarrow \frac{5}{6}\pi r^2$

Putting the values of radius = 3.2 cm

We have

$\frac{5}{6}\times \frac{22}{7}\times \frac{32}{10}\times \frac{32}{10}$

$\Rightarrow \frac{5\times 22\times 32\times 32}{6\times 7\times 100}$

$\Rightarrow \frac{3604480}{4200}$

$\Rightarrow 858209$

$\Rightarrow 858.21c{m}^2$