44
The perimeter of a certain sector of a circle of radius $5.6$ m is $20.0$ m. Find the area of the sector.
Show SolutionHide Solution↓
$2r + \frac{2\pi r \theta}{360} = 20$
$2(5.6) + 2 \times \frac{22}{7} \times 5.6 \times \frac{\theta}{360} = 20$
$11.2 + 2 \times \frac{22}{7} \times 5.6 \times \frac{\theta}{360} = 20$
Solving, we get $\theta = 90^\circ$
$\therefore$ Area of sector $= \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 5.6 \times 5.6$
$= 24.64$ m$^2$
$2(5.6) + 2 \times \frac{22}{7} \times 5.6 \times \frac{\theta}{360} = 20$
$11.2 + 2 \times \frac{22}{7} \times 5.6 \times \frac{\theta}{360} = 20$
Solving, we get $\theta = 90^\circ$
$\therefore$ Area of sector $= \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 5.6 \times 5.6$
$= 24.64$ m$^2$