Question

Match the following with the area of shaded regions in $c{m}^2$.

• A. $\frac{250\pi }{3}$
• B. $\frac{50\pi }{3}-25\sqrt{3}$
• C. $\frac{50\pi }{3}$

Answer 1

Here, $r=10$ cm and $\theta ={60}^{\circ }.$

For the given figure:

$Areaofminorsector$
$=\frac{\theta }{360}\times \pi r^2$
$=\frac{60}{360}\times \pi \times 10\times 10$
$=\frac{50\pi }{3}c{m}^2$

For the given figure:

$Areaofmajorsector=Areaofcircle-Areaofminorsector$

$Areaofcircle$
$=\pi r^2$
$=\pi \times {10}^2$
$=100\pi c{m}^2$

$\Rightarrow Areaofmajorsector$
$=100\pi -\frac{50\pi }{3}$
$=\frac{250\pi }{3}c{m}^2$

For the given figure:

$Areaofthesegment=Areaofminorsector-Areaof△OAB$
$Areaof△OAB$
$=\frac{\sqrt{3}}{4}\times 10\times 10$
$=25\sqrt{3}c{m}^2$
[$△OAB$ is an equilateral triangle with side 10 cm.]
$\Rightarrow$ Area of the segment
$=\frac{50\pi }{3}-25\sqrt{3}c{m}^2$