Question

If a chord of a circle of radius $5\text{cm}$ subtends ${120}^{\circ }$ at the centre of the circle, then the area of the corresponding segment of the circle is

• A. $25(\frac{\pi }{3}-\frac{1}{4}){\text{cm}}^2$
• B. $25(\frac{\pi }{3}-\frac{\sqrt{3}}{4}){\text{cm}}^2$
• C. $25(\frac{\pi }{3}-\frac{1}{2}){\text{cm}}^2$
• D. $25(\frac{\pi }{3}-\frac{\sqrt{3}}{2}){\text{cm}}^2$

Answer 1

The correct option is B $25(\frac{\pi }{3}-\frac{\sqrt{3}}{4}){\text{cm}}^2$

Draw perpnedicular bisectors $OC$ and $AB$.
$\angle AOC=\angle BOC=\frac{\angle AOB}{2}={60}^{\circ }$

In $\Delta ACO$,
$\cos {60}^{\circ }=\frac{OC}{AO}$
$\Rightarrow \frac{1}{2}=\frac{OC}{5}$
$\Rightarrow OC=\frac{5}{2}\text{cm}$
$\sin {60}^{\circ }=\frac{AC}{AO}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{AC}{5}$
$\Rightarrow AC=\frac{5\sqrt{3}}{2}\text{cm}$
Since, $AC=CB=\frac{AB}{2}$
$\therefore AB=2AC=5\sqrt{3}\text{cm}$
$\therefore$ Area of $\Delta AOB=\frac{1}{2}\times AB\times OC$
$=\frac{1}{2}\times 5\sqrt{3}\times \frac{5}{2}=\frac{25\sqrt{3}}{4}{\text{cm}}^2$
Area of sector $AOB=\frac{\theta }{360}\times \pi r^2$
$=\frac{{120}^{\circ }}{{360}^{\circ }}\times \pi \times 5^2$
$=\frac{25\pi }{3}{\text{cm}}^2$
$\therefore$ Area of corresponding segment $=$ Area of sector $AOB-$ Area of $\Delta AOB$
$=\frac{25\pi }{3}-\frac{25\sqrt{3}}{4}$
$=25(\frac{\pi }{3}-\frac{\sqrt{3}}{4}){\text{cm}}^2$

Hence, the correct answer is option b.