Question

Area of the segment $ABC$ given below is:

• A. $\frac{2\pi }{3}-\sqrt{3}{\text{cm}}^2$
• B. $\frac{2\pi }{3}+\sqrt{3}{\text{cm}}^2$
• C. $\frac{\pi }{3}-2\sqrt{3}{\text{cm}}^2$
• D. $\frac{\pi }{3}+2\sqrt{3}{\text{cm}}^2$

Answer 1

The correct option is A $\frac{2\pi }{3}-\sqrt{3}{\text{cm}}^2$


Given, radius$\left(r\right)=2\text{cm},\theta ={60}^{\circ }$

Step 1: $\text{Area of the sector}AOBC\text{=}\frac{\theta }{{360}^{\circ }}\times \pi r^2$

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

$=\frac{1}{6}\times \pi \times 4$

$=\frac{2\pi }{3}{\text{cm}}^2$

Step 2: Let $OD\perp AB,$ where $D$ is the mid point of $AB$

In $△ODB,\angle DOB=\frac{{60}^{\circ }}{2}={30}^{\circ }$
Using $\cos$ formula:
$\cos {30}^{\circ }=\frac{\text{Adjacent}}{\text{Hypotenuse}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{OD}{OB}(\because \cos {30}^{\circ }=\frac{\sqrt{3}}{2})$
Cross multiplying, we get
$\Rightarrow OD=\frac{\sqrt{3}}{2}\times OB$
$\Rightarrow OD=\frac{\sqrt{3}}{2}\times 2$
$\Rightarrow OD=\sqrt{3}$

Similarly, $\sin {30}^{\circ }=\frac{\text{Opposite}}{\text{Hypotenuse}}$
$\Rightarrow \frac{1}{2}=\frac{DB}{OB}$
Cross multiplying, we get
$\Rightarrow DB=\frac{1}{2}\times OB$
$\Rightarrow DB=\frac{1}{2}\times 2$
$\Rightarrow DB=1$

Area of the $△OAB=\frac{1}{2}\times AB\times OD$
$=\frac{1}{2}\times 2\times DB\times OD$
$=DB\times OD$
$=1\times \sqrt{3}$
$=\sqrt{3}$

Step 3: $\text{Area of the segment}ABC\text{= Area of the sector}AOBC-\text{Area of the}△OAB$
$=\frac{2\pi }{3}-\sqrt{3}{\text{cm}}^2$

Option (a.) is the correct choice.