Question

Find the perimeter of the minor segment $ACB$ shown below.

• A. $2\pi +3\text{cm}$
• B. $2\pi +6\text{cm}$
• C. $\pi +12\text{cm}$
• D. $\pi -6\text{cm}$

Answer 1

The correct option is B $2\pi +6\text{cm}$

Given, the radius of a circle $\left(r\right)=OB=6\text{cm}$
and the angle subtentded by segment $ACB$ at the centre of the circle$\left(\theta \right)={90}^{\circ }.$

Step 1: Let's find the length of $AB:$
In $△ODB,$ where $OD\perp AB$ and $D$ is the mid point of $AB$
$\angle BOD=\frac{{60}^{\circ }}{2}={30}^{\circ }$

Using $\sin$ formula:
$\sin {30}^{\circ }=\frac{\text{Opposite}}{\text{Hypotenuse}}$
$\Rightarrow \frac{1}{2}=\frac{BD}{OB}(\because \sin {30}^{\circ }=\frac{1}{2})$
Cross multiplying, we get
$\Rightarrow BD=OB\times \frac{1}{2}\Rightarrow BD=6\times \frac{1}{2}\Rightarrow BD=3\text{cm}$

$\therefore AB=2\times BD=2\times 3=6\text{cm}$

Step 2: Length of the arc
$\stackrel{⌢}{BCA}=\frac{\theta }{{360}^{\circ }}\times \pi \times d=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times 2\times 6$
$=\frac{1}{6}\times \pi \times 12$
$=2\pi$

Step 3: Perimeter of the segment $ACB$
$=\stackrel{⌢}{BCA}+AB$
$=2\pi +6$

Hence, the perimeter of the minor segment $ACB$ is $2\pi +6\text{cm}.$
Option (b.) is the correct choice.