$A B C$ is a triangle with $\angle A=30^{\circ}$ and $B C=10 \mathrm{~cm}$. The area of the circum circle of the triangle is

• A

  $5 \mathrm{sq} \mathrm{cm}$
• Correct answer
  B

  $100 \pi \mathrm{sq} \mathrm{cm}$
• C

  $\frac{100 \pi}{3} \mathrm{sq} \mathrm{cm}$
• D

  $25 \mathrm{sq} \mathrm{cm}$

Solution

$$ \bar{Z}=-1-i $$ We know that, $\angle B O C=2 \angle B A C=2 \times 30^{\circ}=60^{\circ}$ Since, $O B=O C$ $\therefore \angle O B C=\angle O C B=\frac{1}{2}\left(180^{\circ}-\angle B O C\right)=60^{\circ}$ So, $\triangle B O C$ is an equilateral triangle. $\begin{aligned} \therefore \quad O B=B C=10 \mathrm{~cm} \\ \text { Therefore, area of circumcircle } &=\pi(O B)^{2}=\pi(10)^{2} \\ &=100 \pi \mathrm{sq} \mathrm{cm} \end{aligned}$

Hence, the correct answer is option: 'B'.