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$A B C$ is a triangle with $\angle A=30^{\circ}$ and $B C=10 \mathrm{~cm}$. The area of the circum circle of the triangle is
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Verified Answer
The correct answer is:
$100 \pi \mathrm{sq} \mathrm{cm}$
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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.
$\begin{aligned} \therefore \quad O B=B C=10 \mathrm{~cm} \\ \text { Therefore, area of circumcircle } &=\pi(O B)^{2}=\pi(10)^{2} \\ &=100 \pi \mathrm{sq} \mathrm{cm} \end{aligned}$
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