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$A B C$ is a triangle with $\angle A=30^{\circ}, B C=10 \mathrm{~cm}$. The area of the circumcircle of the triangle is
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Verified Answer
The correct answer is:
$100 \pi \mathrm{sq} \mathrm{cm}$
In $\triangle A B C, \angle A=30^{\circ}$
$B C=10 \mathrm{~cm}$
$O$ is the centre of circle.
$$
\begin{aligned}
&\therefore \quad \angle B O C=60^{\circ} \text { and } O B \text { and } O C \text { are the radius } \\
&\therefore \quad \angle O B C=\angle O C B=60^{\circ} \\
&\Rightarrow \triangle O B C \text { is an equilateral triangle. } \\
&\therefore \text { Radius of circle is } O B=O C=B C=10 \mathrm{~cm} \\
&\text { Now, area of the circumcircle is } \pi r^{2} \\
&\qquad=\pi(10)^{2}=100 \pi \mathrm{sq} \mathrm{cm}
\end{aligned}
$$
$B C=10 \mathrm{~cm}$
$O$ is the centre of circle.
$$
\begin{aligned}
&\therefore \quad \angle B O C=60^{\circ} \text { and } O B \text { and } O C \text { are the radius } \\
&\therefore \quad \angle O B C=\angle O C B=60^{\circ} \\
&\Rightarrow \triangle O B C \text { is an equilateral triangle. } \\
&\therefore \text { Radius of circle is } O B=O C=B C=10 \mathrm{~cm} \\
&\text { Now, area of the circumcircle is } \pi r^{2} \\
&\qquad=\pi(10)^{2}=100 \pi \mathrm{sq} \mathrm{cm}
\end{aligned}
$$
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