KCET · Maths · Straight Lines
\(A B C\) is a triangle with \(\angle A=30^{\circ}, B C=10 \mathrm{~cm}\). The area of the circumcircle of the triangle is
- A \(100 \pi \mathrm{sq} \mathrm{cm}\)
- B \(5 \mathrm{sq} \mathrm{cm}\)
- C \(25 \mathrm{sq} \mathrm{cm}\)
- D \(\frac{100 \pi}{3} \mathrm{sq} \mathrm{cm}\)
Answer & Solution
Correct Answer
(A) \(100 \pi \mathrm{sq} \mathrm{cm}\)
Step-by-step Solution
Detailed explanation
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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