If \( \Large \triangle ABC \) is an equilateral triangle of area \( \Large 36\sqrt{3} cm^{2} \), then area of the inscribed circle is


A) \( \Large 48 \pi cm^{2} \)

B) \( \Large 36 \pi cm^{2} \)

C) \( \Large 24 \pi cm^{2} \)

D) \( \Large 12 \pi cm^{2} \)

Correct Answer:
D) \( \Large 12 \pi cm^{2} \)

Description for Correct answer:

Let a be the side of the equilateral triangle.

Therefore, Its area = \( \Large \frac{a^{2}\sqrt{3}}{4} \)

\( \Large \therefore \frac{a^{2}\sqrt{3}}{4} = 36\sqrt{3} \)

=> \( \Large a^{2} = 36 \times 4 \)

=> \( \Large a = 6 \times 2 = 12 cm\)

Let r be the radius of the inscribed circle.

Therefore, \( \Large r = \frac{\triangle}{s}=\frac{36\sqrt{3}}{\frac{1}{2} \left(12+12+12\right) } = 2\sqrt{3} cm \)

Area of inscribed circle = \( \Large \pi r^{2}= \pi \times \left(2\sqrt{3}\right)^{2}=12 \pi cm^{2} \)


Part of solved Mensuration questions and answers : >> Aptitude >> Mensuration








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