A) \( \Large 48 \pi cm^{2} \) |
B) \( \Large 36 \pi cm^{2} \) |
C) \( \Large 24 \pi cm^{2} \) |
D) \( \Large 12 \pi cm^{2} \) |
D) \( \Large 12 \pi cm^{2} \) |
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 = \( \Large \pi r^{2}= \pi \times \left(2\sqrt{3}\right)^{2}=12 \pi cm^{2} \)