A) \( \Large 12 cm^{2} \) |
B) \( \Large \frac{3\sqrt{3}}{4}cm^{2} \) |
C) \( \Large \frac{9\sqrt{3}}{4}cm^{2} \) |
D) \( \Large 3 \sqrt{3}cm^{2} \) |
D) \( \Large 3 \sqrt{3}cm^{2} \) |
since, \( \Large \triangle's \) AOB,BOC,COD are equilateral.
Sides = 2 cm
Now, total area = \( \Large 3\times \)\( \Large \frac{\sqrt{3}}{4}(Side)^{2} \)
= \( \Large 3\times \)\( \Large \frac{\sqrt{3}}{4}\times (Side)^{2}=3\sqrt{3} cm^{2} \)