CAT 2003 Slot 1 — QA Question 30

Consider the diagram below, as per the given conditions.

Let r and R be the radii of the inner circle and outer circle respectively.

As area of the outer circle is 4 times the area of the inner circle, we have, R = 2r

In âˆ†OAM, sin θ = $\frac{OM}{OA}=\frac{r}{2r}=\frac{1}{2}$

∴ ∠OAM = θ = 30°

Similarly,

∠OAM = ∠OBM = ∠OAC = ∠OCA = 30°

∠OBC = ∠OCB = 30°

∴ ∠BAC = ∠ACB = ∠CBA

∴ ∆ABC is an equilateral triangle.

AB = 2 × $\sqrt{(O{A}^2-O{M}^2)}=2\sqrt{3}r$

Area of âˆ†ABC = $\frac{\sqrt{3}}{4}$ × (AB)² = $\frac{\sqrt{3}}{4}$ × 4 × 3 × r² = $r\sqrt{3}{r}^2$ ...(i)

But, area of the outer circle = π(2r)² = 4πr² = 12

∴ r² = $\frac{3}{\pi }$

∴ Area of âˆ†ABC = $3\sqrt{3}\times \frac{3}{\pi }=\frac{9\sqrt{3}}{\pi }$

Hence, option (c).