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Question:

There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. Let A, B and C be three distinct points on the perimeter of the outer circle such that AB and AC are tangents to the inner circle. If the area of the outer circle is 12 square centimetres then the area (in square centimetres) of the triangle ABC would be

Explanation:

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 ∆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 ∆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 ∆ABC = $3\sqrt{3}\times \frac{3}{\pi }=\frac{9\sqrt{3}}{\pi }$

Hence, option (c).