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
Choose one:
A) π√​12 B) 9/π C) 9√3/π D) 6√3/π E) None of these
Explanation :   Outer circle:   Since πR² = 12, R = √(12/π) = 2√(3/π).   Inner circle:   Since the area is 1/4 the area of the outer circle, πr² = 3 and r = √(3/π)   AO is the Radius of the outer circle and OD is the radius of the inner circle.   A line tangent to a circle and a radius drawn to the point of tangency form a right angle.   Thus, ∆AOD is a right triangle.   Hypotenuse AO is twice the length of leg OD.   A right triangle whose hypotenuse is twice the length of a leg is a 30-60-90 triangle.   Thus, ∆AOD is a 30-60-90 triangle.   In a 30-60-90 triangle, the sides are proportioned x : x√3 : 2x.   Thus, AD = √3 * √(3/π) * = 3/√π.   Using the same reasoning for ∆OCD, we can deduce that CD = 3/√π.   Thus, AC = 2 * 3/√π = 6/√π.   The formula for the area of an equilateral triangle = (s²√3)/4.   Since AC = 6/√π, the area of ∆ABC = (6/√π)² * √3/4 = 36/π * √3/4 = (9√3)/π.   Answer : 9√3/π
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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