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 centimeters then the area (in square centimeters) of the triangle ABC would be

• A. \(\pi\sqrt{12}\)
• B. \(9/\pi\)
• C. \(9\sqrt{3}/\pi\)
• D. \(6\sqrt{3}/\pi\)

Hint:

Find the radius ratio of the circles first. It tells you the angle between the two tangents at A, and so the shape of triangle ABC.

Answer 1

The Correct Option is C

Step 1: Set up the circles:
Let the outer circle have radius \(R\) and the inner circle radius \(r\), with the same centre \(O\). The area of the outer circle is four times the inner area, so \(\pi R^2 = 4\pi r^2\), which gives \(R = 2r\).
The outer area is 12, so \(\pi R^2 = 12\) and \(R^2 = \frac{12}{\pi}\).

Step 2: Find the angle at A:
A lies on the outer circle, so \(OA = R\). AB and AC are tangents from A to the inner circle. The distance from \(O\) to each tangent is \(r\).
The line AO bisects the angle between the two tangents. Calling that half-angle \(\theta\), we have
\[ \sin\theta = \frac{r}{OA} = \frac{r}{2r} = \frac{1}{2} \Rightarrow \theta = 30^{\circ} \]
So \(\angle BAC = 2\theta = 60^{\circ}\).

Step 3: Show triangle ABC is equilateral:
By symmetry about the line AO, AB = AC. A triangle with two equal sides and a \(60^{\circ}\) angle between them is equilateral. So ABC is an equilateral triangle inscribed in the outer circle.

Step 4: Find the side:
For a triangle inscribed in a circle of radius \(R\), the sine rule gives side \(= 2R\sin 60^{\circ}\).
\[ \text{side} = 2R \cdot \frac{\sqrt{3}}{2} = R\sqrt{3} \]

Step 5: Find the area:
Area of an equilateral triangle \(= \frac{\sqrt{3}}{4}\,(\text{side})^2\).
\[ \text{Area} = \frac{\sqrt{3}}{4} \cdot 3R^2 = \frac{3\sqrt{3}}{4}R^2 \]
Put \(R^2 = \frac{12}{\pi}\):
\[ \text{Area} = \frac{3\sqrt{3}}{4} \cdot \frac{12}{\pi} = \frac{9\sqrt{3}}{\pi} \]

Step 6: Check the options:
Option 1, \(\pi\sqrt{12}\), is about 10.9. A triangle inside a circle of area 12 can never have an area that large (the biggest possible is about 4.96). So it is wrong.
Option 2, \(\frac{9}{\pi}\), is about 2.86, which is smaller than the area of the inner circle (3). The inner circle lies inside the triangle, so the triangle must be larger than 3. Wrong.
Option 4, \(\frac{6\sqrt{3}}{\pi}\), is about 3.31, which is smaller than our correct value \(\frac{9\sqrt{3}}{\pi} \approx 4.96\). Wrong.

Final Answer:
The area of triangle ABC is \(\frac{9\sqrt{3}}{\pi}\) square centimetres, option 3. \[ \boxed{\text{Option 3: } \dfrac{9\sqrt{3}}{\pi}} \]