The vertices of an equilateral triangle lie on a circle with radius 2 . The area of the triangle is
(A) $3 \sqrt{3}$
(B) $4 \sqrt{3}$
(C) $6 \sqrt{3}$
(D) $5 \sqrt{3}$
(E) $2 \sqrt{3}$

## Part C: Each correct answer is worth 8.
Suppose that a circle with centre $O$ has radius 2 and that equilateral $\triangle P Q R$ has its vertices on the circle.

Join $O P, O Q$ and $O R$.

Join $O$ to $M$, the midpoint of $P Q$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-055.jpg?height=447&width=458&top_left_y=817&top_left_x=888)

Since the radius of the circle is 2 , then $O P=O Q=O R=2$.

By symmetry, $\angle P O Q=\angle Q O R=\angle R O P$.

Since these three angles add to $360^{\circ}$, then $\angle P O Q=\angle Q O R=\angle R O P=120^{\circ}$.

Since $\triangle P O Q$ is isosceles with $O P=O Q$ and $M$ is the midpoint of $P Q$, then $O M$ is an altitude and an angle bisector.

Therefore, $\angle P O M=\frac{1}{2} \angle P O Q=60^{\circ}$ which means that $\triangle P O M$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Since $O P=2$ and is opposite the $90^{\circ}$ angle, then $O M=1$ and $P M=\sqrt{3}$.

Since $P M=\sqrt{3}$, then $P Q=2 P M=2 \sqrt{3}$.

Therefore, the area of $\triangle P O Q$ is $\frac{1}{2} \cdot P Q \cdot O M=\frac{1}{2} \cdot 2 \sqrt{3} \cdot 1=\sqrt{3}$.

Since $\triangle P O Q, \triangle Q O R$ and $\triangle R O P$ are congruent, then they each have the same area.

This means that the area of $\triangle P Q R$ is three times the area of $\triangle P O Q$, or $3 \sqrt{3}$.

Answer: (A)

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