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## Thursday, 25 August 2011

### Math GRE - #24

What is the greatest possible area of a triangular region with one vertex at the centre of a unit circle and the other two vertices on the circle?

1. $\frac{1}{2}$
2. $1$
3. $\sqrt{2}$
4. $\pi$
5. $\frac{1+\sqrt{2}}{4}$

Solution :

The area of any triangle with sides $a$, $b$, and angle $\theta$ between $a$ and $b$ is: $A = \frac{1}{2}ab\sin\theta$
Since the sides have the same length as the radius of the circle, $a=b=1.$
Therefore, $A=\frac{1}{2}\sin\theta$ which equals $\frac{1}{2}$ at the maximum of $\theta=\frac{\pi}{2}$.
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