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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 :

Choice 1 is the answer.

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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