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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}.

2 comments:

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[hide] Anonymous said...

You left out the radius of the circle until the solution. :(

on 26 August 2011 at 03:59
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[hide] Paul Liu said...

Oops! Sorry! Added now :P

on 26 August 2011 at 04:02
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