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Gmat Prep Q#2

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by jangojess » Sun Oct 14, 2007 10:12 pm
given that O is the centre and so AC is the diameter..AC = 2 as radius = 1. BC = 1, so AB = sqrt3. Now area of ABC = 1/2 * BC * AB = (sqrt3)/2..so ans is B
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by dpatwa » Mon Oct 15, 2007 6:37 am
I don't understand what leads one to assume that this is a right triangle? To have the a:a*sqrt(3):2a relationship, it must be a right triangle.
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by jangojess » Mon Oct 15, 2007 6:49 am
it is given that O is the centre of the circle and AC is the diameter of the circle.
Rule - any triangle inscribed in a circle with one side as the diameter of the circle and the other two sides of the traingle joining from the ends of the diameter to any point on the circle would be a right triangle or in short any triangle inscribed in a semicircle wud be a right triangle...u need to remember this rule...
now radius is provided which gives AC and the other side is provided. Find the 3rd side using pythagoras theorem and finally get the area of a right triangle...
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