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GMATPrep: Triangle inscribed in circle

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Junior | Next Rank: 30 Posts
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by tariqal » Sun Oct 30, 2011 7:22 am
Sorry if this is a repost, but it's hard to search for questions in the forums. Can anyone help with the below?


Image

it seems I'm getting hung up on trying to find the height of the triangle, which forms one of the sides of a right-angled triangle for which the hypotenuse is BC. Thanks!

EDIT: making this a bit more searchable for others as well by retyping the question here:

"In the figure above, the radius of the circle with center O is 1 and BC = 1. What is the area of the triangular region ABC?"
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by tpr-becky » Sun Oct 30, 2011 8:32 am
the trick to this problem is to know that the two ends of a diameter that are brought together at the edge of the circle to create a triangle create a right triangle. Thus ABC is a right triangle and AC is the hypotenuse. AC is 2 and BC = 1 thus AB^2 + 1^2 = 2^2 or AB^2 = 3 thus AB = sqrt 3.

In a right triangle the two legs are the base and height. Thus the areas is 1/2(sqrt3)(1) = sqrt3/2
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by tariqal » Sun Oct 30, 2011 1:12 pm
Thanks, Becky, that was clear. These tricks from the grade school days come swooshing back when they're pointed out!
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