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what is the greatest possible area of a triangular region with one vertex on center and two points on the circle? 1) PI/2 2) 1/2 3) SQRT3/4 4) 1 5) SQRT 2

Definitely some piece of information is missing here...Is the radius of the circle given...????

But if the radius of the circle is 1. Then the greatest area would be possible when angle at the center is right angle, with legs as the radius of the circle. Hence area = 1/2 *B * H = 1/2 * 1*1= 1/2.

radius is needed. I will solve as r as radius. Area of that triangle is \(r*cos x * r*sin x /2\) (x is half of the center angle). This makes \([m]r^2\)/2 * (sinx*cosx)[/m] taking derivative of that. we get. \(r^2/2*\) (\(cos^2\)x - \(sin^2\)x) it must be zero when the area is greatest. simplfying that we get \(cos^2\)x - \(sin^2\)x = 0 so sinx = cosx. That is x is 45 degrees. And this triangle is a right triangle. So the area is \(r*cos x * r*sin x /2\) = r * \sqrt{2} * r * \sqrt{2}/2 = \(r^2\) If r = 1 than area is 1.

The way you have answered it, I think the whole question is messed up. In my personal opinion, such solutions are not required at GMAT. Kindly correct me if some one thinks I am wrong!!

sdrandom seems to have correct answer!!
_________________

Definitely some piece of information is missing here...Is the radius of the circle given...????

But if the radius of the circle is 1. Then the greatest area would be possible when angle at the center is right angle, with legs as the radius of the circle. Hence area = 1/2 *B * H = 1/2 * 1*1= 1/2.

What is the OA?

Yes, the answer is 1/2. could you please help me understand why the greatest area is possible when the angle at the center is right angle? thanks