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Geometry PS problem

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Geometry PS problem [#permalink]

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New post 25 Jun 2009, 12:38
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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

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Re: Geometry PS problem [#permalink]

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New post 25 Jun 2009, 13:13
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?

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Re: Geometry PS problem [#permalink]

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New post 26 Jun 2009, 00:30
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.

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Re: Geometry PS problem [#permalink]

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New post 26 Jun 2009, 01:45
@maliyeci:

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!!
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Re: Geometry PS problem [#permalink]

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New post 26 Jun 2009, 02:13
I made a mistake in solution (the sides are not r*root 2 but r * root 2 / 2 so the area must be 2* r^2*root 2 ^2 * 1/4). Yes the answer will be 1/2.

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Re: Geometry PS problem [#permalink]

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New post 26 Jun 2009, 02:46
sdrandom1 wrote:
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

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Re: Geometry PS problem [#permalink]

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New post 26 Jun 2009, 11:42
See the following thread which discusses the same problem

triangular-circle-area-problem-79634.html

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Re: Geometry PS problem [#permalink]

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New post 26 Jun 2009, 12:18
Nice problem/solution! I got really scared when I saw the cos and sin solution. yikes!

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Re: Geometry PS problem   [#permalink] 26 Jun 2009, 12:18
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