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Director
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What is the greatest possible area of a triangular region [#permalink]
 22 Jul 2007, 21:30
What is the greatest possible area of a triangular region with one vertex at the center of circle of radius 1 and the other two on the circle?
sqrt3/4, 1/2, pi/4, 1, sqrt2
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Re: Area of the triangular region [#permalink]
22 Jul 2007, 21:47
bewakoof wrote: What is the greatest possible area of a triangular region with one vertex at the center of circle of radius 1 and the other two on the circle?
sqrt3/4, 1/2, pi/4, 1, sqrt2
Triangle with largest area is an equilateral triangle. Since, one vertex is at the centre and the other 2 on the circle, side of the triangle is 1.
Hence, area = sqrt(3/4). A.
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Director
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Should be '1/2'
The greatest area will be of right angle triangle, with base and height equal to radius.
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Director
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Its not 1/2..
and pls show some work..
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"Triangle with largest area is an equilateral triangle."
Is this a theorem or smth? It makes sense, but how do you know?
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Senior Manager
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Hayabusa wrote: "Triangle with largest area is an equilateral triangle."
Is this a theorem or smth? It makes sense, but how do you know?
I remember reading it somewhere, but cannot recollect where. Of course, I may be wrong. 
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I agree it is 1/2
sqrt(3)/4 is less than 1/2
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Manager
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Is the answer root of 2? If it is, I'll explain.
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Director
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bewakoof wrote: Its not 1/2..
and pls show some work..
It should be 1/2. I am 100% sure.
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sumande wrote: Hayabusa wrote: "Triangle with largest area is an equilateral triangle."
Is this a theorem or smth? It makes sense, but how do you know? I remember reading it somewhere, but cannot recollect where. Of course, I may be wrong. Just guessing on this (based on sketches) - possibly the largest triangle that can be inscribed in a circle is equilateral, but if one vertex is the center of the circle, rather than on the circumference, then I agree it has to be a right triangle. So the area would be 1/2.
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Manager
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By using Heron's formula, you can maximize the area of a triangle.
I asked a math wiz about it and he came up with root of 2.
I think he used differentiation.
Not sure if we are likely to see that on the GMAT.
Then again, I can be wrong.... 
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Director
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I have to dig my error log for the correct answer but I promise I'll do it tommorow.. I chose 1/2 and i got it wrong so i knew 1/2 wasnt the answer
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bewakoof wrote: I have to dig my error log for the correct answer but I promise I'll do it tommorow.. I chose 1/2 and i got it wrong so i knew 1/2 wasnt the answer but it should be 1/2. make sure you do not have incorrect answer.
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Himalayan wrote: bewakoof wrote: I have to dig my error log for the correct answer but I promise I'll do it tommorow.. I chose 1/2 and i got it wrong so i knew 1/2 wasnt the answer but it should be 1/2. make sure you do not have incorrect answer. Sunil
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The perimeter is not set so the theorem doesn't hold. Could someone please explain why the answer is sqrt2?
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Area = 1/2 * 1 * 1 * sin90 = 1/2
Value of sin is max at 90 (i.e. 1)...so max area should be 1/2.
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