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sampolo · Tue Mar 06, 2007 1:01 pm

Guys,

I encountered the following problem in my test. Can anyone suggest the solution and the approach?

"What is the maximum area of a triangle with one of its vertex at the center of a circle with radius 1.

I can not use any geometrical formula to calculate the maximum area.

Any clue will be very helpful.

Thanks,
Sam

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ssiva · Wed Mar 07, 2007 3:43 pm

If one of the vertex is on the center of the circle, for maximum area of triangle, the height of the triangle should also be the radius.

Area = (1/2)(1)(1) = 1/2

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sampolo · Thu Mar 08, 2007 12:13 am

Thanks Ssiva for the reply.

How do you get to the conclusion that the height has to be the radius?

Area of triangle = 1/2 * Base * Height.

In this case, since one vertex of the triangle is at the center, b and h are inversely proportional.

Can you tell me how you reach your conclusion about the height and the base?

Thanks,[/quote]

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Stacey Koprince · Sun Mar 11, 2007 10:57 pm

Hi, Sampolo

Can you please post the entire text of the question (and the answer choices)? What you have posted only says that one vertex of a triangle is at the center of a circle with radius 1. The lengths could be anything. Are the other two vertices on the circle itself or something like that?

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sampolo · Tue Mar 13, 2007 10:51 pm

I am sorry.. I thought that I mentioned it. You are correctc stacy, the other two points are on the circle.

I encountered the problem during the test, and forgot to write down all the choices.

I wanted to know what should be the strategy to slve this one.

Thanks

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Stacey Koprince · Tue Mar 13, 2007 11:44 pm

Got it. So one vertex is at the center and the other two are on the circle itself. If we call the point at the center A and the other two B and C, AB will be 1 and AC will be 1 b/c both are radii of the circle.

As a rule, for any particular dimensions, you will maximize the area if you make the triangle a right triangle. This is because the two legs will then be the base and the height of the triangle (which matches the triangle formula: 1/2 * B * H). For a non-right triangle, the height will always be less than the length of the leg which would have been the height if this were a right triangle.

Try drawing some triangles, some right and some non-right. Anything other than 90 will result in the height equaling something less than 1. If we want to maximize, we want to use the largest numbers for B and H, hence we draw a right triangle and B and H are both equal to 1.

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sampolo · Wed Mar 14, 2007 7:37 pm

Hi Stacy,

I agree mostly with your point that the height has to be maximum and if we draw the triangle as right triangle, we get that. Unfortunately, I have a different theory

Area = 1/2 * B * H

Now, you mentioned that if it is not a right triangle, the height will be less than 1. This is correct. But, I can draw a triangle where the base is more than 1 and the angle is not 90. Thus, the base can compensate for the height.

Am I missing something here?

Thanks,

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Stacey Koprince · Thu Mar 15, 2007 12:02 am

Yes, you can draw triangles that way too, but the areas will still be less than the area of the right triangle.

You can follow the below link if you want to read more. Generally, though, just know the rule that to maximize the area of an isosceles triangle, you need a right triangle. (This is no different than knowing, say, the Pythagorean theorem. You don't need to prove it on the test - you just need to know. The only different with this problem is that the rule is a bit more obscure.)

http://my.nctm.org/eresources/view_media.asp?article_id=2073

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