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Geometry

by vish150783 » Tue Dec 02, 2008 10:10 am
Faced this in GMATPrep

What is the greatest possible area of a triangular region with one vortex at teh centre of teh circle or raidus 1 and the other two vertices on the circle?

A. (Sqrt 3)/4
B. 1/2
C. pi/4
D. 1
E. sqrt 2
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by reneerenee » Tue Dec 02, 2008 5:39 pm
i think its C. the triangle that you get when the angle is 90 is half of a 1x1 square and it's area is 1/2.

there is a similar question like this up at the gmat and gre math blog (gmatgremath.blocked) He or she or they or whoever runs that blog has really great geometric pictures to explaing the problems.

(also, i hope you meant vertex instead of vortex! :) )
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by ifairo » Tue Dec 02, 2008 5:57 pm
Answer "1/2" means choice B.
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by anayeri » Tue Dec 02, 2008 6:04 pm
Answer: B.

The maximum area of a triangle, must be a right triangle. So (1*1)/2 = 1/2.

But, why do we necessarily need a right triangle to maximize area? Well, we know for sure that our base is 1 (pitcure a clock, with the base being drawn from 3:00 to the centre of the clock). The right triangle would have the following sides: 3:00-to centre, centre to 12:00, 12:00 to 3:00.

But, if we had a triangle drawn from 10:00 to centre, centre to 3:00, and 3:00 to 10:00, then the base would be the same, but the height would be smaller than 1 (because 10:00 is lower than 12:00, seeing as 12:00 is the apex of the clock). Same if you did it the other way, instead of 10:00, say the other vertex is at 1:00. Seeing as 1:00 is lower than 12:00, we still wouldn't be maximizing the area of our triangle within the clock.

So, the largest triangle inside a circle must go from the centre to two dots that are "3 hours" apart from one another, hence a right triangle.

Hope that helps!
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by reneerenee » Tue Dec 02, 2008 6:05 pm
:D whoops! sorry about that! I totally meant 1/2, and 1/2 = B. you're right. I'm gonna get killed on the gmat if i can't fill in the right letters!
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by vish150783 » Tue Dec 02, 2008 6:51 pm
Hey thanks a lot for your post. the OA is 1/2.

I was getting confused as to what the triangular region meant. When I saw this problem, I could not really picture a triangle. I was looking at a sector.

So, I began with a theta/360 * pi r^2 formula.

I guess wording plays a role in PS as well...yiikeees..

But I guess the general concept I learnt from all your posts is that a given a constant base, the right angled triangle gives the maximum area.

Am I right?
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by anayeri » Tue Dec 02, 2008 7:47 pm
Vish, that's a great way to put it!

Let's hope that comes up on both our exams!
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by anayeri » Tue Dec 02, 2008 8:22 pm
Vish, that's a great way to put it!

Let's hope that comes up on both our exams!
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