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 Post subject: GMAT Geometry (Data Sufficiency)
PostPosted: Mon Jul 13, 2009 2:51 pm 
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Joined: Sun Apr 19, 2009 6:56 pm
Posts: 32
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In the figure above, three segments are drawn to connect opposite vertices of a hexagon, forming six triangles. All three of these segments intersect at a point A. What is the area of the hexagon?

(1) One of the triangles has an area of 12
(2) All sides of the hexagon are of equal length

My question:
On the gmat, will it specifically state whether the hexagon is a regular
hexagon or not? I assumed that when you mentioned hexagon it was a regular
hexagon.


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 Post subject: Re: Test 1, question 25
PostPosted: Mon Jul 13, 2009 2:54 pm 
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Joined: Mon Apr 06, 2009 5:44 pm
Posts: 81
With the problem at hand, AEK, you don't need to know if the hexagon is regular or not. All of the necessary information is provided. However, on the GMAT when you actually encounter a problem that involves a geometric figure that is regular, and, it is required that you know the latter in order to solve the problem, then yes, the GMAT question will provide you with that information.

Take care,
Steve


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 Post subject: Re: Test 1, question 25
PostPosted: Mon Jul 13, 2009 3:00 pm 
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Joined: Sun Apr 19, 2009 6:56 pm
Posts: 32
A regular hexagon has all its sides equal to each other. The
one shown in your example does not have all its sides equal and hence
doesnot justify your explanation.


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 Post subject: Re: Test 1, question 25
PostPosted: Mon Jul 13, 2009 3:02 pm 
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It looks like (2) states that all sides of the hexagon are the same length. :geek:


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 Post subject: Re: Test 1, question 25
PostPosted: Mon Nov 14, 2011 1:13 pm 
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I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!


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 Post subject: Re: Test 1, question 25
PostPosted: Mon Nov 14, 2011 1:55 pm 
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questioner wrote:
I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!
No. You can see that if you draw various cases of 3 arbitrary segments that intersect in one point, then you just need to connect the ends of the segments to create various hexagons.

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