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by vaivish » Sun Aug 17, 2008 10:27 am
The shaded region is a regular hexagon, created by linking the midpoints of the sides if the large regular hexagon. If the area of the large hexagon is 54*root3, what is the area of the small hexagon?
(A) 27*root3
(B) 36*root3
(C) 40.5*root3
(D) 42.5*root3
(E) 45*root3

OA is C.
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Source: — Problem Solving |

by bourne159 » Sun Aug 17, 2008 10:47 am
The answer is C

First figure out length of each side of the large hexagon:
Area = 3 root(3)/2 * side^2 = 54 root(3)
Side = 6

Now for the smaller hexagon:
If we figure out the side of the smaller hexagon we get the area.
Each of the small triangles can be partitioned into two equal right angled triangles (30-60-90 right triangles) The side ratios will be 1:root(3):2
Using that you can get length of each side of the smaller hexagon to be 3 root(3)

Use the area of the hexagon formula above to get 40.5 root(3)
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by pepeprepa » Sun Aug 17, 2008 11:01 am
Can you give the source of the question?
I find it strange you need to use the area formula of an hexagon
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by parallel_chase » Sun Aug 17, 2008 11:08 am
It is a strange question. The method posted by BOURNE 159 is absolutely correct, but it requires one to assume that all the sides are of equal length.
Which assures that this is not a real GMAT question.
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by sudhir3127 » Sun Aug 17, 2008 11:11 am
parallel_chase wrote:It is a strange question. The method posted by BOURNE 159 is absolutely correct, but it requires one to assume that all the sides are of equal length.
Which assures that this is not a real GMAT question.
i second chase and pepeprepas thoughts!! do let us know the source of the question..
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by pepeprepa » Sun Aug 17, 2008 11:27 am
Chase concerning the fact that all sides are equal, the question tells it by "regular hexagon". All internal angles are 120° each and all its sides are the same.
And the formula of the area is (3*sqrt(3)/2)*s^2 as bourne used it. For a non regular hexagon this formula is false.
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by youngwolf » Mon Aug 18, 2008 9:07 pm
There is no need to remember the formula for regular hexagon. You break down the regular hexagon in six and work with 6 identical equilateral triangles.
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