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In the figure above, three segments are drawn to connect
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Updated on: 29 Jul 2015, 03:15
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17% (01:30) correct 83% (01:16) wrong based on 165 sessions
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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 point A. What is the area of the hexagon? 1) One of the triangles has an area of 12. 2) All the sides of the hexagon are of equal length. OFFICIAL SOLUTION:Attachment:
hexagon1.gif [ 3.85 KiB  Viewed 4112 times ]
(E) Statement (1) is not sufficient because we do not know whether the hexagon is a regular hexagon (meaning that all 6 sides have the same length). We could only calculate the area if the hexagon could be divided into six congruent triangles. Statement (2) is also insufficient because we are not given any numbers. So, it is impossible to calculate the area. Combined, the two statements are still insufficient. A hexagon could have six equal sides, yet not be a regular hexagon. Imagine pushing opposite sides of a regular hexagon closer to each other: the hexagon would flatten, while all the sides would remain the same length. One of the triangles within this flattened hexagon could have an area of 12, but not all of the triangles would have the same area. Therefore, we can't calculate the area, even by combining the statements, and the correct answer is choice (E).
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Originally posted by coelholds on 28 Jul 2009, 14:28.
Last edited by Bunuel on 29 Jul 2015, 03:15, edited 6 times in total.
Edited the question.



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Re: In the figure above, three segments are drawn to connect
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28 Jul 2009, 14:46
A If the hexagon is a true hexagon and all sides equal, all angles equal, etc, then drawing three lines between the verticies should create 6 equal triangles, therefore, knowing the area of one, means we can determine the area of all 6 and thereby determine the area of the hexagon. B is insufficient because even if the sides are equal, it's not new information. That's generally understood unless this is not a perfect hexagon where all sideas and angles are equal. If we cannot assume this is a pure hexagon, then neither is sufficient and they're not sufficient together, so the answer would be E. Otherwise, my answer is A. coelholds wrote: 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 point A. What is the area of the hexagon?
1) One of the triangles has an area of 12. 2) All the sides of the hexagon are of equal length.
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Re: In the figure above, three segments are drawn to connect
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28 Jul 2009, 15:34
jallenmorris wrote: A
If the hexagon is a true hexagon and all sides equal, all angles equal, etc, then drawing three lines between the verticies yes but they did not said that it was a "true one, with all side equal" Answer should be C. we can only solve the pb if we know that it is a regular hexagone. Attachment:
hex.GIF [ 2.49 KiB  Viewed 6903 times ]
exemple of an hexagone that is not regular...



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Re: In the figure above, three segments are drawn to connect
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28 Jul 2009, 16:04
IMO C,
Only the question stem will not suffice that it's a regular hexagon.
1) We dont know yet if its a regular hexagon yet.  NS 2) Says that it is a regular hexagon, but doesn't give us any other info.  NS 1+2) Regular hexagon with each triangle of area 12 . So Area = 12* 6 = 72 square units. Suff



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Re: In the figure above, three segments are drawn to connect
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28 Jul 2009, 16:54
As I said before, if we do not know that the sides are equal length AND the angles are all equal the answer should be E. Here is why: We are still missing a piece of necessary information if we combine #1 and #2. I think we can all agree at this point that each independent statement is insufficient and the choices really come down to C or E. If we want to use the information in #1 to multiply by 6 for the entire area of the hexagon, then we must be able to determine that all triangles created are equal, but in order to know that all triangles are equal, we must know that the sides of the hexagon are equal AND the angles are equal as well. Statement #2 only gives us the length of the sides as equal, and not the angles too. See the picture attached to my answer. The lines in the picture are all equal, but the triangles created in the interior are not equal, so by knowing the area of 1, we could not find the area of the others. Answer E should be correct. Attachment:
Hexagon.jpg [ 568.41 KiB  Viewed 6915 times ]
coelholds wrote: 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 point A. What is the area of the hexagon?
1) One of the triangles has an area of 12. 2) All the sides of the hexagon are of equal length.
Attachments
Hexagon.jpg [ 568.41 KiB  Viewed 6902 times ]
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Re: In the figure above, three segments are drawn to connect
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28 Jul 2009, 17:35
You certainly do have a point there.
What is the OA and explaination?



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Re: In the figure above, three segments are drawn to connect
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28 Jul 2009, 19:58
Will go with E.. nothing is given about the 3 line segments intersecting each other.
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Re: In the figure above, three segments are drawn to connect
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28 Jul 2009, 20:11
I would go with E too. What is the OA?



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Re: In the figure above, three segments are drawn to connect
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Updated on: 29 Jul 2009, 05:26
What I wanted to point out is that If all side of an hexagone are equal, then all angle are also equal. hence we can solve this question.
Like a triangle, If a triangle has all sides equals, then the angles are equals too.
am I wrong? if this assumption is wrong therefore answer is E
what is OA?
Originally posted by madeinafrica on 29 Jul 2009, 04:03.
Last edited by madeinafrica on 29 Jul 2009, 05:26, edited 1 time in total.



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Re: In the figure above, three segments are drawn to connect
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29 Jul 2009, 04:53
Yes guys OA is E That is why I have posted this question here. This took me a while to believe. The fact is that you need to visualize that figure that jallenmorris have attached. Congrats!!!!



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Re: In the figure above, three segments are drawn to connect
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29 Jul 2009, 06:06
coelholds wrote: Yes guys OA is E Thanks guy Could some one telle me where i'm wrong? on wiki i can find this: "The internal angles of a regular hexagon (where all of the sides are the same) are all 120° and the hexagon has 720 degrees T."So I still think that if all side of an hexagon are the same, then all angles are equal. hence I do not understant why answer is not C.



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Re: In the figure above, three segments are drawn to connect
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29 Jul 2009, 06:34
madeinafrica, no problem. You are just stuck in the same problem that I was. Pay attention in what you wrote: "The internal angles of a regular hexagon..." ops!!! Did the question say that it is a REGULAR hexagon? No, and that is why we can not assume that it is. Even with a such beautiful and perfect image in the question. Remember, in DATA SUFFICIENCY questions, the image is not on scale. So, what about an hexagon as jallenmorris has posted? Imagine the extremities even more close until forming almost a square. Calculate the area of the triangles, and you will see they are different. That is because it is not a REGULAR hexagon. So, you can have a hexagon with equal legs, but different angles. BUT, you can not have a hexagon with the equal angles and different legs! It is clear now? If not, ask again, then I will try a more detail explanation.... PS.: If you liked the post, consider a kudo. I need just one more to access the GMATClub tests!!! Thank you



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Re: In the figure above, three segments are drawn to connect
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29 Jul 2009, 13:31
Here is another example of a hexagon with equal sides but unequal angles. Only with regard to a triangle can we assume equal angles with equal sides. No other polygon fits into this situation.
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Re: In the figure above, three segments are drawn to connect
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30 Jul 2009, 00:39
jallenmorris wrote: Here is another example of a hexagon with equal sides but unequal angles.
Only with regard to a triangle can we assume equal angles with equal sides. No other polygon fits into this situation. Thanks. Kudos to you.



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Re: In the figure above, three segments are drawn to connect
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15 Feb 2012, 05:52
Taking statement 2 into consideration,the hexagon is regular.If its regular with divided by 3 line segments intersecting with their ends separated equally will lead equal triangle areas.that way we can get 6*12 as the area.When have i gone wrong.Please help



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Re: In the figure above, three segments are drawn to connect
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15 Feb 2012, 08:08
gmatnerd wrote: Taking statement 2 into consideration,the hexagon is regular.If its regular with divided by 3 line segments intersecting with their ends separated equally will lead equal triangle areas.that way we can get 6*12 as the area.When have i gone wrong.Please help What if I squeeze some air out of the hexagon, it still has equal sides but the inside diagonals will differ, and nowhere does the question say that the segments are equal. Think, Think ! Its a nice question. Even I was a bit apprehensive about it, but I guess the diagram is misleading, it makes you think that the segments are equal in length. If you still don't get it, look at the digram below. Attachment:
Hexagons.jpg [ 85.57 KiB  Viewed 6053 times ]



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Re: In the figure above, three segments are drawn to connect
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15 Feb 2012, 19:13
Thanks a lot for the eplination.I was still not convinced(may be because i dont have the math to prove either way) so went and checked in some math forum.2 hexagons are congruent IFF both sides and all internal angles are equal (which means its possible to construct more than one hexagon with equal sides).This means that the area cannot be estimated from just knowing the side length and all of them are equal



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Re: In the figure above, three segments are drawn to connect
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05 May 2018, 23:17
TAKEAWAY Only triangle is the polygon where equal sides proves regular polygon."If all sides of a triangle are equal, triangle is equilateral with all angles equal. It is a regular polygon." "For polygons (other than triangle), equality of sides doesn't make it a regular polygon."For example in quadrilateral, all sides equal proves Quadrilateral to be rhombus, not square. we need additional information to prove the same.
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