Please explain - Hexagon problem

[sharmeenkhan81]

I came across this problem in one of the 800score.com practice tests.

There's a picture of a seemingly regular hexagon with the three diagonals drawn out that create six triangles.

Question: In the figure above, three segments are drawn to connect the opposite vertices of a hexagon, forming six triangles. All of the three 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.

I think the answer should be (C) because statement (2) essentially tell us that its a regular hexagon. Therefore doesn’t it mean that the three diagonals split the regular hexagon in six triangles with equal areas? Since statement (1) gives us the area of one of the triangles we can get the area of the hexagon by multiplying 12 by 6?

According to 800score.com the result is (E). Please advise. Thanks.

[ms]

G'day Sharmeen,
A polygon is defined as a 'regular polygon' only if it is equiangular (all angles are equal) and equilateral (all sides are equal).
1 and 2 combined don't give us complete info to say that the hexagon described is a regular hexagon (because we can't say yet that the all internal angles of this hexagon are equal). Hence the answer is E.
Cheers

[RonPurewal]

hi -

you can make hexagons that are equilateral (all sides equal) but NOT regular. here's an example of such a hexagon:
http://www.drking.plus.com/hexagons/misc/hex-area2.png
(image posted jan. 28 - i don't know how long this will stay up, the internet being the internet)

note that this is just a regular hexagon that is "flattened" a bit.

go ahead and sketch that one out (or flatten it out even more, just for emphasis). if you do so, then all the triangles won't have the same area anymore.

for the record, i wouldn't really worry about this problem. it's very gimmicky - a "gotcha" style problem; the test really doesn't do a lot of that.