Hi Prachich1987,

Nice question ... Thanks !

Pls find attached the explanation as to why (1) is not suff.

Hope this helps. Thanks.

4GMAT_Mumbai wrote:Hi Prachich1987,

Nice question ... Thanks !

Pls find attached the explanation as to why (1) is not suff.

Hope this helps. Thanks.

HI! 4gmat_mumbai, Its nice to see that statement 1 is not sufficient.

But why statement 2 on its own its sufficient.

Hexagon has in total 9 diagonals , 3 of length 2a and 6 of length of length sqrt 3 a ,

and those 3 diagonals bisect each other, I am not able to draw a figure where 3 diagonals of a hexagon each in equal length , bisect each other and still not a regular hexagon...

Shovan has given a very good explanation on Hexagon , I am posting the link , Hope you like it.
https://www.beatthegmat.com/hexagon-t70405.html#318024

Ok so what we need to prove here is that this hexagon consists of 6 equilateral triangles, and by this we need to prove either all the triangle sides are equal and/or that its interior angles each = 60

Using info. in B "all diagonals are equal and bisect each other", this means that the triangles formed by them each have 2 equal sides i.e isosceles and that the two angles opposite these sides are equal.

Using info. in A, "all hexagon sides are equal", this means that the angles opposite these sides (the ones around the center) are also equal and = 60 (cause 360 (sum of interior angles around center) /6= 60)

Now you proved that two sides and angles of these triangles are equal, and that one angle of these triangles =60 making the other two =60 as well, then they're all equilateral triangles.

So, OA is B?

C cause you use both info.

Please help me is B or C??

Any expert???

Anurag@Gurome wrote:Only if the hexagon is a regular hexagon, then all the six triangles formed by the three diagonals of the hexagon will be equilateral triangles.

Statement 1: All sides of the hexagon are of the same length.
This doesn't ensure that the

Not sufficient


Statement 2: The three diagonals of a hexagon are equal in length & bisect each other.
This also doesn't ensure that hexagon is a regular hexagon. See the figure below.

Not sufficient

1 & 2 Together: Now we cannot bend the properties as we did while analyzing the above two statements separately. Thus we will always get a regular hexagon.

Sufficient

The correct answer is C.

Nicely explained Anurag !
I liked the images you have posted. That is how I also try to solve some of these problems, they save significant amount of time and the concept becomes clearer (as we can see what is going on).