What assumption do I have to make to solve this problem? This is an SAT math problem that has been really bugging me lately.

I know that each side of the hexagon is six units long. This is as far as I got (forgive me for not being able to do anything else).
 A: Since each side of the hexagon is $6$, that means every side of the large equilateral triangles is $6$. Do you see how two of the sides of each large equilateral triangle is divided into $3$ pieces? Two bold lines, one dashed? Since all the sides are equal and the small triangles are also equilateral, those $3$ pieces must be equal, which means the length of each is $2$. Once we have that, we can see that we have $6 \cdot 8 =48$ sides, so the perimeter of the whole figure is $48 \cdot 2 = 96$
A: Each side of hexgin has been repalced by 2 sides of equilateral triangle. 36 *2 = 72. Each side of the big equilateral triangele has been deleted by 1/3 that 2 = 6/2. The deleted portion is pacthed up by two sides of small equilateral triangle. We have used one side to cover up the deleted portion, and need to couter one ly side of the small triangle. The extra length is 6 *2 *2 = 24.
So the total is 72 +24 = 96.
A: This question throws you off by getting you to focus on the hexagon. Instead look at one of the large triangles. With some work you can determine that the perimeter of a large triangle is 54. Notice that when you remove the middle third of each side and replace it with two sides of the appropriate equilateral triangle, the perimeter increases by one third. This happens twice, resulting in the final figure. Increase 54 by one third and you get 72. Do this one more time and you get 96. The answer is C.  
