Question 509333
Let's call the base b.

Imagine we have some sort of perpendicular bisector in our equilateral triangle.

This cuts our equilateral triangle into two 30,60,90 triangles.

We know that the 30 degree side is going to be b/2.

If we look at the perpendicular bisector, it corresponds to 60 degrees, and its relationship to b/2 will be sqrt(3) * (b/2).

So, our base is b and our height is (sqrt(3)*b)/2.

Then our area formula will be  (1/2)bh  =  (sqrt(3)/4)* b^2

So, (sqrt(3)/4) *b^2  =  316

sqrt(3)*b^2 = 1264

b^2 =  1264 /sqrt(3)

Rationalize the denominator.

b =  sqrt((1264*sqrt(3)) / 3)  

So the perimeter is   3*sqrt((1264*sqrt(3))/3)cm^2 or 81.04cm^2.

Please e-mail me at onlyyourmathtutor@yahoo.com if this does not make sense and I will be able to visually show you what I'm talking about.