CU...it might be easier if we just considered this :
{BTW....this is not a "square " ...it's a rhombus]
Note that triangles ABC and FGC are similar
And the height of ABC = sqrt (3)
And note that FG is 1 unit above AB
So....the height of FGC must just be [ sqrt (3) - 1]
So
height of FGC 1/2 base of FGC
___________ = _______________
height of ABC 1/2 base of ABC
[sqrt(3) - 1 ] (1/2 ) base FGC
__________ = _____________
sqrt (3) 1
So (1/2) base FGC = [ sqrt (3) - 1] / sqrt (3)
So...its area must be its height * (1/2) its base =
[ sqrt (3) - 1 ] [sqrt (3) - 1 ] / sqrt (3) = [ 3 - 2sqrt (3) + 1 ] / sqrt (3) = [ 4 - 2 sqrt (3)] / sqrt (3)
So...by symmetry.....the area of the rhombus must be twice this = [ 8 - 4sqrt(3)] / sqrt (3) units^2