Square with triangles question thingy

If the side length is 2...the center of the square is   1,1     ....is that what you are looking for????

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

Thank you CPhill! That is the explanation I am looking for!