Just assume the triangles are equilateral and that it isn't squiggly because I drew it.

Solve for the center square thingy

I got the answer of \(\frac{1}{2}\) but apparently it was wrong...

Please explain!!!

CalculatorUser Apr 9, 2019

#1**0 **

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

ElectricPavlov Apr 9, 2019

#2**+2 **

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

CPhill Apr 9, 2019