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avatar+2551 

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!!!

 Apr 9, 2019
 #1
avatar+19789 
0

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

 Apr 9, 2019
 #2
avatar+106519 
+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

 

cool cool cool

 Apr 9, 2019
 #3
avatar+2551 
+1

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

 Apr 9, 2019

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