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# Square with triangles question thingy

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3 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...

Apr 9, 2019

#1
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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
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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   Apr 9, 2019
#3
+1

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

Apr 9, 2019