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Right triangle \(\triangle ABC\) has side lengths AB=3, BC=4, and AC=5. Square XYZW is inscribed in \(\triangle ABC\) with X and Y on \(\overline{AC}\), W on \(\overline{AB}\), and Z on \(\overline{BC}\). What is the side length of the square?

 Oct 13, 2019
edited by xXxTenTacion  Oct 13, 2019
 #1
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Let the side of the square be  =  S

 

Because XW = YZ.....then WZ  is parallel to AC

Triangle ABC  is right with angle ABC = 90°

 

So.... 

angle BZW  = angle BCA 

angle BWZ  = angle BAC

And angle ZYC  = angle WXA = angle ABC  

 

 

Triangle YCZ is similar to triangle BCA  by AA congruency

So

YC / ZY  = BC/BA

YC / S  = 4/3

YC  = (4/3)S

 

And triangle XAW  is similar to triangle BAC  by AA congruency

So

AX / XW  =  BA / BC

AX / S = 3 /  4

AX  = (3/4)S

 

And  

AX + XY +  YC  =  5

So

(3/4)S  + S  +  (4/3)S  =  5

S [ 3/4 + 4/3 + 1 ]  = 5

S [ 9/12 + 16/12 + 12/12 ] / 12  = 5

S [ 37/12]  = 5

S  =  5 (12) /37   =  60 / 37

 

 

cool cool cool

 Oct 13, 2019

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