hectictar

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Usernamehectictar
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 #3
avatar+9466 
+5

Here's a graph you can play with:  https://www.desmos.com/calculator/wy7dptwaij

 

To start with, both angles are set at  38°  so both triangles are congruent (by SAS congruency).

 

Try changing the slider for  b  to make that angle bigger and see what happens to the length of  YX

 

And note that the lengths of  WX  and  WZ  stay constant.

 

*edit*

 

Here is a new graph with calculation for the lengths of  ZY and  XY  and a slider for the length of  WY  (in the folder):

 

https://www.desmos.com/calculator/xfphvzq2jt

Jul 25, 2019
 #1
avatar+9466 
+5

(This question reminds me of this solution smiley)

 

m∠EFH  =  m∠EGF   because they are both right angles

m∠FEH  =  m∠GEF   because they are the same angle

 

So by the AA similarity theorem,  △EFH ~ △EGF

 

EF / EH  =  EG / EF

                                  Multiply both sides of the equation by  EF

EF2 / EH  =  EG

                                  Multiply both sides of the equation by  EH

EF2  =  (EG)(EH)

                                  Take the square root of both sides.

EF  = \(\sqrt{\text{(EG)(EH)}}\)

 

By the way, welcome to the forum! smiley

Jul 25, 2019