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Find AC.

 

 May 6, 2022
 #1
avatar+124595 
+1

NVM .....Guest has the correct answer  !!!

 

cool cool cool

 May 6, 2022
edited by CPhill  May 6, 2022
 #2
avatar
+1

Draw CF ⊥ DA ..( F on DA)
Let z= x^2
Applying Pythagoras’s theorem in triangles ABD, BCE and ACF we get
BE^2=x^2-121=z-121
BD^2=z-169
AC^2=CF^2+AF^2 => (BE+BD)^2+4=z
Or 2.BE.BD=z-BE^2-BD^2-4= 286-z
4.BE^2.BD^2=4(z-121)(z-169)=(286-z)^2
Simplify this equation we get 3.z^2-588.z= 0
We get 2 solutions z= 0 and z= 196
Only z=196 is acceptable so x= 14

 May 6, 2022
 #3
avatar+124595 
0

THX, Guest....I realized later that mine was wrong  !!!!

 

 

cool cool cool

CPhill  May 6, 2022
 #4
avatar+2440 
+1

Here is the link to the original problem, if anyone is interested in other solutions: https://gogeometry.blogspot.com/2016/03/geometry-problem-1194-equilateral.html

 May 6, 2022

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