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$ABC$ is an equilateral triangle such that vertices $B,C$ lie on two parallel lines at a distance of 6 units. $A$ lies between the parallel lines at a distance 4 units from one of them.  Find the side length of the equilateral triangle.

 

 Dec 29, 2020
 #1
avatar+836 
+2

 ABC is an equilateral triangle such that vertices B,C lie on two parallel lines at a distance of 6 units. A lies between the parallel lines at a distance of 4 units from one of them.  Find the side length of the equilateral triangle.

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The side length of a triangle ABC is ≈ 6.11 units

 Dec 29, 2020
 #2
avatar+114171 
+2

Imgaine that we  have a rectangle with a height of  6

 

From C, draw a perpendicular to the base of the rectangle

 

We have  three right triangles 

 

sqrt (S^2  - 2^2)   is the width of the  rectangle

And  sqrt (S^2  - 6^2)  +  sqrt (S^2 - 4^2)   also represents the width

 

So

 

sqrt (S^2  -  4)  = sqrt ( S^2 - 36) + sqrt ( S^2 - 16)        square both sides

 

S^2 - 4 =  S^2 - 36  + S^2 -16   + 2 sqrt [ (S^2  -36) ( S^2 - 16) ]

 

48 - S^2  =  2sqrt  [(S^2 - 36) (S^2 -16)  ]        square both sides again

 

S^4  - 96S^2  + 2304  = 4 S^4 - 208S^2  + 2304

 

3S^4 - 112S^2  =  0

 

S^2  ( 3S^2  - 112)   = 0

 

The second factor will produce a positive result for S

 

S^2  =   112/3

 

S =  sqrt  (112 / 3)  =   sqrt  (16 * 7) / sqrt (3)  =  4sqrt (7/3) ≈  6.11   (as jugoslav found  !!!)

 

 

cool cool cool

 Dec 29, 2020

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