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avatar+1117 

A 30-60-90 triangle is drawn on the exterior of an equilateral triangle so the hypotenuse of the right triangle is one side of the equilateral triangle. If the shorter leg of the right triangle is 6 units, what is the distance between the two vertices that the triangles do not have in common? Express your answer in simplest radical form.

 

Thanks!

AnonymousConfusedGuy  May 18, 2018
 #1
avatar+86859 
+1

For convenience, position the lower left vertex of the equilateral triangle at (0,0)

 

The side of this triangle is twice the length of the side opposite the 30 degree angle in the other triangle  =  2 * 6   = 12

 

And the height of the other triangle  is  √3  times the side opposite the 30 degree angle  =   6√3  units =  √108  units

 

So...the vertex  at the top right of the figure has the coordinates  (12, √108 )

 

And the distance between this vertex  and  (0, 0)  is

 

√ [ 12^2  + (√108)^2  ]  =

 

√ [144  + 108 ] =

 

√252  =

 

√ [ 36 * 7 ]  =

 

6√7  units

 

 

cool cool cool

CPhill  May 18, 2018
 #2
avatar+1117 
+1

Thanks!

AnonymousConfusedGuy  May 18, 2018

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