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In convex quadrilateral ABCD, AB=BC=13, CD=DA=24, and angle D=60 degrees. Points X and Y are the midpoints of segment BC and segment DA respectively. Compute XY^2 (the square of the length of XY). 

 Jun 9, 2019
 #1
avatar+129849 
+5

It's easiest if we lay it out like this

 

 

D = (0,0)   A = (-12, 12sqrt(3) )  B = (12, 12sqrt(3) )

 

We can find C  by constructing a circle with a radius of 13 centered at B  and letting x = 0

 

(0 - 12)^2 + (y - 12sqrt(3))^2 = 169

12^2 + (y - 12sqrt (3))^2 = 169

144 + (y -12sqrt (3))^2 = 169

(y - 12sqrt(3))^2 = 25

y - 12sqrt (3)  = 5

y = 5 + 12sqrt (3)

 

So    C  = (0 , 5+12sqrt(3))

 

The midpoint of DA  =  (-6, 6sqrt(3) )  = Y

 

The midpoint of  BC  =  [   6,  [5 + 12sqrt (3) + 12sqrt(3) / 2 )  =  (6 , 2.5 + 12sqrt(3) ) = X

 

 

So   XY^2  =

 

(-6 -6)^2   + (2.5 + 12sqrt(3) - 6sqrt(3) )^2  =

 

(-12)^2  +  ( 2.5 + 6sqrt(3) )^2  =

 

144 + ( 2.5 + 6sqrt(3) )^2  =

 

144 + 6.25 + 30sqrt (3) + 108  ≈ 

 

310.21 units

 

 

cool cool cool

 Jun 9, 2019
 #2
avatar+88 
+2

thank you so much!!!!! 

 Jun 9, 2019

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