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).
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