+87 In convex quadrilateral abcd,ab=bc=13 , cd=da=24, and \(\angle D=60^\circ\). Points X andY are the midpoints of \(\overline{BC}\) and \(\overline{DA}\) respectively. Compute XY^2 (the square of the length of XY).
+129852 See the image below
Let D = (0,0)
Let C = (24 sin60°, 24cos60°) = (12, 12√3)
Let A = (-12, 12√3)
And we can find B thusly
The distance from C to A = 24
And, by symmetry, EC = 12
So triangle EBC is a 5-12-13 Pythagorean Right Triangle
So B = (0, 12√3 + 5 )
Y is the midpoint of DA = (-6, 6√3)
X is the midpoint of BC = ( [ 0 + 12] / 2 , [ 12√3 + 5 + 12√3] / 2 ) =
(6, 12√3 + 2.5)
So....the square of the distance from X to Y is given by :
(-6 - 6)^2 + ( [ 12√3 + 2.5 ] - [6√3] )^2 =
(-12)^2 + [ 6√3 + 2.5 ]^2 =
144 + [ √108 + 2.5 ]^2
144 + 6.25 + 108 + 5√108 =
258.25 + 5√108 units ≈ 310.21 ≈ (17.61)^2 units
Thank you for trying though
