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).

thunderkirby
Nov 21, 2018

#2**+1 **

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

CPhill
Nov 21, 2018