In convex quadrilateral ABCD, AB=BC=13, CD=DA=24, and \(\angle D=60^\circ\). Points X and Y are the midpoints of \(\overline{BC}\) and \(\overline{DA}\) respectively. Compute \(XY^2\)(the square of the length of XY).
+130071 Here's the image :
Let D = (0,0)
Let C = ( 24sin 60, 24 sin 60) = (12, 12sqrt3 )
And by symmetry, angle BDC = 60/2 = 30°
We can find sin DBC as follows
sin BDC /13 = sin BCD/24
sin30/13 = sinDBC/24
1/26 =sinDBC/24
24/26 = sin DBC
12/13 = sin DBC
So
5/13 = cos BCD
And we can find BD using the Law of Cosines
DC^2 = BD^2 + BC^2 - 2(BD*BC) cosBCD
24^2 = BD^2 +13^2 - 2(BD *13)(5/13)
576 -169 = BD^2 - 10BD
407 = BD^2 - 10BD
BD^2 - 10BD - 407 = 0
The solution to this is
BD = 5 + 12sqrt3
So B = (0 , 5 + 12sqrt 3 )
So X is the midpoint of B and D =
[ 12/2, [ 5 +12sqrt(3) + 12sqrt(3)] /2 ] =
[ 6 , 2.5 + 12sqrt(3) ]
So....by symmetry... A is (-12, 12sqrt3)
So....Y is the midpoint of D and A = (-6, 6sqrt(3) )
So....the distance^2 X to Y =
[ 6 - (-6) ]^2 + [ 2.5 + 12sqrt (3) - 6sqrt(3) ]^2 =
12^2 + [ 2.5 + 6sqrt(3) ]^2 =
144 + 6.25 + 30sqrt(3) +108 ≈
310.21 units
