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# help

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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).

Dec 20, 2018

#1
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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   Dec 21, 2018