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Geometry help plz!

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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).

Nov 21, 2018

#1
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I tried to do it, but that is not enough information.

Nov 21, 2018
#2
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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

Nov 21, 2018
edited by CPhill  Nov 21, 2018
#3
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I am sorry Cphill, but that is incorrect  Thank you for trying though

thunderkirby  Nov 28, 2018
edited by thunderkirby  Nov 28, 2018