Question 1138627
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Please take the time to look at your post before you submit it.  The trapezoid does not have a vertex D; there is no segment ZD.  It is clear that you must mean ZX... but we shouldn't have to figure that out.<br>
Here is a rough picture, approximately to scale....<br>
{{{drawing(400,200,-15,15,-2,10,
line(0,0,12,0),line(12,0,4.5,6.6),line(4.5,6.6,-12,6.6),line(-12,6.6,0,0),line(0,0,4.5,6.6),
locate(0,-.5,Z),locate(12,-.5,Y),locate(4.5,8,X),locate(-12,8,W),
locate(6,-.5,12),locate(9,3.3,10),locate(-4.5,8,x),locate(-6.5,3,17),locate(2.2,3.3,8)
)}}}<br>
Angles YZX and WXZ are congruent -- parallel lines WX and YZ cut by transversal ZX.<br>
By the law of cosines in triangle XYZ,<br>
{{{10^2 = 8^2+12^2-2*8*12*cos(theta)}}}<br>
{{{cos(theta) = 108/192 = 9/16}}}<br>
By the law of cosines in triangle WXZ,<br>
{{{17^2 = x^2+8^2-2*x*8*cos(theta)}}}
{{{189 = x^2+64-16x(9/16)}}}
{{{x^2-9x-125 = 0}}}<br>
The quadratic expression does not factor; the quadratic formula gives us<br>
{{{x = (9 + sqrt(81+500))/2}}}<br>
which, to a few decimal places, is x = 16.552.<br>
ANSWER: The length of base WX is approximately 16.552.