Hmmm, I couldn't find a geometric way to solve this within the 1 minute I looked at this problem. I am very inpatient!
So let us use coordinates!
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Find ST
ST = 40/3 by simple algebra I think you can do at your skill level judging by the difficult of this problem.
Let QR = \(a\)
Slope of PT: \(-\frac{\frac{40}{3}}{a}\)
Y-intercept of PT: 20
Equation: \(y=-\frac{\frac{40}{3}}{a}x+20\)
Slope of QS: (20/a)
Y intercept of QS: 0
Equation: \(y=\frac{20}{a}x\)
Substitute
\(-\frac{\frac{40}{3}}{a}x+20=\frac{20}{a}x\)
\(20 =\frac{\frac{100}{3}}{a}x\)
\(20a=\frac{100x}{3}\)
\(60a=100x\)
\(3a=5x\)
\(\frac{5}{3}x=a\)
Interpret this:
\(x\) is the x-coordinate of the solution of we solved for the location of the intercept at point U. That means \(x\) is the length of QV. We know \(a\) is the length of QR.
That means QV is three-fifths the length of QR.
We know that QUV is similar to QSR by AA similarity through a series of proofs.
Since we know the proportion of the sides, we can solve for UV:
Solve:
20 * (3/5) = 12
Ta-da! Mathz! 
