Find the lenth

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!