The perimeter of $\triangle UVZ$ equals $UV+UZ+VZ$.
We know that $UV=20$. We need to calculate $UZ$ and $VZ$.
Let $O$ be the point on $XW$ directly underneath $Z$.
Since $Z$ is the highest point on the semi-circle and $XW$ is the diameter, then $O$ is the centre of the semi-circle.
We join $UO, VO, UZ$, and $VZ$.
Since $UVWX$ is a rectangle, then $XW=UV=20$ and $UX=VW=30$.
Since $XW$ is a diameter of the semi-circle and $O$ is the centre, then $O$ is the midpoint of $XW$ and so $XO=WO=10$.
This means that the radius of the semi-circle is 10, and so $OZ=10$ as well.
Now $\triangle UXO$ and $\triangle VWO$ are both right-angled, since $UVWX$ is a rectangle.
By the Pythagorean Theorem, $UO^{2}=UX^{2}+XO^{2}=30^{2}+10^{2}=900+100=1000$ and $VO^{2}=VW^{2}+WO^{2}=30^{2}+10^{2}=1000$.
Each of $\triangle UOZ$ and $\triangle VOZ$ is right-angled at $O$, since the semi-circle is vertical and the rectangle is horizontal.
Therefore, we can apply the Pythagorean Theorem again to obtain $UZ^{2}=UO^{2}+OZ^{2}$ and $VZ^{2}=VO^{2}+OZ^{2}$.
Since $UO^{2}=VO^{2}=1000$, then $UZ^{2}=VZ^{2}=1000+10^{2}=1100$ or $UZ=VZ=\sqrt{1100}$.
Therefore, the perimeter of $\triangle UVZ$ is $20+2 \sqrt{1100} \approx 86.332$.
Of the given choices, this is closest to 86.