In the diagram, $U V W X$ is a rectangle that lies flat on a horizontal floor. A vertical semi-circular wall with diameter $X W$ is constructed. Point $Z$ is the highest point on this wall. If $U V=20$ and $V W=30$, the perimeter of $\triangle U V Z$ is closest to
(A) 95
(B) 86
(C) 102
(D) 83
(E) 92

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-041.jpg?height=333&width=561&top_left_y=381&top_left_x=1151)
The perimeter of $\triangle U V Z$ equals $U V+U Z+V Z$.

We know that $U V=20$. We need to calculate $U Z$ and $V Z$.

Let $O$ be the point on $X W$ directly underneath $Z$.

Since $Z$ is the highest point on the semi-circle and $X W$ is the diameter, then $O$ is the centre of the semi-circle.

We join $U O, V O, U Z$, and $V Z$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-070.jpg?height=453&width=542&top_left_y=763&top_left_x=867)

Since $U V W X$ is a rectangle, then $X W=U V=20$ and $U X=V W=30$.

Since $X W$ is a diameter of the semi-circle and $O$ is the centre, then $O$ is the midpoint of $X W$ and so $X O=W O=10$.

This means that the radius of the semi-circle is 10 , and so $O Z=10$ as well.

Now $\triangle U X O$ and $\triangle V W O$ are both right-angled, since $U V W X$ is a rectangle.

By the Pythagorean Theorem, $U O^{2}=U X^{2}+X O^{2}=30^{2}+10^{2}=900+100=1000$ and $V O^{2}=V W^{2}+W O^{2}=30^{2}+10^{2}=1000$.

Each of $\triangle U O Z$ and $\triangle V O Z$ 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 $U Z^{2}=U O^{2}+O Z^{2}$ and $V Z^{2}=V O^{2}+O Z^{2}$.

Since $U O^{2}=V O^{2}=1000$, then $U Z^{2}=V Z^{2}=1000+10^{2}=1100$ or $U Z=V Z=\sqrt{1100}$.

Therefore, the perimeter of $\triangle U V Z$ is $20+2 \sqrt{1100} \approx 86.332$.

Of the given choices, this is closest to 86 .