A point $P$ is selected at random from the interior of the pentagon with vertices $A=(0,2)$, $B=(5,0)$, $C=\left(2\pi+\frac54,0\right)$, $D=\left(2\pi+\frac54,4\right)$, and $E=(0,4)$. What is the probability that $\angle APB$ is obtuse?

1. $\angle APB$ is obtuse iff $P$ lies inside the circle with diameter $AB$.
2. $AB=\sqrt{5^2+2^2}=\sqrt{29}$, so the semicircle with diameter $AB$ on the pentagon's side has area $\frac12\pi\left(\frac{\sqrt{29}}2\right)^2=\frac{29\pi}{8}$. Its center is $\left(\frac52,1\right)$ and its radius $\frac{\sqrt{29}}2<2.7$, so its highest point is below $y=4$ and its rightmost point is left of $x=2\pi+\frac54$: it lies entirely inside the pentagon.
3. $[ABCDE]=4\left(2\pi+\frac54\right)-\frac12(2)(5)=8\pi+5-5=8\pi$.
4. Probability $=\dfrac{29\pi/8}{8\pi}=\boxed{\dfrac{29}{64}}$.