A point $ P$ is selected at random from the interior of the pentagon with vertices $ A \equal{} (0,2)$, $B \equal{} (4,0)$, $C \equal{} (2 \pi \plus{} 1, 0)$, $D \equal{} (2 \pi \plus{} 1,4)$, and $ E \equal{} (0,4)$. What is the probability that $ \angle APB$ is obtuse?
[asy]
size(150);
pair A, B, C, D, E;
A = (0,1.5);
B = (3,0);
C = (2 *pi + 1, 0);
D = (2 * pi + 1,4);
E = (0,4);
draw(A--B--C--D--E--cycle);
label("$A$", A, dir(180));
label("$B$", B, dir(270));
label("$C$", C, dir(0));
label("$D$", D, dir(0));
label("$E$", E, dir(180));
[/asy]
$ \displaystyle \textbf{(A)} \ \frac {1}{5} \qquad \textbf{(B)} \ \frac {1}{4} \qquad \textbf{(C)} \ \frac {5}{16} \qquad \textbf{(D)} \ \frac {3}{8} \qquad \textbf{(E)} \ \frac {1}{2}$
To solve this problem, we need to determine the probability that the angle \( \angle APB \) is obtuse when point \( P \) is randomly selected from the interior of the pentagon with vertices \( A = (0,2) \), \( B = (4,0) \), \( C = (2\pi + 1, 0) \), \( D = (2\pi + 1, 4) \), and \( E = (0,4) \).

1. **Understanding the Geometry**:
   - The pentagon is defined by the vertices \( A, B, C, D, \) and \( E \).
   - We need to find the region within the pentagon where \( \angle APB \) is obtuse.

2. **Condition for Obtuse Angle**:
   - An angle \( \theta \) is obtuse if \( 90^\circ < \theta < 180^\circ \).
   - For \( \angle APB \) to be obtuse, the point \( P \) must lie in a specific region relative to the line segment \( AB \).

3. **Using the Perpendicular Bisector**:
   - The perpendicular bisector of \( AB \) will help us determine the region where \( \angle APB \) is obtuse.
   - The coordinates of \( A \) and \( B \) are \( (0,2) \) and \( (4,0) \) respectively.
   - The midpoint \( M \) of \( AB \) is \( \left( \frac{0+4}{2}, \frac{2+0}{2} \right) = (2, 1) \).
   - The slope of \( AB \) is \( \frac{0-2}{4-0} = -\frac{1}{2} \).
   - The slope of the perpendicular bisector is the negative reciprocal, which is \( 2 \).

4. **Equation of the Perpendicular Bisector**:
   - Using the point-slope form of the line equation, the perpendicular bisector passing through \( M(2,1) \) with slope \( 2 \) is:
     \[
     y - 1 = 2(x - 2)
     \]
     Simplifying, we get:
     \[
     y = 2x - 3
     \]

5. **Region Determination**:
   - For \( \angle APB \) to be obtuse, point \( P \) must lie above the line \( y = 2x - 3 \).
   - We need to find the area of the pentagon above this line.

6. **Area Calculation**:
   - The total area of the pentagon can be calculated using the vertices.
   - The vertices form a simple polygon, and we can use the Shoelace Theorem to find the area.
   - The coordinates of the vertices are \( (0,2) \), \( (4,0) \), \( (2\pi + 1, 0) \), \( (2\pi + 1, 4) \), and \( (0,4) \).

7. **Shoelace Theorem**:
   - The area \( A \) of a polygon with vertices \((x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\) is given by:
     \[
     A = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|
     \]
   - Applying this to our vertices:
     \[
     A = \frac{1}{2} \left| 0 \cdot 0 + 4 \cdot 0 + (2\pi + 1) \cdot 4 + (2\pi + 1) \cdot 4 + 0 \cdot 2 - (2 \cdot 4 + 0 \cdot (2\pi + 1) + 0 \cdot (2\pi + 1) + 4 \cdot 0 + 4 \cdot 0) \right|
     \]
     Simplifying, we get:
     \[
     A = \frac{1}{2} \left| 8\pi + 4 + 8\pi + 4 - 8 \right| = \frac{1}{2} \left| 16\pi \right| = 8\pi
     \]

8. **Area Above the Line**:
   - The line \( y = 2x - 3 \) intersects the pentagon, dividing it into two regions.
   - The area above the line can be approximated by considering the geometry and symmetry of the pentagon.

9. **Probability Calculation**:
   - The probability that \( \angle APB \) is obtuse is the ratio of the area above the line to the total area of the pentagon.
   - Given the symmetry and the nature of the problem, this probability is approximately \( \frac{1}{2} \).

The final answer is \( \boxed{\frac{1}{2}} \)