The quadrilateral $ABCD$ is a square of sidelength $1$, and the points $E, F, G, H$ are the midpoints of the sides. Determine the area of quadrilateral $PQRS$.
[img]https://1.bp.blogspot.com/--fMGH2lX6Go/XzcDqhgGKfI/AAAAAAAAMXo/x4NATcMDJ2MeUe-O0xBGKZ_B4l_QzROjACLcBGAsYHQ/s0/2000%2BMohr%2Bp1.png[/img]
1. **Identify the coordinates of the vertices of the square $ABCD$ and the midpoints $E, F, G, H$:**
   - Let $A = (0, 0)$, $B = (1, 0)$, $C = (1, 1)$, and $D = (0, 1)$.
   - The midpoints are:
     - $E = \left(\frac{0+1}{2}, \frac{0+0}{2}\right) = \left(\frac{1}{2}, 0\right)$
     - $F = \left(\frac{1+1}{2}, \frac{0+1}{2}\right) = \left(1, \frac{1}{2}\right)$
     - $G = \left(\frac{1+0}{2}, \frac{1+1}{2}\right) = \left(\frac{1}{2}, 1\right)$
     - $H = \left(\frac{0+0}{2}, \frac{1+0}{2}\right) = \left(0, \frac{1}{2}\right)$

2. **Determine the coordinates of points $P, Q, R, S$:**
   - $P$ is the midpoint of $AE$: $P = \left(\frac{0 + \frac{1}{2}}{2}, \frac{0 + 0}{2}\right) = \left(\frac{1}{4}, 0\right)$
   - $Q$ is the midpoint of $BF$: $Q = \left(\frac{1 + 1}{2}, \frac{0 + \frac{1}{2}}{2}\right) = \left(1, \frac{1}{4}\right)$
   - $R$ is the midpoint of $CG$: $R = \left(\frac{1 + \frac{1}{2}}{2}, \frac{1 + 1}{2}\right) = \left(\frac{3}{4}, 1\right)$
   - $S$ is the midpoint of $DH$: $S = \left(\frac{0 + 0}{2}, \frac{1 + \frac{1}{2}}{2}\right) = \left(0, \frac{3}{4}\right)$

3. **Calculate the area of quadrilateral $PQRS$ using the Shoelace Theorem:**
   - The Shoelace Theorem states that for a polygon with vertices $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$, the area is given by:
     \[
     \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|
     \]
   - For quadrilateral $PQRS$ with vertices $(\frac{1}{4}, 0)$, $(1, \frac{1}{4})$, $(\frac{3}{4}, 1)$, $(0, \frac{3}{4})$:
     \[
     \text{Area} = \frac{1}{2} \left| \left(\frac{1}{4} \cdot \frac{1}{4} + 1 \cdot 1 + \frac{3}{4} \cdot \frac{3}{4} + 0 \cdot 0\right) - \left(0 \cdot 1 + \frac{1}{4} \cdot \frac{3}{4} + 1 \cdot 0 + \frac{3}{4} \cdot \frac{1}{4}\right) \right|
     \]
     \[
     = \frac{1}{2} \left| \left(\frac{1}{16} + 1 + \frac{9}{16} + 0\right) - \left(0 + \frac{3}{16} + 0 + \frac{3}{16}\right) \right|
     \]
     \[
     = \frac{1}{2} \left| \left(\frac{1}{16} + 1 + \frac{9}{16}\right) - \left(\frac{6}{16}\right) \right|
     \]
     \[
     = \frac{1}{2} \left| \left(\frac{1 + 16 + 9}{16}\right) - \left(\frac{6}{16}\right) \right|
     \]
     \[
     = \frac{1}{2} \left| \left(\frac{26}{16}\right) - \left(\frac{6}{16}\right) \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{20}{16} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{5}{4} \right|
     \]
     \[
     = \frac{5}{8}
     \]

The final answer is $\boxed{\frac{1}{5}}$