Let $ABCD$ be a trapezoid such that $|AC|=8$, $|BD|=6$, and $AD \parallel BC$. Let $P$ and $S$ be the midpoints of $[AD]$ and $[BC]$, respectively. If $|PS|=5$, find the area of the trapezoid $ABCD$.
1. **Identify the given information and construct the necessary points:**
   - Trapezoid \(ABCD\) with \(AD \parallel BC\).
   - Diagonals \(AC\) and \(BD\) with lengths \(|AC| = 8\) and \(|BD| = 6\).
   - Midpoints \(P\) and \(S\) of \(AD\) and \(BC\) respectively, with \(|PS| = 5\).

2. **Construct parallelograms \(CADM\) and \(BDMN\):**
   - Extend \(AD\) and \(BC\) to form parallelograms \(CADM\) and \(BDMN\).
   - Since \(P\) and \(S\) are midpoints, \(PS\) is parallel to both \(AD\) and \(BC\).

3. **Analyze the properties of the trapezoid and the midsegment:**
   - The midsegment \(PS\) of a trapezoid is parallel to the bases and its length is the average of the lengths of the bases.
   - Let \(AD = a\) and \(BC = b\). Then, \(|PS| = \frac{a + b}{2}\).

4. **Use the given length of the midsegment to find the relationship between \(a\) and \(b\):**
   - Given \(|PS| = 5\), we have:
     \[
     \frac{a + b}{2} = 5 \implies a + b = 10
     \]

5. **Determine the height of the trapezoid using the diagonals:**
   - Since \(AC \perp BD\), the diagonals intersect at right angles.
   - The area of the trapezoid can be found using the formula for the area of a right-angled triangle formed by the diagonals:
     \[
     \text{Area} = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 8 \times 6 = 24
     \]

6. **Verify the area calculation:**
   - The area of the trapezoid is indeed given by the product of the diagonals divided by 2, confirming the calculation.

\(\blacksquare\)

The final answer is \( \boxed{ 24 } \).