Compute the area of the trapezoid $ABCD$ with right angles $BAD$ and $ADC$ and side lengths of $AB=3$, $BC=5$, and $CD=7$.
1. **Identify the shape and given dimensions:**
   - The trapezoid \(ABCD\) has right angles at \(BAD\) and \(ADC\).
   - Given side lengths are \(AB = 3\), \(BC = 5\), and \(CD = 7\).

2. **Determine the height of the trapezoid:**
   - Since \(BAD\) and \(ADC\) are right angles, \(AB\) and \(CD\) are perpendicular to \(AD\).
   - Let \(P\) be a point such that \(BP \perp DC\). This means \(BP\) is the height of the trapezoid.

3. **Use the Pythagorean Theorem to find \(BP\):**
   - In the right triangle \(BCP\), where \(BC = 5\) and \(CP = CD - AB = 7 - 3 = 4\):
   \[
   BP^2 + CP^2 = BC^2 \implies BP^2 + 4^2 = 5^2 \implies BP^2 + 16 = 25 \implies BP^2 = 9 \implies BP = 3
   \]

4. **Calculate the area of the trapezoid:**
   - The area \(A\) of a trapezoid is given by:
   \[
   A = \frac{1}{2} \times (b_1 + b_2) \times h
   \]
   where \(b_1\) and \(b_2\) are the lengths of the parallel sides, and \(h\) is the height.
   - Here, \(b_1 = AB = 3\), \(b_2 = CD = 7\), and \(h = BP = 3\):
   \[
   A = \frac{1}{2} \times (3 + 7) \times 3 = \frac{1}{2} \times 10 \times 3 = 15
   \]

The final answer is \(\boxed{15}\)