Line segment $\overline{AB}$ has perpendicular bisector $\overline{CD}$, where $C$ is the midpoint of $\overline{AB}$. The segments have lengths $AB = 72$ and $CD = 60$. Let $R$ be the set of points $P$ that are midpoints of line segments $\overline{XY}$ , where $X$ lies on $\overline{AB}$ and $Y$ lies on $\overline{CD}$. Find the area of the region $R$.
1. **Identify the coordinates of points \(A\), \(B\), \(C\), and \(D\):**
   - Let \(A = (-36, 0)\) and \(B = (36, 0)\) since \(AB = 72\) and \(C\) is the midpoint of \(\overline{AB}\).
   - Therefore, \(C = (0, 0)\).
   - Since \(\overline{CD}\) is the perpendicular bisector of \(\overline{AB}\) and \(CD = 60\), \(D\) lies on the y-axis. Let \(D = (0, 30)\) and \(-D = (0, -30)\).

2. **Determine the coordinates of \(X\) and \(Y\):**
   - \(X\) lies on \(\overline{AB}\), so \(X\) can be any point \((x, 0)\) where \(-36 \leq x \leq 36\).
   - \(Y\) lies on \(\overline{CD}\), so \(Y\) can be any point \((0, y)\) where \(-30 \leq y \leq 30\).

3. **Find the coordinates of the midpoint \(P\) of \(\overline{XY}\):**
   - The midpoint \(P\) of \(\overline{XY}\) is given by:
     \[
     P = \left( \frac{x + 0}{2}, \frac{0 + y}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right)
     \]
   - Therefore, \(P\) can be any point \(\left( \frac{x}{2}, \frac{y}{2} \right)\) where \(-36 \leq x \leq 36\) and \(-30 \leq y \leq 30\).

4. **Determine the range of \(P\):**
   - Since \(x\) ranges from \(-36\) to \(36\), \(\frac{x}{2}\) ranges from \(-18\) to \(18\).
   - Since \(y\) ranges from \(-30\) to \(30\), \(\frac{y}{2}\) ranges from \(-15\) to \(15\).

5. **Identify the region \(R\):**
   - The set of points \(P\) forms a rectangle with vertices at \((-18, -15)\), \((18, -15)\), \((18, 15)\), and \((-18, 15)\).

6. **Calculate the area of the region \(R\):**
   - The length of the rectangle is \(18 - (-18) = 36\).
   - The width of the rectangle is \(15 - (-15) = 30\).
   - Therefore, the area of the rectangle is:
     \[
     36 \times 30 = 1080
     \]

The final answer is \(\boxed{1080}\).