Let $ABCD$ be a rectangle in which $AB + BC + CD = 20$ and $AE = 9$ where $E$ is the midpoint of the side $BC$. Find the area of the rectangle.
1. Let \( AB = x \) and \( BC = 2y \). Since \( E \) is the midpoint of \( BC \), we have \( BE = EC = y \).
2. Given that \( AB + BC + CD = 20 \), and since \( AB = CD \) (opposite sides of a rectangle are equal), we can write:
   \[
   AB + BC + CD = x + 2y + x = 2x + 2y = 20
   \]
   Simplifying, we get:
   \[
   x + y = 10 \quad \text{(Equation 1)}
   \]
3. We are also given that \( AE = 9 \). Since \( AE \) is the diagonal from \( A \) to \( E \), we can use the Pythagorean theorem in triangle \( ABE \):
   \[
   AE^2 = AB^2 + BE^2
   \]
   Substituting the given values, we get:
   \[
   9^2 = x^2 + y^2
   \]
   Simplifying, we get:
   \[
   x^2 + y^2 = 81 \quad \text{(Equation 2)}
   \]
4. We need to find the area of the rectangle, which is \( AB \times BC = x \times 2y = 2xy \).
5. To find \( 2xy \), we square Equation 1:
   \[
   (x + y)^2 = 10^2
   \]
   Expanding, we get:
   \[
   x^2 + 2xy + y^2 = 100
   \]
6. We already have \( x^2 + y^2 = 81 \) from Equation 2. Subtracting Equation 2 from the expanded form, we get:
   \[
   x^2 + 2xy + y^2 - (x^2 + y^2) = 100 - 81
   \]
   Simplifying, we get:
   \[
   2xy = 19
   \]
7. Therefore, the area of the rectangle is:
   \[
   2xy = 19
   \]

The final answer is \(\boxed{19}\).