9. Rectangle $A B C D$ has an area of 2011 square centimeters. The vertices of trapezoid $A F G E$ are such that $F$ is on $B C$, and $D$ is the midpoint of the leg $E G$. Find the area of trapezoid $A F G E$.
【Analysis】According to the problem, connect $D F$. Triangle $A D F$ and rectangle $A B C D$ have the same base and height, so the area of triangle $A D F$ is half the area of rectangle $A B C D$. Since point $D$ is the midpoint of $E G$, and $A E$ is parallel to $F G$, triangle $A D F$ is also half the area of trapezoid $A F G E$. Because point $D$ is the midpoint of line segment $E G$, the areas of triangle $A D E$ and triangle $D G F$ are half the area of trapezoid $A F G E$, meaning the area of the trapezoid equals the area of the rectangle. According to this, we can solve the problem.
【Solution】Solution: As shown in the figure, connect $D F$.
The area of triangle $A D F = 2011 \div 2 = 1005.5$ (square centimeters),
Since point $D$ is the midpoint of $E G$,
The area of triangle $A E D +$ the area of triangle $D F G = 1005.5$ (square centimeters),
The area of trapezoid $A F G E$: $1005.5 + 1005.5 = 2011$ (square centimeters),
Answer: The area of trapezoid $A F G E$ is 2011 square centimeters.