【Analysis】We can use the principle of equal-area transformation, translating $\triangle D E F$ towards point $B$. The shape and size of $\triangle D E F$ remain unchanged. After translation, $D F$ of $\triangle D E F$ overlaps with $A B$. At this point, the area between the isosceles right triangle $A B C$ and the isosceles right triangle $D E F$ remains unchanged. However, the length of $E C$ changes from 4 to 6. At this point, it is not difficult to calculate the area of triangle $A B C$.

【Solution】Solution: According to the analysis, using the principle of equal-area transformation, we translate $\triangle D E F$ towards point $B$. The shape and size of $\triangle D E F$ remain unchanged. After translation, $D F$ of $\triangle D E F$ overlaps with $A B$. At this point, the area between the isosceles right triangle $A B C$ and the isosceles right triangle $D E F$ remains unchanged.

At this point, the length of $E C$ changes from 4 to 6, as shown in the figure:
Draw $E G \perp A C$ intersecting $A C$ at $G$. In the right triangle $\triangle E G C$, it is not difficult to see that $E G = G C = \frac{1}{2} \times \sqrt{2} \times 6 = 3 \sqrt{2}$.
Since the area between the isosceles right triangle $A B C$ and the isosceles right triangle $D E F$ is 20, i.e., the area of trapezoid $A C E F$ is 20,
\[
\therefore (E F + A C) \times E G \times \frac{1}{2} = (E F + A G + G C) \times E G \times \frac{1}{2} = (2 \times E F + 3 \sqrt{2}) \times 3 \sqrt{2} \times \frac{1}{2} = 20 \Rightarrow E F = \frac{11 \sqrt{2}}{6}
\]
Thus, $B F = \frac{11 \sqrt{2}}{6}$, the area of $\triangle B E F = \frac{1}{2} \times B F \times E F = \frac{1}{2} \times \frac{11 \sqrt{2}}{6} \times \frac{11 \sqrt{2}}{6} = \frac{121}{36}$, and the area of triangle $A B C = \text{area of } \triangle B E F + 20 = \frac{121}{36} + 20 = \frac{841}{36}$.
Therefore, the answer is: $\frac{841}{36}$.
11. (10 points) As shown in the figure, the area between isosceles right triangle $ABC$ and isosceles right triangle $DEF$ is $20, BD=$ 2, $EC=4$, find the area of triangle $ABC$.

