80
【Solution】According to the analysis, as shown in the figure, draw $D F \perp B C$ intersecting $B C$ at $F$. In the isosceles right triangle $B D E$, it is obvious that $F B=F D=F E$. In $\triangle A B C$, when $A B \perp A C$, $A B: A C=1: 2$. Similarly, $D F \perp F C$, so $D F$ : $F C=1: 2$,

Also, $\because D F=F E, \therefore D F=E C$, i.e., $F D=F B=F E=E C, \quad \therefore B E=\frac{2}{3} \times B C$, hence $\mathrm{BE}^{2}=\frac{4}{9} \times \mathrm{BC}^{2}=\frac{4}{9}\left(12^{2}+24^{2}\right)=320$
Therefore, the area of the shaded part $S=\frac{1}{2} \times \mathrm{BE} \times \mathrm{DF}=\frac{1}{2} \times \mathrm{BE} \times\left(\frac{1}{2} \times \mathrm{BE}\right)=\frac{1}{4} \times \mathrm{BE}^{2}=80$ (square centimeters).
The answer is: 80.
11. (12 points) As shown in the figure, in right triangle $A B C$, the length of $A B$ is 12 cm, and the length of $A C$ is 24 cm. Points $D$ and $E$ are on $A C$ and $B C$ respectively. The area of isosceles right triangle $B D E$ is ______ square centimeters.