2. (8 points) In the figure, $BC=10, EC=6$, the area of right triangle $EDF$ is 5 less than the area of right triangle $FAB$, then the area of rectangle $ABCD$ is $\qquad$ .
【Solution】Solution: According to the analysis, the area of right triangle $E D F$ is 5 less than the area of right triangle $F A B$,
so we have: $\left(S_{\triangle A B F}+S_{\text {quadrilateral } D F B C}\right)-\left(S_{\triangle E D F}+S_{\text {quadrilateral } D F B C}\right)=5$
$\Rightarrow$ The area of rectangle $A B C D$ - the area of triangle $E B C$ $=5$;
Also, $\because$ the area of triangle $E B C$ $=\frac{1}{2} \times 10 \times 6=30$.
$\therefore$ The area of rectangle $A B C D$ $=$ the area of triangle $E B C$ $+5=30+5=35$.
Therefore, the answer is: 35.