9. (3 points) In the figure, $ABC$ is an obtuse triangle. $BC=6$ cm, $AB=5$ cm, and the height $AD$ from $BC$ is 4 cm. If this triangle moves upwards along the direction of $DA$ at a speed of 3 cm per second. After 2 seconds, the area swept by this triangle is $\qquad$ square cm.
【Analysis】When a figure is translated, each point of the figure moves in the same direction and the same distance. If a triangle moves upward along its height at a speed of 3 cm per second for 2 seconds, it means it has moved $3 \times 2=6 \mathrm{~cm}$ along the height. The area swept by the triangle should be the area of a square plus the area of the triangle, plus the area of parallelogram $A B F E$.
【Solution】The area swept should be the area of square $B C D F$ plus the area of the upper triangle, plus the area of parallelogram $A B F E$,

Given $A D=4, A B=5$,
and $A B^{2}-A D^{2}=B D^{2}$,
which is $5^{2}-4^{2}=3^{2}$,
so $D B=3$ cm,
$3 \times 2=6$ cm,
$$
\begin{array}{l}
6 \times 6+6 \times 4 \div 2+6 \times 3, \\
=36+12+18, \\
=66 \text { (square cm). }
\end{array}
$$

Answer: The area swept by this triangle is 66 square cm.
The answer is: 66.