3. As shown in the figure, there is a square and an isosceles right triangle on the same straight line. The side length of the square is 12 cm, and the hypotenuse of the triangle is 18 cm. They are 13 cm apart. Now the square is moving to the right at a speed of 3 cm per second, and the triangle is moving to the left at a speed of 2 cm per second. Question: How many seconds will it take for the overlapping part of the two to be 63 square cm?
【Analysis】When two figures start to overlap, consider the following:
(1) When $D F$ coincides with point $G$ on $C D$ (point $F$ has not reached point $B$), the maximum overlapping area is $9 \times 9 \div 2=40.5 \mathrm{~cm}^{2}<63 \mathrm{~cm}^{2}$
(2) When point $G$ enters the interior of the square to the point where $F$ coincides with $B$, the overlapping area continuously increases, and the maximum area (as shown in the figure) is: $18 \times 9 \div 2-6 \times 6 \div 2=63 \mathrm{~cm}^{2}$. At this point, the movement has taken $(12+13) \div(3+2)=5$ seconds.
(3) When point $F$ coincides with $B$ to the point where $G$ reaches the symmetry axis of the square, the overlapping area also continuously increases, but the overlapping area is not equal to $63 \mathrm{~cm}^{2}$.
According to symmetry, the only other time the overlapping area is $63 \mathrm{~cm}^{2}$ is when point $E$ coincides with point $C$. At this point, the movement has taken $(13+18) \div(3+2)=6.2$ seconds.

In summary, the overlapping area of the two figures is $63 \mathrm{~cm}^{2}$ after 5 seconds or 6.2 seconds.
【Answer】 5 or 6.2