63. As shown in the figure, in the rectangular paper piece $A B C D$, $A B=4, B C=6$. By cutting out 3 isosceles right triangles from this rectangular paper piece, the minimum area of the remaining part among all cutting methods is $\qquad$ .
Answer: 2.5
Solution: As shown in the figure, the cutting steps are as follows:
(1) Draw the angle bisectors of $\angle B, \angle C$, intersecting at point $E$ (since the distance from point $E$ to side $BC$ is 3, it is known that point $E$ is inside the rectangle $ABCD$);
(2) Extend $BE$ to intersect $AD$ at point $F$;
(3) Draw a perpendicular line from point $E$ to $CD$, and let the foot of the perpendicular be $G$.

Obviously, $\triangle EBC, \triangle ABF, \triangle CEG$ are all isosceles right triangles, with areas respectively as
$$
S_{1}=\frac{1}{2} \times 3 \times 6=9, \quad S_{2}=\frac{1}{2} \times 4 \times 4=8, \quad S_{3}=\frac{1}{2} \times 3 \times 3=4.5,
$$

Therefore, the remaining part $S_{\text {quadrilateralEGDF }}=4 \times 6-\left(S_{1}+S_{2}+S_{3}\right)=24-21.5=2.5$,