5. (20 points) Two identical rectangular pieces of paper, each with an area of 48 square centimeters. As shown in the figure, when they are stacked together, the covered area is 72 square centimeters. Given that one diagonal $B D$ of the overlapping quadrilateral $A B C D$ is 6 centimeters, what is the length of each rectangular piece of paper in centimeters?

---

The translation maintains the original text's format and line breaks.
【Analysis】It is known that $A B C D$ is a parallelogram, and its area is $48+48-72=24$ square centimeters.
Let the width of the rectangular paper be $h$, then $h$ is also the distance between the two sets of parallel lines of the parallelogram, so $A D \times h=$ $\square A B C D$'s area $=D C \times h$,

we get $A D=D C$, therefore $A B C D$ is a rhombus, the diagonals $A C$ and $B D$ of the rhombus are perpendicular to each other at $O$, so $A C \times 6$ $\div 2=24$, thus $A C=8$.

In the right triangle $C O D$, $O D=3$ cm, $O C=4$ cm, so $D C=5$ cm.
Therefore, $5 \times h=24$, so, $h=4.8$ cm.
Since the area of the rectangle is 48 square centimeters, the length of the rectangle is 10 cm.
【Solution】Solution: The area of the parallelogram $A B C D$ is:
$48+48-72=24$ square (centimeters).
Let the width of the rectangular paper be $h$, then $h$ is also the distance between the two sets of parallel lines of the parallelogram, so $A D \times h=\square A B C D$'s area $=D C \times h$,

we get $A D=D C$, therefore $A B C D$ is a rhombus, the diagonals $A C$ and $B D$ of the rhombus are perpendicular to each other at $O$, so $A C \times 6 \div 2=24$, thus $A C=8$ cm.

In the right triangle $C O D$, $O D=3$ cm, $O C=4$ cm,
so $D C=\sqrt{3^{2}+4^{2}}=5$ (cm).
Therefore, $5 \times h=24$, so, $h=24 \div 5=4.8$ (cm)
$48 \div 4.8=10$ (cm)
Answer: The length of each rectangular paper is 10 cm.