53. As shown in the figure, in the rectangular paper piece $\mathrm{ABCD}$, $\mathrm{AB}=4, \mathrm{BC}=6$. The paper is folded so that point $\mathrm{B}$ coincides with point $\mathrm{D}$, and the fold line is $\mathrm{EF}$. The area of the part of the rectangle that does not overlap is $\qquad$ -
Answer: $\frac{20}{3}$
Solution: Let $\mathrm{CE}=\mathrm{x}$, then $\mathrm{BE}=\mathrm{ED}=6-\mathrm{x}$,
In RT $\triangle \mathrm{CDE}$, by the Pythagorean theorem, we have
$$
x^{2}+4^{2}=(6-x)^{2},
$$

Solving for $x$, we get
$$
x=\frac{5}{3},
$$

By the fold, we know that
$\triangle \mathrm{CDE} \cong \triangle \mathrm{A}^{\prime} \mathrm{DF}$,

Therefore, the area of the non-overlapping part in the rectangle is
$$
2 \times \frac{1}{2} \times 4 \times \frac{5}{3}=\frac{20}{3} \text { . }
$$