4. As shown in Figure 1, in the rectangular paper piece $A B C D$, $A B=3 \mathrm{~cm}, B C$ $=4 \mathrm{~cm}$. Now the paper is folded and pressed flat so that $A$ coincides with $C$, and the crease is set as $E F$. Then the area of the overlapping part $\triangle A E F$ is equal to ( ).
(A) $\frac{73}{8}$
(B) $\frac{75}{8}$
(C) $\frac{73}{16}$
(D) $\frac{75}{16}$
4.D.

Let $CE=x$, then $AE=AF=x, BE=4-x$. In the right triangle $\triangle ABE$, by $AE^2=AB^2+BE^2$, we get $x^2=3^2+(4-x)^2$.

Solving for $x$ gives $x=\frac{25}{8}$.
Therefore, $S_{\triangle AEF}=\frac{1}{2} AF \cdot AB=\frac{1}{2} \times \frac{25}{8} \times 3=\frac{75}{16}$.