4. As shown in Figure 1, ABCD and CEFG are two squares with side lengths of 5 cm and 4 cm, respectively. Erase the side GC, leaving an outline, then connect AE and BF intersecting at point H, and connect BG with AE intersecting at point I (as shown in Figure 2). What is the area of the shaded part in Figure 2 in square centimeters?
【Analysis】Observing the hourglass model $\mathrm{AEFB}$, we have $\frac{A H}{H E}=\frac{A B}{E F}=\frac{5}{4}$. Since $\frac{B C}{C E}=\frac{5}{4}=\frac{A H}{H E}$, it follows that $\mathrm{HC} / / \mathrm{AB}$, meaning $\mathrm{H}$ is on $\mathrm{CD}$. Given $\mathrm{HC} / / \mathrm{AB}$, we have $\frac{H C}{A B}=\frac{E C}{E B}=\frac{4}{9} \Rightarrow H C=\frac{20}{9}$, thus $G H=4-\frac{20}{9}=\frac{16}{9}$. Observing trapezoid AGHB, $S_{A G H B}=\frac{1}{2} \times\left(5+\frac{16}{9}\right) \times 5=\frac{305}{18}$. Also, $G H: A B=\frac{16}{9}: 5=16: 45$,
By the butterfly model, we know $S_{\triangle G H I}: S_{A G H B}=16^{2}:(16+45)^{2}=16^{2}: 61^{2}$,
Therefore, $S_{\triangle G H I}=\frac{16^{2}}{61^{2}} \times S_{A G H B}=\frac{16^{2}}{61^{2}} \times \frac{305}{18}=\frac{640}{549}$.