8. As shown in Figure 5, the area of rectangle $A B C D$ is $35 \mathrm{~cm}^{2}$, the area of shaded $\triangle A B E$ is $5 \mathrm{~cm}^{2}$, and the area of shaded $\triangle A D F$ is $7 \mathrm{~cm}^{2}$. What is the area of $\triangle A E F$ in square centimeters?
8.15.5.

As shown in Figure 13, draw rectangle $A D F P$ and rectangle $A B E T$. Let the intersection of $P F$ and $T E$ be $O$. Given $S_{\triangle B E}=5 \mathrm{~cm}^{2}$, we know

Thus, $S_{\text {rectangleECor }}=35-10=25\left(\mathrm{~cm}^{2}\right)$.
Given $S_{\triangle A D F}=7 \mathrm{~cm}^{2}$, we know
$S_{\text {rectangleDDPP }}=14 \mathrm{~cm}^{2}$.
Thus, $S_{\text {rectangle } B \text { CPP }}=35-14=21\left(\mathrm{~cm}^{2}\right)$.
Therefore, $\frac{S_{\text {rectangleSCFP }}}{S_{\text {rectangleECFO }}}=\frac{B C}{E C}=\frac{S_{\text {rectangleECDA }}}{S_{\text {rectangleECor }}}$, which means $\frac{21}{S_{\text {rectangleECro }}}=\frac{35}{25}$.
So, $S_{\text {rectangleECFO }}=21 \times 25 \div 35=15\left(\mathrm{~cm}^{2}\right)$, $S_{\triangle S C F}=\frac{1}{2} S_{\text {rectangle } E C F O}=7.5 \mathrm{~cm}^{2}$.
Thus, $S_{\triangle A E F}=S_{\text {rectangle } A B C D}-S_{\triangle A B E}-S_{\triangle A D F}-S_{\triangle E C F}$ $=35-5-7-7.5=15.5\left(\mathrm{~cm}^{2}\right)$.