14. As shown in Figure 4, point $A$ is on $B G$, quadrilaterals $A B C D$ and $D E F G$ are both squares, with areas of $7 \mathrm{~cm}^{2}$ and $11 \mathrm{~cm}^{2}$, respectively. Then the area of $\triangle C D E$ is $\qquad$
14. $\sqrt{7} \mathrm{~cm}^{2}$.

Draw $E H \perp C D$, intersecting the extension of $C D$ at point $H$.
Since $\angle E D H=\angle G D A, D E=D G$, therefore,
Rt $\triangle E D H \cong$ Rt $\triangle G D A$.
Thus, $E H=A G, D H=A D$.
From the areas of square $A B C D$ and square $D E F G$ being $7$ and $11$ respectively, we get
$$
A D=\sqrt{7}, G D=\sqrt{11}, A G=2 .
$$

Therefore, $S_{\triangle C D E}=\sqrt{7}$.