12. As shown in the figure, with the midpoint $O$ of the line segment $A B$ of length $4 \mathrm{~cm}$ as the center and a radius of $2 \mathrm{~cm}$, draw a circle intersecting the perpendicular bisector of $A B$ at points $E$ and $F$. Then, with $B$ and $A$ as centers and a radius of $4 \mathrm{~cm}$, draw circular arcs intersecting the ray $A E$ at point $C$ and the ray $B E$ at point $D$. Finally, with $E$ as the center and $D E$ as the radius, draw the circular arc $D C$. Find the area of the "oval" $A F B C D A$ formed by these 4 real curve arcs. (Use $\pi$ for the value of pi, do not use an approximate value)
$$
\begin{array}{l}
S_{\text {semicircle } \widehat{A F B}}=\frac{1}{2} \cdot \pi \cdot 2^{2}=2 \pi, S_{\text {sector } B A C}=S_{\text {sector } A B D}=\frac{1}{2} \cdot \frac{\pi}{4} \cdot 4^{2}=2 \pi, \\
S_{\text {sector } E C D}=\frac{1}{2} \cdot \frac{\pi}{2} \cdot(4-2 \sqrt{2})^{2}=(6-4 \sqrt{2}) \pi, S_{\triangle A B E}=\frac{1}{2} \cdot 4 \cdot 2=4,
\end{array}
$$

Therefore, the area of the "egg shape" $A F B C D A$ is $(12-4 \sqrt{2}) \pi-4\left(\mathrm{cm}^{2}\right)$.