2. In the given parallelogram $A B C D$, point $E$ is the midpoint of side $B C$ and point $F$ lies inside side $A B$. The area of triangle $A F D$ is $15 \mathrm{~cm}^{2}$ and the area of triangle $F B E$ is $14 \mathrm{~cm}^{2}$. Determine the area of quadrilateral $F E C D$.
Another solution. Triangles $B E F$ and $E C F$ have the same height from vertex $F$ and congruent bases $B E$ and $E C$. Therefore, the areas of both triangles are the same. From Fig. 2, we see that the area of triangle $C D F$ is half the area of parallelogram $A B C D$ (both figures have the common base $C D$ and the same height), the other half is formed by the sum of the areas of triangles $A F D$ and $B C F$. Therefore, $S_{F E C D}=S_{E C F}+S_{C D F}=S_{E C F}+\left(S_{A F D}+S_{B C F}\right)=S_{A F D}+3 S_{F B E}=$ $=57 \mathrm{~cm}^{2}$.