13. As shown in the figure, the area of triangle $S_{\triangle B E C}$ is 30 square centimeters, $E$ is the point on side $A B$ closer to $B$, dividing $A B$ into four equal parts, and the area of parallelogram $D E C F$ is $\qquad$ square centimeters.
Analysis: Personally, I believe this problem lacks conditions to be solved, as the problem does not specify what type of quadrilateral ABCD is. If judging from the diagram, I can only assume it to be a parallelogram. The solution below assumes quadrilateral $\mathrm{ABCD}$ to be a parallelogram. From the diagram, we know that the area of triangle DCE is half the area of parallelogram $\mathrm{ABCD}$, and also half the area of parallelogram DECF. That is, the area of parallelogram DECF equals the area of parallelogram $\mathrm{ABCD}$. Since $\mathrm{E}$ is the quarter point of $\mathrm{AB}$, we have $S_{\triangle A E D}: S_{\triangle E B C}=A E: E B=3: 1, S_{\triangle A E D}=30 \times 4=120\left(\mathrm{~cm}^{2}\right)$, the area of parallelogram $\mathrm{ABCD}$ is $120 \times 2=240$ (square centimeters), so the area of parallelogram DECF is also 240 square centimeters.