【Solution】Solution: Connect $AC$. Since $E$ is the midpoint of $BC$, according to the principle that triangles with equal bases and heights have equal areas, we have
$S_{\triangle ACE}=S_{\triangle ABE}=\frac{1}{4} S_{\text{parallelogram } ABCD}$
Also, $DF=2FC$
$S_{\triangle AFC}=\frac{1}{3} S_{\triangle ADC}=\frac{1}{6} S_{\text{parallelogram } ABCD}$
$\frac{1}{4} S_{\text{parallelogram } ABCD}+\frac{1}{6} S_{\text{parallelogram } ABCD}=10$
$\frac{5}{12} S_{\text{parallelogram } ABCD}=10$
$S_{\text{parallelogram } ABCD}=24$
Answer: The area of the parallelogram is 24.
Therefore, the answer is: 24.
7. (5 points) As shown in the figure, in parallelogram $A B C D$, point $E$ is the midpoint of $B C$, and $D F=2 F C$. If the area of the shaded part is 10, then the area of parallelogram $A B C D$ is $\qquad$