6. (8 points) As shown in the figure, the area of trapezoid $ABCD$ is 20. Point $E$ is on $BC$, and the area of triangle $ADE$ is twice the area of triangle $ABE$. The length of $BE$ is 2, and the length of $EC$ is 5. Therefore, the area of triangle $DEC$ is $\qquad$
【Solution】Since the area of triangle $A D E$ is twice the area of triangle $A B E$, we have $A D=2 B E$
$$
\begin{array}{l}
=2 \times 2, \\
=4,
\end{array}
$$

The height of trapezoid $A B C D$: $20 \times 2 \div(4+2+5)$
$$
\begin{array}{l}
=40 \div 11 \\
=\frac{40}{11},
\end{array}
$$

The area of triangle $D E C$: $5 \times \frac{40}{11} \div 2$
$$
\begin{array}{l}
=\frac{200}{11} \div 2 \\
=9 \frac{1}{11}
\end{array}
$$

Answer: The area of triangle $D E C$ is $9 \frac{1}{11}$.
Therefore, the answer is: $9 \frac{1}{11}$.