6. As shown in the figure, $P$ is a point outside the parallelogram $A B D C$. It is known that the area of $\triangle P C D$ is equal to 5 square centimeters, and the area of $\triangle P A B$ is equal to 11 square centimeters. Then the area of parallelogram $A B C D$ is $\qquad$

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[Answer] 12
[Solution] Draw $P F \perp A B$, since $A B / / D C$, so $P F \perp C D$. It is easy to know $\left\{\begin{array}{l}S_{\triangle P A B}=\frac{1}{2} A B \cdot P F \\ S_{\triangle P C D}=\frac{1}{2} C D \cdot P E\end{array}\right.$, since $A B=C D$, therefore
$$
\begin{aligned}
S_{\triangle P A B}-S_{\triangle P C D} & =\frac{1}{2} A B \cdot P F-\frac{1}{2} C D \cdot P E \\
& =\frac{1}{2} A B \cdot P F-\frac{1}{2} A B \cdot P E \\
& =\frac{1}{2} A B \cdot(P F-P E) \\
& =\frac{1}{2} A B \cdot E F
\end{aligned}
$$

The area of parallelogram $A B D C$ is $S_{A B D C}=A B \cdot E F$, so $S_{A B D C}=2\left(S_{\triangle P A B}-S_{\triangle P C D}\right)=12$