5. As shown in Figure 2, $P$ is a point inside $\square A B C D$, and $S_{\triangle P A B}=5$, $S_{\triangle P A D}=2$. Then $S_{\triangle P A C}$ equals ( ).
(A) 2
(B) 3
(C) 4
(D) 5
5.(B).
$$
\begin{array}{l}
\text { Because } S_{\triangle P A B}+S_{\triangle P C D}=\frac{1}{2} S_{\triangle A B C D}=S_{\triangle D M C}, \\
\text { Therefore, } S_{\triangle P A C}=S_{\triangle D A C}-S_{\triangle A D P}-S_{\triangle D P C} \\
\quad=S_{\triangle P A B}-S_{\triangle P A D}=3 .
\end{array}
$$