4. In the pyramid $P$ - $A B C$, it is known that $P A \perp$ plane $A B C, A B \perp A C$, and $A C=a, A B=2 a, P A=3 a$. A section is made through $A B$ intersecting $P C$ at point $D$. Then the minimum area of the section $A B D$ is
A. $\frac{3}{\sqrt{10}} a^{2}$
B. $\frac{a^{*}}{\sqrt{10}}$
C. $\frac{3}{\sqrt{5}} a^{2}$
D. $\frac{a^{2}}{\sqrt{5}}$
4. $\mathrm{A}$ As shown in the figure, $P A \perp$ plane $A B C$, so $P A \perp A B$. Also, $A B \perp A C$, hence $A B \perp$ plane $P A C$, thus $A B \perp A D$.
Therefore, $S_{\triangle A B D}=\frac{1}{2} A B \cdot A D=a \cdot A D$.
Notice that, when $A D \perp P C$, $A D$ is the shortest, at this time, $S_{\triangle A B D}$ has the minimum area. Therefore, the minimum area of $S_{\triangle A B D}$ is $a \cdot \frac{P A \cdot A C}{P C}=\frac{3}{\sqrt{10}} a^{2}$.