4. In the right triangular pyramid $P-ABC$, it is known that $PA \perp$ plane $ABC$, $AB \perp AC$, and $AC=a$, $AB=2a$, $PA=3a$. A section is made through $AB$ intersecting $PC$ at point $D$. Then the minimum area of the section $ABD$ 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.
Thus, the minimum area of $S_{\triangle A B D}$ is $a \cdot \frac{P A \cdot A C}{P C}=\frac{3}{\sqrt{10}} a^{2}$.