9. In parallelogram $A B C D$, $\overrightarrow{A B} \cdot \overrightarrow{B D}=0$, fold along $B D$ to form a right dihedral angle $A-B D-C$, and $4 A B^{2}+2 B D^{2}=1$, then the surface area of the circumscribed sphere of the tetrahedron $A-B C D$ is
A. $\frac{\pi}{2}$
B. $\frac{\pi}{4}$
C. $\frac{\pi}{48}$
D. $\frac{\sqrt{2}}{24} \pi$
Solve:
Construct a rectangular cuboid, $AC$ is the diagonal of the cuboid,
$$
(2 R)^{2}=AC^{2}=AB^{2}+BD^{2}+DC^{2}
$$
$=2 AB^{2}+BD^{2}=\frac{1}{2}$, so the surface area of the circumscribed sphere of the tetrahedron $A-BCD$
is $\frac{\pi}{2}$, choose $A$.