9. Figure 1 shows the three views of a geometric solid, where the front view is an equilateral triangle.
Then the ( ) of this geometric solid is:
(A) The radius of the circumscribed sphere is $\frac{\sqrt{3}}{3}$
(B) The volume is $\sqrt{3}$
(C) The surface area is $\sqrt{6}+\sqrt{3}+1$
(D) The surface area of the circumscribed sphere is $\frac{16 \pi}{3}$
9. D.

It is easy to know that the required geometric body is a triangular pyramid with one side perpendicular to the base. Then
$$
\begin{array}{l}
V_{\text {triangular pyramid}}=\frac{1}{3} \times \frac{1}{2} \times 2 \times 1 \times \sqrt{3}=\frac{\sqrt{3}}{3}, \\
S_{\text {triangular pyramid}}=\frac{1}{2} \times 2 \times \sqrt{3}+\frac{1}{2} \times 2 \times 1+2 \times \frac{1}{2} \times \sqrt{2} \times \sqrt{\frac{7}{2}} \\
=\sqrt{7}+\sqrt{3}+1 .
\end{array}
$$

Therefore, the radius of the circumscribed sphere is $\frac{2 \sqrt{3}}{3}$, and the surface area of the circumscribed sphere is
$$
\frac{16 \pi}{3} .
$$