1. Let the right quadrilateral prism $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ have a rhombus as its base, and the area of the base is $2 \sqrt{3}, \angle A B C=60^{\circ}, E, F$ are points on the edges $C C^{\prime}, B B^{\prime}$ respectively, and $E C=B C=2 F B$, then the volume of the quadrilateral pyramid $A-B C E F$ is
A. $\sqrt{3}$
B. $\sqrt{5}$
C. 6
D. 9
1. On the base $ABCD$, take the midpoint $K$ of $BC$, and connect $AK$. Then $AK \perp BC$. Since $ABCD-A'B'C'D'$ is a right prism, $BB' \perp$ plane $ABCD$, thus $AK \perp BB'$. Therefore, $AK \perp$ plane $BCEF$.
Because $S_{\text {rhombus } ABCD}=BC^{2} \sin 60^{\circ}=2 \sqrt{3}$,
so $BC=AB=AC=2, AK=\sqrt{3}, CE=2, BF=1$.
Thus $\quad V_{A-BCFE}=\frac{1}{3} S_{\text {trapezoid EFBC }} \cdot AK=\sqrt{3}$.