1. Given that $A B C D$ is a unit square, and $P$ is the midpoint of side $A B$. By folding $A B C D$ along $P C$ and $P D$ to form the tetrahedron $P A C D$ (where $A$ coincides with $B$), the volume of the tetrahedron $P A C D$ is
A. $\frac{\sqrt{3}}{30}$
B. $\frac{\sqrt{3}}{24}$
C. $\frac{\sqrt{3}}{20}$
D. $\frac{\sqrt{3}}{12}$
1. B As shown in the figure, after folding, let $A$ and $B$ coincide at point $E$. Since $P E \perp E C, P E \perp E D$, it follows that $P E \perp$ plane $E C D$. Therefore, $V_{P-E C D}=\frac{1}{3} S_{\triangle E C D} \cdot P E=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 1^{2} \times \frac{1}{2}=\frac{\sqrt{3}}{24}$.