2. Let the volume of a tetrahedron be $V_{1}$, and the volume of a convex polyhedron formed by the midpoints of its edges be $V_{2}$. Then $\frac{V_{2}}{V_{1}}$ is $(\quad$.
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) uncertain
2.A.

The volume of the convex polyhedron $V_{2}$ is the hexahedron obtained by cutting off 4 small tetrahedra from the original tetrahedron, i.e.,
$$
V_{2}=V_{1}-4 V_{1} \times\left(\frac{1}{2}\right)^{3}=\frac{V_{1}}{2} .
$$