9 The number of spatial quadrilaterals with vertices at the vertices and midpoints of the edges of a tetrahedron is ( ).
(A) 141
(B) 144
(C) 423
(D) 432
(9) $\mathrm{C}$ Hint: Method 1: A tetrahedron has 4 vertices and 6 edges, with each edge having 1 midpoint, making a total of 10 points. The number of ways to choose 4 non-coplanar points from these 10 points is as follows:
(1) The 4 points are all vertices, and there is only 1 way to choose them;
(2) The 4 points include 3 vertices, and there are $4 \times 3=12$ (ways);
(3) The 4 points include 2 vertices, and there are $6 \times\left(\mathrm{C}_{5}^{2}-2\right)=48$ (ways);
(4) The 4 points include 1 vertex, and there are $4 \times\left(\mathrm{C}_{6}^{3}-3\right)=68$ (ways);
(5) The 4 points include no vertices, and there are $\mathrm{C}_{6}^{4}-3=12$ (ways).
In summary, the number of ways to choose 4 non-coplanar points from these 10 points is
$$
1+12+48+68+12=141 \text { (ways). }
$$

Each set of 4 non-coplanar points can form 3 different spatial quadrilaterals, so the number of spatial quadrilaterals with these 10 points as vertices is $141 \times 3=423$ (quadrilaterals).

Method 2: From these 10 points, any 4 can be chosen, and there are $\mathrm{C}_{10}^{4}=210$ (ways) to do so. Among these, the 4 chosen points are coplanar in the following 3 cases:
(1) Choosing any 4 points from the 6 points on each face are coplanar, and there are $4 \mathrm{C}_{6}^{4}=60$ (ways);
(2) The 3 points on each edge and the midpoint of the opposite edge are coplanar, and there are 6 such ways;
(3) Removing 1 pair of opposite edges, the midpoints of the remaining 4 edges are coplanar, and there are 3 such ways.

In summary, the number of ways to choose 4 coplanar points is $60+6+3=69$ (ways). Therefore, the number of ways to choose 4 non-coplanar points is $210-69=141$ (ways), and the rest is the same as Method 1. Hence, the answer is $\mathrm{C}$.