4. If four points in space $A, B, C, D$ satisfy $A B=C D=8, A C$ $=B D=10, A D=B C=13$, then the number of such tetrahedra $A B C D$ is ( ).
(A) 0
(B) 1
(C) 2
(D) More than 2
4. A. Construct $\triangle A B C$, such that $A B=8$, $B C=13$, $A C=10$. Using $B C$ as one side, continue to construct $\triangle B D C$. Clearly, when $\triangle A B C$ and $\triangle B C D$ are in the same plane, $A D$ has its maximum value, at which point quadrilateral $A B C D$ is a parallelogram, thus we have
$$
\begin{array}{l}
A D^{2}+B C^{2}=2\left(A B^{2}+A C^{2}\right), \\
A D^{2}=2\left(8^{2}+10^{2}\right)-13^{2}=159<13^{2} .
\end{array}
$$

Therefore, the maximum value of $A D$ is less than 13, which contradicts $A D=13$. Hence, such a tetrahedron does not exist.