4 Tetrahedron $ABCD$ has six edges of lengths $7, 13, 18, 27, 36, 41$. If $AB=41$, then the length of $CD$ is
A. 7
B. 13
C. 18
D. 27
4 B. If $C D=7$, then one of the other four edges is 36. Let's assume $A D=36$, then regardless of what $A C$ is, it always holds that $A D>A C+C D$, which is impossible.
If $B D=7$, then $A D=36$. If $A C=18$, then $A C+C D<36$ or $A C+B C<$ 41. Therefore, $A C=27, B C=18, C D=13$.