4. Given a tetrahedron $ABCD$ with the lengths of its 6 edges being $4$, $7$, $20$, $22$, $28$, and $x$ (unit: $\mathrm{cm}$). Then the integer part of the minimum value that $x$ can take is ( ).
(A) 10
(B) 9
(C) 8
(D) 7
4.C.

When $x \leqslant 4$, the two side triangles with a side length of $28 \mathrm{~cm}$ must simultaneously satisfy the condition that the sum of any two sides is greater than the third side. Only $22+7>28$, $22+20>28$. At this time, the edge with a length of $22 \mathrm{~cm}$ cannot appear on both sides that have an edge of $28 \mathrm{~cm}$.

When $x>4$, the tetrahedron as shown in Figure 4 can make $x$ reach its minimum.
In $\triangle A B C$ and $\triangle A B D$, we have
$$
\left\{\begin{array}{l}
28-20<x<28+20, \\
7-4<x<7+4 .
\end{array}\right.
$$

Solving this, we get $8<x<11$.
Thus, $[x]=8,9,10$, so $[x]_{\text {min }}=8$.