7. Given a tetrahedron $ABCD$ with the lengths of its 6 edges being $4, 7, 20, 22, 28, x$, then the minimum possible value of $[x]$ is $\qquad$ (here, $[x]$ denotes the greatest integer less than or equal to $x$).
7. 8 .

When $x \leqslant 4$, the two side triangles with a side length of 28 must both satisfy the condition that the sum of any two sides is greater than the third side, which is only true for $22+7>28, 22+20>28$. At this time, the edge with a length of 22 cannot appear on both sides that have an edge of length 28.
When $x>4$, the tetrahedron as shown in Figure 4 may allow $x$ to 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 $[x]_{\text {min }}=8$.