4. Given the lengths of the three face diagonals of a rectangular prism are 5, $4, x$. Then the range of values for $x$ is ( ).
(A) $(2, \sqrt{41})$
(B) $(3,9)$
(C) $(3, \sqrt{41})$
(i)) $(2,9)$
4. (C).

Solution 1: As shown in Figure 1, let $A_{1} D_{1}=a, A_{1} B_{1}=b$, $A A_{1}=c$, then we have
$$
\left\{\begin{array}{l}
a^{2}+c^{2}=4^{2} \\
b^{2}+c^{2}=5^{2}
\end{array}\right.
$$

Thus, $0<c<4$,
and $x=\sqrt{a^{2}+b^{2}}=\sqrt{4^{2}+5^{2}-2 c^{2}}$.
Substituting (3) into (2), we get
$$
3<x<\sqrt{41} \text {. }
$$

Solution 2: As shown in Figure 4, draw the altitude $A_{1} E$ of the right triangle $\triangle A_{1} A B_{1}$ from $A_{1}$, then $E$ lies on the segment $A B_{1}$; connect $D_{1} E$, then $A_{1} D_{1}$ is the perpendicular from the plane $A_{1} A B_{1}$, $D_{1} E$ is the slant line, and $A_{1} E$ is the projection, so $D_{1} E \perp A B_{1}$. In the right triangles $\triangle A D_{1} E$ and $\triangle B_{1} D_{1} E$, $\angle D_{1} A B_{1}$ and $\angle D_{1} E_{1} . A$ are both acute angles. Similarly, $\angle A D, B_{1}$ are also acute angles.

To determine the range of the third side $B_{1} D_{1}$ of the acute triangle $\triangle A E_{1} D_{1}$, we draw Figure 5, where on the quarter circle with radius 4, there is a tangent line $B_{1} P$ and a secant line $B_{1} Q$, with $A B_{1}=5$, then $\triangle A P B_{1}$ and $\triangle A Q B_{1}$ are both right triangles. For $\triangle A B_{1} D_{1}$ to be an acute triangle, $D_{1}$ should lie on the arc $\overparen{P Q}$, so we have
$$
B_{1} P<B_{1} D_{1}<B_{1} Q \text {, }
$$

i.e., $3<x<\sqrt{41}$.