10. Let the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ be $1, \alpha$ be the plane passing through the line $B D_{1}$. Then the range of the area of the section cut by $\alpha$ on the cube is $\qquad$
10. $\left[\frac{\sqrt{6}}{2}, \sqrt{2}\right]$.

As shown in Figure 3, let the section be $B M D_{1} N$, and the distance from point $M$ to line $B D_{1}$ be $h$. Then the area of the section is
$$
S=2 S_{\triangle B M D_{1}}=B D_{1} h .
$$

Obviously, the minimum value of $h$ is the distance between the skew lines $A A_{1}$ and $B D_{1}$.

Notice that, when $M$ is the midpoint of $A A_{1}$ and $O$ is the midpoint of $B D_{1}$, $M O$ is the common perpendicular segment of $A A_{1}$ and $B D_{1}$.
Thus, $h_{\min }=\frac{\sqrt{2}}{2}$, and at this time the area is $\frac{\sqrt{6}}{2}$.
Obviously, when point $M$ coincides with $A\left(A_{1}\right)$, $h$ is maximized. At this time, the section is $A B C_{1} D_{1}\left(A_{1} B C D_{1}\right)$, and its area is $\sqrt{2}$.