7. Given a square $A B C D$ with side length $a$, vertex $A$ lies in plane $\beta$, and the other vertices are on the same side of plane $\beta$. The distances from points $B$ and $D$ to plane $\beta$ are 1 and 2, respectively. If the angle between plane $A B C D$ and plane $\beta$ is 30 degrees, then $a=$
7. $2 \sqrt{5}$

Analysis: Let $A D=A B=a, B, D$ have projections $B_{1}, D_{1}$ on plane $\beta$, so $A B_{1}=\sqrt{a^{2}-1}$, $A D_{1}=\sqrt{a^{2}-4}$, $B D=\sqrt{2} a$, we can get $B_{1} D_{1}=\sqrt{2 a^{2}-1}$. By the cosine rule, $\cos \angle D_{1} A B_{1}=\frac{a^{2}-4+a^{2}-1-\left(2 a^{2}-1\right)}{2 \sqrt{a^{2}-4} \cdot \sqrt{a^{2}-1}}=\frac{-2}{\sqrt{a^{2}-4} \cdot \sqrt{a^{2}-1}}$, thus $\sin \angle D_{1} A B_{1}=\frac{\sqrt{a^{4}-5 a^{2}}}{\sqrt{a^{2}-4} \cdot \sqrt{a^{2}-1}}$. Therefore, $S_{\triangle} A D_{1} B_{1}=\frac{1}{2} \sqrt{a^{4}-5 a^{2}}$. According to the problem, $\cos 30^{\circ}=\frac{\sqrt{3}}{2}=\frac{S_{\triangle} A D_{1} B_{1}}{S_{\triangle} A D B}=\frac{\sqrt{a^{4}-5 a^{2}}}{a^{2}}$, solving this gives $a=2 \sqrt{5}$.