Example 7
An equilateral triangle $E F G$ with side length $a$ covers a square $A B C D$ with side length 1, find the minimum value of $a$.
We can assume that each side of $\triangle E F G$ has at least one vertex of the square $A B C D$, otherwise, we could translate this side, let's say $F G$ (Figure 1.9), inward to $F^{\prime} G^{\prime}$, such that it contains a vertex of the square $A B C D$, thus obtaining a smaller equilateral triangle $E F^{\prime} G^{\prime}$.
At this point, there are two scenarios:
(1) All four vertices of the square $A B C D$ are on the sides of $\triangle E F G$ (Figure 1.10).
In this case, $a=E D+D F=1+\frac{2}{\sqrt{3}}$.
(2) One vertex of the square $A B C D$ is not on the sides of $\triangle E F G$ (Figure 1.11). In this case,
$$
\begin{aligned}
a & =A F+A E=\frac{\sin \alpha}{\sin 60^{\circ}}+\frac{\sin \beta}{\sin 60^{\circ}} \\
& =\frac{2}{\sqrt{3}}(\sin \alpha+\sin \beta)=\frac{4}{\sqrt{3}} \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} .
\end{aligned}
$$

Since $\alpha+\beta=360^{\circ}-60^{\circ}-60^{\circ}-90^{\circ}=150^{\circ}$,
and $\alpha>60^{\circ}, \beta>60^{\circ},-30^{\circ}<\frac{\alpha-\beta}{2}<30^{\circ}$, we have
$$
\begin{aligned}
a & \leq \frac{4}{\sqrt{3}} \sin \frac{150^{\circ}}{2} \cos \frac{30^{\circ}}{2} \\
& =\frac{2}{\sqrt{3}}\left(\sin 90^{\circ}+\sin 60^{\circ}\right)=1+\frac{2}{\sqrt{3}} .
\end{aligned}
$$

Combining (1) and (2), we know that the side length is minimized if and only if $\triangle E F G$ is as shown in Figure 1.10, with the minimum side length being $1+\frac{2}{\sqrt{3}}$.