(7) As shown in the figure, $AB$ is the diameter of semicircle $O$, and $C, D$ are two moving points on the semicircle, with $CD \parallel AB$. If the radius of the semicircle is 1, then the maximum value of the perimeter of trapezoid $ABCD$ is . $\qquad$
Method 1: As shown in Figure 1, connect $A C$, and draw $C H \perp A B$ at $H$. Let $\angle A B C=\theta\left(0<\theta<\frac{\pi}{2}\right)$, then
$$
A D=B C=A B \cos \theta=2 \cos \theta,
$$

thus
$$
B H=B C \cos \theta=2 \cos ^{2} \theta,
$$

so
$$
C D=A B-2 B H=2-4 \cos ^{2} \theta .
$$

Therefore, the perimeter of trapezoid $A B C D$ is
$$
\begin{aligned}
l & =A B+B C+C D+D A \\
& =4+4 \cos \theta-4 \cos ^{2} \theta \\
& =5-4\left(\cos \theta-\frac{1}{2}\right)^{2} .
\end{aligned}
$$

Thus, when $\cos \theta=\frac{1}{2}$, i.e., $\theta=\frac{\pi}{3}$, $l_{\text {max }}=5$.
Method 2: As shown in Figure 2, the points $C$ and $D$ are symmetric to points $C^{\prime}$ and $D^{\prime}$ about the diameter $A B$ on the lower half of the circle.

Since $A B=2$ is a constant, according to symmetry, the perimeter of trapezoid $A B C D$ is maximized when and only when the perimeter of the inscribed hexagon $A D C B C^{\prime} D^{\prime}$ is maximized.

It is easy to see that when $A D C B C^{\prime} D^{\prime}$ is a regular hexagon, its perimeter is maximized, at which time
$$
B C=C D=D A=\frac{1}{2} A B=1 .
$$

Therefore, the maximum perimeter of trapezoid $A B C D$ is 5.