7. As shown in Figure $2, A B$ is the diameter of the semicircle $\odot O$, and $C, D$ are two moving points on the semicircle, with $C D / / A B$. If the radius of the semicircle $\odot O$ is 1, then the maximum value of the perimeter of trapezoid $A B C D$ is $\qquad$ .
7.5.

In Figure 2, connect $A C$, and draw $C H \perp A B$ at point $H$. 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, $B H=B C \cos \theta=2 \cos ^{2} \theta$.
Therefore, $C D=A B-2 B H=2-4 \cos ^{2} \theta$.
Hence, the perimeter of trapezoid $A B C D$ is
$$
\begin{aligned}
l & =A B+B C+C D+D A \\
& =4+4 \cos \theta-4 \cos ^{2} \theta \\
& =5-4\left(\cos \theta-\frac{1}{2}\right)^{2} .
\end{aligned}
$$

Therefore, when $\cos \theta=\frac{1}{2}$, i.e., $\theta=\frac{\pi}{3}$, $l_{\max }=5$.