Question 7 As shown in Figure 9, the three sides $AB$, $BC$, and $CD$ of the isosceles trapezoid $ABCD$ are tangent to the graph of the function $y=-\frac{1}{2} x^{2}+2, x \in[-2,2]$ at points $P$, $Q$, and $R$. Find the minimum value of the area of trapezoid $ABCD$.
From the appearance of the graph, question 6 and question 7 have very similar aspects. Here is the solution: $\square$

Let the area of the isosceles trapezoid $A B C D$ be $S$, and the coordinates of point $P$ be $\left(t, -\frac{1}{2} t^{2} + 2\right) (0 < t \leqslant 2)$. According to the problem, the coordinates of point $Q$ are $(0, 2)$, and the equation of line $B C$ is $y = 2$.

Since $y = -\frac{1}{2} x^{2} + 2$, we have $y' = -x$, and thus $\left.y'\right|_{x=t} = -t$.
Therefore, the equation of line $A B$ is
$$y - \left(-\frac{1}{2} t^{2} + 2\right) = -t(x - t)$$

which simplifies to $y = -t x + \frac{1}{2} t^{2} + 2$.
Setting $y = 0$, we get $x = \frac{t^{2} + 4}{2 t}$, so $A\left(\frac{t^{2} + 4}{2 t}, 0\right)$.
Setting $y = 2$, we get $x = \frac{1}{2} t$, so $B\left(\frac{1}{2} t, 2\right)$.
Thus, $S = \frac{1}{2} \times \left(\frac{1}{2} t + \frac{t^{2} + 4}{2 t}\right) \times 2 \times 2 = 2\left(t + \frac{2}{t}\right) \geqslant 4 \sqrt{2}$.

The equality holds if and only if $t = \frac{2}{t}$, i.e., $t = \sqrt{2} \in (0, 2]$.

Therefore, when $t = \sqrt{2}$, the area $S$ of the isosceles trapezoid $A B C D$ reaches its minimum value, which is $4 \sqrt{2}$.