Example 3 As shown in Figure 3, in trapezoid $A B C D$, $\angle D=$ $90^{\circ}, M$ is the midpoint of $A B$. If $C M=6.5, B C+C D+$ $D A=17$, then the area of trapezoid $A B C D$ is ( ).
(A) 20
(B) 30
(C) 40
(D) 50
(2007, Shandong Province

Junior High School Mathematics Competition)
Solve As shown in Figure 3, extend $C M$ to intersect the extension of $D A$ at point $E$. Then $\triangle B C M \cong \triangle A E M$.
Therefore, $C E=2 C M=13, A E=B C$.
Let the area of trapezoid $A B C D$ be $S$. Then
$$
\left\{\begin{array}{l}
D E^{2}+C D^{2}=C E^{2}, \\
\frac{1}{2} D E \cdot C D=S .
\end{array}\right.
$$

From $C E=13$ we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
D E^{2}+C D^{2}=169 \\
2 D E \cdot C D=4 S .
\end{array}\right. \\
\text { (1) + (2) gives }(D E+C D)^{2}=169+4 S \text {. } \\
\text { Also, }(D E+C D)^{2}=(A D+B C+C D)^{2}=17^{2}=289 \\
\Rightarrow 169+4 S=289 \Rightarrow S=30 \text {. } \\
\end{array}
$$