8. As shown in Figure 2, 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
8. B.

As shown in Figure 9, extend $C M$ to intersect the extension of $D A$ at point $E$, then $\triangle B C M \cong \triangle A E M$. Therefore, $C E = 2 C M = 13$, and $A E = B C$. 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
$$
\left\{\begin{array}{l}
D E^{2} + C D^{2} = 169, \\
2 D E \cdot C D = 4 S .
\end{array}\right.
$$
$$
\begin{array}{l}
\text { (1) + (2) gives }(D E + C D)^{2} = 169 + 4 S . \\
\text { Also, }(D E + C D)^{2} = (A D + B C + C D)^{2} \\
= 17^{2} = 289,
\end{array}
$$

Therefore, $169 + 4 S = 289$.
Thus, $S = 30$.