[Solution] If $E$ is the tangency point of $BC$ and $\odot O$, connect $OE$, and draw $DF \perp AB$.
Let $AB = a, \quad CD = b$.
Then $AD = 2OE = AB + CD = a + b$.
Also, $BC^2 = DF^2 = AD^2 - AF^2 = (a + b)^2 - (a - b)^2 = 4ab$, thus $BC = 2\sqrt{ab}$.
Then $S_{ABCD} = \frac{1}{2}(AB + CD) \cdot BC$
$$
\begin{array}{l}
= \frac{1}{2}(a + b) \cdot 2\sqrt{ab} \\
= (a + b) \sqrt{ab}.
\end{array}
$$

Clearly, among the given sets of values, only when $AB = 9, CD = 4$ is $\sqrt{ab}$ an integer.
Therefore, the answer is $(D)$.
20.95 As shown in the figure, $A B \perp B C, B C \perp C D, B C$ is tangent to the circle $O$ with diameter $A D$. In which of the following cases is the area of $A B C D$ an integer?
(A) $A B=3, C D=1$.
(B) $A B=5, C D=2$.
(C) $A B=7, C D=3$.
(D) $A B=9, C D=4$.
(E) $A B=11, C D=5$.
(39th American High School Mathematics Examination, 1988)