2001 AMC 10 Problem 24

Drop a perpendicular from $B$ to $CD,$ meeting it at $E.$ Then $BE=AD=7$ and $CE=CD-AB.$ By the Pythagorean theorem, $BC^2=BE^2+CE^2.$ Since $BC=CD+AB,$ $$\begin{aligned} &(CD+AB)^2 \\ &\quad {}-(CD-AB)^2 \\ &\quad =BE^2=49. \end{aligned}$$ The left side equals $4\cdot AB\cdot CD,$ so $AB\cdot CD=\dfrac{49}{4}=12.25.$ Thus, the correct answer is B.