In trapezoid $ABCD$, $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$, with $AB+CD=BC$, $AB<CD$, and $AD=7$. What is $AB \cdot CD$? Express your answer in decimal form.
Let $E$ be the foot of the perpendicular from $B$ to $\overline{CD}$. Then $AB = DE$ and $BE =
AD = 7$. By the Pythagorean Theorem, \begin{align*}
AD^2 = BE^2 &= BC^2 - CE^2\\
&= (CD+AB)^2 - (CD - AB)^2\\
&=(CD+AB+CD-AB)(CD+AB-CD+AB)\\
&=4\cdot CD \cdot AB.
\end{align*}Hence, $AB \cdot CD = AD^2/4=7^2/4=49/4=\boxed{12.25}$.

[asy]
pair A,B,C,D,I;
A=(0,0);
B=(0,5);
C=(7,7);
I=(7,5);
D=(7,0);
draw(A--B--C--D--cycle);
draw(B--I);
label("$A$",A,W);
label("$B$",B,W);
label("$C$",C,E);
label("$E$",I,E);
label("$D$",D,E);
[/asy]