5. As shown in Figure 1, in the convex quadrilateral $A B C D$, $A B=4, B C$ $=3, C D=\frac{5}{2}$, and $\angle A D C=$ $\angle A B C=90^{\circ}$. Then $B C \cdot A D$ equals ( ).
(A) $3 \sqrt{3}+\frac{25}{4}$
(B) $3 \sqrt{3}+\frac{27}{4}$
(C) $3 \sqrt{3}+8$
(D) $3 \sqrt{3}+\frac{29}{4}$
5.B.

As shown in Figure 4, by the Pythagorean theorem, we have
$$
\begin{array}{l}
A C=\sqrt{4^{2}+3^{2}}=5 \\
=2 \times \frac{5}{2}=2 C D,
\end{array}
$$

and $\angle A D C=90^{\circ}$, thus
$$
\angle C A D=30^{\circ} \text {. }
$$

Since $\angle A D C=\angle A B C$
$=90^{\circ}$, therefore, points $A, B, C, D$ are concyclic.
Connecting $B D$, we have $\angle A B D=\angle A C D=90^{\circ}-30^{\circ}=60^{\circ}$.
Let $\angle B A C=\alpha$ (where $\alpha$ is an acute angle), then
$$
\sin \alpha=\frac{3}{5}, \cos \alpha=\frac{4}{5}\left(0^{\circ}<\alpha<60^{\circ}\right) .
$$

Construct rectangle $C B A F$, then
$$
A F=B C, \angle F A D=90^{\circ}-\left(\alpha+30^{\circ}\right)=60^{\circ}-\alpha \text {. }
$$

Therefore, $B C \cdot A D=A F \cdot A D=|A F| \cdot|A D| \cos \angle F A D$
$$
\begin{aligned}
& =3\left(\frac{A B}{\sin \angle A D B} \cdot \sin \angle A B D\right) \cos \left(60^{\circ}-\alpha\right) \\
= & 3 \times \frac{4}{\sin \left(90^{\circ}-\alpha\right)} \cdot \sin 60^{\circ}\left(\frac{1}{2} \cos \alpha+\frac{\sqrt{3}}{2} \sin \alpha\right) \\
= & 3 \sqrt{3}+\frac{27}{4} .
\end{aligned}
$$

Editor's Note: This problem can be solved more concisely using complex numbers.