If the area of a cyclic quadrilateral ABCD is $\frac{3\sqrt{3}}{4}.$. The radius of its circumcircle is 1. If AB = 1, BD = $\sqrt{3}$ then BC.DC =

$\begin{array}{l}\mathrm{R}=1\text{. In }{\mathrm{\Delta }}^{\mathrm{le}}\mathrm{ABD}\Rightarrow \frac{\mathrm{AB}}{\sin \lfloor \mathrm{ADB}}=\frac{\mathrm{BD}}{\sin \mathrm{A}}=2\\ \Rightarrow \sin ⌊\mathrm{ADB}=\frac{1}{2}\Rightarrow ⌊\mathrm{ADB}={30}^{\circ }\\ \sin \mathrm{A}=\frac{\mathrm{BD}}{2}\Rightarrow \frac{\sqrt{3}}{2}\\ \Rightarrow ⌊\mathrm{A}={60}^{\circ }\text{ or }{120}^{\circ }\end{array}$
Case -1: $\angle A$=60^o
$\begin{array}{l}\frac{1}{2}\times \mathrm{BA}\times \mathrm{BD}\times \sin {90}^{\circ }+\frac{1}{2}\mathrm{BC}\times \mathrm{DC}\times \sin 120=\frac{3\sqrt{3}}{4}\\ \text{ Case-2: If }⌊\mathrm{A}={120}^{\circ }\Rightarrow ⌊\mathrm{C}={60}^{\circ }\\ \Rightarrow \frac{\sqrt{3}}{\sin 120}=\frac{1}{\sin \lfloor \mathrm{ADB}}=\mathrm{\angle }\mathrm{ADB}={30}^{\circ }\\ \Rightarrow \mathrm{\angle }\mathrm{ADB}={30}^{\circ }\end{array}$
Then area of $\mathrm{ABCD}=\frac{3\sqrt{3}}{4}\Rightarrow \mathrm{BC}\times \mathrm{DC}=2$