$\text{ In a }\mathrm{△}\mathrm{ABC},\mathrm{a}=6,\mathrm{b}=3\text{ and }\cos \mathrm{C}=\frac{\sqrt{3}}{2},\text{ then the area of triangle = }$

$\begin{array}{l}\text{ Given }\cos C=\frac{\sqrt{3}}{2}\Rightarrow \sin C=\frac{1}{2}\\ \text{ Area of triangle }=\frac{1}{2}\mathrm{absin}\mathrm{C}\\ =\frac{1}{2}\times 6\times 3\times \frac{1}{2}=\frac{9}{2}\end{array}$