Two angles of a triangle are $\frac{\pi }{6}$ and $\frac{\pi }{4}$ and included side is $\sqrt{3}+1\mathrm{cm}$ then area of triangle is

Suppose that $\angle A=30°,\angle B=45°$

The measure of angle $C$ is $180°-75°=105°$

The length of the side included between $A$ and $B$ is $c=\sqrt{3}+1$

Use the sine rule $\frac{c}{SinC}=\frac{b}{SinB}$ to find the other side

$\begin{array}{rcl}\frac{\sqrt{3}+1}{\sin 105°}& =& \frac{b}{\sin 45°}\\ b& =& \left(\sqrt{3}+1\right)\frac{1}{\sqrt{2}}\left(\frac{2\sqrt{2}}{\sqrt{3}+1}\right)\\ & =& 2\end{array}$

Therefore, the area of the triangle is 

   $\begin{array}{rcl}∆& =& \frac{1}{2}bc\sin A\\ & =& \frac{1}{2}\left(2\right)\left(\sqrt{3}+1\right)\sin 30\\ & =& \frac{1}{2}\left(\sqrt{3}+1\right)\end{array}$