G8.2 If the area of the equilateral triangle $P Q R$ is $6+b \sqrt{3}$, find the value of $b$.
$$
P Q=2+2 \tan 60^{\circ}=2+2 \sqrt{3}
$$

Area of $P Q R=\frac{1}{2}(2+2 \sqrt{3})^{2} \sin 60^{\circ}=2(1+2 \sqrt{3}+3) \cdot \frac{\sqrt{3}}{2}$
$$
\begin{array}{l}
6+b \sqrt{3}=6+4 \sqrt{3} \\
b=4
\end{array}
$$