SG. 3 The area of an equilateral triangle is $100 \sqrt{3} \mathrm{~cm}^{2}$. If its perimeter is $p \mathrm{~cm}$, find $p$.
$\begin{array}{l}\text { Each side }=\frac{p}{3} \mathrm{~cm} \\ \frac{1}{2} \cdot\left(\frac{p}{3}\right)^{2} \sin 60^{\circ}=100 \sqrt{3} \\ p=60\end{array}$