The sides of an equilateral triangle are increasing at the rate of 2 cm/sec.

 Let x be the side and A be the area of the equilateral triangle at any time t. Then,

$A=\frac{\sqrt{3}}{4} x^{2}$

$\Rightarrow \frac{d A}{d t}=2 \times \frac{\sqrt{3}}{4} x \frac{d x}{d t}$

$\Rightarrow \frac{d A}{d t}=\frac{\sqrt{3}}{2} \times 2 \times 10$

$\Rightarrow \frac{d A}{d t}=10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$