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 an equilateral triangle at any time t.

It is given that

$\frac{d x}{d t}=2 \mathrm{~cm} / \mathrm{sec}$

Area of the equilateral triangle, $A=\frac{\sqrt{3}}{4}(\text { Side })^{2}=\frac{\sqrt{3}}{4} x^{2}$

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

Differentiating both sides with respect to t, we get

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

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

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

When $x=10 \mathrm{~cm}$ and $\frac{d x}{d t}=2 \mathrm{~cm} / \mathrm{sec}$, we get

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

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

Thus, the area of the equilateral triangle increases at the rate of $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$

The sides of an equilateral triangle are increasing at the rate of $2 \mathrm{~cm} / \mathrm{sec}$. The rate at which the area increases, when the side is $10 \mathrm{~cm}$, is $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$