The side of an equilateral triangle is increasing at the rate of $2\ \text{cm/s}$. The rate at which area increases when the side is $10\ \text{cm}$ will be —

For an equilateral triangle, area $A = \dfrac{\sqrt{3}}{4}s^2$ Differentiate with respect to time $t$: $\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2}s\dfrac{ds}{dt}$ Given $\dfrac{ds}{dt} = 2$ and $s = 10$, $\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times 10 \times 2 = 10\sqrt{3}$ Hence, rate = $10\sqrt{3}$ cm$^2$/s