The side of an equilateral triangle is increasing at the rate of 2 cm/s. The rate at which area increases when the side is 10 is

Assume that x be the side of an equilateral triangle and A be its area.
Hence, $\mathrm{A}=(\sqrt{3}/4){\mathrm{x}}^2$ square units
Differentiate both sides with respect to t, we get
$\frac{\mathrm{dA}}{dt}=\left(\frac{\sqrt{3}}{4}\right)2\mathrm{x}\left(\frac{dx}{dt}\right)\dots \left(1\right)$
Given that $\mathrm{x}=10\mathrm{cm}$ and $\mathrm{dx}/\mathrm{dt}=2\mathrm{cm}/\mathrm{s}$
Now, substitute the values in (1), we get
$\frac{\mathrm{dA}}{dt}=\left(\frac{\sqrt{3}}{4}\right)2\left(10\right)\left(2\right)$
Therefore, $\frac{\mathrm{dA}}{dt}=10\sqrt{3}{\mathrm{cm}}^2/\mathrm{s}$