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

$10\sqrt{3}cm^2/s$

The correct answer is Option (3) → $10\sqrt{3}cm^2/s$

Let the side of the equilateral triangle be $x$ cm.

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

$⇒\frac{dA}{dt}=\frac{\sqrt{3}}{4}×2x×2$

Now,

$\left.\frac{dA}{dt}\right|_{x=10}=\frac{\sqrt{3}}{4}×2×10×2=10\sqrt{3}cm^2/s$