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

$10\sqrt{3}\, cm^2/sec$

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

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

Area of an equilateral triangle: $A = \frac{\sqrt{3}}{4}x^2$

Differentiate both sides with respect to time $t$:

$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt} = \frac{\sqrt{3}}{2}x \cdot \frac{dx}{dt}$

Given: $x = 10$ cm, $\frac{dx}{dt} = 2$ cm/sec

$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot 10 \cdot 2 = 10\sqrt{3}$ cm²/sec