If the sides of an equilateral triangle are increasing at the rate of $2 \text{ cm/s}$ then the rate at which the area increases, when side is $10 \text{ cm}$, is

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

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

Let the side of an equilateral triangle be $x \text{ cm}$.

$∴\text{Area of equilateral triangle, } A = \frac{\sqrt{3}}{4}x^2 \quad \dots(i)$

Also, $\frac{dx}{dt} = 2 \text{ cm/s}$

On differentiating Eq. $(i)$ w.r.t. $t$, we get

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

$= \frac{\sqrt{3}}{4} \cdot 2 \cdot 10 \cdot 2 \quad \left[ ∵x = 10 \text{ and } \frac{dx}{dt} = 2 \right]$

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