The side of an equilateral triangle is increasing at the rate of 2 cm/sec. At what rate its area increasing when the side of the triangle is 20 cm?

$20\sqrt{3}\, cm^2/s$

The correct answer is Option (2) → $20\sqrt{3}\, cm^2/s$

Let a cm be the side of an equilateral triangle and A be its area at any time t seconds, then

$A = \frac{\sqrt{3}}{4}a^2$   ...(i)

Differentiating (i) w.r.t. t, we get $\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2a \cdot \frac{da}{dt} = \frac{\sqrt{3}}{2}a \cdot \frac{da}{dt}$   ...(ii)

But $\frac{da}{dt}=2\,cm/sec$ (given)

∴ From (ii), $\frac{dA}{dt} =\frac{\sqrt{3}}{2}a.2=\sqrt{3}a$

When $a=20\,cm,\frac{dA}{dt} =\sqrt{3}×20=20\sqrt{3}$

Hence, the area is increasing at the rate of $20\sqrt{3}\, cm^2/sec$ when the side of equilateral triangle is 20 cm.