The side of an equilateral triangle increases at the rate of 3 cm/sec, then the rate at which the area of the triangle increases when the side is 4 cm, is :

$6\, \sqrt{3}cm^2/sec$

The correct answer is Option (2) → $6\, \sqrt{3}cm^2/sec$

s → side of equal lateral triangle

$\frac{ds}{dt}=3cm/sec$

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

$\frac{dA}{dt}=\frac{\sqrt{3}}{2}s\frac{ds}{dt}=\frac{\sqrt{3}}{2}×4×3$

$=6\sqrt{3}cm^2/\sec$