The side of an equilateral triangle is increasing at the rate of 3 cm/sec. The rate at which the area of the triangle is increasing. when the side is 4 cm is :

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

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

side of equilateral Δ = x = 4 cm

$\frac{dx}{dt}=3$

area = $\frac{\sqrt{3}}{4}side^2$

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

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

$=\frac{\sqrt{3}}{2}×4×3=6\sqrt{3}cm^2/sec$