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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(6\sqrt{3}\,\text{cm}^2/\text{sec}\)
Let the side be x cm at time t. The area of an equilateral triangle is
\( \displaystyle A = \frac{\sqrt{3}}{4}\,x^2 \)
Differentiating with respect to t (chain rule):
\( \displaystyle \frac{dA}{dt} = \frac{\sqrt{3}}{4}\cdot 2x\,\frac{dx}{dt} = \frac{\sqrt{3}}{2}\,x\,\frac{dx}{dt} \)
With x = 4 cm and \(\frac{dx}{dt}\) = 3 cm/sec:
\( \displaystyle \frac{dA}{dt} = \frac{\sqrt{3}}{2}\times 4\times 3 \)
\(= 6\sqrt{3}\ \text{cm}^2/\text{sec}\)