The side of an equilateral triangle is increasing at the rate of 2 cm/sec. The rate at which the area of the triangle is increasing, when the side of the triangle is 5 cm, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5\sqrt{3} \) cm2/sec
For an equilateral triangle of side s, the area is \( A=\frac{\sqrt{3}}{4}s^2 \). Differentiating with respect to t (chain rule):
\( \displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{4}\cdot 2s\,\frac{ds}{dt} \)
\( \displaystyle =\frac{\sqrt{3}}{2}\,s\,\frac{ds}{dt} \)
With s = 5 cm and \( \frac{ds}{dt}=2 \) cm/sec:
\( \displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{2}\times 5\times 2 \)
\( \displaystyle =5\sqrt{3}\ \text{cm}^2/\text{sec} \)