Question

The sides of an equilateral triangle are increasing at the rate of 4 cm/sec. The rate at which its area is increasing, when the side is 14 cm is

• A. $42c{m}^2/sec$
• B. $10\sqrt{3}c{m}^2/sec$
• C. $14c{m}^2/sec$
• D. $28\sqrt{3}c{m}^2/sec$

Answer 1

The correct option is D $28\sqrt{3}c{m}^2/sec$

Let $x$ be the side of the equilateral triangle.
Then, Area$\left(A\right)=\frac{\sqrt{3}}{4}{x}^2$
Given, $\frac{dx}{dt}=4$
$\frac{dA}{dt}=\frac{\sqrt{3}}{4}\cdot 2x\cdot \frac{dx}{dt}$
$\therefore \frac{dA}{dt}{|}_{x=14}=\frac{\sqrt{3}}{4}\times 2\times 14\times 4=28\sqrt{3}c{m}^2/sec$