Question

The sides of an equilateral triangle are increasing at the rate of $2cm/s$. The rate at which the area increases when the side is $10cm$, is

• A. $\sqrt{3}{cm}^2/s$
• B. $10{cm}^2/s$
• C. $10\sqrt{3}{cm}^2/s$
• D. $\frac{10}{\sqrt{3}}{cm}^2/s$

Answer 1

Answer: (C)

$10\sqrt{3}{cm}^2/s$

Let $x$ be the side of an equilateral triangle and $A$ be the area.

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

On differentiating both sides with respect to $t$, we get

$\frac{dA}{dt}=\frac{\sqrt{3}}{4}2x\frac{dx}{dt}$

Given, $x=10cm$

and $\frac{dx}{dt}=2cm/s$

$\therefore \frac{dA}{dt}=\frac{\sqrt{3}}{4}2\times \left(10\right)\times 2$

$=10\sqrt{3}{cm}^2/s$