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.

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

Answer 1

Answer: (B)

Let $x$ be the side of an equilateral triangle and $A$ be the area.
$<br/>\therefore A=\frac{\sqrt{3}}{4}{x}^2<br/>$
On differentiating both sides with respect to $t$, we get
$<br/>\frac{dA}{dt}=\frac{\sqrt{3}}{4}2x\frac{dx}{dt}<br/>$
Given, $x=10cm$
and $\frac{dx}{dt}=2cm/s$
$<br/>\therefore \frac{dA}{dt}=\frac{\sqrt{3}}{4}2\times (10)\times 2<br/>$
$<br/>=10\sqrt{3}c{m}^2/s<br/>$