Question

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

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

Answer 1

Answer: (C)

Area of Equilateral triangle, $A=\frac{\sqrt{3}}{4} x^{2}$
$\frac{d x}{d t}=2 \,cm / s , x=10\, cm $
$\frac{d A}{d t}=\frac{\sqrt{3}}{4} 2 x \frac{d x}{d t} $
$=\frac{\sqrt{3}}{4}(20) \cdot 2 $
$=10 \sqrt{3} \,cm ^{2} / s$