Q. 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) $\frac{10}{\sqrt{3}}{cm}^{2}/s$
Explanation
(c)
If $x$ is the side of an equilateral triangle and $A$ is its area, then
$A = \frac{\sqrt{3}}{4}x^{2}\ \ \ \ \Rightarrow \ \ \ \frac{dA}{dt} = \frac{\sqrt{3}}{4}2x\frac{dx}{dt}$
Here, $x = 10$cm and $\frac{dx}{dt} = 2$ cm/s
$\therefore\ \ A = \frac{\sqrt{3}}{4}2(10)2 = 10\sqrt{3}cm^{2}/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.