Question

An equilateral triangle’s sides increase at the rate of $2 \, cm/sec.$ If the area of its incircle increases at a rate of $k \, cm^{2}/sec$ (when the length of the side is $\frac{6}{\pi }cm$ ), then the value of $k$ is

Answer 1

$\frac{r}{\frac{a}{2}}=tan\left(\left(30\right)^{o}\right)$
$\frac{2 r}{a}=\frac{1}{\sqrt{3}}$
$\Rightarrow r=\frac{a}{2 \sqrt{3}}$
$\therefore $ Area of incircle, $A=\pi r^{2}$ $=\frac{\pi a^{2}}{12}$
$\therefore \frac{d A}{d t}=\frac{2 \pi a}{12}\frac{d a}{d t}$
$=\frac{\pi }{6}\cdot \frac{6}{\pi }\cdot 2$
$=2c m^{2} / s e c.$