Question

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

Answer 1

Answer: 2

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