Question

A point is selected at random inside an equilateral triangle whose side length is $3$ units. If the probability that its distance from any corner of the triangle is greater than $1$ unit is $1-\frac{2\pi }{k\sqrt{3}}$, then $k$ is

Answer 1

Area of the sector
$=\frac{\theta }{2\pi }\times \pi r^2=\frac{r^2\theta }{2}=\frac{1\times \frac{\pi }{3}}{2}=\frac{\pi }{6}$

Now, favourable area
$=\frac{\sqrt{3}}{4}{a}^2-3\times \frac{\pi }{6}=\frac{\sqrt{3}}{4}\times 3^2-3\times \frac{\pi }{6}$
Therefore, the required probability
$=\frac{\frac{\sqrt{3}}{4}\times 3^2-3\times \frac{\pi }{6}}{\frac{\sqrt{3}}{4}\times 3^2}=1-\frac{2\pi }{9\sqrt{3}}\Rightarrow k=9$