Question

We shoot at a target whose shape is a regular hexagon. Suppose that the hit is
uniformly distributed on the target. What is the probability that we hit the inside
of the circle inscribed in the hexagon? (So that the sides of the hexagon are tangent
to the circle.)

Answer 1

i need somebody to check my answer, please
is it pi(r^2)/ (square root of 3)/2*3r^2 ?

Answer 2

\[\text{desired prob}=\frac{\text{area of circle}}{\text{area of hexagon}}\]
Notice that the radius of the circle (denoted \(r\)) is equal to the apothem of the hexagon. The area of a regular polygon with apothem \(r\) is \(\dfrac{rp}{2}\), where \(p\) is the perimeter of the polygon.
\[\text{desired prob}=\frac{\pi r^2}{\frac{rp}{2}}=\frac{2\pi r}{p}\]
Let \(s\) be the side length of the hexagon (and so \(p=6s\)). The apothem to side length ratio of the hexagon is \(\dfrac{s}{2}\) to \(\dfrac{s\sqrt3}{2}\), which means the side length in terms of the apothem is \(s=\dfrac{2r}{\sqrt3}\).

So,
\[\text{desired prob}=\frac{\sqrt3\pi }{6}\]

Answer 3

thanks