Consider a n-sided polygon.

In order for a diagonal to form we need to connect 2 vertices out of the n vertices on the polygon.
We can do this is \(^nC_2\) ways

But there are \(n\) such possibilities out of these which are sides of the polygon, we just need to subtract those.

Hence number of diagonals in a n-sided polygon = \(^nC_2 - n = \frac{n(n-1)}{2*1} - n = \frac{n(n-1) - 2n}{2} = \frac{n^2 - 3n}{2} = \frac{n(n-3)}{2}\)

Hence number of diagonals in a n-sided polygon (Hexagon) = \(\frac{n(n-3)}{2} = \frac{6(6-3)}{2} = \frac{6*3}{2} = 3*3 = \textbf{9}\)

Hence, Answer is D