Question

A polygon of n sides has $\frac{n(n-3)}{2}$ diagonals. If a polygon has 9 diagonals, find the number of sides of the polygon.

Answer 1

$\frac{n(n-3)}{2}=9$

$\Rightarrow n^2-3n=18$

$\Rightarrow n^2-3n-18=0$

$\Rightarrow n^2-6n+3n-18=0$

$\Rightarrow n(n-6)+3(n-6)=0$

$\Rightarrow (n+3)(n-6)=0$

So $n=6$ or $n=-3$

Since, no. o sides cannot be negative. Hence $n=6$.