Question

A polygon with n sides has n(n-3)/3 diagonals. So how many sides for a polygon with 104 diagonals

Answer 1

There is polygon with exactly 9 diagonals has 6 sides

solution

no of diagonals=n(n-3)/2

given no of diagonals=9

by problem

n(n-3)/2=9

n(n-3)=18

nsquare -3n =18

nsquare -3n-18=0

nsquare +3n-6n-18=0

n(n+3)-6(n-3)=0

(n+3)(n-6)=0

n=-3 or 6

side can't be negative so it has 6sides.