Question

A polygon has number of sides n and diagonals n(n-3)/2

Answer 1

Answer:

Step-by-step explanation:

There is polygon with exactly 9 diagonals has 6 sides

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

Mark as brainliest please

Answer 2

Answer:

Step-by-step explanation:

Polygon Diagonals

The number of diagonals in a polygon = n(n-3)/2, where n is the number of polygon sides.

For a convex n-sided polygon, there are n vertices, and from each vertex you can draw n-3 diagonals, so the total number of diagonals that can be drawn is n(n-3). However, this would mean that each diagonal would be drawn twice, (to and from each vertex), so the expression must be divided by 2.