Number of diagonals of a polygon – Solved Examples

Number of diagonals of a polygon 

A line segment connecting any two vertices (non-adjacent) of a polygon is known as the diagonal of the polygon. A polygon needs to have at least 4 sides to have a diagonal. Triangle having three sides doesn’t have a diagonal because all the three vertices are adjacent to each other.

The number of diagonals of a polygon having “a” number of sides is equal to\frac{a(a-3)}{2} where (a > 3).

From each vertex (a – 3) diagonals can be drawn. So, from “a “number of vertices a total of “a (a – 3)” diagonals can be drawn. But the diagonals drawn from opposite sides should be counted as one. Hence by discarding the duplicates, a total number of \frac{a(a-3)}{2} diagonals can be drawn in a polygon having “a” sides.

Let us find the number of diagonals of some polygons using the above formula.

1. Quadrilateral

Number of sides (a) = 4

\text { Number of diagonals }=\frac{a(a-3)}{2}=\frac{4(4-3)}{2}=\frac{4 \times 1}{2}=2

2. Pentagon 

Number of sides (a) = 5

\text { Number of diagonals }=\frac{a(a-3)}{2}=\frac{5(5-3)}{2}=\frac{5 \times 2}{2}=5

3. Hexagon 

Number of sides (a) = 6

\text { Number of diagonals }=\frac{a(a-3)}{2}=\frac{6(6-3)}{2}=\frac{6 \times 3}{2}=9

4. Heptagon 

Number of sides (a) = 7

\text { Number of diagonals }=\frac{a(a-3)}{2}=\frac{7(7-3)}{2}=\frac{7 \times 4}{2}=14

Given Below is the table showing the number of diagonals for different types of polygon.

Solved Examples

1. How many diagonals does a nonagon have?

Nonagon is a polygon having 9 sides

Number of sides (a) = 9

Number of diagonals =\frac{a(a-3)}{2}=\frac{9(9-3)}{2}=\frac{9 \times 6}{2}=27

Hence there are 27 diagonals in a nonagon.

2. A polygon has 54 diagonals. Find the number of sides of this polygon.

Number of diagonals  =\frac{a(a-3)}{2}

\Rightarrow \quad 54=\frac{a(a-3)}{2}

\Rightarrow a(a-3)=54 \times 2=108

\Rightarrow \quad a=12

Hence the polygon having 54 diagonals have 12 sides.