The number of diagonals in a polygon of n sides is
a. n(n-1)/2
b. n(n-2)/2
c. n(n-3)/2
d. n(n-3)

Solution:

We have to find the number of diagonals in a polygon of n sides.

A polygon of n sides also has n vertices.

By joining any two vertices of a polygon, we obtain a diagonal of that polygon.

The number of line segments obtained by joining the vertices of a n sided polygon taken two points at a time.

The number of ways of selecting 2 points at a time from n number of points is given as ⁿC₂.

We know, ⁿC_r = n!/r!(n-r)!

So, ⁿC₂ = n!/2!(n-2)!

= n(n-1)(n-2)/2!(n-2)!

= n(n-1)/2!

= n(n-1)/2

On subtracting the number of edges,

Number of diagonals = n(n-1)/2 - n

= [n(n-1) - 2n]/2

= n(n-3)/2

Therefore, the number of diagonals is n(n-3)/2

✦ Try This: The number of diagonals in a regular hexagon is a. 7, b. 9, c. 2, d. 4

☛ Also Check: NCERT Solutions for Class 8 Maths

NCERT Exemplar Class 8 Maths Chapter 5 Solved Problem 1

The number of diagonals in a polygon of n sides is, a. n(n-1)/2, b. n(n-2)/2, c. n(n-3)/2, d. n(n-3)

Summary:

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

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