Question

number of diagonals in a polygon of n sides =​

Answer 1

Answer:

The number of diagonals of a polygon of n sides is given by the formula d=n(n-3)/2

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Answer 2

$\large\underline{\sf{Solution-}}$

Given a polygon of n sides.

So, it means polygon have n - vertices.

We know, by joining any two vertices, we get number of lines which together form sides and diagonals of a polygon.

So, it implies

Number of lines formed by joining 2 points out of n points in a a plane = Number of diagonals + Number of sides

So,

Number of diagonals = Number of lines formed by joining2 points out of n points in a plane - Number of sides of polygon.

We know,

The number of ways in which 'r' objects can be chosen from 'n' distinct objects is

$\boxed{ \rm{ \: ^nC_r \: = \: \frac{n!}{r! \: (n \: - \: r)!}}}$

So,

The number of lines formed by joining 2 points out of n points in a plane

$\rm \:  =  \:  \: ^nC_2$

$\rm \:  =  \:  \: \dfrac{n!}{2! \: (n - 2)!}$

$\rm \:  =  \:  \: \dfrac{n \times (n - 1) \times (n - 2)!}{2 \times 1 \times \: (n - 2)!}$

$\rm \:  =  \:  \: \dfrac{n(n - 1)}{2}$

So,

$\red{\rm :\longmapsto\:Number_{diagonals} = Number_{lines} - Number_{sides \: of \: polygon}}$

$\rm :\implies\:Number_{diagonals} = \dfrac{n(n - 1)}{2} - n$

$\rm :\implies\:Number_{diagonals} = \dfrac{n(n - 1) - 2n}{2}$

$\rm :\implies\:Number_{diagonals} = \dfrac{n(n - 1- 2)}{2}$

$\bf :\implies\:Number_{diagonals} = \dfrac{n(n - 3)}{2}$

Additional Information :-

If there are n points in a plane, none of them are collinear except m points are collinear, then

$\boxed{ \rm{ Number_{(lines)} = \: ^nC_2 - \: ^mC_2 + 1}}$

$\boxed{ \rm{ Number_{(triangles)} = \: ^nC_3 - \: ^mC_3}}$

If m set of parallel lines are intersected by n set of parallel lines, then

$\boxed{ \rm{ Number_{(parallelogram)} = \: ^nC_2 \times \: ^mC_2}}$