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How many diagonals are there in a polygon with n sides?

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Last updated date: 22nd Jun 2024
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Answer
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Hint: Before attempting this question, one must have prior knowledge about polygon and also remember that the number of selecting two vertices out of n sides of a polygon can be given as; $^n{C_2}$, use this information to approach the solution.

Complete step-by-step answer:
As the given polygon is of n sides so eventually it will have n vertices. Now if we choose any two vertices out of these n vertices then we can have a side or a diagonal for the polygon. So, the number of line segment obtained by joining the vertices of a n sided polygon taken two at a time that is
Numbers of all straight lines formed by joining of two vertices at a time can be given as; $^n{C_2}$
Now we know that ways of choosing r entities out of n entities will be $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
As we know that $n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)..........\left( {n - p} \right)!$ where p < n
Therefore, $^n{C_2} = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{2!\left( {n - 2} \right)!}}$
So, on simplification we get
$^n{C_2} = \dfrac{{n\left( {n - 1} \right)}}{{2!}}$
We know that $2! = 2 \times 1!$ and ${\text{1! = 1}}$
Therefore, $^n{C_2} = \dfrac{{n\left( {n - 1} \right)}}{2}$
So, as these straight lines includes n sides of the polygon
Therefore, the numbers of diagonals = $\dfrac{{n\left( {n - 1} \right)}}{2} - n$
$ \Rightarrow $Numbers of diagonals = $\dfrac{{{n^2} - n - 2n}}{2}$
$ \Rightarrow $Numbers of diagonals $ = \dfrac{{{n^2} - 3n}}{2}$or $\dfrac{{n\left( {n - 3} \right)}}{2}$
Therefore, numbers of diagonals of the polygon with n sides will be $\dfrac{{n\left( {n - 3} \right)}}{2}$.

Note: Whenever we face such type of problem, we always need to remember that how many selections out of the given vertices will eventually leads to formation of that particular required quantity, then by using the basics of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$we can reach to the right answer.