Question

A polygon has 44 diagonals. Find the number of its sides.

Answer 1

Hint: A polygon having n sides $\left( {}^{n}{{C}_{2}}-n \right)$ number of diagonals. Given are the number of diagonals i.e. 44, put this value equal to the formula stated above thus, find the number of sides.

Complete step by step answer:
Before proceeding to the question, let us first understand clearly what is a combination in mathematics because we will solve this problem very easily with the help of a combination formula.
So, in mathematics, a combination is a selection of items from a collection such that the order of selection does not matter.
Suppose, there are total n objects and we are choosing r objects from n objects, then the number of r objects chosen from n objects would be:
\[\begin{align}
  & {}^{n}{{C}_{r}},\text{ n = total objects} \\
 & \text{r = choosen objects} \\
 & \text{C = symbol of combination} \\
\end{align}\]
${}^{n}{{C}_{r}}$ can be expanded as:
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!},\text{ ! = factorial sign}\]
For example, \[6!=6\times 5\times 4\times 3\times 2\times 1=720\]
Similarly,
\[\begin{align}
  & n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right).......\left( n-\left( n-1 \right) \right) \\
 & r!=r\times \left( r-1 \right)\times \left( r-2 \right)\times \left( r-3 \right).......\left( r-\left( r-1 \right) \right) \\
 & \left( n-r \right)!=\left( n-r \right)\times \left( n-r-1 \right)\times \left( n-r-2 \right).......\left( n-r-\left( n-r-1 \right) \right) \\
\end{align}\]
We have a combination formula for a polygon having n sides, the number of total diagonals is given by:
\[{}^{n}{{C}_{2}}-n\]
Given here, the diagonal are 44, suppose n are the sides of polygon then,
\[\begin{align}
  & {}^{n}{{C}_{2}}-n=44 \\
 & \dfrac{n!}{2!\left( n-2 \right)!}-n=44\Rightarrow \dfrac{n\times \left( n-1 \right)\times \left( n-2 \right)!}{2\times 1\times \left( n-2 \right)!}-n=44 \\
 & \Rightarrow \dfrac{{{n}^{2}}-n}{2}-n=44\Rightarrow \dfrac{{{n}^{2}}-n-2n}{2}=44 \\
 & \Rightarrow {{n}^{2}}-2n-n=88\Rightarrow {{n}^{2}}-3n-88=0 \\
\end{align}\]
We will solve this quadratic equation with the factorization method. In factorization, we will split the middle term such that the product of the two terms after splitting is equal to the product of the first term and the constant term.
Therefore, we will split ─3n as ─11n + 8n.
\[\begin{align}
  & \Rightarrow {{\text{n}}^{2}}-11\text{n}+8\text{n}-88=0 \\
 & \Rightarrow \text{n}\left( \text{n}-11 \right)+8\left( \text{n}-11 \right)=0 \\
 & \Rightarrow \left( \text{n}+8 \right)\left( \text{n}-11 \right)=0 \\
\end{align}\]
Thus, either n = ─8 or n = 11.
So, n = 11, since negative roots here are not possible for sides.

Therefore, the sides of a polygon having 44 diagonals are 11.

Note: There is one more method of solving this question without going to the concept of combination. There is a predefined formula:
\[\dfrac{n\left( n-3 \right)}{2}=\text{ diagonals}\] where n are sides of polygon