NIMCET 2021 Question Paper with solutions

Let the number of sides of polygon be $\mathit{n}$
Then total numbers of diagonal will be $={}^{\mathit{n}}{\mathit{C}}_2-\mathit{n}$
So, Given that
Total number of diagonal is $=44$
$\therefore {}^{\mathit{n}}{\mathit{C}}_2-\mathit{n}=44$
$\Rightarrow \frac{\mathit{n}\left(\mathit{n}-1\right)}{2}-\mathit{n}=44$
$\Rightarrow {\mathit{n}}^2-\mathit{n}-2\mathit{n}=88$
$\Rightarrow {\mathit{n}}^2-3\mathit{n}-88=0$
$\Rightarrow {\mathit{n}}^2-11\mathit{n}+8\mathit{n}-88=0$
$\mathit{n}\left(\mathit{n}-11\right)+8\left(\mathit{n}-11\right)=0$
$\left(\mathit{n}+8\right)\left(\mathit{n}-11\right)$
$\mathit{n}=11,-8$
We know that number of sides can not be negative.
Hence total number of sides are 11 .