Question

A polygon has $44$diagonals, then the number of its sides are$A.$ $11$B. 7$C.$ $8$$D.$ $None{\text{ }}of{\text{ }}these$

Hint: Here we have to select two such points on joining them we create a diagonal .So for selection We Use combination to find no. of diagonals.
Let the polygon have n sides.
This means that the polygon has n vertices.
No. of diagonals of the polygon $=$ No. of ways two vertices could be connected by a line segment $-$
No. of sides of the polygon
$= c\left( {n,2} \right) - n{\text{ - }}\left( 1 \right)$

We know that,
$c\left( {n,r} \right) = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} \\ \Rightarrow c\left( {n,2} \right) = \dfrac{{n!}}{{\left( {n - 2} \right)!2!}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!2!}} = \dfrac{{n\left( {n - 1} \right)}}{2} \\$
Using the above obtained formula in equation $\left( 1 \right)$
No. of diagonals of the polygon $= c\left( {n,2} \right) - n = \dfrac{{n\left( {n - 1} \right)}}{2} - n = \dfrac{{n\left( {n - 3} \right)}}{2}$
Given in the problem the polygon has $44$diagonals.
Using the above formula, we get,
$44 = \dfrac{{n\left( {n - 3} \right)}}{2} \\ \Rightarrow 88 = {n^2} - 3n \\ \Rightarrow {n^2} - 3n - 88 = 0 \\$
Factoring the above quadratic equation, we get,
${n^2} - 11n + 8n - 88 = 0 \\ \Rightarrow n\left( {n - 11} \right) + 8\left( {n - 11} \right) = 0 \\ \Rightarrow \left( {n + 8} \right)\left( {n - 11} \right) = 0 \\$
From the above equation either $n = - 8$ or $n = 11$.
Since the no. of sides of a polygon is a positive real integer, we neglect the value $n = - 8$
$\Rightarrow n = 11$
Therefore, the number of sides of the polygon with $44$diagonals is $11$.
Hence option A is correct.

Note: Try to analyse the problems of above type using permutations and combinations. Sides of the polygon are to be excluded from the combination as mentioned in the above solution.