$\begin{array}{1 1}(A)\;9\\(B)\;10\\(C)\;11\\(D)\;12\end{array} $

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- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

Let the number of sides of polygon =n

$\therefore$ Number of angular points =n

$ \therefore$ Number of straight lines joining any two of these n points =$nC_2$

Now the number of sides of the polygon =n

$\therefore$ Number of diagonals =$nC_2-n$

But it is given the number of diagonals =44

$nC_2-n=44$

$\large\frac{n(n-1)}{2}$$-n=44$

$n^2-n-2n=88$

$n^2-3n-88=0$

$(n-11)(n+8)=0$

$n=11$ or $n=-8$

Rejecting negative quantity,

$n=11$

Hence the required number of sides =11

Hence (C) is the correct answer.

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