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(D) Let the number of sides of the polygon be $n$.The number of diagonals in a polygon is given by the formula $\frac{n(n-3)}{2}$.Given that the numbe

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$11$

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(D) Let the number of sides of the polygon be $n$.The number of diagonals in a polygon is given by the formula $\frac{n(n-3)}{2}$.Given that the number of diagonals is $44$,we have:$\frac{n(n-3)}{2} = 44$$n(n-3) = 88$$n^2 - 3n - 88 = 0$Factoring the quadratic equation:$(n - 11)(n + 8) = 0$Since the number of sides $n$ must be a positive integer $(n \in \mathbb{N})$,we have $n = 11$.Thus,the polygon has $11$ sides.

If a polygon has $44$ diagonals,then it has ...... sides.

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