If the number of diagonals of a regular polyg | vedclass

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(B) The number of diagonals of a polygon with $n$ sides is given by the formula: $\frac{n(n-3)}{2}$.Given that the number of diagonals is $104$,we hav

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$16$

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(B) The number of diagonals of a polygon with $n$ sides is given by the formula: $\frac{n(n-3)}{2}$.Given that the number of diagonals is $104$,we have:$\frac{n(n-3)}{2} = 104$$n(n-3) = 208$$n^2 - 3n - 208 = 0$Solving this quadratic equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-208)}}{2(1)}$$n = \frac{3 \pm \sqrt{9 + 832}}{2}$$n = \frac{3 \pm \sqrt{841}}{2}$$n = \frac{3 \pm 29}{2}$Since $n$ must be positive,$n = \frac{3 + 29}{2} = \frac{32}{2} = 16$.Thus,$n = 16$.

If the number of diagonals of a regular polygon of $n$ sides is $104$,then $n=$

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