If a polygon of n sides has 275 diagonals,the | vedclass

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(A) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.Given that the number of diagonals is $275$,we have

Vedclass · Accepted Answer

$25$

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(A) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.Given that the number of diagonals is $275$,we have:$\frac{n(n-3)}{2} = 275$$n(n-3) = 550$$n^2 - 3n - 550 = 0$Solving the quadratic equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-550)}}{2(1)}$$n = \frac{3 \pm \sqrt{9 + 2200}}{2}$$n = \frac{3 \pm \sqrt{2209}}{2}$$n = \frac{3 \pm 47}{2}$Since $n$ must be positive,$n = \frac{3 + 47}{2} = \frac{50}{2} = 25$.Thus,the number of sides $n$ is $25$.

If a polygon of $n$ sides has $275$ diagonals,then $n$ is

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