The interior angles of a polygon with n sides | vedclass

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(C) The sum of interior angles of a polygon with $n$ sides is $(n-2) 	imes 180^{\circ}$.Let the angles be in an $A.P.$ with first term $a$ and common

Vedclass · Accepted Answer

$20$

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(C) The sum of interior angles of a polygon with $n$ sides is $(n-2) 	imes 180^{\circ}$.Let the angles be in an $A.P.$ with first term $a$ and common difference $d = 6^{\circ}$.The sum of the $A.P.$ is $\frac{n}{2}[2a + (n-1)d] = (n-2) 	imes 180^{\circ}$.Given the largest angle $a + (n-1)d = 219^{\circ}$,so $a = 219^{\circ} - 6(n-1) = 225^{\circ} - 6n$.Substituting $a$ into the sum formula:$\frac{n}{2}[2(225 - 6n) + (n-1)6] = (n-2)180$$n[225 - 6n + 3n - 3] = (n-2)180$$n[222 - 3n] = 180n - 360$$222n - 3n^2 = 180n - 360$$3n^2 - 42n - 360 = 0$Dividing by $3$: $n^2 - 14n - 120 = 0$$(n - 20)(n + 6) = 0$Since $n$ must be positive,$n = 20$.

The interior angles of a polygon with $n$ sides are in an $A.P.$ with a common difference of $6^{\circ}$. If the largest interior angle of the polygon is $219^{\circ}$,then $n$ is equal to . . . . . .

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