Let P be a closed polygon with 10 sides and 1 | vedclass

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(C) For a polygon with $n$ sides,the sum of interior angles is given by $(n-2) 	imes 180^{\circ}$.For $n = 10$,the sum of interior angles is $(10-2)

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) For a polygon with $n$ sides,the sum of interior angles is given by $(n-2) 	imes 180^{\circ}$.For $n = 10$,the sum of interior angles is $(10-2) 	imes 180^{\circ} = 8 	imes 180^{\circ} = 1440^{\circ}$.Let $k$ be the number of reflex angles (angles $> 180^{\circ}$) and $m$ be the number of non-reflex angles (angles $\le 180^{\circ}$).We have $k + m = 10$.The sum of the $k$ reflex angles is less than $k 	imes 360^{\circ}$ but must be greater than $k 	imes 180^{\circ}$.The sum of the $m$ non-reflex angles is greater than $0^{\circ}$ and less than or equal to $m 	imes 180^{\circ}$.The total sum is $1440^{\circ}$.If $k = 8$,the sum of reflex angles would be at least $8 	imes 180^{\circ} = 1440^{\circ}$,which leaves $0^{\circ}$ for the remaining $2$ angles,which is impossible for a polygon.If $k = 7$,the sum of $7$ reflex angles is $> 7 	imes 180^{\circ} = 1260^{\circ}$. The remaining $3$ angles must sum to $1440^{\circ} - (	ext{sum of } 7 	ext{ reflex angles})$. Since the sum of $7$ reflex angles can be slightly more than $1260^{\circ}$,the remaining $3$ angles can be positive and sum to less than $180^{\circ}$.Thus,the maximum value of $k$ is $7$.

Let $P$ be a closed polygon with $10$ sides and $10$ vertices (assume that the sides do not intersect except at the vertices). Let $k$ be the number of interior angles of $P$ that are greater than $180^{\circ}$. The maximum possible value of $k$ is

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