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A regular polygon with 9 sides (a nonagon) has a perimeter of 72 inches. What is the area of this polygon? Provide mathematical evidence.

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The perimeter of the regular nonagon is 72 inches, the length of each side can be determined as,


\begin{gathered} P=9s \\ 72=9s \\ s=8\text{ inches} \end{gathered}

The diagram can be drawn as,

The value of apopthem a can be determined as, where n is the number of sides,


\begin{gathered} a=(s)/(2\tan ((180^(\circ))/(n))) \\ =(8)/(2\tan20^(\circ)) \\ =10.98\text{ in} \end{gathered}

The area can be determined as,


\begin{gathered} A=(P* a)/(2) \\ =\frac{72\text{ inches}*10.98\text{ inches}}{2} \\ =395.63in^2 \end{gathered}

Thus, the required area of the polygon is 395.63 square inches.

A regular polygon with 9 sides (a nonagon) has a perimeter of 72 inches. What is the-example-1
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