Question

1.
4m
Find the area of the polygon

Answer 1

Explanation:

Area = 1/2 apothem × perimeter

Area = 1/2 apothem × perimeter A= 1/2 r cos(20) × N × S

Area = 1/2 apothem × perimeter A= 1/2 r cos(20) × N × S we need to find radius so

Area = 1/2 apothem × perimeter A= 1/2 r cos(20) × N × S we need to find radius so s = 2rsin(180/90)

Area = 1/2 apothem × perimeter A= 1/2 r cos(20) × N × S we need to find radius so s = 2rsin(180/90)4 = r ( 2 sin(20) )

Area = 1/2 apothem × perimeter A= 1/2 r cos(20) × N × S we need to find radius so s = 2rsin(180/90)4 = r ( 2 sin(20) )r = 4/ ( 2 sin(20)

Area = 1/2 apothem × perimeter A= 1/2 r cos(20) × N × S we need to find radius so s = 2rsin(180/90)4 = r ( 2 sin(20) )r = 4/ ( 2 sin(20)r = 5.847 approximate to 6

when we go back

A = 1/2( 6 × cos(180/n) )×9×4

A = 1/2( 6 × cos(180/n) )×9×4A = 1/2( 6 × cos(180/9) )×9×4

A = 1/2( 6 × cos(180/n) )×9×4A = 1/2( 6 × cos(180/9) )×9×4A = 101.486m2 app|ozximate to 102 m2

Answer 2

`\underset{ \textit{angle in degrees} }{\textit{area of a regular polygon}}\\\\ A=\cfrac{ns^2}{4}\cot\left( (180)/(n) \right) ~~ \begin{cases} n=\stackrel{sides'}{number}\\ s=\stackrel{side's}{length}\\[-0.5em] \hrulefill\\ n=9\\ s=4 \end{cases}\implies A=\cfrac{(9)(4)^2}{4}\cot\left( (180)/(9) \right) \\\\\\ A=36\cot(20^o)\implies A\approx 98.91~m^2`

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