set up trigonometric ratios for three separate regular pentagons with different givens:

Solution
From the taingle
Using trigonometric ratio it follow
The length ofa sde of the pentagon is
[tex]\begin{gathered} 2\times5tan36 \\ = \end{gathered}[/tex]Therefore the perimeter of the pentagon is \:
[tex]\begin{gathered} 5\times2\times5\times\tan36 \\ =50tan36 \\ =36.32 \end{gathered}[/tex]Therefore the area of the perimter is
[tex]\begin{gathered} A=\frac{1}{2}\times36.32\times5 \\ A=90.8 \end{gathered}[/tex]