What’s the answer for this question??
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The area of the regular pentagon is equal of 5 times the area of the triangle.
The formula of an area of a triangle:
[tex]A_\triangle=\dfrac{bh}{2}[/tex]
We have:
[tex]b=11.8\ in\\h=8.1\ in[/tex]
Substitute
[tex]A_\triangle=\dfrac{11.8\cdot8.1}{2}=47.79\ in^2[/tex]
The area of the regular pentagon:
[tex]A=5\cdot47.79=238.95\ in^2[/tex]