To answer this question, we will use the following diagram as reference:
From the above diagram, we want to determine the length of HI and L. Now, we know that:
[tex]HB=9mm\text{.}[/tex]Since triangle HIB is a right triangle and α=360/7 degrees, then:
[tex]\begin{gathered} \sin \frac{\alpha}{2}=\frac{\frac{L}{2}}{HB}\text{.} \\ \sin (\frac{360}{14})=\frac{L}{18\operatorname{mm}}\text{.} \end{gathered}[/tex]Therefore:
[tex]L=\sin (\frac{360}{14}^{\circ})18\operatorname{mm}.[/tex]To determine HI we use the cosine function:
[tex]\begin{gathered} \cos (\frac{\alpha}{2})=\frac{HI}{HB}=\frac{HI}{9\operatorname{mm}}, \\ HI=\cos (\frac{360}{14}^{\circ})9\operatorname{mm}. \end{gathered}[/tex]Finally, recall that the area of a heptagon is given by the following formula
[tex]\text{Area}=\frac{apothem\cdot perimeter}{2}.[/tex]Substituting the values we found we get:
[tex]A=\frac{\cos (\frac{360}{14}^{\circ})9\operatorname{mm}(\sin (\frac{360}{14}^{\circ})18\operatorname{mm}\times7)}{2}\approx221.649mm^2.[/tex]Answer:
[tex]A=221.649mm^2.[/tex]
