Given:
A regular heptagon with 7 sides is inscribed in a circle.
The radius is 10 mm.
The inscribed angle in the regular heptagon is 51.43 degrees.
Consider,
[tex]\begin{gathered} a^2=10^2+10^2-2(10)(10)\cos 51.43^{\circ} \\ a^2=200-200(0.6235) \\ a=8.68 \end{gathered}[/tex]The area of the heptagon is,
[tex]\begin{gathered} A=\frac{7}{4}a^2\cot (\frac{180^{\circ}}{7}) \\ =\frac{7}{4}(8.68)^2\cot (\frac{180^{\circ}}{7}) \\ =273.7\operatorname{mm} \end{gathered}[/tex]
Answer: option a) 273.641 mm² ( approximately)
