Find the area of the pentagon formed in the plane with the fifth roots of unity as its vertices

Question

Find the area of the pentagon formed in the plane with the fifth roots of unity as its vertices.

is there any formula to solve this type of problem?

Answer 1

Consider the lines joining the vertices of the pentagon to the center. Each of these triangles are isosceles, with side length 1 and vertex angle $\frac {2 \pi}{5}$.

Hence, the area of the pentagon is $\frac {5}{2} \sin \frac {2\pi}{5}$, which we can evaluate to be $ \frac {5}{4} \sqrt{ \frac {5+ \sqrt{5}}{2} } $.

In general, the area of the n-gon is $\frac {n}{2} \sin \frac {2\pi}{n}$.

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If you have to use complex numbers to approach this question, then since the cross product uses $\sin \theta$, hence the area of one of these triangles will be $\frac {1}{2} \left \| 1 \times \omega \right \| = \frac {1}{2} \sin \frac {2 \pi}{5}$.