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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?

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Do you have to use complex numbers? Can you use trigonometry? – Calvin Lin Jan 13 '13 at 4:15

Consider the lines joining the vertices of the circle 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 $\frac {n}{2} \sin \frac {2\pi}{n}$.

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 $\frac {1}{2} \left \| 1 \cdot \omega \right \| = \frac {1}{2} \sin \frac {2 \pi}{5}$.

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Did you lose a $1/2$ in your general formula for an $n$-gon's area? – Benjamin Dickman Jan 13 '13 at 4:32
@B.D I did. Thanks for finding it! – Calvin Lin Jan 13 '13 at 4:39

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