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ellipse described about the circle in which a regular pentagon is constructed mapped on an ellipseenter image description here

enter image description here

The surface can be calculated from my formula


Total area will be an ellipse

enter image description here

Area n work will be

$An=\frac{b.\sin\alpha.a \cos\alpha}{2}+\frac{b.\sin\alpha.a(1-\cos\alpha)}{2} $

$An=\frac{a.b}{2}(\sin\alpha\cos\alpha+\sin\alpha-\sin\alpha\cos\alpha) $

$An=\frac{a.b}{2}\sin\alpha=\frac{a.b}{2}\sin(\frac{360}{n}) $

How is this n part of it multiplied by n

$ A=a.b.\pi $

$ \frac{n}{2}\sin(\frac{360}{n})$ look

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It is $\frac{1}{5}$ of the area of the ellipse, which is $\frac{\pi}{5}ab$. The mapping from the circle to the ellipse by scaling along the $y$-axis maps the points from the pentagon, to the points on the ellipse. This mapping preserves relative areas, and since the pentagon divides the circle into $5$ equal sectors, the mapped points divide the ellipse into $5$ equal sectors.

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True to the method in the division area of ​​ellipse – user14319 Aug 8 '11 at 19:38

The above is exactly the method suggested in one of the answers to your earlier question about dividing an ellipse into an odd number of parts. But the picture certainly improves the presentation.

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Sorry, I'd not seen the earlier question. I will peruse posters' recent questions more carefully. – robjohn Aug 8 '11 at 21:21

The area of an ellipse sector from $0$ to $\theta$ degrees is $a b \theta/2$. The angle $\theta$ is measured along the circle with radius $a$. If a regular n-gon in inscribed into this circle then the angle of one sector is $\theta = 2\pi/n$. Hence, the area of an ellipse sector corresponding to one n-gon sector is $\pi a b / n$, for a pentagon $\pi a b / 5$.

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