Regular n-gons: area and perimeter

The table draws a triangle, a square, a pentagon, and a hexagon. Four names look like four area formulas. They are one regular $n$-gon with $n$ frozen. Regular means every side has the same length and every interior angle is equal — that extra freeze sits on top of the interior sum $(n-2)\pi$ from measuring angles. Write $r$ for the distance from the centre to any vertex. Then $n$ and $r$ fix the figure.

This math handbook writes the central angle $\phi$ in radians; a full turn is $2\pi$.

$n$ copies of one isosceles triangle

$n$ isosceles triangles pack the polygon, each with two sides $r$ and included angle $\phi$. The interior angle $\alpha$ at a vertex is the supplement of that central angle:

$\phi =\frac{2\pi }{n},\alpha =\pi -\phi .$

Half of $\phi$ cuts a right triangle whose opposite side is $a/2$, so the side, the perimeter $C$, and the area $A$ are

$a=2r\sin \frac{\phi }{2},C=na,A=\frac{1}{2}n{r}^2\sin \phi .$

The last formula is $n$ copies of $\frac{1}{2}{r}^2\sin \phi$, one per isosceles triangle. Drop equal sides or equal angles and this packing fails; the boards below refuse that drop.

The smallest pack

Freeze $n=3$. Then $\phi =2\pi /3$ and $\alpha =\pi /3$. That is the equilateral triangle. Drag $B$; $C$ follows so every side stays $a$. Formulas that start from the side $a$ rather than from $r$ sit with the other triangles.

A right central angle

Keep the same packing and freeze one more vertex: $n=4$. Then $\phi =\alpha =\pi /2$. The regular $4$-gon is a square. Drag $B$; $C$ and $D$ follow. Formulas that start from the side $a$ sit with the other quadrilaterals.

The angles stay put

Freeze $n=5$. Then $\phi =2\pi /5$ and $\alpha =3\pi /5$. Drag $B$; $r$, $a$, $C$, and $A$ scale together, while the two angles stay put. That is the general regular freeze made visible: $n$ is locked, $r$ is the only scale.

The side catches the radius

Freeze $n=6$. Then $\phi =\pi /3$ and $\alpha =2\pi /3$. Then $\sin (\phi /2)=\sin (\pi /6)=1/2$, so $a=r$: each side of a regular hexagon equals the circumradius. That identity is this freeze, not a new polygon. Drag $B$; the remaining vertices follow.

Circles, circular sectors, and ellipses live under circles, sectors, and ellipses.