I am wondering if it is always the case that the area of any regular polygon of side length $s$ must include $s^2$ in it. Obviously for a square and hexagon it holds (area of square: $s^2$; area of hexagon: $\dfrac{3\sqrt{3}}{2}s^2$), but is it always true?
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$\begingroup$Suppose you have a regular $n$-gon with side length $s.$ By drawing segments from each vertex to the center of the $n$-gon, one splits the $n$-gon onto $n$ congruent isosceles triangles with a base of $b=s$, where the angle opposite the base is $2\pi/n$ radians (i.e.: one $n$th of $360^\circ$). Some trigonometry shows that the height of each triangle is $$h=\cfrac{\frac12s}{\tan\left(\frac\pi n\right)}=\frac1{2\tan\left(\frac\pi n\right)}s,$$ so the area of each triangle is $$\frac12bh=\frac1{4\tan\left(\frac\pi n\right)}s^2,$$ and so the total area is $$\frac{n}{4\tan\left(\frac\pi n\right)}s^2.$$
$\endgroup$ $\begingroup$Yes, it is always true.
Since the basic feature that can be known about any regular polygon can be expressed in terms of the side-length $s$, as you have mentioned, the expression for area of the polygon can be written in terms of $s$ and other real numbers.
But the physical quantity "area" has a dimension of $M^0L^2T^0$ i.e. $L^2$ or in other words, product of $2$ lengths. Hence the area can be represented as some number times $s^2$ for all regular polygons.
$\endgroup$ $\begingroup$Do you mean that the formula for the area must be of the form $c\cdot s^2$ for some constant $s$? In that case, the answer is yes. Say that the regular polygon with side length $1$ has area $c$. Since area is $2$-dimensional, area scales by the square of the side length. So the regular polygon with side length $s$ will have area $c\cdot s^2$. Finding the constant $c$ is another question, but it can be found using trigonometry.
To see why the area scales with the square of the side-length, imagine dividing the $n$-gon into $n$ congruent isosceles triangles by drawing lines from the center of the $n$-gon to the vertices. Now these triangles have base $s$, and the angle opposite the base is $\theta=2\pi/n$ radians or $360/n$ degrees. To find the height $h$ of this triangle, drop a perpendicular from the vertex opposite the base to the base. This bisects the angle $\theta$, so now we have two congruent right triangles. One leg has length $\frac{s}{2}$, the other leg has length $h$, and the angle opposite the leg of length $\frac{s}{2}$ is $\theta/2=\pi/n$ radians or $180/n$ degrees. Thus $\tan(\theta/2) = \frac{s/2}{h}$, and $h=\frac{s/2}{\tan(\theta/2)}$. So the area of each isosceles triangle (using $\text{Area}=\frac{1}{2}(\text{base})(\text{height})$) is $\frac{1}{2}\cdot s\cdot \frac{s/2}{\tan(\pi/n)}= \frac{1}{4\tan(\pi/n)} s^2$. Hence the area of the $n$-gon is $\frac{n}{4\tan(\pi/n)} s^2$. This tells you the constant $c$ from above is $\frac{n}{4\tan(\pi/n)}$.
More generally, area always scales with the square of a linear scale factor. To see this for more general shapes you need to get into the calculus of measuring area.
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