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Say you were drawing n-sided regular polygons on a square grid (first side you drew being flat always). You wanted any polygon you drew to be 100 units high, aka the uppermost point being y = 100. With a square, this would mean you would want the perimeter to be 400, because if each side was 100 then from top to bottom it would be 100.

I know a rough approximation of some answers, so for # of sides, 4 = 100, 5 = ~65, 6 = ~58, 7 = ~45, 8 = ~41.5, 9 = ~35, etc.

I'm hoping there is some sort of formula for this, where given the desired total height and # of sides, you could come up with a measure for the perimeter and thus each side length.

I tried playing with Pythagorean theorem, triangle areas, apothems, etc. but I don't have a firm enough grasp on that stuff to put something together.

EDIT: I think I got it, perimeter = n*height*tan(180/n)

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    $\begingroup$ Draw line segments from the center to the vertices. You then have $n$ isosceles triangles. Each of these has central angle $\theta=(360/n)^\circ$ and side lengths $r$, the "radius" of the polygon. A bit of trig will get you the third side length: $2r\sin(\theta/2)$. $\endgroup$ – David Mitra Feb 9 '14 at 9:53
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Let $r$ be the radius of the circumcircle of the polygon. From this you get the perimeter as

$$s=2nr\sin\tfrac\pi n$$

This is becuse you can divide your polygon into $n$ isosceles triangles, and dividing each gives $2n$ congruent right triangles, each of which has a central angle of $\alpha=\frac{360°}{2n}$ and a hypothenuse of length $r$. This is the green triangle in the following image:

Illustration

But what is the height of the polygon, in terms of $r$? For even $n$ you'd get

$$h=2r\cos\tfrac\pi n$$

since from bottom edge to midpoint as well as from midpoint to top edge, you have one of the legs of such a right triangle. So for even $n$ you get

$$s = 2n\frac{h}{2\cos\tfrac\pi n}\sin\tfrac\pi n=nh\,\tan\tfrac\pi n$$

This is the result you had yourself already. But for odd $n$, the length from center to top vertex is $r$, so your total height is

$$h = r\left(1+\cos\tfrac\pi n\right)$$

and the perimeter is correspondingly

$$s=2n\frac{h}{1+\cos\tfrac\pi n}\sin\tfrac\pi n$$

Wolfram Alpha will help you find a shorter formula for that as well:

$$s=2n\frac{h}{1+\cos\tfrac\pi n}\sin\tfrac\pi n=2nh\cot\tfrac\pi{2n}$$

I guess you can find a geometric explanation for that formula if you consider the regular $2n$-gon, but I'll leave it at that for now.

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