Proof for area of any regular polygon

This note is a follow up for my circle area proof here

In this note I will prove that the area of any regular polygon can be written as ap2\frac{ap}{2} where a is the anothem and p is the perimeter. I will try my best to make this understandable but it is much easier to understand this proof through video.

In the picture above, we see that all regular polygons can be divided into congruent triangles. The area of a triangle is of course bh/2. However, the triangle's height is equal to the apothem of the original polygon. Thus we have a(b/2).

Now, if we look at all the bases of the triangles, they add up to the perimeter of the original polygon. So the total area of the original polygon is the total area of all the triangles which is abn2a\frac{bn}{2} where n is the number of sides.

Finally, since bn= the perimeter of the polygon, we arrive at the conclusion that ap2\frac{ap}{2} is the area of the original polygon.

Thank you for the challenge @JubayerNirjhor: In my next note, I will prove that the area of any regular polygon can be represented as

14na2cotπn\dfrac{1}{4} na^2\cot\dfrac{\pi}{n}. Where n is the number of sides and a is the side length. You can find it here

Note by Trevor Arashiro
5 years, 1 month ago

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In most circumstances, people don't know the length of the apothem, but the length of a side. So it's useful to have an area formula with side length and number of sides argument.

If nn is the number of sides and aa is the length of one side, then the formula is 14na2cotπn\dfrac{1}{4} na^2\cot\dfrac{\pi}{n}. (Prove it)

Jubayer Nirjhor - 5 years ago

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I will try. But trig is not my strong suit.

Trevor Arashiro - 5 years ago

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Got it! It was a fun proof to do, took me about 10 mins because I was thinking of it the wrong way (trying to use identities). I'll post the proof in just a sec.

Trevor Arashiro - 5 years ago

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@Trevor Arashiro Can you add this discussion (along with part 2) to the Regular Polygons - Area wiki page? Thanks!

Calvin Lin Staff - 5 years ago

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