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 \(\frac{ap}{2}\) 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 \(a\frac{bn}{2}\) where n is the number of sides.

Finally, since bn= the perimeter of the polygon, we arrive at the conclusion that \(\frac{ap}{2}\) 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

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

## Comments

Sort by:

TopNewestIn 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 \(n\) is the number of sides and \(a\) is the length of one side, then the formula is \(\dfrac{1}{4} na^2\cot\dfrac{\pi}{n}\). (Prove it) – Jubayer Nirjhor · 3 years ago

Log in to reply

– Trevor Arashiro · 3 years ago

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.Log in to reply

– Trevor Arashiro · 3 years ago

I will try. But trig is not my strong suit.Log in to reply

@Trevor Arashiro Can you add this discussion (along with part 2) to the Regular Polygons - Area wiki page? Thanks! – Calvin Lin Staff · 2 years, 11 months ago

Log in to reply