Here is the link to my previous proof for the area of any quadrilateral being \(\frac{ap}{2}\) where a is the apothem and p is the perimeter. These two notes are quite closely related and the other note helps you to get a better understanding of this note but it is not at all necessary.

In this note, I will prove that the area of any polygon is \(\dfrac{1}{4} na^2\cot(\dfrac{\pi}{n})\) where n is the number of sides and a is the side length.

We start by drawing lines to each of the vertices of the polygon. Each of the angles will be congruent and of course add up to 360 deg. Thus since there are n angles (n being the number of sides), each angle is \(\frac{360}{n}\).

Next, we draw the perpendicular bisectors of each side. Because each of the interior triangles is isosceles, the bisector will bisect both the angle and the side length. This makes the base of each "half" triangle \(\frac{a}{2}\) and the angle closest to the polygon's in-center \(\frac{180}{n}\Rightarrow \frac{\pi}{n}\)).

Next, we find the height of the triangle (which is equivalent to the apothem. In this case, because we are given the angle closest to the incenter of the polygon and the corresponding opposite side length \((\frac{a}{2})\), we must multiply the base by \(\cot\dfrac{\pi}{n}\). By doing this, we are left with the height since cot represents (in this case) the \(\frac{adjacent}{base}\).

Then, we get the area of each triangle \(\dfrac{b\times h}{2}\Rightarrow \dfrac{a\times (\dfrac{a}{2} \times (\cot\dfrac{\pi}{n}))}{2}\Rightarrow \dfrac{1}{4} a^2\cot(\dfrac{\pi}{n})\).

Finally, because there are as many triangles as there are sides, we multiply our formula by n (the number of sides and are left with \(\boxed{\dfrac{1}{4} na^2\cot(\dfrac{\pi}{n})}\).

## Comments

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TopNewestWrite a comment or ask a question... – Vishwesh Agrawal · 2 years, 10 months ago

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– Trevor Arashiro · 2 years, 10 months ago

Haha. Good use of the back ground text.Log in to reply

Good work. (y) – Jubayer Nirjhor · 2 years, 10 months ago

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you cheat !!!!!!!(to sumit sakarkhar – Vishwesh Agrawal · 2 years, 9 months ago

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toooooooooo good – Sumit Sakarkar · 2 years, 9 months ago

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thanks a lot – Vishwesh Agrawal · 2 years, 10 months ago

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awesome!!!!!!!!! well said :) – Shubham Dwivedi · 2 years, 10 months ago

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45 – Joseph Pulapaka · 2 years, 10 months ago

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awesome work, – Lorenz Tazan · 2 years, 10 months ago

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awesome work. – Lorenz Tazan · 2 years, 10 months ago

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This is great work. I have yet to understand trigonometry to this extent. – Nastacio Tafoya · 2 years, 10 months ago

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gr8 work – Gopalgoel Goel · 2 years, 10 months ago

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Absolutely brilliant!!!!!!!! well said – Vishwesh Agrawal · 2 years, 10 months ago

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absolutely brilliant well said – Vishwesh Agrawal · 2 years, 10 months ago

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Shouldn't the formula be n (a^2)tan(pi/n) paragraph 2 If "a" is the radius, then the area converges to pi at n=infinity – M Lin · 2 years, 10 months ago

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– Trevor Arashiro · 2 years, 10 months ago

Well, if we try a triangle of side length 6, the area is \(9\sqrt3\). By your formula, it's supposedly \(3(36)tan(60)=108\sqrt3\)Log in to reply

– M Lin · 2 years, 10 months ago

"a" is the radius, not the side length. If it was the side length, then your formula is correct. Your diagram made it confusing.Log in to reply

Could we say the triangles formed are equilateral? For example a hexagon would have 6 equilateral triangles area would be 6

a^2root(3)/4 ?? – Somesh Chadda · 2 years, 9 months agoLog in to reply