Proof for area of any Regular Polygon 2!

Here is the link to my previous proof for the area of any quadrilateral being ap2\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 14na2cot(πn)\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 360n\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 a2\frac{a}{2} and the angle closest to the polygon's in-center 180nπn\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 (a2)(\frac{a}{2}), we must multiply the base by cotπn\cot\dfrac{\pi}{n}. By doing this, we are left with the height since cot represents (in this case) the adjacentbase\frac{adjacent}{base}.

Then, we get the area of each triangle b×h2a×(a2×(cotπn))214a2cot(πn)\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 14na2cot(πn)\boxed{\dfrac{1}{4} na^2\cot(\dfrac{\pi}{n})}.

Note by Trevor Arashiro
5 years, 2 months ago

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Good work. (y)

Jubayer Nirjhor - 5 years, 2 months ago

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Write a comment or ask a question...

vishwesh agrawal - 5 years, 1 month ago

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Haha. Good use of the back ground text.

Trevor Arashiro - 5 years, 1 month 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 - 5 years, 2 months ago

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Well, if we try a triangle of side length 6, the area is 939\sqrt3. By your formula, it's supposedly 3(36)tan(60)=10833(36)tan(60)=108\sqrt3

Trevor Arashiro - 5 years, 2 months ago

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

M Lin - 5 years, 1 month ago

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absolutely brilliant well said

vishwesh agrawal - 5 years, 1 month ago

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Absolutely brilliant!!!!!!!! well said

vishwesh agrawal - 5 years, 1 month ago

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gr8 work

gopalgoel goel - 5 years, 1 month ago

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This is great work. I have yet to understand trigonometry to this extent.

Nastacio Tafoya - 5 years, 1 month ago

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awesome work.

Lorenz Tazan - 5 years, 1 month ago

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awesome work,

Lorenz Tazan - 5 years, 1 month ago

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45

Joseph Pulapaka - 5 years, 1 month ago

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awesome!!!!!!!!! well said :)

Shubham Dwivedi - 5 years, 1 month ago

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thanks a lot

vishwesh agrawal - 5 years, 1 month ago

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toooooooooo good

sumit sakarkar - 5 years, 1 month ago

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you cheat !!!!!!!(to sumit sakarkhar

vishwesh agrawal - 5 years, 1 month ago

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Could we say the triangles formed are equilateral? For example a hexagon would have 6 equilateral triangles area would be 6a^2root(3)/4 ??

Somesh Chadda - 5 years, 1 month ago

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