Waste less time on Facebook — follow Brilliant.
×

Proof for area of any Regular Polygon 2!

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 \(\boxed{\dfrac{1}{4} na^2\cot(\dfrac{\pi}{n})}\).

Note by Trevor Arashiro
2 years, 8 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Write a comment or ask a question... Vishwesh Agrawal · 2 years, 8 months ago

Log in to reply

@Vishwesh Agrawal Haha. Good use of the back ground text. Trevor Arashiro · 2 years, 8 months ago

Log in to reply

Good work. (y) Jubayer Nirjhor · 2 years, 8 months ago

Log in to reply

you cheat !!!!!!!(to sumit sakarkhar Vishwesh Agrawal · 2 years, 7 months ago

Log in to reply

toooooooooo good Sumit Sakarkar · 2 years, 7 months ago

Log in to reply

thanks a lot Vishwesh Agrawal · 2 years, 8 months ago

Log in to reply

awesome!!!!!!!!! well said :) Shubham Dwivedi · 2 years, 8 months ago

Log in to reply

45 Joseph Pulapaka · 2 years, 8 months ago

Log in to reply

awesome work, Lorenz Tazan · 2 years, 8 months ago

Log in to reply

awesome work. Lorenz Tazan · 2 years, 8 months ago

Log in to reply

This is great work. I have yet to understand trigonometry to this extent. Nastacio Tafoya · 2 years, 8 months ago

Log in to reply

gr8 work Gopalgoel Goel · 2 years, 8 months ago

Log in to reply

Absolutely brilliant!!!!!!!! well said Vishwesh Agrawal · 2 years, 8 months ago

Log in to reply

absolutely brilliant well said Vishwesh Agrawal · 2 years, 8 months ago

Log in to reply

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, 8 months ago

Log in to reply

@M Lin Well, if we try a triangle of side length 6, the area is \(9\sqrt3\). By your formula, it's supposedly \(3(36)tan(60)=108\sqrt3\) Trevor Arashiro · 2 years, 8 months ago

Log in to reply

@Trevor Arashiro "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 · 2 years, 8 months ago

Log in to reply

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 · 2 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...