My Math Forum Math for Fun, Area of a circle

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 March 22nd, 2018, 05:02 AM #1 Member   Joined: Oct 2017 From: Japan Posts: 62 Thanks: 3 Math for Fun, Area of a circle If anyone is interested I came across an old and simple method to get the area of a circle and made a lesson out of it. Do you know that you can easily get the formula of the area of a circle from very basic math, without using calculus? In this lesson I will show you how. Please subscribe if you like and let me know if you have any question.
 March 22nd, 2018, 08:45 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Well, I think this is a matter of "sweeping the hard part under the rug". That is, it uses the fact that the circumference of a circle is $\displaystyle 2\pi r$ - and introducing $\displaystyle \pi$ itself is the hard part! Last edited by skipjack; March 22nd, 2018 at 10:46 AM.
 March 22nd, 2018, 10:58 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions When I was doing my physics degree in 2002, I cam up with the following: Consider a regular polygon with side length $\displaystyle a$ and number of sides $\displaystyle n$. Then split up the regular polygon into triangles by drawing a line from each vertex to the centre of the shape. Each of those triangles can now be split into 2 right-angled triangles by drawing a bisecting line (of length $\displaystyle h$) from the centre of the shape to the middle of each side of the polygon. You then have the following: The angle, $\displaystyle \theta$, between two of those straight lines at the centre of the polygon is $\displaystyle \theta = \frac{2 \pi}{2n} = \frac{\pi}{n}$ but $\displaystyle \tan \theta = h / (a/2) = \frac{2h}{a}$ Therefore $\displaystyle h = \frac{a}{2} \tan \theta = \frac{a}{2} \tan \left(\frac{\pi}{n}\right)$ The area of the triangle is $\displaystyle A_{tri} = \frac{1}{2} base \times height = \frac{1}{2} \times \frac{a}{2} \times h = \frac{a^2}{8}\tan \left(\frac{\pi}{n}\right)$ Therefore, the area of the regular polygon is this area times the number of triangles, which is $\displaystyle 2n$, therefore $\displaystyle A = 2n A_{tri} = \frac{a^2n}{4}\tan \left(\frac{\pi}{n}\right)$ This is valid for $\displaystyle n \ge 3$. Let's check: $\displaystyle n = 3, A = \frac{\sqrt{3}}{4} a^2$ $\displaystyle n = 4, A = a^2$ etc... Alternatively, if we don't have $\displaystyle a$ but we do have $\displaystyle h$, we can resubstitute to get: $\displaystyle A = h^2n\cot \left(\frac{\pi}{n}\right)$ We can now consider the limit $\displaystyle n \rightarrow \infty$. With some nice maths, you can obtain $\displaystyle \pi r^2$ I thought it was original until I found a super old 1950s textbook with a full description there Thanks from jonah and Sebastian Garth
 March 22nd, 2018, 11:51 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 What does "some nice maths" mean? Thanks from Benit13
March 23rd, 2018, 02:12 AM   #5
Senior Member

Joined: Apr 2014
From: Glasgow

Posts: 2,161
Thanks: 734

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
 Originally Posted by skipjack What does "some nice maths" mean?
L'Hôpital's rule and a couple of small angle approximations. I'll put it here if I get time.

Last edited by skipjack; March 23rd, 2018 at 04:32 AM.

 March 23rd, 2018, 06:11 AM #6 Member   Joined: Oct 2017 From: Japan Posts: 62 Thanks: 3 I agree on that but still find the method interesting nonetheless.

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