Understanding why πr^2 works for different area calculations

  • #1
paulb203
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A kind of personal 'proof' to help me grasp πr^2
I know there are videos etc explaining why but I thought I would try to find a way to understand this myself.

Imagine a version of πr^2, but instead of being for the area of a circle, it’s for the area of a square.
Call it sqi ar ^2
sqi = the ratio of the ‘diameter’ of the square to the square’s ‘circumference’ (the ‘diameter’ of the square being a vertical or horizontal line through the square’s centre, and the ‘circumference’ being its perimeter).
So, sqi=4 (like π=3.14...)
ar = the ‘radius’ of the square (half it’s ‘diameter’, just like with a circe and it’s radius)

Now take a square 4 cm by 4cm
And apply sqi ar ^2
Which gives us 4x2^2
Which = 16cm^2
Which matches with the 16cm^2 we would get from the conventional way of finding the area of the square.

This helps make sense of πr^2, for me at least
Any thoughts?
 
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  • #2
That is a start in the understanding. Of course, one example does not make a proof. Try some other examples, such as regular hexagon, octagon, etc.

See how the ratio changes as the limit of number of sides increases.
 
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  • #3
The coincidence you are wondering about may be due to the similarity of the figure in the X and Y axis. If you consider the area of a rectangle or an ellipse, the pattern is broken.
 
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  • #4
Your proof has determined that the value of the perimeter/diameter ratio (aka π for circumferences) for a square (which "radius" has been defined by you) is 4.

Please, see:
https://www.mathsisfun.com/geometry/circle.html

https://www.mathsisfun.com/geometry/circle-area-lines.html

circle-area-lines.svg



circle-area-vs-square.svg


Area circle-square.jpg
 
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  • #5
scottdave said:
That is a start in the understanding. Of course, one example does not make a proof. Try some other examples, such as regular hexagon, octagon, etc.

See how the ratio changes as the limit of number of sides increases.
Thanks, Dave.

First challenge, for me; draw a regular hexagon. I had to dig out my protractor, and think about how to go about that. Got there in the end.
Drew a regular hexagon, sides 6cm.
Wondered what would count as the 'diameter'. Thinking corner to furthest away corner, so 12cm (that was interesting; twice the length of the sides).
Then realised I'd never considered what the area of a hexagon might be, what the formula for that might be. Found 2 online. 3√3/(2)(a^2) where a=side
This gave the area as approx 93.53
And,
A=0.5ap where a=apothem and p=perimeter
Once I'd Googled 'apothem' I applied the formula and got 90, which sounded right as I had approximated the apothem as 5

So now to apply a version of sqi ar ^2
Call it hxi ar ^2
hxi being the ratio of the perimeter of the hexagon (36 in this case) to the 'diameter' of the hexagon (12 in this case, if I'm correct in thinking that corner to furthest corner counts as the 'diameter'?).
And 36:12 = 3:1
So hxi ( 'pi for a hexagon' )= 3?

I'm knackered after that :)
 
  • #6
paulb203 said:
the ‘diameter’ of the square being a vertical or horizontal line through the square’s centre,
paulb203 said:
if I'm correct in thinking that corner to furthest corner counts as the 'diameter'?).
You are being inconsistent here. You should either use the side midpoints or the corners. One method approaches pi from above, the other from below, as you increase the number of sides.
 
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  • #7
A.T. said:
You are being inconsistent here. You should either use the side midpoints or the corners. One method approaches pi from above, the other from below, as you increase the number of sides.
Thanks, A.T.
I'll use the side midpoints as that's what made sense for me with a square.
And now I've found out that is the apothem I can dispense with calling it 'ar' and call it 'a' for apothem instead
But I've got two different answers.
Using [3√3/(2)](a^2) for the area (where a=length of side) which gives 93.5 I get 3.09 for 'hxi'
hxi(a^2)=93.5 (a=apothem) (apothem = 5.5)
hxi=93.5/5.5^2
hxi=3.09

And, using Area=0.5aP where a=apothem and P=perimeter I get 99 for the area (?I?!)
Which gives me;
hxi(a^2)=99
hxi=99/5.5^2
hxi=3.27 (the 27 recurring)

?I?!
 
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