# Why do shapes with the same area have different perimeter?

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For example, Consider two shapes; a circle and rectangle.

Both these shapes have same area but the perimeter of circle is less than that of rectangle. Why?

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Consider a rectangle that is 100m long by 1cm: its area will be 1 m2. and its perimeter is about 200m. This is the same as a square 1m x 1m, but that has a perimeter of only 4m.

Why? Because one is long and thin and the other is square.

A circle is even more "efficient" than a square and so has a shorter (the shortest) perimeter for a given area.

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Perimeter is a one dimensional attribute, while area is  two dimensional.
IOW, you need one piece of information for one and two for the other.

That allows for considerable ( infinite ) variation.

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14 hours ago, MigL said:

Perimeter is a one dimensional attribute, while area is  two dimensional.
IOW, you need one piece of information for one and two for the other.

That allows for considerable ( infinite ) variation.

what do you mean by considerable (infinite) variation?

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2 hours ago, King E said:

what do you mean by considerable (infinite) variation?

Look at my rectangle example: you can make the ratio of perimeter to area as large as you want

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There is no definite relationship between the perimeter and the area of a plane figure.

There is a maxiumum area that can be enclosed by a given perimeter, given by the circular shape as already noted, but there is no minimum area.

But you might like to look at this curiosity.

In the plane the perimeter of a figure based on the curves 1/x2 and -1/x2 and the axes is infinite
yet the area of that figure is finite.

There is an even more impressive 3D version called Gabriel's Horn or trumpet.

Here the surface area is infinite yet the volume enclosed is finite.

The converse situation ie a finite boundary enclosing an infinite area or volume is not possible.

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You can even have something that's got a finite area, and an infinite perimeter, but that can be enclosed in a finite "box" (or drawn on a finite piece of paper).

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23 hours ago, studiot said:

The converse situation ie a finite boundary enclosing an infinite area or volume is not possible.

What about if you draw a circle on a plane, and consider the exterior region outside that circle? Isn't that an example of an infinite area bounded by a finite boundary? Granted, the boundary here doesn't 'enclose' the area...

Edited by Markus Hanke
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1 minute ago, Markus Hanke said:

What about if you draw a circle on a plane, and consider the exterior region outside that circle? Isn't that an example of an infinite area bounded by a finite boundary?

Very good

But only if the universe is infinite. 🙂

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2 hours ago, Markus Hanke said:

What about if you draw a circle on a plane, and consider the exterior region outside that circle? Isn't that an example of an infinite area bounded by a finite boundary? Granted, the boundary here doesn't 'enclose' the area...

Hello, Marcus, I see you are thinking outside the box again. Quite right, any complete curve/line divides the plane into two regions, either both infinite or one finite if the curve forms a loop.

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4 hours ago, Markus Hanke said:

Isn't that an example of an infinite area bounded by a finite boundary?

It isn't actually 'bounded', though, is it ?
The area outside the circle is not technically 'bounded', although it has a ( interior ) boundary.
( confusing choice of words ? )

For example, is the set of natural numbers excluding 1 to 10, 'bounded' ? It still goes off to infinity.
Is there some other mathematical term that implies a boundary, but is still 'unbounded' ?
Or am I simply misapplying the term 'unbounded' ?

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2 hours ago, MigL said:

It isn't actually 'bounded', though, is it ?
The area outside the circle is not technically 'bounded', although it has a ( interior ) boundary.
( confusing choice of words ? )

For example, is the set of natural numbers excluding 1 to 10, 'bounded' ? It still goes off to infinity.
Is there some other mathematical term that implies a boundary, but is still 'unbounded' ?
Or am I simply misapplying the term 'unbounded' ?

Hi, MigL, the Natural numbers is the wrong choice of set for that example.

Bounded sets and subsets (such as the natural numbers) of real numbers are characterised by two bounds

An upper bound or maximum number and a lower bound or minimum number. Neither bound need be part of the set, but the maxima and minima are part of the set.

In the case of the natural numbers, they have no upper bound and no maximum but they are 'bounded below' by any number (integer)less than 1.

The set in your example is really two sets  -

The set of negative integers which is bounded above by any integer >= 0

and the set of the positive integers which is bounded below by any integer <= 0

Finally none of these sets can set up a circle as described by Marcus. That requires complex numbers in the complex plane, which has different boundary requirements or alternatively the cartesian product set R x R.

But it was a good question though. +1

Edited by studiot
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OK, I might have used a bad example, but I'm still uncertain about the term 'bounded'.

Lets go back to one of the original examples Markus used.
Consider an infinite plane that is bisected by an infinitely long line.
Both resulting planes have a boundary, say one on the 'left', and the other on the 'right'.
But they are still both infinite planes, so I would consider them 'unbounded'.

My question is in regards to the words 'bounded' and 'unbounded'. Does the fact that they have a ( any ) boundary mean they are 'bounded' ?
Or does the fact that they are still infinite mean they are 'unbounded'.
I'm not sure of the ( confusing ) usage of the words.

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