Following a discussion on Reasonable Deviations. I was prompted to write on this.

The above is an illustration on the generation of the Sierpinski Gasket. For simplicity we assume that the area of the initial triangle is 1. We split this triangle into four triangles by joining the mid-points of the sides of the triangle. These smaller triangles as shown in figure 2 have equal areas. We then remove the middle triangle. We adopt the convention that we will only remove the middle triangle and not a triangle at the edge.

In each of the three remaining triangles we repeat the process and remove the middle triangles. This is where **self-symmetry** comes in. If we pretend to see only one of the three small triangles then we are actually doing the same thing as we did to the original triangle. Albeit on a smaller scale. The above figure shows the third and fourth iterates of the original triangle. This process repeated *ad infinitum *gives rise to the Sierpinski gasket.

The L-System representation of the process is:

**variables** : A B

**constants** : + −

**start** : A

**rules** : (A → B−A−B),(B → A+B+A)

**angle** : 60°

Anyway, now what is interesting about this figure is its area and the perimeter.

Continuing with our earlier assumption, suppose the initial triangle had an area

A_{0}= 1;

Now in the first iteration we remove one of the four equal areas in that triangle and keeping the other three that remain.

Therefore the total area of the first iterate will be equal to

A_{1 }= (3/4) x 1;

Similarly in the second iteration, we repeat the process as I have already noted above. Thus,

A_{2 }= (3/4)(3/4) x 1;

For *n *iterations the area will be given by:

A_{n }= (3/4)^{n} x 1;

Now if *n* is arbitrarily large then the area, it would follow will be **ZERO.**

Now finding out the perimeter is a similar exercise. The length of the boundary of the n^{th} iterate of the original triangle is the total length of the boundaries of all the shaded small triangles in the n^{th} iterate. It can be shown that this gets arbitrarily large as n gets arbitrarily large. Therefore we conclude that a Sierpinski gasket has infinite perimeter!

And that it has **zero area inside and infinite perimeter**.

This is against the Koch Snowflake, which is a figure having a** finite area inside an infinite perimeter**. This goes against the way we think and according to geometric intuition that we have. But it actually is characteristic of many shpaes in nature. For example (i really like this example :D ) if we take all the arteries, veins and capillaries in the human body, then they occupy a relative small fraction of the body. Yet if we were able to lay them out end to end, we would find that the total length would come to over 60000 kilometres. This example i am quoting from an article i had copied years ago.

**Related Article:**

on August 2, 2013 at 4:11 am |Being a Triangle and a Verb | Musings of Dennis and Jack[…] Fractal Info: https://onionesquereality.wordpress.com/2008/02/27/a-fractal-with-zero-area-and-infinite-perimeter/ […]

on August 2, 2017 at 9:51 pm |RaheelKoch Snowflake has infinite perimeter and a finite area, what i studied, either rectify your statement or ratify my statement

on August 4, 2017 at 10:57 pm |Shubhendu TrivediOfcourse. But no such claim is made (infact a sentence about the koch snowflake says exactly that). The zero area and infinite perimeter bit is for the sierpinski gasket.

on November 20, 2019 at 6:28 pm |eJournal (Final Entry 2): “Eat, Sleep, Maths and Repeat” – Life in Numbers[…] Trivedi, S. (2008). A Fractal with Zero Area and an Infinite Perimeter?. Retrieved 20 November 2019, from https://onionesquereality.wordpress.com/2008/02/27/a-fractal-with-zero-area-and-infinite-perimeter/ […]