Cynthia Lanius

The Koch Snowflake

What about the AREA?

 
Table of Contents
   Introduction

  Why study fractals?
    What's so hot about
    fractals, anyway?

  Making fractals
    Sierpinski Triangle
         Using Java
         Math questions
         Sierpinski Meets Pascal
    Jurassic Park Fractal
         Using JAVA
         It grows complex
         Real first iteration
         Encoding the fractal
         World's Largest
    Koch Snowflake
         Using Java
         Infinite perimeter
         Finite area
         Anti-Snowflake
            Using Java

  Fractal Properties
    Self-similarity
    Fractional dimension
    Formation by iteration

  For Teachers
    Teachers' Notes
    Teacher-to-Teacher

  Comments
    My fractals mail
    Send fractals mail

  Fractals on the Web
    The Math Forum

  Other Math Lessons
    by Cynthia Lanius

   
In the Koch Snowflake, an infinite perimeter encloses a finite area. The perimeter of the Koch Snowflake gets bigger and bigger with each step. But what about the area? Imagine drawing a circle around the original figure. No matter how large the perimeter gets, the area of the figure remains inside the circle. Could we actually calculate the area?

If you'd like to draw and experiment with the figures, you may print and use this triangle grid paper to help you with the drawing.

Remember the process:

  1. Divide a side of the triangle into three equal parts and remove the middle section.

  2. Replace the missing section with two pieces the same length as the section you removed.

  3. Do this to all three sides of the triangle.

Let's investigate the area below.

Notice the second iteration of the Koch Snowflake above. Notice that the original triangle (yellow) is still contained in the Koch Snowflake with three smaller triangles (red) added in the first iteration, and twelve even smaller triangles (blue) added in the second iteration. So finding the area of the Koch Snowflake is just an addition problem. You find the area of the original triangle, add the area of the three red triangles for the first iteration, and add the twelve blue triangles for the second iteration.

Let's use the triangles of our grid to measure the area. The original yellow triangle has 81 triangles inside. So we'll say its area is 81. What is the area of each red and each blue triangle? Let's organize all our data into a table.

Iteration No. Area of 1 triangle No. of triangles added Amount of area added Total Area
--
    
81
1
9
3
27
108
2
1
12
12
120
Each "Area of one triangle" is 1/9 the previous one. Each "No. of triangles added" is 4 times the previous one. Let's predict the next steps from the rule we notice above.
Iteration No. Area of 1 triangle No. of triangles added Amount of area added Total Area
--
     
81
1
9
3
27
108
2
1
12
12
120
3
1/9
48
5.33
125.33
4
1/81
192
2.37
127.7
5
1/729
768
1.05
128.75
6
1/6561
3072
.4682
129.21
7
1/59049
12288
.2081
129.43
8
1/531441
49152
.0924
129.522
Observe what is happening. The Snowflake is still growing in area but more and more slowly. It is converging on some number, getting closer and closer to that number, but will never go above it.

lanius@rice.edu

You may obtain a print version of this page.

Copyright 1996-2007 Cynthia Lanius
URL http://math.rice.edu/~lanius/frac/koch3.html