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