Maths investigation - Essay Example

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Maths investigation

1/3*1/3=1/9. Thus the number of sides is increased by 4 times i.e. 4*12=48. Length of the side is decreased by 3 times. The perimeter is calculated directly as number of sides in this stage multiplied by the length of the side in the same stage i.e.48*1/9. Now each side in the previous stage gives rise to one equilateral triangle in this stage with side 1/3 of the side in the previous stage i.e.1/9. Therefore the number of new triangles added in this stage is equal to the number of sides in the previous stage i.e. 12. Therefore the area in this stage is calculated as the area in the previous stage plus the sum of the areas of the newly developed triangles.
In the above graph the stage number is taken on x-axis and the no. of sides are taken on y-axis. It shows that the no. of sides increases with the stage number. As n-> Nn ->, since in the limiting situation the fractal does not have an edge but is bound by a smooth curve. ...Show more

Summary

At stage 1 each side of the equilateral triangle is replaced by the zigzag line as shown on the page 7 against the label stage 1 with each segment of the zigzag line being of length 1/3. At this stage for each side of the equilateral triangle in the stage 0 there are 4 sides and thus the total number of sides is 4*3=12 and perimeter is calculated by the number of sides times the length of a side in this stage i.e…
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