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Maths investigation - Essay Example

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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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Maths investigation
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Download file to see previous pages 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.
In general for any n>0 ,
Nn = 3*4n = N0*4n










the graph shows that the length of side decreases with the stage no. and finally converges to 0. i.e. the limiting fractal will not have any edges but will be bound by a smooth curve.
In general
In = 1/3n



The graph shows that the perimeter increases with the stage no. but it cannot tend to infinity as the area is bounded by 0.7 as can be seen from the subsequent graph.
in general
Pn = Nn*In
























Q7 part 2) The graph shows that area increases with stage number but it does not exceed 0.7 as the curve is tangential to the line y=0.7. in fact as n-> the area tends to 0.7.
In general
An = A0(1+3(k=0 to k=n-1, j=1 to j=n-1[4k*1/32*j]))
= A0(1 + 3(40*1/32 + 41*1/34 + 42*1/36 +..+ ...Download file to see next pagesRead More
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