# Koch Snowflake Fractal Creation - Pinterest

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8 Oct 2010 Continue this construction: the Koch curve is the limiting curve obtained the area of the region inside the snowflake curve and its perimeter. The resulting shape is highly complex, has a large perimeter and is roughly similar to natural fractals like coastlines, snowflakes and mountain ranges. You can A closed figure with an infinitely long perimeter what? One of the classic questions in the field of fractal geometry is: How long is the coast of Britain? Repeat! The Koch Snowflake - Perimeter. Question: If the perimeter of the equilateral triangle that you start with is 27 units What is the perimeter of the larger square?

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If you remember from the snowflake the three segments became four. The equation to get the perimeter for this iteration is. P n = P 1 (5/3)^n-1.

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period† von Koch snowflake sub. Kochkurva, snö-.

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https:// As all the sides are equal, perimeter = side length * number of sides. So, the perimeter of the nth polygon will be: 4^(n - 1) * (1/3)^(n - 1) = (4/3)^(n - 1) In each successive polygon in the Von As a result, the perimeter Pn of the Koch snowflake is calculated as follows (11) P n = N n · L n = 4 3 N n − 1 · L n − 1 = 4 3 P n − 1 = 4 n 3 n − 1 l. Therefore, if we start our computations from the original triangle from Fig. 1, after ①steps we have the snowflake having the infinite number of segments N ① = 3 · 4 ①. Also that after a segment of the equilateral square is cut into three as an equilateral square is formed the three segments become five. If you remember from the snowflake the three segments became four. The equation to get the perimeter for this iteration is.

The Koch snowflake (also known as the Koch curve, Koch star, or Koch island [1] [2]) is a fractal curve and one of the earliest fractals to have been described. It is based on the Koch curve, which appeared in a 1904 paper titled "On a Continuous Curve Without Tangents, Constructible from Elementary Geometry" [3] by the Swedish mathematician Helge von Koch . This is then repeated ad infinitum.

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If we just look at the top section of the snowflake. You can see that the iteration process requires taking the middle third section out of each line and replacing it with an equilateral triangle (bottom base excluded) with lengths that are equal to the length extracted. Koch Snowflake Investigation-Alish Vadsariya The Koch snowflake is a mathematical curve and is also a fractal which was discovered by Helge von Koch in 1904. It was also one of the earliest fractal to be described. 5.

Von Koch’s Snowflake is named after the Swedish mathematician, Helge von Koch. He was the one who described the Koch curve in the early 1900s.

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### 3x^2 = 4.35, what is x? : maths - Reddit

p = n*length. p = (3*4 a )* (x*3 -a) for the a th iteration. Again, for the first 4 iterations (0 to 3) the perimeter is 3a, 4a, 16a/3, and 64a/9.

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613-984-7598 Koch Etkinhostdns jagong. 613-984-3107 Personeriasm | 928-536 Phone Numbers | Snowflake, Arizona. 613-984- whiney snowflakeMånad sedan Alain KochMånad sedan Install the best perimeter fence you can and use single strand temporary electric fence on interior Perimeter of the Koch snowflake Each iteration multiplies the number of sides in the Koch snowflake by four, so the number of sides after n iterations is given by: N n = N n − 1 ⋅ 4 = 3 ⋅ 4 n . {\displaystyle N_{n}=N_{n-1}\cdot 4=3\cdot 4^{n}\,.} PERIMETER (p) Since all the sides in every iteration of the Koch Snowflake is the same the perimeter is simply the number of sides multiplied by the length of a side. p = n*length. p = (3*4 a )* (x*3 -a) for the a th iteration. Again, for the first 4 iterations (0 to 3) the perimeter is 3a, 4a, 16a/3, and 64a/9.

the outer perimeter of the shape formed by the outer edges when the process Investigate the increase in area of the Von Koch snowflake at successive stages.