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    Koch Snowflake

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    Part I: Here we will look at the Koch snowflake which is constructed as follows:

    a. Start by drawing an equilateral triangle: we will call this T1 as it is the first step in the process.

    b. Next, on each of the three sides of the equilateral triangle, identify the middle third of that side, erase it, and then put an equilateral triangle here (facing outward). Call the new shape T2.

    c. Next, on each of the sides of the new shape T2, repeat step b, that is erase the middle third of each side and replace it with a new equilateral triangle facing outward. Call this new shape T3.

    d. We can continue the procedure indefinitely, producing shapes T4, T5, . . .. If we were able to repeat this infinitely many times, we would get Tâ?? which is the Koch snowflake.

    Part II: Then Let A denote the area of T1 and P the perimeter of T1.

    a. Find the area of T2,T3,...

    b. What is the area of T?

    c. Find the perimeter of T2, T3, . . .

    d. What is the perimeter of T?

    e. In what ways do you see self similarity in T?

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    https://brainmass.com/math/triangles/drawing-koch-snowflake-435785

    Solution Preview

    I have attached a document and a diagram that should help you understand the problem. The Koch diagram document is for part 1, and the Koch document is for part 2.

    Part II: Then Let A denote the area of T1 and P the perimeter of T1.
    a. Find the area of T2,T3,...
    b. What is the area of T∞?
    c. Find the perimeter of T2, T3, . . .
    d. What is the perimter of T∞?
    e. In what ways do you see selfâ€"similarity in T∞?
    Area
    Taking s as the side length, the original triangle area is . This is the formula for calculating area of an equilateral triangle. This is equal to T1.
    The side length of each successive small triangle is 1/3 of those in the previous iteration
    The following formula is to calculate area in a Koch snowflake.

    For T2, this simplifies to (see ...

    Solution Summary

    Drawing the Koch diagram is advised in the solution.

    $2.19