Chapter 8.3, Problem 64E

To determine

To find: The perimeter and area of first four iteration of Koch snowflake.

Explanation of Solution

Given information:

Below diagram represents first four iteration of Koch Snowflake.

Calculation:

Below diagram represents first four iteration of Koch Snowflake.

From above diagram, one side of the figure from the previous stages becomes four sides in the following stage. First stage there are three sides. i.e. $a_1=3,r=4$ .

So, $a_n=3{\left(4\right)}^{n-1}$ , where $a_n\text{ is number of sides in }nth\text{ stage}\text{.}$

Again, in every iteration length of a side is $\frac{1}{3}$ of length of side of previous stage. And length of side of first triangle is one. i.e. $x_1=1,r=\frac{1}{3}$ .

  $\text{So, }{x}_n=1{\left(\frac{1}{3}\right)}^{n-1}$ , where $x_n\text{ is the side length at }nth\text{ stage}\text{.}$

Thus perimeter will be given as number of sides multiplied by side length.

  $\begin{array}{l}{P}_n=a_n\times x_n\\ \Rightarrow P_n=3{\left(4\right)}^{n-1}\times {\left(\frac{1}{3}\right)}^{n-1}\\ \Rightarrow P_n=3{\left(\frac{4}{3}\right)}^{n-1}\end{array}$

  $\begin{array}{l}\text{Stage 1: }{P}_1=3{\left(\frac{4}{3}\right)}^{1-1}=3\\ P_2=3{\left(\frac{4}{3}\right)}^{2-1}=4\\ P_3=3{\left(\frac{4}{3}\right)}^{3-1}=\frac{16}{3}\\ P_1=3{\left(\frac{4}{3}\right)}^{4-1}=\frac{64}{9}\end{array}$

For Koch snowflake, side of each triangle added in the second stage has length $\frac{1}{3}$ of the side length of first stage, therefore newly added triangles have area $\frac{1}{9}$ of first triangle. Similarly, each triangle added in third stage has $\frac{1}{9}$ area of the triangles added in second stage, and so forth. Suppose first triangle has area of $1$ .

The total area of snowflake will be −

  $\begin{array}{l}A=1+3\left(\frac{1}{9}\right)+12{\left(\frac{1}{9}\right)}^2+48{\left(\frac{1}{9}\right)}^3+\mathrm{...}\\ A_1=1\\ A_2=1+\frac{3}{9}=\frac{4}{3}\\ A_3=1+\frac{3}{9}+\frac{12}{81}=\frac{40}{27}\\ A_4=1+\frac{3}{9}+\frac{12}{81}+\frac{48}{729}=\frac{376}{243}\end{array}$

Perimeter of Koch snowflake forms a geometric sequence but area does not form a geometric sequence. Although, area excluding the first term does form a geometric sequence.