Koch's+Snowflake

[|Koch's Snowflake Assignment] Begin with an equilateral triangle. Then remove the inner third of each side of the triangle and construct an equilateral triangle where the segment was removed. By repeating this process multiple times, we create an image known as Koch's Snowflake. The link below will direct you to the Logo website where you can enter the commands to create Koch's Snowflake. [|Logo] =Koch Snowflake= || pen on loop 3 forward 243 right 120 sleep 500 loopend || || math \text{Perimeter} = L* S\\ L=\text{length of each line segment}\\ S=\text{number of line segments}\\\\\\ \text{Level 0}\\ L=243\\ S=3\\ \text{Perimeter}= 729 math || pen on loop 6 forward 81 left 60 forward 81 right 120 sleep 500 loopend || || math \text{Perimeter} = L* S\\ L=\text{length of each line segment}\\ S=\text{number of line segments}\\\\\\ \text{Level 1}\\ L= 81\\ S= 3*4= 12\\ \text{Perimeter}= 972 math || pen on loop 6 forward 27 left 60 forward 27 right 120 forward 27 left 60 forward 27 left 60 forward 27 left 60 forward 27 right 120 forward 27 left 60 forward 27 right 120 loopend || || math \text{Perimeter} = L* S\\ L=\text{length of each line segment}\\ S=\text{number of line segments}\\\\\\ \text{Level 2}\\ L= 27\\ S=3*4*4=3*4^{2} = 48\\ \text{Perimeter}=1296 math || \text{Perimeter of n-th Koch Snowflake}\\ =L*S\\ =x(\frac{1}{3})^{n}*(3*4^{n})\\\\ \text{L=length of each line segment}\\ \text{x=the original length of each line segment}\\ \frac{1}{3}^{n}=\text{factor by which each length decreases}\\\\ \text{S=the number of line segments}\\ \text{3=the number of sides in a triangle}\\ 4^{n}=\text{the factor by which the S increases} math ||
 * = =Level= ||= =Logo Commands= ||= =Image of Koch's Snowflake= ||= =Perimeter of n-th=
 * ===0=== || pen up gotoxy -120 90
 * ===1=== || pen up gotoxy -120 90
 * ===2=== || pen up gotoxy -120 90
 * ===n=== ||  ||   || math

What is the perimeter of the "last" level of Koch's snowflake?
If we take our perimeter equation from above with x= 243 and n=100th level, we have the following: math 243(\frac{1}{3})^{100}*(3*4^{100})= 2.27*10^{15} math If we have n continually increasing, the perimeter will also increase substantially. Let's try infinity as our n value. Taking the limit of our perimeter then gives us: math \lim_{n\to \infty}({243(\frac{1}{3})^{\infty}*(3*4^{\infty}))= \infty math Therefore the perimeter of the "last" level of Koch's snowflake is infinite.

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