Koch Snowflake

A Koch snowflake begins with an equilateral triangle with a side length of 1 (stage 0). Trisect each side and attach a smaller equilateral triangle to the center of each side (stage 1). Repeat for each triangle to obtain stage 2. Find the perimeter and area of the stage 2 snowflake.

Source : NCTM Mathematics Teacher 2008

SOLUTION
P_0 = perimeter stage 0
=3\cdot 1
=3
A_0 = area stage 0
=\dfrac{1}{2}\cdot 1\cdot\dfrac{\sqrt{3}}{2}
=\dfrac{\sqrt{3}}{4}

In stage 1, each side is replaced by 4 smaller sides each of length \dfrac{1}{3}.
P_1=3\cdot 4\cdot\dfrac{1}{3}
=4
3 smaller equilateral triangles are added. By virtue of similarity the smaller triangle’s area is proportional the original triangle’s area. The areas’ ratio is equal to the square of the lengths’ ratio.
\dfrac{area}{A_0}=\left (\dfrac{1}{3}\right )^2
area=\dfrac{A_0}{9}
A_1=A_0+3\cdot\dfrac{A_0}{9}
=\dfrac{4A_0}{3}

In stage 2 each side is replaced by 16 smaller sides each of length \dfrac{1}{9}.
P_2=3\cdot 16\cdot\dfrac{1}{9}
=\dfrac{16}{3}
12 smaller equilateral triangles are added.
\dfrac{area}{A_0}=\left (\dfrac{1}{9}\right )^2
area=\dfrac{A_0}{81}
A_2=A_1+12\cdot\dfrac{A_0}{81}
=\dfrac{4A_0}{3}+\dfrac{4A_0}{27}
=\dfrac{40A_0}{27}
=\dfrac{40}{27}\cdot\dfrac{\sqrt{3}}{4}
=\dfrac{10\sqrt{3}}{27}

Answer: perimeter = \dfrac{16}{3} units; area = \dfrac{10\sqrt{3}}{27} square units

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , . Bookmark the permalink.

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