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## Geometry (all content)

### Course: Geometry (all content)>Unit 8

Lesson 8: Koch snowflake fractal

# Area of Koch snowflake (1 of 2)

Starting to figure out the area of a Koch Snowflake (which has an infinite perimeter). This is an advanced video. Created by Sal Khan.

## Want to join the conversation?

• this is a little confusing can you give me a small summery? •   Sure. He starts with an equilateral triangle, and pastes smaller triangles to the sides. This makes new sides. Then he adds even smaller triangles. For some crazy reason, he's never going to stop. :)

He wants to know how big his snowflake is. It gets a little bigger each time. Two ways it gets bigger are
Area: Adds up the area of of all the triangles.
Perimeter: This is a little trickier. When he pastes new triangles, they cover some of the old perimeter. He would have to subtract the edges that are now inside, and add the new edges. Here, Sal keeps track of the number of triangles but does not calculate the perimeter.

So the process is fairly simple: paste some triangles, add up the perimeter and area. Paste some more, add up the perimeter and area.

The interesting part is that the perimeter keeps on getting bigger... if you give me a big number and say, "the perimeter can never be longer than this", I can tell you exactly how many triangles I need to add, to make the perimeter bigger than that.

....but the same is not true for area. The snowflake will never reach the edges of the Sal's blackboard. You can draw a box around the snowflake and the area will never be as large as that box.

So there area is limited. Limited to what number? Sal is trying to figure out that number.

Adding up infinite sums isn't easy. Is that the part that's confusing?

Hope that helps.
• Why is the area divided by 4?! • If it goes on forever wouldn't tha area also be infinite • No; for the area to be infinite, the snowflake would have to take up an infinite amount of space. It doesn't take up and infinite amount of space, so the area can't be infinite.

A more mathematical way to explain it is this: as the snowflake continues growing, the extra parts get smaller and smaller; they get small so fast that the area of the snowflake slows down, and the area never gets larger than a certain amount.
• At or so, I get REALLY puzzled. I was having a hard time following before that, but when Sal started tossing fours all over the place he lost me. If one multiplies by 4 and then divides by 4, doesn't one end up right back where one started? If so, what's the point? • This is a common algebraic move that Sal used to "clean up" his equation, and it works because you are absolutely right: multiplying by 4 and dividing by 4 leaves the value of the equation unchanged. What's happening is that by leaving the division alone for the moment (Sal leaves it out front as 1/4), he can use the multiplication by 4 where he knows he wants it, which is distributed out to the terms of the summation so he can raise the power of the 4's. Sal does mention what he's doing, even though he doesn't go into it in detail. If you're interested, try watching some videos on algebra, or try writing down the equation Sal had and see if you can follow along his reasoning.
• So the area is finite and the perimeter is infinate? • what are some applications of the koch snowflake fractal? • When he starts factoring at , why doesn't the square (the 2 on the blue section) get factored out with the rest of the equation? Is it because it's not exactly s^2 like the original yellow equation? • ``sqrt(3)•s^2/4 + 3•sqrt(3)•(s/3)^2/4 + 12•sqrt(3)•(s/9)^2/4 + 48•sqrt(3)•(s/27)^2/4 ...sqrt(3)•(s^2/4 + 3•(s/3)^2/4 + 12•(s/9)^2/4 + 48•(s/27)^2/4 ...)sqrt(3)/4•(s^2 + 3•(s/3)^2 + 12•(s/9)^2 + 48•(s/27)^2 ...)sqrt(3)/4•(s^2 + 3•s^2•(1/3)^2 + 12•s^2•(1/9)^2 + 48•s^2•(1/27)^2 ...)sqrt(3)•s^2/4•(1 + 3•(1/3)^2 + 12•(1/9)^2 + 48•(1/27)^2 ...)``
(1 vote)
• It irked me that the third simplified equation used a blue addition sign rather than it being pink.

Anyways....

Isn't something to the first power pointless? Math hurts my brain sometimes. • So, is this like the Koch curve?
(1 vote) • At approximately , why are we allowed to divide the "outside bit" by four, and multiply the "inside bit" by four?
(1 vote) • You can multiply both the outside and the inside because any number is itself multiplied by 1, and 1 = 4 * (1/4)

You can hence multiply an expression by 4 * (1/4) while keeping its value. Using the property of multiplication m * (a * b) = (m * a) * b = a * (m * b) you can write that as 1/4 times the outside multiplied by 4 times the inside.

I hope this helps.

--Phi φ
(1 vote)