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## Geometry (all content)

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

Lesson 8: Koch snowflake fractal

# Area of Koch snowflake (2 of 2)

Summing an infinite geometric series to finally find the finite area of a Koch Snowflake. This is an advanced video. Created by Sal Khan.

## Want to join the conversation?

• I'd like to see a proof that the perimeter of Koch Snowflake is infinite.
• Are you left with just 4/9 when you subtract S-4/9S?

Aren't you left with 4/9 - (4/9)^(infinity +1) which reduces to 4/9
• S-(4/9)S = (9/9)S -(4/9)S = (5/9)S
• I sort of follow the working through of the proof, but I can't get my head around the idea that as the perimeter increases with each pass, the area remains finite. Are there other cases of this? (besides Koch snowflake)
• It sounds like you may be confusing terms where the area has a finite limit but it is not constant. As the perimeter increases the area increases as well, but because the triangles are getting smaller and smaller in a logarithmic fashion the amount of area added becomes less and less significant. At the same time you would be increasing the number of sides of the overall snowflake in a greater and greater quantity.
• at : just wonder why did you time S with 4/9???
• It's not intuitive, but he multiplies the series by 4/9 in order create a finite difference between S and 4/9S, so that he can solve for S. That is, he can't solve by adding up all the numbers in an infinite series, because it would take an infinite number of steps, but he can assert that the difference between the infinite series S and the infinite series 4/9S is 4/9, which allows him to solve for S. It totally blows my mind, too.
• Cool problem. My question, though, is about the perimeter. If we keep adding and adding more, it's reasonable that it's going to be infinite, but if we try to measure it in a known unit, e.g. centimeters, without being allowed to zoom in, we would be forced to come up with a finite answer. Even if we were allowed to zoom in, we would just get measurements that are in the range of μm or nm so they would not affect our final result. Is there a logical fallacy in what I'm proposing?
• For it to be infinite, it must be proportional to the zoom. As in, what you see is the length you get... so "enlargement" would be a better term than "zoom".
• How can an infinite sum be equal to 4/5 ?
• An infinite sum can be equal to almost anything. For example, the infinite sum that begins
0.3 + 0.03 + 0.003 + 0.0003 + 0.00003 ....
• imho: as i said before, saying that the area is finite is the same as saying you can continues to add sides, and make new triangles, but at some point you'll stop adding new areas. It's a very interesting thing going on here: the drawing shows you a frozen Koch snowflake, but we are asked to imagine a continuous creation. So, saying that the perimeter is infinite is saying that you can keep breaking a line, which is true... But then when you talk about the final area you want to draw a line around the fractal and say, see the area does not go across this point, therefore it is finite. But that treats area as a frozen line around the figure, whereas the figure, the snowflake was IN MOTION constantly growing and adding sides. So treat area as a dynamic creation of new space as well... As new sides are added, and they form triangles, new area is added too, even though it never crosses a boundary.
Sorry for the length....
• An interesting point. Rather valid, if you ask me. Unfortunately, I must be the one to point this out...how is this a question?
• So, when you do the trick with the sums, and 4/9S, if you had a finite series, say if the series ends at 4/9to the 3rd, and you try to multiply s with 4/9S you can't do the trick with the subtraction, and 'cancellation of the infinite tail' So if this is a finite figure, that is a snowflake with a finite number of sides, then each iteration has a bigger area than the one before. Only if you assume a figure with an infinite number of sides, then you can do the whole trick with the sums.... So when they show you an animation of the Koch snowflake and tell you, this is a figure with infinite perimeter and finite area, that is not true, because they always show you a finite figure, and ask you to imagine what if you were to take this to infinity... But they never actually show you a geometrical like they claim.
• i dont know if anybody cares but the equation is false. Whe nKhan writes out ()S= the third number is (4/3)^3 which is incorrect. It might be just my bad eyes but i am pretty sure its there.
(1 vote)
• You are correct (unless my eyes fail me as well).
Sal wrote (4/3)^3 instead of (4/9)^3, but the later evaluation uses it as (4/9)^3, so the final result is correct.

--Phi φ
• At , why does Khan leave a blank space?

shouldn't it be; (assume "///" is just a space for formatting purpose)
///////////////////// (4/9)^1 + (4/9)^2 + (4/9)^3 + . . .
///////////////////// (4/9)^2 + (4/9)^3 + (4/9)^4 + . . .
//////////////////// ----------------------------------------- -
not
/////////////////////////// (4/9) + (4/9)^2 + (4/9)^3 + ...
///////////////////////////////////// (4/9)^2 + (4/9)^3 + (4/9)^4 + ...
///////////////////// ----------------------------------------- -
``(4/9)^1 + (4/9)^2 + (4/9)^3 + ...(4/9)^2 + (4/9)^3 + (4/9)^4 + ...``
``(4/9)^1 +   (4/9)^2 +   (4/9)^3 +   ...            (4/9)^2 +   (4/9)^3 +   (4/9)^4 + ...(4/9)^1 + 2*(4/9)^2 + 2*(4/9)^3 + 2*...``