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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.(15 votes)
- 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(12 votes)- S-(4/9)S = (9/9)S -(4/9)S = (5/9)S(2 votes)

- 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)(10 votes)
- 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.(14 votes)

- at1:57: just wonder why did you time S with 4/9???(6 votes)
- 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.(14 votes)

- 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?(8 votes)
- 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".(2 votes)

- How can an infinite sum be equal to 4/5 ?(2 votes)
- 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 ....

Adds up to ⅓(12 votes)

- 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....(3 votes)- An interesting point. Rather valid, if you ask me. Unfortunately, I must be the one to point this out...how is this a question?(2 votes)

- 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.(3 votes)
- i dont know if anybody cares but the equation is false. Whe nKhan writes out (1:50)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(5 votes)

- At2:03, 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 + ...

///////////////////// ----------------------------------------- -(2 votes)- Use code blocking for monospace formatting. (Click the "Formatting tips" hyperreference below the entry block)
`(4/9)^1 + (4/9)^2 + (4/9)^3 + ...`

(4/9)^2 + (4/9)^3 + (4/9)^4 + ...

You want the addition to be applied to equivalent exponents:`(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*...(2 votes)

## Video transcript

In the last video we got as far as figuring out that the area of this Koch snowflake This thing that has an infinite perimeter, can be expressed as this infinite sum over here So our job in this video is to try to simplify this, and hopefully get a finite value Let's do our best to actually simplify this thing right over here So the easiest part of this thing to simplify is this right over here So let's just focus on that Then if we can get a value for this part that I am bracketing off Then we can just place that value here and simplify the rest of it So what I've just bracketed off can be re-written as three times four ninths plus four ninths squared plus four ninths to the third power And you can go on and on and on Plus four ninths to every other power, all the way through infinity! Lucky for us, there is a way to figure out this infinite (geometric) series There's a way to figure this out, and I've done several videos where we prove the general thing But I'll just do it by hand this time, just so that we don't have to resort to some magical formulas So let's say that we define some sum, this one over here (let's call it S) Let's say that S is equal to what we have in parentheses over here It's going to be equal to four ninths, plus four ninths squared, plus four ninths to the third all the way to infinity Now let's also say that we multiply S by four ninths What's four ninths S going to look like? So then, I'm just essentially multiplying every term here by four ninths So if I take this first term and multiply it by four ninths, what am I going to get? Well I'm going to get four ninths squared If I take the second term and multiply it by four ninths, I'm going to get four ninths to the third power And we are going to go all the way to infinity So this is interesting When I multiply four ninths times this I get all of the terms here except for this first four ninths Now, this is kind of the magic of how we can actually find the sum of an infinite geometric series We can subtract this term right over here (this pink line) from this green line If we do that, clearly this is equal to that and this is equal to that So if we subtract this from that its equivalent to subtracting the pink from the green So we get S minus four ninths S is equal to... Well, every other term, this guy minus this guy is going to cancel out And that's going to happen all the way to infinity and on the right hand side you're only going to be left with this four ninths over here Then this four ninths, we can (S is the same thing as nine over nine) write this as nine over nine S minus four ninths S is equal to four ninths So nine over nine minus four over nine of something gives us five over nine So this becomes five ninths S is equal to four ninths Then to solve for S (and this is kind of magical but it's actually quite logical) Multiply both sides times the inverse of this, so times nine fifths on both sides These guys cancel out, and we get S is equal to four fifths That's really neat! We've just shown that this whole thing over here is equal to four fifths So this entire bracket that we did over here is equal to three times four fifths This entire bracket is equal to twelve over five Now let's go to our original expression, so we don't lose track of what we are doing We have this square root of three times S squared over sixteen Then we have this four here plus the entire thing in brackets (which simplified to twelve fifths) Just to add these two together we can rewrite four as twenty fifths and then twenty over five plus twelve over five is thirty-two over five Let me write that down over here This is the home stretch now, this is very exciting! We are about to find the finite area of something that has an infinite perimeter! So it's going to be square root of three times S squared over sixteen Times thirty-two over five We can divide the thirty-two in the numerator by the sixteen there, which gives two We are left with the area of a Koch snowflake (where the initial equalateral triangle that we started with has each of its sides as length S) is, two times the square root of three times S squared all over five For example, if that first equalateral triangle had a side-length of one Then the area of this crazy thing (that has an infinite perimeter) would just be two square roots of three over five Anyway, I think that's kind of cool