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## 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.

## 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