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### 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?(3 votes)
- 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.(59 votes)

- Why is the area divided by 4?!(7 votes)
- The value needs to stay constant so your to divide and the multiply. I got it !!!(12 votes)

- If it goes on forever wouldn't tha area also be infinite(2 votes)
- 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.(8 votes)

- At9:40or 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?(2 votes)
- 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.(6 votes)

- So the area is finite and the perimeter is infinate?(2 votes)
- yes

watch http://www.khanacademy.org/video/koch-snowflake-fractal?playlist=Geometry

this clarifies on that point(4 votes)

- what are some applications of the koch snowflake fractal?(3 votes)
- It's more of a math theory- it wouldn't actually work in the real world. It is a nice way to show how fractals work without a SUPER complicated shape. Edit: Well, you could do it to a point in the real world, but it would not be infinite. In fact, you see fractals A LOT in nature.(1 vote)

- When he starts factoring at6:29, 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?(2 votes)
`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.(2 votes) - So, is this like the Koch curve?(1 vote)
- Yes, but instead of one side, he uses the three sides of a triangle.(1 vote)

- At approximately8:59, 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)

## Video transcript

We now know how to find the
area of an equilateral triangle. What I want to do
in this video is attempt to find the
area of a-- and I know I'm mispronouncing it-- a
"kawch," or "coach" snowflake. And the way you
construct one is you start with an
equilateral triangle. And then on each of the sides
you split them into thirds and in the middle third you
put another smaller equilateral triangle. And that's after one pass. And on the next pass, you do
that for all of the sides here. So a little one here, here,
here, here, here, here, here, here, here. I think you get
the general idea. That's the next pass. And on the next pass you do
it for all of these sides. And what's really
neat about this-- and we showed this
in a previous video-- is that you have a figure here
that has an infinite perimeter. But we're about to see
in this video it actually has a finite area, which is kind
of interesting to think about. So let's start with a
clean equilateral triangle right over here. We're going to assume that each
of the sides have length s. So it's going to be a
clean equilateral triangle. Each of the sides-- let me
draw that a little bit neater. Each of the sides have length s. And so what I'm
going to do is I'm going to keep track
of two things. I'm going to keep track of the
sides on this triangle as it turns into a snowflake . I'm going to keep track
of the number of sides, and I'm going to keep
track of the area after each pass of adding
more smaller triangles. So this is going to be
our count of the area. Actually, let me give myself
a little bit more real estate, just because I have a feeling
I might need to use it. So this is the sides. I'm going to write up here. And then this is our running
count of the area down here. So right when we
start, we have 3 sides. And our area, we already
figured out in a previous video, is going to be-- if we
assume each of the sides are length s-- is going to
be square root of 3s squared over 4. Fair enough. That was just a simple
equilateral triangle. Now we're going to take
each of these sides, divide them into thirds, and
then in that middle third, we're going to add another
smaller equilateral triangle. So it'll look like that on
that side right over there. I want you to think about what
we're doing to each side right here. So before I did this,
this was just one side. Then I split it into thirds. And that middle
third, I essentially put two sides in there. I put an equilateral triangle. So one side is now turned into
one, two, three, four sides. So every time we do a pass
of making the snowflake more intricate, each side will
turn into four sides. So you could imagine if we
do this on all three sides, we have 4 times 3, which
is now twelve sides. So if you multiply
this times 4, this gets us to twelve sides now. We can count them
out just to make sure where our logic is correct. 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12 sides. And now what is the area, now? Well it's going to be the
area of our original yellow equilateral triangle, plus the
area of each of these smaller ones. And what's the area of
each of these smaller ones? Well first of all, we
have three of them. There are three of each
of these smaller ones and then we use the
formula for the area of an equilateral
triangle again. So it's going to be square
root of 3 times s squared. But now the length of
each of the sides of each of these smaller
equilateral triangles-- they aren't s anymore,
they are s over 3. Remember, this length right
over here is s over 3. So this is going to
be s over 3, as well. Every pass, the sides of
the equilateral triangle become 1/3 of the previous pass. So it's not going to
be s squared anymore. It's going to be
s over 3 squared. And then all of that over 4. And then let's do another pass. So we're going to add these
triangles right over here. Going to add these
right over there. And this is the last
pass where I'll actually attempt to draw all of
the triangles over there. So how many sides
am I going to have, first of all, after
I do another pass? Well, the previous
pass, I had 12 sides. Each of those 12 sides are now
going to turn to 4 new sides when I add these little
orange bumps there. So I'm going to multiply
it times 4 again. I'm going to
multiply it times 4. So now I'm going
to have 48 sides. And how many new triangles? So what's the area? Well it's going to
be the yellow area plus the blue area
plus the orange area. So how many new orange
triangles do I have.? Well I'm adding a
new orange triangle to each of the sides
for the previous pass. In the previous
pass I had 12 sides. So now I'm going to add
12 orange triangles. And actually let me write that. I'll just write 12
orange triangles. But it's really I just
multiplied it times 4. And then I'm going to have
times the square root of 3. And now this isn't going
to be s over 3 any more. These are now going
to be s over 9. These have 1/3 of the dimensions
of these blue triangle. So this is going to be s over 9. s over 9 squared over 4. And so I think you might
start to see the pattern building if we do another
pass after this one. Move to the right a little bit. What will that look like? Let me do this in a different
color that I haven't used yet. Let me see I haven't
used this pink, yet. So now I'm going to have the
previous number of sides, that's my number of
new triangles, 48 times the square root
of 3 times s over-- now these are going to be even
1/3 of this-- s over 27 to the second power. All of that over 4. And I'm going to keep adding
an infinite number of terms of this to get the area
of a true koch snowflake. So I'm just going to keep doing
this over, and over again. So the trick really is
finding this infinite sum and seeing if we get a
finite number over here. So the first thing I want to
do, just to simplify, well, let me just rewrite it a little
bit differently over here. So the first thing
that might be obvious is that we can factor
out a square root of 3s squared over 4. So let me factor that out. So if we factor out
a square root of 3s squared over 4 from
all of the terms. Then this term right over
here will become a 1. This term right over here
is going to become a 3. Let's see, we factored
out a square root of 3. We factor out a four. And we factored
out the s squared. We factor out only
the s squared. So now it's going to have
plus 3 times 1/3 squared. That's all we have left here. We have the 1/3 squared. And then we have this 3. And I'm not simplifying
this on purpose, so that we see a pattern emerge. And then this next term, right
over here, plus-- so this 12 is still going to be there. But I'm going to write
that as 3 times 4. Let's see, we're factoring
out the square root of 3, we're factoring out the 4, we're
factoring out the s squared. And so we're going to
be left with 3 squared. That's what this is
down here, squared. So this is 1/3 squared. And then that, squared. So that's what we're left
with, with that orange term. And then we go to the pink term. This pink term. 48 is just 3 times 4 times 4. Three times 4. I'll write 4 squared here,
because each time we're going to multiply
it times 4 again. So the next one's going
to be 4 to the third, because we're really-- each
side turns into four sides. That's where that came from. 4 squared, we're factoring
out the square root of 3, we're factoring
out the four, we're factoring out the s squared. And all we're left is 1 over
3 to the third power, squared. So times 1 over 3 to the
third power, squared. And we're just going to keep
going like that forever. So on each step, we're
multiplying by 4, and we're also multiplying--
I guess we say, the power of this
4 is incrementing. So there's really 4 to
the 0-th power here. We have a 1 here you
can imagine, implicitly. The 4 to the first
power, 4 squared, then it'll go 4 to the third. This power is also
incrementing-- 3 to the first, 3 to the
second, 3 to the third. But we see that this power
is always one more than that. And it'll be much easier to
calculate this infinite-- what's going to turn into an
infinite-- geometric series, if those are actually
the same power. So what I want to do is I want
to increase the power of 4 in all of those. But I can't just willy-nilly
multiply everything by 4. If I'm going to multiply
everything by 4, I also need to divide
everything by 4. So what I'm going to do in
this step right over here is I'm going to multiply
and divide everything by 4. So if we divide by 4, I
can do that on the outside. So let me multiply 1/4
times this right over here. And so I'm dividing
by 4 out here. And then I'm going to
multiply this by 4. And so I'm not
going to be changing the value of the actual thing. This is going to be 4 plus
3 times 4 plus 3 times 4 to the third. And so what was cool about
this is now that the power of 4 and the power of
this 3 down here are going to be the same power. But it still seems
a little weird because we're taking
this 1 over 3 squared and then we're squaring it. And here we just
have to realize-- so this is always
going to be squared. And this is the thing
that's incrementing. But in general, if I
have 1 over 3 to the n, and I'm squaring it,
this is equal to 1 over 3 to the 2n power. So I'm just multiplying it by 2. If I'm raising something
to the exponent, then raising that
to an exponent, that's just raising it to
the n times 2 exponent. And this is the exact same thing
as 1 over 3 squared, raised to the nth power. So we can actually switch
these two exponents in a very legitimate way. And then let me
rewrite everything. Because I don't want to
do too much on this one step right over here. So this part right
over here gives us square root of 3s
squared over 16. And then that's
going to be times-- I'll open and close
the parentheses. So then we have 4 plus. Then in blue, I'll write 3
times 4 to the first power. And then I can rewrite
this as times 1/3. We could view this
is 1/3 squared. We could view this as 1 over
3 to the first power, squared, or we could view this as 1 over
3 squared to the first power. And I'm going to
write it that way. So times 1/9 to the first power. And then plus 3 times 4 squared. And then this we can write as,
times 1/9 to the second power. And then this one
we could write, plus 3 times 4 to
the third times-- we could write this is 1
over 27 to the second power. But we could also write this
based on what we saw over here. We could write this as 1 over
3 squared to the third power, based on this right over here. Let me make this clear. 1 over 3 to the third,
to the second power. This is the same thing as 1 over
3 squared to the third power. That's what we showed
right over here. So this is the equivalent
to 1/9 to the third power. And now we start to
see the pattern is starting to clean
up a little bit. And let me just
do one more step. And then we'll finish
this in the next video. So this is equal to square root
of 3s squared over 16 times 4 plus 3 times, this is
4/9, plus the next term is three times 4/9 squared. And then we have plus 3 times
4 over 9 to the third power. And then we're just
going to keep going on, and on, and on, and
on taking the 3 times 4/9 to the successively
larger, and larger powers. So this is what we have to find
the sum of to find our area. And we're going to do
that in the next video. And we're going to use
some of the tools we've used to find the sums of
infinite geometric series. But we're going to
re-do it in this video just so that you
don't have to remember that formula or that proof.