Therefore:
3 x 4/3 = 4 which is Stage no. 1
3 x 4/3 x 4/3 = 3 x (4/3)² = 5 1/3 which is Stage no. 2
3 x 4/3 x 4/3 x 4/3 = 3 x (4/3)³ = 7 1/9 which is Stage no. 3
General Rule for Perimeter is:
n = stage no. P = Perimeter of shape
Pn = 3 x (4/3) ⁿ
Area: Area of a triangle = ½ abSinC
A0 = ½ x 1 x 1 x Sin60º
= ½ x 1 x ( 3)/2
= ( 3)/4
A1 = 3 (½ x 1/3 x 1/3 x Sin60º) + ( 3)/4
= 3 (½ x 1/9 x ( 3)/2) + ( 3)/4
= 3( 3)/36 + 9( 3)/36
= 12 ( 3)/36
= ( 3)/3
A2 = 12 (½ x 1/9 x 1/9 x Sin60º) + ( 3)/3
= 12 (½ x 1/9² x ( 3)/2) + ( 3)/3
= 12 ( 3)/324 + ( 3)/3
= ( 3)/27 + ( 3)/3
= ( 3)/27 + 9 ( 3)/27
= 10 ( 3)/27
A3 = 48 (½ x 1/27 x 1/27 x Sin60º) + 10 ( 3)/27
= 48 (½ x 1/27 x 1/27 x ( 3)/2) + 10 ( 3)/27
= 48 (½ x 1/9³ x ( 3)/2) + 10 ( 3)/27
= 48 ( 3)/2916 + 10 ( 3)/27
= 48 ( 3)/2916 + 1080 ( 3)/2916
= 1128 ( 3)/2916
= 94 ( 3)/243
The following is a table showing the area at each stage in a geometric series
A = ( 3)/4
3 x 4 = 12 12 x 4 = 48
IT MULTIPLIES BY 4 EVERY SUCCESSEIVE TERM.
1/9 x 1/9 = 1/9² 1/9² x 1/9 = 1/9³
IT DIVIDES BY 9 EVERY SUCCESSIVE TERM.
This is a geometric series with the first added term being: (3 x A/9) = A/3, the common ratio being 4/9 and the number of added terms (and the index of 9) are equal to n.
In order to find the rule for this geometric sequence I must use this equation:
= a [(1-r ⁿ)/ (1-r)]
n = number of terms r = common ratio a = first term
Area = A/3 [(1-(4/9) ⁿ)/ (1-(4/9))]
This is equal to:
Area = 3A/5 (1-(4/9) ⁿ)
I must now adapt this equation in order for it to suit my sequence. The first shape’s area of ( 3)/4 is irregular, because earlier in my project, I used an odd result to calculate this number. Due to this, I must use the first added term as my first term. My common ratio is 4/9 as I worked out earlier. These have been substituted into the equation below:
Area = 3 ( 3)/20 (1-(4/9)ⁿ)
This equation is incorrect. However, I will now add on ( 3)/4, as I had discarded it previously for being an exception to the pattern:
An = ( 3)/4 + 3 ( 3)/20 (1-(4/9)ⁿ)
This equation works and can be used to find the area at any Stage No.
e.g. A2 = ( 3)/4 + 3 ( 3)/20 (1-(4/9)²)
= ( 3)/4 + 13( 3)/108
= 27 ( 3)/108 + 13 ( 3)/108
= 40 ( 3)/108
= 10 ( 3)/27
At Stage No. 4: n=4
N4 = N0 x r^4
N4 = 3 x 4^4
N4 = 3 x 256
N4 = 768
L4 = L0/r^4
L4 = 1/3^4
L4 = 1/81
P4 = 3 x (4/3) ^4
P4 = 3 x (256/81)
P4 = 768/81
P4 = 9 13/27
A4 = ( 3)/4 + 3 ( 3)/20 (1-(4/9) ^4)
A4 = ( 3)/4 + 3 ( 3)/20 (1 – (256/6561))
A4 = ( 3)/4 + 3 ( 3)/20 (6305/6561)
A4 = ( 3)/4 + 18915 ( 3)/131220
A4 = ( 3)/4 + 1261 ( 3)/8748
A4 = 2187 ( 3)/8748 + 1261 ( 3)/8748
A4 = 3448 ( 3)/8748
A4 = 862 ( 3)/2187
Step 5)
Sum to infinity
As n increases the perimeter of the Koch snowflake continues to increase with every stage, however after reaching a certain stage the area remains the same from there onwards. Therefore, this shows that the Koch snowflake has a finite area surrounded by an infinite perimeter. This also shows that the area series is convergent.
As n →
Sum = a/(1-r)
In order for this to work with my series I must add on the area at Stage 0.
= ( 3)/4 + { ( 3)/12] / [(1-(4/9)]}
= ( 3)/4 + 9( 3)/60
= 15( 3)/60 + 9( 3)/60
= 24( 3)/60
= 2( 3)/5
= 0.69282032302755091741097853660235
This answer is consistent with my previous results as the area gets closer and closer to this figure and will continue to do so, but without ever reaching it.
Proof by Induction
An = 3( 3)/20 [1-(4/9)ⁿ] + ( 3)/4
Firstly. I will prove that this general equation is true for n = 1
A1 = 3( 3)/20[1-(4/9)^1] + ( 3)/4
= 15( 3)/180 + ( 3)/4
= ( 3)/12 + ( 3)/4
= ( 3)/12 + 3( 3)/12
= 4( 3)/12
= ( 3)/3
The answer is correct.
Now, I will assume that it is true for n = k
Ak = 3( 3)/20 [1-(4/9) ^k] + ( 3)/4
Then, I must show that it is true for n = k + 1
A (k+1) = 3( 3)/20 [1-(4/9) ^ (k+1)] + ( 3)/4
This shows the (k+1)th term in the form of:
An = 3( 3)/20 [1-(4/9)ⁿ] + ( 3)/4
( 3)/4 → The initial area
( k+1) → The +1 is the inductive step
Instructions to draw fourth stage of Koch snowflake in Microsoft Windows Logo
to side :x :y → Type into commander box
if :y=0 [fd :x stop] → enter into To Mode box the click ok (applies to each of the next instructions)
side :x/3 :y-1
lt 60 side :x/3 :y-1
rt 120 side :x/3 :y-1
lt 60 side :x/3 :y-1
end → do no type this – click cancel to end
to vonkoch1 (Same process as above)
;superimpose 6 stages
cs pu bk 300 lt 90 fd 200 rt 90 pd
vonkoch2 500 0
vonkoch2 500 1
vonkoch2 500 2
vonkoch2 500 3
vonkoch2 500 4
end
to vonkoch2 :x :y (same process as above)
;draws single curve size :x stage :y at current cursor position
repeat 3 [side :x :y rt 120]
end
Now just type vonkoch1 in the commander box
