NCERT Solutions Ganita Prakash (Part 2) Chapter 4 Koch Snowflake — Figure it Out
Book page 73 Updated on2026-09-05
Q1.
Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.
Answer
Start with an equilateral triangle. Replace every straight side by the four-segment ‘bump’: divide the side into three equal parts, build an equilateral triangle on the middle part, and rub out that middle part.
The generating rule. The dashed middle third is removed, so 1 side of length s turns into 4 sides of length s⁄3.
Step 0
An equilateral triangle — 3 sides.
Step 1
The six-pointed star (Star of David outline) — each of the 3 sides has become 4, giving 12 sides.
Step 2
A small bump is raised on each of those 12 sides, giving 48 sides — the outline already looks lacy.
Tip: Draw Step 0 with a side of 9 cm. Then Step 1 has sides of 3 cm and Step 2 has sides of 1 cm, so you can measure every bump with an ordinary ruler.
Q2.
Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.
Answer
Sn = 3 × 4n.
Every side is replaced by 4 sides, so Sn+1 = 4 Sn S0 = 3 S1 = 4 × 3 = 12 S2 = 4 × 12 = 48 S3 = 4 × 48 = 192 In general, Sn = 3 × 4n
Why 4 and not 3: the middle third is removed but two new sides of the raised triangle take its place. So of the three thirds, two survive and the removed one is replaced by two — 2 + 2 = 4 sides where there was 1.
Q3.
Find the perimeter of the shape at the nth step of the sequence. Take the starting equilateral triangle to have a sidelength of 1 unit.
Answer
Pn = 3 × (4⁄3)n units.
Length of one side at Step n = (1⁄3)n Number of sides at Step n = 3 × 4n Pn = 3 × 4n × (1⁄3)n = 3 × (4⁄3)n
Step n
0
1
2
3
4
Sides
3
12
48
192
768
Side length
1
1⁄3
1⁄9
1⁄27
1⁄81
Perimeter
3
4
16⁄3 ≈ 5.33
64⁄9 ≈ 7.11
256⁄27 ≈ 9.48
Why it happens: four times as many sides, each one-third as long, means the total length is multiplied by 4 × 1⁄3 = 4⁄3 at every step. Since 4⁄3 > 1 the perimeter grows without limit — P20 is already more than 900 units. Yet the whole snowflake always stays inside a small circle drawn round the first triangle. An unlimited boundary enclosing a limited region is one of the surprises fractals hold.
Compare: for the two Sierpinski fractals the multiplier (8⁄9 and 3⁄4) was less than 1 and the area died away. Here the multiplier 4⁄3 is greater than 1 and the perimeter blows up. Same kind of rule, opposite behaviour — the multiplier decides everything.