Figure it Out — The Koch Snowflake — Exploring Some Geometric Themes

1. 12 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 73

   Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.

   Hint. The rule replaces every straight side by a four-piece 'bump'.

   The rule. Start with an equilateral triangle. Then, on every side: (i) divide the side into 3 equal parts, (ii) raise an equilateral triangle on the middle part, pointing outward, and (iii) remove the middle part itself.

   The effect is that each straight side is replaced by a bump made of four segments, each one third of the original side. Then repeat the rule on every side of the new shape.

   Step 0. An equilateral triangle — 3 sides.

   Step 1. Each of the 3 sides becomes 4 shorter ones, giving 12 sides. The shape is the familiar six-pointed Star of David outline.

   Step 2. Each of those 12 sides becomes 4, giving 48 sides. The outline now has small bumps on the sides of the bigger bumps.

   Step 3, for the pattern. 192 sides, and the outline starts to look furry.

   Drawing tips. · Start with a side length that divides neatly by 3 twice over — 9 cm or 18 cm works well, since Step 2 then needs 1 cm or 2 cm segments. · Mark all three thirds on a side before drawing anything, so the bumps sit evenly. · The new triangles always point outward, away from the centre — pointing inward gives the Koch anti-snowflake, a different shape.

   The contrast with the Sierpinski fractals. Those two removed material at every step and lost area. Here material is added at every step, so the shape grows outward — which is why the perimeter behaves so strangely, as the next questions show.

   ✦ Step 0 is a triangle with 3 sides; Step 1 has 12 sides and looks like a six-pointed star; Step 2 has 48 sides — each step replaces every side by a four-segment bump.

2. 23 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 73

   Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.

   Hint. Count how many sides one side turns into.

   Count what happens to a single side. The side is divided into 3 equal parts. The middle part is removed, but two new sides appear as the raised triangle's other two sides. So the pieces are: · the first third — 1 side · the two sides of the raised equilateral triangle — 2 sides · the last third — 1 side That is 4 sides where there was 1. So the count is multiplied by 4 at every step:

   Sₙ₊₁ = 4Sₙ, with S₀ = 3

   Unroll it. S₀ = 3 S₁ = 3 × 4 = 12 S₂ = 3 × 4² = 48 S₃ = 3 × 4³ = 192 so Sₙ = 3 × 4ⁿ.

   Step n	0	1	2	3	4	5
   Sides	3	12	48	192	768	3072

   Why 4 and not 5. It is easy to think the middle third is still there, giving 5 pieces. It is not — the rule explicitly removes it and replaces it with the two sloping sides of the new triangle. Counting 5 would give 3 × 5ⁿ and every later answer would be wrong.

   ✦ Sₙ = 3 × 4ⁿ sides at step n — 3, 12, 48, 192, 768, …

3. 34 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 73

   Find the perimeter of the shape at the nth step of the Koch Snowflake sequence. Take the starting equilateral triangle to have a sidelength of 1 unit.

   Hint. Perimeter = (number of sides) × (length of one side). You already have the first factor.

   Two factors to track.

   Number of sides — from the previous question, Sₙ = 3 × 4ⁿ.

   Length of one side — each step divides a side into three, so every side is one third as long as at the step before. Starting from 1: Lₙ = (1/3)ⁿ

   Multiply. Perimeterₙ = Sₙ × Lₙ = 3 × 4ⁿ × (1/3)ⁿ Pₙ = 3 × (4/3)ⁿ

   Step n	0	1	2	3	4	5
   Sides	3	12	48	192	768	3072
   Side length	1	1/3	1/9	1/27	1/81	1/243
   Perimeter	3	4	16/3 ≈ 5.33	64/9 ≈ 7.11	256/27 ≈ 9.48	1024/81 ≈ 12.64

   What the formula says. Each step multiplies the perimeter by 4/3, a factor bigger than 1, so the perimeter grows without limit. There is no number it settles down to — go far enough and the outline of a snowflake that fits on your page is longer than a kilometre.

   And yet the area does not. The shape always stays inside a circle drawn round the original triangle, so its area stays bounded — in fact the true Koch Snowflake has area exactly 8/5 of the starting triangle. A finite area with an infinite boundary is the signature strangeness of fractals, and it is the reason coastlines have no well-defined length: measure with a shorter ruler and you get a longer coast.

   ✦ Pₙ = 3 × (4/3)ⁿ units — 3, 4, 16/3, 64/9, … The perimeter is multiplied by 4/3 at every step and therefore grows without bound, even though the shape stays inside a fixed circle.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 2, Reprint 2026-27 (hegp204.pdf), where this is Chapter 4 (pages 70-102) — the eleventh chapter of the Class 8 course and the longest in the book. Like the rest of Part 2 it carries NO printed answer key, so every formula and count was derived and then independently recomputed in Python before being written: the Sierpinski Carpet recurrences (R_n = 8^n, H_n = (8^n - 1)/7), the Sierpinski Triangle counts (3^n and (3^n - 1)/2), the areas (8/9)^n and (3/4)^n, the Koch side count 3 x 4^n and perimeter 3 x (4/3)^n, and the face/edge/vertex formulas for prisms and pyramids (checked against Euler's relation for every case). THIS IS A HEAVILY VISUAL CHAPTER, so figure-only items are handled in one of two ways and never guessed. (1) MEASURED FROM THE RENDERED PAGE: the cube-stack count on page 97 was settled by rendering the figure at 400 dpi and observing that each row sits one step BACK as well as one step up (every bottom cube shows its full top face), which makes it a square-layered step pyramid of 16 + 9 + 4 + 1 = 30 cubes rather than the ten visible; and the three letters in the page-96 puzzle were read off the printed pixel glyphs at 700-900 dpi as C (front), A (top) and F (side). (2) FLAGGED AND ANSWERED BY METHOD: the six candidate cube nets, the projection-matching sets, the cube-combination views, the isometric figures to copy, the rolling ball and the impossible triangle are all printed diagrams; each solution gives the full method and reasoning and says plainly that the diagram is not reproduced. ONE ITEM IS LEFT OPEN BY THE BOOK ITSELF and is reported as such: the 30 x 12 x 12 shortest-path Try This on page 87, where the book computes 42 cm and 40 cm for two unfoldings (24^2 + 32^2 = 1600 verified) and then says all unfoldings must be listed to find the answer — so the solution establishes only that the shortest path is at most 40 cm. The tetracube count in the page-100 exercise was verified by exhaustive computer enumeration: 8 arrangements up to rotation, 7 up to rotation and reflection, of which 5 are flat.. Questions are referenced from the NCERT textbook for identification.