In-text — Fractals and the Sierpinski Carpet — Exploring Some Geometric Themes

1. 13 marksGanita Prakash Cl-8 Part 2, Section 4.1, pages 70 and 74-75

   What makes a shape a fractal? Give examples of self-similarity from nature and from human art, as described in the chapter.

   Hint. Look for the same pattern reappearing at smaller and smaller scales inside itself.

   The defining idea — self-similarity. A fractal is a shape that contains smaller copies of itself, and those copies contain smaller copies again, and so on without end. The same pattern reappears at smaller and smaller scales. A shape is not a fractal merely because it is complicated; it is a fractal because its complication repeats as you zoom in.

   In nature. · The fern is the chapter's opening example — the whole frond has leaves shaped like smaller ferns, and those leaves have sub-leaves shaped the same way again. · A tree has a trunk with limbs, limbs with branches, branches with branchlets — the same branching rule at every scale. · Clouds, coastlines, mountains and lightning all show the same behaviour: a small piece of a coastline, magnified, looks much like a longer stretch of coastline.

   In human art — and India has some of the oldest examples. · The Kandariya Mahadev Temple at Khajuraho, Madhya Pradesh, completed around 1025 CE, is built as a tall structure made of smaller copies of the whole structure, which themselves carry still smaller copies. Similar fractal-like patterns appear in temples at Madurai, Hampi, Rameswaram and Varanasi. · Nigerian Fulani wedding blankets carry diamond patterns containing smaller diamond patterns inside them. · The Dutch artist M.C. Escher is the modern master of the form; his print 'Smaller and Smaller' repeats an identical lizard pattern at ever-decreasing scales.

   In mathematics. The chapter builds three exactly-defined fractals — the Sierpinski Carpet, the Sierpinski Gasket (Triangle) and the Koch Snowflake. Each is produced by taking a simple shape and applying one geometric rule over and over, generating a sequence of shapes that approaches the fractal.

   ✦ A fractal is a shape that is self-similar — it contains smaller copies of itself at every scale. Nature gives ferns, trees, clouds, coastlines, mountains and lightning; art gives the Kandariya Mahadev Temple at Khajuraho, Fulani wedding blankets and Escher's Smaller and Smaller; mathematics gives the Sierpinski Carpet, the Sierpinski Gasket and the Koch Snowflake.

2. 24 marksGanita Prakash Cl-8 Part 2, Section 4.1, pages 71-72

   In the Sierpinski Carpet, a square is broken into 9 smaller squares and the central one is removed, and the process repeats. Let Rn be the number of remaining squares and Hn the number of holes at step n. Find a formula for each.

   Hint. Write down how each quantity is built from the step before, then unroll the recurrence.

   Set up the two recurrences from the construction.

   Remaining squares. Every square that survives at step n is cut into 9 and loses its centre, so it produces 8 surviving squares at step n + 1: Rₙ₊₁ = 8Rₙ

   Holes. Every surviving square at step n produces exactly one new hole at step n + 1, and all the old holes are still there: Hₙ₊₁ = Hₙ + Rₙ

   Unroll the first recurrence. Starting from R₀ = 1 (the whole square, no holes yet): R₀ = 1 R₁ = 8 × 1 = 8 R₂ = 8 × 8 = 8² R₃ = 8³ so in general Rₙ = 8ⁿ.

   Unroll the second. Starting from H₀ = 0: H₁ = 0 + R₀ = 1 H₂ = 1 + R₁ = 1 + 8 = 9 H₃ = 9 + R₂ = 1 + 8 + 64 = 73 H₄ = 1 + 8 + 64 + 512 = 585 Each Hₙ is the sum of all the earlier R values: Hₙ = 1 + 8 + 8² + … + 8ⁿ⁻¹

   A closed form for Hₙ. That is a geometric sum. Multiply it by 8 and subtract: 8Hₙ − Hₙ = 8ⁿ − 1, so 7Hₙ = 8ⁿ − 1, therefore Hₙ = (8ⁿ − 1) ÷ 7 Check: n = 3 gives (512 − 1) ÷ 7 = 511 ÷ 7 = 73 ✓, and n = 4 gives (4096 − 1) ÷ 7 = 585 ✓

   n	0	1	2	3	4
   Rₙ	1	8	64	512	4096
   Hₙ	0	1	9	73	585

   Worth noticing. Both counts explode, but the squares always outnumber the holes — the remaining squares are also shrinking, so what the eye sees is a shape full of holes at every scale.

   ✦ Rₙ = 8ⁿ, and Hₙ = 1 + 8 + 8² + … + 8ⁿ⁻¹ = (8ⁿ − 1) ÷ 7.

3. 33 marksGanita Prakash Cl-8 Part 2, Section 4.1, page 72

   Show that by joining the midpoints of an equilateral triangle, we divide it into 4 identical equilateral triangles. [Hint: note that the corner triangles are isosceles.]

   Hint. Look at one corner triangle first: what are its two sides, and what is the angle between them?

   Set up the names. Let the equilateral triangle be ABC with each side of length 2a, and let D, E and F be the midpoints of BC, CA and AB. Joining D, E and F creates four small triangles: the three corner triangles AFE, BDF and CED, and the middle triangle DEF.

   Step 1 — the corner triangles are equilateral. Take triangle AFE. Since F and E are midpoints, AF = ½ AB = a and AE = ½ AC = a. So AFE is isosceles with AF = AE, which is the hint. Its apex angle is ∠A = 60°, because ABC is equilateral. The two base angles of an isosceles triangle are equal, so each is (180° − 60°) ÷ 2 = 60°. All three angles are 60°, so triangle AFE is equilateral with side a. The same argument at B and at C shows BDF and CED are equilateral with side a as well.

   Step 2 — the middle triangle is equilateral too. Since AFE is equilateral with side a, its third side FE also has length a. Likewise DF = a from triangle BDF and ED = a from triangle CED. So DEF has all three sides equal to a, making it equilateral with side a.

   Step 3 — all four are identical. Every one of the four triangles is equilateral with side a, so by SSS they are all congruent to one another.

   Check with areas. Four identical triangles must each be a quarter of the original. And indeed halving every side of an equilateral triangle divides its area by 4, since area scales as the square of the side — which is exactly the fact behind the "half the side is a quarter of the area" idea from Chapter 9.

   ✦ Each corner triangle has two sides of length a with a 60° angle between them, so it is isosceles with both base angles 60° and is therefore equilateral of side a; the middle triangle then also has all sides a, so all four are congruent equilateral triangles, each a quarter of the original.

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.