Figure it Out — The Sierpinski Triangle — Exploring Some Geometric Themes

1. 12 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 72

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

   Hint. One rule, applied to every surviving triangle: join the midpoints and take out the middle.

   The rule, stated once. Start with an equilateral triangle. Join the midpoints of its three sides — this cuts it into 4 identical equilateral triangles — and remove the middle one. Then apply the same rule to every triangle that is still there.

   Step 0. One equilateral triangle, complete and solid.

   Step 1. Join the midpoints and remove the central triangle. What remains is 3 equilateral triangles, one at each corner, each half the side of the original. There is now 1 upside-down hole in the middle.

   Step 2. Repeat inside each of those 3 triangles. Each one loses its own central triangle, so 3 new holes appear and 9 triangles remain, each a quarter of the original side.

   Step 3, for the pattern. 27 triangles remain and there are now 13 holes altogether, each triangle an eighth of the original side.

   Drawing tips. · Start large — at least 16 cm along the base — because by Step 2 you are drawing triangles a quarter of that, and by Step 3 an eighth. · Mark midpoints with a ruler rather than by eye; small errors compound at every step. · Shade the holes rather than erasing, so the pattern of what has been removed stays visible.

   What to look for in your own drawing. At Step 2 the picture already contains three shrunken copies of the Step 1 picture — that is the self-similarity, appearing as soon as the second step.

   ✦ Step 0 is one solid triangle; Step 1 has 3 triangles and 1 hole; Step 2 has 9 triangles and 4 holes — each step joins the midpoints of every surviving triangle and removes its middle.

2. 24 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 72

   Find the number of holes, and the number of triangles that remain, at each step of the shape sequence that leads to the Sierpinski Triangle.

   Hint. Set up the same two recurrences as for the Carpet, but with 3 survivors instead of 8.

   Set up the recurrences. Let Tₙ be the number of triangles remaining and Hₙ the number of holes at step n.

   Every surviving triangle is cut into 4 and loses its middle, so it leaves 3 survivors: Tₙ₊₁ = 3Tₙ, with T₀ = 1 Every surviving triangle also creates exactly one new hole, and old holes stay: Hₙ₊₁ = Hₙ + Tₙ, with H₀ = 0

   Solve the first. T₀ = 1, T₁ = 3, T₂ = 9, T₃ = 27, so Tₙ = 3ⁿ.

   Solve the second. H₁ = 0 + 1 = 1 H₂ = 1 + 3 = 4 H₃ = 4 + 9 = 13 H₄ = 13 + 27 = 40 so Hₙ = 1 + 3 + 3² + … + 3ⁿ⁻¹.

   Summing the geometric series: 3Hₙ − Hₙ = 3ⁿ − 1, so 2Hₙ = 3ⁿ − 1 and therefore Hₙ = (3ⁿ − 1) ÷ 2 Check: n = 4 gives (81 − 1) ÷ 2 = 40 ✓

   Step n	0	1	2	3	4	5
   Triangles Tₙ = 3ⁿ	1	3	9	27	81	243
   Holes Hₙ = (3ⁿ−1)/2	0	1	4	13	40	121

   A neat relation. Notice Hₙ = (Tₙ − 1) ÷ 2, so the holes are always just under half the triangles. Compare the Carpet, where Hₙ = (Rₙ − 1) ÷ 7 — the same structure with 8 in place of 3, because the branching number changed.

   ✦ Tₙ = 3ⁿ triangles remain and Hₙ = (3ⁿ − 1) ÷ 2 holes have been made: 1 and 0, then 3 and 1, then 9 and 4, then 27 and 13, and so on.

3. 34 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 72

   Find the area of the region remaining at the nth step of each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit.

   Hint. At each step, work out what fraction of the area survives — then apply that fraction n times.

   Sierpinski Carpet. Each square is cut into 9 equal pieces and 1 is removed, so 8/9 of the area survives each step. Applying that factor n times: Areaₙ = (8/9)ⁿ

   Alternatively, count directly: at step n there are 8ⁿ surviving squares, each of area (1/9)ⁿ, giving 8ⁿ × (1/9)ⁿ = (8/9)ⁿ — the same answer, which is a useful cross-check.

   n	0	1	2	3	4
   Area	1	8/9 ≈ 0.889	64/81 ≈ 0.790	512/729 ≈ 0.702	≈ 0.624

   Sierpinski Triangle. Each triangle is cut into 4 identical triangles and 1 is removed, so 3/4 of the area survives: Areaₙ = (3/4)ⁿ

   Counting directly: 3ⁿ surviving triangles, each of area (1/4)ⁿ, giving (3/4)ⁿ ✓

   n	0	1	2	3	4
   Area	1	3/4 = 0.75	9/16 ≈ 0.563	27/64 ≈ 0.422	81/256 ≈ 0.316

   What happens in the long run. Both areas are being multiplied by a fixed fraction less than 1 at every step, so both shrink towards zero. The triangle shrinks faster, since 3/4 is a smaller factor than 8/9.

   This is the surprise the fractals are built to deliver: the number of pieces grows without limit while the area they cover falls to nothing. The true Sierpinski fractals — the shapes these sequences approach — have zero area, even though something is always left to see.

   ✦ Carpet: (8/9)ⁿ square units. Triangle: (3/4)ⁿ square units. Both tend to zero as n grows.

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.