MathsCBSE Class 8 · Exploring Some Geometric Themes

Fractals and Self-similarity for Class 8

Updated 17 October 2026 · Fact-checked

A fractal is a self-similar shape: the same pattern repeats at smaller and smaller scales. In the Sierpinski carpet, step n has 8^n squares and (8^n - 1)/7 holes, with area (8/9)^n of the start; in the Sierpinski triangle, 3^n triangles and (3^n - 1)/2 holes with area (3/4)^n. The Koch snowflake has 3 x 4^n sides and perimeter 3 x (4/3)^n for a unit triangle.

Understand Fractals and Self-similarity

Self-similarity. A fractal is a shape that shows the same or a similar pattern again and again at smaller and smaller scales. A fern has leaves that are smaller copies of the whole fern, with sub-leaves that are smaller copies again. Trees, clouds, coastlines, mountains and lightning show similar self-similarity. The chapter also notes fractals in human art, for example the tall structure of the Kandariya Mahadev Temple at Khajuraho (completed around 1025 CE) made of smaller copies of itself, and patterns in Fulani wedding blankets.

Sierpinski carpet. Start with a square. Break it into 9 equal squares and remove the central one (a hole). Repeat the procedure on each of the 8 remaining squares, and so on. Let $R_{n}$ be the number of remaining squares and $H_{n}$ the number of holes at step $n$ (step 0 is the original square).

Every remaining square gives 8 remaining squares in the next step: $R_{n} = 8 R_{n - 1}$ , so $R_{n} = 8^{n}$ ( $1, 8, 64, 512, \dots$ ).
Every remaining square gives one new hole, and old holes stay: $H_{n} = H_{n - 1} + R_{n - 1}$ . So $H_{n} = 1 + 8 + 8^{2} + \dots + 8^{n - 1} = \frac{8 ^{n} - 1}{7}$ ( $0, 1, 9, 73, 585, \dots$ ).
Each remaining square has $\frac{1}{9}$ of the area of its parent square and there are 8 of them per parent, so the remaining area multiplies by $\frac{8}{9}$ each step: area at step $n$ is $(\frac{8}{9})^{n}$ if the start has area 1.

Sierpinski triangle (gasket). Start with an equilateral triangle, join the midpoints of the sides to cut it into 4 identical equilateral triangles (the corner triangles are isosceles with a $6 0^{\circ}$ angle, hence equilateral), and remove the central one. Repeat on the 3 remaining triangles. Then $R_{n} = 3^{n}$ triangles, $H_{n} = H_{n - 1} + 3^{n - 1} = \frac{3 ^{n} - 1}{2}$ holes ( $0, 1, 4, 13, 40, \dots$ ), and the remaining area is $(\frac{3}{4})^{n}$ of the start.

Koch snowflake (named after the Swedish mathematician von Koch, who described it in 1904). Start with an equilateral triangle. Divide each side into 3 equal parts, raise an equilateral triangle over the middle part and remove the middle part. Each side is replaced by 4 sides, each one third as long. If the first triangle has side 1: the number of sides at step $n$ is $3 \times 4^{n}$ , each of length $(\frac{1}{3})^{n}$ , so the perimeter is $3 \times (\frac{4}{3})^{n}$ , growing by $\frac{4}{3}$ at each step.

Patterns for growth. The counts $8^{n}$ and $3^{n}$ show exponential growth: each step multiplies by a fixed factor. The areas $(8/9)^{n}$ and $(3/4)^{n}$ keep shrinking (a factor less than 1 each step) but never reach zero.

Nets of solids, shortest paths on a cuboid and projections are on the next topic page of this chapter.

Fractals and Self-similarity: formulas to remember

Carpet: squares remaining: $R_{n} = 8^{n}$; Each remaining square gives 8.
Carpet: holes: $H_{n} = H_{n - 1} + R_{n - 1} = \frac{8 ^{n} - 1}{7}$; Holes remain and each old square adds one.
Carpet: area remaining: $A_{n} = (\frac{8}{9})^{n}$; Start with area 1.
Triangle: triangles, holes, area: $R_{n} = 3^{n}, H_{n} = \frac{3 ^{n} - 1}{2}, A_{n} = (\frac{3}{4})^{n}$; Start with area 1.
Koch snowflake: $sides = 3 \times 4^{n}, perimeter = 3 (\frac{4}{3})^{n}$; Starting side 1 unit; multiply by the side for other sizes.

How to solve Fractals and Self-similarity questions

Use the generating rule of the fractal to count and measure.

1Write the rule for one step: what happens to one remaining square or side.
2Convert the rule into a recurrence: R_n = k x R_(n-1), or H_n = H_(n-1) + R_(n-1).
3Compute step 1, 2 and 3 in a table to see the pattern.
4Write the general expression using powers (8 to the n, 3 to the n, 4 to the n).
5For area or length, track the scale factor for the pieces and the number of pieces.
6Check against the table for small n.

Fractals and Self-similarity in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Koch border

Suppose a designer makes a decorative border in the Koch snowflake pattern starting from an equilateral triangle of side 12 cm.

Find the perimeter at steps 0, 1, 2 and 3.

Try it first, then show the solutionSolution

Step 0: $3 \times 12 = 36$ cm.
Each step multiplies the perimeter by $\frac{4}{3}$ : step 1 $= 36 \times \frac{4}{3} = 48$ , step 2 $= 48 \times \frac{4}{3} = 64$ , step 3 $= 64 \times \frac{4}{3} = \frac{256}{3} \approx 85.33$ cm.

Answer: 36 cm, 48 cm, 64 cm and about 85.33 cm

Memory hook: More sides, each one third as long: four thirds of the old length.

Counting squares

Imagine a tile pattern in which a square is cut into 9 equal squares with the centre one removed, and each remaining square is treated the same way.

How many squares and holes after 3 steps?

Try it first, then show the solutionSolution

Squares $= 8^{3} = 512$ .
Holes $= \frac{8 ^{3} - 1}{7} = \frac{511}{7} = 73$ .

Answer: 512 squares and 73 holes

Memory hook: Squares multiply by 8; holes add up the earlier squares.

Tricks to remember Fractals and Self-similarity

Short memory aids. Say each one aloud once, then use it on a question the same day.

Carpet: times 8 squares, add holes
Each square gives 8 squares and a new hole: squares x 8, holes + previous squares.
Use it when: Building the table for the carpet.
Triangle: times 3
Each triangle gives 3 triangles and a new hole.
Use it when: Counts for the Sierpinski triangle.
Koch: four thirds
Each side becomes 4 sides, a third as long, so the perimeter becomes four thirds of itself.
Use it when: Perimeter of the snowflake.
Area ratio is pieces times piece scale
Carpet: 8 pieces, each one ninth of the area: 8/9. Triangle: 3 pieces, each one quarter: 3/4.
Use it when: Remaining area.

Fractals and Self-similarity: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Squares at step 3 · 1 mark · MCQBoard level

The number of squares that remain at Step 3 of the Sierpinski carpet is: (a) 24 (b) 64 (c) 512 (d) 729

Try it first, then show the solutionSolution

$R_{3} = 8^{3} = 512$ .

Answer: (c) 512

Example 2: Holes and squares at step 2 · 2 marks · Short answerBoard level

Find the number of holes and of remaining squares at Step 2 of the Sierpinski carpet.

Try it first, then show the solutionSolution

$R_{2} = 8^{2} = 64$ . (1 mark)
$H_{2} = H_{1} + R_{1} = 1 + 8 = 9$ . (1 mark)

Answer: 64 squares and 9 holes

Example 3: Remaining area · 3 marks · Short answerHarder board level

The starting square of a Sierpinski carpet has area 1 sq unit. Find the area still remaining after 3 steps (as a fraction and as a decimal to 3 places) and the area removed.

Try it first, then show the solutionSolution

The area is multiplied by $\frac{8}{9}$ each step, so after 3 steps it is $(\frac{8}{9})^{3} = \frac{512}{729}$ . (1 mark)
$\frac{512}{729} \approx 0.702$ . (1 mark)
Removed $= 1 - \frac{512}{729} = \frac{217}{729} \approx 0.298$ . (1 mark)

Answer: Remaining 512/729 (about 0.702); removed 217/729 (about 0.298)

Example 4: Koch snowflake at step 3 · 3 marks · Short answerHarder board level

A Koch snowflake starts from an equilateral triangle of side 1 unit. Find the number of sides and the perimeter at step 3, and explain why the perimeter grows by the same factor each step.

Try it first, then show the solutionSolution

Sides: $3 \times 4^{3} = 192$ . (1 mark)
Each side has length $(\frac{1}{3})^{3} = \frac{1}{27}$ , so the perimeter is $192 \times \frac{1}{27} = \frac{64}{9} \approx 7.11$ units. (1 mark)
Each side is replaced by 4 sides of one third the length, so the perimeter is multiplied by $\frac{4}{3}$ at every step. (1 mark)

Answer: 192 sides; perimeter 64/9 (about 7.11); factor 4/3 per step.

Example 5: Sierpinski triangle at step 4 · 4 marks · Long answerHarder board level

For the Sierpinski triangle, find the number of triangles and holes at Step 4, and the area that remains if the starting triangle has area 1. Show that H_4 = H_3 + R_3.

Try it first, then show the solutionSolution

$R_{n} = 3^{n}$ , so $R_{4} = 81$ . (1 mark)
$H_{1} = 1$ , $H_{2} = 1 + 3 = 4$ , $H_{3} = 4 + 9 = 13$ , $H_{4} = 13 + 27 = 40$ , so $H_{4} = H_{3} + R_{3}$ with $R_{3} = 27$ . (1 mark)
Check with the formula: $\frac{3 ^{4} - 1}{2} = \frac{80}{2} = 40$ . (1 mark)
Remaining area $= (\frac{3}{4})^{4} = \frac{81}{256} \approx 0.316$ . (1 mark)

Answer: 81 triangles, 40 holes, area 81/256 (about 0.316)

Where students lose marks in Fractals and Self-similarity

The slips examiners see most often, and what to write instead.

Mistake: Counting the holes as equal to the squares
Holes accumulate over steps; squares multiply by 8.
Wrong working
"Step 2: 64 squares and 64 holes."
Right working
$H_{2} = 1 + 8 = 9$ .
Do this: Use H_n = H_(n-1) + R_(n-1).
Usually costs: 1 mark.
Mistake: Starting the numbering at step 1 with the original square
The original square is step 0, so step n has 8^n squares.
Wrong working
"Step 1 has 1 square."
Right working
Step 0 has 1 square; step 1 has 8.
Do this: Say "Step 0" for the starting figure.
Usually costs: The whole table.
Mistake: Forgetting that the sides get shorter in the Koch snowflake
More sides does not mean a perimeter of 4 times as large; each new side is one third as long.
Wrong working
"Step 1 has 12 sides, so the perimeter is 12 x 9."
Right working
$12 \times 3 = 36$ cm (each side is 3 cm).
Do this: Write the side length at every step.
Usually costs: 2 marks.
Mistake: Using 8/9 for the triangle and 3/4 for the carpet
The carpet keeps 8 of 9 pieces; the triangle keeps 3 of 4.
Wrong working
"Triangle: area after n steps = (8/9)^n."
Right working
Triangle: $(3/4)^{n}$ .
Do this: Write pieces kept over pieces made.
Usually costs: 1 mark.

Exam tips

Build a small table for steps 0 to 3 first; it shows the pattern and gives you a check on the formula.
State the rule for one square (or side) in a sentence before using symbols.
Express areas as fractions of the start so no units are needed.
Write the number of sides and the length of each side separately in Koch questions.
Give decimals to the number of places asked.

Practice questions on Fractals and Self-similarity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonBoard level
Assertion (A): At Step 1 of the Sierpinski carpet there are 8 remaining squares and 1 hole.
Reason (R): Each remaining square at one step gives 8 remaining squares and 1 new hole at the next step.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Start from one square at Step 0.
Answer
(a). Step 0 has 1 square and no hole; applying the rule once gives 8 squares and 1 hole.
Q2 · 2 marks · Very short answerBoard level
A Koch snowflake starts from an equilateral triangle of side 9 cm. Find the number of sides and the perimeter at steps 0, 1 and 2.
Hint
Each side becomes 4 sides, each a third as long.
Answer
Step 0: 3 sides of 9 cm, perimeter 27 cm. Step 1: 12 sides of 3 cm, perimeter 36 cm. Step 2: 48 sides of 1 cm, perimeter 48 cm.
Q3 · 3 marks · Short answerHarder board level
After how many steps does the remaining area of a Sierpinski carpet (starting area 1) first fall below one half? Give the area at that step and the step before.
Hint
Compute (8/9) to the power n for n = 1, 2, ... until it is below 0.5.
Answer
$(8/9)^{5} = \frac{32768}{59049} \approx 0.555$ , which is still above 0.5. $(8/9)^{6} = \frac{262144}{531441} \approx 0.493$ , which is below 0.5. So it takes 6 steps.
Q4 · 4 marks · Long answerHOTS / competency
For the Sierpinski carpet, (i) explain why R_n = 8 R_(n-1) and H_n = H_(n-1) + R_(n-1); (ii) complete a table of R_n and H_n for n = 0 to 4; (iii) check that H_4 equals (8^4 - 1)/7.
Hint
Every remaining square makes a hole in the next step.
Answer
(i) Every remaining square is divided into 9 and the centre removed, so it gives 8 squares and 1 hole; holes from earlier steps stay. (ii) $R$ : 1, 8, 64, 512, 4096. $H$ : 0, 1, 9, 73, 585 (each $H_{n} = H_{n - 1} + R_{n - 1}$ ). (iii) $\frac{8 ^{4} - 1}{7} = \frac{4095}{7} = 585$ , which matches.
Q5 · 4 marks · Case-basedHOTS / competency
Case study: a designer makes a Koch snowflake from a triangle of side 12 cm.
(i) Find the perimeter at steps 0, 1, 2 and 3. (2 marks)
(ii) At step 3, find the number of sides and the length of each side. (1 mark)
(iii) Will the perimeter ever stop growing? Explain. (1 mark)
Hint
Each step multiplies the perimeter by 4/3.
Answer
(i) $36$ , $48$ , $64$ and $\frac{256}{3} \approx 85.33$ cm. (ii) Sides $= 3 \times 4^{3} = 192$ , each $\frac{12}{27} = \frac{4}{9} \approx 0.44$ cm (check $192 \times \frac{4}{9} = \frac{256}{3}$ ). (iii) No: the perimeter is multiplied by $\frac{4}{3} > 1$ at every step, so it keeps increasing and has no largest value.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a fractal?Show answer
A self-similar shape that shows the same pattern at smaller and smaller scales.
2.How many squares remain at step n of the carpet?Show answer
8 to the n.
3.How many holes are there at step n of the Sierpinski triangle?Show answer
(3 to the n - 1) / 2.
4.By what factor does the Koch perimeter grow each step?Show answer
4/3.
5.What fraction of area remains after one step of the carpet?Show answer
8/9.
6.What is the Sierpinski triangle also called?Show answer
The Sierpinski gasket.

Fractals and Self-similarity: frequently asked questions

Does the area of the carpet ever become zero?

No. The area is multiplied by 8/9 each step, which gets smaller but never reaches zero.

Why are the corner triangles equilateral when midpoints are joined?

They are isosceles with a 60 degree angle at the corner, so the other angles are also 60 degrees; the chapter hints at this reasoning.

Is the Koch snowflake in Grade 6?

The chapter says we have already met this fractal in Grade 6 Ganita Prakash.

Can I use the formulas without proving them?

In exams you are expected to explain the rule for one step; the formula then follows from the pattern.