Class 9 - Ganita Manjari Chapter 8 - Part 17
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Class 9 Ganita Manjari Chapter 8: Sequences and Progressions
Part 17
FRACTALS IN SQUARES Exercise 8.3 - The SierpiΕ„ski Carpet Stage 0 1 Square Stage 1 8 Squares Stage 2 64 Squares Geometric Progression: \(1, 8, 64 \dots\)
We previously explored the SierpiΕ„ski Triangle. Now, let's look at another famous fractal: The SierpiΕ„ski Carpet.

Instead of a triangle, we start with a solid square. We divide it into a $3 \times 3$ grid (9 smaller squares) and remove the central one. We then repeat this process infinitely for every remaining square. The mathematics of counting these squares and calculating their shrinking area relies perfectly on Geometric Progressions! Let's solve Exercise 8.3 Q7.
1 SierpiΕ„ski Carpet: Square Count
Ganita Manjari - Pg 194
Exercise 8.3 - Q7(a): In the construction of a SierpiΕ„ski Carpet, we start with 1 solid square (Stage 0). In Stage 1, we divide it into 9 equal squares and remove the central one. In Stage 2, we repeat this for the remaining solid squares.

Write the sequence for the number of solid squares at Stage 0, Stage 1, Stage 2, and Stage 3. Find the common ratio and write the explicit formula \(t_n\).
Concept:
1
At each stage, every single solid square is split into 9 pieces, and 1 is removed. This means every 1 square turns into 8 squares.
2
This creates a Geometric Progression where we continually multiply by 8. Because we start at Stage 0 (\(n=0\)), the explicit formula simplifies to \(t_n = a \cdot r^n\).
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Hint:
1
Stage 0: 1 square.
2
Stage 1: \(1 \times 8 = 8\).
3
Stage 2: \(8 \times 8 = 64\). Multiply by 8 one more time for Stage 3.
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Mistakes to Avoid:
!
Using \((n-1)\) for Stage 0: The formula \(t_n = a \cdot r^{n-1}\) is strictly for sequences that start at position 1. Since the textbook defines the start as Stage 0, the exponent is exactly \(n\).
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Solution:
Calculating the sequence of solid squares:
Stage 0 (\(t_0\)) = 1
Stage 1 (\(t_1\)) = \(1 \times 8 = 8\)
Stage 2 (\(t_2\)) = \(8 \times 8 = 64\)
Stage 3 (\(t_3\)) = \(64 \times 8 = 512\)

Sequence and Formula:
The sequence is \(1, 8, 64, 512 \dots\)
Common Ratio (\(r\)) = \(\frac{8}{1} = 8\).
Ans: The explicit formula is \(t_n = 8^n\)
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2 SierpiΕ„ski Carpet: Area Decay
Ganita Manjari - Pg 194
Exercise 8.3 - Q7(b): If the initial area of the solid square at Stage 0 is \(1\) unit, what is the sequence for the area of the remaining solid squares at Stage 1, Stage 2, and Stage 3?

Find the common ratio and write the explicit formula for the area at Stage \(n\).
Concept:
1
At each stage, the total area is divided into 9 pieces, and 1 piece is removed. This means the remaining area is exactly \(\frac{8}{9}\) of the previous stage.
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Hint:
1
Stage 0 Area = 1.
2
Stage 1 Area = \(1 \times \frac{8}{9}\).
3
Stage 2 Area = \(\frac{8}{9} \times \frac{8}{9} = \frac{64}{81}\).
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Mistakes to Avoid:
!
Confusing number of squares with area: While the number of squares multiplies by 8 (growing infinitely), the size of those squares shrinks by 9! The combined area ratio is \(\frac{8}{9}\).
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Solution:
Calculating the sequence of the Area:
Stage 0 (\(A_0\)) = 1
Stage 1 (\(A_1\)) = \(1 \times \frac{8}{9} = \frac{8}{9}\)
Stage 2 (\(A_2\)) = \(\frac{8}{9} \times \frac{8}{9} = \frac{64}{81}\)
Stage 3 (\(A_3\)) = \(\frac{64}{81} \times \frac{8}{9} = \frac{512}{729}\)

Sequence and Formula:
The sequence is \(1, \frac{8}{9}, \frac{64}{81}, \frac{512}{729} \dots\)
Common Ratio (\(r\)) = \(\frac{8}{9}\).
Ans: The explicit formula is \(A_n = \left(\frac{8}{9}\right)^n\)
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Select the correct option:
In the SierpiΕ„ski Carpet, each solid square is divided into 9 smaller squares, and the central one is removed. What is the common ratio (\(r\)) for the sequence of the number of solid squares?
Concept:
1
If 1 item produces \(X\) new items, the sequence of the total count forms a GP where the common ratio \(r = X\).
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Hint:
1
Stage 0 = 1 square. Stage 1 = 8 squares. \(\frac{8}{1} = ?\)
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Mistakes to Avoid:
!
Confusing counting with measuring: The number of squares is an integer count (1, 8, 64). The area of the carpet is a fractional measurement (1, 8/9, 64/81).
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Solution:
Because each square yields 8 new smaller squares, the number of squares multiplies by 8 at each stage.
Therefore, \(r = 8\).
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Select the correct option:
What is the common ratio (\(r\)) for the sequence of the remaining area in the SierpiΕ„ski Carpet?
Concept:
1
If a fraction \(X\) of an object is removed, the remaining fraction is \(1 - X\).
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Hint:
1
1 out of 9 squares is removed.
2
\(1 - \frac{1}{9} = ?\)
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Mistakes to Avoid:
!
Choosing the removed fraction: The sequence tracks the area that is LEFT, not the area that was taken away! Don't select 1/9.
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Solution:
At every step, the area retains \(\frac{8}{9}\) of its previous size.
Therefore, \(r = \frac{8}{9}\).
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Select the correct option:
What happens to the total area of the SierpiΕ„ski Carpet as the number of stages (\(n\)) approaches infinity?
Concept:
1
If the common ratio of a GP is a positive fraction less than 1 (\(0 < r < 1\)), the sequence exhibits exponential decay toward zero.
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Hint:
1
Calculate \((\frac{8}{9})^{1000}\). The result is incredibly small.
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Mistakes to Avoid:
!
Confusing limits: The sequence starts at 1 and drops to 8/9, then 64/81. It does not stop at 8/9. It keeps getting smaller forever.
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Solution:
Because the common ratio \(\frac{8}{9}\) is less than 1, multiplying the area by it repeatedly causes the value to shrink infinitely towards 0.
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Select the correct option:
If you want to find the number of solid squares in the SierpiΕ„ski Carpet at Stage 3 using the explicit formula \(t_n = 8^n\), what is the correct calculation?
Concept:
1
Because Stage 0 defines the starting position, the exponent is simply \(n\). Substitute \(n=3\).
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Hint:
1
Calculate \(8^3\).
2
\(8 \times 8 \times 8 = ?\)
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Mistakes to Avoid:
!
Multiplying base by exponent: \(8^3\) is NOT \(8 \times 3 = 24\). An exponent means multiplying the base by itself!
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Solution:
\(t_3 = 8^3 = 8 \times 8 \times 8 = 512\).
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Select the correct option:
What is the defining difference between the SierpiΕ„ski Triangle and the SierpiΕ„ski Carpet?
Concept:
1
Both are fractals with an area that approaches 0. The difference is their geometric construction and their specific GP ratios.
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Hint:
1
Triangle Area Ratio: \(\frac{3}{4}\). Carpet Area Ratio: \(\frac{8}{9}\).
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Mistakes to Avoid:
!
Assuming they behave differently at infinity: Because both use a ratio less than 1, BOTH of their areas shrink to zero at infinity.
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Solution:
The Sierpinski Triangle uses a triangle base and an area ratio of 3/4.
The Sierpinski Carpet uses a square base and an area ratio of 8/9.
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Chapter 8: Sequences & Progressions

Part 1: Intro to Sequences & Patterns Ganita Manjari Pgs 174-175 Part 2: Explicit Rules & Predicting Terms Ganita Manjari Pgs 176-177 Part 3: Recursive Rules & Fibonacci Ganita Manjari Pgs 178-179 Part 4: Exercise 8.1 - Mastery (Q1-Q3) Ganita Manjari Pg 179 Part 5: Exercise 8.1 - Recursive (Q4-Q6) Ganita Manjari Pg 179-180 Part 6: Intro to AP & Common Difference Ganita Manjari Pgs 180-182 Part 7: AP Formulas & Taxi Fare Math Ganita Manjari Pgs 182-183 Part 8: Sum of First n Natural Numbers Ganita Manjari Pgs 183-185 Part 9: Exercise 8.2 - AP Basics (Q1-Q3) Ganita Manjari Pg 185 Part 10: Exercise 8.2 - Missing Terms (Q4-Q5) Ganita Manjari Pgs 185-186 Part 11: Exercise 8.2 - AP Word Problems Ganita Manjari Pg 186 Part 12: Intro to GP & Common Ratios Ganita Manjari Pgs 186-188 Part 13: Fun with Fractals: SierpiΕ„ski Ganita Manjari Pgs 188-190 Part 14: GP Graphs & Bouncing Balls Ganita Manjari Pgs 190-193 Part 15: Exercise 8.3 - GP Basics Ganita Manjari Pg 193 Part 16: Exercise 8.3 - Rebounds & Roots Ganita Manjari Pgs 193-194 Part 17: Exercise 8.3 - SierpiΕ„ski Carpet Ganita Manjari Pg 194 Part 18: End of Chapter - AP Logic Ganita Manjari Pgs 194-195 Part 19: End of Chapter - Divisibility Ganita Manjari Pg 195 Part 20: End of Chapter - Consecutive Sums Ganita Manjari Pg 195 Part 21: End of Chapter - Dual Equations Ganita Manjari Pg 195 Part 22: End of Chapter - GP Proofs Ganita Manjari Pg 195 Part 23: End of Chapter - Complex Logic Ganita Manjari Pg 195 Part 24: End of Chapter - Recursive Chains Ganita Manjari Pg 195 Part 25: Chapter Summary & Finale Ganita Manjari Pg 196
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