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Fractals and geometric progressions · 5 marks

Observe the Sierpiński triangle and try to answer the following questions

  1. (a) How many black triangles are there in Stages 0 to 3 of the figure?
  2. (b) Can you predict the number of black triangles at Stages 4 and 5?
  3. (c) Can you find a rule for the number of black triangles at the nth stage?
  4. (d) Suppose the area of the triangle (that is, the black region) in Stage 0 is 1 square unit. What is the area of the black region in Stages 1, 2 and 3? What will be the area of the black region in Stages 4 and 5? Find a rule for the area of the black region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?
Stage 0Stage 1Stage 2Stage 3
Answer: (a) 1, 3, 9, 27 (b) 81 and 243 (c) 3n (d) 34, 916, 2764; 81256, 2431024; area = (34)n, which gets closer and closer to 0 as n increases.

Step-by-step solution

Idea: At each stage every black triangle is split into 4 equal smaller triangles and the middle one is removed. So the number of black triangles is multiplied by 3, while the black area is multiplied by 34. Both give geometric progressions.

(a) How many black triangles are there in Stages 0 to 3 of the figure?

  1. Stage 0: 1. Each black triangle becomes 3 black triangles at the next stage.½ mark
  2. Stage 1: 3; Stage 2: 3 × 3 = 9; Stage 3: 9 × 3 = 27.½ mark
1, 3, 9, 27

(b) Can you predict the number of black triangles at Stages 4 and 5?

  1. Stage 4: 27 × 3 = 81.½ mark
  2. Stage 5: 81 × 3 = 243.½ mark
81 and 243

(c) Can you find a rule for the number of black triangles at the nth stage?

  1. 1 = 30, 3 = 31, 9 = 32, 27 = 33: the power of 3 equals the stage number.½ mark
  2. So tn = 3n (a GP with common ratio 3). As a recursive rule: start with 1 at Stage 0, and each stage is 3 × the previous stage.½ mark
tn = 3n

(d) Suppose the area of the triangle (that is, the black region) in Stage 0 is 1 square unit. What is the area of the black region in Stages 1, 2 and 3? What will be the area of the black region in Stages 4 and 5? Find a rule for the area of the black region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?

  1. Each stage keeps 3 of the 4 equal parts of every black triangle, so the black area is multiplied by 34 each time.½ mark
  2. Stage 1: 34; Stage 2: 34 × 34 = 916; Stage 3: 2764 square units.½ mark
  3. Stage 4: (34)4 = 81256; Stage 5: (34)5 = 2431024 square units.½ mark
  4. Rule: sn = (34)n. Since 34 < 1, each stage makes the area smaller: about 0.75, 0.56, 0.42, 0.32, 0.24, … So as n increases, the black area gets closer and closer to 0 (it never becomes exactly 0).½ mark
34, 916, 2764; 81256, 2431024; (34)n; it approaches 0
(a) 1, 3, 9, 27 (b) 81, 243 (c) 3ⁿ (d) 3/4, 9/16, 27/64; 81/256, 243/1024; (3/4)ⁿ; the area keeps decreasing towards 0 as n increases.

Check: Stage 3 area = 27 small triangles, each (14)3 = 164 of the original: 27 × 164 = 2764 ✓.

Answer to write in the exam

(a)

Each black triangle → 3 black triangles at the next stage

∴ Stages 0 to 3: 1, 3, 9, 27

(b)

Stage 4 = 27 × 3 = 81

Stage 5 = 81 × 3 = 243

(c)

1 = 30, 3 = 31, 9 = 32, 27 = 33

∴ tn = 3n

(d)

Area at each stage = 34 × area at previous stage

Stage 1 = 34, Stage 2 = 916, Stage 3 = 2764 sq. units

Stage 4 = 81256, Stage 5 = 2431024 sq. units

∴ sn = (34)n

34 < 1 ∴ the area decreases and gets closer and closer to 0 as n increases.

Common mistakes that cost marks

  • Counting the white (removed) triangles as well. Count only the black ones: 1, 3, 9, 27.
  • Writing the area at Stage 2 as 34 − 14 = 12. The area is multiplied by 34 again, not reduced by a fixed amount: 916.
  • Writing the rule as 3n−1 by habit. The stages start at 0, and Stage 0 has 30 = 1 triangle, so the rule is 3n.

How this can come in the exam

MCQ (1 mark)

In the Sierpiński triangle, if Stage 0 has area 1 square unit, the black area at Stage 6 is

  1. 7294096
  2. 1824
  3. 324
  4. 2434096
Show answer

(A) 7294096
(34)6 = 7294096.

Case-based (4 marks)

An art student makes a Sierpiński triangle from a black card of area 64 cm2, following the same steps.
(i) How many black triangles are there at Stage 3? (ii) What is the area of one black triangle at Stage 3? (iii) What is the total black area at Stage 3? (iv) At which stage is the total black area first less than 20 cm2?

Show answer(i) 33 = 27 (1 mark). (ii) Each stage quarters the size: 64 ÷ 43 = 1 cm2 (1 mark). (iii) 27 × 1 = 27 cm2 (1 mark). (iv) Areas: 64, 48, 36, 27, 20.25, 15.1875, … first below 20 at Stage 5 (1 mark).

Try one yourself

For the Sierpiński triangle, how many black triangles are there at Stage 7, and what fraction of the original area is still black?

Show answer

37 = 2187 triangles; black area = (34)7 = 218716384 of the original (about 0.13).

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