If the black region at Stage 0 is 1 square unit, then the black region at Stage 1 is 34 square units. This process is repeated on Stage 1 to arrive at Stage 2. Hence, the black region at Stage 2 will be 34 of the black region of Stage 1, which is equal to 34 × 34 = (34)2. Can you explain why the area of the black region at Stage n will be (34)n?
Step-by-step solution
Idea: The same step (keep 34 of what is black) is repeated at every stage, so the areas form a GP with first term 1 and common ratio 34.
- Going from any stage to the next, each black triangle is split into 4 equal triangles and 1 is removed. So every black triangle keeps 34 of its area, and the whole black region keeps 34 of its area.1 mark
- Stage 0 = 1, Stage 1 = 1 × 34, Stage 2 = (34)2, Stage 3 = (34)3, … After n stages we have multiplied by 34 exactly n times, so the area is (34)n.1 mark
Answer to write in the exam
Each stage: every black triangle → 4 equal parts, 1 removed
∴ Black area at each stage = 34 × black area at previous stage
Stage 0 = 1, Stage 1 = 34, Stage 2 = (34)2, …
∴ Stage n = 1 × 34 × 34 × … (n times) = (34)n
Common mistakes that cost marks
- Subtracting 14 at each stage (1, 34, 12, 14, 0). Only 14 of the current black area is removed, so the area is multiplied by 34.
- Writing (34)n−1. Stage 0 is the starting point, so Stage n has n multiplications.
- Thinking the removed middle triangle is 13 of the area. Joining midpoints makes 4 equal triangles, so it is 14.
How this can come in the exam
A shape of area 1 square unit is cut into 5 equal parts and 1 part is removed; the same is done to every remaining part at each stage. The area left at Stage 2 is
- 1625
- 35
- 425
- 810
Show answer
(A) 1625
45 of the area is kept each stage: (45)2 = 1625.
Try one yourself
At each stage, 25 of a shape’s shaded area is removed. If Stage 0 has area 1, write the area at Stage n and at Stage 3.
Show answer
35 is kept each time, so area = (35)n; Stage 3: 27125.
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