G19: Similar Shapes
Compare the areas and volumes of similar shapes; apply relationships between lengths, areas and volumes in similar figures
📋 Key Concepts
Similar shapes have the same shape but different sizes. All corresponding angles are equal, and corresponding sides are in proportion.
Scale factor: The ratio of corresponding lengths in similar shapes.
📝 Length Scale Factor
Length scale factor (k): The ratio of corresponding lengths.
k = (length in larger shape)/(length in smaller shape)
Example 1
Two similar triangles have corresponding sides 4 cm and 12 cm. Find the scale factor.
Solution:
Scale factor k = 12/4 = 3
The larger triangle is 3 times bigger than the smaller one.
📝 Area Scale Factor
Important: If the length scale factor is k, the area scale factor is k².
Area scale factor = k²
Example 2
Two similar shapes have length ratio 2:5. Find the ratio of their areas.
Solution:
Length scale factor k = 5/2 = 2.5
Area scale factor = k² = 2.5² = 6.25
Area ratio = 1 : 6.25 or 4 : 25
Example 3
A small shape has area 8 cm². A similar larger shape has lengths 3 times bigger. Find the larger area.
Solution:
Area scale factor = 3² = 9
Larger area = 8 × 9 = 72 cm²
📝 Volume Scale Factor
Important: If the length scale factor is k, the volume scale factor is k³.
Volume scale factor = k³
Example 4
Two similar solids have corresponding heights 3 cm and 9 cm. The smaller volume is 20 cm³. Find the larger volume.
Solution:
Length scale factor k = 9/3 = 3
Volume scale factor = 3³ = 27
Larger volume = 20 × 27 = 540 cm³
Example 5
Two similar shapes have volumes in ratio 8:125. Find the ratio of their lengths.
Solution:
Volume ratio = 8 : 125
Length ratio = ∛8 : ∛125 = 2 : 5
Length scale factor = 5/2 = 2.5
📝 Summary Table
| Scale Factor | Length | Area | Volume |
|---|---|---|---|
| k | × k | × k² | × k³ |
Working backwards:
- Area → length: take the square root
- Volume → length: take the cube root
- Volume → area: take the cube root, then square
📝 Finding Scale Factors
Example 6
Two similar shapes have areas 12 cm² and 48 cm². Find the length scale factor.
Solution:
Area scale factor = 48/12 = 4
Length scale factor = √4 = 2
Example 7
Two similar solids have volumes 27 cm³ and 216 cm³. Find the ratio of their surface areas.
Solution:
Volume scale factor = 216/27 = 8
Length scale factor = ∛8 = 2
Area scale factor = 2² = 4
Surface area ratio = 1 : 4
🧮 Worked Examples — Exam Style
Similar shapes: lengths scale by k, areas by k², volumes by k³. Identify WHICH type of quantity you are scaling before touching a number.
Two similar triangles have lengths ratio 2 : 5. The small triangle has area 8 cm². Find the large area.
Length k = 5/2. Area factor k² = 25/4
Area = 8 × 25/4 = 50 cm²
M1 for k² = 25/4, A1 for 50. Squaring the FRACTION (not the decimal) keeps it exact.
Two similar bottles hold 500 ml and 2000 ml. The small one is 15 cm tall. Find the height of the large one.
Volume ratio = 500 : 2000 = 1 : 4 → length k = ∛4 ≈ 1.587
Height = 15 × 1.587 = 23.8 cm (1 d.p.)
M1 for cube-rooting 4, A1 for 23.8. Cube roots of awkward numbers need the calculator — this is Paper 2/3 material.
Exam wording to expect: "The shapes are mathematically similar", "Find the volume/area of …", or reverse (find a length from volumes).
❓ Practice Questions
Q1: Two similar shapes have length ratio 4:7. Find the area ratio.
Q2: A shape has area 15 cm². A similar shape has lengths 4 times bigger. Find the larger area.
Q3: Two similar solids have volume ratio 27:64. Find the length ratio.
Q4: Two similar shapes have areas 9 cm² and 144 cm². Find the length scale factor.
Q5: A solid has volume 40 cm³. A similar solid has heights 0.5 times the original. Find the new volume.
Q6: Two similar cylinders have heights 4 cm and 10 cm. The smaller has volume 50 cm³. Find the larger volume.
Q7: Similar shapes have area ratio 25 : 49. Write the length ratio.
✅ Answers
- 16 : 49
- 15 × 16 = 240 cm²
- 3 : 4 (∛27 : ∛64)
- √(144/9) = √16 = 4
- 40 × 0.5³ = 40 × 0.125 = 5 cm³
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. Area ratio = k² = 4² : 7² = 16 : 49
Q2. k = 4 → k² = 16 → 15 × 16 = 240 cm²
Q3. Length ratio = ∛27 : ∛64 = 3 : 4
Q4. k² = 144/9 = 16 → k = 4
Q5. k = 0.5 → k³ = 0.125 → 40 × 0.125 = 5 cm³
Q6. k = 10/4 = 2.5 → volume factor = 15.625 → 50 × 15.625 = 781.25 cm³
Q7. Length ratio = √25 : √49 = 5 : 7
📌 Key Facts to Memorise
- Lengths: × k · Areas: × k² · Volumes: × k³
- Going backwards: length ratio = square-root the area ratio, cube-root the volume ratio
- k > 1 enlarges; k < 1 shrinks; both preserve shape and angles
- Volume ratio (a : b) → length ratio (∛a : ∛b)
- Area ratio 1 : 16 → volume ratio 1 : 64 (square then cube the length factor 1 : 4)
🗺️ In the Exam
Similar shapes: 3–5 marks on Papers 2/3 (calculator needed for cube roots). Choosing k, k² or k³ correctly IS the whole assessment.
Links to: proportion (R7), area and volume (G16), circles (G17).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Square the length scale factor for areas
- Cube the length scale factor for volumes
- Work backwards by taking roots
- Always check if you need length, area, or volume ratio
- Read carefully: are shapes getting bigger or smaller?
- Units must be consistent before comparing
🧠 Problem-Solving Strategies
Problem-Solving
Similar shapes have equal angles and proportional sides. Linear scale factor = big length / small length. Area scale factor = (linear scale factor) squared. Volume scale factor = (linear scale factor) cubed. To find a length from an area ratio, take the square root. To find a length from a volume ratio, take the cube root.
Multi-Step Problem
Two similar cylinders have heights 4 cm and 10 cm. The smaller has volume 60 cm cubed. Find the volume of the larger cylinder.
Solution: Linear SF = 10/4 = 2.5. Volume SF = 2.5 cubed = 15.625. Larger volume = 60 x 15.625 = 937.5 cm cubed.
⚠️ Common Errors
Watch Out!
1. Wrong: Using the linear scale factor for area: if sides double, area doubles too Correct: If sides double (SF = 2), area multiplies by 2 squared = 4 and volume by 2 cubed = 8.
2. Wrong: Multiplying area by the scale factor instead of squaring the scale factor Correct: Area scales as SF squared. If SF = 3, area multiplies by 9, not 3.
3. Wrong: Confusing similar shapes with congruent shapes Correct: Similar shapes have the same shape but different sizes (SF not 1). Congruent shapes are identical in size and shape (SF = 1).
✍️ 6-Mark Exam Question
Extended Answer
6 marks: Two similar cones have heights 6 cm and 15 cm. The smaller cone has surface area 60 pi cm squared. (a) Find the linear scale factor. (b) Find the surface area of the larger cone. (c) The smaller cone has volume 96 pi cm cubed. Find the volume of the larger cone.
(a) Linear SF = 15/6 = 2.5.
(b) Area SF = 2.5 squared = 6.25. Larger surface area = 60 pi x 6.25 = 375 pi cm squared.
(c) Volume SF = 2.5 cubed = 15.625. Larger volume = 96 pi x 15.625 = 1500 pi cm cubed.
Mark scheme: M1 linear SF, A1 2.5, M1 area SF, A1 375 pi, M1 volume SF, A1 1500 pi
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Two similar rectangles have areas 20 cm squared and 45 cm squared. The smaller has width 4 cm.
(a) Find the width of the larger rectangle.
(b) A student says "The lengths are in the same ratio as the areas." Is this correct?
(c) A third similar rectangle has area 80 cm squared. Find its width.
Answers: (a) Area ratio = 45/20 = 2.25. Linear SF = root 2.25 = 1.5. Larger width = 4 x 1.5 = 6 cm. (b) No — lengths are in the ratio root(area ratio) = 1.5, not the area ratio of 2.25. The student confuses linear and area scaling. (c) Area ratio from small = 80/20 = 4. Linear SF = root 4 = 2. Width = 4 x 2 = 8 cm.
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🎬 Video Resources
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