The idea, step by step
Count dimensions
Area multiplies two lengths. Scaling each by k therefore scales area by k². A volume multiplies three lengths, so similar solids have volume factor k³. Surface area, like any area, has factor k².
The objects must be similar. If only the height of a box changes, its volume may change with height, but there is no single similarity factor for the whole object.
| Length factor k | Area factor k² | Volume factor k³ |
|---|---|---|
| 2 | 4 | 8 |
| 3 | 9 | 27 |
| 1.5 | 2.25 | 3.375 |
Recover a length factor
An area ratio of 9:25 gives a length ratio of 3:5. A volume ratio of 8:27 gives a length ratio of 2:3. Use positive roots because lengths are positive.
Keep the direction consistent: smaller to larger uses the reciprocal of larger to smaller. Write the ratio in words before calculating if the direction is unclear.
A common trap
Using the length factor directly on a volume.
Instead: Similar volumes use k³, not k.
See the thinking
Enlarging a solid
A cube has surface area 24 cm² and volume 8 cm³. Every edge is enlarged by a factor of 3. Find the new surface area and volume.
- Surface area factor: 3²=9. New surface area=24×9=216 cm².
- Volume factor: 3³=27. New volume=8×27=216 cm³.
- Check: the original edge is 2 cm and the new edge is 6 cm, giving 6×6²=216 cm² and 6³=216 cm³.
Recovering scale
Similar containers hold 64 L and 125 L. A 12 cm length on the smaller matches what larger length?
- Volumes are 4³:5³, so lengths are 4:5.
- 12×5/4=15 cm.
Check your understanding
Similar figures have area ratio 4:49. Find the length ratio.
Show answer
2:7
Take positive square roots.
Your turn
6 questions · 13 marks · Work at your own pace.
Read the questions online
- Similar solids have length factor 2. Give area and volume factors.2 marks
- A similar enlargement has k=1.5 and original area 24 cm². Find new area.2 marks
- A solid of volume 40 cm³ is similarly enlarged with k=3. Find volume.2 marks
- Similar areas are 18 and 50 cm². Find the length ratio.2 marks
- Similar volumes are 27 and 216 cm³. Find the small-to-large length factor.2 marks
- A similar solid doubles its surface area. Give its exact volume factor.3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Similar solids have length factor 2. Give area and volume factors.2 marks
2²=4; 2³=8.
4 and 8 - A similar enlargement has k=1.5 and original area 24 cm². Find new area.2 marks
24×1.5²=54.
54 cm² - A solid of volume 40 cm³ is similarly enlarged with k=3. Find volume.2 marks
40×27=1080.
1080 cm³ - Similar areas are 18 and 50 cm². Find the length ratio.2 marks
18:50=9:25.
Square roots give 3:5.
3:5 - Similar volumes are 27 and 216 cm³. Find the small-to-large length factor.2 marks
216/27=8; cube root 8=2.
2 - A similar solid doubles its surface area. Give its exact volume factor.3 marks
Length factor .
Volume factor .
Curriculum reference & lesson scope
Australian Curriculum v9 and SA Curriculum
Surface area of composite solids; volume of composite solids; proportion/scaling of objects; related length/area/volume factors; measurement error; the effect of rounding and repeated approximations; practical design models; evaluating assumptions and suitable precision.
Official curriculum source ↗