Similar Shapes: Length, Area and Volume
Similar shapes have the same shape but not necessarily the same size. Corresponding angles are equal and corresponding lengths are in a constant ratio. At Higher tier, that linear scale factor connects directly to area and volume scale factors.
Higher — area and volume scale factors. The opening length recap is useful at both tiers.
Linear scale factor
If two shapes are similar, every corresponding length is multiplied by the same linear scale factor k.
Worked example: corresponding sides are 6 cm and 15 cm. The scale factor from the smaller shape to the larger is 15÷6 = 2.5. A corresponding 8 cm side therefore becomes 8×2.5 = 20 cm.
Area scale factor
When every length is multiplied by k, area is multiplied by k². This is because area contains two length dimensions.
Worked example: two similar shapes have linear scale factor 3. If the smaller area is 14 cm², the larger area is 14×3² = 14×9 = 126 cm².
In reverse, if the area factor is known, take its positive square root to recover the linear factor. An area factor of 49 gives linear factor √49 = 7.
Volume scale factor
For similar solids, a linear scale factor k produces a volume scale factor of k³.
Worked example: similar solids have linear scale factor 2.5. Their volume factor is 2.5³ = 15.625. If the smaller volume is 32 cm³, the larger is 32×15.625 = 500 cm³.
In reverse, take the positive cube root. A volume factor of 125 gives linear factor ∛125 = 5.
Finding missing areas and volumes
Keep the direction of the scale factor consistent. If k describes small → large, then multiply by k, k² or k³. To travel large → small, divide by the corresponding factor.
Worked example: two similar solids have corresponding lengths 4 cm and 10 cm, so k = 2.5 from small to large. The larger volume is 750 cm³. The smaller volume is 750÷2.5³ = 750÷15.625 = 48 cm³.
Frustums and similar solids
When a cone or pyramid is cut parallel to its base, the removed smaller solid is similar to the original. Corresponding lengths can therefore establish k; then k³ connects their volumes and k² connects corresponding surface areas.
A common strategy is: identify the similar full solids, find the linear factor, calculate the missing volume, then subtract to obtain the frustum.
Worked example — find a frustum volume
A cone has height 15 cm and base radius 6 cm. A smaller cone of height 5 cm is cut off parallel to the base. The small-to-large length factor is 5/15=⅓, so the removed cone has radius 2 cm. Full volume=⅓π×6²×15=180π cm³. Removed volume=(⅓)³×180π=20π/3 cm³. Frustum volume=180π−20π/3=520π/3 cm³. Use corresponding full-cone heights, not the frustum height of 10 cm, to find the scale factor.
Think Like a Mathematician
Before calculating, write a three-line scale-factor chain: length k, area k², volume k³. Then mark the direction of travel. This avoids applying the right factor in the wrong direction.
Common Mistakes
Using k for everything: lengths use k, areas use k² and volumes use k³.
Reversing the ratio: decide whether your scale factor is small → large or large → small before using it.
Taking the wrong root: square-root an area factor; cube-root a volume factor.
Practice — similar shapes
1. Similar shapes have corresponding sides 5 cm and 12 cm. Find the small-to-large linear scale factor.
k = 12÷5 = 2.4.
2. The linear scale factor is 4. What is the area scale factor?
4² = 16.
3. The linear scale factor is 3. What is the volume scale factor?
3³ = 27.
4. Similar shapes have area factor 64. Find the linear factor.
√64 = 8.
5. Similar solids have volume factor 216. Find the linear factor.
∛216 = 6.
6. The smaller of two similar shapes has area 18 cm². The linear scale factor to the larger is 2.5. Find the larger area.
Area factor = 2.5² = 6.25. Area = 18×6.25 = 112.5 cm².
7. The larger of two similar solids has volume 686 cm³. The linear factor from smaller to larger is 3.5. Find the smaller volume.
Volume factor = 3.5³ = 42.875. Smaller volume = 686÷42.875 = 16 cm³.
8. A small cone and a large similar cone have heights 6 cm and 15 cm. The small cone volume is 32π cm³. Find the large cone volume.
k=15÷6=2.5. Volume factor=2.5³=15.625. Large volume=32π×15.625=500π cm³.
Frequently Asked Questions
Why does area use k²?
Area has two length dimensions, so scaling both by k multiplies the area by k×k = k².
Why does volume use k³?
Volume has three length dimensions, so scaling each by k multiplies the volume by k×k×k = k³.
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