Volume Factor of Similar Shapes

Use the cube of the length scale factor to find volumes of similar solids, and move between length, area and volume factors.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 35 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Three Scale Factors

Similar solids have the same shape, and every length is multiplied by the same number, $k$ .

Areas grow faster than lengths, and volumes grow faster again.

What you compare	Multiply by
Lengths (height, radius)	$k$
Areas (surface area)	$k^{2}$
Volumes (capacity, mass)	$k^{3}$

Rule

Always find the length factor $k$ first. Then square it for area, or cube it for volume.

Volume From Two Lengths

The 2-mark question gives two matching lengths and one volume. Two similar jars are 6 cm and 9 cm tall. The small one holds 160 cm³.

Worked example

1
$k = \frac{9}{6} = 1.5$ — The two heights are lengths, so they give the length factor $k$ : big height over small height. 9 ÷ 6 = 1.5.
2
$k^{3} = 1.5^{3} = 3.375$ — Volume grows by $k$ cubed, because it grows in three directions. 1.5 × 1.5 × 1.5 = 3.375.
3
$160 \times 3.375 = 540 cm^{3}$ — Multiply the small jar's volume by the volume factor to get the big jar's volume.

Common mistake

Multiplying 160 by 1.5. That only works for a length. A volume uses $k^{3}$ .

From Areas to Volumes

The 3-mark question gives two surface areas and one volume. Go back to the length factor first.

Surface areas 75 cm² and 108 cm². The small solid has volume 250 cm³.

Worked example

1
$\frac{108}{75} = 1.44$ — Big surface area over small surface area gives the area factor. 108 ÷ 75 = 1.44.
2
$k = 1.44 = 1.2$ — The area factor is $k^{2}$ , so undo the square with a square root to get the length factor. 1.2 × 1.2 = 1.44.
3
$k^{3} = 1.2^{3} = 1.728$ — Now we need volume, which uses $k^{3}$ . 1.2 × 1.2 × 1.2 = 1.728.
4
$250 \times 1.728 = 432 cm^{3}$ — Multiply the small solid's volume by the volume factor to get the big solid's volume.

Common mistake

Multiplying the volume by the area factor 1.44. Areas and volumes never use the same factor.

Key idea

The other way round: volumes 128 and 250 give $\frac{250}{128} = 1.953125$ . Cube root gives 1.25, then square it: area factor 1.5625.

Volumes to a Length

The 3-mark question often gives two volumes and one height, and asks for the other height. The cube works backwards here, so you use a cube root.

Formula

\frac{big length}{small length} = k = 3 \frac{big volume}{small volume}

Here $k$ is the length factor. Work out the volume factor first, then take its cube root, then multiply the small height by $k$ .

Worked example

1
$\frac{250}{54} = \frac{125}{27}$ — Big volume over small volume gives the volume factor. Divide the top and bottom by 2 so the cube roots come out exactly.
2
$k = 3 \frac{125}{27} = \frac{5}{3}$ — The volume factor is $k^{3}$ , so undo the cube with a cube root: $3125 = 5$ and $327 = 3$ .
3
$6 \times \frac{5}{3} = 10 cm$ — A height is a length, so multiply it by $k$ , not by the volume factor. The small jar is 6 cm tall.
4
$(\frac{10}{6})^{3} = \frac{1000}{216} = \frac{125}{27}$ — Check: cube the ratio of the heights and you should get the volume factor back. It does.

Common mistake

Multiplying the height by the volume factor. Only volumes use the cube. A height uses $k$ .

Mass works the same way, because the objects are made of the same material.

Worked example

1
$k = \frac{20}{8} = 2.5$ — A model is 8 cm tall and has mass 400 g. The statue is 20 cm tall. The heights give $k$ .
2
$k^{3} = 2. 5^{3} = 15.625$ — Same material, so mass grows with volume, which uses $k^{3}$ . 2.5 × 2.5 × 2.5 = 15.625.
3
$400 \times 15.625 = 6250 g$ — Multiply the model's mass by the volume factor.
4
$6250 g = 6.25 kg$ — Divide by 1000 to change grams into kilograms.

Key idea

Ratios follow the same rules. Lengths in the ratio 2 : 3 give areas 4 : 9 and volumes 8 : 27. Going back, volumes 8 : 27 give lengths 2 : 3. Show every line: the volume factor and the cube root each earn their own mark.

When You Know the Difference

The 4-mark question gives the difference between the two volumes, not either volume. Call the small volume $V$ .

Heights 8 cm and 12 cm. The big volume is 1330 cm³ more than the small one.

Worked example

1
$k = \frac{12}{8} = 1.5$ — The heights give the length factor: big over small. 12 ÷ 8 = 1.5.
2
$k^{3} = 1.5^{3} = 3.375$ — Volume uses $k^{3}$ . 1.5 × 1.5 × 1.5 = 3.375, so the big volume is 3.375 lots of $V$ .
3
$3.375 V - V = 1330$ — The big volume minus the small volume is the difference we were told.
4
$2.375 V = 1330$ — 3.375 lots of $V$ take away 1 lot of $V$ leaves 2.375 lots.
5
$V = 1330 \div 2.375 = 560 cm^{3}$ — Divide both sides by 2.375 to get the small volume.
6
$560 \times 3.375 = 1890 cm^{3}$ — Check: the big volume is the small volume times the volume factor.
7
$1890 - 560 = 1330$ — The difference matches the question, so 560 is right.

Common mistake

Dividing 1330 by 3.375. The difference is 2.375 lots of the small volume, not 3.375.

Percentages and Chains

The 4-mark question can give a percentage. "72.8% more volume" means the volume factor is 1.728.

Worked example

1
$100% + 72.8% = 172.8% = 1.728$ — “72.8% more” means the original 100% plus 72.8% extra. Divide by 100 to turn the percentage into the volume factor.
2
$k = 3 1.728 = 1.2$ — The volume factor is $k^{3}$ , so undo the cube with a cube root. 1.2 × 1.2 × 1.2 = 1.728.
3
$k^{2} = 1.2^{2} = 1.44$ — Surface area uses $k^{2}$ . 1.2 × 1.2 = 1.44.
4
$1.44 = 144%$ — Multiply by 100 to turn the area factor back into a percentage.
5
$144% - 100% = 44% more$ — Take away the original 100% to see how much extra there is. The surface area is 44% more.

Common mistake

Saying the area is also 72.8% more. Each quantity has its own factor.

Key idea

Three solids in a chain? Find $k$ for each pair, then multiply the two length factors.

Which Method? Read the Question

Look at what you are given: lengths, areas or volumes.

You see	You do
Two lengths	cube the ratio
Two areas	square root, then cube
Two volumes, find an area	cube root, then square
Two volumes, find a length	cube root of the volume factor
A sum or difference	call the small one V
A percentage more or less	turn it into a factor first
A cone cut off the top	Frustum of a Cone
Areas only	Area Factor of Similar Shapes

Key idea

Mass and capacity behave like volume: they also use $k^{3}$ .

Practise volume factor of similar shapes now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on volume factor of similar shapes

Each with a full worked solution.

All 35 questions on this topic →

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.