Related length/area/volume factors | Year 10

01 / LEARN

Understand the idea.

Count dimensions

For similar objects, every corresponding length is multiplied by $k$ . An area contains two length factors, so it scales by $k^{2}$ ; a volume contains three, so it scales by $k^{3}$ . Surface area follows the area law.

These laws require the same factor in every direction. Doubling only a box's height does not make a similar solid, so a volume factor of $2^{3}$ would be unjustified in that situation.

L : k, A : k^{2}, V : k^{3}

Corresponding factors for similar objects
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

To recover a length factor from an area factor, take its positive square root. From a volume factor, take its positive cube root. Positive roots are required because the scale compares physical lengths.

Keep the comparison direction consistent throughout. If the volume ratio is smaller to larger, derive the length factor in that same direction before applying it to the smaller object's measurement. Reversing direction takes the reciprocal.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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.

Worked solution

1
Find the area and volume factors.
$k = 3, k^{2} = 9, k^{3} = 27$
Every edge is multiplied by the same factor, so the cubes are similar.
2
Scale the surface area.
$A_{new} = 24 (9) = 216 {cm}^{2}$
Surface area has two length dimensions.
3
Scale the volume.
$V_{new} = 8 (27) = 216 {cm}^{3}$
Volume has three length dimensions; equal numerical results do not make the units interchangeable.

Answer

Surface area 216 cm²; volume 216 cm³.

EXAMPLE 02

Recovering scale

Similar containers hold 64 L and 125 L. A 12 cm length on the smaller matches what larger length?

Worked solution

1
Find the volume factor from smaller to larger.
$k^{3} = \frac{125}{64}$
The corresponding containers are similar and their capacities compare interior volumes.
2
Recover the length factor.
$k = \sqrt[3]{\frac{125}{64}} = \frac{5}{4}$
Take the positive cube root of the volume factor.
3
Scale the corresponding length.
$L = 12 \cdot \frac{5}{4} = 15 cm$
All corresponding lengths use $k$ , rather than $k^{3}$ .

Answer

15 cm

03 / YOUR TURN

Try it yourself.

Check your understanding

Similar figures have area ratio 4:49. Find the length ratio.

Show answer & reasoning

2:7

The length ratio is the positive square root of the area ratio: $\sqrt{4} : \sqrt{49} = 2 : 7$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

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
$2^{2} = 4$ ; $2^{3} = 8$ .
4 and 8
A similar enlargement has $k = 1.5$ and original area 24 cm². Find new area.
2 marks
$24 \times {1.5}^{2} = 54$ .
54 cm²
A solid of volume 40 cm³ is similarly enlarged with $k = 3$ . Find volume.
2 marks
$40 \times 27 = 1080$ .
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
$\frac{216}{27} = 8$ ; $\sqrt[3]{8} = 2$ .
2
A similar solid doubles its surface area. Give its exact volume factor.
3 marks
Length factor $\sqrt{2}$ .
Volume factor ${(\sqrt{2})}^{3} = 2 \sqrt{2}$ .
$2 \sqrt{2}$

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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 ↗

In this lesson

Understand the idea.

Count dimensions

Recover a length factor

See the method in action.

Enlarging a solid

Find the area and volume factors.

Scale the surface area.

Scale the volume.

Recovering scale

Find the volume factor from smaller to larger.

Recover the length factor.

Scale the corresponding length.

Try it yourself.

Check your understanding

Keep practising