Similar Figures — AQA GCSE Mathematics

SPEC G19 (H), R12 · GEOMETRY & MEASURES

⬣ HIGHER TIER ONLY ⬣

Area & Volume Scale Factors

The k, k², k³ rule — how areas and volumes grow when you enlarge a similar figure

What you will learn

By the end of this lesson you will:

Recall the length scale factor k for similar figures (from Lesson 4.07)

State and apply the rule: area SF = k 2 , volume SF = k 3

Work backwards — find the length SF from an area SF (square root) or volume SF (cube root)

Solve multi-step problems combining length, area and volume — the R12 ratio-notation link

Avoid the most common exam trap: applying a linear SF directly to an area or volume

Why this matters

Recent examiner reports flag problems involving ratio, volume and surface area as topics where students struggle. The single most common error is using a length scale factor where a squared or cubed factor is needed. Lock this in and you've secured a chunk of high-value Higher marks.

Quick Recap: Length Scale Factor

From Lesson 4.07: two figures are similar when all corresponding angles are equal and all corresponding sides are in the same ratio.

LENGTH SCALE FACTOR

k = new length ÷ original length. Every pair of corresponding lengths is related by the same k.

The question this lesson answers

If two similar shapes have length SF k, what happens to their areas and their volumes? Each quantity scales by a different factor — and getting the factor wrong wastes every mark on the question.

The k , k ², k ³ Rule

If the length scale factor between two similar figures is k, then:

length SF = k → area SF = k2 → volume SF = k3

LENGTH

× k

1-dimensional

AREA

× k2

2-dimensional

VOLUME

× k3

3-dimensional

Intuitively

The exponent matches the dimension. Length has 1 dimension (k1 = k). Area has 2. Volume has 3. The next two slides show why.

Why k ² for Area?

Consider a 1 × 1 square next to a 2 × 2 square. Their length SF is k = 2. How many copies of the small square fit inside the big one?

THE RULE EXPLAINED

Area = length × length. Scaling the length by k scales each factor by k, so area is multiplied by k × k = k2. Double the length → quadruple the area. Triple the length → 9× the area.

Why k ³ for Volume?

Now move to three dimensions. A 1 × 1 × 1 cube next to a 2 × 2 × 2 cube. How many copies of the unit cube fit inside the big cube?

THE RULE EXPLAINED

Volume = length × length × length. Scaling length by k scales each factor by k, giving k × k × k = k3. Double the length → 8× the volume. Triple the length → 27× the volume.

Forward Direction: Length → Area or Volume

When the question gives you a length SF and asks for an area or volume in the other shape:

GivenTo findOperationQuick example ( k = 4)

Length SF kArea SFSquare k : k 24 2 = 16 → area is 16× bigger

Length SF kVolume SFCube k : k 34 3 = 64 → volume is 64× bigger

Mini-example

Two similar triangles have length SF k = 5. The small one has area 12 cm2. Large area = 12 × 52 = 12 × 25 = 300 cm 2.

Watch the exponent

Squaring and cubing k is where marks are won or lost. k = 3 → area 9× (not 3×); volume 27× (not 3×). Small mistakes here sink the whole question.

Worked Example 1

01Volume from Length 3 marks

Two similar cylinders have radii of 2 cm and 6 cm. The volume of the smaller cylinder is 50 cm 3. Find the volume of the larger cylinder.

M

Use the radii to find the length scale factor, then cube it to get the volume scale factor, then multiply.

C

Length SF: k = 6 ÷ 2 = 3. Volume SF: k3 = 33 = 27. Larger volume = 50 × 27 = 1350 cm 3.

E

Volume of the larger cylinder = 1350 cm 3 . Sanity check: the large cylinder has 3× every length, so its volume should be 27× bigger — 50 × 27 = 1350 ✓.

Reverse Direction: Area → Length

When the question tells you the area SF (or gives you two corresponding areas) and you need the length SF, square-root it.

length SF = (area SF)

Area SFOperationLength SF

49497

10010010

2.252.251.5

16 / 9(16/9)4 / 3

Why square root?

Because area SF = k2. To undo the square and recover k, you take the square root. If area SF = 49 then k = 49 = 7, and every length is 7× the original.

Reverse Direction: Volume → Length

When the question gives you the volume SF (or two corresponding volumes), take the cube root to get the length SF.

length SF = ∛(volume SF)

Volume SFOperationLength SF

8∛82

27∛273

125∛1255

125 / 8∛(125/8) = ∛125 / ∛85 / 2

Why cube root?

Because volume SF = k3. To undo the cube and recover k, take the cube root. Calculator button: ∛ — on most calculators it's SHIFT + x3 or SHIFT + power button.

Worked Example 2

02The Chain: Volume → Length → Area 4 marks

Two similar water bottles have volumes of 250 cm 3 and 2000 cm 3. The smaller bottle has a surface area of 200 cm 2. Find the surface area of the larger bottle.

M

The question gives volumes and asks for area. There's no direct volume → area rule — you route via length: volume SF → cube root → length SF → square → area SF → multiply.

C

Step 1: Volume SF = 2000 ÷ 250 = 8. Step 2: Length SF = ∛8 = 2. Step 3: Area SF = 22 = 4. Step 4: Larger SA = 200 × 4 = 800 cm 2.

E

Surface area of the larger bottle = 800 cm 2 . A common wrong answer here is 200 × 8 = 1600 (linear SF to area) or 200 × 2 = 400 (forgetting to square). Neither gives 800.

The Full Operation Map

Any similar-figures problem reduces to finding k first. From k, you can jump to any quantity.

You're givenGet length SF k by...Then for area, ×Then for volume, ×

A pair of lengthsDirect: k = new/originalk 2k 3

A pair of areas(area ratio)area ratio directly( (area ratio)) 3

A pair of volumes∛(volume ratio)(∛(vol ratio)) 2volume ratio directly

Master move

Whatever the question, aim to get k first. Once you have k, every other quantity is one arithmetic step away — square for area, cube for volume.

R12 ratio notation link

Spec point R12 asks you to compare lengths, areas and volumes in ratio form. A length ratio of 2 : 3 gives an area ratio of 4 : 9 and a volume ratio of 8 : 27. Same rule, just written with colons.

Examiner Alert: The Classic Error

⚠️ EXAMINER ALERT — linear SF applied to area or volume.This is the single most common error on similarity questions at Higher Tier. Students multiply an area by k or a volume by k2, losing every method mark.

Wrong answerWhy it's wrongCorrect

Area × kUsed linear SF directly on areaArea × k 2

Volume × kUsed linear SF directly on volumeVolume × k 3

Volume × k 2Used area SF instead of volume SFVolume × k 3

Length = volume SFForgot to cube-rootLength = ∛(volume SF)

Anti-trap habit

Before writing an answer, ask: am I changing dimensions? If I'm going from length to area, I need k2. From length to volume, k3. From volume to length, cube root. One check saves three marks.

Application: Area from Length

Quick Application 3 marks

A small flag is similar to a large flag. The small flag has height 6 cm and area 48 cm 2. The large flag has height 18 cm. Find the area of the large flag.

M

Length SF from the pair of heights, then square for area SF, then multiply.

C

k = 18 ÷ 6 = 3. Area SF = 32 = 9. Large area = 48 × 9 = 432 cm 2.

E

Large flag area = 432 cm 2 . Verify: small flag is 6 × 8 = 48 cm2 ✓. Large is 18 × 24 = 432 cm2 ✓.

Worked Example 3

Length Ratio from Volume Ratio 2 marks03

Two similar shapes have volumes in the ratio 125 : 8. Find the ratio of their corresponding lengths.

M

Volume ratio = k3 ratio. To recover the length ratio, take the cube root of each side of the ratio.

C

Length ratio = ∛125 : ∛8 = 5 : 2.

E

Length ratio = 5 : 2. Check by cubing: 53 : 23 = 125 : 8 ✓.

In ratio form

The same rule works symmetrically across ratios: lengths a : b → areas a 2 : b 2 → volumes a 3 : b 3. Going backwards — for area-to-length, ∛ for volume-to-length — undoes the powers term by term.

Common Errors

Students often writeCorrect versionWhy it matters

k = 3, so area SF = 3, volume SF = 3k 2 = 9 for area; k 3 = 27 for volumeZero marks — dimensions ignored

Volume ratio 8:27 → length ratio 8:27Length ratio = ∛8 : ∛27 = 2 : 3Forgot to cube-root

Area ratio 4:9 → length ratio 2:3 → volume ratio 2:3Length ratio 2:3 → volume ratio 8:27Used length ratio as volume ratio

∛125 = 12.5∛125 = 5Mixed up cube root with dividing by 10

Large volume = small volume × kLarge volume = small volume × k 3Applied linear SF to volume

⚠️ Calculator check. If you type 125 into your calculator and press the cube-root button and get 12.5, you hit a wrong button (probably ÷ 10). ∛125 = 5 exactly. The cube root button is usually SHIFT + x 3.

Exam Technique: The SF Chain

Every similar-figures problem at Higher follows the same 4-step structure. Writing the steps out secures method marks even if the final arithmetic slips.

The 4-step chain

Identify given. Two lengths? Two areas? Two volumes? A mix?

Find k (the length SF). Direct / / ∛ as appropriate.

Convert k to the scale factor for the target dimension (k2 for area, k3 for volume).

Apply to the given quantity and write the final answer with units.

Units reminder

Lengths: cm, m, mm. Areas: cm2, m2. Volumes: cm3, m3. Finishing with the wrong units — or forgetting them — costs the A1 final mark.

SIGNATURE HIGHER QUESTION

AQA loves to mix dimensions — give you volumes, ask for an area or vice versa. The problem is harder only because you must route through length. Once you spot that, the method is always: ∛ or to get k, then square or cube to get the target.

Summary

The one rule to memorise

length SF = k → area SF = k2 → volume SF = k3

Forward direction

From length k, square for area (k2), cube for volume (k3). Every length-to-area or length-to-volume question.

Reverse direction

From area SF, square-root to get k. From volume SF, cube-root to get k. Then convert to target quantity.

Top exam tip

Always find k (the length SF) first. It's the hub of every similar-figures problem. Once you have k, any other dimension is one power (square or cube) or one root ( or ∛) away.

⚠️ Final reminder: the single biggest Higher-tier trap is applying a linear SF directly to an area or volume. Always check: am I changing dimensions? If yes, square or cube k.