G5, G6, G19Geometry & MeasuresFoundation & HigherGrades 4–8

Similarity & Congruence

Congruent shapes are identical in shape and size. Similar shapes have the same shape but may be different sizes — one is an enlargement of the other. Similarity gives a powerful method for finding unknown lengths, and at Higher tier it extends to areas and volumes, where the scale factor is squared or cubed.

45 min Video by Maths Genie AQA GCSE Maths

The big picture

In similar shapes, corresponding angles are equal and every pair of corresponding lengths is in the same ratio. That ratio is the linear scale factor $k$ , and one pair of matching sides is enough to find it.

Area is two lengths multiplied, so it scales by $k^2$ . Volume is three lengths multiplied, so it scales by $k^3$ . A model car at $\frac{1}{20}$ scale has $\frac{1}{400}$ of the paint area and $\frac{1}{8000}$ of the volume.

Similar ShapesCurated by Kepler, not produced by us. Watch it, then read the notes below.

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What you'll learn

Identify congruent triangles using SSS, SAS, ASA and RHS
Identify similar shapes and find the scale factor
Find missing lengths in similar shapes, including overlapping triangles
Use area and volume scale factors (Higher)

Congruent triangles

Two triangles are congruent if any one of these conditions holds: $\textbf{SSS}$ (three sides equal), $\textbf{SAS}$ (two sides and the included angle), $\textbf{ASA}$ (two angles and a corresponding side, sometimes written AAS), or $\textbf{RHS}$ (right angle, hypotenuse and one other side).

AAA is $\textbf{not}$ a congruence condition — three equal angles only guarantee similar triangles.

Example 1Triangle

ABC

has

AB = 7

BC = 9

and angle

B = 50^\circ

. Triangle

PQR

has

PQ = 7

QR = 9

and angle

Q = 50^\circ

. Are they congruent?

1Two sides and the angle between them match.

Answer

Yes, by SAS.

Tip — In a congruence proof, state each pair of equal sides or angles with a reason, then name the condition at the end.

Similar shapes and missing lengths

Find the linear scale factor from one pair of corresponding sides: $k = \frac{\text{length in larger}}{\text{matching length in smaller}}$ . Multiply to go from small to large; divide to go back.

Match sides carefully by their position relative to equal angles — redraw the shapes the same way round if necessary.

The triangles are similar.

k = \frac{15}{6} = 2.5

, so

x = 4 \times 2.5 = 10

cm.

Example 1Two similar rectangles: the smaller is 8 cm by 5 cm; the larger has length 20 cm. Find its width.

k = \frac{20}{8} = 2.5

2Width

= 5 \times 2.5 = 12.5

Answer

12.5

Your turnSame method, new numbers — try it before moving on.

Two similar triangles have corresponding sides of 9 cm and 6 cm. Another side of the larger triangle is 13.5 cm. Find the matching side of the smaller triangle, in cm.

Your answer

Show hint

Going from large to small, divide by the scale factor.

Show worked solution

Solution

1 $k = \frac{9}{6} = 1.5$ .
2 $13.5 \div 1.5 = 9$ cm.

Answer: $9$ cm

Example 2In triangle

ADE

BC

is parallel to

DE

, with

B

AD

and

C

AE

AB = 4

BD = 6

BC = 5

. Find

DE

1Triangles

ABC

and

ADE

are similar (parallel lines give equal corresponding angles).

AD = 4 + 6 = 10

, so

k = \frac{10}{4} = 2.5

DE = 5 \times 2.5 = 12.5

Answer

DE = 12.5

Your turnSame method, new numbers — try it before moving on.

In triangle $PST$ , $QR$ is parallel to $ST$ with $Q$ on $PS$ and $R$ on $PT$ . $PQ = 3$ cm, $QS = 9$ cm and $ST = 16$ cm. Find $QR$ , in cm.

Your answer

Show hint

Use the whole side $PS = 12$ , not $QS$ .

Show worked solution

Solution

1 $PS = 3 + 9 = 12$ , so the scale factor from $PQR$ to $PST$ is $\frac{12}{3} = 4$ .
2 $QR = 16 \div 4 = 4$ cm.

Answer: $4$ cm

Separate the triangles

In overlapping triangles, the scale factor uses whole sides from the shared vertex. Using $BD = 6$ instead of $AD = 10$ is the most common error, and it gives an impossible answer — so always sketch the two triangles separately.

Area and volume scale factors (Higher)

If the linear scale factor is $k$ : area scale factor $= k^2$ and volume scale factor $= k^3$ .

To go backwards, take the square root (from areas) or cube root (from volumes) to find $k$ first.

Example 1Two similar cylinders have heights 6 cm and 9 cm. The smaller has volume 80 cm³. Find the volume of the larger.

k = \frac{9}{6} = 1.5

2Volume scale factor

= 1.5^3 = 3.375

80 \times 3.375 = 270

Answer

270

cm³

Your turnSame method, new numbers — try it before moving on.

Two similar shapes have areas 20 cm² and 180 cm². The smaller has a side of 4 cm. Find the matching side of the larger, in cm.

Your answer

Show hint

Area scale factor is 9, so the length scale factor is $\sqrt{9}$ .

Show worked solution

Solution

1Area scale factor $= 180 \div 20 = 9$ .
2 $k = \sqrt{9} = 3$ , so the side is $4 \times 3 = 12$ cm.

Answer: $12$ cm

Example 2Two similar bottles have volumes 250 ml and 2000 ml. The smaller has a label of area 60 cm². Find the label area on the larger bottle.

1Volume scale factor

= 8

, so

k = \sqrt[3]{8} = 2

2Area scale factor

= 2^2 = 4

60 \times 4 = 240

Answer

240

cm²

Your turnSame method, new numbers — try it before moving on.

Two similar solids have surface areas 50 cm² and 200 cm². The smaller has volume 30 cm³. Find the volume of the larger, in cm³.

Your answer

Show worked solution

Solution

1Area factor $4$ , so $k = 2$ .
2Volume factor $2^3 = 8$ : $30 \times 8 = 240$ cm³.

Answer: $240$ cm³

Think like an examiner

Write the scale factor as a separate line — it is usually worth a method mark.
In congruence proofs, give a reason for each equal pair ("common side", "vertically opposite angles") and name the condition.
Check which quantity you are given — length, area or volume — before choosing $k$ , $k^2$ or $k^3$ .

Scale factors

k = \frac{\text{new length}}{\text{original length}}

Linear scale factor.

\text{area} \times k^2

Areas scale by the square.

\text{volume} \times k^3

Volumes scale by the cube.

Watch out for these

Using AAA to prove two triangles congruent.

AAA shows similarity only; congruence needs a side.

Doubling lengths doubles the area.

Area scale factor is

2^2 = 4

In overlapping triangles, using the partial side

BD

Use the whole side

AD

from the shared vertex.

From an area ratio of 9, using

k = 9

for lengths.

k = \sqrt{9} = 3

Stretch yourself

A cone of height 12 cm is filled with water to a depth of 8 cm (measured from the point, which faces downwards). What fraction of the cone's volume is filled?

Hint — The water forms a smaller cone similar to the whole cone.

Reveal solution

SolutionThe water is a similar cone with

k = \frac{8}{12} = \frac{2}{3}

. Volume scale factor

= \left(\frac{2}{3}\right)^3 = \frac{8}{27}

. So

\frac{8}{27}

of the cone is filled — less than a third, even though the water is two-thirds of the way up.

\blacksquare

Questions students ask

Are all circles similar? All squares?

Yes — every circle is similar to every other circle, and likewise every square. Rectangles, however, are not all similar.

What is the difference between ASA and AAS?

Both use two angles and a side. If two angles match, the third must too, so any corresponding side works — AQA accepts both.

Do I need to show triangles are similar before using scale factors?

If the question states they are similar, no. Otherwise, give a reason, such as equal angles from parallel lines.

Key takeaways

Congruent: SSS, SAS, ASA, RHS. Similar: equal angles, sides in proportion.
Linear scale factor from matching sides: $k = \frac{\text{new}}{\text{original}}$ .
Overlapping triangles: separate them and use whole sides.
Area scale factor $k^2$ ; volume scale factor $k^3$ .
Work back to $k$ with square or cube roots.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 42 marks

Triangles $ABC$ and $DEF$ are similar. $AB = 6$ cm, $BC = 8$ cm, $DE = 9$ cm.

Work out the length of $EF$ .[2]
Show mark scheme
- M1Scale factor $\frac{9}{6} = 1.5$
- A1 $EF = 12$ cm
Answer: $12$ cm

Question 2Grade 63 marks

$ABCD$ is a parallelogram. The diagonal $AC$ divides it into two triangles.

Prove that triangle $ABC$ is congruent to triangle $CDA$ .[3]
Show mark scheme
- B1 $AB = CD$ and $BC = DA$ (opposite sides of a parallelogram are equal)
- B1 $AC$ is a common side
- B1Congruent by SSS

Question 3Grade 83 marks

Two similar solid statues are made from the same metal. The smaller is 15 cm tall and has mass 1.2 kg. The larger has mass 9.375 kg.

Work out the height of the larger statue.[3]
Show mark scheme
- M1Mass (volume) scale factor $= \frac{9.375}{1.2} = 7.8125$
- M1Linear scale factor $= \sqrt[3]{7.8125} = 1.984\ldots$ (or $\sqrt[3]{\frac{125}{16}}$ )
- A1 $29.8$ cm (29.76…)
Answer: $29.8$ cm

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Which is NOT a condition for congruent triangles?
- AAA
- SSS
- SAS
- RHS
Show answer and explanation
Answer: AAA
AAA only gives similarity.
A shape is enlarged by scale factor 3. A 4 cm side becomes…
- 12 cm
- 7 cm
- 36 cm
- 1.3 cm
Show answer and explanation
Answer: 12 cm
$4 \times 3$ .
Similar triangles have sides 5 cm and 20 cm. The scale factor is…
- $4$
- $15$
- $25$
- $\frac{1}{4}$
Show answer and explanation
Answer: $4$
$20 \div 5$ .
Lengths are multiplied by 2. Area is multiplied by…
- $4$
- $2$
- $8$
- $6$
Show answer and explanation
Answer: $4$
$2^2$ .
Lengths are multiplied by 3. Volume is multiplied by…
- $27$
- $9$
- $3$
- $6$
Show answer and explanation
Answer: $27$
$3^3$ .
Similar shapes have areas 12 cm² and 75 cm². Linear scale factor?
- $2.5$
- $6.25$
- $6.25^2$
- $63$
Show answer and explanation
Answer: $2.5$
$\sqrt{75 \div 12} = \sqrt{6.25}$ .
Similar solids have volumes 16 cm³ and 54 cm³. Area scale factor?
- $2.25$
- $1.5$
- $3.375$
- $38$
Show answer and explanation
Answer: $2.25$
$k = \sqrt[3]{3.375} = 1.5$ ; $k^2 = 2.25$ .
A 1 : 50 model has a door of area 8 cm². The real door area is…
- 2 m²
- 400 cm²
- 20 m²
- 0.2 m²
Show answer and explanation
Answer: 2 m²
$8 \times 2500 = 20{,}000$ cm² $= 2$ m².

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