OCR GCSE Mathematics: Similarity and enlargement (J560)

This study guide covers similarity and enlargement for the OCR GCSE (9-1) Mathematics specification (J560), Version 2.1 (August 2026). It follows content references 9.04a, 9.04b and 9.04c in content area 9, Congruence and similarity. Most of the unit is for both tiers. Two parts are Higher tier only, and each is labelled “Higher tier only” where it appears. The specification states that content can appear on calculator and non-calculator papers alike, so every method here is set out so that you can do it by hand.

Use this page with the revision notes and the practice questions. Course pages: the OCR GCSE Mathematics hub, the printable checklist and the free diagnostics. The related unit on congruence is worth reading first, and ratio supports the scale factor work.

What this unit covers

Reference	Skill	Tier
9.04a	Identify similar triangles	Both
9.04a	Prove that two triangles are similar	Both
9.04b	Enlarge a simple shape from a given centre using a whole number scale factor, and identify the scale factor	Both
9.04b	Identify the centre and scale factor (including fractional) of an enlargement, and enlarge by a fractional scale factor	Both
9.04b	Perform and recognise enlargements with negative scale factors	Higher tier only
9.04c	Compare lengths, areas and volumes using ratio notation and scale factors	Both
9.04c	Calculate unknown lengths in similar figures	Both
9.04c	Understand the relationship between lengths, areas and volumes of similar shapes	Higher tier only

9.04a Similar triangles

Two shapes are similar when one is an enlargement of the other. Matching angles are equal, and matching sides all share one ratio, called the scale factor. Similar shapes can be turned or reflected, so check which sides correspond before you divide.

Three tests for triangles, and you only need one of them:

• All three pairs of corresponding sides are in the same ratio.
• Two pairs of corresponding sides are in the same ratio, and the angles between them are equal.
• Two pairs of corresponding angles are equal. The third pair then has to be equal too, because angles in a triangle add to 180°.

Worked example 1: identifying similar triangles

Triangle P has sides 6 cm, 10 cm and 12 cm. Triangle Q has sides 9 cm, 15 cm and 18 cm. Triangle R has sides 9 cm, 14 cm and 18 cm. Which of these are similar to P?

Match shortest to shortest, and so on.
P to Q:  9/6 = 1.5,  15/10 = 1.5,  18/12 = 1.5   equal, so P and Q are similar
P to R:  9/6 = 1.5,  14/10 = 1.4,  18/12 = 1.5   not equal, so R is not similar to P

P and Q are similar, with scale factor 1.5 from P to Q.

Worked example 2: proving triangles similar

In triangle ABC, D lies on AB and E lies on AC. DE is parallel to BC. AD = 6 cm, DB = 3 cm and DE = 8 cm. Prove that triangles ADE and ABC are similar, then find BC.

Angle DAE = angle BAC           (common angle)
Angle ADE = angle ABC           (corresponding angles, DE parallel to BC)
Angle AED = angle ACB           (corresponding angles, DE parallel to BC)
All three pairs of angles are equal, so triangles ADE and ABC are similar.

AB = 6 + 3 = 9 cm
Scale factor from ADE to ABC = AB/AD = 9/6 = 1.5
BC = DE x 1.5 = 8 x 1.5 = 12 cm

A proof needs a reason with every statement. Name the triangles with matching letters, as above: A with A, D with B, E with C.

9.04b Enlargement

An enlargement changes size and keeps shape. It needs a scale factor and a centre of enlargement. Every image point sits on the straight line from the centre through its original point, and its distance from the centre is the original distance times the scale factor. Lengths change by the scale factor, but angles do not change.

Whole number scale factors

1. Work out the horizontal and vertical steps from the centre to a vertex.
2. Multiply both steps by the scale factor.
3. Count those new steps from the centre to find the image vertex.
4. Repeat for each vertex and join them.

Worked example 3: enlarging from a centre

Triangle ABC has A(3, 1), B(5, 1) and C(5, 4). Enlarge it by scale factor 2 with centre (1, 2).

Steps from the centre (1, 2):
A: right 2, down 1   -> right 4, down 2   -> A' = (5, 0)
B: right 4, down 1   -> right 8, down 2   -> B' = (9, 0)
C: right 4, up 2     -> right 8, up 4     -> C' = (9, 6)

Check: AB is 2 units long and A’B’ is 4 units long, so the lengths have doubled.

Identifying the scale factor

Divide a length on the image by the matching length on the original. Here A’B’/AB = 4/2 = 2. If the image is the smaller shape, the scale factor is a fraction less than 1.

Fractional scale factors

Using a fraction smaller than 1 as the scale factor shrinks the shape. The method does not change: multiply the steps from the centre by the fraction.

Worked example 4: a fractional scale factor

Shape P has vertices (4, 2), (10, 2) and (10, 6). Enlarge it by scale factor 1/2 with centre (2, 0).

Steps from (2, 0):
(4, 2):   right 2, up 2   -> right 1, up 1   -> (3, 1)
(10, 2):  right 8, up 2   -> right 4, up 1   -> (6, 1)
(10, 6):  right 8, up 6   -> right 4, up 3   -> (6, 3)

The base of P is 6 units and the image base is 3 units, so the scale factor 1/2 checks out.

Worked example 5: finding the centre and the scale factor

Triangle K has vertices (4, 1), (6, 1) and (4, 3). Its image K’ has vertices (5, 2), (9, 2) and (5, 6). Describe the enlargement fully.

Scale factor: base of K = 2, base of K' = 4, so scale factor = 4/2 = 2.
Centre: draw a straight line through each pair of matching vertices.
  (4, 1) to (5, 2) is the line y = x - 3
  (6, 1) to (9, 2) is the line y = (x - 3)/3
  These meet where x - 3 = (x - 3)/3, so x = 3 and y = 0.

On a grid you can simply draw the lines and read off where they meet. Check with a vertex: from (3, 0) to (4, 3) is 1 right and 3 up, and from (3, 0) to (5, 6) is 2 right and 6 up, which is twice as far. The answer is an enlargement, scale factor 2, centre (3, 0).

Two lines through matching vertices are enough to find the centre, and a third line is a useful check.

When you describe a transformation, give three things: the word enlargement, the scale factor and the centre. Without the centre the description is not complete.

Negative scale factors (Higher tier only)

With a negative scale factor, the image appears across the centre from the original and is turned through a half turn. The number without its sign gives the size, so −2 gives an image twice as big, on the far side of the centre.

Worked example 6: scale factor −2 (Higher tier only)

Enlarge triangle ABC with A(1, 3), B(3, 3) and C(3, 4) by scale factor −2, centre (2, 1).

Steps from (2, 1), then multiply by -2 (so the direction reverses):
A: left 1, up 2    -> right 2, down 4   -> A' = (4, -3)
B: right 1, up 2   -> left 2, down 4    -> B' = (0, -3)
C: right 1, up 3   -> left 2, down 6    -> C' = (0, -5)

The same result comes from enlarging by scale factor 2 and then turning the image through a half turn about the centre. To recognise a negative scale factor, look for an image across the centre that has been turned through 180°. A scale factor of −1/2 gives a smaller image on the far side.

9.04c Similar shapes

Ratio notation and scale factors

A ratio such as 2 : 5 compares lengths on the original with lengths on the image. Divide the second number by the first to get the scale factor, here 5/2 = 2.5. A scale factor from the larger shape back to the smaller one is the reciprocal, 2/5.

Worked example 7: comparing lengths

A photograph measures 8 cm by 12 cm. It is enlarged, and its shorter side becomes 20 cm.

Ratio of shorter sides   8 : 20 = 2 : 5
Scale factor             20 / 8 = 2.5
Longer side              12 x 2.5 = 30 cm

The enlarged photograph is 20 cm by 30 cm. The lengths are in the ratio 2 : 5.

Unknown lengths in similar figures

1. Pick a pair of sides that you know on both shapes, and divide to find the scale factor.
2. To find a length on the larger shape, multiply by the scale factor. To find a length on the smaller shape, divide.
3. Check that the answer is bigger or smaller, as it should be.

Worked example 8: finding sides

Quadrilaterals ABCD and PQRS are similar. AB = 12 cm matches PQ = 18 cm. BC = 14 cm, and RS = 27 cm matches CD. Find QR and CD.

Scale factor from ABCD to PQRS = 18/12 = 1.5
QR = 14 x 1.5 = 21 cm
CD = 27 / 1.5 = 18 cm   (CD is on the smaller shape, so divide)

Areas and volumes of similar shapes (Higher tier only)

If the length scale factor is k, then the area scale factor is k² and the volume scale factor is k³. In ratio terms, lengths 3 : 5 give areas 9 : 25 and volumes 27 : 125.

At Foundation tier you can still compare areas and volumes by working them out, then writing the ratio. The shortcut below is the Higher tier only part.

Worked example 9: areas and volumes (Higher tier only)

A label has area 72 cm². A similar label has lengths in the ratio 3 : 5. Find the area of the larger label.

Area ratio = 3² : 5² = 9 : 25
Scale factor for area = 25/9
Larger area = 72 x 25/9 = 200 cm²

Now a reverse problem. Two similar solids have volumes 54 cm³ and 128 cm³. The smaller has a height of 9 cm. Find the height of the larger.

Volume ratio = 54 : 128 = 27 : 64
Length ratio = cube root of 27 : cube root of 64 = 3 : 4
Larger height = 9 x 4/3 = 12 cm

For a longer chain, change in steps. Two similar containers have surface areas 100 cm² and 225 cm². The smaller holds 40 cm³. The area ratio is 100 : 225 = 4 : 9, so the length ratio is 2 : 3 and the volume ratio is 8 : 27. The larger holds 40 x 27/8 = 135 cm³.

Common errors

• Matching sides in the wrong order. List the sides of each triangle from shortest to longest before dividing.
• Dividing the wrong way. A length on a bigger shape is found by multiplying by a scale factor greater than 1.
• Describing an enlargement as just “scale factor 2”. The centre is needed too.
• Counting steps from the shape instead of from the centre.
• Using the length ratio for areas (Higher tier only). For area use the square of the length scale factor, and for volume use its cube.
• Using a square root on a volume ratio when a cube root is needed (Higher tier only).
• Giving a proof of similarity with angles but no reasons, such as parallel lines or a common angle.

Next steps

Work through the practice questions, then use the revision notes to keep the methods fresh.

Official syllabus

Aligned to the OCR GCSE (9-1) in Mathematics (J560) specification, Version 2.1, August 2026, published by OCR, for first assessment in 2017. Content references 9.04a (Similar triangles), 9.04b (Enlargement) and 9.04c (Similar shapes).