OCR GCSE Mathematics: Similarity and enlargement (J560) – Revision Notes

These revision notes condense similarity and enlargement for the OCR GCSE (9-1) Mathematics specification (J560), Version 2.1 (August 2026), content references 9.04a to 9.04c. Both tiers are covered. Two items carry the label “Higher tier only”: negative scale factors in 9.04b, and the relationship between lengths, areas and volumes of similar shapes in 9.04c. For full explanations and longer worked examples, read the study guide.

Then test yourself with the practice questions. Course pages: the OCR GCSE Mathematics hub, the checklist and the free diagnostics.

Key definitions

Term	Meaning
Similar shapes	Same shape, different size. Corresponding angles are equal and corresponding sides are in the same ratio
Scale factor	The number you multiply a length by to get the matching length on the image
Centre of enlargement	The fixed point that every image point is measured from
Image	The shape after the enlargement
Corresponding sides	Sides in the same position on the two shapes, such as shortest to shortest

9.04a Similar triangles

Tests for similar triangles (any one is enough):

Test	What to show
Sides	All three pairs of sides in the same ratio
Sides and angle	Two pairs of sides in the same ratio, with equal angles between them
Angles	Two pairs of equal angles, because the third pair must then be equal as well

Method to prove similarity, in steps:

1. Name the triangles with matching letters.
2. State each pair of equal angles, with its reason (common angle, corresponding angles, alternate angles, vertically opposite angles).
3. Conclude: “the triangles are similar” and, if needed, write the scale factor.

Worked reminder: triangle ABC has D on AB and E on AC, with DE parallel to BC. Angle A is common, and the corresponding angles ADE and ABC are equal, so triangles ADE and ABC are similar. Parallel lines in a triangle are the most common source of similar triangles.

9.04b Enlargement

Skill	Rule
Enlarge from a centre	Take the steps from the centre to a vertex, multiply them by the scale factor, then count from the centre
Find the scale factor	Image length divided by original length
Find the centre	Draw lines through pairs of matching vertices. They meet at the centre
Fractional scale factor	A fraction less than 1 shrinks the shape, and the same method works
Describe fully	The word enlargement, the scale factor and the centre

Worked reminder: enlarge the point (6, 4) with scale factor 1/4 and centre (2, 0). The steps from the centre are right 4 and up 4, which become right 1 and up 1. The image is (3, 1).

Worked reminder for finding a centre by calculation: a vertex at (2, 1) maps to (5, 4) and the scale factor is 3. Call the centre (a, b). Then 5 = a + 3(2 − a), so a = 1/2, and 4 = b + 3(1 − b), so b = −1/2. Check: the steps from (1/2, −1/2) to (2, 1) are 1.5 right and 1.5 up, and tripling gives 4.5 and 4.5, which lands on (5, 4). On a grid you would simply draw lines through matching vertices and read off where they meet.

Going the other way: if a shape is mapped to a smaller image by scale factor 1/3, then the image maps back to the original by scale factor 3 about the same centre.

Negative scale factors (Higher tier only)

• The image lies across the centre from the original and has been turned through 180°.
• The size of the number gives the size of the image. The sign gives the side.
• Worked reminder: the point (8, 6) with centre (2, 2) and scale factor −1/2. The steps are right 6, up 4. Multiplying by −1/2 gives left 3, down 2. The image is (−1, 0).

9.04c Similar shapes

Matching lengths on similar shapes share one ratio. Use these steps to find an unknown length.

1. Find the scale factor from a pair of known matching sides.
2. Multiply by it to go from the smaller to the larger shape. Divide to go from the larger to the smaller.
3. Check that the size of the answer makes sense.

Worked reminder: two similar triangles have matching sides 5 cm and 8 cm. A side of 12 cm on the first triangle matches 12 x 8/5 = 19.2 cm on the second.

Lengths, areas and volumes (Higher tier only)

If the ratio of lengths is	Then the ratio of areas is	And the ratio of volumes is
a : b	a² : b²	a³ : b³
3 : 7	9 : 49	27 : 343
3 : 5	9 : 25	27 : 125

To work backwards, take the square root of an area ratio or the cube root of a volume ratio. Always reduce the ratio first, since it is easier to take roots of small numbers.

Worked reminder: the lengths of two similar shapes are in the ratio 3 : 5, and the smaller has an area of 18 cm². The larger has an area of 18 x 25/9 = 50 cm². Two similar solids with the same length ratio have the larger volume 500 cm³. The smaller has 500 x 27/125 = 108 cm³.

At Foundation tier, compare areas and volumes by calculating each one and then writing the ratio.

Must-know distinctions

• Similar and congruent: similar shapes can differ in size, while congruent shapes cannot. See congruence.
• Scale factor and ratio: ratio 3 : 5 gives scale factor 5/3 from first to second, and 3/5 from second to first.
• Enlargement and similarity: an enlargement fixes the position of the image using the centre, while similarity only says two shapes have the same shape.
• Scale factor and centre: the scale factor changes the size, and the centre changes where the image lands.
• Length, area and volume scale factors (Higher tier only): k, k² and k³.

Quick self-test

1. Triangle A has angles 40° and 75°. Triangle B has angles 75° and 65°. Are they similar?
2. Triangle C has sides 3 cm, 4 cm and 6 cm. Triangle D has sides 9 cm, 12 cm and 20 cm. Are they similar?
3. A length of 6 cm on a shape becomes 21 cm on its image. What is the scale factor?
4. Enlarge the point (3, 2) by scale factor 3, centre (1, 1). Where is the image?
5. Enlarge the point (6, 4) by scale factor 1/4, centre (2, 0). Where is the image?
6. The point (3, 1) is mapped to (5, 5) by an enlargement with scale factor 2. Find the centre.
7. Two similar shapes have matching sides of 5 cm and 8 cm. What does a side of 12 cm on the first become?
8. A model has scale 1 : 40. A real object is 3.2 m long. How long is the model, in centimetres?
9. (Higher tier only) Enlarge the point (4, 3) by scale factor −2, centre (1, 1).
10. (Higher tier only) The lengths of two similar shapes are in the ratio 3 : 4, and the smaller has area 36 cm². Find the larger area.

Answers

1. Third angles are 65° and 40°. Triangle A has 40°, 75°, 65° and triangle B has 75°, 65°, 40°. Yes, similar.
2. 9/3 = 3 and 12/4 = 3, but 20/6 is not 3. No.
3. 21 / 6 = 3.5.
4. Steps right 2, up 1. Times 3 gives right 6, up 3. Image (7, 4).
5. Steps right 4, up 4. Times 1/4 gives right 1, up 1. Image (3, 1).
6. Centre = 2 x (3, 1) − (5, 5) = (1, −3). Check: steps from (1, −3) to (3, 1) are (2, 4), doubled is (4, 8), giving (5, 5).
7. 12 x 8/5 = 19.2 cm.
8. 3.2 m = 320 cm, and 320 / 40 = 8 cm.
9. Steps right 3, up 2. Times −2 gives left 6, down 4. Image (−5, −3).
10. Area ratio is 9 : 16, so 36 x 16/9 = 64 cm².

Where marks are usually lost

• Writing “similar” with no stated reason such as “corresponding angles” or “common angle”.
• Pairing sides in the wrong order when the triangles are turned or reflected.
• Dividing when you should multiply, so the enlarged length is smaller than the original.
• Counting steps from a vertex of the shape instead of from the centre.
• Describing an enlargement with only a scale factor and no centre.
• Treating a scale factor of 1/2 as 2, or the reverse, when going between big and small.
• Using the length scale factor for area or volume (Higher tier only).
• Applying a square root to a volume ratio, or a cube root to an area ratio (Higher tier only).

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).