OCR GCSE Mathematics: Similarity and enlargement (J560) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover OCR GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), content references 9.04a, 9.04b and 9.04c. Questions marked (Higher tier only) cover negative scale factors and the link between lengths, areas and volumes. The mark allocations are set by Marlbridge for practice. Read the study guide first, then use the revision notes as a reminder. Course pages: the OCR GCSE Mathematics hub, the checklist and the free diagnostics.

Questions

1. Triangle ABC has angles of 48° and 67°. Triangle DEF has angles of 65° and 48°. Show that the triangles are similar. [2]

2. (a) Triangle P has sides 4.5 cm, 6 cm and 7.5 cm. Triangle Q has sides 7.5 cm, 10 cm and 12.5 cm. Show that they are similar and state the scale factor from P to Q. [2] (b) Triangle R has sides 5 cm, 9 cm and 12 cm. Triangle S has sides 10 cm, 18 cm and 25 cm. Explain why R and S are not similar. [2]

3. Triangle S has vertices (1, 2), (2, 2) and (2, 4). Enlarge S by scale factor 3 with centre (0, 1). Give the coordinates of the image. [3]

4. Triangle M has vertices (2, 5), (10, 5) and (10, 9). Its image N has vertices (0, 3), (4, 3) and (4, 5). Give the full description of the one transformation that maps M to N. [3]

5. In triangle ABC, D lies on AB and E lies on AC. DE is parallel to BC. AD = 5 cm, DB = 3 cm, DE = 7.5 cm and AE = 4 cm. (a) Prove that triangles ADE and ABC are similar. [3] (b) Find BC. [2] (c) Find EC. [2]

6. Parallelograms ABCD and EFGH are similar. AB = 7.5 cm matches EF = 12 cm. BC = 10 cm, and EH = 20.8 cm matches AD. (a) Find the scale factor from ABCD to EFGH. [1] (b) Find FG. [2] (c) Find AD. [2]

7. A model of a bridge is made to a scale of 1 : 250. (a) The span of the model is 36 cm. Find the span of the real bridge, in metres. [2] (b) A tower on the real bridge is 120 m high. Find the height of the tower on the model, in centimetres. [2]

8. (Higher tier only) Triangle ABC has A(2, 2), B(5, 2) and C(5, 3). (a) Enlarge ABC by scale factor −2 with centre (3, 1). Give the coordinates of the image. [3] (b) Write down the length of A’B’, where A’ and B’ are the images of A and B. [1]

9. (Higher tier only) Two similar bottles have heights 12 cm and 18 cm. (a) The smaller bottle holds 400 ml. How much does the larger hold? [2] (b) The label on the smaller bottle has area 150 cm². Find the area of the label on the larger bottle. [2]

10. (Higher tier only) Two similar solids have surface areas 96 cm² and 150 cm². The larger has volume 250 cm³. Find the volume of the smaller solid. [5]

11. A vertical pole 2.4 m high casts a shadow 3.6 m long on level ground. At the same time a tree casts a shadow 15 m long. (a) Explain why the triangles formed by the pole and the tree are similar. [2] (b) Find the height of the tree. [3]

12. (Higher tier only) A solid cone has base radius 8 cm, height 24 cm and volume 512π cm³. It is cut by a plane parallel to the base, 9 cm from the apex. This removes a small cone from the top. (a) Explain why the small cone is similar to the whole cone. [1] (b) Find the radius of the circular cut face. [2] (c) Find the volume of the small cone, in terms of π. [2] (d) Find the volume of the remaining solid, in terms of π. [1] (e) Find the ratio of the area of the cut face to the area of the base. [1]

Answers

1. Third angle of ABC = 180° − 48° − 67° = 65° [1]. DEF has angles 65°, 48° and 180° − 65° − 48° = 67°. The angles of both are 48°, 65° and 67°, so the triangles are similar [2] [1] Examiner insight: work out the missing angles and list all three for each triangle. Two angles alone can look like a guess.

2. (a) 7.5 ÷ 4.5 = 10 ÷ 6 = 12.5 ÷ 7.5 = 5/3 [1]. The sides are in the same ratio, so P and Q are similar, with scale factor 5/3 [2] [1] (b) 10 ÷ 5 = 2 and 18 ÷ 9 = 2 [1], but 25 ÷ 12 is about 2.08 [1]. The ratios are not all equal, so R and S are not similar [2] Examiner insight: in (b) quote the three ratios. Saying “the numbers don’t match” with no working gives the reader nothing to follow.

3. Steps from (0, 1): (1, 2) is right 1, up 1; (2, 2) is right 2, up 1; (2, 4) is right 2, up 3 [1]. Multiply by 3: right 3 up 3, right 6 up 3, right 6 up 9 [1]. Image: (3, 4), (6, 4), (6, 10) [3] [1] Examiner insight: count from the centre, not from the original vertex. A sketch with the centre marked helps you spot a slip.

4. Base of M is 8 and base of N is 4, so the scale factor is 4 ÷ 8 = 1/2 [1]. Lines through (2, 5) and (0, 3), and through (10, 5) and (4, 3), meet at (−2, 1) [1]. The transformation is an enlargement with that scale factor and centre [3] [1] Examiner insight: name the transformation, the scale factor and the centre. A description missing any one is incomplete. Check that the third pair of vertices also lines up with the centre.

5. (a) Angle A is common to both triangles [1]. Angle ADE = angle ABC, corresponding angles with DE parallel to BC [1]. Angle AED = angle ACB, corresponding angles, so the triangles are similar [3] [1] (b) AB = 5 + 3 = 8 cm, so the scale factor from ADE to ABC is 8/5 [1]. BC = 7.5 × 8/5 = 12 cm [2] [1] (c) AC = 4 × 8/5 = 6.4 cm [1]. EC = 6.4 − 4 = 2.4 cm [2] [1] Examiner insight: the scale factor uses AB, not DB. A frequent slip is to divide by the shorter part of the side.

6. (a) 12 ÷ 7.5 = 1.6 [1] [1] (b) FG = 10 × 1.6 [1] = 16 cm [2] [1] (c) AD is on the smaller shape, so divide: 20.8 ÷ 1.6 [1] = 13 cm [2] [1] Examiner insight: check the direction. The smaller parallelogram must give the smaller length, so AD should be less than EH.

7. (a) Real length = 36 × 250 = 9000 cm [1] = 90 m [2] [1] (b) 120 m = 12 000 cm [1]. 12 000 ÷ 250 = 48 cm [2] [1] Examiner insight: write the units at each step. The scale is for lengths in the same unit, so convert metres and centimetres before dividing.

8. (a) Steps from (3, 1): A is left 1, up 1; B is right 2, up 1; C is right 2, up 2 [1]. Multiply by −2: A right 2, down 2; B left 4, down 2; C left 4, down 4 [1]. Image: A’(5, −1), B’(−1, −1), C’(−1, −3) [3] [1] (b) AB = 3, so A’B’ = 3 × 2 = 6 [1] [1] Examiner insight: the image should land on the opposite side of the centre. If it ends up on the same side, the sign has been ignored. You can check with A’B’ = 6.

9. (a) Length ratio 12 : 18 = 2 : 3, so the volume ratio is 8 : 27 [1]. 400 × 27/8 = 1350 ml [2] [1] (b) Area ratio is 4 : 9 [1]. 150 × 9/4 = 337.5 cm² [2] [1] Examiner insight: write the ratio you are using next to the working. Mixing up the area and volume factors is the usual way to lose marks here.

10. Area ratio 96 : 150 = 16 : 25 [1]. Length ratio = 4 : 5 [1]. Volume ratio = 64 : 125 [1]. The smaller volume is 250 × 64/125 [1] = 128 cm³ [5] [1] Examiner insight: say which solid each number belongs to. Because the larger volume is given, you scale down by 64/125, and the answer must be smaller than 250.

11. (a) The sun’s rays are parallel, so the angle of elevation is the same for the pole and the tree [1]. The pole and tree are both vertical, so the right angles are equal, so the triangles are similar [2] [1] (b) Scale factor = 15 ÷ 3.6 [1]. Height = 2.4 × 15/3.6 [1] = 10 m [3] [1] Examiner insight: give the geometric reason for the equal angle. Saying “they are similar” without a reason is a conclusion, not an explanation.

12. (a) The cut is parallel to the base, so the small cone has the same angles as the whole cone and its lengths are all in the same ratio [1] [1] (b) Scale factor = 9/24 = 3/8 [1]. Radius = 8 × 3/8 = 3 cm [2] [1] (c) Volume ratio = 3³ : 8³ = 27 : 512 [1]. Volume = 512π × 27/512 = 27π cm³ [2] [1] (d) 512π − 27π = 485π cm³ [1] [1] (e) Area ratio = 3² : 8² = 9 : 64 [1] [1] Examiner insight: cube for volume in (c) and square for area in (e). Check (e) by direct calculation: 9π to 64π reduces to 9 : 64.

Where marks are usually lost

• Naming a pair of triangles as similar without giving a reason for each equal angle.
• Pairing sides of a turned triangle in the wrong order.
• Multiplying when the answer should be smaller, or dividing when it should be larger.
• Using DB instead of AB as the base for the scale factor when a smaller triangle sits inside a larger one.
• Counting steps from the shape instead of from the centre of enlargement.
• Leaving out the centre, or the scale factor, when asked to describe an enlargement.
• Ignoring the minus sign in a negative scale factor, so the image is on the wrong side of the centre (Higher tier only).
• Using the length scale factor for area or volume, or taking the wrong root when working backwards (Higher tier only).

Next steps

• Review the revision notes and the study guide.
• Go to the OCR GCSE Mathematics hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

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).