Every similarity question from 13 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 4.4. 14 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/12/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Geometry past paper questions.
0580 Similarity past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q13(a)
A trapezium T is drawn on a 1 cm² grid. Its parallel sides are 2 cm and 4 cm and its height is 2 cm. On the grid, draw a different trapezium that is mathematically similar to trapezium T.
Show solution
- Similar shapes have all lengths multiplied by the same scale factor, and the angles stay the same.
- Use scale factor 2: parallel sides 2 × 2 = 4 cm and 4 × 2 = 8 cm, height 2 × 2 = 4 cm.
- Draw a trapezium with parallel sides 4 cm and 8 cm, height 4 cm, with the sloping side in the same direction as in T.
May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q3(a)
The diagram shows shapes A to E and a shaded shape drawn on a 1 cm² grid. Write down the letter of the shape that is congruent to the shaded shape.
Show solution
- Congruent shapes are exactly the same shape and size. They may be turned or flipped over.
- Compare each shape’s side lengths and angles with the shaded shape.
- The only exact match is shape D
May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q23
ABCD is mathematically similar to PQRS. AD = 5 cm, PS = 8 cm and PQ = 14 cm. Calculate AB.
Show solution
- AD and PS are matching sides. Scale factor from PQRS to ABCD = 5⁄8.
- AB matches PQ, so AB = 14 × 5⁄8 = 70⁄8 = 8.75 cm
Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q24
Triangles ABC and PQR are mathematically similar. AC = 12 cm, PR = 28 cm and PQ = 42 cm. Calculate AB.
Show solution
- AC and PR are matching sides. Scale factor from PQR to ABC = 12⁄28 = 3⁄7.
- AB matches PQ, so AB = 42 × 3⁄7 = 6 × 3 = 18 cm
- Check: 18⁄42 = 3⁄7 ✓
Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q29(a)
Two similar triangles, A and B, have corresponding sides 9.1 cm and 1.4 cm. Another side of A is 20.8 cm and the matching side of B is x cm. Calculate the value of x.
Show solution
- Matching sides are 9.1 cm on A and 1.4 cm on B, so the scale factor from A to B = 1.4⁄9.1.
- x = 20.8 × 1.4⁄9.1 = 20.8 × 0.1538… = 3.2
- Check: 9.1⁄1.4 = 6.5 and 20.8⁄3.2 = 6.5 ✓
May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q23
Rectangle A measures 8 cm by 2 cm. Rectangle B measures x cm by 1.5 cm. Rectangle A is mathematically similar to rectangle B. Work out the value of x.
Show solution
- In similar shapes, every length is multiplied by the same scale factor.
- The short sides match: 2 cm in A and 1.5 cm in B. Scale factor = 1.5 ÷ 2 = 0.75.
- The long side of B = 8 × 0.75 = 6, so x = 6
0580 Similarity past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q23
Two solids are mathematically similar. The volume of the larger solid is 64 cm³ and the volume of the smaller solid is 27 cm³. The height of the larger solid is 10 cm and the height of the smaller solid is h cm. Find the value of h.
Show solution
- For similar solids, volume scale factor = (length scale factor)³.
- Volume scale factor from large to small = 27⁄64.
- Length scale factor = ∛27⁄64 = ∛27⁄∛64 = 3⁄4.
- h = 10 × 3⁄4 = 7.5
May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q20
Two bottles are mathematically similar. The volume of the larger bottle is 216 cm³ and the volume of the smaller bottle is 27 cm³. The height of the larger bottle is 10 cm and the height of the smaller bottle is h cm. Find the value of h.
Show solution
- For similar solids, volume scale factor = (length scale factor)³.
- Volume scale factor = 216⁄27 = 8.
- Length scale factor = ∛8 = 2, so the larger bottle is twice as tall.
- h = 10 ÷ 2 = 5
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q20
The two cylinders, A and B, are mathematically similar. The surface area of A is 80 cm² and the surface area of B is 180 cm². The height of A is h cm and the height of B is 12 cm. Find the value of h.
Show solution
- For similar shapes, area scale factor = (length scale factor)².
- Area scale factor from B to A = 80⁄180 = 4⁄9.
- Length scale factor = √4⁄9 = 2⁄3.
- h = 12 × 2⁄3 = 8
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q8(b)(i)
AB and CD are parallel lines, and AD and BC are straight lines that intersect at X. Triangle ABX is similar to triangle DCX. CX = 5 cm, XD = 8 cm and AD = 20 cm. Find BX.
Show solution
- Find AX first: AX = AD − XD = 20 − 8 = 12 cm.
- In the similar triangles, AX matches DX and BX matches CX.
- Scale factor = AX⁄DX = 12⁄8 = 1.5.
- BX = 1.5 × CX = 1.5 × 5 = 7.5 cm
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q14
Two bottles are mathematically similar. The larger bottle has a capacity of 1.35 litres and a height of 24 cm. The smaller bottle has a capacity of 0.4 litres. Calculate the height of the smaller bottle.
Show solution
- The capacities are volumes. Volume scale factor = small⁄large = 0.4⁄1.35 = 8⁄27.
- Length scale factor = cube root of the volume scale factor: ∛8⁄27 = 2⁄3.
- Height of the smaller bottle = 24 × 2⁄3 = 16 cm
- Tip: Never use the volume ratio directly on a length. Take the cube root first.
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q14(b)
AC and BD are chords of a circle that intersect at X. Triangle CXD is mathematically similar to triangle BXA. DX = 8.0 cm, BX = 2.7 cm and AX = 4.0 cm.
(i) Work out the length of CX. (ii) Complete the statement: area of triangle CXD : area of triangle BXA = … : …
Show solution
- Match the sides of the similar triangles: DX matches AX, and CX matches BX.
- Scale factor from BXA to CXD = DX⁄AX = 8.0⁄4.0 = 2.
- CX = BX × 2 = 2.7 × 2 = 5.4 cm
- For similar shapes, area scale factor = (length scale factor)² = 2² = 4.
- Area CXD : area BXA = 4 : 1
Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q11
S lies on PQ and T lies on PR. ST is parallel to QR.
(a) Explain why triangle PQR is mathematically similar to triangle PST. Give a geometrical reason for each statement you make.
(b) ST = 3 cm, QR = 9 cm and PS = 5 cm. Work out PQ.
(c) The area of triangle PST is 2k cm². Find, in terms of k, the area of quadrilateral QRTS.
Show solution
- To show two triangles are similar, show that all three pairs of angles are equal.
- ST is parallel to QR, and PQ crosses both lines: Angle PQR = angle PST (corresponding angles)
- ST is parallel to QR, and PR crosses both lines: angle PRQ = angle PTS (corresponding angles)
- Both triangles share the angle at P: angle QPR = angle SPT (common angle)
- All three angles match, so the triangles are similar.
- Matching sides: ST matches QR. Scale factor = QR⁄ST = 9⁄3 = 3.
- PS matches PQ, so PQ = 5 × 3 = 15 cm
- For similar shapes, area scale factor = (length scale factor)² = 3² = 9.
- Area of PQR = 9 × 2k = 18k cm².
- The quadrilateral QRTS is triangle PQR with triangle PST removed: 18k − 2k = 16k cm²
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q22
(a) Two mathematically similar triangles have corresponding sides 10 mm and 14 mm. The side 45 mm in the first triangle matches x mm in the second. Find the value of x.
(b) The surface areas of two mathematically similar containers are 124 cm² and 279 cm². The capacity of the smaller container is 56 ml. Find the capacity of the larger container.
Show solution
- The matching sides 10 mm and 14 mm give the scale factor: 14 ÷ 10 = 1.4.
- x = 45 × 1.4 = 63
- Area scale factor = 279⁄124 = 2.25.
- Length scale factor = √2.25 = 1.5.
- Volume scale factor = 1.5³ = 3.375.
- Capacity of the larger container = 56 × 3.375 = 189 ml
- Tip: Go from area to length with a square root, then from length to volume by cubing.
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