Every surface area and volume question from 22 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 5.4. 27 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/13/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 Mensuration past paper questions.
0580 Surface Area and Volume past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q14(b)
A right-angled triangular prism has a triangular cross-section with perpendicular sides 3 cm and 4 cm, and length 4 cm. Find the volume of the triangular prism.
Show solution
- Volume of a prism = area of cross-section × length.
- Cross-section is a right-angled triangle: area = 1⁄2 × 3 × 4 = 6 cm².
- Volume = 6 × 4 = 24 cm³
May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q15
(a) A cuboid measures 8 cm by 5 cm by 3 cm. Calculate the surface area of the cuboid.
(b) A sphere has a radius of 6.4 cm. Calculate the volume of the sphere.
Show solution
- A cuboid has three pairs of identical faces: 8 × 5, 8 × 3 and 5 × 3.
- Areas: 8 × 5 = 40, 8 × 3 = 24, 5 × 3 = 15.
- Surface area = 2 × (40 + 24 + 15) = 2 × 79 = 158 cm²
- Volume of a sphere = 4⁄3πr³.
- V = 4⁄3 × π × 6.4³ = 4⁄3 × π × 262.144 = 1098.06…
- Round to 3 significant figures: 1100 cm³
May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q11
A cuboid has length 10.8 cm, width 3.7 cm and height 4.6 cm. Work out the volume of the cuboid. Give your answer correct to 1 decimal place.
Show solution
- Volume of a cuboid = length × width × height.
- V = 10.8 × 3.7 × 4.6 = 183.816 cm³.
- Round to 1 decimal place: 183.8 cm³
May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q24
A sphere has radius 6 cm. Work out the surface area of the sphere. Give the units of your answer.
Show solution
- Surface area of a sphere = 4πr².
- 4 × π × 6² = 4 × π × 36 = 144π = 452.38…
- Round to 3 significant figures and give units of area: 452 cm²
May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q26
The radius of a sphere is 3.6 cm. A cone has radius r cm and slant height 8 cm. The surface area of the sphere is equal to the curved surface area of the cone. Find the value of r.
Show solution
- Surface area of the sphere = 4πr² = 4 × π × 3.6² = 4 × π × 12.96 = 51.84π cm².
- Curved surface area of the cone = πrl = π × r × 8 = 8πr cm².
- They are equal: 8πr = 51.84π.
- Divide both sides by 8π: r = 51.84 ÷ 8 = 6.48
- Tip: Leave π in both expressions. It cancels, so the answer is exact.
Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q12
A cuboid has length 10 cm, width 3 cm and height 4 cm. Work out the surface area of the cuboid.
Show solution
- A cuboid has three pairs of identical faces: 10 × 3, 10 × 4 and 3 × 4.
- Areas: 10 × 3 = 30, 10 × 4 = 40, 3 × 4 = 12.
- Surface area = 2 × (30 + 40 + 12) = 2 × 82 = 164 cm²
Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q17
(a) A cone has a radius of 7 cm and a height of 8 cm. Work out the volume of the cone.
(b) A trapezium has parallel sides 6 m and 20 m and perpendicular height 10 m (slanting side 15 m). Work out the area of the trapezium. Give the units of your answer.
Show solution
- Volume of a cone = 1⁄3πr²h.
- V = 1⁄3 × π × 7² × 8 = 1⁄3 × π × 49 × 8 = 410.50…
- Round to 3 significant figures: 411 cm³
- Area of a trapezium = 1⁄2(a + b)h, using the parallel sides and the perpendicular height.
- Area = 1⁄2 × (6 + 20) × 10 = 13 × 10 = 130.
- The lengths are in metres, so the area is in square metres: 130 m²
- Tip: The 15 m sloping side is not needed for the area.
Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q19
A cylinder has radius 5 cm and height 12 cm. A cube has edges of x cm. The cylinder has the same volume as the cube. Find the value of x.
Show solution
- Volume of the cylinder = πr²h = π × 5² × 12 = 300π = 942.47… cm³.
- Volume of the cube = x³, and the volumes are equal: x³ = 942.47…
- Take the cube root: x = ∛942.47… = 9.804…
- 9.80
Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q23(b)
A candle is in the shape of a cone. The volume of the candle is 900 cm³. The height of the candle is 8 cm. Calculate the radius of the candle.
Show solution
- Volume of a cone = 1⁄3πr²h, so 1⁄3 × π × r² × 8 = 900.
- Multiply both sides by 3: 8πr² = 2700.
- Divide both sides by 8π: r² = 2700 ÷ (8π) = 107.43…
- Square root: r = √107.43… = 10.36… = 10.4 cm
May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q11
Calculate the volume of a cube with side length 3 cm.
Show solution
- Volume of a cube = side × side × side.
- Volume = 3 × 3 × 3 = 27 cm³
May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q10
A cuboid has length 5 cm, width 2 cm and height 3 cm.
(a) Draw a net of the cuboid on the 1 cm² grid.
(b) Work out the volume of the cuboid. Give the units of your answer.
Show solution
- A cuboid has 6 faces in 3 matching pairs: two 5 × 2, two 5 × 3 and two 2 × 3 rectangles.
- Draw a strip of four faces (5 × 3, 5 × 2, 5 × 3, 5 × 2) that wraps round the cuboid.
- Add the two 2 × 3 end faces on opposite sides of the strip, so the net folds into a closed box.
- A correct net of six faces: two 5 cm × 2 cm, two 5 cm × 3 cm and two 2 cm × 3 cm.
- Volume of a cuboid = length × width × height.
- Volume = 5 × 2 × 3 = 30. The lengths are in cm, so the units are cubic centimetres.
- 30 cm³
May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q8
The diagram shows two faces of a net of a cuboid on a 1 cm² grid: a 3 cm by 3 cm square joined to a 5 cm by 3 cm rectangle.
(a) Complete the statement. The dimensions of the cuboid are ….. cm by ….. cm by ….. cm.
(b) On the grid, complete a net of the cuboid.
(c) Work out the volume of the cuboid.
Show solution
- The square face is 3 cm by 3 cm and the rectangle is 5 cm by 3 cm.
- So the cuboid’s edges are 3 cm, 3 cm and 5 cm: 3 cm by 3 cm by 5 cm
- This cuboid has 6 faces: two 3 × 3 squares and four 5 × 3 rectangles.
- Two faces are drawn, so add one more 3 × 3 square and three more 5 × 3 rectangles.
- Join them so they fold into a closed box with no overlaps.
- A correct net with two 3 × 3 squares and four 5 × 3 rectangles.
- Volume = length × width × height = 3 × 3 × 5 = 45 cm³
May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q9
The diagram shows the net of a square-based pyramid. The perpendicular height of each triangle is 4 cm. The base has side length 5 cm.
(a) Calculate the surface area of the pyramid.
(b) Write down the number of edges of the square-based pyramid.
Show solution
- The surface area is the total area of all the faces: 4 triangles and 1 square.
- Area of one triangle = 1⁄2 × base × height = 1⁄2 × 5 × 4 = 10 cm².
- Four triangles = 4 × 10 = 40 cm².
- Square base = 5 × 5 = 25 cm².
- Surface area = 40 + 25 = 65 cm²
- The base has 4 edges, and there are 4 sloping edges up to the top point.
- Number of edges = 4 + 4 = 8
0580 Surface Area and Volume past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q21(a,b)
A solid cuboid ABCDEFGH has BF = 2 cm, BC = 6 cm and CD = 9 cm.
(a) Calculate the volume of the cuboid. (b) Calculate the surface area of the cuboid.
Show solution
- Volume of a cuboid = length × width × height = CD × BC × BF.
- Volume = 9 × 6 × 2 = 108 cm³
- A cuboid has three pairs of identical faces: 9 × 6, 9 × 2 and 6 × 2.
- Areas: 9 × 6 = 54, 9 × 2 = 18, 6 × 2 = 12.
- Surface area = 2 × (54 + 18 + 12) = 2 × 84 = 168 cm²
May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q2
The diagram shows an accurate net of an open box in the shape of a cuboid. All the dimensions are whole numbers of centimetres.
(a) Find the total area of the net. (b) Find the volume of the open box.
Show solution
- Measure the net with a ruler. The edges are 2 cm, 3 cm and 5 cm.
- The open box has a 3 × 5 base, two 2 × 3 end faces and two 2 × 5 side faces. It has no lid.
- Add the areas of the five faces.
- Base: 3 × 5 = 15. Ends: 2 × (2 × 3) = 12. Sides: 2 × (2 × 5) = 20.
- Total area = 15 + 12 + 20 = 47 cm²
- Volume = length × width × height = 5 × 3 × 2 = 30 cm³
May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q16
The volume of a sphere is 500 cm³. Using the approximation of 3 for π, find an estimate for the radius of the sphere.
Show solution
- Volume of a sphere = 4⁄3πr³. Use 3 for π.
- 4⁄3 × 3 × r³ = 500. The 3s cancel, so 4r³ = 500.
- Divide by 4: r³ = 125.
- Cube root: r = ∛125 = 5 cm
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q14
A sphere has volume 36π cm³. Find the radius of the sphere.
Show solution
- Volume of a sphere = 4⁄3πr³.
- 4⁄3πr³ = 36π. Divide both sides by π: 4⁄3r³ = 36.
- Multiply by 3⁄4: r³ = 36 × 3⁄4 = 27.
- r = ∛27 = 3 cm
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q16(a,b)
ABCDE is a square-based pyramid. EC and BD intersect at F. AF is perpendicular to the base BCDE. BC = 16.5 cm and AF = 28 cm.
(a) Write down the number of planes of symmetry of the pyramid.
(b) Calculate the volume of the pyramid.
Show solution
- A square-based pyramid has two planes through the diagonals of the base and two through the midpoints of opposite sides.
- 4
- Volume of a pyramid = 1⁄3 × base area × height.
- Base area = 16.5 × 16.5 = 272.25 cm². Height AF = 28 cm.
- V = 1⁄3 × 272.25 × 28 = 2541 cm³
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q21
The diagram shows a frustum. The radius of the top is 8 cm, the radius of the base is 12 cm and the height is 9 cm. Calculate the volume of the frustum.
Show solution
- A frustum is a large cone with a small cone cut off the top. Find the height of the small cone first.
- Let the small cone have height h. The large cone then has height h + 9.
- The two cones are similar, so h⁄8 = h + 9⁄12.
- Cross-multiply: 12h = 8(h + 9) = 8h + 72, so 4h = 72 and h = 18. The large cone has height 18 + 9 = 27 cm.
- Volume of large cone = 1⁄3 × π × 12² × 27 = 1296π.
- Volume of small cone = 1⁄3 × π × 8² × 18 = 384π.
- Frustum = 1296π − 384π = 912π = 2865.1…
- 912π = 2870 cm³ (3 s.f.)
Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q18
A cone has radius r cm and sloping edge 12 cm. A hemisphere has radius 4 cm. The total surface area of the cone is equal to the total surface area of the hemisphere.
(a) Show that r² + 12r − 48 = 0.
(b) Find the exact value of the radius of the cone. Give your answer as a surd in its simplest form.
Show solution
- Total surface area of the cone = circular base + curved surface = πr² + πrl = πr² + 12πr.
- Total surface area of the hemisphere = curved surface + flat circle = 1⁄2 × 4π × 4² + π × 4².
- That is 32π + 16π = 48π.
- Set them equal: πr² + 12πr = 48π.
- Divide by π and rearrange: r² + 12r − 48 = 0 ✓
- Complete the square: r² + 12r = (r + 6)² − 36.
- So (r + 6)² − 36 − 48 = 0, which gives (r + 6)² = 84.
- Square root both sides: r + 6 = ±√84. Simplify √84 = √(4 × 21) = 2√21.
- r = −6 ± 2√21. A radius must be positive, so r = 2√21 − 6
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q9
A tank in the shape of a sphere has radius 40 cm.
(a) Show that the capacity of the tank is 268 litres, correct to the nearest litre.
(b) The tank contains 268 litres of water. The water is emptied into a cylindrical container at a rate of 300 ml per second. (i) Calculate the time taken for the tank to be completely emptied. Give your answer in minutes and seconds, correct to the nearest second. (ii) The cylindrical container has radius 50 cm and height 40 cm. Calculate the percentage of the container that is filled with water.
Show solution
- Volume of a sphere = 4⁄3πr³.
- V = 4⁄3 × π × 40³ = 4⁄3 × π × 64 000 = 268 082.5… cm³.
- 1 litre = 1000 cm³, so divide by 1000: 268.08… litres.
- To the nearest litre this is 268 litres ✓
- Change litres to millilitres: 268 litres = 268 000 ml.
- Time = 268 000 ÷ 300 = 893.33… seconds.
- Change to minutes: 893.33… ÷ 60 = 14 remainder 53.33…, so 14 minutes and 53.33… seconds.
- 14 min 53 s
- Volume of the cylinder = πr²h = π × 50² × 40 = 100 000π = 314 159.2… cm³.
- The water is 268 litres = 268 000 cm³.
- Percentage filled = 268 000⁄314 159.2… × 100 = 85.30…
- 85.3%
Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q3
A cuboid has length 6 cm, width 5 cm and height 2.5 cm. Work out the volume of the cuboid.
Show solution
- Volume of a cuboid = length × width × height.
- 6 × 5 = 30, then 30 × 2.5 = 75.
- Volume = 75 cm³
Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q23
The diagram shows a major sector, AOB, of a circle. The sector angle is 240° and the radius is 12 cm.
(a) Show that the length of the major arc AB is 16π cm.
(b) OA is joined to OB to form a cone. Work out the volume of the cone. Give your answer in the form (a√b)π⁄3, where a is an integer and b is a prime number.
Show solution
- Arc length = angle⁄360 × 2πr = 240⁄360 × 2 × π × 12.
- 240⁄360 = 2⁄3 and 2 × 12 = 24, so arc = 2⁄3 × 24π.
- Arc length = 16π cm ✓
- When the sector is rolled into a cone, the arc becomes the circumference of the base.
- 2πr = 16π, so the base radius r = 8 cm.
- The radius of the sector, 12 cm, becomes the slant height of the cone.
- Height of the cone by Pythagoras: √(12² − 8²) = √(144 − 64) = √80 = √(16 × 5) = 4√5 cm.
- Volume = 1⁄3πr²h = 1⁄3 × π × 64 × 4√5 = 256√5π⁄3.
- 256√5π⁄3 cm³
Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q29
In this question all measurements are in centimetres. Solid A is made from a cylinder and a hemisphere, both of radius 2R. The cylinder has height 5R. Solid B is a cone of radius R and sloping edge x. The total surface area of solid A is equal to the total surface area of solid B. Find R in terms of x.
Show solution
- Solid A has three outside surfaces: the circular base of the cylinder, the curved side of the cylinder and the curved hemisphere on top.
- Base: π(2R)² = 4πR². Cylinder side: 2π × 2R × 5R = 20πR². Hemisphere: 1⁄2 × 4π(2R)² = 8πR².
- Total for A = 4πR² + 20πR² + 8πR² = 32πR².
- Solid B (cone): base πR² plus curved surface πRx.
- Set them equal: 32πR² = πR² + πRx.
- Divide every term by πR: 32R = R + x.
- Subtract R: 31R = x, so R = x⁄31
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q14
A cube contains a solid metal sphere. The sphere touches all the faces of the cube. The side length of the cube is 8 cm.
(a) Show that the volume of the sphere is 256⁄3π cm³.
(b) Calculate the percentage of the cube that is not occupied by the sphere.
(c) The density of the metal is 7.86 g/cm³. Calculate the mass of the sphere in kilograms. (Density = mass ÷ volume)
(d) The sphere is melted down and made into a solid cylinder with radius 3.1 cm. Calculate the total surface area of the cylinder.
Show solution
- The sphere touches all six faces, so its diameter equals the side of the cube: 8 cm.
- Radius r = 8 ÷ 2 = 4 cm.
- Volume of a sphere = 4⁄3πr³ = 4⁄3 × π × 4³ = 4⁄3 × 64 × π
- = 256⁄3π cm³ ✓
- Volume of the cube = 8 × 8 × 8 = 512 cm³.
- Volume of the sphere = 256⁄3π = 268.08… cm³.
- Space not filled = 512 − 268.08… = 243.91… cm³.
- Percentage not filled = 243.91…⁄512 × 100 = 47.6%
- Rearrange density = mass ÷ volume to get mass = density × volume.
- Mass = 7.86 × 268.08… = 2107.1… g.
- Change grams to kilograms (÷ 1000): 2.11 kg
- Melting does not change the volume, so the cylinder also has volume 256⁄3π = 268.08… cm³.
- Volume of a cylinder = πr²h, so π × 3.1² × h = 268.08…
- h = 268.08… ÷ (π × 3.1²) = 268.08… ÷ 30.190… = 8.879… cm.
- Total surface area = two circles + curved surface = 2πr² + 2πrh.
- 2 × π × 3.1² = 60.38… cm² and 2 × π × 3.1 × 8.879… = 172.95… cm².
- Total = 60.38… + 172.95… = 233.3… = 233 cm²
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q13
The diagram shows solid A and solid B. Solid A is made from a hemisphere and a cone each with radius 6 cm. The cone has sloping edge 10 cm. Solid B is a cylinder with radius 4 cm and height h cm. The total surface area of solid A is equal to the total surface area of solid B.
(a) Work out the value of h.
(b) Work out the height of solid A.
Show solution
- Solid A has two curved surfaces and no flat faces, because the cone and hemisphere are joined.
- Curved surface of the cone = πrl = π × 6 × 10 = 60π.
- Curved surface of the hemisphere = half of 4πr² = 1⁄2 × 4π × 6² = 72π.
- Total surface area of A = 60π + 72π = 132π.
- Solid B (cylinder): two circular ends = 2πr² = 2 × π × 4² = 32π. Curved surface = 2πrh = 8πh.
- Total surface area of B = 32π + 8πh.
- Set them equal: 32π + 8πh = 132π. Divide by π: 32 + 8h = 132.
- 8h = 100, so h = 12.5
- The height of the cone comes from Pythagoras, using the slant edge and the radius: √(10² − 6²) = √64 = 8 cm.
- The hemisphere adds its radius below the cone: 6 cm.
- Height of solid A = 8 + 6 = 14 cm
May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q16
The diagram shows a solid made by joining a hemisphere to a cylinder. The radius of both the hemisphere and the cylinder is 6 cm. The height of the cylinder is 5 cm. Find the total surface area of the solid. Give your answer in terms of π.
Show solution
- The surface of the solid has three parts: the curved top of the hemisphere, the curved side of the cylinder and the flat circle at the bottom.
- The flat face where the hemisphere joins the cylinder is inside the solid, so do not count it.
- Hemisphere curved surface = half of 4πr² = 1⁄2 × 4π × 6² = 72π.
- Cylinder curved surface = 2πrh = 2 × π × 6 × 5 = 60π.
- Circular base = πr² = π × 6² = 36π.
- Total = 72π + 60π + 36π = 168π cm²
Browse IGCSE Maths questions by topic
Book a free trial class
Tell us a little about your requirement.