Edexcel International GCSE Maths Surface Area and Volume Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Geometry and Trigonometry

Surface Area and Volume: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every surface area and volume question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 10 questions worth 41 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

13 marksHigherJanuary 2023 · Paper 1HR (4MA1/1HR) · Q2

The diagram shows a solid triangular prism. The cross section is a right-angled triangle with sides 4.8 cm, 3.6 cm and 6 cm. The length of the prism is 7 cm. Work out the total surface area of the triangular prism. Give your answer correct to 3 significant figures.

4.8 cm3.6 cm6 cm7 cm
Show solution
  1. Two triangular ends: 2 × ½ × 4.8 × 3.6 = 17.28 cm2
  2. Three rectangles, each 7 cm long: (4.8 + 3.6 + 6) × 7 = 14.4 × 7 = 100.8 cm2
  3. Total = 17.28 + 100.8 = 118.08 cm2

Answer: 118 cm2

24 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q3

The diagram shows a container for water in the shape of a prism of length 125 cm. Its cross-section is 85 cm wide; it is 60 cm tall for the first 30 cm of its width and 40 cm tall for the rest (so the top 20 cm is only 30 cm wide). The rectangular base of the prism is horizontal. The container is completely full of water. Tuah is going to use a pump to empty the water from the container so that the volume of water in the container decreases at a constant rate. The pump starts to empty water from the container at 10 30 and at 12 00 the water level in the container has dropped by 20 cm. Find the time at which all the water has been pumped out of the container.

30 cm20 cm40 cm60 cm85 cmlength 125 cm
Show solution
  1. The top 20 cm of water sits only in the narrow part: 30 cm wide, 20 cm deep and 125 cm long. Volume pumped out in the first 1.5 hours = 30 × 20 × 125 = 75 000 cm3
  2. Cross-section area = 85 × 40 + 30 × 20 = 3400 + 600 = 4000 cm2, so the total volume = 4000 × 125 = 500 000 cm3
  3. The pump removes 75 000 cm3 every 90 minutes. Time for all the water = 500 000⁄75 000 × 90 = 600 minutes = 10 hours
  4. 10 30 + 10 hours

Answer: 20 30 (8.30 pm)

34 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q5

The diagram shows a cuboid and a cylinder. The dimensions of the cuboid are x cm by 12 cm by 5 cm. The volume of the cuboid is 270 cm3. The radius of the cylinder is x cm. The height of the cylinder is 2x cm.

(a) Work out the volume of the cylinder. Give your answer correct to the nearest whole number. (3)

(b) Change 1 m3 to cm3 (1)

Show solution

Part (a)

  1. Volume of the cuboid: 12 × 5 × x = 60x = 270, so x = 4.5
  2. Cylinder: radius 4.5 cm and height 2 × 4.5 = 9 cm. Volume = π × 4.52 × 9 = 572.55…

Answer: 573 cm3

Part (b)

  1. 1 m = 100 cm, so 1 m3 = 100 × 100 × 100 cm3

Answer: 1 000 000 cm3

44 marksHigherJune 2024 · Paper 1HR (4MA1/1HR) · Q6

The diagram shows a solid wooden cylinder with radius 8 cm and height h cm. The volume of the cylinder is 1208 cm3.

8 cmh cm

(a) Work out the value of h. Give your answer correct to the nearest whole number. (2)

The density of the wood is 1.25 g/cm3

(b) Work out the mass of the cylinder. Give your answer in kilograms. (2)

Show solution

Part (a)

  1. Volume of a cylinder = πr2h, so 1208 = π × 82 × h = 64πh
  2. h = 1208 ÷ (64π) = 6.008…

Answer: h = 6

Part (b)

  1. Mass = density × volume = 1.25 × 1208 = 1510 g
  2. There are 1000 g in 1 kg, so divide by 1000.

Answer: 1.51 kilograms

Tip: Use the exact volume 1208 cm3 from the question, not a volume worked out from the rounded height.
53 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q7

Here is a solid cylinder. The radius of the cylinder is 8 cm and the height is h cm. The volume of the cylinder is 3892 cm3. Work out the value of h. Give your answer correct to one decimal place.

Show solution
  1. Volume of a cylinder = πr2h, so 3892 = π × 82 × h = 64πh
  2. h = 3892 ÷ (64π) = 19.357…

Answer: h = 19.4

63 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q10

The diagram shows two water containers. One is a cuboid measuring 35 cm by 28 cm by 20 cm. The surface of the water in the cuboid is 9 cm above the base of the cuboid. The other is a cylinder with a radius of 10 cm and a height of 33 cm. The cylinder is completely full of water. Izzy is going to pour all the water from the cylinder into the cuboid. Show that the cuboid will not be completely full of water.

Show solution
  1. Volume of water in the cylinder = πr2h = π × 102 × 33 = 10 367.2… cm3
  2. Empty space left in the cuboid = 35 × 28 × (20 − 9) = 35 × 28 × 11 = 10 780 cm3
  3. The water from the cylinder (10 367 cm3) is less than the empty space (10 780 cm3).

Answer: 10 367 cm3 < 10 780 cm3, so the cuboid will not be full

Tip: Another way: water in the cuboid is 35 × 28 × 9 = 8820 cm3, and 8820 + 10 367 = 19 187 cm3, which is less than the cuboid’s full volume of 19 600 cm3.
75 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q20

The diagram shows a frustum of a cone and a sphere. The frustum is made by removing a small cone from a large cone. The cones are similar. The height of the small cone is h cm. The height of the large cone is 2h cm. The radius of the base of the large cone is r cm. The radius of the sphere is r cm. Given that the volume of the frustum is equal to the volume of the sphere, find an expression for r in terms of h. Give your expression in its simplest form.

Show solution
  1. The small cone has half the height of the large cone, so (being similar) it has half the radius: r⁄2.
  2. Large cone: ⅓πr2(2h) = 2⁄3πr2h
  3. Small cone: ⅓π(r⁄2)2h = 1⁄12πr2h
  4. Frustum = 2⁄3πr2h − 1⁄12πr2h = 8⁄12πr2h − 1⁄12πr2h = 7⁄12πr2h
  5. Set equal to the sphere: 7⁄12πr2h = 4⁄3πr3. Divide both sides by πr2: 7⁄12h = 4⁄3r
  6. r = 7⁄12h × 3⁄4 = 7⁄16h

Answer: r = 7h⁄16

85 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q22

A solid is made from a cone and a hemisphere. The circular plane face of the hemisphere coincides with the circular base of the cone. The radius of the hemisphere and the radius of the circular base of the cone are both 20 cm. The curved surface area of the cone is 580π cm2. The volume of the solid is kπ cm3. Work out the exact value of k.

20 cm
Show solution
  1. Curved surface area of a cone = πrl, where l is the slant height: π × 20 × l = 580π, so l = 29 cm
  2. The height, radius and slant height form a right-angled triangle: h = √292 − 202 = √841 − 400 = √441 = 21 cm
  3. Volume of the cone = ⅓πr2h = ⅓ × π × 400 × 21 = 2800π
  4. Volume of the hemisphere = 2⁄3πr3 = 2⁄3 × π × 8000 = 16 000⁄3π
  5. Total = 2800π + 16 000⁄3π = 8400 + 16 000⁄3π = 24 400⁄3π

Answer: k = 24 400⁄3 (= 81331⁄3)

95 marksHigherJune 2023 · Paper 1H (4MA1/1H) · Q23

A frustum is made by removing a small square-based pyramid from a similar large square-based pyramid. The height of the small pyramid is 15 cm. The height of the large pyramid is 45 cm. The square base of the large pyramid has side length 39 cm. Work out the total surface area of the frustum. Give your answer correct to the nearest whole number.

39 cm39 cm15 cm45 cm
Show solution
  1. The small pyramid is 15⁄45 = 1⁄3 the size of the large one, so the top square has side 39 ÷ 3 = 13 cm.
  2. Each sloping face of the frustum is a large triangle minus a small triangle. We need the slant height of a face (the height of the triangle up the middle of the face).
  3. Large pyramid: the slant height goes from the middle of a base edge (19.5 cm from the centre) to the top (45 cm up): l = √452 + 19.52 = 49.043… cm. The small pyramid’s slant height is a third of this: 16.347… cm.
  4. Area of one sloping face = ½ × 39 × 49.043… − ½ × 13 × 16.347… = 956.34… − 106.26… = 850.08… cm2
  5. Total = base + top + 4 faces = 392 + 132 + 4 × 850.08… = 1521 + 169 + 3400.3…

Answer: 5090 cm2

Tip: “Total” surface area includes the square top and the square base, not just the four sloping faces.
105 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q23

A solid shape is made by removing a hemisphere from a cone. The radius of the hemisphere is 2x cm, the radius of the base of the cone is 5x cm and the vertical height of the cone is 6x cm. The volume of the solid shape is 6948π cm3. Work out the total surface area of the solid hemisphere that has been removed from the cone. Give your answer correct to the nearest integer.

Show solution
  1. Volume of cone = ⅓πr2h = ⅓ × π × (5x)2 × 6x = 50πx3
  2. Volume of hemisphere = ½ × 4⁄3πr3 = 2⁄3 × π × (2x)3 = 16⁄3πx3
  3. Volume of solid = 50πx3 − 16⁄3πx3 = 134⁄3πx3
  4. Set equal to 6948π and cancel π: 134⁄3x3 = 6948, so x3 = 6948 × 3 ÷ 134 = 155.55…
  5. x = ∛155.55… = 5.378… so the hemisphere radius 2x = 10.756… cm
  6. A solid hemisphere has a curved surface (half a sphere, 2πr2) and a flat circular face (πr2). Total = 3πr2.
  7. 3 × π × 10.756…2 = 1090.39…

Answer: 1090 cm2

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