3D Shapes and Volume · Study Notes · Edexcel IGCSE 4MA1 Higher

Volumes and surface areas of cuboids, prisms, cylinders, pyramids, cones, spheres. Memorise the formulae.

At a glance

Cuboid: $V = lwh$ .
Prism: $V =$ cross-section area $\times l$ .
Cylinder: $V = \pi r^{2} h$ .
Pyramid/cone: $V = \dfrac{1}{3} \times$ base × height.
Sphere: $V = \dfrac{4}{3}\pi r^{3}$ , $SA = 4\pi r^{2}$ .
Cone curved surface: $\pi r l$ (slant height).
Volume in cubes ( $\text{cm}^{3}$ , $\text{m}^{3}$ ).

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.5 — Find volumes of cuboids, prisms, cylinders, pyramids, cones, spheres.
4.5 — Find surface areas (including formulae and net method).
4.5 — Use Pythagoras to find dimensions in 3D.

Prisms (including cuboids and cylinders)

▼

$V =$ cross-section area × length. Uniform along length.

Prism = uniform cross-section.

General formula: $V =$ area of cross-section $\times l$ .

Specific:

Shape	Volume
Cube	$s^{3}$
Cuboid	$lwh$
Triangular prism	$\dfrac{1}{2}b h_{\triangle} \times l$
Cylinder	$\pi r^{2} h$

Surface area of cylinder: $2$ circles + curved (rectangle when unrolled).

$SA_{\text{cylinder}} = 2\pi r^{2} + 2\pi r h$ .

The curved surface unrolls to a rectangle: width = circumference 2πr, height = h.

Worked qualitative. Why is the curved surface of a cylinder a rectangle?

Imagine cutting and unrolling.
Width: circumference $2\pi r$ .
Height: $h$ .
Area: $2\pi r \times h = 2\pi r h$ .

Edexcel tip. Always state UNITS for volume (cm³, m³). Mark schemes deduct for using cm².

$V =$ cross-section × length.
Cuboid: $lwh$ .
Cylinder: $\pi r^{2} h$ .
Cylinder SA: $2\pi r^{2} + 2\pi r h$ .

Pyramids and cones

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$V = \dfrac{1}{3} \times$ base × height. Cone curved surface uses slant.

Pyramid volume: $V = \dfrac{1}{3} \times$ base area $\times h$ .

The $\dfrac{1}{3}$ is universal for pyramids/cones — exactly $\dfrac{1}{3}$ of the bounding prism/cylinder.

Cone-specific:

$V_{\text{cone}} = \dfrac{1}{3}\pi r^{2} h$

$SA_{\text{cone}} = \pi r^{2} + \pi r l$

where $h$ = perpendicular height, $l$ = SLANT height.

$h, r, l$ form a right triangle: $h^{2} + r^{2} = l^{2}$ (Pythagoras).

Perpendicular height h, radius r and slant height l form a right triangle.

Worked example. Cone with $r = 5$ , $l = 13$ .

Find $h$ : $h^{2} = 169 - 25 = 144$ , $h = 12$ .
$V = \dfrac{1}{3}\pi \times 25 \times 12 = 100\pi$ .
Curved surface: $\pi \times 5 \times 13 = 65\pi$ .
Total SA: $25\pi + 65\pi = 90\pi$ .

Worked qualitative. Why is cone volume $\dfrac{1}{3}$ of cylinder volume (same $r, h$ )?

Cylinder: $\pi r^{2} h$ .
Cone fits inside it.
Calculus shows the cone is exactly $\dfrac{1}{3}$ .
Geometrically: cones with the same base and apex tapered uniformly.

Edexcel tip. Pyramids/cones use PERPENDICULAR height for VOLUME, SLANT height for CURVED SURFACE. Don't confuse.

$V = \dfrac{1}{3} \times$ base × height.
Cone: $\dfrac{1}{3}\pi r^{2} h$ .
Curved surface: $\pi r l$ .
Use Pythagoras for $h, r, l$ .

Spheres

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$V = \dfrac{4}{3}\pi r^{3}$ , $SA = 4\pi r^{2}$ .

Sphere formulas:

$V = \dfrac{4}{3}\pi r^{3}$ $SA = 4\pi r^{2}$

Both use the radius $r$ .

A hemisphere is exactly half a sphere; its surface adds a flat circular base.

Hemisphere (half a sphere):

$V_{\text{hemi}} = \dfrac{2}{3}\pi r^{3}$ .
$SA_{\text{hemi}} = 2\pi r^{2}$ (curved) plus $\pi r^{2}$ (circular base) = $3\pi r^{2}$ .

Worked example. Sphere with $r = 6$ .

Volume: $\dfrac{4}{3}\pi \times 216 = 288\pi$ cm³.
Surface: $4\pi \times 36 = 144\pi$ cm².

Worked qualitative. Why is sphere SA $4\pi r^{2}$ , exactly $4 \times$ area of equator circle?

Beautiful result from calculus.
Beyond IGCSE proof — but memorise.

Edexcel tip. When the question says 'in terms of $\pi$ ', leave answer as $288\pi$ , not the decimal.

$V = \dfrac{4}{3}\pi r^{3}$ .
$SA = 4\pi r^{2}$ .
Hemisphere: half + circular base.
Both use $r$ .

3D Pythagoras

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Used for finding diagonals of cuboids and other 3D distances.

3D Pythagoras for the long diagonal of a cuboid:

If a cuboid has sides $a, b, c$ , the long diagonal $d$ satisfies:

$d^{2} = a^{2} + b^{2} + c^{2}$

Why? First, the floor diagonal: $d_{\text{floor}}^{2} = a^{2} + b^{2}$ (2D Pythagoras).

Then, considering the long diagonal as the hypotenuse of a right triangle with sides $d_{\text{floor}}$ and $c$ (height):

$d^{2} = d_{\text{floor}}^{2} + c^{2} = a^{2} + b^{2} + c^{2}$ .

Pythagoras twice: first the floor diagonal, then combine with height c.

Worked example. Cuboid $3 \times 4 \times 12$ . Long diagonal?

$d^{2} = 9 + 16 + 144 = 169$ .
$d = 13$ .

Worked qualitative. What's the diagonal of a unit cube ( $1 \times 1 \times 1$ )?

$d^{2} = 1 + 1 + 1 = 3$ .
$d = \sqrt{3} \approx 1.732$ .

Edexcel tip. 3D Pythagoras applies wherever you can find a right triangle in 3D. Often combined with cones or other shapes.

$d^{2} = a^{2} + b^{2} + c^{2}$ for cuboid diagonal.
Apply 2D Pythagoras twice.
Find right triangles within the 3D shape.
Common in cone problems too ( $h, r, l$ ).

Quick recap

Prisms: $V =$ cross-section × length.
Pyramids/cones: $V = \dfrac{1}{3} \times$ base × height.
Sphere: $V = \dfrac{4}{3}\pi r^{3}$ , $SA = 4\pi r^{2}$ .
Cone curved surface: $\pi r l$ (slant).
3D Pythagoras: $d^{2} = a^{2} + b^{2} + c^{2}$ .

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Cuboid: $V = lwh$ .
Cylinder: $V = \pi r^{2} h$ .
Pyramid/cone: $V = \dfrac{1}{3} \times$ base × height.
Sphere: $V = \dfrac{4}{3}\pi r^{3}$ , $SA = 4\pi r^{2}$ .
Cone curved surface: $\pi r l$ .

How it’s examined

3D shapes appear every Higher Tier paper (5-8 marks). Often combined with Pythagoras or scale factors. Examiner reports flag (1) using slant for volume, (2) forgetting $\dfrac{1}{3}$ for pyramids/cones, (3) wrong units (cm² vs cm³).

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on 3d shapes and volume, written exactly the way a tutor would explain them at the board.

1Volume of a cuboid
Foundation• cuboid
▼
Question
Cuboid: $5 \times 3 \times 4$ cm. Volume?
Step-by-step solution
1. Step 1
  $V = \text{length} \times \text{width} \times \text{height}$ .
2. Step 2
  $V = 5 \times 3 \times 4 = 60$ cm³.
Answer
$60$ cm³
2Cylinder volume and surface area
Higher• Adapted from 4MA1/1H May/Jun 2024 Q13• cylinder
▼
Question
Cylinder: radius $4$ cm, height $10$ cm. Find (a) volume, (b) total surface area in terms of $\pi$ .
Step-by-step solution
1. Step 1
  (a) $V = \pi r^{2} h = \pi \times 16 \times 10 = 160\pi$ cm³.
2. Step 2
  (b) Surface area: 2 circles + curved surface (rectangle when unrolled).
3. Step 3
  Two circles: $2 \times \pi r^{2} = 2 \times 16\pi = 32\pi$ .
4. Step 4
  Curved: $2\pi r h = 2\pi \times 4 \times 10 = 80\pi$ .
5. Step 5
  Total: $32\pi + 80\pi = 112\pi$ cm².
Answer
(a) $V = 160\pi$ cm³ (b) Total SA $= 112\pi$ cm²
3Cone volume and surface area
Higher• cone
▼
Question
Cone: radius $5$ cm, slant height $13$ cm. Find (a) perpendicular height, (b) volume, (c) total surface area. In terms of $\pi$ .
Step-by-step solution
1. Step 1
  (a) Pythagoras: $h^{2} + r^{2} = l^{2}$ . $h^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$ . $h = 12$ .
2. Step 2
  (b) $V = \dfrac{1}{3}\pi r^{2} h = \dfrac{1}{3}\pi \times 25 \times 12 = 100\pi$ cm³.
3. Step 3
  (c) Surface: base circle $\pi r^{2} = 25\pi$ . Curved: $\pi r l = 5 \times 13 \times \pi = 65\pi$ .
4. Step 4
  Total: $25\pi + 65\pi = 90\pi$ cm².
Answer
(a) $h = 12$ cm (b) $V = 100\pi$ (c) SA $= 90\pi$ cm²
Examiner tip
Cone volume uses PERPENDICULAR height. Curved surface uses SLANT height. Edexcel mark schemes are precise about this.
4Sphere
Higher• Adapted from 4MA1/2H Jan 2024 Q14• sphere
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Question
Sphere of radius $6$ cm. Find (a) volume, (b) surface area. In terms of $\pi$ .
Step-by-step solution
1. Step 1
  (a) $V = \dfrac{4}{3}\pi r^{3} = \dfrac{4}{3}\pi \times 216 = 288\pi$ cm³.
2. Step 2
  (b) $SA = 4\pi r^{2} = 4\pi \times 36 = 144\pi$ cm².
Answer
(a) $V = 288\pi$ (b) SA $= 144\pi$ cm²
5Prism volume
Foundation• prism
▼
Question
Triangular prism: triangle base $6$ cm × height $4$ cm; prism length $10$ cm. Volume?
Step-by-step solution
1. Step 1
  Cross-sectional area (triangle): $\dfrac{1}{2} \times 6 \times 4 = 12$ cm².
2. Step 2
  Volume = cross-section × length: $12 \times 10 = 120$ cm³.
Answer
$120$ cm³
6Edge of a cube from its volume [1 mark]
Getting started• Paper 1H/2H style — 1 mark• cube, volume
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Question
A cube has a volume of $343$ cm³.

Write down the length of one edge of the cube. [1]
Step-by-step solution
1. Step 1
  The volume of a cube is the edge length cubed, so the edge is the cube root of the volume.
  $e^{3} = 343$
2. Step 2
  B1 $7 \times 7 \times 7 = 343$ .
  $e = \sqrt[3]{343} = 7$
Answer
$7$ cm
Examiner tip
Examiner tip: Learn the cubes up to $10^{3} = 1000$ . They appear in volume questions without a calculator prompt.

Common mistake: Dividing by $3$ to get $114.3$ , or taking the square root to get $18.5$ . The inverse of cubing is the cube root.
7Area of the label on a tin [2 marks]
Getting started• Paper 1H/2H style — 2 marks• cylinder, surface area
▼
Question
A tin is a cylinder with diameter $7.4$ cm and height $10.5$ cm. A paper label covers the whole curved surface of the tin, with no overlap.

Calculate the area of the label.

Give your answer correct to $3$ significant figures. [2]
Step-by-step solution
1. Step 1
  M1 The label unrolls into a rectangle whose width is the circumference. The curved surface area $2\pi r h$ is on the formula sheet; the radius is $7.4 \div 2 = 3.7$ cm.
  $2\pi \times 3.7 \times 10.5$
2. Step 2
  A1
  $244.10\ldots \approx 244$
Answer
$244$ cm²
Examiner tip
Examiner tip: A label covers the curved surface only. Read the question to decide which surfaces are wanted before you reach for a formula.

Common mistake: Using the diameter as the radius, which doubles the answer to $488$ cm², or adding the two circular ends when the label does not cover them.
8Filling conical cups from a bottle [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• cone, capacity, litres
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Question
A paper cup is in the shape of a cone with radius $3.5$ cm and perpendicular height $9$ cm.

A bottle holds $2$ litres of water.

Work out the greatest number of cups that can be filled completely from the bottle. [3]
Step-by-step solution
1. Step 1
  M1 Volume of one cup, from $V = \tfrac{1}{3}\pi r^{2} h$ on the formula sheet.
  $\tfrac{1}{3}\pi \times 3.5^{2} \times 9 = 115.45\ldots$
2. Step 2
  M1 $2$ litres $= 2000$ cm³. Divide by the volume of one cup.
  $2000 \div 115.45\ldots = 17.32\ldots$
3. Step 3
  A1 Only complete cups count, so round down.
  $17$
Answer
$17$ cups
Examiner tip
Examiner tip: Decide the direction of rounding from the context. "Filled completely" means the eighteenth cup cannot be filled, so round down.

Common mistake: Leaving out the $\tfrac{1}{3}$ , which gives $346.3\ldots$ cm³ and $5$ cups, or dividing $2$ by the volume without converting litres to cm³.
9Mass of a metal pipe [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• cylinder, density
▼
Question
A metal pipe is a hollow cylinder of length $60$ cm. The outer radius is $5$ cm and the inner radius is $4$ cm.

The density of the metal is $7.9$ g/cm³.

Work out the mass of the pipe in kilograms.

Give your answer correct to $3$ significant figures. [4]
Step-by-step solution
1. Step 1
  M1 The cross-section is a ring: the outer circle minus the inner circle.
  $\pi \times 5^{2} - \pi \times 4^{2} = 25\pi - 16\pi = 9\pi$
2. Step 2
  M1 Volume of metal = cross-section $\times$ length.
  $9\pi \times 60 = 540\pi = 1696.46\ldots$
3. Step 3
  M1 mass $=$ density $\times$ volume, in grams.
  $7.9 \times 1696.46\ldots = 13\,402.0\ldots$
4. Step 4
  A1 Divide by $1000$ to change grams to kilograms.
  $13\,402.0\ldots \div 1000 \approx 13.4$
Answer
$13.4$ kg
Examiner tip
Examiner tip: Subtract the areas, not the radii. The ring has area $\pi(5^{2} - 4^{2})$ , which is not $\pi \times 1^{2}$ .

Common mistake: Using $\pi \times (5 - 4)^{2} \times 60$ , which gives a volume about nine times too small, or giving $13\,400$ when the question asks for kilograms.
10A sphere that fits exactly in a cylinder [5 marks]
Stretch• Paper 1H/2H style — 5 marks• sphere, cylinder, algebra
▼
Question
A sphere of radius $r$ fits exactly inside a closed cylinder. The sphere touches the curved surface and both ends of the cylinder.

(a) Show that the volume of the sphere is $\dfrac{2}{3}$ of the volume of the cylinder. [3]

(b) The radius of the sphere is $5$ cm. Work out the volume of the space inside the cylinder that is not filled by the sphere. Give your answer correct to $3$ significant figures. [2]
Step-by-step solution
1. Step 1
  B1 (a) The cylinder has the same radius as the sphere, and its height is the diameter of the sphere, $2r$ .
  $V_{\text{cyl}} = \pi r^{2} \times 2r = 2\pi r^{3}$
2. Step 2
  M1 Divide the volume of the sphere by the volume of the cylinder.
  $\dfrac{\tfrac{4}{3}\pi r^{3}}{2\pi r^{3}}$
3. Step 3
  A1 Cancel $\pi r^{3}$ to complete the argument.
  $\dfrac{4}{3} \div 2 = \dfrac{4}{6} = \dfrac{2}{3}$
4. Step 4
  M1 (b) The space is $\tfrac{1}{3}$ of the cylinder, which has volume $2\pi \times 5^{3} = 250\pi$ .
  $250\pi - \tfrac{4}{3}\pi \times 5^{3} = \tfrac{1}{3} \times 250\pi$
5. Step 5
  A1
  $\tfrac{250}{3}\pi = 261.79\ldots \approx 262$
Answer
(a) Shown: $\dfrac{4}{3}\pi r^{3} \div 2\pi r^{3} = \dfrac{2}{3}$ (b) $262$ cm³
Examiner tip
Examiner tip: In a "show that" question the final mark is for a complete argument. State the height of the cylinder as $2r$ and show the cancelling; writing $\dfrac{2}{3}$ at the end with no algebra scores nothing.

Common mistake: Taking the height of the cylinder as $r$ , which makes the sphere larger than the cylinder. In part (b), subtracting the radii or rounding $250\pi$ and $\tfrac{500}{3}\pi$ before subtracting.

Question 1
Paper 1H/2H style — 1 mark1 mark
Change $45\,000$ mm³ into cm³. [1]
Model answer
$1\ \text{cm}^{3} = 10 \times 10 \times 10 = 1000\ \text{mm}^{3}$

$45\,000 \div 1000 = 45$

Answer: $45$ cm³
Why this scores
Mark scheme:
- B1 for $45$ cao
Why this scores full marks: The candidate uses $1$ cm $= 10$ mm in each of the three directions, so $1$ cm³ $= 10^{3} = 1000$ mm³, and divides because a cm³ is the larger unit. Weaker scripts divide by $10$ and give $4500$ , divide by $100$ and give $450$ , or multiply.
Question 2
Paper 1H/2H style — 2 marks2 marks
A solid pyramid has a square base of side $6$ cm. The vertical height of the pyramid is $11$ cm.

Work out the volume of the pyramid. [2]
Model answer
$V = \tfrac{1}{3} \times 6^{2} \times 11 = \tfrac{1}{3} \times 396 = 132$

Answer: $132$ cm³
Why this scores
Mark scheme:
- M1 for $\tfrac{1}{3} \times 36 \times 11$ oe
- A1 for $132$ cao
Why this scores full marks: The candidate uses the area of the base, $6^{2}$ , and the vertical height, then multiplies by $\tfrac{1}{3}$ . The pyramid formula is not on the formula sheet, so it has to be remembered. Weaker scripts give $396$ by forgetting the $\tfrac{1}{3}$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
A block of wood is a cuboid measuring $8$ cm by $5$ cm by $4$ cm. The mass of the block is $112$ g.

Work out the density of the wood. [2]
Model answer
$V = 8 \times 5 \times 4 = 160\ \text{cm}^{3}$

$\text{density} = \dfrac{112}{160} = 0.7$

Answer: $0.7$ g/cm³
Why this scores
Mark scheme:
- M1 for $112 \div (8 \times 5 \times 4)$ oe
- A1 for $0.7$ oe
Why this scores full marks: The volume is found first, then density $=$ mass $\div$ volume. Weaker scripts divide the wrong way, giving $1.43$ , or divide by one face area.
Question 4
Paper 1H/2H style — 3 marks3 marks
A closed box is a cuboid measuring $12$ cm by $7$ cm by $4.5$ cm.

Work out the total surface area of the box. [3]
Model answer
$12 \times 7 = 84 \qquad 12 \times 4.5 = 54 \qquad 7 \times 4.5 = 31.5$

$2 \times (84 + 54 + 31.5) = 2 \times 169.5 = 339$

Answer: $339$ cm²
Why this scores
Mark scheme:
- M1 for the area of one face, e.g. $12 \times 7$
- M1 for $2 \times (84 + 54 + 31.5)$ oe, all six faces
- A1 for $339$ cao
Why this scores full marks: The three different faces are listed, then doubled, because opposite faces of a cuboid are equal. Weaker scripts add only three faces ( $169.5$ ) or work out the volume, $378$ cm³.
Question 5
Paper 1H/2H style — 3 marks3 marks
A cylinder has height $12$ cm and volume $850$ cm³.

Calculate the diameter of the cylinder.

Give your answer correct to $3$ significant figures. [3]
Model answer
$\pi r^{2} \times 12 = 850$

$r^{2} = \dfrac{850}{12\pi} = 22.546\ldots \qquad r = \sqrt{22.546\ldots} = 4.7483\ldots$

$d = 2 \times 4.7483\ldots = 9.4967\ldots$

Answer: $9.50$ cm
Why this scores
Mark scheme:
- M1 for $\pi r^{2} \times 12 = 850$ oe
- M1 for $r = \sqrt{\dfrac{850}{12\pi}}$ oe ( $4.748\ldots$ )
- A1 for $9.49$ to $9.50$
Why this scores full marks: The equation is set up from $V = \pi r^{2} h$ on the formula sheet and solved one step at a time, keeping the unrounded $r$ before doubling it. Weaker scripts stop at the radius, $4.75$ , when the question asks for the diameter, or leave out $\pi$ and get $\sqrt{850 \div 12} = 8.42$ for the radius.
Question 6
Paper 1H/2H style — 3 marks3 marks
A sphere has volume $500$ cm³.

Calculate the radius of the sphere.

Give your answer correct to $3$ significant figures. [3]
Model answer
$\tfrac{4}{3}\pi r^{3} = 500$

$r^{3} = \dfrac{500 \times 3}{4\pi} = 119.36\ldots$

$r = \sqrt[3]{119.36\ldots} = 4.923\ldots$

Answer: $4.92$ cm
Why this scores
Mark scheme:
- M1 for $\tfrac{4}{3}\pi r^{3} = 500$
- M1 for $r^{3} = \dfrac{1500}{4\pi}$ oe
- A1 for $4.92$ to $4.924$
Why this scores full marks: The equation is rearranged one step at a time, and the cube root is taken of the unrounded value $119.36\ldots$ . Weaker scripts take a square root, giving $10.9$ , or divide by $\tfrac{4}{3}$ and forget the $\pi$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
A bowl is a hemisphere with internal radius $9$ cm. The bowl is full of water.

All the water is poured into an empty cylindrical jug. The jug has internal radius $6$ cm and height $12$ cm.

Show that the water overflows, and work out the volume of water that overflows. Give your answer correct to $3$ significant figures. [4]
Model answer
$\text{Bowl} = \tfrac{1}{2} \times \tfrac{4}{3}\pi \times 9^{3} = 486\pi$

$\text{Jug} = \pi \times 6^{2} \times 12 = 432\pi$

$486\pi > 432\pi$ , so the water overflows.

$486\pi - 432\pi = 54\pi = 169.64\ldots$

Answer: $170$ cm³
Why this scores
Mark scheme:
- M1 for $\tfrac{1}{2} \times \tfrac{4}{3}\pi \times 9^{3}$ oe
- M1 for $\pi \times 6^{2} \times 12$ oe
- M1 for a comparison of the two volumes with a conclusion, or for $486\pi - 432\pi$
- A1 for $169.6$ to $170$
Why this scores full marks: Both volumes are kept in terms of $\pi$ , so the comparison $486\pi > 432\pi$ is exact and the conclusion is written in words. Weaker scripts use the whole sphere, $972\pi$ , or compare numbers without saying what they show.
Question 8
Paper 1H/2H style — 4 marks4 marks
A closed cylindrical tank has radius $1.2$ m and height $3$ m. The whole outside surface of the tank, including both ends, is to be painted.

One litre of paint covers $8$ m². Paint is sold in tins of $2.5$ litres.

Work out the least number of tins of paint needed. [4]
Model answer
$2\pi \times 1.2 \times 3 + 2 \times \pi \times 1.2^{2} = 7.2\pi + 2.88\pi = 10.08\pi = 31.667\ldots\ \text{m}^{2}$

$31.667\ldots \div 8 = 3.958\ldots \text{ litres}$

$3.958\ldots \div 2.5 = 1.58\ldots$

Answer: $2$ tins
Why this scores
Mark scheme:
- M1 for $2\pi \times 1.2 \times 3$ or $2 \times \pi \times 1.2^{2}$
- M1 for a complete method for the total surface area, $31.66\ldots$
- M1 for their area $\div 8 \div 2.5$ oe
- A1 for $2$ from correct working
Why this scores full marks: Both ends and the curved surface are included, and the number of tins is rounded up because $1$ tin is not enough. Weaker scripts leave out the ends, or give $1.58$ or $1$ as the answer.
Question 9
Paper 1H/2H style — 5 marks5 marks
A solid cube has edges of length $10$ cm. A hemisphere of radius $3$ cm is cut out of the cube. The flat face of the hemisphere lies in the top face of the cube, with its centre at the centre of the top face.

(a) Work out the volume of the solid that is left. Give your answer correct to $3$ significant figures. [2]

(b) Work out the total surface area of the solid that is left. Give your answer correct to $3$ significant figures. [3]
Model answer
(a)

$10^{3} - \tfrac{1}{2} \times \tfrac{4}{3}\pi \times 3^{3} = 1000 - 18\pi = 943.45\ldots$

Answer: $943$ cm³

(b) The top face loses a circle of radius $3$ cm, and the curved inside of the hemisphere is added.

$6 \times 10^{2} - \pi \times 3^{2} = 600 - 9\pi$

$\text{Curved surface of hemisphere} = \tfrac{1}{2} \times 4\pi \times 3^{2} = 18\pi$

$600 - 9\pi + 18\pi = 600 + 9\pi = 628.27\ldots$

Answer: $628$ cm²
Why this scores
Mark scheme:
- M1 for $10^{3} - \tfrac{1}{2} \times \tfrac{4}{3}\pi \times 3^{3}$ oe
- A1 for $943$ to $943.5$
- M1 for $6 \times 10^{2} - \pi \times 3^{2}$ oe
- M1 for $\tfrac{1}{2} \times 4\pi \times 3^{2}$ oe added to a surface area of the cube
- A1 for $628$ to $628.3$
Why this scores full marks: The candidate treats the two parts differently. Cutting the hole removes volume, but for the surface it removes a flat circle and adds a curved bowl that is twice as large, so the surface area goes up. Weaker scripts subtract the hemisphere in part (b) as well, giving $600 - 18\pi = 543.5\ldots$ , or forget to remove the circle and give $600 + 18\pi = 656.5\ldots$ .
Question 10
Paper 1H/2H style — 6 marks6 marks
A solid hemisphere and a solid cone each have radius $r$ cm.

The total surface area of the hemisphere is equal to the total surface area of the cone.

The volume of the cone is $72\sqrt{3}\,\pi$ cm³.

Find the value of $r$ . Show clear algebraic working. [6]
Model answer
Hemisphere: $\tfrac{1}{2} \times 4\pi r^{2} + \pi r^{2} = 3\pi r^{2}$

Cone, slant height $l$ : $\pi r^{2} + \pi r l$

$\pi r^{2} + \pi r l = 3\pi r^{2} \quad \Rightarrow \quad \pi r l = 2\pi r^{2} \quad \Rightarrow \quad l = 2r$

$h = \sqrt{(2r)^{2} - r^{2}} = \sqrt{3r^{2}} = \sqrt{3}\,r$

$\tfrac{1}{3}\pi r^{2}(\sqrt{3}\,r) = 72\sqrt{3}\,\pi \quad \Rightarrow \quad \dfrac{\sqrt{3}}{3} r^{3} = 72\sqrt{3} \quad \Rightarrow \quad r^{3} = 216$

Answer: $r = 6$
Why this scores
Mark scheme:
- M1 for $3\pi r^{2}$ oe for the hemisphere
- M1 for $\pi r^{2} + \pi r l = 3\pi r^{2}$ oe
- A1 for $l = 2r$
- M1 for $h = \sqrt{(2r)^{2} - r^{2}}$ oe
- M1 for $\tfrac{1}{3}\pi r^{2} \times$ their $h = 72\sqrt{3}\,\pi$
- A1 for $r = 6$ cao
Why this scores full marks: The flat circular face of the hemisphere is included, so its total surface is $3\pi r^{2}$ , not $2\pi r^{2}$ . Brackets go round $(2r)^{2}$ , and $\pi$ and $\sqrt{3}$ cancel so the equation reduces to $r^{3} = 216$ . Weaker scripts use $4\pi r^{2}$ for the hemisphere, or write $2r^{2}$ for $(2r)^{2}$ and cannot reach $r = 6$ .

4MA1 3D Shapes and Volume

At a glance

What you’ll learn

Prisms (including cuboids and cylinders)

Pyramids and cones

Spheres

3D Pythagoras

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Volume of a cuboid

2Cylinder volume and surface area

3Cone volume and surface area

4Sphere

5Prism volume

6Edge of a cube from its volume [1 mark]

7Area of the label on a tin [2 marks]

8Filling conical cups from a bottle [3 marks]

9Mass of a metal pipe [4 marks]

10A sphere that fits exactly in a cylinder [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10