Properties of 3D shapes - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 20 exam-style questions on Properties of 3D shapes for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

20 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Write down the number of faces, the number of edges and the number of vertices of a pentagonal pyramid.

Model answer

Faces: 1 pentagon + 5 triangles = 6
Edges: 5 round the base + 5 to the apex = 10
Vertices: 5 on the base + 1 apex = 6

Explanation

A pyramid has a base and one triangular face for each side of the base, all meeting at the apex. For an n-sided base there are n + 1 faces, 2n edges and n + 1 vertices; with n = 5 this gives 6, 10 and 6.

Mark scheme

1 mark
B1 for 6 faces and 6 vertices
correct answer with no working scores full marks
1 mark
B1 for 10 edges

Question 2

3 marks

A polyhedron has 20 faces.
Every face is a triangle.
Each edge is shared by exactly two faces.

For any polyhedron, the numbers of faces F, vertices V and edges E satisfy
\[F + V - E = 2\]

Work out the number of edges and the number of vertices of the polyhedron.

Model answer

Each of the 20 faces has 3 edges, but each edge belongs to 2 faces:
\(E = \dfrac{20 \times 3}{2} = 30\)
\(F + V - E = 2\)
\(20 + V - 30 = 2\)
V = 12
30 edges and 12 vertices.

Explanation

Counting 3 edges for each face counts every edge twice, because each edge is shared by two faces, so divide by 2. Then use Euler's formula to find V. This solid is a regular icosahedron. A common mistake is to give 60 edges.

Mark scheme

1 mark
M1 for \(E = \dfrac{20 \times 3}{2}\)
correct answer with no working scores full marks
1 mark
A1 for E = 30
1 mark
B1 for V = 12
ft their E: V = 2 + E − 20

Question 3

3 marks

A sphere of radius r cm fits exactly inside a cylinder.
The sphere touches both ends of the cylinder and the curved surface of the cylinder.

Show that the curved surface area of the cylinder is equal to the surface area of the sphere.

(Surface area of a sphere \(= 4\pi r^2\))

Model answer

The sphere touches the curved surface, so the cylinder has radius r.
The sphere touches both ends, so the height of the cylinder is the diameter, 2r.
Curved surface area of cylinder \(= 2\pi r \times h = 2\pi r \times 2r = 4\pi r^2\)
Surface area of sphere \(= 4\pi r^2\)
So the two areas are equal.

Explanation

The curved surface of a cylinder unrolls into a rectangle of length \(2\pi r\) (the circumference) and width h. Here the sphere fixes the radius as r and the height as 2r. This surprising result was discovered by Archimedes.

Mark scheme

1 mark
B1 for the cylinder has radius r and height 2r
1 mark
M1 for curved surface area of cylinder \(= 2\pi r \times 2r\)
1 mark
C1 for \(= 4\pi r^2\), which is the surface area of the sphere
must state the conclusion

Question 4

4 marks

A sphere has radius 6 cm.
A solid cylinder has the same total surface area as the sphere.
The height of the cylinder is equal to its diameter.

Work out the radius of the cylinder.
Give your answer correct to 3 significant figures.

(Surface area of a sphere \(= 4\pi r^2\))

Model answer

Surface area of sphere \(= 4 \times \pi \times 6^2 = 144\pi\) cm²
Let the radius of the cylinder be r cm; its height is 2r.
Total surface area of cylinder \(= 2 \times \pi r^2 + 2\pi r \times 2r = 2\pi r^2 + 4\pi r^2 = 6\pi r^2\)
\(6\pi r^2 = 144\pi\)
\(r^2 = 24\)
\(r = 4.8989...\)
Radius = 4.90 cm (3 s.f.)

Explanation

A solid cylinder has three surfaces: two circular ends and one curved surface. With height 2r its total surface area is \(6\pi r^2\). Setting this equal to the sphere's area lets \(\pi\) cancel. A common mistake is to use only the curved surface, which gives r = 6.

Mark scheme

1 mark
P1 for the surface area of the sphere \(= 4 \times \pi \times 6^2\) (= \(144\pi\) = 452.3...)
correct answer with no working scores full marks
1 mark
P1 for the total surface area of the cylinder \(= 2\pi r^2 + 2\pi r \times 2r\) (= \(6\pi r^2\))
must include both circular ends
1 mark
P1 for \(6\pi r^2 = 144\pi\) or \(r^2 = 24\)
1 mark
A1 for 4.90 cm
awrt 4.90 (4.898...); accept \(\sqrt{24}\) or \(2\sqrt{6}\)

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