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Spheres and cones

Free Printable Spheres and cones PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 5)

Download free spheres and cones worksheets (PDF), GCSE maths past paper questions and worked solutions for Spheres and cones (5S). This is a Grade 5 geometry topic from the GCSE questions by topic collection for Foundation and Higher tier crossover GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Maths4Everyone, Maths4EveryOne, Corbett Maths, 1st Class Maths, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our spheres and cones revision notes and worked examples with diagrams, try real GCSE past paper spheres and cones questions with step-by-step solutions, or download a PDF of all spheres and cones worksheet links.

Free Spheres and cones PDF Worksheets & Answers

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GCSE Maths Grade 5 Revision Notes & Method

Guide to Spheres and cones: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

For a sphere of radius r: Volume = 43πr3 and Surface Area = 4πr2. For a hemisphere, halve the volume (23πr3) and add the flat circular base for total surface area (2πr2 + πr2 = 3πr2).

For a cone of radius r, perpendicular height h, and slant height l: Volume = 13πr2h and Total Surface Area = πrl + πr2. Use Pythagoras (l2 = r2 + h2) to switch between h and l.

Example 1: Find the volume of a sphere with diameter 12 cm. Give your answer in terms of π and to 3 s.f.

Solution:

Radius r = 12 ÷ 2 = 6 cm
V = 43π(63) = 288π cm3
Ans: 288π (905 cm3)
r = 6 d = 12 cm

Example 2: A solid cone has base radius 5 cm and vertical height 12 cm. Find its total surface area in terms of π.

Solution:

l = √(52 + 122) = 13 cm
Curved = π(5)(13) = 65π
Total = 65π + 25π = 90π cm2
12 5 cm l = 13
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Spheres and cones Questions & Answers

Past Paper Example 1 IGCSE Jun 2024 1H Q23 5 marks
A solid shape is made by removing a hemisphere, shown shaded, from a cone as shown in the diagram.
Diagram NOTaccurately drawn
The radius of the hemisphere is 2x cm
The radius of the base of the cone is 5x cm
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.
Worked Solution & Mark Scheme (5 marks)
Volume of cone = ⅓π(5x)²(6x) = 50πx³ M2
Volume of hemisphere = ½ × (4/3)π(2x)³ = (16/3)πx³ M1
Remaining volume = (50 − 16/3)πx³ = (134/3)πx³ = 6948π ⇒ x³ = 6948 × 3 ÷ 134 = 10422/67 ≈ 155.552 ⇒ x = 5.378 cm M1
Total surface area of solid hemisphere = 3πR² = 3π(2x)² = 12πx² = 12π(5.378)² ≈ 1090.35... ≈ 1090 cm² A1
Answer: 1090 cm²
Past Paper Example 2 GCSE Jun 2022 2H Q17 5 marks
A solid cone is joined to a solid hemisphere to make the solid **T** shown below.
y cm7 cmVolume of sphere =43πr3rVolume of cone =13πr2hrlh
The diameter of the base of the cone is 7 cm.
The diameter of the hemisphere is 7 cm.
The total volume of **T** is 120π cm3
The total height of **T** is y cm.
(a) Calculate the value of y.
Give your answer correct to 3 significant figures.
[4 marks]
(b) explain the effect this would have on your answer to part (a). [1 mark]
Worked Solution & Mark Scheme (5 marks)
Part (a) Solution:
r = 3.5 cm M1
Hemisphere volume = (2/3)π(3.5³) = (343/12)π M1
Cone volume = (1/3)π(3.5²)(y – 3.5) = (49/12)π(y – 3.5) M1
Total: 343/12 + 49/12(y – 3.5) = 120 ⇒ y – 3.5 = 1097/49 ≈ 22.388 ⇒ y ≈ 25.888 → 25.9 A1
Answer (a): y = 25.9
Part (b) Solution:
The value of y (the total height) would decrease
Answer (b): The value of y (the total height) would decrease
Past Paper Example 3 IGCSE Nov 2024 1H Q22 3 marks
The diagram shows a bowl in the shape of a hemisphere.
The bowl is made from metal.
9 cm12 cmDiagram NOTaccurately drawn
The outer radius of the bowl is 12 cm
The inner radius of the bowl is 9 cm
The thickness of the bowl is uniform.
Work out the volume of the metal.
Give your answer correct to the nearest whole number.
Worked Solution & Mark Scheme (3 marks)
Volume = ½ × (4/3)π(12³ − 9³) = (2/3)π(1728 − 729) = (2/3)π(999) = 666π = 2092.3007... ≈ 2092 cm³ M2 A1
Answer: 2092 cm³

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