Spheres and cones Quiz (Grade 5)

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Score: 0 / 10

Spheres and cones — Notes & Worked Examples Show / Hide

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)

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

One Page Guide (PDF) Spheres and cones Worksheets

Question 1 AQA Nov 2024 3F Q26

[2 marks]

Hemisphere radius 7.5 cm. Ria writes Volume = 4/3 × 7.5 × 3 ÷ 2 = 15. Write down two mistakes she has made. [2 marks]

@@TABULAR₀@@
Ria works out the volume of this **hemisphere** in terms of π
Here is her work.
Volume of a hemisphere = 43 × 7.5 × 3 ÷ 2 = 15
Write down **two** mistakes she has made.

Worked Solution & Answer

1. She omitted π from the formula. M1
2. She calculated 7.5 × 3 instead of 7.5³. C1

Answer: Mistake 1: She did not include π in her calculation
Mistake 2: She multiplied 7.5 by 3 instead of cubing 7.5

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Question 2

[2 marks]

The diagram shows a hemisphere with diameter 8.4 cm.

Work out the volume of the hemisphere.
Give your answer correct to 3 significant figures.

cm³

Worked Solution & Answer

Radius r = 8.4 ÷ 2 = 4.2 cm M1
Volume = 12 × 43 × π × 4.2³ = 155.169... = 155 cm³ (3 s.f.) A1

Answer: 155 cm³

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Question 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.

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.

cm³

Worked Solution & Answer

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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Question 4 AQA Jun 2019 2H Q21

[3 marks]

A solid shape is made by joining two cones.
Each cone has the same radius.
One cone has slant height = 2 × radius
The other cone has slant height = 3 × radius
The total surface area of the shape is 57.8π cm²

Work out the radius.

Curved surface area of a cone = π r l where r is the radius and l is the slant height

Answer cm

Worked Solution & Answer

The total surface area consists of the curved surface areas of both cones (bases are joined): M1
Total SA = π r (2r) + π r (3r) = 2π r² + 3π r² = 5π r²
5π r² = 57.8π M1
5r² = 57.8 ⇒ r² = 11.56
r = √(11.56) = 3.4 cm A1

Answer: Answer 3.4 cm

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Question 5

[4 marks]

The diagram shows a solid shape.
The shape is a cone on top of a hemisphere.

The height of the cone is 10 cm.
The base of the cone has a diameter of 6 cm.
The hemisphere has a diameter of 6 cm.
The total volume of the shape is kπ cm³, where k is an integer.
Work out the value of k.

k =

Worked Solution & Answer

r = 6 ÷ 2 = 3 cm M1
Volume of cone = 1/3 × π × 3² × 10 = 30π M1
Volume of hemisphere = 2/3 × π × 3³ = 18π M1
Total volume = 30π + 18π = 48π ⇒ k = 48 A1

Answer: k = 48

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Question 6

[4 marks]

Here is a sphere.

38 of the surface area of this sphere is 75π cm²
Find the diameter of the sphere.
Give your answer in the form a√(b) where a is an integer and b is a prime number.

Worked Solution & Answer

Total surface area = (75π ÷ 3) × 8 = 200π cm² M3
4πr² = 200π ⇒ r² = 50 ⇒ r = 5√2 ⇒ diameter = 2r = 10√2 cm A1

Answer: 10√(2) cm

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Question 7 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.

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.

cm²

Worked Solution & Answer

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

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Question 8

[5 marks]

Shape **S** is one quarter of a solid sphere, centre O.

The volume of **S** is 576π cm³
Find the surface area of **S**.
Give your answer correct to 3 significant figures.
You must show your working.

cm²

Worked Solution & Answer

Volume of S = 1/4 × (4/3 πr³) = 1/3 πr³ = 576π ⇒ r³ = 1728 ⇒ r = 12 cm M4
Total surface area of S = 1/4(4πr²) + 2 × (1/2 πr²) = 2πr² = 2 × π × 12² = 288π ≈ 905 cm² (3 s.f.). A1

Answer: 905 cm²

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Question 9

[5 marks]

A solid cone is joined to a solid hemisphere to make the solid **T** shown below.

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]

y =

(b) explain the effect this would have on your answer to part (a). [1 mark]

Worked Solution & Answer

(a)

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

(b)

The value of y (the total height) would decrease

Answer (b): The value of y (the total height) would decrease

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Question 10 IGCSE Nov 2021 2H Q22

[6 marks]

The diagram shows a sphere of diameter x cm and a pyramid ABCDE with a horizontal rectangular base BCDE.

The vertex A of the pyramid is vertically above the centre O of the base so that AB = AC = AD = AE.
BC = x cm, CD = 2x cm and AO = 5x cm.
The volume of the sphere is 288π cm³
Calculate the total surface area of the pyramid.
Give your answer correct to the nearest cm²

cm²

Worked Solution & Answer

Volume of sphere = (4/3)π(x/2)³ = 288π ⇒ (x/2)³ = 216 ⇒ x/2 = 6 ⇒ x = 12 M2
Base dimensions: 12 cm by 24 cm (area = 288 cm²), height AO = 60 cm M1
Slant height to BC (and ED) = √(60² + 12²) = √3744 = 12√26 cm M1
Slant height to CD (and BE) = √(60² + 6²) = √3636 = 6√101 cm M1
Total surface area = 288 + 2(½ × 12 × 12√26) + 2(½ × 24 × 6√101) = 288 + 144√26 + 144√101 = 288 + 734.2588... + 1447.1903... = 2469.449... = 2469 cm² A1

Answer: 2469 cm²

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Score: 0 / 10 (0 / 39 marks)

Spheres and cones Worksheets Next: Similar shapes (lengths) →

Grade 5 Quizzes

25 Topics

5A - Combining ratios	5B - Direct and inverse proportion
5C - Reverse percentages	5D - Standard form
5E - Speed	5F - Changing the subject
5G - Quadratic expressions	5H - Solving quadratic equations
5J - Drawing quadratic graphs	5K - Cubic/reciprocal graphs
5L - Simultaneous equations	5M - Sector areas and arc lengths
5N - Sim. equ. graphically	5P - Midpoint of a line segment
5Q - Gradient of a line	5R - Equation of a line
5S - Spheres and cones	5T - Similar shapes (lengths)
5U - Trigonometry (SOHCAHTOA)	5V - Exact trig. values
5W - Probability trees	5X - Vectors
5Y - Venn diagrams	5Z - Density
5AA - Pressure