Volume of a frustum Quiz (Grade 8+)

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Volume of a frustum — Notes & Worked Examples Show / Hide

Explanation:

A frustum is the solid remaining when a smaller cone (or pyramid) is sliced off the top of a larger cone parallel to its base. Because the removed top cone and the original full cone are mathematically similar, the ratio of their heights equals the ratio of their radii: hH = rR.

Use similar triangles first if the height of the removed top cone is not given, then calculate: Volume of frustum = Volume of large cone − Volume of small cone = 13πR²H − 13πr²h.

Example 1: A cone of radius 9 cm and height 15 cm has a smaller cone of radius 3 cm and height 5 cm removed from its top. Find the exact volume of the frustum.

Solution:

V_large = 13π(9)²(15) = 405π; V_small = 13π(3)²(5) = 15π

V_frustum = 405π − 15π = 390π cm³

Example 2: A frustum has top radius 4 cm, base radius 10 cm, and vertical height 9 cm. Find the volume of the frustum (3 s.f.).

Solution:

hh + 9 = 410 ⇒ 10h = 4h + 36 ⇒ h = 6 cm (H = 15 cm)

V = 13π(10²×15 − 4²×6) = 468π = 1470 cm³ (3 s.f.)

One Page Guide (PDF) Volume of a frustum Worksheets

Question 1

[1 mark]

Work out
A frustum is made by removing a small cone of base radius 4 cm and height 6 cm from a similar large cone of base radius 8 cm and height 12 cm.
Find the exact volume of the frustum in terms of π.

Worked Solution & Answer

V = 13π(8² × 12 - 4² × 6) = 224π cm³ B1

Answer: 224π cm³

Question 2

[1 mark]

Work out
A frustum is made by removing a small cone of base radius 3 cm and height 6 cm from a similar large cone of base radius 6 cm and height 12 cm.
Find the exact volume of the frustum in terms of π.

Worked Solution & Answer

V = 13π(6² × 12 - 3² × 6) = 126π cm³ B1

Answer: 126π cm³

Question 3

[4 marks]

Work out
A frustum F is made by removing a cone of height 10 cm from a cone of height 15 cm and base diameter 24 cm. Solid S is made by removing F from a hemisphere of diameter 24 cm. Calculate the volume of solid S to 3 s.f.

The hemisphere has diameter 24 cm.
Calculate the volume of solid **S**.
Give your answer correct to 3 significant figures.

cm³

Worked Solution & Answer

Small cone radius = (10/15) × 12 = 8 cm M1
Volume of F = (1/3)π(12² × 15 – 8² × 10) = (1/3)π(2160 – 640) = (1520/3)π ≈ 1591.74 cm³ M1
Hemisphere volume = (2/3)π(12³) = 1152π ≈ 3619.11 cm³ M1
Volume of S = 3619.11 – 1591.74 = 2027.37 cm³ ≈ 2030 cm³. A1

Answer: 2030 cm³

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Question 4 AQA Jun 2017 1H Q28

[4 marks]

Volume of cone = 13π r² h where r is the radius and h is the perpendicular height.

A cone has a horizontal base of radius 5 cm and height of 15 cm.
The cone contains water to a depth of 9 cm.
Work out the volume of the water, in cm³
Give your answer in terms of π.

Answer cm³

Worked Solution & Answer

The water forms a frustum at the base of the cone M1
Total volume of cone:
V_total = 13 π (5²)(15) = 125π cm³
Height of empty cone at the top: 15 - 9 = 6 cm M1
By similar triangles, radius of empty cone:
r_top = 5 × 615 = 2 cm
Volume of empty cone: M1
V_empty = 13 π (2²)(6) = 8π cm³
Volume of water:
V_water = 125π - 8π = 117π cm³ A1

Answer: Answer 117π cm³

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Question 5 AQA Nov 2023 3H Q14

[4 marks]

A solid trophy consists of a stand and a player.
The stand is made by removing a small pyramid from a large pyramid.
• **Large pyramid** Square base, edge 8 cm, Perpendicular height 16 cm
• **Small pyramid** Square base, edge 5 cm, Perpendicular height 10 cm
@@TABULAR₀@@

(a) Show that the volume of the **stand** is 258 cm³ [2 marks]

(b) The trophy is made from a metal of density 7.5 grams per cm³
The **total** mass of the trophy is 2340 grams.
Work out the volume of the **player**. [2 marks]

Answer cm³

Worked Solution & Answer

(a)

Volume of large pyramid = 1/3 × 8² × 16 = 1024/3 cm³ M1
Volume of small pyramid = 1/3 × 5² × 10 = 250/3 cm³
Volume of stand = 1024/3 – 250/3 = 774/3 = 258 cm³ C1

Answer (a): Large = 1/3 × 8² × 16 = 341.33..., Small = 1/3 × 5² × 10 = 83.33...
Stand = 1024/3 – 250/3 = 774/3 = 258 cm³

(b)

Total volume of trophy = 2340 ÷ 7.5 = 312 cm³ M1
Volume of player = 312 – 258 = 54 cm³ A1

Answer (b): Answer 54 cm³

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Question 6 IGCSE Nov 2023 2H Q23

[5 marks]

Here is a frustum of a cone.
The frustum is made by removing a small cone from a similar large cone.

The height of the large cone is 15 cm.
The radius of the base of the large cone is 6 cm.
The radius of the base of the small cone is x cm.
Given that the volume of the frustum is 421225π cm³
work out the value of x
Show clear algebraic working.

x =

Worked Solution & Answer

Volume of large cone = (1/3)π(6²)(15) = 180π cm³ M2
Volume of small cone = 180π − (4212/25)π = (288/25)π = 11.52π cm³ M2
By similarity, (x/6)³ = 11.52π / 180π = 288 / 4500 = 8 / 125 ⇒ x/6 = 2/5 ⇒ x = 2.4. A1

Answer: x = 2.4

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Question 7 AQA Nov 2022 3H Q23

[5 marks]

Alec makes a bowl for dog food from a solid wooden cone.
The sketches show how the bowl is made.
The cone has radius 9 cm and perpendicular height 30 cm
A smaller cone, with radius 6 cm, is removed.

@@TABULAR₀@@
A hemisphere with radius 6 cm is then removed.
@@TABULAR₁@@
Work out the volume of the remaining wood that forms the bowl.

Answer cm³

Worked Solution & Answer

By similar cones: height of removed cone = 30 × (6/9) = 20 cm M1
Volume of large cone = 1/3 × π × 9² × 30 = 810π cm³ M1
Volume of small cone = 1/3 × π × 6² × 20 = 240π cm³ M1
Volume of hemisphere = 2/3 × π × 6³ = 144π cm³ M1
Remaining volume = (810 – 240 – 144)π = 426π = 1338.3 cm³. A1

Answer: Answer 1338 cm³

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Question 8

[5 marks]

A solid frustum is made by removing a small cone from a large cone as shown in the diagram.

The slant height of the small cone is 6 cm.
The slant height of the large cone is 10 cm.
The radius of the base of the large cone is 3 cm.
Calculate the total surface area of the frustum.
Give your answer correct to 3 significant figures.

cm²

Worked Solution & Answer

Radius of top circular face = (6/10) × 3 = 1.8 cm M2
Curved SA = π(3)(10) – π(1.8)(6) = 30π – 10.8π = 19.2π M2
Total SA = 19.2π + π(3²) + π(1.8²) = (19.2 + 9 + 3.24)π = 31.44π = 98.77... → 98.8 cm² A1

Answer: 98.8 cm²

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Question 9 IGCSE Jun 2023 1H Q23

[5 marks]

A frustum is made by removing a small square-based pyramid from a similar large square-based pyramid as shown in the diagram.

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.

cm²

Worked Solution & Answer

Scale factor = 15/45 = 1/3, so top square has side 39 ÷ 3 = 13 cm M1
Slant height of large pyramid L = √(45² + 19.5²) = 49.0433... cm M1
slant height of small pyramid l = L/3 = 16.3477... cm M1
Lateral area = 4 × (1/2 × 39 × 49.0433... – 1/2 × 13 × 16.3477...) = 3825.38 – 425.04 = 3400.34 cm² M1
Total surface area = 3400.34 + 39² + 13² = 3400.34 + 1521 + 169 = 5090.34 -> 5090 cm². A1

Answer: 5090 cm²

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Question 10 Eduqas Jun 2023 2H Q21

[7 marks]

An inverted cone is held in a stand.
The radius of the cone is 20 cm and the height of the cone is 50 cm.

Jake pours some water into this cone.
When he stops, the radius of the surface of the water is 6 cm.
The surface of the water is parallel to the base of the cone.
Rhian then pours water into the cone until it is completely full.
She pours the water at a rate of 1 litre per 10 seconds.
How many seconds does it take Rhian to fill the cone?

Worked Solution & Answer

Total volume of cone: M1
V_total = 13π (20²)(50) = 200003π ≈ 20943.95 cm³ M1
Height of water poured by Jake: M1
By similar triangles, h50 = 620 ⇒ h = 15 cm
Volume of water = 13π (6²)(15) = 180π ≈ 565.49 cm³ M1
Volume Rhian needs to add: M1
V_add = 20943.95 - 565.49 = 20378.46 cm³ = 20.3785 litres M1
Time taken at 1 litre per 10 seconds:
Time = 20.3785 × 10 ≈ 203.8 seconds ≈ 204 seconds A1

Answer: 204 seconds (or 203.8 seconds)

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Score: 0 / 10 (0 / 41 marks)

Volume of a frustum Worksheets

Grade 8+ Quizzes

14 Topics

8A - Quad. sim. equations	8B - Transforming graphs y=f(x)
8C - Algebraic proof	8D - Completing the square
8E - Quadratic inequalities	8F - Quadratic sequences
8G - Velocity time graphs	8H - Circles and tangents
8J - Circle theorem proof	8K - Vector proof
8L - Probability equation questions	8M - Invariant points
8N - Spheres, cones and cylinders	8P - Volume of a frustum