Three-Dimensional Geometrical Shapes: Mathematics questions

Section A: objective questions

11 marks

Choose the correct answer.

A pyramid has a hexagonal base. How many faces does it have, including the base?

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A hexagon has 6 sides, so there are 6 triangular faces. With the hexagonal base that makes 7 faces. 6 forgets the base; 12 is the number of edges.

For the teacher: Understanding, low difficulty, standard 6.1.1

Textbook p. 101

Which statement about a cuboid is correct?

It is a prism because it has a uniform rectangular cross section and two congruent parallel bases
It is a pyramid because it has a flat base
It is not a prism because all its faces are rectangles
It has one curved surface

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A cuboid has the same rectangular cross section from end to end and two congruent parallel bases, so it is a prism. A pyramid has triangular side faces that meet at an apex.

For the teacher: Understanding, low difficulty, standard 6.1.1

Textbook p. 101

Diagram 1 shows a net made from a circle and a sector. Which solid can be made from the net?

A sector of radius 5 cm with a circle of radius 3 cm attached below the curved edge — Diagram 1

Cylinder
Pyramid
Prism
Cone

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A circle for the base and a sector for the curved surface make a cone. The radius of the sector (5 cm) becomes the slant height. A cylinder needs two circles and a rectangle.

For the teacher: Understanding, low difficulty, standard 6.2.1

Textbook p. 103

The net of a triangular prism is made up of ______.

3 triangles and 2 rectangles
4 triangles and 1 square
2 triangles and 3 rectangles
2 circles and 1 rectangle

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A triangular prism has two triangular bases and three rectangular side faces. 4 triangles and 1 square is the net of a square-based pyramid; 2 circles and 1 rectangle is a cylinder.

For the teacher: Remembering, low difficulty, standard 6.2.1

Textbook p. 103

Diagram 2 shows a cuboid. Find its surface area.

Cuboid with length 8 cm, width 5 cm and height 3 cm — Diagram 2

158 cm²
120 cm²
79 cm²
240 cm²

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Surface area = 2(8 × 5 + 8 × 3 + 5 × 3) = 2(40 + 24 + 15) = 158 cm². 120 is the volume (8 × 5 × 3). 79 is only half of the surface.

For the teacher: Applying, low difficulty, standard 6.3.1

Textbook p. 107

A closed cylinder has a radius of 7 cm and a height of 10 cm. What is its surface area? (Use π = 227)

594 cm²
748 cm²
1 540 cm²
440 cm²

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2πr² = 2 × 227 × 49 = 308. 2πrh = 2 × 227 × 7 × 10 = 440. Total = 748 cm². 594 uses only one circle; 1 540 is the volume; 440 is the curved surface only.

For the teacher: Applying, low difficulty, standard 6.3.1

Textbook p. 108

Diagram 3 shows a cone. Find its total surface area. (Use π = 227)

A cone with base radius 7 cm and vertical height 24 cm — Diagram 3

550 cm²
682 cm²
154 cm²
704 cm²

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The slant height s = √(7² + 24²) = √625 = 25 cm. Surface area = πr² + πrs = 154 + 227 × 7 × 25 = 154 + 550 = 704 cm². 682 uses h = 24 instead of s; 550 is the curved surface only.

For the teacher: Analysing, medium difficulty, standard 6.3.1

Textbook p. 108

What is the surface area of a sphere with a radius of 21 cm? (Use π = 227)

5 544 cm²
2 772 cm²
38 808 cm²
1 386 cm²

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Surface area = 4πr² = 4 × 227 × 21² = 5 544 cm². 2 772 is 2πr² (half). 38 808 is the volume 43πr³.

For the teacher: Applying, low difficulty, standard 6.3.2

Textbook p. 109

A cone and a cylinder have the same base and the same height. The volume of the cone is ______ of the volume of the cylinder.

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Three cones of sago fill the cylinder, so the cone is 13 of the cylinder: V = 13πr²h.

For the teacher: Understanding, low difficulty, standard 6.4.1

Textbook p. 112

A pyramid has a rectangular base 9 cm by 6 cm and a height of 10 cm. Find its volume.

540 cm³
180 cm³
90 cm³
60 cm³

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V = 13 × 9 × 6 × 10 = 180 cm³. 540 forgets the 13.

For the teacher: Applying, low difficulty, standard 6.4.2

Textbook p. 113

Find the volume of a hemisphere with a radius of 3.5 cm. (Use π = 227)

134.75 cm³
179.67 cm³
269.50 cm³
89.83 cm³

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V = 23πr³ = 23 × 227 × 3.5³ = 89.83 cm³. 179.67 cm³ is the whole sphere. 134.75 is πr³ and 269.50 is 2πr³.

For the teacher: Applying, medium difficulty, standard 6.4.2

Textbook p. 114

Section B: structured questions

29 marks

Answer all questions. Show your working.

Name the three-dimensional shape that has each geometrical characteristic.

[1 mark] (a)One curved surface, and every point on it is the same distance from the centre.

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Sphere [1]1

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[1 mark] (b)One apex and a polygon for its base; the other faces are triangles.

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Pyramid [1]1

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[1 mark] (c)Two congruent parallel bases that are polygons, with a uniform cross section.

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Prism [1]1

Tick each point you got.

For the teacher: Remembering, low difficulty, standard 6.1.1

Textbook p. 101

Diagram 4 shows the net of a solid.

Net of a prism: three rectangles 3 by 10, 4 by 10 and 5 by 10 in a row, with a right-angled triangle above and below the middle rectangle — Diagram 4

[1 mark] (a)Name the solid that is formed.

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Triangular prism [1]1

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[2 marks] (b)How many edges and how many vertices does the solid have?

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9 edges [1]1
6 vertices [1]1

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For the teacher: Understanding, low difficulty, standard 6.2.1

Textbook p. 103

A closed tin is a cylinder with a radius of 3.5 cm and a height of 12 cm. (Use π = 227)

[1 mark] (a)Find the area of one circular end.

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227 × 3.5² = 38.5 cm² [1]1

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[1 mark] (b)Find the area of the curved surface.

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2 × 227 × 3.5 × 12 = 264 cm² [1]1

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[1 mark] (c)Find the total surface area of the tin.

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2 × 38.5 + 264 = 341 cm² [1]1

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For the teacher: Applying, low difficulty, standard 6.3.1

Textbook p. 108

Diagram 5 shows a solid made of a cube of side 8 cm and a pyramid with a square base of the same size. The height of the pyramid is 3 cm.

A cube with side 8 cm and a square-based pyramid of height 3 cm on top — Diagram 5

[2 marks] (a)Calculate the slant height of the pyramid.

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Half of base = 4 cm, s² = 3² + 4² [1]1
s = 5 cm [1]1

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[2 marks] (b)Calculate the total surface area of the solid, in cm².

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5 × (8 × 8) + 4 × (½ × 8 × 5) [1]1
320 + 80 = 400 cm² [1]1

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[1 mark] (c)Express the surface area in m².

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400 ÷ 10 000 = 0.04 m² [1]1

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For the teacher: Applying, medium difficulty, standard 6.3.3

Textbook p. 109

Diagram 6 shows a triangular prism. The cross section is a right-angled triangle with sides 6 cm, 8 cm and 10 cm. The prism is 15 cm long.

Triangular prism with a right-angled triangle cross section: legs 6 cm and 8 cm, hypotenuse 10 cm; length of prism 15 cm — Diagram 6

[1 mark] (a)Find the area of the cross section.

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½ × 6 × 8 = 24 cm² [1]1

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[1 mark] (b)Find the volume of the prism.

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24 × 15 = 360 cm³ [1]1

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[2 marks] (c)Find the total surface area of the prism.

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2 × 24 + (6 + 8 + 10) × 15 [1]1
48 + 360 = 408 cm² [1]1

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For the teacher: Applying, medium difficulty, standard 6.4.2

Textbook p. 112

Diagram 7 shows an ice-cream cone of vertical height 12 cm with a hemisphere on top. The radius of both is 7 cm. (Use π = 227)

A cone pointing down with a hemisphere on top; radius 7 cm, cone height 12 cm — Diagram 7

[2 marks] (a)Calculate the volume of the cone.

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13 × 227 × 7² × 12 [1]1
616 cm³ [1]1

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[2 marks] (b)Calculate the volume of the hemisphere.

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23 × 227 × 7³ [1]1
718.67 cm³ [1]1

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[2 marks] (c)How many such ice-creams can be made from 10 litres of ice cream? Give your answer as a whole number. (1 litre = 1 000 cm³)

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Total volume = 616 + 718.67 = 1 334.67 cm³ [1]1
10 000 ÷ 1 334.67 = 7.49, so 7 ice-creams [1]1

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For the teacher: Analysing, high difficulty, standard 6.4.3

Textbook p. 114

A factory wants a closed tin that holds 1 540 cm³ of syrup. The tin is a cylinder with a radius of 7 cm. (Use π = 227) A cuboid box with the same volume is also being considered.

[2 marks] (a)Find the height of the cylindrical tin.

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227 × 7² × h = 1 540, so 154h = 1 540 [1]1
h = 10 cm [1]1

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[1 mark] (b)A cuboid box 14 cm long, 11 cm wide and 10 cm high is proposed. Show that its volume is the same.

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14 × 11 × 10 = 1 540 cm³ [1]1

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[2 marks] (c)Which container needs less material to make? Support your answer with calculations.

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Cylinder: 2 × 154 + 2 × 227 × 7 × 10 = 748 cm²; cuboid: 2(154 + 140 + 110) = 808 cm² [1]1
The cylinder uses less material (748 is less than 808) [1]1

Tick each point you got.

For the teacher: Evaluating, medium difficulty, standard 6.4.3

Textbook p. 115

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Section B: structured questions

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18