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Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

GET DIRECTORATE

MATHEMATICS LESSON

GRADE 7

2. CONCEPTS AND SKILLS TO BE ACHIEVED

By the end of the lesson, learners should know and be able to:

2.1. use appropriate formulae to calculate the surface area, volume and

capacity of a cube

2.2. use appropriate formulae to calculate the surface area, volume and

capacity of rectangular prisms

2.3. describe the interrelationship between surface area and volume of cubes

2.4. describe the interrelationship between surface area and volume of

rectangular prism

2.5. solve problems involving surface area, volume and capacity

2.6 use and convert between appropriate SI-units, including:

π‘šπ‘š2 ↔ π‘π‘š2, π‘π‘š2 ↔ π‘š2, π‘šπ‘š3 ↔ π‘π‘š3and π‘π‘š3 ↔ π‘š3

2.7. use equivalence between units when solving problems:

1 cm3 ↔ 1 ml and 1 m3 ↔ 1 kl

TOPIC: SURFACE AREA AND VOLUME OF 3-D OBJECTS

Lesson: Surface area and volume

Page 2 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

3. RESOURCES DBE workbook 1, textbooks. Sasol-Inzalo learner book

ONLINE RESOURCES

https://drive.google.com/open?id=1Qw6gZzmSxQ-

ypsHmqx1LHnVbA2HsKX79

https://www.thelearningtrust.org/asp-treasure-box

4. PRIOR KNOWLEDGE

Area, Volume and capacity of the rectangular prisms done in previous

grades.

5. INTRODUCTION (Suggested time: 10 Minutes)

Consider the cube below which is made from small cubes of 1π‘π‘š by 1π‘π‘š by 1π‘π‘š:

One face of the cube

1. Use the Cube above to answer the following questions.

a) How many small cubes make up the big cube?

b) What is the volume of the big cube?

c) How many small cubes make up the first layer of the big cube?

d) How many cubes make up the length of each edge of the big cube?

e) Use the formula to calculate the area of the bottom face (base) of the big cube.

f) How many layers (height of the big cube) does the big cube have?

g) Is there no other way that we can use to find the volume of the prism without counting the number

of cubes which makes up the prism?

2. Calculate the area of each face of the big cube.

A cube has six (6) identical faces, see the net alongside

Page 3 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

6. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)

Activity 1:

Calculate the volume of the following cubes.=

a)

b)

Note: After learners’ calculations and teacher’s feedback no (b) is emphasised as a formula for

calculating volume of any given cube.

Activity 2: Calculate the surface area of the cubes below.

a)

b)

A formula for calculating surface area of any given cube, is:

7. CLASSWORK (Suggested time: 15 minutes) Learners complete the exercises:

a) Use the cube alongside and calculate using appropriate formulae its: i. Volume

ii. surface area

b) The dimensions of a 3-D object are: 7 π‘π‘š Γ— 7 π‘π‘š Γ— 7 π‘π‘š. Calculate its

i. Volume

ii. surface area

5 π‘π‘š

Volume of a cube = SΒ³

3 π‘π‘š

𝑠

Surface area of a cube= 6SΒ²

𝑠

3 π‘π‘š

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Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

8. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)

Emphasise that:

Volume of an object is the amount of space it occupies.

Formula thereof is

Surface area of an object is the total area of the areas of the faces.

Formula thereof is

9. Homework: The solutions can be found at the end of the lesson.

COMPLETE THE EXERCISES BEOFE YOU WORK THROUGH THE SOLUTIONS.

1. Use the cube alongside and calculate using formulae its: i. Volume

ii. surface area

2. The dimensions of a 3-D object are: 2,5 π‘π‘š Γ— 2,5 π‘π‘š Γ— 2,5 π‘π‘š. Calculate its:

i. Volume

ii. surface area

MEMORANDUM : DAY 1:

ACTIVITY SOLUTION

5. INTRODUCTION 1(a) 27 cubes

(b) 27 cm3

(c) 9 cubes

(d) 3 cubes

(e) Area of base = 𝑠2 = (3 π‘π‘š)2 = 9 π‘π‘š2

(f) 3 lae

(g) Yes, Area of the cube x number of layers = Area of

base x height

= 𝑠2 Γ— 𝑠

= 9 π‘π‘š2 Γ— 3 π‘π‘š

= 27 π‘π‘š3

Volume of a cube = SΒ³

Surface area a cube= 6SΒ²

4 π‘π‘š

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Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

2. Surface area of a cube = Total of Area of all faces

= 9 π‘π‘š2 + 9 π‘π‘š2 + 9 π‘π‘š2 + 9 π‘π‘š2 + 9 π‘π‘š2 +

9 π‘π‘š2

= 6 Γ— 9 π‘π‘š2

= 54 π‘π‘š2

Hence the surface area of a cube = 6 Γ— 𝑠2

6.PRESENTATION/DEVELOPMENT

Aktivity1(a) Volume = 𝑠3

= (3 π‘π‘š)3

= 27 π‘π‘š3

(b) Volume = 𝑠 Γ— 𝑠 Γ— 𝑠

= 𝑠3

Aktivity 2(a) Surface area = 𝑠2 Γ— 6

= (3 π‘π‘š)2 Γ— 6

= 9 π‘π‘š2 Γ— 6

= 54 π‘π‘š2

(b) Surface area = 𝑠 Γ— 𝑠 Γ— 6

= 6 𝑠2

7. CLASSWORK

(a) i V= 125 cm3

ii BO = 150 cm2

(b)i V = 343 cm3

ii BO = 294 cm2

9.HOMEWORK

1 (i) V= l x b x h = 4 cm x 4 cm x 4cm = 64 cm3

(ii) SA = 6(l x l) = 6 x 16 cm2 = 96 cm2

2 (i) V = l x l x l; 2,5 cm x 2,5 cm x 2,5 cm = 15,625 cm3

(ii) SA = 6(l x l); 6(2,5 cm x 2,5 cm) = 37,5 cm2

Page 6 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

DAY 2

PRIOR KNOWLEDGE Area, volume and capacity of the rectangular prisms done in previous

grades.

INTRODUCTION (Suggested time: 10 Minutes)

Consider the following rectangular prism made from small cubes of 1 π‘π‘š Γ— 1 π‘π‘š Γ— 1 π‘π‘š and its

faces.

N.B.: Remember each of these faces has a replica.

1. Use the rectangular prism above to answer the following questions.

a) How many small cubes make up the rectangular prism?

b) What is the volume of the rectangular prism?

c) How many small cubes make up the first layer of the rectangular prism?

d) How many cubes make up the length and breadth of the first layer (base)?

e) Use the formula to calculate the area of the bottom face (base) of the rectangular prism.

f) How many layers (height of the rectangular prism) does the prism have?

g) Is there no other way that we can use to find the volume of the prism without counting

the number of cubes which makes up the prism?

2. Calculate the area of each pair of faces of the rectangular prism.

3. What is the total surface area of the rectangular prism?

10. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)

Activity 1: What is the volume of these rectangular prisms?

(a) (b) (c) h

8cm 3cm

3cm 2cm

5cm 2cm

l b

(i) (iii) (ii)

Page 7 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

a) 3 cm Γ— 2 cm Γ— 8 cm

b) 5 cm Γ— 2 cm Γ— 3 cm

c) l Γ— b Γ— h =

After learners’ calculations and teacher’s feedback no (c) is emphasised as a formula for calculating

any given rectangular prism.

Activity 2: What is the surface area of these rectangular prisms?

(a) (b) (c) h

8cm 3cm

3cm 2cm

5cm 2cm

l b

(c) is emphasised as a formula for calculating any given rectangular prisms

11. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)

a) Emphasise that:

Volume of a rectangular prism is 𝑙 x 𝑏 x β„Ž

Surface area of a rectangular prism is 2(π‘™β„Ž + π‘β„Ž + 𝑙𝑏)

b) Homework: Educators must choose appropriate exercises from the textbook.

MEMORANDUM: DAY 2

ACTIVITY SOLUTION

11. INTRODUCTION

1(a) 210 cubes

(b) 210 π‘π‘š3

(c) 30 cubes

(d) Length = 6 cubes and the breadth = 5 cubes

(e) Area of the base = 𝑙 Γ— 𝑏 = 6 π‘π‘š Γ— 5 π‘π‘š = 30 π‘π‘š2

(f) 7 layers

(g) Yes,

Volume of a rectangular prism = l x b x h

Surface area of a rectangular prism = 2(π‘™β„Ž + π‘β„Ž + 𝑙𝑏)

Page 8 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

Volume of the rectangular prism =Area of the base Γ—

number of layers

=Area of the base Γ— height

= 𝑙 Γ— 𝑏 Γ— β„Ž

= 6 π‘π‘š Γ— 5 π‘π‘š Γ— 7

= 210 π‘π‘š3

2(a) 2(7π‘π‘š Γ— 5π‘π‘š)

= 2(35 π‘π‘šΒ²)

= 70 π‘π‘š2

(b) 2(6π‘π‘š Γ— 5π‘π‘š)

= 2(30 π‘π‘šΒ²)

= 60 π‘π‘š2

(c) 2(7π‘π‘š Γ— 6π‘π‘š)

= 2(42 π‘π‘šΒ²)

= 84 π‘π‘š2

3. 70π‘π‘šΒ² + 60π‘π‘šΒ² + 84π‘π‘šΒ²= 214 π‘π‘šΒ²

12.LESSON

PRESENTATION/DEVELOPMENT

Activity 1(a) 48 cmΒ³

(b) 30 cm3

(c) β–‘ cubic units

Activity 2

(a)

2(8 π‘π‘š x 3 π‘π‘š) + 2(8 π‘π‘š x 2 π‘π‘š) + 2(3 π‘π‘š Γ— 2 π‘π‘š)

= 2(24 π‘π‘šΒ²) + 2(16 π‘π‘šΒ²) + 2(6 π‘π‘šΒ²)

= 48 π‘π‘šΒ² + 32 π‘π‘šΒ² + 12 π‘π‘šΒ²

= 92 π‘π‘šΒ²

(b) 2(5π‘π‘š x 3π‘π‘š) + 2(3π‘π‘š x 2π‘π‘š) + 2(5π‘π‘š Γ— 2π‘π‘š)

= 2(15 π‘π‘šΒ²) + 2(6 π‘π‘šΒ²) + 2(10 π‘π‘šΒ²)

= 30 π‘π‘šΒ² + 12 π‘π‘šΒ² + 20 π‘π‘šΒ²

= 62 π‘π‘šΒ²

(c) 2(𝑙 x β„Ž) + 2(𝑏 Γ— β„Ž) + 2(𝑙 Γ— 𝑏)

= 2(π‘™β„Ž + π‘β„Ž + 𝑙𝑏)

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Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

DAY 3

12. PRIOR KNOWLEDGE Surface area and volume of a cube

13. INTRODUCTION (Suggested time: 10 Minutes)

Learners must revise the following work done:

Determine the surface area and the volume of the following 3-D shapes in the table.

Name Object Volume Surface Area

Cube

14. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)

Activity 1:

a) Complete the table: .

Size of Cube Volume Surface Area

1 π‘š x 1 π‘š x 1 π‘š

2 π‘š x 2 π‘š x 2 π‘š

3 π‘š x 3 π‘š x 3 π‘š

4 π‘š x 4 π‘š x 4 π‘š

5 π‘š x 5 π‘š x 5 π‘š

8 π‘š x 8 π‘š x 8 π‘š

b) Does the surface area increase or decrease as the length of the side of the cube increases?

c) Does the volume increase or decrease as the length of side of the cube increases?

d) Does volume or surface area increase more rapidly when the length of the side of the cube

increases?

3cm

Page 10 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

15. CLASSWORK (Suggested time: 15 minutes) Consider the table above (i.e. in Activity 1)

a) Which length(s) of the side of the cube will make the volume and surface area the same?

b) Which lengths of the side of the cube will make the volume less than the surface area?

c) Which lengths of the side of the cube will make the surface area less than the volume?

16. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)

Emphasise that: (Below, in red, are also the answers to no. 17 a, b and c above).

a). A cube that has length of 6 units will have the same volume as the surface area.

b). All lengths smaller than 6 units will have the volume smaller than the surface area.

c). All lengths bigger than 6 units will have the volume bigger than the surface area.

MEMORANDUM: DAY 3

ACTIVITY SOLUTION

15. INTRODUCTION 𝑉 = 𝑠3

= 33π‘π‘š3

= 27π‘π‘š3

𝑆𝐴 = 6 Γ— 𝑠2

= 6 Γ— 32π‘π‘š2

= 6 Γ— 9π‘π‘š2

= 54π‘π‘š2

16. a) Size of Cube Volume Surface Area

1 π‘š x 1 π‘š x 1 π‘š 1π‘š3 6 Γ— (1π‘š)2 = 6π‘š2

2 π‘š x 2 π‘š x 2 π‘š 8π‘š3 6 Γ— (2π‘š)2 = 24π‘š2

3 π‘š x 3 π‘š x 3 π‘š 27π‘š3 54π‘š2

4 π‘š x 4 π‘š x 4 π‘š 64π‘š3 96π‘š2

5 π‘š x 5 π‘š x 5 π‘š 125π‘š3 150π‘š2

8 π‘š x 8 π‘š x 8 π‘š 512π‘š3 384π‘š2

b).Yes, the surface area of the cube increases as the length of the

side increases.

c) Yes, the volume increase as the length of the side of the cube

increases.

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Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

d) The volume increases more rapidly when the length of the side of

the cube increases.

17.CONSOLIDATION/ CONCLUSION & HOMEWORK

a). A cube that has length of 6 units will have the same volume as the

surface area.

b). All lengths smaller than 6 units will have the volume smaller than

the surface area.

c). All lengths bigger than 6 units will have the volume bigger than the surface area

Page 12 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

DAY 4

17. PRIOR KNOWLEDGE Volume and capacity of a cube and

Volume and capacity rectangular prisms

18. INTRODUCTION (Suggested time: 10 Minutes)

Revise the following work.

Determine the surface area and the volume of the following 3-D object in the table.

Name Object Volume Surface Area

Rectangular

Prism

19. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)

Activity: Consider the 3-D objects below.

(i) (ii)

a) Calculate the volume of each of the rectangular prism?

b) Calculate the surface area of each of the rectangular prism?

c) Compare the two rectangular prisms in terms of volume and surface area?

Size of Rectangular Volume Surface Area

1 π‘š x 1 π‘š x 16 π‘š

1 π‘š x 2 π‘š x 8 π‘š

1 π‘š x 4 π‘š x 4 π‘š

2 π‘š Γ— 2 π‘š Γ— 4 π‘š

Note: 1 m x 1 m x 16 m has a one side with the greatest dimension

𝑏 = 2 π‘π‘š 𝑙 = 4 π‘π‘š

β„Ž = 3 π‘π‘š

6 π‘π‘š

2 π‘π‘š

2 π‘π‘š 1 π‘π‘š

1 π‘π‘š

24 π‘π‘š

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Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

20. CLASSWORK (Suggested time: 15 minutes) a) Calculate the volume of each of the rectangular prism with the following dimensions:

i. 6π‘š Γ— 2π‘š Γ— 3π‘š

ii. 4π‘š Γ— 3π‘š Γ— 3π‘š

iii. 9π‘š Γ— 2π‘š Γ— 2π‘š

b) Arrange the rectangular prisms from the one with smallest surface area to the one with biggest surface area.

21. CONSOLIDATION AND CONCLUSION (Suggested time: 5 minutes)

Emphasise that:

Rectangular prisms with the same volume and different dimensions have different surface area.

Page 14 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

MEMORANDUM: DAY 4

ACTIVITY SOLUTION 20. INTRODUCTION

Name Object Volume Surface Area

Rectangular

Prism

𝑉 = 𝑙 Γ— 𝑏 Γ— β„Ž

= 4π‘π‘š Γ— 2π‘π‘š Γ—

3π‘π‘š

= 24π‘π‘š3

𝑆𝐴 = 2π‘™β„Ž + 2π‘β„Ž + 2𝑙𝑏

= 2(4 π‘π‘š Γ— 3 π‘π‘š) + 2(2 cmΓ— 3 π‘π‘š) +

2(4 π‘π‘š Γ— 2 π‘π‘š)

= 24 π‘π‘š2 + 12 π‘π‘š2 + 16 π‘π‘š2

= 52π‘π‘š2

21. LESSON PREPARATION/DEVELOPMENT

Activity: (a).

(i) 𝑉 = 𝑙 Γ— 𝑏 Γ— β„Ž (ii) 𝑉 = 𝑙 Γ— 𝑏 Γ— β„Ž

= 6 π‘π‘š Γ— 2 π‘π‘š Γ— 2 π‘π‘š = 24 π‘π‘š Γ— 1 π‘π‘š Γ— 1 π‘π‘š

= 24 π‘π‘šΒ³ = 24 π‘π‘šΒ³

(b). (i) π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π‘Žπ‘Ÿπ‘’π‘Ž = 2𝑙𝑏 + 2π‘™β„Ž + 2π‘β„Ž

= 2 Γ— 6 π‘π‘š Γ— 2 π‘π‘š + 2 Γ— 6 π‘π‘š Γ— 2 π‘π‘š + 2 Γ— 2 π‘π‘š Γ— 2 π‘π‘š

= 56 π‘π‘šΒ²

(ii) π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π‘Žπ‘Ÿπ‘’π‘Ž = 2𝑙𝑏 + 2π‘™β„Ž + 2π‘β„Ž

= 2 Γ— 24 π‘π‘š Γ— 1 π‘π‘š + 2 Γ— 24 π‘π‘š Γ— 1 π‘π‘š + 2 Γ— 1 π‘π‘š Γ— 1 π‘π‘š

= 98 π‘π‘šΒ²

(c). If the volume of different rectangular prisms is the same, then

the rectangular prism that has the greatest one dimension of

all other dimensions have the greatest surface area.

Size of Rectangular Volume Surface Area

1 π‘š x 1 π‘š x 16 π‘š 16 π‘š3 66 π‘š2

1 π‘š x 2 π‘š x 8 π‘š 16 π‘š3 52 π‘š2

1 π‘š x 4 π‘š x 4 π‘š 16 π‘š3 48 π‘š2

2 π‘š Γ— 2 π‘š Γ— 4 π‘š 16 π‘š3 40 π‘š2

22. CLASSWORK

(a). i V= l Γ—b Γ—h = 36 m3 (a). ii V= l Γ—b Γ—h = 36 m3

𝑏 = 2 π‘π‘š 𝑙 = 4 π‘π‘š

β„Ž = 3 π‘π‘š

Page 15 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

(a). iii V= l Γ—b Γ—h = 36 m3

(b). (ii), (i), (iii)

DAY 5

22. PRIOR KNOWLEDGE formulae for surface area of a cube and rectangular prism

formulae for volume of a cube and rectangular prism

23. INTRODUCTION (Suggested time: 10 Minutes)

Revise the following formulae for calculating the surface area and volume of the 3-D objects in the

table below. Learners must name each 3-D object and give its formula.

NAME OF 3-D

OBJECT

SURFACE AREA /

VLOLUME FORMULAE

Surface area

Volume

Surface area

Volume

NOTE:

The surface area of any 3-D object is the sum of the areas of all its faces.

The volume of any 3-D object is given by area of the base Γ— heght.

s

s s

h

b

Page 16 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

24. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)

Example 1: Consider the following rectangular prism with inside measurements as shown:

a) Calculate its surface area.

b) Calculate its volume.

c) What is the capacity of rectangular prism in π‘šπ‘™?

Example 2: Consider the following cube:

a) Calculate its surface area.

b) Calculate its volume.

25. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)

a) Emphasise that:

the volume of prism = area of the base Γ— height

the surface area of the prism = the sum of the area of all its faces

the volume of a cube = 𝑠3 or 𝑙3

the volume of a rectangular prism = 𝑙 Γ— 𝑏 Γ— β„Ž

b) Homework:

Educators choose appropriate exercises from the textbook.

Page 17 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

MEMORANDUM: DAY 5

ACTIVITY SOLUTION 25. INTRODUCTION

NAME OF 3-D OBJECT

SURFACE

AREA /

VLOLUME

FORMULAE

CUBE

Surface area

π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 6 Γ— π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘œπ‘›π‘’ π‘“π‘Žπ‘π‘’

= 6 Γ— 𝑠2

Volume

𝑉 = 𝑠3

RECTANGULAR PRISM

Surface area

π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π‘Žπ‘Ÿπ‘’π‘Ž = 2 Γ— 𝑙𝑏 + 2 Γ— π‘™β„Ž + 2 Γ— π‘β„Ž

= 2(𝑙𝑏 + π‘™β„Ž + π‘β„Ž)

Volume

𝑉 = 𝑙 Γ— 𝑏 Γ— β„Ž

26. LESSON

PRESENTATION/DEVELOPMENT

Example 1: (a).

Surface area = 2(𝑙𝑏 + π‘™β„Ž + π‘β„Ž) = 2(8 π‘π‘š Γ— 5 π‘π‘š + 8 π‘π‘š Γ— 2 π‘π‘š + 5 π‘π‘š Γ—2 π‘π‘š) = 2(40 π‘π‘š2 + 16 π‘π‘š2 + 10 π‘π‘š2) = 2(66 π‘π‘š2) = 132 π‘π‘š2

(b). Volume = 𝑙 Γ— 𝑏 Γ— β„Ž = 8 π‘π‘š Γ— 5 π‘π‘š Γ— 2 π‘π‘š

= 80 π‘π‘š3

(c). Capacity = 80 π‘šπ‘™

Example 2 (a).

Surface area = 6 Γ— 𝑠2

= 6 Γ— (5 π‘π‘š)2

= 6 Γ— 25 π‘π‘š2

= 150 π‘π‘š2

(b). Volume = 𝑠3

= (5 π‘π‘š)3

= 125 π‘π‘š3

s s

s

h

b

Page 18 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

26. PRIOR KNOWLEDGE

Volume and capacity of a cube and rectangular prisms done in the

previous lessons.

substitution

27. INTRODUCTION (Suggested time: 10 Minutes)

Revise the following formulae for calculating the surface area and volume of the 3-D objects on the

table below. Ask learners to name each 3-D object and give its formulae (previous lesson).

NAME OF 3D

OBJECT

SURFACE AREA /

VLOLUME FORMULAE

Surface area

s s

s

Page 19 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

Volume

Surface area

Volume

NOTE:

The surface area of any 3-D object is the sum of the areas of all its faces.

The volume of any 3-D object is given by area of the base Γ— heght.

28. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)

Teaching activities

h

b

Page 20 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

Present the following examples in activity 1 to learners:

Activity 1:

Example 1:

a) The volume of a prism is 350 π‘š3. What is the height of the prism if its length is 10 π‘š and its breadth is 5 π‘š?

b) Calculate the volume of a prism with a surface base of 48 π‘š2 and a height of 4 π‘š.

Example 2

a) Calculate the capacity of a rectangular prism with the following inside measurements: length = 3 π‘š,

breadth 2 π‘š and height = 1,5 π‘š

b) A water tank has a square base with internal edge lengths of 15 π‘π‘š. What is the height of the tank

when the maximum capacity of the tank is 11 250 π‘π‘š3?

Activity 2

The volume of a cube is 64 π‘π‘š3. a) Determine the length of each side face. b) Determine the surface area of the cube.

29. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)

Page 21 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

Emphasise that:

the volume of prism = area of the base Γ— height

the surface area of the prism = the sum of the area of all its faces

the volume of a cube = 𝑠3 or 𝑙3

the volume of a rectangular prism = 𝑙 Γ— 𝑏 Γ— β„Ž

the amount of space inside the prism is called its capacity

the amount of space occupied by a prism is called its volume

MEMORANDUM: DAY 6

ACTIVITY SOLUTION Activity 1, example

1(a). 𝑙 Γ— 𝑏 Γ— β„Ž = 𝑉

10 π‘š Γ— 5 π‘š Γ— β„Ž = 350 π‘š3

50 π‘š2 Γ— β„Ž = 350 π‘š3

β„Ž = 7 π‘š

(b). 𝑉 = 𝑙 Γ— 𝑏 Γ— β„Ž

= 48 π‘š2 Γ— 4 π‘š

= 192 π‘π‘š3

Activity 1, example

2(a). πΆπ‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦ = 𝑙 Γ— 𝑏 Γ— β„Ž

= 3 π‘š Γ— 2 π‘š Γ— 1,5 π‘š

= 9 π‘š3

(b). 𝑙 Γ— 𝑏 Γ— β„Ž = πΆπ‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦

15 π‘π‘š Γ— 15 π‘π‘š Γ— β„Ž = 11 250 π‘π‘š3

225 π‘π‘š2 Γ— β„Ž = 11 250 π‘π‘š3

β„Ž = 50 π‘π‘š

Activity 2

2(a) Side length of one face = √64 π‘π‘š33= 4 π‘π‘š

(b) Surface area of the cube = 6 Γ— 𝑠2

= 6 Γ— (4 π‘π‘š)2

= 6 Γ— 16 π‘π‘š2

= 96 π‘π‘š2

π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π‘π‘Žπ‘ π‘’ = 𝑙 Γ— 𝑏 = 48 π‘š2

Solve by inspection by asking: 224 multiplied by what will be 11 250

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Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

30. PRIOR KNOWLEDGE

Volume and capacity of a cube and rectangular prisms done in the

previous lessons.

Conversion between appropriate units done in area and perimeter

of 2-D shapes.

31. INTRODUCTION (Suggested time: 10 Minutes)

Revise the conversion between appropriate SI- units as shown in the table below:

Hence, we see from the table that:

1 π‘π‘š2 = 100 π‘šπ‘š2 1 π‘šπ‘š2 = 0,01 π‘π‘š2

1 π‘š2 = 10 000 π‘π‘š2 1 π‘π‘š2 = 0,0001 π‘š2

POSING A PROBLEM: How many π‘šπ‘š3 would fit into π‘π‘š3?

How many π‘π‘š3 would fit into π‘š3?

To convert Do this To convert Do this

π‘π‘š π‘‘π‘œ π‘šπ‘š Γ— 10 π‘šπ‘š π‘‘π‘œ π‘π‘š Γ· 10

π‘š π‘‘π‘œ π‘π‘š Γ— 100 π‘π‘š π‘‘π‘œ π‘š Γ· 100

π‘π‘š2 π‘‘π‘œ π‘šπ‘š2 Γ— 100 π‘šπ‘š2 π‘‘π‘œ π‘π‘š2 Γ· 100

π‘š2 π‘‘π‘œ π‘π‘š2 Γ— 10 000 π‘π‘š2 π‘‘π‘œ π‘š2 Γ· 10 000

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Grade 7 Lesson: Term 2

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33. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)

Emphasise that:

To convert Do this To convert Do this

π‘π‘š3π‘‘π‘œ π‘šπ‘š3 Γ— 1 000 π‘šπ‘š3 to π‘π‘š3 Γ· 1 000

π‘š3π‘‘π‘œ π‘π‘š3 Γ— 1 000 000 π‘π‘š3 to π‘š3 Γ· 1 000 000

32. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)

Learners must complete example 4, below.

Example 1: Convert π‘π‘š3 to π‘šπ‘š3

Example 2: Work out how many π‘šπ‘š3 are equal to 1 π‘π‘š3?

Example 3: Convert π‘šπ‘š3 to π‘π‘š3

Example 4:

1. Write the following volumes in π‘π‘š3:

a) 3000 π‘šπ‘š3

b) 50 π‘šπ‘š3

c) 4 450 π‘šπ‘š3

d) 2,23 π‘š3

2. Write the following volumes in π‘š3:

a) 500 000 π‘π‘š3

b) 350 000 π‘π‘š3

c) 4 000 000 π‘π‘š3

Page 24 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

MEMORANDUM: DAY 7

ACTIVITY SOLUTION 34. LESSON PRESENTATION/ DEVELOPMENT

Example 1 Solution: To convert 1 π‘π‘š3 to π‘šπ‘š3 is the same as

finding out how many π‘šπ‘š3 would fit into 1 π‘π‘š3?

Consider the sketch below which shows cube A with a length edge of 1 π‘š. Also shown is cube B with an

edge length of 1 π‘π‘š.

How many small cubes can fit into the large cube?

100 small cubes can fit along the length of the

base of cube A (because there are 100 π‘π‘š in 1

π‘š).

100 small cubes can fit along the breadth of the

base of cube A.

100 small cubes can fit along the height of cube A.

The total number of 1 π‘π‘š3 cubes in 1 π‘š3 = 100 Γ—

100 Γ— 10 =

1000 000

1 π‘š3 = 1 000 000 π‘π‘š3

Example 2 1π‘π‘š3 = 1 π‘π‘š Γ— 1 π‘π‘š Γ— 1 π‘π‘š

= 10 π‘šπ‘š Γ— 10 π‘šπ‘š Γ— 10 π‘šπ‘š

= 1 000 π‘šπ‘š3

Example 3 Solution: 1 π‘šπ‘š3 = 1 π‘šπ‘š Γ— 1 π‘šπ‘š Γ— 1 π‘šπ‘š

= 0,01 π‘π‘š Γ— 0,01 π‘π‘š Γ— 0,01 π‘π‘š

= 0,001 π‘π‘š3

Example 4 1. Write the following volumes in π‘π‘š3:

π‘Ž) 3000 π‘šπ‘š3 = 3 π‘π‘š3

𝑏) 50 π‘šπ‘š3 = 0,05 π‘π‘š3

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Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

c) 450 π‘šπ‘š3 = 4,45 π‘π‘š3

d) 2,23 π‘š3 = 2 230 000 π‘π‘š3

2. Write the following volumes in π‘š3:

π‘Ž) 500 000 π‘π‘š3 = 0,5 π‘š3

𝑏) 350 000 π‘π‘š3 = 0,35 π‘š3

c) 4 000 000 π‘π‘š3 = 4 π‘š3

Page 26 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

34. PRIOR KNOWLEDGE Conversion between appropriate SI-units done in the previous

lesson

35. INTRODUCTION (Suggested time: 10 Minutes)

Activity 1: Learners revise the definition of the following concepts:

Volume: The amount of space occupied by a 3-D object.

Capacity: The amount of space inside a 3-D object.

Activity 2: Learners must see the equivalence between units of volume and capacity by presenting

the following scenario:

Page 27 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

36. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)

Present the following examples to learners:

Examples:

1. Write the following volumes in π‘šπ‘™:

a) 2 000 π‘π‘š3

b) 2,5 𝑙

c) 1 𝑙

2. Write the following volumes in π‘˜π‘™:

a) 6 500 π‘š3

b) 20 𝑙

c) 1 423 000 π‘π‘š3

3. A glass can hold up to 250 π‘šπ‘™ of water. What is the capacity of the glass:

a) in π‘šπ‘™?

b) in π‘π‘š3?

4. A glass tank has the following inside measurements: length = 250 π‘šπ‘š, breadth =

120 π‘šπ‘š and height = 100 π‘šπ‘š.

Calculate the capacity of the tank in millilitres.

37. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)

a. Emphasise that:

1 𝑙 = 1 000 π‘šπ‘™ and 1 π‘šπ‘™ = 0,001 𝑙

1 π‘˜π‘™ = 1 000𝑙 and 1 𝑙 = 0,001 π‘˜π‘™

1 𝑙 = 1 000 π‘π‘š3

1 π‘π‘š3 = 1 π‘šπ‘™ and 1 π‘š3 = 1 π‘˜π‘™

b. Homework:

The primary purpose of Homework is to give each learner an opportunity to demonstrate mastery

of mathematics skills taught in class. Therefore, Homework should be purposeful and the principle

of β€˜Less is more’ is recommended, i.e. give learners few high quality activities that address variety

of skills than many activities that do not enhance learners’ conceptual understanding. Carefully

select appropriate activities from the DBE workbooks and/or textbooks for learners’ homework. The

selected activities should address different cognitive levels.

Page 28 of 28

Grade 7 Lesson: Term 2

SURFACE AREA AND VOLUME OF 3-D OBJECTS

MEMORANDUM: DAY 8

ACTIVITY SOLUTION

38. LESSON PRESENTATION/

DEVELOPMENT

Write the following volumes in π‘šπ‘™:

d) 2 000 π‘π‘š3 2 000 π‘šπ‘™

e) 2,5 𝑙 2 500 π‘šπ‘™

f) 1 𝑙 1 000 π‘šπ‘™

2. Write the following volumes in π‘˜π‘™:

π‘Ž) 6 500 π‘š3 6 500 π‘˜π‘™

𝑏) 20 𝑙 0,02 π‘˜π‘™

𝑐) 1 423 000 π‘π‘š3 1,423 π‘˜π‘™

3.

a) in π‘šπ‘™? 250 π‘šπ‘™

b) in π‘π‘š3? 250 π‘π‘š3

4. Solution:

Capacity = 250 π‘šπ‘š Γ— 120 π‘šπ‘š Γ— 100 π‘šπ‘š

= 3 000 000 π‘šπ‘š3

= 3000 π‘π‘š3

= 3000 π‘šπ‘™

Or

Capacity = 250 π‘šπ‘š Γ— 120 π‘šπ‘š Γ— 100 π‘šπ‘š

= 25 π‘π‘š Γ— 12 π‘π‘š Γ— 10 π‘π‘š

= 3000 π‘π‘š3

= 3000 π‘šπ‘™

Divide by 1 000

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