mathematics lesson
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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:
ππ2 β ππ2, ππ2 β π2, ππ3 β ππ3and ππ3 β π3
2.7. use equivalence between units when solving problems:
1 cm3 β 1 ml and 1 m3 β 1 kl
TOPIC: SURFACE AREA AND VOLUME OF 3-D OBJECTS
Lesson: Surface area and volume
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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 1ππ by 1ππ by 1ππ:
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
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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 learnersβ calculations and teacherβs 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 ππ Γ 7 ππ Γ 7 ππ. Calculate its
i. Volume
ii. surface area
5 ππ
Volume of a cube = SΒ³
3 ππ
π
Surface area of a cube= 6SΒ²
π
3 ππ
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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 ππ Γ 2,5 ππ Γ 2,5 ππ. 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 = π 2 = (3 ππ)2 = 9 ππ2
(f) 3 lae
(g) Yes, Area of the cube x number of layers = Area of
base x height
= π 2 Γ π
= 9 ππ2 Γ 3 ππ
= 27 ππ3
Volume of a cube = SΒ³
Surface area a cube= 6SΒ²
4 ππ
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SURFACE AREA AND VOLUME OF 3-D OBJECTS
2. Surface area of a cube = Total of Area of all faces
= 9 ππ2 + 9 ππ2 + 9 ππ2 + 9 ππ2 + 9 ππ2 +
9 ππ2
= 6 Γ 9 ππ2
= 54 ππ2
Hence the surface area of a cube = 6 Γ π 2
6.PRESENTATION/DEVELOPMENT
Aktivity1(a) Volume = π 3
= (3 ππ)3
= 27 ππ3
(b) Volume = π Γ π Γ π
= π 3
Aktivity 2(a) Surface area = π 2 Γ 6
= (3 ππ)2 Γ 6
= 9 ππ2 Γ 6
= 54 ππ2
(b) Surface area = π Γ π Γ 6
= 6 π 2
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
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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 ππ Γ 1 ππ Γ 1 ππ 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)
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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 learnersβ calculations and teacherβs 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 π x π x β
Surface area of a rectangular prism is 2(πβ + πβ + ππ)
b) Homework: Educators must choose appropriate exercises from the textbook.
MEMORANDUM: DAY 2
ACTIVITY SOLUTION
11. INTRODUCTION
1(a) 210 cubes
(b) 210 ππ3
(c) 30 cubes
(d) Length = 6 cubes and the breadth = 5 cubes
(e) Area of the base = π Γ π = 6 ππ Γ 5 ππ = 30 ππ2
(f) 7 layers
(g) Yes,
Volume of a rectangular prism = l x b x h
Surface area of a rectangular prism = 2(πβ + πβ + ππ)
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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
= π Γ π Γ β
= 6 ππ Γ 5 ππ Γ 7
= 210 ππ3
2(a) 2(7ππ Γ 5ππ)
= 2(35 ππΒ²)
= 70 ππ2
(b) 2(6ππ Γ 5ππ)
= 2(30 ππΒ²)
= 60 ππ2
(c) 2(7ππ Γ 6ππ)
= 2(42 ππΒ²)
= 84 ππ2
3. 70ππΒ² + 60ππΒ² + 84ππΒ²= 214 ππΒ²
12.LESSON
PRESENTATION/DEVELOPMENT
Activity 1(a) 48 cmΒ³
(b) 30 cm3
(c) β‘ cubic units
Activity 2
(a)
2(8 ππ x 3 ππ) + 2(8 ππ x 2 ππ) + 2(3 ππ Γ 2 ππ)
= 2(24 ππΒ²) + 2(16 ππΒ²) + 2(6 ππΒ²)
= 48 ππΒ² + 32 ππΒ² + 12 ππΒ²
= 92 ππΒ²
(b) 2(5ππ x 3ππ) + 2(3ππ x 2ππ) + 2(5ππ Γ 2ππ)
= 2(15 ππΒ²) + 2(6 ππΒ²) + 2(10 ππΒ²)
= 30 ππΒ² + 12 ππΒ² + 20 ππΒ²
= 62 ππΒ²
(c) 2(π x β) + 2(π Γ β) + 2(π Γ π)
= 2(πβ + πβ + ππ)
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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 x 1 π x 1 π
2 π x 2 π x 2 π
3 π x 3 π x 3 π
4 π x 4 π x 4 π
5 π x 5 π x 5 π
8 π x 8 π x 8 π
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
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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 π = π 3
= 33ππ3
= 27ππ3
ππ΄ = 6 Γ π 2
= 6 Γ 32ππ2
= 6 Γ 9ππ2
= 54ππ2
16. a) Size of Cube Volume Surface Area
1 π x 1 π x 1 π 1π3 6 Γ (1π)2 = 6π2
2 π x 2 π x 2 π 8π3 6 Γ (2π)2 = 24π2
3 π x 3 π x 3 π 27π3 54π2
4 π x 4 π x 4 π 64π3 96π2
5 π x 5 π x 5 π 125π3 150π2
8 π x 8 π x 8 π 512π3 384π2
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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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
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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 x 1 π x 16 π
1 π x 2 π x 8 π
1 π x 4 π x 4 π
2 π Γ 2 π Γ 4 π
Note: 1 m x 1 m x 16 m has a one side with the greatest dimension
π = 2 ππ π = 4 ππ
β = 3 ππ
6 ππ
2 ππ
2 ππ 1 ππ
1 ππ
24 ππ
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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 Γ 2π Γ 3π
ii. 4π Γ 3π Γ 3π
iii. 9π Γ 2π Γ 2π
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.
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MEMORANDUM: DAY 4
ACTIVITY SOLUTION 20. INTRODUCTION
Name Object Volume Surface Area
Rectangular
Prism
π = π Γ π Γ β
= 4ππ Γ 2ππ Γ
3ππ
= 24ππ3
ππ΄ = 2πβ + 2πβ + 2ππ
= 2(4 ππ Γ 3 ππ) + 2(2 cmΓ 3 ππ) +
2(4 ππ Γ 2 ππ)
= 24 ππ2 + 12 ππ2 + 16 ππ2
= 52ππ2
21. LESSON PREPARATION/DEVELOPMENT
Activity: (a).
(i) π = π Γ π Γ β (ii) π = π Γ π Γ β
= 6 ππ Γ 2 ππ Γ 2 ππ = 24 ππ Γ 1 ππ Γ 1 ππ
= 24 ππΒ³ = 24 ππΒ³
(b). (i) ππ’πππππ ππππ = 2ππ + 2πβ + 2πβ
= 2 Γ 6 ππ Γ 2 ππ + 2 Γ 6 ππ Γ 2 ππ + 2 Γ 2 ππ Γ 2 ππ
= 56 ππΒ²
(ii) ππ’πππππ ππππ = 2ππ + 2πβ + 2πβ
= 2 Γ 24 ππ Γ 1 ππ + 2 Γ 24 ππ Γ 1 ππ + 2 Γ 1 ππ Γ 1 ππ
= 98 ππΒ²
(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 x 1 π x 16 π 16 π3 66 π2
1 π x 2 π x 8 π 16 π3 52 π2
1 π x 4 π x 4 π 16 π3 48 π2
2 π Γ 2 π Γ 4 π 16 π3 40 π2
22. CLASSWORK
(a). i V= l Γb Γh = 36 m3 (a). ii V= l Γb Γh = 36 m3
π = 2 ππ π = 4 ππ
β = 3 ππ
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Grade 7 Lesson: Term 2
SURFACE AREA AND VOLUME OF 3-D OBJECTS
(a). iii V= l Γb Γh = 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
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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 ππ?
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 = π 3 or π3
the volume of a rectangular prism = π Γ π Γ β
b) Homework:
Educators choose appropriate exercises from the textbook.
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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
ππ’πππππ π΄πππ = 6 Γ π΄πππ ππ πππ ππππ
= 6 Γ π 2
Volume
π = π 3
RECTANGULAR PRISM
Surface area
ππ’πππππ ππππ = 2 Γ ππ + 2 Γ πβ + 2 Γ πβ
= 2(ππ + πβ + πβ)
Volume
π = π Γ π Γ β
26. LESSON
PRESENTATION/DEVELOPMENT
Example 1: (a).
Surface area = 2(ππ + πβ + πβ) = 2(8 ππ Γ 5 ππ + 8 ππ Γ 2 ππ + 5 ππ Γ2 ππ) = 2(40 ππ2 + 16 ππ2 + 10 ππ2) = 2(66 ππ2) = 132 ππ2
(b). Volume = π Γ π Γ β = 8 ππ Γ 5 ππ Γ 2 ππ
= 80 ππ3
(c). Capacity = 80 ππ
Example 2 (a).
Surface area = 6 Γ π 2
= 6 Γ (5 ππ)2
= 6 Γ 25 ππ2
= 150 ππ2
(b). Volume = π 3
= (5 ππ)3
= 125 ππ3
s s
s
h
b
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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
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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
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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 π3. What is the height of the prism if its length is 10 π and its breadth is 5 π?
b) Calculate the volume of a prism with a surface base of 48 π2 and a height of 4 π.
Example 2
a) Calculate the capacity of a rectangular prism with the following inside measurements: length = 3 π,
breadth 2 π and height = 1,5 π
b) A water tank has a square base with internal edge lengths of 15 ππ. What is the height of the tank
when the maximum capacity of the tank is 11 250 ππ3?
Activity 2
The volume of a cube is 64 ππ3. a) Determine the length of each side face. b) Determine the surface area of the cube.
29. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)
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Grade 7 Lesson: Term 2
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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 = π 3 or π3
the volume of a rectangular prism = π Γ π Γ β
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). π Γ π Γ β = π
10 π Γ 5 π Γ β = 350 π3
50 π2 Γ β = 350 π3
β = 7 π
(b). π = π Γ π Γ β
= 48 π2 Γ 4 π
= 192 ππ3
Activity 1, example
2(a). πΆππππππ‘π¦ = π Γ π Γ β
= 3 π Γ 2 π Γ 1,5 π
= 9 π3
(b). π Γ π Γ β = πΆππππππ‘π¦
15 ππ Γ 15 ππ Γ β = 11 250 ππ3
225 ππ2 Γ β = 11 250 ππ3
β = 50 ππ
Activity 2
2(a) Side length of one face = β64 ππ33= 4 ππ
(b) Surface area of the cube = 6 Γ π 2
= 6 Γ (4 ππ)2
= 6 Γ 16 ππ2
= 96 ππ2
ππ’πππππ πππ π = π Γ π = 48 π2
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 ππ2 = 100 ππ2 1 ππ2 = 0,01 ππ2
1 π2 = 10 000 ππ2 1 ππ2 = 0,0001 π2
POSING A PROBLEM: How many ππ3 would fit into ππ3?
How many ππ3 would fit into π3?
To convert Do this To convert Do this
ππ π‘π ππ Γ 10 ππ π‘π ππ Γ· 10
π π‘π ππ Γ 100 ππ π‘π π Γ· 100
ππ2 π‘π ππ2 Γ 100 ππ2 π‘π ππ2 Γ· 100
π2 π‘π ππ2 Γ 10 000 ππ2 π‘π π2 Γ· 10 000
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33. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)
Emphasise that:
To convert Do this To convert Do this
ππ3π‘π ππ3 Γ 1 000 ππ3 to ππ3 Γ· 1 000
π3π‘π ππ3 Γ 1 000 000 ππ3 to π3 Γ· 1 000 000
32. LESSON PRESENTATION/DEVELOPMENT (Suggested time: 20 minutes)
Learners must complete example 4, below.
Example 1: Convert ππ3 to ππ3
Example 2: Work out how many ππ3 are equal to 1 ππ3?
Example 3: Convert ππ3 to ππ3
Example 4:
1. Write the following volumes in ππ3:
a) 3000 ππ3
b) 50 ππ3
c) 4 450 ππ3
d) 2,23 π3
2. Write the following volumes in π3:
a) 500 000 ππ3
b) 350 000 ππ3
c) 4 000 000 ππ3
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Grade 7 Lesson: Term 2
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MEMORANDUM: DAY 7
ACTIVITY SOLUTION 34. LESSON PRESENTATION/ DEVELOPMENT
Example 1 Solution: To convert 1 ππ3 to ππ3 is the same as
finding out how many ππ3 would fit into 1 ππ3?
Consider the sketch below which shows cube A with a length edge of 1 π. Also shown is cube B with an
edge length of 1 ππ.
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 ππ in 1
π).
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 ππ3 cubes in 1 π3 = 100 Γ
100 Γ 10 =
1000 000
1 π3 = 1 000 000 ππ3
Example 2 1ππ3 = 1 ππ Γ 1 ππ Γ 1 ππ
= 10 ππ Γ 10 ππ Γ 10 ππ
= 1 000 ππ3
Example 3 Solution: 1 ππ3 = 1 ππ Γ 1 ππ Γ 1 ππ
= 0,01 ππ Γ 0,01 ππ Γ 0,01 ππ
= 0,001 ππ3
Example 4 1. Write the following volumes in ππ3:
π) 3000 ππ3 = 3 ππ3
π) 50 ππ3 = 0,05 ππ3
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c) 450 ππ3 = 4,45 ππ3
d) 2,23 π3 = 2 230 000 ππ3
2. Write the following volumes in π3:
π) 500 000 ππ3 = 0,5 π3
π) 350 000 ππ3 = 0,35 π3
c) 4 000 000 ππ3 = 4 π3
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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:
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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 ππ:
a) 2 000 ππ3
b) 2,5 π
c) 1 π
2. Write the following volumes in ππ:
a) 6 500 π3
b) 20 π
c) 1 423 000 ππ3
3. A glass can hold up to 250 ππ of water. What is the capacity of the glass:
a) in ππ?
b) in ππ3?
4. A glass tank has the following inside measurements: length = 250 ππ, breadth =
120 ππ and height = 100 ππ.
Calculate the capacity of the tank in millilitres.
37. CONSOLIDATION/CONCLUSION & HOMEWORK (Suggested time: 5 minutes)
a. Emphasise that:
1 π = 1 000 ππ and 1 ππ = 0,001 π
1 ππ = 1 000π and 1 π = 0,001 ππ
1 π = 1 000 ππ3
1 ππ3 = 1 ππ and 1 π3 = 1 ππ
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 βLess is moreβ is recommended, i.e. give learners few high quality activities that address variety
of skills than many activities that do not enhance learnersβ conceptual understanding. Carefully
select appropriate activities from the DBE workbooks and/or textbooks for learnersβ homework. The
selected activities should address different cognitive levels.
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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 ππ:
d) 2 000 ππ3 2 000 ππ
e) 2,5 π 2 500 ππ
f) 1 π 1 000 ππ
2. Write the following volumes in ππ:
π) 6 500 π3 6 500 ππ
π) 20 π 0,02 ππ
π) 1 423 000 ππ3 1,423 ππ
3.
a) in ππ? 250 ππ
b) in ππ3? 250 ππ3
4. Solution:
Capacity = 250 ππ Γ 120 ππ Γ 100 ππ
= 3 000 000 ππ3
= 3000 ππ3
= 3000 ππ
Or
Capacity = 250 ππ Γ 120 ππ Γ 100 ππ
= 25 ππ Γ 12 ππ Γ 10 ππ
= 3000 ππ3
= 3000 ππ
Divide by 1 000