SK Year 5 Mathematics DLP

Name Mass
SUBTRACTION OF UNITS OF MASS
Khairul 34.9 kg
Divesh 41.2 kg
1 Calculate the difference in mass Khairul Divesh
between Khairul and Divesh.

41.2 kg – 34.9 kg = kg
10
3 0 12
4 1 . 2 kg
− 3 4 . 9 kg
6 . 3 kg

41.2 kg – 34.9 kg = 6.3 kg

The difference in mass between Khairul and Divesh is 6.3 kg.


2 How much more is the mass of butter than

Butter Cake Recipe the mass of sugar? State the answer in g.
Ingredients A 1 kg – 0.1 kg = g
1 4
• kg of butter
4 Method 1
• 0.1 kg of sugar
1 1 250
• 5 yolks kg – 0.1 kg = ( × 1 000) g – (0.1 × 1 000) g
4 4
• 230 g of 1
wheat flour = 250 g – 100 g

= g
Method 2


0.25 kg 1 kg – 0.1 kg = 0.25 kg – 0.10 kg
4 1.00 kg 4
1 −0 = 0.15 kg
Convert kg
4 1 0 = (0.15 × 1 000) g
to decimal kg − 8
first. 20 = g
− 20
0 Are the answers
for method 1 and
method 2 the same?


The mass of butter is g more than the mass of sugar.


• Carry out activities such as to find the difference in mass between two friends. 193
5.2.3 • Encourage pupils to subtract decimals using vertical form to ensure that the
units are the same and the decimal points are aligned.

3 6 042 g – 1.07 kg – 980 g = kg
Convert unit of g to kg Subtracting consecutively


6 042 g = (6 042 ÷ 1 000) kg 9 18
5 1014 3 8 17
= 6.042 kg 6 .0 4 2 kg 4 .9 7 2 kg
– 1 .0 7 0 kg – 0 .9 8 0 kg
980 g = (980 ÷ 1 000) kg 4 .9 7 2 kg 3 .9 9 2 kg

= 0.980 kg


6 042 g – 1.07 kg – 980 g = 3.992 kg

4
4 10 kg – 1 503 g – 3.96 kg = g
5

Convert unit of kg to g Subtracting consecutively

4 4 1 0 8 0 0 g
10 kg = 10 kg + kg
5 5 4 200 – 1 5 0 3 g
= (10 × 1 000) g + ( × 1 000) g g
5
1 – 3 9 6 0 g
= 10 000 g + 800 g
g
= 10 800 g


3.96 kg = (3.96 × 1 000) g 1 p
= 3 960 g 1 kg – 800 g = kg
q
5
Find values of p and q.



1 Calculate.
3
a 7.1 kg – 2.54 kg = kg b kg – 0.69 kg = g
7 4
c 5 120 g – 1 kg – 2.093 kg = kg
10
1
d 10 kg – 4 860 g – 3.72 kg = g
2
1
2 a Calculate the difference between 38.92 kg and 60 kg.
5
b How much more is 13.051 kg than 7 360 g?
2
c Subtract 5 kg from 9.018 kg. State the answer in g.
5

• Carry out a question and answer session involving subtraction of mass
194 using words such as calculate the balance, find the difference, and how
5.2.3 much more is a particular mass from another mass.

May I help you,
MULTIPLICATION OF UNITS OF MASS madam?


1 Calculate the mass of 7 packets of hazelnuts.
7 × 0.25 kg = kg

1 3
0 . 2 5 kg
× 7 kg
1 . 7 5 kg

7 × 0.25 kg = 1.75 kg I want 7 packets of
The mass of 7 packets of hazelnuts is 1.75 kg. hazelnuts.


2 What is the mass of 15 packets of crisps?
1
15 × kg = g
2
1 1 500
15 × kg = 15 × ( × 1 000) g
2 2
1 2
= 15 × 500 g 1 5
× 5 0 0 g
= 7 500 g
1 7 5 0 0 g 1
15 × kg = 7 500 g kg
2 2
The mass of 15 packets of crisps is 7 500 g.

3
3 10 × 40 kg = g
5
Method 1 Method 2

3 3 3 3
40 kg = 40 kg + kg 10 × 40 kg = 10 × (40 kg + kg)
5 5 5 5
3 200 = 10 × (40 kg + 0.6 kg)
= (40 × 1 000) g + ( × 1 000) g
5 = 10 × 40.6 kg
1
= 40 000 g + 600 g = 406 kg

= 40 600 g
406 kg = (406 × 1 000) g
10 × 40 600 g = 406 000 g = 406 000 g

3
10 × 40 kg = 406 000 g
5



• Encourage pupils to do quick calculation when multiplying any 195
5.2.4
mass by 10, 100 or 1 000.

4 No. Number of packets Brand of cat food Mass per unit
1. 100 Meow Meow 1.2 kg
2. 1 000 Comel 0.85 kg

3. 1 000 Anggun 480 g
Based on the table above, calculate the total mass of:

a Meow Meow. b Comel.

100 × 1.2 kg = kg 1 000 × 0.85 kg = g

100 × 1.2 kg = 120 kg 1 000 × 0.85 kg = 850 kg

100 × 1.2 kg = 120 kg = (850 × 1 000) g
= 850 000 g
The total mass of Meow Meow
is 120 kg. 1 000 × 0.85 kg = 850 000 g

The total mass of Comel
is 850 000 g.




Calculate the total mass,
in kg, of Anggun.








1 Calculate.

a 9 × 1.07 kg = kg b 16 × 4.5 g = kg
1
c 24 × 3 kg = g d 50 × 6.78 kg = g
10
2 Quick calculation.
4
a 10 × 2.06 kg = kg b 10 × kg = g
5
1
c 100 × 1.9 kg = g d 100 × 1 kg = g
4
7
e 1 000 × 0.003 kg = g f 1 000 × 3 kg = kg
10





196 • Drill multiplication using flash cards. Encourage pupils to calculate
5.2.4
mentally when multiplying any mass by 10, 100 or 1 000.

DIVISION OF UNITS OF MASS



1 Calculate the mass of one chocolate bar.
0.35 kg ÷ 7 = kg The mass of 7

0.0 5 kg chocolate bars
7 0.3 5 kg is 0.35 kg.
−0
0 3
− 0
3 5
− 3 5
0

0.35 kg ÷ 7 = 0.05 kg
The mass of one chocolate bar is 0.05 kg.



1
2 12.33 kg ÷ 18 = g 3 16 kg ÷ 10 = g
2
0 0.6 8 5 kg 1
1
18 1 2.3 3 0 kg 16 kg = 16 kg + kg
2
2
−0 = 16 kg + 0.5 kg
1 2
− 0 = 16.5 kg
1 2 3
− 1 0 8 16.5 kg ÷ 10 = 1.65 kg
1 5 3
− 1 4 4 = (1.65 × 1 000) g
9 0 = 1 650 g
− 9 0
0 1
16 kg ÷ 10 = 1 650 g
2
0.685 kg = (0.685 × 1 000) g
= 685 g
Try another method by converting
12.33 kg ÷ 18 = 685 g 1
16 kg to g first, then divide.
2








• Guide pupils to do a group activity. Pupils will weigh suitable items 197
5.2.5 and divide the number of packets of item according to the same
mass by giving the answers in the required units.

4 5 1.4 kg ÷ 1 000 = g

1.4 kg ÷ 1 000 = ( × ) g ÷ 1 000

= g ÷ 1 000

The mass of 100 pieces of = g
1
macaroon is 1 kg.
2
Calculate the mass, in g, of one Try to solve it.
piece of macaroon.
1
1 kg ÷ 100 = g
2
1 3 500
1 kg ÷ 100 = ( × 1 000) g ÷ 100
2 2
1
1 500
= g 0 4 2 1 3
100
= 15 g Use all the number cards above to

1 complete the empty boxes below.
1 kg ÷ 100 = 15 g
2
The mass of one piece of . kg = . kg ÷
macaroon is 15 g.





1 Calculate.


a 9.02 kg ÷ 5 = kg b 10.2 kg ÷ 6 = kg
3
c 13.51 kg ÷ 14 = g d 7 kg ÷ 50 = g
4
4
e 21.1 kg ÷ 10 = g f 36 kg ÷ 100 = kg
5
7
g 6.8 kg ÷ 1 000 = g h 9 kg ÷ 1 000 = g
10
2 Based on the diagram on the left,
complete the empty boxes below and

Granulated solve it.
Granulated
Sugar
Sugar
4.5 kg kg ÷ = g
4.5 kg

198 • Guide pupils to apply cancellation method or shift the decimal point
5.2.5
when dividing any mass by 10, 100 and 1 000.

CONVERT UNITS OF MILLILITRE AND LITRE



1 a Convert 0.3 to m .

0.3 = m

This bottle 0.3 = (0.3 × 1 000) m
contains This box
1
0.3 of juice. contains = 300 m
2
of juice.
0.3 = 300 m



1
1 Is 0.5 equal to ?
2
b = m Explain.
2
1 1 500
= ( × 1 000) m 1.0 1 000 m
2 2
1
= 500 m 0.5 500 m

1
= 500 m
2



2 2 750 m =


in decimal in fraction


2 750 m 2 750 m = 2 000 m + 750 m

= (2 750 ÷ 1 000) = (2 000 ÷ 1 000) + ( 750 )

= 2.75 75 ÷ 25 1 000
= 2 + ( )
100 ÷ 25
3
= 2 +

4 1 = 1 000 m
3
= 2 × 1 000
4
m

3
2 750 m = 2.75 or 2 ÷ 1 000
4



• Ask pupils to measure the volume of coloured liquid using 199
5.3.1 measuring apparatus and convert the units into fractions or
decimals.

3 42 9 m = 4 700 m =
9 700
7

42 9 m = 42 + ( ) 700 m = ( ) Is equal to
1 000 1 000 10
= 42 + 0.009 7 0.7 ? Discuss.
=
= 42.009 10
42 9 m = 42.009 700 m = 7
10


3
5 6 025 m = 6 27 = m
6 025 5
6 025 m = ( ) 3 3
1 000 27 = 27 +
6 000 25 5 5
= ( + ) = (27 × 1 000) m
1 000 1 000
25 ÷ 25 3 200
= 6
1 000 ÷ 25 + ( × 1 000) m
5
1
= = m + m

= m




1 Convert to m or vice versa.

a b c d e
Pomegranate
Juice




0.35 1.2 15 m 200 m 1 400 m

2 State the answers in fractions.



a 800 m = b 1 200 m = c 3 900 m =
3 State the answers in decimals.




a 5 m = b 10 080 m = c 26 40 m =
4 Solve these.
1 7
a 6 = m b 5 = m
8 10

200 • Use fractions with denominators 2, 4, 5, 8 and 10 to reinforce
5.3.1 pupils’ understanding involving the conversion of units of volume of
liquid.

ADDITION OF UNITS OF VOLUME OF LIQUID


1












1
5 1.875 125 m
2
a Calculate the total volume of blue and yellow paints.
1
1.875 + 5 =
2 0 5.5
1 11 2 1 1 .0
1.875 + 5 = 1.875 +
2 2 −0
1 1
= 1.875 + 5.5 − 1 0
1 0
= 7.375 1
1 .8 7 5 − 1 0
+ 5.5 0 0 0
7 .3 7 5
1
1.875 + 5 = 7.375
2
The total volume of blue and yellow paints is 7.375 .

b Find the total volume of blue and red paints.


1.875 + 125 m = m


1.875 + 125 m = (1.875 × 1 000) m + 125 m
= 1 875 m + 125 m
= 2 000 m

1.875 + 125 m = 2 000 m

The total volume of blue and red paints is 2 000 m .


The total volume of the three paints is more
than 8 . Is this statement true? Prove it.



201
5.3.2 • Stress that the unit of volume must be the same before
performing addition.

2

2 9.27 + 13 500 m + 10 =
5
13 500 m = 13 500 m 2 2
10 = 10 + 0.4


13 500 5 5 5 2.0
= ( ) = 10 + 0.4

1 000 −0
= 13.5 = 10.4 2 0
−2 0
1 1
9 .2 7 0
1 3 .5 0
+ 1 0 .4 0

3 3 . 1 7
2
9.27 + 13 500 m + 10 = 33.17

5

9

3 40.08 + 11 + 76 m = m
10
Method 1 9 9 Method 2
11 = 11 +
10 10 76 m = (76 ÷ 1 000)
= 11 + 0.9 0.076
= 11.9 =
1
1
= (11.9 × 1 000) m 40.0 8 5 1 .9 8 0
= 11 900 m + 1 1.9 0 + 0 .0 7 6
40.08 = (40.08 × 1 000) m 5 1.9 8 5 2 .0 5 6

= 40 080 m 52.056 = (52.056 × 1 000) m

4 0 0 8 0 m = m
1 1 9 0 0 m
+ 7 6 m
Are both methods above
m correct? Discuss.




Calculate.
3
a 3.8 + 9.204 = b 2 + 7.265 =
4
9

c 10.46 + 578 m = m d + 12 304 m =
10
1 1

e 3 + 680 m + 0.2 = f 8 645 m + 12 + 4.1 =
5 2
202
5.3.2 • Identify pupils’ weaknesses and provide more exercises to enhance
their understanding.

SUBTRACTION OF UNITS OF VOLUME OF LIQUID



1 What is the difference in volume of the
liquid in the syringes?

5 m – 3.3 m = m
4 10
5. 0 m
.
− 3. 3 m
.
.
1. 7 m
5 m – 3.3 m = 1.7 m

The difference in volume of the liquid in the syringes is1.7 m .


What is the balance of the volume of
2 I have filled up
2.5 juice the juice in the container?

into a jug. 1
10 – 2.5 =
4

Step 1 Step 2

1 1 9
10 = 10 + 0 10 12
Full 4 4
volume = 10 + 0.25 1 0 .25
1 − 2 .50
10
4 7 .75
= 10.25
1
10 – 2.5 = 7.75
4
The balance of the volume of the
juice in the container is 7.75 .


3 12.09 − 780 m = m


12.09 − 780 m = (12.09 × 1 000) m − 780 m 1 10
1 2 0 9 0 m
= 12 090 m − 780 m − 7 8 0 m

= 11 310 m 1 1 3 1 0 m


12.09 − 780 m = 11 310 m

203
5.3.3 • In groups, carry out an activity of subtracting two volumes using
a game of Dominoes.

2
4 8 – 1 360 m – 4.91 =
5
2
8 – 1 360 m – 4.91
5
1 360
= (8 + 0.4 – ( ) – 4.91 1 = 0.2 or 200 m
)
1 000 5

= 8.4 – – 4.91 2 = 0.4 or 400 m



= 5
1
5 20 35 m – 0.948 – 16 = m
8
Convert unit Subtracting consecutively

0.948 = (0.948 × 1 000) m m m

= 948 m 1 9 1 0 3 5 1 8 1 0 8 7
2 0 3 5 1 9 8 7
1 1
16 = 16 + − 9 4 8 − 1 6 1 2 5

8 8
1 125 1 9 8 7 2 9 6 2

= 16 + ( × 1 000) m
8
1

= 16 + 125 m
= 16 125 m

1
20 35 m – 0.948 – 16 = 2 962 m

8



1 Calculate.
1

a 10 m – 4.5 m = m b 13 – 1.85 =

2
3
c 7.025 – 629 m = m d 9 – 2 084 m = m

5
3
e 12 – 960 m – 8.47 =

4
1
f 8 – 3 640 m – 1.02 = m

10
2 Solve these.
1
a 15.24 – 6 – 120 m = m

8
1

b 6 320 m – 4.5 – 1 = m
2
204 • Carry out a question and answer session involving subtraction of
1
5.3.3 volume of liquid such as “Subtract 0.2 from equals


how many m ?”. 2

MULTIPLICATION OF UNITS OF VOLUME OF LIQUID


1 Calculate the total volume of 5 bottles
of cultured milk. Each bottle


5 × 0.08 = contains 0.08 .
4
0. 0 8
× 5
0. 4 0


5 × 0.08 = 0.4
The total volume of 5 bottles of cultured milk is 0.4 .


2 What is the total volume of water in the box? 48 x 230 m

48 × 0.23 = m

Method 1
1 Method 2
• Multiply. 1 2
0 .2 3 • Convert 0.23 to m
.
× 4 8 (0.23 × 1 000) m = 230 m

1 1
1 8 4 1
+ 0 9 2 0 • Multiply. 2
1 1 .0 4 2 3 0 m
× 4 8

• Convert 11.04 to m . 1
(11.04 × 1 000) m = 11 040 m 1 8 4 0

+ 9 2 0 0
48 × 0.23 = 11 040 m 1 1 0 4 0 m

The total volume of water in the box is 11 040 m .


3
3 10 × 1 =
5
3 2 8
10 × 1 = 10 ×
5 5
1
= 16
State the answer in m .
3
10 × 1 = 16
5


• Explain the meaning of cultured milk, which is the dairy products 205
5.3.4 produced when sterilised milk, pasteurised milk, or skimmed milk is
added with specifically cultured bacteria that helps in the digestion
process.

4 BBB

HAND
CLEAN WORLD FACTORY HAND 1
SANITIZER 50 m SANITISER 0.4
B Sanitiser 2
BB
,
a Calculate the volume, in m of 100 bottles of BB hand sanitiser.
100 × 0.4 = m



Method 1 Method 2
0 .4 40 = (40 × 1 000) m 0.4 = (0.4 × 1 000) m
× 1 0 0 = 40 000 m = 400 m
4 0 .0
100 × 400 m = 40 000 m
100 × 0.4 = 40 000 m

The volume of 100 bottles of BB hand sanitiser is 40 000 m .


b What is the volume, in m , of 1 000 bottles of BBB hand sanitiser?

500 1
1 000 × = do cancellation
2 Calculate the
1 = ( × ) m convert unit volume, in of 10
,
= m bottles of
B hand sanitiser.

The volume of 1 000 bottles of BBB hand sanitiser

is m .






1 Solve these.

a 16 × 0.39 = b 13 × 5.7 = m c 25 × 0.416 = m
1 7
d 48 × 2 = m e 100 × 6 = m f 1 000 × 7.8 m =
4 10
2 Complete these.
3
1 000 × 75 m 10 × 0.75 100 ×
4
as as
m



206 • Vary questions such as using a mind map to reinforce
5.3.4 pupils’ understanding.
• Instil moral values such as taking care of hygiene and health.

DIVISION OF UNITS OF VOLUME OF LIQUID



1 What is the volume of a bowl of mutton soup
based on the picture given?

2.5 ÷ 4 = I poured 2.5 of mutton soup
0 .6 2 5 equally into 4 bowls.
4 2 .5 0 0
− 0
2 5
− 2 4
1 0
− 8
2 0
− 2 0
0

2.5 ÷ 4 = 0.625
The volume of a bowl of mutton soup is 0.625
.



2 6.93 ÷ 15 = m


Method 1 Method 2


0.4 6 2 6.93 = (6.93 × 1 000) m
15 6.9 3 0 = 6 930 m
−0 0 4 6 2 m
6 9
−6 0 15 6 9 3 0 m
1 Solve these. 9 3 − 0
6 9
a 16 × 0.39 = b 13 × 5.7 = m c 25 × 0.416 = m − 9 0 − 6 0
3 0
1 7 9 3
d 48 × 2 = m e 100 × 6 = m f 1 000 × 7.8 m = − 3 0
4 10 − 9 0
0
2 Complete these. 3 0
0.462 = (0.462 × 1 000) m − 3 0
= 462 m 0

6.93 ÷ 15 = 462 m



207
5.3.5 • Guide pupils to divide using various strategies to reinforce
their knowledge.

0. 1 4 8
2
3 7 ÷ 50 = 50 7 . 4 0 0
5 −0
2 37
7 ÷ 50 = ÷ 50 7 4
5 5 −5 0
= 7.4 ÷ 50 2 4 0
= 0.148 −2 0 0
4 0 0
2
7 ÷ 50 = 0.148 − 4 0 0
5 0



4 What is the volume, in m , of a mini bottle of honey?

2.5 ÷ 100 = m HONEY HONEY HONEY
HONEY HONEY
HONEY
HONEY HONEY HONEY
2.5 = (2.5 × 1 000) m 2 500 m = 25 m
= 2 500 m 100 PARTY SOUVENIRS

2.5 ÷ 100 = 25 m contains 100
mini bottles of honey
The volume of a mini bottle of honey is 25 m .



5 48 ÷ 1 000 = State the
answer in m .
48 = 0.048
1 000
1
48 ÷ 1 000 = 0.048 ÷ 100 = ÷ 10
5
What is the value in ?





1 Calculate.

a 1.2 ÷ 4 = b 9.6 m ÷ 8 = m
1 4
c 25 ÷ 50 = m d 60 ÷ 32 = m
4 5
7
e 19 ÷ 100 = f 43.9 ÷ 100 = m
10
2 a Divide 31 800 m by 1 000. Give the answer in m .

b Calculate 33.36 divided by 60. State the answer in m .




208
5.3.5 • Vary calculation strategies to find the answers.

GROUP ACTIVITY


Tools/Materials product catalogues, manila cards, glue, pens


Task


1 Gather information regarding 2 Cut and paste the information
length, mass, or volume of liquid on the manila card.
from product catalogues (printed
or downloaded from websites)
or through online shopping sites.

Potato Crisps






1.5
130 g


3 Construct questions on unit conversion or addition, subtraction,
multiplication, and division operations. Solve them. For example:


Convert 1.5 to: Calculate the total
a m . mass, in kg, of 2 cans

b in fraction. of potato crisps.
Potato Crisps
a 1.5 = (1.5 × 1 000) m 2 × 130 g = kg
= 1 500 m 1 3 0 g

1.5 b 1.5 = 1 + 0.5 × 2
5 ÷ 5 130 g 2 6 0 g
= 1 +
10 ÷ 5
1 260 g = (260 ÷ 1 000) kg
= 1
2 = 0.26 kg

4 Present the work and discuss. 5 Display the work at the
mathematics corner.





5.1, • Assess pupils’ mastery in terms of knowledge, communication 209
5.2, 5.3 skills, thinking skills, soft skills including attitude and values while
performing the activity.

SOLVE THE PROBLEMS


1 Cliff wants to form a right-angled triangle.
He uses three ropes and 12 cm 1 m
the measurements are as shown. 5

a Calculate the total length, in cm, of the
ropes.
0.16 m
b What is the difference, in m, between
the longest rope and the shortest rope?

1
Understand • The length of the ropes are 12 cm, 0.16 m, and m.
5
the problem • Calculate the total length of the ropes in cm.
• Find the difference between the longest rope and the
shortest rope in m.

Plan the strategy 1
12 cm m 0.16 m
5
total length
Solve

1 1
a 12 cm + m + 0.16 m = cm b m – 12 cm = m
5 5
1 = 20 cm – 12 cm
12 cm + m + 0.16 m
5 = 8 cm
1 20
= 12 cm + ( × 100) cm + (0.16 × 100) cm = (8 ÷ 100) m
5
1 = 0.08 m
= 12 cm + 20 cm + 16 cm

= 48 cm

Check b 0.08 m + 12 cm

a 4 8 cm 3 2 cm = 0.08 m + (12 ÷ 100) m
– 1 6 cm – 2 0 cm = 0.08 m + 0.12 m
3 2 cm 1 2 cm = 0.2 m

1 1
12 cm + m + 0.16 m = 48 cm m – 12 cm = 0.08 m
5 5
The total length of the ropes The difference between the longest

is 48 cm. rope and the shortest rope is 0.08 m.


210 • Guide pupils to find keywords in the questions and check
5.4.1
the answers using a calculator.

1
8 km
Ashley’s 4 school
home
2 Jeera’s
home
Based on the diagram above, the distance of Ashley’s home to school is

3 times the distance of Jeera’s home to school. What is the distance, in m,
from Jeera’s home to school?

1
Solution The distance of 8 km Draw the
Ashley’s home 4 diagram. Write
the number
The distance of sentence.
Jeera’s home ?
1
8 km ÷ 3 = m
4
1 250
Convert 8 km 1 33
4 8 km = ( × 1 000) m
to m. 4 4
1 2 7 5 0 m
1 3 8 2 5 0 m
1 − 6 Check
2 5 0 m 2 2 2 1
× 3 3 − 2 1 2 7 5 0 m
1 1 5 × 3
7 5 0 − 1 5
+ 7 5 0 0 0 0 8 2 5 0 m

1 8 2 5 0 m − 0
8 km ÷ 3 = 2 750 m
4 0
The distance from Jeera’s home to school is 2 750 m.



3 In conjunction with the recycling campaign, Underline the
5 Meteor pupils collected 4.67 kg of cans. The mass important information.
of paper collected is 12 times the mass of cans.
Calculate the mass, in g, of paper collected.
1 1
Solution 12 × 4.67 kg = g 4 6 7 0 g
× 1 2
Convert 4.67 kg = (4.67 × 1 000) g 1 1
4.67 kg to g. = 4 670 g 9 3 4 0

+ 4 6 7 0 0
5 6 0 4 0 g
The mass of paper collected is g.


• Show various problem-solving strategies such as constructing 211
5.4.1 a table or simulation.
• Discuss the method to check the answer for example 3.

4 The table on the right shows the Product Volume
volume of some products. Puan Ani
bought two products. She bought date milk 200 m
10 bottles of each product. The total
pomegranate
volume of the products bought juice 0.25
.
is 7 What are the two possible
products bought by Puan Ani? 1
olive oil
2


hand gel 450 m


Solution

Convert all volume in decimal Calculate the total
.
200 m = (200 ÷ 1 000) volume of 10 bottles.

= 0.2 Identify two
1 10 × 0.2 = 2.0 volume with the
= 0.5
2 total of 7
.
450 m = (450 ÷ 1 000) 10 × 0.25 = 2.5

= 0.45 2.0 + 5.0 = 7


10 × 0.5 = 5.0
2.5 + 4.5 = 7

10 × 0.45 = 4.5



The two possible products bought by Puan Ani are date milk and

olive oil or pomegranate juice and hand gel.



Puan Ani wants to fill up one bottle of pomegranate juice equally into
2 glasses. Which glass should she choose? Explain.





P Q
120 m 130 m






212 • Guide pupils to calculate mentally and estimate efficiently to identify
5.4.1
the total of a certain quantity.

Solve the following problems.

a Puan Anita bought a piece of white cloth and batik cloth. The length
3
of the white cloth is 20 m. The length of the batik cloth is 1.5 m more
4
than the length of the white cloth. Calculate the length, in cm, of the
batik cloth bought by Puan Anita.

b The diagram below shows the location of some places on a straight
road.
Azam’s Ajay’s recreational swimming
home home park complex


1
36 km
2
Ajay’s home is located 19.7 km away from Azam’s home and

9 km 80 m from the recreational park.
i Calculate the distance, in km, from the recreational park to the
swimming complex.
ii Azam goes to Ajay’s home twice a week for a group study.
Calculate the total distance of a round-trip, in m, taken by Azam.

c
2.1 kg


The diagram above shows some rubber balls of the same type and

size on a balanced weighing scale. Calculate the mass, in g, for one
rubber ball.

d The table on the right shows the volume of three
Water
drinking water containers prepared by Lay Ting. container Volume
,
i Calculate the total volume, in of drinking P 3.18
water in the three containers. Q 750 m
ii The drinking water in container R was filled 1
equally into 30 glasses. Each glass contains R 3
2
150 m of drinking water. Is it a true statement?

Prove it.






• Construct more questions as in (a) to (d) and ask pupils to solve the 213
5.4.1
questions in groups.

1 Convert the measurements to the required units.
1
a 0.8 cm = mm b 3 m = cm c 9 002 m = km
4
d 17 m 3 cm = m e 126 mm = cm f 45 km 9 m = km
2 Calculate.
3
a 10.9 cm + 8.21 cm = cm b 6 m + 0.78 m + 94.1 cm = cm
5
1 1
c 9 m – 65 cm = cm d 13 km – 6.03 km – 40 m = km
10 8
4
e 8 × 1 m = m f 100 × 0.46 km = m
5
g 9.1 km ÷ 26 = m h 76 m 9 cm ÷ 1 000 = cm

3 Solve these.
a Add 24.6 cm and 37 cm. State the answer in mm.

1
b Is 18 × 7 cm equal to 5.4 m ÷ 4? Show the calculation.
2
4 Complete the unit conversion.

Measurement In decimal In fraction
a 4 200 g kg kg

b g 3.9 kg kg

c 600 m
d m 8.7

5 Calculate.
1 3
a kg + 0.15 kg = kg b 47.07 kg + 13 080 g + 9 kg = g
2 4
1 9
c 7 kg – 0.26 kg = kg d 12 kg – 6.98 kg – 1 120 g = g
4 10
7
e 18 × 4.5 g = kg f 2 kg ÷ 100 = g
8
6 Find the answers.
1

a 0.375 + 900 m = m b 6 + 1 140 m + 3.9 =
4
1

c 15.01 – 860 m = d 40.308 – 17 – 580 m = m
2
1
e 27 × 3 = m f 8 92 m ÷ 1 000 = m

5
214 5.1, • Encourage pupils to use various calculation methods to find
5.2, 5.3
the answers.

7 Solve these.
a How much should be subtracted from 3 206 g to become 1.507 kg?
3
b Find the quotient of 16 and 25. State the answer in .
4
8 Solve the following problems.
1
a Aishah used 2 m of red ribbon and 85 cm of blue ribbon to make
4
a handicraft.
i Calculate the length, in m, of the blue ribbon.
ii Calculate the total length, in m, of the two ribbons.
b Based on the diagram on the right, 2 school
the distance from the school to the 1 km
5
bookshop is 4 times the distance
from Adam’s home to the school. Adam’s
Solve these. home
Adam cycles from his home to the
a Add 24.6 cm and 37 cm. State the answer in mm.
bookshop passing by the school. 5.09 km
1
b Is 18 × 7 cm equal to 5.4 m ÷ 4? Show the calculation. Then, he returned home by the shortest route.
2
i Calculate the distance, in m, from the school
to the bookshop. Bookshop
bookshop
ii Calculate the total distance, in km, taken by Adam.
c Encik Sulaiman donated two boxes that contained 96 bottles of
mineral water altogether to the flood victims. The total volume

of water is 24 . What is the volume, in , for one bottle of
mineral water? State the answer in fraction.
d The table shows the mass of three types of
Cake Mass
cake that is sold in a shop. 1
i What is the mass, in kg, for one Chocolate 1 kg
2
strawberry cake? 50 g more than
ii The shopkeeper sold 6 kg of cake in one Strawberry carrot cake
afternoon. State the number and types of Carrot 1.2 kg
cakes that might have been sold by him.

e I used 125 mm of
1 wire to form a regular I used 0.5 cm less than

a 0.375 + 900 m = m b 6 + 1 140 m + 3.9 = the length of wire you
4 pentagon.
1 used to form a square.

c 15.01 – 860 m = d 40.308 – 17 – 580 m = m
2 State the length of sides, in cm, for:
1
e 27 × 3 = m f 8 92 m ÷ 1 000 = m Jacky i pentagon. ii square.
5 Farah
5.2.3, • In groups, carry out quizzes to solve “Try It Again” questions. 215
5.3.5, 5.4
• Add simple questions to improve problem-solving skills.

Tools/Materials coins, markers , game cards, papers, pens



Participants 2 players and 1 referee



S & A GAME SOLVE AND ARRANGE


1 7
0.84 cm + 37 mm + cm m – 34 cm – 5.6 cm 1
2 8 70 × 3 km = m
= mm = cm 2

marker station
1
2 kg ÷ 8 = g 1 000 × 17.8 =
5


3 Divide 9.018 km by 9.
Multiply kg by 15. Give
4
the answer in g. Give the answer in m.


A piece of robe needs Aida drinks 14.7 of
3
3 m of cloth. What The mass of 1 teabag is water in one week. In
4
is the length, in cm, 0.002 kg. The mass of one day, she drinks
100 teabags is g.
for 6 pieces of robes? m of water.




How to play

1 Put all the questions cards face down.
2 Toss the coin to decide the first player.
3 The first player opens a question card and answer the question.

4 Put a marker on any circles in the marker station if the answer is correct.
Only one marker is to be placed for each correct answer.
5 Take turns. Repeat steps 3 and 4 until all the question cards are answered.
6 The first player to place all the markers in these forms
, , or , wins.



216 5.1, 5.2,
5.3, 5.4 • Modify questions on the game cards to suit pupils’ ability.

6 6 SPACE
SP
ACE





REGULAR POLYGONS



1




Let’s identify the
characteristics of
Wow! This mural is regular polygons.
beautiful.










There is a Yes, there are regular
pentagon! polygons and
irregular polygons.




a b
straight side

teacher friends
diagonals
all straight sides
are of equal length


closed all angles
corner interior angle shape are of equal
characteristics size
Characteristics of a of regular
the number polygons
regular pentagon of angles the number of
• 5 straight sides are of equal length equals the symmetrical
number of axes equals
• 5 interior angles corners the number of
• 5 corners reference straight sides
• 5 diagonals books websites

• 5 symmetrical axes


• Carry out polygon-shaped folding activities for pupils using pieces of
paper or card to explore the characteristics of regular polygons. 217
6.1.1 • Explain to pupils that corners are synonymous with vertices. Review the
characteristics of regular polygons. Explain the meaning of interior angle.

2 Name this regular polygon. Explain the
characteristics of this regular polygon.
A diagonal is a
corner corner straight line drawn
from one corner to
the opposite corner.
diagonal diagonal





corner corner




1 Name each polygon based on its characteristics.


a • 6 straight sides b • 3 straight sides c
which are closed of equal length • 7 corners
• all straight sides of • 3 equal interior • all straight sides
equal length angles of equal length
• 6 symmetrical axes • 0 diagonal • 14 diagonals



2 How many diagonals a b
are there in each
of these regular
polygons?


3 Complete the table below.

Number Number Number of Number
Regular of straight of symmetrical Number of
polygon of angles
sides corners axes diagonals
equilateral 3 3
triangle
square 4 4

pentagon 5 5

hexagon 6 6 9
heptagon 7 7 14

octagon 8 8 20



218 • Create a mobile project (dangling ornament) of regular polygons with their
6.1.1 characteristics and display them at the mathematics corner.
• Guide pupils to identify the number of diagonals for regular polygons.

MEASURING INTERIOR ANGLES


1 The size of all interior angles in a regular pentagon are equal.
What is the value of angle a ?

A protractor is used to Ways to measure the value of angle a:
measure the value of
angle a. 1 Place the centre of 2 Make sure line
the protractor at AE overlaps the
corner A. baseline of the
B C protractor.

3 Read the inner
scale from 0°
(line AE) to the
inner scale B side (line AB).
D
4 The value of a
outer scale o
is 108 or
o o o
110 − 2 = 108 .
o
108
a The value of angle a is one

A E hundred and eight degrees.
baseline centre of protractor o
108



2 State the value of angle p.
angle p corner P

P
Make sure the centre
of the protractor
p
overlaps the corner
of the angle that is
to be measured, p.




The value of angle p is .


Can you state the types of angles for a and p? Why?


• Introduce the characteristics of protractor to pupils such as the
inner scale, outer scale, baseline, and centre of protractor. 219
6.2.1 • Demonstrate how to measure angles using a protractor in detail.

S
3 a b




projection
line
b


p
a






R
The value of angle a is .

The value of angle b is .
Outer scale reading in S: 0°
Outer scale reading in R: 120°

The value of angle p = −
=
Name the polygon which has a right
angle 90° and acute angle.








Copy the regular polygon diagrams below.
Measure and write down the value of the interior angles.

a b Q c
q





p


w
p w




220 • Guide pupils to solve the ‟Try These” question while ensuring that they
6.2.1 measure the value of the angles using the correct technique.
• Carry out a project of measuring angles of a polygon mask to reinforce
pupils’ understanding.

PERIMETERS OF COMPOSITE SHAPES




1

Indung, we
have made
this bookmark Shanti, let’s decorate
out of these the outline of the
two regular bookmark with this
polygons. green-coloured
paper.



How do we determine the length of the green-coloured paper

that is needed?
The length of the 9 cm
green-coloured 9 cm
paper is the length
of the outer sides or
the perimeter of the 9 cm
composite shape.
9 cm






1 Measure all the 9 cm
outer sides of 9 cm 9 cm
the composite
shape. Perimeter = (9 + 9 + 9 + 9 + 9 + 9 + 9) cm
2 Total up the
length of all the = 63 cm
outer sides of the OR
composite shape. Perimeter = 7 × 9 cm

= 63 cm
The length of the green-coloured paper

9 cm needed is 63 cm.


Is the perimeter of the composite shape of this
regular polygon 63 cm as well? Discuss.





• Provide each group with a few regular polygon shapes. Each group must 221
6.3.1 form a composite shape from two polygons and determine the perimeter
of the composite shape.

4 m
2
4 m 4 m
end 10 m


4 m start
Determine one
4 m 10 m side as a starting
4 m point to calculate.
4 m 6 m


10 m
Method 1 Perimeter = (10 + 10 + 6 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 10) m

= 64 m

+
Method 2 Perimeter = Perimeter of the outer Perimeter of the outer
sides of the octagon sides of the square

= (7 × 4 m) + (10 m + 10 m + 10 m + 6 m)
= m + m
= m



3 Determine the perimeter of the shaded region.

start
Perimeter of the shaded region Total up the
6 cm surrounding sides of
= (6 + 6 + 5 + 6 + 5 + 6 + 6) cm the shaded region.

= cm 6 cm

5 cm
5 cm


4 m
4
Discuss the error in
1 m calculation of this perimeter.


Perimeter of the shaded region
2 m 5 m
(4 + 5 + 2 + 1 + 1) m = 13 m





222 • Carry out exploration activities of two composite shapes using regular
6.3.1 polygons including triangles and quadrilaterals to find perimeter.

MY DESIGN


Tools/Materials MS Word, A4 papers, printer, rulers, pens

Participants 4 pupils in a group


Task
1 Launch MS Word.

2 Click Insert. Click Shapes. Select two regular polygon EXAMPLES OF
COMPOSITE SHAPES
shapes up to six sides including right-angled triangles,
isosceles triangles, and rectangles.
3 Make a few composite shapes from any two regular
polygons using your own creativity by adjusting the
size, sequence, and colour for each shape.

4 Print and measure the perimeter of each shape that
you have made.
5 Present your completed work through Gallery Walk.






1 Calculate the perimeter for each of the composite shapes below.
a b 4 m
3 cm 3 cm 3 cm
3 cm


3 cm 5 m
12 m
3 cm 3 cm
3 cm
3 cm
2 Find the perimeter of the shaded region.
a b


7 m 10 cm
6 cm
5 cm



5 m 3.5 cm 4.5 cm



• Carry out activity in groups to answer the ‟Try These” questions. 223
6.3.1 • Each group presents their work and other groups can suggest other ways
when appropriate.

AREA OF COMPOSITE SHAPES



1 Calculate the area of the composite shape coloured by Chin.


Write down the
4 units measurement of its
length, breadth,
4 units A B base, and height.


2 units
4 units


Area of the square, A:
4 units × 4 units = 16 units 2

Area of the right-angled triangle, B:
Area of a square 1 1
= length × breadth 2 × 2 units × 4 units = 4 units 2
1
Area of a triangle
2
2
1 Area of the composite shape = 16 units + 4 units
= × base × height
2 = 20 units 2
The area of the composite shape coloured by Chin is 20 units .
2


2 2 m
What is the area of the painted wall?


3 m The area of the painted wall

= surface A + surface B
A
= (3 m × 2 m) + (3 m × 5 m)

2
= 6 m + 15 m 2
3 m B = 21 m 2


2
The area of the painted wall is 21 m .
5 m




• Discuss the role of the inner sides of the composite shape to calculate
224 the area of the composite shape.
6.3.2
• Carry out group activities to calculate areas based on two given
composite shapes. Each group presents their work.

10 cm
3 Area of the composite shape


= area + area
60 cm = ( × 60 cm × 40 cm) + ( × cm × 100 cm)
40 cm
1
1
10 cm 100 cm = 2 cm + cm 2 2
2

= cm 2


4 Dad planted grass on the shaded region as shown. Find the area of
the grassy compound.
6 m
Area of the grassy compound
= area of the rectangle – area of the square

4.8 m = (6 m × 4.8 m) – (2 m × 2 m)
2
2 m = m – m 2
= m 2
2 m






1 Calculate the area of each of the composite shapes below.

a 7 m b
5 m
4 m 10 cm
7 m
6 cm 6 cm
2 cm 6 cm
1 m
2 Calculate the area of the shaded region. 18 cm

a b



100 m
80 m
3 cm


50 m 50 m 5 cm


• Relate the activity of calculating the area of the composite 225
6.3.2 shapes to daily life situations such as the area of group table’s
surface, the area of playground, and the area of tiled floor.

VOLUME OF COMPOSITE SHAPES



1
1 unit
We will learn how to
determine the volume 1 unit
of composite shapes. 1 unit
The volume of
the small cube
is 1 cubic unit.









What is the volume of the composite shape?


Method 1 Count the number of cubes
of 1 cubic unit.



Volume of the
6 cubes 54 cubes composite shape
3
1 cubic unit 1 cubic unit = 6 units + 54 units 3
6 units 3 54 units 3 = 60 units 3
Method 2
2 units Volume of the composite shape
1 unit
= Volume + Volume
3 units
= (2 units × 1 unit × 3 units)
+ (9 units × 2 units × 3 units)
3 units
3
= 6 units + 54 units
3
2 units 3
9 units = 60 units
The volume of the composite shape is 60 units .
3



Build a combination of cube and cuboid from 59 cubic units. For each of
the composition of shapes, state the volume of the cuboid and the cube.


• Emphasise that the volume of each of the composite shape is the
226 same even though they are in different arrangements.
6.4.1
• Provide cubes and cuboids of various sizes for pupils to explore
various compositions of two shapes to find its volume.

2 Volume of the cuboid = 10 cm × 8 cm × cm
= cm
3
Volume of the cube = cm × cm × cm

= cm 3

17 cm 8 cm Volume of the composite shape
3
= cm + cm 3
8 cm = cm 3
Complete the
8 cm
10 cm 8 cm calculation.



3 Calculate the volume of the remaining block after the middle part
is removed.

The volume of the remaining block 6 m

= (4 m × 5 m × 6 m) – (2 m × 4 m × 3 m)
3 m 5 m
3
3
= 120 m – 24 m
3
= 96 m 4 m
3
The volume of the remaining block is 96 m . 2 m 4 m






Calculate the volume of the following blocks.
a b 8 cm

2 m

4 cm 8 cm
4 cm

10 m
3 m


3 m 8 cm
3 m 12 m






• Guide pupils to calculate the unknown length of a side in
‟Try These” question. 227
6.4.1
• In groups, conduct a Think-Pair-Share session on the volume of
two composite shapes.

SOLVE THE PROBLEMS


1 Zariq designed an ornamental fish pond by
composing two shapes. He will build a fence 1 m 8 m 1 m
around the pond. What is the length of the fence?


20 m

Understand the problem
• The composition of a rectangle and regular octagon.
• The pond measurements as in the plan. 10 m

• Find the length of the fence around the pond.




Plan the strategy the length
of the outer
• Mark the length of the outer sides of the pond. sides of the
• Add the length of all outer sides to find the 8 m pond
length of the fence.



20 m
Solve

The length of the fence 10 m
= length of the outer sides + length of the outer sides of
of the rectangular pond the regular octagon pond
= (1 m + 20 m + 10 m + 20 m + 1 m) + (7 × 8 m)
= 52 m + 56 m
= 108 m

Check
Perimeter of the pond
Perimeter of the pond = (2 × 20 m) + (2 × 10 m)
= 8 × 8 m = 40 m + 20 m

= 64 m = 60 m
The perimeter of the outer sides of the pond = 64 m + 60 m − (2 × 8 m)
= 124 m − 16 m
= 108 m

The length of the fence is 108 m.

228 • Ask pupils to individually create an object such as a picture frame and
6.5.1 a greeting card using a composition of two regular polygons. Ask them
to measure the perimeter of the object.

14 cm


2 Jafri sketches the parts of a
wooden block to be cut. The
sketch would be in the shape of 14 cm
a cuboid. Calculate the volume of 18 cm
the wooden block after being cut. 16 cm
23 cm


Understand the problem 14 cm
• Identify the length, breadth, and

height of the original wooden block
and the shape that has been cut. 14 cm
• Cuboid-shaped cuts. 18 cm

• Find the volume of the wooden 16 cm

block after being cut.
23 cm
Plan the strategy

• The volume of the original wooden block = 23 cm × 16 cm × 18 cm.
• The volume of the cuboid to be cut = 14 cm × 16 cm × 14 cm.
• The volume of the wooden = The volume of the – The volume of the
block after being cut original wooden block cuboid to be cut

Solve


The volume of the original The volume of the cuboid to be cut
wooden block 2 1
1 5 6 1 4 cm 2 2 4cm 2
2 3 cm 3 68 cm 2 × 1 6 cm × 1 4cm
× 1 6 cm × 1 8 cm 1 1 1
1 3 8 1 1 8 4 8 9 6
+ 2 3 0 2 9 44 + 1 4 0 + 22 4 0
3 6 8 cm 2 + 3 6 80 2 3 1 3 6cm 3
6 6 24 cm 3 2 2 4 cm

11 Check
The volume of the 5 1 14 1 1
wooden block 6 6 2 4 cm 3 3 4 8 8 cm 3 3
after being cut − 3 1 3 6 cm 3 + 3 1 3 6 cm
3 4 8 8 cm 3 6 6 2 4 cm 3

3
The volume of the wooden block after being cut is 3 488 cm .

229
• Guide pupils to find the volume of a single shape before calculating the
6.5.1
volume of the composite shape.

2 m


3 Dad wants to build a brick footpath 4 m
footpath in the house compound (B)
as shown in the diagram. He has grass grass
allocated RM4 000 for this. The
cost of building 1 m brick footpath
2
is RM100. Is there enough money?
footpath (A) 4 m





Solution 6 m


Identify the length Find the Calculate Conclude

and the breadth of total area the total cost whether the
footpaths A and B. of footpaths for the areas allocation of
Calculate the area of A and B. of footpaths RM4 000 is
footpaths A and B. A and B. enough.




Area of footpath A = 6 m × 4 m
= 24 m 2

Area of footpath B = 2 m × 4 m
= 8 m 2


Total area of footpaths A and B = Area of footpath A + Area of footpath B
= 24 m + 8 m
2
2
= 32 m 2

Total cost of the footpath = Total area of footpaths A and B
2
× The cost of 1 m brick footpath
= 32 × RM100
= RM3 200



Conclusion

Yes. No. The allocation of RM4 000 is enough not enough .




230 • Get each group of pupils to give their opinions on the method of calculation.
6.5.1 • Instil entrepreneurship and the value of making a budget before
making decisions.

10 m
Solve the problems below.

parit
a Uncle Naim dug a ditch ditch
around his farm as shown
in the diagram. Calculate 10 m 6.7 m
the length of the ditch.
2.5 m
4.8 m
3.5 m
10.2 m
b A card with a rectangular

surface has been cut as shown
in the diagram.
i Find the perimeter of the 40 cm 50 cm
card that has not been cut. 50 cm
ii What is the area of the card
that has not been cut?
30 cm 30 cm





c The shape of Seng Huat’s
15 m
house compound is
shown in the diagram. He
planted grass in the whole 15 m 15 m 15 m
compound. What is the area
of the grassy compound? 25 m




d In a design project, a cuboid was cut 5 cm
and removed from a cube.
5 cm
i Calculate the volume of the 10 cm
cuboid removed. 20 cm
ii Is the volume of the remaining
3
cube 7 750 cm ? Prove it. 20 cm
20 cm





• Get pupils to complete the ‟Try These” questions in groups. 231
6.5.1 • Next, carry out activities on a station basis. Each group will show the
steps of their calculations.

1 Copy the following pictures of regular polygons. Write down
the characteristics and names of the regular polygons in the
empty boxes.
a b
















2 Measure the interior angles of these regular polygons. Write down
the value of the angles.
a b













r
r


3 Calculate the perimeters of these composite shapes.

a b
5 cm





1 cm 6 m

3 m

7 cm 7 cm 2 m





232 6.1.1, 6.2.1,
6.3.1

4 Determine the area of these composite shapes.
a 2 m b
80 cm


40 cm

6 m



2 m 1 m 60 cm


20 mm
5 Calculate the volume of the blocks below.
a 14 cm b
14 cm
35 mm
18 mm
18 mm

14 cm
8 cm 16 mm

18 mm
6 cm
10 cm 36 mm


6 Solve these problems. Dad cuts off the middle part of a

a 6 cm 6 cm cuboid-shaped block. The middle part is
cube-shaped as shown in the diagram.

i What is the volume of the cube?
6 cm 12 cm ii Calculate the volume of the original
cuboid.
6 cm iii Is the volume of the remaining
6 cm 3
18 cm block 936 cm ? Prove it.
15 m
b Pak Ali’s cultivated paddy field is
shaded as shown in the diagram.
i Calculate the area of Pak Ali’s 13 m
m

13
paddy field. 12 m
ii What is the length of the irrigation
canal surrounding Pak Ali’s paddy irrigation 5 m
field? canal

6.3.2, 6.4.1, 233
6.5.1

SPACE BUNTING


5 task cards, 5 manila cards, pens, a 1 m roll of
Tools/Materials
ribbon, a paper puncher, polygons, composite
shapes of cubes and cuboids

Participants 5 groups of pupils Example of
completed work
Steps

1 Each group is given one task card
to complete. For example:


Card 1 Card 2
Choose a regular Choose a regular
polygon and paste polygon and paste
it on a manila card. it on a manila card.
Write down the Label all the values of
characteristics and the interior angles.
the name of the
regular polygon.

Card 3 Card 4
Combine and Combine two regular
paste two regular polygons, or one regular
polygons. Calculate polygon and a triangle.
the perimeter of the Calculate the area of the
composite shape. composite shape.

Card 5
Choose a composite shape of cube
and cuboid. Calculate the volume of the
composite shape.

2 Carry out a Goldfish Bowl session.
Each group will present their work
while members of other groups will
ask questions to obtain the required
information.
3 Collect the work of all groups to be
turned as a bunting.
4 Display the bunting at the
mathematics corner.

6.1.1, 6.2.1,
234 6.3.1, 6.3.2,
6.4.1

7 7 COOR DIN A TES, RA
COORDINATES, RATIO, TIO,

AND PROPORTION AND PROPORTION




DISTANCE BETWEEN TWO COORDINATES


Horizontal distance and vertical distance from the origin
The Cartesian plane shows several areas in a zoo.

Ticket counter is at the origin.
y
The distance of (0, 5)
each grid is 1 unit. 5



4
(1, 4)

vertical axis 3 (5, 3)




2

The distance (5, 2)
along the
y-axis is 1
vertical
distance.
(4, 0)
ticket x
counter O 1 unit 1 2 3 4 5 6
horizontal axis

The distance along the x-axis is horizontal distance.

The horizontal distance
a The horizontal distance of the parrot
from the origin is x unit.
from the ticket counter is 4 units.
b The vertical distance from the ticket ( x, y )

counter to the jetty is 5 units.
c The horizontal distance from The vertical
the origin to the lion is 5 units. distance from the
The vertical distance is . origin is y unit.


• Start the lesson by recalling the coordinates of a point in the first
quadrant of the Cartesian plane. 235
7.1.1
• Discuss the horizontal distance and vertical distance of other places
in the Cartesian plane above.

Horizontal distance and vertical distance between two coordinates

The Cartesian plane shows the interior of Rashwin’s home.

a What are the horizontal distance and vertical distance
from the kitchen to room 2?
y


6
toilet (4, 6)
room 2 (6, 5)
5
room 1 (2, 5)
4
Start moving from
the kitchen and
3 stop at room 2.
living room (3, 3) 5 units

2


1
dining area
(0, 1) x
O 1 2 3 4 5 6
kitchen (4, 0)

2 units


To go to room 2 Place Distance from origin
from the kitchen, Horizontal distance Vertical distance
move 2 units to the kitchen 4 units 0 unit
right and 5 units up.
room 2 6 units 5 units

the difference

of distances 6 units – 4 units = 2 units

the difference
of distances 5 units – 0 unit = 5 units

The distance from the kitchen to room 2 is
2 units horizontally and 5 units vertically.
Rashwin wants to go to the toilet from the
living room. Describe the distance he has to travel.

236 • Emphasise the importance of the starting and ending points to
7.1.1
determine the exact horizontal distance and vertical distance.

b Calculate the horizontal distance and the vertical distance from room 1
to the dining area.
Horizontal distance

2 units – 0 unit = 2 units From room 1,
move 2 units to
Room 1 coordinate (2, 5) Dining area coordinate (0, 1) the left and
units downwards
Vertical distance to the dining area.
units – 1 unit = units


The distance from room 1 to the dining area is
2 units horizontally and units vertically.

State the horizontal distance and vertical distance between:
a room 1 and room 2.

b the living room and dining area.







The horizontal distance and vertical distance from point L (a, b) to point

M (1, 8) is 2 units horizontally and 3 units vertically. What are the values
of a and b?



y
Based on the Cartesian plane on the
1
6 left, state the horizontal distance and
B K
5 vertical distance:
Q
4 a of point A from the origin.
A b of point B from the origin.
3
c of point C from the origin.
2
P d from point P to point T.
1
C T x e from point Q to point P.
O 1 2 3 4 5 6 f from point P to point K.
2 Calculate the horizontal distance and vertical distance from:

a point L (3, 6) to point J (4, 9). b point M (1, 8) to point N (5, 7).



237
7.1.1

RATIO BETWEEN TWO QUANTITIES


1 The picture shows the uniformed units’ camping activity.

The Malaysian Red
Crescent Society (MRCS)
has 5 members. How
many scouts are here?



Ratio is the comparison of
measurements with other
measurements or values
with other values.

We have 3 members.






a What is the ratio of the number of scouts to the number of
MRCS members?

scouts MRCS
the ratio of three to five
:
3 5
The ratio of the number of scouts to the number of MRCS
:
members is 3 5.
b State the ratio of the number of MRCS members to the total number of
MRCS members and scouts.





:
5 8
The ratio of the number of MRCS members to the total number of
:
MRCS members and scouts is 5 8.
c The ratio of the total number of scouts and MRCS members to the
:
number of scouts is .

State the ratio of the number of boys to the
number of girls in the picture above.


• Emphasise how to write and pronounce the correct ratio.
238 7.2.1 For example, 5 : 8 pronounced as the ratio of five to eight.
(i), (ii), (iii)
• Prepare appropriate questions to state ratios and ensure that the
answers cannot be simplified.

2 The following is part of the recipe to make dodol by Puan Maslina.



10 kg of glutinous flour
1 kg of rice flour We use 11 kg of
11 kg of granulated sugar granulated sugar
2 kg of brown sugar and 10 kg of
4 kg of coconut milk glutinous flour.










a What is the ratio of the mass of granulated sugar to the mass of

glutinous flour? the mass of granulated sugar the mass of glutinous flour


1 kg 1 kg
the ratio of eleven to ten
:
11 10
The ratio of the mass of granulated sugar to the mass of glutinous
:
flour is 11 10.
b The ratio of the mass of glutinous flour to the total mass of flour
:
is 10 .
c The ratio of the total mass of sugar to the mass of brown sugar
:
is .


3






Alif Raudah Whose answer
is correct?
Why?











• Emphasise that units of measurement must be the same to state
7.2.1 239
(i), (ii), (iii) the ratio of two quantities.
• Get other recipes from websites for activities to state ratios.

The mass of each watermelon is

the same. State the ratio of:
a the mass of a watermelon to
the mass of a pineapple.
b the mass of a pineapple to the
total mass of the fruits.
4 kg 3 kg






Based on the picture, state the ratio of:
1
a the number of roosters to the
number of hens.
b the number of roosters to the total
number of chickens.

c the total number of chickens to the
number of hens.


2 The table shows the length of fabric bought by Raysha’s mother.

Fabric colour Yellow Green Blue

Length 2 m 300 cm 4 m
State the ratio of:
a the length of yellow fabric to the length of green fabric.

b the length of blue fabric to the total length of all fabrics.
c the total length of all fabrics to the length of the yellow fabric.

3 The picture on the right shows the volume of liquids 15 m 11 m 9 m 10 m
in four test tubes. State the ratio of:
a the volume of liquid in test tube A
to the volume of liquid in test tube B.
b the volume of liquid in test tube B to the total
volume of liquids. A B C D




240 7.2.1 • Guide pupils to carry out exercises and if necessary, repeat the
(i), (ii), (iii) learning activities using manipulative materials for them to master
the skills.

PROPORTION TO FIND A VALUE


1 I got the recipe for
a batik cake, mum.
• 600 g Marie biscuits
• 200 g butter
• 1 cup of cooking chocolate
• 1 cup of condensed milk
1
• cup of cocoa powder
2
1
• cup of boiled water
2
• 2 eggs
• Vanilla essence


a I want to make two batik cakes.
What is the mass of butter for
2 cups of cooking chocolate?

1 cup of cooking chocolate 200 g butter

2 cups of cooking chocolate 2 × 200 g = 400 g
The mass of butter for 2 cups of cooking chocolate
is 400 g.

b What is the mass of Marie biscuits needed when 300 g of butter
is used?
200 g butter for 600 g of Marie biscuits.

Divide by 2 to find the mass of Marie biscuits
for 100 g butter. Total up the mass of 200 g

100 g 300 g Marie biscuits and
200 g 600 g 100 g butter.
2 butter for 2 Marie biscuits.
1 1
300 g butter The mass of Marie

biscuits needed
200 g + 100 g 600 g + 300 g = 900 g

The mass of Marie biscuits needed when
300 g of butter is used is 900 g.


• Emphasise the use of arrows in proportional representations.
• Construct other appropriate questions from the recipe above to 241
7.3.1
enhance pupils’ understanding.

2






1 kg GREEN 1 kg
RM8.50 BEANS BARLEY RM7.30




a Calculate the price for 2 kg 400 g of green beans.

2 kg 400 g = 2 kg + 400 g


Step 1 Step 2

1 kg RM8.50 1 000 g RM8.50

2 kg 2 × RM8.50 100 g RM8.50 ÷ 10 = RM0.85
= RM17.00 400 g 4 × RM0.85 = RM3.40


Step 3

2 kg 400 g 2 kg 400 g


RM17.00 + RM3.40 = RM20.40

The price for 2 kg 400 g of green beans is RM20.40.

b Is RM30 enough to buy 4 kg of barley?


Method 1 RM7.30 + RM7.30 + RM7.30 + RM7.30 = RM29.20

1 kg + 1 kg + 1 kg + 1 kg = 4 kg


Method 2 RM7.30 1 kg
2 × RM7.30 = RM14.60 2 × 1 kg = 2 kg
2 × RM14.60 = RM29.20 2 × 2 kg = 4 kg

Yes, RM30 is enough to buy 4 kg of barley.



Find the price for 3 kg 500 g of barley.



242 • Vary questions involving daily life situations such as the relationship
7.3.1
between the volume of petrol and its price.