NOMBOR BULAT

Nombor0, 1, 2, 3, ..., 50 ..., 8614, ... yang digunakan dalam proses pengiraandipanggil 'nombor bulat' (whole numbers).

Nombor bulat terdiri daripada angka 0, 1, 2, 3, 4, 5, 6, 7, 8 dan 9 yang bolehditulis dalam perkataan atau angka.

Nombor 1, 2, 3, 4, ... dipanggil nombor tabii / asli (natural numbers). Set tiga titik dibelakang nombor-nombor tersebut menunjukkan bahawa corak terus-menerusnya yang juga dipanggil elipsis (ellipsis).

Setiap1,2,3,4.... bulat mewakili nilai tertentu mengikut kedudukan nilaiangka dalam nombor tersebut. Sebagai contoh, nilai bagi setiap angka dalamnombor 4,128,925 adalah seperti berikut;

Juta - angka 4
Ratus ribu - angka 1
Puluh ribu - angka 2
Ribu - angka 8
Ratus - angka 9
Puluh -angka 2
Unit - angka 5
Nilai bagi setiap angka dalam nombor boleh ditunjukkan dengan menggunakanjadual. Kedudukan angka dalam sesuatu nombor menunjukkan nilainya.

Nombor
Nilai angka 5
Nilai angka 3

5 643
5 000
3

2 358
50
300

.
Menghitung nombor bulat

Hitung dalam gandaan sepuluh dari 20 hingga 50
(Jwb: 20, 30, 40, 50)

Hitung dalam gandaan dua dari 4 hingga 20
(Jwb: 4, 6, 8, 10, 12, 14, 16, 18, 20)

Hitung dalam pengurangan lima dari 30 hingga 10
(Jwb: 30, 25, 20, 15, 10)

Hitung dalam pengurangan seratus dari 1100 hingga 500
(Jwb: 1100, 1000, 900, 800, 700, 600, 500)
Menulis nombor bulat dalam perkataan

1 020 (Jwb: Seribu dua puluh)
19 503 (Jwb: Sembilan belas ribu lima ratus tiga)
634 202 (Jwb: Enam ratus tiga puluh empat ribu dua ratus dua)
1 007 032 (Jwb: Satu juta tujuh ribu tiga puluh dua)
Menulis nombor bulat dalam bentuk angka

Lima ratus tiga puluh empat (Jwb: 534)
Dua belas ribu tujuh puluh (Jwb: 12 070)
Lima ratus dua puluh empat ribu, dua ratus empat puluh dua (Jwb: 524 242)
Dua puluh satu juta, tujuh ribu empat puluh lapan (Jwb: 21 007 048)
Mengenal pasti nilai tempat

24 763

2 - puluh ribu
4 - ribu
7 - ratus
6 - puluh
3 - unit
374 125

3 - ratus ribu
7 - puluh ribu
4 - ribu
1 - ratus
2 - puluh
5 - unit
Mengenal pasti nilai angka dalam nombor bulat

Kenal pasti nilai angka 6 dalam nombor 36 052
Jwb: Nilai tempat angka 6 adalah ribu. Oleh itu nilai angka 6 adalah 6000.

Kenal pasti nilai angka 7 dalam nombor 472 351
Jwb: Nilai tempat angka 7 adalah puluh ribu. Oleh itu nilai angka 7 adalah 70 000.
Nombor bulat boleh di'bundarkan' (rounded off) kepada nilai tempat tertentu mengikut ketepatan yang diperlukan.

Kaedah untuk pembundaran (rounding off) nombor bulat kepada nilai tempat yang diberi adalah seperti berikut;

Cari angka di sebelah kanan nilai tempat yang diberi.
Jika angka ini adalah kurang dari 5, biarkan angka pada nilai tempat tersebut tidak berubah.
Jika angka ini adalah 5 atau lebih, maka tambahkan 1 pada angka dalam nilai tempat yang diberi dan gantikan setiap angka di sebelah kanan dengan 0.
Pembundaran nombor bulat

Bundarkan 37 268 yang nilai yang terhampir

Ribu

Nilai tempat yang diberi, ribu = 7.
Angka disebelah kanan nilai tempat yang diberi adalah 2 (2<5 ).
Biarkan angka 7 tidak berubah dan gantikan kesemua angka di sebelah kanan angka 7 dengan 0.Oleh itu pembundaran 37 268 kepada nilai ribu yang terhampir adalah 37 000.

Puluh ribu

Nilai tempat yang diberi, puluh ribu = 3.
Angka disebelah kanan nilai tempat yang diberi adalah 7 (7>5).
Tambahkan 1 pada angka 3 dan gantikan kesemua angka di sebelah kanan angka 3 dengan 0.Oleh itu pembundaran 37 268 kepada nilai puluh ribu yang terhampir adalah 40 000.

Bab 2 Pola dan Turutan Nombor

Corak/Pola Nombor dan Urutannya

Nombor-nombor yang disusun dalam corak tertentu dikenali sebagai urutan (sequence).

Corak urutan nombor (number sequence) boleh ditentukan dengan menambah, menolak, mendarab atau membahagikan ‘nombor dalam urutan yang sebelumnya’, dengan bilangan/nombor-nombor tertentu.

Urutan Fibonacci (Fibonacci Sequence)

Ahli matematik telah mengkaji corak selama berabad-abad. Corak nombor 1, 1, 2, 3, 5, 8, …dipanggil urutan Fibonacci.

Urutan ini bermula dengan 1, 1 dan setiap selepas sebutan (term) yang kedua, diperolehi dengan menambah dua sebutan (term) sebelumnya yang terdapat dalam urutan.

Menggambarkan corak/pola urutan nomborGambarkan corak setiap urutan nombor berikut:

5, 12, 19, 26, …
Jwb:

Corak urutan nombor 5, 12, 19, 26, … diperolehi dengan menambah (adding) 7 pada nombor sebelumnya (nombor dalam urutan).
1, 4, 16, 64, …
Jwb:

Corak urutan nombor 1, 4, 16, 64, … diperolehi dengan mendarabkan (multiplying) nombor dalam urutan sebelumnya dengan 4.
40, 35, 30, 25, …
Jwb:

Corak urutan nombor 40, 35, 30, 25, … diperolehi dengan menolakkan (subtracting) 5 dari nombor sebelumnya (nombor dalam urutan).
144, 72, 36, 18, …
Jwb:

Corak urutan nombor 144, 72, 36, 18, … diperolehi dengan membahagikan (dividing) nombor dalam urutan sebelumnya dengan

Bab 3 Pecahan

Pecahan

Apabila satu unit dari keseluruhan kuantiti dibahagikan kepada bahagian yang sama, mana-mana bahagian dalam kuantiti tersebut dipanggil sebagai pecahan unit (fractions of the unit).

Jadual dibawah menunjukkan bagaimana pecahan dibaca.

Pecahan adalah nombor yang mewakili satu atau lebih bahagian yang sama dibahagikan daripada keseluruhannya. Contohnya, Dalam rajah dibawah, setiap bahagian adalah 1 daripada 6 bahagian yang sama.

Pecahan juga boleh digunakan untuk menamakan sebahagian daripada koleksi atau satu set objek atau kumpulan. Contohnya,

Setiap pensil di atas adalah 1 daripada sekumpulan 4 pensel.

Pecahan boleh diwakili oleh gambar rajah. Contohnya,

Bahagian-bahagian yang berlorek dalam gambar rajah adalah mewakili 8 bahagian daripada 9 bahagian yang sama. Dengan kata lain, 8/9 daripada gambar rajah.

Menulis pecahan.

Pecahan ditulis dalam bentuk

, dimana a adalah pengangka (numerator) dan b adalah penyebut (denominator). Contohnya,

Bilangan 1 adalah mewakili semua bahagian keseluruhannya.

Contoh 1:

Apakah pecahan yang diwakili oleh bahagian-bahagian yang berlorek dalam setiap gambar rajah di bawah?

Jwb: 5/8
Jwb: 1/4
Jwb: 3/6 = 1/2

Bab 4 Nombor Perpuluhan

Nombor Perpuluhan dan Pecahan

Mewakilkan pecahan 1/10 dan 1/100 sebagai nombor perpuluhan dan sebaliknya.

Nombor perpuluhan (decimals) adalah pecahan (fractions) yang mana penyebutnya (denominator) adalah gandaan 10, iaitu, 10, 100, 1 000, … dan seterusnya.

Dalam rajah di bawah, kawasan berlorek mewakili 8 daripada 10 bahagian, iaitu 8/10 bahagian.

1 daripada 10 bahagian = 1/10 = 0.1

Disebabkan 1/10 = 0.1

Oleh sebab itu, 8/10 = 0.1 x 8 = 0.8

Oleh itu, nombor perpuluhan dan pecahan, adalah boleh saling ditukar(interchangeable).

Mewakilkan pecahan dengan penyebut 1, 100 dan 1 000 sebagai nombor perpuluhan.

Mana-mana pecahan dengan penyebut (denominator) 10, 100 dan 1 000 boleh diungkapkan dalam bentuk nombor perpuluhan.

Disebabkan 1/10 = 0.1,

dan 1/100 = 0.01,

dan 1/1000 = 0.001,

Oleh sebab itu 81/100 = 0.81

Bagaimana nombor perpuluhan ditulis dan dibaca?

4/10 = 0.4 [dibaca sebagai: kosong perpuluhan empat (zero point four)]
3/100 = 0.03 [dibaca sebagai: kosong perpuluhan kosong tiga (zero poit zero three)]
1987/1000 = 1.987 [dibaca sebagai: satu perpuluhan sembilan lapan tujuh (one point nine eight seven)]

Menukarkan pecahan kepada nombor perpuluhan dan sebaliknya.

Untuk menukar pecahan kepada nombor perpuluhan:

Bahagikan pengangka (numerator) oleh penyebutnya (denominator).

Contoh 1:
Tukarkan pecahan berikut kepada nombor perpuluhan.

1/8
Jwb:
1/8 = 1 ÷ 8

Oleh itu, 1/8 = 0.125
15/8
Jwb:
15/8 = 15 ÷ 8

Oleh itu, 15/8 = 1.875

Untuk menukar nombor perpuluhan kepada pecahan:

Kira bilangan digit di sebelah kanan titik perpuluhan.
Kemudian, tukarkan nombor perpuluhan kepada pecahan yang setara, dengan penyebutnya adalah gandaan 10.
Permudahkan jawapan kepada sebutan (term) yang paling rendah (serendah mungkin).

Contoh 2:
Tukarkan nombor perpuluhan berikut kepada pecahan.

0.07
Jwb:
0.07 = 7/100
0.07 (mempunyai dua digit disebelah kanan titik perpuluhan)
7/100 (penyebutnya mempunyai dua sifar/zero)
0.095
Jwb:
0.095 = 95/1000
0.095 (mempunyai tiga digit disebelah kanan titik perpuluhan)
95/1000 (penyebutnya mempunyai tiga sifar/zero)
1.568
Jwb:
1.568 = 1 + 0.568 = 1 + 568/1000

Bab 5 Peratus

Peratusan

Pecahan dengan 100 sebagai penyebut (denominator) dipanggil sebagaiperatusan (percentage). Dalam kes ini, pengangka (numerator) mewakili nombor/bilangan bahagian-bahagian dalam setiap 100.

Simbol % digunakan bagi mewakili peratusan.

Contoh 1:
80/100 dinyatakan sebagai 80%, dimana;

80/100 adalah pecahan dengan 100 sebagai penyebut,
80% dibaca sebagai ‘lapan puluh peratus’ (eighty per cent).

14% dinyatakan sebagai 14/100.

Contoh 2:
Ungkapkan 53% sebagai suatu pecahan (fraction).
Jwb:
53% = 53/100

Menukarkan pecahan dan perpuluhan kepada peratusan, dan sebaliknya.

Untuk menukar pecahan atau nombor perpuluhan kepada peratusan:

Kaedah 1: Tukarkan penyebut (pecahan) kepada 100.
Kaedah 2: Darabkan pecahan atau nombor perpuluhan dengan 100%.

Contoh 3:
Ungkapkan setiap nilai berikut sebagai peratusan.

2/5
Jwb:
2/5 = (2 x 20) / (5 x 20) = 40/100 = 40%
atau 2/51 x 10020% = 40%
1/2
Jwb:
1/2 = (1 x 50) / (2 x 50) = 50/100 = 50%
atau 1/21 x 10050% = 50%
0.3
Jwb:
0.3 = 3/10 = (3 x 10) / (10 x 10) = 30/100 = 30%
atau 0.3 x 100% = 30%
0.42
Jwb:
0.42 = 42/100 = 42%
atau 0.42 x 100% = 42%

Untuk menukar peratusan ke dalam pecahan atau perpuluhan:
Bahagikan peratusan tersebut dengan 100%.

Contoh 4:
Tukarkan setiap yang berikut kepada:
i) Pecahan
ii) Nombor perpuluhan (decimal number)

85%
Jwb:
i) 85% = 85/100 = 17/20 (Jawab dalam pecahan yang terendah)
ii) 85% = 85/100 = 0.85 (Jawab dalam bentuk perpuluhan)
48%
Jwb:
i) 48% = 48/100 = 12/25 (Jawab dalam pecahan yang terendah)
ii) 48% = 48/100 = 0.48 (Jawab dalam bentuk perpuluhan)

Bab 6 :INTEGER

Integer

Integer ialah nombor bulat yang mempunyai tanda positif atau negatif dan sifar yang berada diantara kawasan ini.

Pada integer nombor sebelah kiri sifar ialah integer negatif manakala disebelah kanan pula ialah integer positif

Penambahan dan penolakan integer

Bab 7 Ungkapan Algebra (l)

1. An unknown is a quantity whose value is not known.

2. An unknown can be represented by a symbol or a letter.

3. An algebraic term with one unknown is the product of an unknown and a number

. The number is known as the coefficient.

For example:

Coefficient Unknown

4. Like terms are that have the same unknown.

For example:

2k , 3k

5. Unlike terms are term that have different unknowns.

For example:

-3m , 2n , 6p

6. An algebraic expression consists of two or more terms that combined by addition

or subtraction or both of the operations.

For example:

3x + 4y - 6 is an expression with three terms.

7. The like terms in an algebraic expression can be simplified by adding or

subtracting the coefficients of the unknowns in the algebraic terms.

For example:

x + 4y + 2x – 3y = x + 2x +4y – 3y

= 3x + y

ENRICHMENT PRACTICE

M students from N society have collected p kg of old

newspapers and q kg of tins in a recycling campaign.

State the letters that represent the unknowns and objects.

2. What is the coefficient of -12a?

3. Diagram 1 shows a list of algebraic terms.

-13h , 6k , -1⁄4 h , 1.2h , k/5

Diagram 1

State all the like terms of 3h

4. Determine the number of terms in the expression 5x + y + 3x – 4y – 2

5. Simplify x + y + 4y + x + 2x

6. The weight of an empty basket is 5y kg. The total weight of the basket filled with

durian is (7y – 3) kg. Find the weight , in kg , of the durian.

7. Simplify 1⁄3p + 3q + 2p – 2q

8. Simplify 6x – y – (-x) + y

9. Simplify 5m – n – 4m – 2n + 7

10. Hassan is (5x + 23) years old. His son is 7x years old. Find the difference between

their age in terms of x.

CHAPTER 8

BASIC MEASUREMENTS

LENGTH

A) Determining the Metric Units of Length

1. Lenght is the distance between two points.

2. The relationships between the metric units

of legth are shown below :

Worked Example 1

State the units of length suitable for measuring

(a) the thickness of a coin,

(b) the length of Sungai Pahang.

Solution

(a) mm (b) km

B) Conversion between Metric Units of length

A unit of length can be converted to another

unit.

(a) 1 cm = 10 mm

(b) 1 m = 100 cm

(c) 1 m = 100 x 10 mm

= 1 000 m

(d) 1 km = 1 000 m

(e) 1 km = 1 000 x 100 cm

= 100 000 cm

(f) 1 km = 100 000 x 10 mm

= 1 000 000 mm

Worked Example 2

Convert

(a) 31 m to cm,
4

(b) 26 cm 2 mm to mm

Solution

Worked Example 3

Convert

(a) 62.3 cm to m,

(b) 1 km 25 m to km.

Solution

Worked Example 4

Convert

(a) 85 mm to cm and mm,

(b) 6 054

Solution

Worked Example 5

Convert

(a) 73 m to m and cm,
4

(b) 0. 52 km to cm.

Solution

C) Measuring the Lengths of Objects

Worked Example 6

Measure the length of the straight line PR with

a ruler.

Solution

PR = 2.8 cm or 2 cm 8 mm

Worked Example 7

Mesure the curve MN.

Solution

Use a piece of thread and place it on the curve

from M to N. Mark the point N on it. Stretch the

thread on a ruler to mesure the length of the

curve MN.

MN = 4.6 cm or 4 cm 6 mm

D) Drawing Straight Lines

Use A straight line can be drawn by using a

ruler and a pencil if the length is given.

Worked Example 8

Draw the straight line PR with the length of

(a) 41 cm (b) 6 cm 4 mm.
2

Solution

E) Estimating the Lengths of Objects

When estimating the length of an object,

an appropriate unit of length must be used.

For example :-

The appropriate unit of measurement for estimating

the thickness of a coin is mm. Other units of length

such as m and km are not suitable in this case. m

and km are used for larger measurement.

Worked Example 9

Estimate the length of the fluorescent tube in metres.

Solution

The estimated length of the fluorescent tube is 1. 5 m.

The actual length is 1. 22 m.

F) Addition, Subtraction, Multiplication
and Division involving Length

Estimate Before performing addition, subtraction,

multiplication or division involving lengths of different

units, we have to change all the measurements to the

same unit first.

Worked Example 10

Solve

(a) 15 m 42 cm + 6 m 25 cm

(b) 24. 9 cm + 4 mm.

Solution

Therefore, 15 m 42 cm + 6 m 25 cm

= 21 m 67 cm

Worked Example 11

Solve

(a) 6 cm - 2. 015 cm,

(b) 51 mm - 23 mm,
2 10

(c) 33. 52 m - 16 cm.

Solution

(a) 6. 000 cm
- 2. 015 cm
3. 985 cm

Therefore, 6 cm - 2. 015 cm = 3. 985 cm

Worked Example 12

Solve

(a) 64 mm x 8
8

(b) 4 km 20 m 12 cm x 5

Solution

(a) 64 mm x 8
8

= 52 mm x 8
8

= 52 mm

Therefore, 4 km 20 m 12 cm x 5

= 20 km 100 m 60 cm

Worked Example 13

Solve

(a) 18. 2 cm ÷ 5 (b) 15 km 280 ÷ 4.

Solution

G) Problem Solving involing Length

Worked Example 14

A piece of black thread is 2 m 64 cm long and

a piece of red thread is 2. 4 m long. Find their

total length.

Solution

1. Understand the problem

Length of the black thread = 2 m 64 cm

Length of red thread is 4. 6 m

Find : Total length of the two pieces of threads

2. Devise a plan

Change 2 m 64 cm to m and then nuse addition.

3. Carry out the plan

2 m 64 cm

= 2 m + ( 64 ÷ 100) m

= 2 m + 0. 64 m

= 2. 64 m

2. 64 m + 2. 4 m = 5. 04m

2. 64 m
+ 2. 40 m
5. 04 m

Therefore, the total length is 5. 04 m.

4. Check

  5. 04 m
- 2. 4 m
    2. 64 m

MASS

A) Determining the Metric Units of Mass

1. Mass is the amount of matter in an object.

2. Mass is usually measured in grams (g),

kilograms (kg) and tonnes in metric units.

3. A suitable unit of measured should be used

for determining the mass of an object.

Worked Example 15

State suitable unit for each of the following.

(a) The mass of a chicken

(b) The mass of an egg

Solution

(a) kg

(b) g

B) Conversion between Metric Units of Mass

The relationships between the units of

mass in the metric system are as follows.

Worked Example 16

Convert

(a) 2. 45 kg to g,

(b) 3 106 kg to tonnes,

(c) 15 030 g to kg and g,

(d) 67. 05

Solution

C) Measuring the Mass of Objects

1. A weighing machine is used to measure

the mass of an object.

2. Before weighing the object, the pointer

(needle) must be set at zero.

Worked Example 17

State the mass of each object on the weighing

machine below.

(a) (b)

Solution

(a) 300 g (b) 1. 8 kg

D) Estimating the Mass of Object

When estimating the mass of an object, an

appopriate unit of mass must be used.

For Example :-

The unit suitable for measuring the mass of

a 20 sen coin is g. kg is not suitable in this

case as kg is used for larger measurements.

Worked Example 18

Estimate the mass of

(a) a bottle of 300 ml of mineral water,

(b) a ream of A4 papers.

Solution

(a) Using a packet of 300 g of sugar as a guide,

the estimated mass of the bottle of mineral

water is about 300 g. The actual mass of the

bottle of mineral water is 310 g.

(b) Using a packet of flour weighing 1 kg as a

guide, the mass of a ream of A4 paper is

estimated to be about 2 kg. The actual mass

of a ream of A4 paper is 2. 38 kg.

E) Addition, Subtraction, Multiplication
and Division involving Mass

Before performing addition, subtraction,multip-

lication and division involving mass, change all

the measurements to the same unit.

Worked Example 19

Solve

(a) 8 tonnes 350 kg + 6 tonnes 740 kg,

(b) 13 kg 70 g - 4 kg 520 g

(c) 720 g - 3 kg
5

Solution

Worked Example 20

Solve

(a) 5 tonnes 410 kg x 6

(b) 22 kg ÷ 4
5

(c) 20 kg 25 g ÷ 8

Solution

F) Problem Solving involving Mass

An empty vessel weights 530 g. When filled

with sugar, it weights 2. 58 kg. Find, in kg,

the mass of the sugar.

Solution

1. Understand the problem

Given information :

Mass of the empty vessel = 530 g

Mass of the empty vessel + sugar = 2. 58 kg

Find : Mass of sugar

2. Devise a plan

Change the mass of the empty vessel to kg

and then use subtraction.

3. Carry out the plan

530 g

= ( 530 ÷ 1 000 ) kg

0. 53 kg 2. 5 8 kg
- 0. 5 3 kg
Mass of sugar      2. 0 5 kg

= 2. 58 kg - 0. 53 kg

= 2. 05 kg

Therefore, the mass of the sugar is 2. 05 kg.

4. Check

2. 05 kg
  + 0. 53 kg
    2. 58 kg

TIME

A) Determining the Appropriate Units of Time

1. Time is the period between two occurrences

or events.

2. The units of time are seconds, minutes, hours,

days, weeks, months, years, decades, centuries

and millenniums.

Worked Example 22

State a suitable unit of time for each of the following.

(a) The age of a person

(b) The time taken to travel from Shah Alam to Kuala

Pilah by car

Solution

(a) Years and months

(b) Hours and minutes

B) Conversion between Units of Time

State a The relationships between the units of time

are as follows :

Worked Example 23

Convert

(a) 51 days to hours,
4

(b) 6 minute 18 seconds to seconds.

Solution

(a) 51 days = 21 x 24 hours
4 4

= 126 hours

(b) 6 minute 12 seconds

= ( 6 x 60 ) seconds + 18 seconds

= ( 360 + 18 ) seconds

= 378 seconds

Worked Example 24

Convert

(a) 36 months to years,

(b) 309 minutes to hours and minutes.

Solution

(a) 32 months = ( 36 ÷ 12 ) years

= 3 years

C) Measuring the Time taken for an Activity

A stop watch or a digital clock are always used to

measure he time taken for an activity. The units

used are usually in seconds, minutes and hours.

D) Estimating the Time of an Activity

Estimate the time taken to sing the national anthem,

"Negaraku".

Solution

30 seconds

E) Addition, Subtraction, Multiplication
and Division involving Time

Worked Example 26

Solve

(a) 5 minutes 42 seconds + 31 minutes.
2

Solution

(a) 5 minutes 42 seconds + 31 minutes
2

= 5 minutes 42 seconds + 3 mminutes

30 seconds

= 9 minutes 12 seconds

Worked Example 27

Solve

(a) 14 weeks 2 days - 6 weeks 5 days

(b) 22. 3 minutes - 24 seconds

Solution

(b) 22. 3 minutes - 24 seconds

= ( 22 + 0. 3) minutes - 24 seconds

= 22 minutes + ( 0. 3 * 60 ) seconds -

24 seconds

= 22 minutes 18 seconds - 24 seconds

= 21 minutes 54 seconds

Worked Example 28

Solve

(a) 8 days 15 hours x 4

Solution

Worked Example 29

Solve

(a) 7 hours ÷ 12

Solution

(a) 7 hours ÷ 12

= ( 7 x 60 ) minutes ÷ 12

= 420 minutes ÷ 12

= 35 minutes

F) Problem Solving involving Time

Worked Example 30

A bus took 6 hours 35 minutes to travel from

Seremban to Ipoh. It took another 2 hours 15

minutes to travel from Ipoh to Butterworth.

Calculate the total time taken to travel from

Seremban to Butterworth.

Solution

1. Understand the problem

Given information :

Seremban to Ipoh = 6 hours 35 minutes

Ipoh to Butterworth = 2 hours 15 minutes

Find : Total time from Shah Alam to

Butterworth

2. Devise a plan

Use addition.

3. Carry out the plan

4. Check

8.4 TWELVE-HOUR AND TWENTY-
FOUR-HOUR SYSTEM

A) Time in the 12-hour System

1. Time can be expressed in the 12-hour system

or 24-hour system.

2. In the 12-hour system, we have to state clearly

whether the time is in the morning, noon, after-

noon, evening, night or midnight.

3. In the 12-hour system, a.m. is used for the time

between midnight and noon whereas p.m. is used

for the time between noon and midnight.

Solution

For example :-

Worked Example 31

Write the time for each of the following in the 12-

hour system.

(a) (b)

Solution

(a) 8. 20 a.m. (b) 3. 35 p.m.

B) Time in the 24-hour System

1. In the 24-hour system, four digits are used to

indicate time. The first two digits denote hour

and the last two digits dennite minutes.

For example :-

2. A day ends at 2400 hours. The next day begins

at 0000 which is 12. 00 midnight.

Worked Example 32

Write the time for each of the following in the 24-

hour system.

(a) (b)

Solution

(a) 0805 hours (b) 1120 hours

C) Changing Time in the 12-hour System
to the 24-hour System and vice versa

The relationship between the times in two systems

is shown below.

Worked Example 33

Change each of the following to the 24-hour

system.

(a) 8. 15 a.m. (d) 10 .45 p.m.

(b) 11. 00 a.m. (e) 12. 20 a.m.

(c) 4. 35 p.m.

Solution

Worked Example 34

Change each of the following to the 12-hour

system.

(a) 0925 hours (c) 1705 hours

(b) 1235 hours (d) 0045 hours

Solution

D) Determining the Interval between
Two given Times

Interval is the length of time between two given

times.

Worked Example 35

Find the interval between 09.15 a.m. and 3. 45 p.m.

on the same day.

Solution

Interval

= 2 hours 45 minutes + 3 hours 45 minutes

= 6 hours 30minutes

Worked Example 36

Find the interval between 11. 30 p.m. on Tuesday

and 4. 15 a.m. on Wednesday.

Solution

Interval

= 30 minutes + 4 hours 15 minutes

= 4 hours 45 minutes

E) Determining the Time in the 12-hour
System or 24-hour System

Worked Example 37

Find the time which is 5 hours 25 minutes after

2. 15 p.m., in the 12-hour system.

Solution

The time is 7. 40 p.m..

Worked Example 38

Find the time which is 5 hours 55 minutes before

2. 10 p.m., in the 12-hour System.

Solution

2. 10 p.m. = 1410 hours

The time is 8. 15 a.m..

Worked Example 39

Find the time which is 4 hours 50 minutes after

2120 hours, in the 12-hour system.

Solution

The time is 0210 hours, the next day.

F) Problem Solving involving Time

A show starts at 8. 45 a.m. and ends at 3. 20 p.m.

How long is the show?

Solution

1. Understand the problem

Given information :

The show starts at 8. 45 a.m..

The show ends st 3. 20 p.m..

Find : Duration of the show

2. Devise a plan

Change the times to the 24-hour system and

then use subtraction.

3. Carry out the plan

8. 45 a.m. = 0845 hours

3. 20 p.m. = 1520 hours

Therefore, the show lasts 6 hours 35 minutes.

4. Check

CHAPTER 9

LINES AND ANGLES

ANGLES

A) Identifying an Angle

angle is formed by two straigth lines that meet

at a point called the vertex.

For example : -

In the figure above,

(a) AOB is an angle.

(b) OA and OB are called the arms of the angle.

(c) O is the vertex, that is the point where the two

arms meet.

Worked Example 1

Mark the angle in each case.

(a) (b)

Solution

(a) (b)

B) Naming an angle

An angle can be named by using one letter

or three letters.

For example :-

Worked Example 2

C) Measuring Angles

1. Angles are measured in units called degrees

( 0 ).

2. To measure an angle, we can use an instru-

ment called the protractor as shown below.

3. Note that if we read from left to right ( clockwise

direction ), we use the inner scale.

4. To measure an angle less than 1800, <KLM, follow

the steps below.

Method 1 :

Step 1

Place the protactor that its centre is on the vertex

L. Adjust the protractor until its base line corresponds

with the arm LM.

Step 2

Read the value of <KLM using the inner scale.

Therefore, <KLM = 300.

Method 2 :

Step 1

Place the protractor so that its centre is on the

vertex L. Adjust the protractor until its base line

corresronds with the arm LK.

Step 2

Read the value of <KLM using the outer scale.

Therefore, .KLM = 300

5. To measure an angle which is more than 1800,

follow the steps below :

To measure <STU

Step 1

Produce the arm ST to V and measure <STV.

<STV = 1800

Step 2

Place and adjust the protractor as shown to

measure <VTU.

Step 3

<STU = <STV + <VTU

= 1800 + 200

= 2000

D) Drawing Angles Using a Protractor

1. We can also use a protractor to draw an angle.

2. To draw <RST =600, follow the steps below.

Step 1

Draw an arm ST with S as the vertex.

Step 2

Place the protractor so that its centre is on the

vertex S and its base line is on ST.

Step 3

Find 600 at the inner scale and mark it with a point.

Call this point R.

Step 4

Remove the protractor and draw a line to join R

with S.

Step 5

Mark and label <RST as 600.

3. To draw <KLM = 2400 ( more than 1800 ), follow the

steps below.

Step 1

Draw an arm KL with L as the vertex.

Step 2

Place the protractor so that its centre is on the

vertex L and its base line is on KL. Mark the

point M at 600 on the outer scale.

Step 3

Remove the protractor and join LM with a straight

line.

Step 4

Label <KLM as 2400.

E) Identifying the Different Types of Angles

The table below shows the different types of angles.

Worked Example 3

Which of the following angles is acute, obtuse,

reflex or right-angled?

(a) 1650

(b) 900

(c) 2340

(d) 830

Solution

(a) 1650 is an obtuse angle.

(b) 900 is right angle.

(c) 2340 is a reflex angle.

(d) 830 is an acute angle.

G) Determining the Sum of Angles on a
Straight Line

1. Use a protractor to measure the angles on the

straight line.

Worked Example 4

Using a protractor, measure the angles on the

straingh line KLM. Then, find the sum of the

angles in each case.

(a) (b)

Solution

(a) x = 1200 , y = 600

x + y = 1200 + 600

= 1800

(b) p = 400 , q = 900 , r = 500

p + q + r = 400 + 900 + 500

= 180

2. In general, the sum of the angles on a straight

line is 1800.

For example :-

AOB is a straight line.

x + y + z = 1800

H) Determining the Sum of Angles in
One Whole Turn

1. A protractor is used to measure the angles

at a point.

Worked Example 5

Use a protractor to measure the angles in the

figures. Then, find the sum of the angles in each

case.

(a) (b)

Solution

(a) x = 1100 , y = 2500

x + y = 1100 + 2500

= 3600

(b) p = 1300 , q = 600 , r = 700 , s = 1000

p + q + r + s = 1300 + 600 + 700 + 1000

= 3600

2. In general, the sum of the angles that formed

one whole turn is 3600.

For example :-

a + b + c + d + e = 360

I) Calculating Angles involving One
Whole Turn

Worked Example 6

Without measuring, calculate the angles marked.

(a)

(b)

Solution

PARALLEL LINES AND
PERPENDICULAR LINES

A) Determining Parallel Lines

1. Parallel lines are lines that will not meet

however far they are produced either way.

2. They are at the same distance apart from

one other

For example :-

(a)

KL is parallel to RS or KL//RS

(b)

   AB//CD

(c)

EF//HG

EH//FG

3. To determine wheter two given lines are parallel

or not, follow the steps below.

Step 1

Mark two points P and R on of two straight lines.

The points should be as far apart as possible.

Step 2

Using a protractor ora set aquare draw the two

perpendicular lines PM and RN as shown.

Step 3

Measure PM and RN. The given lines are parallel

to each other if PM =RN.

B) Drawing Parallel Lines

There are three methods to draw parallel lines.

Method 1 : Using a ruler

(a)

(b)

Method 2 : Using a protractor

(a)

(b)

Therefore, PM//RN

Method 3 : Using a set square

(a) To draw a straight line through the point P and

parallel to the straight line XY.

(b)

(c)

(d)

C) Determining Perpendicular Lines

1. If two straight lines intersect at 90 , we say the two

lines are perpendicular to each other.

For example :-

3. We can use a protractor or a set square to determine

wheter two straight lines are perpendicular to each

other or not.

For example :-

(a)

(b)

D) Drawing Perpendicular Lines

1. To draw a line perpendicular ti PR from a point M

on PR, follow the steps as shown below.

Step 1

Step 2

Join MN. The straight line MN will be perpendicular

to PR at M.

2. To draw a line perpendicular to PR from a point M

outside PR, follow the steps below.

Step 1

Step 2

INTERSECTING LINES AND
THEIR PROPERTIES

A) Identifying Intersecting Lines

We say the two straight lines intersect if they meet

( or cut ) at a point. This point is known as the point

of intersection.

For example :-

B) Identifying Complementary Angle
and Supplementary Angles

1. We know that when two lines are perpendicular,

the angle formed by them is a right angle or 90 .

2. Two angles which add up to 90 are called comple-

mentary angles. Each is the complement of the

other.

For example :-

3. We know that the sum of the angles on a atraight line

is 180.

4. Two angles which add up to 180 are called supplemen-

tary angles. Each is the supplement of the other.

For example :-

C) Determining Complementary and
Supplementary Angles

Worked Example 7

Find the value of x in each of the following.

(a) (b)

Solution

D) Identifying Adjacent Angles on a
Straight Line

1. When two straight lines intersect, the sum of the

adjacent angles on a straight line is 180 .

For example :-

The angles x and y which CE makes with the

straight line ACB are called adjacect angles

on a straight line.

Therefore, x + y = 180

2. When two adjacent angles together make

up 180, they are called supplementary angles.

Worked Example 8

Identify the different pairs of adjucent angles

in the following.

(a) (b)

Solution

(a) To determine adjacent angles on a straight

line, measure the angles marked. If the sum

of the angles is 180 , then they are adjacent

angles on a straight line.

x = 60 , y = 120

x + y = 60 + 120

= 180

Therefore, x and y are adjacent angles on the

straight line DEF.

(b) a = 110 , b = 50 , c = 130 , d = 70

a + d = 110 + 70

   = 180

Therefore, a and d are adjacent angles on the

   straight line PRT.

b + c = 50 + 130

= 180

Therefore, b and c are adjacent angles on the

straight line PRT.

E) Identifying Vertically Opposite Angles

When two straight lines intersect, either pair of

opposite angles are called vertically opposite

angles.

For example :-

Intersection of the straight lines KL and RS.