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Mathematics Grade 9 Flipbook PDF
Mathematics Grade 9
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GRADE
9
Extended Programme
Student’s Book
Mauritius Institute of Education under the aegis of
GRADE 9 Student's Book
MATHEMATICS
Professor Vassen Naëck - Head Curriculum Implementation,Textbook Development and Evaluation
MATHEMATICS PANEL Dr Khemduth Singh Angateeah - Coordinator, Lecturer, MIE
- Associate Professor, MIE Mr Sooryadev Purdasseea - Lecturer, MIE Miss Dilshad Beebee Codobaccus - Educator Mr Vinayesingh Sookarah - Educator Mrs Beeneswaree Sonah - Educator Mr Joel Descubes - Educator Miss Hanna Khodabocus - Educator Mr Baboojeesing Bhantooa - Educator Mr Zaheer Hissoob - Educator Dr Rajeev Nenduradu
Design Kunal Sumbhoo Karnesh Ramful
- Graphic Designer, MIE - Graphic Designer, MIE
© Mauritius Institute of Education (2020) ISBN: 978-99949-53-54-7
Acknowledgements The Mathematics textbook panel wishes to thank the following: - Mathematics Education Department (MIE) for vetting. - Joseph Richard Constant Perrine (Educator from Rodrigues) for vetting. - Laxmi Devi Angateeah (Educator) for vetting. - Dr Yesha Devi Mahadeo-Doorgakant (Lecturer, MIE) for proof reading.
Consent from copyright owners has been sought. However, we extend our apologies to those we might have overlooked. All materials should be used strictly for educational purposes.
FOREWORD This textbook is designed for Year 3 of the Extended Programme (EP). It is based on the Nine Year Continuous Basic Education (NYCBE) curriculum for Grades 7, 8 and 9 which is accessible through the MIE’s website, www.mie.ac.mu. The textbook builds upon the competencies acquired in the first two years of the Extended Programme. The content and pedagogical approaches are adapted to the cognitive level and profile of the students, while also paving the way for them to acquire essential skills and knowledge to move to the next level. The fact that the material is contextualised will certainly make it highly appealing to the learners. The writing of the textbooks involved several key contributors, namely academics from the MIE and educators from Mauritius and Rodrigues, as well as other stakeholders. It took into account feedback obtained during community-of-practice sessions to ensure relevance and adequacy. This was a highly innovative approach which promoted a valuable partnership with EP educators during and for the development of educational resources. We are especially appreciative of the insight brought by educators whose suggestions emanated from long-standing experience and practice in the field. The development of textbooks has been a very challenging exercise for the writers and the MIE. We had to ensure that the learning experiences of our students are enriched through approaches which appeal to them, without compromising on quality. I would, therefore, wish to thank all the writers and contributors who have produced content of high standard, thereby ensuring that the objectives of the National Curriculum Framework are skilfully translated through the textbooks. Every endeavour involves several dedicated, hardworking and able staff whose contribution needs to be acknowledged. Professor Vassen Naëck, Head Curriculum Implementation and Textbook Development and Evaluation provided guidance with respect to the objectives of the NCF, while ascertaining that the instructional designs are appropriate for the age group targeted. I also acknowledge the efforts of the graphic designers who put in much hard work to maintain the quality of the MIE publications. My thanks also go to the support staff who ensured that everyone receives the necessary support and work environment which is conducive to a creative endeavour. I am equally thankful to the Ministry of Education, Human Resources, Tertiary Education and Scientific Research for actively engaging the MIE in the development of the textbooks for the reform project. I wish enriching and enjoyable experiences to all users of the new set of Grade 9 textbooks.
Dr O Nath Varma Director Mauritius Institute of Education
i
PREFACE The Grade 9 Mathematics Student’s Book and Teacher’s Book for the Extended Programme have been developed to meet the philosophical principles of the New Curriculum Framework (Secondary) as proposed by the Nine-Year Basic Education Reform. The focus of the current mathematics curriculum for secondary education is on developing the mathematical proficiency of all learners. This Grade 9 textbook adopts similar pedagogical approaches to those used in Grade 7 and Grade 8 textbooks. The textbook embraces a student-centered and an activity-based learning approach. In line with the spiral mathematics curriculum, concepts learnt in lower grades are reviewed and new concepts are embedded therein. The same core curriculum of the 3-year mainstream cycle (Grade 7 to 9) are adapted to match the learning disposition of the Grade 9 students in the Extended stream. To foster the development of mathematical proficiency among Grade 9 students, the textbook advocates the use of the concrete-pictorial-abstract (CPA) strategy, through meaningful activities and an inductive approach for the development of concepts. It is envisaged that the activities will encourage students’ active participation in class and will also enrich their understanding and communication so that they can solve a range of problems in mathematics. The book also lays strong emphasis on problem representations and connections with real life. The textbook provides diverse tasks to engage students in the effective learning of mathematics. Examples provided are followed by carefully graded exercises to consolidate the concepts learned. The continuous assessment component gauges the extent of their learning. The IT corners integrated provide opportunities for pupils to enhance their understanding of the concepts using technology. Educators are encouraged to make optimal use of the ‘Note to Teacher’ in their books to promote the learning of Mathematics. They are urged to act as facilitators and support students to learn key ideas and concepts presented in the Student’s Book. Educators have the freedom to choose classroom activities to progressively assess the competencies of learners. The Mathematics Panel hopes that students, teachers and parents will make optimum use of the textbook to drive up the achievement in mathematics.
Mathematics Panel
ii
Table of Contents 1. Numbers
1.1 Numeration 1 1.2 Types of Numbers 6 1.3 Integers 16 1.4 Indices 32 1.5 Order of Operations 41 1.6 Factors and Multiples 47 1.7 Fractions 57 1.8 Decimals 66 1.9 Real Numbers 82 1.10 Sequences 93 1.11 Percentage 99 1.12 Ratio and Proportion 112
2. Geometry 2.1 Pythagoras' Theorem 125 2.2 Angles 134 2.3 Polygons 155 2.4 Coordinates 177 2.5 Geometrical Constructions 195 2.6 Symmetry 207 2.7 Reflection 215
3. Measurement 3.1 Area 225 3.2 Speed 244
4. Algebra
4.1 Algebraic expressions 249
5. Statistics
5.1 Statistics 263
iii
NUMBERS
1
1.1 NUMERATION By the end of this topic, you should be able to: • Read numbers from 0 to 1 000 000. • Count numbers up to 1 000 000. • Write numbers from 0 to 1 000 000.
My father wants to buy a new car. There is a variety of choice at different price. What is the price of some of the cars?
FLASH CAR SALE
Special offer
AS FROM Rs 450 000 normal price Rs 500 000
Special offer
AS FROM Rs 500 000 normal price Rs 560 000
Special offer
AS FROM Rs 1 000 000 normal price Rs 1 200 000
Special offer
AS FROM Rs 535 000 normal price Rs 565 000
Special offer
AS FROM Rs 590 000 normal price Rs 630 000
With the help of your teacher, try to answer the following questions: 1.
What can you see on the advert?
2.
Which car is the cheapest?
3.
Which car is the most expensive?
1
1. NUMBERS RECALL
1. Write in figures.
(a) Two thousand seven hundred and fifty four
=
(b) Six thousand three hundred and one
=
(c) Eighty two thousand five hundred and sixty one
=
(d) Ninety three thousand seven hundred and fourteen
=
(e) Twenty four thousand three hundred and twelve =
2. Write the following in words.
(a) 585
(b) 802
(c) 8 521
(d) 5 961
(f ) 85 356
(e) 63 123
2
Breaking down six digit numbers For example, consider the number 426 015, 4 2 6 0 1 5 is four hundred and twenty six thousand and fifteen. 5 units 1 ten 0 hundreds 6 thousands 2 ten thousands 4 hundred thousands Similarly, 235 914 is two hundred and thirty five thousand nine hundred and fourteen.
Hundred thousand as a place value For example, consider the number 264 537 on an abacus, HTH
TTH
TH
H
T
U
2
6
4
5
3
7
2 has the value of 2 hundred thousands = 200 000 6 has the value of 6 ten thousands
= 60 000
4 has the value of 4 thousands
= 4 000
5 has the value of 5 hundreds
= 500
3 has the value of 3 tens
= 30
7 has the value of 7 units
=7
hundred thousands
ten thousands
thousands
hundreds
tens
units
2
6
4
5
3
7 3
1. NUMBERS EXERCISE 1.1.1
1. Complete the following with the missing numbers.
(a) 569 230 = _____ hundred thousands + _____ ten thousands + _____ thousands + _____ hundreds + _____ tens + _____ units
(b) 309 415 = _____ hundred thousands + _____ ten thousands + _____ thousands + _____ hundreds + _____ tens + _____ units
(c) 158 006 = _____ hundred thousands + _____ ten thousands + _____ thousands + _____ hundreds + _____ tens + _____ units
(d) 560 317 = _____ hundred thousands + _____ ten thousands + _____ thousands + _____ hundreds + _____ tens + _____ units
2. Write in figures.
(a) Two hundred and thirteen thousand seven hundred and fifty four = _______________
(b) Six hundred and seven thousand three hundred and one = ________________
(c) Nine hundred and eighty two thousand five hundred and sixty one = ____________
(d) Five hundred and fifty three thousand seven hundred and fourteen = ____________
3. Write the following in words.
4
(a) 241 958 = _________________________________________________________
(b) 156 235 = _________________________________________________________
(c) 365 614 = _________________________________________________________
(d) 958 250 = _________________________________________________________
One million Do you know which whole number comes just after 999 999?
+1
999 997
+1
999 998
+1
999 999
Count in hundred thousands.
700 000
800 000
What will 1000 times thousand be?
1 000 times
1 000 000
900 000
1 000 thousands
1 million
Continuous Assessment 1.1 1. Complete the following with the missing numbers. (a) 629 352 = _____ hundred thousands + _____ tens thousands + _____ thousands + _____ hundreds + _____ tens + _____ units (b) 934 715 = _____ hundred thousands + _____ tens thousands + _____ thousands + _____ hundreds + _____ tens + _____ units 2. Write in figures. (a) Four hundred and seventy thousand five hundred and forty-two = _______________
(b) Eight hundred and twelve thousand nine hundred and fifty-seven = _____________
3. Write the following in words.
(a) 742 598 = (b) 453 231 = 5
1. NUMBERS 1.2 TYPES OF NUMBERS By the end of this topic, you should be able to: • Identify triangular and square numbers. • Distinguish between triangular and square numbers. • Find the square root of perfect squares. • Find the cube of a number. • Find the cube root of a cubic number.
What are you doing?
I am trying to make some triangles.
Let’s do that.
1
Did you count the number of bottle caps you have used?
2 3 4 5 6
Observe the pictures carefully. Have you ever tried this? Discuss with your friends what the girl is building. How many bottle caps is she adding in each row? Using bottle caps, try to build one on your own. 6
Triangular Numbers A triangular number counts objects arranged in an equilateral triangle.
Let’s try this activity.
Activity - Stacking counters in rows
For this activity, counters such as bottle caps, plastic cups or coins will be needed. 1. Observe the following figures. Count the total number of counters used. Figure 1
Figure 2
Figure 3
1
1+2
1+2+3
3
Figure 1 2 3 4 5 6 7 8 9 10
Figure 4
Figure 5
1+2+3+4
1+2+3+4+5
10
15
6
Number of dots 1 1+2 1+2+3 1+2+3+4 1+2+3+4+5
Triangular number 1 3 6 10 15 Complete the table up to Figure 10 (that is, up to the 10th triangular number).
We say that 1, 3, 6, 10 and 15 are triangular numbers. Write down the other triangular numbers that you obtained after completing the above table. Can you find two more triangular numbers? 7
1. NUMBERS Square numbers Perfect squares can be obtained by forming squares with counters. Square numbers are obtained by multiplying a number by itself.
Activity - Placing paper strips so that they form squares. For this activity, paper strips will be cut out into squares.
Let’s try this activity.
1. Cut papers into small squares.
2. Using these small squares, form bigger squares as illustrated.
Number of small squares used: 1 4 9 Figure 1 Figure 2 Figure 3
16 Figure 4
______ Figure 5
3. Count the number of small squares in each figure. Then verify by multiplying the length and the width. 4 3 2 1 1 1×1=1
8
2 2×2=4
4
3 3×3=9
4 × 4 = 16
___ × ___ = ____
Figure 1 2 3 4 5 6 7 8 9 10
Number of small squares used 1 4 9
Number of squares 1 × 1 = 12 = 2 × 2 = 22 = 3 × 3 = 32 = 4 × 4 = 42 =
1 4 9 16
Square numbers 1 4 9 16
Complete the table up to the 10th row (Figure 10).
Note that: 3 × 3 = 32 32 is read as 3 square
We say that 1, 4, 9 and 16 are square numbers. Can you write down the other square numbers that you obtained after completing the above table. Can you find two more square numbers which are more than 100?
EXERCISE 1.2.1
1. Complete the table below. Number of rows
Number of dots Triangular number
1
2
3
1
1+2
1+2+3
1
3
5
10
12
...... .
2. Write down all triangular numbers between 0 and 100. _______________________________________________________________ 3. Which of the following are not triangular numbers? 4, 6, 9, 15, 22,
28, 36, 65, 80, 100
9
1. NUMBERS
4. Write down all the square numbers between 0 and 150. _______________________________________________________________ 5. Circle the square numbers in the list below: 0, 2, 10, 16, 24, 46, 64, 81, 90, 100 6. Evaluate the following:
(a) 92
(b) 132 (c) 252 7. Which triangular numbers are also square numbers (between 0 and 100)?
8. Complete each of the following:
10
(a) 81 =
9 × 9
(b) 144 =
(c) 400 =
(d) 900 =
❑ ×❑ ❑ ×❑ ❑ ×❑
Square root Let’s consider the square number 16. Which number multiplied by itself gives 16?
1
2
3
4 multiplied by itself gives 16.
4
4 × 4 = 16 or 42 = 16
2
square, 42
3 4
4
16
square root, 16
We say that 42 = 16 Or the square root of 16 is 4, that is The square root (
16 = 4
) of a number ( 16 ) is the number ( 4 ) which
when multiplied by itself give that number ( 16 ). Let’s consider the square number 4. Which number multiplied by itself gives 4? 2 multiplied by itself gives 4.
1
2 × 2 = 4 or 22 = 4
2
square, 22
2
2
4
square root, 4
We say that 22 = 4 Or the square root of 4 is 2, that is 4 = 2 Example: 1 × 1 = 12 = 1
→
1 = 1
2 × 2 = 22 = 4
→
4 = 2
3 × 3 = 32 = 9
→
9 = 3
4 × 4 = 42 = 16
→
16 = 4
(square root of 4 is 2)
EXERCISE 1.2.2
Find the square root of the following numbers. (a) 9
(b) 25
(c) 36
(d) 64
(e) 100 11
1. NUMBERS Square root of bigger numbers Example: Find the square root of: (i) 4
(ii) 16
(iii) 36
Solution: 4 =
2×2=2
(ii)
16 =
2×2×2×2
= =
(iii) 36 = (iv) (v)
= = 64 = = = 100 = = =
(vi) 900 =
= =
2 × 4
×
2 36 2 18 3 9
3
3 3
2×2×2×2×2×2 2 × 8
(vi) 900
2
2×2×3×3 2 6
(v) 100
To find the square root of a number, we express the number as a product of its prime factors.
(i)
(iv) 64
2
×
1
2
2×2×5×5 2 × 10
5
2×2×3×3×5×5 2 × 30
3
×
5
2 900 2 450 3 225 3 75 5 25 5 5 1
EXERCISE 1.2.3
Find the square root of the following numbers. (a) 36
12
(b) 121
(c) 196
(d) 225
(e) 400
(f ) 784
Cube of a number The cube of a number is obtained by multiplying the number by itself three times. For example:
Note: 2 × 2 × 2 = 23 23 is read as 2 cube.
The cube of 2 is 2 × 2 × 2 or 23.
4 3
2
1 1
1
2
1×1×1=1 Figure 1
4 2
2×2×2=8 Figure 2
3
4
3
3 × 3 × 3 = 27 Figure 3
4 × 4 × 4 = 64 Figure 4 Complete the table up to the 10th row (Figure 10).
Figure 1 2 3 4 5 6 7 8 9 10
Number of small cubes used 1 8 27 64
Number of cubes 1 × 1 × 1 = 13 = 1 2 × 2 × 2 = 23 = 8 3 × 3 × 3 = 33 = 27 4 × 4 × 4 = 43 = 64
Cubic number 1 8 27 64
We say that 1, 8, 27 and 64 are cubic numbers. Write down the other cubic numbers that you obtained after completing the above table. Can you find one more cubic number which is more than 1 000? 13
1. NUMBERS Cube root The cube root ( 3 ) of a number ( 8 ) is the number ( 2 ) which when multiplied by itself three times give that number ( 8 ). Example 1: 1 × 1 × 1 = 13 = 01
➞
3
1
= 01
cube root of 01
is 1
2 × 2 × 2 = 23 = 08
➞
3
8
= 02
cube root of 08
is 2
3 × 3 × 3 = 33 = 27
➞
3
27
= 03
cube root of 27
is 3
4 × 4 × 4 = 43 = 64
➞
3
64
= 04
cube root of 64
is 4
5 × 5 × 5 = ___ = ___
➞
3
= ___
cube root of ____ is ____
6 × 6 × 6 = ___ = ___
➞
3
= ___
cube root of ____ is ____
7 × 7 × 7 = ___ = ___
➞
3
= ___
cube root of ____ is ____
8 × 8 × 8 = ___ = ___
➞
3
= ___
cube root of ____ is ____
9 × 9 × 9 = ___ = ___
➞
3
= ___
cube root of ____ is ____
10 × 10 × 10 = 103 = 1 000 ➞
3
1 000 = 10
cube root of 1 000 is 10
Cube root of bigger numbers Example 2: Find the cube root of: Solution: (i) 3
8 = =
3
(i) 8
(ii) 216 (ii) 3
2×2×2 2
2 216 2 108
216 =
3
2×2×2×3×3×3
=
2
=
6
×
3
2 54 3 27 3 9 3 3 1
EXERCISE 1.2.4
Find the cube root of: (a) 64 14
(b) 512
(c) 1 728
(d) 2 744
(e) 3 375
(f ) 1 000 000
Continuous Assessment 1.2 1. Write down all triangular numbers between 0 and 100.
2. (a) Which of the following are triangular numbers? 6, 11, 15, 24, 28, 36, 70, 87, 100
(b) Illustrate your answer using diagrams.
3. Write down all the square numbers between 0 and 150.
4. Circle the square numbers in the list below: 4, 8, 9, 25, 30, 49, 60, 81, 96 5. Find the square root of:
(a) 64
(b) 100
(c) 144
(d) 256
(e) 625
6. Circle the cubic numbers in the list below: 4, 8, 18,
27,
36, 64, 125, 150, 216
7. Find the cube root of:
(a) 64
(b) 125
(c) 729
(d) 1 000
(e) 1 728
8. Which number is a square number and a cubic number? 4,
9,
27,
36,
64,
81,
125
15
1. NUMBERS 1.3 INTEGERS By the end of this topic, you should be able to: • Demonstrate an understanding of integers. • Represent positive and negative integers on a number line. • Compare integers using symbol > or ) and less than ( < ) The terms “greater than” and “less than” can be represented by symbols > and < as follows: > : is greater than
< : is less than
Consider 2 and 5 on a number line. Smaller
Greater
0
1
2
3
4
5
We observe that, 5 is greater than 2 (5 is on the right 2). “5 is greater than 2” is denoted as 5 > 2. On the other hand, we can say that 2 is less than 5. “2 is less than 5” is denoted as 2 < 5. Example 1 : Using the number line, fill in the blanks with “”. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
(a) 3 ____ 2
(b) –3 ____ 2
(c) –2 ____ 3
(d) –2 ____ –3
(e) –5 ____ 2
(f ) 2 ____ –5
(g) 5 ____ 2
(h) –5 ____ –2
Answers: (a) > (b) < (c) < (d) > (e) < (f ) > (g) > (h) < Example 2: Fill in the blanks with “”. (a) 55 ____ 63 (b) 23 ____ 17 (c) –20 ____ 15 (d) 35 ____ – 40 (e) –10 ____ –30
(f ) –50 ____ –25
Answers: (a) < (b) > (c) < (d) > (e) > (f )
0 and n and m are integers. Example 2: (a) 73 × 710 = 7(3 + 10) = 713 (b) 57 × 58 = 5(7 + 8) = 515
EXERCISE 1.4.5
Simplify, giving your answers in index form.
36
(a) 25 × 24
(b) 32 × 33
(c) 83 × 87
(d) 92 × 96
(e) 711 × 75
(f ) 1030 × 1020
(g) 522 × 534
(h) 4100 × 450
Division involving powers (indices) Example 1: Simplify, giving your answer in index form. (a) 37 ÷ 32 37 ÷ 32 =
37 3×3×3×3×3×3×3 = 32 3×3 37 ÷ 32 = 35
=3×3×3×3×3 = 35 (b) 58 ÷ 54 58 ÷ 54 =
5×5×5×5×5×5×5×5 58 = 5×5×5×5 54 =5×5×5×5
58 ÷ 54 = 54
= 54
Based on the above examples, express the following in index form (without expanding). (i) 225 ÷ 25 = _________ (ii) 330 ÷ 320 = _________ (iii) 2 m ÷ 2 n = _________ Thus, am ÷ an = a(m – n) where a > 0 and m and n are integers. Example 2: Simplify, giving your answer in index form. (a) 27 ÷ 23 = 2(7 – 3) = 24 (b) 39 ÷ 35 = 3(9 – 5) = 34
EXERCISE 1.4.6
Simplify, giving your answers in index form. (a) 23 ÷ 21
(b) 64 ÷ 62
(c) 89 ÷ 85
(d) 76 ÷ 73
(e) 107 ÷ 102
(f ) 1218 ÷ 1216
(g) 230 ÷ 210
(h) 550 ÷ 525 37
1. NUMBERS Power Law Consider (34)2 If 32 = 3 × 3, then (34)2 = (34) × (34) Example: Simplify, giving your answer in index form. (a) (34)2 (34)2 = 34 × 34 = (3 × 3 × 3 × 3) × (3 × 3 × 3 × 3) (34 )2 = 38
= 38 (b) (23)4 (23)4 = 23 × 23 × 23 × 23
= (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) (23 )4 = 212
= 212 (c) (75)2 (75)2 = 75 × 75
(75 )2 = 710
= 7(5 + 5) = 710 Thus,
Using multiplication law
(am)n = a(m × n) = amn
EXERCISE 1.4.7
Simplify, giving your answers in index form. Example: (86)5 = 8(6 × 5) = 830
38
(a) (72)3
(d) (38)3
(b) (24)5
(e) (57)5
(c) (96)2
(f ) (63)4
where a > 0 and n and m are integers
Zero Power Consider 23 ÷ 23 (i) Using Division law 23 ÷ 23 = 23 – 3 = 2o (ii) Using Expansion 23 2×2×2 = =1 3 2 2×2×2 From (i) and (ii), we observe that
23 ÷ 23 =
20 = 1.
Similarly, 30 = 1 40 = 1 Thus,
a0 = 1, a ≠ 0
EXERCISE 1.4.8
Evaluate: (a) 70
(b) 90 (c) 1000 (d) 40 + 42
(d) 60 × 67
(f ) a0 (g) 3b0 (h) 5(30)
39
1. NUMBERS Continuous Assessment 1.4 1. State the base and index in each of the following.
(i) 23
(ii) 57
2. Write in expanded form.
(a) 17
(b) 63
(e) 205
(f ) 1002
(c) 85
(d) 114
3. Write in index form:
(a) 3 × 3
(d) 5 × 5 × 5 × 5 × 5 × 5
(b) 2 × 2 × 2 × 2
(e) 4 × 4 × 4 × 4 × 4
(c) 9 × 9 × 9
(f ) 10 × 10 × 10 × 10 x × 10 × 10 × 10 × 10
4. Express the following numbers in index form.
(a) 27
(b) 49
(c) 64
(d) 125
5. Express the following numbers in index form.
(a) 48
(b) 75
(c) 72
(d) 196
6. Simplify, giving your answer in index form.
(a) 23 × 26
(b) 52 × 53
(c) 62 × 68
(d) 35 × 33
(e) 92 × 98
(f ) 740 × 720
7. Simplify, giving your answer in index form.
(a) 36 ÷ 32
(b) 49 ÷ 45
(c) 57 ÷ 53
(d) 106 ÷ 103
(e) 89 ÷ 82
(f ) 6150 ÷ 650
8. Simplify, giving your answer in index form.
(a) (42)4
(b) (34)5
(c) (78)3
9. Write down the value of:
(a) 10
10. Evaluate:
40
(a) 50 + 51 × 52
(b) 23 × 20
(c) (60)7 × 62
(d) 30 + 31 + 32
(b) 160
(c) 10000
(d) 2a0
1.5 ORDER OF OPERATIONS
By the end of this topic, you should be able to: • Demonstrate an understanding of order of operations. • Perform arithmetic operations using the BODMAS convention. • Use the commutative, associative and distributive laws.
Khem and Zara, find out the answer for: 6+4÷2
6+4÷2=? I will start with division first: 4÷2=2 then I continue with addition: 6+2=8
6+4÷2=? I will start with addition first: 6 + 4 = 10 then I continue with division: 10 ÷ 2 = 5
Observe and discuss the picture. Who has the right answer? In which order would you carry out each operation?
41
1. NUMBERS BODMAS When we have two or more operations, we use the BODMAS convention to find the correct answer. Order of Operations
Brackets Orders Division Multiplication Addition Subtraction
Note: operators ➜ + , – , × , ÷ operations ➜ process
Consider
6 + 4 ÷ 2
addition
division
Division comes before addition.
B O
6 + 4 ÷ 2 =6 + 2 =
Division: 4 ÷ 2
D M
Addition: 6 + 2
A S
8
Work out the following. Multiplication comes before subtraction.
10 – 3 × 2
Example 1:
multiplication
subtraction
10 – 3 × 2 = 10
–
=
4
6
Multiplication: 3 × 2 = 6 Subtraction: 10 – 6 = 4
Example 2:
We carry out the operation in the brackets first.
3 + (12 – 5) × 2 brackets addition
multiplication
3 + (12 – 5) × 2 = 3 + 7 × 2
42
= 3
+
=
17
14
B O D M A S
Brackets: (12 – 5) = 7 Multiplication: 7 × 2 = 14 Addition: 3 + 14 = 17
B O D M A S
Example 3: 15
÷
3
+ addition
division
15
÷
3
= 15
÷
=
+
We evaluate the number in the index form (order) first.
22 order
B
22
Order: 2 = 4
O
3 +
4
Division: 15 ÷ 3 = 5
+
4
Addition: 5 + 4 = 9
D M A
5
=
2
S
9
Example 4: 8
+
6
+
= 8
+
=8
+
=
2
×
division
addition
8
÷
6
÷
We carry out division first and then we multiply.
5
B O
multiplication
2
3
× × 15
5
Division: 6 ÷ 2 = 3
D
5
Multiplication: 3 × 5 = 15
M
Addition: 8 + 15 = 23
A S
23
EXERCISE 1.5.1
1. Work out the following:
(a) 12 + 5 × 7
(b) 8 ÷ (9 – 7)
(c) 26 – 4 × 5
(d) 18 + 12 ÷ 6
(e) 18 ÷ 9 + 23
(f ) (9 + 6) ÷ 5
2. Work out the following:
(a) 8 × (7 + 3) – 1
(b) 10 + 6 ÷ 3 × 4
(c) 9 + 12 ÷ 6 × 4
(d) 8 – 2 + 3 × 1
(e) 4 × (5 – 3) – 1 + 7
(f ) 4 × (9 – 7) + 6 ÷ 3
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1. NUMBERS COMMUTATIVE, ASSOCIATIVE AND DISTRIBUTIVE LAWS
Description Commutative law We can add in any order (Addition)
+
=
Example 4+5=5+4=9 Similarly
+
6+3
a+b=b+a
3+6
Commutative law We can multiply in any order
4 × 5 = 5 × 4 = 20
(Multiplication)
Similarly,
= 4×2
2×4 Associative law
a×b=b×a
When adding, we can group in any order
5
(Addition) +
=
(6 + 3) + 4
Associative law
(2 + 3) + 4 = 2 + (3 + 4) + 4 = 2 + 7 9 = 9
+
Similarly,
6 + (3 + 4)
When multiplying, we can group in any order
(a + b) + c = a + (b + c)
(2 × 3) × 4 = 2 × (3 × 4) 6
(Multiplication)
× 4 = 2 × 12 24 = 24
=
Similarly, (a × b) × c = a × (b × c)
(2 × 4) × 3
2 × (4 × 3)
Distributive law
2 (3 + 4) = 2 × 3 + 2 × 4 2 (7) = 6 + 8
= 3 × (2 + 4)
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14 = 14
3×2 +
3×4
Similarly, a(b + c) = ab + ac
NOTE: The Commutative Law does not work for subtraction or division. Example 1: 5 – 3 = 2 whereas 3 – 5 = – 2 (subtraction is not commutative) 1 (division is not commutative) 6 ÷ 2 = 3 whereas 2 ÷ 6 = 3 NOTE: The Associative Law does not work for subtraction or division. Example (subtraction)
(9 – 4) – 3 = 5 – 3 = 2
But
9 – (4 – 3) = 9 – 1 = 8
Example (division) 24 ÷ (8 ÷ 4) = 24 ÷ 2 = 12, but 3 (24 ÷ 8) ÷ 4 = 3 ÷ 4 = 4 Uses: Sometimes it is easier to add or multiply in a different order. Example 2: (i) What is 18 + 33 + 7 ?
18 + 33 + 7 = 18 + (33 + 7)
(we group 33 + 7 first as it is easier)
= 18 + 40 = 58 (ii) What is 2 × 13 × 5?
2 × 13 × 5 = (2 × 5) × 13
= 10 × 13 = 130
(we can rearrange, with 2 × 5 together) (as multiplication by 10 is easier)
(iii) 3 lots of (2 + 4) is the same as 3 lots of 2 plus 3 lots of 4
That is 3 × (2 + 4) = 3 × 6 = 18
And
Thus, we have 3 × (2 + 4) = 3 × 2 + 3 × 4 = 18
3 × 2 + 3 × 4 = 6 + 12 = 18
Example 3: (i) What is 6 × 207 ?
6 × 207 = 6 × (200 + 7)
= 6 × 200 + 6 × 7
= 1200 + 42
= 1242
(207 = 200 + 7) (apply distributive law)
(i) What is 17 × 6 + 17 × 4?
17 × 6 + 17 × 4
= 17 × (6 + 4)
(apply distributive law in the reverse)
= 17 × 10 = 170 45
1. NUMBERS We can use distribution law in subtraction as well: Example 4: Calculate 27 × 3 – 22 × 3. 27 × 3 – 22 × 3 = (27 – 22) × 3
= 5×3
= 15
Continuous Assessment 1.5 1. Work out the following:
(i) 4 ÷ 6 × 2
(ii) 4 + 6 ÷ 2
(iv) 10 ÷ 2 + 3 × 4
(v) 4 × (5 – 3) + 6 ÷ 2
(iii)
5 + (6 – 2) × 3
2. Fill in the blanks with ‘=’ or ‘≠’.
(i) 2 + 3 ___ 3 + 2
(ii) 2 × 3 ___ 3 × 2
(iii) 2 ÷ 3 ___ 3 ÷ 2
(iv) 2 – 3 ___ 3 – 2
3. Fill in the blanks with ‘=’ or ‘≠’.
(i) (3 + 4) + 5 ___ 3 + (4 + 5)
(ii) (3 – 4) – 5 ___ 3 – (4 – 5)
(iii) (3 × 4) × 5 ___ 3 × (4 × 5)
(iv) (3 ÷ 4) ÷ 5 ___ 3 ÷ (4 ÷ 5)
4. Evaluate
(i) 13 + 24 + 7
(ii) 5 × 17 × 2
5. Fill in the blank.
46
7 × 42 = 7 × (40 + ___ )
= 7 × 40 + 7 × ___
= ____ + ____
= _____
(iii) 98 + 46
1.6 FACTORS AND MULTIPLES By the end of this topic, you should be able to: • Find the factors of a number. • Find the prime factors of a number. • Identify common factors of numbers. • Find the Highest Common Factor (H.C.F.) of numbers. • Find multiples of a number. • Find common multiples of numbers. • Find the Least Common Multiple (L.C.M.) of numbers.
I have one 20-rupee coin.
Can I pay with two 10-rupee coins?
Yes and I have four 5-rupee coins
Ice cream Rs 20
Discuss with your friend different ways in which we can pay Rs 20 exactly. What can we say about the numbers 1, 2, 4, 5, 10 and 20? What are the different pairs of numbers that when multiplied make 20?
47
1. NUMBERS FACTORS Activity Let us find the factors of 12 by using counters. How can we group 12 counters? (i) 1 row of 12 counters
1 × 12 = 12
(ii) 2 rows of 6 counters
2 × 6 = 12
(iii) 3 rows of 4 counters
3 × 4 = 12
We say that 1, 2, 3, 4, 6 and 12 are factors of 12.
Factors of a number (e.g., 12) are all numbers (e.g., 1, 2, 3, 4, 6, 12) that can exactly divide the number (e.g. 12) or that can be multiplied to obtain that same specific number (e.g., 12). 2 × 6 = 12 Factors of 12 Therefore: 1, 2, 3, 4, 6, 12 are factors of 12. Note that a number is a factor of itself.
48
A prime number is a number having exactly two different factors (one and itself ). For example: 2 = 2 × 1 2 is a prime number as it has exactly two different factors, namely
Note: 1=1×1 Thus 1 is not a prime number as it does not have 2 different factors.
1 and 2. Examples of other prime numbers are: 3, 5, 7, 13, 19, … Let’s find the prime factors of the number 12. Method 1: Repeated division by prime factors Divide 12 by using only prime numbers. 2 12
Therefore 12 = 2 × 2 × 3
2
6
3
3
1
2 and 3 are prime numbers.
12
Method 2: Factor tree
2
2
×
6
×
2
×
Note: Index form 2 × 2 = 22 3 × 3 = 32 2 × 2 × 2 × 2 = 24 2 × 2 × 3 × 3 = 22 × 32
3
Expressing 12 as a product of its prime factors. 12 = 2 × 2 × 3 12 = 22 × 3
expanded form index form
Example: Find the prime factors of (a) 28
(b) 72
(a) 28
(b) 72
Using the factor tree 28
2
Using repeated division
2 72 2 36
2
×
14
×
2
×
2 18 7
28 = 2 × 2 × 7 28 = 22 × 7
72 = 2 × 2 × 2 × 3 × 3 72 = 23 × 32
3
9
3
3
1
EXERCISE 1.6.1
1. Find the factors of:
(a) 8
(b) 9
(c) 18
(d) 24
(e) 48
(f ) 64
2. Express each of the numbers as a product of its prime factors:
(a) 6
(b) 30
(c) 54
(d) 250
49
1. NUMBERS Consider the numbers 18 and 30. Factors of 18 are 1, 2, 3, 6, 9 and 18. Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. We observe that 1, 2, 3, and 6 are common factors to both 18 and 30. Therefore, the Highest Common Factor of 18 and 30 is 6. Highest Common Factor (H.C.F.) (e.g., is the largest factor TheThe Highest Common Factor (H.C.F.) (e.g., 6) is6)the largest factor which is common to both numbers (e.g.,30). 18 and 30). which is common to both numbers (e.g., 18 and We now discuss three methods to find the H.C.F. of two numbers. Method 1: Using expanded form Write down all the prime factors of 18 and 30. Using expanded form: 18 = 2 × 3 × 3 30 = 2 × 3
× 5
REMEMBER
2 18
2 30
3
9
3 15
3
3
5
5
1
1
Index / power 2 × 2 × 2 = 23 base
The Highest Common Factor or the H.C.F. of 18 and 30 is 2 × 3 = 6. Method 2: Using index form 18 = 21 × 32 30 = 21 × 31 × 51 H.C.F. = 21 × 31
=6
2 18
2 30
3
9
3 15
3
3
5
5
1
1
Take common factors with least index.
The Highest Common Factor or the H.C.F. of 18 and 30 is 21 × 31 = 6. Method 3: Using repeated division Divide 18 and 30 by common prime numbers 2 18 , 30 3
9 , 15
Divide the two numbers by common prime factors of the 2 numbers
3 , 5 The Highest Common Factor or the H.C.F. of 18 and 30 is 2 × 3 = 6. 50
EXERCISE 1.6.2
Find the Highest Common Factor (H.C.F.) of the following numbers. (a) 8 and 12
(b) 16 and 24
(c) 21 and 35
(d) 30 and 45
(e) 39 and 78
(f ) 45 and 60
(g) 24 and 108
(h) 64 and 144
WORD PROBLEMS INVOLVING H.C.F. Read the following word problem carefully.
Example: The lengths of a red ribbon is 12 m and a blue ribbon is 16 m. The ribbons are to be cut into pieces of the same length. Find the maximum length of each piece. 16 m 12 m Solution: To find the maximum length of each piece to be cut, we find the H.C.F. of 12 and 16. 12 = 2 × 2 × 3
or
16 = 2 × 2 × 2 × 2
12 = 22 × 3 16 = 24
H.C.F. = 2 × 2 = 4 Therefore, the maximum length of each piece will be 4 m. 4m
4m 4m 16 m
4m
4m 12 m
4m
4m
NOTE: Observe that each piece is of the same length.
EXERCISE 1.6.3
The lengths of a green rope is 18 m and a black rope is 15 m. The ropes are to be cut into pieces of equal length. Find the maximum length of each piece. 51
1. NUMBERS Activity Consider sets of 5 bottle tops. Step 1: Find the number of bottle tops in each set. 1 set of 5
1×5=5
2 sets of 5
2 × 5 = 10
3 sets of 5
3 × 5 = 15
Step 2: Write down the number of bottle tops for: 4 sets of 5 5, 10, 15, … are called multiples of 5. Can you name other multiples of 5? All multiples of 5 are exactly divisible by 5. Question: Is 20 a multiple of 5?
5 20
Is 32 a multiple of 5?
4
20 is exactly divisible by 5. 20 is a multiple of 5.
5 32
6
32 is not exactly divisible by 5. 32 is not a multiple of 5.
Remainder 2
Multiples (e.g., 5, 10, 15…) of a number (e.g., 5) are obtained by multiplying the number (e.g., 5) by whole numbers such as 1, 2, 3 and so on. Example: Multiples of 5 are 5, 10, 15, …
Multiples of 3 are 3, 6, 9, …
EXERCISE 1.6.4
1. Find the first five multiples of the following numbers.
(a) 3
(b) 4
(c) 12
(d) 15
(e) 25
2. In the list below, circle the numbers which are multiples of 7. 52
7, 35, 44, 63, 100, 24, 49, 87
LEAST COMMON MULTIPLE (L.C.M.) OF 2 NUMBERS
To find the L.C.M. of 2 and 3. Consider the multiples of 2 and 3. Multiples of 2 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ... Multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, … Common multiples for 2 and 3 are 6, 12, 18, … Note that the least common multiple of 2 and 3 is 6. The Least Common Multiple or the L.C.M. of 2 and 3 is 6. Example: Find the L.C.M. of 6 and 20. Method 1: Using expanded form
The common prime factors are used only once.
6 = 2
2 6
2 20
20 = 2 × 2
3
3
2 10
1
5
5
1
×3 ×5
L.C.M. = 2 × 2 × 3 × 5 = 60
The L.C.M. of 6 and 20 is 2 × 2 × 3 × 5 = 60 Method 2: Using index form
6 = 2×3
20 = 22
×5
L.C.M. = 22 × 3 × 5
2 6
2 20
3
3
2 10
1
5
5
1
= 4 × 3 × 5 = 60 We select all prime factors of the 2 numbers with the highest power. E.g., 22 is greater than 21.
The Least Common Multiple (L.C.M.) of 6 and 20 is 60. The Least Common Multiple (L.C.M.) of two numbers (e.g., 6, 20) is the smallest multiple (e.g., 60) which is common to both numbers. 53
1. NUMBERS Method 3: Dividing by prime factors Step 1: Find a prime number by which 6 and 20 are divisible. 2 6 , 20
3 , 10
Step 2: We repeat the process until we obtain 1 as remainder. 2 6 , 20 3
3 , 10
2
1 , 10
5
1 , 5
1 , 1
Since 10 is not divisible by 3, we rewrite it in the next line. When we obtain the number 1 we do not divide further.
The Least Common Multiple (L.C.M.) of 6 and 20 is 22 × 31 × 51 = 60.
EXERCISE 1.6.5
Find the Least Common Multiple (L.C.M.) of the following numbers.
54
(a) 2 and 3
(b) 5 and 7
(c) 6 and 9
(d) 8 and 12
(e) 24 and 32
(f ) 18 and 27
(g) 64 and 96
(h) 45 and 60
WORD PROBLEMS INVOLVING L.C.M.
Two neon lights, A and B, are turned on at the same time. One blinks every 25 seconds and the other blinks every 30 seconds. After how many seconds will they next blink at the same time? Solution: Light A 0
25
50
75
100
125
150
seconds
150
seconds
Light B 0
30
60
90
120
We need to find the time at which they will next blink together. The times at which the lights will blink are multiples of 25 and 30: Light A : 25, 50, 75, 100, 125, … Light B : 30, 60, 90, 120, … To find when the lights will blink together, we calculate the L.C.M.
25 =
52
30 = 21 × 31 × 51 L.C.M. = 21 × 31 × 52
= 150
Light A and light B will next blink together after 150 seconds.
EXERCISE 1.6.6
Boxes that are 12 cm tall are being placed next to boxes that are 18 cm tall. What is the shortest height at which the two piles of boxes will be the same height?
55
1. NUMBERS Continuous Assessment 1.6 1. Find the factors of
(i) 6
(ii) 20
(iii) 36
2. Find the prime factors of
(i) 5
(ii) 20
(iii) 32
3. Find the common factors of 12 and 18. 4. Find the Highest Common Factor (H.C.F.) of
(i) 9 and 15
(ii) 16 and 28
(iii) 18 and 42
(iv) 30 and 45
5. Find the first five multiples of
(i) 3
(ii) 7
(iii) 8
6. Find the Least Common Multiple (L.C.M.) of the following numbers.
(i) 4 and 8
(ii) 9 and 15
(iii) 8 and 3
(iv) 2 and 6
7. The lengths of a blue rope is 20 m and a red rope is 24 m. The ropes are to be cut into pieces of equal length. Find the maximum length of each piece. 8. Bell A rings every 3 hours and bell B rings every 4 hours. At 08 00, the two bells ring together. After how many hours will the bells next ring together?
56
1.7 FRACTIONS By the end of this topic, you should be able to: • Recall addition and subtraction of fractions. • Perform multiplication and division on fractions. • Solve word problems involving fractions. A person cutting each pizza. Pizza A
Pizza B
Everyone will not have the same size of pizza.
Observe and discuss the way in which the pizzas have been cut. What can you say about the shares in both pizzas? In which pizza the shares are equal?
A fraction is a part of a whole. It is used to describe how an object has been divided into equal parts. 1 Pizza A has been cut into 8 equal parts. Each share represents of the pizza. 8 The shares in pizza B are not equal. 57
1. NUMBERS FRACTIONS IN REAL LIFE 1 2
0.5
PRIC E SAL E
One third of the teddies is dressed in blue.
I am halfway through reading my books. One third of my apples is green.
Half this bus is painted green.
One third of the peppers is orange.
Half the apples are red.
Here are some examples of fractions. Can you name them? Can you tell some examples where fractions are used?
INTERESTING FACT
From as early as 1800 BC, the Egyptians were writing fractions. The Egyptians wrote all their fractions using what we call unit fractions. A unit fraction has 1 as its numerator (top number). They put a mouth picture (which meant part) above a number to make it into a unit fraction.
58
1 6
RECALL ADDITION AND SUBRACTION OF FRACTIONS Do you remember that we can add simple fractions and mixed numbers? For example: +
(a) 1 + 4 7 7
1 7
1 + 4 = 5 7 7 7
(b) 1 – 2 2 5 ×5
= 4 7
5 7
×2
1 5 = 2 10
2 4 = 5 10
×2 ×2 5 – 4 = 1 10 10 10
5 1 = 10 2
4 2 = 10 5 remove
4 10
5 10
1 1 (c) 2 + 1 2 3
We express the mixed number into improper fraction
2
1 1 +1 2 3
=
5 4 − 2 3
= 15 − 8 6 6
= 7 6
=11 6
1 + 1 + 1 = 5 halves 2 1 = 5( ) 2
1 =21 2 2
1 10
1
+ 1 =11 3 3
1 = 4 thirds 3 1 = 4( ) 3
2
1
= 5 2 1 5 2 = 2 2
= 4 3 1 4 1 = 3 3
RECALL
When fractions have the same denominator, we can add or subtract the numerator. When fractions have different denominators, we express them as equivalent fractions with a common denominator and then proceed with addition or subtraction.
NOTE
Equivalent Fractions 5 = 15 2 6 4 = 8 3 6
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1. NUMBERS EXERCISE 1.7.1
1. Work out the following. (a) 3 + 2 (b) 3 + 1 7 7 8 4 2. Work out the following.
(a) 4 – 1 5 5
(b) 1 – 2 2 10
(c) 6 + 2 11 5 (c) 2 – 1 3 4
3. Work out the following.
(a) 1 2 + 1 3 3
(b) 1 1 + 2 1 2 3
4. Work out the following.
(a) 1 1 – 3 2 4
(b) 2 1 – 1 1 3 4
MULTIPLYING FRACTIONS Consider (a) 1 × 2 and (b) 1 × 2 2 4 3 3 Solution: (a) 1 × 2 3 2
1 2
Observe that and
2 3
1 of 2 1 × 2 1× 2 = 2 3 6 2
2 3 2 = 2 = 1 3 6 3
(b) 1 × 2 4 3
1 4
2 3
1 of 2 4 3 1 × 2 = 2 4 3 12
1 ×2 = 2 12 4 3
2 5 What is 3 × 7 ?
[can you guess ! !]
To multiply two fractions ( 1 × 2 ), we multiply the numerators (1 × 2) and the denominators 2 3 (2 × 3). That is, 1 × 2 = 1 × 2 = 2 2 × 3 6 2 3 60
Example: Work out the following: (a) 2 × 3 5 7 Solution
(b) 2 × 3 5 8
(c) 2 × 4 9
(d) 1 1 × 2 1 2 3
(a) 2 × 3 = 2 × 3 = 6 7 5 7×5 35
(b) 2 × 3 = 2 × 3 = 6 = 3 5×8 8 5 40 20 (c) 2 × 4 = 2 × 4 = 2 × 4 = 8 9×1 9 9 9 1 (d) 1 1 × 2 1 = 3 × 7 = 3 × 7 = 21 = 7 = 3 1 6 2 3 2 3 2×3 2 2
EXERCISE 1.7.2
Work out the following. 1. (a) 2 × 3 4 3 2. (a)
4 × 5 9
3. (a) 1
1 3 × 2 4
(b) 1 × 7 5 8 (b)
2 × 7 3
(b) 1
2 1 × 2 3 2
(c) 3 × 1 7 4
(d) 1 × 3 2 5
3 × 6 4
(d) 2 × 3 5
(c)
(c) 2
1 1 × 1 4 3
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1. NUMBERS DIVIDING FRACTIONS What does 3 ÷ 1 mean? 2
This means how many halves ( 1 ) do we have in 3 wholes. 2 1 2 1 2
1 2 1 2
1 2 1 2
3 wholes = 6 halves 3÷ 1 =6 2
Thus 3 ÷
Similarly, 2 ÷ 1 4 1 4
Note:
1 =6 2
[that is, there are 6 halves in 3 wholes]
1 1 means how many do we have in 2 wholes. 4 4 1 4 1 4
1 4 1 4
Reciprocal of
Reciprocal of 2 is
1 4 1 4
2 wholes = 8 quarters 2÷
Thus, 2 ÷
1 =8 4
1 =8 4
[there are 8 quarters in 2 wholes]
Observe that 3 ÷
2 1 =3× =6 1 2
and 2 ÷
4 1 =2× =8 1 4
now, 4 ÷
1 = 3
Thus, when we divide by a fraction ( That is
62
2 3 2 4 8 ÷ = × = 5 4 5 3 15
2 3 is 3 2
[can you guess!!]
3 4 ), we multiply by its reciprocal ( ) 4 3
1 2
Example: Work out the following: 2 3 2 ÷ (b) ÷ 4 7 5 3
(a)
(c) 1
1 1 ÷2 2 3
Solution (a)
3 2 5 10 2 ÷ = × = 5 7 3 21 7
(b)
2 2 4 2 1 1 2 ÷4= ÷ = × = = 12 3 1 3 4 6 3
(c) 1
1 3 7 3 3 9 1 ÷2 = ÷ = × = 3 2 3 2 7 14 2
EXERCISE 1.7.3
Work out the following. 1. (a)
3 2 ÷ 4 3
(b)
1 3 ÷ 5 8
(c)
2 1 ÷ 7 3
2. (a)
2 ÷ 5 3
(b)
1 ÷ 7 2
(c)
3 ÷ 5 4
3. (a) 1
1 3 ÷ 2 4
(b) 1
2 1 ÷ 3 2
(c) 2
1 1 ÷ 1 4 3
63
1. NUMBERS Word Problems Example: 1. There are 24 eggs in a crate. One quarter of them is broken. How many eggs are unbroken? Solution: 1 4
Fraction of broken eggs =
Fraction of unbroken eggs = 1−
Number of unbroken eggs =
2. Raj ate
Unbroken
Broken
3 1 = 4 4
3 4
1 4
3 × 24 = 18 4
1 1 of a cake and gave of the remaining to his sister Laxmi. What fraction of 3 4
the whole cake did Laxmi receive?
Ate
Solution:
Fraction of cake Raj ate =
Fraction of cake left = 1 −
Fraction of whole cake Laxmi received =
Left
1 3 1 3
2 1 = 3 3
2 3
1 of what is left 4
=
2 1 of 3 4
=
2 2 1 1 × = = 3 12 6 4
EXERCISE 1.7.4
1. If
2 of 40 pupils are boys, how many are girls? 5
2. Sam spent
If there are Rs 3000 left, what is his salary?
3. Kevin ate 64
1 1 of his salary on food and on transport. 3 2
1 2 of a cake and gave of the remaining to his son Adi. 4 5
What fraction of the whole cake did Adi receive?
Continuous Assessment 1.7 Work out the following 1. (a)
3 2 + 7 7
2. (a)
2 1 × 4 3
(b)
3 × 2 5
(c) 1
1 1 × 2 4
3. (a)
3 1 ÷ 4 5
(b)
3 ÷ 2 5
(c) 1
2 1 ÷ 3 3
4. If
(b)
5 3 − 8 8
(c)
2 1 − 3 5
(d) 2
2 1 −1 3 5
3 of 32 pupils are boys, how many are girls? 8
1 3 of his salary on food and on transport. 5. Ali spent 3 7 If there are Rs 2000 left, what is his salary?
6. Nasreen ate
1 1 of a cake and gave of the remaining to her daughter Arifa. 3 4
What fraction of the whole cake did Arifa receive?
65
1. NUMBERS 1.8 DECIMALS By the end of this topic, you should be able to: • Demonstrate understanding of decimals. • Review knowledge of place value. • Convert fractions into decimals. • Convert decimals into fractions. • Compare and order decimals. • Perform arithmetic operations on decimals. • Add and subtract decimals. • Multiply a decimal by 10, 100, 1000. • Multiply a decimal by a multiple of 10, 100, 1000. • Multiply a decimal by a whole number. • Multiply a decimal by another decimal. • Divide a decimal by 10, 100, 1000. • Divide a decimal by a multiple of 10, 100, 1000. • Divide a decimal by a whole number. • Divide a decimal by another decimal • Solve word problems involving decimals.
Observe and discuss the picture. Do you recognize this athlete? What can you say about the number written on the board? What does the dot between 9 and 58 represent? Is 9.58 greater or smaller than 10? Why?
66
DECIMALS IN REAL LIFE Decimals are another way to write fractions. They show values less than 1.
Money exchange rates
Digital kitchen scale
Here are some examples where you can encounter decimals. Can you name other examples?
Decimal comes from the Latin word decimus, meaning tenth, which comes from the word decem or 10.
While numbers play an extremely important role in our everyday life, decimals give more meaning and accuracy to them. The denominator will always be 10, 100, 1000… Example: 0.1 =
1 10
2.29 = 2
29 951 0.951 = 100 1000
67
1. NUMBERS PLACE VALUE AND READING OF DECIMALS Place value refers to the value of each digit in a number. We can use visuals to represent decimals. Example: Let us represent 2.346 Here the unit is the 1000-cube. One small cube is
2 units
1 . 1000
3 tenths
4 hundredths
6 thousandths
2.346 3 tenths
2 units
4 hundredths
6 thousandths
Remember, the decimal point separates the whole numbers on the left from the decimal digits on the right. We can also represent 2.346 on an abacus:
Note that decimal fractions are smaller than 1 unit. For example: 0.1 or
1 is smaller than 1 10
1 is smaller than 1 100
0.01 or
0.001 or 2
.
3
4
6
We write 2.346 in expanded form as follows: 2.346 = (2 × 1) + (3 ×
1 1 1 ) + (4 × )+(6× ) 100 1000 10
2.346 = (2 × 1) + (3 × 0.1) + (4 × 0.01) + (6 × 0.001) 68
1 is smaller than 1 1000
CONVERTING A FRACTION INTO A DECIMAL One way to convert a fraction into a decimal is to express it into an equivalent fraction with denominator either in 10, 100, 1 000. We can also use division to convert a fraction into a decimal. Example: Write (a)
3 1 and (b) into decimals. 4 2
Method 1 (a)
5 1 is equivalent to 10 2
5 1 = 2 10
5 = 0.5 10
(b)
75 3 is equivalent to 100 4
3 75 1 = = 75 × 4 100 100
75 = 0.75 100
1 2
3 4
Method 2 (a)
=
=
5 10
75 100
3 1 (b) 4 2 1
2 1.0 0.5
3 2
4 3.00 0.75
69
1. NUMBERS We use equivalent fractions with denominator either in 10, 100, 1 000 to convert a fraction into decimal. Example: Write as decimal: (a)
3 5
(b)
5 8
(c)
9 25
Method 1: (a)
3 3×2 6 = = 5 5 × 2 10
(b)
= 0.6
5 5 × 125 = = 625 8 8 × 125 1000
= 0.625
Method 2: 5 2 4
3
5 3.0 8 5.000 (a) (b) 0.6 0.625
EXERCISE 1.8.1
Convert the following fractions into decimals.
70
(a)
3 10
(b)
3 9 (c) 8 100
(d)
20 25
(e)
17 20
(f )
89 200
(c)
9 9×4 36 = = 25 25 × 4 100 = 0.36
CONVERTING A DECIMAL INTO A FRACTION
To convert a decimal into a fraction, we express it into a fraction with denominator either in 10, 100, 1000... . Example: Convert the following into fraction. (a) 0.6 (b) 0.27 (c) 0.125 Solution (a) 0.6 is 6 tenths
0.6 =
6 10
(b) 0.27 is 27 hundredths
0.27 =
(c) 0.125 is 125 thousandths
27 125 0.125 = 100 1000
3
There is no common
5
simplify further.
1 5 25
Divide both 6 and 125 6 factor for 27 and 10 by a common = = 100, we cannot 1000 10 factor 2
0.6 =
200 40 8
1 3 0.125 = 8 5
EXERCISE 1.8.2
Convert the following decimals into fractions. (a) 0.8
(b) 0.35
(c) 0.225
(d) 0.625
71
1. NUMBERS ARITHMETIC OPERATIONS AND DECIMALS Recall: Adding and Subtracting decimals For addition, we arrange the decimals so that the place value (units, tenths, hundredths) are in the same column. Example: Work out
4 5 3
• •
4 3
+
7 7
thousandths
hundredths
•
10 1 10 2 0
tenths
–
7 5 8
(c) 2.104 + 23.71
units
2 • 3 + 4 • 5 6 • 8
(b) 58.2 – 4.43
tens
(a) 2.3 + 4.5
2
•
1 0 4
2 3
•
7 1 0
2 5
•
8 1 4
EXERCISE 1.8.3
Work out the following. 1. (a) 5.3 + 2.4 (c) 3.504 + 12.81
(b) 12.5 + 3.4 (d) 113.25 + 198.96
2. (a) 11.28 + 3.5 (c) 123.5 + 80.07
(b) 10.7 + 1.04 (d) 23.302 + 14.14
3. (a) 4.5 – 2.3 (c) 72.3 – 4.52
(b) 6.2 – 3.7 (d) 43 – 7.41
4. (a) 27.8 – 15.2 (c) 274.05 – 38.79
(b) 138.82 – 125.17 (d) 258.42 – 127.51
5. Aditi weighs 50.26 kg and Hesha weighs 3.75 kg more. How much does Hesha weigh? 6. Shaun has Rs 125.75. He buys a set of copybooks at Rs 80.50. How much money has he left? 7. Find the perimeter of the rectangle below. 7.3 cm 4.1 cm 72
MULTIPLYING DECIMALS Multiplying by 10, 100, 1000 We know that 2 can be written as 2.0, 2.00, 2.000 And,
2 × 10 = 2.0 × 10 = 20.
2 × 100 = 2.00 × 100 = 200.
2 × 1 000 = 2.000 × 1 000 = 2000. Observe that:
What do you observe about the position of the decimal point when multiplying by 10, 100, 1 000?
2.0 × 10 = 2.0 × 10 = 20.
2.00 × 100 = 2.00 × 100 = 200.
2.000 × 1 000 = 2.000 × 1 000 = 2 000. We can extend the above example to decimal places. Example 1: Evaluate (a) 2.1 × 10
= 2.1 × 10
(b) 3.1 × 100
× 10 is equivalent to shifting the decimal point to 1 digit to the right. × 1 000 is equivalent to shifting the decimal point to 3 digits to the right.
(c) 4.5 × 1 000
= 3.10 × 100
= 4.500 × 1 000
= 21 = 310 = 4 500 Example 2: Evaluate (a) 0.3 × 10
(b) 4.49 × 100
(c) 5.234 × 1 000
0.3 × 10 4.49 × 100 5.234 × 1 000 = 3 = 449 = 5 234 EXERCISE 1.8.4
Evaluate the following. (a) 0.4 × 10
(b) 9.1 × 100
(c) 0.185 × 100
(d) 44.75 × 10
(e) 1.456 × 1000
(f ) 23.86 × 1000
73
1. NUMBERS Multiplying by a multiple of 10, 100, 1000 Example: Calculate (i) 0.3 × 20
(ii) 2.34 x 300
(iii) 1.456 x 2 000
= 0.3 × 10 × 2
= 2.34 × 100 × 3
= 1.456 × 1 000 × 2
=
3
× 2
=
234
× 3
=
1456
=
6
=
702
=
2 912
×2
EXERCISE 1.8.5
Calculate the following.
74
1. (a) 2.1 × 30
(b) 15.3 × 20
(c) 0.84 × 50
(d) 125.3 × 40
2. (a) 3.2 × 200
(b) 51.4 × 500
(c) 0.41 × 300
(d) 40.01 × 600
3. (a) 4.5 × 2 000
(b) 23.9 × 3 000
(c) 0.145 × 5 000
(d) 0.0015 × 7 000
MULTIPLYING A DECIMAL BY A WHOLE NUMBER Example: Given 123 × 4 = 492 Evaluate (i) 12.3 × 4
Observe the decimal point in the examples.
(ii) 1.23 × 4
Solution: We know that 123 × 4 = 492 (i) 12.3 × 4
(ii) 1.23 × 4
12.3 is approximately 12
1.23 is approximately 1
and 12 × 4 = 48
and 1 × 4 = 4
12.3 × 4 = 49.2
1.23 × 4 = 4.92
Position of decimal point
Position of decimal point
The value of 12.3 × 4 is close to 48, that is 49.2
The value of 1.23 × 4 is close to 4, that is 4.92
From the above example, we observe: 12.3 × 4 = 49.2 1.23 × 4 = 4.92 0.123 × 4 = . . . . . . . . . .
Can you guess this one?
When we multiply a decimal (e.g., 1.23) by a number (e.g., 4), we first multiply the numbers (123 × 4 = 492) then adjust the decimal point (4.92)
EXERCISE 1.8.6
1. Given 2 341 × 7 = 16 387, evaluate:
(a) 2.341 × 7
(b) 234.1 × 7
(c) 23.41 × 7
(d) 0.2341 × 7
(c) 3 520 × 6
(d) 0.352 × 6
(b) 4512.3 × 2
(c) 4.5123 × 2
(d) 451.23 × 2
(b) 5.9 × 4
(c) 25.3 × 9
(d) 120.1 × 7
2. Given 352 × 6 = 2 112, evaluate:
(a) 3.52 × 6
(b) 35.2 × 6
3. Given 45 123 × 2 = 90 246, evaluate:
(a) 45.123 × 2
4. Evaluate:
(a) 1.7 × 5
75
1. NUMBERS MULTIPLYING A DECIMAL BY A DECIMAL Consider 4.1 × 0.3 Observe that 4.1 =
41 3 and 0.3 = 10 10
Now, 4.1 × 0.3 =
41 3 41 × 3 × = 10 10 10 × 10
=
123 100
= 1.23
That is, to evaluate 4.1 × 0.3, we first consider 41 × 3 = 123 digit the decimal 1 digit 2 digit Then we 1adjust point.
That is
4.1
×
1 digit
0.3
=
1.23
1 digit
2 digits
Example: Evaluate (i) 2.3 × 1.2
23
(ii) 1.24 × 2.1
We consider 23 × 12 = 276
×
12
Then we adjust the decimal
46
That is, 2.3 × 1.2 = 2.76 1 digit
1 digit
+ 2 3 0
276
2 digits
We consider 124 × 21 = 2 604 × 21 124 Then we adjust the decimal + 2480 1.24 × 2.1 = 2.604 2604 2 digits
1 digit
EXERCISE 1.8.7
Complete the following.
76
124
(a) 5.2 × 0.3
(b) 0.7 × 0.7
(c) 2.15 × 0.05
(d) 13.1 × 1.8
(e) 30.1 × 0.2
(f ) 1.44 × 0.27
3 digits
DIVIDING DECIMALS Dividing a decimal by 10, 100 or 1 000. Example: Calculate (i) 12.5 ÷ 10
÷ 10 is equivalent to shifting the decimal point to 1 digit to the left. ÷ 1 000 is equivalent to shifting the decimal point to 3 digits to the left.
12.5 ÷ 10 = 1.25
(ii) 235.4 ÷ 100
235.4 ÷ 100 = 2.354
(iii) 425.6 ÷ 1 000
425.6 ÷ 1 000 = 0.4256
(iv) 35.1 ÷ 1 000
35.1 ÷ 1 000 = 035.1 ÷ 1 000
= 0.0351
EXERCISE 1.8.8
1. Calculate
(a) 45 ÷ 10
(b) 47.2 ÷ 10
(c) 5.8 ÷ 10
(d) 0.85 ÷ 10
(b) 25.7 ÷ 100
(c) 4.97 ÷ 100
(d) 0.147 ÷ 100
(b) 1678.4 ÷ 1 000
(c) 541.7 ÷ 1 000
(d) 27.59 ÷ 1 000
2. Calculate
(a) 253 ÷ 100
3. Calculate
(a) 4957 ÷ 1 000
77
1. NUMBERS Dividing a decimal by a multiple of 10, 100, 1000 Example: Calculate the following.
(i) 4.6 ÷ 20 =
4.6 20
=
4.6 2 × 10
=
0.46 2
= 0.23
Dividing by 20 is the same as dividing by 10 and then dividing by 2.
2 0.46 0.23
(ii) 630.9 ÷ 300 =
630.9 300
=
630.9 3 × 100
=
6.309 3
= 2.103
Dividing by 300 is the same as dividing by 100 and then dividing by 3. 3 6.309 2.103
906.3 3000
(iii) 906.3 ÷ 3 000
=
=
906.3 3 × 1000
=
0.9063 3
= 0.3021
We divide by 1 000 and then by 3.
3 0.9063 0.3021
EXERCISE 1.8.9
1. Calculate
(a) 40.6 ÷ 20
(b) 123.6 ÷ 30
2. Calculate
(a) 639.6 ÷ 300
(b) 122.4 ÷ 400
3. Calculate 78
(a) 2468.4 ÷ 2 000
(b) 21.35 ÷ 7 000
Dividing a decimal by a whole number Example: Calculate the following. (iii) 25.05 ÷ 5 hundredths
units
2 0 • 8 0 • 4
tenths
(ii) 6.39 ÷ 3 tenths
units
(i) 0.8 ÷ 2
2
5 2 5 • 0 5 0 5 • 0 1
3 6 • 3 9 2 • 1 3
0.8 ÷ 2 = 0.4
6.39 ÷ 3 = 2.13
25.05 ÷ 5 = 5.01
EXERCISE 1.8.10
Calculate the following. (a) 3.2 ÷ 2
(b) 4.8 ÷ 6
(c) 19.6 ÷ 8
(d) 99.99 ÷ 9
(e) 0.216 ÷ 3
(f ) 252.8 ÷ 4
79
1. NUMBERS Dividing a decimal by a another decimal Example : Calculate
(i)
24.86 0.2
We make the denominator a whole number.
24.86 0.2 × 10
24.86 248.6 = 0.2 2
Multiply both numerator and denominator by 10.
× 10
(ii)
2 248.6 124.3
= 124.3
9.126 0.03 9.126 912.6 = 0.03 3
Multiply both numerator and denominator by 100. 1
3 912.6 304.2
= 304.2
EXERCISE 1.8.11
Calculate the following. 1. (a) 4.8 ÷ 0.4
(b) 0.64 ÷ 0.8
(c) 4.97 ÷ 0.07
(d) 12.55 ÷ 0.5
2. Devi has 5 identical marbles. The total mass of the 5 marbles is 11.5 g. What is the mass of each marble? 3. Seema shares Rs 12.80 equally among her 4 children. How much money does each child receive?
80
Continuous Assessment 1.8 Complete the following. 1. (a) 2.1 + 0.3
(b) 7.25 + 5.206
(c) 4.704 + 23.81
2. (a) 9.2 – 2.1
(b) 8.21 – 3.01
(c) 6.324 – 3.85
3. (a) 0.7 × 10
(b) 8.2 × 100
(c) 0.328 × 1000
4. (a) 14.25 × 20
(b) 5.023 × 300
(c) 21.12 × 4 000
5. (a) 1.9 ÷ 10
(b) 7.4 ÷ 100
(c) 0.123 ÷ 1000
6. (a) 9.63 ÷ 30
(b) 32.4 ÷ 400
(c) 2.34 ÷ 2 000
7. (a) 1.2 × 4
(b) 4.5 × 7
(c) 35.2 × 8
8. (a) 4.8 ÷ 4
(b) 4.9 ÷ 7
(c) 22.5 ÷ 5
9. Jonny has 7 pencils. Each pencil has a mass of 5.3 g. What is the total mass of the 7 pencils? 10. Ali shares Rs 15.60 equally among his 3 children. How much money does each child receive?
81
1. NUMBERS 1.9 REAL NUMBERS By the end of the topic, you should be able to: • Recall natural numbers, whole numbers and integers. • Demonstrate an understanding of rational, irrational numbers and real numbers. • Represent real numbers on a number line. • Compare and order real numbers. • Convert a fraction into a decimal to identify the recurring part of non-terminating decimals. • Approximate a given number to required decimal places.
Types of Numbers Natural number Natural numbers are counting numbers. The set of Natural Numbers is denoted by letter N where N = {1, 2, 3, 4, ...}. They can be represented on a number line as follows.
1 2 3 4 5 6 7 8 9 10
Whole number Whole numbers consist of positive numbers and zero. The set of Whole Numbers is denoted by the letter W, where W = {0, 1, 2, 3, 4, ....}. The number line given below illustrates whole numbers.
0 1 2 3 4 5 6 7 8 9 10
Integer (Z) Integers consist of positive and negative whole numbers, and zero. The set of Integers is denoted by the letter Z, where Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}. It can be represented on a number line. 82
-3 -2 -1 0 1 2 3
1. NUMBERS Decimal decimal
non-terminating
terminating
recurring
non-recurring
Terminating decimal A terminating decimal contains a finite number of digits. Example : 1.5, 0.25, 3.142, 0.025, ...
Non- terminating decimals Consider the following decimals. (i)
0.17546321357....
(non-recurring) .
(ii) 0.33333333333.......
(recurring)
Observe that in both cases, the decimal goes on and on, thus they are called non-terminating decimal. In the first case (0.1754...), the digits do not repeat themselves. Hence the decimal 0.1754... is a non-recurring decimal. However, in the second case (0.333...), the digit 3 repeats itself indefinitely. The decimal 0.333… is called a recurring decimal.
The dot notation A dot is placed on the digit(s) that recur.
.
For example, 0.333.... = 0.3
83
1. NUMBERS Study the following recurring decimals.
.
(i) 2 = 0.6666…
(ii) 99 = 0.434343…
(iii) 999 = 0.437437437…
. . (iv) 1 = 0.142857142857… = 0.142857 (a dot on the first and last recurring digits)
3
43
437
= 0.6
..
= 0.43
(a dot on both recurring digits)
. .
= 0.437
(a dot on the first and last recurring digits)
7
EXERCISE 1.9.1
Use dot notation to write the recurring decimals
(a) 0.22222….
(b) 0.77777….
(c) 0.292929…
(d) 0.535353…
(e) 0.203203203…
(f) 0.126126126…
(g) 0.92222…
(h) 0.573333…
Expressing Fraction as recurring Decimal Example Use dot notation to write the recurring decimals.
(i)
4 3
(ii)
5 6
Solution
84
(i)
(ii) 3 4.101010
6 5.50 20 20
4 3
1. 3 3 3.....
.
= 1. 333... = 1.3
5 6
0. 8 3 3 .....
.
= 0.8333... = 0.83
1. NUMBERS EXERCISE 1.9.2
Use dot notation to express the following fraction into decimal. 2
1
5
3
(i) (ii) (iii) (iv) 9 3 6 7
Rational numbers (Q) p
A rational number is a number that can be expressed in the form q where p and q are integers and q ≠ 0. 1 For example, 2 , – 3 , 4 , 9 7 1 5
Activity Discuss whether the following numbers are rational numbers. (i) 3
(ii) 0.5
(iii) 0.1234
(iv)
1
23
Solution 3
(i) 3
= 1
(ii) 0.5
=
(iii) 0.1234 = 1
(iv) 2 3
=
5 10
1234 10000 7 3
p
is in the form q , thus 3 is a rational number. p
is in the form q , thus 0.5 is a rational number. p
is in the form q , thus 0.1234 is a rational number. p
1
is in the form q , thus 2 3 is a rational number.
Do you think √2 a rational number ?
85
1. NUMBERS EXERCISE 1.9.3
Are the following numbers rational? Justify your answer. 1
(a) 7
(b) 3
(c) 0.25
(d) 1 5
3
(f ) 0
2
(e) 0
Irrational numbers p
Irrational numbers are numbers that cannot be written in the form q , where p and q are integers and q≠0. 3
For example, 0 , π, √2, √3, 0.347591… EXERCISE 1.9.4
Circle the irrational numbers in the following. 5 4 5, 0 , – 4 , π, 0.7238, √ 5
Summary Decimals
Terminating 0.83 (rational)
Non-terminating
recurring 0.333... (rational)
86
Non-recurring √ 2 = 1.414... (irrational)
1. NUMBERS Real Number (R) Rational and irrational numbers together form the set of real numbers. Thus, any number on a number line is a real number.
Example 1 Represent the following real numbers on a number line. 1 2
Solution
, 2.1, √3, 2.9, – 0.5
1 – 0.5 2
NOTE
√3 2.1 2.9
-1 0 1 2 3 Example 2 Find a real number between (i)
2.1 and 2.2
(ii) –0.7 and –0.6
There are other possible solutions. E.g., a real number between 2.1 and 2.2 may be any of: 2.14, 2.15 or 2.16.
Solution –0.67
2.13 2.10
–0.7
2.20
A real number between 2.1 and 2.12 is 2.13
–0.60
A real number between – 0.7 and – 0.6 is – 0.67
EXERCISE 1.9.5 1 7
1.
Represent the real numbers 0.5, √8, 2.52, 3 , 3 , –1, π on a number line
2.
Find a real number between
(i)
–3 and 3
(ii) 4 and 5
(iii) 2.1 and 2.3
(iv) –2 and –1
87
1. NUMBERS Summary
Real Number Rational 2 3 –3
Integer Whole Natural 1, 2, 3...
88
Irrational
–2.65 –19 0
1 6 4
π √2
1. NUMBERS Approximation Rounding off to 1 decimal place (1 d.p.) Example
(a) Round off 4.23 correct to 1 decimal place (c.t. 1 d.p.).
Solution
We first represent the number 4.23 on a number line. 4.23 4.20
4.25
4.30
Since 4.23 is closer to 4.20, thus 4.23 is 4.2 to 1 d.p.
That is 4.23 = 4.2 c.t. 1 d. p.
(b) Round off 4.27 correct to 1 decimal place (c.t. 1 d.p.).
Solution
4.27 4.20
4.25
4.30
Since 4.27 is closer to 4.30, thus 4.27 is 4.3 to 1 d.p.
That is, 4.27 = 4.3 c. t. 1 d. p.
(c) Round off 4.25 correct to 1 decimal place (c.t. 1 d.p.).
Solution
4.25 4.20
4.25
The number 4.25 is between 4.20 and 4.30
By convention 4.25 is 4.3 to 1 d.p.
That is, 4.25 = 4.3 c. t. 1 d. p.
4.30
89
1. NUMBERS EXERCISE 1.9.6
Round off the following numbers to one decimal place (1 d.p.).
(a)
8.47
= .....
(b) 5.59
(c)
21.68
= .....
(d) 58.23 = .....
(e)
0.198
= .....
(f ) 25.58 = .....
(g) 53.63
= .....
(h) 60.98 = .....
(i)
= .....
(j) 95.02 = .....
82.75
= .....
Rounding off to 2 decimal places Example
(a) Round off 3.123 correct to 2 decimal place (c.t. 2 d.p.).
Solution
The number 3.123 is between 3.120 and 3.130. 3.123 3.10
3.125
3.130
Since 3.123 is closer to 3.120, thus 3.123 = 3.12 (c. t. 2 d. p.)
(b) Round off 3.127 to 2 decimal place.
Solution
The number 3.127 is between 3.120 and 3.130. 3.127 3.120
3.125
3.130
Since 3.127 is closer to 3.130, thus 3.127 = 3.13 (c. t. 2 d. p.)
(c) Round off 3.125 to 2 decimal place.
Solution The number 3.125 is the mid-point of 3.120 and 3.130. 3.125 3.120 90
By convention, 3.125 = 3.13 (c. t. 2 d. p.).
3.130
1. NUMBERS EXERCISE 1.9.7
Round off the following numbers to 2 decimal places (2 d.p.).
(a) 3.237
(b)
10.983
(c)
5.597
(f ) 66.202
(g)
58.937
(h) 28.128
(d) 13.345
(e) 13.101
(i)
(j) 45.360
73.998
Rounding off Based on the previous examples, we observe the following rules for rounding off decimals. 1.
Find the place you are rounding to.
2.
If the digit on the right:
a. Is 5 or more, add 1 to the place you are rounding to.
b. Is less than 5, add nothing.
3.
Then, drop down all the digits on the right of the decimal place you are looking for.
Example Round off
(i) 4.2385 correct to 1 decimal place
(ii) 4.2385 correct to 2 decimal places
(iii) 4.2385 correct to 3 decimal places
Solution
(i) 4.2385 correct to 1 decimal place
Rounding off to 1 d. p.
4.2385 1st d.p.
less than 5, add nothing and drop all digits
... 4.2385 = 4.2 c .t. 1 d. p.
(ii) 4.2385 correct to 2 decimal places
Rounding off to 2 d. p. +1
4.2385
2nd d.p.
reater than 5, add 1 to the 2nd decimal place and g drop all digits.
... 4.2385 = 4.24 c. t. 2 d. p. 91
1. NUMBERS (iii) 4.2385 correct to 3 decimal places
Rounding off to 3 d. p. +1
4.1385 3rd d.p.
... 4, 1385 = 4. 139 c. t. 3 d. p.
equal to 5, add 1 to the 3rd decimal place and drop all digits.
Continuous Assessment 1.9 1.
Write down (i) two whole numbers
2.
From the following numbers
(ii) two integers
(iii) two natural numbers.
1
– 2 , 0. 5, 1 2 , √5, 3, π
Write down
(i) the integers
(ii) the irrational numbers
(iii) the rational numbers.
3.
Express the given recurring decimals in dot notation.
(a) 0.5555…
(b) 1.222…
(c) 4.1313…
4.
Use dot notation to express the following fractions into decimal 3
4
2
8
(i) (ii) (iii) (iv) 7 9 7 3 5. Represent the following real numbers on a number line 2
5
0.75, √5, 2.48, 3 , 3 , –2, π 6.
Find a real number between
(i) 1 and 2
(ii) 0.7 and 0.8
7. Express each of the following numbers correct to number of decimal places indicated in brackets.
92
(a) 2.31278 (2 d.p.)
(b) 5.65445 (1 d.p.)
(c) 12.0678 (2 d.p.)
(d) 15.4796 (3 d.p.)
1.10 SEQUENCES By the end of the topic, you should be able to: • Identify and complete number patterns and figures. • Recognise and use Fibonacci sequence. • Complete sequence of ordered pairs.
In this game, you will jump onto squares containing counting numbers. The purpose of the game is to skip count by 2. For example, if you skip count by 2, you will jump on 2, then on 4, then on 6 …
2
5
6
3
4
1 Put a tick on the next two numbers on which I should jump.
10 6
9 8
7
Complete the following: (i)
2, 4 , 6 , _______ , ________
(ii)
30 , 40 , 50 , _______ , ________
93
1. NUMBERS A sequence is a pattern that involves a mathematical rule.
Sequences Involving Addition Example: Write down the next two terms of the following sequences:
(a)
2, 5, 8, 11, ...., …..
(b)
1, 5, 9, 13, …., ….. +4
+3
1 2 3 4 5 6
1 2 3 4 5 6
The rule is: add 3 2, 5, 8, 11, 14, 17
The rule is: add 4 1, 5, 9, 13, 17, 21
EXERCISE 1.10.1
For each of the following sequences, write down the next two terms and the rule.
(i)
3, 5, 7, 9, …. , ….
The rule is ……………………..
(ii)
1, 6, 11, 16, …. , ….
The rule is ………………………
(iii) –10, – 8, – 6, –4, …. , ….
The rule is ………………………
(iv) 1.2, 1.5, 1.8, 2.1, ….. , ….
The rule is ………………………
(v) 5 2 , 6, 6 2 , 7, 72 , …. , ….
1
1
1
The rule is ………………………..
Sequences Involving Subtraction Example: Write down the next two terms in the following sequences:
(a)
15, 13, 11, 9, .... , …. -2
94
-2
15 14 13 12 11 10 The rule is: subtract 2 15, 13, 11, 9, 7, 5
(b) 13, 10, 7, 4, …. , …. -3
-3
13 12 11 10 9 8 7 6 The rule is: subtract 3 13, 10, 7, 4, 1, –2
EXERCISE 1.10.2
For each of the following sequences, write down the next two terms and the rule.
(i)
100, 90, 80, 70, ….. , ….
The rule is ………………....
(ii)
12, 10, 8, 6, ….. , ….
The rule is ………………....
(iii) 20, 15, 10, 5, …. , ….
The rule is ………………....
5 11 9 (iv) , , 7 , 17 , …. , …. 17 17 17
The rule is ………………....
The rule is ………………....
(v)
0.9, 0.7. 0.5, 0.3, …. , ….
Sequences Involving Multiplication Example: Write down the next term and the rule for the following sequences. ×2 (a)
(b) × 2
×2
3 , 6, 12, 24, ....
×3
×3
×3
1, 3, 9, 27, …..
The next term is: 48
The next term is: 81
The rule is : multiply by 2
The rule is : multiply by 3
EXERCISE 1.10.3
Write down the next term and the rule for the following sequences.
(a)
2, 4, 8, ….
The rule is ………………………………………
(b) 5, 20, 80, ……
The rule is ………………………………………
(c)
6, 18, 54, ….. The rule is ………………………………………
1 , 4 , 16 , ….. (d)
The rule is ………………………………………
The rule is ………………………………………
5
(e)
5
5
10, 100, 1000, .......
95
1. NUMBERS Sequences Involving Division Example: Write down the next term and the rule for the following sequence.
(a)
÷2
÷2
÷2
64, 32, 16, 8 , ......
(b)
÷3
81,
÷3 27,
÷3 9,
3, ……
The next term is: 4
The next term is: 1
The rule is: divide by 2
The rule is : divide by 3
EXERCISE 1.10.4
Write down the next term and the rule for the following sequences.
(a)
125, 25, 5, …..
The rule is …………………………………………………
(b)
0.8, 0.4, 0.2, …..
The rule is …………………………………………………
(c)
27, 9, 3, ….
The rule is …………………………………………………
(d)
16, 8, 4, …..
The rule is …………………………………………………
(e)
64, 8, 1, ……..
The rule is …………………………………………………
Growing Patterns Study carefully the given figures and then shade Fig. 4 appropriately.
Fig. 1
Fig. 2
Fig. 3
Fig. 4
Based on the above figures, complete the table given below. Figure number Number of red squares Number of white squares 96
1 1 8
2 4 12
3 9
4
5
20
24
6 36
7
8
32
36
9 81
The Fibonacci Sequence Observe the following sequences.
(a)
0, 1, 1, 2, 3, 5, 8, ...
(b)
1, 3, 4, 7, 11, 18, ...
(c)
4, 1, 5, 6, 11, 17, ...
Can you find out the pattern in each case? For such patterns, the next term is found by adding the two previous consecutive terms. Example: 0, 1, 1, 2, 3, 5, ...
EXERCISE 1.10.5
1.
Identify the Fibonacci sequence in the following.
(a)
0, 2, 4, 6, 8, 10, ...
(b)
1, 2, 3, 5, 8, 13, ...
(c)
3, 2, 5, 7, 12, 19, ...
(d)
0, 1, 1, 2, 3, 5, 8, ...
(e)
1, 3, 6, 9, 12, 15, ...
2.
Write down the next two terms in the following Fibonacci sequences.
(a)
0, 2, 2, 4, 6, …, …...
(b)
1, 2, 3, 5, 8, …., …..
(c)
0, 5, 5, 10, 15, …., ……
(d)
2, 5, 7, 12, 19, …., ….
(e)
2, 6, 8, 14, 22, …., …….
Sequence of Ordered Pairs Consider the following sequence and complete it with the help of your teacher.
(1,2), (2,4), (3,6), (4,8), (...,...), (...,...), (....,....)
This sequence is an example of a sequence of ordered pairs. Observe that the first number from each pair form the sequence 1, 2, 3, 4, ... and the second numbers form the sequence 2, 4 ,6, 8, .... 97
1. NUMBERS EXERCISE 1.10.6
Write down the next two terms in each of the following sequence of ordered pairs.
(a)
(0, 1), (1, 2), (2,3), (...,...), (...,...)
(b)
(1,3),(3, 5),(5, 7), (...,...), (...,...)
(c)
(5,20), (10,30), (15,40), (...,...), (...,...)
Continuous Assessment 1.10
1.
Write down the next term and the rule for the following sequences.
(i) 2, 7, 12, 17, ….
The rule is ………………………………………
(ii) 51, 47, 43, 39, ...
The rule is ………………………………………
(iii) 2, 6, 18, ...
The rule is ………………………………………
(iv) 64, 32, 16, ...
The rule is ………………………………………
1 , 1 , 3 , 1 , ... (v)
The rule is ………………………………………
(vi) 3, 2, 5, 7, ……
The rule is ………………………………………
Write down the next two terms in the following sequences.
8
2.
4
8
2
(i) (4, 1), (3, 2), (2,3), (...,...), (...,...)
(ii) (16,3), (8, 6), (4, 12), (...,...), (...,...)
Triangular numbers may be formed by building triangular patterns as follows.
3.
6
3
1 1 Figure 1
1+2 2
1+2 + 3+4 4
(i) Draw the next figure.
(ii) Based on the triangles, complete the table below.
Figure number Number of blue triangles Number of white triangles 98
1+2 + 3 3
1 1 0
2 3 1
3 6
4 6
5
6
7
8
9
1.11 PERCENTAGES By the end of this topic, you should be able to: • Recognise the use of percentages in real life situations. • Represent a percentage on a 100-square grid and percent bar. • Convert a percentage into a fraction and/or decimal and vice versa.
Mira, there is a sale in the new shopping mall. All articles are 25 % off.
What does 25% off mean?
Observe the above picture. What do you understand by 25%? Where else have you seen the symbol % ?
99
1. NUMBERS Percentages in real life context Observe and discuss the images. What do you observe?
1.5%
Discount offer
3%
0%
Showing amount of fat in milk
Interest on home loan
What is common in the images? What does the symbol “%” represent? Percentage is another way of representing fractions and decimals. It is often denoted using the percent sign “%”. Example: Consider the 100 - square grid. The shaded region represents : 1 100
= 0.01 = 1 % Decimal
Fraction shaded
100
percentage
Representation of percentage We use a 100-square grid to represent percentage. Example 1 In the 100-square grid, 20 squares are shaded. 20
Fraction shaded is 100 . 20 100 20 100
can also be written as 0.20. = 20%
A percent is equivalent to a fraction in which the denominator is 100. 25
Example: 25 % = 100 A percent is also equivalent to a decimal number in hundredths. Example: 25% = 0.25 Example 2 In the 100-square grid, 50 squares are shaded. 50
Fraction shaded is 100 . 50 100
can also be written as 0.50.
50 100
= 50% 50 % 50 100
1
= 2
0.5
Example 3 In the 100-square grid, 45 squares are shaded. 45
Fraction shaded is 100 . 45 100
can also be written as 0.45.
45 100
= 45%
45 % 45 100
9
= 20
0.45 101
1. NUMBERS Try the exercises below.
EXERCISE 1.11.1
1.
Complete the table below. An example has been done for you. Representation
Fraction shaded
7 100
102
Decimal
0.07
Percentage
7%
Colour the grid appropriately. 2.
Using the hundred grid, represent the given percentages:
(a)
2%
(b) 18 %
(c)
29 %
(d) 100 %
103
1. NUMBERS Percent bar We can also represent percentages using a percent bar. The whole bar represents 100 %. 0%
100%
0 10 20 30 40 50 60 70 80 90 100
50 % 100 %
0 %
0 10 20 30 40 50 60 70 80 90 100
0 %
25 % 100 %
0 10 20 30 40 50 60 70 80 90 100
EXERCISE 1.11.2
1.
What is the percentage represented on each percent bar?
(a)
0 %
100 %
(b)
0 %
100 %
(c) 104
0 %
100 %
2. Represent the following on a percent bar. (a) 90% 0%
100 %
0%
100 %
0%
100 %
(b) 60%
(c) 15 %
Converting a fraction into a percentage Consider the following examples. 1
Example 1 : Convert 2 into percentage. Method 1 : Using equivalent fractions. 1 2 1 2
= x 50
=
? 100
2 100 50
50 100
x 50
50 100
= 50%
Method 2 : Multiply by 100 % 1 2
1 2
= 50%
x 100 % = 50 % 3
Example 2 : Convert 4 into percentage.
Remember, a percentage is a fraction whose denominator is 100.
Method 1 : Using equivalent fractions. x 25
3 4
= x 25
75 100
3 4
=
75 100
75 100
=
75%
4 100 25
Method 2 : Multiply by 100 % 3 4
Note that to convert fraction into percentage, we use equivalent fractions if the denominator is a factor of 100. For example, 1, 2, 4, 5, 10, 20, 25, 50 and 100.
x
3 4
= 75%
100 % = 75 %
Observe that to convert a fraction into percentage, we associate equivalent fractions and decimals, or we multiply by 100 %. 105
1. NUMBERS EXERCISE 1.11.3
1.
Convert the following into percentage. 1
(a) 100
7 10
(b)
3 5
(c)
8
20
9
(d) 20 (e) 25 (f ) 50
2
2.
Laura cleans 5 of the lawn. What percentage of the lawn does she clean?
3.
Dave eats 10 of a pizza. What percentage of the pizza does he eat?
3
1
In previous section, we have seen that 2 is 50 %. Now, we convert fractions whose denominators are not a factor of 100. 1
Example 1 : Convert 3 into percentage. 1 3
x 100 % =
1 3
100
x 1
100
1 1
3 100 33 R 1
= 3
1 1 3 = 33 3 %
1
= 33 3 %
= 33.33 %
11 11
3 100.00 33.33
1 1 3 = 33 3 % or 33.33 %
4
Example 2 : Convert 7 into percentage. 4
We multiply 7 by 100 4 7
x 100 % =
4 7
100
x 1
4 7
400
= 7
= 57.14 %
106
= 57.14 %
5 1 3
7 400.00 57.14
4 5 7 = 57 7 % or 57.14 %
EXERCISE 1.11.4
Convert the following into percentage. 1
2
(a) 2
(b) 3
(c)
5 5
3
4
5
(d) 7 (e) 9
(f ) 6
Converting a mixed number into a percentage (exceeding 100%) 1
3
A mixed number (e.g., 1 2 , 2 4 ) can be converted into percentage. 1
Example : Convert 1 2 into percentage. Method 1 : using diagram
Method 2 : multiply by 100% 1
Rewrite 1 2 as an improper fraction. 1
1 2
1 = 100%
1
1 2
= 1 +
3
1 2 = 2
= 50%
Then, multiply by 100 %
1 2
= 100 % + 50 %
= 150 %
3 2
50
x 100 % = 150 %
1
3
Example : Convert 2 4 into percentage. Method 2 : multiply by 100%
Method 1 : using diagram
3
Rewrite 2 4 as an improper fraction. 3
1 = 100%
1 = 100%
3 4
3 4
2
= 2 +
= 200 % + 75 %
= 275 %
3 4
11
2 4 = 4
= 75%
Thus, 11 4
25
x 100 % = 275 %
1
Fractions greater than one represent percentages exceeding 100. 107
1. NUMBERS EXERCISE 1.11.5
Convert the following into percentage.
(a) 2 2
1
1
1
(b) 1 4 (c) 6 2
3
(d) 4 10
1
(e) 2 5
4
(f ) 5 50
Converting a percentage into fraction. To convert a percentage into a fraction, we rewrite the percentage as a fraction with denominator 100.
Therefore, 50 % = 1 2
Example 1: Convert 50 % into fraction. Rewrite 50 % as a fraction with denominator 100. 50
50 % = 100 50 100
1
= 2
50 % = 1 2
1
Example 2 : Convert 33 3 % into fraction. 1
Rewrite 33 3 % as a fraction. 33
1
1 3
%=
33 3
100 1
= 33 3 ÷ 100
=
= 3 x 100
100 3
÷ 100
100
1
1
1
33 3 % = 3
1
= 3
Example 3: Convert 350 % into fraction. Rewrite 350 % as a fraction over 100. 350
350 % = 100
108
=
1
350% = 3 2
35 10
NOTE
7
= 2 =3
1 2
Therefore, 350 % = 3 1 2
To convert a percentage into a fraction, we divide the percentage by 100
EXERCISE 1.11.6
Convert the following percentage into fraction. 1.
(a) 1 % 2
2. (a) 66 3
(b) 15 %
(c) 20 %
(d) 35 % (e) 75 %
1
(b) 87 2 % (c) 225 % (d) 435 % (e) 840 %
Converting a percentage into decimal number. To convert a percentage into a decimal number, we simply divide the percentage by 100. Example 1 : Convert 31 % into a decimal number. 31
31 % = 100
= 0.31
Example 2 : Convert 6 % into a decimal number. 6
6 % = 100
= 0.06
Example 3 : Convert 515 % into a decimal number. 515
515 % = 100
= 5.15 Note: when dividing a decimal number by 100, we move the decimal point 2 digits to the left.
EXERCISE 1.11.7
Convert the following percentages into decimal. 1.
(a) 25 %
(b) 43 %
(c) 80 %
(d) 70 %
(e) 9 %
(f ) 5 %
2.
(a) 647 %
(b) 412 %
(c) 308 % (d) 205 % (e) 760 % (f ) 910 %
109
1. NUMBERS Converting a decimal number into a percentage To convert a decimal number into a percentage, we multiply by 100%.
(i)
Convert 0.75 into a percentage.
0.75 = 0.75 x 100% = 75 %
(ii)
Converting 0.4 into a percentage.
0.4 = 0.4 x 100%
= 40 %
Add 0 after 4 to move the decimal point 2 digits to the right.
(iii) Converting 2.38 into a percentage.
2.38 = 2.38 x 100%
Move the decimal point 2 digits to the right.
= 238 %
EXERCISE 1.11.8
Convert the following decimal numbers into percentages.
110
1. (a) 0.6 (b) 0.5
(c) 0.1
(d) 0.75
(e) 0.25
(f ) 0.08
2. (a) 3.2 (b) 8.1
(c) 5.08
(d) 2.09
(e) 1.15
(f ) 6.71
Continuous Assessment 1.11 1.
Represent the given percentage on the following hundred grid.
(a) 15 %
2.
Represent the given percentages on the following percent bar.
(a) 20%
(b) 63 %
0% 3.
100 %
(b) 85%
100 %
0% Convert the following fractions into percentage. 1
6
1
9
(b) (c) (d) (a) 100 10 4 5
1
(e) 2 2
4.
Convert the following percentages into fraction.
(a) 25 %
5.
Convert the following decimals into percentage.
(a) 0.17
6.
Convert the following percentages into a decimal.
(a) 75 %
7.
Cinthia eats 4 of a cake. What percentage of the cake does she eat?
8.
Salim cleans 5 of a room. What percentage of the room does he clean?
9.
Fill in the table below.
(b) 45 %
(b) 0.04
(b) 50 %
(c) 3 %
(c) 0.5
(c) 1%
2
(d) 200 % (e) 66 3 %
(d) 7.25
(d) 35 %
(e) 4.01
(e) 245 %
3
4
Percentage Fraction
2 %
18 %
2 100
Decimal
0.02
7 10
0.43
225%
1 5
0.95
111
1. NUMBERS
1.12 RATIO AND PROPORTION By the end of this topic, you should be able to: • Demonstrate an understanding of ratio and direct proportion. • Compare two quantities in terms of a ratio and proportion. • Solve word problems involving ratio and direct proportion.
Observe and discuss the scenario in the picture. How many cups of water is needed for the dough? How many cups of flour is needed for the dough? What will happen if the baker uses 4 cups of water and 1 cup of flour?
Here, we are comparing two quantities, that is, water and flour. The baker used 2 cups of water and 3 cups of flour to prepare the dough.
That is for every two cups of water we need three cups of flour. Or for every three cups of flour we need two cups of water.
112
1. NUMBERS
1 boat
3 planes
For each boat there are 3 planes. For every 3 planes there is 1 boat.
Or
Observe that we can compare two quantities in different order.
diluted into We can say for each :
1 bottle we have 3 glasses.
3 glasses we have 1 bottle.
Or
The comparison of two quantities is known as ratio. Example 1
There are 3 boys and 2 girls. The ratio of boys to girls is 3 to 2, denoted by 3 : 2. Ratio of boys to girls = 3 : 2 Ratio of girls to boys = 2 : 3 Example 2 Consider the chairs and tables and find :
(i)
the ratio of tables to chairs.
(ii)
the ratio of chairs to table.
Solution
(i) The ratio of tables to chairs = 2 : 5
(ii) The ratio of chairs to tables = 5 : 2 113
1. NUMBERS EXERCISE 1.12.1
1.
Consider the following pencils.
(a)
The ratio of red pencils to blue pencils is ______________.
(b)
The ratio of blue pencils to red pencils is ______________.
2.
Consider the following fruits.
(a)
The ratio of oranges to apples is ______________.
(b)
The ratio of apples to oranges is ______________.
Example Sam
Jane
50 kg
27 kg
What is the ratio of the masses of Sam to Jane ? The ratio of the masses of Sam to Jane is 50 : 27
114
EXERCISE 1.12.2
1. Mr James sold 7 kg of potatoes and 3 kg of tomatoes. Find the ratio of the masses of tomatoes to potatoes. 2.
A shopkeeper sold two wires of length of 8 cm and 3 cm. What is the ratio of the lengths of the shorter wire to the longer wire?
3. Mrs Chin had two plants in her garden. The mango plant was 5 m high and the guava plant was 2 m high. What is the ratio of the heights of the mango plant to the guava plant? 4.
A painter poured 25 cL of white paint in a mug and 13 cL of blue paint in a second mug. What is the ratio of the capacity of blue paint to white paint?
Simplifying ratios Example 1 Write down 3 : 15 in its simplest form. Solution We divide both numbers by 3. ratio 1 : 5 divide by 3
ratio 1 : 5 ratio 1 : 5
ratio 3 : 15 ÷ by 3.
3 : 15 1:5
÷ by 3
Ratio 3 : 15 = 1 : 5 Example 2 Write down 12 : 18 in its simplest form. Solution
÷2
÷3
12 : 18 6:9 2:3
÷2 ÷3
Ratio 12 : 18 = 6 : 9 = 2 : 3 115
1. NUMBERS EXERCISE 1.12.3
Rewrite the following ratios in their simplest form.
(a) 6 : 9
(b) 4 : 20
(c)
24 : 6
(d) 10 : 30
(e) 25 : 75
(f )
40 : 100
Expressing ratios in terms of fractions A ratio can be expressed in terms of fractions.
The ratio of yellow marbles to green marbles is 4 : 8. 4
4 : 8 can also be written as 8 . 4 8
can also be expressed in its lowest term.
4 8
= 1
÷4
2
4:8 1:2
÷4 Fraction reduced to its simplest term.
Fraction reduced to its lowest term. Example 1 : Express the following ratio as fraction. (a) 5 : 18 (b) 6 : 12 Solution 5 6 1 (a) 5 : 18 = 18 (b) 6 : 12 = 12 = 2 Example 2 : Express the following fraction as ratio. 2 1 (a) (b) 3 5 Solution 2 (a) =2:3 3 116
1
(b) 5 = 1 : 5
EXERCISE 1.12.4
1.
Express each of the following fractions as a ratio in the simplest form. 5
(a) 7
12 8
(b)
(c)
15 75
2.
Express the following ratios as fractions in the simplest form.
(a) 10 : 6
(b) 20 : 30
(c)
(d)
18 : 72
(d)
16 32
15 : 45
Solving word problems Example 1 A picture frame is 18 cm long and 12 cm wide.
18 cm
(a) Find the ratio of length to width.
12 cm
(b) Find the ratio of length to perimeter. Solution: (a) The ratio of length to width = 18 : 12
= 3:2
÷6
(b) The ratio of length to perimeter.
The perimeter = 18 + 12 + 18 + 12 = 60 cm
The ratio of length to perimeter = 18 : 60
= 3 : 10
÷6
Example 2 Jame shares Rs 60 between his two children, Sam and Bob, in the ratio 2 : 3 respectively. Find (a) the total number of shares. (b) amount of money Sam receives. (c)
amount of money Bob receives.
117
1. NUMBERS
Solution: (a) Sam : Bob = 2 : 3 Total number of shares = 2 + 3 = 5 (b) Sam’s Share is: 2 out of 5 shares 2 Amount of money Sam receives = 5 x 60 = R 24 (c)
Bob’s share is: 3 out of 5 shares 3 Amount of money Bob receives = 5 x 60 = R 36
EXERCISE 1.12.5
1. Annelise has 20 sweets. She shares them with Riya and Chin in the ratio 1 : 3. Find the number of sweets Riya gets. 2. Arav and Priscila shared 50 marbles between them in the ratio 3 : 5. Find the number of marbles: (a) Arav got. (b) Priscila got. 3. A shopkeeper sold 50 kg of rice to two clients A and B in the ratio 3 : 7 respectively. Find (a) mass of rice sold to client A. (b) mass of rice sold to client B.
118
Direct Proportion Kenny wants to buy some cupcakes for his birthday party. He goes to a pastry shop.
How much do the cupcakes cost?
Seller replying: Each cupcake costs Rs 15.
Let's discuss How much must Kenny pay if he wants to buy:
(a) 1 cupcake?
________
(b) 2 cupcakes?
________
(c)
3 cupcakes?
________
(d) 5 cupcakes?
________
(e) 10 cupcakes?
________
Justify your answer.
119
1. NUMBERS We observe that as the number of cupcakes increases, the cost also increases. 1 cupcake
1 x Rs 15
= Rs 15
2 cupcakes
2 x Rs 15
= Rs 30
3 cupcakes
3 x Rs 15
= Rs 45
10 cupcakes
10 x Rs 15 = Rs 150
We say that the number of cupcakes and the cost are in direct proportion. That is, the cost of cupcake is directly proportional to the number of cupcakes and vice versa.
Let's work out. Example 1 One pencil costs Rs 7. Find the cost of:
(a) 2 pencils.
(b) 15 pencils.
Solution :
2 pencils cost 2 x Rs 7 = Rs 14
The cost of 2 pencils is Rs 14.
120
(a) 1 pencil costs Rs 7
(b) 1 pencil costs Rs 7
15 pencils cost 15 x Rs 7 = Rs 105
The cost of 15 pencils is Rs 105.
1. NUMBERS Example 2 One banana costs Rs 4.
(a) Find the cost of 12 bananas.
(b) How many bananas can be bought from Rs 84?
Solution:
(a) 1 banana costs Rs 4.
12 bananas cost 12 x Rs 4 = Rs 48
The cost of 12 bananas is Rs 48.
(b) With Rs 4, we bought 1 banana.
1
Rs 1
banana
4
x 84 = 21 bananas.
1
Rs 84
4
21 bananas can be bought from Rs 84.
EXERCISE 1.12.6
1.
If 1 apple costs Rs 6, find the cost of:
(a) 8 apples. (b) 20 apples.
2.
One box contains 30 glasses. How many glasses can 18 boxes contain?
3.
One ruler costs Rs 9.
(a) Find the cost of 3 rulers.
(b) How many rulers can be bought from Rs 108?
Let's work out. Example 3 3 oranges cost Rs 18. Find the cost of 8 oranges. Solution: 3 oranges cost Rs 18 18
1 orange
Rs 3
8 oranges
= Rs 6
8 x Rs 6 = Rs 48
The cost of 8 oranges is Rs 48. 121
1. NUMBERS Example 4 To make 10 pancakes, Yesha needs 3 cups of flour. How many cups of flour will she need to make 50 pancakes? Solution For 10 pancakes, Yesha needs 3 cups of flour 3
1 pancake
10
50 pancakes
10 x 50 cups of flour = 15 cups of flour
Yesha needs 15 cups of flour.
cups of flour
3
EXERCISE 1.12.7
1.
Noah buys 5 kg of potatoes for Rs 55. How much will he pay for 12 kg of potatoes?
2.
If 13 pens cost Rs 78, find the cost of 10 pens.
3. A factory produces 224 bags in 7 days. How many bags can the factory produce in 15 days? 4. A shoemaker takes 5 hours to make 3 pairs of shoes. Find out how long he will take to make 15 pairs of shoes.
Continuous Assessment 1.12
1.
(a) Write down the ratio of triangles to stars
sss HHHHH
(b) Write down the ratio of blue to purple circles.
2.
Rewrite the following ratios in their simplest forms.
(a) 60 : 6 = ____________.
3.
Express each of the following fractions as a ratio in the simplest form. 18
(a) 3 122
12
(b) 30
(c)
(b) 25 : 15 = ____________. 60 15
1. NUMBERS 4.
Express the following ratios as fractions in the simplest form.
(a) 10 : 20
(b) 8 : 30
(c) 18 : 10
5. In a basket, the ratio of peaches to kiwis is 4 : 7. If there are 55 fruits in the basket, find out the number of peaches. 6, There are 60 marbles in a bag. 20 of them are red, 15 are blue and the rest are yellow. Find, in its simplest form, the ratio of:
(a) blue to red marbles.
(b) red to yellow marbles.
7.
Louise runs 1 km every 15 minutes. How long will she take to run 7 km?
8.
One pear costs Rs 6.
(a) Find the cost of 14 pears.
(b) How many pears can be bought from Rs 114?
9.
A car travels 60 km in 40 minutes. What distance will it cover in 120 minutes?
123
NOTES
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124
GEOMETRY
2
2.1 Pythagoras' Theorem By the end of this topic, you should be able to: • Identify the hypotenuse of a right-angled triangle. • Use Pythagoras’ theorem to find an unknown side in a right-angled triangle. • Apply Pythagoras’ theorem to real life situations.
Study the above picture carefully. What is the width of the canal? What is the height at which the man is seeking help? Can you guess the length of the ladder needed? How will you calculate it?
125
2. GEOMETRY Chapter 2 - Binomial expressions
Recall – Right-Angled Triangles Right-angled triangle
Properties
Example in real life set squares
acute
A right-angled triangle has one right angle.
Activity Investigating the sides of a right-angled triangle Material needed: ruler 1. Consider the right-angled triangle given below. A
B
C
2. Measure and write down the length of all the three sides of the right-angled triangle ABC.
AB = ___________________ cm
BC = ___________________ cm
AC = ___________________ cm
3. Which is the longest side of the right-angled triangle ABC? ____________________ 4. Which side of the right-angled triangle ABC is opposite to the angle of 90°? _________ 126
The hypotenuse of a right-angled triangle From the previous activity, it can be noted that the longest side, AC, of the right-angled triangle ABC is opposite to the angle of 90°. In fact, the longest side of any right-angled triangle is always opposite to the angle of 90° and is called the hypotenuse. Example Which side is the hypotenuse in each of the following right-angled triangles? (a) (b) X Q
R
Z
Y
Solution (a) XZ (opposite 90°)
P
(b) PR (opposite 90°)
EXERCISE 2.1.1
Tick the hypotenuse in the following right-angled triangles.
(a)
(b)
(c)
(d)
127
2. GEOMETRY Chapter 2 - Binomial expressions Discovering the formula connecting the sides of a right-angled triangle 1.
Observe carefully the following right-angled triangle.
5 cm
4 cm
3 cm 2. A square is drawn on each of the sides of the right-angled triangle. The area of each square is given. 5 cm 4 cm
4 cm
Area = 5 x 5 = 52 = 25 cm2
Area = 4 x 4 = 42 = 16 cm2
3 cm
3.
5 cm
Area = 3 x 3 = 32 = 9 cm2
3 cm Find the sum of the areas of the two smaller squares.
_________________________________________________________ 4. Compare your answer with the area of the biggest square. What do you observe? _________________________________________________________ 5.
Hence, give a formula connecting the squares of the sides of the right-angled triangle.
_________________________________________________________ 128
Pythagoras’ Theorem
AN INTERESTING FACT
From the previous activity, you must have observed that the area of the biggest square is equal to the sum of the areas of the two smaller squares, that is, 25 = 9 + 16 52 = 32 + 42 Pythagoras' theorem states that : In a right-angled triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the two other sides.
a
c2 = a2 + b2
c
Pythagoras (569 – 495 B.C.E.) was born on the island of Samos in Greece. He studied mathematics among other things during his numerous travels. Though the famous Pythagoras’ theorem had been named after him, it is most probable that the Babylonians had already some proof of the theorem a thousand years before.
b Example Find the unknown side in the following right-angled triangles. (a)
(b)
y 8
8
x
17
6
Solution (a) x2 = 62 + 82 (b) 82 + y2 = 172 2 x = 36 + 64 64 + y2 = 289 2 x = 100 y2 = 289 – 64 = 225 x = 100 y = 225 x = 10 y = 15 129
2. GEOMETRY Chapter 2 - Binomial expressions
EXERCISE 2.1.2
Find the unknown sides in the following right-angled triangles.
(a)
(b)
4
p
x
24 7
3
(c) (d)
5
u 15
y 17
130
13
Applications of Pythagoras’ Theorem The Pythagoras' theorem is generally used in architecture, carpentry and land surveying.
Laying out a right-angle correctly during a construction project.
Use of Pythagoras' theorem in carpentry.
Example A ladder, 13 m long, leans against a wall. The foot of the ladder is 5 m from the base of the wall. How far up the wall does the ladder reach?
13m
5m Solution The situation can be represented on a right-angled triangle as follows. h2 + 52 = 132 h2 + 25 = 169 h2 = 169 - 25 h2 = 144 h = 144 h = 12
✓
h
13
5
The ladder reaches 12 m up the wall. EXERCISE 2.1.3
Try these. 1. PQRS is a rectangle of length 8 cm and width 6 cm. Find the length of the diagonal QS.
P
Q 6 cm
S
8 cm
R
131
2. GEOMETRY Chapter 2 - Binomial expressions 2.
Find the value of x. 13 m
xm 8m
12 m
3. Kingston Town is 15 km north and 8 km east of Queens Town. How far are the two towns apart? 4. Given a triangle ABC where AB = 12 cm, BC = 5 cm and AC = 13 cm, determine whether it is a right-angled triangle.
Continuous Assessment 2.1 1.
Tick the hypotenuse in the following right-angled triangles.
(a)
2.
(b)
Find the unknown sides in the following right-angled triangles.
(a)
(b) 8
5
12
x 6
132
w
(c)
(d) z 6
y
15
17
3.
10
Find the value of x. 10 m
24 m
25 m
xm
4. A ladder, 17 m long, leans against a wall. The foot of the ladder is 8 m from the base of the wall. How far up the wall does the ladder reach?
17m
8m
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2. GEOMETRY Chapter 2 - Binomial expressions
2.2 Angles By the end of this topic, you should be able to: • Recognise and use fundamental geometrical terms such as point, line, line segment, ray, plane, angle, vertex, parallel and perpendicular lines and transversal. • Distinguish among different types of angles (acute, obtuse, right-angled, reflex, straight angle, complete turn). • Measure and construct angles using geometrical instruments. • Identify complementary and supplementary angles. • Identify parallel lines and transversals. • Identify vertically opposite, corresponding, alternate and interior angles. • Find unknown angles using the notions of complementary, supplementary, corresponding, alternate and co-interior angles.
Points, Lines and Planes Name and Shape Properties Real life examples Point A point has Trace left by the • no length, tip of a pencil on • no width, and paper • no thickness.
Line
134
A line has length. It continues endlessly in both directions.
Horizon
Name and Shape
Properties
Line segment
A line segment has length and two endpoints.
Real life examples Curtain rail
Ray
A ray is a half-line. It has length. It continues endlessly in one direction only.
Parallel lines
Parallel lines are at a fixed distance apart. They never meet each other.
Railway track
Perpendicular lines meet each other at right angles, that is, at 90°.
Lug wrench
Transversal
A transversal is a line that cuts two parallel lines.
Top bar of a goal post
Plane
A plane has both length and width.
Surface of a table
Perpendicular lines
A ray from the sun
135
2. GEOMETRY Chapter 2 - Binomial expressions Angles An angle is formed where two lines meet. The point where they meet is called the vertex of the angle.
angle vertex
Types of angles
Properties
acute angle
• Smaller than a right angle • Less than 900
Right angle
• Quarter Turn • Exactly 900
Obtuse angle
Straight angle
• Bigger than a right angle • Smaller than a straight angle • Between 90° and 180°
Real life examples Foot tongs
Set squares
Poolside chair
Roof top
• Half Turn • Exactly 1800 Bridge and horizon
136
Types of angles
Properties
Real life examples
• Bigger than a straight angle • Smaller than a complete turn • Between 180° and 360°
Reflex angle
Lantern and house
• Complete Turn or Full Turn • Exactly 360°
Angle at a point
Wheel
How to denote angles?
A
The angle on the right can be denoted as: ABC, B, ABC, B, CBA or CBA. B
Example Which of the following angle is obtuse ?
C R
L
M
N
P
Q
Answer PQR is an obtuse angle.
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2. GEOMETRY Chapter 2 - Binomial expressions
EXERCISE 2.2.1
1.
Fill in the boxes with the appropriate angles from the list: 32°, 104° and 318°.
(a)
2.
(c)
State whether each of the following angles is acute, obtuse or reflex.
(a) 340
(b) 1100
(c) 860
(d) 325°
(e) 63°
(f ) 216°
(g) 142°
(h)
(i)
3.
Which one of the following angles is acute?
(a)
B
A
138
(b)
191°
98°
(b) F
D C
E
How to measure angles? A protractor is used to measure or construct angles. The most commonly used protractor is semi-circular in shape and it can measure angles up to 1800. inner scale Outer scale
outer scale centre
A protractor
baseline
inner scale
Material needed: protractor. 1. Consider the angle ABC given below.
A
B
C
139
2. GEOMETRY Chapter 2 - Binomial expressions 2. Place the centre of your protractor on the vertex of the angle ABC such that the right baseline of the protractor lies exactly on the line segment BC. A
C B zero - use inner scale 3. The point C indicates that you have to use the inner scale. So, starting at 0° on the inner scale, move round the protractor in an anticlockwise direction until you reach the other line segment AB. 4. 5.
Now, take the reading from the inner scale and hence fill in the blanks below. ABC is an _______________ angle and is equal to ______________. Next, consider the angle DEF given below. F
D
140
E
6. Place the centre of your protractor on the vertex of the angle DEF such that the left baseline of the protractor lies exactly on the line segment DE. F
D
EE zero -outer scale
7. The point D indicates that you have to use the outer scale. Therefore, starting at 0° on the outer scale, move round the protractor in a clockwise direction until you reach the other line segment EF. 8. Take the reading from the outer scale and hence write down the size of the angle DEF. DEF is an _______________ angle and is equal to ______________.
141
2. GEOMETRY Chapter 2 - Binomial expressions
Example Write down the size of each of the following angles. (a) ABC
C
A
B
(b) DEF D
E
Solution (a) ABC = 400
142
(b) DEF = 1200
F
EXERCISE 2.2.2
1.
Write down the size of each of the following angles.
(a) ABC
(b)
DEF
C D
A
(c) KLM
(d)
PQR
M
K
F
E
B
P
R
Q
L
2. Use a protractor to measure each of the following angles. (a) (b)
(c)
(d)
(f )
(e)
143
2. GEOMETRY Chapter 2 - Binomial expressions NOTE
If the line segments are too short, use a pencil to extend them to help you read the scale on the protractor.
Constructing an angle of 60° Materials needed: pencil, ruler, protractor. To construct angle ABC = 60°: 1. Using a ruler, draw and label a line segment AB. A
B
2. Place the centre of the protractor on B with the left baseline exactly on the line segment AB.
A
B
3. Starting at 0° on the outer scale (on the left), move round the protractor in a clockwise direction until 60° is reached. Mark this position ‘C’.
C
A
4.
B
Join point B to point C to form the line segment BC and label angle ABC. C
A
144
600
B
Constructing an angle of 200° Materials needed: pencil, ruler, protractor. To construct angle DEF = 200°: 1. Using a ruler, draw and label a line segment DE of length 6 cm. D
E
2. Place the centre of the protractor on E with the left baseline exactly on the line segment DE.
D E
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2. GEOMETRY Chapter 2 - Binomial expressions
3. Starting at 0° on the outer scale (on the left), move round the protractor in a clockwise direction until 160° is reached. Mark this position ‘F’.
F D
4.
E
Join point E to point F to form the line segment EF and label reflex angle DEF 200°. F
E D 2000
EXERCISE 2.2.3
Use a ruler and a protractor to construct the following angles.
146
(a)
30°
(b)
90°
(c)
54°
(e)
227°
(f )
160°
(g) 270°
(d)
105°
(h)
333°
Discovering Complementary Angles 1. Consider the following angles and fill in the blanks. (i) (ii)
150
300 600
750
The sum of the angles in case (i) is ______ ° + ______ ° = ______ °. The sum of the angles in case (ii) is ______ ° + ______ ° = ______ °.
2.
What can you deduce about the angles in case (i) and in case (ii)?
Complementary Angles Two angles are complementary if their sum is equal to 90°.
a
a + b = 90° b
NOTE :
a is the complement of
b and vice-versa.
Example Find the unknown angle in each of the following. (a) (b)
y 380
x 700
Solution (a) x = 900 – 700 = 200 (b) y = 900 – 380 = 520 147
2. GEOMETRY Chapter 2 - Binomial expressions
Discovering Supplementary Angles 1.
Consider the following angles and fill in the blanks.
(i) (ii) 510
1400 400
1290
The sum of the angles in case (i) is ______ ° + ______ ° = ______ °.
The sum of the angles in case (ii) is ______ ° + ______ ° = ______ °.
2.
What can you deduce about the angles in case (i) and in case (ii)?
_______________________________________________________
Supplementary Angles a
Two angles are supplementary if their sum is equal to 180°. a + b = 180° Note :
a is the supplement of
b and vice-versa.
Example Find the unknown angle in each of the following. (a)
(b)
x
530
y
1480
Solution (a) x = 1800 – 1480 = 320 (b)
148
y = 1800 – 530 = 1270
b
EXERCISE 2.2.4
1.
(a) Find the complement of (i) 34° (ii) 88°
(iii) 51° (iv) 65° (v) 27°
(b) Find the supplement of (i) 129° (ii) 91°
(iii) 33° (iv) 148° (v) 12°
2.
Find the unknown angle in each of the following.
(a)
(b)
(c)
150 c
a
460 720
b
(d)
(e)
(f )
d 380
f 40
0
29
0
e
410
360
3.
Find the unknown angle in each of the following.
(a)
1500
u
(d) 1360
(b)
(c) 780
v
(e)
710 y
x
w
590
(f )
470
z 460
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2. GEOMETRY Chapter 2 - Binomial expressions
Angles Properties Angles Properties Vertically opposite angles
Examples
a
a = b
b = 500
Angle at a point
b
b a
c
1200
800
d
a + b + c + d = 3600
Corresponding angles
b
500
b
1000 b + (800 + 1200 + 1000) = 3600
b + 300 = 3600 b = 3600 – 3000 = 600
a
a b
700
a = b
a = 700
Alternate angles x
a
700
b
a = b
Co-interior angles
x = 700
a b
700 x
a + b = 1800
x + 700 = 1800 x = 1800 – 700
x = 1100
150
EXERCISE 2.2.5
Find the unknown angles. 1. (a)
(b) 1200
800 x
x
2. (a) 500
(b) 600
1100
x
1200
x
3. (a)
1100
(b)
e
g
a
a f
d
700
c
4. (a)
b
(b) 60
0
1200
x
x
5. (a)
(b) 1150
800 x
x
151
2. GEOMETRY Chapter 2 - Binomial expressions Continuous Assessment 2.2 1.
Match the following from Column A to Column B.
Column A
Column B
.
plane
ray
point
line
2.
segment Fill in the correct terms: Perpendicular, Transversal, Parallel L
M
R
P
Q
152
(a)
Lines L and M are _____________
(b) Lines P and Q are _____________
(c)
R is a ____________
3.
Match the following from Column A to Column B and Column C.
Column A a
Column B
Column C
Reflex angle
900
Right angle
1200
obtuse angle
600
acute angle
3000
b
4.
c
d
Fill in the blank with the correct angle.
a b c
e d
(a)
Angle _____ and angle ____ are complementary.
(b)
Angle _____ and angle ____ are supplementary.
153
2. GEOMETRY Chapter 2 - Binomial expressions
5.
Find the unknown angle in each of the following.
(a)
(b)
(c) 1450
b
a
c
680
6.
520
Find the unknown angle in each of the following.
(a)
(b) p
y
x
420
39
0
7.
q
r
Find the unknown angle in each of the following.
(a)
(b)
(c)
m
k
n 36
0
62
0
8.
1170
Measure and write down the following angles.
(a)
(b) a
154
b
9.
Use a ruler and a protractor to construct the following angles.
(a) 750
(b) 1280
2.3 Polygons By the end of this topic, you should be able to: • Identify and name polygons (up to decagon), including regular polygons. • Differentiate among scalene, isosceles and equilateral triangles in terms of lengths and angles. • Find unknown angles in triangles. • Differentiate among different types of quadrilaterals (rectangle, square, parallelogram, rhombus, kite, trapezium, arrowhead). • Find unknown angles in quadrilaterals. • Demonstrate an understanding of properties of regular polygons. • Identify and determine the interior and exterior angles of a polygon geometrically and using formulae. • Solve problems involving regular polygons.
Polygons
Football Beehive
Black panel
Stop sign Cell
What can you say about the above pictures? How many sides does: (a) the black panel of a football have? (b) each cell of a beehive have? (c) a stop sign have? Where else have you seen shapes? What type?
155
2. GEOMETRY Chapter 2 - Binomial expressions
In general, a 2-D shape having many straight sides is called a polygon.
A regular polygon is a 2-D shape in which all the sides are of the same length and all the angles are equal.
An irregular polygon is a 2-D shape in which the angles are not all of the same size and the sides are not all of the same length.
156
Regular Polygons Regular Polygons
Number of Sides
Examples in Real Life
Equilateral Triangle
3
Square
4 Window
Pentagon
5
Hexagon
6 Blanket
Road Sign
Art Work
Heptagon 7 7-Day Pill Box
Octagon
8 Clock
Nonagon 9 Austrian 5 Euro Coin
Decagon 10 Umbrella Top View
157
2. GEOMETRY Chapter 2 - Binomial expressions
EXERCISE 2.3.1
1.
158
Name the following polygons and state whether they are regular.
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
(j)
(k)
(l)
2.
Draw the following polygons :
(a) pentagon
(b) hexagon
(c) heptagon
(d) octagon
(e) nonagon
(f ) decagon
Side-Named Triangles Types of triangles
Properties An isosceles triangle has two equal sides. An equilateral triangle has all 3 sides equal. A scalene triangle has no equal sides.
Examples in real life earrings furniture
pool triangle
cookie cutter
boat sails
Angle-Named Triangles Types of triangles
acute
acute
An acute-angled triangle has all angles less than 90°.
acute
A right-angled triangle has one right angle.
obtuse
acute
Properties
An obtuseangled triangle has one obtuse angle.
Examples in real life pendant
set squares
roof facade
159
2. GEOMETRY Chapter 2 - Binomial expressions
EXERCISE 2.3.2
160
1.
Circle the triangles.
2.
Name the following triangles according to their (i) lengths, (ii) angles.
(a)
(b)
(c)
(d)
(e)
(f )
Activity Finding sum of angles in a triangle
Materials needed: rectangular piece of bristol paper, ruler, pencil, scissors, glue.
1. Using a ruler and a pencil, draw a triangle on your rectangular piece of bristol paper and cut it out (as shown in Fig. 1).
Fig. 1 2.
Name the angles in your triangle a, b and c (as shown in Fig. 2). a
b
c Fig. 2
3.
Using a ruler and a pencil, draw a horizontal line in your copybook.
4. Cut out the three angles from your triangle and place them along the horizontal line in your copybook (as shown in Fig. 3).
a $ b
c
c
a
b
Fig. 3 5.
Stick the angles in your copybook.
6.
Can you deduce the sum of the angles in your triangle?
161
2. GEOMETRY Chapter 2 - Binomial expressions Sum of angles in a triangle From the previous activity, it can be noted that the three angles in a triangle form a straight angle. a
c
b
Hence, the sum of the angles in a triangle is 1800. Example Find the unknown angles x, y and z in the following triangles. (a)
(b)
(c)
x
400
350 700
500
y
Solution (a) x = 1800 – (700 + 500) (b) y = 1800 – (900 + 350)
z
(c) 2z + 400 = 1800
= 1800 – 1200
= 1800 – 1250
2 z = 1800 – 400
= 600
= 550
2 z = 1400
162
z
0 z = 140
2
z = 700
EXERCISE 2.3.3
1.
Find the unknown angles in the following triangles.
(a) a
(b) 0 65
(c) b
740
480
c
360
(d)
(e)
(f ) 220
1300
570
d
e
180
f
2. An equilateral triangle is a regular 2-D shape. It has three equal sides and three equal angles. Given the equilateral triangle below, find x.
x
x
3.
x
Below is an isosceles right-angled triangle. Find y.
y
y
163
2. GEOMETRY Chapter 2 - Binomial expressions Recall – Quadrilaterals Types of quadrilaterals
Properties
A square has 4 equal sides and 4 right angles.
Examples in Real Life Carom board
A rectangle has Football 2 pairs of equal pitch sides and 4 right angles.
A parallelogram has 2 pairs Tiles of equal and parallel sides A rhombus is a parallelogram Carpet with 4 equal design sides. A trapezium has at least one pair of parallel Roof sides. A kite has 2 Kite pairs of equal adjacent sides.
Arrowhead Types of arrowheads
Properties
An arrowhead has at least one reflex pair of equal adjacent sides and it has one reflex reflex angle.
164
Examples in Real Life Tip of an arrow
EXERCISE 2.3.4
Circle the quadrilaterals and write down their names.
165
2. GEOMETRY Chapter 2 - Binomial expressions Activity Finding sum of angles in a quadrilateral Materials needed: rectangular piece of bristol paper, ruler, pencil, scissors, glue 1. Using a ruler and a pencil, draw a quadrilateral on your rectangular piece of bristol paper and cut it out (as shown in Fig. 4).
Fig. 4 Name the angles in your quadrilateral a, b, c and d (as shown in Fig. 5).
2.
a
d
b c Fig. 5 3.
Using a pencil, make a dot in your copybook.
4.
Cut out the four angles from your quadrilateral and place them around the dot in your copybook (as shown in Fig. 6).
$ c b
d Fig. 6
166
5.
Stick the angles in place in your copybook.
6.
Can you deduce the sum of the angles in your quadrilateral ?
a
Sum of angles in a quadrilateral As it can be observed from the previous activity, the four angles in a quadrilateral form a complete turn. c d
a
b
a + b + c + d = 3600 Hence, the sum of the angles in a quadrilateral is 3600. Example Find the unknown angles in the following quadrilaterals. (a)
(b)
P
Q a
b S
(c)
550
L
B
A c
e
1300 K
1250 500
R
a
M
e
d C
D
400
1000
N
Solution PQRS is a parallelogram. (a)
(c)
ABCD is an isosceles trapezium.
b = 180° – 125°
AD = BC
= 55° (Co-interior angles)
Thus, c = 180° – 50°
a = 180° – 55° = 125° (Co-interior angles)
It is symmetrical about KM. Sum of angles in a quadrilateral = 360° e + e + 40° + 100° = 360°
= 130°
2e + 140 = 360°
(Co-interior angles)
2e = 360° – 140°
d = 180° – 130° = 50°
(b) KLMN is a kite.
2e = 220° 2e = 220° 2
e = 110°
(Co-interior angles)
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2. GEOMETRY Chapter 2 - Binomial expressions
EXERCISE 2.3.5
1.
Find the unknown angles in each of the following quadrilaterals.
(a)
(b)
a 720
(c) 1240
580
1020 360
b c
(d)
(e)
1160 1020
(f )
540
f
e 1260
d
1150
2. A square is a regular 2-D shape. It has four equal sides and four right angles. Given the square below, find the unknown angles x and y. x y
3.
Find the unknown angles m and n in the following rectangle. 360 m n
168
Sum of interior angles of a polygon The sum of the interior angles of an n-sided polygon is given as (n - 2) x 1800 where n - 2 is the number of triangles formed. E.g.
Quadrilateral 2 triangles 2 × 180° = 360°
Pentagon 3 triangles 3 × 180° = 540°
Hexagon 4 triangles 4 × 180° = 720°
The size of one interior angle of a regular n-sided polygon is given as (n-2) x 1800 n since the interior angles of a regular polygon are all equal in size. Example 1 Find the sum of the interior angles of a heptagon. Solution 5 triangles are formed. A heptagon has 7 sides. Thus, n = 7. Sum of the interior angles of a heptagon = (7 – 2) × 180° = 5 × 180° = 900° Example 2 The interior angles of a hexagon are 90°, 110°, 135°, 124°, 191° and x°. Find x. Solution A hexagon has 6 sides. Thus, n = 6. Sum of the interior angles of a hexagon = (6 – 2) × 180° = 4 × 180° = 720° Sum of the 5 given angles of the hexagon = (90° + 110° + 135° + 124° + 191°) = 650° The 6th angle = 720° – 650° = 70°. Hence, x = 70° 169
2. GEOMETRY Chapter 2 - Binomial expressions
Example 3 Find the size of one interior angle of a regular nonagon. Solution A nonagon has 9 sides. Thus, n = 9. Size of one interior angle of a regular nonagon =
(9-2) x 1800 = 7 x 1800 9 9
= 7 x 200 = 1400
EXERCISE 2.3.6
1.
Find the sum of the interior angles of:
(i)
2.
a pentagon,
The interior angles of a heptagon are 92°, 121°, 143°, 164°, 175°, 108° and x°. Find x.
(ii) an octagon
1750 x0
920 1210
1080 1640
170
1430
3.
The interior angles of an octagon are 133°, 156°, 202°, 117°, 304°, x°, 2x° and 3x°. Find x.
4.
Find the size of one interior angle of:
(i)
a regular pentagon,
(ii) a regular decagon.
Exterior angles of a polygon
a
The exterior angle of a polygon is the angle formed when one side of the polygon is extended as shown in the diagram on the right.
h
e
Interior angle
d
Since the word ‘exterior’ means ‘outside’, the exterior angle is thus found on the outside of the polygon.
g
b f
exterior angle
c
Angles e, f, g, h are exterior angles. Angles a, b, c, d are interior angles.
NOTE : While extending the sides of the polygon, it should be done in one direction only, that is, either clockwise or anti-clockwise but not in both directions as shown in the diagrams below. anti-clockwise
clockwise a
x
p z
b q
c
y
r
This representation is not possible since it has mixed directions.
Exterior angles of a triangle d
c
a
c + d = 180° b
c
d c
a
d b
d=a+b • Angles c and d form a straight angle and therefore they are supplementary. • Angles a and b add up to give angle d. The exterior angle of any triangle is equal to the sum of the two opposite interior angles. 171
2. GEOMETRY Chapter 2 - Binomial expressions
Example Find the exterior angle x in the following triangles. (a)
(b)
700
840
600
500
x
350
x
Solution x = 1800 – 500 = 1300 or
x = 840 + 350 = 1190
x = 700 + 600 = 1300 EXERCISE 2.3.7
Find the exterior angle marked in each of the following triangles. (a)
(b)
680
620
370
640
480
(c)
x
810 (d)
y
290
700 42 0
172
w
870
y
Sum of exterior angles of a polygon The sum of the exterior angles of a polygon is 360°. The size of one exterior angle of a regular n-sided polygon is given as 360° since the n exterior angles of a regular polygon are all equal in size.
Example The exterior angles of a hexagon are 40°, 60°, 75°, 24°, 85° and x°. Find angle x.
Solution Sum of the exterior angles of a hexagon = 360° Sum of the 5 given exterior angles of the hexagon = (40° + 60° + 75° + 24° + 85°) = 284° The 6th exterior angle = 360° – 284° = 76°. Hence, x = 76°.
x0
EXERCISE 2.3.8
720
1. The exterior angles of a quadrilateral are 72°, 88°, 124° and x°. Find x. 880 124
0
2. The exterior angles of a pentagon are 92°, 64°, 56°, 108° and x°. Find x.
3. The exterior angles of a heptagon are 54°, 46°, 35°, 81°, x°, 2x° and 3x°. Find x. 4.
Find the size of one exterior angle of:
(i) a regular hexagon,
(ii)
a regular nonagon.
5. Given the size of one interior angle of a regular polygon is 140°, find the number of sides of that regular polygon. 173
2. GEOMETRY Chapter 2 - Binomial expressions Continuous Assessment 2.3 1.
Name each of the following polygons.
(a)
2.
Draw the following polygons:
(a)
a quadrilateral,
3.
Match each shape in Column A to its name in Column B.
(b)
(b)
a heptagon,
Column A
(c)
(c)
a nonagon.
Column B
Isosceles triangle
Equilateral triangle
174
Scalene triangle
(d)
4.
Match each shape in Column A to its name in Column B.
Column A
Column B
Acute-angled triangle
Right-angled triangle
Obtuse-angled triangle
5.
Find the unknown angles in the following triangles.
(a)
(b) (c) a 320
1240 b
400
c
150
6.
Find the unknown angles in the following quadrilaterals.
(a)
(b) (c)
550
1360
x
y
z
1220 980
175
2. GEOMETRY Chapter 2 - Binomial expressions 7.
Find the size of one interior angle of:
(i)
a regular hexagon,
8.
The interior angles of an octagon are 154°, 135°, 197°, 120°, 306°, x°, 2x° and 3x°. Find x.
9.
Find the exterior angle marked in each of the following triangles.
(a)
(ii) a regular nonagon.
(b) 320 720
580
850
m n
10. Given the size of one interior angle of a regular polygon is 108°, find (a)
the size of one exterior angle,
(b) the number of sides of that regular polygon.
176
2.4 Coordinates By the end of this topic, you should be able to: • Identify the axes in a Cartesian plane (x-y plane). • Locate and plot points in the Cartesian plane. • Determine the equation of lines parallel to the x and y axes. • Generate and plot points with integer coordinates and draw lines. • Find point of intersection of two lines graphically.
Locating the position of a student in class Let’s ask Mr. Owen.
I want to know where Mark is sitting.
Pam
Quin
Aditi
Sonu
Tom
Jerry
Krish
Khem
Mark
Nina
Elias
Farrah
Gina
Hema
Ivan
Mr.Owen
Alex
Benny
Carl
Dev
Discuss how you would find your way to Mark. Have you realized that you need a starting point? If Mr Owen is the starting point, how would you find your way to Mark?
177
2. GEOMETRY Chapter 2 - Binomial expressions To locate Mark from Mr Owen’s desk, we first move three desks to the right, then two desks forward, as shown on the diagram below.
3
Pam (__,__)
2
Jerry (__,__)
1
Elias (__,__)
0 Mr.Owen (__,__)
Quin
Aditi
Sonu
Tom
(__,__)
(__,__)
(__,__)
Krish
Khem
Mark
Nina
(__,__)
(__,__)
3, 2 (__,__)
(__,__)
Gina
Hema
Ivan
Farrah (__,__)
(__,__)
(__,__)
Alex
Benny
Carl
(__,__)
(__,__)
(__,__)
(__,__)
(__,__) Dev (__,__)
0 1 2 3 4 AN INTERESTING FACT
The Cartesian Plane
The plane below is known as a Cartesian plane. y - axis
0
x - axis
The French mathematician Rene Descartes (1596 - 1650) was once lying on his bed when he noticed a fly on the ceiling. He devised a way to locate the fly by using two perpendicular axes, the x and y axes.
Observe that it consists of two intersecting lines. The horizontal number line is called the x-axis. The vertical number line is called the y-axis. The point where the two axes meet is called the origin and is denoted by the letter O. 178
Locating the position of an object on a map Pirate John is going on a treasure hunt on Orchid Island. He leaves his ship at Pirates' Bay and follows the trail on his treasure map. He walks by a coconut tree, a waterfall and finally reaches the spot where the treasure is buried. y x
x
x Pirates' Bay
x
x
Study the map carefully and answer the following questions. 1.
Taking Pirates' Bay as the origin, write down the positions of :
(a)
the cocunut tree,
(
,
)
(b)
the waterfall,
(
,
)
2.
Give the position of the spot where the treasure was buried.
(
3.
On the map, draw a hill with position (2, 3)
,
)
179
2. GEOMETRY Chapter 2 - Binomial expressions Coordinates of a point The position of any point on the Cartesian plane is described by the ordered pair (x, y) where l x is the horizontal position of the point from the origin, O, l y is the vertical position of the point from O, l x is called the x-coordinate and y is called the y-coordinate of the point. To plot any point (x, y), we start form the origin, O, and we move x steps horizontally, then y steps vertically. Note that if: l x is positive we move x steps to the right of the origin, O, l x is negative we move x steps to the left of the origin, O, l y is positive we move y steps upwards, l y is negative we move y steps downwards. Consider the following graphs. y B(-3, 2)
A (3, 2)
+2 -3 -3
+2
+3
x
0
-2 D(-3, -2)
C(3, -2) y L
+2
-3 0
N(0, -1)
180
To plot B(–3, 2), starting from O, move 3 steps (x = –3) to the left and 2 steps (y = +2) upwards. To plot C(3, –2), starting from O, move 3 steps (x = +3) to the right and 2 steps (y = –2) downwards. To plot D(–3, –2), starting from O, move 3 steps (x = –3) to the left and 2 steps (y = –2) downwards.
(0, 4)
+4 M(-3, 0)
To plot A(3, 2), starting from O, move 3 steps (x = +3) to the right and 2 steps (y = +2) upwards.
K (2, 0)
x
To plot K(2, 0), starting from O, move 2 steps (x = +2) to the right. To plot L(0, 4), starting from O, move 4 steps (y = +4) upwards. To plot M(–3, 0), starting from O, move 3 steps (x = –3) to the left. To plot N(0, –1), starting from O, move one step downward.
Example Write down the coordinates of the points given on the Cartesian plane below. y A G H F
B
x
0 C
E D Solution
A (1, 3)
B (2, 0)
E (-1, -2)
F (-2 2 , 0)
1
C (3, -1)
D (0, -3)
G (-2, 2)
H (0, 1)
EXERCISE 2.4.1
1. Write down the coordinates of the points in the spaces given on the Cartesian plane below. y H( , ) B( A(
G(
, )
, )
C(
, )
0 F(
, )
x
, ) D(
, )
181
2. GEOMETRY Chapter 2 - Binomial expressions 2.
Write down the coordinates of the points given on the following x-y plane. y B( K( J(
, )
L(
, )
, )
, ) A( I(
, ) C(
, )
, )
x
0
H(
E(
, )
, ) D(
F(
, )
, )
Coordinates in real life Coordinates are used on maps to describe the position of objects or places. For example, air traffic controllers assign coordinates to airplanes so as to locate them easily on their radar. Meteorologists use of coordinates to map cylones on weather charts.
Agalega
Tromelin
Intense Tropical Cyclone Gelena
St Brandon
Mauritius
Rodrigues
Réunion
Radar screen showing the positions of several aeroplanes 182
Weather Chart with cyclone track
Example 1 Plot the following points on a Cartesian plane. (a) A (1, 0) (b) B (3, 4) (c) C (0, 2) (e) E(2, -3) (f ) F (0, -4) (g) G (-2, 3)
(d) D (-3, 0) (h) H (-4, -2)
y
Solution
B G C
D
A
x
0 H E F
Example 2 Plot the following points on an x-y plane. (a) (i) A(–2, 2) (ii) B(3, 2) (iii) C(3, –1) (iv) D(–2, –1) (b) Join A to B, B to C, C to D and D to A using a pencil and a ruler. Name the 2-D shape obtained. y Solution (a) A
B
x
0 0 D
C
(b) ABCD is a rectangle. 183
2. GEOMETRY Chapter 2 - Binomial expressions
EXERCISE 2.4.2
Answer the whole of this question on a sheet of graph paper. 1. Plot the following points on a Cartesian plane.
(a)
A(5, 0)
(b)
B(4, 5)
(c) C(0, 4)
(d) D(–4, 0)
(e)
E(3, –7)
(f)
F(0, –6)
(g) G(–2, 6)
(h) H(–5, –3)
2. Answer the whole of this question on a sheet of graph paper.
(i)
On an x-y plane, draw the triangle ABC with vertices:
(a)
On the same graph, draw the trapezium DEFG with vertices:
(ii)
(a)
A(–1, –1) D(–4, –1)
(b) (b)
B(2, 0)
E(–3, –3) (c)
C (–1, 4) F (3, –3) (d) G (4, –1)
(iii) Name the figure formed by the triangle ABC and the trapezium DEFG.
3.
Answer the whole of this question on a sheet of graph paper.
(i)
On a well labelled Cartesian plane, draw the square ABCD with vertices:
(a)
On the same graph, draw the rectangle EFGH with vertices:
(ii)
(a)
A (–3, 2) E(–2, –1)
(b) (b)
B (3, 2) F(0, –1)
(c) (c)
C (3, –4) G(0, –4)
(d) D (–3, –4) (d) H(–2, –4)
(iii) On the same graph, draw the trapezium KLMN with vertices:
(a)
K(–2, 4)
(b)
L(2, 4)
(c)
M(4, 2)
(d) N(–4, 2)
(iv) Name the figure formed by the square ABCD, the rectangle EFGH and the trapezium KLMN altogether.
184
(c)
Recall A line can be vertical, horizontal or inclined.
Types of Lines
Property
Vertical
Examples in Real Life
A lamp post
The horizon Horizontal
Inclined
An inclined ramp
185
2. GEOMETRY Chapter 2 - Binomial expressions Equation of lines parallel to the x axis
(a)
Observe carefully the horizontal line given below. y 3 A
2
C
B
D
1
–2
0
–1
x 1
2
3
5
4
–1 –2 –3
(b)
Write down the coordinates of the points A, B, C and D.
A( ___ , ___ ), B( ___ , ___ ), C( ___ , ___ ), D( ___ , ___ )
What do you observe about the y-coordinates of the four points?
(c)
________________________________________________________
(d)
Where does the horizontal line cut the y-axis?
NOTE: Equation of horizontal line y
In general, any horizontal line parallel to the x-axis has equation y = k where k is the y-intercept. K
0
186
y=k
x
Equation of lines parallel to the y axis
(a)
Observe carefully the vertical line given below. y P
3 2
Q
1
`R 0
–1
–2
1
x 2
3
4
5
–1 –2
S
–3
(b)
Write down the coordinates of the points P, Q, R and S.
P( ___ , ___ ), Q( ___ , ___ ), R( ___ , ___ ), S( ___ , ___ )
What do you observe about the x-coordinates of the four points?
(c)
________________________________________________________
(d)
Where does the vertical line cut the x-axis?
________________________________________________________ NOTE: Equation of vertical line
y
In general, any vertical line parallel to the y-axis has equation x = h where h is the x-intercept.
0
h
x
x= h 187
2. GEOMETRY Chapter 2 - Binomial expressions Example 1 Given the equation of each of the vertical and horizontal lines given below. (a) y
(b)
y
3 2 1 –1 0 –1
1
2
3
4
x
–1
1
2
3
4
x
–2
–2 (c) y
–2
0 –1
0
–1
(d)
1
2
3
y
x
–1 –2 –4
–3
–3
–4
–2 –1 0 –1
1
–2
Solution
(a)
x = 1
(b) y = 2
(c)
y = –3
(d)
x = –2
(c)
y = – 4
(d)
x = –6
Example 2 On the same x-y plane, draw the lines : (a) x = 5
188
(b) y = 1
x
Solution
y
y=1 x
0
y = –4 x = –6
x=5
EXERCISE 2.4.3
1.
Give the equation of each of the vertical and horizontal lines given below. y y (a) (b)
x
0 x
0
(c)
(d)
y
y
0
0
x
x
189
2. GEOMETRY Chapter 2 - Binomial expressions 2.
On the same x-y plane, draw the lines:
(a)
x = 5
(d) y = 4
3.
(b) y = 7
(c)
x = –2
(e) x = -4 (f ) y = –5
Give the equation of the y-axis and the x-axis.
Point of intersection of a vertical line and a horizontal line 1.
Study the graph below. y
A
B
x
0
D
x = -1
190
y=2
C
x=4
y = –3
2.
Copy and complete the following table.
Equation of vertical line
Equation of horizontal line
Point of intersection
x = -1
y = 2
A(
,
)
x = 4
y = 2
B(
,
)
x = 4
y = -3
C(
,
)
x = -1
y = -3
D(
,
)
Note: The point where two lines meet is called the point of intersection.
3.
What do you observe ?
Point of intersection of a vertical line and a horizontal line From the previous activity, you must have observed that the point of intersection of a vertical line x = h and a horizontal line y = k has coordinates (h, k). y (h, k)
y= k
x
0
x= h
191
2. GEOMETRY Chapter 2 - Binomial expressions Example Write down the coordinates of the point of intersection of the vertical line and horizontal line in each of the following. (a)
(b)
y
y
x
0
x
0
Solution (a) (1, 2)
(b) (– 2, – 3)
EXERCISE 2.4.4
Write down the coordinates of the point of intrsection of the vertical line and horizontal line in each of the following. y
y
0
x
0
192
x
Continuous Assessment 2.4
1. Label the x and y axes
2. Write down the coordinates of the following points.
3. Plot the following points. P(1, 3), Q(0, 2), R(-2, 2), S(-3, 0), T(-3, -1), U(0, -2), V(2, -3), W(2, 0)
193
2. GEOMETRY Chapter 2 - Binomial expressions 4.
Write down the equation of the following lines. (a)
(c)
(b)
(d)
5. Write down the equation of the (i) x-axis (ii) y-axis. 6. The diagram below shows the lines x = -2, x = 3, y = -1 and y = 2. R, S T and U are points of intersection of the lines. Find the coordinates of the points R, S, T and U.
7. Write down the point of intersection of the lines
194
x = 2 and y = -3
(a)
(b) x = - 5 and y = 15
(c)
x-axis and y = 6
2.5 GEOMETRICAL CONSTRUCTION By the end of this topic you should be able to: • Measure the length of a line segment. • Draw a line segment of given length. • Draw parallel lines. • Bisect a line segment. • Bisect an angle.
This building is beautiful. What a shape !
A B
D
C
Observe the picture carefully. What can you say about the length of the line segments AB and DC? Are they parallel? Which sides are equal? How would you measure the length of line AB? How do you draw parallel lines?
195
2. GEOMETRY Chapter 2 - Binomial expressions Geometrical construction in real life Geometry is present all around us. Some examples are given below.
Geometry in Architecture
Geometry in students' playground
196
Geometry on the road
Geometrical instruments The table below shows some geometrical instruments that are needed to draw and measure line segments and parallel lines. Instruments needed for accurate drawing are ruler, set square, protractor, pair of compasses and divider.
Rulers
Set squares
Pair of compasses
A ruler is used to measure and draw line segments.
2 types of set squares are used to draw parallel lines.
A pair of compasses is used to draw lines and bisect angles.
Activity Measure and write down the length of the line segments below. (a)
A
(b)
B C
AB = _____ cm CD = _____cm I
D (c)
E
FE = ______cm
F (d)
(e) G
H
GH = ______cm
J IJ = ______cm 197
2. GEOMETRY Chapter 2 - Binomial expressions
RECALL
Construction of a line segment To draw a line segment, we either use a ruler or a pair of compasses together with a pencil.
A line is straight and extends in both directions without ending. It has no thickness. A line segment is part of a line and has two end points. It has a length. P
Q
NOTE: Line and Segment are used interchangeably in this topic.
Example Draw a line segment of length 3.5 cm.
Method 1
Method 2
Step 1: Mark a point P in your copy book.
Step 1: Draw a straight line and mark a point P at the beginning of the line as shown below.
P
∙
P
Step 2: Using a ruler, measure a distance of 3.5 cm and mark the point as Q. 0
3.5 cm
P
Step 2: Place the pointer of the pair of compasses on the mark 0 of the ruler and measure 3.5 cm as shown below.
Q
Step 3: Join the points P and Q to form a line 3.5 cm 0 segment. Step 3: Now place the pointer of the pair of compasses on the point P and draw an arc to cut the line as indicated below. 3.5 cm P Q P Line segment PQ = 3.5 cm
Step 4: Mark this cut as Q. You get a line segment of 3.5 cm as shown below. 3.5 cm P
Q
EXERCISE 2.5.1
1. 2. 198
Using a pair of compasses, construct the following line segments. (a) 5 cm (b) 6.4 cm (c) 12.5 cm Construct a line segment AB of length 75 mm using a pair of compasses. (a) Mark the point X where AX is 2.5 cm. (b) Measure and write down the length of XB.
Construction of a line parallel to a given line. To construct a line segment ST, parallel to a given line GH.
Step 1 Consider the line segment GH below.
G
Step 2 Place the set square on the line GH and the ruler perpendicular to the line GH as shown below.
RECALL Parallel lines are lines that never meet. Example:
H
G
H
Step 3 Step 4 Slide the set square along the ruler Draw a line ST parallel to the line to the position where you need to GH as shown below. draw the parallel line. See below.
Slide
G
H
S
T
G
H
Thus, we get line segment ST parallel to the line GH. S
T
G
H
199
2. GEOMETRY Chapter 2 - Binomial expressions EXERCISE 2.5.2
(a) Use set square and ruler to construct a line parallel to each of the given lines. (a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
Construction of the perpendicular bisector of a line. A perpendicular bisector (e.g., PQ) is a line that divides a line segment (e.g., AB) into two equal parts and makes a right angle to the line segment.
P
A
B
Perpendicular bisector of AB
Q
200
Steps to construct the perpendicular bisector of line AB are as follows: Consider the line AB. A
B
Step 1 Open the pair of compasses greater than half the length of the line AB. Using A as centre draw two arcs (one above the line AB and one below the line AB)
Step 2 Keep the same measurement set on the pair of compasses at step 1. Use B as the centre and draw two other arcs such that they cut the first two arcs formed. Mark their points of intersection as C and D. C
A
B
A
B
D
Step 3 Use a ruler to join the points C and D. Mark the intersection of the two lines with the letter O.
Step 4 Measure angle COB and angle COA. Measure the length of AO and OB. What do you notice? C
C
A
O
B
D
A
O
B
D
CD cuts AB into two equal parts and is perpendicular to AB. CD is called the perpendicular bisector of AB.
201
2. GEOMETRY Chapter 2 - Binomial expressions EXERCISE 2.5.3
1. Use a pair of compasses to construct the perpendicular bisector of each of the given lines. (a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
2. Draw a line segment FG of length 5 cm. Construct the perpendicular bisector of FG. 3. Draw a line segment NM of length 8 cm. Construct the perpendicular bisector of NM.
Construction of an angle bisector. An Angle Bisector is the line segment that divides an angle into two equal parts. A
D
B
C
Line BD is the angle bisector which divides angle ABC into two equal parts. Follow the steps below to draw the bisector of an angle. Construct angle ABC = 600 and its angle bisector.
202
Step 2 Place the pointer of the pair of compasses on point B and draw arcs on BA and on BC. Label the points of intersection of the arcs on lines AB and AC as T and S respectively.
Step 1 Use the protractor to draw angle ABC (600).
A A T S
B
C
C
B
Step 3 Step 4 (a) Place the pointer of the pair of compasses Join the point Z to point B. on T and draw an arc between BA and BC. (b)Repeat the process with the pointer on S and label the resulting point of intersection of the arcs as Z. A
part (a) A
Angle bisector of
Z
T
C B
Z
T
S
B
ABC
C
S
part (b) Step 5 Measure angle ABZ and angle CBZ. What do you notice? The line BZ divides the angle ABC into 2 equal parts. That is, BZ bisects angle ABC.
A Z T 300
Angle bisector of
ABC
30
0
B
S
C
203
2. GEOMETRY Chapter 2 - Binomial expressions EXERCISE 2.5.4
Draw the angle bisector of each of the following angles.
204
(a)
(b)
(c)
(e)
(f )
(g)
Continuous Assessment 2.5 1.
Measure and write down the length of the following line segment.
A
(a)
B
AB = _______ cm D
(b) C
CD = _______ cm
2.
Using a pair of compasses, construct the following line segments.
(a) 4 cm
3.
Construct a line segment RQ of length 60 mm using a pair of compasses. (a) Mark the point G where RG is 2.5 cm. (b) Measure the length GQ.
4.
Use set square and ruler to construct a line parallel to each of the given lines.
(b) 8 cm
(c) 10.5 cm
(a)
(b)
(c)
(d)
5.
Draw the perpendicular bisector of the following line segments.
(a)
(b)
(c)
205
2. GEOMETRY Chapter 2 - Binomial expressions 6. (a)
(d)
206
Draw the angle bisector of each of the following angles.
(b)
(c)
2.6 SYMMETRY By the end of this chapter, you should be able to: • Determine the number of lines of symmetry in plane shapes. • Locate and draw the line(s) of symmetry in a given shape. • Complete plane figures given line(s) of symmetry (horizontal, vertical and slant).
RECALL - Symmetry
If a rectangular piece of bristol paper is folded in half along a vertical line or a horizontal line as shown in the following diagrams, the two halves fit exactly on top of one another.
Fold
Fold When a 2-D shape is divided along a line such that the two halves fit exactly on top of each other, the shape is said to have line symmetry. The line is known as the line of symmetry or the mirror line and is represented by a dotted line.
Isosceles triangle
Rectangle
Letter H
207
2. GEOMETRY Chapter 2 - Binomial expressions RECALL - Symmetry
A 2-D shape may have more than 2 lines of symmetry which can be vertical, horizontal or inclined. The number of lines of symmetry of a regular polygon is equal to its number of sides. Examples of regular polygon are equilateral triangle and square.
An equilateral triangle has 3 lines of symmetry.
A square has four lines of symmetry.
Example Locate and draw the line(s) of symmetry, if any, of the following shapes. (a)
(b) (c)
Solution (a)
(b) (c)
No line of symmetry
208
EXERCISE 2.6.1 1.
Locate and draw the line(s) of symmetry of the following regular polygons.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
209
2. GEOMETRY Chapter 2 - Binomial expressions 2.
Locate and draw the line(s) of symmetry, if any, of the following shapes.
(a)
(b)
(c)
(d)
(e)
(f)
RECALL
It is easy to complete symmetrical figures using gridlines. Two examples are given below. (a) (b)
210
Solution (a) (b)
Example Complete each of the symmetrical shapes given below. (a) (b)
Solution (a) yy Points on the mirror line do not move. yy The mirror line is inclined. yy We measure the perpendicular distance (number of diagonals) from each point to the mirror line. yy We count the same number of diagonals on the opposite side of the line of symmetry to obtain the image of each point. yy We complete the symmetrical figure.
211
2. GEOMETRY Chapter 2 - Binomial expressions (b)
yy Points on the line of symmetry do not move. yy The line of symmetry is inclined. yy We measure the perpendicular distance (number of diagonals) from each point to the line of symmetry. yy We count the same number of diagonals on the opposite side of the line of symmetry to obtain the image of each point. yy We complete the symmetrical figure.
EXERCISE 2.6.2 Complete each of the symmetrical shapes given below.
212
(a) (a)
(b) (b)
(c) (c)
(d) (d)
Continuous Assessment 2.6 1. State the number of line(s) of symmetry, if any, of following plane shapes. (a)
(b)
(c)
2. Draw the lines of symmetry of the following regular polygons. (a)
(b)
(c)
3. Draw the line(s) of symmetry, if any, of the following figures. (a)
(b)
(c)
(e)
(f)
(d)
213
2. GEOMETRY Chapter 2 - Binomial expressions 4. Complete each of the symmetrical shapes given below. (a)
(b)
5. Complete each of the symmetrical shapes given below. (a)
(b)
6. Complete each of the symmetrical shapes given below. (a)
(b)
7. State the number of line(s) of symmetry of: (a) a square,
214
(b) a parallelogram,
(c) a kite,
(d) a rhombus.
2.7 REFLECTION By the end of this topic, you should be able to: • Demonstrate an understanding of the notion of reflection. • Reflect points, line segments and polygons in lines of reflection. • Reflect figures and shapes in lines (horizontal, vertical and inclined) of reflection. • Locate the line of reflection given the object and the image (through construction).
Dad! What is this word? I cannot read it.
Have a look in the rearview mirror. AMBULANCE
ECNALUBMA
Discuss what you see in the picture. Why is the word “AMBULANCE” written as “ What happened to the word in the mirror?
”?
Activity Reflection in a mirror (Group Activity) Materials needed: pencil, ruler, scissors, square mirror (one for each group), 2 rectangular pieces of bristol paper (for each group), sticky stuff. 1. Use a pencil and a ruler to draw a scalene triangle and a parallelogram on the first piece of bristol paper.
2. Cut out each 2-D shapes using a pair of scissors.
215
2. GEOMETRY Chapter 2 - Binomial expressions Activity 3. Now, draw a straight-line AB in the middle of the second piece of bristol paper using a pencil and a ruler. A
Making groups of 10
B 4. Stick your triangle on one side of the line AB on the bristol paper using blu tack. Next, place the mirror along AB facing your 2-D shape and take a look at the image in the mirror. What do you observe? A
Mirror
B 5. Repeat steps 4 and 5 using your parallelogram and answer the following questions in each case.
(a) Which shape do you see in the mirror?
_____________________________________________________
(b) Compare the size of your 2-D shape and that of its image in the mirror?
_____________________________________________________
(c) What can you say about the orientation of your 2-D shape and that of its image in the mirror?
_____________________________________________________
(d) Compare the distance of your 2-D shape and that of its image from the line AB.
_____________________________________________________
216
Note: Orientation is how an object is placed in the space it occupies.
DID YOU KNOW
Every day we encounter reflection almost everywhere. For example, in rear view mirrors, lakes, glass doors and so on.
Reflection in a rear view mirror
Reflection in glass doors
Reflection in a lake
Reflection is a process which flips an object over a mirror line or a line of reflection resulting in a mirror image.
Image
Object
The object and its image have the same size and shape though the image is laterally inverted (i.e., it faces the opposite direction). The perpendicular distance of the object from the mirror line is equal to the perpendicular distance of the image from the mirror line.
217
2. GEOMETRY Chapter 2 - Binomial expressions Example Reflect the point A(3, 2) in the y-axis. Solution yy We draw our x - y plane and plot the point A(3, 2).
y
A x
O y
yy We draw a dotted line from the point A such that it cuts our mirror line, the y-axis, at right angle, say, at the point M.
M
yy We extend this dotted line beyond the mirror line.
x
O y
yy We count the number of squares from A to M. yy We have 3 squares. yy We count the same number of squares from M on the extended dotted line and label this point A1.
x
O y
A1
M O
218
A
M
yy The point A1 (–3, 2) is the image of A (3, 2) under a reflection in the y-axis. yy The perpendicular distance of the point A from the mirror line is equal to the perpendicular distance of its image A1 from the mirror line.
A
A x
EXERCISE 2.7.1
1. Answer the following questions on separate sheets of graph paper. (a) Reflect triangle ABC in the x-axis.
(b) Reflect trapezium KLMN in the y-axis. y
y
L
B K
A
C x
O
x
O N M
(c) Reflect kite PQRS in the line x = 2. y
(d) Reflect rectangle EFGH in the line y = 3. y
x=2 Q P
E
F
H
G
R x
O
y=3
S O
x
2. Answer the following questions on separate sheets of graph. (a) Reflect square ABCD in the y-axis given A(–3, 3), B(–1, 3), C(–1, 1) and D(–3, 1). (b) Reflect parallelogram PQRS in the x-axis given P(1,0), Q(4,0), R(3,2) and S(0,2). (c) Reflect triangle EFG in the line y = 2 given E(–3, 2), F(1, 3) and G(3, 5). (d) Reflect kite KLMN in the line x = 3 given K(2, –2), L(3, –1), M(1, –1) and N(2, 3).
219
2. GEOMETRY Chapter 2 - Binomial expressions Example Reflect the triangle ABC in the line x = –3 given A(–5, 4), B(–5, 0) and C(–2, 2). y
A
y
A
Solution Triangle A1 B1 C1 is the image of triangle ABC under a reflection in the line x = -3 y A 1 A A
C
C
C1
x = –3
B
x
O
B
C
C1
C
B
B1 O
B
x
O
y
A1
x
B1 O
x = –3
x
x = –3
x = –3 EXERCISE 2.7.2
1. Answer the following questions on separate sheets of graph paper. (a) Reflect KLMN in the line x = –4. x = –4
x = –4
L K
K
y
(b) Reflect rectangle EFGH in the line y = –3. y
y
y
F
O
L
EE
M M
x
F
O
x
G G –3 yy == –3 HH
N N
O O
x x
2. Answer the following questions on separate sheets of graph
(a) Reflect triangle LMN in the line y = –4 given L(1, –3), M(2, –5) and N(5, –3). (b) Reflect kite ABCD in the line x = –3 given A(–3, 5), B(–2, 6), C(–1, 5) and D(–2, 2).
220
Reflecting a triangle in an inclined line Example Reflect the triangle ABC in the line PQ given A(0, 1), B(1, 1) and C(1, 3). Solution y
yy We draw our x-y plane together with our triangle ABC and the line PQ.
Q
C
yy The point B is on the mirror line and so it does not move. A
B
x
O P
yy The mirror line is inclined.
y
yy Instead of counting the number of squares, we count the number of diagonals.
Q C
yy We count one diagonal along the dotted line from the point C to the mirror line.
A
yy So, we move one diagonal along the extended dotted line to locate the image of the point C.
B
C1
x
O P y
yy We label it C1. yy We count half diagonal along the dotted line from the point A to the mirror line.
Q C
yy We move half diagonal along the extended dotted line to locate the image of the point A.
A B
yy We label it A1.
O
C1
x
A1
P
y
Q
yy We draw the triangle A1 BC1 which is the image C
of the triangle ABC under a reflection in the line PQ. A
B
O
A1
C1
x
221
y O
2. GEOMETRY Chapter 2 - Binomial expressions
Reflect the triangle KLM in the line AB given K(–2, 2), L(1, 3) and M(3, 3). y
Q
x
C
Example
A
A1
L
M
Solution
A
B
O
A1
y
A
K
C1
L
x
M
K x
O y
A
L
K
M
L1 A
B
M1 K x
O
O y
x L
M B x
O L1
B
EXERCISE 2.7.3
B
M1
y questions on separate sheets of graph paper. Answer the following y=x C y = -x
(a) Reflect triangle ABC in the line y = x.
R
S
y
Q T (b) Reflect figure PQRST in the line y = - x.
B
A
P y O A
y=x x
C
y = -x
R
S
Q
T
y O
x
B P
O
222
x
O
x
Locating the mirror line using construction To locate the line of reflection or the mirror line, we draw the perpendicular bisector of the dotted line segment connecting the object and its image. y
Example
P1
Given the triangle PQR has been mapped onto the triangle P1 Q1 R1 under a reflection as shown in the diagram on the right, locate the mirror line and write down its equation.
R1 y
R P
P1
O
R1
Solution yy We draw, say, the dotted line RR1.
P
O
yy We draw the perpendicular bisector of the dotted line RR1.
O
y (a) Draw the mirror line in each of the following. A
B y B1 A
B
A1
B1
C
Q
P
y G y
(b)
D
CO
D1 x D
C1
E1
E
G
(c)
(c)
y
(d)
(d)
Q Q11
R
O
O
R
F
K
K
QQ
P1
P1
O
M
M
x
x
11 1
L1
11 1
L1
M1
y
y L
L
x
x
O
y
PP
F
F1
x
O
E
F1
1
x
x
(b)
A1
C1
Q
y=2
E1
D
Q1
Q1
P
R
EXERCISE 2.7.4
x
y=2
R
R1
O
x
R1 P1
The equation of the mirror line is y = 2.
Q
P1
y
yy The perpendicular bisector of the dotted line RR1 is the mirror line.
Q
Q1
y
R
(a)
Q1
M1
O
x
O
x 223
2. GEOMETRY Chapter 2 - Binomial expressions Continuous Assessment 2.7 1. Which of the following shows reflection?
AA
B
B
2. Answer the following questions on separate sheets of graph paper. (a) Reflect line segment AB in the y-axis. y
(b) Reflect triangle UVW in the line. x = –3. y
x = –3 U
A
x
O
V
W
x
O B
(c) Reflect flag PQRST in the line y = 1. y
R
S
Q
T
P
(d) Reflect rectangle KLMN in the line y = x. y
y=1
y=x
K
L
N
M
x
O
x
O
2. Write down the equation of the line of reflection in each of the following cases. (a) (b) y y A
P
B
D
C O C1
D1 A1
224
B1
x
S O
Q
R
Q1
P1
R1
S1 x
MEASUREMENT
3
3.1 AREA By the end of this topic, you should be able to: • Recall area of triangle, rectangle, parallelogram and trapezium. • Calculate area of kite and rhombus. • Convert metre square to centimetre square and vice versa. • Convert centimetre square to millimetre square and vice versa. • Convert kilometre square to metre square and vice versa. • Convert hectare into metre square. • Differentiate between hectare, arpent,perche and toise.
225
3. MEASUREMENT Chapter 2 - Binomial expressions In grades 7 and 8 we calculated area of quadrilaterals (Squares, Rectangles, Trapeziums and Parallelograms). In this topic we will calculate area of Rhombus and Kite. Area is the amount of surface inside a 2D shape. It is measured in unit square (e.g., cm2, km2). Area helps to calculate how many tiles are needed to cover a surface. Shapes
Formula
Rectangle
Area = length × breadth =lxb b
Examples l = 10 cm
b = 5 cm
Area = length x breadth = 5 × 10 = 50 cm2
l
Paralellogram
Area = base × height
A
=b×h
B
10 cm
h
C Base
Triangle
40 cm
D
Area = 40 × 10 = 400 cm2 Area =
× base × height ×b×h
height
10 cm
=
1 2 1 2
30 cm
Area = 12 × 30 × 10 = 150 cm2
Base
Trapezium
a
h
Area =
1 2
=
1 2
sum of parallel sides
6 cm
× height
h=8 cm
(a + b) h
10 cm b
Area =
1 2
× (6 + 10 ) x 8
= 12 x 16 × 8 = 64 cm2 226
EXERCISE 3.1.1
Find the area of the following shapes a)
7 cm
b)
8 cm
c)
6 cm
3 cm
7 cm
8 cm
d) e)
f) 4 cm
9 cm 3 cm
6 cm 8 cm
13 cm
4 cm 13 cm
8 cm
10 cm
KITE Properties of a kite A kite has 2 pairs of equal adjacent sides. It is symmetrical about the longer diagonal. B AB and BC are adjacent and are equal in length. A
C
AB = BC AD and DC are adjacent and are equal in length. AD = DC.
D
227
3. MEASUREMENT Chapter 2 - Binomial expressions Area of a Kite Consider the following kite. B
B
A
C
B
A
C
A
C
D
D
D
Observe that the kite can fit in only half of the rectangle ABCD. Area of kite = half Area of rectangle. = =
1 2 1 2
× AC × BD x product of diagonals of kite. B product of diagonals
Thus Area of kite =
1 2
×
=
1 2 1 2
× AC × BD
=
A
h
×a×h D
Example Find the area of the following shapes. (i) B
Q
(ii)
6m
8m
A
C
P
R 10 m
S
20 m
D Solution Area = 228
1 2
× 8 × 20
= 4 × 20 = 80 m2
Area =
1 2 1 2
× PR × QS
= × 10 × 6 = 30 m2
a
C
EXERCISE 3.1.2
1. Find the area of the following kites. B
(a)
A
(b)
5m
(c) Q
T
4m
C P
14 m
R W S
8m
U
9m
D
V
8m
2. Given that the area of each of the following kite is 36 cm2, find x. (a) (b)
x cm
12 m
8m
x cm
Area of Rhombus A rhombus has the following features: All sides are equal, that is, AB = BD = DC = AC yy Opposite sides are equal and parallel. yy Opposite angles are equal but not necessarily 900. yy Diagonals bisect each other at right angles (AD and BC are diagonals). A
B
A
B
height C
D
C
base
D
229
3. MEASUREMENT Chapter 2 - Binomial expressions Area of Rhombus A
B
A
B
8 cm
D
X
8 cm
C
10 cm
D
10 cm
C
Y
If we extract the triangle ADX and stick it on the right side of the rhombus, we will obtain a rectangle with base 10 cm and height 8 cm Therefore, area of Rhombus = base × height
= 10 × 8 = 80 cm2 Note: A rhombus is also a special type of kite, we may thus apply the formula for kite to find area of rhombus. B A
B
Rotate
A
C
Area = D
1 2
× AC × BD
C D
Thus, a rhombus may be considered as a kite or a parallelogram.
Example Find the area of rhombus ABCD. (a) (b) A B
30 cm
A
B
40 cm
10 cm
D
X
60 cm
C
D
Area = base × height
Area =
= DC × AX
=
= 60 × 10
=
= 600 cm2
=
230
C 1 2 1 2 1 2 1 2
(product of diagonals) × AC × BD × 30 × 40 × 1200
= 600 cm2
EXERCISE 3.1.3
1. Calculate the area of the following rhombuses. A
B
A
B
5 cm
D
C
7 cm
D
C
AC = 10 cm and DB = 13 cm 2. Calculate area of the following rhombuses. a)
b)
c)
12 cm
d) 8 cm
10 cm
8 cm
5 cm
15 cm
10 cm
6 cm
Conversion of units of Area RECALL
Measurement of length The metre rule
10 cm
0
20 cm
30 cm
40 cm
50 cm
60 cm
70 cm
80 cm
90 cm
100 cm
With the help of your teacher, answer the following questions:(a) How many centimetres are there in the metre? (b) On the metre ruler identify the following lengths:
20 cm, 50 cm, 75 cm, 43 cm and 92 cm.
(c) From the above metre rule, we observe that there are 100 cm in 1 metre. 1m = 100 cm 231
3. MEASUREMENT Chapter 2 - Binomial expressions Converting metre square ( m2) to centimetre square (cm2) Consider a square of size 1 m or 100 cm. Note: Area is measured in m2, cm2, mm2, km2 1m (100 cm)
(100 cm) 1m
Area = 1 × 1 = 1 m2 or Area = 100 × 100
= 10 000 cm2
Observe that the area of the square is calculated in different units. Therefore 1 m2 = 10 000 cm2
Example 1 : Convert 6 m2 into cm2. 6 m2 can be represented as follows: 1m
1m
1m
1m
1m
1m
1 m 10000 cm2 1 m 10000 cm2 1 m 10000 cm2 1 m 10000 cm2 1 m 10000 cm2 1 m 10000 cm2
Area = 6 m2 = 6 × 10000 cm2 = 60000 cm2
Example 2 : Convert 5 m2 into cm2. 5m
1m
232
5 m2 = 5 × 10000 cm2
= 50000 cm2
Converting centimetre square (cm2) to metre square ( m2) 1 cm
RECALL
1 cm2
1 cm
Recall: - 1 m = 100 cm 100 cm = 1 m 1
Therefore 1 cm = 100 m
Area = 1 × 1 cm2
= 1 cm2
Area = =
1 1 × 100 m2 100 1 m2 10000
Note: 1 cm2 = 1
10000
m2
Example 1 : Convert 20000 cm2 into m2. 1 cm2 =
1 10000
m2
20000 cm2 = 20000 ×
1 10000
m2
= 2 m2
Note: To convert cm2 to m2, we divide by 10000 or 1 multiply by 10000 .
The following diagram shows how to convert m2 to cm2 and vice versa. × 10000 1 m2
10000 cm2
×
1 10000
Example 2 : Convert the following into cm2. (a) 4 m2 (b) 7.5 m2 1 m2 = 10 000 cm2 1 m2 = 10 000 cm2 Then 4 m2 = 4 × 10 000 cm2 then 7.5 m2 = 7.5 × 10 000 cm2 = 40 000 cm2 = 75 000 cm2
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3. MEASUREMENT Chapter 2 - Binomial expressions Example 3 : Convert the following into m2. (a) 80 000 cm2
1 cm2 =
(b) 105 000 cm2
1 10 000
m2
80 000 cm2 = 80000 × = 8 m2
1 10 000
m2
1 10 000
1 cm2 =
105 000 cm2 = 105 000 × = 10.5 m2
m2 1 10 000
m2
EXERCISE 3.1.4
1. Circle the correct answer. (a) 3 m2 = __________ cm2
A 300
B 30
C 0.00003
D 30 000
(b) 9.7 m2 = _________ cm2
A 9700
B 97000
C 970
D 0.00097
(c) 73500 cm2 = _________m2
A 735
B 73.5
C 7.35
D 0.0735
(d) 115000 cm2 = _______ m2
A 11.5
B 1150
2. Convert the following into cm2. (b) 13 m2 (a) 7 m2 (e) 24.5 m2 (d) 16.8 m2
C 11500
D 115
(c) 22 m2 (f ) 1 m2 2
3. Convert the following into m2.
(a) 90 000 cm2
(b) 120 000 cm2
(c) 145 000 cm2
(d) 55 000 cm2
(e) 175 000 cm2
(f ) 2500 cm2
4. The length of a football playground is 90 m and its width is 70 m. Find the area of the playground in 90 m (a) m2 (b) cm2
234
70 m
Converting centimetre square (cm2) to millimetre square (mm2) X
0
1 cm
2 cm
3 cm
4 cm
With the help of your teacher observe the above picture and answer the following: (a) How many small intervals can you see from 0 cm to 1 cm? (b) What is the reading (in mm) pointed at X ? Observe that each 1 cm is divided into 10 equal intervals and each interval is 1 mm. Thus 1 cm = 10 mm
1 cm (10 mm) 1 cm (10 mm)
Area
= 1 cm × 1 cm
= 1 cm2
or Area = 10 mm × 10 mm 1 cm
2
= 100 mm2
That is, 1 cm2 = 100 mm2
The following diagram shows how to convert cm2 to mm2 and vice versa. × 100 cm2
mm2
1 x 100
Example : Convert the following into mm2. (a) 5 cm2 (b) 4.3 cm2 1 cm2 = 100 mm2 1 cm2 = 100 mm2 Then 5 cm2 = 5 × 100 mm2 then 4.3 cm2 = 4.3 × 100 mm2 = 500 mm2 = 430 mm2 235
3. MEASUREMENT Chapter 2 - Binomial expressions Converting millimetre square (mm2) to centimetre square (cm2) Example : Convert the following into cm2. (a) 900 mm2
(b) 1850 mm2
1 1 mm2 = 100 cm2
1 900 mm2 = 900 × 100 cm2
1 mm2 =
1 100
cm2
1 1850 mm2 = 1850 × 100 cm2
= 9 cm2 = 18.5 cm2
EXERCISE 3.1.5
1. Circle the correct answer. (a) 6 cm2 = __________ mm2.
(b) 8.9 cm2 = _________ mm2.
A 60
A 89000
B 600
C 6000
B 89
C 890
D 60 000
D 8900
(c) 700 mm2 = _________ cm2. A 7
B 0.7
C 0.07
D 70
C 69
D 6.9
(d) 690 mm2 = _______ cm2. A 0.069
B 0.69
2. Convert the following into mm2.
(a) 4 cm2
(b) 11 cm2 (c) 17 cm2
(d) 5.8 cm2
(e) 12.5 cm2 (f )
3 4
cm2
3. Convert the following into cm2.
(a) 800 mm2 (b) 1300 mm2 (c) 2200 mm2 (d)1450 mm2 (e) 2270 mm2 (f ) 25 mm2
4. A birthday card has a length of 20 cm and a breadth of 10 cm. Find the area of the card in
(a) cm2
(b) mm2
20 cm
10 cm
236
Converting kilometre square (km2) to metre square (m2) A man walks from La Croisette shopping mall to Super U shopping mall in Grand Bay. The distance between these two places is 1 km apart. We can also say that the man has walked 1000 m, as 1 km is equal to 1000 m. Super U shopping mall
1 km m 1000
La Croisette shopping mall
1 km= 1000 m
Consider a square of length 1 km that is, 1000 m. 1km (1000 m)
Area
= 1 km × 1 km
= 1 km2
= 1 000 000 m2
1km 1km2 or Area = 1000 m × 1000 m (1000 m)
Thus 1 km2 = 1 000 000 m2
The following diagram shows how to convert km2 to m2 and vice versa. 1 km2
× 1 000 000 x
1 000 000 m2
1 1000 000
Example 1 : Convert the following into m2. (a) 6 km2 (b) 3.2 km2 1 km2 = 1000 000 m2 1 km2 = 1000 000 m2 Then 6 km2 = 6 × 1000 000 m2 then 3.2 km2 = 3.2 × 1000 000 m2 = 6 000 000 m2 = 3 200 000 m2
Example 2 : Convert the following into km2. (a) 10 000 000 m2
1 m2 =
(b) 9 400 000 m2 1 1000000
km2
1 m2
=
1 1000000
km2
10 000 000 m2 = 10 000 000 × 1 0001 000 km2 9 400 000 m2 = 9 400 000 × 1 0001 000 km2 = 9.4 km2 = 10 km2 237
3. MEASUREMENT Chapter 2 - Binomial expressions EXERCISE 3.1.6
1. Circle the correct answer. (a) 2 km2 = __________ m2
A 2 00 000
B 2 000 000
C 20 000
D 2000
(b) 6.3 km2 = _________m2
A 6 300 000
B 6300
C 630
D 63
(c) 1 500 000 m2 = _________ km2 A 1500
B 150
C 1.5
D 15 000
(d) 4 250 000 m2 = _______ km2
A 4250
B 425
C 42500
D 4.25
2. Convert the following into m2.
(a) 8 km2
(e) 7.1 km2
(b) 4 km2
(d) 5.8 km2
(c) 17 000 000 m2
(d) 1 400 000 m2
(f ) 2 15 km2
3. Convert the following into km2. (a) 8 000 000 m2 (b) 13 000 000 m2 (f) 3 125 000 m2 (e) 2 250 000 m2 4.
(c) 12 km2
The picture shows a residential land of length 8 km and width 6 km. Calculate the area of the land in (a) km2 (b) m2.
HECTARE (ha) An hectare is a unit of area. It is widely used for the measurement of large areas of land.
1 ha
100 m
100 m
The area of a square of side 100 m is one hectare. 238
1 ha = 100 × 100 m2 = 10 000 m2
thus
1 ha = 10 000 m2
The following diagram shows how to convert hectare to m2 and vice versa. × 10000 1 ha 10000 m2 ×
1 10 000
Example 1 : Convert the following into m2. (a) 5 ha (b 7.5 ha 1 ha = 10 000 m2 7.5 ha = 7.5 × 10000 m2 5 ha = 5 × 10000 m2 = 75 000 m2 = 50000 m2
Example 2 : Convert the following into ha. (a) 50 000 m2
(b) 75 000 m2
=
1 ha 10 000
1 m2
50 000 m2 = 50 000 ×
1 ha 10 000
1 m2
=
1 ha 10 000
75000 m2 = 75 000 ×
= 5 ha
1 ha 10 000
= 7.5 ha
EXERCISE 3.1.7
1. Circle the correct answer.
(a) 6 ha = __________ m2
A 600
(b) 9.1 ha = _________m2
A 910
(c) 80000 m2 = _________ ha
A 8
B 6 000
B 91 000
B 80
(d) 122000 m2= _______ ha A 0.122 B 1.22
C 60 000
C 9100
C 800
C 122
D 600 000
D 910 000
D 0.8
D 12.2 239
3. MEASUREMENT Chapter 2 - Binomial expressions 2. Convert the following into m2. (a) 4 ha (b)11 ha (c)17 ha
(d) 3.9 ha
(e) 10.8 ha
(f ) 1 ha 2
3. Convert the following into ha. (a) 90000 m2 (b) 130000 m2 (c) 65000 m2 (d)113000 m2 (e) 167000 m2 (f ) 5000 m2 4. The picture shows the top view of SSR international airport. Calculate the area of the land in (b) ha (a) m2 3000 m
2500 m
Other units of measure DID YOU KNOW 1 millimetre
1 centimetre
The foot was a common unit of measurement throughout Europe. It is Henry I of England that passed the law that the foot be as long as a person's own foot.
1 inc 1 foot (Not to scale)
FOOT
1 foot is approximately 30 cm (12 inch) The area of the tile is 1 ft Area = 1ft × 1ft = 1ft2
1 foot ≈ 30 cm (12 inch)
1 ft 240
TOISE It is a unit of area. It is used in measuring small plot of land.
1ft 1ft
Area = 6ft × 6ft = 36 ft2 PERCHE 1 perche = 11 1 toise 2 It is used to find the area of a plot of land. It is bigger than a toise.
Plot of land ARPENT 1 arpent = 1111 toise An arpent is bigger = 100 perche than a perche.
Sugar cane field
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3. MEASUREMENT Chapter 2 - Binomial expressions Continuous Assessment 3.1 1)
Calculate the area of the following shapes. (a) (b) 3 cm 4 cm
(c)
3 cm
4 cm 2 cm
4 cm
2) Find the following area. B A
C cm 10
cm
15
D BD = 25 cm and AC = 20 cm
3) Circle the correct answer i)
5 m2 = …………….. cm2
(a) 500
(b) 50
(c) 0.00005
(d) 50 000
ii) 125000 cm2 = ……….. m2
(a) 12.5
(b) 1250
(c) 12500
(c) 11000
(d) 125
iii) 11 cm2 =…………….. mm2
(a) 110
(b) 1100
(d) 111000
Iv) 820 mm2 = ……………cm2
(a) 0.082
(b) 0.82
(c) 82
(c) 920
(d) 8.2
v) 9.2 m2 =…………. cm2
(a) 9 200000
(b) 9200
(d) 92
vi) 8.6 ha = ………….. m2
242
(a) 860
(b) 86000
(c) 8600
(d) 860 000
4) Find the following area. (a) 3 km
(b)
Area = ______km2
= ______m2
(c)
4m
2 km
5m
5 cm
Area = ______ m2
8 cm
Area = ______ cm2
= ______cm2
= ______ mm2
5) Match the area in column A with its most appropriate unit in column B.
Column A
Column B
Arpent
Perche
Toise
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3. MEASUREMENT Chapter 2 - Binomial expressions 3.2 SPEED By the end of this topic you should be able to: • Demonstrate an understanding of the terms speed and average speed. • Solve real life problems involving speed (including average speed).
The Bugatti Chiron has recently broken the top speed record for a road legal car. This car can go up to 490 km/h.
Road sign for Speed limit
Observe and discuss the pictures. What is the top speed of the car? What distance can the car cover in 1 hour? What is the meaning of 60 on the road sign? Do you use speed in your daily life activities? Where?
244
DID YOU KNOW
Speed Movement
Consider a car travelling a distance of 10 metres (m) in every second. 10m
1s
10m
1s
10m
1s
Speed (m/s)
Walking casually
10m
1s
2.0
Aeroplane taking off
83
Sound waves in air
330
Light waves in vacuum
300 000 000
Since the car is covering a distance of 10 m every second, we say that the car is moving at a speed of 10 metre per second, denoted by 10 m/s or 10 ms-1. Units
Speed is a measure of the distance travelled per unit time.
Distance Kilometre -km Metre – m Centimetre – cm
Example 1 (i)
A lorry covers a distance of 25 m every second
Its speed is 25 m/s.
Speed m/s metre per second km/h kilometre per hour
[read as 25 metre per second]
(ii) A car covers a distance of 80 km every hour
Time h hour s second
Its speed is 80 km/h. [read as 80 kilometre per hour]
(iii) A snail covers a distance of 10 cm every hour
Its speed is 10 cm/h. [ read as 10 centimetre per hour] Consider a car travelling at a speed of 60 km/h In 1 hour, distance travelled = ( 60 × 1 ) km In 2 hours, distance travelled = ( 60 × 2 ) km In 3 hours, distance travelled = ( 60 × 3 ) km
Note: D S
Distance = Speed × Time Distance Time = Distance Speed
Distance = Speed × Time T
D S
T
Time =
Distance Speed
Speed =
Distance Time
Speed = Time
D S
T
Example 2 A car travels at a constant speed of 50 km/h for 3 hours. What is the distance travelled? Solution Speed = 50 km/h Time = 3 hours Distance = Speed × time = 50 × 3 km = 150 km
Constant Speed Speed stays the same for the whole journey
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3. MEASUREMENT Chapter 2 - Binomial expressions EXERCISE 3.2.1
1.
Calculate the distance travelled in the following:
a) Speed = 50 km/h, Time taken = 4 hours.
b) Speed = 10 m/s, Time taken = 5 minutes.
c) Speed = 24 km/h, Time taken = 1 hours 20 minutes.
Average Speed Consider a bus travelling 32 km from Rose Hill to Flacq in 1 hour 15 minutes. As the bus goes by, its speed changes. It can go faster and slower and even stop at lights.
1h15 mins 32 km
In this case we calculate the average speed of the bus, that is, how far it travels over a period of time. Average speed =
total distance travelled total time taken
Thus, Distance travelled by bus = 32 km Time taken by bus = 1 h 15 mins = 11h
4 5 h = 4
total distance travelled total time taken = 32 ÷ 5 km/h
Thus, average speed =
246
4 = 32 × 4 km/h 5
= 25.6 km/h
The average speed 25.6 km/h implies that: if the bus was to travel at a constant speed (without stopping) from Rose Hill to Flacq in 1 hour 15 minutes, that speed would have been 25.6 km/h.
Example 1 John runs 12 km in 40 minutes. What is the average speed in km/h? Solution Distance = 12 km Time
= 40 minutes
= 40 hour 60
= 23 hour
Average Speed =
Distance Time
= 12 ÷ 23
= 12 × 32
= 18 km/h
Example 2 The metro travels from Rose Hill to Port Louis at an average speed of 50 km/h. If the railway track from Rose Hill to Port Louis is 15 km, what is the time taken, in minutes, by the metro to complete the journey? Solution Speed
= 50 km/h
Distance = 15 km Time
= Distance Speed
= 15 50 hour
15
= 50 × 60 minutes = 18 minutes
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3. MEASUREMENT Chapter 2 - Binomial expressions EXERCISE 3.2.2
1.
Calculate the missing items in the following table: Distance Travelled
Time Taken
Average Speed
(i)
30 km
60 km/h
(ii)
1500 m
5 m/s
(iii)
2h
42 km/h
(iv)
15 s
10 m/s
(v)
12 km
0.5 h
(vi) 600 m
120 s
2. 3. 4.
A car covered 72 km in 1 hour 30 minutes. Find its average speed in km/h. Karl walked for 4 hours at an average speed of 5 km/h. Find the distance he walked. Laxmi travelled 60 km at an average speed of 40 km/h. Find the time taken, in hour, to complete her journey.
Continuous Assessment 3.2 1.
Calculate the missing items in the following table: Distance Travelled
Average Speed
(i)
72 km
24 km/h
(ii)
600 m
4 m/s
(iii)
3h
24 km/h
(iv)
50 s
8 m/s
(v)
10 km
(vi) 200 m
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Time Taken
0.25 h 20 s
2.
A motorcyclist travelled 45 km in 1 hour 15 minutes. Find its average speed in km/h.
3. 4.
Ali cycled 2 hours at an average speed of 24 km/h. Find the distance he cycled. Sarah travelled 64 km at an average speed of 32 km/h. Find the time taken, in hour, to complete her journey.
ALGEBRA
4
4.1 Algebraic expressions By the end of this topic, you should be able to: • Use letters to represent unknown quantities. • Recognise algebraic terms, coefficients and expressions. • Perform arithmetic operations on algebraic expressions. • Expand m(x +y), where m is a rational number. • Evaluate algebraic expressions. • Solve linear equations of the form ax + b = c.
History of Algebra The word “algebra” is derived from ‘Aljebar w’al almugabalah’. It was written by the mathematician Mohammed Ibn Al Khowarizmi of Baghdad at around 825 AD.
+3=8 What is the missing number?
A. 3
A. 4
A. 5
A. 6
Rs Rs Rs Rs Rs Rs Rs Rs Rs Rs Rs Rs Rs Rs Rs
According to you, what is the correct answer? How did you calculate it? Can we replace the empty box by a letter? Why?
249
4. ALGEBRA Chapter 2 - Binomial expressions What is algebra? Algebra is a part of mathematics in which signs and letters are used to represent numbers.
With the help of your teacher solve the puzzle on your right.
Investiagtion puzzles Puzzle - Think of a number - Add three to the number - Double the result - Subtract 4 from it. - Divide the result by 2 - Subtract your original number. - Your result is….1!
Use letters to represent unknown quantities What is the missing number? − 1 = 5. The answer is 6, right? Because 6 − 1 = 5. In Algebra we do not use blank boxes.
We use a letter (usually an x or y, but any letter is fine). Thus we can write the above statement as
x − 1 = 5.
x is called an unknown as we do not know its value. Example
1. Krishtee thinks of a number and subtracts 10 from it. What is the result? Let y be the number
The result is
y − 10.
1. Zyan had some stickers. He buys 15 more. How many stickers does he have now? Let x be the number of stickers Zyan had at the beginning.
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He now has x + 5 stickers.
EXERCISE 4.1.1
Use letters to represents the unknown quantities. 1. Yuthika had some pens. For her birthday, she received 8 more as gift. How many pens does she have in all now? 2. John had some marbles. He gave 5 to his brother. How many marbles does John have now? 3. For the marathon this year, there are three times the number of runners that participated last year. What is the number of runners this year? 4. Zaina had some sweets. She shared them equally among 4 friends. How many sweets does each friend get? 5. Think of a number and square it. 6. Think of two numbers then multiply them.
ALGEBRAIC TERM Term
What is a term?
Algebraic term Consists of a coefficient and an unknown
Constant term Is simply a number Example: 5, -8, 12
unknown
Example: 3x, 5y2, 15xy2 coefficient
What are the different types of terms?
Terms
Like Terms Examples of like terms are:
Unlike Terms Examples of unlike terms are:
(a) 3x, -8x, 14 x, 0.5x
(a) 3x2, 4y, 7z (b) 3x2, -6y2, 3z2
(b) 15y2, -7y2, 12 y2, 0.3y2 (c) 2xy, 34
xy, 0.2 xy,-3xy
NOTE: Like terms have same unknown
(c) 2xy, 7wy, -5ab, 15 cd NOTE: Unlike terms have different unknown
251
4. ALGEBRA Chapter 2 - Binomial expressions EXERCISE 4.1.2
Write down 3 like terms and 3 unlike terms. Like terms :- ……………………………………………………… Unlike terms:- ……………………………………………………
ALGEBRAIC EXPRESSIONS An algebraic expression is the collection of terms.
Example of an algebraic expression. Coefficient
Note : An algebraic expression consists of terms connected by operators (+, -).
3x + 5 unknown
constant
Examples of algebraic expression: (a) 3x + 5
(b) –3y –7
(c) 2a + 3b
Activity Group the pens in box A and erasers in box B.
A All the pens are called like terms, of one type. If we denote pen by p and eraser by e, We have 5 pens plus 4 erasers, that is 5p + 4e
252
B
Group the like terms from the list below.
Example: 3x, 4y, -5x, 8x, 9y Unknown x: 3x, -5x, 8x Unknown y: 4y, 9y
EXERCISE 4.1.3
Group the like terms from the following. (a) 2x, 3y, 7x, 5xy. (b) 9xy, 8x, -3xy, y
(c) 12x2, 10y, -5x2, 7y, 2x
Performing arithmetic operations on algebraic expressions. Addition and subtraction of algebraic expressions Example 1: Simplify (a) 2a + 8a
Note : • Like terms can be added or subtracted • Unlike terms cannot be added or subtracted
(b) 9b - 5b
(c) 8a - 2a + 3a
(b) 9b - 5b = 4b
(c) 8a - 2a + 3a = 9a
Solution (a) 2a + 8a = 10a
Example 2: Simplify (a) 12v + 8 + 6v
(b) 7x + 3y - 2x - 5y
Solution (a) 12v + 8 + 6v (b) 7x + 3y - 2x - 5y = 12v + 6v + 8 = 7x - 2x + 3y – 5y = 18v + 8 = 5x - 2y
253
4. ALGEBRA Chapter 2 - Binomial expressions EXERCISE 4.1.4
1. Simplify the following expressions.
(a) 2x + 3x
(e) – 3f – 8f
(b) 7y + 6y
(c) 7p – 3p
(f ) –5m + 2m
(d) 5z – 8z
(g) –20v – 15v
(h) – 15w + 20w
2. Simplify the following expressions.
(a) 5x – 3y + 5y
(b) 8a + 5 – 2a
(d) 8 + 5t + 10
(e) –3 – 9s +4
(c) 6v + 8 w – 4 v (f ) 15h – 4 + 6h
3. Simplify the following expressions.
(a) 7xy + 3xy - 2xy
(b) 5x2 + 2x2 – 6x2
(c) –15vw + 7vw + 3 vw
Multiplication and division of algebraic expressions Example: Simplify the following expressions. (a) 5 × 3a
(b) 8 × 5 mn
(c) 15v ÷ 3
(d) 24b ÷ 8
Solution (a) 5 × 3a = 5 × 3 × a (b) 8 × 5mn = 8 × 5 × mn (c) 15v ÷ 3 = 15v = 15a
= 40 mn
(d) 24b ÷ 8 = 24b
3 15 = × v 3
8 24 = ×b 8
= 5v
= 3b
Note : 2x × 5 = 10x And 5 × 2x = 10x Thus, 2x × 5 = 5 × 2x EXERCISE 4.1.5
1.
Simplify the following expressions.
(a) 6 × 4a
(b) 7cd × 4
(c) 3 × 12t2
(d) 7 × 8 m3
2. Simplify the following expressions.
254
(a) 32a ÷ 8
(b) 63b2 ÷ 9
(c) 28ef ÷ 7
(d) 48v3 ÷ 12
Expanding algebraic expressions To expand an expression, we have to remove the brackets.
Example 1: Expand the following algebraic expressions. (a) 3(x + 5)
(b) 5(y – 2)
Method 1 Method 2 (a) 3(x + 5) is equivalent to looking for the area of a rectangle of sides 3 and (x + 5).
x 3 3x
3(x + 5)
(a)
= (3 × x) + (3 × 5) = 3x + 15
+5 +15
Area = 3(x + 5) = 3x + 15 (b) 5 (y - 2)
(b)
y -2
5
5y
= (5 × y) + (5 × -2) = 5y – 10
–10
5 (y – 2)
Area = 5(y -2) = 5y – 10
EXERCISE 4.1.6
Expand the following expressions. (a) 2(x + 8)
(b) 7(y + 3)
(c) 4(v – 6)
(d) 8(z – 1)
(e) 5(v - 8)
(f ) 9(3m - 2v)
(g) 3(-2t + 4v)
(h) 6(-5k - 6h)
Example 2: Expand the following algebraic expressions. (a) -2(v + 4)
(b) -6 (w - 3)
Solution (a) -2(v + 4) = (-2 × v) + (-2 × 4)
(b) -6(w -3) = (-6 × w) + (-6 × -3)
= -2v + (- 8) = -2v - 8
= -6 w + (+18) = -6w + 18 255
4. ALGEBRA Chapter 2 - Binomial expressions EXERCISE 4.1.7
Expand the following expressions. (a) –3(a + 5)
(b) –1(c + 9)
(c) –5 (d – 2)
(d) –9(e – 2)
(e) –7(3+ w)
(f ) –12(– 9 + w)
(g) –2(–3 – 4b) (h) –2(10 – 3m)
Evaluate algebraic expressions Example : Evaluate the following expressions for the given values. (a) x + 4, when x = 6
(b) x – 3, when x = 5
(d) x , when x = 6 3
(c) 2 x, when x = 3
(e) 3x + 1, when x = 4.
Solution (a) x + 4, when x = 6 (replace the value of x by 6) x +4= 6+4 = 10
(b) x – 3, when x = 5 (replace the value of x by 5) x –3=5–3 =2
(d) x when x = 6 3 (replace the value of x by 6) x = 6 3 3 =2
(e) 3x + 1, when x = 4 (replace the value of x by 4)
(c) 2 x, when x = 3 (replace the value of x by 3) 2x = (2 × 3) =6
3x + 1 = (3 × 4) + 1 = 12 + 1 = 13
EXERCISE 4.1.8
1. Evaluate the following expressions for the given values. (a) x + 3 when x = 5
256
(d) 5z when z = 8
(b) 9 - a when a = 2 (e) 10 when p = 2
p
(c) x when x = 4 2
Example : Given a = 2 and b = 4, evaluate the following expressions.
(a) a + b
(d)
b a
(b) a - b = 2 - 4 (c) 6a = 6 x 2
(d)
b 4 a = 2
(b) a – b
(c) 6a
Solutions (a) a + b = 2 + 4
= 6
= -2
= 12
(e) 3a + 2b
= 2
(e) 3a + 2b = 3 × 2 + 2 × 4
=6+ 8
=14
Additive inverse Find the missing value. =0 (b) (a) 2 +
-2 =0
Observe that 2 + -2 = 0 , We say that -2 is the additive inverse of 2. Similarly
2 - 2 = 0 , the additive inverse of -2 is 2 Study the table below.
Number
Additive inverse
-2
2
5
-5
1 2
-
3
-3
-3
3
1 2
EXERCISE 4.1.9
Find the missing additive inverse. (a) (c) 10 +
+5 =0
(b)
=0
(d) 12 +
-8= 0 =0 257
4. ALGEBRA Chapter 2 - Binomial expressions Multiplicative inverse Find the missing value × 2 = 1 (a)
(b
× -2 = 1
Observe that 1 2
1 2
× 2 = 1 we say that the multiplicative inverse of 2 is
And - 12 × -2 = 1, the multiplicative inverse of -2 is -
1 2
.
.
EXERCISE 4.1.10
1. Find the missing multiplicative inverse (a)
× 5 = 1
(c) 15 ×
=1
(b) (d) - 14 ×
2. Complete the following table Number
Multiplicative inverse
-2
2 -
1 2
8
258
-3 3
3
× 8 = 1 = 1
ALGEBRAIC EQUATIONS An equation is a mathematical statement containing ‘equal to’ (=).
What is an equation?
Activity In the scale, container B holds 5 marbles and container A holds 3 marbles and a box containing some marbles.
A
B
An equation has 2 sides, left hand side and right hand side. It can be seen as the balance where the "Left hand side = Right hand side". + 3 = 5
x + 3 = 5
Solving Linear Equation To solve an equation implies looking for the unknown.
Example: Solve the following equations. (a) x + 2 = 5
(b) a – 3 = 10
x is the unknown value here.
To obtain x, we eliminate 2 on both sides.
x + 2 - 2 = 5 – 2 (additive inverse of 2 is –2)
a is the unknown value here.
To obtain a, we eliminate –3 on both sides
a – 3 + 3 = 10 + 3 (additive inverse of –3 is 3)
x + 0 = 3 x = 3
a + 0 = 13 a = 13
EXERCISE 4.1.11
Solve the following equations. (a) x + 4 = 6
(d) m – 20 = 28
(b) x + 16 = 20
(e) b – 16 = 24
(c) x + 8 = 12
(f ) 15 + z = 35 259
4. ALGEBRA Chapter 2 - Binomial expressions Example 1: Solve the following equations. (b) x = 3 4
(a) 2x = 8 Solution
x = 3 4 1 x 2x = 8 × 1 multiplicative 4 x 1 . x = 3 × 4 4 2 2 inverse of 2 is 12 1. x = 4 1. x = 12
2x = 8
multiplicative inverse of 1 is 4 4
x = 4 x = 12 EXERCISE 4.1.12
Solve the following equations. (a) 5x = 15
(b) -8y = 24
(c) 12a = -60
(d) b = 6 8
c
(e) -6 = -4
Example 2: Solve the following equations. (a) 6 + 2 x = 10
(b) x + 3 = 8 2
Solution 6 + 2x = 10 6 - 6 + 2x = 10 - 6 (use -6 additive inverse of 6, on both sides) 2x = 4 1 × 2x = 4 × 1 2 2 (multiplicative inverse of 2 is 1 ) 2 4 x 2 = 2 2 Hence, x = 2 (a)
(b)
x + 3 = 8 2 x + 3 - 3 = 8 - 3 (use -3, additive inverse of 3, on 2 both sides.) x = 5 2 x × 2 = 5 × 2 (use 2, multiplicative inverse of 1 2 2 on both sides.) Thus, x = 10
EXERCISE 4.1.13
Solve the following equations. (a) 3 x + 20 = 29
(d) 12 - t = 18 3
(b) v + 4 = 6 2 (e) -3d -18 = 9
(c) -4m – 6 = 22 CHECK THIS LINK http://www.educationworld.com/a_ tsl/archives/07-1/lesson016.shtml
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Continuous Assessment 4.1 1. Use letters to represent the unknown quantities in the following. (a) Aditi has some balls and she received 5 more as gift. How many balls does she have in all? (b) Samy has some bottle caps, he gave 6 to his brother. How many bottle caps has he left now? (c) Koreshia had some dresses. For her birthday she got more dresses. She now has three times the number of dresses she had previously. How many dresses does she have now? (d) Arav had some toys. He shared it equally among 5 friends. How many toys does each friend get? 2. In each of the following, state the unknown, the coefficient and the constant term. (a) 2 x + 3 Unknown……… Coefficient……… Constant………
(b) -3a + 5 Unknown……… Coefficient……… Constant………
(c) -8c - 6 Unknown……… Coefficient……… Constant ………
3. Circle the like terms in the list below.
(a) 6x, 4y, -8 x, 11y (b) 2x, -3xy, 6z , 16xy (c) 3x2, 9y , 17xy, 13x2, -8y
4. Simplify the following expression.
(a) 8x + 9x
(b) 5a - 3a (c) -2c - 7c
(d) 6 × 4x
(e) 24 m ÷ 3
5. Expand and simplify the following.
(a) 2 (x + 4)
(b) 2 (a – 3) (c) –3 (d + 5)
(d) –5( –m – 2)
6. Evaluate the following expressions. (a) x + 5, when x = 3 (b) b - 6 , when b = 10 (d) 2x , when x = 6 (e) 3m2, when m = 2 3
7. Solve the following equations: (a) x + 8 = 1
(b) x – 6 = 2
(c) 3 x = 12
(c) 3d, when d = 4
(d) x = 5 4
(e) 2x + 1 = 5
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NOTES
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STATISTICS
5
5.1 REPRESENTING AND INTERPRETING INFORMATION By the end of this topic you should be able to: • Demonstrate an understanding of the term data. • Collect and classify statistical data. • Construct and use frequency table. • Calculate the mean, median and mode of set of raw data. • Draw a pie chart.
PARKING
Observe the above picture carefully. How many red cars are there? How many yellow cars are there? How can we represent the number of cars in a diagram?
263
5. STATISTICS Chapter 2 - Binomial expressions DATA What are data? Data are facts or observations collected in terms of numbers or words. For example: Data (numbers) Data (words) • Age: 12 , 13, 14, 15 years Colours: red, blue, yellow • Height: 5 m, 100 cm,150 cm Country: Mauritius, India, France • Mass: 250 mg, 300 g, 50 kg Sports: Football, Volleyball, Tennis
A survey is conducted to collect data. For example: • What are the heights of students in your class? • What is the favourite colour of students in your class? • What is the mode of transport that students in your class use to travel to school?
Collect and Classify Statistical Data
Activity 1 : Tossing a Die 1. Toss a die 30 times and record the scores obtained in the grid below.
264
Tally marks can be used to classify the data collected. Example of tallying
Each Tally mark ( | ) represents 1 unit. A horizontal Tally mark ( — ) is used to cross over four Tally marks, making it five units ( |||| ). For example: |||| || represent 7 units.
Yellow
||||
Red
||||
Blue
|||| |
Green
|
Pink
||||
4 5 6 1 4
2. Use tally marks to classify the data collected in question 1. Then, answer the following questions. Tossing of die Tally Frequency
Score 1 2 3 4 5 6
Frequency is the number of times an event occurs, or an item appears. E.g., if the score ‘4’ is obtained ten times, the frequency is 10.
Total (a) Fill in the frequency table to represent the above data. Score Frequency
1
A Frequency table shows the number of occurrences of different observations.
2
(b) What is the highest frequency? __________ (c) Are there similar frequencies? If yes, what are they? ___________ (d) Which score is most popular? ________ EXERCISE 5.1.1
1.
Cars in a Parking Lot
Maria noted the colour of cars in a parking lot as follows. silver silver grey brown brown
grey grey grey brown grey
silver grey brown brown brown
silver silver grey brown silver
grey grey brown brown grey
grey silver grey brown grey 265
5. STATISTICS Chapter 2 - Binomial expressions Use the given data on colour of cars and complete the table below. Colour of car
Tally
Frequency
Silver Grey Brown Total Answer the questions below. (a) How many brown cars were there in the parking? _______ (b) How many silver and grey cars were there altogether in the car park? ________ (c) What is the total number of cars in the parking? ___________
2. Favourite Vegetable
In a survey, patients in a clinic were asked about their favourite vegetable. Their responses are listed below. tomato carrot carrot cucumber tomato
lettuce cucumber brocoli lettuce brocoli brocoli lettuce tomato brocoli brocoli
brocoli cucumber tomato carrot tomato tomato cucumber lettuce brocoli lettuce cucumber tomato tomato carrot cucumber carrot carrot carrot carrot carrot
Fill in the table below with the given data. Then answer the following questions.
Vegetables Tomato Lettuce Broccoli Cucumber Carrot 266
Favourite Vegetables Tally
(a) Fill in the frequency table below. Favourite Vegetables
Tomato
Lettuce
Broccoli
Cucumber
Carrot
Frequency (b) How many patients preferred Lettuce? ________ (c) Which was the most favourite vegetable? ____________
A. carrot
B. tomato
C. brocoli
D. lettuce
(d) How many more patients preferred tomato than cucumber? __________ (e) How many patients were there in all? ____________
MEAN, MODE & MEDIAN MEAN Let us find out what is the Mean of a set of data.
Three children, Kevin, Vinay and Anna, each have the amount of money as shown below.
Kevin
Vinay
Anna
If the amount of money was distributed equally among them how much money would each of them have?
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5. STATISTICS Chapter 2 - Binomial expressions Now redistribute it equally among the three children.
Collect all the coins together. Complete the sharing.
Kevin
Vinay
Anna
Each child will get Rs _____. This amount is also known as the mean amount.
The mean amount of money received by each child is equal to the amount of money each child receive. From the above example: Mean: Sum of money = 12 = 4 3 Number of children
Observe the picture below.
Kevin
Vinay
Anna Before redistribution.
After redistribution, each of them recieved 4 coins. Mean = 4 (equal sharing) 268
Example 1 The temperature in 4 regions (Rose Hill, Port Louis, Curepipe and Mahebourg) were as follows; 260 C, 310 C, 220 C and 290 C. What is the mean temperature? Mean = Sum of temperature Number of regions = 26 + 31 + 22 + 29 4 = 108 4 = 270 C Therefore, the mean temperature is 270 C.
Example 2
Example 3
Find the mean of the following data set. 2, 3, 5, 6, 5
A die is tossed 7 times and the scores are as follows 4, 6, 2, 3, 5, 2, 6
Mean =
sum of values Number of values =2 +3+5+6+5 5 = 21 5 = 4.2
The mean is 4.2.
Find the mean score. Mean =
sum of scores Number of tosses
= 4+6+2+3+5+2+6 7 = 28 7 =4 The mean score is 4.
EXERCISE 5.1.2
1.
Find the mean of each of the following sets of data. a) 1, 2, 9 b) 3, 3, 4, 2, 3 c) 1, 5, 3, 4,1, 5, 2 d) 2, 2, 3, 1, 1, 2 e) 7, 9, 3, 0, 4, 0, 5
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5. STATISTICS Chapter 2 - Binomial expressions 2. Ken has 6 friends. Their ages, in years, are as follows 16, 12, 8, 14, 5, 11. Find the mean age of Ken’s friends. 3. The mean mass of 4 boys is 150 kg. a. What is the total mass of the 4 boys? b. If the mass of three of the four boys are 100 kg, 120 kg and 140 kg, what is the mass of the fourth boy?
MODE What is the Mode?
Let us discover together.
Khem tosses a die 10 times. The scores are given below. 5
3
3
4
1
3
2
3
6
6
Let us count how many times each score appeared. In ascending order: 1 2 3 3 3 3 4 5 6 6
It will be easier to count if we put the scores in order first.
The score 3 appears four times. The score 6 appears twice. The remaining scores appear only once. The score that occurs most frequently is 3. The most frequent number is known as the mode. Thus, the mode is 3.
Mode is the number that occurs the most.
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Yes, the mode is the most popular (frequent) number. To find the mode, we arrange the data set in ascending or descending order. Then choose the number having the highest frequency.
Example: Find the mode of the following data set. (a) 14, 5, 7, 5, 4, 5, 3, 10
(b) 20, 30, 35, 20, 40, 20, 30, 20
Solution: Arrange in ascending order. 3, 4, 5, 5, 5, 7, 10, 14.
Solution: Arrange in ascending order. 20, 20, 20, 20, 30, 30, 35, 40
Mode = 5 as it occurs the most.
Mode = 20 as it occurs the most.
c) 4, 7, 5, 10, 12, 3 Solution: Arrange in ascending order. 3, 4, 5, 7, 10, 12
No mode All number occur equally once.
NOTE: If each value appears only once, then there is no mode.
CAUTION: If there is no mode, this does not imply that, mode is equal to 0.
EXERCISE 5.1.3
1 Circle the correct answer.
(a) Find the mode of 3, 7, 1, 6, 4, 7.
A. 1
(b) What is the mode of 23, 12, 6, 7, 21?
A. 0
2
Find the mode of each of the following.
(a) 2, 9, 0, 9, 7, 0, 6, 0, 1
(b) 8, 6, 7, 4, 2, 2, 2
(c) 3, 3, 3, 3, 3, 3, 3
(d) 3, 2, 4, 4, 5, 5, 4, 5
(e) 4, 5, 6, 7, 8, 9, 3
3
Gracy tossed a die several times.
Her scores are as follows; 1, 5, 1, 5, 6, 5, 4.
Find her modal score.
4
If the mode of the given data set is 3, find the missing value.
2, 3, 7, 2, 5, 5, ____, 3, 7
B. 4
B. no mode
C. 6
C. 7
D. 7
D. 2
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5. STATISTICS Chapter 2 - Binomial expressions MEDIAN How to find the Median? We first rearrange the given data in ascending or descending order and pick out the middle value.
Median is the middle number in an ordered list
Example 1 Find the median of the following data set. 4, 6, 7, 3, 3, 3, 9, 8, 3 Solution Method 1 Ascending order. 3
3
3
3
Method 2 Descending order. 4
6
7
8
9
Middle number. 3
3
3
3
9
8
7
6
4
3
3
3
3
Middle number. 4
6
7
8
9
9
8
7
6
4
3
3
3
3
4 is the number in the middle of the 4 is the number in the middle of the list, so the Median is 4. list, so the Median is 4.
Example 2 Aditi has six cats with the following masses in kg. Find the median mass of her cats. 3, 5, 2, 6, 3, 5. Solution Step 1: Arrange in ascending order. 2, 3, 3, 5, 5, 6 step 2: Middle number. 2, 3, 3, 5, 5, 6 step 3: Median here is the value between 3 and 5. Median = mean of 3 and 5 = 3+5 2 = 4 Therefore, the median mass of the cats is 4 kg.
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If there are two middle values, the median is the mean of the two values.
Note: We cut out each pair of extreme values to obtain the middle value(s)
EXERCISE 5.1.4
1. Circle the correct answer.
(a) Find the median of 5, 9, 3, 8, 6, 9, 2.
A. 2 B. 5 C. 6 (b) What is the median of 17, 20, 13, 22, 17, 19 ? A. 18 B. 17 C. 20 2.
D. 8
D. 22
Find the median of each of the following set of data. (a) 5, 4, 3, 3, 1 (b) 8, 8, 8, 6, 1, 2, 2 (c) 1, 3, 8, 1 (d) 3, 2, 1, 4, 3, 4 (e) 7, 6, 4, 4, 3, 2, 1, 1, 4
3. Trisha noted her assessment marks in her journal as follows. 67, 55, 69, 61, 70, 57, 80, 81, 62, 78. What is the median mark? CHECK THIS LINK https://www.mathsisfun.com/median.html https://www.mathsisfun.com/mean.html
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5. STATISTICS Chapter 2 - Binomial expressions PIE CHART A pie chart consists of a circle divided into sectors. Each sector represents a part of the whole data set. The angle of each sector represents the size of the data.
Sector 60O
In grade 8, you interpreted information from a pie chart. In this section, you will learn how to draw a pie chart. RECALL
Car
Try this: The pie chart shows the vehicles owned by a company. (a) Find x. (b) If the company owns 30 Lorries, how many vehicles the company have? (c) How many cars does the company have?
Van
80o
xo 120o Lorry
Drawing a pie chart To draw a pie chart, we calculate the angle for each item to be represented,
Example: In a group of 12 passengers flying to Rodrigues, there are 3 men, 4 women and 5 children. Draw a pie chart to represent the given information. Solution We find the angle that represent men, women, and children.
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Number of passengers
Fraction
Angle
Men
3
3 12
3 × 360° = 90° 12
Women
4
4 12
4 × 360° = 120° 12
Children
5
5 12
5 × 360° = 150° 12
Then, we draw the pie chart as follows.
Men Children
150o 120o Women
EXERCISE 5.1.5
1. Jenny has 10 fish, 6 birds and 2 dogs. Draw a pie chart to represent the information. 2. In the school parking, there are 10 red cars, 19 white cars and 7 blue cars. Draw a pie chart to represent the information. CHECK THIS LINK https://www.mathplayground.com/piechart.html interpreting information in a pie chart.
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5. STATISTICS Chapter 2 - Binomial expressions Continuous Assessment 5.1 1. The favourite hobby of Grade 9 Red students are as follows. Football
Painting
Football
Football
Painting
Football
Painting
Painting
Football
Painting
Painting
Painting
Dancing
Painting
Dancing
Dancing Dancing
Dancing
Dancing
Dancing
Dancing Painting
Dancing
Football
Painting
(a) Use the above given data to complete the table below. Hobbies
Tally
Frequency
Football Painting Dancing Total (b) Hence complete the following frequency table. Hobbies Frequency
Football
Painting
Dancing
2. The table below shows the number of pencils that Kushi has. Pencils Frequency
Blue 12
Black 10
Red 18
Green 5
Study the above table and answer the following questions. (a) How many black pencils does Kushi have? B. 18 B. 12
C. 10
D. 5
(b) How many more red pencils were there than blue pencils? __________ (c) How many pencils did Kushi have in all? ____________ (d) What fraction of the total number of pencils does black represent? _________
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3.
Find the mean of each of the following set of data. a) 2, 3, 10 b) 5, 3, 6, 2, 4 c) 1, 4, 2, 5,1, 6, 2, 8
4. Ryan has 4 friends. Their heights are as follows 160 cm, 120 cm, 130 cm and 140 cm. Find the mean height of Ryan’s friends. 5.
Find the mode of each of the following. (a) 3, 4, 5, 1, 2, 2, 6 (b) 9, 6, 8, 5, 3, 4, 6 (c) 1, 8, 0, 8, 6, 0, 5, 0, 2
6.
Find the median of each of the following set of data. (a) 6, 5, 4, 3, 1 (b) 7, 6, 7, 6, 2, 3, 3 (c) 2, 7, 5, 1, 9, 8, 12, 3
7. The mass of a group of friends are as follows 55 kg, 60 kg, 61 kg, 57 kg, 60 kg, 61 kg, 62 kg, 63 kg and 62 kg. What is the median mass? 8. Mrs Meera sold several jewelries for Christmas. The frequency table below provides details of her sales. Jewelries Number sold
Silver 10
Gold 18
Plattine 8
Draw a pie chart to represent the above information.
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NOTES
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© Mauritius Institute of Education - 2020 ISBN: 978-99949-53-54-7