FOCUS-ON TEXTBOOK MATHS 9 Singapore Maths Approach WƌŽŐƌĞƐƐŝǀĞWƌĂĐƟĐĞƐ 21st Century Learning Skills &ŽƌŵĂƟǀĞΘ^ƵŵŵĂƟǀĞƐƐĞƐƐŵĞŶƚƐ Digital Resources Enhancements ©Praxis Publishing_Focus On Maths
FOCUS-ON MATHS TEXTBOOK 9 First Published 2023 6001 Beach Road, #14-01 Golden Mile Tower, Singapore 199589. E-mail: [email protected] ISBN 978-981-17293-4-8 Distributed by PT. Penerbitan Pelangi Indonesia Ruko the Prominence, Block 38G No. 36, Jl. Jalur Sutera, Alam Sutera, Tangerang, 15143, Indonesia. Tel: [021]29779388 Fax: [021]30030507 Email: [email protected] Printed in Malaysia by The Commercial Press Sdn. Bhd. Lot 8, Jalan P10/10, Kawasan Perusahaan Bangi, Bandar Baru Bangi, 43650 Bangi, Selangor Darul Ehsan, Malaysia. 2023 All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, photocopying, mechanical, recording or otherwise, without the prior permission of ©Praxis Publishing_Focus On Maths
II FOCUS-ON MATHS is an exciting new series that has been developed to match the latest Indonesian Mathematics syllabuses (Phase D) for Grade 7 to Grade 9. The topic coverage in each grade is arranged to address all the learning achievements (Capaian Pembelajaran) as prescribed by the Indonesian Ministry of Education. The series adopts the Singapore Maths method which is a world-class maths teaching approach. FOCUS-ON MATHS is a complete mathematics programme that comprises the Textbook, Workbook, Teacher’s Guide and online resources. It is a comprehensive, task-based and learner-centred programme designed to cultivate students’ interest in the learning of Mathematics, equip them with an in-depth understanding of Mathematical concepts and help them to achieve their fullest potential in mathematics. The content is structured to develop 21st Century Skills and Higher Order Thinking Skills in students. Students are challenged with problem-solving tasks and problems in real-word contexts. This enables them to think independently and build foundations for many of the advanced applications of mathematics that are relevant to today’s world. A collection of STEM activities is also integrated in this series to foster inquiring minds, logical reasoning and collaboration skills in students. FOCUS-ON MATHS aims to develop a greater awareness of the nature of mathematics and bring NYLH[LYJVUULJ[PVUZIL[^LLUKPɈLYLU[[VWPJZPUTH[OLTH[PJZHZ^LSSHZIL[^LLUTH[OLTH[PJZ and other subjects, creating a deeper and more robust understanding of mathematics. PREFACE 21st Century Learning Skills Indonesian Maths Syllabuses s Singapore Maths Method FOCUS-ON MATHS Formative and Summative Assesment Digital Resources Enrichment Project - Based Learning STEM Activities ©Praxis Publishing_Focus On Maths
III KEY FEATURES • Introduces the chapter in a real-world setting, allow students to realise the relevance and utility of mathematics in our daily lives. • Poses mind-stimulating questions for students to think creatively, logically and analytically. • Presents the important mathematical terms in the chapter. • Outlines the learning outcomes that will be covered in the upcoming lesson. • Provide an overview of the chapters in the book which sets the learning pace. • Provide a the chapt which set pace. • Introduce setting, al relevance our daily l • Recounts the history and development of mathematics and the contributions of great mathematicians. the learning s that will be in the upcoming • Outlines outcomes covered i lesson. APPLICATION OF THIS CHAPTER CONCEPT MAP LEARNING OUTCOMES KEY TERMS MATHS HISTORY CREATIVE THINKING ©Praxis Publishing_Focus On Maths
IV • Direct students to websites with extra learning materials or resources to enhance the learning experience. • Challenges students with questions that promote critical DQGUHÁHFWLYHWKLQNLQJ • Prepares various types of activities aimed at involving students individually, in pairs or in groups inside or outside the classroom. • A quick assessment of VWXGHQWV·SULRUNQRZOHGJH of the concepts learnt previously. • Demonstrates the steps in solving mathematical problems and techniques in answering questions. k assessment of QWV·SULRUNQRZOHGJH concepts learnt usly. Demonstrates the steps in solving mathematical problems and techniques in answering questions. • SCAN ME LET’S EXPLORE FLASHBACK WORKED EXAMPLE CRITICAL THINKING ©Praxis Publishing_Focus On Maths
V LET’S INVESTIGATE • Encourages students to explore the learning contents by themselves and involves them in active discussions during the lesson. ©Praxis Publishing_Focus On Maths
VI • Summarise important WHUPVGHÀQLWLRQVDQG mathematical properties to be used as helpful reinforcements. • Provides extra PDWKHPDWLFDOVFLHQWLÀF or historical facts and information related to the topic. • Remind students of key information or support that is useful to tackle an exercise, or simply useful to know. MATHS TIPS MATHS INFO ©Praxis Publishing_Focus On Maths
VII • Helps pupils practice answering questions that promote higher order thinking skill. HOTS CHALLENGE • Provides questions ZLWKGLIIHUHQWGLIÀFXOW\ levels that allow students to practise methods that have just been introduced. These range from VLPSOH¶UHFDOODQGGULOO· activities to applications and problem-solving tasks. PRACTICE ©Praxis Publishing_Focus On Maths
VIII • Presents different methods of working in a practical and easy-to-follow way. ALTERNATIVE METHOD • Provides students opportunities to work in small groups to explore mathematical concepts. • Provides students opportunities to work in small groups to explore mathematical concepts. INTERACTIVE ZONE ©Praxis Publishing_Focus On Maths
IX CALCULATOR CORNER • 6KRZVWKHXVHRIVFLHQWLÀF calculators in calculations • Provides opportunities to students to explore how certain mathematical concepts can be connected or related to other concepts. MATHS LINK ©Praxis Publishing_Focus On Maths
X • Prepares various types of activities aimed at involving students individually, in pairs or in groups inside or outside the classroom. SUMMARY • Sums up and highlights important concepts and formulae for quick revision. TEAM WORK ©Praxis Publishing_Focus On Maths
XI MASTERY PRACTICE • Provides a further exercise WRWHVWVWXGHQWV·PDVWHU\ of the concepts and skills learnt in each chapter. Students can download the free GeoGebra Classic software WYVNYHT[VVWLU[OLYLSH[LKÄSLZPU . https://www.geogebra.org/download ©Praxis Publishing_Focus On Maths
XII CONTENTS CHAPTER 1 Indices XIV ϭ͘ϭ /ŶĚĞdžEŽƚĂƟŽŶ 3 ϭ͘Ϯ DƵůƟƉůŝĐĂƟŽŶ ŽĨEƵŵďĞƌƐŝŶ/ŶĚĞdžEŽƚĂƟŽŶ 6 ϭ͘ϯ ŝǀŝƐŝŽŶ ŽĨEƵŵďĞƌƐŝŶ/ŶĚĞdžEŽƚĂƟŽŶ 8 ϭ͘ϰ ZĂŝƐŝŶŐEƵŵďĞƌƐĂŶĚůŐĞďƌĂŝĐdĞƌŵƐŝŶ/ŶĚĞdžEŽƚĂƟŽŶ ƚŽĂ WŽǁĞƌ 9 ϭ͘ϱ ĞƌŽ/ŶĚĞdžĂŶĚEĞŐĂƟǀĞ/ŶĚŝĐĞƐ 12 CHAPTER 2 Pythagoras’ Theorem 18 Ϯ͘ϭ ^ƋƵĂƌĞƐĂŶĚ^ƋƵĂƌĞ ZŽŽƚƐ 21 Ϯ͘Ϯ ZĞůĂƟŽŶƐŚŝƉ ĞƚǁĞĞŶ ƚŚĞ^ŝĚĞƐ ŽĨĂ ZŝŐŚƚͲŶŐůĞĚdƌŝĂŶŐůĞ 22 Ϯ͘ϯ ^ŽůǀŝŶŐ WƌŽďůĞŵƐ/ŶǀŽůǀŝŶŐ WLJƚŚĂŐŽƌĂƐ͛dŚĞŽƌĞŵ 32 CHAPTER 3 Circles 38 ϯ͘ϭ WĂƌƚƐ ŽĨĂ ŝƌĐůĞ 41 ϯ͘Ϯ ŝƌĐƵŵĨĞƌĞŶĐĞĂŶĚƌĞĂ ŽĨĂ ŝƌĐůĞ 52 CHAPTER 4 Geometry 70 ϰ͘ϭ 'ĞŽŵĞƚƌŝĐĂů WƌŽƉĞƌƟĞƐ ŽĨdŚƌĞĞͲŝŵĞŶƐŝŽŶĂů ^ŚĂƉĞƐ 73 ϰ͘Ϯ EĞƚƐ ŽĨdŚƌĞĞͲŝŵĞŶƐŝŽŶĂů ^ŚĂƉĞƐ 77 ϰ͘ϯ ^ƵƌĨĂĐĞƌĞĂ ŽĨ ƵďŽŝĚƐ͕ ƵďĞƐ͕ WƌŝƐŵƐĂŶĚ WLJƌĂŵŝĚƐ 81 ϰ͘ϰ ^ƵƌĨĂĐĞƌĞĂ ŽĨ LJůŝŶĚĞƌƐ͕ ŽŶĞƐĂŶĚ^ƉŚĞƌĞƐ 88 ϰ͘ϱ sŽůƵŵĞ ŽĨ ƵďŽŝĚƐ͕ ƵďĞƐ͕ WƌŝƐŵƐĂŶĚ WLJƌĂŵŝĚƐ 96 ϰ͘ϲ sŽůƵŵĞ ŽĨ LJůŝŶĚĞƌƐ͕ ŽŶĞƐĂŶĚ^ƉŚĞƌĞƐ 104 ©Praxis Publishing_Focus On Maths
XIII CHAPTER 5 Cartesian Coordinate System 118 ϱ͘ϭ ĞƚĞƌŵŝŶŝŶŐ >ŽĐĂƟŽŶ 121 ϱ͘Ϯ ĂƌƚĞƐŝĂŶ ŽŽƌĚŝŶĂƚĞƐ 123 CHAPTER 6 dƌĂŶƐĨŽƌŵĂƟŽŶƐ 142 ϲ͘ϭ dƌĂŶƐĨŽƌŵĂƟŽŶƐ 145 ϲ͘Ϯ dƌĂŶƐůĂƟŽŶ 147 ϲ͘ϯ ZĞŇĞĐƟŽŶ 158 ϲ͘ϰ ZŽƚĂƟŽŶ 165 ϲ͘ϱ ffiŶůĂƌŐĞŵĞŶƚ 173 CHAPTER 7 Probability 188 ϳ͘ϭ /ŶƚƌŽĚƵĐƟŽŶ ƚŽ WƌŽďĂďŝůŝƚLJ 191 ©Praxis Publishing_Focus On Maths
1INDICES Applications of this chapter Indices are used in a lot of areas in our modern technological world. Indices are used in computer game physics, pH and Richter Measuring Scales, science, engineering, economics, accounting and many other disciplines. Indices are critically important in modern internet-based VDOHV DQGPDUNHWLQJ LQYHVWPHQW DQG ÀQDQFH,QGLFHV DUHWKH EDVLV RI demographics and also part of food technology and microbiology. What profession uses indices? XIV ©Praxis Publishing_Focus On Maths
Zero Index Index Notation Negative Index Indices Application of the Concept of Exponents in Problem-Solving Indices • Index notation • Index • Base • Repeated multiplication • Law of indices Concept Map Key Terms Maths History Samuel Jeake (1623-1690), is an astrologer in the 16th century. His principal mathematical work was Logisticelogia, or Arithmetick Surveighed and Reviewed published in four books in 1696. The table of powers from A Compleat Body Learning Outcomes • Understand and use the concept of real numbers expressed in index notation. • Multiply and divide real numbers in index notation. • Perform computations involving negative indices. • Understand and use the concept of real numbers expressed in exponential notation with integral indices. • Explain results of expression in exponential notation in integral numbers, ratios and decimals. • Multiply and divide real numbers in the form of indices with the same base and integral indices. 1 of Arithmetic is one of the work written by Samuel Jeake in 1671. His son, Samuel Jeake, the younger (1652–1699), also an astrologer, had introduced the term 'indices' in 1696. ©Praxis Publishing_Focus On Maths
CHAPTER 1 Indices 2 Flashback 1. 2 × 2 = 2. 4 × 4 = 3. 3 × 3 = 4. 22 = 5. 5 × 5 × 5 = As I was going to Jalan Jawa I met a mother with seven children, Each child had seven bags; each bag had seven chicks, Each chick had seven seeds: seeds, chicks, bags and children, How many were going to Jalan Jawa? (a) How many seeds, chicks, bags and children are suggested by the rhyme? (b) In the rhyme the number seven is used; how many seeds, chicks, bags and children would be suggested by the rhyme if the number nine replaces the number seven? 1 Suppose the number seven is replaced by a number that makes the number of seeds suggested by the rhyme become 14 641. What is the number that replaced seven? Critical Thinking 6. 72 = 7. 6 × 6 × 6 × 6 = 8. 2 × 2 × 3 × 3 = 9. 82 = 10. 42 × 52 = ©Praxis Publishing_Focus On Maths
3 Indices CHAPTER 1 1.1 Index Notation Index notation is used to shorten the way products of numbers or pronumerals are written. Discuss with your friends and complete the table below. Index form Expanded form Value 21 2 2 22 2 = 2 23 2 = 2 = 2 24 25 26 27 28 29 2 A Represent repeated multiplication in index notation an When a number is multiplied by itself a certain number of times, it can be written in index notation as an , a × a × a × … × a = an n factors where a is known as the base and n is power that known as index. n also known as exponent. Conversely, any number in index notation can be written as a repeated multiplication. Can every integer, other than 0, be expressed in index notation? Explain. Critical Thinking ©Praxis Publishing_Focus On Maths
CHAPTER 1 Indices 4 An index, or power, is used to show that a quantity is repeatedly multiplied by itself. EXAMPLE 1 Express each of the following repeated multiplication in index notation. (a) 3 × 3 × 3 × 3 × 3 (b) 0.7 × 0.7 × 0.7 × 0.7 × 0.7 × 0.7 (c) – 2 5 × – 2 5 × – 2 5 Solution: (a) 3 × 3 × 3 × 3 × 3 = 35 (b) 0.7 × 0.7 × 0.7 × 0.7 × 0.7 × 0.7 = 0.76 (c) – 2 5 × – 2 5 × – 2 5 = – 2 5 3 Index notation an means the value of a is multiplied repeatedly by n times. an is read as a raised to the nth power. B Rewrite a number in index notation and vice versa We can express a number in index form by performing the following steps: Write the number in the form of repeated multiplication. Write the product using base and index. For example, 32 = 2 × 2 × 2 × 2 × 2 = 25 2 as the base that is multiplied by 5 times. :HFDQÀQGWKHYDOXHRIDQXPEHULQLQGH[IRUPE\SHUIRUPLQJWKHIROORZLQJVWHSV Write the number in the form of repeated multiplication. Find the product. For example, 74 = 7 × 7 × 7 × 7 = 2401 EXAMPLE 2 Express each of the following numbers in index notation in the base indicated in brackets. (a) 25 (base 5) (b) –100 000 (base –10) (c) 27 125 base 3 5 Solution: (a) 25 = 5 × 5 = 52 7 is multiplied by 4 times. ©Praxis Publishing_Focus On Maths
Indices CHAPTER 1 5 (b) –100 000 = –10 × –10 × –10 × –10 × –10 = (–10)5 (c) 27 125 = 3 5 × 3 5 × 3 5 = 3 5 3 EXAMPLE 3 Find the value of each of the following. (a) 46 (b) (– 6)3 (c) – 2 3 3 Solution: (a) 46 = 4 × 4 × 4 × 4 × 4 × 4 = 4096 (b) (–6)3 = –6 × –6 × –6 = –216 (c) – 2 3 3 = – 2 3 × – 2 3 × – 2 3 = – 8 27 Certain numbers can be expressed in more than one index notation. For example, 16 = 24 or 42 . Practice 1.1 Basic Intermediate Advanced Express each of the following in index notation. (a) 2 × 2 × 2 × 2 (b) –3 × (–3) × (–3) × (–3) × (–3) × (–3) (c) 0.6 × 0.6 × 0.6 (d) – 4 3 × – 4 3 × – 4 3 × – 4 3 × – 4 3 Express each of the following as a repeated multiplication. (a) 72 (b) 54 (c) (– 4)5 (d) 0.36 (e) – 3 4 3 (f) 1 1 2 4 Find the value of each of the following. Then check your answer with a calculator. (a) 27 (b) 54 (c) (–3)5 (–a)n is a negative value if n is an odd number. (–a)n is a positive value if n is an even number. Technology Computers rely on algebraic formulae to enhance their algorithms and ensure that devices function correctly. Exponential functions SOD\ D VLJQL¿FDQW UROH LQ WKHVH IRUPXODH ZLWK PDQ\ DOJRULWKPV UHO\LQJRQSRZHUWHUPVWKDWLQYROYHLQGLFHV6SHFL¿FDOO\FRPSXWHU processors frequently utilise powers of 2 in their calculations, since binary is the fundamental language of computation and uses 2 as its base. By incorporating these formulae and exponential functions, computers can perform complex calculations and execute programs with remarkable speed and precision. Maths LINK Press ( (–) 2 ab/c 3 ) ^ 3 = –8/27 ©Praxis Publishing_Focus On Maths
6 CHAPTER 1 Indices 1.2 Multiplication of Numbers in Index Notation A Multiplication of numbers and algebraic terms in index notation with the same base Objective: To correlate the multiplication of numbers in index notation that has the same base as repeated multiplication. Instruction: Do this activity in groups of four. 1. Copy and complete the following table. Multiplication of number in index notation Repeated multiplication Multiplication result in index notation Addition in index 22 × 25 (2 × 2) × (2 × 2 × 2 × 2 × 2) 27 22+5 34 × 32 ( × × × ) × ( × ) 3 3 + 53 × 54 ( × × ) × ( × × × ) 5 5 + a3 × a4 ( × × ) × ( × × × ) a a + 2. Based on the pattern from the table, what is the correlation between the multiplication of numbers in an index notation that has the same basis as repeated multiplication? 3. What generalisations can be made about am × an provided that m and n are positive integers? 1 Multiplication of numbers or algebraic terms expressed in index notation with the same base FDQEHVLPSOLÀHGE\DGGLQJXSWKHLQGLFHVZKLOHNHHSLQJWKHEDVHXQFKDQJHGWKDWLV am × a n = a m + n . (d) (– 4)4 (e) 0.33 (f) (– 0.2)6 (g) 1 3 4 (h) – 2 5 5 (i) 1 2 3 3 Express each of the following numbers in index notation in the base indicated in brackets. (a) 16 (base 2) (b) 81 (base 3) (c) 625 (base 5) (d) –216 (base –6) (e) 0.064 (base 0.4) (f) 125 64 (base 5 4 ) Given that a × a × a × a = 9n, where a and n are positive integers. State the values of a and n. Given that k7 ²flÀQGWKHYDOXHRIk. The product of the repeated multiplication of a number by n times is 7776. Determine the value of n. ©Praxis Publishing_Focus On Maths
Indices CHAPTER 1 7 EXAMPLE 4 Simplify each of the following. (a) 54 × 52 (b) 4a3 × 5a4 × (– 3a) Solution: (a) 54 × 52 = 54 + 2 (b) 4a3 × 5a4 × (– 3a) = 56 = 4 × 5 × (– 3) × (a3 × a4 × a) = – 60a3 + 4 + 1 = – 60a8 B Multiplication of numbers and algebraic terms in index notation with different bases When multiplying numbers or algebraic terms in index notation with different bases, simplify it as follows: Group the numbers or the algebraic terms with the same base together. Apply the law of indices am × an = a m + n to each group of numbers or algebraic terms with the same base. EXAMPLE 5 Simplify each of the following. (a) 2 × 33 × 24 (b) 2x 2 × 5y × 4x 3 × y 5 Solution: (a) 2 × 33 × 24 = (2 × 24 ) × 33 = 21 + 4 × 33 = 25 × 33 (b) 2x 2 × 5y × 4x 3 × y 5 = (2 × 5 × 4) × (x 2 × x 3) × (y × y 5 ) = 40 × x 2 + 3 × y1 + 5 = 40x 5 y 6 Numbers in index notation with different bases are separated by the multiplication sign ×. For example, 36 × 42 . Algebraic terms in index notation with different bases are not separated by any multiplication sign ×. For example, 24p5 q4 . The bases in algebraic terms are usually expressed in alphabetical order. For example, 24p5 q4 , and not 24q4 p5 . Simplify each of the following. (a) 2 × 24 (b) 32 × 34 × 3 (c) 42 × 43 × 4 × 44 (d) 7 × 72 × 73 (e) 82 × 8 × 85 × 83 (f) 112 × 113 × 11 × 114 × 115 Simplify each of the following. (a) 37 × 34 (b) 185 × 182 (c) 56 × 5 × 58 (d) k 9 × k10 × k 3 Simplify the following. (a) a × a5 (b) f 2 × f 3 × f 4 (c) 2m × 3m3 (d) 5q 2 × 2q 3 × 4q (e) –2r × 5r 3 × 6r 4 × 3r 2 (f) –y × 2y 3 × (–5y 2 ) Practice 1.2 Basic Intermediate Advanced ©Praxis Publishing_Focus On Maths
8 CHAPTER 1 Indices 1.3 Division of Numbers in Index Notation A Division of numbers and algebraic terms expressed in index notation with the same base Objective: To correlate the division of numbers in index notation that has the same base as repeated multiplication. Instruction: Do this activity in groups of four. 1. Copy and complete the following table. Division of number in index notation Repeated division Division result in index notation Subtraction in index 47 ÷ 42 47 42 = 4 × 4 × 4 × 4 × 4 × 41 × 41 14 × 41 45 47 – 2 65 ÷ 63 65 63 = × × × × × × 6 6 – 78 ÷ 75 78 75 = 7 7 – a8 ÷ a5 a8 a5 = a a – 2. Based on the pattern from the table, what is the correlation between the division of numbers in an index notation that has the same basis as repeated multiplication? 3. What generalisations can be made about am ÷ an provided that m and n are positive integers? 2 Division of numbers or algebraic terms expressed in index notation with the same base can EHVLPSOLÀHGE\VXEWUDFWLQJWKHLQGLFHVZKLOHNHHSLQJWKHEDVHXQFKDQJHGWKDWLV am ÷ a n = a m – n . Simplify each of the following. (a) 22 × 34 × 3 (b) 42 × 5 × 53 × 43 (c) 5 × 63 × 53 × 62 (d) 72 × 23 × 24 × 62 × 73 × 65 Simplify each of the following. (a) 6x 2 × 2x 4 (b) m3 n × m7 n5 (c) 4ab3 × 7a9 b2 (d) p2 q6 × p5 r 4 × q 9 r 10 Express each of the following in the simplest index notation. (a) b2 × a2 × b (b) g × h2 × h3 × g2 (c) 2q × r 3 × 3r × q3 (d) 2x 2 × y × 5z × 4z 2 × (–2y3 ) ©Praxis Publishing_Focus On Maths
Indices CHAPTER 1 9 EXAMPLE 6 Simplify each of the following. (a) 26 ÷ 22 (b) 15e7 ÷ 3e4 Solution: (a) 26 ÷ 22 = 26 – 2 = 24 (b) 15e7 ÷ 3e4 = 15 3 × e7 – 4 = 5e 3 1.4 Raising Numbers and Algebraic Terms in Index Notation to a Power A :PTWSPÄJH[PVUVMU\TILYZHUKHSNLIYHPJ[LYTZL_WYLZZLKPUPUKL_UV[H[PVUYHPZLK[V a power Numbers in index form raised to a power can be related by repeated multiplication as follows. (42 ) 3 = 42 × 42 × 42 (42 ) 3 = (4 × 4) × (4 × 4) × (4 × 4) (42 ) 3 = 4 × 4 × 4 × 4 × 4 × 4 (42 ) 3 = 46 (42 )3 = 42 × 3 Numbers in index form raised to a power can be related by repeated multiplication as follows. am n = a m × n where m and n are positive integers. Repeated multiplication of 42 by 3 times Indices are multiplied, 2 × 3 = 6 Simplify each of the following. (a) 26 2 (b) 511 57 (c) 76 ÷ 72 (d) 1312 ÷ 135 Simplify each of the following. (a) 76 ÷ 74 (b) 2913 ÷ 295 (c) 128 123 (d) y16 ÷ y7 Express each of the following in its simplest index notation. (a) f 10 ÷ f 3 (b) p12 p 7 (c) 12h 5 h 3 (d) –48q 7 ÷ 36q 4 Simplify each of the following. (a) 18h10 ÷ 3h4 (b) a4 b17 ÷ ab3 (c) k 9 h5 k 2 h3 (d) 128x 20y7 32x 12y Practice 1.3 Basic Intermediate Advanced ©Praxis Publishing_Focus On Maths
CHAPTER 1 Indices 10 EXAMPLE 7 Simplify each of the following. (a) (67 )4 (b) (p3 )5 (c) (3x 2 )4 Solution: (a) (67 ) 4 = 67 × 4 = 628 (b) (p3 ) 5 = p 3 × 5 = p15 (c) (3x 2 )4 = 34 × x 2 × 4 = 81x 8 EXAMPLE 8 Simplify the following. (a) (32 × 7)4 (b) (2p2qr 3) 4 Solution: (a) (32 × 7)4 = 32 × 4 × 71 × 4 = 38 × 74 (b) (2p2qr 3)4 = 24 p2 × 4q1 × 4r 3 × 4 = 16p8q4r 12 EXAMPLE 9 Simplify each of the following. (a) 24 32 2 (b) (p4 ÷ q 2 )3 B Multiplication and division of numbers and algebraic terms expressed in index notation with different bases raised to a power When the multiplication of numbers or algebraic terms expressed in index notation with different EDVHVLVUDLVHGWRDFHUWDLQSRZHULWFDQEHVLPSOLÀHGE\PXOWLSO\LQJWKHLQGH[RIHDFKWHUP in the multiplication to the power, that is, am × bn p = amp × bnp . When the division of numbers or algebraic terms expressed in index notation with different EDVHVLVUDLVHGWRDFHUWDLQSRZHULWFDQEHVLPSOLÀHGE\PXOWLSO\LQJWKHLQGH[RIHDFKWHUP in the brackets to the power, that is, am bn p = amp bnp . ©Praxis Publishing_Focus On Maths
Indices CHAPTER 1 11 Solution: (a) 24 32 2 = 2 4 × 2 32 × 2 (b) (p4 ÷ q 2)3 = (p4 )3 ÷ (q 2 )3 = 28 34 = p12 ÷ q 6 = p12 q 6 C Combined operations involving multiplication, division and raised to a power on numbers or algebraic terms The combined operations involving multiplication, division and raising to a power on numbers RUDOJHEUDLFWHUPVLQLQGH[QRWDWLRQFDQEHVLPSOLÀHGE\XVLQJ laws of indices. EXAMPLE 10 Simplify each of the following. (a) 76 × (52 )4 55 (b) (3g 3 h2)2 9g 2 h Solution: (a) 76 × (52 ) 4 55 = 76 × 52×4 55 = 76 × 58 55 = 76 × 58–5 = 76 × 53 (b) (3g 3 h2 )2 9g 2 h = 32 g 3×2h2×2 9g 2 h = 9g 6 h4 9g 2 h = g 6–2h 4–1 = g 4 h 3 Simplify. (a) (32 )3 (b) (53 )4 (c) (75 )2 (d) (103 )0 (a) (d 3 ) 2 (b) (e 2 ) 4 (c) (p 5 )6 (d) (q7 )4 (a) ( j 2 k) 3 (b) (2f 3 g2 )3 (c) (5m2 pq 3 )4 (d) (4tu 3 vw 2 )5 (a) (2 = 32 )2 (b) (42 = 53 )4 (c) (34 = 5 = 72 ) 5 (d) (56 = 7 = 92 )3 (e) (74 = 9 = 113 = 135 ) 3 (a) 23 5 2 (b) 43 72 4 (c) (32 ÷ 7)5 (d) (113 ÷ 134 ) 2 (a) b2 c 2 (b) h3 g2 4 (c) (m3 ÷ n) 5 (d) (x3 ÷ y 5 )6 (a) 25 ÷ (22 ) 2 = 23 (b) 47 = (63 ) 2 ÷ (42 ) 3 (c) (32 ÷ 54 )2 ÷ (52 )3 (d) (32 )4 = (72 )3 72 (e) 212 92 ÷ 23 9 4 (f) (33 )3 = 87 (82 ) 2 = 33 (a) (a3 d 2 )3 ÷ a5 d 4 (b) (m4 n2 )2 m5 n3 (c) (p4 q5 )3 (p2 q)4 (d) (2r 2 s 3 ) 2 = rs 2 4r 3 s 5 (e) 24e5 = (f 6 )2 ÷ (2e2 f 3 )2 (f) (4x 2 y 5 ) 3 ÷ (2x 2 y 4 = 8xy 3 ) am × an = am + n am ÷ an = a m – n (am)n = amn Practice 1.4 Basic Intermediate Advanced ©Praxis Publishing_Focus On Maths
CHAPTER 1 Indices 12 1.5 Zero Index and Negative Indices A Verify that a0 = 1 and a–n = 1 an Objective: To verify a0 = 1 and a–n = 1 a n . Instruction: Do this activity in groups of four. 1. Copy and complete the following table. Division with repeated multiplication am ÷ an = a m – n 25 ÷ 25 = 2 1 × 21 × 21 × 21 × 21 21 × 21 × 21 × 21 × 21 = 25 ÷ 25 = 25 – 5 = 2 25 ÷ 26 = = 25 ÷ 26 = 2 – = 2 25 ÷ 27 = = 25 ÷ 27 = 2 – = 2 25 ÷ 28 = = 25 ÷ 28 = 2 – = 2 2. Based on the pattern from the table above, what conclusions can be drawn about a0 and a–n ? 3 From the activity above, we can simplify an ÷ an as follows: an ÷ an = an – n or an ÷ an = an an = a0 = 1 7KHUHIRUH LW LV YHULÀHG WKDW a0 = 1 where a ʒ For example, 90 = 1, 7 8 0 = 1, (– 0.6)0 = 1. For a0 ÷ an , we can simplify as follows: a0 ÷ an = a0 – n or a 0 ÷ an = a0 an a0 = 1 = a –n = 1 an 1 1 ©Praxis Publishing_Focus On Maths
Indices CHAPTER 1 13 7KHUHIRUH LW LV YHULÀHG WKDW a–n = 1 an where a ʒ For example, 3–1 = 1 3 , z –5 = 1 z 5 . Why are 40 and (– 4)0 equal to 1, while –40 is equal to –1? Critical Thinking EXAMPLE 11 State each of the following in the form 1 an . (a) 2–1 (b) 7–3 Solution: (a) 2–1 = 1 2 (b) 7–3 = 1 73 EXAMPLE 12 State each of the following in index notation. (a) 1 8 (b) 1 106 Solution: (a) 1 8 = 8–1 (b) 1 106 = 10– 6 EXAMPLE 13 Express (a) 2 9 –1 in the form of positive index. (b) x y 4 in the form of negative index. Solution: (a) 2 9 –1 = 1 2 9 a–n = 1 an = 1 ÷ 2 9 = 9 2 ©Praxis Publishing_Focus On Maths
CHAPTER 1 Indices 14 (b) x y 4 = x 4 y 4 = 1 ÷ y 4 x 4 = 1 ÷ y x 4 = 1 y x 4 1 an = a–n = y x – 4 B Combined operations involving negative indices on numbers and algebraic terms EXAMPLE 14 Evaluate. (a) 3–3 × 3–2 (b) 28 × (2–2 × 3–1) 3 Solution: (a) 3–3 × 3–2 = 3–3 + (–2) = 3–5 = 1 35 = 1 243 EXAMPLE 15 Simplify each of the following. (a) e2 ÷ (e–3) 4 (b) (–3x 2 y) 3 ÷ xy –2 Solution: (a) e2 ÷ (e –3)4 = e 2 ÷ e –3 × 4 = e 2 ÷ e –12 = e 2 – (–12) = e14 (b) 28 × (2–2 × 3–1) 3 = 28 × 2–2 × 3 × 3–1 × 3 = 28 × 2–6 × 3–3 = 28 + (–6) × 1 33 = 22 × 1 27 = 4 27 (b) (–3x 2 y)3 ÷ xy –2 = (–3)3 x 2 × 3y 3 ÷ xy –2 = –27x 6 y 3 ÷ xy –2 = –27x 6 – 1 y 3 – (–2) = –27x 5 y 5 ©Praxis Publishing_Focus On Maths
15 Indices CHAPTER 1 Basic Intermediate Advanced Practice 1.5 State each of the following as a fraction. (a) 4–1 (b) p –1 (c) 5–2 (d) 12–3 (e) g – 4 (f) z –7 State each of the following in index notation. (a) 1 5 (b) 1 d (c) 1 94 (d) 1 243 (e) 1 f 4 (f) 1 m10 Evaluate each of the following. (a) 2–2 = (23 ) 2 (b) 52 ÷ (5–1) 2 (c) 2–4 = (2–3)–2 = 24 (d) 42 = (4–2)3 4–5 (e) 2–3 3 2 = 3 24 –2 (f) (23 = 3)3 (2–4 = 3)–2 Express (a) in the form of positive index. (i) 1 5–1 (ii) 1 10–2 (iii) p q –3 (b) in the form of negative index. (i) 63 (ii) 3 8 (iii) x y 6 Simplify. (a) a 4 = (a2 ) 3 ÷ a –3 (b) (d 2 ) –3 ÷ (d –2) 2 = d (c) (m–1n2)–2 m2 n–3 (d) (p3 r 2 )2 ÷ (pr 3 )–2 (e) (3s 2)2 = (t 2)–3 s 6 t –5 (f) (2x 3 y – 4z) 2 ÷ 4x 5 y –5 = 8yz 2 Summary Summary Spot and correct any mistakes made by a student as shown in the diagram. Explain how you think the error occurred. team work Indices Index notation: Index an = a × a × a × … × a n factors Law of indices 1. Multiplication of indices am × an = am + n 2. Division of indices am ÷ an = a m – n 3. Indices raised to a power (am)n = amn 4. Negative indices a –n = 1 an 5. a0 = 1 ©Praxis Publishing_Focus On Maths
16 CHAPTER 1 Indices Section A 1. (–3)4 = A 12 B –12 C 81 D –81 2. 1 1 2 3 = A 1 1 8 B 3 3 8 C 3 1 6 D 1 3 8 3. Which of the following values is not equal to 256? A 28 B 44 C 83 D 162 4. 23 × 3 × 2 × 36 = A 23 × 36 B 24 × 37 C 628 D 611 5. 810 ÷ 85 = A 8–5 B 82 C 85 D 815 6. Simplify (9pq 3 )2 . A 18p2 q 6 B 81p2 q 6 C 81p3 q5 D 18p3 q5 7. Simplify 2p3 × 4q2 × 3p × q5 . A 12p2 q 7 B 24p3 q 10 C 24p4 q7 D 24p5 q6 8. Simplify 6(m2 n3 )3 ÷ 2m 3n. A 3m3n5 B 3m2 n9 C 3m3n8 D 3m9 n10 9. 52 ÷ (5 × 2–1) 3 = A 1000 B 40 C 5 8 D 8 5 10. Simplify 2pq–1 × (3p 2 q –3)2 ÷ 6p3q –5. A 3p 2 q2 B p2 q2 C 3p q2 D p q2 Section B 1. Evaluate (a) 53 ÷ (52 × 5–3) (b) 3 2 2 × 2 3 – 2 2. Simplify (a) (g 2) 3 ÷ g –9 × g3 (b) 6p2 q3 ÷ 2(pq –1)–2 × 4pq 3. Evaluate. (a) (–3)4 (b) 42 × 4 ©Praxis Publishing_Focus On Maths
17 Indices CHAPTER 1 (c) 24 42 2 (d) 52 ÷ (5 × 2–1)3 4. Simplify. (a) (9pq3 ) 2 (b) 6(m2 n3 ) 3 ÷ 2m3 n (c) 2pq–1 × (3p2 q – 3) 2 ÷ 6p3 q – 5 5. (a) Given that 32x – 1 = 27, calculate the value of x. (b) Simplify (g2 ) 2 × (4fg2 )3 ÷ f 2g7 . 6. Simplify. (a) 3(92x + 1) (b) 4[22m × (2 1 —3 )–3] 7. Given that 16 × 2p = 29 ÀQGWKHYDOXHRIp. 8. Given that 22x – 1 = 8 4 ÀQG WKH YDOXH RI x. 9. Given that 2(162m + 1 flÀQGWKHYDOXHRI m. 10. (a) Simplify (p3)5 ÷ p7 × 1 p 2 . (b) Given that 3n × 27 (3n )2 fl Ànd the value of n. (c) Given that x y n – 1 = y x n – 5ÀQGWKHYDOXH of n. 11. (a) Simplify m4 n × (mn3)2 mn 5 —2 . (b) Simplify 18n + 1 × 31 – n 2n – 1 . (c) Given that 16(25x) = 125(32y ÀQG WKH values of x and y. ©Praxis Publishing_Focus On Maths
2 Applications of this chapter Applications of this chapter 7KH 3\WKDJRUDV· WKHRUHP H[SODLQV WKH UHODWLRQVKLS EHWZHHQ WKH WKUHH VLGHVRIDULJKWDQJOHGWULDQJOH7KLVFRQFHSWLVZLGHO\XVHGLQWKHÀHOGV RIDUFKLWHFWXUHFRQVWUXFWLRQODQGVXUYH\LQJQDYLJDWLRQDQGRWKHUÀHOGV 0RVW DUFKLWHFWV XVH WKH FRQFHSW RI WKH 3\WKDJRUDV· WKHRUHP WR ÀQG XQNQRZQ GLPHQVLRQV7KH IDFH UHFRJQLWLRQ IHDWXUH LQ VHFXULW\ FDPHUDV XVHVWKH VDPH FRQFHSW 3HRSOH ZKRWUDYHO E\ VHD XVHWKLV FRQFHSW WR ÀQGWKHVKRUWHVWGLVWDQFHDQG URXWHWRSODFHVWKH\FDUHDERXW,WLVDOVR XVHG LQ LQWHULRU GHVLJQ DQGWKH FRQVWUXFWLRQ RIKRXVHV DQG EXLOGLQJV PYTHAGORAS' THEOREM +DYH\RXHYHUXVH3\WKDJRUDV·WKHRUHPLQ\RXUGDLO\OLIHDSSOLFDWLRQV" 18 ©Praxis Publishing_Focus On Maths
Hypotenuse of a Right-Angled Triangle Types of Triangles Lengths of Sides of Geometric Shapes Length of the Unknown Side of a Right-Angled Triangle Pythagoras’ Theorem Pythagorean Triples Usage of Pythagoras’ Theorem Square and Square Roots Relationship Between the Sides of a Right-Angled Triangle Pythagoras’ Theorem Maths History • Right-angled triangle • Hypotenuse • Pythagoras· theorem • Pythagorean triple • Converse of Pythagoras· theorem • Acute angle • Obtuse angle • ,GHQWLI\WKHK\SRWHQXVHRIDULJKWDQJOHGWULDQJOH • 'HWHUPLQHWKHUHODWLRQVKLSEHWZHHQWKHVLGHVRIDULJKWDQJOHGWULDQJOHDQGKHQFH H[SODLQWKH3\WKDJRUDV·WKHRUHP • 8QGHUVWDQGDQGGHWHUPLQHWKH3\WKDJRUHDQWULSOHV • $SSO\WKHFRQFHSWRI3\WKDJRUDV·WKHRUHPWRVROYHPDWKHPDWLFDOSUREOHPVDQG UHDOOLIHSUREOHPV Concept Map Key Terms Learning Outcomes Pythagoras (569 BC - 475 BC) was a Greek philosopher and mathematician. Although Pythagoras introduced and popularised the theorem, WKHUH LV VXIÀFLHQW HYLGHQFH SURYLQJ LWV existence in other FLYLOLVDWLRQVyears before Pythagoras was born. The oldest known HYLGHQFHGDWHVEDFNWREHWZHHQWKHWKWR WKcentury BC in the Old Babylonian period. 19 ©Praxis Publishing_Focus On Maths
CHAPTER 2 Pythagoras’ Theorem 20 D 8VHDSURWUDFWRUWRPHDVXUHWKHDQJOHVRIHDFKWULDQJOH E 8VHDFHQWLPHWUHUXOHUWRPHDVXUHWKHVLGHVRIHDFKWULDQJOH F &ODVVLI\WKHWULDQJOHVDERYHEDVHGRQVLGHVDQGDQJOHV How can you tell if a triangle is a right-angled triangle? Critical Thinking 2. Find the value of each of the following. (a) 9 (b) 36 (c) 121 (d) 225 Flashback 1. Find the value of each of the following. (a) 42 (b) 72 (c) 132 (d) 182 1 ©Praxis Publishing_Focus On Maths
Pythagoras’ Theorem CHAPTER 2 21 5 = 5 = 25 and (–5) = (–5) = 25. So, 25 RU ² Usually, only the positive value is given. In many applications, the negative square root does not make sense or is not meaningful, so we only consider the positive value. (J 36 = 6 EXAMPLE 1 )LQGWKHVTXDUHVRIWKHIROORZLQJQXPEHUV D E 3 5 F G ² Solution: D 2 = ffi E fl 5 2 = fl 5 = fl 5 = 64 25 F 2 = G ² 2 = (–5) = (–5) = 25 EXAMPLE 2 )LQGWKHVTXDUHURRWRIHDFKRIWKHIROORZLQJQXPEHUV (a) E ffi F fl G 144 Solution: (a) 16 = 4 = 4 = 4 E 49 = = F fl 9 = 9 = 9 G 144 = 12 = 12 = 12 EXAMPLE 3 8VLQJDFDOFXODWRUÀQGWKHYDOXHVRIWKHIROORZLQJ*LYHWKH DQVZHUVFRUUHFWWRGHFLPDOSODFHV (a) E 526 F fl G ffi 2.1 Squares and Square Roots A Square of a number ,IDQXPEHULVPXOWLSOLHGE\LWVHOIWKH UHVXOWLVFDOOHGWKHsquareRIWKHQXPEHU7KHVTXDUH RIDQXPEHUnLVZULWWHQDVn2 )RUH[DPSOH 4 × 4 = 42 LVWKH VTXDUH RI 7KHQXPEHU2 FDQEHUHDGDV¶VTXDUHRI·RU¶VTXDUHG·RU¶WRWKHSRZHURI· B Square root of a number The square root RI D QXPEHU LVWKH QXPEHUWKDWZDVPXOWLSOLHG E\ LWVHOIWR JHWWKH VTXDUH QXPEHU7KHVTXDUHURRWV\PERO LVFDOOHGDUDGLFDO )RUH[DPSOH 16 = 4 = 16 LVUHDGDV¶VTXDUHURRWRIVL[WHHQ· 7KHVTXDUHDQGVTXDUHURRWRIDQXPEHUDUHLQYHUVH RSHUDWLRQV 6TXDUH 5 25 6TXDUHURRW A negative number has no VTXDUH URRW :H FDQQRW ¿QG the square root of a negative number because the product of two numbers with the same sign is always positive. E.g. –9 = (–3) = 3 (The multiplicand and multiplier are not the same.) ©Praxis Publishing_Focus On Maths
CHAPTER 2 Pythagoras’ Theorem 22 Solution: (a) Press 1 7 = GS E 3UHVV 5 2 6 = ffiflffiflfl 526 ffi GS F 3UHVV 0 · 2 8 = ffi fl ffi GS G 3UHVV 0 · 0 9 4 = ffiffi ffi GS ab = a = b. a b = a b Where a 0 and b > 0 Practice 2.1 Basic Intermediate Advanced 1 )LQGWKHVTXDUHVRIWKHIROORZLQJQXPEHUV D fl E 4 F G H ² I ² 1 2 )LQGWKHVTXDUHURRWRIHDFKRIWKHIROORZLQJ SHUIHFWVTXDUHVZLWKRXWXVLQJDFDOFXODWRU (a) E F 64 G 100 (e) I 169 )LQGWKHYDOXHRIHDFKRIWKHIROORZLQJDQG JLYHWKHDQVZHUWRGHFLPDOSODFH (a) E F 130 8VLQJDFDOFXODWRUÀQGWKHYDOXHRIHDFKRI WKHIROORZLQJDQGJLYHWKHDQVZHUFRUUHFWWR GHFLPDOSODFHV (a) E fl 2.2 Relationship Between the Sides of a Right-Angled Triangle /RRNDWWKHVHWULDQJOHV D :KLFKWULDQJOHLVDULJKWDQJOHGWULDQJOH" E :KLFKWULDQJOHLVDQREWXVHWULDQJOH" F :KLFKWULDQJOHLVDQDFXWHWULDQJOH" +RZGLG\RXGHFLGH" 2 ©Praxis Publishing_Focus On Maths
Pythagoras’ Theorem CHAPTER 2 23 A Identifying hypotenuse of a right-angled triangle Objective: 7RLGHQWLI\WKHK\SRWHQXVHRIDULJKWDQJOHGWULDQJOH Instruction: 'RWKLVDFWLYLW\LQJURXSVRIIRXU Materials: 3HQFLOUXOHUSDSHU 1. (DFKPHPEHURIWKHJURXSGUDZVDULJKWDQJOHGWULDQJOH 2. /DEHOWKHULJKWDQJOHIRUWKHWULDQJOHGUDZQ 3. 0HDVXUHWKHOHQJWKVRIWKUHHVLGHVRIWKHWULDQJOHGUDZQ 4. 'LVFXVVZLWK\RXUPHPEHUVRQWKHIROORZLQJTXHVWLRQV D :KLFKVLGHLVWKHORQJHVW" E ,VWKHORQJHVWVLGHRSSRVLWHWRWKHULJKWDQJOH" 5. 3UHVHQW\RXUÀQGLQJVLQFODVV 1 ,QDULJKWDQJOHGWULDQJOHWKHORQJHVWVLGHLVFDOOHGDhypotenuse 7KHK\SRWHQXVHLVWKHVLGHWKDWLVDOZD\VRSSRVLWHWRWKHULJKWDQJOH How can you identify the hypotenuse of a right-angled triangle? Critical Thinking Hypotenuse EXAMPLE 4 ,GHQWLI\DQGVWDWHWKHK\SRWHQXVH LQ WKH GLDJUDP Solution: TQ LV WKH K\SRWHQXVH RIWULDQJOHPQT. TQ LVWKH ORQJHVW VLGH RI WULDQJOHPQT. SR LV WKH K\SRWHQXVH RIWULDQJOHQRS. SR LVWKH ORQJHVW VLGH RI WULDQJOHQRS. T S R Q P The hypotenuse is only found in right-angled triangles. Discuss with your classmates and present to the class why the side labelled as c is not the hypotenuse. INTERACTIVE ZONE 95° c c ©Praxis Publishing_Focus On Maths
CHAPTER 2 Pythagoras’ Theorem 24 B Pythagoras’ theorem Objective:7RGHWHUPLQHWKHUHODWLRQVKLSEHWZHHQWKHVLGHVRIDULJKWDQJOHGWULDQJOH Instruction:'RWKLVDFWLYLW\LQJURXSVRIIRXU Material:6TXDUHJULGVFLVVRUVJOXH A B Square R Square Q Square P C a c b 1. 'UDZ D ULJKWDQJOHG WULDQJOH ACB ZLWK OHQJWK RI VLGH a XQLWV DQG OHQJWK RI VLGH b XQLWVRQDVTXDUHJULG 2. 'UDZWKUHHVTXDUHVPQDQGRRQVLGHVabDQGcUHVSHFWLYHO\ 3. &RORXUVTXDUHPDQGVTXDUHQDVVKRZQDERYHVRWKDW\RXFDQVWLOOVHHWKHJULGOLQHV 4. &XWVTXDUHQ7KHQFXWVTXDUHPLQWRVPDOOVTXDUHV 5. 3DVWH WKH FXWRXW RI VTXDUHQ RQ VTXDUH R DQG SDVWH WKH VPDOO VTXDUHV RIP RQ WKH UHPDLQLQJVSDFHRIVTXDUHR 6. 2EVHUYHZKHWKHUVTXDUHQDQGWKHVPDOOVTXDUHVFXWRXWIURPVTXDUHPFRPSOHWHO\ÀOO VTXDUHR 7. :KDW LV WKH UHODWLRQVKLS EHWZHHQ WKH DUHDV RI WKH WKUHH VTXDUHV" 0DNH D FRQFOXVLRQ DERXWWKHUHODWLRQVKLSEHWZHHQWKHOHQJWKVRIWKHVLGHVRIWKHULJKWDQJOHGWULDQJOHACB 8. 5HSHDW6WHSVWRZLWKWKHOHQJWKRIVLGHa XQLWVDQGOHQJWKRIVLGHb XQLWV RQWKHVTXDUHJULGV 9. 3UHVHQW\RXUÀQGLQJVLQFODVV 2 )URP$FWLYLW\ ZH IRXQGWKDW $UHDRIVTXDUHR $UHDRIVTXDUHP$UHDRIVTXDUHQ c 2 = a2 + b2 The Pythagoras’ theoremVWDWHVWKDWIRUDULJKWDQJOHGWULDQJOHWKHVTXDUH RIWKHK\SRWHQXVHLVHTXDOWRWKHVXPRIWKHVTXDUHVRIWKHRWKHUWZRVLGHV c 2 = a 2 + b2 c b a ©Praxis Publishing_Focus On Maths
Pythagoras’ Theorem CHAPTER 2 25 EXAMPLE 5 6WDWH WKH UHODWLRQVKLS EHWZHHQ WKH OHQJWKV RI VLGHV RI WKH IROORZLQJULJKWDQJOHGWULDQJOHV (a E R P Q Solution: (a) r 2 = p 2 + q 2 E PR 2 = PQ 2 + QR 2 r q p Pythagoras’ theorem can only be applied to a right-angled triangle. C Pythagorean triples A Pythagorean tripleLVDVHWRISRVLWLYHLQWHJHUVabDQGcWKDWVDWLVÀHVWKH3\WKDJRUDV· WKHRUHP c 2 = a 2 + b2 EXAMPLE 6 'HWHUPLQHZKHWKHUWKHIROORZLQJLVD3\WKDJRUHDQWULSOH D flDQG E DQG Solution: (a) Let a flb c c 2 = a2 + b2 2 fl2 + 152 flffi flffi flffiLVDWUXHVWDWHPHQW fl LV D 3\WKDJRUHDQWULSOH EHFDXVHWKH VHW RI LQWHJHUVVDWLVÀHVWKH3\WKDJRUDV·WKHRUHPc 2 = a2 + b2 E /HWa b c = 15 c 2 = a2 + b2 152 2 + 132 225 = 49 + 169 flLVDIDOVHVWDWHPHQW LVQRWD3\WKDJRUHDQWULSOHEHFDXVHWKHVHW RI LQWHJHUV GRHV QRW VDWLVI\ WKH 3\WKDJRUDV· WKHRUHP c 2 = a2 + b2 c LVWKH ORQJHVW VLGH Common Pythagorean triples in a right-angled triangle are: (3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (20, 21, 29) ©Praxis Publishing_Focus On Maths
CHAPTER 2 Pythagoras’ Theorem 26 EXAMPLE 7 :ULWH WKH YDOXH RI x LQ HDFK RI WKH IROORZLQJ ULJKWDQJOHG WULDQJOHV D E x 4 cm 3 cm x 15 cm 17 cm Solution: (a) x 2 = 32 + 42 x = 32 + 42 = 25 FP E 2 = x 2 + 152 x 2 2 – 152 x = 2 – 152 = 64 flFP D Using Pythagoras’ theorem (I) Finding the length of the unknown side of a right-angled triangle 7KH3\WKDJRUDV·WKHRUHPFDQEHXVHGWRÀQGWKHOHQJWKRIWKHXQNQRZQVLGHRIDULJKWDQJOHG WULDQJOH ,QWKHGLDJUDPVKRZQ c 2 = a2 + b2 so a2 + b2 = c 2 a2 = c 2 – b2 or b2 = c 2 – a2 Work with your team members. Two numbers in a Pythagorean triple are 20 and 29. Find the third number. How many solutions are possible? team work c a b The length of the hypotenuse of a right-angled triangle can EHGHWHUPLQHGXVLQJDVFLHQWL¿F calculator. For example, in example 7(a), press Pol ( 3 , 4 ) = 5 Engineering A housing contractor and a civil engineer use the Pythagoras’ theorem to solve problems involving right angles in the construction of buildings. Engineers use the Pythagoras· theorem to ensure that the corners’ angles are correct while creating a foundation. The Pythagoras’ theorem allows you to compute the diagonal length joining two straight lines. Maths LINK ©Praxis Publishing_Focus On Maths
Pythagoras’ Theorem CHAPTER 2 27 +RZFDQ\RXXVHWKH3\WKDJRUDV¶WKHRUHPWR¿QGWKHOHQJWKRIWKHGLDJRQDOLQDUHFWDQJOH" Critical Thinking (II) Finding the lengths of sides of geometric shapes EXAMPLE 8 ,QWKHGLDJUDPVKRZQABCLVDVWUDLJKW OLQH)LQGWKHOHQJWKRICD Solution: BD = 102 – 62 = 64 flFP CD = 152 fl2 = flffi FP EXAMPLE 9 7KHGLDJUDPVKRZVDQHTXLODWHUDOWULDQJOH PQR)LQGWKHKHLJKWRSRIWKHWULDQJOH *LYH \RXU DQVZHU FRUUHFW WR GHFLPDO SODFHV Solution: PR = PQ = QR FP PS FP¸ FP RS = 62 – 32 = FP EXAMPLE 10 7KH GLDJUDP VKRZV D ULJKWDQJOHG WULDQJOHKLM)LQGWKHOHQJWKRIKM*LYH \RXUDQVZHUFRUUHFWWRGHFLPDOSODFHV Solution: LM = KL FP KM = 52 + 52 = 50 FP A B C D 10 cm 6 cm 15 cm P S Q R 6 cm 60° L M K 5 cm 45° M fl ï ffi ï VR 6KLM LV DQ LVRVFHOHV WULDQJOH ZLWK KL = LM. ©Praxis Publishing_Focus On Maths
CHAPTER 2 Pythagoras’ Theorem 28 EXAMPLE 11 7KH GLDJUDP VKRZV D UHFWDQJOH ABCD )LQG WKH OHQJWK RI WKH GLDJRQDOAC Solution: AC = $%ɧ2 + %&ɧ 2 = 122 + 52 = 169 FP A D C B 5 cm 12 cm The diagram shows a right-angled triangle ABC. A C 4 cm B Find the length of AC and AB. 30° Hint: • The interior angles of a triangle add up to 180°. • Each interior angle of an equilateral triangle is 60°. EXAMPLE 12 P Q T S R 6 cm 5 cm 9 cm 7KH GLDJUDP VKRZV D NLWH PQRS )LQG WKH OHQJWK RI WKH GLDJRQDOPR*LYH\RXUDQVZHUFRUUHFWWRGHFLPDOSODFH Solution: ST ¸ FP PR = PT + TR = 36 ɧ2 ï67 ɧ2 + 65ɧ2 ï67 ɧ2 = 52 ï2 + 92 ï2 = 16 + fl FP EXAMPLE 13 7KH GLDJ UDP VKRZ V D FXERLG ABCDEFGH)LQGWKHSHULPHWHURIWKH VKDGHG SODQH ACGE FRUUHFW WR VLJQLÀFDQWÀJXUHV P T S R 3 cm 5 cm 9 cm E H D A F G C 5 cm 7 cm 6 cm B a b d2 d1 The length of diagonal, d2 , can be determined using the formula: d2 = a2 í 1 2 d1 2 + b2 í 1 2 d1 2 where a = length of the shortest side b = length of the longest side d1 = length of the diagonal which is bisected by another diagonal ©Praxis Publishing_Focus On Maths
Pythagoras’ Theorem CHAPTER 2 29 Solution: AC = $%ɧ2 + %& ɧ2 = 62 2 = fl ffiFP 3HULPHWHURIVKDGHGSODQHACGE ffi ACGE LV D UHFWDQJOH fl flFP (III) Identifying the types of triangles The FRQYHUVHRI3\WKDJRUDV·WKHRUHPFDQEHXVHGWRVKRZWKDWDWULDQJOHLVDULJKWDQJOHG WULDQJOH 7KH FRQYHUVHRI3\WKDJRUDV·WKHRUHP VWDWHVWKDW LQ DWULDQJOH LIWKH VTXDUH RIWKH ORQJHVW VLGHLVHTXDOWRWKHVXPRIWKHVTXDUHVRIWKHRWKHUWZRVLGHVWKHQWKHDQJOHRSSRVLWHWKH ORQJHVWVLGHLVDULJKWDQJOH Let cEHWKHORQJHVWVLGH D ,Ic 2 = a2 + b2 WKHQWKHWULDQJOHLVULJKWDQJOHGDWC E ,Ic 2 a2 + b2 WKHQWKHWULDQJOHLVREWXVHDQJOHGDWC F ,Ic 2 a2 + b2 WKHQWKHWULDQJOHLVDFXWHDQJOHGDWC 6ABC LV D ULJKWDQJOHG WULDQJOH C A B a b c EXAMPLE 14 'HWHUPLQHZKHWKHUWULDQJOHVZLWKWKHIROORZLQJVLGHVLQFP DUHULJKWDQJOHGWULDQJOHV D fl E ffi Solution: D 2 flffi 6TXDUH RIWKH ORQJHVW VLGH fl2 + 152 flffi 6XP RI VTXDUHV RIWKH RWKHU WZR VLGHV 2 fl2 + 152 ,WLVDULJKWDQJOHGWULDQJOH E 2 = 256 92 + 122 = 225 162 92 + 122 ,WLVQRWDULJKWDQJOHGWULDQJOH EXAMPLE 15 'HWHUPLQHZKHWKHUWKHWULDQJOHVZLWKWKHIROORZLQJVLGHVDUH ULJKWDQJOHGREWXVHDQJOHGRUDFXWHDQJOHGWULDQJOHV D FPFPDQGFP E FPFPDQGflFP F FPffiFPDQGFP C is a right angle (90°). C is an obtuse angle (90°). C is an acute angle (90°). C A B C A B C A B ©Praxis Publishing_Focus On Maths
CHAPTER 2 Pythagoras’ Theorem 30 1DPH WKH K\SRWHQXVH RI HDFK RI WKH IROORZLQJULJKWDQJOHGWULDQJOHV D E r p q x z y F G R P Q TU S :ULWHWKHUHODWLRQVKLSEHWZHHQWKHVLGHVRI HDFKRIWKHIROORZLQJULJKWDQJOHGWULDQJOHV D E x y z v u w F G J K L A B C 'HWHUPLQHZKHWKHUHDFKRIWKHIROORZLQJLV D3\WKDJRUHDQWULSOH D flDQG E DQG )LQGWKH OHQJWK RIWKH K\SRWHQXVH LQ HDFK RIWKHIROORZLQJULJKWDQJOHGWULDQJOHV D E A C B 7 cm 24 cm 12 cm 5 cm E G F F G A 3 cm 4 cm B C L 20 cm 21 cm K M H I P 8 cm 15 cm R Q T S 40 cm 9 cm U )LQGWKHYDOXHRIxLQHDFKRIWKHIROORZLQJ ULJKWDQJOHGWULDQJOHV FRUUHFWWR GHFLPDO SODFHVZKHUHQHFHVVDU\ D E x 5 cm 6 cm x 10 cm 9 cm F G x 12 cm 13 cm x 40 cm 41 cm H I x 12 cm 8 cm x 5 cm 7 cm )LQGWKHYDOXHRIxLQHDFKRIWKHIROORZLQJ GLDJUDPVFRUUHFWWRGHFLPDOSODFHVZKHUH QHFHVVDU\ D E x P T Q S R 12 cm 8 cm 6 cm D A B C 15 cm 12 cm 5 cm x Practice 2.2 Basic Intermediate Advanced Solution: (a) 52 = 25 32 + 42 = 9 + 16 = 25 52 = 32 + 42 ,WLVDULJKWDQJOHGWULDQJOH E fl2 = 64 52 + 62 = 25 + 36 = 61 fl2 52 + 62 ,WLVDQREWXVHDQJOHGWULDQJOH F 2 = 100 2 + 92 ffifl 102 2 + 92 ,WLVDQDFXWHDQJOHGWULDQJOH ©Praxis Publishing_Focus On Maths
Pythagoras’ Theorem CHAPTER 2 31 F x T S Q P R 3 cm 5 cm 3 cm 25 cm 'HWHUPLQH ZKHWKHU WKH IROORZLQJ WULDQJOHV DUHULJKWDQJOHGWULDQJOHV (a) 26 cm 24 cm 10 cm E 8 m 6 m 13 m F 2 cm 2.5 cm 1.5 cm G 0.3 m 1.2 m 0.5 m 'HWHUPLQH ZKHWKHU WULDQJOHV ZLWK WKH IROORZLQJ VLGHV DUH ULJKWDQJOHG REWXVH DQJOHGRUDFXWHDQJOHGWULDQJOHV D FPFPFP E flFPFPFP F PflPP G ffiPPP 24 cm 15 cm 8 cm S R P Q ,QWKHGLDJUDPDERYHPQRSLVDWUDSH]LXP ZKHUH PS DQG QR DUH WZR SDUDOOHO VLGHV )LQGWKHOHQJWKRIPQ )LQGWKHOHQJWKRIWKHXQNQRZQGLDJRQDORI HDFKRIWKHIROORZLQJ D E 9 cm 12 cm 24 cm 16 cm 37 cm 7KHOHQJWKRIRQHVLGHRIDVTXDUHLVflFP )LQG WKH OHQJWK RI LWV GLDJRQDO *LYH \RXU DQVZHUFRUUHFWWRGHFLPDOSODFHV S R P Q ,Q WKH GLDJUDP PQRS LV D UKRPEXV ZLWK VLGH ffi FP HDFK *LYHQ WKDW WKH OHQJWK RI GLDJRQDO QS LV fl FP ÀQG WKH OHQJWK RI GLDJRQDOPR B C 7 cm E F H G A D 7KH GLDJUDP DERYH VKRZV D FXERLG )LQG WKHOHQJWKRIAF*LYH\RXUDQVZHUFRUUHFW WRGHFLPDOSODFHV F E K G C D J 8 cm 10 cm 9 cm H 7KHGLDJUDPDERYHVKRZVDFXERLG)LQG D WKHOHQJWKRIEJ E WKHOHQJWKRIFJ F WKHSHULPHWHURIVKDGHGSODQHEFJ *LYH \RXU DQVZHUV FRUUHFW WR GHFLPDO SODFHV P Q S R 45° 45° 4 cm 7KHGLDJUDPDERYHVKRZVDVTXDUHPQRS )LQG WKH OHQJWK RI PQ *LYH \RXU DQVZHU FRUUHFWWRGHFLPDOSODFH 7KH OHQJWK RI RQH VLGH RI UKRPEXVPQRS LV FP ZKLOHWKH OHQJWK RI LWV GLDJRQDO LV flFP*LYHQWKDWPQR ÀQG D WKHSHULPHWHURIWKHUKRPEXV E WKHDUHDRIWKHUKRPEXV ©Praxis Publishing_Focus On Maths
CHAPTER 2 Pythagoras’ Theorem 32 2.3 Solving Problems Involving Pythagoras’ Theorem EXAMPLE 16 7KH GLDJUDP VKRZV D WUDSH]LXP PQRS DQG D ULJKWDQJOHG WULDQJOH PTQ&DOFXODWHWKHDUHDLQFP2 RIWKH VKDGHGUHJLRQ Solution PQ = 62 fl2 = 100 FP $UHDRIVKDGHGUHJLRQ $UHDRIWUDSH]LXPPQRS²$UHDRIWULDQJOHPTQ = 1 2 (14 + 10) = 12 – 1 2 = 6 =fl = 144 – 24 FP2 EXAMPLE 17 ,Q WKH GLDJUDP WQRS DQG VWTU DUH VTXDUHV ,I WKH DUHD RI VWTU LV FP2 FDOFXODWH WKH DUHD LQ FP2 RIWQRS Solution VQ 2 = PQ 2 + PV 2 = 242 2 = 625 7KHUHIRUH VQ = 625 FP *LYHQWKDWWKHDUHDRIVWTULVFP2 7KHUHIRUHVW = 400 FP WQ = VQ – VW = 25 – 20 FP 7KHDUHDRIWQRS = 5 × 5 FP2 S 14 cm R 12 cm 8 cm 6 cm P Q T R V S U T W P 24 cm 7 cm Q ©Praxis Publishing_Focus On Maths
Pythagoras’ Theorem CHAPTER 2 33 EXAMPLE 18 7KHGLDJUDPVKRZVDFXERLG)LQG D WKHOHQJWKRIAN E WKHDUHDRIVKDGHGSODQHANM *LYH\RXUDQVZHUVFRUUHFWWRGHFLPDO SODFHV Solution (a) $1ɧ2 = $'ɧ2 + '1ɧ2 $1ɧ2 2 + 42 $1ɧ2 = 65 AN = 65 flFP E $UHDRIVKDGHGSODQHANM = 1 2 =fl= 9 ANM LV D ULJKWDQJOHG WULDQJOH FP2 S M R P Q D C B N 7 cm 4 cm 6 cm 9 cm A EXAMPLE 19 $VKLSVDLOHGflNPGXHQRUWKWKHQ FKDQJHG LWV FRXUVH DQG VDLOHG NP GXH ZHVW +RZ IDU LV WKH VKLSIURPLWVVWDUWLQJSRLQW" Solution x 2 fl2 + 152 x 2 = 64 + 225 x 2 flffi x = flffi 7KHVKLSLVNPDZD\IURPLWVVWDUWLQJSRLQW x km 15 km 8 km N S W E The diagram shows a ladder AB with the length of 3 m leaning against the wall. Given that point B at the foot of the ladder is 1.8 m from the wall. If the ladder slides down and point A on the ladder moves 0.2 m downwards, calculate the new distance of point B from the wall. 1.8 m B A ©Praxis Publishing_Focus On Maths
CHAPTER 2 Pythagoras’ Theorem 34 EXAMPLE 20 $VWXGHQWZDVJLYHQWKUHHVWLFNVRIOHQJWKVFPFPDQG FPUHVSHFWLYHO\7KHWHDFKHUDVNHGKLPWRXVHWKHVWLFNV WRIRUPDULJKWDQJOHGWULDQJOH&DQKHGRLW" Solution 252 = 625 2 + 242 = 625 252 2 + 242 <HVKHFDQIRUPDULJKWDQJOHGWULDQJOHZLWKWKHVWLFNV 12 cm 13 cm 15 cm S T R P Q ,QWKHGLDJUDPDERYHPQRSLVDUHFWDQJOH &DOFXODWHWKHDUHDRIWKHVKDGHGUHJLRQ $ODGGHURIOHQJWKPOHDQVDJDLQVWDZDOO 7KHIRRWRIWKHODGGHULVPDZD\IURPWKH EDVHRIWKHZDOO+RZIDUXSWKHZDOOGRHV WKHODGGHUUHDFK"*LYH\RXUDQVZHUFRUUHFW WRVLJQLÀFDQWÀJXUHV $ SLORW ÁHZ DQ DLUSODQH NP GXH VRXWK WKHQ WXUQHG DQG ÁHZ fl NP GXH ZHVW +RZIDULVWKHSLORWIURPKLVVWDUWLQJSRLQW" *LYH \RXU DQVZHU FRUUHFW WR GHFLPDO SODFHV K N M L S R Q 10 cm 12 cm 8 cm P 7KH GLDJUDP VKRZV D FXERLG )LQG WKH DUHDRIWKH VKDGHG UHJLRQLPR*LYH \RXU DQVZHUVFRUUHFWWRGHFLPDOSODFHV 7.5 m 12.5 m 10 m 10 m 7.5 m 7KH GLDJUDP DERYH VKRZV D SLHFH RI ZRRGHQ SODQN 'HWHUPLQH ZKHWKHU WKH ZRRGHQSODQNLVLQWKHVKDSHRIDUHFWDQJOH Practice 2.3 Basic Intermediate Advanced Junita wants to make a right-angled triangle out of 3 pieces of wood measuring 36 cm, 48 cm and 64 cm. Will she be able to form the right-angled triangle? Explain. Critical Thinking ©Praxis Publishing_Focus On Maths
35 Pythagoras’ Theorem CHAPTER 2 Summary Pythagoras’ Theorem Pythagoras’ theorem a b c In a right-angled triangle, a2 = b2 + c2 where a is the hypotenuse. Converse of Pythagoras’ theorem a A C b c B If a2 = b2 + c 2 where a is the longest side, then the triangle is right-angled at A. Section A 1. P Q R T S U ,QWKHGLDJUDPDERYHWKHK\SRWHQXVHRIWKH ULJKWDQJOHGWULDQJOHRSULV A ST C UR B RS D SU 2. :KLFKRIWKHIROORZLQJLVnotD3\WKDJRUHDQ WULSOH" A fl C B ffi D 3. :KLFKRIWKHIROORZLQJLVnotDULJKWDQJOHG WULDQJOH" A C 12 cm 13 cm 5 cm 16 cm 30 cm 34 cm B 8 cm D 24 cm 25 cm 16 cm 12 cm 20 cm 4. ,QWKH GLDJUDPPQR LVDWULDQJOHRLV A DULJKWDQJOH B DQREWXVHDQJOH C DQDFXWHDQJOH D DUHÁH[DQJOH 5. 9 cm 15 cm A BCD ,QWKH GLDJUDPBCD LV D VWUDLJKW OLQH DQG BC = CD )LQG WKH OHQJWK LQ FP RI AC FRUUHFWWRGHFLPDOSODFHV A ffi C B ffifl D fl 6. P Q R S 17 cm 8 cm 10 cm 7KH GLDJUDP VKRZV WZR ULJKWDQJOHG WULDQJOHV PQR DQG PRS )LQG WKH OHQJWK LQFPRIPQFRUUHFWWRGHFLPDOSODFHV A C B fl D 9 cm 7 cm 8 cm P Q R ©Praxis Publishing_Focus On Maths