Teacher's Guide - Pearson

IntroductionThis Teacher’s Guide will assist you in interpreting and presenting the activities inthe accompanying Student’s Book. It begins with an introduction to the RevisedJunior Secondary School syllabi.Introduction to the Revised Junior Secondary School syllabi• y The Revised Junior Secondary programme has not been drastically changed.• y The revised syllabi aim to convey the ideals reflected in the Revised NationalPolicy on Education as well as Vision 2016.• y They are still based on the ten-year basic education philosophy and equipstudents with knowledge and skills that are relevant in today’s world.• y Students are encouraged to excel within their own capabilities. Teachers areencouraged to use participatory teaching and diverse learning approaches asindividual talents, needs and learning styles are recognised.• y This inclusive approach urges teachers to accommodate all children regardless oftheir physical, intellectual, social, emotional, linguistic or other conditions.• y The syllabi also focus on the inculcation of attitudes and values that need to benurtured.• y Emerging issues help students to understand and cope with the challenges anddevelopments happening around them. These emerging issues, such asenvironmental education, HIV/AIDS education, gender equality, and the world ofwork, are infused into the syllabi.• y Assessment is a crucial part of the teaching and learning process, and should alsoconsider those students with special needs.Features of the Student’s Book• y The Student’s Book is divided into chapters. The first page of each chapter lists theobjectives that will be covered in the chapter.• y At the start of the chapter, you will also find a list of key skills that students willacquire while engaging with the activities.• y Activities are used to explore, experiment, research and discover, while exercisesreinforce the work the students have learnt.• y We have provided additional activities for students of mixed abilities. There areadditional activities for students who need additional stimulation. They are calledextension activities. There are also activities for students who need extra time toreflect on concepts, and practise what they have learnt. They are called supportactivities.• y Emerging issues boxes link current social and economic issues to the environmentof students and make the learning more practical.• y There are also case studies, investigations and homework exercises to reinforcelearning.• y Summaries help the students reflect on the work they have learnt.Introductionv

c h a p t e r1 NumbersSyllabus: General objectiveThis chapter deals with numbers and number concepts. Students must have a clearunderstanding of all the different types of numbers.Specific objectives• y Classify numbers into the categories in which they belong.• y Use a calculator to deal with numbers.Students have to develop the skills to do the four basic mathematical operations –multiply, divide, add and subtract – with very large and very small numbers inscientific notation and normal notation on their calculators. Students should knowhow to convert between the different modes and be able to interpret numbers inscientific notation and normal notation.Help students develop a ‘feel’ for significant numbers. For example, the budget ofBotswana is measured in billions of pula. This means that amounts of P500 orP1 000 do not have any significance when compared to such a large figure.However, values of P1.00 and P0.50 are significant when referring to, for example,the price of petrol.AnswersActivity 1.1 (Student’s Book page 4)1. a) –4 ∈ Z, IQ, IRb)1__ ​ ∈ IQ, IR3c) –2 ∈ Z, IQ, IR2. a) Trueb) Falsec) TrueNote: π is a transcendental number, because it does not originate from an integervalue. The value of ​√ ___13 ​is irrational because it is non-recurring and 13 is an integer.For the purpose of this book, students should regard π as irrational.Chapter 1: Numbers1

Extension Activity 1.1 (Student’s Book page 4)1. True2. False3. TrueSupport Activity 1.1 (Student’s Book page 6)1. a) 0.2b) 0.428571428…c) 2.666666666…d) 0.625e) 0.818181818…2. a) 4b) 3.605551275…c) 3.141592654…3. a)1__5 ​, ​ 5__ ​and ​√___ 16 ​8b)3__7 ​, 2​ 2__3 ​ and ___ ​ 911 ​c) ​√ ___13 ​ and πActivity 1.2 (Student’s Book page 6)1. a) 0.2b) 0.428571428…c) 2.666666666…d) 0.625e) 0.818181818…f) 0.111111111…g) 0.8h) 0.87i) 1.058823529…j) 0.944444444…2. a) 4b) 3.605551275…c) 3.141592654…d) 1.570796327…e) 0.666666666…f) 2.53. a)1__5 ​, ​ 5__8 ​, ___ ​ 1215 ​, ____ ​ 522​, ​√___ 16 ​,600​√ __4__9 ​ ​and ___ 5522 ​b)3__7 ​, 2​ 2__3 ​, ___ ​ 911 ​, ​ 1__9 ​, ___ ​ 1817 ​; ___ ​ 1718 ​and √ __ ​ 4__ ​9c) ​√ ___13 ​, π and π__2 ​2 EXPLORING MATHEMATICS FORM 3

​​​​​​1.Support Activity 1.2 (Student’s Book page 7)1__5 ​ ∈ IQ​3__7 ​ ∈ IQ2​2__3 ​ ∈ IQ​5__8 ​ ∈ IQ___ 911 ​ ∈ IQ​1__9 ​ ∈ IQ___ 1215 ​ ∈ IQ____ 522600 ​ ∈ IQ___ 1817 ​ ∈ IQ___ 1718 ​ ∈ IQ​√ ___16 ​ ∈ IQ​√ ___13 ​ ∈ IQ'π ∈ IQ'​π__2 ​ ∈ IQ'​√ __4__ ​ ​∈ IQ9___ 5522 ​ ∈ IQ2. IR IQ IQ' 0Z Undefineda) 5 ✗ ✗b) ​√ __5 ​ ✗ ✗ ✗ ✗ ✗c)2__0 ​ ✗ ✗ ✗ ✗ ✗ ✗ d) –4 ✗ ✗ ✗ ✗e) 1.1 . 23 . ✗ ✗ ✗ ✗ ✗f) π 2 ✗ ✗ ✗ ✗ ✗g)7__9 ​ ✗ ✗ ✗ ✗ ✗h) ​√ __9 ​ ✗ ✗i) ​ 3 √ ____–27 ​ ✗ ✗ ✗ ✗Chapter 1: Numbers3

Activity 1.3 (Student’s Book page 8)1. See Support Activity 1.2 above.2. IR IQ IQ' 0Z Undefineda) 5 ✗ ✗b) ​√ __5 ​ ✗ ✗ ✗ ✗ ✗c)2__0 ​ ✗ ✗ ✗ ✗ ✗ ✗ d) –4 ✗ ✗ ✗ ✗e) 1.1 . 2 . 3 . ✗ ✗ ✗ ✗ ✗f) π 2 ✗ ✗ ✗ ✗ ✗g)7__9 ​ ✗ ✗ ✗ ✗ ✗h) ​√ __9 ​ ✗ ✗i) ​ 3 √ ____227 ​ ✗ ✗ ✗ ✗j) ______ (6 2 6)​ ✗ ✗ ✗6k) 4.5 ✗ ✗ ✗ ✗ ✗l) 20.00098 ✗ ✗ ✗ ✗ ✗m) 0.012301230123 ✗ ✗ ✗ ✗ ✗n) 3.1415926535… ✗ ✗ ✗ ✗ ✗3. No. ___ ​22​is only an approximation of the value of π. π is irrational, so we cannot7express it as a fraction in the form Z__Z ​.Exercise 1.1 (Student’s Book page 10)1. a) 1​1__9 ​b)4__9 ​287c) 3​ ____999 ​d)5__9 ​10e) 100​ ___99 ​101f) 100​ ____999 ​1 001g) 100​ _____9 999 ​7 331h) 4​ _____9 999 ​2. Write the fraction without the decimal point: 1 239. Subtract 1 from this numberto get 1 238. Divide 1 238 by 999 to get the original decimal fraction of 1.2 . 39 . .4 EXPLORING MATHEMATICS FORM 3

Exercise 1.3 (Student’s Book page 16)1. a) 4.97b) 0.24c) 21.11d) 1.01e) 10.25Extension Activity 1.3 (Student’s Book page 17)0.557Revision Exercise (Student’s Book page 17)1. a) Irrationalb) Rationalc) Rationald) Rationale) Rationalf) Irrational2. a) 0.97 . 8.85 ​ ___9 ​ 5 ___ ​ 8890 ​ 5 ___ ​ 4445 ​b) 0.1 . 27 . 5 0.127127127… 5 ____ ​127999 ​c) 5.3 . 5 . 5 5.353535… 5 ____ 530​99 ​d) 0.001 . 5 0.00111… 5 ______ ​11199 900 ​3. a) 0.8 . 1 .b) 1.3 . 11 .c) 5.71 . 3 .d) 0.0 . 01 .4. a) Trueb) False (Fractions – common and decimal – and irrational numbers are alsoreal numbers.)c) Trued) False (​√ ___16 ​= 4 ∈ IQ)e) False (π is an irrational and real number, but not a natural number.)f) False (​0__ ​is undefined.)0g) Trueh) False (7 < ​√ ___50 ​< 8)i) False (A prime number has two different factors, namely, 1 and itself.)j) False (All mixed fractions are rational.)5. a) 1.806 or 1.782b) 0.978c) 3.7856 EXPLORING MATHEMATICS FORM 3

2c h a p t e rPatterns and sequencesSyllabus: General objectiveThis chapter deals with number patterns and sequences. Students must be able toidentify number patterns, and make connections between patterns in nature andarithmetic sequences.Specific objectives• y Describe in words the pattern of a linear sequence.• y Express the pattern of a linear sequence in algebraic form.• y Generate a sequence of numbers.• y Use the patterns of sequences to represent given situations and use them to solvereal-life problems.Teach students the skill of looking for number patterns or patterns that aremathematically significant. Remind students to be aware of the world around themand that mathematical patterns appear frequently.AnswersActivity 2.1 (Student’s Book page 21)1. T 55 11 Sequence starts with 3. Add 2 to the previous term.2. T 55 8 Sequence starts with 24. Add 3 to the previous term.3. T 55 29 Sequence starts with 7. Subtract 4 from the previous term.4. T 55 25 Sequence starts with 1. Square the number of the term.5. T 55 225 Sequence starts with 25. Subtract 5 from the previous term.6. T 55 3 125 Sequence starts with 5. Multiply the previous term by 5.7. T 55 232 Sequence starts with 22. Multiply the previous term by 22.8. T 55 275 Sequence starts with 23. Multiply 23 by the square of the numberof the term.Chapter 2: Patterns and sequences7

Activity 2.2 (Student’s Book page 21)Pattens 1 (12), 2 (13), 3 (24) and 5 (25) are linear.Exercise 2.1 (Student’s Book page 22)1. Linear. Add 20.3 to the previous term.2. Not linear.2. Not linear.4. Linear. Add 2 to the previous term.5. Not linear.6. Linear. Add 23.5 to the previous term or subtract 3.5.Activity 2.3 (Student’s Book page 23)1. T n5 23 1 (n 2 1) 3 3 5 23 1 3(n 2 1) 5 23 1 3n 2 3 5 3n 2 6;T 105 3(10) 2 6 5 30 2 6 5 242. T n5 n 2 1​1__2 ​; T 5 8​ 1__10 2 ​3. T n5 0.3n 2 1; T 105 24. T n5 6 2 2n; T 105 214Exercise 2.2 (Student’s Book page 23)1. Sequence starts with 214. Add 14 to the previous term; T n5 14n 2 28; T 85 842. Sequence starts with 7. Add 2 to the previous term; T n5 2n 1 5; T 85 213. Sequence starts with 8. Subtract 3 from the previous term; T n5 1 2 3n; T 85 2134. Sequence starts with 24. Subtract 5 from the previous term; T n5 1 2 5n;T 85 2395. Sequence starts with 1__3 ​. Add ​ 1__3 ​to the previous term; T 5 ​ 1__n 3 ​n; T 5 2​ 2__8 3 ​6. Sequence starts with 1 1__4 ​. Subtract ​ 1__2 ​from the previous term; T 5 1​ 3__n 4 ​ 2 ​ 1__2 ​n;T 85 22​1__4 ​Support Activity 2.1 (Student’s Book page 24)1. T 205 40; T 505 1002. 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; …3. T 505 160Activity 2.4 (Student’s Book page 25)1. a) 2; 5; 10; 17 b) 3; 7; 11; 15 c) 33; 34; 35; 362. a) 401 b) 79 c) 523. b) and c)8 EXPLORING MATHEMATICS FORM 3

Exercise 2.3 (Student’s Book page 25)1. 23; 21; 1; T 105 15 (linear)2. 1; 23; 27; T 105 235 (linear)3. 12; 13; 14; T 105 21 (linear)4. 27​1__2 ​; 28; 28​ 1__2 ​; T 5 212 (linear)105. 4; 6.5; 9; T 105 26.5 (linear)6. 0; 6; 16; T 105 198Exercise 2.4 (Student’s Book page 27)1. a) T n5 13 1 5nb) T 405 2132. a) T 305 100b) T 505 180, so there are 50 terms (n 5 50).Extension Activity 2.1 (Student’s Book page 27)1. T n5 n 2 3 (21) n2. T n5 ______ n​(n 1 1) ​3. T n5 (n 2 )(3 n )4. T n5 4n 2 75. T n5 _______ (2n 2 1)​(3n 2 2) ​Exercise 2.5 (Student’s Book page 28)1. a) d 5 1​1__2 ​b) T n5 1​1__2 ​n 1 3​ 1__2 ​c) T 1005 153​1__2 ​d) n 5 1502. 3; 7; 113. a) 96 is not a term in this sequence.b) T n5 6n 1 10200 5 6n 1 10∴ n 5 31.6 .∴ 200 is not a term in this sequence because n ∈ .4. a) T 415 18.7b) T 45 0.2 ∴ T 4is closest to zero.c) T 775 36.7d) 23 terms are less than 10.Chapter 2: Patterns and sequences9

Extension Activity 2.2 (Student’s Book page 28)1. The first difference for this sequence is 2; 1; 0; 21; …. Therefore, the constantsecond difference is 21; 21; 21; 21; …, because the pattern in the first differenceis to subtract 1 from the previous term.2. T 15 a(1) 2 1 b(1) 5 3T 25 4a 1 2b 5 5Solve simultaneously:a 5 2 ​1__ ​; b 5 3.52∴ T n5 2 ​1__2 ​n2 1 3.5n3. T 65 3; T 75 0; T 105 2154. No, the general term shows us that this is a quadratic sequence.Exercise 2.6 (Student’s Book page 30)1. a) 17 logsb) 10 rows2. a) 15 stonesb) 9 rows3. T n5 25n 1 25∴ 350 5 25n 1 25∴ n 5 13∴ They will swim 350 m in the 13 th session.4. a) P2 200b) Year 75. a) A 5 660n 1 6 000b) T 45 P8 640; T 55 P9 300; T 65 P9 960; T 75 P10 620; T 85 P11 280c) T 205 P19 200d) 35 yearsInvestigation (Student’s Book page 32)1. T 15 1 2 2 1 1 11 5 11 T 45 23T 25 2 2 2 2 1 11 5 13 T 55 31T 35 3 2 2 3 1 11 5 17Yes, the conjecture seems to be correct, because these five numbers are all primenumbers.2. The first five terms are prime numbers, but not all the prime numbers between11 and 31 are generated: 19 and 29 are missing.3. The conjecture does not hold water for all values of n because n 5 11 generates aperfect square. Using a spreadsheet we find that, among the first 40 terms, thefollowing terms are not prime: T 11; T 12; T 15; T 25; T 28; T 33; T 34; T 38; … .10 EXPLORING MATHEMATICS FORM 3

Extension Activity 2.3 (Student’s Book page 34)1. 13; 6; 21; 28; …2. c)Revision Exercise (Student’s Book page 35)1. a) No, because substituting the value of n into the rule does not give the valueof the number at that position in the sequence.b) T n5 3n 1 3c) T 1005 3032. a) 2; 1; 0; 21; 22b) 7; 9; 11; 13; 15c) 22; 3; 8; 13; 18d) 1​1__4 ​; 1​ 3__4 ​; 2​ 1__4 ​; 2​ 3__4 ​; 3​ 1__4 ​e) 14; 33; 52; 71; 90f) 28; 211; 214; 217; 2203. a) 13; 15; 17; T n5 2n 1 3; T 505 103b) 211; 215; 219; 223; 227; T n5 9 2 4n; T 505 2191c) 21; 25; 29; 33; 37; T n5 4n 1 1; T 505 201d) 0.3; 0; 20.3; 20.6; 20.9; T n5 1.8 2 0.3n; T 505 213.2e) 275; 350; 425; 500; 575; T n5 75n 2 25; T 505 3 725f) 21​3__4 ​; 2​ 3__4 ​; 2​ 1__4 ​; ​ 1__4 ​; T 5 0.5n 2 3​ 1__n 4 ​; T 5 21​ 3__50 4 ​4. a) T n5 5n 1 1b) T 2005 1 001c) 120 hexagons5. a) 14b) 4; 8; 12; …c) There is a difference of 4 between terms in the same position.d) T n5 4n 1 36. a) P7 000; P6 550; P6 100; …b) P5 200c) 14 years7. a) 8 000; 7 966; 7 932; 7 898; 7 864b) 119 daysChapter 2: Patterns and sequences11

3re t p a h cIndicesSyllabus: General objectiveThis chapter deals with indices. Students must acquire knowledge on indices andtheir role in real-life problems.Specific objectives• y Interpret integral and positive fractional indices.• y Simplify expressions that contain indices.• y Solve problems that involve integral and positive fractional indices.• y Order numbers 2 in ascending order and descending order 2 expressed instandard form.• y Solve problems that involve numbers expressed in standard form.Teach students how to write and work with powers of numbers. Ensure that they canuse exponential notation to write very large or very small numbers in a moreconvenient way.AnswersActivity 3.1 (Student’s Book page 39)1. x 42. d 103. x 3 y 74. x 9 y 215. 8a 96. 81x 127. 227x 128. 8a 9 b 15 c 129. ​x__4 ​10. 224t 1311. a 4b12. ​ ___5 ca ​ 212 EXPLORING MATHEMATICS FORM 3

Homework Exercise (Student’s Book page 40)Yes. (2 2 ) 3 5 64, but ​2​ (23 )​5 2 8 5 256. The correct answer is 64.Support Activity 3.1 (Student’s Book page 41)1. 3 2. 10 3. 8 4. 5Activity 3.2 (Student’s Book page 41)1. a) 3 b) 10 c)4__d) 29e) 10 f) 4 g) 1 000 h) 8i) 52. a) Undefinedb) Defined: ​ 3 √ ____28 ​ 5 3 ​√ ______(22) 3 ​ 5 22c) Defined: ​ 3 √ _____216 ​ 22.523. a) False: ​√ _______9 1 16 ​ 5 ​√ ___25 ​ 5 5 3 1 4; square and add first, then find the squareroot.b) True: ​√ _______9 1 16 ​ 5 ​√ ____144 ​ 5 12 5 3 3 4Exercise 3.1 (Student’s Book page 42)1. a) x1​ __ ​ 2__ 5​ ​ 3b) ac) 6x 4d) 2a 3 b 2e) ​3​ 9x3 ​or ​27​ 3x3 ​f) Stays unchanged.2. a) 3 √ __a 2 ​b) 3​ 5 √ __x 3 ​c) 4x 2d) a 2e) 3​ 5 √ __x 2 ​f) 8Exercise 3.2 (Student’s Book page 43)1. a) 27b) 32c) 32a 10d) 9a 22. a) 5 ​√ __4 ; 4 ​√ __3 ​ ​b) ​ 3 √ __7 ; 4 ​√ ___12 ​ ​; ​√ __3 ​Chapter 3: Indices13

Activity 3.3 (Student’s Book page 43)1. The size of the flea population doubles every five days. Therefore, the exponentshould be expressed as a multiple of 5.2. 240 fleas3. 20 daysActivity 3.4 (Student’s Book page 44)1. Growth 5 P10 000 1 P10 000 3 ____ 14.5​100 ​5 P10 000(1 1 0.145)5 P10 000 3 1.145∴ Penny’s money is multiplied by 1.145 each year.2. P10 0003. a) A 5 P10 000 3 1.145 25 P13 110.25b) A 5 P10 000 5 1.145 55 P19 680.11c) A 5 P10 000 5 1.145 105 P38 730.664. No, the investment’s value after 5 years is P19 680.11. Therefore, the investmentwill double in the 6 th year.5. P10 000 3 1.145 n 5 P338 000∴ 1.145 n 5 33.8∴ n 26 yearsExercise 3.3 (Student’s Book page 44)1. a) The number of bacteria in the culture doubles every minute.b) 180 bacteriac) t 0 1 2 3 4 5N 5 180 3 2 t 180 360 720 1 440 2 880 5 760d) 180 3 2 t > 40 000∴ 2 t > 222.22∴ t > 7.8∴ t 8 min2. a) 100% 2 5.5% 5 94.5%, which represents the volume of air that remains aftereach day.94.5b) V air5 4 200 3 (​ ____100 ​)t5 4 200 3 0.945 t , where t is the number of daysc) 2 671.18 cm 314 EXPLORING MATHEMATICS FORM 3

d) 4 200 3 0.945 t < 1 000∴ t > 25.4∴ t 26 daysNote: Students should use the trial-and-error method. The inequality sign changesaround because the bases on either side of the equation are fractions. The followingexample may help students to understand it better:If (​1__2 ​)t < (​1__4 ​), then (​ 1__2 ​)t < (​1__2 ​)2 . This is possible only for t > 2.3. a) P 5 2 000 000 3 (1 1 ____ 1.9​100 ​)t5 2 3 10 6 3 1.019 tb) P 5 2 3 10 6 3 1.019 t 2 197 358 peoplec) 1.019 t 5 1.1∴ t 5.1 yearsProject (Student’s Book page 45)1. _____ 42 500 ​ 3 ____ ​ 1001 ​ 5 0.16% On the first day, this doesn’t seem much, but he should be concerned because theplant grows exponentially (the next day it covers double the surface area than theprevious day).2. It is exponential, because the growth doubles every day from the day before, notfrom the first day. A 5 4 3 2 t 2 1 , where t is the number of days starting from1 April. Alternatively,A 5 2 3 2 t 5 2 t 2 1 , where t is the number of days starting from 1 April.3.Surface area (m 2 )Number of days4. t 9.3 days (determined from the graph), so on 9 April it will cover half.5. t 10.3 days (determined from the graph), so on 10 April it will cover the damcompletely.6. Kill all the plants before 3 April, when it will still be manageable (16 m 2 ).Chapter 3: Indices15

Support Activity 3.2 (Student’s Book page 46)1. a) 7.89 3 10 5b) 2.0785 3 10 6c) 2.84 3 10 27d) 5 3 10 272. a) 23 000 000b) 0.000752Exercise 3.4 (Student’s Book page 47)1. a) 4.56 3 10 5b) 8.09 3 10 6c) 9.89 3 10 27d) 1 3 10 27e) 4.50 3 10 25f) 1.23 3 10 8g) 4.35 3 10 21h) 2 3 10 92. a) 87 000 000b) 23 400c) 0.0000000275d) 0.0004Activity 3.5 (Student’s Book page 47)1. 7.5 3 10 3 > 1.22 3 10 3 > 7.05 3 10 2 > 3.45 3 10 22 > 4.53 3 10 232. 9.34 3 10 23 < 8.76 3 10 22 < 9.8101 3 10 1 < 2.0237 3 10 2 < 2.35098 3 10 2 < 3.4 3 10 6Exercise 3.5 (Student’s Book page 48)1. a) 7.5 3 10 8b) 2.6 3 10 3c) 2.1 3 10 25d) 3.4 3 10 112. a) 2.077 3 10 9b) 1.152 3 10 215c) 7.651 3 10 8d) 2.736 3 10 22Activity 3.6 (Student’s Book page 48)1. 319 450 or 3.1945 3 10 5 people2. 122 988 or 1.22988 3 10 5 people16 EXPLORING MATHEMATICS FORM 3

3. Gaborone: P 5 1.86 3 10 5 3 (1 1 ____ 0.85​100 ​)95 2.007 3 10 6 peopleMahalapye: P 5 1.0445 3 10 5 3 (1 1 ____ 0.85​100 ​)95 1.127 3 10 5 peopleLobatse: P 5 2.9 3 10 4 3 (1 1 ____ 0.85​100 ​)95 3.130 3 10 4 peopleExercise 3.6 (Student’s Book page 49)1. 2 3 10 6 2 1.785 3 10 65 2.15 3 10 5 people5 215 000 people2. P 5 1.785 3 10 6 3 (1 1 ____ 0.85​100 ​)95 1.926 3 10 6 peopleThis is not accurate, because the answer should be 2 3 10 6 . The growth rate wasactually closer to 1.27%. Find this value by the following calculation:2 5 1.785(1 1 i) 9i 5 9 ​√ __________ 21.785 2 1 0.0127∴ r 1.27%3. Number of births 5 _____ 24.8​1 000 ​ 3 (2 3 106 )5 49 600 birthsNumber of deaths 5 _____ 13.9​1 000 ​3 (2 3 106 )5 27 800 deaths4. CO 2emissions 5 (2 3 10 6 ) 3 2.45 4.8 3 10 6 units of CO 25. Required total calories 5 (2 3 10 6 ) 3 (2.16 3 10 3 )5 4.32 3 10 9 calories6. Number of people who do not have water 5 0.3 3 (2 3 10 6 )5 6 3 10 5 or 600 000 peopleExtension Activity 3.1 (Student’s Book page 50)1. a) Population density 5 ___________(2 3 10​6 )(5.696 3 10 5 ) ​5 3.5 people/km 2b) Arable land 5 (5.696 3 10 5 ) 3 0.025 1.1392 3 10 4 km 25 11 392 km 2Chapter 3: Indices17

3. Bank A: A 5 P5 000(1 1 0.1) 105 P12 968.71Bank B: A 5 P5 000(1 1 ______ 0.0975​5 P13 203.7212 ​)120Bank B is the better option because it yields P235.01 more than Bank A. However,point out to students that a difference of P235.01 over a 10-year period is notsignificant. Small differences in interest rates in this range only becomesignificant when investing large sums of money.4. a) P220 400 5 P150 000(1 1 i 3 5)∴ i 0.0939 9.39%b) P220 400 5 P150 000(1 1 i) 5∴ i 5 0.085 8%c) P220 400 5 P150 000(1 1 ___ i​12 ​)60 0.0772 7.72%5. a) Option 1: P24 645.48Option 2: P24 812.29Option 3: P24 890.13b) Option 1 is the cheapest. Point out to students that a lower interest rate doesnot necessarily mean that it is cheaper, but that it also depends on thenumber of times the interest is compounded per annum.Exercise 4.3 (Student’s Book page 62)1. P 5 ______ P2 355​1.12 ​5 P2 102.68VAT 5 P2 102.68 3 ____ ​12100 ​5 P252.322. a) A 5 P35 240 3 1.125 P39 468.80b) A 5 P39 468.80 3 1.35 P51 309.44c) A 5 P51 309.44 3 (1 2 0.3)5 P35 916.61P35 916.61d) Profit 5 ​ _________ ​ 2 P35 2401.125 2P3 171.605 P3 171.60 lossNote: Point out to students that a mark-up of 30% followed by a discount by thesame margin on the marked-up price results in a loss. This occurs because we first22 EXPLORING MATHEMATICS FORM 3

multiply by 1.3 and then by 0.7, but 1.3 3 0.7 1. Rather, 1.3 3 0.7 5 0.91. Thisindicates a loss of 1 2 0.91 5 0.09, which is equivalent to 9%.e) % loss 5 ________ P3 171.60​P35 240 ​ 3 ____ ​ 1001 ​5 9%Note: This is the same result as that found in question d).3. a) No, because a 20% reduction would result in a price of:P120 000 3 0.8 5 P96 000.P100 000b) Purchase price 5 ​ ________ ​1.185 P84 745.76Note: It is essential that students understand the difference in calculation betweenmarked-up pricefinding the original price from a mark-up (Original price 5(​ ___________________(1 1 ________________percentage mark-up​​) ) ​100and reducing the price of an article by a certain percentage(reduced price 5 price 3 ​( 1 2 _________________percentage discount​100 )​).4. You can buy from either shop, because both values work out P1 008.Exercise 4.4 (Student’s Book page 65)1. a) A 5 P180 000(1 2 5 3 0.13)5 P63 000b) A 5 P180 000(1 2 0.13) 55 P89 715.772. a) The computers depreciate at P10 000 per annum until their value is P0.P50 000b) _______P10 000 ​ 5 5 yearsNote: Point out that when an article depreciates according to the straight-linemethod, its value eventually becomes zero. In contrast, when an article depreciatesaccording to the reducing-balance method, its value can never become zero. You canintroduce the asymptotic effect here.3. a) P800 5 P(1 2 5 3 0.15)∴ P 5 P3 200b) P800 5 P(1 2 0.15) 2∴ P P1 8034. a) P800 5 P4 500(1 2 6 3 i)∴ i 0.1962 19.6%b) P800 5 P4 500(1 2 i) 6∴ i 5 1 2 6 ​√ ____________ P800P4 500 25%Chapter 4: Money23

5. P8 000 5 P(1 2 0.15) 6∴ P 5 P21 211.756. A 5 P40 000(0.88) 105 P11 140.047. Depreciation 5 ________ P270 000​9 years ​5 P30 000 per annumRate of depreciation 5 ________ P30 000​P270 000 ​ 3 ____ ​ 1001 ​ 11.1%Extension Activity 4.1 (Student’s Book page 65)n 5 12 yearsExercise 4.5 (Student’s Book page 66)1. a) Selling price 5 P850 000 3 1.165 P986 000b) A 5 P986 000(1 1 0.095) 55 P1 552 1992. a) A 5 P110 000(1 1 0.075) 12 P261 996b) A 5 P20 000(1.06) 12 P40 244c) First year: P40 244Second year: P42 659Third year: P45 218Total: P128 121Phembi would have enough money.3. Note: Appreciation on property is compounded annually.Property value 5 P50 000(1.075) 20 P212 393Shares’ value 5 P50 000(1 1 0.12 3 20)5 P170 000Precious should take the property option because the property investment at alower interest rate is more profitable.4. P1 000 000 5 P222 000(1 1 i) 15∴ i 0.11259 11.26%5. 2 5 1(1 1 i) 8i 0.0905 9.05%24 EXPLORING MATHEMATICS FORM 3

6. A 5 P1 500 000(1 1 0.089) 35 P1 937 202Exercise 4.6 (Student’s Book page 67)1. Inflation rate 5 ___________(39.98 2 36)​​ 3 ____ 100​36 1 ​5 11.1%2. P35 890 5 P25 000(1 1 i) 5∴ i 0.07499 7.5%3. A 5 7.89 3 1.0435 P8.23P29.104. Inflation rate 5 2​ _______P289.45 ​ 3 ____ ​ 1001 ​5 210.1%Note: Point out that negative inflation means that the price of consumer goodsdecreases. Although it is unlikely to happen in present times, it is not impossible.5. A 5 P135(1 1 0.073) 25 P155.43Revision Exercise (Student’s Book page 71)1. The National Bank: A 5 P12 000(1 1 0.105) 55 P19 769Invest Bank: A 5 P12 000(1 1 _____ 0.102​2 ​)105 P19 734The National Bank’s deal is better by P35, which is insignificant over a 5-yearperiod.2. P78 600 5 P49 000(1 1 i) 3∴ i 0.1706 17.1%3. a) P1 320 000 5 P682 000(1 1 i) 5∴ i 0.1411 14.1%b) P141 429 2 P73 071 5 P68 358 more VAT4. P8 200 5 P5 600(1 1 __ i​4 ​)1212​∴ i 5 4 3 (​√ ______________ P8 200 ​P5 600 ​ 2 1) 0.1291 12.9%Chapter 4: Money25

5. a) P5.50 3 1.033 7 5 P6.90b) P5.50 5 P3.25(1 1 i) 3∴ i 0.1916 19.2%6. Simple interest: A 5 1(1 1 1 000 3 0.105)5 P106Compound interest: A 5 1(1 1 0.105) 1 0005 P2.3029 3 10 437. a) Interest 5 P50 000 3 _____ 0.095​4 ​5 P1 187.50b) A 5 P50 000(1 1 _____ 0.095​4 ​)125 P66 266.958. Depreciated value 5 P150 000(0.85) 55 P66 555.80New car price 5 P150 000(1.12) 55 P264 351.25Patrick will need P264 351.25 2 P66 555.80 5 P197 795.45.26 EXPLORING MATHEMATICS FORM 3

c h a p t e r5 MatricesSyllabus: General objectiveThis chapter deals with matrices and is an extension of the work covered in Form 2.In this chapter, students learn how to work with 4 3 4 matrices and some of theoperations that are applicable to matrices.Specific objectives• y Multiply matrices that comprise up to 4 columns and 4 rows.• y Investigate the properties of associativity, commutativity and identity of 2 3 2matrices in arithmetic computations.Ensure that students can manipulate matrices algebraically and use a calculator toperform calculations that involve matrices.AnswersSupport Activity 5.1 (Student’s Book page 76)1. Monday(2 3 1Tuesday 1 1 0Thursday 3 2 2Saturday 2 1 32. Meat(42.00Milk 7.50Bread 8.503. No, they are not in the same order.((Extension Activity 5.1 (Student’s Book page 76)1. 0 5 no incident of influenza passing from one person to another1 5 one incident of influenza passing from one person to another2. Column A: from whom person A caught influenzaColumn B: from whom person B caught influenzaColumn C: from whom person C caught influenzaColumn D: from whom person D caught influenzaChapter 5: Matrices27

Row A: to whom person A passed on influenzaRow B: to whom person B passed on influenzaRow C: to whom person C passed on influenzaRow D: to whom person D passed on influenza3. The diagonal represents the values for the relationship of the same person,i.e. A to A, B to B, C to C and D to D.4. For example:Exercise 5.1 (Student’s Book page 78)1. A: 3 3 1 column matrixB: 3 3 3 square matrixC: 2 3 2 square matrixD: 1 3 4 row matrixE: 2 3 1 column matrixF: 1 3 2 row matrixG: 2 3 3 row matrix2. a) Equalb) Not equal3. You can only add and subtract matrices of the same order.4. a) 8 145 13((b) 4 910 210c) (4 14 16)d)(14228242249(e) x 4x8x 6x3x 2xf)(9 0 15132 63 290212 27 233(((((28 EXPLORING MATHEMATICS FORM 3

(g)(6 13211 187 21Activity 5.1 (Student’s Book page 81)(1. Product equals R 13 C 1R 13 C 2for 2 3 2 matrices.R 23 C 1R 23 C 2(In other words, multiply each element in row 1 by each element of the sameposition in column 1, and so on.Note: Ensure that students understand how to add and subtract matrices and why itis important to know this skill. You can introduce the use of a calculator here, butensure that students first understand the mechanism of multiplication.2. DE: (3 3 0) 1 (6 3 21DF: (1 3 2) 1 (5 3 5)(3 3 2) 1 (6 3 5)KF: (3 3 1) 1 (1 3 3); (3 3 2) 1 (1 3 5)(1 3 2) 1 (3 3 5)(22 3 1) 1 (1 3 3); (22 3 21) 1 (1 3 2)3. The first row and the second column give the order. A 2 3 2 times a 2 3 2 yieldsa 2 3 2 matrix; a 4 3 2 times a 2 3 3 yields a 4 3 3 matrix.Activity 5.2 (Student’s Book page 83)1. a) Yes; 2 3 4b) Yes; 3 3 6c) Yes; 1 3 1d) Noe) Yes; 1 3 3f) Nog) Noh) Noi) Yes; 10 3 12j) No2. a) Any two matrices where the number of columns in matrix 1 is equal to thenumber of rows in matrix 2.b) Any two matrices where the condition in question 2a) is not met.Exercise 5.2 (Student’s Book page 84)1. No product is possible.2. 1 3 2; (23 15)Chapter 5: Matrices29

3. 3 3 4;(88 268 245 16225 2130 250 10059 254 276 44. 2 3 3; 219 220 226211 233 815. 1 3 3; (120 132 236)6. No product is possible.7. 2 3 2; 3 522 3((Note: Ensure that students are aware of the result when multiplying the identitymatrix. We shall deal with this concept formally later in the chapter.((((8. 1 3 4; (14 211 14 3)9. 4 3 4;(1 8 5 43 12 7 55 9 27 32 5 24 2Activity 5.3 (Student’s Book page 85)1. B 3 A2.(3 300)3802203. The total number of Economy seats in their fleet is 3 300, the total number ofBusiness seats is 380 and the total number of First class seats is 220.Exercise 5.3 (Student’s Book page 86)1. Wednesday: P206.50Thursday: P254.50Friday: P303.00Saturday: P442.00Total: P1 206.002. a) (550 180)b) Store A: P8 200Store B: P9 840Store C: P6 380Total: P24 42030 EXPLORING MATHEMATICS FORM 3

c) Trued) Truee) Truef) Falseg) Trueh) False3. a) T 2 V 5( ​ ​22 2 ​1 2 ) ​V 2 T 5 ​( 2 22 ​21 22 ) ​∴ T 2 V V 2 Tb) After removing the brackets, you can see that V 2 T 2 H V 2 T 1 H.f) VH 5 ​ 13 18 ​( 27 10 ) ​HV 5 ​ ​4 21 ​( 0 21 ) ​∴ VH HVh) V 1 I 25( ​ 4 1 ​22 21 ) ​ 1 ( ​ ​1 0 ​0 1 ) ​5( ​ 5 1 ​22 0 ) ​ VHomework Exercise (Student’s Book page 92)1. For real numbers a and b, a 3 b 5 0 means either a 5 0 or b 5 0.AB 5 ( ​ ​ 1 2 ​ 1 ​0 ) ​ ( ​ ​ 0 3 ​ 0 ​0 ) ​ 5 ( ​ ​ 0 0 ​ 0 ​0 ) ​ A matrix is made of different entries.2. For real numbers, if a da 5 dc, then a 5 c.DA = ​ ( ​ 8 4 ​ ​0 0 ) ​DC = ​ ( ​ 8 4 ​ ​0 0 ) ​Multiplying matrix A and C give the same results when matrix A is NOT equal tomatrix C.Revision Exercise (Student’s Book page 93)1. a) ​ ( ​ .710 ​ ​36 35 ​ ) ​b) ​ ( ​ 22.22 ​ ​24 211 ​ ) ​c) ​ ( ​ 7018 ​ ​20 16 ​ ) ​32 EXPLORING MATHEMATICS FORM 3

​d)(​ .2 ​ ​ .4 ​ ​ .6​1816 ​ ​12.8 ​ ​10​ ) ​20 ​e) Not possible​ ​ ​ ​ ​ ​ ​f)(​0​ ​ 8​–7​ ​–3​)​​ 5​16 ​ ​ ​ –3​–1​–2 3 –15 11 5 4 –22. a) Trueb) Falsec) Falsed) False3. AC BC CA4. When adding matrices you can change the order of the matrices.5–7. Check own answers.Chapter 5: Matrices33

Assessment 1 Chapters 1–5​Question 11.1 T n5 26 1 3(n 2 1) = 3n 2 9 ✓✓✓ (3)1.2 T 535 150 ✓✓ (2)1.3 T 65 26 1 3(6 1 1) 5 9 ✓; 26 1 (23) 1 0 1 3 1 6 1 9 5 9 ✓✓ (3)1.4 n 5 150 ✓✓✓ (3)[11]Question 22.1 a) (2a 2 ) 3 or 8a 6 ✓✓✓ (3)b) 8a 6 ✓✓✓ (3)c) (2x 3 ) 2 or 4x 6 ✓✓✓✓ (4)2.2 a) 1.23 3 10 26 ✓✓ (2)b) 5.6 3 10 6 ✓✓ (2)2.3 1.275 3 10 211 ✓✓ (2)2.4 a) ___ 71✓✓✓✓99(4)b) It is rational. ✓ (1)[21]Question 33.1 a) P21 735.51 ✓✓✓✓ (4)b) P20 282.30 ✓✓ (2)3.2 P20 282.30 ✓✓✓✓✓ (5)3.3 P5.73 ✓✓✓ (3)[14]Question 44.1 a) X: 3 3 3; Y: 3 3 3 ✓ (1)(b) i) ​1 ​ –2 ​ ​5​2 ​ ​10 ​ ​6 ​​) ​ ✓✓✓✓ (4)1 6 2ii) 7 ​ 0 ​ ​7(​21 ​ ​10 ​ ​6 ​​) ​ ✓✓✓ (3)3 8 74.2 a) A 1 B 5 B 1 A 5 ​ ( ​4 2 ​ 21 ​21​) ​ ✓✓ (2)b) AB BA è ( ​ ​ 122 ​ 28 ​23​) ​ ( ​ ​ 26 5 ​ ​ 421 ​ ) ​ ✓✓ (2)c) A 31 0( 5 2 00 1 ( ✓ 0 22B 31 0( 5 0 00 1 ( ✓ (2)0 1[14]Total: 60((((34 EXPLORING MATHEMATICS FORM 3

6r e t p a h c TrianglesSyllabus: General objectiveThis chapter deals with the properties of triangles.Specific objectives• y Derive the Midpoint theorem by investigation.• y Derive the converse of the Midpoint theorem by investigation.• y Solve problems involving the Midpoint theorem.• y Solve problems involving the converse of the Midpoint theorem.• y Calculate unknown sides and angles using properties of congruent triangles.• y Calculate unknown sides using properties of similar triangles.• y Solve problems involving properties of congruent triangles and similar triangles.Properties of triangles are of great importance in all aspects of mathematics andmathematics-based disciplines, particularly where geometry is concerned. Manygeometrical problems involve triangles, and knowledge of properties of triangles canbe used to solve a variety of problems, both in further mathematics courses andreal-life situations or professions such as architecture.AnswersActivity 6.1 (Student’s Book page 101)1–3. In triangle MNO below, A is the midpoint of side MN, line AR is parallel toside NO and it cuts side MO at B so that is MB 5 BO.Chapter 6: Triangles35

4. A parallelogram, AR||NO and RO||MN.5. AN 5 AM (constructed A as Midpoint of MN)AN 5 RO (opposite sides of parallelogram)6. MAB and BOR are the same shape and size (AM 5 RO; opp s at B;MAB 5 BRO)7. MB and BO are equal.8. B9. The converse of the midpoint theoremSupport Activity 6.1 (Student’s Book page 102)KL 5 6.4 cmExercise 6.1 (Student’s Book page 103)1. 66 mm2. a)b) MBA (MB 5 MA) and MTR (MT 5 MR)3. GK||LO (G, K midpoints, so GK||LM)KO||GL (K, O midpoints, so KO||LP)∴ LGKO is a parallelogram (2 pairs of opposite sides parallel)36 EXPLORING MATHEMATICS FORM 3

Extension Activity 6.1 (Student’s Book page 103)1. 23 cm2. a) 39.8 cmb) 6.64 cmc) 229.1 cm 2Support Activity 6.2 (Student’s Book page 104)1. GXY5.2 cmIH2. YG 5 5.2 cm (Midpoint theorem)3. GH 5 5.2 1 5.2 5 10.4 cmExercise 6.2 (Student’s Book page 104)1. The straight line drawn through the midpoint of one side of a triangle parallel toanother bisects the third side.2.a) U is the midpoint of EF and S is theEmidpoint of ED, therefore US||FD.b) EP 5 PT (US bisects a line drawn fromE to FD)c) 20 cm, as it is given that the triangleUPsides are 40 cm, and that U and S areSmidpoints on the sides. The Midpointtheorem states that a line joining themidpoints of two sides of a triangle ishalf as long as the third side, soFDTUS = 1__2 ​ FD3. She needs to pull a line parallel to the first floor until she meets the third side ofthe roof. That point will be the bisector of that side.4. a) Since D is the midpoint of CF and DE is parallel to FG, then according to theconverse of the midpoint theorem, E is the midpoint of CG and bisects CG.b) B divides CA into two equal parts. Therefore CB is 1__ ​AC = 3.1 cm.2Chapter 6: Triangles37

5. a) Angle ADE = 80º (Corresponding angles)b) Angle FEC = 40º (Corresponding angles)c) EF = 5 cm. D is the midpoint because E is the midpoint of AC and ED isparallel to BC. Hence DB = EF because of a parallelogram.Extension Activity 6.2 (Student’s Book page 106)Scale factor of length in the triangles is half. Therefore the scale factor of the areas isone quarter.∴ Area BAC 5 20 cm 2∴ Area BKMC 5 20 cm 2 2 5 cm 2 5 15 cm 2Activity 6.2 (Student’s Book page 106)SSS: EnoughAAA: Not enoughSAS: EnoughSSA: Not enoughSAA: EnoughRSH: EnoughSupport Activity 6.3 (Student’s Book page 108)1. a) FJ 5 7.5 cmb) QJ 5 5.2 cmc) RQ 5 6.8 cm2. FJQ3. The formed alternating angles are equal.Exercise 6.3 (Student’s Book page 109)1. PQM and PSM (RSH)PQR and PSR (SSS)QMR and SMR (RSH)2. MO 5 10 cm, N 5 60°3. a) XOY and WOZ are congruent.b) They are congruent because of the SAS rule.c) XY 5 WZ because XOY WOZ.38 EXPLORING MATHEMATICS FORM 3

Support Activity 6.4 (Student’s Book page 110)5. Ratios are equalHomework Activity (Student’s Book page 112)1. AAA (3 pairs of corresponding angles are equal)Exercise 6.4 (Student’s Book page 113)1. a) EMP and JFKb) JFc) ___ EM​JF ​ = ​ 1__3 ​d) i) EP 5 3 cmii) FK 5 12.9 cm2. ACB 5 RCT (Vertically opposite angles)CAB 5 CRT (Alternating angles and parallel lines)ABC 5 RTC (Alternating angles and parallel lines)∴ by the AAA property the triangles are similar3. b) i) 6 cm ii) 4.8 cm iii) 9 cm iv) 4 cm v) 5.6 cm4. a) Yes, AAA 5.x__3 ​ = ___ ​ 1.24.5 ​b) ABR|||YXR ∴ x = 0.8 mc) ___ ARRY ​ = ___ ​ 169 ​d) AB = ________ 1.3 3 3.2​ ​1.8= 2.3 mExercise 6.5 (Student’s Book page 116)​ ​​ 1. a) x 5 7.2 cm b) p 5 70° and y 5 40° c) z 5 6.3 cm and q 5 40°2. a) SAS8b) (​ ___4.8 ​) 5 ________ (k 1 4.2)∴ k 5 2.8 cm4.2g(​ __6 ​) 5 (​ ___ 8​) ∴ g 5 10 cm4.83.43. a) Ratios: ___5.1 5 2__3 and ____ 4.56.75 5 2__3 ​The two corresponding sides are equal and the angle included, so the trianglesare similar.b) The scale factor of reduction is 2__3 ​.4. a) DE||BF and AD = DB∴ DE bisects ACb) DE = 1__ ​BC2(D and E are midpoints ofAB and AC)∴ AE = EC∴ 2DE = BCChapter 6: Triangles39

Extension activity 6.3 (Student’s Book page 118)1. EI 5 7.2 cm2. If P is the midpoint of AB, Q the midpoint of BC and R the midpoint of AC, then:In RQC and APR: RC 5 AR (Midpoint theorem)AP 5 RQ (Midpoint theorem)PR 5 QC (Midpoint theorem)By SSS, the two triangles RQC and APR are congruent.The same reasoning works for PBQ with APR and/or RQC.For QRP and PBQ: BQ 5 RP (Midpoint theorem)BP 5 RQ (Midpoint theorem)PQ is commonTherefore by SSS the two triangles are congruent.It follows that all four triangles in ABC are congruent.3. Ratio of areas PQR:PST:PVW(Remember: ratio of areas is the square of length of sides.)Ratio of PR:PT:PW 5 2:4:6 5 1:2:3Ratio of PQR:PST:PVW 5 1:4:9Revision Exercise (Student’s Book page 119)1. a) ZY 5 6.2 cmb) XY 5 9.6 cm2. a) PA 5 5.9 cmb) PH 5 4.7 cmc) APH 5 60°d) PHY 5 95°3. a) A and C (SAS and ratios of corresponding sides are equal)b) D and F (SAS and ratios of corresponding sides are equal)c) All triangles are similar4. a) p 5 9.3 cmb) k 5 2.4 cmc) m 5 5.07 cm5. BD 5 6 cm6. The triangles are similar because of AAA. Scale factor is 3__4 ​.40 EXPLORING MATHEMATICS FORM 3

7c h a p t e rThe PythagoreantheoremSyllabus: General objectiveThis chapter deals with the Pythagorean theorem.Specific objectives• y Derive the Pythagorean theorem by investigation.• y Calculate the unknown sides of a right-angled triangle using thePythagorean theorem.• y Use the Pythagorean theorem to solve problems related to real-life situations.The Pythagorean theorem is a very powerful and useful theorem that can be appliedto calculate sizes and length (or distances) in right-angled triangular shapes andobjects. It can be employed to minimise and maximize resources where necessary invarious practical situations involving right-angled triangles.Therefore, it is important to expose students to this theorem and its importance,since they always encounter situations in their life in which understanding of thetheorem will be handy.The trigonometric ratios are also very important, as they can be used to achieveaccuracy in angle sizes and lengths of sides of right-angled triangular shapes andobjects.AnswersInvestigation (Student’s Book page 125)1. a) STb) 9 cm 2 , 16 cm 2 and 25 cm 2c) 25 5 9 1 16; the area of the square on the longest sides is equal to the sum ofthe areas of the other squares.2. a) 169b) 25 and 144c) 169 5 144 1 25; again the area of the square on the largest side of the triangleequals the sum of the areas of the other squares.Chapter 7: The Pythagorean theorem41

3. Diagram Area of square A Area of square B Area of square C Sum of areas ofsquares A and B(a) 144 25 169 169(b) 16 9 25 25(c) 64 225 289 2894. Students should have observed that in a right-angled triangle, the area of thesquare on the hypotenuse is always equal to the sum of the areas of the squareson the other two sides of the triangle.Note: This is also true for semi-circles, equilateral triangles and numerous otherequilateral shapes. Let the students investigate this further.Support Activity 7.1 (Student’s Book page 130)1. 15 cm2. 12 cm3. 12 cm4. 4.9 cmExercise 7.1 (Student’s Book page 131)1. b 5 28 cm, c 5 7.5 cm2. s 5 2.5 cm, t 5 1.4 cm3. b 5 4, a 5 154. b 5 4.9 cm, a 5 10 cm, c 5 14.9 cm5. 5 cm6. AC 5 12 cmExtension Activity 7.1 (Student’s Book page 132)1 cm​√ __2 ​cm1 cmHomework Activity (Student’s Book page 132)One side is 1 cm. The other isde is the hypotenuse of the previous triangle.42 EXPLORING MATHEMATICS FORM 3

Exercise 7.2 (Student’s Book page 133)1. No2. Yes, T^ = 90°3. Yes, T^ = 90°Exercise 7.3 (Student’s Book page 135)1. 18 sec2. Height 5 15 m3. H 5 3.9 m4. 21.8 km5. 1.7 m6. 12.1 mRevision Exercise (Student’s Book page 137)1. a) 4 cmb) 2.3 cmc) 12.46 cm2. d 5 14.48 cm3. AB 5 2.5 m4. 7.07 cm5. 17 cm6. XY 5 25.3 mChapter 7: The Pythagorean theorem43

8c h a p t e rTrigonometrySyllabus: General objectiveThis chapter deals with trigonometric ratios.Specific objectives• y Identify opposite, adjacent and hypotenuse sides of a right-angled triangle withreference to the given angle.• y Calculate the three trigonometric ratios of sine, cosine and tangent in a rightangledtriangle.• y Use the three trigonometric ratios and a calculator to calculate the unknown angleand side in a right-angled triangle.• y Use both the trigonometric ratios and the Pythagorean theorem to solve problemsrelated to real-life situations.AnswersSupport Activity 8.1 (Student’s Book page 141)1. a) Ab)TadjacentxRoppositehypotenuseoppositehypotenuseBadjacentxCc)xVQhypotenuseadjacentWoppositeP44 EXPLORING MATHEMATICS FORM 3

Exercise 8.1 (Student’s Book page 141)Triangle Hypotenuse Opposite Adjacent(a) 5 cm 3 cm 4 cm(b) 13 cm 5 cm 12 cm(c) 10 cm 6 cm 8 cm(d) z y x(e) p r qSupport Activity 8.2 (Student’s Book page 144)1. Triangle Hypotenuse Opposite Adjacent(a)​√ ___34 ​ 3 5(b) 10 6 82. a) sin θ 5 325cos θ 5 tan θ 5 ​3__34345 ​6b) sin θ 5 ​ ___10 ​ cos θ 5 ​ ___ 810 ​ tan θ 5 ​ 6__8 ​Activity 8.1 (Student’s Book page 144)1. a) The length of hypotenuse is 13 cm.The opposite side is 5 cm.The adjacent side is 12 cm.b) The length of hypotenuse is 41 cm.The opposite side is 9 cm.The adjacent side is 40 cm.c) The length of hypotenuse is 5 cm.The opposite side is 4 cm.The adjacent side is 3 cm.2. a) sin α = ___ 941 ​; cos α = ___ 125​; tan α =13 ___12 ​b) sin α = ___ 941 ​; cos α = ___ 409​; tan α =41 ___40 ​c) sin α = 4__ 3__​; cos α = ​; tan α =4__5 5 3 ​d) sin α = ___ 125​; cos α =13 ___ ​; tan α =1213 ___5 ​e) sin α = _____ 7​; cos α = _____ 11​; tan α = ___ 7170 ​ 170 ​ 11 ​​√ ____​√ ____Chapter 8: Trigonometry45

Exercise 8.2 (Student’s Book page 146)1. a)___ ACAB ​, ___ ​ CBAB ​, ___ ​ ACCB___ NTKT ​, ___ ​ KNKT ​, ___ ​ NTKNb)___ FEHE ​, ___ ​ HFHE ​, ___ ​ FEHF ​ c) ____ XYWY ​, ____ ​ WXWY ​, ____ ​ XYWX___ SUST ​ or ___ ​ TUST ​, ___ ​ SUST ​ or ___ ​ TUST ​, ___ ​ SUTU ​d)e)2 a) sin/ 5 ​3__5 ​, cos/ 5 ​ 4__5 ​, tan/ 5 ​ 3__4 ​ b) sin/ 5 ___ ​ 1213 ​, cos/ 5 ___ ​ 513 ​, tan/ 5 ___ ​ 125 ​c) sin/ 5 ___ ​1​√ ​, cos/ 5 ___ ​12 ​ ​√ ​, tan/ 5 1 d) sin/ 5 ​4__2 ​5 ​, cos/ 5 ​ 3__5 ​, tan/ 5 ​ 4__3 ​e) sin/ 5 ___ 15​17 ​, cos/ 5 ___ ​ 817 ​, tan/ 5 ___ ​ 158 ​3. There are alternative answers such as the following :a) BC 5 10 sin A, BC 5 6 cos B, BC 5 6 tan Ab) EF 5 13 cos G, EF 5 5 tan G, EF 5 13 cos Fc) KL 5 ____ ​√__ 3 ​sin k ​Exercise 8.3 (Student’s Book page 150)1. a) 22.6˚ b) 29˚ c) 60˚ d) 20.9˚ e) 61˚ f) 77˚2. a) 1.33 b) 0.74 c) 0.866 d) 1 e) 0.60 f) 0.213. a) 6.69 b) 4.62 c) 2.65 d) 11.2 e) 5.36 f) 12.8g) 6.39 h) 6 i) 11.7Extension Activity 8.1 (Student’s Book page 151)/BAD = 97.6°Activity 8.4 (Student’s Book page 151)1. 2. Yes 3. 30˚1 250 m625 mExercise 8.4 (Student’s Book page 152)1. 4.3 cm 2. 4.3 m 3. a) 4.56 m b) 37.9°Extension Activity 8.2 (Student’s Book page 153)1. cos 60˚ 5 ___ BC​8 ​BC 5 8 cos 60˚5 4BD 5 ​√ _______4 2 1 4 2 ​5 5.66 m46 EXPLORING MATHEMATICS FORM 3

2. tan θ 5 ​1__ ​, where θ is the angle that the road makes with the horizontal.5∴ θ 5 11.3˚Revision Exercise (Student’s Book page 154)1. a) /A 5 cos 21 (​4__7 ​) 5 55.2˚, /C 5 sin21 (​4__ ​) 5 34.8˚, and BC 5 5.74 cm7b) /A 5 180 2 (90 1 70) 5 20˚, AB 5 4.5 tan 70 5 12.4 cm, andAC 5 ​√ ____________12.4 2 1 4.5 2 ​ 5 13.2 cm2. sin θ 5 ___ 1.5​ ​, where θ is the angle made by the door AB and the roof2.5∴ θ 5 sin 21 1.5(​ ___2.5 ​)∴ θ 5 36.9˚153. tan b 5 ​ ___ ​, where b is the angle at F8∴ b 5 tan 21 15(​ ___8 ​)∴ b 5 61.9˚4. a) sin 26˚ 5 ___ PS​ ​, so PS 5 4 sin 26˚ 5 1.75 m4b) cos 26˚ 5 ___ ​AP​, so AP 5 4 cos 26˚ 5 3.6 m4AB 5 7 5 AP 1 PQ 1 QB∴ 7 5 AP 1 PS 1 QB, since PQ 5 PS∴ 7 5 4 cos 26˚ 1 4 sin 26˚ 1 QB∴ QB 5 1.65 mc) RB 2 5 QB 2 1 PS 2∴ RB 5 ​√ __________QB 2 1 PS 2 ​5 ​√ ______5.785 ​5 2.41 md) QRB 5 _______ 4 sin 26˚​ ​RB5 46.6˚5. tan VXY 5 ___ ​425 ​∴ /VXY 5 9.09˚3606. a) /AOB 5 ​ ____5 ​ 5 72˚ and /OAB 5 ________ ​ 180 2 72​ 5 54˚2b) Area nAOB 5 ​1__2 ​ 3 20 3 ______ ​ 10tan 36 ​5 137.6 cm 2c) Area pentagon 5 137.6 3 55 688.2 cm 2Chapter 8: Trigonometry47

9c h a p t e rCoordinate geometrySyllabus: General objectiveThis chapter deals with the use of coordinates in a Cartesian plane.Specific objectives• y Calculate the distance between two points.• y Calculate the coordinates of the midpoint of a line segment given the coordinatesof its endpoints.• y Calculate the coordinates of the endpoint of a line segment given the coordinatesof the midpoint and one end point.• y Solve problems involving the applications of distance between two points andmidpoint of the two end points of a line segment.AnswersActivity 9.1 (Student’s Book page 158)1. A(1;2), B(6;8)4. C(6; 2)5. AC 5 6 2 1 5 5 units; BC 8 2 2 5 6 units6. AB 5 ​√ _______5 2 1 6 2 ​ 5 7.8 units7. AB 5 (x 22 x 1) 2 1 (y 22 y 1) 2√Exercise 9.1 (Student’s Book page 160)1. a) Distance 5 ​√ _____________________(1 2 9) 2 1 (24 2 14) 2 ​5 ​√ ______________(28) 2 1 (218) 2 ​5 √ 64 1 3245 19.7 unitsb) Distance 5 10 unitsc) Distance 5 5.7 unitsd) Distance 5 9.06 units2. a) AB 5 4 units; AC 5 ​√ ___29 ​ unitsBC 5 ​√ ___29 ​ units è Isosceles48EXPLORING MATHEMATICS FORM 3

) AB 5 17.1 units; AC 5 12 units;BC 5 17.1 units è Isoscelesc) AB 5 6.3 units; AC 5 8.5 unitsBC 5 10.4 units è Not isoscelesd) AB 5 4.1 units; AC 5 4.1 units;BC 5 2 units è IsoscelesExtension Activity 9.1 (Student’s Book page 161)1. AD 5 3 units; DB 5 4 units2. 3 units in y-direction and 4 units in x-direction.3. AB 5 5 units4. AB 5 ​√ ____________________________(0 2 4) 2 1 (3 2 0) 2 1 (0 2 0) 2 ​5 5 units5. E is in the y-z-plane which means zero units in the x-direction, but 3 units in they- and 2 units in the z-direction.6. AE 5 ​√ ___29 ​ 5 5.4 units7. AE 5 ​√ __________AB 2 1 BE 2 ​5 ​√ _______5 2 1 2 2 ​5 5.4 unitsActivity 9.2 (Student’s Book page 162)1.–3.7654321yA(2;3)C(5;5)2 4 6 8B(8;7)4. x 5 _____ 2 1 8​2 ​ 5 5; y 5 _____ ​ 3 1 7​ 5 5, so c 5 (5;5)25. Depends on students' drawing.6. x 5 x ______1 x 1 2​, y 5 ______y​11 y 2​227. Students’ own answers.xExercise 9.2 (Student’s Book page 162)1. a) (5.5;5.5)b) (4;3.2)Chapter 9: Coordinate geometry49

​​c) (20.5;3)d) (21;0.5)2. a) 20 unitsb) 62.8 unitsc) (8;9)Support Activity 9.1 (Student’s Book page 163)24 1 x1. ______ ​ 5 0; [ x 5 42_____ 7 1 y​ 5 3; [ y 5 21228 1 x2. ______ ​ 5 23; [ x 5 22______ 26 1 y​ 5 4; [ y 5 142Exercise 9.3 (Student’s Book page 164)1. a) (4;21)b) (2;14)c) (28;2)d) (23;215)2. b) C(3;8)c) D(3;4)Extension Activity 9.2 (Student’s Book page 164)1. (1;4); (1;0) and (5;0)2. Gradient AB 5 ______ ​11 2 74 2 2 ​ 5 2Gradient CB 5 ______ ​11 2 14 1 1 ​ 5 2So A, B and C are in a straight line.3. Gradient AB 5 2​4__3 ​Gradient of line is 3__ ​ (product of the gradients of perpendicular lines is 21).4Line goes through the midpoint of AB which is (1;2).Therefore: y 2 y 15 m(x 2 x 1)∴ y 2 2 5 ​3__ ​(x 2 1)4∴ y 5 ​3__4 ​x 1 ​ 5__4 ​50 EXPLORING MATHEMATICS FORM 3

1.Activity 9.3 (Student’s Book page 166)B(8;4)M(3;21)θA(22;26)135˚2. Inclination 5 tan θ 5 ________ 4 2 (26)​8 2 (22) ​ 5 1θ 5 45˚3. a) Strut supports at (3;21)b) 135˚ with the ‘positive x-axis’c) Girder mass 5 6 3 ​√ ____200 ​ 5 84.9 kgExercise 9.4 (Student’s Book page 166)1. Diameter 5 13 units; Radius 5 6.5 units2. a) AB 5 2.8 unitsb) M(1;4)c) AC 5 7.2 units; BC 5 7.2 units.So, ABC is isoceles.d) No, 7.2 2 7.2 2 1 2.8 2e) i) Perimeter 5 17.2 unitsii) MC 5 ​√ ___50 ​ 5 7.1 unitsiii) MC is an altitude sincem MC3 m AB5 21Area nABC 5 ​1__ ​ 3 AB 3 MC25 ​1__ ​ 3 2.8 3 7.125 9.9 units 23. a) M(0.5;6)b) P is halfway between A and M.So, P(22.75;4)4. AP 5 5.8 units, so A is not on the circle.BP 5 5 units, so B is on the circle.Chapter 9: Coordinate geometry51

Revision Exercise (Student’s Book page 168)______ ______y 11 y 22 2 )​Distance 5 √ ____________________;​ (x 12 x 2) 2 1 (y 12 y 2) 2 ​1. Midpoint 5 ​( x 1 1 x 22. a) Length 5 19.7 unitsMidpoint (5;5)b) Length 5 10 unitsMidpoint (24;23)c) Length 5 5.7 unitsMidpoint (1;1)d) Length 5 9.1 unitsMidpoint (2.5;24.5)3. a) Not isoscelesb) Isoscelesc) Isoscelesd) Isosceles4. Radius 5 5 units5. a) M(2;1); N(3;9)b) MN 5 2.236 unitsBC 5 4.472 unitsc) Midpoint theorem as BC 5 2MN52 EXPLORING MATHEMATICS FORM 3

10c h a p t e rPlans, elevations andbearingsSyllabus: General objectiveThis chapter deals with the practical applications of plans, elevations and bearings.Specific objectives• y Draw the front, side and plan elevations of the same shape and house.• y Find a point’s position using direction/bearings, distances and/or angles in a givenjourney.• y Use the Pythagorean theorem and trigonometric ratios to solve problemsinvolving angle of elevation, angle of depression and bearings.AnswersExercise 10.1 (Student’s Book page 173)1. a)frontsideb)planplanfrontsidec)plan front sideChapter 10: Plans, elevations and bearings53

d)planfrontside2. a) b)3. a)plan front sideb)frontsidec)planfrontsided)planfrontside54 EXPLORING MATHEMATICS FORM 3

e)planf)frontsideplanfrontsideActivity 10.3 (Student’s Book page 176)2. a) Due North (000˚)b) 045˚c) 108.4˚d) 206.6˚Activity 10.4 (Student’s Book page 176)Remember to always draw north linesparallel to each other.North080°3 kmNorthSchool2 kmNorthVillage health clinicSadi’s homeThe bearing of the village health clinic from Sadi’s home measures to be 080˚.The bearing of the direct route from the clinic 5 360˚ – 100˚, using the property ofthe sum of interior angles formed along a transversal line cutting parallel lines andthe area around a point.So, the bearing of the direct route 5 260˚.Chapter 10: Plans, elevations and bearings55

Exercise 10.2 (Student’s Book page 177)2. a) 050˚b) 080˚c) 340˚d) 220˚e) 050˚f) 330˚3. a)CN55˚75˚ B50˚Ab) i) 050˚ii) 305˚iii) 230˚ (180˚ + 50˚)Homework Activity (Student’s Book page 178)1. 270˚2. 180˚3. 000˚ (or 360˚)Extension Activity 10.1 (Student’s Book page 179)1. Let the bearing from B to A be x.Thus the bearing from A to B would be 180˚ 1 x[ 180˚ 1 x 5 4xx 5 60˚Bearing from B to A is 60˚ and the bearing from A to B is 180 o 1 60˚ 5 240˚2. 180˚ 1 x 5 7xx 5 30˚B è A is 30˚ and A è B is 210˚Activity 10.5 (Student’s Book page 180)Elevation: Line of sight is at an angle upwards from the horizontal (look up fromthe horizontal).Depression: Line of sight is at an angle downwards from the horizontal (cast youreyes downward from the horizontal).56 EXPLORING MATHEMATICS FORM 3

Activity 10.6 (Student’s Book page 181)1.NNH9 kmP10 kmQ2. 48˚3. 222˚4. 13.5 km5. sin 48˚ 5 ____ 10​13.5 ​0.74 5 0.74 (confirmed)Exercise 10.3 (Student’s Book page 182)1. a) 15 kmb) 56.3˚c) 326.3˚2. 20.6˚3. a) angle of depressionb) 2.87˚4. 23.4˚Extension Activity 10.2 (Student’s Book page 183)AC 1 CB 5 ______ 210​cos 54˚​ 1 ______ ​ 2105 574.7 kmsin 75˚​Chapter 10: Plans, elevations and bearings57

Revision Exercise (Student’s Book page 184)1. a)planfrontsideb)planfrontsidec)plan front side2.3. a)plan front side58 EXPLORING MATHEMATICS FORM 3

)planfrontsidec)planfront and side5. b) i) 60˚ii) 240˚iii) 215˚6. h 5 150 tan 35˚ 5 105 m7. Distance 5 37.3 metres8. tan 37˚ 5 ___ 1.7​ x ​; x 5 2.26 mBlind distance 5 x 2 0.85 5 1.4 m9 a) a 5 60 cos 30˚ < 52 mb) b 5 60 cos 20˚ < 56.4 mc) b 2 a 5 4.4 mChapter 10: Plans, elevations and bearings59

11c h a p t e rTransformation geometrySyllabus: General objectiveThis chapter deals with understanding and using combined transformations.Specific objectives• y Draw combined transformations up to three different types.• y Identify and describe fully combined transformations involving up to threedifferent types.• y Solve problems involving combined transformations.AnswersActivity 11.1 (Student’s Book page 188)1. a) Flip in a mirror line.b) Shift or slide through a point (usually (0;0)) over a certain distance, i.e. by avector ​( ​ x y ​ ) ​.c) Rotated through a certain angle (usually 90˚ or 180˚) relative to a referencepoint.d) A shape is made bigger or smaller by a constant factor (scale factor).Exercise 11.1 (Student’s Book page 191)1. a) Translation by vector ​ ( ​ 4 6 ) ​b) Reflection in y 5 xc) Reflection in y 5 2d) Reflection in y 5 0e) Rotation about centre (0,0) through 90˚2. a) Image F: (22.5;2.5), (21;2.5) and (21;1.5)Note: an enlargement of 1__ ​is actually a reduction. To prevent confusion amongst2students, it is safer to talk of a reduction by a factor 1__ ​, so just point this out.2b) Image G: (25;2), (23;2) and (25;5)60 EXPLORING MATHEMATICS FORM 3

Exercise 11.2 (Student’s Book page 193)1. Image triangle has vertices (1;6), (4;6) and (1;5)2. Image triangle has vertices (2;1), (3;22) and (3;1)3. Image triangle has vertices (5;3), (5;4) and (8;3)4. Image triangle has vertices (4;1), (4;2.5) and (8.5;1)Exercise 11.3 (Student’s Book page 195)1. a) Image triangle has vertices (25;25), (23;25) and (25;28).b) Rotation about centre (2;23) through 180˚ followed by translation by vector ( ​ ​ 0 0 ) ​or a rotation about (0;0) through 180° followed by a translation ( ​ ​ 4–6 ​ ) ​.c) Rotation through 290˚ about centre (0;0) followed by reflection in x 5 3 andthen translation by vector ​ ( ​ 2 0 ) ​.d) Image triangle has vertices (22;24), (22;210) and (2;24).2. a) H: (5;22); (5;27); (2;27)b) K: (7;22); (7;25); (2;25)c) Reflection in y 5 0 (x-axis)3. b) Enlargement, about centre (17;0) with scale factor 1.5.Extension Activity 11.1 (Student’s Book page 196)1. yA 2A 4A 3A 1x2. (4;1)​ ( ​ 210 ​ ​0 21 ​ ) ​ 5 (24;21) (A 2 ), (4;1)​ ( ​ 0 1 ​ ​1 0 ) ​ 5 (1;4) (A 3 ),(4;1)​ ( ​ 1 0 ​ ​0 1 ) ​ 5 (4;1)3. The matrix ​ ( ​ 210 ​ 21 ​0 ​ ) ​, as shown by point A . 24. ( ​ 1 0 ​ 1 ​0 ) ​, as this is the identity matrix for multiplication.5. The matrix ​ ( ​ 0 1 ​ ​1 0 ) ​6. ​ ( ​ 0 21 ​ ​1 0 ) ​Chapter 11: Transformation geometry61

Activity 11.4 (Student’s Book page 197)1. a) Reflection x 5 0 or y-axis and translation ( ​ ​0211 ​ )b) Cʹ(0;–4)2. Reflection in the y-axis or the line x 5 0Exercise 11.4 (Student’s Book page 197)1. a) Translation vector ​ ( ​ 253 ​ ) ​.b) B and C, reflection in y 5 xc) Triangle with vertices (20.5;21), (0.5;21) and (20.5;22.5)2. Reflection in y 5 2x and translation by vector ( ​ ​ 322 ​ ) ​.3. A is point (22;25).4. Rotation about the origin through 180˚5. Reflection in the y-axis, rotation about (2;2) through 90˚ clockwise, reflection inthe x-axis, enlargement about the origin with scale factor 1__2 ​.Revision Exercise (Student’s Book page 202)1. c) nEFG: (22;24), (21;0) and (3;21)2. c) Rotation about the origin through 90˚ anticlockwise.3. Reflection in y 5 x.Enlargement by a scale factor 2 through the point (27;1).4. Rotation by 290˚ about (0, 0)(Rule: (x;y) è (y;2x))5. Enlargement with scale factor 1__ ​and centre (12;0).26. a) P'Q'R': (22;21), (22;1), (0;1)c) The original object was translated by the vector ( ​ ​ 2626 ​ ) ​to give the final image.Assessment 2 Chapters 6 – 11Question 11.1 a) NZ 5 60✓ mm (N midpoint since MN||ZY)✓✓ (3)b) MN 5 ​1__ ​ZY 5 40 mm ✓✓2ML 5 40 1 24 5 64 mm✓ (3)c) YK 5 2ML✓ 5 128 mm✓ (2)1.2 a) (2x 2 3) 2 5 x 2 1 9✓✓ (2)b) 4x 2 2 12x 1 9 5 x 2 1 9✓∴ 4x 2 2 x 2 2 12x 5 9 2 9✓62 EXPLORING MATHEMATICS FORM 3

∴ 3x 2 2 12x 5 0✓∴ 3x(x 2 4) 5 0✓∴ x 5 0 or x 5 4 ∴ x 5 4 units✓ (4)1.3 a) AC 5 ​√ _________40 2 1 40 2 ​✓ 5 56.56 mm✓ (2)b) Area 5 3 200 mm 2 ✓ (2)[18]Question 22.1 tan 27° 5 h ✓ ∴ h 5 150 tan 27°✓ 5 76.4 m✓150(3)2.2 a) BE 5 12 tan 58.6˚✓✓ 5 19.7 m✓ (3)b) AE 5 12 tan 63˚✓✓ 5 23.6 m✓ (3)c) AB 5 AE 2 BE 5 3.9 m✓ (1)[10]Question 33.1 x 5 21✓✓y 5 22✓✓ (4)3.2 a) Perimeter 5 XY 1 XZ 1 YZ✓5 10✓ 1 ​√ ___34 ​✓ 1 ​√ ____146 ​✓b) 2.92 units✓Question 45 27.9 units✓✓ (6)Midpoints: (21;5)✓✓ and ( 1__2 ​; 5__ ​)✓✓ (5)2[15]4. Plan: Front: Side:✓ ✓✓ ✓✓Question 5y5.1 – 5.45Object4✓✓✓3D2✓✓✓1EB ✓✓✓x27 26 25 24 23 22 21-11 2 3 4 5 6 7-2C23 ✓✓✓24[5][12]Total: 60Chapter 11: Assessment 263

12c h a p t e rArea and perimeterSyllabus: General objectiveThis chapter deals with the application of ratio and enlargement/reduction on areaand perimeter.Specific objectives• y Solve problems involving practical applications of ratio on area and perimeter oftwo-dimensional shapes.• y Solve problems involving practical applications of scale factors of enlargement/reduction on area and perimeter of two-dimensional shapes.In this chapter students will learn the skill to manipulate two-dimensional shapeswith regard to their area and perimeter by changing their dimensions by a certainratio or scale factor. It is important that students have a clear understanding ofthe effect that a change to dimensions of a shape has on the area and perimeterof the object.AnswersSupport Activity 12.1 (Student’s Book page 209)Area 1 () 5 l 3 b 5 14 3 6 5 84 cm²Area 2 () 5 ​1__2 ​b 3 h 5 ​ 1__ ​ 3 4 3 6 5 12 cm²2Area 3 ( ) 5 ​1__2 ​πr²5 ​1__2 ​ 3 π 3 165 25.13274122 cm²Total Area 5 84 1 12 1 25.13 5 121.13 cm²Perimeter 5 18 1 6 1 6 1 ​√ _______4 2 1 6 2 ​ 1 1__2 (2π.4)5 32 1 ​√ ___52 ​+ 4π5 32 1 7.21 + 12.575 51.78 cm64EXPLORING MATHEMATICS FORM 3

Exercise 12.1 (Student’s Book page 210)1 a) Perimeter 5 2 3 30 1 2 3 605180 cmb) h 15 ​√ __________(30 2 225 2 ) ​ 5 16.58 cmh 25 ​√ ___________(60 2 2 25 2 ) ​ 5 54.54 cm∴ h 11 h 25 71.12Area 5 ​1__ ​(a 3 b)25 ​1__ ​ (50 3 71.12)25 1 778 cm 2c) Length 5 71.12 1 2(25) cm5 121.12 cmd) and e) Cost 5 ____ ​122100 ​ 3 ____ ​ 8.50 ​ 1 ______ ​ 1 778 ​ 3 121 10 0005 P12.50Note: It is important for students to understand that a length of 121.12 cm willmost likely be rounded off to 122 cm (in some cases hardware stores may cut onlyhalf-metre lengths and Thsepo might have to buy 1.5 metres, but for the purpose ofthis calculation we assume that he can buy 1.22 metres). Students must also realisethe importance of converting the measurements to metres and m 2 , and of coursethey must know how to do such conversions, i.e. why we divide by 100 in the onecase but by 10 000 in the next.2. a) Perimeter 5 2πr5 2π 3 10 62.8 mb) Cost 5 62.8 3 55.50 5 P3 485.40Investigation (Student’s Book page 211)AC1. ___DF ​ 5 ___ ​ 105 ​ 5 2 ___ ABDE ​ 5 ​ 6__3 ​ 5 2 ___ BCEF ​ 5 ​ 8__4 ​ 5 2ACTherefore ​ ___DF ​ 5 ___ ​ ABDE ​ 5 ___ ​ BCEF ​ 5 22. Perimeter of triangle DEF 5 12 cm and perimeter of ABC 5 24 cmperimeter ABCThe ratio 5 2. The ratios of sides and perimeters are the same.perimeter DEF3. Area of DEF 5 6 cm 2 and area of ABC 5 24 cm 2area ABCThe ratioarea DEF 5 ___ ​ 246 ​ 5 4 5 22 . The ratio of the area is the square of theratio of the side lengths.Chapter 12: Area and perimeter65

4.R915S12TRSRatio of sides___DE ​ 5 ​ 9__3 ​ 5 3 ___ STEF ​ 5 ___ ​ 124 ​ 5 3 ___ RTDF ​ 5 ___ ​ 155 ​ 5 3perimeter RSTRatio of perimeters ______________perimeter of DEF ​ 5 ___ ​ 3612 ​ 5 3area of RSTRatio of areas__________area of DEF ​ 5 ___ ​ 546 ​ 5 95. The ratios of the lengths of the sides are equal, because the triangles are similar.The ratios of the perimeters are the same, because the perimeters are the sum ofthe sides.The ratio of the areas is the same, because the ratio of the perimeters are equal;and the ratio of the areas is a square, because length 3 length is a square.Activity 12.1 (Student’s Book page 213)​ ​ ​ 1. Perimeter of rectangle ABCD 5 2(3 1 10) 5 26 cmArea of rectangle ABCD is 3 3 10 5 30 cm 2Perimeter of MNOP 5 2(4.2 1 14) 5 36.4Area of MNOP 5 58.8 cm 2Ratio of length 5 ___ 147__ 7__5 ​10 5Ratio of side 5 ___ 1410 5 7__5 ​ Ratio of perimeters ____ 36.426____ 364260Ratio of areas 58.85 ____30 5 ____ 588300___ 4921 (​ 7__5 ​)22. a) TrianglesRatio of the sides is 1:3 Ratio of perimeters is 1:3 Ratio of areas is 1:9b) RectanglesRatio of sides is 2:3 Ratio of perimeters is 2:3 Ratio of areas is 4:9c) ArrowsRatio of sides 7:4 Ratio of perimeters is 7:4 Ratio of areas is 49:16Extension Activity 12.1 (Student’s Book page 214)Ratio of perimeters 5 3:4, therefore ratio of areas is 9:16.The sum of the areas is 75 cm 2 .9The area of the smaller triangle is 75 3 (​ ___25 ​) 5 27 cm2 .16The area of the larger triangle is 75 3 (​ ___25 ​) 5 48 cm2 .66 EXPLORING MATHEMATICS FORM 3

Exercise 12.2 (Student’s Book page 214)1. a) 6:5 and 36:25b) 4:9 and 4:92. a) 1__2 ​bh ∴ h 5 ____ ​ Area1__​ 5 ___ ​ 60​ 5 12 cm2 ​b 5b) h 5 ​2__3 ​ 3 ___ ​ 121 ​ 5 8 cm; b 5 ​ 2__3 ​ 3 ___ ​ 101 ​ 5 ___ ​ 203 ​ = 6​ 2__3 cmArea DEF 5 ​4__9 ​ 3 ___ ​ 601 ​ 5 ____ ​ 2409 ​ 5 26​ 6__9 5 26​ 2__3 cm2c) Area 5 ​1__ ​bh = ​1__2 2 ​ 3 ___ ​ 203 ​ 3 ​ 8__1 ​ 5 ____ ​ 1606 ​ 5 26​ 4__6 ​3. a)9__4 ​b)9__4 ​ 3 ____ ​ 1001 ​ 5 225%c) Area 1: 1__ ​ 3 60 3 40 5 1 200 cm22New dimensions: 3__2 ​ 3 60 5 90 cm 3__ ​ 3 40 5 60 cm2Area 2: 1__ ​ 3 90 3 60 5 2 700 cm22This is the same as 2.25 3 1200.BC4. a) ___FH ​ 5 ___ ​ DAGE ​b) Ratio ABCD:EFHG5 ​6__4 ​ 5 3:2c)x__8 ​ 5 ​ 2__3 ​ ∴ x 5 ___ ​ 16​ cm 5 5.3 cm3y__5 ​ 5 ​ 3__2 ​ ∴ y 5 ___ ​ 15​ 5 7.5 cm2d) Perimeter ABCD:Perimeter EFHG 5 3:2Area ABCD:Area EFHG 5 9:4e) Area EFHG 5 ​4__9 ​ 3 ___ ​ 81​ 5 36 cm215. a) One window:Frame perimeter 5 2(2 1 1.5) 5 7 mLength of slats 5 (4 3 1.5) 1 (2 3 2) 5 10 mCost per window:Frame 5 7 3 20.50 5 P143.50Slats 5 10 3 15.25 5 P152.50Total 5 P296/windowCost of 9 windows: P2 664b) 4:25c) Cost 5 ​2__ ​ 3 296 5 P118.405Chapter 12: Area and perimeter67

Extension Activity 12.2 (Student’s Book page 216)Perimeter of ABCD 5 50 cm Perimeter of A 1B 1C 1D 15 50 3 (​5__ ​) 5 125 cm2Area of ABCD 5 150 cm 2 and Area of A 1B 1C 1D 15 150 cm 2 .Ratio of perimeters is 5:2 and Ratio of area is 1:1.The two triangles are not similar because the ratio of area is not a square of the ratioof perimeters.Investigation (Student’s Book page 216)1. Side Perimeter Area1 cm 4 cm 1 cm 22 cm 8 cm 4 cm 23 cm 12 cm 9 cm 24 cm 16 cm 16 cm 25 cm 20 cm 25 cm 26 cm 24 cm 36 cm 2a) Perimeter doubles in each case.b) Area increases by a factor of 4.perimeterc) ________ ​ 5 4 ____ area​ 5 1, then 2, then 3, then 4, etc. So the perimeter increasessidesidelinearly and the area increases exponentially.d) Tapiwa is right. If you increase the sides by a constant factor, the areaincreases by more than that constant factor.2. Side Perimeter Area8 cm by 4 cm 24 324 cm by 2 cm 12 82 cm by 1 cm 6 23. When you increase the dimensions by scale factor k, the perimeter increases byscale factor k and area increases by scale factor k 2 .Activity 12.2 (Student’s Book page 218)1. 2; 42.1__5 ​; ___ ​ 125 ​3. 4; 44. 2; 468 EXPLORING MATHEMATICS FORM 3

Extension Activity 12.3 (Student’s Book page 218)For a square, if the side s 5 x: P 5 4x A 5 x 2If the sides double, so s 5 2x: P 5 4(2x) A 5 (2x) 2For a rectangle with sides x and y: P 5 2(x + y) A 5 xyIf the sides double, so 2x and 2y: P 5 4(x + y) A 5 4xyExercise 12.3 (Student’s Book page 218)1. a) x 5 4 cm; perimeter 5 16 cmb) Sides 5 8 cm; scale factor 2c) 2 cm2. a) Area 5 144 mm 2 ; perimeter 5 48 mmb) i) Perimeter Y 5 1.5 3 48 5 72 mmArea Y 5 1.5 2 3 144 5 324 mm 2ii) Perimeter Z 5 1.5 3 72 5 108 mmArea Z 5 1.5 2 3 324 5 729 mm 2c) Side length of Y 5 1.5 3 12 5 18 mmSide length of Z 5 1.5 2 3 1251.5 3 18 5 27 mmd) 18 2 5 324 mm 2 ; 27 2 5 729 mm 24 3 18 5 72 mm; 4 3 27 5 108 mme) 48:72:108 5 4:6:9f) Area X:Area Y5 144:3245 1:2.25Area Y:Area Z5 324:7295 1:2.25X:Y:Z 5 144:324:729 5 16:36:813. a) ___ 125 ​b) 2πr; ​2__5 ​πr; ___ ​ 225 ​πr; ____ ​ 2125 ​πrRatio 5 ​1__5 ​Exercise 12.4 (Student’s Book page 220)1. a) Volume increases by a scale factor of 36. He will therefore use 360 ml of paint.b) Perimeter increases by a scale factor of 6. He will therefore use 6 times morebraiding and it will cost him 6 3 2.50 5 P15.00.2. a)1__ ​ 3 60 5 15 m of fencing3b)1__ ​ 3 450 5 P509Chapter 12: Area and perimeter69

3. a) _____ 11 000 ​b) _____ 11 000 ​4. a) The length and the width must each increase by a factor ​√ _____1.21 ​ 5 1.1.So, width 5 27.5 m; length 5 35.2 mProject (Student’s Book page 221)1. 960 000 ha.32. ​ ____178 ​3. 150 000 3 1.045 5 156 7504. 1 003 2005. Percentage increase in land over five years is 1.0456. 4.5%7. It is not sustainable, as the elephants will be taking all the land in a few years tocome. Elephants will have to be reduced, for example, relocated.Revision Exercise (Student’s Book page 222)1. a) 5__2 ​; ___ ​ 254 ​b) halve; be a quarter of the original.c) 4; 4d) 3:2; 9:42. a) 4 cmb) Area ABC 5 ​1__2 ​ 3 4 3 3 5 6 cm2c) ____ 12.5 ​ 5 ____ ​ 12 6.25 ​d) Area DEF 5 Area ABC 3 6.255 6 3 6.255 37.5 cm 2e) DE 5 2.5 3 3 5 7.5 cmEF 5 2.5 3 4 5 10 cmDF 5 2.5 3 5 5 12.5 cmArea DEF 5 ​1__ ​ 3 10 3 7.5 5 37.5 cm223. a) Shaded area 5 18 3 20 2 π 3 9 25 105.5 m 2π 3 9b) Ratio 5​ ______2105.5 ​ 5 _____ ​ 254.5105.5 ​ 5 ____ ​ 509211 ​ 5 ___ ​ 2.41 ​c) Shaded area 5 ​1__ ​ 3 105.5 5 26.4 m24Ratio 5​ _______ π 3 4.5 2​ 5 ___ ​ 2.426.4 1 ​70 EXPLORING MATHEMATICS FORM 3

Note: It can be pointed out to students that the ratio stays the same since bothareas changed by the same factor.d) Shaded area 5 18 3 20 2 2 3 π 3 4.5 25 232.8 m 2Note: It is important for students to understand that two circles with a diameter of9 metres each do not have the same area as one circle with diameter 18 metres,i.e. 2 3 4.5 2 9 2 .4. a) Shaded area 5 π 3 40 2 2 2 3 π 3 20 25 2 513.3 m 2b) Ratio 5 ​1__ ​ 5 1:1 (i.e. the areas are equal)1c) Perimeter small circles 5 2(2 3 π 3 20)5 251.33 mPerimeter big circle 5 2 3 π 3 405 251.33 mRatio 5 1:1d) Diameters and perimeters double.5. a) ​√ ________(5 2 2 3 2 ) ​ 5 4 cmb) Area 5 ​1__ ​(sum || sides) 3 ⊥height25 ​1__ ​ (18 1 12) 3 425 60 cm 2c) Area 5 3 2 3 60 5 540 cm 2d) Ratio 5 9:16. a) 0.75 mb) Area door 5 3.9 m 2Perimeter door 5 7.9 mArea windowc) ___________ ​ 5 ___ 16​Area door 25 ​ 5 0.64 : 1_______________Perimeter window​ 5 ​4__Perimeter door 5 ​d) Area window 5 2.5 m 2Perimeter window 5 6.3 mChapter 12: Area and perimeter71

13c h a p t e rSurface area and volumeSyllabus: General objectiveThis chapter deals with applications of ratio, proportion, enlargement and reductionon surface area and volume of three-dimensional objects. It is important thatstudents have a clear understanding of the effect that a change to dimensions of a3-D object has on the surface area and volume of the object.This chapter lends itself to teaching students the skill to manipulate threedimensionalobjects with regard to their surface area and volume through changingtheir dimensions by a certain ratio or scale factor.Specific objectives• Solve problems that involve the application of ratio and proportion and scalefactors of enlargement and reduction on surface area and volume of threedimensionalobjects.AnswersInvestigation (Student’s Book page 227)1. Surface area (SA 1) 5 6 3 1.2 m 3 1.2 m 5 8.64 m 22. Volume (V 1) 5 1.2 m 3 1.2 m 3 1.2 m 5 1.728 m 33. Surface area (SA 2) 5 6 3 2.4 m 3 2.4 m 3 2.4 m 5 34.56 m 24. Volume (V 2) 5 2.4 m 3 2.4 m 3 2.4 m 5 13.824 m 3SA5. ___ 2​ 5 4 and __V 2​ 5 8SA 1V 16. The surface area changed by a scale factor of 4 5 2 2 and the volume changed by ascale factor of 8 5 2 3 .Exercise 13.1 (Student’s Book page 229)1. a) V a5 113.1 cm 3 ; SA a5 113.1 cm 2b) V b5 3 053.6 cm 3 (5 27 3 V a)SA b5 1 017.9 cm 2 (5 9 3 5)Vc) __ b​ 5 27 5 3V 3aSAd) ___ b​ 5 9 5 3SA 2a72 EXPLORING MATHEMATICS FORM 3

2. a) V a5 3 10 2 3 30 5 9 424.8 cm 3SA a5 2r 2 1 2rh 5 2 513.3 cm 2b) V b5 ​1__8 ​ 3 V 5 1 178.1 cm3aSA b5 ​1__ 5 628.3 cm2c)d)__V b​ 5 ​1__V a8 ​___SA b​ 5 ​1__SA a4 ​4 ​ 3 SA aSupport Activity 13.1 (Student’s Book page 229)1. 4:9; 8:272. 3:2; 27:83. 2:5; 4:25Activity 13.1 (Student’s Book page 230)1. a) Diameters: ___ 8​24 ​ 5 ​ 1__3 ​Surface areas: 1__9 ​Volumes: ​ ___ 127 ​Note: In the formula V 5 ​1__3 ​r2 h, both r and h are multiplied by the scale factor ​ 1__3 ​,therefore the scale factor by which the volume decreases works out to be (​ 1__3 ​)2 3 ​1__3 ​ 5 ___ ​ 127 ​.b) and c) Truncated cone surface area5 SA full cone2 SA small cone1 3 4 25 240 2 ​1__ ​ 3 240 1 1695 229​1__3 ​ ​( 8__ cm2 9 ​ 3 240 1 16 )​∴ Ratio 5 229​ 1________ 3 ​240 ​ 5 ____ ​ 688720 ​ 5 ___ ​ 4345 ​Truncated cone volume5 V full cone2 V small cone5 2304 2 ___ ​127 ​ 3 2304265 ​ ___27 ​ 3 23045 2 218 2__ ​ cm332 218​2__∴ Ratio 5 _______ 3​​2304 ​ 5 _____ ​ 6 6566 912 ​ 5 ___ ​ 2627 ​OR:Ratio 5 1 2 ​ ___ 127 ​ 5 ___ ​ 2627 ​Chapter 13: Surface area and volume73

Exercise 13.2 (Student’s Book page 231)1. a) V 5 324 cm 3SA 5 342 cm 2SAb) _____ small​ 5 ​4__SA big9 ​; ____ ​ l small​ 5 ​2__l big3 ​c) V small5 ___ 8​ ​ 3 324 5 96 cm327SA small5 ​4__ ​ 3 342 5 152 cm29d) Dimensions5 (​2__3 ​ 3 3) 3 (​ 2__3 ​ 3 12) 3 (​ 2__3 ​ 3 9)5 2 3 8 3 6 cme) V 5 96 cm 3SA 5 2[(2 3 8) 1 (2 3 6) 1 (8 3 6)]5 152 cm 22. a) V 5 729 cm 3 ; SA 5 6 3 81 5 486 cm 2b) V 5 (​4__3 ​)3 3 729 5 1 728 cm 3SA 5 (​4__3 ​)2 3 486 5 864 cm 2c) Side 5 ​4__ ​ 3 9 5 12 cm3 12 cm 3 12 cm 3 12 cmd) V 5 12 3 5 1 728 cm 3SA 5 6 3 12 3 12 5 864 cm 23. a)H__h ​ 5 ​ 3__2 ​ 5 ​ R__r ​SA bigb) _____ ​ 5 ​9__SA small4 ​; V big5 ___ ​27V small8 ​c) SA big5 ​9__ ​ 3 400 5 900 cm24d) V small5 ___ 8​ ​ 3 108 5 32 cm327Investigation (Student’s Book page 232)1. If the sides increase by scale factor 3, the surface area will increase by scalefactor 9. The area will increase by scale factor 27.2. If you halve the sides, the surface area decreases by scale factor 1__ ​. The area4decreases by scale factor 1__8 ​.3. Conclusion: The surface area increases or decreases by the square of the sides. Thearea decreases by the factor of the power of three of the side.4. a) Side x Surface area is 6x 2 Volume is x 3b) Side 3x Surface area 6(3x) 2 Volume is (3x) 3c) Side ( 1__ ​)x2Surface area 6(1__2 ​x)2 Volume is ( 1__2 ​x)374 EXPLORING MATHEMATICS FORM 3

Activity 13.2 (Student’s Book page 233)1. 4; 82.1__2 ​; ​ 1__8 ​3. 3; 27Exercise 13.3 (Student’s Book page 234)1. a) SA 5 2r 2 1 2rh5 534.1 cm 2V 5 942.5 cm 3b) SA 5 6 3 8 2 5 384 cm 2V 5 83 5 512 cm 32. a) SA 5 9 3 534.1 5 4 806.9 cm 2V 5 27 3 942.5 5 25 447.5 cm 3b) SA 5 9 3 384 5 3 456 cm 2V 5 273 512 5 13 824 cm 33. a) SA 5 4r 2 5 314.2 cm 2 5 0.0314 m 2b) Number of spheres 5 ______ 5​0.0314 ​ 5 159c) Scale factor 5 (​1__2 ​)2 5 ​1__4 ​Number of spheres 5 4 3 159 5 636Note: The actual answer is 636.9, but since 0.9 is not a full sphere, the number ofspheres that can be fully covered by the available material is only 636.4. a) V 5 5 301.4 m 3b) SA 5 r 2 1 2rh 5 1 590.4 m 2c) Cost 5 1 590.4 3 10.75 5 P3 419.435Note: Realistically, paint gets bought in 20-litre containers, especially for a job ofthis magnitude. It would therefore be more realistic to work out how many 20-litrecontainers are needed ( 1 590.4 4 20 5 15.9 16 containers), and then calculate the5cost of it, in this case 16 3 20 3 10.75 5 P3 440.d) i) V 5 ​1__ ​ 3 5 301.4 5 662.7 m38ii) Cost 5 ​1__ ​ 3 3 419.43 5 P854.86 (P860)45. a) _________ volume Bvolume A ​ 5 _____ ​ 1 02416 ​ 5 64length of base Bb) ______________ ​ 5 cube root of 64 5 4length of base Asurface area of Bc) ______________ ​ 5 square of length 5 16surface area of Ad) Base area of pyramid B is 8 3 16 5 128 cm 2Chapter 13: Surface area and volume75

6. If the second block of wood is half the size, the cost of the second block will beP320 3 (​1__4 ​) = P80.00. He will need 1.2 litres 3 (​ 1__ ​) 5 0.15 litres of varnish.8Extension Activity 13.1 (Student’s Book page 236)1. a) s 5 ​√ ________12 2 1 6 2 ​ 5 13.4 cmb) SA 5 2r 2 1 rs5 2 3 6 2 1 3 6 3 13.416407…5 479.1 cm 22. SA 5 ​9__ ​ 3 479.08800… 5 1 077.9 cm24Project (Student’s Book page 237)Diameter in metres Height in metres Volume in m 31 2.5 1.963751 3 2.35651.5 2.5 4.41843751.5 3 5.3021252 1.75 5.49852 2 6.2841. 1.5 m in diameter and height 3 m is enough.2. The tank with the smallest surface area is the one 1 metre in diameter andheight 2.5 m.3. SA 5 3.142 3 1 3 (0.5 1 2.5) 5 9.426 m 2 .The amount of metal to the nearest metre will be 10 m 2 .Cost: 10 m 2 3 180 + ____ ​ 18 ​ 3 1 800 = P2 124100Revision Exercise (Student’s Book page 238)1. a) 9:4; 27:8b) 2; 82. a) V 5 162 cm 3 ; SA 5 198 cm 2b) 1:4; 1:2c) V 5 20.25 cm 3 ; SA 5 49.5 cm 23. a) SA per ball 5 4r 2 5 254.5 cm 2Leather required 5 3 053.63 cm 2b) Leather required 5 (​2__3 ​)2 3 3053.6351 357.17 cm 24. a) 1:25b) 1:5c) 9 cm76 EXPLORING MATHEMATICS FORM 3

165. a) ___49 ​ and ____ ​ 64343 ​b) i) 210 mmii)4__ ​ 3 210 5 120 mm76. Dimensions increase by a factor of 3 ​√ ______ 278 5 ​ 3__2 ​37.5 m2.25 m4.5 m12 m7. a) V 5 (R 2 2 r 2 ) 3 h5 (4 2 2 2.5 2 ) 3 0.155 4.6 m 3 concreteb) Cost 5 4.6 3 150 5 P690c) SA 5 30.6 m 2d) Cost 5 P78.95 3 30.6 5 P2 415.87e) He decreased the dimensions by factor 2__3 ​.f) i) V 5 (​2__3 ​)3 3 4.6 5 1.36 m 3 concreteii) Savings 5 690 3 ___ 19​27 ​ 5 P485.56(1 2 (​2__3 ​)3 5 ___ 19​27 ​)Note that cost 5 (​2__3 ​)3 3 6905 P204.44iii) SA 5 (​2__3 ​)2 3 30.6 5 13.6 m 2iv) Savings 5 (1 2 (​2__3 ​)2 ) 3 2 415.875 P1 342.15Chapter 13: Surface area and volume77

14c h a p t e rTravel graphsSyllabus: General objectiveThis chapter deals with applications of distance, time and speed to real-lifesituations.Specific objectives• Draw and interpret displacement-time graphs and solve problems involving these.• Draw and interpret velocity-time graphs and solve problems involving these.This chapter teaches students the skill to draw, interpret and use displacement-timeand velocity-time graphs in real-life situations.AnswersSupport Activity 14.1 (Student’s Book page 245)1. A. Gradient 5 0; Velocity 5 0 m/sObject stationary.B. Gradient 5 constant and positive Displacement uniform Velocity 5 positive, so the object moves at constant velocity away from thereference point, so in a straight line.C. Gradient NOT constant acceleration, so velocity increases with time and sodisplacement per second increases. Gradient gets steeper with time, BUT: it isstill a CONSTANT acceleration, so there is even growth in steepness of graph.D. Gradient NOT constant acceleration (although negative), but the velocitydecreases with time, so the displacement per second decreases. Gradient getsflatter with time, although it is still a CONSTANT deceleration.Note: A negative acceleration is a deceleration.2. a) Ab) Dc) Bd) C78EXPLORING MATHEMATICS FORM 3

Activity 14.1 (Student’s Book page 248)2. a) and b)c) Phako has stopped to look at a flower bush.d) 3 seconds.Activity 14.2 (Student’s Book page 249)1. a) (0;0) is at P, the starting point. It then moves away from P, stops at B (turningpoint/maximum point on curve) comes back towards P (displacementdecreases) until it goes past P (cuts time axis) in the opposite direction (belowtime axis) until it reaches D where it stops.b) Displacement is positive above the time axis (when the ball moves between Pand B) and negative below it (when the ball moves between P and D).c) Velocity positive from P to B and negative from B back to P through to D.Note: The velocity is zero at B since it is a turning point.d) At B and D.e) 3 m to the left of P.2. It travels upwards while it slows down uniformly until it stops (at B); it thencomes back at a uniform acceleration (under the influence of gravity) and movespast the point from where it was launched (at P) until it hits the floor (at D). No,the ball travels in a straight line that goes vertically upwards and then along thesame line vertically downwards until it reaches the floor. The graph is arepresentation of the displacement of the ball relative to the reference point (P).3. No. Displacement changes as time goes on and the horizontal axis is the timeaxis, so the graph is bound to 'follow the time' from left to right.Chapter 14: Travel graphs79

Activity 14.4 (Student’s Book page 252)2. a) The ball goes up in the air to 3 metres above the point from where it wasthrown (reference point) in 1 second, after which it stops and then falls back,past the reference point to a point 9 metres below the reference point in thenext 2 seconds. The ball falls under a constant acceleration, which is why thecurve has a parabolic shape (there is a quadratic relationship between thedisplacement and the time).b) Displacement positive for 0 t 2.c) After 1 second (the velocity changes direction).d) 3 m/s upwards.Exercise 14.1 (Student’s Book page 253)1. a) 1.5 metres1.5b) ___0.8 ​ 5 1.875 sc) 3.375 sd)e) Total displacement 5 0 m2. a) AB: Object stationary for 2 seconds at 1 metre from the reference point.BC: Object accelerates at a constant rate (velocity increases) for 2 seconds andis displaced 3 metres from the starting point (4 metres from the referencepoint).CD: Object moves at constant velocity (0.5 m/s) for 2 seconds and is displacedby another metre.DE: Object moves back to the reference point at a constant velocity (2 5m/s)in 1 second.b) 21 mc) Distance 5 3 1 1 1 5 5 9 md) Velocity 5 ____________displacement ​ 5 2​5__ ​ 5 25 m/stime 1e) Average velocity 5 2​1__ ​ 2 0.14 m/s780 EXPLORING MATHEMATICS FORM 3

3.f) Object stopped and turned back on the same path.g) Average velocity 5 ​3__ ​ 5 1.5 m/s24. a) AB: Moves at constant velocity for 3 seconds, displaced 3 metres; stops at B.BC: Turns back and moves at constant velocity for 2 seconds in oppositedirection; displaced 22 metres.CD: Moves at increased velocity until 1 metre past reference point, all in1 second. Displaced 22 metres; stops at D.DE: Turns back on same path and moves at constant velocity for 2 seconds,1 metre past reference point. Displaced 2 metres.b) Stopped and turned back at B and D. Accelerated at C.c) V AB5 1 m/s in a certain directionV BC5 21 m/s in the opposite directionV CD5 22 m/s in the opposite directionV DE5 1 m/s in the original directiond) V AB5 3.6 km/h in a certain directionV BC5 23.6 km/h; opposite directionV CD5 27.2 km/h; opposite directionV DE5 3.6 km/h in the original directione) i) 1 metreii) 9 metresiii) Average speed 5 ________ distance​time ​ 5 ​ 9__ ​ 5 1.125 m/s8f) At reference point A.g) At 0 s and at 7 s.Chapter 14: Travel graphs81

1.Extension Activity 14.1 (Student’s Book page 255)2. a) Same for both, since they reach the end point at exactly the same time.b) Ra Sam: 14.3 m/s; 7 m/s; 4.7 m/s; 4 m/s; 3.3 m/s all in a northern direction.Note: Only the average velocity per 3-second interval can be calculated since RaSam’s velocity changes constantly (broken-line graph). The instantaneous velocitycan be found at a given time by finding the gradient of the curve at that time. (Thistechnique falls outside the syllabus for Form 3.)Ra Buthi: velocity 5 ____ ​100​ 5 6.7 m/s North15 He maintained a constant velocity (let’s assume it to be possible) frombeginning to end.c) Ra Sam was at 43 m, 64 m, 78 m, 90 m and 100 m from the starting point. Hecovered 43 m in the first 3 seconds, then 21 m in the next 3 s, 14 m in thenext, 12 m and then 10 m. Ra Buthi was at 20 m, 40 m, 60 m, 80 m and 100 m from the starting point,so he covered 20 metres every 3 seconds throughout the race.d) No winner, they both finished at the same time.3. a) 23 metresb) 25 metresc) 18 metresd) 10 metrese) 0 metres4. a) Average velocity 5 ____ 100​ ​ 5 6.7 m/s North15b) Ra Sam: 14.3 m/s North;7 m/s North;4.7 m/s North; 4 m/s North;3.3 m/s NorthRa Buthi: Velocity 5 ​ ____ 100​ 5 6.7 m/s North155. Graph (b)6. No – they are too slow. Olympic athletes run 100 m in under 10 s.82 EXPLORING MATHEMATICS FORM 3

Project (Student’s Book page 257)1. AB: Constant velocity for 2 minutes, away from A in a positive direction.Displacement is a straight line, which indicates a constant velocity. BC is wherethe object stands still, so the velocity is zero. CD is displacement in a directionopposite to the original direction, therefore it is a negative velocity (since it is inthe opposite direction) but still at a constant velocity, hence the straight line fordisplacement. Note that the time on the velocity-time graph has been convertedto hours in order to express the velocity in km/h.2. The velocity stays constant, i.e. acceleration is zero.3. Area A 5 2 kmArea B 5 26 kmThe areas represent the displacement since km/h 3 h 5 km.4. The displacement is zero in that time interval.5. Yes, the object ends up 4 km from its starting point, but in a negative direction.Activity 14.5 (Student’s Book page 259)1. Velocity 5 4 m/s East after 2 seconds2. Velocity stays constant at 12 m/s East for 14 seconds.3. The scooter stops.4. 0 – 6 seconds5. BC: 20 – 24 seconds6. It accelerates constantly from a standstill for 6 seconds, then travels at a constantvelocity (in a straight line) for 14 seconds, after which it slows down at aconstant rate for 4 seconds until it stops.7. 6 seconds8. Displacement 5 area under velocity-time graph.∴ Displacement 5 ​1__2 ​ 3 4 3 8 5 16 m9. Total area 5 total displacement5 ​1__2 ​ 3 6 3 12 1 14 3 12 1 ​ 1__2 ​ 3 4 3 125 228 metres10. 222 metresChapter 14: Travel graphs83

Exercise 14.2 (Student’s Book page 260)1. a) An object moving at a constant (uniform) velocity, for example, a car on astraight road.b) An object slowing down at a uniform rate until it stops, for example, a scooter.c) An object slowing down at a constant rate until it comes to a standstill, andthen it accelerates in the opposite direction at the same constant rate.d) An object slowing down at a non-uniform rate until it stops. A car, truck,train, etc.2. a) 22.5 m; 67.5 m; 157.5 mb) The car slowed down uniformly.c) The car stopped.d) 213.5 m (13.5 m in direction opposite to the initial direction of movement)e) 157.5 mf) 0 m/s 2g)1.Extension Activity 14.2 (Student’s Book page 261)Radius of orbit 5 1 609 1 6 4005 8 009 kmDistance 5 2 3 8 009 5 50 322 km2. 0 km84 EXPLORING MATHEMATICS FORM 3

3. Speed 5 ________ distance​time​50 322 km5118,41hours6050 322 km51.9735 h≈ 25 500 km/h4. Velocity 5 0 since displacement 5 0Revision Exercise (Student’s Book page 262)1. a)b)c)d)___ 333 ​ 5 ___ ​ 1​ 5 0.09 m/s away from A.11___ 8​ 5 20.1 m/s away from A, i.e. 0.1 m/s towards A.802. a)3__ ​ 5 0.6 m/s5b) 0 m/sChapter 14: Travel graphs85

c)3. a) Velocity 5 40 3 3.65 144 km/hNo, it is not safe and it is over the speed limit anyway.b)c) Distance 5 Area under velocity-time graph 1 5 metresDistance 5 40 3 15 1 5 3 30 1 ​1__2 ​(5 3 10) 1 10 3 30 1 ​ 1__ ​(4 3 30) 1 525 1 140 metres (5 1.14 km)d) Distance 5 40 3 34 5 1 360 metresThe car would have been 220 metres past the elephant.4. a) A: The potato was launched.B: The potato reached its maximum height, after 2 seconds, where it stoppedand turned back.C: The potato reached the point from where it was launched, after 4 seconds.b) It starts at v 5 30 m/s upwards, slows down to v 5 0 m/s within 2 secondsand then accelerates uniformly in the opposite direction until it reachesv 5 30 m/s downwards after 2 seconds (or v 5 230 m/s upwards).86 EXPLORING MATHEMATICS FORM 3

c)d) 0 metrese) 60 metresf) Speed 5 ________ distance​time ​ 5 ___ ​ 60​ 5 15 m/s = 54 km/h4Chapter 14: Travel graphs87

15c h a p t e rWorking with variablesSyllabus: General objectiveThis chapter deals with knowledge on working with variables.Specific objectives• Expand and/or simplify algebraic expressions of up to 4 terms and of degree 3.• Express as a single fraction, algebraic fractions with two terms and of degree 3,with both numerator and denominator having two terms.• Factorise quadratic expressions and include the concept of difference of twosquares.The use of algebra and algebraic concepts play a huge role in problem solving inMathematics, and the skills dealt with here form an integral part of the biggerpicture in Mathematics. One often hears students ask: “… but where will I use it inlife one day?”. This section of work lends itself to sketching the broader picture tostudents and to help them understand that Mathematics is like a tapestry frombehind: it often seems like a lot of loose ends with no purpose, but when iteventually all comes together one day, it makes the most beautiful picture!AnswersActivity 15.1 (Student’s Book page 266)1. a) x 2 ; 6x; 24; 4xb) Total area 5 x 2 1 6x 1 24 1 4x5 x 2 1 10x 1 24 units 2Exercise 15.1 (Student’s Book page 268)1. a) 2a 1 2ab 1 4bb) x 2 1 2x 2 3c) x 2 2 5x 2 14d) x 3 2 4x 2 1 3xe) x 3 1 2x 2 1 2x 2 388EXPLORING MATHEMATICS FORM 3

2. a) x 2 1 5x 1 6b) 23x 2 1 3c) 22x 3 1 x 2 1 6x 2 3d) 23a 3 2 3a 2 1 9ae) 23a 3 1 8a 2 2 a 2 6f) 4pk 2 8p 2 12k 1 24g) 3x 3 1 15x 2 1 18xh) p 3 2 4p 2 1 p 1 63. a) Perimeter 5 10x 1 4 cmb) Area 5 6x 2 1 8x 2 8 cm 2Extension Activity 15.1 (Student’s Book page 268)1. (a 1 b) 1 5 a 1 b(a 1 b) 2 5 a 2 1 2ab 1 b 2(a 1 b) 3 5 a 3 1 3a 2 b 1 3ab 2 1 b 3(a 1 b) 4 5 a 4 1 4a 3 b 1 6a 2 b 2 1 4ab 3 1 b 42. 11 2 11 3 3 11 4 6 4 13. Add the two numbers above to get the next row:1 5 10 10 5 11 6 15 20 15 6 11 7 21 35 35 21 7 11 8 28 56 70 56 28 8 1Support Activity 15.1 (Student’s Book page 271)1. 4(x 1 2y) 1 a(x 1 2y)5 (x 1 2y)(4 1 a)2. 3a(1 2 2a) 2 2b(1 2 2a)5 (1 2 2a)(3a 2 2b)3. 3xy(8x 2 3y) 2 2y(8x 2 3y)5 y(8x 2 3y)(3x 2 2)4. 6(2m 1 n) 2 2n(2m 1 n)5 2(2m 1 n)(3 2 n)Chapter 15: Working with variables89

Exercise 15.2 (Student’s Book page 271)1. a) 2b) 8xc) 11x 2d) 7mne) 8xyf) 2x 2 mg) 3c 2 12. a) 3(a 1 b)b) 7mn(m 2 n 2 8)c) (3c 2 1)(2a 2 b)d) (2r 2 1)(3x 1 2y)e) (x 1 5)(x 1 3)f) (x 1 5)(2x 1 3)g) (x 1 6)(x 2 6)h) x(x 2 3) 1 3(x 2 3)5 (x 1 3)(x 2 3)3. a) 2(3 2 y) 2 x(3 2 y)5 (3 2 y)(2 2 x)b) (x 2 1)(x 2 5)c) x(x 2 2) 2 8(x 2 2)5 (x 2 2)(x 2 8)d) (x 1 4)(x 1 1)e) x(x 1 3) 1 7(x 1 3)5 (x 1 3)(x 1 7)f) (2x 2 3)(2x 1 3)g) a(x 1 7)(x 2 3)h) 2x(2x 1 3) 2 7(2x 1 3)5 (2x 2 7)(2x 1 3)Activity 15.2 (Student’s Book page 272)1. 602. 403. 8a 44. a 2 (a 1 1)5. a 2 b 3Exercise 15.3 (Student’s Book page 273)1.y__5 ​90 EXPLORING MATHEMATICS FORM 3

2. _______ 3a 1 4b​ ​ab2xy 2 x3. _______6y ​ 3b 1 34. ________2b(1 2 b) ​5. ___________ 22b(b 1 1)(b 1 2) ​6.2x 1 8 _______x(x 2 2) ​Exercise 15.4 (Student’s Book page 274)1. a)b)c)d)e)f)g)___ 3x 32z___ a 22b ​________ 12a 2 (3 2 b) ​_____ 23 2 b ​1__x ​a__2 ​___________2x(35x 1 48)​6x 2 (x 2 4)2. Perimeter 5 2(l 1 b)5 2​( ____ 12ba ​1 ___ ​ a 38b )​5 ________ 2​(​[ 96b2 1 a 48ab ] ​ )​5 ​ ________ a 4 1 96b 2​4abArea 5 l 3 b5 ____ ​12ba ​ 3 ___ ​ a 38b ​3a5 ​ ___22 ​Extension Activity 15.2 (Student’s Book page 274)xxxx + 1x + 1x + 2x + 2x + 2xSmaller volumes: x 2 (x 1 2) 1 x(x 1 1)(x 1 2)+ x(x 1 2) 2 1 (x 1 1)(x 1 2) 2= 4x 3 1 14x 2 1 14x 1 4Big volume: (2x + 1)(x + 2)(2x + 2)= 4x 3 + 14x 2 + 14x + 4This equals the sum of the smaller volumes.Chapter 15: Working with variables91

Activity 15.3 (Student’s Book page 275)1. a) 1b) x 1 2c) 24xd) x 2 1e) x 2 5f) (x 2 3)(x 2 1)g) x 2 1 2x 2 8h) (x 2 4)(x 1 1)i) (x 1 4)(x 2 1)Activity 15.4 (Student’s Book page 277)1. a) x 2 1 3x 1 3x 1 9b) x 2 2 4x 2 3x 1 12c) x 2 2 5x 1 x 2 5d) x 2 1 4x 2 2x 2 82. a) (x 1 3)(x 1 3)b) (x 2 4)(x 2 3)c) (x 2 5)(x 1 1)d) (x 1 4)(x 2 2)Exercise 15.5 (Student’s Book page 277)1. (x 1 5)(x 1 5)2. (x 1 5)(x 2 1)3. (y 1 3)(y 2 1)4. (x 1 7)(x 2 1)5. (x 2 5)(x 1 4)6. (x 2 12)(x 2 2)7. (k 2 3)(k 2 3)8. (x 2 8)(x 2 1)9. (x 2 30)(x 2 1)Exercise 15.6 (Student’s Book page 278)1. (5x 2 4)(x 1 1)2. (3a 1 1)(a 1 8)3. (2d 2 1)(d 1 3)4. (7x 2 3)(2x 1 5)5. (4y 2 5)(2y 2 3)92 EXPLORING MATHEMATICS FORM 3

Activity 15.5 (Student’s Book page 279)6. a) (x 1 1) 2b) (x 1 2) 2c) (x 1 6) 2d) (2x 2 1) 2Exercise 15.7 (Student’s Book page 280)1. a) (a 1 2) 2b) (y 1 7) 2c) (x 1 y) 2d) (b 2 8) 2e) (2x 1 y) 22. a) (k 2 9)b) (7a 2 5b)c) (3a 1 2b)Investigation (Student’s Book page 281)1. a) AB 5 3, BC 5 4, AC 5 5b) AB 5 8, BC 5 6, AC 5 10c) AB 5 5, BC 5 12, AC 5 132. Yes. The sides satisfy the equation a 2 1 b 2 5 c 2 .3. For right-angled triangle ABC:AC 2 5 AB 2 1 BC 2= (m 2 2 n 2 ) 2 1 (2mn) 2= m 4 2 2m 2 n 2 1 n 4 1 4m 2 n 2= m 4 2 2m 2 n 2 1 4m 2 n 2 1 n 4= m 4 1 2m 2 n 2 1 n 4= (m 2 1 n 2 ) 2= AC 2Extension Activity 15.3 (Student’s Book page 281)Hypotenuse 5 ​√ ________________(n 2 2 4) 2 1 (4n) 2 ​5 ​√ ____________________n 2 2 8n 1 16 1 16n 2 ​5 ​√ _______________17n 2 1 8n 1 16 ​Chapter 15: Working with variables93

Exercise 15.8 (Student’s Book page 283)1. (x 2 9)(x 1 9)2. (2x 2 1)(2x 1 1)3. (j 2 13y)(j 1 13y)4. (3m 2 5)(3m 1 5)5. 2(t 2 2)(t 1 2)6. 5m(m 2 3n)(m 1 3n)7. 2(6 2 x)(6 1 x)8. [5 2 (b 1 3)][5 1 (b 1 3)]5 (2 2 b)(8 1 b)9. (23 2 d)(9 2 d)10. 3(5 2 x)(5 1 x)11. (4z 2 3)(4z 1 3)12. 3ab 2 (9a 2 2 4)5 3ab 2 (3a 2 2)(3a 1 2)13. 3x 3 y 3 z(x 2 4yz 2 )(x 1 4yz 2 )Exercise 15.9 (Student’s Book page 284)1. a)_________4b 2 (1 1 2b)​ 3 ___________(b 2 1)(b 1 1)​​b 1 1b5 4b(1 1 2b)(b 2 1)b) 4(b 2 2 3)(2b 1 3)(b 2 1)c) _____________(2b 2 3)(2b 1 3) ​2 _______ ​ 3(b 2 3) ​3bb5 ​ __________2 2 8b 1 92b(2b 2 3) ​ OR _________ ​8b – b2 – 9b(2b – 3) ​2. P 5 ___________(b 1 1)(b 1 3)​1 ___________​ (b 2 2)(b 1 2)2​[ 3(b 1 1) 3(b 1 2) ] ​5 4b 1 23A 5 ___________​ (b 2 2)(b 1 2) ​ 3 ___________(b 1 3)(b 1 1)​​3(b 1 3) 3(b 1 1)5 ​ _____ b 2 2 4​9Revision Exercise (Student’s Book page 285)1. a) b 2 (b 2 1) 1 3b(b 2 1)5 b(b 2 1)(b 1 3)b) a 2 1 a 2 6c) x 3 2 3x 2 2 10xd) x 2 2 8x 1 16e) 4a 2 1 12ab 1 9b 294 EXPLORING MATHEMATICS FORM 3

f) 2x(4x 2 1 4x 1 1)5 8x 3 1 8x 2 1 2x2xy 2 x2. a) _______6y ​ 3b)c)22b ___________(b 1 1)(b 1 2) ​__________________x 2 1 x 2 (x 2 2)(x 1 4)​x(x 2 2)5 ​ _______ 2x 1 8 ​ x(x 2 2)3abd) ______2 c 2 d​2e)y__2 ​d 2 3f) _____3 ​3. a) (p 1 q) 2b) (6x 1 1) 2c) (t 2 7)(t 1 7)d) [7 2 (g 1 6)][7 1 (g 1 6)]5 (1 2 g)(13 1 g)e) (x 1 5)(x 2 4)f) 2(x 2 2)(x 2 1)g) (3x 1 1)(2x 2 1)h) (3x 2 2)(2x 1 3)i) (x 1 1)(2 1 y)j) (x 2 1)(3x 1 4y)Chapter 15: Working with variables95

16c h a p t e rFormulaeSyllabus: General objectiveThis chapter deals with knowledge of formulae.Specific objectives• Change the subject of a formula to a specified variable by involving the use offactorising, square roots and cube roots.• Solve problems involving changing the subject of a formula as applied inpractical situations.AnswersActivity 16.1 (Student’s Book page 289)1. a) Area of a circleb) Speed 5 _______ distance​time​c) Compound interestd) n th term of arithmetic sequencee) Distance between points (x 1;y 1) and (x 2;x 2)Extension Activity 16.1 (Student’s Book page 289)1. V esc5 ( 2 3 6.673 3 10220 3 5.9742 3 10 24 1__2)​6.371 3 10 31_5 (125.5739)​ 25 11.19 km/s3. V esc5 ( 2 3 6.673 3 10220 3 7.36 3 10 22 1__21.737 3 10)​35 2.3780 km/s2.3780Factor 5 ______ ​11.19 ​ 5 0.296EXPLORING MATHEMATICS FORM 3

Homework Activity (Student’s Book page 291)Yes, –40 °C = –40 °FInvestigation (Student’s Book page 292)1. 1, 3, 5, 7, 9, 11, 13, 15, 17, etc.2. 4, 9, 16, 25, 36, etc.3. The resulting sums are perfect squares.4. n 2 , where n is the position of the number, e.g. the sum of the first three oddnumbers is 9, which is 3 2 .Exercise 16.1 (Student’s Book page 293)1. h 5 ​V__lb ​2. t 5 d S3. P 5 _______ ​A(1 1 i) ​ n4. n 5 ______T​n2 ba ​5. x 25 2x 2 x 1Exercise 16.2 (Student’s Book page 295)1. yz 1 5z 5 3k∴ z(y 1 5) 5 3k∴ z 5 ​ _____ 3k ​ y 1 52. x(5 1 y 1 d 2 2) 5 1∴ x 5 ​ ___________ 15 1 y 1 d 2 2 ​3. 3d 2 5 3v∴ d 5 ±​√ __ v ​4. t 5 ___ ​EI 2 R ​__5. r 5 √ ​A__6. I 2 5 __ ​ E ​ Rt√(r always > 0)I 5 ​E__ ​ (currently always > 0)Rt___a ct 2( is a constant value, so it does not make sense in reality to make a4rconstant the subject of a formula, but for the purpose of the exercisewe can accept it.)7. 5 ​√ ___Chapter 16: Formulae97

8. T 2 5 A 2 3 20d∴ d 5 ​ ____ T 2​20A 29. H 2 5 4 2 ___ L9.8 ​∴ L 5 _____ ​ 9.8H 2​4 210. r 5 3 ​√ ___P__C ​ ​2 l11. b 5 3a(​ 3 √ __k ​2 x)Activity 16.3 (Student’s Book page 296)1. a) 6 m 3b) V 5 lbhc) h 5 ​V__lb ​d) h 5 ______ 6​ ​5 2.5 m1.2 3 22. h 5 1 500(26 2 T) 5 1 500(26 1 52)5 117 000 mExtension Activity 16.2 (Student’s Book page 297)1. Cost 5 1.15 3 10 1 3.52x2. P 5 ​ __________________________________________0.04 1 1.7y 1 1.5hhd 2 0.13rf 2 2.1hr 1 0.53gov 2 Q​0.17Exercise 16.3 (Student’s Book page 297)1. a) p 5 2A qb) p 5 _______ ​ 2 3 100 ​5 8 cm252. R 5 ​E__I ​a) E 5 IRb) I 5 E Rc) E 5 1.5 3 25 5 37.5 voltsd) I 5 ___ ​ 12 ​5 0.8 amperes__________ 153. d 5 ​ √8h​( R 2 ​h__2 )___4. v 5 √ ​ ___ 2Em ​ ​5 √ ____​ ____ E6005. a) Cost 5 300 1 45pb) p 5 _______ C 2 300​ ​5 ________ 560 – 300​ ​ 5 5.845 45c) Kabelo can take 4 people plus himself (5 people in total).98 EXPLORING MATHEMATICS FORM 3

6. a) A 5 6x m 2288b) ____ ​5 24 m212c) x 5 4 m7. t = 4.1 sRevision Exercise (Student’s Book page 299)___1. a) v 5 ±​ √___F crmb) a 5 ​√ _______2(s2ut) ​4c) t 5 √____ ​ ____2 ra cd) T 2 5 ____ ​ 4 2 Lg​____∴ g 5 ​( 42 LT 2 )​e) s 5 ​√ ___2gt ​f) c 2 5 ​E__m ​∴ m 5 ​E__c ​ 2 abg) h 5√a 3 1 b 32. a) n 5 3 ​√ ___8( ​T n2 3)5 2​ 3 √ ______T n2 3 ​b) n 5 2 3 3 ​√ _______30 2 3 ​ 5 6, so it is the 6 th term3. b 5 ______ v​2D 1 3h ​5 ___________​ (2.5) 2​5 0.11550 1 3 3 1.54. a) C 5 50x 1 25yb) 800 5 50 3 8 1 25y∴ y 5 16 chickensChapter 16: Formulae99

Assessment 3 Chapters 12 – 16Question 11.1 Shaded area 5 153.5 m 2 ✓ (4)1.2 Unshaded area 5 300 2 153.5✓5 146.5 m 2 ✓ (2)1.3 Perimeter 1 circumference 5 66.4 m✓Cost 5 66.4 3 25✓✓ 5 P1 660✓ (4)1.4 a) Cost 5 ​1__ ​ 3 66.4 3 25✓ 5 P830✓2b) Shaded area 5 (​1__2 ​)2 3 153.5✓5 38.4 m 2 ✓ (4)[14]Question 22.1 h 5 ​√ ________10 2 2 6 2 ​ ✓✓5 8 cm✓ (3)2.2 Area 5 12 3 12 3 8✓ 5 48 cm 2 ✓ (2)2.3 Volume 5 area base 3 height5 48 3 20✓5 960 cm 2 ✓ (2)2.4 TSA 5 220 310 1 12 3 20 1 2 3 48✓✓5 736 cm 2 ✓✓ (4)2.5 V 5 27 3 960✓ 5 25 920 cm 3 ✓TSA 5 9 3 736✓ 5 6 624 cm 2 ✓ (4)[15]Question 33.1 The object moves 1 metre away from A to B (left) in 1 second✓, stopsfor 1 second, moves 4 metres in one second to the right toD✓ and then turns back and moves 2 metres left in 2 seconds to E.✓ (3)3.2 Displacement 5 1 m to the right.✓ (3)3.3 v AB5 1 m/s left; v BC5 0 m/s✓✓; v CD5 4 m/s right✓✓; v DE5 1 m/s left✓✓ (6)3.4Velocity (m/s)543✓212122✓1 2 3 4 5 6✓Time (s)23 (4)[16]✓100 EXPLORING MATHEMATICS FORM 3

Question 44.1 8x 3 1 64y 3 (3)4.2 218x 3 1 60x 2 − 50x (3)[6]Question 55.1 a) ___ 3y 22 ​✓ ✓ (2)3x 1 5b) ______ ​✓2 ✓ (2)5.2 a) (x 1 2)(a 2 b)✓✓✓ (3)b) (x 2 2y)(x 1 2y)✓✓ (2)c) (2x 1 3)(x 2 4)✓✓ (2)d) (2x 1 y)(2x 2 y 1 6)✓✓✓ (3)5.3 a) x 5 1 or x 5 22✓✓✓ (3)b) x 5 0 or x 5 5✓✓✓ (3)c) 2x 2 2 3x 2 9 5 0✓∴ (2x 1 3)(x 2 3) 5 0✓x 52 ​3__ ​ or x 5 3✓✓ (4)2[24]Question 6___6.1 g 5 6​ √___ 12tm ✓✓✓ (3)m6.2 x 5 ​ _______ ​✓✓ (2)3(a 1 b)[5]Total: 80Chapter 16: Formulae101

172c h a p t e rSyllabus: General objectiveThis chapter deals with the understanding and application of quadratic equations.Specific objectives• Solve quadratic equations by factorisation.• Solve problems involving quadratic equations as applied in practical lifesituations.• Use a spreadsheet to solve quadratic equations.AnswersSupport Activity 17.1 (Student’s Book page 305)1. x 5 152. x 5 ___ 60​ y ​3. x 5 04. x IR if y 5 05. x IR6. x 5 0; x 5 27. x 5 0; x 5 218. x 2 2 2x 2 4 5 0∴ x 5 2 6 √ 20 5 3.24 or 21.2429. x(x 2 1) 5 0∴ x 5 0 or x 5 1Exercise 17.1 (Student’s Book page 306)1. x 5 1 or x =222. x 5 1 or x 5 273. x 5 5 or x 5 24. k 5 27 or k 5 55. m 5 2​ ___ 11 ​ or m 5 326. x 5 23 or x 5 ​3__​ Quadratic equations102EXPLORING MATHEMATICS FORM 3

7. j 5 258. w 5 ​7__4 ​9. n 5 ±810. x 5 ​2__5Exercise 17.2 (Student’s Book page 307)1. (k – 4)(k – 2) 5 0∴ k 5 4 or k 5 22. (x 1 1)(x 1 4) 5 0∴ x 5 21 or x 5 243. (2m 1 1)(m 1 5) 5 0∴ m 5 2​1__ ​ or m 5 2524. (3a 1 2)(a 2 2) 5 0∴ a 5 2​ 2 __3 ​ or a 5 25. (w 2 11)(w 1 11) 5 0∴ w 5 ±11Exercise 17.3 (Student’s Book page 308)1. x 2 1 7x 2 30 5 0∴ (x 1 10)(x 2 3) 5 0∴ x 5 210 or x 5 32. c 2 112c 2 64 5 0∴ (c 116)(c 2 4) 5 0∴ c 5 216 or c 5 43. a 2 2 14a 2 32 5 0∴ (a 216)(a 1 2) 5 0∴ a 516 or a 5 224. b 2 5 36∴ b 5 ±​√ ___36 ​ 5 ±65. x 2 2 49 5 0∴ (x 2 7)(x 1 7) 5 0∴ x 5 ±76. y 2 2 y 2 12 5 0∴ (y 2 4)(y 1 3) 5 0∴ y 5 4 or y 5 237. g 5 528. m(m 2 5) 5 0∴ m 5 0; m 5 5Chapter 17: Quadratic equations103

9. j 2 2 6j 1 8 5 0∴ (j 2 4)(j 2 2) 5 0∴ j 5 4 or j 5 2Extension Activity 17.1 (Student’s Book page 308)1. Multiply each term by x first: 2x 2 5 2 2 3x∴ 2x 2 1 3x 2 2 5 0∴ (2x 2 1)(x 1 2) 5 0∴ x 5 ​1__ ​ or x 5 2222. Multiply each term by x first: 3x 2 2 3x 5 2 1 x 2∴ 3x 2 2 3x 2 2 2 x 2 5 0∴ 2x 2 2 3x 2 2 5 0∴ (2x 1 1)(x 2 2) 5 0∴ x 52​ __ 1 ​ or x 5 22Activity 17.2 (Student’s Book page 311)1. l 5 x 1 52. x(x 1 5) 5 143. x 2 1 5x 2 14 5 0∴ (x 1 7)(x 2 2) 5 0∴ x 5 27 or x 5 24. x 5 25. Substitute x 5 2 into the first expression to get the length, so l 5 2 1 5 5 7.Therefore width is 2 m and length is 7 m.Exercise 17.4 (Student’s Book page 311)1. 7x 1 2x 2 5 4∴ 2x 2 1 7x 2 4 5 0∴ (2x 2 1)(x 1 4) 5 0∴ x 5 ​√ __2 ​or x 5 24The answer must be an integer, so the answer is 24.2. x(x 2 5) 5 24∴ x 2 2 5x 2 24 5 0∴ (x 2 8)(x 1 3) 5 0∴ x 5 8 or x 5 23The length cannot be negative, so l 5 8 cm and w 5 3 cm.3. ∴ x 2 2 23x 5 210∴ x 2 2 23x 2 210 5 0104 EXPLORING MATHEMATICS FORM 3

∴ (x 2 30)(x 1 7) 5 0∴ x 5 30 m (length)area∴ Width 5length 5 21030 5 7 m4. a)x 1 1x 1 2xb) (x 1 2) 2 5 (x 1 1) 2 1 x 2c) x 2 1 4x 1 4 5 x 2 1 2x 1 1 1 x 2∴ 0 5 x 2 2 2x 2 3∴ x 2 2 2x 2 3 5 0∴ (x 2 3)(x 1 1) 5 0∴ x 5 3 is the solution (measurements are positive)Measurements are: 3, 4 and 5 metres.5. a) 48 cm 2c) (6 1 x)(8 1 x) 5 96∴ x 2 1 8x 1 6x 1 48 2 96 5 0∴ x 2 1 14x 2 48 5 0∴ x 5 2b6 √ b2 2 4ac5 214± √ ____388 ​/22ad) x 2.85 cm to be added to each side.6. a) 65 m (when t 5 0)b) When h 5 0 ∴ 5t 2 2 65 5 0∴ 5t 2 5 65∴ t 2 5 13∴ t 5 3.6 s (only positive value is possible)7. a) 200b) t 2 2 24t 1 200 5 56∴ t 2 2 24t 1 144 5 0∴ (t 2 12) 2 5 0∴ t 5 12 yearsExercise 17.5 (Student’s Book page 314)1. x 5 0.73 or x 5 22.732. x 5 20.24 or x 5 4.24Chapter 17: Quadratic equations105

Project (Student’s Book page 316)1. d 5 71.5 m2. 0.007v 2 1 0.015v 2 20 5 0∴ v 5 52.4 km/h 21 (for 20 metres, from formula)0.007v 2 1 0.0015v 2 10 5 0∴ v 5 36.7 km/h 21 (for 10 metres)0.007v 2 1 0.0015v 2 5 5 0∴ v 5 25.7 km/h 21 (for 5 metres)Revision Exercise (Student’s Book page 317)1. a) (x 2 7)(x 2 5) 5 0∴ x 5 7 or x 5 5b) (x 1 10)(x 2 3) 5 0∴ x 5 210 or x 5 3c) y 5 0 or y 5 6d) b 2 2 b 2 20 5 0(b 2 5)(b 1 4) 5 0∴ b 5 5 or b 5 24e) h 5 ±kf) x 2 1 7x 2 30 5 0∴ (x 1 10)(x 2 3) 5 0∴ x 5 210 or x 5 3g) a 5 16 or a 5 22h) x 5 25 or x 5 33. a)40 1 2xx20 1 2xxb) l 5 40 1 2xw 5 20 1 2xc) Area walkway5 (40 1 2x)(20 1 2x) 2 8005 120x 1 4x 2d) 4x 2 1 120x 5 544∴ x 2 1 30x 2 136 5 0∴ (x 2 4)(x 1 34) 5 0∴ x 5 4 m (the answer can’t be negative)106 EXPLORING MATHEMATICS FORM 3

4. ​1__ ​(2 1 x)x 5 242∴ x 2 1 2x 2 48 5 0∴ (x 1 8)(x 2 6) 5 0∴ x 5 6 (only solution)Height 5 6 m; Base 5 8 m5. a) x 2 63 33 3b) 3 cmc) V 5 (x 2 6) 2 3 35 3(x 2 2 12x 1 36)∴ 3x 2 2 36x 1 108 5 48∴ x 2 2 12x 1 36 5 16∴ x 2 2 12x 1 20 5 0∴ (x 2 10)(x 2 2) 5 0∴ x 5 10d) Cardboard is 10 cm 3 10 cmChapter 17: Quadratic equations107

18c h a p t e rQuadratic graphsSyllabus: General objectiveThis chapter deals with the use of graphs in quadratic equations.Specific objectives• Use graphical methods to solve quadratic equations.• Solve problems involving the use of graphical methods in quadratic equations asapplied in practical situations.• Use a spreadsheet to solve quadratic equations by a graphical method.AnswersSupport Activity 18.1 (Student’s Book page 321)x 24 23 22 21 0 1 2x 2 16 9 4 1 0 1 45x 220 215 210 25 0 5 106 6 6 6 6 6 6 6y = x 2 1 5x 1 6 2 0 0 2 6 12 202. (24;2) (23;0) (22;0) (21;2) (0;6) (1;12) (2;20)3.y201510y 5 x 2 1 5x 1 65x24 23 22 21 0 1 24. (23;0) and (22;0)108EXPLORING MATHEMATICS FORM 3

Activity 18.1 (Student’s Book page 324)1.–3. The graph cuts the x-axis at 22 and 12.y4225 22 0 2 522244. 4 2 x 2 5 0∴ x 2 5 4∴ x 5 6 4 5 12 or 22√Activity 18.2 (Student’s Book page 325)2. For a < 0: maximum functionFor a > 0: minimum function3. Fold along the axis of symmetry.4. The graph turns on the fold line (axis of symmetry).5. The fold line is halfway between the x-intercepts.Activity 18.3 (Student’s Book page 327)1. It has a shape, so a maximum function; AS: x 5 1; turning point is at (1;4);x-intercepts: x 5 3 and x 5 21; y-intercept: (0;3).2. y4221 0 352224Chapter 18: Quadratic graphs109

Extension Activity 18.1 (Student’s Book page 327)1. (1;4)2. x 5 13. Axis of symmetry4. (–1; 0); x = 21Exercise 18.1 (Student’s Book page 327)1. a) Minimum function; a = 22x-intercepts: –3 and –1Turning point (22;21)y-intercept (0;3)b) Minimum function; TP (25;0); AS: x 5 25;y-intercept (0;25)c) Minimum function; TP (0;29); AS: x 5 0;Roots: x 5 ±3; y-intercept (0;29)d) Maximum function; TP (0;1); AS: x 5 0;Roots: x 5 ±1; y-intercept (0;1)y2. a) 4b) y30220250x102224210 250xc)20yd) y515105250525x25 0255210x210215220110 EXPLORING MATHEMATICS FORM 3

Extension Activity 18.2 (Student’s Book page 328)1. a) x 5 22 or 21b) x 5 23c) No real roots2. 2, 1 and 03. b 2 2 4ac 5 0 è one root(turns on the x-axis)b 2 − 4ac < 0 è no roots(turns before x-axis)b 2 − 4ac > 0 è two roots(cuts x-axis at two points)4. Can be 2 (maximum), 1 or zero (minimum).Exercise 18.2 (Student’s Book page 329)1. Solutions: (23;26); (22;26)2. Solutions: (22;4); (1;1)3. Solutions: (20.83;4); (4.83;4) (the quadratic formula is needed for this)4. Solutions: (4;16); (24;16)5. Solutions: (4;12); (23;12)Activity 18.4 (Student’s Book page 331)y2.521.510.50x0.5 1 1.5Let the equation be y 5 k(x 2 0)(x 2 1).The TP is at ( 1__ ​;2), so substitute this into the equation:2∴ 2 2 k( 1 2 2 0)(1 2 2 1)5 k( 1 2 )(21 2 )5 2 1 4 k∴ k 5 28The correct equation is y = 28x(x 2 1)Chapter 18: Quadratic graphs111

Investigation (Student’s Book page 331)1.6 2 xx3. (1;5), (2;8), (3;9), (4;8), (5;5), (6;0)4.–5. y10(3;9)987(2;8) (4;8)65(5;5)4321(6;0) x0 1 2 3 4 5 6 76. x-interceps: 0 and 6. It tells us that the area 5 0 when x 5 0 or x 5 6.7. AS: x 5 38. (3;9)9. x 5 310. 9 units 2Activity 18.5 (Student’s Book page 332)1. There will be a minimum number of bacteria (it is a minimum function).2. y201816AS1412108642x-4 -3 -2 -1 0 1 2 3 4(1;5)4. Minimum at T 5 ​1__2 ​ oC5. TP (​1__2 ​;5​ 3__4 ​)112 EXPLORING MATHEMATICS FORM 3

6. 5.75 3 100 5 575 bacteria7. T 2 2 T 1 6 > 8T 2 2 T 2 2 > 0(T − 2)(T 1 1) > 0T < −1 ˚C or T > 2 ˚CExercise 18.3 (Student’s Book page 332)1. a) P 5 100l 2 5l 2550500450400350300250200150100500yb) i) I 5 2100 5 10 amp210ii) P max5 500 W5 10 15 20c) P 5 375 Wd) 100I 2 5I 2 5 20I 2 2 20I 1 4 5 0I 5 19.8 amp or 0.2 amp (quadratic formula used)2. a) Length 5 (x 2 4) cmb) Cut-out lengths are 2 cm each.c) Volume 5 area base 3 heightd) and e) y555045403530252015105f) 2(x 2 4) 2 5 50∴x 5 9 cm3. a) (15;45)0x1 2 3 4 5 6 7 8 9 10xChapter 18: Quadratic graphs113

) y 5 kx(x 2 30)∴ 45 5 k(15)(15 2 30)∴ k 5 20.2∴ y 5 20.2x(x 2 30) (third equation)c) y 5 20.2 3 10(10 2 30) 5 40 mExercise 18.4 (Student’s Book page 337)1. a) x 5 23 or x 5 21b) x 5 25c) x 5 3d) x 5 1Revision Exercise (Student’s Book page 339)1. a) x 5 3 or x 5 22b) x 5 26 or x 5 2c) No solutiond) x 5 5 or x 5 23e) No solution3. a)h5045403530252015105t0 1 2 3 4 5 6 8 9 10 11 12 14 15b) i) 12 mii) After 12 secondsiii) 2t 2 1 11t 1 12 5 6∴ t 5 ​ 11 ± ​√ ____________ 145 ​​ 5 11.5 seconds (t cannot be negative)2iv) Maximum height at t 5 5.5 seconds (axis of symmetry).Max h 5 2(5.5) 2 1 11(5.5) 1 125 42.25 m114 EXPLORING MATHEMATICS FORM 3

19r e t p a h c Processing data andpresentationsSyllabus: General objectiveThis chapter deals with understanding different ways of processing data andcompiling presentations.Specific objectives• Present data in a cumulative frequency table.• Draw a cumulative frequency curve.• Estimate the median from a cumulative frequency curve.• Interpret data from a cumulative frequency curve.AnswersExtension Activity 19.1 (Student’s Book page 345)6 childrenSupport Activity 19.1 (Student’s Book page 345)2. a) 11 plants were of a height less or equal to 69 mm.b) 80–89mmc) 60d) 42e) Height in mm Frequency Height in mm Cumulative frequency50–59 9 50–59 960–69 11 50–69 2070–79 6 50–79 2680–89 16 50–89 4290–99 8 50–99 50100 + 10 50–100 + 60f) 20 plants (of 60 plants) were of a height less than or equal to 69, which meansthey were all less than 70 mm.Chapter 19: Processing data and presentations115

Exercise 19.1 (Student’s Book page 346)1. a) 3b) 23c) 100d) Number of computers, n Frequency Cumulative frequencye) 12. a) 12.9%b) Number of daysabsentc) 458 students0 5 5n 1 35 40n 2 23 63n 3 14 77n 4 18 95n 5 5 100FrequencyNumber of daysabsent, nCumulativefrequency0 80 0 801 45 n 1 1252 125 n 2 2503 68 n 3 3184 140 n 4 4585 86 n 5 5446 34 n 6 5787 40 n 7 6188 2 n 8 6003. a) Weight (kg) Frequency Cumulativefrequencyb) 63c) 37%0 < w 0.5 2 20.5 < w 1.0 10 121.0 < w 1.5 23 351.5 < w 2.0 28 632.0 < w 2.5 25 882.5 < w 3.0 12 100116 EXPLORING MATHEMATICS FORM 3

4. a) Age last birthday Frequency Age lastbirthdayb) 110 peoplec) 20–39d) 73 peopleCumulativefrequency0–19 28 0 < a 19 2820–39 45 20 < a 39 7340–59 30 40 < a 59 10360–79 7 60 < a 79 110Homework Activity (Student’s Book page 348)Age Frequency Cumulativefrequency0–5 2 26–10 2 411–15 2 616–20 2 821–25 8 1626–30 0 1631–35 4 20x 5 16Activity 19.1 (Student’s Book page 351)1. Distance Cumulative frequency2. 1813. 201–5 56–10 4611–15 12316–20 18121–25 22026–30 23731–35 240Chapter 19: Processing data and presentations117

4. Table of ordered pairs:x 5 10 15 20 25 30 35x 5 46 123 181 220 237 2405. Distance travelled to cattleposts300Cumulative frequency250200150100500510 15 20 25 30 35Distance (km)Exercise 19.2 (Student’s Book page 352)1. Weight to the nearest kilogram Cumulative frequency35304 45 126 157 268 319 32Cumulative frequency curve for weight to the nearest kgCumulative frequency25201510504 5 6 7 8 9Weight (kg)118 EXPLORING MATHEMATICS FORM 3

2. a) 200b) 31–40 years was the most common age group, and 71–80 was the leastcommon age groupc) 129 peopled) Age upper boundary (years) Cumulative frequency10 520 2430 7140 15050 18260 19770 19980 200250Ages of people visiting a museumCumulative frequency20015010050010 20 30 40 50 60 70 80Age (years)3. a) i) Number of drinks Frequency0–4 35–9 410–14 815–19 1220–24 1525–29 1230–34 435–39 1Chapter 19: Processing data and presentations119

ii) Upper number of drinks Cumulative frequency4 39 714 1519 2724 4229 5434 5839 59b)7060Drinks sold at lunchtime in a schoolCumulative frequency504030201000–4 5–9 10–14 15–19 20–24 25–29 30–34 35–39Number of drinks4. Marks Frequency Mark upper limit Cumulative frequency0–5 0 5 06–10 0 10 011–15 1 15 116–20 3 20 421–25 4 25 826–30 4 30 1231–35 6 35 1836–40 3 40 2141–45 5 45 2646–50 4 50 30120 EXPLORING MATHEMATICS FORM 3

35Marks scored in a Mathematics test by Form 3 students30Cumulative frequency25201510500–5 6–10 11–15 16–20 21–25 26–30 31–35 36–40 41–45 46–50MarksActivity 19.3 (Student’s Book page 355)1. Age last birthday Frequency Age last birthday Cumulative frequency0–19 12 0–19 1220–39 25 0–39 3740–59 35 0–59 7260–79 16 0–79 8880–99 3 0–99 91Chapter 19: Processing data and presentations121

2. Median 5 middle value of number of people, so at 47:Cumulative Frequency100908070605040302010019 39 59 79 99The middle value (median age) is estimated at 44 years.Exercise 19.3 (Student’s Book page 356)1. 6.1 kg, 34 years, 15–19 drinks; 30 marks2. a) Spectators wanted to be on time for the start of the game so not manyarrived too late or after the start.b) Minutes before thegame startedFrequencyLower boundary ofclass intervalCumulative frequency60 T > 50 35 50 3550 T > 40 62 40 9740 T > 30 95 30 19230 T > 20 120 20 31220 T > 10 54 10 36610 T > 0 34 0 400122 EXPLORING MATHEMATICS FORM 3

c) Cumulative frequency of time of arrival by spectatorsbefore a game450400350Cumulative frequency30025020015010050050 40 30 20 10 0Time of arrival before a game (minutes)d) Median: 30 minutesExercise 19.4 (Student’s Book page 358)1. a) 46 kgb) 84 studentsc) 44 kg2. a) 55%b) 65%c) 30 studentsChapter 19: Processing data and presentations123

Revision Exercise (Student’s Book page 360)1. a) Month Money saved Cumulative frequency(Pula)January 300 300February 400 700March 600 1 300April 900 2 200May 450 2 650June 750 3 400b) P3 400c) 4 monthsd) He saved more money during the second half of the time, because P1 300 is lessthan half of P3 400 (after 3 months, end of March).2. a) Number of school Number of Upper boundary of Cumulative frequencydays missed students number of schoolmissed limits1–5 460 5 4606–10 740 10 1 20011–15 392 15 1 59216–20 205 20 1 79721–25 95 25 1 89226–30 66 30 1 95831–35 64 35 2 02236–40 8 40 2 030b) Number of days missed by students due to HIV/AIDSCumulative frequency2 5002 0001 5001 00050001–5 6–10 11–15 16–20 21–25 26–30 31–35 36–40Number of days missed124 EXPLORING MATHEMATICS FORM 3

c) About 1 800 studentsd) Out of 2 030 students who missed full days of school, most missed less thanthe median days, so a small percentage of students missed more than themedian days due to HIV/AIDS-related illness. We also see that 230 studentsmissed a lot of school due to HIV/AIDS-related illness.3. Marks of 210 students in a test250Cumulative frequency20015010050010 20 30 40 50 60 70 80 90 100Marks %a) 210 candidatesb) Estimated median mark is 53%.c) 70 studentsd) 45% is the pass mark (since 150 students scored above 45%)Chapter 19: Processing data and presentations125

20re t p a h cMeasures of centraltendency and dispersionSyllabus: General objectiveThis chapter deals with knowledge on measures of central tendency and measures ofdispersion.Specific objectives• Find the modal class, median class, and mean class from a frequency tablecomprising of grouped data.• Find modal class and mean class from a histogram comprising of grouped data.• Calculate range and inter-quartile range from a distribution and cumulativefrequency curve.• Estimate quartiles and percentiles from a cumulative frequency curve.AnswersExercise 20.1 (Student’s Book page 367)1. a) i) Modal class is 61–70.ii) Median class is 61–70.iii) Percentage mark Frequency Middle value Middle value 3 frequency31–40 3 35.5 106.541–50 4 45.5 18251–60 16 55.5 88861–70 26 65.5 1 70371–80 14 75.5 1 05781–90 8 85.5 68491–100 9 95.5 859.5Total 80 5 480Mean 5 _____ 5 480​80 ​ 5 68.5So the mean class is 61–70.b) 16 1 26 1 14 1 8 1 9 5 73 students2. a) i) 138–142ii) 138–142126EXPLORING MATHEMATICS FORM 3

iii) Mass in kg Frequency Middle value Middle value 3 frequency128–132 8 130 1 040133–137 12 135 1 620138–142 13 140 1 820143–147 7 145 1 015148–152 6 150 900153–157 4 155 620Total 50 7 015Mean 5 _____ 7 015​50 ​ 5 140.3So the mean class is 138–142.b) 10 lambsActivity 20.2 (Student’s Book page 369)1. The value in the middle position is 16.2. It is the median of the values to the left of 16.3. It is the median of the values to the right of 16.4. The range 5 20 2 125 85. The inter-quartile range 5 18.5 2 13.55 5Activity 20.3 (Student’s Book page 371)1. Inter-quartile values are like medians in that they occupy the middle positionwhen arranged in order in the lower and in the upper half of the distribution.Lower quartile is the median of the lower half of the distribution and upperquartile is the median of the upper half of the distribution.2. a) n 1 1 = 1012 2= 50.5. Estimated median weight for this value is 46 kg.b) The lower quartile value is 41.The upper quartile value is 54.c) Inter-quartile range is 54 2 41 5 13. (Note: mark according to the students’answers in (c)).Chapter 20: Measures of central tendency and dispersion127

Exercise 20.2 (Student’s Book page 372)1. Marks of 210 students in a test250Cumulative frequency20015010050010 20 30 40 50 60 70 80 90 100Marks %a) 43%b) 65%c) 65 2 43 5 222. Distance travelled by students to school300Cumulative frequency2502001501005005 10 15 20 25 30 35Distance (km)a) Median 5 15 kmb) Lower quartile 5 11.5 kmc) Upper quartile 5 20 kmd) Inter-quartile range 5 20 2 11.5 5 8.5 km128 EXPLORING MATHEMATICS FORM 3

Exercise 20.3 (Student’s Book page 375)1. Weights of 100 students120Cumulative frequency100806040202.029 39 49 59 69 70 80Weight (kg)a) i) 46 kgii) 40 kgiii) 54 kgiv) 54 2 40 5 14b) i) 23 rd percentile 5 39 kgii) 60 th percentile 5 48 kgiii) 89 th percentile 5 64 kgc) About 87%300Distance travelled by students to schoolCumulative frequency2502001501005005 10 15 20 25 30 35Distance (km)a) i) 15 kmii) 11.5 kmiii) 20 kmiv) 20 2 11.5 5 8.5 kmb) i) 10 kmii) 22 kmChapter 20: Measures of central tendency and dispersion129

Revision Exercise (Student’s Book page 377)1. Mean is 6.933 Mode is 10 Median is 82. a) Amount in Pula Middle value Frequency Middle value 3 FrequencyP0 – P1 000 P500 3 P500 3 3 5 P1 500P1 001 – P2 000 P1 500.50 3 P1 500.50 3 3 5 P4 501.50P2 001 – P3 000 P2 500.50 9 P2 500.5 3 9 5 P22 504.50P3 001 – P4 000 P3 500.50 18 P3 500.50 3 18 5 P63 009P4 001 – P5 000 P4 500.50 13 P4 500.5 3 13 5 P58 506.50P5 001 – P6 000 P5 500.50 10 P5 500.50 3 10 5 P55 005P6 001 – P7 000 P6 500.50 2 P6 500.50 3 2 5 P13 001P7 001 – P8 000 P7 500.50 2 P7 500.5 3 2 5 P15 001b) i) Modal class is P3 001 2 P4 000ii) Median class is P3 001 2 P4 000Total 60 P233 028.50iii) Mean P233 028.50 /60 5 P3 883.80. Mean class is P3 001 2 P4 000.3. a) Median value is P19 600b) Range of the distribution is P36 000 2 P11500 5 P24 500c) Lower quartile is P14 600.d) Upper quartile is the middle number in the upper half of this distribution.e) Inter-quartile range is P27 000 2 P14 600 2 P12 400.130 EXPLORING MATHEMATICS FORM 3

21c h a p t e rScatter graphsSyllabus: General objectiveThis chapter deals with the understanding and application of scatter graphs inpractical situations.Specific objectives• Use a spreadsheet to draw scatter graph.• Use a spreadsheet to draw a line of best fit on a scatter graph.AnswersExercise 21.1 (Student’s Book page 382)1.908070Length against mass of a certain plant’s leavesMass (g)60504030201000 2040 60 80 100Length (mm)Chapter 21: Scatter graphs131

2. Height against weight for 12 patients in a clinic66656463Weight (kg)62616059585756154 156 158 160 162 164 166Height (cm)3.200Mother’s height against son’s height150Son’s height (cm)1005000 50 100 150 200Mother’s height (cm)132 EXPLORING MATHEMATICS FORM 3

Exercise 21.2 (Student’s Book page 385)1.90807060Length against mass of a certain plant’s leavesMass (g)504030201000 2040 60 80 100Length (mm)2. Height against weight for 12 patients in a clinic66656463Weight (kg)62616059585756154 156 158 160 162 164 166Height (cm)Chapter 21: Scatter graphs133

3.180160140Mother’s height against son’s heightSon’s height (cm)12010060402000 50 100 150 200Mother’s height (cm)Revision Exercise (Student’s Book page 387)1. Hours worked against hours watching sports40Hours watching sports353025201510500 10 20 30 40Hours worked134 EXPLORING MATHEMATICS FORM 3

2. a) Water used against average growth8Average growth (cm)64200 1 2 3 4 5 6Water used (kl)b) There is a strong, positive correlation, i.e. more water used resulted in moregrowth.3. a) Goals scored against goals conceded by 12 teams10080Goals conceded60402000 20 40 60 80 100Goals scoredb) There is a weak, negative correlation.Chapter 21: Scatter graphs135

22C H A P T E RProbabilitySyllabus: General objectiveThis chapter deals with the understanding and application of probability tableson real-life situations.Specific objectives• Represent possible outcomes of two events involving real-life situations usingprobability tables/sample space diagrams.• Use probability tables/sample space diagrams to calculate the probabilities of twoevents involving real-life situations.AnswersActivity 22.1 (Student’s Book page 392)1. Green spinner 3 (1,3) (2,3) (3,3)2 (1,2) (2,2) (3,2)1 (1,1) (2,1) (3,1)1 2 3Red spinnerThere are 9 possible outcomes.2. Coin 1 H (H, H) (H, T)T (T, H) (T, T)H TCoin 2There are 4 possible outcomes.136EXPLORING MATHEMATICS FORM 3

Exercise 22.1 (Student’s Book page 392)1. Using SU for Scripture Union, TD for Traditional dance , MC for MathematicsClub, CH for Chess, GC for Guidance and Counselling, MU for Music, DM forDrama, AT for Art, and DB for Debating:SU (SU, GC) (SU, MU) (SU, DM) (SU, AT) (SU, DB)TD (TD, GC) (TD, MU) (TD, DM) (TD, AT) (TD, DB)Monday MC (MC, GC) (MC, MU) (MC, DM) (MC, AT) (MC, DB)CH (CH, GC) (CH, MU) (CH, DM) (CH, AT) (CH, DB)GC MU DM AT DBWednesday2. a) Taking TE for Tea, JU for Juice, WA for Water, ML for Milk, FD for Fizzy drink,BS for Biscuits, CK for Cake, and MP for meat pie:DrinkTE (TE, BS) (TE, CK) (TE, MP)JU (JU, BS) (JU, CK) (JU, MP)WA (WA, BS) (WA, CK) (WA, MP)ML (ML, BS) (ML, CK) (ML, MP)FD (FD, BS) (FD, CK) (FD, MP)BS CK MPb) There are 15 possible outcomes.3. a) A (A, R) (A, L) (A, D)Consonants E (E, R) (E, L) (E, D)I (I, R) (I, L) (I, D)R L DVowelsb) 9 choicesActivity 22.2 (Student’s Book page 394)ChickenBeefVegetarianchipssaladchipssaladchipssaladChapter 22: Probability137

The possible combinations are:Chicken and chipsChicken and saladBeef and chipsBeef and saladVegetarian and chipsVegetarian and saladActivity 22.3 (Student’s Book page 395)1. The first slice and the second slice eaten by the child.2. Because the same slice cannot be taken twice (once eaten it cannot be put backon the plate).3. There are 20 possible choices.4. 6 choices35. a) ___ ​ b) ​110 ___ ​ c) ​3__10 5 ​Activity 22.4 (Student’s Book page 397)1.1__2 ​​2__3 ​ G1​2__2 ​​2__2​1__​3 ​ R 0G2RG2R2. a) 0 b) 1__ ​ c)2__3 3 ​Exercise 22.2 (Student’s Book page 397)1. a)2. a)___ 1202__5​​b)2__ ​5c)1__5 ​b) ___ 212 ​138 EXPLORING MATHEMATICS FORM 3

3. a)0.50.5HTH0.5TH0.50.5T0.50.50.50.50.50.50.5HTHTHTHTb) i) 0.125 ii) 0.125 iii) 0.875Revision Exercise (Student’s Book page 399)1. a) G 5R 3G 5R 2G 5R 1G 5G 1G 5G 2G 5G 3G 5G 4G 5G 4R 3G 4R 2G 4R 1G 4G 1G 4G 2G 4G 3G 4G 5G 4G 3R 3G 3R 2G 3R 1G 3G 1G 3G 2G 3G 4G 3G 5G 3G 2R 3G 2R 2G 2R 1G 2G 1G 2G 3G 2G 4G 2G 5G 2G 1R 3G 1R 2G 1R 1G 1G 2G 1G 3G 1G 4G 1G 5G 1R 1R 3R 1R 2R 1G 1R 1G 2R 1G 3R 1G 4R 1G 5R 1R 2R 3R 2R 1R 2G 1R 2G 2R 2G 3R 2G 4R 2G 5R 2R 3R 2R 3R 1R 3G 1R 3G 2R 3G 3R 3G 4R 3G 5R 3R 3R 2R 1G 1G 2G 3G 4G 5b)4__7 ​​5__8 ​ G​3__7 ​​5__7​3__​8 ​ R2__7 ​GRGRc) 56d) i)___ 3​ ii) ​2__ ​ iii) ​1521 7 ___28 ​Chapter 22: Probability139

2. a) dicedisc1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 8b) i) ___ 712 ​ ii) ___ 7​ iii) ​3__12 4 ​3. a)3__5 ​W​4__6 ​ W​2__5 ​​4__5​2__​6 ​ B1__5 ​BBW140 EXPLORING MATHEMATICS FORM 3

23c h a p t e rResearch skills: Data analysisand interpretationSyllabus: General objectiveThis chapter deals with the application of knowledge, skills and processes learnt toidentify and solve real-life problems.Specific objectives• Gather data to support or refute the conjectures/hypothesis.• Analyse and interpret data.• Generalise results.• Evaluate the product.AnswersActivity 23.3 (Student’s Book page 406)1. Pie chart or bar chart2. Pie chart3. Bar chartActivity 23.4 (Student’s Book page 407)1. a) South Africa and Mauritius followed the world trend: slight increase from1995–2000, then remained constant. Tanzania and Botswana: sharp increase from 1995–2000, then remainedconstant. Outperformed the world trend. Uganda had a continuous increase from 1995 until 2002. Outperformed theworld trend.Zimbabwe and Kenya: industry deteriorated, below world trend.b) South Africa and Mauritius both have well-developed tourism industries thatare running almost at capacity, i.e. there is not much room for further growth.Botswana and Tanzania both have fast-growing tourism industries that are notat full capacity yet, so there’s still a lot of growth possible, and they utilise it.Uganda’s tourism industry is growing rapidly and they use all the means thatare available to develop it further.Chapter 23: Research skills: Data analysis and interpretation141

Zimbabwe and Kenya are losing tourists for various reasons, and students candiscuss the possible reasons why these two African countries are goingbackwards in their tourism industries while the rest of the African countriesstudied here are growing.2. a) Just below 8%; 8%.b) July & October highest; May lowest.c) South Africa has a higher visitor rate in the summer months, probably due toits long coastal line. Many tourists visit the South African beaches. Botswanahas a higher visitor rate in June-July, the winter months, when their climate isfar more moderate than South Africa’s.Exercise 23.1 (Student’s Book page 409)2. The infant mortality rate was still on the increase from 2000 to 2004(62–70/1 000), but since then there was a sharp decrease from about70 per 1000 to only 15 per thousand. It might be due to improved healthcare in the rural areas; better after-birth care; young mothers being educatedon caring for their infants; better community involvement in looking afterthe babies; decline in teenagers falling pregnant.3. More teenage mothers in rural areas than urban areas; an increase in teenagemotherhood from 1971 to 1988, and there after a significant decrease to 2006.There are half as many urban mothers as there are rural mothers. Reason: mightbe that the urban teenagers are more street-wise or that they are more careful bynot having unprotected sex, better protected by their social groups that theymove in, whereas the rural teenagers might be more exposed to the risks ofunhealthy relationships which might contribute to early pregnancies.142 EXPLORING MATHEMATICS FORM 3

Assessment 4Question 11.1 a) Two widths 5 2a and length 5 b; fourth side of pen is formed by the river. ü(1)b) b 5 100 2 2a ü A 5 a(100 2 2a) ü(2)c) a(100 2 2a) 5 800 ü 2a 2 2 100a 1 800 5 0 ü a 5 40 (not applicable) ü or a 5 10 ü So a 5 10 and b 5 80 ü(5)d) Dimensions: 80 metres long, 10 metres wide. üü(2)1.2 a) (x 2 5)(x 1 1) 5 0 ü x 5 5 ü or x 5 21 ü(3)b) x 5 5 metres is where the water jet hits the ground. üü(2)c) 5 m üü(2)d) 2 m üü(2)e) 9 m (substitute x 5 2 into y 5 2x 2 1 4x 1 y) ü(1) [20]Question 22.1 a) 26,7 kg üü(2)b) Q 15 30 kg üü; Q 35 22 kg üü(4)c) 3 rd : 14.5 kg üü; 10 th : 18 kg üü: 80 th : 31.5 kg üü(6)d) 75 lambs are below 37 kg. üü(2)e) 15 heaviest lambs: 32 2 40 kg üü(2)2.2 a)Classes Frequency (f) Class midpoints (x) f 3 x10 < x < 15 3 12.5 37.515 < x < 20 11 17.5 192.520 < x < 25 17 22.5 382.525 < x < 30 28 27.5 770.030 < x < 35 12 32.5 390.035 < x < 40 9 37.5 337.5üüf 5 n 5 80 üfx 5 2 110 ü(4)b) Model class: 25 < x < 30 üü(2)c) Median class: 25 < x < 30 üü(2)d) Estimated mean = ___ fxn ​2 110= ​ _____80 ​ ü= 26.375 ü(2)Chapter 23: Research skills: Data analysis and interpretation143

e) It is based on class mid-values, which is a rough approximation of theaverage for every class interval. It is the mean obtained for grouped dataand is not based on the raw data. (1)[27]Question 33.1(7)3.2 y = 26 + 0.62x üü(2)3.3 Strong ü (r = 0.81) ü(2)3.4 Language proficiency enhances Mathematical proficiency for obvious reasons:if students understand what they read, and they can interpret questions well,then they can also find a way to solve Mathematical problems. üü(2)[13]Question 34.1 düü4.2 büü4.3 aüü4.4 cüü4.5 büü[10]Question 45.1 a) Student could have drawn EITHER a probability table OR a tree diagramHeads (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6)Coin Tails (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)1 2 3 4 5 6Die üüü (3)144 EXPLORING MATHEMATICS FORM 3

OR üüü (3)b) ___ 212 ​ = ​ 1__6 ​ üü(2)5.2 a) Student could have drawn EITHER a probability table OR a tree diagram.The outcomes are:R1; R2; R3; R4; R5; R6B1; B2: B3; B4; B5; B6Y1; Y2: Y3; Y4; Y5; Y6 à 18 outcomes üüü (3)Y1 + Y3 + Y5b) ___________ ​ ü = 318___18 ​ = ​ 1__6 ​ ü(2)[10] Total: 80Chapter 23: Research skills: Data analysis and interpretation145

Sample Examination Paper 1 Answers: MathematicsMultiple-choice questions Marks: 40AnswersBBCBBACBCBACAABADBDCBCDCBBBBDACCCBBDBBBD146EXPLORING MATHEMATICS FORM 3

Sample Examination Paper 2 Answers: MathematicsSECTION AQuestion 11.1 4.8 üü(2)1.2 üü (Segments correct)Station Aü (Labels correct)45ºStation E120ºStation C90ºStation D30º Station B(3)75º[5]Question 22.1 21 ü(1)2.2 Triangle ü (1)[2]Question 33.1 90 ü(1)3.2 (Line of best fit) üNumber of accidents100908070605040302010’00 ’01 ’02 ’03 ’04 ’05 ’06 ’07 ’08 ’09[2]YearsSample Examination Paper147

4.1 10.5 cm üü(2)4.2 DE ü (1)4.3 44º ü(1) [4]Question 5rP(n + 1)5.1 c = _______ ​ü (1)2n5.2 P32.50 üü(2) [3]Question 66.1 6d 2 + 22d + 12 ü(1)6.2 2 ü(1) [2]Question 77.1 South Africa ü(1)7.2 P817.22 (to two d.p.) ü (1)7.3 P837.65 ü(1) [3]Question 8y = x + 1 ü(1)[1]Question 99.1 P22 000 ü(1)9.2 P7 000 ü(1)[2]Question 1020.8 km/h üü(2)[2]148 EXPLORING MATHEMATICS FORM 3

Question 1111.1 Figure L is the reflection of K under the line m. ü (1)KmLH11.2 ​( 7 1 )​ ü(1)[2]Question 1212.1 (60º angle correct) ü(4)12.2 (Bisection correct) ü(1) 60°[2][30]ACSample Examination Paper149

Question 13SECTION B13.1 üüü (3)13.2 384 750 m 2 üü(4)13.3 3 415.9 m üü(3)13.4 86 rolls üü(2) [12]Question 1414.1 110 ü(1)14.2 1976, 1977 ü(1)14.3 45 cases in 1977 üü(2)14.4 42.6 üüü(3) [7]Question 1515.1 10 m/s üü(1)15.2 2 seconds ü (1)15.3 36 km/h üü(1) [3]Question 1616.1 448 cm 2 üü(2)16.2 480 cm 3 üü (2)16.3 Surface area is halved, so 224 cm 2 ü; volume divided by 4 = 120 cm 3 ü(2)[6]Question 1717.137°37° üü (2)17.2 4.33 m üüü(3)[5]150 EXPLORING MATHEMATICS FORM 3

18.1 x 22 21 0 1 2x 2 4 1ü 0ü 1 4x 2 2 2 2 21ü 22ü 21ü 2 (5)18.2 (Coordinates and parabola correct) üü (2)18.3 (0;22) üü (2)18.4 (2;2) ü (22;2) ü (2)Question 19The church looks like this in 3-D.[11]Sample Examination Paper151

Plan of the church:Question 1919.1 T 5 100 ü (1)P 5 30 units ü (1)19.2 12 units ü (1)19.3 T 5 n 2 ü (1)P 5 3n ü (1)19.4 n 5 ​P__3 ​when T 5 9 ü n 5 3 ü (2)Question 2020.1 y 5 ​x__5 ​ ü (1)20.2SECTION C[4][50][7]425 10 15222426üü (2)[3]152 EXPLORING MATHEMATICS FORM 3

Question 2121.1 ü (1)21.2 132 rpm üü (2)Question 2222.1[3]üü (2)22.2üü (2)[4]Question 2361 2534üü (3)[4]Sample Examination Paper153