Preface. Enhanced Learning

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Rafe Bell
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1 Preface This book is just what you are looking for Secondary 2 Mathematics made easy and comprehensible so you need not struggle to make sense of all the new and unfamiliar concepts. Specially written to help you ease into secondary level mathematics, this book will be a great companion to all textbooks. So what can you expect to find in this book? Important Formulae and Notes These are included to help you recap essential concepts that you must know for each topic. These summarise what you learn in school and will save you precious time to get started on the practice questions. Worked Examples These show detailed and specific steps to solving a problem. Learning is also reinforced through similar question types following the examples. Useful Tips These are the extra information that can help you better understand the concepts and problem-solving skills so that you can work out the solutions faster. Review Exercises These consolidate what you have learnt in each topic and give a comprehensive overview of the concepts covered. Fully Worked Solutions These are the step-by-step, worked solutions to all the questions in the book. Not only do they serve as a quick reference for checking of your answers, they also provide learning for those questions that you were not able to solve. Finally, practise, practise, practise! Work through these questions and work towards achieving distinction in Mathematics. Enhanced Learning resources available Additional practice questions can be found online. They are carefully chosen questions to help develop higher order thinking processes and skills. The online Mid-Year Examination Specimen Papers cover the first 6 topics so you can assess your learning progress. The online Final-Year Examination Specimen Papers cover all 11 topics and you can assess your overall understanding of the school year s work. The Editorial Team

2 Contents Chapter 1 Congruence and Similarity...1 Chapter 2 Direct and Inverse Proportion...13 Chapter 3 Expansion and Factorisation of Algebraic Expressions...27 Chapter 4 Algebraic Manipulation and Formulae...45 Chapter 5 Simultaneous Linear Equations...59 Chapter 6 Trigonometry and Pythagoras Theorem...71 Chapter 7 Volume and Surface Area of Pyramids, Cones and Spheres...83 Chapter 8 Graphs of Linear Equations in Two Unknowns...93 Chapter 9 Graphs of Quadratic Equations Chapter 10 Statistical Diagrams and Averages of Statistical Data Chapter 11 Probability Mid-Year Examination Specimen Paper 1 and Paper 2 Final-Year Examination Specimen Paper 1 and Paper 2 Refer Specimen Papers for assessment of your learning progress. Download from Solutions...S1 S41

3 Chapter 1 Congruence and Similarity 1.1 Congruent Figures and Objects Two figures or objects are congruent if they have exactly the same shape and size. They may not be identical; they can be different in colour or texture. The symbol means is congruent to. Example 1 Given that ABCD KLMN, copy and complete the following: A L M D B N C (a) AB = (e) ABC = (b) BC = (f) BCD = (c) CD = (g) CDA = (d) AD = (h) BAD = K Since ABCD is congruent to KLMN, then (a) AB = KL (e) ABC = KLM (b) BC = LM (f) BCD = LMN (c) CD = MN (g) CDA = MNK (d) AD = KN (h) BAD = LKN Example 2 Given that PQR XYZ, write down the missing measurements. P Y 4 cm X Q 3 cm 53º R Z 5 cm 1

4 Since PQR is congruent to XYZ, then PQ = XY = 4 cm P = X = 180º 90º 53º = 37º QR = YZ = 3 cm Q = Y = 90º PR = XZ = 5 cm R = Z = 53º Example 3 In the diagram, two parallelograms, EFGH and ELKJ, are congruent. Given that KL = 6 cm, EF = 8 cm and EJK = 70º, find (a) the length of JK, (b) the length of HL, (c) HEF. J E F K H G (a) Since EFGH is a parallelogram, HG = EF = 8 cm Since EFGH ELKJ, JK = HG = 8 cm (b) Since EFGH ELKJ, Since EFGH is a parallelogram, Since ELKJ is a parallelogram, L GF = KL = 6 cm EL = JK = 8 cm HE = GF = 6 cm HL = EL HE = 8 6 = 2 cm (c) JEL = 180º EJK (int. s, JK // EL) = 180º 70º = 110º Since EFGH ELKJ, HEF = JEL = 110º Example 4 Given that RST UVW such that UV = 4 cm, VW = 7 cm, RT = 6 cm, V = 59º and T = 35º. Find (a) U, (b) the length of RS, (c) the length of UW. 2

5 (a) Since RST UVW, W = T = 35º U = 180º V W ( s sum of ) = 180º 59º 35º = 86º (b) RS = UV = 4 cm (c) UW = RT = 6 cm 1.2 Similar Figures and Objects Two figures or objects are similar if they have exactly the same shape but not necessarily the same size. When two figures are similar, then 1. all the corresponding angles are equal, 2. all the corresponding sides are proportional. All congruent figures or objects are similar but not all similar figures or objects are congruent. Example 5 In the diagram, pentagons ABCDE and PQRST are similar. Copy and complete the following: A P B E Q T R S C D (a) PQ AB = BC = RS = DE = PT (b) A = (d) C = (f) E = (c) B = (e) D = Since ABCDE is similar to PQRST, 1. all the ratio of the corresponding sides are equal: (a) PQ AB = BC QR = CD RS = DE ST = AE PT 2. all the corresponding angles are equal: (b) A = P (c) B = Q (d) C = R (e) D = S (f) E = T 3

6 Example 6 In the diagram, EFG is similar to XYZ. Write down the missing measurements. E 30º 6 cm F 14 cm Z X 7 cm 131º Y 4.7 cm G Since EFG and XYZ are similar, X = E = 30º F = Y = 131º G = Z = 180º 131º 30º ( s sum of ) = 19º EF XY = EG XZ FG YZ = XZ EG XY 6 = 7 14 FG 4.7 = 14 7 XY = FG = = 3 cm = 9.4 cm Example 7 Given that two cylindrical cans, A and B, are similar such that the ratio of the diameter of can A to the diameter of can B is 5 : 9. If the height of can A is 12.5 cm, calculate the height of can B. Height of can B = Diameter of can B Height of can A Diameter of can A Height of can B 12.5 = 9 5 Height of can B = = 22.5 cm 4

7 1.3 Similarity and Enlargement Under an enlargement, a figure and its image are similar. An enlargement with: (i) a scale factor greater than 1 produces an enlarged image (ii) a scale factor between 0 1 produces a diminished image (iii) a scale factor of 1 produces a congruent image Example 8 In the diagram, KLM is mapped onto PQR by an enlargement with centre O and scale factor 3. If KL = 1.5 cm and PR = 6.9 cm, find (a) the length of KM, (b) the length of PQ. P R Q M K L O (a) Since KLM is similar to PQR under the enlargement, KM PR = 1 (KM < PR, ratio < 1) scale factor KM 6.9 = 1 3 KM = = 2.3 cm (b) PQ KL = scale factor (PQ > KL, ratio > 1) PQ 1.5 = 3 PQ = = 4.5 cm 5

8 Review Exercise 1 1. (a) = (b) = (a) HCF of 30 and 45 = 3 5 = LCM of 30 and 45 = = 90 (b) HCF of 15, 20 and 28 = LCM of 15, 20 and 28 = = (a) = = (b) (i) HCF of 90 and 132 = 2 3 = 6 LCM of 90 and 132 = = 1980 (ii) 132 = ( ) = (a) (5 10) (b) (3 10) (c) (d) (e) (7 100) (f) (2 10) (g) (h) (a) 2377 (b) 1 (c) 32 (d) Base area = volume height n Least value of n = = 2178 = = 256 m 2 Length of the box = base area = 256 = = 16 m The length of the box is 16 m. 7. (a) Area of the graph paper = = mm 2 Area of each square box = = 400 mm 2 Length of the side of each square box = 400 = = 20 mm The length of the side of each square box is 20 mm. BE GREA 1 T! MATHEMATICS Secondary 1 S Review Exercise 1

CHAPTER 5 : THE STRAIGHT LINE

EXERCISE 1 CHAPTER 5 : THE STRAIGHT LINE 1. In the diagram, PQ is a straight line. P 4 2 4 3 2 1 0 1 2 2 2. Find (a) the -intercept, (b) the gradient, of the straight line. Q (5,18) Q Answer :a).. b) 3