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Hassanein, Ismaiel and Mohd Nor, Shamsudin (2015) The Importance of learning Arabic Vowel systems among non-native speakers in determining the Lexical meaning of Arabic words. Jawab ď † = BA = 0,8 x 2 x 10-4 weber = 1,6 x 10-4 weber = 0,16 mili weber Contoh 2.3 Fluksi magnet 0,06 weber mengalir melewati cela udara suatu motor listrik.
OXFORD FAJAR ADVISORY BOARDThe board consists of a team of experienced teacherswho review our titles to ensure that the contents arein line with the current syllabus and examinationrequirements as set by the Examination Syndicate,Ministry of Education Malaysia.Success Mathematics SPM e-book was reviewed by• Elaine Yek• Chong Geok Chuan• Rosaimi Abd. Wahab• Ng Kai Che• Long C. Y.*****************************************Oxford Fajar Sdn. Bhd. (008974-T)(Formerly known as Penerbit Fajar Bakti Sdn. Bhd.)4 Jalan Pemaju U1/15, Seksyen U1Hicom-Glenmarie Industrial Park40150 Shah AlamSelangor Darul Ehsan© Oxford Fajar Sdn. Bhd. (008974-T) 2013First published 2013ISBN 978 983 47075 8 3All rights reserved.No part of this publication may be reproduced,stored in a retrieval system, or transmitted in any form orby any means, electronic, mechanical, photocopying,recording or otherwise, without the prior permission ofOxford Fajar Sdn. Bhd. (008974-T)Text set in 10 point ITC Giovanni byUG Graphic Services Sdn. Bhd., Kuala LumpurWord Power ii
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ContentsSPM Analysis Table (2003–2011) v CHAPTER 6 91Aim for Success vi 92Zoom-in (Forms 1–3) P1 Statistics III 98 6.1 Class Intervals 101 FORM 4 1 6.2 Mode and Mean of Grouped Data 105 2 6.3 Histograms 110CHAPTER 1 4 6.4 Frequency Polygons 112 9 6.5 Cumulative Frequencies 120Standard Form 6.6 Measures of Dispersion 1.1 Significant Figure SPM Exam Practice 6 1241.2 Standard Form 125SPM Exam Practice 1 CHAPTER 7 126 129CHAPTER 2 12 Probability I 134 13 7.1 Sample Space Quadratic Expressions and Equations 14 7.2 Events 1362.1 Quadratic Expressions 18 7.3 Probability of an Event 1372.2 Factorisation of Quadratic Expressions 20 SPM Exam Practice 7 1402.3 Quadratic Equations 25 1442.4 Roots of Quadratic Equations CHAPTER 8 148SPM Exam Practice 2 Circles III 151CHAPTER 3 8.1 Tangents to a Circle 8.2 Angles between Tangents and Chords 152Sets 28 8.3 Common Tangents 163 SPM Exam Practice 8 1673.1 Sets 29 CHAPTER 9 1723.2 Subsets, Universal Sets and Complement 173 Trigonometry II 176of a Set 32 9.1 Values of Sine, Cosine and Tangent 1803.3 Operations on Sets 35 of an Angle 183 9.2 Graphs of Sine, Cosine and Tangent 184SPM Exam Practice 3 47 SPM Exam Practice 9 192 198CHAPTER 4 CHAPTER 10Mathematical Reasoning 53 Angles of Elevation and Depression 10.1 Angles of Elevation and Depression 4.1 Statements 54 10.2 Solving Problems SPM Exam Practice 10 4.2 Quantifiers ‘All’ and ‘Some’ 55 CHAPTER 114.3 Operations on Statements 57 Lines and Planes in 3-Dimensions 4.4 Implications 61 11.1 Angle between a Line and a Plane 11.2 Angle between Two Planes 4.5 Arguments 63 SPM Exam Practice 11 4.6 Deduction and Induction 66SPM Exam Practice 4 68CHAPTER 5The Straight Line 725.1 Gradient of a Straight Line 735.2 Gradient of a Straight Line in CartesianCoordinates 745.3 Intercepts 775.4 Equation of a Straight Line 795.5 Parallel Lines 83SPM Exam Practice 5 86 iii
Persamaan Transistor Ba 388 3
FORM 5 202 CHAPTER 6 310 203CHAPTER 1 213 Gradient and Area Under a Graph 311 6.1 Quantity Represented by the GradientNumber Bases 3191.1 Numbers in Base Two, Eight and Five of a Graph 328SPM Exam Practice 1 6.2 Quantity Represented by the Area UnderCHAPTER 2 a Graph SPM Exam Practice 6 Graphs of Functions II 2.1 Graphs of Functions 215 CHAPTER 72.2 Solution of Equations by the 216 Probability II 331 Graphical Method 2252.3 Regions Representing Inequalities 7.1 Probability of an Event 332 231 in Two Variables 236 7.2 Probability of the Complement of an Event 336SPM Exam Practice 2 7.3 Probability of a Combined Event 339 7.4 Solving Problems using Tree Diagrams 345 SPM Exam Practice 7 350CHAPTER 3 CHAPTER 8Transformations III 240 Bearing 3553.1 Combination of Two Transformations 241 8.1 Bearing 3563.2 Solving Problems involving SPM Exam Practice 8 363Transformations 258SPM Exam Practice 3 261 CHAPTER 9CHAPTER 4 Earth as a Sphere 366Matrices 265 9.1 Longitudes 3674.1 Matrices 266 9.2 Latitudes 3704.2 Equal Matrices 268 9.3 Location of a Place 373 9.4 Distance between Two Points Along a4.3 Addition and Subtraction of Matrices 270 Great Circle 3764.4 Multiplication of a Matrix by a Number 273 9.5 Distance between Two Points Along a4.5 Multiplication of Two Matrices 277 Parallel of Latitude 3814.6 Identity Matrices 282 9.6 Shortest Distance between Two Points 3834.7 Inverse Matrices 284 9.7 Solving Problems 3854.8 Solving Simultaneous Linear Equations SPM Exam Practice 9 388using Matrices 286SPM Exam Practice 4 289 CHAPTER 10CHAPTER 5 Plans and Elevations 393 10.1 Orthogonal Projections 394Variations 292 10.2 Plans and Elevations 400 SPM Exam Practice 10 4145.1 Direct Variation 2935.2 Inverse Variation 297 SPM Model Test 4215.3 Joint Variation 301SPM Exam Practice 5 306 Answers 431 Word Power 468 iv
SPM Analysis Table (2003–2011) Topics 2003 2004 2005 2006 2007 2008 2009 2010 2011 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2FORMS 1 – 3 1 Linear Equations 1–1–1–1–1–1–1–1–1– 2 Simultaneous Linear Equations –1–1–1–1–1–1–1–1–1 3 Transformations I & II 3–2–2–2–2–2–2–2–2– 4 Polygons 2–1–2–1–2–1–2–2–1– 5 Statistics I & II 2–2–1–2–2–2–2–3–2– 6 Arcs and Sectors –1–1–1–1–1–1–1–1–1 7 Indices 2–2–2–1–2–2–2–2–2– 8 Algebraic Expressions 1–1–1–1–1–2–2–1–1– 9 Algebraic Formulae 1–1–1–1–1–1–1–1–1–10 Linear Inequalities 2–2–1–1–1–1–2–2–1–11 Volumes and Surface Areas –1–1–1–1–1–1–1–1–1FORM 4 1 Standard Form 4–4–3–4–3–4–4–4–4– 2 Quadratic Expressions and Equations 1 1 1 1 1 1 1 1 1 1 – 1 – 1 1 1 1 1 3 Sets 3–313–313–312–2131 4 Mathematical Reasoning –1–1–1–1–1–1–1–1–1 5 The Straight Line 112121212121212121 6 Statistics III –1–11111–1–111–1–1 7 Probability I 2–2–2–1–1–1–1–1–1– 8 Circles III 1–1–1–1– 1–1–1–1–1– 9 Trigonometry II 2–3–3–3–3–3–2–2–3–10 Angles of Elevation and Depression 1 – 2 – 1 – 2 – 1 – 2 – 2 – 2 – 2 –11 Lines and Planes in 3-Dimensions 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1FORM 5 1 Number Bases 2–2–2–2–2–2–2–2–2– 2 Graphs of Functions II 121112111211121111 3 Transformations III –1–1–1–1–1–1–1–1–1 4 Matrices 311121212121212121 5 Variations 2–2–3–3–3–3–2–2–3– 6 Gradient and Area Under a Graph –1–1–1–1–1–1–1–1–1 7 Probability II –111–1111111111111 8 Bearing 1–1–1–1–1–1–1–1–1– 9 Earth as a Sphere 11112111211111111110 Plans and Elevations –1–1–1–1–1–1–1–1–1TOTAL QUESTIONS 40 16 40 16 40 16 40 16 40 16 40 16 40 16 40 16 40 16 v
WTMds(mahapeknashsiliisetnidlahglaasisnesPmectemceahautdnpreiiameosedcttfnrkioeccol2Arsbiofsratenqqtiombgumuctiaopenieeslpassisigacttgtfwaiirseooosiuissknnttgerheivssel.ssiliSiisttmStinuh–ertupxdeduethxioqecdriienruenicct)tniaS.osrtnetIPemhswttdMsewcptittisaeSphioEnlt?sPleatxeanMfhnaoxlcmsespryowMomianieusnnaraoessttetsSwsihhtotPteeiuhnpsMrmdii.ossneIuapPtgnstbuaeitisjspccleatsaetcirolrotps.n1otahpeer Overview of SPM How to use it? 2004 — 2011 Glance through Aim for Success section. Get an overview of the importance of the Forms 1–3, Form questions trends 4 and Form 5 topics in the SPM examination. Paper 1 35% 25% 40% Forms 1–3 Form 4 Form 5Read and practice the Read and practice the Paper 2questions in the Form 4 questions in the Form 5section of the book. section of the book.No No 28.5% 12.5% 59%Answer the questions in the Aim for Success section.Do you manage to answer all the questions correctly?(You may refer to our Companion Website for the fullyworked solutions to the questions.) Yes Form 4 Form 5We wish you success in your SPM examination. Forms 1–3 Practice Makes Perfect.SPM Mathematics Exam Format Time Number of Questions 1 hour 15 minutes Paper 40 multiple-choice questions (Answer all questions) 2 hours 30 minutesPaper 1 (40 marks) Section A: 11 questionsObjective Test (Answer all questions)Paper 2 (100 marks) Section B: 5 questionsSubjective Test (Answer 4 questions) vi
SPM Mathematics Exam TipsA Answer according to the instructions given in the C Do not waste time struggling questions. If not, you will lose marks unnecessarily. with the question(s) that you find difficult. Instead, you should Example: attempt other easier questions first. After you have finished with the rest, you will regain yourA1 Matrices: confidence. Then, only you retry the question(s) thatUsing the matrix method, find the values of x and y you skipped just now.which satisfy the simultaneous linear equations. [Advice:Do not solve the simultaneous linear equations using the D What is the implication of each of the followingsubstitution or elimination method. You should use the instructions? inverse matrix method, i.e. AX = Y ⇒ X = A–1Y.] A2 Lengths of Arcs and Areas of Sectors: Instruction Implication Using π = 22 , calculate the perimeter and Calculate, Obtain the answer by showing 7 the area of the shaded region. [Advice: You find, determine proper steps of working. should not use π = 3.142 nor the value of π Write down, Write down the answer straight from the scientific calculator.] state away without having to show the steps of working. A3 Volumes of Solids: Calculate the volume of the solid, giving your answer Express Find an unknown in terms ofcorrect to two decimal places. [Advice: You should not other unknown(s).state your final answer in less than or more than two decimalplaces.] Solve Find the root(s) of a given equation.A4 Probability:List the sample space and list all the outcomes of the Prove, show The answer is given. Show theevent and hence, calculate the probability that ... proper steps of working to obtain the given answer.[Advice: You should not solve the probability problem usingthe reasoning method. You should use the formula Estimate A certain range of answers is acceptable.P(A) = n(A) .] n(S) A5 Plans and Elevations: Draw Prepare a table of values, plot You are not allowed to use a graph paper to answer (a graph) the points on a graph paperthis question. [Advice: You should not draw the plans using a suitable scale and thenand elevations on a piece of graph paper. You should draw join the points to form a straightthe plans and elevations on the empty space provided in the line or a smooth curve.answer space.] Sketch A graph paper is not required. (a graph) Marks are given based on theB Remember to bring a scientific calculator, a shape of the graph and the transparent long ruler (for drawing histograms, position of the graph with plans and elevations), a flexible curve rule (for respect to the axes. drawing an ogive), a pair of compasses (for drawing a sphere to represent the Earth in the topic Earth as Draw to full Draw the plan and elevations a Sphere) and a set-square (for drawing plans and scale of a solid according to the exact elevations) into the examination hall. measurements. vii
FORMS 1–3 Centre of enlargement Scale factor A (4, –1) —21 1 In the diagram, PQRST is a regular 2P1 pentagon. USW, TRW and ZTS are straight B (3, 0) —23 C (3, –3) —32 lines. D (3, –3) US y° W 50° R T x° Q 4 Given that 4(p – 1) = 2p – 2, find the value ofZP p. 3 P1Find the value of x + y. A 1 C 3 5 5A 68° C 98°B 86° D 122° B 2 D 4 5 5 2 The diagram shows five points plotted onP1 a Cartesian plane. 5 Express 5 – h–2 as a single fraction in its 6e 2eh y P1 simplest form. 4 BA A 2h + 1 C h–3 3eh 3eh 2 Q 2h – 1 h+3 B 3eh D 3eh–4 –2 O x 6 Given that h + 3 = t, express h in terms of t. 24 P1 A h = 4t2 – 2 C h = 2t2 – 3 D C 3 –2Which of the points, A, B, C or D, is B h = 4t2 + 3 D h = (2t – 3)2the image of Q under an anticlockwiserotation of 90° about the centre (1, 1)? 7 Simplify (3m2n–1)2 ÷ (m 3n2)3. P1 A 3m–5n–8 C 3m5n8 3 The diagram shows two triangles, drawn B 9m–5n–8 D 9m5n8P1 on a Cartesian plane. Triangle Q is the 1 2 8 2 –3 image of triangle P under an enlargement. P1 5 Find the centre and the scale factor of the = enlargement. 1 2A 2 —32 y 5 1 2B 2 –—23 5 6 4P 1 2C 5 —32 2Q 2 1 2D 5 –—32 2–2 0 x 9 List all the integers x which satisfy the –2 2 468 –4 P1 inequalities 3x – 2 ≤ x ≤ 4 + 5x. A 0, 1 C –1, 0, 1, 2 B –1, 0, 1 D –2, –1, 0, 1, 2 COMPANION WEBSITE Fully Worked Solutions viii
10 The pie chart shows the distribution of the (a) the perimeter, in cm, for the wholeP1 different grades of eggs sold by a shop. diagram, (b) area, in cm2, of the shaded region. Grade C Grade D Answers 2. A 3. C 4. C 100° 1. B 6. A 5. D 7. B 8. A 60° 9. B 10. A 11. 610 2 cm3 Grade B Grade A 7 12. (a) 57 11 cm (b) 122 10 cm2 21 21Given that the sum of the number of the gradeA and grade C eggs sold is 240, calculate the FORM 4 Chapter 1sum of the number of the grade B and grade Deggs sold. Standard FormA 300 C 800 1 Evaluate 10.32 – 14.62 ÷ 0.2 and roundB 540 D 1440 P1 off the answer correct to one decimal 11 The diagram shows a composite solid formed place.P2 by joining a right pyramid with a half-cylinder A –62.78 C –21.5 at the rectangular plane DEFG. It is given that DE = 6 cm, EF = 16 cm and the height of the B –62.8 D 21.5 pyramid is 12 cm. 2 Express 0.000000049 in standard form. V P1 A 4.9 × 107 C 4.9 × 10–7 B 4.9 × 108 D 4.9 × 10–8 3 2.7 × 10–7 – 9.1 × 10–8 = P1 A 6.4 × 10–8 C 1.79 × 10–8 B 6.4 × 10–7 D 1.79 × 10–7 DG EF 4 Given that 1 MB = 1.02 × 106 bytes, the 22 P1 capacity of a computer thumb drive with a 7Using π = , calculate the volume, in cm3, of storage capacity of 512 MB, in bytes, isthe composite solid. A 5.13 × 105 C 5.22 × 107 D 5.22 × 108 12 The diagram shows two sectors of two circles Answers 2. D 3. D 4. DP2 ORQ and OPS, with centre O. 1. B R 12 cm FORM 4 Chapter 2 150° O 7 cm Quadratic Expressions and PQ Equations S 1 (m – n)2 + (m + n)(n – m) = P1 A 2n2 – 2mn C 2n2 – 2m B 2m2 – 2mn D 2m2 – 2nUsing π = 22 , calculate 2 Solve the following quadratic equation by 7 P2 factorisation:P1 Paper 1 P2 Paper 2 (y – 4)2 = 2y – 5 ix
3 Solve the following quadratic equation by Given that n(ξ) = 280, n(G) = 210, n(H) = 70, n(G H) = 30, calculate theP2 factorisation: number of students who do not like Geography nor History. m2 + 1 = 5m A 10 2 3 B 20 C 30Answers 2. 3 or 7 3. 1 or 3 D 40 1. A 3 4 The Venn diagrams show the sets A, B and C FORM 4 Chapter 3 P2 such that the universal set,Sets 1 The diagram shows a Venn diagram with ξ = A B C. On the diagram, shadeP1 the universal set, ξ = P Q. (a) the set A B, Q B C •g P A •f (b) the set (A B) C’. List all the subsets of set Q. B A {f }, {g} C B { }, {f }, {g} C {f }, {g}, {f, g} A D { }, {f }, {g}, {f, g} Answers 2. D 3. C 2 The diagram shows a Venn diagram with 1. DP1 the universal set, ξ = X Y Z. 4. (a) B C XY A A BC ZD (b) B C Which region, A, B, C or D, represents the A set (X Y)’ Z? 3 The diagram shows a Venn diagram whichP1 represents the number of students in the sets G and H. It is given that set G = {students who like Geography} and set H = {students who like History}. ξG H COMPANION WEBSITE Fully Worked Solutions x
FORM 4 Chapter 4 A – 53 C 8 5Mathematical Reasoning B 3 D 8 1 (a) State whether the following sentence is a 5 3P2 statement or not. Explain why. 2 The gradient of a straight line is 2 . If the “All prime numbers are odd numbers.” P1 y-intercept of the straight line is 3 (b) Write down the converse of the following –6, find implication and state whether the its x-intercept. converse is true or false. “If k is a multiple of 10, then k is a A –9 C 4 multiple of 5.” (c) Make a general conclusion by induction B –4 D 9 for the number sequence 11, 26, 51, 86, … which follows the following pattern: 3 The diagram shows a trapezium PQRS 11 = 5(1)2 + 6 P2 drawn on a Cartesian plane such that PS 26 = 5(2)2 + 6 51 = 5(3)2 + 6 is parallel to QR. The point Q and S lie on 86 = 5(4)2 + 6 the x-axis. PQ is parallel to the y-axis. 2 (a) State whether the following statement is yP2 true or false. P (–3, 4) “1 is a prime number and 4 is a factor of QO S 20.” x (b) Write down two implications based on R (6, –3) the following statement. Find “x > y if and only if x – 2 > y – 2.” (a) the equation of the straight line PS, (c) Complete Premise 2 in the following (b) the x-intercept of the straight line PS. argument. Answers Premise 1: 0007If h is an even number, then 1. A = 2. D 3 h + 1 is an odd number. 3. (a) y – 31 x + Premise 2: ___________________________ Conclusion: h is not an even number. (b) 9 Answers FORM 4 Chapter 6 1. (a) Statement, because it is false. Statistics III (b) If k is a multiple of 5, then k is a multiple of 10. 1 The data in the diagram show the marks P2 obtained by 45 students in a Science pre- False (c) 5n2 + 6, where n = 1, 2, 3, 4, … test. 2. (a) False (b) If x > y, then x – 2 > y – 2. 36 12 15 22 18 If x – 2 > y – 2, then x > y. 39 9 20 28 22 (c) h + 1 is not an odd number. 33 36 36 34 43 22 10 23 5 10 FORM 4 Chapter 5 18 8 33 43 11 20 40 30 14 10The Straight Line 15 23 28 24 18 28 28 9 36 22 1 The gradient of the straight line 18 17 12 18 28P1 3x + 5y = 8 is xi
(a) Based on the given data, complete the (b) State the size of the uniform class interval following table. used in the table in (a). Marks Midpoint Frequency (c) Using a scale of 2 cm to 10 marks on the 5–9 74 horizontal axis and 2 cm to 5 students10 – 14 on the vertical axis, draw an ogive to15 – 19 represent the given data.20 – 2425 – 29 (d) From your ogive, answer the following30 – 34 questions.35 – 3940 – 44 (i) Find the third quartile. (ii) Hence, explain briefly the meaning of the third quartile.(b) State the size of the uniform class Answers interval used in the table in (a). 1. (a)(c) Calculate the mean mark. Marks Midpoint Frequency(d) Using a scale of 2 cm to 5 marks on 5–9 7 4 10 – 14 12 7 the x-axis and 2 cm to 1 student on 15 – 19 17 8 the y-axis, draw a frequency polygon 20 – 24 22 9 for the data. 25 – 29 27 5(e) Based on your frequency polygon in 30 – 34 32 4 (d), state the modal class. 35 – 39 37 5 40 – 44 42 3 2 The following data show the marks of 40P2 students in a History test. 61 53 70 49 30 (b) 5 75 92 64 82 40 (c) 22.67 37 56 81 58 65 (d) 74 36 58 66 78 48 78 83 57 60 Frequency 41 66 63 73 95 84 67 86 53 46 9 79 65 37 69 53 8 7(a) Based on the given data, complete the 6 following table. 5 4 Marks Upper Frequency 3 boundary 0 220 – 29 130 – 39 29.5 0240 – 4950 – 59 7 12 17 22 27 32 37 42 4760 – 69 Marks70 – 7980 – 89 (e) 20 – 2490 – 99 COMPANION WEBSITE Fully Worked Solutions xii
2. (a) Upper Frequency Find the value of x. boundary A 157 Marks 0 B 113 29.5 4 C 67 20 – 29 39.5 5 D 23 30 – 39 49.5 7 40 – 49 59.5 10 2 In the diagram, APB is a tangent to the 50 – 59 69.5 7 P1 circle at P. 60 – 69 79.5 5 70 – 79 89.5 2 R QA 80 – 89 99.5 90 – 99 112° 35°(b) 10 P(c) Cumulative y° frequency S Find the value of y. B A 33 40 B 35 35 C 55 D 68 30 Answers 2. A 25 1. D 20 FORM 4 Chapter 9 15 Trigonometry II 10 5 1 In the diagram, RQT is a straight line. 0 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5 Marks P1 75 R(d) (i) 75 (ii) 43 (i.e. 30) of the students have P 7 cm marks of 75 and below.FORM 4 Chapter 8Circles III 1 In the diagram, LSK is a tangent to the circle 6 cm Q 8 cmP1 with centre O, at S. x° K TS T Given that sin x° = 4 , find the length, x° 5 P 67° S in cm, of TS. O A 18 B 15 Q C 12 L D 10 xiii
90° 180° –1 2 In the diagram, QRST is a parallelogram DyP1 and PQR is a straight line. 1 TS O 90° 180° –1 x P QRGiven that sin ∠PQT = 12 , find the value 4 Given that cos θ = –0.4067 such thatof cos ∠TSR. 13 P1 0° ≤ θ ≤ 360°, find the values of θ.A 5 A 114° and 246° 13 B 114° and 294° C 156° and 246°B 5 D 246° and 294° 12C – 153D – 12 Answers 2. C 3. B 4. A 13 1. B 3 Which of the following graphs represents FORM 4 Chapter 10P1 y = cos x for 0° ≤ x ≤ 180°? Angles of Elevation and Depression 1 The diagram shows the vertical side view of a A y P1 chair on a horizontal floor. 1 S Ox TR 90° 180° –1By 180° x 1 O 90° –1C 90° 180° x PQ y Name the angle of elevation of point S from 1 point T. O A ∠RTS –1 B ∠RST C ∠QTS y D ∠QPS 1 COMPANION WEBSITE Fully Worked Solutions xiv
2 The diagram shows two vertical poles, AB and 2 The diagram shows a right prism with aP1 CD, on a horizontal plane. P1 horizontal rectangular base ABCD. D H E G F B D C A 30 m B 18 m C What is the angle between the plane A AHGB and the plane ABFE?Given that the angle of depression of point A HAD B HAEB from point D is 52°, calculate the angle of C GBF D GBCelevation of point D from point A.A 51°57’ 3 The diagram shows a right prism with aB 59°2’ P2 horizontal square base PQSR. TrapeziumC 70°45’D 71°15’ HPQL is the uniform cross-section of the prism.Answers K 5 cm J 1. A 2. D FORM 4 Chapter 11 H L RLines and Planes in 3-Dimensions T S 1 The diagram shows a pyramid with a 12 cmP1 horizontal rectangular base JKLM. Vertex N is P 10 cm Q vertically above point M. Given that PT = 3 PH, name and calculate N 4 the angle between the plane TQR and the JM base PQRS. KL Answers 1. AThe angle between the line KN and the plane 2. B JMN is 3. ∠TQP = 41°59’A ∠KNJB ∠KNMC ∠KNLD ∠NLM xv
Smd Transistor Ba
7 x O FORM 5 Chapter 1 DyNumber Bases 1 1100102 – 111012 = O7 xP1 A 101012 1 B 110002 C 1101012 2 On the graph in the answer space, shade the D 1011012 P2 region which satisfies the three inequalities 2 Given that (53 × 4) + (52 × 3) + 5k = 43205, x + y ≤ 9, 3y ≥ x + 3, y < x.P1 find the value of k. Answer:A 0 C 3 yB 2 D 4 3 If 1010001102 = x8, then the value of x is 8P1 A 568 C 5038 x+y=9B 4068 D 5068 6 4 3y = x + 3 4 Express 1235 as a number in base 8. 2P1 A 388 C 648B 468 D 1738Answers O 2 4 6 8x 1. A 2. B 3. D 4. B 3 (a) Complete the following table for the P2 equation y = – 8 . x FORM 5 Chapter 2Graphs of Functions II x –5 –4 –3 –2 –1.6 –1 y 2 2.7 4 5 8 1 Which of the following graphs representsP1 y = 7 – 3x2? x 1 1.6 2 3 4 5 Ay y –8 –4 –2.7 –2 –1.6 7 (b) By using a scale of 2 cm to 2 units on both axes, draw the graph of y = – 8 for –5 ≤ x ≤ 5. x x O x (c) From the graph in (b), findBy (i) the value of y when x = 3.4, O 7 (ii) the value of x when y = 4.4. (d) Draw a suitable line on your graph to find all the values of x which satisfy the 8 equation x + 1 – x = 0 for –5 ≤ x ≤ 5. State these values of x.Cy Answers 1. A 7 x O y COMPANION WEBSITE Fully Worked Solutions xvi
2. y y C 8 y=x 14 6 B 4 3y = x + 3 2 x+y=9 12 A D 10 8 KQ P 6 R 4L O 2 4 6 8x 2 S 3. (a) x = –5, y = 1.6 M x = 1.6, y = –5 O 2 4 6 8 10 x (b) y (a) Transformation W is a reflection in 8 1 2the straight line x = 3. 6 2 . Transformation T is a translation –4 4.4 State the coordinates of the image 4 of point D under the combined 2 transofrmations. 2.4 3.4 (i) WT, x (ii) TW. 6 –6 –4 –2 O2 4 (b) (i) PKLM is the image of –3.4 –1.8 ABCD under a combined –2 transformation VU. Describe in –2.4 –4 full, the transformations y = –x – 1 (a) U, –6 (b) V. (ii) If the area of the quadrilateral –8 ABCD is 160 cm2, find the area, in cm2, of the shaded region.(c) (i) y ≈ –2.4 Answers (ii) x ≈ – 1.8(d) Draw y = –x – 1 1. (a) (i) (0, 7) x ≈ – 3.4 or 2.4 (ii) (4, 7) (b) (i) (a) Anticlockwise rotation of 90° about the centre FORM 5 Chapter 3 (10, 9)Transformations III (b) Enlargement with centre 1 The diagram shows three quadrilaterals, P(8, 7) and a scale factorP2 ABCD, PQRS and PKLM, drawn on a Cartesian 3 of 2 . plane. (ii) 200 cm2 xvii
FORM 5 Chapter 4 AnswersMatrices 1. A 2. D 1 21 2 5 1 1 2 3. (a) 3 – 5 2 2 1 2 3 1 = –1 2P1 –3 0 –5 1 4 5 21 x 2 1 6 2 1 2 0 (b) 2 3 y = 8 A –13 x = –11, y = 10 –3 4. (a) h = –2, k = 14 1 2 x –1 –4 6 y 18 B (0 –13 –3) 1 21 2 1 2(b) = 1 2 5 1 x = –3, y = 1 C 2 –15 FORM 5 Chapter 5 –3 0 Variations 5 2 –3 1 The table shows the corresponding values of x 1 –15 0 P1 and y. 1 2D 1 2 1 2 1 2 1 2 P21 Gfinivdenthtehvaatlu–e47of–t2+25=24, xpr t. 1 –2 –7 y q —23 A –2 p q B –3 If x varies directly as y and = 6, find the value of r. C 2 D 3 A 2 C 4 4 5 B 3 D 5 2 31 2 3 It is given that matrix A = . 2 Given that f varies inversely as the square rootP2 (a) Find the inverse matrix of A. P1 of g and f = 1 when g = 49, find the value of f (b) Write the following simultaneous when g = 64. linear equations as matrix equation: A 5 8 4x + 5y = 6 2x + 3y = 8 B 3 4 Hence, using the matrix method, find the value of x and of y. C 1 2 1 2 –4 61 2 4 It is given that matrix P= and D 7 8P2 6 h 4 1 1 2matrixQ = 1 such that k 1 0 3 The table shows some values of the variables 0 1 P1 A, B and C. 1 2PQ = . (a) Find the value of h and of k. A 2m (b) Write the following simultaneous B48 linear equations as a matrix equation. x + 2y = –1 C39 6y – 4x = 18 Hence, using matrices, find the value of x and of y. COMPANION WEBSITE Fully Worked Solutions xviii
Given that A ∝ B2 , find the value of m. 2 The diagram shows the speed-time graph C P2 of a particle for a period of 15 s.A 8 Speed (m s–1) 3 23B 4 3C 8 9D 2 k 3 5 O 6 10Answers 2. D 3. A 15 Time (s) 1. C FORM 5 Chapter 6 CalculateGradient and Area Under a Graph (a) the distance, in m, the particle moves 1 The diagram shows the distance-time graph ofP2 the journey of a bus. with constant speed, (b) the rate of change of speed, in m s–2, Distance (km) in the first 6 s, 22 (c) the value of k if the total distance travelled in the first 15 s is 139 m. Answers 1. (a) 12 km (b) 60 km h–1 (c) T = 25 2. (a) 20 m (b) –3 m s–2 (c) k = 9 10 FORM 5 Chapter 7 O 10 17 T Probability I & II Time (minutes) 1 The table shows the scores and theThe bus leaves station A for station B. After P1 number of times Fadillah obtains thestopping for 7 minutes at station B, it travels tostation C. scores in a computer game.Calculate(a) the distance, in km, from station B to Score Number of times 2000 – 2999 175 station C, 3000 – 3999 145(b) the speed of the bus, in km h–1, from 4000 – 4999 100 5000 – 5999 80 station A to station B,(c) the value of T if the speed of the bus from If she plays the game for total 500 times, station B to station C is 90 km h–1. how many times is she expected to achieve a score of at least 4000? A 180 C 220 B 200 D 240 xix
2 A group of visitors to Zoo Negara (a) List the sample space. (b) List all the outcomes of the events andP1 consists of 480 adults and a few children. find the probability thatIf a visitor is chosen at random, the (i) the code begins with the letter G, (ii) the code consists of two vowels or twoprobability that the visitor is a child is 3 . 7 consonants.Calculate the number of children among Answersthe visitors to the zoo. 1. A 2. C 3. B 4. D 5. (a) {(A, N), (A, G), (A, L), (A, E),A 1120 (N, A), (N, G), (N, L), (N, E),B 840 (G, A), (G, N), (G, L), (G, E), (L , A), (L, N), (L, G), (L, E), (E, A), (E, N),C 360 (E, G), (E, L)} (b) (i) {(G, A), (G, N), (G, L), (G, E)}D 206 51 (ii) {(A, E), (N, G), (N, L), (G, N), 3 A box contains 6 white ping-pong ballsP1 and a number of yellow ping-pong balls. (G, L), (L, N), (L, G), (E, A)} 25 The number of yellow balls is twice the number of white balls. After 4 more FORM 5 Chapter 8 white balls have been added to the box, Bearing a ball is selected at random from the box. 1 The diagram shows the positions of the points Calculate the probability of selecting a P1 J, K and M on horizontal ground such that yellow ball. JK = JM.A 5 11 KB 6 J 84° 11C 13 22D 7 11 4 A box contains 50 marbles. There areP1 blue marbles and red marbles. A marbleis chosen at random from the box. Theprobability that a red marble is chosenis 2 . How many red marbles need to be 5added to the box so that the probability 3that a red marble is chosen becomes 5 ?A 10 C 20B 15 D 25 5 The diagram shows five cards labelled MP2 with letters of the word ‘ANGLE’. Given that point K lies due north of point J, find the bearing of point J from point M. ANG L E A 084° B 096° C 264° All these cards are put into a box. A two- D 276° letter code is to be formed by using any two of these cards. Two cards are picked COMPANION WEBSITE at random, one after the other, without replacement. Fully Worked Solutions xx
2 Skukri left point P to point Q which is at 3 R(40°N, 80°W), P and T are three points P2 on the surface of the earth. RP is theP1 a bearing of 310° from point P. Then he diameter of a parallel of latitude 40°N.continued his journey to point R which is at T is 6600 nautical miles due south of R. (a) State the longitude of P.a bearing of 200° from point Q. Given that (b) Find the latitude of T. (c) Calculate the shortest distance, in∠QRP is 35°, find the bearing of point R from nautical miles, from R to P measuredpoint P. along the surface of the earth. (d) A plane flew from P to R along theA 235° C 125° common parallel of latitude and then flew due south to T. The total timeB 145° D 055° taken for the journey was 20 hours. Calculate the average speed of theAnswers 2. A plane for the whole journey. 1. C Answers FORM 5 Chapter 9 1. C 2. C Earth as a Sphere 3. (a) 100°E 1 In the diagram, N is the North Pole, S is the (b) 70°SP1 South Pole and NOS is the axis of the earth. (c) 6000 n.m. (d) 743.7 knots N 50º O Q P FORM 5 Chapter 10 30ºE S Plans and ElevationsFind the longitude of Q. 1 (a) Diagram (i) shows a solid right prism.A 20°E C 160°E P2 Trapezium ABCD is its uniform cross-B 20°W D 160°W section of the prism. The base ADEF is on a horizontal 2 In the diagram, N is the North Pole, S is theP1 South Pole and NOS is the axis of the earth. plane. BA, CD, HE and GF are vertical edges. The plane BCHG is inclined. N CH X 7 cm G 65º B O D 52º Equator 4 cm E A 6 cm X 15ºW S F 4 cmFind the position of point X. Diagram (i)A (65°N, 37°E) C (25°N, 37°E)B (65°N, 67°E) D (25°N, 67°E) xxi
Draw to full scale, the elevation of the (b) (i) solid right prism on a vertical plane parallel to AF as viewed from X. C/K/D H/E(b) A half-cylinder with a radius of 3 cm 6 cm and a height of 6 cm is joined to the solid in Diagram (i) at the vertical plane ABLKD to form a combined solid as shown in Diagram (ii). CH KML 7 cm B G M/B/A 4 cm G/F D 6 cm E (ii) Y MA 4 cm F 2 cm Diagram (ii) G/B H/C 4 cm 1 cm Draw to full scale, (i) the plan of the combined solid, F/A LK (ii) the elevation of the combined 6 cm solid on a vertical plane parallel to EF as viewed from Y.Answers H 6 cm E/D 1. (a) C 3 cmBG 4 cmA/D 4 cm F/E COMPANION WEBSITE Fully Worked Solutions xxii
A Linear Equations I – Linear Equations (a) Work on the brackets first, if there areform 2 in One Unknown brackets. (b) Group the unknowns on the left-hand side 1 A linear equation in one unknown is an equation which shows the relationship between numbers of the equation and the numbers on the and a linear term with one unknown, e.g. right-hand side. (a) —12 x – 3 = 7 (c) Solve the equation using combined (b) 2y – 5 = 3(5 – y) operations of addition, subtraction, 2 The steps to solve a linear equation in one multiplication and division. unknown are: 2 SPM SPM Clone Clone ’07 ’08 (Paper 1/Compulsory question/1 mark) 1 CSlPoMne CSlPoMne Given that —7k–+2––1–0– = 4 + k, then k = ’06 ’10 A – —31 C – —65 (Paper 1/Compulsory question/1 mark) B – —25 D —52Given that 2 – —3(—p2–—1)– = —3p , find the value of p. Solution —7k–+2––1–0– = 4 + kA —171– C —75 7k + 10 = 2(4 + k)B —2111– D —251– 7k + 10 = 8 + 2k 7k – 2k = 8 – 10Solution 5k = –2 k = – —252 – —3(—p2–—1)– = —p3 2p = 3(7 – 3p) Answer: B —4 –—3—(2p—– 1—) = —p3 —4 —– 3—2p +—3– = —p3 2p = 21 – 9p Key Terms —7 —–23—p = —p3 2p + 9p = 21 Linear equation – Persamaan linear Unknown – Anu 11p = 21 p = —2111– Answer: BMultiple-choice Questions Paper 1 A —13 C —34 4 Given that —32 x + —89 = 1, then x = 1 Given that —32 m + 4 = 13, then m = B —23 D —83 A —61 C —227–A 2 C 6 —31 u B —34 D —1276– C B 4 D 9 3 Given that 1 – = 4, then u = A – —16 –6 2 Given that 2h – —13 = 1, then h = B –9 D 9 P1 Zoom-in (Forms 1–3)
5 Given that —2p—5+—3 = 2, then p = C 4 D 10 14 Given that 2(3 – k) = —8k—2– 5–, find the value of k.A —27 C —123– 10 Given that v – 3(2 + v) = 10, then v = CSlPoMne A –10 C 8 ’08 A —1127– C —167–B —72 D 7 B –8 D 10 B —1107– D —147– 6 Given that n = —n +—3 8–, then n = 11 Given that —2h = —53 , then h = —1 –—24p– CSlPoMne A —130– C —130– 15 Given that = 3(p – 2), ’05 then p = CSlPoMneA 2 C 4 B —65 D —65 ’09 A —130– C —56B 3 D 5 7 Given that –3(2u – 3) = 8 – 2u, 12 Given that —32 – 5k = –2(4 – k), find B —87 D —1103–then u = CSlPoMne the value of k.A – —14 C —14 ’06 A —1121– C —171– Given that —y –—510– – —y –—3 7– = 1,B – —16 D —16 B —1149– D —131– 16 calculate the value of y. SPM A –1 C –3 Clone ’11 8 Given that 3r + 4 = 20 – (2 – r), C’S 1 0lPo7Mn3e GvAai lvu—e52eno t fhta.t —t –3—1 C= 2t—45– 1, find the B –2 D –5then r =A 5 C 7B 6 D 8 Answers: 1 C 2 B 3 B 4 C 9 Given that 2(3k – 5) – (k + 3) = 7, B —35 D –—45 5 B 6 C 7 C 8 C 9 C 10 B 11 C 12 B then k = 1 3 A 14 A 15 D 16 DA 2 B 3 B Linear Equations II – Simultaneous Linear Equations in Two Unknownsform 3Solving two simultaneous linear equations is a process of finding a pair of values for the unknownswhich satisfy both the equations. 3 CSlPoMne CSlPoMne ’07 ’10 (Paper 2/Compulsory question/4 marks)Calculate the value of h and of k that satisfy the Alternative Method (Substitution Method)following simultaneous linear equations:2h – k = 7 2h – k = 7 ... ①h + 3k = –7 h + 3k = –7 ... ②Solution From ②: …③ h = –3k – 7 2h – k = 7 ... ① Substituting ③ into ①:h + 3k = –7 ... ② 2 (–3k – 7) – k = 7 2h – k = 7 ... ①– 2h + 6k = –14 ... ② × 2 –6k – 14 – k = 7 –7k = 21 Multiplying the second –6k – k = 7 + 14 k = —–271– equation throughout by k = –3 2 so that the coefficients –7k = 21From equation ①: of h in both equations are k = —–217– the same to enable the k = –3When k = –3, 2h – (–3) = 7 elimination process to be 2h + 3 = 7 carried out. From ③: h = –3(–3) – 7 = 2 2h = 7 – 3 2h = 4 h = 2Zoom-in (Forms 1–3) P2
Persamaan Transistor Ba 388 2
Key TermsSimultaneous linear equations – Persamaan serentak linear 4 CSlPoMne CSlPoMne ’05 ’08(Paper 2/Compulsory question/4 marks)Calculate the value of p and of q that satisfy the 2p – 12q = 36 ... ③ × 2simultaneous equations: – 2p + 5q = 2 ... ②—31 p – 2q = 6 –17q = 34 Next, balance the coefficients of p2p + 5q = 2 q = —–314–7– in both equations by multiplying q = –2 equation ③ throughout by 2.Solution—13 p – 2q = 6 ... ① The coefficient of p in the first From equation ②: ... ② equation is a fraction. Multiply When q = –2, 2p + 5(–2) = 2 2p + 5q = 2 the equation throughout by 3 2p – 10 = 2 to eliminate the fraction. 2p = 2 + 10 2p = 12 p – 6q = 18 ... ① × 3 p = —122– p = 6 p – 6q = 18 ... ③ Label the equation which is free 2p + 5q = 2 ... ② from fractions as equation ③.Subjective Questions Paper 2 5 Calculate the value of e and of 9 Find the value of k and of m that 1 Calculate the value of x and of f that satisfy the simultaneous CSlPoMne satisfy the following simultaneous y that satisfy the simultaneous equations: equations: ’07 linear equations: 4k – 3m = —129– x – 3y = 11 k + 2m = 1 2x + y = 8 3e + —21 f = 4 2e – f = 8 2 Calculate the value of r and of 6 Calculate the value of d and of 10 Find the value of h and of k which t that satisfy the simultaneous SPM satisfy the following simultaneous equations: e that satisfy the simultaneous r – 11 = 4t Clone 2r – 7t = 20 equations: d – —41 e = 3 3d + e = 2 ’08 linear equations: 3 Calculate the value of h and of h + —23 k = –4 4h – k = 12 k that satisfy the simultaneous equations: 7 Calculate the value of h and of 11 Calculate the value of m and of 3h + 4k = –7 CSlPoMne k that satisfy the simultaneous 5h – 3k = 27 ’05 equations: CSlPoMne k that satisfy the simultaneous ’09 equations: —12 h + k = –4 h – 4k = 10 4m + k = 9 2m – 3k = 15 8 Calculate the value of p and 12 Calculate the value of m and 4 Calculate the value of v and of CSlPoMne of q that satisfy the following CSlPoMne of n that satisfy the following simultaneous equations: ’11 simultaneous linear equations: w that satisfy the simultaneous ’06 m + 2n = 10 —23 m – n = 7 equations: p + 2q = 12 v – —31 w = –5 —23 p – q = –2 5v + 2w = –36 d = 2, e = –4 5 e = 2, f = –4 10 h = 2, k = –4 9 k = 2, m = – 12— 2 r = 3, t = –2 Answers:12 m = 6, n = 2 11 m = 3, k = –3 8 p = 2, q = 5 1 x = 5, y = –2 3 h = 3, k = –4 4 v = –3, w = 6 7 h = –2, k = –3 P3 Zoom-in (Forms 1–3)
C Transformations I, II – Translation, _ + 2 A translation is described in the form h , where kform 2,3 Reflection, Rotation and Enlargement (a) h represents the movement to the rightC1 Translation (positive) or to the left (negative) which is 1 Translation is a form of transformation where all parallel to the x-axis, points on a given plane are moved along a straight line in the same direction and by the same (b) k represents the movement upwards distance. Example: (positive) or downwards (negative) which is P´ Q´ parallel to the y-axis. 3 If P'(x', y') is the image of an object P(x, y) under a 1 2translationh k , then PQ (a) the coordinates of the image P' are given by R´ P' = (x + h, y + k) R (b) the coordinates of the object P are given by 1 2Translation 5 P = (x' – h, y' – k) 2 5 CSlPoMne ’03(Paper 1/Compulsory question/1 mark)y In the diagram, point Q is the Solution In the diagram, if PQ = 10 units and TP = 6 units,Q image of point P(6, 2) under y then using Pythagoras’ 1 2a translationh Q Theorem, k . The 8 10 QT = √102 – 62 distance of PQ is 10 units. T6 = √ 64 1 2The translation h is O k = 8 unitsP (6, 2) Answer: A Hence, the translation1 2 1 2O x –6 –6 A 8 C 10 P(6, 2) 1 hk 2 = _ –86+. x B 1 68 2 D 1 –1062 1 In the diagram, D is the image of C under a certain y translation. If point K(2, 1) is the image of point H 6 under the same translation, the coordinates of the 4 point H are D2 A (–3, 7) C (–3, –5)–6 –4 –2 O –2 B (7, 7) D (7, –5) –4 K (2, 1) Solution –5 246 6 x 1 2C is mapped onto D under a translation . C 8 1 2H –5 Translation 6 K(2, 1) ∴ H = (2 – (–5), 1 – 6) = (7, –5) Answer: D Key TermsTranslation – TranslasiZoom-in (Forms 1–3) P4
C2 Reflection Key Terms 1 Reflection is a form of transformation where all Reflection – Pantulan points on an object are reflected in a line known Axis of reflection – Paksi pantulan as the axis of reflection. Example: C3 Rotation 1 A rotation is a transformation in which all points Axis of reflection on a plane are turned about a fixed point (known E E´ as centre of rotation) through the same angle and in the same direction (either clockwise or D D´ anticlockwise). 2 If the object and its image under a rotation are 2 The distance of the object and the distance of the given, the centre of rotation can be determined. image from the axis of reflection are the same. 7 CSlPoMne 3 The straight line which joins the object to the ’06 image is perpendicular to the axis of reflection. (Paper 1/Compulsory question/1 mark) 6 CSlPoMne CSlPoMne ’03 ’10 (Paper 1/Compulsory question/1 mark) y 2The diagram shows two quadrilaterals, KLMN andABCD, drawn on a square grid. ABCD is the image A Sof KLMN under a reflection. –4 –2 O x W D –2 2 46 L U B S C M Q N –4 C K The above diagram shows five points plotted on a Cartesian plane. Which of the points, A, B, C or D, D is the image of point S under a clockwise rotation of 90° about the centre (3, –2)? P AB Solution R Point C is the image of point S under a clockwise T rotation of 90° about the centre (3, –2), as illustrated V in the following diagram.The axis of reflection is the straight lineA PQ C TUB RS D VWSolution U y 2 L –4 S x –2 O 246 M Answer: C –2 (3, –2) N –4 C C K D AB TThe axis of reflection is the straight line TU.Answer: C P5 Zoom-in (Forms 1–3)
8 CSlPoMne ’08 (Paper 1/Compulsory question/1 mark) • Join point L to point L'. Using a ruler and a protractor, draw the perpendicular bisector of y In the diagram, ∆K'L'M' the line segment LL'.6 KL is the image of ∆KLM under an anticlockwise The perpendicular bisector of the line4 M segment LL’ is the straight line that2 L´ ABC D rotation of 90°. Which of passes through the midpoint of LL’ and is the points, A, B, C or D, perpendicular to the line segment LL’. M´ is the centre of rotation? • Join point M to point M'. Draw the K´ perpendicular bisector of the line segmentO 24 6 x MM'.Solution • The intersection point of the perpendicular bisectors is the centre of rotation, which is theObject K L M point C.Image K' L' M' Answer: C y Key Terms 6 KL Rotation – Putaran Centre of rotation – Pusat putaran 4 MC Perpendicular 2 L´ M´ bisectors K´ 4 x O2 6C4 Enlargement 3 When an object and its image under an enlargement are given, its centre of enlargement 1 Enlargement is a form of transformation that involves an invariant point (or fixed can be determined by the following steps. point) known as the centre of enlargement. All the points on the plane (except the centre (a) Draw any two non-parallel lines passing of enlargement) will move from the centre of through two pairs of the corresponding enlargement to form an image according to a points on the object and the image. certain ratio, known as the scale factor. Example: (b) The intersection point of these two lines represents the centre of enlargement. Example: P P´ O P´ Construction OR R´ P lines Centre of Q enlargement Q Q´ R R´ Q´ 4 For an enlargement with a scale factor of k: Area of image = k2 × Area of objectEnlargement with centre O and a scale factor of 2 2 (a) Scale factor, k = Length of one side of image Length of corresponding side of object ( b) S=caDDleiissfttaaancntccoeer,ookffcpoorirnetsopnonimdianggepforoinmt oOn , object from O where O is the centre of enlargementZoom-in (Forms 1–3) P6
9 CSlPoMne CSlPoMne Key Terms ’04 ’10 Enlargement – Pembesaran Centre of enlargement – Pusat pembesaran(Paper 1/Compulsory question/1 mark) Invariant point – Titik tetap Scale factor – Faktor skalaThe diagram shows five triangles, Q, A, B, C and D,drawn on a square grid. 2 B P A Q Q CD The above diagram is drawn on a square grid. Q is the image of P under an enlargement. If the area of P is 30 cm2, calculate the area of the region Q which is not shaded.Which of the triangles, A, B, C or D, is not the A 60 cm2 C 240 cm2image of the triangle Q under an enlargement? B 90 cm2 D 270 cm2Solution Solution The scale factor of Centre of enlargement this side is 3 . 2The scale factor of A Peach side is 1 . 2 A´ B B Q A Q The scale factor of B´ this side is 2. —AA'BB—' —26 C Scale factor, k = = = 3 D The scale factor of Area of Q = k2 × Area of P each side is 2. = 32 × 30 Im age = 270 cm2 ObjectEach side of triangle B is not enlarged by an equalscale factor. Hence, triangle B is not the image of the ∴ The area of the region Q which is not shaded triangle Q under an enlargement. = 270 – 30Answer: B = 240 cm2 Answer: C P7 Zoom-in (Forms 1–3)
Multiple-choice Questions Paper 1 4 In the diagram, point P' is the C2 ReflectionC1 Translation 7 The diagram below shows five 1 In the diagram below, N is the image of point P under a 1 2translation3 points plotted on a Cartesian plane. image of M under a translation. –2 . y N y P 4 ABC P´ 4 2 2D −4 −2 O x O 24 x 24 6 EQ M Find the coordinates of point P. A (–4, 5) C (2, 5) Which of the points, A, B, C or D, is the image of point E under aWhat is the translation? B (–4, 1) D (2, 1) reflection in the straight line PQ?1 2A –8 C 1 –58 2 –5 5 In the diagram, point R is the image 8 The diagram below shows five of point Q under a translation.B 1 –58 2 D 1 –85 2 triangles drawn on a Cartesian plane. y y 6 2 The diagram below shows two pentagons drawn on a square grid. N4 DA2 2 CB R 2 x –2 O M A –2 −6 −4 −2 O 2x Q −2 B Which of the triangles, A, B, C or D, is the image of triangle M underThe translation that maps A onto If point N is the image of point M a reflection in the line y = x?B is C 1 –47 2 under the same translation, the 9 The diagram below is drawn on a1 2A –7 square grid. –4 coordinates of the point M are A (–5, 7) C (1, 7) R B (–5, –1) D (1, –1)1 2B –4 D 1 –74 2 6 The diagram shows five points on a P –7 B CSlPoMne Cartesian plane. 3 In the following diagram, point AQ Q(6, 6) is the image of point P(2, 3) ’05 under a translation. y S DC y 10 A D Shahrir draws a polygon Q as the image of a polygon P under a 8 reflection in the line RS. Which of the points, A, B, C or D, is incorrect? Q(6, 6) 6 Q 10 The following diagram is drawn on a Cartesian plane. 4 P(2, 3) 2C y 4 D(4, 1) B ABC Ox E2 D O xThe coordinates of the image 2 4 6 8 10of point D(4, 1) under the sametranslation are Point Q is the image of point P xA (0, –2) C (7, 5)B (1, –3) D (8, 4) 1 2under a translation 3 . Which of –2 O 246 –4 –2 E´ the points, A, B, C or D, represents –4 M point P?Zoom-in (Forms 1–3) P8
Point E' is the image of point E 14 The diagram below shows a few C4 Enlargement isosceles triangles. 18under a reflection. The axis of Mreflection is J´A MA C MCB MB D MD11 The diagram is drawn on a O J Cartesian plane. Point P is the image of point M under a reflection AD B L´ K´ in a certain straight line. C y If the shaded triangle, M, is rotated BC LK through 180° about the point O, AD which of the triangles, A, B, C or6 P D, is the image of triangle M? ∆ J'K'L' is the image of ∆ JKL Q under a certain enlargement.4 15 The diagram is drawn on a Which of the points, A, B, C or D,2M Cartesian plane. Figure Q is the image of figure P under an is the centre of the enlargement? anticlockwise rotation.O x 19 The diagram is drawn on a square 246 8 y grid. ∆KLM is the image of ∆PQRThe coordinates of the image of 6 under a certain enlargement. 4Qpoint Q under the same reflection are 2PA (3, 0) C (4, 0) ACB (3, 1) D (4, 1) BD12 The diagram shows five objects, K LCSlPoMne F, A, B, C and D, drawn on a R’11 Cartesian plane. O x 2468 y The coordinates of the centre of PQ 6 rotation are Which of the points, A, B, C or D, A 4F represents the vertex M? A (3, 1) C (4, 1) 2 B B (3, 2) D (4, 2) −6 −4 −2−O2 24 6 x 16 In the diagram, pentagon M is 20 The diagram is drawn on a square D C −4 CSlPoMne the image of pentagon N under a CSlPoMne grid. ’08 clockwise rotation of 90°. K ’06 L −6 Which of the objects, A, B, C or D, N AM MN is the image of the object F under a B P reflection in the straight line y = –x? C QR DC3 Rotation Which of the points, A, B, C or D, Quadrilateral KPQR is the image13 The diagram shows three points on is the centre of rotation? of quadrilateral KLMN under an a Cartesian plane. enlargement. Which of the following 17 The diagram shows five triangles gives the correct centre and the scale y factor of the enlargement? 2 Q CSlPoMne drawn on a Cartesian plane. x −4 −2 O ’09 y −2 24 Centre Scale factor AK H2 BK 3 AB x CM 4 P −4 Q´ −4 −2 O 24 DM 3 −2 4Point Q' is the image of point Q CD 21 The diagram below shows fiveunder a certain rotation about theorigin. What are the coordinates of Which of the triangles, A, B, C or CSlPoMne pentagons drawn on a square grid.point P under the same rotation? D, is the image of triangle H under Which of the pentagons, A, B, CA (–3, 2) C (3, –2) a clockwise rotation of 90° about ’07B (–2, 3) D (2, –3) the point (–1, –1)? or D, is the image of pentagon M CSlPoMne under an enlargement? ’08 P9 Zoom-in (Forms 1–3)
M y Which of the rectangles, A, B, C A 12 or D, is the image of the shaded rectangle under an enlargement C B 10 A with centre O(0, 0) and a scale D factor of —12 ? 8 6 4 23 ΔPQR is the image of ΔABC 2B CSlPoMne under an enlargement with a scale ’11 factor of k. If the area of ΔABC and ΔPQR are 6 cm2 and 54 cm222 The diagram shows five rectangles −4 −2 O 2 4x −2 C respectively, find the value of k. drawn on a Cartesian plane. A —19 −4 D C 3CSlPoMne B —31 ’09 D 9 23 C 1 3 A 14 C 15 D 16 D 17 A 18 A 19 D 20 B 21 C 22 B 1 C 2 A 3 D 4 A 5 B 6 A 7 B 8 D 9 D 10 B 11 B 12 D Answers: D Polygons II – Interior and Exterior 10 CSlPoMne CSlPoMneform 3 Angles ’06 ’08 1 The sum of all the interior angles of an n-sided (Paper 1/Compulsory question/1 mark) polygon = WU x° T (n – 2) × 180° 40° S 2 Each interior angle of a regular n-sided polygon = V —(n—– —2)n—× 1—80—° y° P QR 3 The sum of the exterior angles of a polygon is In the above diagram, UVQRST is a regular hexagon. 360°. WVQ and PQR are straight lines. The value of x + y = 4 Each exterior angle of a regular n-sided polygon = A 95 C 105 —36—n0°– B 100 D 110 5 Properties of a parallelogram: p° R Solution q° S x° The sum of all the interior angles of a regular hexagon y° = (6 – 2) × 180° = 720° Hence, each interior angle of a regular hexagon = —726—0° 1 x = 180 –120 = 30 2 q°P p° y° = 120° W U T x° x° 120° 3 y = 180 – 40 – 60 Q x°(a) Opposite sides are equal and parallel. = 80 40° V(b) Opposite angles are equal. S 6 Properties of a rhombus: y° y° x° x° y° 60° 120° R P Q 2 ∠WQP = 180° – 120° = 60° x°x° y° ∴ x + y = 30 + 80 = 110 y° Answer: D(a) All sides are equal in length.(b) Diagonals bisect each other at right angles.Zoom-in (Forms 1–3) P10
Key Terms 12 CSlPoMne CSlPoMnePolygon – Poligon ’04 ’10Interior angle – Sudut pedalamanExterior angle – Sudut peluaran (Paper 1/Compulsory question/1 mark) In the diagram, PQRST is a pentagon and RST is an isosceles triangle. VQP and STU are straight lines. Y R 78° y° Q 11 CSlPoMne ’07(Paper 1/Compulsory question/1 mark) 114° SThe diagram below shows an octagon ABCDEFGH.BAJK is a straight line. B A JK P x° TC 115° x° H UDG Find the value of x + y. C 135 A 125 D 140 B 130 EF SolutionFind the value of x. Y RA 145 y°B 150C 155 Q 78°D 160 180° – y° 33°Solution 114° SThe sum of all the interior angles of a regularoctagon 33°= (8 – 2) × 180° 147° – x°= 1080° P x° THence, each interior angle of a regular octagon U = —10—880—°= 135° In ΔRST, ∠SRT = ∠STR = —18—0°—–2—11—4°– = 33°∴ ∠BAH = 135°Since BAJK is a straight line, then ∠QPT = 90°∠HAJ = 180° – 135° = 45° ∠PQR = 180° – y°∴x = ∠JHA + ∠HAJ ∠PTR = 180° – x° – 33° = 147° – x° = 115 + 45 = 160 Sum of interior angles in quadrilateral PQRT = 360°Answer: D 90 + (180 – y) + 78 + (147 – x) = 360 x + y = 90 + 180 + 78 + 147 – 360 = 135 Answer: C P11 Zoom-in (Forms 1–3)
Multiple-choice Questions Paper 1 V T S 8 In the diagram, PQVW is a W 80° R 1 In the diagram, U x° U CSlPoMne parallelogram and PQR is a y° QPQRST is P ’07 straight line. Ua regular y° CSlPoMne 120° 75° Tpentagon. UTP ’09 W V 2x° y° S 80°and VRQ are S 110° V 110° 50°straight lines. T x° RThe value of The value of x + y = PQ Rx+y= A 70 C 90 Find the value of x + y.A 132 PQ B 80 D 100 A 135 C 145B 142 B 140 D 150C 152 5 In the following diagram, PQRST 9 The diagram shows a regular is a regular pentagon. VQR is an CSlPoMne pentagon PQRST and anD 162 isosceles triangle. UPT is a straight ’08 equilateral triangle PUV. QUR is a line. 2 In the diagram, ABCDEG is a straight line. Q regular hexagon. FED and FGB VR S are straight lines. 70° y° U 130° Q P x° RF ED 29° x° x° V TS UP T Find the value of x. G C The value of x + y = y° A 46 C 67 A 82 C 117 B 60 D 75 B 89 D 120 A B 10 In the diagram, PQRST is a regular 6 In the PThe value of x + y = CSlPoMne pentagon and PJKLMT is a regular ’09 hexagon.A 150 C 170 CSlPoMne diagram, 30° ’05 PQRTU isB 160 D 180 Q U x° J K a regular QR 3 In the diagram, PQRSV is a regular pentagon pentagon and STUVWX is a regular hexagon. and RST is a k° PL y° P straight line. R S T S Q x° Find the value of k. WV A 36 C 78 TM B 70 D 72 The value of x + y = A 20 C 42 7 The diagram shows a regular B 30 D 48 CSlPoMne hexagon PQRSTU and a X y° U ’06 parallelogram PTVW. TUV is a 11 In the diagram, PQRSTU is anR straight line. CSlPoMne irregular polygon. PUS, QUT and ’11 RST are straight lines. STThe value of x + y = T UV P T S x° 38°A 138 C 150 80° U P x°B 148 D 158 W 4 In the diagram, PQRSTU is a 120° S regular hexagon. VTS and VUQ are straight lines. The straight lines RQ QR VW and TU are parallel to each other. Find the value of x. Given that ∠TQR = 2 × ∠PQT, A 108 C 144 find the value of x. B 120 D 150 A 89 C 80 B 82 D 7211 A Answers: 1 B 2 D 3 A 4 A 5 C 6 C 7 D 8 B 9 C 10 D Zoom-in (Forms 1–3) P12
E Statistics I, II – Pictograms, Bar Charts, 14 CSlPoMneform 2,3 Pie Charts, Mode, Median and Mean ’10E1 Pictograms (Paper 1/Compulsory question/1 mark)A pictogram is a representation of data using pictures The diagram is a bar chart showing the number ofor symbols showing the frequency of the given blouses sold by a salesgirl from January to June 2010.information. The symbols used can represent one ormore than one unit of the given data. Number of blouses sold 24 20 13 CSlPoMne CSlPoMne 16 12 ’06 ’10 8 (Paper 1/Compulsory question/1 mark) 4 0 Jan Feb Mac Apr May Jun MonthThe diagram is a pictogram showing the sales ofguidebooks in a particular month at a bookshop. The number of blouses sold from January to June is Science 20% more than the number of blouses sold from JulyMathematics to December. The commission earned by the salesgirlHistory for each blouse sold was RM50. Calculate the total commission earned by the salesgirl in that year. represents 10 copies of guidebooks A RM3250 C RM5720 The number of History guidebooks sold is not B RM3900 D RM7150shown. Science guidebooks make up 50% of the Solution The number of blouses sold from January to Junetotal sales that month. Calculate the number of = 18 + 16 + 14 + 8 + 10 + 12 = 78 The number of blouses sold from January to JuneHistory guidebooks sold. is 20% more than the number of blouses sold from July to December (y), i.e.A 30 C 50 —1102—00 y = 78 y = —1120—00 × 78 = 65B 40 D 60SolutionNumber of Science guidebooks sold = 9 × 10 = 90Number of Mathematics guidebooks sold= 6 × 10 = 60Let N = Total number of guidebooks sold The total number of blouses sold for the whole yearGiven that Science guidebooks make up 50% of the is 78 + 65 = 143total sales, Hence, the total commission earned—150—00 N = 90 = 143 × 50 = RM7150 N = 90 × —1500—0 Answer: D N = 180Therefore, the total number of guidebooks sold is Key Terms180.Hence, the number of History guidebooks sold Pictogram – Piktogram/Piktograf= 180 – Number of Science guidebooks – Number Bar chart – Carta palang of Mathematics guidebooks E3 Pie Charts= 180 – 90 – 60 = 30 1 A pie chart represents data in relative quantitiesAnswer: A using the areas of sectors in a circle such that the quantities are proportional to the size of theE2 Bar Charts angles of the sectors.A bar chart is a representation of data using 2 In a pie chart, the angle of a sector whichhorizontal or vertical bars. The width of each bar represents a particular quantity =is the same throughout and the height of each bar isproportional to the frequency. Quantity represented by the sector × 360° Total quantity P13 Zoom-in (Forms 1–3)
15 CSlPoMne ’07 (Paper 1/Compulsory question/1 mark) Solution Angle of the sector which represents the amountThe pie chart shows the division of the expenditure spent on foodof the family of Madam Tan in a particular month. = —2547—6000– × 360° = 150° Clothing Entertainment Angle of the sector which represents the amount spent on entertainment 30° = 360° – 60° –30° – 150° = 120° Bills 60° Hence, the ratio of the amount spent on entertainment Food to the amount spent on food = —1152—00The total monthly expenditure in that month is = —45 =4:5RM5760. The amount spent on food is RM2400. Answer: CFind the ratio of the amount spent on entertainmentto the amount spent on food.A 2 : 3 C 4 : 5B 3 : 4 D 5 : 6E4 Mode,Median and Mean 4 In a set of data, 1 Mode is the value of data which occurs most Mean = Sum of all the values of data frequently in a set of data. Number of data 2 In a frequency table, the mode is the value of data 5 When the data is given in a frequency table, which has the highest frequency. Mean = Sum of (value × frequency) 3 Median is the value located at the centre of a set Sum of frequencies of data after the values of data are arranged in an ascending order. Key Terms Pie chart – Carta pai 16 CSlPoMne 3 ’04 Based on the bar chart provided in ‘Exam Watch 16’, find the median score. (Paper 1/Compulsory question/1 mark) A 2.5 B 3 The diagram shows the scores of a group of students C 3.5in a Mathematics quiz. D 4 Number of students Solution Write the scores presented in the bar chart in the 6 form of a ‘set of data’, as shown below. 5 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5 4 3 2 1 O1 2 34 5 ScoreThe modal score is C 5A 2 D 6 Centre valueB 3 Score with the highest Median = 3Solution frequency (highest bar) Answer: BModal score = 3Answer: BZoom-in (Forms 1–3) P14
4Based on the bar chart provided in ‘Exam Watch Mean16’, calculate the mean score. = Sum of (score × frequency) Sum of frequenciesA 2.6 C 2.8B 2.7 D 2.9 = —(1–×—3—) +—(2—×—35)+—+5(—+3 ×6—+6—)5+—+(42—×—5)—+ —(5—× 2—)Solution = —2611–A frequency table can be constructed to represent = 2.9the distribution of scores presented in the barchart, as shown below.Score Number of students (frequency) Answer: D 1 3 2 5 3 6 4 5 5 2 Key Terms Median – Median Mean – MinMode – Mod Multiple-choice Questions Paper 1 2 The following pictogram shows January CSlPoMne the number of scooters sold by a FebruaryE1 Pictograms ’05 company in five particular months. March 1 The pictogram shows the types of April January books Encik Fadzli has in his home February represents 600 cars library. March April The number of cars sold in March Science and May technology 2009 is twice the number of cars Novel sold in April 2009. Given that the Autobiography represents x scooters total number of cars sold in the represents 5 books first four months of 2009 is 12 600 The number of autobiography The mean number of scooters sold units, calculate the number of cars books he has is not shown. Novels make up 35% of Encik Fadzli’s per month is 18. Calculate the sold in March 2009. books. Calculate the number of autobiography books he has. number of scooters sold in March. A 1800 C 3000 A 15 B 20 A 1 C 12 B 2400 D 3600 C 35 D 45 B 6 D 18 4 The pictogram in the diagram 3 The pictogram shows the number CSlPoMne shows the number of students ’09 taking tuition for four subject at SPM of cars sold by a car dealer in Clone a tuition centre. The numbers of ’06 January and February 2009. The students taking Mathematics and numbers of cars sold in March and English tuitions are not shown. April 2009 are not shown. P15 Zoom-in (Forms 1–3)
Mathematics A 48° C 72° The ratio of the number of mango B 96° D 120° trees to the number of durian Physics E3 Pie Charts trees is 5 : 4. If a pie chart is 7 The pie chart drawn to represent all the given Chemistry shows the information, find the angle of the modes of sector representing the number of transportation English to school of School Walking starfruit trees. 1200 students. bus 40°1264°° A 24° C 48° represents 5 students B 36° D 60° The ratio of the number of Bicycle E4 Mode,Median and Mean 10 The table below shows the students taking English tuition The number Motorcycle Car distribution of scores of 36 to the number of students taking of students students in a health quiz. Mathematics tuition is 3 : 4. Given who take a school bus is 700. that the total number of students Calculate the ratio of the number taking tuition for the four subjects of students who walk to school to Score 1234 is 135, calculate the number of the number of students who take a Frequency 5 18 10 3 students taking English tuition. school bus to school. If the above data is represented A 25 C 35 A 1 : 2 C 2 : 3 using a pie chart, then the angle of B 30 D 40 B 1 : 3 D 3 : 5 the sector of the modal score isE2 Bar Charts 8 Diagram (i) shows a bar chart A 150° C 210° CSlPoMne which represents the number of 5 The double bar chart shows the ’07 students from three schools, P, Q B 180° D 240° number of male and female students CSlPoMne and R, who were chosen for the in three schools, P, Q and R. ’08 ‘National Service’ programme. 11 The bar chart shows the distances, correct to the nearest km, from the School Diagram (ii) shows a pie chart house to the school, of 40 students showing the number of students in a class.P according to gender.Q 14Number of students 10R Number of students 8 12 6 4 Number of students 10 2 Male students Female students 8 School Q has 540 female students. 6 Calculate the difference between 4 the total number of male students 2 x° Female O1 23 4 5 0 P Q R School Male Distance from house and the total number of female to school (km) students in the three schools. Male Female Calculate the mean distance, in km. A 180 C 360 Diagram (i) Diagram (ii) A 2.5 C 2.7 B 270 D 450 Find the value of x. B 2.6 D 2.8 6 The diagram is a bar chart showing A 207 C 225CSlPoMne the time spent for five activities in a’11 week. B 216 D 234 12 The table shows the sales of books CSlPoMne of four subjects on a certain day at Minute 9 The diagram is an incomplete bar ’08 a bookshop. CSlPoMne chart which shows the number of 60 ’09 fruit trees in an orchard. The bar Subject Frequency 50 40 for durian trees is not shown. Mathematics 10 30 20 Number of fruit trees Physics 5 10 100 Chemistry p 0 Reading Swim- Yoga Internet Ping Activity 80 ming pong Biology 6 60 If the data is represented by a pie If the mode is Mathematics, find chart, calculate the maximum 40 difference in the sector angles the maximum value of p. between any two of the activities. 20 A 8 C 10 0 Papaya Mango Durian Starfruit Types of fruit trees B 9 D 11 Answers: 1 A 2 B 3 D 4 B 5 C 6 B 7 B 8 B 9 C 10 B 11 C 12 BZoom-in (Forms 1–3) P16
F Circles I – Length of an Arc and Area of 2 Area of a sector of a circle = —An—gl—e36a—0t °c—en—tre– × Area of circleform 2 a Sector of a Circle = —An—gl—e36a—0t °c—en—tre– × πr2 1 Length of an arc of a circle A= —An—gl—e36a—0t °c—en—tre– × Circumference r= —An—gl—e36a—0t °c—en—tre– × 2πr A Area of the sector OAB O θ ° Sector Arc = —3θ6—0°° × πr2 r r B BLength of the arc AB O θ°= —3θ6—0°° × 2πr r 17 CSlPoMne CSlPoMne ’06 ’08(Paper 2/Compulsory question/6 marks)The diagram shows two sectors, OQR and OTS, with Area of region QTSRthe same centre O and a quadrant, PTQ, with centre T. = Area of sector OTS – Area of sector OQR = —37600– × —272– × 212 – —37600– × —272– × 72 P = 269 —21 – 29 —1178– T = 239 —59 cm2 Q Hence, the area of the shaded regionO 70° = Area of sector PTQ + Area of region QTSR = 154 + 239 —95 R = 393 —59 cm2 S (b) Length of arc PQ = —39600– × 2 × —272– × 14It is given that OS = 21 cm and PT = 14 cm. = 22 cmUsing π = —272–, calculate Length of arc TS(a) the area of the shaded region, = —37600– × 2 × —272– × 21 = 25 —23 cm(b) the perimeter of the whole diagram.Solution – 3—6θ—0 × 2πr(a) Area of sector PTQ = —39600– × —272– × 142 3– —6θ—0 × πr2 = 154 cm2 P Hence, the perimeter of the whole diagram 14 cm T Q = Length of arc PQ + PT + Length of arc7O2c1mc7m0R° TS + OS + OQ —32 = 22 + 14 + 25 + 21 + 7 = 89 —23 cm S Key Terms Area of sector – Luas sektorLength of arc – Panjang lengkok P17 Zoom-in (Forms 1–3)
18 CSlPoMne CSlPoMne ’07 ’10 (Paper 2/Compulsory question/6 marks)The diagram below shows a sector PQR of a circle 1 2 1 2 =—3246–00– × —272– × 142 – —14 × —272– × 72with centre O and a quadrant OSM. – 38—12 410—32 – 3—6θ—0 × πr2 Q = 372—61 cm2 O = S 120° P (b) Perimeter of the shaded region M = OP + Length of arc PQR + RM + SM + Length R of arc SOGiven that OP = 14 cm, OM = 7 cm and taking = 000714 + —2364–00– × 2 × —272– × 14 + 7 + 7π = —272–, calculate 1 2 (a) the area, in cm2, of the shaded region,(b) the perimeter, in cm, of the shaded region. 1 2 19+47—+—23415c8×m—232 ×+—21724–+×171 3– —6θ—0 × 2πrSolution(a) Area of the shaded region = Area of the sector OPQR – Area of the quadrant OSMSubjective Questions Paper 2 S 4 In the diagram, AF is an arc of a circle, with centre B. DEF and 1 The diagram shows two sectors of T 120° R BCD are semicircles. The points B, two concentric circles, ABC and U D and F lie on a straight line such DEF, with the same centre O. OGA that BD = 2DF. is a semicircle with centre D. OFC P 28 cm Q is a straight line and OA =14 cm. It A is given that ∠COA = 90˚. U(as) intgheπa=re—a272o–f, calculate 21 cm the whole diagram, 60° E DFB (b) the perimeter of the whole B E D A diagram. O Using π = —272–C, calculate FG 3 In the diagram, RNS is an arc of a (a) the area of the whole diagram, C quadrant, with centre P. RQPT is (b) the perimeter of the whole an arc of a circle, with centre O. Using π = —272–, calculate ROP is a straight line. diagram. (a) the perimeter, in cm, of the R 5 In the diagram, QR and TU are the whole diagram, arcs of two different circles, with (b) the area, in cm2, of the shaded QO N centre O. QPOU and RSTO are 250° T straight lines and QU = 21 cm. region. P 28 cm S R 2 In the diagram, PQRS is a S quadrant, with centre P and PU Using π = —272–, calculate T is an arc of a circle, with centre T. (a) the perimeter of the whole The points S, T and P are collinear Q POU and PQ = 2ST. diagram, (b) the area of the shaded region.Zoom-in (Forms 1–3) P18
Using π = —272–, calculate (b) the area, in cm2, of the shaded 10 In the diagram, PQR and ST are the(a) the area of the shaded region, region. CSlPoMne arcs of two circles, both with centre(b) the perimeter of the whole ’09 O. PUO is a semicircle and S is the 8 The diagram shows the quadrant diagram. CSlPoMne OAB and semicircle RST, both with midpoint of OR. POT is a straight ’07 centre O. It is given that OA = 14 cm line. It is given that OR = 14 cm 6 The diagram Z Y and ∠SOT = 50°. V X and S is the midpoint of OA.CSlPoMne shows two arcs QR’05 of two circles, AVWX and ZY,both with centre O S USO. OVZ and OXY 70° Tare straight lines 50° P O Tsuch that OX = W OB XY = VX = 14 cm. R Using π = —272–, calculateUsing π = —272–, calculate Using π = —272–, calculate (a) the perimeter, in cm, of the(a) the perimeter of the shaded (a) the area, in cm2, of the shaded whole diagram, region, region, (b) the area, in cm2, of the shaded (b) the perimeter, in cm, of the(b) the area of the whole diagram. whole diagram. region. 7 In the diagram, OSN and OPQ 11 In the diagram, PMQL is a sector CSlPoMne of a circle with centre P and OPRQCSlPoMne are two sectors of two concentric 9 The diagram shows the arcs RS and ’11 is a semicircle with centre O.’06 circles, both with centre O. ONRM CSlPoMne PQ of two concentric circles, both ’08 with the same centre O. OPR is ais a quadrant, also with centre O. straight line.ORQ and OMP are straight lines. R Q OM is the midpoint of OP. It is given S 21 cm Pthat OP = 14 cm, ∠SON = 45° and Q∠QOP = 60°. Q 120° O PR M L 14 cm NR Given that MP = 14 cm and using π = —272–, calculate S It is given that ∠POQ = 54° and —272–, 45° ∠ROS = 120°. Using π = (a) the perimeter, in cm, of the 60° calculate M P whole diagram, O (a) the perimeter, in cm, of theUsing π = —272–, calculate sector ORS, (b) the area, in cm2, of the shaded(a) the perimeter, in cm, of the (b) the area, in cm2, of the shaded region. whole diagram, region. Key Terms Collinear – Segaris 496 92— cm2 (b) 3 (a) 116 cm 2 (a) 513 13— cm2 (b) 114 23— cm (b) 369 35— cm2 9 (a) 86 cm 89 65— cm2 (b) 7 (a) 51 56— cm 6 (a) 87 31— cm (b) 924 cm2 4 (a) 288 43— cm2 (b) 76 cm 333 23— cm2 (b) 11 (a) 86 32— cm 10 (a) 65 89— cm (b) 166 65— cm2 5 (a) 52 12— cm2 (b) 55 21— cm 8 (a) 124118— cm2 (b) 56 94— cm (b) 423 21— cm2 Answers: 1 (a) 102 cm G Indices 3 (am)n = amn 1 2e.g. 42 ÷ 32 = —34 2 e.g. (32)3 = 36form 3 6 a0 = 1, a ≠ 0 4 am × bm = (ab)m e.g. 60 = 1Laws of Indices e.g. 23 × 43 = (2 × 4)3 = 83 1 am × an = am + n 7 a–n = —a1n , a ≠ 0 1 2 5 am ÷ bm =—ab m e.g. 3–2 = —312 = —19 e.g. 32 × 34 = 32 + 4 = 36 , b ≠ 0 Zoom-in (Forms 1–3) 2 am ÷ an = am – n e.g. 34 ÷ 32 = 34 – 2 = 32 P19
8 a—n1 = n√a , a . 0, n . 0 9 a—mn = (n√a )m = n√am , a . 0, e.g. 125—13 = 3√125 = 5 n.0 e.g. 64—23 = (3√64)2 = 42 18 CSlPoMne 20 CSlPoMne CSlPoMne CSlPoMne CSlPoMne ’07 ’07 ’09 ’10 ’11 (Paper 1/Complusory question/1 mark) (Paper 1/Compulsory question/1 mark)Simplify (x12 y–8)—14 × y–4z4. C x3y–2z4 √1 2 —98 –5 =A x3y–12z4 1 2A —98 —52 1 2C —89 —52B x3y–6z4 D x3y2z4Solution 1 2B —98 – —25 1 2D —89 – —25(x12 y–8)—14 × y–4z4 = (x12)—41 × (y–8)—14 × y–4z4 = (x3y–2) × y–4z4 (am)n = amn Solution = x3y–6z4Answer: B x3y–2 + (–4)z4 √1 2 1 2 √1 2 1 2 —98 –5 = ———891 –5– = —89 5 = —98 —52 19 CSlPoMne CSlPoMne Answer: A a–n = —a1–n ’08 ’09 (Paper 1/Compulsory question/1 mark) 21 CSlPoMne CSlPoMne CSlPoMne ’08 ’10 ’11Given that 24—58 = x√24y , state the value of x and of y. (Paper 1/Compulsory question/1 mark)A x = 5, y = 8 C x = 5, y = —18 Simplify —(pp—3–q8q4–)2—–2 .B x = 8, y = 5 D x = 8, y = —15 A p–2q10 C p–14q–6Solution B p2q–10 D p14q624—85 = x√24y Solution p q3 × (–2) 4 × (–2)24—85 = 8√245∴ x = 8, y = 5 a—mn− = n√am —(pp—3–q8q4–)2—–2 = —pp—––68qq––2—8Answer: B = p–6 – q(–8) –8 – 2 = p2q–10 Answer: BMultiple-choice Questions Paper 1 2 (64x)–—31 = √1 2 3 —185– –7 A —64—13√x– B —43√—1 x– ——1 – = 43√x —2 C – SPM 1 Given that 25 3 = p√25q , state the Clone —185– —72 1 2C —185– —27 value of p and of q. ——1 – 1 2 ’09 1 2D —185– – —72 643√x A p = 2, q = 3 C p = 2, q = —13 D – A CSlPoMne ’10 1 2B —185– – —27 B p = 3, q = 2 D p = 3, q = —21Zoom-in (Forms 1–3) P20
4 (216–2) —1 = A —1y6–4z—42 C —y26–4z—2 12 Evaluate (6–3 ÷ 22) × (68 ÷ 2–12)—14 . 3 B —14–z4—2xy–6 D —24z–x2—y6CSlPoMne A 36 C – —316– A 3 C —13 D –36’11 C – —641–x—2 B —316– D – —161–x—2 1 2 9 Simplify —4x––4—y–——23 3 × y2w3. B —136– D —112– –w 5 – (64x3) –—32 = ’C 1S0lPo3M6ne A—(3 34—–3×3×—6 62)——12 = B – —31 CSlPoMne A 64x2 A —64y–x4—12 C – —64y–x4—12 B —x6142– D – —x6142–’07 C —13 D 3 B —16–1x—2 Simplify —–6–pr—32q–——16 —11 2 6 3 q 2 r3. Simplify —((61–42—m5m6—)6—65n—1×5)n——329 .CSlPoMne × 10 14 Evaluate 81—34 ÷ 27–—34 × 3–2.’10 CSlPoMne A —21r–63—p9 C – —18r–p3—9q A 39 C 35 ’11 A —3225–m—n B —18r–p3—9q D – —21r–63—p9 C —25mn– B 37 D 3–3 7 (m–9 k12)—32 × m9 k6 = B —325–mn—3 D 2—5mn–3 15 (23 × 3–2) ÷ (16 × 3–3) = 3 14 C m–15 k14 11 Evaluate (46 × 3–3)—31 ÷ (15 × 2–6). A —4816– C —23A m kCSlPoMne D m–15 k18 —16 D —32’07 B m3 k18 A —140–25—4 C —64– B 451 2 8 Simplify 1—2x–z–—3—41 y—–3 2 × x—12 z4. B —1156– D —54 Answers: 1 B 2 B 3 A 4 B 5 D 6 D 7 A 8 A 9 D 10 A 11 A 12 C 13 C 14 C 15 D H Algebraic Expressions III – Subtraction 23 CSlPoMne CSlPoMneform 3 of Algebraic Fractions ’09 ’11When subtracting algebraic fractions, we must first (Paper 1/Compulsory question/1 mark)find the LCM of their denominators. E xpr ess –6—+p—p ÷ –3—p–2—p as a single fraction in its 22 SPM SPM simplest form. Clone Clone –6—3p –+—pp–2 C –63—+–—pp ’08 ’10 A (Paper 1/Compulsory question/1 mark) Express –k—–k—4 – –3—(6k—–2 k–) as a single fraction in its B –63—+–—pp2 D –3—–6—psimplest form. A –k—2 –—kk2—– 1—8 C –k—2 –—kk2—– 2—2 SolutionB –k2—–—3kk2—– —18 D –k2—–—3kk2—– —22 –6—+p—p ÷ –3—p–2—p = –6—+p—p × –3—p–2—pSolution = –p—(36—–+p—p)––k—–k—4 – –3—(6k—–2 k–) = –k—2 k–—24k– – –(1—8k—–2 3–k–) –6—3p –+–p—p2 = –k—2 –—4k—–k2—18—+ —3k = = –k—2 –—kk2—– 1—8 Answer: AAnswer: A P21 Zoom-in (Forms 1–3)
Multiple-choice Questions Paper 1 4 Express —82+–a–a– – —4b–a—–b a– as a 8 Express —8cd– – —c2 –––1–2– as a single single fraction in its simplest form. 24cd 1 Express —57+–k–k ÷ —28–1—–4k1—21k– as a CSlPoMne A —a +–a—b4b– C —22–+a—ba– CSlPoMne fraction in its simplest form. ’10 A —c2+–d—2 —c2 –+—3 ’07 6cd single fraction in its simplest form. C A —10–k—+ –2—k2 C —10–—+ 2–k—2 B —2b–2–—b a– D —2–2+—bb– B —c122––c—d6 D —c122–+c—d6 28 – 11k 28 – 11k B —10–k—+–k—2 D —1–0—+ –k—2 5 Express —35m– – —91–2–m—7m2– as a single 9 Express —65h– – —112–8––h–2h– as a single 28 – 11k 28 – 11k 2 Express —4(–xx—+y 2–)– – —12–3—–y x– as a CSlPoMne fraction in its simplest form. CSlPoMne fraction in its simplest form. A —9m4–m—–23– C —91m2–m—+21– A —8h9–h—–26– C —h18––h—32 ’05 ’08 single fraction in its simplest form. B —131–m2—m–2–9– D —9m4–m—–24– B —7h9–h—–26– D —h18––h—72 —x2––—24–x—+—24– A 3xy B —x2–+—2–4– C S’1lPo61Mne EAfrxa cp—6trqie(o(— qsnqs2 i—+–—nqpp6i–+–t2)ps–)–p–s im÷ Cp—q l2pe—––6sqq–tpp(—–f(p2oqa—+r–smqap–.))–single 10 Express —24+–a–a– + —32––a–a– as a single 3xy B —6(—qq–—p)– D —6p—(qq—– p–)– fraction in its simplest form. C —x2–+—8– A —8 +–a—a C —84+–a—a 3xy D —–x–2 —+ 2–4– B —8 ––a—a D —8 ––—a xy 4a Express – —81––6—p2qq–– + —81p– as a 7 Express —4xx––y–y– ÷ —82––y–y– as a single 11 Expre ss —65x– – —3 1––2–x5–2x– as a single 3 single fraction in its simplest form. fraction in its simplest form. A —p1q– C —q4–p–—q2 SPM fraction in its simplest form. A —51x–2––x–23– C —51x–2––x–21– Clone A —1x C —88xx—––—2xyy– ’09 B —q8–p–—q2 D – —p1q– B —8—x –—y– D —88x—x––—2yy– B —5x4–x––2–3– D —5x4–x––2–1– 8x – xy Answers: 1 A 2 B 3 C 4 D 5 A 6 B 7 C 8 D 9 A 10 D 11 D I Algebraic Formulae – Expressing a 24 CSlPoMne CSlPoMneform 3 Specified Variable as the Subject of a ’09 ’10 Formula (Paper 1/Compulsory question/1 mark)In the formula A = πr2, A is the subject of the formula. Given that m = 5n2 + 6, express n in terms of m.The subject is on the left-hand side of the formula. r √A n = –m—5+–6– C n = √–m—5+––6–can also be made the subject of the formula as shown,i.e. A = πr2 √ B n = –m—5––6– D n = √–m—5–––6–—Aπ = r2 Solution r2 = —A π m = 5n2 + 6√ r = —A 5n2 + 6 = m π 5n2 = m – 6 n2 = –m—5–––6–Zoom-in (Forms 1–3) P22
√ n = –m—5–––6– SolutionAnswer: B 1√ 2 E = 5 —P +–1—Q √ –E—5 = —P +–1—Q 25 CSlPoMne CSlPoMne Square both sides to –E2—52 = —P +–1—Q eliminate the square root ’08 ’11 on the right-hand side. P + Q = –E2—52 (Paper 1/Compulsory question/1 mark) P = –E2—52 – Q Apply cross multiplication1√ 2Given that E = 5 —P +–1—Q , express P in terms of Eand Q.A –E—52 – Q C –E2—52 – E 2Q Answer: BB –E2—52 – Q D –E—52 – E 2QMultiple-choice Questions Paper 1 5 Given that —5m–—– 2– = —27m–, then k = 8 Given that v = √ u2 + 2as, then a = 3√k 1 Given that p = 7 + 5q2, then q = CSlPoMne A —(v –2–—su)–2 C —v2–2–—su2–CSlPoMne ’08’10 A √—p–+5—7– √C —p–+5—7– A —49–(3—56mm––—2 2—)2 B —v ––—u D —v2–+—u2– √D —p––5—7– 2s 2s B √—p––5—7– B —7(–5—m6m––—2)—2 9 Given that —5–w— = —3w–2—+ 4–, then x = 2√x C —7(–53—m6m––—22)—2 2 Given that 3x = —–4—–, then y = A —(3–2w—5+w–42—)2 C —(3–w5—w+2–4—)2 √y + 1CSlPoMne —49–(—5m––—2—)2 B —3w–5—w+ –4– D —9w2–25—+w21–6– A —91x6–2 + 1 1 2C —4– + 1 2 D 6m’05 3x —34x– – 1 D —91x6–2 – 1 6 Given that 2p = —11 +––—qq , then q = 1 2B 2 10 Given that 2y = 4x2 – 8, then x = CSlPoMne ’11 A —12p+–—–21p– C —21p––+—2p1– 3 Given that —p2–5+—a– = —b3 , then p = √CSlPoMne —y +–2—8 √C —y +–2—4 ’09 A √ √A —b1––5—a C —5b–3–—a B —1 ––2—2p– D —1 +–2—2p– √ √B —y ––2—8 D —y ––2—4 √B —5b–—–3 3— a D √5b – 3a 7 Given that c = 3b + —25 a√ t , then t = 11 Given that —2p–24—+ 1– = q – 1, then p = CSlPoMne 1 2CSlPoMne —4(c–—– 3–b—)2 ’07 5a ’06 A A —q ––4—1 + 1 —12 —4b–2—+ c– 4 Given that 6a2 = 3 , then b = B —2(c–—– 3–b—)2 1 2B —q ––2—1 – —21 —12 5a 1 2 1 2A —9a–2 —12 —9a–2 —4c —12 2 – c C 2 – C —2(c–2—–5a3–2b—)2 —–4— —1 —12 q–1 2 1 2C – B (18a2 – c)—12 D (9a2 – 4c)—12 D —4(c–2—–5a3–2b—)2 1 2D —q ––4—1 – 1 —12 P23 Zoom-in (Forms 1–3)
12 Given that F = —G–M— + 5, then r = 13 Given that y = —2√–x—+—2 , then x = √A —3Vπ–h– √C —3πV–h– 2√r 3√x – 2 √B —3Vπ–h– √D —3πV–h– A —(G–M4—F+2–5—)2 —3 +–2—y2y– —2 +–—2y– 2 3y – 2 1 2 1 2A 2 C —3 –2–—y2y– —32y––+—22y– B —G–—2M–2— 1 2 1 2B 2 D 2 15 The surface area, A, of a cylinder 2F – 10 of radius r and height h, is given by A = 2πr2 + 2πrh. Express h in terms of A, π and r. C —2(–FG—–M–5—)2 14 The volume, V, of a cone of —r13adπiru2shr. A —2Aπ–r– – 1 C —2Aπ–r– – r D —4(G–F—2–M–52—)2 and height h, is given by V = B —2Aπ–r– – r D —A2–π––rr– Express r in terms of V, π and h. Answers: 1 D 2 B 3 B 4 C 5 A 6 C 7 D 8 C 9 A 10 C 1 1 B 12 D 13 C 14 D 15 C J Linear Inequalities – Solving Linear (b) x < 0 and x . –3form 3 Inequalities in One Unknown 1 If x . y, then −4 −3 −2 −1 0 1 (a) x + z . y + z (d) x—xzz ,y—zyz, z, ,0 0 ∴ –3 , x < 0 ; x = –2, –1, 0 (b) x – z . y – z (e) z. (c) x .1 and x > –2 (c) xz . yz, z . 0 (f) —xz , —yz , z , 0 2 If x . y and y . z, then x . z. −2 −1 0 1 2 3 The number line can be used to show the x . 1; x = 2, 3, 4, ... solutions of linear inequalities. For example, (a) x < 1 and x < –2 27 CSlPoMne ’08 −3 −2 −1 0 1 2 (Paper 1/Compulsory question/1 mark) ∴ x < –2; x = ..., –3, –2 Which number line represents the solution of the simultaneous linear inequalities 7x + 8 > –6 and 26 CSlPoMne CSlPoMne CSlPoMne CSlPoMne 10 – 3x . –5? A ’06 ’09 ’10 ’11 −5 −4 −3 −2 −1 0 1 2 3 4 5 B (Paper 1/Compulsory question/1 mark) −5 −4 −3 −2 −1 0 1 2 3 4 5 CList all the integer values of x for which 2x . 56 and −5 −4 −3 −2 −1 0 1 2 3 4 564 – 2x > 2. D A 29, 30 C 29, 30, 31 −5 −4 −3 −2 −1 0 1 2 3 4 5B 28, 29, 30 D 28, 29, 30, 31Solution Solution 7x + 8 > –6 2x . 56 64 – 2x > 2 7x > –14 10 – 3x . –5 –3x . – 15 x . –5—26 64 –22x > 2x x > –2 x , –––—135 Reverse the < 62 inequality sign x . 28 x , 5 when a division x < 31 ∴ –2 < x , 5 by a negative number is∴ x = 29, 30, 31 Answer: B carried out.Answer: CZoom-in (Forms 1–3) P24
28 CSlPoMne 29 CSlPoMne CSlPoMne ’05 ’07 ’09 (Paper 1/Compulsory question/1 mark) (Paper 1/Compulsory question/1 mark)Find all the integer values of a that satisfy the Solve the inequality 6x – 7 , 2 – —32x–.inequality 12 , 5a + 2 < 37. A x , – —190– C x , —65A 2, 3, 4, 5 C 2, 3, 4, 5, 6B 3, 4, 5, 6, 7 D 2, 3, 4, 5, 6, 7 B x , – —32 Solution D x , 2 12 , 5a + 2 < 37 Solution 12 – 2 , 5a < 37 – 2 10 , 5a < 35 6x – 7 , 2 – —32x– —150 , a < —355– 2 , a < 7 12x – 14 , 4 – 3x × 2∴ a = 3, 4, 5, 6, 7 12x + 3x , 4 + 14 15x , 18 x , —65Answer: B Answer: CMultiple-choice Questions Paper 1 6 Find all the integer values of k that 11 Find all the integer values of q that 1 List all the integer values of p for satisfy the inequality CSlPoMne satisfy the inequality ’05 2q – 1 , 17 < 4q – 3.CSlPoMne which 16p . 96 and 10 , 3k – 2 < 19.’06 7 – 12p > –101. A 6, 7, 8 C 6, 7, 8, 9 A 5, 6 C 5, 6, 7 B 5, 6, 7, 8 D 5, 6, 7, 8, 9 A 7, 8, 9, 10 C 6, 7, 8 B 6, 7, 8, 9 D 7, 8, 9 B 4, 5, 6 D 4, 5, 6, 7 7 Given that m is an integer and 12 Find the smallest integer k which 2 Find all the integer values of x that 0 , m2 < 45, find the greatest satisfies the inequality satisfy the inequality value of m, such that m2 is a perfect –10 , 3k – 7 , –1. 4x – 5 , 51 < 6x – 9. square. A –1 C 1 A 10, 11, 12 B 11, 12, 13 A 36 C 6 B 0 D 2 C 10, 11, 12, 13 D 10, 11, 12, 13, 14 B 25 D 5 13 Which of the following is 3 Find all the integer values of m that 8 List all the integer values of y for CSlPoMne equivalent to 7 + —4x . 5 + 2x? satisfy the inequality which –8 < y , –3 and 3 , 2y + 15. ’07 A —72x– . 4 C —–4x– . 4 – x A –5, –4 19 < 12m – 5 , 67. B –6, –5, –4 B —74x– . 4 – x D —74x– , 2 A 2, 3, 4, 5, 6 C 3, 4, 5, 6 C –6, –5, –4, –3 B 2, 3, 4, 5 D 4, 5, 6 D –8, –7, –6, –5, –4 4 Find all the integer values of x that 9 Find the smallest prime number p 14 Which of the following inequalities, satisfy the inequality for which 2p + 1 > 21. I 2x . 10 –11 < 5 – 2x < – 5. A 19 C 13 II x + 2 > 7 A 5, 6, 7 C –8, –7, –6 III x – 9 . –5 B 5, 6, 7, 8 D –8, –7, –6, –5 B 17 D 11 has / have the possible solution 5 Given that x < 12—14 , find the 10 List all the integer values of y for x = 5, 6, 7, ...? largest possible value of x if x is a A III only CSlPoMne which 7y , 61 and 25 – 7y < –3. B I and II onlyprime number. ’11 A 5, 6, 7 C II and III only D I, II and IIIA 7 C 12 B 4, 5, 6, 7 C 5, 6, 7, 8B 11 D 13 D 4, 5, 6, 7, 8 P25 Zoom-in (Forms 1–3)
15 5, 6 and 7 are the common values 17 The number line representing the 18 Find all the integer values of x that of CSlPoMne solution of –1 < 3p – 7 , 17 is A p > 5 and p , 8 ’08 A CSlPoMne satisfy both of the inequalities B p . 5 and p , 8 ’10 3x < 15 and 19 – 4x , 11. C p . 5 and p < 8 −1 17 D p > 5 and p < 8 A 4, 5, B 8 B 2, 3, 416 C 3, 4, 5 2 D 2, 3, 4, 5 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 C 19 The solution of 4 – 7x > 8 + 2x is Which of the following simultane- ous inequalities is represented by 2 8 CSlPoMne A x > – —49 the above number line? A –4 , q , 4 ’09 B –4 < q and q . 4 C –4 , q < 4 D 24 B x < – —49 D –4 , q and q > 4 6 C x > —49 D x < —49 Answers: 1 D 2 C 3 B 4 B 5 B 6 C 7 C 8 A 9 D 10 D 1 1 B 12 B 13 D 14 C 15 A 16 C 17 C 18 C 19 B K Solid Geometry II, III – Surface Areas Solids Formulae 5 VVo=lu—13me×, (Base area) × hform 2,3 and Volumes of Solids Total surface area, h A = Base area + Sum ofSolids Formulae Pyramid areas of all slanting1 Volume, V = l3 6 faces l Surface area, A = 6l2 hl Volume, V = —31 πr2h l Volume, V = l × b × h r Curved surface area l Surface area, = πrl A = 2(lb + lh + bh) Cone Total surface area, A Cube 7 = πrl + πr22 Volume, V = πr2h Curved surface area r Volume, V = —34 πr3 h = 2πrh Surface area, A = 4πr2 b Total surface area Sphere l = 2πrh + 2πr2 8 Volume, V = —23 πr3 Total surface area, Cuboid Volume, V r A = 2πr2 + πr23r =Area of cross-section × l = 3πr2 Total surface area, Hemisphere h A = (Perimeter of cross- Cylinder section) × l + 2(Area4 of cross-section) lArea ofcross-section PrismZoom-in (Forms 1–3) P26
30 CSlPoMne CSlPoMne ’09 ’10 (Paper 2/Compulsory question/4 marks)The diagram shows a solid = —17–32—48– cm3hemisphere. A solid cone is Volume of the cone = —13 πr2hhollowed out.The radius of the hemisphere is = —31 × —272– × 92 × 714 cm. The radius and height of = —17–38–2– cm3the cone are 9 cm and 7 cm respectively. Volume of the remaining solid = —17–32—48– – —17–38–2–Using π = —272–, calculate the volume, in cm3, of theremaining solid.Solution hemisphere = —32 πr3 = —15–34—66–Volume of = —23 × —272– × 143 = 5155—31 cm3 31 CSlPoMne CSlPoMne ’08 ’11 (Paper 2/Compulsory question/4 marks)The diagram shows a composite solid formed by Solutionthe combination of a right prism and a half circular (a) Volume of half circular cylinder Radiuscylinder at the rectangular plane PQUT. Right- = —12 πr2h = —Dia—m2—et—er —21 × —272– = —14—2cm—angled triangle SUT T S = × 7 × 7 × 12 = 924 cm3 = 7 cmis the uniform cross- V 12 cmsection of the prism. U R (b) Volume of right prism WP = Volume of s olid – Volume of half Q composite circular cylinderThe diameter of the half circular cylinder is 14 cm and = 2436 – 924 = 1512 cm3the volume of the composite solid is 2436 cm3. Using —12 × TU × US × SR = 1512 cm3π = —272–, calculate(a) the volume, in cm3, of the half circular cylinder, —21 × 14 × US × 12 = 1512 cm3 US = 18 cm(b) the length, in cm, of US.Subjective Questions Paper 2 2 The diagram shows four identical 3 The diagram shows a solid formed spheres with a radius of 4 cm in a CSlPoMne by combining a cylinder and a 1 The diagram cuboid. ’07 cuboid. The height of the cylindershows a solid X 7 cm 16 cm is 8 cm and its diameter is 7 cm.comprising 8 cm EFtwo cones, 16 cmX and Y. 6 cm Y Calculate the volume, in cm3, ofThe height the empty space. [Use π = 3.142] Dratio of X to Y is 2 : 1. H G C 9 cmCalculate the volume, in cm3, of 20 cm3 4the solid. Use π = —272– A 10 cm B P27 Zoom-in (Forms 1–3)