PELANGI BESTSELLER
MATHEMATICS SPM
FORM
4∙5
KSSM
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Official Publish D ual L anguage
P rogramme
m5Textbook
Ng Seng How er NEW SPM ASSESSMENT
Ooi Soo Huat KSSM
Samantha Neo Ma FORMAT 2021
Yong Kuan Yeoh thematics For
CONTENTS
Mathematical Formulae iv 6Chapter Linear Inequalities in Two
Form 4 Variables 85
1Chapter Quadratic Functions and
Equations in One Variable 1
1.1 Quadratic Functions and Equations 2
SPM Practice 1 13
Penerbitan Pelangi Sdn Bhd. All Rights Reserved6.1 Linear Inequalities in Two Variables 86
6.2 System of Linear Inequalities in Two
Variables 91
SPM Practice 6 97
7Chapter
Graphs of Motion 101
2Chapter 7.1 Distance-Time Graphs 102
7.2 Speed-Time Graphs 106
Number Bases 16
2.1 Number Bases 17 SPM Practice 7 113
SPM Practice 2 24
8Chapter Measures of Dispersion for
3Chapter
Ungrouped Data 118
Logical Reasoning 26
8.1 Dispersion 119
8.2 Measures of Dispersion 121
SPM Practice 8 135
3.1 Statements 27
3.2 Argument 36 9Chapter Probability of Combined
SPM Practice 3 45 Events 139
4Chapter 9.1 Combined Events 140
9.2 Dependent Events and
Operations on Sets 48 Independent Events 141
4.1 Intersection of Sets 49 9.3 Mutually Exclusive Events and
4.2 Union of Sets 53 Non-Mutually Exclusive Events 147
4.3 Combined Operation on Sets 59 9.4 Application of Probability of
SPM Practice 4 64 Combined Events 154
SPM Practice 9 157
5Chapter Network in Graph Theory 69 1 0Chapter Consumer Mathematics: 161
Financial Management
5.1 Network 70
SPM Practice 5 81 10.1 Financial Planning and
Management 162
SPM Practice 10 172
ii
Form 5 6Chapter Ratios and Graphs of
1Chapter Trigonometric Functions 260
Variation 175 6.1 The Value of Sine, Cosine and Tangent
for Angle q, 0° < q < 360° 261
1.1 Direct Variation 176 6.2 The Graphs of Sine, Cosine and
1.2 Inverse Variation 181 Tangent Functions 270
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
1.3 Combined Variation 185 SPM Practice 6 277
SPM Practice 1 188 7Chapter Measures of Dispersion for
2Chapter Grouped Data 283
Matrices 191 7.1 Dispersion 284
2.1 Matrices 192 7.2 Measures of Dispersion 294
2.2 Basic Operation on Matrices 194
SPM Practice 2 206 SPM Practice 7 301
3Chapter Consumer Mathematics: 8Chapter
Insurance 209 Mathematical Modeling 305
8.1 Mathematical Modeling 306
3.1 Risk and Insurance Coverage 210 SPM Practice 8 316
SPM Practice 3 219 SPM Model Paper 318
4Chapter Consumer Mathematics: Answers 335
Taxation 222
4.1 Taxation 223
SPM Practice 4 232
5Chapter Congruency, Enlargement and
Combined Transformations 234
5.1 Congruency 235
5.2 Enlargement 239
5.3 Combined Transformation 247
5.4 Tessellation 254
SPM Practice 5 257
iii
4Chapter Learning Area: Discrete Mathematics
Form 4
Operations on Sets
Penerbitan Pelangi Sdn Bhd. All Rights ReservedWhere should number 2 be written on a Venn diagram to
show the relation between set of prime numbers and set of
even numbers?
• Complement of a set – Pelengkap bagi suatu set Concept
• Intersection of sets – Persilangan set map
• Sets – Set
• Union of sets – Kesatuan set
Iatnu•ro noVdSueecexnnirpdenlntoduhcrisaeeeg,cgaraarlnimonvduin–bpmgesGaacnotmlhifapibsnuesaglvirafsetrinaeeadj.anrhTeauhVcmecceblonanenrsurdssmii,nfbirgeeedartol2antciushcmeoaibrrdepcrirnosim,gmewmthnooounlmetphbnreeouirrp.meTcbrotheirimesssmnosouronmrtabhpteairrootncptaaehlnretnyiaeuslams.robeMebmaresnu.cyclhatsehsaiinfsigieesdr
48
4.1 Intersection of Sets Mathematics SPM Chapter 4 Operations on Sets
(c) If B A, then A B = B.
ξ
BA
A Determining and describing the
intersection of sets
1. The intersection of two sets A and B is a set
where the elements of the set are the common
elements of set A and set B. The intersection 5. The intersection of three sets A, B and C,
of sets is denoted using the symbol . The A B C, is a set where the elements of the
intersection of set A and set B is written as set are the common elements of set A, set B and
A B. also set C.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 4 For example, A = {1, 2, 3, 4, 5} 6. The intersection of three sets can be represented
B = {3, 4, 5, 6, 7} using a Venn diagram.
Therefore, A B = {3, 4, 5} ξA
2. The intersection of two sets can be represented C B The shaded region
using a Venn diagram. represents set
ξ A B C.
A
B
7. The intersection of three sets obeys the
The shaded region
represents set associative law, (A B) C = A (B C),
A B. for any three sets.
1
3. The intersection of two sets obeys the
Given the universal set,
commutative law, A B = B A, for any two = {x : x is a whole number not more than 15},
sets A and B. A = {x : x is a whole number less than 10} and
B = {x : x is an odd number}.
4. Given set A and set B, observe the following
cases. (a) Determine set A B by using
(a) If A B and B A, then A B A and (i) descriptions,
A B B. (ii) listing,
(iii) set builder notation.
ξ B
A
(b) Represent set A B using a Venn diagram.
Solution
(b) If A B, then A B = A. (a) (i) A B is a set of odd numbers less than 10.
ξ (ii) A = {1, 2, 3, 4, 5, 6, 7, 8, 9} List the common
AB B = {1, 3, 5, 7, 9, 11, 13, 15} elements of
A B = {1, 3, 5, 7, 9} set A and set B.
(iii) A B = {x : x is an odd number and
x 10}
49
Mathematics SPM Chapter 4 Operations on Sets
(b) ξ A B Solution
(a) P Q = {–2, –1, 3, 4, 5} {–1, 0, 1, 2, 3, 4, 5}
•2 •1 • 11 • 10 = {–1, 3, 4, 5}
•4 •3 • 13 • 12
•5 • 15 • 14 (b) P Q R = {–1, 3, 4, 5} {–2, –1, 0, 2, 4}
•6 •7 = {–1, 4}
•8 •9
4
The Venn diagram below shows the universal set,
and three sets P, Q and R.
2Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Given the universal set, ξP Q
= {x : x is a colour of rainbow},
Form 4 A = {Blue, Green, Yellow, Red, Orange} and •d ••me •j •b
B = {Yellow, Indigo, Green}. •g •l
•h •f •a
(a) List all the elements of set A B and represent
set A B using a Venn diagram. •n
•k
(b) Given C = {Blue, Red, Yellow} •c R
D = {Red, Purple}
Determine the set List the elements of set
(i) D C, (a) P Q,
(ii) A C and state the relation between set A (b) Q R,
(c) P Q R.
and set C. Solution
(a) P Q = {e, m, f }
Solution
(a) A B = {Yellow, Green} ξP •d ••me Q
•g
ξA •j •b
•l
B • h •f •a
•k •n
•c R
• Red • Yellow • Indigo (b) Q R = {f, a}
• Blue • Green
• Orange
ξP •d ••me Q
•g
• Purple •j •b
(b) (i) D C = {Red} •l
(ii) A C = {Blue, Red, Yellow} • h •f •a
= C •k •n
Therefore, C A.
•c R
(c) P Q R = {f }
ξP •d ••me Q
•g
3 •j •b
Given the universal set,
= {x : x is an integer and –5 x 5}, • h •f •a •l
P = {–2, –1, 3, 4, 5},
Q = {–1, 0, 1, 2, 3, 4, 5} and •k •n
R = {–2, –1, 0, 2, 4}. •c R
Determine the set
(a) P Q, Try Questions 1 – 6 in Try This! 4.1
(b) P Q R.
B Determining the complement of the
intersection of sets
1. The complement of set A B is a set where
the elements of the set are all elements in the
universal set which are not the elements of A B.
50
2. The complement of set A B is written as Mathematics SPM Chapter 4 Operations on Sets
(A B)ʹ. The set (A B)ʹ can be represented by
using a Venn diagram. Solution
(a) n(B) = 9 + 8
= 17
ξA B ξ B
A
The shaded region 6 89
represents set
(A B)ʹ.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved 5
Form 4
(b) n(A B)ʹ = 6 + 9 + 5
= 20
5
Given the universal set, ξ B
= {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}, A
P = {1, 5, 9, 15, 19},
Q = {3, 5, 9, 11, 13, 17, 19} and 6 89
R = {7, 9, 11, 15, 17}.
5
Determine the set
(a) (P Q)ʹ,
(b) (Q R)ʹ.
Try Questions 7 – 9 in Try This! 4.1
Solution C Solving problems involving the
(a) P Q = {1, 5, 9, 15, 19} {3, 5, 9, 11, 13, 17, 19} intersection of sets
= {5, 9, 19}
(P Q)ʹ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
= {1, 3, 7, 11, 13, 15, 17}
7
Remove the elements in the universal set The Venn diagram below shows the days Ali performed
which are elements of set P Q. two sports activities in a week. Set A represents cycling
activity and set B represents running activity.
(b) Q R = {3, 5, 9, 11, 13, 17, 19} {7, 9, 11, 15, 17}
= {9, 11, 17}
ξ B
(Q R)ʹ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} A
= {1, 3, 5, 7, 13, 15, 19} • Sunday
• Monday
• Friday • Saturday
• Wednesday
6 • Thursday
The Venn diagram below shows the number of • Tuesday
elements in the universal set and two sets A and B.
(a) How many days in a week does Ali miss cycling
ξA B activity?
689 (b) State the days Ali performed both sports
5 activities.
Determine Solution
(a) n(B), (a) Aʹ = {Tuesday, Thursday, Saturday}
(b) n(A B)ʹ. n(Aʹ) = 3
Therefore, there are 3 days in a week Ali misses
cycling activity.
51
Mathematics SPM Chapter 4 Operations on Sets (a) Determine set P Q by using
(i) descriptions,
(b) A B = {Wednesday, Friday} (ii) listing,
Therefore, Ali performed both sports activities (iii) set builder notation.
on Wednesday and Friday. (b) Represent set P Q using a Venn diagram.
8 2. Given the universal set,
Class 4 Amanah has 35 pupils. 18 pupils are members = {x : x is a day in a week},
of Mathematics Society and 22 pupils are members of A = {Sunday, Thursday, Friday, Saturday} and
English Language Society. Given that each pupil is the B = {Thursday, Monday, Tuesday, Wednesday,
member of at least one of these two societies. Find the
number of pupils who are members of
(a) both societies,
(b) only one society.
Solution
Let P = {Members of Mathematics Society}
Q = {Members of English Language Society}
x = Number of pupils who join both societies.
Therefore, n(P Q) = x
PQ
18 – x x 22 – x
Penerbitan Pelangi Sdn Bhd. All Rights Reserved Saturday}.
Form 4 (a) List all the elements of set A B and represent
set A B using a Venn diagram.
(b) Given C = {Thursday, Friday, Saturday}
D = {Friday, Saturday}
Determine the set
(i) D C,
(ii) A C.
(c) Determine the relation between set A C and
set D C.
(a) Therefore, (18 – x) + x + (22 – x) = 35 3. Given the universal set,
= {x : x is an integer and –10 x 3},
P = {–9, –7, –5, 0, 1},
Q = {–8, –7, 0, 1, 2, 3} and
R = {–6, –7, 1, 2}.
Determine the set
(a) P Q,
(b) Q R,
(c) P Q R.
40 – x = 35 4. Given the universal set,
x = 5 = {a, b, c, d, e, f, g, h, i, j, k, l },
n(P Q) = 5 A = {a, c, e, f, h},
B = {a, e, f, g, j, k, l},
Therefore, there are 5 pupils who join both C = {b, d, g, i, k} and
societies. D = {a, e, f, k}.
(b) n(P Q)ʹ = (18 – x) + (22 – x) Determine the set (a) A B,
(b) A C,
n(P Q)ʹ = 40 – 2x (c) A B D.
= 40 – 2(5) 5. ξP
= 30
•1 •5
Therefore, there are 30 pupils who join only one • 11 Q
society.
•2 •4 •3 • 10 • 15
•6 •7
Try Questions 10 – 11 in Try This! 4.1 • 8 • 12 • 14
R
• 9 • 13
Try This! 4.1 The Venn diagram above shows the universal set,
and three sets, P, Q and R. List the elements of set
1. Given the universal set, (a) P Q,
= {x : x is an integer and 70 x 90}, (b) Q R,
P = {x : x is a prime number} and (c) P R,
Q = {x : x 80}. (d) P Q R.
52
6. In each of the following Venn diagrams, shade the Mathematics SPM Chapter 4 Operations on Sets
stated region.
(a) P Q (b) (P Q R)ʹ
ξP ξ
Q
P
Q
R
R
Q 10.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved ξA
Form 4(b) Q RB
ξP • Grape • Papaya • Kiwi
• Mango
• Watermelon • Orange • Guava
• Apple • Banana
R The Venn diagram above shows the types of fruits
sold by two stalls A and B.
(a) Which fruits are sold by both stalls?
(b) State the fruits sold by only one of the stalls.
7. Given the universal set,
= {x : 12 x 25}, 11. In a survey about the types of vehicles owned by
P = {14, 15, 16, 17, 23, 25}, 40 families in a housing area, it was found that
Q = {12, 14, 15, 16, 17, 23, 25} and 25 families owned cars and 28 families owned
R = {14, 15, 16, 17, 18, 19, 20, 21, 23}. motorcycles. Given that each family owned at least
Determine the set one type of vehicle.
(a) (P Q)ʹ, (a) Using
(b) (Q R)ʹ. K = {Families that owned cars} and
M = {Families that owned motorcycles}
8. ξ B represent the information above using a Venn
A C
diagram
7 12 5 9 11
(b) Find the number of families that
3 (i) owned both types of vehicles,
(ii) owned only one type of vehicle.
The Venn diagram above shows the number of 4.2 Union of Sets
elements in the universal set, , and three sets, A, B
and C. Determine A Determining and describing the union
(a) n(A B)ʹ, of sets
(b) n(B C)ʹ,
(c) n(A C)ʹ.
9. In each of the following Venn diagrams, shade the 1. The union of two sets A and B is a set where the
stated region. elements of the set are the elements of set A or set
(a) (Q R)ʹ
ξ Q B. The union of sets is represented by the symbol
P . The union of set A and set B is written as
A B.
For example, A = {1, 2, 3, 4, 5}
B = {3, 4, 5, 6, 7} The
R common
Therefore, A B = {1, 2, 3, 4, 5, 6, 7} elements
are written
only once.
53
Mathematics SPM Chapter 4 Operations on Sets
2. The union of two sets can be represented using 5. The union of three sets A, B and C, A B C,
a Venn diagram. The shaded region in the Venn is a set where the elements of the set are elements
diagram below represents set A B. of set A or set B or set C.
ξ B 6. The union of three sets can be represented using
A a Venn diagram. The shaded region in the Venn
diagram below represents set A B C.
ξA
BC
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 4 For the case A B ≠ φ
ξ B
A 7. The union of three sets obeys the associative
law, (A B) C = A (B C), for any three
sets, A, B and C.
9
For the case A B = φ Given the universal set,
= {x : x is a whole number and 5 x 20},
3. The union of two sets obeys the commutative P = {16, 17, 18, 19, 20} and
law, A B = B A for any two sets A and B. Q = {x : x is an even number}.
(a) Determine set P Q by using
4. Given set A and set B, observe the following (i) descriptions,
cases. (ii) listing,
(iii) set builder notation.
(a) If A B and B A, then A A B and (b) Represent set P Q using a Venn diagram.
B A B.
Solution
ξA B (a) (i) P Q is a set of even numbers or numbers
not less than 16.
(b) If A B, then A B = B. (ii) P = {16, 17, 18, 19, 20}
Q = {6, 8, 10, 12, 14, 16, 18, 20}
ξ P Q = {6, 8, 10, 12, 14, 16, 17, 18, 19, 20}
AB
Combine the elements of set P and set Q. The common
elements (elements of P Q) are written only once.
(c) If B A, then A B = A.
(iii) P Q = {x : x is an even number or x 16}
ξ BA
(b) ξ PQ
• 17 • 16 •6 •5 Observe that
• 19 • 18 •8 •7 PQ
• 20 • 10 •9 = {16, 18, 20}.
• 12 • 11
• 14 • 13
• 15
54
Mathematics SPM Chapter 4 Operations on Sets
10 (a) P Q
(b) Q R
Given the universal set, (c) P Q R
= {x : x is a state in Peninsular Malaysia}, Solution
A = {x : x is the name of a state that begins with (a) P Q = {a, b, c, d, e, f, h, j, m, n}
letter P} and
B = {x : x is the name of a state that begins with letter ξ P •d Q
K}. •e •f
(a) List all elements of set A B and represent set •m
A B using a Venn diagram. •b • j
(b) State the relationship between A B and A.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved •h •a •n
Form 4•c
•k
•l •i •g R
Solution
(a) A = {Perlis, Penang, Perak, Pahang} (b) Q R = {a, c, f, g, h, i, j, l, m, n}
B = {Kedah, Kelantan}
A B = {Perlis, Penang, Perak, Pahang, Kedah, ξ P Q
•f
Kelantan} • •d •m
e
ξ • Johor • Selangor •b • j
A • Perlis B
•h •a •n
•c
• Penang • Kedah •k
• Pahang • Kelantan •l •i •g R
• Perak Observe that
AB=φ
• N.Sembilan • Melaka
• Terengganu
(b) A A B (c) P Q R = {a, b, c, d, e, f, g, h, i, j, l, m, n}
ξ P •d Q
•e •f
11 •m
Given the universal set, •b • j
= {a, b, c, d, e, f, g, h, i, j, k, l},
P = {b, e, h, i, j}, •h •a •n
Q = {c, d, g, h, e, k} and •c
R = {e, f, h, k, l}. •k
Determine the set •l •i •g R
(a) P Q,
(b) P Q R. Try Questions 1 – 5 in Try This! 4.2
B Determining the complement of the
union of sets
Solution 1. The complement of set A B is a set where
(a) P Q = {b, e, h, i, j} {c, d, g, h, e, k} the elements of the set are all the elements in
= {b, c, d, e, g, h, i, j, k} the universal set which are not the elements of
A B.
(b) P Q R
= {b, c, d, e, g, h, i, j, k} {e, f, h, k, l} 2. The complement of set A B is written as
= {b, c, d, e, f, g, h, i, j, k, l} (A B)ʹ. Set (A B)ʹ can be represented using
a Venn diagram. The shaded region in the Venn
12 diagram below represents set (A B)ʹ.
ξ P •d Q
•e •f
•b •m ξ B
•j A
•h •a •n
•c
•k
•l •i •g R
The Venn diagram above shows the universal set, For the case A B ≠ φ
and three sets, P, Q and R. List the elements of set
55
Mathematics SPM Chapter 4 Operations on Sets
ξ B Solution
A (a) n(Aʹ) = 15 + 6
= 21
ξ B
A
10 5 15
6
For the case A B = φ
Form 4 Penerbitan Pelangi Sdn Bhd. All Rights Reserved13 (b) n(A B)ʹ = 6 B
Given the universal set,
= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}, ξ
P = {2, 6, 12, 16, 20}, A
Q = {4, 8, 10, 12, 16, 22, 24} and
R = {2, 6, 14, 18}. 10 5 15
Determine the set 6
(a) (P Q)ʹ,
(b) (Q R)ʹ.
Solution Try Questions 6 – 8 in Try This! 4.2
(a) P Q = {2, 6, 12, 16, 20} {4, 8, 10, 12, 16, 22, 24}
= {2, 4, 6, 8, 10, 12, 16, 20, 22, 24}
C Solving problems involving the union
(P Q)ʹ = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24} of sets
= {14, 18}
Remove the elements in 15
the universal set which are
the elements of set P Q.
(b) Q R = {4, 8, 10, 12, 16, 22, 24} {2, 6, 14, 18} ξA B
= {2, 4, 6, 8, 10, 12, 14, 16, 18, 22, 24} • Susita •Devi • May Lin
• Hua An •Jefri • Sabri
(Q R)ʹ = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24} • Rajoo
= {20} • Firdaus • Subra • Anjung
14 B The Venn diagram above shows the participation
of ten pupils in two contests. Set A represents the
ξ pupils who participated in drawing contest and set
A B represents the pupils who participated in essay
writing contest.
10 5 15 (a) State the pupils who did not participate in both
6 contests.
(b) State the number of pupils who participated in
The Venn diagram above shows the number of
elements in the universal set, and two sets A and B. one contest only.
Determine
(a) n(Aʹ) Solution
(b) n(A B)ʹ (a) (A B)ʹ = {Firdaus, Subra, Anjung}
Therefore, pupils who did not participate in both
contests are Firdaus, Subra and Anjung.
56
Mathematics SPM Chapter 4 Operations on Sets
(b) A B = {Susita, Hua An, Rajoo, Jefri, Devi, (ii) n(P Q) – n(P Q) = 42 – 18
May Lin, Sabri} = 24
A B = {Jefri, Devi} Therefore, 24 customers like one type of
Let P = {Pupils who participate in one contest food only.
only} Try Questions 9 – 10 in Try This! 4.2
ξA B
Try This! 4.2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 4 1. Given the universal set,
= {a, c, e, f, i, j, k, l, m, n, o, p},
n(P) = n(A B) – n(A B) A = {k, e, c, i, l } and
= 7 – 2 B = {x : x is a vowel}.
= 5
Therefore, there are 5 pupils who participated in (a) Determine set A B by using
(i) descriptions,
one contest only. (ii) listing,
(iii) set builder notation.
(b) Represent set A B using a Venn diagram.
16 2. Given the universal set,
= {x : x is a regular polygon that has less than 10
A food stall only sells nasi lemak and roti canai. A
survey was carried out on 50 customers about their sides},
favourite food. The result of the survey showed that A = {square, regular pentagon, regular hexagon}
32 customers like nasi lemak, 28 customers like roti
canai and 18 customers like both types of food. and
(a) Represent the given information using a Venn B = {x : x is a regular polygon that has not more
diagram. than 7 sides}.
(b) Hence, find the number of customers who (a) List the elements of set A B and represent
(i) do not like both types of food.
(ii) like one type of food only. the set A B using a Venn diagram.
(b) State the relationship between set A B and
A.
Solution 3. Given the universal set,
Let P = {Customers who like nasi lemak} = {x : x is a positive integer not more than 18},
Q = {Customers who like roti canai} P = {2, 4, 7, 9, 10},
Therefore, n() = 50 Q = {1, 2, 3, 4, 5, 6, 7, 8} and
n(P) = 32 R = {6, 9, 12, 17, 18}.
n(Q) = 28 Determine the set
n(P Q) = 18
(a) P Q,
(b) Q R,
(c) P Q R.
4. ξP
(a) ξ P Q •r •p •q •n
•x •w
•u •y •z Q
•m •s •t •v
14 18 10
R
8 The Venn diagram above shows the universal set,
(b) (i) n(P Q)ʹ = 8 and three sets, P, Q and R. List the elements of the
Therefore, 8 customers do not like both set
(a) P Q,
types of food. (b) Q R,
(c) P Q R.
57
Mathematics SPM Chapter 4 Operations on Sets (b) (P Q)ʹ
5. In each of the following Venn diagrams, shade the ξQ
stated region.
(a) P Q P
ξ
P
Q R
Penerbitan Pelangi Sdn Bhd. All Rights Reserved(b) Q RR 9. ξ A • Nleamsai k B
Q • Fried • Tom yam
Form 4 ξ • Soup
P noodle
Chicken •Fried
•Curry egg
noodle •Fried
rice
• Fried noodle
• Burger C
R The Venn diagram above shows the types of food
sold by three food stalls, A, B and C.
(a) State the types of food sold by stall A or stall B.
(b) State the types of food which are not sold by
6. Given the universal set, stall B or stall C.
= {x : 10 x 20 and x is an integer},
P = {10, 12, 17, 18, 19}, 10. A survey was carried out to study the favourite
Q = {10, 11, 12, 17, 20} and subjects of 50 pupils. The result of the survey
R = {10, 12, 14, 16, 18, 20}. showed that 25 pupils like Science, 29 pupils like
Determine the set History, 12 pupils like Mathematics and Science,
13 pupils like Mathematics and History, 4 pupils like
(a) (P Q)ʹ, Science only and 5 pupils like all three subjects.
(b) (P R)ʹ, Given that there are no pupils who dislike all three
(c) (Q R)ʹ. subjects.
(a) By using
7. ξ M = {Pupils who like Mathematics}
S = {Pupils who like Science} and
A 8 C J = {Pupils who like History},
5 B
12 represent the given information using a Venn
36 10 diagram.
8 (b) Hence, find the number of pupils who
(i) like Mathematics or Science,
The Venn diagram above shows the number of (ii) like History only,
elements in the universal set, and three sets, A, B (iii) dislike History or Science.
and C. Determine
(a) n(A B)ʹ,
(b) n(A C)ʹ,
(c) n(B C)ʹ.
8. In each of the following Venn diagrams, shade the
stated region.
(a) (Q R)ʹ
ξ Q
P
R
58
Mathematics SPM Chapter 4 Operations on Sets
4.3 Combined Operations on Sets (b) (P Q) (P R)
ξP Q
A Determining and describing the
combined operations on sets
R
Solution
17 (a) ξ
Given the universal set,
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12},
P = {1, 2, 3, 4, 5, 6},
Q = {4, 5, 6, 7, 8, 9} and
R = {7, 8, 9, 10, 11}.
Determine the set
(a) (P Q) R,
(b) (P Q) R,
(c) (P Q) (Q R).
Penerbitan Pelangi Sdn Bhd. All Rights Reserved P
Form 4
Q R
ξP
ξP
Solution Q RQ R
(a) (P Q) = {1, 2, 3, 4, 5, 6} {4, 5, 6, 7, 8, 9}
= {4, 5, 6} Mark all the regions Mark all the regions
of P Q with /. of P R with X.
(P Q) R = {4, 5, 6} {7, 8, 9, 10, 11} Therefore, (P Q) (P R) is the regions
= {4, 5, 6, 7, 8, 9, 10, 11} that contain / or X.
(b) (P Q) = {1, 2, 3, 4, 5, 6} {4, 5, 6, 7, 8, 9} (b) ξ
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
P Q
(P Q) R = {1, 2, 3, 4, 5, 6, 7, 8, 9}
{7, 8, 9, 10, 11}
= {7, 8, 9}
(c) (P Q) = {4, 5, 6} R
(Q R) = {4, 5, 6, 7, 8, 9} {7, 8, 9, 10, 11} ξP Q ξP Q
= {7, 8, 9}
(P Q) (Q R) = {4, 5, 6} {7, 8, 9} RR
= {4, 5, 6, 7, 8, 9}
Mark all the regions Mark all the regions
of P Q with /. of P R with X.
18 Therefore, (P Q) (P R) is the regions that
Shade the region that represents the given combined contain / and X.
set in the following Venn diagram.
(a) (P Q) (P R) SPM Tips
ξP The combined operations on sets obey the distributive
law where for any three sets, P, Q and R,
Q R (a) P (Q R) = (P Q) (P R)
(b) P (Q R) = (P Q) (P R)
Therefore, in Example 18(a), the shaded region P (Q R)
are the same as (P Q) (P R). Likewise, in Example
18(b), the shaded region of P (Q R) are the same
as (P Q) (P R).
Try Questions 1 – 2 in Try This! 4.3
59
Mathematics SPM Chapter 4 Operations on Sets (b) Pʹ (Qʹ R)
B Determining the complement of ξ
combined operations on sets P
Q
19 R
Given the universal set,
= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24},
P = {2, 6, 12, 16, 20}, Solution
Q = {4, 8, 10, 12, 16, 22, 24} and
R = {2, 6, 14}.
Determine the set
(a) (P Q)ʹ R,
(b) P (Q R)ʹ,
(c) (P R)ʹ (Q R)ʹ.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved (a) ξ P
Form 4
Solution Q R
(a) (P Q) = {2, 6, 12, 16, 20} {4, 8, 10, 12, 16, ξP
22, 24}
= {2, 4, 6, 8, 10, 12, 16, 20, 22, 24}
ξP
(P Q)ʹ = {14, 18}
(P Q)ʹ R = {14, 18} {2, 6, 14} QR QR
= {14}
Mark all the regions of Mark all the regions of
(P R) with /. Q with X.
(b) (Q R) = {4, 8, 10, 12, 16, 22, 24} {2, 6, 14}
= {2, 4, 6, 8, 10, 12, 14, 16, 22, 24} Therefore, (P R) Q is the regions that
contain / or X.
(Q R)ʹ = {18, 20}
P (Q R)ʹ = {2, 6, 12, 16, 20} {18, 20} (b) ξ
= {20}
P
(c) (P R) = {2, 6, 12, 16, 20} {2, 6, 14} Q
= {2, 6, 12, 14, 16, 20}
(P R)ʹ = {4, 8, 10, 18, 22, 24} R
(Q R) = {2, 4, 6, 8, 10, 12, 14, 16, 22, 24}
(Q R)ʹ = {18, 20}
(P R)ʹ (Q R)ʹ = {4, 8, 10, 18, 22, 24 } ξ Q ξ Q
{18, 20} P P R
= {18} R
20 Mark all the regions of Mark all the regions
Shade the region that represents the given combined (Q R) with /. of P with X.
set in the following Venn diagram.
(a) (P R)ʹ Q Therefore, P (Q R) is the regions that
contain / and X.
ξP
Try Questions 3 – 4 in Try This! 4.3
Q R
60
Mathematics SPM Chapter 4 Operations on Sets
C Solving problems involving the 22
combined operations on sets
A survey was carried out on 90 families in a housing
21 area about the newspaper they subscribe. The result
of the survey showed that 35 families subscribe
ξG P newspaper A, 45 families subscribe newspaper B, 36
families subscribe newspaper C, 12 families subscribe
65 9 newspaper A and newspaper B, 18 families subscribe
3 8 newspaper A and newspaper C, 10 families subscribe
newspaper B and newspaper C and 8 families do not
2 subscribe any newspaper.
(a) Use x to represent the number of families that
subscribe all three newspapers, represent all
the information above using a Venn diagram.
Hence, find the value of x.
(b) Based on the Venn diagram, find the number of
families who subscribe
(i) newspaper B only,
(ii) two newspapers only,
(iii) newspaper A but do not subscribe
newspaper C.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved5y
Form 4V
The Venn diagram above shows the number of pupils
in a music class which has 40 pupils. Given
G = {Pupils who play guitar},
P = {Pupils who play piano} and
V = {Pupils who play violin}.
(a) Find the value of y. Solution
(b) Find the number of pupils who play
(i) both guitar and violin, (a) ξ A B
(ii) piano or guitar,
(iii) one type of musical instrument only,
(iv) both guitar and piano but do not play violin.
Solution 5+x 12–x 23+x
(a) 6 + 5 + 9 + 2 + 3 + 8 + 5 + y = 40 18–x x 10–x
38 + y = 40
y = 2 8+x
C 8
(b) (i) Number of pupils who play both guitar and
violin = 2 + 3 (5 + x) + (12 – x) + (23 + x) +
= 5 ξG P (18 – x) + (10 – x) + (8 + x) + 8 + x = 90
65 9 84 + x = 90
23 8
x = 6
V5 2
(ii) Number of pupils who play piano or guitar (b) (i) Number of families who subscribe
=6+5+9+2+3+8 newspaper B only = 23 + x
= 33 ξG P = 23 + 6
65 9 = 29
23 8
V5 2 (ii) Number of families who subscribe two
newspapers only
(iii) Number of pupils who play one type of
musical instrument only = (12 – x) + (10 – x) + (18 – x)
=6+9+5 ξG P = 40 – 3x
= 20 = 40 – 3(6)
65 9 = 22
23 8
V5 2
(iv) Number of pupils who play both guitar and (iii) Number of families who subscribe
piano but do not play violin newspaper A but do not subscribe
= 5 ξG newspaper C
P
= (5 + x) + (12 – x)
65 9 = (5 + 6) + (12 – 6)
23 8 = 17
V5 2
61
Mathematics SPM Chapter 4 Operations on Sets Solution
SPM Highlights Based on the Venn diagram below,
x = n(R) – 8
(a) Given that set P = {Even numbers less than 40} P
and Q = {6, 18, 34}. Complete the following Venn = 18 – 8 Q
diagram to show the relation between set P and z y
set Q. = 10
42
P y =n(Q) – 42 – 8
= 54 – 42 – 8 8
=4 x
z =n(P) – y R
= 24 – 4
= 20
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 4 (b) The Venn diagram below shows set P, Q and R Number of pupils who do not like curry mee
= n(Qʹ)
where the universal set, = P Q R. Write a set =x+z
involving set P, Q and R to represent the shaded = 10 + 20
region. = 30
PQ Answer : A
R
Solution SPM Highlights
(a) P •4 •2 Q • 30 The table below shows the favourite colours selected
• 8 • 14 • 28 • 32 by 40 pupils.
• 10 • 16 •6 • 26 • 36 Observe that
• 18 Q , P. Colour Number of pupils
• 34
Qʹ R.
• 12 • 20 • 22 • 24 R Blue 26
(b) P Q Red 23
Green 25
Red and green 16
Blue and red 14
Blue only 8
The set representing the shaded region is Red only 4
P (Qʹ R).
Given that set B = {Pupils who select blue colour},
set M = {Pupils who select red colour} and set
H = {Pupils who select green colour}. Which of the
SPM Highlights following Venn diagrams represents the information in
P the table?
Q A C
BB
42 88
8 34 25
11 11
4 55 HM 4 65 H
M
R B
D 8
The diagram above shows a Venn diagram with three B
sets, P, Q and R. Given the universal set, = P Q R,
set P = {Pupils who like laksa}, set Q = {Pupils who B
like curry mee} and set R = {Pupils who like roti canai}.
Given n(P) = 24, n(Q) = 54 and n(R) = 1. Find the 8
number of pupils who do not like curry mee. 33 35
11 11
A 30 C 22 4 74 4 67
M H M H
B 26 D 20
62
Solution Mathematics SPM Chapter 4 Operations on Sets
Based on the given information, it is found that
n(B) = 26 2. Shade the region that represents the given combined
n(M) = 23 operation on sets in the following Venn diagram.
n(H) = 25 (a) P (Q R)
n(M H) = 16
n(B M) = 14 ξP
n(M H)’ = 8
n(B H)’ = 4 QR
(b) (Q R) (P R)
Penerbitan Pelangi Sdn Bhd. All Rights ReservedLet n(B M H) = xB
Form 4The information in the table8ξQ
is as shown in the Venn P
diagram on the right. 14 – x
x
4 16 – x H R
M
Therefore, 4 + (14 – x) + x + (16 – x) = n(M)
= 23 3. Given the universal set,
34 – x = 23 = {a, b, c, d, e, f, g, h, i, j, k, l, m, n},
x = 11 P = {a, c, e, h, j, k, m},
Q = {b, c, e, f, g, k, m, n} and
The Venn diagram can be B R = {f, g, h, k, l}.
updated as in the diagram 8
on the right. Determine the set
3y (a) (Rʹ P) Q,
Therefore, M 11 H (b) (Q R)ʹ P,
8 + 3 + 11 + y = n(B) (c) R (P R)ʹ,
= 26 4 5z (d) (P Q)ʹ (Q R)ʹ.
y = 4
4. Shade the region that represents the given combined
Hence, 11 + 5 + y + z = n(H) operation on sets in the given Venn diagram.
= 25 (a) (Q R)ʹ P
11 + 5 + 4 + z = 25
z = 5 ξ
P
Therefore, the complete B Q R
Venn diagram is shown in 8
the diagram on the right.
34 (b) Rʹ (P Q)
11 Q
ξ R
4 55 P
M H
Answer : A
Try Questions 5 – 6 in Try This! 4.3
5.
Try This! 4.3 ξM B
7 4 8
4 5
1. Given the universal set,
= {x : 30 x 50 and x is an integer}, 9
P = {32, 36, 39, 40, 45, 47, 49},
Q = {x : x is an odd number} and Ex 5
R = {x : x is a multiple of 3}.
Determine the set The Venn diagram above shows the number of
pupils in a class who passed in three subjects in an
(a) (P R) Q, examination where
(b) (Q R) P,
(c) (P Q) (Q R).
63
Mathematics SPM Chapter 4 Operations on Sets who master Chinese and Bahasa Melayu only, 5
pupils who master English and Chinese only,
M = {Pupils who passed Mathematics} 7 pupils who master Bahasa Melayu and English
B = {Pupils who passed Bahasa Melayu} only, 4 pupils who master all three languages.
E = {Pupils who passed English} (a) By using
M = {Pupils who master Bahasa Melayu},
(a) Given that there are 20 pupils who passed E = {Pupils who master English} and
English, find the value of x. C = {Pupils who master Chinese},
represent all the information above using a
(b) Find the number of pupils who passed
(i) Mathematics, Venn diagram.
(ii) Mathematics and Bahasa Melayu, (b) Based on the Venn diagram, find the number of
(iii) any two subjects only,
(iv) Mathematics or English but failed Bahasa pupils who master
Melayu. (i) Chinese only,
(ii) any two languages only,
6. In a survey about the languages mastered by 40 (iii) any one language only,
pupils, it was found that there were 25 pupils who (iv) English and Bahasa Melayu only.
master Bahasa Melayu, 23 pupils who master
English, 20 pupils who master Chinese, 8 pupils
Form 4 Penerbitan Pelangi Sdn Bhd. All Rights Reserved
SPM Practice 4
PAPER 1 ξ take any of the subject, Physics and Biology is 6.
1. P Find the number of pupils who do not take Physics.
A 56
Q B 62
C 68
In the diagram above, is the universal set. Given D 72
n() = 20, n(P) = 8, n(Q) = 12 and n(P Q) = 6.
Therefore, n(Pʹ) = 4. R S
x+2 x 2x+1
A 4 C 12
B 8 D 16 The Venn diagram above shows the number of
participants of two contests. Given the universal set,
2. Given that = R S, set R = {Participants of calligraphy
writing contest} and set S = {Participants of
Universal set, = {C, O, M, P, U, T, E, R, S}, drawing contest}. Given that there are 17 persons
set P = {M, O, U, S, E} and participated in calligraphy writing contest only, find
set Q = {M, E, P, S }, the total number of participants.
A 59
B 61
C 63
Find n(P Qʹ). D 65
A 1 C 3
B 2 D 4
3. ξ
FB
5. A survey was carried out to study the favourite
SPM
2018 drinks of 50 pupils. The result showed that 32 pupils
like coffee and 28 pupils like tea. Find the number of
The diagram above shows a Venn diagram with pupils who like both coffee and tea.
the universal set, = {Form 5 pupils}, set F =
{Pupils who take Physics} and set B = {Pupils who A 4
take Biology}. Given that n(F) = 34, n(B) = 68,
n(F B) = 12 and the number of pupils who do not B 6
C 8
D 10
64
Mathematics SPM Chapter 4 Operations on Sets
6. The diagram below shows a Venn diagram with Given that n(Q) = n(P R)ʹ, find n().
SPM universal set , set U, set V and set W.
2019 A 62
B 64
jU V C 66
D 68
W
9 9. R P
A
Set U = {Pupils who like badminton} Q
Set V = {Pupils who like tennis} B CD
Set W = {Pupils who like table tennis}
Given that n() = 321, n(U) = 215, n(V) = 85,
n(W) = 55 and n(U V) = 35, find the difference
Penerbitan Pelangi Sdn Bhd. All Rights Reserved The Venn diagram above shows the universal set,
Form 4between the number of pupils who do not like the = P Q R. Which of the regions A, B, C and D,
represents the set (Q R)ʹ Pʹ?
three games and the number of pupils who like
badminton only. 10. P
A 160 Q
B 170
C 180
D 190
7. Given P, Q and R are three different sets. If a R
is an element such that a P, a P Q and
a P R. Which of the following shows the correct The Venn diagram above shows three sets P, Q and
position of a? R. Given the universal set, = P Q R. Which
of the following sets represents the shaded region?
A P Q R A P Q R
B (P Q)ʹ R
a C (P Q)ʹ (P R)
D (P Q) (P R)
B P 11. Given that = {x : 55 < x < 65}
2S 0P1M9 M = {x : x is a multiple of 4}
N = {x : x is a whole number which
aQ gives a remainder of 3 when
R
divided by 4}
State the elements of (M N)ʹ.
C P Q A {55, 59, 63}
R B {57, 61, 62}
a
C {57, 58, 61, 62, 65}
D {57, 59, 60, 61, 63, 65}
D P R 12. Given that set M = {2, 4, 6, 8, 10}, K = {1, 3, 5, 7}
and L = {2, 3, 7}. List all the elements of
a
Q (K L) M.
8. The Venn diagram below shows the number of A {3, 7}
elements in the universal set, , set P, set Q and B {2, 3, 4, 7}
set R. C {2, 4, 6, 8, 10}
D {2, 3, 4, 6, 7, 8, 10}
ξ 13. Which of the following Venn diagrams represents
P the set P Q R?
x–1 A ξ P Q
4x R
2
x+2 7 6
QR
65
Mathematics SPM Chapter 4 Operations on Sets
B ξ P Q 17. S
R R 40 T
C ξ P 6
Q
The diagram above shows a Venn diagram with
three sets R, S and T. Given the universal set,
= R S T, set R = {Pupils who like walking
activity}, S = {Pupils who like running activity}
and set T = {Pupils who like cycling activity}.
Given that n(R) = 26, n(S) = 50 and n(T ) = 20.
Find the number of pupils who do not like running
activity.
A 36
B 20
C 10
D 6
Form 4 Penerbitan Pelangi Sdn Bhd. All Rights ReservedRQ
D ξ P
R
14. P Q R
18. ξ G
M
SPM
2017
The diagram above shows a Venn diagram with the 46
universal set, = P Q R. Which of the following S x 12 5x
80
operations on sets is an empty set?
A (Q R)ʹ P C (Q R)ʹ Q
B (Q R)ʹ R D (Q R)ʹ R
The diagram above shows a Venn diagram with
15. In a survey about electric cooking appliances owned set M = {Members of Mathematics club}, set S =
{Members of History club} and set G = {Members
by 500 families, it was found that 36% of the families
of Geography club}. Given the number of members
have microwave ovens, 24% have convection ovens, 1
4
12% have air fryers. It was also found that 8% of of History club is of the number of members of
the families have microwave ovens and convection Mathematics club. Calculate the number of members
ovens, 5% have microwave ovens and air fryers, of Geography club. HOTS
4% have convection ovens and air fryers whereas A 150 Applying
2% have all the three types of appliances. Find the B 182
number of families that have microwave ovens only. C 242
A 100 C 136 HOTS D 310
B 125 D 142 Analysing
16. P 19. ξ
P
SPM Q
2018 R
QR The Venn diagram above shows the universal set,
, set P, set Q and set R. Which of the following
The diagram above shows a Venn diagram with represents the shaded region?
three sets P, Q and R. Given that the universal set, A Qʹ (P R)
= P Q R. Which of the following represents B Qʹ (P R)
the shaded region? C Q (Pʹ R)
A (Pʹ Q) R D Q (Pʹ R)
B (Pʹ Q) R
C (P Q) Rʹ
D (P Q) Rʹ
66
Mathematics SPM Chapter 4 Operations on Sets
20. The table below shows the sports activities selected PAPER 2
SPM 1.
2018 by 60 pupils. ξ
P
Sports activity Number of pupils Q
Badminton 32
Table tennis 33
Tennis 37 The diagram above shows a Venn diagram
with the universal set, = {Pupils of School M},
set P = {Pupils who join Drama club} and set Q
= {Pupils who join Culture club}.
Given that n(P) = 234, n(Q) = 338, n(P Q) = 56 and
the number of pupils who do not join both clubs are
98 pupils. Find the total number of pupils in School M.
2. Given that the universal set, = {x : x is an integer
and 1 x 15}, set P = {2, 3, 5, 6, 7, 11} and set
Q = {x : x is a prime number}. Find n(Pʹ Q).
Penerbitan Pelangi Sdn Bhd. All Rights ReservedTable tennis and tennis20
Form 4
Badminton and table tennis 18
Badminton only 10
Table tennis only 6
Given that set B = {Pupils who select badminton}, 3. P
set P = {Pupils who select table tennis} and set T 1
= {Pupils who select tennis}. Which of the following R
Venn diagrams represents the information in the 4x
table above? 3
A B 9–x 2 4
Q
10
The Venn diagram above shows the number of
83 elements in each region in set P, Q and R. Given the
11 universal set, = P Q R and n(Qʹ) = n(P R)ʹ.
Find the value of x.
6 8 14
4. The following Venn diagram shows the universal
PT set, , set P, set Q and set R where = P Q R.
Shade each of the following sets in the diagram
B B given.
(a) P Q
10
PQ
64
11 R
(b) P (Rʹ Q) Q
6 10 12
P
PT
C B R
10
5. The following Venn diagram shows the universal
74
11 set, , set E, set F and set G where = E F
G. Shade each of the following sets in the diagram
6 9 13 given.
PT (a) F G
D B F
10 E
93 G
11
6 7 16
PT
67
Mathematics SPM Chapter 4 Operations on Sets
(b) G (Fʹ E) 9. The following Venn diagram shows the universal
F SPM set, , set K, set M and set N where = K M N.
E 2017 Shade each of the following sets in the diagram
G given.
(a) K M
M
K
6. The Venn diagram below shows the universal set,
, set M, set N and set S where = M N S. N
Shade the set M (N S) on the diagram given.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
M N
S (b) M (N Kʹ)
Form 4
M N
K
7. The following Venn diagram shows the universal
set, , set M, set N and set S where = M N S.
Shade each of the following sets in the diagram 10. (a) Given that set P = {Multiples of 3 larger than
given. SPM 30} and Q = {36, 42, 48}. Complete the Venn
(a) N S 2018 diagram below to show the relationship between
MN
S
set P and set Q.
P
N
(b) (M S)ʹ N (b) The following Venn diagram shows three sets P,
Q and R where the universal set = P Q R.
M Write a set which involves set P, Q and R to
represent the shaded region.
S
Q
8. The following Venn diagram shows the universal PR
set, , set F, set G and set H where = F G H.
Shade each of the following sets in the diagram
given. 11. (a) The diagram below shows a Venn diagram with
(a) G H SPM set M and set N. Given that the universal set
2019 = M N. By using symbol of set, write the
F
H
G
relationship between set M and set N.
NM
(b) (G Hʹ) F
H
F
(b) Given that set X = {x : x is a multiple of 6},
G set Y = {x : x is a factor of 3} and set Z =
{x : x is a factor of 9} such that the universal set
= X Y Z. Based on the information given,
draw a Venn diagram to show the relationship
between set X, set Y, and set Z.
68
5Chapter Learning Area: Measurement and Geometry
Form 5
Congruency, Enlargement and
Combined Transformations
Penerbitan Pelangi Sdn Bhd. All Rights Reserved• Congruency – Kekongruenan Concept
• Enlargement – Pembesaran map
• Reflection – Pantulan
• Rotation – Putaran
• Scale factor – Faktor skala
• Similarity – Keserupaan
• Tessellation – Teselasi
• Transformation – Transformasi
• Translation – Translasi
Aotrhfcehtrictaoencnstcfseopartmlwoaatfyiostnrausnscesadfnotrhbmeeatcciooonnmc?beiWpnthedaotftiostrptahrnoesdftuoracrnme sauftnoioirqmnuaettoisohdnaepaseipgspn.liWeadhb?autilbduinilgd.inTgheindMifaflearyesnita types
apply
234
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
5.1 Congruency (a) A 6 cm C R
A Differentiate between congruent and 8 cm 53° 10 cm
non-congruent shapes based on sides 37° P 8 cm Q
and angles B K 5 cm N
C
1. Congruent shapes are the shapes that have (b)
the same size and shape regardless of their
orientation. D
2. A pair of congruent polygons have the same 5 cm 8 cm
measurement for the corresponding sides and
angles.
3. The diagram below shows two quadrilaterals,
ABCD and EFGH, that are congruent.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved A 9 cm B M
Form 5L
(c) H Z
7 cm 54° 66° X
K JY 7 cm
D 10 cm C H (d) D 8 cm C U 12 cm V
50° 119° 5 cm
50° 7 cm
7 cm 15 cm
10 cm 15 cm
123° 68° 123° E 98°
A 8 cm B 82°
A 12 cm B X 8 cm W
G 119° 8 cm
5 cm 68° Solution
(a) BC = √62 + 82
F = 10 cm
= QR
From the diagram above, the same pairs of
corresponding sides and angles are shown in the ∠ACB = 180° – 90° – 37°
table below.
Corresponding Corresponding = 53°
sides angles = ∠PRQ
The measurements of all corresponding sides
AB = EF ∠ABC = ∠EFG and angles are equal. Hence, both shapes are
BC = FG ∠BCD = ∠FGH congruent.
CD = GH ∠CDA = ∠GHE
AD = EH ∠DAB = ∠HEF (b) Although all the corresponding angles are equal,
AB ≠ KL and CD ≠ MN.
Hence, both shapes are not congruent.
SPM Tips (c) ∠HJK = 180° – 54°
63° 2
• If there is one pair of the corresponding sides or angles =
are different sizes, then both of the shapes are not ≠ ∠XYZ
congruent. Hence, both shapes are not congruent.
• The arrangement of the shapes does not affect the (d) ∠ADC = 180° − 82°
congruency between the shapes. = 98°
= ∠UXW
1 The measurements of all corresponding sides
Determine whether each pair of the following shapes and angles are equal. Hence, both shapes are
are congruent. congruent.
Try Questions 1 – 2 in Try This! 5.1
235
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
B Making and verifying the conjecture of triangle congruency based on sides and angles
1. Triangle congruency can be determined from the specific properties of a pair of triangles.
2. Two congruent triangles satify the properties of triangle congruency.
3. The properties of triangle congruency are as follows.
(a) Side-Side-Side (SSS)
(b) Side-Angle-Side (SAS)
(c) Angle-Side-Angle (ASA)
(d) Angle-Angle-Side (AAS)
(e) Angle-Angle-Angle (AAA)
(f) Side-Side-Angle (SSA)
4. The table below shows the explanation of the properties of triangle congruency using two congruent
triangles, ABC and PQR.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Properties of triangle congruency Situation of triangle congruency
Side-Side-Side (SSS)
• Each pair of the corresponding sides are of equal AP
length. B CQ R
• AB = PQ, BC = QR, AC = PR A
P
Side-Angle-Side (SAS)
• Two pairs of the corresponding sides are B CQ R
A
equal in length and the sizes of the pair of the P
corresponding subtended angle between the two
sides are equal. B CQ R
• AB = PQ, ∠ABC = ∠PQR, BC = QR A
Angle-Side-Angle (ASA) P
• Two pairs of the corresponding angles are equal
Form 5 and the lengths of the pair of the corresponding B CQ R
side between the two angles are equal. A
• ∠ABC = ∠PQR, BC = QR, ∠BCA = ∠QRP P
Angle-Angle-Side (AAS)
• Two pairs of the corresponding angles are equal B CQ R
and the lengths of a pair of the corresponding A
sides which do not lie between the two angles are P
equal.
• ∠ABC = ∠PQR, ∠BCA = ∠QRP, AC = PR B CQ R
Angle-Angle-Angle (AAA)
• All the three corresponding angles are equal.
• The areas of the pair of triangles must be equal.
• ∠ABC = ∠PQR, ∠BCA = ∠QRP, ∠CAB = ∠RPQ
Side-Side-Angle (SSA)
• Two pairs of the corresponding sides are equal
in length and a pair of the corresponding angles
which are not subtended between the two sides
are equal.
• The areas of the pair of triangles must be equal.
• AB = PQ, BC = QR, ∠BCA = ∠QRP
236
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
SPM Tips 3
The diagram below shows a triangle PQR.
AP
R
B CQ R P
Right angle-Hypotenuse-Side (RHS) is a special triangle Q
congruency for right-angled triangle. If two right-angled
triangles have the same length of the hypotenuse It is given that another triangle CDE is congruent to
and a pair of the other corresponding sides that is not the triangle PQR. State the triangle congruency used
the hypotenuse are the same, then both the triangles to determine both triangles are congruent if
are congruent. This triangle congruency is based on (a) PQ = CD, PR = CE and ∠QPR = ∠DCE
Side-Side-Angle (SSA). (b) QR = DE, ∠PQR = ∠CDE and ∠PRQ = ∠CED
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Form 52SolutionR
The diagram below shows two congruent triangles (a)
which satisfy the conditions of Angle-Angle-Side
(AAS). P
CZ Q
E
Y
A BX C
Complete the table below based on the triangle D
congruency of Angle-Angle-Side (AAS).
Two pairs of the corresponding sides and a pair
Corresponding angles Corresponding sides of the corresponding subtended angle between
∠ACB = ∠YZX the two sides are given. Hence, the triangle
congruency is Side-Angle-Side (SAS).
Solution
(b) R
Corresponding angles
∠ACB = ∠YZX Corresponding sides P
∠CBA = ∠ZXY AB = YX
Q
or Corresponding sides E
AB = YX
Corresponding angles C
∠ACB = ∠YZX
∠BAC = ∠XYZ D
REMEMBER! Two pairs of the corresponding angles and a
pair of the corresponding side between the two
Identify the corresponding vertices before determining angles are given. Hence, the triangle congruency
the corresponding sides and angles, for instance, A = Y, is Angle-Side-Angle (ASA).
B = X and C = Z.
Try Questions 3 – 4 in Try This! 5.1
237
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
C Solving problems involving congruency Solution:
4 x cm 8 cm Analyse the
The diagram below shows a parallelogram ABCD. diagram given
16 cm to obtain the
D QC measurement
So, x = 8 ÷ 2
=4
Area = 3 × [(4 × 8) + (8 × 8)]
= 288 cm2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
A PB Try this HOTS Question
The diagram below shows the plan of an office.
P and Q are midpoints of the sides AB and CD
respectively. Determine whether the triangle APD is 20 m
congruent with
(a) triangle QCP. 5m
(b) triangle CQB.
The office is divided into two congruent parts.
Solution If the total area of the office is 550 m2, find the
(a) D Q C perimeter for one congruent shape.
Answer: 84 m
A PB
Try This! 5.1
It is found that AP = QC and AD = QP but
∠PAD ≠ ∠CQP and PD ≠ PC. 1. Determine whether each pair of the following shapes
are congruent.
Hence, triangles APD and QCP are not
congruent. (a)
(b) D Q C
8 cm 14 cm
A PB 5 cm
It is found that AP = CQ, AD = CB and (b) 14 cm 13 cm
∠PAD = ∠QCB, so they satisfy triangle
Form 5 congruency Side-Angle-Side. Hence, triangles 13 cm 14 cm
APD and CQB are congruent. 72° 98°
Try Questions 5 – 7 in Try This! 5.1 (c)
Example of HOTS Question 10 cm 10 cm
The diagram below shows a shape consisting of a 4 cm 6 cm
combination of three congruent shapes.
7 cm 3 cm
8 cm (d)
16 cm 110°
105°
Find the area, in cm2, of the shape.
238
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
2. Determine the pairs of the congruent shapes in the 6. The diagram below shows two congruent right-
diagram below. angled triangles, KLM and PQR.
M Q
P
P QR S 37° 20 cm
R
16 cm
T U VW KL
Find
(a) ∠PRQ.
(b) the area, in cm2, of triangle PQR.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
3. The diagram below shows two congruent triangles, Form 5 7. The diagram below shows a shape consisting of a
ABC and EFG. combination of three congruent arrows.
CF P
AE 3 cm
BG 5 cm
It is given that AB = EF and ∠ABC = ∠EFG. State Q
another condition if the triangle congruency involved
is It is given that PQ is the axis of symmetry of
(a) Angle-Side-Angle the shape and the height of each arrow is 9 cm,
(b) Side-Angle-Side calculate the area, in cm2, of the shape.
4. The diagram below shows a triangle PQR. 5.2 Enlargement
R
PQ A Explaining the meaning of similarity
of geometric objects
It is given that another triangle XYZ is congruent
with the triangle PQR. State the triangle congruency 1. Similarity of geometric objects means that the
involved if objects have the same shape regardless of their
(a) PQ = XY, QR = YZ and PR = XZ size and orientation.
(b) QR = YZ, ∠PQR = ∠XYZ and ∠PRQ = ∠XZY
2. Two similar objects have the same corresponding
5. The diagram below shows a trapezium ABCD. angles and fixed ratio of the corresponding sides.
DC 3. The diagram below shows two similar triangles,
ABC and PQR.
CR
AB A BP Q
It is given that AD = BC, determine whether each • Each pair of the corresponding angles are
pair of the following triangles are congruent. equal, namely ∠A = ∠P, ∠B = ∠Q and
(a) ∆ACD and ∆BDC ∠C = ∠R.
(b) ∆ABC and ∆ACD
• All the ratios of the corresponding sides are
constant, namely PQ = QR = RP
AB BC CA
239
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
5 point from a particular point on the image and
from the corresponding point on the object is
Determine whether each pair of the following constant.
geometric objects are similar. 2. The fixed point is known as centre of
enlargement.
(a) A BL 3. The ratio of the distance of centre of enlargement
from a point on the image to the distance of
50° M centre of enlargement from the corresponding
point on the object is known as scale factor.
120° 4. In an enlargement,
D (a) the corresponding angles between the
object and image are equal.
(b) the ratios of the corresponding sides are
constant.
5. Both object and image in an enlargement are
similar.
6. The diagram below shows an enlargement.
AЈ
Penerbitan Pelangi Sdn Bhd. All Rights ReservedC50°100°
K N
(b) Z 21 cm
R
14 cm 10 cm Y
9 cm X
P 15 cm
6 cm
Q
Solution A Object Image
(a) ∠B = 360° – 50° – 120° – 90° CЈ
= 100° Centre of C BЈ
= ∠N enlargement O
∠L = 360° – 50° – 100° – 90° mB
= 120° n
= ∠D
∠A = ∠K = 50° 7. Scale factor, k, of an enlargement can be
∠C = ∠M = 90° determined as follows.
All the corresponding angles are equal. Hence,
distance of point of image from O n OB9
quadrilateral ABCD and quadrilateral KLMN are distance of point of object from O m OB
similar.
(b) PQ = 6 = 3 k= = =
YX 10 5
or
9 3
QR = 15 = 5 k= length of side of image = A'B9
XZ length of side of object AB
PR = 14 = 2 8. For scale factor, k, within the range 0 , k , 1,
YZ 21 3 the size of the image formed is smaller than the
object as shown in the diagram below.
Form 5 The ratio of the corresponding sides PR and
YZ is different with the ratio of the other
corresponding sides. Hence, triangle PQR and
triangle XYZ are not similar.
Try Questions 1 – 2 in Try This! 5.2 Centre of Image Object
enlargement O
10 cm
B Making a connection between 20 cm
similarity and enlargement, hence
describing enlargement using Scale factor, k = 10 cm = 1
representation 20 cm 2
1. Enlargement is a type of transformation where 9. When the scale factor is a negative value, the
the image is formed based on a fixed point image formed under enlargement is located on
such that the ratio of the distance of the fixed the opposite side of the object as shown in the
diagram below.
240
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
Image Centre of Object 6
10 cm
enlargement In each of the following diagrams, triangle A9B9C9 is
O the image of the object ABC under an enlargement.
(a) y
20 cm
CЈ 10
Scale factor, k = – 20 cm = –2 8 BЈ
10 cm AЈ
Penerbitan Pelangi Sdn Bhd. All Rights Reserved 6
10. The following table shows the size of image Form 5x
and the position of image for the scale factor in C4 24
different range. B
2
A
–6 –4 –2 O
Scale factor, Size of image Position of image (b) y
k
to the centre of C 6
enlargement O B
larger than the Be on the same 4
size of object side as the object
k>1 CЈ 2
BЈ
A x
2
smaller than Be on the same –8 –6 –4 –2 O
0<k<1 the size of side as the object AЈ –2
object
–4
smaller than Be on the opposite Describe the enlargement in each diagram.
the size of side as the object
−1 < k < 0 object Solution
(a)
k < −1 larger than the Be on the opposite y
size of object side as the object CЈ 10
8 BЈ
AЈ x
k = 1 or equal in size to When k = 1, the C 6 24
k = −1 the object image is on the
same side as the P(–4, 1) 4
object. –6 –4 B
When k = −1, the
image is on the 2
opposite side as A
the object. –2 O
Scale factor = PA9 The ratio of the
PA distance of point on
image from P to the
= 6 units distance of point on
2 units object from P
SPM Tips
= 3
The centre of enlargement can be determined from
the intersection point between all the straight lines that A9B9C9 is the image of ABC under an enlargement
connecting each pair of the corresponding points. at centre P(−4, 1) with scale factor 3.
241
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
(b) y Scale factor = – distance of S from centre
distance of H from centre
C 6
B = – 3 units Negative sign shows
2 units that the image is on the
4
A CЈ 2BЈ =– 3 opposite side as the object
2
–8 –6 –4 –2 O x
AЈ 2 PQRS is the iwmitahgescoafleEfFaGctHoru–nd32er. an enlargement at
centre (2, 3)
–2
P(0, –3)
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
–4
Scale factor = B9C9 Try Questions 3 – 4 in Try This! 5.2
BC
The ratio of the
= 3 units length of side of C Determining the image and object of
6 units image to the length an enlargement
of side of object
= 1 The flow map below shows the steps to determine the
2 image or object of an enlargement.
A9B9C9 is Pth(0e,im−3a)gewoitfhAsBcCaluenfadcetroarn21en. largement Draw the
at centre image or object
Identify Draw the which has the
7 Identify the the projection similar shape
In the diagram below, PQRS is the image of EFGH centre of scale line from according to
under an enlargement. enlargement factor the centre the scale factor
of
enlargement
y 8
8F
The diagram below shows four similar triangles drawn
6 E on the Cartesian plane.
G
S4 H y
Rx
2 8
246
P
–4 –2 O 6
4P
–2
Q–4
Describe the enlargement. 2
Form 5 Solution –8 –6 –4 A–2 –2 O 2 4 6 8 x
10
y –4 Q
8F –6
6 E –8 R
G
S4 H –10
Rx
2 Determine the image of triangle A under the
246
P
–4 –2 O enlargement at centre (−8, 4) with scale factor 5 .
3
–2
Q
–4
242 242
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
Solution y Solution
8 (a) When k = 3, the distance of each corresponding
(–8, 4)
6 vertex of image from O is 3 times the distance
4P of the vertex of object from O on the same
direction.
2
x
8 10
Penerbitan Pelangi Sdn Bhd. All Rights Reserved–8 –6 –4 –2 A O 2 4 6
Form 5–2
–4 Q O
–6
R (b) When k= − 1 , the distance of each corresponding
–8 vertex of 2 1 times the distance
2
–10 image from O is
The image of A is Q. of the vertex of object from O on the opposite
REMEMBER! direction.
When the scale factor, k = 5 , the distance of each
3
corresponding vertex of image from centre of enlargement
is 5 times the distance of the vertex of object from centre O
3
of enlargement on the same direction.
9 SPM Tips
Draw the image for each of the following objects
under the enlargement at centre O with the given We can use the horizontal distance and vertical distance
scale factor. to determine the ratio of a point from the centre of
(a) Scale factor, k = 3 enlargement as follows.
It is given that point P is object and point P9 is its image,
O is the centre of enlargement and k = 3.
k = 3 means that the horizontal distance and vertical
distance of P9 from O to the horizontal distance and
vertical distance of P from O follow the ratio 3 : 1.
PЈ
O 3 × 2 = 6 units P
2 units
1 5 units
(b) Scale factor, k = – 2 3 × 5 = 15 units O
10
O Draw the object for each of the following images
under the enlargement at centre O with the given
scale factor.
243
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
(a) Scale factor, k = 4 D Making and verifying conjecture on
3 the relation between area of the
image and area of the object of an
O enlargement
1. The diagram below shows the enlargement of a
right-angled triangle at centre of enlargement O
with scale factor k.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved(b) Scale factor, k = −2 O
A y cm q cm
x cm AЈ
p cm
O Since scale factor = k, so the ratio of the length
of corresponding sides of image to the length of
Solution 4 sides of object = k,
(a) When k = 3
, the distance of each vertex of object which is p = k and q = k
x y
from O is 3 times the distance of corresponding p = kx q = ky
4
vertex of image from O in the same direction. 1
2. Hence, area of the object = 2 xy
O and area of the image = 1 pq
2
1
= 2 (kx)(ky)
= 1 k2xy
2
(b) vfWreorhmteenxOkoif=sim−212a,tgitemhfeerdos mitshteaOndciinestotahfneecaeocpohpfvocesorirttreeexdsopirfoeoncbtdijioenncgt. 3. From the ratio of area of the image to area of the
object,
O
area of the image = 1 k2xy = k2
area of the object 2
1
2 xy
Form 5 4. Generally, the area of the image of an enlargement
can be determined by the following relation.
SPM Tips
We can use the inverse scale factor to determine the 11
object of an image under an enlargement. For example, It is given that the area of a geometric object is 16 cm2.
if A9 is image of A under the enlargement at centre O Calculate the area of its image under an enlargement
with scale factor k, then we can perform the enlargement with scale factor,
at centre O with scale factor 1 on the image A9 to (a) k = 5
determine the object A. k
(b) k= 3
4
Try Questions 5 – 7 in Try This! 5.2
244
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
Solution Solution
(a) Area of the image = k2 × area of the object AB
= 52 × 16 (a) Scale factor, k = PB
= 400 cm2 = 8 + 12
12
1 2(b) Area of the image = 3 2 × 16 5
4 = 3
= 9 cm2 (b) Area of the image = k2 × area of the object
Penerbitan Pelangi Sdn Bhd. All Rights Reserved12 1 2 5 2
Form 53
The area of the image of object A under an enlargement Area 100 ×
is 54 cm2. Find 100 = × area of the object
(a) the area of the object if the scale factor of the object = 9
of the 25
enlargement is 3,
(b) the scale factor if the area of the object is 24 cm2. = 36 cm2
Area of the shaded region = 100 – 36
= 64 cm2
Solution 14
(a) Area of the image = k2 × area of the object The diagram below shows the measurement of a small
54 = 32 × area of the object Jalur Gemilang.
54
Area of the object = 69cm2
=
(b) 54 = k2 × 24 90 cm
k2 = 54 1.8 m
24
9 A school intends to make a large Jalur Gemilang
= 4 with an area of 103.68 m2 by using the concept of
enlargement. Find the length and width of the Jalur
k = + 23 or – 3 Gemilang that the school intends to make.
2
Try Question 8 in Try This! 5.2 Solution
Area of the small Jalur Gemilang = 1.8 × 0.9 = 1.62 m2
E Solving problems involving and area of the large Jalur Gemilang = 103.68 m2
enlargement
k2 = Area of the image
13 Area of the object
In the diagram below, triangle ABC is the image of
triangle PBQ under an enlargement at centre B. = 103.68 m3
1.62 m2
C = 64
Q k = 8
Length of Jalur Gemilang = 1.8 × 8 = 14.4 m
Width of Jalur Gemilang = 0.9 × 8 = 7.2 m
A 8 cm P 12 cm B Try Questions 9 – 10 in Try This! 5.2
Find Try This! 5.2
(a) the scale factor of the enlargement.
(b) the area of the shaded region, if the area of 1. Determine whether each pair of the following
geometric objects are similar.
triangle ABC is 100 cm2.
245
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
(a) (b) y
76° 10
42°
8
76°62° 6
AЈ 4 A
2
(b) 14 cm 10.5 cm
9.3 cm
140° 12.4 cm x
12 cm 80° 24 68
16 cm
Penerbitan Pelangi Sdn Bhd. All Rights Reserved –8 –6 –4 –2 O
(c) 14 cm (c) y
7 cm 10
4 cm
8 cm 8
6A
AЈ 4
(d) 9 cm 2
–10 –8 –6 –4 –2 O
6 cm x
24 6
5 cm 5 cm 8 cm 8 cm
5 cm 5 cm 8 cm 8 cm 4. Describe the enlargement in the following diagram
6 cm where P9 is the image of P.
9 cm
2. Determine similar triangles in the following diagrams. PЈ
A C P
B O
6 cm 4 cm
E 5. Based on the given scale factor, determine the object
F and image of each of the following enlargement.
(a) Scale factor, k = 2
D
P
G Q
Form 5 3. Describe the enlargement in each of the following
diagrams where A9 is the image of A.
(a) y (b) Scale factor, k = 1
3
4
2A
–10 –8 –6 –4 –2 O x
AЈ –2 246
–4 Y
X
–6
–8
246
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
6. Copy each of the following diagrams. Draw the 10. The diagram below shows two right-angled triangles
where ΔQST is the image of ΔPQR under an
image of each of the diagrams under an enlargement
at centre O. enlargement. HOTS
Analysing
(a) Scale factor = 3 (b) Scale factor = 4 R
2
3 cm Q S
P
O
O 12 cm
T
(a) Describe the enlargement.
(b) Given that the area of the whole diagram is
102 cm2, find the perimeter, in cm, of the whole
diagram.
(c) Penerbitan Pelangi Sdn Bhd. All Rights ReservedScale factor=–1 (d) Scale factor = −2
Form 53
O 5.3 Combined transformation
O
A Determining the image and object of
a combined transformation
7. Copy each of the following diagrams. Draw the object 1. The table below shows the image A9 of the object
A under four types of transformations.
for each of the diagrams under an enlargement at
centre O.
3
(a) Scale factor = 2 (b) Scale factor = – 2 Translation Reflection
Reflection on the line
Translation 1 5 2 y=3
–2
y y
6 5 units
O 6
A 2 units A
4 AЈ
O 4
2
2 y=3
AЈ
8. The table below shows the values of the area of the x x
object, the area of the image and the scale factor O 246 O 246
under the different enlargement. Complete the table.
Area of the Area of the Scale factor Rotation Enlargement
object image Rotation of 90° Enlargement at centre
clockwise at centre (0, 6) with scale
(a) 10 cm2 90 cm2 (1, 2) factor 3
(b) 80 cm2 2 y y
(c) 5
6 6
7 unit2 – 1 A
2 4A
4
9. In the diagram below, trapezium APQR is the image
of trapezium ABCD under an enlargement. Vertex A 2 AЈ 2 AЈ
is the centre of the enlargement. (1, 2) x
O x
DC O2 46 246
RQ 2. Generally, the combination of two
transformations, A and B, can be performed in
A PB different orders, that are transformation AB or
12 cm 8 cm transformation BA.
It is given that the area of the shaded region is
80 cm2. Find the area, in cm2, of trapezium APQR.
247
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
3. Combined transformation AB means y
transformation B occurred then followed by QЉ (–2, 7) 8
transformation A.
6
4. Combined transformation AA means 4 P(3, 2)
transformation A occurred two times in a row 2 246
and is written as A2.
–2 O
5. The diagram below shows the steps to determine x
the image or object of a combined transformation 8
AB.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Transformation Image under Transformation (a) Determine the image of point P under the
B transformation A combined transformation AB.
B (b) If Q0 is the image of point Q under the
transformation AC, determine the coordinates
Object Image of point Q.
Object of the Object under Solution
Object under image under transformation (a) Combined transformation AB means
transformation transformation
A transformation B followed by transformation A.
BA P9 is the image of P under the transformation B
and P0 is the image of P9 under the transformation
A.
SPM Tips y
8
The table below shows the ways to determine the object P Љ(–4, 6) y=x
of an image under a transformation. 6
Type of transformation Transformation 4 P Ј(2, 3)
performed on the image
to determine the object 2 P(3, 2)
Translation 1 x 2 Translation 1 –x 2 –4 –2 O x
y –y 246
Reflection on the line y = x Reflection on the line y = x The image of point P is P 0(−4, 6).
Rotation in clockwise Rotation in anticlockwise
direction at centre (x, y) direction at centre (x, y)
(b) Combined transformation AC means
Enlargement at centre Enlargement at centre O transformation C followed by transformation
O with scale factor k A. Q9 is the object of image Q0 under the
with scale factor 1 transformation A and Q is object of image Q9
k under transformation C.
Form 5
15 y Q(8, 8)
Q Ј(4, 4) (8, 4)
It is given that transformations 8
QЉ (–2, 7)
A = translation 1 –6 2
3 6
B = reflection on the line y = x 4
2
C = rotation of 90° anticlockwise at centre (8, 4) –2 O x
2468
The diagram below shows the point P and point Q0 on The coordinates of point Q are (8, 8).
a Cartesian plane.
248
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
REMEMBER! (b) Combined transformation PQ means
transformation Q followed by transformation P.
When determining the object of an image under a combined A9 is the image of A under the transformation
transformation AB, the order of the transformations is Q and A0 is the image of A9 under the
reversed, that is determine the object of the image under transformation P.
transformation A followed by transformation B.
y
6
16
The diagram below shows a triangle A drawn on a
Cartesian plane.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved 4A
Form 5
2
y –10 –8 –6 –4 –2 O2 4 x
6 6
–2
4A AЈ –4
2 –6
AЉ
x –8
246
O –10
It is given that transformations
P = translation 1 1 2 REMEMBER!
–4
Q = rotation of 180° at the origin For the rotation of 180°, its image is the same either in
R = enlargement at centre (4, 3) with scale factor 2 clockwise or anticlockwise direction.
Determine the image of triangle A under the (c) Combined transformation QR means
combined transformation transformation R followed by transformation Q.
(a) P2 A9 is the image of A under the transformation
(b) PQ R and A0 is the image of A9 under the
(c) QR transformation Q.
Solution y
(a) Combined transformation P2 means 8
transformation P is performed twice in a row. 6
A9 is the image of A under the transformation AЈ
P and A0 is the image of A9 under the 4
transformation P.
2
y
–8 –6 –4 –2 O
6 –2 A
(4, 3)
4 A x
2 2 AЈ 4
24 6 8
–6 –4 –2 O
–2 x –4 AЉ
6 –6
–4 AЉ –8
–6
249
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
17 (b) Combined transformation CB means
The diagram below shows some similar pentagons transformation B followed by transformation
drawn on a Cartesian plane. C. Y is the object of image P under the
transformation C and X is the object of image Y
y under the transformation B.
8
6R y x=2
(–1, 3) 4
P
2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved4
PQ –8 –6 –4 –2 Y O x
–2 2 4X 6
2
x –4
–8 –6 –4 –2 O 2468
–2
U
–4 V Hence, the object of image P under the combined
T –6 transformation CB is U.
S –8
Try Questions 1 – 4 in Try This! 5.3
It is given that transformations B Making and verifying the conjecture
about commutative law in combined
A = enlargement at centre (−2, 3) with scale factor transformation
1
− 2 1. A combined transformation AB satisfies
the commutative law if the images under the
B = reflection on the line x = 2 combined transformations AB and BA are the
same.
C = rotation of 90° clockwise at centre (−1, 3)
2. A combined transformation AB does not satisfy
Determine the object of image P under the combined the commutative law if the images under the
transformation combined transformations AB and BA are not
(a) AB the same.
(b) CB
Form 5 Solution 18
(a) Combined transformation AB means The diagram below shows a quadrilateral K drawn on
a Cartesian plane.
transformation B followed by transformation
A. Y is the object of image P under the
transformation A and X is the object of image Y
under the transformation B.
y x=2 y
8 10
X6 Y 8
(–2, 3) 4 K
P
6
2
4
–8 –6 –4 –2 O
x 2
24 6 8
Hence, the object of image P under the combined O x
transformation AB is Q. 2 4 6 8 10
250
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
It is given that transformations For combined transformation CB, transformation
1 3 2 B performed first followed by transformation C.
A = translation 2
y
B = reflection on the line y = 5
C = rotation of 180° at centre (6, 5) 10
Determine whether each of the following combined 8
transformations satisfies the commutative law.
(a) Combined transformation AB K KЉ
(b) Combined transformation BC 6 y=5
(6, 5)
4 KЈ
SolutionPenerbitan Pelangi Sdn Bhd. All Rights Reserved 2
(a) For combined transformation AB, transformation Form 5
O x
B performed first followed by transformation A. 2 4 6 8 10
y The images of the combined transformations
10 BC and CB are the same. Hence, the combined
8 transformation BC satisfies the commutative
K law.
6 KЉ Try Question 5 in Try This! 5.3
y=5
4 KЈ
2
O x C Describing combined transformation
2 4 6 8 10 1. The diagram below shows key points to describe
For combined transformation BA, transformation a transformation.
A performed first followed by transformation B.
Object
y
10
KЈ
8
K
6
4 y=5 Similar image Congruent
but different size image
2 KЉ
O x Enlargement Same Inverted
2 4 6 8 10 orientation orientation
The images of the combined transformations
AB and BA are not the same. Hence, the
combined transformation AB does not satisfy
the commutative law. Reflection
Distances Distances
(b) For combined transformation BC, transformation between each between each
C performed first followed by transformation B. corresponding corresponding
y point are point are
equal different
10
8 Translation Rotation
K KЉ
6
4 (6, 5) y=5 2. When describing a combined transformation
KЈ AB, we need to follow the order, which is
transformation B comes first and followed by
2 transformation A.
O x
2 4 6 8 10
251
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
19 (ii) y
In the diagram below, B is the image of object A 10
under a combined transformation PQ while C is the AЈ x = –1 8 AЉ
image of A under a combined transformation RS. A
6
y (–7, 4)
10 4
2
8 x
246
Penerbitan Pelangi Sdn Bhd. All Rights Reserved6C –8 –6 –4 –2 O
A
For transformation RS, transformation S
4 performed first followed by transformation
R,
B2
–8 –6 –4 –2 O x which is A S A9 R A0,
246
It is given that transformations where transformation S = enlargement
Q = rotation at centre (−7, 4) with scale factor 2;
S = enlargement at centre (−7, 4) transformation R = reflection on the line
(a) Describe in full, the combined transformation x = −1.
(i) PQ (ii) RS
(b) Describe a single transformation which is Hence, object A experienced the
enlargement at centre (−7, 4) with scale
equivalent to the combined transformation PQ. factor 2 followed by the reflection on the
line x = −1.
Solution
(a) (i) y (b) The single transformation which is equivalent to
10 the combined transformation PQ is the rotation
of 90° anticlockwise at centre (–2, 6).
8 Try Questions 6 – 7 in Try This! 5.3
A (–5, 4) 6 D Solving problems involving combined
4 transformation
AЈ
AЉ 2
–8 –6 –4 –2 O x SPM Highlights
246
In the diagram below, pentagon ABCDE is the plan of
Form 5 For transformation PQ, transformation Q a shopping complex. Pentagon QSTUV is the plan of
performed first followed by transformation Taman Sarjana with PQRWX is a commercial region
P, and the shaded region is the residential region.
which is A Q A9 P A0, y
where transformation Q = rotation of 90° AE 8 S
anticlockwise at centre (−5, 4); D 6T
1 2 transformation P = translation 5 U 4 W R
–1 BC X
Hence, through combined transformation 2 P Q
PQ, object A experienced the rotation of V x
90° anticlockwise at centre (−5, 4) followed –6 –4 –2 O 246
1 2 5 (a) Pentagon QSTUV is the image of pentagon
–1 ABCDE under a combined transformation LH.
by the translation . Describe in full, the transformation
(i) H (ii) L
(b) It is given that the area of the residential region
is 12 600 m2. Calculate the area of the shopping
complex.
252
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
Solution y (a) Determine the image of point A under the
(a) (i) 8 combined transformation
AE 6 (i) K 2
D (ii) LM
(iii) MK
BC 4 W R (b) It is given that K is the image of a point
2 X
under the transformation KL. Determine the
coordinates of the point.
2. The diagram below shows some quadrilaterals
drawn on a Cartesian plane.
y
8
Penerbitan Pelangi Sdn Bhd. All Rights Reserved–6 –4 –2 OP Q
Form 5x
246
H = rotation of 90° anticlockwise at (1, 8)
(ii) Scale factor, k= QS = 25 6
QR IV
L = enlargement at centre Q(5, 1) with
I4
scale factor 5 2
2
–8 –6 –4 –2 O x
(b) Area of ABCDE = area of PQRWX –2 2468
Area of QSTUV = 1 5 22 × area of ABCDE II –4 P
2
=245 × area of ABCDE –6
Area of shaded region III
= area of QSTUV − area of PQRWX –8
12 600 = 25 × area of ABCDE − area of ABCDE –10
4
21
12 600 = 4 × area of ABCDE It is given that transformations
A = reflection on the x-axis
Area of ABCDE = 2 400 m2
B = rotation of 180° at the origin
Hence, the area of shopping complex is 3
C = enlargement at centre (9, 2) with scale factor 2
2 400 m2.
Determine the image of quadrilateral P under the
Try Question 8 in Try This! 5.3 combined transformation
(a) BA (b) CA
Try This! 5.3 3. The diagram below shows an isosceles triangle A
drawn on a Cartesian plane.
1. The diagram below shows some points drawn on a
Cartesian plane. y
y 8
10 A6
S 4
P8 2
Q
–8 –6 –4 –2 O x
6R
K4 It is given that transformations
A2 T P = translation 1 –3 2
–1
x Q = reflection on the line x = 2
–6 –4 –2 O 2468
R = enlargement at centre (2, 4) with scale factor
It is given that transformations – 32
K = translation 1 5 2 Copy the diagram and draw the image of the isosceles
2 triangle A under the combined transformation
L = reflection on the line y = 3
M = rotation of 90° anticlockwise at centre (−1, 6) (a) PQ (b) QR
253
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
4. The diagram below shows some triangles drawn on 7. In the diagram below, A9 is the image of A under the
a Cartesian plane. combined transformation KL.
y y
8
8C D
6 6A
H4 4
2A B 2 AЈ
Penerbitan Pelangi Sdn Bhd. All Rights Reserved–4 –2 O x O x
2468 2468
It is given that transformations It is given that transformations
P = translation 1 –4 2 K = translation 1 –2 2
2 –4
Q = rotation of 90° clockwise at centre (1, 7) Describe
R = reflection on the y-axis
(a) the transformation L.
Determine the object of H under the combined
(b) a single transformation which is equivalent to
transformation the combined transformation KL.
(a) PQ 8. In the diagram below, trapeziums A and B are
two congruent gardens. Trapezium PQRS is a
(b) RP construction region.
5. The diagram below shows a right-angled triangle y
drawn on a Cartesian plane.
6 R
y A
8 4
6 2
4 S x
–8 –6 –4 –2 O 2468
2
–2
–2 O x –4
2 4 6 8 10 B
P –6
It is given that transformations
–8
A = translation 1 –2 2
1 –10
Q
B = rotation of 90° clockwise at centre (6, 1)
C = enlargement at centre (6, 1) with scale factor 2
Determine whether each of the following combined It is given that B is the image of A under a rotation.
Trapezium PQRS is the image of trapezium A under
transformations satisfies the commutative law. the combined transformation UV.
(a) Describe the
Form 5 (a) Combined transformation AB (i) transformation V.
(ii) transformation U.
(b) Combined transformation BC (b) Given that the area of A is 120 m2, calculate the
6. In the diagram below, C0 is the final image of object area of the shaded region.
C under the combined transformation MN.
y 5.4 Tessellation
8
6C
CЈ
4
2 CЉ A Explaining the meaning of tessellation
1. Tessellation is a pattern of recurring shapes that
O x
2468 fills a plane without leaving empty spaces or
overlapping.
Describe the transformation M and transformation N.
254
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
2. The diagram below shows several examples of 2. The table below shows that how a tessellation
tessellations. can be designed using transformations.
Type of Tessellation Explanation
transformation
x • B is the image
involved of A under the
Translation
Tessellation Tessellation Tessellation yA B translation 1 x 2.
consisting of consisting of consisting of 0
triangles only. a combination a combination
of squares and of equilateral
trapeziums. triangles, squares
Penerbitan Pelangi Sdn Bhd. All Rights Reservedand regular DC • C is the image
Form 5hexagons.xof B under the
20 translation 1–0y2.
• D is the image
of C under the
Determine whether each of the following is a translation 1–0x2.
tessellation.
Reflection • B is the image
(a) (b) B of A under the
PQ reflection on
AC the side PQ.
(c) (d)
SR • C is the image
D of A under the
reflection on
Rotation AB the side QR.
60°
Solution P CQ • D is the image
(a) It is a tessellation consisting of quadrilaterals of A under the
D reflection on
only. the side RS.
(b) It is a tessellation consisting of a combination of
• B is the image
equilateral triangles and regular hexagons. of A under the
(c) It is not a tessellation because it consists of the rotation of 60°
clockwise at
shape which is not recurring, which is pentagon. centre Q.
(d) It is a tessellation consisting of two recurring
• C is the image
patterns without overlapping. of A under the
rotation of 60°
Try Question 1 in Try This! 5.4 clockwise at
centre P.
B Designing tessellation involving
isometric transformation • D is the image
of C under the
1. We can design a tessellation by using the rotation of 60°
isometric transformations such as translation, anticlockwise at
reflection and rotation. centre Q.
3. Escher tessellation is a type of tessellation
created by Maurits Cornelis Escher (1898-1972).
Escher tessellation consists of a combination of
congruent patterns.
255
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
4. The diagrams below show the examples of Escher Solution
tessellation. (a) C is the image of A under the rotation of 150°
5. The diagram below shows one of the simple anticlockwise at centre Q.
ways to produce an Escher tessellation, starting D is the image of A under the rotation of 150°
from a piece of square cardboard.
anticlockwise at centre P.
(b) E is the image of B under the reflection on the
line PR.
F is the image of B under the rotation of 150°
clockwise at centre R.
Try Question 2 in Try This! 5.4
Penerbitan Pelangi Sdn Bhd. All Rights Reserved
Try This! 5.4
1. Determine whether each of the following is a
tessellation.
(a)
Cut two Combine the Create more identical
small small poligons shapes with different (b)
polygons with another colours and combine
from two two sides of them to form an Escher
sides the square. tessellation. (c)
of the
square.
(d)
SPM Tips
We can use more than one type of transformations or
combined transformation to design a tessellation.
Form 5 21
The diagram below shows a tessellation consisting of
equilateral triangles and squares. 2. The diagram below shows a tessellation consisting
of squares and regular octagons.
AQC 5 units B
B A PQ C
P E RF
E
D D
Based on the diagram, describe the transformation Based on the diagram, describe the transformation
used to produce the tessellation from used to produce the tessellation from
(a) object A to the images C and D. (a) object A to images C and D.
(b) object B to the images E and F. (b) object B to image E.
256
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
SPM Practice 5
PAPER 1 5. The diagram below shows a rectangle.
1. The diagram below shows two congruent triangles.
15 cm
9 cm
Penerbitan Pelangi Sdn Bhd. All Rights Reserved46° 46°
Form 5
70° 70° Which of the following is correct?
9 cm 9 cm
Triangle congruency shown on the diagram is Scale factor Image
A
A Side-Side-Angle 5 cm
B Side-Side-Side 3 3 cm
C Angle-Angle-Side
D Angle-Side-Angle
2. The diagram below shows a trapezium, PRST. B 10 cm
2
S
4.5 cm
T U C 10 cm
64°
2
PQ R 3 6 cm
It is given that PRT and RSQ are two congruent D 5 cm
triangles. Find the ∠TUS. 1 4.5 cm
A 26° C 90° 2
B 64° D 104°
3. The diagram below shows two similar parallelograms.
18 cm 6. Transformation P is a translation 1 6 2 and
–1
12 cm 16 cm
transformation Q is a rotation of 90° anticlockwise at
centre (0, 3). State the coordinates of the image of
The value of x is x cm point (3, 1) under the combined transformation PQ.
A 20 C 28 A (1, 3) C (7, 2)
B 24 D 32
B (6, 8) D (8, 5)
4. In the diagram below, H9 is the image of H under an 7. It is given that transformation R is a translation 1 –2 2
–3
SPM enlargement. and transformation S is a reflection on the line x = 4.
2018
y Which of the following is the image of T under the
8 combined transformation SR?
6 HЈ y
H
8 T
4 A
2 6
x 4C
2 4 6 8 10
O 2B D
The centre of enlargement is x
24 68
A (4, 3) C (5, 4) O
B (4, 5) D (5, 6)
257
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
8. It is given that the areas of shape P and shape (b) The diagram below shows the geometric
Q are 36 cm2 and 16 cm2 respectively. If Q is the shapes ABCD, PQRS and PQTS drawn on a
image of P under an enlargement, determine the Cartesian plane.
scale factor of the enlargement.
y
A 2 C 3 AD 8 SP
3 4
B 2 D 1 6
2 BQ
9. The diagram below shows that Y is the image of X C4 R
under a combined transformation AB.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved 2
y T x
6
8 –8 –6 –4 –2 O 2 4
6 Y (i) PQTS is the image of ABCD under the
4X
combined transformation XY. Describe in
2 full, the transformation
O x (a) Y
24 68 (b) X
(ii) It is given that PQRS represents a region
It is given that transformation A is a translation 1 1 2.
–5 which has an area of 48 m2. Calculate the
area, in m2, of the shaded region.
Which of the following is transformation B? 2. (a) The diagram below shows two points, A and B,
SPM on a Cartesian plane where B is the image of A
A Rotation of 90° clockwise at centre (5, 4) 2019 under the combined transformation PQ.
B Rotation of 90° clockwise at centre (3, 1)
C Rotation of 90° anticlockwise at centre (5, 4) y
A8
D Rotation of 90° anticlockwise at centre (3, 1)
10. A tessellation cannot be produced from 6
4B
A equilateral triangles
B rectangles 2
C regular pentagons
D regular hexagons
PAPER 2 –6 –4 –2 O x
246
1. (a) The diagram below shows a point K on a It is given that transformation P is a rotation
SPM and transformation Q is a reflection on the line
2018 Cartesian plane. y = 6. Describe in full,
(i) transformation P.
y (ii) a single transformation which is equivalent
to the combined transformation PQ.
4
(b) In the diagram below, PQRSTU is the image
2 of ABCDEF under the combined transformation
HK.
Form 5 –6 –4 –2 O x
–2 24 y
P4
–4 K
UQ
It is given that transformations 2
A = translation 1 –5 2 –6 –4 –2 S O 2 E4 F6 x
3 T –2 R 8
B = enlargement at centre (3, −4) with scale
factor 3 –4 D A
State the coordinates of the image of point –6 C B
K under each of the following combined
(i) Describe in full, transformation
transformations. (a) K
(i) A2 (b) H
(ii) AB
258
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
(ii) It is given that PQRSTU represents 4. (a) The diagram below shows a tessellation
a region which has an area of 76 m2. consisting of equilateral triangles and squares.
Calculate the area, in m2, of the shaded
region. 3 units
A PC
3. (a) The diagram below shows two similar BD
trapeziums.
Q
120° y cm 16 cm
x
73°
24 cm
Penerbitan Pelangi Sdn Bhd. All Rights Reserved107° 12 cm Based on the diagram, describe the
Form 5transformation used to produce the tessellation
Find the value of from
(i) x (i) object A to image C.
(ii) y (ii) object B to image D.
(b) In the diagram below, pentagon PQRST is the
image of pentagon ABCDE under the combined (b) The diagram below shows two artificial lakes, X
transformation UV. and Y, and an industrial region Z.
y
y 4
10 A X2 Z
x
8 –8 –6 –4 –2 O
Y –2 246
ST
6 FB
P E
RQ 4 In the diagram, triangle Z is the image of
KG triangle X under the combined transformation
PQ.
2 (i) Describe in full, the transformation
(a) Q
–8 –6 –4 D H C (b) P
–2 O 24 x (ii) It is given that the area of each lake is
280 m2. Calculate the area, in m2, of the
6 industrial region.
It is given that the area of the shaded region is
126 cm2.
(i) Describe in full, the transformation
(a) V
(b) U
(ii) Calculate the area, in cm2, of pentagon
PQRST.
259
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Focus SPM 2022 - Mathematics
Discover the best professional documents and content resources in AnyFlip Document Base.
Search