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CC038233
FORM KSSM FOCUS
4∙5
Mathematics SPM
FOCUS SPM KSSM Form 4 • 5 – a complete and precise series of reference books Mathematics SPM
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A great resource for every student indeed!
• Neo Geok Kee (Textbook Writer)
REVISION REINFORCEMENT EXTRA Mathematics • Ng Seng How (Textbook Writer)
› Comprehensive Notes & ASSESSMENT FEATURES • Ooi Soo Huat
› Example & Solution › Try This! › Example of • Yong Kuan Yeoh
› SPM Tips › SPM Practices HOTS Questions
› Remember › SPM Model Paper › SPM Highlights
› Complete Answers › QR code
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CC038233
ISBN: 978-629-7537-21-4
Based on the
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Format: 190mm X 260mm TP Focus SPM 2023 Mat BI version _pgi CRC
Mathematics SPM
FORM
4∙5 KSSM
• Neo Geok Kee (Textbook Writer)
• Ng Seng How (Textbook Writer)
• Ooi Soo Huat
• Yong Kuan Yeoh
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CONTENTS
Mathematical Formulae iv Chapter
6 Linear Inequalities in Two 85
Variables
FORM 4
6.1 Linear Inequalities in Two Variables 86
Chapter 6.2 System of Linear Inequalities in Two
1 Quadratic Functions and Variables 91
Equations in One Variable
1
SPM Practice 6 97
1.1 Quadratic Functions and Equations 2
Chapter
SPM Practice 1 13
7 Graphs of Motion 101
Chapter 7.1 Distance-Time Graphs 102
2 Number Bases 16 7.2 Speed-Time Graphs 106
SPM Practice 7 113
2.1 Number Bases 17
SPM Practice 2 24 Chapter
8 Measures of Dispersion for 118
Ungrouped Data
Chapter
3 Logical Reasoning 26 8.1 Dispersion 119
8.2
121
Measures of Dispersion
SPM Practice 8 135
3.1 Statements 27
3.2 Argument 36 Chapter
SPM Practice 3 45 9 Probability of Combined 139
Events
Chapter 9.1 Combined Events 140
4 Operations on Sets 48 9.2 Dependent Events and
Independent Events 141
4.1 Intersection of Sets 49 9.3 Mutually Exclusive Events and
4.2 Union of Sets 53 9.4 Non-Mutually Exclusive Events 147
Application of Probability of
4.3 Combined Operation on Sets 59 Combined Events 154
SPM Practice 4 64 SPM Practice 9 157
Chapter
Chapter Consumer Mathematics:
5 Network in Graph Theory 69 10 Financial Management 161
5.1 Network 70 10.1 Financial Planning and
SPM Practice 5 81 Management 162
SPM Practice 10 172
ii
00 Content Focus SPM Math.indd 2 15/02/2023 11:21 AM
FORM 5 Chapter Ratios and Graphs of
6
Chapter Trigonometric Functions 260
1 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
1.3 Combined Variation 185 SPM Practice 6 277
SPM Practice 1 188
Chapter
Chapter Measures of Dispersion for
2 Matrices 191 7 Grouped Data 283
2.1 Matrices 192 7.1 Dispersion 284
2.2 Basic Operation on Matrices 194 7.2 Measures of Dispersion 294
SPM Practice 2 206 SPM Practice 7 301
Chapter Chapter
3 Consumer Mathematics: 209 8 Mathematical Modeling 305
Insurance
3.1 Risk and Insurance Coverage 210 8.1 Mathematical Modeling 306
SPM Practice 3 219 SPM Practice 8 316
Chapter SPM Model Paper 318
4 Consumer Mathematics: 222
Taxation
Answers 335
4.1 Taxation 223
SPM Practice 4 232
Form 1 - 3 Intensive Revision
Chapter Congruency, Enlargement
5 and Combined
234
Transformations
https://qr.pelangibooks.com/?u=FSPM2023MMIntRev123
5.1 Congruency 235
5.2 Enlargement 239
Form 1 - 3 Intensive Revision Answers
5.3 Combined Transformation 247
5.4 Tessellation 254
https://qr.pelangibooks.com/?u=FSPM2023MMIntRev123A
SPM Practice 5 257
iii
00 Content Focus SPM Math.indd 3 15/02/2023 11:21 AM
Mathematical Formulae
– 2
1. n(A B) = n(A) + n(B) – n(A B) 21. Variance, σ = Σf(x – x) = Σfx 2 – x –2
2
Σf Σf
2. n(A) = n() – n(A)
22. Standard deviation,
Σx
Σ(x – x)
2
– 2
=
– x
–2
3. n(A B) = n(A B) σ =
N
N
4. n(A B) = n(A B) 23. Standard deviation,
Σfx
Σf(x – x)
2
– 2
=
– x
–2
Σf
5. P(A) = n(A) σ =
Σf
n(S)
6. Complement of event A, P(A) = 1 – P(A) 24. A = 1 3 d –b 4
–1
ad – bc –c a
7. P(A and B) = P(A B)
×
25. Premium = Face value of policy Premium rate
RMx
per RMx
8. P(A B) = P(A) × P(B)
26. Amount of required insurance
9. P(A or B) = P(A B) = Percentage of × Insurable value
co-insurance of property
10. P(A B) = P(A) + P(B) – P(A B)
27. Chargeable income
11. Σd(v) = 2E; v V = Total annual – Tax – Tax relief
income exemption
Vertical distance
12. Gradient, m = 28. Property assessment tax
Horizontal distance = Property assessment tax rate × Annual value
y – y
2
13. m = x – x 1 1 29. Quit rent
2
y-intercept = Quit rent rate per unit area × Total land area
14. m = –
x-intercept 30. Scale factor, k
Distance Distance of corresponding point of
15. Speed =
Time = image from O
Total distance travelled Distance of point of object from O
16. Average speed =
Total time taken = Length of corresponding side of image
Change of speed Length of side of object
17. Acceleration =
Change in time 31. Area of image = k × Area of object
2
–
18. x = Σx
N
32. Range = Midpoint of the – Midpoint of the
–
19. x = Σfx highest class lowest class
Σf
– 2
20. Variance, σ = Σ(x – x) = Σx 2 – x 33. Interquartile range
–2
2
= Third quartile, Q − First quartile, Q
N N 3 1
iv
0b Math Formulae Focus SPM Math.indd 4 15/02/2023 11:22 AM
Learning Area: Numbers and Operations Form 4
Chapter
2 Number Bases
1 foot = 12 inches.
1 foot = 12 inch.
The length of this 2 feet = 24 inches.
The length of this
2 foot = 24 inches.
board is two feet. Oh, the length of the
board is two feet
Oh, the length of the
board is 24 inches.
board is 24 inches.
Access to
i-Study SPM
• Convert – Tukar
• Digit value – Nilai digit Access to
• Number base – Asas nombor INFOGRAPHIC
• Place value – Nilai tempat
• Repeated division – Pembahagian berulang
Different number bases have different purposes and usages. We often use base ten system
in our daily lives, for example, in measurement of length and weight. The Maya and Babylon
civilisation used base-20 system and base-60 system respectively. These system may be easier
to perform certain Mathematics operations at that time, for example the calculation involving
circles. Nowadays, computers use binary system (base two system). Do you know how binary
system works in a computer?
16
02 Focus SPM Maths F4.indd 16 17/02/2023 9:39 AM
Mathematics SPM Chapter 2 Number Bases
8. In the number system in base m, we count
2.1 Number Bases something by arranging them in group of m,
then group of (m × m) and so on.
A Representing and explaining numbers 9. Each digit in a number base has a certain value
in various bases according to the place value. The place value of
a number is m where m is the base and n is the
n
1. The base for a number consists of base 2, 3 and power, n = 0, 1, 2, ….
so on without limit. For example, place value of each digit in the
numbers 147 and 101101 2 is as shown below:
2. The number of digits used in the formation of FORM
the number system depends on the base. Number in base 10 1 4 7
3. The table below shows the examples of number Place value 10 2 10 1 10 0 4
bases and digits used.
Number in base 2 1 0 1 1 0 1
Number base Digits used
Two 0, 1 Place value 2 5 2 4 2 3 2 2 2 1 2 0
Three 0, 1, 2 10. The value of a digit is the product of the digit
Four 0, 1, 2, 3 with its place value.
For example, the value of digit 3 in number 3212 4
Five 0, 1, 2, 3, 4 can be determined by the following methods.
Six 0, 1, 2, 3, 4, 5 (a) Multiplication of digit and place value
Seven 0, 1, 2, 3, 4, 5, 6 3212 4
Eight 0, 1, 2, 3, 4, 5, 6, 7 Number 3 2 1 2
Nine 0, 1, 2, 3, 4, 5, 6, 7, 8 Place value 4 3 4 2 4 1 4 0
Ten 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Digit value 3 × 4 2 × 4 1 × 4 2 × 4
0
3
1
2
= 192 = 32 = 4 = 2
4. The number that we use daily is in base ten. The
ten digits used in base ten numbers are 0, 1, 2, 3, (b) Block method
4, 5, 6, 7, 8 and 9. For example, 147 is a number 3212 4
in base ten. Number 3 2 1 2
5. Numbers in base eight use eight digits, which Place value 4 3 4 2 4 1 4 0
are 0, 1, 2, 3, 4, 5, 6 and 7. For example, 37 8 is a Digit value
number in base eight.
6. Numbers in base five use five digits, which are
0, 1, 2, 3 and 4. For example, 234 5 is a number in
base five.
7. Numbers in base two use only two digits, which
are 0 and 1. For example, 1101 2 is a number in
base two. 192 32 4 2
SPM Tips 11. The value of a number is the sum of all digit
values.
234 5 is read as “two three four base five”. For example, the number value of 1204 5 can be
determined by the following methods.
17
02 Focus SPM Maths F4.indd 17 17/02/2023 9:40 AM
Mathematics SPM Chapter 2 Number Bases
(a) Collecting process 3
1204 5
Find the value for each of the following numbers.
Number 1 2 0 4 (a) 121 3
Place value 5 3 5 2 5 1 5 0 (b) 405 6
(c) 10101 2
Digit value 1 × 5 2 × 5 0 × 5 4 × 5 (d) 1023 4
0
1
3
2
Number (1 × 5 ) + (2 × 5 ) + (0 × 5 ) + (4 × 5 ) Solution
0
3
2
1
value = 125 + 50 + 0 + 4 (a) 121 3
2
0
1
= 179 = (1 × 3 ) + (2 × 3 ) + (1 × 3 )
= 9 + 6 + 1
FORM
(b) Adding digit value using blocks = 16 Number 1 2 1
1204 5 Place value 3 2 3 1 3 0
4
1
0
2
Number 1 2 0 4 (b) 405 6 = (4 × 6 ) + (0 × 6 ) + (5 × 6 )
= 144 + 0 + 5
Place 5 3 5 2 5 1 5 0 = 149
value
(c) 10101 2 = (1 × 2 ) + (0 × 2 ) + (1 × 2 ) + (0 × 2 )
1
3
4
2
+ (1 × 2 )
0
= 16 + 0 + 4 + 0 + 1
Digit = 21
value
(d) 1023 4 = (1 × 4 ) + (0 × 4 ) + (2 × 4 ) + (3 × 4 )
3
1
0
2
= 64 + 0 + 8 + 3
= 75
Number 125 + 50 + 0 + 4
value = 179 Try Questions 1 – 3 in Try This! 2.1
1 SPM Highlights
State the place value of the underlined digit for each Given 4 × 5 + 1 × 5 + 5p = 4130 5 , find the value of p.
2
3
of the following numbers. A 0 C 5
(a) 221 3 (b) 406 7 B 3 D 6
(c) 5324 9 (d) 34 5 Solution
3
2
1
Solution 4130 5 = 4 × 5 + 1 × 5 + 3 × 5 + 0 × 5 0
= 4 × 5 + 1 × 5 + 5p
2
3
(a) 3 Number 2 2 1 Therefore, 5p = 15
2
(b) 7 1 Place value 3 2 3 1 3 0 p = 3
(c) 9 2 Answer: B
(d) 5 0
2
Find the value of the underlined digit in each of the SPM Highlights
following numbers. State the value of digit 5 in the number 2567 8 , in base
(a) 357 8 (b) 11011 2 ten.
(c) 412 5 (d) 6012 7 A 40 C 125
B 64 D 320
Solution
(a) 5 × 8 = 40 Number 3 5 7 Solution
1
4
(b) 1 × 2 = 16 Place value 8 2 8 1 8 0 Digit value 5 = 5 × 8 2
(c) 2 × 5 = 2 = 320
0
2
(d) 0 × 7 = 0 Answer: D
18
02 Focus SPM Maths F4.indd 18 17/02/2023 9:40 AM
Mathematics SPM Chapter 2 Number Bases
B Converting numbers from one base to 11 101 2
another (1 × 2 ) + (1 × 2 ) (1 × 2 ) + 0 + (1 × 2 )
0
1
2
0
1. Numbers in base m can be converted to base n 3 5
by using place value and base value. 11 can be written as 011 in the group of three digits.
2. Number in base ten can be converted to other 5. The table below shows the number in base two
bases by the following methods. For example, and its equivalent number in base eight.
convert 341 to number in base six.
(a) Division using place value Number in base two Number in base eight FORM
000 0
Place 6 = 1 296 6 = 216 6 = 36 6 = 6 6 = 1
0
4
2
1
3
value 001 1 4
1 3 2 5 010 2
Step 216 ) 341 36 )125 ) 6 17 ) 1 5 011 3
1 296 > 341 – 216 – 108 – 12 – 5
125 17 5 0 100 4
Base 6 1 3 2 5 101 5
341 = 1325 6 110 6
(b) Division using base value 111 7
6 341 Remainder
6 56 – 5 6. To change a number in base eight to a number
in base two, change each digit of the number in
6 9 – 2 The digits are read base eight to the equivalent three digits in base
from the bottom
6 1 – 3 upwards. two.
0 – 1 For example, 42 8 = 100010 2 .
The division is continued
until digit zero. 4 2 8
341 = 1325 6 100 010
3. Numbers in base m can be converted to other
base, n, by using the following steps. 4
Convert each of the following numbers to a number
Number in base m in the given base.
Find number value (a) 479 [base five]
(b) 11101 2 [base eight]
Number in base ten (c) 1256 8 [base three]
(d) 53 6 [base four]
Division (e) 341 5 [base eight]
Number in base n Solution
(a) 5 479 Remainder
4. To change a number in base two to a number 5 95 – 4 The digits are read from
in base eight, group every three digits of the 5 19 – 0 the bottom upwards.
number in base two from right to left. Then, 5 3 – 4
replace each group with the equivalent number 0 – 3
in base eight. The division is continued
until digit zero.
For example, 11101 2 = 35 8
479 = 3404 5
19
02 Focus SPM Maths F4.indd 19 17/02/2023 9:40 AM
Mathematics SPM Chapter 2 Number Bases
(b) 11101 2 = (1 × 2 ) + (1 × 2 ) + (1 × 2 ) + (0 × 2 ) 5
3
1
2
4
+ (1 × 2 )
0
= 16 + 8 + 4 + 0 + 1 Convert
= 29 (a) 1101111 2 to a number in base eight.
8 29 Remainder (b) 275 8 to a number in base two.
8 3 5
0 3 Solution
5
6
(a) 1101111 2 = (1 × 2 ) + (1 × 2 ) + (0 × 2 ) +
4
∴ 11101 2 = 35 8 (1 × 2 ) + (1 × 2 ) + (1 × 2 ) + (1 × 2 )
3
0
1
2
(c) 1256 8 = (1 × 8 ) + (2 × 8 ) + (5 × 8 ) + (6 × 8 ) = 64 + 32 + 0 + 8 + 4 + 2 + 1
1
0
3
2
= 512 + 128 + 40 + 6 = 111
FORM
= 686
4
3 686 Remainder 8 111 Remainder
3 228 2 8 13 7
3 76 0 8 1 5
3 25 1 0 1
3 8 1 ∴ 1101111 2 = 157 8
3 2 2
0 2
∴ 1256 8 = 221102 3 Alternative Method
1 101
(d) 53 6 = (5 × 6 ) + (3 × 6 ) 111 2
1
0
= 30 + 3 1 5 7
= 33
∴ 1101111 2 = 157 8
4 33 Remainder
4 8 1 (b) 275 8 = (2 × 8 ) + (7 × 8 ) + (5 × 8 )
0
2
1
4 2 0 = 128 + 56 + 5
0 2 = 189
∴ 53 6 = 201 4
2 189 Remainder
(e) 341 5 = (3 × 5 ) + (4 × 5 ) + (1 × 5 ) 2 94 1
2
0
1
= 75 + 20 + 1 2 47 0
= 96
2 23 1
8 96 Remainder 2 11 1
8 12 0 2 5 1
8 1 4 2 2 1
0 1 2 1 0
∴ 341 5 = 140 8 0 1
∴ 275 8 = 10111101 2
C
C C A L C U L A T O R Corner
Scientific calculator can be used to covert a number in Alternative Method
certain base to a number in other base. 2 7
Steps: 5 8
1. Set the calculator in BASE mode. 10 111 101
2. Key in the number as follow: ∴ 275 8 = 10111101 2
BIN 1 1 1 0 1 = OCT
Final display: 35 8 Try Questions 4 – 8 in Try This! 2.1
∴ 11101 2 = 35 8
20
02 Focus SPM Maths F4.indd 20 17/02/2023 9:40 AM
Mathematics SPM Chapter 2 Number Bases
SPM Highlights Subtraction
Express 436 8 as a number in base five. Subtract If the first digit is smaller Repeat the
A 2121 5 C 3421 5 the digits than the digit being subtraction
B 1120 5 D 1320 5 from subtracted, borrow 1 from process
right to next place value which for all the
left. has the value of m. digits.
Solution
436 8 = (4 × 8 ) + (3 × 8 ) + (6 × 8 )
2
0
1
= 256 + 24 + 6 (b) Conversion of number in base m to base
= 286 ten
5 286 Remainder Convert Perform Convert the FORM
5 57 1 the addition answer in
5 11 2 number in or base ten to 4
5 2 1 base m to subtraction base m
base ten
0 2
∴ 436 8 = 2121 5
Answer: A 6
Find the value of
(a) 23 5 + 14 5
SPM Highlights (b) 111 2 + 11 2
(c) 102 3 – 21 3
Express 5(5 + 5 + 2) as a number in base five.
2
1
(d) 235 8 – 37 8
A 3210 5 C 1120 5
B 3012 5 D 1040 5 Solution
Solution (a) 1
5(5 + 5 + 2) = 5 + 5 + 2(5) 2 3 5
1
2
3
2
= (1 × 5 ) + (1 × 5 ) + (2 × 5 ) + (0 × 5 ) + 1 4 5
3
2
0
1
= 1120 5 4 2 5
Write 2 in the answer
Answer: C space and 1 is carried
3 + 4 = 7 10 = 12 5 forward to the next
1 + 2 + 1 = 4 10 = 4 5 place value.
Therefore, 23 5 + 14 5 = 42 5
C Performing computations involving
addition and subtraction of numbers Alternative Method
in various bases 23 5 + 14 5 = 13 + 9 5 22 Remainder
= 22 5 4 – 2
1. Addition and subtraction in number base m can 0 – 4
be carried out using the following methods. Therefore, 23 5 + 14 5 = 42 5
(a) Using vertical form (b) 1 1 1
1 1 1 2
Addition + 1 1 2 Write 0 in the
answer space and 1
1 0 1 0 2 is carried forward to
The sum If the number next place value.
Add the of the in base m is Repeat 1 + 1 = 2 10 = 10 2
more than
digits digits one digit, the 1 + 1 + 1 = 3 10 = 11 2 Write 1 in the
from in base carry forward addition 1 + 1 = 2 10 = 10 2 answer space and 1
right to ten is the second process is carried forward to
left. converted for all the next place value.
to base digit to next digits.
m. place value. Therefore, 111 2 + 11 2 = 1010 2
21
02 Focus SPM Maths F4.indd 21 17/02/2023 9:40 AM
Mathematics SPM Chapter 2 Number Bases
Alternative Method 2 18 Remainder
111 2 + 11 2 = 7 + 3 2 10 Remainder 2 9 0
= 10 2 5 – 0 2 4 1
2 2 – 1 2 2 0
2 1 – 0 2 1 0
0 – 1 0 1
Therefore, 111 2 + 11 2 = 1010 2 ∴ 101 2 + 23 5 = 10010 2
(b) 211 3 – 14 8 = 22 – 12
(c) 2 1
0 3 Borrow 1 from the next 2 2 211 3 = (2 × 3 ) + (1 × 3 )
place value which has + (1 × 3 )
0
1 0 2 3 the value of 3. – 1 2 = 22
FORM
– 2 1 3 1 0 14 8 = (1 × 8 ) + (4 × 8 )
0
1
4
= 12
1 1 3
5 10 Remainder
5 2 0
2 – 1 = 1 3
3 – 2 = 1 3 0 2
Therefore, 102 3 − 21 3 = 11 3 ∴ 211 3 – 14 8 = 20 5
Alternative Method Try Questions 9 – 10 in Try This! 2.1
102 3 – 21 3 = 11 – 7 3 4 Remainder
= 4 3 1 – 1 SPM Highlights
Therefore, 102 3 – 21 3 = 11 3 0 – 1
(d) 8 Borrow 1 from the next 101011 2 + 1011 2 =
A 101100 2
1 2 8 place value which has
2 3 5 8 the value of 8. B 110101 2
C 110110 2
– 3 7 8 D 111010 2
1 7 6 8
Solution
1 1 1
8 + 5 – 7 = 6 8
8 + 2 – 3 = 7 8 + 1 0 1 0 1 1 2
Therefore, 235 8 − 37 8 = 176 8 1 0 1 1 2
1 1 0 1 1 0 2
Alternative Method Answer: C
235 8 – 37 8 = 157 – 31 8 126 Remainder
= 126 8 15 – 6 SPM Highlights
8 1 – 7
Therefore, 235 8 – 37 8 = 176 8 0 – 1 11101 2 – 111 2 =
A 10101 2
7 B 10110 2
C 10011 2
Find the value of D 10001 2
(a) 101 2 + 23 5 , give the answer in base two. Solution
(b) 211 3 – 14 8 , give the answer in base five.
2
Solution 0 0 2
(a) 101 2 + 23 5 = 5 + 13 1 1 1 0 1 2
5 101 2 = (1 × 2 ) + 0 + (1 × 2 ) – 1 1 1 2
2
0
= 5
+ 1 3 23 5 = (2 × 5 ) + (3 × 5 ) 1 0 1 1 0 2
1
0
1 8 = 13 Answer: B
22
02 Focus SPM Maths F4.indd 22 17/02/2023 9:40 AM
Mathematics SPM Chapter 2 Number Bases
D Solving problems involving number Try This! 2.1
bases
1. State the place value of the underlined digit in each
8 of the following numbers.
(a) 11011 2
Given X is the biggest three-digit number in base six. (b) 302 4
State X as a number in (c) 4653 7
(a) base four, (b) base eight. (d) 25 6
Solution 2. Find the value of the underlined digit in each of the
(a) The biggest three-digit number in base six = 555 6 following numbers. FORM
555 6 = (5 × 6 ) + (5 × 6 ) + (5 × 6 ) (a) 4121 5
2
0
1
= 180 + 30 + 5 (b) 111011 2 4
(c) 8457 9
= 215 (d) 2032 4
4 215 Remainder 3. Find the number value for each of the following.
4 53 3 (a) 67 8
4 13 1 (b) 235 6
4 3 1 (c) 1021 3
(d) 155 7
0 3
4. Convert each of the following numbers to a number
∴ 555 6 = 3113 4 in base two.
(b) (a) 32 5
(b) 476 8
8 215 Remainder (c) 2211 3
8 26 7
8 3 2 5. Convert each of the following numbers to a number
in base five.
0 3 (a) 110011 2
∴ 555 6 = 327 8 (b) 553 7
(c) 121 4
Try Questions 11 – 12 in Try This! 2.1
6. Convert each of the following numbers to a number
in base six.
Example of HOTS (a) 40 5
HOTS Question
Given 42 5 Y 103 5 where Y is an odd number in (b) 321 4
base ten. List all the possible values of Y. (c) 2367 8
Solution: 7. Convert each of the following numbers to a number
in base eight.
42 5 Y 103 5
42 5 = (4 × 5 ) + (2 × 5 ) (a) 100101 2
1
0
= 22 (b) 58 9
(c) 422 5
103 5 = (1 × 5 ) + (0 × 5 ) + (3 × 5 )
2
1
0
= 28 8. Convert
22 Y 28 (a) 10011101 2 to a number in base eight.
(b) 111011100 2 to a number in base eight.
∴ Y = 23, 25, 27
(c) 156 8 to a number in base two.
(d) 347 8 to a number in base two.
Try this HOTS Question
Given 61 7 K 112 7 where K is a number in 9. Find the value of
base ten. Find the biggest difference between (a) 21 4 + 33 4
values of K. (b) 240 5 + 132 5
(c) 543 6 + 441 6
Answer: (d) 1010011 2 – 1110 2
57 – 44 = 13 (e) 75 8 – 47 8
(f) 352 9 – 81 9
23
02 Focus SPM Maths F4.indd 23 17/02/2023 9:40 AM
Mathematics SPM Chapter 2 Number Bases
10. Find the value of 11. Given X is the biggest four-digit number in base
(a) 222 3 + 323 5 , give the answer in base three. three. State X as a number in
(b) 625 8 + 1101001 2 , give the answer in base five. (a) base six,
(b) base nine.
(c) 235 7 – 310 4 , give the answer in base six.
(d) 334 6 – 128 9 , give the answer in base eight. 12. Given 203 4 Y 50 8 where Y is an integer in base
five. List all the possible values of Y.
FORM
SPM Practice 2 Full
solution
4
PAPER 1 6. The value of digit 8 in the number 5847 9 is 8 × 9 .
m
State the value of m.
1. State the value of digit 5 in the number 23561 8 in A 0
base ten. B 1
A 1080 C 2
B 320 D 3
C 500
D 560 7. Given that 257 8 = K 5 , find the value of K.
A 1114
B 1200
2. Express 2 + 2 + 2 + 1 as a number in base two. C 1014
4
5
A 110010 2 D 1002
B 110011 2
C 110101 2 8. Given that 1m13 5 = 183 10 , find the value of m.
A 2
D 11011 2
B 1
3. The table below shows the equivalent value of C 3
numbers in base 5 and base 2. D 4
Base 5 Base 2 9. Convert 10011101 2 to a number in base eight.
1 001 A 215 8
2 010 B 245 8
3 011 C 225 8
4 100 D 235 8
Find the equivalent value of 23 5 in base 2. 10. Given 10110 2 = P 8 , where P is an integer, find the
A 1110 2 value of P.
B 1101 2 A 26
B 22
C 1011 2
D 1111 2 C 16
D 12
2
4. Express 5(5 + 4) as a number in base five. 11. Given that P 8 = 8 + (3 × 8 ) + (5 × 8 ), then P =
2
4
0
A 140 5 A 135
B 1035
B 301 5
C 13005
C 104 5
D 1040 5 D 10305
5. Express 6 + 6 + 12 + 4 as a number in base six. 12. Express 2212 3 as a number in base seven.
2
4
A 10114 6 A 104 7
B 214 7
B 10110 6
C 10124 6 C 140 7
D 224 7
D 402124 6
24
02 Focus SPM Maths F4.indd 24 17/02/2023 9:40 AM
Mathematics SPM Chapter 2 Number Bases
13. Given that 25k 6 = 1100111 2 , find the value of k. 3. Convert each of the following numbers to a number
A 5 in base seven.
B 3 (a) 1221 4
C 2 (b) 57 8
D 1
4. Find the number value for each of the following.
14. 110101 2 – 11011 2 = 1m01n 2
A m = 1, n = 1 (a) 566 7
B m = 1, n = 0 (b) 110011 2
C m = 0, n = 1 (c) 77 8
D m = 0, n = 0 (d) 134 5
15. The table below shows the number of cakes 5. Express 2 × 6 + 1 × 6 + 3 × 6 + 3 as a number in FORM
1
4
5
produced in a bakery from January to April. base six.
January February March April 4
m
6. (a) The value of digit 4 in the number 427 8 is 4 × 8 .
2132 4 X 5 235 7 305 8
State the value of m.
Calculate the value of X if the number of cakes (b) State the biggest three-digit number in base
produced in February is twice the number of cakes three.
produced in March.
A 443 7. (a) Given 1011101 2 = P 5 = Q 8 . Find the value of P
B 444 and of Q.
C 1443 (b) Given 1100 2 R 25 6 where R is an integer in
D 1444
base five. List all the possible values of R.
16. Find the value of P in the equation 301 4 + P 4 = 1230 4 , HOTS
where P is an integer. Applying
A 122 8. (a) Given 13 5 X 22 5 . List all the possible
B 323 values of X.
C 321 (b) Given 24 5 = T 10 = U 8 . Find the value of U – T.
D 210
17. Find the value of Q in the equation 225 7 – Q 7 = 32 7 , 9. P is the smallest three-digit number in base seven.
where Q is an integer. State P as a number in
A 113 (a) base two,
B 153 (b) base nine.
C 163
D 123 10. Given 2H5 8 is a three-digit number in base eight.
Find the value of H if
18. 321 5 + 244 5 =
A 1220 5 (a) 2H5 8 = 10111101 2 .
(b) 2H5 8 = 141 10 .
B 1230 5
C 1120 5
D 1110 5 11. (a) Given 2 m – 1 – 1 = 1111 2 . Find the value of m.
19. Given 415 7 – 2201 3 = X 5. Find the value of X. (b) Given 5 n – 1 + 1 = 11010 2 . Find the value of n.
A 1020 HOTS Analysing
B 135 12. Given 5P2 7 is a three-digit number in base seven.
C 120 Determine the value of P if
D 1121
(a) 5P2 7 = 5(7 ) + 3(7 ) + 2(7 ).
1
2
0
PAPER 2 (b) 5P2 7 = 376 8 .
1. Find the value of the underlined digit in each of the 13. The diagram below shows butters of Brand P and
following numbers. Brand Q.
(a) 2101 3
(b) 65 8
RM110 3 RM22 5
(c) 431 5
1.25 kg 1.15 kg
(d) 5411 6
Brand P Brand Q
2. Convert each of the following numbers to a number
in base three. Which brand of butter is sold at a more economical
(a) 10111 2 price? Justify your answer.
(b) 246 9
25
02 Focus SPM Maths F4.indd 25 17/02/2023 9:40 AM
Learning Area: Measurement and Geometry Form 5
Chapter
5 Congruency, Enlargement and
Combined Transformations
• Congruency – Kekongruenan
• Enlargement – Pembesaran Access to
• Reflection – Pantulan i-Study SPM
• Rotation – Putaran
• Scale factor – Faktor skala
• Similarity – Keserupaan Access to
• Tessellation – Teselasi INFOGRAPHIC
• Transformation – Transformasi
• Translation – Translasi
Architects always used the concept of transformation to design a building. The different types
of transformations can be combined to produce unique shapes. What building in Malaysia apply
the concept of transformation? What is the transformation applied?
234
05 Focus SPM Math F5.indd 234 17/02/2023 9:48 AM
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
5.1 Congruency (a) A 6 cm C R
8 cm 53° 10 cm
37°
A Differentiate between congruent and P 8 cm Q
non-congruent shapes based on sides B
and angles (b) K 5 cm N
D C
1. Congruent shapes are the shapes that have
the same size and shape regardless of their 5 cm 8 cm
orientation. A 9 cm B
L M
2. A pair of congruent polygons have the same
measurement for the corresponding sides and (c) H Z
angles. 54°
7 cm X
3. The diagram below shows two quadrilaterals, 66°
ABCD and EFGH, that are congruent. K J 7 cm
Y
D 10 cm H (d) 8 cm C 12 cm
50° C D U V
119° 50° 7 cm
7 cm 5 cm 15 cm
123° 68° 10 cm 123° E 15 cm
A 8 cm B 82° 98°
A 12 cm B X 8 cm W
119° 8 cm
G
68° Solution
5 cm
2
2
F (a) BC = √6 + 8
From the diagram above, the same pairs of = 10 cm
corresponding sides and angles are shown in the = QR
table below. ∠ACB = 180° – 90° – 37°
= 53°
Corresponding Corresponding = ∠PRQ
sides angles 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° FORM
2
= 63°
• If there is one pair of the corresponding sides or angles ≠ ∠XYZ 5
are different sizes, then both of the shapes are not Hence, both shapes are not congruent.
congruent.
• The arrangement of the shapes does not affect the (d) ∠ADC = 180° − 82°
congruency between the shapes. = 98°
= ∠UXW
The measurements of all corresponding sides
1 and angles are equal. Hence, both shapes are
congruent.
Determine whether each pair of the following shapes
are congruent. Try Questions 1 – 2 in Try This! 5.1
235
05 Focus SPM Math F5.indd 235 17/02/2023 9:48 AM
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.
Properties of triangle congruency Situation of triangle congruency
Side-Side-Side (SSS) A P
• Each pair of the corresponding sides are of equal
length.
• AB = PQ, BC = QR, AC = PR
B C Q R
Side-Angle-Side (SAS) A P
• Two pairs of the corresponding sides are
equal in length and the sizes of the pair of the
corresponding subtended angle between the two
sides are equal. B C Q R
• AB = PQ, ∠ABC = ∠PQR, BC = QR
Angle-Side-Angle (ASA) A P
• Two pairs of the corresponding angles are equal
and the lengths of the pair of the corresponding
side between the two angles are equal.
• ∠ABC = ∠PQR, BC = QR, ∠BCA = ∠QRP B C Q R
Angle-Angle-Side (AAS) A P
• Two pairs of the corresponding angles are equal
and the lengths of a pair of the corresponding
sides which do not lie between the two angles are
equal. B C Q R
• ∠ABC = ∠PQR, ∠BCA = ∠QRP, AC = PR
Angle-Angle-Angle (AAA) A P
FORM
• All the three corresponding angles are equal.
5
• The areas of the pair of triangles must be equal.
• ∠ABC = ∠PQR, ∠BCA = ∠QRP, ∠CAB = ∠RPQ
B C Q R
Side-Side-Angle (SSA)
• Two pairs of the corresponding sides are equal A P
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. B C Q R
• AB = PQ, BC = QR, ∠BCA = ∠QRP
236
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
SPM Tips 3
The diagram below shows a triangle PQR.
A P
R
B C Q R P
Right angle-Hypotenuse-Side (RHS) is a special triangle
congruency for right-angled triangle. If two right-angled Q
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 hypotenuse are the same, then both the triangles the triangle PQR. State the triangle congruency used
are congruent. This triangle congruency is based on to determine both triangles are congruent if
Side-Side-Angle (SSA).
(a) PQ = CD, PR = CE and ∠QPR = ∠DCE
(b) QR = DE, ∠PQR = ∠CDE and ∠PRQ = ∠CED
2 Solution
(a)
The diagram below shows two congruent triangles R
which satisfy the conditions of Angle-Angle-Side
(AAS). P
C Z Q
E
Y
C
A B X
D
Complete the table below based on the triangle
congruency of Angle-Angle-Side (AAS). Two pairs of the corresponding sides and a pair
of the corresponding subtended angle between
Corresponding angles Corresponding sides the two sides are given. Hence, the triangle
∠ACB = ∠YZX congruency is Side-Angle-Side (SAS).
(b) R
Solution
Corresponding angles Corresponding sides P
∠ACB = ∠YZX AB = YX
∠CBA = ∠ZXY Q
E
or FORM
Corresponding angles Corresponding sides C 5
∠ACB = ∠YZX AB = YX D
∠BAC = ∠XYZ
Two pairs of the corresponding angles and a
pair of the corresponding side between the two
REMEMBER! angles are given. Hence, the triangle congruency
is Angle-Side-Angle (ASA).
Identify the corresponding vertices before determining
the corresponding sides and angles, for instance, A = Y,
B = X and C = Z. Try Questions 3 – 4 in Try This! 5.1
237
05 Focus SPM Math F5.indd 237 17/02/2023 9:48 AM
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
C Solving problems involving congruency Solution:
Analyse the
4 8 cm diagram given
x cm to obtain the
The diagram below shows a parallelogram ABCD. 16 cm measurement
D Q C So, x = 8 ÷ 2
= 4
Area = 3 × [(4 × 8) + (8 × 8)]
= 288 cm
2
Try this HOTS Question
A P B
P and Q are midpoints of the sides AB and CD The diagram below shows the plan of an office.
respectively. Determine whether the triangle APD is
congruent with
(a) triangle QCP. 20 m
(b) triangle CQB.
Solution 5 m
(a) D Q C
The office is divided into two congruent parts.
If the total area of the office is 550 m , find the
2
perimeter for one congruent shape.
Answer: 84 m
A P B
It is found that AP = QC and AD = QP but
∠PAD ≠ ∠CQP and PD ≠ PC.
Hence, triangles APD and QCP are not Try This! 5.1
congruent.
1. Determine whether each pair of the following shapes
(b) D Q C are congruent.
(a)
8 cm 14 cm
5 cm
A P B
It is found that AP = CQ, AD = CB and
∠PAD = ∠QCB, so they satisfy triangle (b) 14 cm 13 cm
congruency Side-Angle-Side. Hence, triangles
APD and CQB are congruent. 13 cm 14 cm
Try Questions 5 – 7 in Try This! 5.1 72° 98°
FORM
Example of HOTS (c)
HOTS Question
5
The diagram below shows a shape consisting of a 10 cm 10 cm
combination of three congruent shapes. 4 cm 6 cm
7 cm 3 cm
8 cm
(d)
16 cm
110°
105°
2
Find the area, in cm , of the shape.
238
05 Focus SPM Math F5.indd 238 17/02/2023 9:48 AM
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 Q S 37° 20 cm
R
16 cm
R P
U V W K L
T Find
(a) ∠PRQ.
(b) the area, in cm , of triangle PQR.
2
3. The diagram below shows two congruent triangles, 7. The diagram below shows a shape consisting of a
ABC and EFG. combination of three congruent arrows.
C F P
3 cm
A E
5 cm
B G
It is given that AB = EF and ∠ABC = ∠EFG. State
another condition if the triangle congruency involved Q
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 cm , of the shape.
2
4. The diagram below shows a triangle PQR.
5.2 Enlargement
R
A Explaining the meaning of similarity
Q of geometric objects
P
1. Similarity of geometric objects means that the
It is given that another triangle XYZ is congruent
with the triangle PQR. State the triangle congruency objects have the same shape regardless of their
involved if size and orientation.
(a) PQ = XY, QR = YZ and PR = XZ 2. Two similar objects have the same corresponding
(b) QR = YZ, ∠PQR = ∠XYZ and ∠PRQ = ∠XZY angles and fixed ratio of the corresponding sides.
3. The diagram below shows two similar triangles,
5. The diagram below shows a trapezium ABCD.
ABC and PQR.
D C FORM
C R 5
A B A B P 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
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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.
(a) A B 2. The fixed point is known as centre of
L enlargement.
50° M
3. The ratio of the distance of centre of enlargement
from a point on the image to the distance of
120° 50° 100° centre of enlargement from the corresponding
C point on the object is known as scale factor.
K N
D
4. In an enlargement,
(b) Z (a) the corresponding angles between the
R 21 cm object and image are equal.
14 cm (b) the ratios of the corresponding sides are
15 cm Y
P 9 cm 10 cm constant.
6 cm
Q X
5. Both object and image in an enlargement are
Solution similar.
(a) ∠B = 360° – 50° – 120° – 90° 6. The diagram below shows an enlargement.
= 100°
= ∠N A
∠L = 360° – 50° – 100° – 90° A
= 120° Centre of C Object Image
= ∠D enlargement O C B
∠A = ∠K = 50° m B
∠C = ∠M = 90° n
All the corresponding angles are equal. Hence, 7. Scale factor, k, of an enlargement can be
quadrilateral ABCD and quadrilateral KLMN are determined as follows.
similar. distance of point of image from O n OB9
(b) PQ = 6 = 3 k = distance of point of object from O = m = OB
YX 10 5 or
QR = 9 = 3 length of side of image A'B9
XZ 15 5 k = length of side of object = AB
PR = 14 = 2
YZ 21 3 8. For scale factor, k, within the range 0 , k , 1,
The ratio of the corresponding sides PR and the size of the image formed is smaller than the
YZ is different with the ratio of the other object as shown in the diagram below.
corresponding sides. Hence, triangle PQR and
triangle XYZ are not similar.
Image Object
FORM
Try Questions 1 – 2 in Try This! 5.2 Centre of
5
enlargement O
10 cm
B Making a connection between 20 cm
similarity and enlargement, hence 10 cm 1
describing enlargement using Scale factor, k = 20 cm = 2
representation
9. When the scale factor is a negative value, the
1. Enlargement is a type of transformation where image formed under enlargement is located on
the image is formed based on a fixed point the opposite side of the object as shown in the
such that the ratio of the distance of the fixed diagram below.
240
05 Focus SPM Math F5.indd 240 17/02/2023 9:48 AM
Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
6
Centre of
enlargement Object In each of the following diagrams, triangle A9B9C9 is
Image
O the image of the object ABC under an enlargement.
20 cm 10 cm
(a) y
C 10
Scale factor, k = – 20 cm = –2 8
10 cm B
6
10. The following table shows the size of image C 4
and the position of image for the scale factor in B 2
different range. A A
–6 –4 –2 O 2 4 x
Position of image
Scale factor, Size of image to the centre of
k (b) y
enlargement O 6
C B
4
larger than the Be on the same
k > 1
size of object side as the object 2
A C B
x
smaller than –8 –6 –4 –2 O 2
0 < k < 1 the size of Be on the same A –2
side as the object
object –4
smaller than Describe the enlargement in each diagram.
−1 < k < 0 the size of Be on the opposite
side as the object
object Solution
(a) y
larger than the Be on the opposite C 10
10
k < −1
size of object side as the object
8
B
When k = 1, the 6
image is on the C 4
same side as the B
k = 1 or equal in size to object. P(–4, 1) 2 A
A
k = −1 the object When k = −1, the –6 –4 –2 O 2 4 x
image is on the
opposite side as PA9 The ratio of the FORM
the object. Scale factor = PA distance of point on
distance of point on
= 6 units image from P to the 5
2 units
object from P
SPM Tips = 3
The centre of enlargement can be determined from A9B9C9 is the image of ABC under an enlargement
the intersection point between all the straight lines that
connecting each pair of the corresponding points. at centre P(−4, 1) with scale factor 3.
241
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
(b) y Scale factor = – distance of S from centre
6 distance of H from centre
C B 3 units Negative sign shows
4 = –
2 units that the image is on the
2 opposite side as the object
C B 3
A = –
x 2
–8 –6 –4 –2 O 2
A –2 PQRS is the image of EFGH under an enlargement at
3
P(0, –3) centre (2, 3) with scale factor – .
–4 2
Try Questions 3 – 4 in Try This! 5.2
Scale factor = B9C9 The ratio of the
BC length of side of
= 3 units image to the length C Determining the image and object of
of side of object
6 units an enlargement
= 1 The flow map below shows the steps to determine the
2 image or object of an enlargement.
A9B9C9 is the image of ABC under an enlargement
1
at centre P(0, −3) with scale factor . Draw the Draw the
2 Identify projection image or object
Identify the the line from which has the
7 centre of scale the centre similar shape
enlargement
according to
In the diagram below, PQRS is the image of EFGH factor of the scale factor
enlargement
under an enlargement.
y
8 F 8
6
G E The diagram below shows four similar triangles drawn
4
S on the Cartesian plane.
2 H
y
P R x
–4 –2 O 2 4 6 8
–2 6
Q
–4 P
4
Describe the enlargement. 2
x
Solution –8 –6 –4 –2 O 2 4 6 8 10
A –2
y
8 –4
F Q
FORM
6 –6
5
G E
4 –8 R
S
2 H –10
P R x
–4 –2 O 2 4 6 Determine the image of triangle A under the
–2
5
Q enlargement at centre (−8, 4) with scale factor .
–4 3
242
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
Solution Solution
(a) When k = 3, the distance of each corresponding
y
vertex of image from O is 3 times the distance
8
of the vertex of object from O on the same
6 direction.
(–8, 4) P
4
2
x
–8 –6 –4 –2 A O 2 4 6 8 10
–2
–4
Q
–6 O
R 1
–8 (b) When k = − , the distance of each corresponding
2
1
–10 vertex of image from O is times the distance
The image of A is Q. 2
of the vertex of object from O on the opposite
direction.
REMEMBER!
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
3 O
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
to determine the ratio of a point from the centre of
scale factor. enlargement as follows.
(a) Scale factor, k = 3 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.
P
3 × 2 = 6 units FORM
O P
2 units 5 units 5
1
(b) Scale factor, k = – O
2 3 × 5 = 15 units
10
O Draw the object for each of the following images
under the enlargement at centre O with the given
scale factor.
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
4
(a) Scale factor, k = 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.
(b) Scale factor, k = −2
O y cm
A q cm
x cm A
p cm
O
Since scale factor = k, so the ratio of the length
of corresponding sides of image to the length of
Solution sides of object = k,
4
(a) When k = , the distance of each vertex of object which is p = k and q = k
3 x y
3
from O is 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 = xy
2
O 1
and area of the image = pq
2
= 1 (kx)(ky)
2
= 1 k xy
2
2
3. From the ratio of area of the image to area of the
(b) When k = −2, the distance of each vertex of object object,
1
from O is times the distance of corresponding 1 2
2
vertex of image from O in the opposite direction. area of the image = 2 k xy = k
2
area of the object 1 xy
2
4. Generally, the area of the image of an enlargement
O can be determined by the following relation.
FORM
5
SPM Tips
We can use the inverse scale factor to determine the 11
object of an image under an enlargement. For example, 2
if A9 is image of A under the enlargement at centre O It is given that the area of a geometric object is 16 cm .
with scale factor k, then we can perform the enlargement Calculate the area of its image under an enlargement
at centre O with scale factor 1 k on the image A9 to with scale factor,
determine the object A. (a) k = 5
3
(b) k =
4
Try Questions 5 – 7 in Try This! 5.2
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
Solution Solution
(a) Area of the image = k × area of the object (a) Scale factor, k = AB
2
= 5 × 16 PB
2
= 400 cm = 8 + 12
2
12
3
(b) Area of the image = 1 2 2 × 16 = 5
4
= 9 cm 3
2
(b) Area of the image = k × area of the object
2
12 100 = 1 2 2 × area of the object
5
3
The area of the image of object A under an enlargement Area of the object = 100 × 9
is 54 cm . Find 25
2
2
(a) the area of the object if the scale factor of the = 36 cm
enlargement is 3, Area of the shaded region = 100 – 36
(b) the scale factor if the area of the object is 24 cm . = 64 cm
2
2
Solution
(a) Area of the image = k × area of the object 14
2
54 = 3 × area of the object The diagram below shows the measurement of a small
2
Area of the object = 54 Jalur Gemilang.
9
= 6 cm
2
90 cm
(b) 54 = k × 24
2
k = 54 1.8 m
2
24
= 9 A school intends to make a large Jalur Gemilang
4 with an area of 103.68 m by using the concept of
2
3
k = + or – 3 enlargement. Find the length and width of the Jalur
2 2 Gemilang that the school intends to make.
Try Question 8 in Try This! 5.2 Solution
Area of the small Jalur Gemilang = 1.8 × 0.9 = 1.62 m
2
E Solving problems involving and area of the large Jalur Gemilang = 103.68 m
2
enlargement Area of the image
2
k =
13 Area of the object
= 103.68 m 3
In the diagram below, triangle ABC is the image of 1.62 m 2
triangle PBQ under an enlargement at centre B. = 64 FORM
k = 8
C
Length of Jalur Gemilang = 1.8 × 8 = 14.4 m 5
Q
Width of Jalur Gemilang = 0.9 × 8 = 7.2 m
Try Questions 9 – 10 in Try This! 5.2
A 8 cm P 12 cm B
Find
(a) the scale factor of the enlargement. Try This! 5.2
(b) the area of the shaded region, if the area of 1. Determine whether each pair of the following
triangle ABC is 100 cm . geometric objects are similar.
2
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
(a) (b) y
10
76°
42°
8
76°
62°
6
A
4
(b) 14 cm A
10.5 cm 2
140° 12.4 cm 9.3 cm
16 cm 12 cm 80° x
–8 –6 –4 –2 O 2 4 6 8
(c) y
(c) 14 cm 10
7 cm 8
8 cm A
4 cm 6
A 4
2
(d) 9 cm
6 cm x
5 cm 5 cm 8 cm 8 cm –10 –8 –6 –4 –2 O 2 4 6
5 cm 5 cm
8 cm 8 cm 4. Describe the enlargement in the following diagram
6 cm
9 cm where P9 is the image of P.
2. Determine similar triangles in the following diagrams.
P
A P
B O
C 6 cm 4 cm
5. Based on the given scale factor, determine the object
E and image of each of the following enlargement.
(a) Scale factor, k = 2
F
D
G
P
Q
3. Describe the enlargement in each of the following
diagrams where A9 is the image of A.
(a) y 1
FORM
(b) Scale factor, k =
4 3
5
2 A
x
–10 –8 –6 –4 –2 O 2 4 6
–2
A Y
–4 X
–6
–8
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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
image of each of the diagrams under an enlargement where ΔQST is the image of ΔPQR under an
at centre O. enlargement.
3 HOTS
(a) Scale factor = (b) Scale factor = 4 Analysing
2 R
3 cm
Q S
P
O
12 cm
O T
(a) Describe the enlargement.
(b) Given that the area of the whole diagram is
1
(c) Scale factor = – (d) Scale factor = −2 102 cm , find the perimeter, in cm, of the whole
2
3 diagram.
5.3 Combined transformation
O
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
for each of the diagrams under an enlargement at
centre O. A under four types of transformations.
(a) Scale factor = 2 (b) Scale factor = – 3 Translation Reflection
2
5
Translation 1 2 Reflection on the line
y = 3
–2
y y
5 units
O 6 6
A 2 units A
4 4
O A
2 2 y = 3
A
8. The table below shows the values of the area of the O x O x
object, the area of the image and the scale factor 2 4 6 2 4 6
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
(a) 10 cm 2 90 cm 2 clockwise at centre (0, 6) with scale
2 (1, 2) factor 3
(b) 80 cm 2
5 y y FORM
(c) 7 unit 2 – 1 6 6
2 A 5
4 4 A
9. In the diagram below, trapezium APQR is the image
of trapezium ABCD under an enlargement. Vertex A 2 (1, 2) 2 A
is the centre of the enlargement. A
O 2 4 6 x O 2 4 6 x
D C
R Q 2. Generally, the combination of two
transformations, A and B, can be performed in
P
A B 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 cm . Find the area, in cm , of trapezium APQR.
2
2
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
3. Combined transformation AB means
transformation B occurred then followed by y
transformation A. Q (–2, 7) 8
6
4. Combined transformation AA means
transformation A occurred two times in a row 4
and is written as A . 2 P(3, 2)
2
5. The diagram below shows the steps to determine x
the image or object of a combined transformation –2 O 2 4 6 8
AB.
Transformation Transformation (a) Determine the image of point P under the
B Image under A combined transformation AB.
transformation (b) If Q0 is the image of point Q under the
B transformation AC, determine the coordinates
Object Image of point Q.
Solution
Object of the
Object under image under Object under (a) Combined transformation AB means
transformation transformation transformation transformation B followed by transformation A.
B A A P9 is the image of P under the transformation B
and P0 is the image of P9 under the transformation
A.
SPM Tips y
The table below shows the ways to determine the object 8
of an image under a transformation. P (–4, 6)
6 y = x
Transformation
Type of transformation performed on the image 4 P (2, 3)
to determine the object 2 P(3, 2)
–x
x
Translation 1 y 2 Translation 1 –y 2 x
–4 –2 O 2 4 6
Reflection on the line y = x Reflection on the line y = x
Rotation in clockwise Rotation in anticlockwise The image of point P is P0(−4, 6).
direction at centre (x, y) direction at centre (x, y)
Enlargement at centre Enlargement at centre O (b) Combined transformation AC means
1
O with scale factor k with scale factor transformation C followed by transformation
k A. Q9 is the object of image Q0 under the
transformation A and Q is object of image Q9
under transformation C.
y
FORM
15 8 Q(8, 8)
5
It is given that transformations Q (–2, 7)
6
–6
A = translation 1 2 4 Q (4, 4) (8, 4)
3
B = reflection on the line y = x 2
C = rotation of 90° anticlockwise at centre (8, 4) –2 O 2 4 6 8 x
The diagram below shows the point P and point Q0 on
a Cartesian plane. The coordinates of point Q are (8, 8).
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
(b) Combined transformation PQ means
REMEMBER! transformation Q followed by transformation P.
A9 is the image of A under the transformation
When determining the object of an image under a combined
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
4 A
The diagram below shows a triangle A drawn on a
Cartesian plane. 2
x
–10 –8 –6 –4 –2 O 2 4 6
y
–2
6
A –4
4 A
–6
2
A
–8
O 2 4 6 x
–10
It is given that transformations
1
P = translation 1 2 REMEMBER!
–4
Q = rotation of 180° at the origin
R = enlargement at centre (4, 3) with scale factor 2 For the rotation of 180°, its image is the same either in
clockwise or anticlockwise direction.
Determine the image of triangle A under the
combined transformation
(a) P (c) Combined transformation QR means
2
(b) PQ transformation R followed by transformation Q.
(c) QR
A9 is the image of A under the transformation
Solution R and A0 is the image of A9 under the
(a) Combined transformation P 2 means transformation Q.
transformation P is performed twice in a row.
A9 is the image of A under the transformation y
P and A0 is the image of A9 under the 8
transformation P.
6
y A
4
6 A (4, 3)
2 FORM
4 A
x
2 –8 –6 –4 –2 O 2 4 6 8 5
–2
x
–6 –4 –2 O 2 A 4 6
–2 –4 A
–6
–4
A
–8
–6
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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
C. Y is the object of image P under the
drawn on a Cartesian plane.
transformation C and X is the object of image Y
under the transformation B.
y
8 y
(–1, 3) 4 x = 2
6 R
P
2
4
P Q x
2 –8 –6 –4 –2 O 2 4 6
Y X
x –2
–8 –6 –4 –2 O 2 4 6 8 –4
–2 U
–4 Hence, the object of image P under the combined
transformation CB is U.
T –6 V
S –8 Try Questions 1 – 4 in Try This! 5.3
It is given that transformations B Making and verifying the conjecture
A = enlargement at centre (−2, 3) with scale factor about commutative law in combined
− 1 transformation
2
B = reflection on the line x = 2 1. A combined transformation AB satisfies
C = rotation of 90° clockwise at centre (−1, 3) the commutative law if the images under the
Determine the object of image P under the combined combined transformations AB and BA are the
transformation same.
(a) AB 2. A combined transformation AB does not satisfy
(b) CB the commutative law if the images under the
combined transformations AB and BA are not
Solution the same.
(a) Combined transformation AB means
transformation B followed by transformation
A. Y is the object of image P under the 18
transformation A and X is the object of image Y The diagram below shows a quadrilateral K drawn on
under the transformation B. a Cartesian plane.
y y
8 x = 2 10
FORM
5
6 8
X Y
K
(–2, 3) 4 6
P
2 4
x 2
–8 –6 –4 –2 O 2 4 6 8
O 2 4 6 8 10 x
Hence, the object of image P under the combined
transformation AB is Q.
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
It is given that transformations For combined transformation CB, transformation
3
A = translation 1 2 B performed first followed by transformation C.
2
B = reflection on the line y = 5 y
C = rotation of 180° at centre (6, 5) 10
Determine whether each of the following combined 8
transformations satisfies the commutative law. 6 K K y = 5
(a) Combined transformation AB 4 (6, 5)
(b) Combined transformation BC K
2
Solution
(a) For combined transformation AB, transformation O 2 4 6 8 10 x
B performed first followed by transformation A. The images of the combined transformations
y BC and CB are the same. Hence, the combined
10
transformation BC satisfies the commutative
8 law.
K
6
K
4 y = 5 Try Question 5 in Try This! 5.3
K
2
C Describing combined transformation
O 2 4 6 8 10 x
For combined transformation BA, transformation 1. The diagram below shows key points to describe
A performed first followed by transformation B. a transformation.
y
10 Object
K
8
K
6
4 y = 5 Similar image Congruent
but different size image
2
K
O x
2 4 6 8 10 Enlargement Same Inverted
The images of the combined transformations orientation orientation
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 FORM
point are point are
y equal different 5
10
8 Translation Rotation
K K
6
4 (6, 5) y = 5 2. When describing a combined transformation
K
2 AB, we need to follow the order, which is
O 2 4 6 8 10 x transformation B comes first and followed by
transformation A.
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
19 (ii) y
10
In the diagram below, B is the image of object A
under a combined transformation PQ while C is the x = –1 8
image of A under a combined transformation RS. A 6 A
A
y 4
10 (–7, 4)
2
8
x
6 C –8 –6 –4 –2 O 2 4 6
A For transformation RS, transformation S
4
performed first followed by transformation
2
B R,
x S R
–8 –6 –4 –2 O 2 4 6 which is A A9 A0,
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 Hence, object A experienced the
(b) Describe a single transformation which is 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
the combined transformation PQ is the rotation
10
of 90° anticlockwise at centre (–2, 6).
8
Try Questions 6 – 7 in Try This! 5.3
6
A (–5, 4) 4 D Solving problems involving combined
A transformation
A 2
x SPM Highlights
–8 –6 –4 –2 O 2 4 6
For transformation PQ, transformation Q In the diagram below, pentagon ABCDE is the plan of
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.
Q P
which is A A9 A0, y
8
where transformation Q = rotation of 90° A E
anticlockwise at centre (−5, 4); D 6 T S
FORM
U 4
5
5
transformation P = translation 1 2 B C 2 X W R
–1
Q
Hence, through combined transformation V P x
PQ, object A experienced the rotation of –6 –4 –2 O 2 4 6
90° anticlockwise at centre (−5, 4) followed (a) Pentagon QSTUV is the image of pentagon
ABCDE under a combined transformation LH.
5
by the translation 1 2 . Describe in full, the transformation
(i) H
(ii) L
–1
(b) It is given that the area of the residential region
is 12 600 m . Calculate the area of the shopping
2
complex.
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
(a) Determine the image of point A under the
combined transformation
Solution (i) K
2
(a) (i) y (ii) LM
8 (iii) MK
A E (b) It is given that K is the image of a point
D
6 under the transformation KL. Determine the
coordinates of the point.
4
B C W
R 2. The diagram below shows some quadrilaterals
2 X drawn on a Cartesian plane.
Q
P x
–6 –4 –2 O 2 4 6 y
8
H = rotation of 90° anticlockwise at (1, 8)
5
(ii) Scale factor, k = QS = 6 IV
QR 2 I 4
L = enlargement at centre Q(5, 1) with
scale factor 5 2
2
x
(b) Area of ABCDE = area of PQRWX –8 –6 –4 –2 O 2 4 6 8
5
Area of QSTUV = 1 2 2 × area of ABCDE –2
2
= 25 × area of ABCDE II –4 P
4 –6
Area of shaded region III
= area of QSTUV − area of PQRWX –8
12 600 = 25 × area of ABCDE − area of ABCDE
4 –10
12 600 = 21 × area of ABCDE It is given that transformations
4
2
Area of ABCDE = 2 400 m A = reflection on the x-axis
B = rotation of 180° at the origin
Hence, the area of shopping complex is 3
2 400 m . C = enlargement at centre (9, 2) with scale factor 2
2
Determine the image of quadrilateral P under the
combined transformation
Try Question 8 in Try This! 5.3 (a) BA (b) CA
3. The diagram below shows an isosceles triangle A
Try This! 5.3 drawn on a Cartesian plane.
y
1. The diagram below shows some points drawn on a 8
Cartesian plane.
y A 6
10 4
S
P 8 2
Q FORM
6 R x
–8 –6 –4 –2 O
K 4 5
It is given that transformations
–3
A 2 T P = translation 1 2
–1
x Q = reflection on the line x = 2
–6 –4 –2 O 2 4 6 8
R = enlargement at centre (2, 4) with scale factor
It is given that transformations – 2
3
5
K = translation 1 2 Copy the diagram and draw the image of the isosceles
2
L = reflection on the line y = 3 triangle A under the combined transformation
M = rotation of 90° anticlockwise at centre (−1, 6) (a) PQ (b) QR
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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 8
C
6 D 6 A
H
4 4
2 A B 2 A
x x
–4 –2 O 2 4 6 8 O 2 4 6 8
It is given that transformations It is given that transformations
–4
–2
P = translation 1 2 K = translation 1 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
(b) RP
two congruent gardens. Trapezium PQRS is a
construction region.
5. The diagram below shows a right-angled triangle
drawn on a Cartesian plane.
y
y 6
8 A
4
R
6
2
4 S x
–8 –6 –4 –2 O 2 4 6 8
2
–2
x
–2 O 2 4 6 8 10 –4
B
It is given that transformations P –6
–2
A = translation 1 2 –8
1
B = rotation of 90° clockwise at centre (6, 1) –10 Q
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.
transformations satisfies the commutative law. Trapezium PQRS is the image of trapezium A under
(a) Combined transformation AB the combined transformation UV.
(b) Combined transformation BC (a) Describe the
(i) transformation V.
6. In the diagram below, C0 is the final image of object (ii) transformation U.
C under the combined transformation MN. (b) Given that the area of A is 120 m , calculate the
2
FORM
area of the shaded region.
y
5
8
C 5.4
6 Tessellation
C
4
2 C A Explaining the meaning of tessellation
O 2 4 6 8 x 1. Tessellation is a pattern of recurring shapes that
Describe the transformation M and transformation N. fills a plane without leaving empty spaces or
overlapping.
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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
transformation Tessellation Explanation
involved
Translation x • B is the image
of A under the
Tessellation Tessellation Tessellation y A B x
consisting of consisting of consisting of translation 1 2 .
0
triangles only. a combination a combination D C • C is the image
of squares and of equilateral x of B under the
trapeziums. triangles, squares
0
and regular translation 1 2 .
hexagons. –y
• D is the image
of C under the
20
–x
translation 1 2 .
Determine whether each of the following is a 0
tessellation. Reflection • B is the image
(a) (b) B of A under the
P Q
A C reflection on
R the side PQ.
S
D • C is the image
of A under the
(c) (d) reflection on
the side QR.
• D is the image
of A under the
reflection on
the side RS.
Solution Rotation • B is the image
A B of A under the
(a) It is a tessellation consisting of quadrilaterals 60°
only. P C Q rotation of 60°
D clockwise at
(b) It is a tessellation consisting of a combination of centre Q.
equilateral triangles and regular hexagons. • C is the image
(c) It is not a tessellation because it consists of the of A under the
shape which is not recurring, which is pentagon. rotation of 60°
(d) It is a tessellation consisting of two recurring clockwise at FORM
patterns without overlapping. centre P.
• D is the image 5
Try Question 1 in Try This! 5.4 of C under the
rotation of 60°
anticlockwise at
B Designing tessellation involving centre Q.
isometric transformation
3. Escher tessellation is a type of tessellation
1. We can design a tessellation by using the created by Maurits Cornelis Escher (1898-1972).
isometric transformations such as translation, Escher tessellation consists of a combination of
reflection and rotation. congruent patterns.
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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°
anticlockwise at centre Q.
D is the image of A under the rotation of 150°
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°
5. The diagram below shows one of the simple clockwise at centre R.
ways to produce an Escher tessellation, starting Try Question 2 in Try This! 5.4
from a piece of square cardboard.
Try This! 5.4
1. Determine whether each of the following is a
tessellation.
(a)
Cut two Combine the Create more identical (b)
small small poligons shapes with different
polygons with another colours and combine
from two two sides of them to form an Escher
sides the square. tessellation.
of the
square.
(c)
SPM Tips
We can use more than one type of transformations or
combined transformation to design a tessellation.
(d)
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.
A Q C B
FORM
B 5 units
P R F P Q
5
E A C
D
E
D
Based on the diagram, describe the transformation
used to produce the tessellation from Based on the diagram, describe the transformation
(a) object A to the images C and D. used to produce the tessellation from
(b) object B to the images E and F. (a) object A to images C and D.
(b) object B to image E.
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Mathematics SPM Chapter 5 Congruency, Enlargement and Combined Transformations
SPM Practice 5 Full
solution
PAPER 1 5. The diagram below shows a rectangle.
1. The diagram below shows two congruent triangles. 15 cm
9 cm
46° 46°
70° 70°
9 cm 9 cm Which of the following is correct?
Triangle congruency shown on the diagram is Scale factor Image
A Side-Side-Angle
B Side-Side-Side A 5 cm
C Angle-Angle-Side 3 3 cm
D Angle-Side-Angle
2. The diagram below shows a trapezium, PRST.
B 10 cm
S
2 4.5 cm
T C
64° 10 cm
U 2
3 6 cm
P Q R
It is given that PRT and RSQ are two congruent
triangles. Find the ∠TUS. D 5 cm
A 26° C 90° 1
B 64° D 104° 2 4.5 cm
3. The diagram below shows two similar parallelograms.
6
18 cm 6. Transformation P is a translation 1 2 and
16 cm –1
12 cm transformation Q is a rotation of 90° anticlockwise at
centre (0, 3). State the coordinates of the image of
x cm point (3, 1) under the combined transformation PQ.
The value of x is A (1, 3) C (7, 2)
A 20 C 28 B (6, 8) D (8, 5)
B 24 D 32
–2
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
–3
enlargement. and transformation S is a reflection on the line x = 4. FORM
y Which of the following is the image of T under the
8 combined transformation SR? 5
6 y
H H 8
4 A T
6
2
4 C
x
O 2 4 6 8 10
2 B D
The centre of enlargement is
A (4, 3) C (5, 4) O x
B (4, 5) D (5, 6) 2 4 6 8
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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 cm and 16 cm respectively. If Q is the shapes ABCD, PQRS and PQTS drawn on a
2
2
image of P under an enlargement, determine the Cartesian plane.
scale factor of the enlargement. y
A 2 C 3 8
3 4 A D S P
B 2 D 1 6
2 B Q
4 R
9. The diagram below shows that Y is the image of X C
under a combined transformation AB. 2
y T x
8 –8 –6 –4 –2 O 2 4 6
(i) PQTS is the image of ABCD under the
6 combined transformation XY. Describe in
X full, the transformation
4
(a) Y
2 Y (b) X
(ii) It is given that PQRS represents a region
x
2
O 2 4 6 8 which has an area of 48 m . Calculate the
2
1
It is given that transformation A is a translation 1 2 . area, in m , of the shaded region.
–5
Which of the following is transformation B? 2. (a) The diagram below shows two points, A and B,
A Rotation of 90° clockwise at centre (5, 4) on a Cartesian plane where B is the image of A
B Rotation of 90° clockwise at centre (3, 1) under the combined transformation PQ.
C Rotation of 90° anticlockwise at centre (5, 4) y
D Rotation of 90° anticlockwise at centre (3, 1) A
8
10. A tessellation cannot be produced from 6
A equilateral triangles
B rectangles 4 B
C regular pentagons
D regular hexagons 2
x
PAPER 2 –6 –4 –2 O 2 4 6
1. (a) The diagram below shows a point K on a It is given that transformation P is a rotation
Cartesian plane. and transformation Q is a reflection on the line
y = 6. Describe in full,
y (i) transformation P.
4 (ii) a single transformation which is equivalent
to the combined transformation PQ.
2
(b) In the diagram below, PQRSTU is the image
x of ABCDEF under the combined transformation
–6 –4 –2 O 2 4 HK.
–2
K y
P
–4 4
FORM
U Q
It is given that transformations 2
5
–5
A = translation 1 2 –6 –4 –2 S O 2 4 6 8 x
3
B = enlargement at centre (3, −4) with scale T –2 R E F
factor 3
–4 D A
State the coordinates of the image of point –6
K under each of the following combined C B
transformations. (i) Describe in full, transformation
(i) A 2 (a) K
(ii) AB (b) H
258
05 Focus SPM Math F5.indd 258 17/02/2023 9:49 AM
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 m . consisting of equilateral triangles and squares.
2
2
Calculate the area, in m , of the shaded
region. 3 units
3. (a) The diagram below shows two similar A P C
trapeziums. B D
Q
120° 16 cm
y cm x
12 cm
73° 107° Based on the diagram, describe the
24 cm transformation used to produce the tessellation
from
Find the value of (i) object A to image C.
(i) x (ii) object B to image D.
(ii) y (b) The diagram below shows two artificial lakes, X
(b) In the diagram below, pentagon PQRST is the and Y, and an industrial region Z.
image of pentagon ABCDE under the combined
transformation UV. y
y 4
10 A X 2 Z
8 x
–8 –6 –4 –2 O 2 4 6
S T
6 –2
P E F B Y
4
R Q In the diagram, triangle Z is the image of
K G
2 triangle X under the combined transformation
D H C PQ.
–8 –6 –4 –2 O 2 4 6 x (i) Describe in full, the transformation
It is given that the area of the shaded region is (a) Q
126 cm . (b) P
2
(i) Describe in full, the transformation (ii) It is given that the area of each lake is
2
2
(a) V 280 m . Calculate the area, in m , of the
(b) U industrial region.
(ii) Calculate the area, in cm , of pentagon
2
PQRST. FORM
5
259
05 Focus SPM Math F5.indd 259 17/02/2023 9:49 AM
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