TABLE OF CONTENTS
CORE THEORY SUMMARIES
Paper 1
Number Systems
Integers
Rational Numbers
Algebra
Exponents
Equations
Finance
Ratio and Rate
Number Patterns
Graph Interpretations
2
5
7
12
16
20
24
27
29
31
Paper 2
Euclidean Geometry: Lines and Angles
Euclidean Geometry: Triangles
Quadrilaterals
Transformations
Measurement
Statistics
Probability
www
33
36
40
44
50
56
61
www
Number Systems
Grade 8 Maths Essentials
REAL NUMBERS (ℝ)
Integers:
= {… − 3, − 2, − 1, 0, 1, 2, 3, …}
Whole Numbers:
= {0, 1, 2, 3, 4, …}
Important Terms:
• Factor: A factor can be divided into a number with no remainder.
e.g. The factors of 20 are F20 = {1, 2, 4, 5, 10, 20}
• Multiple: The multiples of a number are found by multiplying the number by only natural numbers.
e.g. The first 5 multiples of 5 are M5 = {5, 10, 15, 20, 25 …}
• Prime Numbers: A prime number has only two factors, 1 and itself.
e.g. 17 is a prime since its only two factors are 1 and 17
• Composite Number: A composite number has more than two factors.
e.g. 8 is a composite number since F8 = {1, 2, 4, 8}
• Universal number: 1 is the universal number (it is neither prime nor composite)
• Square number: A number multiplied by itself is square.
e.g. 16 is a square number since 4 × 4 = 16
LEARN THIS!
Irrational Numbers:
Non-recurring and nonterminating decimals
i.e. 5 or π
Rational Numbers:
a
Can be written as if b ≠ 0
b
(i.e. fractions)
SCIENCE CLINIC 2022 ©
Natural Numbers:
= {1, 2, 3, 4, …}
EXAMPLE 1.1
1) Using the set of {1, 2, 3, 4, …, 25}, list the following:
a) Factors of 24
1 × 24
2 × 12
3×8
4×6
∴ F24 = {1, 2, 3, 4, 6, 8, 12, 24}
b) Prime factors of 24
PF24 = {2; 3}
Hint: only use the factors from
(a) that are prime numbers
c) Multiples of 6
M6 = {6, 12, 18, 24}
Each composite number can be written as a product of its prime factors e.g. 20 = 4 × 5 = 2 × 2 × 5
For bigger numbers: use the tree or ladder methods to find their products. (Hint: always start with the smallest prime
factors)
EXAMPLE 1.2
1). Write 248 as a product of its prime factors using the:
a) Tree Method
b) Ladder method
2) Write 300 as a product of its prime factors using the:
a) Tree Method
b) Ladder Method
248
300
2 300
2 248
d) Prime Numbers
{2, 3, 5, 7, 11, 13, 17, 19, 23}
2
2 124
124
2
2 150
150
3 75
2 62
2) Write down the:
a) Factors of 30
1 × 30
2 × 15
3 × 10
5×6
F30 = {1, 2, 3, 5, 6, 10, 15, 30}
2
62
2
31 31
1
31
31
1
2
75
3
5 25
∴ 248 = 2 × 2 × 2 × 31
5 5
25
5
1
5
5
b) First five multiples of 30
M30 = {30, 60, 90, 120, 150, …}
c) Prime factors of 30
PF30 = {2, 3, 5}
Writing numbers as products of prime factors:
1
∴ 300 = 2 × 2 × 3 × 5 × 5
Hint: only use the factors from
(a) that are prime numbers
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Page 2
Number Systems
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
SQUARES, CUBES AND ROOTS
Using prime factors to find HCF and LCM:
HCF: Highest Common Factor
LCM: Lowest Common Multiple
A number SQUARED: a number multiplied by itself
e.g. 6 × 6 = 62 = 36
* For smaller numbers you can write out a list of the factors and multiples to find the HCF and LCM
A number CUBED: a number multiplied by itself twice
e.g. 5 × 5 × 5 = 53 = 125
e.g. Find the HCF and LCM of 24 and 18
F24 = {1, 2, 3, 4, 6, 8, 12, 24}
F18 = {1, 2, 3, 6, 9, 18}
M24 = {24, 48, 72, 96, 120 . . . }
M18 = {18, 36, 54, 72, 90...}
A SQUARE ROOT: 4 means ?2 = 4
and since 2 × 2 = 4 then 4 = 2
HCF = 6 since it is the
highest factor on both lists
LCM = 72 since it is the
lowest multiple on both lists
A CUBE ROOT: 3 27 means ?3 = 27
and since 3 × 3 × 3 = 27 then 3 27 = 3
* For larger numbers, use prime factors.
HCF: use the pairs common to both lists
LCM: use the pairs and ‘leftovers’
EXAMPLE 1.4
1) 13
= 1×1×1
= 1
EXAMPLE 1.3
Use prime factors to find the HCF and the LCM of:
2) 52
= 5×5
= 25
2) 25 and 135
1) 36 and 68
2 36
2 68
2 18
2 34
3 9
17 17
3 3
1
135
25
5
3
5
5
1
45
3
1
(Circle the matching pairs)
36 = 2 × 2 × 3 × 3
68 = 2 × 2 × 17
(Only the pairs)
HCF = 2 × 2
= 4
(Pairs and leftover factors)
LCM = 2 × 2 × 3 × 3 × 17
= 612
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3)
15
3
5
(Circle the matching pairs)
25 = 5 × 5
135 = 3 × 3 × 3 × 5
(Only the pairs)
HCF = 5
(Pairs and leftover factors)
LCM = 5 × 5 × 3 × 3 × 3
= 675
4)
5
1
5)
4 + 3 125
= 2+ 5
= 7
64 + 36
=
100
= 10
25 × 9
=
25 ×
= 5×3
= 15
Alternatively:
25 × 9
=
225
= 15
Remember BODMAS
(2 × 2 = 4 and 5 × 5 × 5 = 125)
(10 × 10 = 100)
9
A root sign acts like a bracket,
so first add or subtract the
terms under the root, then
find the root
Multiplication makes one term
so you can root each factor
separately as it can make the
sum easier
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Page 3
Number Systems
Grade 8 Maths Essentials
LEARN THIS!
ORDER OF OPERATIONS: BODMAS
SCIENCE CLINIC 2022 ©
Note:
There are different ways of writing multiplication and division
Brackets
Order (Exponent)
Division (Left to Right)
Multiplication (Left to Right)
Addition (Left to Right)
Subtraction (Left to Right)
MULTIPLICATION
DIVISION
5 × 3 = 15
7÷7= 1
4(2) = 8
12
= 6
2
(2) × (6) = 12
(8)(1) = 8
1
10 ⋅ 5 = 50 (different from 10.5 = 10 )
2
EXAMPLE 1.5
1) 8 + 3 − 1 + 4
= 14
+ and − from L → R
2) 8 + 3 × 1 + 4
= 8+ 3+ 4
= 15
× first
+ from L → R
3) 6 × 2 ÷ 3(4)
= 6 × 2 ÷ 12
= 1
6
+ 2
3
= 6 −2 + 2
= 6
4) 6 −
5) 1 + 6 × (8 − 4) ÷ 3
= 1+ 6×4÷3
= 1+ 8
= 9
6)
8 −7 ÷ 1 −1
16 ÷ (12 − 4)
Brackets
× and ÷ from L → R
8)
16 ÷ 8 × 5
10 − 5 × 2
8 −7 −1
16 ÷ 8
− from L → R (top); ÷ (bottom)
=
10
10 − 10
=
0
2
÷
Remember:
0 ÷ □= 0
=
10
0
7) 3 × 7 − 15 ÷ 5
= 21 − 3
= 18
+ and − from L → R
9)
=
= undefined
= 0
OF first
× and ÷ from L → R
÷ the fraction
÷ (top); brackets (bottom)
16
100 − 5 × 10
−
10 − 8
20 ÷ 2
× and ÷ from L → R
× and ÷
=
16 100 − 50
−
2
10
− in the numerator
=
16 50
−
2
10
÷ each fraction
× and ÷ from L → R
Remember:
□ ÷ 0 = undefined
10) 15 ÷ 5 × 0
= 0
× and ÷ from L → R
11) 4 + 2(9 − 5)2
= 4 + 2(4)2
= 4 + 2(16)
= 4 + 32
= 36
Bracket
Exponent
× into the bracket
Remember:
□ × 0 = 0
= 8 −5
= 3
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Page 4
Integers
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
SUBTRACTING INTEGERS
TERMINOLOGY
Integers (ℤ) include whole numbers which are both positive or negative or zero.
ℤ = { . . . − 3, − 2, − 1, 0, 1, 2, 3 . . . }
On a number line, integers look like this:
-3
-2
-1
0
1
2
3
4
When you subtract a positive ( + ) integer:
Move left on the number line.
When you subtract a negative ( − ) integer:
Move right on the number line.
NOTE:
Subtracting a negative ( − ) integer is the same as adding its additive inverse. The
additive inverse of −5 is + 5 and the additive inverse of −3 is + 3.
EXAMPLE 1.7
ADDING INTEGERS
When you add a positive ( + ) integer: Move right on the number line.
1) 3 − 5
= −2
When you add a negative ( − ) integer: Move left on the number line.
-2
-1
-0
1
2
3
Subtract the + integer
∴ move left
4
EXAMPLE 1.6
-3 -2
-1
0
1
2
3
4
Add the + integer
∴ move right
End
Start
End
-5
2) −1 − (−5)
= −1 + 5
= 4
-2
2) −1 + (−5)
= −6
-6
Start
End
1) −3 + 7
= 4
-1
0
1
Start
-4 -3
-2
-1
Start
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0
Add the − integer
∴ move left
2
3
4
5
Additive Inverse
Subtract the − integer
∴ move right
End
3) −1 + (−4) − 4 + (+ 1)
= −5 −4 + 1
= −9 + 1
= −8
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BODMAS ∴ add and
subtract from L → R
Page 5
Integers
Grade 8 Maths Essentials
LEARN!
MULTIPLYING INTEGERS
(+ )×(+ )= (+ )
( + ) × ( −) = ( −)
( −) × ( −) = ( + )
( −) × ( + ) = ( −)
EXAMPLE 1.10
1)
EXAMPLE 1.8
−3 ⋅ 4 × (−2)
−5 − 3 + 9
24
=
1
BODMAS (top: × L → R bottom: + & − ; L → R)
(+ )÷(+ )= (+ )
= 24
1) (+ 2) × (3)
= 6
(+ )×(+ )= (+ )
2) (+ 2) × (−3)
= −6
( + ) × ( −) = ( −)
3) −8 × 5 × (−2)
= − 40 × (−2)
= 80
BODMAS ∴ L → R
( + ) × ( −) = ( −)
( −) × ( −) = ( + )
Remember:
0 × □ = 0
4) 0 × (−10)
= 0
2) (−3 + 27 − 12) ÷ 3
(+ )÷(+ )= (+ )
( + ) ÷ ( −) = ( −)
( −) ÷ ( −) = ( + )
( −) ÷ ( + ) = ( −)
= 4
3) [3 + (6)(−4) − (8 + 1)] + 15
= − 15
4)
top & bottom: multiply
− from L → R
( −) ÷ ( + ) = ( −)
5) 3 − (−10) + 4 ÷ (−2)
( + ) ÷ ( −) = ( −)
= 3 − (−10) − 2
Subtract [ − ] is the same as [ + ] its additive inverse
= 3 + 10 − 2
+ &−; L→R
= 11
INTEGERS WITH ROOTS
( −) ÷ ( + ) = ( −)
3) (−24) ÷ (−2) ÷ (−6)
= 12 ÷ (−6)
= −2
÷L→R
( −) ÷ ( −) = ( + )
( + ) ÷ ( −) = ( −)
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6(3) − 10(3) + (−4)
10 × 3 − 14
18 − 30 − 4
=
30 − 14
−16
=
16
= −1
(+ )÷(+ )= (+ )
2) −15 ÷ 3
= −5
4) (−16) ÷ 4 ÷ 0
= −4 ÷ 0
= undefined
() First
− from L → R
= [3 − 24 − 9] + 15
EXAMPLE 1.9
10
1)
5
= 2
Brackets (+ & − ; L → R)
(+ )÷(+ )= (+ )
= 12 ÷ 3
DIVIDING INTEGERS
LEARN!
SCIENCE CLINIC 2022 ©
EXAMPLE 1.11
1)
3
1 − 64
= 1−4
= −3
2)
3
−8 × (−5)2
= (−2) × (25)
= − 50
(−2) × (−2) × (−2) = 8
(−5) × (−5) = 25
( −) × ( + ) = ( −)
Remember:
□ ÷ 0 = undefined
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3)
81 +
9
24 ÷ (−2)
9+ 3
=
−12
12
=
−12
= −1
( + ) ÷ ( −) = ( −)
( + ) ÷ ( −) = ( −)
Page 6
Rational Numbers
Grade 8 Maths Essentials
TERMINOLOGY
Numerator
Denominator
3
4
SCIENCE CLINIC 2022 ©
EQUIVALENT FRACTIONS
These are fractions which have the same value but look different. They are found by either
multiplying or dividing both the numerator and denominator by the same number.
7
Improper Fraction: The numerator is bigger than the denominator. e.g.
5
Mixed Number: A whole number with a fraction. e.g. 1
1
1
which means 1 +
3
3
Changing mixed numbers into improper fractions:
EXAMPLES 1.12
3 2
×
4 2
6
=
8
3
6
So =
4
8
12 3
÷
15 3
4
=
5
12
4
=
So
15
5
1)
2)
STEPS:
ADDING AND SUBTRACTING FRACTIONS
1) Multiplying the whole number by the denominator
2) Add the numerator
STEPS:
1) Convert mixed numbers to improper fractions
2) Find the LCD (Lowest Common Denominator)
3) Add and subtract the numerators (Remember to use BODMAS)
4) Simplify if possible (don't convert back to mixed numbers)
3) Write it above the original denominator
4) Keep the original sign
EXAMPLE 1.13
3
4
4×1+ 3
=
4
7
=
4
1) 1
2)
4
2
5
5×2+ 4
=
5
14
=
5
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3) −1
1
2
EXAMPLE 1.14
2×1+ 1
2
3
= −
2
= −
1
4) −3
4
1)
Keep the original sign
2)
4×3+ 1
= −
4
13
= −
4
1 2
+
4 4
3
=
4
1 1
+
2 6
1 3 1
= × +
2 3 6
3 1
= +
6 6
4
=
6
2
=
3
Denominators
are the same
LCD = 6
3)
1 1
+
8 4
1
= +
8
1
= +
8
1
=−
8
−
1 2 1 4
× − ×
4 2 2 4
2 4
−
8 8
1
1
1
+ 2 −3
3
4
6
4 9 19
−
= +
3 4
6
16 27 38
=
+
−
12 12 12
5
=
12
4) 1
Simplify ÷ 2
1
2
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LCD = 8
BODMAS: L → R
Convert to
improper fraction
LCD = 12
BODMAS: L → R
Page 7
Rational Numbers
Grade 8 Maths Essentials
MULTIPLYING FRACTIONS
DIVIDING FRACTIONS
STEPS:
STEPS:
1) Convert to improper fractions
1) Convert to improper fractions
2) Simplify (cross cancel): Top with bottom
2) Tip and Times
3) Multiply numerators
4) Multiply denominators
5) Simplify
EXAMPLE 1.15
1)
1 4
×
8 7
=
1
41
×
82
7
=
1×1
2×7
=
1
14
3
2) 2 ×
16
21
3
=
×
1
168
=
3) 3
3
8
1
3
1
×
×
4 26 9
=
13
3
1
×
×
4
26 9
=
131
31
1
×
×
4
262 93
=
1×1×1
4×2×3
=
1
24
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EXAMPLE 1.17
1 7
1
− ×1
2 8
21
5 7 22
=
− ×
2 8 21
5 71 2211
=
−
×
2 84
213
5 11
=
−
2 12
30 11
=
−
12 12
19
=
12
1) 2
(i.e. Invert the divisor and multiply)
3) Simplify
EXAMPLE 1.16
1)
1 1
÷
2 3
1 3
= ×
2 1
1×3
=
2×1
3
=
2
1
1
2)
÷1
4
16
1 17
= ÷
4 16
1 16
= ×
4 17
1
164
=
×
41
17
4
=
17
3)
1 2 21
÷ ÷
3 7
8
1 7
8
= × ×
3 2 21
1 71
84
= ×
×
3 21 213
4
=
9
SCIENCE CLINIC 2022 ©
Tip & Times
2)
1 1
1 1
− ÷
−
5 3 (5 3)
1 1
3
5
− ÷
−
5 3 ( 15 15 )
1 1
−2
=
− ÷
5 3 ( 15 )
Mixed no. to improper fraction
Multiply first
Cancel
LCD = 12
LCD = 15
=
Mixed no. to
improper fraction
Tip & Times
1
1
−155
−
×(
)
5 31
2
1 5
=
+
5 2
2
25
27
=
=
+
10
10 10
=
Cancel
3)
Tip & Times twice
1 1
−15
− ×
5 3 ( 2 )
=
Divide (Tip & Times)
Multiply
Cancel
( −) × ( −) = ( + )
Add (LCD = 10)
−10
1
2
=
− 34
−10
2
4
− 34
−1
( 4 )
−10
−4
=
×
( 1 )
1
40
=
1
= 40
LCD = 4
= − 10 ÷
Cancel
Divide (Tip & Times)
( −) × ( −) = ( + )
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Page 8
Rational Numbers
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
COMMON FRACTIONS, DECIMALS AND PERCENTAGES
REMEMBER: A percentage is a fraction out of 100, e.g. 50 % =
50
1
=
= 0,5
100
2
DECIMAL TO PERCENTAGE
STEPS:
1) Multiply by 100 (so move comma two places to the right)
COMMON FRACTION TO DECIMAL
STEPS:
1) Convert denominator to smallest power of 10 (10, 100, 1000...etc)
2) If you can't, do division
EXAMPLE 1.18
EXAMPLE 1.20
Convert to decimals (without a calculator) and round off to 3 decimal places where necessary
3. 1,25
= 125 %
1. 0,64
= 64 %
1)
=
2 decimals ∴ denominator = 100
3)
=2
326
2
÷
1000 2
=2
163
500
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4)
25
(100 = 4 × 25)
25
Mixed no. to improper fraction
×
2
(10 = 5 × 2)
2
1 zero ∴ 1 dec places
Can't make 7 a power of 10
0,4285...
= 7 3,0000...
= 0,4285
≈ 0,429
3 decimals ∴ denominator = 1000
Simplify
3
7
= 3÷7
= 7 3
Simplify
13
20
2) 2,326
326
=2
1000
2 zeros ∴ 2 dec places
2) 1
EXAMPLE 1.19
65
5
÷
100 5
=
2
5
7
=
5
14
=
10
= 1,4
DECIMAL TO COMMON FRACTION
STEPS:
1) Use denominators of 10, 100, 1000... etc.
2) Simplify Fractions
=
×
75
100
= 0,75
2. 0,053
= 5,3 %
1) 0,65
65
=
100
3
4
2
9
= 9 2
Remember 3 = 3,0000...
Round up to 3 dec places
Can't make 9 a power of 10
Remember 2 = 2,0000...
0,2222...
= 9 2,0000...
= 0,2222
≈ 0,222
Round down to 3 dec places
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Page 9
Rational Numbers
Grade 8 Maths Essentials
COMMON FRACTION TO PERCENTAGE
STEPS:
1) Convert to decimal
2) Multiply by 100 to make it into %
PERCENTAGE TO DECIMALS AND COMMON FRACTIONS
Remember:
□
% means
100
EXAMPLE 1.21
5
1)
8
=
625
1000
= 0,625
EXAMPLE 1.22
125
×
(1000 = 8 × 125)
125
3 zeros ∴ 3 decimal places
× 100 (comma moves 2 right)
= 62,5 %
2)
1) 20 %
20
=
100
5
(100 = 20 × 5)
5
=
2 zeros ∴ 2 decimal places
2) 36 %
36
=
100
× 100
Decimal = 0,36
Mixed no. to improper fraction
3) 115 %
115
=
100
= 5%
1
3) −1
6
= −
7
6
= −6 7
1,1666...
= − 6 7,0000...
·
= − 1,1666
Common fraction =
1
5
Common fraction =
9
25
Common fraction =
23
20
or = 0,2
×
= 0,05
(÷20)
Decimal = 0,20
1
20
5
100
SCIENCE CLINIC 2022 ©
7 = 7,0000...
Decimal = 1,15
(÷4)
(÷5)
( −) ÷ ( + ) = ( −)
× 100
= − 116,67 %
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Page 10
Rational Numbers
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
SQUARE ROOTS & CUBE ROOTS OF FRACTIONS & DECIMALS
KEY TO REMEMBER
STEPS:
1) Write as common fraction (if a decimal)
The root of a fraction can be split into the
2) Convert any mixed numbers to improper fractions
root of the numerator over the root of the
3) Take root of numerator & denominator
denominator
4) Convert back to a decimal (if necessary)
EXAMPLE 1.23
4
9
1)
4)
=
4
=
9
=
36
1
=
10
Simplify (÷2)
4
=
50
2
=
25
a
3
b
non-real
(−x)2 = non-real
Note:
3
125000
64
125000
Simplify (÷2)
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1
100
=
3
125 × 1000
3
0,027
=
3
=
3
10
27
1000
(− × − = + )
2 decimal places
2 zeros
1 zero ∴ 1 decimal place
= 0,1
64
125000
3
3
The square root of a negative number is
1 zero ∴ 1 decimal place
=
6)
=
1000
64
3
a
=
b
8
0,01
5)
8
6
4
=
3
3
3 zeros
3
b
(−x) × (−x) ≠ (−x)2
=
3)
3
3 decimal places
= 0,2
Mixed no. to improper fraction
64
36
=
8
1000
3
2
=
10
28
36
1
0,008
3
=
2
=
3
2)
3
a
a
=
b
BUT
The cube root of a negative number is
negative
(−x)3 = − x
3 decimal places
(−x) × (−x) × (−x) = (−x)3
(− × − × − = −)
3 zeros
1 zero ∴ 1 decimal place
= 0,3
= 5 × 10
= 50
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Page 11
Algebra
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
TERMINOLOGY:
Variable: A lower (small) letter of the alphabet used in algebra to indicate an "unknown"
SUBSTITUTION
Always substitute using brackets in place of the variable
Remember to use BODMAS
•
•
Constant: A term with a fixed value (it does not contain a variable). For example, 2 or − 3
Terms: These are separated by + or - signs
e.g. x + 2y − 3 (contains 3 terms), while x + 2y × 3 (contains only 2 terms)
Coefficient: The numbered sign that a variable is multiplied by
1
1
e.g. in the term "4a" the coefficient of a is 4 and in the term - b the coefficient of b is 4
4
Degree: The highest power of the variable in the expression
e.g. the degree of x in the expression x 2 − 3x 4 + 5x is the 4th degree
Algebraic Expression: A sum with one or more terms of which at least one is available,
e.g.: 8x, 3a b c, 5(a)(b)(c) = 5a b c are called monomials
7x + y, 3a − 4 are called binomials
3x 2 + 6x − 4, 3a − 7b + c are called trinomials
All the above examples, as well as expressions with more that three terms, are called polynomials.
NOTES:
1. 4(a) means 4 × a or 4a
2. x means 1x and −y means −1y
EXAMPLE 2.1
EXAMPLE 2.2
1) Determine the value of 5x + 10 if x = − 2
5x + 10
= 5(−2) + 10
= −10 + 10
=0
2) If y = 3, calculate the value of
Given x = − 2
BODMAS ∴ M first
15y − 5
2
15y − 5
2
=
15(3) − 5
2
=
45 − 5
2
=
40
2
Given y = 3
Multiply
= 20
For the expression:
x3
x5 + x2 −
−4 + x
2
3) Find the value of 3x + y − x y if x = 5 and y = − 1
Write down the:
1) Variable
x
4) Degree of x
5th degree (x 5 is the highest power)
2) Number of terms
5 terms
5) Constant term
−4 (term with no variable)
3
3) Coefficient of x
1
−x 3
has the coefficient −
2
2
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3x + y − x y
= 3(5) + (−1) − (5)(−1)
= 15 − 1 + 5
= 19
Given x = 5; y = − 1
4) Calculate a 2 + 2b if a = − 2 and b = − 3
a 2 + 2b
= (−2)(−2) + 2(−3)
= (−2)(−2) + 2(−3)
= −2
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Given a = − 2; b = − 3
Page 12
Algebra
Grade 8 Maths Essentials
WRITING ALGEBRAIC EXPRESSIONS
SCIENCE CLINIC 2022 ©
LIKE & UNLIKE TERMS
Terms in algebraic expressions can only be added or subtracted if they are "like terms".
EXAMPLE 2.3
To be like terms both the variable and its power must be identical e.g. 5x and −2x are like
terms but 5y and −2y 2 are not like terms (we call them unlike terms)
1) A certain number is subtracted from 5.
2) Three times a certain number squared is added to 10.
3) A certain number is divided by 6 and then added to that same number multiplied by 3.
NOTE:
5a b = 5b a
Therefore 4a b and −2b a are like terms
4) 10 added to a certain number is then subtracted from that same number cubed.
EXAMPLE 2.4
SOLUTIONS
1) 5 − x
(start at the word "from")
1. 6a + 2a − 4a
= 4a
BODMAS ∴ + & − from L → R
2. 7p − q + 2q + p
= 7p + p − q + 2q
= 8p + q
Group like terms
7p + p = 8p and −1q + 2q = + q
(unlike terms ∴ stop here)
3. 2b c − 4cb + 6ca
= −2b c + 6ca
BODMAS ∴ + & − from L → R
4. 5m × n+ 2m(n) − 6n × m
= 5m n+ 2m n− 6m n
= m n (or 1m n)
BODMAS ∴ × & ÷ from L → R
5. 6x × y + x y − 20x y ÷ 2
= 6x y + x y − 10x y
= −3x y
BODMAS ∴ × & ÷ from L → R
6. 5x 2 + 4x − 8x 2 − 4x
= 5x 2 − 8x 2 + 4x − 4x
= −3x 2 + 0
= −3x 2
Group like terms (x 2 and x are unlike terms!)
2) 10 + 3p 2
3)
c
+ 3c
6
4) y 3 − (y + 10)
NOTE:
1. Any variable could be used in these examples
2. Remember to start at from & use brackets to group an operation
HINT:
When doing calculations,
always arrange the terms alphabetically. Eg.: Write “a b”
instead of “ba”, write “x yz”
instead of “z x y”. It really
helps to identify like
terms!
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Page 13
Algebra
Grade 8 Maths Essentials
ADDING & SUBTRACTING ALGEBRAIC EXPRESSIONS
•
•
•
When adding: Signs remain the same
When subtracting: Use brackets & signs change start at the word "from"
Group like terms and simplify
EXAMPLE 2.5
ADDING & SUBTRACTING ALGEBRAIC EXPRESSIONS WITH FRACTIONS
5x + 2y + x − 3y
Group like terms
1)
= 5x + x + 2y − 3y
= 6x − y
6a − 4b + 7b − 5a + 2c
2)
= a + 3b + 2c
3) From 2a + 6c subtract 2a − 3c
Start at "from"
2a + 6c − (2a − 3c)
Use brackets
= 2a + 6c − 2a + 3c
Signs change
= 2a − 2a + 6c + 3c
Group like items
= 0 + 9c
3)
= 9c
4) Subtract 7p − 8 from 10 + 17p
Start at "from"
10 + 17p − (7p − 8)
Use brackets
= 10 + 17p − 7p + 8
Signs change
= 10 + 8 + 17p − 7p
Group like terms
= 18 + 10p
REMEMBER!
A “−” (minus) in front of a bracket means “−1” must be distributed in
Bracket → Distribute (signs will change)
Add like terms
No brackets → no distribution
Add like terms
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2x
2
LCD = 2
2x ÷ 2 = 1x or x
=x
Group like terms
= 6a − 5a − 4b + 7b + 2c
Bv.2: a + b − a + b
= a −a + b + b
= 2b
x
x
+
2 2
=
2) To 6a − 4b add 7b − 5a + 2c
= a + b − 1(a + b)
= a + b − 1a − 1b
= 0
Group like terms
Find the LCD for like terms
Add/Subtract numerators of like terms only!
Simplify if possible
•
•
•
•
EXAMPLE 2.6
1) Add 5x + 2y and x − 3y
Eg.1: a + b − (a + b)
SCIENCE CLINIC 2022 ©
4)
a 2a
+
3
5
=
5a 6a
+
15 15
=
11a
15
y 2y
−
4
5
=
5y 8y
−
20 20
=
−3y
20
2x 5y x
+
−
3
2
9
LCD = 15
LCD = 20
Group like terms
=
2x x 5y
− +
3
9
2
For x terms: LCD = 9
=
6x x 5y
− +
9
9
2
Unlike terms ∴ no LCD
=
5x 5y
+
9
2
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Page 14
Algebra
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
DISTRIBUTIVE DIVISION
DISTRIBUTION LAW FOR MULTIPLICATION
•
•
4(a) means 4 × a
6(x + y) means 6 × (x + y) and so each term in the bracket should be multiplied by 6
STEPS:
1) Apply BODMAS
2) Distribute & multiply
3) Group like terms
4) Simplify
STEPS:
1) Divide each term in the numerator by the divisor
2) Simplify, if possible
EXAMPLE 2.8
1)
EXAMPLE 2.7
1) 6(x + y)
= 6(x) + 6(y)
=
Distribute
Multiply
= 5(a) + 5(3)
Distribute
Multiply
2)
= 5a + 15
3) −3(y − 2)
= − 3(y) − 3(−2)
6a 3b
+
3
3
Divide each term
Simplify
= 2a + b (or 2a + 1b)
= 6x + 6y
2) 5(a + 3)
6a + 3b
3
Distribute
Multiply − × − = +
5x − 2y
2
=
5x 2y
−
2
2
=
5x
5x
− y (or
− 1y)
2
2
Divide each term
Simplify
= − 3y + 6
4) 5(a + c) − 2(a − b)
Distribute
= 5(a) + 5(c) − 2(a) − 2(−b)
Multiply
= 5a + 5c − 2a + 2b
Group like terms
= 5a − 2a + 5c + 2b
= 3a + 5c + 2b
5) x (5y + 3) − y (2 − 3x)
Unlike terms
Distribute
= x (5y) + x (3) − y (2) − y (−3x)
Multiply
= 5x y + 3x − 2y + 3x y
Group like terms
3)
4(−2x + 5c)
−5
Distribute
=
4(−2x) + 4(5c)
−5
Multiply
=
−8x + 20c
−5
Divide each term
=
−8x 20c
+
−5
−5
Simplify (− ÷ − = + and + ÷ − = −)
=
8x
− 4c
5
= 5x y + 3x y + 3x − 2y
= 8x y + 3x − 2y
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Unlike terms
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Page 15
Exponents
Grade 8 Maths Essentials
exponent
an
base
a n = a × a × a × a … n times
power
MULTIPLICATION OF POWERS
2
3 ×3
So we can see that if we add the exponents we would get 35
LAW:
p a × p b = p a+ b
Simplify the following:
4 6
5) 3 b × 3b × 3 b
4+ 1+ 2
= 3
= 212
= 37 b 14
3
2) 3 × 3 × 3
Note: 3 =
31
= 38+ 1+ 3
3) 25 × 5 × 23 × 54
= 25+ 3 × 51+
3
6) 2a × 4a
×b
6+ 1+ 7
4
= 28 × 55
4) m 2 × m 3 × m 5
= m 2+ 3+ 5
2) m × m × m × m = m 4
EXAMPLE 3.4
Simplify the following:
1)
= 6a
3
Note the difference when
multiplying and adding
powers (Examples 6 and 7)
5)
=
= 32 (not necessary)
57
515
m 25
m 35
=
= 25
2)
6)
1
m 35−25
1
m 10
38b 5
3b 8
=
7×7×7
7×7×7×7×7×7
=
=
=
1
73
1
515−7
=
1
58
38−1
b 8−5
=
37
b3
3
7) 2a 3 + 4a 3
211
26
= 211−6
For this one we would have also subtracted the
exponents but in the denominator
= 8a 6
= 312
2) p 7 = p × p × p × p × p × p × p
73
76
2 7
= 2
8
1) 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 = 59
So we can see that if we subtract the exponents
we would also get 53
EXAMPLE 3.3
6+ 4+ 2
1) 35 = 3 × 3 × 3 × 3 × 3
= 53
Or
1) 2 × 2 × 2
Simplify the following:
5×5×5×5×5
5×5
=
2
Expand the following:
5
52
= 35
4
EXAMPLE 3.2
5
= 3×3×3×3×3
6
EXAMPLE 3.1
DIVISION OF POWERS
3
SCIENCE CLINIC 2022 ©
3)
36 × 74
34 × 78
LAW:
=
36−4
78−4
pa
= p a−b; if a > b
pb
=
32
74
pa
1
= b−a ; if a < b
pb
p
4)
7)
12a 8b 7
3a 3b 4
=
12 a 8−3 × b 7−4
×
3
1
= 4a 5b 3
x 22
x 14
= x 22−14
= x8
= m 10
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Page 16
Exponents
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
RAISING A POWER TO A FURTHER POWER
Note:
(23)4
= 23 × 23 × 23 × 23
= 23+ 3+ 3+ 3
= 212
a3
a3
Note if we multiplied the exponents, we would have got the same answer
LAW:
( p a )b = p a×b
LAW:
( p a × q b )c = p a×c × q b×c
EXAMPLE 3.5
EXAMPLE 3.6
Simplify the following:
Simplify the following:
1) (25)10
1) (34 × 55)7
= 25×10
= 34×7 × 55×7
50
= 2
= 328 × 535
2) (k 8)7
2) (a 4 b 6 )6
= k
8×7
= a
= k 56
×b
6×6
a×a×a
a×a×a
= 1
BUT
a3
= a 3−3
a3
= a0
Therefore a 0 = 1
= a 24 b 36
3) (22 × 54 )3
2
4
2
4
2
3) (3a 5b 4 )3
4
= (2 × 5 ) × (2 × 5 ) × (2 × 5 )
2+ 2+ 2
= 2
6
4×6
=
= 31×3 × a 5×3 × b 4×3
4+ 4+ 4
×5
= 33a 15b 12
12
= 2 ×5
Or = 27a 15b 12
Note: both base exponents were multiplied by the
exponent on the outside of the bracket
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Page 17
Exponents
Grade 8 Maths Essentials
EXPONENT LAWS:
EXAMPLE 3.7
p a × p b = p a+ b
LEARN THESE LAWS!
SCIENCE CLINIC 2022 ©
Simplify the following leaving answers in the positive exponential form. (Remember BODMAS)
a
p
= p a−b; if a > b
pb
pa
1
= b−a ; if a < b
pb
p
a b
(p ) = p
6)
1) 2x (x 3) − 3x 2(2x 2 )
= 2x 4 − 6x 4
= − 4x 4
a×b
2) (−1)4 = 1
( p a × q b )c = p a×c × q b×c
0
a = 1
a = a
4) −8x y (x y 2 )3
= − 8x y × x 3y 6
KNOW THE DIFFERENCE!
NOTE HOW THE PLACEMENT OF THE
MINUS SIGN CHANGES THE ANSWER
First raise power to
remove the bracket
7)
= − 8x 4 y 7
2
2 = 2×2 = 4
5)
2
−2 = − (2 ) = − (2 × 2) = − 4
(−2)2 = (−2 × −2) = 4
24x 10 y 3
36x 4 y 8
−23 = − (23) = − (2 × 2 × 2) = − 8
(−2)3 = (−2 × −2 × −2) = − 8
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24 × 38
22 × 310
=
24−2
310−8
=
22
or
32
2x 6
3y 5
4
9
−10d e f (−2d e 2 )3
20d 4 e 7
First raise power to
remove the bracket
=
−10d e f × −8d 3e 6
20d 4 e 7
Multiply
=
80d 4 e 7 f
20d 4 e 7
Divide
= 4d 0 e 0 f
e 0 = 1 and d 0 = 1
= 4f
24 x 10−4
=
× 8−3
36
y
=
First raise power to
remove the bracket
=
3) (−1)7 = − 1
1
2
(2 ⋅ 32 )4
22 ⋅ 310
8)
6x 8 y 4
( 18x y 3 )
3
Simplify inside the
bracket first
=
1x 7 y
( 3 )
=
1x 21y 3
or
33
3
x 21y 3
27
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Page 18
Exponents
Grade 8 Maths Essentials
ROOTS
SQUARE ROOTS:
16 =
SCIENTIFIC NOTATION
This is a method used to write exceptionally large numbers in simpler
method using the powers of 10
4×4 = 4
121 =
11 × 11 = 11
Note: you must know all the squared numbers from 1 - 13: {1; 4; 9; 16; 25; 36; 49; 64; 81; 100; 121; 144; 169}
x2 =
x×x= x
x6 =
x 2 × x 2 × x 2 = x × x × x = x 3 OR =
m6 × m6 = m6
CUBE ROOTS:
3
3
8=
3
64 =
4×4×4 = 4
Note: you must know all perfect cubed numbers from 1 - 10: {1; 8; 27; 64; 125; 216; 343; 512; 729; 1000}
3
x3 =
3
x×x×x= x
3
x6 =
3
x 3 × x 3 = x × x = x 2 OR
3
m 15 =
3
3
x6 =
3
x2 × x2 × x2 = x2
m 3 × m 3 × m 3 × m 3 × m 3 = m × m × m × m × m = m 5 OR
3
m 15 =
3
m5 × m5 × m5 = m5
25k 10
3)
52 × k 5 × k 5
3
=
27h 12
3
33h 4 × h 4 × h 4
= 3h 4
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3
=
= 5k 5
2)
c) The distance from the sun to the earth: 150,95 million km
150 950 000
= 1,51 × 108 km
d) 14,45 × 107
= 1,45 × 108
Simplify the following:
=
a) 1 200 000,0
= 1,2 × 106 (the comma moves 6 places forward to behind the 1)
b) 122 350 000 000
= 1,22 × 1011
EXAMPLE 3.8
1)
EXAMPLE 3.9
1) Write the following in scientific notation, correct to 2 decimal
places:
2×2×2 = 2
3
100 = 10 2
1000 = 10 3
10 000 = 10 4
100 000 = 10 5
Etc…
For correct scientific notation only 1 significant figure can be in front
of the decimal comma. A significant figure is a NON ZERO number.
x3 × x3 = x3
m 2 × m 2 × m 2 × m 2 × m 2 × m 2 = m × m × m × m × m × m = m 6 OR =
m 12 =
SCIENCE CLINIC 2022 ©
729f 6 g 18
3
93 f 2 × f 2 × f 2 × g 6 × g 6 × g 6
= 9f 2 g 6
49a 6 b 18
4)
=
72 a 3 × a 3 × b 9 × b 9
(Note: this is not correct scientific notation as there
are 2 significant figures in front of the comma
2) Write the following in decimal notation:
a) 7,34 × 10 4
= 73 400
(decimal moves 4 places back, i.e. so 2 digits
behind 4 and then add 2 zeros)
b) 3,578 × 1010
= 35 780 000 000
(move comma 10 places)
= 7a 3b 9
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Page 19
Equations
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
An equation is a statement in maths where LHS (left hand side) = RHS (right hand side), for example: 5 + 3 = 1 + 7
Solve for x:
An equation can also be thought of as a scale, where the
weights on each side must be equal in order for the scale to
balance.
x+ 3= 8
x = 5
You can solve for x by inspection or using a flow diagram
Flow Diagram
+3
x
8
Work backwards
from 8. The
opposite operation
of addition is
subtraction.
In order to maintain the balance, whatever is added to/subtracted from the one side, must be added to/subtracted
from the other side. If there are 21 marbles on the right of
the scale, then there must also be 21 marbles on the left of
the scale in order for it to balance.
LINEAR EQUATIONS
a x1 + b = 0
x will always be to the power of 1
Let’s apply this new knowledge of equations (a flow diagram
takes too long!)
Solve for x
1) x − 5 = 10
x = 10 + 5
x = 15
-3
Although you can solve these just by looking at them, in maths
you need to solve them as an equation.
2) x + 5 = − 10
∴x+ 3= 8
x = 8 −3
x = − 10 − 5
x = 5
x = − 15
This is called a linear equation, where only 1 answer/
solution will make this statement true.
3) 2x = 10
5+ 3= 8
LHS = RHS
GOLDEN RULE!
Whatever is done to one side of
the equation, MUST be done to
the other side
e.g. Solve the following: 3x + 2 = − 10
Step 1: Subtract 2 from both sides
3x + 2 − 2 = − 10 − 2
3x = − 12
Step 2: Divide both sides by 3
3x
−12
=
3
3
On the left of the scale, the 21 marbles are made up of 5
visible marbles and 2 bags with an unknown number of
marbles in each. To determine how many marbles are in the
bags, we can remove 5 marbles from each side. This leaves
us with 2 bags of marbles on the left and 16 marbles on the
right.
Let’s consider a flow diagram: the opposite of × 2 is ÷ 2
✖2
x
x = −4
2x = 16
x = 8
(−2)x = 10
∴ 2x = 10
x =
10
2
x =
✖ (-2)
x
10
−2
x = −5
x = 5
2x + 5 = 21
2x = 21 − 5
10
➗2
Both bags are marked as x, so we know that both bags have
an equal number of marbles. The total marbles in the two
bags on the left is equal to the 16 marbles on the right. We
can divide 16 equally between the two bags to give us a
total of 8 marbles in each bag. This can be represented in
an equation as follows
Step 3: Simplify
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NOTE: integers are
important here
10
REMEMBER:
the opposite
operation of × is ÷
➗ (-2)
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Page 20
Equations
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
SOLVING LINEAR EQUATIONS
WORD PROBLEMS
Remember to perform the same operation on both sides of the equation to isolate the variable
In this section, we solve real life problems using equations
EXAMPLE 4.1
EXTENSION
Solve the following equations
Using the equation from Q3: 8b + 6 = 3b + 1
1)
3m − 1 = 20
3m = 20 + 1
21
m=
3
2)
3)
Get the variables on one
side and the numbers
on the other side.
Is b = − 1 a solution to the above equation?
LHS : 8(−1) + 6
= −2
5y + 2 = 4y − 8
RHS : 3(−1) + 1
5y − 4y = − 8 − 2
= −3 + 1
y = − 10
= −2
8b − 3b = 1 − 6
5b = − 5
−5
b=
= −1
5
4) 2(x + 1) = 3x − 4
2x + 2 = 3x − 4
2x − 3x = − 4 − 2
−1x = − 6
−6
x =
−1
x = 6
5) 5( p − 1) − (1 − 2p) = 8
5p − 5 − 1 + 2p = 8
7p = 14
p= 2
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• addition
• more than
• subtract
• less than
• sum
• add
• minus
• difference
• deduct
×
∴ LHS = RHS
∴ b = − 1 is true
8b + 6 = 3b + 1
−
• plus
= −8 + 6
m= 7
+
÷
• times
• triple
• divide
• multiply
• product
• quotient
• double
• twice as many
• half
• a third
REMEMBER:
BODMAS (Brackets)
• Consecutive numbers (integers): x ; x + 1; x + 2
Use the Distributive Law
• Consecutive EVEN numbers (integers): 2x ; 2x + 2; 2x + 4
• Consecutive ODD integers: If 2x represents an even number then 2x + 1 represents odd number. An odd number is
always 1 more than an even number. eg. 3 = 2 + 1 ∴ consecutive odd ℤ’s = 2x + 1; 2x + 3; 2x + 5
• A number squared: x 2
• Total cost = (price of item) × no. of items
• Rectangle
l
b
Ar ea : A = l × b
Per i m e t er : P = 2(l + b)
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Page 21
Equations
Grade 8 Maths Essentials
SOLVING WORD PROBLEMS
EXAMPLE 4.2
STEPS:
Set up an equation and solve
1) Read the context and look for NB word.
2) Let the unknown = x
3) Set up your equation.
4) Solve for x
5) Answer the question
1) When twice a certain number is subtracted from 15, the result is equal
to 3 more than the number. What is the number?
Let the number = x
∴ 15 − 2x = x + 3
15 − 3 = x + 2x
12 = 3x
4= x
∴ the number is 4
2) Sara made twice as many mask as her friend Amara. Together they
made 156 masks. How many did Sara make?
Let the no. of masks Amara made = x ∴ Sara made 2x
A + S = 156
x + 2x = 156
3x = 156
x = 52
∴ Sara made 104
3) a) The perimeter of your garden is 122 meters. The length of your
garden is 1 meter more than double the width of your garden.
W= x
L = 2x + 1
∴ 2L + 2W = p
2(2x + 1) + 2x = 122
4x + 2 + 2x = 122
6x = 120
x = 20m
∴ b = 20m
L = 2(20) + 1
= 41m
b) You want to put instant lawn down. If it costs R12.50/m 2, how much
will it cost you to get this instant lawn for your garden?
Area =
=
=
∴ cost =
=
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SCIENCE CLINIC 2022 ©
EXAMPLE 4.3
1) At a soccer match, the tuck shop sold three times as many
hamburgers as boerewors rolls.
Hamburgers = R30
Boerie Rolls = R20
The tuck shop counted their money at the end of the day to be
R5500. How many boerewors rolls did they sell?
Let no. of B.rolls sold = x Let no. of Hamburgers sold = 3x
∴ money received from B rolls = 20 × x and Hamburgers = 30 × 3x
∴ B + H = 5500
20x + 90x = 5500
110x = 5500
x = 50
∴ they sold 50 boerewors rolls
2) Thandi is 11 years older than Fred. In 5 years time, she will be
twice Fred’s ages. How old is Fred now?
Now: Fred: = x
Thandi: = x + 11
5 years time: Fred: = x + 5
Thandi: = (x + 11) + 5 = x + 16
∴ Thandi + 2Fred
x + 16 = 2(x + 5)
16 − 10 = 2x − x
6= x
Fred is 6 years old
Did you notice how you used SUBSTITUTION
to help solve these word problems?
l ×b
41 × 20
820m 2
820 × R12,50
R10 250
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Page 22
Equations
Grade 8 Maths Essentials
EQUATIONS WITH MORE THAN ONE VARIABLE
SCIENCE CLINIC 2022 ©
EXTENSION WORK: EQUATIONS WITH EXPONENTS
Lets look at a different type of equation:
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
30 = 1
31 = 3
32 = 9
33 = 27
34 = 81
50 = 1
51 = 5
52 = 25
53 = 125
25 = 32
y = 2x − 1
y = double (x) subtract 1
INPUT
OUTPUT
-1
-3
x
y
0
2
-1
-1
✖
1
1
y = 2x − 1
If x = − 1 then y = 2(−1) − 1
= −2 −1
= −3
(−1; − 3)
NOTE:
We can write the values
of x and y as co-ordinates,
in the format
(x value; y value)
in order to show the pairs
If x = 0 then y = 2(0) − 1
= 0 −1
= −1
(0; − 1)
32 × 32
9×9
EXAMPLE 4.4
Solve for x:
1) 2x = 8
2 x = 23
x = 3
2) 3x = 27
3x = 33
x = 3
Substitution
26 = 64
(27 = 33)
3) 3x+ 1 = 27
3x+ 1 = 33
x+ 1= 3
x = 2
4) 5x = 1
x = 0
Try this:
If x = 1 then y = 2(1) − 1
= 1
(1 : 1)
5) x 2 = 25
The opposite of squaring is
And 25 = 5
∴x = 5
∴ x = 5 or x = − 5
Table
y = 2x − 1
(5)2 = 25
BUT (−5)2 = 25
Note:
x 1 + 2 = 7 (1 answer) - Linear equation
x 2 = 25 (2 answers) - Quadratic equation (you’ll learn about these in Grade 10)
x
-1
0
1
2
y
-3
-1
1
3
(x; y)
(-1;-3)
(0;-1)
(1;1)
(2;3)
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(square rooting)
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Page 23
Finance
Grade 8 Maths Essentials
WORKING WITH PERCENTAGES
REMEMBER: % (percent) means per 100. This is revision of Grade 7 work.
INCREASING OR DECREASING BY A PERCENTAGE
Increasing by a % =
100 + percent
× given value
100
Decreasing by a % =
100 − percent
× given value
100
EXAMPLE 5.1
1) If a theatre can seat 260 people, what percentage of the theatre would
be occupied if 104 of the seats were occupied?
EXAMPLE 5.2
104
× 100 = 40 %
260
1) Goods cost R525 excluding VAT. What will they cost with VAT(15%)?
115
R525 ×
= R603,75
100
2) What percentage is R126 of R210?
126
× 100 = 60 %
210
2) What will a computer costing R7 500 cost next year if inflation is 6,5%?
106,5
R7 500 ×
= R7 987,50
100
3) Calculate 35% of R700.
35
× 700 = R245
100
4) JJ earns a gross salary of R12 500 per month. He pays 20% in taxation,
5% for medical aid and 1% UIF. How much is each deduction and what
is his net salary?
Tax:
20
× R12 500 = R2 500
100
Med aid:
UIF:
SCIENCE CLINIC 2022 ©
3) Shoes marked at R325 are marked down by 25% on a sale. What would you pay for them on the sale?
100 − 25
R325 ×
100
75
= R325 ×
100
= R243,75
4) Mankwe bought a pair of shoes for R274 during a 25% sale. What was the original price of the shoes?
100 − 25
x×
= R274
100
100
x = R274 ×
75
5
× R12 500 = R625
100
1
× R12 500 = R125
100
Net Salary: R12 500 − R2 500 − R625 − R125 = R9 250
= R365,33
5) Determine the price before VAT (15%) if the goods cost R456,99 with VAT.
100 + 15
x×
= R456,99
100
100
x = R456,99 ×
115
= R397,38
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Page 24
Finance
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
% increase or decrease =
second amount − first amount
× 100
first amount
EXAMPLE 5.3
1) Calculate the percentage mark-up if goods costing
R45,50 were sold for R69.
R69 − R45,50
% mark up =
× 100
R45,50
= 51,65 %
2) A book marked at R148,50 is sold for R118,77.
What percentage discount was given?
% discount =
R148,50 − R118,77
× 100
R148,50
= 20,02 %
3) Determine the inflation rate if a car cost R250 000 last
year and now costs R268 750.
% inflation =
R268 750 − R250 000
× 100
R250 000
= 7,5 %
PROFIT AND LOSS
KNOW THE DIFFERENCE!
PERCENTAGE INCREASE OR DECREASE
INCREASING/DECREASING BY A
PERCENTAGE: is to add/subtract a
percentage to a given value. It represents
Profit = selling price - cost price
= a positive value
Loss = selling price - cost price
= a negative value
the portion to be added/subtracted.
The answer will be an amount.
PERCENTAGE INCREASE/DECREASE
is the percentage by which a certain amount
% Profit =
% Loss =
Profit
× 100
Cost
Loss
× 100
Cost
has gotten more/less. The answer will be a
percentage.
EXAMPLE 5.4
1) Thabiso bought 10 tickets for a soccer match for R90 each.
He sells 4 for R110 each and the rest for R70 each.
a) Determine if he made a profit or a loss.
Selling price – cost price
= (R110 × 4 + R70 × 6) − (R90 × 10)
= − R40
∴ Thabiso made R40 loss
b) Determine the percentage profit or loss
R40
% Loss =
× 100
R900
= 4,44 %
2) Eggs are bought by SJ Wholesalers for R16,20 per dozen
(12 eggs). They sell them for R17,80 per dozen. What is
their profit and percentage profit?
Profit = R17,80 − R16,20
= R1,60
% Profit =
R1,60
× 100
R16,20
= 9,88 %
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Page 25
Finance
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
SIMPLE INTEREST
Simple Interest is used to calculate the interest for a
loan taken or a principal amount deposited. The interest
is dependent on the total initial value, the interest rate
and the time period over which the loan is to be repaid
or the deposited amount remains in the bank.
The formula can be manipulated as follows :
Simple interest (SI) =
P × r × t
100
Where P = Principle amount deposited
r = Interest rate (%)
t = Time in years
Time (t) =
SI × 100
P ×r
Rate (r) =
SI × 100
P ×t
Present value (p) =
SI × 100
t ×r
EXAMPLE 5.5
EXAMPLE 5.6
1) Joe deposits R5000 in an account offering 11% p.a. (per annum) simple interest for 6 years.
1) Mary earns R594,75 interest after three years. How much did she invest (present value)
if the interest rate was 13% p.a.?
a) How must interest does he earn?
b) What is his final bank balance?
a) SI =
SI =
P=
P × r × t
(p = R5000, r = 11%, t = 6)
100
R5 000 × 11 × 6
100
R594,75 × 100
3 × 13
= R1 525
SI = R3 300
2) If Gina invests R1 800 from 1 January 2017 to 31 December 2019 and makes R432
interest, what was the rate p.a.? (Note: 1 January 2017 to 31 December 2019 is 3 years)
b) Final Balance = P + SI
= R5 000 + R3 300
= R8 300
r=
2) How much simple interest would Keke pay on a loan of R1 900 for 3 years at 9,5% p.a.
simple interest?
P × r × t
SI =
(p = R1900, r = 9,5%, t = 3)
100
SI =
=
SI × 100
t × r
=
SI × 100
P × t
R432 × 100
R1 800 × 3
= 8%
R1 900 × 9,5 × 3
100
SI = R541,50
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Page 26
Ratio and Rate
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
RATIO
COMMON CONVERSIONS
A ratio is a way of comparing two quantities that are the same. Some real life examples are:
a scale on a map
•
measurements when baking
•
•
•
1 kg = 1000 g
1 kl = 1000 l
1 m = 100 cm
1 g = 1000 mg
1 l = 1000 ml
1 cm = 10 mm
COMPARING AND SIMPLIFYING RATIOS
1 hour = 60 min
1 km = 1000 m
R1 = 100 c
Ensure all units are the same (or that all fractions have an LCD)
Divide each part of the ratio by the largest number that still leaves the answers with whole numbers.
1 min = 60 sec
1 m = 1000 mm
EXAMPLE 6.1
INCREASING OR DECREASING
IN A GIVEN RATIO
1) Simplify:
a) R2,20 : 60 c
R2,20 : 60 c
220 c : 60 c
11 : 3
× 100 to get same units
÷ 20
No units in final answer
b) 3,4 kg : 34 mg
3,4 kg : 34 mg
3 400 g : 34 mg
3 400 000 mg : 34 mg
× 1000 (kg)
× 1000 (g)
1
h : 4 seconds
2
1
2 h : 4 sec
2
150 min : 4 sec
9 000 sec : 4 sec
3
8
3
8
3
8
3
8
1
4
1
:4
4
17
:
4
34
:
8
:4
Convert to improper fraction
Leave out denominator
5 200 l : 120 l
× 1000 (kl)
÷ 40
÷4
EXAMPLE 6.2
5
= R125
4
Note: Numerator larger than denominator
Check:
125
5
=
1
4
240 ml ×
3
= 90 ml
8
Note: Numerator smaller than denominator
2) Simplify:
× 1000 (km)
5 mm : 1 200 m
× 1000 (m)
5 mm : 1 200 000 mm
÷5
1 : 240 000
When decreasing in a given ratio the numerator
needs to be smaller than the denominator.
2) Decrease 240ml in the ratio 3:8.
× 60 (min)
2 250 : 1
a) 5 mm : 1,2 km
•
R100 ×
130 : 3
× 60 (h)
When increasing in a given ratio the numerator
needs to be bigger than the denominator.
1) Increase R100 in the ratio 5:4.
e) 5,2 kl : 120 l
5,2 kl : 120 l
•
LCD = 8
3 : 34
÷ 34
100 000 : 1
c) 2
d)
b) 500 mm : 1 cm
500 mm : 10 mm
× 10 (cm)
÷ 10
Check:
90
3
=
240
8
50 : 1
1mm on map represents
240 000 mm in real life
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Page 27
Ratio and Rate
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
RATIOS AND SHARING
RATE
If sweets are shared between you and your friend in the ratio 2:3, it means that for every two sweets
that you get your friend gets 3. So the total shared is 5 (2+3=5 parts).
A rate is a way of comparing two different quantities. Some real life examples are:
The price of meat (R/kg)
•
The speed of a car (km/h)
•
EXAMPLE 6.3
EXAMPLE 6.4
1) Divide 30 sweets between Thabo and Jenny in the ratio 2:3.
Thabo : Jenny
} Given ratio
2:3
So total parts = 2 + 3 = 5
2
∴ Thabo gets × 30 = 12 sweets
5
3
Jenny gets × 30 = 18 sweets
5
1) Express each of these as a rate:
a) 5 kg of chicken for R179,99
R179,99
5k g
= R36,00/ kg
b) Traveling 275 km in 2
2) Share 69 apples between the horses, pigs and goats on a farm in the ratio 15:5:3.
horses : pigs : goats
} Given ratio
15 : 5 : 3
Total parts = 15 + 5 + 3 = 23
15
× 69 = 45 apples
∴ Horses get
23
5
pigs get
× 69 = 15 apples
23
3
goats get
× 69 = 9 apples
23
(check 45 + 15 + 9 = 69)
3) In a bag of marbles there are blue and pink marbles in the ratio 2:7. If there are 16 blue marbles,
how many marbles are there in total?
Blue : pink
} Given ratio
2 : 7
pink
blue
known/unknown values
x = 56 pink marbles
∴ Total = 16 + 56 = 72 marbles
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= 12 trees/row
= 35,998 ≈ 36,00
(check 12 + 18 = 30)
16 : x
x
7
=
16
2
7
x =
× 16
2
c) 180 apple trees planted in 15 rows
180 trees
15 rows
275k m
1
hours
2
d) 1250 l of paint used in 53 hours
2 12
1250l
53h
= 23,58l / h
= 110km / h
2) Which deal is cheaper? 30 milk chocolate bars for R270 or 42 mint chocolate
bars for R320.
Milk
R270
30
Mint
R320
42
= R9/ bar
= R7,62/ bar
∴ the mint bars are cheaper
3) A 200 ml can of orange juice costs R8,20. Determine the cost per litre.
200 ml = 0,2 l
NOTE:
FRACTIONS are a common
way of writing ratios
Price per litre =
R8,20
0,2l
= R41/ litre
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Page 28
Number Patterns
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
You must be able to identify and expand different kinds of number patterns. For example 4; 7; 10; 13; 16; . . .
T4 = 13
You have to identify 2 key components in any number pattern in order to describe that pattern in your own words:
1)
2)
Where the pattern starts
How it changes from term to term
This pattern starts at 4 and then 3 is added to every term after that. Each number in the pattern is called a term. The 1st
term is 4 and the 2nd term is 7 etc.
This number represents
the position of the
term.
This is the numerical
value of the term.
We use the following notation: T1 = 4; T2 = 7; T3 = 10; . . .
PATTERNS WITH A CONSTANT DIFFERENCE
Continuing with the previous example: 4; 7; 10; 13; 16; . . .
To find the general rule:
When we have a constant difference, we can find the general
rule as follows: Use the constant difference (+3)
multiplied by the term number. (i.e. for this pattern, you
are actually working with multiples of 3)
Term 1 (T1):
T2:
T3:
T4:
(3 × 1) ±
(3 × 2) ±
(3 × 3) ±
(3 × 4) ±
□=
□=
□=
□=
4
7
10
13
PATTERNS WITH A CONSTANT RATIO
EXAMPLE 7.1
Observe the following pattern:
2; 6; 10; 14; . . .
3) Write the general term (Tn) for this pattern:
T1 : 4(1) − 2 = 2
T2 : 4(2) − 2 = 6
T3 : 4(3) − 2 = 10
T4 : 4(4) − 2 = 14
∴ Tn = 4n− 2
We can now write the general rule (Tn or nth term) using the
above information.
4) Find the 20th term in this pattern
T20 = 2(20) − 2
= 78
Tn = 3 × n+ 1
5) Which term has the value of 126?
Tn = 4n− 2
126 = 4n− 2
4n = 126 + 2
n = 128 ÷ 4
∴ n = 32
NOTE:
This method is just to make sure your formula is
correct, you do not have to do this every time.
The more you practice, the easier it will become
to determine the formula mentally.
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1; 3; 9; 27; . . .
1) State the next 3 terms
18; 22; 26
2) Describe the pattern in words.
The pattern starts at 2 and then adds 4 to every term
after that.
Look at T1. What must be added in the block to get an
answer of 4?
T1: (3 × 1) ± □
1 = 4
1 = 7
T2: (3 × 2) ± □
T3: (3 × 3) ± □
1 = 10
T4: (3 × 4) ± □
1 = 13
We call this number pattern a geometric number pattern.
This pattern occurs when we multiply or divide every term
with the same number.
×3 ×3 ×3
We describe this number pattern in words as:
Start at 1, and multiply every term with 3.
EXAMPLE 7.2
Write down the next three terms and the describe the rule
in your own words for each of the patterns:
1) 1; 10; 100; 1 000; . . .
10 000; 100 000; 1 000 000; . . .
Start the pattern at 1 and then multiply every term
by 10
2) 64; 32; 16; . . .
8; 4; 2; . . .
Start the pattern at 64 and then divide every term
by two.
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Page 29
Number Patterns
Grade 8 Maths Essentials
OTHER TYPES OF PATTERNS
DESCRIBING PATTERNS IN A DIFFERENT WAY
1) 1; 4; 9; 16; 25; . . .
These are all square numbers. The next three terms
will be:
36; 49; 64; . . .
The pattern below is made from squares:
Fig. 1
You start at 4 and + 1 to get the next term, +2 to
get the term after that and + 3 to get the term
after that, etc. (I.e. you will increase your
difference with +1 every time). The next three
terms will be:
25; 32; 40; . . .
3) 1; 1; 2; 3; 5; 8; . . . This is the Fibonacci series :)
To get the next term of this pattern, you need to
find the sum of the two previous terms. The next
three terms will be:
13; 21; 34; . . .
Fig.2
Fig. 3
Fig. 4
1) How many squares are there in fig 1 – fig 4?
1; 4; 7; 10
2) How many squares will there be in fig 5?
13
So, we have a constant difference of +3. From here we can find
the nth term and subsequently the value of any term number.
Tn = 3n− 2
2) How many blocks will be needed to create shape 17?
(3 × 17) + 2 = 53
Using tables to represent patterns
Consider the following pattern:
Term Number
1
2
3
4
Term Value
2
7
12
5
7
10
97
Determine the general rule for this pattern: 2; 7; 12; . . .
We have a constant difference of +5 and the pattern starts
at 2.
These are all cube numbers. The next three terms
will be:
125; 216; 343; . . .
∴ Tn = 5n− 3
Fig. 1
Fig. 2
Fig. 3
Fig. 4
The key is to convert the shape pattern into a number pattern:
1; 3; 6; 10
+ 2 + 3 + 4
Now it is easier to determine the next three terms:
15; 21; 28
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1) How many blocks will be needed to create the next
shape?
(3 × 4) + 2 = 14
3) Write a general term for this pattern:
Tn = 3 × n+ 2
Tn = 3n+ 2
Study the following pattern:
4) 1; 8; 27; 64; . . .
EXAMPLE 7.3
We sometimes make use of shapes to describe a number pattern,
instead of words.
2) 4; 5; 7; 10; 14; 19; . . .
+ 1 + 2 + 3 + 4 + 5
SCIENCE CLINIC 2022 ©
Complete the table:
T 7 = 5(7) − 3 = 32
T 10 = 5(10) − 3 = 47
97 = 5n− 3
97 + 3 = 5n
100 = 5n
n = 20
Term Number
1
2
3
4
5
7
10
20
Term Value
2
7
12
17
22
32
47
97
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Page 30
Grade 8 Maths Essentials
Graph Interpretations
SCIENCE CLINIC 2022 ©
Graphs are mathematical diagrams that gives us a visual picture of the relationship between two sets of data. This is normally two variable quantities and each variable is
measured along one axis on the Cartesian plane. They represent different types of relationships which you should be able to identify.
Continuous relationships:
This relationship is shown on a graph in the form of a continuous line or curve.
Discrete relationships:
This relationship is shown on a graph in the form of discrete or separate points.
The two variables we are working with are always an independent variable and a dependent variable.
Just as the words describe, one variable’s value will depend on the others value. The independent
variable will go on the x-axis and the dependent variable will go on the y-axis.
HINT:
To help you remember
which variable belongs on
which axis, try to remember
“DRY MIX”
Dependent variable is the Result
and goes on the Y-axis. The
Manipulated variable is the
Independent variable and
goes on the X-axis
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Page 31
Graph Interpretations
Grade 8 Maths Essentials
Trends can be determined by looking at the graphs. The trends in a continuous or discrete
graph can be linear or non-linear. Graphs with a linear trend follow a straight line pattern.
Graphs that are non-linear follow a curved pattern.
You
•
•
•
need to be able to tell whether the line is:
Linear or non-linear
Discrete or continuous
Increasing, decreasing or constant (a horizontal line)
SCIENCE CLINIC 2022 ©
EXAMPLE 8.2
The graph shows the water usage of a family for 2019. Study the graph and answer the
questions that follows:
You need to read information from the graph like the minimum or maximum values and
then compare two or more lines to discuss the differences.
EXAMPLE 8.1
The following graph shows the temperature change on a certain day of the year. The
temperature is taken every hour, on the hour.
36˚
34˚
Temperature
32˚
28˚
●
24˚
20˚
●
16˚
4˚
●
●
●
●
●
12˚
8˚ ●
●
●
●
●
●
●
0˚
05:00 06:00 07:00 08:00 09:00 10:00 11:00 12:00 13:00 14:00 15:00 16:00 17:00 18:00
Time of Day
1) Is this data discrete or continuous?
Discrete – temperatures we taken once every hour.
1) Is this graph discrete or continuous?
Continuous
2) Is this graph linear or non-linear?
Non-linear
3) What is the minimum water usage for this period?
50kl in June
2) Is this trend linear or non-linear?
It is non-linear.
3) At what time of day was the temperature at its maximum?
At 14:00 – 28°C
4) Give a reason for the minimum usage.
It is winter and the garden did not need watering, or the swimming pool was covered and
did not need refilling.
4) Describe the decreasing and increasing trend of this data.
The first reading was done at 05:00. The temperature decreased until 06:00 where it
started increasing until 14:00. It started decreasing then until it reached 16°C at 18:00.
5) Is the graph increasing or decreasing in the period from June to December?
Increasing
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Page 32
www
Euclidean Geometry: Lines and Angles
Grade 8 Maths Essentials
REMINDERS FROM GRADE 7:
Alphabet letters (A,B,C...) are used to indicate points or lines in figures/sketches
(i.e. AB describes line AB or part AB of a bigger line/shape
SCIENCE CLINIC 2022 ©
PARALLEL LINES:
Lines that are always the same (equal) distance from each other. They never cross or touch!
Lines that are the same length are depicted as follows:
PERPENDICULAR LINES:
̂ = 90∘ ∴ A D ⊥
Lines that cut/cross each other at exactly 90∘. ∴ ACD
BC
The following geometry symbols are important for you to memorise:
Â
= angle A
LEARN!
= Parallel lines
NAMING ANGLES:
= Perpendicular line (cut each other at 90∘)
̂
B1̂ = A BC
̂
̂
B2 = C BD
̂
B1̂ + B2̂ = A BD
Note: B1+̂ 2 ≠ B1̂ + B2̂
Horizontal line (parallel to the ground or horizon)
Vertical line (runs from top to bottom, at 90∘ with the ground)
USEFUL TERMINOLOGY:
•
•
•
•
•
Complimentary angles: add up to 90∘
Supplementary angles: add up to 180∘
Parallel Lines (ll): are always the same distance apart (so they never meet/touch).
Perpendicular lines: meet or intersect at 90∘(a right angle).
Adjacent Angles: are next to each other and share a common vertex, but must be
on opposite sides of the common side.
COMPLIMENTARY ANGLES:
Angles that add up to 90∘.
SUPPLEMENTARY ANGLES:
Angles that add up to 180∘.
KEEP IN MIND:
•
•
•
•
Each new statement needs a reason.
Once you have calculated an angle size you can use that value for other calculations.
There are often multiple ways to solve a geometry problem.
Set up your geometry solutions in a statement/reason table.
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B1̂ + B2̂ + B3̂ = 90∘
C1̂ + C2̂ + C3̂ + C4̂ + C5̂ = 180∘
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Page 33
Euclidean Geometry: Lines and Angles
Grade 8 Maths Essentials
TYPES OF ANGLES
SCIENCE CLINIC 2022 ©
BASIC ANGLES
Statement
∘
∘
x + 30 = 90 x = 90 − 30
x = 60∘
Complementary angles add up to 90
Smaller than 90∘
ACUTE
∘
Reason
∘
Compl. ∠ 's
30º
x
Exactly 90
RIGHT
∘
Vertically opposite angles are equal
y
Adjacent angles on a straight line are
supplementary (add up to 180)
Exactly 180∘
STRAIGHT
Angles around a point add up to 360
Between 180∘ and 360∘
REFLEX
vert. opp. ∠ 's =
a + 160∘ = 180∘a = 180∘ − 160∘
a = 20∘
∠ 's on str. line
150∘ + 30∘ + 35∘ = 360∘
215∘ = 360∘
360∘ − 215∘
145∘
∠ 's round a pt
64º
Between 90∘and 180∘
OBTUSE
y = 64∘
160º
a
b
b
b
b
35º 30º
150º
b
+
+
=
=
Exactly 360∘
REVOLUTION
EXAMPLE 9.1
Determine, giving reasons, the unknowns
Statement
1)
a
b
a = 62∘
b + 62∘ = 180∘
b = 118∘
Reason
Statement
2)
vert. opp. ∠ 's =
p
∠ 's on str. line
40º
62º
q
220º
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p
p
q
q
q
+
=
+
+
=
40∘ = 90∘
50∘
220∘ + 90∘ = 360∘
310∘ = 360∘
50∘
Reason
Statement
3)
compl. ∠ 's
∠ 's round a pt
x
y
3x
3x = 90∘
x = 30∘
x + 3x + y = 180∘
30∘ + 90∘ + y = 180∘
y + 120∘ = 180∘
y = 60∘
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Reason
Given
∠ 's on str. line
Page 34
Euclidean Geometry: Lines and Angles
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
ANGLES FORMED BY PARALLEL LINES
“FUN” Angles
Remember to name the parallel lines with reason
Statement
Corresponding
angles are equal
A
B
x = 56∘
Reason
Corresp. ∠ 's A B ∥ CD
EXAMPLE 9.2
Determine, giving reasons, the unknowns
x
C
Co-interior angles
are supplementary
(add up to 180∘)
56º
P
Q
1)
t + 91∘ = 180∘
t = 180∘ − 91∘
t = 89∘
R
r
W
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R
Y
y = 26∘
Corresp. ∠ 's PQ ∥ R S
∘
x + 10 + y = 180
x+10º S
26º
∘
Co-int. ∠ 's PQ ∥ R S
x + 10∘ + 26∘ = 180∘
x + 36∘ = 180∘
2)
Alt. ∠ 's V W ∥ X Y
r = 30∘
q r
p
A
X
50º
Q
y
x = 144∘
S
V
P
Co-int. ∠ 's PQ ∥ R S
t 91º
r = 50∘
Alternate angles are
equal
Reason
Statement
D
C
30º
B
p = 30
Corresp. ∠ 's A B ∥ CD
∘
p + q = 180
Alt. ∠ 's A B ∥ CD
∘
∠ 's on str. line
30∘ + q = 180∘
D
q = 180∘ − 30∘
q = 150∘
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Page 35
Euclidean Geometry: Triangles
Grade 8 Maths Essentials
CLASSIFYING TRIANGLES
SCIENCE CLINIC 2022 ©
PROPERTIES OF TRIANGLES
ANGLES
STATEMENT
60º
All angles are equal
60º
Equilateral triangle
60º
The sum of the interior angles
of a triangle is 180∘
a
b
Two angles are equal
70º
Isosceles triangle
70º
One angle is 90∘
REASON
The exterior angle of a triangle
is equal to the sum of the
interior opposite angles
a + b + c = 180∘
int. ∠ 's of △
x = y+ z
ext. ∠
p = 60∘
equilat. △
t = 50∘
∠ 's opp = sides
AC = BC
sides opp = ∠ 's
c
y
z
Right-angled triangle
of △
x
The interior angles of an
equilateral triangle are each 60∘
SIDES
p
All sides are equal
Two sides are equal
Equilateral triangle
In an isosceles triangle the
angles opposite the equal sides
are equal
In an isosceles triangle the sides
opposite the equal angles are
equal
No sides are equal
A
75º
Scalene triangles
75º
C
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t
50º
Isosceles triangle
B
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Page 36
Euclidean Geometry: Triangles
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
EXAMPLE 10.1
Determine the values of the unknowns and give reasons
1)
38º
a
b
Statement
Reason
a + 38∘ + 90∘ = 180∘
a + 128∘ = 180∘
a = 52∘
int . ∠ ′s of △
(Note: there are 2
options to calculate b)
i) a + b = 180∘
52∘ + b = 180∘
b = 128∘
ii) b = 90∘ + 38∘
b = 128∘
Statement
3)
p
q = 60∘
q = 180∘
60∘ = 180∘
120∘
equilat . △
∠ ′s on str . line
q r
∠ ′s on str . line
ext ∠
p=
r+
r+
r =
Reason
of △
4)
2)
Statement
x = y
x + y + 25∘ = 180∘
x + y = 155∘
∴ x = y = 77,5∘
z = y = 77,5∘
25º
x
y
Reason
r
p F
23º
E
∠ ′s opp = sides
int ∠ ′s of △
G
vert . opp . ∠ ′s =
z
65º
q
Statement
H
p
q
q
q
r
=
+
+
=
=
65∘
23∘ + 65∘ = 180∘
88∘ = 180∘
92∘
q = 92∘
Reason
alt ∠ ′s EF ∥ GH
int ∠ ′s of △
corresp ∠ ′s EF ∥ GH
(This problem has multiple solutions: only one is shown)
NOTE:
If there are ∥ lines,
remember to look for the
”FUN” angles.
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Page 37
Euclidean Geometry: Triangles
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
THEOREM OF PYTHAGORAS
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
NOTE:
EXAMPLE 10.2
Determine, giving reasons, the values of x, y, z
The Theorem of Pythagoras is only valid for
right-angled triangles
1)
Statement
REMINDER: In a right-angled triangle, the hypotenuse is the side directly opposite
the right-angle (and it is always the longest side).
x
4
hypotenuse
3
hypotenuse
2)
HOW TO USE THE THEOREM OF PYTHAGORAS:
AC 2 = A B 2 + BC 2
42 + 32
16 + 9
25
25
5
Statement
5
The square of the length of the hypotenuse is placed on the left of the equation and is
equal to the sum of the squares of the lengths of the other two sides of the triangle.
Statement
x2 =
x2 =
x2 =
x =
x =
12
x
Reason
x2 =
x2 =
x2 =
x =
x =
52 + 122
25 + 144
169
169
13
Reason
Pyth.
Reason
Pyth.
Pyth.
NOTE:
The hypotenuse
always stands
alone on the left of
the equal sign.
3)
4
6
5
3
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Statement
x
10
10 2 = x 2 + 62
100 = x 2 + 36
64 = x 2
x =
64
x = 8
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Reason
Pyth.
Page 38
Euclidean Geometry: Triangles
Grade 8 Maths Essentials
EXAMPLE 10.3
EXAMPLE 10.4
Determine, giving reasons, the values of x and y
Determine, giving reasons, the values of x, y, z
A
A
6
z
x
D
B
4
y
⎷ 23cm
△
x2 =
x2 =
x2 =
x =
x =
ABD
62 + 82
36 + 64
100
100
10
△ BCD
x 2 = y 2 + 42
10 2 = y 2 + 42
100 = y 2 + 16
84 = y 2
y=
84
Reason
Pyth.
Pyth.
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Statement
1cm
x
C
△ ABD
2
2
z =
23 +
z 2 = 23 + 2
z =
25
z = 5
Both triangles are right-angled
•
The bottom triangle is also an
isosceles triangle.
∠ ′s opp =
int ∠
of △
sides
Pyth.
Pyth.
23 + y 2
2
•
Reason
△ BCD
x + x + 90∘ = 180∘
2x = 90∘
x = 45∘
y 2 = 12 + 12
y2 = 2
y=
2
z2 =
What should you notice?
y
B
C
Statement
D
8
SCIENCE CLINIC 2022 ©
2
2
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Page 39
Quadrilaterals
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
1) Parallelogram
4) Square
6) Trapezium
Definition: A quadrilateral with both pairs of opposite sides parallel.
Definition: A square is a rectangle with 2 adjacent sides equal.
Definition: A trapezium is quadrilateral with 1 pair of
opposite sides parallel.
Properties:
Properties:
• 2 pairs opposite sides equal
P
Properties:
S
• All sides equal
• 2 pairs opposite angles equal
• Diagonals bisect one another
• Diagonals bisect one another
• All angles are 90∘
• Diagonals bisect area of
parallelogram
• Diagonals are equal
Q
2) Rectangle
Definition: A rectangle is a parallelogram with a 90∘ angle
R
• Diagonals cut at 90∘
• 2 pairs opposite sides equal
• 2 pairs opposite angles equal
• Diagonals bisect one another
• All angles are 90∘
• Diagonals are equal (PR=QS)
D
• 1 pair opposite
sides are parallel
C
B
• Diagonals bisect the angles
forming 45∘
Angles of a Quadrilateral
5) Kite
Definition: A kite is a quadrilateral with two pairs of adjacent sides
equal.
Properties:
A
Properties:
A1̂ + B̂ + Ĉ1 = 180∘ (sum ∠ ′s of △)
A2̂ + D̂ + Ĉ2 = 180∘ (sum ∠ ′s of △)
 + B̂ + Ĉ + D̂ = 360∘ (sum ∠ ′s of □)
• Diagonals cut at 90∘
A
• One diagonal bisect the other
(BO=OD)
2
1
• Diagonal AC bisects the angles
3) Rhombus
Definition: A rhombus is a parallelogram with 2 adjacent sides equal
• Angles opposite diagonal AC
are equal (B̂ = D̂ )
B
D
Properties:
P
S
• All sides equal
2
1
• 2 pairs opposite sides parallel
2
1
• 2 pairs opposite angles equal
• Diagonals bisect one another at 90∘
Q
R
• Diagonals bisect the angles
̂ = P̂ 2 etc)
(P1
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Page 40
Grade 8 Maths Essentials
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Quadrilaterals
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SCIENCE CLINIC 2022 ©
Page 41
Quadrilaterals
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
“Exclusive clubs in the quadrilateral society”
The quadrilateral
Rule: Must have four sides
The trapesium
Rule: Must have only one
parallel side
The parallelogram
Rule: Must have two pairs
of parallel sides
The rhombus
Rule: Sides have equal lengths.
The Rectangle
Rule: Must
have four 90º
angles.
The
Square*
The Kite
Rule: two pairs
of adjacent are
sides equal
*The Square is part of every exclusive club! It is a rectangle, a rhombus, a parallelogram and a quadrilateral*
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Page 42
Quadrilaterals
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
EXAMPLE 11.1
EXAMPLE 11.2
Calculate the sizes of angles indicated by letters, give reasons for answers:
Determine with reasons, values of variables:
A
75º
1)
1)
D
a
C
c
d
5cm
10º
B 80º
a
 + B̂ + Ĉ + D̂ = 360∘
75∘ + 80∘ + a + 10∘ = 360∘
a = 360∘ − 165∘ = 195∘
a=
b=
c2 =
c=
d =
B
A
3c
(sum ∠ ′s of quad)
D
30º
b
m
90∘
30∘
52 − 32
16 = 4c m
3c m
(diag . rhom.)
(diag . rhom.)
(Pythagoras)
(diag . rhom.)
C
2)
M
2)
c
22
º
N
58º
P
d
e 80º
̂ = 180∘
O c + N̂ + MON
c
c
e
d
d
+
=
=
=
=
(sum ∠ ′s of △)
58∘ + 80∘ = 180∘
180∘ − 138∘ = 42∘
42∘
(alt ∠ ′s , MN ∥ PO)
360∘ − 22∘ − 42∘ − 58∘ − 80∘ − 42∘ (sum ∠ ′s of quad)
116∘
2x + 20∘ = 5x − 40∘
2x − 5x = − 40∘ − 20∘
−3x = − 60∘
x = 20∘
y = 180∘ − (2x + 20∘ )
y = 180∘ − 60∘
y = 120∘
(opp . ∠ ′s ∥ gram)
GI ⊥ HJ
H M = MJ = 8m
i 2 = 82 + 62
i=
100 = 10m
k = 25∘
j = 180∘ − 90∘ − 25∘
j = 65∘
(diag . kite)
(diag . kite)
(Pythagoras)
(co − int ∠ ′s DE ∥ GF)
3)
3)
2x + 10º
x
2x
2x + 10º
 + B̂ + Ĉ + D̂ = 360∘
2x + 2x + 10∘ + x + 2x + 10∘ = 360∘
7x = 340∘
x = 48,57∘
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(sum ∠ ′s of quad)
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(diag . kite)
(sum ∠ ′s of △)
Page 43
Transformations
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
In Grade 8, we focus on 3 types of transformations: 1) Translation 2) Reflection 3) Enlargement
NOTE:
The original (given) point or
shape is called the
OBJECT.
The transformation is called the
IMAGE.
TRANSLATION
Translation is the horizontal movement left or right, or the vertical up or down movement.
Plot the following points on the Cartesian plane and label with the appropriate coordinate pairs:
5
A(5; 1), B(−3; − 2), C(2; 3)
4
5
C’
3
2
4
C
1
A’
3
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
2
A
-1
1
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
B’
-2
-3
-1
B
-4
-2
-5
-3
-4
The “new” coordinates will now be denoted as A′, B′ and C′.
-5
As you know, a coordinate is written as (x ; y). The left or right movement will be the change to x
Now apply the following changes to all three points. Move 1 unit down and 2 units left.
Let’s see what changes happened in the coordinate: A(5; 1) → A′(3; 0)
Do you see the x is 2 units less and y is 1 unit less. In other words, we subtracted in both instances.
So we can describe the change to the coordinate like this: (x ; y) → (x − 2; y − 1)
The general rule for translation is:
(x; y) → (x + a; y + b)
a will represent the left/right (horizontal) movement
b will represent the up/down (vertical) movement
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Page 44
Transformations
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
EXAMPLE 12.1
EXAMPLE 12.2
Given the object ABC below:
2)
5
4
4
A
A
3
2
2
C
B
1
-5
-4
-3
-2
-1
P(1; 4) → P′(−13; 25)
Q(−6; 7) → Q′(−20; 28)
R(−1; 0) → R′(−15; 21)
3
C
B
-6
1) Without drawing the image, give the coordinates of △ PQ R [P(1; 4), Q(−6; 7),
R(−1; 0)] if it is translated by the following
rule: (x ; y) → (x − 14; y + 21)
5
A’
1
0
1
2
3
4
5
6
-6
-5
-4
-3
-2
B’
-1
-1
0
-1
-2
-2
-3
-3
-4
-4
-5
-5
1
2
3
4
C’
5
6
2) Write the rule for the translation of 4 units
right and 3 units up.
(x ; y) → (x + 4; y + 3)
3) Write the rule for the translation of 8 units
up and 3 units left.
(x ; y) → (x − 3; y + 8)
EXAMPLE 12.3
1) Give the coordinates of object ABC.
A(−4; 3), B(−5; 1) and C(−1; 1)
2) Draw the image of object ABC if it is translated 4 units to the right
and 2 units down and label it A′B′C′.
3) Write the coordinates of A′B′C′.
A′(0; 1), B′(−1; − 1) and C′(3; − 1)
4) Give the general rule for this translation.
(x ; y) → (x + 4; y − 1)
5) Find the area of the triangle using the area formula
1
Area △ ABC =
× base × ⊥ height
2
1
=
×4×2
2
= 4 units2
The area of △ A′B′C′ will be the same.
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NOTE:
Looking at the translation, you should see
that the size and shape of the triangle
stayed the same. The shape just shifted.
Describe the translation that has taken place
(in words and using the rule) according to the
following points:
1) (4; 8) → (7; 10)
Translation of 3 units right and 2 units up.
(x ; y) → (x + 3; y + 2)
2) (−2; 7) → (1; 4)
Translation of 3 units right and 3 units
down.
(x ; y) → (x + 3; y − 3)
3) (0; − 2) → (−2; 4)
Translation of 2 units left and 6 units up.
(x ; y) → (x − 2; y + 6)
4) (7; 7) → (−3; − 3)
Translation of 10 units left and 10 units
down.
(x ; y) → (x − 10; y − 10)
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Page 45
Transformations
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
REFLECTIONS
When we talk about reflection, you must visualise looking at a reflection in a mirror. In this case, your x-axis or y-axis will be your “mirror”.
We can reflect both points and images like this
and the name for these “mirrors” over which
the the points/images are reflected is the axis
of symmetry.
The x-axis is the horizontal axis of symmetry
and the y-axis is the vertical axis of symmetry.
Reflection in the y-axis
Reflection in the x-axis
The general rule for reflection in the y-axis is:
The general rule for reflection in the x-axis is:
(x; y) → (−x; y)
(x; y) → (x; − y)
An easy way to remember the rule: Reflection in y-axis → y-value stays
the same x-value changes sign
An easy way to remember the rule: Reflection in x-axis → x-value stays
the same y-value changes sign
5
4
A’
A
3
Observe the object △ A BC with coordinates A(−4; 3), B(−5; 1) and C(−1; 1)
Image △ A′′B′′C′′ is the reflection of the object in the x-axis and the coordinates
for this image will be: A′′(−4; − 3), B′′(−5; − 1) and C′′(−1; − 1)
2
△ABC
B
Image △ A BC is the reflection of the object in the y-axis and the coordinates for
this image will be: A′(4; 3), B′(5; 1) and C′(1; 1)
△A’B’C’
C
C’
1
-6
-5
-4
-3
-2
-1
0
1
B’
2
3
4
5
6
C"
B"
△A”B”C”
-1
-2
A"
-3
-4
-5
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Page 46
Transformations
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
EXAMPLE 12.4
Study the object ( △ A) and answer the questions below:
2)
5
5
4
4
B
B
B’
3
3
C
A
2
C2 C’
A
1
-6
-5
-4
-3
-2
-1
A1
B
1
0
1
2
3
4
5
6
-6
-5
-4
A
-1
-3
-2
-1
0
1
2
3
-1
-2
-2
-3
-3
-4
-4
-5
-5
4
5
6
A’
1) Give the coordinates of the vertices of △ A
(−4; − 1), (−1; 3) and (0; 2)
2) Draw the reflection of △ A in the y-axis and label it △ B.
3) Give the coordinates of the vertices of △ B
(4; − 1), (1; 3) and (0; 2)
4) Give the rule that describes the transformation from △ A 𝒕𝒐 △ B.
(x ; y) → (−x ; y)
5) Without drawing the image, give the coordinates of △ C if this is a reflection of △ A in the x-axis.
(−4; 1), (−1; − 3) and (0; − 2)
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Page 47
Transformations
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
ENLARGEMENTS/REDUCTIONS
An enlargement is the exact “copy” of the shape, just larger. The reduction will be smaller. The shapes will be similar. We enlarge the image by a
“scale factor” and we work with the origin as the centre of enlargement (0; 0).
The following general rule will apply for enlargements:
(x; y) → (k x; k y)
k will represent the scale factor with which the image will be enlarged or reduced.
In the diagram below is a triangle with the vertices A(1; 1), B(1; 3) and C(4; 1) on the Cartesian plane. △ A BC has been enlarged with a scale factor of 3. The centre of enlargement
is the origin (0; 0).
10
B’ (3;9)
You will notice that the coordinates of the object are A(1; 1), B(1; 3) and C(4; 1)
9
Because of the enlargement with a scale factor of 3, the coordinates of the image
will be A′(3; 3), B′(3; 9) and C′(12; 3)
8
7
So the rule we applied for this transformation is: (x ; y) → (3x ; 3y)
6
When we look at △ A BC we can find the area of the triangle:
5
Area △ ABC =
1
× base × ⊥ height
2
4
B (1;3)
1
=
×3×2
2
= 3 units2
A’ (3;3)
3
2
A (1;1)
C (4;1)
1
Now we need to determine the area for △ A′B′C′
1
× base × ⊥ height
2
1
=
×9×6
2
= 27 units2
-1
C’ (12;3)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Area △ ABC =
IN SUMMARY:
If an object is enlarged by a scale factor of k, then
• the sides will be k times longer
• the perimeter will be k times more and
• the area will be k 2 times bigger.
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Page 48
Transformations
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
EXAMPLE 12.5
10
B’
C’
9
8
7
6
5
4
A’
B
C
D’
3
2
A
1
-1
0
1
D
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
7) Calculate the area of A BCD
1) Write down the co-ordinates of A BCD
A = l ×b
A(1; 1), B(1; 3), C(4; 3) and D(4; 1)
= 2×3
2) Write down the co-ordinates of A′B′C′D′
= 6 units2
A(3; 3), B(3; 9), C(12; 9) and D(12; 3)
3) What is the scale factor for this enlargement?
k = 3
8) Calculate the area of A′B′C′D′
A = l ×b
= 6×9
= 54 units2
4) Calculate the perimeter of A BCD
P = 2(2 + 3) = 10 units
9) Compare the area of A BCD and A′B′C′D′. What do you notice?
The area of A′B′C′D′ is 9 times bigger than the area of A BCD.
5) Calculate the perimeter of A′B′C′D′
So, Area of A′B′C′D′ = Area of A BCD = 32
P = 2(6 + 9) = 30 units
54 = 6 × 9
6) Compare the perimeter of A BCD and A′B′C′D′. What do you notice?
So, the area of A′B′C′D′ = Area of A BCD × k 2
The perimeter of A′B′C′D′ is 3 times bigger. 10 × 3 = 30
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Page 49
Area and Perimeter of Polygons and Circles
KNOW THESE SYMBOLS!
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
POLYGONS: rectangles, squares, triangles
P: Perimeter of a polygon
A: Area of polygon or circle
NOTE:
When calculations are done for
area and perimeter,
all measurements
must have the same units!
b: length of base of polygon
h: Perpendicular (⊥ ) height of base of polygon
C: Circumference of circle
r: Radius of circle
d: diameter of circle
DIFFERENCE BETWEEN PERIMETER AND AREA
• Perimeter measures the total distance around a shape
• Area measures the total square units within a shape.
• Perimeter is calculated by adding all the sides of a polygon.
• Area is calculated by multiplying the perpendicular dimensions of
the shape. The result of the product is a two-dimensional area.
4
2
2
2
4
4
Perimeter = 4 + 2 + 4 + 2
= 12 units
Area = 4 × 2
= 8 square units
Units :
Square units:
✖1000
km
➗1000
✖100
m
✖10
cm
➗100
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✖10002
mm
➗10
km2
➗10002
✖1002
m2
✖102
cm2
➗1002
mm2
➗102
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Page 50
Area and Perimeter of Polygons and Circles
Grade 8 Maths Essentials
RECTANGLE
TRIANGLE
SQUARE
CIRCLE
a
base (b)
Perimeter = sum of 4 sides
Perimeter = sum of 4 sides
c
a
d
r
b
a
base
height
b
base (b)
a
a
height (h)
a
height (h)
a
height
a
b
b
base
b
∴P= a+ a+ a+ a
∴ P = 4a
Area = base × height
Area = base × height
Perimeter = sum of 3 sides
∴P= a+ b+ c
∴ A = b×a
∴ A = a ×a
∴ A = a2
Area = area of rectangle ÷ 2
EXAMPLE 13.1
3cm
P = 5cm + 5cm + 3cm + 3cm
= 16cm
3m
EXAMPLE 13.3
P = 3m + 3m + 3m + 3m
= 12m
20mm ÷ 10 = 2cm
b×h
1
=
b ⋅h
2
2
NOTE:
The base is not always the horizontal
side of the triangle but it is the side on
which the ⊥ height is indicated.
5cm
A = 5cm × 3cm
= 15cm2
base
∴A=
EXAMPLE 13.2
1)
3cm
A = 3m × 3m
= 9m2
2c
m
P = 3+ 4+ 6
= 13cm
6cm
6×2
2
= 6cm2
A=
4cm
mal digits. We round off the value of 𝜋 to 3,14 when
using calculations or we can use the value that is programmed into our calculators for better accuracy.
Circumference (C) of a circle:
c
= π
d
c = πd
c = π × 2r (d = 2r)
∴ c = 2π r
5×3
2
= 7,5cm2
A = π r2
EXAMPLE 13.4
1)
2)
3mm
cm
m
𝜋 is a decimal fraction with an infinite number of deci-
16
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9c
3cm
A = 2cm × 4cm
= 8cm2
P = 5+ 4+ 9
= 18cm
5cm
4cm
P = 4cm + 4cm + 2cm + 2cm
= 12cm
2)
What is π?
When any circle’s circumference (C) is divided by its
diameter (d), it always gives the same constant value
known as 𝜋.
C
= π
d
Area (A) of a circle
1)
4cm
20mm
c
c
a
∴P= a+ a+ b+ b
∴ P = 2a + 2b
2)
SCIENCE CLINIC 2022 ©
A=
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C=
=
=
=
2×π ×r
2×π ×3
6π
18,85mm
A=
=
=
=
π × r2
π × 32
9π
28,27mm
C=
=
=
=
2×π ×r
2×π ×8
16π
50,27cm
A=
=
=
=
π × r2
π × 82
64π
201,01cm2
Page 51
Area and Perimeter of Polygons and Circles
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
PERIMETER AND AREA OF COMPOSITE OR PARTIAL FIGURES
EXAMPLE 13.5
1) Calculate the perimeter of the following polygon:
4) Determine the area and perimeter of the semi-circle with diameter 16cm correct
to two decimal places.
3cm
P = 3+ 3+ 1+ 1+ 2+ 3+ 4+ 7
3cm
3cm
4cm
1cm
7cm
NOTE:
To calculate the perimeter, fill in the
missing side lengths and add all the
side lengths together.
2) Calculate the area of the polygon:
= 16cm2
5 × 2 = 10cm2
NOTE:
Divide the polygon into smaller shapes
and add all the areas together.
3×2=
3cm
6cm2
3cm
2cm
4cm
P = 3+ 3+ 5+ 2+ 4+ 2+ 3
= 22cm
5cm
½×4×3
= 6cm2
4 × 2 = 8cm2
A = 9+ 6+ 8
= 23cm2
2cm
1
c+ r+ r
4
1
=
(2 × π × 4) + 4 + 4
4
= 2π + 8
= 14,3cm
1
circle
4
1
=
(π × 42 )
4
= 4π
= 12,6cm2
P=
A=
6) Calculate the area and perimeter of the following figure correct to 2 decimal
places.
3) Calculate the area and perimeter of the polygon:
3×3=
9cm2
16cm
∴ r = 8cm
A = 10 + 6
1cm
3cm
1
circle
2
1
=
× π r2
2
1
=
× π 82
2
= 100,53cm2
A=
5) Calculate the area and perimeter of the quarter circle with radius 4cm.
(Correct to 1 decimal place.)
5cm
2cm
1
c+ d
2
1
=
(2 × π × 8) + 16
2
= 41,13cm
P=
= 24cm
2cm
cm
P = 10 + 8 + 6 +
10
1
(2 × π × 6)
4
= 33,42cm
= 24 +
0,08m
6cm
1
c
4
6×8 1
+ (π × 62 )
2
4
1
= 24 + (36π)
4
= 52,27cm2
A=
NOTE:
Only the exterior sides are added.
4cm
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Page 52
Volume and Surface Area - Prisms
SCIENCE CLINIC 2022 ©
PRISMS
B: Area of the base of the prism
A Prism is a three-dimensional object with two polygonal bases that are congruent (identical) and parallel. The faces that
connects the two bases are rectangles, squares or parallelograms.
P: Perimeter of the base of the prism
H: Height between the two bases of the prism
(distance between the bases)
base
V: Volume of a prism
TSA: Total surface area of a prism
Take note: A prism does not necessarily stand
on its base.
Base of a prism: The two parallel, congruent
(the same) faces of a prism
height
KNOW THESE SYMBOLS!
Grade 8 Maths Essentials
Height of a prism : The distance between the
two bases.
base
Capacity of a Prism: Capacity describes the amount of fluid that a 3-dimensional shape can hold.
Relationship between capacity and volume:
1 litre = 1000ml
1 kilolitre = 1000 litre
Volume: Volume measures the space that is taken by a shape in three dimensions and it is the result
of the product of 3 perpendicular (orthogonal) dimensions.
Cubic units:
✖1003
✖10003
km3
➗10003
m3
✖103
cm3
➗1003
mm3
➗103
Total Surface Area (TSA): The surface area of the unfolding of a prism in 2 dimensions. The TSA is
calculated by adding up the surface areas of all the sides/surfaces of the prism.
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Page 53
Volume and Surface Area - Prisms
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
VOLUME OF A PRISM
General formula for the volume of all prisms:
Volume = area of base x height
V= B×H
VOLUME OF A CUBE
VOLUME OF A RECTANGULAR PRISM
VOLUME OF A TRIANGULAR PRISM
The base of a cube is a square.
The base of a rectangular prism is a rectangle.
The base of a triangular prism is a triangle.
H
a
a
a
a
V= B×H
V = (a × a) × a
V = a3
EXAMPLE 14.1
a
b
V= B×H
V = (a × b) × H
V= B×H
1
b×h
V=
× H or V =
(b × h) × H
2
2
EXAMPLE 14.3
2cm
5cm
5cm
4cm
5m
3m
6cm
4m
V = 5cm × 5cm × 5cm
= 125cm3
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V = 6cm × 4cm × 2cm
= 48cm3
h
b
EXAMPLE 14.2
5cm
H
c
4m × 3m
V=
× 1,5m
2
= 6m2 × 1,5m
= 9m3
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1,5m
Page 54
Volume and Surface Area - Prisms
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
TOTAL SURFACE AREA OF A PRISM
TSA = (2 × B) + (P × H)
General formula for the TSA of all prisms:
TSA = (2 x area of base) + (perimeter of base x H)
TSA OF A CUBE
TSA OF A RECTANGULAR PRISM
TSA OF A TRIANGUALR PRISM
Unfolding a cube:
Unfolding of a rectangular prism:
Unfolding of a triangular prism:
c
a
a
a
H
a
a
a
a
a
a
H
b
a
a
a
b
b
a
b
H
a
H
We can think of the surface area of a cube as the sum of 6
identical squares each with an area of a 2.
TSA = 6 × a
TSA = 6a 2
2
OR
When we apply the general rule:
TSA
TSA
TSA
TSA
=
=
=
=
2×B+ P + H
2(a × a) + (a + a + a + a) × a
2a 2 + 4a × a
2a 2 + 4a 2 = 6a 2
We can split the rectangular prism into 3 pairs of rectangles that
are identical and thus have the same area.
TSA = 2 ×
TSA = 6(5cm)2
= 150cm2
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b×h
+ (a + b + c) × H
2
EXAMPLE 14.6
EXAMPLE 14.5
4cm
5cm
a
TSA = (2 × B) + (P × H)
OR
When we apply the general rule:
TSA = (2 × B) + (P × H)
TSA = 2ab + (2a + 2b) × H
5m
3m
2cm
5cm
b
TSA = 2ab + 2aH + 2bH
EXAMPLE 14.4
5cm
c
6cm
TSA = 2(6cm × 4cm) + 2(4cm × 2cm) + 2(6cm × 2cm)
TSA = 48cm2 + 16cm2 + 24cm2
TSA = 88cm2
OR
TSA = 2(6cm × 4cm) + (6cm + 4cm + 6cm + 4cm) × 2cm
TSA = 48cm2 + 40cm2
TSA = 88cm2
4m
1,5m
3m × 4m
+ (3m + 4m + 5m) × 1,5m
2
TSA = 12m2 + (12m × 1,5m)
TSA = 12m2 + 18m2
TSA = 30m2
TSA = 2 ×
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Page 55
Statistics
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
Statistics consists of the following parts:
SORTING DATA
• Collecting data
• Sorting data
We can sort data by using a frequency table or stem-and-leaf displays.
• Presenting data
• Analysing and interpreting data
EXAMPLE 15.1
Information is referred to as data, and we use information to make informed decisions and predictions.
COLLECTING DATA
The pupils in a class have been numbered from 1 to 30 in order to determine their
participation in winter sport.
R: Rugby
S: Soccer
N: Netball
H: Hockey
C: Cross Country O: No Sport
We collect data using some of the following tools: Surveys, questionnaires and interviews.
Raw data is data that has not been organised in any way. After collection it must be sorted in a
meaning full way and presented in such a way so that we can analyse and interpret the information
collected.
Discrete data is data that can be counted. The data values are all whole numbers. For example, if
you count people, or goals scored, or shoe sizes.
Continuous data is data that is measured. The data values are rational numbers. For example, if you
work with lengths of people, or times taken by athletes running races, or different weights of people
attending a gym.
We want to study the shoe sizes of woman in South Africa. A population is made up of the entire
group of people studied, in this case, ALL the woman in South Africa. You will agree it is close to
impossible to collect the data of ALL the woman in South Africa. That is the reason why will only use a
sample of a population. A sample is a smaller selection of that population.
The following conditions for selecting a sample must be adhered to:
• The sample must represent the whole population.
• The sample must be chosen at random to not influence the results. The sample must be
unbiased.
1
S
6
O
11
S
16
O
21
H
26
H
2
R
7
H
12
R
17
H
22
N
27
C
3
S
8
N
13
N
18
N
23
N
28
N
4
N
9
C
14
N
19
H
24
C
29
S
5
H
10
R
15
S
20
R
25
S
30
S
Let’s set up a frequency table
SPORT
TALLY
FREQUENCY
R
||||
4
N
|||| |||
8
H
|||| |
6
S
|||| ||
7
C
|||
3
O
||
2
The raw data is now sorted and we can follow on to the next step.
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Page 56
Statistics
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
EXAMPLE 15.2
In the previous example, there were only a few data items. When there are many different data items, it is easier to group the data in class intervals to make tallying more efficient.
Let’s set up a stem-and-leaf display (The “tens” in the numbers form the stem and the
AGES OF TEACHERS
“units” form the leaves)
24
27
36
47
39
38
64
30
36
38
Stem
35
62
29
49
51
2
4,7,7,9,5
36
58
49
25
52
49
27
45
32
41
3
8,5,6,6,0,6,2,9,8
4
9,9,5,7,9,1
5
8,1,2
6
4,2
Let’s set up a frequency table
CLASS INTERVALS
TALLY
10-19
FREQUENCY
Leaf
0
You record all the ages as you read them, including the repeated values and zero’s. (30).
After you have done this, you now need to go and order you stem-and-leaf display:
20-29
||||
5
30-39
|||| ||||
9
40-49
|||| |
6
50-59
|||
3
60-69
||
2
Stem
Leaf
2
4,5,7,7,9
3
0,2,5,6,6,6,8,8,9
4
1,5,7,9,9,9
5
1,2,8
6
2,4
Now your collected data has been organised and you are ready to go on to the next step.
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Page 57
Statistics
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
PRESENTING DATA
Once the data has been collected and sorted, we can represent the data graphically using bar graphs, histograms, pie charts and broken-line graphs.
This will give you a visual picture of the data.
Bar graphs:
Histograms:
A bar graph has vertical or horizontal bars and usually represent discrete data. The bars have
spaces between them.
Histograms are used to represent grouped data graphically. You also use bars to represent the
data, but there are no spaces between the bars. The x-axis normally shows continuous measurement and is labelled with the class intervals.
We are going to use the same information used in EXAMPLE 15.1 to draw our bar graph.
For the histogram, we are going to use the information in EXAMPLE 15.2.
SPORT
TALLY
FREQUENCY
CLASS INTERVALS
R
||||
4
10-19
N
|||| |||
8
20-29
||||
5
H
|||| |
6
30-39
|||| ||||
9
S
|||| ||
7
40-49
|||| |
6
C
|||
3
50-59
|||
3
O
||
2
60-69
||
2
9
8
Frequency
Frequency
7
6
5
4
3
2
1
Rugby
Netball
Hockey
Soccer
Cross Country
FREQUENCY
0
Ages of teachers in school
Winter sport participation
0
TALLY
No Sport
Sport
10
9
8
7
6
5
4
3
2
1
0
0-9
10-19
20-29
30-39
40-49
50-59
60-69
Ages
You can now clearly see that Netball and Soccer are the two most popular sports.
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From this histogram we can see that the most teachers fall within the age-range of between
30 and 40 years.
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Page 58
Statistics
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
Pie Charts:
Broken-line Graphs:
A pie chart is a circle that is divided into sectors. Whenever we need to represent information in percentage, we will use
a pie chart.
A broken-line graph is made up of straight lines connecting points
that have been plotted. The y-axis normally represents the frequency
and it does not have to start at zero.
The following background knowledge is necessary to enable you to draw a pie chart:
A circle consists out of 360∘, thus if you have a sector of 45∘ you need to:
45∘
1
=
of the total
∘
360
8
1) Express it as a fraction:
2) Write it as a percentage:
45º
The following table shows the maximum temperature in °C as
measured at 14:00 every day for the first ten days of September.
1
× 100 = 12,5%
8
Draw a graph for the data below.
EXAMPLE 15.3
The table below shows the number of favourite fruit chosen by Grade 3 learners:
Favourite Fruit
Banana
Apple
Naartjie
Orange
Grapes
No. of learners
7
3
5
4
2
1)
How many learners participated in this survey?
7 + 3 + 5 + 4 + 2 = 21
2)
In the table, calculate:
i) The fraction of pupil choosing a fruit
ii) The fraction as a percentage.
iii )The size of the angle representing each sector.
Fruit
Banana
Apple
Naartjie
Orange
Grape
Fraction
7
21
3
21
5
21
4
21
2
21
Percentage
Angle size
33%
120∘
14%
51∘
24%
86∘
19%
69∘
10%
34∘
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EXAMPLE 15.4
Day
1
2
3
4
5
6
7
8
9
10
Temp (C)
27
25
18
22
28
28
31
30
29
32
Now draw a pie chart representing the
information.
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Page 59
Statistics
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
ANALYSING AND INTERPRETING DATA
You will now study the graphs and look for trends in the data. We use measures of central tendency (mean, median and mode) as well as
measures of dispersion (range and outliers) to investigate different aspects of the data.
Measures of Central Tendency:
EXAMPLE 15.5
Mean:
The average of all the data values
Below is the percentages achieved by 40 pupils who wrote a Maths exam:
sum of all the values
number of values
40
61
43
29
95
74
63
49
74
63
54
39
86
89
85
66
48
95
28
63
67
69
84
83
94
54
83
29
54
83
84
85
63
87
96
73
59
53
72
Complete the frequency table.
% Intervals
Tally
Frequency
20-29
||||
4
30-39
|
1
40-49
||||
4
50-59
||||
5
60-69
|||| |||
8
Mode:
This is the number that occurs most often.
70-79
||||
4
80-89
|||| ||||
10
Measures of Dispersion:
90-99
||||
4
Median:
Once the data is sorted in numerical order,
you find the middle value. If there are an
even number of values, the median lies
between the two middle values.
Range:
Range is the spread of the data that has
been collected.
40
Represent the data on a stem-and-leaf display.
Stem
Range = Highest Value − Lowest Value
Outliers:
These are values that are significantly
higher or lower than the other data values.
They “stand out”. Outliers can affect the
mean and are sometimes excluded from
the data.
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1) How many pupils obtained more than 90%?
4 pupils
2) How many pupils failed Maths?
5 pupils
3) Determine the mean of this data.
2641
Mean =
= 66%
40
4) What is the median percentage?
Median = 66,5% (between 66% and 67%)
5) Give the mode of this data.
Mode = 63% (appears 4 times)
6) Draw a histogram of the Maths marks.
Leaf
2
5, 8, 9, 9
3
9
4
0, 3, 8, 9
5
3, 4, 4, 4, 9
6
1, 3, 3, 3, 3, 6, 7, 9
7
2, 3, 4, 4
8
3, 3, 3, 4, 4, 5, 5, 6, 7, 9
9
4, 5, 5, 6
Results of the Maths exams
Frequency
Mean =
25
10
9
8
7
6
5
4
3
2
1
0
0-9
10-19
20-29
30-39
40-49
50-59
60-69
70-79
80-89 90-100
Marks in %
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Page 60
Probability
Grade 8 Maths Essentials
SCIENCE CLINIC 2022 ©
Probability is all about chance! What is the chance that something might or might not happen?
One of the probabilities that you should have come across is the weather. If you have ever planned a birthday party outside, surely you would have watched the weather on the news to see
what the chances of rain are?
Relative frequency: This is the actual number of times a specific event occurred
out of the total number of trials done.
KNOW THESE TERMS!
Using the example of rolling a die:
Experiment: The experiment is the actual rolling of the die.
Trial: Rolling the die a certain number of times
Relative Freq . =
Outcome: The outcome is the result of the roll.
number of times the event happens
the total number of trials
Sample Space: This is a set of ALL possible outcomes. In
this example there will be six (6) possible outcomes.
The table shows the results of physically rolling the die 10 times
Probability Scale: All probabilities will lie between 0 and 1.
impossible
0
0,1
0,2
high chance
0,3
Probability =
0,4
0,5
0,6
1
2
3
4
5
6
7
8
9
10
Result
6
4
2
5
3
3
2
6
4
1
certain
even chance
low chance
Trial
0,7
0,8
So the relative frequency of rolling a 3 is:
0,9
1
Relative Frequency =
2
1
=
10
5
number of times the event occurs
the number of possible outcomes
Remember that the probability can be expressed in 3 ways.
What is the probability of rolling a die, and landing on a 2?
1
• Fraction: P(2) = 6
• Decimal Number: P(2) = 0,17
• Percentage: P(2)=17%
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Page 61
Grade 8 Maths Essentials
Probability
EXAMPLE 16.1
EXAMPLE 16.3
Fifteen cards numbered from 1 to 15 are placed in a box. One card is drawn from the box.
Determine the following:
A card is drawn from a fair pack of cards excluding the jokers.
NOTE:
A pack of cards has 52 cards (excluding jokers);
13 of each suit (hearts, diamonds, spades or
clubs), 12 picture cards (Jack, Queen and King);
4 letter cards (Aces)
1) P(odd number)
8
=
15
2) P(even number)
7
=
15
Determine the probability of drawing:
1) A heart
3) P(prime number)
6
2
=
=
15
5
P(Heart)
13
=
52
1
=
4
4) P(number > 10)
5
1
=
=
15
3
2) A jack of clubs
P(Jack of clubs)
1
=
52
5) P(number ≤ 3)
3
1
=
=
15
5
6) P(number between 7 and 8)
0
=
15
3) A king or queen
EXAMPLE 16.2
A die is rolled 3000 times. Predict how many times you would expect to roll the number 6.
Prediction = probability × no of trials
=
1
× 3000
6
= 500 times
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SCIENCE CLINIC 2022 ©
P(King or Queen)
8
=
52
2
=
13
4) Neither a spade nor a heart
P(Neither a Spade or a Heart)
26
=
52
1
=
2
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Page 62