GRADE 10
SETS
1. Given that Set A has 5 elements and B has 128 subsets.
(a) Find the number of subsets of A
(b) Find the number of elements of Set B.
Solutions
a) Number of subsets i s given by 2n where n is the number of elements in a given set.
Number of subset
= 2n
A has 5 elements
= 25
= 32, A has 32 subsets
b) No. of subset = 2n
128 = 2n
27 = 2n
n = 7, B has 7 elements
2. If E = { Natural numbers less than 13}
P = {x: x is a prime number}
O = {x:x is an old number}
S = x:x is a square number}
List Sets E, P, O and S and hence find the following:
(a) P′
(b) (P ∩ O)′
(c) (P ∪ S)′
(d) (P ∪ S ∪ O)′
Solutions
E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
P = {2, 3, 5, 7, 11}
O = {1, 3, 5, 7, 9, 11}
S = {1, 4, 9}
a) P′ elements in the universal set that are not in P ∴ P′ = {1, 4, 6, 8, 9, 10, 12}
b) (P ∩ O)′ elements in the universal set that are not in P ∩ O = {3, 5, 7, 11}
∴ (P ∩ O)′ = {1, 2, 4, 6, 8, 9, 10, 12}
c) P ∪ S elements of both Sets P ∪ S = {1, 2, 4, 5, 7, 9, 11} and common elements
should be listed once.
Page 1 of 173
d) First list P ∪ S ∪ O = {1, 2, 3, 4, 5, 7, 9, 11} then list elements of universal that are
not P ∪ S ∪ O ∴ (P ∪ S ∪ O)′ = {6, 8, 10, 12}
3. The diagram below shows three intersecting sets A, B and C. It is given that n (A) =
50, n (B) = 42 and n (C) = 62
a) By considering Sets B and C, show that
E
a + b = 38 – x
(i)
B
A
a + 2b = 58= x
(ii)
b) Hence or otherwise, find the value of b
S
a
b
c) Given that
S = b + 10 find the values of S, a and x.
4
x
b
a
2b
C
Solutions
a) From Sets B, a + b + 4 + x = 42
and from C,
a + 2b + 4 + 4 + x = 62
a + b = 42 – 4 – x
a + 2b = 62 – 4 – x
a + b = 38 – x
a + 2b = 58 – x
∴ a + b = 38 – x
a + 2b = 58 – x
(i)
(ii) Hence shown
b) Value of b can be solved by showing the two equations above simultaneously.
𝑎 + 𝑏 = 38 − 𝑥
−(
)
𝑎 + 2𝑏 = 58 − 𝑥
−𝑏 = −20
∴ 𝑏 = 20
Value a from Set A
Value of x from Set B
S + 2a + 4 = 50
a + b + x + 4 = 42
S = 20 + 10
30 + 4 + 2a = 50
8 + 20 + x + 4 = 42
S = 30
34 + 2a = 50
x = 42 – 8 – 20 – 4
2a = 56 – 34
x = 10
c) S = b + 10
2𝑎
𝑎
=
16
𝑎
a=8
Page 2 of 173
∴ a = 8, s = 30 and x = 10
4. At Hillcrest Technical Secondary School, a group of 70 take optional subjects as
illustrated in the Venn diagram below.
E
History
Commerce
RE
x
3x – 5
2x
9
(i)
(ii)
Calculate the value of x
Find the value of Girls who take
a) History only
b) Commerce
Solutions
(i)
3x – 5 + x + 2x + 9 = 70
3x + x + 2x – 5 + 9 = 70
6x + 4 = 70
6x = 70 – 4
6𝑥
6
=
(ii) (a) History only = 3x – 5
3 (11) – 5
33 – 5
28 History only
66
6
Value of x = 11
(b) Commerce = x + 2x + 9
= 11 + 2(11) + 9
= 11 + 22 + 9
= 42
5. In the diagram below shows three Sets A, B and C
C
A
10
B
X
3
C
3X
5
i.
Give that ∩ (A ∪ B ∪ = 50 𝑓𝑖𝑛𝑑
(i) The value of x
(ii) ∩ (A ∪ B)
(iii) ∩ (B ∪ 𝐶)′
(iv) ∩ (A′ ∩ C′)
4
Page 3 of 173
Solution
(i)
10 + x + 3 + 3x + 5 + 4 = 50
10 + 3 + 5 + 4 + x + 3x = 50
22 + 4x = 50
4x = 50 – 22
4𝑥
4
=
28
4
x=7
(ii)
∩ (A∪ 𝐵) = 10 + 𝑥 + 3 + 3𝑥
= 10 + 7 + 3 +3(7)
= 10 + 7 + 3 + 21
= 41
(iii)
(B ∪ 𝐶)′ = 10 + 4
= 10 + 4
= 14
(iv)
∩ (A′ ∩ C′) = 3 + 4
=7
6. On the Venn diagram below, shade the Set (A′ ∩ B) ∩ C
Solution
E
Hint
A
B
Shade A′ and B to give the region
common to (A′ and B)
- Shade C to give the region common
to (A′ ∩ B) ∩ C
C
7. A survey was conducted on 60 women connecting the types of Sim cards used in their
cell phones for the past 2 years. Their responses are given in the diagram below.
E
Cell Z
Airtel
a
4
14
2
3
a) Given that 23 women have used Cell
Z Sim cards, find the values of a and
b
b) How many women have used only
two different Sim cards.
b
10
MTN
Page 4 of 173
c) If a woman is selected at random from the group, what is the probability that
i.
She has no cell phone
ii.
She used only type of a Sim card
d) How many women did not use MTN and Cell Z Sim Cards?
e) How many women used either Airtel or MTN Sim Cards but not Cell Z.?
Solutions
a) a + 4 + 2 + 3 = 23
a + 9 = 23
a = 23 – 9
a = 14
b) a + b + 4 + 2 + 3 + 10 + 14 + 8 = 60
14 + b + 4 + 2 + 3 + 10 + 14 + 8 = 60
b + 55
= 60
b
= 60 – 55
b=5
∴ 𝑎 = 14 𝑎𝑛𝑑 𝑏 = 5
c) (i) No cell phone
(iii)
8
60
=
2
15
Only one type of Sim card = a + 14 + 10
= 14 + 14 + 10
= 38
∴ 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =
38
60
=
19
30
d) (Cell Z ∪ 𝑀𝑇𝑁)′ = 14 + 8
= 22
e) 10 + 5 + 2 + 3 + 14 + 4 = 38 used either Airtel or MTN but not Cell Z
Page 5 of 173
8. If P = {-2, -1, 0, 1, 2,3, 4 …….}, express
a) P in set builder notation
P = {x : x ≥ - 2; x ∈ 𝑍)
(The inequality ≥ is used since elements are building
from 2 (inclusive) to the positive side).
b) Write Set A in listed form,
A = (x : 2 ≤ x < 8
x < 8, 𝑥 ≥ 2
(By splitting the set builder notation)
(Using the number line)
2
3
4
5
6
∴ 𝐴 = {2, 3, 4, 5, 6,7}
7
8
(2 is included in the solution set when 8 is not included)
9. Using the information given in the Venn diagram below
E
A
B
.b
.k
.f
.m
.c
.g
a) List the Set (A ∩ B)
(A∩B) = {f, m}
b) List (A ∩B)′
(A∩B)′ = {k, b, c, g}
(Complement of a set are elements outside the given
sets (A∩B)
Page 6 of 173
10. E (Universal Set) = {2, 4, 6, 8, 10, 12, 14, 18, 20}
A = {2, 4, 6. 8}
B = {4, 8, 12, 14} C = {2, 4, 12, 18, 20}
i.
Illustrate the above information on a Venn diagram.
E
A
B
.6
.8
.14
4
.2
.10
20
12
18
20
C
.16
20
ii.
iii.
List the Set
(A ∩ B ∩ C) = {4}
(The intersection of three sets)
(A ∩ B′) ∪ 𝐶
A ∩ B′ = {2, 4, 6, 8} ∩ {2, 6, 10, 16, 18, 20)
(Dealing with what is inside the
brackets by listing Set A and Set
B′)
(A ∩ B′) = {2, 6}
Set C = {2, 4, 12, 18, 20}
∴ (𝐴 ∩ 𝐵 ′ ) ∪ 𝐶 = {2, 6} ∪ {2, 4, 12, 18, 20}
(A∩ B′) ∪ 𝐶 = {2, 4, 6, 12, 18, 20} (Do not repeat elements in the union of sets)
iv.
Find the value of ∩ ( A ∪ 𝐵 ∪ 𝐶)′
(A∪ 𝐵 ∪ 𝐶)′ = {10, 16}
∴ ∩ (A∪ 𝐵 ∪ 𝐶)′ = 2
v.
By listing
(There are two elements in that Set)
In the Venn diagram, shade the region
(A′∪ 𝐵) ∩ 𝐶
(A′ ∪ 𝐵) = {12, 14, 18, 20, 10, 16) ∪ {4, 8, 12, 14}
(A′ ∪ 𝐵) = {4, 8, 10, 12, 14, 16, 18, 20}
(A′ ∪ 𝐵) ∩ 𝐶 = {4, 8, 10, 12, 14, 16, 18, 20} ∩ {2, 4, 12, 18, 20}
(A′ ∪ 𝐵) ∩ 𝐶 = {4, 12, 18, 20} Shade the region where these elements are lying.
Page 7 of 173
11. 70 learners at Mansa Secondary School were asked to mention their favourite subjects
between Maths and Science.
The results are shown in a Venn diagram below
E
Maths
Science
25
30
10
i.
How many learners like Maths only? 25 learners
ii.
How many learners do not like Maths nor Science?
70 – (25 + 30 + 10)
70 – (65)
70 – 65
= 5 learners do not like Maths nor Science
INDEX NOTATION
5 −2
Q1 a) Evaluate ( )
3
1
9 2
(𝑡 6)
b) simplify
2𝑥 3 𝑦
c) simplify
6𝑥𝑦 2
solution
5 −2
a) ( )
3
9
3 2
𝟗
=( ) =
5
𝟐𝟓
1
1
(3)2 2
1
2
3 2 2
𝟑
b) ( 6) =[(𝑡 3)2] =[( 3) ] = 𝟑
𝑡
𝑡
𝒕
c)
2𝑥 3 𝑦
6𝑥𝑦 2
=
2𝑥 3−1
3𝑦
=
2−1
𝒙𝟐
𝟑𝒚
Q2 a) Evaluate 4−2
Page 8 of 173
1
b) Simplify [
9𝑥𝑦 6 2
𝑥3𝑦2
]
solution
a) 4−2 =
b) [
1
42
1
9𝑥𝑦 6 2
𝑥3𝑦
𝟏
=
] =[
2
𝟏𝟔
1
1
9𝑦 4 2
𝑥
3𝑦 2
] = [(
2
𝑥
2 2
) ] =
𝟑𝒚𝟐
𝒙
Q3 Work out the value of
a) 30
b)5−1
c)6−2
2 −2
d)( )
5
solutions
a) 30 = 1
𝟏
b)5−1 =
c)6−2 =
𝟓
1
62
2 −2
d)( )
5
=
=
𝟏
𝟑𝟔
1
5 2
( )
2
=
𝟐𝟓
𝟒
Q4 a)Evaluate 50 − 5−1
b)Simplify(5𝑥 3 )2
1
c)Simplify
16 2
[ 16]
𝑛
Solution
1
𝟒
5
𝟓
a) 50 − 5−1 = 1 − =
b) (5𝑥 3 )2 = 𝟐𝟓𝒙𝟔
Page 9 of 173
1
1
c)
16 2
[ 16]
𝑛
4 2 2
𝟒
= ⌈( 8 ) ⌉ = 𝟖
𝑛
𝒏
Q5a) The population of a country is 3.2 X 106. There are 8 x 105 children.
What fraction of the whole population are children ? Give your answer in its
simplest form
b) Simplify 25𝑥 2 ÷ 5𝑥 −4
solution
a)
8×105
3.2×106
=
8×105
32×105
=
8
32
=
𝟏
𝟒
b) 25𝑥 2 ÷ 5𝑥 −4 = 5𝑥 2−(−4) = 𝟓𝒙𝟔
Q6 Evaluate
a) 170
5
b)42
c)(0.2)−2
solution
a) 170 = 1
5
5
b)42 = (22 )2 = (2)5 = 𝟑𝟐
2 −2
c)(0.2)−2 = ( )
10
10 2
= ( ) = (5)2 = 𝟐𝟓
2
1 −2
Q7 a) Evaluate ( )
4
2
b) Simplify 643
1
c) Simplify[
4𝑥 2 𝑦 9 2
𝑥4𝑦
]
solution
1 −2
a) ( )
4
4 2
= ( ) = 𝟏𝟔
1
Page 10 of 173
2
2
b) 643 = (43 )3 = 42 = 𝟏𝟔
1
c) [
4𝑥 2 𝑦 9 2
𝑥4𝑦
] =[
1
1
4𝑦 8 2
𝑥
] =
2
42𝑦
8×
1
2×
𝑥 2
1
2
=
𝟐𝒚𝟒
𝒙
Q8 a)simplify 𝑝2 (𝑝3 − 3𝑝−2 )
1
b) simplify (27𝑥 6 )3
solution
a) 𝑝2 (𝑝3 − 3𝑝−2 ) = 𝑝2+3 − 3𝑝−2+2 = 𝒑𝟓 − 𝟑
1
1
1
b) (27𝑥 6 )3 = [33 (𝑥 2 )3 ]3 =[(3𝑥 2 )3 ]3 = 𝟑𝒙𝟐
Q9 a) Find the value of a when 3𝑎 ÷ 34 = 32
b) Find the value of b when 82 = 2
solution
3𝑎 ÷ 34 = 32
3𝑎 = 34 × 32
∴𝑎 =2+4=𝟔
b) 8𝑏 = 2
(2)3𝑏 = (2)1 ∴ 3𝑏 = 1𝒃 =
𝟏
𝟑
Q10
Page 11 of 173
Q 11
Page 12 of 173
PRACTICE QUESTIONS:
(1) EVALUATE 42 + 41 + 40.
(2) (a) Find the value of 2n − 𝑛2 .
(i)
when n = 0,
(ii) when n = 3.
1 -2
3
(b) Find the value of
(3) Find a, b and c when
(a) 3𝑎 ÷ 35 = 27,
(b) 125𝑏 = 5,
(c) 10𝑐 = 0.001.
ANSWERS:
(1) 21
(2) (a) (i) 1
(b) 9
(3) (a) 8
(b)
(ii) -1
1
3
(c ) -3
Type equation here.
FRACTIONS:
1. Find the value of
3
2
1
(a) 3 4 × 2 5 ÷ 1 2
5
1
(b) 6 − 6
1
3
1
(c) 2 2 + 4 ÷ 18
Solution
3
2
1
1.(a) 3 4 × 2 5 ÷ 1 2
Page 13 of 173
=
15 12 3
×
÷
4
5 2
=
15 12 2
×
×
4
5 3
=𝟔
5
1
(b). 6 − 6
=
1
3
5−1
6
=
4
6
=
𝟐
𝟑
1
(c) 22 + 4 ÷ 18
=
5 3 18
+ ×
2 4 1
=
5 9
+
2 2
=
5+3
2
=
8
2
=4
2
2. Convert 15 as a decimal fraction.
Solution
2
7
15 = 5
=𝟏. 𝟒
RATIO AND PROPORTION:
1.The ratio of boys to boys at Mimi Secondary school is 4:5. If there are 800 boys, find
(a) The number of girls in the school.
(b) The total number of pupils in the school.
Solution
Page 14 of 173
(a)Let 𝑥 be the number of girls in school
4 ……….𝑥𝑔𝑖𝑟𝑙𝑠
5……….800 boys
5𝑥 = 4 × 800
4 × 800
5
𝑥=
𝑥 = 640
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑖𝑟𝑙 𝑖𝑠 𝟔𝟒𝟎.
(b) Total number of pupils is 800 + 640
= 1440
2.Bwalya is 1.8 m and Gondwe is 1.2 m tall. On a bright and sunny afternoon, they were
standing next to each facing the direction of the sun. If Gondwe’sshadow was 6 m long, how
long was Bwalya’s shadow?
Solution
Let 𝑥 𝑏𝑒 𝑡ℎ𝑒 𝑠ℎ𝑎𝑑𝑜𝑤 𝑜𝑓 𝐵𝑤𝑎𝑙𝑦𝑎
Object
image
1.8m ………….𝑥
1.2m ………...6m
𝑥=
=
6 × 1.8
1.2
6 × 1.8 × 10
1.2 × 10
=
6 × 18
12
=𝟗
3.In an election, 80 000 people voted. The candidates A, B and C got were in the ratio 9:5:2
respectively. How many votes did candidate B received?
Solution
Total =9 +5+2
= 16
Page 15 of 173
5
Candidate B = 16 × 80 000
= 𝟐𝟓 𝟎𝟎𝟎votes
4. Express the ratio 1.5 km to 25 cm in its simplest form.
Solution
1.2 km : 25 cm
𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑘𝑚 𝑡𝑜 𝑐𝑚
1.2× 1000 × 100 ∶ 25
120000 : 25
4800 : 1
5. The scale of a map is 1 : 2500, calculate the length, in centimeters ,on the map that
represents an actual distance of 1.2 km.
Solution
𝑙𝑒𝑡 𝑥 𝑏𝑒 𝑎𝑐𝑡𝑢𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
1 : 2500
𝑥 : 120000
𝑥=
120000
2500
𝑥 = 𝟒𝟖 𝒄𝒎
6. One painter would take 24 hours to paint a classroom. How long would 3 painters take to
paint the classroom
Solution
𝑙𝑒𝑡 𝑥 𝑏𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 ℎ𝑜𝑢𝑟𝑠
1 painter…….24hours
3 painters……𝑥
𝑥=
1 × 24
3
𝑥=𝟖
So 3 painters would take 8 hours to paint the classroom.
Page 16 of 173
PYTHAGORUS THEOREM
1.The area of triangle PQR = 24 𝑐𝑚2 . Given that PQ = 8 cm. Calculate
(a) RQ
(b) PR
Solution
1
2
1.(a) A= 𝑏ℎ
24 =
1
×8×ℎ
2
24 = 4ℎ
ℎ = 𝟔𝒄𝒎
(b) 𝑃𝑅 2 = 𝑃𝑄 2 + 𝑄𝑅 2
= 82 + 62
= 64 + 36
𝑃𝑅 = √100
𝑃𝑅 = 𝟏𝟎𝒄𝒎
2.The diagram below shows two straight roads AB and BC which join the main road A and C. The
road AB meets the road BC at right angles.
Page 17 of 173
Given that AB = 8 km and AC = 10 km, find the length of the road BC.
Solution
𝐵𝐶 2 = 𝐴𝐶 2 − 𝐴𝐵2
= 102 − 82
= 100 − 64
= 36
𝐵𝐶 = √36
BC= 6 km
ALGEBRA
1. Solve the equation
7
1

4
x  2 x 1
[3 MARKS]
2. Peter cuts a square out of a rectangular piece of metal.
Page 18 of 173
2x + 3
Diagram NOT
accurately drawn
x+2
x+4
x+2
The length of the rectangle is 2x + 3.
The width of the rectangle is x + 4.
The length of the side of the square is x + 2.
All measurements are in centimetres.
The shaded shape in the diagram shows the metal remaining.
The area of the shaded shape is 20 cm2.
(a)
Show that x2 + 7x – 12 = 0
(4)
Page 19 of 173
(b)
(i)
Solve the equation x2 + 7x – 12 = 0
Give your answers correct to 4 significant figures.
(ii)
Hence, find the perimeter of the square.
Give your answer correct to 3 significant figures.
7r + 2 = 5(r – 4)
3. Solve
4. Simplify fully
(i)
(p3)3
3q4  2q5
q3
(ii)
5. The force, F, between two magnets is inversely proportional to the square of the
distance, x, between them.
When x = 3, F = 4.
6.
(a)
Find an expression for F in terms of x.
(b)
Calculate F when x = 2.
(c)
Calculate x when F = 64.
40 – x
=4+x
3
(a)
Solve
(b)
Simplify fully
4x2 – 6x
4x2 – 9
7. A van can carry a maximum load of 400 kg. It carries boxes weighing 20 kg and 40 kg. It
carries at least 7 boxes weighing 40 kg. The number of boxes weighing 40 kg is not more
than twice the number of 20 kg boxes.
Let x represent the number of 20 kg boxes and y the number of 40 kg boxes.
a) Write down three inequalities involving x and y .
b) Illustrate the three inequalities by a suitable diagram on graph paper. Let 2 cm represent 1
box on both axes.
c) From the diagram determine the least weight the van carries.
d) What combinations give the greatest weight?
8. The dimensions of a rectangle are such that its perimeter is greater than 20 metres and less
than 30 metres. One side must be greater than the other. The larger side must be less than
twice the size of the smaller side.
Let x represent the length of the smaller side and y the length of the larger one.
a) Write down four inequalities involving x and y.
b) On graph paper, illustrate these inequalities using a scale of 2 cm to represent 2 metres on
each axis, clearly showing the area containing the solution.
c) What whole number dimensions will satisfy these three inequalities?
Page 20 of 173
5𝑝−4𝑞
9. Given that 3𝑝 = 2𝑟−3𝑞 , express q in terms of p and r. Find the value of q when p =2 and
r = -5.
10. The surface area of a solid cone is given by the formula
A = 𝜋𝑟 2 + 𝜋𝑟𝑙.
(i) Factorise fully the expression A = 𝜋𝑟 2 + 𝜋𝑟𝑙.
(ii) Rearrange the formula to express l in terms of 𝜋, r and A.
11. The formula used in connection with the mirror is
1
𝑢
(i)
(ii)
1
1
+𝑣 = 𝑓
Given that v = 9, and f = 5, find u.
Express v in terms of u and f.
FUNCTIONS
-
When two members of the two sets are connected, it is called a relationship
-
A relation is a collection of ordered pairs. A function is a special type of relation
-
Functions and relations can be represented by:
(i) a mapping (ii) a table
(iii) an ordered pair (iv) an algebraic sentence (v) a graph
QUESTIONS
1. A relation from set A =
2, 4, 6, 8
to set B =
1, 3, 5, 7, 9
is given as “is
one more than”
(a)Draw an arrow diagram to show the relation
Answers
A
(i)
is one than
2
1
4
3
6
5
8
7
9
7
9
Page 21 of 173
(b) the type of relationship
one – to – one relationship
2. Study the mapping below
(a)
3.
.1
4.
.2
5.
.7
Complete the following
(i)
9 and ….
3 is mapped into …..
Answer
(ii)
4 is mapped into………
Answer
(iii)
1 and 2
2
5 is mapped into…….
Answer 7
(a)List the set of:
(i)
The domain
Answer - Domain =
(ii)
The range
Answer - Range =
3. Set D =
3, 4, 5
1, 2, 7
(2,4), (2, 6), (2,8), (2,10), (3, 6), (3, 9), (4,4), (4,8), (5, 10)
(a) Illustrate this information on an arrow diagram.
2
4
3
6
4
8
5
9
10
4. Complete the following table
10
10
Page 22 of 173
i.
ii.
iii.
iv.
Input
1
0
2
f: x
f:x
f:x
3x + 1
3 (1) + 1
3 (0) + 1
Output
4
Ordered pair
1,4)
Answer
i.
ii.
iii.
iv.
input
1
0
-2
-3
f: x
f: x
f: x
f: x
f: x
3x + 1
3(1) + 1
3(0) + 1
3 (-2) + 1
3(-3) + 1
Output
4
1
-5
-8
Ordered pair
(1,4)
(0,1)
(-2, -5)
(-3, -8)
5. A function f is such that f )t) = 2t – 2. Find:
(a) f (0)
(a)
f (t) = 2t – 2
f (0) = 2 (0) -2
−1
(b) f (2)
(c) (
(b) f (t) = 2t – 2
(c) f (t) = 2 (t) -2
f (2) = 2 (2) -2
2
)
=2(
−1
2
) -2
= 0–2
= 4–2
= -1-2
= -2
= 2
= -3
6. Given that the ordered pairs (m, 25) and (n, -10) belong to the mapping h:x
x+4
Find the values of M and n.
Answer
(M, 25)
h:x
x+4
(n, -10)
h (x) = x + 4
h:x
x+4
h(m) = M = 4
h(x)
x+4
25 = M + 4
h(n) = n + 4
25 – 4 = M
-10 = n + 4
21 = M
-10 – 4 = n
-14 = n
7. In the first year, Grace made K800 selling cellphones. She increased her earnings
by K50 each year for the next four years
(a) Draw up a table
(b) Draw up a linear graph
Page 23 of 173
Year
1
2
3
4
Answer
Earnings
K800
K850
K900
K900
Answer
1000
950
900
850
800
0
(c) Write the co-ordinate pairs
Answer
(x, y)
(1, K800)
(2, K850)
(3, K900)
(4, K950)
(a) Give the domain and range
Domain = {1, 2, 3, 4}
Range = {K800, K850, K900, K950}
1
2
3
4
(d) Write an algebraic sentence
Answer
Algebraic Sentence
Y = x + K50
PRACTICE QUESTIONS
1. If f : x
2
3𝑥
+ 5, find
a) f(2)
b) x when f(x) =7
c) f-1(x)
1
2. A function h is defined as h(x) = 2x – 5, find
a) h(-4)
b) the value of x for which h(x)= 3
c) h-1(x)
3. Given that f(x) =
3𝑥−5
2
and g(x) =
𝑥−4
6
, find
Page 24 of 173
a) f(-9)
b) f-1(x)
c) the value of x for f(x) = 3g(x)
4. If h(x) = 3x – 5, find
a) h(3)
b) h(x) = 10
c) h-1(x)
EXPECTED ANSWERS
2
1
1. a) 56 or 52
1
b) X = 3
2
c) f-1(x) = 3𝑥−15
2. a) h (-4) = -7
b) X = 16
c) h-1(x) = 2x + 10
3. a) f(-9) = -16
b) f-1(x) =
2𝑥+5
3
1
c) X = 2
4. a) h(3) = 4
b) X = 5
c) h-1(x) =
𝑥+5
3
MATRICES
1. Given that A
, find
(a)
(b)
Solutions
(a)
Page 25 of 173
(b)
2. Given that
(i)
(ii)
(iii)
(i)
, and
, find
The inverse of matrix A.
3A – B
AB
Solutions
Determinant of A = ( 5 x 0) – (2 x 1 )
= 0–2
=-2
=
(ii)
3A –B
=
=
=
(iii)
AB =
=
2
3.If P= (1
8
4 1
1 0
)and
Q
=(
3 1
2 1
2 6
1 0
4
1) ,evaluate PQ.
1
Page 26 of 173
PQ =
11 4
=( 8 3
18 2
13
8)
40
4.Solve the following simultaneous equations using the matrix method.
Solution
)
Find the determinant of the matrix of coefficient of
Det = (3 x 3) – (2x2)
=9–4
=5
Find inverse of matrix of coefficient of
Inverse =
)
( )
5. If matrix A
, is a singular matrix , find the value of
Solution
A singular matrix have the determinant equal to zero
(
=0
- 24 = 0
Page 27 of 173
SIMILARITY AND CONGRUENCE
1. SIMILARITY
Two objects are said to be similar if:
i.
The corresponding angles are equal
ii.
The ratio of the corresponding sides is the same or equal.
1.1 SIMILARITY IN TRIANGLES
For two triangles to be similar, they need to satisfy any of the three cases:
i.
Three pairs of corresponding angles are equal (AAA)
ii.
The ratio of corresponding sides is the same (SSS)
iii.
Two pairs of corresponding sides are proportional and the included angles are
equal (SAS).
2. CONGRUENCY
Two objects are congruent if they have the same shape and size.
QUESTIONS
1) State the two triangles in the diagram below which are similar. Give the reason why.
Answer: ΔABC and ΔADE are similar.
Since DE and BC are parallel, ˂𝐴𝐵𝐶 = ˂𝐴𝐷𝐸 𝑎𝑛𝑑 ˂𝐴𝐶𝐵 = ˂𝐴𝐸𝐷
( Corresponding Angles). ˂A is common to both triangles, hence satisfying AAA.
2) Determine whether or not the two rectangles below are similar
Solution
2:3 ≠ 3:5
Therefore XYWZ and RSTQ are not similar.
Page 28 of 173
a) In the diagram, DE is parallel to AB, DE = 3cm, AB = 9cm and CD = 4cm
b) Find the ratio of corresponding sides.
c) Find the value of x
Solution
a) 3: 9 = 1: 3
b)
3
4
=
9 4+𝑥
3(4+x) = 9×4
12 + 3x = 36
3x = 24
X = 8cm
3) Name all the pairs of congruent triangles in the figure below:
Solution
ΔEOF = ΔGOF
ΔEOH = ΔGOH
ΔFEH = ΔFGH
4) Show that ΔABC and ΔPQR below are congruent
Page 29 of 173
Answer:
˂ABC = ˂PQR,
˂BAC = ˂RPQ = 30˚ and
˂BCA = ˂PRQ = 40˚
5) Find the length of YU in the diagram below
Solution:
˂UVY = ˂XVZ (Vertically opposite angles)
UV = XV
YV = VZ
Therefore UY = XZ = 7cm.
6) A wall, which is 4m high, is built next to a street light that is 8m high. The shadow of the
wall is 5m long. How far is the wall from the street light?
Solution
4
8
5
= 5+𝑥
4(5+x) = 8×5
20+ 4x = 40
X = 5m
NOTE: If two figures are similar and the lengths of their corresponding sides are in the
ratio
𝒑∶ 𝒒,
then the ratio of their area is 𝒑𝟐 ∶ 𝒒𝟐 ,
and the ratio of their volumes is 𝒑𝟑 ∶ 𝒒𝟑 .
PRACTICE QUESTIONS:
Page 30 of 173
(1) A model of a tanker is made using a scale of 1: 20.
(a) The length of the tanker is 15 m. Calculate the length , in centimetres, of the
model.
(b) The model holds 12 litres of liquid. Calculate the number of litres the tanker will
hold.
(2) (a)
Are triangles ABC and DEF similar? Explain your answer clearly.
(b)
Triangle LMN and PQR are similar. Calculate the value of x.
(3) In the diagram, ABCD is a quadrilateral with BA parallel to CD.
AC and BD meet at X where CX = 8 cm and XA = 10 cm.
(a) Given that BD =27 cm , find the length BX.
(b) Find the ratio area of triangle BXC : area of triangle AXD.
(4) Two pots are geometrically similar. The height of the smaller pot is 5 cm.
The height of the bigger pot is 15 cm.
Page 31 of 173
(a) The diameter of the base of the larger pot is 7 cm. Find the diameter of the base of
the smaller pot.
(b) Find the ratio of the volume of the smaller pot to that of the larger. Give your
answer in the form 1 : n.
(5) The ratio of the areas of the bases of two geometrically similar buckets is 4 : 9.
(a) The area of the top of the smaller bucket is 480 cm2.
What is the area of the top of the larger bucket?
(b) Write down the ratio of the heights of the two buckets.
(c) Both buckets are filled with sand. The mass of sand in the larger bucket is 36 kg.
Find the mass of sand in the smaller bucket.
ANSWERS:
(1) (a) 75cm ,
(b) 96000 litres
(2) (a) not similar, because not all the 3 corresponding sides are proportional.
(b) 7
(3) (a) 15 cm ,
(b) 1 : 1
1
(4) (a) 23 cm,
(b) 1 : 27
(5) (a) 1080 cm2 ,
(b) 2 : 3 ,
2
(c) 103 kg .
KINEMATICS,
Problems involving distance, time, speed(velocity) and acceleration are given the
name of kinematics (kinema= motion)
UNITS USED:
 Distance travelled- (metres) or (m)
Page 32 of 173



Time taken- (seconds) or (s)
Velocity- (metres/second) or (m/s)
Acceleration- (metres/second/second) or (m/s2 )

Velocity(speed)=

Acceleration =
NOTE:
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒
m/s
m/s2
GRADIENT GRAPHS
If motion of an object is given in a graph by a straight line then the object travels
at
constant speed or uniform speed determined by the gradient of the line.
If the graph is a curve, then the object concerned has different speeds at each instant.
The gradient of the tangent to the curve at that point gives the speed of the
object.
TRAVEL GRAPHS
(i)
DISTANCE TIME GRAPH
constant rate of change = constant speed.
Page 33 of 173
varying rate of change = varying speed
EXAMPLE 1:
The distance time graph a shows an object starting from a point O, travelling 15m in
2 s, is stationary for another 2s and finally travels back to O in 1s.
Graph can be interpreted as follows:
The total distance travelled is 30 m (going+returning)
From O to A: constant speed =
15𝑚
2𝑠
= 7.5 m/s
From A to B: The car is stationary, speed = 0 m/s
From B to O: The speed is constant =
15𝑚
1𝑠
= 15 m/s
(ii) SPEED (VELOCITY) TIME GRAPH
hints:
 If the graph is a straight line parallel to the time axis (x-axis), then the object
is said to travel at constant speed, i.e. there is no acceleration.
 The total distance travelled in t sec is given by the area under speed time
graph for that time.
Page 34 of 173
NOTE:
 From O to A, Speed is constantly changing, hence there is constant(uniform)
acceleration.
 From A to B, Speed is constant (not changing), hence there is no acceleration.
 From B to C, Speed is decreasing uniformly, hence there is deceleration.
EXAMPLE 2
The diagram shows the velocity - time graph of a particle during a period
of t seconds.
Calculate
a) the acceleration of the particle in the first 10
seconds,
b) the value of t, if it travelled 50m from the 20𝑡ℎ
second,
c) the average speed of the particle for the whole
journey.
EXPECTED ANSWER:
Page 35 of 173
(a) 𝑎 =
𝑣−𝑢
𝑡
, 𝐵𝑢𝑡 𝑢 = 30𝑚/𝑠 , 𝑣 = 10𝑚/𝑠 , 𝑡𝑖𝑚𝑒 𝑡 = 10𝑠
𝑎=
10 − 30
2
𝑎 = −𝟐𝒎/𝒔𝟐
(b) Distance traveled is equal to the area under the graph.
𝐴𝑟𝑒𝑎 =
50 =
1
×𝑏 ×ℎ
2
1
× (𝑡 − 20) × 10
2
50 = 5𝑡 − 100
5𝑡 = 50 + 100
5𝑡 150
=
5
5
∴ 𝑡 = 𝟑𝟎𝒔
(c) 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 =
𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑
𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ
∴ 𝐷 = 𝑎𝑟𝑒𝑎 𝐴 + 𝑎𝑟𝑒𝑎 𝐵 + 𝑎𝑟𝑒𝑎 𝐶.
𝐷=
1
1
× 10 × 20 + 10 × 20 + × 10 × 10
2
2
𝐷 = 100 + 200 + 50
𝐷 = 350𝑚 𝑎𝑛𝑑 𝑡𝑖𝑚𝑒 𝑡 = 30𝑠
∴ 𝐴𝑣. 𝑆𝑝𝑒𝑒𝑑 =
350𝑚
30𝑠
=
35
3
𝟐
𝑚/𝑠 = 𝟏𝟏 𝟑 𝒎/𝒔.
EXAMPLE 3
The diagram below is the speed-time graph of a bus which leaves a bus stop and accelerates
uniformly for 10 seconds over a distance of 100m.
It then maintains the speed it has attained for 30 seconds and finally retards uniformly to rest
at the next bus stop. The whole jouney takes t seconds.
Page 36 of 173
If the two bus stops are 1 kilometre apart, find
(i)
the value of V,
(ii)
the acceleration in the first 10 seconds,
(iii)
the total time (t) taken for the whole journey.
EXPECTED ANSWER
(i)
distance= 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ
1
× 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
2
1
100𝑚 = × 10𝑠 × 𝑉 𝑚/𝑠
2
=
∴𝑉=
100
5
m/s
∴ 𝑉 = 𝟐𝟎m/s
(ii) 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑠𝑝𝑒𝑒𝑑
𝑡𝑖𝑚𝑒
m/s2
∴ 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
20𝑚/𝑠
10𝑠
∴ 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝟐 m/s2
(iii) 𝑛𝑜𝑡𝑒: 1𝑘𝑚 = 1000𝑚,
∴ 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
1
(𝑎 + 𝑏)ℎ
2
𝑏𝑢𝑡 𝑑𝑖𝑠𝑡 = 1000𝑚; 𝑎 = 40 − 10 = 30𝑠; 𝑏 = 𝑡; ℎ =
20𝑚
,
𝑠
1
(40 + 𝑡)20
2
∴ 100 = (40 + 𝑡)
∴ 1000 =
∴ 𝑡 =100−40
∴ 𝑡 = 𝟔𝟎𝒔𝒆𝒄𝒐𝒏𝒅𝒔
∴ 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 = 𝟔𝟎𝒔𝒆𝒄𝒐𝒏𝒅𝒔
LEARNER ACTIVITY:
Page 37 of 173
The diagram below is the speed-time graph of a particle. The particle accelerates uniformly
from a speed of vm/s to a speed of 5v m/s in 20 seconds.
(a) find the expression in terms of v, for acceleration.
(b) the distance travelled by the object from 0 seconds to 20 seconds is 80m. Find the
value of v.
(c) Find the speed at t = 15seconds.
EXPECTED ANSWERS
(a) acceleration=
𝑣−𝑢
𝑡
but u=v and v= 5v
∴acceleration=
𝟓𝐯−𝐯
𝟐𝟎
𝟒𝐯
m/s2= 𝟐𝟎 𝒎/𝒔2
𝟏
∴acceleration= 𝟓 𝐯 𝒎/𝒔2
(𝑏)𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
1
(𝑎 + 𝑏)ℎ
2
but𝑎=v, b=5v, h=20 and dist=80m
∴ 80 =
1
(v + 5v)20
2
∴ 6v=8
8
∴v= 6 𝑚/𝑠
𝟒
∴v= 𝟑 𝒎/𝒔
(𝑐 )
𝟏
8
4
∴acceleration= 𝟓 × 6 𝑚/𝑠2 = 15 𝑚/𝑠2 ,
𝒗−𝒖
𝒕
4
𝑙𝑒𝑡 𝑠𝑝𝑒𝑒𝑑 𝑎𝑡 𝑡 = 15 s 𝑏𝑒 = 𝑦= final velocity, 𝑎𝑛𝑑 𝑣 = 3 𝒎/𝒔 beinitialvelocity.
∴ 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑠𝑝𝑒𝑒𝑑 𝑎𝑡 𝑡 = 15𝑠, 𝑤𝑒 𝑢𝑠𝑒 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 =
∴
4
15
2
𝒚−
4
3
𝑚/𝑠 = 𝟏𝟓
Page 38 of 173
∴ 15 × 4 = 15 × 𝑦 − 15 ×
4
3
4
3
4
∴𝑦 =4+
3
𝟏𝟔
∴ 𝑦 = 𝟑 m/s = speed at t =15s
∴4=𝑦−
LEARNER ACTIVITY:
The diagram represents the speed-time graph of a moving object.
(i)
(ii)
(iii)
Calculate the speed of the object when t= 4s.
Calculate the distance travelled in the first 15 seconds.
Given that the rate at which the object slows down after t=15 is equal to half the
rate
at which the object accelerates during the first 6 seconds, calculate the time at
which
it stops.
EXPECTED ANSWERS:
(i)
(ii)
(iii)
speed at t= 4s is equal to 32m/s,
distance travelled= 528m,
time at which it stops= t=35s.
PRACTICE QUESTION:
(Q ) The diagram is the speed- time graph of a car which is uniformly retarded from u m/s to
20 m/s in 10 seconds.
The car is then uniformly retarded at a different rate until it finally comes to rest after a
further 40 seconds.
Page 39 of 173
Calculate
(a) the speed of the car after 20 seconds,
(b) the retardation during thefinal 40 seconds of its motion,
(c ) the value of u , if the distance travelled in the first 10 seconds is 275 metres.
ANSWERS
(a) 15 m/s,
(b)
1⁄ m/s2 ,
2
(c)
35 m/s .
SOCIAL AND COMMERCIAL ARITHMETIC
1. Mr. Baldwin bought 500 shares of a company at k3100 per share. The nominal value of a
share was K1000.
(a) What did he pay for the 500 shares?
(b) What is the total nominal value of the shares?
SOLUTIONS
Cost of shares = number of shares x cost per share
= 500 x K3100
= K1550000
(b) Total nominal value=number of shares x nominal value of a share
= 500x K1000
= K500000
Q2. The director of Mukuba pensions decide to pay a total dividend of K978000 on 1250000
shares
Page 40 of 173
(a)Calculate the dividend per share.
(b)Mukuyu Trust holds 15000 shares in the company how much is paid out in dividends to
the company?
SOLUTIONS
(a)
dividend =
=
total dividend amount
Number of shares issued
K978, 000
1250,000 = K0.78
(b) Total dividend = dividend x number of shares
= K0.78 x 15000
= K11, 700
Q3. The value of shares that Mr. ZIBA bought increased .When the market value per share
was
K30.52, he sold his 7500 share.
(a)If the dividend per share was k1.05 when he sold his shares, what was the total dividend
he received when he sold all his shares?
7500 shares x K1.05
=K 7875
(b)What would Mr Ziba have received for his shares if he had sold them at K23.23 per
share?
23.23 X 7500
=K 174225
(c)How much profit did he make when he sold his shares at k28.23?
28.23 X 7500
=K211725 therefore profit will be equal to
= K211725 – K7875
Page 41 of 173
= K203850
BEARING AND SCALE DRAWING
1). BEARING: Bearing refers to the direction of a movement.
We use three methods to show direction.
i. COMPAS BEARING (Nautical bearing)
 There are four cardinal points on the compass.
N
W
E
S


There are other points half way between the cardinal points e.g. North West
(NW), South East (SE) etc.
Nautical bearings are measured as acute angles from the North or South to the
East or West.
N
N
A
Bearing of F from O is 064°
45°
O
ii. THREE FIGURE BEARING
 These are given in three figures
 They are always measured in the clockwise direction starting from the North.
N
Bearing of F from O is 064°
F
064°
O
2). SCALE DRAWING
When making a scale drawing;
i. Make a rough drawing (sketch diagram) to give a general idea of what the
final diagram should look like.
ii. Choose a suitable scale to give a large diagram.
 The larger the diagram the more accurately you can read the
lengths and angles.
Page 42 of 173

Draw parallel lines from North to South (North lines) if you are
making a scale diagram to show bearings at the points from where
you are reading the bearings.
7.
N
A
B
30°
C
In the diagram above, B is due East of A and due North of C. Angle BAC is 30°, find the
bearing of A from C.
Solution
Find find <ACB
<ABC+<BAC+<ACB=180°
(angles in a triangle)
90°+30°+<ACB=180°
Therefore, the bearing of A from C =360°-60°
<ACB=180°-120°
=300°
<ACB=60°
N
N
A
B
30°
60°
C
8).
N
N
A
110°
B
C
D
In the diagram above, 𝐶 is due south of 𝐴, 𝐴𝐵 = 𝐵𝐶 and the bearing of 𝐵 fro 𝐴 is 110°. Find
the bearing of 𝐶 from 𝐵.
Solution
< 𝐵𝐴𝐶 =< 𝐴𝐶𝐵 = 70° (Base angles on an isosceles triangle)
< 𝐴𝐶𝐵 =< 𝐶𝐵𝐷 = 70°
∴the bearing of 𝐶 from 𝐵
= 180° + 70°
= 250°
Page 43 of 173
9). Three towns 𝑃, 𝑄 and 𝑅 are such that Q is 45𝑘𝑚 from P on a bearing of 145°. Make an
accurate scale drawing to show the positions of P,Q and R. From the scale drawing, find the
distance and bearing of R from P.
Solution
Sketch
N
N
Q 145°
45km
60km
030°
P
R
Accurate drawing:
Q
4.5cm
030°
P
6cm
R
From the drawing, 𝑃𝑅 = 7.6𝑐𝑚
7.6 × 10 = 76𝑘𝑚
The bearing of R from P is
1. B. Show this information in a diagram in a diagram. SW is 225° from N, therefore: A
is on a bearing of 225°.
SOLUTION
At B, draw a line showing N and measure an angle of 225° clockwise from N. A is
south west of point
N
B
A
A
Page 44 of 173
2. Figure 2 shows the bearing of P from Q.
Find the bearing of Q from P.
N
P
Q 300°
The bearing of Q from P is the angle between North and the direction
from Q to P as parallel to NX shown in figure 3,
It can be found in the following way:
At P, draw line NY.
Solution
N
N
< 𝑌𝑃𝑄 = 𝑁𝑄𝑃 – 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠
< 𝑁𝑄𝑃 = 360⁰ − 300⁰ = 60⁰
𝑃
Hence < YPQ = 60°
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, < 𝑁𝑃𝑄 = 180° – 60° = 120°
𝑇ℎ𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑜𝑓 𝑄 𝑓𝑟𝑜𝑚 𝑃 𝑖𝑠 120⁰.
Q
300°
Y
X
3. Points A, B, C and D lie on level ground.
The pint D is due North of A.
DÂC = 140°, CÂB = 90° and AB̂C = 75°. Find the bearing of:
North
a) A from C
b) B from A
D
c) C from B
140°
A
75°
B
C
Solution
D
a)
Y = 180° - 140° (interior opposite <s)
= 40°
D
140°
A
∴ Bearing of A from C is 040°
x
b) X =360° - (140° +90°)=130°
75°
C
C
B
∴ Bearing B from A is 130°
c) Bearing of C from B
<x+<z=180° (Interior opposite angles)
130°+<z=180°
<z=50°
∴ Bearing of C from B
Page 45 of 173
=360° - (75°+ 50°)
=235°
B
4.
A
2.2m
1.9m
42°
C
D
The diagram shows a frame work 𝐴𝐵𝐶𝐷, 𝐴𝐷 = 2.2𝑚
𝐵𝐷 = 1.9𝑚 𝑎𝑛𝑑 𝐵𝐶𝐷 = 45⁰. 𝐴𝐷𝐶 = 90⁰
Calculate i) 𝐴𝐷𝐵
ii) 𝐵𝐶
b) A vertical flagpole, 18m high, stands on horizontal ground,
Calculate the angle of elevation of the top of the flagpole
from a point, on the ground, 25m from its base.
Solutions
a) i) 𝐼𝑛 ∆ 𝐴𝐷𝐵,
1.9
𝐶𝑜𝑠 < 𝐴𝐷𝐵 = 2.2
< 𝐴𝐷𝐵 = 𝐶𝑜𝑠 −1
1.9
2.2
= 30.27°
= 30.3° (1 𝑑. 𝑝)
𝑖𝑖) 𝐼𝑛 ∆ 𝐵𝐷𝐶,
1.9
𝑆𝑖𝑛 42° = 𝐵𝐶
𝐵𝐶 = 2.839
= 2.84𝑚 (3𝑠. 𝑓).
b)
1.8m
Q
25m
𝑇𝑎𝑛 𝑄 =
18
25
𝑄 = 35.8⁰ (1 𝑑. 𝑝)
∴ < 𝑜𝑓 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 35.8⁰.
5. The diagram shows the position of a harbour, H, and three islands A, B
and C.
C is due North of H.
Page 46 of 173
C
A
128°
31km
54km
B
62°
H
The bearing of A from H is 062° and HÂB = 128° HA = 54km and AB = 31 km.
a) Calculate the distance HB
b) Find the bearing of B from A
c) The bearing of A from C is 133°. Calculate the distance AC.
Solution
a) Using Cosine Rule
𝐻𝐵 = 54 + 31 – 2(54)(31) 𝐶𝑜𝑠 128
= 77.059
= 77.1𝑚 (3𝑠. 𝑓)
b).
N
𝑋° = 62° (𝑎𝑙𝑡 < 𝑠)
128° − 62° = 66°
180° − 66° = 114°
∴Bearing of B from A is 114°
A
𝑥
128°
31km
62°
B
c). < 𝐻𝐶𝐴 = 180⁰ − 133⁰
Using Sine Rule;
𝑆𝑖𝑛𝑒47⁰
54
=
𝐴𝐶 =
𝑆𝑖𝑛 62⁰
𝐴𝐶
54 𝑆𝑖𝑛 62⁰
𝑆𝑖𝑛 47⁰
= 65.19
= 65.2𝑚 (3𝑠. 𝑓)
1. The diagram below shows three points P, Q and R on the map. Given
that the bearing of Q from P is 035°, the bearing R from Q is 110° and <
QRP = 33°
N
Page 47 of 173
Q 110°
𝑥 𝑦
N
N
w
33°
R
35°
V
P
Find:
(a)
(b)
(c)
(d)
the bearing of P from Q
the bearing of Q from R
the bearing of P from R
the bearing of R from P
Solution:
a) P from Q
𝑋 = 35° (𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠)
𝑌 = 180° − 110
𝑌 = 70°
The bearing of P from Q = X + Y + 110°
= 35° + 70° + 110°
= 215°
b) Q from R
110° + W + 180 (interior angles)
W = 180° - 110°
W = 70°
The bearing of Q from R = 360° - 70°
= 290°
c) P from R
360° − (70° + 33°) 𝑠𝑖𝑛𝑐𝑒 𝑊 = 70
= 257°
∴ The bearing of P from R is 257°
d) R from P
𝑉 = 180° − (𝑥 + 𝑦 + 33) (interior angles)
𝑉 = 180° − (35° + 70° + 33°)
𝑉 = 180° − 138° = 42°
∴ The bearing of R from P is 42° + 35° = 077°
PRACTICE QUESTIONS:
(1) The diagram below shows an equilateral triangle ABC. A is due North of B and
CN is
parallel to BA.
Page 48 of 173
Find (a)
(b)
BCN ,
the bearing of C from A.
Answer: (a)………………….
(b)………………….
(2) The diagram below shows triangle PQR in which R is due east of Q, angle PQR
=135° and angle QPR=33°.
Calculate (a) angle PRQ,
(b) the bearing of P from R.
Answer: (a)………………….
(b)…………………
(3) The bearing of a point B from A is 129°. What is the bearing of A from B?
Page 49 of 173
(4) The bearing of B from A is 072°.
(a) Find the bearing of A from B.
(b) C is due South of B and BA = BC. Find the bearing of A from C.
(5) A, B and C are three towns. C is equidistant from A and B.
The bearing of C from A is 132° and angle BAC = 75°.
Find
(a) (i) the acute angle ACB,
(ii) the reflex angle ACB,
(b) the bearing of A from C,
(c) the bearing of A from B.
ANSWERS
(1) (a) angle BCN = 120° ,
(b) bearing of C from A = 120°
(2) (a) 12°,
(b) bearing of P from R =256°
(3)
bearing of A from B = 309°
(4) (a) bearing of A from B = 252°
(b) bearing of A from C =306°
(5) (a) (i) angle ACB = 30°
Page 50 of 173
(ii) the reflex angle ACB = 330°
(c) bearing of A from C = 312°
(d) bearing of A from B = 27°
Definition of Bicimal Numbers
The word bicimal comes from a combination of the words binary and decimal.
This entails that we are dealing with numbers in base two (2) called binary numbers
and decimal numbers.
A bicimal is the base two analog of a decimal, it has a bicimal point and bicimal
places.
-
Examples of bicimals
- Note : The places in a bicimal can either be terminating or repeating.
1001.011
0.1101
10101
1111.001
 Reminder on converting from base 10 ( denary) to base 2 (binary)
2
17
when converting 17ten to base two
2
8r 1
2
4r 0
2
2r 0
2
1r 0
2
0r 1
17ten in base two is 10001two
Converting from binary and denary
Reminder on Converting from binary (base two) and denary ( base ten)
Convert 10101two (binary) to base ten (Denary)
4
3
2
1
0
2
2
2
2
2
1
0
1
0
1
0
1x2 =1
1
0x2 =0
2
1x2 =4
3
0x2 =0
4
1 x 2 = 16
21ten
Page 51 of 173
Converting decimal numbers in base 10 to bicimal
Example
30.375ten to base two (2)
2
30
then 0.375 x 2= 0.75
2
15 r 0
0.75 x 2 = 1.5
2
7 r 1
0.5 x 2 = 1.0
2
3 r 1
2
1 r 1
2
0 r 1
therefore: 0. 011two
11110
Then 30.375ten is 11110.011two
Converting numbers in Bicimal to base 10
Example
Convert 0.1101two to base 10
-1
2
-2
2

-3 -4
2 2
11 1 1
2 4 8 16
1
1
0
1
•
(2 × 1) + (4 × 1) + (8 × 0) + (16 × 1)
1
1
1
1
1 1
1
+ +0+
2 4
16
8+4+0+1
16
𝟏𝟑
𝟏𝟔
or 0.8125ten
ACTIVITY
CONVERT THE FOLLOWING TO BICIMAL NUMBERS
A. 19.125
B. 22.375
C. 8.125
CONVERT THE FOLLOWING BICIMALS TO BASE TEN
A. 111.01
B. 0.11101
C. 10101.110
Page 52 of 173
GRADE 8 & 9 SYLLABI ON THE TOPIC
COMPUTERS
What is a computer?
A computer is an Electronic device that can:
-
Accept data, as input
Process the data
Store data and information
Produce information, as output.
BASIC ELEMENTS OF A COMPUTER
INPUT
PROCESS
OUTPUT
STORAGE
FLOW CHARTS
-
A computer carries out all its tasks in a logical way.
A set of logical steps that need to be followed in order to solve a problem are also
referred to as FLOW CHART.
How does one construct a flow chart?
-
To construct a flow chart one needs to firstly master the symbols used and their
meaning.
BASIC FLOWCHART SYMBOLS
SYMBOL
MEANING
BEGIN/END OR
START/ STOP
INPUT/OUTPUT
ORENTRY/DISPLAY
PROCESS
DECISION
PROGRAM FLOW
Page 53 of 173
BASIC FOUR OPERATORS
SYMBOL
MEANING
ADDITIONAL
+
‒
*
/
SUBTRACTION
MULTIPLICATION
DIVISION
EXAMPLE:
- Construct a flow chart program to calculate the perimeter of a square, given its
length.
- Since the formula is:
Perimeter = 4l
- Data needed for input is the length
BEGIN
ENTER length
IS length
≥0?
NO
ERROR: length
MUST BE ≥ 0
YES
PERIMETER = 4*length
Page 54 of 173
DISPLAY PERIMETER
Carry out the following activities:
Construct a flow chart program on how to find Mr. Mwansa’s age given his year of
1.
birth as y.
2.
Prepare two questions using the well-known formulae or mathematical concepts on
which a flow chart can be used to arrive at its solutions.
Topic Study –Mathematics Computer & Calculator
FUNCTIONS ON A CALCULATOR
-
A relevant component of the syllabus but with most challenges when it comes to
teaching arising from the fact that :
1.
there is no standard or specific type of a calculator to be used in our
schools.
2. the subtopic is perceived as obvious by both the learners and teachers.
3. the supposition that the subtopic is taken care of by the manual the
calculators come with.
-
Hence, this subtopic goes un attended to in most cases.
Thus far it is advised that we try to find resolutions to these
- Challenges and start teaching this subtopic.
BASIC COMPONENTS OF A COMPUTER
A computer is an Electronic device that can:
- Accept data, as input
- Process the data
- Store data and information
- Produce information, as output.
- Parts or devices that serve in each of the processes below:
INPUT
PROCESS
OUTPUT
STORAGE
-
Input parts/devices
Examples include Keyboard, mouse etc.
Process part/device
For example CPU or Systems Unit.
Output parts/ devices
Examples include Monitor, Printer etc.
Storage parts/devices
Page 55 of 173
For example Flash disk, Memory card etc.
ALGORITHMS.
-
A flow chart as understood from the previous presentation can also be considered as
an example of an Algorithm
Definition
-
An algorithm is generally a set of logical steps that need to be followed in order to
solve a problem.
For example the solving of a malaria problem using any anti-malaria drug.
METHODS OF IMPLEMENTING AN ALGORITHM.
- There are two basic methods of implementing an algorithm and these are:

-
(i)
FLOW CHARTS
(ii)
PSEUDO CODE
FLOW CHARTS
The underlying factors of any flow chart are the use of the correct symbols for each
step and the correct operation symbols.
For example; calculate how far point A is from Mufulira, given point A.
BEGIN
ENTER Point A
Is Point A
within
Mufulira
YES
ERROR: Point A
MUST BE OUTSIDE
MUFULIRA
NO
Measure the distance
between them.
DISPLAY
DISTANCE
Page 56 of 173

PSEUDO CODE
-
This is a derivative of a flow chart and its underlying factors are the correct extraction
of statements inside a flow chart symbol, listing them vertically and preservation of
the logical steps also known as dentation.
-
For example; An equivalent Pseudo code to the flow chart above is:
EQUIVALENT PSEUDO CODE
BEGIN
ENTER Point A,
IF Point A is within Mufulira THEN
DISPLAY Error message,
ENTER Point A,
ELSE
Measure Distance between them,
ENDIF
DISPLAY Distance,
END.
GROUP ACTIVITIES
Working in pairs
(i) Identify a problem whose solution is dependant on one parameter.
(ii) Demonstrate how you would get to this solution by using both methods of
implementing an algorithm.
(iii)
Each one of the members in a group to present each of the methods.
CONCLUSION
This topic on computer and calculator in mathematics does not replace the need for one to be
computer literate via the learning of computer studies but among others:
-
Provide for the knowledge on stages of problem solving (define a problem, analysis
method of solution) and knowledge on how to write a computer program.
Page 57 of 173
GRADE 11
APPROXIMATION
Specific outcome: work with relative and absolute error
Relative error
1) Find the absolute error, the upper limit and the lower limit for 8h
Solution
Least unit of measurement = 1h
Absolute error =
Upper limit =
Lower limit
2) Find the relative error of 5.5l
Solution
Absolute error = 0.05
Relative error
3) Find the limits between which the areas of the following shapes must lie.
a) A square of side 3cm
b) A right-angled triangle with hypotenuse 5m and the other sides 3m and 4m
long
Solution
a). least unit of measurement = 1cm
Absolute error = 0.5
Upper limit is
Lower limit is
Maximum area
Minimum area
Therefore, the limits are
b). least unit of measurement = 1m
Absolute error = 0.5
Upper limit for the height=4+0.5=4.5m
Lower limit for the height= 4-0.5=3.5m
Upper limit for the base= 3+0.5=3.5m
Page 58 of 173
Lower limit for the base= 3-0.5= 2.5m
Area of a triangle
Therefore, maximum area
Minimum area
The limits are
1. The length and breadth of a rectangle, given to the nearest centimeter, are 15cm and 10cm
respectively. Find :
a) The shortest possible length and shortest possible breadth of the rectangle.
b) The longest possible length and the longest possible breadth of the rectangle
c) The limits between which the area must lie.
Ans: 15cm, given to the nearest centimeter, lies between 14.5cm and 15.5cm. 10cm lies
between 9.5cm and 10.5cm
a) The shortest possible length = 14.5 and the shortest possible breadth = 9.5cm
b) The longest possible length =15.5cm and the longest possible breadth is 10.5cm
c) The smallest possible Area= 14.5 9.5
= 137.75cm2
The longest possible Area= 15.5
10.5
=162.75cm2
So, the area lies between 137.75cm2 and 162.75cm2 or (137.75cm2
162.75cm2)
2. A car is driven a distance of 30km, measured to the nearest km, in 20 minutes, measured to
the nearest min. between what limits will the average speed lie?
Ans: 30km to the nearest km means.
29.5km
distance
30.5km
20min to the nearest min means
19.5min
time 20.5min
Greatest average speed=
Page 59 of 173
Least possible speed =
=
= 1.44km/min = 86km/h
Hence, the average speed = ( 90 4) km/h
3. The true value of the length or a rectangle is 5.5m. If this is recorded as 5m, find
A] The absolute error
B] The relative error
C] The percentage error
SOLUTIONS.
True value is 5.5m recorded value is 5m.
A] Absolute error =[recorded value –true value]
=[5-5.5] m =[-0.5]= 0.5m.
B] Relative error=
=
=0.09
C]The percentage error = relative error
= 0.09
100%
100% = 9%
4.The length of a square is given as 10cm correct to the nearest centimeter, calculate
A)
B)
C)
D)
E)
F)
The lower bound length.
The upper bound length.
The smallest possible perimeter.
The largest possible perimeter.
The smallest possible area.
The largest possible area.
Solutions
The length of a square is 10cm.
Error =
= 0.5cm
a) The lower bound length = actual length +error
Page 60 of 173
=10cm
0.5cm
=9.5cm
b) The upper bound length = actual length + error
= 10cm + 0.5cm = 10.5cm
c) Perimeter of a square = 4 length
Therefore , smallest possible perimeter = 4 lower breadth length
= 4 9.5cm = 38cm
d) Largest possible perimeter = 4 upper bound length
= 4 10.5cm = 42cm
e) Area of a square = L2
Therefore the smallest possible area = (lower bound length)2
= (9.5cm)2 = 90.25cm2
f) Largest possible area = (upper bound length)2
= (10.5cm)2 = 110.25cm2
ARITHMETIC AND GEOMETRIC EXPRESSIONS
Sequence
Sequence is a set of numbers listed in a well defined order with a specific rule that can be
used to state the next numbers in that set.
1. Write the next three terms of each of the sequences below
(a) 1, 2,4,8,……….
Answers
16, 32,64
(b)-4,-1,2,5,8,11
Answers
14,17,20
Series
A series is the sum of all the terms of a sequence e.g
Page 61 of 173
(i) 1+2+4+8+16+,…….
Answers
14,17,20
1. . For the AP, 2+5+8+………… find
(i)
The 10th term
(ii)
Answers
a=2, first term
d=5-2
D=3
T10 = a+ (n-1)d
= 2+ (10-1)3
= 29
(iii)
The 51 st term
Answers
T51 = 2+ (51 - 1)3
- = 2+ 50 x 3
= 152
(iv)
(v)
The nth term
Answers
Tn = a + (n - 1) d
= 2+ (n - 1)3
= 2 + 3n - 3
= 3n - 1
3. find the number of terms in the AP 3 + (-1) + (-5) +….+ (-53).
Answers
a = 3, d = -4, Tn = -53
Tn = a + (n - 1)d
-53 = 3 + (n - 1) -4
Page 62 of 173
-53 = 3 - 4n + 4
4n = 53 + 7
1
1
 4 n  60 
4
4
n = 15
4. The 10th term of an AP is 37 and the 16th term is 61, for this AP find:
(i)
The common difference
Answers
Tn = a + (n - 1)d
T10 = a + 9d
37 = a + 9d……………eqn 1
and
T16 = a + (16 - 1)d
16 = a+15d…………..eqn 2 and solve the equations simultaneously.
a + 9d = 37
-(a+15d = 61)
6
 24

4
4
d  4

(ii) The first term
Answers
First term
a+ 9d = 37
a+ 9(4) = 37
a+ 36 = 37
a = 37 - 36
a=1
(iii) the 30th term
Answers
Tn = a + (n - 1)
T30 = 1 + (30 - 1)4
T30 = 117
Page 63 of 173
5. The nth term (Tn) of an AP is given by Tn = 1/2(4n - 3).
(a) State (i) the 5th term (ii) the 10th term
Answers
The 5th term
T5 = 1/2(4n - 3)
T5 = 1/2(4 x 5n - 3
T5 = 1/2(17)
T5 = 8.5
(iii) the 6th term
Answers
T10 = 1/2(4 x 10 - 30)
T10 = 1/2(40 - 3)
T10 = 18.5
T6 = 1/2(4 x 6 - 3)
T6 = 1/2(24 - 3)
T6 = 1/2(21)
T6 = 10.5
(b) the common difference
Answers
d = T6 - T5
d = 10.5 - 8.5
d=2
Therefore, the common difference is 2.
6. If x + 1, 2x - 1 and x + 5 are three consecutive terms, find the value of x.
Answers
X+1
T1
,
2x - 1
T2
,
x+5
T3
For an AP,
Common difference, d = T2 - T1 = T3 - T1
(2x - 1) - (x + 1) = (x + 5) - (2x - 1)
2x - 1 - x - 1 = x + 5 - 2x
2x - x - 1 - 1 = x - 2x + 5 + 1
X - 2 = -x + 6
X+x=2+6
2x = 8
X=4
7. (i) if the numbers 3,m,n and 8 are three consecutive terms of an AP, find the values of m
and n.
Answers
M-3=n-m
M+m=n+3
2m = n + 3
and
n - m = 18 - n
n + n = 18 + m
2n = 18 + m
Page 64 of 173
n  3
2
m=
…………..eq1
m = 2n - 18…………….eq2
Equate m = m
3
2
n+
= 2n - 18/1
n + 3 = 2(2n - 18)
n + 3 = 4n - 36
n-4n = -36 -3
-3n = -39
3n =
 39
3
 3
n = 13
for m
m=
n 3
2
13 3
2
m=
m=
16
2
m=8
Therefore, m = 8 and n = 13
(ii)
The numbers m - 1, 4m + 1 and 5m - 1 are three consecutive terms of an
AP, find the numbers.
(iii)
Answers
m - 1, 4m + 1, 5m - 1 and 4m + 1 is an arithmetic mean between m - 1 and
5m - 1
b=
a  c
2
4m + 1 =
m  1  5m  1
2
2(4m + 1)=m + 5m-1-1
8m+2=6m - 2
8m - 6m = -2 - 2
Page 65 of 173
2m = -4
M = -2
Substitute for m = 1 in the series we get, -3, -7 and -11
8.(i) Find the arithmetic mean of the first 6 terms of 3 + 8 +………
Answers
First term
a=3
Common difference d = 8-3
d=5
Therefore, the 6 terms are 3,8,13,18,23,28.
Arithmetic mean = 3 + 8 + 13 + 18 + 23 + 28
6
93
Arithmetic mean =
6
Arithmetic mean = 15.5
Or
Arithmetic mean = median
=
13  18
2
= 15.5
(ii) Find the arithmetic mean and the geometric mean of 4 and 64.
Answers
Given 4 and 64
Arithmetic mean =
=
4  64
2
68
2
= 34
Geometric mean = square root of 4 and 64
=2x8
Page 66 of 173
= 16
9. An arithmetic progression has a 1st term to be 2 and common difference of 2, show that
the sum of the first nth terms of the AP is given by Sn = n2 + n. hence find the sum of the 21st
terms of an AP.
Answers
A=
2, d = 2
n
Sn =
(2a  (n 1)d)
2
=
=
=
n
(2x2  (n 1)2)
2
n
(4  2n  2)
2
n
(2  2n)
2
= n + n2
Sn = n2 + n is required
The sum of the first 21st terms
S21 = 212 + 21
S21 = 441+ 21
= 462
10. The sum Sn of the first n terms of an AP is given by Sn = n2 + n, find (i) the first
term (ii) common difference (iii) the formula for the sum of the first n - 1 terms
Answers
(i)
Sn = n2 + 2n
To find the first term we put n = 1 in the given sum
S1 = 12 + 2(1)
Page 67 of 173
(ii)
(iii)
a=3
The common difference
d = S2 - 2S1
=8-6
=2
Sn = n2 + 2n
Sn-1 = (n - 1)2 + 2(n - 1)
= n2 - n - n + 1 + 2n - 2
= n2 - 2n + 2n + 1 - 2
= n2 - 1
Geometric Progression (GP)
A geometric progression (GP) is a sequence in which each term is formed by multiplying the
previous term by a constant amount.
nth term of a Geometric Progression
The nth term of a GP with first term a and common ratio r is: Tn = arn - 1
1. For a GP, 2 + 6 + 18 + …………, find (i) the tenth term (ii) the 17 th term
Solution
First term (a) = 2
Common ratio r =
T 2
6

T 1
2
r  3
n = 10
Tn  arn1
T10  2 3101
T10  39366 So the 10th term is 39 366
(ii) T17 = 2 X 317 - 1
= 2 x 316
= 86 093 442
2. The third term of a GP is 9 and the tenth term is 19 683, find;
(i) the common ratio
(ii) the 8th term
Solutions
Page 68 of 173
(i) T3 = ar2
ar2 = 9………….eqn (i)
T10 = ar9
Ar9 = 19 683……..eqn (ii)
Dividing equation (i) by equation (ii)
ar9 19683

ar2
9
7 7
r  7 2187
r 3
Therefore the common ratio is 3
(ii) ar2 = 9
A x (3)2 = 9
9a 9

9
9
a 1
The first term is 1
(iii) Tn = arn -1
T8 = 1 x 38 - 1
= 37
T8 = 2187 the 8th term is 2 187
3. Given that x +2, x + 3 and x + 6 are the first three terms of a GP, find
(a) the value of x
(ii) the 5th term of the GP.
Solutions
(i) Common ratio (r) =
T
T
2
1

T
T
3
2
x 3 x 6

x  2 x 3
(x + 3)(x + 3) =(x + 2)(x + 6)
X2 + 3x + 3x + 9 = X2 + 6x + 2x + 12
6x + 9 = 8x + 12
8x - 6x = 9 - 12
2 x
 3

2
2
x   1 13
Page 69 of 173
(ii) First term (a) = x + 2
3
2
2
3 4

2

a 
Common ratio (r) =
1
2
x 3
x2
3  3
 23 12
2 1
3 3
3 2
      
 2 1  2 1 
3 1
 
2 2
3
 2
2
r 3
T5  arn1
1
  34
2
81
 or40.5
2
The nth term (Tn) of a GP is given by Tn = 29 - n. Find
(i) the first term
(ii) the common ratio
(iii) the sum of the first 9 terms.
Solutions
(i)
T n = 29 - n
T1 = 2 9 - 1
T1 = 28
T1 = 256
Page 70 of 173
a = 256
(ii) To find the common ratio, first calculate the second term (T2)
T2 = 29-2
= 27
= 128
Common ratio (r)
(iii) Sum
T2
T1
128

256
1

2

a(1 r n )
1 r
256[1 ( 12 )9 ]

1 12
256(0.998046875)

1

2
255.5
0.5
Sum 511

Sum of a GP
6. Calculate, correct to three significant figures, the sum of the first 8 terms of the
GP 12, 8, 5 13 ..........
Solutions
8
12
3
r  or0.75
4
r
First term a = 12
Page 71 of 173
a(1 rn )
1 r
12 (1 ( 34 )8

1 34
12(0.899837085)

1 34
10.79864502

0.25
 43.19458008
S8 


= 43.2 correct to 3 significant figures
1 1 1
, , ,......... ..
8 4 2
7. Work out the sum of the first 10 terms of
Solution
Common ratio (r ) =
1 1

4 8
1
r  8
4
1
8
n
a(r 1)
S10 
r 1
10
1 (2 1)
8
2 1
1 (10241)
8
1
1
 (1023)
8
S10  127.875
r  2anda
Geometric Mean
8. Find the geometric Mean of 4 and 64.
Solution
Page 72 of 173
 4  64
 4  64
 2  8
 16
9. The sum of infinity of a certain GP is 28. if the first term is 37, find r
Sum to infinity
a
 28
1 r
a  37
a
 28
1 r
37
 28
1 r
28(1 r)  37
28  28r  37
 28r  37  28
28r
9

or  032
28  28
S 
10. Write down the number of terms in the following GPs
(i) 2 + 4 + 8 + …………..+512
(ii) 81 + 27 + 9 + ……….. +
1
27
Solution
First term a = 2, r =2
Last term = 512
L  arn1
512 2  2n1

2
2
n1
256  2
Page 73 of 173
28 = 2n - 1
8=n-1
n=8+1
n=9
Factorising 256
2 128
2
64
The GP has 9 terms
2 32
2 16
2
8
2
4
2
2
2
1
(ii) 1 + 27 + 9 + ……….. +
27
1
1
, last 
3
27
L
log a
n  1
logr
a  81, r 
1
1
log 27
81
 1
log 13
1
1
log(27
 81
)
 1
1
log 3
log(13 )3  ( 13 )4
n  1
log(13 )
log(13 )7
n  1
log(13 )
7log(13 )
n  1
log(13 )
n  1 7
n 8
The GP has 8 terms
Page 74 of 173
ARITHMETIC PROGRESSION
PRACTICE QUESTIONS
1. The fourteenth term of an AP is 5.2 and the twenty fifth term of the AP is 105.find the sum to the
first 79 term of AP?
T14=a+(14-1)d=58.2
therefore the value of a+24d=105
T25=a+(25-1)d=105
a+24(4.25)=105
a +13d=58.2
a=105-102
a+24d=105
a=3
subtract= -11d=-46.8
-11
therefore the sum is S79=79/2[2(3)+(79-1)4.25]
-11
S79=13,331.25
D=4.25
2. The twenty first of an AP is 184.9 and the twelfth is 104.8 calculate the sum of 81.
T21=a+(21-1)d=184.9
therefore the value of a is a+11d=104.8
T12=a +(12-1)d=104.8
a+11(8.9)=104.8
=a+20d =184.9
Subtract
a+11d=104
a=6.9
therefore the sum is S81=81/2[2(6.9)+(89-1)8.9]
=9d/9=80.1/9
=35,466.5
D=8.9
3.the nineth term of AP is 70.7 and the sixteenth term of the AP is 59.3 .find the T39
T19=a+18d=70.7
therefore a+18d=70.7
T16=a+15d=59.3
3d/3=11.4/3
D=3.8
T39=2.3+(39-1)3.8
a+18(3.8)=70.7
T39=2.3+144.4
a=70.7-68.4
T39=146.7
a=2.3
4. the twenty third term of an AP is 159.5 and the eighteenth term of the AP is 123.5.find the sum of
18.
T23= a +(23-1)d=159.5 a+22d=159.5
T18=a +(18-1)d=123.5 a+22(7.2)=159.5
S18=18/2[2(1.1)+(18-1)7.2]
=1,121.4
Page 75 of 173
a +22d=159.5
a+17d=123.5
a=159-158.4
a=1.1
5d/5=36/5
D=7.2
5.the twenty first term of an AP is 42.6 and the nineteeth term of the AP is 39.find T 46
T21=a+(21-1)d=42.6
a+20d=42.6
T19= a+(19-1)=39
a+20(1.8)=42.6
2d/2=3.6/2
D=1.8
T46=a+(n-1)
= 6.6+(46.1)1.8
a=42.6-81
=87.6
a=6.6
GEOMETRICAL PROGRESSION
1. The third term of a G.P is 4096 and the fifth term is 1024. Calculate
T3=ar3-1
T5= ar5-1
S12= 16384×4095/4096/1/2
T3=ar2= 4096
T5= ar4= 1024
S12= 32760
Tr/T3= 1024/4096
ar3= 4096
Ar2= ¼
a(1/2)2= 4096
√r2=√1/4
a= 4096/(1/2)2
r= ½
a=16384
2. The seventh term of a G.P is 729 and the fourth term is 19683. Find the fifth term.
T7= ar7-1
T4=ar4-1
ar6= 729
=ar3= 19683
T7/T4=r6/r3
r=
=1/27
r3=729/19683
Page 76 of 173
CO-ORDINATE GEOMETRY
1. Find the length of AB if A (2,3) and B(4,5)
Y
B (4,5)
5
A(2,3
)
3
0
2
4
X
A(2,3) and B(4,5)
Using Pythagoras theorem
AQ =2 units
QB = 2 units
Therefore, AB2 = (QA)2 = (BQ)2
AB2 = (4-2)2 + (5-3)2
AB2 = (2)2 + (2)2
AB2 = 4+4
AB2 = 8
AB = √8
From above, we can say that the distance along the x-axis = (x2 – x1) and
distance along the y-axis = (y2 – y1)
AB2 = (X2 – X1)2 + (Y2 –Y1)2
AB2 = (4-1)2 + (5-3)2
AB2 = (2)2 + (2)2
AB2 = 4+4
Page 77 of 173
AB = √8
Therefore, distance between two points = √(𝑋2 – 𝑋1)2 + (𝑌2 – 𝑌1)2
2. Find the midpoint of A and B
X1+ X2 , Y1 +Y2
(halfway x1 and x2 and halfway y1 and y2)
2
2
x1, y1
x2, y2
A (2,3) and B (4,5)
= 2+4, 3+5
2
2
6
8
Midpoint= ( + )
2
2
Midpoint = ( 3,4 )
3. Calculate the gradient of the line C(-5,-3) and D(-2,6) and E(1,2) and (3,-1)
.D (-2,6)
6
5
4
3
2
.E(1,2)
1
-5
0
-4
-3
-2
-1
0
-1
1
2
3
. F (3, -1)
-2
. C (-5,-3)
M = ∆y = M = Y1-Y2
∆x
x1-x2
-3
C = (-5, -3) and D (-2, 6)
Page 78 of 173
MCD = 6-(-3)
-2-(-5)
MCD = 6+3
-2+5
MCD = 9
3
MCD = 3
TIP - since the line slops to the right, the gradient should be positive
X1,Y1 , X2 , Y2
(ii) MEF =∆ Y = MEF = Y2 -Y1
EF = (1,2) and (3,-1)
∆X
X2 –X1
MEF = -1-2
3-1
MEF = -3
2
Since the line is sloping to the left, the gradient should be negative
4. (i) Find the equation of the straight line through (-2,-3) with a gradient 2.
Y – Y1 = M (X – X1)
(equation to use when a point on the
line
Y-(-3)=2(X-(-2))
is given and gradient) M=Y2-Y1
Y+3=2 (X+2)
X 2-X1
Y+3=2X +4
Y=2X +4 -3
Y=2X+1
(ii) Find the equation of the straight line through (2,4) and (-2, -4)
M = Y-Y1
X-X1
Y-Y1 = M (X-X1)
Since we do not have M then we find M as y2 –
y1
Page 79 of 173
Y – Y1 = y2 –y1
x1
x2 –x1
Y -4 = -4-4
-2-2 (X-2)
equation
x2 –
using the point (2,4) to substitute in to the
Y – 4= 2 (X-2)
Y=2X-4+4
Y = 2X
(iii) Find the equation of the straight line through the point (4,-2) and the
origin
i.e (4,-2) and (0,0)
origin has co-ordinates(0,0)
Y- Y1 = M (X-X1)
Y-Y1 = Y2-Y1 ,(X-X1)
X2-X1
Y-0 = 0-(-2) (X-0)
using the point (0,0)
0-4
Y = -2X OR 4Y = -2X
4
(iv) Find the equation of the straight line with the gradient 3 and y – intercept
of 6
Y = MX +C , M is the gradient, C is the y-intercept, the point
Where the line cuts the y-axis
Therefore, Y = 3X + 6
By substituting the values of M and C
(V) Find the equation of the straight line with the gradient -2/9 and x-intercept
3
Y = MX+C
axis. Along
Y=
−2
9
Y=
(3) +C
−2
3
+C
X-intercept is the point where the line cuts the xx-axis all the y- co-ordinates are zero(0)
therefore, the point is (3,0)
Page 80 of 173
0=
−2
3
C=
+C
2
3
The equation of the straight line is
y=
−2
2
9
3
x+
or 9y = -2x +6
,by multiplying through by 9.
(vi) Find the equation of the straight line with an x- intercept 4 and y- intercept
6.
i.e (4,0) and (0,6)
M = Y2+Y1
X2+Y1
M = 6-0
0-4
M=
−6
4
=
−3
2
Using the point (0,6) i.e C=6
Y = MX +C
Y=
−3
2
x+6. Or
2y = -3x = 12
multiplying throughout by 2.
5.(i) Find the equation of the straight line that passes through the point (3,8) and
is
parallel to y = 2x -9
- The gradient of the new line is 2 since parallel lines have the same
gradient
i.e M1 = M2.
Using the point (3,8)
Y = MX +C
8 = 2(3) + C
8 = 6 +C
2=C
Y = 2X +2
Page 81 of 173
(ii) Find the equation of the line that is perpendicular to 3y – 2x = 4 and passes
through the point (-7,4)
3y = 2x +4
making y the
subject of
y = 2x + 4
the formula
3
3
Therefore, M = 2
3
- Gradient of parallel lines M2× M2 = -1
2⁄
3 × 𝑀2 = −1
2M2 = -3
M2 =
−3
2
Using the point (-7,4) to substitute into the equation
Y = MX + C,
4=
−3 (−7)
2
+C
4 = 21 +C
2
4 -21 = C
1 2
8-21 = C
2
-13 =C
2
The equation perpendicular to 3y-2x=4 which passes through point (-7,4) is
Y=
−3𝑥
2
-
13
2
or
2y = -3x - 13
PRACTICE QUESTIONS:
(1) Find the gradient of the straight line whose equation is 3y + x = 5.
(2) Find the equation of the straight line passing through (-4, 4) and is
𝑥
perpendicular to the straight line whose equation is 𝑦 + = 1.
7
(3) In the diagram below, the points A and B are (4,0) and (0,8) respectively.
Page 82 of 173
Find the equation of AB.
(4) In the diagram, B is the point (0,16) and C is the point (0,6). The sloping line
through B and the horizontal line through C meet at the point A.
(a) Write down the equation of the line AC.
(b) Given that the gradient of the line AB is 2, find the equation of the line
AB.
(c) Calculate the coordinates of the point A.
(d) Calculate the area of the triangle ABC.
ANSWERS:
(1) m = −
1
3
(2) y = 7x + 32
(3) y = -2x + 8
(4) (a) y = 6
(b) y = 2x + 16
(c ) A(-5, 6)
Page 83 of 173
(d ) area = 25 units2
QUADRATIC FUNCTIONS
1. (a) Make a table of values of the function f (x) = x2 – 2x, with the domain -2 ≤ x ≤ 4,
XER, and sketch the graph. Use a scale of 1cm to 2 units on the Y-axis and 1cm to 1
unit on
the X-axis.
(b) Write down the range
(c) Find the turning point and state whether it is maximum or minimum
(d) Write down the equation of the line of symmetry.
SOLUTION
(a) f(x) = x2 – 2x
x
X2
-2x
F(x)
-2
4
4
8
-1
1
2
3
0
0
0
0
1
1
-2
-1
2
4
-4
0
3
9
-6
3
4
16
-8
8
Points are: (-2,8), (-1,3), (0,0), (1,-1), (2,0), (3,3), (4,8). Graph on graph paper.
(b) From table: the smallest value f(x) is -1 and largest value is 8.
Therefore the range -1 ≤ f(x) ≤ 8, f(x) ER
(c) Turning point (1,-1)
Minimum
(d) Line of symmetry x = 1
2. Answer the whole of the question on a sheet of graph paper.
The valuables x and y are connected by the equation y = x2 – 4x + 3.
Some of the corresponding values of x and y correct to one decimal place where
necessary are given
in the table below.
Page 84 of 173
X
y
0
3
0.2
2.2
0.5
r
0.8
0.4
1
0
1.5
-0.8
2
-1
2.3
-0.9
2.5
-0.8
2.8
-0.4
3
0
(a) Calculate the value of r
(b) Using a scale of 4cm to represent 1 unit on the horizontal axis and 2cm to represent 1
unit on the vertical axis, draw the graph of y = x2 – 4x + 3 for 0 ≤ x ≤ 3.
(c) Showing your method clearly, use your graph to find values of x which satisfy the
equation
x2 – 4x + 3 = -
1
2
(d) By drawing a suitable straight line on the same axes, use your graph to find the values of
x which satisfy the equation x2 – 4x + 3 = x + 1
(e) By drawing a suitable tangent, find the gradient of the curve at the point where x = 1.5
Solution on graph paper
3. A solution of the equation x2 + Kx + 9 = 0 is x = 3. Find the value of K
Solution
x2 + Kx + 9 = 0
x = (3)2 + K(3) + 9 = 0
9 + 3K + 9 = 0
3K = -18
K=
−18
3
K = -6
3.
12
A
x
B
x
Q
8
S
x
D
R
x
C
In the diagram, ABCD is a rectangle
AB = 12cm and BC = 8cm
Page 85 of 173
AP = BQ = CR = DS = x centimetres
(a) Find an expression, in terms of x for
(i)
the length of QC
(ii)
the area of triangle CRQ
(b) Hence show that the area in square centimeters, of the quadrilateral PQRS is
2x2 – 20x + 96
(c) When the area if quadrilateral PQRS is 60cm2 form an equation in x and show
that it simplifies to x2 – 10x + 18 = 0
(d) Solve the equation x2 – 10x + 18 = 0, giving each answer correct to 2 decimal
places
Solution
(b) (i) QC = (8 – x) cm
(ii) Area of
CQR =
=
=
(c) In
1
2
1
2
𝟏
𝟐
X RCXQC
x (8 – x)
(8x – x2) cm2
BQP, PB = 12 - x
1
∴ Area of BQP = 2 x (12 – x)
1
= 2 (12x – x2) cm2
Total area of four triangle
= Area of
= 2(
BQP + area of
BQP) + 2 (
1
DSR + Area of
APS + Area of
CRQ
CRQ)
1
=2(2 (12x – x2) + 2 (2 (8x – x2))
= 12x – x2 + 8x – x 2
= (20x – 2x2) cm2
Area of rectangle ABCD = 12 x 8 = 96cm2
∴ area of PQRS = 96 – (20x – 2x2)
=96 – 20x + 2x2
=2x2 – 20x + 96cm2
shown
(d) Area of PQRS = 60cm2
2x2 - 20x + 96 = 60
2x2 – 20x + 96 – 60 = 0
2x2 – 20x + 36 = 0
x2 – 10x + 18 = 0
shown
Page 86 of 173
(e) x2 - 10x + 18 =0
−𝑏±√𝑏 2 −4𝑎𝑐
x=
x=
2𝑎
−(−10)±√𝑏(−102 −4 𝑥 1 𝑥 18
2 (1)
10±√100−72
x=
x=
x=
x1=
2
10±√28
2
10 ±502915
2
10 +5.2915
x1 =
10 − 5.2915
or
2
15.2915
2
x2 =
2
= 7.6457
=
4.7085
2
4.7085
2
= 2.3547
∴ x = 7.65
or
2.35
QUADRATIC EQUATIONS
1. Solve the following quadratic equations :
(b)
(c)
(giving your answer correct to 2 decimal places for b)
2. The area of a rectangle is
and its perimeter is
.Find the length and width
of the rectangle.
Solutions
A quadratic equation is suppose to be expressed in standard form ie
1. (a)
(b)
Page 87 of 173
(x
=7
= 4.65 ,
Or
,
,
,
3.Let be the length of a rectangle and
A=
be the breadth.
and P=
………………….1
Taking (2)
……….3
Substituting (3) in (1)
Page 88 of 173
– 15 = 0 ,
=15 ,
4=0
4
When
So length= 15
.
VARIATION
Direct variation: A quantity 𝑦 is said to vary directly as another quantity 𝑥 if the ratio 𝑦 to
𝑥=
𝑦
𝑥
is always constant. If this is the case then 𝑦 is said to be directly proportional to 𝑥.
𝑇ℎ𝑒 symbol ∝ denotes ‘proportional to’. Thus the statement 𝑦 is proportional to 𝑥 is
expressed mathematically as 𝑦 ∝ 𝑥.
To find the equation connecting 𝑦 and 𝑥, we introduce a constant of proportionality k, say.
Hence the equation connecting 𝑦 and 𝑥 is
𝑦
= 𝑘 𝑜𝑟 𝑦 = 𝑘𝑥
𝑥
Inverse variation: A quantity 𝑦 varies inversely as another quantity 𝑥 if the ratio 𝑦 𝑡𝑜
1
𝑥
=
𝑦
1
𝑥
1
= 𝑥𝑦 is constant for all pairs(𝑥, 𝑦). The statement 𝑦 varies inversely as 𝑥 is written as 𝑦 ∝ 𝑥
𝑘
and the formula linking 𝑦 and 𝑥 is 𝑦 = 𝑥 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. If two quantities 𝑥 and 𝑦
vary in such a way that when one increases the other decreases, or vice versa, then the two are
said to be showing inverse variation.
Joint variation: A variation in which one variable depends on two or more other variables.
Partial variation: Variation as the sum of parts.
Direct variation:
Question 1
Answer:
Page 89 of 173
a) 𝑦 ∝ 𝑥 2
𝑦 = 𝑘𝑥 2
1=
𝑘
4
𝑘=4
∴ 𝑦 = 4𝑥 2
b) 9 = 4𝑥 2
𝑥2 =
9
4
∴𝑥=±
3
2
Question 2
Answer:
a) 𝑦 = 𝑘√𝑥
12 = 𝑘√36
12 = 6𝑘
𝑘=2
𝑦 = 2 √𝑥
b) 10 = 2√𝑥
√𝑥 = 5
𝑥 = 25
Inverse variation:
Question 1
Page 90 of 173
Answer
(i)
−1 =
(ii)
64 =
𝑘
23
−8
∴ 𝑘 = −8
𝑥3
3
64𝑥 = −8
𝑥3 =
−8
𝑥3 =
−1
𝑥=
64
8
−1
2
Question 2
Answer
a) 𝑝 𝑣𝑎𝑟𝑖𝑒𝑠 𝑖𝑛𝑣𝑒𝑟𝑠𝑒𝑙𝑦 𝑎𝑠 𝑎 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑜𝑓 𝑞.
b) 4 =
12
√𝑞
4√𝑞 = 12
√𝑞 = 3
𝑞 = 32 ∴ 𝑞 = 9
Joint variation:
Question 1
1
a. Given that y varies directly as x and z and that y = 9 when x = 6 and 𝑧 = 2 ,find;
i.
k (the constant of variation )
ii.
the value of y when x=4 and z = 3
iii.
the value of x when y = 4 2 and z = 5
1
b. It is given that 𝑦 varies inversely as 𝑥. Some corresponding values of 𝑥 and 𝑦 are
given in the table below.
Page 91 of 173
𝑥
0.6
0.9
𝑏
𝑦
30
𝑎
9
i. Find the equation connecting 𝑥 and 𝑦
ii. Find the values of 𝑎 and 𝑏
Soln
a. .
i. 𝑦 ∝ 𝑥𝑧
𝑦 = 3𝑥𝑧
ii.
𝑦 = 𝑘𝑥𝑧
iii.
1
𝑦 = 3(4)(3)
1
9 = 𝑘(6) (2)
𝑦 = 3𝑥𝑧
4 2 = 3𝑥(5)
9
𝒚 = 𝟑𝟔
2
9 = 3𝑘
= 15𝑥
30𝑥 = 9
𝟑
3=𝑘
𝒙 = 𝟏𝟎
𝒌=𝟑
b. .
i.
1
𝑦∝𝑥
ii.
𝑘
𝑦=𝑥
30 =
30 =
𝑘
0.6
𝑘
6
10
𝒚=
𝟏𝟖
𝒚=
𝒙
18×10
𝑎 = 0.9×10
𝑎=
180
9
𝒂 = 𝟐𝟎
18 = 𝑘
𝒚=
𝟏𝟖
10 was
multiplie
d to in
order to
do away
of a
decimal
in 0.9
9=
𝟏𝟖
𝒙
18
𝑏
9𝑏 = 18
𝒃=𝟐
𝒙
Partial variation:
Question 1
The resistance (R Newtons) to the motion of the motor vehicle is partly
constant and partly varies directly as the square of the velocity (v m/s). Write
down a formula for r in terms of v.
Answer:
R = a + bv2
Question 2
The velocity v m/s of a body moving with constant acceleration is given by v = u
+ at where t seconds is the time the body is in motion and then u and a are
Page 92 of 173
constants. If v = 30 when t = 10 and v = 40 when t =15, find the values of u and
a.
Answer:
30 = u + 10a ………………………………………………………………………………………..( i)
40 = u + 15a ……………………………………………………………..…………………………(ii)
Solving these equations simultaneously we get
a= 2 and u = 10.
PRACTICE QUESTIONS:
(1) Given that y is proportional to 𝑥 3 and that y =250 when 𝑥 = 10, find
(a) the value of the constant 𝑘,
(b) y when 𝑥 = 4,
(c) 𝑥 when y = 54.
(2) It is given that 𝑤 varies directly as the square of 𝑥 and inversely as y.
(a) Write an expression for 𝑤, in terms of 𝑥, 𝑦 and a constant 𝑘.
(b) If 𝑥 = −6, 𝑦 = 12 and 𝑤 = 15, find 𝑘.
(c) Find the value of 𝑦 when 𝑥 = 8 and 𝑤 = 20.
(3) Two variables 𝑝 and 𝑞 have corresponding values as shown in the table
below.
𝑝
3
𝑞
6
14
5
5
5
7
6
Given that 𝑞 varies directly as 𝑝, find
(a) the constant of variation, 𝑘,
(b) the value of 𝑞 when 𝑝 = 5,
(c) the value of 𝑝 when 𝑞 = 6.
(4) Given that 𝑥 varies as 𝑦 and inversely as 𝑧 2 and that 𝑥 = 12 when 𝑦 =
3
Page 93 of 173
and 𝑧 = 2, 𝑓𝑖𝑛𝑑
(a) the equation connecting 𝑥, 𝑦 and z,
(b) the value of 𝑥 when 𝑦 = 3 and 𝑧 = 4,
(c) the values of z when x =4 and 𝑦 = 25.
ANSWERS:
1
(1) (a) 𝑘 = ,
(b) y = 16,
4
𝑤=𝑘
(2)
(a)
(3)
(a) 𝑘 =
(4)
(a) 𝑥 = 16
2
5
𝑥2
𝑦
,
(b)
(c ) 𝑥 = 6.
𝑘 =5,
(b) 𝑞 = 2 ,
,
𝑦
𝑧2
(b)
𝑥 =3,
(c)
𝑦 = 16.
(c) 𝑝 = 15 .
(c)
𝑧 = 10 .
CIRCLE THEOREM
hint:
- Discuss circle properties such as Radius, Diameter, Circumference, Sector, Chords,
Segment, and Tangent properties. Then move on to angles in a circle.
Theorems of angles in a Circle.
1. Angle at the centre theorem:
- Angle subtended by an arc at the centre is twice the angle subtended at the
circumference.
2. Angle in the same segment:
- Angles subtended by the same segment of a circle are equal.
Page 94 of 173
3. Angle in a semi-circle:
- Angle in a semi-circle= 90°.
4. Angles associated with a cyclic-quadrilateral:
-
The opposite angles of a cyclic-quadrilateralare supplementary.
The exterior angle of a cyclic-quadrilateral is equal to the opposite interior angle.
5. Alternate Segment theorem:
- The angle between a chord and a tangent at the point of contact is equal to any angle
in the Alternate Segment.
Example
In the diagram, A, B, C, D and E lie on the circumference of a circle.
Page 95 of 173
DE is parallel to CB. Angle ACE = 48°, angle CED = 65° and angle CBE = 73°.
Calculate;
a) 𝑎𝑛𝑔𝑙𝑒𝐴𝐵𝐸 b) 𝑎𝑛𝑔𝑙𝑒𝐴𝐶𝐵,
c) 𝑎𝑛𝑔𝑙𝑒𝐵𝐸𝐶, d) 𝑎𝑛𝑔𝑙𝑒𝐶𝐷𝐸
EXPECTED ANSWERS:
1) 𝐷𝐸 ⃦ 𝐶𝐵
𝐺𝑖𝑣𝑒𝑛
< 𝐴𝐶𝐸 = 48°
Given
< 𝐶𝐸𝐷 = 65°
𝐺𝑖𝑣𝑒𝑛
< 𝐶𝐵𝐸 = 73°
Given
(a) < 𝐴𝐵𝐸 =< 𝐴𝐶𝐸 = 48° (angles in the same segment)
∴<ABE=48°
(b) < 𝐴𝐶𝐵 =< 𝐵𝐶𝐸−< 𝐴𝐶𝐸
𝐻𝑖𝑛𝑡: 𝐵𝑢𝑡 < 𝐵𝐶𝐸 =< 𝐷𝐶𝐸 = 65°
(𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙)
∴< 𝐴𝐶𝐵 = 65° − 48°
∴< 𝐴𝐶𝐵 = 17°
(c) < 𝐵𝐸𝐶+< 𝐵𝐶𝐸+< 𝐶𝐵𝐸 = 108°
(𝑠𝑢𝑚 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝑎 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒)
∴< 𝐵𝐸𝐶 = 180° − (< 𝐵𝐶𝐸+< 𝐶𝐵𝐸)
= 180° − (65° + 73°
= 180° − 138°
∴< 𝐵𝐸𝐶 = 42°
(d)< 𝐶𝐷𝐸+< 𝐶𝐵𝐸 = 180° (opposite angles of a cyclic quadrilateral)
∴< 𝐶𝐷𝐸 = 180° − (< 𝐶𝐵𝐸)
< 𝐶𝐷𝐸 = 180° − 73°
∴< 𝐶𝐷𝐸 = 107°
ACTIVITY 2
In the diagram below, ABC is a tangent to the circle BDEF at B, angle DFB= 40°,
angle EBF=15°
Page 96 of 173
and angle DBE= 45°.
Find (a) angle DEB,
(b) angle DEF,
(c) angle ABF.
EXPECTED ANSWERS:
(a) angle DEB= angle DFB= 40°
(angles in the same segment)
(b) angle DEF+angle DBF=180°
∴ 𝑎𝑛𝑔𝑙𝑒 𝐷𝐸𝐹 + 60° =180°
(opposite angles of a cyclic quadrilateral)
∴ 𝑎𝑛𝑔𝑙𝑒 𝐷𝐸𝐹 =180° − 60°
∴ 𝑎𝑛𝑔𝑙𝑒 𝐷𝐸𝐹 = 𝟏𝟐𝟎°
(c) angle ABF= angle BDF
=180° − (45° + 15° + 40°)
∴ angle ABF = 𝟖𝟎°.
(alternate segment theory)
ACTIVITY 3
In the diagram below, A,B,C and D lie on the circumference of the circle, center o.
BO is parallel to CD, angle BAD=62° and BCE is a straight line.
Calculate:
(a)
angle t,
(b)
angle BCD,
(c)
angle OBC,
(d)
angle DCE.
EXPECTED ANSWERS:
Page 97 of 173
angle t= 2× 𝑎𝑛𝑔𝑙𝑒𝐵𝐴𝐷
∴angle t= 2× 62°
∴angle t=124°.
(a)
(b)
(c )
(angle at the centre = 2×angle on the
circumference)
angle BCD + angle BAD= 180°
∴ angle BCD =180° − 62°
∴ angle BCD = 118°.
angle OBC= 180° − 118°
∴angle OBC= 𝟔𝟐°
(d) angle DCE= 𝑎𝑛𝑔𝑙𝑒𝐵𝐴𝐷 = 𝟔𝟐°
opposite interior angle)
(opp. angles of a cyclic quad.)
(angles associated with parallel lines)
(exterior angle of cyclic quad=
PRACTICE QUESTIONS:
(1) O is the centre of the circle through A,B,C and D. Angle BOC= 100° and angle
OBA=62°.
Calculate
(i)
BAC,
(ii)
OCB,
(iii)
ADC.
(2) In the diagram, AB is a diameter of the circle, centre O. P and Q are two points on
the circle and APR is a straight line.
Given that angle QBA = 67° and angle PAQ = 32° ,
calculate
(a) angle QAB,
(b) angle RPQ,
(c) angle POB.
Page 98 of 173
(3) BT is a diameter of a circle and A and C are points on the circumference. The tangent
to the circle at the point T meets AC produced at P.
Given that angle ATB = 42° and angle CAT = 26°,
calculate
(i)
angle CBT,
(ii)
angle ABT,
(iii)
angle APT.
ANSWERS:
(1) (i) 50°,
(2) (a) 23° ,
(3) (i) 26° ,
(ii) 40° ,
(b) 67° ,
(ii) 48° ,
(iii) 78° .
(b) 110° .
(iii) 22°
Page 99 of 173
Page 100 of 173
Page 101 of 173
Page 102 of 173
VECTORS
1. (a) The position vector of a point A is(
(b)(i) Find the column vector m such that (
and AB =(
, find the coordinates of B.
−8
−5
)−𝑚 =( )
6
2
(ii) Hence find |𝑚|
−1
(c).Given that PQ= ( ), find QP in component form.
9
Solutions
1.(a) 𝐴𝐵 = 𝐴𝑂 + 𝑂𝐵
(
−3
−2
) = ( ) = 𝑂𝐵
2
1
−3
−2
𝑂𝐵 = ( ) − ( )
2
1
−1
=( )
1
𝑩(−𝟏, 𝟏)
−8
−5
(b)(i)( ) − 𝑚 = ( )
6
2
(
−8
−5
)−( )= 𝑚
6
2
𝟑
𝑚=( )
−𝟒
(ii)|𝑚| = √(3)2 + (−4)2
= √9 + 16
= √25
=5
−1
(c).𝑃𝑄 = ( )
9
𝟏
𝑄𝑃 = ( )
−𝟗
15
6
−9
2.Given that 𝑢 = ( ), 𝑣 = ( ) and 𝑤 = ( ), find
𝑝
−8
10
(a) (i) |𝑢|
(ii) 2𝑢 + 𝑣
(b).Given that vector 𝑤 is parallel to vector 𝑢, calculate the value of 𝑝.
Solution
Page 103 of 173
2(a)(i) |𝑢| = √(6)2 + (−8)2
= √36 + 64
= √100
=10 units
6
−9
(ii) 2𝑢 + 𝑣 = 2 ( ) + ( )
−8
10
12
−9
=(
)+( )
−16
10
= (
𝟑
)
−𝟔
(iii) 𝑢 = 𝑘𝑤 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 then
(
15
6
) = 𝑘( )
𝑝
−8
15𝑘
6
( )=(
)
𝑘𝑝
−8
6 = 15𝑘
2
𝑘=5,
−8 = 𝑘𝑝, 𝑏𝑢𝑡 𝑘 =
2
5
2
−8 = 𝑝
5
2𝑝 = −40
𝑝 = −𝟐𝟎
3.In the diagram, 𝑂𝐴 = 2𝑝, 𝑂𝐵 = 3𝑞 and 𝐵𝑋 = 𝑝 − 𝑞. The lines OX and AB intersects at L.
(i)
Express as simply as possible in terms of 𝑝 and / or 𝑞
(a) 𝑂𝑋
Page 104 of 173
(b) AB
(ii)
Given that 𝐴𝐿 =h𝐴𝐵, express 𝐴𝐿 in terms of 𝑝, 𝑞 and h.
(iii)
Hence show that 𝑂𝐿 = (2 − 2ℎ)𝑝 + 3ℎ𝑝
Solution
3. (i)(a) 𝑂𝑋 = 𝑂𝐵 + 𝐵𝑋
= 3𝑞 + 𝑝 − 𝑞
= 3𝑞 − 𝑞 + 𝑝
=𝟐𝒒 + 𝒑
(b) 𝐴𝐵 = 𝐴𝑂 + 𝑂𝐵
= −2𝑝 + 3𝑞
= 3𝑞 −2p
(ii).𝐴𝐿 =h𝐴𝐵
𝐴𝐿 =h(3𝑞 − 2𝑝)
=𝟑𝒉𝒒 − 𝟐𝒉
(iii) 𝑂𝐿 = 𝑂𝐴 + 𝐴𝐿
=2𝑝 + 3ℎ𝑞 − 2ℎ𝑝
=2𝑝 − 2ℎ𝑝 + 3ℎ𝑞
= (−𝟐𝒉)𝒑 + 𝟑𝒉𝒒
4. In the diagram, 𝑂𝐴 = 2𝑏, 𝑂𝐶 = 3𝑎 and 𝐴𝐵 = 2𝑎. The lines OB and AC intersects at X..
Express as simply as possible in terms of 𝑎 and / or 𝑏.
(a) 𝑂𝐵
(b) BC
(ii)
Given that 𝐶𝑋 = ℎ𝐶𝐴, express 𝐶𝑋 in terms of 𝑎, 𝑏 and ℎ.
(iii)
Hence show that 𝑂𝑋 = (3 − 3ℎ)𝑎 + 2ℎ𝑏
(i)
Solutions
4(i)(a) 𝑂𝐵 = 𝑂𝐴 + 𝐴𝐵
= 2𝑏 + 2𝑎
(b) 𝐵𝐶 = 𝐵𝑂 + 𝑂𝐶
Page 105 of 173
= −(2𝑎 + 2𝑏) + 3𝑎
= −2𝑎 − 2𝑏 + 3𝑎
= 3𝑎 − 2𝑎 − 2𝑏
= 𝒂 − 𝟐𝒃
(ii) 𝐶𝑋 = ℎ𝐶𝐴𝑏𝑢𝑡 𝐶𝐴 = 𝐶𝑂 + 𝑂𝐴
=2𝑏 − 3𝑎
𝐶𝑋 = ℎ(2𝑏 − 3𝑎)
=𝟐𝒉𝒃 − 𝟑𝒉𝒂
(iii) 𝑂𝑋 = 𝑂𝐶 + 𝐶𝑋
= 3𝑎 + 2ℎ𝑏 − 3ℎ𝑎
= 3𝑎 − 3ℎ𝑎 + 2ℎ𝑏
= (𝟑 − 𝒉)𝒂 + 𝟐𝒉𝒃shown.
PRACTICE QUESTIONS
(1)
OABC is a parallelogram .
The point X on AC is such that AX = 1⁄5 𝐴𝐶. The point Y on AB is such that AY
= 1⁄4 𝐴𝐵. Given that OA = 20p and OC= 20q, express in terms of p and q
(i)
AC,
(ii) AX ,
( iii) OX ,
(iv) OY.
What do the results of (iii) and (iv) tell you about O, X and Y?
Page 106 of 173
(2) In the diagram XVZ is a straight line ,XY = 8p , XZ = 4p + 9q and YV =-6p +
cq.
(i)
Express XV in terms of p,q and c.
(ii)Given that XV = h XZ, form an equation involving p, q , h and c.
(iii)
The point K is outside triangle XYZ and is such that
XK = -4p + 3q.
Is XK parallel to YV? Justify your answer.
(3) In the triangle ORS , the point A on OR is such that OA = 2AR. B is the midpoint of OS,
X is the midpoint of AB and OX produced meets RS at Y.OA = 2p and OB =2q.
(a) Express in terms of p and/or q
(i)
AB,
(ii)
AX,
(b) Given that RY = k RS ,
(iii)
OX,
(iii)
RS
Express RY in terms of p,qand h.
(c) Hence show that OY = 3(1- h)p +4hq.
(d) Given also that OY =k OX , Express OY in terms of p, q and k.
Page 107 of 173
(e) Using these two expressions for OY, find the value of h and the value of k.
(f) Find the ratio RY : YS.
(g) Express XY in terms of pand q.
(4) p = (−𝟐𝟓),
q = (−𝟑
) , 𝒓 = (𝟕𝒙 ) ands = (𝟒𝒚).
𝟏
(a) Find (i)
2p + 3q,
(ii)
p− q.
Given that 2p = r + s calculate the value of 𝑥 and the value of y.
( b)
(5) OA = (−2
)and OB = (34).
1
(a)
Given that OC = OA + OB, express oc as a column vector.
(b) Express AB as a column vector.
(c )
4
If LM = (−2
), what is the special name given to the quadrilateral ALMO?
𝟏
(6) p = (−𝟒
) ,
q = (−𝟑
)and r = (𝒎
).
𝟒
𝟐
(a) Find q .
(b) Express 2p–q as a column vector.
( c) Given that p is parallel to r, find m.
ANSWERS:
(1) (i) 20(q–p),
(ii) 4( q – p)
(iii) 4(4p+ q) (iv) 5(4p+ q)
The points O,X and Y lie on the same line.
(2) (i) 2p + cq ,
(ii) h= 1⁄2 ,
c= 41⁄2 ,
(ii)
YV= 3⁄2 (−4𝑷 + 3𝒒) ,
XK isparallel to YV.
(3) (a) (i) 2p −2q ,
( b)
(f)
h(4q – 3p ) ,
3:4
,
(ii) q – p ,
(iii) p+ q ,
(d) k( p+ q ) ,
(g)
(e )
(iv) 4q – 3p
.
h= 3⁄7 , k = 1 5⁄7.
5⁄ (𝑷 + 𝒒)
7
Page 108 of 173
(4)
(a) (i)
(5) (a) OC = (𝟏𝟓) ,
(6)
(a) 5 ,
1
(−1
) ,
(ii)
( b)
8
(−3
),
AB =(53) ,
5
(−12
) ,
(b)
(b)
x= 6 , y =−11.
(c)
(c)
Trapezium.
- 1⁄2 .
TRIGONOMETRY QUESTIONS
Sine, cosine, tangent
The trigonometric ratios sine, cosine and tangent are defined in terms of the hypotenuse,
opposite and adjacent side of a right-angled triangle.
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑆𝑖𝑛𝑒 𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Opposite
Hypotenuse
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Cosine= 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝜃
𝑇𝑎𝑛𝑔𝑒𝑛𝑡 𝜃 =
Adjacent
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
Adjacent
Sine rule
Sine rule is used when you are given
i.
two angle and one side
ii. two side and a non included angle
In obtuse-angled triangle 𝑠𝑖𝑛 𝜃 = 𝑠𝑖𝑛 (180° − 𝜃)
Cosine rule
Cosine rule is used when you are given;
i.
two side and an included angle
ii. three side only
In obtuse-angled triangle 𝑐𝑜𝑠 𝜃 = −𝑐𝑜𝑠(180° − 𝜃)
Pythagorus
In a right angle triangle, the square on the hypotenuse is equal to the sum of the
the two adjacent sides.
squares on
TRIGONOMETRY ON THE CARTESIAN PLANE
1. Determining the signs of the three trig ratios in the quadrants.
2
2 quadrant
nd
180°
90°
1
1 quadrant
st
360°
Page 109 of 173
3
3 quadrant
4
4 quadrant
rd
th
270°
The quadrants are numbered 1 to 4 in an anti-clock wise direction.
i.
angles between 0 and 90 fall in the first quadrant
ii. angles between 90 and 180 fall in the second quadrant
iii. angles between 180 and 270 fall in the third quadrant
iv.
angle between 270 and 360 fall in the fourth quadrant
We can find trig ratios on Cartesian plane given a point
y
P
i)
r
𝜃
0
y
x
X
y
P
r
ii) 𝑦
𝜃
𝑥
𝑂
𝑋
y
y
iii)
Type equation here.
𝑥
𝜃
X
0
y
r
P
y
iv)
Type equation here.
𝜃
𝑥
0
Type equation here.
X
Page 110 of 173
r
y
P
 P(x,y) is a point in each of the four quadrants of the Cartesian plane.
The length 𝑂𝑃 = 𝑟, where 𝑟 > 𝑜 and < 𝑃𝑂𝑋 = 𝜃
𝑦
𝑥
𝑆𝑖𝑛 𝜃 = 𝑟 , 𝐶𝑜𝑠 𝜃 = 𝑟

and 𝑇𝑎𝑛 𝜃 =
𝑦
𝑟
The signs of x and y differ from quadrant to
quadrant so the values of sin, cos and tan will differ as well.

𝜃 is the angle between the line OP and the positive 𝑥 − 𝑎𝑥𝑖𝑠
TRIGONOMETRIC EQUATIONS
Sin𝜃, cos𝜃 and tan𝜃 can be positive or negative depending on the quadrant within
which 𝜃 fall.
The cartesian diagram
Sine +ve
S
All positive
A
Tan +ve
T
Cosine +ve
C



The cast diagram is very useful when solving trig equations
A trigonometric equations will usually have two solutions
To solve equations of the form 𝑠𝑖𝑛𝜃 = 𝑘, 𝑐𝑜𝑠𝜃 = 𝑘 𝑜𝑟 𝑡𝑎𝑛 𝜃 = 𝑘
i.
find the reference angle ∝ in the 1st quadrant
ii. determine in which two quadrants 𝜃 will lie
iii. find the corresponding angle in the two quadrant
In the diagram below P has coordinates (12,5)
y
P(12, 5)
r
0
y
𝑥
Type equation here.
Page 111 of 173
Find the value of;
a) < sin <XOP
b) cos <XOP
c) tan XOP
Solutions
First find side ‘r’ by using the Pythagoras theorem.
𝑟2 = 𝑦2 + 𝑥2
𝑟 2 = 52 + 122
𝑟 2 = 25 + 144
𝑟 2 = 169
√𝑟 2 = √169
𝑟 = 13𝑢𝑛𝑖𝑡𝑠.
𝑂
𝐴
𝑂
∴ (a) 𝑠𝑖𝑛 < 𝑋𝑂𝑃 = 𝐻
(𝑏) 𝑐𝑜𝑠 < 𝑋𝑂𝑃 = 𝐻
(𝑐) 𝑡𝑎𝑛 < 𝑋𝑂𝑃 = 𝐴
𝑦
𝑥
𝑦
=
=
=
𝑟
𝑟
𝑥
5
12
5
=
=
=
13
13
12
5. Solve the equation sin 𝜃=0.766 for 0° ≤ 𝜃 ≤ 180°.
Solution
𝑆𝑖𝑛 𝜃 = 0.766
∝= 𝑠𝑖𝑛−1 0.766
= 49.99603866°
= 50°
Sin is positive in the 1st and 2nd quadrant
(i) 1st quadrant
𝜃 = 50°
(ii) 2nd quadrant
𝜃 = 180° − 50°
𝜃 = 30°
∴ 𝜃 = 50° 𝑎𝑛𝑑 𝜃 = 130°
6. Solve the equation 𝑡𝑎𝑛𝜃 = −5.67 𝑓𝑜𝑟 0° ≤ 𝜃 ≤ 360°.
Solution
𝑡𝑎𝑛𝜃 = −5.67
∝ = 𝑡𝑎𝑛−1 5.67
= 79.99°
= 80°
Tan is negative in the 2nd and 4th quadrants.
2nd quadrant
4th quadrant
𝜃 = 180° − 80°
𝜃 = 360° − 80°
= 100°
= 280°
∴ 𝜃 = 100° 𝑎𝑛𝑑 𝜃 = 280°
Page 112 of 173
A kite (k) being flawn is such that its vertical height HK is 6m and the angle formed between
the vertical height and the string Skit is attached to is 60° as shown in the diagram below.
K
60°
6m
S
H
If 𝑠𝑖𝑛 60° = 0.866, 𝑐𝑜𝑠 60° = 0.5 and 𝑡𝑎𝑛 60° = 1.73, calculate the length of the string
SK.
Solution
𝐴
𝐶𝑜𝑠 𝜃 =
𝐻
6
𝐶𝑜𝑠 60° =
𝑆𝐾
6 = 𝑆𝐾𝑐𝑜𝑠 60°
6
𝑆𝐾 =
= 12𝑚.
0.5
In the diagram below, AC=10cm, BC=5cm and <ACB=60°. Given that
𝑠𝑖𝑛60° = 0.866, 𝑐𝑜𝑠60° = 0.5 𝑎𝑛𝑑 𝑡𝑎𝑛60° = 1.73. Calculate the value of (𝐴𝐵)2 .
B
5cm
A
60°
10cm
C
Solution
(𝐴𝐵)2 = (𝐴𝐶)2 + (𝐵𝐶)2 − 2𝐴𝐵 × 𝐵𝐶 × cos 𝐶
= 102 + 52 − 2(10)(5)(0.5)
= 100 + 25 − 50
2
(𝐴𝐵) = 75𝑐𝑚
The figure below shows triangle ABC in which AC=5cm. Given that
𝑠𝑖𝑛 𝐵 = 0.5, 𝑐𝑜𝑠𝐵 = 0.9 𝑎𝑛𝑑 𝑡𝑎𝑛𝐵 = 0.6. Calculate the length of BC;
C
Page 113 of 173
5cm
/
A
B
Solution
𝑂
𝑆𝑖𝑛 𝜃 = 𝐻
𝐴𝐶
𝑆𝑖𝑛 𝐵 = 𝐵𝐶
5
0.5 = 𝐵𝐶
0.5𝐵𝐶 = 5
𝐵𝐶 = 5 ÷ 0.5
∴ 𝐵𝐶 = 10𝑐𝑚
2. PQ and R are fishing camps along the banks of lake kaliba joined by straight paths PQ, QR
and RP. P is 7.6km from Q and Q is 13.2km from R and <PQR=120°.
Type equation here.
Q
120°
13.2km
R
P
7.6km
a) Calculate;
i. The distance PR
ii. The area of triangle PQR
iii. Find the shortest distance from Q to PR
b) A fisherman takes 30 minutes to move from R to P. calculate his average
speed in km/h.
Solution
a) (i) (𝑃𝑅)2 = (7.6)2 + (13.2)2 − 2(7.6 × 13.2 × cos 120°)
= 232 − 2(−50.16)
2
(𝑃𝑅) = 332.32
𝑃𝑅 = 18.2𝑘𝑚 (3 𝑠. 𝑓)
1
(ii) Area of trianglePQR=2 × 7.6 × 13.2 sin 120°
= 43.4𝑘𝑚2
1
(iii) in the formula, 𝐴 = 2 𝑏 ℎ ( ℎ is a shortest distance)
A=43.4𝑘𝑚2 , 𝑏 = 𝑃𝑅 = 18.2𝑘𝑚
1
43.4=2 × 18.2 × ℎ
ℎ =
43.4
9.1
ℎ = 4.773𝑘𝑚
ℎ = 4.8𝑘𝑚
∴ 𝑡ℎ𝑒 𝑠ℎ𝑜𝑟𝑡𝑒𝑠𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑠 4.8𝑘𝑚
Page 114 of 173
b) 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 =
=
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
30
𝐷 = 18.2𝑘𝑚 , 𝑇 = 60 = 0.5ℎ𝑟𝑠
𝑇𝑖𝑚𝑒
18.2𝑘𝑚
0.5ℎ𝑟𝑠
= 36.4𝑘𝑚/ℎ
4) 𝐴𝐵𝐶 is a triangle in which angle 𝐴𝐶𝐵 = 90°, D is a point on 𝐴𝐶,
𝐴𝐵 = 20𝑐𝑚, 𝐵𝐶 = 12𝑐𝑚, 𝐶𝐷 = 5𝑐𝑚, 𝐵𝐷 = 13𝑐𝑚 and 𝐷𝐴 = 11𝑐𝑚. Giving each answer
as a fraction, find;
a) Tan <CDB
b) Cos <CAB
c) Sin <ADB
C
5cm
D
12cm
11cm
13cm
A
20 cm
B
Solution
𝐵𝐶
a) 𝑡𝑎𝑛 < 𝐶𝐷𝐵 = 𝐷𝐶
c) 𝑠𝑖𝑛 < 𝐴𝐷𝐵 =?
12
𝐷𝐶 2 = 132 − 122
𝑠𝑖𝑛 < 𝐵𝐷𝐶 = 13
12
= 169 − 144
𝑠𝑖𝑛 < 𝐴𝐷𝐵 = − 13
𝐷𝐶 2 = 25
𝐷𝐶 = 5𝑐𝑚
∴ 𝑡𝑎𝑛 < 𝐶𝐷𝐵 =
b) 𝑐𝑜𝑠 < 𝐶𝐴𝐵 =
=
12
5
16
20
4
5
5) In the diagram below = 1.8 𝑢𝑛𝑖𝑡𝑠, 𝑄𝑅 = 2.5 𝑢𝑛𝑖𝑡𝑠, find the size of the angle marked 𝜃.
P
1.8units
Q
2.5units
𝜃
Page 115 of 173
R
Solution
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
1.8
𝑆𝑖𝑛 𝜃 =
2.5
−1 1.8
𝜃 = 𝑠𝑖𝑛 (2.5)
𝑆𝑖𝑛 𝜃 =
𝜃 = 46°
6)
C
A
4cm
5cm
120°
B
D
ABC is a triangle with AB=5cm., BC=4cm and angle ABC=120°. AB is produced to D and
angle BCD=90°. Using as much information in the table below as necessary;
sin
cos
tan
0.87
-0.5
-1.73
120°
Calculate,
a) The area of the triangle ABC
b) The length of BD
Solution
1
a) Area of triangle 𝐴𝐵𝐶 = 2 × 5 × 4 × 𝑠𝑖𝑛 120°
1
= 2 × 20 × 0.87
= 8.7𝑐𝑚2
b) 𝐵𝐷 =? 𝑐𝑜𝑠 < 𝐶𝐵𝐷 =
𝐵𝐶
𝐵𝐷
4
𝑐𝑜𝑠 60° = 𝐵𝐷
4
0.5 = 𝐵𝐷
∴ 𝐵𝐷 = 8𝑐𝑚.
PRACTICE QUESTIONS:
(1) In the right angled triangle ABC, P is a point on the side AB. Given that AP= 4cm,
PB = 5cm, BC = 12cm and PC = 13cm, calculate
Page 116 of 173
(a)
(b)
(c)
(d)
AC,
cos BPC,
tan PAC,
sin APC.
(2) ABCD represents a building with a vertical flagpole AP on the roof.
The point O is on the same level as C and D.
The angle of elevation of A from O is 15°, OA = 60 metres and
(i)
(ii)
POA =7°
Calculate
(a) the height AD of the building,
(b) the height of the flagpole, AP.
Given also that AB =10 metres, calculate the angle of elevation of P from B.
(3) In the diagram, ABC represents a horizontal triangular field and AD represents
a vertical tree in the corner of the field. A path runs along the edge BC of the
field.
AB =83m, AC= 46m and angle BAC= 67°.
(a) The angle of elevation of the top of the tree when viewed from B is 14°.
Calculate the height of the tree.
(b) Calculate the length of the path BC.
Page 117 of 173
(c) Calculate the area of the field ABC.
(d) Calculate the shortest distance from A to the path BC.
(e) Calculate the greatest angle of elevation of the top of the tree when viewed
from any point on the path.
(4) The diagram shows a triangular field PQR.
Calculate (a) angle QPR,
(b ) the area of triangle QPR.
ANSWERS:
(1) (a) AC = 15cm,
(b) cos
(2) (i) (a) 15.53m
(b) 7.89m
(ii )
5
BPC= 13 ,
(c) tan
4
PAC= 3 , (d) sin
APC=
12
13
.
38.3° .
(3) (a) 20.7m ,
(4) (a) 101.7°,
(b) 77.6m ,
(c )
1760m2 ,
(d) 45.3m ,
(e)
24.6° .
(b) 1660.76m2 .
MENSURATION
Specific outcome:




Calculate the area of a sector.
Calculate surface area of three dimensional figures.
Calculate volume of prisms.
Solve problems involving area and volume.
1
1. [Volume of a cone= 3 × 𝑏𝑎𝑠𝑒 𝑎𝑟𝑒𝑎 × ℎ𝑒𝑖𝑔ℎ𝑡]
The diagram shows a plant pot. The open end of the plant pot is a circle of radius
10cm. The closed end is a circle of radius 5cm. The height of the pot is 12cm. The
plant pot is part of the right circular cone of height 24cm.
Page 118 of 173
I.
II.
Calculate the volume of the plant pot. Give your answer in litres.
How many of these plant pots can be completely filled from a 75litre bag of
compost?
Solutions
i.
1
Volume of larger cone= 3 × 𝜋𝑟 2 × ℎ
1
= 3 × 3.142 × 102 × 24
=2513.6𝒄𝒎𝟑
1
Volume of the smaller cone= 3 × 3.142 × 52 × 12
=314.2𝒄𝒎𝟑
Volume of plant pot= 2513.6 − 314.
= 2199.4𝑐𝑚3
1 litre → 1000𝑐𝑚3
𝑥 → 2199.4𝑐𝑚3
𝑥=
2199.4
1000
= 2.1994
=2.20 litres
Page 119 of 173
ii.
75
The number of pots =2.1994 = 34.100
Therefore the number of completely filled = 34
2.
Diagram 1
Diagram 2
Diagram 1 shows a hollow cone whose sloping edge is of length t cm. The radius of
the circular top is r cm. The cone is cut along its sloping edge and laid flat to form the
sector OPQ of a circle of radius t cm as shown in Diagram 2.
1). Find an expression in terms of r, for the length of the arc PQ
2). It is given that t = 5r
a). calculate POQ
b). given also that 𝑡 = √40, calculate the area of the sector OPQ, expressing your
answer as a multiple of π.
Solutions
1). Length of arc PQ= circumference of circular top
Arc PQ= 2πr
𝜃
2).a).length of arc= 360 × 2𝜋𝑟
2𝜋𝑟 =
𝑃𝑂𝑄
360
× 2𝜋 × 5𝑟, since t= 5r
720 = 𝑃𝑂𝑄
𝑃𝑂𝑄 = 𝟕𝟐𝟎
𝜃
b). Area of sector = 360 × 𝜋𝑟 2
72
= 360 × 𝜋 × (√40)2
72
= 36∅ × 𝜋 × 4∅
= 𝟖𝝅
3. The base of a pyramid is a square with diagonals of length 6cm. The sloping faces are
isosceles triangles with equal sides of length 7cm. the height of the pyramid is √𝑙.
calculate 𝑙
Page 120 of 173
Solutions
Hint
Construct a right-angled triangle and use the Pythagoras theorem to find the height.
Applying the Pythagoras theorem
72 = ℎ2 +32
72 − 32 = ℎ2
49 − 9 = ℎ2
(√40)2 = (√ℎ)2
40 = ℎ
ℎ = 40
Therefore, 𝑙 = 40
4. A solid cuboid measures 7cm by 5cm by 3cm
3
7
I.
II.
Calculate the total surface area of the cuboid.
A cube has the same volume as the cuboid. Calculate the length of an edge of
this cube
Solutions
i).The total surface area =2(7x5) + 2(3x5) + 2(7x3)
= 70 + 30 + 42
= 142cm2
ii). Volume of the cuboid = 7 x 5 x 3
= 105𝑐𝑚3
Page 121 of 173
Length of edge = √105
= 4.72cm (to 3s.f)
5. A closed container is made by joining together a cylinder of radius 9cm and a
hemisphere of radius 9cm as shown in diagram 1
The length of the cylinder is 18cm.
The container rests on a horizontal surface and is exactly half full of water.
a. Calculate the surface area of the inside of the container that is in
contact with the water.
b. Show that the volume of the water is 972𝜋𝑐𝑚3
Solutions
1
a. Surface area =2 𝜋𝑟 2 +
1
1
2
(2𝜋𝑟ℎ) +
1
1
4
[4𝜋(9)2 ]
1
= 2 𝜋(9)2 + 2 [2𝜋(9)(18)] + 4 [4𝜋(9)2
81
= 2 𝜋 + 162𝜋 + 81𝜋
81
= 𝜋 ( 2 + 162 + 81)
= 283.5 x 3.142
= 891𝒄𝒎𝟐 𝑡𝑜 3𝑠𝑓.
1 2
b. Volume of water = 2 [3 𝜋𝑟 3 + 𝜋(81)(18)]
1 2
= 2 [3 𝜋(9)3 + 𝜋(81)(18)]
1
= 2 (486𝜋 + 1458𝜋)
1
= 2 (1944𝜋)
=972𝝅𝒄𝒎𝟑 ℎ𝑒𝑛𝑐𝑒 𝑠ℎ𝑜𝑤𝑛.
5. A car’s windscreen wiper left a part of a windscreen unwiped, as shown in the diagram below
Page 122 of 173
a) Calculate the length of arc BB1
b) Calculate the length of arc AA1
c) Calculate the Area of the shaded part of the windscreen.
Ans; a) BB1 =
2
=
2
= 54.977 = 54.98cm
b) AA1 =
2
=
2
= 36.6519
= 36.65cm
c)A = area of sector OBB1 – area of sector OAA1
=
2
=
2
2
-
2
= 577.267 – 256.563
= 320.704
= 320.70cm
6.
O
Page 123 of 173
In the diagram, O is the centre of a circle of radius 8cm. PQ is a chord and POQ= 150 .
The minor segment of the circle formed by the chord PQ is shaded. (Take
as 3.142),
calculate
i)the length of triangle the minor arc PQ
ii)the area of triangle OPQ
iii)the shaded area
Ans: i) minor arc PQ=
2
=
2
=20.9cm
ii) area of triangle OPQ =
ab sin
=
8
8 sin 150
= 16cm2
iii) area of sector =
=
2
3.142
= 83.79cm
Shaded area = area of sector – area of triangle
= 83.29cm2 – 16cm2
= 67.8cm2
7. OAB is the sector of a circle of radius r cm,
AOB = 60
A
O
Page 124 of 173
Find, in its simplest form, an expression in terms of r and
for
a) The area of the sector
b) The perimeter of the sector
a) Area of sector =
=
=
b) Arc length AB =
x2 r
=
x2 r
=
r
Perimeter of the sector = OA + OB + AB
=r+r+
r
= 2r +
= r(
r
)
8. In the diagram below, MN is an arc of a circle whose centre is O and radius 21cm
Given that
[ Take
MON = 120 , calculate the area of the sector MON
to be
]
Page 125 of 173
Ans:
x
x 212 = 462cm2
8. The net of a square based pyramid is shown below
40cm
Calculate the total surface area of the pyramid
Ans: TSA= S(2L +S)
= 40( 2 x 32.02 + 40)
= 4161.6cm2
9. The area of the base of a square- based pyramid is 100mm2 and its slant height is
25mm
a) Calculate the total surface area of the pyramid
b) Express the relationship between the total surface area of the pyramid and the
area of its base as a ratio in its simplest form
ans: S2 = 100mm2
s= 10mm
a) TSA= s(2l + s)
= 10( 50 + 10)
= 10(60)
Page 126 of 173
= 600mm2
b) Ratio
=600: 100
= 6:1
10. The volume of a cone is x base area x height.The area of the curved surface of a cone of
radius r and slant L πrl.
A solid cone has a base radius of 8cm and a height of 15cm.
Calculate
(i). its volume
(ii).its slant height
(iii). its curved surface area
(iv).its total surface area
Solutions
(i)
Volume = πr2h
= x x 82 x 15
=1005.3cm2
(ii)
(iii)
Slant height =
= 17cm
Curved surface area = πrl
= π x 8 x 17
= 427.256
Page 127 of 173
= 427.26 cm2
(iv)
Total surface area =
+
= π (8)2+ 427.26
= 201.09 + 427.26
= 628.348 cm2
11. The volume of a cone =
x base area x height. The diagram shows a plant pot. The open
end of the plant pot is a circle of radius 10cm. the closed end is a circle of 5cm. The height of
the pot is 12cm. The plant pot is part of a right circular cone of height 24cm.
(i). calculate the volume of the plant pot. Give your in litres
(ii). A smaller plant pot is geometrically similar to the original plant pot. The open end of this
plant pot is a circle of radius 5cm.
How many of these plants pots can be completely filled from a 75 litre bag of compost?
Solution
(i) Volume of larger cone
=⅓ × πr2 × h
=2513.6 cm3
Volume of smaller cone
=⅓ × 3.142 × (5)2 × 12
Page 128 of 173
=314.2cm2
Volume of plant pot
=2513.6 314.2
=2199.4cm3
=1 litre = 1000cm3
=2.1994 litres
=2.20 litres
(i)
number of pots =
= 34.100
(ii)
number of pots completely filled = 34
=
3
3
=
=
=
Volume of small pot =
×2.1994
=0.2749 litres
No of pots =
= 272.826
No of pots completely filled= 272
The base of a pyramid is a square with diagonals of length 6cm.
The sloping faces are isosceles triangles with equal sides of length 7cm
The height of the pyramid is
cm. Calculate L
Page 129 of 173
Solution
Applying Pythagoras theorem
72=h2+32
49=h2+9
h2=40
h=
2
=40
Or L=40cm.
Paper 2 – STATISTICS
The Cumulative Frequency Curve
A cumulative frequency curve, also called the Orgive curve can be used to find the mean,
quartiles (lower, upper, interquartiles, semi-interquartiles) of a given distribution. To find the
cumulative frequency, find the accumulated totals and plot them against the data or score
values. The cumulative frequency is formed by joining the points with a smooth curve.
Question
Answer the whole of this question on a sheet of graph paper.
The waiting time for 55 passengers at the power tools bus station in Kitwe for them to board
a Lusaka bound bus on a particular day were as follows:Waiting time
(in minutes)
Number of
Passengers
1≤ 𝑥 ≤ 3
4≤𝑥 ≤6
7≤ 𝑥 ≤ 9
10≤ 𝑥 ≤ 12
13 ≤ 𝑥 ≤ 15
6
11
20
13
5
≤ 12
≤ 15
a) Calculate the estimate of the mean waiting time.
b) Copy and complete the cumulative frequency table below.
Waiting time
(in minutes)
≤3
≤6
≤9
Page 130 of 173
Number of
Passengers
6
17
55
c) Using a horizontal scale of 2cm to represent 2 minutes for times from 0 to 15 minutes and
a vertical scale of 2cm to represent 10 passengers.
Draw a smooth cumulative frequency curve.
d) Showing your method clearly, use your graph to estimate the
i.
Median
ii.
Lower quartile
iii.
Upper quartile
iv.
Interquartile range
v.
Semi-interquartile range
vi.
60th percentile
e) Find the number of passengers who waited for more than 6 minutes.
f) (i) If a passenger was chosen at random, find the probability that he waited for less than 9
minutes.
g) If two passengers were chosen at random. Find the probability that they both waited for
more than 12 minutes.
Relative Cumulative Frequency Table
The Table below shows a frequency table of the marks obtained by 120 pupils in a
Mathematics Test.
Marks
0 -4
5–9
10 – 14 15 – 19 20 – 24 25 – 29 30 – 34 35 – 39
40 – 44 45 – 49
Frequency
0
4
6
9
i.
ii.
10
14
24
28
19
Construct the relative cumulative frequency curve for the above mentioned data.
From the curve, estimate the 74th percentile.
a) Relative cumulative frequency =
Mark
𝐶𝑙𝑎𝑠𝑠 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
Frequency
0–4
0
5–9
4
𝑇𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
Cumulative
Frequency
0
4
Relative Cumulative
Frequency
0
=0
120
4
= 0.03
120
Page 131 of 173
6
10 – 14
6
10
10
= 0.08
120
15 – 19
10
20
20
= 0.17
120
20 – 24
14
34
34
= 0.28
120
25 – 29
24
58
58
= 0.48
120
30 – 34
28
86
86
= 0.72
120
35 – 39
19
105
105
= 0.88
120
40 – 44
9
114
114
= 0.95
120
45 – 49
6
120
120
= 1.00
120
1.00
0.9
0.8
74th Percent
Frequency
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Page 132 of 173
Variance and Standard Deviation
1.1
Ungrouped Data
Variance
This is a measure of spread that measures the distances or spread of data about the
mean. It is found by the formula.
Variance = ∑𝑛𝑖=1
(𝑥 − 𝑥)
𝑛
Where x is the mean of x1, x2, x3, …..., xi and R is the number of observations.
An alternative method that can be used to calculate the variance for ungrouped is
𝑥2
𝑛
Variance = ∑
1.2
− (𝑥)
2
Standard Deviation
The standard deviation is the positive square root of the variance.
Standard deviation = √𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒
(S.D)
Page 133 of 173
It can be calculated by the formula
Standard deviation = √∑(𝑥 − 𝑥) 2
𝑛
Alternatively, the standard deviation of the ungrouped data can be found by the
formula
Where 𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛
2
2
S.D = √∑ 𝑥 − (𝑥)
𝑛
Question 1
Calculate the variance and standard deviation of the following set of data.
8, 5, 10, 25 and 32.
Solution
Mean 𝑥 = ∑𝑥
𝑛
Mean 𝑥 =
8+5+10+25+32
80
5
x= 5
x = 16
Method 1
Using
Score
x
8
5
10
25
32
Ʃ 𝑥 = 80
Variance = ∑
(𝑥−𝑥)2
𝑛
Deviation
x–x
-8
-11
-6
9
16
Ʃ (𝑥 − 𝑥) = 0
(𝒙 − 𝒙)²
64
121
36
81
256
Ʃ (𝑥 − 𝑥)² = 558
2.0. GROUPED DATA
2.1 THE MEAN OF GROUPED DATA
Page 134 of 173
The mean of grouped data also called the estimate of the mean is denoted by x is calculated using the
formula
Mean (𝑥) = ∑𝑓𝑥
∑𝑓
2.2 THE STANDARD DEVIATION OF GROUPED DATA
The formula used to calculate the standard deviation of the grouped data is
Standard deviation = √∑ 𝑓(𝑥 − 𝑥) 2
𝑛
Where
𝑛 = ∑𝑓
𝑥 = ∑𝑓𝑥
∑𝑓
The alternative formula for calculating the standard deviation is
Standard deviation = √∑ 𝑓𝑥2 − 𝑥 2
∑𝑓
where
𝑥 = ∑𝑓𝑥
∑𝑓
Variance ==Mean (𝑥) = ∑(𝑥 − 𝑥)2
𝑛
= 558
5
=111.6 (Id.p)
S.D
= √𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒
= √∑(𝑥 − 𝑥)2
𝑛
= √558
5
=10.56 (2d.p)
Standard deviation = √∑ 𝑥 2 − (𝑥) 2
𝑛
Page 135 of 173
𝑛
8
64
5
25
10
100
25
625
𝒙 = 𝟏𝟔
32
1024
= 80
5
∑𝑥 = 80 ∑𝑥2=1838
Standard deviation = √∑ 𝑥2 − (𝑥)2
𝑛
= √1838 − 16 2
5
= √111.6
= 𝟏𝟎. 𝟓𝟔
Question
An intelligence quotient test that was taken by pupils at Kitwe Boys Secondary School in Kitwe
Showed the following results.
Time taken
(Minutes)
Frequency
0<x<1
1<x<2
2<x<3
3<x<5
5<x<10
10
15
25
40
25
Find
(a) The estimate of the mean
(b) The standard deviation
Solution
To calculate the estimate of the mean and standard deviation, we use the mid-internal or midpoint x
Time
0 < 𝑥 <1
1
.. < 𝑥 <2
𝒙
0.5
𝑓
10
𝑓𝑥
5
𝑓𝑥
2.5
2
Page 136 of 173
1.5
2 < 𝑥 <3 2.5
3 < 𝑥 <5 4
5 < 𝑥 <10 7.5
15
25
40
25
22.5
62.5
160
187.5
∑𝑓 = 115 ∑𝑓𝑥 = 437.5
33.75
156.25
640
1406.25
∑𝑓𝑥 = 2238.75
(a)estimate of the mean
𝑥 = ∑𝑓𝑥
∑𝑓
𝑥 = 437.5
115
𝒙 = 𝟑. 𝟖𝟎
(b)S.D
S.D= √∑ 𝑓𝑥 2 − (𝑥)2
∑𝑓
S.D= √2238.75 − 3.80 2
115
𝑆. 𝐷 = 2.2 (𝐼𝑑. 𝑝)
(c) (i) Median = ½ Q
Q2 = ½ x 55
=27.5
Q2 =7.8 Minutes
(ii) Lower Quartile = ¼ Q
Q1 = ¼ x55
= 13.75
Q1 = 5.2 Minutes
(iii) Upper quartile = ¾ Q
Q3 =3/4 x55
= 41.25
=9.7 Minutes
(iv) Interquartile range = Upper quartile – Lower quartile
= Q3 - Q1
9.7- 5.2
4.5 Minutes
(v) Semi – interquartile range = 1.2 x interquartile range
½ x 4.5 Minutes
2.25 Minutes
th
(Vii) 60 Percentile = 60 X 56
100
60% of Q = 33.6
= 8.45 Minutes
Page 137 of 173
(d) 55-17
=38 Pupils
P(> 6 Minutes) = 38
55
(e) (i) P(<9 Minutes ) = 37
55
(ii) P(< 12 Minutes and < 12 Minutes )
P(> 2Minutes) x P(< 12 Minutes )
5 X 4 =2
55
54 297
a)
Mean (𝑥) = ∑𝑓𝑥
𝑥 =
2x 6 + 5x11 + 8x20 + 11x12 + 14+5
∑𝑓
55
𝑥 = 12+55+160+132+70
= 429
55
55
𝑥 = 7.8 Minutes (I.d.p)
Waiting time (mm)
Frequency
<3
16
<6
17
<9
37
<12
50
<15
55
Number of Passengers
60
50
Upper Quartile
(Q )
60th Percentile
40
Median (Q2)
30
20
Lower Quartile
10
0
Page 138 of 173
2
4
6
8
10
12
14
16
GEOMETRICAL TRANSFORMATIONS
1. DEFINITIONS
A geometrical transformation moves an object(a point or a shape) from one position to
another. During a transformation some points will move while others may not move.
The new position is known as the image of the original object.
Those points which do not move are known as invariant points and the line on which
these points lie is the line of invariant points. At this level, there are seven types of
transformations that you are expected to know in preparation for your examinations.
These are: translation, reflection, rotation, dilatation(enlargement or reduction), stretch
and shear.
2. TRANSLATION
A translation is represented by a
21 column matrix or vector T  ba .
 
A translation is completely described by stating the column vector of T.
Finding the image under a Translation
The image of the point (x, y) is (x1, y1) and is defined by
 x1    x    a .
 y   y   b
 1
Examples
Page 139 of 173
(a) Find the image of the point (3,  2) under the translation T  6 .
2
 
(b) The image of the point (9, 1) is (4, 5) . Find the translation vector T.
(c) Under the translation T    7 , the image of the point A is (2, 5) .
5


Find the coordinates of the point A.
Solutions
(a)
 x1    x    a
 y   y   b
 1
(b)
 x1    x    a
 y   y   b
 1
 x1    x    a
 y   y   b
 1
(c)
 x1    3   6
 y    2  2
 1  
 4   9   a
 5  1  b 
 2    x    7
  5  y   5
   
 x1   
y  
 1 
 a  
 b 
 x    2     7
 y    5  5 
9
0
4   9
5  1
 a    5
 b  6 
The image is (9, 0) .
 T   65
 
 x    9 
 y   10
 A is
(9, 10)
Exercise 1
(a) Find the image of the point (8,  4) under the translation T  5 .
4
 
(b) The image of the point (9, 3) under a translation T is (7, 5) .
Find the translation vector T.
(c) Under the translation T   2 , the image of the point B is (2, 5) . Find the
3
 
coordinates of the point B.
(d) The image of the point (0, 1) under a translation T is (8, 8) .
Find the translation vector T.
(e) Under the translation T   1 , the image of the point C is (11, 5) . Find the
4


coordinates of the point C.
Page 140 of 173
Answers
(a) (13, 0)
(b) (16, 8)
(c) (0, 8)
(e) (12, 9)
(d) (8, 9)
----------------------------------------------------------------------------------------------------------------
3. REFLECTION
A reflection is completely described by giving the equation of the line of reflection.
The following table gives the matrices associated with the given line of reflection.
Line of Reflection
Matrix
The line y = 0
(or the x – axis)
Mx  1
0
The line x = 0
(or the y – axis)
My   1
 0
0
1
The line y = x
Myx   0
1
1
0
The line
y  x
0
1
Myx   0
 1
1
0 
Finding the image under a Reflection
The image (x1, y1) of the point (x, y) under a reflection is defined by
where M is the
22 matrix associated with the given
 x1   M x  ,
y 
 y
 1
reflection.
Page 141 of 173
Examples
(a) Find the image of the point (2, 5) under reflection in
(i) the line x = 0
(ii) the line y = x
(b) The point (2, 8) is the image of the point Q under reflection in the line
y  x .
Find the coordinates of Q.
Solutions
(a) (i)
 x1    1 0 2 
 y   0 1  5
 
 1
(ii)
 x1    2
 y   5
 1
 x1   0 1 2
 y  1 0  5
 
 1
 x1    5
 y   2
 1
The image is (2, 5)
The image is (5, 2)
  2   0 1 x     2    y 
  8  1 0 y 
  8   x 
(b)
2   y  y  2
8   x  x  8
The coordinates of Q are (8, 2) .
Exercise 2
(a) Find the image of the point (2, 5) under reflection in
(i) the line y = 0
(ii) the line
y  x
(iii) the y – axis
(b) The point (3, 12) is the image of the point R under reflection in the line
y x .
Find the coordinates of R.
(c) Under a reflection in the line x = 0, the point B is mapped onto the point (6, 15) .
Find the coordinates of B.
(d) Find the image of the point (4, 3) whenit is reflectedin theline y 2.
Page 142 of 173
Answers
(a)(i) (2, 5)
(ii) (5,  2)
(iii ) (2, 5)
(b) (12, 3)
(c) (6, 15)
(d) (4, 1)
4. ROTATION
A rotation is completely described by giving the centre, the angle and the
direction(clockwise or anticlockwise) of rotation.
The following table gives the matrices associated wih Rotations.
Rotation, Centre (0, 0)
Matrix
90° anticlockwise
(positive quarter turn)
R90  0
1
1
0
90° clockwise
(negative quarter turn)
R90   0
 1
1
0
180°
(Half Turn)
270°
(three quarter turn)
R180   1 0 
 0 1
R2700   0
 1
1
0
Page 143 of 173
Finding the image under a Rotation
(i) When the centre is the origin (0, 0)
The image (x1, y1) of the point (x, y) under a rotation, centre (0, 0), is defined by
 x1   R x  , where
 y
y 
 
 1
R is the
22 matrix associated with the given
rotation.
(ii) When the centre is (r, s), where r ≠ 0 and s ≠ 0
The image (x1, y1) of the point (x, y) under a rotation, centre (r, s), is defined
by
 x1   R x r  
y 
 y  s


 1
r  , where
 s
R is the
22 matrix associated with the
given rotation.
Finding the Centre and Angle of a Rotation
Examples
(a) Find the image of the point (4, 1) under a 90° negative quarter turn, centre (0, 0)
(b) Find the image of the point (2, 5) under a clockwise rotation, centre (2, 4).
Page 144 of 173
(c) The point (2, 8) is the image of the point Q under an anticlockwise rotation of 90°,
centre (1, 3). Find the coordinates of Q.
Solutions
(a)  0
1 4   1

1
0 1   4

The image is (1,  4)
(c)  1
 0
0  a  1   1   2
1 b  3 3   8
  a  1   1   2
  b  3 3   8


 a2 2  a4
(b)
0
1

  0
1
1 2  2    2
0  5  4   4
1 0    2
0  9   4

  9    2
4
0


 
  11 
4
The image is (11, 4)
b68  b14
 Q is (4, 14)
Exercise 3
(a) Find the image of the point (7, 1) under a 90° positive quarter turn, centre (0, 0).
(b) Find the image of the point (9, 3) under a half turn rotation, centre (2, 4).
(c) The point (4, 5) is the image of the point D under an anticlockwise rotation of 90°,
centre (1, 3). Find the coordinates of D.
5. DILATATION
Page 145 of 173
A Dilatation can either be an Enlargement (where an object is enlarged in size) or
a Reduction (where an object is reduced in size).
Enlargement
An Enlargement is represented by the
22 matrix E  k0 k0 .


The linear scale factor is k, where k ˃ 1 and the area scale factor is k2, which is
the determinant matrix E . The object and its image are in the ratio
1 : k 2 . This means
if the area of the object is A cm2, the area of the image will be k2 A cm2.
If the image of the line AB is A1B1 , then the linearscalefactork 
A1B1
.
AB
If k ˂ 0 (negative), the image is turned round. The transformation is equivalent to
a combination of an enlargement followed by a 180° rotation or vice-versa.
If k = 1, the object will remain where it is since, in this case, we shall have
the identity matrix 1
0 .
0 1
An enlargement is completely described by stating the centre of enlargement and
the linear scale factor.
Reduction
This has all the properties of Enlargement except that the lengths are reduced
by the same linear scale factor k, where
1 ˂ k ˂ 1.
Page 146 of 173
Finding the Centre of an Enlargement
Finding the image under an Enlargement
(i) When the Centre is the origin (0, 0)
The image (x1, y1) of the point (x, y) under an enlargement, centre (0, 0), is defined
by
 x1   k 0 x , where
 y   0 k  y
 1
 k
0
0
k  is the matrix associated with the enlargement
and k as the linear scale factor.
(ii) When the Centre is (r, s), where r ≠ 0 and s ≠ 0
The image (x1, y1) of the point (x, y) under an enlargement, centre (r, s), is defined
by
 x1   k 0 x r    r  , where
 y   0 k  y  s   s 


 1
 k
0
0
k  is the matrix associated with the
enlargement and k as the linear scale factor.
Examples
(a) Find the image of the point (4, 1) under an enlargement, scale factor 2, centre (0, 0) .
(b) Find the image of the point (2, 5) , under an enlargement, scale factor 3, centre (2, 4).
(c) The point (2, 8) is the image of the point Q under an enlargement scale factor
 3,
centre (1, 3). Find the coordinates of Q.
Solutions
Page 147 of 173
(a)  2
0
0 4   8
21  2 
The image is (8, 2)
(c)   3
 0
0  a  1   1   2
 3 b  3 3   8
  3a  3   1   2
  3b  9  3   8


3a4 2  a2
(b)
3
0

0  2  2    2
3  5  4   4
  3 0 0    2
 0 3  9   4 

  0    2
  27  4
  2 
  23
The image is (2,  23)
 3b 128  b 6 23
 Q is (2, 6 23 )
Exercise 4
(a) Find the image of the point (1,  2), under an enlargement, scale factor 2, centre (3, 1).
(b) Find the image of the point (4, 0) under an enlargement, scale factor 4, centre (0, 0) .
(c) The point (7, 10) is the image of the point C under an enlargement scale factor
2,
centre (5, 0). Find the coordinates of C.
6. STRETCH
Page 148 of 173
A Stretch with the x- axis as the invariant line is represented by the matrix S  1
0
0
k .
A Stretch with the y- axis as the invariant line is represented by the matrix S   k
0
0
1.


The stretch factor

distanceof image frominvariantline
distanceof objectfrominvariantline
.
A Stretch is completely described by stating the invariant line, the linear scale factor
and the direction.
----------------------------------------------------------------------------------------------------------------
7. SHEAR
A Shear with the x- axis as the invariant line is represented by the matrix S  1
0
k .
1
A Shear with the y- axis as the invariant line is represented by the matrix S   1
k
0
1.


The shear factor

distancemovedby point
distanceof thatpointfromthe invariantline
.
A Shear is completely described by stating the equation of the invariant line, the linear
scale factor and the direction.
Page 149 of 173
Exercise 5 : Miscellaneous
 1 0
 , B =
0 1
1.(a) A = 
 1 0

 , C =
 0 1
 6 
  , D =
 1
 1 0

 , F =
 0 1
 0

 1
1
.
0 
(i) Name the transformation that each of matrix A, B, C, D and F represents.
(ii) Find the image of P(2, 3) under
(a) a reflection in the line y = 0
(b) a positive quarter turn
(c) the translation given above
0 
 6 x
 represents an enlargement.
8  2 y 4  x 
(b) The matrix L = 
(i) Find the value of x and the value of y.
(ii) Write down the area scale factor.
-----------------------------------------------------------------------------------------------------------2. (a)
Triangle V is the image of ∆XYZ, with vertices X(1, 2), Y(4, 2) and Z(1, 3),
 1 0 
 .
 0 2
under the transformation given by the matrix A  
(i) Draw and label triangle V using a scale of 1 unit to 1 cm on each axis.
Take
 5 ≤ x ≤ 5 and  7 ≤ y ≤ 7 .
(ii) Triangle W is the image of triangle V under a rotation of 180˚ about
the origin O. Draw and label triangle W .
(iii) Determine the matrix representing the single transformation which maps
triangle V onto triangle W .
(b) The point (x, y) is mapped onto the point (x1 , y1 ) by the transformation D
 1  2
 2
 x1 
 x
 and B  
  A   B, where A  
 0 1
 5
 y
 y1 
described by 

 .

(i) Find the coordinates of the point P1 , the image of P(3, 2) under D.
Page 150 of 173
(ii) Q1 (1, 6) is the image of Q(a, b) under the transformation D .
Find the value of a and the value of b .
---------------------------------------------------------------------------------------------------------------3. Answer the whole of this question on a sheet of graph paper
Triangle A has vertices (4, 1), (4, -1) and (5, 1)
Triangle B has vertices (1, 4), (-1, 4) and (1, 5)
(i) Using a scale of 1 cm to represent 1 unit on each axis, draw axes for values
of x and y in the range -6 ≤ x ≤ 6 and -6 ≤ y ≤ 6 . Draw and label the
triangles A and B .
(ii) Describe fully the single transformation which maps triangle A onto
triangle B
 0 1
 maps triangle A
 1 0 
(iii) The transformation represented by the matrix 
onto triangle C. Draw and label triangle C .
(iv) Write down the matrix representing the transformation which maps
triangle B onto triangle C .
(v) Given also that triangle C is mapped onto triangle D by a translation given
 5
 3
by 

 , draw and label triangle D .

---------------------------------------------------------------------------------------------------------4. The points A(1,1) , B( 2, 3) and C(3, 2) are vertices of
ABC.
(a) Using a scale of 1 cm for 1 unit on each axis, draw and label
ABC.
 2 1
 .
S  
 0 1
ABC is transformed to A1B1C1 , where A1 , B1 and C1 are respectively the
images of A, B and C under the transformation with matrix S.
(b) (i) Find the coordinates of A1 , B1 and C1 .
Page 151 of 173
(ii) Draw and label A1B1C1 .
 0 2
 .
T  
 1 1
A1B1C1 is transformed to A2 B2C2 , where A2 , B2 and C2 are respectively
the images of A1 , B1 and C1 under the transformation with matrix T.
(c)
Draw and label A2 B2C2 .
An enlargement, centre O, followed by a rotation about O transforms
ABC
onto A2 B2C2 .
(d)
Find
(i) the scale factor of the enlargement,
(ii) the angle of rotation.
(e) Find the matrix of the transformation that maps
ABC onto A2 B2C2 .
----------------------------------------------------------------------------------------------------------5.(a)
Find the matrix for the stretch S parallel to the y – axis if the x – axis is invariant
and the point P(1, 2) is mapped onto Pʹ(1, 6).
(b) (i) Plot the points A(2, 1), B(3, 5) and C(5, 1).
(ii) If S is a stretch such that the y – axis is invariant and the point (1, 0) is mapped
onto (3, 0), plot the image of triangle ABC under under S.
(iii) Find the matrix S and state its determinant.
(iv) Find the ratio of the areas of the the two triangles.
--------------------------------------------------------------------------------------------------------------
TOPIC: INTRODRODUCTION TO CALCULUS
Calculus is a branch of mathematics which was developed by Newton (1642-1727)
and Leibnitz (1646-1716) to deal with changing quantities.
Page 152 of 173
EXPECTED OUTCOMES:
A: Differentiation.
1. Differentiate functions from first principles.
2. Differentiate functions using the formula
3. Calculate equations of tangents and normals
B: Integration
Find indefinite integrals
Evaluate simple definite integrals
Find the area under the curve
A: DIFFERENTIATION.
1. DIFFERENTIATING FUNCTIONS FROM FIRST PRINCIPLES.
EXAMPLES:
1. If
f  x  2x 5, f ' (x) from
first principle.
SOLUTION
f '(x)  lim
ho
f (x  h)  f (x)
h
DATA:
f (x)  2x 5
f (x h)  2(x h) 5[Plug in these functions in the formula above]
f '(x)  lim
ho
f (x  h)  f (x)
h
2(x  h)  5  (2x  5)
h
2x  2h  5  2x  5
 lim
ho
h
2h
 lim
ho h
 lim
2
ho
f '(x)  lim
ho
f '(x)  2
2. Find
dy from first principle for the function y  2x2 .
dx
SOLUTION:
Page 153 of 173
dy
f (x  h)  f (x)
 lim
dx ho
h
DATA.
f (x)  2x2
f (x  h)  2(x  h)2
[plug in these in the formula above]
f '(x)  lim
ho
f (x  h)  f (x)
h
dy
2(x  h)2  2x2
 lim
dx ho
h
2(x2  2xh  h2 )  2x2
 lim

ho
h
2
2x  4xh  2h2  2x2
 lim
ho
h
2
4xh  2h
 lim
ho
h
 lim4
x  2h
ho
dy
 4x (notethat ash o,2h  0)
dx
EXERCISE:
1. Find f '(x) for each of the following functions by first principle.
(a) f (x)  5x  4
(b) f (x)  x2 1
(c) f (x)  20x2  6x  7
Expected Answers:
(a) f '(x)  5
(b) f '(x)  2x
(c) f '(x)  40x  6
2. DIIFFERENTIATING FUNCTIONS USING THE FORMULA:
A. The Derivate of axn
Page 154 of 173
Given a function
y  f (x)  axn then it follows that dy  f '(x)  anxn1 .
dx
Examples
1. Given that y  7 , find
dy .
dx
SOLUTION:
y  7isthesameas y  7x0 sin cex0 1
y  7x0
dy
 (0)7x01
dx
dy
0
dx
NOTE THAT: The derivate of any constant is Zero (0)
2. Given that y  5x , find
dy .
dx
SOLUTION:
y  5xisthesameas y  5x1
y  5x1
dy
1(5)x11
dx
dy
1(5)x0
dx
dy
5
dx
3. Find the derived function of
y  2x4  5x3  x2  2
SOLUTION:
Page 155 of 173
y  2x4  5x3  x2  2
dy
 4(2)x41  3(5)x31  2x21  0
dx
dy
 8x3 15x2  2x1
dx
dy
 8x3 15x2  2x
dx
B. The Derivate of
(ax  b)n
The derivate of the function
y  (ax  b)n is given by the formula
dy
 n(ax  b)n1 a
dx
EXAMPLE:
1. If y  (3x  5)4 , find
dy .
dx
SOLUTION:
y  (3x  5)4
dy
 4(3x  5)41 3
dx
dy
12(3x  5)3
dx
C. The Derivate of a product. (Product Rule)
If
y  (ax  b)n (cx  d)m we can let u  (ax b) andv  (cx d) . From it follows that, the
derivative of a product is given by the formula
dy
du
dv
 v u
dx
dx
dx
Example:
1. Given that y  (3x 1)
2
(2x 5)3 , find dy .
dx
SOLUTION:
y  (3x 1)2 (2x 5)3
We let ---
Page 156 of 173
u  (3x 1)2 and v  (2x 1)3
du
dv
 2(3x 1) 3 and  3(2x 1)2  2
dx
dx
du
dv
 6(3x 1)
 6(2x 1)2
dx
dx
From the above it follows that =
dy du dv
 v u
dx dx dx
dy
 (2x 1)3 6(3x 1)  (3x 1)2 6(2x 1)2
dx
 6(2x 1)3 (3x 1)  6(3x 1)2 (2x 1)2
 6(2x 1)2 (3x 1)[2x 1 3x 1]
 6(2x 1)2 (3x 1)(5x)
 30x(2x 1)2 (3x 1)
C: THE DERIVATE OF A QUOTIENT
If y  f (x) is a ratio of functions u andv where u and v are also functions of x , the
derivative of the function y with respect to x is given by the formula
du dv
dy v dx  u dx vu 'uv '


dx
v2
v2
Example:
1. Differentiate
(x  3)2
.
(x  2)2
SOLUTION:
SINCE y 
We let
(x 3)2
(x  2)2
u  (x 3)2 and v  (x  2)2
Page 157 of 173
u '  2(x  3) and v '  2(x  2)
dy vu ' uv '

dx
v2
dy (x  2)2 2(x  3)  (x  3)2 2(x  2)

2
dx
 x  22 


2
2(x  2) (x  3)  2(x  3)2 (x  2)

(x  2)4
2(x  2)(x  3)[(x  2)  (x  3)]

(x  2)4
2(x  3)(x  2  x  3)

(x  2)3
2(x  3)(5)

(x  2)3
dy 10(x  3)

dx (x  2)3
EXERCISE:
1. Differentiate
y  6x2  3x2  x  9
2. Differentiate
(x3  4x)7
3. Differentiate
f (x)  (x 1)2 (x  2)3
EXPECTED ANSWERS:
1.
dy
 12x3  6x 1
dx
2.
dy
 7(3x2  4)(x3  4x)7
dx
3.
f '(x)  (x 1)(5x  7)(x  2)2
Page 158 of 173
TANGENTS AND NORMALS
If y  f (x) is a curve, we can find the gradient at any point on the curve. This gradient is
equal to the gradient of the tangent to the curve at that point.
If the gradient of the tangent is m1 and that of the normal line is
m2 it follows that
m1 m2 1
The tangent and the normal are perpendicular to each other at the point of contact.
Example (FINDING THE GRADIENT OF THE TANGENT AND THENORMAL)
1. Find the gradient of the tangent and the normal to the curve
y  3x2  4x 1 at the
point where x  4 .
SOLUTION
y  3x2  4x 1is the equation of the curve.
The gradient of the tangent is
dy = m
1
dx
m1  6x  4 and since the value of x  4
m1  6(4)  4
m1  24  4
m1  28
Page 159 of 173
To find the gradient of the normal, recall that
m1  m2  1
28 m2  1
1
m2 
28
The gradient of the tangent,
m2 
m1  28and the gradient of the normal,
1
28
Example (FINDING THE EQUATION OF THE TANGENT AND THE NORMAL)
1. Find the equation of the tangent and the normal to the curve
point where
y  3x2  4x 1at the
x  4.
SOLUTION
We have the value for x. So let’s find the corresponding value for y.
y  3x2  4x 1, x  4
y  3(4)2  4(4) 1
y  3(16) 16 1
The point is (4,65)
y  48 16 1
y  65
The gradient of the tangent is m1  28and that of the normal is m2 
1 at the point
28
(4,65)
Equation of the tangent
y  m1x  c
65  28(4)  c
65 112  c
c  47
y  28x  47
Equation of the normal
y  m2 x  c
1
65  (4)  c
28
1
65   c
7
1
65   c
7
456
c
7
1 456
y  x
28
7
EXERCISE
Page 160 of 173
1. Find the equation of the tangent and the normal to the curve
x2  4y at the point
(6,9).
Expected Answers: y  3x 9 for the tangent and y 
1
x 11for the normal.
3
INTEGRATION
The inverse of differentiation or the reverse of differentiation is called integration.
Since integration is the reverse of differentiation the following steps must be taken:1. Increase the power of the variable by 1.
2. Divide the term (variable term) by the new power.
3. Then finally add the arbitrary term C
Example (INDEFINITE INTEGRALS)
An indefinite integral must contain an arbitrary constant (C). An integral of the form
 f (x) dx is called an indefinite integral.
1. Integrate the following gradient functions
(a)
dy
 3x
dx
(b)
f '(x)  6x3  2x2  x
SOLUTION:
dy
 3x  3x1
dx
3x11
y
c
(a)
2
3x2
y  2
2
3
f '(x)  6x  2x2  x
f '(x)  6x3  2x2  x1
6x31 2x21 x11
f (x) 


c
4
3
2
6x4 2x3 x2
f (x) 

 c
4
3 2
3
2
1
f (x)  x4  x3  x2  c
2
3
2
(b)
EXERCISE
1. Integrate the following gradient functions
Page 161 of 173
(a) 5x2  x 1
(b) x6 3x4  2x2 1
EXPECTED ANSWERS:
5
3
1
2
1(a) y  x3  x  x  c
1
7
3
5
2
3
(b) y  x7  x5  x3  x +c
DEFINITE INTEGRALS
b

A definite integral is an integral performed between the limits. Thus A  f (x) dx is
a
an integral performed between the limiting values a and b for x.
NOTE that y 
dy
 dx   f '(x) dx
Example
1. Evaluate the definite integral of 4x3 1between x 1 and x  3
f (x)  4x3 1
3
3
1
1
 f (x) dx  (4x3 1) dx
4
 [ x4  x]13
4
 [x4  x]13
 (34  3)  (14 1)
 (81 3)  (11)
 78  0
 78
Integrals can be used to compute the area under a given curve.
Example
1. Find the area of the region bounded by the curve y  x
2
1, the ordinates x=1 and
x=2 and the x- axis.
Page 162 of 173
2
A   (x2 1) dx
1
x3
 x]12
3
23
13
 (  2)  ( 1)
3
3
8
1
 (  2)  ( 1)
2
3
1
A  3 units2
3
[
EXERCISE:
1. Find the area under the curve
y  x  x2 between x=1 and x=3 Expected Answer
2
12 units2
3
2. Find the area enclosed by the x – axis, the curve
y  3x2  2 and the straight lines
x=3 and x= 5. Expected Answer. 102 squared units.
INTEGRATION
Integration is the reverse of differentiation.
Since integration is the reverse process of differentiation, the standard integrals listed in
table 1 may be deduced and readily checked by differentiation.
Table 1. Standard integrals
(i) ∫ 𝑎𝑥 𝑛 𝑑𝑥 =
𝑎𝑥 𝑛+1
+𝑐
𝑛+1
𝑒𝑥𝑐𝑒𝑝𝑡 𝑤ℎ𝑒𝑟𝑒 𝑛 = −1
1
(ii) ∫ 𝑐𝑜𝑠𝑎𝑥 𝑑𝑥 = 𝑠𝑖𝑛 𝑎𝑥 + 𝑐
𝑎
1
(iii) ∫ 𝑠𝑖𝑛𝑎𝑥 𝑑𝑥 = − 𝑎 𝑐𝑜𝑠 𝑎𝑥 + 𝑐
(iv) ∫ 𝑠𝑒𝑐 2 𝑎𝑥 𝑑𝑥 =
1
𝑎
𝑡𝑎𝑛 𝑎𝑥 + 𝑐
1
(v) ∫ 𝑐𝑜𝑠𝑒𝑐 2 𝑎𝑥 𝑑𝑥 = − 𝑎 𝑐𝑜𝑡 𝑎𝑥 + 𝑐
1
(vi)∫ 𝑒 𝑎𝑥 𝑑𝑥 = 𝑎 𝑒 𝑎𝑥 + 𝑐
1
(vii) ∫ 𝑥 𝑑𝑥 = 𝐼𝑛 𝑥 + 𝑐
Page 163 of 173
QUESTIONS
1. Determine (a) ∫ 5𝑥 2 𝑑𝑥
(b) ∫ 2𝑡 3 𝑑𝑥
Solutions
(a) ∫ 5𝑥 2 𝑑𝑥 =
=
5𝑥 2+1
2+1
𝟓𝒙𝟑
𝟑
+𝑐
(b)
+𝒄
∫ 2𝑡 3 𝑑𝑥 =
=
ans.
3
2. Determine (𝑎) ∫ 𝑥 2 𝑑𝑥
2𝑡 3+1
3+1
𝟐𝒕𝟒
𝟒
+𝑐
+𝒄
𝒂𝒏𝒔.
(𝑏) ∫ 3√𝑥 𝑑𝑥
Solutions
3
(a) ∫ 𝑥 2 𝑑𝑥 = ∫ 3𝑥
−2
3𝑥 −2+1
𝑑𝑥 =
−2+1
+𝑐
(b) ∫ 3√𝑥 𝑑𝑥 = ∫ 3𝑥
1
2
1
𝑑𝑥 =
3𝑥 2
+1
+𝑐
1
+1
2
3
= −3𝑥
=
−𝟑
𝒙
−1
+𝑐
=
3𝑥 2
3
2
+𝑐
𝟑
= 2𝑥 𝟐+𝒄
+ 𝒄 𝒂𝒏𝒔.
= 𝟐√𝒙𝟑 + 𝒄
3. Determine (a) ∫ 4𝑐𝑜𝑠 3𝑥 𝑑𝑥
(b) ∫ 5𝑠𝑖𝑛2𝜃𝑑𝜃
Solutions
1
(a) ∫ 4𝑐𝑜𝑠 3𝑥 𝑑𝑥 = (4) (3) 𝑠𝑖𝑛3𝑥 + 𝑐
1
(b) ∫ 5𝑠𝑖𝑛2𝜃𝑑𝜃 = 5 (− 2) 𝑐𝑜𝑠2𝜃 + 𝑐
𝟒
𝟓
= 𝟑 𝒔𝒊𝒏𝟑𝒙 + 𝒄
= − 𝟐 𝒄𝒐𝒔𝟐𝜽 + 𝒄
4. Determine (a) ∫ 7𝑠𝑒𝑐 2 4𝑡 𝑑𝑡
(b) ∫ 3𝑐𝑜𝑠𝑒𝑐 2 2𝑥 𝑑𝑥
Solutions
1
(a) ∫ 7𝑠𝑒𝑐 2 4𝑡 𝑑𝑡 = (7) (4) tan 4𝑡 + 𝑐
1
(b) ∫ 3𝑐𝑜𝑠𝑒𝑐 2 2𝑥 𝑑𝑥 = (3) (− 2) 𝑐𝑜𝑡2𝑥 + 𝑐
𝟕
𝟑
= 𝐭𝐚𝐧 𝟒𝒕 + 𝒄
= − 𝒄𝒐𝒕𝟐𝒙 + 𝒄
𝟒
5. Determine (a) ∫ 5𝑒 3𝑥 𝑑𝑥
𝟐
2
(b) ∫ 3𝑒 4𝑡 𝑑𝑡
Page 164 of 173
Solutions
1
2
(a) ∫ 5𝑒 3𝑥 𝑑𝑥 =5 (3) 𝑒 3𝑥 + 𝑐
2
2
1
(b) ∫ 3𝑒 4𝑡 𝑑𝑡 = ∫ 3 𝑒 −4𝑡 𝑑𝑡 = (3) (− 4) 𝑒 −4𝑡 + 𝑐
𝟓
𝟏
= 𝟑 𝒆𝟑𝒙 + 𝒄
= − 𝟔 𝒆−𝟒𝒕 + 𝒄
𝟏
= − 𝟔𝒆𝟒𝒕 + 𝒄
3
6. Determine ∫ 5𝑥 𝑑𝑥
Solution
3
3 1
∫ 𝑑𝑥 = ∫ ( ) ( ) 𝑑𝑥
5𝑥
5 𝑥
𝟑
= 𝟓 𝑰𝒏𝒙 + 𝒄
ans.
Application of Integration
𝑑𝑦
7. Find 𝑦 given that 𝑑𝑥 = 2𝑥 − 3 and that 𝑦 = −4 𝑤ℎ𝑒𝑛 𝑥 = 1.
Solution
𝐼𝑓
𝑑𝑦
= 2𝑥 − 3, then 𝑦 = ∫(2𝑥 − 3)𝑑𝑥 = 𝑥 2 − 3𝑥 + 𝑐
𝑑𝑥
𝑤ℎ𝑒𝑛 𝑥 = 1 , 𝑦 = 1 − 3 + 𝑐 = −4 𝑠𝑜 𝑐 = −2
𝐻𝑒𝑛𝑐𝑒 𝑦 = 𝑥 2 − 3𝑥 − 2
8. The gradient of the tangent at a point on a curve is given by 𝑥 2 + 𝑥 − 2. Find the equation
of the curve if it passes through (2,1).
Solution
𝑑𝑦
Gradient = 𝑑𝑥 = 𝑥 2 + 𝑥 − 2
Then 𝑦 = ∫( 𝑥 2 + 𝑥 − 2)𝑑𝑥 =
𝑥3
3
+
𝑥2
2
− 2𝑥 + 𝑐
𝑤ℎ𝑒𝑛 𝑥 = 2,
𝑦=
1
8 4
+ −4+𝑐 = 1
3 2
𝐻𝑒𝑛𝑐𝑒 𝑐 = 3.
𝑻𝒉𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒊𝒔 𝒚 =
𝒙𝟑
𝟑
+
𝒙𝟐
𝟐
𝟏
− 𝟐𝒙 + 𝟑 𝒐𝒓 𝟔𝒚 = 𝟐𝒙𝟑 + 𝟑𝒙𝟐 − 𝟏𝟐 + 𝟐.
EARTH GEOMETRY
( i) THE EARTH
Page 165 of 173
hints:

The Earth is very nearly spherical.

Mean Radius of the Earth is 6 370 km or 3 437nm.

Longitude: The meridian that passes through Greenwich is called the prime
meridian. All longitudes run through the north pole and the south pole. All longitudes
are measured west or east of the Prime (Greenwich) Meridian.

Latitude: Latitude marks the distance (in degrees) given to a place north orsouthof
the equator

All latitudes parallel to the equator.

Among the latitudes, the equator is the only great circle.
LEARNER ACTIVITY:
In the diagram below,
Page 166 of 173
(i)
state the longitude of the points A, B, C, T and R.
(ii)
find the difference in longitude between A and T; A and R; E and N; B
and C.
(iii)
find the difference in longitude between A and B; B and T; Tand C; Cand
R.
(iv)
state the latitude of the points C, D, E, M, N and R.
(v)
find the difference in latitude between C and D; D and E; C and E; M and
N; Rand N.
DISTANCE BETWEEN POINTS ALONG THE SAME CIRCLE OF LONGITUDE.
NOTE:

If the two given points are in the same hemisphere,
𝜽 = 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒊𝒏 𝒍𝒂𝒕𝒊𝒕𝒖𝒅𝒆.

If the two given points are in different hemispheres,
𝜽 = 𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒊𝒓 𝒂𝒏𝒈𝒍𝒆𝒔 𝒐𝒇 𝒍𝒂𝒕𝒊𝒕𝒖𝒅𝒆𝒔.
i) Distance in nautical miles

along a Great circle, 1° of arc = 60 nm
∴distance between two points= 𝜽 × 𝟔𝟎𝑛𝑚.
where𝜃 is the difference in latitude between the given
points.
𝜽
ordistance between two points = 𝟑𝟔𝟎 × 𝟐𝝅𝑹 𝑛𝑚.
where θ is the difference in latitude and R= 3 437nm
is the radius of the Earth.
ii) Distance in kilometres (km)

Radius of the Earth in kilometres = 6 370km
Page 167 of 173
∴ 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒘𝒐 𝒑𝒐𝒊𝒏𝒕𝒔 =
𝜽
× 𝟐𝝅𝑹 km
𝟑𝟔𝟎
with R= 6370 km.
iii) distance between any two points on the equator:
NOTE: Distance is calculated as in i) and ii), but
𝜽is the difference in Longitudebetween the
two given points for points on the same side
of the Prime Meridian, OR the sum of their
Longitudes, if they are on opposite sides of
the Prime Meridian.
iv) distance between points on the same circle of Latitude
other than the Equator.

Fomulae:
i) distance = 𝜽 × 𝟔𝟎𝑛𝑚 × 𝒄𝒐𝒔𝜶 𝑛𝑚, where 𝜽is the difference
OR sum in longitude and 𝜶is the
angleof Latitude.
𝜽
ii) distance= 𝟑𝟔𝟎 × 𝟐𝝅𝒓,but r=R cos 𝜶
𝜽
∴distance== 𝟑𝟔𝟎 × 𝟐𝝅𝑹𝒄𝒐𝒔𝜶
NOTE: In nm use R= 3437nm; in km use R= 6370km. and let 𝝅=
𝟐𝟐
𝟕
or 3.142.
ACTIVITY 1.
Find the distance along a circle of latitude between P(40°N, 30°E) and
Q(40°N, 50°E).
EXPECTED ANSWERS:
-
difference in longitude=50° − 30° = 20°
Page 168 of 173
-
latitude= 40°
distance= 𝜃 × 60 × 𝐶𝑂𝑆𝛼
∴distance= 20° × 60 × 𝐶𝑂𝑆40°
∴distance=919 nm.
ACTIVITY 2.
The diagram below shows a wire model of the earth, The circle of latitude in the north is
80°𝑁 and the circle of latitude 0°. The meridian 𝑁𝑃𝑅𝑆 is 60°𝐸 and meridian 𝑁𝑄𝑇𝑆 is
directly opposite 𝑁𝑃𝑅𝑆.
i) State the position of the point 𝑄.
ii) Find the distance along the circle of latitude 80°𝑁 between 𝑄 and 𝑃 in 𝑘𝑚.
iii) Calculate the shortest distance between 𝑃 and 𝑄
iv) If the time at 𝑅 is 20: 00ℎ𝑟𝑠, what is the time at 𝑇
EXPECTED ANSWERS:
i.
∴ 𝑄(80𝑁, 120𝑊)
ii.
180
× 2 × 3.142 × 6370 × 𝑐𝑜𝑠 80
360
= 3475.5𝑘𝑚
= 𝟑𝟒𝟖𝟎𝒌𝒎 (3 𝑠. 𝑓)
=
iii. SHORTEST DISTANCE BETWEEN P AND Q.
𝜃
𝐷=
× 2𝜋𝑅
360
Page 169 of 173
20
× 2 × 3.142 × 6370
360
𝐷 = 2224𝑘𝑚
𝐷 = 𝟐𝟐𝟐𝟎𝒌𝒎(3 𝑠. 𝑓)
𝐷=
iv.
180
15
= 12 ℎ𝑜𝑢𝑟𝑠
∴ 20: 00 − 12: 00 = 𝟎𝟖: 𝟎𝟎 𝒉𝒐𝒖𝒓𝒔.
ACTIVITY 3.
The diagram below is a sketch of the earth and on it are the points
P(20°N,80°E), Q(40°S, 80°E) and R(40° S, 30° E). [Use 𝜋 = 3.142 𝑎𝑛𝑑 𝑅 =
6370𝑘𝑚]
(i)
(ii)
Calculate the distance QR in kilometres.
An aeroplane starts from P and flies due west on the same latitude
covering a distance of 1 232km to point T.
(a) Calculate the difference in angles between P and T,
(b) Find the position of T.
EXPECTED ANSWERS:
(i)
Dist.𝑃𝑄 =
𝜃
360°
DISTANCE OF PQ:
-difference in longitude= 80° − 30°
= 50° = 𝜃,
𝑃𝑄 𝑖𝑠 𝑜𝑛 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 = 40°𝑆 = 𝛼,
× 2𝜋𝑅𝑐𝑜𝑠𝛼
50°
× 2 × 3.142 × 6370𝑐𝑜𝑠40°
360°
∴ Distance PQ = 4258.8964 km
≈ 𝟒𝟐𝟓𝟖. 𝟗𝒌𝒎 (1 𝑑. 𝑝. )
∴ Dist. 𝑃𝑄 =
(ii)
𝜃
(a) Distance = 360° × 2𝜋𝑅𝑐𝑜𝑠𝛼, but Dist= 1232km,
R= 6370km,
𝛼 = 20°.
∴ 1232 =
𝜃
× 2 × 3.142 × 6370𝑐𝑜𝑠20°
360°
Page 170 of 173
104.4861975
1232
∴ 𝜃 = 11.79103106
≈ 𝟏𝟏. 𝟖°
∴ difference in angles between P and T≈ 𝟏𝟏. 𝟖°
∴ 𝜃=
(𝑏)Position of T is:
𝑻(𝟐𝟎°𝑵, 𝟔𝟖. 𝟐°𝑺)
80° − 𝟏𝟏. 𝟕𝟗𝟏𝟎° = 𝟔𝟖. 𝟐𝟎𝟗°
∴
ACTIVITY 4.
In the diagram below, A(65°N, 5°E), B(65°N, 45°W) and Care three
points on the surface of the model of the earth and O is the centre of the
model. The point C due south of A, is such that AOC = 82° .
[𝜋 = 3.142, 𝑹 = 3437nm]
(i)
(ii)
(iii)
State the longitude of A,
Calculate the latitude of C,
Calculate, in nautical miles, the shortest distance
(a) between A and C and measured along the common longitude,
(b) betweenA and B measured along the circle of latitude.
EXPECTED ANSWERS:
(i)
(ii)
(iii)
Longitude of A = 𝟓° 𝑬
Latitude of C = 82° − 65° = 17° 𝑆
(a)𝟏° 𝒐𝒇 𝒂𝒓𝒄 = 𝟔𝟎𝒏𝒎,
difference in latitude= 82° = 𝜃,
∴ Shortest distance between A and C:
𝜃 × 60𝑛𝑚 = 82° × 60𝑛𝑚 = 𝟒𝟗𝟐𝟎 𝒏𝒎.
(b)difference in longitude between A and B
=45°𝑊 + 5°𝐸 = 𝟓𝟎°.
∴ distance along circle of latitude 65°𝑵, between A and B
= 𝜃 × 60𝑛𝑚 × 𝑐𝑜𝑠𝛼
= 50° × 60𝑛𝑚 × 𝑐𝑜𝑠65°
Page 171 of 173
= 𝟏𝟐𝟔𝟕. 𝟖𝟓𝟒𝟕𝟖𝟓 ≈ 𝟏𝟐𝟔𝟕. 𝟗𝒏𝒎(1 d.p.)
ACTIVITY 5
In the diagram below, the points P and Q lie on the same latitude, O is the Centre of
the earth and angle NOQ = 60°. (𝑇𝑎𝑘𝑒 𝜋 = 3.142 𝑎𝑛𝑑 𝑅 = 6370𝑘𝑚)
(i)
(ii)
(iii)
State the latitude where the points P and Q are lying.
Find the distance between P and T.
Given that the point P is on longitude 18°𝑊 and the time difference
between P and Q is 5hours, calculate the longitude on which Q lies.
EXPECTED ANSWERS:
(i)
Latitude where P and Q lie = 90° − 60°
= 𝟑𝟎° N.
(ii)
(iii)
difference in latitude = 30° − 0°
= 30° = 𝜃.
𝜃
∴ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑷 𝑎𝑛𝑑 𝑸 =
× 2𝜋𝑅
360°
30°
=
× 2 × 3.142 × 6370𝑘𝑚
360°
= 𝟑𝟑𝟑𝟓. 𝟕𝟓𝟔𝟔𝟔𝟕𝒌𝒎
= 𝟑𝟑𝟑𝟓. 𝟕𝟔𝒌𝒎 (2 𝑑. 𝑝. )
Note: 𝟏𝒉𝒓 = 𝟏𝟓°.
5hr = y°
∴y° = 5ℎ𝑟𝑠 𝑡𝑖𝑚𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝟕𝟓°
∴ 𝒍𝒐𝒏𝒈𝒊𝒕𝒖𝒅𝒆 𝑜𝑓 𝑸 = 75° − 18°
∴ 𝒍𝒐𝒏𝒈𝒊𝒕𝒖𝒅𝒆 𝑜𝑓 𝑸 = 57°.
ACTIVITY 6
The diagram below shows a wire model of the earth. The circle of latitude in the north
is 50°𝑵and the circle of latitude in the south is 60°𝑺. A and C are on longitude 55°W
while B and D are on longitude 50°E.
(Take 𝜋 = 3.142 𝑎𝑛𝑑 𝑅 = 3437𝑛𝑚)
Page 172 of 173
(i)
(ii)
(iii)
(iv)
Write the positions, using longitudes and latitudes, of the points A and D.
Calculate the difference in longitudes between A and B.
Given that the time at town D is 09 20 hours, what would be the time at town C?
Calculate the distance BD along the longitude 50°E in nautical miles.
EXPECTED ANSWER:
(i)
A(50°N, 55°W) and D(60°S, 50°E)
(ii)
Difference in longitudesbetween A and B= 55° + 50° = 𝟏𝟎𝟓°
(iii) TIME AT C=
note: C(60°S, 55°W), D(60°S, 50°E)
Difference in longitudesbetween C and D=55° + 50° = 𝟏𝟎𝟓°
note: 1hr = 15°
xhrs= 𝟏𝟎𝟓°
∴ xhrs =
𝟏𝟎𝟓°
15°
= 7hrs time difference,
∴ TIME AT C= 09 20hrs −7ℎ𝑟𝑠(C is West of D, so subtract)
∴ TIME AT C= 02 20hrs.
(iv)
Difference in latitudes= 50° +60° = 110°,
∴ 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝑩 𝒂𝒏𝒅 𝑫 = 110° ×60nm= 𝟔𝟔𝟎𝟎𝒏𝒎.
Page 173 of 173