Mathematics lesson note for SS1 Third Term is now available for free. The State and Federal Ministry of Education has recommended unified lesson notes for all secondary schools in Nigeria, in other words, all private secondary schools in Nigeria must operate with the same lesson notes based on the scheme of work for Mathematics.

Mathematics lesson note for SS1  Third Term has been provided in detail here on schoolings.org

For prospective school owners, teachers, and assistant teachers, Mathematics lesson note is defined as a guideline that defines the contents and structure of Mathematics as a subject offered at SS level. The lesson note for Mathematics for SS stage maps out in clear terms, how the topics and subtopics for a particular subject, group works and practical, discussions and assessment strategies, tests, and homework ought to be structured in order to fit in perfectly, the approved academic activities for the session.

To further emphasize the importance of this document, the curriculum for Mathematics spells out the complete guide on all academic subjects in theory and practical. It is used to ensure that the learning purposes, aims, and objectives of the subject meant for that class are successfully achieved.

Mathematics Lesson note for SS1 carries the same aims and objectives but might be portrayed differently based on how it is written or based on how you structure your lesson note. Check how to write lesson notes as this would help make yours unique.

The SS1 Mathematics lesson note provided here is in line with the current scheme of work hence, would go a long way in not just helping the teachers in carefully breaking down the subject, topics, and subtopics but also, devising more practical ways of achieving the aim and objective of the subject.

The sudden increase in the search for SS1 Mathematics lesson note for Third Term is expected because every term, tutors are in need of a robust lesson note that carries all topics in the curriculum as this would go a long way in preparing students for the West African Secondary Examination.

This post is quite a lengthy one as it provides in full detail, the Mathematics-approved lesson note for all topics and sub-topics in Mathematics as a subject offered in SS1.

Please note that Mathematics lesson note for SS1 provided here for Third Term is approved by the Ministry of Education based on the scheme of work.

I made it free for tutors, parents, guardians, and students who want to read ahead of what is being taught in class.

SS1 Mathematics Lesson Note (Third Term) 2025

SS1 THIRD TERM: E-LEARNING NOTES

 

SCHEME THIRD TERM

WEEK	TOPIC	CONTENT
1	LOGICAL REASONING	(a) Simple statements. (b) Meaning of simple statement – (i) True or false (ii) negation of simple statements. (c) Compound statements (i) Meaning (ii) Conjunction (iii) Disjunction (iv) Implication (v) bi-implication. (d) Logical operators and symbols. (i) list of logical operators and symbols (ii) Truth value of logical operators.
2	TRIGONOMETRY 1	(a) Basic Trigonometric Ratios – (i) sine (ii) cosine (iii) tangent with respect to right-angled triangles. (b) Trigonometric ratio of; – (i) Angle 300 (ii) Angle 450 (iii) Angle 600.
3	TRIGONOMETRY 2	(a) Application of trigonometric ratios (angle of elevation and depression; bearing). (b) Trigonometric ratios related to the unit circle. (c) Graphs of sines and cosines.
4	MENSURATION 1	(a) Length of arcs of circle. (b) Perimeter of sectors and segments. (c) Areas of sectors of a circle. (d) Areas of segments of a circle.
5	MENSURATION 2	(a) Relation between the sector of a circle and the surface area of a cone. (b) Surface area and volume of solids; (i) Cube, cuboids (ii) Cylinder (iii) Cone (iv) Prisms (v) Pyramids.
6	MENSURATION 3	(a) Surface areas and volume of frustum of a cone and pyramid. (b) Surface area and volume of compound shapes.
7	MID-TERM BREAK.
8	DATA PRESENTATION	(a) Revision on collection, tabulation and presentation of data. (b) Frequency distribution (grouped data).
9	DATA PRESENTATION	(i) Line graph, (ii) Bar graph. (iii) Histograms. (iv) Pie Chart. (v) Frequency polygon. (vi)Deductions and Interpretations.
10	REVISION
11	EXAMINATION

WEEK 1 Date……………………………………

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: LOGICAL REASONING

CONTENT:

• (a) Simple statements.
• (b) Meaning of simple statement – (i) True or false (ii) negation of simple statements.
• (c) Compound statements (i) Meaning (ii) Conjunction (iii) Disjunction                                  (iv) Implication (v) bi-implication.
• (d) Logical operators and symbols. (i) List of logical operators and symbols (ii) Truth value of logical operators.

 

• SIMPLE STATEMENTS

Logic is the science of thinking about or explaining the reason for something. It is a particular method or system of reasoning which arrive at conclusions by way of valid evidence.

Mathematical logic can be defined as the study of the relationship between certain objects such as numbers, functions, geometric figures etc.  .

Example: The following are logical statements;

1. Nigeria is in Africa
2. The river Niger is in Enugu
3. 2 + 5 = 3
4. 3 ≤ 7

(N.B The educator should ask the students to give their examples)

Example: The following are not logical statements because they are neither true nor false.

1. What is your name?
2. Oh what a lovely day
3. Take her away
4. Who is he?
5. Mathematics is a simple subject (note that this statements is true or false depending on each individual, so it is not logical).

Educator to ask the students to give their own examples.

• MEANING OF SIMPLE STATEMENT.

Statements are verbal or written declarations or assertions. The fundamental (i.e logical) property of a statement is that it is either true or false but not both. So logical statements are statements that are either reasonably true or false but not both

True or False statements:                                                                                                                                                                       To determine the truth or falsity of a simple statement, one requires pre-knowledge and/or definition of the main concepts related to the statements. For example, the simple statement ‘it is hot’ is true if ‘it’ refers to a hot object or weather. Otherwise the statement is false. A true statement is said to have a truth value T while a false statement is said to have a truth value F.

Example: indicate T or F for the truth value of each statement.

1. is equal to 10
2. Green is one of the colours on the Nigerian flag
3. How far is Abuja from here?
4. 3 {2, 4, 6, 8, …}
5. The perimeter of a room 2.5m by 3.5m is 6m

Solution:

1. F
2. T
3. Not applicable
4. F
5. F

(note: educator to explain closed statements and open statements as in question 3)

NEGATION OF SIMPLE STATEMENTS                                                                                                                          The opposite of a statement is called the negation of the statement. Given any logical statement P, the negation (or the contradiction or the denial) of P is written symbolically as P

Examples: write the negation of each of the following statements.

1. I am a Mathematician
2. 2 > 4
5. The sky is the limit

Solution:

1. I am not a mathematician
2. 2≯4
5. The sky is not the limit

 

 

Truth table for P:

 

P	P
T	F
F	T

 

Class Activity:

Write the negation of the following statements:

1. All polygons are quadrilaterals
2. It is a sunny day
3. XYZ is an isosceles right angled triangle
4. The figure is a cube
5. X is not a prime number

Compound statements:

       When two or more simple statements are combined, we have a compound statement. To do this, we use the words: ‘and’, ‘or’, ‘if … then’, ‘if and only if’, ‘but’. Such words are called connectives.

Conjunction (or ˄) of logical statements:

Any two simple statements p,q can be combined by the word ‘and’ to form a compound (or composite) statement ‘p and q’ called the conjunction of p,q denoted symbolically as p˄q.

Example: 1. Let p be “The weather is cold” and q be “it is raining”, then the conjunction of p,q written as p˄q is the statement “the weather is cold and it is raining”.

2. The symbol ‘˄’ can be used to define the intersection of two sets A and B as follows;

 

The truth table for p˄q is given below;

P	Q	P˄Q
T	T	T
F	F	F
T	F	F
F	T	F

Class Activity:

Form compound statements using ‘and’, and express the following compound statements in symbol form.

1. P: It is cold.

Q: It is wet.

2. P:

Q: = -3

3. P: f(= ² + 2

Q: f(1) = 7

4. P: ² is a perfect square.

Q: when x = 1, (+2)² = 9

Disjunction (or ˅) of logical statements.                                                                                                                                         Any compound statement formed by using the word ‘or’ to combine simple statements is called a disjunction. The symbol ‘˅’ stands for ‘or’.

Examples

1. Let `p ‘ be “Bola studied Mathematics”, and `q’ be “Ngozi studied French”. Then the disjunction of p, q (p˅q) is the statement “Bola studied Mathematics or Ngozi studied French”.
2. p: You will read your notes .                                                                                                                                          q: You will fail                                                                                                                                                                                p˅q: You will read your notes or fail
3. P:

Q:

P ˅ Q :

4. P: The solution of

Q: The solution of -3.

P˅Q: The solution of

 

The truth table for p˅q is illustrated below

P	Q	P˅Q
T	T	T
F	F	F
T	F	T
F	T	T

Class Activity:                                                                                                                                                                              Express the compound statements in symbolic form.

1. P: is a rational number   , Q:  is an even number.
2. P: the trade union is stubborn, Q: the workers strike will soon be ended.
3. P: 5 is a prime number, Q: 7 is an even number.
4. P: a person who has taken physics can go to geophysics, Q: a person who has taken geology can go for geophysics.

Implications (conditional statements)                                                                                                                                   When the connective ‘if…then’ is used to combine simple statements, the result is called an implicative or conditional proposition. We denote implication symbolically by  i.e pq means if p is true, then q is true. (or p implies q or p only if q, etc.)

Examples: Form compound statements using ‘if … then’

1. P: The triangle is an equilateral triangle

Q: The angles are equal

P Q: if the triangle is equilateral then the angles are equal.

2. P:

Q:

P Q: ifthen

3. P: Isa is a Mathematician .

Q: He is intelligent.

P Q: if Isa is a mathematician then he is intelligent.

N.B : Educator to explain antecedent and consequent and Examples should be given.

The truth or falsity of the implication P Q: is shown below;

P	Q	P Q
T	T	T
T	F	F
F	T	T
F	F	T

 

Class Activity:

Form compound statements using ‘if … then ’

1. P: y = 2

Q:

2. P: A student reads Mathematics

Q: the student reads science

3. P: Damilola is a youth corper

Q: she has a degree

4. Identify the antecedent and consequent in the statement below;

If Mathematics teachers work very hard then they will be compensated.

Bi-implication or Bi-conditional statement (equivalence).                                                                                                             Another common statement in Mathematics is of the form “p if and only if q”. This statement is actually the combination of two conditional statements and so it is called bi-conditional or equivalence and is denoted by  or sometimes p iff q (if and only if) i.e implies and is implied by.

Examples:                                                                                                                                                                                                      1.  Let p be “he is a handsome man” and q be “10 > 6” then  is the statement “he is a handsome man if and    only if 10 > 6”, then

p

2. P: A number is divisible by 3

Q: the sum of the digits of the number is divisible by 3

A number is divisible by 3 iff the sum of its digits is divisible by 3

The truth table for  is shown below;

P	Q
T	T	T
T	F	F
F	T	F
F	F	T

 

 

Logical operators and symbols

The word ‘not’ and the four connectives ‘and’, ‘or’, ‘if … then’, ‘if and only if’ are called logic operators. They are also referred to as logical constants. The symbols adopted for the logic operators are given below.

                       Logic Operators      Symbols

‘not’

                                          ‘and’        

‘or’         ˅

‘if … then’

‘if and only if’    ↔

When the symbols above are applied to propositions p and q, we obtain the representations in the table below:

Logic operation          Representation

‘not p’                       p or

‘P and q’                       p˄q

‘p or q’                         p˅q

‘if p then q’                     pq

‘p if and only if q’                    p↔q

 

PRACTICE EXERCISE:

1. What are the Truth values of this compound statement? (P ˄ Q)
2. Determine the truth value of the compound statement
3. Use a truth table to prove that; is a Tautology.
4. Copy and Complete the table below;

P	Q	Q	P ⇒ Q	P ˅Q	(P⇒Q)↔(P˅Q)
T	T
T	F
F	T
F	F

 

5. If P and Q are two logical statements, copy and complete the following truth table

P	Q	P˅Q	(P˅Q)	(P˅Q) ˄ Q	(P˅Q) ⇒ P

 

ASSIGNMENT:

1. If P and Q are two logical statements, copy and complete the following truth table

P	Q	P ˅ Q	(P ˅ Q)	(P ˅ Q) ˄ P	(P ˅ Q) ⇒ P ˄Q)

 

2. Study the following;
3. Antecedent
4. Consequent

• Converse, inverse and contrapositive statements

1. Tautology and contradiction

WEEK 2

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: TRIGONOMETRY 1

CONTENT: 

• (a) Basic Trigonometric Ratios – (i) sine (ii) cosine (iii) tangent with respect to right-angled triangles. (b) Trigonometric ratio of; – (i) Angle 30⁰(ii) Angle 45⁰                 (iii) Angle 60⁰.

Basic trigonometric ratios (i) Sine  (ii) Cosine  (iii) Tangent with respect to right-angled triangles.

These trigonometric ratios are applicable to right – angled triangle. A right – angle triangle is 90⁰. Thus the remaining two angles add up to 90⁰ since every triangle contains two right angles.

B

 

c      a

 

A          b               C

In ∆ABC, B + C = 90⁰.

Such angles whose sum is 90⁰ are said to be complementary angles. While capital letter are used for angles, small (lower case) letters are used for sides. Notice that the side opposite A is labelled a, the one opposite B is labelled b etc.

The side opposite the right angle is called the hypotenuse. Every right – angled triangle obeys the Pythagoras theorem. This theorem states that the square of the hypotenuse of any right angled triangle is equal to the sum of the square of the other two sides.                                                                                         Thus in the above triangle,

 

Apply Pythagoras theorem to the right – angled triangles below to find the lettered sides.

12

5 x            y 17

 

15

There are six basic trigonometric ratios viz: sine, cosine, tangent, cosecant, secant and cotangent. The first three are commonly used.

They are applicable only to right – angled triangles. Their short forms are: Sin, Cos, tan, cosec, sec, and cot respectively.

B

 

c       a

 

A         b               C

in the figure above, sin B =

sin C =

In short, sine =

Similarly, cos B =

Cos C =

In short, Cosine =

Also Tan B =

What is Tan c ?

In short, tan =

Using the first letter of these three words of three formulae, we have SOH CAH TOA

SINE

The trigonometric ratio sine, is opposite divided by hypotenuse. Its reciprocal is cosecant.

Example 1:

 

x          17

15

Find sin  in the above figure.

Solution

Sin =

The opposite side of  is not given. By using Pythagoras theorem, it can be found.

Let the opposite side to  be x

By Pythagoras theorem

x² + 15² = 17²

x² + 225 = 289

x² =  289 – 225

x² = 64

x = 8

Sin  =

Example 2:

 

X           10m

22⁰

 

Find x.

Solution

Since the opposite of the given angle is known and the hypotenuse of the triangle, one can use sine ratio.

Sin 22⁰ =

x = 10m x sin 22⁰

= 10m x 0.3746

= 3. 746m

Class Activity:

1. Find SinB, SinC in the figure below.

B

 

15

A           9             C

(2) In right – angled triangle XYZ, with Z = 90⁰. If l XY l = 5m and l YZ l = 3m. Find (i) Sin x  (ii) y

(3)  If a ladder of length 2m leans against a wall and makes 30⁰ with the floor, how high above the

floor does the ladder reach on the wall?

(4) New General Maths SS1 Ex. 11a;  No. 1, 2a, b, d, 3a, b

COSINE

The trigonometric ratio cosine is adjacent divided by hypotenuse. Its reciprocal is secant.

B

a

c

A   b          C

 

In the triangle above, B and C are complementary Sin B = b/a = Cos C

Also Sin C = c/a = Cos B

For complementary angles, the sine of one is the cosine of the other i.e

Sin  = Cos (90 – )

Example 1:

Solve sin2x = cos3x

Solution

Since Sin  = cos (90 – )

So, sin2x = cos (90 – ) (i)

But we are given that

Sin2x = cos3x (ii)

From the right hand sides of equation (i) and (ii)

We conclude that

90⁰ – 2x  = 3x

90⁰  = 5x

So x = 90/5 = 18⁰.

Example 2.

The angle of elevation of the top of a tree is 60⁰. If the point of observation is 4m from the foot of the tree. How far is the point from the top of the tree?

Solution

x

 

4m

Let the point of observation from the top of the tree be x metres.

Cos = 4/x

0.5 = 4/x

0.5x = 4

x = 4  0.5

= 8m.

Example 3

Given that Cos x = 0.7431, 0 < x < 90⁰, use tables to find the values of (i) 2sinx (ii) tan x/2

Solution

Cos x = 0.7431

X = cos^-1 0.7431

= 42⁰.

• 2Sinx = 2Sin42⁰

= 2(0.6691)

= 1.3382

• Tan x/2 = tan 42/2

= tan 21⁰

                    = 0.3839

 

Class Activity:

1. Find the unknown sides of the following triangles.

y                                         21⁰

12       68⁰

10               15       z

60^{0                                     x}                                                         

2. If sin = 0.3970 use the tables to find (i) cos   (ii) tan

TANGENT

 

Tan =

Example 1:

Find y in the figure

 

 

y

32⁰

6

Tan 32 = y/6

y = 6tan 32

= 6 x 0.6249

= 3.75

 

 

Example 2:

Find x in the figure  below

 

 

x

35⁰

x + 15

Tan 35⁰=

0.7002 =

x = 0.7002 (x+15)

x(1 – 0.7002) = 0.2998x

x       = 10.503

x =

= 35.03

Example 3.

If sin x = 3/5 and x is an acute angle, find cosx + tanx

Solution

 

3 5

x

4

Since sin x = 3/5

Opposite = 3, hypotenuse = 5,

Let the adjacent be a, by Pythagoras theorem,

a² +  =

a² = 25 – 9

a² = 16

Hence a = 4

so, cos x + tan x   = 4/5 + ¾

= (16+15)/20

= 31/20

= 1¹¹/_20.

Class Activity:

1. If Cos P = 4/5 and P is an acute angle, what is the value of tan P?
2. Given that tan x = 8/15. What is the value of sin x + cos x?

 

 

 

 

 

 

Angles 30⁰ and 60⁰.

To consider the trigonometric ratios for the special angles 30⁰ and 60⁰, we shall consider an equilateral triangle of side 2units. If the triangle is bisected we shall have

30⁰

 

2                      2                         2           h

60⁰               60⁰

1 1            1

If the value of the altitude h, can be obtained using Pythagoras theorem.

h² +

h² = 4 – 1

h=

consequently, Sin30⁰ = opp/hyp = ½

Cos30⁰ = adj/hyp

= /2

Tan30⁰ = opp/adj

=

Sin 60⁰ = /2

Cos 60⁰ = ½

Tan 60⁰ = /1

=

 

 

 

Example 1

Find x and y in the figure below

y

 

4           x

60⁰                     

Look at the figure with respect to the given 60⁰, the known side is adjacent. The unknown side x is the hypotenuse, the trigonometric ratio that connects the known (adjacent) and the unknown (hypotenuse) is cosine. Thus we write

Cos 60⁰ = ⁴/_x

But cos 60 = ½

So           ½ = ⁴/_x

Hence     x = 8.

To find y, y is the opposite side and 4 is the adjacent. Hence we use tangent

Tan 60⁰ = ^y/₄

But tan 60⁰ =

,

hence y = 4

Class Activity:

Do not use tables but leave your answer in surds form

1.  Find the values of the lettered sides;

(a)                                               (b)                                                               (c)

30⁰                                                      60⁰                                                              60⁰ 30⁰

10              y                      x               3             x

 

X                                                   y                                                       y                     z

2.New General Mathematics SS 1, Ex 11d. Nos 3 – 8

 

ANGLE 45⁰.

Consider a square one unit.                                                                      1

1

 

1

Suppose the length of its diagonal is x.                                      1

Using Pythagoras theorem,

x² = 1² + 1²

x² = 2

x =

from this isosceles triangle obtained from above

 

 

1

 

1

Sin 45⁰ = ^opp/_hyp

= 1/

Cos 45⁰ = adj/hyp

= 1/

tan 45⁰ = opp/adj = = 1/

Example 1:

Find x and y in the figure below:

X

45

Y                6

To find x, use cosine

Cos 45⁰ = adj/hyp

Cos 45⁰ = x/6

1/ = x/6

x = 6

x = 6/

Rationalize the denominator to get

x = (6)/(

= /2

= 3

likewise Sin 45⁰ = y/6

1/ =y/6

y = 6

x = 6/

= 3

 

 

Example 2.

Find x and y below

 

 

4

 

45                        60

Solution

Sin 60⁰ = x/

/= x/

2x =

x = /

= 3

= 6

To get y, use Sine

Sin 45⁰ = x/y

1/ = 6/y

y = 6

Class Activity:    Find the marked sides in the triangles below. Do not use tables.

• x                                                             (2)

 

y                  2                                   y                   x

45                                              45                        60

 

Below is the following summary of trigonometric ratio for 30⁰ 45⁰ and 60⁰.

	300	450	600
Sin	½		/2
Cos	/2		1/2
Tan		1

 

PRACTICE EXERCISE:

Find the marked angles below:

 

5 9

X                                                              Y

• 3                                                                     (2)

5

 

 

 

ASSIGNMENT

1. New General Mathematics for SS 1   Ex. 11a, Page 135, No.  6, 7, 8,9, 14
2. New General Mathematics  for SS 1  Ex. 11b, Page 139, No.  5, 15, 19
3. New General Mathematics for SS 1   Ex. 11c, Page 142, No.  14, 15
4. New General Mathematics for SS 1,  Ex 11d,  Page 146, No.  11 – 20
5. New General Mathematics for SS 1,  Ex 11e,  Page  147, No.  1, 2, 7, 13

 

WEEK 3 

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: TRIGONOMETRY 2

CONTENTS:

• (c) Application of trigonometric ratios (angle of elevation and depression; bearing).(d) Trigonometric ratios related to the unit circle. (e) Graphs of sines and cosines.

Application of trigonometric ratios (angle of elevation and depression; bearing).

                               Angle of depression 

 

                                    Angle of elevation

 

Angle of elevation:

Example 1:

The angle of elevation of a point P on a tower from a point Q on the horizontal ground is 60⁰. If /PQ/=74m, how high is P above the ground?

Solution:                          P

74m             Tower = x

60⁰

The relevant sides to 60⁰ are Opp and Hyp (SOH)

sin60⁰ =

=

 

 

Example 2:

A man 1.8m tall observes a bird on top of a tree. If the man is 21m away from the tree and his angle of sighting the bird is 30⁰, calculate the height of the tree.

Solution:

Tan 30⁰ =

 

 

Thus height of the tree =  + 1.8m = 12.12m + 1.8m = 13.92m

Angle of depression:

Example 3:

A boat can be sighted at the sea 71.5m from the foot of a cliff which is 26m high. Calculate the angle of depression of the boat from the top of cliff, correct to 2 sig. figures.

Top of cliff

 

                                                     26m

                            

          Boat              71.5m           Foot of cliff

Tan =

 

= 19.98

= 20⁰ (2 s.f)

Example 4:

The pilot of an air craft 2,000m metres above the sea level observes at an instance that the angles of depression of two boats which are in direct straight line are 58⁰ and 72⁰. Find correct to the nearest metres, the distance between the two boats.

 

Solution:

A  Pilot

72⁰  58⁰

 

2000m

 

 

B                                   C           x             D

The distance between the two boats is

In triangle ABC, tan 72⁰ =

BC  =

=  649.84m

In triangle ABD, tan 58⁰ =

BD  =

= 1249.74m

But,

=  1249.74m  –  649.84m

=  599.9m

=  600m  (to the nearest metres)

Bearing:

(a)  A boy sets out to travel from A to C via B. From A he travels a distance of 4km on a bearing  to B.  From B he travels a further 3km due east. Calculate how far is C

(i) North of A

(ii) East of A

(b)  Hence, or otherwise, calculate the distance AC correct to 1 decimal place.

Class Activity:

OBJECTIVE QUESTIONS

1. A ladder 9m long leans against a vertical wall making an angle of 64⁰with the horizontal ground. Calculate correct to one decimal place. How far the foot of the ladder is from the wall?
2. 4.0m B. 5.8m C. 7.1m D. 8.1m E. 18.5m

 

2. When an Airplane is 900m above the ground, its angle of elevation from a point P on the ground is 30⁰. How far is the plane from P by time of right?
3. 400m B. 800m C. 1500m D. 1600m E. 1700m

 

3. The angle of elevation of X from Y is 30⁰. If /XY/ = 40m, how high is X above the level of Y?
4. 10m B. 20m C. 30m D. 40m E. 50m

 

4. If the shadow of a pole 7m high is ½ its length, what is the angle of elevation of the sun, correct to the nearest degree?
5. 90⁰ B. 63⁰ C. 60⁰ D. 26⁰ E. 0⁰
6. From the top of a building 10 m high, the angle of depression of a stone lying on the horizontal ground is 69⁰. Calculate, correct to 1 decimal place, the distance of the stone from the foot of the building
7. 3.8m B. 6.0m C. 9.3m D. 26.1m

THEORY QUESTIONS

1. A ladder of length 4.5 m leans against a vertical wall making an angle of 50⁰with the horizontal, if the bottom of a window is 4m above ground what is the distance between the top of the ladder and the bottom of the window?    [Answer correct to the nearest m].
2. From a horizontal distance of 8.5 km, a pilot observes that the angle of depression of the top and base of a control tower are 30⁰and 33⁰ respectively. Calculate, correct to 3 significant figures.

(a) The shortest distance between the pilot and the base of the control tower;

(b) The height of the control tower.

 

 

Trigonometric ratios related to the unit circle

1. Angles between 0⁰and 360⁰.

Consider a circle of radius one unit and centre at O, the origin, in the XY plane.

y-axis

p(x,y)

1            y

0          x                                  x-axis

 

 

The circle is called a unit circle. Point P has co-ordinate (x,y) and lOPl = 1 unit.

Line Op makes an angle of  with Ox.

In the right – angled triangles,

Sin = y/1 = y

Cos = x/1 = x

The Y – axis and x – axis divides the plane into 4 parts (or 4 quadrants)

Y – axis

 

2^nd quadrant 1^st quadrant

 

X – axis

3^rd quadrant             4^th quadrant

 

Starting from the OX line, positive angles are measured in the anticlockwise direction while the negative angles are measured in the clockwise direction

y-axis

 

 

70

30           x-axis

 

 

 

Notice that +70⁰ = -290⁰ (same direction)

-30⁰ = +330⁰

In what quadrant is (i) +170⁰  (ii) -170⁰ (iii)  -45⁰  (iv) -260⁰ (v)235⁰.?

What positive angle is equivalent to

(i) -160⁰  (ii) -180⁰ (iii)  -270⁰?

What negative angle is equivalent to

(i) 167⁰  (ii) 202⁰ (iii)  285⁰

For angles in the first quadrant, 0⁰ <  < 90⁰.

P(x,y)

 

r                    y

 

x

 

Sin   = y/x

Cos   = x/r

tan   = y/r

All the trigonometric ratios are positive

For angles – the second quadrant, 90⁰ < <  180⁰.

P(x,y)

r

y

 

 

 

 

In this quadrant, x is negative, y is positive.

r is always taken to be positive r  = x² + y²)

Sin = y/r = Sin () which is positive

Cos = -x/r = -Cos () which is negative

Tan = y/-x = -tan () which is negative

For angles in the third quadrant, 180⁰ < < 270⁰.

 

-x

-y    r

 

Sin = -y/r = -Sin   – 180⁰). This is negative

Cos = -x/r = -Cos   – 180⁰). This is negative

Tan = -y/-x = tan   – 180⁰). This is positive

For angles negative, the fourth quadratic, 270⁰ < 360⁰.

x

360⁰ –

 

 

X is positive

Y is negative

Sin = -y/r = -Sin 360⁰ – ). This is negative

Cos = x/r = Cos360⁰ – ). This is positive

Tan = -y/x = – tan360⁰ – ). This is negative

Class Activity:

Use tables to find the values of the following.

• Cos 130⁰
• Tan (-130⁰)
• Sin 111⁰
• Sin (-320⁰)

2. Fill the table below

	00	300	600	900	1200	1500	1800	2100	2400	2700	3000	3300	3600
Sin
Cos
Tan

 

Graphs of Trigonometric function of sine and cosine

The graph of y = Sin can be drawn using the unit circle.

One can plot values of y against  to get a wave-like curve.

For instance when = 360⁰, the corresponding y value is obtained by drawing dotted horizontal line from mark 30⁰ on the unit circle to meet the y – axis.

When  = 90⁰, the dotted horizontal line for 90⁰ on the circle will meet the y – axis where y = 1

 

 

 

1

 

0

 

-1

 

 To draw the graph of Cos x, use the corresponding values of x and .

For instance on the circle, when  = 0⁰, that corresponds to x = 1, when  = 60⁰, dotted line from point 600 on the unit circle vertically downwards will meet the x-axis at x = ½

(i.e when = 60⁰, x = ½).

Plotting values of x against corresponding values of  gives the graph of Cos . it is another wave-shaped curve. As  increases beyond 360⁰, the curves of Sin and Cos repeat themselves.

Class Activity:

1. Draw the graph of for
2. Draw the graph of for

PRACTICE EXERCISE:

1. At a point 500m from the base of a water tank the angle of elevation of the top of the tank is 45⁰, find the height of the tank.
2. 250m     B. 353m    C. 354m    D. 433m    E. 500m (SSCE 1993)

 

2. A ladder 6m long leans against a vertical wall so that it makes an angle of 60⁰with the wall. Calculate the distance of the foot of the ladder from the wall.
3. 3m    B. 6m    C. 2√3m D. 3√3m    E. 6√3m     (SSCE 1994)

 

3. The angle of elevation of the top X of a vertical pole from a point P on a level ground is 600, the distance from a point P to the foot of the pole is 55 m, without using tables, find the height of the pole.
4. 50/3m    B. 50m    C. 55√3m    D. 60m    E. 65cm (SSCE 1996)

 

4. A boat is on the same horizontal level as the foot of a cliff, and the angle of depression of the boat from the top of the cliff is 30⁰. If the boat is 120m away from the foot of the cliff, find the height of the cliff correct to 3 significant figures. (SSCE 1992)
5. 5. A simple measuring device is used at points X and Y on the same horizontal level to measure the angle of elevation of the peak P of a certain mountain. If X is known to be 5,200m above sea level,

/XY/ = 4000m and the measurement of the angles of elevation of P at X and Y are 15⁰ and 35⁰ respectively, find the height of the mountain. (Take tan 15⁰ = 0.3 and tan 35⁰ =0.7)   (SSCE 1993)

 

6. Draw the graph of for  at intervals of 30⁰. Using the graph find the solution set of the equation

ASSIGNMENT:

1. The angle of elevation of a point T on a tower from a point U on the horizontal ground is 30⁰, if

TU = 54 cm, how high is T above the horizontal ground?

46. 108m   B. 72m    C. 46.3m    D. 31.2m    E. 27m (SSCE 1997)

 

2. A ladder 5m long rest against a wall such that its foot makes an angle of 30⁰with the horizontal. How far is the foot of the ladder from the wall.
3. 5√2/3m B. 2½m    C. 5√3/2m   D. 10√3/3m    E. 10√3m (SSCE 1998)
4. The angles of depression of the top and bottom of a building are 51⁰ and 62⁰ respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures. (SSCE 2004)
5. From two points on opposite sides of a pole33m high, the angles of elevation of the top of the pole are 53⁰ and 67⁰. If the two points and the base of the pole are on the same horizontal level, calculate, correct to three significant figures, the distance between the two points. (SSCE 2007)

 

5. (i) Copy and complete the following table for f:x

x	00	300	600	900	1200	1500	1800	2100	2400     2700    3000    3300       3600
f(x)							-9

(ii)  Draw the graph of f for

• Use your graph to estimate; the maximum and minimum values of f correct to 1 decimal place, stating nearest degree for which they occur.
• The truth set of the equation

WEEK 4 

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: MENSURATION

CONTENTS:

(a) Length of arcs of circle.

(b) Perimeter of sectors and segments.

(c) Areas of sectors of a circle.

(d) Areas of segments of a circle.

 

DEFINITION AND MEANING, LENGTH OF ARCS OF CIRCLE

• AN ARC: an arc of a circle is a part of the circumference of the circle.

Major arc

 

 

A                                            B

Minor arc

Hence, an arc is a length or a distance along the circumference of a circle. It is never an area.

• A SECTOR: a sector is a part or a fraction of a circle bounded by an arc and two radii.

 

 

 

 

A                                                         B

 

 

Hence, an arc is a length whereas the sector covers an area of a circle.

• A SEGMENT

The segment of a circle is the part cut off from the circle by a chord. A chord is the line segment AB.

Major segment

 

 

A                                                     B

                                                                                                        Minor segment

LENGHT OF ARCS OF CIRCLES

If 5 sectors are cut off from 5 different circles and the lengths of the arcs l, radii r and angles measured and compared.

Then

i/2π = /360 or L = /360 x 2 πr = 2 πr/360

hence, in a circle of radius r, the length l of an arc that subtends angle  at the centre is given by

L =  

 

 

A                                                B

l               L

 

Example 1:                                                   

Find the length of an arc of a circle of radius 5.6cm which subtends an angle of 60⁰ at the centre of the circle (Take π = 22/7)

Solution

Length of arc = /360 x 2 πr

Given  = 60⁰, r = 5.6cm, π = 22/7

Substituting into the formula,

Length of arc AB = /360 x 2  x 22/7 x 5.6

= 1/6 x 2 x 22/7 x 5.6

= 1/6 x 2 x 22/7 x 0.8

= 17.6/3

= 5.8667cm

= 5.87cm (2 decimal places)

Example 2;

What angle does an arc 6.6cm in length subtends at the centre of a circle of radius 14cm. Use π = 22/7)

Solution

Length of arc xy = /360 x 2 πr

6.6 = /360 x 2 x 22/7 x 14

x 2 x 22 x 14 = 6.6 x 360 x 7

=

= (33 x 18)/11×2

= 3 x 9

= 27⁰

= 27⁰ , the angles subtend by the arc

Example 3: An arc of length 12.57cm subtends an angle of 60⁰ at the centre of a circle. Find

The radius of the circle

The diameter of the circle.

Solution

Arc = /360 x 2 πr

12.57 = /360 x 2 x 22/7 x r/1

12.57 x 360 x 7 = 60 x 2 x 22 x r

=

r =

r = 527.94/44

r = 11.99

r = 12cm

Example 4:

An arc of a circle of diameter 28m subtends an angle of 108⁰ at the centre of the circle. Find the length of the major arc.

Solution

Minor  arc angle = 108⁰

Major  arc angle = 360⁰ – 108⁰

= 252⁰.

Arc = = /360 x 2 πr

= /360 x 2/1  x 22/7 x 14/1

=

=

= 61.6m

The length of major arc = 61.6m

Class Activity:

1. Find the radius of a circle which subtends an angle of 120⁰at the centre of the circle and is of length 2.8cm (π = 22/7)
2. In terms of π, what is the length of an arc of a circle of radius 3¹/₂m?

 

• PERIMETER OF SECTORS AND SEGMENTS

PERIMETER OF SECTORS

The word perimeter simply means the distance round an object. So the perimeter of a sector of a circle is the distance round the circle.

 

 

A

L

 

B

hence the perimeter of a sector AOB is the sum of two radii (2r) and length of arc l, where r is radius and l = length of arc.

Perimeter of sector AOB =  

Example 

Find the perimeter of the sector of radius 3.5cm which subtends an angle of (i) 45⁰ (ii) 315⁰.

Solution

Length of arc =  x πr

Here  = 45⁰, r = 3.5cm, π =

Length of arc =  x  x  x 3.5

=  x x  x

= 22/8

= 2.75cm

The perimeter of the sector is 2r + l, here r = radius which is 3.5cm and l = 2.75cm.

Perimeter = 2r + length of arc

= 2r + 2.75cm

= (2 x 3.5) + 2.75cm

= 7.0 + 2.75cm

= 9.75cm

(ii)  = 315⁰, π =  , r = 3.5cm

Length of arc =   x  x  x 3.5cm

=   x  x  x  cm

=   x  x  x  cm

= 77/4 = 19.25cm

Perimeter = 2r + 19.25cm

= (2 x 3.5) + 19.25

= 7 + 19.25

= 26.25cm

Class Activity:

1. Calculate the perimeter of a sector of a circle of radius 14cm, where the sector angle is 60⁰.       Take π =
2. The perimeter of a sector is 61.43cm. If the angle subtended by the sector at the centre is 120⁰. Find the radius of the sector.

PERIMETER OF SEGMENT

The perimeter of a segment = length of arc + length of chord.

To find the length of chord AB, we bisect <AOB

Using trigonometric ratios

= Sin

AD = r Sin

But, AD = DB

Hence

AB chord AB = AD + DB

= r Sin  + r Sin

= 2r Sin units

Length of Chord = 2r Sin units

Thus,

Perimeter of segment =  x πr + 2r Sin

Example :

AB is a chord of a circle with centre o and radius 4cm, ÐAOB = 120⁰. Calculate the perimeter of the minor segment (π = ).

Solution

A                       B                 B

Chord AB = rSin , r = 4cm,  = 120⁰.

=  = 60⁰.

So chord AB = 2 x 4 Sin 60⁰

= 8 x sin 60⁰

= 8 x

= cm

= 6.92cm

Length of arc AB =  x 4 x

=  x x

=

= 8.38cm (2 d.p)

Perimeter of minor segment = length of arc + chord

= 8.38 + 6.92cm

= 15.30cm

AREA OF SECTORS OF A CIRCLE

The area of a sector =  x

  Area of a sector   =  x r²    

where  is the angle formed at the centre by the arc of the circle.

Example 1:

Find the area of the sector of a circle of radius 4.8cm which subtends an angle of 135⁰ at the centre.(Take  = 3.142)

Solution

The area of a sector =  x r²

Here  = 135⁰, r = 4.8cm,  = 3.142

The area of a sector =  x 4.8)²

=  x 4.8 x 4.8

= 27 x 3.142 x 0.4 x 0.8

= 27.14688cm²

Area of sector = 27.15cm² (decimal places)

Example 2:

AB is an arc of a circle of length 9.2cm with centre 0 and the radius is 4.6cm. Find the area of the sector AOB.

Solution

Length of arc AB =   x 2πr

r = 4.6cm, length of arc = 9.2cm

Length of arc AB =  x 2 x π x 4.6

9.2 =  x 2 x π x 4.6

9.2 x 360 =  x 2 x π x 4.6

=

=

Area of sector AOB

=  x πr²

=  x πr²

=  x  x π x (4.6)²

= (4.6)²

= (4.6) x (4.6)

= 21.16cm²

Class Activity:

1. Calculate the area of a sector of a circle which subtends an angle 45⁰at the centre of the circle, radius 14cm.
2. A sector of 80⁰is removed from a circle of radius 12cm. What area of the circle is left? Use.
3. New General Mathematics for senior secondary school, Book 1 page 158 exercise 12d Nos 1 – 6.

AREA OF SEGMENTS OF CIRCLES

Area of segments = area of sector – area of triangle

Area of segment =  x πr²  r² Sin

Example:

The arc AB of a circle, radius 6.5cm, subtends an angle of 45⁰ at the centre O. Find the area of the minor segment cut off by the chord AB (Take π = )

Solution:

Area of segment = Area of sector AOB – Area of

Area of sector AOB =  x πr²

= 45⁰, r = 6.5, π =

Area of sector AOB =  x (6.5)²cm²

=  x 6.5 x 6.5

=  x 42.25

= 16.598cm²

 

Area of  =  X OB Sin45⁰

=  X 0.5 x 6.5 Sin45⁰

=  x

=

=

= 14.935375cm²

Area of segment = 16.598 – 14.935

1.66cm² (2 d.p)

Class Activity:

1. An arc AB of a circle radius 4.8cm subtends an angle of 158⁰at the centre O. Find
2. The area of the sector AOB
3. The area of the minor segment cut off by the chord AB (π = 3.142)
4. A sector of a circle radius 16cm subtends an angle of 84⁰at the centre O. Calculate the area of the shaded segment of the circle. ( Take  )

 

PRACTICE EXERCISE:

1. A chord of a circle subtends an angle of 60⁰at the centre of a circle of radius 14cm. Find the length of the chord.
2. A rope of length 18m is used to form a sector of a circle of radius 2.5m on a school playing field. What is the size of the angle of the sector? Correct to the nearest degree?
3. An arc of length 21.34cm subtends an angle 101⁰at the centre of a circle. Find the diameter of the circle.
4. The perimeter of a sector is 75.43cm. If the angle – subtended by the sector at the centre is 135⁰. Find the radius of the sector.
5. XY is a chord of a circle centre O and radius 7cm. The chord XY which is 8cm long subtends an angle of 120⁰at the centre of the centre of the circle. Calculate the perimeter of the minor segment.

(Take  )

ASSIGNMENT:

XOY is a sector of a circle centre O of radius 3.5cm which subtends an angle of 144⁰ at the centre. Calculate, in terms of, the area of the sector.

In the diagram, XY is a chord of a circle of radius 5cm. The chord subtends an angle 96⁰ at the centre. Calculate, correct to 3.s.f; the area of the minor segment cut-off by XY.

In the diagram, C is the centre of the circle of radius 5cm, and /PR/ = 8cm. Find the area of the shaded region.

WEEK 5 

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: MENSURATION

CONTENTS:

(e) Relation between the sector of a circle and the surface area of a cone.

(f) Surface area and volume of solids; (i) Cube, cuboids (ii) Cylinder (iii) Cone

(iv)  Prisms (v) Pyramids.

Relation between the sector of a circle and the surface area of a cone.

If a sector of a circle AOB is cut and folded into a come as shown in the diagram below.

The arc AB becomes the circumference of the base of the cone. The radius R becomes the slant edge l of the cone.

Arc AB = 2πr

From the above diagram.

The area of the sector = area of the curved surface of the cone

Length of arc AB = circumference of the circular base of the cone

Curved surface area of cone =  x l²

Also,  x l =

=

Total surface area of cone = rl + r²

= r(l + r).

Example 1:  Calculate in terms of , the total surface area of a cone of base diameter 12cm and height 10cm.

Solution

 

 

 

Using Pythagoras rule,

l² = 10² + 6²

l² = 100 + 36

= 136

l=

Total surface area

= rl + r²

= r(l + r).

= ( + 6)cm².

 

 

 

 

 

Example 2:

A 216⁰ sector of a circle of radius 5cm is bent to form a cone. Find the radius of the base of the cone and its vertical angle.

Solution

 

 

 

 

Radius = r

Vertical angle = 2

Circumference of base of cone = length of arc of sector

Substituting values

x  =

r =  x  x

r = = 3

r = 3cm

Sin =  = 0.6000

= 36.87⁰

= 36.87⁰ x 2

= 73.74⁰

Radius of base = 3cm

Vertical angle = 73.7⁰ (to 0.1⁰)

Class Activity:

Find the curved surface and total surface areas of a closed cone of height 4cm and base radius 3cm.              (Take )

SURFACE AREA OF CUBES, CUBOIDS AND CYLINDER

CUBE: since a cube is solid with six faces

Area of one face is a  x a = a².

Total area = a  x a x 6 = 6a².

Surface Area = 6a² units

Example 1:

What is the surface area of a cube of edges 12cm.

Solution

A = 6 x 12 x 12

= 6 x 144

= 864cm²

Example 2: Calculate the surface area of a cube of edge 11cm.

Solution

A = 6a²

= 6 x 11 x 11

= 6 x 121

= 726cm²

CUBOIDS

Surface area of a cuboids = 2(lb + bh + lh)

Example 3: Given a cuboid of edges 3cm, 5cm, 8cm. calculate the surface area.

Solution

Surface area of a cuboids = 2(lb + bh + lh)

= 2(8 x 5 +8 x 3 + 5 x 3)

= 2(40 + 24 + 15)

= 2 x 79

= 158cm²

CYLINDER

Area of cylinder with one end closed =  2πrh + π

= πr(2h + r)

If two ends are closed.

Area = 2πrh + 2πr²

= 2πr(h + r), where r = radius of cylinder, h = height of cylinder

Example 4:

Find the surface area of a cylinder with height 14cm and base radius of 7cm, consider all the three cases namely;

• Open ended cylinder
• One end open cylinder
• Both ends closed cylinder

Solution

• Open ended cylinder

Area = 2πrh

r = 7cm, h = 14cm

A = 2 x  x 7 x 14

= 44 x 14

= 616cm²

• One end open cylinder

Area = 2πrh + πr²

= (616 +  x 7² ) cm²

= (616 + 154) cm²

= 770cm².

• Both ends closed cylinder

Area = 2πr(h + r)

= 2 x  x 7(14 + 7)cm²

=  x 7 x 21

= 924cm².

Class Activity:

1. Calculate the surface area of a hollow cylinder which is closed at one end, if the base radius is 3.5cm and the height is 8cm. Take (Take )
2. A Cylindrical container, closed at both ends, has a radius of 7cm and height 5cm. Find the total surface area of the container (Take )

SURFACE AREA OF PRISMS AND PYRAMIDS

PRISMS: The total surface area is the sum of the surface of the five faces as shown below

 

 

 

 

Total surface area of prism = Areas of (ABCD + AFED + BCEF) + Areas of ()

Example 1:

Find the total surface area of the prism shown below

Solution

Area of ABCD

=20 x 10cm²

= 200cm²

Area of AFED = 20 x 6cm²

= 120cm²

Area of BCEF = 20 x 8cm²

= 160cm²

Area of  = area of

=  X 8 X 6cm²

= 24cm²

Total surface area of prism = (200 + 120 + 160 + 24 + 24)cm²

= 528cm²

In the case of a rectangular prism or cuboid, the total surface area is determined by finding and summing up the areas of the four rectangular faces and the two end faces.

 

Examples 2: Find the total surface area of the solid.

Solution

The net of the solid gives a clearer picture of the shape

Area of A = 5 x 3cm² = 15cm²

Area of B = 8 x 5cm² = 40cm²

Area of C = 8 x 3cm² = 24cm²

Area of D= 8 x 5cm² = 40cm²

Area of E = 8 x 3cm² = 24cm²

Area of F = 5 x 3cm² = 15cm²

Total surface area = (15 + 40 + 24 + 40 + 24 +  15)cm²

= 158cm².

 

 

PYRAMIDS: The total surface area of a pyramid is found by summing up areas of the common shapes that make up the pyramid.

Example 3:

Find the total surface area of a right pyramid with a rectangular base 6cm by 10cm, a height of 8.3cm and a slant edge of 10.14cm.

Solution

The total surface area is the sum of the surface areas of the five faces:

ABCD, VAB, VDC, VBC and VAD

ABCD is a rectangle so AB = DC and AD = BC

If AB = 10cm, then DC = AB = 10cm

If AD = 6cm, then BC = AD = 6cm

Hence, area of ABCD = 10 x 6 = 60cm²

Area of  = Area of  and Area of  = Area of

Area of  =  x AB x VP

By using Pythagoras rule

VP² = VB² – PB²

= (10.14)² – 5²

= 102.8 – 25

= 77.8

VP =

= 8.82cm

Area of  =  x AB x VP

=  x 10 x 8.82

= 5 x 8.82

= 44.10cm²

Area of = Area of

 

 

 

 

 

 

 

 

 

 

 

 

 

Area of  =  x AB x VF

But VF² = VC² – FC²

= (10.14)² – 3²

= 102.8 – 9

= 93.8

VF =

= 9.69cm

Area of  =  x BC x VF

=  x 6 x 9.69

= 3 x 9.69

= 29.07cm²

Area of = Area of

Area of rectangular base ABCD = 10 x 6 = 60cm².

Total surface area of the pyramid = [(41.10 + 41.10 + 29.07 + 29.07 + 60)]

= (82.20 + 58.14 + 60) cm²

= 200.34cm²

Class Activity:

1. Find the length of the slant edge of a right pyramid with
2. A rectangular base 3cm by 5cm and a height of 4.2cm
3. A rectangular base 6cm by 4.5cm and a height of 3.4cm
4. Find the surface area of a triangular prism 10.8cm long and having a triangular face of dimensions 8.8cm by 5.7cm by 6.8cm.

 VOLUME OF CUBES, CUBOIDS, CYLINDER, CONE, PRISMS AND PYRAMIDS

The Volume of cube = s² x s = s³

Example 1:

Find the volume of a cube whose edges are 7cm each.

Solution

Volume = s² x s = s³

= 7 x 7 x 7cm³

= 49 x 7cm³

= 343cm³

Volume of cuboids

The volume of a cuboid = Area x height

= lb x h

= lbh cm³

Example 2:

What is the volume of a cuboid if

• Height = 6cm, breadth = 10cm, length = 13cm?
• Area of cross – section is 105m²and height = 5cm?
• Area of cross-section has the same numerical value with the height = 16cm?

Solution

• Volume (V) = lbh = 13 x 6 x 10 = 780cm³
• Volume = Ah = 105 x 5 = 525cm³
• V = A x h

= 16 x 16

= 256cm³

Volume of cylinder

With the Cylinder the cross section is a circle.

Area = πr², where is the radius of the circular base. Height = h

Volume = Area x height

=  πr² x h

= πr²hcm3

Example 4:

What is the volume of a cylinder if the radius is 5cm and height is 20cm?

Solution

Volume = area of circular base x height

= πr² x h

=  X 5² X 20cm²

=  X 5 X x 5 x 20cm²

= 1571.43cm³

Volume of Cone

The volume of cone =  x volume of cylinder

=  πr²h where r = radius of the cone and h = height of the cone.

Example 5:

Find the volume of a cone if the perpendicular height is 9cm and radius 4cm.

Solution

Volume of cone =  πr²h

r = 4cm, h = 9cm

V =  x  x 4 x 4 x 9cm³

= cm³

= 150.86cm³.

Volume of prisms

Volume of prisms = area of cross – section x distance between the end faces.

Example 6:

Find the volume of the triangular prism in the diagram below.

Solution

Volume of prisms = area of cross – section x distance between the end faces.

Here, end face = CDE

Area of CDE = CE x DG

= x 5 x 2.6cm²

= cm²

= 6.75cm²

Volume of prism = Area of CDE x BC

= 6.75 x 4.5cm³

30.375cm³

Volume of triangular prism

= 30.4cm3

Volume of pyramids

Volume of pyramid =  x base area x perpendicular height

 

Example 7:

Find the volume of right pyramid with vertex V and a rectangular base measuring 5.4cm by 4cm and a height of 8cm.

Solution

Volume of pyramid =  x base area x perpendicular height

= Volume of pyramid =  x 5.4 x 4 x 8cm³

Volume of pyramid =  x 5.4 x 32cm3

Volume of pyramid = cm³

= 57.6cm³

 

PRACTICE EXERCISE:

1. Find the total surface area of a solid circular cone with base radius 3cm and slant height 4cm

[Take π = ]

1. 37cm² B. 75cm² C. 66cm² D. 88cm² E. 78cm² (SSCE 1995)
2. 2. A hollow sphere has a volume of k cm³and a surface area of k cm³. Calculate the diameter of the sphere.
3. 3cm   B. 6cmnd base C. 9cm    D. 12cm   E. more information is needed

(SSCE 1995)

3. 3. Calculate the total surface area of a solid cone slant height 15cm and base radius 8cm in terms of π.
4. 64πcm² B. 120πcm² C. 184πcm² D. 200πcm² E. 320πcm² (SSCE 1997)
5. 4. The cross-section of a prism is a right-angled triangle 3 cm by 4 cm by 5 cm. the height of the prism is 8cm. Calculate its volume.
6. 48cm³ B. 60cm³ C. 96cm³ D. 120cm³ E. 240cm³ (SSCE 1997)
7. 5. Find the curved surface area of a cone of radius 3cm and slant height 7cm. [Take π =]
8. 22cm² B. 44cm² C. 66cm² D. 132cm² E. 198cm² (SSCE 1998)
9. 6. The height of a pyramid on a square base is 15cm. If the volume is 80cm, find the area of the square base.
10. 8cm² B. 9.6cm² C. 16cm² D. 25cm² (SSCE 2000)
11. 7. A right pyramid is on a square base of side 4cm. The slanting side of the pyramid is Calculate the volume of the pyramid.
12. 5cm² B. 10cm² C. 16cm² D. 32cm² (SSCE 2001)
13. 8. The height of a right circular cone is 4cm. The radius of its base is 3cm. Find its curved surface area.
14. 9πcm² B. 15πcm² C. 16πcm² D. 20πcm² (SSCE 2001)
15.   A sector of a circle of radius 7 cm subtending an angle of 270⁰ at the centre of the circle, issued to form a cone.
16. Find the base radius of the cone
17. Calculate the area of the base of the cone correct to the nearest square centimeter.

(Take π =) (SSCE 1988)

 

10.  A solid metal cone of height 20cm and radius 12cm is melted down to form a cylinder of the same height. What is the radius of the cylinder?

ASSIGNMENT:

1. A right pyramid has a square base of side 8cm. The height of the pyramid is half the side of the square. Find the length of the sloping edge.
2. A rectangular tank is 76cm long, 50cm wide and 40cm high. How many litres of water can it hold? (WAEC)
3. Calculate in terms of π the total surface area of a cylinder of radius 3cm and height 4cm. (WAEC)
4. A conical container has radius 7cm and height 5cm. Calculate the volume of the container

(WAEC)

5. The volume of a tank is 4.913cm³, what is the length of its edge.
6. What is the edge of a cube whose volume is 729cm³
7. A rectangular tank is 76cm long, 50cm wide and 40cm height. How many litres of water can it hold? (WAEC)
8. Find the volume of a triangular prism of height 10cm whose cross-section is an equilateral triangle of side 4cm.
9. An open cylindrical container has base radius of 3.5cm. if the ratio of the area of its base to that of its curved surface is 1:6, what is the height of the container?
10. A cylindrical container of radius 10cm and height 20cm is filled to the brim with water. A steel ball of radius 6cm is now dropped into the cylinder so that the excess water flows out. How much water is left in the container? (leave your in terms of )

 

 

 

WEEK 6

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: MENSURATION

CONTENTS:

(g) Surface areas and volume of frustum of a cone and pyramid.

(h) Surface area and volume of compound shapes.

1. Surface area of frustum of a cone and pyramids
2. Volume of frustum of a cone and pyramid
3. Surface area and volume of compound shapes.

TOTAL SURFACE AREA OF FRUSTUM OF CONE AND PYRAMIDS

A frustum is the remaining part of cone or pyramid when the top part is cut off as shown below. Daily examples of frustums are buckets, lamps shades e.t.c

Frustum of a cone

Rectangular pyramid

Frustum of a pyramid

For Surface area of the frustrum of a pyramid, we sum up all areas of the faces that make up the frustum.

For Surface area of the frustum of a cone,

Total surface area of a Closed frustum = π(height x sum of radii) + area of top and base circles.

Total surface area of a Open frustum (bucket) = π(height x sum of radii) + area of circle.

Example 1:

Find the total surface area of a bucket 36cm in diameter at the top and 24cm at the bottom.

The depth of the bucket is 30cm.

Solution

The total surface area of bucket = sum of curved part + area of bottom circle

= π (height x sum of radii) + area of bottom

= π x 30 x (18 + 12) + π12²

= π x 30 x 30 + π(144)

= 900π + 144π

= 1044πcm²

Example 2:

Find in cm², the area of material required for a lamp shade in the form of a frustum of a cone of which the top and bottom diameters are 20cm and 30cm respectively and the vertical height is 12cm.

Solution

y similar triangles

=

5x = 24

 

By Pythagoras’s theorem, y = 26 and z = 13

i.e from

 

 

z² = 12² + 5²

= 144 + 25

= 169

Z =

= 13

Surface area of frustum

= π x 15 x 39 – π x 10 x 26cm²

= 13π(45 – 20)cm²

= 13π x 25cm²

= 1021cm²

Area of material required = 1021cm². (3 s. f)

Class Activity:

A bucket is 12cm in diameter at the top, 8cm in diameter at the bottom and 4cm deep. Calculate its volume in cm³ in terms of π. (JAMB)

 SURFACE AREA OF FRUSTRUM OF PYRAMIDS

Example:

A frustum of a pyramid consists of a square base of length 10cm and a top square of length 7cm, height of the frustum is 6cm. Calculate to the nearest whole number, the surface area of frustum of the pyramid.

Solution

Original Frustum

Completed to pyramid

1. Diagonals of the top and bottom

IPRI² = IPQI² + IQRI²

= 10² + 10²

= 100 + 100

= 200

IPRI =

= 14.14

IYRI =

= 7.07cm

= 7.1cm

Diagonal AC

IACI² = IABI² + IBCI²

= 7² + 7

= 49 + 49

= 98

IACI² =

= 9.89

BUT CX =

= 4.945

= 4.95cm

 

To calculate the height of the added pyramid.

Let the heights of the added pyramid be h, by similar

=

7.07h = 4.95h + 29.7

7.07h – 4.95h = 29.7

2.12h = 29.7

=

= 14cm

height of bigger pyramid

14 + 6cm

20cm

To calculate the slant side of the smaller and bigger s

Bigger

IVRI² = 7.1² + 20²

= 50.41 + 400

= 450.41

IVRI² =

21.22cm

21cm

Smaller :

 

IVCI² = ICXI² + h²

= 4.95² + 14²

= 24.50 + 196

= 220.50

IVCI =

= 14.8493cm

= 14.85cm

To find the surface area of frustum of pyramids using Hero’s formula for finding the area of the triangular face.

A =

Where s = (a + b + c) and a, b and c are sides of the triangle

s = (21 + 21 + 10)

=  x 52

= 26cm

Area of a ∆face of big pyramid =

A =

=

=

= 101.98

= 102cm²

Area of 4 ∆faces of the pyramid

= 4 x 101.98cm²

= 407.92cm²

= 408cm²

Area of 4 ∆faces of the small pyramid

s = (a + b + c)

s = (14.85 + 14.85 + 7)

s =   x 36.7

= 18.35cm

Area of a face =

=

=

=

= 50.51cm²

Area of a 4face = 4 x 50.51

= 202.04cm²

= 202cm²

Area of the base of the frustum = l x b

= 10 x 10

= 100cm²

Area of the top of the frustum = l x b

= 7 x 7

= 49cm²

Total surface area of the frustum = [(Area of the 4∆faces of the big pyramid – Area of the 4∆faces of the small pyramid )] + Bottom area + Top area

= [(408 – 202) + 100 + 49]

=  206 + 149

= 355cm²

Class Activity:

A pyramid is on a square base of 25m side and 25m high. The top of the pyramids 10m high was cut off. Find the surface area of the frustum formed.

VOLUME OF FRUSTUM OF CONE AND PYRAMIDS

Volume of frustum = volume of the full cone/pyramid – volume of the part cut off.

Example:

A circular cone 24cm high and 18cm in diameter at its base is cut off at half of its height. Calculate the volume of the remaining frustum. Leave π in your answer.

Solution

(i)

 

 

 

 

 

(ii)

 

 

Volume of original cone = πr²h

= π x 9² x 24

= π x 81 x 24

= π x 27 x 24

= 648πcm³

To find the radius r of the base of the cut off part, we use similarity of ∆s =  =

r =  =  = 4¹/₂cm

Volume of small cone = π()² x 12

= 81πcm³

Hence, Volume of frustum (i) – (ii)

= 27 x 24π – 81π

= (81 x 8) π – 81π

= 81π(8 – 1)

= 567 πcm³

 

 

 

 

Example 2:

Find the volume of the frustum of a pyramid with 30cm square top and 50cm square base and height 20cm.

Solution

 

 

 

 

 

 

 

 

 

 

If x is the height of the cut – off part and 15cm is half the side of its square base,

Then

=

i.e 25x = 300 + 15x

10x = 300

x = 30cm.

Volume of entire pyramid

=

= (base area x height)

= (50 x 50) x (20 + 30)

= (2500 x 50)

= (125,000)cm³

But volume of top pyramid

= (base area x height)

= (30 x 30 x 30)

= (27,000)

= Volume of frustum = (125,000 – 27,000)

= (98,000)cm³

=32,666 cm³

Class Activity:

What is the capacity of a bucket that is 42cm deep and inner radii of the base and topmost part of the bucket are 12cm and 20cm respectively?

Further Example:

A machine part is made up of a cylinder and a cone on each end, the dimension of which are shown below. Calculate the total surface area of the machine.

 

 

 

 

Solution

Surface area of 1^st cone = πrl

By Pythagoras rule.

L² = (6² + 4²)cm²

= (36 + 16)cm²

= 52cm

l =

the surface area of the come

= π x 4cm x cm

= 4πcm²

Surface area of 2^nd cone = πrl

l² = 9² + 4²

= 81 + 16

= 97cm

L = cm²

Area = cm²

Therefore total surface area of the machine = surface area of cylinder + surface area of 1^st cone + surface area of 2^nd cone

= cm² + cm² + cm²

= cm²

Class Activity:

Find the surface area of the bullet in the figure below, using those dimensions

PRACTICE EXERCISE:

1. A cone of radius 7cm is 42cm deep. If the cone is filled with water. How deep is the water in the cone (Take π = )
2. A square base of a pyramid of side 3cm has height 8cm. If the pyramid is cut into two parts by a plane parallel to the base midway between the base and the vertex, calculate the volumes of the two sections to the nearest centimetre.
3. What is the height of a plastic bucket whose two radii are 20cm and 29cm and a slant height of 41cm?
4. A cone of height 6cm and radius of base 4cm has its top cut-off by plane parallel to its base and 4cm from it. Find the volume of the remaining frustum.
5. A bucket full of water is in the form of a frustum of a cone. The bottom and top radii of the frustum are 18cm and 28cm respectively and the vertical depth is 30cm. If the water in the bucket is then poured into an empty cylindrical container with base radius 20cm, find the depth of the water in the container.  (Take π = )

ASSIGNMENT:

1. The internal and external radii of a water pipe are 9cm and 10cm respectively. If the pipe is 35cm long, find, in cm³, the volume of material used in making it. (Take π = )
2. The internal and external radii of a cylindrical bronze pipe are 1.5cm and 2cm respectively. If the pipe is 10cm long, calculate the volume of bronze used.  (Take π = )
3. A cylindrical pipe is 28metres long. Its internal radius is 5cm. calculate :
   • The volume of water, in litres that the pipe can hold when full;
   • The volume in cm³of metal used in making the pipe. (Take π = )
4. A Cylindrical vessel open at one end is made of metal. The internal diameter is 7cm, the internal depth 10cm and the thickness of the metal is 0.5cm. Calculate:  (a) the internal volume of the vessel     (b) the volume of the metal.  (Take π = )
5. A rectangular block of metal, 6cm long by 3cm wide and 3cm high has a cylindrical hole of radius 1.2cm and depth 2.5cm bored out in the centre of the top surface. Calculate the volume of the remaining part of the block.

 

WEEK 7

MID-TERM BREAK

 

WEEK 8

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: DATA PRESENTATION

CONTENTS:

(a) Revision on collection, tabulation and presentation of data.

(b) Frequency distribution (grouped data).  

Revision on collection, tabulation and presentation of data.

Statistics can be considered as a way or method used in the collection and organisation of data in order to interprets, predict and get other information required out of the data.

This data could be information about some people e.g.. ages, weights, heights, examination scores, etc

Collection of Data

Let us remind ourselves that there are two kinds of data namely:

• Discrete data (b) continuous data

Discrete data are those which are obtained by direct counting e.g numbers of persons born on a particular days of the week, month of the year, etc.

Continuous data are those which require measurement before counting e.g number of persons of some age, height, weight etc.

Class Activity: 

Collect and tabulate data on the days of the Week each member of the class was born.

Tabulation and Presentation of data.

For easy access to information, data are normally presented using frequency tables. This table marches each data with the number of times it appeared. The frequency table is prepared as follows:

1. Draw the columns
2. Represent the data given in the first column in ascending or descending order.
3. Represent each data with the use of a tally.

Raw data and frequency tables

When data is first collected and has not been organised in any way, it is called Raw Data.

Example 1:

In a game, a die was thrown several times. Below are the results of the scores.

2	3	4	4	2	1	3	2	6	5
3	2	1	1	2	5	2	1	4	4
6	5	6	1	6	5	4	5	4	3
6	5	5	3	5	2	1	4	5	2
4	5	4	6	3	1	5	6	6	5

The above is an example of raw data. One way we can organise the above data is to present it in a frequency distribution table (or frequency table for short) as shown below.

No of throws	Tally	Frequency
1	1111 11	7
2	1111 111	8
3	1111 1	6
4	1111 1111	9
5	1111 1111 11	12
6	1111 111	8
	TOTAL	50

 

Note – Frequency table can be given with or without the tally column

Example 2

In a test marked out of 10, a group of pupils obtained the following marks.

3	4	6	3	4	3	5	6	7	6
8	9	5	9	10	7	8	2	6	5
4	10	5	6	7	3	8	9	4	2

 

Prepare a frequency table for the distribution

Solution

Marks x	Tally	Frequency
2	11	2
3	1111	4
4	1111	4
5	1111	4
6	1111	5
7	111	3
8	111	3
9	111	3
10	11	2
	Total	30

 

Class Activity:

The weight to the nearest kilogram, of a group of 50 students in a college of technology is given below. Prepare a grouped frequency table with class intervals                                        45 – 49, 50 – 54, 55 – 59, e.t.c

65	70	60	46	51	55	59	63	68	53
47	53	72	58	67	62	64	70	57	56
73	56	48	51	58	63	65	62	49	64
53	59	63	50	48	72	67	56	61	64
66	52	49	62	71	58	53	69	63	59

 

 

PRACTICE EXERCISE:

1. The score of 50 students in a mid-term test were as shown below. Construct a frequency table using the class intervals 1 – 5, 6 – 10, 11 – 15, e.t.c

11	20	30	24	13	28	33	40	23	28
40	8	30	13	15	34	8	34	32	22
26	21	25	18	26	10	19	3	27	18
18	24	26	25	27	29	28	13	35	24
9	24	14	28	27	38	40	32	33	34

 

 

 

 

 

 

2. The age in years of 50 teachers in a school are given below. Form a frequency distribution of the data using the internals 21 – 25, 26 – 30, 31 – 35, e.t.c

21	37	49	27	49	42	26	33	46	40
50	29	23	24	29	31	36	22	27	38
30	26	42	39	34	23	21	32	41	46
46	31	33	29	28	43	47	40	34	44
26	38	34	49	45	27	25	33	39	40

 

3. Using class interval 30 – 34, 35 – 39, 40 – 44, …… Construct the frequency distribution table.

62	54	53	44	46	55	46	56	68	63
59	61	66	54	39	48	47	53	59	57
50	35	40	30	46	44	36	49	54	51
57	56	45	33	38	41	40	45	53	58
51	45	48	34	36	46	43	49	63	52

WEEK 9

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: DATA PRESENTATION

CONTENTS:

(i) Line graph, (ii) Bar graph (iii) Histograms (iv)  Pie Chart (v) frequency polygon                                 (vi) Deductions and Interpretations.

LINE GRAPH

Line graphs are used to show trends over a period of time and have the advantage that they can be extended. To draw a line graph, plot the given data as a series of points and then join the points together by straight lines. The lines can be drawn vertically or horizontally and have no thickness.

Example 1

A comparison of traffic densities in a number of countries produced the following results.

Country	Number of Vehicles
U. S.A	61
U.K	51
West Germany (W.G)	51
Italy	49
Belgium	36
France	26
Japan	20

Present the data in a line graph

Solution

 

 

 

 

 

 

 

 

Example 2:

The table below shows the number of eggs laid in one month by 5 hens. Draw a horizontal line graph for the data.

Hen	A	B	C	D	E
No of Eggs	24	21	27	30	18

`Solution

 

 

 

 

The number of eggs laid in one month by 5 hens can be presented in the following horizontal line graph.

 

A

B

C

D

E

0         3        6        9        12        15        18        21        24        27         30

                                                                                                                  No. of eggs

Class Activity:

1. The student population of the Federal Government college Kano was recorded as follows.

Class	1	2	3	4	5
No of Students affected	30	29	30	28	27

 

Plot the line graph of the above data.

2.   The table below shows the expenditure of a company in Benue State in a Certain year.

Description	Expenditure/N1000
Wages and Salaries	25
Fuel and Power	15
Raw Materials	65
Maintenance	5
Miscellaneous	10

 

Represent the information on a line graph

BAR GRAPH/CHART

Bar charts are rectangular shapes of equal widths but different lengths drawn to represent the frequency. It is uniform in thickness. Bars can be drawn vertically or horizontally.

Example 1:

The table below shows the number of babies born to a number of women within a given age range to a number of women within a given age range.

Women ages	24	25	26	27	28
No. of babies	1	4	2	3	2

Draw a bar chart to illustrate the above distribution.

Solution

 

 

 

 

 

Example 2:

The number of bottles of soft drinks sold in a restaurant one evening is given by the data in the table below.

Type of soft drink	No of bottles
Coke	12
Fanta	10
Sprite	6
Lemon	4
Pepsi	8

Draw a bar chart to display this information

Solution

 

 

 

 

HISTOGRAMS

Histogram is a rectangular block graph representing the frequency distribution of the analysed items. The blocks are joined to one another.

Example 1: The marks scored by 40 students in a particular subject are as follows:

38	74	28	32	10	31	49	34	50	19
30	92	50	42	38	64	24	65	91	77
18	35	12	87	41	27	8	90	22	21
42	43	52	59	72	70	90	91	29	28

• Prepare a frequency table, using class interval: 1 – 20, 21 – 40, 41 – 60, …….
• Use the table to draw a histogram.

 

 

 

 

 

 

 

 

Solution

(a)

Class interval                             Tally                         Frequency

1 – 20                            1111   11                            7

 

21 – 40                         1111   1111   111               13

 

41 – 60                          1111   1111                         9

 

61 – 80                           1111   1                              6

 

81 – 100                           1111                                 5

 

(b)

Class Activity:

1. Drawing histogram using class boundary, draw a histogram for the frequency distribution in the table below. (Use class boundaries to plot against the frequency).

NB: The use of class boundaries to plot against the frequency is important in using histogram to estimate the mode.

Class	1 – 5	6 – 10	11 – 15	16 – 20	21 – 25
Frequency	2	4	6	5	3

 

 

 

2. The table below shows the marks obtained by forty pupils in a Mathematics test.

Marks                     0 – 9	10 – 19	20 – 29	30 – 39	40 – 49	50 – 59
No of pupils               4	5	6	12	8	5

Draw a histogram for the mark distribution

PIE CHART.

A pie chart is a circular shape which is divided into sections whose angles are proportional to the frequencies of the items.

Steps to Drawing a  Pie Chart

Step 1.

Calculate the angles of sector of the pie chart using  x  360⁰. Where x is the frequency of the total frequencies of all the items.

Step 2.

Using a pair of compasses, draw a circle with a suitable radius.

Step 3.

Partition it into the various angles of sectors obtained in step 1 using a protractor.

Step 4.

Label each sector showing the information which relates to the sector.

Step 5

Give a title to your pie chart, by stating the information the pie chart is representing.

Example 1:

The table below shows the expenditure of a Bachelor in a given month

Items	Expenditure N1000
Food	10
Rent	5
Bills	3
Clothing	2
Savings	6
Miscellaneous	4

 

Draw a pie chart to illustrate the information

Solution

First, we shall calculate the angle of sector for each item as follows.

Items	Expenditure in N1000	Angle of Sectors
Food	10	X = 1200
Rent	5	X = 600
Bills	3	X = 360
Clothing	2	X = 240
Savings	6	X = 720
Miscellaneous	4	X = 480
Total	30	=3600

 

 

Title: Pie chart showing the expenditure of Bachelor in a certain month.

Example 2:

In a certain school, the lesson periods for each week are as itemised below.

English 10, Mathematics 7, Biology 3, Statistics 4, Ibo 3, others 9.

Draw a pie chart to illustrate this information

Solution

Subjects	Periods	Sector (degrees)
English	10	X = 1000
Mathematics	7	X = 700
Biology	3	X = 300
Statistics	4	X = 400
Ibo	6	X = 300
others	9	X = 900
Total	36	=3600

 

 

 

Title : pie Chart showing the lesson periods for each week in a certain school.

Example 3:

The pie chart illustrates the amount of private time a student spends in a week studying various subjects. Find the value of k.

 

Solution

Since the total angles of all sectors in a given circle is 360⁰.

105⁰ + 75⁰ + (2k)⁰ + k⁰ + (3k)⁰ = 360⁰

105⁰ + 75⁰ + (6k)⁰ =360⁰

1800 + 6k =360⁰

6k = 360⁰ -180⁰

6k = 180⁰

Divide through by 6

6k/6 = 180⁰/6

K = 30⁰.

Example 4:

The pie chart represents the fruits on display in grocery shop. If there are too oranges on display. How many apples are there?

 

Solution

The only sector that has useful information is that of oranges.

Number of oranges = 60

Sector angle of orange = 360 – (60 + 100 + 120)

= 360⁰ – 280⁰

= 80⁰.

But  x   = 80⁰.

x   = 80⁰

= 80/1

Cross multiplying

21600 = 80(total fruit)

Dividing both sides by 80

=

= 270

Total number of fruit is 270.

Let the number of apples be x, then;

x   = 120⁰

=

= 90⁰

Example 5:

The Pie chart below shows the distribution of students in a certain school into some major ethnic groups in Nigeria.

 

 

• In its simplest form, what fraction of the students are Igbos?
• What percentage of the students are Hausas?
• What is the ratio of the Binis to Hausas in the school in its simplest terms.

 

Solution

• The angle of the sector representing Igbos is 70⁰.

= 70/360 of the students are Igbos.

= 70/360 = 7/36

• The angle of sector for Hausa is x⁰and this can be obtained using

110⁰ + x⁰ + 70⁰ + 60⁰ + 90⁰ = 360⁰

330⁰ + x⁰ = 360⁰

x = 30⁰.

Taking the percentage of the fraction, the students who are Hausas use

= 30/ 300 x 100/1

= 1/12 x 100/1

= 100/12

8.3%

 

• The angle of sector for binis = 90⁰

The angle of sector for Hausas= 30⁰

ratio of Binis to  Hausas 90⁰ : 30⁰

= 3 : 1

Example 6:

The Pie chart below shows the weekly sales of a motor dealer in Lagos in 1999.

 

 

• What fraction of the cars were Toyota?
• What percentages of the Cars were Datsun?
• If the dealer sold 16 Peugeots, How many BMW did he sell in a week?

Solution

• The angle of the sector representing Toyota is 120⁰.

=  120/ 360

= 12/36

= 1/3

 

• Taking the percentage of the fraction of the Datsun Cars

=  90/ 360 x 100/1

= 1/4 x 100/1

= 25%

• To get the total number of Cars, we will first get the angle of sector for Peugeot.

70 + 120 + 90 + x + 360

280 + x = 360

X = 360 – 280

x = 800

Assuming the total number of Cars to be y, then

16/y x 360/1 = 80⁰

5760/y =80/1

Cross multiplying

5760 = 80y

Divide both sides by 80

5760/80=80y/80

y = 72⁰.

To get the number of BMW sold in a week, let the number of the BMW sold be m

m/72 x 360/1 = 70⁰

360m/72 = 70

Cross multiplying

360m = 5040

Dividing both sides by 360

360m/360 = 5040/360

m = 14

 

Class Activity:

• The pie chart below shows the allocation of money to the different departments in a secondary school.

 

 

If applied Science Department were allocated the sum of N120,000.00. What was the total allocation to Mathematics Department?

2.   In a certain year, government bought 240 Cars; 110 were Peugeots, 54 were Datsuns, 35 were Fords,       28 were Hondas and the rest were Volkswagens. Draw a pie chart to represent the above information.

Frequency polygon

The frequency polygon is obtained by joining the midpoints of the tops of the histogram rectangles with straight lines. Since the polygon is a closed figure, we take one interval below the lowest internal on the base axis. Then we join the ends of the polygon to the midpoints of the new intervals as shown below. Class Activity:

The table shows the distribution of the masses of fifty logs exported in September 1981 by a timber and plywood company.

Mass (kg)	No of Logs
150 -154	1
155 – 159	4
160 – 164	8
165 – 169	13
170 – 174	12
175 – 179	8
180 – 184	3
185 – 189	1

Construct the frequency polygon for the distribution

PRACTICE EXERCISE:

1. The body temperatures of some patients in a hospital ward one morning are as follows:   Illustrate this data in a bar chart.

No. of beds	P	Q	R	S	T
Temperature	980	1000	101.10	780	1050

2. A dice is thrown 27 times. The table below shows the distribution of the number of times each number showed up.  Present this data in a bar chart.

Face	1	2	3	4	5	6
Frequency	4	6	8	5	3	1

3. A class of students had a test. The following table gives the number f of students obtaining various marks x

	0	1	2	3	4	5	6	7	8	9	10
	2	3	3	4	6	8	5	4	3	2	0

Illustrate the information on a histogram.

4. The table below shows how a company’s sales manager spent his 2017 annual salary:

Food	30%
Rent	18%
Car Maintenance	25%
Savings	12%
Taxes	5%
Others	10%

• Represent this information on a pie chart.
• Find his savings at the end of the year if his annual salary was #60,000.00

5. The student intake at a University for a particular year was distributed among its faculties as follows:

• Illustrate the data below on a pie chart
• What is the percentage of students admitted into the faculty of science?

Faculties	Numbers
Agriculture	200
Arts	372
Law	56
Science	540
Technology	272

 

 

ASSIGNMENT:

1. The pie chart represents the fruits on display in a grocery shop. If there are 60 oranges on display, how many apples are there?

 

 

 

 

 

 

 

2. The table below shows how a man spends his income in a month.

Items	Amount spent
Food	N4500
House rent	N3000
Provisions	N2500
Electricity	N2000
Transportation	N5000
Others	N3000

• Represent this information on a pie chart.

 

• What percentage of his income is spent on transportation?

 

3. The following data shows the marks of 40 students in a history examination.

41   52   37   56   63   48   65   46

54   32   51   66   74   23   35   61

58   44   49   53   45   57   56   38

59   28   50   49   67   56   36   45

76   68   43   56   26   47   55   71

1. Form a grouped frequency table with the class intervals 20 – 29, 30 – 39, 40 – 49 etc.
2. Draw the histogram of the distribution.

 

 

4. The distribution of junior workers in an institution is as follows;

Clerk         78

Drivers         36

Typists         44

Messengers        52

Others                 30

Represent the above information by a pie chart.

5. The table below shows the marks obtained by forty pupils in a mathematics test.

Marks	0-9	10-19	20-29	30-39	40-49	50-59
Number of pupils	4	5	6	12	8	5

 

• Draw a histogram for the mark distribution
• Use your histogram to estimate the mode

Hope you got what you visited this page for? The above is the lesson note for Mathematics for SS1 class. However, you can download the free PDF file for record purposes.

If you have any questions as regards Mathematics lesson note For SS1 class, kindly send them to us via the comment section below and we shall respond accordingly as usual.