Mathematics Lesson Note For SS1 (Second Term) 2024

Mathematics lesson note for SS1 Second 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 Second Term has been provided in detail here on schoolgist.ng

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 Second 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 Second 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 (Second Term) 2024

SS1 SECOND TERM: E-LEARNING NOTES

SCHEME SECOND TERM

WEEK	TOPIC	CONTENT
1	MODULAR ARITHMETIC 1	(a) Revision of addition, subtraction, multiplication and division of integers. (b) Concept of module arithmetic.
2	MODULAR ARITHMETIC 2	(a) Addition, subtraction, multiplication and division operations in module arithmetic. (b) Application to daily life.
3	QUADRATIC EQUATION 1	(a) Revision of factorization of quadratic expressions. (b) Solution of quadratic equation of the form: ab = 0 i.e. a = 0 or b = 0. (c) Formation of quadratic equation with given roots.
4	QUADRATIC EQUATION 2	(a) Drawing quadratic graph. (b) Obtain roots from a quadratic graph. (c) Application of quadratic equation to real life situations.
5	CONSTRUCTIONS 1	Revision of (i) Construction of triangles with given sides. (ii) Bisection of an angle; 30⁰, 45⁰, 60⁰ and 90⁰.
6	CONSTRUCTIONS 2	Construction of (i) An angle equal to a given angle. (ii) 4-sided plane figure given certain conditions. (iii) Locus of moving points equidistance from 2 lines, 2 points, and constant distance from a point, etc.
7	MID-TERM BREAK
8	PROOFS OF SOME BASIC THEOREMS 1	Proofs of (i) sum of a triangle is 180⁰ (ii) The exterior angle of a triangle is equal to the sum of two interior opposite angles. (iii) Congruency and similarity of triangles.
9	PROOFS OF SOME BASIC THEOREMS 2	Riders including – (i) angles of parallel lines (ii) angles in a polygon (iii) congruent triangles (iv) properties of parallelogram (v) intercept.
10	REVISION
11	EXAMINATION

WEEK1 DATE……………………

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: Modular Arithmetic

CONTENT:

Revision of addition and subtraction of integers
Revision of multiplication and division of integers
Concept of modular arithmetic/Cyclic events

Revision of addition, subtraction, multiplication and division of integers

Recall: Integer is a counting whole numbers, either positive or negative. Examples

These numbers can be added, subtracted, divided or multiplied.

Addition of integers;
Subtraction of integers;
Multiplication of integers;
Division of integers;

Class Activity:

Solve the following;

Concept of Modular Arithmetic

The word Modular implies consisting of separate parts or units which can be put together to form something, often in different combinations.

Arithmetic the science of numbers involving adding, subtracting, multiplying and dividing of numbers

Modular Arithmetic is the type of arithmetic that is concerned with the remainder when an integer is divided by a fixed non-zero integer. The word remainder as used in definition practically refers to the excess in number, after full cycles have been completed.

Examples;

Reduce 65 to its simplest form in:

modulo 3 (b) modulo 4 (c) modulo 5 (d) modulo 6

Solution

If 20 oranges are to be put into bags that can contain a maximum of eight oranges in each , calculate
The number of bags that will be filled with the oranges
The number of oranges in the bag with some space left.

Solution

2 bags will be filled with oranges
4 oranges will be in the bag with space left

Class Activity:

Reduce 72 to its simplest form

Modulo 3
Modulo 4
Modulo 5
Modulo 6
Modulo 7

Cyclic Events: Cyclic means happening in cycles.

Just as you ride your bicycle, the wheel rotates from a point to another. There are events that have constant ice day’s interval of three days, four, five or a week.

Examples: If ice cream is served every three days. If you are served on Thursday, the next serving will be

Find the number which results from the following additions on the number cycle below of ice cream

Class Activity:

Use the number cycle 5 to simplify

3+35

PRACTICE EXERCISE

Thirty nine oranges are to be put into bags that have a capacity of 7 oranges. If the bags are to be filled in turn, how many oranges would be required to fill the last bag. How many bags will be needed to contain all the oranges
State the quotient on division of
6 by 3
15 by 7
Use the cyclic 4 to simplify the following
3 +13
1 + 3
Reduce 35 to simplest form
Module 3
Module 4
Arrange the days of the week, Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday on a circle using the number code 0 for Sunday, 1 for Monday, 2 for Tuesday , 3 for Wednesday, 4 for Thursday, 5 for Friday and 6 for Saturday. If Thursday, which day will it to be in,
5 day’s time
10 day’s time

ASSIGNMENT

Arrange the days of the week, Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday on a circle using the number code 0 for Sunday, 1 for Monday, 2 for Tuesday , 3 for Wednesday, 4 for Thursday, 5 for Friday and 6 for Saturday. If Thursday, which day will it to be in,
7 day’s time
39 day’s time
Use the cyclic 4 to simplify the following
0 + 12
2 +32
State the quotient on division of

6 by 8
-13 by 5

15 by -7

Reduce 35 to simplest form
Module 5
Module 6
Find the simplest form of the following in given moduli
-5(mod 6)
-52(mod 11)

-75 (mod 7)

-50(mod 4)

KEYWORDS: Arithmetic, Modular, cyclic, events e.t.c

WEEK2 DATE…………………….

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: Modular Arithmetic

CONTENT:

Addition, Subtraction, Multiplication and Division operations in module arithmetic
Application to daily life

ADDITION AND SUBTRACTION:

In modular arithmetic, addition and subtraction are symbolized by ⊕ and ⊖ respectively. In the table we enter only the remainders when a pair of elements in the ⊕ on the set {0,1,2,3,4} modulo 5

⊕	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	0
2	2	3	4	0	1
3	3	4	0	1	2
4	4	0	1	2	3

Addition ⊕ in mod 5

Simplify Find 39 ⊕ 29(mod 6)

Solution: 39 ⊕ 29= 68

= (6×11+2)

= 2(mod 6)

N.B

Calculate the following in the given moduli (a) 12Θ5(mod 4) (b) 38 Θ42(mod 7)

Solution: (a) 12Θ5 = 7

7 = 4 + 3

= 3(mod 4)

(b) 38 Θ 42 = 4

4 = 7 + 3

= 3(mod 7)

Class Activity:

Find the following additions modulo 5
3 9
65 32
41 52
8 17
Find the simplest positive form of each of the following numbers modulo 5

Multiplication of modulo

Examples: Evaluate the following modulo 4

2 ⊕ 2
5 ⊕7
6 ⊕73

Solution:

2 ⊕2 = 4

= 4 + 0(mod 4)

= 0(mod 4)

5 ⊕ 7 = 35

= 4 x 8 + 3

= 3(mod 4)

6⊕73 = 438

= 4 x 109 + 2

= 2(mod 4)

Class Activity:

Find the values in the moduli written beside them

16 7(mod 5)
21 18(mod 10)
8 25(mod 3)
27 4(mod 7)
80 29(mod 7)

DIVISION OF MODULO

Examples: Find the values of the following;

23(mod 4)
72(mod 5)
22(mod 4)

Solution:

If 23 =

⇒

Cross-multiply ,

Add 4 to RHS

3 = 2 + 4(mod 4)

3 = 6(mod 4)

Divide both sides by 3

= 2(mod 4)

72 =

2 = 7(mod 5)

2 = (5 x 1) + 2(mod 5)

2 = 2

= 1

22 =

2 = 2(mod 4)

Divide both sides by 2

= 1

2 = 2 + 4(mod 4)

2 = 6(mod 4)

= 3(mod 4)

N.B If 32 = , then 2 = 3

No multiple of 4 can be added to 3 to make it exactly divisible by 2. There are no values of 32 in modulo 4.

Class Activity:

Calculate the following division in modulo 5

N.B Educators should also solve various examples.

PRACTICE EXERCISE :

Copy and complete the table for addition (mod 5)

	0	1	2	3	4
0					4
1
2		3
3
4	4

Copy and complete the table for subtraction modulo 6

Θ	0	1	2	3	4	5
0
1
2
3
4

Complete the multiplication modulo 5 in the table below

	0	1	2	3	4	5
0	0	0	0	0
1	0
2	0
3	0		1
4					1
5	0				0

Simplify the following

8-5+2 in mod 6

Evaluate

APPLICATION OF MODULAR ARITHMETICS TO DAILY LIFE

Time plays crucial role in indicating how often events occur or qualities vary. Whenever the time rate of occurrence of events is constant, the order of occurrence is repeated.

Lets consider these basic facts:

1 rotation of hour of hand of the clock or record 12hrs (half day)
1 rotation of the minute hand of the clock or watch record 60mins or 1hr

1 rotation of the second hand of the clock or watch records 60 secs or 1 minute.

1 rotation of 7days record 1 week
1 rotation of 24 hours record is 1 day

Etc.

Examples

If the hour hand of a clock is at 2:00 a.m. What time of the day will it indicate after 20 rotations

Solution

We apply mod 12 since the movement is that of the hour hand

We then have

Therefore the day is 8hrs from 2’0 clock into its second phase

The require time 2 + 8= 10pm

The market in a village holds every 6days. If the current market is on Wednesday when will the next market held?

Solution

We operate in mod 7

Then Wednesday

Next market holds 6 days after

Total number of days from Sunday

i.e 9

the next market is 2 days after Sunday

therefore the expected market day is Tuesday

ASSIGNMENT:

Construct an addition table in mod 5. Use it to evaluate:
Hint: 3
3 + 2

4 + 3 + 3

The minute of a stop watch is 3. Where will it be if it were round
rotations clockwise
rotation anti-clockwise

rotations anti-clockwise

Find the complete set of solutions to the following
Dayo attends a sports club as a member every five days. If he made his fifteenth attendance on a Thursday, when did he first attend the club as a member.
Construct an operation table for multiplication and another for addition in module 7

KEYWORDS: Arithmetic, Modular, cyclic, events e.t.c

WEEK 3 Date……………………….

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: QUADRATIC EQUATIONS

CONTENT:

Revision of linear and quadratic expressions
Solution of quadratic expression of the form ab=0, a=0 or b=0
Formation of quadratic equation with given roots

REVISION OF LINEAR AND QUADRATIC EXPRESSIONS

Any expression in which highest power of the unknown is 1 is called a linear expression. Some examples of linear expressions are; (a) x + 1 (b) 2y + 3 (c) p

In general, linear expressions are expressions of the form , where a & b are constants and x is a variable.

A quadratic expression is that whose highest power of the unknown is 2. Examples are; (a) (b)

Factorization of quadratic expressions;

Examples; (i) Factorize the following quadratic expressions

Solution:

, is a common factor of the terms . Hence can be written as isolating common factors, we have
, the common factor of the terms is

can be written as , hence we obtain

(ii) Factorize the following;

(a) (b) (c)

Solution: (a) , find the product of the first and last terms . Find two terms such that their product is and their sum is

Factors of	sum of factors
(a)
(b)
(c)
(d)
(e)
(f)

Of these, only (d) gives the required result. Replace with in the given expression. Then factorize by grouping the terms.

, 3 is common factor , first take out the common factor.

Factors of	Sum of factors
+6a and +a	+2a
+3a and +2a	+5a

, find the product of the first and last terms i.e

Find two terms such that their sum is and their product is . Since the middle term is negative, consider negative factors only. The terms are , replace with in the given expression.

Class Activity:

Factorize the following quadratic expressions

SOLUTION OF QUADRATIC EXPRESSION OF THE FORM ab=0, a=0 OR b=0

If the product of two numbers is 0, then one of the numbers (or possibly both of them) must be zero. For example, , .

In general, if

Examples 1. Solve the equation

Solution:

If , then either

⇒

Solve the equation

If , then either

⇒

Solve the equations (i) (ii)

Solution: (i) if

Then,

⇒

(ii) , then any one of the four factors of LHS may be 0

i.e

⇒

Class Activity:

Solve the following equations;

(8 – v)( 8 – v)

FORMATION OF QUADRATIC EQUATION WITH GIVEN ROOTS

The roots of a quadratic equation are the solutions of that equation. Suppose the roots of a quadratic equation in are , then we can write;

Examples; (1) Find the quadratic equation whose roots are

Solution: let , then

On careful expansion, we obtain

Find the quadratic equation whose roots are

Solution: If the roots are

Let

⇒

Class Activity:

Form the quadratic equations whose roots are;

3 and – 4

PRACTICE EXERCISE

Find the quadratic equation whose roots are
²²²²

² (SSCE 1988)

Find the roots of the equation ²
¹/₂ B.

¹/₂ or 2 (SSCE 1988)

Solve the following equation: ²
x=¹/₂or x=-2¹/₂ B. x=¹/₃ or x=-2¹/₂ C. x=1²/₃or x=-¹/2 D. x=-1²/₃ or x=¹/₂
x=⁵/₆or x=-1 (SSCE 1989)

Solve for ²

(SSCE 1989)

Solve the quadratic equation 3x²+ 4x + 1 = 0

ASSIGNMENT

Factorize ²²
(SSCE 1996)

Factorize ²

(SSCE 2001)

Which of the following is not a quadratic expression?

²²

Factorize

(SSCE 2002)

One of the factors of ²is.The other factor is:
(SSCE 2011)KEYWORDS: EXPRESSION, EQUATION, QUADRATIC, LINEAR, ROOTS

WEEK 4 Date……………………………………..

SUBJECT: MATHEMATICS

CLASS: SS 1

TOPIC: QUADRATIC EQUATION

CONTENT:

Revision of linear graph and drawing quadratic graph
Obtaining roots from a quadratic graph
Finding an equation from a given graph
Application of quadratic equation to real life situations

LINEAR GRAPHS

Recall that any equation whose highest power of the unknown is 1 is a linear equation. To draw the graph of a linear equation, we need to

Make a table of value for the equation
Plot the graph of the linear equation

Example: Draw the graph of

Solution; y = x – 1

-2	0	2
-1	-1	-1
-3	-1	1

Scale: 2cm to 1unit on both axes

y-axis

Y=x-1 graph of y=x-1

x-axis

-1

-2

DRAWING QUADRATIC GRAPH: To draw a quadratic graph, we need to also follow the same process of drawing linear graph

Example: Draw the graph of

Solution; since , we shall now make a table for the values of x & y.

-3	-2	-1	0	2	3
9	4	1	0	4	9
-6	-4	-2	0	4	6
1	1	1	1	1	1
4	1	0	1	9	16

Note: when plotting the graph,

we choose a scale such that our graph is as large as possible and also occupies the centre of the graph sheet. This will enable us to obtain the point where the graph cuts the x-axis more easily.
We join the points in the graph by a smooth curve

Class Activity:

Draw the graph of

OBTAINING ROOTS FROM A QUADRATIC GRAPH

To obtain the roots of a quadratic equation form a quadratic graph, we need to first plot the graph of the expression and then obtain the roots by reading the two values of x where the graph cuts or touches the x-axis, i.e where y=0

Example: Draw a graph to find the roots of the equation

Solution;

0	1	2	3	4	5
0	4	16	36	64	100
0	-20	-40	-60	-80	-100
25	25	25	25	25	25
25	9	1	1	9	25

From the graph it is clear that the curve does not cut the x-axis. It appears to touch the x-axis where x=2.5. this result can be checked by factorisation.

i.e

(2x-5)(2x-5)=0

Note: when the curve touches the x-axis, the roots are said to be coincident

Class Activity:

Use the table of values below to solve the equation graphically for

	-4	-3	-2	-1	0	1	2	3
				-8				4

FINDING AN EQUATION FROM A GIVEN GRAPH

It is possible to find the equation of a curve from its graph. The graph of cuts the x-axis (i.e the line y=0) at the points x=-1 and x=3. This implies that -1 and 3 are the roots of the equation . Therefore in general if a graph cuts the x-axis at points a & b, it satisfies the equation (x-a)(x-b)=0

Example 1: Obtain the equation of the graph below

y-axis

2 P

0 1 2 x-axis

Solution; From the graph when y=0, x=1 and x=2,

Then (x-1)(x-2)=0

.at point P, y=2 when x=0

Example 2: obtain the equation of the graph below (WAEC)

y-axis

-1 -0.5 0 1 2 x-axis

-2

Solution: In the graph above, where y=0, x= and x=2

Second, in the curve above, at point P, y=-2 when x=0. However the constant term in equation (i) is only -1. So we multiply both sides of the equation (i) by 2

Equation (ii) satisfies and the requirement that the constant term should be -2

The equation of the curve is

Class Activity:

Find the equation of the graphs below

Y-axis Y-axis

-2.5 -1 0 1 2.5 3.5 -2 -1.5 -1 0 1 1.5 2 x-axis

-1

APPLICATION OF QUADRATIC EQUATION TO REAL LIFE SITUATIONS

Example 1: The area of a rectangle is The length is 11cm more than the width. Find the width.

Solution:

Let the width be , length =

.length will be cm.

The length is 11cm more than the width gives

Simplifying;

i.e

factorizing completely we have,

. the width is 4cm since it cannot be negative.

Check: length = = 15 = 4 + 11 and 15 x 4 = 60

Class Activity:

When 11 times a certain integer is subtracted from twice the square of the integer, the result is 21. What is the integer?
A rectangular lawn is 4cm longer than its width. If its area is 165, calculate its width
Musa is 60years old and Joy is 25years old. How many years from now will the product of their ages be 2244years?

PRACTICE EXERCISE:

Solve the following equations graphically and obtain the least value of y

The sum of the ages of a mother and her child is 63. If the product of their ages four years ago was 484. What are their ages now?
Find the equations of the graphs below

Y-axis (b) y-axis

-5 0 1 x-axis -1 0 3 x-axis

-3

The following is a graph of a quadratic function. Use it to answer question 3 and 4.

Find the co-ordinates of point
(0, 4) (1, 4) C. (0, -4) D. (-4, 0) (SSCE2009)
Find the values of when
1, 3 B. 1, 4 C. 2, 3 D.1, 6 (SSCE 2009)

(SSCE 2006)

The figure is the graph of a quadratic equation. Use the information to answer questions 1 and 2.

What are the roots of the equation?
-1 and 1 4 and -1 C. 1 and 4 D. 4 and 4 (SSCE 2005)

What is the equation of the curve?
C. D.

ASSIGNMENT

Draw the graph of 25 and for.

Use the graph(s) to:

find the roots of the equation 25 and
determine the line of symmetry of the curve25.

(a) Copy and complete the table of values for the relation for.

x	-3	-2	-1	0	1	2	3
y		-4		2			-4

(b)Using scales of 2cm to 1 unit on the x-axis and 2cm to 2 units on the y-axis.

Draw a graph of the relation

minimum value of y
roots of the equation

iii. gradient of the curve at (SSCE 2010)

KEYWORDS: EXPRESSION, EQUATION, QUADRATIC, LINEAR, ROOTS, MINIMUM VALUE, MAXIMUM, INTERCEPT ETC

WEEK 5 DATE……………………….

Subject: Mathematics

Class: SS 1

TOPIC: Constructions

Content:

Guidelines for constructions
Constructions of basic angles
Construction of triangles with given sides and angle.
Bisection of basic angles

GUIDELINES FOR CONSTRUCTIONS

When making constructions, the following guidelines should be followed.

A short pencil of about 3 inches should be fixed on the pair of compasses when constructing to avoid any obstruction when turning your compass round to draw arcs.
Ensure that the pivot of your pair of compasses is tight to avoid unwanted shift when carrying out your constructions.
To ensure that your lines and points are as fine and accurate as possible make use of a hard pencil with a sharp point.
Before making the actual construction, make a rough sketch of the problem under consideration. This will make the construction of the actual problem easy.
Leave all your arcs and construction lines visible. Do not clean any arc that leads you to your final result.
Double lines and arcs in constructions are not allowed, hence clean up all double arcs and lines neatly and re-draw.

Bisecting Of Angles

To bisect the angle ABC,

Step 1: –Measure a length of about 2.5cm with your pair of compasses. Fix pin at B and draw an arc to cut the two line AB and BC that formed the angle. Cut the arc at x and y.

Step 2: – Fix pin at x and draw an arc and at y and draw another arc both to meet at z.

Step 3: – Join the point zB. \ABz = zBC

z y

θ^◦

A θ^◦ B

Bisecting A Line AB:

To bisect the line AB

Step 1: – Measure the line AB with your ruler.

Step 2: – Measure about ¾ of the given line with your compasses, use for step 3 and step 4.

Step 3: – Fix pin at A and draw an arc above the line AB and below the line AB.

Step 4: – Fix pin also at B and draw an arc above the line AB to cut the arc drawn in step 3 at x and draw another arc below to cut the other arc drawn in step 3 at y.

Step 5: – Join the points xy with your ruler and produce both ways.

A 8cm B

Class Activity:

What are the basic guidelines for constructions?

CONSTRUCTIONS OF THE BASIC ANGLES

To construct the basic angles, we shall make use of the line AB of length 8cm for each case.

NOTE:

Angles constructed at the point A are normally read from the right side of A to left using the protractor and angles constructed at point B are normally read from the left side of B to the right.

Constructions Of The Basic Angles

To construct the basic angles, we shall make use of the line AB of length 8cm for each case.

Note:

Angles constructed at the point A are normally read from the right side of A to left using the protractor and angles constructed at point B are normally read from the left side of B to the right.

Angle 60⁰

Step 1: – Measure a length of about 2.5cm with your pair of compasses

Step 2: – Put pin at A and draw an arc in form of a semi-circle to cut the line AB at x.

Step 3: – Using the same length as in step 1 draw an arc putting pin at x cut the arc in step2 at y.

Step 4: – Join Ay and produce. Angle yAB = 60⁰.

60⁰

A B

Angle 120⁰ (Obtained by constructing 60⁰ + 60⁰)

The steps for constructing angle 60⁰ above is followed from step 1 to step 3.

Step 5: –Using the same arc length as in step 1 above put pin at point y and draw an arc to cut the arc in step2 at point z.

Step 6: – Join Az and produce

Angle zAB = 120⁰

z y

120⁰

A x B

Class Activity

Construct all the basic angles shown above at point B on the line AB. Where AB = 8cm.

Angle 90⁰ (Obtained by constructing 60⁰ + 30⁰)

The steps for constructing angle 120⁰ above are followed from step 1 to step 5.

Step 6: – Bisect the angle formed by arc yz to have 30⁰ added to angle yAx = 60⁰. This is done by drawing an arc with compass pin at point z and another arc with pin at point y, both to meet at point p.

Step 7: – Join Ap produced. Angle PAB = 90⁰

P P₁

90⁰

z y y₁z₁ A x x₁ B

NB: – To construct angle at point B as shown above, the first arc is drawn by fixing pin at B and cutting arc from x₁.

Angle 75⁰ (obtained by constructing 60⁰ + 15⁰)

The steps for constructing angle 90⁰ above are followed from step 1 to step 7.

As for the step 7, the line Ap is drawn using broken line, since the required angle is not 90⁰. The angle, 90⁰ will only aid us in constructing 75⁰.

Step 8: – Fix pin at point where Ap cut arc yz draw an arc and also fix pin at y and draw an arc both to meet at q.

Step 9: – Join Aq and produce \ qAB = 75⁰

z y

75⁰

A x B

Angle 105⁰ (obtained by constructing 90⁰ + 15⁰)

The steps for constructing angle 90⁰ above are followed from step 1 to step 7.

As for step 7, the line Ap is drawn using broken line, since the required angle is not 90⁰.

The angle 90⁰ will only aid us in constructing 105⁰.

Step 8: – Fix pin at point where Ap cut arc yz draw an arc and also fix pin at z and draw an arc both to meet at q.

Step 9: – Join Aq and produce \ qAB = 105⁰

z y

105⁰

A x B

Class Activity

Construct all the basic angles shown above at point B on the line AB. Where AB = 8cm.

Angle 135⁰ (obtained by constructing 90⁰ + 45⁰)

The steps for constructing angle 90⁰ above are followed from step 1 to step 7.

As for step 7, the line Ap is drawn using broken line, since the required angle is not 90⁰. The angle 90⁰ will only aid us in constructing 135⁰.

Step 8:– Produce BA to cut the 1^st arc at point n.

Step 9:–Fix pin at point n and draw an arc, fix pin also at the point where Ap cut arc yz and draw an arc both to meet at q.

Step 10:– Join Aq and produce \ qAB = 135⁰

q z y

135⁰

n A x B

Angle 45⁰(obtained by constructing 90⁰¸2)

The steps for constructing angle 90⁰ above are followed from step 1 to step 7.

As for step 7, the line Ap is drawn using broken line, since the required angle is not 90⁰.

Step 8: – Bisect angle pAB (90⁰) to get 45⁰.

i.e. Fix pin at x and draw an arc and fix pin at the point where pA cuts arc zy and draw another arc both to meet at q.

Step 9: –Join Aq and produce.

\ qAB = 45⁰

z y

45⁰

A x B

Angle 30⁰:

This is obtained by bisecting angle 60⁰.

Step 1:- Construct angle 60⁰ with broken line since the required angle is not 60⁰.

Step2: –To bisect the angle, fix pin at x and draw an arc and also fix pin at y and draw another arc both to meet at p.

Step3: –Join Ap and produce. \pAB =30⁰

30⁰

B x A

Note that:

Other angles can be obtained by bisecting these basic angles that have been drawn already.

e.g. – 15⁰ can be obtained by bisecting 30⁰

22½ can be obtained by bisecting 45º
52½ can be obtained by bisecting 105⁰
67½ can be obtained by bisecting 135⁰
37½ can be obtained by bisecting 75⁰

In practical construction, the letter x, y, z, p and q are not indicated at the points where the arcs are drawn. We did this in this text to enable you understand the steps in constructing the basic angles. Also small letters are not used to label points in Geometry.
These basic angles can also be constructed at the point B on the line AB. In this case the arcs x, y, z are drawn from the left side of B. The angles on the protractor are also read from the left side of B to the right.

Class Activity

Construct all the basic angles shown above at point B on the line AB. Where AB = 8cm.
Using a ruler and a pair of compasses only, construct the following angles at the point B on the line AB = 6.5cm.

(a) 60⁰ (b) 120⁰ (c) 90⁰ (d) 45⁰ (e) 135⁰ (f) 75⁰ (g) 105⁰ (h) 52½⁰ (i) 22½⁰ (j) 37½⁰ (k) 67½⁰ (l) 30⁰ (m) 15⁰ (n) 150⁰

CONSTRUCTION OF TRIANGLES:

Case 1: When only the three sides are given.

Example 1: C

Construct a triangle ABC such that

AB = 8cm, BC = 7cm and AC = 6cm.

Solution:

Step 1: – Draw line AB = 8cm with your ruler.

Step 2: – Measure 7cm from your ruler with 6cm 7cm

your pair of compasses, fix pin at B and

draw an arc at the suspected position of C.

Step 3: – Measure 6cm from your ruler

With your compasses, fix pin at A and A 8cm B

draw an arc to cut the arc drawn in step 2 at C.

Step 4: – Join AC and BC and label the vertices and sides.

Case 2: When given two sides and the included angle.

Example 2:

Construct a triangle PQR such that

PQ = 7.5cm PQR = 60⁰ and QR = 6cm.

Solution:

Step 1: – Draw the line PQ = 7.5cm R

with your ruler.

Step 2: – Construct the angle 6cm

60⁰ at the point Q.

Step 3: – Measure 6cm from your ruler

with your pair of compasses. Fix pin at

point Q and cut an arc on QR at R. 60⁰

Step 4: – Join PR to form a triangle. P 7.5cm Q

Class Activity

(1) Using a ruler and a pair of compasses only, construct the following triangles.

(a) D ABC, such that ôABô = 8cm, ôBCô = 6cm and ôACô = 7cm. Measure CAB

(b) D ABC, such that ôABô = 9cm, ôBCô = 8cm and ôACô = 7cm. Construct also a circle passing through A, B and C. Find the radius of the circle.

(d) D PQR, such that ôPQô = 7.5cm, PQR = 75⁰ and QR = 6cm. Measure PR.

(e) D XYZ, such that YZ = 7cm, XYZ = 45⁰ and XY = 10cm. Measure XZ.

Case 3:When given one side and two angles.

Example 3:

Construct a triangle XYZ such that XY = 8cm, XYZ = 60⁰ and ZXY = 45⁰

Solution:Step 1: – Draw the line XY = 8cm using your ruler.

Step 2: – Construct angle 60⁰ at point Y.

Step 3: – Construct angle 45⁰ at point X.

Step 4: – Produce the line for angle 60⁰ from Y and the line for angle 45 from X to meet at point Z Z

45⁰ 60⁰

X 8cm

Case 4: When given two sides and an angle.

Example 4:

Construct a triangle ABC such

that AB = 7.5cm, ABC = 75⁰ and

AC = 9cm.

C

Solution:

Step 1: – Draw the line AB = 7.5cm

using your ruler.

Step 2: – Construct angle 75⁰ at B.

Step 3: – Measure 9cm from 9cm

your ruler with a pair of compasses,

fix pin at point A and cut

an arc on the line BC at C.

Step 4: – Join the line AC 75⁰

to form the triangle. A 7.5cm B

Class Activity

(i) D ABC, such that ôABô = 6.8cm, ABC = 60⁰ and CAB = 75⁰. Measure AC.

(j) D XYZ, such that XY = 9cm, XYZ = 45⁰ and ZXY = 75⁰. Construct also a circle passing through the points X, Y and Z. Measure the radius of the circle.

(k) D PQR, such that PQ = 6.5cm, PQR = 75⁰ and ôPRô = 9cm. Measure QR.

(l) D XYZ, such that XY = 7cm, ZXY = 60⁰ and ôYZô = 10cm. Measure XZ.

(m) D PQR, such that ôPQô = 10cm PQR = 37½ and RPQ = 60⁰. Measure QR.

(n) D PQR, such that ôPQô = 7cm, PQR = 52½ and QR = 6cm. Measure PR.

PRACTICE EXERCISE

DPQR, such that ôPQô = 8.4cm, PQR = 90⁰ and QR = 6.5cm. Construct a circle also passing through P, Q and R. What is the radius of the circle?
Construct the DABC such that AB= 6cm ,BC = 5.5cm and 0
Using a ruler and a pair of compasses only, construct;

a triangle PQR such that /PQ/ = 10cm, /QR/ = 7cm and PQR = 90⁰