Mathematics at a glance for Beginners Volume 1

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MATHEMATICS AT A GLANCE FOR BEGINNERS VOLUME 1

Kennedy Samson

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Table of Contents MATHEMATICS AT A GLANCE FOR BEGINNERS VOLUME 1 ........................................ 1 Kennedy Samson ................................................ 1 PREFACE .................................................................. 7 This book is useful in providing the fundamental background knowledge of Mathematics most especially in high schools and other institutions where Basic General Mathematics is one of the core courses to be passed before graduation. It is an invaluable companion for students, its approach and methodology is well simplified to the understanding of the students. The presentation is elaborate and in simple English. Also, topics are developed from the simple to the more complex aspect of the concept treated, thereby elimination the ambiguity and fear of Mathematics. ............................................................ 7 The book covers a wide range of topics in Geometry, Trigonometry, Statistics and Probability with practical examples and exercises for students to practice. The topics treated include, Unit of Measurement, Plane and Solid Geometry, Areas of Plane Shapes, Perimeter of Regular and Irregular Shapes, Volumes of Solids, Statistics and Probability. ....................................... 7 All materials consulted in the preparation of this book are acknowledge in the form of References 8 3

Kennedy Samson .....................................................8 DEDICATION ...........................................................9 This book is dedicated to God Almighty, my entire house hold and to all who applies Mathematics in all aspects of life. ..........................9 CHAPTER ONE.......................................................10 THE UNIT OF MEASUREMENT ..........................10 INTRODUCTION: ...............................................10 TYPES OF MEASUREMENTS ..............................11 THE STANDARDIZED MEASUREMENT .......11 THE NON-STANDARDIZED MEASUREMENT .12 THE METRIC SYSTEM .........................................12 PREFIX USED FOR MEASUREMENT ...............13 UNIT OF MEASUREMENT FOR LENGTH .....13 LINEAR MEASURE ...........................................15 RULES FOR CONVERSION ..............................17 UNIT OF MEASUREMENT FOR AREA ..........20 METRIC CONVERSION FACTORS FOR AREA ..............................................................................22 Exercises ...............................................................24 THE UNIT OF MEASUREMENT FOR VOLUME .............................................................24 THE LINEAR MEASURE FOR CAPACITY .....26 THE UNIT OF MEASUREMENT FOR MASS AND WEIGHT.....................................................30 LINEAR MEASURE FOR MASS AND WEIGHT ..............................................................................31 CONVERSION TO GRAMMES .............................32 THE UNIT OF MEASUREMENT OF TIME .33 CONVERSION TABLE FOR THE UNIT OF TIME ....................................................................34 THE ALGEBRA OF TIME .................................34 THE UNIT OF MEASUREMENT FOR MONEY. ..............................................................................43 4

MISCELLANEOUS EXERCISES ...................... 49 CHAPTER TWO ..................................................... 51 GEOMETRY ........................................................... 51 INTRODUCTION ............................................... 51 ELEMENTS OF GEOMETRY ........................... 52 CHAPTER THREE.................................................. 58 SOLID GEOMETRY ............................................... 58 PROPERTIES OF SOLIDS ................................. 59 CHAPTER 4 ............................................................ 66 PLANE SHAPES AND THEIR PROPERTIES ...... 66 CONCEPT OF LINE AND ANGLES ................. 66 PROPERTIES OF ANGLES ............................... 71 CHAPTER 5 ............................................................ 80 PROPERTIES OF A TRIANGLE ........................... 80 PROPERTIES OF TRIANGLE ........................... 80 ANGLES PROPERTIES OF TRIANGLE .......... 86 CONGRUENT TRIANGLES .............................. 95 Condition for Congruency.................................... 95 CHAPTER SIX ...................................................... 106 POLYGONS .......................................................... 106 CHAPTER 7 .......................................................... 117 AREAS OF PLANE SHAPES ............................... 117 CHAPTER 8 .......................................................... 133 PERIMETER OF PLANE SHAPES...................... 133 CHAPTER NINE ................................................... 145 AREA OF SOLIDS ................................................ 145 SURFACE AREAS: .......................................... 145 CHAPTER TEN ..................................................... 159 VOLUMES AND AREAS SOLIDS GEOMETRY ................................................................................ 159 CHAPTER 11 ........................................................ 172 ARCS, SECTORS AND SEGMENTS OF A CIRCLE ................................................................. 172 REFERENCES................................................... 195 5

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PREFACE This book is useful in providing the fundamental background knowledge of Mathematics most especially in high schools and other institutions where Basic General Mathematics is one of the core courses to be passed before graduation. It is an invaluable companion for students, its approach and methodology is well simplified to the understanding of the students. The presentation is elaborate and in simple English. Also, topics are developed from the simple to the more complex aspect of the concept treated, thereby elimination the ambiguity and fear of Mathematics. The book covers a wide range of topics in Geometry, Trigonometry, Statistics and 7

Probability with practical examples and exercises for students to practice. The topics

treated

include,

Unit

of

Measurement, Plane and Solid Geometry, Areas of Plane Shapes, Perimeter of Regular and Irregular Shapes, Volumes of Solids, Statistics and Probability. All materials consulted in the preparation of this book are acknowledge in the form of References Kennedy Samson

8

DEDICATION This book is dedicated to God Almighty, my entire house hold and to all who applies Mathematics in all aspects of life.

9

CHAPTER ONE THE UNIT OF MEASUREMENT INTRODUCTION: Measuring is part of man’s way of life. Everyday, we measure either directly or indirectly. So, man is regarded as a measuring animal, irrespective of the age or profession. Teachers, Traders, Businessmen, lawyers, Doctors, Engineers, Bricklayers, Carpenters and many more, measure. But the more experienced or educated a man is, the more “accurate” he measures. Measure can be done when we want to find out; (i) How much time is allowed to write a given paper in an examination. (ii) Which of two dressed turkey weighs more than the other? (iii) Which tank contains more fuel? (iv) Which is shorter or taller of two students in a given class? 10

(v) Which is shorter or longer of two given distances? The solution to these question and other similar questions result in measurement.

TYPES OF MEASUREMENTS There are two basic types; viz;

THE STANDARDIZED MEASUREMENT The standardized measurements are universal or international. In a standardized measurement, the unit is mostly used internationally to mean unity (one) of a quantity. Examples include 1 metre (m), 1 gramme (g), 1 litre (l) etc. There is the international standard unit of measurement. The most common is the International System of measurement usually abbreviated as S.I unit. The S.I basic unit of time is second (s); of length is metre (m); of mass is grammes (g); of volume (capacity) is litre (l). 11

THE NON-STANDARDIZED MEASUREMENT Man makes use of any instrument available to him, including parts of his body and his sense of judgment to measure. Examples include the use of ropes, sticks, footsteps which are used to measure length and distance; cups, buckets, sack, bowl etc. to measure volumes / capacity. Hands are used to compare weights (mass) of any two items. For time, man make use of his shadow, stick or sometimes cockcrows are used to predict time. For Money, cowry shells or trade by Barter are used.

THE METRIC SYSTEM The metric system of weights and measures is an International System of units (S.I). The metric system is a system of measurement that has basic units of measure for length, width, area, volume, 12

money, time and weight that are in decimal relationship to each other i.e. the metric system is based on base 10. PREFIX MEASUREMENT Prefix Kilo Hecto Deca

USED

Symbol k h da

Deci

d

Centi

c

Milli

m

FOR

Meaning 1000 100 10 1 = 0 .1 10 1 or 0.01 100

1 or 0.001 1000

UNIT OF MEASUREMENT FOR LENGTH In measuring length, we are interested in comparing two or more measurement to determine which is longer, or shorter or taller. The use of foot steps, foot strides, number of sticks or 13

number of ropes are examples of the non – standard unit of measuring length. The basic unit of length in the metric system is the metre (m). Although, the metre is the basic unit of measurement of length in the metric system, many lengths (and thickness) are measured in terms of parts of metre such as decimeter (dm), centimeter (cm), and millimeter (mm). The kilometer is the most commonly used unit for longer lengths.

For example, all races in the Olympic Games are described in metres; 100 metre dash, 200 metre dash, 400 metre dash, 800 metre run, 1500 metre run etc, but the distance between two cities is measured in kilometres.

The table below summarizes the units related to the metre.

14

The chart below helps to find equivalent measure of length.

LINEAR MEASURE 10 millimetre (mm) = 1 centimetre (cm) 10 centimetre (cm) = 1 decimetre (dm) 10 decimetre (dm) = 1 metre (m) 10 metre (m) = 1 decametre (dam) 10 decametre (dam) = 1 hectometre (hm) 10 hectometre (hm) = 1 kilometre (km). The millimetre is not normally used to measure distance; it is used in the 15

industries. The centimetre replaces inches in measurement while feet and yards were replaced by the metre (m) and the kilometre replaced the miles.

The relationship between these forms of measurement is presented below. 1 metre (m)  3.281 feet (ft) 1 metre (m)  1.094 yards 1foot  0 .3048 metre 1 yard  0.9144 metre 1 foot  0.3333 yards 1 centimetre  0.3937 inches 1 inch  2.54 cm 1 kilometre  0.6214miles 1 mile  1.6093 kilometre 1 foot  30 centimetre 1 yard (yd)  0.9 metres, 1 millimetre  0.04 inches 1 metre (m)  3.3 feet 1 metre (m)  1.1 yards

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RULES FOR CONVERSION 1. To convert from metre to kilometre divide by 1000 2. To convert from kilogram to metre, multiply by 1000 3. To convert from centimetre to metre, divide by 100 4. To convert from metre to centimetre, multiply by 100 5. To convert from millimetre to litre, divide by 1000 6. To convert from metres to millimetre, multiply by 1000 etc. Example 1 1. Convert (i) 22m to cm (ii) 314mm to cm (iii) 15mm to dm (iv) 4.2km to m (v) 77dm to mm Solution (i) 1m = 10  10cm = 100cm Therefore, 22m = 22  100cm = 2200cm (ii) 10mm = 1cm  1mm = 1 = 0.1cm 10 Thus 314mm = 314  0.1cm 17

= 3.14cm (iii) 1 dm = 10cm = 10  10mm = 100mm 1 dm = 0.01dm 100 Thus, 15mm = 15  0.01dm = 0.15dm 1m = 10  10  10 or 1000mm Thus 4.1m = 4.1  1000m = 4100mm

Therefore, 1mm = (iv) (v) (vi)

1 km = 1000m Thus 4.2km = 4.2  1000m = 4200m 1 hm = 1000dm Therefore, 1dm = 11000 hm = 0.001hm Thus, 77dm = 77  0.001hm = 0.077hm.

Example 2 Convert each given measurement to the indicated measurement (a) 10 miles to km (b) 25 feet to metres (c) 105 niches to cm (d) 25 yards to feet. Solution a. Since 1 mile  1.6093km, then 10 miles  10  1.6093km = 16.093km  0.3048m, then 25 feet b. I foot = 25  0.3048m = 7.62 m

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= 2.54 cm, = 105 2.54 cm = 266.7 cm

them 105inches

c.

I inch

d.

I yard = 3 feet, then 25 yards = 3  25 feet = 75 feet

Example 3 A boy was asked to buy 30 yards of electrical cable. Find the equivalent in (i) metres (ii) feet Solution (a) 1 yard  0.9144 metres, thus, 30 yards = 30 0.9144 metres = 27.43 metres (b) I yard = 3 feet. Thus, 30 yards = 3  30 feet = 90 feet Example 4 Convert each speedometer reading to the indicated measurement (a) 30 mile / hr to km/h (b) 100 km/hr to mile / hr. 19

Solution (a)

1 mile  1.6093km.Thus,

30 mile / hr = 30 x 1.6093= 48.3km / hr

(b)

1km  0.6214 mile. Thus

100 km / hr = 100  0.6214 mile = 62.14 mile / hr.

Exercises 1. Complete each of the following

(a ) (b ) (c )

4.2m = 5km = 13.24km =

( ( (

)mm )mm )mm

2. If the distance between town A and town A is 1070km. Find the distance in miles 3. The speed limit in city is 55miles per hour. What is the speed limit in kilometer per hour?

UNIT OF MEASUREMENT FOR AREA An area is the amount of surface covered by a figure; examples of such 20

figures are rectangles, triangles, circles, pyramids etc. Areas are measured in square units. For example the floor of a room that measures 8ft by 10ft has an area of 8  10 = 80 square feet. (80 sq. ft ). A square whose 1cm by 1cm is 1sq . measurements is 2 centimeter (1cm ) . Square centimeter is used to find the area of a relatively small region such as area of the page of a note book. Larger areas are measured in terms of metre square (m 2 ). Since 1m = 100cm, (1m2 ) = 100 then 1 square metre 100cm2 = 10,000cm 2 (Read ten thousand square centimetres). The areas of very large region is measured in hectares (10,000square metres) . Land measured in terms of areas can also be measured in hectares, since 1 hectare = 10,000 sq . metres. 1 hectare is the area of a square that measures 100metres on each side.

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METRIC CONVERSION FACTORS FOR AREA

( ) = 1 sqaure feet (1ft ) = 1 sqaure Yard (1yd ) = 1 sqaure mile (1mi ) =

1 square inches 1m 2 2

2

6.5cm 2 0.09m 2 0.8m 2

2

2.6km 2 1 arce = 0.4 hectres 1 square millimeter (mm 2 ) = 0.000001m 2

1 square centimeter (cm 2 ) = 0.0001m 2 1 square kilometre (km2 ) = 1 000000m 2 1 Hectare (ha) = 10,000m 2 1 sqaure centimetre

=

1 sqaure metre m 2 1 sqaure kilometre 1 hectares (ha) 1 inch

= = = =

1 ft 2

= 0.0929m2

1yd 2

= 0.8361m2

1 mi 2

= 2.5899 km 2

( )

0.16 square inches 1.2 square yards 0.4 square mile 2.5 acres. 6.4516 square centimetre

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Worked Examples: 1.

Express 2.5m 2 in mm 2

Solution 1m 2 = 1000000m 2

Thus 2.5m 2 = 1000000  2.5m 2 = 250, 000mm 2

2. Tom bought 6m2 piece of cloth, while Dave bought 6000cm2 of another cloth, who bought the bigger piece of cloth? Solution 1m 2 = 10, 000cm 2

So that 6m 2 = 6  10, 000 = 60, 000cm 2 . Thus, Tom bought the bigger piece of cloth since 6m2 = 60, 000cm 2 is greater than 6000cm 2 3. Sum up the following areas and leave your answer in metres square. 1000cm 2 + 1m 2 + 3 hectares. Solution 1m 2 = 10,000cm 2 , so that 1cm 2 = 0.0001m 2 1 hectare = 10,000m 2 so that 3 hectares 3  10,000m 2 = 30,000m 2

23

Thus, 1000cm + 1m 2 + 3 hectares gives 0.1m 2 + 1m 2 + 30,000m 2 = 30,001.1m 2

Exercises 1 Find the difference between 10,000m 2 and 1m 2 2. Add up 2.5m 2 ,1.5km2 and 12 hectares leave your answer in square metre. 3.

Which of these is greater

1

1 2

1 2

hectares and 1 km 2 ? 4. 5.

Express 1.2m 2 in square centimeter Express 1.5km 2 in square metres

THE UNIT OF MEASUREMENT FOR VOLUME The volume is the measure of how much a container can hold, i.e. its capacity. The basic unit used to measure capacity is the litre. A litre is defined as the volume of a cubic decimeter. In other words, litre is the capacity of a cube (box) which is 1 decimetre long, 1 decimetre wide and 1 decimetre deep. The most 24

commonly used unit is the millilitres, centilitres and litres 1dcm 1dcm 1dcm

1 litre is equal to 1 cubic decimetre and is equal to 1000 cubit centimetres. The notation is “cc”. The millimeter is a small unit of volume measure. For example, 5 millilitres (ml) = 1 teaspoon, and 15 millilitres is equal to 1 tablespoon. Most liquid prescriptions obtained at the pharmacies are sold in millilitres. A cubic metre (m 3 ) is used to measure larger volumes. One cubic metre is equivalent to 1.3 cubic yards. The millimeter is the same as cubic centimeter (cm 3 ) . The litre replaces the “Pint”, in measurement. The cubic metre (m 3 ) is used in business and industry for larger volumes.

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Symbol kl hl dal l dl cl ml

Word kilolitres hectolitres decalitres litres decilitres centilitres millilitres

1000cm 3 i.e. 1cm3

= 1000ml

= 1ml

Meaning 1000 litres 100 litres 10 litres 1 litre 0.1 litres 0.1 lites 0.001 litres

= litre

= 0.001l.

THE LINEAR MEASURE FOR CAPACITY 10 millilitres 10 centilitres 10 decilitres 10 litres 10 decalitres 10 hectolitres

= = = = = =

1 centilitres 1 decilitres 1 litre 1 decalitres 1 hectolitres 1 kilolitres

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CONVERSION CHART FOR LITRES

Examples 1. Complete each of the following

(i ) (ii )

642cl = 74.3l =

( )l ( ) cl

Solution

(i )

1 litre =

1ml

=

1000ml , so that

1 = 0.001litres 1000

Thus = 0.001 x 642 litres = 0.642 litres (ii ) 1 litre = 100cl Thus, 74.3l = 74.3 x 100 cl = 7430 cl 642cl

2. How many litres will an aquarium hold if it is 70cm long, 0.5m wide and 500mm high? 27

Solution Convert all measurements centimetres Length 70 cm, width = 0.5m = 0.5  100

to

= 50cm, height = 500mm = 500  10 = 50cm

But volume = Length  Width  Height = 70  50  50 = 175000cm 3 .

But 1 litre = 100cm 3 Thus, volume = 175000cm 3 = 175000 0.0001 litres = 175 litres.

3. Find the volume of a box that is 1 metre long, 40cm wide and 50cm high? Leave your answer in centimetres Solution 1m = 100cm  length = 1m = 1 x 100cm = 100cm. widt h = 40cm and height = 50cm. But volume (V) = Length (L)  Width (W)  Height (H) = 100  40  50 cm 3 = 200 0000cm 3 .

4.

Complete each of the following 28

(i)

60cl to litres

(ii)

120l to

( ) cl (iii)

120ml to (

) cl

Solution

(i )

60mcl

to

( )l

1 l = 0.01 litres 100 = 60  0.01 litres = 0.6 litres.

1 litre = 100cl , then 1cl = Thus 60cl

(ii )

1litre = 100cl thus, 120 litres = 120  100 cl = 12, 000cl

(iii)

1 cl = 10ml. Then 1ml = 0.1cl Thus 120ml = 120  0.1cl = 12cl

Exercises 1.

2.

3. 4.

Calculate the volume of a tank (in cm 3 ) if the capacity of the tank is 50l . Calculate in litres the capacity of a cuboid of length 6cm, by 9cm by 10cm . Convert 145cm 3 to mm 3 Johnny was asked by a physician to take 2 table spoon of a certain drug 3 29

times a day. How many millilitres is he expected to take for a particular day? 5. A swimming pool has sides of 8cm by 3cm by 10cm, what is the capacity of the pool?

THE UNIT OF MEASUREMENT FOR MASS AND WEIGHT Mass is the quantity of matters contained in a body which is always constant. Weight on its part is the gravitational force acting on a body. It is important to note here that it changes from one place to another depending on the force of gravity. Mass and weight measurements take place in our daily activities. These are carried out in our various schools, hospitals, markets and so on. Scales are usually used to measure weights. The basic standard unit of mass and weight is the gramme (g). A gramme is a small unit of mass. The milligramme is used to measure small masses such as dosages of medicine. 30

The gramme replaces ounces in measurement. The kilogram replaced the pounds in measurement. The metric tonne is used to measure very large masses such as a truck load of cassava, oranges cocoa, timber and many more.

LINEAR MEASURE FOR MASS AND WEIGHT 10 milligrammes (mg) = 1 centigramme (cg) 10 centigrammes (cg) = 1 decigramme (dg) 10 decigrammes (dg) = 1 gramme (g) 10 grammes (g) = 1 decagrammes (dag) 10 decagrammes (dag) =1 hectogrammes (hg) 10 hectogrammes (hg) = 1 kilogramme (kg) 1000 grammes = 1 kilogramme 1000 kilogrammes = 1 tonne (t). 31

CONVERSION TO GRAMMES 1 metric tonne (1000kg) = 1,000, 000kg 1 killogramme (kg) = 1000g 1hectogramme (hg) = 100g 1decagramme (dag) = 10g 1 = 0.1g 1 decigramme (dg) = 10 1 centigramme (cg)

=

1 milligramme (mg)

=

1 = 0.01g 100

1 = 0.001g. 1000

Examples 1. Complete each of the following

(i )

50kg to

( )g

(ii)

4317g to

( )kg (iii)

5.234g

Solution i. 1kg = 1000g , thus 50kg = 50  1000g = 50, 000g

ii. 1 gramme = 1 kg = 0.001kg, thus 4317g = 1000 4317  0.001kg = 4.317kg 32

iii. 1 cg = 10mg , thus 5.234cg = 5.234  10mg = 52.34mg .

Exercises Express in grammes (g) the following masses (i ) 1.60kg (ii) 0.035tones (iii) 14.36kg (iv) 0.235t (iv) 28360kg .

THE UNIT OF MEASUREMENT OF TIME The uses of time by man cannot be over emphasized. Man needs time for him to perform his duties effectively. Before now, various means were used by man to estimate (measure) time, these includes his shadows or shadows of stick stand out in the sun during the day or the crow of the cock during the early morning hours. But over time, these means of estimating time were replaced by wrist watches, clocks and even calendars to record dates of events. The Standard International (S.I) unit of time is the 33

seconds (s). Others are minutes, hours, days, weeks, months, years and many more.

CONVERSION TABLE FOR THE UNIT OF TIME 60 seconds (s) (mm) 60 minutes (mm) 24 hours (hr) 7 days (d) 4 week (wk) (mth) 12 months (mth) 365 days (d) 52 days (d)

=

1

minute

= = = =

1 hour (hr) 1 day (d) 1 week (wk) 1 month

= = =

1 year (yr) 1 year (yr) 1 year (yr).

THE ALGEBRA OF TIME The four basic arithmetic operations of addition, subtraction, multiplication and division can be performed on time. When calculating time, seconds and minutes are in base 60 or other bases as stated in the conversion table above. 34

Worked Examples 1. Add the following (i) 5minutes 28 seconds, 3minutes 42seconds, 4minutes 20second. (ii) 6weeks 4days, 5days, 2weeks 4days and 1 week 2 days (iii) 3 months 1week, 2days, 1month 2weeks 3days, 2months 2week, 3days. Solution +

Min 5 3 4 13

S 28 42 20 30

Explanation: Add 28 seconds to 42 seconds to 20seconds to get 90seconds which is 1-minute 30seconds. (Since 60 seconds gives 1 minute). Put 30 in the column of second and add the 1 minute to the result of adding 5minutes, 3minutes and 4minutes to get 13 minutes. Thus, the required result is 13 minutes, 30 second.

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(ii) Wk 6 + 2 1 11

d 4 5 4 2 1

Note: 7days gives 1 week, so that 15days gives 2weeks 1day.

(iii) Mth 3 1 2 7

Wk 1 2 2 2

d 2 3 3 1

The required result is 7months 2weeks and 1day. Example 2 Simplify the following 36

(a )

Mth - 9 4

(c)

3

Wk 1 3

d 3 6

(b ) 

Mth 2

Wk 1

d 3 4

4mth 3wk 2d.

Solution

Explanation: Take 1week from the existing 1week i.e. 7days, add the 7days to 3days to get 10days. Subtract 6days from the 10days to get 4days. Take 1month which is 4weeks from the existing 9months, subtract 3week from 4weeks to get 1week, finally subtract 4 months from the remaining 8months to get 4 months. Thus, the result is 4month, 1week and 4days. 37

(b )

Mth 2  2

Wk 1 1

d 3 4 5

Thus, we have 9mth 1wk 5days

3 4mth 3wk 2d 3. Convert all to days

4months = 4  4 weeks = 16 weeks 16 weeks = 16  7 days = 112 days 3 weeks = 3  7days = 21days 2days = 2days Total number of days = 135days

Therefore, 3

=

4mth 3wk 2d

3

135 days

= 45 days.

Since 7 days = 1 week 45 days gives 6 wks 3days 4weeks = 1 month, Thus 6 weeks = 1 month 2 weeks. Therefore 45 days = 1month, 2week 3days.

Example 3 If today is Monday, what day of the week will it be in (i) 10days (ii) 40days (iii) 500days (iv) a year’s time 38

Solution

(i )

7 days = 1 week 10 days = 1 week 3 days

Therefore in 1 week time, it will be on a

Monday, the 3 days are Tuesday, Wednesday, Thursday. Thus, the day will be Thursday, (i) 40days = 5week and 5days In 5weeks time, it will be on a Monday, the remaining 5days will be a Saturday. Thus, the day will be on Saturday. (ii) 500days = 71week and 3days. In 71week’s time, it will be on a Monday, the remaining 3days will be a Thursday. Thus, the day will be on Thursday. (iv) 1 year is 365days 365 days = 32weeks and 1 day. In 32week’s time, it will be on a Monday, the remaining 1 day will be a Tuesday, and thus, the day will be a Tuesday. The 12 – Hour Clock System: This uses the digits 1- 12. This is the type that is usually used mainly by all. 39

The 24 – Hour Clock System: The 24 – hour clock system uses the digits 1 -24 in its operations. This is used by some countries of the world. The relationship between 12–hour clock system and the 24 – hour clock system is presented below. 12 - hours

24 – hours

1.am 2am 3am 4am 5am 6am 7am 8am 9am 10am 11am 12noon 1 pm 2pm 3pm 4pm 5pm 6pm 7pm

1 am 2 am 3am 4am 5am 6am 7am 8am 9am 10am 11am 12noon 13hour /pm 14hours 15pm 16pm 17pm 18pm 19pm 40

8pm 9pm 10pm 11pm 12midnight

20pm 21pm 22pm 23pm 24pm or 00 hours.

Note: Pm means post meridian. That is, from noon to midnight which is between 12noon and midnight. am means ante – meridians (before noon) which is the period between midnight and noon). Examples 1. Convert 15.40Pm to 12-hour clock. Solution 15.40Pm is 15.40 – 12.00 = 3.40Pm. 2. The mathematics examination ended at 3.35Pm. If the paper was taken for 3hours 15minutes, when did the paper commenced. Solution Convert 3.35Pm to 24hour clock. i.e. 3.35Pm = 12.00 + 3.35Pm = 15.35Pm 41

Since the paper lasted for 3hours 15 (3.15), then we have that the paper started at 15. 35 − 3.15 = 12.20

Thus, the paper started at 12.20Pm. 3. A drama programme aired on the radio started at 10.25am and ended at 2.40Pm. How long was the programme?

Solution 2.40Pm is 24hour clock is 12.00 + 2.40Pm = 14.40Pm. If the programme started at 10.25am and ended at 2.40Pm = 14.40Pm, then it lasted for 14.40 − 10.25 = 4.25 i.e. 4 hours 25 minutes.

Exercises 1. How many hours are in each of the following times (i) 221 minutes (ii) 180 seconds? 2. Find the sum of 365 second, 14 minutes and 9 hours leaving your answer in minutes. 3. How many years is (i) a decade (ii) a millennium. 42

4.

A child spends 6 years in the primary school. if the child finished his primary school in 1991, which year did he start his primary school. 5. Suppose today is Friday, what day of the week will be in a. 12 days’ time b. 22 days’ time c. 300 days’ time

6.

An examination is to last for / hour 20 minutes. The paper started at 11.00am, when will the end.

THE UNIT OF MEASUREMENT FOR MONEY. Money has been variously defined by different schools of thought. But it is generally seen as “any acceptable medium of exchange”. Trade by Barter (direct exchange of goods for goods) was the first means of exchange. But over time, various forms of money came into being. Examples include cowry shell. But today every independent state has a unique unit 43

of money. In Nigeria, we make use of the naira (N) and Kobo (K). The table below gives names of some countries and their currencies.

44

OPERATIONS WITH MONEY The four basic arithmetic operations of addition subtractions, multiplications and division can also be performed with money. DIVISION OF MONEY In the world all over, the conversion factor of the currency used within a country is 100. In Nigeria, 1 Naira (N) = 100 Kobo (k) USA 1 Dollar ($) = 100 cent (c) UK 1 pound (€) = 100 pence (p) Ghana: 1 cedi (C) = 100 pesewa (p) Gambia: 1 Dalasi (D) = 100 butut (b) Sierra Leone: 1 Leone (Le) = 100 cent (c) and so one. Worked Examples 1. Complete each of the following a) N 3,000 to ( ) kobo b) $ 1,000 to ( ) cent c) 40,00kobo to ( ) naira.

45

Solution 1. N 1.00 = 100k (a) Thus N 3000 = 3000 x 100k = 30 00 00 kobo (b) $1 = 100cent, thus $1000 = 1000  100cent = 10 0000 cents (c) N 1 = 100k then 1kobo = 1100 = N (0.01)

2. a)

Thus 40,000 kobo give N (40, 000 x 0.01) = N 400. 00. Simplify the following

N 346 + 78

b) c)

:

K 26 28

N5, 600.00 ÷ 3 € 265.02  75.

Solution a)

N 346 + 78 424

:

K 26 28 54

N 424. 54K . 46

b)

1866.66 3

5,600.00 3 26 24 20 18 20 18 2

Thus N 5600.00  3 = N1,866.67k c) € 265. 02  75 = € 19876. 50 3. The exchange rate of $ to the naira is N185 to a dollar. Find (a) The value in naira of $12,000 (b) The value is $ of N100 000 000. Solution (a) $1.00 = N 185, thus 12, 00 = N (12,000  185) = N222 0000 = N2, 220, 000 If $1.000 is N185. 000 then N1.00 = $ 1185 = $0.00541.

(

)

Thus N100, 000 000 000  0.00541) =

=

$

(100,000

$ 540, 540. 54. 47

Exercise 1. Express the following money in kobo (a) N165 .30 (b) 2 fifty naira notes and 6 ten Naira notes 2. Find the cost of 4 tonnes of rice at N 120.00 per tonne. 3. Adamu ordered for 6 cars from United States of America at the cost of N 580, 000 each, if the exchange rate is N 150.00 to $1.00. find the cost of all the cars in dollars. 4. Add N 2.40, N13.65 and 656k leaving your answer in Naira 5. A man traveled to the United Stated with N 3,000,000.00. He spent ten thousand in America, and left for Japan where he bought a car worth 5,000 ten. He returned back to Nigeria with the remaining amount all in dollars. Calculate the balance money in Naira. If $1 is equivalent to $ 150,00 and 1 yen is equivalent to $0.009. 48

MISCELLANEOUS EXERCISES 1.

Express the following in meters (i) 0.165km (ii) 33.33cm (iii) 3330cm (iv) 13265mm 2. Express the following in millimeter (i) (ii) 0.56km 3.55cm 3. Add the following leaving your answer in metre (a) 2.3m, 5.6cm, 2.1km (b) 587m 76cm, 605mm (c) 77m 0.36cm 0.365m 4. Express the following capacities in milliliters (a) 1.9 litres (b) 0.025 liters (c)180 litres 5. If 6 litres of water are poured into a tank with a volume Vcm2 a length 50cm and a width 40cm. find the height of the water in the tank 8. Calculate the volume of air in a room whose volume is 2080com2. 9. Express to the nearest second 49

(a)

10.

1.412 hours (b) 0.14 hours

How many hours are in each of the following times (a) 221 mins (b) 180 seconds

50

CHAPTER TWO GEOMETRY INTRODUCTION The term geometry is derived from the Greek word geometria, meaning "to measure the Earth." In its most basic sense, then, geometry was a branch of mathematics originally developed and used to measure common features of Earth. Most people today know what those features are: lines, circles, angles, triangles, squares, trapezoids, spheres, cones, cylinders, and the like. Humans have probably used concepts from geometry as long as civilization has existed. But the subject did not become a real science until about the sixth century B.C. At that point, Greek philosophers began to express the principles of geometry in formal terms. The one person whose name is most closely associated 51

with the development of geometry is Euclid (c. 325–270 B.C.), who wrote a book called Elements. This work was the standard textbook in the field for more than 2,000 years, and the basic ideas of geometry are still referred to as Euclidean geometry.

ELEMENTS OF GEOMETRY Axiom: A mathematical statement accepted as true without being proved. Construction: A geometric drawing that can be made with simple tools, such as a straight edge and a compass. Euclidean geometry: A type of geometry based on certain axioms originally stated by Greek mathematician Euclid. Line: A collection of points with one dimension only—that of length. Line segment: A portion of a line. Non-Euclidean geometry: A type of geometry based on axioms other than those first proposed by Euclid. 52

Plane geometry: The study of geometric figures that can be represented in two dimensions only. Point: A figure with no dimensions. Proposition: A mathematical statement that can be proved or disproved. Proof: A mathematical statement that has been demonstrated logically to be correct. Solid geometry: The study of geometric figures that can be represented in three dimensions. 2. 2 PLANE GEOMETRY Plane geometry is that aspect of geometry that deals with two (2) dimensional objects. They have only lengths and width or breadth. But no thickness. Examples are rectangles, squares kites parallelograms and so on.

53

PROPERTIES OF PLANE SHAPES Quadrilaterals: A quadrilateral is any plane figure bounded by four straight lines. Examples are squares, rectangles, parallelograms, rhombuses, trapeziums. PROPERTIES OF A SQUARE

1. All the four sides are equal 2. The diagonals bisect the angles 3. The diagonals bisect each other at right angles 2. (i) (ii) (iii) (iv) (v)

Rectangles It has four sides The opposite sides are equal All the angles are right angles Pairs of opposite sides are parallel The diagonals are equal and they bisect each other.

54

3.

Parallelograms

(i) The opposite sides area equal (ii) Opposite sides are parallel (iii) The opposite angles are equal (iv) The diagonals bisect each other. (v) Each diagonal bisects the parallelogram into two congruent triangles (vi) It is a special rectangle. 4.

Rhombus

A rhombus is a quadrilateral in which (i) All sides are equal in length (ii) The opposite angles are equal 55

(iii) The diagonal are lines of symmetry (iv) The diagonal bisect each other at right angle. It is a special square.

5.

Kite

A Kite is a quadrilateral which has (i) Two pairs of sides of the same length (ii) One pair of opposite angles are of equal size (iii) The diagonals are of different lengths (iv) One line of symmetry 6.

Trapezium

56

A trapezium is a quadrilateral which has (i) (ii) (iii) (iv)

One axis of symmetry A pair of equal sides The diagonals equal 2 pair of equal angles

Exercises 1.

List 4 properties of the following (i) Square (ii) Rectangle (iii) Parallelogram 2.

List the difference (s) between (i) Square and Rhombus (ii) Rectangles and parallelogram (iii) Kite and Trapezium.

57

CHAPTER THREE SOLID GEOMETRY Solids are three (3) dimensional objects. This is because they have length, breadth and height. Some solid have regular shaper while other have irregular shapes. For example, a book has a regular shape while a piece of stone has irregular shape. Some of the common solid found at home include chairs, tables, beds, boxed, pots, cups, tubers of yam, bag of rice and so on. Some of these solids have regular shapes such as boxes. Others have irregular shapes such as tubers of yam. Outside the homes, we find such solids as rock, buildings, trees, basin of garri, grains of rice. While the buildings have regular shapes, the rocks have irregular shapes. In the classroom, we can also find solid such as chalk boxes, wooden tables, chairs, desks, pencils, pen, books, walls, tin cans, play balls and so on. All these solids have regular shapes. Some of the common solid are cuboids, cubes, cylinders, prisms, cone, 58

pyramids, spheres, hemispheres, frustum, compound solids.

PROPERTIES OF SOLIDS Before we enumerate the properties of solids let us first explain some common concept used. (i) Face is one outer view of a solid. (ii) Surface is the total outer view of a solid. That is, the faces of a solid taken together made up its surface (iii) An edge is that part of a surface where two faces of a solid meet. An edge may be straight or curve. (iv) A vertex is a point or corner where three or more edges meet. The point or corner is usually sharp. Vertices is the plural form vertex. The properties of solid include: 1. All solid have faces or surfaces separated by edges 2. Most solid have edges which join the faces 3. Most solids also have an inside surface if they have an opening. 59

Vertex

Edge Face

PROPERTIES SOLIDS 1.

OF

SOME

COMMON

Cube

A cube has (i) 6 faces which are made up of squares (ii) 12 edges that are equal (ii) 8 vertices 2.

Cuboid

60

A cuboid has; (i) 6 faces which are made up of rectangles (ii) 12 edges, but only pairs of them are equal (iii) 8 vertices 3,

Sphere

A sphere has (i) A round surface (ii) No vertex (iii) No edge 4.

Cone 61

A cone has (i) A curved surface (ii) A circular flat face (iii) 1 vertex (iv) 1 edge

5.

Cylinder

A cylinder has (i) 1 round face (ii) 2 flat circular faces (iii) 2 edges (iv) No vertices 6.

Pyramid

62

Pyramid has a number of faces depending on the shape of the base. Some examples are outlined below. (i) Rectangular Pyramid has five faces, four of which are triangular and one is rectangular. (ii) A Square Pyramid has five faces, four of which are triangular and one is a square. Remarks: It is the shape of the base of the pyramid that determines the number of faces of the pyramid. 7.

Prism

A prism has a number of faces depending on the shape. A triangular prism has three rectangular faces, and two triangular faces. The number of edges and vertices are determined by the shape of the solid. A triangular prism has an edge, 10 vertices and 7 faces. 8.

Tetrahedron:

63

A tetrahedron has (i) 4 triangular faces (ii) 6 edges (iii) No vertex 9.

Ellipsoid

A solid ellipsoid has (i) An oral surface (ii) No vertex (iii) No edge 10.

Hexagonal Wedge

A hexagonal wedge has (i) 6 vertices (ii) 9 edges (iii) 5 faces 64

Exercise: 1. List five solids found at home. 2. What are the differences between a cube and a cuboid 3. What is the difference between a cuboid and a rectangular prism? 4. Draw (i) a cube (ii) a cuboid (iii) a square pyramid (iv) Basin full of garri.

5.

Copy and complete the table below

Name of Vertices

Edges

Faces

solid Cube

12

Cuboid Sphere

6 0

Cylinder

2

Prism

7

Ellipsoid

1

Cone

1 65

CHAPTER 4 PLANE SHAPES AND THEIR PROPERTIES CONCEPT OF LINE AND ANGLES 1. Point: A point is the smallest shape in geometry. It is considered to have no size. It is normally represented by a dot (.) or a cross (x).

66

2.

Lines: These are series of points joined together, they can lie on a straight line or they may not, this means that a line can be straight or may not be straight.

3.

Line segment: A line segment is part of a line with some definite length. It has two endpoints. This is sometimes written as AB or AB , this is represented below A

4.

B

Angles:

When two straight lines meet, an angle is formed, the points where two lines meet is also known as a vertex. The figure below shows angles formed between two lines 67

NAMING ANGLES An angle is a measure of the region between two intersecting lines. Consider the figure below. B

R

Q

A

(b)

(a) P C

In fig (a) the two straight edges which meet at the point P are PR and PQ, the 68

angle between PR and PQ is called angle RPQ written as  RPQ or  QPR OR Pˆ , In fig (b) the two straight edges which meet at the point A are AB and AC, the angle between AB and BC is called ˆ.  CAB OR  BAC OR A

Parallel lines: If two or more straight line does not meet (intersect) in a plane, such lines are said to be parallel to each other. They maintain the same distance between then. The figure below gives some sets of parallel lines

69

If the line AB is parallel to another line CD, we write this as AB//CD. A

B

C

D

Perpendicular lines Two lines are said to be perpendicular if the angle formed at the point of contact (meeting point) is rightangle (i.e. 90o). In the figure below, AB is perpendicular to PQ. The is written as AB ⊥ PQ

B

P

A

70

Q

PROPERTIES OF ANGLES 1.

2.

3.

Complementary Angles: Two angles are said to be complementary if their sun is 90o. If A is an angle and B another angles, if their sum i.e. Aˆ + Bˆ = 90o , then the two angles are said to be complementary. Examples are 30o and 60o, 20o and 70o, 50o and 40o and so on Supplementary Angles: Angles that add up to 180o are said to be supplementary. If a, b, c are angles and their sum a + b + c = 180o , then the angles are said to be supplementary. Vertically Opposite Angles: Consider the figure bellows, where two lines AB and CD intersects (meet). A

D doa b c

C

B 71

Four angles are form. They are