prescribed by the national curriculum and textbook board as afor this reason the study of algebra...

Prescribed by the National Curriculum and Textbook Board as aTextbook for class VII from the academic year 1998

Junior Secondary Mathematics[ For class VII ]

Written by

Suruj Uddin Ahmed

A M M Ahsan Ullah

Translated by

Md. Wahiduzzaman

Kemal Raza

Edited by

Dr. Anower Hossain

NATIONAL CURRICULUM AND TEXTBOOK BOARD, DHAKA

Published byNational curriculum and Textbook Board

69-70, Motijheel Commercial Area, Dhaka

[ All rights reserved by the Publisher ]

First Edition : 1998

Reprint : 2007

Reprin : 2010

Reprint :

Computer Compose Perform Color Graphics (Pvt.) Ltd

Cover Design

Nasir Biswas

Design

NCTB, Dhaka

For free distribution from academic year 2010 by the Government of Bangladesh

Printed by :

Preface

Education is the key to development. A progressively improved education system largely determines the pace and the quality of national development. To reflect the hopes and aspirations of the people and the socio-economic and cultural reality in the context of the post independent Bangladesh, new textbooks were introduced in the beginning of the 1980s folIowing the recommendations of the National Curriculum and Textbook Committee.

In 1994, in accordance with the need for change and development, the textbooks of lower secondary, secondary and higher secondary were revised and modified. The textbooks from classes VI to IX were written in 1995. In 2000, almost all the textbooks were rationally evaluated and necessary revision were made. In 2008 the Ministry of Education formed a Task Force for Education. According to the advice and guidance of the Task Force, the cover, spelling and information in the textbooks were updated and corrected.

The study of arithmetic has been limited to class VI and more importance has been given to the study of algebra. So that application and use of Mathematics can be made simple and easy. For this reason the study of algebra has been introduced from class six. It is hoped that students can easily solve mathematical problems through algebraic formulas. It is necessary to develop mathematical skill in the learners so that they can study science in a better manner. Therefore, new techniques and methods has been presented in a simple and concrete way. As a result learners can themselves solve their problems without depending fully on teachers. Mathematics is a subject where practice is needed. It cannot be memorised. Exercises after each chapter, therefore, includes both traditional practice questions as well as creative questions.

This book of Junior Secondary Mathematics for class VII is the English Version of the original textbook entitled 'Nimna Madhyamik Ganit' written in Bangla.

We know that curriculum development is a continuous process of which textbooks are written. Any logical and formative suggestions for improvement will be considered with care. On the event of the golden jubilee of the Independence of Bangladesh in 2021, we want to be a part of the ceaseless effort to build a prosperous Bangladesh.

In spite of sincere efforts in translation. Editing and printing some inadvertent errors and omissions may be found in the book. However our efforts to make it more refined and impeccable will continue. Any constructive suggestion towards its further improvement will be gratefully considered.I thank those who have assisted us with their intellect and efforts in the writing, editing and rational evaluation of this book. We hope that the book will be useful for the students

for whom it is written.Prof. Md. Mostafa Kamaluddin

ChairmanNational Curriculum and Textbook Board, Dhaka.

Index

Chapter Contents Pages

Arithmetic

First Chapter Square Root 03

Second Chapter Ratio, Proportion 18

Third Chapter Unitary Method 34

Fourth Chapter Measurement and Unit 47

Algebra

First Chapter Addition, Subtraction, Multiplication and

division of algebraic Expression 60

Second Chapter First Four Formulae with its application 77

Third Chapter L.C.M. and H.C.F of Algebraic

Expressions 89

Fourth Chapter Algebraic Fraction 96

Fifth Chapter Simple Equations and Applications 104

Geometry

First Chapter Primary Discussion and Preliminary

Concepts 118

Second Chapter Theorem Concerning Triangles 127

Third Chapter Problems On Triangles 150

Answers 159

Junior Secondary Mathematics

Arithmetic

Chapter One

Square Root

1.1. Explanation Of Square Roots

We know, a square is a geometrical figure. A square is a rectangle whose sides are

equal to one another and the space enclosed by the square is called the area of the

square. The area of a square is the product of its length and breadth. If the length of a

square be x units, then its breadth is also x units. In that case the area of the square

will be (x × x) square units or x2 square units. Conversely, we may say if the area of a

square is x2 square units, then the length of its each side is x units.

Let us assume that marbles are arranged

(figure alongside) in 4 rows so that each

row contains 4 marbles kept at equal

distance apart. Then the total number of

marbles is 4 × 4 = 42 = 16.

[Since the number of marbles in each row is equal to the number of rows, the figure

will be a square]. Again let us suppose that 16 marbles are arranged in rows so that

each row contains a number of marbles equal to the number of rows. In that case the

number of rows is 4 and the number of marbles is also 4. Here we say, the square of 4

is 16 and the square root of 16 is 4.

When any number is multiplied by itself once the product is called the square of that

number and the number itself is called the square root of the product. Thus. 3 × 3 or 9

is the square of 3 and 3 is the square root of 9, 12 × 12 or 144 is the square of 12 and

12 is the square root of 144. etc.

Perfect Square :

A number whose square root can be expressed exactly either by a whole number or by a

fraction is called a perfect square. For example, 4, 9, 16, 4/9, 9/16, 16/25, 144, 256, 0.09.

Below is a list of few perfect squares with their square roots :Perfect Square Perfect Square Perfect square

square root square root square root

1 1 64 8 225 15

4 2 81 9 256 169 3 100 10 289 1716 4 121 11 324 1836 6 169 13 400

441202149 7 196 14

If we carefully notice the table, we understand,

a) If the digit in the unites' place of a number is 1 or 9, the digit in the units' place of

its square is 1.

For example,. the square of 11 is 121, the square of 9 is 81, the square of 19 is 361.

b) If the digit in the units' place of a number is 2 or 8,. the digit in the units' place of

its square is 4.

For example, the square of 2 is 4. the square of 12 is 144, the square of 8 is 64, the

square of 18 is 324.

c) If the digit in the units' place of a number is 3 or 7, the digit in the units' place of its

sqare is 9.

For example. the square of 3 is 9, the square of 7 is 49, the square of 13 is 169, the

square of 17 is 289.

d) If the digit in the units' place of a number is 4 or 6, the digit in the units' place of its

square is 6.

For example, the square of 4 is 16, the square of 6 is 36, the square of 14 is 196, the

square of 16 is 256,

e) If the digit in the units' place of a number is 5, the digits in the units' place of its

square is 5.

For example, the square of 5 is 25, the square of 15 is 225.

f) If the digit in the units' place of a number is 0, the digits in the units' and tens place

of its square root are also 0.

For example, the square of 10 = 100, the square of 20 = 400.

g) The digit in the units' place of a perfect square is 0,1,4,6, or 9.

h) The digit in the units' place of a perfect square cannot be 2,3, 7, or 8

Sign of A Square Root:

To express a square root two symbols are used. Square root of 25 is written as

25 or (25) We know, 5 × 5 = 25. again (_5) x (_5) = 25. As such, the square root of

25 is 5 or _5. However, in arithmetic 25 is 5. Also (25) may be 5 or _5.

The sign written to the left of a number expresses the square root of the number

4 Junior Secondary Mathematics

1_2

1_2

Surd: Number which are not perfect squares, their square roots cannot be determined exactly, such square roots are called surds or irrational numbers.

For example: 2, 3, 10, Although the value of a surd cannot be determined exactly, we may find the value of the surd to the required decimal places.

1.2 Square Root With The Help Of Factor. (Or Factor Method)

This method is applicable to perfect squares only.

Rule: (1) First, the given number is to be expressed in its prime factors.

(2) To make groups, each consisting of 2 identical factors.

(3) To Write one factor from each group.

(4) The product of the successive multiplication of the written factors gives the

required square root.

Point To Remember:

One Factor Out of Every Pair Of Factors is The Square Root

Example 1. Find the square root of 1024

Solution :

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= (2 × 2) × (2 × 2) × ( 2 × 2) × (2 × 2) × (2 × 2)

Therefore, 1024 = 2 × 2 × 2 × 2 × 2 = 32

2 1024

2 512

2 256

2 128

2 64

2 32

2 16

2 8

2 4

2 2

1

Square Root 5

25

Example 2 : What is the least mumber by which 2450 must be multiplied in order to

become a perfect square?

Solution: Here, 2450 = 2 × 5 × 5 × 7 × 7 = 2 × (5×5) × (7 × 7) .

It is seen that the factor 2 is not in pair, If 2 is found in pair then the number would

have been a perfect square. Therefore, if"the number is multiplied by 2 the number

will be a perfect square.

∴ the required number = 2

Example 3 : What is the least number by which 4608 must be divided in order to

become a perfect square?

Solution: Here, 4608 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

= (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × 2 × ( 3 × 3)

It is seen that the factor 2 is onlywithout pair. Therefore the number will be a perfect

square when divided by 2.

The required least number = 2

1.3 To Find The Square Root By Division.

An example illustrates the method to determine the square root of a number.

Example 4 : Find the square root of 2304 by division.

(1) Write down the number

(2) Divide the number into pairs or periods of two digits each beginning

from the extreme right digit. Give bar over every pair /period.

(3) Draw a vertical line to the right of the number, as in division.

(4) The first pair or period is 23. The nearest perfect square before 23

is 16, whose square root is 16 or 4: now write 4 to the right of

the vertical line and 16 just below 23, Draw a horizontal line

below 16.

(5) Subtract 16 from 23

(6) Put the next pair or period 04 to the right side of the difference

7. Draw a vertical line (sign of division) to the left of 704.

(7) Twice the quotient is 4 × 2 or 8, put it to the left of 704

separated by the vertical line. Keep a space for inserting a

digit between 8 and the vertical line.

2304

2304

2304

2304 4

16

2304

2304 4

16

7

2304 4

16

2304 4

16

704

8 704


(8) Now find out a digit and place it to the right of 8 so that

the number so formed when multiplied by that digit

equals 704 or is less than 704.

In this case it is 8:

Put 8 to the right of 4 in the quotient

Note: 88 × 8 = 704

(9) 48 is obtained in the quotient.This is the required square root.

∴ 2304 = 48

Explanation To Find The Square Root Of A Number By Division.

We know from the formula in Algebra that,

(a+b)2 = a2 + 2ab + b2

= a2 + (2a + b)b

So, (30 +6)2 = 302 + 2 × 30 × 6 + 62

= 302+(2 × 30 + 6) × 6

And 125 = (120 + 5)2

= 1202 + 2 × 120 × 5 + 52

= 1202 + (2 × 120 + 5) × 5

Let us find the square root of 7569

Now, 802 = 6400

and 902 = 8100

Then, clearly the square root of 7569 is greater than 80 and less then 90.

So, the digit in the tens place of the square root will be 8. If we now subtract 802 or

6400 from 7569, the difference will be 1169 and that will be equal to (2 × 80 + digit in

the units' place) × (that very digit in the units' place). By trial it is found that the digit

in the unit place is 7. For (2 × 80 + 7) × 7 = 1169. The method is in the next page.

2304 48

16

88 704

704

0

7569 80 + 7

6400

1169

1169

0

2 × 80 + 7 = 167

Square Root 7

In practice the work is shortened by omitting zeroes thus:

Example 5 : Find by the method of divison, the square root of 92416.

Solution :

∴ the. reqired square root = 304

Note: while forming pairs, 9 is taken into account and is considered as pair by itself.

After the pair 24 is brought down any digit placed to the right of 6 makes the trial

divisor greater than 24.For Example 61 × 1 = 61 So, placing 0 in the quotient, the pair

16 is set.

Example 6 : what is the least number which must be subtracted from 7428 in order

to get a perfect square? what is that perfect squre ?

Solution :

Therefore, 7428 is not a perfect square. If 32 is subracted from 7428 the difference

will be a perfect Square.

∴ the required perfect square is (7428 _ 32) or, 7396

Example 7 : What is the least number to be added to 651201 in order to obtain a

perfect square?

Solution :

9 24 16 304

9

604 2416

2416

0

74 28 86

64

166 1028

996

32

7569 87

64

167 1169

1169

0

65 12 01 806

64

1606 11201

9636

1565


Therefore, 651201 is not a perfect square. The least number which when added to it

will make the sum a perfect square and then its square root will be 806 + 1 = 807

∴ the required least number = 807 × 807_ 651201

= 651249 _ 651201

= 48

that is, 48 is the least number to be added

Ans. Least number 48. square root = 807

Important Points About Perfect Squares And Square Roots :

(1) For any number, one factor out of every two factor is its square root.

(2) If the digit in the extrame right of a number, that is the digit in the unit's place is

2 or 3 or 7 or 8, then it is not a perfect square.

(3) If odd number of zeroes are found to the right of a number, it cannot be a perfect

square.

(4) If the digit in the units'place of a number is 1 or 4 or 5 or 6 or 9 it may be a perfect

square. Again, if even number of zeroes are found to the right of a number, it may

be a perfect square.

For example, 81, 64, 25, 49, 100 etc,

(5) If a point be placed over every second digit in any number beginning with the

units place, the number of points will be the same as the number of digits in the

square root.

For example,

81 = 9 (one digit, here one dot over the number)

100 = 10 (two digits, here two dots over the number)

47089 = 217 (Three digits, here three dots over the number)

Exercise 1.1

1. Find by factors the square roots of:

(a) 361 (b) 784 (c) 1296 (d) 2025

(e) 3969 (f) 7396 (g) 8100 (h) 9216

Square Root 9

2. Find by division method, the square root of:

(a) 729 (b) 961 (c) 6561 (d) 49284

(e) 54756 (f) 717409 (g) 193600 (h) 1002001

3. What are the least numbers by which the following numbers be multiplied in order

to become a perfect square?

(a) 432 (b)2187 (c) 23805 (d) 23104

4. What are the least numbers by which the following numbers be divided in order

to become a perfect square?

(a) 2205 (b) 2187 (c) 1922 (d) 3072 (e) 4056

5. What is the least number that must be added to 7245 so that the sum is a perfect

square?

6. What is the least number that must be subtracted from 49289 so that the

difference is a perfect square?

7. Find the least number that must be subtracted from 6558 so that the difference is

a perfect square.

1.4 Square Root or Decimal Fractions

To find the square root of a decimal fraction we proceed as in the case of a whole

number or an integer.

Rule:

(a) The integral part is divided into pairs or periods beginning from the digit

immediately preceding the decimal point as in the case of a whole number. That

is, marking off the periods of two digits each starting from the digit in the units,

place.

(b) The decimal part is marked off in pairs or periods moving to the right of the

decimal point.

If, however, it is found that only one digit remains at the extreme right end then

annexe a zero after that digit and mark over the two digits.

(c) In the usual process of extracting square root when the work is over for the

integral part, the decimal point should, be put in the square, root before bringing

down the pair following the decimal point.


(d) For every pair of zeroes in the decimal part of a number, one zero is put in the

square root after the decimal point.

Example 8 : Find the square root of 644.1444

Solutions :

∴ the required square root = 25.38

Example 9 : Find the square root of 0.001936

Solution :

∴ the required square root = 0.044

Example 10 : Find the square root of 25.462 to two places of decimals.

Solution :

∴ the required square root = 5.05 (to two places of (decimals)

Remarks : Since the third place after decimal is 5 so 1 is added to the digit in the

second place to determine the approximate value of the square root to two decimal

places, which is 5.05.

644.1444 25.38

4

45 244

225

503 1914

1590

5068 40544

40544

0

0.00 19 36 .044

16

84 336

336

0

25.46 20 00 5.045

25

1004 46 20

40 16

10085 6 04 00

5 04 25

99 75

Square Root 11

Rule :

(1) If asked to find the square root to two decimal places. obtain the value up to three

decimal places.

(2) If asked to find the square root of a number up to three decimal places, take at

least 6 digits after the decimal point in the number. If necessary, annexe

required number of zeroes after the extreme right digit of the number. This does

not change the value of the number.

(3) If the third digit after the decimal in a square root is 0, 1, 2, 3 or 4, then there will

be no change in the second digit after the decimal in the square root.

(4) If the third digit after the decimal in a square root is 5, 6, 7, 8 or 9, then add 1 to

the digit in the second place after the decimal in the square root.

Example 11 : Find the square root of 3rd to 2nd places of decimals.

Solution :

∴ the required square root = 1.73 ( to two places of decimals)

Exercise l.2

Find the square root of the following numbers: (Question 1-9)

1. 0.49 2. 1.21 3. 0.0025. 4. 4.7089 5. 153.76.

6. 0.005329 7. 0.00005625 8. 644.1444 9. 1.002001

Find the square root to 3rd place of decimals (Question 10-19)

10. 5 11. 8.12 12. 21 13. 0.5 14. .03 15. 0.527

16. 2.7 17. .3 18. 3.36 19. 0.037

3.00 00 00 1.732

1

27 200

189

343 1100

1029

3462 7100

6924

176


1.5 Square Root of Fraction

The square root of a fraction reduced to its lowest term is the square root of its

numerator divided by the square root of its denominator. If the denominator be 'not a

perfect square, it is convenient to make it So by multiplication.

Perfect Square Fraction :

When a fraction is reduced to its lowest term, the numerator and the denominator are

both perfect squares, then the fraction is said to be a perfect square fraction.

Example 12: Find the square root of and

Solution : Square root of = =

Square root of = = = (Reduced to lowest terms)

Example 13 : Find the square root of 210 and 25

Solution : Square root of 210 = 210 = = = 14

Square root of 25 = 25 = = = = 5

Example 14 : Find the square root of and to there places of decimals.

Solution : = = = = = 0.527

= = = = = 0.764

Exercise 1.3

1. Find the square root of the following numbers:

(a) (b) (c) (d) (e) 3

(f) 83 (g) 38 (h) 104

2. Find the square root of three places of decimals.

(a) (b) (c) (d) 4 (e) 8 (f) (g) 4

1.6 Practical Application Of Square RootExample 15 : A general wishing to arrange 63009 soldiers into a solid square found

that these is an excess of 8 soldiers. How many soldiers were there in each row?

144289

144289

192363

192363

64121

242288

8414

242288

242288

7442288

3721144

9121

144289

44110000

811

144289

1217

14

14

14

292

6112

518

518

712

149

181

1125

4081

2249

57

518

712

512

1718

532

313

2136

√216

1036

√106

5×2

18×27×3

12×3

4.6826

3.1626

712

112

12

192363

Square Root 13

Solution: Since 8 soldiers were excess, so if 8 is subtracted from 63009 the difference

will be a perfect square.

Ans. There were 251 soldiers in each row.

Example 16 : A battalion of soldiers can be arranged in rows of 8, 10 and 12. They

may again be arranged squared form, What is the least number of soldiers in the

battalion ?

Solution : The battalion can be arranged in row of 8,10 and 12. So the number of

soldiers in that battalion should be divisible by 8 , 10 and 12.

Such least number is the L. C. M of 8, 10 & 12.

∴ the L.C. M = 2 × 2 × 2 × 5 × 3 = 120

The battalion consiting of the L. C. M = 120 can be arranged in rows of 8, 10 and 12

but cannot be arranged in a square; become 120 is not a perfect square.

To make 120 or 2 × 2 × 2 × 5 × 3 a perfect square, we should multiply this number by

at least 2 × 3 × 5.

∴ the least number of soldiers which can be arranged in rows of 8, 10 and 12 and into

a squared form = 2 × 2 × 2 × 3 × 5 × 2 × 3 × 5 = 3600

Ans. 3600 soldiers.

63009

(_) 8

63001 251

4

45 230

225

501 501

501

0

2 8, 10, 12

2 4, 5, 6

2, 5, 3


Exercise 1.4

1. In a certain garden there were 8 rows of trees. To plant 8 trees in each row, how

many trees are required ?

2. In a certain garden there are 1024 coconut trees. If the number of trees in rows

along the length and this breadth of the garden be same then how many trees are

there in each row ?

3. What least number of soldiers should be withdrawn from 9220 soldiers so that the

soldiers may be arranged in a squared form.

4. What least number of soldiers should be subtracted from or added to 56728

soldiers so that the soldiers can be arranged in solid square ?

5. The product of two numbers is 2688 and their quotient is 6/7. What are the numbers?

6. What is the least perfect square divisible by 9, 15 and 25 ?

7. All the pupils in a class subscribed as many five paisa coins as there were pupils:

the whole subscription raised being Tk. 125.00. How many pupils were there in

that class ?

8. The difference of the squares of two consecutive numbers is 91. Find the numbers.

9. The Product of two numbers is 2250 and their quotient is 9/10, what are the

numbers? .

10. All the members in a boarding subscribe as many 4 times, as there are members,

the whole subscription raised being Tk. 4900. How many members are there?

11. The monthly expenditure for each student is 10 times as the member of the

students in a hostel.There monthly total expenditure is Tk. 62500. How many

students are in the hostel ?

Square Root 15

Multiple Choice Questions:

1. How much is the square root of ?

(a) (b)

(c) (d)

2. How much is the square root of 1.21 ?

(a) 1.01 (b) 1.10

(c) 1.11 (d) 0.110

3. What is the square of 0.03 ?

(a) 0.0900 (b) 0.9000

(c) 0.0090 (d) 0.0009

Answer the questions 4 _ 6 from the following informations:

The difference of square of two consecutive integers is 11.

4. If one number is 6, then how much is the other number?

(a) 4 (b) 5

(c) 7 (d) 8

5. How much is the sum of the square of the two numbers?

(a) 16 (b) 17

(c) 61 (d) 71

6. How much should, be subtracted from the product of the two numbers so that the

result would be a whole square?

(a) 5 (b) 6

(c) 7 (d) 9

7. Observe the following statement

(i) 64 is a whole square.

(ii) 112 is a whole square.

(iii) The square root of 81 is 9

which one of the following is correct?

(a) i and ii (b) ii and iii

(c) i and iii (d) i,ii and iii


144289

12289

14417

1712

1217

CREATIVE QUESTIONS

1. An army team can be arranged in 4, 5, 9. rows but they cannot be arranged in a

square shape.

a) What are the factors of 9?

b) By which smallest number the total number of the soldiers should be

multiplied to arrange the army in a square?

c) At least how many soldiers should join the troop to arrange them in a square?

2. The monthly expenditure of each student of a hostel is ten times of the number of

students' living in that hostel. Monthly expenditure is Tk. 9000 in that hostel.

a) Consider the number of students is x, express the monthly expenditure in

terms as x.

b) Find the number of students of that hostel.

c) The students can not be arranged in a square after admission of 12 students

in to the hostel. At least how many students should be left to arrange them in

a square?

3. A team of 655 members of Bharoteswary homes arrives at Dhaka Stadium for

the independent day display. There are 6 teachers in the team.

a) How many students of homes are present in the stadium?

b) At least how many students should be removed from the team to arrange

them in a square?

c) At least how many students should join the team to arrange them in a square?

4. A farmer has 535 mango trees and 1156 coconut trees. He wants to plant equal

number of trees along the length and the width of the garden.

a) How many trees does have?

b) If he plants coconut trees in his garden, find the number of coconut trees in

each row.

c) How much more mango trees would he require to plant equally in each row

along length and width?

Junior Secondary Mathematics 17

Chapter Two

Ratio, Proportion2.1 Ratio

In the above pictures there are four circles with their respective radii. The circles are arranged in ascending order of their sizes from the left. On the other hand it can be seen that the cirles arranged in discending order of their sizes from the right.

Now if we compare the first circle with the fourth one, we can say that,

(a) The diametre of the fourth circle is a four times than that of the 1st

circle or times. In other words,

(b) The diametre of the 1st circle is th of the diametre of the 4th circle.

In case of (a) we can say that the ratio between the diametres of the 4th to the 1st circle is .

Similarly in case of (b) the ratio of the diametres between the 1st and the

4th circles is .

So the term 'ratio' is used to compare between the two quantities.

When two quantities of same category are compared to ascertain how many times the first contains the other, this kind of comparison is known as the ratio between the two quantities.

The ratio in case of 'a' can be written as 4t1 and is read as 4 is to 1. Again in case of (b) the ratio is written as 1 t 4 and read as 1 is to 4. The mathematical symbol of ratio is represented by the sign:

41

14

41

14

1cm2cm

3cm4cm

e.g. the ratio between Tk. 5 and Tk. 3 can be expressed by 5 t 3

Observation : The ratio between 1 cm and 4 cm is written as

Therfore, 1 t 4 =

Hence, the ratio is a pure number and a fraction.

2.2 The antecedent and the consequent of simple ratio.

Simple ratio :

The ratio having two quantities is called simple ratio. The 1st quantity of simple is called antecedent and the 2nd quantity is called consequent e.g. in the ratio 7 t 5, here the quantity 7 is called antecedent and 5 is called consequents.

Ratio of less Inequality : The ratio in which the antecedent is smaller than the consequent is called the Ratio of less Inequality.

Ratio of greater Inequality: The ratio in which the antecedent is greater that consequent is called the ratio of greater Inequality.

Unit Ratio: The ratio in which the antecedent and the consequent are equal is called unit ratio.

2.3 General Rules regarding Ratios :

(a) Value of any ratio remain unchange if antecedent and consequent of a ratio multiply or divide by same number except 0.

e.g. 5t 4 = (5 × 10) t (4 × 10) = 50t 40

50t 40 = (50 ÷ 10) t (4 ÷ 10) = 5 t 4(b) Ratio can express as smallest form like fraction.

e.g. 50 t 40 = 5t 4 [Divide Antecedent and Consquent by 10]

2.4 Different types of ratios.a) Inverse ratio: The ratio formed by the cosequent of the given ratio as its antecedent and the antecedent of the given ratio as its consequent then this ratio is called the Inverse ratio of the given ratio. For example the inverse ratio of 16 t 5 is 5 t 16.

(b) Mixed or Compound Ratio: The ratio obtained by successive product 'of the antecedents of given simple ratios as its antecedent and the successive product of the consequents of the given ratios as its consequent then this is called the compound ratio.

Example 1. Find the compound ratio of 4 t 3, 7t12, 9t 5Solution : The antecedent of the given ratios are 4, 7, 9 and their product 4 × 7× 9 = 252

14

14

Ratio, Proportion 19

The consequents of the given ratios are 3, 12, 5 and their product = 3 × 12 × 5 = 180 .

∴ The required compound ratio 252 t 180 or 7t 5 [After simplification]

Ans. 7 t 5

c) Duplicate Ratio: The ratio formed by squaring the antecedent of the given ratio as its antecedent and by squaring the consequents of the given ratio as its consequent then this ratio is called duplicate ratio of the given ratio. e.g. the duplicate ratio of 7 t 5 is 72 t 52 or 49 t 25

d) Sub-Duplicate ratio : The ratio formed by taking the square roots of both the

antecedent and consequent of the given ratio is said to be the sub-duplicate of the given ratio. e.g. the sub--duplicate ratio of 25 t 9 is 25 t 9 = 5 t 3.

2.5 The double rule of three :

Let the length, breadth and height of a house are 17. 15, and 10 metres respectively.

The ratio of length, breadth and height = 17, 15 t 10

In short,

Length: Breadth: Height = 17 t 15 t 10.The above ratio is known as double rule of three. Similarly we can form a ratio for more than three quantities also.

Example 2. The length and breadth of a garden are 15m and 10m 25 cm respectively. Find the ratio of length and breadth of the garden.

Solution: Length of the garden = 15 m = (15 ×100) cm = 1500 cm.

Breadth of the garden = 10m 25cm = (10 × 100 + 25) cm. = 1025 cm

∴ Lengtht Breadth =1500cm t 1025cm = 1500 t1025 = 60 t 41 (Dividing by 25)

Ans. 60 t 41

Remark: Both the quantities of the ratio should be converted into the same units.

Example 3. Find the Inverse ratio, Duplicate ratio and sub-duplicate ratios of the following Ratios:

a) 7 t 16 b) 4 t 25

Solutions: a) The inverse ratio of 7 t16 is 16 t 7

The duplicate ratio of 7 t 16 is 72 t 162 or 49 t 256. The sub-duplicate ratio of 7 t 16 is

7 t 16 or 7 t 4


b) The inverse ratio of 4 t 25 is 25 t 4. The duplicate ratio of 4 t 25 is 42 t 252 or 16 t 625. The sub -duplicate ratio of 4 t 25 is 4 t 25 or 2 t 5

Example 4. The ratio of the ages of father and the son's age is 11 t 4. The age of the son is 16 yrs. Find the age of the father.

Solution : Age of father t Age of son = 11 t 4

Therefore, the age of father is times than that of son.

Therefore, the age of father = × age of son = ( × 16) ytars = 44 years

Ans. 44 years.

Exercise 2 .1

1. Express the first quantity as the ratio of the second quantity.

a) 7.5 and 3.5 b) 8 and 5

c) 5 hrs 10 minutes and 3 hrs. d) 10 m and 5m 7cm

e) 5 kg and 7kg 9 gram f) 2 years. 6 months and 5 years.

2. Express into mixed ratiot

a) 7 t 5, 4t 13, 11t 7 b) 5 t 9, 4 t 17, 3 t 11

c) 6 t 13, 7t 12, 26t 3 d) 8 t 7, 14 t 23, 46 t 9

3. The ratio between the circumference and diameter of a circle is 22 t 7. The diameter of the circle is 4m 20 cm. What is the circumference of that circle.

4. The ratio of the age of Mr. Azmal and his son is 9 t 2 If the age of the son is 12 years, what is the age of Mr. Azmal ?

5. Which one is the biggest ratio among the following ratios 3 t 7, 5 t 9 and 7t 11.

6. The ratio of the prices of two books is 3 t 5. If the price of the 1st book is Tk. 19. 20 then what is the price of the 2nd one ?

7. An ornament weighing 30 gm is made by mixing gold and silver. The weight of the gold is 25gm. What is the ratio of the weight of silver and gold ?

8. The ratio of the ages of Monowara and shipra is 7 t 4. If the age of Monowara is 35 years, find the age of shipra ?

114114

114

12

12


9. Out of the three coloured balls, red, yellow and white, the ratios of the weight of red to yellow balls is 5 t 6. The weight of yellow and white is 5 t 4, The yellow is 180 gm. find the ratio of the weights of the white to the red balls.

10. The ratio of the price of two televisions is 4 t 3. The price of the 2nd television is Tk. 21,000.00. If the price of the 1st television is 1,000.00 taka higher than the price of the 2nd one what is the ratio of prices of the two televisions ?

11. The sum of the two quantities is 450 and their ratio is 8 t 7. Find the number.

12. Asish covers 16 kilometres in 3 hrs and Babu cover 18 kilometres in 4 hrs.

Find the ratio between their speeds.

2.6 Successive Ratios :

The two ratios A t B and B t C can be written as A t B t C. This is called

Successive Ratio. Any two ratios can be express into Successive Ratio.

Observation: The consequent of the 1st ratio A t B is B and the antecedent of the 2nd ratio B t C is B.

If the consequet of 1st ratio is equal to the antecedent of 2nd ratio then they can be written as successive ratios.

Example 5. Express 5 t 7 and 9 t 11 as successive ratio.

Solution : 1st ratio = 5 t 7 = 5 × 9 t 7 × 9 t 45 t 63

2nd ratio = 9 t 11 = 9 × 7 t 11 × 7 = 63 t 77

∴ The successive ratios of the given two ratios are 45 t 63 t 77.

Ans. 45 t 63 t 77

Rule: The two quantities of 1st ratio should be multiplied by the antecedenr of the 2nd ratio. Again both quantities of the 2nd ratio must be multiplied by the consequent of 1st ratio.

2 .7 Mixture/CompoundA mixture is made by mixing more than one kind of things. The things which are mixed together are called the elements of the mixture. A mixture car, be produced by mixing elements in different ratios. For example, a mixture may be made by mixing 2 litres of water with 10 litres of syrup. In that case we say that this is a water mixed syrup. In this mixture the ratio of syrup and water = 10 litre t 2 litre = 10 t 2 = 5 t 1

12

910

12


We can produce new material by combining two or more matarial or things in a definite ratio by Chemical reaction, e.g. If we mix up zinc with oxygen at the ratio 130.34 t 32 then we get zinc oxide.

Example 6. A compound manure is produced by mixing up 6 kg of ammonia 8 kg of potash and 20 kg of cowdung. In this compound manure find the ratio of amonia, potash and cowdung.

Solution : The weight of amonia : weight of Potash : weight of cowdung

= 6 kg t 8 kg t20 kg

= 6 t 8 t20 = 3 t 4 t 10 [ Dividing each of the quantity by 2 ]

Ans. 3 t 4 t 10

2.8 Proposition

Suppose there are 4 quantities 6 kg, 8kg.Tk.24 and Tk.32.

Now, I st quantityt 2nd quantity = 6 kg t 8 kg = 6 t 8 = 3 t 4And 3rd quantityt 4th quantity = Tk. 24 t Tk. 32 = 24 t 32 = 3 t 4Therefore, 1st quantity t 2nd quantity = 3rd quantity : 4th quantity

∴ The two ratios are equal.

In this case usually it can be said that the four quantities 6kg, 8kg 24 Taka and 32 Taka form a proportion. These four quantities are said to be proportionals.

If the ratio of the 1st quantity an 2nd quantity is equal to the ratio of the 3rd quantity and the fourth quantity of given four quantities then these four quantities form a proportion. The four quantities are called proportionals.

Remarks: The 4 quantities of the proportionality may not be of the same units. But each of quantities of the ratio must be of the same units.

The fourth quantity of the proportionality is said to be the 4th proportional.

Extreme quantities: The 1st and the 4th quantities/terms of the proportion are called extreme element or simply extremes.

Mid-element /middle term : The 2nd and 3rd quantities of the proposition are called mid element/middle terms.

The proportion 6 t 8 = 24 t 32 is generally written as 6 t 8 t t 24 t 32 . We use the symbol 't t' in place of the symbol (=).


In the example we observe t 1st quantities/lst term 2nd quantity/2nd term

6 x 32 = 8 x 24

4th quantity/4th term 3rd quantity /3rd term

Generally,

In the proportional, I st quantity × 4th quantity = 2nd quantity × 3rd quantity

Example 7 : Find the fourth proportional of 5, 8, 15

Solution : When the four quantities are in proportion

then 1st quantity × 4th quantity = 2nd quantity × 3rd quantityHere, 1st quantity = 5, 2nd quantity = 8, 3rd quantity = 15

∴ 5 x 4th quantity = 8 × 15

∴ 4th quantity = = 24

Ans. 24.

2.9 Continued proportion :

Let us take the three quantities 3kg, 6kg and 12kg respectively. Here two ratios can be formed out of these three quantities. These are 3 t 6 and 6 t 12 Clearly, 3 t 6 t t 6 t 12

This type of proportionality is called continued proportion, The three quantities 3kg, 6kg and 12 kg are of continued proportion.

When the ratio of the 1st to the second is the same as the ratio of the 2nd to the third of the three given quantities of taken units are said to be in continued proportion.

The second quantity is said to be the mean proportional of the 1st and the 3rd quantities. The third is called the 3rd proportional of Ist and 2nd quantities.

Remarks : The three quantities of continued proportion must be of the same kind we observe the following in the continued proportion.

3rd quantity

3 × l2 = (6)2

1st quantity Second quantity

8 × 155


Generaly,

In the continued proportion, 1st quantity × 3rd quantity = (2nd quantity)2

Example 8 : The 1st and 3rd quantities of a continued proportional quantities are 6 and 24 respectively. Find the mid-term of the continued proportionality and write down the continued proportion. .

Solution : We know in any continued proportion

(middle quantity)2 = 1st quantity × 3rd quantity.

Here, 1st quantity = 6, 3rd quantity = 24

∴ (Middle quantity)2 = 6 × 24 =144

That means middle quantity = 144 = 12

Again, the required continued proportion is 6 t 12 t t 12 t 24

Ans. Middle quantity = 12, Continued proportion = 6 t 12 t t12 t 24

2.10 The rule of three

It has been discussed earlier that in case four quantitis in a proportion.1st quantity × 4th quantity = 2nd quantity × 3rd quantity

Now, the 1st quantity, the 2nd quantity and the third quantity are 7. 14 and 15 respectively.

∴ 7 × 4th quantity =14 × 15

∴ 4th quantity = = 30.

Therefore, if the three quantities of a proportion are known then we can easily find the 4th one. This process of finding the 4th quantity is called the rule of three.

Example 9. If the price of 8 kg of rice is Tk. 120.00 then what is price of 12 kg of rice?

Solution: The ratio in which the quantity of rice increases the price of rice also increases in the same ratio.

That is the ratio of the quantity of rice = the ratio of the price of rice.

∴ 8 kg t 12kg = 120 Taka t Required price or 8 t 12 = 120 Taka t Required price

Here, 1st quantity = 8, 2nd quantity = 12, 3rd quantity = 120 Taka.

14 × 157


∴ 8 x Required price =12 × 120 Taka

∴ Required price = Taka = 180 Taka,

Ans. 180 Taka.

Remarks : The method of finding the 4th proportion as described is said to be simple rule of three.

In solving this type of problem the following things should be considered:

(1) The quantity which is similar in nature with the required quantity is cal1ed 3rd quantity of the proportion and the required is the 4th quantity.

(2) The other two quantities should be placed in the 1st and 2nd quantities respectively. These two should be without any unit [Since in the example we have seen ,8kg t 12kg = 8 t 12]

(3) The units of 3rd & the rquired quantity must be of the same units.

Example 10. A work can be done in 10 days by 8 persons. How many days would it require to finish the work by 20 persons.

Solution: The ratio of increase or decrease of manpower is the same as the ratio of decrease or increase of days required. In other words we can say that the inverse ratio of manpower is equal to the simple ratio of days required.

Here No. of persons days

8 persons 10 days

20 persons Required days.

∴ 20 t 8 = 10 days t Required days.

∴ 20 × Required days = 10 × 8 days

∴ the required days = days = 4 days

Ans. 4 days.

Remarks : The method described earlier to find the 4th proponional is said to be an

Inverse Rule, of three.

12 × 1208

10 × 820


Exercise 2.2

I. The extreme quantities of three consecutive quantities are given. Find the

proportion.

a) 5, 45 b) 25, 81

c) , 1 d) 1.5, 13.5

2. Put down the exact quantity in the box :

a) 12 t 16 t t t 20 b) 4t tt 9 t 18

c) 2.6t1.3tt 1.1: d) t tt t

3. Find the fourth proportional of the following :

a) 5, 10, 8 b) 24, 32, 36

c) 4, 4 , 2 d) 10, 40, 50 e) 3, 4.5, 8

4. 5 kg of sugar is mixed with 50 kg of milk. Find the ratio between the sugar and

the milk in the mixture of sugar and milk.

5. A farmer has 40 quintal of cowdung and 16 quintal of Amonia manure. The farmer

purchased same potash and mixed it with cowdung and Amonia. In this mixture the ratio

among Amonia, cowdung and potash is 2 t 5 t 3. Find the quantity of potash purchased.

6. Out of three quantities-of continued proportionals the product of the two quantities

is 36. If one of the quantities is 12 then find the other quantities continued

proportional.

7. The cost of 12 kg of pulses is 480 Taka. Find the cost of 25 kg of pulses.

8. The cost of 5 books is Tk. 126.25. How many books are available by Tk 429.25.

9. A work can be done in 20 days by 18 persons. How many days will be required

to do the same work by 27 persons ?

10. There is a proyision of food for 50 students for 20 days in a boarding. After 3

days 16 students left the boarding: How many days the food can be served the

remaining students ?

11. A factory can produce 14.600 bulbs in 365 days. How many bulbs can the factory

produce in 140 days ?

12. The ratio of the salaries of Mr. Jamal and Mr. Samaresh is 6 t 5. The total amount

of their salaries is Tk. 11,000.00. The salaries of Mr Jamal and Mr Samaresh are

increased by Tk 100.00 and Tk.75.00 respectively after one year. Find the ratio of

their salaries after one year.

35

13

12

15

310

35


13. In a mixture the ratio of weight between the powder milk and water is 5 t 6. In

another mixture the ratio of the weights between the water and the sugar is 7 t 4.

If the mixtures are combined together then find the ratio among the milk, water

and sugar.

14. The ratio of the weights of acid and water in 30 litre of mixtures is 7 t 3. How

much water should be mixed to get the ratio 3 t 7 between the Acid and the

water.

2. 11 Proportional Division :

Suppose we like to divide 30 metre length of a tape into two parts, the ratio 7t3. To

divide the tape at the ratio 7 t 3 means that we are to divide the tape into (3+7) =10

equal portion .and then take 7 portion and 3 portion out of 10

portions. In other words the 1st part of the tape will be equal to parts of the tape.

The length of 1st portion = × 30 metre = 21 metre

and the length of 2nd portion = × 30 metres = 9 metre .

Therefore, one portion = × The given quantity

To divide a given quantity by the process described above is called proportional

division.

To divide a given quantity into two or more parts in a given number of portion is

called proportional division.

Example 11. An amount of Tk. 8,000.00 is divided among A, B and C in the ratio 7 t 5 t 4. How much amount will each of A, B and C get?

Solution: The sum of the proportion = 7 + 5 + 4 = 16

A will get × 8000 Taka = 3500 Taka.

B will get × 8000 Taka = 2500 Taka.

C will get × 8000 Taka = 2000 Taka.

Ans. 3500.00 Taka. 2500.00 Taka. 2,000.00 Taka.

2.12 Partnership business :If two or more people own a business then that business is called partnership business. And

each of them is the owner of that business. They invest money in the business according some

agreed agreement thus the business runs.

716

710

710

310

516416

The respective part of the proportion

The sum of the proportion


The partners after through discussion among themselves distribute the profit and loss

of the business under same agreement. If nothing is mentioned in the agreement about

the share of profit and loss then the profit will be distributed proportional to their

money invested in the business. Similarly, the partners also share the loss accordingly.

For example A and B invested Tk. 2000 and Tk. 3000 respectively in a business. The

ratio of their profit will then be 2000 t 3000 = 2 t 3.

Example 12. Mr Jamal. Mr Azmal and Mr Suresh invested an amount of Tk 6400.00,

Tk 8000.00 and Tk. 9600.00 respectively in a joint stock company.

The profit comes to Tk 3600.00 in one year. How much profit each of them will get ?

Solution : Capita1 of Mr Jamal : the capital of Mr Azmal t the capital of Mr. Suresh =

6400 t 8000 t 9600 = 4 t 5 t 6

∴ The sum of the ratios = 4+5+6 =15

∴ The amount of profit of Mr Jamal = × 3600 Taka = 960 Taka

The amount of profit of Mr. Azmal = × 3600 Taka = 1200 Taka

The amount of profit of Mr Suresh = × 3600 Taka = 1440 Taka

Ans. 960.00 Taka, 1200.00 Taka, 1,440.00 Taka.

Example 13. Rani and Jani started a business jointly with the capitals 5000 Taka and

6000 Taka respectively. Papi joins the business with the capital 8000 Taka as a partner

after 6 months. The amount of profit is taka 3600.00 after one year. How much

amount of profit will each of them get?

Solution : In this business Rani, Jani and Popi invested their capitals in different time

period. Therefore. the amount of profit for a particular and for equivalent amount of

capital of each of them and then distribute the profit proportionately. The amount of

profit obtained by investing Tk.5000.00 for 12 months is equivalent to the amount of

profit by investing (5000 × 12) Taka 1 month.

∴ The equivalent capital of Rani for one month = (5000 × 12) Taka. Similarly, the

equivalent capital of Joni for 1 month = (6000 × 12) Taka.

Since, Popi has invested the capital for 6 months, The equivalent capital for 1 month =

(8000 × 6) Taka.

:. The equivalent capital of Ranit The equivalent capital of Joni t The amount of

capital of Popi.

515

515

615


= (5000 × 12) t (6000 × 12) t (8000 × 6)

= 10 t 12 t 8 [Dividing (1000 × 6)]

=5 t 6 t 4

The sum of the ratios = 5 + 6 + 4 = 15

Therefore, the profit of Rani = × 3600 Taka = 1206 Taka

The profit of Joni = × 3600 Taka = 1440 Taka

The profit of Popi = × 3600 Taka = 960 Taka

Ans. 1200.00 Taka, 1440.00 Taka and 960.00 Taka.

Exercise 2.3

1. If At B = 4 t 5, B t C = 7 t 11 then find A . B t C.

2. The ratio of iron and carbon in a steel plate is' 49t 1. How much amount of carbon

contained in 200kg of steel?

3. The weight of a iron and copper mixed solid is 156 gm. The ratio of copper and

iron is 7 t 6. How much gram of iron should be mixed with copper so that the

ratio of copper and iron would be 6 t 7 ?

4. A farmer uses bio-manure, chemical manure in his field at the ratio 9t2. He

requires 22 sacs of mixed manure per hector of land. How much manure of each

kind does he require to use 8 hectors of land ?

5. The ratio of Acid and water in a mixture is 88.9 t 11.1 How much quantity of

acid and water contained in 100 kilogram of mixture?

6. The amount of Tk. 2160.00 is distributed among Jahanara. Ayesha and Minati at

the ratio 3t 7t 14. How much amount of money each of them will get ?

7. Tk 570 has been distributed among three boys as per their age ratio. The ages of

them are 10,13 and 15 respectively. How much money each of them will get ?

8. Some amount of money has been distributed among Azmal, Kader and Rodricks,

at the ratio 7 t 5 t 4. Amal gets an amount of Tk. 350.00. Find the total amount of

money.

9. The ratio of successful and unsuccessful students of a school of 132 students is 9:

2. If the number of successful students is increased by 4 more then what will be

the ratio of successful and unsucessful students ?

415

615

515


10. The monthly salary of Mr. Atik is Tk.5600.00. He saves an amount of Tk.500.00

per month and spend the rest for house rent, monthly expenditure and educational

expenditure of his children at the ratio 5t 11t 1. How much amount does he

expend for the educational expenditure of his children ?

11. The ratio of the incomes of Panir and Monir is 5t 4.The ratio of income of Monir

and Tapan is 3 t 4. The income of Panir is Tk 120.00. Find the income of Tapan.

12. The two friends Nitai and Tanim started business by investing an amount of

money 1500 Taka and 2500 Taka respectively. After one year, the amount of

profit stands at the amount of Tk 849.00. What is the amount of profit of Nitai

and Tanim ?

13. Anwar, Uttam and Jalal started a business investing the amount Tk. 5000, Tk.

6000 and Tk. 5500 respectively. After one year the profit stands at Tk. 2475.00.

How much profit will each of them get ?

14. Ishaque, Ramesh and Zaman invested Tk.1500, Tk. 1200 and Tk. 1700

respectively in a joint cloth business. After one year the loss stands at an amount

of Tk. 220.00. How much amount of loss will each of them share ?

15. Ashiq and Jafar purchased a Television at the cost of Tk. 25000.00. and sells it at

an amount ofTk 2625.00. Ashiq paid an amount of 1 times than that of Jafar

during purchase of the Television. How much money will Ashiq get from the

profit ?

16. At the begining of the year Zafar and Kashem invested an amount of Tk. 12000

and Tk.15000 respectively in a joint business. After 5 months Jamir invested an

amount of Tk. 2000.00 in the same business. After the end of the year they

profited of Tk.6760.00.How much each of Zamir and Kashem will get out of this

profit ?

17. The three labours Jahed, Maqbul and Rashid established a lath Machine business

by investing Tk. 6500.00; Tk. 9100.00 and Tk. 5200.00 respectively. At the end

of the year the profit stands at an amount of Tk. 5200.00. How much profit will

each of them get ?

18. An amount of Tk. 315.00 is divided among Lily, Mally and Pinkey, Here the

amount of Lily is times as the amount of Molly the amount of Molly is twice

the amount of Pinky. Find the amount of Lily, Molly and Pinky.

35

12


19. An ornament is made by mixing copper. Zinc and Silver. The ratio of copper and

Zinc in that ornament is 1 t 2, the ratio of Zinc and Silver is 3 t 5. How much

quantity of Zinc will there be when the ornament weight 38 gram ?

20. Suman and Jamal invested an amount of Tk.5000 and Tk. 4000 respectivly in a

business. After 3 'months Suman invested Tk. 1000.00 more and Dilip invested

Tk.7000.00 in the same business as a new partner. After one year the profit stands

at an amount of Tk.3600.00. How much profit will each of them earn ?

21. Three glass of same size are full of sugar juice. The ratio of water and syrup is 3 t 1 in the 1st glass, 5 t 3 in the 2nd glass and 9 t 7 in the 3rd glass. The Sarbat of

three glass was poured into another big container. Find the ratio of water and

syrup in the new container.

Multiple Choice Ouestions: 1 . What is ratio?

(a) An integer (b) a fraction

(c) A prime number (d) an even number

2. Which one of the following is the sub-duplicate ratio of 4 t 9 ?

(a) 2 t 3 (b) 4 t 9(c) 9 t 4 (d) 16 t 81

3. Observe the following informations:

i) 5 t 2 is the ratio of greater inequality.

ii) Three quantities of succession proportion are not homogeneous.

iii) If a t b.= 2 t 3 and b t c = 3 t 4, then a t b t c = 2 t 3 t 4Which one of the following is correct ?


(c) i and iii (d) i, ii and iii

Siam and Fatema started a jointventure business and their earned profit is Tk.

256. The dividend of Fatema is Tk. 160.

Answer the questions 4-6 in view of above informations:

4. Which one of the following is ratio of profit of Siam and Fatema?

(a) 8 t 5 (b) 3 t 8(c) 3 t 5 (d) 5 t 3


5. If the capital of Siam is Tk. 3000, how much taka is the capital of Fatema?

(a) 11 000 (b) 8000

(c) 5000 (d) 3000

6. If Fatema received dividend Tk. 156, which one of the following is the ratio of

Profit earned by Siam and Fatema?

(a) 3 t 5 (b) 25 t 64

(c) 64 t 39 (d) 25 t 39

CREATIVE QUESTIONS

1. The weight of a red ball and a green ball are 72 gm and 96 gm respectively. The

ratio of weight of a green ball and a black ball is equal to the ratio of weight of a

red ball and a green ball.

(a) Find the ratio of weight of a red ball and a green ball.

(b) Find the weight of a black ball.

(c) If the ratio of buying prices of red, green, and black ball is equal to the ratio

of their weight, find the buying price of red and green ball. (Here the total

price of a red, a green and a black ball is Tk. 296).

2. Mustafiz and Ayesha reached Dhaka travelling 50 km is 4 hours and 52 km in

3 hours respectively for business purpose. They earned 5% profit by investing

Tk. 22000 in business.

(a) Find the speed of Mustafiz.

(b) Determine the ratio of speed of Mustafiz and Ayesha.

(c) If the profit of Mustafiz and Ayesha is equal to the ratio of their speed, find

their profit.

3. The ratio of monthly salary of Arefin and Amena is 4 t 4 . The total of

their monthly salary is Tk. 29000. The salary of Arefin and Amena are increased

by Tk. 575.75 and Tk. 650.50. respectively after one year.

(a) Express the ratio of salary of Arefin and Amena in the form of a:b where a. b

are integer.

(b) Find the salary of Arefin and Amena.

(c) Find the ratio of salary of Arefin and Amena After one year.

12

12

15

12


Chapter Three

Unitary Method

3.1 Revision Of Previous Lessons.

We know, unitary method is to find the cost, weight, measure ete. of one thing trom

the cost, weight, measure etc. of any "number of things and from the cost weight,

measure etc. of one thing, the cost, .weight, measure etc. of particular number of thing

of the same kind its found out.

Let, the cost of 10 oranges be TK. 45.00, we can easily say that the cost of onc orange

is Tk. (45 ÷10)

That is, the cost of one orange is Tk. 45 ÷ 10. Now from the cost of one orange we

can find the cost of any number of oranges. For example,

Cost of 8 oranges = Tk. x 8 = Tk. 36

Rules for Solving the Problem

(1) The cost of oranges reduces if the number of oranges is less. For that reason, the

cost of 10 oranges is divided by 10 to obtain the cost of 1 orange. (2) The cost of

orange increase if the number of oranges increase. Therefore, the cost of 1 orange is

multiplied by 8.

Again let us assume that 10 people can do a piece of work in 7 days. In this case we

may say that 1 person will require 10 times the total time to do that work alone. That

is, 1 person can do the work in 7x10 days. Now if 5 people can do this work, they will

require less time than 1 person.

∴ 5 people can do it in days or, 14 days,

Rules for Solving the Problem,

(1) When number of people decreases the time required enhances. For this, the time

required by 10 people is multiplied by 10 to obtain the time required by 1 person.

(2) Again, when the number of people increases. the time required diminishes. So, the

time required by1 person is divided by 5.

4510

9

2

4

7x 105

2

Example 1. If the cost of 50 kg of sugar is Tk. 1475, what is the cost of 63 kg of

sugar?

Solution: Cost of 50 kg sugar is Tk. 1475

∴ " " 1 Tk.

∴ " " 63 Tk.

or, Tk.

or, Tk. 1858.50

Ans: TK. 1858.50

Example 2. Working 8 hours a day, 20 workers reap crop of a field in 15 days.

Working 6 hours a day, in how many days will 25 workers reap the field ?

Solution: In 15 days the workers work (8 x 15) hours or 120 hours.

That is 20 workers reap in 120 hours

∴ 1 worker reaps in 120 × 20 hours

∴ 25 workers reap in hours = 96 hours

Since the workers work 6 hours daily, so they will reap field in (96 ÷ 6) days, or 16 days.

Ans : 16 days.

Remark : We have taken into consideration that the workers have equal capability to

work.

Exercise 3.1

(1) If the cost of 105 kg. of dal is Tk. 4331.25 what is the cost of 70 kg. of dal (pulse)?

(2) If the weight of 275 silver coins is 2 kg. what is the weight of 325 silver coins?

(3) 7 kg. of coal boils (steams) 52 kg of water. How many kilogrammes of coal is

required to steam (boil) water weighing 915 kg.

(4) A shop owner purchases 80 kg of rice at Tk. 18.50 per kg and another 100 kg of

rice at Tk. 15.00 per kg. By selling both the kinds of rice he gains Tk. 170.00.

What is the average selling price per kg of rice?

147550

37172

1475 × 6350

120 × 2025

34

12

Unitary, Method 35

(5) A man earns Tk. 252 by investing a capital of Tk. 7000. How much will he earn

if he invests a capital of Tk. 10,500? .

(6) Arif Shaheb deposits some money in a bank. He receives a profit to Tk. 585 after

a year. For every Tk. 100 if he receives Tk.6.50 profit in I year, how much money

he had deposited?

(7) 55 people can do a piece of work in 15 days. How many people is required to

finish the work in 25 days?

(8) If 3 quintals of rice last 9 people 30 days, how long would they last 25 people ?

(9) 42 workers build 5000 bricks in 30 days. How long will it take 18 men to build

those bricks ?

(10) The cost of 5 radios and 4 television together is Tk. 89,500. If a television cost

Tk. 19.500. how much will 15 radios cost?

(11) A hostel of 30 students has provisions for 15 days. Some new students were

admitted to the hostel as such the provision lasted 10 days. How many new

students were admitted?

(12) 25 people can dig a canal in 17 days, working 8 hours a day. In how many days

can 34 people dig the same canal working 6 hours a day ?

( 13) 4 men or 8 women can do a piece of work in 10 days. In how many days 2 men

and 12 women could do it ?

(14) Rafiq covers a distance of 480 km in 12 days. walking 10 hours a day. In how

many days will he cover a distance of 360 km, walking 9 hours a day?

(15) A contractor contracts to build 20 km of road in 45 days. He employed 150

workers and could finish half the work in 30 days. What additional bumber of

workers must he employ to complete the work within the stipulated time?

(16) 15 men can do a piece of work in 8 days, working 6 hours a day. How many

hours a day must 10 men work to do the same in 9 days?

(17) 4 men and 10 boys can do a piece of work in 18 days. In how many days can 12

men and 30 boys do it?

3.2 Problems Relating to Time and Work

In problems relating to time and work, two or three different kinds of quantities are

associated. These are:


(a) Required time

(b) Amount of work

(c) The bumber of workers doing the work

The Tips Given Below Should Be Kept In Mind While Solving Problems Relating To Time And Work By Unitary Method:

(1) Time increases if quantity of work increases keeping the number of workers

constant. Again time reduces if the quantity of work decreases.

(2) Quantity of work increases if the number of worker increases keeping the time

fixed. Again the quantity of work decreases when the number of wrker decreases.

(3) Time reduces if the number of worker increases keeping the quantity of work

fixed. Again time increases if the number of worker decreases.

Example 3.

5 workers keep a jute field measuring 8 hectares in 4 days. In how many days could

25 workers keep 70 hectares?

Solution :

5 workers reap 8 hectares in 4 days

∴ 1 " " 8 " " 4 × 5 days

∴ 1 " " 1 " " days

∴ 25 " " 1 " " days

∴ 25 " " 70 " " days or, 7 days

Ans. 7 days.

Note : When some people together do a piece of work in a certain time. We consider

that each of them has equal capability to do the work.

Solution of problems in unitary method relating to constant amount or work done by two or more people taking different amount of time :

Example 4. Shafi can do a piece of work in 24 days.Ratan can do it in 12 days. In

how many days could Shafi and Ratan together do it ?

4 × 58

4 × 5

8 × 254 × 5 × 70

8 × 25

Unitary, Method 37

Solution : In 24 days Shafi can do a whole work or 1 work

∴ 1 day " " " " " of the work or works

Again, in 12 days Ratan can do 1 work

∴ in 1 day Ratan can do works .

So, Shafi and Ratan together can do ( + ) works.

or, works or, works

∴ Shafi and Ratan together can do of the works in 1 day

∴ Shafi and Ratan together can do 1(whole work in(1 ÷ ) days = 8 days

That is, Shaft and Ratan together can do the work in 8 days.

Ans. 8 days.

Example 5. Zakir and Monir do a piece of work in 20 and 30 days respectively. Zakir

and Monir together worked for 7 days and left. Ashish finished the remaining work in

10 days. In how many days could Ashish do the whole work?

Solution: In 20 days Zakir does 1 work

∴ 1 day Zakir does work

In 30 days Monir does 1 work

∴ 1 day Monir does work

∴ + = =

∴ In 1 day Zakir and Monir together can do works

∴ In 7 days Zakir and Monir together can do works

The work remaining is (1_ ) works or, works

That is, Ashish does works in 10 days

∴ Ashish does 1 (whole) work in (10 ÷ ) days

= (10 × ) = 24 days

Ans: 24 days.

124

124

124

112

112

324

18

18

120

130

120

130

560

112

112

712

712

512

512

512

125

18


3.3 Problems Concerning Time And Distance.

Example 6. A train of length 120 metres will cross a bridge of length 330 metres. If

the speed of the train is 30 km per hour, how much time the train will require to cross

the bridge?

Solution : The train has to cross the length of the bridge and its own length.

That is, it crosses, (330 + 120) metres or 450 metres.

Here, 30 km = 30 × 1000 metres,

and I hour = 60 × 60 seconds

The train travels 30 × 1000 m in 60 × 60 seconds

∴ " " 1 " " seconds

∴ " " 450 " '' seconds

Ans. 54 seconds.

Explanation :

The above figure shows the process of crossing starts when the front left end of the

bridge. From the figure at the bottom that the process of crossing ends when the end

of the train just leaves the right end of the bridge. So we find that to cross the bridge

the train has to cross the length of bridge and its own length i.e. (length of the bridge

+ length of the train).

Example 7. The distance between two places 'A' and 'B' is 37 km. 3 hours after

Ashish starts from A, his friend Rahman starts from the same place by a motor cycle.

If the speeds of Ashish and Rahman are 10 km per hour and 50 km per hour

respectively where and after what time will they meet?

Solution: In 1 hour Ashish covers a distance of 10 km

∴ In 3 hours Ashish covers a distance of 10 × 3 or 30 km

60 × 60 × 450

30 × 1000

60 × 60

30 × 1000

12

Unitary, Method 39

So, when Rahman starts, the two friends are at a distance of 30 km apart. The

difference of speeds of Rahman and Ashish in 1 hour is (50-10)km or 40 km. That is,

the distance of 40 km decreases between Rahman and Ashish in 1 hour is (50-10) km

or 40 km,

That is, the distance of 40 km decreases between Rahman and Ashish in 1 hour

∴ the distance of 1 km decreases between Rahman and Ashish in hours.

∴ the distance of 30 km decreases between Rahman and Ashish in hours.

or, hours.

So, the two meet hours after Rahman starts.

In hours Rahman travels km or 37 km.

both of them meet 37 km away from 'A' that is at 'B'.

Ans. The two meet at B, hours or 45 minutes after Rahman starts.

3.4 Problems Relating, to Pipe and Cistern/TankCistern/tank or such things requires pipe to fill water or liquid in it. Again to bring

out water or liquid, pipe is required. Due to the difference in the dimensions of a

pipe there is a difference in time to fill the cistern/tank. If there is leak in the cistern

/tank, water or. liquid leaks through it.

Example 8. A water tank has 3 pipes. The tank can be filled by the first and the

second pipe in 20 and 30 minutes respectively. The full tank is emptied in 60

minutes by the third pipe. In what time the tank be filled if all the three be opened at

once? .

Solution :

In 20 minutes the first pipe fills 1 (whole) tank

∴ In 1 minute the first pipe fills tanks

In 30 minutes the second pipe fills 1 (whole) tank

∴ In 1 minutes the second pipe fills tanks

140

34

34

34

34

120

130

12

1 × 3040

50 × 34


In 60 minutes the third pipe empties 1 (whole) tank

∴ In 1 minute the third pipe empties tanks

∴ When the three pipes are opened at-once they will fill in 1 minute ( + _ ) tanks

Now, ( + _ ) = = =

∴ tanks filled by the three pipes in 1 minute

∴ 1 (Whole) tank is filled by the three pipe in (1÷ ) minutes

= 15 minutes

Ans. 15 minutes.

3.5 Problems Related to Boat and Flow of StreamThe speed of a boat in still water is its actual speed. The speed of the boat along with

the current and against the current is its effective speed.

Effective speed of a boat along with = Actual speed of the boat + speed

the current (downstream). of the river current.

Effective speed of a boat against = Actual speed of the boat - speed

current (upstream). of the river current.

A boat travels in a river with its effective speed.

Example 9. In still water the speed of a boat is 7 km per hour. Such a boat takes 3

hours to move along with the stream a distance of 33 km. what time will it require to

return ?

In 3 hours the boat moves along with the current a distance of 33 km.

∴ In 1 hour the boat moves alongwith the current a distance of or, 11 km

That is, actual speed of the boat and speed of the river current is 11 km per hour. or, 7

km per hour + speed of the river current = 11 km per hour.

∴ the speed of the river current = ( 11 _ 7) km per hour or 4 km per hour.

When the boat returns, the effective speed

= Actual speed of the boat - speed of the river current.

= 7 km per hour - 4 km per hour

= 3 km per hour.

120

460

115

115

333

115

130

160

120

130

160

160

3+2_160

Unitary, Method 41

The boat returns a distance of 3 km in 1 hour.

∴ boat returns a distance of 1 km in hours

∴ boat returns a distance of 33 km in hours or, 11 hours

Ans. 11 hours.

Example 10. A boatman can go against the river current a distance of 32 km in 8

hours. He requires 4 hours to travel that distance. Find the speed of the river current

and the actual speed of the boat.

Solution : In 8 hours the boat moves against the current a distance of 32 km.

∴ In 1 hour the boat moves against the current a distance of km = 4 km.

In 4 hours the boat moves alongwith the current a distance of 32 km.

∴ In 1 hour the boat moves alongwith the current a distance of km= 8 km.

We know, the effective speed of a boat against the current = Actual speed of the boat-

speed of the current.

∴ actual speed of the boat + speed of the current = 8 km/hour

and actual speed of the boat - speed of the current = 4 km/hour

adding, 2 x Actual speed of the boat = (8 + 4) km / hour

∴ actual speed, of the boat = km / hour = 6 km hour

and the speed of the current = (8 _ 6) km / hour = 2 km / hour

Ans. 2 km/ hour, 6 km/hour

Remark : We may solve the problem much easily by forming algebraic equations.

Exercise 3.2

1. 3 Workers, working 7 hours a day, can white-wash 6504 square metres of wall in

12 days. How many workers can white-wash 4336 square metres of wall in 14

days working 6 hours a day?

2. The time required to make an almirah is three times that required to make a bench.

6 carpenters make 36 benches and 5 almirahs in 12 days. How long will 10

carpenters take to make 61 benches and 8 almirahs ?

13

328

324

8 + 42

1 × 333


3. 30 masons can build a house in 25 days. After to days, work could not be

continued for 5 days due to bad weather. How many more masons have to be

employed to finish the work in time?.

4. A contractor contracts to build a 1920 metres long embankment in 120 days and

immediately employs 160 labourers. After 24 days it was found that only 240

metres had been constructed. How many extra labourers should be employed to

finish the work in time?

5. 3 men or 4 boys working 8 hours a day and earn tk. 9750 in 65 days. In how

many days could 7 men and 8 boys, working 10 hours a day, earn Tk. 24375 ?.

6. Ishaque Shaheb uses 4 cows to plough 40 hectares of land in 7 days. How much

hectares of land could he get ploughed by 6 cows in 10 days? .

7. In a garments factory, 150 workers make 6750 shirts in 15 days. In how many days

could 65 workers of that factory make 3315 shirts?

8. Lily alone can do a work in 10 hours. Hamida alone can do it in 12 hours. In how

many hours could lily and Hamida together do the work?

9. Three friends Zahir, Mihir and Shishir can do a piece of work in 6 days. 8 days

and 12 days respectively. The three friends worked at it for 2 days and then Zahir

and Mihir left. In how many days could shishir do the remaining work?

10. 2 men do as much work as 3 women 4 men 9 women can do a piece of work in

30 days. In how many days could 5 men and 15 women finish that work ?

11. Aziz and Kamal together could build a wall in 20 days. After working together

for 12 days, Aziz finish the remaining work in 10 days. In what time could each

of them alone build the wall ?

12. A and B can do a piece of work in 12 days, B and C in 15 days and A and C in 20

days. In what time could each of them do the whole work alone?

13. 1 man, 1 woman and 1 boy together can do a piece of work in 5 days. But 5

women and 5 boys together can do of the work in 1 day. In how many days

could 1 man complete the work?

14. Two buses start from Gabtali bus depot towards Aricha at a speed of 20 km per

hour. After reaching Savar one bus stopped. But the other continue to move. After

half an hour the bus which stopped at savar started moving at a speed of 25 km

per hour. At what distance from savar would both the buses meet ?

Unitary, Method 43

12

72

15. Distances between two places A and B is 35 km. Monowar starts from A and

Salam from B at the same time. If the speeds of Monowar and salam be 6 per

hour and 8 km per hour respectively, at what time after starting will they meet?

16. In a 12 km cycle race Shuvra beats kajal by 5 seconds. If Kajal's speed be 18 km

per hour, then what is the speed of shuvra ?

17. A 50 metres long train travels at a speed of 36 km per hour. In how many seconds

the train will pass a pole at the way side?

18. The speed of a 100 metres long train is 48 km per hour. That train crosses a

bridge in 21 seconds. What is the length of the bridge?

19. A train 150 metres long, takes 30 seconds to cross a bridge 250 metres long.

What time will the train take to cross a platform of 130 metres long?

20. A cistern has two pipes. The first and the second pipe can fill the empty cistern in

6 hours and 12 hours, respectively. If both the pipes be opened at once in what

time will the empty cistern be filled?

21. A water tank on the roof of a house can be filled by a pipe in 25 minutes.lf the

pipe used for supplying water to the house be opened, the full tank is emptied in

50 minutes. If both the pipes be in action when the tank is half full, how long will

it take to fill it ? .

22. A cistern can be filled by two pipes in 20 minutes and 30 minutes, respectively.

Both the pipes were opened at once when the tank was empty. When will the first

pipe be closed so that the whole cistern is filled in 18 minutes ?

23. A cistern has 3 pipes connected to it. It can be filled by two pipes in 10 minutes

and 30 minutes respectively. Water flows out through the third. When the three

pipes be opened at a time the cistern fill in 15 minutes. In what time the third

pipe empties the water-filled cistern ?

24. A boat covers a distance of 40 km in 4 hours along the current. If the speed of

that boat in still water be 8 km per hour, what is the speed of the river current ?

25. In still water a boat moves 6 km per hour. But in the upstream it takes 3 times the

time to cover 6 km. What time will it take to move 50 km down stream.

26. An anchored boat on a quay snapped the rope because of the high tides and was

carried away to a distance of 7.5 km in 2 hours. After wards the boat man rowed

it back to that quay in 3 hours. What was the actual speed of the boat when

rowed ?.


27. A steamer left Sadarghat, Dhaka to travel downsteam a certain distance in 10

hours. It takes 20 hours to return to Sadarghat against the current. If the actual

speed of the steamer be 21 km per hour, what is the distance between Sadarghat

and the place of destination?

28. With the help of oar a boat can travel 1.5 km in 6 minutes along the current and

against the stream it travels 1.25 km in 15 minutes. Find the speed of the boat and

the current ?

29. The ratio of doing work by a man, a woman and a boy is 3 t 2 t 1. 4 men, 3

women and 6 boys, working 10 hours a day, can finish a piece of work in 14

days. In how many days could 4 men, 4 women and 15 boys working 6 hours a

day, finish another piece of work twice as great as that ?

Multiple Choice Question:

1. If the number of labour increased to double to finish a particular work, how much

time of the preceding time is required to do the work?

(a) 4 time (b) time .

(c) 2 time (d) time

2. If the velocity of stream is 2 km per hour and the velocity of a boat is 5 km per

hour in favour of stream. How much is the velocity of the boat in still water?

(a) 7 km (b) 5 km

(c) 3 km (d) 1 km

3. The price of 15 goats is equal to the price of 3 cows. How many cows can be

bought in exchange of 20 goats?

(a) 4 (b) 5

(c) 6 (d) 10

4. The length of a platform is 100 meter. How much distance does a 150'meter long

train have to travel to cross the platform?

(a) 50 meter (b) 100 meter

(c) 150 meter (d) 250 meter.

Unitary, Method 45

14

12

5. Olsesve the following statements :

(i) The effective velocity of a boat in favour of stream

= Exact velocity of boat + velocity of stream.

(ii) The effective velocity of the boat against the stream

= Velocity of stream --- Exact velocity of boat.

(iii) The velocity of boat in still water is the exact velocity of boat. Which one of

the following is correct?

(a) i and ii (b) i and iii

(c) ii and iii (d) i, ii and iii

CREATIVE QUESTIONS

1. The distance of Magura from Jessore is 56 km. Rahman, a friend of Roushan

starts from Jessore 3 hours after Rowshan started. The speed of Rowshan and

Rahman are 15 km and 75 km per hour respectively:

(a) How much distance was between them before Rahman starts?

(b) In how much time the distance between Rahman and Rowshan decreases by l km?

(c) Where and when the two friends will meet?

2. Sohel and Masum can harvest a paddy field in 20 and 30 days respectively. They

left the work after working together 7 days. Raton finished the remaining work

by 10 days.

(a) How much does Sohel can do of the work in one day?

(b) How much do Sohel and Masum can do together of the work in one day?

( c) In how many days Raton can finish the work alone?

3. There is a water tank on the roof of your school. There are three pipes with the

tank. The tank fill up by 1st and 2nd pipes in 20 and 30 minutes respectively. The

tank become empty by the 3rd pipe in 60 minutes.

(a). How much part of the tank vacates in one minute by the 3rd pipe?

(b) How much part of the. tank will be filled by the 1st and 2nd pipes separately in

1 minute?

(c) In how much time the tank will be filled up if the three pipes get open at the

same time?


14

Chapter Four

Measurement and Unit

4.1 Linear measurement or Length.

Generally we say Let us purchase 4 metres of clothes from the market or the

distance of our school is 2 km away from my residence or "The length of the rod is 5

feet" etc. In the above example kilometre, metre, foot etc. are the units of linear

measurement or simply unit of length. The word linear or length derives from the

word line.The Units Yard, Foot, Inch etc., were first introduced in England. The Units

Yard, Foot and Inch are called British system of measurement of length .

Now a days most of the countries of the

world use Metric system. 1 tenth million Part

of the length from North pole to the equator

alongth any latitude is considered to be one

metre.

Throughout the world the metal scale made

of the metal admixture of Platinum Uramium

and is considered as the 'ideal' or 'standard'

scale. It is kept in French Science Meusium.

When needed any country of the world can

get sample from the standard scale.

1 metre = one croreth part of the distance from the north pole to the equator,

along any longitude.

N.B. For measuring length and weights, Bangladesh introduced the International

standard or system of International units in 1982.

The main unit of length is 'metre' The main Unit of weight is 'gram'.

The list of British system and the Metric system for measuring length are given below.

12 inch = 1 foot

3 feet = 1yard

1760 yards = 1 mile

10 millimetre (mm) = 1 centimetre (cm)

100 Centimetre (mm) = 1 metre (m)

1000 metres (m) = 1 kilometre (km)

4.2 Units of Area

An enclosed surface of a plane is called a space.The quantity or the measurement of

the space is called area. The unit of a square is generally taken as the area of a square

having its side as 1 unit. The unit of the area is written as square unit.The area of a

square having its side as 1 inch is equivalent to 1 sq inch. Similarly if the length of

the side of a square is 1 foot its area will be 1 sqft. Again if the length of square be 1

cm then its area will be 1 sq. cm.

In measuring the area of a space one needs to get the number of square units

contained in it

For Example : Suppose the length of the side of the above squares is 1 yrd.

Therefore, the area of this square is 1 sq.yrd.

British system

(for measuring lands)

1 inch = 2.54 em (approx)

39.37 inch (app) = 1 metre

1 mile = 1.61 k.m(app)

1 k.m =0.62 Mile(app)

7.92 inch = 1 Link

25 Links = 1 Rod

4 Rods = 1 Chain

10 Chains = 1 Farlong

8 Farlongs = 1 mile

Relation between, British and

Metric systems.

1 sq. inch1 inch 1 inch

1 inch

1 inch

1 cm 1 sq cm

1 cm

1 cm

1 cm

lsq.ft

l ft

l ft 1 sq.ft lsq. ft

lsq. ft Isq.ft 1 sq.ft

I sq.ft Isq.ft lsq .ft

l ft l ft

l ft

l ft


Each of the sides of the square is divided into three equal parts. Then join the

opposite sides of the squares. The main square is then gets divided into several

smaller squares and the area of each of the square becomes 1 sq. ft. In the figure we

can find three squares in each of the three rows. i.e In the main square contains 3 × 3

= 9 smaller squares.

Therefore, 9sq. ft = 1 sq. yrd.

Similarly, it can be shown that 1 sq. ft = 12 × 12 =144 sq. inches by dividing a 1 unit

square in to 12 equal smaller.

Squares and joining the points of sides actually in the same way as done in the earlier

example.

Since 12 inch = l ft, ∴ 12 × 12 or 144 inch = 1sq. ft

and Since 3 ft = 1 yrd ∴ 3 × 3 or 9 sq. ft = 1 sq.yrd.

Remarks : 4 metre square and 4 square metre do not represent the same thing. The

statement 4 metre square means a square having 4 metres length as its side and

whose area will be equal to 4 × 4 = 16 sq metres. On the other hand 4 square metres

means a square having its sides equals to 2 metres.

4.3 Measurement of land

4.4 Measurement of weights/weighting

Every solid thing has its own weight. Different countries of the world has different

kinds of uints to measure the weight. We have been using the International standard

kilogram as unit for weighting things in Bangladesh since 1982.

British system

144 sq. inch = 1 sq. ft.

9 sq.ft = 1 sq. yrd.

4840 sq.yrd = 1 acre.

Relationship between British and local system

4 'Ganda' = 1 sq.yrd.

5 sq. yrd = 1 'satak' .

80 sq. yrds = 720 sq. ft.=1 'katha'

1600 sq. yrds = 1 'Bigha'

640 'Acres' = 1 sq. mile.

Local system

1sq 'hand' = 1Ganda'

20 'Ganda' = 1satak'

16 'satak' = 1katha'

20 'kathas'= 1 'Bigha'

Measurement and Unit 49

Before 1982 the weighting systems in Bangladesh were as follows:

The main unit used for measuring weight is gram. The weight of left water at

4˚celcius is equal to 1 gram. At this temperature the weight of water is highest.

Relations between different units of weight.

1 gram = 0.0022 Pound (approx.)

= 0.086 Tola (approx.)

1 Pound = 453.60 gram (approx.)

1 Tola = 11.63 gram (approx.)

1 kilogram (kg) = 2.20 Pound (approx.)

= 1.07 Seer (approx.)

1 Seer = 0.93 kilogram (approx.)

1 Mound = 37.32 kilogram (appro x.)

1 Handar = 1 mourid, 14 seers 7 chataks (approx.)

= 50.8 kilo gram (approx.)

1 ton (Bitish) = 27 Mound 9 Seer (approx.)

= 1 metricton 16 Kilograms, (approx.)

4.5 Measuring of SolidsVolume : A solid has its length, breadth and height. We should find out the amount of

the space occupied by a solid.

The volume of the solid occupying the space containing 1 cm length, 1 cm breadth &

1 cm height is equal to 1 cubic centimetre.

British System:

16 Ounce = 1 pound

28 pounds = 1 quarter

4 quarter-= 1 Hander

20 Handar = 1 British ton

Local system

5 Tolas = 1 ' chatak'

16 'chataks'= 1'seer'

40 'seers' = 1 'mound'

Metric system

1000 milligram = 1 gram

1000 grams = 1 kilogam (kg)

100 kiloram = 1 quintal


The solid having equal length, breath and height is called a cube.

The volume of a rectangular solid = Length × Breadth × Heigth

The above formula is used in finding the volume of liquid substance.

4.6 Volume of Liquid substance

Make a box of thin paper having 10 cm

length, 10cm breadth and 10cm height.

Then the Volume of the (inner) box will

be (l0 × 10 × 10) cubic metre. 1 litre is

equivalent to1000 cu.cm. The amount of

water 10 cm contained in the box is equal

to 1000 cu.cm or 1 litre.

Remarks: Cubic centimetre is abriviated by c.c.

1 cubic cm(c.c)= 1 milritre 1 cubic inch = 16.39 cubic cm ( approx)

4.7 The relation between Metric system and decimal.We know in decimal number system the value of each digit increases ten times as it

moves one place to the left. In the reverse order the value decreases ten times as each

digits moves to the right. This same relation also holds between the successive units of

lengths, weights & volumes. So it can easily be expressed in any units by using

decimal in measuring length, weight and volumes in Metric system.

Thousands hundreds tens unit tenths hundredths thousandths

1000 100 10 1 .1 .01 .001

Metre

or

Kilo Hecto Deca Gram Deci Centi Milli

or Litre


Breadth

Length

Height

10 cm

10 cm

10 cm

Exercise 4.11. Express in ecntimetres and milimetres :

a) 8 Metre b) 7.4 Metre c) 13.96 Metre

d) 0.007 Metre e) 0.0479 Metre

2. Express in Metres.

a) 493 mm b) 756.5 cm c) 23.8 km

d) 8 km. 9 cm e) 25 km. 3 cm.

3. Express in Kilometres:

a) 40390 cm b) 5359 mm c) 97865 mm

d) 9 metre 7cm 8 mm e) 75 metre 250 mm

4. Express in Metre, Decimetre, Centimetre and Milimetres:

a) 3.5 metre b) .764 metre

c) 5.37 Deeimetre d) .0056 Decimetre

4.8 Problems relating to measurement and its solution.

Some formula to measure the areas of space of different shape are given below:

Area = Length x Breadth

b = 1 × b

Rectangle

Area = Length × Breadth

= 1 × h

Parallelogram

Area = (Base × Altitude)

= (a × h)

Triangle

Example 1. The length and breadth of a rectangle are 9 metres and 3 metres

respectively Find the area of the rectangle. .

1

12

12


h

1

h

a

Solution : The required area = (9 × 3) sq. metre

= 27 sq. metre.

Exampie 2. There are 3 doors and 6 windows of a house. The length and breadth of

each doors are 2 metre and 1.25 metres respectively. The length and breadth of each

windows are 1.25 metres and I metre respectively. Given that the length and breadth

of each of the planks are 5 metres and 0.60 metre respectively. Find the number of

planks required to prepare those doors and windows.

Solution: The area of three doors = (2 × 1.25) × 3 sq.metres

= 7.5 sq. metre.

The area of 6 windows = (1.25 × l) × 6 sq. metre

= 7.5 sq.metre.

The area of a plank = (5 × 0.6) sq.metre

= 3 sq. metre.

Required no.of plank

= The sum of the areas of the doors and the windows ÷ the area of the plank.

= (7.5 + 7.5) ÷ 3

= 15 ÷ 3 = 5.

Ans: 5 Nos.

Example 3. A land having length and breadths equal to 70 metres and 60 metres

respectively. There is a pond inside that land /field. If the breadth.of each of banks of

pond is 4 metres then find the area of the banks of the pond.

Solution :


60 m

70 m

4 m

4 m

4 m

4 m

The area of the field including the pond = 70 × 60 sq. metre. = 4200 sq. metre.

The area of the field excluding the pond = {70 _ (4 × 2)} {6o _ (4 × 2)} sq. metre.

= (62 × 52) sq. metre = 3224 sq. metre.

The area of the bank of the pond = (4200 _ 3224) sq. metre.

= 976 sq. metre.

Ans. 976 sq. metre.

Exercise 4.2

Find the area of each of the triangle whose bases heights are given below:

1. Base 10 metre and height 4m.

2. Base 25 cm and height 12 cm.

3. Base 23 m and height 10m.

4. Base 110m and height 25 m.

5. The length and breadth of a rectangular garden are 60 metre and 40 metre

respectively. There is a road of breadth 2 metre inside the garden. Find the area of

the road.

6. The. amount of length of a house is three times the breadth. The cost to cover the

floor with a carpet is Tk. 1102.50 at the rate of Tk. 7.50 per sq.metre. Find the

length and breadth of the floor

7. The length of one of edges of a cube is 5m. Find the volume of the cube.

8. The volume of a mug is 1.5 litre. How many mugs of water will be there in 450

litre of water?

9. The length and heigth of a parallelogram are respectively 75 cm and 40 cm. Find

the area of the parallelogram.

10. The base and height of a parallelogram are respectively 40 metre and 30 metre.

Find its area.

11. The length of a square is 16 metre. Find its area.

12. The length, breadth and height of a golden bar are respectively 20 cm, 1 5 cm.

and 4 cm Gold is 19.3 times heavier than water. Find the weight of the golden bar

in kilogram.


13. The length, breadth and height of a rectangular cistern are respectively 5.5 metre,

4 metre and 2 metre Find the volume in litre and weight in kilogram when the

cistern is filled with water. [weight of 1 litre water is 1 kilogram].

14. A family requires 1.25 litre milk daily. Price of 1 litre milk is Tk. 12.00. Find the

cost of milk of that family for 30 days.


I. 1 Mound = How much kilogram?

(a) 37.32 (b) 32.37

(c) 39.37 (d) 37.39

2. (i) Volume= length × Width × Height.

(ii) Area of rectangle = Length × width.

(iii) Area of parallelogram = Length × Width.

Which one of the following is correct?



Answer the questions 3-4 on the basis of thc following informations.

The Length, width and height of a box are 10 cm, 10 cm and 10 cm respectively.

3. How much cubic cm.of volume does the box has?

(a) 10 (b) 100

( c) 1000 (d) 10000

4. How much would the weight of the water if the box is filled by water?

(a) 1 kg (b) 10 kg

(c) 100 kg (d) 1000 kg

5. Obsesve the following statments:

(i) The basic unit of length measurement is meter.

(ii) The basic unit of weight measurement is gram. (gm)

(iii) The basic unit of volume measurement of liquid is kg..



(a) i (b) ii

(c) i and ii (d) ii and iii

6. 0.007 meter = How much cm?

(a) 0.7 (b) 7

(c) 0.07 (d) 70

7.

How much is the area of the region ABCD drawn in the figure if each of the

smallest square is of 1 square inch?

(a) 3 (b) 4

(c) 12 (d) 16

CREATIVE QUESTIONS

1. There is a rectangular garden of length 80 meter and of width 70 meter. There is

a road of width 5 meter along the four sides inside the garden.

(a) Find the length of the garden excluding the road.

(b) Find the area of the garden excluding the road.

(c) How much would be the cost for grafting grass on the road at the rate of Tk.

64 per square meter?

A

D

B

C


2. There are 4 doors and 8 windows in a hall room of a school. Each door is of length

2.5 meter and of width 1.5 meter, each window is of length 2 meter and of width

1.5 meter.

(a) Find the area of each door.

(b) Find the total area of the windows.

(c) How many plank of length 5 meter and of width 0.60 meter would require to

make the doors and windows of that hall room?

3. The amount of length is 3 times of the width of a room. It cost total Tk. 1102.50

to cover the floor of the room with a carpet.

(a) Express the length in terms of x if the width of the room is x meter.

(b) Find the area of the room if the cost is Tk. 7.5 per square meter.

(c) Find the length and width of the room.

4. The length, width and height of a rectangular tank are 5.5. meter, 400 cm and 200

cm respectively.

(a) How much cm is the length of the tank?

(b) Find the volume of the tank.

(c) How much is the weight of the water of the tank?

(The weight of 1 liter .water is 1 kilogram)


Junior Secondary Mathematics

Algebra

Chapter One

Addition, Subtraction, Multiplicationand division of algebraic expression

4.1. Revision.

Algebraic Expression :5a ÷ b + 3c _ 2d × f is an algebraic expressions in brief an expression; Any

meaningful arrangement of literal and arithmetical numbers alongwith arithmetical

operations is called an algebraic expression or an expression.

Term Of An Expression:

The parts of an expression connected by addition (+) or subtraction (_) signs are

called terms of the expression. For example, there are 4 terms in

5a ÷ b + 3c _ 2d × 9e + f. They are 5a ÷ b, 3c, 2d × 9e and f.

An expression containing only one term is called a Mononomial. For example, 3a ÷

5a2 × 3c is a mononomial. Expression containing two terms is a Binomial and

containing three terms is a Trinomial. Generally a trinomial is called a multinominal.

An expression containing 3 or more terms is called a Multinomial or Polynomial.

For example, x3 + 3x2y + x2 + y3 is a multinomial.

Coefficient:

If any number or expression is multiplied by 1, no change takes place. For that reason

the numerical coefficient to the left of every number and expression is 1.

Index Or Exponent :

We know a × a × a × a × a is written in short as a5. Here a5 is the power of a. 5 is called

the Index or Exponent of the power of a. The second and third power of a number are

respectively called the square and cube of that number.

Additive Invene :

Two numbers a, b are called Additive Inverse to each other, if a + b = 0. The additive

inverse of a number a is written as _ a.

For example, the additive inverse of 5 is _5 and the additive inverse of _5 is _ (_5) = 5.

Like and Unlike terms:

Those terms which are contained in one or more algebraic expressions and differ only in

their numerical coefficients are called like terms. However numerical coefficients in

like terms may be same. For example:

Like Terms: (i) 3abx2, 5x2ab (ii) 5x2yz, 7yx2z and 5zx2y (iii) 9a3b3c3, 8b3c3a3 and

8b3a3c3

Unlike Terms: (i) 2x2y and 3xy2, (ii) 2x2y2 and 5x2y2z

1.2 Commutative Law:

a + b = b + a is called commutative law of Addition and a × b = b × a is called

commutative law of multiplication.

1.3 Associative Law :

(a + b) + c = a + (b + c) is called associative law of addition and

(a × b) × c = a × (b × c) is called associative law of multiplication.

1.4 Addition of Algebraic Expressions :

Example 1. Add 3a+5b, 2a

Solution: (3a + 5b) +2a

= 3a + (5b + 2a) [Associative Law]

= 3a + (2a + 5b) [Commutative Law]

= (3a + 2a) + 5b [Associative Law]

= 5a + 5b

Note : Like Terms have been taken adjacent by associative law and commutative 1aw.

Example 2. Add 3a + 2b+5c, 4a _ 3b _ 2c

Solution: (3a + 2b + 5c) + (4a _ 3b _ 2c) .

= (3a + 4a) + (2b _ 3b) + (5c _ 2c) [Like terms have been taken by

associative law and commutative law.]

= 7a _ b + 3c

Example 3. add:

3a2b _ 7ab2 + 8b3, -10b3 + 9ab2 + 10a2b

Solution :

The required sum = (3a2b _ 7ab2 + 8b3) + (_ l0b3+9ab2 + l0a2b)

=3a2b _ 7ab2 + 8b3_ 10b3 + 9ab2 + 10a2b

Addition, Subtraction, Multiplication and division of algebraic expression 61

= (3a2b + 10a2b) + (_ 7ab2 + 9ab2) + (8b3 - 10b3)

= 13a2b + 2ab2 _ 2b3 [By applying the associative and commutative law]

Note: 13= 3 + 10 2 = _7 + 9 _ 2 = 8 _ 10.

So, it is seen that sum of the numerical coefficients of like terms have been found. The

related symbols have been placed besides the sum so obtained. In this way the sum of

all the terms so obtained is the required sum.

Alternative Method

3a2b _ 7ab2 + 8b3

10a2b + 9ab2 _ 10b3

13a2b + 2ab2 _ 2b3

∴ the required sum 13a2 + 2ab2 _ 2b3.

Rules : (i) We have to first arrange the terms so that the like terms are placed, one

below the other.

(ii) Now the numerical coefficients of the like terms should be added according

to the rule of addition of numbers with signs. Then the symbols of the

related like terms should be placed beside the sum so obtained.

(iii) Sum of all the like terms and unlike terms is the required sum. .

Example 4. Add:

x3 + 3x2y _ 5x + 4, 2x3 _ 6x2 + 7x _ 8.

x3 + 7x2 _ 2xy2 + 9 and 5x2 + 2.

Solution : Arranging the like terms one below the other, we obtain

x3 + 3x2y _ 5x + 4

2x3 + 7x _ 8 _ 6x2

x3 + 9 + 7x2 _ 2xy2

+ 2 + 5x2

4x3 + 3x2y + 2x + 7 + 6x2 _ 2xy2

∴ the required sum = 4x3 + 3x2y _ 2xy2 + 6x2 + 2x + 7


Exercise 1.1

Add: (Question. 1 _ 12)

1. _ k + 3m + 5n and 5k _ 4m _ 6n.

2. _ a + 2b + 5c and 7a _ 3b _ 8c

3. _ 3x + 5y _ 9z, 5x _ 3y + 7z and _ 2y + z

4. 3a _ 2b + 7c _ 8d, 2c + 6d _ 5a, 3b + d _ 10c, and _ 7d + 5b

5 8x3 _ 7x2 + 13x + 8, _ 6x3 _ 7x + 9 and 8x2 + 9x _ 7

6. x3 _ 3x2y + 3xy2 _ y3, x3 + 3x2y + 3xy2 + y3 and _ 2x3 + 5

7. 5x3y _ 3x2y2 + 9xy3 _ y4, _ 2x3y _ 9x2y2 + y4 and 5x2y2 + 2xy3

8. a3 _ b3, 3a2b + 3ab2 + b3, _ 2a3 _ 2b3 and _ 5a2b + 5b3

9. 8a2xy2 + 3a2x2y2 _ 8, _ 5a2xy2 _ 6a2x2y2 and 7a2xy2 + y2

10. p3q3r3 _ 3p2q2r2 + 5p4q4r4, _ 7p4q4 + 8 and 7p3r3q3 _ 10

11. x + y _ z, z _ y + x and z _ x

12. a2x _ ax2 _ a3, a3 _ a2x + ax2 and ax2 _ a3

If a = 1, b = 2, x = 3 then find the value of: (Question 13 - 18)

13. (a3 + 3a2b + 3ab2 + b3) + (_ 3a2b + 5ab2 _ b3) + (2a3_ 8ab2)

14. (x2y2 + 3x _ 5) + (6x _y3 + 2) + (y3_ x2y2 + 3)

15 (3x4 + 5x3 + 8x2 - 6x + 5) + (7+6x _ 9x2 _ 5x3 _ 3x4)

16. (a2x2 _ 3a3x _ 25a4) + (_ 5a3x + l1a2x2 _ 12a4) + (8a3x _ 12a2x2 _ 14a4)

17. (a2 + 4ab + 4b2) + (4b2 _ 4ab + a2) + (a2 _ 8b2)

18. (a2b2x2 + 3a3b3x2 _ 6a5xy) + (2a5xy _ 4x2a3b3) + (b3a3x2 + 4a5xy)

1.5 Subtraction of Algebraic ExpressionIn arithmetic we subtract smaller number from greater number as. 5 - 3 = 2. Now, we

can find the sum {( + 5) + (_ 3))} with the help of the number line.

12

18

58

34

18

12

34

12

14

12

13

14

13

12

12


From the above number line we may say, (+5) + (_ 3) = 2.

5_3 = (+5) + (_3).

That is, to subtract 3 from 5 implies adding (_3), the ivnerse of (+3),with 5. In general,

a _ b = a + (_ b).

To subtract one quantity from another is same as adding the inverse of the second

to the first.

Example 5. Subtract 2x2y2 from 5x2y2

Solution : 5x2y2 _ 2x2y2

= 5x2y2 + (_ 2x2y2)

= 3x2y2

Note: The sign of the subtrahend is changed to plus sign.

Example 6: Subtract 4y2 + 4xy _ 2x2 from 2y2 _ 3xy + x2

Solution : By taking the inverse terms of every terms in the subtrahend that is by

changing the sign of the terms of the subtrahend, we obtain _ 4y2 _ 4xy +2x2. We now

have to add the first expression and the subtrahend so changed. By arranging the like

terms of these two expressions one below the other we obtain.

2y2 _ 3xy + x2

_ 4y2 _ 4xy + 2x2

Adding, _ 2y2 _ 7xy + 3x2

∴ the required difference = _ 2y2 _ 7xy + 3x2

Example 7: Subtract 2x3 _ 4x2 + 7x + 5 From x3 _ 3x2 + 6x

Solution: x3 _ 3x2 + 6x

2x3 _ 4x2 + 7x + 5

(_) (+) (_) (_)

_ x3 + x2 _ x _ 5

∴ the required difference = _ x3 + x2 _ x _ 5


-2 -1 0 1

-2

2 3 4 5 6

_3

Rules :

(i) The minuend expression at the top and the subtrahend expression below the

minuend be so placed that the like terms are one below the other.

(ii) We have to add by changing the signs of the terms of the subtrahend. If

necessary, the changed signs may be written below.

(iii) The sum of the numerical coefficient of like terms should procede the related

literal number. In this way the sum of all like terms and unlike terms is the

required difference.

Exercise 1.2Subtract the second expression from the first: (Question 1- 6)

1. 2x + 3y _ 4z, _ x+y _ z

2. 4a2 _ 6a + 5, 7a2 + 3a _ 2

3. a2 _ 5ab _ 8b2, _ 3a2 + 2ab _ 7b2

4. 3ax _ 7by _ 7cz, 7cz _ 6by + 2ax.

5. 10m2 + 7mn _ 12n2, 4n2 + 8m2 _ 3mn.

6. 4x2 _ 9x2y _ 10xy2 + 7y3, 2x3 _ 5x2y + 5xy2 _ 2y3.

Subtract after arranging like terms one under another (7 - 13)

7. 5a4 _ 8a _ 13a2 + 9, 4a _ 3 _ 11a2 + 6a4

8. a3 _ 3a2b + 3ab2 _ b3, a3 + 3a2b + 3ab2 + b3

9. 9x3 + 6x2y + 5xy2 + y3, 3x3 + y3

10. 6x2y _ 8xy2, 10x3 + 3y3 _ 5x2y + 3xy2

11. a3 + b3, 2a3 + 2b3 _ 3a2b + 3ab2

12. 3x5 + 6x4y _ 3x2y3 + 9y5, 3y5 + 3x4y _ 3x2y3 + x5

13. x2 + y2 + z2, x2 + y2 + z2 _ yz _ zx _ xy

14. By how much is 20a2 + 9ab _ 10b2 greater than 7a2 _ 5ab + 5b2?

15. What must be added to x3 _ 6x2 _ 6x + 1 so that the sum is 3x3 _ 6x2 _ 13x + 11 ?

16. What must be added to the sum of (x3 + y3) and (7x3 _ 8y3) so that the sum is

l3x3 + l3y3 ?

17. What must be subtracted from x3 + 3x2y + 5xy2 _ 2y3 so that the difference is 9x3?

18. If A = 5x3 _ 3a2x2 + 4a3

B = 4x3 _ 3a2x2 + 3a3, then prove that A _ B = x3 + a3.


1.6 Multiplication Of Algebraic Expression

In arithmetic only positive numbers are used. But in algebra, besides positive

numbers, negative numbers and literal numbers are used. For this we need to have a

clear understanding of few points while multiplication and division. Meanwhile we

discussed about the commutative and associative laws of multiplication.

Further discussions are given below;

Multiplication of Numbers With Signs:

We know, 1 + 1 + 1 + 1 + 1 = 5 that is, if 1 is taken 5 times the particular number is 5.

Similarly, if 3 is taken 5 times it becomes (3 + 3 + 3 + 3 +3) or 15.

The operation which transforms 1 to 5 when applied on 3 is called 3 is multiplied by 5

That is 3 × 5 = 3+3 + 3 + 3 + 3=15.

In general, a × b = ab

So, (_3) × 5 = (_3) + (_3) + (_3) + (_3) + (_3) = _15

In general, (_ a) × b = _ ab

Again (_1) + (_1) + (_1) = _3

That is, if the inverse of 1 is taken 3 times the particular number becomes (_ 3).

∴ 5 × (_3) = (_5) + (_5) + (_5) = _ 15 [since the inverse of 5 is (_ 5) ]

In general, a × (_ b) = _ (a × b) = _ ab

(_ p) × (_ q) = _{(_ p) × q} [ a × (_ b) = _ (a × b)]

= _ { _ (p × q)} [ (_ a × b) = _ (a × b)]

= p × q [ The positive inverse of a is _ a]

That is (_ p) × (_ q) = p × q = pq

The product of two like signed numbers will be preceded by (+) sign.

The product of two unlike signed numbers will be preceded by (_) sign.

. ... ... ..


Rules Relating To Exponent Or Index ;

We know,

a × a = a2, a × a × a = a3

a2 × a3 = (a × a) × (a × a × a) = a5 = a2+3

In general, am × an = am+n (m and n are each natural number)

Again, (a2)3 = a2 × a2 × a2 = a6 = a2×3

In general, (am)n = amn

Distributive law :

We know, 3(a + b) = (a + b) + (a + b) + (a+ b) = 3a + 3b.

In general, x (a + b + c + d + .......................)

= xa + xb + xc + xd +.....................

This rule is called the Distributive Law Of Multiplication.

Example 8. Multiply 5a3b2 × 4a4b3

Solution : ( 5a3b2 × 4a4b3)

= (5 × 4 ) × (a3 × a4) × (b2 × b3), [by applying the commutative and

associative laws of multiplication]

= 20a7b5 [ by rules of indices ]

∴ the required product = 20a7b5

Example 9. Multiply _ 3a2b2c by 4a2b2

Solution: (_3a2b2c) × 4a2b2

= (_3) × 4a2 × a2 × b2 × c

= _12 × a4 × b4 × c

= _12a4b4c

∴ the required product = _ 12a4b4c

Example 10: Multiply _3x2y2z2 by _ 5a2x3y3

Solution: (_3x2y2z2) × (_5a2x3y3)

= (_3) × (_5) × a2 × x2 × x3 × y2 × y3 × z2

= 15a2 x5y5z2

∴ the required product = 15a2x5y5z2


Example 11: Multiply -2m3 + 3m2n - 5mn2 by 4mn

Solution: (_2m3 + 3m2n _5mn2) × 4mn

= _2m3 × 4mn + 3m2n × 4mn _5mn2 × 4mn. [by distributive law]

= _ 8m4n + 12m3n2 _ 20m2n3

∴ the required product = _ 8m4n + 12m3n2 _ 20m2n3

Polynomial Multiplied By Polynomial: (a + b) × (c + d) = (a + b) x p [ let, p = c + d)

= a × p + b × p. [ by distributive law]

= a × (c + d) + b × (c + d) [ we assumed, p = c + d)

= a × c + a × d + b × c + b × d

That is, the terms of the second expression are separately multiplied by the terms

of the first expression. In this way, the algebraic sum of the products so obtained

gives the required product.

Example 12 : Multiply a2 + 2ab + b2 by a + b.

Solution: a2 + 2ab + b2 (multiplicand)

a + b (multiplier)

a3 + 2a2b + ab2 (multiplying by a)

a2b + 2ab2 +b3 (multiplying by b)

Adding, a3 + 3a2b + 3ab2 + b3

∴ the required product = a3 + 3a2b + 3ab2 + b3

Rules : (i) At first we multiply every term of the multiplicand by the first term of the

multiplier.

(ii) Then the terms of the multiplicand are multiplied by the second term of

the multiplier. This product should be so written that the like terms of

both the product lie one below the other.

(iii) The algebraic sum of the two products is the required product.

. ..


Exercise 1.3 Multiply the first expression by the second expression (Question 1-26)

1. 2a2, 3ab

3. 5a2x2, 2ax5y

5. _ 3bx2, 10b3x2yz

7. _ 4abxy, _ 8a3x4by2

9. 7a3xy5z2, _ 2z5x3a2y4

11. 2a2 _ 3b2 _ c2, abc

13. a3 _ b3+ 3abc, a4b

15. a + b, a _ b

17. 5x _ 8y, 2x _ 7y

19. a2 _ ab + b2, 3a + 4b

21. x4 + y4, x4 _ y4

23. 4p2 _ 4pq + q2, 2p _ q

25. a2 + ab + b2, b2 _ ab + a2

27. If A = x2 + xy + y2 and B = x _ y, then prove that AB = x3 _ y3

28. Show that (x +1) (x _ 1) (x2 + 1) = x4 _ 1

1.7 Division of Algebraic ExpressionDivision of numbers with signs:

Divisor Quotient Dividend

We know, if a × b = c

then c ÷ a = b, c ÷ b = a

In multiplication of numbers with signs a × (_ b) = _ ab

∴ (_ ab) ÷ a = _ b and (_ ab) ÷ (_ b) = a

Again, (_ a) × (_ b) = ab

∴ ab ÷ (_ a) = _ b and ab ÷ (_ b) = _ a.

The quotient of two like signed numbers will be preceded by (+) sign.

The quotient of two unlike signed numbers will be preceded by (_) sign.

2. 6ab, _ 9b2

4. 2ax2, _ 7a3x

6. _ 3p2q3, _ 5p5q4

8. _14xy4, _ 5x4yz

10. 2x _ 3y, 4xy

12. _ 2c2d + 3d3c _ 5cd2, _ 6c2 _ d4

14. x + 5, x + 4

16. a + b _ c, b + c

18. x2 + 2x _ 3, x + 4

20. a2 _ 2ab + b2, a _ b

22. x2 + 2ax + a2, x + a

24. x2-x + 1, 1 + x + x2

26. x + y + z, x + y + Z


Remarks: If both the numbers are preceded by (+) sign or by (_) sign the numbers are

called mutually like signed numbers, Again if one number is preceded by (+) sign' and

the other by (_) sign then they are called mutually unlike signed numbers.

Division Rule of Exponents:

a5 ÷ a3 = = = a × a [cancelling the common factors from

the numerator and the denominator]

That is, a5 ÷ a3 = a × a = a2 = a (5_3)

In general, am ÷ an = am _ n, where m and n are natural numbers and m>n.

Corollary : am ÷ am = am_

m = a0

Also, am ÷ am = = 1

∴ a0 = 1

Example 13. Divide 15a4b7 by 5a2b4

Solution : The required quotient

=

= × ×

= 3a4_2b7

_4

= 3a2b3

Example 14. Divide _12 x13y9z3 by _ 3x4y5z2

Solution : The required quotient =

= × x13_

4 × y9_5 × z3

_2

= 4x9y4z

Division of A Multinomial by A Mononomial

We know, (5 + 15 + 20) ÷ 5 = 40 ÷ 5 = 8

Since 5 ÷ 5 = 1, 15 ÷ 5 = 3, 20 ÷ 5 = 4

∴ 5 ÷ 5 + 15 ÷ 5 + 20 ÷ 5 = 1 + 3 + 4 = 8

so, (5 + 15 + 20) ÷ 5 = (5 ÷ 5) + (15 ÷ 5) + (20 ÷ 5)

In general, (a + b + c) ÷ d = (a ÷ d) + (b ÷ d) + (c ÷ d)

a × a × a × a × aa × a × a

a5

a3

am

am

155

a4

a2b7

b4

15a4b7

5a2b4

_12 x13y9z3

_3x4y5z2

_12_3


That is, = + +

This is called the distributive law of division

Example 15. Divide 6a3x2-10a2x3 + 8ax4 by 6ax,

Solution : The required quotient =

= + + [by the distributive

law of division] = 3a2x _ 5ax2 + 4x3

Division of a multinomial by a mononomial :

By the commutative law of addition 5a2x+ 3ax3_2a3x2 may be written as

3ax3_2a3x2 + 5a2x

Again 5a2x +3ax3 _ 2a3x2 may be written as 5a2x _ 2a3x2 + 3ax3

Note : The exponents or indices of the powers of x in the second rearranged case is 1.2.3

In the first case we say the expression has been arranged in the descending order of

the powers of x. In the second case it is arranged in the ascending order of the powers

of x.

Rules :

(i) The dividend and the divisor are both arranged in ascending order of the powers

of their common algebraic symbol.

Then divison is done by placing them as in arithmetic.

(ii) The quotient so obtained by dividing the first term of the dividend by the first

term of the divisor is the first term of the required quotient.

(iii) The product so obtained by multiplying all the terms of the divisor by the term of

the required quotient is subtracted from the dividend.

(iv) The difference is the new dividend. The difference should be written in ascending

or descending order as before.

(v) The quotient so obtained by dividing the first term of the new dividend by the first

term of the divisor is the second term of the required quotient.

6a3x2_10a2x3 + 8ax4

6ax

a+b+cd

ad

bd

cd

6a3x2

6ax

_10a2x3

6ax8ax4

6ax


Example 16. Divide l0x2 + 29x_21 by 2x + 7

Solution : Both the expressions are arranged in descending order of powers of

∴ the required quotient = 5x_3

Example 17. Divided a16_3 a8 + 1 by (a8 + a4_1)

Solution : Both the expressions are arranged in descending order of powers of a

∴ the required quotient = a8 + a4 _1.

Remarks : The new dividend should be arranged in ascending or descending order of

the powers of a as before.

Exercise 1.4

Divide the first expression by the second: (Question 1-35)

1. 16x4, 4x. 19. 9m3n4 _ 21m2n5 _12m4n3, 3m2n3

2. _27a6, 3a3 20. x3 + 3x4y5 _ 5x3y4z2, 5x3

3. _15x12, _ 5x6 21. a3b2 + 2a2b3, a + 2b

4. _30a4x3, 6a2x 22. x2 _ 9x + 14, x _7

5. _18a3z3y3, _ 6ayz 23. 12a2 _ 8a _ 32, 4a _ 8

6. 26p2q2r2, _13pq 24. 2x2 _7xy + 6y2, x _ 2y

2x + 7 ) 10x2 + 29x _21 (5x _ 3 10x2 + 35x _ 6x _ 21 _ 6x _ 21 0

10x2 ÷ 2x = 5x_ 6x ÷ 2x = _3

a16 ÷ a8= a8

a16 ÷ a8 = a8

_ a8 ÷ a8 = _1

a8 _ a4 _1) a16 _ 3a8 +1 (a8 + a4_1

a16 _ a8 _ a12

a12 _ 2a8 + 1

a12 _ a8_ a4

_ a8 + a4 + 1

_ a8 + a4 + 1

0


7. 5x12y9q4, _ x9y2q4 25. x2 - 8xy + 16y2, x- 4y

8. _ 22x2yz, _ 2xyz 26. a2 + 4axyz + 4x2y2z2, a + 2xyz

9. 50x4y6z, _ 10x4zy5 27. x4 _ 1, x2 + 1

10. _ 35xy3a5b4, _ 7a4b2xy2 28. x4y4 _ 1, x2y2 + 1

11. 3a3b2 _ 2a2b3, a2b2 29. 25x2_ 9y2, 5x _ 3y

12. 6a5b3 _ 9a3b4, 3a2b2 30. x2 _ 8abx +15a2b2, x _ 3ab

13. a3b4 _ 3a7b7, _ a3b3 31. x4 + 8x2 + 15, x2 + 5

14. x2yz + zy2z + xyz2, _ xyz 32. 2a2b2 + 5abd + 3d2, ab + d

15, 15a3x _ 20a2x2 + 10ax3, _ 5ax 33. x6 + x4 + 1, x2 _ x + 1

16. 8p4q2 _16p3q3 _ 20p2q4, _ 4p2q2 34. 4a4 + b4 _ 5a2b2, 4a2 _ b2

17. 3x6 _ 9x2y + 12xy3, _3x 35. 81p4 + q4 _ 22p2q2, 9p2 + 2pq _ q2

18. 21a2b2c _12a4b4c4 _ 9a2b6c2, _ 3ab2

1.8 Brackets and Removal of Brackets of Algebraic Expression :We know, 5 + (7 _ 2) = 5 + 5 =10 and 5 + 7_ 2 = 12 _ 2 =10

∴ 5 + (7_ 2) = 5 + 7 _ 2

In general, a + (b _ c) = a + b _ c

Note: * (+ ) sign precedes the bracket

*After removal of the bracket there were no change in the sign of the terms

inside the bracket.

So If (+) sign precedes a bracket there will be no change in the signs of the terms

inside the bracket when the bracket is removed.

Again, 10 _ (8 _ 2) = 10 _ 6 =4 and 10 _ 8+2 = 2 + 2 = 4

∴ 10 _ (8 _ 2) = 10 _ 8 + 2

In general, a _ (b _ c) = a _ b + c

Note: * (_) sign precedes the bracket.

*After removal of the bracket the signs of the terms inside the bracket changed

to opposite sign.


So. If (_) sign preceds a bracket due to removal of the bracket the signs preceding

the terms inside the bracket change. That is, the terms preceded by (+) signs

change to (_) signs and the terms preceded by (_) signs change to( +) signs.

We have learnt before, a + (b-c) = a + b _ c

That is, a + b _ c = a + (b _ c)

Similarly, a _ b + c = a _ (b _ c)

Note: * In the first case the ( + ) sign preceds the bracket to enclose the terms within

the bracket. Here the sign preceding the terms remained intact.

* In the second case the (_) sign preceds the bracket to enclose the terms. Here

the signs preceding the terms changed.

So Terms of an expression which are enclosed within bracket With (+ )sign

preceding the bracket, the signs of each terms remain intact. Again when the

terms of an expression are enclosed within the bracket with (_) sign preceding

the bracket their signs require to be changed.

Example 18. Simplify: [b_{c _ d _ (e _ f + g)}]

Solution: a + [b _{c _ d _ (e _ f+ g)}]

= a + [b _ {c _ d _ e + f _ g}]

= a + [b _ c + d + e _ f + g]

= a + b _ c + d + e _ f + g

Remark: As in arithmetic, first, second and third brackets are removed successively.

Example 19. Simplify: 2x _ [3x + {4y _ (2x _ y) + 5x _ 7y}]

Solution: 2x _ [3x + {4y _ (2x _ y) + 5x _ 7y}]

= 2x _ [3x + {4y _ 2x + y + 5x _ 7y}]

= 2x _ [3x + (3x _ 2y)]

= 2x _ [3x + 3x _ 2y]

= 2x _ [6x _ 2y]

= 2x _ 6x + 2y

= _ 4x + 2y



1. Which one of the following is the sum of

3a+ 2b _ c and 2c _ 2a _ b?

(a) a + 2b + c (b) a + b + c

(c) 2a + b + c (d) a + b + 2c

2. Which one of the following is the result of subtraction of

3ax _ 6by + 2cz from 4ax _ 5by + 3cz?

(a) ax + by + cz (b) 7ax _ ll by + 5cz

(c) ax _ 11by + cz (d) ax _ by + cz

3. Which one of the following is the quotient if of 18a5b2 divided by 3a3?

(i) 6ab2 (ii) 6a2b

(iii) 6a2b2 (iv) 6a3b2

4. Which one of the following is the product of 2a2b and _5ab2?

(i) _10a2b2 (ii) _10a3b2

(iii) _10a2b3 (iv) _10a3b3

5. Which one of the following is the correct value of a3 +2a2 _1?

When a = _1

(a) _ 2 (b) _ 1

(c) 0 (d) 1

6. Which one of the following is the simplified result of a _ {b _ (c _ d)}?

(i) a _ b + c + d (ii) a _ b + c _ d

(iii) a _ b _ c _ d (iv) a _ b _ c + d

7. Which one of the following is the quotient if 8a6b5c4 is divided by _ 2a2b2c2?

(i) _ 4a4b2c3 (ii) _ 4a4b3c2

(iii) _ 4a3b3c3 (iv) _ 4a4b3c3

8. Which one of the following is the co-efficient of xy in the expression xy?

(a) 1 (b) 0

(b) x (d) y


9. Which one of the following is the additive inverse of x?

(i) x (ii) _x

(iii) (iv) _1

10. Which one of the following pairs is similar?

(i) 5abx2, 3x2ab (ii) 2x2y, 3xy2

(iii) 5ab2, 7ab (iv) 10a3b, 10ab3

CREATIVE QUESTIONS

1. (l5x2+7x_2) and (5x_1) are two algebraic expressions

i) Find the sum of the two expressions.

ii) Write the quotient of (15x2+7x_2) divided by (5x_l)

iii) Find the product of the two expressions.

2. (a_3), (a+3) and (a2+2) are three expressions.

(a) Subtract the 2nd expression from the 1st expression.

(b) Find the sum of 3rd expression with the product of 1st and 2nd expressions.

(c) If a = 3 then find the value of the product of the three expressions.

3. 2x + y, 3x_z and x_4y_3z+2 are three algebraic expressions.

(a) In case of 3rd expression close first two terms by bracket beginning with (+)

sign and close the next terms by bracket beginning with (_) sign.

(b) Subtract the 3 times of 2nd expression from 2 times of the 1st expression.

(c) Simplify 7 + (12x + y) _ {(3x _ 7}_ (x _ 4y) _ 3z + 2( + l0}] .

4. 5a2b _ 15a3b and 20a4b _ 10a3b2 + 15a2b4 are three algebraic expressions.

(a) Multiply the 2nd expression by the 1st expression.

(b) Divide the 3rd expression by lst expression.

(c) a =1 and b =2 the find the summation of the above three expressions.

5. a6_ a3_ 2 and l5 + 2a12 _ 10a6 are two expressions formed using a.

(i) Arrange the 2nd expression in descending order of power of a

(ii) Multiply the 1st expression by (a2)3.

(iii) Find quotient and remainder of 2nd expression divided by 1st expression.

1x


Chapter Two

First four formulae with its application

2.1 Defination of algebraic formulaeIf we divide a number with another number then we get Dividend = Divisor ×

Quotient. For all numbers it is true. So it is a postulate. Now let us denote dividend,

the divisor and the quotient by the letters N, D and Q respectively to get N = DQ.

The above relation is an algebraic formulae.

Any deduction expressed by algebraic symbols is called an algebraic formulae or

Simply a formula.

2.2 First four rules.In this chapter the first four rules are presented

Rule/Formula 1. (a + b)2 = a2 +2ab + b2

Proof. (a + b)2 is equal to (a + b) (a + b)

∴ (a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b)

= a2 + ab + ab + b2

= a2 + 2ab + b2

i.e. The square of the sum of two quantities or two numbers = the square of 1st

number + 2 × 1st number × 2nd number + the square of the Second number.

Corollary 1. a2 + b2 = (a2 + 2ab + b2) _ 2ab

= (a + b)2_2ab.

Rule 2. (a _ b)2 = a2 _ 2ab + b2

Proof: (a _ b)2 = (a _ b)( a _ b)

= a (a _ b) _ b( a _ b)

= a2 _ ab _ ab + b2

= a2 _ 2ab + b2

i.e. The square of the sum of two quantities = the square of the first quantity _ 2 ×

first quantity × second quantity + the square of the second quantity.

Remarks: (a _ b)2 = {a + (_ b)}2

= a2 + 2a(_ b) + (_ b)2

= a2 _ 2ab + b2

i.e, the second rule can be obtained by using the first rule.

Corollary 2: a2 + b2 = (a2 _ 2ab +b2) + 2ab

= (a _ b)2 +2ab

Corollaries from the 1st two rules :Corollary 3: (a + b)2 = a2 + 2ab + b2

= a2 + b2 _ 2ab + 4ab

= (a _ b)2 + 4ab [ 2ab = _ 2ab + 4ab]

Corollary 4: (a _ b)2 = a2 _2ab + b2

= a2 + 2ab + b2 _ 4ab

= (a + b)2 _ 4ab

Corollary 5 : (a + b)2 + (a _ b)2

= a2 + 2ab + b2 + a2 _ 2ab + b2

= 2a2 + 2b2

= 2(a2 + b2)

Corollary 6 : (a + b )2 _ (a _ b )2

= (a2 + 2ab + b2) _ (a2 _ 2ab + b2)

= a2 + 2ab + b2 _ a2 + 2ab _ b2

= 4ab.

Example 1. Find the square of 5x + 3y

Solution: (5x + 3y)2

= (5x)2 + 2.5x.3y + (3y)2

= 25x2 + 30xy + 9y2

Example 2. Find the square of 9a _ b

Solution : (9a _ b)2

= (9a)2 _ 2.9a.b + b2

= 81a2 _18ab + b2

Example 3. Find the square of 999

Solution :

999 = 1000 _ 1

∴ (999)2 = (1000 _ 1)2

= (1000)2 _ 2 × 1000 × 1 + (1)2

= 1000000 _ 2000 + 1

= 998001

. ..


. ..

Example 4. Find the square of a + b + c.

Solution: (a + b + c)2

= {a+(b+c)}2

= a2 + 2a(b + c) + (b + c)2

= a2 + 2ab + 2ac + b2 + 2bc + c2

= a2 + b2 + c2 + 2ab + 2bc + 2ac

Remarks/Conclusion: a + b + c

Following the rule showing in the above picture the value of (a + b+c)2 is equal to the

grand total of twice the products of two numbers as per above picture & the sum of

the squares of each of the numbers. To find the solution of this complex type of

problems we can apply the above rule.

Example 5. Find the square of 3x _ 5y + 2z

Solution:

3x _ 5y + 2z = 3x _ (5y _ 2z)

∴ (3x _ 5y + 2z)2

= {3x _ (5y _ 2z)}2

= (3x)2 _ 2(3x)(5y _ 2z) + (5y _ 2z)2 [ 1st quantity = 3x &

2nd quantity = 5y _ 2z ]

= 9x2 _ 30xy + 12xz + (5y)2 _ 2.5y.2z + (2z)2

= 9x2 _ 30xy + 12xz + 25y2 _ 20yz + 4z2

= 9x2 + 25y2 + 4z2 _ 30xy + 12zx _ 20yz

Example 6. Simplify: (5x + 4y)2 _ 2(5x + 4y)(3y + 5x) + (3y + 5x)2

Solution : Let 5x + 4y = a and 3y + 5x = b

∴ The given expression = a2 _ 2ab + b2

= (a _ b)2

= {(5x + 4y) _ (3y + 5x)}2

= (5x + 4y _ 3y _ 5x)2

= (y)2

= y2

First four formulae with its application 79

. ..

. ..

Example 7. Given a = 15 and b = 6, Fine the value of 9a2 _ 48ab + 64b2

Solution : The given expression = 9a2 _ 48ab + 64b2

= (3a)2 _ 2.3a.8b + (8b)2 = (3a _ 8b)2

= (3×15 _ 8 × 6)2

= (45 _ 48)2

= (_3)2

= _3 × _ 3

= 9

Example 8. Given a + b = 7 and ab = 24, Find the value of a2 + b2

Solution: a2 + b2 = (a + b )2 _ 2ab

= (7)2 _ 2 × 24

= 49 _ 48

= 1

Example 9. If x _ = 9 then Prove that x2 + ( )2 = 83

Sobition : x2 + ( )2 = 83

=(x_ )2 + 2.x.

= (9)2 + 2

= 81 + 2

= 83 (Proved)

Exercise 2.1

Find the squares of the following: (Question 1 - 19)

1. 2a + 3 . 2. 5x + 2y 3. 9x + 5a

4. 3m + 7x 5. 5a _7 6. 3x _ 7y

7. ax + by 8. 3a _ 11xy 9. 5a2 + 9m2

10. 9x2 _ lly2 11. 99 12. 1005

13. 997 14. a _ b + c 15. a + 2b + 3c

16. 2x + y _ z 17. bc + ca + ab 18. x2 + y2 _ z2

19. a2 + 2b _ c2

1x

1x

1x

1x

1x


Simplify : (Question 20 - 24)

20. (a + b)2 _ 2(a + b)(a _ b) + (a _ b)2

21. (2x + 3y)2 + 2(2x + 3y)(2x _ 3y) + (2x _ 3y)2

22. (2a _ 3b + 4c)2 + (2a + 3b _ 4c)2 + 2(2a _ 3b + 4c)(2a + 3b _ 4c)

23. (5a _ 7b)2 + 2(5a _ 7b)(9b _ 4a) + (9b _ 4a)2

24. (2a + 5b + 3c)2 + (5b + 3c _ a)2 _ 2(2a + 5b + 3c)(5b + 3c _ a)

Find the values of the following: (Question 25 - 28)

25. 9x2 + 30x + 25, when x = _ 2

26. 4x2 _ 28xy + 49y2, when x = 7 & y = 2

27. 16a2 _ 40ab + 25b2 when a = 7 & b = 6

28. 81x2 + 18xz+ z2, when x = 7 & z = _ 67

29. If m + = 4, prove that m2 + = 14

30. If a_ = 5, prove that a2 + = 27

31. Find the value of m2 + if m + = p

32. If a + b = 5 and ab = 12, then prove that a2 + b2 = 1.

33. Find the value of ab when a + b = 8 and a _ b = 4

34. If x + = 3, show that x4 + ( )4 = 47

35. Find the value of a2 + b2 + 5ab if a + b = 7 and ab = 10.

Rule 3. (a + b)(a _ b) = a2 _ b2

Proof: (a + b)(a _ b)

= a(a _ b) + b(a _ b)

=a2 _ ab + ab _ b2

= a2 _ b2

i.e. Sum of two quantities × its differencs = the defference of the squares of these

two quantities

Ruls 4. (x + a)(x + b) = x2 + (a + b)x + ab

1m

1m

1x

1x

1a

1

m2

1

m2

1

a2


Proof: (x + a)(x + b)

= (x + a)x + (x + a)b

= x2 + ax + bx + ab

=x2 + (a + b)x + ab

i.e. (x + a)(x + b)

= x2 + (algebraic sum of a & b) x + (the product of a and b).

Corollary 1. (x + a)(x _ b)

= x2 + (a _ b)x + a(_b)

= x2 + (a _ b )x _ ab

Corollary 2. (x _ a)(x + b)

= x2 + (_a + b)x + (_a)(b)

= x2 + (b _ a)x _ ab.

Corollary 3. (x _ a)(x _ b)

= x2 + (_ a _ b)x + (_ a)(_b)

= x2 _ (a + b)x + ab

Example 10. Using an appropriate formula multiply 2x + 3 by 2x _ 3.

Solution: (2x + 3)(2x _ 3)

= (2x)2 _ (3)2

= 4x2 _ 9.

Example 11. Find the value of (2x + 3y + 1) (2x _ 3y + 1)

Solution: (2x + 3y + 1) (2x _ 3y + 1)

= {(2x + 1) + 3y}{(2x + 1) _ 3y}

= (2x + 1)2 _ (3y)2

= (2x)2 + 2.2x.l + 12 _ 9y2

= 4x2 + 4x+1 _ 9y2

= 4x2 _ 9y2 + 4x + 1

Example 12. Multiply x + 3a by x + 7a


Solution : The first term of both the expressions is x & the second terms are 3a and 7a

respectively.

∴ (x + 3a)(x + 7a)

= x2 + (3a + 7a)x + (3a)(7a)

= x2 + 10ax + 21a2

Example 13. Multiply 3a2 _ 7 by 3a2 _ 9.

Solution : (3a2 _ 7)(3a2 _ 9)

= (3a2)2 + (_7 _ 9)3a2 + (_7)(_ 9)

= 9a4 _ 48a2 + 63

Exercise 2.2

Multiply the 1st expression by the second one using the formulae given above: (Question 1-15)

1. 4x + 3, 4x _ 3

3. 4a + 7b, 4a _7b

5. ax2 + b, ax2 _ b

7. x2 _ x+ l, x2 + x +1

9. 1 + 4x4 _ 2x2, 4x4 + 2x2 + 1

11. x + 16, x + 9

13. 2x + 5, 2x + 8

15. x+ , x+

Find the products with the help of rules: (Question 16-18)

16. (x_ a)(x + a) 17. ( x + y) ( x _ y)18. (x + 1)(x _1)(x2 + 1)

2.3. Factors of algebraic expressions.

We know, 15 = 3 x 5

Here we say that 3 and 5 are the two factors of 15.

From the formulae/rule 3. We know that a2 _ b2 = (a + b)(a _ b).

Here (a + b) and (a_b) are the two factors of the algebraic expression of a2 _ b2

23

13

13

13

15

16

15

16

2. 5a + 8b, 5a _ 8b

4. 13 _ 12p, 13 + 12p

6. a _ b _ c, a + b + C

8. 4x2 + 6x + 9, 4x2 _ 6x + 9

10. a4 + 3a2x2 + 9x4, 9x4_ 3a2x2 + a4

12. Px + 3, Px _ 5

14. q2x2 + r, q2x2 + 3r


When an expression is a product of two or more expressions, each of these two or more expressions is termed as a factor.

Factorize the algebraic expression' means that one is to find the factors of that

expression. By applying associative, distributive and reflexible rules in case of

multiplication and also by applying the algebraic formulae we can express any

algebraic quantity as the product of two or more factors.

Application of Associative, Distributive and Reflexible rules in finding the factors of any algebraic expression :

Example 14. Factorize the following expression: ax + ay

Solution : ax + ay

= a.x + ay

= a (a + y) [Distributive law]

Example 15. Factorize: x3 + 5x2 +3x +15

Solution : x3 + 5x2 + 3x + 15

= x2.x + x2.5 + 3x + 3 _ 5

= x2(x + 5) + 3(x + 5) [Distributive law]

= (x + 5)(x2 + 3) [ Distributive law]

Remarks: You are to select two quantities in such a way so that by applying the

distributive law you can find a common factor between the two quantities. Use of the

formula a2 _ b2 = (a + b)(a _ b) in finding the factors :

Example 16. Factorize : 81x2 _ 25

Solution: 81x2 _ 25

= (9x)2 _ (5)2

= (9x + 5) (9x _ 5)

Example 17. Find the factors of the following : ax4 _ 4a

Solutions: ax4 _ 4a

= a.x4 _ 4.a

= a(x4 _ 4) [By distributive law]

= a{(x2)2 _ 22}

= a(x2+ 2)(x2 _ 2)


Remarks : If necessary first find the factors by using distributive law.

Example 18. Factorize 25(3a _ 2b)2 _ 16 (2a + 3b)2

Solution: Let 3a _ 2b = x and 2a + 3b = y

Then the given quantity = 25x2 _ 16y2

= (5x)2 _ (4y)2

= (5x + 4y)(5x _ 4y)

= {5 (3a _ 2b) + 4 (2a + 3b)} {5(3a _ 2b) _ 4(2a + 3b)}

= (15a _ l0b + 8a + 12b)(15a _ l0b _ 8a _ 12b)

(Putting the values of x and y)

=(23a + 2b)(7a _ 22b)

Example 19. Resolve into factors: x4 + x2 + 1

Solutions: x4 + x2 + 1

= (x2)2 + x2 + (1)2

= (x2)2 + (1)2 + 2x2 1_ 2x2 + x2

= (x+1)2 _ x2

= (x2 + 1 + x)(x2 + 1_ x)

= (x2 + x + l)(x2 _ x + 1) [After arrangement]

Rule : (i) Twice of the product of the given two quantities should be added or

subtracted with twice the sum of the squares of the given two quantities. (ii) The

quantity which is added or subtracted should, respective, be added with or

subtracted from the given quantity.

Example 20. Resolve into factors : a2 _ 2bc _ 2ab _ c2

Solution : a2 _ 2bc _ 2ab _ c2

= a2 _ 2ab + b2 _ b2 _ 2bc _ c2 [ adding and subtracting b2 ]

= (a2 _ 2ab + b2) _ (b2 + 2bc + c2)

= (a _ b)2 _ (b + c)2

= {(a _ b) + (b + c)}{(a _ b) _ (b + c)

= (a _ b + b + c)(a _ b _ b _ c)

= (a + c)( a _ 2b _ c)

Note from the above example:

* b2 is added to a2 _ 2ab. By this we get a perfect square.

* Since, we added b2, we subtracted b2.


This resulted the given quantity unchanged.

Example 21. Resolve into factors: x2 + 7x + 10

Solution : x2 + 7x+ 10

=x2+(5+2)x+10

= x2 + 5x + 2x+ 1 0

= x(x + 5) + 2(x + 5)

= (x + 2)(x + 5)

Rule : The co-efficient of x should be divided into two parts such that the product of

these two must be equal to the third term. That is if the two quantities be a and b then

a + b = 7 and ab = 10

Example 22 . Resolve into factors: x2 + 5x _ 6

Solution : Here, if a = 6 and b = _ 1

then a + b = 6 _ 1= 5 and ab = 6(_1) = _ 6,

∴ x2 + 5x _ 6

= x2 +(6 _1) x + 6(_1)

= x2 + 6x _ x _ 6 .

= x(x + 6) _ 1 (x + 6)

= (x + 6)(x _ 1).

Remarks : The algebraic expression should be resolve into factors as mentioned in

the above example without the direct application of the formula.

Exercise 2.3

Resolve into factors(Question 1_ 34)

1. xy + a2x 2. a2bc + ab2c + abc2

3. a2 + bc + ca + ab 4. x2 _ ax _ bx + ab

5. 1 + a + a2 + a3 6. ab(px + qy) + a2qx + b2py

7. 4ab_10bc + 4ad _ 10cd 8. x2 _ 25

9. 4x2 _ y2 10. a2 _ 81b2

11. 49 _ a2b2 12. a2 _ (x + y)2

13. x2 _ (b _ c)2 14. 80x3 _ 45x

15. p2 _ 16. pq3 _ p3q1

9


17. 16x4 _ 81y4 18. x2 _ 3

19. 25x4 _ 36y4 20. 9(2a + 5x)2 _ 16(3a _ x)2

21. (x2 + 1)2 _ (y2 + 1)2 22. 144x7 _ 25x3a4

23. 64 _ 9(x _ y)2 24. 4x4 + 81

25. a2 _ 2ab + b2 _ c2 26. a2 _ b2 + 2bc _ c2

27. 4x2 _ 4xy + y2 _ z2 28. x2 + 5x + 6

29. x2 + llx + 30 30. x2 _ 14x + 45

31. a2 + 3a _ 40 32. m2 + m _ 30

33. a2 _ 6a _ 91 34. m2 _ 15m + 56.


1. If p + = 2 then which one of the following is correct value of

(a) 1 (b) 2

(c) _1 (d) _2

2. If x + = 1 and x2 _ = 0, which one of the fol1owing is correct

value of ( x_ ) ?(i) _1 (ii) 0

(i i i) 1 (iv) 2

3. If a+b = 3 and a2+b2 =1, which one of the following is correct value of ab?

(i) 1 (ii) 3

(iii) 4 (iv) 8

4. i) It is good not to write (1000 _1 )2 for the square of 999.

(ii) 4xy = (x+y)2 _ (x _ y)2

(iii) The product of (x +1) and (x _1) can be written in the form x2 +(_1+l)x+(_1)(1)


(i) i and ii (b) i and iii

(c) ii and iii (d) i. ii and iii


13

1p

1x

1x

1x2

1p ?

Answer the questions 5 _ 7 on the basis of the following informantion's.

Observe the algebraic expressions:

(i) x2_25 (ii) x2 + 4x _ 5

5. Whieh one of the following is the difference of i and ii putting x = 1?

(a) 0 (b) _ 24

(c) 24 (d) 25

6. Which one of the following is the factorization of (ii) ?

(a) (x_1) (x+5) (b) (x+l) (x_5)

(c) (_x+1) (x+5) (d) (_x+l) (5_x)

7. Which one of the following is the common factor of i and ii ?

(a) (x _ 5) (b) (x _ 1)

(c) (x + 1) (d) (x + 5)

CREATIVE QUESTIONS

l. If x + = 2, then

(a) Show that, x2 _ 2x + 1= 0

(b) Prove that, x2 _ = 0

(c) Find the value of x4 +

2. If x + y = 4 and xy = 3, then

(a) Find the square of (x _ y)

(b) Show that, x2 _ y2 =10

(c) Find the value of (x_ y)2 + (x2 _ y2)

3. Observe the following algebraic expressions :

(i) x2 + 5x

(ii) x2 + 7x + 10

(iii) ax4 _ 625a

(a) Factorize the expression (i).

(b) Using the formula, show that the product of (x + 5) and (x+2) is equal to the

expression (ii)

(c) Show that (x_5) is a factor of the expression (iii)


1x

1

x2

1

x4

Chapter Three

L.C.M. and H.C.F. of Algebraic

Expressions.

3.1. Conception about Dividend, Divisor and Multiple. a ÷ b = c

Dividend Divisor Quotient

In the above example you can find the process of Division. We already know that the

quantity which divides another quantity or the dividend is called divisor. If the

quotient C in the above example is an integer then a is divisible by b. Here a is called

a multiple of b.

If a quantity (Dividend) is divisible by another quantity (Divisor) then the

Dividend is a multiple of the Divisor.

3.2. H.C.F or G.C.M.The factors of abd are a,b, and d;

The factors of 5a are 5 and a.

The factors of apq are a, p and q.

∴ The Common factor of these quantities abd, 5a and apq is 'a'.

The quantity which is Common to two or more quantities then the first quantity is

a factor of each of them and that quantity is called common factor.

From Arithmetic we get

12 = 2 × 2 × 3= 22 × 3

24 = 2 × 2 × 2 × 3 = 23 × 3

The H.C.F. of 12 and 24 = 2 × 2 × 3 = 22 × 3

Note: * Each of Prime factors 2 and 3 is a common factor of both 12 and 24

* Here 2 comes twice i.e. the highest power of 2 is 2 and the highest power of 3 is I.

Therefore, the highest Common factor (H.C.F) = 22 × 3

The product of the highest number of prime factors which are common to two or more number or quantities is called the Highest Common Factors (H.C.F) of those quantities.

Rules of finding H.C.F.i) The H.C.F of the co-efficients should be determined by applying the rules of

Arithmetic.

ii) The prime factors should be found of those algebraic quantites.

iii) The product of the highest number of common factors of algebraic quantity and

the H.C.F of the co-efficient is the H.C.F of the given quantities.

Example 1. Find the H.C.F of 6a2bc and 8a3b2c2

Solution : 6a2bc = 6 x a2 × b × c

8a3b2c2 = 8 × a3 × b2 × c2

Here the H.C.F of the Co-efficient 6 and 8 is 2

Here a with highest power common to both the quantities. = a2

b. " " " " " " = b

c " " " " " " = c

∴ The required H.C.F = 2 × a2 × b × c = 2a2bc

Remarks : Here a comes twice means that the highest power of a is 2 i.e the value = a2

Example 2. Find the H.C.F of 3 (a2 _ b2) and (a2 + 2ab + b2)

Solution: Here the 1st quantity = 3(a2 _ b2) = 3 (a + b)(a _ b)

The 2nd quantity = (a2 + 2ab + b2) = (a + b)2

The H.C.F of the co-efficient 3 and 1 is 1

The highest power of the prime Common factor a + b is 1 i.e. the value = a + b

∴ the required, H.C.F = a + b.

Example 3. Find the H.C.F of x2 _ 9, x2 + 7x + 12 and 3x + 9.

1st quantity = x2 _ 9

= (x + 3)(x _ 3)

2nd quantity = x2 + 7x + 12

= x2 + (4 + 3)x + 3 x 4

= x2 + 4x + 3x + 12

= x (x + 4) + 3(x + 4)

= (x + 3)(x + 4)

3rd quantity = 3x + 9

= 3(x + 3)


The H.C.F of the Co-efficients, 1, 1 and 3 is 1

The highest power of the Common factor (x + 3) is 1 i.e. the value = (x + 3)

∴ The required H.C.F = 1 x (x + 3) = x + 3

Example 4. Find the H.C.F of 16x2 _ 9y2, 4x2 + 3xy and 16x2 + 9y2 + 24xy.

Solution :

1st quantity = 16x2 _ 9y2

= (4x)2 _ (3y)2

= (4x + 3y)(4x _ 3y)

2nd quantity = 4x2 + 3xy

= x(4x + 3y)

3rd quantity = 16x2 + 9y + 24xy

= (4x)2 + (3y)2 + 2.4x.3y

= (4x + 3y)2

∴ The required H.C.F = 4x + 3y

3.3. L.C.MWe know, x2y2 ÷ xy2 = x

x2y2 ÷ x2y= y

i.e. x2y2 is divisible by both the quantities xy2 and x2y. therefore x2y2 is , a common

multiple of xy2 and x2y.

Similarly, x3y2 , x3y3, x2y3 etc are also the common multiple of xy2 and x2.

Again, xy2 = x × y2

x2y = x2 × y

i.e.The common prime factors of xy2 and x2y are x and y. Now the value of the

common factor x be at least equal to x2 then it will be,divisible by both x2 and x.

Similiarly if we can have the common factor y twice if the value of the common factor

be equal to y2 then must be divisible by both y and y2 i.e The minimum common

factors between this two quantities must be x2 and y2

∴ The L.C.M of these two quantities = x2y2

The common multiple which has the minimum number of factors, present then

that common multiple is the least common multiple i.e. L.C.M of those quantities.


Ramarks : This definition of L.C.M is also applicable for more than two quantities.

By this time we have seen that the L.C.M of x2y and xy2 is equal to x2y2.

Example 5. Find the L.C.M of a2b, b2c and abc3

Solution : a2b = a2 × b

b2c = b2 × c

abc3 = a × b × c3

In the above three quantities a, b, c are the three factors. The highest value of a is a2,

the highest value of b is b2 and the highest value c is c3.

∴ The required L.C.M = a2b2c3

Example 6. Find the L.C.M of 6a2bcd, 15ab3c2d3 and 3b2c2d3.

Solution: The L.C.M of the Co-efficients 6, 15, 3 is 30.

The highest values of the factors a, b, c, d are a2, b3, c2, d5 respectively.

∴ The required L.C.M = 30a2b3c2d5

Remarks: The L.C.M of the Co-efficients must be shown seperately.

Example 7. Find the L.C.M of 3x2 + 9, x4 _ 9 and x4 + 6x2 + 9

Solution: 1st quantity = 3x2 + 9 = 3(x2 + 3)

2nd quantity = x4 _ 9

= (x2)2 _ (3)2

= (x2 + 3)(x2 _ 3)

3rd quantity = x4 + 6x2 + 9

= (x2)2 + 2x2.3 + (3)2

= (x2 + 3)2 .

The L.C.M of the Co-efficients 3, 1, 1 = 3

The highest values of the factors (x2 + 3), (x2 _ 3) are (x2 + 3)2, x2 respectively.

∴ The required L.C.M = 3(x2 + 3)2 (x2 _ 3)

Example 8. Find the L.C.M of x2 _ 8x + 15, x2 _ 25 and x2 + 2x _15

Solution: 1st quantity = x2_ 8x + 15.

= x2 + (_5 _3)x + (_5)(_3)

= x2 _ 5x_ 3x + 15

= x(x _ 5) _ 3(x _ 5)

= (x _ 3)(x _ 5).


2nd quantity = x2 _ 25

= (x)2 _ (5)2

= (x + 5)(x _ 5)

3rd quantity = x2 + 2x _ 15

= x2 + (5 _ 3)x + (5)( _ 3)

= x2 + 5x _ 3x _ 15

= x (x + 5) _ 3(x + 5)

= (x _ 3)(x + 5)

∴ The required L..C.M = (x _ 5)(x + 5) (x _ 3)

= (x2 _ 25)(x _ 3)

Exercise - 3

Find the H.C.F for the following (Question 1-12)

1. x3y5, xy4 2. 3a3b2, 2a2b3

3. 4a4x7, 6a5x4 4. 3xyz2, 2yz2x2, zx2y2

5. 16a3x4y, 40a2y3x, 28ax3 6. a2 + ab, a2 _ b2

7. a3 _ ab2, a4 + 2a3b + a2b2 8. 4a4x _ 9a2x3, 4a2x2 + 6ax3

9. x2 + 7x + 12, x2 + 9x + 20 10. x2 _ x _ 20, x2 _ 9x + 20

11. x2 + 3x + 2, x2 _ 4 12. xy _ y, x3y _ xy, x2 _ 2x + 1

Find the L.C.M of the following (Question 13-24)

13. 2a2b, 3ab2c 14. 4a2x3c, 6ac4x2

15. 13c3d2, 26acd4, 39c5d 16. 4ab, 9a2c,12b3

17. 2a3b2, 3x3a, 5ax2y2z2 18. 3x2, 4x3 + 8x

19. x2 + 2x, x2 + 3x + 2 20. x2 + l0x + 21, x4 _ 49x2

21. 9x2 _ 25y2, 15ax _ 25ay 22. x2 _ 3x _ 10, x2 _ l0x + 25

23. ax2 + 2a, x4 _ 4, x4 + 4x2 + 4

24. a2 _ 7a + 12, a2 + a _ 20, a2 + 2a _ 15.


Multiple Choice Questions :

1. Obsesve the following statements:

(i) The full expression of L.C.F. is Least Common Factor.

(ii) The full expression of G.C.F is Greatest Common Factor.

(iii) Common multiple is a multiple of each of the expressions.

Which one of the following is corrrect ?



2. Which one of the following is the G.C.F of x3y5 an xy4?

(a) xy (b) xy4

(c) x3y5 (d) x4y9

3. Which one of the following is the L.C.M. of 4a2cx3 and 6ac4x2 ?

(a) 2acx (b) 12a2c4x3

(c) 24a3c5x5 (d) 3ac3x

4. If a quantity is divisible by another quantity, then what is called dividend in

respect of divisor ?

(a) Remainder (b) Multiple

(c) Factor (d) Quotient

Answer the questions 5 _7 on the basis of the following information's. a2 _ ab, a2 _ b2

are two algcbraic expressions.

5. Which one of the following is the G.C.F. of the two expressions?

(i) b (ii) a2

(iii) a + b (iv) a _ b

6. How much is the multiple of (a+b) in the 1st expression?

(i) a (ii) a _ b

(iii) a(a+b) (iv) a(a _b)

7. Which one of the following is the L.C.M. of the two expressions?

(i) a2 (ii) a(a+b)

(iii) a(a _ b) (iv) a(a+b) (a _ b)


CREATIVE QUESTIONS

1. a2b, a3b+3a2b+2ab, a4+2a3+a2 are three algebraic expressions.

(a) How much is the multiple of b in the 1st expressions ?

(b) Factorize the 3rd expression.

(c) Find the G.C.F. of the three expressions.

2. 4(a2 _ b2), 2(a2+2ab+b2), 6(a2b+ab2), 2(a4 _ b4) are four algebraic expressions.

(a) How much is the G.C.F. of the numerical co-effecients of the four expressions ?

(b) Factorize the 1st and 2nd expression.

(c) Find the G.C.F of the four expression.

3. x4 _ 16, x2 _ 3x _ 10 are two algebraic expressions.

(a) Which formula has to use to factorize the 1st expressions?

(b) Factorize the 2nd expression.

(c) Find the L.C.M. of the two expressions.


Chapter Four

Algebraic Fraction

We already have discussed on fractions in Arithmetic. is a fraction of which 5 is

Numerator and 16 is Denominator. In place of the number 5 and 16 if we replace them

by algebraic symbol like a and b then we have the fraction like . Then is called an

algebraic fraction. And again if 16 is replaced by b only then we shall get the fraction in

the form , which is also an algebraic fraction. Like wise if the Denominator or

numerator of any fraction be algebraic quantity, the fraction is an algebraic fraction,

e.g. ; ; etc. are the examples of algebraic fractions.

Remarks : The expression (a ÷ b) can be written in the form of an algebraic fraction, like

4.1 Change of A Fraction or Its Lowest Term :

If we multiply or divide both the numerator and the denominator of a fraction by same

quantity except O then the value of the fraction remains unchanged.

Example 1. Express in its lowest fraction,

Solution . = =

Remarks: Both the numerator and the denominator is divided by their H.C.F

Example 2. Express in its lowest form of fraction.

Solution : Numerator = a2 + 2a = a(a + 2)

Denominator = a2 _ 4 = (a)2 _ (2)2 = (a + 2)(a _ 2)

the given fraction =

= [ Dividing both the Numerator and the

Denominator by the common factor (a + 2)]

ab

ab

5b

ab

x2+2x+2

a2+2a

a2_4

a(a+2)

(a+2)(a _ 2)

4a2b3c3

10ab4c3

4a2b3c3

10ab4c32ab3c3 × 2a

2ab3c3 × 5b

1x+1

x+53

516

2a5b

aa_2

Exercise 4.1

Express the following in the smallest form :

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

11. 12. 13.

4.2 Change of a fraction in the form of Common Denominator

Let and be the two fractions.

Here the L.C.M of the Denominators b and d = bd.

Now, = [Since bd ÷ b = d]

=

and = [Since bd ÷ d = b]

=

Here the Denominator of both these fractions is bd.

Rule : (i) L.C.M of the denominators should be determined.

(ii) The L.C.M should be divided by each of the denominator. Then multiply

both the numerator and the denominator by the respective quotient.

Example 3. Express the fractions and as the lowest fraction of which the

denominator is same or equal.

Solution : The L.C.M of the denominators x + 1 and x_1= (x+1)(x_1)

∴ = [ (x+1)(x_1) ÷ (x+1) = (x_1)]

and = [ (x+1)(x_1) ÷ (x_1) = (x+1)]

Example 4. Express and in their smallest term such that the

denominators are equal.

. ..

. ..

ab

ab

cd

axdbxd

cxbdxb

xx+1

xx+1

yx_1

xx2_1

x(x_1)(x+1)(x_1)

y(x+1)(x_1)(x+1)

1

x2+3x+2

yx_1

adbd

bcbd

cd

38k2p3m4

57k3p3m3

x3y2z2

x2y3z33abc

15a2b2c

3a6ab

46x4y5z9

69x2y3x12

4x2 _ 9a2

4x2 + 6ax

x2+7x=12x2+4 _12

x2+4x+4

x2_4

a2+2a_15

a2+9a+20x2_5x_14x2_4x_21

4_81x3

9x4_2x3

x2_9ax+3a

3x2 _ 12xy

48y2 _ 3x2

Algebraic Fraction 97

Solution : The denominator of the 1st fraction = x2 + 3x + 2

= (x+1)(x+2)

The denominator of the 2nd fraction = x2_1

=(x+1)(x_1)

The L.C.M of the denominators = (x+1)(x_1)(x+2)

Dividing the L.C.M by each of the denominators we get (x_1) & (x+2) respectively.

∴ =

and = =

Exercise 4.2

Reduce the following fractions into their lowest term with common denominator: (Q1-10)

1. 2. 3. 4.

5. 6.

7. 8.

9. 10

4.3. Addition, Subtraction and Simplification of fractions.

By distribution law, we have

+ _

= a ÷ x + (b ÷ x) _ (c ÷ x)

=(a + b _ c) ÷ x

=

Observation: The denominators of each of the above three factors are x i.e the

fractions are of equal denominators. Again the algebraic sum of the three factions is

also a fraction.

1x2+3x+2

x

x2_1

x_1

(x+1)(x+2)(x_1)

x(x+2)

(x+1)(x_1)(x+2)x2+2x

(x+1)(x+2)(x_1)

ax

a + b _ cx

bx

cx

a2b

3c4d

abc

bca

cab

ab+c

bc+a

x

x_y

yx+y

zx(x+y)

x2

a2+2ab

3

a2_a_2

x

x2_a

y

x2+2x_15

1

4x2_25y2

y

2ax+5ay

z

2x+5y

5

a2+a_6

2x

a2(a+x)

3y

b2(a_x)4z

c2(a2_x2)

y2

a_2b

1

a2+3a

b

a2+5a+6

c

a2_a+2

,

,

,

, , ,

,

, ,

,

, , , , ,


Rules of addition and subtraction of algebraic fractions.

i) Express the fraction such that their denominators are equal.

ii) The sum or difference of numerators is the numerator of the required

fraction. The common denominator is the denominator of the required

fraction.

Example 5. Add with

Solution. + = =

Example 6. Add With

Solution : The L.C.M of the denominators of the given fraction = (a+b)(a _ b)

1st fraction = =

2nd fraction = =

∴ = + = +

= =

Example 7. Subtract, from

Solution :

2x

3+2x

a_ba+b

a_ba+b

a+ba_b

a+ba_b

(a+b)2

(a_b)(a+b)

(a+b)2

(a+b)(a_b)

2(a2+b2)

a2_b2

(a+b)2+(a_b)2

(a+b)(a_b)

(a_b)2

(a+b)(a_b)

a_b

a+b

a+ba_b

x+ax_a

4ax

x2_a2

(a_b)2

(a+b)(a_b)

3x

2x

3x

5x

x+ax_a

4ax

x2_a2_

x+ax_a

4ax

(x+a)(x_a)_

=

(x+a)2

(x_a)(x+a)

4ax

(x+a)(x_a)_

=

(x+a)2_ 4ax

(x_a)(x+a)=

(x_a)2

(x_a)(x+a)= [ By formula]

(x+a)

(x_a)= [ Simplest formula]


Simplification of fraction :

Two or more fractions connected by mathematical operational symbols like +, -, etc.

can be converted into one quantity or one fraction. This is called the simplification of

the fractions.

Example 8. Simplify :

Solution : The L. C. M of the denominator = ab(a_b)

The given quantity

Exercise 4.3

Simplify : (Question 1-20)

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

xa

3x3

a

a2

b

a+bb

3a+3

xx+y

yx_y

2a+2

a_b_ca

a+b+ca

a_ba

b2

a

2yb

y

b

x2

ab(x+a)2

a(a_b)

(x+b)2

b(a_b)+

=

=

_

x2

ab(x+a)2

a(a_b)

x2(a_b)

ab(a_b)

=a2b_ab2

ab(a_b)= = 1

ab(a_b)

ab(a_b)

=x2(a_b)+b(x+a)2_a(x+b)2

ab(a_b)

=ax2_bx+bx2+2abx+a2b_ax2_2abx_ab2

ab(a_b)

+b(x+a)2

ab(a_b)a(x+b)2

ab(a_b)

(x+b)2

b(a_b)+ _

_

+

xy

y

x+

+

+

+

_

_

_x+yx

1a+1

1x_2

x_yy+

x2+y2

xyx2_y2

xy_

_

+

1a+3

1a2_9


13. 14.

15. 16.

17. 18.

19. 20.


1. Which one of the following is a fraction expressed in the least form ?

(i) (ii)

(iii) (iv)

2. Which one of the following is the value of ( ) ? (a) _ (b)

(c) (d)

3. How much is the value of ?

(a) 1 (b) 0

(c) (d) a2 _ b2

4. Which one of the following is the L.C.M. of the denominators

of the fractions and ?

(i) 2(a2_b2) (ii) a2 _ b2

(iii) 1 (iv) (a2+b2) (a_b)

b_cbc

aa_b

4x+1

1

x2+7x+12

3

x2_ 4x_5

2a

a2_ 4b2

a

2b2+ab

ba_b

1x_1

a2+b2

a2_b2

5

x2_6x+5

3a

4x2_9y2

a_b

2(a+b)

1

2x+3y

c_aca

a_bab+

_

+

+

+

_

_

+

a2+aa+1

x2y3

x3y2

x2_9ax+3a

3x

1x

6x

14x

1x

4x

2x

a+ba_b

_ +

ab

ba

_

a2+b2

a2_ b2

a2 _ b2

ab

a_b2(a+b)


5. How much would be if it is expressed in the least form ?

(a) (b)

(c) (d)

6. If is expressed in the least form then it would be -

(i) (ii)

(iii) (iv)

7. How much would be the result if is from subtracted from ?

(a) x (b) y

(c) 1 (d) 0

CREATIVE QUESTIONS

1. Two algebraic fractions are and

(a) Factorize the denominator of the 1st fraction

(b) Find the L.C.M. of the denominators of the two fractions.

(c) Express the fraction in to the fraction with least common denominator.

2. and are three algebraic fraction.

(a) Factorize the denominator of 2nd fraction

(b) Find the L.C.M. of the denominators of the fractions.

(c) Subtract the 3rd fraction from the summation of 1st and 2nd fraction.

46x4y3z9

69x7y2z6

4a2b_9b3

4a2b+6ab2

1

x2+3x+2

xx2_1

x2

ab(x+a)2

a2_ab(x+b)2

ab_b2

23xyz3

34x3z2

3y2z3

4x3

2yz2

4x3

2a+3b2ab

2a+3b2a

2a_3b2ab

2a_3b2a

2yz3

3x3

y

x_y

y

x_y

,


3. and are three algebraic fractions.

(a) How much is the index of a in the denominator of 1st fraction ?

(b) Factorize the denominators of 2nd and 3rd factions.

(c) Express the three fractions with the least common denominator.

1

a(a+3)b

a2+5a+6c

a2_a_12,


Chapter Five

Simple Equations and Applications

5.1 Revision On Previous Lessons.

Equation: The relation formed between two expression of which at least one has to

be an algebraic expression when connected by the equal sign (=) is called an equation.

The expression to the left of the equal sign (=) in an equations called the left hand side

and that to the right is called the right hand side. for example, x + 5 = 1 4, x + 5x +3=

0, 2x + 3y = 5 are equiations.

Unknown Quantity: In an equation the algebraic symbol or literal number whose one or

more certain values equate both the sides of the equation is called the unknown qantity.

Root of an Equation: The value of the unknown quantity which makes the left hand

side of an equation equal to the right hand side is said to satisfy the equation and is

known as the root of the equation.

Simple Equation: Equation in a' single unknown quantity of first degree (linear) is

called an simple equation For exqmple, x _ 13 = 0, 3x + 5 = x + 7 etc. are equations.

Solution of an Equation : The process which enables to find the root of an equation

is called the solution of the equation.

Remarks: The word equation is used extensivly. If an equation is satistied for particular

values of the unknown quantity, it is called an equation. If, however, the equation is

satisfied for any values of the unkown quantity, it is called an identity. For example

(x + 2)(x _ 2) = x2 _ 4. This is an identity, for it will be satisfied for any values of x.

5.2 The Following Axioms Are Used ln Solving Equations

(i) If same quantities are added to equal quantities their sums will also be equal to

each other.

(ii) If same quantities are subtracted from equal quantities, the difference will also

be equal to each other

(iii) If equal quantities are multiplied by the same quantities, the products will also be

equal to each other.

(iv) If equal quantities are divided by the same quantities excepting zero, the quotients

wi1l also be equal to each other.

5.3. Laws Of Equation:

Law of Transposition :

The given Equation Derived Equation Axiom

3x +5_5 = 2x_5 We subtract 5 from both the

3x+5 = 2x or, 3x = 2x _5 sides of the given equation.

3x + 9 +2x = 5_2x+2x We add 2x to both the sides

3x+9=5_2x or, 3x + 9 + 2x =5 the given equation.

Note: * In the first case 5 is transposed from the left hand side to the right hand side

by changing sign .

* In the second case -2x is transposed from the right hand side to the left hand

side by changing sign.

Any term of an equation may be transposed (moved) from one side to the other

provided that the sign of the term is changed, This is called the Law of Transposition

Law Of Cancellation Of Addition

The Given Equation Derived Equation Axiom.

3x+5_5= x +5_5 We subtract 5 from

3x+5 = x+5 or, 3x = x both the sides of the equation

2x_3+ 3 = 9x _3 + 3 We add 3 to both the sides

2x_3 = 9x_3 or, 2x = 9x.. of the equation

We may cancell same terms from both the sides of an equation, provided they are

preceded by the same sign. This is called the Law Of Cancellation of Addition

Law of Cancellation of Multiplication :

The Given Equation Derived Equation Axiom

3(x+ 3) = 3 (2x+5) x+3 = 2x+5 Both the sides have been

divided by the common factor 3

Simple Equations and Applications 105

Note : The common factor 3 is cancelled from both the sides

Common factor may be cancelled from both the sides. of an equation. This is called

the Law Of Cancellation Of Multiplication:

Law of Cross Multiplication :

Let, =

∴ x bd = x bd [ Both the sides are multiplied by bd ]

or, ad = bc.

That is Numerator of the left hand side × Denominator of the right hand side =

denominator of the left hand side × numerator of the right hand side. This

is called the Law of Cross Multiplitcation

The Following Rules Are Followed In Solving Equation :(i) The unknown quantity is usually retained in the left hand side of an equation,

(ii) All the terms of the left hand side of an equation can be transposed to the

right side and simultaneously all the terms of the right hand side can be

transposed to the left side without changing their signs.

For example,

∴ 5x _ 9 = 2x+3

2x + 3 = 5x_9 [By law of transposition]

or, _5x + 9 = _2x_3

or, _(5x_ 9) = _ (2x + 3)

or, (_1)(5x_9) = (_1)(2x + 3)

∴ 5x _ 9 = 2x+3 [_1 is cancelled from both the sides by

the law of cancellation of multiplication]

This is called the law of transposition of equation.

5.4 Solution Of Simple EquationExample 1. Solve: 5x _ 13 = 12

Solution : The given equation, 5x _ 13 = 12

or, 5x = 12 + 13 [Law of transposition]

or, 5x = 25

a

ba

b

c

d

c

d


∴ x= 5, [Dividing both the sides by 5]

∴ the root of the equation = 5

Checkup : L.H.S = 5x_13

= 5×5_13 = 25_13=12

and R.H.S = 12

∴ L.H.S = R.H.S

∴ the solution of the equations is correct.

Example 2. Solve : 5(1_x) +3 (2_x)= _29

Solution : The given equations 5(1_x) + 3(2_x)= _29

or, 5_5x+6 _3x= _29

or, _5x_3x = _29 _ 5 _ 6

or, _ 8x= _ 40

or, x =

∴ x = 5

∴ the required root = 5

Check Up : L.H.S = 5(1_x)+3(2_x)

=5(_5) + 3(2_5)

=5(_ 4) + 3(_3)

= _ 20 _ 9 = _29

and R.H.S = _29

∴ L.H.S = R.H.S

So, the solution of the equation is corrct.

Example 3. Sove: _ = _ + 7

Solution : The given equation is : _ = _ + 7

or,

or,

or,

or,

_ 40_ 8

x6

x5

x15

x3

x6

x5

x15

x3

x6

x5

x15

x3

_ _ + = 7

= 7

5x _ 6x _ 2x + 10x 30

15x _ 8x30

=7x30

71


= 7

or, 7x=7 × 30 [ by cross multiplication]

∴ x=30 [ by cancelling 7 from both the sides]


Remark : check should be done separately.

Alternative Method:The denominators of the terms of the equation are 6, 5, 15 and. 3; Their L.C.M. is 30.

We multiply both the sides by 30 to obtain,

5x _ 6x = 2x _ 10x + 210

or, 5x _ 6x _ 2x + 10x = 210

or, 15x _ 8x = 210

or, 7x = 210

∴ x = = 30

the required root = 30

Example 4. Solve:

Solution : The given equation is

Here the denominators of the terms are 3,4 and 12;. their L.C.M. is 12. Multiplying

both sides of the equation by 12. we obtain.

4.2x + 3.(5x_2) = 11 x + 6

or, 8x + 15x_6 = 1lx + 6

or. 8x + I5x _ 11x = 6 + 6

or, 23x _ 11 x = 12

or, 12x =12

∴ x = = 1


Example 5. Solve:

Solution : The given equation,

2107

2x3

5x_24

11x+612

+ =

2x3

1212

5x_24

x_25

x+29

2x+13

11x+612

+ =

_

_

= 5 _

x_25

x+29

2x+13

= 5 _


Here the denominators are 5, 9 and 3; their L.C.M is 45.

Multiplying both sides by 45, we obtain,

9(x _ 2) _5(x +2) = 5 x 45 _ 15(2x +1)

or, 9x _ 18 _ 5x _ 10 = 225 _ 30x _ 15

or, 4x _ 28 = 210 _ 30x.

or, 4x+30x=210+28

or, 34x = 238

∴ x = = 7


Exercise 5.1

Solve: (Question 1-30)

1. 4x + 3 = 2x + 5 2. 3x + 2 = x + 6

3. 5x _ 6 = 2x + 3 4. 15x _ 9 = 1lx _ 25

5. 19 _ 3x = 5x+ 35 6. 13x _ 5 = 3 _ 2x

7. 2x _ 9 = 13x _ 7 8. 15x _ 9 = 11 x _ 35

9. = _ 4 10. +5 = + 7

11. 12. _ = +

13. 14 + = 3 _

15. 4(2_x)+2(3 _2x) = 30 16. 7(3_2x) + 5(x_l) = 34

17. 3(x_2) + 5(2x_3) = 5(1_2x) _59 18. 8(2x_7) _ 9(3x_14) _15 = 0

19. 5(x+ ) _ 6(2x + ) = 7 20. (x+l) + (x+ 2)+ (x + 3) =16

x7

x2

x3

x3

12

x4

12

x4

12

12

12

13

13

14

35

34

x5

23834

x2

x3

x5

16

_ = _

x5

27

5x7

45

_ = _


21. 22.

23. 24.

25. 26.

27. 28.

29. (x+1)(x+2) = (x+4)(x_2) 30. (2x_1)(x+3) = 2x(x+1)

5.5 Application of Equations.

Many arithmetical problems can be solved easily with the help of equations. The

required quantity is represented by literal number or algebraic symbol. Then equation

is formed with the help of the condition given in the problem and the solution of the

equation is determined. The value obtained in the solution of the equation is the

required value of the quantity .

Example 6. Raihan obtained 185 marks all together in General Mathematics and

Higher Mathematics. He obtained 5 marks less in Higher mathematics than General

Mathematics. What is his marks in each subject.

Solution : Let in General Mathematics Raihan obtain x marks

So, he obtains in Higher Mathematics (x_5) marks [since he obtained 5 marks less in

Higher Mathematics]

So, according to the given condition, x + (x _ 5) =185

or, x + x _ 5 = 185

or, 2x = 185 + 5

or, 2x = 190

or, x = = 95

∴ x = 95

∴ Raihan obtains in General Mathematics 95 marks, and in Higher mathematics

(95_5) marks or 90 marks.

3x+15

2x_73

=

=

=

x+64

x+26

x_ 43

x_25

x_23

x_35

x_315

x2

_ _

= 3x +

_

_

= 3 _

= x _x_24

1902

2x_13

+3x+7

4

2x+139

5x_47

12

13

_x+12

76

_

+3x+ 1_

2

3x+1

5

= 2

2x+35

13x+710

+5x_1

4=


Example 7. The sum and difference of the two numbers are 320 and 20 respectively,

Find the two numbers.

Solution : Let the greater number be x

Since sum of the two numbers is 320, the other number is (320 _ x)

So, according to the given condition.

x _ (320 _ x) = 20 [Since the difference of the two numbers is 20]

or, x _ 320 + x = 20

or, 2x = 20 + 320

or, 2x = 340

or, x =

∴ x = 170

∴ the greater number is 170 and the other number is (320 _ 170) or, 150

∴ the required two numbers are 170 and 150.

Example 8. Zaman started from Dhaka at a speed of 20 kilometres per hour and

reached his home. Again he returned to Dhaka from his home at a speed of 30

kilometres per hour. It took him in all 10 hours. What is the distance of his home from

Dhaka?

Solution: Let the distance from Dhaka to Zaman's home be x km.

The time required to travel to his home from Dhaka is hours.

Again the time required to return to Dhaka from his home is hours.

So, according to the given condition,

= 10

or, 3x + 2x = 600 [Multiplying both sides by 60]

or, 5x = 600

or, x =

∴ x =120

the distance from Dhaka to Zaman's home is 120 km.

x20

x30

x20

x30

6005

+


340

2

Exercise 5.2

Form equation and solve: (Question 1-15)

1. 5 is added to which number, so that the sum is 38 ?

or

Find the number which when added to 5 makes the sum 38.

What is the number to which 5 is added so that the sum is 38 ?

2. What is the number from which 15 is subtracted, so that the difference is 21 ?

3. Sum of the ages of Azmal and Shafique is 35 years. If Azmal is 5 years older than

Shafique; find their ages.

4. The sum and difference of two numbers is 80 and 24 respectively; what are the

two numbers ?

5. The sum of three consecutive numbers is 135. Find the three numbers.

or

Find three consecutive numbers whose sum is 135 ?

6. What is the number whose one third when added to one fourth of it gives the

sum 35 ?

7. What is the number so that when 63 is deducted from 11 times of that number,

the difference is 190 ?

or

When 63 is subtracted from 11 times a certain number, the difference is 190.

Find the number.

8. Jamal, Kamal and Haripada altogether have Tk. 177, Jamal has Tk. 5 less than

Kamal and Haripada has Tk. 7 more than Jamal. Find how much taka each have.

9. Of the total number of fruits a fruit seller has apples, mangoes, oranges

and 165 lichis. How many fruits he possessed ?

10. The distance between Dhaka and Sreepur is 48 Kilometres. Two people start at

the same time from Dhaka and Sreepur and meet after 6 hours, if the speed of the

first is 3 times that of the second find the speed of each person.

18

16

14


11. A train travelling at 30 km per hour reach its destination If the speed of the train

is 27 km per hour it would take 20 minutes longer to reach the destination. Find

the distance the train travels.

12. The difference of two numbers is 25. If 4 times the greater number equals 5 times

the smaller. What are the two numbers?

13. A boatman rows down a river 45 kilometres in 5 hours with the stream. If the

speed of the boat in still water is 7 km per hour; find the speed at which the

stream flows per hour.

14. A boat travels with the stream at a spead 5 times the speed at which it travels

against the stream, if the speed of the boat in still water be 6 km per hour, what is

the speed at which the stream flows per hour?

15. A pen and a book together cost Tk 85. If the cost of the pen is Tk. 15 more and

the cost of the book Tk 14 less, the cost of the pen will be twice that of the book.

What is the cost of pen ?


1. How many unknown quantity are there in a simple equation?

(a) 1 (b) 2

(c) 3 (d) 4

2. Which one of the following indicates an equation?

(a) xo =1 (b) 4x2 = 4x.x

(c) (x+1 )2 = x2+2x+1 (d) x+5=14

3. Rahim has obtained total 180 marks in Mathematics and Higher Mathematics. He

obtained 10 marks less in Higher Mathematics than Mathematics. If he obtained

x in Mathematics, then according to the above information which one of the

following relation is correct?

(a) x_ ( x_ l0) =180 (b) x_ (x _ l0) =180

(c) (x_ l0) _ x =180 (d) 180 _ x =10 _ x


4. If the summation of two consecutive integers is 15, then which one of the

following is the correct value of the two numbers?

(a) (4, 11) (b) (10, 5)

(c) (7, 8) (d) (9, 6)

5. Observe the following informations :

(i) The unknown quantity is always kept in the left side in an equation.

(ii) (x + y)2 = (x _ y)2 + 4xy is an identify.

(iii) Common factors can be omitted from both the sides of and quation

Which one of the following is correct in the basis of above information ?



Answer the questions 6 _7 in the basis of the information given below.

Sapnil has 25 more pencils than that of Shithil has. The 4 times of the number of

pencils of Sapnil is eqqual to 8 times of the number of pencils of Shithil.

6. If the number of pencil of Sapnil is x and that of Shithil is. y, which one of the

following relation is correct?

(a) x + y + 25 = 0 (b) x + 25 = y

(c) x + y = 25 (d) x = y + 25

7. Which of the following is the number of pencils of Shithil ?

(a) 25 (b) 50

(c) 75 (d) 100

CREATIVE QUESTIONS

1. Observe the two following equations:

(i)

(ii)

(a) Find the L.C.M. of denominators of equation No. i.

(b) Solve the equation No. i

(c) Show that the root of equation i is the root of the equation ii.

x2

x3

x5

16

2x15

_ _=

=

+

2x3

5x_25

11x + 612

+


2. The total price of a pen and a book is Tk. 190. If the price of a pen is Tk. 30 more

and pricc of a book is Tk. 25 less then the price of a pen become double of the

price of a book.

(Let price of a pen be Tk. x and price of a book be Tk. y)

(a) Express the total pricc of a pen and a book in an equation.

(b) Express x in terms of y.

(c) Determine the price of a pen and a book separately.

3. The distance of Gaforgaon from Dhaka is 72 km. Keya and Sumi started from

Dhaka and Gaforgaon at the same time and they met after 6 hours. The speed of

Keya is 3 times of the speed of Sumi.

(a) If the speed of Sumi is x km per hour then find the sum of the speed of both

of them.

(b) Find the speed of Sumi.

(c) If the speed of both is increased by 2 km then find that how much more time

would required by Sumi than the time of Keya to reach a fixed destination.


Junior Secondary MathematicsGeometry

Chapter OnePrimary Discussion and Preliminary

Concepts.

1.1. Geometry

'Geo' means land and 'Metry' means the method of measurement, This branch of

science has been developed from the method of measuring land.

1.2. The symbols used in Geometry.

Symbol Meaning Symbol Meaning

+ Addition || is Parallel to

= Equal to angle

> is Greater than is Perpendicular to

< is less than Δ triangle

~= is Congruent to/is equal Circle

in all respect to ∴ Therefore

Since

1.3. Geometrical argumentProposition : The subjects discussed in Geometry is generally called proposition.

Problem : The proposition in which one has to Construct & Prove the

Construction by arguments/reasons is Called Problem.

1) Data: The given facts in the Problem.

. ..

Proposition

Problems

Data Construction Proof Particular

Enunciation

General

Enunciation

Construction

Theorems

2.) Construction: The drawings which are made to solve the problems.

3) Proof: Justification of the construction by argument.

Theorem : The proposition which is established for some geometrical statement by

argument is called a theorem.

The theorem consists of following parts :

a) General Enunciation : This is a preliminary statement describing in general terms

the purpose of the proposition.

b) Particular Enunciation: This repeats in special terms the statement already made,

and refers it to a diagram, which enables the reader to fol low the reasoning more easily.

c) Construcdons: The additional drawing made to prove the truth of a problem.

d) Proof: The proof shows that the object proposed in a problem has been

accomplished, or that the property stated in a Theorem is true.

e) Corollary: This is a statement the truth of which follows readily from an established

proposition as an inference or deduction, which usually requires no further proof.

f) Converse theorem: If we interchange the hypothesis and conclusion of a theorem,

we enunciate a new theorem which is called the Converse of the 1st. The theorems

which were discussed in books of lower levels are described below:

Theorem - 1

The adjacent angles which are straight line makes with another straight line on one

side of it are together equal to two right angles.

Let the straight line OC makes with the

straight line AB at the point O, the

adjacent ∠AOC and ∠BOC.

Therefore, ∠AOC + ∠BOC = 2 right

angles.

Primary Discussion and Preliminary Concepts 119

A O B

C

Theorem - 2

If at a point in a straight line, two other straight lines, on opposite sides of it, make the

adjacent angles together equal to two right angles, then these two straight lines lie on

one and the same straight line.

In the figure OA and OB are on the opposite sides of OC and make the adjacent

∠AOC and ∠BOC together equal to two right angles, Therefore. OB lies on the same

straight line as OA.

Theorem - 3

If two straight lines cut one another, the vertically opposite angles are equal.

Let the straight lines AB and CD cut one another at the point 'O'

Therefore, ∠AOD = vertically opposite ∠BOC

and ∠AOC = vertically opposite ∠BOD.


A O B

C

O

A D

C B

Theorem - 4

If a straight line cuts two parallel lines, it makes

a) the alternate angles equal to one another.

b) the corresponding angles equal to one another.

c) the two interior angles on the same side together equal to two right angles.

In the figure AB || CD and PQ cuts the lines AB and CD at the points E and F respectively.

Hence, a. (i) ∠BEF = alternate ∠CFE

(ii) ∠AEF =. alternate ∠EFD

b. (i) ∠BEF = corresponding ∠DFQ

(ii) ∠PEB = Corresponding ∠EFD

(iii) ∠AEF = Corresponding ∠CFQ

(iv) ∠PEA = Corresponding ∠EFC

c. (i) ∠BEF + ∠EFD = 2 right angles,

(ii) ∠AEF + ∠EFC = 2 right angles.

Theorem - 5

If a straight line cuts two other straight lines so as to make a) the alternate angles

equal or b) the corresponding angles equal or c) the interior angles on the same side

equal to two right angles, then in each case the two straight lines are parallel.


P

B

D

E

A

C F

Q

In the figure PQ cuts the straight lines AB and CD at the points E and F respectively and

a) ∠AEF = The alternate ∠EFD

or. b) ∠PEB = Corresponding ∠EFD

or. c) ∠BEF + ∠EFD = 2 right angles,

Hence AB and CD are Parallel to one another.

Remarks : Theorem 5 is the reverse proposition of theorem 4.

Theorem - 6The straight lines which are parallel to the same straight lines are parallel to one

another.

In the figure AB is parallel to EF, CD is parallel to EF.

Hence, AB and CD are parallel to each other.


P

B

D

E

A

C F

Q

P

B

D

E

A

C

F

Q

Exercise 1.11. Find the magnitude of the directed angles in the adjacent figure.

2. Find the reasons for the following sentences to be true in the adjaceut figure.

(a). a + a' = two right angles.

(b). c' > a

(c). b = x

(d). c' > b

(e). c = y

(f). a + b = c'

(g). a' = b + c

(h). b + b' = two right angles.


p

qr

55

z

y

x

105°

110°u

19w

a

a'

bb'

x

cc'

y


1.

In the figure: if ∠PQR=55˚, ∠LRN = 90˚ and PG || MR then which one of the

following is the value of ∠MRN?

(a) 35˚ (b) 45˚

(c) 55˚ (d) 90˚

2. In the figure: if PQ || SR, ∠QPR = 55˚ and ∠PRQ = 50˚ then which one of the

following is the value of ∠LRS?

(a) 40˚ (b) 50˚

(c) 55˚ (d) 75˚

3. In the isosceles triangle ABC the line EF is parallel to base BC and intersected

AB and AC at the. points E and F.


PN M

LRQ

P S

LR

Q

If ∠B= 52˚ the which one of the following is the value of ∠A+ ∠F?

(a) 76˚ (b) 104˚

(c) 128˚ (d) 156˚

AB || CD || EF

4. Which one of the following is the correct value of ∠x?

(a) 28˚ (b) 32˚

(c) 45˚ (d) 58˚

5. Which one of the following is the value of ∠Z?

(a) 58˚ (b) 103˚

(c) 122˚ (d) 148˚

6. Which one of the following is the correct value of y-2x?

(a) 58˚ (b) 71˚

(c) 103˚ (d) 122˚

7. Obsesve the following informations:

(i) Two adjacent angles situated on the same line can be equal to one-another.

(ii) The bisectors of the vertically opposite angles are situated on the same

straight line.


A

Y

E Z

X

C

B

D

F320

450

Answer the questions 4-6 on the basis of the informations given below:




CREATIVE QUESTIONS:

In the figure, AB || CD, ∠BPE = 60˚ and PQ = PR

(a) Show that ∠APE = 120˚

(b) Find the value of ∠CQF

(c) Prove that, PQR is an equilateral triangle.


A

C

E

P

Q

B

DR

F

Chapter two Theorem Concerning Triangles

2.1 Triangle

The boundary line of a closed area formed by the parts of three straight lines is called

a triangle and the parts of the lines are called side. The common point of the two sides

is vertex. The angle formed at the vertex formed by two sides is an angle of the

triangle. There are three sides and three angles in a triangle. The sum of the length of

the sides is called parameter. The area enclosed by the sides of the triangle is called

triangle area.

ABC is triangle in the adjent figure.

A,B,C are vertices. AB, BC', CA are its

three sides and ∠A,∠B,∠C are its three

angles. AB+BC+CA is the parameter of

the triangle. If three sides of triangle are

equal then it is called equilateral triangle.

If two sides of a triangle are equal then it

is called isosceles triangle .

Congruency :

If two triangles coincide in all

respects when one is applied on the

other, then the triangles are congruent.

If ΔABC and ΔDEF are congruent

and vertices A,B,C fall respectively

on D,E,F then AB = DE, AC = DF,

BC = EF ∠A = ∠D, ∠B = ∠E, ∠C =

∠F sometimes congruency of two

triangles ABC, DEF is denoted by

ΔABC≅ ΔDEF.

Exterior and interior angle of triangle :The angle formed by extending a side of a triangle is called an exterior angle of the

triangle. Except the adjacent angle of this angle, the other two angles of the triangle is

called opposite interior angle of this exterior angle.

A D

B C

A

B C

E F

(a) In the adjacent figure side BC of

ΔABC is extended up to D, ∠ACD is an

exterior angle ∠ABC, ∠BAC and ∠ACB

are three interior angle of ΔABC.

∠ACB is called adjacent angle with

respect to ∠ACD ∠ABC and ∠BAC both

are opposite interior angle with respect to

∠ACD.

(b) In, the figure, AB||CE

∴ ∠BAC = ∠ACE [alternate angle] and

∠ABC = ∠ECD.

Therefore, ∠BAC + ∠ABC = ∠ACE

+∠ECD = ∠ACD

ie, ∠ACD is equal to sum of ∠BAC and

∠ABC

∴ ∠ACD is greater than each ∠BAC and

∠ABC

ie, The exterior angle formed by

extending any side of a triangle is greater

than each opposite interior angle. This

proposition will be use in the next

theorems. .

Theorem - 7

If two triangles have two sides of the one .equal to two sides of the .other, each to

each, the angles included by those sides are also equal, then the triangles are. equal in

all respects.


B

A C

E

D

B

A

C

E

D

B

A

C D

F

Particular Enunciation : Let ΔABC and ΔDEF be two triangles in which AB = DE,

AC = DF and the included ∠BAC= the included ∠EDF. It is required to prove that

ΔABC ≅ ΔDEF.

Proof : Apply ABC to the DEF So that the point A falls on the point D, the side AB

along the side DE and C falls on the same side of DE as F.

Then because AB = DE

∴ the point B must coincide with the point E

And again because AB falls along DE and ∠BAC = ∠EDF

∴ AC must fall along DF, Now since AC = DF

∴ the point C must coincide with the point F.

Then since B coincides with E and C with F

the side BC must coincide with the side EF

Hence the ΔABC coincides with the ΔDEF.

ΔABC ≅ ΔDEF (Proved).

A special proposition : (This proposition

will be used to prove next theorems) If a

side of a triangle is extended, the exterior

angle so formed is greater than each of

the two interior opposite angles. Suppose

the side BC of the ΔABC is extended up

to D. It is required to prove that the

exterior angle ∠ACD is greater than each

of the two interior opposite angles ∠CAB

and ∠ABC.

Construction: We join B and M, mid point of AC and extend BM up to E so that ME

= BM.

Proof: In the triangles ΔMCE and ΔMAB

MC = MA (∴ M is the mid point of AC)

ME = MB ( According to construction)

∠CME = ∠AMB ( opposite angle)

ΔMCE and ΔMAB are congruent.

∴ ∠MCE =∠MAB

i.e. ∠ACE = ∠CAB ..........(i)

Theorem Concerning Triangles 129

A

M

E

BC D

But they are alternative angle of EC ar.d AB with respect to their intersector.

Hence, EC||AR

So, ∠ECD = ∠ABC............. (ii)

But both ∠ACE and ∠ECD are part of ∠ACD

Hence, ∠ACD> ∠ACE and ∠ACE> ∠ECD

∴ Front (i) and (ii) we get

∠ACD> ∠CAB and ∠ACD > ∠ABC (proved)

Example 1. In the figure, AO = OB, CO = OD

Prove that, ΔAOD ≅ ΔBOC

Proof: In the ΔAOD and ΔBOC AO = OB,

CO = OD ( given) and the included ∠AOD =

the included ∠BOC (vertically opposite

angles are equal to each other)

ΔΑOD ≅ ΔBOC (theorem - 7) proved.

Example 2. In the figure AD is a line

segment and B and C lie on the line

AD.

Again, ∠EBC = ∠ECB

BE = CE

AB= CD.

Prove that ΔABE ≅ ΔDCE.

Proof: The line segment BE meets

the line AC at the extreme point B.

∴ ∠EBA + ∠EBC = 2 right angles ....................(l) (Theorem-1)

And the line CE meets the line BD at the extreme point C

∴ ∠ECB + ∠ECD = Two right angles ................(2)

We get from (l) and (2)

∠EBA + ∠EBC = ∠ECB + ∠ECD

or ∠EBA + ∠EBC = ∠EBC + ∠ECD [Since ∠EBC = ∠ECB (given)] Cancelling

∠EBC from both the sides, ∠EBA = ∠ECD.

Now from the ΔΑBE and ΔDCE

BE = CE, AB = CD and the included EBA = included ECD

∴ ΔABE ≅ ΔDCE [theorem. 7] (proved).


D

A

O

C

B

E

A B C D

Exercise .2.1

1. In the figure, CD is the

Perpendicular bisector of AB, prove

that, ΔADC ≅ ΔBDC.

2. In the figure, CD = CB and ∠DCA=

∠BCA, prove that AB = AD

3. In the figure. XY = XZ and XM is the

bisector of ∠YXZ. Prove that, the

line segment XM is the bisector of

∠YLZ

4. In the figure the diagonals AC and BD

of the Parallelogram ABCD bisect

each other at the point O. Prove that,

ΔAOD ≅ ΔCOB and ΔDOC ≅ ΔAOB

5. Prove that any point on the perpendicular bisector of a line segment is equidistance

from the extreme points of the line segment.

6. In the figure, AD and BC are the two

diametres of the circle ABDC, Prove

that AB = CD.

7. In the figure C is the mid Point of the

line segment BD.

x = y, AC = CE,

Prove that AB = ED.


C

B

O

A

D

C

B

O

A

D

Z

M

Y

LX

AD B

C

A

D

B

C

B C D

A

x y

E

8. In the figure ∠BAC = ∠ACD and

AB = DC.

Prove that AD = BC.

Theorem - 8

If two sides of a triangle are equal, then the angles opposite the equal sides are also

equal.

Particular enunciation :

Suppose in the ΔABC, AB = AC. It is

required to prove that, ∠ABC = ∠ACB

Construction : We construct the bisector

AD of ∠BAC, which meets

BC at D.

Proof: In the ΔABD and ΔACD.

AB = AC (by hypothesis), AD is common to both the triangles and the included

∠BAD = the included ∠CAD (by construction).

So, ΔΑΒD ≅ ΔACD (Theorem-7)

∴ ∠ABD = ∠ACD

That is, ∠ABC = ∠ACB ( Proved).

Example 1 : If the equal sides of an isosceles triangle are produced the exterior

angles so formed are equal.

Suppose in the ΔABC, AB = AC. AB and

AC are produced to D and E respectively.

It is required to prove that ∠DBC =

∠ECB

Proof : The line-segment BC meets the

straight line AD at the point B. Therefore,

∠ABC + ∠DBC = 2 right angles .....................(1) [Theorem-1]

Similarly, ∠ACB + ∠ECB = 2 right angles .........................(2)


DC B

A

D C

BA

D

B

A

C

E

So, ∠ABC + ∠DBC = ∠ACB + ∠ECB

or. ∠ABC + ∠DBC = ∠ABC + ∠ECB, [∴ AB = AC ; ∠ABC = ∠ACB]

Cancelling, ∠ABC from both the sides,

∴ ∠DBC = ∠ECB. [Proved]

Example 2 : In the figure, AB = AC and BD = CE in the ABC, Prove that AD = AE.

Proof: In the ΔABC, AB = AC, so,

∠ACB = ∠ABC. (Theorem-8) Now in the

ΔABD and ΔACE, AB = AC, BD=CE and

∠ABD = ∠ACE Therefore, ΔADB ≅

ΔACE [Theorem- 7]

So, AD = AE [Proved]

Exercise 2.2

1. If the base of an isosceles triangles are produced both ways, show that the exterior

angles so formed are equal.

2. In the figure, AD= AE, BD= CE.

Prove that AB = AC.

3. In the figure, ΔABC and ΔDBC

are both isosceles triangles.

Prove that, ΔABD ≅ ΔACD

4. Show that the medians drawn from the extremities of the base of an isosceles

triangle to the opposite sides are equal to one another.


A

B D E C

A

B D E C

A

B

D

C

5. In the figure, AB and CD bisect

each other at O and OA = OC.

Prove that, ∠ODB =∠OBD.

6. In the figure ABC is a triangle in

which ΔABC, AB = AC, BD = DC

and DE = DF Prove that ∠EDB

=∠FDC.

7. In the quadrilateral ABDC, AB = AC, AD is the bisector of ∠BAC. Prove that BD

= CD.

8. Prove that the angles of an equilateral triangle are equal to one another.

Theorem - 9

If two angles of a triangle are equal, then the sides opposite to the equal angles

are equal.

Particular enunciations : Let ABC be

a triangle in which the ∠ACB = the

∠ABC. It is required to prove that AB

= AC.

Proof: If AC and AB are not equal, suppose that AB>AC.

From BA cut off BD equal to CA, join CD.

Now in the ΔABC and ΔDBC, AC = BD, BC is common and the included.

∠ACB = the included ∠DBC.

∴ ΔABC ≅ ΔDBC [Theorem-7]

∴ ∠DCB=∠ABC But ∠ABC=∠ACB (given)

∴ ∠DCB=∠ACB

That is, area of ΔABC = area of ΔDBC. But this is absurd. because a part can not be

equal to the whole.


D

A B

C

O

A

E

B

F

CD

A

B C

D

∴AB is not greater than AC. Similarly AC is not greater than AB.

∴AB = AC (Proved)

Example 1 : In the figure, ABC is a triangle in which DE and DF are two

perpendiculars drawn on AB and AC respectively. DE = DF and BE = CE Prove that

AB = AC

Proof: In ΔBDE and ΔCFD

DE = DF. BE = CF (given)and included ∠BED = included ∠CFD [Since each of them

is equal to one right angle]

∴ ΔBDE =ΔCFD [Problem-7 ]

∴ ∠B = ∠C [Conesponding angles of

congruent triangles]

∴ AC=AB [Problem-9]

Example 2. In the figure ABCD is a quadrilateral in which AD = AB, ∠ADC

∠ABC. It is required to prove that BC = CD .

Proof: In ADB, AD= AB (given)

∠ADB = ∠ABD [Theorem-8]

Again, ADC = ABC (given)

or, ∠ADB + ∠CDB = ∠ABD + ∠CBD.

or, ∠ADB + ∠CDB = ∠ADB + ∠CBD

[Since ∠ADB = ∠ABD]

or, ∠CDB = ∠CBD

(Cancelling ∠ADB from both the sides)

∴ BC = CD [Theorem : 9] (proved)

Exercise 2.3

1. In the ΔABC, BD = CD and AD⊥ BC, Prove that ABC is an isosceles triangle.

2. In the ΔABC, ∠A = ∠C, AC = BC, Prove that ABC is an equilateral triangle.


A

CD

B

A

F

CD

B

E

3. In the ABC, D,E and F are points on BC, AB and AC respectively, BD = DC, ED

= FD and ∠EDB = ∠FDC, Prove that AB = AC.

4. In the given figure, BO and CO

bisect the ∠ABC and ∠ACB

respectively and BO = CO, Prove

that AB = AC

5. In the figure ∠B = ∠C and BD =

EC, Prove that ΔADE is an

isosceles triangle.

6. Prove that if the three angles of a

triangle are equal to one other then

its sides are also equal.

7. If the bisector of the vertical angle of a triangle bisects the base, then prove that it

is an isosceles triangle.

Theorem - 10

If two triangles have the three sides of the one equal to the three sides of the other,

each to each, then they are equal in all respects.

Particular Enunciation:

In the ΔABC and ΔDEF, AB = DE, AC = DF and BC=EF,

It is required to prove that ΔABC ≅ ΔDEF.

Proof: Let us assume that BC and EF are the largest sides of the ΔABC and ΔDEF

respectively.

We now apply the ΔABC to the ΔDEF in such a way that the point B falls on the point

E and the side BC falls along the side EF but the point A falls on the side of EF

opposite the point D. Let the point G be the new position of the point A. Since BC =

EF, the point C falls on the point F.


A

B

O

C

D EB C

A

D

E F

G

B C

A

So, ΔGEF is the new position of the ΔABC.

That is EG = BA. FG = CA and ∠EGF = ∠BAC.

We join D and G Now, in the ΔEGD, EG = ED [since EG = BA = ED].

Therefore, ∠EDG = ∠EGD [Theorem- 8]

Again, in the ΔFGD, FG = FD, [since FG = CA = FD]

Therefore, ∠FDG = ∠FGD [Theorem-8]

So, ∠EDG = ∠FDG =∠EGD + ∠FGD

or, ∠EDF = ∠EGF.

that is, ∠BAC = ∠EDF

So, in the ΔABC and ΔDEF, AB = DE, AC =DF and the included

∠BAC = the included ∠EDF.

Therefore, ΔABC = ΔDEF [Theorem-7] (proved).

Example 1 : Prove that the diagonals of a rectangle bisect it into two congruent

triangles.

Particular Enunciation:Suppose in the rectangle ABCD, BD be the

diagonal. It is required to prove Δ ABD ≅ Δ BCD

Proof: In ΔΑBD and ΔBCD.

AB = DC

AD = BC

[ the opposite sides of a rectangle are equal] and BD is common to both the

triangles.

Therefore, ΔABD ≅ ΔBCD

[Theorem- 10] (proved).

Exercise - 2.4

1. In the ΔABC, AB = AC and O is a point such that OB = OC Prove that ∠AOB = ∠ AOC.

2. OP is the bisector of the chord AB of a circle centre O Prove that ∠OPA = ∠OPB.

or

In a Circle with centre at O, OP bisects the chord AB. Prove that, ∠OPA =∠OPB

3. In the ΔABC, D and E are point on AB and AC respectively such that BD = CE

and BE = CD. Prove that ∠ABC = ∠ACB.

4. Two isosceles triangles ABC and DBC drawn on the same base BC and on the

. ..

A

B C

D


same side of it. Prove that the straight line joining A and D bisects both the

vertical angles.

5. In the quadrilateral ΑBCD, AB = CD and the diagonals BD and AC are equal and

intersect each other at O. Prove that, BOC is an isosceles triangle.

6. In a circle with centre O, the chords BC and ED are equal in length. Prove that

∠BOC = ∠EOD.

Theorem - 11

If one side of a triangle is greater than another, then the angle opposite the greater side

is greater than the angle opposite the less.

Particular Enunciation: Let ABC be a triangle, in which the AC >AB. It is required

to prove that ∠ABC > ∠ACB.

Construction: From AC we cut off AD equal to AB. B,D is joined.

Proof: In the ΔABD, AB = AD.

∴ the ∠ADB = the ∠ABD [Theorem-8]

But ∠ADB is the exterior angle of the

ΔBDC

∴∠ADB > ∠BCD Because if a side of a

triangle is extended then an external

angle is produced which is greater than each of the two opposite internal angles.

∴ the ∠ADB > the ∠ABD

∴ the ∠ABD> the ∠BCD

or the ∠ABD> the ∠ACB

But the ∠ABC > the ∠ABD [Since the ∠ABD is a part of the ∠ABC] Therefore,

∠ABC > ∠ACB.

Remarks: The exterior angle formed by extending a side of a triangle is greater than each of the two interior opposite angles.

Theorem - 12

If one angle of a triangle is greater than another, then the side opposite the greater

angle is greater than the side opposite the less.

Particular Enunciation : Let ABC be a

triangle in which ∠ABC > ∠ACB. It is

required to prove that AC>AB.

Proof: If AC is not greater than AB, it

must be either equal or less than AB.

A

B

D

C

A

B C


i.e. AC=AB or AC<AB.

Now if AC were equal to AB then the ∠ABC would be equal to the ∠ACB. [Theorem -8]

But by hypothesis, this is not true.

Again, if AC were less than AB then

the ∠ABC would be less than the ∠ACB. [Theorem-11]

But, by Hypothesis this is not true. That is AC is neither equal to, nor less the an AB.

:. AC>AB Proved.

Remarks/Conclusion : Theorem 12 is the converse of theorem 11.

Example 1. ABCD is a quadrilateral in which AB = AD, BC = CD and BC>AB. It is

required to prove that the ∠DAB > the ∠DCB


Let ABCD be a quadrilateral in which AB =AD, BC = CD & BC >AB. It is required

to prove that the ∠OAB> the ∠OCB

Proof: In the ΔABC, BC >AB,

Therefore, the ∠BAC> the

∠BCA .............................(1) [Theorem- 11]

Again, from the ΔADC, CD>AD.

Therefore. the ∠DAC> the

∠DCA ...................(2) [Theorem-11]

Adding (1) and (2) we have the ∠BAC+the ∠DAC > the ∠BCA + the ∠DCA,

That is, the ∠DAB> the ∠DCB (proved).

Exercise 2.51. In the ΔABC, AB>AC and the bisectors of the ∠B and ∠C intersects at the point

P, Prove that PB>PC

2. ABC is an isosceles triangle and AB=AC.. BC is extended up to E. Prove that

AD>AB.

3. In the quadrilateral ABCD, AB = AD, BC = CD and CD > AD. Prove that

∠DAB> ∠BCD.

4. In the ΔABC, AB=AC and D is any point on BC. Prove that AB>AD.

5. In the ΔΑBC, AB ⊥ AC and D is any point on AC. Proved that BC>BD.

6. Prove that the hypotenuse is the greatest side of a right angled triangle.

A

B C

D


7. Prove that the angle opposite the greatest side of a triangle is also the greatest

angle of that triangle.

8. ABCD is a quadrilateral in which the ∠B = the ∠C and AB> CD. Prove that the

∠D> the ∠A.

Theorem - 13

Any two sides of a triangle are together greater than' the, third side.

Particular Enunciation: Let ABC be a triangle, it is required to prove that any two of

its sides are together greater than. the third side. Let, BC to be the greatest side. Then

it is enough to prove that AB+AC>BC

Construction: BA is produced to D so that. AD = AC, We join C and D

Proof: In the Δ ADC, AD = AC

Therefore, ∠ACD=∠ADC [Theorem-8]

But, ∠BCD>∠ACD [since, ∠ACD is a part of ∠BCD]

Therefore, ∠BCD>∠ADC

That is, in the ΔBCD, ∠BCD>∠BDC

Therefore, BD>BC [ Theorem-12 ]

But, BD= AB + AD = AB + AC [Since AC = AD]

Therefore, AB + AC > BC (proved)

Example 1: In the ΔABC, AB= AC and

D is a point on AC produced. Prove that,

BD>CD.


In the ΔABC, AB = AC and D is a point

on AC produced. It is required to prove

that BD>CD.

D

C

A

B

A

B C

D


Proof: In the Δ ABD,

AB + BD>AD [Theorem-13]

or, AB + BD>AC + CD

or, AB + BD > AB + CD [AB = AC, by hypothesis]

or, BD > CD [Cancelling AB from both sides] (proved).

Theorem - 14Of all line segments that can be drawn to a given straight line from a point outside it,

the perpendicular is the shortest,

Pitrticular Enunciation : Let AB be a

straight line and O, a point outside it. It is

required to prove that of all line segments

that can be drawn from O to AB, the

perpendicular is the shortest. Let OM

drawn from O to AB. Then it is enough to

prove that OP< OM.

Proof: In the Δ OPM ∠OPM =1 right angle = ∠OPA

But ∠OPA>∠OMP. [If one side of a triangle is produced the exterior angle that is

greater than the opposite two interior angle]

Therefore, ∠OPM> ∠OMP

So, OM > OP. [Theorem-12]

That is, OP<OM (proved)

Exercise-2.6

1. Prove that the difference of any two sides of a triangle is less' than the third side.

2. D is any point within the ΔABC. Prove that, AB + AC > BD + CD.

3. The diagonals AC and BD of a .quadrilateral ABCD intersect at O. Prove that.

AC+ BD>BC+AD

4. O is any point within the ΔABC. Prove that, AB+ BC + CA>OA+ OB + OC

5. O is any point within the ΔABC and AB > OC. Prove that AO+BO>CO.

6. Prove that in any triangle, the sum of the medians is less than its perimeter

O

A p M B


7. Prove that, the sum of the two diagonals of a quadrilateral is less than its perimeter.

8. Prove that, of all the line segments drawn from the vertex of an isosceles triangle

to its base, the median is the shortest.

Theorem - 15The three angles of a triangle are together equal to two right angles.

Particular Enunciation: Let ABC be a triangle.

It is required to prove that ∠BAC+ ∠ABC +∠ACB = Two right angles.

Construction:

We produce the side BC to any point D and a line CE is drawn parallel to the line BA

through C.

Proof: Because BA and CE are parallel and AC meets them.

∴ the ∠BAC = the alternate ∠ACE

Again because BA and CE are parallel and BD meets them

∴ the ∠ABC = the corresponding ∠ECD.

Therefore, the ∠BAC + the ∠ABC = the ∠ACE + the ∠ECD = the ∠ACD ........... (1)

Adding the ∠ACB to both the sides,

we have, the ∠BAC+ the ∠ABC+ the ∠ACB = the ∠ACD + the ∠ACB.

But the ∠ACD +the ∠ACB = 2 right angles [theorem-l]

∴ the ∠BAC + the ∠ABC + the ∠ACB = 2 right angles (proved).

Corollary 1: If a side of a triangle is produced then the exterior angle so formed is

equal to the sum of the two opposite exterior angles. .

Corollary 2: If a side of a triangle is produced then the exterior angle so fomed is

greater then each of the two interior opposite angles.

Corollary 3: The acute angles of a right angled triangle are complementary to one another.

Corollary 4: In an equilateral triangle each angle measures 60˚.

A

B

E

DC


Theorem - 16If two triangles have two angles of one equal to two angles of the other, each to each,

and any side of the 1st equal to the corresponding side of the other then

the triangles are equal in all respects.

Particular Enunciation: Let ABC and DEF be two triangles in which the ∠A = the

∠D, the ∠B = the ∠E .

and the side BC = the corresponding side EF.

It is required to prove that the ΔABC ≅ the ΔDEF

Proof: ABC is a triangle in which the ∠A + the ∠B + the ∠C = 2 right angles

[Theorem-15].

From the ΔDEF

the ∠D +the ∠E + the ∠F = 2 right angles [Theorem-15]

Since, the ∠A = The ∠D and

the ∠B = the ∠E

∴ the ∠C = the ∠F

Apply the ΔABC to the ΔDEF, so that B falls on E, BC along EF and A falls on the

side of EF as D.

Then because BC = EF.

∴ C coincides with F

Again because the ∠B = the ∠E

∴ BA must fall along DE and because the ∠C = the ∠F

∴ CA must fall along FD.

∴ the point A; which falls both on ED and FD must coincide with D.

That is, the ΔABC coincides with the ΔDEF

∴ the ΔABC ≅ the ΔDEF

Example 1: Prove that if an angle of a triangle is equal to the sum of the other two

angles then the triangle is right angled.

Proof: We know that in any ΔABC, the ∠A + the ∠B + the ∠C = 180° [Given, the

∠A = the ∠B + the ∠C ]

A

B C

D

E F


∴ 2∠A=180°

∴ the ∠A = 90°

Therefore, the ΔABC is a right angled triangle.

Example 2: If the bisector of the vertical angle of a triangle is perpendicular to the

base then prove that it is an isoceles triangle.

Particular Enunciation :In the figure, let

ΔABC be triangle.The bisector AD of the

vertical angle A is perpendicular at the

point D on the base BC of the triangle

ABC.

It is required to prove that, AB =AC

Proof: In the ΔABD and the ΔACD

the ∠BAD=∠CAD [ AD is the bisector of the ∠BAC]

the ∠ADB = the ∠ADC [ AD is perpendicular on BC]

and AD is common

Therefore, the ΔABD ≅ the ΔACD [ Theorem-16]

Hence, AB = AC (proved)

Theorem - 17If two right angled triangles have their hypotenuses equal and one side of one equal to

one side of the other then they are equal in all respects.

Particular Enuncition :

Let ABC and DEF be two right angled triangles, in which the hypotenuse AC=

hypotenuse DF and AB = DE It is required to prove that ΔABC ≅ ΔDEF

. ... ..

A

B D C

A

B CG E F

D


Proof: Apply the ΔΑBC to the ΔDEF so that the point B falls on the point E, BA falls

on the side ED and the point C on the side of ED opposite to 'F' Let, G be the point on

which C falls since AB = DE

∴ A falls on. D. Thus ΔDEG represents the ΔABC is its new position.

Therefore, DG = AC = DF, ∠DEF = ∠ABC = one right angle

and∠DGE =∠ACB

since ∠DEF + ∠DEG = 1 right angle + 1 right angle = 2 right angle

∠GEF is a straight line.

Now, since in the ΔDGF, DG=DF

∴ ∠DFE = ∠DGE [Theorem -8]

Therefore, ∠DFE = ∠ACB

Now in the ΔABC and ΔDEF

∠ABC = ∠DEF [ each is equal to one right angle]

∠ACB = ∠DFE and side AB = side DE

Therefore, ΔABC ≅ ΔDEF [ Theorem -16] (Proved)

Exercise -2.7

1. In the figure, ABC is a triangle in

which ∠ABC = 90˚, ∠BAC = 48˚ and

BD perpendicular to AC. Find of the

remaining angles.

2. The vertical angle of an isosceles triangle is 50˚. Find the other two angles.

3. Prove that the sum of the angles of a quadrilateral equal to 4 right angles.

4. The two line segments PQ & RS intersects at the point O and LM ⊥ RS, EF ⊥ PQ. Prove that ∠MLO = ∠FEO

. ..


A

B

D

Czyx

48°

5. In the ΔABC, the internal bisectors of the ∠B and ∠C intersect at the point D.

Prove that ∠BDC = 90˚ + ∠A.

6. The ΔABC is right angled at C, E is a point on. AC produced. The perpendicular

ED on AB is drawn to meet BC at O.

Prove that ∠CEO = ∠DBO.

7. In the ΔABC, the external bisectors of the ∠B and ∠C intersect at the point O.

Prove that ∠BOC = 90˚ _ ∠A.

8. If the bisector of the vertical angle of any mangle is perpendicular to the base

then prove that the triangle is isosceles.

9. In the figure ABC is a triangle in

which AB = AC and ∠BAD =

∠CAE ; Prove that BD = CEO

10. In the quadrilateral ABCD. AC is the bisector of the ∠BAD and ∠BCD. Prove

that ∠B = ∠D.

11. In the figure ABCD is a quadrilateral

in which AB is equal and parallel to

CD and the diagonals AC and BD

intersect at the point O. Prove that

AD =BC.

12. In the figure, ABC is a triangle in

which AB = AC, BC is produced on

both sides to D and E and ∠DAB =

∠EAC proved that AD = AE

13. Proved that the perpendiculars from the end points of the base of an isosceles

triangle to the opposite sides are equal.

14. Proved that, If the perpendiculars from the end points of the base of a triangle to

the opposite sides are equal then the triangle isosceles triangle.

15. In the quadrilateral ABCD. AB = AD and ∠B = ∠D = 1 right angle. Prove that

ΔABC ≅ ΔADC.


A

D B C E

A

D B C E

A

D

B

O

C

12

12


Answer the questions 1_3 on the basis of the following information:

In the figure, the side BC of ΔABC is extended to D and CE is

the bisector of ∠ACD. AB || CE and ∠ECD = 60˚

1. Which one of the following is the value of ∠BAC

(a) 30˚ (b) 45˚

(c) 60˚ (d) 120˚

2. Which one of the following is the value of ∠ACD?

(a) 60˚ (b) 90˚

(c) 120˚ (d) 180˚

3. What kind of triangle is ΔABC is?

(a) obtuse-angled (b) isosceles

(c) equilateral (d) right-angled.

4. If ∠A=70˚ and ∠B = 40˚ in ΔABC then what type of triangle is ΔABC?

(a) obtuse-angled (b) right_angled

(c) equilateral (d) isosceles.

5 Two sides of a triangle are of length 5 cm. and 4 cm. Which one of the following

is the measurement of the other side of the triangle?

(a) 1 cm (b) 4 cm

(c) 9 cm. (d) 10 cm.


A

B CD

E

60°

6. The equal sides of an isosceles triangle are extended and if one of the external

angles is 120˚ then how much is the other external angle?

(a) 120˚ (b) 90˚

(c) 60˚ (d) 30˚

7. If one of the two acute angles of a right angled triangle is 40˚, then which of the

following is the value of other acute angle?

(a) 40˚ (b) 45˚

(c) 50˚ (d) 60˚

8, If the sum of two angles is equal to the other angle of a triangle, what type of

triangle is it?

(a) Equilateral (b) acute-angled

(c) right-angled (d) abtuse-angled

CREATIVE QUESTIONS

1.

In the figure, AB = AC in ΔABC, ∠ACB=60˚, BO and CO are external bisectors of

∠B and ∠C respectively.

(a) How much is the value of ∠ABC ?

(b) Find the value of ∠BAC and show that AB=BC

(c) Prove that ∠BOC= 90˚ _ ∠A


Α

Β

Ε Ο

C

F

2.

In the figure, ∠QPM = ∠RPM; PM ⊥ QR and ∠QPM ,= 90˚

(a) Find the value of ∠QPM.

(b) How much are the values of ∠PQM and ∠PMR?

(c) If PQ =6 cm, find the value of PR.

3. The sides AB and AC of ΔABC are equal to one another. AD is perpendicular to

BC from A.

(a) Draw the figure on the basis of the above informations.

(b) Show that AD < AC

(c) Prove that ΔABD ≅ ΔACD

4. AB > AC in ΔABC and the bisectors of ∠A, ∠B and ∠C intersected at the point O.

(a) Draw the figure on the basis of above informations.

(b) Show that, OB > OC

(c) Prove that, (AB+BC+CA) <2(DB+OA+OC)

5. QR of ΔPQR is extended to S. RT || QP and PQ||RT.

(a) Draw the figure on the basis of the above informations.

(b) Show that ∠PQR + ∠QPR=∠PRS

(c) Joining P with T, prove that ΔPQR ≅ ΔPRT.


P

Q RM

Chapter Three

Problems On Triangles

Every triangle has (a) three sides, and (b) three angles.

Two triangles can be equal in all respects or congruent if three parts of one are equal

to corresponding parts of the other. If only those parts are given the shape of the

triangle can be determined and the triangle can be constructed or drawn. If one of the

following data of a triangle is known, a particular triangle may be drawn :

(1) Two sides and the angle included by those sides.

(2) Three sides, (sum of the length of any two sides is greater than the length of the

third side).

(3) Two angles and one side.

(4) Two sides and an angle opposite to one of the two sides.

Problem - 9To construct a triangle having given the lengths of the three sides.

Let a, b, c be the given length of the three sides of a triangle. We are to construct the

triangle.

Construction: We cut off BC equal to a, from any line segment BD. With center at B

and C and radius equal to c and b respectively, we draw two arcs on the same side of

BC. The two arcs intersect each other at A. We join A with B and C. Then, ABC is the

required triangle.

Proof: By construction, in the ΔABC, BC = a, AC = b and AB = c.

∴ ABC is the required triangle.

A

B Ca

a

bb cc

Remarks : According to Theorem -13 any two sides of a triangle are together Greater

than the third side. So the given sides should be such that the sum of the lengths of

any two sides is greater than the length of the third side. It is only then that it is

possible to construct the triangle.

Problem - 10 To construct a triangle having given two sides and an angle included by those sides.

Let a and b be the two given sides and ∠C the given angle included by those sides. We

are to draw the triangle.

Construction: We cut off BC equal to a from any line-segment BD. We draw ∠BCE

equal to ∠C at the point C. We cut off CA equal to b from the line segment CE We

join A and B. Then ABC is the required triangle.

Proof: By construction, in the ΔABC, BC = CA= b and ∠ACB= ∠C

Therefore, ABC is the required triangle.

Problem - 11To construct a triangle having given one side and two angles adjacent to it.

Problems On Triangles 151

a

b

C

E

A

B C D

F EA

BB C

a

C D

Let a be a given side and ∠B and ∠C, the two given angles adjacent to it. We are to

construct the triangle.

Construction: We cut off BC equal to a, from any line-segment BD. We construct

∠CBE = ∠B at B and ∠BCF = ∠C at C on the line-segment BC. BE and CF intersect

each other at A. Then, ABC is the required triangle.

Proof: By construction, in the ΔABC, BC= a, ∠ABC=∠B and ∠ACB = ∠C.

Therefore ABC is the required triangle.

Remarks : By theorem- 15 the sum of the three angles of a triangle is equal to

two right angles, so the two given angles be such that their sum should be less than

two right angles, otherwise no triangle can be drawn.

Problem - 12To construct a triangle having given two angles and side opposite to one of them.

Let ∠A and ∠B be the two given angles and a the given length of the side opposite to

∠A. We are to construct the triangle.

Construction: We cut off BC equal to a, from any line-segment BD. We draw∠CBF

and ∠DCE equal to ∠B at B and C respectively. Again on the line segment CE at C

we draw ∠ECO equal to ∠A on the side of CE opposite to the angle drawn equal to

∠B. The straight line CG intersects BF at A.

Then ABC is the required triangle.

Proof: By construction, ∠ABC = ∠ECD. Since these are corresponding angles.

So BA|| CE

Now BA || CE and AC intersects them.


GA

A

a

FE

DCB

B

Therefore, ∠BAC = the alternate ∠ACE = ∠A,

So in ΔABC, ∠BAC = ∠A, ∠ABC = ∠B and BC = a

Therefore, ΔABC is the required triangle.

Problem - 13To construct a triangle having given two sides and an angle opposite to one of them

Let b and c be the two given sides and B the given angle opposite the side b. We are to


Construction: We draw ∠DBE equal to ∠B at B on any line seagment BD. We cut

off BA equal to the side c from BE. Now with centre at A and radius equal to the side

b, we draw an arc cutting line segment BD at C and C.We join A with C and C'. Then

both the ΔABC and ΔABC are the required triangles.

Proof: By construction in the ΔABC, BA = c, AC = b and ∠ABC = ∠B and in the

ΔABC', BA=c AC' = b and ∠ABC' =∠B

Therefore, both ABC and ABC' are the required triangles.

Problem - 14To construct a right angled triangle having given hypotenuse and one side.

Let a be the given hypotenuse of a right angled triangle and b the given side. We are to



AE

C' DCB

c

b

B

Ab

a

E

B C D

Construction: We cut off BC equal to b from any line segment BD, We draw a

perpendicular BE to BD centre at A. With centre at C and radius equal to a, we draw

an arc-cutting BE at A. We join, A and C. Then ABC is the required tringle.

Proof : By construction, the hypotenuse AC= a, BC = b and ∠ABC= 1 right angle.

Therefore ABC is the required triangle.

Problem - 15To construct a right angled triangle having given the hypotenuse and one angle

Let a be the given hypotenuse of a right angled triangle and B the given angle. We are

to construct the triangle.

Construction: We cut off BD = a from any line segment. BE. We find the middle

point O of the line segment BD. Now with centre at O and radius equal to OB we

draw a semicircle. We draw ∠DBA= ∠B at B on BD. The straight line BA meets the

semicircle at A. We join A and D. Then ABD is the required triangle.

Proof : We join A and O.

Since OB = OA. [being radius of the same circle]

Therefore, ∠OAB = ∠OB A [Theorem-8]

Again, OA = OD, [being radius of the same circle]

Therefore, ∠OAD = ∠ODA

Therefore. ∠OAB + ∠OA D = ∠OBA + ∠ODA

or, ∠BAD = ∠ABD + ∠BDA = 1 right angle.

Now. BD = the hypotenuse a ∠ABD= ∠B and ∠BAD = 1 right angle

So, ADB is the required triangle.


a

B

B O D E

A

Exercise - 3.1

1. Construct a triangle whose sides are:

(a) 2 cm, 3 cm and 4 cm

(b) 3 cm, 4 cm and 5 cm

(c) 3.5 cm, 2.7 cm and 4.8 cm

2. (a) Construct an isosceles triangle having given the base of the triangle and any one side.

(b) The length of a side of an equilateral triangle is 3.8 cm. Draw the triangle.

3. Construct a triangle above two sides and the angle included by these sides are:-

(a) 2 cm, 3.5 cm and 60˚

(b) 3.8 cm, 4.5 cm and 45˚

4. Construct a triangle in which a side and two angles adjacent to it are:-

(a) 3 cm, 30˚ and 45˚

(b) 3.5 cm, 50˚ and 60˚

5. Construct a triangle in which two angles and opposite to the first angle are:-

(a) 120˚, 30˚ and 5cm

(b) 60˚, 30˚ and 3cm

6. Construct a triangle in whieh two sides and an angle opposite to the first side are:-

(a) 5.4 cm, 6 cm and 60˚

(b) 4 cm, 5 cm and 30˚

7. Construct a triangle in which the length of the hypotenuse and one side adjacent

to it are:-

(a) 7.6 cm and 4.5 cm

(b) 3 cm and 2.8 cm

8. The length of a hypotenuse of an isosceles right angle triangle is 6 cm. Construct

the triangle.

9. The length of a particular side of a right angled triangle is 5 cm and one of the

acute angles is 40˚, Construct the triangle.



1. If two sides of a triangle and one angle opposite to one of that sides are given,

then how many triangles can be drawn?

(a) 1 (b) 2

(c) 3 (d) 4

2. In which case it is possible to draw a triangle when the length of the sides are

respectively_

(a) 1 cm, 2 cm, 3 cm (b) 3 cm. 4 cm, 5 cm.

(c) 2 cm, 4 cm, 6 cm. (d) 3 cm, 4 cm, 7 cm.

3. (i) If two sides of a triangle and the included angle are given then the triangle

can be drawn.

(ii) If the sum of two sides of a triangle is greater than third side then the

triangle can be drawn.

(iii) There may exist more than one abtuse angle in a triangle. Which one of the

following is correct?


(c) i and iii (d) i, ii and iii.

Answer the questions 4-5 according to the following figure

4. To draw a line parallel to AB at the point C, angle equal to which angle has to draw?

(a) ∠ABC (b) ∠ACB

(c) ∠BAC (b) ∠CAD

5. Which one of the following is equal to ∠CAD?

(a) ∠BAC+∠ACB (b) ∠ABC+ ∠ACB

(c) ∠ABC + ∠ACB +∠BAC (d) ∠ABC+∠BAC


A

D

CB

CREATIVE QUESTION

1. There are three points A, B and C so that they are not colinear.

(a) Draw a triangle through the three points.

(b) Draw perpendicular to the base from the vertex of the drawn triangle. .

(c) If the base of the drawn triangle is the hypotenuse of right angled isosceles

triangle then draw the triangle.

2.

(a) Which is the hypotenuse of the triangle in the figure above?

(b) Measure the hypotenuse in centimeter and draw a angle equalizing the angle

∠ACB.

(c) Draw a right-angled triangle whose hypotenuse is 2 cm greater than that of

drawn triangle and an angle is equal to ∠ACB

3.

In figure, lines BA and CE are parallel to one-another.

(a) Write the relation between ∠BAC and ∠ACE

(b) Show that ∠BAC+ ∠ABC = ∠ACD

(c) Draw a triangle whose two angles are equal to the angles ∠ECD and ∠ABC

and one side is equal to AB.


A

B C

A

BC

D

E

4. There are two sides a = 3.2 cm, b= 4.5 cm and an angle ∠B=30˚ of a triangle.

(a) Draw an angle equal to ∠B.

(b) Draw a triangle whose two sides are equal to a and b and included angle is

equal to ∠B

(c) Draw such a triangle whose one side is b and the opposite side of ∠B is a.

5. The length of one side is 4 cm of a triangle and two angles are 37˚ and 46˚

adjacent to the side.

(a) How much is the amount of the other angle?

(b) What type of triangle it is and why?

(c) Draw the triangle.


Answers

ArithmeticExercise 1.1

1. (a) 19 (b) 28 (c) 36 (d) 45 (e) 63 (f) 87 (g) 90 (h) 96

2. (a) 27 (b) 31 (c) 81 (d) 222 (e) 234 (f) 847 (g) 440 (h) 1001

3. (a) 3 (b) 3 (c) 5 (d) 4

4. (a) 5 (b) 3 (c) 2 (d) 3 (e) 6

5. 3 6. 5 7. 20

Exercise 1.2

1. 0.7 2. 1.1 3. .05 4. 2.17 5. 12.4 6. .073

7. .0075 8. 25.38 9. 1.001 10. 2.236 11. 2.828 12. 3.494

13. .707 14. .173 15. .726 16. 1.643 17. .548 18. 1.833

19. .192

Exercise 1.3

1. (a) (b) (c) (d) (e) 1 (f) 9 (g) 6 (h) 10

2. (a) .845 (b) .527 (c) .764 (d) 2.101 (e) 2.991 (f) .395 (g) 2.057

Exercise 1.4

1. 64 tress 2. 32 trees 3. 4 4. 84,393 5. 48,56 6. 225 7. 50 pupils

8. 45,46 9. 45,50 10. 35 people 11. 25 people

Exercise 2.1

1. a) 15� 7 b) 5000 � 7009 c) 31� 18 d) 1000 � 507 e) 1� 2 f) 5 � C

2. a) 44�56 b) 20�561 c) 7 � 3 d) 32 � 9

3. 13m 20cm 4. .54 years 5. 7 � 11 6.32 � 00 Taka 7. 1 � 5

17

67

15

19

29

311

1217

21100

8. 20 years 9. 24 � 25 10. 29 � 21 11. 240, 210

12. Sodium 78.67 gm (Appx.), Chlorine 121.33 gm 13. 55 � 42

Exercise 2.21. a) 1 � 3 � 9 b) 25 � 45 � 81 c) 9 � 1� 25 d) 1 � 3 � 9

2. a) 15 b) 8 c) 0.55 d)

3. a) 16. b) 48 c) 2 d) 200 e) 12.

4. 1 � 10 5. 24 quintal 6. 3 � 6 �12 7. 4 gm 8. 1000.00 Taka

9. 17 Nos. 10. 13 days 11. 25 days 12. 5,600 Nos 13. 244 � 203

14. 35 � 42 � 24 15. 40 litres.

Exercise 2.31. 28 � 35 � 55 2.4kg 3. 26gm

4. Biomanure 144 sacs, Chemical fertilizer 32 sacs.

5. Oxygen 88.9 kg. Hydrogene 11.1 kg. 6. 270 Taka. 630 Taka, 1260 Taka

7. Tk.150.00 Tk.195'00 Tk.225.00 8. Tk 800,00 9. 14 � 3 10. Tk. 300.00

11. Tk. 128.00 12. Tk. 315.00, Tk 525.00

13. Tk.750. Tk. 900, Tk. 825 14. Tk. 75 Tk. 60, Tk. 85

15. Tk.750, Tk.500 16. Tk. 3160, Tk. 3600.

17. Tk.1625, Tk.2275, Tk.1300

18. Lily 90 Taka, Moly 150 Taka, Pinky 75 Taka.

19. 12 gm. 20. Male 500 Taka, Female 200 Taka, Boys 120 Taka. .

21. Suman 1380 Taka, Jamal 960 Taka. Dilip 1260 Taka,

22. 31 � 17

Exercise 3.11. Tk.2887.50 2. 3 kg. 3. 122 kg 4. Tk.17.50 5. Tk.378,

6. Tk.9000 7.33 People 8. 10 days 9. 70 days 10. Tk 34500

12

14

14

45

13


11. 15 Students 12. 16 13. 5 days 14. 10 days 15. 150 workers

16. 8 hours 17. 6 days,

Exercise 3.21. 2 workers 2. 12 days 3. 15 masons 4. 120 labourers 5.30 days

6. 90 hectors. 7. 17 days 8. 5 hours 9. 3days 10.20 days

11. Aziz in. 25 days , Kamal in 100 days,

12. A in 30 days, B in 20 days, C in 60 days.

13. In 12 days. 14. 50 Km 15. 2 hours 30 minutes

16. 18.04 km per hour (Approx.) 17. 5 seconds

18. 180 metres. 19. 21 seconds 20. 4 hours 21. 25 minutes

22. After 8 minutes. 23. 15 minutes 24. 2 Km per hour

25. 6 hour 26. 6.2 Km per hour 27. 280 Km

28. Speed of the boat Km per hour, speed of the current Km per hour.

29. 32 days.

Exercise 4.11. (a) 800 cm, 8000 mm (b) 740 cm, 7400mm. (c) 1396 em, 13960mm.

(d) 0.7 cm, 7mm (e) 4790cm, 47900mm.

2. (a) 0.493m (b) 7.565 m (c) 23800m (d) 8000.09m (e) 25000.03m

3. (a) 0.4039km. (b) 0.005359 km (c) 0.097865 km (d) .009078 km

(e) 0.07525km

4. (a) 3m 5 decim (b) 7 decim 6cm 4mm c) 53m 7 decim d) 5cm 6mm

Exercise 4.2

1. 20 sq.m 2; 150 sq.cm 3. 115 sq m. 4.1375 sq.m

5. 384sq.m 6. 21 m,7m 7. 125 cm 8. 300 mgs

9. 3000 sq.cm. 10. 1200 sq. metre 11. 250 sq.metre 12. 23.16 kg.

13. 44,000 litre and 44,000 kg. 14. Tk. 450.00

23

511

16

112

Answers 161

AlgebraAnswer

Exercise 1.1

1. 4K_m_ n. 2.6a_b _3c 3.2x_ z 4. _ 2a+ 6b_ c_ 8d

5. 2x3 + x2 + 15x + 10 6. 6xy2 + 5 7. 3x3y _7x2y2 + ll xy3

8. _a3 + 3b3_2a2b + 3ab2 9. l0a2xy2-3a2x2y2 + y3-8.

10. 8p3q3r3 _ 3p2q2r2 _ 2p4q4r4 _2 11. x _ y + z

12. _ a3 + ax2 13.3 14. 27 15.3 16. _1 17. 3. 18. 36.

Exercise 1.2

1. 3x + 2y _3z 2. _ 3a2_9a+ 7

3. 4a2 _7ab _ b2 4. ax_by_14cz

5. 2m2+ 10mn _16n2 6. 2x3_ 4x2y _15xy2+ 9y3

7. _ a4 _ 2a2 _12a +12 8. _ a3+ 3b3 _ 2a2b +3ab2

9. 6x3+ 6x2y + 5xy2 10. llx2y _ llxy2_10x3 _ 3y3

11. _ a3_b + 3a2b _3ab2 12. 2x5+3x4y + 6y5

13. yz+zx + xy 14. 13a2+14ab _18b2

15. 2x3_7x+ 10 16. 5x3 + 20y3

17. _8x3 + 3x2y + 5xy2 _2y3

Exercise 1.3

1. 6a3b. 2. _54ab3 3. 10a3x7y 4. _14a4x3

5. _30b4x4yz, 6. 15p7q7 7. 32a4b2x5y3 8. 70x5y5z

9. _14a5 x4y9z7 10.8x2y_12xy2 11. 2a3bc _3ab3c _ abc3

12. 12c4d5-18c3d7 + 30c3d6 13. a7b _ a4b4+ 3a5b2c.

14. x2 + 9x + 20 15. a2 _ b2 16. ab+ac+b2_ c2

17. 10x2-51xy + 56y2 18. x3+6x2+5x_12

14

58

512

16

712


19. 3a3 + a2b _ ab2 + 4b3 20. a3_3a2b + 3ab2_b3 21. x8_y8

22. x3 + 3x2a + 3xa2 + a3 23. 8p3 _ 12p2q + 6pq2 _ q3 24. x4 + x2+ 1

25. a4+a2b2+b4 26. x2+y2+z2+2xy+2yz+2zx

Exercise 1.4

1. 4x2 2. _ 9z3 3. 3x6 4. _5a2x2 5. 3a2z2y

6. _2pqr2 7. _5x3y7 8. 11x 9. _5y. 10. 5yab2

11. 3a_2b. 12 2a3b _ 3ab2 13. _b+ 3a4b4 14. _x_y_z

15. _3a2 +4ax_2x2 16. _2p2 + 4pq +5q2 17. _x5+3xy_4y3

18. _7ac+4a3b2c4+3ab4c2 19. 3mn_7n2_4m2 20. + xy5_y4z2

21. a2b2 22. x_2 23. 3a +4 24. 2x_3y 25. x_ 4y

26. a +2xyz 27. x2_1 28. x2y2 _ 1. 29. 5x + 3y 30. x_ 5ab.

31. x2+3. 32. 2ab +3d. 33. x2 + x + 1. 34. a2_b2 35. 9p2_2pq_ q2

Exercise 1.5

1. a_b+c_d 2. 2x_2a. 3. 5y_5x 4. 3x+4y_z + b + 2c.

5.7b_2a. 6. 0 7.5y_2x+7z. 8.6a_6b_18c.

9. 5a_b+llc. 10. _10x+14y_18z. 11. 2a+3b+28c. 12. _ 9.

13. x_(_y +a_b) 14. _(a_b+c) + (d_c)

15.3x_{4y+(8z_5)} 16. 10 _[7+ {_3+(4_8)}], 10

Exerctse 2.1

1. 4a2 +12a+9 2. 25x2 +20xy + 4y2

3. 8lx2 + 90ax+25a2 4. 9m2 + 42mn + 49n2

5. 25a2 _ 70 a+49 6. 9a2 _ 42xy + 49y2

7. a2x2 _ 2abxy+b2y2 8. 9a _ 66axy + 121x2y2

9. 25a4+90a2m2+81m4 10. 81x4 _ 198x2y2 + 121y4

11. 9801 12. 1010025

35

15

Answers 163

13. 994009 14. a2+b2 +c2_2ab + 2ca _2bc

15. a2+4b2+9c2+4ab+6ca+12bc 16. 4x2 + y2 +z2+ 4xy_ 4zx_2yz

17. b2c2 +c2a2+a2b2 + 2abc2+ 2ab2c + 2a2bc

18. x4+y4_z4 + 2x2y2_2z2x2_2yz2

19. a4+ 4b2 + c4+ 4a2b _2c2a2_4bc2

20.4b2 21.16 x2 22.16a2 23. a2 + 4ab + 4b2 24. 9a2 25.1

26. 0 27. 4 28.16 31. P2_2 33. 12 35. 79.

Exercise 2.21. 16x2 _ 9 2. 25a2 _ 64b2 3. 16a2 _ 49b2 4. 169 _ 144p2

5. a2x4 _ b2 6. a2 _ b2 _ c2 _ 2bc 7. x4 + x2 + 1

8. 16x4 + 36x2 +81 9. 1 + 4x4 + 16x8 10. a8 + 9a4x4 + 81a8

11. x2 + 25x + 144 12. p2x2 _ 2px _ 15 13. 4x2 + 26x +40

14 q4x4 + 4rq2x2+ 3r2 15. x2+x+ 16. x2 _ a2

17. x2 _ y2 18. x4 _1

Exercise 2.3

1. x(y+a2) 2. abc(a+b+c) 3. (c+a)(a+b)

4.(x_a) (x_b) 5. (1+a) (1+a2) 6. (ax +by) (bp+ aq)

7. 2(b+d) (2a _5c) 8. (x+5)(x_5) 9. (2x+y)(2x_y)

10. (a+9b) (a_9b) 11.(7+ab)(7_ab) 12. (a+x+y) (a_x_y)

13. (x+b_c) (x_b+c) 14. 5x(4x+3) (4x_3) 15. (p+ ) (p_ )

16. pq(p+q) (p_ q) 17. (2x +3y) (2x_3y) (4x2 +9y2) 18. (x+3)(x_3)

19. (5x2+6y2)(5x2_6y2) 20. (18a+llx)(19x_6a)

21. (x+y)(x_y)(x2+y2+2) 22. x3(12x2+5a2)(l2x2_5a2)

23. (8+3x_3y)(8_3x+3y) 24. (2x2+6x+9)(2x2_6x+9)

25. (a_b+c)(a_b_c) 26. (a+b_c)(a_b+c)

29

125

136

13

13

13

19


27. (2x_y+z)(2x_y_z) 28. (x+3)(x+2)

29. (x+5)(x+6) 30. (x_9)(x_5)

31. (a+8)(a_5) 32. (m+6)(m_5)

33. (a_13)(a+7) 34. (m_8)(m_7)

Exercise 3

1. xy4 2. a2b2 3.2a4x4 4. xyz

5.4ax 6. (a+b) 7. a(a+b) 8. ax(2a+3x)

9. (x+4) 10. (x_5) 11. (x+2) 12. (x_1)

13.6a2b2c 14. 12a2x3c4 15. 78ac5d4 16. 36a2b3c

17. 30a3b2x3y2z2 18. 12x2(x+2) 19. x(x2+3a+2) 20. x2(x2_49)(x+3)

21. 5a(9x2_25y2) 22. (x_5)2(x+2) 23. a(x2_2)(x2+2) 24. (a2_7a+12)(a_5)

Exercise 4.1

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

11. 12. 13. _

Exercise 4.2

1. , 2. 3.

4. 5.

6.

7. 8.

9.

10.

2ad

4bd

x2+x2y

x(x2_y2)

x2y_xy2

x(x2_y2)

zx_yz

x(x2_y2)

ax2_2bx2

a(a+2b)(a_2b)

a2y2+2aby2

a(a+2b)(a_2b)

2xb2c2(a_x)

a2b2c2(a2_x2)

3a+9

(a+1)(a+3)(a-2)

a2_2a_8

a(a+3)(a+2)(a_4)

ax

a(2x+5y)(2x_5y)

3yc2a2(a+x)

a2b2c2(a2_x2)

3bc

4bd

a2

abc

12b

15ab

xyz

2m3k

2x2y2

3z

xx+4y

2x_3a2x

x+3x_3

x+2x_2

x_3a

a_3a+4

x+2x+3

2+9x

x3

,

, ,,

,

,

,

, 2xy-5y2

a(2x+5y)(2x_5y)

2axz+5ayz

a(2x+5y)(2x_5y),

a2_ 4ab

a(a+3)(a+2)(a_4)

a2c+2ac

a(a+3)(a+2)(a_4),

5a+5

(a+1)(a+3)(a-2)

x2+5x

(x+3)(x_3)(x+5)

xy+3y

(x+3)(x_3)(x+5),

4za2b2

a2b2c2(a2_x2),

b2

abc, c2

abc, ac+a2

(b+c)(c+a)

b2+bc

(b+c)(c+a),

Answers 165

Exercise 4.3

Exercise 5.11. 1 2. 2 3. 3 4. _ 4 5. _ 2 6. 7. 2 8. _ 6

9. _ 28 10. 12 11. 5 12. 12 13. 1 14. 7 15. _2 16. _ 2

17. _ 1 18. 5 19. _ 1 20. 13 21. 38 22. 1 23. _ 2 24. _ 7

25. 5 26. _ 6 27. 28. 41 29. _ 10 30. 1

Exercise 5.21. 33 2. _ 6 3. Azmal 20 years, Shafique 15 years.

4. 52, 28. 5. 44, 45, 46 6. 60 7. 23

8. Kamal Tk. 60, Jamal Tk. 55, Haripada Tk. 62: 9. 360 fruits,

10. 2km/hour and 6Km/hour 11. 90 Km

12. Greater 125 Smaller 100. 13. 2 Km/hour

14. 4 Km/hour 15. Tk. 49

Geometry

Exercise 1.11. x = 60°, y = 120°, z = 60°

p = 135°, q = 55° r = 135°

u = 50°, v = 130°, w = 50°

Exercise 2.71. x = 42°, y = 48°, z = 42°

2. 65°, 65°

bx+ay

ab

3x3b+2ay

xy

x2_y2

xya3_b3

ab

x2+y2

xy

815

411

12

23


2. 3. 4. 5.1.

a2+b2

ab

2y

x2x_1

x2_x_2

a

a2+5a+67. 8. 9. 10.6.

a_ 4

a2_9

x2+y2

x2_y2

a+b

2a_7b12. 13. 14. 15.11.

1

x+3

x

x2_ 6x+5

23_4x

x2_ 4x+5

5x_3y

4x2_9y2

a3

b(a2_4b2)17. 18. 19. 20.16.

2

0 1

prescribed by the national curriculum and textbook board as afor this reason the study of algebra...

Documents