i i i Government of Karnataka MATHEMATICS English Medium 7 SEVENTH STANDARD Part - I KARNATAKA TEXT BOOK SOCIETY (R) 100 Feet Ring Road, Banashankari 3rd stage, Bengaluru-85 ©KTBS Not to be republished
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Government of Karnataka
MATHEMATICS English Medium
7SEVENTH STANDARD
Part - I
KARNATAKA TEXT BOOK SOCIETY (R)100 Feet Ring Road, Banashankari 3rd stage,
Bengaluru-85
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PrefaceThe Textbook Society, Karnataka, has been engaged in producing new textbooks according to the new syllabi which in turn are designed on NCF - 2005 since June 2010. Textbooks are prepared in 12 languages; seven of them serve as the media of instruction. From Standard 1 to 4 there is the EVS, mathematics and 5th to 10th there are three core sub-jects, namely, mathematics, science and social science.NCF - 2005 has a number of special features and they are: connecting knowledge to life activities. learning to shift from rote methods. enriching the curriculum beyond textbooks. learning experiences for the construction of knowledge. makingexaminationsflexibleandintegratingthemwithclassroom
experiences. caring concerns within the democratic policy of the country. making education relevant to the present and future needs. softening the subject boundaries integrated knowledge and
the joy of learning. the child is the constructor of knowledge. The new books are produced based on three fundamental ap-proaches namely, Constructive approach, Spiral approach and Inter-grated approach. The learner is encouraged to think, engage in activities, master skills and competencies. The materials presented in these books are integrated with values. The new books are not examination oriented in their nature. On the other hand they help the learner in the all round development of his/her personality, thus help him/her become a healthy member of a healthy society and a productive citizen of this great country, India. Mathematics is essential in the study of various subjects and in real life. NCF 2005 proposes moving away from complete calcula-tions, construction of a framework of concepts, relate mathematics to real life experiences and cooperative learning. Many students have a maths phobia and in order to help them overcome this phobia, jokes,
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puzzles, riddles, stories and games have been included in textbooks. Each concept is intoduced through an activity or an interesting story at the primary level. The contributions of great Indian mathematicians are mentioned at appropriate places. We live in an age of Science and Technology. During the past fivedecadesmanhasachievedgreatthingsandrealizedhisdreamsand reached pinnacle of glory. He has produced everything to make life comfortable. In the same way he has given himself to pleasures and reached the stage in which he seems to have forgotten basic sci-ences. We hope that at least a good number of young learners take to science in higher studies and become leading scientists and contrib-ute their share to the existing stock of knowledge in order to make life prosperous. Ample opportunity has been given to learners to think, read, discuss and learn on their own with very little help from teach-ers. Learning is expected to be activity centered with the learners doing experiments, assignments and projects. 6th standard Social Science Textbook has been prepared based on the prescribed syllabus. And all the features of NCF 2005 and KCF 2007 have been included in the Text Book. This new Text Book has given importance to enhance the creativity of students by including ac-tivities. Many projects are included to help students to gain knowledge. This Text Book has been written in such a way that students need not memorise historical dates and other information. The Textbook Society expresses grateful thanks to the Chairper-sons, Writers, Scrutinisers, Artists, Staff of DIETs and CTEs and the Members of the Editorial Board and Printers in helping the Textbook Society in producing these textbooks.
Prof. G.S. MudambadithayaCoordinator
Curriculum Revision and TextbookPreparation
Karnataka Textbook Society®Bengaluru, Karnataka
NagendrakumarManaging Director
Karnataka Textbook Society®Bengaluru, Karnataka
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Chairperson's LetterDear All,NCF 2005 intends that the aim of learning Mathematics to be of higher value rather than
mere learning algorithms and this is stated as learning of Mathematics is 'Mathemetisation. To achieve this objective our educational system, that is administration, class, school, society and persons in these agencies has to provide opportunities to the child to gain excellence experiences and fecilitate to construct knowledge himself.
The logical thinking adopted to solve a problem is to be of higher significance rather than knowing the solution to the problem. This develops rational and logical thinking among learners and enables them to explore their own method or approach of finding the solution and keeps them to be active participants. To make it possible, teachers have to become friendly facilitators to provide learning opportunities to the learners; as well as encourage pupils to learn with co - operation, in peer groups.
'Mathematics' is a challenge at 7th standard because it has to be nearer to the experiences and environment of learner and the learner has to comprehend the logical process and the abstract idea/concept.
Scope for exploration and creation is required in Mathematics instead of the rote problems and complicated calculations. Students have to be encouraged to solve the problems in different ways. To achieve this, previous learning is linked with present topics of learning Model activities, activities that can be done by the students and activities that have to be performed by the students are given to the possible extent Illustration, figures, laws/ principles and worked (model) problems are given to facilitate better comprehension of the subject. Additional information / issues are included here and there to enrich learning. To make learning easier and meaningful simple language along with wider scope and social situations are used. For reinforcement of each unit exercises, based on knowledge, understanding, skills and application- are given as far as possible. Our Committee is grateful to the chief co-ordinator, the managing director, the joint director and programme co-ordinator of karnataka textbooks society for providing an opportunity and responsibility of producing this textbook through which we have the pleasure to reach and serve larger educational community. We thank the editorial board and the scrutinizers for the valuable guidance and suggestions the members and DIETS involved in 'Try - out' for the feedback. All of this is attended to improve the text.
I am thankful to the members, translators, artists of this committee, technical assistants and other departmental, non departmental persons who have assisted me in this service .
It is tried to make this text to be free from doubts, confusions, ambiguity and printing mistakes. In spite of it, I request you kindly to communicate with me for any such things. Any suggestions for the improvement will be acknowledged whole heartedly. Please, share your opinion regarding the quality of the textbook with us and the department.
Yours truly,N. Kaleshwara Rao
Chair Person,7th Std Maths Text book
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Text Book Committee
Chair Person :Sri Kaleshwar Rao Bagoor # 3 Santhasa, 1st Cross, Aravinda Marga, J.P. Nagar, IInd
Phase, Bengaluru.Members :Smt B. Keerthi Principal, Siliconcity P. U. College, Konanakunte, Bengaluru
-62.Sri Subramanya Bhat Assistant Teacher, K.V.S.M High school, Kanchana, Puttur (T)Sri Sadananda Kumar Government Girls P. U. College, Hampi Road, Hospete Ballari
District- 583218.Sri Pramod G. Kulkarni V. B. Darbar P. U. College, Vijayapura.Sri C. L. Bhaskar Assistant Professor Vijaya Teachers College, Jayanagar 4th
Block Bengaluru - 56001.Smt Renuka Assistant Teacher, Goverment composite P. U. College
Kamalapur Taluk Hospete. Ballari District.Sri Jangi Art Teacher. DSERT Bengaluru.Scrutinizers :Prof. Prabhakar R. V. Deen Vijaya College Jayanagar and Assistant Secretary B. H.
S. H. E. S. Jayanagar 4th Block. BengaluruDr R. Latha Kumari Principal, Sanjaya Gandhi College of Education, cholanagar,
Hebbal, Bengaluru.Editorial Committee :Dr Ravindra Former NCERT Director Arehalli BSK 3rd stage Bengaluru - 61. Dr Upadya B. S. Lecturer and Head, Department of Mathematics, Education
RIE, Mysuru.Dr Prasad S. V. Lecturer RIE Mysuru.Dr Sharad Sure Assistant Professor Azim Premji university, PES School of
Engineering Campus Konappana Agra Hara Bengaluru.Translation Committee :Smt Vasanthi Rao Retired Teacher, Chord Road, IInd Stage Bengaluru.
Sri Sadananda Kumar Government Girls P. U. College. Hampi Road, Hospet Ballari District 583218.
Sri Ravikumar T. R. Assistant Teacher, A. L. S. High School, Basaweshwara Nagar, Bengaluru.
Chief - Coordinators :Sri G. S. Mudambaditaya Co-ordinator, Curriculum revision and Text Book Preparation.
Guidance :Sri Nagendra Kumar Managing Director, Karanataka, Text Book Society.Smt. C. Nagamani Deputy Director, Karnataka, Text Book Society.Programme Coordinators :Smt Vijaya Kulkarni Assistant Director, Karanataka Text Book Society.Smt Prema B. R. Technical Assistant ,Karanataka Text Book Society.
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About the Revision of Textbooks
Honourable Chief Minister Sri Siddaramaiah who is also the Finance Minister of Karnataka, in his response to the public opinion about the new textbooks from standard I to X, announced, in his 2014-15 budget speech of constituting an expert-committee, to look into the matter. He also spoke of the basic expectations there in, which the textbook experts should follow: “The textbooks should aim at inculcating social equality, moral values, development of personality, scientific temper, critical acumen, secularism and the sense of national commitment”, he said.
Later, for the revision of the textbooks from class I to X, the Department of Education constituted twenty seven committees and passed an order on 24-11-2014. The committees so constituted were subject and class-wise and were in accordance with the standards prescribed. Teachers who are experts in matters of subjects and syllabi were in the committees.
There were already many complaints, and analyses about the textbooks. So, a freehand was given in the order dated 24-11-2014 to the responsible committees to examine and review text and even to prepare new text and revise if necessary. Eventually, a new order was passed on 19-9-2015 which also give freedom even to re-write the textbooks if necessary. In the same order, it was said that the completely revised textbooks could be put to force from 2017-18 instead of 2016-17.
Many self inspired individuals and institutions, listing out the wrong information and mistakes there in the text, had sent them to the Education Minister and to the Textbook Society. They were rectified. Before rectification we had ex-
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changed ideas by arranging debates. Discussions had taken place with Primary and Secondary Education Teachers’ As-sociations. Questionnaires were administered among teach-ers to pool up opinions. Separate meetings were held with teachers, subject inspectors and DIET Principals. Analytical opinions had been collected. To the subject experts of sci-ence, social science, mathematics and languages, textbooks were sent in advance and later meetings were held for dis-cussions. Women associations and science related organi-station were also invited for discussions. Thus, on the basis of all inputs received from various sources, the textbooks have been revised where ever necessary.
Another very important aspect has to be shared here. We constituted three expert committees. They were constituted to make suggestions after making a comparative study of the texts of science, mathematics and social science subjects of central schools (NCERT), along with state textbooks. Thus, the state text books have been enriched based on the com-parative analysis and suggestions made by the experts. The state textbooks have been guarded not to go lower in stan-dards than the textbooks of central school. Besides, these textbooks have been examined along side with the textbooks of Andhra Pradesh, Kerala, Tamil Nadu and Maharashtra states.
Anotherclarificationhastobegivenhere.Whateverwehave done in the committees is only revision, it is not the total preparation of the textbooks. Therefore, the structure of the already prepared textbooks have in no way been affected or distorted. They have only been revised in the background of gender equality, regional representation, national integrity, equality and social harmony. While doing so, the curriculum frames of both central and state have not been transgressed. Besides, the aspirations of the constitution are incorporated
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carefully. Further, the reviews of the committees were once given to higher expert committees for examination and their opinions have been inculcated into the textbooks.
Finally, we express our grateful thanks to those who strived in all those 27 committees with complete dedication and also to those who served in higher committees. At the same time, wethankallthesupervisingofficersoftheTextbookSocietywho sincerely worked hard in forming the committees and managed to see the task reach its logical completion. We thank all the members of the staff who co-operated in this venture. Our thanks are also due to the subject experts and to the associations who gave valuable suggestions.
Narasimhaiah
Mangaging Director Karnataka Textbook Society
Bengaluru.
Prof. Baraguru Ramachandrappa Chairman-in-Chief
Textbook Revision Committee Bengaluru.
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Text Books Revision Committee
Chairman-in-chief.Prof. Barguru Ramchandrappa, State Revision Committee, Karnataka
textbooks Society®, Bengaluru. Revision Committee
ChairpersonDr. Narasimhamurthy S.K. Professor and Chairman, Department
of Mathematics , Kuvempu University, Shankaraghatta-577 451. Shivamogga
MembersDr. B . Chaluvaraju, Professor, Department of Mathematics,
Bengaluru University, Bengaluru.Sri. B. K. Vishwanath Rao, Rtd., Principal, No.94, ''Prashanthi'', 30th
Cross, BSK 2nd Stage, Bengaluru.Sri Narasimha murthy G.N., ‘Beladingalu’ No.23/1,5th cross, Hosalli,
Bengaluru.Sri Shankarmurthy M.V. Rtd Headmaster, Sarvodaya High-school,
BengaluruSri H.N.Subbarao, Headmaster, Sadvidya Highschool,
N.S.Road, Mysuru.Smt S.S. Thara, Headmistress, Govt. High School,
Mavattur, K.R. Nagar taluk, Mysuru Dist,Smt Sushma Nagaraj Rao, High School Teacher, Govt. Higher Primary
School, RamnagarSri Shrinath Shastri, Kannada Ganak Parishat, Chamrajpete,
Bengaluru.High Power CommitteeDr.Kashinath Biradar, Plot No.7, Gangasiri, Jayanagar,
Kalburgi - 585 105.Smt. L. Padmavati, Vice-principal, Empress Girls
High-school, Tumkur.Sri T Gangadharaiah, Associate Professor, Department of
Mathmetics, Govt. women’s college, Kolar Chief Advisors Sri Narasimaiah, Managing Director, Karnataka Textbooks
Society®, Banashankari 3rd stage, Bengaluru-85.
Smt Nagamani C. Karnataka Textbooks Society®, Banashankari 3rd stage, Bengaluru-85.
Programme co-ordinator:Smt. Vijaya Kulkarni, Asst.Director, Karnataka Textbooks
Society®, Banashankari 3rd stage, Bengaluru-85.
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CONTENTSPart - I
Sl. No Chapters Pages
1 Integers 1 - 25
2 Fractions 26 - 51
3 Rational Numbers 52 - 79
4 Algebraic Expressions 80 - 111
5 Pair of angles 112 - 135
6 Pair of lines 136 - 159
7 Properties of triangles 160 - 181
8 Symmetry 182 - 203
Answers 204 - 214
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CHAPTER– 1
INTEGERS
After studying this chapter you : multiply positive integer by positive integer as well as
by negative integer, multiply negative integer by negative integer, compare integers and identify smaller, greater among
given integers, follow proper method in multiplication, as well as relate
multiplication and division of numbers, divide an integer by another integer, understand why an integer can not to be divisible by
zero, understand the properties of integers: commutative,
associative and distributive, with respect to fundamental operations.
Suhasini saw a rose plant having 5 roses. She plucked 3 roses. Find how many roses are left in the plant.
Solution:
Total number of roses in the plant = 5
Number of roses plucked = 3
Number of roses left in the plant = 5 – 3 = 2.
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Positive integers, negative integers along with zero are termed as integers. We know how to add and subtract integers on the number line.Example 1 : Rita wanted to play a game. She jumped 3 steps to her right and then she decided to jump 5 steps to her left.How can you show her jumping on the number line ?
If we consider the first position of Rita as zero, then the jumps can be shown on the number line, where does she reach ?
0 1 2 3
-2 2-1 1-3 30
0
3 + (–5) = –2
Remember : Counting numbers/Natural numbers are positive
integers. Zero is not a natural number. Natural numbers and "0" together termed as "Whole
numbers". Integers include whole numbers as well as negative
integers.
Exercise 1.1
I. Simplify the following .
1) 8 + 5 2) 2 + (–7) 3) (–3) + (4) 4) (– 7) + (– 2)
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Multiplication of integers. Multiplication of a positive integer by a positive integer:
If a sign is not attached to a number, it is considered to be positive.
What is multiplication ?
We know that multiplication is repeated addition. Let us observe this from the following example.Example 1:
Bharath bought 3 pencil boxes from a shop. If each box contains 6 pencils, find the total number of pencils Bharatha has
6 + 6 + 6 = 18 (addition)
How to mark this on a number line?
0 6 12 18
Look at the arrow marks shown on the number line. There are 3 arrow marks on the number line, each arrow covering ‘6’. The third arrow reaches at 18.
Therefore, 6 × 3 = 18. This is represented on the number line.Example 2 :
5 friends went to a shop. Each one bought 12 mangoes. How many mangoes did they buy altogether?
× =
Know this : Z is used to represent
Integers. Zahlen is the German word for Integers. Integers are also represented by ‘I’.
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Multiplication of a positive integer by a negative integer :Nidhi was fond of Almonds. Her mother kept some almonds
in a container. Nidhi ate 2 almonds each day. Every day the number of almonds is reduced by 2. After three days the container was empty!
How many almonds were there in the container?Let us denote the reduced number
of almonds by a negative integer –2
–2, –2, –2. Now add these integers. (–2)+(–2)+(–2) = –6
∴ 3 × (–2) = –6
Nidhi ate 6 almonds in 3 days.
So, there were 6 almonds in the box.
Let us mark this on the number line.
-6 -5 -4 -3 -2 -1 0-7
(–2) × 3 = –6Example 1 : Find 5 × (–4)
We can simply take "5 time of −4", like this:-20 -16 -12 -8 -4 0 4 8
1×-4=-4 2×-4=-8 3×-4=-12 4×-4=-16 5×-4=-20
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Example 2 : Let us try this; 4 × (–5) = ?
-20 -15 -10 -5 0 5 10
4 × (–5) = –20
If we multiply a positive integer by a negative integer we get a negative integer.Example 3 : (+5) × (–6) =?
+5 × (–6) =–30
This means 5 times –6
5 × (–6)=(–6) + (–6) + (–6) + (–6) + (–6)= –30
I have neither positive sign, nor negative sign. I am free from sign
Have I any sign?
Ha ha ha
...No sign
Multiplication of a negative integer by a positive integer:What happens if we multiply a negative integer by a
positive integer?Look at this pattern , 5 × 4 = 20 4 × 4 = 16 = 20 – 4 3 × 4 = 12 = 16 – 4 2 × 4 = 8 = 12 – 4 1 × 4 = 4 = 8 – 4 0 × 4 = 0 = 4 – 4–1 × 4 = 0 – 4 = – 4–2 × 4 = –4 – 4 = –8–3 × 4 = –8 – 4 = –12–4 × 4 = –12 – 4 = –16–5 × 4 = –16 – 4 = –20
Thus we get (–5) × 4 = –20
Do you know which is the
largest negative integer?
yes, it is -1
my friend
Know this
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When we multiply a negative integer and a positive integer, we multiply them as whole numbers and put a negative sign to the product. We get a negative integer.Example 1: (–6) × 7 =? Multiply the absolute values.
6 × 7 = 42
5 × 7 = 35 = 42 – 74 × 7 = 28 = 35 – 73 × 7 = 21 = 28 – 72 × 7 = 14 = 21 – 71 × 7 = 7 = 14 – 70 × 7 = 0 = 7 – 7 –1 × 7 = 0 –7 = –7
–2 × 7 = –7 – 7 = –14
–3 × 7 = –14 – 7 = –21
–4 × 7 = –21 – 7 = –28
–5 × 7 = –28 – 7 = –35
–6 × 7 = –35 –7 = –42 ∴ (–6) × 7 = –42
If we multiply two integers having unlike signs then, their product will be always negative.
Example 2 : (–8) × 2 = –16
Write in the pattern as shown above.Example 3:
A diver dives into the sea. He dives at a speed of 9m per minute. Find out the position of the diver in the water after 6 min if he maintains the same speed.
(The person who dives in the sea is called scuba diver)
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If we consider the water level as zero (0), under water diver’s position can be taken as negative number.
Here 9 is negative number.
Therefore, in 6 minutes divers position will be = –9 × 6 = –54
The diver is 54 m below the water level or he will be at a distance of –54 metre.
Multiplication of an integer by '0'
Observe the following pattern
6 × 2 = 12 = 12 - 0
5 × 2 = 10 = 12 – 2
4 × 2 = 8 = 10 – 2
3 × 2 = 6 = 8 – 2
2 × 2 = 4 = 6 – 2
1 × 2 = 2 = 4 – 2
0 × 2 = 0 = 2 - 2
0 × 2 can be written as 2–2
∴ 0 × 2 = 0
When an integer is multiplied by zero we get zeroExample :
i) 5 × 0 = 0
ii) –10 × 0 = 0
iii) 0 × 0 = 0
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Multiplication of a negative integer by a negative integer: Read this:
I eat.I do not eat.I fast.I do not fast.
Which integer is greater than any negative integer and less than any positive integer?
Hey it's zero my friend.
The first and the last sentences give the same meaning.And the second and third give the same meaning..If I say, ‘I do not disagree’ that means ‘I agree’.We can conclude that any sentence with double negative,
gives positive meaning.Consider x = 3 × 5
x + 0 = 3 × 5 + 0 Let y = 3 × (–5) and its additive inverse (–3) ×(–5)
x + y – y = 3 × 5 + 3 × (–5) – [3 × (–5)]
= 3 [5 + (–5)] – [3 × (–5)] (x + y is simplified)
= 3 × 0 – [3 × (–5)]
= 0 – [3 × (–5)]
– 3 × (–5)
∴ 3 × 5 = –3 × –5
15 = –3 × –5
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By generalising this a × b = a × b + 0 = a × b + a (–b) – a (–b) = a ( b + –b) – a (–b) = a (0) – a (–b) = 0 –a (–b) ∴ ab = (–a) (–b) = –a × –b –a × –b = a × b
The product of two negative integers is a positive integer.Let us multiply the following numbers using a pattern. (–3) × (– 6) = ? Let us write this in the form of a pattern, The product of two negative integers is a positive integer.
We multiply the absolute values and then assign positive sign to the product.
– 3 × 6 = –18 – 3 × 5 = –15 = –18 – (– 3) – 3 × 4 = – 12 = –15 – (– 3)
(recall the additive inverse that you have studied in
6th Standard)– 3 × 3 = –9 =–12 – ( – 3)– 3 × 2 = – 6 = –9 – ( – 3) – 3 × 1 = – 3 =–6 – (– 3) – 3 × 0 = 0 =–3 – (– 3) - 3 × -1 = ?
-3 × -1 = 0 - (- 3) = 0 + 3 = 3 -3 × -2 = 3 - (- 3) = 3 + 3 = 6 -3 × -3 = 6 - (- 3) = 6 + 3 = 9 -3 × - 4 = 9 - (- 3) = 9 + 3 = 12 -3 × -5 = 12 –(-3) = 12 + 3 = 15 -3 × -6 = 15 –(-3) = 15 + 3 = 18
We know that any number multiplied by '0', the product is '0' only. -3 × 0 = 0 = -3 - (-3)
If two integers have like sign then, their product is always positive.
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Example 1 : (–4) × (–3) = ?Multiply the absolute numbers. 4 × 3 = 12But the product of two negative integers is a positive
integer. Thus ( – 4) × ( – 3) = 12
Multiplication of more than two integers:Example 1: 3 × 5 × 6 = ?
3 × 5 × 6 = 15 × 6 = 90Example 2: 3 × 5 × (–6) =?
3 × 5 × (–6) = 15 × (–6) = –90Example 3: 2 × 3 × 5 × 6 = ?
2 × 3 × 5 × 6 = 6 × 30 = 180Example 4: 2 × 3 × (–5) × (–6) =?
2 × 3 × (–5) × (–6) = 6 × 30 = 180Example 5: (–2) × (–3) × 5 × (–6) = ?
(–2) × (–3) × 5 × (–6) = 6 × (–30) = –180Example 6: (–2) × (–3) × (–5) × (–6) = ?
(–2) × (–3) × (–5) × (–6) = +6 × + 30 = +180 = 180
Know this?Step 1 : Find the product of the numerical values of the given numbers.Step 2 : Count the number of negative integers in the given number.Step 3 : If the number of negative integers counted in the step 2 is even, the product is just the product from step 1, if the number of negative integers is odd, the product is the product from step 1 with a negative sign. Verbal problems:Example 1:
Nandan went to a fruit shop.
Do you know?One Dozen = twelve
He bought a dozen bananas. If the cost of each banana is ` 3, how much has he to pay ?
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Solution: One dozen = 12
The cost of one banana = ` 3 To find the amount to pay, multiply 12 by 3 Total amount Nandan has to pay = 12 × 3 = 36.So, Nandan has to pay ` 36 to the shop keeper.
Example 2: A submarine is submerging from the surface at the rate of 15m/minute. At what depth is the submarine after 5 minutes?Solution: Note that moving downwards is negative.Change in position of submarine in one minute = –15 m∴The position of submarine after 5 minutes = –15 × 5 = –75 m
So, the submarine is at depth of 75 m from the surface after 5 minutes.
Exercise 1.2
I. Let us play with numbers and enjoy1) Multiply each number in the row by the numbers in
the column and write the product as shown.
× –3 6 11 –5
7 –21
4 44
–8
–2
12
–9
–4
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2) Multiply the number in the outer circle by the number in the inner circle and fill in the appropriated boxes given outside the circle by their product.
Now select any numbers from the 8
5-5
-67-2
1wheel and multiply by other number or numbers and enjoy.
3) Let us play one more game. i) Take a board marked from (– 50) to 50 as shown in the
figure
–50 –49 –48 –47 –46 –45 –44 –43 –42 –41
–31 –32 –33 –34 –35 –36 –37 –38 –39 –40
–30 –29 –28 –27 –26 –25 –24 –23 –22 –21
–11 –12 –13 –14 –15 –16 –17 –18 –19 –20
–10 –9 –8 –7 –6 –5 –4 –3 –2 –1
9 8 7 6 5 4 3 2 1 0
10 11 12 13 14 15 16 17 18 19
29 28 27 26 25 24 23 22 21 20
30 31 32 33 34 35 36 37 38 39
50 49 47 46 45 44 43 42 41 40
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ii) Take a bag containing two blue dice and two red dice. Let the
number of dots on the blue dice indicate the positive integers and number of dots on the red dice indicate the negative integers.
Rules of the game:Every player has to place the pawn at '0' in the beginning .Every player will take out two dice at a time from the bag
and throw them.After every throw, the player has to multiply the numbers
marked on the dice.The player will move his/her pawn to the product what he/
she will obtain. The player who reaches the greater number in 15 min is the winner.
Exercise 1.3
I. Find the products of the following integers :
1) 5, (– 3) 2) (– 3), 8
3) 7, (– 34) 4) (–3), (– 12)
5) (– 21), 6 6) (–2), (3), –(4)
7) (–3), 4, (–5), (–1)
II. Fill in the blanks appropriately and write the product
1) 4 × + 7 = + 28
2) 3 × + 5 = – .......
3) + 9 × – 7 = .......
4) 6 × – 7 = + .......
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III. Tick the correct answer in the following situations :1) An open tank filled with water is on the terrace of a
building. Evaporation causes the height of the water level to change by 2cm each day.
If the first day the water level is x cm, then after 6 days the level is
a) x cm + (2 × 6) cm b) x cm – (2 × 6) cm
c) x cm – (2 × –6) cm d) 6 × x cm – 2 cm
2) During the past 8 weeks Mr. Girish has deposited ` 7500 each week to his bank account. He has a bank balance of ` 62000 now. So his bank balance 8 weeks ago if no interest is added up in this period was,
a) ̀ (62000 – 8 × 7 × 7500) b) ` (62000 – 8 × 7500)
c) ` (62000 + 8 × 7500) d) ` (8 × 7500 – 62000)
Think: i) Can you write the greatest positive integer and the
smallest negative integer?ii) Which is the smallest positive integer and the biggest
negative integer
Division of integers.Situation 1 :Latha and team : Happy Birthday, Uma.Uma : Thank you .Hey friends,here is a box of sweets.
Let us share these equally among ourselves.Rabia : Ok, I will help you. (counts and says) There are 20
sweets in the box,and we are 5 members including Uma.
Shrinidhi : Ok, fine, divide 20 by 5.
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Latha : So let us take 4 sweets each.Jose : Do you know Latha that 5 ×4 is also equal to 20.Shrinidhi : Multiplication and division are related. Latha, Jose ,Shrinidhi, Rabia: Any way thank you Uma.Uma : Friends, if I have not taken the share, you could
have got more sweets .Latha : yes, 20 ÷ 4 = 5.Situation 2 :Jack and Jill were playing in a village. After some time they were thirsty. They wanted to drink some water. They saw a well and a bucket with a rope. They decided to fetch the water from that well. After drinking water their conversation was like this.Jack : Jill, What may be the depth of this well?Jill : Hmm. Hey, we will ask that uncle. What is the depth of this well uncle?Raju : It is approximately 35 feet from the ground
level dear kids.Jack : What may be the water level uncle?Raju : 28 feetJill : Thank you uncle. Hey Jack, let us assume that
ground level as zero.Jack : Yes, then 28 feet of water means, the water level
is 35–28 = 7 feet below the ground level.Jill : Let us denote it by –7. Jack : Since water is below the ground level, we can
denote the measure (depth) of the water as –28 feetJill : If a bucket fell into the well and reaches the bottom
in 4 min from the top of the water level what will be the speed of the bucket Jack ?
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Jack : Good question Jill. Let us calculate. We have studied speed in Science.
Jill : Yes, speed = distance divided by time.Jack : So speed = 28 ÷ 4 = 7 feet per minJill : But we have considered the depth of the water as–28.Jack : And 4 × ? = –28
Jill : It is –7. Therefore –28 ÷ 4 = –7. The speed of the bucket is –7 feet per min.
Negative sign indicates the movement of the bucket downwards.
Jack : Enough Jill. Let us continue playing. (Both Jack and Jill went back to play).We know the relation between multiplication and division.Example : i) 6 × 2 = 12 = 2 × 6
So 12 ÷ 6 = 2 and 12 ÷ 2 = 6
ii) (– 4) × 5 = – 20 using method applied in (i) we can write (– 20) ÷ 5 = – 4 and (– 20) ÷ – 4 = 5.
Each multiplication statement has two division statements. Fill up the table with your answers.
Multiplication statement Corresponding division statement
( – 8) × 2 = – 16 –16 ÷ 2 = – 8, –16 ÷ – 8 = 2
(– 5 ) × (– 4 ) = 20 20 ÷ – 4 = – 5, ––––––––––––––––
9 × (– 4) = – 36 –36 ÷ – 4 = 9, – 36 ÷ 9 = – 4
( – 7)× (– 6) = 42 –––––––––––––––, 42 ÷ – 7= – 6
3 × (– 10) = – 30 – 30 ÷ – 10 = 3, –––––––––––––––
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Observations from the above table are
• Division involving two integers with the same signs always results in a positive integer.
• Division involving two numbers with different signs always results in a negative integer.
What happens if we divide an integer by zero?
Know thisS. Ramanujan, a great Indian Mathematician asked this question in the class, when he was a student.
Let us see this by an activity : Take 8 paper sheets of equal size.
that is, 8, 1 time = 8Divide 8 into 2 equal groups. 4 + 4 that is, 4, 2 times = 8Divide these each groups into 2 further equal groups. 2 + 2 + 2 + 2 that is, 2, 4 times = 8Again, divide these each groups into 2 further equal groups.
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 that is, 1, 8 times = 8
Cut each of these 8 sheets into 2 pieces.
21+ 2
1+ 21+ 2
1+ 21+ 2
1+ 21+ 2
1+......+ 21+ 2
1+ 21+ 2
1+ 21
that is, 21 , 16 times = 8
Again cut each of these 16 pieces into 2 pieces.
41 + 4
1 + 41 + 4
1 + 41 + 4
1 + 41 + 4
1 +....+ 41 + 4
1 + 41 + 4
1 + 41
that is, 41 , 32 times = 8
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or .0 258 =32 0.25+0.25+………+0.25 (0.25,32 times) = 8
similarly .0 18 = 80 0.1+0.1+……….+0.1 (0.1,80 times) = 8
.0 018 =800 0.01+0.01+……+0.01 (0.01, 800 times) = 8
08= ? How many zeroes add up to 8?
That is no group of zeroes can be found to make eight. So
we cannot divide any integer by zero.(Note: Division of any number by zero is not defined)
Exercise 1.4
I. Simplify the following:
1) (– 40) ÷ (– 10) 2) 34 ÷ (– 2) 3) (– 44) ÷ 4
4) (– 28) ÷ (– 7) 5) 0 ÷ (– 8)
II. Fill in the blanks:
1) (– 45) ÷ ––––– = – 45 2) (– 27) ÷ (– 27) = –––––
3) 30 ÷ –––– = – 15 4) –––– ÷ 4 = – 3
5) –––– ÷ ( – 3) = 10
III. Fill this table:
Division of Integers Quotient
24 ÷ 12
24 ÷ (–12)
(–24) ÷ 12
(–24) ÷ (–12)
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IV. Patil purchased 8 packets of Dharwad peda. Each packet contains equal number of pedas. There were 320 pedas in total. Calculate the number of pedas in each packet.
Properties of integers.a) Commutative property:Addition: 2 + (– 5) = (–3) and (–5) + 2 = (–3).
For any two Integers a and b, if a+ b = b +a then, this property is called commutative property.
When we add integers, order doesn't matter, we get the same answer.
The Integers satisfy the commutative property under addition.Multiplication: 2 × −5 = −5 × 2 = –10
For any two Integers a and b, a × b = b × aOrder doesn't matter when we multiply integers, we get
the same answer.The Integers satisfy the commutative property under
multiplication.Subtraction: 18 – 15 = 3 and 15 – 18 = –3
18 – 15 ≠ 15 – 18When the order changes in subtraction then, the answer
also changes.Therefore the commutative property does not satisfy with
respect to subtraction of integers.Division: 12 ÷3 = 4 and 3 ÷12 = 12
341=
12 ÷3 ≠ 3 ÷12If dividend and divisor are interchanged then the quotient
also changes.Therefore the commutative property does not satisfy with
respect to division of Integers.
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b) Associative PropertyAddition: (4 + −2) + −5 = 2 + –5 = –3.
4 + (−2 + −5) = 4 + –7 = –3.For any three integers a,b and c, (a+b)+c = a+(b+c) When we add three integers, it doesn't matter if we start
adding the first pair or the last pair; the answer is the same.So the Integers satisfy the associative property under
addition.Multiplication: (4 × −2) × −5 = –8 × –5 = 40
4 × (−2 × −5) = 4 × 10 = 40
For any three integers a, b, c; a × (b × c) = (a × b) × cWhen we multiply three integers, it doesn't matter if we
start multiplying the first pair or the last pair; the answer is the same.
The Integers satisfy the associative property under multiplication.
Subtraction: 6 – (3 – 7) = 6 – (–4) = 10
(6 – 3) – 7 = 3 – 7 = –4
6 – (3 – 7) ≠ (6 – 3) – 7It is clear that the Integers do not satisfy the associative
property under subtraction.Division: 8 ÷ (4 ÷ 2) = 4
(8 ÷ 4) ÷ 2 = 1 8 ÷ (4 ÷ 2) ≠(8 ÷ 4) ÷ 2It is clear that the Integers do not satisfy the associative
property under division.c) Additive Identity
(−5) + 0 = (−5)
0 + (−5) = (−5)
For any Integer a, a + 0 = a = 0 + a.
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Zero is the identity element for addition. By adding zero on either side, the number will not change.d) Multiplicative Identity
(−3) × 1 = (–3) = 1 × (−3) = −3
For any Integer a, a × 1 = a = 1 × aOne is the identity element for multiplication. By multiplying
1 on either side, the number remains same.e) Distributive Property
Observe the multiplication of the numbers inside the brackets by the number outside, given below.Example 1 : 3 × (2 + (−4))= (3 × 2) + (3 × (−4))Example 2 : [(–2) × (5 − 7)] = [(–2) × 5)] – [(–2) × (–7)]
For any three integers a, b and c a × (b + c) = (a × b) + (a × c)
a × (b – c) = (a × b) – (a × c)
This is called Distributive property
Exercise 1.5
I. Simplify the following using appropriate property.
1) 5 × [(–4) + 2] 2) (–3) ×( 8 – 5) 3)[(4 ×(–2)]+[(5 ×(–2)]
4) (5 + (–3))–2 5) (–7) + (8 – 3) 6) [7 ×(–2)] + (4 –7) –6 (2 ×(–3)
Verbal problems. Example 1 : Ramya went to a textile shop to buy trouser pieces and shirt pieces to her 3 brothers. The cost of a trouser piece is ` 450 and the cost of a shirt piece is ̀ 320. What is the amount she has to pay in the textile shop for 3 pairs of dresses?
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Solution: The cost of a trouser piece = ` 450
Therefore, the cost of 3 trouser pieces = 3 × 450
= ` 1350
The cost of a shirt piece = ` 320
Therefore, the cost of 3 shirt pieces = 3×320
= ` 960
Therefore, the total amount spent by Ramya in the textiles = 1350+960
total amount = ` 2310.
Example 2 : Sona had ` 1020 in her bank account. She deposited ` 200 on Monday, withdrew ` 500 on the same day, and deposited ` 580 on Wednesday. On Saturday, if she deposited ̀ 1000, find her balance in the account on Saturday.
Solution:
Here depositing the amount is considered as positive and withdrawing as negative.
The amount Sona had in the bank account = ` 1020
On Monday =1020 + 200 – 500 =1220 – 500 = ` 720
On Wednesday = 720 + 580 = ` 1300
On Saturday = 1300 + 1000 = ` 2300
Her balance in the account on Saturday is ` 2300.
Example 3: A company gains profit of `12 per each piece of pen and loss of ` 8 per each box of pencil. The company sells 2500 pieces of pen and 4000 boxes of pencil in a month. What is the profit or loss?
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Solution:The profit by selling the pen = `12
Number of pens sold = 2500 Total profit by selling pen = 2500 × 12
= `30000
The loss by selling pencils = ` 8
Number of boxes of pencils sold = 4000
Total loss by selling pencils = 4000 × 8
= ` 32000
= 32000 – 30000
The total loss by selling pen and pencils = ` 2000Example 4: A quiz question paper contains 12 questions, each carrying 5 marks. Scheme of evaluation is +5 marks for correct answer and for incorrect answer (–2)
i) Bharathi attempts all the questions in which 9 are correct and 3 are incorrect.
ii) Manjula attempts only 10 questions in which 6 are correct and the remaining are incorrect.
Who is the winner in that quiz ?
Solution:
i) Marks given for one correct answer = 5
Marks for 9 correct answer = 5 × 9 = 45
Marks given for one incorrect answer = –2
Marks for 3 incorrect answer = (–2) × 3= (–6)
Therefore, Bharathi's total marks = 45 + (–6)= 39
ii) Marks for 6 correct answer = 5 × 6 = 30
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Marks given for one incorrect answer = –2
Marks for 4 (10 – 6) incorrect answer = –2 × 4= –8
Therefore, Manjula's total marks = 30 + (–8) = 22
39 > 22, Bharathi scores more
∴ Bharathi is the winner.
Exercise 1.6
I. Solve the following Verbal Problems.1) Monika had 5 boxes of chocolates with her. If each box
contains 25 chocolates, find the total chocolates she had.
2) Aftab observed that the normal temperature of Bengaluru during a month was 210C. On a particular day, the changes in temperature as compared with the normal temperature were as follows:
Morning : 5 am: –50 C ; 10 am: + 50 C;
After noon : 12 noon: + 70 C
Evening : 5 pm: + 20 C
Help Aftab to find the actual temperature at 5 am, 10 am, 12 noon and 5pm.
3) Rashmi had `12350 in her bank account. If she withdrew `200 from her account 3 times, What is the balance amount in her account ?
4) The cost of a pen is `8 and the cost of a pencil is `5. Find the cost of a dozen pens and half a dozen pencils.
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5) There are 38 students in a class. The cost of a set of books for one student is `1235. Find the cost of books for 38 sets.
6) In a container 45kg sugar was there. Three years old Meera spilt 750gm of sugar. How many packets of 250gm of sugar can be made from the remaining sugar?
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CHAPTER - 2
FRACTIONS
After studying this chapter you : multiply a given fraction by a whole number,
multiply a fraction by another fraction,
understand the use 'of ' in proper order,
write and use reciprocal of a fraction,
divide a fraction by a whole number and a whole number by a fraction,
divide a fraction by another fraction,
solve statement problems involving multiplication and division of fractions.
You have learnt the meaning of a fraction, types of fractions, equivalent fractions and also addition and subtraction of fractions. Recall them.
Classification of fractions
1) Let us classify the following into proper, improper and mixed fractions.
, ,2 , , , ,2 , , ,43152
5347
911
65
811361315
61
Observe each fraction and write in respective row.
a) Proper fractions : , , ,315265136
b) Improper fractions : , , ,47911
1315
c) Mixed fractions : , ,2 53 2 8
1 461
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2) Improper fractions in list A are converted into mixed fractions and mixed fraction in list B are converted into Improper fractions. Have a keen look at it.
, , ,29
512
615
1112
AList, , ,2 4
3 5 61 4 7
2 6 53
BList
29 4 2
1
512 2 5
2
615 2 6
3
1112 111
1
=
=
=
=
2 43
411
561
631
4 72
730
6 53
533
=
=
=
=
Example 1 : Mohammed has 86 parts of chocolate and Manya
has 43 parts of chocolate. Who has got more chocolate?
What is your conclusion?
a) b)
Both have equal size of chocolates
Hence, 86
43= are equivalent fractions.
Note : )
)
a
b
86
8 26 2
43
43
4 23 2
86
''
##
= =
= =
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Example 2 :
Shaila decorated 125 part of a school hall and Renu
decorated 63 part of the same hall. Find out total part of the
hall decorated by them?
The part of the hall decorated by Shaila = 125
The part of the hall decorated by Renu = 63
Total part of the hall decorated 125
63
125
6 23 2
125
126
125 6
1211
##
= +
= +
= +
= + =
by Shaila and Renu together
Total parts of the hall decorated by both=1211
Example 3 :There was 12
8 kg of sugar in a container. 8
3 kg of sugar is utilized. Find the remaining quantity of sugar in the container?
The sugar in the container 128= kg
The sugar used up 83= kg
The L.C.M. of denominators
12,8 6,4 3,2
22
2×2×3×2=24
Remaining sugar
128
83
12 28 2
8 33 3
2416
249
2416 9
247
##
##
= -
= -
= -
= - =
∴Remaining sugar = 247 kg
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Exercise 2.1
I Classify the following into proper, improper and mixed
fraction.
2 , , , 6 , ,61
127
1618
92
45
203
II Write two equivalent fractions for each of the following.
1) 63 2) 5
4 3) 107
III Reduce the following fractions to the lowest form.
1) 128 2) 30
18 3) 5628
IV Convert the following into mixed fractions.
1) 815 2) 12
25 3) 497
V Convert the following into improper fractions.
1) 3 43 2) 8 2
1 3) 465
VI Simplify.
1) 43
35+ 2) 3 2
143
52+ +
3) 159
52- 4) 4 5
3 265-
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Multiplication of Fractions :A class teacher gave pictorial problems to solve. Sanvi and
Puneeth solved these problems as follows.
1) There are 5 parts each of which is 4
1 . Join them and write
14 + 1
4 + 14 + 1
4 + 14
=(1+1+1+1+14
)= 54
= 54 = 1 1
4
41
41
41
41
41
41 1 1 1 1
45
45 1 4
1
+ + + +
= + + + + =
= =
` j41
41
41
41
41
41 1 1 1 1
45
45 1 4
1
+ + + +
= + + + + =
= =
` j41
41
41
41
41
41 1 1 1 1
45
45 1 4
1
+ + + +
= + + + + =
= =
` j41
41
41
41
41
41 1 1 1 1
45
45 1 4
1
+ + + +
= + + + + =
= =
` j41
41
41
41
41
41 1 1 1 1
45
45 1 4
1
+ + + +
= + + + + =
= =
` j
2) There are 3 parts each of which is 4
3 . Join them and write
43
43
43
43 3 3
49
49 2 4
1
+ +
= + +
=
= =
43
43
43
43 3 3
49
49 2 4
1
+ +
= + +
=
= =
43
43
43
43 3 3
49
49 2 4
1
+ +
= + +
=
= =
43
43
43
43 3 3
49 2 4
1
+ +
= + +
= =
Shreya who is observing the above, posed a question to the teacher.
The above two sums have repetitive addition of fractions. Can we multiply these fractions as we do for whole numbers?
Yes, of course, you are right! You can solve the above problems using multiplication process.
Now, let us learn multiplication operation of fractions.
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Multiplication of fractions by whole numbers :
Example 1 : Observe the figure given above. How much parts of each circle are shaded?
Each circle has 41 part shaded
∴Total parts shaded 41
41
41
41 3#= + + =
Now the total of 3 shaded parts can be represented as below. That is 4
3 part. Hence 4
1 3 41 3
43# #= = .
The numerator is multiplied by the whole number and written as the numerator and denominator remains the same.
Know this : A whole number has 1 in the denominator. Hence, 3 ,5 ,81
315
18
41 3 4
113
4 11 3
43# #
##
= = =
= = =
Example 2 : ?52 2# =
To find the product of this join the strips of 52 . (two times)
52
+ =
52
54
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252
52
12
5 12 2
54# #
##= = =
Product of Whole number and numerator
Product of denominators =
Example 3 : 29
x 3 = ?
+ + =
39
292
13
9 12 3
96# #
##= = =
Product of the numerators →Product of the denominators →
Observe the reduced form
33
96
32
32
#
#= =
Alternate method 29
23
62
93
3×
==
Example 4 :
22
125 2
125
12
12 15 2
1210
2 62 5
65
65
#
###
##
#
#
= =
= =
= =
← Common factor of 10 and 12 is 2
← Answer is in the reduced form
Do it yourself : a) 73 2# b) 10
3 3# c) 2 94# d) 5 15
3#
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Multiplication of improper fraction by a whole number.
Example 1 : Look at this picture convert the fractions in each block into improper fraction.
41
41
41
41
41
41
41
41
41
41
41
41
41
41
41
In each group there are [ 54
14
×5 = ]Total parts 34
5
45
13
4 15 3
415
#
###
=
= = =
[This can be written in mixed form]
∴ 415 3 4
3= ∴ Total parts = 3 43
Example 2 : 3 ?79
79 3 7
913
7 19 3
727 3 7
6
#
# ###
=
= = = =
Example 3 : ?10 56
10 56
110
56
1 510 6
560
5 15 12 12
#
# ###
#
#
=
= = = = =
Do it yourself : a) 58 3# b) 6
13 2# c) 6 35# d) 12 7
8#
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Multiplication of whole numbers by mixed fractions. Observe these examples
Example 1 : How many parts are given in total ?
1 1 41 + 1 1 4
1 + 1 1 41
?2 41 2 4
1 2 41+ + =
Hence 2 41 of 3.
2 341
49
13
4 19 3
427 6 4
3
#
###
=
= =
= =
Step 1 : Convert the mixed fraction into improper fraction.
Step 2 : Then multiply. The total parts = 6 4
3
Example 2 : 2 7 ?54
514
17
5 114 7
598
19 53
#
###
=
= = =
=
Do it Yourself: a) 5 21 7# b) 3 6
2 3# c) 8 2 41# d) 12 3 2
1#
Fundamental operation related to fraction (of)Example 1 : Rekha shaded half part of each circle by red.
19598
548453
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What is the total parts shaded by Rekha?
We say this as 21 of 3. Let us learn the method of using the
"of " in fractionsof2
1 3 21
13
2 11 3
23 1 2
1
#
##
=
= = =
'of 'means multiplication,
consider 'of ' as multiplication use the sign '×'
Alternate method
3 of 21 =
3 21
13
21
23 1 2
1
#
#= = =
∴ Total parts shaded = 1 21
Example 2 : Milk business is carried out in Ravi's house. One day Ravi filled 8 bottles each with 5
4 litre of milk. Find out how many litres of milk did he used to fill 8 bottles
54 l
54 l 5
4 l 54 l 5
4 l 54 l 5
4 l 54 l
Total quantity of milk in 8 bottles
Alternate Method
8 of 54
8
l54
1 58 4
532 6 5
2
#
##
=
= = =
of
l
54 8
54 8 5
418
5 14 8
532
6 52
# #
##
= =
= =
=
Quantity of milk filled by Ravi = 652 Litres
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Example 3 : In a class of 42 Students, 72 of them practised
volley ball, 73 of them practised karate and the remaining
practiced Kabbadi. a) Find the number of students who practiced Volleyball? b) Find the number of students who practiced Karate? c) Find the number of students who practiced Kabbadi? a) The number of students of7
2 42
72 42
72 42
121
6# #
=
= =
=
who practiced Volley ball
∴ Number of students who practiced volley ball = 12
b) The number of students of73 42
73 42
73
142 18
6
# #
=
= = =
who practiced Karate.
∴ No of students who practiced karate = 18c) The number of students who practiced Kabbadi
= Total students - (Students who practice Volley ball + Karate) = 42 - (12 + 18)
= 42 - 30
= 12
∴ Number of students who practiced Kabbadi = 12
Do it yourself :1) a) of43 9 b) of6
2 15 c) of75 4 d) 20of6
5
2) There are 45 students in a class. Of them 53 are girls.
Find out the number of girls and number of boys in this class
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Exercise 2.2
I. Write the sum of the parts mention in figures given below in multiplication form.
Example : 34
64× 2 =+ =
1) 41
41
41
41
41+ + + + = ___× ___
2) = ___× ___+21 +2
121
3)
4) 31
31 + 3
1
31 + 3
1
31 + 3
1
31 = _× _
II. Multiply and write the product in reduced form (simplified form).
A. 1) 72 2# 2) 5
4 3# 3) 92 5# 4) 8
7 6#
5) 9 21# 6) 10 3
2# 7) 15 53# 8) 13 4
1#
B. 1) 2 41 5# 2) 3 2
1 8# 3) 6 53 4# 4) 9 2 2
1#
5) 3 4 53# 6) 2 110
3# 7) 14 2 71# 8) 30 310
7#
+ = ___× ___
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III. Simplify and write the answer in its lowest form.
4) 1) 6 of 32 2) 10 of 5
3 3) 1 of6 81 4) 28 of 7
2
5) 20of43 6) 6of8
2 5 7) 6of107 8) 4of9
5
3) 1) Madhu scored 53 of 50 marks in maths test. Find the
marks scored by her
2) A school arranged an excursion for 60 students. The students were asked to bring lunch for themselves. 12
3 of them brought Chapathis, 5
2 of them brought Pulao and 20
7 of them brought Chitranna. Find the
a) number of students who brought Chapathis
b) number of students who brought Pulao
c) number of students who brought Chitranna
Multiplication of a fraction by another fraction
Reshma brought 43 kg of sugar from a shop. 2
1 of this was taken by her neighbour. Find how much sugar the neighbour took?
The quantity of sugar brought by Reshma is 43 kg.
The quantity taken by the neighbour of kg
kg
21
43
21
43
83
#
=
=
=
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There are two fractions in this problem and we will learn to find out the product of it. Let us understand the method of multiplying a fraction by another fraction.
Example 1 :
What is the 41 part of shaded region in figure?
Shaded region 21=
41 of the shaded region is = of4
121
It means divide the shaded region into 4 equal parts. Take
one part from it. of41
21 is 8
1 the part of the whole thing.
Multiply the numerators of both the fractions and write as the numerator
Multiply denominator of both the fractions and write as the
denominator.
of
of
41
21
41
21
41
21
81
41
21
81
#
#
`
=
= =
=
Let us solve the problem of Reshma discussed above The quantity of sugar brought by Reshma = 4
3 kg
The quantity of sugar taken by the neighbour = of21
43
43
21
4 23 1
83
#
##
=
= = kg
numeratordenominator
∴The quantity of sugar taken by the neighbour 83= kg
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Verify : of43
21 and of2
143 . Are they equal?
Do it yourself : a) 31
41# b) 5
331#
c) 65
72# d) 8
352#
In our daily life, we have to multiply proper, improper and mixed fractions. Let us learn it now.
Example 1 :
Veena bought 4 21 m cloth to prepare a doll. She used 3
1 of the cloth to prepare one doll. Find the length of cloth she used to prepare a doll?
The length of the cloth bought by Veena 4 21= m
The length used for preparing one doll of31 4 2
1
31 4 2
1
31
29
3 29 1
69
2 33 3
23 1 2
1
#
#
##
#
#
=
= =
= = =
[convert 421 into improper fractions]
[taking common factor]
∴ Cloth used for making one doll = 1 21 m
Example 2 :
The rate of 1 metre ribbon is ` 4 53 . Gowthami purchased
2 21 metre length of the ribbon. Find the cost of the ribbon she
purchased
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The cost of 1m ribbon = ` 4 53
The cost of 2 21 m ribbon = ` 4 5
3 2 21# ←[Here there are two
mixed fractions]
5
5232
1
1
#= ← [Both converted into improper fractions and multiplied]
232 = 11 1
2
∴Cost of the ribbon is ` 11 12
Exercise 2.3
I. Multiply and write the product in reduced form.
1) 65
32# 2) 8
743# 3) 5
497# 4) 5
935#
5) 31
97# 6) 3
11103# 7) 7
459# 8) 2
11213#
II. Multiply and write the product in its lowest form. 1) 4
3 251# 2) 5
4 365# 3) 3
5 2 21# 4) 7
3 4 31#
5) 4 53 2 2
1# 6)5 41
23# 7) 3 9
5 5# 8) 6 41 35
1#
III. Do it yourself:
1) 3 52 4# 2) 4 3
253# 3) 5 4
3 2 21# 4) 25 3 2
1#
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IV. Solve the problems.
1) A school van travels 8 km per litre of diesel. If it has 8 43
litres of diesel, find the distance it could travel.
2) The cost of one litre of milk is ` 34 21 . Find the cost of 12
litres of milk.
3) Roshani started reading a historical novel at the rate of 1 4
3 hour a day. She completed reading the book in 6 days. Calculate the number of hours she took to read novel completely.
4) The cost of 1m zip is ` 5 41 . Find the cost of 8 5
4 m zip?
5) A thin rectangular sheet of metal has 3 21 m length and
2 21 m breadth. Calculate its area.
[ Area of a rectangle = length × breadth]
Division of fractionsRadha distributed 4 biscuits among her friends such that
each got half of a biscuit. How many friends of Radha got 21
biscuits. This is shown below.
Total Biscuits When they are cut into halves
When distributed
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The 4 biscuits are distributed among 8 friends then each will get half bisect.
It can be written as 4 21 8' =
That means 21 8 4# =
Let us study the operation used here.When 4 is divided by 2
1 means 4 12# getting 8 halves
i.e, 4 is multiplied by 2
4
421
12
18
8
'
#= =
=
By inter changing the numerator and denominator of 2
1 we get 12 .
This is the inverse of 21
Reciprocal form (or) Inverse form
When the numerator and the denominator of fractions are inter changed we get inverse form of the original fractions. Observe the example :-
Reciprocal of 53 is 3
5 Reciprocal of 87 is 7
8
Reciprocal of 49 is 9
4 Reciprocal of 32 is 2
3
• What is the inverse of 2 43 ?
First, convert the mixed fraction into improper fraction. Find the inverse of it.
Improper of 2 43 = 4
11
Inverse of 411 is 11
4
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• What is the inverse of 8?
8 is written as 81 (8= 8
1 )The inverse of 8 is 8
1
Observe the product when a fraction is multiplied by its inverse.
What is your inference?
Note: Any number except zero is multiplied by its inverse the product is equal to 1.
Do it yourself: Write the Inverse of the following a) 4
3 b) 95 c) 13 d) 3 4
1
Division of a whole number by a fractionExample 1 :
Simplify : 8 43
8 34
18
34
332 10 3
2
'
#
#
=
=
= =
Step-1 Convert division sign as multiplication sign and write the reciprocal of divisor .
Step-2 Continue multiplication process .
1
1
1
41
14
44
11
52
25
1010
11
32
23
66
11
5 51
55
11 1
#
#
#
#
= = =
= = =
= = =
= = =
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Example 2 : How many 41 part can be obtained from three
circular discs?
From three discs we can get 12 equal parts each of which are 4
1 of the circle.
This can be written as
3 41' Step 1 : convert the division symbol into
multiplication symbol by writing the reciprocal of the divisor
3 14#= Step 2 : then proceed with the multiplication
process
1 13 4
112 12
##= = =
∴ Number of 41 parts got from 3 circular discs = 12
Example 3 : Simplify : 12 265'
When the divisor is a mixed fraction convert it into improper fraction and then solve.
12 2
12
12
65
617
176
1772 417
4
'
'
#
=
=
= =
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Do it yourself: a) 6 53' b) 8 4
1' c) 2 53'
d) 9 73' e) 15 2 2
1' f) 12 3 41'
Dividing a fraction by a whole number
Example 1 : 53 2'
Here divisor is 2, multiply dividend by the inverse of 2.
53 2 5
321
103' #= =
Similarly
1) 587
87
51' #= =
2) 1065 ' = =
Example 2 : Observe multiplication of a mixed fraction by a whole number. What would be done to divide mixed fraction? Convert mixed fraction into improper fraction and proceed.
5 41 9
421 9 4
2191
3621
'
' #= = =Similarly
1) 3 5 521
27
27
51' ' #= = =
2) 4 732 ' = = =
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Division of a fraction by a fraction
Example 1 : 41
53' , 4
1 is divided by 53 Here 4
1 is to be multiplied by the reciprocal of the divisor 5
3 .
Then 41
53
41
35
125' #= =
Example 2 : 32
83'
32
83
32
38
916 1 9
7' #= = =
Similarly 1) 107
31
107
13' #= = =
2) 65
73' = =
Try this : a) 98
72' , b) 5
3109' , c) 2 4
3125' , d) 11
521'
Exercise 2.4
I. Simplify.
1) 9 21' 2) 13 4
3' 3) 15 61'
4) 20 73' 5) 15 2 3
1' 6) 10 3 72'
II. Write the inverse of these.
1) 52 2) 9
7 3) 121
4) 261 5) 9 6) 4 3
2
III. Simplify.
1) 32 5' 2) 8
7 3' 3) 76 13'
4) 59 4' 5) 2 4
3 7' 6) 3 21 14'
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IV. Simplify.
1) 85
31' 2) 6
572' 3) 9
458'
4) 54 1 2
1' 5) 3 43 1 3
2' 6) 3 72 15
1'
V. Solve the problems.
1) The cost of 6 chocolates is ̀ 10 21 . What is the cost of one
chocolate?
2) A school was provided with 12 43 l of milk for a day. Each
child was given 203 l of milk. Find out how many children
got the milk?
3) Two students together purchased 15 note books. If the total cost of 15 note books is ` 142 2
1 , find the cost of each note book.
4) How many packets are required to fill 10 54 l of curd so
that each packet contains 52 l ?
Mixed OperationsWe come across many instances in our daily life,
where we use different operations on fractions.
Let us take some examples :
Example 1 : Asma had ̀ 3 21 . She bought 1 pencil for ̀ 2 4
3 and one rubber of ` 2
1 .What is the amount left with her?
Solution : You have to subtract the total amount spent for pencil and a rubber from `3 2
1 . Observe the method to solve the different operations.
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3 12 - (2 3
4 + 12 )
= 72 - (11
4 +1 ×22 ×2)
= 72 - (11
4 + 24 )
= 7 ×22 ×2 - 13
4
= 144 - 13
4 =14 -134 = 1
4
← L.C.M of the denominators of fractions which are to be added
← Equalising the Denominator
∴ Remaining amount with Asma = ` 41
BODMAS means divide first, then multiply, then add and at last substract.Example 2 : A tailor bought 12 pieces of cloth, each measuring
52m. He stitched two curtains each measuring 2 8
1m. Calculate the length of the remaining cloth.
Solution : Total length of cloth 52 12#=
length of the cloth used for stitching two curtains 2 281 #=
Bought cloth - used cloth Remaining length of the cloth = ( 2
5 ×12) - (2 18 ×2)
Observe how BODMAS rule is used here. First we have to multiply and then subtract.
= ( 25 ×12) - (2 1
8 ×2)= ( 2
5 ×121 ) - (17
8 × 21 )
= 245 -17
4
= 24 ×45 ×4 -17 ×5
4 ×5
= 9620 -85
20
= 96 -8520 = 11
20
∴Remaining length of the cloth = 1120 m
← The denominators are equalized
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Example 3 : Simplify : 2 21
21
41
52
21# '+ -
In the above sum there are multiplication, addition, division and subtraction.
RememberAccording to BODMAS rule we have to do division first then multiplication, addition and finally subtraction.
Solution: 2 12 × 1
2 + ( 14 ÷ 2
5 ) - 12
Step 1 41
52
41
25
85' #= =2 1
2 × 12 + ( 5
8 ) - 12
= (2 12 × 1
2 ) + 58 - 1
2
Step 254 + 5
8 - 12
= ( 54 + 5
8 ) - 12
2 21
21
25
21
45# #= =
Step 3815
21-
815
2 41 4
815
84
815 4
811
##-
= -
= - =
45
85
4 25 2
85
810 5
815
##+ = +
= + =
Step 4 1811
83=
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Exercise 2.5
I. Simplify. 1) 8
741
21#- 2) 2 5
353 1 2
1103#+ - 3) 2 4
143
41 2 2
1# '-
4) 4 21
41
51
53'- + 5) 2 5
353
27#+ 6) 15
354
31
41' #+
II. Solve these problems.
1) A school bought 15 m ribbon for a function. Out of
which m441 , is used for arch (toran), 410
7 m is used
for badges, remaining length is used for decorating the
manuscript magazine. Find the length of the ribbon
used for decorating the manuscript magazine?
2) Organiser of a function bought 3 21 kg of sugar 4 times.
In that, 9 43 kg of sugar was used. What is remaining
quantity of sugar?
3) Raju needs 2 21 m length of cloth for stitching a shirt.
The shop keeper has ' 21 m' scale. How many times has
he to use this scale to measure 2 21 m cloth?
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CHAPTER - 3
RATIONAL NUMBERS
After studying this chapter you will: know the meaning, standard form and equivalents of
rational numbers, develop the skill of representing the rational numbers
on a number line, know the comparison of rational numbers, method of
finding the rational numbers between any two rational numbers,
know about operations on rational numbers (Addition, Subtraction, Multiplication, Division) and also know the method of solving verbal problems related to them,
know the method of writing rational numbers in decimal form,
know the multiplication, division operation of decimal numbers. Also know the method of solving problems related to them,
understand the method of converting measuring units (about measurement of length and mass).
Meaning of Rational numbers:You already know about natural numbers (N),whole
numbers (W), integers (Z). N = { 1, 2, 3, 4...............}
W = { 0, 1, 2, 3..............}
Z = {...................-4, -3, -2, -1, 0, 1, 2, 3........................}
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In the beginning only natural numbers were there in use '0' (zero) is invented to represent the non existence of an object. In this way whole numbers came into existence. Integers set developed because negative numbers also occur, while finding the difference of two natural numbers.
(Example : 3-5=-2, 10-15=-5), a new set of numbers like - 5 is obtained, these are called negative integers. So, due to the necessity, the development of integers set came to existence.
Know this : A well defined, related objects, ideas, or numbers represents a set.
In the same way, while dividing one integer by another integer, a new set will be obtained. Example : ,5
3710- . These
numbers are not in integer set. So the set of rational numbers (Q) came into existence.
We have already learnt in the previous class about fractions. Fraction should be in the form p
q , where q ≠ o and p, q are natural numbers. But while subtracting fraction of the form
pq possibility of obtaining negative numbers is as
shown in the example. So, a new set of numbers is obtained in which -5
10 is neither fraction nor integer.
All the numbers of the form pq are rational numbers.
In pq , q ≠ o and p,q are integers.
Set of rational numbers is represented by a letter 'Q'.Q = {contains all the numbers of the form p
q , q ≠ 0 and p, q are intergers.}
Example : , , , , , , ...etc32453074
715
94103- - -
-
We can write all natural numbers, integers and whole numbers as rational numbers.
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Example : 3 ,12 ,8
0 ,0 ,0
, ,
13
112
18
10
20
100
5 15 7 1
7 10 110
= = =
= = =
- = - - = - - = -
The set of rational numbers includes natural N
WZQ
number, whole number and integers.
Therefore, it is the biggest set.
Think! All fractions are rational numbers. But all rational numbers are not fractions? Give reason.
Know this : rational number is derived from the word ratio p
q = p : q
Standard form (Simplest form) of rational numbers.Observe these rational numbers. , , , ,5
341107
73114-
In these numbers the common factor of numerator and denominator is 1 (or H.C.F is 1). These rational numbers are said to be in standard (simplest) form.
Consider 104 (Divide both numerator and
denominator by 2)
104
52` = Standard form of is10
452
2515 = 25
1553
5
3
= . (Divide both numerator and denominator by 5)
53 is standard form of 25
15
3627
43- = - (Standard form) 36
2743- = - (Divide both numerator
and denominator by 9)
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Standard form of is3627
43- -
Standard (simplest) form of rational numbers is obtained by dividing both numerator and denominator of rational number by their HCF
Example : a) 5522
52- = - b) 30
16158
-- = c) 39
2632- = - d) 63
392113=
Equivalent rational numbers :
Recall the method of obtaining equivalent fractions. In the same way equivalent rational numbers can be obtained by multiplying or dividing both numerator and denominator of a rational number by the same non-zero integer.
Example 1 :Write 4 equivalent rational numbers of 32 .
a) 3 22 2
64
## = b) 3 3
2 396
## = c) 3 4
2 4128
## = d) 3 5
2 51510
## =
Example 2 : Write 4 equivalent rational numbers of 75-
a) 7 25 2
1410
##- = - b) 7 3
5 32115
##- = -
c) 7 45 4
2820
##- = - d) 7 5
5 53525
##- = -
Example 3 : Write 4 rational numbers equivalent to 8040 .
a) 80 240 2
4020
'' = b) 80 4
40 42010
'' =
c) 80 540 5
168
'' = d) 80 10
40 1084
'' =
Know this: If p1
q1 and
p2
q2 are equivalent rational numbers.
Then p1 × q2 = p2 × q1.
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Representing (locating) rational numbers on a number Line :
Already, you know the method of representing whole numbers and integers on a number line:
On the number line, positive integers are on the right side and negative integers are on the left side of zero at equal distance.
Example 1 : Represent 53 on a number line.
To represent a rational number on the number line, divide the unit length on the number line into the number of parts as in the denominator of the rational number. Then mark the number of parts as in the numerator of the rational number.
0
0 +1-1
53+
Note that the distance between 0 and 1 is divided into 5 equal parts
Example 2 : Represent 25 on a number line
0
-3 -2 -1 0 +1 +2 +3
25
Here each unit is divided into 2 equal parts.
Example 3 : Represent 47- on a number line. Here each unit
in divided into 4 equal parts.-2 -1 0 +1 +2
047-
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Exercise 3.1
I. State whether the following are true or false, if false correct it and write.
1) - 12 is a rational number.
2) 03 is a rational number.
3) 43- is a fraction.
4) 75- is a not rational number.
5) 40 is a rational number.
6) 85-- is a positive rational number.
7) 2812 is in the simplest form.
II. Write these rational numbers in standard form.
1) 2028 2) 150
120 3) 4015- 4) 56
32--
III. Write 4 equivalent rational numbers to each of the following.
1) 92- 2) 10
3 3) 54- 4) 96
80
IV. Represent these rational numbers on a number line. (Use different lines for each set)
1) , ,73
7475- 2) , ,4
74549-
3) , ,35
3238- 4) , ,5
65758-
V. Group the following rational numbers as positive and negative rational numbers.
, , , , , ,54
67
35
710
97
152
611-
- -- -
--
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VI. Fill up the blanks. 1) 4
3 12= 2) 65
24- = 3) 28
12 3- = -
4) 2015
4= 5) 18 149- = 6) 12 2
5- = -
Operations on rational numbersYou have learnt addition, subtraction, multiplication
and division of integers and fractions. Now, let us do these operations with rational numbers.Addition : Rule used in the addition of fractions is to be used here also. If denominator is same, then write the same denominator and add the numerator and write as numerator.
Example 1 :
If the denominators are different, take the L.C.M of the denominator, convert them to have the same denominator and then add.
a) 32
31
32 1
33 1+ = + = =
b) 52
51
52 1
53+ = + =
c) 87
82
87 2
85- + = - + = -
d) ( )107
103
107 3
104
52+ - = + - = =
Example 2 :
a)32
41
128 3
1211+ = + = LCM of 3 and 4 = 12
b) (-35)+ 7
10
-6 + 77 =
110
LCM of 5 and 10 = 10
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c) (-56)+(-4
5)
ú25 + 2430 =
4930
LCM of 6 and 5 = 30
d)35
21
43
1220 6 9
1235
+ +
= + + =
LCM of 3,2,4 = 12
2 × 3 × 2 = 12 3,2,42
32
3,1,21,1,21,1,1
Example 3 : Shanthamma bought 21 kg beans, 4
1 kg green
chillies, 23 kg potatoes, 10
1 kg ginger from a vegetable shop.
What is the total weight of vegetables she bought?
Solution:Weight of beans = 2
1 kg
Weight of green chillies = 41 kg
Weight of potato = 23 kg
Weight of ginger = 101 kg
∴Total weight
kg
21
41
23
101
2010 5 30 2
2047
2 207
= + + +
= + + +
=
=
LCM of 2, 4, 2, 10 = 20 2,4,2,10
1,2,1,51,1,1,51,1,1,1
225
2 × 2 × 5 = 20
Example 4 : 45m shirting , 3
2m trouser cloth and 41 m cap cloth
is required to stitch a dress for a child. How many metres of cloth is required for this ?
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Solution:
Cloth required to stitch a shirt = 45 m
Cloth required to stitch a pant = 32 m
Cloth required to stitch a cap = 41 m
∴ Total cloth required LCM of 4, 3, 4
2
m
45
32
41
1215 8 3
1226
122
261
= + +
= + +
=
=
=
4,3,41,3,11,1,1
43
LCM 4 × 3 = 12
Additive inverse of a rational numberWe know additive inverse in integers. In the same way, all
rational numbers also have additive inverse.Example: Additive inverse of 3 is -3 Additive inverse of 1
2 is - 1
2
Additive inverse of -35
is +35
Additive inverse of - 710
is 710
Know this: The sum of a number and its additive inverse is '0' (zero); a + (-a) = a - a = 0, 0 is called additive identity.
Additive inverse of a rational number is the same number having opposite sign in numerator.Additive inverse of a
b is -a
b Additive inverse of -x
y is -x
y
Subtraction :In integers, when we subtract 5 from 8, we write 8 -5, That
can be written as 8 + (-5), which means the additive inverse
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of the number to be subtracted is added.To subtract -4 from 7 means -4 is subtracted from 7.i.e., 7 + 4 = 11 ( Additive inverse of -4 is + 4)
In the subtraction of rational numbers, same method to be followed.
Example 1 : a) Subtract 71 from 7
5 .
= 57 +(- 1
7 ) = 57 - 1
7 = 5 -17 = 4
7 b) Subtract 8
1- from 87 .
= 78 -(- 1
8 ) = 78 + 1
8 = 7 +18 = 81
81 = 1
To subtract rational numbers having different denominator convert their denominator to the same denominator and proceed.Example 2 : Subtract from3
265
65
32- = 6
5 461= - =
LCM of 6 and 3 = 6,32,11,1
32
3 × 2 = 6
Example 3 : Subtract from43
711
711
43- =
2844 21
2823
= -
=
LCM of 7 and 4 = 28
7,41,41,1
74
7 × 4 = 28
Example 4 : Subtract 185 from 9
4-
= -49
- 518
= - 8 - 518
= -1318
LCM of 9 and 18 = 18
9,181,21,1
92
9 × 2 = 18
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Example 5 : A teacher brought 13 43 kg sweets to distribute
among the children, on the occasion of Gandhi Jayanthi. If he distributes 12 8
7 kg sweets to the children, what is the weight of remaining sweets ?Solution :Total weight of sweet brought = 13 4
3 kg
Weight of sweet distributed among the children = 12 87 Kg
∴ Weight of remaining sweet 13 43 12 8
7
455
8103
8110 103
= -
= -
= -
4,81,21,1
42
4 × 2 = 8
∴ Weight of remaining sweet 87= kg
Example 6 : Rahim spent 21 of his income on food, 5
1 part for children's education, 4
1 part for other items from his salary and saved the remaining part. What is the part of salary saved by him?
Solution :
Part of the salary used for food by Rahim in a month 21=
Part of the salary used for children's education 51=
Part of the salary used for other expenses = 41=
∴ Total part of salary spent21
51
41
2010 4 5
2019
= + +
= + +
=
2,5,41,5,21,1,21,1,1
252
2 × 5 × 2 = 20
LCM of 2, 5, 2 = 20
Let total salary be 1
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∴ Remaining part of the salary 1 2019
2020 19
201
= -
= -
=
∴ Rahim saves 201 part of his salary in a month.
Exercise 3.2
1) Find the sum of the following.
a) 52
107+ b) 3
292+ c) 5
365- + d) 7
531
23+ +
2) Subtract.
a) 85 from 4
1 b) 65 from 6
1-
c) 53 from 15
4 d) 87 from 4
3
3) Usha bought 5 43 kg of pulses and 2 3
2 kg of vegetables from the market. What was the total weight of pulses and vegetables she bought?
4) As "Kshira Bhagya" plan a school got 15 43 kg milk powder
per week. 1451 Kg of milk powder is used, what is the
quantity of remaining milk powder?
5) Ravi bought 8 75 m of cloth from a textile shop. In that 4 3
1 m of cloth for his son and 34
3 m cloth for his daughter is used for stitching dress. What is the length of remaining cloth?
6) There are four cows in Leela's house. One day 20 53 l of
milk is collected from them. If Leela sold 18 32 l out of this,
what is the quantity of milk remaining?
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Multiplication of rational numbers :
Rational numbers can be multiplied in the same way as fractions.
Example 1 :
a) 43
13
49# =
Note : While multiplying rational numbers, multiply the numerator by the numerator and the denominator by the denominator b) 3
234
98# =
c) -34 × 5 = -15
4
d) -21
5 ×(- 342
) = 310
e) 43
21
65
4 2 21 5
165
1
2= =# ## ##
Example 2 : If a student reads 52 part of a book containing
100 pages in a day, then how many pages are read by him?
Solution :
Total number of pages = 100
Part of it read by a student = 52 of 100
5
1002 401
20#= = pages.
Example 3 : If a metre of cloth costs ` then, what is the
total cost of 14 21 m of cloth at the same rate?
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Solution : Cost of 1 m of cloth ` 3034
Cost of 14 21 m of cloth = 30
34 � 14
12
= 1234 �
292
= 3567
8 = 445 78
Therefore, the cost of 1412 m cloth = `445
78
Multiplicative inverse (Reciprocal) of rational numbers:
To obtain the reciprocal (multiplicative inverse) of a rational number, interchange the numbers in the numerator and denominator of that rational number.Example : 1) Multiplicative inverse of 4
3 is 34
2) Multiplicative inverse of 25 is 5
2
3) Multiplicative inverse of 103- is 3
10-
4) Multiplicative inverse of 4 is 41
5) Multiplicative inverse of a is a1 . (where a≠0)
Know this : No change in the sign while writing a multiplicative inverse. The product of a number and its reciprocal is always 1 (Identity element of multiplication is 1 i.e; a a
1 1# = ).
Think ! : '0' has no multiplicative inverse. Why?
Division of rational numbers :We know that dividing a fraction by another fraction
means, multiplying the dividend by the reciprocal of the divisor. Division of rational numbers to be done in the same
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way.
Example 1 : a) 43
21' means
43
12
23
2
1=#
b) 52
103'- means
52
310
34
1
2- = -#
c) 65
34
65
43
85
1
2- - = -
-=+' #
d) 83
76
83
67
167
1
2= =' #
e) -45 ÷ 2
3 = -45 × 3
2 = -65
2
f) 1512 10 15 10
112756
252
5
6
25
2- =
-=-
' # = -
2) Find the quantity of milk shared by each student,
if 11 41 l of milk is distributed equally among 45 students.
Solution : Quantity of milk available 11 4
1 l Total number of students = 45
∴ Quantity of milk shared by each student 11 45
45
4545
41
445
41
41
1
1
'
'
#
=
=
=
=
Each student gets 41 l of milk
3) Find the weight of apple got by each, if 3 53 kg of apple
is shared equally by 3 friends.
Solution :Total weight of apple = 3 5
3 kg.
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Number of friends who shared the apple = 3∴ Weight of apple each gets 3 5
3 3
518 3
518
31
56
151
1
6
=
=
=
=
'
'
#
=
∴ Each of the friends gets 151 kg of apple
Exercise 3.4
I) Multiply the following: 1) 4
5103# 2) 8
7425#- 3) 15
7143#-
4) 65
152#- - 5) 9
786
72# #- - 6) 10
935
32# #
II) Divide the following: 1) 9
173' 2) 8
767' 3) 3
295' 4) 5
4107'- -
III) Write the additive inverse of the following:
1) 53- 2) 4
9 3) -10 4) 2
7 5) 51- 6) 0.
IV) Write the multiplicative inverse (reciprocal) of the following.
1) 109 2) 6
5- 3) 1813- 4) 7
5-
V) A box has 15 43 kg sugar. How many packets of each
containing 1 43 kg can be made?
VI) A student requires 2 43 m cloth for an uniform. Find the
total length of cloth required to stitch uniform for 44 such students?
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VII) A rectangular garden has length 351 m and breadth 2 4
3m What is its area?
VIII) In a school 20 54 l of milk is brought for distribution.
If 51 l of milk is given to each student, find the number of
students in the school?To write rational numbers as decimals.
You know already, the method of writing fractions in decimal form. In the same way, rational numbers can be written in decimal form.
Decimal numbers have two parts.In 61.35, 61 is the integer part and 35 is the decimal part. Method of reading 61.35 → Sixty one point three five. Similarly, method of reading 27.834 → Twenty seven point
eight three four.
Activity: Put some flash cards with different decimal numbers in a box. Ask each student to take one card from it and read the decimal number written in the proper way and to say integer part as well as decimal part in it and to cite the place value of each digit.
You know the method of writing fractions in decimal form.Example 1 :
a) .107 0 7= b) .10
27 2 7=
c) .53 0 6= d) .4
3 0 75=
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e) 0.002001 5= f) .3
1 0 3333= ...
g) .65 0 8333= h) .11
1 0 090909= ...
In the same way, any rational number can be expressed in decimal form.
If we observe the above examples while writing a rational number in decimal form its digit may be stopped in one digit or more than one digit or may be repeating.
In examples a , b , c , d and e digits in the decimal part ends. Therefore these are called terminating decimals.
But in example f , g , h one or more than one digits in the decimal part are repeated. Even if the division continues the same digits will be recurring in decimals. We put dots to show that the digits are repeated in the decimal part. These are called, repeating decimals or recurring decimals.
1) Terminating decimals :
a) 3.57 b) 15.035 c) 0.0004321 d) 10.23576
2) Repeating or recurring decimals :
a) 2.544 ................ b) 10.272727 ................
c) 15.145614561456 ................ d) 18.321321321.............
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Think ! All rational numbers can be written as terminat-ing or recurring decimals. But if digits in the decimal part neither end nor repeat then what type of numbers are they. Can they be written in the form b
a ?
Examples of converting rational numbers into decimals
A) .72 0 285714285714= b) 1 1.9
5 555= ...
.914 1 555= ...
c) .4152
1562 4 133= = ...
d) 0.1253 024=
Exercise 3.5
I) Write these rational numbers in decimal form.
1) 54 2) 8
3 3) 165 4) 14
3
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II) Identify terminating, non terminating and recurring decimals. 1) 8
5 2) 121 3) 9
2 4) 161
3) Classify the following decimals as terminating, non-terminating, recurring and not belonging to any of this group of decimals. a) 8.751 b) 2.5444 ...
c) 0.0303... d) 9.2874
e) 2.456731456731456731 ... f) 10.56173824931685...
g) 3.147521896397...
To write the decimals as rational numbers :To convert a decimal into a rational number, write the
numbers given as it is without decimal point as numerator. Then write 1 in the denominator followed by zeroes equal to number of digits in decimal parts.
Example : a) .0 3 103= b) 0.12 100
12= c) 1 .104
1014 1 4= =
Multiplication of decimal numbers :While multiplying decimal numbers find the product of the
numbers without taking decimal point for consideration, later put the decimal point in that product by leaving the digits to right side equal to the number of decimal places both in multiplicand and multiplier. Suppose, there is no sufficient number of digits in the product write zeros instead of that.Example 1 : a) 0.2 × 0.3 = 0.06
b) 0.12×0.6 = 0.072
c) 1.2×1.5 = 1.80
d) 0.003×0.12 = 0.00036
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e) 0.13×0.0005 = 0.000065
f) 3.41×2.678 = 9.13198
Example 2 : What is the area of a rectangle, if its length is 4.7 cm and breadth is 3.5 cm?
Length of a rectangle = 4.7 cm
Breadth of a rectangle = 3.5 cm
∴ Area of a rectangle = Length × Breadth = 4.7 × 3.5 = 16.45 sq. cm
Example 3 : Find the cost of 3.20 m of cloth, if the cost of 1 m of cloth is ` 98.75
Cost of 1 m of cloth = ` 98.75
∴ Cost of 3.20 m of cloth = 98.75 × 3.20
Cost of 3.20 m of cloth = ` 316
Example 4 : For mid - day meal 0.035 kg of dal is given to a child per day. What is the quantity of dal required for 38 children?
Dal given to a child per day = 0.035 kg
∴ Quantity of dal given for 38 children = 0.035×38
= 1.330 kg
Division of decimal numbers :
To divide decimal numbers by integers:Example 1 :
a) . . .3 5 7 73 5 0 5' = = b) . .4
6 4 1 6=
c) . .88 24 1 03= d) . .25
15 625 0 625=
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To divide decimal number by a decimal:
Example 1 :
) ..
.
.
a 0 48 4
0 4 108 4 10
484 2121
##= = =
) ..
..
.
b 1 83 24
1 8 1003 24 100
180324
1018 1 8
10
18##=
= = =
Note: Multiply both numerator and denominator by the same integer to remove decimal points and then divide.
) ..
..
c 12 56 25
12 5 1006 25 100
1250625
21
1
502
25
##= = =
) . .
..
..
d 0 729 1 80
1 800 729
1 80 10000 729 1000
1800729
20081
600
243
200
81
'
##
=
= = =
Observe :To make an integer a number having 1 decimal place must be multiplied by 10, 2 decimal place by 100 ... continued.
Example 2 :
If a bus covers a distance 48.4 km in 1.1 hrs then, find its speed per hour?.
Total distance covered by bus = 48.4 km ..
1 1 1048 4 10
1148444##
=Time taken = 1.1 hrs
Speed= Distance/ Time = .
.
/km hr1 148 4
44
=
=
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Example 3 :
If Kavitha purchased 12.5 m cloth for ̀ 427.50 to stitch the dress for her children, then find the cost of cloth per meter?
Quantity of cloth purchased by Kavitha = 12.5 m
Total cost of cloth = ` 427.50
∴ Cost of cloth per metre = 427.50 ÷ 12.5
12.50 100427.50 100
05171
34.20`
12505
4275 171= =
##
∴ Cost of cloth per metre =
know thisIf two rational numbers of the form b
a , where a < b are multiplied then the product obtained is less than both the rational numbers. If one number is divided by the other, then the quotient obtained is greater than both the rational numbers.
Exercise 3.6
I. Multiply.
1) 8.6×4 2) 3.75×2
3) 4.105×9 4) 2.56×1.3
5) 0.03×1.456 6) 11.2×0.15
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II. Divide.
1) 0.42÷6 2) 0.144÷12
3) 4.97÷10 4) 6.75÷0.25
5) 2.86÷1.3 6) 68.8÷0.16
III. Solve the following Problems.
1) If the cost of sugar per kg is ` 35.45 then find the cost of 20.25 kg of sugar.
2) If a bus travels 4.25 hrs at a speed of 28.25KM per hour then find the distance travelled by it.
3) If a car travels 21.5 km distance per litre of petrol then find the number of litres of petrol required to travel 219.30 km distance.
4) If Sheela purchased 12.5 m coloured ribbon for `257.50 then find the cost of ribbon per m?
Conversion of units of measurementsQuantity of measurement gives numerical value. Quantities
can be compared by units. We have accepted a few unit of measurements internationally. These measurements are called Standard units. Small and big measurements are derived from fundamental units (basic unit).International fundamental(basic) unit of length is metre (m)
International fundamental(basic) unit of weight is kilo gram (kg).
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know this1 inch = 2.54 cm
1 foot = 30.48 cm
1 yard = 0.914 m
12 inches = 1 foot
3 feet = 1 yard
220 yards = 1 furlong
8 furlongs = 1 mile
1 mile = 1.61 kilometer
1 km = 0.62 mile
1 ounce = 28.35 gram
1 pound = 0.45 kg
1 kg = 2.204623 pound
1 kilo metre = 1000 metre
1 metre = 100 cm
1 cm = 10 mm
1 kilo gram = 1000 gram
1 gram = 1000 milli gram
Do you know this?Kilo metreHecto metreDeca metreMetreDeci metreCenti metreMilli metre
÷ 10× 10
That means,1 kilo metre = 10 Hecto metreSimilarly,1 cm = 10 mm
Example 1 : 1) Convert these into centimeter. a) 45 m = 45 × 100 = 4500 cm
b) 2.7 m = 2.7 × 100 = 270 = 270 cm
c) 4 43 m = 4
19 100 19 25 475# #= = cm
d) 3.02 m = 3.02 × 100 = 302.00 = 302 cm
Note : To convert metre into cm it should be multiplied by 100 a 1m = 100 cm 1 cm = 100
1 m
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Example 2 : Convert these into metres.
a) 20 cm 10020
51= = m
b) 2 cm 1002
501= = m
c) 415 cm 4.15100415= = m
d) 0.003 cm 0.003 1001
10003
1001
1000003# #= = = m
e) 4.5 km = 4.5 × 1000 = 4500 m
f) 25 km = 25 × 1000 = 25000 m
g) 2.1 km = 2.1 × 1000 = 2100 m
Example 3 : Write these in cm.a) 6mm
= 6 101
53# = cm
b) 23 mm
= .1023 2 3= cm
c) 276 mm
= .10276 27 6= cm
Example 4 : Write these in km.a) 105 m
1051000
12002121
200#= = km
b) 725 m
7251000
1402529
40#= = km
Example 5 : Write these in kg.a) 720 g
1000720
10072
2518
25
18= = = kg
b) 3150 g
3.1510003150
100315= = = kg
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Example 6 : Write these in grams.a) 1.4 kg
1.4×1000
= 1400 g
b) 10 21 kg
10 21 kg =
221 1000 10500
1
500# = g
c) 3.42 kg
3.42 kg
= 3.42 × 1000 = 3420 g
Example 7 : Write these in grams.a) 740 mg
00
10074
10074
5037= = g
b) 210 mg
00
10021
10021= g
Example 8 : Write 125 mg in kg.gm kg
mg kg
mg kg
kg
125 10000001
400005
1000 11000000 1
1 10000001
80001
540000
1
8000# =
==
=
=
Example 9 : A bus travels 16.2 miles in one hour. Express it in km.
1 mile = 1.61 km
∴ 16.2 miles = 16.2 × 1.61
= 26.082 km
Example 10 :Find the length of the room in cm if its length is 4.5 feet .
1 foot = 30.48 cm
∴ 4.5 feet = 30.48 × 4.5
= 137.160 cm
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Exercise 3.7
1) Express these in Grams
a) 2.51 kg b) 72.5 mg
2) Express these in kilograms
a) 625 gm b) 10825 mg
3) Write these in kilo metres
a) 1450 m b) 17.3 m c) 21350 cm
4) Write these in metres
a) 2.7 km b) 7525 cm c) 1.58 km
5) Write these in cm
a) 12.5 m b) 4.7 km
6) Convert 7.2 miles into km.
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CHAPTER– 4
ALGEBRAIC EXPRESSIONSAfter studying this chapter you : come to know algebraic expressions, definemeaningofalgebraicexpression, obtain algebraic expressions by combining variables
with themselves or with other variables, convert verbal statements into algebraic expression. comprehend themeaning of degree of algebraic
expression,
applyfundamentaloperations to algebraic expressions.
Introduction : Algebraic expressions are core concepts in algebra. These
are powerful tools to solvemany problems in daily life.Expressions enable us to find out solutions for unknownquantitieswhilesolvingequations.Elementary concepts in AlgebraLiteral Numbers:
SumaandMadhuwereplayingwithmatchsticks.TheydecidedtomakethelettersoftheEnglishalphabetbyusingmatch sticks.
Sumatookthreematchstickstoformtheletter'C'
ThenMadhupicked threematchsticks to formanotherletterCandputitnexttothe'C'madebySuma.
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SumaandMadhuwentonformingtheletter'C'byusingmatchsticksandputthemoneaftertheother.
Finally they thought of preparing
a table which shows relation between numberofsticksrequiredandnumberof'C's
No.of'C'formed
1 2 3 4 5 ... 20 ... 25 ... n
No.ofsticksused
3 6 9 12 15 ... 60 ... 75 ... 3n
1×3 2×3 3×3 4×3 5×3 ... 20×3 ... 25×3 ... n×3
Whilewritingthetable,theyrealisedthatthenumberofmatchsticks required is thrice thenumber of 'C's formed,that isNumberofmatchsticksrequired=3×numberof'C'stoform'c'. If 'n' stands for number of 'C's formed, then, number ofmatchsticksrequired= 3 × n = 3n
Weknowthat'n'isaliteralnumberThus,For n=1,thenumberofsticksrequired= 3 × 1 = 3For n=2,thenumberofsticksrequired= 3 × 2 = 6For n=3,thenumberofsticksrequired= 3 × 3 = 9
Here,nisthenumberof'C'sinthepatternandntakesvalues1, 2, 3, 4,....... in the table.The valueofn goesonchanging.Asaresult,thenumberofmatchsticksrequiredalso goes on changing.
nisanexampleofavariable.Literalnumbersareusedas variables also
It can take any value1, 2, 3, 4, ......................
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The word "variable" means some thing that can vary ie., change.Thevalueofavariableisnotfixed.Itcantakedifferentvalues.someexamplesforvariable,
Yah, a,b,c,....x,y,z areusedtorepre-sent variables as wellasunknowns.Youknowthat;
Areaofrectangle=Length× Breadth A = l × b
In the following table different rectangle of same area is listed.Nameofrectangle
Measuresof
length (l) cm
breadth(b) cm
Area A = l × b sq cm
ABCD 20 cm 3 cm 60 sq cm
PQRS 15 cm 4 cm 60 sq cm
KLMN 12 cm 5 cm 60 sq cm
WXYZ 10 cm 6 cm 60 sq cm
The above table shows 'l'hasdifferentvaluesfordifferentrectangles and 'b' also has different values for differentrectangles.Thustheliteralnumberslike'l' and 'b' are called variables.
A variable is a literal number or any symbol which attains different values according to the situation.
Thesymbolsthatareusedtorepresentunknownnumbersare also called variables.
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Know this :Historical Background : Around 300 BC, use of letters to denote unknowns
and forming expressions from them was quite common in India. Many great mathematicians – Aryabhatta (476 AD), Brahmagupta (598 AD), Mahavira (who lived around 850 AD) and Bhaskara II (1114 AD) contributed a lot to the study of algebra. They name dvariables as beeja, varna. The Indian name for algebra is Beejaganit.
Francois Viete, one of the great mathematicians of 16th century, was the first person who used letters to describe general arithmetic patterns.
Constant :Canthenumberofsidesofatrianglebemorethan3 ?
Definitelynot,hence,thenumberofsidesofatriangleisafixednumber;itisconstant.
Canthenumberofverticesofaquadrilateralbelessthan4 ?
Definitely not, hence the number of vertices of aquadrilateralisafixednumberandthusitisconstant.
Whenwewritethenumeral4,itrepresentsnumber4
Whenwewritethesymbol5,itrepresentsnumber5 (fivein english).
A constant is a number or other symbol that represents only one value
5, 9, 10, 3/4, –8, 0.01.......... are constants
CircumferenceDiameter
= π which is a constant
c = 3 × 108 m/s is a constant i.e., (velocityoflightinvacuum)
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Algebraic Term
We have already learnt
2xy, 4pq, x y43 2- , 10,
m are algebraic terms
Algebraic terms have numbers,variables&bothnumbersand
variablesthattheproductof6 and 8 i.e., 6 × 8 = 48 Similarly,thequotientobtained by dividing 18 by 2 i.e., 18 ÷ 2 = 9
Canyoufindtheproductofx and 3 ?
Yes,theproductis3x
Canyoufindthequotientobtainedbydividingyby4 ?
Yes,thequotientis y4These 3x, y4 etc are called algebraic terms. Some more
examplesforalgebraictermsare3x , –6x y, 2x2y, y/2, –2x /3
Notice the constant and the Think x-2, x/y, 1/x are
algebraic terms
variables in these algebraic terms In the algebraic term
3x, 3 is constant and x is variable
In the algebraic term –6x y, -6 is constant and x and y are variables
In the algebraic term 2x2y, 2 is constant and x and y are variables etc.
An algebraic termiseitheraconstantortheproductorquotientobtainedbymultiplicationofdivisionbyconstants.
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Exercise 4.1
I. Make a list of the variable or variables in each of the following algebraic terms 1) 5x 2) –3a 3) 7x y 4) 4
3 x 2y5) 0.8a2b 6) m2n2 7) xyz 8) –5m2np
2. Make a list of the numerical constant in each of the following algebraic terms 1) 2x 2) –5x 2y 3) m2n 4) –8p5) 9p2qr 6) 4
3 x 2y 7) 0.5pqr2 8) 0.008mn
Algebraic ExpressionsHow are mathematical expressions
ALGEBRAISFUN!EXPERIENCE THE THRILLOFFINDING
'UNKNOWNS'
formed?
Considerthefollowingmathematical statements. Thesumofanumberandfiveis59
+ 5 = 59
i.e., 59 = 54 + 5Thereforetherequirednumberis54
Similarly,Theproductofanumberand8 is 88
8 × = 88
i.e., 8 × 11 = 88
Thereforetherequirednumberis11.
If we have toomany unknown quantities, it becomesdifficult to createmathematical statement. To simplifywedenoteunknownquantitiesbythevariablesx, y, z and so on.
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Consider the following exampleA girl is now 10 years old. How old will she be1) 5 years later ? 2) 3 years ago ? 3) x years later ? 4) y years ago ?
Solution : The girl becomes 1 year older every year.1) 5 years later she will be (10 + 5) = 15 years old. 2) 3 years ago she was (10 – 3) = 7 years old. 3) 'x' years later she will be (10 + x) years old. 4) 'y' years ago she was (10 – y) years old. The expressions (10 + x), (10 – y) which contain variables
are called algebraic expressions.
Look at the following expressions
i) x2 ii) 2y2 Know thisThese symbols are read as
:-IIIly- Similarly
i.e. - that is ∴ Thereforebcongruent a because
iii) 3x + 2 iv) 2pq + 7
The expression x2 means, multiplying x by itself. i.e., x × x = x2
IIIly 2 × y × y = 2y2
How is 3x + 2 obtained ?The expression 3x + 2 is
obtainedbymultiplyingx by 3 and adding 2totheproduct.
i.e., (x × 3) + 2 = 3x + 2
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How is 2pq + 7 obtained ?The expression 2pq + 7isobtainedasfollows;Bymultiplyingpandqwegetpq,thenmultiplyingitby2
to get 2pq and 7isaddedtotheproducttogettheexpression.All the above expressions x2, 2y2, 3x + 2, 2pq + 7 are called
algebraic expressions.Look at the Table;
Statement Algebraic expression
8 is added to x x + 8
5issubtractedfromy y – 5
pismultipliedby6 6 × p
'm' is divided by n m ÷ n
Here, we observe according to the statement, operations ofaddition,subtraction,multiplicationordivisionareappliedonvariablesandconstantsresultinginanexpression.
"Algebraicexpressionareobtainedbydoingfundamentaloperations with algebraic terms"
Terms of an algebraic expressionLook at the following expressions
Algebraic Expression
Terms of theExpression
No. of Terms
2x + 3y 2x, + 3y 2
x2 – 4x + 3 x2, – 4x, + 3 3
3ab – 4bc – 6cd 3ab, – 4bc, 6cd 3
–2p2q + 3q2r – rp + 4pqr – 2p2r, + 3q2r, – rp, 4pqr 4
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Now consider the expression, 2x + 3yIn this expression there are two variables x and y and
twonumbers2 and 3. 2xisonetermoftheexpressionand3y is another term. Similarly identify the terms in otherexpressions.Product, Factor and Coefficient :
We know that 4 × 5 = 20
Here 20isproduct,4 and 5arefactorsof20
Whentwovariablesaremultiplied,whatistheproduct?Suppose,wemultiply'x'and'y',whatistheproduct?Wewritetheproductof'x' and 'y' as (x × y) or xy
Similarly,Theproductof2 and p = 2p
Note: The variables written side by side, indicates their prod-uct. 2andparefactorsof2p
Theproductof5 × m × n = 5mn
5,mandnarefactorsof5mn
Theproductof7 x × x × y = 7x2y
7, x,yarethefactorsof7x2y
Diagram of an algebraic expression and its terms
2
8x2y
8 5x x xy y
+5xy -8
Expression :
Terms :
Factors :
8x2 + 5 xy - 8
-4
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Coefficient :Letusconsidertheproduct
of x and 8, 8x. In this 8 is a number,x is a variable. 8x means + 8x, 8 is the numerical factor. It is alsocalled arithmetical factor or numerical coefficient ofx. Similarly, x is called the variable factor or variablecoefficientof8.
x = 1 x
y = + y
z = +1z
When a variable is written withoutitsnumericalcoefficientitsnumericalcoefficientis1
Look at some more examples.
Product Co–efficientNumericalcoefficient
Literalcoefficient
in it
8xy = 8 × x × y
co–efficientofyis8xco–efficientofxis8yco–efficientof8isxy
8
8
1
xyxy
–21x2y=–21x × x × y
co–efficientofyis–21x2
co–efficientofxyis–21xco–efficientofx2 is–21yco–efficientof21 is –x2y
–21
–21
–21
-1
x2
xy
x2y
Exercise 4.2
I. Classify the following expressions into numerical expressions and literal numerical expressions (algebraic expressions):
1) 8 + 5 – 3 2) 3×–8 3) (7 × 6) – 4m
4) 3p + 4q 5) (20 × 7) – (5 × 10) – 45 6) 2y + 6 – 4z
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II. a) Write the terms of the following algebraic expressions: 1) 3x + 4y 2) 2pq – 8qr 3) 3 x 2 – 3x + z
4) ab + bc – ca 5) 9m + 6n 6) – 3xy + x – y
b) Write the diagram of the terms and factors of the following algebraic expressions:
1) 2xy + 5 2) x2 + xy + 4 3) 3p – 5y2 4) 2ab + bc + ca
III. Write the following statements as algebraic expressions:
1) 8 is added to 'x'
2) 7issubtractedfrom'y'
3)pismultipliedby12
4) q is divided by 5
5) 4 times x is added to 3 times y
6) 5timesyissubtractfrom10
7)Theproductofpandqisaddedto3x
8) 3 times 'l'issubtractedfrom5 times 'm'
9)yismultipliedby10 and then 15isaddedtotheproduct
10)Theproductofx and y is divided by z
IV. Write the numerical co–efficient of x in the following
1) 3x 2) –4x 3) x43
4) 10.5x 5) 168x 6) x98-
V. Write the variable coefficient in the following
1) 3m 2) –9xy 3) 16pqr
4) 10c 5) mn45- 6) 0.8 x 2 y
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VI.Writethecoefficientsof27 xyzasdirected
i)Thenumericalcoefficient of xyz is ______
ii)Theliteralcoefficientofxy is ______
iii)Theliteralcoefficientof27z is ______
iv)Theliteralcoefficientof27 is ______
LikeandUnlikeTermsFour Students Rama, Suma, Salma and Robert are
discussing in the class room.
Rama says : 20 pens + 5 pens = 25 pens
Suma says : 50 note books – 5 note books = 45 note books
Salma says : 20 pens + 15 note books = 35
Robert asked whether 35 notebooks or 35 pens? You cannot add like this. Salma reacted why she could not add like this ?
Robert : They are different things, we cannot add or subtract things of different types.
Salma : Yes, we can't add 3 kg of sugar with 2 kg of salt.
While friends are discussing, teacher enters the class room and hangs a chart on board and asked them to
observe the Box –1 and Box –2 in the chart.
Box - 11) x, 3x, -6x2) 2ab, -3ba, 2
1 ab3) y2, 3y2, -52 y2
Box - 21) x, 3y, -6z2) 2ab, -3bc, 2
1 ca
3) x, x2, -2x3
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Teacher : WhattypeoftermsdoesBox1 contain?
Students: Box –1containsthegroupofalgebraictermswiththesameliteralfactorsandthesameexponent.
Teacher : What does Box 2 contain?
Students: Box –2containsthegroupofalgebraictermswitheitherdifferentliteralfactorsorthesameliteralfactorswithdifferentexponents.
Therefore –Groups1, 2, 3 in box 1representgroupofliketerms.Groups1, 2, 3 in box 2 representgroupofunliketerms.
Algebraic terms having the same literal factors with the same exponent are called like terms. Algebraic terms having different literal factors or same literal factors with different exponents are called unlike terms.
Classification of Algebraic ExpressionsObservethefollowingboxes;
Box–1 Box–2 Box–3 Box–4
–3x, 2xy, 7x2y,4pqr, – mn
710-
x+y, 2ab+c,4p2+3y, m–7
x+y+z,2x2–3x+4,ab+bc+ca
5x4+4x3–3x2+2x+8,x5–3x2+7x+8,9p2+q2–rq+pq–8
Box 1: contains expressions having Mono →stands for one Bi →is for twoTri →is for three Poly →is for more
only one term.Box 2: contains expressions having two terms.Box 3: contains expressions having three terms.Box 4: contains expressions having many terms.
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The classification is done in the following manner i)Monomial : An expression having only one term ii) Binomial :Anexpressionhavingtwounliketermsiii) Trinomial :Anexpressionhavingthreeunliketermsiv) Polynomial :Anexpressionhavingmoreunliketerms.
but inAlgebra, irrespective of numberof terms, the expression is known asa polynomial. In general all algebraic expressions with terms having positive exponents are called polynomials.
Note : Inalgebra, liketermscanbeaddedorsubtracted a + 3a = 4a, 3ab + 5ab + 6ab = 14ab ∴ a + 3a is not a binomial, it is monomial. Similarly, 3ab + 5ab + 6ab is not trinomial, it is monomial.
Algebraic Expression
Constant Monomial Binomial Trinomial
Polynomial
Know this :+ and - sign separates algebraic terms.× and ÷ sign do not separate algebraic terms.
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Some special features of a polynomial in Algebra1) Any expression with one or more terms is called a
polynomial. Hence a monomial, a binomial and a trinomial are polynomials.
2) The index or exponent of the variable in each termshouldbewholenumber.
3) Terms like x½, 5x – 2, 1/x, p1/3 are not polynomials.Activity: Identificationofmonomial,binomialandtrinomialexpressions.
Step 1 : Place 3 open boxes A, B, C and label them as monomial, binomial and trinomial respectively.
Step 2 : Placeflashcardscontainingalgebraicexpressionsofdifferenttypesonstoolinfacedownposition.
Step 3 : Call students one by one and ask them to takeonecard,readaloudtheexpressionandputitinappropriate box.
Step 4 : Countthenumberofcardsineachboxandeliminatethecardsthathavebeenwronglyplaced.Makealistoftheobservationsinthetable.
Group No.ofCards No.ofTerms TypeofExp.
A ___________ One MonomialB ___________ Two BinomialC ___________ Three Trinomial
Have you understood on what basis the algebraicexpressions are classified?What is the reason fornaminglike this?
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Powered Numbers :We have already learnt that the Area ofasquare= side × sideIf the lengthofeachsideofasquareis 2 cm then area of the square = 2 × 2 = 22 sq cm = 4 sq cmIfthelengthofeachsideis'x' cm, then
Areaofsquare= x cm × x cm = x2 sq cm
Similarly, Volume o f a cube = length × breadth ×heightIflength,breadthandheight are '2'cmthen, Volume ofcube
= (2 × 2 × 2)cubiccm = 23cubiccm
= 8cubiccm
Ifthelength,thebreadthandtheheightofacubeare'y'cmthen,volumeofthecube = (y × y × y)cubiccm = y3cubiccmorccm. Thus,x2, y3, p4, q6, d7 and m9etc.arecalledpowerednumbersorexponentialnumbers.
Howtoreadpowerednumbers?
• a2isreadas'asquared'or'atothepower2'
• x3 is read as 'xcubed'or'x to the power 3'
• p4 is read as 'p to the power 4'
• d7 is read as 'd to the power 7'
•m9 is read as 'm to the power 9'
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Terms associated with powered numbers :In the powered numbers
Index/Exponent/Power
base
x2
such as x2, y3, n7, r4 etc, the numbers2, 3, 7, 4 written on thetopofrighthandcorneroftheliteralnumbersarecalledindices or exponents or powers.
Thenumbersx, y, m, k are known as bases. The index or exponentindicatesthenumberoftimesthebaseistobeusedasafactor.
a2 = a × a (base'a'isusedasfactortwice)x3 = x × x × x (base xisusedasafactorthrice) (ab)4 = ab × ab × ab × ab (baseabisusedasafactor4 times).
Note : Thesizeofindexshouldbesmallerascomparedtothesizeofbase.
Degree of algebraic expression :We have already learnt about the terms of algebraic
expressionandthepowerednumbersLetusconsideranexpressionx 4 + x 2 – 3x + 4Thetermsofexpressionarex 4, x 2, –3x, 4Whichtermofthisexpressionhashighestexponent?Yes, the term x 4 has the highest exponent i.e., 4Thus,4iscalleddegreeofexpressionx 4 + x 2 – 3x + 4Hencetheexpressioniscalledfourthdegreeexpression.In the expression a2 + 5a3 – 4a – 8 the term 5a3 has highest
exponent 3. Hence, the expression is called third degree expression.
In polynomial having only one variable, the highest exponent of the variable among the terms of the expression is the degree of that expression.
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More about degree of polynomials
consider the expression
3x4 – 4x3y2 + 8xy + 7
Take each term and test its power
In 3x4, 4 is the index and it is 4th degree
In – 4x3 y2,addindexofxtotheindexofy
ie.. 3+2=5. It is 5th degree
In the above expressions the highest degree term is – 4x3y2 and its degree is 5,Thereforethedegreeofthispolynomialis 5. Hence,
The degree of polynomial with more than one variable is the highest sum of indices of the variables of the terms of that expression.
Excercise 4 .3
I. Classify these expressions as monomial, binomial or trinomial expressions 1) 2x + y 2) 2xy 3) 5 + 6a + 4b
4) 3x2 + 5x 5) xyz 6) ab – bc
7) 2y z x37- + 8) 3xp ÷ q 9) a2 – 3ab + c
10) x2 + 4 – 3x
II. Which of the following expressions are polynomials 1) x + 2yq 2) x3 + x – 2 + 1
3) y2 + xy½ – 6 4) m3 + 2m2 + 3m – 4
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III. Mark the group of like terms by 'L' and unlike terms by 'U'.
1) 3x, 5x, 8x 2) x3, –3x2, 8x
3) –8p2, 6p2, 10p2 4) 2ab, 6ba, 8ab
5) 3a2b, – 2ab2, 7a2b2 6) – a3, 2a2, – 8a
IV. Group the like terms together
1) x, y3, 2x, – 7y3, –8x, 23 y3, 6x, –y3
2) 7ab, 6bc, –8ba, 2ca, – 3ab, 2abc, 4ab, 2a2b
3) 7p, 8pq, –5pq, 2p and 3p
V. Write the numerical coefficient of the following
1) 3xy 2) ab32- 3) 0.3p
4) 24xyz 5) –18p2q 6) m np119 2-
VI. Write the coefficient of
1) xy in –3axy 2) ab in 4a2b 3) z2 in p2yz2
4) xy in 10xy 5) 15 in –15p2 6) mn in –mn
VII. Write the degree of each of the following
1) 3x2 2) 3 – p2 + p3 3) 7m2n
4) z – p 5) a2 + 2ab + abc 6) m2 + 2mn2 + n2
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VIII. Complete the following table
Powerednumbers Base Exponent / Index
x5
(ab)6
(13p)9
(–y)10
(xyz)7
(0.59)20
IX.Write the following in exponential form a) p × p × p × p × p b) m × m × m × m
c) ab × ab × ab × ab × ab × ab d) z × z × z × z × z × z × z × z
e) abc × abc × abc f) b×b×b×......10timesBasic operations in algebraic expressions
As inarithmetic,allbasicoperationscanbeperformedonalgebraicexpressionsandtheyobeyallthebasicrulesofaddition, subtraction,multiplication anddivision just likenumbers.Thesignsusedarealsothesameasthatofintegers.
Addition and subtraction of algebraic expressionsLetusconsider3 baskets containing some apples, some
mangoes and some oranges.
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Letthe1st basket contain 8 apples and 8 mangoes,
the 2nd basket contain 6 apples and 4 oranges,
and the 3rd basket contain 5 mangoes and 6 oranges.
Howdowespeakabout the fruits, if the fruitsareputtogether in a new basket?.
Whenthese fruitsareput together inonebasket, thenthere are 8 + 6 =14 apples,
8 + 5 = 13 mangoes and
and 4 + 6 = 10 oranges.
Thesameprocedureisfollowedinadditionandsubtractionofalgebraicexpressions.Weaddorsubtractliketerms.
Additionandsubtractionofalgebraicexpressions,followtherulesofadditionandsubtractionofintegers.
Recall
1) Thesumofpositiveintegersispositive.
2) Thesumofnegativeintegersisnegative.
3) The sumof apositive integer andanegative integerdependsonabsolutevalueoftheinteger.
Thatistosayittakessignofbiggerabsolutenumber.
Addition of algebraic expressionsa) Addition of monomials
i) only like terms can be added
ii) add theirnumericalcoefficientstofindtheirsum
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Example 1 : Add 4x, 8x and 5x
Solution 4x + 8x + 5x All are like terms= (4 + 8 + 5)x Add numerical coefficients of
variable x= 17x And put variable ‘x’ with this
numericalcoefficient
Example 2 : Add 3a, 5b, 8a and 2bSolution 3a + 5b + 8a + 2b
= 3a + 8a + 5b + 2b
= (3+8)a + (5+2)b
= 11a + 7b
There are two variables a and bAddnumerical coefficient of aandput“a”withthenumericalcoefficient and similarly forvariable b.
Example 3 : Add 3x2y, 2xy2 and 7yx2
Solution 3x2y + 2xy2 + 7yx2
= 3x2y + 7yx2 + 2xy2 = (3+7)x2y + 2xy2
= 10x2y + 2xy2
Note : i) x2y = yx2
ii) x2y ≠ xy2
b) Addition of Binomial and Trinomials
i) Row method :Alltermsofexpressionsarewritteninahorizontal line, like terms are written together and are added.
Example 1 : Add (3a + 2b) and (5a + 7b)
Solution : Row method
(3a + 2b) + (5a + 7b)
= 3a + 2b + 5a + 7b
= 3a + 5a + 2b + 7b
= 8a + 9b
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ii) Column method :Expressionsarewritteninsuchthat a way their like terms are arranged one below the other inacolumn.Thenadditionofthetermsisdonecolumnwise.Solution:
3a + 2b ... Addend
5a + 7b ... Addendum
8a + 9b ... Sum
Example 2 : Add (14m2n–10n2) and (10m2n+6n2)
Solution: Row Method
(14m2n –10n2) + (10m2n + 6n2)
= 14m2n – 10n2 + 10m2n + 6n2
= 14m2n + 10m2n –10n2 + 6n2
= 24m2n – 4n2
Column Method 14m2n–10n2
10m2n+6n2
24m2n – 4n2
Example 3 : Findthesumof(4x+3y+5z), (2x+5y–2z) and (3x–4y–z)
Solution: Row Method
(4x + 3y + 5z) + (2x + 5y –2z) + (3x – 4y – z)
= 4x + 3y + 5z + 2x + 5y – 2z + 3x – 4y – z
= (4x + 2x + 3x) + (3y + 5y – 4y) + (5z – 2z – z)
= 9x + 4y + 2z
Column method
4x + 3y + 5z
+2x + 5y – 2z
+3x –+4y – z
9x + 4y + 2z
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Excercise 4.4
I. Find the sum of:1) 4a, 7a, 9a, 21a 2) 6p2, 4p2, –10p2, 12p2
3) 6y, 7x, –3y, –4y, –4x 4) 2a2, a, 3b2, 4a2, 11a, –4b2
II. Simplify:1) 5x + 7x – 3x + 2x – 6x 2) 5a2 + 2a2–a2 + 7a2
3) 5xy + 6xy – 9xy – xy + 10xy – 2xy4) 2x 2 + 4y – 3z + x2 – y + z + 4x2 + 3y + z
III. Find the sum of the following binomials:1) (3x + 4y) and (5x + 7y)
2) (13a – 4b) and (4a – 6b) 3) (xy – 7z) and (6z –10xy)
4) (20a2bc – 3ab2c) and (10ab2c – 3a2bc)
IV. Find the sum of the following trinomials:1) (4x + 3y + 5z) and (2x – 6y – 2z)
2) (p2 + 3q2 + 5r2) and (6p2 – 2q2 + 6r2)
3) (4a + b – 2c) and (3b + 2a + 5c)
4) (4x2y – 5y2z + 6z2x) and (7z2x – 2x2y + 3y2z)
V. Add the following expressions:1) (5mn – 3p), (mn + p), (6mn – 2p)
2) (x + y – z), (2x + 3y + 2z) and (5x – 6y + 3z)
3) (5a2 + 2a–3), (2a2 + 3a + 1) and (3a2 – 4a + 5) VI. Add all terms
5x2, 2x2 + y, 3 x2 + 3y + 7, y2+ 5z + 6y + 4
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Subtraction of algebraic expressions a) Subtraction of monomials
i) Onlyliketermscanbesubtractedii) Subtract their numerical coefficient to find their
differenceExample 1: Subtract5xfrom12x.
Solution: (12x) – (5x)
= (12 – 5) x
= 7x
Example 2 : Subtract5xyfromthesumof7xy and 3xySolution: (7xy + 3xy) – 5xy
= 10 xy – 5 xy
= 5xy b) Subtraction of binomials and trinomials i) Row method :
All expressions are written in a horizontal line and then the termsarearrangedtocollectallgroupsofliketermstogether.Change the sign (from+ to –orfrom– to +)ofeachtermintheexpressionwhichistobesubtractedandthenaddthetwoexpressions. Example 1:Subtract(5x+3y)from(9x+7y) Solution: Row method
(9x + 7y) – (5x + 3y) = 9x + 7y – 5x – 3y = 9x – 5x + 7y – 3y = (9 – 5)x + (7 – 3)y = 4x + 4y
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ii) Column method :
In this method expressions are written in a separate row suchthattheirliketermsarearrangedonebelowtheotherinacolumn.Changethesignofeverytermintheexpressiontobesubtractedbelowtheoriginalsignofeachterm.Thenadd like terms.
Example 1 :Subtract(5x+3y)from(9x+7y)
9x + 7y minuend
(–)5x + (–) 3ysubtrahend
4x + 4y → Difference
Example 2: Subtract(ax + by)from(2ax + 5by)
Solution: Row method Column method
(2ax + 5by) – (ax + by) 2ax + 5 by minuend
= 2ax + 5by – ax – by (–)ax + (–) by subtrahend
= (2 – 1)ax + (5 – 1) by ax + 4 by difference
= ax + 4by
Example 3:Subtract(2a + b + 6c)from(10a – 5b + 12c)
Solution: Row method Column method
(10a – 5b + 12c) – (2a + b + 6c) : 10a – 5b + 12cminuend
= 10a – 5b + 12c – 2a – b – 6c : (–)2a +(–)b +(–)6cubtrahend
= (10a – 2a) + ( – 5b – b) + (12c – 6c) : = 8a – 6b + 6c : 8a – 6b + 6cdifference
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Excercise 4.5
I. Subtract the first term from the second term in the following: 1) 8x, 15x
2) 6a2b, 14a2b
3) –6ab, 10ab
4) 8x2y, –14x2y
II. Subtract :
1) (7x + 4y)from(12x + 6y)
2) (13a – 6b)from(17a – 4b)
3) (– 6m2n – mn2)from( – 10m2n + 4mn2)
4) (7pqr – 8q + 6)from(10 – 4pqr + 3q)
III. 1)Byhowmuch(6x + 10y) is greater than (2x + 6y)
2)Byhowmuch(2d – c) is greater than (5c – 2d)
3)Whatshouldbeaddedto(3x + 4y + 6) to get (4x – 2y – 8)
4)Whatshouldbeaddedto(9m2n2 – 12xy + 10) to get (15m2n2 – 10xy)
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Multiplication of algebraic expressionsLookattheshelf
5 glasses
5 glasses
5 glasses
5 glasses
The glasses are arranged in 4 rows. In each row there are 5glasses.Whatisthetotalnumberofglassesinfourrows?
5 + 5 + 5 + 5 = 20 glasses
Wehave already learnt thatmultiplication is repeatedaddition.
5 × 4 = 20 glasses
Supposeeachrowhas'2x'numberofglasses
Then,totalnumberofglassesintheshelfisgivenby
2x + 2x + 2x + 2x = 8x
2x × 4 = 8x glasses
Recall the signs used while multiplying integers
Positive integer (+a) × Positive integer (+b) = Positive integer(+ab)
Positive integer (+a) × Negative integer (–b) = Negative integer(-ab)
Negative integer (–a) × Positive integer (+b) = Negative integer(-ab)
Negative integer (–a) × Negative integer (–b) = Positive integer(+ab)
Multiplication of monomialsExample 1 :Multiply3x by 5y
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Solution : Multiplythenumericalcoefficientfirstandthentheliteralcoefficient. Note :
3×5≠35But,a × b = abIfvariablesarewritten side by side it indicates theirproduct.
(3x) (5y) = (3 × x) × (5 × y) = (3 × 5) × ( x × y) = 15xyExample 2 :Multiply(–20x) by (+3y)
Solution : (–20x) (+3y) = (–20 × x) (3× y)
= (–20 × 3) (x × y)
= –60xyExample 3 : Findtheproductof(-2x), (–3x), (4x)
Solution : (–2x) (–3x) (4x) = (–2 ×–3 × 4) (x × x × x)
= +24x 3
Example 4 : Findtheproductof3x 2 and 5x 3
Solution : (3x 2) (5x 3)
= (3 × 5) (x2 × x3)
= 15x 5
Multiplication of a binomial by a monomialLetusconsider theproduct 5 × 108 = 540
Wemaywritethisintheform5 × (100+8)
Letuswrite5 ×108 = 5 (100+8)
= (5 × 100) + (5 × 8)
= 500 + 40
= 540
i.e., a (b + c) = (a × b) + (a × c)
RecallLawsofexponent am × an = am+n
Recallthedistributivepropertyofintegers
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Example 1 : Multiply(3x + 2y) by 5y
Solution : 5y (3x + 2y)
= (5y × 3x) + (5y × 2y)
= 15xy + 10y2
Example 2 : Findtheproductof(8y – 3) and 4x
Solution : 4x × (8y – 3)
= (4x × 8y) + (4x × – 3)
= 32xy – 12x
Example 3 : Find theproductof(8x –17y) and ( – 2x)
Solution : (8x – 17y) ( – 2x)
= (8x × –2x) + (–17y ×– 2x)
= –16x2 + 34xy
Multiplication of a binomial by a binomial.
Aswithmultiplying amonomial by a binomial, usingdistributiveproperty,wecanmultiplyabinomialbyabinomial.
Letusconsidertheproductofbinomials.Example 1 : Multiply(x+1) and (x+2)
(x + 1) (x+ 2) = x (x + 2) + 1(x + 2)
= x2 + 2x + x +2
= x2 + 3x + 2
WecanalsouseFOILmethodtomultiplybinomialbyabinomial.
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(x + 1) (x + 2)
F – MultiplytheFIRSTterms x × x = x2
O – MultiplytheOUTERterms x × 2 = 2x
I – MultiplytheINNERterms 1 × x = x
L – MultiplytheLASTterms 1 × 2 = 2
so, (x + 1) (x +2)
= x2 + 2x + x + 2
= x2 + 3x +2
Example 2 :Multiply (2a + 5) and (a – 7)
Solution : (2a + 5) (a – 7)
= 2a (a–7) + 5 (a–7)
= 2a2 – 14a + 5a – 35
= 2a2 – 9a – 35
Example 3 :Findtheproductof(n – 11) and (n – 5)
(n – 11) (n – 5) = n (n – 5) – 11 ( n – 5)
= [(n × n) – (n × 5)] – [(11 × n) – (11 × 5)]
= [(n × n) – (n × 5)] – [(11 × n) – (11 × 5)]
= n2 – 5n – (11n – 55)
= n2 – 5n – 11n + 55
= n2 – 16n + 55
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Exercise 4.6
I. Multiply: 1) 7x , 8y, 5z 2) –2p, 59, –4r 3) (6m) , (–8n) , (+3p)
4) (–3a) , (–b) , (–6c) 5) –3x , 4x , 5x 6) (–2p) (+3p2) (–3p3)
II. Find the product of:1) 6xy, 2yz and 7xz 2) –7ab, 5bc and –6ca3) 4x2y, –3y2z and x2 4) –3a2, –2b3 and –4e4
III. Multiply: 1) (6x + 7y) by 2z 2) (3p – 5q) by –5p
3) (–18x – y) by 2z 4) (2ab + 3bc) by 6abc
IV. Multiply 1) (x + 4) (x + 8) 2) (5n + 2) (n –3)
3) (3a –b) (2a + b) 4) (5x – 2p) (5x + 2p)
5) (2x – 7) (x – 3)
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CHAPTER- 5
PAIR OF ANGLESAfter studying this chapter you : understand the meaning of adjacent angles,
complementary angles, supplementary angles, and vertically opposite angles,
identify adjacent angles, complementary angles, supplementary angles, linear pair and vertically opposite angles,
solve the problems based on adjacent angles, complementary angles, supplementary angles, linear pair and vertically opposite angles.
In our day to day activities the knowledge of geometry is useful. The concepts of geometry such as line and angle are essential in many occupations like carpentry, engineering and tailoring. We come across situations which involve these concepts in our daily life. For example, corners formed by walls, slant position of ladder against the wall, inclination of hill. The angles formed in these examples can be determined.
In your previous class, you have learnt different lines and angles.
Now, identify the types of lines from the following :
(i) (ii) (iii)
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Example : figure (i) is a horizontal line
Figure (ii) -------------- Figure (iii) --------------
Identify the type of angles formed in the following figures?
O
A
B
Fig iOP
Q
Fig ii O
K
L
Fig iii
O
X
Y
Fig iv
Example : In the above figure AOB is an acute angle Some letters in the English alphabet are given below. Mark the angles that you can notice and try to name them as acute, obtuse and right angle.
Example : In a Letter ''T'' the marked angle is a right angle.Remember• A Point is the simplest of all the geometrical figures
Conceptually, a point has no size, but has ''position". A fine dot made by a sharp pencil on a plane sheet of paper is the physical representation of a point.
• A Line has no beginning point or end point. Imagine it is continuing indefinitely in both the directions. It is indicated by small arrows on both the ends.
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• A Line Segment has a beginning point and an end point. It is a part of the line.
X
M
K
L O
N
PQY
All the sides of the triangle KLO are line segments.• A ray has a beginning point but no end point. Think
of Sun's rays. They start from Sun and move in all directions.
Sun
• An angle is formed when two rays meet at a common point. The two rays of the angle are called the arms and the point where they meet is called the vertex.
The size of an angle depends upon the slope
O
A
B
or inclination of one arm on the other arm of an angle.
In the given figure OA and OB are two rays starting from O. The angle between horizontal ray OB and vertical ray OA is right angle. That is AOB 900=
(Then, What do you call those rays?)
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Know this :Measurement of angles : An angle is also defined as the amount of the rotation made by a ray from its initial position to the terminal position. The rotation may be clockwise or anticlockwise.
Initial position
Terminal Position
Initial position
Terminal Position
anticlockwiseclockwise
Note : An angle formed by clockwise or anti clockwise direction, the value of the angle remains the same.
Congruent [Equal] Angles :Look at the following figures, What is the measure of each
angle?Plane - 1
400400
400
Plane - 2
300 300
Plane - 3
500
500
In the Plane -1, the measure of each angle is 400. In the Plane 2, it is 300 and in the plane 3, it is 500.
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The angles having the same measurement on a same plane are called equal ( Congruent ) angles.
CONGRUENT is derived from a Latin word, CONGRUENTEM which means "to agree" this word is used for 'equal'
Pair of angles :In geometry, certain pair of angles can have special
features. Our knowledge of acute, right, and obtuse angles helps to know the relation between pairs of angles.
Adjacent angles :
Look at the given figure, identify the angles, which are placed side by side.
• How many angles are formed in the figure ? Name them.• Identify the angles which are placed side by side in the
figure.• Write the name of the vertex and arms of the angles
placed side by side.Here, ''O'' is the vertex OB and OC are the arms of an angle
BOC . similarly OA and OB are the arms of the another angle AOB .
That is to say;i) They have one common vertex.
ii) They have one common arm separating the two arms.The pair of angles which have one common vertex and one common arm (side), which separates the angles are called adjacent angles.
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In the given figure, AOB and BOC are adjacent angles.
Activity - 1 :
In the given figure, locate the three pairs of
BC
O
A
adjacent angles.
Think• Can two obtuse angles be adjacent angles? Justify• Can acute angle be adjacent to obtuse angles? Justify?
Think : Look at the angles AOB and
AOC Are they adjacent angles ?
Exercise 5.11. Look at these figures and fill up the table given below :
(i)(ii)
(iii)
(iv) (v)(vi)
(vii)
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Figure Angles Arms Common Vertex
Common Arm
Angles are adjacent or
not
1 2 3 4 5
i ,AOB BOC OA, OB, OC O OByesAOB
& BOC
ii
iii
iv
v
vi
vii
2. In the given figure, identify the pairs of adjacent angles, mention its common vertex and common side.
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Angles
Nam
e of
th
e ve
rtex
Nam
e of
th
e ar
ms
Com
mon
V
erte
x
Com
mon
A
rm
Pair
s of
ad
jace
nt
angl
es (Y
es/N
o)
AOB and BOD
AOD and BOC
AOC and BOC
BOC and AOB
AOD and AOB
Group a few more pairs of angles and examine whether they are adjacent angles ?
3. Observe the angles marked with one arc and two arc in the given figure. Are they pairs of adjacent angles ? Why ?
A
DO 1
2 C
F
E
B
R
Q
PO 1
2
X
X
Z
O
1
2
W
O L
J
21
K
R S
P
21
Q
B1 2
C
A D M
O
P
NQ
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Complementary Angles :Observe the figures given below in pairs on the same plane.
plane (i)
AC
Q
RPB
700
200
plane (ii)
A B650
250
Q
PRC
plane (iii)
A
B
C
800
Q
P
100
In the given figure; i), ii) and iii) Find the sum of the measures of pairs of angles.
If the sum of two angles in the same plane is equal to 900 then they are called complementary angles.
That is by adjoining the angles we get a new angle which measure 900
In the above figure; ABC PQR 900+ = . Here, ABC is complementary to PQR and PQR is complementary to ABC .
Note : Whenever two angles are complementary, each angle is said to be complement to the other angle.
Think : If one of the angles is x0 then what would be the complement of this angle ?
Example 1 : Find the complementary angle of 350. Let the complementary angle for 350 is x0. We know that , 350 + x0 = 900
x0 = 900 - 350 x0 = 550 The complementary angle of 350 is 550
R
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Example 2 : If an angle is 5 times of its complementary angle. Find the
measure of pair of angles. Let an angle be x0
Five times of an angle x0 = 5x0
x0 + 5x0 = 900
6x0 = 900
x0 = 900
6 x0 = 150
5x0 = 5 × 15 = 750 The pair of angles is 750 and 150 .Therefore, the complementary angle for 150 is 750 .
Think: * What is the measure of complementary angle of a right angle ?* What is the measure of complementary angle of 600 ?
Think : Can two acute angles be complement to each other ? Can two obtuse angles be complement to each other ? Can two right angles be complement to each other ?
Activity : Take a rectangular sheet of paper. Mark a point 'O' on BC . Fold the sheet along DO and crease them, unfold so as to have figure (III) along the
Fig (i) Fig (ii) Fig (iii)
o o o
vertex as shown in the figure, then measure marked angles formed at the folded part.From this, What do you conclude ?
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Activity 1 :Draw two circles of same convenient radius on a hard
card board sheet.Label the circles as A and B respectivelyDraw a radius to the circumference of a circle ''A''.Mark 100, 200, 300, 400, 500, 600, 700 and 800, angles
respectively on a circle as shown in the figure, using a protractor and a scale.
Cut the sectors of each angle separately, so that you get 8 sectors.
Mark 900 on the sheet having circle B.
403020
8070
60
50
10
A
y
O
Lx
B
Now place one of these 8 sectors, in such a way that one arm of the sector coincides with OX as shown in the figure. Let the sector cover LOX Now try to adjust another sector that coincide LOY
Repeat this activity with other pair of sectors. Then, make a list of 2 sectors such that they can cover the sector XOY
List the complementary angles that you have used in your activity.
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Exercise 5.2
1. In the following figure each pair of angles are complementary to each other, one of the angles is given. Find the other angle.
(i) (ii) (iii) (iv)
2) Find the complement of each of the following angles. a) 220 b) 840 c) 20.50 d) 40.50
3) If an angle is 3 times its complement, then find the measure of the pair of angles.
Supplementary angles :Look at the following pairs of angles on the same plane,O
500 1300
S P Q R
M
fig -i fig -ii
O
7201080
S P Q R
M
fig -iii
O 800
1000
S
P Q R
M
In the given figure i), ii) and iii) find the sum of the pair of angles SPO and RQM
If the sum of angles on the same plane is a straight angle, then, they are called supplementary angles.
Supplementary angles are pairs of angles on a plane whose sum is a straight angle or 1800
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In the above figure; SPO MQR 1800+ =
Hence SPO is supplement to ,MQR and ,MQR is supplement to SPO .
Note : When two angles are supplementary, each angle is said to be the supplement of the other.
Think : If one of the angle is x0 then what would be the supplement of this angle ?
Example 1 : Find the supplementary angle for 130.50 .Let the supplementary angle for 130.50 be x0
We know that the sum of 2 angles in the above case is 1800 i.e. x0 + 130.50 = 1800
x0 = 1800 - 130.50
x0 = 49.50
Example 2 : One angle is supplementary to the other angle. The measure of the bigger angle is 900 more than the smaller angle. Find the measure of each angle.Let the smaller angle be x0 and the bigger angle is x0 + 900
i.e. We know that x0 + (x+90)0 = 1800
x0 + x0 + 900 = 1800
2x0 = 1800 - 900
x0 = 2900
x0 = 450
Therefore , The smaller angle is 450.The bigger angle is x0 + 900 = 450 + 900 =1350
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Think : Whether the following angles are supplementary?
RO
1450450
P
Q
Think : Can two acute angles be supplement to each other ?
Justify. Can two obtuse angles be supplement to each other ?
Justify. Can two right angles be supplement to each other ?
Justify.
Note : Supplementary angles need not be adjacent angles.
Exercise 5.3
1) In the following figure, Which of the following angles together make supplementary angles?
A O B
C
D
E
600700
200
300
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2) Verify the following pair of angles; Are they Supplementary ?
O
700 1100
P
QR
fig. iO
1350
450
P
M
Nfig. ii
O
750 1050
A C
B
fig. iii
A
200 1600
EBC F
D
fig. ivO
8001100
Q
P
S
R
T
fig. v
3) Write the supplementary angle of each of the following angles.
a) 950 b) 1100 c) 120.50 d) 125.50
4) Of the two supplementary angles the measure of the larger angle is 500 more than the measure of the smaller angle. Find the measure of each angle.
5) If an angle is four times of its supplement, then find that angle.
6) In the given figure; ABC 350= and DBE 900= then find the measure of
i) Supplementary angle of CBD
B350900
A
C
D
E
ii) Supplementary angle of EBC
iii) Supplementary angle of ABD
iv) Supplementary angle of EBD
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Linear pair of angles :Look at the measures of adjacent angles, in the given
figures.
O3001500
B
C
A
fig. i
O900 900
B
C
A
fig. ii
O600 1200
B
C
A
fig. iii
Find their sum. Then enter in the tabular form;
Figure AOC BOC AOC BOC+
i. 1500 300 1500 + 300 = 1800
ii. 900 - -
iii. 600 - -
Are AOC and BOC adjacent angles ?In the figures i, ii, and iii the sum of adjacent angles is 1800
and OA and OB rays form a straight line.What do you call such pair of angles ?They are called linear pair of angles.
Two angles are said to be linear pair if. They are adjacent angles. The sum of the measurement of the angles is 1800.
Note : A pair of supplementary angles form a linear pair when they are placed adjacent to each other.
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Exercise 5.4
1. Observe the angles AOB and BOC drawn in the figure. Do AOB and BOC form a linear pair? Discuss and Justify.
O
650
1350
B
C
A
O600
1300B
C
A
2. Find which of the following pairs of angles form a linear pair, when both the angles are placed side by side.
Q
B
O
A
1250
550
R
P
fig-i
Q700
R
1100
O B
AP
fig-ii
PA
O BQ
R
300
1500
fig-iii
A
O B
R
P Q900
900
fig-iv
A
O BR
PQ
900
900
fig-v
P
QR
1400
A
B
O400
fig-vi3) Find the type of another angle for the following types,
if they have to form a linear pair of angles.A) An acute angle B) An obtuse angle C) A right angle
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GameTake two drawing sheets, cut each sheet to get 40 pieces
and draw the angles of the following measurement on each piece as given below. Make two sets of the same. The angles are 10, 20, 30, 40, 50, 60, 70, 80, 90,100, 110, 120, 130, 140, 150, 160, 170, 180 and 25, 35, 45, 55, 65, 75, 85, 95, 105 115, 125, 135, 145, 155, 165, 175 (you may choose any measurement, but think of supplementary or complementary angles). Then at the back of each card mark one of the following numbers 1, 2, 3, 4, 5 and 6. Place all the cards, so that you can see the numbers on them.
Take a dice having 1 to 6 numbers or dots. (two or more friends can play this game)
Rule : Each member has to throw the dice and take the card of that number. After all the cards are over, that is cards are with the players, exchange one to one card, so that you can make a complementary or supplementary angle with that.
The person who holds maximum pairing cards is the winner.
Vertically Opposite Angles :Two sticks are tied across one another as shown in the
figure. Let AB and CD be the sticks and O be the point where they are banded.
If AB and CD are line segments A
B
D
C
O
then, 'O' is vertex. , , ,AOC BOC AOD BOD are angles.How many adjacent angles are there to BOC ? Name them
How many adjacent angles are there to AOC ? Name them
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Now, observe BOC and AOD
IIIrly, observe AOC and BOD
Are they adjacent ?
How are they related ?
BOC and AOD are on either side of the vertex. IIIrly AOC and BOD are also on either side of the vertex.
What do you call ? such pairs of angles ?
The angles opposite to each other at the point of intersection of two lines are called vertically opposite angles.
Note : The vertically opposite angles have a common vertex.
To find the relation between the pair of vertically opposite angles, let us do the following activity.
Acitivity -1
A
B
D
C
O
A
B
D
C
O
Draw two lines, say ↔AB and ↔CD on a sheet of paper
intersecting at a point O. Then mark the four angles by , ,COB BOD AOD and AOC as shown in the figure.
Now place a tracing paper or a transparent sheet over it. Trace the lines AB and CD on this transparent sheet.
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Now put the traced copy on the original in such away that the line ↔AB drawn on the sheet coincide with the line ↔AB on the tracing paper.
Now fix a pin at the point of intersection of the line. Rotate the traced paper in such a way that line
↔AB on
traced copy lies on ↔BA opposite to the line ↔AB drawn on the original sheet of paper.
What do you observe from the above activity? We observe that,
1) COB drawn on original sheet covers completely the AOD drawn on the tracing sheet.
Which means COB coincides with AOD
2) AOD drawn on original sheet covers completely the COB drawn on the tracing sheet.
Which means AOD coincides with COB
3) BOD drawn on the tracing paper completely covers the AOC drawn on the original sheet.
Which means BOD coincides with AOC
4) AOC drawn on the tracing paper completely covers the BOD drawn on the white sheet.
Which means AOC coincides with BOD
Thus we can conclude that
COB AOD= and BOD AOC=
If two lines intersect each other, a pair of vertically opposite angles formed are equal.
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Activity - 2 :Take a sheet of white paper. Fold it length wise thrice and crease. Unfold the paper. Then fold the paper diagonally and crease, then unfold it, you get a number of intersecting points mark them as A, B, C........Measure the angles formed at each of the vertex calling them A1, A2, A3.............. and so on. Write the measurements on a tabular form in such a way that measure of vertically opposite angles lie side by side.
Sl. No. Vertex 1st pair of vertically
opposite angles2nd pair of vertically
opposite angles
i. AA1 = A3 =
A2 =A4 =
ii.
Give some situations from real life where you can see vertically opposite angles.
Exercise 5.5
1) In the given figure, AOC 600= Find out the other angle.
A
B
D
C
600O
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2) In the given figure, Identify the pairs of vertically opposite ang l e s , name them , I f , BOE 1400= and OC⊥EO then,
calculate the measurement of all the angles.
E
A B1400
C
D
O
3) In the given figure if AOC 300=find the measurement of the remaining angles.
A
B
D
C
300 O
Think : Out of 2 pairs of vertically opposite angles, one pair has
the following condition. What would be the nature of other pair of vertically opposite angle ? One pair of acute angle. One pair of right angle. One pair of obtuse angle.
Exercise : 5.6
I. What is the type of angle that forms a linear pair with.
1) acute angle
2) obtuse angle
3) right angle
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II. Find the value of 'x' and measure of each angle from the following.
x 750
fig. i
3x x+20
fig. ii
x2x x+50
fig. iii
x+15 2x+15x+10
fig. iv
III. In the given figure 9AOC 00= PQ is the bisector of the angle BOD and RS is the bisector of the angle AODFind the other angles.
Q
D
R
A
900P
C
S
B
O
IV. Find the value of x, y and z from the following.
B
D
2z 3z
450
A
C E
Ox
y
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V. In the given figure OE ⊥AB, (OE is perpendicular to AB) then find the following pairs.
i) Linear pair.
B
D
A
CE
Oii) Supplementary angles.
iii) Vertically opposite angles.
iv) Complementary angles.
VI. Fill in the blanks :
i) If two angles are complementary, then the sum of their measure is ____________ .
ii) If two angles are supplementary, then the sum of their measure is ____________ .
iii) Two angles forming a linear pair are ____________ .iv) If two adjacent angles are supplementary, they form a
____________ .v) If two lines intersect at a point, and one pair of vertically
opposite angles are acute then the other pair of vertically opposite angles are ____________.
VII. In the given figure KL = PQ then find
1) Equivalent supplementary angles.
2) Vertically opposite angles.
Q
S
K
450
P
R
L
O
3) Unequal supplementary angles.
4) Adjacent Complementary angles.
5) Vertically opposite obtuse angles.
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CHAPTER- 6PAIR OF LINES
After studying this chapter you : define the meaning of pair of lines,
define the meaning of intersecting lines,
understand the meaning of parallel lines and transversal,
identify parallel lines and transversal lines in day to day situations,
identify the angles made by a transversal; corresponding angles, alternate angles, interior angles, angles on the same side of the transversal,
solve the problems related to the angles.
Observe the photograph of window grill, lines drawn in your note book, etc.
Look at the figures below; you observe that each line in a pair lie on the same plane.
fig (a)
fig (b) fig (c)
fig (d)
Count the number of lines drawn in each figure. How many lines are there in each figure? Yes, there are two lines in each figure. Observe these figures. Can you identify the lines crossing each other in fig (a) and fig (c) ?
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Any two lines on the same plane, irrespective of type or position, are called pair of lines.
Note: Any two or more lines or line segments on same plane are called coplanar lines or coplanar line segments.
Intersecting Lines
i) Choose a point on a plane, and draw two rays. Name them.
ii) Draw a line segment PQ and mark a point R on it. P QR
Draw another line segment RT.
T
QP R
Let RT stands on PQ. Extend the TR to get TS.
P R
S
Q
T
Look at the above figure, SR and TR meet the line segment PQ at 'R'. ST crosses PQ through a point 'R'. The line segment ST is called intersection line segment of PQ and R is the intersecting point.
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Look at these figures formed by rays and lines.
fig. I fig. II
fig. IIIIn the fig (i) The rays BA and BC start from point B. In the
figure (ii) and (iii) lines AB and CD cross each other at a point O.AB and CD are called intersecting lines and 'O' is called the
intersecting point.
Any two lines which cross through a common point are called intersecting lines and the common point is called the point of intersection.
In the given figure PQ and RS intersect P S
QR
O
each other. Where do the lines PQ and RS intersect each other ?
Which point is common to both the line? What do you call this common point?
O
Activity 1 : In the given figure, identify intersecting line segments. For example, AC and BD are intersecting at O. Try to find out other pairs of intersecting line segments and make a list of them.
Think : Does any pair of lines or line segments always intersect ?
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Parallel Lines Activity 2 : Draw a line segment AB. Mark points K,L,M and N on the line segment AB. Mark 900 at M then draw an angular line and mark a point 'P 'on that line. Now join PK , PL , and PN as shown in the figure. Measure the lengths PK , PL , PM and PN using scale and measure , ,PMB PMA PNM and PKL by using a protractor. Compare the lengths of PK , PL , PM and PN . Which line appears to be the least in length ? Name the shortest length which stands on AB.
The least distance between the given line segment from an external point is PM. The angle at M is 900. Then, the line PM is called perpendicular to AB. So, Among all the line segment drawn from an external point to a given line segment perpendicular has the least length.
Note: The symbol used to denote perpendicular is ⊥. PM is perpendicular to AB. It is represented by PM AB= .
Parallel LinesScale
Post Card
©
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Look at the boundaries of the scale and post card.Draw lines along the length wise boundaries of the scale
them. Name as PQ and RS as shown below. Now extend the lines PQ and RS on either side.
P
R
Q
S
Do the two lines meet each other ?Measure the perpendicular distance between the lines at
different points.What do you infer from them ?What name is given to such lines ?Here, PQ and RS are at same distance at all corresponding
points. So, we call such lines as ''parallel lines''.
A pair of straight lines are said to be parallel, if the distance between corresponding points of the lines are always the same.
Note : 1) Two parallel lines can be represented by marking the
arrows on both the lines in the same direction. ⇒2) If two parallel lines are extended in both directions, they
never meet each other on the same plane.
In the given figures, line AB and CD are two lines drawn in different positions and the distance between them are same.
Why do these lines not meet or cross ?
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The distance between two lines does not change. Hence, the parallel lines do not meet anywhere.
Activity 3 : Observe the lines drawn in your graph book and identify;
i) Parallel lines. ii) Perpendicular lines.What difference do you observe between them ?
Activity 4 : List at least 5 examples from your surroundings where you see ; (i) Parallel lines. ii) Perpendicular lines.
Note :
1) If two lines AB and CD are parallel, we write AB || CD 2) If they are perpendicular, we write AB ⊥ CD .
Transversal Lines Activity 5 :
Draw a line segment AB and another line segment PQ as shown in the figure, Where do they meet each other ? Now, let us draw another line segment CD above the line segment AB which intersects PQ as shown in the figure. Then mark the point of intersection as L.
A
A B
D
L
KK
PP
C
AB B
What do you observe ? PQ intersectsAB at K and CD at L.
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Observe the following figures
fig - i fig -ii
C
AB
D
P
KLM
Qfig - iii
A C
B D
LKQP
fig - ivIn the above figures,Name the line that intersects the lines AB and CD in each
case ?Mark the points of intersection as K, L as shown in the
figure (i).In the above figures, PQ intersects at least two lines; such
an intersecting line is called transversal.The line that intersects at least two lines on a same
plane is called a transversal line.
Activity 6 : Take a sheet of white paper. Make two or three folds length wise and crease it, then
unfold the paper. Now fold the paper other than length wise and crease,
then unfold the paper. Mark the lines caused by folding using a pencil. Name the lines and intersecting points. Make a list of
intersecting lines and transversal lines.
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Activity 7 : In the figures given below, name the transversal and justify.
A
B
C
D
P
Q
L
K
AB
CD
P
Q
L
K
A
B
C
D
P QLK
Think : In the given figure, 'EF' is not a transversal all though it cuts the line 'AB' and 'CD'?
By observing the above, under what condition is a line called transversal ?
Angles formed by a transversal line Look at the figure;
Name the line segment, transversal and the point of intersection made by the transversal.
AB, CD and PQare lines, in which PQ is the transversal; K and L are distinct points of intersection. PKB , AKQ are vertically opposite angles,
find out other opposite angles formed by the point K and L ,PKB AKP are supplementary angles. Identify other pairs of angles formed by the intersecting point K and L
There are eight different angles formed at the intersecting point K and L
How can these angles further be classified ?
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Interior angles
C
A
D
BP
2K
143
LQ
In the figure AB, CD and PQare three straight lines.
Which is the transversal line ?
Which are the angles formed by PQ at K and L intersecting points.
Observe the following four angles ,AKQ BKQ formed at K, CLP and DLP formed at L.
What commonality do we observe among these four angles ?
We observe that these angles which lie between the lines 'AB' and 'CD' ,'KL' is the interior part of transversal 'PQ'.Since these four angles are formed interior to the pair of lines. They are named as interior angles.
The angles formed by a line segment of the transversal and between pair of lines are called interior angles.
Therefore, in the above figure AKL , KLC and BKL , KLD are interior angles.Exterior angle; [External angle]
A
CL
Q
D
BK
P In the figure AB, CD and PQ are three straight lines Which is the transversal line?
Now, name the interior angles on the same side of a transversal PQ.Observe the pair of angles outside the
pair of lines on the same side of the transversal.
Here AKP at point K and CLQ at point L are pair of angles.Similarly, PKB at K, and QLDat L are pair of angles.What are the similarities between these pair of angles ?
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Notice that these pair of angles lie outside the parallel lines AB and CD on same side of the transversal .
What name do you suggest for such pair of angles ?Certainly these angles are outside the lines other than the
transversal and hence, they are termed as External angles or Exterior angles.
Therefore, the pair of angles which lie outside the pair of lines, but on the same side of the transversal are called Exterior angles or External angles.
Activity 8 : Identify the pair of interior angles A
C
ED
B
F Gand exterior angles from the following figure and make a list of them.
Consecutive Interior Angles In the given figure, observe the pair of
A
BC
D
P
Q
L
Kangles AKL and KLC and another pair BKL and KLD with respect to a transversal PQ.
Observe that, AKL and KLC are pair of interior angles on the same side of the transversal PQ. ly the pair of angles BKL and KLD are on the right side of the transversal. Such pair of angles are called consecutive interior angles.
In the above figure AKL and KLC form one pair of consecutive interior angles. Can you name other pair of consecutive interior angles in the same figure ?
The interior angles on the same side of a transversal are called "Consecutive Interior angles".
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Corresponding AnglesObserve the angles marked in each of the figures.
AB
D
K
LCQ
P
fig ifig ii
fig iii
A B
D
K
LC
Q
P
fig iv
In the above figures, PQ is a transversal. It cuts AB and CD at K and L respectively.
In the given figure (i) AKP and CLK are marked on left side of the transversal PQ and AKP lies above AB and CLKlies above CD.
In the given figure (ii) pair of angles AKL and CLQ are on left side of the transversal PQ and are below the lines ABand CD.
In the given figure (iii) pair of angles PKB and KLD are at right side of the transversal PQ and lie above the lines AB and CD.
In the given figure (iv) pair of angles BKL and DLQ are at right side of the transversal PQ and lie below the lines AB and CD.
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What can we observe from the above ?The pair of angles in each figure,a) have different vertices,b) are on the same side of the transversal,c) are above the pair of parallel lines in fig (i) and(ii) in fig
(iii) and fig (iv) below the parallel lines. Such pair of angles are termed as Corresponding Angles.
The pair of angles which have different vertices, lie on the same side of transversal, and are located in respective positions of the pair of lines are called corresponding angles.
In the figures (i,ii,iii & iv) the pair of angles (i) ( AKP and CLK ) (ii)( AKL and CLQ ) (iii)( PKB and KLD ) & (iv) ( BKL and DLQ
are corresponding angles.
Activity 9 : Fold a sheet of paper and crease it to get parallel lines and transversal lines as shown in the figure. Name the lines and intersecting points. Write pairs of corresponding angles in each case.
Think : The letter 'F' helps you to remember corresponding angles. How ?
Alternate Interior Angles :Observe the angles marked in each of the given figures.
Fig i Fig ii
AB
D
K
LCQ
P
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In the figure (i) AKL lies to the left of transversal PQ and below the line AB while the KLD lies to the right of the transversal PQ and above the line CD .
In the figure (ii) BKL is on the right side of transversal PQ and lies below the line AB while KLC is on left side of the transversal PQ and lies above the line CD.
Make a list of what conclusions do you draw from the above information ? Yes, the pair of angles in each figure,
(a) have different vertices
(b) are at opposite sides of the transversal
(c) are in the interior side of the parallel lines
(d) are in different position of the parallel lines that is if one angle is above the line, the another angle is below the line,
So, such pair of angles are termed as Alternate Interior angles.
In the above figures, pair of Angles ,AKL KLD and ,BKL KLC are Alternate Interior Angles.
The pair of angles whichi) have different verticesii) lie on opposite sides of the transversal, andiii) lie in the interior of parallel lines andiv) located in different positions of parallel lines are called
Alternate interior angles.
Activity 10: Make paper folding as in previous activity mark lines, then identify alternate interior angles.
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Think :How are the letter N and Z useful to identify the alternative angles ?
Transversals to parallel lines :Observe the following pictures that you have seen in your daily life.
fig i fig ii fig iii
In the above figure; (i) picture of a ladder, (ii) photograph of a window, and (iii) photo of railway track. Identify which of them looks like (a) parallel lines and (b) transversal line ?
Activity 11 : Take a sheet of paper. Draw two parallel lines AB and CD on it. Then draw a
transversal PQ to the lines.Mark the pair of corresponding angles as shown below.
Fig i Fig ii
Fig iii Fig iv
Place a tracing paper over the figure (i), trace figure (i) correctly. Now slide the tracing paper on vertex L, in such a way that the line segment AB coincides with the line CD and the vertex K at vertex L.
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What do you observe from the above diagrams ?
We observe that the traced angle AKP in each case completely covers the angle. KLC at vertex L.
What kind of angles are AKP and KLC ?
AKP and KLC are corresponding angles. From this activity we notice that AKP = KLC .
Now, mark PKB on a trace sheet in figure (iii) Place the marked tracing sheet coincide with the figure(iii). Then slide the tracing paper such that AB coincides with CD and K coincides with L.
Observe PKB and KLD . What is your inference ?
Yes, PKB KLD=
,LKB QLD AKL CLQ= = Therefore we conclude that,.
Relation between a pair of alternate angles made by a transversal to parallel lines.
Activity 12 : Draw two parallel lines AB and CD and also draw a transversal PQ on a sheet of paper, then mark the pair of alternate angles as shown below.
Fig (i) Fig (ii) Fig (iii) Fig (iv)
Trace figure (i) on transparent paper. Fold the traced figure and crease such away that line AB exactly coincides with the line CD and also the vertex K coincides with L.
Then, unfold the transparent paper and mark the point of intersection on the line segment PQ. Mark that point as 'x' .
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Then place a transparent paper over the corresponding figure drawn on a sheet of paper in such a way that the line AB on the transparent sheet coincide with the line AB on the sheet as in fig (iii).
Then fix a drawing pin at the point 'x' on the transparent paper and mark AKL . Now rotate the transparent sheet, in such away that the line AB of the transparent sheet coincides with the line CD drawn on the sheet of paper, but directed in the opposite direction of CD. Observe the AKL that coincide exactly with KLD .
What do you infer from this activity? yes we infer that AKL KLD= .
Now mark BKL as shown in the figure (iii). Then rotate on the axis 'x'. Now can you justify with the pair of angles; BKL KLC= .
Therefore, we conclude from the above activity that;If transversal is drawn to a pair of parallel lines, the pair of alternate angles are always equal.
Interior angles on the same side of the Transversal :
Activity 13:Draw two parallel lines AB and CD¸and a transversal PQ on a sheet of paper.
Fig (i) Fig (ii)
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Now, trace figure (i) on the tracing sheet; then place the tracing sheet on the sheet of paper in such away that the line AB of tracing paper coincide with the line CD such that A coincides with D.
Note : The direction of arrow of line AB of tracing sheet and direction of arrow of the line CD on the sheet is opposite.
Notice that AKL KLD= Now you measure the angles BKL and KLD and find the sum of these two angles.
We know that 180AKL BKL 0+ = , because they are adjacent angles.
Now substitute AKL for KLD then KLD BKL 1800+ =
Similarly, BKL CLK= , therefore AKL CLK 1800+ = .
Hence we conclude that,
Let us observe the relation with pair of angles and line seg-ments with this activity.
Step 1 : Take a sheet of paper. Draw a line AB. Then mark a point K on it.
A BK
Step 2: Construct an angle of 1000 at K. Draw an angular line and produce it A BK
1000
Step 3: Take a tracing paper, draw a line CD by using marker, then mark a point L on it.
C DL
Step 4: Construct an angle of 1000 at L. Draw an angular line and produce as you did it earlier C DL
1000
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Step 5: Now you place a tracing paper below or above the line AB . Adjust the position of AB and CD by sliding up and down in such away that lines drawn from K and L coincide exactly with one another, so that the line is a transversal.
A
C D
B
P
Q
K
L
Observe the pair of angles AKP and CLK . Which type of angles are they?
They are corresponding angles and they are equal. Now measure the height between ,AB CD at different positions. Are they same? Then, What relation do you observe between the pair of lines AB and CD ?
When a transversal cuts two lines in such away that each pair of corresponding angles remain equal, then the lines are parallel.
Think : When a transversal cuts two lines in such away that
each pair of alternate angles is equal, then what would be the nature of lines ?
When a transversal lines cuts two line segments, the corresponding angles formed are equal, then what would be the nature of lines ?
When a transversal cuts two lines in such away that the sum of the interior angles on the same side of the transversal are supplementary, then what would be the nature of lines ?
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Think : 1) Observe the shapes given below. How can you identify
the interior angles on the same side of the transversal lines ?
2) If a transversal cuts two line segments, if the sum of the interior angles are supplementary, then what do you call those line segments ?
Activity 14: Draw two non - parallel lines AB and CD and draw a transversal PQ, which cuts the parallel lines at K and L respectively, then measure a pair of corresponding angles, a pair of alternate angles, and interior angles on the same side of the transversal. Write your conclusions. A simple procedure to verify whether a given pair of lines are parallel or not.
Draw two pair of lines on the same sheet of paper as in the fig - (i) and fig - (ii).
fig-i fig- ii
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A
C
B
D
A
C
B
D
Name the two given lines as AB and CD. Put a ruler (scale) in such away that its edges cross both the lines 'AB' and 'CD'. Mark lines along the edges of the scale.
Remove the scale and name the points P and Q where the lines marked by the scale which meet on the given line 'AB '. Also name the points as 'R' and 'S' where lines marked by the scale meet the line 'CD '. Measure the length of the segments PQ and RS in each case.
fig (i) fig (ii)
In the given figure (i) we observe that length of PQ = length RS. We conclude, that lines AB and CD are parallel to each other where as in the figure (ii) the length of PR ≠ QS we can say that the lines AB and CD are not parallel.
And hence, we conclude that, PQ and RS are perpendicular to line AB and CD then, they are parallel.
If the perpendicular distance between two lines are equal, then the lines are parallel.
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Example 1 : In the given figure
ABIICDIIEF. GF=EF and BGD = 500
Find the remaining angles
A B
D500
FE
C G
+ = 180O [ CD║EF]
= 180O
- 90O
= 90O
ABG and BGD are alternate
angles.
= = 50O
= 50O
Here CGB and BGD form a
linear pair,∴ CGB + BGD = 180O
CGB = 180O - 50O
CGB = 130O
Example 2 : In the given figure. Find the measure of marked angle. CFE + EFG = 180O A
C DF
E
400
G
B
CFE + 40O = 180O
CFE = 180O + 40O
CFE = 140O
The marked angles is FEB
,CFE FEB are alternate angle
Since ABIICD
CFE FEB=
FEB 1400=
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Example 3 : Find the marked
angle p, q, r and s from the given
figure.
i) ABC 600=
Since DEIIBC Corresponding angles are equal
ADF ABCp ABC
ABCp
6060
0
0`
====
ii) AFD and EFC are the vertically opposite angles
AFD EFC=700 = qq = 700
iii) EFC and ACBACB EFC
rr
7070
0
0
`
`
===
are alternate angles.
AFD + CFD = 1800 [ linear angles]700 + CFD = 1800
∴ CFD = 1800 - 700
∴ CFD = 1100
CFD and AFE are vertically opposite angles
[ DE║BC and AC transversal line]∴ CFD = AFEs = 1100
∴ s = 1100
A
C
D
r
700
EF
600
B
pq?
s
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Exercise 6.1
1. Write down the angle that corresponds to the coloured angle?
a) b) c)
d) e) f)
2) Find the measure of the angles of each marked letter in each figure.
a)
A
C
F H
K DL
I J B
GE700
b) c)
3) In the given figure PQRS is a parallelogram then find the measure of all interior angles.
P
Q R
S
T1300
4) In the figure MN||BC. Find the measure of all interior angles of triangle ABC
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5) Decide whether AB is parallel to CD in the given figure
AC
F
D
BG 1260
440H
E
750750
AC
F
DB
E
6) In the given figure, the arms of the two angles are parallel, if 70ABC 0= then find, DGC and DEF .
7) Find the measure of marked angles a, b, c and d from the given figures.
1) 2) 3)
4) 5)
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CHAPTER- 7
TRIANGLES
After studying this chapter you : find through activities the sum of the interior angles
of a triangle,
establish the sum of the interior angles of a triangle is 1800 through paper folding (origamy),
prove the sum of interior angles of a triangle is 1800
using the properties of parallel lines,
Identify the difference between logical proof and activity based proof,
establish the relationship between exterior angle of a triangle and interior opposite angles,
understand that the sum of any two sides of a triangle is always greater than the third side,
understand the Pythagorean theorem through activities.
Introduction
Triangle is one of the most familiar geometrical shapes. You observe many triangles in the above given figures.
Poles tied in the triangular shape attain more stability. We observe many triangular objects and patterns in our surroundings. Hence, it is necessary to learn the properties of triangles.
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Angle sum property of a triangle
Activity 1 : A
B Cfig (i)
A
BC fig (ii)
Draw a triangle as shown in fig (i) Name its vertices as A, B, C Measure the interior angles of the triangle using
protractor. (as shown in the fig ii) Write the measures in the table given below. For example
in the fig (1) the measures of angles are written. Find the sum of the three angles.
Interior angles of the
triangles
Triangle 1 above series
Triangle 2 Triangle 3 Triangle 4
Triangle 5
angle 1 640
angle 2 560
angle 3 600
sum of the three angles 1800
Similarly, construct different types of triangles (eg :- equilateral, scalene, acute angle, obtuse angle) Measure each interior angles of these triangles.
Write these measures in the given table. Find the sum of three angles of each triangle.From the above activity, what do you observe about the
sum of the three interior angles of a triangle?
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Let us do another activity.
Activity 2 : Take a thick paper. Draw a triangle
of any measurement. Mark the three angles as 1, 2 and 3. Cut angle 1, angle 2 and angle 3 as
shown in fig (1).
1
2 3
fig (1)
Rearrange the cut pieces in such
fig (2)
away that the vertices coincide at a point as shown in fig (2).
What is your observation about the sum of the three angles of triangle?
∴ What do you observe about the sum of the interior angles of a triangle?
Let us do another activity (without cutting the paper)Activity 3: Draw a triangle ABC on a thick
sheet as shown in the figure. Name the angles as 1, 2 and 3. Mark the mid points of AB and
AC as P and Q respectively. as shown in the fig (1).
Fold the triangle APQ on the line PQ. Now, vertex A touches BC at M.
Fold the vertex B so that BP coincides with PM.
Fold the vertex C so that QC coincides with QM
Now you will find ,1 2 and 3 makes a straight angle.
fig (1)
132
MB
A
P Q
C
fig (2)
From the above three activities we establish that the sum of the three interior angles of a triangle is 1800.
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Exercise 7.1
1) Take a rectangular sheet of paper. Cut a right angle triangle at one of its corner. Cut two angles other than the right angle. Place the other two angles at the vertex where the right angle is formed in such away that they all coincide with each other. State your opinion about the sum of the three angles of a triangle.
2) Draw an obtuse angled triangle as
B
A
C
shown in the adjacent figure. Cut this triangle without cutting the three angles. Verify the sum property of the triangle.
Think: In all these activities we found that the sum of the
interior angles of a triangle is 1800. Why is it not 1500 or 2000?
There are innumerable triangles of different shapes and sizes. Do all these triangles have their sum of the interior angles as 1800?
Is there any triangle in which sum of the interior angles is not equal to 1800?
To find the sum of the interior angles of triangle logically.
All the activities conducted indicate that the sum of the interior angles of a triangle is 1800. But there is no definiteness of the property which is applicable to all the triangles, Even after measuring the angles of thousands of triangles, one cannot arrive at the exactness of the angle sum of property. This is the limitation for the method in which examples are taken. This limitation can be over come by some other method.
To generalise the inference, there is a method which is known as logical method. To prove the mathematical statements, logical method is applied.
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Let us use the logical method to prove the statement ''sum of the three interior angles of a triangle is 1800''.
Let us apply logical proof to derive the sum
B C
Aof the angles ,A B and C of OABC in figure is equal to 1800.
Is it possible to have a construction to bring all the three angles of the triangle ABC at any point (say A, B or C) on a straight line? Yes, let us draw a straight line RS passing through A.
In this figure, we have two pairs of alternate
B C
A
R
S
∗
∗
angles. They are
i) ABC and BAR (represented by ∗)
ii) ACB and SAC (represented by )
But these alternate angles are not equal.
In the above triangle ABC, if there could be one pair of parallel lines, there would be alternate angles which are equal.
We already know that, if a pair of parallel lines are cut by a transversal, the alternate angles so formed are equal.
So, draw a line PQ through A parallel to BC
B C
AP Q∗
∗
Here PQ || BCand AB is the transversal.
Now observe the figure
Do you find any pairs of angles equal?
PAB ABC= ........(i) [alternate angles marked as ∗]
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Similarly PQ || BC and AC is the transversal.
Do you find any pairs of angles which are equal ?
QAC ACB= ........(ii) [alternate angles marked as ]
Observe the lines PA , AB and AC
PAB BAC QAC+ + = Straight angle = 1800 → (iii)
Substitute the RAB and SAC by ABC and ACB respectively in equation (iii)
we get ABC BAC ACB 1800+ + =
These are the interior angles of a triangle.
Hence, the sum of the interior angles in a triangle is 1800.
The logical proof discussed is true of all shapes and sizes of triangles. Hence the conclusion is applied to all triangles.
Calculation of the angles of a triangle by logical methodSolved problem 1:
In the adjoining fig PQ || RS. A is a point on PQ.
AB and AC are drawn from the point A.
50 70PAB Q AC0 0= = . F i n d the measure of all the angles of triangle ABC logically.
Observe only parallel lines PQ and RS and another line AB.
What type of angles are PAB and ABC ?
They are alternate angles. Hence, ABC PAB 500= =
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Now observe the lines PQ, RSand AC.What type of angles are QAC and ACB ?They are alternate angles.Hence. ACB QAC 700= =
Now observe only the lines PQ, AB, AC What type of angles are ,PAB QAC and BAC ?
Are they supplementary angles ?Then ∠PAB + ∠BAC + ∠QAC = 1800
500 + ∠BAC + 700 = 1800
∴ ∠BAC = 1800 - 500- 700
= 600
Solved Problem 2 : In the figure, the two angles
of triangle PQR is given(400 and 300). Find the third angle of the triangle logically.
Draw a line through P parallel to QR.
Mark the alternate angles formed at the point P as 400 and 300 in the figure.
P
400 300
?
Q R
In the adjacent triangle PQR Q R40 300 0= = .Sum of the
three angles at point P is 1800.So, the third angle =1100-400-300=1800
P
400
400 300
300
?
Q R
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Example 1 : In the figure ,A B60 500 0= = Find the measure of C logically. A B C 1800+ + = The sum of the three angles of triangle is 1800
600 + 500 + ∠C = 1800
1100 + ∠C = 1800
∠C = 1800 - 1100
∠C = 700
Example 2 :Find the value of x in the adjacent figure.
180A B C 0+ + = The sum of the three angles of triangle is 1800
600 + x0 + 80 = 1800 [Substituted the measure of A and C ] 1400 + x 0 = 1800
x 0 = 1800 - 1400
x 0 = 400
Exercise 7.2
1. Find the value of x logically in the following figures:
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2. Two angles of the triangle ABC are given. Find the third angle and fill up the blanks in the table:
∠A ∠B ∠C
1 600 700
2 1000 350
3 650 650
4 1100 360
3. Three angles are given in the table. Find the sum of these three angles. Can these become the angles of a triangle?
A B C Sum Can it form a triangle Yes/No
1 500 600 700
2 600 700 800
3 650 750 550
4 560 640 600
5 570 640 790
4) In a triangle ABC, if ,A B45 650 0= = then find the measure of C .
5) Construct the angle ABC with the following measure AB =5cm 5PAB 00= and QBA 600=. Measure the ACB and verify the measure of ACB logically.
6) In a right angle triangle if one angle measures 350. Find the measure of the other two angles.
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7) Construct an equilateral triangle having the side 4 cm and find out the measure of each angle.
8) In a triangle ABC A 350= and B is greater than A by 150 then find C .
9) In a right angle triangle, one of the other two angles is twice the other. Find the measure of the other angle which is not a right angle.
Exterior angle of the angle
A triangle has three sides,
adjacent interior angle
exteriorif any one of the sides of the triangle is extended, an angle which is formed out side the triangle at the vertex and this angle is called an exterior angle.
The angle which is adjacent to the exterior angle is called adjacent interior angle.
adjacent interior angle exterior angle
adjacent interior angle
exterior angle
adjacent interior angleexterior angle
adjacent interior angleexterior angle
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If a side of any triangle is
adjacent interior angle exterior angle
extended, an exterior angle and interior angles are formed. what is the sum of an exterior angle and an interior angle adjacent to it ?
These two angles are supplementary angles or linear pair angles.
Hence, the sum of an interior angle and an exterior angle is 1800.Interior Opposite angle
The angle opposite to the adjacent interior angle of a triangle is called interior opposite angle.
interior opposite angle 2
interior opposite angle1 adjacent opposite angle exterior angle
There are two interior opposite angles and one adjacent interior angle to every exterior angle of a triangle.
Is there any relation between the exterior angle and the Interior opposite angles of a triangle ?
Activity 1 : As shown in the fig (1)construct
any triangle on a thick sheet of paper.
Extend the base of this triangle. Name the three angles of the
triangle as 1, 2 and 3. Name the exterior angle as 4.
fig(i)
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As shown in the fig (2) cut the angles (1) and (2) arrange these two angles on the exterior angle 4 as shown in the fig.
What is your conclusion? angle 1+ angle 2= exterior angle 4. This activity establishes that
exterior angle of a triangle is equal to the sum of opposite interior angles.
fig(ii)
Let us examine some examples related to the above properties.Example 1 : In the figure 700
500
C D
A
Bexternal angle=?
given below, the two angles of the triangle A B C a r e g i v e n . Calculate the measure of exterior angle.∠A + ∠B + ∠C = 1800 (Sum of the three angles of triangle = 1800) 700 + 500 + ∠C = 1800 (Substitute the measure of A and B ). 1200 + ∠C = 1800
∠C = 1800 - 1200 = 600
∴∠ACB = 600 Interior angle
ACB Exterior angle ACD 1800+ = (∴ The sum of one adjacent interior and exterior angle = 1800)
∴ ∠ACB + ∠ACD = 1800
600 + ∠ACD = 1800
∠ACD = 1800 - 600 = 1200
Exterior angle 12ACD 00= sum of interior opposite angle = 700 + 500 = 1200
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Hence, the exterior angle is equal to sum of opposite interior angles.
Example 2 : In the given figure
400
550RS
P
Qexterior anglethe two angles of triangle PQR are
given. Find the measure of exterior angle.
∠P + ∠Q + ∠R = 1800 (Sum of the three angles of the angles in 1800)400 + 550 + ∠R = 1800 (Substituting the values of P and Q 950 + ∠R = 1800
∠R = 1800 - 950 = 850
∴ ∠PRQ = 850 Interior angle∠PRQ Interior angle and ∠PRS exterior angle are complimentary
∴ ∠PRQ + ∠PRS = 1800
850 + ∠PRS = 1800
∠PRS = 1800 - 850 = 950
(One interior angle + One adjacent exterior angle = 1800)In the above example exterior angle = 950. The sum of interior opposite angles = 400 + 550 = 950
Hence exterior angle of a triangle is equal to the sum of the opposite interior angles.
Example 3 : In the figure if ∠BAC = 500 and
A
500
850
DCB
∠ABC = 850 then find the measure of ACB and ACD
∠BAC+∠ABC +∠ACB=1800 (Sum of the angles of the triangle is = 1800)500 + 850 + ∠ACB = 1800 (Substituting the values of A and B
1350 + ∠ACB = 1800
∠ACB = 1800 - 1350 = 450
∠ACB + ∠ACD = 1800(Supplementary angles)
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450 + ∠ACD = 1800
∠ACD = 1800 - 450 = 1350
Exercise 7.3
1. Find the value of x and y in the following figures.
a)600
400 x Y
c)
x y
550
550
b)
1450 700
X
Y
d)500
650X Y
2) In the fig. if 45BAC 0= and
ABC 750= Find the measure
of ACB and ACD .
A
450
750
DCB
3) In fig if PQR 500= PRS 1300= ,
find the measure of QPR500 1300
R S
P
Q
4) In the fig if NKL 350= KNL 300= and LNM 750= then find NLM and NML .
300 750
MLK
N
350
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5) In triangle ABC, BC is produced to D. A and D are joined. If 60 , 45BAC ABC0 0= = find ACD .
Sum of any two sides of a triangle
Activity 1 :
Mark three non- collinear points A, B A
B C
and C in the courtyard. Join AB, BC and
CA with chalk powder. These three lines
represent three sides of the triangle
ABC. Ask your friend to walk from A to
C using any route.
There are two routes to reach from A to C. One is directly
from A to C, another is from A to B then B to C.
The first route is short and everybody follows that route.
It is a longer distance to walk through A to B and then B to C.
In triangle ABC
AB + BC > AC → (1)
There are two routes to reach from B to A. One is directly
from B to A, another is from B to C and C to A.
then BC + CA > BA → (2)
There are two routes to reach from B to C. One is directly
from B to C, another is from B to A and Ato C.
∴ BA + AC > BC → (3)
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Activity 2 : Take sticks of length 3cm, 5cm, 7cm, 9cm, 11cm, 13cm, 15cm, 17cm, 19cm, 21cm. Select any three sticks to form a triangle. Similarly take different length of sticks to form a triangle. Is it possible to construct a triangle with the sides as 3cm, 4cm and 6cm?
For example, take two sticks of length 5cm and 7cm. You need a third stick to complete the triangle. What should be the length of the third side?. That stick should be more than 7-5 = 2cm and less than 7 + 5 = 12cm.
Hence, to form a triangle the sum of the lengths of two sticks should be greater than the third stick.
From the above activity 1 and 2 we conclude that,
The sum of any two sides of a triangle should be greater than the third side.
Example 1 : Let us verify whether the sum of any two sides of a triangle greater than third side.
1) 3cm + 4cm > 6cm sum of any two sides of a triangle is greater than the third side
2) 4cm + 6cm > 3cm sum of any two sides of a triangle is greater than the third side
3) 6cm + 3cm > 4cm sum of any two sides of a triangle is greater than the third side
∴ A triangle can be drawn with these measures
Example 2 : Is it possible to construct a triangle with sides 3cm, 4cm, and 7cm? Verify?
Let us verify whether the sum of two sides of a triangle is greater than the third side.
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a) 3cm + 4cm = 7cm The sum of the two sides of a triangle is not greater than the third side.b) 3cm + 7cm > 4cm The sum of the two sides of a triangle is greater than the third side.c) 4cm + 7cm > 3cm The sum of the two sides of a triangle is greater than the third side.
In one occasion (a), the sum of 2 sides is not greater than the third side. Hence, it is not possible to construct a triangle with the given sides.
Exercise 7.4
I. Verify whether a triangle can be drawn with following measures.
1) 3cm, 4cm, 5cm 2) 3cm, 2cm, 6cm3) 2cm, 6cm, 6cm 4) 5cm, 3cm, 8cm
II. M is any point inside the triangle ABC. Are the following statements True/False A
MB C
1) MA + MB > AB?
2) MB + MC < BC?
3) MC + MA > CA?
Pythagoras Theorem (Bodhayana sutra)Pythagoras was a greek philosopher. He introduced
an important property of a right angled triangle in 600 BC. Hence, the law related to right angled triangle is named after him.
If one of the angles of a triangle is 900, it is called as right angled triangle. The side opposite to right angle is called hypotenuse. Hypotenuse is the greatest side of a right angled triangle.
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Activity 1 : Take a white rectangular sheet of paper. As per the measurements indicated in the figure, draw 4 right angled triangles at the corners of the sheet. Triangles are ∆ABC, ∆KLM, ∆PQR and ∆XYZ. Fold along the hypotenuse of these triangle such as AC, KH, PR and XZ.
A X 12 cm
12 cm8 cm
3cm
5cm
4cm
9cm6c
m
Y
ZR
QPML
KC
B
Tabulate the lengths of the sides of the triangles in the following table. Follow the information written in case of ∆ABC.
Triangle∆
Measureof
hypot-enuse
Squareof the hypot-enuse
Lengthof one side
Squareof it
Lengthof
another side
Squareof it
Sumof the
squares on two sides
∆ABC 5 25 3 9 4 16 9+16=25
∆KLM
∆PQR
∆XYZ
In ∆ABC AB = 3cm BC = 4cm. Find AC, is AC = 5cm?
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Let us calculate the sum of the squares of two sides.
AB2 + BC2 = 32 + 42
= 9 + 16
= 25
Now find the square of AC, AC2 = 25
∴ AC2 = AB2 + BC2
52 = 32 + 42
25 = 9 + 16
= 25
From the above calculation we understand, in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This is universally accepted as Pythagoras therom.
Pythagoras theorem
"In a right angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides".
square on one side
AB2
square on another BC
sideBC2
square on the
hypotenuse AC 2
In triangle ABC if ∠B = 900 then AB2 + BC2 = AC2
A
B C
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Know this
The "shulba sutra" composed during the period
800-600 BC by "Bodhayana" states
which means "The diagonal of a rectangle produces both
areas in which length and breadth of the rectangle produce
separately".
Example 1:
In the adjacent figure ∆ABC, AB = 6cm, BC = 8cm and
B 900= calculate AC.
Solution
As per the Pythagoras theorem A
900
6cm
8cmB C
AC2 = AB2 + BC2
= 62 + 82
= 36 + 64
= 100
= 102
AC = 10
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Example 2: A 13 m length ladder is placed slantly on the wall. If the top of the ladder reaches a height of 12m from the floor on the wall, calculate the distance between foot of the ladder and the wall.
Solution As per the pythagoras theorem A
wall13 m
?B C
12 m
floor
ladder
AB2 + BC2 = AC2
122 + BC2 = 132
144 + BC2 = 169
BC2 = 169 - 144
BC2 = 25
BC = 5
∴ The distance from foot of the ladder to the wall = 5m
Exercise 7.5
1) A flag staff is to be erected vertically on
86?
the ground. A rope is tied to the flag staff to pegs on the ground. The peg is 6m away form the bottom of the flag staff. If the rope is tied to the flag staff at a height of 8m from the ground, find the length of the rope.
2) In triangle ABC, AB = 5cm, BC = 12cm ABC 900= calculate the length of AC.
3) A 15m ladder reaches the wall at a height of 12m. Find how far is the foot of the ladder from the wall.
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4) Given below are the sides of a triangle. In which of these cases are the triangles right angled?
a) 2cm, 2cm, 5cm b) 1.5cm, 2cm, 2.5cm
c) 6cm, 8cm, 10cm
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CHAPTER- 8
SYMMETRY
After studying this chapter you : identify the symmetry and reflection symmetry,
define rotational symmetry of 2D figures,
identify rotational symmetry in the case of 900, 1200, 1800 rotation,
manipulate rotation through 900, 1800 to show symmetry,
operate the figures with both rotation and reflection symmetry,
identify the figures that have reflection and rotation symmetry.
Symmetry
You have studied symmetrical figures and reflection symmetry in your previous class
What are symmetrical figures ?
The figures with balanced proportion with reference to a line in terms of shape and size are called symmetrical figures. Many monuments, architectural constructions and gold ornaments are beautiful because of their symmetrical constructions. Many pictures of living and non- living things exhibit symmetry.
Ravi and Mohan are friends.Ravi : Do you like nature, Mohan ?
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Mohan : Yes Ravi, very much.Ravi : I clicked a photograph see.
Mohan : Wow! Beautiful Ravi. Sky is seen in the water. Wonderful!!!
What speciality do you observe in the figure ?Do you see the same image on either side ?We have a special name for this, that is symmetry.We can see many things in nature which seem to be
symmetrical, but symmetry in mathematics is a particular concept.
Look at this picture. Is it beautiful to see ?What aspect of geometry makes it beautiful?It is the symmetry.
Try to identify symmetrical parts.
Activity 1: Spray ink on a sheet of paper, fold the paper equally to half of its size in lengthwise and press and crease. Open the sheet and look at the pattern formed. Mark the line of symmetry. Think why and how symmetry is caused here.
If a line divides the figure into two identical halves then the figure is said to be symmetrical figure and it has reflection symmetry. The line is the axis of symmetry.
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Observe the following figures.
The dotted line represents line of symmetry. It is called the axis of symmetry.
Symmetry provides a balance and perfection to the objects.
Reflection symmetry and mirror reflection are related to each other. In reflection symmetry the half side is the mirror image of the other.
Activity 2: Draw the line/lines of symmetry for the follow-ing figures.
Recall: The line that divides
a figure into two symmetrical parts is called
the axis of symmetry.
Do you know ? Our national flag is one of the best examples for
symmetry.
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Activity 3: Nitya likes origami. She folded a
i) ii)
sheet of paper lengthwise into two equal halves, creased it and then cut it off along the edges as shown in the figure.
Then she unfolded the paper. What did she observe ?
Try to answer seeing the picture (ii).Yes, she observed that the figure is divided into two equal
parts in size and shape along the fold. Recall the name of the folded line.The folded line is called the line of symmetry or axis of
symmetry.
Activity 4: Take a sheet of paper which is in a triangular shape with PQ = PR as in the figure.
Q S R
PWhen the half of a figure
is the mirror image of the other, it is said to have reflection symmetry.
Let us fold this sheet through PS such that PR coincides with PQ, crease it and then unfold. We can see the mirror image of ∆PQS as ∆PRS. Can we draw another axis of symetry? Isn’t it? No. This triangle has only one axis of symmetry.
Isosceles triangle which has reflection symmetry along the axis of symmetry. (Can we draw another axis of symmetry to this triangle ?)
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Activity 5: Now take a rectangular shaped sheet of paper PQRS as shown in the figure.
S R
P Q
Fold this sheet in such away that QR coincides with PS. crease it and unfold the sheet.Now fold the sheet in such away that PQ coincides with SR. crease it and unfold it.How many folded lines can you see?
Two lines can be seen. The rectangle has two lines of symmetry.
Activity 6: Let us draw an equilateral triangle on a sheet of paper. Mark the mid points of each side. Join each midpoint of the side to each opposite vertex.
B
F
A
E
CD
ABC is an equilateral triangle. ΔADC is the mirror image of ΔADB. We can notice that AD, BE and CF are the axes of symmetry. That is, there are 3 axes of symmetry in an equilateral triangle. Find the remaining mirror images.
Activity 7: In a square PQRS, all the sides are equal.
SN
K M
R
P QL
Let us draw lines of symmetry.PR, NL are two axes of symmetry, name the other two. How many axes of symmetry are there?There are four axes of symmetry.There are two diagonals and two perpendicular bisectors for each pair of opposite sides.
For different figures, we can draw different number of axes of symmetry.
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Note : • An equilateral triangle has 3 lines of symmetry • A square has 4 lines of symmetry • A regular pentagon has 5 lines of symmetry • A regular hexagon has 6 lines of symmetry.The number of lines of symmetry of a regular polygon is
equal to its number of sides.
Can you find the number of lines of symmetry for a circle?Observe the figures given below.
Think!!!! A circle has unlimited lines of symmetry. Why?
Note: We can see a reflection of a person in a mirror. The image of that person in the mirror is definitely not the reflection symmetry. Reflection of an object and reflection symmetry are not one and the same.
Exercise 8.1
I. Which of the following figures have line of symmetry?a) b) c) d)
II. Draw as many lines of symmetry as possible for the following letters.
a) A b) H c) X
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III. Whether the dotted line on each shape is a line of symmetry? Write yes or no and justify your answer.
a) b) c) d)
e) E f) g) h)
IV. Write the number of lines of symmetry in the following
a) An isosceles triangle b) An equilateral triangle
c) A square d) A rhombus
e) A circle f) A parallelogram
g) A regular pentagon h) A regular hexagonV. Match the following English Alphabet given below with their mirror images.
Alphabets Mirror Images
B E
C B
D E
E C
R D
R
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RotationIn our daily life we see many objects which show rotational
movement.
Eg: i) Rotation of wings of a fan.
ii) Rotation of wheels of vehicles.
iii) Rotation of hands of a clock
You have seen the movement o f hands of a clock. This type of movement is called clockwise movement. The move-ment opposite to the movement of hands of a clock is called anti-clockwise movement.
In rotational movements, there are two types:
i) Clockwise movement ii) Anti-clockwise movement
Look at this clock. The two hands of the clock rotate around a fixed point in the middle of the clock. This fixed point is called centre of rotation.
Observe the figures. consider the wheel and observe a dot on the wheel.
In both the figures the shape and the size of the wheel do not change and the wheel turns around a fixed point. The dot implies the rotation.
One more example of rotation.
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John is a naughty boy. On Christmas vacation his mother bought a star for him. He played with that star by turning it as shown in the figure.
That is he rotated the star.
A
B
C
D
D
A
B
CD
A
B
C
D
A
B
C
He rotated that in all possible directions.Then he rotated it from left to right and right to left.He rotated the star, both ways clockwise and anti clockwise
direction.The rotation is a process in which an object turns about
a fixed point in any direction in the plane. This fixed point is called Centre of rotation or Point of
rotation.Activity 8: Fix a small nail to a cardboard and
90 90
9090
hang a thin strip of card board with an arrow mark on one side as shown in the figure. Here, nail is the centre of the rotation. Now rotate the strip 900, 1800, 2700 and finally one full round. What do you observe?
After completing 3600 or one full rotation the strip comes to its original position.
So, all figures after completing 3600 or one full rotation, come back to their original position.
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What else do you understand from the above activity ?
We can observe that after a quarter turn, the strip covers 900, half turn, it covers 1800, after three quarter turns, it covers 2700 and after one complete rotation, it covers 3600 .Observe the rotation of the following figures:
1)
2)
3)
B
CD
A C
DA
B D
AB
C A
BC
D B
CD
A
4)
5)
B
CD
A
DA
B CD
AB
C
BC
D AB
CD
A
Activity 9: Find the image of a rectangle OABC when rotated about O through an angle of .
(i) 90° ii) 180° iii) 270° iv) 360°
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C''
C'
C
BA
O A'
B'
B''
B'''
A''
C'''
A'''
Step 1: Draw a rectangle OABC and let the point of rotation be O.
Step 2: Rotate the rectangle through 900 . Let OA take the place of OA'.
Step 3: Clearly the image of OABC will be OA'B'C' under rotation through 900.
Step 4: Rotate it further through 900 and the rectangle takes the position of OA''B''C''.
Step 5: Rotate it, further through 2700 from the original position and the rectangle takes the position of OA'''B'''C'''.
Step 6: A complete rotation through 3600 brings back the rectangle to its original position.
Rotational symmetry
Example 1: Can you rotate this ? as shown in the figure
If you rotate two times it takes the original shape. Isn’t it?
Note: Rotational symmetry is always considered with a relative reference
Example 2: Do you know what this is ? Yes, you are right. That is a wind mill.
It can also rotate at its point of centre.
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Example 3:Generally flowers are liked by all. Is it not beautiful ? Apeksha wanted to rotate a flower by 900 in clockwise direction.
The flower is rotated in the clockwise direction by 900, 1800, 2700 and 3600.Example 4: Consider this triangular lamina. Let us rotate this by 900 in anti-clockwise direction.
900
1800
2700
3600
When, we rotate this in the anti-clockwise direction by 900, 1800, 2700 and 3600 the figure takes the above positions. After one complete rotation, it takes the original position.
Example 5:
1200 2400 3600
Here, the figure is rotated clockwise by 1200, 2400, and 3600. Observe that in all these positions; the figure exactly looks like the original figure.
This figure is symmetrical. When the object rotates in few angles through rotational centre, there will not be any change. Such figures have rotational symmetry.
The angle by which the object rotates to take exactly the
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same shape of the original is called the angle of rotation.In example 5, O is the point of rotation and the angle of
rotation is 1200.A figure is said to have rotational symmetry, if it looks
exactly same as the original figure, when it is rotated at a fixed point with a particular angle of rotation. A figure appears exactly same as original after it rotates 360°.
Make a paper windmill and turn it. Compare it with the rotation of the figure shown below.
Rotate the following figure completely one turn.
OQ
RS
P
OR
SP
Q
OS
PQ
R
OP
QR
S
OQ
RS
PIn a full turn ie, when we rotate it completely 3600 there
are 4 positions when the figure looks exactly the same.Then, we say, it has rotational symmetry of order 4.In a complete turn, the number of times an object looks
exactly the same is called the order of rotational symmetry.In the above example 1 order of rotation is 2. We rotated
that figure by 900.In the above example 2 order of rotation is 3. In 3 rotations
of 1200 we get the original position.In the above example 3 order of rotation is 4. In this case
after 4 positions it takes the original position. In the above example 4 order of rotation is 4.In the above example 5 order of rotation is 3.Now, let us consider the following figure.
P
O
RQ
S O
S
QP
R O
R
PS
Q O
Q
SR
P O
P
RQ
S
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This figure matches itself 4 times as it is rotated through a complete turn about the point O. Four times it looks like the same, when it has been rotated through 900, 1800, 2700 and 3600. So, it is said to have rotational symmetry of order 4.
Consider a parallelogram PQRS. By rotating this 1800 each time we get the following figures.
S R
QP
S
R Q
P S R
QP
• Parallelogram has rotational symmetry of order 2.
Think!
What is the order of rotational symmetry of a circle?
Take a sheet which is a regular pentagon ABCDE. Fix a pin the number of its centre and rotate. Observe how many times it appears exactly as the first position after one complete rotation.
A B
CE
D
B C
DA
E
C D
EB
A
D E
AC
B
E A
BD
C
A B
CE
D
• Regular pentagon has rotational symmetry of order 5.
∴Order of rotational symmetry Angle of rotation A360 360c
cc= = .
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Activity 10: Take a piece of paper and fix it to a card board. Draw an equilateral triangle of each side 6 cm on the paper. Mark the midpoints of the sides and join them to the opposite vertices and get the point O. Now take a transparent piece of paper and draw a triangle identical to the orig-inal triangle by placing a transparent paper over
the original triangle and also draw the lines joining the mid-points of the sides to its vertices and get the point O in new triangle.
Now place the new triangle on the first triangle so that O and O and all other sides coincides each other. Now fix a pin into the shape at the point where O and O1 coincide each other. Turn the transparent paper in clockwise direction and observe how many times the shape coincides with the original shape in one full round. It coincides 3 times. So the order of rotational symmetry is 3 and angle of rotational symmetry is 1200.
Try this activity with a parallelogram and a rhombus.
If we know the order of rotational symmetry, we can find the angle of rotation using the following formula.From the above activity, Angle of rotation =Order of Rotation
3600 Angle of rotation = 3
3600
= 1200 And the order of rotation = Angle of rotation
3600 =
120360
0
0
= 3
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Every rotating figure occupies the same position after a rotation of 360°. Its order of rotation is one.
Do you know, what this is ? This is a circular disc. Its centre is fixed on a nail, so that, the disc can be easily rotated.
This is an example for rotational symmetry.
Can you find the order of this ?
Yes it is the number of black or white triangles on the board. That is, the order of this disc is 10.
A figure which looks exactly same as the original figure more than once during a complete rotation is said to have different order of rotational symmetry.
• A figure has rotational symmetry when angle of rotation is less than or equal to 1800.
• All figures when rotated through 3600 come back to their original position.
• All figures have rotational symmetry of at least order 1.
• Cut out 2 squares of same size from a sheet of paper.• Draw the diagonals and locate the centre.• The point where the diagonals intersect is the centre.• Place the sheets one upon the other, so that, the two
figures match exactly.• Pierce a pin through the centre point to hold both the
sheets.• Turn the top sheet only and note, how many times the
top figure matches with the bottom figure.What is the angle of rotation and order of rotation? Would
it change, if you change the direction of rotation?
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Example 1 : If we rotate the figure through 900, 1200, 1800 about O, then at which angle does the image look like the original one?. Also find the order of rotational symmetry.
Solution: When we rotate above figure about O through 900, we get shape. We get rotation through 1200, we get (not symmetry)
Rotation through 1800 we get,
We can have the rotational symmetry if it rotates about 1800.
∴Order of rotational symmetry 180360 20
0
= =
Example 2 :
Find the order of rotational symmetry of the equilateral triangle ABC.
Solution: When an equilateral triangle is rotated about its central point O, it attains its original form at 1200.
∴Order of rotational symmetry 120360 3cc= =
O
A
CB
There are some figures, which have both
reflection and rotational symmetry.
Figure A Figure B
Figure B is the reflection of Figure A.
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Rotation of this figure around the point of rotation through the angle 900 are as shown in the following.
The square has both reflection and rotational symmetry.Now you identify the reflection of Figure 1 from the following
figures.
Fig 1 (a) (b) (c)
The equilateral triangle has both reflection and rotational symmetry.
Consider a regular pentagon.
Fig :A Fig :B
Figure B is the reflection of Figure A. Now rotate it through 720 around the point of rotation.We get the following figures.The fifth figure is coinciding the original
figure after 4 rotations.
Fig (1) Fig (2) Fig (3) Fig (4) Fig (5)
The regular pentagon has both reflection and rotational symmetry.
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Try This: Identify some letters of the English alphabet which have both reflection and rotational symmetry.
Cut figures of H, N, O, S, X, Z and visualize it. Look at the table given below which is very informative.
Shapes Polygon sides lines of
Symmetry
Angle of
rotation
Order of rotational symmetry
Equilateral Triangle 3 3 1200 3
Rectangle 4 4 900 4
Rhombus 4 4 900 4
Regular Pentagon 5 5 720 5
Regular Hexagon 6 6 600 6
Note the following from this table.Number of sides, the number of lines of symmetry and the
order of rotational symmetry are same in each regular polygon.
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Try this : Observe the giant wheel. Try to find the order of rotational symmetry of giant wheel.
….Oops I lost the count…, I cannot count....
Activity 12: Locate an item in the room that has symmetry and ask a friend to try and guess which object you have selected. Offer 10 clues and allow the partner to guess after each clue. If the partner guesses the object then it is his/her turn to locate an object that has symmetry and offer you the clues. Points to be remembered:
• Rotation may be clockwise or anti-clockwise.• Every figure will have rotational symmetry of at least
order 1.• The point at which the figure is rotated is called the
point of rotation.• The angle of turning during rotation is called the angle
of rotation.• Number of times the figure matches with the original
figure in one complete rotation is called the order of rotation.
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Exercise 8.2
1) Consider the letters of the alphabet S and I . Mark the point of symmetry and also find the order of rotational symmetry.
2) Write three letters of the English alphabet which have reflection symmetry but not rotational symmetry.
3) Write three letters of English alphabet which have both reflection and rotational symmetry.
II. 1) What is the order of rotational symmetry of a ceiling fan with (i) 3 blades (ii) 4 blades ?
2) Write some numbers which have rotational symmetry.3) This star is made up of black and white
equilateral triangles: What is the order of rotational symmetry of
the star ?
III. 1) Draw an equilateral triangle ABC with side 2 cms, rotate the triangle through 900 and 1800 with B as the point of rotation and find the image.
2) Take an isosceles triangle OAB in which OA = AB. O be the fixed point, rotate it through 900, 1800 and 2700. Draw the diagrams to show the rotations.
IV. 1) Identify the rotational symmetry and reflection symmetry in the following figure.
E E E EE E(a) (b) (c)
(a) (b) (c)
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10) Find the order of rotational symmetry in i) rectangle ii) rhombus
11) Fill in the table given below.
Alphabet Letters
Reflection symmetry Yes /No
No. of lines of
symmetry
Rotational symmetry Yes /No
Order of rotational symmetry
I
B
O
H
E
G
Z
S
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Chapter - 1 Integers
Exercise 1.1 I. 1) 13, 2) -5, 3) 1 4) - 9
Exercise 1.21. x -3 6 11 -5
7 -21 42 77 -354 -12 24 44 -20-8 +24 -48 -88 +40-2 +6 -12 -22 +1012 -36 72 132 -60-9 +27 -54 -99 +45-14 +42 -84 -154 +70
2.
85
-5-6
7-2
1-25
+3010
-5
-40
-35
Exercise 1.3 I. 1) -15, 2) -24, 3) -238, 4) 276, 5) -126, 6) 24,7) -60
II. 1) 4 7 28#+ + = +
2) 3 5 15#- + = -
3) 9 7 63#+ - = -
4) 6 8 48#- - = +
III. 1) b) A cm - 2 × 6 cm 2) ` 62000 - 8 × 7500
Exercise 1.4 I. 1) + 4, 2) -17, 3) -11, 4) 4 , 5) 0 IV. 40
II. 1) 1, 2) 1, 3) -2, 4) -12, 5) -30 III. Integer Quotient+ 24 ÷ + 12 + 2+ 24 ÷ -12 - 2- 24 ÷ +12 - 2- 24 ÷ -12 + 2
Exercise 1.5 I. 1) -10, 2) -9, 3)
-18 4) 0, 5) -2, 6) -19
Exercise 1.6 I. 1) 125, 2) 5am
16oC, 10am 26oC, 12noon 28oC, 3pm 23oC 3) ` 11,750, 4) ` 126,
5) ` 46930, 6) 177
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Chapter -2 Fractions
Exercise 2.1 : I. Proper Fractions : ,127
203
Improper Fractions : ,1618
45 Mixed Fractions : ,2 6
1 6 92
II 1) ,126189 , 2) ,10
81512 , 3) ,20
143021 III 1) 3
2 , 2) 53 , 3) 2
1
IV 1) 1 87 , 2) 212
1 , 3) 24 41 , V 1) 4
15 , 2) 217 , 3) 6
29
VI 1) 1229 2) 20
93 3) 51 4) 1 30
23
Exercise 2.2 : I 1) 41 5# , 2) 2
1 3# , 3) 83 2# , 4) 3
2 4#
II A. 1) 74 , 2) 5
12 , 3) 910 , 4) 4
21 , 5) 29 , 6) 3
20 , 7) 9, 8) 143 ,
B. 1) 11 41 , 2) 28, 3) 26 5
2 , 4) 22 21 , 5) 13 5
4 , 6) 2 53 , 7) 30, 8) 111
III A. 1) 4, 2) 6, 3) 2, 4) 8, 5) 15, 6) 14, 7) 521 , 8) 9
20
B. 1) 30, 2) a) 15, b) 24, c) 21
Exercise 2.3 : I 1) 95 ,2) 32
21 , 3) 4528 4) 3 5) 27
7 , 6) 1011 7) 35
36 8) 2413
II 1) 1 2013 , 2) 315
1 , 3) 461 , 4) 1 7
6 , 5) 11 21 , 6) 7 8
7 , 7) 17 97 8) 20
III 1) 2) 3) 4)
IV. 1) 70 k.m. 2) ` 414, 3) 10 21 hour, 4) ` 465
1 , 5) 8 43 sq.cm.
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Exercise 2.4 : I. 1) 1) 18, 2) 352 , 3) 90, 4) 3
140 , 5) 6 73 , 6) 3 23
1
II) 1) 25 , 2) 7
9 , 3) 12, 4) 136 , 5) 9
1 , 6) 43 III) 1) 15
2 2) 247 3) 91
6 4) 209
5) 2811 6) 4
1 IV) 1) 815 2) 12
35 3) 185 4) 15
8 5) 2 41 6) 2 42
31
V 1) ` 1 43 2) 85 3) ` 9 2
1 4) 27
Exercise 2.5 : I 1) 43 2) 310
2 3) 47801 4) 2 20
13
II 1) m6201 2) kg4 4
1 3) 5m
Chapter -3 Rational Numbers
Exercise 3.1 : I 1) Yes 2) No 3) Yes 4) No 5) Yes 6) Yes 7)No
II 1) 57 2) 5
4 3) 83- 4) 7
4 III 1) , , ,184276368
4510- - - - 2) , , ,20
6309
4012
5015
3) , , ,108
1512
2016
2520
- - - - 4) , , ,192160
288240
384320
480400 V , , ,6
7710
97
611
--
--
, ,54
35152-
-- VI 1) 16 2) -20 3) 7 4) 3 5) -28 6) 30
Exercise 3.2 : I. 1) 1011 2) 9
8 3) 307 4) 42
107
II. 1) 83 2) -1 3) 3
1- 4) 81
III.1) 8 kg125 2) 1 kg20
11 3) m8453 4) 1 l15
14
Exercise 3.3 : I 1) 83 2) 48
5- 3) 101- 4) 9
1 5) 61 6) 1 II 1) 27
7 2) 43 3) 5
6
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4) 78 III.1) 5
3 2) 49- 3) 10 4) 2
7- 5) 51 6) 0 IV 1) 9
10 2) 56- 3) 13
18- 4) 57-
V.9 VI 121 m VII 8 54 sq.m VIII 104
Exercise 3.4 :I. 1) 0.8 2) 0.375 3) 0.3125 4) 0.2142857142...
Never ending numbers = 121 , 16
1
Ending numbers = 85 , 9
2
Exercise 3.5 :I. 1) 34.4 2) 7.5 3) 36.945 4) 3.328 5) 0.04368
6) 1.68 II) 1) 0.07 2) 0.012 3) 0.497 4) 27 5) 2.2 6) 430
III) 1) ` 717.86 2) 120.06 km 3) 10.2l 4) ` 20.6m
Exercise 3.6 :
1) a) 2510 g b) 0.0725 g
2) a) 0.625 kg b) 0.0108 kg
3) a) 1.45 km b) 0.173 km c) 0.2135 km
4) a) 2700 m b) 75.25 m c) 1580 m
5) a) 1250 cm b) 470000 cm
6) 11.59 kmChapter - 4 Algebraic Expression
Exercise 4.1 1) x 2) a 3) xy 4) x2y 5) a2b 6) m2n2 7) xyz 8) m2np
II 1) 2 2) -5 3) 1 4) -8 5) 9 6) 43 7) 0.5 8) 0.008
Exercise 4.2 I Variable number : 1) 8 + 5 - 3 2) 3x - 8 5) (20 ×
7) - (5 × 10) - 45
Algebraic term :3) (7 × 6) -4m 4) 3p + 4q 6) 2y + 6 - 4z II 1) 3x, + 4y
2) 2pq - 8qr 3) 3x2, -3x, +z 4) ab, + bc, -ca 5) 9m, + 6n 6) 6, -3xy, +x, -y
2)1) xy, 2, 5 2) x2, 1, xy, 4 3) p, 3, y2, -5 4) ab, 2, bc, 1, ca, 1
III 1) x + 8 2) y - 7 3) p 12 = 12p 4) q5 5) 4x + 3y 6) 10 - 5y
7) pq + 3x 8) 5m - 3l 9) 10y + 15 10) zyx IV 1) 3 2) -4 3) 4
3 4) 10.5
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5) 168 6) q8- V) 1) m 2) -xy 3) pqr 4) c 5) mn 6) x2y
VI 1) 27 2) z 3) xy 4) xyz
Exercise 4.3 I Monomial: 2xy, xyz, 3xp ÷ q Binomial: 2x + y,
3x2 + 5x, ab - bc
Trinomial: 5 + 6a + 4b, 2y z x37- + , a2 - 3ab + c, x2 + 4 - 3x
II) x3 + x - 2 + 1, m3 + 2m2 + 3m - 4, x + 2yzIII Like terms (L) Unlike terms (U)
3x, 5x, 8x-8p2, 6p2, 10p2
2ab, 6ba, 8ab
x3, - 3x2, 8x3a2b, -2ab2, 7a2b2
-a3, 2a2, -8aIV 1) y3, -7y3, 23y3, -y3 2) 7ab, -8ba, -3ab, 4ab 3) 7p, 2p, 3p
V 1) 3 2) 32- 3) 0.3 4) 24 5) -18 6) 11
9- VI 1) 3a 2) 4a3) p2y 4) 10
5) -p2 6) -1 VII 1) 2, 2) 3, 3) 3, 4) 1, 5) 3,6) 3
VIII Base Exponentx 5
ab 613p 9-y 10
xyz 70.59 20
IX i) p5 ii) m4 iii) (ab)6
iv) z8 v) (abc)3 vi) b10
Exercise 4.4 I 1) 41 a 2) 12 p2 3) -1y + 3x 4) 6a2 - b2 + 12a
II 1) 5x 2) 13a2 3) 9xy 4) 7x2 + 6y - z III 1) 8x + 11y 2) 17a - 10b
3) -9xy-z 4) 17a2bc + 7ab2c IV 1) 6x - 3y + 3z 2) 7p2 + q2 + 11r2
3) 6a + 4b + 3c 4) 2x2y - 2y2z + 13z2x V 1) 12mn - 4p 2) 8x - 2y - 4z
3) 10a2 + a + 3 VI 10x2 + 10y + y2 + 5z + 11
Exercise 4.5 I 1) 7x 2) 8a2b 3) 16ab 4) -22x2y II 1) 5x + 2y 2) 4a + 2b
3) -4m2n + 5mn2 4) -11pqr + 11q + 4 III 1) 4x + 4y 2) -6c + 4d
3) x-6y - 14 5) 2xy - 10 VII 1) 27 2) Z 3) xy 4) xyz
Exercise 4.6 I 1) 280 xyz 2) 40 pqr 3) -144 mnp 4) -8abc 5) -60x3
6) +18 p6 II 1) 84 x2y2z2 2) + 210 a2b2c2 3) -12x4y3z 4) -24a2b3c4
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III1) 12xz + 14yz 2) -15p2 + 25pq 3) -36xz - 2yz 4) 12a2b2c + 18ab2c2
IV 1) x2 + 12x + 32 2) 5n2 - 13n - 6 3)6a2+ab-b2
4) 25x2 - 4p2 5) 2x2 - 13x + 21Chapter - 5 Pair of Angles
Exercise 5.1 Look at these figures and fill up the table given below
Figure Angles Arms Common Vertex
Common Arm
Angles are adjacent or
not
i ,AOB BOC OA, OB, OC,
O OB Yes ,AOB BOC
ii , ,PQR RQSSQJ
QP,QR,QS,QJ
Q QRQS
No
iii ,LMN KLM LK, LM, MN
- LM No
iv ,DEF FEG ED, EF, EG
E DE Yes ,DEF FEG
v ,EOG GFH OE, OG, FG, GH
- - No
vi ,OAB DOC AB, OA,OC, OD
- OC Yes
vii ,HOG GOJ OH, OG, OJ
O OG Yes
2. In the given figure, which of the following are pairs of adjacent angles. Mentioning the common vertex and common side.
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AnglesN
ame
of
the
vert
ex
Nam
e of
th
e ar
ms
Com
mon
V
erte
x
Com
mon
A
rm
Pair
s of
ad
jace
nt
angl
es (Y
es/
No)
,AOBBOD
0 , ,OA OB OD Yes Yes OB
Yes
,AODBOC
0 , , ,OA OD OB OC Yes No No
,AOCBOC
0 , ,OA OB OC Yes Yes OC
Yes
,BOCAOB
0 , ,OA OB OC Yes Yes OB
Yes
,AODAOB
0 , ,OA OB OD Yes Yes OB
Yes
3. Adjacent angles : i) DOC and COE ii) POR and POQ
iv) JOK and KOL vii) QOM and MON ,
Exercise 5.2 I. 1) 500, 2) 700, 3) 350 4) 520 II. a) 680, b) 600 c)
69.50 d) 49.50 III. 22.50 and 67.50
Exercise 5.3 1.a) AOD BOD AOB+ = b)
AOE EOB AOB+ =
c) AOC COB AOB+ = 2. (i), (ii), (iii), (iv) Supplementary angle
3. a) 850 b) 700 c) 59.50 d) 54.50 4. 650 5. 360 and 1440 6. i) 1250,
ii) 350 iii) 900 iv) 900
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Exercise 5.4 1) It is not a supplementary angle. 2) (ii), (iv),
(v) and (vi) Pair of angles 3) a) Obtuse angle b) Acute angle c)
Right angleExercise 5.5 1) ,
,AODBODBOC
12060120
0
0
0
===
2) ,,
AOEBOCEOD
405090
0
0
0
===
3)
BOCBOD
15030
0
0
==
Exercise 5.6 I 1) Obtuse Angle 2) Acute Angle
3) Right Angle II i) 1050, ii) 400 iii) 32.50 iv) 350
III
90454545
AOC BODAOD COBAOR SOBDOR COSPOC DOQAOP BOQ
90
45
0
0
0
0
0
0
`
= == == == == == =
IV x = 450, z = 27, y = 1350
V i) &AOD AOCii) a) &AOE BOE b) &COA AOD iii) &AOC BOD iv) &AOC COEVI) 1) 900 2) 1800 3) Supplementary angles 4)Pair of angles 5) Obtuse angle
Chapter - 6 Pair of Lines
Exercise 6.1 1.a) RNK b) QMN c) MNR d) SNK e) MNR f) QMN
2. a) EIJEIJ AIKEIJ IKLIKL CKFJIK BJLAIE JIK
70707070110110
0
0
0
0
0
0
== == == == == =
[Opposite angles][Corresponding angles]
[ Opposite angles][Corresponding angles][Opposite angles
b) [Opposite angles][Corresponding angles][Opposite angles]
GIB AIJAIJ CJKDJKDJK FKH
6060
180 60 120120
0
0
0 0 0
0
= == == - == =
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c)
EHB = GHF = 50o [Opposite anglesGHF = AGD = 50o Corresponding angles
3. Alternative anglesCorresponding angles
PARPST QPSPST QRS
50130130
0
0
0
== == =
4. , ,BAC ABC ACB80 45 550 0 0= = =
5. a) AB # CD b) AB CD
6. 70DGCGEF 70
0
0
==
7. a b c d
(i) 800 500 1000 1300
(ii) 700 800 700 -
(iii) 500 1300 1300 -
(iv) 400 2600 600 -
(v) 1300 700 1300 1100
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Chapter - 7 Properties of Triangles
Exercise 7.1 1800
Exercise 7.2 1. a) x = 500 b) x = 800 c) x = 200 c) x=620
2. 1) C = 500 2) B = 450 3) C = 500 4) C = 340
3. 1) 500 + 600 + 700 = 1800 Yes 2) 600 + 700 + 800 = 2100 No
3) 650 + 750 + 550 = 1950 No 4) 560 + 640 + 600 = 1800 Yes
5) 570 + 640 + 790 = 2000 No
4. C = 700 5. C = 700 6. x = 550 x =900
7. x = 600 8. C = 950
Exercise 7.3 a) x = 800, y = 1000 b) y = 700, x = 1100
c) y = 350, x = 750 d) x = 650, y = 1150
2. ∠ACB = 600 ∠ACD = 1200 3. ∠QRP = 500, ∠RPQ = 800
4. ∠KLN = 1150, ∠NLM = 650, ∠LMN = 400 5. ∠ACD = 1050
Exercise 7.5 1. AC = 10, 2. AC = 13 cms, 3. AB = 9 m 4. b) c)
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Chapter - 8 SymmetryExercise 8.1
I. a c d
II. a) 1 b) 2 c) 2III a) b) c) d) e) f) g)
No Yes Yes Yes Yes No Yes h) i) j) k) l)
No No Yes Yes Yes IV a) 1 b) 3 c) 4 d) 4 e) Infinity
f) 4 g) 4 h) 5 i) 6
V C
D
ER
A - B
i) C -
ii) D -
iii) E -
iv) R -
Exercise 8.2 I. 1) S I (2,2)
2) M W E C U
3) O N Z
II. 1) 1200, 3 2) 900, 4
III 2) 0,8,1,6,9 3) 3
11) English
Alphabets Reflection Symmetry Yes / No
No of lines of
Symmetry
Rotational Symmetry Yes / No
Order of Rotational Symmetry
I Yes 1 Yes 4B Yes 1 Yes 4O Yes Many Yes ManyH Yes 2 Yes 2E Yes 1 Yes 4G No - Yes 4Z Yes - Yes 2S Yes - Yes 2
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