9em Mat PDF
9em Mat PDF
9em Mat PDF
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CHILDREN! THESE DIINSTRUCTIONS FOR YOU...
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♦ For each and every conceptual understanding, a real life context with appropriate
illustrations are given in the textbook. Try to understand the concept through keen
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reading of context along with observation of illustration.
♦ While understanding the concepts through activities, some doubts may arise.
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Clarify those doubts by through discussion with your friends and teachers,
understand the mathematical concepts without any doubts.
♦ ''Do this/Do these" exercises are given to test yourself, how far the concept has
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been understood. If you are facing any difficulty in solving problems in these
exercises, you can clarify them by discussing with your teacher.
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♦ The problems given in "Try this/try these", can be solved by reasoning, thinking
creatively and extensively. When you face difficulty in solving these problems,
you can take the help of your friends and teachers.
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♦ The activities or discussion points given "Think and disicuss" have been given
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for extensive understanding of the concept by thinking critically. These activities
should be solved by discussions with your fellow students and teachers.
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♦ Different typs of problems with different concepts discussed in the chapter are
given in an "Exercise" given at the end of the concept/chapter. Try to solve these
problems by yourself at home or leisure time in school.
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♦ The purpose of "Do this"/do these", and "Try this/try these" exercises is to solve
problems in the presence of teacher only in the class itself.
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♦ Wherever the "project works" are given in the textbook, you should complete
them in groups. But the reports of project works should be submitted individually.
♦ Try to solve the problems given as homework on the day itself. Clarify your
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doubts and make corrections also on the day itself by discussions with your
teachers.
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♦ Try to collect more problems or make new problems on the concepts learnt and
show them to your teachers and fellow students.
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♦ Try to collect more puzzles, games and interesting things related to mathematical
concepts and share with your friends and teachers.
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♦ Do not confine mathematical conceptual understanding to only classroom. But,
try to relate them with your suroundings outside the classroom.
♦ Student must solve problems, give reasons and make proofs, be able to
communicate mathematically, connect concepts to understand more concepts &
solve problems and able to represent in mathematics learning.
♦ Whenever you face difficulty in achieving above competencies/skills/standards,
you may take the help of your teachers.
MATHEMATICS
CLASS - IX
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TEXTBOOK DEVELOPMENT & PUBLISHING COMMITTEE
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Chief Production Officer : Sri. A. Satyanarayana Reddy,
Director, SCERT, Hyderabad.
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Director, Govt. Text Book Press, Hyderabad.
Organising Incharge :
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Dr. Nannuru Upender Reddy,
Prof. & Head, Curriculum & Text Book Department,
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SCERT, Hyderabad.
Chairperson for Position Paper and Mathematics Curriculum and Textbook Development
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Prof. V.Kannan,
Department of Mathematics and Statistics,
Hyderabad Central University, Hyderabad
Chief Advisors
Sri Chukka Ramaiah Dr. H.K.Dewan
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Published by
The Government of Telangana, Hyderabad
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© Government of Telangana, Hyderabad.
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All rights reserved.
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writing of the publisher, nor be otherwise circulated
in any form of binding or cover other than that in
which it is published and without a similar condition
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including this condition being imposed on the sub-
sequent purchaser.
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The copy right holder of this book is the Director
of School Education, Hyderabad, Telangana
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Printed in India
at the Telangana Govt. Text Book Press,
Mint Compound, Hyderabad,
Telangana.
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Text Book Development Committee
Writers
Sri. Tata Venkata Rama Kumar Sri. Gottumukkala V.B.S.N. Raju
H.M., ZPPHS, Mulumudi, Nellore Dt. SA, Mpl. High School, Kaspa, Vizianagaram.
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PGT. APTWRS, Chandrashekarapuram, Nellore SA, ZPHS,Thakkasila, Alampur Mandal Mahabubnagar Dt.
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PGT.APTWR School of Excellence, Srisailam. SGT, MPUPS,Chamallamudi, Guntur Dt.
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SA,GHS, Zamisthanpur, Manikeshwar Nagar, Hyd. Lecturer, Govt D.I.E.T., Vikarabad, R.R. Dt.
Dr. A. Rambabu
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H.M.,St. Peter’s High School, R.N.Peta, Nellore. Lecturer, Government CTE, Warangal
Editors
Dr. S Suresh Babu Prof. N.Ch.Pattabhi Ramacharyulu (Retd.) Sri. K Brahmaiah
Professor, Dept. of Statistics, National Institute of Technology, (Retd.)
SCERT, Hyderabad Warangal. Prof., SCERT, Hyderabad
Co-ordinators
Sri Kakulavaram Rajender Reddy Sri K.K.V Rayalu
Resource Person, SCERT, Hyderabad Lecturer, IASE, Masab Tank, Hyderabad
Sri Inder Mohan Sri Yashwanth Kumar Dave Sri Hanif Paliwal Sri Asish Chordia
Vidyabhawan Society Resource Centre, Udaipur
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Foreword
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studied upto the upper primary stage to the study of mathematics as a discipline. The logical
proofs of propositions, theorems etc. are introduced at this stage. Apart from being a specific
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subject, it is to be treated as a concommitant to any subject involving analysis as reasoning.
I am confident that the children in our state of Telangana learn to enjoy mathematics,
make mathematics a part of their life experience, pose and solve meaningful problems,
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understand the basic structure of mathematics by reading this text book.
For teachers, to understand and absorb critical issues on curricular and pedagogic
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perspectives duly focusing on learning in place of marks, is the need of the hour. Also coping
with a mixed class room environment is essentially required for effective transaction of curriculum
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in teaching learning process. Nurturing class room culture to inculcate positive interest among
children with difference in opinions and presumptions of life style, to infuse life into knowledge
is a thrust in the teaching job.
The afore said vision of mathematics teaching presented in State Curriculum Frame
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work (SCF -2011) has been elaborated in its mathematics position paper which also clearly
lays down the academic standards of mathematics teaching in the state. The text books make
an attempt to concretize all the sentiments.
The State Council for Education Research and Training Telangana appreciates the hard
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work of the text book development committee and several teachers from all over the state
who have contributed to the development of this text book. I am thankful to the District
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Educational Officers, Mandal Educational Officers and head teachers for making this possible.
I also thank the institutions and organizations which have given their time in the development
of this text book. I am grateful to the office of the Commissioner and Director of School
Education (T.S.) and Vidya Bhawan Society, Udaipur, Rajastan for extending co-operation in
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developing this text book. In the endeavor to continuously improve the quality of our work,
we welcome your comments and suggestions in this regard.
The Government of Telangana has decided to revise the curriculum of all the subjects
based on State Curriculum Frame work (SCF - 2011) which recommends that children’s life
at schools must be linked to their life outside the school. Right to Education (RTE - 2009)
perceives that every child who enters the school should acquire the necessary skills prescribed
at each level upto the age of 14 years. The introduction of syllabus based on National
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Curriculum Frame Work - 2005 is every much necessary especially in Mathematics and
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Sciences at secondary level with a national perspective to prepare our students with a strong
base of Mathematics and Science.
The strength of a nation lies in its commitment and capacity to prepare its people to
meet the needs, aspirations and requirements of a progressive technological society.
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The syllabus in Mathematics for three stages i.e. primary, upper primary and secondary
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is based on structural and spiral approaches. The teachers of secondary school Mathematics
have to study the syllabus of classes 8 to 10 with this background to widen and deepen the
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understanding and application of concepts learnt by pupils in primary and upper primary
stages.
The syllabus is based on the structural approach, laying emphasis on the discovery and
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l The syllabus has been divided broadly into six areas namely, Number System, Algebra,
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situations.
The text book attempts to enhance this endeavor by giving higher priority and space to
opportunities for contemplations. There is a scope for discussion in small groups and
activities required for hands on experience in the form of ‘Do this’ and ‘Try this’. Teacher’s
support is needed in setting the situations in the classroom.
Some special features of this text book are as follows
l The chapters are arranged in a different way so that the children can pay interest to all
curricular areas in each term in the course of study.
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Care has been taken to see that every theorem is provided initially with an activity for easy
understanding of the proof of those theorems.
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l Continuous Comprehension Evaluation Process has been covered under the tags of ‘Try
this’ and ‘Think, Discuss and Write’. Exercises are given at the end of each sub item of the
chapter so that the teacher can assess the performance of the pupils throughout the chapter.
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l Entire syllabus is divided into 15 chapters, so that a child can go through the content well in
bit wise to consolidate the logic and enjoy the learning of mathematics.
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l Some interesting and historical highlights are given under titles of Brain teasers, Do you
child can visualise the rational and irrational numbers by the representation of them on number
line. Some history of numbers is also added e.g value of to create interest among students. The
representation of real numbers on the number line through successive magnification help to
visualise the position of a real number with a non-terminating recurring decimal expansion.
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Chapter (2) Polynomials and Factorisation under the area algebra dealt with polynomials in one
variable and discussed about how a polynomial is diffierent from an algebraic expression.
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Facrtorisation of polynomials using remainder theorem and factors theorem is widely discussed
with more number of illustrations . Factorisation of polynomials were discussed by splitting the
middle term with a reason behind it. We have also discussed the factorisation of some special
polynomials using the identities will help the children to counter various tuypes of factorisation.
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Chapter (3) Linear equations in two variables under the same area will enable the pupil to
discover through illustative examples,the unifying face of mathematical structure which is the
ultimate objective of teaching mathematics as a system. This chapter links the ability of finding
unknown with every day experience.
There are 7 chapters of Geometry i.e (3 ,4,7,8,11,12, and 13 ) were kept in this book. All these
chapters emphasis learning geometry using reasoning , intutive understanding and insightful
personal experience of meanings. It helps in communicating and solving problems and obtaining
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new relations among various plane figures. Development geometry historically through
centuries is given and discussed about Euclid’s contribution in development of plane geometry
through his collection “The Elements” . The activities and theorem were given on angles,
triangles, quadrilaterals , circles and areas. It will develop induction, deduction, analytical
thinking and logical reasoning. Geometrical constructions were presented insuch a way that
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the usage of an ungraduated ruler and a compass are necessary for a perfect construction of
geometrical figures.
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Chapter (5) deals with coordiante geometry as an alternate approach to Euclidean geometry
by means of a coordinate system and associated algebra. Emphasis was given to plot
ordered pairs on a cartesian plane ( Graph ) with a wide variety of illustratgive examples.
Chapter (9) statistics deals with importance of statistics , collection of statistical data i.e
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grouped and ungrouped , illustrative examples for finding mean, median and mode of a
given data was discussed by taking daily life sitution.
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Chapter (14) Probability is entirely a new chaper for secondary school students was
introduced with wide variety of examples which deals with for finding probable chances of
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success in different fields. and mixed proportion problems with a variety of daily life situations.
Chapter (10) surface areas and volumes we discussed about finding curved (lateral) surface
area, total surface area and volume of cylinder, cone and sphere. It is also discussed the
relation among these solids in finding volumes and derive their formulae.
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Chapter (15) Proofs in mathematics will help ;the students to understand what is a
mathematical statement and how to prove a mathematical statement in various situations.
We have also discussed about axiom , postulate, conjecture and the various stages in proving
a theorem with illustrative examples. Among these 15 chapters the teacher has to Real
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lines and angles, Statistics, Surface as a part of are volume, Circles Geometrical constructions
and probability under paper - II.
The success of any course depends not so much on the syllabus as on the teacher and the
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teaching methods she employs. It is expected that all concerned with the improving of
mathematics education would extend their full co operation in this endeavour.
Mere the production of good text books does not ensure the quality of education, unless
the teachers transact the curriculum the way it is discussed in the text book. The involvement
and participation of learner in doing the activities and problems with an understanding is
ensured. Therefore it is expected that the teachers will bring a paradigm shift in the classroom
process from mere solving the problems in the exercises routinely to the conceptual
understanding, solving of problems with ingenity.
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Highlights from History
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One day in an Arithmetic class on
division the teacher said that if three = 1 + (2 × 4)
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bananas were given to three boys, each
boy would get a banana and he generalised = 1 + 2 16
this idea. Then Ramanujan asked “Sir, if
Ramanujan = 1 + 2 1 + 15
no banana is distributed to no
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1 1 student will every one still get a
= 1 + 2 1 + (3 × 5)
+ 2 = 1
4 2 banana?”
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4
1
+ (2 × 3) = 2
2
2
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Ramanujan’s math ability
and so on ...
taught, he feasted on “Loney’s Trigonometry” at the age of 13, and at the age of 15, his senior
friends gave him synopsis of Elementary results in pure and Applied mathematics by George
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Carr. He used to write his ideas and results on loose sheets. His filled note books are now
famous as “Ramanujan’s Frayed note books”. Though he had no qualifying degree, the
university of Madras granted him a monthly Scholarship of Rs. 75 in 1913. He had sent papers of
120 theorems and formulae to great mathematican G.H. Hardy (Combridge University, London).
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They have recognised these as a worth piece and invited him to England. He had worked with
Hardy and others and presented numerical theories on numbers, which include circle method in
number theory, algebra inequalities, elliptical functions etc. He was second Indian to be elected
fellow of the Royal Society in 1918. He became first Indian elected fellow of Trinity college,
Cambridge. During his illness also he never forget to think about numbers. He remarked the taxi
number of Hardy, 1729 is a singularly unexceptional number. It is the smallest positive integer
that can be represented in two ways by the sum of two cubes; 1729 = 13+123 = 93+103.
Unfortunately, due to tuberculosis he died in Madras on April 26, 1920. Government of India
recognised him and released a postal stamp and declared 2012 as “Year of Mathematics” on
the eve of his 125th birth anniversary.
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2 Polynomials and Factorisation June, July 27-58
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3 The Elements of Geometry July 59-70
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5 Co-Ordinate Geometry December 107-123
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Linear Equation in Two variables August, September 124-147
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7 Triangles October, November 148-173
Jana-gana-mana-adhinayaka, jaya he
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Bharata-bhagya-vidhata.
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Punjab-Sindh-Gujarat-Maratha
Dravida-Utkala-Banga
Vindhya-Himachala-Yamuna-Ganga
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Uchchala-Jaladhi-taranga.
Tava shubha name jage,
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Tava shubha asisa mage,
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Gahe tava jaya gatha,
Jana-gana-mangala-dayaka jaya he
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Bharata-bhagya-vidhata.
Jaya he, jaya he, jaya he,
Jaya jaya jaya, jaya he!
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PLEDGE
- Pydimarri Venkata Subba Rao
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1
Real Numbers
01
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1.1 I NTRODUCTION
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Let us have a brief review of various types of numbers.
Consider the following numbers.
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1 3 13 −5 213 −69 22
7, 100, 9, 11, -3, 0, − , 5, 1, , −1, 0.12, − , 13.222 ..., 19, , , , ,
4 7 17 3 4 1 7 5.6
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John and Sneha want to label the above numbers and put them in the bags they belong to.
Some of the numbers are in their respective bags..... Now you pick up rest of the numbers and
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put them into the bags to which they belong. If one number can go in more than one bag then
copy the number and put them in the relevent bags.
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Q
Z 30, 7, 100,
-9,
0,-10,
7, , 9, 7,
W 7 -3, 37
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7, 7,
0, 9, −3,3, 7
0, 1,
7,7,8, N 10,100
11, ... -2,-3
100 100
100, 101
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You have observed bag N contains natural numbers. Bag W contains whole numbers.
Bag Z contains integers and bag Q contains rational numbers.
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The bag Z contains integers which is the collection of negative numbers and whole numbers.
It is denoted by I or Z and we write,
Z = {... −3, −2, −1, 0, 1, 2, 3, …}
p
Similarly the bag Q contains all numbers that are of the form where p and q are integers
q
and q ≠ 0.
You might have noticed that natural numbers, whole numbers, integers and rational numbers
p
can be written in the form , where p and q are integers and q ≠ 0.
q
2 IX-CLASS MATHEMATICS
−15
For example, -15 can be written as ; here p = -15 and q = 1. Look at the Example
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1 2 10 50
= = = ... and so on. These are equivalent rational numbers (or fractions). It
2 4 20 100
means that the rational numbers do not have a unique representation in the form p/q, where p
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and q are integers and q ≠ 0. However, when we say p/q is a rational number or when we
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represent p/q on a number line, we assume that q 0 and that p and q have no common factors
other than the universal factor ‘1’ (i.e., p and q are co-primes.) There are infinitely many
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fractions equivalent to , we will choose i.e., the simplest form to represent all of them.
2 2
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You know that to represent whole numbers on number line. We draw a line and mark
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a point ‘0’ on it. Then we can set off equal distances on the right side of the point ‘0’ and
label the points of division as 1, 2, 3, 4, …
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0 1 2 3 4
The integer number line is made like this,
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− 5 − 4 − 3 −2 − 1 0 1 2 3 4 5
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To recall this, let’s first take the fraction and represent it pictorially as well as on number
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line.
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We know that in , 3 is the numerator and 4 is the denominator.
4
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Which means that 3 parts are taken out of 4 equal parts from a given unit.
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Here are few representations of .
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1234
4444
-1 0 1 1 2
3 3 2
4 (Pictorially) 4 (Number line)
5 5
Example-1. Represent and − on the number line.
3 3
Solution : Draw an integer line representing −2, −1, 0, 1, 2.
-7 -2 -5 -4 -1 -2 -1 0 1 2 1 4 5 2 7
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3 (= -6 ) 3 3 ( = -3 ) 3 3 3 3 (= 3 ) 3 3 (= 6 ) 3
3 3 3 3
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Divide each unit into three equal parts to the right and left sides of zero respectively.
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Take five parts out of these. The fifth point on the right of zero represents and the fifth
3
−5
one to the left of zero represents .
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DO THIS
1. Represent
−3
on number line.
N2. Write 0, 7, 10, -4 in p/q form.
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3. Guess my number : Your friend chooses an integer between 0 and 100.
You have to find out that number by asking questions, but your friend can
answer only in ‘yes’ or ‘no’. What strategy would you use?
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Example-2. Are the following statements True? Give reasons for your answers with an example.
i. Every rational number is an integer.
ii. Every integer is a rational number
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Solution :
Method-I : We know that the rational number that lies between two rational numbers
a+b
a and b can be found using mean method i.e. .
2
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a+b
Here a = 3 and b = 4, (we know that is the mean of two integers ‘a’, ‘b’ and it lie
2
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between ‘a’ and ‘b’)
(3 + 4) 7 7
So, = which is in between 3 and 4. 3 < < 4
2 2 2
7
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If we continue the above process, we can find many more rational numbers between 3 and
2
3 + 72 6+2 7 132 13 13
= = = =
2
13 7
3< < < 4
2 2 2× 2 4
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4 2
Method-II : The other option to find two rational numbers in single step.
Since we want two numbers, we write 3 and 4 as rational numbers with denominator 2 + 1 = 3
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3 6 9 4 8 12 16
i.e., 3 = = = and 4 = = = =
1 2 3 1 2 3 4
10 11
Then you can see that , are rational numbers between 3 and 4.
3 3
9 10 11 12
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Now if you want to find 5 rational numbers between 3 and 4, then we write 3 and 4 as
rational number with denominator 5 + 1 = 6.
18 24 18 19 20 21 22 23 24
i.e. 3 = and 4 = 3= < , , , , < =4
6 6 6 6 6 6 6 6 6
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From this, you might have realised the fact that there are infinitely many rational numbers
between 3 and 4. Check, whether this holds good for any other two rational numbers? Thus we
can say that , there exist infinite number of rational numbers between any two given rational
numbers.
DO THIS
i. Find five rational numbers between 2 and 3 by mean method.
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ii. Find 10 rational numbers between − and .
11 11
7 10
Example-4. Express , and 2 in decimal form.
16 7 3
Solution : 0.4375 1.428571 0.666
16 7.00000 7 10 3 2.0000
0 7 18
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70 30 20
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64 28 18
60 20 20
48 14 18
120 60 2
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112 56
80
80 N 40
35
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0 50 2
10 49 ∴ = 0.666 = 0.6
7 ∴ =1.428571 3
∴ = 0.4375 7
16 10 is a non-terminating
is a non-terminating
recurring decimal
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DO THIS
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1 1
Find the decimal form of (i) (ii)
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p
Example -5. Express 3.28 in the form of (where p and q are integers, q ≠ 0 ).
q
328
Solution : 3.28 =
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100
328 ÷ 2 164
= =
100 ÷ 2 50
164 ÷ 2 82
= = (Numerator and denominator are co-primes)
50 ÷ 2 25
82
∴ 3.28 =
25
p
Example-6. Express 1.62 in form where q ≠ 0 ; p, q are integers.
q
Solutions : Let x = 1.626262..... (1)
multiplying both sides of equation (1) by 100, we get
100x = 162.6262... (2)
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Subtracting (2) from (1) we get
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100x = 162.6262...
x = 1.6262...
- -
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99x = 161
161
x =
161
99
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∴1.62 =
99
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TRY THESE
I. Find the decimal values of the following:
1 1 1 1
i. ii. iii. iv.
2 22 5 5× 2
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3 27 1 7
v. vi. vii. viii.
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10 25 3 6
5 1
ix. x.
12 7
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Can you tell the special character of denominator which makes the fraction as terminating
or non-terminating recurring decimals.
Write prime factors of denominator of each rational number.
What did you observe from the results?
EXERCISE - 1.1
1. (a) Write any three rational numbers
(b) Explain rational number in your own words.
2. Give one example each to the following statements.
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i. A number which is rational but not an integer
ii. A whole number which is not a natural number
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iii. An integer which is not a whole number
iv. A number which is natural number, whole number, integer and rational number.
v. A number which is an integer but not a natural number.
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3. Find five rational numbers between 1 and 2.
3 2
4. Insert three rational numbers between
−8
N 5
and
3
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8
5. Represent and on a numberline
5 5
6. Express the following rational numbers in decimal form.
242 354 2 115
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7. Express each of the following decimals in form where q ≠ 0 and p, q are integers
q
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9. Without actually dividing find which of the following are terminating decimals.
3 11 13 41
(i) (ii) (iii) (iv)
25 18 20 42
1.2 I RRATION
RRATION AL N UMBERS
TIONAL
Let us take a look at the number line again. Are we able to represent all the numbers on the
number line? The fact is that there are infinite numbers left on the number line.
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3
–16 –9 –4 0 3 8 14
To understand this, consider these equations
(i) x2 = 4 (ii) 3x = 4 (iii) x2 = 2
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For equation (i) we know that value of x for this equation is 2 and −2. We can plot
2 and −2 on the number line.
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3x 4 4
For equation (ii) 3x = 4 on dividing both sides by, 3 we get = ⇒ x = . We can
3 3 3
plot this on the number line.
When we solve the equation (iii) x2 = 2, taking square root on both the sides of the
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equation ⇒ x2 = 2 ⇒ x = ± 2 . Let us consider x = 2.
Can we represent 2 on number line ?
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What is the value of 2 ? To which numbers does 2 belong?
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Let us find the value of 2 by long division method.
1.4142135
1 2. 00 00 00 00 00 00 00 Step 1 : After 2, place decimal point.
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2824 11900
11296
28282 60400
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56564
282841 383600
282841
2828423 10075900
8485269
28284265 159063100
141421325
∴ 2 = 1.4142135 …
28284270 17641775
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But decimal number for 2 is non-terminating and non-recurring decimal. Can you
represent this using bar? No we can’t. These type of numbers are called irrational numbers and
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they can’t be represented in p/q form. That is 2 ≠ p/q (for any integers p and q, q ≠ 0).
Similarly 3 = 1.7320508075689.....
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5 = 2.2360679774998.....
These are non-terminating and non-recurring decimals. These are known as irrational
numbers and are denoted by ‘S’ or ‘Q1’.
Examples of irrational numbers N
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(1) 2.1356217528..., (2) 2, 3, π , etc.
In 5th Century BC the Pythagoreans in Greece, the followers of the famous
mathematician and philosopher Pythagoras, were the first to discover the numbers
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which were not rationals. These numbers are called irrational numbers. The Pythagoreans
proved that 2 is irrational number. Later Theodorus of Cyrene showed that
3, 5, 6, 10, 11, 12, 13, 14, 15 and 17 are also irrational numbers.
There is a reference of irrationals in calculation of square roots in Sulba Sutra
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(800 BC).
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Now can you identify which are rational and which are irrational?
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THINK DISCUSS AND WRITE
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2 p
Kruthi said 2 can be written as which is in form. So 2 is a rational number.
1 q
Do you agree with her argument ?
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Know About π
c
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π is defined as the ratio of the circumference (C) of a circle to its diameter (d). i.e. π =
As π is in the form of ratio, this seems to contradict the fact that π is irrational. The
d
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circumference (C) and the diameter (d) of a circle are incommensurable. i.e. there does not
exist a common unit to measure that allows us to measure the both numerator and denominator.
If you measure accurately then atleast either C or d is irrational. So π is regarded as irrational.
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The Greek genius Archimedes was the first to compute the value of π . He showed
the value of π lie between 3.140845 and 3.142857. (i.e., 3.140845 < π < 3.142857)
Aryabhatta (476-550 AD), the great Indian mathematician and astronomer, found the
value of π correctly upto four decimal places 3.1416. Using high speed computers and
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advanced algorithms, π has been computed to over 1.24 trillion decimal places .
π =3.14159265358979323846264338327950 ….. The decimal expansion of π is non-
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22
terminating non-recurring. So π is an irrational number. Note that, we often take as
7
22
an approximate value of π , but π ≠ .
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We celebrate March 14th as π day since it is 3.14 (as π = 3.14159 ....). What a
coincidence, Albert Einstein was born on March 14th, 1879!
TRY THESE
Find the value of 3 upto six decimals.
A
numbers only. Is it true? Can’t you represent 2 on number line? Let us discuss and locate
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irrational numbers such as 2 , 3 on number line.
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By Pythagoras theorem OB = 12 + 12 = 2
N C
2
B
1
2
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O
1 A1
-7 -3 -5 -2 -3 -1 -1 0 1 K3 2 5 3 7
2 2 2 2 2 2 2 2
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Fig. (i)
We have seen that OB = 2 . Using a compass with centre O and radius OB, draw an arc
on the right side to O intersecting the number line at the point K. Now K corresponds to 2 on
the number line.
T
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C B
2
1 3
O A
1
-7 -3 -5 -2 -3 -1 -1 0 1 1 K3 L 2 5 3 7
2 2 2 2 2 2 2 2
Fig. (ii)
Using a compass, with centre O and radius OD, draw an arc which intersects the number
line at the point L right side to 0. Then ‘L’ corresponds to 3 . From this we can conclude that
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many points on the number line can be represented by irrational numbers also. In the same way,
we can locate n for any positive integers n, after n − 1 has been located.
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TRY THESE
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Locate 5 and − 5 on number line. [Hint : 52 = (2)2 + (1)2]
p
numbers that cannot be written in the form , where p and q
q
19
are integers and are called irrational numbers. If we represent , 7, 8, -2,
2007
all rational numbers and all irrational numbers and put these on
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the number line, would there be any point on the number line -7, 100, 2 , 5 ,
that is not covered? π , 9, 999...
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on the number line. We can say that every real number is represented by a unique point on the
number line. Also, every point on the number line represents a unique real number. So we call this
as the real number line.
Here are some examples of Real numbers
1 22
−5.6, 21, −2, 0,1, , , π, 2, 7, 9, 12.5, 12.5123..... etc. You may find that both
5 7
rational and Irrationals are included in this collection.
1 2
Example-9. Find any two irrational numbers between and .
5 7
1
Solution : We know that = 0.20
5
2
A
= 0.285714
7
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1 2
To find two irrational numbers between and , we need to look at the decimal form of
5 7
the two numbers and then proceed. We can find infinitely many such irrational numbers.
Examples of two such irrational numbers are
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0.201201120111..., 0.24114111411114…, 0.25231617181912..., 0.267812147512 …
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Can you find four more irrational numbers between
1
5
2
and ?
7
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Example-10. Find an irrational number between 3 and 4.
Solution :
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If a and b are two positive rational numbers such that ab is not a perfect square of a
rational number, then ab is an irrational number lying between a and b.
= 3×2 = 2 3
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(i) (3 + 3 ) + (3 − 3 ) ( )(
(ii) 3 + 3 3 − 3 )
( )
10
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2
(iii) (iv) 2 +2
2 5
Solution :
(i) (3 + 3 ) + (3 − 3 ) = 3+ 3 +3− 3
= 6, which is a rational number.
(ii) (3 + 3 )(3 − 3 )
We know that ( a + b )( a − b ) ≡ a 2 − b2 is an identity.
( )(
Thus 3 + 3 3 − 3 = 32 − ) ( 3)
2
= 9 − 3 = 6 which is a rational number.
10 10 ÷ 2 5 5× 5
(iii) = = = = 5 , which is an irrational number.
2 5 2 5÷2 5 5
A
( ) = ( 2)
2 2
(iv) 2 +2 + 2. 2 .2 + 22 = 2 + 4 2 + 4
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= 6 + 4 2 , which is an irrational number.
EXERCISE - 1.2
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1. Classify the following numbers as rational or irrational.
(i) 27
(iv) 7.484848…
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(ii) 441
(v) 11.2132435465
(iii) 30.232342345…
(vi) 0.3030030003.....
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2. Give four examples for rational and irrational numbers?
5 7
3. Find an irrational number between and . How many more there may be?
7 9
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ACTIVITY
Constructing the ‘Square root spiral’.
Take a large sheet of paper and construct the ‘Square root spiral’ in the following manner.
A
Step 1 : Start with point ‘O’ and draw a line segment OP of S 1
1 R
1 unit length. T
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1
4 3 Q
Step 2 : Draw a line segment PQ perpendicular to OP of
2
unit length (where OP = PQ = 1) (see Fig) 1
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Step 3 : Join O, Q. (OQ = P
2) O 1
Step 7 : Continue in this manner for some more number of steps, you will create a beautiful
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spiral made of line segments PQ , QR , RS , ST , TU ... etc. Note that the line
segments OQ , OR , OS , OT , OU ... etc. denote the lengths 2, 3, 4, 5, 6
respectively.
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In the previous section, we have seen that any real number has a decimal expansion.
Now first let us see how to represent terminating decimal on the number line.
Suppose we want to locate 2.776 on the number line. We know that this is a terminating
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-4 -3 -2 -1 0 1 2 3 4
2 3
2.1 2.2 2.3 2.4
2.5 2.6 2.7 2.8 2.9
Fig.(i)
So, let us look closely at the portion of the number line between 2 and 3. Suppose we
divide this into 10 equal parts as in Fig.(i). Then the markings will be like 2.1, 2.2, 2.3 and so on.
To have a clear view, let us assume that we have a magnifying glass in our hand and look at the
portion between 2 and 3. It will look like what you see in figure (i).
Now, 2.776 lies between 2.7 and 2.8. So, let us focus on the portion between 2.7 and 2.8
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(See Fig. (ii). We imagine that this portion has been divided into ten equal parts. The first mark
will represent 2.71, the second is 2.72, and so on. To see this clearly, we magnify this as shown
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in Fig(ii).
2 3
2.1 2.2 2.3 2.4
2.5 2.6 2.7 2.8 2.9
2.7
N 2.752.76
G 2.8
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2.71 2.72 2.73 2.74 2.77 2.78 2.79
Fig.(ii)
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2.77 2.78
2.775
2.771 2.772 2.773 2.774 2.776 2.777 2.778 2.779
Fig.(iii)
T
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Again 2.776 lies between 2.77 and 2.78. So, let us focus on this portion of the number line
see Fig. (iii) and imagine that it has been divided again into ten equal parts. We magnify it to see
it better, as in Fig.(iii).
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The first mark represents 2.771, second mark 2.772 and so on, 2.776 is the 6th mark in
these subdivisions.
We call this process of visualization of presentation of numbers on the number line through
a magnifying glass, as the process of successive magnification.
Now let us try and visualize the position of a real number with a non-terminating recurring
decimal expansion on the number line by the process of successive magnification with the following
example.
Example-12. Visualise the representation of 3.58 on the number line through successive
magnification upto 4 decimal places.
Solution: Once again we proceed with the method of successive magnification to represent
3.5888 on number line.
3 4
Step 1 : 3.5 3.6 3.7 3.8 3.9
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3.1 3.2 3.3 3.4
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3.5 3.6
3.51 3.52 3.53 3.54
3.553.56 3.57 3.58 3.59
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Step 2 :
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3.58 3.59
3.585
3.581 3.582 3.583 3.584 3.586 3.587 3.588 3.589
Step 3 :
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3.588 3.589
3.5881 3.5883
3.5885 3.5887 3.5889
Step 4 :
T
3.58
3.5888
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E XERCISE - 1.3
1. Visualise 2.874 on the number line, using successive magnification.
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A
7
= 1 . Here 1 is a rational number.
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7
What do you observe? The sum, difference, quotients and products of irrational numbers
need not be irrational numbers.
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So we can say irrational numbers are not closed with respect to addition, subtraction,
multiplication and divisioin.
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Let us see some problems on irrational numbers.
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5
Example-13. Check whether (i) 5 2 (ii) (iii) 21 + 3 (iv) π + 3 are irrational numbers
2
or not?
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Solution : (3 5 − 7 3) − (5 3 + 7 5)
= 3 5 −7 3 − 5 3−7 5
= −4 5 − 12 3
= − (4 5 + 12 3)
We now list some properties relating to square roots, which are useful in various ways.
Let a and b be non-negative real numbers. Then
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(i) ab = a b
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a a
(ii) = ; if b ≠ 0
b b
(iii) ( a + b )( a − b ) = a − b
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(iv) ( a + b )( a − b ) = a − b 2
(v) ( a + b )( c + d ) = ac + Nad + bc + bd
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( a + b ) = a + 2 ab + b
2
(vi)
(vii) a + b + 2 ab = a + b
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2
(iii) 5− 2 5+ 2
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Solution :
(i) ( 3 + 3 )( 2 + 2 ) = 6 + 3 2 + 2 3 + 6
( 2 + 3 )( 2 − 3 ) = 2 − ( 3 ) = 4 - 3 = 1
2
2
(ii)
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( 5 + 2 ) = ( 5 ) + 2 5 2 + ( 2 ) = 5 + 2 10 + 2 = 7 + 2
2 2 2
(iii) 10
(iv) ( 5 − 2 )( 5 + 2 ) = ( 5 ) − ( 2 ) = 5 − 2 = 3
2 2
= 3+ 2 + 2⋅ 3 ⋅ 2 ∵ a + b + 2 ab = a + b
= 3+ 2
A
2
How do we find the value? As 2 = 1.4142135..... which is neither terminating nor
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repeating. Can you divide 1 with this?
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Let us try to change the denominator into a rational form.
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To rationalise the denominator of 1 , multiply the numerator and the denominator of
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2
1 by
2 , we get
2
1 1 2 2
× =
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= . Yes, it is half of 2 .
2 2 2 2
2
Now can we plot on the number line ? It lies between 0(zero) and 2.
2
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is the rationalising factor (R.F) of the other. Also notice that the R.F. of a given irrational number
is not unique. It is convenient to use the simplest of all R.F.s of given irrational number.
DO THIS
1 3 1
Find rationalising factors of the denominators of (i) (ii) (iii) .
2 3 5 8
1
Example-18. Rationalise the denominator of
4+ 5
( )(
Solution : We know that a + b a − b = a 2 − b )
A
1
Multiplying the numerator and denominator of by 4 − 5
4+ 5
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1 4− 5 4− 5 4− 5 4− 5
× = = =
4+ 5 4− 5
( 5) 16 − 5
2 11
42 −
1
Example-19. If x = 7 + 4 3 then find the value of x +
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x
Solution : Given x = 7 + 4 3
Now
1
=
1
×
7−4 3
=
7−4 3 N ( )
=
7−4 3
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x 7+4 3 7−4 3 72 − 4 3
2
49 − 16 × 3
7−4 3
= = 7−4 3
49 − 48
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1
∴x+ = 7 + 4 3 + 7 − 4 3 = 14
x
1 1
Example-20. Simplify +
7+4 3 2+ 5
T
1 1
= +
7+4 3 2+ 5
1 7−4 3 1 2− 5
= × + ×
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7+4 3 7−4 3 2+ 5 2− 5
7−4 3 2− 5
= 2 + 2
7 − (4 3) 2
2 − ( 5) 2
7−4 3 2− 5
= +
49 − 48 (4 − 5)
7−4 3 2− 5
= +
1 (−1)
= 7−4 3 − 2+ 5 = 5−4 3 + 5
a m−n if m > n
am
i) a .a = a
m n m+n ii) (a ) = a
m n mn
iii) n = 1 if m = n
A
a n1−m if m < n
a
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1
iv) a m b m = (ab) m v) = a−n vi) a0 = 1 (a ≠ 0)
an
Here a, b ‘m’ and ‘n’ are integers and a ≠ 0. b ≠ 0, a, b are called the base and m, n are
the exponents.
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For example
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i) 73.7 −3 = 73+ ( −3) = 7 0 = 1 ii) (23 ) −7 = 2 −21 =
1
221
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23−7
iii) 4
= 23−7− 4 = 23−11 iv) ( 7 )
−13
. ( 3)
−13
= ( 7 × 3)
−13
= ( 21)
−13
23
Suppose we want to do the following computations
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4 1
2 1 1 35 1 1
i) 2 3 .2 3 ii) 5 7 iii) iv) 717 .1117
1
33
How do we go about it? The exponents and bases in the above examples are rational
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numbers. Thus there is a need to extend the laws of exponents to bases of positive real numbers
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and to the exponents as rational numbers. Before we state these laws, we need first to understand
what is nth root of a real number.
i.e., 2
9 =3
25 = ( 25) 2 = 52 ( )
1
1
2× 12
If 52 = 25 then 25 = 5 i.e., 25 = 5 moreover =5 =5
2 2 2
( )
1
If 23 = 8 then 8 = 8 3 = 23 =2
1
8 = 2 (cube root of 8 is 2);
3 3 3
16 = (16 ) 4 = 24 ( )
1 1
32 = ( 32 ) 5 = 25 ( )
1 1
25 = 32 then =2
th
32 = 2 (5 root of 32 is 2);
5 5 5
64 = ( 64 ) 6 = 26 ( )
1 1
26 = 64 then =2
th
64 = 2 (6 root of 64 is 2);
6 6 6
.............................................................................................................
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b = (b) n = an ( )
1 1
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Let a > 0 be a real number and ‘n’ be a positive integer.
If bn = a, for some positive real number ‘b’, then b is called nth root of ‘a’ and we write
n
a = b . In the earlier discussion laws of exponents were defined for integers. Let us extend the
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laws of exponents to the bases of positive real numbers and rational exponents.
Let a > 0 be a real number and p and q be rational numbers then, we have
i) a p .a q = a p + q N
ii) (a ) = a
p q pq
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ap p −q
iii) q = a
1
iv) ap.bp = (ab)p v) n
a = an
a
Now we can use these laws to answer the questions asked earlier.
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Example-21. Simplify
4 1
2 1 1 35 1 1
i) 2 3 .2 3 ii) 5 7 iii) iv) 717 .1117
1
33
T
2 1
( 23 + 13 ) = 2 33 = 21 = 2
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Solution : i) 2 3 .2 3 = 2
DO THIS
4
17 4
=5
7
ii) 5 Simplify:
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(16 ) (128)
1 1
i. 2
ii. 7
( 1 − 13 )
1
35 3− 5 −2 1
iii)
1
=35 = 3 15 = 315 = 32 /15
( 343)
1
33 iii. 5
1 1 1 1
iv) 717 .1117 = ( 7 × 11)17 = 7717
S ur
urdd
If ‘n’ is a positive integer greater than 1 and ‘a’ is a positive rational number but not nth
power of any rational number then n a (or) a1/n is called a surd of nth order. In general we say
the positive nth root of a is called a surd or a radical. Here a is called radicand , n is called
A
radical sign. and n is called the degree of radical. Forms of Surd
Here are some examples for surds. Exponential form a 1n
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3
2, 3, 9, ..... etc Radical form n a
1
Consider the real number 7 . It may also be written as 7 2 . Since 7 is not a square of any
rational number, 7 is a surd.
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Consider the real number 3 8 . Since 8 is a cube of a rational number 2, 3
8 is not a surd.
DO THIS
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57 17 6 55 142 2
EXERCISE - 1.4
SC
i) ( 5 + 7 )( 2 + 5 ) (
ii) 5 + 5 5 − 5 )( )
iii) ( 3+ 7 )
2
iv) ( 11 − 7 )( 11 + 7 )
2. Classify the following numbers as rational or irrational.
( )
2
i) 5 − 3 ii) 3+ 2 iii) 2 −2
( )( )
2 7 1
iv) v) 2π vi) vii) 2 + 2 2 − 2
7 7 3
3. In the following equations, find whether variables x, y, z etc. represent rational or irrational
numbers
A
i) x 2 = 7 ii) y = 16 iii) z = 0.02
2 2
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17
iv) u = v) w2 = 27
2
vi) t4 = 256
4
4. Every surd is an irrational, but every irrational need not be a surd. Justify your answer.
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5. Rationalise the denominators of the following:
1 1 1 6
i) ii) iii) iv)
3+ 2
N
7− 6 7 3− 2
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6. Simplify each of the following by rationalising the denominator:
6−4 2 7− 5 1 3 5− 7
i) ii) iii) iv)
6+4 2 7+ 5 3 2 −2 3 3 3+ 2
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10 − 5
7. Find the value of upto three decimal places. (take 2 = 1.414 and
2 2
5 = 2.236 )
T
8. Find:
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1 1 1
i) 64 6 ii) 32 5 iii) 625 4
3 2 −1
iv) 16 2 v) 243 5 vi) (46656) 6
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10. If ‘a’ and ‘b’ are rational numbers, find the value of a and b in each of the following
equations.
3+ 2 5+ 3
i) = a+b 6 ii) = a − b 15
3− 2 2 5 −3 3
11. Find the square root of 11 + 2 30
A
p
2. A number which cannot be written in the form , for any integers p, q and q ≠ 0 is
AN
q
called an irrational number.
3. The decimal expansion of a rational number is either terminating or non-terminating
recurring.
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4. The decimal expansion of an irrational number is non-terminating and non-recurring.
5. The collection of all rational and irrational numbers is called Real numbers.
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6. There is a unique real number corresponding to every point on the number line. Also
corresponding to each real number, there is a unique point on the number line.
LA
q
7. If q is rational and s is irrational, then q+s, q-s, qs and are irrational numbers.
s
8. If n is a natural number other than a perfect square, then n is an irrational number.
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( a+ b )( )
a − b =a−b ( a + b )( a − b ) = a −b
T
2
iii) iv)
( )
2
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v) a+ b = a + 2 ab + b vi) a + b + 2 ab = a + b
11. Let a > 0, b > 0 be a real number and p and q be rational numbers. Then
ap
i) a p .a q = a p + q ii) (a p )q = a pq iii) q
= a p −q
a
iv) a p .b p = (ab) p
12. If ‘n’ is a positive integer > 1 and ‘a’ is a positive rational number but not nth power of
1
any rational number then n
a or a n is called a surd of nth order.
A
2.1 I NTRODUCTION
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There are six rows and each row has six plants in a garden. How many plants are there in
total ? If there are ‘x’ plants, planted in ‘x’ rows then how many plants will be there
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in the garden? Obviously it is x2.
The cost of onions is `10 per kg. Inder purchased
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p kg., Raju purchased q kg. and Hanif purchased r kg.
How much each would have paid? The payments would
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be `10p, `10q and `10r respectively. All such examples
show the use of algebraic expression.
We also use algebraic expressions such as ‘s2’ to
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polynomials. Note that, all algebraic expressions we have considered so far only have non-
negative integers as exponents of the variables.
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Can you find the polynomials among the given algebraic expressions:
1 3
x2, x 2 + 3, 2x2 − + 5; x2 + xy + y2
x
SC
1 1
From the above x 2 + 3 is not a polynomial because the first term x 2 is a term with an
1 3
exponent that is not a non-negative integer (i.e. ) and also 2x2 − + 5 is not a polynomial
2 x
because it can be written as 2x2 − 3x−1 + 5. Here the second term (3x−1) has a negative
exponent. (i.e., −1). An algebraic expression in which the variables involved have only non-
negative integral powers is called a polynomial.
28 IX-CLASS MATHEMATICS
A
1
(v) 3x 2 + 5 y (vi) + 1 (x ≠ 0) (vii) x (viii) 3 xyz
x
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We shall start our study with polynomials in various forms. In this chapter we will also
learn factorisation of polynomials using Remainder Theorem and Factor Theorem and their use
in the factorisation of polynomials.
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2.2 P OLYNOMIAL
OLYNOMIAL
YNOMIALSS IN ONE V ARIABLE
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Let us begin by recalling that a variable is denoted by a symbol that can take any real
value. We use the letters x, y, z etc. to denote variables. We have algebraic expressions
LA
3 s
Such as 2x, 3x, −x, x .... all in one variable x. These expressions
4
are of the form (a constant) × (some power of variable). Now,
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5 cm P = 4 × 5 = 20 cm
10 cm P = 4 × 10 = 40 cm
Here the value of the constant i.e. ‘4’ remains the same throughout this situation. That is,
the value of the constant does not change in a given problem, but the value of the variable (s)
keeps changing.
Suppose we want to write an expression which is of the form ‘(a constant) × (a variable)’
and we do not know, what the constant is, then we write the constants as a, b, c ... etc. So
these expressions in general will be ax, by, cz, .... etc. Here a, b, c ... are arbitrary constants.
You are also familiar with other algebraic expressions like x2, x2 + 2x + 1, x3 + 3x2 − 4x + 5.
All these expressions are polynomials in one variable.
DO THESE
A
• Write two polynomials with variable ‘x’
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• Write three polynomials with variable ‘y’
• Is the polynomial 2x2 + 3xy + 5y2 in one variable ?
• Write the formulae of area and volume of different solid shapes. Find out the variables
and constants in them.
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2.3 DEGREE OF THE POLYNOMIAL
POLYNOMIAL
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Each term of the polynomial consists of the product of a constant, (called the coefficient
of the term) and a finite number of variables raised to non-negative integral powers. Degree of a
term is the sum of the exponent of its variable factors. And the degree of a polynomial is the
largest degree of its variable terms.
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Lets find the terms, their coefficients and the degree of polynomials:
(i) 3x2 + 7x + 5 (ii) 3x2y2 + 4xy + 7
In the polynomial 3x2 + 7x + 5, each of the expressions 3x2, 7x and 5 are terms. Each
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You know that the degree of a polynomial is the highest degree of its variable term.
As the term 3x2 has the highest degree among all the other terms in that expression, Thus
the degree of 3x2 + 7x + 5 is ‘2’.
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Now can you find coefficient and degree of polynomial 3x2y3 + 4xy + 7.
The coefficient of x2y3 is 3, xy is 4 and x0y0 is 7. The sum of the exponents of the
variables in term 3x2y3 is 2 + 3 = 5 which is greater than that of the other terms. So the degree
of polynomial 3x2y3 + 4xy + 7 is 5.
Now think what is the degree of a constant? As the constant contains no variable, it can be
written as product of x0. For example, degree of 5 is zero as it can be written as 5x0. Now that
you have seen what a polynomial of degree 1, degree 2, or degree 3 looks like, can you write
down a polynomial in one variable of degree n for any natural number n? A polynomial in one
variable x of degree n is an expression of the form
anxn + an–1xn–1 + . . . + a1x + a0
where a0, a1, a2, . . ., an are constants and an ≠ 0.
A
In particular, if a0 = a1 = a2 = a3 = . . . = an = 0 (i.e. all the coefficients are zero), we get
the zero polynomial, which is denoted by ‘0’.
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Can you say the degree of zero? It is not defined as we can’t write it as a product of a
variable raised to any power.
DO THESE
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1. Write the degree of each of the following polynomials
(i) 7x3 + 5x2 + 2x − 6
5p −
N (ii) 7 − x + 3x2
(v) −5xy2
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(iii) 3 (iv) 2
2. Write the coefficient of x2 in each of the following
3
Zero Constant polynomial −12; 5; etc
4
SC
A
2 Binomial 3x + 5 3x, 5
3 Trinomial 2x2 + 5x + 1 ...........................
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More than 3 Multinomial ........................... 3x3, 2x2, −7x, 5
Note : A polynomial may be a multinomial but every multinomial need not be a polynomial.
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A linear polynomial with one variable may be a monomial or a binomial.
Eg : 3x or 2x − 5 N
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THINK, DISCUSS AND WRITE
How many terms a cubic (degree 3) polynomial with one variable can have?
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Give examples.
If the variable in a polynomial is x, we may denote the polynomial by p(x), q(x) or r(x)
etc. So for example, we may write
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TRY THESE
p(x) = 3x2 + 2x + 1
ER
t(z) = z2 + 5z + 3
EXERCISE - 2.1
1. Find the degree of each of the polynomials given below
(i) x5 − x4 + 3 (ii) x2 + x − 5 (iii) 5
A
3
2. Which of the following expressions are polynomials in one variable and which are not ?
AN
Give reasons for your answer.
2
(i) 3x2 − 2x + 5 (ii) x2 + 2 (iii) p2 − 3p + q (iv) y +
y
G
(v) 5 x+x 5 (vi) x100 + y100
3. Write the coefficient of x3 in each of the following
(i) x3 + x + 1 (ii) 2 − x3 + x2N (iii) 2 x3 + 5 (iv) 2x3 + 5
LA
π 2
(v) x3 + x (vi) − x3 (vii) 2x2 + 5 (vi) 4
2 3
4. Classify the following as linear, quadratic and cubic polynomials
TE
5. Write whether the following statements are True or False. Justify your answer
T
(vi) π r2 is a monomial.
6. Give one example each of a monomial and trinomial of degree 10.
A
= 0+0+4 = 1−5+4
AN
= 4 =0
Can you find the value of p(−4)?
G
s(y) = 4y4 − 5y3 − y2 + 6
s(1) = 4(1)4 − 5(1)3 − (1)2 + 6
N
= 4(1) − 5(1) − 1 + 6
= 4−5−1+6
LA
= 10 − 6
= 4
TE
DO THIS
Find the value of each of the following polynomials for the indicated value of
T
variables:
(i) p(x) = 4x2 − 3x + 7 at x = 1
ER
A
polynomial equation in x . We observe that ‘−1’ is the root of the polynomial f(x) in the above
example So we say that ‘−1’ is the zero of the polynomial x + 1, or a root of the polynomial
AN
equation x + 1 = 0.
l Now, consider the constant polynomial 3.
TRY THESE
Can you tell what is its zero ? It does not Find zeroes of the following
have a zero. As 3 = 3x0 no real value of x polynomials
G
gives value of 3x0. Thus a constant polynomial 1. 2x − 3 2. x2 − 5x + 6
has no zeroes. But zero polynomial is a 3. x + 5
N
constant polynomial having many zeros.
Example-1. p(x) = x + 2. Find p(1), p(2), p(−1) and p(−2). What are zeroes of the
LA
polynomial x + 2?
Solution : Let p(x) = x + 2
replace x by 1
TE
p(1) = 1 + 2 = 3
replace x by 2
p(2) = 2 + 2 = 4
replace x by −1
T
p(−1) = −1 + 2 = 1
ER
replace x by −2
p(−2) = −2 + 2 = 0
Therefore, 1 , 2, −1 are not the zeroes of the polynomial x + 2, but −2 is the zero of the
SC
polynomial.
1
So, − is a zero of the polynomial 3x + 1.
3
A
As 2x − 1 = 0
AN
1
x = (how ?)
2
1
Check it by finding the value of P
2
G
Now, if p(x) = ax + b, a ≠ 0, a linear polynomial, how will you find a zero of p(x)?
N
As we have seen to find zero of a polynomial p(x), we need to solve the polynomial
equation p(x) = 0
LA
Which means ax + b = 0, a ≠ 0
So ax = −b
−b
TE
i.e., x =
a
−b
So, x = is the only zero of the polynomial p(x) = ax + b i.e., A linear polynomial
a
in one variable has only one zero.
T
DO THIS
ER
Polynomial
x+a −a
x−a -------------
ax + b -------------
b
ax − b
a
A
also replace x by 1
AN
p(1) = (1)2 − 3(1) + 2
=1−3+2
=0
G
Hence, both 2 and 1 are zeroes of the polynomial x2 − 3x + 2.
Is there any other way of checking this?
N
What is the degree of the polynomial x2 − 3x + 2 ? Is it a linear polynomial ? No,
it is a quadratic polynomial. Hence, a quadratic polynomial has two zeroes.
LA
Example-5. If 3 is a zero of the polynomial x2 + 2x − a. Find a?
Solution : Let p(x) = x2 + 2x − a
TE
9+6−a =0
15 − a = 0
ER
−a = −15
or a = 15
SC
EXERCISE - 2.2
1. Find the value of the polynomial 4x2 − 5x + 3, when
1
(i) x = 0 (ii) x = −1 (iii) x = 2 (iv) x =
2
2. Find p(0), p(1) and p(2) for each of the following polynomials.
(i) p(x) = x2− x +1 (ii) p(y) = 2 + y + 2y2 − y3
(iii) p(z) = z3 (iv) p(t) = (t − 1) (t + 1)
(v) p(x) = x − 3x + 2
2
A
3. Verify whether the values of x given in each case are the zeroes of the polynomial or not?.
1 −3
p(x) = 2x + 1; x = − (ii) p(x) = 5x − π; x =
AN
(i)
2 2
(iii) p(x) = x2 − 1; x = +1 (iv) p(x) = (x - 1)(x + 2); x = − 1, −2
b
(v) p(y) = y2; y = 0 (vi) p(x) = ax + b ; x = −
G
a
1 2 1 −1
(vii) f(x) = 3x2 − 1; x = −
3
,
N 3
(viii) f (x) = 2x - 1, x = ,
2 2
LA
4. Find the zero of the polynomial in each of the following cases.
(i) f(x) = x + 2 (ii) f(x) = x − 2 (iii) f(x) = 2x + 3
(iv) f(x) = 2x − 3 (v) f(x) = x2 (vi) f(x) = px, p ≠ 0
TE
A
3x3 + x2 + x as x(3x2 + x + 1)
AN
What are the factors of 3x3 + x2 + x ?
(iii) Consider another example (2x2 + x + 1) ÷ x 2x + 1
x 2x + x + 1
2
2 x2 x 1
Here, (2x2 + x + 1) ÷ x = + +
G
x x x −2 x 2
1 x +1
= 2x + 1 +
Is it a polynomial ? N x −x
1
LA
1 1
As one of the term has a negative integer exponent (i.e. = x−1)
x x
1
TE
∴ 2x + 1 + is not a polynomial.
x
We can however write
(2x2 + x + 1) = [x × (2x + 1)] + 1
By taking out 1 separately the rest of the polynomial can be written as product of two
T
polynomials.
ER
Here we can say (2x + 1) is the quotient, x is the divisor and 1 is the remainder. We must
keep in mind that since the remainder is not zero, ‘x’ is not a factor of 2x2 + x + 1.
DO THESE
SC
−2 x 3 x 2 + 3x
Step 3 : Divide = −2 , it becomes the 2nd term in the quotient. − −
x
− 2x −1
Step 4 : Multiply (x + 1)(−2) = −2x − 2,
− 2x − 2
A
Subtract it from −2x − 1, which gives ‘1’. + +
+1
AN
Step 5 : We stop here as the remainder is 1, a constant.
(Can you tell why a constant is not divided by a polynomial?)
This gives us the quotient as (3x − 2) and remainder (+1).
Note : The division process is said to be complete if we get the remainder 0 or the degree of
G
the remainder is less than the degree of the divisor.
Now, we can write the division fact as
N
3x2 + x − 1 = (x + 1) (3x − 2) + 1
i.e. Dividend = (Divisor × Quotient) + Remainder.
LA
Let us see by replacing x by −1 in p(x)
p(x) = 3x2 + x − 1 It is observed that the value of p(−1)
TE
+ −
2 x4
= 2 x3 − 2 x 2 − 3x − 1
x
Now multiply (x − 1) (2x3) = 2x4 − 2x3 − 2x2 + 2x
+ −
Then again see the first term of the remainder that − 5x − 1
is −2x3. Now do the same. − 5x + 5
+ −
−6
A
=2−4−3−1
AN
= −6
Is the remainder same as the value of the polynomial f(x) at zero of (x − 1) ?
From the above examples we shall now state the fact in the form of the following theorem.
G
It gives a remainder without actual division of a polynomial by a linear polynomial in one
variable.
Remainder Theorem : N
LA
Let p(x) be any polynomial of degree greater than or equal to one and let ‘a’
be any real number. If p(x) is divided by the linear polynomial (x − a), then the
remainder is p(a).
Let us now look at the proof of this theorem.
TE
Proof : Let p(x) be any polynomial with degree greater than or equal to 1.
Further suppose that when p(x) is divided by a linear polynomial g(x) = (x − a), the
quotient is q(x) and the remainder is r(x). In other words, p(x) and g(x) are two polynomials
such that the degree of p(x) > degree of g(x) and g(x) ≠ 0 then we can find polynomials q(x)
T
and r(x) such that, where r(x) = 0 or degree of r(x) < degree of g(x).
ER
By division algorithm,
p(x) = g(x) . q(x) + r(x)
∴ p(x) = (x − a) . q(x) + r(x) ∵ g(x) = (x − a)
SC
Since the degree of (x − a) is 1 and the degree of r(x) is less than the degree of (x − a).
∴ Degree of r(x) = 0, implies r(x) is a constant, say K.
So, for every real value of x, r(x) = K.
Therefore,
p(x) = (x − a) q(x) + K
If x = a, then p(a) = (a − a) q(a) + K
=0+K
=K
Hence proved.
Let us use this result in finding remainders when a polynomial is divided by a linear polynomial
without actual division.
A
Solution : Here p(x) = x3 + 1
is −1 (x + 1 = 0 ⇒ x = −1)
AN
The zero of the linear polynomial x + 1
So replacing x by −1
p(−1) = (−1)3 + 1
G
= −1 + 1
=0
N
So, by Remainder Theorem, we know that (x3 + 1) divided by (x + 1) gives 0 as the
remainder.
LA
You can also check this by actual division method. i.e., x3 + 1 by x + 1.
Can you say (x + 1) is a factor of (x3 + 1) ?
TE
= 8 − 2(4) − 10 + 4
= 8 − 8 − 10 + 4
SC
= − 6.
As the remainder is not equal to zero, the polynomial (x − 2) is not a factor of the given
polynomial x3 − 2x2 − 5x + 4.
Example10. Check whether the polynomial p(y) = 4y3 + 4y2 − y − 1 is a multiple of (2y + 1).
Solution : As you know, p(y) will be a multiple of (2y + 1) only, if (2y + 1) divides p(y) exactly.
−1
We shall first find the zero of the divisor , 2y + 1, i.e., y = ,
2
−1
Replace y by in p(y)
2
3 2
−1 −1 −1 −1
p = 4 + 4 − −1
2 2 2 2
A
−1 1 1
= 4 + 4 + −1
8 4 2
AN
−1 1
= +1+ −1
2 2
=0
G
So, (2y + 1) is a factor of p(y). That is p(y) is a multiple of (2y + 1).
N
Example-11. If the polynomials ax3 + 3x2 − 13 and 2x3 − 5x + a are divided by (x − 2) leave
the same remainder, find the value of a.
LA
Solution : Let p(x) = ax3 + 3x2 − 13 and q(x) = 2x3 − 5x + a
∵ p(x) and q(x) when divided by x − 2 leave the same remainder.
TE
∴ p(2) = q(2)
a(2)3 + 3(2)2 − 13 = 2(2)3 − 5(2) + a
8a + 12 − 13 = 16 − 10 + a
8a − 1 = a + 6
T
8a − a = 6 + 1
ER
7a = 7
a=1
SC
EXERCISE - 2.3
1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following
Linear polynomials :
1
(i) x+1 (ii) x − (iii) x (iv) x + π
2
(v) 5 + 2x
2. Find the remainder when x3 − px2 + 6x − p is divided by x − p.
3. Find the remainder when 2x2 − 3x + 5 is divided by 2x − 3. Does it exactly divide the
polynomial ? State reason.
2
4. Find the remainder when 9x3 − 3x2 + x − 5 is divided by x −
3
5. If the polynomials 2x3 + ax2 + 3x − 5 and x3 + x2 − 4x + a leave the same remainder
A
when divided by x − 2, find the value of a.
6. If the polynomials x3 + ax2 + 5 and x3 − 2x2 + a are divided by (x + 2) leave the same
AN
remainder, find the value of a.
7. Find the remainder when f (x) = x4 − 3x2 + 4 is divided by g(x) = x − 2 and verify the
result by actual division.
G
8. Find the remainder when p(x) = x3 − 6x2 + 14x − 3 is divided by g(x) = 1 − 2x and
verify the result by long division.
N
9. When a polynomial 2x3 +3x2 + ax + b is divided by (x − 2) leaves remainder 2, and
LA
(x + 2) leaves remainder −2. Find a and b.
As we have already studied that a polynomial q(x) is said to have divided a polynomial
p(x) exactly if the remainder is zero. In this case q(x) is a factor of p(x).
For example. When p(x) = 4x3 + 4x2 − x − 1 is divided by g(x) = 2x + 1, if the remainder
is zero (verify)
T
Factor Theorem : If p(x) is a polynomial of degree n > 1 and ‘a’ is any real
number, then (i) x − a is a factor of p(x), if p(a) = 0 (ii) and its converse
“if (x − a) is a factor of a polynomial p(x) then p(a) = 0.
Let us see the simple proof of this theorem.
Proof : By Remainder Theorem,
p(x) = (x − a) q(x) + p(a)
(i) Consider proposition (i) If p(a) = 0, then p(x) = (x − a) q(x) + 0.
= (x − a) q(x)
A
for some polynomial q(x)
Let p(x) = ax3 + bx2 + cx + d (a ≠ 0) and
∴ p(a) = (a − a) q(a) (x + 1) is a factor of p(x) ⇒ p(–1) = 0
AN
= 0 ⇒ b+d=a+c
i.e. the sum of the coefficients of even
∴ Hence p(a) = 0 when (x − a) is a factor
power terms is equal to the sum of the
of p(x)
coeffients odd power terms then (x + 1) is
G
Let us consider some more examples. a factor.
=0
So, by the Factor Theorem, x + 2 is a factor of x3 + 2x2 + 3x + 6.
Example-13. Find the value of K, if 2x − 3 is a factor of 2x3 − 9x2 + x + K.
T
3
If (2x − 3) = 0 then x =
2
3
∴ The zero of (2x − 3) is
2
3
SC
27 − 81 + 6 + 4K = 0
−48 + 4K = 0
4K = 48
So K = 12
A
Example-14. Show that (x − 1) is a factor of x10 − 1 and also of x11 − 1.
Solution : Let p(x) = x10 − 1 and g(x) = x11 − 1
AN
To prove (x − 1) is a factor of both p(x) and g(x), it is sufficient to show that p(1) = 0
and g(1) = 0.
Now
G
p(x) = x10 − 1 and g(x) = x11 − 1
p(1) = (1)10 − 1 and g(1) = (1)11 − 1
=1−1 N =1−1
TRY THESE
LA
=0 =0
Thus by Factor Theorem, Show that (x - 1) is a
(x − 1) is a factor of both p(x) and g(x). factor of xn - 1.
TE
We shall now try to factorise quadratic polynomial of the type ax2 + bx + c, (where
a ≠ 0 and a, b, c are constants).
Let its factors be (px + q) and (rx + s).
T
b = ps + qr
c = qs
This shows that b is the sum of two numbers ps and qr,
whose product is (ps) (qr) = (pr)(qs)
= ac
Therefore, to factorise ax2 + bx + c, we have to write b as the sum of two numbers
whose product is ac.
A
(1, 18), (2, 9), (3, 6) of these pairs, 2 and 9 will satisfy p + q = 11
So 3x2 + 11x + 6 = 3x2 + 2x + 9x + 6
AN
= x(3x + 2) + 3(3x + 2)
= (3x + 2) (x + 3).
G
DO THESE
Factorise the following
1. 6x2 + 19x + 15
N
2.10m2 - 31m - 132 3. 12x2 + 11x + 2
LA
Now, consider an example.
x2 − 3x + 2 = x2 − 2x − x + 2
T
= x(x − 2) − 1(x − 2)
= (x − 2) (x − 1).
ER
= 32 − 48 + 12 + 6 − 2
= 50 − 50
=0
= 2−6+3+3−2
A
= 8−8
AN
= 0
G
As both (x − 2) and (x − 1) are factors of p(x), then their product x2 − 3x + 2 is also a
factor of p(x) = 2x4 − 6x3 + 3x2 + 3x − 2.
So (x − 1) is a factor of p(x)
EXERCISE - 2.4
1. Determine which of the following polynomials has (x + 1) as a factor.
(i) x3 − x2 − x + 1 (ii) x4 − x3 + x2 − x + 1
A
(iii) 3) x + 3
2. Use the Factor Theorem to determine whether g(x) is factor of f(x) in each of the
AN
following cases :
(i) f(x) = 5x3 + x2 − 5x − 1, g(x) = x + 1
(ii) f(x) = x3 + 3x2 + 3x + 1, g(x) = x + 1
G
(iii) f(x) = x3 − 4x2 + x + 6, g(x) = x − 2
(iv) f(x) = 3x3 + x2 − 20x + 12, g(x) = 3x − 2
N
(v) f(x) = 4x3 + 20x2 + 33x + 18, g(x) = 2x + 3
LA
3. Show that (x − 2), (x + 3) and (x − 4) are factors of x3 − 3x2 − 10x + 24.
4. Show that (x + 4), (x − 3) and (x − 7) are factors of x3 − 6x2 − 19x + 84.
1
TE
Identity III : (x + y) (x − y) ≡ x2 − y2
Identity IV : (x + a)(x + b) ≡ x2 + (a + b)x + ab.
Geometrical Proof :
For Identity (x − y)2
y (x-y) × y (y×y)
Step-I Make a square of side x.
A
Step-II Subtract length y from x.
(x-y) × y
Step-III Calculate for (x − y)2 x (x-y)2
AN
(x-y)
= x2 − [(x − y) y + (x − y) y + y2]
= x2 − xy + y2 − xy + y2 − y2
(x-y) y
= x2 − 2xy + y2 x
G
TRY THIS
(i) ( x + y ) 2 ≡ x 2 + 2 xy + y 2
N
Try to draw the geometrical figures for other identities.
(ii) ( x + y )( x − y ) ≡ x 2 − y 2
LA
(iii) ( x + a )( x + b) ≡ x 2 + ( a + b) x + ab
(iv) ( x + a )( x + b)( x + c ) ≡ x 3 + (a + b + c ) x 2 + ( ab + bc + ca ) x + abc
TE
DO THESE
Find the following product using appropriate identities
(i) (x + 5) (x + 5) (ii) (p − 3) (p + 3) (iii) (y − 1) (y − 1)
(iv) (t + 2) (t + 4) (v) 102 × 98 (vi) (x + 1) (x + 2) (x + 3)
T
Identities are useful in factorisation of algebraic expressions. Let us see some examples.
ER
Example-18. Factorise
(i) x2 + 5x + 4 (ii) 9x2 − 25
(iii) 25a2 + 40ab + 16b2 (iv) 49x2 − 112xy + 64y2
SC
Solution :
(i) Here x2 + 5x + 4 = x2 + (4 + 1)x + (4) (1)
Comparing with Identity (x + a) (x + b) ≡ x2 + (a + b)x + ab
we get (x + 4) (x + 1).
(ii) 9x − 25 = (3x)2 − (5)2
2
A
Using Identity I, (x + y)2 ≡ x2 + 2xy + y2
we get 25a2 + 40ab + 16b2 = (5a + 4b)2
AN
= (5a + 4b) (5a + 4b).
(iv) Here 49x2 − 112xy + 64y2 , we see that (x + y)2 + (x − y)2 = 2(x2 + y2)
G
49x2 = (7x)2, 64y2 = (8y)2 and (x + y)2 − (x − y)2 = 4xy
112 xy = 2(7x) (8y)
Thus comparing with Identity II,
(x − y)2 ≡ x2 − 2xy + y2,
N
LA
we get, 49x2 − 112xy + 64y2 = (7x)2 − 2(7x) (8y) + (8y)2
= (7x − 8y)2
TE
DO THESE
T
9 2 y2
2
(i) 49a + 70ab + 25b 2
(ii) x −
16 9
(iii) t2 − 2t + 1 (iv) x + 3x + 2
2
SC
So far, all our identities involved products of binomials. Let us now extend the identity I to
a trinomial x + y + z. We shall compute (x + y + z)2.
Let x + y = t, then (x + y + z)2 = (t + z)2
= t2 + 2tz + z2 (using Identity I)
= (x + y)2 + 2(x + y) z + z2 (substituting the value of ‘t’)
= x2 + 2xy + y2 + 2xz + 2yz + z2
By rearranging the terms, we get x2 + y2 + z2 + 2xy + 2yz + 2zx
Alternate Method :
You can also compute (x + y + z)2 by regrouping the terms
[(x + y) + z]2 = (x + y)2 + 2(x + y) (z) + (z)2
= x2 + 2xy + y2 + 2xz + 2yz + z2 [From identity (1)]
A
= x2 + y2 + z2 + 2xy + 2yz + 2xz
AN
In what other ways you can regroup the terms to find the expansion ? Will you get the same
result ?
So, we get the following Identity
G
Identity V : (x + y + z)2 ≡ x2 + y2 + z2 + 2xy + 2yz + 2zx
N
Example-19. Expand (2a + 3b + 5)2 using identity.
Solution : Comparing the given expression with (x + y + z)2,
LA
we find that x = 2a, y = 3b and z = 5
Therefore, using Identity V, we have
TE
(2a + 3b + 5)2 = (2a)2 + (3b)2 + (5)2 + 2(2a)(3b) + 2(3b) (5) + 2(5) (2a)
= 4a2 + 9b2 + 25 + 12ab + 30b + 20a.
A
DO THESE
AN
(i) Write (p + 2q + r)2 in expanded form.
(ii) Expand (4x − 2y − 3z)2 using identity
(iii) Factorise 4a2 + b2 + c2 − 4ab + 2bc − 4ca using identity.
G
So far, we have dealt with identities involving second degree terms. Now let us extend
Identity I to find (x + y)3.
We have N
LA
(x + y)3 = (x + y)2 (x + y)
= (x2 + 2xy + y2) (x + y)
= x(x2 + 2xy + y2) + y(x2 + 2xy + y2)
TE
= x3 + y3 + 3xy(x + y).
ER
TRY THESE
SC
Let us see some examples where these identities are being used.
A
(2a + 3b)3 = (2a)3 + (3b)3 + 3(2a)(3b) (2a + 3b)
AN
= 8a3 + 27b3 + 18ab (2a + 3b)
= 8a3 + 27b3 + 36a2b +54 ab2
= 8a3 + 36a2b + 54ab2 + 27b3.
G
(ii) Comparing the given expression with (x − y)3, we find that x = 2p and y = 5
So, using Identity VII , we have
N
(2p − 5)3 = (2p)3 − (5)3 − 3(2p)(5) (2p − 5)
LA
= 8p3 − 125 − 30p (2p − 5)
= 8p3 − 125 − 60p2 + 150p
= 8p3 − 60p2 + 150p − 125.
TE
= 1000000 + 27 + 900(103)
= 1000000 + 27 + 92700
= 1092727.
(ii) We have (99)3 = (100 − 1)3
Comparing with (x − y)3 ≡ x3 − y3 − 3xy(x − y) we get
= (100)3 − (1)3 − 3(100)(1) (100 − 1)
= 1000000 − 1 − 300 (99)
= 1000000 − 1 − 29700
= 970299.
A
Solution : The given expression can be written as
8x3 + 36x2y + 54xy2 + 27y3 = (2x)3 + 3(2x)2 (3y) + 3(2x) (3y)2 + (3y)3
AN
Comparing with Identity VI, (x + y)3 ≡ x3 + 3x2y + 3xy2 + y3,
G
= (2x + 3y) (2x + 3y) (2x + 3y).
DO THESE N
LA
1. Expand (x + 1)3 using an identity
2. Compute (3m − 2n)3.
3. Factorise a3 − 3a2b + 3ab2 − b3.
TE
= x3 + xy2 + xz2 − x2y − xyz − x2z + x2y + y3 + yz2 − xy2 − y2z − xyz + x2z
+ y2z + z3 − xyz − yz2 − xz2
SC
A
= 8a3 + b3 + c3 − 6abc
AN
Example-26. Factorise a3 − 8b3 − 64c3 − 24abc
Solution : Here the given expression can be written as
a3 − 8b3 − 64c3 − 24abc = (a)3 + (−2b)3 + (−4c)3 − 3(a)(−2b)(−4c)
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Comparing with the identity VIII,
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x3 + y3 + z3 − 3xyz ≡ (x + y + z) (x2 + y2 + z2 − xy − yz − zx)
we get factors as
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= (a − 2b − 4c) [(a)2 + (− 2b)2 + (− 4c)2 − (a) (− 2b) − (− 2b) (− 4c) − (− 4c) (a)]
= (a − 2b − 4c) (a2 + 4b2 + 16c2 + 2ab − 8bc + 4ca).
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DO THESE
1. Find the product (a − b − c) (a2 + b2 + c2 − ab + bc − ca) without
actual multiplication.
T
Example-27. Give possible values for length and breadth of the rectangle whose area is
2x2 + 9x −5.
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∴ length = (x + 5)
breadth = (2x − 1)
Let x = 1, l = 6, b = 1
x = 2, l = 7, b = 3
x = 3, l = 8, b = 5
A
..............................
..............................
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Can you find more values ?
EXERCISE - 2.5
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1. Use suitable identities to find the following products
(i) (x + 5) (x + 2) (ii) (x − 5) (x − 5) (iii) (3x + 2)(3x − 2)
(iv)
2 1 2 1
x + 2 x − 2
x x
N (v) (1 + x) (1 + x)
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2. Evaluate the following products without actual multiplication.
1 1
(i) 101 × 99 (ii) 999 × 999 (iii) 50 × 49
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2 2
(iv) 501 × 501 (v) 30.5 × 29.5
3. Factorise the following using appropriate identities.
(i) 16x2 + 24xy + 9y2 (ii) 4y2 − 4y + 1
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y2
4x2 − (iv) 18a2 − 50
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(iii)
25
(v) x2 + 5x + 6 (vi) 3p2 − 24p + 36
4. Expand each of the following, using suitable identities
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5. Factorise
(i) 25x2 + 16y2 + 4z2 − 40xy + 16yz − 20xz
(ii) 9a2 + 4b2 + 16c2 + 12ab − 16bc − 24ca
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(i) 8a3 + b3 + 12a2b + 6ab2 (ii) 8a3 − b3 − 12a2b + 6ab2
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12 2 6 1
(iv) 8 p − p + p−
3
(iii) 1 − 64a3 − 12a + 48a2
5 25 125
9. Verify (i) x3 + y3 = (x + y) (x2 − xy + y2) (ii) x3 − y3 = (x − y) (x2 + xy + y2)
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using some non-zero positive integers and check by actual multiplication. Can you call
these as identites ?
10. Factorise (i) 27a3 + 64b3
N (ii) 343y3 − 1000 using the above results.
11. Factorise 27x3 + y3 + z3 − 9xyz using identity.
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1
12. Verify that x3 + y3 + z3 − 3xyz = ( x + y + z )[( x − y ) 2 + ( y − z ) 2 + ( z − x) 2 ]
2
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(i)
3 3 3
1 1 5
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15. Give possible expressions for the length and breadth of the rectangle whose area is given
by (i) 4a2 + 4a − 3 (ii) 25a2 − 35a + 12
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16. What are the possible polynomial expressions for the dimensions of the cuboids whose
volumes are given below?
(i) 3x3 − 12x (ii) 12y2 + 8y − 20.
17. If 2(a2+b2) = (a+b)2, then show that a = b
WHAT
HAT WE HAVE DISCUSSED
HAVE
A
respectively the coefficients of x0, x1, x2, .... xn and n is called the degree of the polynomial
if an ≠ 0. Each anxn ; an−1xn−1 ; .... a0, is called a term of the polynomial p(x).
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2. Polynomials are classified as monomial, binomial, trinomial etc. according to the number
of terms in it.
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3. Polynomials are also named as linear polynomial, quadratic polynomial, cubic polynomial
etc. according to the degree of the polynomial.
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4. A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. In this case, ‘a’ is also called
a root of the polynomial equation p(x) = 0.
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5. Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial
has no zero.
6. Remainder Theorem : If p(x) is any polynomial of degree greater than or equal to 1 and
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p(x) is divided by the linear polynomial (x − a), then the remainder is p(a).
7. Factor Theorem : If x − a is a factor of the polynomial p(x), then p(a) = 0. Also if
p(a) = 0 then (x − a) is a factor of p(x).
8. Some Algebraic Identities are:
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Brain teaser
A
3.1 INTRODUCTION
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You may have seen large structures like bridges, dams, school buildings, hostels, hospitals
etc. The construction of these structures is a big task for the engineers.
Do you know how we estimate the cost of the construction? Besides wages of the
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labour, cost of cement and concrete. It depends upon the size and shape of the structure.
N
The size and shape of a structure include the foundation, plinth area, size of the walls,
elevation, roof etc. To understand the geometric principles involved in these constructions, we
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should know the basic elements of geometry and its application.
We also know that geometry is widely used in daily life activities such as paintings,
handicrafts, laying of floor designs, ploughing and sowing of seeds in fields. So in other words,
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The great construction like the Pyramids in Egypt, the Great wall of China, Temples,
Mosques, Cathedral, Tajmahal, Charminar and altars in India, Eifel tower of France etc. are
some of the best examples of application of geometry.
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In this chapter, we will look into the history to understand the roots of geometry and the
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different schools of thought that have developed the geometry and its comparison with modern
geometry.
3.2 H IST OR
ISTOR
ORYY
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The domains of mathematics which study the shapes and sizes of structures are described
under geometry. The word geometry is derived from the Greek ‘geo’ means earth and ‘metrein’
means measure.
The earliest recorded beginnings of geometry can be traced to early people, who discovered
obtuse angled triangles in the ancient Indus valley and ancient Babylonia. The ‘Bakshali manuscript’
employs a handful of geometric problems including problems about volumes of irregular solids.
Remnants of geometrical knowledge of the Indus Valley civilization can be found in excavations
60 IX-CLASS MATHEMATICS
The ‘Sulba Sutras’ in Vedic Sanskrit lists the rules and geometric principles involved in
the construction of ritual fire altars. The amazing idea behind the construction of fire altars is that
they occupy same area although differ in their shapes. Boudhayana (8th century B.C.) composed
A
the Boudhayana Sulba Sutra, the best-known Sulba Sutra which contains examples of simple
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Pythagorean triples such as (3,4,5), (5,12,13), (8,15,17)… etc. as well as a statement of
Pythagorean theorem for the sides of a rectangle.
Ancient Greek mathematicians conceived geometry as the crown jewel of their sciences.
They expanded the range of geometry to many new kinds of figures, curves, surfaces and solids.
G
They found the need of establishing a proposed statement as universal truth with the help of logic.
This idea led the Greek mathematician Thales to think of deductive proof.
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Pythagoras of Ionia might have been a student of Thales and the theorem that was named
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after him might not have been his discovery, but he was probably one of the mathematicians who
had given a deductive proof of it. Euclid (325-265B.C) of Alexandria in Egypt wrote 13 books
called ‘The Elements’. Thus Euclid created the first system of thought based on fundamental
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Euclid thought geometry as an abstract model of the world in which they lived. The notions
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of point, line, plane (or surface) and so on were derived from what was seen around them. From
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studies of the space and solids in the space around them, an abstract geometrical notion of a solid
object was developed. A solid has shape, size, position and can be moved from one place to
another. Its boundaries are called surfaces. They separate one part of the solid from another,
and are said to have no thickness. The boundaries of the surfaces are curves or straight lines.
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These lines end in points. Consider the steps from solids to points (solids-surfaces-lines-point)
Observe the figure given in the next page. This figure is a cuboid (a solid) [fig.(i)]. It has
three dimensions namely length, breadth and height. If it loses one dimension i.e. height then it will
have only two dimensions which is rectangle. You know that a rectangle has two dimensions
length and breadth [fig.(ii)]. If it further loses another dimension i.e. breadth then it will leave with
only line segment [fig.(iii)] and if it has to lose one more dimension, there remain only the points
[fig.(iv)]. We may recall that a point has no dimensions. Similarly when we see the edge of a table
or a book, we can visualise it as a line. The end point of a line or the point where two lines meet
is a point.
A
AN
solids → surfaces/curves → lines → points
3-D 2-D 1-D no dimension
These are the fundamental terms of geometry. With the use of these terms other terms like
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line segment, angle, triangle etc. are defined.
Based on the above observatioin, Euclid defined point, line, plane.
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In Book 1 of his Elements, Euclid listed 23 definitions. Some of them are given below.
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• A point is that which has no part
• A line is breadthless length
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Father of Geometry
• A plane surface is a surface which lies evenly with the straight lines on itself
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In defining terms like point, line and plane, Euclid used words or phrases like ‘part’,
‘breadth’, ‘evenly’ which need defining or further explanation for the sake of clarity. In defining
terms like plane, if we say ‘a plane’ occupies some area then ‘area’ is again to be clarified. So to
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define one term you need to define more than one term resulting in a chain of definitions without
an end. So, mathematicians agreed to leave such terms as undefined. However we do have a
intuitive feeling for the geometric concepts of a point than what the “definition” above gives us.
So, we represent a point as a dot, even though a dot has some dimension. The Mohist philosophers
in ancient China said “the line is divided into parts and that part which has no remaining part is a
point.
A similar problem arises in definition 2 above, since it refers to breadth and length, neither
of which has been defined. Because of this, a few terms are kept undefined while developing any
course of study. So, in geometry, we take a point, a line and a plane (in Euclid’s words a
plane surface) as undefined terms. The only thing is that we can represent them intuitively, or
explain them with the help of ‘physical models.’
Euclid then used his definitions in assuming some geometric properties which need no
proofs. These assumptions are self-evident truths. He divided them into two types: axioms and
A
postulates.
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3.3.1 Axioms and Postulates
Axioms are statements which are self evident or assumed to be true with in context of a
particular mathematical system. For example when we say “The whole is always greater than the
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parts.” It is a self evident fact and does not require any proof. This axiom gives us the definition
of ‘greater than’. For example, if a quantity P is a part of another quantity C, then C can be
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written as the sum of P and some third quantity R. Symbolically, C > P means that there is some
R such that C = P + R.
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Euclid used this common notion or axiom throughout the mathematics not particularly in
geometry but the term postulate was used for the assumptions made in geometry. The axioms are
the foundation stones on which the structure of geometry is developed. These axioms arise in
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different situations.
Some of the Euclid’s axioms are given below.
• Things which are equal to the same things are equal to one another
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• If equals are subtracted from equals, the remainders are also equal.
• Things which coincide with one another are equal to one another.
• Things which are double of the same things are equal to one another
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• Things which are halves of the same things are equal to one another
These ‘common notions’ refer to magnitudes of same kind. The first common notion could
be applied to plane figures. For example, if the area of an object say A equals the area of another
object B and the area of the object B equals that of a square, then the area of the object A is also
equals to the area of the square.
Magnitudes of the same kind can be compared and added, but magnitudes of different
kinds cannot be compared. For example, a line cannot be added to the area of objects nor can
an angle be compared to a pentagon.
TRY THIS
Can you give any two axioms from your daily life.
A
1. Mark two distinct points A and B on a sheet of paper.
Draw a straight line passing through the points
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A and B. How many such lines can be drawn through
A B
point A and B? We can not draw more than one
distinct line through two given points.
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Euclid’s first postulate gives the above
concept. Postulate is as follows-
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Postulate-1 : There is a unique line that passes through the given two distinct points.
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In Euclid’s terms, “To draw a straight line from any point to any point”.
P Q
How far the line segment PQ can be extended both sides? Does it have any end points?
We see that the line segment PQ can be extended on both sides and the line PQ has no end
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Postulate-2 : A line segment can be extended on either side to form a straight line.
In Euclid’s terms ‘To produce a finite straight line continuously in a straight line’ Euclid
used the term ‘terminated line’ for ‘a line segment’.
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3. Radii of four circles are given as 3 cm, 4 cm, 4.5 cm and 5 cm. Using a compass, draw
circles with these radii taking P, Q, R and S as their centres.
If the centre and radius of a circle are given can you draw the circle? We can draw a
circle with any centre and any radius. (See chapter-12 Circle)
Euclid’s third postulate states the above idea.
(To describe a circle with any centre and distance)
Postulate-3 : We can describe a circle with any centre and radius.
A
4. Take a grid paper. Draw different figures which represent a right angle. Cut them along
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their arms and place all angles one above other. What do you observe?
A C B B C
B
N G
A C
B C A A
You observe that both the arms of each angle fall on one above the other, (i.e.) all right
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angles are equal. This is nothing but Euclids fourth axiom. Can you say this for any angle? Euclid
take right angle as a reference angle for all the other angles and situation which he stated further.
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Now we shall look at the Euclid’s fifth postulate and its equivalent version.
Postulate-5 : If a straight line is falling on two straight lines makes the interior angles on the same
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side of it taken together is less than two right angles, then the two straight lines, if produced
infinitely, meet on that side on which the sum of the angles is less than two right angles.
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C Q
eventually intersect on the left side of PQ. D
This postulate has acquired much importance as many mathematicians including Euclid
were convinced that the fifth postulate is a theorem. Consequently for two thousand years
mathematicians tried to prove that the fifth postulate was a consequence of Euclid’s nine other
axioms. They tried by assuming other proposition (John Play Fair) which are equivalent to it.
A
n
may be drawn to the given line. (John Play Fair – 1748-1819) P
Let l be a line and P be a point, not on l. So through P, there k
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exists only one line parallel to l. This is called Play Fair’s axiom.
G
2
two right angles. (Legendre)
∠1 + ∠2 + ∠3 =1800
N 1 3
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• There exists a pair of lines everywhere equidistant from one another. (Posidominus)
l
m t
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p s
q
T
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• If a straight line intersects any one of two parallel lines, then it will intersect the other
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also.(Proclus)
• Straight lines parallel to the same straight line are parallel to one another.(Proclus)
If any one of these statements is substituted for Euclid’s fifth postulate leaving the first four
the same, the same geometry is obtained.
So after stating these five postulates, Euclid used them to prove many more results by
applying deductive reasoning and the statements that were proved are called propositions or
theorems.
Sometimes a certain statement that you think is to be true but that is an educated guess
based on observations. Such statements which are neither proved nor disproved are called
conjectures (hypothesis). Mathematical discoveries often start out as conjectures (hypothesis).
(“Every even number greater than 4 can be written as sum of two primes” is a conjecture
(hypothesis) stated by Gold Bach.)
A
A conjecture (hypothesis) that is proved to be true is called a theorem. A theorem is
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proved by a logical chain of steps. A proof of a theorem is an argument that establishes the truth
of the theorem beyond doubt.
Euclid deducted as many as 465 propositions in a logical chain using defined terms, axioms,
postulates and theorems already proven in that chain.
G
Let us study how Euclid axioms and postulates can be used in proving the results.
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Example-1. If A,B,C are three points on a line and B lies between A and C, then prove that
AC - AB = BC.
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A B C
Solution : In the figure, AC coincides with AB+BC
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Euclid’s 4th axiom says that things which coincide with one another
are equal to one another. Therefore it can be deducted that
AB + BC = AC,
Substituting this value of AC in the given equation AC - AB = BC
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AB + BC - AB = BC
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Note that in this solution, it has been assumed that there is a unique line passing through
two points.
Propostion -1. Prove that an equilateral triangle can be constructed on any given line segment.
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P Q
P Q
P Q
From Euclid’s 3rd postulate, we can draw a circle with any centre and any radius. So, we
can draw a circle with centre P and radius PQ. Draw another circle with centre Q and radius QP.
The two circles meet at R. Join ‘R’ to P and Q to form ∆ PQR.
Now we require to prove the triangle thus formed is equilateral i.e., PQ = QR = RP.
PQ = PR (radii of the circle with centre P). Similarly, PQ = QR (radii of the circle with
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centre Q)
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From Euclid’s axiom, two things which are equal to same thing are equal to each another,
we have PQ = QR = RP, so ∆PQR is an equilateral triangle. Note that here Euclid has
assumed, without mentioning anywhere, that the two circles drawn with centre P and Q will meet
each other at a point.
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Let us now prove a theorem.
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Example-3. Two distinct lines cannot have more than one point in common.
l
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Given : Two lines l and m.
Required to Prove (RTP): l and m have only m
A B
one point in common.
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Proof: Let us assume that two lines intersect Note that Euclid state this for straight lines
in two distinct points say A and B. only, not for the curved lines. Where ever line
is written always assume that we are talking
Now we have two lines passing through A and
about straight line.
B. This assumption contradicts with the
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pass through two distinct points. So we can conclude that two distinct lines cannot have more
than one point in common.
X
B
Example-4. In the adjacent figure, we have AC = XD, C and D
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EXERCISE - 3.1
1. Answer the following:
i. How many dimensions a solid has?
ii. How many books are there in Euclid’s Elements ?
A
iii. Write the number of faces of a cube and cuboid.
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iv. What is sum of interior angles of a triangle ?
v. Write three un-defined terms of geometry.
2. State whether the following statements are true or false? Also give reasons for your answers.
a) Only one line can pass through a given point.
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b) All right angles are equal.
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c) Circles with same radii are equal.
d) A line segment can be extended on its both sides endlessly to get a straight line.
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A C B
A B C D E F G H
1
4. If a point Q lies between two points P and R such that PQ = QR, prove that PQ = PR.
2
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7. Mark two points P and Q. Draw a line through P and Q. Now how many lines are
parallel to PQ, can you draw?
n
8. In the adjacent figure, a line n falls on lines l and m such that
l
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the sum of the interior angles 1 and 2 is less than 180°, then
1
what can you say about lines l and m. 2
m
3 9. In the adjacent figure, if ∠1 = ∠3,
4 ∠2 = ∠4 and ∠3 = ∠4, write the relation between
1 ∠1 and ∠2 using an Euclid’s postulate. B
2
1 1
10. In the adjacent figure, we have BX = AB, BY = BC and X Y
2 2
AB = BC. Show that BX = BY
A C
N ON -E UCLIDIAN G EOMETR
EOMETRYY
A
is true or some contrary postulate can be substituted A B
for it. If substituted with other, we obtain, Geometry A C
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different from Euclid’s Geometry, hence called non- (i) (ii)
Euclidian Geometry.
If plane is not flat what happens to our theorems?
Let us observe.
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Take a ball and try to draw a triangle on it? What difference do you find between
triangle on plane and on a ball. You observe that lines of a triangle on paper are straight but
not on ball.
N
See in figure (ii), the lines AN and BN (which are parts of great circles of a sphere) are
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perpendicular to the same line AB. But they are meeting at N, even though the sum of the
angles on the same side of line AB is not less than two right angles (in fact, it is 90° + 90° =
180°). Also, note that the sum of the angles of the triangle NAB on sphere is greater than 180°,
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as ∠A +∠ B = 180°.
We call the plane on a sphere as a spherical plane. Can any parallel lines exist on a
sphere? Similarly by taking different planes and related axioms new geometries arise.
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WHAT
HAT WE HAVE DISCUSSED
HAVE
ER
• The three building blocks of geometry are Points, Lines and Planes, which are undefined
terms.
• Euclid developed a system of thought in his “The Elements” that serves as the foundation
for development of all subsequent mathematics.
• Things which are equal to the same things are equal to one another
• If equals are subtracted from equals, the remainders are also equal.
• Things which coincide with one another are equal to one another.
• Things which are double of the same things are equal to one another
A
• Things which are halves of the same things are equal to one another
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Postulate - 1: To draw a straight line from any point to any point
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Postulate - 3: To describe a circle with any centre and radius
Brain teaser
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2. If the diagonal of a square is ‘a’ units, what is the diagonal of the square, whose area is
doubel that of the first square?
A
AN
4.1 I NTRODUCTION
Reshma and Gopi have drawn the sketches of their school and home respectively. Can
you identify some angles and line segments in these sketches?
D
G
T S
U X
V E C
W
N R
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P
Q A B
(i) (ii)
In the above figures (PQ, RS, ST, ...) and (AB, BC, CD, ...) are examples of line
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segments. Where as ∠UPQ, ∠PQR, ... and ∠EAB, ∠ABC, ... are examples of some angles.
Do you know whenever an architect has to draw a plan for buildings, towers, bridges
etc., the architect has to draw many lines and parallel lines at different angles.
In science say in Optics, we use lines and angles to assume and draw the movement of
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light and hence the images are formed by reflection, refraction and scattering. Similarly while
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finding how much work is done by different forces acting on a body, we consider angles between
force and displacement to find resultants. To find the height of a place we need both angles and
lines. So in our daily life, we come across situations in which the basic ideas of geometry are in
much use.
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DO THIS
Observe your surroundings carefully and write any three situations of your daily life
where you can observe lines and angles.
Draw the pictures in your note book and collect some pictures.
72 IX-CLASS MATHEMATICS
A
part of a line. It begins at a point and goes on endlessly in a specified direction. While line can be
extended in both directions endlessly.
AN
A part of a line with two end points is known as line segment.
We usually denote a line segment AB by AB and its length in denoted by AB. The ray
AB is denoted by AB and a line is denoted by AB . However we normally use AB , PQ for
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etc. lines and some times small letters l, m, n etc. will also be used to denote lines.
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If three or more points lie on the same line, they are called collinear points, otherwise
they are called non-collinear points.
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Sekhar marked some points on a line and try to count the line segments formed by them.
(Note PQ and QP represents the same line segment)
S.No. Points on line Line Segments Number
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1. P R Q PQ , PR , RQ 3
2. P S R Q PQ , PR , PS , SR , SQ , RQ 6
3. P S T R Q .............................................
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Do you find any pattern between the number of points and line segments?
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Take some more points on the line and find the pattern:
No. of points 2 3 4 5 6 7
on line segment
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O Initial position
The angle is formed by rotating a ray
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from an initial position to a terminal position.
rm
Te
ina
rm
lp
in
The change of a ray from initial position
os
al
iti
po
on
to terminal position around the fixed point ‘O’ is
sit
io
O
n
called rotation and measure of rotation is called Initial position
A
angle.
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One complete rotation gives 3600. We also draw angles with compass.
An angle is formed when two rays
originate from the same point. The rays
making an angle are called arms of the
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angle and the common point is called
vertex of the angle. You have studied
different types of angles, such as acute
angle, right angle, obtuse angle, straight
N
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acute angle : 0°< x <90° right angle : y = 90°
angle and reflex angle in your earlier
classes.
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obtuse angle : 90° < z < 180° straight angle : s = 180° reflex angle : 180° < t < 360°
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A
called point of concurrence.
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THINK, DISCUSS AND WRITE
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EXERCISE - 4.1
1. In the given figure, name:
(i) any six points
N E G
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X M P
(ii) any five line segments A B
11 12 1 C
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10 2
9 A 3
8 4 B
7 6 5
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A
(a) 9’O clock (b) 6’O clock (c) 7:00 PM
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4.3 P AIRS OF A NGLES
Now let us discuss about some pairs of angles.
Observe the following figures and find the sum of angles.
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40°
N
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60°
30° 50°
(i) (ii)
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What is the sum of the two angles shown in each figure? It is 900. Do you know what do
we call such pairs of angles? They are called complementary angles.
If a given angle is x0, then what is its complementary angle? The complementary angle of
x0 is (900- x0).
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Example-1. If the measure of an angle is 62°, what is the measure of its complementary angle?
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Solution : As the sum is 90°, the complementary angle of 62° is 90° - 62° = 28°
Now observe the following figures and find the sum of angles in each figure.
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What is the sum of the two angles shown in each figure? It is 1800. Do you know what
do we call such pair of angles? Yes, they are called supplementary angles. If the given angle is x0,
then what is its supplementary angle ? The supplementary angle of x0 is (180° - x°).
Example-2. Two complementary angles are in the ratio 4:5. Find the angles.
Solution : Let the required angles be 4x and 5x.
Then 4x + 5x = 900 (Why?)
9x = 900 ⇒ x = 100
A
Hence the required angles are 400 and 500.
Now observe the pairs of angles such as (120°, 240°) (100°, 260°) (180°, 180°) (50°,
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310°) ..... etc. What do you call such pairs? The pair of angles, whose sum is 360° are called
conjugate angles. Can you say the conjugate angle of 270°? What is the conjuage angle of x°?
DO THESE
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1. Write the complementary, supplementary and conjugate angles for the following angles.
(a) 450 (b) 750N (c) 54° (d) 300
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(e) 600 (f) 90° (g) 0°
2. Which pairs of following angles become complementary or supplementary angles?
TE
120°
T
30° 60°
C
2
1 3
B (iii)
3 1 2 2
1
2 1
O A
(i) (ii) (iv)
In figure (i) we can observe that vertex ‘O’ and arm ‘ OB ’ are common to both ∠ 1 and
∠ 2. What can you say about the non-common arms and how are they arranged? They are
arranged on either side of the common arm. What do you call such pairs of angles?
A
In fig.(ii), two angles ∠1 and ∠2 are given. They have neither a common arm nor a
AN
common vertex. So they are not adjacent angles.
TRY THIS
G
(i) Find pairs of adjacent and non-adjacent angles in the above figures (i, ii, iii & iv).
From the above, we can conclude that pairs of angles which have a common vertex, a
common arm and non common arms lie on either side of common arm are called adjacent angles.
Observe the given figure. The hand of the athlete
is making angles with the Javelin. What kind of angles
T
ACTIVITY
Measure the angles in the following figure and complete the table.
A
1 2
2 1
AN
(i) 1 2 (iii)
(ii)
Figure ∠1 ∠2 ∠1 + ∠2
G
(i)
(ii)
(iii) N
LA
4.3.1 Linear pair of angles axiom
Axiom : If a ray stands on a straight line, then the sum of the two adjacent angles so formed is
TE
180°.
When the sum of two adjacent angles is 180°, they
are called a linear pair of angles. 1
2
In the given figure, ∠1 + ∠2 = 180 0
T
Let us do the following. Draw adjacent angles of different measures as shown in the fig.
ER
Keep the ruler along one of the non-common arms in each case. Does the other non-common
arm lie along the ruler?
C C
SC
A
A
C 60°
80° 150° 70°
60°
30° O B O B
O B (ii)
(i) (iii)
A
A
125°
55°
= 180°. In other figures it is not so. O
B A
AN
(iv)
G
non-common arms of the angles form a line. This is the
converse of linear pair of angle axiom. 1 2
N
Angles at a point : We know that the sum of all the angles
around a point is always 360°.
5
4
3
LA
In the given figure ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 3600
Draw any two intersecting lines and label them. Identify the linear
P S
pairs of angles and write down in your note book. How many pairs are a
formed? d b
O
c
In the figure, ∠POS and ∠ROQ are opposite angles with same
T
R Q
vertex and have no common arm. So they are called as vertically opposite
ER
How many pairs of vertically opposite angles are there? Can you find them? (See figure)
ACTIVITY :
SC
Measure the four angles 1, 2, 3, 4 in each of the below figure and complete the table:
2 1 2
1 2 3
1
4 3 4 3 4
Figure ∠1 ∠2 ∠3 ∠4
(i)
(ii)
(iii)
A
What do you observe about the pairs of vertically opposite angles? Are they equal? Now
AN
let us prove this result in a logical way.
Theorem-4.1 : If two lines intersect each other, then the pairs of vertically opposite angles thus
formed are equal.
Given: AB and CD be two lines intersecting at O D B
G
O
Required to prove (R.T.P.)
(i) ∠ AOC = ∠ BOD
(ii) ∠ AOD = ∠ BOC. N A C
LA
Proof:
Ray OA stands on Line CD
Therefore, ∠ AOC + ∠ AOD = 180° [Linear pair angles axiom] .... (1)
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∠ AOD = ∠ BOC
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Do it on your own.
DO THIS
SC
1. Classify the given angles as pairs of complementary, linear pair, vertically opposite and
adjacent angles.
a
b
b
a
(i)
(ii)
a
a
b
b
(iii) (iv)
A
AN
2. Find the measure of angle ‘a’ in each figure. Give reason in each case.
G
a a
209°
43° 63°
96°
a
(ii)
N (iv)
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(iii)
50°
(i)
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(2x
A O B
all the angles on AB at a point O is 1800.
∴ (3x + 7)° + (2x - 19)° + x = 180° (Linear angles)
⇒ 6x - 12 = 180 ⇒ 6x = 192 ⇒ x = 32°.
SC
5
Therefore, ∠ POR = × 180 = 75°
12
7
Similarly, ∠ ROQ = × 180 = 105°
12
Now, ∠ POS = ∠ ROQ = 105° (Vertically opposite angles)
A
and ∠ SOQ = ∠ POR = 75° (Vertically opposite angles)
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Example-5. Calculate ∠ AOC, ∠ BOD and ∠ AOE in the adjacent figure given that
∠ COD = 90o, ∠ BOE = 72o and AOB is a straight line,
C D
Solution : Since AOB is a straight line, we have :
∠ AOE + ∠ BOE = 180o
G
90°
x° y°
= 3x° + 72° = 180° A O B
72°
⇒ 3x° = 108° ⇒ x = 36°.
We also know that
N 3x°
LA
∴ ∠ AOC + ∠ COD + ∠ BOD = 1800 (∵ straight angle) E
⇒ x° + 90° + y° = 180°
TE
Find ∠ ROT. T
Solution : Ray OS stands on the line PQ. P O Q
Therefore, ∠ POS + ∠ SOQ = 180° (Linear pair)
SC
Let ∠ POS = x°
Therefore, x° + ∠ SOQ = 180° (How?)
So, ∠ SOQ = 180° – x°
Now, ray OR bisects ∠ POS, therefore,
1
∠ ROS = × ∠ POS
2
1 x
= × x=
2 2
1
Similarly, ∠ SOT = × ∠ SOQ
2
1
= × (180° – x)
2
x°
A
= 90° –
2
Now, ∠ ROT = ∠ ROS + ∠ SOT
AN
x° x°
= + 90° −
2 2
G
= 90°
OS are four rays. Prove that
N
Example-7. In the adjacent figure OP , OQ , OR and
P
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∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°.
Solution : In the given figure, you need to draw opposite
ray to any of the rays OP , OQ , OR or OS O
TE
T Q
Draw ray OT so that TOQ is a line. Now, ray OP R
S
stands on line TQ .
Similarly, ray OS stands on line TQ .
ER
EXERCISE - 4.2
E
1. In the given figure three lines AB , CD and EF
A D
intersecting at O. Find the values of x, y and z it is
y°
being given that x : y : z = 2 : 3 : 5 z° O
A
2. Find the value of x in the following figures. x°
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A C B
B
F
B
(x-24)°
G
(3x+18)° 93° 29° C
A C O
O
(i)
N 296°
(ii)
A
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B C B
(6x+2)°
O C
D
A
(iii) (iv)
T
C
E
ER
D
P
4. In the given figure lines XY and MN intersect M
at O. If ∠ POY = 90° and a: b = 2 : 3, find c.
a
X b Y
c O
S T
A
Q R
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6. In the given figure, if x + y = w + z, then prove that
B
AOB is a line. x
y
G
O
w
z
N A
LA
7. In the given figure PQ is a line. Ray OR is
R
perpendicular to line PQ . OS is another ray lying S
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8. It is given that ∠ XYZ = 64° and XY is produced to point P. A ray YQ bisects ∠ ZYP.
Draw a figure from the given information. Find ∠ XYQ and reflex ∠ QYP.
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l
meets the other lines m and n? Line l meets the lines at two 1
P 2
distinct points. What do we call such a line? It is a transversal. m
4
It is a line which intersects two distinct lines at two distinct 3
points. Line ‘l ’ intersects lines ‘m’ and ‘n’ at points ‘P’ and
5
‘Q’ respectively. So, line l is a transversal for lines m and n. Q 6
n 8
Observe the number of angles formed when a 7
transversal intersects a pair of lines.
A
one is interior and the other is exterior, are called corresponding angles.
AN
From the given figure.
(a) What are corresponding angles?
(i) ∠ 1 and ∠ 5 (ii) ∠ 2 and ∠ 6
(iii) ∠ 4 and ∠ 8 (iv) ∠ 3 and ∠ 7, So there are 4 pairs of corresponding angles.
G
(b) What are alternate interior angles?
(i) ∠ 4 and ∠ 6 (ii) ∠ 3 and ∠ 5, are two pairs of alternate interior angles.(Why?)
(c) N
What are alternate exterior angles?
(i) ∠ 1 and ∠ 7 (ii) ∠ 2 and ∠ 8, are two pairs of alternate exterior angles. (Why?)
LA
(d) What are interior angles on the same side of the transversal?
(i) ∠ 4 and ∠ 5 (ii) ∠ 3 and ∠ 6 are two pairs of interior angles on the
TE
Exterior angles on the same side of the transversal are also referred as consecutive
exterior angle or co-exterior angles or allied exterior angles?
What can we say about the corresponding angles formed when the two lines l and m are
parallel? Check and find. Will they become equal? Yes, they are equal.
SC
Axiom of corresponding angles: If a transversal intersects a pair of parallel lines, then each
P
pair of corresponding angles are equal.
What is the relation between the pairs of alternate
Q
B
interior angles (i) ∠ BQR and ∠ QRC A
1
(ii) ∠ AQR and ∠ QRD in the figure? 2
C R D
Can we use corresponding angles axiom to find
the relation between these alternative interior angles.
S
A
And ∠ PQA = ∠ BQR ..... (2) (Why?)
So, from (1) and (2), you may conclude that ∠ BQR = ∠ QRC.
AN
Similarly, ∠ AQR = ∠ QRD.
This result can be stated as a theorem as follows:
Theorem-4.2 : If a transversal intersects two parallel lines, then each pair of alternate interior
G
angles are equal.
In a similar way, you can obtain the following theorem related to interior angles on the
same side of the transversal.
N
Theorem-4.3 : If a transversal intersects two parallel lines, then each pair of interior angles on
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the same side of the transversal are supplementary.
DO THESE
TE
1. Find the measure of each angle indicated in each figure where l and m are
parallel lines intersected by transversal n. n
n
T
x°
l l
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84°
y°
m m
110°
SC
l 53°
100° l
z°
m s°
m
75°
l
l
A
60°
(11x - 2)°
m (8x -4)°
AN
m
G
n
(14x-1)°
N l
l
LA
(13x-5)°
m
(12x+17)° m
(17x+5)°
TE
ACTIVITY
T
DO THIS
Draw a line AD and mark points B and C on
Q S
it. At B and C, construct ∠ABQ and ∠BCS equal to
E
F
each other as shown. Produce QB and SC on the
A
other side of AD to form two lines PQ and RS.
Q S
AN
A B D
C
G
A C D P R
B
N
Draw common perpendiculars EF and GH for the two lines PQ and RS. Measure the
lengths of EF and GH. What do you observe? What can you conclude from that? Recall that
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if the perpendicular distance between two lines is the same, then they are parallel lines.
Axiom-1 : If a transversal intersects two lines such that a pair of corresponding angles are equal,
then the two lines are parallel to each other.
TE
the mason concludes that the wall is vertical to the ground. Think,
Wall
how he has come to this conclusion?
ER
∠ PQA = ∠ QRC
S
Theorem-4.4 : If a transversal intersects two lines such that a pair of alternate interior angles are
A
equal, then the two lines are parallel.
AN
4.4.1 Lines Parallel to the Same Line
t
If two lines are parallel to the same line, will they 1
be parallel to each other? l
2
G
Let us check it. Draw three lines l, m and n such m
3
that m || l and n || l. n
N
Let us draw a transversal ‘t’ on the lines, l, m and n.
Now from the figure ∠ 1 = ∠ 2 and ∠ 1 = ∠ 3
LA
(Corresponding angles axiom)
So, ∠ 2 = ∠ 3 But these two form a pair of corresponding angles for the lines m & n.
TE
TRY THIS
t
ER
?
n
R
A
Again, EF || AB and AE is the transversal.
105°
AN
A C x°
25
∠ BAE + ∠ AEF = 180 [∵ Co-interior angles]
o
°
E
⇒ 105o + ∠ AEC + ∠ CEF =180o
⇒ 105o + 25o + (180o - xo) = 180o
G
⇒ 310 - x° = 180°
Hence, x = 130°.
N
LA
Example-9. In the adjacent figure, find the value of x, y, z and a, b, c.
bo
⇒ xo + yo = 180o (Linear pair) o
a
o xo zo
⇒ xo + 110o = 180o c
o
110
o y o
65
⇒ x = (180 - 110 ) = 70 .
o o o o
T
co = 65o (How?)
ao + co = 180o [Linear pair]
⇒ ao + 65o = 180o
SC
E G
∠PQS + ∠SQB = 180° (Why?)
A
R S
x° = C D
7
4x°
AN
= 25°
F H
Example-11. In the given figure PQ || RS, ∠ MXQ = 135o and ∠ MYR = 40o, find ∠ XMY.
Solution : Construct a line AB parallel to PQ, through the point M.
G
Now, AB || PQ and PQ || RS. X
P Q
Therefore, AB || RS
Now, ∠ QXM + ∠ XMB = 180o N 135°
LA
(AB || PQ, Interior angles on the same A M B
side of the transversal XM)
So, 135° + ∠ XMB = 180° R
40°
S
Y
TE
Solution : In the given Figure a transversal AD intersects two lines PQ and RS at two points
B and C respectively. Ray BE is the bisector of ∠ ABQ and ray CF is the bisector of ∠ BCS;
and BE || CF.
We have to prove that PQ || RS. It is enough to prove any one of the following pair:
i. Corresponding angles are equal.
ii. Pair of interior or exterior angles are equal.
iii. Interior angles same side of the transversal are supplementary.
From the figure, we try to prove the pairs of corresponding angles to be equal.
Since, it is given that ray BE is the bisector of ∠ ABQ. A
1
∠ ABE = ∠ ABQ. ... (1)
2 E
Similarly, ray CF is the bisector of ∠ BCS.
1
A
P Q
Therefore, ∠ BCF = ∠ BCS ... (2) B F
2
But for the parallel lines BE and CF; AD is a transversal.
AN
R S
C
Therefore, ∠ ABE = ∠ BCF
(Corresponding angles axiom) ... (3)
From the equation (1) and (2) in (3), we get D
G
1 1
∠ ABQ = ∠ BCS
2 2
∴ ∠ ABQ = ∠ BCS
N
But, these are the corresponding angles made by the transversal AD with lines PQ and
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RS ; and are equal.
Therefore, PQ || RS (Converse of corresponding angles axiom)
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Example-13. In the given figure AB || CD and CD || EF. Also EA ⊥ AB. If ∠ BEF = 55°,
find the values of x, y and z. G
Solution : Extend BE to G. A C
z E
Now ∠GEF = 180° - 55° (Why?) 55°
T
= 125° D
y
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3. Showing a pair of interior angles on the same side of the transversal are supplementary.
EXERCISE - 4.3
1. It is given that l || m to prove ∠1 is supplement to ∠8. Write reasons for the
statement.
Statement Reasons
A
2 1
i. l || m ______________
l
AN
3
ii. ∠1 = ∠5 ______________ 4
6 5
iii. ∠5 + ∠8 = 180° ______________ m
7 8
iv. ∠1 + ∠8 = 180° ______________
G
v. ∠1 is supplement to ∠8 ______________
2.
y : z = 3 : 7, find x.
N
In the adjacent figure AB || CD; CD || EF and
A
x
B
LA
y
C D
z
E F
TE
C E D
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point R.] R
r s
2a°
p
6. Find the value of a and b, given that p || q and r || s.
b°
A
80°
q
AN
c d
7. If in the figure a || b and c || d, then name the angles
that are congruent to (i) ∠1 (ii) ∠2.
G
2 1 10 9
a
3 4 11 12
N b
6 5 14 13
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x° 7 8 15 16
y°
8. In the figure the arrow
59° head segments are parallel. find
TE
105°
ER
(3y+6)°
x°
)°
+5
(3y
A P B
13. In the given figure, if AB || CD, ∠ APQ = 50° 50°
and ∠ PRD = 127°, find x and y. y°
x° 127°
C Q R D
A
14. In the adjacent figure PQ and RS are two mirrors P B Q
AN
placed parallel to each other. An incident ray AB
strikes the mirror PQ at B, the reflected ray moves A
D
along the path BC and strikes the mirror RS at C
G
R S
and again reflected back along CD . Prove that C
AB || CD.
N
[Hint : Perpendiculars drawn to parallel lines are also parallel.]
E
LA
15. In the figures given below AB || CD. EF is the
transversal intersecting AB and CD at G and H 2x° G
C D
respectively. Find the values of x and y. Give y
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reasons 3x°
A H B
E E
F (i)
T
2x +15° G
C D C D
G
ER
4x- 23°
3x-20° 3x°
A B A H B
H
(iii)
SC
F (ii)
F t
2 1
A E B
16. In the adjacent figure, AB || CD, ‘t’ is a 3 4
transversal intersecting E and F respectively. If 6 5
C F D
∠ 2 : ∠ 1 = 5 : 4, find the measure of each 7 8
marked angles.
P z°
80° y°
A B
A
y° Q R
C D
AN
z°
18. In the adjacent figure AB || CD. Find the values
70° of x, y and z.
90°
x° x°
A B
E
G
19. In each of the following figures AB || CD. Find the value of x in each case.
B
A
104°
B
NC
35°
D 35° M
75°
D
LA
x° E x°
E x°
116° 65°
A B
C D
(i) (ii) (iii)
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A C
ACTIVITY
ER
A
through ‘C’ draw a line CE parallel to BA
Proof :
AN
BA||CE [By construction]
∠ ABC= ∠ ECD .....(1) [By corresponding angles axiom.]
∠ BAC = ∠ ACE .....(2) [Alternate interior angles for the parallel lines
G
AB and CE]
∠ ACB = ∠ ACB .....(3) [Same angle]
∠ ABC + ∠ BAC + ∠ ACB =
∠ ECD + ∠ ACE + ∠ ACB
N [Adding the above three equations]
LA
But ∠ ECD + ∠ ACE + ∠ ACB = 1800 [Sum of angles at a point on a straight line]
∴ ∠ ABC + ∠ BAC + ∠ ACB = 1800
∠ A + ∠ B + ∠ C = 1800
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You know that when a side of a triangle is produced there forms an exterior angle of the
triangle
When side QR is produced to point S, ∠ PRS is P
T
Q S
Also, see that R
Example-14. The angles of a triangle are (2x)°, (3x + 5)° and (4x − 14)°.
A
Find the value of x and the measure of each angle of the triangle.
AN
Solution : We know that the sum of the angles of a triangle is 180o.
∴ 2x° + 3x° + 5° + 4x° - 14° = 180° ⇒ 9x° − 9° = 180°
⇒ 9x° = 180° + 9° = 189°
G
189°
⇒ x= = 21.
9°
N
∴ 2x° = (2 × 21)° = 42°, (3x + 5)° = [(3 × 21 + 5)]° = 68°.
(4x − 14)° = [(4 × 21) − 14]° = 70°
LA
Hence, the angles of the triangle are 42o, 68o and 70o.
Example-15. In the adjacent figure, AB || QR, ∠ BAQ = 142o and ∠ ABP = 100o.
TE
P
Find (i) ∠ APB (ii) ∠ AQR and (iii) ∠ QRP,
Q
⇒ 142o = 100o + xo R
A
B
Let ∠ DAE = p , ∠ BAE = q , ∠ DCE = z and ∠ ECB = t . Since the exterior
° ° ° °
AN
angle of a triangle is equal to the sum of the interior opposite angles, we have :
z° = p° + 26° A D
p° 26°
° °
t = q + 38 ° q° z°
C
∴ z° + t° = p° + q° + (26 + 38)° = p° + q° + 64° t° E
G
But, p° + q° = 46. (∵ ∠ DAB = 46o) 38°
So, z° + t° = 46 + 64 = 110°.
Hence x° = z° + t° = 110°. N B
LA
Example-17. In the given figure ∠ A = 40°. If BO and CO are the bisectors of ∠ B and
∠ C respectively. Find the measure of ∠ BOC.
TE
Example-18. Using information given in the adjacent figure, find the values of x and y.
A
Solution : Side BC of ∆ABC has been produced to D.
x°
Exterior ∠ ACD = ∠ ABC + ∠ BAC x°
∴ 100° = 65o + xo
⇒ xo = (100o - 65o) = 35o.
100°
∴ ∠ CAD = ∠ BAC = 35o 65° y°
D
B C
In ∆ACD, we have :
∠ CAD + ∠ ACD + ∠ CDA = 180o (Angle sum property of triangle)
⇒ 35o + 100o + yo = 180o
⇒ 135o + yo = 180o
⇒ yo = (180o - 135o) = 45o
A
Hence, x = 35°, y = 45°.
AN
Example-19. Using information given in the adjacent figure, find the value of x and y.
A
Solution : Side BC of ∆ABC has been produced to D.
∴ Exterior angle ∠ACD = ∠BAC + ∠ABC 30°
E y°
G
⇒ xo = 30o + 35o = 65o.
Again, side CE of ∆DCE has produced to A. 35°
x°
45°
N
∴ Exterior angle ∠ DEA = ∠ EDC + ∠ ECD
⇒
B C
D
LA
y = 45 + xo = 45o + 65o = 110o.
Hence, x = 65° and y = 110°.
Example-20. In the adjacent fig. if QT ⊥ PR, ∠ TQR = 40° and ∠ SPR = 30°, find x and y.
TE
Solution : In ∆TQR, P
x°
BO and CO of ∠ CBE and ∠ BCD respectively meet at point
1
O, then prove that ∠ BOC = 90° – ∠ BAC.
2
Solution : Ray BO is the bisector of ∠ CBE. y° z°
1 B C
Therefore, ∠ CBO = ∠ CBE
2
1 E D
= (180° – y°)
2
y°
= 90° – ...(1)
2 O
A
In ∆BOC, ∠ BOC + ∠ BCO + ∠ CBO = 180° ...(3)
AN
Substituting (1) and (2) in (3), you get
z° y°
∠ BOC + 90° – + 90° – = 180°
2 2
z° y°
So, ∠ BOC = +
G
2 2
1
or, ∠ BOC = (y° + z°) ... (4)
But,
2
N
x° + y° + z° = 180° (Angle sum property of a triangle)
LA
Therefore, y° + z° = 180° – x°
Therefore, (4) becomes
1
TE
= 90° – ∠ BAC
2
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EXERCISE 4.4
1. In the given triangles, find out x, y and z.
A
SC
E
50° S
60°
y° R
x°
60° F 70°
B C D G
(i) z°
H (ii) 35° 45°
P Q
(iii)
2
1 3 4
R T
B
S
A
3x°
D 3. In the given figure AB || CD; BC || DE then
105°
AN
A
24° y°
E find the values of x and y.
C
C
G
CD⊥DA then prove that m∠1 ≅ m∠3. 3
1
30°
5y°
A
D N D E
A
LA
5. Find the values of x, y for which the lines AD
2x° (x-y)°
and BC become parallel.
B C
TE
S 30°
45°
P 135°
P 53°
D E
A
95°
AN
R
40° S and RS intersect at point T, such that
T 75° ∠ PRT = 40°, ∠ RPT = 95° and
∠ TSQ = 75°, find ∠ SQT.
Q A
N G
12. In the adjacent figure, ABC is a triangle in which
70°
∠ B = 50° and ∠C = 70°. Sides AB and AC are produced. B 50°
C
LA
If ‘z’ is the measure of the angle between the bisectors of x° y°
x° y°
the exterior angles so formed, then find ‘z’.
z
TE
P Q
x° O
28°
13. In the given figure if PQ ⊥ PS, PQ || SR,
y° ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x
T
65° and y.
S T A
R
ER
A
D B E
C
34°
AN
17. Using information given in the figure, calculate the value
y° of x and y.
E
G
D 24° x° 62°
B
C
WHAT
HAT WE HAVE DISCUSSED
HAVE
N
LA
• Linear pair axiom: If a ray stands on a straight line, then the sum of the two adjacent
angles so formed is 180°.
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• Axiom of corresponding angles: If a transversal intersects two parallel lines, then each
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• Theorem: If a transversal intersects two lines such that a pair of interior angles on the same
side of the transversal are supplementary, then the two lines are parallel.
• Theorem: Lines which are parallel to a given line are parallel to each other.
• Theorem: The sum of the angles of a triangle is 180º.
A
• Theorem: If a side of a triangle is produced, then the exterior angle so formed is equal to
the sum of the two interior opposite angles.
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Do You Know?
The Self-generating Golden Triangle
G
The golden triangle is an isosceles triangle with base
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angles 72o and the vertex angle 36o. When both of
these base angles are bisected the two new triangles
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produced are also golden triangles. This process can
be continued indefinitely up the legs of the original
golden triangle, and an infinite number of golden
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Co-Ordinate Geometry
05
A
5.1 I NTRODUCTION
AN
The minimum and the maximum temperatures of Kufri in Himachal Pradesh on a particular
day in the month of December were - 6oC and 7oC. Can you represent it on a number line?
G
in oC
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
N
Here the numberline acts as a reference scale
HG F ED C B A
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to indicate the status of temperature on a particular
day.
Let us observe the situation as shown in the
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In the above example can you say how many references did you consider? What are they?
Let us discuss one more situation.
A teacher asked her students to mark a point on a
sheet of paper. The hint given by the teacher is “the point
should be at a distance of 6 cm from the left edge.” Some
A
6 cm. A
of the students marked the point as shown in the figure.
AN
In the figure which point do you suppose is correct? 6 cm.
B
Since each point A,B,C and D is at a distance of 6 cm
from the left edge, no point can be denied. To fix the 6 cm.
C
exact position of the point what more information is
G
6 cm. D
needed? To fix its exact position, another reference, say,
the distance from the edge of the top or bottom has to be
given. N
LA
Suppose the teacher says that the point is at a
distance of 6 cm from the left edge and at a distance of 8
6 cm. P cm from the bottom edge, now how many points with
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DO THIS
Describe the seating position of any five students in your classroom.
A
Clip .......... .......... ..........
Teddy .......... .......... ..........
AN
Soap .......... .......... ..........
Is the object in 3rd column and 4th row is same as 4th column and 3rd row?
G
The representation of a point on a plane with idea of two references led to development
of new branch of mathematics known as Coordinate Geometry.
N
Rene Descartes (1596-1650), a French mathematician and
philosopher has developed the study of Co-ordinate Geometry. He found
an association between algebraic equations and geometric curves and figures.
LA
In this chapter we shall discuss about the point and also how to plot the
points on a co-ordinate plane.
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EXERCISE 5.1
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1. In a locality, there is a main road along North-South direction. The
W E
S map is given below. With the help of the picture answer the following
T
questions.
(i) What is the 3rd object on the left side in
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street no. 3?
(ii) Find the name of the 2nd house which is in
right side of street 2.
SC
We use number line to represent the numbers by marking points on the line at equal
distances. Observe the following integer line.
Negative Integer Positive Integer
A
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
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It is observed that distances marked on either side from a fixed point is called origin on
number line and denoted by ‘O’. All positive numbers are shown on the right side of zero and all
negative numbers on its left side.
We take two number lines, perpendicular to each other in a plane. We locate the position
G
of a point with reference to these two lines. Observe the following figure.
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LA
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The perpendicular lines may be in any direction as shown in the figures. But, when we
choose these two lines to locate a point in a plane in this chapter, for the sake of convenience we
take one line horizontally and the other vertically as in fig. (iii). We draw a horizontal number line
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and a vertical number line meeting at a point perpendicular to each other. The point of intersection
is denoted as origin. The horizontal number line XX1 is known as X-axis and the vertical number
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3 3
2 2
Origin 1 1
Origin
1
X X 0 X
1
X
-5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 O 1 2 3 4 5
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
Y1 Y1
A
Y-axis respectively. Also OX1 and OY1 are X
1
X
called the negative directions of the X-axis and O
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the Y-axis respectively. We can observe that
the axes (plural of axis) divide the plane into Quadrant III Quadrant IV
four parts. These four parts are called the Q3 Q4
G
quadrants and are denoted by Q1, Q2, Q3 and
Y1
Q4 in anti clockwise direction. The plane here
is called the cartesian plane (named after Rene
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Descartes) or co-ordinate plane or XY-plane. The axes are called the coordinate axes.
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5.2.1 Locating a Point
Now let us see how to locate a point in the coordinate system. Observe the following
graph. Two axes are drawn on a graph paper.
TE
Y
A and B are any two points on it. Can you name 5
Q• 4
the quadrants to which the points A and B belong (5, 3)A
to? D3
2
T
(iii) The perpendicular distance of the point B from the Y-axis measured along the negative
direction of X-axis is OE=BF= 4 units. i.e. at −4 on X-axis. We call this as X-coordinate
of ‘B’.
(iv) The perpendicular distance of the point B from the X-axis measured along the negative
direction of Y-axis is OF = EB = 3 units. i.e. at −3 on Y-axis. We call this as
Y-coordinate of ‘B’ and (−4, −3) are coordinates of ‘B’.
A
Now using these distances, how can we locate the point? We write the coordinates of a
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point in the following method.
(i) The x-coordinate of a point is the distance from origin to foot of perpendicular on
X-axis.
The x-coordinate is also called the abscissa.
G
The x-coordinate (abscissa) of P is 2.
The x-coordinate (abscissa) of Q is −3.
(ii)
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The y-coordinate of a point is, the distance from origin to foot of perpendicular on
Y-axis.
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The y-coordinate is also called the ordinate.
The y-coordinate or ordinate of P is −2.
The y-coordinate or ordinate of Q is 4.
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Hence the coordinates of P are (2, −2) and the coordinates of Q are (−3, 4).
So the coordinates locate a point in a plane uniquely.
5.2.2 Origin
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1. The intersecting point of X-axis and Y-axis is called origin. We take origin as a reference
point to locate other points in a plane.
ER
Example 1. State the abscissa and ordinate of the following points and describe the position of
each point (i) P(8,8) (ii) Q (6,−8).
Solution : (i) P (8,8)
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Solution : 1. Draw a perpendicular line to X-axis from the point P. The perpendicular line
touches X-axis at 4 units. Thus abscissa of P is 4. Similarly draw a perpendicular
line to Y-axis from P. The perpendicular line touches Y-axis at 3 units. Thus
ordinate of P is 3. Hence the
A
Y
Cordinates of P are (4, 3). Q 5
4
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2. Similarly, the abscissa and ordinate of P
3
the point Q are −4 and 5 respectively.
2
Hence the coordinates of Q are
1
(−4, 5). 1
X
G
X
-5 -4 -3 -2 -1 O 1 2 3 4 5
3. As in the earlier case the abscissa and -1
ordinate of the point R are −2 and −4 -2
3
3 units from the Y-axis and at a distance zero
2
units from the X-axis. Therefore the x
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1
coordinate of A is 3 and y-coordinate is 0. 1
E D C B A
X X
Hence the coordinates of A are (3,0).So think -5 -4 -3 -2 -1 O 1 2 3 4 5
-1
and discuss.
-2
SC
DO THIS
Among the points given below some of the points lie on X-axis. Identify them.
(i) (0,5) (ii) (0,0) (iii) (3,0)
(iv) (-5,0) (v) (-2,-3) (vi) (-6,0)
A
(vii) (0,6) (viii) (0,a) (ix) (b,0)
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Example-4. Write the coordinates of the points marked in graph.
Solution :
G
5 P
from the X-axis and at a distance zero
4
from the Y-axis. Therefore the Q
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x-coordinate of P is 0 and y-coordinate
is 5. Hence the coordinates of P are
3
2
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(0,5). 1 R
1
X X
So think & discuss that- -5 -4 -3 -2 -1 O 1 2 3 4 5
-1
(ii) The coordinates of Q are (0, 3.5), why? -2 S
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-3
(iii) The coordinates of R are (0,1), why?
-4
(iv) The coordinates of S are (0, −2), why? -5 T
1
Y
(v) The coordinates of T are (0, −5), why?
T
Since every point on the Y-axis has no distance from the Y-axis, therefore the x-coordinate
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of the point lying on Y-axis is always zero. Y-axis is denoted by the equation x = 0.
zero. Also it lies on X-axis. Its distance from X-axis is zero. Hence its y-coordinate is zero.
TRY THESE
1. Which axis the points such as (0, x) (0, y) (0,2) and (0,−5) lie on? Why ?
A
6
D
5
AN
4
3
2
1
G
1
X X
-9 -8 -7 -6 -5 -4 -3 -2 -1 O 1 2 3 4 5 6 7 8 9
-1
A
-2
-3
N I
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-4
-5
N
-6
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T
-7
-8
-9
1
Y
T
E 3 7 E (3,7) Q1 (+, +)
From the above table you may have observed the following relationship between the signs
of the coordinates of a point and the quadrant of a point in which it lies.
Y
5
4 Y
A
Q2 3
Q1
(-, +) (+, +)
2 x < 0 (negative) x > 0 (positive)
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1 y > 0 (positive) y > 0 (positive)
1
X X X
1
X
-5 -4 -3 -2 -1 O 1 2 3 4 5 O
-1
G
-2 x < 0 (negative) x > 0 (positive)
Q3 Q4 y < 0 (negative) y < 0 (negative)
-3
(-, -) (+, -)
-4
-5
Y1
N Y1
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EXERCISE 5.2
1. Write the quadrant in which the following points lie?
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3. Which of the following points lie on the axes? Also name the axis.
i) The ordinate of L
A
9
8
M Q
AN
7
6
5
4
G
3
2
X
1 N
-9 -8 -7 -6 -5 -4 -3 -2 -1 O
1
N
1 2 3 4 5 6 7 8 9
X
LA
-1
-2
R(-2, -2) -3
-4 P(5, -4)
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-5
-6
L -7
-8
T
-9
1
Y
ER
iii. The point which lies on both the axes is called origin.
iv. The point ( 2, −3 ) lies in the third quadrant.
v. (−5, −8 ) lies in the fourth quadrant.
vi. The point (−x , −y) lies in the first quadrant where x < 0 , y < 0.
6. Plot the following ordered pairs on a graph sheet.What do you observe?
i.. (1, 0), (3 , 0), (−2 , 0 ), (−5, 0), (0, 0), (5, 0), (−6, 0)
ii. (0, 1), (0 , 3), (0 , −2), (0, −5), (0, 0), (0, 5), (0, −6)
So far we have seen how to read the positions of points marked on a Cartesian plane.
Now we shall learn to mark the point if its co-ordinates are given.
A
For instance how do you plot a point (4, 6).
Can you say in which quardrant the point P lies?
AN
We know that the abscissa (x-coordinate) is 4 and y-coordinate is 6.
Y
9
8
G
7
6 P (4,6)
N 5
4
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3
2
1
A
TE
1
X X
-9 -8 -7 -6 -5 -4 -3 -2 -1 O 1 2 3 4 5 6 7 8 9
-1
-2
-3 Y′
T
• Draw two number lines perpendicular to each other meeting at their zeroes on a graph
paper. Name the horizontal line as X-axis and the vertical line as Y-axis and locate the
meeting point of both the lines as Origin ‘O’.
SC
• Keep the x-coordinate in mind, start from zero, i.e. from the Origin.
• Move 4 units along positive part of X-axis i.e. to its right side and mark the point A.
• From A move 6 units upward along a line parallel to positive part of Y-axis.
• Locate the position of the point ‘P’ as (4, 6).
The above process of marking a point on a Cartesian plane using their co-ordinates is
called “plotting the point”.
6 Y
A
5
M (-2,4) 4
AN
3
2
1
G
X
1 A X
-9 -8 -7 -6 -5 -4 -3 -2 -1 O 1 2 3 4 5 6 7 8 9
-1
-2 N
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-3
A (-5, -3)
-4
-5
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-6 N (1, -6)
-7 Y′
(i) Can you say in which quadrant the point M lies?
It lies in the second quadrant. Let us now locate its position.
T
M (−2,4) : start from the origin, move 2 units from zero along the negative part of
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Move 5 units from zero to its left side that is along the negative part of X-axis.
From there move 3 units along a line parallel to negative part of Y-axis i.e. downwards.
(iii) N (1, −6): start from zero, the Origin.
The point N lies in the fourth quadrant, start from zero the origin.
Move 1 unit along positive part of X-axis i.e. to the right side of zero.
From there move 6 units along a line parallel to negative Y-axis i.e.
downwards.
DO THIS
Plot the following points on a Cartesian plane.
A
Example 8 : Plot the points T(4, −2) and V(−2, 4) on a cartesian plane. Whether these
two coordinates locate the same point?
AN
Y
Solution : In this example we plotted two 5
points T (4, −2) and V(−2, 4) V(-2, 4) 4
Are the points (4, −2) and (−2, 4) 3
G
distinct or same? Think. 2
We see that (4, −2) and (−2, 4) are at 1
-5 -4 -3 -2 -1 O
-1
1 2 3 4 5
X
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B(−5 , 4) and say whether the point (x, y ) is -2
T(4, -2)
different from (y, x ) or not ? -3
-4
From the above plotting it is evident
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-5
that the position of (x, y) in the Cartesian Y1
plane is different from the position of (y, x).
i.e. the order of x and y is important in (x, y).
Therefore (x, y) is called an ordered pair.
T
DO THIS
(i) Write the coordinates
Y A
of the points A, B, C, 9 B
D, E. 8
7
A
(ii) Write the coordinates 6 C
of F, G, H, I, J. 5
AN
4
3 D
E
2
1
1
X X
G
-9 -8 -7 -6 -5 -4 -3 -2 -1 O 1 2 3 4 5 6 7 8 9
-1
F
-2
-3
J
N G
-4
-5
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-6
I H -7
-8
-9
1
Y
TE
E XERCISE 5.3
1. Plot the following points in the Cartisian plane whose x , y co-ordinates
T
are given.
x 2 3 −1 0 −9 −4
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y −3 −3 4 11 0 −6
(x, y)
Are the positions of (5, −8) and (−8, 5) is same? Justify your answer.
SC
2.
3. What can you say about the position of the points (1, 2), (1, 3), (1, −4), (1, 0), and
(1, 8). Locate on a graph sheet .
4. What can you say about the position of the points (5, 4), (8, 4), (3, 4), (0, 4),
(−4, 4), (−2, 4)? Locate the points on a graph sheet. Justify your answer.
5. Plot the points (0, 0) (0, 3) (4, 3) (4, 0) in graph sheet. Join the points with straight lines
to make a rectangle. Find the area of the rectangle.
6. Plot the points (2, 3), (6, 3) and (4, 7) in a graphsheet. Join them to make it a
triangle. Find the area of the triangle.
7. Plot at least six points in a graph sheet, each having the sum of its coordinates equal
to 5.
A
Hint : (−2, 7) (1, 4) .............
AN
L, M. N, P, O and Q.
Y
5 B
G
A L 4 C
D
N 3 P
J K M
1 N 2
Q E F
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1
X O X
-5 -4 -3 -2 -1 1 2 3 4 5
-1 H
I G
-2
TE
-3
-4
-5
1
Y
T
9. In a graph Sheet Plot each pair of points, join them by line segments
ER
i. (2, 5), (4, 7) ii. (−3, 5), (−1, 7) iii. (−3, −4), (2, −4)
iv. (−3, −5), (2, −5) v. (4, −2), (4, −3) vi. (−2, 4), (−2, 3)
vii. (−2, 1), (−2, 0) viii. (4, 7), (4, –3) ix. (4,–2), (2, –4)
SC
x. (4, –3), (2,–5) xi. (2, 5), (2, –5) xii. (–3, 5), (–3, –5)
xiii. (–3, 5), (2, 5) xiv. (–1, 7) (4, 7)
(1, 0), (0, 9); (2, 0), (0, 8); (3, 0) (0, 7); (4, 0) (0, 6);
(5, 0) (0, 5); (6, 0) (0, 4); (7, 0) (0, 3); (8, 0) (0, 2); (9, 0) (0, 1).
ACTIVITY
Study the positions of different cities like Hyderabad, New Delhi,
Chennai and Vishakapatnam with respect to longitudes and latitudes on a globe.
A
CREATIVE ACTIVITY
AN
Take a graph sheet and plot the following pairs of points on the axes and join
them with line segments.
(−9, 0), (−6, 4), (−2, 5), (2, 4), (5, 0) (−2, 0),
(−2, −8), (−3, −9), (−4, −8).
G
What do you notice ?
N
LA
WHAT WE HAVE DISCUSSED
• We need two references to locate the exact position of a point in a plane.
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• A point or an object in a plane can be located with the help of two perpendicular number
lines. One of them is horizontal line (X-axis) and the other is vertical line (Y-axis).
• The representing of points in the plane in the form of coordinates ‘x’ and ‘y’ are called
Cartesian Coordinates.
T
• The ordered pair (x, y) is different from the ordered pair (y, x).
Brain teaser
Look at the cards placed below you will find a puzzle The white card pieces must change
12345 12345 12345
places with the black pieces while
12345 12345 12345
12345 12345 12345 following these rules : (1) pieces of the
12345 12345 12345
12345 12345 12345
12345 12345 12345
12345
12345
12345
12345
12345
12345 same colour cannot jumpone another (2)
12345 12345 12345
12345 12345 12345
12345
12345
12345
12345
12345
12345 move one piece one space or jump at a
time. Find the least number of moves.
Minimum number of moves is 15. Can you do better?
To make the game more challanging, increase the number of pieces of cards
A
6.1 INTRODUCTION
AN
We have come across many problems like
(i) If five pens cost ` 60, then find the cost of one pen.
G
(ii) A number when added to 7 gives 51. Find that number.
Here, in situation (i) the cost of the pen is unknown, while in situation (ii) the number is
N
unknown. How do we solve questions of this type? We take letters x, y or z for the unknown
quantities and write an equation for these situations.
LA
For situation (i) we can write
5 × cost of a pen = 60
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Likewise we can make an equation for situation (ii) and find the unknown number. Such
type of equations are linear equations.
ER
44
-20 -10 0 10 20 30 40 50
A
these.
Kavya did not know the cost of the note book and the
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pen separately. Now can you express this information in
the form of an equation ?
Here, you can see that the cost of the single note
G
book and also of the pen is unknown, i.e. there are two
unknown quantities. Let us use x and y to denote them.
N
So, the cost of a note book is ` x and the cost of a pen
is ` y.
LA
We represent the above information as an
equation in the form 4 x + 2 y = 100,
Have you observed the exponents of x and y in the equation ?
TE
Thus the above equation is in linear form with variables ‘x’ and ‘y’.
If a linear equation has two variables then it is called a linear equation in two
variables.
T
s t
p + 3q = 50, 3u + 2v = 11, − = 5 and 3 = 5x − 7y are examples of linear equation
2 3
in two variables.
SC
Note that you can put the above equations in the form of p + 3q - 50 = 0,
3u + 2v − 11 = 0, 3S– 2t – 30 = 0 and 5x − 7y − 3 = 0 respectively.
Example 2. Hema’s age is 4 times the age of Mary. Write a linear equation in two variables to
represent this information.
A
Solution : Let Hema’s age be ‘x’ years and of Mary be ‘y’ years,
If Mary’s age is y then Hema’s age is ‘4y’.
AN
According to the given information we have x = 4y
⇒ x − 4y = 0 (how?)
G
Example 3. A number is 27 more than the number obtained by reversing its digits. If its unit’s
and ten’s digits are x and y respectively, write the linear equation representing the above statement.
N
Solution : Units digit is represented by x and tens digit by y, then the number is 10y + x
LA
If we reverse the digits then the new number would be 10x + y (Recall the place value
of digits in a two digit number).
Therefore according to the given condition the equation is
TE
Example 4. Express each of the following equations in the form of ax + by + c = 0 and write the
values of a, b and c.
i) 3x + 4y = 5 ii) x−5= 3y
SC
x y 1
iii) 3x = y iv) + =
2 2 6
v) 3x − 7 = 0
Solution : (i)3x + 4y = 5 can be written as
3x + 4y − 5 = 0.
1.x − 3y − 5 = 0.
A
(iii) The equation 3x = y can be written as
AN
3x − y + 0 = 0.
Here a = 3, b = −1 and c = 0.
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x y 1
(iv) The equation + = can be written as
2 2 6
x y 1
+ − =0 ;
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LA
2 2 6
1 1 −1
a= ,b= and c =
2 2 6
TE
3x + 0. y − 7 = 0.
a = 3, b = 0; c = −7
T
ER
Example-5. Write each of the following in the form of ax + by + c = 0 and find the values
of a, b and c
i) x = −5
ii) y=2
SC
iii) 2x = 3
iv) 5y = −3
Solution :
A
a b c
1 x = −5 1.x + 0.y + 5 = 0 1 0 5
AN
2 y=2 0.x + 1.y − 2 = 0 0 1 −2
3 2x = 3 --- --- --- ---
4 5y = −3 ---- ---- --- ---
N G
TRY THIS
LA
1. Express the following linear equations in the form of ax + by + c = 0 and indicate the
values of a, b, c in each case?
i) 3x + 2y = 9 ii) − 2x + 3y = 6 iii) 9x − 5y = 10
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x y
iv) − −5 =0 v) 2x = y
2 3
T
EXERCISE - 6.1
ER
1. Express the following linear equation in the form of ax+by+c=0 and indicate
the values of a, b and c in each case.
i) 8x + 5y − 3 = 0 ii) 28x − 35y = − 7 iii) 93x = 12 − 15y
SC
x y −3
iv) 2x = − 5y v) + =7 vi) y = x
3 4 2
vii) 3 x + 5 y = 12
2. Write each of the following in the form of
ax + by + c = 0 and find the values of a, b and c
y −14
i) 2x = 5 ii) y − 2 = 0 iii) =3 iv) x =
7 13
A
(iv) The cost of a pencil is ` 2 and a ball point pen is ` 15. Sheela pays ` 100
for the pencils and pens she purchased.
AN
(v) Yamini and Fatima of class IX together contributed ` 200/- towards the Prime
Minister’s Relief Fund.
(vi) The sum of a two digit number and the number obtained by reversing the order of its
G
digits is 121. If the digits in unit’s and ten’s place are ‘x’ and ‘y’ respectively.
6.3 S OLUTION
OLUTION OF A N
L INEAR E QUATION
QUA IN TW
TWOO VARIABLES
LA
You know that linear equation in one variable has a unique solution.
What can we say about the solution of this linear equation in two variables? Do we have
only one value in the solution or do we have more ? Let us explain.
T
We get 3 (3) − 2y = 5
9 − 2y = 5
SC
i.e., Still we cannot find the solution of the given equation. So, to know the solution,
besides the value of ‘x’ we also need the value of ‘y’. we can get value of y from the above
equation 9 − 2y = 5. ⇒ 2y = 4 or y = 2
The values of x and y which satisfy the equation 3x − 2y = 5, are x = 3 and y = 2. Thus
to statisfy, a linear equation in two variables we need two values, one value for ‘x’ and one value
for y.
Therefore any pair of values of ‘x’ and ‘y’ which satisfy the linear equation in two variables
is called its solution.
We observed that x = 3, y = 2 is a solution of 3x − 2y = 5. This solution is written as an
ordered pair (3, 2), first writing the value for ‘x’ and then the value for ‘y’. Are there any other
solutions for the equation? Pick a value of your choice say x = 4 and substitute it in the equation
A
3x - 2y = 5. Then the equation reduces to 12 − 2y = 5. Which is an equation in one variable.
AN
On solving this we get.
12 − 5 7
= , so 4, is another solution, of 3x − 2y = 5
7
y=
2 2 2
Do you find some more solutions for 3x − 2y = 5? Check if (1, −1) is another solution?
G
Thus for a linear equation in two variables we can find many solutions.
N
Note : An easy way of getting two solutions is put x = 0 and get the corresponding value
of ‘y’. Similarly we can put y = 0 and obtain the corresponding value of ‘x’.
LA
TRY THIS
Find 5 more pairs of values that are solutions for the above equation.
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Example 6. Find four different solutions of 4x + y = 9. (Complete the table wherever necessary)
Solution :
S.No. Choice of a Simplification Solution
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x or y
⇒ 4x = 9 9
, 0
⇒ x = 9/4 4
3. x=1 4x + y = 9 ⇒ 4 × 1+y=9
⇒ 4+y=9 ——
⇒ y=5
4. x= −1 ——— ( − 1, 13)
9
∴ (0, 9), , 0 , (1, 5) and ( − 1, 13) are some of the solutions for the above equation.
4
A
Solution : We know that if we get LHS = RHS when we substitute a pair in the given equation,
then it is a solution.
AN
The given equation is x + 2y = 4
G
LHS and Solution
RHS
1. (0, 2) N
x + 2y = 0 + (2 × 2) ∴LHS=RHS ∴(0, 2) is a
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=0+4=4 4 Solution
2. (2, 0) x + 2y = 2 + (2 × 0) ...... (0, 2) is a Not
=2+0=2 4 a Solution
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3. (4, 0) x + 2y = 4 + (2 × 0)
=4+0=4 4 LHS = RHS ___
4. ( 2, − 3 2) x + 2y = 2 + 2( −3 2) ( 2, − 3 2)
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= −5 2 ___ Solution
5. (1, 1) ___ 4 LHS ≠ RHS (1, 1) Not a
Solution
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6. ___ x + 2y = − 2 + (2 × 3) (−2, 3) is a
= −2+6=4 4 LHS = RHS Solution
A
⇒ 1=k
∴ k=1
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The resultant equation is 5x − 7y = 1.
Example-9. If x = 2k + 1 and y = k is a solutions of the equation 5x + 3y − 7 = 0, find the value
of k.
G
Solution : It is given that x = 2k + 1 and y = k is a solution of the equation 5x + 3y − 7 = 0
by substituting the value of x and y in the equation we get.
⇒ 5(2k + 1) + 3k – 7 = 0
⇒ 10k + 5 + 3k − 7 = 0
N
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⇒ 13k − 2 = 0 (this is the linear equation in one variable).
⇒ 13k = 2
2
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∴k=
13
EXERCISE - 6.2
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i) 3x + 4y = 7 ii) y = 6x iii) 2x − y = 7
iv) 13x − 12y = 25 v) 10x + 11y = 21 vi) x + y = 0
2. If (0, a) and (b, 0) are the solutions of the following linear equations. Find ‘a’ and ‘b’.
SC
i) 8x − y = 34 ii) 3x = 7y − 21 iii) 5x − 2y + 3 = 0
3. Check which of the following is solution of the equation 2x − 5y = 10
1
i) (0, 2) ii) (0, –2) iii) (5, 0) iv) (2 3, − 3) v) , 2
2
A
them?
AN
6.4 G RAPH OF A LINEAR EQUATION IN TW
EQUA TWOO VARIABLES
We have learnt that each linear equation in two variables has many solutions. If we take
possible solutions of a linear equation, can we represent them on the graph? We know each
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solution is a pair of real numbers that can be expressed as a point in the graph.
Consider the linear equation in two variables 4 = 2x + y. It can also be expressed as
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y = 4 − 2x. For this equation we can find the value of ‘y’ for a particular value of x. For
example if x = 2 then y = 0. Therefore (2, 0) is a solution. In this way we find as many solutions
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as we can. Write all these solutions in the following table by writing the value of ‘y’ against the
corresponding value of x.
Table of solutions: Scale :
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Y X-axis : 1 cm = 1 unit
x y = 4 – 2x (x, y) 6 Y-axis : 1 cm = 1 unit
0 y = 4 – 2(0) = 4 (0, 4) 5
2 y = 4 – 2(2) = 0 (2, 0) 4 A(0, 4)
1 y = 4 – 2(1) = 2 (1, 2) 3
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2 C(1, 2)
3 y = 4 – 2(3) = –2 (3, –2)
1
ER
Pick up any other point on this line AD and check if its coordinates satisfy the equation or
not?
Now take any point not on the line AD say (1, 1). Is it satisfy the equation?
Can you find any point that is not on the line AD but satisfies the equation?
A
Let us list our observations:
1. Every solution of the linear equation
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represents a point on the line of the
equation.
2. Every point on this line is a solution
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of the linear equation.
3. Any point that does not lie on this
line is not a solution of the equation
and vice a versa. N
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4. The collection of points that give the
solution of the linear equation is the
graph of the linear equation.
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Steps :
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4. Write the values of x and its corresponding value of y as coordinates of x and y respectively
as (x, y) form.
5. Plot the points on the graph paper.
6. Join these points.
Thus line drawn is the graph of linear equation in two variables. However to check the
correctness of the line it is better to take more than two points. To find more solutions take
different values for ‘x’ substitute them in the given equation and find the corresponding
values of ‘y’.
TRY THESE
Take a graph paper, plot the point (2, 4), and draw a line passing through it.
Now answer the following questions.
1. Can you draw another line that passes through the point (2, 4).
A
2. How many such lines can be drawn?
AN
3. How many linear equations in two variables exist for which (2, 4) is a solution?
Example-10. Draw the graph of the equation y − 2x = 4 and then answer the following.
(i) Does the point (2, 8) lie on the line? Is (2, 8) a solution of the equation? Check by
G
substituting (2, 8) in the equation.
N
(ii) Does the point (4, 2) lie on the line? Is (4, 2) a solution of the equation? Check algebraically
also.
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(iii) From the graph find three more solutions of the equation and also three more which are not
solutions.
Solution : Given y − 2x = 4 ⇒ y = 2x + 4
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Table of Solutions
x y = 2x + 4 (x, y) Point
0 y = 2(0) + 4 = 4 (0, 4) A(0, 4)
T
Plotting the points A, B and C on the graph paper and join them to get the straight line BC
SC
as shown in graph sheet. This line is the required graph of the equation y − 2x = 4.
(i) Plot the point (2, 8) on the graph paper. From the graph it is clear that the point (2, 8) lies
on the line.
Checking algebraically: On substituting (2, 8) in the given equation, we get
LHS = y − 2x = 8 − 2x2 = 8 − 4 = 4 = RHS, So (2, 8) is a solution
Scale :
Y
X-axis : 1 cm = 2 units
Y-axis : 1 cm = 2 units
8 (2, 8)
A
6 C(1, 6)
AN
4 A(0, 4)
2 (4, 2)
B(−2, 0)
X' X
−8 −6 −4 −2 0 2 4 6 8
−2
G
−4
−6
−8
N
LA
Y'
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(ii) Plot the point (4, 2) on the graph paper. You find that (4, 2) does not lie on the line.
Checking algebraically: By substituting (4, 2) in the given equation we have
LHS = y − 2x = 2 − 2 × 4 = 2 − 8 = − 6 ≠ RHS, so (4, 2) is not a solution.
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(iii) We know that every point on the line is a solution of the given equation. So, we can take
any three points on the line as solutions of the given equation. Eg:(-4, -4). And we also
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know that the point which is not on the line is not a solution of the given equation. So we
can take any three points which are not on the line as not solutions of y - 2x = 4.
eg : (i) (1, 5); ........; .........
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x −3
Solution : We have x − 2y = 3 ⇒ y =
2
Table of Solutions
x −3
x y= (x, y) Point
2
3−3
3 y= =0 (3, 0) A
2
A
1−3
1 y= = − 1 (1, − 1) B
2
AN
−1−3
−1 y= = − 2 ( − 1, − 2) C
2
Plotting the points A, B, C on the graph paper and on joining them we get a straight line as
G
shown in the following figure. This line is the required graph of the equation x − 2y = 3
N Y
Scale :
X-axis : 1 cm = 2 units
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Y-axis : 1 cm = 2 units
10
8
6
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4
2 A(3,0)
X'
−8 −6 −4 −2 0 2 4 6 8 10 X
P(-1,-2)
-3
D 0,
-2 2
T
-4
-6
ER
-8
Y'
(i) We have to find a solution (x, y) where x = − 5, that is we have to find a point which lies
on the straight line and whose x-coordinate is ‘ − 5’. To find such a point we draw a line
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parallel to y-axis at x = − 5. (in the graph it is shown as dotted line). This line meets the
graph at ‘P’ from there we draw another line parallel to X-axis meeting the Y-axis at
y = − 4.
The coordinates of P = ( − 5, − 4)
Since P( − 5, − 4) lies on the straight line x − 2y = 3, it is a solution of x − 2y = 3.
(ii) We have to find a solution (x, y) where y = 0.
Since y = 0, this point (x, 0) lies on the X-axis. Therefore we have to find a point that lies
on the X-axis and on the graph of x − 2y = 3.
A
−3
From the graph it is clear that 0, is this point.
AN
2
−3
Therefore, the solution is 0, .
2
G
EXERCISE - 6.3
i) 2y = −x + 1 ii) –x + y = 6
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1. Draw the graph of each of the following linear equations.
iii) 3x + 5y = 15 iv)
x y
− =3
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2 3
2. Draw the graph of each of the following linear equations and answer the following question.
i) y = x ii) y = 2x iii) y = −2x iv) y = 3x v) y = −3x
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i) Are all these equations of the form y = mx, where m is a real number?
ii) Are all these graphs passing through the origin?
iii) What can you conclude about these graphs?
T
3. Draw the graph of the equation 2x + 3y = 11. Find the value of y when x = 1 from the
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graph.
4. Draw the graph of the equation y − x = 2. Find from the graph
i) the value of y when x = 4
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7. Rajiya and Preethi two students of Class IX together collected ` 1000 for the Prime
Minister Relief Fund for victims of natural calamities. Write a linear equation and draw a
graph to depict the statement.
8. Gopaiah sowed wheat and paddy in two fields of total area 5000 square meters. Write
A
a linear equation and draw a graph to represent the same?
9. The force applied on a body of mass 6 kg. is directly proportional to the acceleration
AN
produced in the body. Write an equation to express this observation and draw the graph
of the equation.
10. A stone is falling from a mountain. The velocity of the stone is given by V = 9.8t. Draw its
G
graph and find the velocity of the stone ‘4’ seconds after start.
N
Example-12. 25% of the students in a school are girls and others are boys. Form an equation
and draw a graph for this. By observing the graph, answer the following :
LA
(i) Find the number of boys, if the
number of girls is 25.
x = 25% of (x + y)
25 1
= of (x + y) = (x + y)
100 4
1
x = (x + y)
4
4x = x + y
3x = y
A
The required equation is 3x = y or 3x - y = 0.
AN
Table of Solutions
x y = 3x (x, y) Point
10 30 (10, 30) A
20 60 (20, 60) B
G
30 90 (30, 90) C
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Plotting the points A, B and C on the graph and on joining them we get the straight line as
shown in the following figure.
LA
Y Scale :
X-axis : 1 cm = 20 units
Y-axis : 1 cm = 20 units
Boys
100
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C(30, 90)
80
60 B(20, 60)
40
A(10, 30)
T
20
X' Girls X
−80 −60−40−20 0 20 40 60 80 100
ER
−20
−40
−60
−80
SC
Y'
Example-13. For each graph given below, four linear equations are given. Out of these find
the equation that represents the given graph.
(i) Equations are
A) y = x Y
Scale :
10
A
B) x + y = 0 X-axis : 1 cm = 2 units
8
Y-axis : 1 cm = 2 units
C) y = 2x 6
AN
D) 2 + 3y = 7x 4
2
B(−1, 1)
X' X
−8 −6 −4 −2 0 2 4 6 8 10
−2 A(1, − 1)
G
−4
−6
C) y = −x + 2 10
8
D) x + 2y = 6
6
4
(−1, 3)
2 (0, 2)
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(2, 0)
X' X
−8 −6 −4 −2 0 2 4 6 8 10
ER
−2
−4
−6
−8
SC
Solution : Y'
(i) From the graph we see (1, −1) (0, 0) (−1, 1) lie on the same line. So these are the solutions
of the required equation i.e. if we substitute these points in the required equation it should
be satisfied. So, we have to find an equation that shoul be satisfied by these pairs. If we
substitute (1, −1) in the first equation y = x it is not satisfied. So y = x is not the required
equation.
Putting (1, −1) in x + y = 0 we find that it satisfies the equation. In fact all the three points
satisfy the second equation. So x + y = 0 is the required equation.
We now check whether y = 2x and 2 + 3y = 7x are also satisfied by (1, −1) (0, 0) and
(−1, 1). We find they are not satisfied by even one of the pairs, leave alone all three. So,
they are not the required equations.
(ii) The points on the line are (2, 0), (0, 2) and (−1, 3). All these points don’t satisfy the
first and second equation. Let us take the third equation y = −x + 2. If we substitute
A
the above three points in the equation, it is satisfied. So required equation is y = −x +
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2. Check whether these points satisfies the equation x + 2y = 6.
E XERCISE - 6.4
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1. In a election 60% of voters cast their
votes. Form an equation and draw the
N
graph for this data. Find the following from
the graph.
LA
(i) The total number of voters, if
1200 voters cast their votes
(ii) The number votes cast, if the
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[Hint: If the number of voters who cast their votes be ‘x’ and the total number of voters
be ‘y’ then x = 60% of y.]
T
2. When Rupa was born, her father was 25 years old. Form an equation and draw a graph
for this data. From the graph find
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Write the linear equation representing this information and draw the graph. With the help
of graph find the distance travelled if the fare paid is ` 55? How much would have to be
paid for 7 kilometers?
4. A lending library has fixed charge for the first three days and an additional charges for
each day thereafter. John paid ` 27 for a book kept for seven days. If the fixed charges be
` x and subsequent per day charges be ` y, then write the linear equation representing the
above information and draw the graph of the same. From the graph, find fixed charges
for the first three if additional charges for each day thereafter is `4. Find additional
charges for each day thereafter if the fixed charges for the first three days of `7.
5. The parking charges of a car in Hyderabad Railway station for first two hours is `
50 and `10 for each subsequent hour. Write down an equation and draw the graph.
Find the following charges from the graph
(i) For three hours (ii) For six hours
A
(iii) How many hours did Rekha park her car if she paid ` 80 as parking charges?
AN
6. Sameera was driving a car with uniform speed of 60 kmph. Draw distance-time
graph. From the graph find the distance travelled by Sameera in
1 1
(i) 1 hours (ii) 2 hours (iii) 3 hours
2 2
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7. The ratio of molecular weight of Hydrogen and Oxygen in water is 1:8. Set up an equation
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between Hydrogen and Oxygen and draw its graph. From the graph find the quantity of
3
Hydrogen if Oxygen is 12 grams. And quantity of oxygen if hydrogen is gms.?
LA
2
[Hint : If the quantities of hydrogen and oxygen are ‘x’ and ‘y’ respectively,
then x : y = 1:8 ⇒ 8x = y]
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8. In a mixture of 28 litres, the ratio of milk and water is 5:2. Set up the equation between
the mixture and milk. Draw its graph. By observing the graph find the quantity of milk in
the mixture.
countries like India, it is measured in Celsius. Here is a linear equation that converts
9
Fahrenheit to Celsius F = C + 32
5
SC
(i) Draw the graph of the above linear equation having Celsius on x-axis and Fahrenheit
on Y-axis.
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) Is there a temperature that has numerically the same value in both Fahrenheit and
Celsius? If yes find it?
6.5 EQUATION
QUA OF LINES PARALLEL TO X- AXIS AND Y- AXIS
Consider the equation x = 3. If this is treated as an equation in one variable x, then it
has the unique solution x = 3 which is a point on the number line
A
−3 −2 −1 0 1 2 3
AN
However when treated as an equation in two variables and plotted on the coordinate plane
it can be expressed as x + 0.y – 3 = 0
G
This has infinitely many solutions, let us find some of them. Here the coefficient of y is zero.
So for all values of y, x becomes 3.
N
Table of solutions
LA
x 3 3 3 3 3 3 …...
y 1 2 3 –1 –2 –3 …..
(x, y) (3, 1) (3, 2) (3, 3) (3, –1) (3, –2) (3, –3) …..
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Points A B C D E F …..
8
Now draw the graph using the 6
above solutions. What do you 4 C(3, 3)
X' −8 −6 −4 −2 0 2 4 6 8 10 X
Is it a straight line? Whether
−2
D(3, −1)
E(3, −2)
−4
is a straight line and is parallel to −6
Y-axis? −8
What is the distance of this Y'
line from the y-axis?
Thus the graph of x = 3 is a line parallel to the y-axis at a distance of 3 units to the right of it.
DO THIS
1. i) Draw the graph of following equations.
a) x=2 b) x = –2 c) x = 4 d) x = –4
ii) Are the graphs of all these equations parallel to Y-axis?
A
iii) Find the distance between the graph and the Y-axis in each case
AN
2. i) Draw the graph of the following equations
a) y=2 b) y = –2 c) y = 3 d) y = –3
ii) Are all these parallel to the X-axis?
G
iii) Find the distance between the graph of the line and the X-axis in each case
N
From the above observations we can conclude the following:
1. The graph of x = k is a line parallel to Y-axis at a distance of k units and passing through
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the point (k, 0)
2. The graph of y = k is a line parallel to X-axis at a distance of k units and passing through
the point (0, k)
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Table of solutions
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x 1 2 3 –1 –2 …...
y 0 0 0 0 0 …..
(x, y) (1, 0) (2, 0) (3, 0) (–1, 0) (–2, 0) …..
SC
Points A B C D E …..
By plotting all these points on the graph paper, we get the following figure. From the
graph what do we notice?
Scale :
Y 10 x-axis : 1 cm = 2 units
y-axis : 1 cm = 2 units
8
6
A
4
AN
(-2,0) 2
(-3,0)
(-1,0)
(1,0)
(3,0)
X' (2,0) X
−8 −6 −4 −2 0 2 4 6 8 10
−2
G
−4
−6
−8
N
LA
Y'
We notice that all these points lie on the X-axis and y-coordinate of all these points is ‘0’.
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Therefore the equation y = 0 represents X-axis. In other words the equation of the X-axis
is y = 0.
TRY THESE
T
EXERCISE - 6.5
1. Give the graphical representation of the following equation.
SC
A
5. Write the equation of the line parallel to Y-axis and passing through the point
AN
i) (–4, 0) ii) (2, 0) iii) (3, 5) iv) (–4, –3)
6. Write the equation of three lines that are
(i) parallel to the X-axis (ii) parallel to the Y-axis.
G
W HAT
HAT WE HA VE DISCUSSED
HAVE
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1. If a linear equation has two variables then it is called linear equation in two
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variables.
2. Any pair of values of ‘x’ and ‘y’ which satisfy the linear equation in two variables
is called its solution.
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8. Equation of X-axis is y = 0.
9. Equation of Y-axis is x = 0.
Triangles
07
A
7.1 I NTRODUCTION
AN
We have drawn figures with lines and curves and studied their properties. Do you
remember, how to draw a line segment of a given length? All line segments are not same in size,
they can be of different lengths. We also draw circles. What measure, do we need and have been
G
used to draw a circle? It is the radius of the circle. We also draw angles equal to the given angle.
A 3 cm. B N
We know if the lengths of two line segments are equal then they are congruent.
P
5 cm.
Q
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C D R 2 cm. S
3 cm.
AB ≅ CD PQ ≅ RS
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(Congruent) (Non-congruent)
P Y D
O B X
(Congruent) (Non-congruent)
From the above examples we can say that to make or check whether the figures are
same in size or not we need some specific information about the measures describing these
figures.
Let’s consider a square : What is the minimum information required to say whether
two squares are of the same size or not?
Satya said- “I only need the measure of the side of the given squares. If the sides of given
squares are equal then the squares are of identical size”.
Siri said “that is right but even if the diagonals of the two squares are equal then we can
say that the given squares are identical and are same in size”.
x
Do you think both of them are right?
Recall the properties of a square. You can’t make two different
x x
squares with sides having same measures. Can you? And the diagonals of 2x
A
two squares can only be equal when their sides are equal. See the given
x
AN
figure:
The figures that are same in shape and size are called congruent
figures (‘Congruent’ means equal in all aspects). Hence squares that have sides with same measure
are congruent and also with equal diagonals are congruent.
G
Note : In general, sides decide sizes and angles decide shapes.
We know if two squares are congruent and we trace one out of them on a paper and
N
place it on other one, it will cover the other exactly.
Then we can say that sides, angles, diagonals of one square are respectively equal to
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the sides, angles and diagonals of the other square.
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m
.
3.5
5 cm. E
m.
Q P I
5c
5c
3.5
m.
4 cm.
4 cm.
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m.
5c
.
m
cm
4c
5c
m.
.
T 5 cm. S
R
(ii) (iii)D (iv) (v)
G
If we trace these triangles from fig.(ii) to (v) and try to cover ∆ABC. We would observe
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that triangles in fig.(ii), (iii) and (iv) are congruent to ∆ABC while ∆TSU in fig.(v) is not congruent
to ∆ABC.
If ∆PQR is congruent to ∆ABC, we write ∆PQR ≅ ∆ABC.
Notice that when ∆PQR ≅ ∆ABC, then sides of ∆PQR covers the corresponding sides
of ∆ABC equally and so do the angles.
That is, PQ covers AB, QR covers BC and RP covers CA; ∠P covers ∠A,∠ Q covers
∠B and ∠R covers ∠C. Also, there is a one-one correspondence between the vertices. That is,
P corresponds to A, Q to B, R to C. This can be written as
P ↔ A, Q ↔ B, R ↔ C
Note that under order of correspondence, ∆PQR ≅ ∆ABC; but it will not be correct
to write ∆QRP ≅ ∆ABC as we get QR = AB, RP = BC and QP = AC which is incorrect for
the given figures.
Similarly, for fig. (iii),
A
FD ↔ AB, DE ↔ BC and EF ↔ CA
and F ↔ A, D ↔ B and E ↔ C
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So, ∆FDE ≅ ∆ABC but writing ∆DEF ≅ ∆ABC is not correct.
Now you give the correspondence between the triangle in fig.(iv) and ∆ABC.
So, it is necessary to write the correspondence of vertices correctly for writing of
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congruence of triangles.
Note that corresponding parts of congruent triangles are equal and we write in
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short as ‘CPCT’ for corresponding parts of congruent triangles.
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DO THIS
1. There are some statements given below. Write whether they are true or false :
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2. Which minimum measurements do you require to check if the given figures are congruent:
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You have learnt the criteria for congruency of triangle in your earlier class.
Is it necessary to know all the three sides and three angles of a triangle to make a unique
triangle?
Draw two triangles with one side 4 cm. Can you make two different triangles with one
side of 4 cm? Discuss with your friends. Do you all get
congruent triangles? You can draw types of triangles if one
side is given say 4 cm.
4 cm. 4 cm.
D
Now take two sides as 4 cm. and 5 cm. and draw as many
triangles as you can. Do you get congruent triangles?
C
5 cm.
We can make different triangles even with these two given .
m
measurements. 5c
A
Now draw triangles with sides 4 cm., 7 cm. and 8 cm. A B
4 cm.
Can you draw two different triangles?
AN
m. You find that with measurement of these three sides, we can make a
7c
4c
m.
unique triangle. If at all you draw the triangles with these dimensions
8 cm. they will be congruent to this unique triangle.
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Now take three angles of your choice, of course The sum of the angles must be 180°.
Draw two triangles for your choosen angle measurement.
Mahima finds that she can make
different triangles by using three angle N 70°
LA
70°
measurement.
∠A = 50°, ∠B = 70°, ∠C = 60° 50° 60° 50° 60°
Now try to draw two distinct triangles with each sets of these three measurements:
i. ∆ABC where AB = 5 cm., BC = 8 cm., ∠C = 30°
ii. ∆ABC where AB = 5 cm., BC = 8 cm., ∠B = 30°
SC
(i) Are you able to draw a unique triangle with the given measurements, draw and check
with your friends.
A A
5 cm.
5 cm.
A' A'
. .
5 cm 5 cm
30° 30° 30°
B 8 cm. C B 8 cm. C B 8 cm. C
A
In the other words you can draw a unique triangle with the measurements given in case(ii).
Have you noticed the order of measures given in case (i) & case (ii) ? In case (i) two
AN
sides and one angle are given which is not an included angle but in case (ii) included angle is given
along with two sides. Thus given two sides and one angle i.e. three independent measures is not
the only criteria to make a unique triangle. But the order of given measurements to construct a
triangle also plays a vital role in making a unique triangle.
G
7.3 C ONGRUENCE OF T RIANGLES
N
The above has an implication for checking the congruency of triangles. If we have two
triangles with one side equal or two triangles with all 3 angles equal, we can not conclude that
LA
triangles are congruent as there are more than one triangle possible with these specifications.
Even when we have two sides and an angle equal we cannot say that the triangles are congruent
unless the angle is between the given sides. We can say that the SAS (side angle side) congruency
rule holds but not SSA or ASS.
TE
We take this as the first criterion for congruency of triangles and prove the other criteria
through this.
Axiom (SAS congruence rule): Two triangles are congruent if two sides and the included
angle of one triangle are equal to the two sides and the included angle of the other triangle.
T
Example-1. In the given Figure AB and CD are intersecting at ‘O’, OA = OB and OD = OC.
Show that
ER
O
Also, since ∠ AOD and ∠ BOC form a pair of vertically opposite
angles, we have
∠ AOD = ∠ BOC. A D
So, ∆AOD ≅ ∆BOC (by the SAS congruence rule)
(ii) In congruent triangles AOD and BOC, the other corresponding parts are also equal.
So, ∠ OAD = ∠ OBC and these form a pair of alternate angles for line segments AD
and BC.
Therefore AD || BC
Example-2. AB is a line segment and line l is its perpendicular bisector. If a point P lies on l,
show that P is equidistant from A and B.
Solution : Line l ⊥ AB and passes through C which is the mid-point of AB
We have to show that PA = PB. l
Consider ∆PCA and ∆PCB. P
A
We have AC = BC (C is the mid-point of AB)
AN
∠PCA = ∠PCB = 90° (Given)
A B
PC = PC (Common) C
G
and so, PA = PB, as they are corresponding sides of congruent triangles.
DO THESE N
LA
1. State whether the following triangles are congruent or not? Give reasons for your answer.
R
F E
A 70° L
TE
.
m
4c
4c
60°
m
.
70° 80°
50° D S T
B C 80° 3 cm.
T
M 3 cm. N
ER
(i) (ii)
D
2. In the given figure, the point P bisects AB and DC. Prove
that A
P B
SC
∆APC ≅ ∆BPD
A
congruence.
AN
Theorem 7.1 (ASA congruence rule) : Two triangles are congruent, if two angles and the
included side of one triangle are equal to two angles and the included side of the other triangle.
Given: In ∆ABC and ∆DEF
G
∠ B = ∠ E, ∠ C = ∠ F and BC = EF
Required To Prove (RTP): ∆ABC ≅ ∆ DEF
N
Proof: There will be three posibilities. The possiblities between AB and DE are either
LA
AB > DE or DE > AB or DE = AB .
We will consider all these cases and see what does it mean for ∆ABC and ∆DEF.
TE
AB = DE (Assumed)
T
∠B = ∠E (Given)
B C E F
ER
BC = EF (Given)
So, ∆ABC ≅ ∆DEF (By SAS congruency axiom)
Case (ii): The second possibility is AB > DE.
SC
A D
So, we can take a point P on AB such that PB = DE.
Now consider ∆PBC and ∆DEF P
PB = DE (by construction)
B C E F
∠B= ∠E (given)
BC = EF (given)
So, ∆PBC ≅ ∆DEF (by SAS congruency axiom)
Since the triangles are congruent their corresponding parts will be equal
So, ∠ PCB = ∠ DFE
But, ∠ ACB = ∠ DFE (given)
So ∠ ACB = ∠ PCB (from the above)
A
Is this possible?
This is possible only if P coincides with A
AN
(or) BA = ED
So, ∆ABC ≅ ∆DEF (By SAS congruency axiom)
G
AB = DE and the two triangles are congruency by SAS rule).
N
Case (iii): The third possibility is AB < DE
A D
LA
M
We can choose a point M on DE such that ME =
AB and repeating the arguments as given in case (ii), we
B
can conclude that AB = DE and so, ∆ABC ≅ ∆DEF. C E F
TE
You know that the sum of the three angles of a triangle is 180°. So if two pairs of angles
are equal, the third pair is also equal (180° − sum of equal angles).
ER
So, two triangles are congruent if any two pairs of angles and one pair of
corresponding sides are equal. We may call it as the AAS Congruence Rule. Let us now take
some more examples.
SC
Example-4. In the given figure, AL || DC, E is mid point of BC. Show that ∆EBL ≅ ∆ECD
Solution : Consider ∆EBL and ∆ECD
D C
∠ BEL = ∠ CED (vertically opposite angles)
E
BE = CE (since E is mid point of BC)
A
∠ EBL = ∠ ECD (alternate interior angles) A
B L
AN
Example-5. Use the information given in the D E
adjoining figure, to prove :
∆DBC ≅ ∆EAC
G
(i)
x x
(ii) DC = EC. A B
C
Solution : Let ∠ ACD = ∠ BCE = x
N
∴ ∠ ACE = ∠ DCE + ∠ ACD = ∠ DCE + x ...... (i)
LA
∴ ∠ BCD = ∠ DCE + ∠ BCE = ∠ DCE + x ...... (ii) E
BC = AC [Given]
D
∠ CBD = ∠ CAE [Given]
T
EXERCISE - 7.1
A
C
1. In quadrilateral ACBD, AC = AD and AB bisects ∠ A
AN
Show that ∆ABC ≅ ∆ABD.
A B
What can you say about BC and BD?
G
D
2. ABCD is a quadrilateral in which AD = BC and A D
P q
4. l and m are two parallel lines intersected by l
A D
another pair of parallel lines p and q . Show
SC
A
(ii) M
(iii) ∆DBC ≅ ∆ACB
AN
B C
1
(iv) CM = AB
2 D C
7. In the adjacent figure ABCD is a square and ∆APB is an
P
equilateral triangle. Prove that ∆APD ≅ ∆BPC.
G
(Hint : In ∆APD and ∆BPC AD = BC, AP = BP and
∠ PAD = ∠ PBC = 90o - 60o = 30o]
N
In the adjacent figure ∆ABC is isosceles as AB = AC, BA
A
P Q
B
LA
8.
and CA are produced to Q and P such that AQ = AP . Show A
that PB = QC
(Hint : Compare ∆APB and ∆AQC)
TE
A
B C
B C
D
ER
10. If the bisector of an angle of a triangle also bisects the opposite side, prove that the
triangle is isosceles. A
11. In the given figure ABC is a right triangle and right angled at
B such that ∠BCA = 2∠BAC.
SC
In the above section you have studied two criteria for the congruence of triangles. Let us
now apply these results to study some properties related to a triangle whose two sides are equal.
ACTIVITY
i. To contruct a triangle using compass, take any measurement and draw a line segment AB.
Now open a compass with sufficient length and put it on point A and B and draw an arc.
Which type of triangle do you get? Yes this is an isosceles triangle. So, ∆ABC in figure is an
A
isosceles triangle with AC = BC. Now measure ∠A and ∠B. What do you observe?
C
AN
G
A B A B
as shown below.
B C
Given: ∆ABC is an isosceles triangle in which AB = AC. D
RTP: ∠ B = ∠ C.
Construction: Let us draw the bisector of ∠ A and let D be the point of intersection of this
SC
Is the converse also true? That is “If two angles of any triangle are equal, can we conclude
that the sides opposite to them are also equal?”
ACTIVITY
1. On a tracing paper draw a line segment BC of length
A
6cm.
AN
2. From vertices B and C draw rays with angle 600 each.
Name the point A where they meet.
3. Fold the paper so that B and C fit precisely on top of B C
each other. What do you observe? Is AB = AC?
G
Repeat this activity by taking different angles for ∠B and
2
LA
Theorem-7.3 : The sides opposite to equal angles of a triangle are
equal. B C
TE
In ∆ACD, AC = CD
⇒ ∠ CAD = ∠ CDA = x
and, ext. ∠ ACB = ∠ CAD + ∠ CDA
= x + x = 2x
A
⇒ ∠ BAC = ∠ ACB = 2x. (∵ In ABC, AB = BC)
∴ ∠ BAD = ∠ BAC + ∠ CAD
AN
= 2x + x = 3x
∠BAD 3 x 3
And, = =
∠ADB x 1
i.e., ∠ BAD : ∠ ADB = 3 : 1.
G
Hence Proved.
Example-9. In the given figure, AD is perpendicular to BC and EF || BC, if ∠ EAB = ∠ FAC,
N
show that triangles ABD and ACD are congruent.
LA
Also, find the values of x and y if AB = 2x + 3, AC = 3y + 1,
BD = x and DC = y + 1. E A F
Solution : AD is perpendicular to EF
TE
⇒ ∠ BAD = ∠ CAD
T
Hence proved.
∠ ABD = ∠ ACD ⇒ AB = AC and BD = CD [By C.P.C.T]
⇒ 2x + 3 = 3y + 1 and x=y+1
⇒ 2x − 3y = −2 and x−y=1
Substituting 2(1+ y) − 3y = −2 Substituting y = 4 in x = 1 + y
x=1+y 2 + 2y − 3y = −2 x=1+4
−y = − 2 − 2 x=5
−y = −4
Example-10. E and F are respectively the mid-points of equal sides AB and AC of ∆ABC
(see figure)
Show that BF = CE
A
Solution : In ∆ABF and ∆ACE,
A
AB = AC (Given)
E F
∠A= ∠A (common angle)
AN
AF = AE (Halves of equal sides)
B C
So, ∆ABF ≅ ∆ACE (SAS rule)
Therefore, BF = CE (CPCT)
G
Example-11. In an isosceles triangle ABC with AB = AC, D and E are points on BC such that
BE = CD (see figure) Show that AD = AE,
Solution : In ∆ABD and ∆ACE,
N
LA
AB = AC (given) ........... (1)
∠ B = ∠ C (Angles opposite to equal sides) ........... (2)
Also, BE = CD
TE
So, BE − DE = CD − DE
A
That is, BD = CE (3)
So, ∆ABD ≅ ∆ACE
T
EXERCISE - 7.2
A
1. In an isosceles triangle ABC, with AB = AC, the bisectors
SC
B D C
A
3. ABC is an isosceles triangle in which altitudes BD and CE
are drawn to equal sides AC and AB respectively (see figure)
Show that these altitudes are equal.
E D
A
B C
A
4. ABC is a triangle in which altitudes BD and CE to sides AC
AN
and AB are equal (see figure) . Show that
E D
(i) ∆ABD ≅ ∆ACE
(ii) AB = AC i.e., ABC
G
B C A
is an isosceles triangle.
5. N
∆ABC and ∆DBC are two isosceles triangles on the
same base BC (see figure). Show that ∠ ABD = B C
LA
∠ ACD.
D
TE
Theorem 7.4 (SSS congruence rule) : Through construction we have seen that SSS congruency
rule hold. This theorem can be proved using a suitable construction.
T
In two triangles, if the three sides of one triangle are respectively equal to the corresponding three
ER
⇒ ∠P = ∠W and PR = WZ (CPCT) X
Similarly, XZ = WZ
In ∆XYW, XY = YW
A
W
⇒ ∠YWX = ∠YXW (In a triangle, equal sides have equal angles opposite to them)
AN
Similarly, ∠ZWX = ∠ZXW
∴ ∠YWX + ∠ZWX = ∠YXW + ∠ZXW
⇒ ∠W = ∠X
G
Now, ∠W = ∠P
∴ ∠P = ∠X
In ∆PQR and ∆XYZ
PQ = XY
N
LA
∠P = ∠X
PR = XZ
∴ ∆PQR ≅ ∆XYZ (SAS congruence criterion)
TE
D C
AB = CD (given)
ER
AD = BC (given)
AC = CA (common side)
∆ABC ≅ ∆CDA (by SSS congruency rule) A B
SC
A
DO THIS
1. In the adjacent figure ∆ABC and ∆DBC are two B C
triangles such that AB = BD and AC = CD .
Show that ∆ABC ≅ ∆DBC.
D
You have already seen that in the SAS congruency axiom, the pair of equal angles has to
be the included angle between the pairs of corresponding equal sides and if not so, two triangles
may not be congruent.
ACTIVITY
Construct a right angled triangle with hypotenuse 5 cm. and one side 3 cm.
long. How many different triangles can be constructed? Compare your triangle with
those of the other members of your class. Are the triangles congruent? Cut them out
A
and place one triangle over the other with equal side placed on each other. Turn the
triangle if necessary what do you observe? You will find that two right triangles are
AN
congruent, if side and hypotenuse of one triangle are respectively equal to the
correseponding side and hypotenuse of other triangle.
Note that the right angle is not the included angle in this case. So we arrive at the following
G
congruency rule.
Theorem 7.5 (RHS congruence rule) : If in two right triangles the hypotenuse and one side of
N
one triangle are equal to the hypotenuse and one side of the another triangle, then the two triangles
are congruent.
LA
Note that RHS stands for right angle - hypotenuse-side.
Let us prove it.
TE
Construction: Produce DE to G B C
So that EG = AB. Join GF.
Proof: G
SC
AC = GF … (2) (CPCT)
Further, AC=GF and AC=DF (From (2) and Given)
Therefore DF = GF (From the above)
So, ∠ D = ∠ G … (3) (Angles opposite to equal sides are equal)
we get ∠ A = ∠ D … (4) (From (1) and (3))
A
Thus, in ∆ABC and ∆DEF ∠ A = ∠ D, (From (4))
∠B= ∠E (Given)
AN
So, ∠ A+ ∠ B = ∠ D + ∠ E (on adding)
But ∠ A + ∠ B + ∠ C = 1800 and (angle sum property of triangle)
∠ D + ∠ E + ∠ F = 1800
G
180 - ∠ C = 180 - ∠ F ( ∠ A+ ∠ B=1800− ∠ Cand ∠ D + ∠ E=1800− ∠ F)
So, ∠ C = ∠ F, … (5) (Cancellation laws)
Now, in ∆ABC and ∆DEF, we have
BC = EF
N (given)
LA
∠C = ∠F (from (5))
AC = DF (given)
∆ABC ≅ ∆DEF (by SAS axiom of congruence)
TE
Example-13. AB is a line - segment. P and Q are points on either side of AB such that each of
them is equidistant from the points A and B (See Fig ). Show that the line PQ is the perpendicular
bisector of AB.
T
Solution : You are given PA = PB and QA = QB and you have to show that PQ is perpendicular
on AB and PQ bisects AB. Let PQ intersect AB at C. P
ER
In these triangles,
SC
AP = BP (Given) Q
AQ = BQ (Given)
PQ = PQ (Common side)
So, ∆PAQ ≅ ∆PBQ (SSS rule)
Therefore, ∠ APQ = ∠ BPQ (CPCT).
Now let us consider ∆PAC and ∆PBC.
You have : AP = BP (Given)
A
and ∠ ACP = ∠ BCP (CPCT)
Also, ∠ ACP + ∠ BCP = 180° (Linear pair)
AN
So, 2 ∠ ACP = 180°
or, ∠ ACP = 90° ........... (2)
From (1) and (2), you can easily conclude that PQ is the perpendicular bisector of AB.
G
[Note that, without showing the congruence of ∆PAQ and ∆PBQ, you cannot show that
∆PAC ≅ ∆PBC even though AP = BP (Given)
PC = PC N (Common side)
LA
and ∠ PAC = ∠ PBC (Angles opposite to equal sides in ∆APB)
It is because these results give us SSA rule which is not always valid or true for congruence of
triangles as the given angle is not included between the equal pairs of sides.]
TE
Example-14. P is a point equidistant from two lines l and m intersecting at point A (see figure).
Show that the line AP bisects the angle between them.
T
Solution : You are given that lines l and m intersect each other at A.
ER
l
PB = PC (Given) B
PA = PA (Common side) m
EXERCISE - 7.3
1. AD is an altitude of an isosceles triangle ABC in which AB = AC.
Show that, (i) AD bisects BC (ii) AD bisects ∠ A.
2. Two sides AB, BC and median
A
A P
AM of one triangle ABC are
respectively equal to sides PQ and
AN
QR and median PN of ∆PQR (See
figure). Show that:
(i) ∆ABM ≅ ∆PQN B
M
C Q
N
R
G
(ii) ∆ABC ≅ ∆PQR
3. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule,
N
prove that the triangle ABC is isosceles.
∆ABC is an isosceles triangle in which AB = AC. Show that ∠ B = ∠ C.
LA
4.
(Hint : Draw AP ⊥ BC) (Using RHS congruence rule)
D
TE
6. ABC is a right angled triangle in which ∠ A = 900 and AB = AC. Show that
∠ B = ∠ C.
ER
ALITIES
NEQUALITIES IN A
So far, you have been studying the equality of sides and angles of a triangle or triangles.
Sometimes, we do come across unequal figures and we need to compare them. For example,
line segment AB is greater in length as compared to line segment CD in figure (i) and ∠ A is
greater than ∠ B in following figure (ii).
C D
A B
A
B
AN
(i) (ii)
Let us now examine whether there is any relation between unequal sides and unequal
angles of a triangle. For this, let us perform the following activity:
G
ACTIVITY
1. N
Draw a triangle ABC mark a point A′ on CA produced (new position of it)
LA
So, A′ C > AC (Comparing the lengths) A"
A'
Join A′ to B and complete the triangle A′ BC.
A
What can you say about ∠ A′ BC and ∠ ABC?
TE
You will observe that as the length of the side AC is increases (by taking different
positions of A), the angle opposite to it, that is, ∠ B also increases.
Let us now perform another activity-
SC
In ∆ABC Figure, BC Z R U
is the longest side and AC is
the shortest side. Q
Also, ∠ A is the largest
Y S T
and ∠ B is the smallest.
A
P
Measure angles and sides of each of the above triangles, what is the relation between a
AN
side and its opposite angle when campared with another pair?
Theorem-7.6 : If two sides of a triangle are C
unequal, the angle opposite to the longer side is
larger (or greater).
G
You may prove this theorem by taking a P
•
point P on BC such that CA = CP as shown in
adjacent figure.
Now, let us do another activity: N A B
LA
ACTIVITY
Measure angles and sides of each triangle given below. What relation you can visualize
for a side and its opposite angle in each triangle.
P
A Y Z
A
AN
B C R X
Q
In this way, we arrive at the following theorem.
Theorem -7.7 : In any triangle, the side opposite to the larger (greater) angle is longer.
G
This theorem can be proved by the method of contradiction.
DO THIS
N
LA
Now draw a triangle ABC and measure its sides. Find the sum of the sides AB + BC,
BC + AC and AC + AB, compare it with the length of the third side. What do you observe?
You will observe that AB + BC > AC,
TE
A
or, AB > AD (AD = AC)
AN
EXERCISE - 7.4
1. Show that in a right angled triangle, the hypotenuse is the longest side.
2. In adjacent figure, sides AB and AC of ∆ABC are A
G
extended to points P and Q respectively.
Also, ∠ PBC < ∠ QCB. Show that AC > AB.
N P
B C
Q
LA
B
D
O 3. In adjacent figure, ∠ B < ∠ A and ∠ C < ∠ D.
A Show that AD < BC.
TE
C
D
4. AB and CD are respectively the smallest and longest sides of a
quadrilateral ABCD (see adjacent figure). A
T
B C
P
6. If two sides of a triangle measure 4cm and 6cm find all possible measurements (positive
Integers) of the third side. How many distinct triangles can be obtained?
7. Try to construct a triangle with 5cm, 8cm and 1cm. Is it possible or not? Why? Give your
justification?
WHAT
HAT WE HAVE DISCUSSED
HAVE
• Figures which are identical i.e. having same shape and size are called congruent figures.
A
• Two triangles are congruent if the sides of one triangle are equal to the sides of another
AN
triangle and the corresponding angles in the two triangles are equal.
G
• In Congruent triangles corresponding parts are equal and we write in short ‘CPCT’ for
corresponding parts of congruent triangles.
• N
SAS congruence rule: Two triangles are congruent if two sides and the included angle of
one triangle are equal to the corresponding two sides and the included angle of the other
LA
triangle.
• ASA congruence rule: Two triangles are congruent if two angles and the included side of
TE
one triangle are equal to two angles and the included side of other triangle.
• SSS congruence rule: If three sides of one triangle are equal to the three sides of another
ER
• RHS congruence rule: If in two right triangles the hypotenuse and one side of one triangle
are equal to the hypotenuse and corresponding side of the other triangle, then the two
SC
• If two sides of a triangle are unequal, the angle opposite to the longer side is larger.
• The sum of any two sides of a triangle is greater than the third side.
Quadrilaterals
08
A
8.1 INTRODUCTION
AN
You have learnt many properties of triangles in the previous chapter with justification. You
know that a triangle is a figure obtained by joining three non-collinear points in pairs. Do you
know which figure you obtain with four points in a plane ? Note that if all the points are collinear,
G
we obtain a line segment (Fig. (i)), if three out of four points are collinear, we get a triangle
(Fig(ii)) and if any three points are not collinear, we obtain a closed figure with four sides (Fig (iii),
(iv)), we call such a figure as a quadrilateral.
A
N A
LA
A
B D
B
D
TE
A B C D B
C D C
(i) (ii) (iii) C (iv)
T
You can easily draw many more quadrilaterals and identify many around you. The
Quadrilateral formed in Fig (iii) and (iv) are different in one important aspect. How are they
ER
different?
In this chapter we will study quadrilaterals only of type
D C
(Fig (iii)). These are convex quadrilaterals.
SC
There are four angles in the interior of a quadrilateral. Can we find the sum of these four
angles? Let us recall the angle sum property of a triangle. We can use this property in finding sum
of four interior angles of a quadrilateral.
A
ABCD is a quadrilateral and AC is a diagonal (see figure).
We know the sum of the three angles of ∆ABC is,
AN
∠CAB + ∠B + ∠BCA = 180 o ...(1) (Angle sum property of a triangle)
Similarly, in ∆ADC, C
D
∠CAD + ∠D + ∠DCA = 180 o ...(2)
G
Adding (1) and (2), we get A B
N
∠CAB + ∠B + ∠BCA + ∠CAD + ∠D + ∠DCA = 180o + 180o
Look at the quadrilaterals drawn below. We have come across most of them earlier. We
will quickly consider these and recall their specific names based on their properties.
T
ER
D C D C
E F
B
A
SC
(i) A B H G
(ii) (iii)
A A
B B
B O D
D C D (vi)
(iv) (v) C
C
We observe that
l In fig. (i) the quadrilateral ABCD had one pair of opposite sides AB and DC parallel
to each other. Such a quadrilateral is called a trapezium.
If in a trapezium non parallel sides are equal, then the trapezium is an isosceles trapezium.
l In fig. (ii) both pairs of opposite sides of the quadrilateral are parallel such a
A
quadrilateral is called a parallelogram. Fig.(iii), (iv) and (v) are also parallelograms.
l In fig. (iii) parallelogram EFGH has all its angles as right angles. It is a rectangle.
AN
l In fig. (iv) parallelogram has its adjacent sides equal and is called a Rhombus.
l In fig. (v) parallelogram has its adjacent sides equal and angles of 90° this is called
a square.
G
l The quadrilateral ABCD in fig.(vi) has the two pairs of adjacent sides equal, i.e.
AB = AD and BC = CD. It is called a kite.
Consider what Nisha says:
N
A rhombus can be a square but all squares are not rhombuses.
LA
Lalita Adds
All rectangles are parallelograms but all parallelograms are not rectangles.
Which of these statements you agree with?
TE
Give reasons for your answer. Write other such statements about different types of
quadrilaterals.
Illustrative examples
T
So in a parallelogram ABCD
∠C = ∠A = 60o and ∠B = ∠D
and the sum of consecutive angles of parallelogram is equal to 180o.
SC
Example-2. In a parallelogram ABCD, ∠DAB = 40o find the other angles of the parallelogram.
Solution :
D C TRY THIS
D C
40°
A
A
B
40°
AN
ABCD is a parallelogram A E
B
∠DAB = ∠BCD = 40° and AD || BC Extend AB to E. Find ∠CBE. What do
As sum of consecutive angles you notice. What kind of angles are
∠CBA + ∠DAB = 180°
G
∠ABC and ∠CBE ?
∴ ∠CBA = 180 − 40°
= 140°
N
From this we can find ∠ADC = 140° and ∠BCD = 40°
Example-3. Two adjacent sides of a parallelogram are 4.5 cm and 3 cm. Find its perimeter.
LA
Solution : Since the opposite sides of a parallelogram are equal the other two sides are 4.5 cm
and 3 cm.
Hence, the perimeter = 4.5 + 3 + 4.5 + 3 = 15 cm.
Example-4. In a parallelogram ABCD, the bisectors of the consecutive angles ∠A and ∠B
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D C
1 1 180 P
∠ A + ∠B =
2 2 2
⇒ ∠PAB + ∠PBA = 90 o
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A B
In ∆ APB,
∠PAB + ∠APB + ∠PBA = 180o (angle sum property of triangle)
E XERCISE - 8.1
1. State whether the statements are True or False.
(i) Every parallelogram is a trapezium ( )
(ii) All parallelograms are quadrilaterals ( )
A
(iii) All trapeziums are parallelograms ( )
(iv) A square is a rhombus ( )
AN
(v) Every rhombus is a square ( )
(vi) All parallelograms are rectangles ( )
2. Complete the following table by writing (YES) if the property holds for the particular
Quadrilateral and (NO) if property does not holds.
G
Properties Trapezium Parallelogram Rhombus Rectangle square
a. Only one pair of opposite YES
b.
sides are parallel
Two pairs of opposite
N
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sides are parallel
c. Opposite sides are
equal
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d. Opposite angles
are equal
e. Consecutive angles
are supplementary
T
f. Diagonals
bisect each other
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right angle
j. Diagonals are per-
pendicular to each
other.
3. ABCD is trapezium in which AB || CD. If AD = BC, show that ∠A = ∠B and
∠C = ∠D .
4. The four angles of a quadrilateral are in the ratio 1: 2:3:4. Find the measure of each angle
of the quadrilateral.
5. ABCD is a rectangle AC is diagonal. Find the nature of ∆ACD. Give reasons.
ACTIVITY
A
Cut-out a parallelogram from a sheet of paper again and cut along one of its
AN
diagonal. What kind of shapes you obtain? What can you say about these triangles?
Place one triangle over the other. Can you place each side over the other exactly.
You may need to turn the triangle around to match sides. Since, the two traingles
G
match exactly they are congruent to each other.
Do this with some more parallelograms. You can select any diagonal to cut along.
N
We see that each diagonal divides the parallelogram into two congruent triangles.
Let us now prove this result.
LA
Theorem-8.1 : A diagonal of a parallelogram divides it into two congruent triangles.
Proof: Consider the parallelogram ABCD.
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D C
Join A and C. AC is a diagonal of the parallelogram.
This means that the two traingles are congruent by A.S.A. rule (angle, side and angle).
This means that diagonal AC divides the parallelogram in two congruent traingles.
A
also AD = BC and ∠DAC = ∠ACB
AN
A B
∠CAB = ∠DCA
∴ ∠ACB + ∠DCA = ∠DAC + ∠CAB
i.e. ∠DCB = ∠DAB
G
We thus have in a parallelogram
i. The opposite sides are equal.
ii. The opposite angles are equal. N
LA
It can be noted that with opposite sides of a convex quadrilateral being parallel we can
show the opposite sides and opposite angles are equal.
We will now try to show if we can prove the converse i.e. if the opposite sides of a
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A D
Consider ∆ABC and ∆CDA
ER
or AB || DC ...(1)
Since ∠ACD = ∠CAB with CA as transversal
We have BC || AD ...(2)
Therefore, ABCD is a parallelogram. By (1) and (2)
You have just seen that in a parallelogram both pairs of opposite sides are equal and
conversely if both pairs of opposite sides of a quadrilateral are equal, then it is a parallelogram.
Can we show the same for a quadrilateral for which the pairs of opposite angles are equal?
Theorem-8.4 : In a quadrilateral, if each pair of opposite angles are equal then it is a parallelogram.
Proof: In a quadrilateral ABCD, ∠A = ∠C and ∠B = ∠D then prove that ABCD is a
parallelogram.
We know ∠A + ∠B + ∠C + ∠D = 360°
D C
360° E
A
∠A +∠B = ∠C + ∠D =
2
AN
i.e. ∠A + ∠B = 180°
A B
Extend DC to E
∠C + ∠BCE = 180° hence ∠BCE = ∠ADC
G
If ∠BCE = ∠D then AD || BC (Why?)
With DC as a transversal
N
We can similarly show AB || DC or ABCD is a parallelogram.
LA
EXERCISE - 8.2
1. In the adjacent figure ABCD is a parallelogram F D E C a n d
ABEF is a rectangle show that ∆AFD ≅ ∆BEC.
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1
Prove that ∠COD = (∠A + ∠B)
ER
8.5 D IA GON
IAGON AL
GONAL
ALSS OF A P ARALLELOGRAM
ARALLELOGRAM
A
Hence we have to check if the converse is also true. Converse is if diagonals of a
quadrilateral bisect each other then it is a parallelogram.
AN
Theorem-8.6 : If the diagonals of a quadrilateral bisect each
D C
other then it is a parallelogram.
Proof: ABCD is a quadrilateral. O
AC and BD are the diagonals intersect at ‘O’,
G
such that OA = OC and OB = OD. A B
Prove that ABCD is a parallelogram.
N
(Hint : Consider ∆AOB and ∆COD. Are these congruent? If so then what can we say?)
LA
8.5.1 More geometrical statements
In the previous examples we have showed that starting from some generalisation we can
find many statements that we can make about a particular figure(Parallelogram). We use previous
results to deduce new statements. Note that these statements need not be verified by measurements
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as ∠A = 90o (given)
∴ ∠B = 180o − ∠A
= 180o − 90o = 90o
Now ∠C = ∠A and ∠D = ∠B (opposite angles of parallelogram)
So ∠C = 90o and ∠D = 90o .
Therefore each angle of a rectangle is a right angle.
Corollary-2 : Show that the diagonals of a rhombus are perpendicular to each other.
Proof : A rhombus is a parallelogram with all sides equal.
ABCD is a rhombus, diagonals AC and BD intersect at O C
A
OA = OC (Diagonals of a parallelogram bisect each other) D O B
AN
OB = OB (common side to ∆AOB and ∆BOC)
AB = BC (sides of rhombus)
A
Therefore ∆AOB ≅ ∆BOC (S.S.S rule)
G
So ∠AOB = ∠BOC
Corollary-3 : In a parallelogram ABCD, if the diagonal AC bisects the angle A, then ABCD is a
rhombus.
ER
A
Hence, ABCD is a rhombus.
AN
Corollary-4 : Show that the diagonals of a rectangle are of equal length.
Proof : ABCD is a rectangle and AC and BD are its diagonals
We want to know AC = BD
G
ABCD is a rectangle, means ABCD is a parallelogram with all its angles equal to right
angle. D C
N
Consider the triangles ∆ ABC and ∆BAD,
AB = BA (Common)
LA
∠B = ∠A = 90o (Each angle of rectangle) A B
BC = AD (opposite sides of the rectangle)
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A B
∠A + ∠B =180o(Consecutive angles of Parallelogram)
1 1
We know ∠BAP = ∠A and ∠ABP = ∠B [Since AP and BP are the bisectors
2 2
of ∠A and ∠B respectively]
1 1 1
⇒ ∠A + ∠B = × 180o
2 2 2
But In ∆APB,
A
= 90o
AN
We can see that ∠SPQ = ∠APB = 90°
Similarly, we can show that ∠CRD = ∠QRS = 90° (Same angle)
But ∠BQC = ∠PQR and ∠DSA = ∠PSR (Why?)
G
∴ ∠PQR = ∠QRS = ∠PSR = ∠SPQ = 90°
Hence PQRS has all the four angles equal to 90°.
N
We can therefore say PQRS is a rectangle.
LA
THINK, DISCUSS AND WRITE
1. Show that the diagonals of a square are equal and right bisectors of each other.
2. Show that the diagonals of a rhombus divide it into four congruent triangles.
TE
Let PQ, RQ, RS and PS are the bisectors of ∠RPB, ∠CRP , ∠DRP and ∠APR
respectively.
∠BPR = ∠DRP (Interior Alternate angles) ...(1)
1
But ∠RPQ = ∠BPR (∵ PQ is the bisector of ∠BPR )
2
...(2)
1
also ∠PRS = ∠DRP (∵ RS is the bisector of ∠DPR ).
2
∴ PQ || RS
A
Similarly
∠PRQ = ∠RPS, hence PS || RQ
AN
Therefore PQRS is a parallelogram ... (3)
We have ∠BPR + ∠CRP = 180 (interior angles on the same side of
o
G
1 1
∠BPR + ∠CRP = 90o
2 2
⇒ ∠RPQ + ∠PRQ = 90o
But in ∆ PQR, N
LA
∠RPQ + ∠PQR + ∠PRQ = 180o (three angles of a triangle)
∠PQR = 180o − ( ∠RPQ + ∠PRQ)
= 180o − 90o = 90o ... (4)
TE
Example-6. In a triangle ABC, AD is the median drawn on the side BC is produced to E such
T
E XERCISE - 8.3
A
1. The opposite angles of a parallelogram are (3x − 2)o and (x + 48)o.
AN
Find the measure of each angle of the parallelogram.
2. Find the measure of all the angles of a parallelogram, if one angle is 24o less than the twice
of the smallest angle.
3. In the adjacent figure ABCD is a D C
G
parallelogram and E is the
midpoint of the side BC. If DE E
and AB are produced to meet at
F, show that AF = 2AB. N A B
F
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D Q C
4. In the adjacent figure ABCD is a parallelogram P and Q are
the midpoints of sides AB and DC respectively. Show that
PBCQ is also a parallelogram.
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A B
P
Q
T
A
B
(iv) ∆ABC ≅ ∆DEF C
AN
D C
8. ABCD is a parallelogram. AC and BD are the Q
diagonals intersect at O. P and Q are the points of O
tri section of the diagonal BD. Prove that CQ || AP
P
and also AC bisects PQ (see figure).
A B
G
9. ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA
N
respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.
LA
8.6 T HE M IDPOINT T HEOREM OF T RIANGLE
We have studied properties of triangle and of a quadrilateral. Let us try and consider
the midpoints of the sides of a triangle and what can be derived from them.
TE
TRY THIS
Draw a triangle ABC and mark the midpoints E and F of two sides of triangle.
T
1
We find ∠AEF = ∠ABC and EF = BC
SC
2 B C
As these are corresponding angles made by the transversal
AB with lines EF and BC, we say EF || BC.
Repeat this activity with some more triangles.
1 A
We have to show that : (i) EF || BC (ii) EF = BC
2
Proof:- Join EF and extend it, and draw a line parallel F D
E
to BA through C to meet to produced EF at D.
In ∆s AEF and ∆CDF
A
B C
AF = CF (F is the midpoint of AC)
AN
∠AFE = ∠CFD (vertically opposite angles.)
G
By A.S.A congruency rule
∴ ∆ AEF ≅ ∆CDF ASA congruency rule
Thus AE = CD and EF = DF N (CPCT)
LA
We know AE = BE
Therefore BE = CD
Since BE || CD and BE = CD, BCDE is a parallelogram.
TE
So ED || BC
⇒ EF || BC
As BCDE is a parallelogram, ED = BC(how ?) (∵ DF = EF)
we have shown FD = EF
T
∴ 2EF = BC
ER
1
Hence EF = BC
2
We can see that the converse of the above statement is also true. Let us state it and then
SC
A
We can not prove the congruence of the B C
AN
triangles as we have not shown any pair of sides in
the two triangles as equal.
To do so we consider EB || DC
G
and ED || BC
Thus EDCB is a parallelogram and we have BE = DC.
Since BE = AE we have AE = DC.
Hence ∆AEF ≅ ∆CFD
N
LA
∴ AF = CF
respectively E
so by Mid-point Theorem,
DE || AC
SC
A
B G E
intersected by the transversals p and q at A, B, C
and D,E, F such that they make equal intercepts m
AN
AB and BC on the transversal p. Show that the
intercepts DE and EF on q are also equal. C F
n
Proof : We need to connect the equality of AB
and BC to comparing DE and EF. We join A to F
G
and call the intersection point with ‘m’ as G.
In ∆ACF, AB = BC (given)
Therefore B is the midpoint of AC.
and BG || CF (how ?)
N
LA
So G is the midpoint of AF (By the theorem).
Now in ∆ AFD, we can apply the same reason as G is the midpoint of AF and GE || AD,
E is the midpoint of DF.
TE
Thus DE = EF.
Hence l, m and n cut off equal intersects on q also.
Example-9. In the Fig. AD and BE are medians of ∆ABC and BE || DF. Prove that
T
1
CF = AC.
4
ER
Proof : If ∆ ABC, D is the midpoint of BC and BE || DF; By Theorem F is the midpoint of CE.
A
1
∴ CF = CE
2
SC
11
= AC (How ?) E
2 2
F
1
Hence CF = AC .
4 B C
D
Example-10. ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB
respectively intersecting at P, Q and R. Prove that the perimeter of ∆PQR is double the perimeter
of ∆ABC.
A
respectively.
1 1 1
∴ AB = PQ; BC = QR and CA = PR (How)
AN
P
2 2 2
(State the related theorem)
Now perimeter of ∆PQR = PQ + QR + PR
= 2AB + 2BC + 2CA
G
= 2(AB + BC + CA)
= 2 (perimeter of ∆ABC).
EXERCISE - 8.4 N
LA
1
1. ABC is a triangle. D is a point on AB such that AD = AB and E is a
4
1
point on AC such that AE = AC. If DE = 2 cm find BC.
TE
4
2. ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA
respectively. Prove that EFGH is a parallelogram.
3. Show that the figure formed by joining the midpoints of sides of a rhombus successively
is a rectangle.
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(ii) MD ⊥ AC
1 D M
(iii) CM = MA = AB .
2
C B
WHAT
HAT WE HAVE DISCUSSED
HAVE
A
quadrilaterals
4. Parallelogram is a special type of quadrilateral with many properties. We have proved
AN
the following theorems.
a) The diagonal of a parallelogram divides it into two congruent triangles.
b) The opposite sides and angles of a parallelogram are equal.
c) If each pair of opposite sides of a quadrilateral are equal then it is a parallelogram.
G
d) If each pair of opposite angles are equal then it is a parallelogram.
N
e) Diagonals of a parallelogram bisect each other.
f) If the diagonals of a quadrilateral bisect each other then it is a parallelogram.
LA
5. Mid point theorm of triangle and converse
a) The line segment joining the midpoints of two sides of a triangle is parallel to the
third side and also half of it.
TE
b) The line drawn through the midpoint of one of the sides of a triangle and parallel
to another side will bisect the third side.
T
Brain teaser
ER
Add two straight lines to the above diagram and produce 10 triangles.
2. Take a rectangular sheet of paper whose length is 16 cm and breadth is 9 cm. Cut it
in to exactly 2 pieces and join them to make a square.
9 cm
16 cm
12 cm
A
9.1 I NTRODUCTION
AN
One day Ashish visited his mathematics teacher at his home.
At that time his teacher was busy in compiling the information which
he had collected from his ward for the population census of India.
G
Ashish : Good evening sir, it seems you are very busy. Can
I help you in your work, Sir?
N
Teacher : Ashish, I have collected the household information
for census i.e. number of family members has, their
LA
age group, family income, type of house they live
and other data.
Ashish : Sir, what is the use of this information?
TE
Teacher : The Census Department compiles this massive data and by using required data
handling tools analyse the data and interpretes the results in the form of information.
ER
Ashish, you must have learnt basic statistics (data handling) in your earlier classes,
didn’t you?
Like Ashish we too come across a lot of situations where we see information in the form of
facts, numerical figures, tables, graphs etc. These may relate to price of vegetables, city temperature,
SC
cricket scores, polling result and so on. These facts or figures which are numerical or otherwise
collected with a definite purpose are called ‘data’. Extraction of meaning from the data is studied
in a branch of mathematics called statistics.
Lets us first revise what we have studied in statistics (data handling) in our previous classes.
The primary activity in statistics is to collect the data with some purpose. To understand
this let us begin with an exercise of collecting data by performing the following activity.
STATISTICS 195
ACTIVITY
Divide the students of your class into four groups. Allot each group the work of
collecting one of the following kinds of data:
i. Weights of all the students in your class.
A
ii. Number of siblings that each student have.
iii. Day wise number of absentees in your class during last month.
AN
iv. The distance between the school and home of every student of your class.
Let us discuss how these students have collected the required information?
G
1. Have they collected the information by enquiring each student directly or by visiting every
house personally by the students?
N
2. Have they got the information from source like data available in school records?
LA
In first case when the information was collected by the investigator (student) with a definite
objective, the data obtained is called primary data (as in (i), (ii), (iv)).
In the above task (iii) number of absentees in the last month could only be known by
TE
school attendance register. So here we are using data which is already collected by class teachers.
This is called secondary data. The information collected from a source, which had already been
recorded, say from registers, is called secondary data.
T
DO THIS
ER
A
So the range = 50 − 7 = 43,
AN
From the above we can also say that our data lies from 7 to 50.
Now let us answer the following questions from the above date.
i. Find the middle value of the given data.
G
ii. Find how many children got 60% or more marks in the mathematics test.
Discussion
(i) N
Ikram said that the middle value of the data is 25 because the exam was conducted for 50
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marks. What do you think?
Mary said that it is not the middle value of the data. In this case we have marks of 15
students as raw data, so after arranging the data in ascending order,
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7, 11, 20, 20, 25, 28, 30, 34, 39, 40, 42, 42, 45, 50, 50
we can say that the 8th term is the middle term and it is 34.
60
(ii) You already know how to find 60% of 50 marks (i.e. × 50 = 30 ).
100
T
You find that there are 9 students who got 60% Marks Tally No of
or more marks (i.e. 30 marks or more).
ER
Marks students
When the number of observations in a data are 1 6
too many, presentation of the data in ascending or 2 6
descending order can be quite time consuming. So we 3 3
SC
A
Sum of all frequencies in the table gives the total number of observations of the data.
AN
As the actual observations of the data are shown in the table with their frequencies, this
table is called ‘Ungrouped Frequency Distribution Table’or ‘Table of Weighted
Observations’.
G
ACTIVITY
Make frequency distribution table of the initial letters of that denotes surnames of your
N
classmates and answer the following questions.
(i) Which initial letter occured mostly among your classmates?
LA
(ii) How many students names start with the alphabat ‘I’?
(iii) Which letter occured least as an initial among your classmates?
TE
Suppose for specific reason, we want to represent the data in three categories (i) how
many students need extra classes, (ii) how many have an average performance and (iii) how
many did well in the test. Then we can make groups as per the requirement and grouped
frequency table as shown below.
T
4-5 (Average) 16
6 - 10 (Well) 19
SC
To classify the data according to the requirement or if there are large number of observations.
We make groups to condense it. Let’s take one more example in which group and frequency
make us easy to understand the data.
Example-2. The weight (in grams) of 50 oranges, picked at random from a basket of oranges,
are given below:
35, 45, 55, 50, 30, 110, 95, 40, 70, 100, 60, 80, 85, 60, 52, 95, 98, 35, 47, 45, 105, 90,
30, 50, 75, 95, 85, 80, 35, 45, 40, 50, 60, 65, 55, 45, 30, 90, 115, 65, 60, 40, 100, 55, 75,
110, 85, 95, 55, 50
To present such a large amount of data and to make sense of it, we make groups like
30-39, 40-49, 50-59, ..... 100-109, 110-119. (since our data is from 30 to 115). These groups
are called ‘classes’ or class-intervals, and their size is called length of the class or class width,
which is 10 in this case. In each of these classes the least number is called the lower limit and the
greatest number is called the upper limit, e.g. in 30-39, 30 is the ‘lower limit’ and 39 is the ‘upper
limit’.
A
(Oranges weight) Tally marks (Number of oranges)
AN
Class interval Frequency
30 - 39 6
40 - 49 8
50 - 59 9
G
60 - 69 6
70 - 79 3
80 - 89
90 - 99 N 5
7
LA
100-109 3
110 - 119 3
Total 50
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Presenting data in this form simplifies and condenses data and enables us to observe certain
important features at a glance. This is called a grouped frequency distribution table.
We observe that the classes in the table above are non-
overlapping i.e. 30-39, 40-49 ... no number is repeating in Classes Class boundaries
T
two class intervals. Such classes are called inclusive classes. 20 - 29 19.5 - 29.5
Note that we could have made more classes of shorter size, 30 - 39 29.5 - 39.5
ER
or lesser classes of larger size also. Usually if the raw data is 40 - 49 39.5 - 49.5
given the range is found (Range = Maximum value − Minimum 50 - 59 49.5 - 59.5
value). Based on the value of ranges with convenient, class 60 - 69 59.5 - 69.5
interval length, number of classes are formed. For instance,
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70 - 79 69.5 - 79.5
the intervals could have been 30-35, 36-40 and so on. 80 - 89 79.5 - 89.5
Now think if weight of an orange is 39.5 gm. then in 90 - 99 89.5 - 99.5
which interval will we include it? We cannot include 39.5 100 - 109 99.5-109.5
either in 30-39 or in 40-49. 110 - 119 109.5-119.5
In such cases we construct real limits (or boundaries) 120 - 129 119.5-129.5
for every class.
Average of upper limit of a class interval and lower limit of the next class interval becomes
the upper boundary of the class. The same becomes the lower boundary of the next class interval.
Similarly boundaries of all class intervals are calculated. By assuming a class interval
before the first class and next class interval after the last class, we calculate the lower boundary
any of the first and upper boundary any of the last class intervals.
Again a problem arises that whether 39.5 has to be included in the class interval 29.5-39.5
or 39.5 - 49.5? Here by convention, if any observation is found to be equivalent to upper
A
boundary of a particular class, then that particular observation is considered under next class, but
not that of the particular class.
AN
So 39.5 belongs to 39.5 - 49.5 class interval.
The classes which are in the form of 30-40, 40-50, 50-60, .... are called over lapping
classes and called as exclusive classes.
G
If we observe the boundaries of inclusive classes, they are in the form of exclusive classes.
The difference between upper boundary and lower boundary of particular class given the length
of class interval. Length class interval of 90 − 99 is (i.e. 99.5 − 89.5 = 10) 10.
N
Example-3. The relative humidity (in %) of a certain city for a September month of 30 days was
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as follows:
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3
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90-92 || 2
92-94 |||| || 7
94-96 |||| | 6
96-98 |||| || 7
98-100 |||| 4
(ii) Range 99.2 - 84.9 = 14.3 (vary for different places).
EXERCISE - 9.1
1. Write the mark wise frequencies in the following frequency distribution table.
A
No of students 5 11 19 31 40 45
AN
2. The blood groups of 36 students of IX class are recorded as follows.
A O A O A B O A B A B O
B O B O O A B O B AB O A
O O O A AB O A B O A O B
G
Represent the data in the form of a frequency distribution table. Which is the most common
N
and which is the rarest blood group among these students?
3. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring
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was noted down as follows;
1 2 3 2 3 1 1 1 0 3 2 1
2 2 1 1 2 3 2 0 3 0 1 2
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3 2 2 3 1 1
Prepare a frequency distribution table for the data given above.
giving options like A – complete prohibition, B – prohibition in public places only, C – not
necessary. SMS results in one hour wereA B A B C B
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A B B A C C B B A B
B A B C B A B C B A
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B B A B B C B A B A
B C B B A B C B B A
B B A B B A B C B A
B B A B C A B B A
Represent the above data as grouped frequency distribution table. How many appropriate
answers were received? What was the majority of peoples’ opinion?
5. Represent the data in the adjacent bar graph as frequency distribution table.
Y Scale : on X-axis 1 cm = 5 Vehicles
90
Cars
Bikes
A
Y
Autos
AN
90
80 Cycles
70
Number of students
X
5 10 15 20 25 30 35 40 45 50
60 Number of Vehicles
50
G
40
6. Identify the scale used on the axes of
30
the adjacent graph. Write the
20
III Class
I Class
V Class
7. The marks of 30 students of a class, obtained in a test (out of 75), are given below:
42, 21, 50, 37, 42, 37, 38, 42, 49, 52, 38, 53, 57, 47, 29
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59, 61, 33, 17, 17, 39, 44, 42, 39, 14, 7, 27, 19, 54, 51.
Form a frequency table with equal class intervals. (Hint : one of them being 0-10)
8. The electricity bills (in rupees) of 25 houses in a locality are given below. Construct a
T
170, 212, 252, 225, 310, 712, 412, 425, 322, 325, 192, 198, 230, 320, 412,
530, 602, 724, 370, 402, 317, 403, 405, 372, 413
9. A company manufactures car batteries of a particular type. The life (in years) of 40
batteries were recorded as follows:
SC
A
Case-2 : Consider the wages of staff at a factory as given in the table. Which salary figure
represents the whole staff:
AN
Staff 1 2 3 4 5 6 7 8 9 10
Salary in ` 12 14 15 15 15 16 17 18 90 95
(in thousands)
G
Case-3 : The different forms of transport in a city are given below. Which is the popular means
of transport?
1. Car 15% N
LA
2. Train 12%
3. Bus 60%
4. Two wheeler 13%
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In the first case, we will usually take an average (mean), and use it to resolve the problem.
But if we take average salary in the second case then it would be 30.7 thousands. However,
verifying the raw data suggests that this mean value may not be the best way to accurately reflect
the typical salary of a worker, as most workers have their salaries between 12 to 18 thousands.
So, median (middle value) would be a better measure in this situation. In the third case mode
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(most frequent) is considered to be a most appropriate option. The nature of the data and its
ER
purpose will be the criteria to go for average or median or mode among the measures of central
tendency.
1. Give 3 situations, where mean, median and mode are separately appropriate and counted.
Considier a situation where fans of two cricketers Raghu and Gautam claim that their star
score better than other. They made comparison on the basis of last 5 matches.
Fans of both the players added the runs and calcuated the averages as follows.
307
Raghu’s average score = = 61.4
5
298
Gautam average score = = 59.6
A
5
Since Raghu’s average score was more than Gautam’s, Raghu fan’s claimed that Raghu
AN
performed better than Gautam, but Gautam fans did not agree. Gautam fan’s arranged both their
scores in descending order and found the middle score as given below:
Raghu 100 76 50 50 31
G
Gautam 100 100 65 23 10
Then Gautam fan’s said that since his middle-most score is 65, which is higher than Raghu
N
middle-most score, i.e. 50 so his performance should be rated better.’
But we may say that Gautam made two centuries in 5 matches and so he may be better.
LA
Now, to settle the dispute between Raghu’s and Gautam’s fans, let us see the three measures
adopted here to make their point.
TE
The average score they used first is the mean. The ‘middle’ score they used in the argument
is the Median. Mode is also a measure to compare the performance by considering the scores
repeated many times. Mode score of Raghu is 50. Mode score of Gautam is 100. Of all these
three measures which one is appropriate in this context?
T
Σx
SC
Sum of observations
Mean x = or x = i
Number of observations n
∑ xi x1 + x2 + x3 + .....xn
Mean x = = Where x1, x2 ..... xn are n observation
n n
4 + 5 + 12 + 3 + 6 + 8 + 0.5 38.5
and x is their mean = = = 5.5 cm.
A
7 7
Example-5. If the mean of 10, 12, 18, 13, P and 17 is 15, find the value of P.
AN
∑ xi
Solution : We know that Mean x =
n
10 + 12 + 18 + 13 + P + 17
G
15 =
6
90 = 70 + P
P = 20. N
LA
9.4.1.2 Mean of Ungrouped frequency distribution
Consider this example; Weights of 40 students in a class are given in the following frequency
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distribution table.
Weights in kg (x) 30 32 33 35 37 41
No of students ( f ) 5 9 15 6 3 2
T
From the table we can see that 5 students weigh 30 kg., each. So sum of their weights is
5 × 30 = 150 kg. Similarly we can find out the sum of weights with each frequency and then their
total. Sum of the frequencies gives the number of observations in the data.
Mean x =
Total number of observations
5 × 30 + 9 × 32 + 15 × 33 + 6 × 35 + 3 × 37 + 2 × 41 1336
So Mean = = = 33.40 kg.
5 + 9 + 15 + 6 + 3 + 2 40
If observations are x1, x2, x3, x4, x5, x6 and corresponding frequencies are f1, f2, f3, f4, f5, f6
then we may write the above expression as
Mean x =
f1 x1 + f 2 x2 + f3 x3 + f 4 x4 + f5 x5 + f 6 x6
=
∑fx
i i
f1 + f 2 + f3 + f 4 + f5 + f 6 ∑f i
∴ x=
∑fx i i
∑f i
A
Example-6. Find the mean of the following data.
AN
x 5 10 15 20 25
f 3 10 25 7 5
G
Solution : xi fi fixi
50
Σfi = 50 Σf i xi =755
Example-7. If the mean of the following data is 7.5 , then find the value of ‘A’.
Marks 5 6 7 8 9 10
T
No. of Students 3 10 17 A 8 4
ER
A
method
AN
Example-8. Find the arithmetic mean of the following data:
x 10 12 14 16 18 20 22
f 4 5 8 10 7 4 2
G
Solution : xi fi fi xi
(i) Simple Method N 10
12
4
5
40
60
LA
Thus in the case of ungrouped frequency distribution,
14 8 112
you can use the formula,
16 10 160
7 18 7 126
∑fx
TE
i =1
i i 20 4 80
622
x =
7
= = 15.55 22 2 44
∑f
i =1
i
40
7 7
∑
i =1
fi =40 ∑ f x =622
i =1
i i
14 8 −2 −16
Sum of frequencies = 40
16 A 10 0 0
Sum of the fi×di products = − 60 + 42
18 7 +2 +14
Σfidi = −18
20 4 +4 +16
Σf d −18
Mean x = A + i i = 16 + 22 2 +6 +12
Σfi 40
= 16 − 0.45 40 −60+42=−18
= 15.55
9.4.2 Median
Median is the middle observation of a given raw data, when it is arranged in an order
(ascending / descending). It divides the data into two groups of equal number, one part comprising
all values greater than median and the other part comprising values less than median.
A
We have already discussed in the earlier classes that median of a raw data with observations,
arranged in order is calculated as follows.
AN
th
n +1
When the data as ‘n’ number of observations and if ‘n’ is odd, median is
2
observation.
th th
n n
G
When n is even, median is the average of and + 1 observations
2 2
TRY THESE N
LA
1. Find the median of the scores 75, 21, 56, 36, 81, 05, 42
2. Median of a data, arranged in ascending order 7, 10, 15, x, y, 27, 30 is 17 and
when one more observation 50 is added to the data, the median has become 18
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Find x and y.
No. of employees 4 18 30 20 15 8 5
N
Find and identify the median class, whose cumulative frequencies just exceeds
2
N
, where N is sum of the frequencies.
2
th th
N N
Here N= 100 even so find and + 1 observations which are 50 and 51
2 2
A
respectively.
From the table corresponding values of 50th and 51st observations is the same, falls in the
AN
wages of 8500. So the median class of this distribution is 8500.
TRY THESE
G
1. Find the median marks in the data.
Marks 15 20 10 25 5
No of students 10 N
8 6 4 1
LA
2. In finding the median, the given data must be written in order. Why?
9.4.3 Mode
Mode is the value of the observation which occurs most frequently, i.e., an observation
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Size 6 7 8 9 10
1. Classify your class mates according to their heights and find the mode of it.
2. If shopkeeper has to place a order for shoes, which number shoes should he order more?
Example-10. Test scores out of 100 for a class of 20 students are as follows:
93, 84, 97, 98, 100, 78, 86, 100, 85, 92, 55, 91, 90, 75, 94, 83, 60, 81, 95
(a) Make a frequency table taking class interval as 91-100, 81-90, .....
(b) Find the modal class. (The “Modal class” is the class containing the greatest frequency).
A
(c) find the interval that contains the median.
Solution :
AN
(a) Test Scores Frequency Greater than
Cumulative frequency
91-100 9 20
G
81-90 6 11
71-80 3 5
61-70 N 0 2
LA
51-60 2 2
Total 20
(b) 91-100 is the modal class. This class has the maximum frequency.
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(c) The middle of 20 is 10. If I count from the top, 10 will fall in the class interval 81-90. If
I count from the bottom and go up, 10 will fall in the class interval 81-90. The class
interval that contains the median is 81-90.
T
9.5 DEVIATION
EVIATION IN VALUES OF CENTRAL TENDENCY
ALUES
ER
What will happen to the measures of central tendency if we add the same amount to all
data values, or multiply each data value by the same amount.
Let us observe the following table
SC
Original Data Set 6, 7, 8, 10, 12, 14, 14, 15, 16, 20 12.2 14 13
Add 3 to each data 9, 10, 11, 13, 15, 17, 17, 18, 19, 23 15.2 17 16
value
Multiply 2 times each 12, 14, 16, 20, 24, 28, 28, 30, 32, 40 24.4 28 26
data value
A
of central tendency will also be affected similarly. If each observation is multiplied by 2, the
mean, mode and median will also be multiplied by 2.
AN
EXERCISE - 9.2
1. Weights of parcels in a transport office are given below.
G
Weight (kg) 50 65 75 90 110 120
No of parcels 25 34 38 40 47 16
N
Find the mean weight of the parcels.
LA
2. Number of families in a village in correspondence with the number of children are given
below:
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No of children 0 1 2 3 4 5
No of families 11 25 32 10 5 1
3. If the mean of the following frequency distribution is 7.2 find value of ‘K’.
ER
x 2 4 6 8 10 12
f 4 7 10 16 K 3
SC
4. Number of villages with respect to their population as per India census 2011 are given
below.
Villages 20 15 32 35 36 7
5. AFLATOUN social and financial educational program intiated savings program among
the high school children in Hyderabad district. Mandal wise savings in a month are given
in the following table.
Mandal No. of schools Total amount saved (in rupees)
Amberpet 6 2154
A
Thirumalgiri 6 2478
Saidabad 5 975
AN
Khairathabad 4 912
Secundrabad 3 600
Bahadurpura 9 7533
G
Find arithmetic mean of school wise savings in each mandal. Also find the arithmetic
mean of saving of all schools.
N
6. The heights of boys and girls of IX class of a school are given below.
LA
Height (cm) 135 140 147 152 155 160
Boys 2 5 12 10 7 1
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Girls 1 2 10 5 6 5
Compare the heights of the boys and girls
[Hint : Find median heights of boys and girls]
7. Centuries scored and number of cricketers in the world are given below.
T
No. of centuries 5 10 15 20 25
ER
No. of cricketers 56 23 39 13 8
Find the mean, median and mode of the given data.
8. On the occasion of New year’s day a sweet stall prepared sweet packets. Number of
SC
9. The mean (average) weight of three students is 40 kg. One of the students Ranga weighs
46 kg. The other two students, Rahim and Reshma have the same weight. Find Rahims
weight.
10. The donations given to an orphanage home by the students of different classes of a secondary
school are given below.
A
VII 7 15
VIII 10 20
AN
IX 15 16
X 20 14
Find the mean, median and mode of the data.
G
11. There are four unknown numbers. The mean of the first two numbers is 4 and the mean
of the first three is 9. The mean of all four number is 15, if one of the four number is 2 find
the other numbers.
N
LA
WHAT
HAT WE HAVE DISCUSSED
HAVE
for how many times, to analyse and to interpret the data easily.
ER
• A measure of central tendency is a typical value of the data around which other
observations congregate.
Σfi xi
• For a ungrouped frequency distribution arithmetic mean x = .
Σ fi
Σf i d i
• By deviation method, arithmetic mean =A + where A is assumed mean and
Σ fi
where Σf i is the sum of frequencies and Σfi di is the sum of product of frequency
and deviations.
A
• Median is the middle observation of a data, when arranged in order (ascending /
descending).
AN
th
n +1
• When number of observations ‘n’ is odd, median is observation.
2
th
G
n
• When number of observations ‘n’is even, median is the average of and
2
th
n
+ 1 observations
2 N
LA
• Median divides the data into two groups of equal number, one part comprising all
values greater and the other comprising all values less than median.
•
TE
Mode is the value of the observation which occurs most frequently, i.e., an observation
with the maximum frequency is called mode.
T
ER
Brain teaser
SC
In a row of students, Gopi is the 7th boy from left and Shankar is the 5th
boy from the right. If they exchange their seats, Gopi is the 8th boy from the
right. How many students are there in the row?
A boy chaitanya carved his name on the bark of a tree at a height of 1.5 m
tall. The tree attains a height of a 6.75 m from the ground at what height
from the ground will chaitanya’s name can be located now?
Give reason to your answer.
A
10.1 I NTRODUCTION
AN
Observe the following figures
(a)
N G
(b)
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Have you noticed any differences between the figures of group (a) and (b)?
From the above, figures of group (a) can be drawn easily on our note books. These figures
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have length and breadth only and are named as two dimensional figures or 2-D objects. In group
(b) the figures, which have length, breadth and height are called as three dimensional figures or
3-D objects. These are called solid figures. Usually we see solid figures in our surroundings.
You have learned about plane figures and their areas. We shall now learn to find the surface
T
areas and volumes of 3-dimensional objects such as cylinders, cones and spheres.
ER
10.2 S URFACE
URFA AREA OF C UBOID
Observe the cuboid and find how many faces it H G
D C
has? How many corners and how many edges it has?
h = 2cm.
SC
If we cut and open the given cuboid along CD, ADHE and BCGF. The figure we obtained
is shown below: D C
l
b (2) b
A
D H G C
b l b
h (3) h (1) h (4) h
AN
b l b
A E F B
b (5) b
G
A l B
D
N h (6)
l
h
C
LA
This shows that the surface area of a cuboid is made up of six rectangles of three identical
pairs of rectangles. To get the total surface area of cuboid, we have to add the areas of all six
rectangular faces. The sum of these areas gives the total surface area of a cuboid.
TE
.....(5)
Area of the rectangle DCBA = l × h = lh .....(6)
ER
= 2(lb + bh + lh)
(1), (3), (4), (6) are lateral surfaces of the cuboid
Lateral Surface Area of a cuboid = Area of (1) + (3) + (4) + (6)
= lh + bh + bh + lh
= 2lh + 2bh
= 2h (l + b)
Now let us find the surface areas of cuboid for the above figure. Thus total surface area is
62 cm.2 and lateral surface area is 32 cm.2.
TRY THIS
Take a cube of edge ‘l’ cm. and cut it as we did in the previous activity and find
total surface area and lateral surface area of cube.
DO THIS
A
1. Find the total Surface area and lateral surface area of the
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Cube with side 4 cm. (By using the formulae deduced
4 cm.
above)
2. Each edge of a cube is increased by 50%. Find the 4 cm.
4 cm.
percentage increase in the surface area.
G
10.2.1 Volume
N
To recall the concept of volume, Let us do the following activity.
LA
Take a glass jar, place it in a container. Fill the glass jar with water up to the its brim.
Slowly drop a solid object (a stone) in it. Some of the water from the jar will overflow into the
container. Take the overflowed water into measuring jar. It gives an idea of space occupied by
a solid object called volume.
TE
T
ER
Every object occupies some space, the space occupied by an object is called its volume.
SC
A
height of the cuboid is ‘h’
Space occupied by the cuboid = Area of plane region occupied by rectangle × height
AN
Volume of the cuboid = l b × h = l bh
∴ Volume of the cuboid = l bh
Where l, b, h are length, breadth and height of the cuboid.
G
a
TRY THESE a
N
(a) Find the volume of a cube whose edge is ‘a’ units. a
a
a
a
a
LA
3 a a
(b) Find the edge of a cube whose volume is 1000 cm . a
a
a
We know that cuboid and cube are the solids. Do we call them as right prisms? You have
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observed that cuboid and cube are also called right prisms as their lateral faces are rectangle and
perpendicular to base.
We know that the volume of a cuboid is the product of the area of its base and height.
Remember that volume of the cuboid = Area of base × height
T
= lb × h
ER
= lbh
In cube = l = b = h = s (All the dimensions are same)
volume of the cube = s2 × s
= s3
SC
Hence volume of a cuboid should hold good for all right prisms.
Hence volume of right prism = Area of the base × height
3 2
In particular, if the base of a right prism is an equilateral triangle its volume is a × h cu.units.
4
Where, ‘a’ is the length of each side of the base and ‘h’ is the height of the prim.
DO THESE
1. Find the volume of cuboid if l = 12 cm., b = 10 cm.
and h = 8 cm.
2. Find the volume of cube, if its edge is 10 cm.
A
3. Find the volume of isosceles right angled triangular 5 cm.
(Fig 1)
prism in (fig. 1).
AN
Like the prism, the pyramid is also a three dimentional solid figure. This figure has fascinated
human beings from the ancient times. You might have read about pyramids of Egypt, which are,
one of the seven wonders of the world. They are the remarkable examples of pyramids on
G
square bases. How are they built? It is a mystery. No one really knows that how these massive
structures were built.
Can you draw the shape of a pyramid?
N
What is the difference you have observed between the prism
and pyramid?
O
LA
What do we call a pyramid of square base? height
ACTIVITY
T
make?
DO THESE
1. Find the volume of a pyramid whose square base is 10 cm. and height 8 cm.
2. The volume of cube is 1728 cubic cm. Find the volume of square pyramid of the same
height.
A
EXERCISE - 10.1
AN
1. Find the later surface area and total surface area of the following right prisms.
(i) (ii)
G
5 cm.
4 cm.
4 cm.
N 8 cm.
LA
2. The total surface area of a cube is 1350 sq.m. Find its volume.
3. Find the area of four walls of a room (Assume that there are no doors or windows) if its
TE
4. The volume of a cuboid is 1200 cm3. The length is 15 cm. and breadth is 10 cm. Find its height.
T
Express in words. Can you find the total surface area of the box if each dimension is raised
to n times?
SC
6. The base of a prism is triangular in shape with sides 3 cm., 4 cm. and 5 cm. Find the
volume of the prism if its height is 10 cm.
7. A regular square pyramid is 3 m. height and the perimeter of its base is 16 m. Find the
volume of the pyramid.
8. An Olympic swimming pool is in the shape of a cuboid of dimensions 50 m. long and 25
m. wide. If it is 3 m. deep throughout, how many liters of water does it hold?
( 1 cu.m = 1000 liters)
ACTIVITY
Cut out a rectangular sheet of paper. Paste a thick
string along the line as shown in the figure. Hold the
string with your hands on either sides of the rectangle
and rotate the rectangle sheet about the string as fast
A
as you can.
AN
Do you recognize the shape that the rotating
rectangle is forming ?
Does it remind you the shape of a cylinder ? F
G
10.3 R IGHT C IRCULAR C YLINDER
Observe the following cylinders:
N
LA
h h h
TE
r r r
(i) (ii) (iii)
(i) What similiarties you have observed in figure (i), (ii) and (iii)?
T
(ii) What differences you have observed between fig. (i), (ii) and (iii)?
ER
Find out which is right circular cylinder in the above figures? Which are not? Give reasons.
You will find it is in rectangular shape. The area of rectangle is equal to the area of curved
surface area of cylinder. Its height is equal to the breadth of the rectangle, and the circumference
of the base is equal to the length of the rectangle.
Height of cylinder = breadth of rectangle (h = b)
Circumferance of base of cylinder with radius ‘r’ = length of the rectangle (2πr = l)
A
Curved surface area of the cylinder = Area of the rectangle
AN
2πr
= length × breadth
h = 2 πr × h
h
= 2πrh
G
πrh
Therefore, Curved surface area of a cylinder = 2π
DO THIS
N
LA
Find CSA of each of following
cylinders
10 cm.
14 cm.
(i) r = x cm., h = y cm.
TE
3 cm.
(ii) d = 7 cm., h = 10 cm.
(iii) r = 3 cm., h = 14 cm. 7 cm.
Do you find that it is a right circular cylinder? What surfaces you have to add to get its total
surface area? They are the curved surface area and area of two circular faces.
Now the total surface area of a cylinder
SC
DO THESE
Find the Total surface area of each of the following cylinders.
(i) (ii)
7 cm.
7 cm.
A
250 cm2
AN
10 cm.
G
10.3.3 Volume of a cylinder
Take circles with equal radii and arrange one over the other.
N
Do this activity and find whether it form a cylinder or not.
LA
In the figure ‘r’ is the radius of the circle, and the ‘h’ is the height up to which the circles
are stacked.
Volume of a cylinder = πr2 × height r
TE
= πr2 × h
= πr2h
r
So volume of a cylinder = πr h 2
T
Example-1. A Rectangular paper of width 14 cm is folded along its width and a cylinder of
22
radius 20 cm is formed. Find the volume of the cylinder (Fig 1) ? Take π =
7
SC
22
= × 20 × 20 × 14
7
= 17600 cm3.
Hence the volume of the cylinder is 17600 cm3.
A
Example-2. A Rectangular piece of paper 11 cm × 4 cm is folded without overlapping to
AN
make a cylinder of height 4 cm. Find the volume of the cylinder.
Solution : Length of the paper becomes the circumference of the base of the cylinder and width
becomes height.
G
Let radius of the cylinder = r and height = h
Circumference of the base of the cylinder = 2πr = 11 cm.
2×
22
7
× r = 11
N 11 cm.
LA
4 cm.
7
∴ r= cm.
4
TE
h = 4 cm
Volume of the cylinder (V) = πr2h
22 7 7
= × × × 4 cm 3
7 4 4
T
= 38.5 cm3.
ER
44 44 × 7
r= = = 7cm.
2 × π 2 × 22
22
= 2× × 7(7 + 18)cm 2
7
= 1100 cm2.
A
Example-4. Circular discs 5 mm thickness, are placed one above the other to form a cylinder
of curved surface area 462 cm2. Find the number of discs, if the radius is 3.5 cm.
AN
5
Solution : Thickness of disc = 5 mm = cm = 0.5 cm
10
Radius of disc = 3.5 cm.
G
Curved surface area of cylinder = 462 cm2.
∴ 2πrh = 462 ..... (i)
Let the no of discs be x N
LA
∴ Height of cylinder = h = Thickness of disc × no of discs
= 0.5 x
22
TE
22
2× × 3.5 × 0.5 x = 462
T
462 × 7
ER
∴ x= = 42 discs
2 × 22 × 3.5 × 0.5
Example-5. A hollow cylinder having external radius 8 cm and height 10 cm has a total
surface area of 338 π cm2. Find the thickness of the hollow metallic cylinder.
SC
A
⇒ (10 × 8) + (r × 10) + 8 − r = 169
2 2
⇒ r − 10r + 25 = 0
2
AN
⇒ (r − 5) = 0
2
∴ r =5
∴ Thickness of metal = R − r = (8 − 5) cm = 3 cm.
G
TRY THESE
N
1. If the radius of a cylinder is doubled keeping its lateral surface area the
LA
same, then what is its height ?
2. A hot water system (Geyser) consists of a cylindrical pipe of length 14 m and
diameter 5 cm. Find the total radiating surface of hot water system.
TE
E XERCISE - 10.2
1. A closed cylindrical tank of height 1.4 m. and radius of the base is 56 cm.
is made up of a thick metal sheet. How much metal sheet is required (Express
T
in square meters)
2. The volume of a cylinder is 308 cm.3. Its height is 8 cm. Find its lateral surface area
ER
4. An overhead water tanker is in the shape of a cylinder has capacity of 61.6 cu.mts.
The diameter of the tank is 5.6 m. Find the height of the tank. r
R
A
in m2.
8. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
AN
(i) its inner curved surface area
(ii) The cost of plastering this curved surface at the rate of Rs. 40 per m2.
9. Find
(i) The total surface area of a closed cylindrical petrol storage tank whose diameter
G
4.2 m. and height 4.5 m.
1
the tank. N
(ii) How much steel sheet was actually used, if
12
of the steel was wasted in making
LA
10. A one side open cylinderical drum has inner radius 28 cm. and height 2.1 m. How much
water you can store in the drum. Express in litres. (1 litre = 1000 cc.)
11. The curved surface area of the cylinder is 1760 cm.2 and its volume is 12320 cm3. Find
TE
its height.
Observe the above figures and which solid shape they resemble?
SC
h h h
r r r
(i) (ii) (iii)
A
In fig.(ii) although it has circular base, but its vertical height is not perpendicular to the
AN
radius of the cone.
Such type of cones are not right circular cones.
In the fig. (iii) although the vertical height is perpendicular to the base, but the base is not in
circular shape.
G
Therefore, this cone is not a right circular cone.
AO is the height of the cone (h) and OB is equal to the radius of the cone (r)
TE
From ∆AOB O r B
2 2 2
AB = AO + OB
h
AB2 = h2 + r2 (AB is called slant height = l ) l
T
l2 = h2 + r2
ER
l = h2 + r2 A
ACTIVITY
Making a cone from a sector
SC
O
Follow the instructions and do as shown in the
figure. A B
(i) Draw a circle on a thick paper Fig(a) (a)
(ii) Cut a sector AOB from it Fig(b).
(iii) Fold the ends A, B nearer to each other slowly A B
and join AB. Remember A, B should not overlap
on each other. After joining A, B attach them
with cello tape Fig(c). (b) (c)
O
A
10.4.2 Curved Surface area of a cone
AN
B
r b1
b2
O b3
h O
l
G
A B
A
(i) (ii) (iii)
N
Let us find the surface area of a right circular cone that we made out of the paper as
discussed in the activity.
LA
While folding the sector into cone you have noticed that OA, OB of sector coincides and
becomes the circumference of
becomes the slant height of the cone, whereas the length of AB
TE
If we find the area of these triangles and adding these, it gives area of the sector. We
know that sector forms a cone, so the area of a sector is equal to curved the surface area of the
ER
Thus, lateral surface area or curved surface area of the cone = πrl
Where ‘l’ is the slant height of the cone and ‘r’ is its radius
If the base of the cone is to be covered, we need a circle whose radius is equal to the
A
radius of the cone.
AN
How to obtain the total surface area of cone? How many surfaces you have to add to get
total surface area?
The area of the circle = πr2
Total surface area of a cone = lateral surface area + area of its base
G
= πrl + πr2
= πr (l + r)
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Total surface area of the cone = πr (l + r)
LA
Where ‘r’ is the radius of the cone and ‘l’ is its slant height.
DO THIS
TE
1. Cut a right angled triangle, stick a string along its perpendicular side, as shown
in fig. (i) hold the both the sides of a string with your hands and rotate it with
constant speed.
What do you observe ?
T
2. Find the curved surface area and total surface area of the each following Right
Circular Cones. P
ER
P
fig. (i)
A O B A O B
OP = 2cm.; OB = 3.5cm. OP = 3.5cm.; AB = 10cm.
SC
Make a hollow cylinder and a hollow cone with the equal radius and equal height and do
the following experiment, that will help us to find the volume of a cone.
h h h
r r r
A
AN
h h h
i. Fill water in the cone up to the brim and pour into the hollow cylinder, it will fill up only
G
some part of the cylinder.
ii. Again fill up the cone up to the brim and pour into the cylinder, we see the cylinder is still
not full.
N
iii. When the cone is filled up for the third time and emptied into the cylinder, observe whether
LA
the cylinder is filled completely or not.
With the above experiment do you find any relation between the volume of the cone and
the volume of the cylinder?
TE
We can say that three times the volume of a cone makes up the volume of cylinder, which
both have the same base and same height.
So the volume of a cone is one third of the volume of the cylinder.
T
1 2
∴ Volume of a cone = πr h
3
ER
where ‘r’ is the radius of the base of cone and ‘h’ is its height.
Example-6. A corn cob (see fig), shaped like a cone, has the radius of its broadest end as 1.4
cm and length (height) as 12 cm. If each 1cm2 of the surface of the cob carries an average of four
SC
grains, find how many grains approximately you would find on the entire cob.
22
= × 1.4 × 12.08 cm 2
7
= 53.15 cm2
= 53.2 cm2 (approx)
Number of grains of corn on 1 cm2 of the surface of the corn cob = 4.
Therefore, number of grains on the entire curved surface of the cob.
= 53.2 × 4 = 212.8 = 213 (approx)
A
So, there would be approximately 213 grain of corn on the cob.
AN
Example-7. Find the slant height and vertical height of a Cone with radius 5.6 cm and curved
surface area 158.4 cm2.
G
Solution : Radius = 5.6 cm, vertical height = h, slant height = l
CSA of cone = πrl = 158.4 cm2
⇒
22
7
× 5.6 × l = 158.4 N A
LA
158.4 × 7 18
⇒ l= = = 9cm
22 × 5.6 2
l
TE
we know l2 = r2 + h2 h
h2 = l2 − r2 = 92 − (5.6)2
= 81 − 31.36
O B
= 49.64
T
h= 49.64
ER
h = 7.05 cm (approx)
Example-8. A tent is in the form of a cylinder surmounted by a cone having its diameter of the
SC
base equal to 24 m. The height of cylinder is 11 m and the vertex of the cone is 5m above the
cylinder. Find the cost of making the tent, if the rate of canvas is `10 per m2.
Solution : Diametre of base of cylinder = diametre of cone = 24m
∴ Radius of base = 12 m
Height of cylinder = 11 m = h1
Height of Cone = 5m = h2
Let slant height of cone be l
l = GD = r 2 + h 2 = 12 2 + 52 = 13m
Area of canvas required = CSA of cylinder + CSA of cone
= 2πrh1 + πrl
G
l = πr (2h1 + l)
5 m.
A
C D 22
F
= × 12(2 × 11 + 13)m 2
AN
7
22 × 12
11 m.
= × 35m 2
7
= 22 × 60 m2
G
A B
E
24 m.
= 1320 m2
Rate of canvas = `10 per m2
∴ Cost of canvas = Rate × area of canvas
N
LA
= `10 × 1320
= `13,200.
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Example-9. A conical tent was erected by army at a base camp with height 3m. and base
diameter 8m. Find;
(i) The cost of canvas required for making the tent, if the canvas cost ` 70 per 1 sq.m.
(ii) If every person requires 3.5 m.3 air, how many can be seated in that tent.
T
d 8
r= = = 4 m.
2 2
height = 3 m.
3 m.
SC
Slant height (l ) = h2 + r 2
8 m.
= 32 + 42
= 25 = 5 m.
22 440 2
= ×4×5= m
7 7
1 2
Volume of the cone = πr h
3
1 22
= × ×4×4×3
3 7
A
352 3
= m
7
AN
(i) Cost of canvas required for the tent
= CSA × Unit cost
440
G
= × 70
7
= `4400
N
(ii) No. of persons can be seated in the tent
LA
Volume of conical tent
=
air required for each
352
= ÷ 3.5
TE
7
352 1
= × = 14.36
7 3.5
= 14 men (approx.)
T
EXERCISE - 10.3
ER
1. The base area of a cone is 38.5 cm2. Its volume is 77 cm3. Find its height.
2. The volume of a cone is 462 m3. Its base radius is 7 m. Find its height.
SC
3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm Find.
(i) radius of the base (ii) Total surface area of the cone.
4. The cost of painting the total surface area of a cone at 25 paise per cm2 is `176. Find the
volume of the cone, if its slant height is 25 cm.
5. From a circle of radius 15 cm., a sector with angle 216° is cut out and its bounding radii
are bent so as to form a cone. Find its volume.
6. The height of a tent is 9 m. Its base diameter is 24 m. What is its slant height? Find the
cost of canvas cloth required if it costs `14 per sq.m.
5 4
7. The curved surface area of a cone is 1159 cm2. Area of its base is 254 cm2. Find its
7 7
volume.
8. A tent is cylindrical to a height of 4.8 m. and conical above it. The radius of the base is
4.5m. and total height of the tent is 10.8 m. Find the canvas required for the tent in square
A
meters.
9. What length of tarpaulin 3 m wide will be required to make a conical tent of height 8m
AN
and base radius 6m ? Assume that extra length of material that will be required for
stitching margins and wastage in cutting is approximately 20 cm (use π = 3.14)
10. A Joker’s cap is in the form of a right circular cone of base radius 7 cm and height 27 cm.
Find the area of the sheet required to make 10 such caps.
G
11. Water is pouring into a conical vessel of diameter 5.2m and slant
height 6.8m (as shown in the adjoining figure), at the rate of 1.8 m3
5.2 m.
N
per minute. How long will it take to fill the vessel?
12. Two similiar cones have volumes 12π cu. units and 96π cu. units. If
LA
m.
the curved surface area of the smaller cone is 15π sq. units, what is
6.8
the curved surface area of the larger one?
TE
10.5 SPHERE
T
ER
Figure (i) is a circle. You can easily draw it on a plane paper. Because it is a plane figure.
A circle is plane closed figure whose every point lies at a constant distance (radius) from a fixed
point (centre)
The remaining above figures are solids. These solids are circular in shape and are called
spheres.
A sphere is a three dimensional figure, which is made up of all points in the space, which
is at a constant distance from a fixed point. This fixed point is called centre of the sphere. The
distance from the centre to any point on the surface of the sphere is its radius.
ACTIVITY
Draw a circle on a thick paper and cut it neatly.
Stick a string along its diameter. Hold the both the
ends of the string with hands and rotate with constant
speed and observe the figure so formed.
A
AN
10.5.1 Surface area of a sphere
Let us find the surface area of the figure with
the following activity.
G
Take a tennis ball as shown in the figure and
wind a string around the ball, use pins to keep the
N
string in place. Mark the starting and ending points
of the string. Slowly remove the string from the surface
LA
of the sphere.
Find the radius of the sphere and draw four circles
of radius equal to the radius of the ball as shown in the
TE
pictures. Start filling the circles one after one with the
string you had wound around the ball.
The string, which had completely covered the surface area of the sphere (ball), has been
used to completely fill the area of four circles, all have same radius as of the sphere.
ER
With this we can understand that the surface area of a sphere of radius (r) is equal to the
four times of the area of a circle of radius (r).
∴ Surface area of a sphere = 4 × the area of circle
SC
= 4 πr2
Surface area of a sphere = 4 πr2 TRY THIS
Where ‘r’ is the radius of the sphere Can you find the surface
area of sphere in any other way?
10.5.2 Hemisphere
Take a solid sphere and cut it through the middle with a plane that passes through its
centre.
Then it gets divided into two equal parts as shown in the figure
Each equal part is called a hemisphere.
A sphere has only one curved face. If it is divided into two equal
parts, then its curved face is also divided into two equal curved faces.
A
What do you think about the surface area of a hemisphere ?
Obviously,
AN
Curved surface area of a hemisphere is equal to half the surface area of the sphere
1
So, surface area of a hemisphere = surface area of sphere
2
G
1
= × 4πr2
N 2
= 2 πr2
LA
∴ surface area of a hemisphere = 2 πr2
The base of hemisphere is a circular region.
Its area is equal to = πr2
TE
Let us add both the curved surface area and area of the base, we get total surface area
of hemisphere.
Total surface area of hemisphere = Its curved surface area + area of its base
T
= 2 πr2 + πr2
ER
= 3 πr2.
πr2.
Total surface area of hemisphere = 3π
SC
DO THESE
1. A right circular cylinder just encloses a sphere of radius r (see figure).
Find : (i) surface area of the sphere
(ii) curved surface area of the cylinder
(iii)ratio of the areas obtained in (i) and (ii)
7 cm.
7 cm.
(i) (ii)
A
AN
10.5.3 Volume of Sphere
To find the volume of a sphere, imagine that a sphere is composed of a great number of
congruent pyramids with all their vertices join at the centre of the sphere, as shown in the figure.
N G
r
LA
Base of
Pyramid
3. Consider a part (pyramid) among them. Each pyramid has a base and let the area of the
base of pyramids are A1, A2, A3.....
ER
The height of the pyramid is equal to the radius of sphere, then the
1
Volume of one pyramid = × Area of the base × height
3
SC
1
= Ar you can take any
3 1
polygon as base of a
4. As there are ‘n’ number of pyramids, then pyramid
1 1 1
Volume of ‘n’ pyramids = A1r + A2r + A3r + ..... n times
3 3 3
1
= r [A1 + A2 + A3 + ..... n times]
3
1 A = A1 + A2 + A3 + ..... n times
= ×Ar
3 = Surface areas of ‘n’ pyramids
5. As the sum of volumes of all these pyramids is equal to the volume of sphere and the sum
of the areas of all the bases of the pyramids is very close to the surface area of the sphere,
(i.e. 4πr2).
A
1
So, volume of sphere = (4πr2) r
3
AN
4 3
= πr cub. units
3
4 3
Volume of a sphere = πr
G
3
Where ‘r’ is the radius of the sphere
N
How can you find volume of hemisphere? It is half the volume of sphere.
1
LA
∴ Volume of hemisphere = of volume of a sphere
2
1 4 3
= × πr
2 3
TE
2
= πr 3
3
[Hint : You can try to derive these formulae using water melon or any other like that]
T
DO THIS
r
ER
Example-10. If the surface area of a sphere is 154 cm2, find its radius.
Solution : Surface area of sphere = 4πr2
22 2
4πr2 = 154 ⇒ 4 × × r = 154
7
154 × 7 7 2
⇒r = 2
=
4 × 22 2 2
7
⇒ r = = 3.5 cm
2
Example-11. A hemispherical bowl is made up of stone whose thickness is 5 cm. If the inner
radius is 35 cm, find the total surface area of the bowl.
Solution : Let R be outer radius and ‘r’ be inner radius Thickness of ring = 5 cm
∴ R = (r + 5) cm = (35 + 5) cm = 40 cm
Total Surface Area = CSA of outer hemisphere + CSA of inner hemisphere + area of the ring.
A
= 2πR2 + 2πr2 + π(R2 − r2)
AN
= π(2R2 + 2r2 + R2 − r2)
35 cm. 5 cm.
22 22
= (3R 2 + r 2 ) = (3 × 40 2 + 35 2 ) cm 2
7 7
G
6025 × 22 2
= cm
7
N
= 18935.71 cm2 (approx).
LA
Example-12. The hemispherical dome of a building needs to be painted (see fig 1). If the
circumference of the base of dome is 17.6 m, find the cost of painting it, given the cost of painting
is Rs.5 per 100 cm2.
Solution : Since only the rounded surface of the dome is to be painted we need to find the
TE
curved surface area of the hemisphere to know the extent of painting that needs to be done.
Now, circumference of base of the dome = 17.6 m Therefore 17.6 = 2πr
7
So, The radius of the dome = 17.6 × m
2 × 22
T
= 2.8 m
ER
22
=2× × 2 .8 × 2 .8 m 2
7
SC
= 49.28 m2.
Now, cost of painting 100 cm2 is Rs.5 fig 1
So, cost of painting 1m2 = Rs. 500
Therefore, cost of painting the whole dome
= Rs.500 × 49.28
= Rs. 24640.
Example-13. The hollow sphere, in which the circus motor cyclist performs his stunts, has a
diameter of 7m. Find the area available to the motor cyclist for riding.
Solution : Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding space
available for the motorcyclist is the surface area of the ‘sphere’ which is given by
22
4 πr 2 = 4 × × 3 .5 × 3 .5m 2
A
7
= 154 m2.
AN
Example-14. A shotput is a metallic sphere of radius 4.9 cm. If the density of the metal is
7.8 g. per cm3 , find the mass of the shotput.
G
Solution : Since the shot-put is a solid sphere made of metal and its mass is equal to the product
of its volume and density, we need to find the volume of the sphere.
Example-15. A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water
it would contain ?
Solution : The volume of water the bowl can contains = Volume of hemisphere
SC
2 3
= πr
3
2 22
= × × 3.5 × 3.5 × 3.5 cm 3
3 7
= 89.8 cm3. (approx).
EXERCISE - 10.4
1. The radius of a sphere is 3.5 cm. Find its surface area and volume.
2
2. The surface area of a sphere is 1018 sq.cm. What is its volume?
7
A
3. The length of equator of the globe is 44 cm. Find its surface area.
AN
4. The diameter of a spherical ball is 21 cm. How much leather is required to prepare 5
such balls.
5. The ratio of radii of two spheres is 2 : 3. Find the ratio of their surface areas and volumes.
G
6. Find the total surface area of a hemisphere of radius 10 cm. (use π = 3.14)
7. The diameter of a spherical balloon increases from 14 cm. to 28 cm. as air is being
N
pumped into it. Find the ratio of surface areas of the balloons in the two cases.
LA
8. A hemispherical bowl is made of brass, 0.25 cm. thickness. The inner radius of the bowl
is 5 cm. Find the ratio of outer surface area to inner surface area.
9. The diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34 g/c3. What is
TE
11. How many litres of milk can a hemispherical bowl of diameter 10.5 cm. hold?
ER
12. A hemispherical bowl has diameter 9 cm. The liquid is poured into cylindrical bottles of
diameter 3 cm. and height 3 cm. If a full bowl of liquid is filled in the bottles, find how
many bottles are required.
SC
WHAT
HAT WE HAVE DISCUSSED
HAVE
1. Cuboid and cube are regular prisms having six faces and of which four are
lateral faces and the base and top.
2. If length of cuboid is l, breadth is ‘b’ and height is ‘h’ then,
Total surface area of a cuboid = 2 (lb + bh + lh)
Lateral surface area of a cuboid= 2 h (l + b)
Volume of a cuboid = lbh
A
1
4. The volume of a pyramid is rd volume of a right prism if both have the same base and
3
AN
same height.
5. A cylinder is a solid having two circular ends with a curved surface area. If the line
segment joining the centres of base and top is perpendicular to the base, it is called right
G
circular cylinder.
6. If the radius of right circular cylinder is ‘r’ and height is ‘h’ then;
N
• Curved surface area of a cylinder = 2πrh
LA
• Total surface area of a cylinder = 2πr (r + h)
• Volume of a cylinder = πr2h
TE
7. Cone is a geometrical shaped object with circle as base, having a vertext at the top. If the
line segment joining the vertex to the centre of the base is perpendicular to the base, it is
called right circular cone.
8. The length joining the vertex to any point on the circular base of the cone is called slant
T
height (l)
ER
l2 = h2 + r2
9. If ‘r’ is the radius, ‘h’ is the height, ‘l’ is the slant height of a cone, then
• Curved surface area of a cone = πrl
SC
A
called a hemisphere.
AN
• Curved surface area of a hemisphere = 2πr2
• Total surface area of a hemisphere = 3πr2
2
• Volume of a hemisphere = πr3
3
N G
LA
Do You Know?
Making an 8 × 8 Magic Square
Simply place the numbers from 1 to 64 sequentially in the square grids, as illustrated
TE
on the left. Sketch in the dashed diagonals as indicated. To obtain the magic square below,
replace any number which lands on a dashed line with its compliment (two numbers of a
magic square are compliments if they total the same value as the sum of the magic’s square
smallest and largest numbers).
T
ER
SC
* A magic square is an array of numbers arrange in a square shape in which any row,
column total the same amount. You can try more such magic squares.
244
Areas
11
A
11.1 INTRODUCTION
AN
Have you seen agricultural fields around your village or town? The land is divided amongst
various farmers and there are many fields. Are all the fields of the same shape and same size?
Do they have the same area? If a field has to be further divided among some persons, how will
G
they divide it? If they want equal area,
what can they do?
How does a farmer estimate the N
LA
amount of fertilizer or seed needed for
field? Does the size of the field have
anything to do with this number?
TE
convenience. In history you may have discussed the floods of river Nile (Egypt) and the land
markings generated later . Some of these fields resemble the basic shapes such as square, rectangle
trapezium, parallelograms etc., and some are in irregular shapes. For the basic shapes, we
follow the rules to find areas from given measurements. We would study some of them in this
SC
chapter. We will learn how to calculate areas of triangles, squares, rectangles and quadrilaterals
by using formulae. We will also explore the basis of those formulae. We will discuss how are
they derived ? What do we mean by ‘area’?
You may recall that the part of the plane enclosed by a simple closed figure is called a
planar region corresponding to that figure. The magnitude or measure of this planar region is its
area.
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A
Triangular Quadrilateral Circular Rectangular Square
region region region region region
AN
A planar region consists of a boundary and an interior region. How can we measure the
area of this? The magnitude of measure of these regions (i.e. areas) is always expressed with a
positive real number (in some unit of area) such as 10 cm2, 215 m2, 2 km2, 3 hectares etc. So,
G
we can say that area of a figure is a number (in some unit of area) associated with the part of the
plane enclosed by the figure.
N
The unit area is the area of a square of a side of unit length. Hence square centimeter
(or 1cm2) is the area of a square drawn on a side one centimeter in length.
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TE
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Area = 1 sq. cm 1m
Area = 1 sq. m 1 km
T
Area = 1 sq. km
ER
The terms square meter (1m2), square kilometer (1km2), square millimeter (1mm2) are to
be understood in the same sense. We are familiar with the concept of congruent figures from earlier
classes. Two figures are congruent if they have the same shape and the same size.
SC
ACTIVITY
Observe Figure I and II. Find the
area of both figures. Are the areas equal?
Trace these figures on a sheet of 2.4 cm 2.4 cm
paper, cut them. Cover fig. I with fig. II.
Do they cover each other completely? 6 cm 6 cm
Are they congruent? I II
(i)
A
sheet of paper. Cut them let III IV
us cover fig. III by fig. IV by conciding their bases (length of same side).
AN
As shown in figure V are they covered completely?
We conclude that Figures I and II are congruent and equal
G
in area. But figures III and IV are equal in area but they are
not congruent.
V
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Q 123456789012345678901234567
B
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X Y Z
You may observe that planar region of figures X, Y, Z is made up of two or more planar
T
Thus the area of a figure is a number (in some units) associated with the part of the plane
enclosed by the figure with the following properties.
(Note : We use area of a figure (X) briefly as ar(X) from now onwards)
(i) The areas of two congruent figures are equal.
If A and B are two congruent figures, then ar(A) = ar(B)
(ii) The area of a figure is equal to the sum of the areas of finite number of parts of it.
If a planar region formed by a figure X is made up of two non-overlapping planar regions
formed by figures P and Q then ar(X) = ar(P) + ar(Q).
If the number of units in the length of a rectangle is multiplied by the number of units in its
breadth, the product gives the number of square units in
D C
the area of rectangle
A
Let ABCD represent a rectangle whose length AB
4 Units
is 5 units and breadth BC is 4 units.
AN
Divide AB into 5 equal parts and BC into 4 equal
parts and through the points of division of each line draw
parallels to the other. Each compartment in the rectangle A B
G
5 Units
represents one square unit (why ?)
EXERCISE - 11.1
A
1. In ∆ABC, ∠ABC = 90 o , AD = DC, AB = 12cm
SC
B C
6.5 cm
2. Find the area of a quadrilateral PQRS in which ∠QPS = ∠SQR = 90°, PQ = 12 cm,
PS = 9 cm, QR = 8 cm and SR = 17 cm (Hint: PQRS has two parts)
S 17 cm
R
9 cm
A
8 cm
AN
P 12 cm Q
3. Find the area of trapezium ABCD as given in the figure in which ADCE is a rectangle.
(Hint: ABCD has two parts)
B
G
3 cm
N
3 cm
E C
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A D
8 cm
4. ABCD is a parallelogram. The D
TE
C
diagonals AC and BD intersect
O
each other at ‘O’. Prove that
ar(∆AOD) = ar(∆BOC). (Hint:
Congruent figures have equal A B
T
area)
ER
We shall now study some relationships between the areas of some geometrical figures
SC
under the condition that they lie on the same base and between the same parallels. This study will
also be useful in understanding of some results on similarity of triangles.
Look at the following figures.
A B P Q A DA B
E F
T U P
D C S R B CD C
(i) (ii) (iii) (iv)
In Fig(i) a trapezium ABCD and parallelogram EFCD have a common side CD. We say
that trapezium ABCD and parallelogram EFCD are on the same base CD. Similarly in fig(ii) the
base of parallelogram PQRS and parallelogram TURS is the same. In fig(iii) Triangles ABC and
DBC have the same base BC. In Fig(iv) parallelogram ABCD and triangle PCD lie on DC so, all
these figures are of geometrical shapes are therefore on the same base. They are however not
A
between the same parallels as AB does not overlap EF and PQ does not overlap TU etc.
Neither the points A, B, E, F are collinear nor the points P, Q, T, U. What can you say about
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Fig(iii) and Fig (iv)?
Now observe the following figures.
A1 B1 E1 F1 P1 T1 Q1 U1 A1 D1 A1 P1 B1
D1 C1 S1 N R1
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B1 C1 D
1
C1
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(v) (vi) (vii) (viii)
What difference have you observed among the figures? In Fig(v), We say that trapezium
A1B1C1D1 and parallelogram E1F1C1D1 are on the same base and between the same parallels
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A1F1 and D1C1. The points A1, B1, E1, F1 are collinear and A1F1 || D1C1. Similarly in fig. (vi)
parallelograms P1Q1R1S1 and T1U1R1S1 are on the same base S1R1 and between the same
parallels P1U1 and S1R1. Name the other figures on the same base and the parallels between
which they lie in fig. (vii) and (viii).
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So, two figures are said to be on the same base and between the same parallels, if they
have a common base (side) and the vertices (or the vertex) opposite to the common base of each
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Which of the following figures lie on the same base and between the same
parallels?
In such a cases, write the common base and the two parallels.
A P B P Q P Q A B P A B Q
M N
T
P
D C S R S R D C S D C R
(a) (b) (c) (d) (e)
11.5 PARALLELOGRAMS
ARALLELOGRAMS ON THE SAME BASE AND BETWEEN THE
SAME PARALLELS
ARALLELS
Now let us try to find a relation, if any, between the areas of two parallelograms on the
same base and between the same parallels. For this, let us perform the following activity.
A
ACTIVITY
AN
Take a graph sheet and draw two parallelograms ABCD and PQCD on it as
show in the Figure-
The parallelograms are on the same base DC
and between the same parallels PB and DC Clearly
G
the part DCQA is common between the two
parallelograms. So if we can show that
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∆DAP and ∆CBQ have the same area then we can
say ar(PQCD) = ar(ABCD).
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Theorem-11.1 : Parallelograms on the same base and between the same parallels are equal in
area.
Proof: Let ABCD and PQCD are two parallelograms on the samebase DC and between the
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in area. Thus the parallelogram with the equal base can be considered to be on the same base for
the purposes of studying their geometrical properties.
Let us now take some examples to illustrate the use of the above Theorem.
Example-1. ABCD is parallelogram and ABEF is a rectangle and DG is perpendicular on AB.
Prove that (i) ar (ABCD) = ar(ABEF) F D E C
A
(ii) ar (ABCD) = AB × DG
Solution : (i) A rectangle is also a parallelogram
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∴ ar(ABCD) = ar(ABEF) ..... (1) A G B
(Parallelograms lie on the same base and between the same parallels)
(ii) ar(ABCD) = ar(ABEF) (∵ from (1))
G
= AB × BE (∵ ABEF is a rectangle)
= AB × DG (∵ DG ⊥ AB and DG = BE )
Therefore ar(ABCD) = AB × DG
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From the result, we can say that “area of a parallelogram is the product of its any side and
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the corresponding altitude”.
Example-2. Triangle ABC and parallelogram ABEF are on the same base, AB as in between
1
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the same parallels AB and EF. Prove that ar(∆ABC) = ar(|| gm ABEF)
2
Solution : Through B draw BH || AC to meet FE produced at H
F C E
∴ ABHC is a parallelogram H
Diagonal BC divides it into two congruent triangles
∴ ar(∆ABC) = ar(∆BCH)
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1
= ar (|| gm ABHC)
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2 A B
But || gm ABHC and || gm ABEF are on the same base AB and between same parallels
AB and EF
∴ ar(|| gm ABHC) = ar(|| gm ABEF)
SC
1
Hence ar(∆ABC) = ar (|| gm ABEF)
2
From the result, we say that “the area of a triangle is equal to half the area of the
parallelogram on the same base and between the same parallels”.
Example-3. Find the area of a figure formed by joining the mid-points of the adjacent sides of
a rhombus with diagonals 12 cm. and 16 cm.
Solution : Join the mid points of AB, BC, CD, DA of a rhombus ABCD and name them M, N,
O and P respectively to form a figure MNOP.
What is the shape of MNOP thus formed? Give reasons?
A
1
∴ ar ∆MNP= ar ABPN .....(i)
2
1
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Similarly ar ∆PON = ar PNCD .....(ii) P N
2
1
and Area of rhombus = × d1d 2 .....(iii)
2
D O C
From (1), (ii) and (iii) we get
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ar(MNOP) = ar(∆MNP) + ar(∆PON)
1 1
=
2
1
2
= ar(rhombus ABCD)
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ar(ABNP) + ar(PDCN)
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2
11
= × 12 × 16 = 48 cm.2
2 2
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EXERCISE - 11.2
D C F E
2
1. The area of parallelogram ABCD is 36 cm .
Calculate the height of parallelogram ABEF
T
if AB = 4.2 cm.
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A B
A B 2. ABCD is a parallelogram. AE is perpendicular
on DC and CF is perpendicular on AD.
F
If AB = 10 cm, AE = 8 cm and CF = 12 cm.
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D C Find AD.
E
A E B
3. If E, F G and H are respectively the
midpoints of the sides AB, BC, CD and AD
H F
of a parallelogram ABCD, show that
1
ar(EFGH) = ar(ABCD) . D C
2 G
4. What type of quadrilateral do you get, if you join ∆APM, ∆DPO, ∆OCN and ∆MNB in
the example 3.
D P C
A
6. P is a point in the interior of a parallelogram
ABCD. Show that A B
AN
1
(i) ar(∆APB) + ar(∆PCD) = ar(ABCD)
2 P
(ii) ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)
D
G
(Hint : Through P, draw a line parallel to AB) C
7. Prove that the area of a trapezium is half the sum of the parallel sides multiplied by the
distance between them.
P A Q
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B 8. PQRS and ABRS are parallelograms and X is
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any point on the side BR. Show that
X
(i) ar(PQRS) = ar(ABRS)
1
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We are looking at figures that lie on the same base and between the
same parallels. Let us have two triangles ABC and DBC on the same base A D
BC and between the same parallels, AD and BC.
What can we say about the areas of such triangles? Clearly there
can be infinite number of ways in which such pairs of triangle between the B C
same parallels and on the same base can be drawn.
ACTIVITY
Draw pairs of triangles one the same base or ( equal bases) and between the
same parallels on the graph sheet as shown in the Figure.
A
Let ∆ABC and ∆DBC be the two triangles lying on the same base BC and between
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parallels BC and AD. Extend AD on either sides and draw CE || AB and BF || CD.
Parallelograms AECB and FDCB are on the same base BC and are between the same
parallels BC and EF.
Thus ar(AECB) = ar(FDCB). (How ?)
F A D E
G
1
We can see ar(∆ABC) = ar(Parallelogram AECB) ...(i)
2 P
1
and ar(∆DBC) = ar(Parallelogram FDCB)
2 N
From (i) and (ii), we get ar(∆ABC) = ar(∆DBC).
.... (ii)(
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You can also find the areas of ∆ABC and ∆DBC by the method of B counting the squares
C
in graph sheet as we have done in the earlier activity and check the areas are whether same.
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F A D E A D
T
ER
B C B C B C
Corollary-1 : Show that the area of a triangle is half the product of its base (or any side) and the
corresponding attitude (height).
Proof : Let ABC be a triangle. Draw AD || BC such that CD =BA.
Now ABCD is a parallelogram one of whose diagonals is AC.
We know∆ABC ≅ ∆ACD
A
So ar∆ABC = ar∆ACD (Congruent triangles have equal area)
1 A D
Therefore, ar∆ABC =
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ar(ABCD)
2
Draw AE ⊥ BC
We know ar(ABCD) = BC × AE
1 1 B C
G
We have ar(∆ABC) = ar(ABCD) = × BC × AE E
2 2
1
So ar∆ABC = × base BC × Corresponding attitude AE.
2
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Theorem-11.2 : Two triangles having the same base (or equal bases) and equal areas will lie
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between the same parallels. A D
Observe the figure. Name the triangles lying on the same
base BC. What are the heights of ∆ABC and ∆DBC?
If two triangles have equal area and equal base, what
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Example 4. Show that the median of a triangle divides it into two triangles of equal areas.
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Draw AE ⊥ BC.
1
Now ar ( ∆ABD) = × base BD × altitude of ∆ADB
2
A
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1
= × BD × AE
2
1
= × DC × AE (∵ BD = DC)
2
1
= × base DC × altitude of ∆ACD
2
B E D C
= ar ∆ACD
Hence ar (∆ABD) = ar (∆ACD)
Example-5. In the figure, ABCD is a quadrilateral. AC is the diagonal and DE || AC and also
DE meets BC produced at E. Show that ar(ABCD) = ar(∆ABE).
Solution : ar(ABCD) = ar(∆ABC) + ar(∆DAC) D
A
∆DAC and ∆EAC lie on the same base AC
A
and between the parallels DE || AC
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ar(∆DAC) = ar(∆EAC) (Why?) B C E
G
Hence ar(ABCD) = ar(∆ABE)
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Example 6. In the figure , AP || BQ || CR. Prove that ar(∆AQC) = ar(∆PBR).
Similarly
ar (∆CQB) = ar (∆RQB) (same base BQ and BQ || CR) ...(2)
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EXERCISE - 11.3
SC
A
1. In a triangle ABC (see figure), E is the
midpoint of median AD, show that
E
(i) ar ∆ABE = ar ∆ACE
1
(ii) ar∆ABE = ar (∆ABC) B D C
4
2. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
C
3. In the figure, ∆ABC and ∆ABD are two
triangles on the same base AB. If line
A O
segment CD is bisected by AB at O, show
B
that
A
ar (∆ABC) = ar (∆ABD). D
AN
A
4. In the figure, ∆ABC, D, E, F are the midpoints of
sides BC, CA and AB respectively. Show that
(i) BDEF is a parallelogram
G
F E
1
(ii) ar ( ∆DEF) = ar (∆ABC)
B
D
C N(iii) ar (BDEF) =
4
1
ar (∆ABC)
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2
A
5. In the figure D, E are points on
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B C
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D C
7. In the figure, diagonals AC and BD of a O
A
P Q
9. In the figure, if ar ∆RAS = ar ∆RBS and [ar (∆QRB) =
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ar(∆PAS) then show that both the quadrilaterals PQSR R S
and RSBA are trapeziums.
G
quadrilateral. The grampanchayat of the village decided to take over some portion of his
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plot from one of the corners to construct a school. Ramayya agrees to the above proposal
with the condition that he should be given equal amount of land in exchange of his land
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adjoining his plot so as to form a triangular plot. Explain how this proposal will be
implemented. (Draw a rough sketch of plot).
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G
Show that
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B Y C
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ar (ACFG)
D X E
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
Can you write the result (vii) in words ? This is a famous theorem of Pythagoras. You
shall learn a simpler proof in this theorem in class X.
WHAT
HAT WE HAVE DISCUSSED
HAVE
A
3. If X is a planer region formed by two non-overlapping planer regions of figures P and
Q, then ar(X) = ar(P) + ar(Q)
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4. Two figures are said to be on the same base and between the same parallels, if they
have a common base (side) and the vertices (on the vertex) opposite to the common
base of each figure lie on a line parallel to the base.
5. Parallelograms on the same base (or equal bases) and between the same parallels are
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equal in area.
6. Area of a parallelogram is the product of its base and the corresponding altitude.
8. If a parallelogram and a triangle are on the same base and between the same parallels,
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then area of the triangle is half the area of the parallelogram.
9. Triangles on the same base (or equal bases) and between the same parallels are equal
in area.
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10. Triangles on the same base (or equal bases) and having equal areas lie between the
same parallels.
DO YOU KNOW?
A PUZZLE (AREAS)
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German mathematician David Hilbert (1862-1943) first proved that any polygon can
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be transformed into any other polygon of equal area by cutting it into a finite number of
pieces.
Let us see how an English puzzlist, Henry Ernest Dudency (1847 - 1930) transforms an
equilateral triangle into a square by cutting it into four pieces.
SC
3 4
3 4
3
4 2 2 1
2
1
1
Try to make some more puzzles using his ideas and enjoy.
Circles
12
A
12.1 I NTRODUCTION
AN
We come across many round shaped objects
in our surroundings such as coins, bangles, clocks,
wheels, buttons etc. All these are circular in shape.
G
You might have drawn an outline along the
edges of a coin, a bangle, a button in your childhood to form a circle.
N
So, can you tell, the difference between the circular objects and the
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circles you have drawn with the help of these objects?
All the circular objects we have observed above have thickness and are
3-dimensional objects, where as, a circle is a 2-dimensional figure, with no
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thickness.
which is the centre of the circle. The length of the fulcrum with
reference to the circle is radius of the circle. Think of some
other examples from your daily life about circles.
In this chapter we will study circles, related terms and properties of the circle. Before
this, you must know how to draw a circle with the help of a compass.
Let us do this.
A
shown in the figure.
If we need to draw a circle of given radius, we do this with the help of a scale.
AN
Adjust the distance between the sharp point of the compass and tip of the pencil equal to
the length of the given radius, mark a point ‘O’(radius of the circle in the figure is 5 cm.) and
draw circle as described above.
C
G
B
5 cm.
N A
O D
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
SCALE E
F
Mark any 6 points A, B, C, D E and F on the circle. You can see that the length of each
line segment OA, OB, OC, OD, OE and OF is 5 cm., which is equal to the given radius. Mark
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some other points on the circle and measure their distances from the point ‘O’. What have you
observed? We can say that a circle is a collection of all the points in a plane which are at a fixed
distance from a fixed point on the plane.
The fixed point ‘O’ is called the centre of the circle and the fixed Start
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the park and completed one round. What do you call the distance covered
by Narsimha? It is the total length of the boundary of the circular park,
and is called the circumference of the park.
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ACTIVITY
Let us now do the following activity. Mark a point on a sheet of
paper. Taking this point as centre draw a circle with any radius. Now
increase or decrease the radius and again draw some more circles
with the same centre. What do you call the circles obtained in this
activity?
Circles having a common centre are called concentric circles.
DO THIS
B A
1. In the figure which circles are congruent to the circle A. D
2. What measure of the circles make them congruent? C E
A
A circle divides the plane on which it lies into three parts. They
AN
X
P Q are (i) inside the circle, which is also called interior of the circle; (ii) on
O T the circle, this is also called the circumference and (iii) outside the circle,
W
which is also called the exterior of the circle. From the above figure,
V
find the points which are inside, outside and on the circle.
G
S
U R
ACTIVITY N
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Take a thin circular sheet and fold it to half and open. Again fold it along any
other half and open. Repeat this activity for several times. Finally when you open it,
what do you observe?
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You observe that all creases (traces of the folds) are intersecting at one point. Do you
remember what do we call this point? This is the centre of the circle.
Measure the length of each crease of a circle with a divider. What do you notice ? They
are all equal and each crease is dividing the circle into two equal halves. That crease is called
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diameter of circle. Diameter of a circle is twice its radius. A line segment joining any two points
on the circle that passes through the centre is called the diameter.
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In the above activity if we fold the paper in any manner not only in half, we see that creases
joining two points on circle. These creases are called chords of the circle.
C
So, a line segment joining any two points on the circle is called D
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a chord.
What do you call the longest chord? Is it passes through the centre? A B
rc
mi
a semicircular arc or a semicircle. In the fig.(ii) ACB
major a
nor
Central
is a semicircle C
arc
C O O Angle
A
B
a semicircle, then the arc is called a major arc. In (ii) (iii)
is a minor arc and ADB
the fig.(iii) ACB is a major arc.
AN
If we join the end points of an arc by a chord,
A A
the chord divides the circle into two parts. The region
Major segment
Minor segment
between the chord and the minor arc is called the minor
se
G
C segment and the region between the chord and the
mi
cir
C happens to be a diameter,
B
D
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then the diameter divides the
r
o
B r
or sect
minor sect
circle into two equal segments.
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t a
m aj
The area enclosed by an arc and the two radii joining the centre to r or
the end points of an arc is called a sector. One is minor sector and
another is major sector (see adjacent figure).
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E XERCISE -12.1
1. Name the following parts from the adjacent figure where ‘O’ is the
centre of the circle.
T
C
(i) AO (ii) AB
(iii) BC
X
ER
X
i. A circle divides the plane on which it lies into three parts. ( )
ii. The area enclosed by a chord and the minor arc is minor segment. ( )
iii. The area enclosed by a chord and the major arc is major segment. ( )
iv. A diameter divides the circle into two unequal parts. ( )
v. A sector is the area enclosed by two radii and a chord ( )
vi. The longest of all chords of a circle is called a diameter. ( )
vii. The mid point of any diameter of a circle is the centre. ( )
mi
c
major ar
nor arc
AO and BO. Angle is made at centre ‘O’ by AO , BO i.e. ∠AOB is O Central
Angle
A
called the angle subtended by the chord AB at the centre ‘O’.
B
What do you call the angles ∠POQ, ∠PSQ and ∠PRQ in the D
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figure?
i. ∠POQ is the angle subtended by the chord PQ at the centre ‘O’ R
ii. ∠PSQ and ∠PRQ are respectively the angles subtended by the
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chord PQ at point S and point R on the minor and major arc. O
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In the figure, O is the centre of the circle and
AB, CD, EF and GH are the chords of the P
S
Q
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circle.
O
H
We can observe from the figure that GH > EF > CD > AB.
G
E F
C D Now what do you say about the angles subtended by these chords at
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A B the centre?
After observing the angles, you will find that the angles subtended by the chords at the
centre of the circle increases with increase in the length of chords.
T
So, now think what will happen to the angle subtended at the centre of the circle, if we take
two equal chords of a circle?
ER
Construct a circle with centre ‘O’ and draw equal chords AB and CD using the compass
and ruler.
Join the centre ‘O’ with A, B and with C, D. Now measure the C
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A
angles ∠AOB and ∠COD. Are they equal to each other? Draw two or O
more equal chords of a circle and measure the angles subtended by them
at the centre. D
B
You will find that the angles subtended by them at the centre are
equal.
Given : Let ‘O’ be the centre of the circle. AB and CD are two equal chords and
∠AOB and ∠COD are the angles subtended by the chords at the centre.
R.T.P. : ∠AOB ≅ ∠COD
A
Construction : Join the centre to the end points of each chord and you get two triangles ∆AOB
and ∆COD.
AN
Proof: In triangles AOB and COD C
A
AB = CD (given) O
G
OB = OD (radii of same circle) D
B
Therefore ∆ AOB ≅ ∆ COD (SSS rule)
N
Thus ∠AOB ≅ ∠COD (corresponding parts of congruent triangles)
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In the above theorem, if in a circle, two chords subtend equal angles at the centre, what
can you say about the chords? Let us investigate this by the following activity.
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ACTIVITY
Take a circular paper. Fold it along any diameter such that the two edges
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coincide with each other. Now open it and again fold it into half along another diameter.
On opening, we find two diameters meet at the
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A 1 C A C
Now take cut-out of the four segments namely
1, 2, 3 and 4
If you place these segments pair wise one above the other the edges of the pairs (1,3)
and (2,4) coincide with each other.
Is AD = BC and AC = BD ?
Though you have seen it in this particular case, try it out for other equal angles too. The
chords will all turn out to be equal because of the following theorem.
A
Note that in given theorem ∠PQR = ∠MQN, then Q
R
∆PQR ≅ ∆MQN (Why?)
AN
Is PR = MN? (Verify) P
EXERCISE - 12.2
G
C
1. In the figure, if AB =CD and ∠AOB = 900 find ∠COD
A
P
N O
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D
Q B
2. In the figure, PQ = RS and ∠ΟRS = 48 . 0
O
S Find ∠OPQ and ∠ROS.
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P S
R
O
3. In the figure PR and QS are two diameters. Is PQ = RS?
Q R
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• Construct a circle with centre O. Draw a chord AB and a perpendicular to the chord
AB from the centre ‘O’.
• Let the point of intersection of the perpendicular on AB be P.
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ACTIVITY
Take a circle shaped paper and mark centre ‘O’
Fold it into two unequal parts and open
it. Let the crease represent a chord AB,
and then make a fold such that ‘A’ coincides
A
with B. Mark the point of intersection of O O
the two folds as D. Is AD = DB?
AN
∠ ODA = ? ∠ ODB = ? Measure the A B
angles between the creases. They are right
angles. So, we can make a hypothesis “the
line drawn through the centre of a circle to bisect a chord is perpendicular to the chord”.
G
TRY THIS
circles. If ‘P’ is a point other than the centre of the circle, then
also we can draw many circles through P.
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circles with centre R, R1, R2 and radii RP, R1P and R2P respectively. Does
these circles also passes through Q (Why?)
If three non-collinear points are given, then how many circles can
C
A
be drawn through them? Let us examine it. Take any three non-
collinear points A, B, C and join AB and BC.
B
Draw PQ and RS the perpendicular bisectors to AB and R O P
C
BC . respectively. Both of them intersect at a point ‘O’(since two A
lines cannot have more than one point in common)
Q B S
Now O lies on the perpendicular bisector of AB , so OA = OB. .....(i)
A
As every point on PQ is at equidistant from A and B
Also, ‘O’ lies on the perpendicular bisectors of BC
R O P
AN
Therefore OB = OC ..... (ii) C
A
From equation (i) and (ii)
Q B S
We can say that OA = OB = OC (transitive law)
G
Therefore, ’O’ is the only point which is equidistant from the points A, B and C so if we
draw a circle with centre O and radius OA, it will also pass through B and C i.e. we have only
one circle that passes through A, B and C.
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The hypothesis based on above observation is “there is one and only one circle that passes
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through three non-collinear points”
Note : If we join AC, the triangle ABC is formed. All its vertices lie on the circle. This circle is
called circum circle of the triangle, the centre of the circle ‘O’ is circumcentre and the
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TRY THIS
If three points are collinear, how many circles can be drawn through these
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points? Now, try to draw a circle passing through these three points.
Example-1. Construct a circumcircle of the triangle ABC where AB = 5cm; ∠B = 750 and
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BC = 7cm
X C
Solution : Draw a line segment AB= 5 cm. Draw BX at B such that
∠B = 750. Draw an arc of radius 7cm with centre B to cut BX at C S
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P
join CA to form ∆ABC, Draw perpendicular bisectors PQ and RS
7c
m.
O
to AB and BC respectively. PQ , RS intersect at ‘O’. Keeping ‘O’ R
75°
as a centre, draw a circle with OA as radius. The circle also passes A 5 cm. B
through B and C and this is the required circumcircle. Q
12.3.2 Chords and their distance from the centre of the circle
A circle can have infinite chords. Suppose we make many chords of equal length in a
circle, then what would be the distance of these chords of equal length from the centre? Let us
check it through this activity.
ACTIVITY
Draw a big circle on a paper and take a
cut-out of it. Mark its centre as ‘O’. Fold it in D D
C
half. Now make another fold near semi-circular C
O
A
edge. Now unfold it. You will get two conguent O
folds of chords. Name them as AB and CD.
AN
Now make perpendicular folds passing through A B A B
centre ‘O’ for them. Using divider compare the
perpendicular distances of these chords from the centre.
Repeat the above activity by folding congruent chords. State your observations as a
G
hypothesis.
“The congruent chords in a circle are at equal distance from the centre of the circle”
TRY THIS
N
LA
In the figure, O is the centre of the circle and AB = CD. OM is C
B
perpendicular on AB and ON is perpendicular on CD . Then prove N
that OM = ON. M
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O
As the above hypothesis has been proved logically, it becomes a D
theorem ‘chords of equal length are at equal distance from the centre A
of the circle.’
T
Example-2. In the figure, O is the centre of the circle. Find the length of CD, if AB = 5 cm.
Solution : In ∆ AOB and ∆ COD, A C
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OA = OC (why?)
m.
OB = OD (why?) °
5c
55° 55
∠AOB = ∠COD B O
D
∴ ∆ AOB ≅ ∆ COD
SC
R.T.P. : AB = CD
Proof : AD is the chord of the bigger circle with centre ‘O’ and OE is perpendicular to AD .
∵ OE bisects AD (The perpendicular from the centre of a circle to a chord bisect it)
A
∴ AE = ED ..... (i)
AN
BC is the chord of the smaller circle with centre ‘O’ and OE is perpendicular to AD.
∵ OE bisects BC (from the same theorem)
∴ BE = CE ..... (ii)
Subtracting the equation (ii) from (i), we get
G
AE - BE = ED - EC
AB = CD
EXERCISE - 12.3
N
LA
1. Draw the following triangles and construct circumcircles for them.
3. If two circles intersect at two points, then prove that their centres
ER
A
lie on the perpendicular bisector of the common chord.
C L
4. If two intersecting chords of a circle make equal angles with E
P O Q
diameter passing through their point of intersection, prove that B M
SC
A B
A
at the centre ‘O’ is the same as the angle subtended D C
A B
by the arc at the centre ‘O’.
AN
(i)
In fig.(ii) AB and CD are two chords of O
G
(ii)
at the centre ‘O’. (Prove
equal to the angle subtended by the arc CD
∆AOB ≅ ∆DOC)
N
From the above observations we can conclude that “Arcs of equal length subtend equal
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angles at the centre”
R R
Consider the circle with centre ‘O’.
in fig. (i) the minor arc, in fig. (ii)
Let PQ
O
P Q
O semicircle and in fig. (iii) major arc.
Take any point R on the circumference. Join R
T
P
Q
with P and Q.
(i) (ii)
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∠PRQ
∠POQ
(iii)
Similarly draw some circles and subtended angles on the circumference and centre of the
circle by their arcs. What do you notice? Can you make a conjecture about the angle made by an
arc at the centre and a point on the circle? So from the above observations, we can say that
“The angle subtended by an arc at the centre ‘O’ is twice the angle subtended by it on the
remaining arc of the circle”.
A
S S
P S Q
Q
AN
Given : Let O be the centre of the circle.
G
)
Let R be a point on the remaining part of the circle (not on PQ
Let us begin by joining the point R with the centre ‘O’ and extend it to a point S (in all
cases)
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For convenience
∠POS is an exterior angle of ∆ ROP
Let ∠ORP = ∠OPR = x
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(construction)
∠POS = ∠1
∠POS = ∠ORP + ∠OPR or 2 ∠ORP ..... (1) ∠1 = x + x = 2x
(∵ exterior angle = sum of opp. interior angles) Let ∠ORQ = ∠OQR = y
SC
Hence the theorem is “the angle subtended by an arc at the centre is twice the angle
subtended by it at any point on the remaining part of the circle.
Example-4. Let ‘O’ be the centre of a circle, PQ is a diameter, then prove that ∠PRQ = 90o
(OR) Prove that angle subtended by semi-circle is 90o.
R
A
Solution : It is given that PQ is a diameter and ‘O’ is the centre of the
circle.
AN
P Q
∴ ∠POQ = 180 [Angle on a straight line]
o
O
G
the angle subtended by it at any other point on circle]
180o
∴ ∠PRQ =
2
= 90o
N
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Example-5. Find the value of x° in the adjacent figure
Solution : Given ∠ACB = 40°
C
By the theorem angle made by the arc AB at the centre 40
o
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Let us now discuss the measures of angles made by an arc in the same segment of a circle.
Consider a circle with centre ‘O’ and a minor arc AB (See figure). Let P, Q, R and S be
points on the major arc AB i.e. on the remaining part of the circle. Now join the end points of the
SC
arc AB with points P, Q, R and S to form angles ∠APB, ∠AQB, ∠ARB and ∠ASB.
A
Therefore ∠APB = ∠AQB =∠ARB = ∠ASB C
Observe that “angles subtended by an arc in the same segment are equal”.
Note : In the above discussion we have seen that the point P, Q, R, S and A, B lie on the same
circle. What do you call them? “Points lying on the same circle are called concyclic”.
Theorem-12.4 : If a line segment joining two points, subtends equal angles at two other points
A
lying on the same side of the line then these, the four points lie on a circle ( i.e. they are concyclic)
AN
You can see the truth of this result as follows:
Given : Two angles ∠ACB and ∠ADB are on the same side of a line segment AB joining two
points A and B are equal.
G
R.T.P : A, B, C and D are concyclic (i.e.) they lie on the same circle.
Construction : Draw a circle passing through the three non colinear point A, B and C.
N
Proof: Suppose the point ‘D’ does not lie on the Circle.
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Then there may be other point ‘E’ such that it will intersect AD (or extension of AD)
D
If points A, B, C and E lie on the circle then
C E C E
∠ACB = ∠AEB
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(Why?) D
EXERCISE – 12.4
C
1. In the figure, ‘O’ is the centre of the circle.
∠AOB = 100° find ∠ADB.
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100°
A B
B
D
A 40°
O 2. In the figure, ∠BAD = 40° then find ∠BCD.
D
3. In the figure, O is the centre of the circle and ∠POR = 120°. Find Q
A
5. In the figure, ‘O’ is the centre of the circle.
S
O OM = 3cm and AB = 8cm. Find the radius of
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the circle
A M B S
G
and OM, ON are the perpendiculars from R
O
the centre to the chords PQ and RS. If
D C N
OM = ON and PQ = 6cm. Find RS. P M Q
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A 7. A is the centre of the circle and ABCD is a square. If BD = 4cm then
B
find the radius of the circle.
D
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ADRILAATERAL D
C
In the figure, the vertices of the quadrilateral A, B, C and D lie on the
same circle, this type of quadrilateral ABCD is called cyclic quadrilateral. A
B
SC
ACTIVITY
Take a circle paper. Mark four points A, B, C
D
and D on the circle paper. Draw cyclic quadrilateral C
A
4
AN
What do you infer from the table?
G
Given : A cyclic quadrilateral ABCD. D
To Prove : ∠A + ∠C = 180° x
∠B + ∠D = 180° N A
O
y
C
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Construction : Join OA, OC
B
1
Proof: ∠D = ∠y (Why?) ..... (i)
2
1
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1
∠D + ∠B = (∠y + ∠x)
ER
2
1
∠B + ∠D = × 360°
2
∠B + ∠D = 180°
SC
Similarly ∠A + ∠C = 180°
A
Theorem-12.6 : If the sum of any pair of opposite angles in a quadrilateral is 180o, then it is
cyclic.
AN
D
E
Given : Let ABCD be a quadrilateral D
E C
such that C
(i) (ii)
∠ABC + ∠ADC = 180o
G
A B A B
∠DAB + ∠BCD = 180 o
But one of these is an exterior angle of ∆ADE and the other is an interior opposite angle.
ER
We know that the exterior angle of a triangle is always greater than either of the opposite interior
angles.
∴ ∠AEC = ∠ADC is a contradiction.
So our assumption that the circle passing through A, B and C does not pass through D is false.
SC
1
Now, ∠CBD = ∠COD (Why?) C
D
2
A
O
Again, ∠ACB = 90° (Why?)
AN
So, ∠BCE = 180° - ∠ACB = 90°
G
EXERCISE 12.5
1. Find the values of x and y in the figures given below.
30o
N
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xo 50o
yo O
x
o o
y 110
o yo
85 xo
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4. For each of the following, draw a circle and inscribe the figure given. If a polygon of the
given type can’t be inscribed, write not possible.
(a) Rectangle
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(b) Trapezium
(c) Obtuse triangle
(d) Non-rectangular parallelogram
(e) Accute isosceles triangle
WHAT
HAT WE HAVE DISCUSSED
HAVE
• A collection of all points in a plane which are at a fixed distance from a fixed point in the
same plane is called a circle. The fixed point is called the centre and the fixed distance is
called the radius of the circle
A
• A line segment joining any points on the circle is called a chord
• The longest of all chords which also passes through the centre is called a diameter
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• Circles with same radii are called congruent circles
• Circles with same centre and different radii are called concentric circles
• Diameter of a circle divides it into two semi-circles
G
• The part between any two points on the circle is called an arc
• The area enclosed by a chord and an arc is called a segment. If the arc is a minor arc then
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it is called the minor segment and if the arc is major arc then it is called the major segment
• The area enclosed by an arc and the two radii joining the end points of the arc with centre
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is called a sector
• Equal chords of a circle subtend equal angles at the centre
• Angles in the same segment are equal
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• The perpendicular from the centre of a circle to a chord bisects the chords. The converse
is also true
ER
A
13.1 I NTRODUCTION
AN
To construct geometrical figures, such as a line segment, an angle, a triangle, a quadrilateral
etc., some basic geometrical instruments are needed. You must be having a geometry box which
contains a graduated ruler (Scale) a pair of set squares, a divider, a compass and a protractor.
G
Generally, all these instruments are needed in drawing. A geometrical construction is the
N
process of drawing a geometrical figure using only two instruments - an ungraduated ruler and
a compass. We have mostly used ruler and compass in the construction of triangles and quadrilaterals
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in the earlier classes. In construction where some other instruments are also required, you may
use a graduated scale and protractor as well. There are some constructions that cannot be done
straight away. For example, when there are 3 measures available for the triangle, they may not be
used directly. We will see in this chapter, how to extract the needed values and complete the
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required shape.
bisector of 30°, 45°, 60°, 90° and 120° or of a given angle, in the lower classes. However the
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reason for these constructions were not discussed. The objective of this chapter is to give the
process of necessary logical proofs to all those constructions.
segment.
P
Example-1. Draw the perpendicular bisector of a given line
segment AB and write justification.
Solution : Steps of construction.
A O B
Steps 1 : Draw the line segment AB
1
Step 2 : Taking A centre and with radius more than , draw
2 AB
an arc on either side of the line segment AB. Q
Step 3 : Taking ‘B’ as centre, with the same radius as above, draw arcs so that they intersect the
previously drawn arcs.
Step 4 : Mark these points of intersection as P and Q.
Join P and Q.
A
Thus the line POQ is the required perpendicular bisector of AB.
AN
How can you prove the above construction i.e. “PQ is the perpendicular bisector of AB”,
logically?
Draw diagram of construction and join A to P and Q; also B to P and Q.
G
We use the congruency of triangle properties to prove the required.
Proof :
Steps
In ∆s PAQ and ∆PBQ
N Reasons
P
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Selected
AP = BP ; AQ = BQ Equal radii A O B
PQ = PQ Common side
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Proved above
OP = OP Common
∴ ∆APO ≅ ∆BPO SAS rule
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180°
We get ∠AOP = ∠BOP = = 90° From the above result
2
Thus PO, i.e. POQ is the perpendicular Required to prove.
bisector of AB
A
A
Step 2 : Taking B as centre and with any radius, draw an arc to
intersect the rays BA and BC , at D and E respectively, as shown in D
AN
the figure.
B E C
A
Step 3 : Taking E and D as centres draw two arcs with equal radii to
G
D intersect each other at F.
A
B E C
N
Step 4 : Draw the ray BF. It is the required bisector of ∠ABC .
D F
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B E C
Let us see the logical proof of above construction. Join D, F and E, F. We use congruency
rule of triangles to prove the required.
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Proof :
Steps Reasons
In ∆s BDF and ∆BEF Selected triangles
A
T
D F
DF = EF Arcs of equal radii
B E C
BF = BF Common
SC
∠ABC
TRY THESE
Observe the sides, angles and diagonals of quadrilateral BEFD. Name the figures given
below and write properties of figures.
A A
F
1. 2.
A
D F D
AN
B C B C
E E
G
ray.
Example-3. Draw a ray AB with initial point A and construct a ray AC such that ∠BAC = 60°.
Solution : Steps of Construction N
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Step 1 : Draw the given ray AB and taking A as centre and some
radius, draw an arc which intersects AB, say at a point D.
A B
D
E
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TRY THIS
Draw a circle, Identify a point on it. Cut arcs on the circle
with the length of the radius in succession. How many parts can
the circle be divided into? Give reason.
A
EXERCISE - 13.1
AN
1. Construct the following angles at the initial point of a given ray and justify the
construction.
(a) 90o (b) 45o
G
2. Construct the following angles using ruler and compass and verify by measuring them by a
protractor.
(a)
(d)
30o
75o
(b) 22 2
(e) 105o
1o
N (c) 15o
f) 135o
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3. Construct an equilateral triangle, given its side of length of 4.5 cm and justify the construction.
4. Construct an isosceles triangle, given its base and base angle and justify the construction.
[Hint : You can take any measure of side and angle]
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13.3
13.3CC ONSTRUCTION OF TRIANGLES (S PECIAL CASES )
We have so far, constructed some basic constructions and justified with proofs. Now we
will construct some triangles when special type of measures are given. Recall the congruency
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properties of triangles such as SAS, SSS, ASA and RHS rules. You have already learnt how to
construct triangles in class VII using the above rules.
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You may have learnt that atleast three parts of a triangle have to be given for constructing
it but not any combinations of three measures are sufficient for the purpose. For example, if two
sides and an angle (not the included angle) are given, then it is not always possible to construct
such a triangle uniquely. We can give several illustrations for such constructions. In such cases
SC
we have to use the given measures with desired combinations such as SAS, SSS, ASA and RHS
rules.
Step 1 : Draw a rough sketch of ∆ABC and mark the given measurements as usual.
(How can you mark AB + AC = 8cm ?)
How can you locate third vertex A in the construction ?
Analysis : As we have AB + AC = 8 cm., extend BA up to D so that D
BD = 8 cm.
A
∴ BD = BA + AD = 8 cm
.
8cm
but AB + AC = 8 cm. (given)
AN
∵ AD = AC
To locate A on BD what will you do ?
60°
As A is equidistant from C and D, draw a perpendicular C
B 5cm.
X
bisector of CD to locate A on BD.
G
How can you prove AB + AC = BD ?
N
Step 2 : Draw the base BC = 5 cm and construct
X
D
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∠CBX = 60 o at B
.
8cm
60°
B 5cm. C
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X
D Step 3:With centre B and radius 8 cm (AB + AC 60°
B 5cm. C
= 8 cm) draw an arc on BX to intersect (meet) at
.
8cm
D.
T
D X
Step 4 : Join CD and draw a perpendicular
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bisector of CD to meet BD at A
.
60°
8cm
B 5cm. C
ABC. 60°
B 5cm. C
Now, we will justify the construction.
A
13.3.2 Construction : To Construct a triangle given its base, a
base angle and the difference of the other two sides.
AN
Given the base BC of a triangle ABC, a base angle say ∠B and the difference of other
two sides AB − AC in case AB>AC or AC-AB, in case AB<AC, you have to construct the
triangle ABC. Thus we have two cases of constructions discussed in the following examples.
Case (i) Let AB > AC
G
Example-5. Construct ∆ABC in which BC = 4.2 cm, ∠B = 30o and AB − AC = 1.6 cm
Solution : Steps of Construction
N
Step 1: Draw a rough sketch of ∆ABC and mark the given
A
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measurements
30°
(How can you mark AB − AC = 1.6 cm ?) B
4.2cm. C
X
TE
A
Analysis : Since AB − AC = 1.6 cm and AB > AC,
mark D on AB such that AD = AC Now
BD = AB − AC = 1.6 cm. Join CD and D
draw a perpendicular bisector of CD to .
.6cm
find the vertex A on BD produced. 1
T
30°
B 4.2cm. C
Join AC to get the required triangle ABC.
ER
.
cm X
1.6
30° A
B 4.2cm. C
D
.
cm
Step 3 : Draw the perpendicular bisector 1.6
30° C
of CD. Let it meet ray BDX at a point A. B 4.2cm.
A
cm
1.6
30° measures by changing the base angle ∠C instead
B 4.2cm. C
of ∠B ? Draw a rough sketch and construct it.
AN
Case (ii) Let AB < AC
Example-6. Construct ∆ABC in which BC = 5cm, ∠B = 45o and AC − AB = 1.8 cm.
G
Solution : Steps of Construction.
Step 1: Draw a rough sketch of ∆ABC and mark the given
measurements. N
Analyse how AC − AB = 1.8 cm can be marked?
45°
LA
Analysis : Since AC − AB = 1.8 cm i.e. AB < AC we have to find D on AB produced such
that AD = AC
Now BD = AC − AB = 1.8 cm (∵ BD = AD − AB and AD = AC)
TE
AB + BD = AC D
So BD = AC − AB
= 1.8 cm
Hence ∆ABC is the required that triangle.
A
AB + BC + CA = 11 cm. A
AN
Solution : Steps of construction.
Step 1 : Draw a rough sketch of a triangle ABC and mark the given
measures 60° 45°
B C
(Can you mark the perimeter of triangle ?)
G
Analysis : Draw a line segment, say XY equal to perimeter of ∆ABC i.e., AB + BC + CA.
Make angles ∠YXL equal to ∠B and L
60° 45° L
X Y
11cm.
M
R
P
Step 5 : Draw perpendicular bisectors of AX
A
and AY to intersect XY at B and C
respectively 60° 45°
X Y
Join AB and AC. B 11cm. C
A
= XY
Again ∠BAX = ∠AXB (∵ XB = AB in ∆AXB) and
TRY THESE
AN
∠ABC = ∠BAX + ∠AXB Can you draw the triangle
(Exterior angle of ∆ABC). with the same measurements in
alternate way?
G
= 2∠AXB
60°
= ∠YXL (Hint: Take ∠YXL = = 30°
= 60o. N and ∠XYM =
45°
2
= 22 12 ° )
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Similarly ∠ACB = ∠XYM = 45 as required
o
2
angle 60°. (Draw major segment Why?) Can you draw a circle without O
a centre?
A 7 cm B
C Analysis: Let ‘O’ be the centre of the
60° circle. Let AB be the given chord and
SC
X
ACB be the required segment of the
Y X circle containing an angle C = 60°.
be the arc subtending the angle at C.
Let AXB
O
A
A 7cm. B
AN
Y X Step-4 : With centre ‘O’ and radius OA or OB, draw the circle.
O C
Step-5 : Mark a point ‘C’
30° 30° 60°
A on the arc of the circle.
G
7cm. B
Join AC and BC. We get Y X
∠ACB = 60°
N
Thus ACB is the required circle segment.
30°
O
30°
LA
A 7cm. B
Let us justify the construction
Proof : OA = OB (radii of circle).
X
∴ ∠OAB + ∠OBA = 30° + 30° = 60°
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∴ ∠ACB = = 60°
2
ER
TRY THESE
SC
What happen if the angle in the circle segment is right angle? What kind of
segment do you obtain? Draw the figure and give reason.
EXERCISE - 13.2
1. Construct ∆ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 12 cm.
2. Construct ∆PQR in which QR = 8 cm, ∠Q = 60° and PQ − PR = 3.5 cm
3. Construct ∆ XYZ in which ∠Y = 30°, ∠Z = 60° and XY + YZ + ZX = 10 cm.
4. Construct a right triangle whose base is 7.5cm. and sum of its hypotenuse and other side
is 15cm.
5. Construct a segment of a circle on a chord of length 5cm. containing the following angles.
i. 90° ii. 45° iii. 120°
A
WHAT
HAT WE HAVE DISCUSSED ?
HAVE
AN
1. A geometrical construction is the process of drawing geometrical figures using only two
instruments - an ungraduated ruler and a compass.
G
2. Construction of geometrical figures of the following with justifications (Logical proofs)
• Perpendicular bisector of a given line segment.
• bisector of a given angle. N
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• Construction of 60° angle at the initial point of a given ray.
3. To construct a triangle, given its base, a base angle and the sum of other two sides.
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4. To construct a triangle given its base, a base angle and the difference of the other two
sides.
5. To construct a triangle, given its perimeter and its two base angle.
T
Brain Teaser
292
Probability
14
A
Probability theory is nothing but common sense reduced to calculation.
AN
- Pierre-Simon Laplace
14.1 I NTRODUCTION
G
Siddu and Vivek are classmates. One day during their lunch they are talking to each other.
Observe their conversation
N
Siddu : Hello Vivek , What are you going to do in the evening today?
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Vivek : Most likely, I will watch India v/s Australia cricket
match.
Siddu : Whom do you think will win the toss ?
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Siddu : No, first I will finish the homework then I will come to your home.
Vivek : Ok.
In many situations we make such statements and use our past experience and logic to take
decisions. For example
It is a bright and pleasant sunny day. I need not carry my umbrella and will take a chance
to go.
A
However, the decisions may not always favour us. Consider the situation. “Mary took her
umbrella to school regularly during the rainy season. She carried the umbrella to school for many
AN
days but it did not rain during her walk to the school. However, by chance, one day she forgot
to take the umbrella and it rained heavily on that day”.
Usually the summer begins from the month of March, but one day in that month there was
a heavy rainfall in the evening. Luckily Mary escaped becoming wet, because she carried umbrella
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on that day as she does daily.
N
Thus we take a decision by guessing the future happening that is whether an event occurs
or not. In the above two cases, Mary guessed the occurrence and non-occurrence of the event
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of raining on that day. Our decision may favour us and sometimes may not. (Why?)
We try to measure numerically the chance of occurrence or non-occurrence of some events
just as we measure many other things in our daily life. This kind of measurement helps us to take
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decision in a more systematic manner. Therefore we study probability to figure out the chance of
something happening.
Before measuring numerically the chance of happening that we have discussed in the above
situations, we grade it using the following terms given in the table. Let us observe the following
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table.
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Equally likely
(Tossing a coin)
A
(Raining in the month of March) (Cold waves in the last week of
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December)
DO THIS
1. Observe the table given in the previous page and give some other example for each term.
G
2. Classify the following statements into the categories less likely, equally likely,
more likely.
N
a) Rolling a die* and getting a number 5 on the top face.
b) Cold waves in your village in the month of November.
LA
c) India winning the next soccer(foot ball)world cup
d) Getting a tail or head when a coin is tossed.
e) Winning the jackpot for your lottery ticket.
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Blue
Blu
d
Bl
Re
Green
Red
en
Red
Red
d
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toss? Can you get a head or tail that you want? In an ordinary coin
that is not possible. The chance of getting either is same and you
cannot say what you would get. Such an experiment known as
‘random experiment’. In such experiments though we know the
possible outcomes before conducting the experiment, we cannot
predict the exact outcome that occurs at a particular time, in advance.
The outcomes of random experiments may be equally likely or may
not be. In the coin tossing experiment head or tail are two possible outcomes.
* A die (plural dice) is a well balanced cube with its six faces marked with numbers from 1 to 6, one number
on each face. Sometimes dots appear in place of numbers.
TRY THESE
1. If you try to start a scooter , What are the possible outcomes?
2. When you roll a die, What are the six possible outcomes?
A B
3. When you spin the wheel shown, What are the possible
A
outcomes?
C
AN
(Out comes here means the possible sector where the pointer
stops)
4. You have a jar with five identical balls of different colours (White,
Red, Blue, Grey and Yellow) and you have to pickup (draw) a
G
ball without looking at it. List the possible outcomes you get.
N
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THINK, DISCUSS AND WRITE
In rolling a die.
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When we toss a coin or roll a die , we assume that the coin and the die are fair and
unbiased i.e. for each toss or roll the chance of all possibilities is equal. We conduct the experiment
many times and collect the observations. Using the collected data, we find the measure of chance
of occurrence of a particular happening.
A coin is tossed several times and the result is noted. Let us look at the result sheet where
we keep on increasing the tosses.
A
60 26 34
AN
70 ...... 30 ...... 40
80 ...... 36 ...... 44
90 ...... 42 ...... 48
G
100 ...... 48 ...... 52
N
We can observe from the above table as you increase the number of tosses, the number of
heads and the number of tails come closer to each other.
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DO THIS
Toss a coin for number of times as shown in the table. And record your findings in the
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table.
No. of Tosses Number of heads No. of tails
10
T
20
ER
30
40
50
SC
This could also be done with a die, roll it for large number of times and observe.
A
50 9 5 12 9 8 7
75 14 10 16 12 10 13
AN
100 17 19 19 16 13 16
125 25 20 24 18 16 22
150 28 24 28 23 21 26
G
175 31 30 33 27 26 28
200 34 34 36 30 32 34
225 37 38 N 40 34 38 38
LA
250 40 40 43 40 43 44
275 44 41 47 47 47 49
300 48 47 49 52 52 52
TE
From the above table, it is evident that rolling a die for a larger number of times, the each
of six outcomes, becomes almost equal to each other.
From the above two experiments, we may say that the different outcomes of the experiment
are equally likely. This means each of the outcome has equal chance of occurring.
T
In the above experiments each toss of a coin or each roll of a die is a Trial or Random
experiment.
Consider a trial of rolling a die,
SC
How many possible outcomes are there to get a number more than 5 on the top face?
It is only one (i.e., 6)
How many possible outcomes are there to get an even number on the top face?
They are 3 outcomes (2,4, and 6).
Thus each specific outcome or the collection of specific outcomes make an Event.
In the above trail getting a number more than 5 and getting an even number on the top face
are two events. Note that event need not necessarily a single outcome. But, every outcome of a
random experiment is an event.
Here we understand the basic idea of the event, more could be learnt on event in higher
classes.
Consider the experiment of tossing a coin once. What are the outcomes? There are only
A
two outcomes Head or Tail and both outcomes are equally likely.
AN
What is the chance of getting a head?
1
It is one out of two possible outcomes i.e. . In other words it is expressed as the
2
1
probability of getting a head when a coin is tossed is , which is represented by
G
2
1
P(H) = = 0.5 or 50%
2
N
What is the probability of getting a tail?
LA
Now take the example of rolling a die. What are the possible outcomes in one roll? There
are six equally likely outcomes 1,2,3,4,5,or 6.
What is the probability of getting an odd number on the top face?
TE
3 1
1, 3 or 5 are the three favourable outcomes out of six total possible outcomes. It is or
6 2
We can write the formula for Probability of an event ‘A’
T
probability of getting atleast one head, (e) probability of getting no heads and (f) probability of
getting only one head.
Solution : (a) The possible outcomes are
Coin 1 Coin 2
Head Head
Head Tail
Tail Head
Tail Tail
A
3
d) Probability of getting atleast one head =
4
AN
[At least one head means getting a head one or more number of times]
1
e) Probability of getting no heads = .
4
G
2 1
e) Probability of getting only one head = = .
4 2
DO THIS N
LA
1. If three coins are tossed simultaneously then write their outcomes.
Example 2 : (a) Write the probability of getting each number on the top face when a die was
SC
rolled in the following table. (b) Find the sum of the probabilities of all outcomes.
Solution : (a) Out of six possibilities the number 4 occurs once hence probability is
1/6. Similarly we can fill the table for the remaining values.
Outcome 1 2 3 4 5 6
A
We can generalize that
Sum of the probabilities of all the outcomes of a random experiment is always 1
AN
TRY THIS
G
Event Favourable Number of Total Number Probability =
outcome(s) favourable possible of total Number of favourable outcomes
N
outcome(s) outcomes possible
outcomes
Number of total possible outcomes
LA
Getting a 5 1 1, 2, 3, 4, 6 1/6
number 5 5 and 6
on the top face
Getting a
TE
number greater
than 3 on the
top face
Getting a prime
number on the
T
top face
Getting a number
ER
less than 5 on
the top face
Getting a number
that is a factor of
6 on the top face
SC
Getting a number
greater than 7
on the top face
Getting a number
that is a Multiple of
3 on the top face
Getting a
number 6 or
less than 6
on the top face
A
b) The probability of an event which is impossible = 0
AN
14.2.5 C ONDUCT YOUR OWN EXPERIMENTS
1. We would work here in groups of 3-4 students each. Each group would take a coin of
the same denomination and of the same type. In each group one student of the group
G
would toss the coin 20 times and record the data. The data of all the groups would be
placed in the table below (Examples are shown in the table).
20 20 20
21 40 − 21 19
2 20 40 14 21 =
40 40 40
3 20 60
T
4 20 80
5 20 100
ER
6 ..... ....
7 .... ....
What happens to the value of the fractions in (6) and (7) when the total number of tosses
SC
of the coin increases? Could you see that the values are moving close to the probability of
getting a head and tail respectively.
2. In this activity also we would work in groups of 3-4. One student from each group would
roll a die for 30 times. Other students would record the data in the following table. All the
groups should have the same kind of die so that all the throws will be treated as the
throws of the same die.
Complete the following table, using the data obtained from all the groups :
A
Number of times Total number of times Number of times
Group(s) 1 turned up a die is rolled 1 turned up
AN
Total number of times
a die is rolled
G
1s t+ 2nd
1st+2nd+3rd
1st + 2nd + 3rd + 4th
N
LA
1st + 2nd + 3rd + 4th + 5th
What do you observe as the number of rolls increases; the fractions in cloumn (4)
TE
1
move closer to . We did the above experiment for the outcome 1. Check the same for the
6
outcome 2 and the outcome 5.
What can you conclude about the values you get in column (4) and compare these with the
T
3. What would happen if we toss two coins simultaneously? We could have either both
coins showing head, both showing tail or one showing head and one showing tail. Would
the possibility of occurrence of these three be the same? Think about this while you do
this group activity.
SC
Divide class into small groups of 4 each. Let each group take two coins. Note that all the
coins used in the class should be of the same denomination and of the same type. Each group
would throw the two coins simultaneously 20 times and record the observations in a table.
20
A
1 st
1st + 2nd
AN
1st + 2nd + 3rd
1st + 2nd + 3rd + 4th
.... .... ....
G
Now we find the ratio of the number of times no head turns up to the total number of
times two coins are tossed. Do the same for the remaining events.
Fill the following table:
N
No. of times No. of times No. of times
LA
Group(s) no head one head two heads
Total tosses Total tosses Total tosses
(1) (2) (3) (4)
TE
Group 1 st
Group 1 + 2 nd
Group 1 + 2 + 3 rd
Group 1 + 2 + 3 + 4 th
T
As the number of tosses increases, the values of the columns (2), (3) and (4) get closer
to 0.25, 0.5 and 0.25 respectively.
SC
Example-3: A spinner was spun 1000 times and the frequency of outcomes was recorded as
in given table:
Find (a) List the possible outcomes that you can see in the spinner (b) Compute the probability
of each outcome. (c) Find the ratio of each outcome to the total number of times that the spinner
spun (use the table)
Solution :
Red
Or
(a) The possible outcomes are 5. They are red, orange, purple, yellow
an
Green
ge
and green. Here all the five colours occupy equal areas in the
P
spinner. So, they are all equally likely.
urp
le
ow
Yell
(b) Compute the probability of each event.
A
Favourable outcomes of red
P(Red) = spinner
AN
Total number of possible outcomes
1
= = 0.2.
5
Similarly
G
1
P(Orange), P(Purple), P(Yellow) and P(Green) is also or 0.2.
5
N
(c) From the experiment the frequency was recorded in the table
No. of outcomes of red in the above experiment
LA
Ratio for red =
Number of times the spinner was spun
185
= = 0.185
1000
TE
Similarly, we can find the corresponding ratios for orange, purple, yellow and green are
0.195, 0.210, 0.206 and 0. 204 respectively.
Can you see that each of the ratio is approximately equal to the probability which we have
obtained in (b) [i.e. before conducting the experiment]
T
Example-4. The following table gives the ages of audience in a theatre. Each person was
ER
given a serial number and a person was selected randomly for the bumper prize by choosing a
serial number. Now find the probability of each event.
Age Male Female
SC
Under 2 3 5
3 - 10 years 24 35
11 - 16 years 42 53
17 - 40 years 121 97
41- 60 years 51 43
Over 60 18 13
Total number of audience : 505
A
Total number of people = 505
67
AN
P(audience of age < 10 years) =
505
b) The probability of female audience of age 16 years or younger
The female audience with age less than or equal 16 years = 53 + 35 + 5 = 93
P(female audience of age < 16 years) = 93/505
G
c) The probability of male audience of age 17 years or above
N = 121 + 51 + 18 = 190
190 38
LA
P(male audience of age > 17 years) = =
505 101
d) The probability of audience of age above 40 years
TE
= 51+43+18+ 13 = 125
125 25
P(audience of age > 40 years) = =
505 101
e) The probability of the person watching the movie is not a male
T
= 5 + 35 + 53 + 97 + 43 + 13 = 246
ER
246
P(A person watching movie is not a male) =
505
A
π(3)2 − π(2)2 Area of a ring = πR 2 − πr 2
=
π(3)2
AN
9π − 4 π
=
9π
5
= 0.556
G
9
TRY THESE
N
LA
From the figure given in example 5.
1. Find the probability of the dart hitting the board in the circular region B (i.e. ring B).
2. Without calculating, write the percentage of probability of the dart hitting the board in
TE
l Meteorological department predicts the weather by observing trends from the data collected
ER
EXERCISE - 14.1
1. A die has six faces numbered from 1 to 6. It is rolled and the number on
the top face is noted. When this is treated as a random trial.
A
a) What are the possible outcomes ?
AN
b) Are they equally likely? Why?
c) Find the probability of a composite number turning up on the top face.
2. A coin is tossed 100 times and the following outcomes are recorded
G
Head:45 times Tails:55 times from the experiment
a) Compute the probability of each outcomes.
N
b) Find the sum of probabilities of all outcomes.
LA
3. A spinner has four colours as shown in the figure. When we spin it once, find
a) At which colour, is the pointer more likely to stop?
Blue
Blu
ue
d
Bl
Re
b) At which colour, is the pointer less likely to stop?
TE
Green
Red
c) At which colours, is the pointer equally likely to stop? Gr Yel
een
G low
re
d) What is the chance the pointer will stop on white?
Re
en
Red
Red
d
4. A bag contains five green marbles, three blue marbles, two red marbles, and two yellow
ER
6. Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights
of flour (in kg):
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of
A
flour.
7. An insurance company selected 2000 drivers at random (i.e., without any preference of
AN
one driver over another) in a particular city to find a relationship between age and accidents.
The data obtained is given in the following table:
Age of Drivers Accidents in one year More than 3
G
(in years) 0 1 2 3 accidents
18-29
30- 50
440
505
N
160
125
110
60
61
22
35
18
LA
Over 50 360 45 35 15 9
Find the probabilities of the following events for a driver chosen at random from the city:
TE
(i) The driver being in the age group 18-29 years and having exactly 3 accidents in
one year.
(ii) The driver being in the age group of 30-50 years and having one or more accidents
in a year.
T
22
(Take π = and express in percentage)
7
WHAT
HAT WE HAVE DISCUSSED
HAVE
• There is use of words like most likely, no chance, equally likely in daily life, are showing
the manner of chance and judgement.
A
• There are certain experiments whose outcomes have equal chance of occurring.
Outcomes of such experiments are known as equally likely outcomes.
AN
• An event is a collection of a specific outcome or some of the specific outcomes of the
experiment.
G
• As the number of trials increases, the probability of all equally likely outcomes come
very close to each other.
Do you Know?
ER
The diagram below shows the 36 possible outcomes when a pair of dice are thrown. It is
interesting to notice how the frequency of the outcomes of different possible numbers (2 through
12) illustrate the Gaussian curve.
SC
This curve illustrate the Gaussian curve, name after 19th century
famous mathematician Carl Friedrich Gauss.
A
AN
15.1 I NTRODUCTION
We come across many statements in our daily life. We gauge the worth of each
statement. Some statements we consider to be appropriate and true and some we dismiss.
G
There are some we are not sure of. How do we make these judgements? In case there is a
statement of conflict about loans or debts. You want to claim that bank owes your money
N
then you need to present documents as evidence of the monetary transaction. Without that,
people would not believe you. If we think carefully we can see that in our daily life we
LA
need to prove if a statement is true or false. In our conversations in daily life we sometimes
do not consider to prove or check statements and accept them without serious examination.
That however will not be accepted in mathematics. Consider the following:
TE
We know, out of these some sentences are false. For example, 4>8 and present New
York is not the capital of USA. Some are correct. These include "sun rises in the east." The
probability............"
The Sun is not stationary......................
Besides those there are some other sentences that are true for some known cases but not
true for other cases, for example x + 2 = 7 is true only when x = 5 and x < y is only true for those
values of x and y where x is less than y.
Look at the other sentences which of them are clearly false or clearly true. These are
statements. We say these sentences that can be judged on some criteria, no matter by what
process for their being true or false are statements.
Think about these:
1. Please ignore this notice..... 2. The statement I am making is false.
A
3. This sentence has some words. 4. You may find water on the moon.
Can you say whether these sentences are true or false? Is there any way to check them
AN
being true or false?
Look at the first sentence, if you ignore the notice, you do that because it tells you to do so.
If you do not ignore the notice, then you have paid some attention to it. So you can never follow
G
it and being an instruction it cannot be judged on a true/false scale. 2nd and 3rd sentences are
talking about themselves. 4th sentence have words that show only likely or possibility and hence
ambiguity of being on both sides.
N
The sentences which are talking about themselves and the sentences with possibility are
LA
not statements.
DO THIS
TE
Make 5 more sentences and check whether they are statements or not. Give
reasons.
We can write infinetely large number of sentences. You can think the kind of sentences you
ER
use and can you count the number of sentences you speak? Not all these however, they can be
judged on the criteria of false and true. For example, consider, please come in. Where do you
live? Such sentences can also be very large in number.
All these the sentences are not statements. Only those that can be judged to be true or
SC
false but not both are statements. The same is true for mathematical statements. A mathemtical
statement can not be ambiguous. In mathematics a statement is only acceptable if it is either true
or false. Consider the following sentences:
1. 3 is a prime number. 2. Product of two odd integers is even.
3. For any real number x; 4x + x = 5x 4. The earth has one moon.
5. Ramu is a good driver. 6. Bhaskara has written a book "Leelavathi".
7. All even numbers are composite. 8. A rhombus is a square.
9. x > 7. 10. 4 and 5 are relative primes.
11. Silver fish is made of silver. 12. Humans are meant to rule the earth.
13. Fo any real number x, 2x > x. 14. Havana is the capital of Cuba.
Which of these are mathematical and which are not mathematical statements?
A
Let us consider some of the above sentences and discuss them as follows:
AN
Example-1. We can show that (1) is true from the definition of a prime number.
Which of the sentences from the above list are of this kind of statements that we can prove
mathematically? (Try to prove).
G
Example-2. “Product of two odd integers is even”. Consider 3 and 5 as the odd integers. Their
product is 15, which is not even.
N
Thus it is a statement which is false. So with one example we have showed this. Here we
are able to verify the statement using an example that runs counter to the statement. Such
LA
an example, that counters a statement is called a counter example.
TRY THIS
TE
Example-3. Among the sentences there are some like “Humans are meant to rule the
earth” or “Ramu is a good driver.”
These sentences are ambiguous sentences as the meaning of ruling the earth is not specific.
T
We therefore recognize that a ‘mathematical statement’ must comprise of terms that are
understood in the same way by everyone.
Example-4. Consider some of the other sentences like
The earth has one Moon.
SC
counter to the statement. In the statement for any real number 2x > x, we can take
1
x = −1 or − .... and disprove the statement by giving counter example. You might have also
2
noticed that 2x > x is true with a condition on x i.e. x belong to set N.
Example-5. Restate the following statements with appropriate conditioins, so that they become
true statements.
A
i. For every real number x, 3x > x.
For every real number x, x2 ≥ x.
AN
ii.
iii. If you divide a number by two, you will always get half of that number.
iv. The angle subtended by a chord of a circle at a point on the circle is 90°.
v. If a quadrilateral has all its sides equal, then it is a square.
G
Solution :
i. If x > 0, then 3x > x.
ii. If x ≤ 0 or x ≥ 1, then x2 ≥ x.
N
iii. If you divide a number other than 0 by 2, then you will always get half of that number.
LA
iv. The angle subtended by a diameter of a circle at a point on the circle is 90°.
v. If a quadrilateral has all its sides and interior angles equal, then it is a square.
EXERCISE - 15.1
TE
1. State whether the following sentences are always true, always false or
ambiguous. Justify your answer.
i. There are 27 days in a month. ii. Makarasankranthi falls on a Friday.
iii. The temperature in Hyderabad is 2°C. iv. The earth is the only planet where life exist.
T
2. State whether the following statements are true or false. Give reasons for your answers.
i. The sum of the interior angles of a ii. For any real number x, x2 ≥ 0.
quadrilateral is 350°.
iii. A rhombus is a parallelogram. iv. The sum of two even numbers is even.
SC
We human beings are naturally curious. This curiosity makes us to interact with the world.
What happens if we push this? What happens if we stuck our finger in that? What happens if we
make various gestures and expressions? From this experimentation, we begin to form a more or
less consistant picture of the way that the physical world behaves. Gradually, in all situations, we
A
make a shift from
AN
‘What happens if.....?’ to ‘this will happen if’
The experimentation moves on to the exploration of new ideas and the refinement of our
world view of previously understood situations. This description of the playtime pattern very
nicely models the concept of ‘making and testing hypothesis.’ It follows this pattern:
G
• Make some observations, Collect data based on the observations.
• Draw conclusion (called a ‘hypothesis’) which will explain the pattern of the observations.
• N
Test out hypothesis by making some more targeted observations.
LA
So, we have
• A hypothesis is a statement or idea which gives an explanation to a series of observations.
Sometimes, following observation, a hypothesis will clearly need to be refined or rejected.
TE
This happens if a single contradictory observation occurs. In general we use word conjecture in
mathematics instead of hypothesis. You will learn the similarities and difference between these
two in the higher classes.
There is often confusion between the ideas surrounding proof, making and testing an
ER
experimental hypothesis which is mathematics, which is science. The difference is rather simple:
• Mathematics is based on deductive reasoning : a proof is a logical deduction from a set
of clear inputs.
• Science is based on inductive reasoning : hypotheses are strengthened or rejected based
SC
A
other side.
AN
A V 8 5
Suppose you are told that these cards follow the rule:
G
“If a card has an odd number on one side, then it has a vowel on the other side.”
What is the smallest number of cards you need to turn over to check if the rule is true?
N
Of course, you have the option of turning over all the cards and checking. But can you
manage with turning over a fewer number of cards?
LA
Notice that the statement mentions that a card with an odd number on one side has a
vowel on the other. It does not state that a card with a vowel on one side must have an odd
number on the other side. That may or may not be so. The rule also does not state that a card
TE
with an even number on one side must have a consonant on the other side. It may or may not.
So, do we need to turn over A ? No! Whether there is an even number or an odd number
on the other side, the rule still holds.
T
What about 8 ? Again we do not need to turn it over, because whether there is a vowel
or a consonant on the other side, the rule still holds.
ER
But you do need to turn over V and 5 . if V has an odd number on the other side, then
the rule has been broken. Similarly, if 5 has a consonant on the other side, then the rule has
been broken.
SC
The kind of reasoning we have used to solve the puzzle is called deductive reasoning. It
is called ‘deductive’ because we arrive at (i.e., deduce or infer) a result or a statement from a
previously established statement using logic. For example, in the puzzle by a series of logical
arguments we deduced that we need to turn over only V and 5 .
Deductive reasoning also helps us to conclude that a particular statement is true, because
it is a special case of a more general statement that is known to be true. For example, once we
prove that the product of two even numbers is always even, we can immediately conclude (without
computation) that 56702 × 19992 is even simply because 56702 and 19992 are even.
A
ii. Some singers are poets. All lyrists are Poets.
Here the deduction based on two statement is wrong. (Why?) All lyricist are poets
AN
(wrong). Because we are not sure about it. There are three posibilities (i) all lyricists could be
poets, (ii) few could be poets or (iii) none of the lyricists is a poet.
You may come to a conclusion that if - then conditional statement comes into deductive
G
reasoning. In mathematics we use this reasoning a lot like if linear pair of angles are 180°. Then
only the sum of angles in a triangle is equal to 180°. Like wise if we are using decimal number
N
system to write a number 5. If we use the binary system we represent the quantity by 101.
Unfortunately we do not always use correct reasoning in our daily life. We often come to
LA
many conclusions based on faulty reasoning. For example, if your friend does not talk to you one
day, then you may conclude that she is angry with you. While it may be true that “if she is angry
at me she will not talk to me”, it may also be true that “if she is busy, she will not talk to me. Why
TE
don’t you examine some conclusions that you have arrived at in your day-to-day existence, and
see if they are based on valid or faulty reasoning?
EXERCISE - 15.2
T
i. Human beings are mortal. Jeevan is a human being. Based on these two
statements, what can you conclude about Jeevan ?
ii. All Telugu people are Indians. X is an Indian. Can you conclude that X belongs to
Telugu people.
SC
iii. Martians have red tongues. Gulag is a Martian. Based on these two statements,
what can you conclude about Gulag?
iv. What is the fallacy in the Raju’s reasoning in the cartoon below?
2. Once again you are given four cards. Each card has a number printed on one side
and a letter on the other side. Which are the only two cards you need to turn over to
check whether the following rule holds?
“If a card has a consonant on one side, then it has an odd number on the other side.”
B 3 U 8
A
AN
3. Think of this puzzle What do you need to find a chosen number from this square?
Four of the clues below are true but do nothing to help in finding the number.
Four of the clues are necessary for finding it.
G
Here are eight clues to use: 0 1 2 3 4 5 6 7 8 9
a. The number is greater than 9. 10 11 12 13 14 15 16 17 18 19
N
b. The number is not a multiple of 10.
c. The number is a multiple of 7.
20 21 22 23 24 25 26 27 28 29
30 31 32 33 34 35 36 37 38 39
LA
40 41 42 43 44 45 46 47 48 49
d. The number is odd.
50 51 52 53 54 55 56 57 58 59
e. The number is not a multiple of 11. 60 61 62 63 64 65 66 67 68 69
f. The number is less than 200. 70 71 72 73 74 75 76 77 78 79
TE
First follow the clues and strike off the number which comes out from it.
ER
Like - from the first clue we come to know that the number is not from 1 to 9. (strike off
numbers from 1 to 9).
After completing the puzzle, see which clue is important and which is not?
SC
A
understanding and experience, i.e., our mathematical intuition. The conjecture may turn out to be
AN
true or false. If we can prove it, then it becomes a theorem. Mathematicians often come up with
conjectures by looking for patterns and making intelligent mathematical guesses. Let us look at
some patterns and see what kind of intelligent guesses we can make.
While studying some cube numbers Raju noticed that “if you take three consecutive whole
G
numbers and multiply them together and then add the middle number of the three, you get the
middle number cubed”; e.g., 3, 4, 5, gives 3 × 4 × 5 + 4 = 64, which is a perfect cube. Does this
N
always work? Take some more consecutive numbers and check it.
Rafi took 6, 7, 8 and checked this conjecture. Here 7 is the middle term so according to
LA
the rule 6 × 7 × 8 + 7 = 343, which is also a perfect cube. Try to generalize it by taking numbers
as n, n + 1, n + 2. See other example:
The dots here arranged in such a way that they form a rectangle. Here T1 = 2,
T2 = 6, T3 = 12, T4 = 20 and so on. Can you guess what T5 is? What about T6? What about Tn?
ER
2 6 12 20 ? .....
Solution :
+4 +6 +8 +10
So, T5 = T4 + 10 = 20 + 10 = 30 = 5 × 6
T6 = T5 + 12 = 30 + 12 = 42 = 6 × 7 ..... Try for T7?
T100 = 100 × 101 = 10, 100
Tn = n × (n + 1) = n2 + n
This type of reasoning which is based on examining a variety of cases or sets of data,
discovering patterns and forming conclusions is called inductive reasoning. Inductive reasoning
is very helpful technique for making conjecture.
Gold bach the renounced mathematician, observed a pattern:
6=3+3 8=3+5 10 = 3 + 7
A
12 = 5 + 7 14 = 11 + 3 16 = 13 + 3 = 11 + 5
AN
From the pattern Gold bach in 1743 reasoned that every even number greater than 4 can
be written as the sum of two primes (not necessarily distinct primes). His conjecture has not been
proved to be true or false so far. Perhaps you will prove that this result is true or false and will
become famous.
G
But just by looking few patterns some time lead us to a wrong conjecture like: in class 8th
Janvi and Kartik while studying Area and Perimeter chapter..... observed a pattern
3 cm. 4 cm.
N 5 cm. 6 cm.
3 cm.
3 cm.
LA 3 cm.
3 cm.
(i) (ii) (iii) (iv)
Perimeter : 12 cm. 14 cm. 16 cm. 18 cm.
TE
2 2 2 2
Area : 9 cm 12 cm 15 cm 18 cm
and stated a conjecture that when the perimeter of the rectangle increases the area will also
increase. What do you think? Are they right? 3 cm.
6 cm.
1 cm.
Do you understand that while making a conjecture we have to look all the possibilities.
SC
TRY THIS
Envied by the popularity of Pythagoras his disciple claimed a different relation
between the sides of right angle triangles. By observing this what do you notice?
25
13 7
5 5
4
3 12 24
(i) (ii) (iii)
Liethagoras Theorem : In any right angle triangle the square of the smallest side
equals the sum of the other sides.
Check this conjucture, whether it is right or wrong.
You might have wondered - do we need to prove every thing we encounter in
mathematics and if not, why not?
A
In mathematics some statements are assumed to be true and are not proved, these are
AN
self-evident truths’ which we take to be true without proof. These statements are called axioms.
In chapter 3, you would have studied the axioms and postulates of Euclid. (We do not distinguish
between axioms and postulates these days generally we use word postulate in geometry).
G
For example, the first postulate of Euclid states:
A straight line may be drawn from any point to any other point.
And the third postulate states:
N
A circle may be drawn with any centre and any radius.
LA
These statements appear to be perfectly true and Euclid assumed them to be true. Why?
This is because we cannot prove everything and we need to start somewhere, we need some
statements which we accept as true and then we can build up our knowledge using the rules of
TE
For example, many of us believe that if a number is added to another number, the result will
be large than the numbers. But we know that this is not always true : for example 5 + (-5) =
ER
4 2
appears bigger.
2
You might then wonder, about the validity of axioms. Axioms
A B
have been chosen based on our intuition and what appears to be self-
evident. Therefore, we expect them to be true. However, it is possible that later on we
discover that a particular axiom is not true. What is a safeguard against this possibility? We
take the following steps:
i. Keep the axioms to the bare minimum. For instance, based only on axioms and five
postulates of Euclid, we can derive hundreds of theorems.
A
Statement-2 : A whole number divided by zero is a whole number.
AN
(Remember, division by zero is not defined. But just for the moment, we assume that it
is possible, and see what happens.)
1
From Statement-2, we get = a, where a is some whole number. This implies that, 1=0.
G
0
But this disproves Statement-1, which states that no whole number is equal to its successor.
iii.
N
A false axiom will, sooner or later, result into contradiction. We say that there is a
contradiction, when we find a statement such that, both the statement and its negation
LA
are true. For example, consider Statement-1 and Statement-2 above once again.
From Statement-1, we can derive the result that 2 ≠ 1.
Let x = y
TE
x × x = xy
x2 = xy
x - y = xy - y2
2 2
(x+y) (x-y) = y (x - y) From Statement-2, we can cancel (x - y) from both the sides.
T
x+y=y
ER
But x=y
so x+x=x
or 2x = x
SC
2=1
So we have both the statements 2 ≠ 1 and its negation, 2 = 1 are true. This is a
contradiction. The contradiction arose because of the false axiom, that a whole number divided
by zero is a whole number.
So, the statement we choose as axioms require a lot of thought and insight. We must
make sure they do not lead to inconsistencies or logical contradictions. Moreover, the
choice of axioms themselves, sometimes leads us to new discoveries.
We end the section by recalling the differences between an axiom, a theorem and a
conjecture. An axiom is a mathematical statement which is assumed to be true without proof; a
conjecture is a mathematical statement whose truth or falsity is yet to be established; and a
theorem is a mathematical statement whose truth has been logically established.
A
EXERCISE - 15.3
AN
1. (i) Take any three consecutive odd numbers and find their product;
for example, 1 × 3 × 5 = 15, 3 × 5 × 7 = 105, 5 × 7 × 9 - .....
(ii) Take any three consecutive even numbers and add them, say,
G
2 + 4 + 6 = 12, 4 + 6 + 8 = 18, 6 + 8 + 10 = 24, 8 + 10 + 12 = 30 and so on.
Is there any pattern can you guess in these sums? What can you conjecture about them?
2. Go back to Pascal’s triangle. N 1
LA
Line-1 : 1 = 110 1 1
Line-2 : 11 = 111 1 2 1
1 3 3 1
Line-3 : 121 = 112
TE
1 4 6 4 1
Make a conjecture about Line-4 and Line-5.
Does your conjecture hold? Does your conjecture hold for Line-6 too?
3. Look at the following pattern:
T
A
5. List five axioms (postulates) used in this book.
In a polynomial p (x) = x2 + x + 41 put different values of x and find p (x). Can you
AN
6.
conclude after putting different values of x that p (x) is prime for all. Is x an element of
N? Put x = 41 in p (x). Now what do you find?
G
15.6 W HAT
HAT IS A M ATHEMATICAL P ROOF ?
THEMATICAL
N
Before you study proofs in mathematics, you are mainly asked to verify statements.
For example, you might have been asked to verify with examples that “the product of two
LA
odd numbers is odd”. So you might have picked up two random odd numbers, say 15 and 2005
and checked that 15 × 2005 = 30075 is odd. You might have done so for many more examples.
Also, you might have been asked as an activity to draw several triangles in the class and
TE
compute the sum of their interior angles. Apart from errors due to measurement, you would have
found that the interior angles of a triangle add up to 180°.
What is the flaw in this method? There are several problems with the process of verification.
While it may help you to make a statement you believe is true, you cannot be sure that it is true
T
in all cases. For example, the multiplication of several pairs of even numbers may lead us to
guess that the product of two even numbers is even. However, it does not ensure that the product
ER
of all pairs of even numbers is even. You cannot physically check the products of all possible
pairs of even numbers because they are endless. Similarly, there may be some triangles which
you have not yet drawn whose interior angles do not add up to 180°.
SC
A
i. First we must understand clearly, what is required to prove, then we should have a
AN
rough idea how to proceed.
ii. A proof is made up of a successive sequence of mathematical statements. Each statement
is a proof logically deduced from a previous statement in the proof or from a theorem
proved earlier or an axiom or our hypothesis and what is given.
G
iii. The conclusion of a sequence of mathematically true statements laid out in a logically
correct order should be what we wanted to prove, that is, what the theorem claims.
N
To understand that, we will analyse the theorem and its proof. You have already studied
LA
this theorem in chapter-4. We often resort to diagrams to help us to prove theorems, and this is
very important. However, each statement in proof has to be established using only logic. Very
often we hear or said statement like those two angles must be 90°, because the two lines look as
TE
if they are perpendicular to each other. Beware of being deceived by this type of reasoning.
Theorem-15.4 : The sum of three interior angles of a triangle is 180°.
A E
Proof : Consider a triangle ABC.
We have to prove that
T
Now, we see how each step has been logically connected in the proof.
Step-1: Our theorem is concerned with a property of triangles. So we begin with a triangle
ABC.
A
to proceed so that to be able to prove the theorem.
Step-3: Here we conclude that ∠CAB = ∠ACE and ∠ABC = ∠DCE, by using the fact that
AN
CE is parallel to BA (construction), and previously known theorems, which states
that if two parallel lines are intersected by a transversal, then the alternate angles
and corresponding angles are equal.
G
Step-4: Here we use Euclid’s axiom which states that “if equals are added to equals, the
wholes are equal” to deduce ∠ABC + ∠BCA + ∠CAB = ∠DCE + ∠BCA + ∠ACE.
N
That is, the sum of three interior angles of a triangle is equal to the sum of angles on
a straight line.
LA
Step-5: Here in concluding the statement we use Euclid’s axiom which states that “things
which are equal to the same thing are equal to each other” to conclude that
TE
Theorem-15.6 : The product of any two consecutive even natural numbers is divisible by 4.
Any two consecutive even number will be of the form 2m, 2m + 2, for some natural
number n. We have to prove that their product 2m (2m + 2) is divisible by 4. (Now try to prove
this yourself).
We conclude this chapter with a few remarks on the difference between how mathematicians
A
discover results and how formal rigorous proofs are written down. As mentioned above, each
AN
proof has a key intiative idea. Intution is central to a mathematicians’ way of thinking and discovering
results. A mathematician will often experiment with several routes of thought, logic and examples,
before she/he can hit upon the correct solution or proof. It is only after the creative phase subsides
that all the arguments are gathered together to form a proper proof.
G
We have discussed both inductive reasoning and deductive reasoning with some examples.
It is worth mentioning here that the great Indian mathematician Srinivasa Ramanujan used
N
very high levels of intuition to arrive at many of his statements, whch he claimed were true. Many
of these have turned out to be true and as well as known theorems.
LA
EXERCISE - 15.4
TE
1. State which of the following are mathematical statements and which are not?
Give reason.
i. She has blue eyes
ii. x + 7 = 18
T
A
repeating these digits in the same order (425425). Your new number is divisible by 7, 11,
and 13.
AN
WHAT W E H AVE D ISCUSSED
HAT
G
1. The sentences that can be judged on some criteria, no matter by what process for their
being true or false are statements.
N
2. Mathematical statements are of a distinct nature from general statements. They can not
be proved or justified by getting evidence while they can be disproved by finding a
counter example.
LA
3. Making mathematical statements through observing patterns and thinking of the rules
that may define such patterns.
A hypothesis is a statement of idea which gives an explanation to a sense of observation.
TE
4. A process which can establish the truth of a mathematical statement based purely on
logical arguments is called a mathematical proof.
5. Axioms are statements which are assumed to be true without proof.
6. A conjecture is a statement we believe is true based on our mathematical intution, but
T
7. A mathematical statement whose truth has been established or proved is called a theorem.
8. The prime logical method in proving a mathematical statement is deductive reasoning.
9. A proof is made up of a successive sequence of mathematical statements.
SC
10. Begining with given (Hypothesis) of the theorem and arrive at the conclusion by means
of a chain of logical steps is mostly followed to prove theorems.
11. The proof in which, we start with the assumption contrary to the conclusion and arriving
at a contradiction to the hypothesis is another way that we establish the original conclusion
is true is another type of deductive reasoning.
12. The logical tool used in establishment the truth of an unambiguious statements to
deductive reasoning.
13. The resoning which is based on examining of variety of cases or sets of data discovering
pattern and forming conclusion is called Inductive reasoning.
328
Answers
A
E XERCISE 1.1
AN
22 −2013
1. a. -5, ,
7 2014
G
p
b. A number which can be written in the form where q ≠ 0; p, q are integers, called
q
a rational number.
3
N
LA
2. (i) (ii) 0 (iii) −5
7
(iv) 7 (v) −3
19 37 77
TE
3 5 11 21 53 , ,
3. , , , , 4.
2 4 8 16 32 30 60 120
−8 8
5. 5 5
-2 -1 0 1 2
T
5 35 4 563
8. (i) (ii) (iii) (iv)
9 9 11 180
9. (i) Yes (ii) No (iii) Yes (iv) No
EXERCISE - 1.2
1. (i) Irrational (ii) Rational (iii) Irrational
(iv) Rational (v) Rational (vi) Irrational
13
2. Rational numbers : -1, , 1.25, 21.8 , 0
7
Irrational numbers : 2 , 7 , π , 2.131415....., 1.1010010001.....
5
3. , infinite solutions
3
A
4. 0.71727374....., 0.761661666..... 5. 5 = 2.236
AN
6. 2.645751 8. 5, 6
9. (i) True (ii) True (iii) True ( 3 ) (iv) True 9
3
(v) True (vi) False
7
G
EXERCISE - 1.4
1. (i) 10 + 5 5 + 2 7 + 35
N (ii) 20
LA
(iii)10 + 2 21 (iv) 4
2. (i) Irrational (ii) Irrational (iii) Irrational (iv) Rational
TE
7
5. (i) (ii) 7+ 6 (iii) (iv) 3 2 + 2 3
7 7
ER
3 2+2 3 9 15 − 3 10 − 3 21 + 14
6. (i) 17 −12 2 (ii) 6 − 35 (iii) (iv)
6 25
1
7. 0.3273 8. (i) 2 (ii) 2 (iii) 5 (iv) 64 (v) 9 (vi) 9. −8
6
SC
−19 5
10. (i) a = 5, b = 2 (ii) a = ,b= 11. 6 + 5
7 7
EXERCISE - 2.1
1. (i) 5 (ii) 2 (iii) 0 (iv) 6
(v) 2 (vi) 1
A
(v) (vi) (vii) 0 (viii) 0
2 3
AN
4. (i) Quadratic (ii) Cubic (iii) Quadratic (iv) Linear
(v) Linear (vi) Quadratic
5. (i) True (ii) False (iii) True (iv) False
(v) True (vi) True
G
EXERCISE - 2.2
1. (i) 3 (ii) 12
N (iii) 9 (iv)
3
LA
2
2. (i) 1, 1, 3 (ii) 2, 4, 4 (iii) 0, 1, 8 (iv) -1, 0, 3
(v) 2, 0, 0
TE
p
−2
ER
5. a = 6. a = 1, b = 0
7
EXERCISE - 2.3
SC
27
1. (i) 0 (ii) (iii) 1
8
−27
(iv) -π 3+3π 2-3π+1 (v)
8
−13
2. 5p 3. Not a factor, as remainder is 5 4. -3 5.
3
−13 21
6. 7. 8 8. 9. a = -7, b = -12
3 8
EXERCISE - 2.4
1. (i) Yes (ii) No (iii) No (iv) No
2. (i) Yes (ii) Yes (iii) Yes (iv) Yes
(v) Yes
A
7. (i) (x − 1) (x + 1) (x − 2) (ii) (x + 1)2 (x − 5)
(iii) (x + 1) (x + 2) (x + 10) (iv) (y + 1) (y + 1) (y − 1)
AN
9. a = 3 10. (y − 2) (y + 3)
EXERCISE - 2.5
G
1. (i) x2 + 7x + 10 (ii) x2 - 10x + 25
1
(iii) 9x2 - 4 (iv) x4 − (v) 1 + 2x + x2
x4
2. (i) 9999 (ii) 998001 N (iii)
9999
4
= 2499
3
4
LA
(iv) 251001 (v) 899.75
y y
3. (i) (4x + 3y)2 (ii) (2y − 1)2 (iii) 2x + 2x −
5 5
(iv) 2 (3a + 5) (3a − 5) (v) (x + 3) (x + 2)
TE
(vi) 3 (P − 6) (P − 2)
4. (i) x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz
(ii) 8a3 − 36a2b + 54ab2 − 27b3
(iii) 4a2 + 25b2 + 9c2 − 20ab − 30bc + 12ac
T
a2 b2 ab a
(iv) + + 1 − − b +
16 4 4 2
ER
4 2 8 3
(v) p3 + 3p2 + 3p + 1 (vi) x3 - 2x2y + xy - y
3 27
5. (i) (-5x + 4y + 2z)2 (ii) (3a + 2b - 4c)2
SC
6. 29
7. (i) 970299 (ii) 10,61,208 (iii) 99,40,11,992 (iv) 100,30,03,001
3
1
8. (i) (2a + b) 3
(ii) (2a − b) 3
(iii) (1 − 4a)3
(iv) 2p −
5
10. (i) (3a + 4b) (9a2 - 12ab + 16b2) (ii) (7y − 10) (49y2 + 70y + 100)
11. (3x + y + z) (9x2 + y2 + z2 − 3xy − yz − 3xz)
−5
14. (i) -630 (ii) 16380 (iii) (iv) −0.018
12
15. (i) (2a + 3) (2a − 1) (ii) (5a − 3) (5a − 4)
16. (i) 3x (x − 2) (x + 2) (ii) 4 (3y + 5) (y − 1)
A
EXERCISE - 3.1
AN
1. (i) 3 (ii) 13 (iii) 6 (iv) 180°
(v) Point, Plane, Line
2. a) False b) True c) True d) True
e) True 7. Infinite 8. Lines intersect on the side of the angle less than 1800
G
9. ∠1 = ∠2
EXERCISE - 4.1 N
LA
2. (i) Reflex angle (ii) Right angle (iii) Acute angle
3. (i) False (ii) True (iii) False (iv) False
(v) True (vi) True (vii) False (viii) True
TE
EXERCISE - 4.2
1. x = 36° y = 54° z = 90°
T
EXERCISE - 4.3
2. x = 126°
3. ∠AGE = 126° ∠GEF = 36° ∠FGE = 54°
4. ∠QRS = 60° 5. ∠ACB = z = x + y
6. a = 40° ; b = 100°
7. (i) ∠3, ∠5, ∠7, ∠9, ∠11, ∠13, ∠15
(ii) ∠4, ∠6, ∠8, ∠10, ∠12, ∠14, ∠16
8. x = 60° y = 59°
9. x = 40° y = 40°
10. x = 60° y = 18°
11. x = 63° y = 11°
13. x = 50° y = 77°
A
15. (i) x = 36°; y = 108° (ii) x = 35° (iii) x = 29°
16. ∠1 = ∠3 = ∠5 = ∠7 = 80° ; ∠2 = ∠4 = ∠6 = ∠8 = 100°
AN
17. x = 20° y = 60° z = 120°
18. x = 55° y = 35° z = 125°
19. (i) x = 140° (ii) x = 100° (iii) x = 250°
G
EXERCISE - 4.4
1.
2.
(i) x = 110°
∠1 = 60°
(ii) z = 130°
3. x = 35°, y = 51°
N (iii) y = 80°
5. x = 50° y = 20°
LA
6. x = 70° y = 40° 7. x = 30° y = 75°
8. ∠PRQ = 65° 9. ∠OZY = 32° ; ∠YOZ = 121°
10. ∠DCE = 92° 11. ∠SQT = 60° 12. z = 60°
TE
EXERCISE - 5.1
1. (i) Water Tank (ii) Mr. ‘J’ house
SC
(iii) In street-2, third house on right side while going in east direction.
(iv) In street 4, first building on right side while going in east direction.
(v) In street 4, the third building on left side while going in east direction
EXERCISE - 5.2
1. (i) Q2 (ii) Q 4 (iii) Q 1 (iv) Q 3
(v) Y-axis (vi) X-axis (vii) X-axis (viii) Y-axis
A
(v) (0, -8) : Y-axis (vi) (7, 0) : X-axis
(vii) (0, 0) : on both the axis.
AN
4. (i) -7 (ii) 7 (iii) R (iv) P
(v) 4 (vi) -3
5. (i) False (ii) True (iii) True (iv) False (v) False (vi) False
G
EXERCISE - 5.3
2.
3.
N
No. (5, -8) lies in Q4 and (-8, 5) lies in Q2
All given points lie on a line parallel to Y-axis at a distance of 1 unit.
LA
4. All points lie on a line parallel to X-axis at a distance of 4 units.
5. 12 Sq.units. 6. 8 Sq. units
EXERCISE - 6.1
TE
1 1
(v) a= b= c = -7
3 4
3
(vi) a= b=1 c=0
2
SC
EXERCISE - 6.2
A
17
2. (i) (0, -34); ( , 0) (ii) (0, 3) ; (-7, 0)
4
AN
3 −3
(iii) (0, ) ; ( , 0)
2 5
3. (i) Not a solution (ii) Solution (iii) Solution
(iv) Not a solution (v) Not a solution
G
8
4. k = 7 5. α= 6. 3
5
EXERCISE - 6.3 N
LA
2. (i) Yes (ii) Yes
3. 3
4. (i) 6 (ii) -5
TE
3
5. (i) ( , 3) (ii) (-3, 6)
2
6. (i) (2, 0) : (0, -4) (ii) (-8, 0) ; (0, 2)
(iii) (-2, 0) ; (0, -3)
T
EXERCISE - 6.4
1. 5x = 3y ; 2000 ; 480 (No. of voters who cast their vote = x,
Total no. of voters = y)
SC
EXERCISE - 6.5
4. (i) y = -3 (ii) y=4 (iii) y = -5 (iv) y=4
5. (i) x = -4 (ii) x=2 (iii) x=3 (iv) x = -4
A
EXERCISE - 7.4
AN
6. 7 7. No.
EXERCISE - 8.1
G
1. (i) True (ii) True (iii) False (iv) True
(v) False (vi) False
2. (a) Yes, No, No, No, No
(c) No, Yes, Yes, Yes, Yes
N (b)
(d)
No, Yes, Yes, Yes, Yes
No, Yes, Yes, Yes, Yes
LA
(e) No, Yes, Yes, Yes, Yes (f) No, Yes, Yes, Yes, Yes
(g) No, No, No, Yes, Yes (h) No, No, Yes, No, Yes
(i) No, No, No, Yes, Yes (j) No, No, Yes, No, Yes.
TE
EXERCISE - 8.3
1. Angles of parallelogram = 73°, 107°, 73°, 107°
T
EXERCISE - 8.4
1. BC = 8 cm.
SC
EXERCISE - 9.1
1. Marks 5 6 7 8 9 10
Frequency (f ) 5 6 8 12 9 5
2. Blood Group A B AB O
Frequency (f ) 10 9 2 15
Most common blood group = O ; Most rarest blood group = AB
3. No. of Heads 0 1 2 3
Frequency (f ) 3 10 10 7
4. Options A B C
Frequency (f ) 19 36 10
A
Total appropriate answers = 65
Majority of people’s opinion = B (Prohibition in public place only)
AN
5. Type of Vehicles Car Bikes Autos Cycles
No. of Vehicles (f) 25 45 30 40
G
on X-axis = 1 cm. = 10 number of students
Class
No. of students (f )
I
40 NII
55
III
65
IV
50
V
30
VI
15
LA
7. Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
(Class interval)
No. of students (f ) 1 4 3 7 7 7 1 0
TE
225 - 300 3
300 - 375 7
ER
375 - 450 7
450 - 525 0
525 - 600 1
SC
600 - 675 1
675 - 750 2
9. Life time (in years) 2-2.5 2.5-3.0 3.0-3.5 3.5-4.0 4.0-4.5 4.5-5.0
(Class Interval)
No. of Batteries 2 6 14 11 4 3
EXERCISE - 9.2
1. x = 85 2. x = 1 .7 1 3. K = 10
4. x = 17.7
5. (i) ` 359, ` 413, ` 195, ` 228, ` 200, ` 837
A
(ii) `444 saving per school.
AN
6. Boy’s height = 147 cm. ; Girl’s height = 152 cm.
7. x = 11.18 ; Mode = 5 ; Median = 10
8. x = 80 ; Median = 75 ; Mode = 50
9. 37 kgs 10. `11.25, Median = ` 10; Mode = ` 10
G
11. 1 = 2 ; 2 = 6 ; 3 = 19 ; 4th = 33
st nd rd
EXERCISE - 10.1
N
LA
1. (i) 64 cm2 , 96 cm2 (ii) 140 cm2, 236 cm2
2. 3375 m2 3. 330 m3 4. 8 cm.
5. (i) 4 times of original area (ii) 9 times of original area (iii) n2 times
TE
EXERCISE - 10.2
1. 6.90 m2 2. 176 cm2; 253 cm2
T
EXERCISE - 10.3
1. h = 6 cm. 2. h = 9 cm.
3. (i) 7 cm. (ii) 462 cm2 4. 1232 cm3
2 3
5. 1018.3 cm3 6. `7920, 15m 7. 3394 cm
7
8. 241.84 m2 (approximate) 9. 63m 10. 6135.8 cm2 11. 24.7 min
12. 18 2π
EXERCISE - 10.4
1. 154 cm2 ; 179.67 cm3 2. 3054.86 cm3
3. 616 cm2 4. 6930 cm2 5. 4 : 9 ; 8 : 27
6
6. 942 cm2 7. 1:4 8. 441 : 400 9. 55 gms or 0.055 kg
A
7
10. 5 cm. 11. 0.303 liters 12. No. of bottles = 9
AN
E XERCISE - 11.1
1. 19.5 cm2 2. 114 cm2 3. 36 cm2
G
E XERCISE - 11.2
1. 8.57 cm 2. 6.67 cm N
LA
E XERCISE - 12.1
1. (i) Radius (ii) Diameter (iii) Minor arc
TE
E XERCISE - 12.2
ER
E XERCISE - 12.4
SC
E XERCISE 12.5
1. (i) x° = 75° ; y° = 75° (ii) x° = 70° ; y° = 95°
(iii) x° = 90° ; y° = 40°
4. (a), (b), (c), (e), (f) = Possible ; (d) = Not possible
E XERCISE - 14.1
1
1. (a) 1, 2, 3, 4, 5 and 6 (b) Yes (c)
3
45 55
2. (a) ; (b) 1
A
100 100
3. (a) Red (b) Yellow (c) Blue and Green (d) No chance
AN
(e) No (It is random experiment)
4. (a) No.
5 1 1 1
(b) P (green) = ; P (blue) = ; P (red) = ; P (yellow) =
G
12 4 6 6
(c) 1
5. (a) P(E) =
5
26
(b) P(E) = N 5
13
(c) 1 (d)
21
26
LA
7
6. P(E) =
11
61 45 261 3.43
7. (i) P= (ii) P= (iii) P= 8.
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EXERCISE - 15.1
1. (i) Always false. There are minimum 28 days in a month. Usually we
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(iii) Ambiguous. At some time in winter, there can be a possibility that Hyderabad have
2°C temperature.
(iv) True, to the known fact, so far we can say this but it can be changed if scientists find
evidances of life on other planets.
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(v) In an equilateral triangle, a median is also an angle bisector.
4. Take any negative number x y
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-2 > -3
x2 = -2 × -2 = 4 (here x2 < y2)
y2 = -3 × -3 = 9
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EXERCISE - 15.2
1. (i) Jeevan is mortal N
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(ii) No, X could be any other state person lke marathi, gujarati, punjabi etc.
(iii)Gulag has red tongue.
(iv) All smarts need not be a president. Here we have given only that all presidents are
smart. There could be some other people like some of the teachers, students who
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E XERCISE - 15.3
1. (i) The possible three conjucture are:
a) The product of any three consecutive odd number is odd.
b) The product of any three consecutive odd number is divisible by 3.
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c) The sum of all the digits present in product of three consecutive odd numbers
is even.
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(ii) The possible three conjuctures are:
a) The sum of any three consecutive number is always even.
b) The sum of any three consecutive number is always divided by 3.
c) The sum of any three consecutive number is always divided by 6.
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4. 1111112 = 12345654321 11111112 = 1234567654321
Conjecture is true
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6. Conjecture is false because you can not find a composite number for x = 41.
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E XERCISE - 15.4
1. (i) No (ii) Yes (iii) No
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4. Let x = 2m and y = 2n
Product xy = (2m) (2n)
= 4 mn
6. (i) Let your original number be n. Then we are doing the following operations:
3n + 9
n → 2n → 2n + 9 → + n = 3n + 9 → = n +3 → n +3+ 4 = n + 7 → n + 7 − n = 7
3
(ii) Note that 7 × 11 × 13 = 1001. Take any three digit number say, abc. Then abc ×
1001 = abcabc. Therefore, the six digit number abcabc is divisible by 7, 11 and 13.
SYLLABUS
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and rational numbers on the number line.
l Representation of terminating / non terminating recurring
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decimals, on the number line through successive
magnification.
l Rational numbers as recurring / terminating decimals.
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l
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as 1.01011011101111—
1.12112111211112—
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and 2, 3, 5 etc.
l Existence of non-rational numbers (irrational numbers)
such as 2 , 3 and their representation on the number
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line.
l Existence of each real number on a number line by using
Pythogorian result.
l Concept of a Surd.
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l Rationalisation of surds
Square root of a surd of the form a + b
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x3 +y3 ≡ (x + y)(x2 − xy + y2)
x3 − y3 ≡ (x − y)(x2 + xy + y2)
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and their use in factorization of polynomials. Simple
expressions reducible to these polynomials.
(ii) Linear Equations in TwoVariables
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l Recall of linear equations in one variable.
l Introduction to the equation in two variables.
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Solution of a linear equation in two variables
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l Graph of a linear equation in two variables.
l Equations of lines parallel to x-axis and y-axis.
l Equations of x-axis and y-axis.
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(i) The Elements of l History – Euclid and geometry in India. Euclid’s method
Geometry of formalizing observed phenomenon onto rigorous
(ii) Lines and Angles
mathematics with definitions, common / obvious notions,
(iii) Triangles
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l (Motivate) Results on corresponding angles, alternate
angles, interior angles when a transversal intersects two
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parallel lines.
l (Motivate) Lines, which are parallel to given line, are
parallel.
(Prove) The sum of the angles of a triangle is 1800.
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Congruence).
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(iv) Quadrilaterals
• (Prove) The diagonal divides a parallelogram into two
congruent triangles.
• (Motivate) In a parallelogram opposite sides are equal
and conversely.
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• (Motivate) In a parallelogram opposite angles are equal
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and conversely.
• (Motivate) A quadrilateral is a parallelogram if one pair
of its opposite sides are parallel and equal.
• (Motivate) In a parallelogram, the diagonals bisect each
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other and conversely.
• (Motivate) In a triangle, the line segment joining the mid
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points of any two sides is parallel to the third side and
(motivate) its converse.
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(v) Area
• Review concept of area, area of planar regions.
• Recall area of a rectangle.
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(vi) Circles
• Through examples, arrive at definitions of circle related
concepts radius, circumference, diameter, chord, arc,
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subtended angle.
• (Prove) Equal chords of a circle subtend equal angles at
the centre and (motivate) its converse.
• (Motivate) The perpendicular from the centre of a circle
to a chord bisects the chord and conversely, the line drawn
through the centre of circle to bisect a chord is
perpendicular to the chord.
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double the angle subtended by it at any point on the
remaining part of the circle.
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• (Motivate) Angles in the same segment of a circle are
equal.
• (Motivate) A line segment joining any two points subtends
equal angles at two other points lying on the same side
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of it then the four points are concyclic.
• (Motivate) The sum of the either pair of the opposite
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angles of a cyclic quadrilateral is 1800 and its converse.
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(vii) Constructions
l Construction of a triangle given its base, sum / difference
of the other two sides and one base angles.
Construction of a triangle when its perimeter and base
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(i) Surface Areas and l Revision of surface area and volume of cube, cuboid.
Surface areas of cylinder, cone, sphere, hemi sphere.
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l
Volumes
l Volume of cylinder, cone, sphere. (including hemi spheres)
and right circular cylinders/ cones.
(i) Statistics
Statistics and Probability
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l
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l Throwing a large number of identical dice/coins together
and aggregating the result of the throws to get large number
of individual events.
l Observing the aggregating numbers over a large number
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of repeated events.Comparing with the data fora coin.
Observing strings of throws, notion of randomness.
Proofs in Mathematics
(5 hrs)
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(i) Proofs in Mathematics
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l Mathematical statements, verifying them.
(i) Proofs in l Reasoning Mathematics, deductive reasoning
Mathematics l Theorems, conjectures and axioms.
What is a mathematical proof.
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l
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Academic Standards
Academic standards are clear statements about what students must know and be able to do.
The following are categories on the basis of which we lay down academic standards
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Problem Solving
Using concepts and procedures to solve mathematical problems
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(a) Kinds of problems:
Problems can take various forms- puzzles, word problems, pictorial problems, procedural
problems, reading data, tables, graphs etc.
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(b) Problem Solving
Reads problems
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Identifies all pieces of information/data
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l Separates relevant pieces of information
l Understanding what concept is involved
l Recalling of (synthesis of) concerned procedures, formulae etc.
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l Selection of procedure
l Solving the problem
l Verification of answers of raiders, problem based theorems.
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(c) Complexity:
The complexity of a problem is dependent on
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l Context unraveling
l Nature of procedures
Reasoning Proof
l Reasoning between various steps (involved invariably conjuncture).
l Understanding and making mathematical generalizations and conjectures
l Understands and justifies procedures· Examining logical arguments.
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l Writing and reading, expressing mathematical notations (verbal and symbolic
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forms)
Ex: 3 + 4 = 7, 3 < 5, n1+n2= n2+n1, Sum of angles = 1800
l Creating mathematical expressions
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l Explaining mathematical ideas in her own words like- a square is closed figure having
four equal sides and all equal angles
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Explaining mathematical procedures like adding two digit numbers involves first adding
the digits in the units place and then adding the digits at the tens place/ keeping in
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mind carry over.
l Explaining mathematical logic
Connections
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