Download KTBS Not tobe republished - Karnataka Textbook Society

TEXT BOOK COMMITTEE
Chairman
Dr. B. J. Venkatachala, Professor, H.B.C.S.E(T.I.F.R, Mumbai),
Department of Mathematics, Indian Institute of Science, Bengaluru.
Members
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1. Dr. G. Sheela, Asst. Professor, Department of Education, Manasagangotri,
Mysore University, Mysuru.
2. Sri T. K. Raghavendra, Lecturer, D.I.E.T., Chickballapur.
3. Sri A. Ramaswamy, Asst. Master, Govt. Empress High School, Tumkuru.
4. Sri Vinay A. Joseph, Asst. Master and P.R.O., St. Xavier’s High School,
Shivajinagar, Bengaluru.
5. Smt. Vasanthi Rao, Retired Teacher, Rajajinagar, Bengaluru.
6. Sri G. M. Jangi, Artist, D.S.E.R.T., Bengaluru.
Scrutinizers
1. Dr. Ashok M. Limkar, Subject Inspector of Mathematics, D.D.P.I., Office,
Vijayapura.
2. Sri A. S. Hanuman, Subject Inspector of Mathematics, D.D.P.I., Office,
Shivamogga.
Editorial Committee Members
1. Dr. K. S. Sameera Simha, Joint Secretary, Vijaya Educational Institutions,
Jayanagar, Bengaluru.
2.Dr. S. Shiva Kumar, Professor, R.V. Engineering College, Bengaluru.
Chief advisors
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1. Sri Nagendra Kumar, Managing Director, Karnataka Text Book Soci-ety,
Bengaluru-85.
2. Sri Panduranga, Deputy Direcor(in-charge), Karnataka Text Book Society,
Bengaluru-85.
Chief Co-ordinator
Prof. G. S. Mudambadithaya, Curriculum review and Text book preparation,
Karnataka Text Book Society, Bengaluru-85.
Programme Co-ordinator
Smt. R. N. Shashikala, Asst. Director, Karnataka Text Book Society,
Bengaluru-85.
Foreword
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The Government of India through NCERT have brought out NCF-2005
to revise the curriculum of schools and suggested all the states to introduce revised textbooks in the schools based on the new curriculum. Accordingly state Governments took up the work and requested respective
DSERTs to start introducing new curriculum and texts. Karnataka Government has suggested to its DSERT to take up the challenge to fulfil the
vision of NCF-2005. DSERT, Karnataka started the process: constituted
committees to revise the syllabi, identified the writers and requested these
people to write texts books based on the new syllabi incorporating the expectations of NCF-2005. Karnataka Text Book Society, took the initiative
and coordinated the whole programme of writing these text books.
The current work, a text book in mathematics for 8-th standard, is a
step taken in this direction. An effort has been made here to look at the
mathematics needed at 8-th standard through a different lens. At first
glance, this may look a totally unconventional approach. Some may feel
that it is hard on the part of 8-th standard students. On the other hand
that is the correct age for the students to learn new concepts and ideas.
Students are receptive to new intellectual challenges. It is the onus of
the teachers to teach new things to the students and prepare them to the
challenges of the ever changing world. This text book is also an effort
to integrate our students with the national mainstream where CBSE has
surged forward and parents think that their wards will be better off by
learning CBSE texts.
We have tried here to tell something new about numbers and number
system. Similarly, some thing new about graphs, postulates of geometry
and congruency of triangles are also introduced with more expectations.
Quadrilaterals have been introduced now itself. There are optional problems at the end to challenge the students.
It is my earnest request to all my teacher friends to take up the new
challenge. Let the parents of our students not feel that their wards are
always in the back seats.
B. J. Venkatachala
Homi Bhabha Centre for
Science Education, TIFR, Mumbai
Table of Contents
Indian Mathematics - A brief introduction
(i)
Chapter 1.
1
Unit 2.
Squares, square-roots, cubes
and cube-roots
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Playing with numbers
27
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Unit 1.
Rational numbers
Unit 4.
Commercial arithmetic
Unit 5.
Statistics
102
Unit 1.
Algebraic Expressions
132
Unit 2.
Factorisation
148
Unit 3.
Linear equations in one variable
156
Unit 4.
Exponents
172
Unit 5.
Introduction to graphs
192
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Unit 3.
53
80
Chapter 2.
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Chapter 3.
Axioms, postulates and theorems
219
Unit 2.
Theorems on triangles
250
Unit 3.
Congruency of triangles
266
Unit 4.
Construction of triangles
291
Unit 5.
Quadrilaterals
306
Mensuration
330
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Unit 1.
Chapter 4.
Unit 1.
Optional problems
341
INDIAN MATHEMATICS - A BRIEF INTRODUCTION
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Indian Mathematics dates back to the Vedic times. The first
significant mathematical texts of Vedic times are Shulva Sutras.
Shulva is a sanskrit word for chord. These contain the details of
construction of sacrificial
altars. These ancient texts introduce
√ √
surds of the type 2, 3, etc. (In fact most of the ancient mathematics was developed because of the interest in Yajna and Yaga
and astrology.) Baudhayana√Sutra and Apastamba Sutra give a
very good approximation to 2 in the form
1+
1
1
1
+
−
,
3 3 × 4 3 × 4 × 34
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which is correct up to 5 decimal places.
The classical Pythagoras’ theorem is stated in the above sutras, far earlier than the Greeks discovered it. Another ancient
unsolved problem known as squaring a circle finds its place in
Shulva sutras. One is required to construct, using only a ruler
and a compass, a square whose area is equal to that of the given
circle. Shulva sutras give approximate methods for constructing such a square. This remained unsolved over two thousand
years, and only in 18th century it was proved that such a construction is impossible.
Indian mathematicians are credited with being the first to
give an approximate value for π . Aryabhata I (476 AD) gave an
approximate value for π as 3.1416; he mentions that a circle of
diameter 20000 units has circumference approximately equal
to 62832 units. It is interesting to note that Aryabhata I clearly
mentions, in the fifth century itself, that π is not rational and he
is using its approximate value. Only in 1761, Lambart proved
that π is an irrational number, and in 1882, Lindeman proved
that π is, in fact, a transcendental number.
Indian Mathematics
(ii)
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The most remarkable contribution for which the entire world
still salutes India is the invention of the decimal system by introducing zero and infinity. The use of decimal system is so
easy that even children can grasp it and use it in their calculations. If you really want to appreciate the simplicity of the decimal system and the concept of place value, you must first study
the prevalent Roman system of representing numbers. According to Florian Cajori, an eminent historian of great repute “ of
all the mathematical discoveries, no one has contributed
more to the general progress of intelligence than Zero.” Using base 10, Indians were able to grasp very large numbers(See
Chapter 2, Unit 4, Exponents for more details).
Ancient Jain contribution to mathematics is another important milestone in the history of Indian Mathematics. Their findings are recorded in famous Jain texts, dating back to 500 BC
to
√ 200 BC. Here again, you see an approximate value for π as
10 and it is calculated up to 13 decimal places.
Ancient India has vastly contributed in the areas: the methods of Arithmetic called Vyakthaganita; The method of Algebra
called Avyakthaganita. Ancient Indian Mathematicians had introduced all the four operations: addition, multiplication, subtraction and division. They also knew how to operate with fractions, solving simple equations, finding square and square-root,
finding cube and cube-root, and also knew about permutation
and combination.
Mahavira(9th century AD), a great Jain mathematician from
Karnataka, gave the well known formula
nC
r=
n!
(n − r)!r!
for the first time in the history of mathematics, in his Ganita
Sara Sangraha. Aryabhata I is one of our greatest mathematicians and astronomers of all times. He is the one who systematically developed mathematics and is called, justifiably, the father of Algebra. He gave tables to trigonometric ratio Sine,
Indian Mathematics
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called jya in Sanskrit, for angles from 0 to 90 degrees at inter3
vals of 3 degrees. He was the first Indian to declare: the Earth
4
is round and that stars appear to move from East to West for a
stationary observer on the Earth.
Aryabhata I was followed by Bhaskara I (6th century AD) who
provided interesting geometrical treatment for many algebraic
formulae and gave a very good rational approximation for sin θ,
even for large values of angle θ. Brahmagupta(628 AD) was
the first to
pgive a formula for the area of a cyclic quadrilateral in
the form (s − a)(s − b)(s − c)(s − d), where a, b, c, d are the sides
of the quadrilateral and s, the semi-perimeter. He is also the
first mathematician to obtain cyclic quadrilaterals with rational
sides.
Another important area, where our ancient mathematicians
made significant contributions, is the solution of equations of
the form ax + by = c, and x2 − N y 2 = 1 in integers, where
a, b, c, N are given integers, and N is generally a square-free positive integer. Such equations are called Diophantine equations
in modern terminology. The second equation is wrongly called
Pell’s equation by the great Euler, but the name still continues. These days, many authors call the equation of the type
x2 − N y 2 = 1 as Brahmagupta-Pell’s equation. Aryabhata I discusses ax + by = c, whereas Brahmagupta made a significant
contribution in the understanding of the equation x2 − N y 2 = 1.
However, later, Bhaskara II(popularly called Bhaskaracharya)
developed a new method, called Chakravala method, for solving x2 − N y 2 = 1 in integers. Here is an interesting thing about
the equation x2 − N y 2 = 1. In 1657, Fermat (who is famous
for his contribution to number theory) proposed the problem of
solving x2 − 61y 2 = 1 for integers x, y , to European mathematicians. But none was able to solve this problem. In 1732, Euler
gave a complete solution to this equation. But surprisingly, as
a matter of divine coincidence, the same equation x2 − 61y 2 = 1
was solved by Bhaskara II(1150 AD)in his Bijaganitam by his
Indian Mathematics
(iv)
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chakravala method more than 5 centuries before. The smallest
solution of x2 − 61y 2 = 1 given by Bhaskara II is x = 226153980
and y = 1766319049. Bhaskara II also introduced calculus, the
concept of derivative, though not rigorously. He clearly mentions equivalent of d(sin x) = cos x dx.
We also see in the works of Bhaskara II and Mahavira very
beautiful and interesting problems, rich in poetic imagination.
Incidentally, Bhaskara II was born in the current Bijapur district of Karnataka state (the exact place of his birth is still a
point of debate among the scholars) and moved to the current
Maharashtra state. Here we give two problems, originally stated
in verses, whose translation runs as follows:
1. (From Lilavati of Bhaskaracharya)
A beautiful maiden asks me which is the number when multiplied by 3, then increased by three-fourths of the product,
divided by 7, diminished by one-third of the quotient, multiplied by itself, diminished by 52, the square-root found,
addition of 8, division by 10 gives the number 2?
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2. (From Ganita Sara Sangraha of Mahavira)
Three merchants find a purse lying on a road. One merchant says, “ If I keep the purse, I shall have twice as much
money as the two of you together.” Give me the purse and I
shall have three times as much" said the second merchant.
The third merchant said, “ I shall be better off than either
of you if I keep the purse; I shall have five times as much as
the two of you together.” How much money is in the purse?
How much money does each merchant have?
The achievements of Indian Mathematicians are remarkable
in certain specific areas: arithmetic; theory of equations; spherical trigonometry and astronomy; geometrical treatment of algebraic equations; plane trigonometry; and mensuration. Unfortunately, after Bhaskara II, because of the invasion, Indian
Mathematics went into hibernation, except for a brief period
Indian Mathematics
(v)
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during which the Kerala school by Nilakanta and Madhava made
some important contributions in the area of series approximation to tan function. With Ramanujan, at the end of 19th century, it recovered its earlier glory. Ramanujan was a prodigious
mathematician. In his brief span of 32 years, he made wonderful contributions to number theory, hyper-geometric series, divergent series, elliptic functions and integrals, and mock-theta
function. Even today, world-wide mathematicians are trying to
understand the depth of his mathematics and trying to prove
the conjectures he made. It is also worthwhile to mention that
Chadrashekhara Samantha of Orissa, at the end of 19th century,
made some important contributions to astronomy.
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(For more details about the Indian mathematics see “Indian
Mathematics and Astronomy: Some landmarks” by Dr S Balachandra Rao, published by Bharatiya Vidya Bhavan, Bengaluru)
CHAPTER 1
UNIT 1
PLAYING WITH NUMBERS
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After studying this unit you learn:
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• to write a given natural number in its general form ( in base 10);
• to formulate some games and puzzles involving numbers;
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• given two positive integers, how to divide one by the other to get the
quotient and the remainder;
• construction of a 3 × 3 magic square;
• divisibility tests for 4,3,9,5,11;
• about some unsolved problems involving numbers.
1.1.1 Introduction
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Numbers have played an important role in the intellectual development
of mankind. This still forms a perfect play-ground for activities of children. One can create simple puzzles which are brain-teasers. They can be
used to play (mental) games among children. We shall explore some nice
properties of numbers which help in engaging children with puzzles and
these puzzles will help in arousing the curiosity of children. On the other
hand the numbers can help us in placing some of the hitherto unproved
conjectures and perhaps those will help the children to further explore the
wonderful universe of numbers.
You write 76 or 315 and say these are natural numbers. For example,
you say that 6 is the digit in the unit’s place of 76, and 7 is the digit in
ten’s place. Similarly, looking at 315, you say that 5 is in the unit’s place,
1 is in the ten’s place and 3 is in the hundred’s place. You understand the
place value of each digit, given a natural number. We explore a little bit
more about these concepts and create puzzles using them.
Looking at the numbers 2, 24, 46, 88 or 122, you immediately say that
these are even numbers and each is divisible by 2. Can you say whether a
number is divisible by 3 by just looking at the number? Can you say that
Unit 1
2
a number is divisible by 4,5, 9 or 11 without actually dividing it by these?
Can you device simple rules which help us to decide whether a number is
divisible by 3,4,5,9,11?
1.1.2 Numbers in general form
Consider the number 45. We write this as
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45 = 40 + 5 = (4 × 10) + (5 × 1).
Similarly, 34 = 30 + 4 = (3 × 10) + (4 × 1). What can we do with 354? Observe
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354 = 300 + 50 + 4 = (3 × 100) + (5 × 10) + (4 × 1).
Activity 1: Write the following numbers in the form described as above:
75, 88, 121, 361, 1024, 2011, 4444, 2345.
Can you see that any natural number can be written in the above form? It
is immaterial how many digits are there in the number. Suppose you have
123456789, a 9-digit number. You may write it as
123456789 = 100000000 + 20000000 + 3000000 + 400000
+50000 + 6000 + 700 + 80 + 9
= (1 × 100000000) + (2 × 10000000) + (3 × 1000000)
+(4 × 100000) + (5 × 10000) + (6 × 1000)
to
+(7 × 100) + (8 × 10) + (9 × 1).
You will learn later that this can be written in a compact form as
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123456789 = (1 × 108 ) + (2 × 107 ) + (3 × 106 ) + (4 × 105 ) + (5 × 104 )
+ (6 × 103 ) + (7 × 102 ) + (8 × 101 ) + (9 × 100 ).
This is called the base 10 representation of the given natural numbers
or the generalised form of the number. This system of representing a
number using base 10 was invented by early Indian mathematicians.
Consider, for example, 136. You write this in the generalised form as:
136 = (1 × 100) + (3 × 10) + (6 × 1).
Can you see that 6 is associated with 1; 3 is associated with 10; and 1 is
associated with 100? This is the reason, 6 is called the digit in the unit’s
Playing with numbers
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place; 3 is the digit in the ten’s place; and 1 is the digit in the hundred’s
place.
Suppose you have a number abcd, with the digits in the unit’s place,
ten’s place’ hundred’s place and thousand’s place respectively as d,c,b and
a. Then its generalised form is
abcd = (a × 1000) + (b × 100) + (c × 10) + (d × 1).
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To avoid the confusion that abcd may represent the product of a, b, c and
d, the number is written in the form abcd. Thus
abcd = (a × 1000) + (b × 100) + (c × 10) + (d × 1).
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Indian mathematics emerged in the Indian subcontinent from 1200 BC
until the end of the 18th century and after that the modern era dawned.
In the classical period of Indian mathematics (400 AD to 1200 AD), important contributions were made by scholars like Aryabhata, Brahmagupta,
and Bhaskara II. The decimal number system in use today and the binary
number system were first recorded in Indian mathematics. Indian mathematicians made early contributions to the study of the concept of zero as a
number, negative numbers, arithmetic, and algebra. These mathematical
concepts were transmitted to the Middle East, China, and Europe and led
to further developments that now form the foundations of many areas of
mathematics.
All mathematical works were orally transmitted until approximately 500
BC; thereafter, they were transmitted both orally and in manuscript form.
The oldest mathematical document produced on the Indian subcontinent
is the birch bark Bakhshali Manuscript, discovered in 1881 in the village
of Bakhshali, near Peshawar (modern day Pakistan).
The representation using base 10 is only a convenient thing. One can use
different bases and represent numbers. For example computers use base
2 representation(called binary codes) and base 16 representation(hexadecimal codes). However, in daily life the use of decimal system(base 10
representation) is the most useful thing and the Indian contribution is forever remembered.
Exercise 1.1.2
1. Write the following numbers in generalised form:
39, 52, 106, 359, 628, 3458, 9502, 7000.
Unit 1
4
2. Write the following in the normal form:
(i) (5 × 10) + (6 × 1); (ii) (7 × 100) + (5 × 10) + (8 × 1); (iii) (6 × 1000) +
(5 × 10) + (8 × 1); (iv) (7 × 1000) + (6 × 1); (v) (1 × 1000) + (1 × 10).
3. Recalling your earlier knowledge, represent 555 in base 5.
1.1.3. Some games and puzzles involving digits
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4. What is the representation of 1024 in base 2?
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Game 1.
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Here we describe some properties of numbers which will help you to
evolve a game to amuse your friends.
You can play a trick with your friend. You do this in several steps.
Step 1. Ask your friend to choose a 2-digit number in his mind and
not to reveal it to you.
Step 2. Tell him to reverse the digits of the number he chose and get
another number.
Step 3. Now tell him to add both the numbers and divide the sum by
11.
Step 4. Surprise him by telling him that the remainder is 0.
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At no stage he reveals you the number or its reversal or their sum. Still
you can conclude that the remainder of the sum when divided by 11 is
zero.
For example suppose your friend chooses 41. The number obtained by
reversing its digits is 14. Their sum is 41 + 14 = 55. When 55 is divided by
11, the remainder is 0.
Are you not curious how it works?
Suppose the two digit number is ab. Then you know that ab = (a × 10) +
(b × 1). The reversed number is ba = (b × 10) + (a × 1). Thus you get the sum
of a number and its reversal as:
ab + ba = (a × 10) + (b × 1) + (b × 10) + (a × 1) = 11(a + b).
Now you see why the remainder when divided by 11 is zero.
Activity 2: You can create a game of your own. Instead of taking the
sum of a 2-digit number and its reversal, suppose you take their positive
Playing with numbers
5
difference. Take several examples, say, 21, 34, 86, 79, 95. Which divisor
is common to all the differences: 21 − 12, 43 − 34, 86 − 68, 97 − 79, 95 − 59?
What is the game you can formulate?
Game 2.
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This time, tell your friend to choose a 3-digit number and to keep it in
his mind. Let him get the number obtained by reversing the digits of the
original number and tell him to find the difference between the original
number and the reversed number. Ask him to divide this difference by 99.
You may surprise him by telling the remainder is zero, even if you do not
know any thing about his choice. For example, if your friend chooses 891,
the reversed number is 198 and their difference is 891 − 198 = 693 = 99 × 7.
Hence the remainder is zero after dividing the difference by 99.
Activity 3: Take several 3 digit numbers, say, 263, 394, 512, 765, 681,
898, 926. Find the difference between each number and the number obtained by its reversal. Find the remainder when the difference is divided
by 99 in each case.
Why does this work in general? If the number chosen is abc, then the
reversed number is cba. Thus
abc − cba = (a × 100) + (b × 10) + (c × 1) − (c × 100) − (b × 10) − (a × 1)
to
= (99 × a) − (99 × c)
= 99(a − c).
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Hence you see that the difference is always divisible by 99.
Game 3.
Now you start with a 3-digit number, say 132. You can get two more
numbers 213 and 321, by cyclically permuting the digits of 132. Add all
of them; you get
132 + 213 + 321 = 666 = 18 × 37.
Repeat this with numbers 196, 225, 308, 446, 589, 678, 846. Do you
observe that in each case the resulting number is divisible by 37? Can
you now formulate a game using this property?
Unit 1
6
Statement 1.
Given a 3-digit number abc, consider two more numbers obtained
by cyclical permutation of its digits, namely, bca and cab. Then 37
divides the sum abc + bca + cab.
321 = (3 × 100) + (2 × 10) + (1 × 1),
Thus we get
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213 = (2 × 100) + (1 × 10) + (3 × 1).
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132 = (1 × 100) + (3 × 10) + (2 × 1),
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The proof is not hard. Let us see how it works for 132. We have
132 + 321 + 213 = 1 × (10 + 100 + 1) + 3 × (100 + 10 + 1) + 2 × (100 + 1 + 10)
= (1 + 3 + 2) × 111
= 6 × 3 × 37.
The same method works for for any number abc. Using the general form,
you get
abc = (a × 100) + (b × 10) + (c × 1),
bca = (b × 100) + (c × 10) + (a × 1),
Hence
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cab = (c × 100) + (a × 10) + (b × 1).
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abc + bca + cab = (a × 100) + (b × 10) + (c × 1) + (b × 100) + (c × 10) + (a × 1)
= 111(a + b + c).
+(c × 100) + (a × 10) + (b × 1)
But 111 = 37 × 3 and hence the number on the right side is divisible by 37.
We may conclude that abc + bca + cab is divisible by 37.
Alpha numerals and puzzles
You can create puzzles involving numbers and letters of an alphabet.
Look at the following examples.
Example 1. Find the digit represented by P in the following addition.
Playing with numbers
7
Solution:
You see that P , being a digit, cannot exceed 9. The
only way you can arrive to 6 from 5 is adding 1.
+
Hence P = 1. Similarly, you get Q = 1. You may
check that 411 + 115 = 526.
Example 2. Find the digits A and C in the following multiplication.
4
Q
5
1 P
1 5
2 6
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Solution:
Solution:
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Here the last digit of 2 × A is 4. Hence either A = 2
or A = 7. Which one to choose?
Suppose A = 2. Then you get the product as 32 ×
3 A
12 = 384. This shows that C = 3. On the other hand,
× 1 2
if A = 7, then the product is 37 × 12 = 444. However,
C 8 4
the digit in the ten’s place of the result must be
8 and not 4. We may therefore reject A = 7. We
conclude A = 2 and C = 3.
Example 3. In the following addition, A, B, C represent different digits.
Find them and the sum.
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Here you observe that the last digit of A + B + C is
C, so that A + B = 10. (Why is that A + B = 0 not
possible?) Since C is a digit, C ≤ 9. Hence the carry
from unit’s place to ten’s place is 1.
A A
Since we are adding only 3 digits, the carry in ten’s
+ B B
place of the sum cannot exceed 2. Hence B cannot
+ C C
be more than 2. Thus you observe that B = 1 or 2.
B A C
The addition of digits in ten’s place gives(along with
carry 1 from unit’s place) A + B + C + 1 = 10 + C + 1
and this must leave remainder A when divided by
10.
If B = 1, then A = 9 and hence C +1 = 9 giving C = 8. We get 99+11+88 =
198, which is a correct answer. If B = 2, you get A = 8 and C + 1 = 8 giving
C = 7. But then 88 + 22 + 77 = 187. This does not fit in as hundred’s place
in the sum is 1 but not 2.
The correct answer is 99 + 11 + 88 = 198.
Unit 1
8
Exercise 1.1.3
1. In the following, find the digits represented by the letters:
3
+ B
7
16
+ 2A
B1
(iv)
(v)
1A
×
1A
1 B A
×
(vi)
2A
A
12 A
3A
×
A
2BA
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1A A
+ 1A A
2 A A
(iii)
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(i)
2. In the adjacent sum, A, B, C are consecutive digits. In the third row, A, B, C appear in some order.
Find A, B, C.
3. Find A, B, C in the adjacent multiplication table.
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4. Is it possible to have the adjacent multiplication
table? Give reasons.
A
+ C
+ –
1 2
B
B
–
4
C
A
–
2
A
×
A C
B
A
6
C
A
C
A
× B
C C
A
B
C
1.1.4. Divisibility and remainders
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One of the important property related to integers is the concept of divisibility. In your earlier class, you have studied how to divide a natural
number by another natural number to get a quotient and a remainder.
If you divide 91 by 13, you see that 13 completely divides 91 and you do
not get any remainder. On the other hand, dividing 85 by 15, you see that
15 × 5 = 75 and 15 × 6 = 90, so that you cannot divide 85 completely by 15.
On actual division, you get 5 as quotient and 10 as remainder.
13
91
7
15
85
5
91
75
—–
—–
00
10
Playing with numbers
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Suppose you divide 304 by 12. The quotient is 25 and the remainder is
4. If you divide 887 by 17, you obtain the quotient 52 and the remainder
3.
12
304
25
17
887
52
24
85
————64
37
60
34
————4
3
We write these divisions in the following form:
91 = (13 × 7) + 0,
85 = (15 × 5) + 10,
304 = (12 × 25) + 4,
887 = (17 × 52) + 3.
Do you observe that 0 < 13, 10 < 15, 4 < 12, 3 < 17? Can you conclude
that the remainder does not exceed the number from which you divide?
Activity 4: Find the quotient and the remainder in each of the following
cases:
(i) 100 divided by 2,3,5,7,11,13,17,23, 29 and 31.
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(ii) 300 divided by 37, 41,43,47,53,59,61,67.
Our observation can be put in a formal way:
Given a non-negative integer a and another integer b > 0, there
exist unique integers q and r such that a = (b × q) + r, where 0 ≤ r < b.
We say b divides a if the remainder is zero, that is, r = 0.
A similar statement can be made when a number is divided by another
non-zero number. You will learn more about these in higher classes. The
above statement can be used as a basis for a nice game which you can
play with your friends.
Game 4.
Ask your friend to choose a number smaller than 1000. Tell him to divide
this number by 7, 11, 13 respectively and ask for the remainders obtained
Unit 1
10
(2 × 715) + (7 × 364) + (11 × 924).
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by these divisions. Using these remainders, you can construct the number
chosen by your friend.
Suppose your friend has chosen 128. Then the remainder when divided
by 7 is 2; the remainder when divided by 11 is 7 and the remainder when
divided by 13 is 11. Now form the sum
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14142 = (1001 × 14) + 128,
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If you simplify this, you get 14142. Divide this by 1001. You see that
so that the remainder is 128. This is the number chosen by your friend.
Are you not thrilled?
These are the steps in this game.
Step 1. Tell your friend to choose a number less than 1000, in his
mind.
Step 2. Tell him to divide the number by 7, 11, 13 and ask him to
give you three remainders.
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Step 3. Now you construct the number he thought of using the three
remainders as follows. Suppose the remainders he gives you are
r1 (remainder when divided by 7), r2 (remainder when divided by 11),
and r3 (remainder when divided by 13); multiply r1 by 715, r2 by 364
and r3 by 924; take care that you are doing the correct multiplication.
Add all three numbers so obtained and divide the resulting number
by 1001. The remainder you obtain is the number chosen by your
friend.
Take another example, say 212. Observe that
212 = (7 × 30) + 2;
212 = (11 × 19) + 3;
212 = (13 × 16) + 4.
Thus r1 = 2, r2 = 3 and r3 = 4. We obtain
(r1 × 715) + (r2 × 364) + (r3 × 924) = (2 × 715) + (3 × 364) + (4 × 924) = 6218.
Divide 6218 by 1001. The remainder is 212, the number you started with.
Playing with numbers
11
You may be wondering how such a game works. Suppose you start with an
arbitrary number a < 1000. Let r1 , r2 , r3 be the remainders when a is divided
by 7, 11, 13 respectively. Then you can write
a = 7q1 + r1 ,
a = 11q2 + r2 ,
a = 13q3 + r3 ,
r1 = a − 7q1 ,
r2 = a − 11q2 ,
r3 = a − 13q3 .
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Hence
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for some integers q1 , q2 , q3 . This shows that
715r1 + 364r2 + 924r3 = 715(a − 7q1 ) + 364(a − 11q2 ) + 924(a − 13q3 )
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= a(715 + 364 + 924) − (7 × 715)q1 − (11 × 364)q2 − (13 × 924)q3 .
However you may notice that
7 × 715 = 7 × 11 × 13 × 5,
11 × 364 = 11 × 7 × 13 × 4,
13 × 924 = 13 × 7 × 11 × 12.
And 1001 = 7 × 11 × 13. Do you now understand why we consider the remainders when divided by 7, 11 and 13? Hence you get
715r1 + 364r2 + 924r3 = a × 2003 − 1001(5q1 + 4q2 + 12q3 ).
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You may also observe that a × 2003 = (a × 1001 × 2) + a. When you divide
715r1 + 364r2 + 924r3 by 1001, you are left with a as all other terms are divisible
by 1001. Since a < 1000, a is indeed the remainder. But that is the number
you have started with.
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Activity 5:
Check the game 4 with some more numbers: 804, 515, 676, 938, 97, 181.
Exercise 1.1.4
1. Find the quotient and the remainder when each of the following number is divided by 13:
8,31, 44, 85, 1220, 2011.
2. Find the quotient and the remainder when each of the following number is divided by 304:
128, 636,785,1038,2236,8858,13765,58876.
3. Find the least natural number larger than 100 which leaves the remainder 12 when divided by 19.
Unit 1
12
4. What is the least natural number you have to add to 1024 to get a
multiple of 181?
5. What is the smallest number you have to add to 100000 to get a
multiple of 1234?
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1.1.5 Magic squares
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Can you arrange the numbers from 1 to 9 in 3 rows and 3 columns
such that the sum of the numbers in each row, each column and each
diagonal are all identical? Look at the following arrangement.(Fig. 1)
8
1
6
6
1
8
3
5
7
7
5
3
4
9
2
2
9
4
Fig. 1
Fig. 2
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You observe that the sum of the numbers in each row is 15, the column
sum is 15 and the diagonal sum is 15. You can also arrange the numbers
as in Fig. 2. Can you see that there is a resemblance between these two
magic squares? The middle column is (1, 5, 9) in both the squares. The
right most column (8, 3, 4) in the first square is the left most column in
the second. Similarly, the left most column (6, 7, 2) in the first square is
the right most column in the second. Thus the second magic square is
obtained by flipping the right most column and the left most column. The
sum 15 is called the magic sum.
Is there a way of constructing such a magic square? Start with the
middle cell of the topmost row and put 1 there. Thus we begin with the
following:
1
Now we follow the following rules:
Playing with numbers
13
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Rule 1. If there is a vacant cell
along the diagonal from left to
right, you fill it with the next num−→
5
ber. (Here there is a vacant cell
4
4
after 4 along the diagonal. Hence
we fill it with 5.)
Rule 2. If there is no vacant cell
1
1
along the diagonal and if there are
−→
columns further, you fill the bot2
tom cell of the next column with
the next number. Follow Rule 1. (Here there is no vacant cell along
the diagonal after 1. Hence we go to the bottom most cell in the next
column and fill with 2.)
Rule 3. If there is no vacant cell
1
1
along the diagonal and if there are
−→
3
no columns further, go to the row
2
2
above the cell you have reached
and fill the left most cell of this row with the next number and follow
Rule 1. (Here there is no vacant cell along the diagonal to move from
2 and we are already in the last column. Hence we move to the row
above and fill the left most cell with 3.)
Rule 4. If at any stage you en1
1 6
counter a cell which has already
−→
3
3 5
been filled, go to the cell below the
4
cell you have reached and then
continue with Rule 1. (Here you cannot move from 3 to the next cell
along the diagonal since you have 1 already there. Hence we go to the
cell below 3 and fill with this 4.)
Rule 5. If you are at the end of
1 6
1 6
the main diagonal, fill the number
−→
3 5
3 5 7
below the last cell in the diagonal
4
4
with the next number. And follow
the appropriate rule further. (Here you have 6 at the end of the main
diagonal. Hence we go to the cell below it and fill it with 7.)
Unit 1
14
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Let us see how it works for a 3 × 3 magic square. We start with the
central cell of the first row and put 1 there. Now apply Rule 2, as we
cannot move diagonally. We fill the bottom cell of the next column, that
is column 3, with number 2. Again, we cannot move diagonally, nor do
we have column further. We apply rule 3 and move to the row above and
fill the left most cell with 3. Now you see that you cannot move along the
diagonal as the cell there is already filled. Hence we use Rule 4 and go to
the cell below the cell we are in. We fill this with 4 and move diagonally
to fill 5 and 6. Again, we cannot move further and we are on the main
diagonal. We use Rule 5 and fill the cell below the last cell on the diagonal
with 7. Now we use Rule 3 and fill the left most cell of the row above with
8. Using Rule 2, we now fill the bottom most cell of the next column with
9. And you have the magic square!
The sequence of operations are shown below
1
1
→ Rule 2 →
2
3
4
1 6
5 7
2
→ Rule 1 →
→ Rule 3 →
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3
4
2
1
3 5
4
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1
1
8
3
4
→ Rule 3 → 3
→ Rule 1 → 3
2
4
1 6
5 7
2
8
→ Rule 2 → 3
4
2
1 6
5
2
→ Rule 4 →
→ Rule 5 →
1 6
5 7
9 2
Activity 6: Using the central cell of the first column as the starting point,
construct a 3 × 3 magic square with numbers from 1 to 9. How does this
compare with the magic square in Fig. 1? What relation is there between
the magic sum and the number in the central cell of the magic square?
Using the numbers from 3 to 11, construct a magic square.
Solution: We use the same sequence of operations as we have used earlier,
but the starting number is 3 instead of 1.
Playing with numbers
15
→ Rule 2 →
5
6
4
5
6
3 8
7 9
4
4
→ Rule 1 →
3
5 7
6
→ Rule 3 →
10 3
5 7
6
→ Rule 3 → 5
→ Rule 1 → 5
4
6
4
3 8
7
4
→ Rule 4 →
→ Rule 5 →
8
10 3 8
9 → Rule 2 → 5
7 9
4
6 11 4
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3
3
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3
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3
Here the magic sum is 21.
Activity 7: Construct a 5 × 5 magic square using the above rules and with
numbers from 1 to 25. What relation is there between the magic sum and
the number in the central cell of the magic square?
Using the five rules, it is possible to construct an m × m magic
square with the numbers from 1 to m2 , for any odd natural number m > 1.
Exercise 1.1.5
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1. Using the numbers from 5 to 13, construct a 3×3 magic square. What
is the magic sum here? What relation is there between the magic sum
and the number in the central cell?
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2. Using the numbers from 9 to 17, construct a 3×3 magic square. What
is the magic sum here? What relation is there between the magic sum
and the number in the central cell?
3. Starting with the middle cell in the bottom row of the square and
using numbers from 1 to 9, construct a 3 × 3 magic square.
4. Construct a 3 × 3 magic square using all odd numbers from 1 to 17.
5. Construct a 5 × 5 magic square using all even numbers from 1 to 50.
Unit 1
16
1.1.6 Divisibility tests
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If a number ends with any of the digits 0, 2,4,6 or 8, you immediately
say the number is divisible by 2. What is your reasoning? You write any
such number a as a = 10k +r, where r is the remainder when divided by 10.
Hence r is one of the numbers 0,2,4,6,8. You now see that 10 is divisible
by 2 and r is also divisible by 2. You conclude that 2 divides a.
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1. Divisibility by 4
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It is natural to think whether such simple tests are available for divisibility by other numbers. We explore some of them here.
If a number is divisible by 4, it has to be divisible by 2(why?). Hence the
digit in the unit’s place must be one of 0,2,4,6,8. But look at the following
numbers: 10,22,34,46, 58. You see the last digit in each of these numbers
are as required, yet none of them is divisible by 4? Thus you may conclude
that it is not possible to decide the divisibility on just reading the last digit.
Perhaps, the last two digits may help.
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If a number has two digits, you may decide the divisibility by actually
dividing it by 4. All you need is to remember the multiplication table for 4.
Suppose the given number is large, say it has more then 2 digits. Consider
the numbers, for example, 112 and 122. You see that 112 = 100 + 12, and
both 100 and 12 are divisible by 4. You may conclude that 112 is divisible
by 4. But 122 = 100 + 22; here 100 is divisible by 4, but 22 is not. Hence
122 is not divisible by 4. We invoke the following fundamental principle on
divisibility:
If a and b are integers which are divisible by an integer m 6= 0,
then m divides a + b, a − b and ab.
How does this help us to decide the divisibility of a large number by 4?
Suppose you have a number a with more than 2 digits. Divide this by 100
to get a quotient q and remainder r: a = 100q + r, where 0 ≤ r < 100. Since
4 divides 100, you will immediately see that a is divisible by 4 if and only
if r is divisible by 4. But r is the number formed by the last two digits of
a. Thus you may arrive at the following test:
Playing with numbers
17
Statement 2
A number a (having more than one digit) is divisible by 4 if and only
if the 2-digit number formed by the last two digits of a is divisible by
4.
Example 4. Check whether 12456 is divisible by 4
Example 5. Is the number 12345678 divisible by 4?
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Solution: Here, the number formed by the last two digits is 56. This is
divisible by 4 and hence so is 12456.
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Solution: The number formed by the last 2 digits is 78, which is not
divisible by 4. Hence the given number is not divisible by 4.
Activity 8:
Ask your friend sitting adjacent to you to give several 4, 5 and 6 digit
numbers. Test divisibility by 4 for them.
Activity 9:
By dividing several 4 and 5 digit numbers by 8, formulate a divisibility test
by 8.
2. Divisibility by 3 and 9
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Consider the numbers 2,23,234,2345, 23456, 234567. We observe that
among these 6 numbers, only 234 and 234567 are divisible by 3. Here,
we cannot think of the number formed by the last two digits, or for that
matter even three digits. Note that 3 divides 234, but it does not divide
34. Similarly, 3 divides 456, but 3 does not divide 23456.
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Activity 10: Write down numbers 1, 11, 21, 31, 41, . . . , 141, 151. (All numbers
from 1 to 151 which ends in 1.) Form the sum of the digits of each number
and tabulate them. Check which numbers are divisible by 3 and whether
the digital sum of that number is also divisible by 3? What do you observe?
Consider the numbers 234 and 234567. The sum of the digits of the
first number is 9 and that of the second is 27. You see that both 9 and 27
are divisible by 3. (In fact, divisible by 9.) Let us explore this in a general
case for a 2-digit, 3-digit and 4-digit numbers. If n = ab is 2-digit number,
then
n = ab = (10 × a) + b = 9a + (a + b).
Unit 1
18
This shows that n is divisible by 3 if and only if a + b is divisible by 3.
Similarly for m = pqr, we have
m = pqr = 100p + 10q + r = (99p + 9q) + (p + q + r),
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and you may observe that m is divisible by 3 if and only if p + q + r is
divisible by 3. Do you observe that you can deduce divisibility test for 9 as
well: 9 divides m if and only if 9 divides p + q + r? Now you do not have any
problem to extend the test for a 4-digit number or a number with more
digits. Observe that the sum of the digits of 234567 is 27. You can check
that 9 divides 234567.
Statement 3
An integer a is divisible by 3 if and only if the sum of the digits of a
is divisible by 3. An integer b is divisible by 9 if and only if the sum
of the digits of b is divisible by 9.
Example 6. Check whether the number 12345321 is divisible by 3. Is it
divisible by 9?
Solution: The sum of the digits is 1 + 2 + 3 + 4 + 5 + 3 + 2 + 1 = 21. Hence the
number is divisible by 3, but not by 9. In fact 12345321 = (9 × 1371702) + 3.
Example 7. Is 444445 divisible by 3?
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Solution: The sum of the digits is 25, which is not divisible by 3. Hence
444445 is not divisible by 3. Here the remainder is 1.
3. Divisibility by 5 and 10
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Activity 9: Take all multiples of 5 from 51 to 100. Tabulate the last digits
of each multiple of 5.
Do you see that 0 or 5 appears as the digit in the unit’s place for every
multiple of 5? Does this observation help you to formulate a divisibility
test for 5 and also for 10?
Statement 4
An integer a is divisible by 5 if and only if it ends with 0 or 5. A
number is divisible by 10 if and only if ends with 0.
Example 8. How many numbers from 101 to 200 are divisible by 5?
Playing with numbers
19
Solution: Write down all numbers from 101 to 200 which end with 0 or
5: 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170,
175, 180, 185, 190, 195, 200. There are 20 such numbers.
Example 9. Is the number 12345 divisible by 15?
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Solution: Note that 15 = 3 × 5. Hence the given number must be divisible
by both 3 and 5.(This is also sufficient to prove that the given number is
divisible by 15. However, a general rule is false. For example, 4 divides
12 and 6 divides 12, but their product 24 does not divide 12. Can you
formulate some rule?) It is obviously divisible by 5, as its last digit is 5.
The sum of the digits is 1 + 2 + 3 + 4 + 5 = 15 and it is divisible by 3. Hence
3 also divides 12345. We conclude that 15 divides 12345.
Example 10. How many numbers from 201 to 250 are divisible by 5, but
not by 3?
Solution: Here again, the numbers divisible by 5 are 205, 210, 215, 220,
225, 230, 235, 240, 245, 250. Now you compute the digital sum of these
numbers: you get 7,3,8,4,9,5,10,6,11,7. Among these, the only numbers
divisible by 3 are 3,6,9. Thus among the 10 numbers divisible by 5, only
three numbers are also divisible by 3. The remaining 7 numbers are not
divisible by 3.
3. Divisibility by 11
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Consider the number 4587. You may check that it is divisible by 11.
(In fact, 4587 = 11 × 417.) We may also write
No
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4587 = (4 × 1000) + (5 × 100) + (8 × 10) + 7
= (4 × 1001) + (5 × 99) + (8 × 11) + (−4 + 5 − 8 + 7)
= (11 × 91 × 4) + (11 × 9 × 5) + (11 × 8) − (4 − 5 + 8 − 7).
Each of the numbers in the first three brackets is divisible by 11. Hence
the divisibility of 4587 by 11 is now related to the divisibility of 4 − 5 + 8 − 7,
which involves only the digits of the given number. The important point to
note here is that the sign alternates with + and −. We also observe that
4 − 5 + 8 − 7 = 0, which is divisible by 11.
Consider a 3 digit number 429. You may easily check hat 429 is divisible by 11: 429 = 11 × 39. On the other hand,
Unit 1
20
429 = (4 × 100) + (2 × 10) + 9
= (4 × 99) + (2 × 11) + (4 − 2 + 9).
= 99a + 11b + (a − b + c).
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n = 100a + 10b + c
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Since 4 − 2 + 9 = 11 is divisible by 11, you may conclude that 11 divides
429, without actually dividing it by 11.
How do you test a general 3-digit or 4-digit number? Suppose n = abc
is a 3-digit number. Then
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Thus n is divisible by 11 if and only if 11 divides a − b + c. If m = pqrs is a
4-digit number, then
n = 1000p + 100q + 10r + s
= 1001p + 99q + 11r − (p − q + r − s).
Hence 11 divides n if and only if 11 divides p − q + r − s. This may be
extended to numbers with any number of digits, with due care.
Statement 5
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Given a number n in decimal form, put alternatively − and + signs
between the digits and compute the sum. The number is divisible
by 11 if and only if this sum is divisible by 11. Thus a number
is divisible by 11 if and only if the difference between the sum of
the digits in odd places and the sum of the digits in even places is
divisible by 11.
Example 11. Is the number 23456 divisible by 11?
Solution: Observe that 2−3+4−5+6 = 4 and hence not divisible by 11. The
test indicates that 23456 is not divisible by 11. In fact 23456 = (11×2123)+4.
A palindrome is a number which reads the same from left to right or
right to left. Thus a palindrome is a number n such that by reversing the
digits of n, you get back n. For example 232 is a 3-digit palindrome; 5445 is
a 4-digit palindrome.
Example 12. Find all 3-digit palindromes which are divisible by 11.
Solution: A 3-digit palindrome must be of the form aba, where a 6= 0 and
Playing with numbers
21
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b are digits. This is divisible by 11 if and only if 2a − b is divisible by 11.
This is possible only if 2a − b = 0 or 2a − b = 11 or 2a − b = −11. Since a ≥ 1
and b ≤ 9, we see that 2a − b ≥ 2 − 9 = −7 > −11. Hence 2a − b = −11 is not
possible. Suppose 2a − b = 0. Then 2a = b; thus a = 1, b = 2; a = 2, b = 4;
a = 3, b = 6; and a = 4, b = 8 are possible. We get the numbers 121, 242, 363,
484. For a = 6, b = 1, we see that 2a − b = 12 − 1 = 11, and hence divisible by
11. Similarly a = 7, b = 3; a = 8, b = 5; and a = 9, b = 7 give the combinations
for which 2a − b is divisible by 11. We get four more numbers: 616, 737, 858,
and 979.
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Thus the required numbers are: 121, 242, 363, 484, 616, 737, 858, 979.
Example 13. Prove that 12456 is divisible by 36 without actually dividing
it.
Solution: First notice that 36 = 4 × 9. Hence it is enough to prove that
the given number is separately divisible by 4 and 9(then it will be divisible
by their LCM which is 36). Consider the number formed by the last two
digits: 56. It is divisible by 4. Hence 12456 is divisible by 4. On the
other hand the sum of the digits is 18 and divisible by 9. Hence 12456 is
divisible by 9 as well. Combining, we get the result.
Exercise 1.1.6
to
1. How many numbers from 1001 to 2000 are divisible by 4?
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2. Suppose a 3-digit number abc is divisible by 3. Prove that abc+bca+cab
is divisible by 9.
3. If 4a3b is divisible by 11, find all possible values of a + b.
4. Prove that a 4-digit palindrome is always divisible by 11.
5. Using the digits 4,5,6,7,8, each once, construct a 5-digit number
which is divisible by 132.
Unit 1
22
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You have seen the first few primes are:
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73. . .
There are infinitely many primes. Among these look at the pairs:
(3,5), (5,7), (11,13), (17,19), (29,31), (41,43), (59,61), (71,73).
These are primes which differ by 2. That is: the difference between them is
2. Such a pair of primes is called a twin prime. It is an outstanding unsolved
problem: Are there infinitely many twin primes?
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Another classical unsolved problem involving numbers is Goldbach’s conjecture.
It dates back to 1742. Consider even numbers greater than 2: observe 4 = 2 + 2,
6 = 3 + 3, 8 = 3 + 5, 10 = 3 + 7, 12 = 5 + 7, 14 = 3 + 11, 16 = 5 + 11, and so on. It
was conjectured by a German mathematician Goldbach that every even number
greater than 2 is a sum of two primes. It has been extensively verified using
computers and there is a strong suspicion that the conjecture is true. But no
mathematical proof is available till today.
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A natural number n is called a perfect number if it is equal to the sum of all
its proper positive divisors.(That is the sum of all its positive divisors excluding
the number itself.) The first perfect number is 6. It has three proper positive
divisors, 1,2,3; and 1 + 2 + 3 = 6. Similarly, 28 has proper divisors, 1,2,4,7,14;
and 1 + 2 + 4 + 7 + 14 = 28. The next numbers are 496 and 8128. All these were
discovered by Euclid. Euclid also proved that 2p−1 (2p − 1) , with p a prime, is a
perfect number whenever 2p − 1 is a prime. Prime numbers of the form 2p − 1
are called Mersenne primes(after a seventh-century mathematician called Marin
Mersenne). It is easy to prove that 2n − 1 is prime only if n is prime. But not all
numbers of the form 2p − 1, where p a prime, are prime numbers. For example,
211 − 1 = 2047 = 23 × 89 and hence not a prime. Using this idea and computers,
one can generate some perfect numbers. The numbers 2p−1 (2p − 1), where p is
equal to
2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281
are the first few perfect numbers. There are some unsolved problems related to
perfect numbers.
1. Are there infinitely many perfect numbers?(Equivalently, are there infinitely many Mersenne primes?)
2. All the perfect numbers so far known are even perfect numbers. This naturally raises the question: Is there any odd perfect number?
Playing with numbers
23
Additional Problems on “Playing with numbers”
1. Choose the correct option:
B. (4 × 100) + (6 × 10) + (5 × 1)
C. (5 × 100) + (4 × 10) + (6 × 1)
D. (6 × 100) + (5 × 10) + (4 × 1)
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(b) Computers use
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(a) The general form of 456 is
A. (4 × 100) + (5 × 10) + (6 × 1)
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A. decimal system B. binary system C. base 5 system D. base
6 system
(c) If abc is a 3-digit number, then the number
n = abc + acb + bac + bca + cab + cba
is always divisible by
A. 8
B. 7 C. 6 D. 5
(d) If abc is a 3-digit number, then
n = abc − acb + bac − bca + cab − cba
is always divisible by
A. 12 B. 15 C. 18 D. 21
A. 1
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(e) If 1K × K1 = K2K, the letter K stands for the digit
B. 2 C. 3 D. 4
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(f) The numbers 345111 is divisible by
A. 15 B. 12 C. 9
D. 3
(g) The number of integers of the form 3AB4, where A, B denote some
digits, which are divisible by 11 is
A. 0
B. 4 C. 7 D. 9
2. What is the smallest 5-digit number divisible by 11 and containing
each of the digits 2,3,4,5,6?
3. How many 5-digit numbers divisible by 11 are there containing each
of the digits 2,3,4,5,6?
Unit 1
24
4. If 49A and A49, where A > 0, have a common factor, find all possible
values of A.
5. Write 1 to 10 using 3 and 5, each at least once, and using addition
and subtraction. (For example, 7 = 5 + 5 − 3.)
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6. Find all 2-digit numbers each of which is divisible by the sum of its
digits.
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7. The page numbers of a book written in a row gives a 216 digit number.
How many pages are there in the book?
8. Look at the following pattern:
1
1
1
1
1
1
2
3
4
1
3
6
1
4
1
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This is called Pascal’s triangle. What is the middle number in the 9-th
row?
9. Complete the
adjoining magic
square.(Hint: In a 3 × 3 magic square,
the magic sum is three times the
central number.)
8
3
7
10. Find all 3-digit natural numbers which are 12 times as large as the
sum of their digits.
11. Find all digits x, y such that 34x5y is divisible by 36.
Playing with numbers
25
12. Can you divide the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 into two groups such
that the product of numbers in one group divides the product of numbers in the other group and the quotient is minimum?
13. Find all 8-digit numbers 273A49B5 which are divisible by 11 as well
as 25.
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14. Suppose a, b are integers such that 2 + a and 35 − b are divisible by 11.
Prove that a + b is divisible by 11.
15. A list of numbers and corresponding
589 724 1306
L
codes are given in the adjacent table.
524 386 9701 2011
Find the number L.
16. In the multiplication table A8 × 3B = 2730, A and B represent distinct
digits different from 0. Find A + B.
17. Find the least natural number which leaves the remainders 6 and 8
when divided by 7 and 9 respectively.
18. Prove that the sum of cubes of three consecutive natural numbers is
always divisible by 3.
Glossary
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Divisibility: an integer a is said to be divisible by another non-zero integer
b, if a = qb for some integer q.
Quotient: if a = qb, for some integers a, b 6= 0 and q, then q is the quotient
upon division of a by b.
Remainder: if a = bq + r, where 0 ≤ r < b for some integer a and natural
number b, then r is the remainder upon division of a by b.
Palindrome: a number which reads the same from left to right and right
to left.
Puzzle: any mind twister.
Alpha numeral: a letter appearing in an equation and taking some numerical value.
Conjecture: a statement which is believed to be true, but without any
substantiating mathematical proof.
Magic square: a square consisting of smaller squares and each smaller
square filled with numbers such that the row sum, the column sum and
Unit 1
26
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Points to remember
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the diagonal sum are all equal.
Perfect number: a natural number whose all positive divisors smaller
than the number add up to the number.
Twin-primes: pairs of prime numbers such that the difference of numbers
in each pair is 2.
Mersenne primes: prime numbers of the form 2p − 1. (If 2p − 1 is a prime,
it forces p is also a prime.)
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• Any natural number can be written in the generalised form using base
10.
• Given any two integers a and b > 0, there exist unique integers q and
r such that a = bq + r, where 0 ≤ r < b.
• A number is divisible by 4 if and only if the number formed by the
last two digits is divisible by 4.
• A number is divisible by 3 or 9 if and only if the sum of the digits is
divisible by 3 or 9 respectively.
• A number is divisible by 5 if and only if it ends in 0 or 5.
to
• A number is divisible by 11 if and only if the difference between the
sum of oddly placed digits and the sum of evenly placed digits is
divisible by 11.
No
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We must think and act like a nation of billion people and not like that
of a million people. Dream, dream, dream!
—–Abdul Kalam
CHAPTER 1
UNIT 2
SQUARES, SQUARE ROOTS, CUBES, CUBE ROOTS
• perfect squares and square root of perfect squares;
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• to recognise the digits in unit’s place of a perfect square;
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After studying this unit, you learn
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• to obtain the remainders when a perfect square is divided by 3 and 4;
• different occasions leading to perfect squares;
• some methods of finding perfect squares and square-roots of perfect
squares;
• perfect cubes and cube-roots of perfect cubes;
1.2.1 Introduction
to
Look at the numbers of the form 1, 4, 9, 16, 25 and so on. What do
you recognise? For example, 4 = 2 × 2, 25 = 5 × 5. You see that each
such number is the product of two equal numbers. Similarly, you see
that 8 = 2 × 2 × 2, 64 = 4 × 4 × 4. Each number is the product of three
equal numbers. These numbers are given special names. In this chapter
we study some properties of such numbers. We also study the reverse
process: whenever a number is the product of two equal numbers or three
equal numbers, can we find these equal numbers?
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1.2.2 Perfect squares
Look at the following diagrams.
How many dots do you find in each figure? You recognise them as 1, 4,
9, 16.
Unit 2
28
D
C
10
Suppose you have a square ABCD
of side-length 10 units. Divide
the square in to smaller unit
squares(as in the adjoining figure)
using lines parallel to the sides.
Can you count that there are 100
unit squares?
9
8
7
6
5
4
3
1
A
B
1
2
3
4
5
6
7
8
9
10
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Activity 1: Repeat this with squares of side-length 8, 12, 15 units and
tabulate your findings.
Observe that 1 = 1 × 1, 4 = 2 × 2, 9 = 3 × 3, 16 = 4 × 4, 100 = 10 × 10.
If a is an integer and b = a × a, we say b is a perfect square.
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Hence 1, 4, 9, 16, 100 are all perfect squares. Since 0 = 0 × 0, we see that
0 is also a perfect square.
If a is an integer, we denote a × a by a2 (we read this as Square of a or
simply a-square). Thus 36 = 62 , 81 = 92 . Thus a perfect square is of the
form m2 , where m is an integer.
Here you may be observing some thing more. For example 4 = 2 × 2
and 4 = (−2) × (−2); in the second representation, you again have equal
integers, but negative this time. There is nothing strange about this. The
property is inherent in the number system: if you multiply two negative
integers, you get a positive integer. Thus for any natural number m, we
get m2 = m × m = (−m) × (−m). This also tells something about the nature
of perfect squares. If m is a natural number, m2 = m × m is also a natural
number and hence m2 is positive. If m = 0, then m2 = 0 × 0 = 0. If m
is a negative integer, then m = −n for some natural number n. Hence
m2 = (−n) × (−n) = n2 , which again is positive.
Thus a perfect square is either equal to 0 or must be a positive
integer. It can never be a negative integer.
We have seen that 1 = 12 and 4 = 22 are perfect squares. Can 2 and
3 also be written as a product of two equal integers? Your intuition tells
Squares and cubes
29
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that this cannot happen. Can we see this mathematically? Suppose m
and n are two natural numbers such that m < n. Then it is easy to see
that m2 < n2 (why?).
If at all 2 is a perfect square, 2 = n2 for some natural number. Then
1 < 2 < 4 gives 1 = 12 < n2 < 22 = 4. This forces 1 < n < 2(why?). Thus n
is a natural number strictly between 1 and 2. But we know that given a
natural number k, there is no natural number between k and its successor
k + 1. Thus no natural number exists between 1 and 2. We conclude that
2 is not a perfect square.
Similarly, you can conclude that 3 is not a perfect square. This may be
extended to prove that any natural number n, such that m2 < n < (m + 1)2 ,
cannot be a perfect square.
Look at the following table:
a
a2
1 2
1 4
3 8 −7
9 64 49
−12 20 −15
144 400 225
Do you see that squares of 2,8,-12, 20 are even numbers, where as the
squares of 1,3, −7, −15 are all odd numbers. What do you infer?
Statement 1. The square of an even integer is even and the square
of an odd integer is odd.
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This is not hard to prove. If m is even, then m = 2n for some integer n
and m2 = (2n) × (2n) = 4n2 is an even integer. If m is odd, then m = 2k + 1
for some integer k, so that
m2 = (2k + 1)(2k + 1)
= (2k + 1) × (2k) + (2k + 1) × 1
= (2k) × (2k) + (1 × 2k) + (2k × 1) + (1 × 1)
= 4k 2 + 2k + 2k + 1
= 4k 2 + 4k + 1,
which is an odd number.
Consider the first ten perfect squares as in the table.
Unit 2
30
12
1
22
4
32
9
42 52 62 72 82
16 25 36 49 64
92 102
81 100
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If you observe the unit’s place in these squares, you see that they are
1,4,9,6,5,6,9,4,1,0 in that order. Thus the only digits which can occupy
the unit’s place in a perfect square are 0,1,4,5,6,9. If you take any number, its unit’s digit is one of 0,1,2,3,4,5,6,7,8,9. Hence the square of that
number(multiply the number with it self in your mind to get the digit in
unit’s place) ends with one of the numbers 0,1,4,5,6,9. Can you see that
the digits 2,3,7,8 can never occur as the last digit of a perfect square? We
can make a formal statement:
Statement 2. A perfect square always ends in one of the digits
0,1,4,5,6,9. If the last digit of a number is 2,3,7 or 8, it cannot be a
perfect square.
Think it over ! If a number ends in 0,1,4,5,6 or 9, then it is not
necessary that the number is a perfect square.
to
In precise mathematical language, we say that a necessary condition
for the given number to be a perfect square is that it should end with one
of the digits 0,1,4,5,6 or 9, but this condition is not a sufficient condition
to ensure that the given number is a perfect square. This helps us to
recognise perfect squares.
Exercise 1.2.2
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1. Express the following statements mathematically:
(i) square of 4 is 16; (ii) square of 8 is 64; (iii) square of 15 is 225.
2. Identify the perfect squares among the following numbers:
1,2,3,8,36,49,65,67,71,81,169,625,125,900,100,1000,100000.
3. Make a list of all perfect squares from 1 to 500.
4. Write 3-digit numbers ending with 0,1,4,5,6,9, one for each digit, but
none of them is a perfect square.
5. Find numbers from 100 to 400 that end with 0,1,4,5,6 or 9, which
are perfect squares.
Squares and cubes
31
1.2.3 Some facts related to perfect squares
There are some nice properties about perfect squares. We study them
here.
(a) Look at the following table:
0
2
2
0
4
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10
20
25
100
300
1000
16 100 400 625 10000 90000 1000000
4
6
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a
a2
The number of
zeros at the
end of a2
What do you observe? The number of zeros at the end of a square is
always an even number(it may be equal to 0, but still an even number).
Moreover the number of zeros at the end of each square is twice the number of zeros at the end of the number whose square is considered. Can
you now formulate this as a statement?
Statement 3. If a number has k zeros at the end, then its square
ends in 2k zeros.
a2
The remainder of a2
when divided by 3
1
1
4
1
9
0
25
1
64
1
121
1
36
0
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Thus, if a number ends in odd number of zeros, it cannot be a perfect
square. This helps us to rule out certain numbers from the list of perfect
squares.
(b) Look at the adjoining table:
1
2
3
5
8
11
−6
The remainder of a2
when divided by 4
1
0
1
1
0
1
0
Do you see that the remainder of a perfect square when divided by 3 is
either 0 or 1? Similarly, the remainder of a perfect square when divided by
Unit 2
32
4 is either 0 or 1. When you divide a number by 3, the possible remainders
are 0, 1 and 2. But when you divide a perfect square by 3, the remainder
is either 0 or 1, but never 2. Similarly, when you divide a number by 4,
the possible remainders are 0,1,2 or 3. However, when you divide a perfect
square by 4, the remainder is either 0 or 1, but it can never be 2 and 3.
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Statement 4. The remainder of a perfect square, when divided by
3, is either 0 or 1, but never 2. The remainder of a perfect square,
when divided by 4, is either 0 or 1, but never 2 and 3.
(c)
Activity 2:
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Think it over !
The remainder of a perfect square, when divided by 8, is either
0 or 1 or 4. It can never be equal to 2,3,5,6,7.
Take any four consecutive natural numbers and form their product. Add
1 to this product. Check whether it is a perfect square. Repeat this with
some more sets of four consecutive natural numbers. Do the same with
some sets of consecutive negative integers. What is your observation if one
of the four consecutive integers is 0 ?
For example:
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(1 × 2 × 3 × 4) + 1 = 24 + 1 = 25 = 52 ;
No
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(8 × 9 × 10 × 11) + 1 = 7920 + 1 = 7921 = 892 .
Statement 5. When the product of four consecutive integers is
added to 1, the resulting number is a perfect square.
(d) Read the following table:
1 = 1 = 12 ,
1 + 3 = 4 = 22 ,
1 + 3 + 5 = 9 = 32 ,
1 + 3 + 5 + 7 = 16 = 42 ,
1 + 3 + 5 + 7 + 9 = 25 = 52 .
Squares and cubes
33
Continue this process for some more rounds, adding the next odd number
to the previous sum. You see that you go on getting perfect squares. A
careful observation also reveals some thing more. The sum of the first 4
odd natural numbers is 42 ; the sum of the first 5 odd natural numbers is
52 . Check this with the sum of the first 8 and 12 odd natural numbers.
Can you formulate this as a new statement?
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Statement 6. The sum of the first n odd natural numbers is equal
to n2 , for every natural number n.
Activity 3:
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(e)
Consider the numbers 11, 101, 1001, 10001 and compute their squares.
Do the same inserting some more zeros. Do you see some pattern?
For example:
112 = 121,
1012 = 10201,
10012 = 1002001,
to
and so on. You see that the middle number of each square is always 2;
on both sides of 2, zeros appear as digits; and the end digits are equal to
1. The number of zeros both sides of 2 are equal and equal to the number
of zeros in the original number. Thus we can formulate a statement as
follows.
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Statement 7. Consider the number N = 1000 · · · 01, where zeros
appear k times. (For example, for k = 6, you get N = 10000001; there
are 6 zeros in the middle.) Then N 2 = 1000 · · · 02000 · · · 01, where the
number of zeros on both sides of 2 is k.
(f) Look at the following patterns:
Unit 2
34
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To generate such a pattern, put a dot in row 1, put 2 dots in row 2,
put 3 dots in row 3, and so on, as shown in the figure. The dots are now
arranged in the shape of a triangle. Count the number of dots in each of
the triangle.(Single dot is considered as a degenerate triangle.) They are
1,3,6, 10, 15, 21, 28, 36 and so on. These numbers are called triangular
numbers (you know the reason). You can see how the triangular numbers
are formed. For n-th triangular number, you form a triangle of dots with
n rows and each row contains as many points as the index of that row. If
you want to find the 8-th triangular number, the number of points in the
8-th triangle is
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36.
Here are the first few triangular numbers:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91.
Take any two consecutive triangular numbers and find their sum. For
example: 10 + 15 = 25 = 52 ; 28 + 36 = 64 = 82 ; 55 + 66 = 121 = 112 ; 36 + 45 =
81 = 92 . You see that the sum of any two consecutive triangular numbers
is a perfect square. You can also observe some thing more. Note that 28
is 7-th triangular number and 36 is 8-th one; their sum is 82 . Similarly
66 is 11-th triangular number and 78 is 12-th one; their sum is 144 = 122 .
Verify this property for some more pairs.
to
Statement 8. The sum of n-th and (n + 1)-th triangular numbers is
(n + 1)2 .
Exercise 1.2.3
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1. Find the sum 1 + 3 + 5 + · · · + 51(the sum of all odd numbers from 1 to
51) without actually adding them.
2. Express 144 as a sum of 12 odd numbers.
3. Find the 14-th and 15-th triangular numbers, and find their sum.
Verify the Statement 8 for this sum.
4. What are the remainders of a perfect square when divided by 5?
1.2.4 Methods for squaring a number
Many times, it is easy to find the square of a number without actually
multiplying the number with itself. Consider 422 . We may write 42 = 40 + 2.
Thus
Squares and cubes
35
422 = (40 + 2)(40 + 2)
= 402 + (40 × 2) + (2 × 40) + 22
= 402 + (2 × 40 × 2) + 22 .
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Here we have used the distributive property of integers. Now it is easy to
recognise 402 = 1600; 2 ×40 ×2 = 160; and 22 = 4. Hence 422 = 1600 + 160 + 4 =
1764. (You can compute 402 , 2 × 40 × 2 and 22 in mind and add.)
Activity 4:
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Note. The basis for this method is the identity (a + b)2 = a2 + 2ab + b2 ,
which you will study later.
Find 892 , 682 , 962 using the above method.
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There is an easy way of computing the square of a number ending with
5. For example, consider 352 . Take the digit in unit’s place, namely 5.
Put 25 (= 52 ) first. Remove the unit’s digit from the given number and
consider the number formed by the remaining digits, which is 3. Consider
the product of 3 and its next number 4; 3 × 4 = 12. Prefix 12 to 25 to get
1225. Check that 352 = 1225.
Take another example, say 1052 . Here the number formed by the remaining digits after the removal of the digit in the unit’s place is 10 and
its successor is 11. Their product is 10 × 11 = 110. Now you may check that
1052 = 11025. You may formulate this as a statement.
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Statement 9. If n = a1 a2 · · · ak 5(represented in base 10), then n2 is
equal to 25 prefixed by
(a1 a2 · · · ak ) × (a1 a2 · · · ak + 1)
Exercise 1.2.4
1. Find the squares of:
(i) 31, (ii) 72; (iii) 37 ;
2. Find the squares of: (i) 85;
(iv) 166.
(ii) 115;
(iii) 165.
3. Find the square of 1468 by writing this as 1465 + 3.
Unit 2
36
1.2.5 Square roots
As you know, if the side-length of a square ABCD is l, then its area is
l2 . Can you reverse the process? Given the area of a square, can we find
its side-length?
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Again consider the following perfect squares:
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Suppose the area of a square is 16 cm2 . To find its side-length, we write
l2 = 16 = 42 and conclude l = 4 cm. Here the square of a number is given
and we have to find the number. Do you see that we are moving in the
opposite direction?
1 = 12 , 4 = 22 , 9 = 32 , 16 = 42 , 49 = 72 , 81 = 92 , 196 = 142 .
In each case the number is obtained by product of two equal numbers.
Here we say 1 is square root of 1; 2 is square root of 4; 7 is square root of
49 and so on.
Suppose N is a natural number such that N = M 2 . The number
M is called a square root of N.
No
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We have seen earlier m2 = m × m = (−m) × (−m) = (−m)2 . Thus m2 has
two square roots: m and −m. Which one should be taken? For example
16 = 42 = (−4)2 . Thus both 4 and −4 are square roots of 16. It is not clear,
which one of these should be taken. Many times, physical context clarify
the matter. As in the above example, if the area of a square is 16 units,
then its side-length is necessarily 4 units(−4 is not admissible as it cannot
be length). However, mathematically both 4 and −4 are acceptable as a
square root of 16. We make the following convention.
Whenever the word square root is used, it is always meant to
be the positive square root. The square root of N is denoted by
√
N.
Activity 5: Fill in the following blanks looking at the similarity of the
Squares and cubes
37
statements:
12 = 1 =⇒
22 = 4 =⇒
52 = 25 =⇒
√
√
√
1 = 1;
4 = 2;
25 = − − −−;
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112 = 121 =⇒ − − −− = 11;
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− − −− = 225 =⇒ − − − − − = 15.
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Activity 6: Fill in the blanks with appropriate word or number looking at
the similarity of the statements:
square
2
square
4
3
9
4
16
square root
square root
square
49
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6
square
We have learnt earlier that the square of a non-zero integer is always a
positive integer. Hence square root is meaningful only for positive integers
or possibly 0 (whose square root is 0).
Square root of a perfect square by factorisation
√
√
We know that 3 = 9 and 4 = 16. However 9 = 3×3 and 16 = 2×2×2×2 =
4 × 4. Thus we can factorise the given perfect square in terms of their
prime factors, combine these prime factors appropriately to write the given
perfect square as a product of two equal integers. This will help us to read
off the square root of the given perfect square.
Unit 2
38
Example 1. Find the square root of 5929.
5929
7
847
11
121
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11
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Solution: We do this in several steps.
Step 1. We express 5929 as a product of prime numbers:
Thus 5929 = 7 × 7 × 11 × 11.
Step 2. We arrange these prime factors suitably to write 5929 = (7 × 11) ×
(7 × 11) = 77 × 77.
Step 3. Since 5929 = 77 × 77, a product of two equal integers, we conclude
√
5929 = 77.
Example 2. Find the square root of 6724.
No
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Solution: We observe that 6724 is even so that 2 is its prime factor.
Thus 6724 = 2 × 3362. Again 3362 is even so that 3362 = 2 × 1681. Thus
6724 = 2 × 2 × 1681. Now there is no easy way of finding prime factors of
1681. We must go on checking whether 1681 is divisible by the primes
in increasing order, starting from 3. We see that it is not divisible by
3,5,7,11,13,17,19,23,29,31,37, but it is divisible by 41 and 1681 = 41 × 41.
Thus we obtain 6724 = 2 × 2 × 41 × 41 = (2 × 41) × (2 × 41) = 82 × 82. We
√
conclude: 6724 = 82.
There is no easy way of finding the prime factors of a given
number. There are some algorithms which can be used on computers to find the prime factors of a given large number. However, these algorithms also use up lot of computer time. The
fact that there is no easy way of factorising a large number is
the basis for modern day security systems used in banks and
other financial institutions.
Squares and cubes
39
Why some numbers are not perfect squares?
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You might have noticed, while finding the square root of a perfect
square by factorising it, that each prime factor of the perfect square occurs even number of times. For example, if you take 1296, you factorise
it as 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3. It has two distinct prime factors,
2 and 3. Now 2 occurs four times and 3 also occurs 4 times. This helps
you to write 1296 = (2 × 2 × 3 × 3) × (2 × 2 × 3 × 3) = 36 × 36, as a product
√
of two equal integers. You conclude that 1296 = 36. The success of this
method depends on the fact that the prime factors can be properly paired
to get the given number as a product of two equal integers. This is possible
because each prime factor occurs even number of times.
Activity 7:
Write all perfect squares between 1000 and 1500. In each case factorise
it as a product of prime numbers. Check that in every case, each of the
prime factor occurs even times in the product.
Now you can see why a number fails to be a perfect square. In its
prime factorisation, some primes may not occur even number of times.
Then there is no way of pairing the factors such that the given number
is equal to a product of two equal integers. However, you can make it a
perfect square on multiplication by a suitable factor or on division by a
suitable factor.
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Suppose a number is not a perfect square. Take for example, 48. We
see that 48 = 2 × 2 × 2 × 2 × 3. Here 2 occurs four times, where as 3 occurs
only once. Hence we cannot properly pair the prime factors. However, if
we multiply 48 by 3, we see that
48 × 3 = 2 × 2 × 2 × 2 × 3 × 3 = (2 × 2 × 3) × (2 × 2 × 3) = 12 × 12,
and we get a perfect square. Of course you may as well multiply 48 with
3 × 2 × 2 and get
48 × 12 = (2 × 2 × 3 × 2) × (2 × 2 × 3 × 2) = 24 × 24,
leading to a perfect square. In fact you can multiply 48 by 3k 2 where k
is any positive integer and get as earlier 48 × 3k 2 = (12k) × (12k), which is
Unit 2
40
a perfect square. However 3 is the least number with which you have to
multiply 48 to get a perfect square.
Example 3. Find the least positive integer whose product with 9408 gives
a perfect square.
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Solution: You can start with 2 and go on dividing by 2 till you get an odd
number: 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 147. Now 147 = 3 × 49 = 3 × 7 × 7. Thus
we get
9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7.
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Here both 2 and 7 occur even number of times, where as 3 occurs only
once. Hence we have to multiply 9408 by 3 to get a perfect square(you get
9408 × 3 = (168)2 ).
Let us go back to 48 = 2 × 2 × 2 × 2 × 3. Instead of multiplying by 3, we
can as well divide by 3:
2×2×2×2×3
48
=
= 4 × 4.
3
3
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Still we get a perfect square. Here again you may divide by 3 × 2 × 2 and
get
48
2×2×2×2×3
=
= 2 × 2,
3×2×2
3×2×2
a perfect square. However 3 is the least number with which you have to
divide 48 to get a perfect square.
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Example 4. Find the smallest positive integer with which one has to divide
336 to get a perfect square.
Solution: We observe that 336 = 2 × 2 × 2 × 2 × 3 × 7. Here both 3 and 7
occur only once. Hence we have to remove them to get a perfect square.
We divide 336 by 3 × 7 = 21 and get
336
= 16 = 42 .
21
The least number required is 21.
Exercise 1.2.5
1. Find the square root of the following numbers by factorisation:
(i)196;
(ii) 256;
(iii) 10404;
(iv) 1156;
(v) 13225.
Squares and cubes
41
2. Simplify:
√
√
√
√
√
√
(i) 100 + 36; (ii) 1360 + 9; (iii) 2704 + 144 + 289;
√
√
√
√
√
25; (v) 1764 − 1444; (vi) 169 × 361.
(iv)
√
225 −
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3. A square yard has area 1764 m2 . From a corner of this yard, another square part of area 784 m2 is taken out for public utility. The
remaining portion is divided in to 5 equal square parts. What is the
perimeter of each of these equal parts?
(ii) 450;
(iii) 1445;
(iv) 1352.
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(i) 847;
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4. Find the smallest positive integer with which one has to multiply each
of the following numbers to get a perfect square:
5. Find the largest perfect square factor of each of the following numbers:
(i) 48;
(ii) 11280;
(iii) 729;
(iv) 1352.
6. Find a proper positive factor of 48 and a proper positive multiple of
48 which add up to a perfect square. Can you prove that there are
infinitely many such pairs?
1.2.6 Perfect squares near to a given number
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Let us start with a non-perfect square, say, 72. Observe that 72 =
2 × 2 × 2 × 3 × 3, so that 2 appears only an odd number of times. We
can multiply 72 by 2 to get 72 × 2 = 144 = 122 . Or we may divide 72
72
= 36 = 62 . But there are more perfect squares between
by 2 to get
2
62 and 122 , namely 72 = 49, 82 = 64, 92 = 81, 102 = 100 and 112 = 121.
Which one is the nearest to 72 among these perfect squares? You see that
82 = 64 < 72 < 81 = 92 and 72 − 64 = 8 < 9 = 81 − 72. Thus we see that 64 is
nearer to 72 than 81. Hence 64 is the nearest perfect square to 72.
In fact, given a non-perfect square, there is a unique perfect square
nearest to it. Suppose N is the given non-perfect square. You can put
it between two consecutive squares; there is a unique n such that n2 <
N < (n + 1)2 .(Can you see why?). Since n and n + 1 are two consecutive
numbers, one of them is even and the other odd. Hence N cannot be
exactly in the middle of n2 and (n + 1)2 ; if N − n2 = (n + 1)2 − N, then
2N = n2 + (n + 1)2 = n2 + n2 + 2n + 1 = 2n2 + 2n + 1 which is impossible since
Unit 2
42
2N is even and 2n2 + 2n + 1 is odd. Hence either n2 is the nearest perfect
square to N or (n + 1)2 is the nearest perfect square. If n2 is the nearest
√
perfect square to N, we say n approximates N; if (n + 1)2 happens to be
√
the nearest square to N, we say n + 1 approximates N . Thus, even if N
√
is not a perfect square(so that N is no more an integer), we can find the
√
√
nearest integer to N and get an integer approximation to N .
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Example 5. If the area of a square is 90 cm2 , what is its side-length,
rounded to the nearest integer?
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Solution: Since A = l2 , we have l2 = 90. But 81 < 90 < 100 and 81 is nearer
√
√
to 90 than 100. Hence the nearest integer to 90 is 81 = 9.
Example 6. A square piece of land has area 112 m2 . What is the closest
integer which approximates the perimeter of the land?
Solution: If l is the side-length of a square, its perimeter is 4l. We know
that l2 = 112. Hence
(4l)2 = 16l2 = (16) × (112) = 1792.
But 422 = 1764 < 1792 < 1849 = 432 and 1764 is nearer to 1792 than 1849.
√
Hence the integer approximation for 1792 is 42. The approximate value
of the perimeter is 42 cm.
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Caution!
√
If you take 112 to the nearest integer you see that it is 11
and you may be tempted to write the approximate value of the
√
perimeter as 44(= 4 × 11) cm. By replacing 112 with 11, you have
already committed an error and when you multiply by 4, the error increases by 4 times. That is the reason we have multiplied
112 by 16 first and then took the square root to the nearest integer. There is nothing strange in this. The nearest integer to
r = 1/4 is 0. But the nearest integer to 3r = 3/4 is 1, not 3 × 0 = 0.
√
Now you may see the limitations of integers. If you want to find 90,
√
you have to take it as 9; if you want 94, then you have to take it as 10.
But neither of them gives you a true picture of what is the square root
of a non-perfect square. The limitation is because: there is no integer
Squares and cubes
43
between n and n + 1. However, this property ceases to be true in the
system of rational numbers. You can find a rational number between any
two rational numbers. This will help in moving further close to the square
root of a non-perfect square. You will learn more about this in your higher
classes.
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Exercise 1.2.6
(i) 232;
(ii) 600;
(iii) 728;
(iv) 824;
(v) 1729.
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1. Find the nearest integer to the square root of the following numbers:
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2. A piece of land is in the shape of a square and its area is 1000 m2 .
This has to be fenced using barbed wire. The barbed wire is available
only in integral lengths. What is the minimum length of the barbed
wire that has to be bought for this purpose?
√
√
3. A student was asked to find 961. He read it wrongly and found 691
to the nearest integer. How much small was his number from the
correct answer?
1.2.7 Perfect cubes
Read the following table:
1 = 1 × 1 × 1;
8 = 2 × 2 × 2;
to
27 = 3 × 3 × 3;
125 = 5 × 5 × 5
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You observe that each number is written as a product of 3 equal integers..
We say that an integer N is a perfect cube if N can be written
as a product of three equal integers. If N = m × m × m, we say N
is the cube of m and write N = m3 (read as cube of m or simply
m-cube).
Consider a few more examples:
(−4) × (−4) × (−4) = −64 = (−4)3 ,
(−5) × (−5) × (−5) = −125 = (−5)3
Unit 2
44
(−8) × (−8) × (−8) = −512 = (−8)3
Do you see that the negative numbers are also perfect cubes? Contrast
this with perfect squares. A non-zero perfect square is necessarily a positive integer. However, perfect cubes can as well be negative.
Example 7. Find the cube of 6.
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Solution: We have 63 = 6 × 6 × 6 = 36 × 6 = 216.
Example 8. What is the cube of 20?
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Solution: Again (20)3 = 20 × 20 × 20 = (400) × 20 = 8000.
You have studied about a solid called cube. It is a solid having equal
length, breadth and height. If l is the side-length of a cube, then its volume
V = l3 cubic units.
Example 9. If a cube has side-length 10 cm, what is its volume?
Solution: We have V = 10 × 10 × 10 = 1000 cm3 .
Example 10. Find the smallest integer larger than 1 which is a perfect
square as well as a perfect cube.
Solution: Start with any number n and multiply it with itself 6 times to
get a number N. We observe that
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N = n × n × n × n × n × n = (n × n) × (n × n) × (n × n)
= (n2 ) × (n2 ) × (n2 ) = (n2 )3 .
Thus N is the cube of n2 . On the other hand you may also observe that
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N = n × n × n × n × n × n = (n × n × n) × (n × n × n)
= (n3 ) × (n3 ) = (n3 )2 .
Hence N is also the square of n3 . Thus N is both a perfect cube and a
perfect square. Taking n = 2, we get the least number: N = 2 × 2 × 2 × 2 ×
2 × 2 = 64. You may verify that 64 = 43 and 64 = 82 .
Example 11. Show that 6 is not a perfect cube.
Solution: We observe that 1 < 6 < 8 so that 13 < 6 < 23 . Since no integer
exists between 1 and 2, 6 cannot be written as a product of three equal
integers. Hence 6 is not a perfect cube.
Squares and cubes
45
Exercise 1.2.7
1. Looking at the pattern, fill in the gaps in the following:
2
3
4
−5
—
23 = 8 33 = — —= 64 — = — 63 = —
8
—
— = — — = −729
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2. Find the cubes of the first five odd natural numbers and the cubes
of the first five even natural numbers. What can you say about the
parity of the odd cubes and even cubes?
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3. How many perfect cubes you can find from 1 to 100? How many from
−100 to 100?
4. How many perfect cubes are there from 1 to 500? How many are
perfect squares among these cubes?
5. Find the cubes of 10, 30, 100, 1000. What can you say about the
zeros at the end?
6. What are the digits in the unit’s place of the cubes of 1,2,3,4,5,6,7,8,9,
10? Is it possible to say that a number is not a perfect cube by looking
at the digit in unit’s place of the given number, just like you did for
squares?
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Activity 8: (More on Hardy-Ramanujan numbers) Express 4104 and
13832 as a sum of two perfect cubes in two different ways. Find some
numbers which can be expressed as a sum of two different perfect squares
in two or more ways. Explore more on this topic.
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1.2.8 Cube root
You know how to find the volume of a cube, given its side-length. Can
you reverse the process? Given the volume of a cube, is it possible to find
its side-length?
Suppose the volume of a cube is 125 cm3 . If l is its side-length, you
write l3 = 125 and conclude l = 5 cm. Here we say 5 is the cube root of 125
√
and write 5 = 3 125.
If N is number and n is another number such that N = n3 , we say
√
n is the cube root of N and write n = 3 N .
Unit 2
46
Remark: The definition of square root and cube root makes sense even for
non-integers(for square root , the number must be non-negative). At this
stage we confine only to integers and do not get involved in generality.
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Contrast the definition of cube root with that of square root. Given a
perfect square, there are two possible square roots; positive and negative.
This is because, the square of a non-zero integer is always positive and
(−n)2 = n2 for any integer n. Such a thing cannot happen for cubes.
If n is positive, then n3 is positive; if n is negative, n3 is also negative.
Hence cube root of a perfect cube is negative or positive depending on the
negativity or positivity of the given perfect cube. This shows that we can,
unambiguously, talk of the cube root of a perfect cube. There is no need
to follow a convention as in the case of square roots.
As in the case of square roots, we can find the cube root of a perfect
cube by prime factorisation.
Example 12. Find the cube root of 216 by factorisation.
Solution: Observe
216 = 2 × (108) = 2 × 2 × (54) = 2 × 2 × 2 × 27 = 2 × 2 × 2 × 3 × 3 × 3
= (2 × 3) × (2 × 3) × (2 × 3)
= 6 × 6 × 6.
√
3
216 = 6.
to
Hence
Example 13. Find the cube root of −17576 using factorisation.
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Solution: Let us first find the cube root of 17576. As earlier, we have
17576 = 2 × (8788) = 2 × 2 × (4394) = 2 × 2 × 2 × (2197)
= 2 × 2 × 2 × 13 × (169)
= 2 × 2 × 2 × 13 × 13 × 13
= (2 × 13) × (2 × 13) × (2 × 13)
= 26 × 26 × 26.
This shows that −17576 = (−26) × (−26) × (−26). Thus
√
3
−17576 = −26.
Squares and cubes
47
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Srinivasa
Ramanujan(1887-1920)
is undoubtedly the greatest Indian
mathematician of all times. He was
self-taught and had an uncanny mathematical manipulative ability. He was
not able to pass his school examinations in India, and had to be content
with a clerical position in the Port Trust
of Madras. However, he continued to
create his own mathematics, obtained
lot of hitherto unknown results. He
sent these results G. H. Hardy who at
once recognized Ramanujan’s intrinsic
mathematical ability and arranged for
him to travel to Cambridge.
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Because of his lack of formal training, Ramanujan sometimes did not differentiate between formal proof and apparent truth based on intuitive or numerical
evidence. His intuition and computational ability allowed him to determine and
state highly original and unconventional results which continued to defy formal
proof until recently.
Ramanujan had an intimate familiarity with numbers, and excelled especially
in number theory. J. Littelewood( a collaborator of G.H.Hardy) exclaimed that
every integer was a personal friend of Ramanujan. His familiarity with numbers may be demonstrated by the following incident. During an illness in England, Hardy visited Ramanujan in the hospital. When Hardy remarked that
he had taken taxi number 1729, a singularly dull number, Ramanujan immediately responded that this number was actually quite remarkable: it is the
smallest integer that can be represented in two ways by the sum of two cubes:
1729 = 13 + 123 = 93 + 103 (Hardy-Ramanujan number).
Unfortunately, Ramanujan’s health deteriorated rapidly in England, due perhaps to the unfamiliar climate, food, and to the isolation which Ramanujan felt
as the sole Indian in a culture which was largely foreign to him. Ramanujan
was sent home to recuperate in 1919, but tragically died the next year at the
very young age of 32.
Example 14. What is the least positive integer with which you have to
multiply 243 to get a perfect cube?
Unit 2
48
Solution: Let us factorise 243. We observe that
243 = 3 × (81) = 3 × 3 × 3 × 3 × 3.
If we multiply 243 by 3, we see that
243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 9 × 9 × 9,
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and we get a perfect cube. The answer is therefore 3.
If you factorise a positive perfect cube, you may observe that the number of times each prime factor occurs is always a multiple of 3(just like it is
a multiple of 2 for perfect squares). Hence to get the least positive integer
whose product with the given integer makes a perfect cube, you have to
see how much each prime factor is deficient in the prime factorisation of
the given number to be away from a perfect cube.
Often, finding the cube root of a given perfect cube may be time consuming. We seek to find easier methods. We can use the behaviour of
the unit’s digit of a cube to fix the cube root. We see that the unit’s digits of the cubes of numbers ending with 1,2,3,4,5,6,7,8,9,0 are uniquely
determined. See the following table:
units’ digit of n
unit’s digit of n3
1
1
2 3 4
8 7 4
5 6 7
5 6 3
8 9
2 9
0
0
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We also tabulate the cubes of first nine numbers:
1
1
2 3
4
8 27 64
5
6
7
8
9
125 216 343 512 729
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n
n3
Let us see how these help us.
Example 15. Find the cube root of 103823.
Solution: Here the units digit of 103823 is 3. If n3 = 103823, then the units
digit of n must be 7. Let us split 103823 as 103 and 823. We observe that
43 = 64 < 103 < 125 = 53 . Hence
403 = 64000 < 103823 < 125000 = 503 .
Hence n must lie between 40 and 50. Since the units digit of n is 7, the
only such number is 47. You may check that 473 = 103823.
Squares and cubes
49
Note: This method works only if you know that the given number is
a perfect cube.
Nevertheless, this helps us in estimating the cube root of a non-perfect
cube. We can squeeze the given number between two perfect cubes and
see which is the nearest one.
Solution: We observe that
203 = 8000 < 12345 < 27000 = 303 .
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Example 16. Find the nearest integer to the cube root of 12345.
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√
Hence 3 12345 must lie between 20 and 30. We do not know whether 12345
is a perfect cube or not. However, we may sharpen the bound: 233 = 12167
√
and 243 = 13824 and hence 3 12345 must be between 23 and 24. Moreover
√
12167 is nearer to 12345 than 13824. Hence the closest integer to 3 12345
is 23.
Exercise 1.2.8
1. Find the cube root by prime factorisation:
(i) 10648;
(ii) 46656;
(iii) 15625.
(ii) 166375;
(iii) 704969.
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2. Find the cube root of the following by looking at the last digit and
using estimation:
(i) 91125;
(iii) 373248.
3. Find the nearest integer to the cube root of each of the following:
(ii) 46656;
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(i) 331776;
Unit 2
50
Additional problems on
“Squares, square roots, cubes and cube roots”
1. Match the numbers in the column A with their squares in the column
B:
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Answers
——–
(1) ——
(2) ——
(3)——
(4)——
(5)——
(6)——
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(a)
(b)
(c)
(d)
(e)
(f)
(g)
B
—–
25
144
36
484
64
4
121
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A
—–
(1)
5
(2)
8
(3)
2
(4) −6
(5) −22
(6)
12
2. Choose the correct option.
(a) The number of perfect squares from 1 to 500 is:
A. 1) B. 16 C. 22 D. 25
(b) The last digit of a perfect square can never be
A. 1
B. 3 C. 5 D. 9
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(c) If a number ends in 5 zeros, its square ends in:
A. 5 zeros B. 8 zeros C. 10 zeros D. 12 zeros
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(d) Which could be the remainder among the following when a perfect
square is divided by 8?
A. 1
B. 3 C. 5 D. 7
(e) The 6-th triangular number is:
A. 6
B. 10 C. 21
D. 28
3. Consider all integers from −10 to 5, and square each of them. How
many distinct numbers do you get?
4. Write the digit in unit’s place when the following numbers are squared:
4,5,9,24,17,76,34,52, 33, 2319, 18, 3458, 3453.
Squares and cubes
51
5. Write all numbers from 400 to 425 which end in 2,3, 7 or 8. Check if
any of these is a perfect square.
6. Find the sum of the digits of (111111111)2.
7. Suppose x2 + y 2 = z 2 .
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(i) if x = 4 and y = 3 find z;
(iii) if y = 15 and z = 17, find x.
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(ii) if x = 5 and z = 13, find y;
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8. A sum of 2304 is equally distributed among several people. Each
gets as many rupees as the number of persons. How much does each
one get?
9. Define a new operation ⋆ on the set of all natural numbers by m ⋆ n =
m2 + n2 .
(i) Is N closed under ⋆?
(ii) Is ⋆ commutative on N?
(iii) Is ⋆ associative on N?
(iv) Is there an identity element in N with respect to ⋆?
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10. (Exploration) Find all perfect squares from 1 to 500, each of which is
a sum of two perfect squares.
11. Suppose the area of a square field is 7396 m2 . Find its perimeter.
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12. Can 1010 be written as a difference of two perfect squares? [Hint:
How many times 2 occurs as a factor of 1010?]
13. What are the remainders when a perfect cube is divided by 7?
14. What is the least perfect square which leaves the remainder 1 when
divided by 7 as well as by 11?
15. Find two smallest perfect squares whose product is a perfect cube.
Unit 2
52
Glossary
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Points to remember
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Perfect square: an integer which is the product of two equal integers.
Triangular numbers: the sum of the first n natural number is called n-th
triangular number.
Square-root: a number a is a square root of b if b = a2 .
Perfect cube: an integer which is the product of three equal integers.
Cube-root: a number c whose cube is d is called a cube-root of c.
Prime factor: a prime number which divides an integer a is a prime factor
of a.
Irrational number: any real number which is not a rational number.
• A perfect square is the product of two equal integers; a perfect cube
is the product of three equal integers.
• A perfect square is always non-negative(0 is also perfect square); a
perfect cube may be negative, equal to 0 or may be positive.
• Given a positive number, there are two square-roots, positive and
negative. But for any number, there is only one cube-root.
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• Given any positive number which is not a square, you can always
squeeze it between two consecutive perfect squares.
CHAPTER 1
UNIT 3
• the concept of fraction and rational numbers;
• to add and multiply rational numbers;
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After studying this unit you learn:
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RATIONAL NUMBERS
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• the properties of rational numbers with respect to addition and multiplication: closure,associativity, commutativity, distributive properties, and existence of identity and inverse;
• representation of rational numbers on the number-line and density
property of rational numbers;
• the gain while moving from integers to rational numbers and the loss.
1.3.1 Introduction
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Earlier, you have studied natural numbers and some of their properties; the numbers {1, 2, 3, . . .}. This is called as the set of all natural numbers and is denoted by N. You have seen that the sum or product of any
two natural numbers is again a natural number: for example 5 + 13 = 18;
12 × 15 = 180. We say that the set of all natural numbers is closed under addition and multiplication(or the closure property of addition and
multiplication of the set of all natural numbers). You have also observed
observed that:
8 + 12 = 12 + 8; 13 + (9 + 21) = (13 + 9) + 21; 15 × 7 = 7 × 15; 3 × (5 × 6) = (3 × 5) × 6.
Here you can take any natural number. Thus you have also seen that for
all natural numbers m, n, p the following hold:
m+n
m + (n + p)
m·n
m · (n · p)
=
=
=
=
n + m,
(commutative property of addition);
(m + n) + p, (associative property of addition);
n · m,
(commutative property of multiplication);
(m · n) · p,
(associative property of multiplication).
Unit 3
54
You have learnt how to combine addition and multiplication and get yet
another property called the distributive property. For example
5 × (7 + 8) = 5 × 15 = 75 = 35 + 40 = (5 × 7) + (5 × 8).
Thus for all natural numbers m, n, p, you may write
(n + p) · m = n · m + p · m,
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From this you also get
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m · (n + p) = m · n + m · p.
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using the commutative property of multiplication and addition. You have
also seen that the natural number 1 satisfies 1 × 8 = 8 × 1 = 8; again 8 is
irrelevant and the relation 1 · m = m · 1 = m holds for all natural numbers
m.
You must have wondered why such a natural number does not exist for
addition; a number u such that m + u = u + m = m for all natural numbers.
This is the reason you have added the number 0 to the set of all natural
numbers and got the set of all whole numbers, W . Thus W = {0, 1, 2, 3, . . .}.
The number 0 satisfies, for example, 8 + 0 = 0 + 8 = 8 and 9 × 0 = 0 × 9 = 0.
Thus the number 0 obeys certain rules:
m + 0 = 0 + m = m,
to
m · 0 = 0 · m = 0,
for all natural numbers m;
for all natural numbers m;
0 + 0 = 0;
No
t
0 · 0 = 0.
If you multiply two non-zero numbers in W , you get a non-zero number:
for example 14 × 6 6= 0. Thus in W , you must have noticed that m · n = 0
is possible if and only if either m = 0 or n = 0(or may be both). Another
important property you have studied about the natural number is that it
is possible to compare any two natural numbers; for example, if you are
asked to compare 12 and 81, you immediately say that 12 is smaller than
81; or 81 is larger than 12. You write 12 < 81 or 81 > 12. Thus given any
two natural numbers m, n, you know that either m < n or m = n or m > n;
and only one of these properties hold. This is called the ordering on the
Rational numbers
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set of all natural numbers; and you write 1 < 2 < 3 < 4 < · · · . Now we place
0 before 1 and get ordering on W ; 0 < 1 < 2 < 3 < 4 < · · · . There is one
fundamental truth about this ordering.
For example, if you consider the set E = {3, 6, 9, . . .}, the set of all multiples of 3, you see that 3 is the smallest element in it. However this set
does not have the largest element. Suppose you take the set of all marks
scored by students in your class in a test. If you write the marks in ascending order, you see that one of the marks is the lowest. Any subset of
N containing at least one element of N is called a non-empty subset of N.
For example, the set of all even natural numbers is a non-empty subset of
N. The set of all numbers which are both even and odd is an empty subset
of N since there is no such number.
Every non-empty subset of natural numbers of N (or W ) has
the smallest element.
This is called the well ordering property of natural numbers.
No
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Activity 1: Write down five finite non-empty subsets of N and find their
smallest elements. Write down two infinite subsets of W and find the least
element of these subsets.
There is one distinct disadvantage inherent in N or W . Consider the
equation x + 5 = 3. You see that there is no natural number m such that
m + 5 = 3. In fact, for any natural number, you know that m + 5 > 5 > 3.
This disadvantage is removed in the set of all integers, Z. You have seen
that you can adjoin to W , another class of numbers called the negative
integers. For each natural number m, you associate another number
−m called the negative of m(or the opposite of m). Thus Z consists of
three parts: the set of all natural numbers; 0; and the set of all negative
numbers. We usually write this as
Z = {· · · − 5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, · · · }.
Here Z is derived from the German word Zahlen(which means number).
You have also seen that you can perform addition and multiplication
on Z. If m and n are two natural numbers, then
(i) (−m) + (−n) = −(m + n);
Unit 3
56
(ii) (−m) + 0 = −m = 0 + (−m);


 −(m − n), if m > n
(iii) (−m) + n =
n − m, if m < n ;


0, if m = n
(v) (−m) · (−n) = m · n;
(vi) (−m) · 0 = 0 · (−m) = 0.
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(iv) (−m) · n = m · (−n) = −(m · n);
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Of course, if m and n are two whole numbers, we retain the same old
addition and multiplication for them.
With this extension of definition of addition and multiplication, Z now
enjoys several nice properties:
1. closure property: for all integers a, b, both a + b and a · b are also
integers;
2. commutative property: for all integers a, b
a + b = b + a and a · b = b · a;
3. associative property: for all integers a, b, c,
to
a + (b + c) = (a + b) + c and a · (b · c) = (a · b) · c;
4. distributive property: for all integers a, b, c,
No
t
a · (b + c) = a · b + a · c;
5. cancellation law: if a, b, c are integers such that c 6= 0 and ac = bc,
then a = b (effectively you can cancel c on both sides).
Note that the cancellation law holds only if c 6= 0. For example, you may be
tempted to write 3 · 0 = 0 = 5 · 0 and try to cancel 0 on both sides to end up
with an absurd conclusion 3 = 5. Many of the fallacious conclusions are
due to such erroneous cancellations.
What is the advantage in moving from N to Z? You see that 0 has a
special status: a + 0 = 0 + a = a for all integers. We say that 0 is the additive
identity of Z(or 0 is the identity with respect to addition). Moreover, for
each integer a, we have another integer −a such that a+(−a) = 0 = (−a)+a;
Rational numbers
57
if a = m is a natural number, −a is the negative integer −m; if a = 0, then
−a = 0; if a is a negative integer, then a = −n for some natural number n
and we take −a = n, so that
a + (−a) = (−m) + m = 0,
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by the way we have defined negative numbers. We say −a is the additive
inverse of a. Every integer has additive inverse. Now you can solve an
equation x + a = b in integers, for any two integers a, b. We can take x =
b + (−a). Then
x + a = (b + (−a)) + a = b + (−a) + a = b + 0 = b;
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we have used associativity of addition, the fact that (−a) is the additive
inverse of a and that 0 is the additive identity.
We can also order the elements of Z. For any natural number n, we
put −n < 0. If m and n are two natural numbers such that m < n, we put
−n < −m. You now see that all the elements of Z can be compared. Every
natural number is an integer and we call the natural numbers as positive
integers.
Here you may be wondering what is the status of subtraction. This is
not introduced as a fundamental operation. You have learnt earlier that
12 − 7 = 5. However, you have learnt now that −7 is the additive inverse of
7. Thus you may think 12 − 7 = 12 + (−7) adding 12 and −7. Here you see
that the subtraction is a convenient name we have given to the extension
of our fundamental operation addition to cover negative integers. This
makes sense even when two integers are negative. Suppose you want to
find −8 − 13. You put this as (−8) + (−13) = −(8 + 13) = −21, by the way
we have defined addition of two negative integers. Now what is 15 − 21?.
Your answer is clear: adding 15 and (−21). The definition shows that
15 + (−21) = −(21 − 15) = −6. What is the additive inverse of −m? Since
m + (−m) = 0, we see that m is the additive inverse of −m; thus you obtain
−(−m) = m. The emphasis here is to the fact that − symbol represents
additive inverse.
It is a universal law that when you gain some thing, you must lose
some thing. In Z, the gain is clear: you are able to solve an equation of the
form x + a = b, whenever a and b are integers. Such an equation cannot be
Unit 3
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solved in N unless b > a, and a, b natural numbers. On the other hand, you
see that in the set of all integers, a non-empty subset need not have the
smallest element: if you take {· · · − 5, −4, −3, −2, −1}, the set of all negative
integers, this does not have the smallest element(why?). But the gain we
have by going to the integers is much more than the loss incurred.
Consider again Z. Here an equation of the form ax = b, where a 6= 0
cannot be solved, in general. (You may solve this provided a divides b.)
Hence Z is also inadequate for our purpose. We look forward for a new
number system in which we can do better things than we are able to do
in Z. We also take care that it has most of the nice properties of Z and
perhaps much more. But remember we have to loose some thing. We shall
see what is the loss and what is the gain?
Exercise 1.3.1
1. Identify the property in the following statements:
(i) 2 + (3 + 4) = (2 + 3) + 4;
(ii) 2 · 8 = 8 · 2;
(iii) 8 · (6 + 5) = (8 · 6) + (8 · 5).
2. Find the additive inverses of the following integers:
6, 9, 123, −76, −85, 1000.
3. Find the integer m in the following:
(i) m + 6 = 8;
m + 25 = 15;
(iii) m − 40 = −26;
4. Write the following in increasing order:
(iv) m + 28 = −49.
to
21, − 8, − 26, 85, 33, − 333, − 210, 0, 2011.
5. Write the following in decreasing order
No
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85, 210, − 58, 2011, − 1024, 528, 364, − 10000, 12.
1.3.2 Rational numbers
In your earlier class, you have learnt about fractions; numbers of the
form p/q, where p and q are natural numbers. For example: 1/2, 1/3, 3/4, 8/3
and such numbers. You have also learnt how to add and multiply such
numbers.
1
8
Example 1. Add and multiply and .
3
5
Solution: We have
Rational numbers
59
1 8
1×8
8
× =
= .
3 5
3×5
15
If you are given a fraction 10/4, you write this in the form
10
5×2
5
=
= ,
4
2×2
2
d
Their product is
1 8
(1 × 5) + (8 × 3)
5 + 24
29
+ =
=
= .
3 5
3×5
15
15
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by canceling one 2 in the numerator with another 2 in the denominator.
In other words, if there are common factors between the numerator and
the denominator, you cancel them for convenience. Thus you do not distinguish between 10/4 and 5/2. You also divide one fraction by another: if
you want to divide 1/3 by 8/5, the result is
1×5
5
1/3
=
= .
8/5
3×8
24
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You have learnt a lot about working with fractions. Can all these be put in
a formal way? Is it possible to include negative fractions just like we have
included negative integers?
p
We define a rational number as a number of the form , where p is
q
an integer and q > 0. Here p is called the numerator and q is called
the denominator of the rational number p/q. Thus the denominator of
a rational number is always a positive integer, where as the numerator
could be positive, negative or possibly 0.
Given two rational numbers a/b and c/d, we say they are equivalent if
a × d = c × b. Thus 10/4 is equivalent to 5/2 since 10 × 2 = 20 = 5 × 4. We say
a rational number a/b is in its lowest form or irreducible form if a and b
do not have any common factors other than 1. Thus 5/2 is its lowest form,
where as 10/4 is not.
Activity 2: Write ten rational numbers equivalent to 3/4. How many rational numbers are there which are equivalent to 3/4?
Thus 3/4, 1/5, 6/7, 7/10, −5/8, −6/11 are all rational numbers. You may
observe that each integer is also a rational number: given an integer a,
you may write it as a/1. Thus we do not distinguish between 7 and 7/1.
Suppose you have a fraction 3/4. Just like we have defined negative
integers using natural numbers(or positive integers), we can also define
Unit 3
60
3
−3
negative of . This is defined as the rational number
. This is denoted
4
4
3
by − .
4
We denote the set of all rational numbers by Q. Thus we can describe
Q as:
p p, q are integers and q > 0, HCF(p,q)=1 .
Q=
q
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Here HCF denotes the highest common factor.
At this point, you may wonder why we are not taking negative integers
p
in the denominator. Suppose you take a number of the form
, where
−q
q > 0 is an integer. You may observe that this is equivalent to the ratio−p
nal number
, since p × q = (−p) × (−q). Thus you are not losing out
q
any rational number by restricting denominators to the set of all positive
integers.
We say a rational number is positive if both its numerator and denominator are positive integers.
The term rational number is derived from the word ratio. A rational
number is a ratio of two integers where the denominator is not equal to
zero.
Exercise 1.3.2
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1. Write down ten rational numbers which are equivalent to 5/7 and the
denominator not exceeding 80.
2. Write down 15 rational numbers which are equivalent to 11/5 and the
numerator not exceeding 180.
3. Write down ten positive rational numbers such that the sum of the
numerator and the denominator of each is 11. Write them in decreasing order.
4. Write down ten positive rational numbers such that numerator − denominator for each of them is −2. Write them in increasing order.
3
5. Is
a rational number? If so, how do you write it in a form conform−2
ing to the definition of a rational number(that is, the denominator as
a positive integer)?
6. Earlier you have studied decimals 0.9, 0.8 . Can you write these as
rational numbers?
Rational numbers
61
1.3.3 Properties of rational numbers
Closure property
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You have learnt that the set of all natural numbers and the set of all integers have addition and multiplication satisfying the closure, commutative,
associative and distributive properties. Can we similarly define addition
and multiplication on the set of all rational numbers having these properties? Our starting point is the addition and multiplication of fractions,
which you have learnt in your lower class.
5
11
Example 1. Let us find the sum of and . It is
6
13
(5 × 13) + (11 × 6)
5 11
+
=
6 13
(6 × 13)
65 + 66
=
78
131
=
.
78
Similarly, the sum of
4
−3
and
is:
7
5
−7
−3
and
is:
4
7
No
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The sum of
to
(4 × 5) + ((−3) × 7)
4 −3
+
=
7
5
7×5
20 + (−21)
=
35
−1
.
=
35
−7 −3
(−7) × 7 + (−3) × 4
+
=
4
7
4×7
(−49) + (−12)
=
28
−61
=
.
28
2
and
11
2
8
× =
11 7
Example 2. The product of
8
is:
7
2×8
16
= .
11 × 7
77
Unit 3
62
−3
−7
and
is:
5
2
−3 −7
(−3) × (−7)
21
·
=
= .
5
2
5×2
10
Similarly, the product of
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Based on these ideas, we can introduce addition and multiplication of
a
c
two rational numbers. Given two rational numbers and , we define
b
d
a c
ad + cb
a c
ac
+ =
;
· = .
b d
bd
b d
bd
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Since b > 0 and d > 0, you see that bd is a natural number. Moreover
ad + cb
ac
ad + cb and ac are integers. Thus you may conclude that
and
bd
bd
are rational numbers. Thus the sum of two rational numbers is again a
rational number. We say that the set of all rationals numbers is closed
under addition.
The set of all rational numbers is closed under addition and
multiplication.
No
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Activity 3: Take ten pairs of rational numbers. Find the sum of the
numbers in each pair. Check that you will always end up with rational
numbers. Thus satisfy your self that the closure with respect to addition
holds. Similarly, multiply the two numbers in each pair and satisfy your
self that the closure property with respect to multiplication also holds.
Associative property You have seen earlier that integers have the properties: p + (q + r) = (p + q) + r and p · (q · r) = (p · q) · r. For example:
3 + (5 + 8) = (3 + 5) + 8 and 3 × (5 × 8) = (3 × 5) × 8. Will such things hold for
rational numbers as well?
Example 3. Consider three rational numbers 1/2, 4/5, −6/7. Observe that
1
4 −6
1
7 × 4 + (−6) × 5
1
28 − 30
+
+
= +
=
+
2
5
7
2
5×7
2
35
1 −2
=
+
2
35
35 × 1 + (−2) × 2
=
70
31
=
.
70
Rational numbers
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On the other hand, you also have
1 4
−6
−6
5+8
13 −6
+
+
+
=
=
+
2 5
7
10
7
10
7
91 − 60
=
70
31
=
.
70
Can you conclude
1 4
−6
1
4 −6
=
+
+
+
+
?
2
5
7
2 5
7
d
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This is true for any three rational numbers a/b, c/d and e/f . You compute
a c e
both
c
e
a
+
+
and
+
+ .
b
d f
b d
f
You get
a
c
e
a cf + ed
adf + (cf + de)b
+
+
= +
=
b
d f
b
df
bdf
adf + cf b + deb
=
.
bdf
c e
ad + cb e
adf + cbf + ebd
+
+ =
+ =
.
b d
f
bd
f
bdf
Now can you see that adf + cf b + deb = adf + cbf + ebd? What properties of
integers have we used here? Thus both the sums are the same.
2 7 11
Similar result is true for multiplication. Consider , , . We have
3 8 13
7 11
2
77
154
2
×
×
= ×
=
;
3
8 13
3 104
312
2 7 11
14 11
154
×
×
=
×
=
;
3 8
13
24 13
312
We conclude that
2 7 11 2 7 11
×
×
=
×
× .
3
8 13
3 8
13
a c
e
For any three rational numbers , and , we obtain
b d
f
a
c e
a ce
ace
=
·
·
·
=
,
b
d f
b df
bdf
a c e
ac e
ace
·
·
=
· =
,
b d
f
bd f
bdf
a
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Similarly,
Unit 3
64
so that
a
·
b
c e
·
d f
=
a c e
·
· .
b d
f
Addition and multiplication are associative on the set of all rational numbers.
Commutative property
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You have studied earlier that addition and multiplication satisfy commutative property on the set of all integers. Given any two integers m and n,
you have m + n = n + m and m · n = n · m; for example 3 + 5 = 5 + 3 and
3 × 5 = 5 × 3. Do we have similar things for rational numbers?
Example 4. Let us take two rational numbers, say 8/11 and −16/9. Observe that
8
−16
8 × 9 + (−16) × 11
+
=
11
9
11 × 9
72 − 176
−104
=
=
.
99
99
Similarly,
8
(−16) × 11 + 8 × 9
−16
+
=
9
11
9 × 11
−176 + 72
−104
=
=
.
99
99
Thus you will get
No
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8
−16
−16
8
+
=
+ .
11
9
9
11
Example 5. Similarly, you may verify that
8 −16
−16 8
·
=
· ,.
11
9
9
11
Can we put this in a more general setting? Take any two rational numbers
a/b and c/d. Observe that
a c
ad + cb
c a
cb + ad
+ =
, and
+ =
.
b d
bd
d b
db
But you know that ad + cb = cb + ad and bd = db(What properties of integers
are used here?). Hence you may conclude that
a c
c a
+ = + .
b d
d b
Rational numbers
65
This gives the commutative property of addition. A similar observation can
be made for multiplication:
a c
ac
ca
c a
· =
=
= · .
b d
bd
db
d b
We obtain the commutative property of multiplication.
he
Distributive property
d
Addition and multiplication are commutative on the set of all
rational numbers.
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Consider the rational numbers
2/3,
1/2 and 1/9. Observe that
2 11
2
1 1
22
11
= ·
·
+
=
= .
3
2 9
3 18
54
27
Similarly, we have
2
2
66
11
2 1 2 1
· + · = +
=
= .
3 2 3 9
6 27
162
27
Note that we have used the property of equivalent fractions. We may con
clude that
2
1 1
2 1 2 1
·
+
= · + · .
3
2 9
3 2 3 9
Activity 4:
to
Take several triples of rational numbers and verify distributive property.
Also prove this as a general statement: if p/q, r/s and u/v are rational
numbers, then
p r u p r p u
·
+
= · + · ,
q
s v
q s q v
using the definition of addition and multiplication of rational numbers and
properties of integers.
No
t
In the set of all rational numbers, multiplication is distributive
over addition.
Think it over!
Given rational numbers a/b, c/d, e/f , can you have
a
c e
a c
a e
+
·
=
+
·
+
?
b
d f
b d
b f
In other words can addition be distributive over multiplication?(That is, can we interchange addition and multiplication
in the above distributive property?)
Unit 3
66
Additive identity
0
Consider the rational number . Observe
1
7 0
7×1+0×8
7
+ =
= .
8 1
8×1
8
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For any rational number a/b, you see that
a 0
a×1+0×b
a
+ =
= .
b 1
b×1
b
he
0 7
7
+ = .
1 8
8
d
Similarly, you may verify that
Similarly, you may easily verify that
0 a
a
+ = .
1 b
b
0
Thus the rational number acts as additive identity. We denote this sim1
ply by 0.
The set of all rational numbers has 0 as additive identity; that
is r + 0 = 0 + r = 0, for all rational numbers r.
to
Multiplicative identity
No
t
Again consider the rational number
1
. We have, for example,
1
11 1
11
× = .
12 1
12
Thus any rational number a/b, we observe that
a 1
a
1 a
· = = · .
b 1
b
1 b
1
Hence the rational number (which again is denoted by 1) is identity with
1
respect to multiplication.
The set of all rational numbers has 1 as multiplicative identity;
that is r · 1 = r = 1 · r, for all rational numbers r.
Rational numbers
67
Additive inverse
8
−8
Take
and
. If we add these two, we get
13
13
8
−8
8 × 13 + (−8 × 13)
0
+
=
=
= 0.
13
13
169
169
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a
This is true for any rational number. For each rational number , consider
b
−a
the rational number
. Let us find their sum:
b
a −a
ab + (−a)b
0
+
=
=
.
b
b
b2
b2
0
0
But the rational number 2 is the same as as they are equivalent frac1
b
−a
a
tions. Thus
is the additive inverse of .
b
b
For each rational number r, there exists a rational number, denoted by −r, such that r + (−r) = 0 = (−r) + r.
Multiplicative inverse
You have seen that an integer may not have multiplicative inverse. For
example, 8 has no multiplicative inverse; 8 × a = 1 is not possible for any
7
integer a. On the other hand, consider . We see that
5
to
35
7 5
× =
= 1.
5 7
35
No
t
This is true for any nonzero rational number.
Take any nonzero rational number a/b. Then a 6= 0 and hence b/a is also
a rational number, Observe that
a b
ab
1
· =
= ,
b a
ba
1
the multiplicative identity. We have used the fact that 1/1 and ab/ba are
equivalent fractions. You observe here that every nonzero rational number
has multiplicative inverse.
For each rational number r 6= 0, there exists a rational number,
denoted by r −1 (or 1/r), such that r · r −1 = 1 = r −1 · r.
Unit 3
68
In integers, you have observed another important property, namely
cancellation law. For example, if 8 × a = 48, you write
8 × a = 8 × 6.
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You cancel 8 both sides and get a = 6. Thus if a, b, c are integers such that
a 6= 0 and ab = ac, then b = c. Effectively, you can cancel equal nonzero
integers on both sides of an equality. This also holds in Q.
For example, we have 4/4 = 2/2, as they are equivalent fractions. But
2
2
1
4
= 2 × , and
=2× .
4
4
2
2
You thus obtain
1
2
=2× .
4
2
Canceling 2 on both sides, you get 2/4 = 1/2, which is true as they are
equivalent fractions.
a
Suppose you have three rational numbers a/b, c/d, e/f such that 6= 0
b
and you have
a e
a c
· = · .
b d
b f
a
Since 6= 0, it has multiplicative inverse b/a. Multiply both sides by b/a:
b
b a c b
a e
= ·
·
·
·
.
a
b d
a
b f
2×
to
Use the associative property to write this as
b a
c
b a
e
·
·
· =
· .
a b
d
a b
f
No
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This gives
c
e
= ,
d
f
and we have canceled a/b both sides.
Now we define two more operations on Q: subtraction and division.
4
12
Consider the rational numbers
and
. We want to give meaning for
13
7
4
12
12
−12
− . Recall, you have the additive inverse of
, which is simply
.
13
7
7
7
We define
4
12
4
−12
−
=
+
.
13
7
13
7
You can simplify this using the definition of addition:
Rational numbers
69
4
−12
4 × 7 + (−12) × 13
−128
+
=
=
.
13
7
13 × 7
91
d
a
In other words subtraction amounts to adding the additive inverse. If
b
c
and are two rational numbers, then
d
a c
a −c
ad − bc
− = +
=
.
b d
b
d
bd
he
a
c
c
and
are two rational numbers and if
is not equal to
b
d
d
c
a
zero,we define the division of by as follows:
b
d
a d
ad
a c
÷ = × = .
b d
b
c
bc
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Similarly, if
a
c
with the multiplicative inverse of , which
b
d
8
exists as this is a non-zero rational number. If you want to divide
by
15
−7
, it is simply
11
8
−7
8
−11
−88
÷
=
×
=
.
15
11
15
7
105
Note that we are multiplying
to
The only fundamental operations are addition and multiplication. The subtraction and division are defined in terms of addition and multiplication.
No
t
Let us look at what we have gained from enlarging our number system
from Z to Q. If a 6= 0 is an integer, then there is no integer b such that
a · b = b · a = 1, unless a = 1 or a = −1. Thus, apart from 1 and −1, no
other integer has multiplicative inverse. On the other hand, every nonzero rational number has its multiplicative inverse in Q.
This helps us to solve an equation of the form rx = s, where r 6= 0 and s
3
5
are rational numbers. Suppose we have to solve the equation x = . You
8
9
solve this for x by multiplying both sides by 8/3. Thus
8 3
8 5
40
× x= × = .
3 8
3 9
27
You get x = 40/27. This you can do for any general equation.
Unit 3
70
Suppose r = a/b and s = u/v, where a, u are integers and b, v are natural
numbers. Since r 6= 0 implies that a 6= 0, r has its multiplicative inverse
b/a. Multiplying both sides by b/a, you get
b a b u
·
·x = · .
a
b
a v
Exercise 1.3.3
d
bu
.
av
he
This gives x =
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1. Name the property indicated in the following:
3 9
27
(i) 315 + 115 = 430; (ii) · = ; (iii) 5 + 0 = 0 + 5 = 5;
4 5
20
8
−8
8
8
22 23
(iv) × 1 = ; (v)
+
= 0; (vi)
·
= 1.
9
9
17
17
23 22
2. Check the commutative property of addition for the following pairs:
−8 23
−7 −18
102 3
, ; (ii)
, ; (iii)
,
.
(i)
201 4
13 27
9 19
3. Check the commutative property of multiplication for the following
pairs:
22 3
−7 25
−8 −17
(i) , ; (ii)
, ; (iii)
,
.
45 4
13 27
9 19
to
4. Check the distributive property for the following triples of rational
numbers:
13
1 1 1
−4 6 11
3
(i) , , ; (ii)
, , ; (iii) , 0, .
8 9 10
9 5 10
8
7
No
t
5. Find the additive inverse of each of the following numbers:
8 6 −3 −16 −4
,
,
,
,
.
5 10 8
3
1
6. Find the multiplicative inverse of each of the following numbers:
6 −8 19
1
2, ,
, ,
.
11 15 18 1000
Rational numbers
71
1.3.4 Representation of rational numbers on the Number
line
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Earlier, you have seen how to represent integers on a line. We choose
an infinite line and fix some point on the line. This is denoted by 0. Fix
a unit of length and on both sides of 0, go on marking points at equal
unit distance. On the right side, at unit distance you get 1. If you move a
further unit distance, you get 2 and so on. If you move to the left by unit
distance, you get −1. If you further move unit distance to the left, you get
−2 and so on. Thus all the integers are represented on the line.
−5 −4 −3 −2 −1 0
1
2 3
4
5
We can also use the same number line to represent rational numbers.
For example, we can represent 1/2 as the mid-point between 0 and 1.
A
−1/2
−1
1/2
0
1
We can obtain A by bisecting the line-segment from 0 to 1. Similarly,
1
we can represent − as the mid-point of the line-segment from −1 to 0.
2
7
How can we get ?
3
Q
to
P
No
t
0
1
P
2
3
4
R
5
6
7
S
8
9
Q
Again consider the line-segment P Q from 0 to 7. We divide P Q in to 3
equal parts: P R = RS = SQ.(Here you need some geometrical constructions, which you will learn later.) Then P R = 7/3. Hence R represents the
7
rational number 7/3. We can get − by locating a point R′ to the left of P
3
such that R′ P = P R.
R’
P
R
−7/3
0
7/3
S
Q
Unit 3
72
Activity 5:
1/8
1/4
3/8
1/2
3/4
1
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1 1 1 4 −3
on it.
Draw a number line and locate the points , , , ,
8 6 3 5 8
In this way, you can represent each rational number by a unique point
on the number line. You may observe that 2/4 and 1/2 are represented
by the same point on the number line. (Can you see why?) All rational
numbers which are equivalent to a given rational number get the same
representation on the number line.
No
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Can you now observe what is happening when you represent the rational numbers on the number line? You have seen that you can locate 1/2
between 0 and 1 as the mid-point of the line-segment from 0 to 1. If you
take the mid-point of 0 and 1/2, you get 1/4; and the mid-point of 1/2 and
1 is 3/4. The mid-point of 1/4 and 1/2 is 3/8. Do you see that the mid-point
of the line-segment joining two points representing two rational numbers
on the number-line is again a rational number? We may observe here that
it is possible to order the rational numbers using the ordering on Z. Supa
c
pose a/b and c/d are two rational numbers. We say < , if ad < bc. Thus
b
d
6
7
−7
−6
< , since 48 < 49. On the other hand
<
, since −49 < −48.
7
8
8
7
2
5
2
5
Suppose you consider and . Then < . Then the average of these
7
8
7
8
two is
2 5
+
7 8 = 51 .
2
112
51
2
5
lies between and . In fact
Now
112
7
8
2
51
5
<
< ,
7
112
8
which may be easily verified.
This is true for any two rational numbers. If we start with two ratioa
c
a
c
nal numbers and such that < , then the mid-point of the points
b
d
b
d
a
c
representing and represents the number
b
d
Rational numbers
73
a c
+
b d = ad + bc .
2
2bd
In fact
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a
ad + bc
<
⇐⇒ a × (2bd) < b(ad + bc)
b
2bd
⇐⇒ 2ad < ad + bc (cancellation of b)
a
c
⇐⇒ ad < bc ⇐⇒ < ,
b
d
d
This is also a rational number. You may observe that
a
c
ad + bc
<
< .
b
2bd
d
which is given. Similarly, you can prove the other inequality. Thus the
ad + bc
a
c
is strictly between and .
rational number
2bd
b
d
Between any two distinct rational numbers, there is another
rational number.
No
t
to
Compare this with the property of integers. Given any integer m, there
is no integer between m and its successor m + 1. This is no longer true for
rational numbers. We can talk of the next integer, but there is nothing
like the next rational number.
This is precisely the loss we have while moving from integers to rational
numbers. Thus the gain is: we can divide one rational number by another
nonzero rational number which helps us in solving an equation of the form
rx = s, where r 6= 0 and s are rational numbers. The loss is: there is no
more the next number in rationals, which we had in integers.
Exercise 1.3.4
1. Represent the following rational numbers on the number line:
−8 3 2 12 45
; ; ; ; .
5 8 7 5 13
2. Write the following rational numbers in ascending order:
3 7 15 22 101 −4 −102 −13
, , , ,
,
,
,
.
4 12 11 19 100 5
81
7
2
3
3. Write 5 rational number between and , having the same denomi5
5
nators.
Unit 3
74
4. How many positive rational numbers less than 1 are there such that
the sum of the numerator and denominator does not exceed 10?
5. Suppose m/n and p/q are two positive rational numbers. Where does
m+p
lie, with respect to m/n and p/q?
n+q
d
6. How many rational numbers are there strictly between 0 and 1 such
that the denominator of the rational number is 80?
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1.3.5 Introducing irrational numbers
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7. How many rational numbers are there strictly between 0 and 1 with
the property that the sum of the numerator and denominator is 70?
You have seen that there is no integer whose square is 2. The argument
used was: 12 = 1 < 2 < 4 = 22 and there is no integer between 1 and 2.
However, you have seen now that there are plenty of rational numbers
between 1 and 2. In fact there are infinitely many rational numbers which
lie between 1 and 2. Hence, it is natural to wonder whether one could get
a rational number between 1 and 2 such that its square is 2. And the
astounding answer is no! There is no rational number r such that r 2 = 2.
The argument is also simple. Suppose, if possible, there is a rational r
such that r 2 = 2. Write r = p/q, such that HCF(p,q)=1 You get the relation
to
p2 = 2q 2 .
No
t
This shows that p2 is even and hence p itself is even; for, if p is odd, p2
must be odd. Hence you may write p = 2a, for some integer a.
Substitution gives
4a2 = 2q 2
which implies q 2 = 2a2 .
But then, q is also even. Thus p and q are both even and must have a
common factor 2. This contradicts what we know: p/q is in its lowest
form. We conclude that no rational number, whose square is 2, exists.
Do you now see that the set of all rational numbers is also inadequate?
You cannot solve an equation of the form x2 = 2 in Q. This is the reason
that the mathematicians started enlarging the rational number system to
a better number system. One can show that for any natural number n,
Rational numbers
75
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which is not a perfect square, there is no rational number r such that
√
r 2 = n. Thus, we need to give a meaning to n, where n is a non-perfect
square. This is taken care in the real number system.
√
The number 2 or such numbers which are not rational numbers are
called irrational numbers. One can also prove that there is no ratio√
nal number whose cube is equal to 2. We denote this by 3 2 called the
cube-root of 2. This again is an irrational number. There is a systematic
construction of the real number system starting from Q, but it involves a
better understanding of the structure of Q.
The set of all real numbers consists of two parts: rational numbers
and irrational numbers. Both parts are infinite. However, in some sense,
the set of all irrational numbers is much larger than the set of all rational
numbers. In this way there are different infinities and a hierarchy among
infinities. You will enter a fascinating world of infinities.
Additional problems on “Rational numbers”
1. Fill in the blanks:
(a) The number 0 is not in the set of —————————-.
(b) The least number in the set of all whole numbers is —————.
to
(c) The least number in the set of all even natural numbers is———
——.
(d) The successor of 8 in the set of all natural numbers is ————–.
No
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(e) The sum of two odd integers is ——————-.
(f) The product of two odd integers is —————-.
2. State whether the following statements are true or false:
(a) The set of all even natural numbers is a finite set.
(b) Every non-empty subset of Z has the smallest element.
(c) Every integer can be identified with a rational number.
(d) For each rational number, one can find the next rational number.
(e) There is the largest rational number.
Unit 3
76
(f) Every integer is either even or odd.
(g) Between any two rational numbers, there is an integer.
3. Simplify:
(i) 100(100 − 3) − (100 × 100 − 3);
(ii) (20 − (2011 − 201)) + (2011 − (201 − 20))
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d
4. Suppose m is an integer such that m 6= −1 and m 6= −2. Which is
m
m+1
larger
or
? State your reasons.
m+1
m+2
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5. Define an operation ⋆ on the set of all rational numbers Q as follows:
r ⋆ s = r + s − (r × s),
for any two rational numbers r, s. Answer the following with justification:
(i) Is Q closed under the operation ⋆?
(ii) Is ⋆ an associative operation on Q?
(iii) Is ⋆ a commutative operation on Q?
(v) What is a ∗ 1 for any a in Q?
(vi) Find two integers a 6= 0 and b 6= 0 such that a ⋆ b = 0.
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to
6. Find the multiplicative inverses of the following rational numbers:
8 12 26 −13 −101
, , ,
,
.
13 17 23 11 100
7. Write the following in increasing order:
10 20 5 40 25 10
, , , , , .
13 23 6 43 28 11
8. Write the following in decreasing order:
21 31 13 41 51 9
, , , , , .
17 27 11 37 47 8
9. (a) What is the additive inverse of 0?
(b) What is the multiplicative inverse of 1?
(c) Which integers have multiplicative inverses?
Rational numbers
77
10. In the set of all rational numbers, give 2 examples each illustrating
the following properties:
(iii) distributivity of multiplica-
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12. Simplify the following:
3
22 36
6
25 12
+ ; (ii)
× ;
(i)
+
+
9
3
5
7
5
7
51 7
3
16 21
15 2
+
+
−
(iii)
÷ ; (iv)
×
.
2
6
5
7
8
3
9
he
11. Simplify the following using distributive property:
5
8
2
1 2
25 32
11 2
; (ii)
; (iii) ×
.
(i) ×
+
×
+
+
5
9 5
12
9
5
9
2
9
d
(i) associativity; (ii) commutativity;
tion over addition.
13. Which is the property that is there in the set of all rationals but which
is not in the set of all integers?
14. What is the value of
1+
to
15. Find the value of
1 1
−
3 4
1
1
1+
1+1
.
?
1 1
−
2 3
.
No
t
16. Find all rational numbers each of which is equal to its reciprocal.
17. A bus shuttles between two neigbouring towns every two hours. It
starts from 8 AM in the morning and the last trip is at 6 PM. On one
day the driver observed that the first trip had 30 passengers and each
subsequent trip had one passenger less than the previous trip. How
many passengers travelled on that day?
18. How many rational numbers p/q are there between 0 and 1 for which
q < p?
19. Find all integers such that
3n + 4
is also an integer.
n+2
Unit 3
78
20. By inserting parenthesis(that is brackets), you can get several values
for 2 × 3 + 4 × 5. (For example (2 × 3) + 4 × 5 is one way of inserting
parenthesis.) How many such values are there?
1
is also in its lowest form.
p+q
lowest form.
14n + 3
is in its
21n + 4
he
22. Show that for each natural number n, the fraction
1
+
q
d
21. Suppose p/q is a positive rational in its lowest form. Prove that
Glossary
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23. Find all integers n for which the number (n+3)(n−1) is also an integer.
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Ordering; the comparability of numbers.
Well ordering property; every non-empty subset has the least element.
Positive integers; the integers which can be identified with natural numbers.
Negative integers; those integers which are additive inverses of positive
integers.
Cancellation law; the law which enables us to cancel two non-zero equal
quantities on both sides of an equality.
Rational number; the numbers which are of the formp/q, where p is an
integer and q is a natural number.
Lowest form or irreducible form of a rational number; that form p/q of
a rational number such that p and q have no common factor.
Additive identity; that number which added to a given number does not
change the given number.
Additive inverse; given a number, that number which added to the given
number gives the additive identity.
Multiplicative identity; that number which multiplied with the given
number does not change the given number.
Multiplicative inverse; given a number, that number which when multiplied with the given number gives the multiplicative identity.
Density property; the inseparable property of numbers; for example rational numbers are dense in number line.
Rational numbers
79
Successor; the next number; integers have this property, where as rationals do not have this property.
Points to remember
d
• A rational number is a ratio of the form p/q, where p is an integer and
q is a natural number.
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• The set of all rational numbers is closed under addition and multiplication.
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• The set of all rational numbers has associative property and commutative property with respect to both addition and multiplication.
Moreover, multiplication is distributive over addition.
• The rational number 0 is the additive identity; 1 is the multiplicative
identity.
• Every rational number has its additive inverse; every non-zero rational number has its multiplicative inverse.
• Between any two distinct rational numbers, there are infinitely many
rational numbers.
No
t
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• Every integer has the next integer, but there is no next rational number for any given rational number.
CHAPTER 1
UNIT 4
COMMERCIAL ARITHMETIC
After studying this unit, you learn to:
• define the term percentage;
• identify profit or loss in commercial transactions;
he
• solve the numerical problems involving percentage;
d
• identify the processes of mathematics in commercial transactions;
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• workout problems on profit, loss, selling price and marked price;
• calculate discount, discount percentage and selling price after discount;
• solve problems on commission and commission percentage;
• define simple interest and other terminologies associated with simple
interest;
• calculate simple interest, principal, time, rate of interest and amount
in numerical problems.
1.4.1 Introduction
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In this unit, we study several aspects of commercial mathematics which
are useful in daily life. You go to a market and buy some essential item.
You pay money for it. But behind this transaction a separate world of
arithmetic exists. The vendor brings the goods from somewhere and he
has to decide how much he should charge for the goods before selling.
Depending on the market strategy, he has to fix his selling prices. Some
times, he may have to sell it for a price lower than the price for which he
has procured the goods. Some other times, he has to announce attractive
incentives to get a good number of customers. When you do some transactions in buying and selling land, houses and cattle, you may have to give
some money for the service you get from the agent. All these things are
generally dealt with money. This is the reason why money is an important
part of human life. Commercial mathematics is an insight into this kind
of transactions in daily life.
Commercial arithmetic
81
1.4.2 Percentage
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You are familiar with the meaning of percent. Per cent means for ev4
ery hundred. Thus, 4 percent (or 4%) means 4 for every hundred or
.
100
Percentage is also a fraction whose denominator is 100. The numerator
of the fraction is called rate percent. Thus 12% means 12 out of hun12
dred or
. The concept of percentage is used in business transactions,
100
calculating interest, comparison of quantities and the like.
Suppose a basket has, say, 6 pineapples and 14 oranges. Then the
number of pineapples and oranges can be compared using the fraction
6
3
3
= . The number of pineapples is times the number of oranges. In
14
7
7
7
the same way, the number of oranges is times the number of pineapples.
3
Comparison can also be done using percentages.
There are 6 pineapples out of
20 (=6+14) fruits. Therefore
percentage of pineapples
30
6
=
= 30%
20
100
to
(here denominator is made
100)
Out of 20 fruits, number of
pineapples is 6. So out of 100
fruits, number of pineapples
is
6
× 100 = 30%.
20
This is called unitary method.
No
t
The basket contains only pineapples and oranges.
Therefore percentage of pineapples + percentage of oranges = 100.
Or 30% + % of oranges =100 or % of oranges = 100 -30= 70.
Thus the basket has 30% pineapples and 70% oranges.
Percentage is a convenient way of comparing quantities.
Example 1. A man spends 78% of his monthly income and saves
What is his monthly income?
1,100.
Solution: Let his monthly income be 100. Then his expenditure is
Therefore, his savings is 100-78 = 22.
78.
Unit 4
82
Let us put this in reverse way. If the savings is
22, then the income
100
is 100. Hence if the savings is 1, then the income is
.
22
100
If the savings is 1,100, then the income is
× 1100 = 5, 000 rupees.
22
Hence his monthly income is 5,000.
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Aliter: Given expenditure is 78%. Therefore savings is (100 − 78) = 22%.
Let the monthly income of the man be x. Then 22% of x is 1,100.
This means
22
x = 1100.
100
Solving for x, we obtain
x = 1100 ×
100
= 5000.
22
5,000.
78
Verification: 78% of 5,000 =
× 5000 = 3900. Hence, his savings is
100
5,000 -3,900 = 1,100.
Hence his monthly income is
Example 2. An athlete won 8 events out of a number of events. If the win
percentage was 40, how many events were there in total?
to
Solution: We are given that win percentage is 40. So,
40 events were won out of 100 events.
100
× 8 = 20 events.
⇒ 8 events were won out of
40
Hence, there were 20 events in total.
Activity 1: Try this in another method.
No
t
Example 3. Ravi’s income is 25% more than that of Raghu. What percent
is Raghu’s income less than that of Ravi?
Solution: Let Raghu’s income be 100. Then, Ravi’s income is 125. Put
this in the reverse way. If Ravi’s income is Rs 125, Raghu’s income is
100
100. Hence if Ravi’s income is Rs 1, Raghu’s income is
. Changing the
125
100
scale to 100, if Ravi’s income is 100, then Raghu’s income is
× 100
125
= 80. Therefore Raghu’s income is 100 - 80 = 20% less than that of Ravi.
Example 4. The salary of an employee is increased by 15%. If his new
salary is 12,650, what was his salary before enhancement?
Commercial arithmetic
83
Solution: Let the salary before enhancement be 100. Since his increment is 15%, his salary after enhancement is 100 + 15 = 115.
Now we reverse the role. If the new salary is
ment is 100.
115, salary before enhance-
he
Exercise 1.4.2
d
100
If the new salary is 12,650, salary before enhancement is
× 12650 =
115
11,000.
Therefore, his salary before enhancement is 11,000.
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1. In a school, 30% of students play chess, 60% play carrom and the rest
play other games. If the total number of students in the school is 900,
find the exact number of students who play each game.
2. In a school function 360 remained after spending 82% of the money.
How much money was there in the beginning? Verify your answer.
3. Akshay’s income is 20% less than that of Ajay. What percent is Ajay’s
income more than that of Akshay?
4. A daily wage employee spends 84% of his weekly earning. If he saves
384, find his weekly earning.
5. A factory announces a bonus of 10% to its employees. If an employee
gets 10,780, find his actual salary.
1.4.3 Profit and Loss
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We purchase goods from shops. The shopkeeper purchases goods either directly from a manufacturer or through a wholesale dealer. The
money paid to buy goods is called Cost Price and abbreviated as C.P.
The price at which goods are sold in shops is called Selling Price abbreviated as S.P. When an article is bought, some additional expenses such
as freight charges, labour charges, transportation charges, maintenance
charges etc., are made before selling. These expenses are known as Overhead Charges. These expenses have to be included in the cost price.
Hence, you may conclude that
the real cost price = total investment
= price for buying goods + overhead charges.
If S.P > C.P, there is a gain or profit. If S.P < C.P, there is a loss.
Therefore,
Unit 4
84
profit = S.P - C.P.
Gain (or profit) or loss on
and
loss = C.P. - S.P.
100 is called as gain percent or loss percent.
Note: Profit or loss is always calculated on cost
price.
gain
× 100;
C.P.
4. Selling Price =
C.P.;
(100-loss%)
×
100
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1. Gain or profit % =
he
d
Remember the following formulae:
loss
× 100;
C.P.
(100+gain%)
3. Selling Price =
×
100
C.P.;
2. Loss % =
5. Cost Price =
100
×
(100+gain%)
S.P.;
6. Cost Price =
100
× S.P..
(100-loss%)
Example 5. The cost price of a computer is 19,500. An additional 450
was spent on installing a software. If it is sold at 12% profit, find the selling
price of the computer.
No
t
to
Solution: Cost price of the computer is
19,500 +
450 (overhead expenses) =
19,950. The
computer is sold at a profit of 12%.
Therefore,
(100+gain%)
× C.P.
100
(100 + 12)
=
× 19950
100
selling price =
112
× 1995
10
= 22344.
=
Thus the selling price is
Aliter: (Unitary method)
Profit = 12%. Hence selling
price is 100 + 12 = 112.
If the cost price is 100, selling price is 112.
If the cost price is 19,950,
112
× 19950
100
= 22344.
selling price =
Hence, the selling price of
computer is 22,344.
22,344.
Commercial arithmetic
100
× S.P.
(100-loss%)
100
=
× 4300
(100 − 14)
100
=
× 4300
86
= 5000.
cost price =
100
× 4300 = 5000.
86
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cost price =
Aliter: (Unitary method)
We know that the loss is 14%.
Therefore selling price =
100 - 14 = 86.
If the selling price is
86,
cost price is 100. If the selling price is 4300,
d
Solution: Selling price of the bicycle is 4,300, and the loss is 14%.
Therefore,
4,300, a dealer loses 14%. For how
he
Example 6. On selling a bicycle for
much should he sell it to gain 14%?
85
Hence the cost price of the bicycle
is 5,000.
Now, the cost price of the bicycle is
be the selling price to get 14% profit.
Expected gain = 14%.
Therefore,
(100+gain%)
× C.P.
100
(100 + 14)
=
× 5000
100
114
=
× 5000
100
= 5700.
No
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to
selling price =
5,000. Let us find out what should
Aliter:(Unitary method)
Expected gain = 14%.
Therefore, selling price
= 100 + 14 = 114.
If the cost price is 100, selling price = 114.
If the cost price is 5,000,
selling price =
Hence, the selling price of the bicycle to gain 14% is
114
×5000 = 5700.
100
5,700.
Example 7. Two cows were sold for 12,000 each, one at a gain of 20%
and the other at a loss of 20%. Find the loss or gain in the entire transaction.
Solution: We first find out the cost price of each cow and then add them
to find the total amount spent. We know the total money received. Com-
Unit 4
86
paring them, we will know whether there is a loss or gain in the whole
transaction.
Second Cow:
First cow:
S.P = 12,000.
S.P. = 12,000.
Gain = 20%. Therefore,
Loss = 20%. Therefore,
d
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Activity 2:
100
× S.P.
(100-loss%)
100
=
× 12000
(100 − 20)
100
=
× 12000
80
= 15000.
Cost price =
he
100
× S.P.
(100+gain%)
100
=
× 12000
(100 + 20)
100
=
× 12000
120
= 10000.
Cost price =
Find the cost price of the cow in both the cases using unitary method.
Let us compute the total cost price and selling price in the combined transaction.
Total C.P. of both cows = (10,000 + 15,000) = 25,000. Total S.P. of both
cows = 12,000 × 2 = 24,000. Here, we observe that S.P. < C.P.
Therefore, loss = (25,000 - 24,000 ) = 1,000. Hence,
to
loss percentage =
loss
1000
× 100 =
× 100 = 4.
C.P.
25000
Therefore, there is a loss of 4% in the whole transaction.
No
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Example 8. If the cost price of 21 cell phones is equal to selling price of
18 cell phones, find the profit percent.
Solution: Let the C.P. of each cell phone be Re. 1. Then the C.P. of 21 cell
phones is 21.
By the given data, S.P. of 18 cell phones = C.P. of 21 cell phones = 21.
21
Therefore, S.P. of 1 cell phone is
. This gives
18
Profit = S.P. -C.P. =
Thus, there is a profit of
21
3
1
−1 =
= .
18
18
6
1/6 on each cell phone. Now we can calculate
Commercial arithmetic
87
profit percentage:
Profit percentage =
1/6
50
2
profit
× 100 =
× 100 =
= 16 .
C.P.
1
3
3
Therefore, the profit percentage is 16 32 %.
d
To find the overall profit or loss, we need to find the combined
C.P. and combined S.P.
85
he
Example 9. A dealer sells a radio at a profit of 8%. Had he sold it for
less, he would have lost 2%. Find the cost price of the radio.
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Solution: Let the cost price of the radio be x. Let us calculate the selling
price with 8% profit and 2% loss, separately.
With 2% loss:
We obtain here
With 8% profit:
We have here
(100+gain%)
× C.P.
100
(100 + 8)
=
× C.P.
100
54
x.
=
50
S.P. =
(100-loss %)
× C.P.
100
(100 − 2)
=
× C.P.
100
49
x.
=
50
S.P. =
to
Activity 3:
Calculate S.P. in both the cases using unitary method.
No
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Thus the difference in the selling price with 8% profit and 2% loss is
54
49
5
x
x− x= x= .
50
50
50
10
But, this difference is given to be equal to
x
= 85. This
10
850.
85, so that
implies that, x = 850. Hence, the cost price of radio is
Exercise 1.4.3
1. Sonu bought a bicycle for 3,750 and spent 250 on its repairs. He
sold it for 4,400. Find his loss or profit percentage.
2. A shop keeper purchases an article for 3,500 and pays transport
charge of 100. He incurred a loss of 12% in selling this. Find the
selling price of the article.
Unit 4
88
3. By selling a watch for 720, Ravi loses 10%. At what price should he
sell it, in order to gain 15%?
4. Hari bought two fans for 2,400 each. He sold one at a loss of 10%
and the other at a profit of 15%. Find the selling price of each fan and
find also the total profit or loss.
d
5. A store keeper sells a book at 15% gain. Had he sold it for 18 more,
he would have gained 18%. Find the cost price of the book.
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6. The cost price of 12 pens is equal to selling price of 10 pens. Find the
profit percentage.
1.4.4 Discount
A reduction on the marked price of articles is called discount. Generally discount is given to attract customers to buy goods or to promote
the sale of goods. The following completely describe the facts related to
discount:
• discount is always given on the marked price of the article;
• discount = marked price - selling price;
• marked price is sometimes called list price;
• discount = rate of discount times the marked price;
to
• net price = marked price - discount.
18,000 was sold at
15,840. Find
No
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Example 10. A computer marked at
the percentage of discount.
Solution: Marked price is 18,000 and selling price is 15,480. Therefore, discount = 18,000 - 15,840 = 2,160. Thus, for a marked price of
18,000, discount is 2,160.
2160
For a marked price of 100, discount =
× 100 = 12%. Therefore the
18000
percentage of discount is 12.
Example 11. A tape recorder is sold at
of 5%. What is its marked price?
5,225 after being given a discount
Solution: We are given that the discount is 5%. This means that for
100, the discount is 5.
Commercial arithmetic
89
Therefore, selling price = 100 - 5= 95.
Thus on a selling price of 95, the marked price is 100.
100
On a selling price of 5,225, the marked price =
× 5225 = 5500.
95
Therefore, marked price of the tape recorder is 5,500.
Solution: Cost price =
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500;
20
profit = 20% of 500 =
× 500 = 100;
100
marked price = cost price + profit = 500 + 100 = 600.
Now the discount given is 12%.
12
× 600 = 72 rupees.
For 600, the discount =
100
Therefore, selling price = cost price - discount= 600 - 72 =
Hence the selling price of the article is 528.
d
Example 12. A shop keeper buys an article for 500. He marks it at 20%
above the cost price. If he sells it at 12% discount, find the selling price.
528.
Example 13. A cloth seller marks a dress at 45% above the cost price and
allows a discount of 20%. What profit does he make in selling the dress?
No
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Solution: Suppose the cost price of the dress material is 100. Since the
seller marks it 45% above the C.P., the marked price would be
100 + 45 = 145.
20
Discount of 20% on this marked price is = 20% of 145 =
× 145= 29.
100
Therefore, selling price = 145 - 29 = 116. We get
profit = S.P. - C.P. = 116 - 100 = 16.
profit
16
Hence, profit percent =
× 100=
× 100 =16.
C.P.
100
Thus the merchant makes a profit of 16% on the marked price. (Observe
it is not 45-20=25 percent.)
Exercise 1.4.4
1. An article marked
800 is sold for
704. Find the discount and
discount percent.
2. A dress is sold at 550 after allowing a discount of 12%. Find its
marked price.
3. A shopkeeper buys a suit piece for 1,400 and marks it 60% above
the cost price. He allows a discount of 15% on it. Find the marked
price of the suit piece and also the discount given.
Unit 4
90
4. A dealer marks his goods 40% above the cost price and allows a discount of 10%. Find the profit percent.
5. A dealer is selling an article at a discount of 15%. Find:
(i) the selling price if the marked price is
(ii) the cost price if he makes 25% profit.
500;
he
d
1.4.5 Commission
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You might have observed some advertisements in news papers regarding the availability of houses, sites, vehicles etc., for sale. Many a times,
these transactions are mediated by a person other than the owner and
the buyer. This mediator who helps in buying and selling is called commission agent or broker. The money that the broker or agent receives in
the deal is called brokerage or commission. Commission is calculated on
the transaction amount in percentage. Commission per hundred rupees
is called commission rate.
Example 14. A real estate agent receives a commission of 1.5% in selling
a land for 1,60,000. What is the commission amount?
to
Solution: Selling price of the land is 1,60,000 and the commission rate
is 1.5%.
If the selling price is 100, commission = 1.5.
1.5
If the selling price is 1,60,000, commission =
× 160000= 2,400.
100
Therefore commission amount is 2,400.
No
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The commission amount can also be calculated directly.
Commission = commission rate × selling price
In the example above,
1.5
× 160000 = 2,400.
100
Example 15. The price of a long note book is 18. A shop keeper sells
410 note books in a month and receives 1,033.20 as commission. Find
the rate of commission.
commission = 1.5% of
1,60,000 =
Solution: Price of one note book is 18.
=⇒ Price of 410 note books = 410 × 18 = 7,380.
Commercial arithmetic
91
Now the commission received for this amount is
100, the commission is
1,033.20. Hence for
1033.20
× 100 = 14.
7380
Therefore, the rate of commission is 14%.
he
d
Example 16. Abdul sold his house through a broker by paying 6,125
as brokerage. If the rate of brokerage is 2.5%, find the selling price of the
house.
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Solution: Brokerage given is 6,125. The brokerage rate is 2.5%.
If the brokerage is 2.5, selling price would be 100.
If the commission is 6,125, selling price is
100
× 6125 = 245000.
2.5
Thus the selling price of the house is
2,45,000.
The selling price can also be calculated directly;
selling price =
In this example, S.P. =
100
× commission.
commission rate
100
× 6125 =
2.5
2,45,000.
to
Exercise 1.4.5
No
t
1. Sindhu sells her scooty for 28,000 through a broker. The rate of
brokerage is 2 12 %. Find the commission that the agent gets and the
net amount Sindhu gets.
2. A share agent sells 2000 shares at 45 each and gets the commission
at the rate of 1.5%. Find the amount the agent gets.
3. A person insures 26,000 through an insurance agent. If the agent
gets 650 as the commission, find the rate of commission.
4. A selling agent gets 10,200 in a month. This includes his monthly
salary of 6000 and 6% commission for the sales. Find the value of
goods he sold.
Unit 4
92
1.4.6 Simple interest
d
People borrow money from banks or financial institutions or money
lenders for various purposes. While returning the borrowed money after a
period of time, they need to pay some extra amount. This extra amount
paid on the borrowed money after a period of time is called interest. In
this context, we define the following terms.
he
1. Principal: the money borrowed is called principal or sum.
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2. Interest: the extra money paid on the principal after a period of time
is called interest.
3. Amount: the total money paid is called amount. Thus Amount =
Principal + Interest.
4. Rate: interest for every
per annum.
100 for one year is known as rate percent
5. Time: time is the duration for which the borrowed money is utilised.
Time is expressed in years or months or days.
6. Simple interest: the interest calculated uniformly on the principal
alone throughout the loan period is called simple interest. In other
words, it is the interest paid on the principal alone.
to
In the world of finance (Bankers rule), time is often
expressed in days also.
Formula to find the simple interest:
No
t
Let P = principal; R = rate of interest per annum; T = time in years; I =
simple interest. These are related by the formula
I=
P ×T ×R
100
From the above formula, we obtain different formulae:
P =
100 × I
,
T ×R
T =
100 × I
,
P ×R
R=
100 × I
.
P ×T
Amount = principal + interest.
Commercial arithmetic
93
Example 17. Calculate the interest on
years.
use the formula for I;
PTR
800 × (7/2) × (13/2)
=
= 2 × 7 × 13 = 182.
100
100
182.
he
Thus the interest is
d
7
13
800; T = 3 21 years = years; R = 6 12 % =
%. We
2
2
Solution: Given: P =
I=
800 at 6 21 % per annum, for 3 12
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Example 18. Find the simple interest on 3,000 at 16% per annum for
the period from 4th February 2010 to 16th June 2010.
Solution: Here the principal is P = 3,000, and the rate of interest R =
16% p.a. However, we are not given time in years; we have only the period,
4th February 2010 to 16th June 2010. We have to convert this to years.
Observe that
February 5th to 28th
March
April
May
June
=
=
=
=
=
24
31
30
31
16
days(2010 is not a leap year)
days
days
days
days.
to
Adding, we get the total time is equal to 132 days. Converting this to years,
132
T =
years. Now we have all the required data to apply formula:
365
I=
(3000 × 132 × 16)
PTR
=
= 173.58.
100
365 × 100
No
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Thus the interest is approximately
174.
Note: 1. For calculating interest, the day on which money deposited is not counted, while the day on which money is withdrawn
is counted.
2. When the time is given in days or months, it is to be expressed in
years.
Example 19. A sum at a simple interest of 12 21 % amounts to
after 3 years. Find the sum, and the interest.
2,502.50
Unit 4
94
2,502.50, and R = 12 21 % =
Solution: Given: Amount =
be x. Using the formula, we have
I=
25
%. Let the sum
2
75x
PTR
x × 3 × 25
3x
=
=
=
.
100
2 × 100
200
8
But we know that
This amount is given to be
d
3x
(8x + 3x)
11x
=
=
.
8
8
8
he
Amount = Principal + Simple interest = x +
2,502.50. Therefore, we obtain 2502.50 =
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Solving for x, we get
11x
.
8
2502.5 × 8
= 1820.
11
1,820. The interest is
2,502.50 - 1,820= 682.50.
x=
Hence, the sum is
Example 20.
800 amounts to
920 in 3 years at a certain rate of
interest. If the rate of interest is increased by 3%, what would the amount
will become?
to
Solution: Recall, interest(I) = amount(A) - principal(P ). Hence,
I = 920 - 800 = 120.
800. Using,
This interest is accrued in T = 3 years, for the principal
PTR
I=
we get,
100
100 × 120
100 × I
R=
=
= 5.
P ×T
800 × 3
No
t
Thus the original rate of interest is 5%. After the increase of interest by
3%, the new rate of interest, which we again denote by R = 5% + 3% =
8%. The principal P = 800 and the period T = 3 years remain the same.
Therefore,
PTR
800 × 3 × 8
I=
=
= 192.
100
100
Therefore, the new amount = 800 + 192 = 992.
Exercise 1.4.6
1. Find the simple interest on
2,500 for 4 years at 6 14 % per annum.
Commercial arithmetic
95
2. Find the simple interest on
165 days.
3. In what period will
simple interest?
3,500 at the rate of 2 12 % per annum for
5,200 amounts to
7,384 at 12% per annum
d
4. Ramya borrowed a loan from a bank for buying a computer. After
4 years she paid 26,640 and settled the accounts. If the rate of
interest is 12% per annum, what was the sum she borrowed?
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1.4.7 Tax
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5. A sum of money triples itself in 8 years. Find the rate of interest.
The Government requires money for its functioning. Money required
for a Government is collected from the public in the form of taxes. One
such method of collecting money is Sales Tax.
Sales Tax is the tax we pay when we buy goods/articles from a shop.
Sales Tax is charged by the Government on the sale of every good/article.
Sales tax is called indirect tax as it is collected from the manufacturer,
wholesaler and retailer (shopkeeper) who in turn collects it from the customer.
Value added tax (VAT)
to
Value Added Tax(VAT) is a revised version of sales tax. Normally an
article, before it reaches consumer, passes through various stages as given
here:
No
t
Manufacturer → Wholesaler → Retailer → Consumer.
The person/company who/which manufactures an article is Manufacturer. Depending on the cost of producing an article, manufacturer marks
the price higher than the cost price. On this marked price manufacturer
has to charge sales tax, which he pays to the Government.
The person who purchases large quantities of articles from the manufacturer is the Wholesaler. He marks a price higher than the price he purchased from the manufacturer(as he has to get his profit). On this marked
price, he also charges sales tax.
Unit 4
96
The person who buys articles in smaller quantities from the wholesaler
is the Retailer. He marks the price higher than his cost price(he again
takes his profit). On this marked price, he charges sales tax.
he
d
The common people who purchase the articles from the shop are Consumers. For the consumer, the cost price is the marked price of the retailer plus the sales tax on the marked price. Thus, VAT is a tax on the
value added at each stage for a product that has to pass through various
stages in the channel of distribution.
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Remember:
No shopkeeper sells any article at loss. Even when a high discount is given, he makes a profit. The discount is given to
attract customers and to expedite sales so that he can reinvest
his principal.
Example 21. Abdul purchases a pair of clothes with a marked price
1,350. If the rate of sales tax is 4%, calculate the amount to be paid by
him.
Solution: Marked price of the item is
to
1,350, and sales tax is 4%, on the
4
× 1350 = 54.
marked price. Hence the total sales tax on the item is
100
Amount to be paid = Marked price + tax = 1,350 + 54= 1,404. Hence,
the amount to be paid by him is 1,404.
Example 22. A blazer marked at
rate of sales tax.
1,600 is billed at
1,696. Find the
No
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Solution: Selling price is 1,696 and marked price is 1,600. Hence
sales tax paid is 96. On 1,600, sales tax is 96. Hence the percentage
of sales tax is
96
× 100 = 6.
1600
Hence, the rate of sales tax is 6%.
Exercise 1.4.7
1. A person purchases the following items from a mall for which the
sales tax is mentioned against:
(a) Stationary materials for
250 and sales tax of 4% there on;
Commercial arithmetic
97
(b) Electronic goods worth 2,580 and sales tax of 10% there on;
(c) Groceries worth 1,200 on which sales tax of 3% is levied;
(d) Medicines worth 200 with sales tax of 6%.
Find the bill amount for each item.
he
d
2. A person buys electronic goods worth 10,000 for which the sales
tax is 4% and other material worth 15,000 for which the sales tax
is 6%. He manufactures a gadget using all these and sells it at 15%
profit. What is his selling price?
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3. A trader purchases 70 kg of tea at the rate of 200 /kg and another
30 kg at the rate of
250/kg. He pays a sales tax of 4% on the
transaction. He mixes both of them and sells the product at the rate
of 240/kg. What is the percentage gain or loss(find approximate
value)?
Additional problems on “Commercial Arithmetic”
1. Four alternative options are given for each of the following statements.
Select the correct option.
700 is:
(a) Nine percent of
A.
63 B.
630 C.
6.3 D.
0.63
to
(b) What percent of 50 metres is 12 metres?
A. 20%
B. 60% C. 24% D. 32%
(c) The number whose 8% is 12 is:
No
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A. 120 B. 150 C. 130 D. 140
(d) An article costing
600 is sold for
750. The gain percentage is:
A. 20 B. 25 C. 30 D. 35
(e) By selling note book for
price of the book is:
A.
18 B.
30 C.
22 a shopkeeper gains 10%. The cost
20 D.
22
(f) The percentage of loss, when an article worth
for 9,000 is:
A. 10 B. 20 C. 15 D. 25
10,000 was sold
Unit 4
98
(g) A radio marked
A.
50 B.
1000 is given away for
100 C.
150 D.
850. The discount is:
200
(h) A book marked
250 was sold for
percentage of discount is:
200 after discount. The
A. 10 B. 30 C. 20 D. 25
A.
185 B.
170 C.
215 D.
d
200. If 15% of discount is
175
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(i) The marked price of an article is
allowed on it, its selling price is:
A.
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(j) One sells his bike through a broker by paying 200 brokerage.
The rate of brokerage is 2%. The selling price of the bike is:
12,000 B.
10,000 C.
14,000 D.
(k) The brokerage amount for a deal of
mission is:
A.
500 B.
250 C.
5,000 D.
12,500
25,000 at 2% rate of com2,500
(l) If 1,600 is the commission at 8% for goods sold through a broker
, the selling price of the goods is:
A.
18,000 B.
20,000 C.
(m) The simple interest on
A.
100 B.
200 C.
22,000 D.
24,000
5,000 at 2% per month for 3 months is:
300 D.
400
to
(n) The time in which simple interest on a certain sum be 0.15 times
the principal at 10% per annum is:
No
t
A. 1.5 years B. 1 year C. 2 years D. 2.5 years
(o) The principal that yields a simple interest of
annum for 8 months is:
A.
10,000 B.
12,000 C.
12,800 D.
1,280 at 16% per
14,000
2. A time interval of 3 minutes and 20 seconds is wrongly measured as
3 minutes and 25 seconds. What is the percentage error?
3. Hari reads 22% of the pages of a book on the first day, 53% on the second day and 15% on the third day. If the number of pages remaining
to be read is 30, find the total number of pages in the book.
Commercial arithmetic
99
4. If 55% of students in a school are girls and the number of boys is 270,
find the number of girls in the school.
5. By selling an article for
price of the article.
920, a shop keeper gains 15%. Find the cost
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6. Amit sells a watch at 20% gain. Had he sold it for 36 more, he would
have gained 23%. Find the cost price of the watch.
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7. On selling apples at 40 per kg, a vendor incurs 10% loss. If he incurs
a total loss of 120, calculate the quantity(in kg) of apples he sold.
8. A dealer allows a discount of 20% and still gains 20%. Find the
marked price of an article which costs the dealer 720.
9. A shop keeper buys an article for 600 and marks 25% above the cost
price. Find (i) the selling price if he sells the article at 10% discount;
(ii) the percentage of discount if it is sold for 690.
10. A retailer purchases goods worth 33, 600 and gets a discount of 14%
from a whole seller. For paying in cash, the whole seller gives an
additional discount of 1.5% on the amount to be paid after the first
discount. What is the net amount the retailer has to pay?
42, 000. If the brokerage
to
11. An old car was disposed through a broker for
is 2 12 %, find the amount the owner gets.
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12. A milk-man sells 20 litres of milk everyday at 22. He receives a commission of 4% for every litre. Find the total commission he receives in
a month of 30 days.
13. A bike was sold for 48, 000 and a commission of
by the dealer. Find the rate of commission.
8, 640 was received
14. In how many years will a sum of money becomes three times at the
rate of interest 10% per annum?
15. In what time will the simple interest on a certain sum be 0.24 times
the principal at 12% per annum?
Unit 4
100
16. Find the amount of 30, 000 from 15-th January, 2010 to 10-th August, 2010 at 12% per annum.
Glossary
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2,50,000. The shop
17. A person purchases electronic items worth
keeper charges him a sales tax of 21% instead of 12%. The purchaser
does not realise that he has over paid. But after some time he finds
that he has paid excess and asks the shop-keeper to return the excess money. The shop-keeper refuses and the purchaser moves the
consumer court. The court with due hearing orders the shop-keeper
to pay the purchaser the excess money paid by the way of sales tax,
with an interest of 12% per annum. If the whole deliberation takes 8
months, what is the money that the purchaser gets back?
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Percentage: it means for every hundred; thus 6% means 6 for hundred.
Rate percent: when the percent is expressed as a fraction with denominator equal to 100, the numerator is called rate percent.
Cost price: the money paid to procure an item is called its cost price and
abbreviated as C.P.
Overhead charges: it is the additional charges on the item before it is
ready for sale, like, labour, transportation, etc.
Selling price: it is the price at which goods are sold; it is abbreviated as
S.P.
Profit: it is the gain in a transaction; it is equal to S.P- C.P. whenever S.P.
> C.P.
Loss: if S.P. is less than C.P., then C.P.-S.P. is the loss.
Discount: it is the reduction on the marked price of an item given by the
seller to attract customers.
Brokerage or commission: it is the money charged by a mediator for the
service provided for a smooth transaction of sales between a seller and a
buyer.
Simple interest: it is the money charged by a lender on the receiver for
the amount lent for a period of time.
Principal: it is the money lent for a period of time by a lender to a receiver.
Rate: it is the money charged as interest for every 100 for one year.
Commercial arithmetic
101
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Time: it is the period for which money is lent.
Amount: principal plus interest is called amount.
Sales tax: on every sales done, the Government levies a certain amount of
money; it is called sales tax; this is a certain percent on the marked price
of an item.
VAT: Value Added Tax; the sales tax goes on adding when there is multiple
sales of goods.
Manufacturer: a company or a factory at which goods originate for sales.
Wholesaler: the purchaser who buys goods in bulk, and in turn sells it to
small vendors.
Retailer: the vendor who buys goods in small quantities and sells it.
Consumer: the ultimate user of the goods.
Points to remember
• Percentage is a method of comparing quantities of the same kind.
• If SP > CP, then SP − CP is the profit; if SP < CP, then CP − SP is the
loss.
• Profit or loss is always calculated on the cost price.
• Discount is the reduction given in the marked price.
• The money that an agent (or broker) receives in a deal is the commission(or brokerage).
to
• While calculating the simple interest, the day on which money is deposited is not counted where as the day on which it is withdrawn is
taken in to account.
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• VAT means value added tax.
CHAPTER 1
UNIT 5
STATISTICS
After learning this chapter, you learn to:
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• explain the terms data, observation, range, frequency, class interval,
exclusive and inclusive class intervals size of class interval, mid-point
of class interval;
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• construct frequency distribution table for exclusive and inclusive class
intervals;
• draw histogram for the given frequency distribution;
• define mean, median and mode;
• calculate mean for grouped and un-grouped data;
• calculate median for grouped and un-grouped data;
• identify mode of un-grouped and grouped data.
1.5.1 Introduction
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Statistics is considered to be a mathematical science pertaining to the
collection, analysis, interpretation and presentation of data.
Statistics is useful in drawing conclusions from numerical data. It is also
useful to predict weather, to obtain information concerning business, import, export, education etc.. All research and investigation require statistical interpretation.
A collection of numerical facts with particular information is called data.
Consider, the marks obtained by 20 students in mathematics in 8th
standard mid-term examination:
56, 31, 44, 78, 67, 74, 38, 60, 56, 59, 87, 73, 38, 77, 84, 80, 49, 60, 60, 71.
The above data is a collection numerical entries. It is called observation.
Such a collection of data is called raw data.
The data can be arranged in ascending or descending order. Arranged in
descending order, we get,
87, 84, 80, 78, 77, 74, 73, 71, 67, 60, 60, 60, 59, 56, 56, 49, 44, 38, 38, 31.
Statistics
103
From this, we can infer that the highest score is 87 and lowest score is 31.
The difference between the highest and the lowest score is called range.
The range of above data is (87 -31) = 56.
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We can observe that the scores 38 and 60 are repeated. The number
38 is repeated twice and 60 is repeated thrice. We say that the frequency
of 38 is 2 and the frequency of 60 is 3. The frequency of rest of the scores
is 1. The number of times a particular observation (score) occurs in a data
is called its frequency. The above data may be represented in a tabular
form, showing the frequency of each distribution. This representation in
tabular form is called Frequency Distribution Table. Tallies are used to
mark the counts; III represents three(3) counts, where as I I I I represents 5
counts.
Example 1. The marks scored by 20 students in a unit test out of 25
marks are given below.
12, 10, 08, 12, 04, 15, 18, 23, 18, 16, 16, 12, 23, 18, 12, 05, 16, 16, 12, 20.
Prepare a frequency distribution table.
Solution: The table looks like:
Marks
No
t
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23
20
18
16
15
12
10
08
05
04
Total
Tally No. of students
marks
(frequency)
II
2
I
1
III
3
IIII
4
I
1
5
IIII
I
1
I
1
I
1
I
1
20
20
Unit 5
104
1.5.2 Grouping Data
Organising the data in the form of frequency distribution table is called
grouped frequency distribution of raw data. Sometimes, we have to deal
with a large data.
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Example 2. Consider the following marks (out of 50) scored in mathematics by 50 students of 8th class:
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41, 31, 33, 32, 28, 31, 21, 10, 30, 22, 33, 37, 12, 05, 08, 15, 39, 26, 41,
46, 34, 22, 09,
11, 16, 22, 25, 29, 31, 39, 23, 31, 21, 45, 47, 30, 22, 17, 36, 18, 20, 22,
44, 16, 24, 10,
27, 39, 28, 17.
Prepare a frequency distribution table.
Solution: If we prepare a frequency distribution table for each observation, then the table would be too long. So for convenience, we make groups
of observations like 0 - 9, 10 - 19 and so on. We obtain a frequency of distribution of the number of observations coming under each group. In this
way, we prepare a frequency distribution table for the above data as below:
Tally marks
III
IIII , IIII
IIII , IIII , IIII , I
IIII , IIII , IIII
IIII , I
No
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Groups
0-9
10 - 19
20 - 29
30 - 39
40 - 49
50 - 59
Total
50
Frequency
03
10
16
15
06
0
50
The data presented in this manner is said to be grouped and the distribution obtained is called grouped frequency distribution. Grouped
frequency distribution table helps us to draw meaningful inferences like:
1. most of the students have scored between 20 and 29;
2. only 3 students have scored less than 10;
3. no student has scored 50 or more than 50.
Statistics
105
In the above table, marks are grouped into 0 - 9, 10 - 19 and the like. No
score overlaps in any group. Each of these groups is called a class interval
or a class. This method of grouping data is called Inclusive Method.
Class Limit: In the class interval, say (10 - 19), 9.5 is called the lower class
limit and 19.5 is called the upper class limit.
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Note: To find the class limit, in inclusive method, subtract 0.5 from
lower score to get lower class limit and add 0.5 to the upper score to
get upper class limit.
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Class Size: The number of scores in the class interval say, (10-19), including 10 and 19, is called the class size or width of the class. In this
example, the class size is 10.
Class Mark: The midpoint of a class is called its class mark (or midpoint
of class interval). It is obtained by adding the two limits and dividing by 2.
For example, the class mark of (10-19) is (10 + 19)/2 = 14.5. The class mark
of (10-20) is (10+20)/2 = 15.
The data in Example 2 can also be grouped in class intervals like 0 10, 10 - 20, 20 - 30 and so on. The frequency distribution table will then
be as follows:
Tally marks
III
IIII , IIII
I I I I , I I I I , I I I I , III
I I I I , I I I I , III
IIII , I
50
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Groups
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
Total
Frequency
03
10
18
13
06
50
Here observe that 10 occurs in both the classes (0 - 10) as well as (10
-20). But it is not possible that an observation (say 10) can belong to two
classes (0 - 10) and (10 - 20) simultaneously. In order to avoid this, we
follow a convention that the common observation (here 10) will belong to
the higher class, that is 10 belongs to (10 - 20) and not to (0 - 10). Similarly
30 belongs to (30 - 40) and not (20 -30). This method of grouping the data
is called Exclusive method.
Unit 5
106
Class Limit: In the class interval (10 - 20), 10 is called the lower limit and
20 is called the upper limit.
Class Size: The difference between the upper limit and the lower limit is
called the class size or width. The width of class in (10 - 20) is 20 - 10 =
10.
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Example 3. Forty candidates from 10th class of a school appear for a
test. The number of questions (out of 60) attempted by them in forty five
minutes is given here.
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52, 42, 40, 36, 12, 28, 15, 37, 35, 22, 39, 50, 54, 39, 21, 34, 46, 31, 10,
09,
13, 24, 29, 31, 49, 58, 40, 44, 37, 28, 13, 16, 29, 36, 39, 41, 47, 55, 52,
09.
Prepare a frequency distribution table with the class size 10 and answer
the following:
(i) Which class has the highest frequency? (ii) Which class has the lowest
frequency?
(iii) Write the upper and lower limits of the class (20- 29).
(iv) Which two classes have the same frequency?
Solution: Let us first prepare the frequency distribution table pertaining
to this data.
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Class-Interval Tally marks Frequency
0-9
II
2
10 - 19
6
IIII , I
20 - 29
7
I I I I , II
30 - 39
11
IIII , IIII , I
40 - 49
8
I I I I , III
50 - 59
6
IIII , I
Total
40
40
Using this table, we can observe:
(i) (30-39) has the highest frequency;
(ii) (0-9) has the lowest frequency;
(iii) upper limit is 29.5 and lower limit is 19.5;
(iv) (10-19) and (50-59) have the same frequency.
Statistics
107
Example 4. The heights of 25 children in centimetre are given below:
174, 168, 110, 142, 156, 119, 110, 101, 190, 102, 190, 111, 172, 140,
136, 174, 128, 124, 136, 147, 168, 192, 101, 129, 114.
Prepare a frequency distribution table, taking the size of the class interval
as 20, and answer the following:
(i) Mention the class intervals of highest and lowest frequency.
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(ii) What does the frequency 6 corresponding to class interval (160-180)
indicate?
(iii) Find out the class mark (or midpoint) of (140-160).
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(iv) What is the range of heights?
Solution: The frequency distribution for the given data is as follows:
Class-Interval Tally marks Frequency
100 - 120
8
I I I I , III
120 - 140
5
IIII
140 - 160
III
3
160 - 180
6
IIII , I
180 - 200
III
3
Total 25
25
Answers:
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(i) Highest frequency: (100-120); lowest frequency: (140-160). and (180200).
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(ii) There are 6 children whose heights are in the range 160 cm to 180
cm.
(iii) Mid-point = (140+160)/2 = 150
(iv) Range = highest score- lowest score= 192 - 101 = 91.
Exercise 1.5.2
1. The marks scored by 40 candidates in an examination (out of 100) is
given below:
75, 65, 57, 50, 32, 54, 75, 67, 75, 88, 80, 42, 40, 41, 34, 78, 43, 61,
42, 46, 68, 52, 43, 49, 59, 49, 67, 34, 33, 87, 97, 47, 46, 54, 48, 45,
51, 47, 41, 43.
Unit 5
108
Prepare a frequency distribution table with the class size 10. Take
the class intervals as (30 - 39), (40 - 49), . . . and answer the following
questions:
d
(i) Which class intervals have highest and lowest frequency?
(ii) Write the upper and lower limits of the class interval 30-39.
(iii) What is the range of the given distribution?
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2. Prepare the frequency distribution table for the given set of scores:
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39, 16, 30, 37, 53, 15, 16, 60, 58, 26, 28, 19, 20, 12, 14, 24, 59,
21, 57, 38, 25, 36, 34, 15, 25, 41, 52, 45, 60, 63, 18, 26, 43, 36, 18,
27, 59, 63, 46, 48, 25, 33, 46, 27, 46, 42, 48, 35, 64, 24. Take class
intervals as (10 -20), (20 - 30), . . . and answer the following:
(i) What does the frequency corresponding to the third class interval
mean?
ii) What is the size of each class interval? Find the midpoint of the
class interval 30 - 40.
iii) What is the range of the given set of scores?
1.5.3 Histogram
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A histogram is a representation of a frequency distribution by means
of rectangles whose widths represent class intervals and whose areas are
proportional to the corresponding frequencies. In a histogram, frequency
is plotted against class interval. Thus, a histogram is a two-dimensional
graphical representation of data. However, if the length of all the class
intervals are the same, then the frequency is proportional to the height of
the rectangle.
Construction of a histogram
We will show how to construct histograms taking some examples.
Example 5. Draw the histogram of the following frequency distribution:
Statistics
109
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Class - Interval Frequency
0-9
5
10 - 19
8
20 - 29
12
30 - 39
18
40 - 49
22
50 - 59
10
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Solution: The given distribution is in inclusive form. It should be converted into exclusive form. This can be done by applying a correction
d
factor , where
2
d = (lower limit of a class) - (upper limit of a class before it)
Here, we have
d
;
2
d
actual lower limit = stated limit - .
2
to
actual upper limit = stated limit +
No
t
For example, consider the class limit 10 - 19. You get
d = lower limit of the class interval -upper limit of class before it =10-9 =1.
d
Hence, d = 1 or = 0.5. Now,
2
d
= 19 + 0.5 =19.5.
2
d
actual lower limit = (stated lower limit) - = 10 - 0.5= 9.5
2
actual upper limit = (stated upper limit) +
Converting into exclusive form, we get the table as below:
Unit 5
110
Y
22
20
d
15
10
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frequency
Actual
Frequency
Class
interval
-0.5 - 9.5
5
9.5 - 19.5
8
19.5 - 29.5
12
29.5 - 39.5
18
39.5 - 49.5
22
49.5 - 59.5
10
5
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Stated
Class
interval
0-9
10 - 19
20 - 29
30 - 39
40 - 49
50 - 59
−0.5
9.5 19.5
29.5 39.5
49.5 59.5
X
class intervals
Construction of histogram:
1. Draw x-axis and y-axis. Choose a proper scale for x and y axes, say
on x-axis: 1cm = 10 and on y-axis: 1cm = 5.
2. Mark the class intervals on x-axis; (0.5 - 9.5), (9.5 - 19.5) and the like.
to
3. Draw a rectangle of height 5cm on the first class interval (0.5 - 9.5).
4. Draw a second rectangle of height 8 cm on the second class interval
and follow the same procedure for the rest of the class intervals and
the corresponding frequencies.
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Then, the histogram takes the form shown above.
In the above histogram, we observe that,
• there are no gaps between rectangles, showing that the distribution
is continuous;
• the heights of rectangles represent the frequencies and the base represents class intervals.
Remember:
• In a bar graph, the height of the bar represents the data. The
bars may be separated or closed.
• In a histogram, the area of each rectangle represents corresponding data (frequency). There should be no gaps between rectangles.
Statistics
111
Example 6. Draw the histogram for the following frequency distribution.
Y
16
d
8
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frequency
12
4
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Class - Interval Frequency
0-5
5
5 - 10
8
10 - 15
15
15 - 20
4
20 - 25
10
5
10
15
20
25
X
class intervals
Solution: The given distribution is in exclusive form. So, we can take the
class-intervals as( 0 - 5), (5 - 10) etc., along the x-axis and frequency along
y-axis.Choosing a proper scale, we can construct a histogram as explained
in the previous example.
Note: Histogram is drawn for class intervals which are exclusive
(continuous). If the class intervals are given in inclusive (discrete), it
should be converted into exclusive form by applying the correction
factor.
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Exercise 1.5.3
No
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1. Draw a histogram to represent
the following frequency distribution.
Class - Interval Frequency
20 -25
5
25 - 30
10
30 - 35
18
35 - 40
14
40 - 45
12
2. Draw a histogram to represent
the following frequency distribution.
Class - Interval Frequency
10 - 19
7
20 - 29
10
30 - 39
20
40 - 49
5
50 - 59
15
Unit 5
112
1.5.4 Mean, Median and Mode
Now we study three important quantities associated with a statistical
data. They give a clear picture of the behaviour of an experiment. They
are generally called measures of central tendencies.
Mean
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Mean is a commonly used as a measure, in a given statistical experiment, to get an idea how the experiment is behaving. This is simply the
average of the numerical data collected during an experiment.
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Mean for an un-grouped data:
It is the sum of the numerical values of all the observations divided by
the total number of observations. If x1 , x2 , x3 , . . . , xN are the values of N
observations, then
(sum of all values of observations) x1 + x2 + x3 + · · · + xN
=
.
(the number of observations)
N
P
P
The sum of N values of x is represented by
x. Here
stands for summation notation. Therefore,
X
x
X=
N
P
Note: Sum is denoted by
and read as
sigma.
Mean is denoted by X. This is read as Xbar.
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Mean =
Example 7. Find the mean of first six even natural numbers.
Solution: The first six even natural numbers are 2, 4, 6, 8, 10, 12. There
are six scores. Therefore, N = 6. The observations are x1 = 2, x2 = 4,
x3 = 6, x4 = 8, x5 = 10, x6 = 12. Hence
X
x = 2 + 4 + 6 + 8 + 10 + 12 = 42
Hence the mean is given by
X=
X
N
x
=
42
= 7.
6
Statistics
113
d
Example 8. Marks scored by Hari in 5 tests (out of 25 marks) are given:
24, 22, 23, 23, 25. Find his average score.
P
Solution: We observe that there are 5 scores, so that
x = 24 + 22 + 23 +
23 + 25 = 117. Hence the mean is given by
X
x
117
X=
=
= 23.4.
N
5
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In the above examples, the number of values is very less. So we could
find the mean easily. If large number of values is given and we need to find
the mean, it is difficult. In these cases, where the given data is more, we
group the data and prepare a frequency distribution table. From frequency
distribution table, we can find the mean.
Mean of a grouped data:
to
Example 9: The number of goals scored by a hockey team in 20 matches
is given here:
4, 6, 3, 2, 2, 4, 1, 5, 3, 0, 4, 5, 4, 5, 4, 0, 4, 3, 6, 4.
Find the mean.
Solution: To find the mean, let us prepare a frequency distribution table
first. We observe that some scores are repeated. So to find the sum of all
scores, we have to multiply each score with its frequency and then find
the sum.
No
t
Scores Tally marks Frequency
0
II
2
1
I
1
2
II
2
3
III
3
4
7
I I I I , II
5
III
3
6
II
2
N = 20
Scores Frequency
fx
(x)
(f )
0
2
0
1
1
1
2
2
4
3
3
9
4
7
28
5
3
15
6
2
12
P
N = 20
f x = 69
To do this, let us denote the scores by x and frequency by f , then multiply
P
f and x and add the product f x. Here
f x denotes the sum of all the
Unit 5
114
products f × x. Now the mean is
sum of the scores
X=
=
number of scores
P
fx
69
= .
N
20
Thus we get X = 3.45.
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Example 10. Find the mean for the given frequency distribution table.
Solution: To find the mean, first we
have to find the mid-point of each class
interval. Mid-point of 0 - 4 = (0+4)/2 =
2; mid-point of 5 - 9 = (5+9)/2= 7 and
the like. Denote the mid- point of the
class interval by x. Write down the frequencies f corresponding to each class
interval.
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Class - Interval Frequency
0-4
3
5-9
5
10 - 14
7
15 - 19
4
20 - 24
6
N = 25
to
Class interval Mid-point of CI (x) Frequency(f )
fx
0-4
2
3
6
5-9
7
5
35
10 - 14
12
7
84
15 - 19
17
4
68
20 - 24
22
6
132
P
N = 25
f x = 325
No
t
P
Multiply f and x to get f x. Add all f x and find out
f x. Now the mean is
calculated using the formula,
X
fx
sum of all scores
X=
=
.
total number of scores
N
Thus we get
X=
X
fx
N
=
325
= 13.
25
Therefore mean is 13.
Activity 1:
Mark the height in centimetres on a wall in your school. (Take the help of
Statistics
115
your teacher). Measure and record the height of 10 of your friends. Find
the mean height.
Median
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Median is the mid-point of the data(raw scores), after being arranged in
ascending or descending order. Median divides the given set of scores into
two equal halves, that is there are as many scores below the median as
above the median.
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The mean depends on the nature of scores. If some scores are
very high(or low), that will influence the mean. For example;
consider the data 5,8, 6, 9, 12, 110,130. If you compute the
mean, it is 40(the sum is 280 and there are 7 scores). But there
are 5 scores below 40 and only two scores above 40. Hence it is
not central. On the other hand the median is 9 and you see that
it is in the centre. Thus mean is influenced by unusually high
scores in the given data and may not be a true representative of
an experiment. In such cases median is preferred.
Median for an un-grouped data:
to
Arrange the given set of scores in ascending or descending order of magnitudes (values). If the total number of scores is odd, then the middle most
score is the median. If the total number of scores is even, then the average
of the two middle most scores is the median.
Example 11. Find the median of the data: 26, 31, 33, 37, 43, 8, 26, 33.
No
t
Solution: Arranging the scores in ascending order, we have 26, 31, 33,
37, 38, 42, 43.
Here the number of terms is 7. The middle term is the 4-th one and it is
37. Therefore, median is 37.
Think it over:
Do you get the same median if the scores are arranged in descending order?
Example 12. Find the median of the data: 32, 30, 28, 31, 22, 26, 27, 21.
Unit 5
116
Solution: Arranging in descending order, we obtain 32, 31, 30, 28, 27,
26, 22, 21.
There are 8 terms. Therefore, median is the average of the two middle
(27 + 28)
= 27.5.
terms, which are 27 and 28. Thus the median is
2
d
Think it over:
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Do you get the same median if the scores are arranged in ascending order?
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Note: When N scores are given, we can use the following method to find
the median:
First arrange the scores in ascending or descending order; (i) If N is odd,
1
(N + 1)
then median is the score at
-th place; (ii) If N is even, median is
2
2
N
N
+ 1 -th place .
the score at -th place + the score at
2
2
Median for a grouped data
to
In the case of an un-grouped data, you could compute its median as the
middle score if the number of scores is odd(or the mean of its two middle
scores in the case of even number of scores). We have to adopt a different
way of computing the median of a grouped data. We describe it through
examples.
Example 13. Find the median for the following grouped data.
Solution: Median is the middle
score. Here actual scores are not
given. So we have to use a different procedure to find the median.
We have N = 20, an even number. Therefore, there are two middle scores: one corresponding to
N/2 = 20/2 = 10-th score and the
other is 11-th score.
We have to locate the median between 10-th and 11-th scores. Since we
are not given individual scores, we have to get some idea about 10-th and
No
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Class - Interval Frequency
1-5
4
6 - 10
3
11 - 15
6
16 - 20
5
21 - 25
2
N = 20
Statistics
117
Class
Frequency(f ) Cumulative
Interval
frequency(fc)
1-5
4
4
6 - 10
3
7
11 - 15
6
13
16 - 20
5
18
21 - 25
2
20
N = 20
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4
4+3=7
7+6=13
13+5=18
18+2=20
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11-th scores. For this it is required to find the cumulative frequency.
Observe the following table, to know how to find cumulative frequency.
Observe that the cumulative frequency corresponding to the last class
interval is equal to N. Counting frequencies from first class interval
downwards, we find that the 10-th score lies in the class interval (11 - 15).
This class interval (11 - 15) is called the median class. The frequency corresponding to this is 6. Its lower real limit (LRL) is 10.5. The cumulative
frequency above this class is 7. Now, knowing:
(a) lower real limit(LRL)=10.5;
(b) frequency of the median class (fm ) = 6;
(c) cumulative frequency above the median class (fc ) = 7; and
(d) size of the class interval(i) = 5;
to
we can find the median, using the formula
No
t
median = LRL+
((N/2) − fc )
× i.
fm
Thus we get
median = 10.5 +
(20/2) − 7
(10 − 7)
× 5 = 10.5 +
×5
6
6
= 10.5 +
Note: Median = LRL+
basic principles.
3
× 5 = 10.5 + 2.5 = 13.
6
(N/2 − fc )
× i. This formula can be derived from the
fm
Unit 5
118
Example 14. Calculate the median for an exclusive(continuous) distribution given below.
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Class
Frequency Cumulative
Interval
(f )
frequency (fc )
10 - 20
11
11
20 - 30
13
24
30 - 40
13
37
40 - 50
9
46
50 - 60
4
50
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Class Interval Frequency
10 - 20
11
20 - 30
13
30 - 40
13
40 - 50
9
50 - 60
4
Solution: We first prepare the cumulative frequency table(see above). Here
you see that the total number of observation is N = 50. Therefore (30 - 40)
is the median class. We also observe that LRL = 30, fc = 24, fm = 13 and
i = 20 − 10 = 10. Now we are in a position to use the formula for median:
median = LRL +
(N/2) − fc
(25 − 24)
10
× i = 30 +
× 10 = 30 +
= 30.77,
fm
13
13
approximately.
Mode
to
There is another measure of central tendency which is used occasionally, called mode. Mode is the score that occurs frequently in a given set
of scores. Mode is the value around which the other scores cluster around
densely.
No
t
Mode for an un-grouped data:
Example 15. Find the mode for the data: 15, 20, 22, 25, 30, 20, 15, 20,
12, 20.
Solution: Here 20 appear maximum times (4 times). Therefore, mode is
20.
Example 16. Find the mode of the data: 5, 3, 3, 5, 7, 6, 3, 4, 3, 5, 8, 5.
Solution: Here 3 and 5 appear 4 times. Therefore modes are both 3 and
5.
Statistics
119
Note: A collection of data can have more than one mode. If the
data has only one mode, we say it has uni-mode, if it has 2
modes, we say it has bi-mode and if it has more than 2 modes,
we say it has multi-mode.
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Mode for a grouped data
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Example 17. Find the mode for the following data.
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For a grouped data, the same score having maximum frequency is
the mode.
Number
12 13
Frequency 7
9
14 15 16 17
6 22 20 19
Solution: Here the maximum frequency is 22. Therefore, the number 15
corresponding to maximum frequency is the mode.
Hence, the mode is 15.
Exercise 1.5.4
1. Runs scored by 10 batsmen in a one day cricket match are given.
Find the average run scored.
23, 54, 08, 94, 60, 18, 29, 44, 05, 86
2. Find the mean weight from the following table.
to
Weight(kg)
29 30 31 32
No. of children 02 01 04 03
33
05
No
t
3. Calculate the mean for the following frequency distribution:
Marks
10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency
3
7
10
6
8
2
4
4. Calculate the mean for the following frequency distribution:
Marks
15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44
Frequency
6
5
9
12
6
2
5. Find the median of the data: 15, 22, 9, 20, 6, 18, 11, 25, 14.
6. Find the median of the data: 22, 28, 34, 49, 44, 57, 18, 10, 33, 41,
66, 59.
Unit 5
120
7. Find the median for the following frequency distribution table:
Class
Intervals
Frequency
110-119
120-129
130-139
140-149
150-159
160-169
6
8
15
10
6
5
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8. Find the median for the following frequency distribution table:
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9. Find the mode for the following data:
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Class-interval 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30
Frequency
5
3
9
10
8
5
(i) 4, 3, 1, 5, 3, 7, 9, 6
(ii) 22, 36, 18, 22, 20, 34, 22, 42, 46, 42
10. Find the mode for the following data:
x
f
5
4
10 12 15 20 30 40
8 11 13 16 12 9
Additional problems on “Statistics”
1. Four alternative options are given for each of the following statements.
Select the correct option.
(a) The size or width of the Class interval (0 - 4) is :
B. 5 C. 3 D. 0
to
A. 4
(b) The midpoint of class interval (10 - 19) is:
No
t
A. 10 B. 14 C. 15 D. 14.5
(c) The difference between highest and lowest score of a distribution
gives:
A. class interval B. class width C. range D. class limit
(d) The number of times a particular observation (score) occurs in a
data is called its:
A. frequency B. range C. class interval D. class limit
(e) In inclusive form, the actual upper limit and lower limit of class
interval (0 - 4) are:
A. -0.5 & 3.5 B. 0.5 & 4.5 C. −1 & 5 D. 1 & 5
Statistics
121
(f) The height of a rectangle in a histogram represents:
A. class interval B. midpoint C. frequency density D. frequency
(g) In a histogram, the width of the rectangle indicates:
d
A. class interval B. midpoint C. frequency density D. frequency
A. 15 B. 13 C. 13.4 D. 14.3
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(i) Class interval grouping of data is done when:
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(h) The mean of scores 10, 15, 12, 15, 15 is:
A. the range of data is small B. the range of data is large
C. the class intervals are small D. class intervals are large
(j) The mean of 6, 4, 7, x and 10 is 8. The value of x is:
A. 10 B. 12 C. 14 D. 13
P
(k) If n = 10 and Mean = 12, then
f x is:
A. 120 B. 1200 C. 12
D. 13
(l) The mean of first three multiples of 5 is :
A. 5
B. 10 C. 15
D. 30
(m) The median of 37, 83, 70, 29, 32, 42, 40 is:
A. 29 B. 30 C. 40 D. 42
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(n) In an inclusive class interval (10 - 14), the lower real limit is:
A. 9.5 B. 10.5 C. 13.5 D. 14.5
No
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(o) In an exclusive class interval (10 - 20), the lower real limit is:
A. 20 B. 10 C. 10.5 D. 20.5
(p) The mode of 2, 3, 3, 5, 3, 5, 7, 3, 5 is:
A. 3
B. 5 C. 3 and 5
D. 3,5,7
(q) For given two values of x, 16, 18 the frequencies are respectively
12 and 20. Then the mode is:
A. 16 B. 18 C. 12 D. 20
(r) A collection of data having more than 3 modes is said to be:
A. uni-mode B. bi-mode
C. tri-mode D. multi-mode
Unit 5
122
2. Prepare a frequency distribution table for the scores given:
42,22,55,18,50,10,33,29,17,29,29,27,34,15,40,42,40,41,35, 27,
44,31,38,19,54,55,38,19,20,30,42,59,15,19,27,23,40,32,28,51.
d
Take the class intervals as 10-20, 20-30, 30-40, 40-50, 50-60. From
the frequency distribution table answer the following questions:
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(i) What does the frequency corresponding to the class interval 2030 indicate?
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(ii) In which class intervals are the scores 10,20 and 30 included?
(iii) Find the range of the scores.
3. The following are the marks scored in a unit test(out of 25). Prepare
a frequency distribution table, taking the class intervals as 0-4, 5-9,
10-14, 15-19, 20-24:
21,14,3,7,23,18,24,16,18,17,20,10,17,18,21,23,19,12,14,9,16,18,12, 14,11.
From the table (i) find the mid-points of each class interval; (ii) find
the class interval having maximum frequency; (iii) find the range of
the scores.
No
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to
4. Draw histogram for the following frequency distribution:
Class Interval Frequency
5-15
2
15-25
8
25-35
14
35-45
14
45-55
12
5. Draw histogram for the following frequency distribution:
Class Interval Frequency
0-10
4
11-20
18
21-30
12
31-40
6
41-50
20
51-60
10
6. The marks obtained by 12 students in a mathematics examination
are given below.
48,78,93,90,66,54,83,58,60,75,89,84.
Find (i) the mean of the marks; (ii) the mean mark of the students if
each student is given 4 grace marks.
Statistics
123
7. If the mean of 8,12,21,42,x is 20, find the value of x.
8. Find the mean for the following distribution:
12,14,10,12,15,12,18,10,15,11,19,20,12,15,19,10,18,16,20,17.
Glossary
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Data: collection of numerical facts with a particular information during
an experiment.
Observation: numerical display of data.
Score: the numerical entries in an observation.
Range: the difference between the highest and the lowest scores in an
observation.
Frequency: the number of times a particular score appears in an observation.
GFD: Grouped frequency distribution; the data is collected in to several
groups and the frequency of scores in each group is recorded.
Class interval: in a grouped frequency distribution, each group is called
a class interval.
Cumulative frequency: the sum of the frequencies up to the current class
interval.
FDT: Frequency Distribution Table; it displays the frequencies of scores
corresponding to various class intervals.
Inclusive method: while grouping, the end points of the groups do not
overlap.
Exclusive method: the end points of consecutive groups overlap.
Class limit: the end points of a class in exclusive method; the end points
of a class with correction factors in inclusive method.
Class size(or width): the difference between the upper class limit and the
lower class limit.
Class mark: the mid point of the class interval; it is equal to the average
between the upper class limit and the lower class limit.
Histogram: a graphical way of representing grouped data using rectangular bars in which the frequency is proportional to the area of the rectangle.
Mean: average of the scores; It is equal to the sum of the scores divided
by the number of scores.
Answers
124
d
Median: the middle score.
Mode: the score which appears maximum number of times in an observation.
Median class: in a grouped data, it is difficult to say which is the median
as scores are not given explicitly; but one can say in which class interval,
the middle score lies. This is called the median class.
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Points to remember
• Statistics is a branch of science which helps us to analyse the collected data in a systematic way.
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• Mean, median and mode are three different measures which, in some
sense, represent the given data. They are called measures of central
tendency.
Answers to Exercises in Chapter 1.
Exercise 1.1.2
to
1. (i) (3 × 10) + (9 × 1); (ii) (5 × 10) + (2 × 1); (iii) (1 × 100) + (6 × 1);
(iv) (3 × 100) + (5 × 10) + (9 × 1);
(v) (6 × 100) + (2 × 10) + (8 × 1);
(vi)
(3 × 1000) + (4 × 100) + (5 × 10) + (8 × 1); (vii) (9 × 1000) + (5 × 100) + (2 × 1);
(viii) (7 × 1000).
2. (i) 56; (ii) 758; (iii) 6058; (iv) 7006; (v) 1010.
3. 555 = (4210)5 = (4 × 125) + (2 × 25) + (1 × 5) + (0 × 1).
4. 1024 =
(10000000000)2 = (1 × 210 ).
Exercise 1.1.3
No
t
1. (i) B = 4; (ii) A = 5, B = 4; (iii) A = 5; (iv) A = 0; (v) two solutions:
A = 0, B = 0 and A = 1, B = 2; (vi) A = 6, B = 1. 2. A = 3, B = 4, C = 5.
3. A = 1, B = 7, C = 9. 4. Not possible. (Hint: AA = 11 × A.)
Exercise 1.1.4
1. If s −→ (q, r) denotes the quotient q and remainder r, when s is divided
by 13, then:
8 −→ (0, 8); 31 −→ (2, 5); 44 −→ (3, 5); 85 −→ (6, 7); 1220 −→ (93, 11).
2. If s −→ (q, r) denotes the quotient q and remainder r, when s is divided
by 304, then
128 −→ (0, 128); 636 −→ (2, 28); 785 −→ (2, 177); 1038 −→ (3, 126); 2236 −→
Answers
125
(7, 108); 8858 −→ (29, 42); 13765 −→ (45, 85); 58876 −→ (304, 204).
3. 107.
4. 62.
5. 46.
Exercise 1.1.5
16 9 14
11 13 15
12 17 10
Magic sum is
27.
Central
number is 9.
We have 27 =
3 × 9.
4.
Magic sum is
39.
Central
number is 13.
We have 39 =
3 × 13.
5.
2 9
7 5
6 1
4
3
8
This is
square.
different
different
you have
only one magic
You can construct
magic squares for
positions of 1, as
seen earlier.
48
10
12
24
36
16
28
40
42
4
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12 5 10
7
9 11
8 13 6
3.
d
2.
he
1.
15 1 11
5
9 13
7 17 3
34
46
8
20
22
2
14
26
38
50
30
32
44
6
18
Exercise 1.1.6
No
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to
1. 250 numbers. 3. The only numbers in the required form and divisible
by 11 are 4939, 4037, 4136, 4235, 4334, 4433, 4532, 4631, 4730. Hence
a + b = 18(in the case of 4939), or a + b = 7.
5. There are 4 numbers:
58476, 48576, 57684, 67584. (Use divisibility by 4 and 11 together.)
Additional problems on “Playing with numbers”
1. (a) A; (b) B; (c) C; (d) C; (e) A; (f) D; (g) D. 2. 24365. 3. 12.
4. 2, 5, 7, 8. 5. 1 = 3+3−5, 2 = 5−3, 3 = 5+5+5−3−3−3−3−3+3,
4 = 3+3+3−5, 5 = 5+5+5+5−3−3−3−3−3, 6 = 5+5+5+3+3−3−3−3−3−3,
7 = 5 + 5 − 3, 8 = 5 + 3, 9 = 5 + 5 + 5 + 3 + 3 + 3 − 3 − 3 − 3 − 3 − 3,
10 = 5 + 5 + 5 + 5 + 5 − 3 − 3 − 3 − 3 − 3. 6. 10, 20, 30, 40, 50, 60, 70, 80,
90, 12, 18, 21, 24, 27, 36, 42, 45, 54, 63, 72, 81, 84 7. 108
pages. 8. 70.
Answers
126
10. Only such number is 108.
11. x = 4,
y = 2 or x = 0, y = 6 or x = 9, y = 6.
12. Take
one group as {1, 2, 3, 5, 8, 7} and the other group as
{4, 6, 10}. Then (1×2×3×5×8×7)/(4×6×10) = 7 and
this is the minimum quotient. 13. 27314925
and 27364975. 15. 8066. 16. 12. 17. 62.
9.
8 9
3 7
10 5
4
11
6
2. 1, 36 = 62 , 49 = 72 , 81 = 92 ,
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1. (i) 42 = 16; (ii) 82 = 64; (iii) 152 = 225.
169 = 132 , 625 = 252 , 900 = 302 , 100 = 102 .
d
Exercise 1.2.2
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3. 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324, 361,
400,441,484. 4. You can take 200,201,204,205,206,209. None of these
is a square as they lie between 196 = 142 and 225 = 152 .
5. 100,121,144,169,196,225,256,289,324,361,400.
Exercise 1.2.3
1. 1 + 3 + 5 + · · · + 51 = 262 = 676. 2. 144 = 122 = 1 + 3 + 5 + · · · + 23.
3. 105 and 120. Their sum is 225 = 152 .
4. 0, 1 or 4.
Exercise 1.2.4
1. (i) 961; (ii) 5184; (iii) 1369;
(iii) 27225. 3. 2155024.
(iv) 27556. 2. (i) 7225;
(ii) 13225;
Exercise 1.2.5
No
t
to
1. (i) 14; (ii) 16; (iii) 102; (iv) 34; (v) 115.
2. (i) 16; (ii) 37;
(iii) 81; (iv) 10; (v) 4; (vi) 247.
3. 56 m. 4. (i) 7; (ii) 2; (iii)
5; (iv) 2.
5. (i) 16; (ii) 16; (iii) 729; (iv) 676.
6. You can take
16 as factor of 48 and 240 as a multiple of 48; 16 + 240 = 256 = 162 . You
can generate infinitely many by taking l = m(3m + 2), m = 1, 2, 3, . . .; 16 as a
factor of 48 and 48l as a multiple of 48. Observe
16 + 48l = 16 + 48m(3m + 2) = 16 + 96m + 144m2 = (4 + 12m)2 .
Exercise 1.2.6
1. (i) 15;
(ii) 24;
(iii) 27;
(iv) 29;
(v) 42.
2. 127 m.
3. 5.
Answers
127
Exercise 1.2.7
1.
4
43 = 64
2.
13
1
33
27
−5
(−5)3 = −125
53
73
93 23
125 343 729 8
43
64
6
8
63 = 216
83 = 512
63
83
103
216 512 1000
−9
(−9)3 = −729
d
3
33 = 27
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2
23 = 8
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The cube of an odd number is odd and the cube of an even number is
even.
3. There are 4 perfect cubes from 1 to 100; there are 9 perfect cubes from
−100 to 100(recall 0 is also a perfect cube). 4. There are 7 perfect cubes
from 1 to 500. Of these 64 is the only perfect square; 64 = 43 = 82 .
5. The number of zeros at the end is always a multiple of 3.
6. Each
digit occurs at the end of some cube. Hence one cannot conclude that
some number is not a cube by looking at the last digit(compare this with
perfect squares).
Exercise 1.2.8
1. (i) 22; (ii) 36;
36; (iii) 72.
(iii) 25.
2. (i) 45;
(ii) 55;
(iii) 89.
3. (i) 69;
(ii)
to
Additional problems on “Squares square roots,
cubes and cube roots”
No
t
2. (a) C; (b) B; (c) C; (d) A; (e) C. 3. 11. 4. 6,5,1,6,9,6,6,4,9,1,4,4,9.
5. None of them is a perfect square. 6. 81. 7. (i) z = ±5; (ii) y = ±12;
(iii) x = ±8. 8. 48. 9. (i) closed; (ii) commutative; (iii) associative;
(iv) no, because m2 + k 2 = m2 implies k = 0 and N does not contain 0.
11. 344 m.
12. If 1010 = a2 − b2 for some integers a and b, then either
both a and b are odd or both even. Hence a2 − b2 is divisible by 4. But
1010 is not divisible by 4. Hence 1010 is not the difference of two perfect
squares. 13. 0,1,6. 14. 1156 = 342 . 15. 4 and 16.
Exercise 1.3.1
1. (i) associative property of addition in Z; (ii) commutative property of
multiplication in Z; (iii) distributivity of multiplication over addition in
Answers
128
Z.
2. −6, −9, −123, 76, 85, −1000.
3. (i) m = 2; (ii) m = −10; (iii)
m = 14; m = −77. 4. −333, −210, −26, −8, 0, 21, 33, 85, 2011. 5. 2011,
528, 364, 210, 85, 12, −58, −1024, −10000.
Exercise 1.3.2
10 15 20 25 30 35 40 45 50 55
,
,
,
,
,
,
,
,
,
.
14 21 28 35 42 49 56 63 70 77
22 33 44 55 66 77 88 99 110 121 132 143 154 165 176
2.Take ,
,
,
,
,
,
,
,
,
,
,
,
,
,
.
10 15 20 25 30 35 40 45 50 55 60 65 70 75 80
10 9 8 7 6 5 4 3 2 1
3.
, , , , , , , , ,
.
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
,
,
.
4. , , , , , , ,
3 4 5 6 7 8 9 10 11 12
3
−3
5. The number
is the same as
, as these are equivalent fractions;
−2
2
recall that the denominator is always positive as a convention.
4
9
8
= .
6. 0.9 = ; 0.8 =
10
10
5
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1. You can take
Exercise 1.3.3
1. (i) closure property of addition; closure property of multiplication; (iii)
0 is the additive
identity;
(iv) 1 is the multiplicative identity.
−8 −6
−3
3 16 4
1 11 −15 18
,
, . 6. ,
,
,
, 1000.
5.
=
, ,
5 10
5
8 3 1
2 6
8
19
to
Exercise 1.3.4
−13
−102
−4
7
3
101
22
15
<
<
<
< <
<
< .
7
81
5
12
4
100
19
11
13 14 15 16 17
3. We can take
, , , , . If you increase both the numerator and
30 30 30 30 30
denominator and use the property of equivalent fractions, you can get any
number of such collections.
1 1 1 1 1 1 1 1 2 2 2 3 3 3 4
4. There are 15 such rational numbers: , , , , , , , , , , , , , , .
2 3 4 5 6 7 8 9 3 5 7 4 5 7 5
m
p
m+p
m
p
m
p
5. If
and are distinct, then
lies between
and . If
= ,
n
q
n+q
n
q
n
q
m+p
m
p
then
=
= . 6. 79. 7. 69.
n+q
n
q
No
t
2.
Additional problems on “Rational numbers”
1. (a) natural numbers; (b) 0; (c) 2; (d) 9; (e) even; (f) odd.
2. (a) false;
Answers
129
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(b) false; (c) true; (d) false; (e) false; (f) true; (g) false.
3. (i) 297; (ii)
39. 4. m/(m + 1) < (m + 1)/(m + 2); do not forget two cases m < −2 and
m > −1. 5. (i) yes; (ii) yes; (iii) yes; (iv) a ∗ 1 = 1; (e) a = 2, b = 2.
13 17 23 −11 −100
10
5
20
25
10
40
6.
,
,
,
,
. 7.
< <
<
<
< .
8 12 26 13
101
13
6
23
28
11
43
21
13
31
9
41
51
8.
>
>
> >
> . 9. (a) 0; (b) 1; (c) 1, −1.
17
11
27
8
37
47
413
225
11825
46
11 (i)
; (ii)
; (iii)
; (iv)
.
13. Every non-zero rational
225
108
6
504
5
number is invertible, but only ±1 are invertible integers. 14. .
3
1
15. .
16. ±1.
17. 140.
18. No rational p/q between 0 and 1 for
2
which q < p. 19. n = 0, −1, −3, −4. 20. 4 values: 26, 46, 50, 70.
23. n = 2, 3, 5, 0, −1, −3.
Exercise 1.4.2
1. chess: 270, carom: 540, other games: 90
4. 2,400. 5. 9,800.
2.
2,000.
3. 25%.
Exercise 1.4.3
1. 10% profit. 2.
3,168. 3.
920. 4. Profit
120 5.
600 6. 20%.
Exercise 1.4.4
1. 12%. 2. 625. 3.
4. 26%. 5. (i) 425.
1904(marked price) and
(ii) 340.
336(discount).
1.
4.
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Exercise 1.4.5
700(commission); Sindhu gets
70,000.
27,300. 2.
1,350. 3. 2.5%.
No
t
Exercise 1.4.6
1.
625.
2.
39.55
3. 3 12 years
4.
18,000.
5. 12 12 %.
Exercise 1.4.7
1.(i) 260; (ii)
3. 7.33% gain.
2,838;
(iii)
1,236.
(iv)
212.
2.
30,245
Additional problems on “Commercial Arithmetic”
1. (a) A.; (b) C.; (c) B.; (d) B. ; (e) C.; (f) A.; (g) C.;
(i) B.; (j) B.; (k) A.; (l) B.; (m) C.; (n) A.; (o) B..
2. 2 12 .
3. 325.
4. 330.
5.
800.
6.
1,200.
(h) C.;
7. 27 kg.
Answers
130
8.
1,080.
9. (i) 675; (ii) 8% .
10 28,462.56.
11. 41,050.
12. 528. 13. 18%. 14. 20 years. 15. 2 years. 16. 32,041.65.
17. 24,300.
Exercise 1.5.2
1.
2.
Class
Interval
10-20
20-30
30-40
40-50
50-60
60-70
4
16
Tally
Frequency
marks
9
I I I I , IIII
12
I I I I , I I I I , II
10
IIII , IIII
9
I I I I , IIII
6
IIII , I
IIII
4
d
Frequency
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Tally
marks
IIII
IIII , IIII ,
IIII , I
I I I I , II
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Class
Interval
30-39
40-49
50-59
60-69
70-79
80-89
90-99
IIII
IIII
III
I
7
5
4
3
1
(i) highest: 40 - 49, lowest: 90 - 99
(ii) 30.5 and 39.5 (iii) 65.
(i) 10 scores between 30 and 40.
ii) 10, 35. (iii) 52.
Exercise 1.5.3
1.
2.
40
40
35
35
30
30
to
25
20
15
10
25
20
15
No
t
10
5
20
25
30
35 40
5
45
9.5
19.5 29.5 39.5 49.5
59.5
Exercise 1.5.4
1. 42.1. 2. 29.73. 3. 42.75. 4. 22.63. 5. 15. 6. 37.5.
7. 136.8.
8. 16.7.
9. (i) 3; (ii) 22 and 42(bi-mode distribution).
10. 20.
Additional problems on “Statistics”
1. (a) B. (b) D. (c) B. (d) A. (e) A. (f) D. (g) A. (h) C.
(k) A. (l) B. (m) C. (n) A. (o) B. (p) A. (q) D. (r) B.
(i) B.
(j) D.
Answers
131
2.
Frequency
(i) The largest number of
scores lie in the class interval 20-30.
8
10
8
8
6
40
d
(ii) 10, 20, 30 are included respectively in the class intervals 10-20, 20-30, 30-40.
(iii) The range of the scores is
59-10=49.
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10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
Total
Tally
marks
I I I I , III
IIII , IIII
I I I I , III
I I I I , III
IIII , I
40
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Class-Interval
3.
Class-Interval
0-4
5-9
10 - 14
15 - 19
20 - 24
Total
to
4.
Tally
Frequency
marks
I
1
II
2
7
I I I I , II
9
I I I I , IIII
6
IIII , I
25
25
No
t
20
14
15
10
14
(i) Mid-points of 0-4, 5-9, 1014, 15-19, 20-24 are respectively 2,7,12,17,22.
(ii) The class interval having
maximum frequency is 1519.
(iii) The range of the scores is
24-3=21.
5.
20
20
18
15
12
8
5
12
10
10
6
5
4
2
5
15 25 35 45 55
0 10 20
6. (i) 73.17(approximately); (i) 77.17(approximately).
30 40 50 60
7. 17.
8. 14.75.
CHAPTER 2
UNIT 1
ALGEBRAIC EXPRESSIONS
After studying this unit, you learn
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• addition and subtraction of polynomials;
d
• the meaning and types of polynomials;
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• multiplication of polynomials: monomials by monomials, binomial by
monomials, binomial by binomial (x + a)(x + b), (a + b)2 , (a − b)2 and
(a + b)(a − b) etc.
2.1.1 Introduction
Let us first review some of the things you have learnt earlier.
A symbol which has a fixed value is called a constant.
√
√
Examples: 5, -7, 2 35 , 5, 2 + 3 etc.
A symbol which does not have any fixed value, but may be assigned value
(values) according to the requirement is called a variable or a literal.
Examples: p, q, x, y, z etc.
Note:
to
1. Combination of a constant and a variable is a variable.
Examples: 3x, (4 + p), 6/x, x/7, x − 4, 9x etc.
2. Combination of two or more variables is either a variable or a constant.
No
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Examples: xy, x/y, (x − y), (y − x), −x, (x + y), xyz, xy/z , 13 + x − y,
14x − y, 10 − xy, 7x/y, 8/xy etc. Observe (4 + x) + (4 − x) = 8 which is a
constant.
A term is a number (constant), a variable, or a combination of (product or
quotient of) numbers and variables.
Examples: 9, x, 3x, 4xy, 7x/15y, 21/xy, yz/x etc.
A single term or a combination of two or more terms connected by additive
(both addition and subtraction) and multiplicative(both multiplication and
division) symbols form an algebraic expression.
Examples: 7 − y, 3x2 − 4y, 6xy, 6 + x2 − 3x, (7x/9) + 4y − 6z etc.
Algebraic Expressions
133
Note: The signs of multiplication and division do not separate terms. For
example: 9x2 4y or 4x2 /7y are single terms.
An algebraic expression which contains only one term is called a monomial.
Examples: 4, 5/ 11, x, 6x, 8xy, 7x2 /y, yz/x, 5x2 y/7z etc.
d
An algebraic expression which contains two terms is called a binomial.
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Examples: 7 + x, xy − 7, (5x/y) − 3x, 3x2 − 6xy, yz 2 + 2z.
An algebraic expression which contains three terms is called a trinomial.
Examples: 4 + x + y, 6x + 15 − y 2 , ax2 + bx + c, ax + by + 2.
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An algebraic expression in which each term contains only the variable(s)
with non negative integral exponent(s) is called a polynomial.
Examples: x2 − 4x, x − 4xy + y 2 , 6 − 5y + xy + x2 y, 4.
x
Note: + 2 is not a polynomial, it is only an algebraic expression.
y
Exercise 2.1.1
1. Separate the constants and variables from the following:
√
√
3, (2/3)xy, (5xy)/2, 7, 7 − x, 6x + 4y,
12 + z, 15, −x/5, −3/7, x,
−7z, (8yz)/3x, y + 4, y/4 and (2x)/(8yz).
2. Separate the monomials, binomials and trinomials from the following:
7xyz, 9 − 4y, 4y 2 − xz, x − 2y + 3z, 7x + z 2 , 8x/y, (8/5)x2 y 2 , 4 + 5y − 6z.
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2.1.2 Algebraic expressions
No
t
Consider 9x. It contains two factors: 9 and x. We say 9 is the numerical
coefficient(or constant or arithmetical coefficient);x is called as variable(or literal factor or literal coefficient). Or consider 9xy. Here there
are two variables x and y. We say 9x is the coefficient of y and 9y is the
coefficient of x. Observe the following tabular columns:
Product
−8xy
Coefficient
Numerical
Literal
Coefficient Coefficient
(constant) (variable)
−8x is the coefficient of y
−8
x
−8y is the coefficient of x
−8
y
xy is the coefficient of −8
–
xy
−8 is the coefficient of xy
−8
–
Unit 1
134
Like and unlike terms
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Note: 1. If the literal factor (variable) has no sign, it is taken as
positive.
2. If the variable has no power, it is taken as 1.
3. If the variable has no numerical coefficient, it is taken as 1.
For example, x means +1x.
4. The highest power of the variable in a polynomial with single
variable is called the degree of the polynomial.
Examples:
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Terms having the same variable with same exponents are called like terms.
5x, 2x, 7x, −9x, (1/3)x etc.;
1
x2 , 2x2 , 6x2 , 9x2 , x2 etc.;
7
1
3
3
3
3
x , 3x , 7x , −9x , x3 etc.
9
Terms having the same variable with different exponents or different
variables with same/different exponents are called unlike terms.
Examples: x, x2 , x3 , x4 , x5 etc.; x, m, n, p etc.; −x, xy, xy 2 etc.
2.1.3 Addition and Subtraction of polynomials
to
Let us first review properties of addition and multiplication in the set of
all integers:
No
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1. Sum of two positive integers is a positive integer; (+7)+(+5) = +7+5 =
+12.
2. Sum of two negative integers is a negative integer; (−7) + (−5) = −7 −
5 = −12.
3. Sum of a positive integer and a negative integer is positive if the absolute value of the negative integer is smaller than the positive integer;
(+7) + (−5) = +7 − 5 = +2.
4. Sum of a positive integer and a negative integer is negative if the absolute value of the negative integer is larger than the positive integer;
(−7) + (+5) = −7 + 5 = −2
5. Product of two positive integers is also a positive integer; (+7) × (+5) =
+35.
Algebraic Expressions
135
6. Product of two negative integers is a positive integer; (−7) ×(−5) = +35
7. Product of a positive integer and a negative integer is a negative integer; (+7) × (−5) = −35.
d
8. Product of a negative integer and a positive integer is a negative integer; (−7) × (+5) = −35.
1. like terms can be added or subtracted;
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2. unlike terms cannot be added or subtracted;
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Make the following rules for addition and subtraction of two polynomials:
3. while adding or subtracting like terms, their numerical coefficients
are added or subtracted;
Example 1. Add 5x2 y, −7x2 y and 9x2 y.
Solution: Recall rule 1. We have
(5x2 y) + (−7x2 y) + (9x2 y) =
=
5 + (−7) + 9 x2 y
5 − 7 + 9 x2 y
= 7x2 y.
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This is horizontal addition. We can also have vertical addition:
+5 x2 y
−7 x2 y
+9 x2 y
7 x2 y
You may observe that we are adding the coefficients and retaining the
variable term as it is.
Example 2. Add: 7x2 − 4x + 5 and 9x − 10.
Solution: Here there are unlike
terms. We can add only like terms.
We write like terms one below other
to facilitate easy addition.
7x2
7x2
−4x
+9x
+5x
+5
−10
−5
Example 3. Add 8xy + 4yz − 7zx, 6yz + 11zx − 6y and −5xz + 6x − 2yx.
Unit 1
136
Solution: Here again there are
many unlike terms. We write
like terms one below other to
facilitate easy addition.
We
are also using the commutative property: xy = yx and xz =
zx.
+8xy
+10yz
−7zx
+11zx
−5xz
−xz
−6y
+6x
+6x −6y
d
−2xy
6xy
+4yz
+6yz
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Example 4. Subtract 2x3 − x2 + 4x − 6 from x3 + 5x2 − 4x + 6.
Solution: We write like terms
+1x3 +5x2 −4x +6 → Minuend
one below other to facilitate
+2x3 −x2
+4x −6 → Subtrahend
easy subtraction. Note that we
(−2) (+1) (−4) (+6)
are subtracting relevant coeffi−1x3 +6x2 −8x +12
cients. Subtracting a negative
number is equivalent to adding the negative of that number. Hence we
have changed the signs of coefficients in the subtrahend and added the
coefficients. You can also do this in a quick way, once you understand the
meaning of subtraction, as follows:
x3 + 5x2 − 4x + 6) − 2x3 − x2 + 4x − 6
= x3 + 5x2 − 4x + 6 − 2x3 + x2 − 4x + 6
No
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to
= (1 − 2)x3 + (5 + 1)x2 + (−4 − 4)x + (6 + 6)
= −1x3 + 6x2 − 8x + 12
= −x3 + 6x2 − 8x + 12.
Exercise 2.1.3
1. Classify into like terms:
1
4x2 , x, −8x3 , xy, 6x3 , 4y, −74x3 , 8xy, 7xyz, 3x2 .
3
2. Simplify:
(i) 7x − 9y + 3 − 3x − 5y + 8;
(ii) 3x2 + 5xy − 4y 2 + x2 − 8xy − 5y 2 .
3. Add:
(i) 5a + 3b, a − 2b and 3a + 5b;
and 3x2 y + 9xy 2 .
(ii) x3 − x2 y + 5xy 2 + y 3 , −x3 − 9xy 2 + y 3 ,
Algebraic Expressions
137
4. Subtract:
(i) −2x2 y + 3xy 2 from 8x2 y;
(ii) a − b − 2c from 4a + 6b − 2c.
2.1.4 Multiplication of Polynomials
Observe the following products: (i) 5x × 6x2 = (5 × 6) × (x × x2 ) = 30x3 ;
(ii) 2x × 6y × 8z = (2x × 6y) × (8z)
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= (2 × 6) × (x × y) × (8z) = 12xy × (8z)
= 12 × 8 × (xy × z) = 96xyz.
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We can also write this in one step: 2x×6y×8z = (2×6×8)×(x×y×z) = 96xyz.
Note: coefficient of the product= the product of the coefficients of
expressions;
algebraic factor of the product= the product of all the algebraic factors.
Example 5. Find the product of 6x and −7x2 y.
Solution: We have (6x) × (−7x2 y) = 6 × (−7) × (x × x2 y) = (−42)x3 y.
to
Note: We are using x × x2 y = (x × x2 )y = x3 y. In other words, we
multiply similar variables and use the law of indices for simplifying
the expression: xm × xn = xm+n for all integers m, n, which you will
study later.
No
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Multiplying a monomial by a monomial
Example 6. Find the product 4x × 5y × 7z.
Solution: We have 4x × 5y × 7z = (4 × 5 × 7) × (x × y × z) = 140xyz.
Example 7. What is the product 2l2 m × 3lm2 ?
Solution: We have 2l2 m × 3lm2 = (2 × 3) × (l2 × l) × (m × m2 ) = 6l3 m3 . We are
using the law of indices: ak × al = ak+l .
Multiplying a monomial by a binomial
Consider the product 9 × 103 = 927. We may also write this in the form
9 × 103 = 9 × (100 + 3) = (9 × 100) + (9 × 3) = 900 + 27 = 927.
Unit 1
138
Do you see that we have used the distributive property of multiplication
over addition? We adopt the same strategy when a binomial is involved.
Thus
2x(3x + 5xy) = (2x) × (3x) + (2x) × (5xy) = 6x2 + 10x2 y.
Example 8. Determine the product (8y + 3) × 4x.
= (4x × 8y) + (4x) × 3
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= 32xy + 12x.
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(8y + 3) × (4x) = (4x) × (8y + 3)
d
Solution: We have
Here we have used the commutativity of the product and right distributivity. We can also use left distributive law and get
(8y + 3) × (4x) = (8y) × (4x) + 3 × (4x)
= 32yx + 12x
= 32xy + 12x,
because xy = yx, the commutative property of the product.
Important points:
No
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We are using all the properties of numbers to the algebraic variables; associativity, commutativity, law of indices, distributive
property. This is because, when we substitute numbers for these
variables, all these properties are true. As an extrapolation, we
expect them to hold for variables as well.
Multiplying a binomial by a binomial
Consider the product of two binomials (4a + 6b) and (5a + 7b). We have
(4a + 6b)(5a + 7b) = 4a(5a + 7b) + 6b(5a + 7b)
= (4a)(5a) + (4a)(7b) + (6b)(5a) + (6b)(7b)
= 20a2 + 28ab + 30ab + 42b2
= 20a2 + 42b2 + 58ab.
Algebraic Expressions
139
Exercise 2.1.4
1. Complete the following table of products of two monomials:
9x2 y
−11x3 y 2
d
−8xy
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4x2
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First → 3x −6y
Second ↓
3x
−6y
4x2
−8xy
9x2 y
−11x3 y 2
2. Find the products:
(i) (5x+ 8)3;
(ii) (−3pq)(−15p3 q 2 −q 3 );
(iii)
6x 3 3
(a −b );
5
(iv) −x(x−15).
3. Simplify the following:
(i) (2x2 y−xy)(3xy−5); (ii) (3x2 y 2 +1)(4xy−6xy 2);
(iv) (2m3 + 3m)(5m − 1).
(iii) (3x2 +2x)(2x2 +3);
2.1.5 Special product
to
Now we study a special product, a product of two binomials. Consider the
product:
No
t
(x + a)(x + b) = x(x + b) + a(x + b) = x2 + xb + ax + ab
= x2 + ax + bx + ab
= x2 + (a + b)x + ab.
We have used commutative property and the distributive property: xb = bx,
(ax + bx) = (a + b)x. We say (x + a)(x + b) = x2 + (a + b)x + ab is an identity.
Thus
(x + 2)(x + 1) = x2 + (2 + 1)x + (2 × 1) = x2 + 3x + 2.
We can also represent the the identity (x + 2)(x + 1) = x2 + 3x + 2, pictorially.
Unit 1
140
1
x
1
1
1 1
1
K
L
x
J
x
(x+1)
x
1
C
1
1
1
G
x
x
x
x
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x2
A
H
d
I
x
he
D
You observe that the
area of the rectangle is
(x + 2)(x + 1). The rectangle is split in to smaller
squares and rectangles:
squares AEKJ, KLHI
and LGCH having areas x2 , 1 and 1 respectively; rectangles EF LK,
F BGL, JKID having areas x, x, x respectively.
Thus the area of ABCD
is x2 + 1 + 1 + x + x + x =
x2 + 3x + 2. We thus obtain
x
E 1
F
1
B
(x+2)
(x + 2)(x + 1) = x2 + 3x + 2.
What is the value of (x + a)(x + b)? If we replace x by y what
happens to the identity: (x + a)(x + b) = x2 + (a + b)x + ab? Can you
see that you again get an identity?
Example 9. Find the product (x + 6)(x + 7).
to
Solution: We observe that (x + 6)(x + 7) = x2 + (6 + 7)x + (6 × 7) = x2 + 13x + 42.
Example 10. Determine the product (x + 8)(x − 4)
No
t
Solution: Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, we get
(x + 8)(x − 4) = x2 + (8 − 4)x + 8 × (−4) = x2 + 4x − 32.
Example 11. Compute (2x + 5)(2x + 3).
Solution: We know that (x + a)(x + b) = x2 + (a + b)x + ab. Thus
(2x + 5)(2x + 3) = (2x)2 + (5 + 3)(2x) + (5 × 3) = 4x2 + 16x + 15,
Example 12. Find the product 103 × 98 using the above identity.
Solution: We observe that
Algebraic Expressions
141
103 × 98 = (100 + 3)(100 − 2)
= (100)2 + (3) + (−2) 100 + 3 × (−2)
= 10000 + (1 × 100) + (−6) = 10094,
Example 13. Find the product (p2 − 5)(p2 − 3).
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Solution: We have
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d
where we have taken x = 100, a = 3 and b = −2 in the expansion: (x + a)(x +
b) = x2 + (a + b)x + ab.
(p2 − 5)(p2 − 3) = (p2 )2 + (−5) + (−3) (p2 ) + (−5) × (−3) = p4 − 8p2 + 15.
Identities
An identity is an equality which is true for every value of the variable in
it.
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For example (x + 3)(x + 2) = x2 + 5x + 6 is an identity. If you give any value
for the variable x, you see that the left side and right side coincide.
No
t
We have some special identities, which are helpful in solving problems.
Consider
(a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ba + b2
= a2 + ab + ab + b2
= a2 + b2 + 2ab.
Observe that we have used the commutative law: ab = ba. We can have a
pictorial proof of this identity using geometrical squares.
Unit 1
142
G
a
D
a
C
ab
a
K
b
F
ab
E
a
b
b
B
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A
b²
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H
Observe that the square
ABCD has area equal to
(a + b)2 , since its side-length
is a + b. Now we divide the
square in to two smaller
squares and two rectangles;
squares HKGD having area
a2 and the square EBF K
having area b2 ; the rectangle
KF CG having area ab and the
rectangle AEKH having area
ab. Thus the area of ABCD is
a2 + b2 + ab + ab = a2 + b2 + 2ab.
d
(a+b)
a²
b
(a+b)
Similarly we obtain
(a − b)2 = a2 + b2 − 2ab.
And its pictorial proof is as follows.
(a−b)
G
(a−b)²
b
C
b(a−b)
to
(a−b)
(a−b)
D
K
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H
b
F
b(a−b)
A
(a−b)
a
b²
E
b
b
B
a
Consider a square ABCD with
side-length a so that its area
is equal to a2 . Now we divide
the square in to two smaller
squares and two rectangles;
squares HKGD having area
(a − b)2 and the square EBF K
having area b2 ; the rectangle KF CG having area b(a − b)
and the rectangle AEKH having area b(a−b). (We assume a is
greater than b.) Now the area of
HKGD is obtained by subtracting the areas of EBKF , KF CG
and AEKH from that of ABCD.
Hence we get
(a − b)2 = a2 − b2 − b(a − b) − b(a − b) = a2 − b2 − ba + b2 − ba + b2
= a2 − 2ab + b2 .
Algebraic Expressions
We also have
143
(a + b)(a − b) = a2 − ab + ba − b2 = a2 − b2 .
These are called standard identities.
Example 14. Find (2x + 3y)2.
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Solution: We use the identity: (a + b)2 = a2 + 2ab + b2 . Taking a = 2x and
b = 3y, we get
Example 15. What is the expansion of (4p − 3q)2 ?
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(2x + 3y)2 = (2x)2 + 2(2x)(3y) + (3y)2 = 4x2 + 12xy + 9y 2.
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Solution: Here we use (a − b)2 = a2 − 2ab + b2 . Taking a = 4p and b = 3q, we
obtain
(4p − 3q)2 = (4p)2 − 2(4p)(3q) + (3q)2 = 16p2 − 24pq + 9q 2 .
Example 16. Compute (4.9)2 .
Solution: We can use identities for this problem. Observe
(4.9)2 = (5 − 0.1)2 = 52 − 2(5)(0.1) + (0.1)2 = 25 − 1 + 0.01 = 24.01.
You may verify this by directly computing (4.9)2 .
Example 17. Compute 54 × 46.
Solution: Here again, identities are useful. We make use of (a + b)(a − b) =
a2 − b2 . Taking a = 50 and b = 4, we see that
Activity 1:
to
54 × 46 = (50 + 4)(50 − 4) = (50)2 − (4)2 = 2500 − 16 = 2484.
No
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Represent the identity (a + b)(a − b) = a2 − b2 pictorially as has been done in
the case of other identities.
Exercise 2.1.5
1. Find the product:
(i) (a+ 3)(a+ 5),
(ii) (3t+ 1)(3t+ 4);
(iii) (a−8)(a+ 2);
(iv) (a−6)(a−2).
2. Evaluate:
(i) 53 × 55;
(ii)102 × 106;
(iii) 34 × 36;
(iv) 103 × 96.
3. Can you imitate the case of the identity (x + a)(x + b) = x2 + (a + b)x + ab,
and get a similar expression for the product (x + a)(x + b)(x + c)?
Unit 1
144
4. Using the identity (a + b)2 = a2 + 2ab + b2 , simplify the following:
(i) (a + 6)2 ,
(ii) (3x + 2y)2 ;
(iii) (2p + 3q)2 ;
(iv) (x2 + 5)2 .
5. Evaluate using the identity (a + b)2 = a2 + 2ab + b2 :
(i) (34)2 ;
(ii) (10.2)2 ;
(iii) (53)2 ;
(iv) (41)2 .
6. Use the identity (a − b)2 = a2 − 2ab + b2 to compute:
(iv) (p2 − q 2 )2 .
(iii) (5a − 4b)2 ;
7. Evaluate using the identity (a − b)2 = a2 − 2ab + b2 :
(i) (49)2 ;
(ii) (9.8)2 ;
(iii) (59)2 ;
(iv) (198)2 .
d
(ii) (3x − 5y)2;
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(i) (x − 6)2 ;
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8. Use the identity (a + b)(a − b) = a2 − b2 to find the products:
(i)
(x + 6)(x
− 6); (ii) (3x + 5)(3x − 5);
2x
2x
+1
−1 .
3
3
9. Evaluate these using identity:
(i) 55 × 45;
(ii) 33 × 27;
10. Find the product:
(iii) 8.5 × 9.5;
(iii) (2a + 4b)(2a − 4b);
(iv)
(iv) 102 × 98.
(i) (x−3)(x+3)(x2 +9); (ii) (2a+3)(2a−3)(4a2 +9); (iii) (p+2)(p−2)(p2 +4);
1
1
1
1
1 2 1
(iv)
m−
m+
m +
; (v) (2x−y)(2x+y)(4x2 +y 2); (vi)
2
3
2
3
4
9
(2x − 3y)(2x + 3y)(4x2 + 9y 2).
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Additional problems on “Algebraic expressions”
1. Choose the correct answer.
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(a) Terms having the same literal factors with same exponents are
called
A. exponents B. like terms C. factors D. unlike terms
(b) The coefficient of ab in 2ab is:
A. ab B. 2 C. 2a D. 2b
(c) The exponential form of a × a × a is:
A. 3a
B. 3 + a
C. a3
D. 3 − a
(d) Sum of two negative integers is:
A. negative B. positive C. zero D. infinite
Algebraic Expressions
145
(e) What should be added to a2 + 2ab to make it a complete square?
A. b2
B. 2ab C. ab D. 2a
(f) What is the product (x + 2)(x − 3)?
A. 2x − 6
B. 3x − 2 C. x2 − x − 6 D. x2 − 6x
(g) The value of (7.2)2 is( use an identity to expand):
d
A. 49.4 B. 14.4 C. 51.84 D. 49.04
D. 4x2 + 9y 2 + 12xy
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C. 2x2 + 3y 2 − 6xy
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(h) The expansion of (2x − 3y)2 is:
A. 2x2 + 3y 2 + 6xy
B. 4x2 + 9y 2 − 12xy
(i) The product 58 × 62 is(use an identity):
A. 4596 B. 2596 C. 3596 D. 6596
2. Take away 8x − 7y − 8p + 10q from 10x + 10y − 7p + 9q.
3. Expand:
(i) (4x + 3)2 ;
(ii) (x + 2y)2 ;
(iii) (x + 1/x)2 ;
4. Expand:
(i) (2t + 5)(2t − 5); (ii) (xy + 8)(xy − 8);
(iv) (x − 1/x)2 .
(iii) (2x + 3y)(2x − 3y).
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5. Expand:
(i) (n − 1)(n + 1)(n2 + 1); (ii) (n − 1/n)(n + 1/n)(n2 + 1/n2 );
(iii) (x − 1)(x + 1)(x2 + 1)(x4 + 1); (iv) (2x − y)(2x + y)(4x2 + y 2 ).
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6. Use appropriate formulae and compute:
(i) (103)2 ; (ii) (96)2 ; (iii) 107 × 93; (iv) 1008 × 992;
7. If x + y = 7 and xy = 12, find x2 + y 2 .
8. If x + y = 12 and xy = 32, find x2 + y 2 .
9. If 4x2 + y 2 = 40 and xy = 6, find 2x + y.
10. If x − y = 3 and xy = 10, find x2 + y 2 .
11. If x + (1/x) = 3, find x2 + (1/x2 ) and x3 + (1/x3 ).
12. If x + (1/x) = 6, find x2 + (1/x2 ) and x4 + (1/x4 ).
(v) 1852 − 1152 .
Unit 1
146
13. Simplify: (i) (x + y)2 + (x − y)2;
(ii) (x + y)2 × (x − y)2.
14. Express the following as difference of two squares:
(ii) 4(x + 2y)(2x + y); (iii) (x + 98)(x + 102);
a+b 2
a−b 2
505 × 495. (Hint. ab =
−
.)
2
2
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15. If a = 3x − 5y, b = 6x + 3y and c = 2y − 4x, find
(i) a + b − c; (ii) 2a − 3b + 4c.
(iv)
d
(i) (x + 2z)(2x + z);
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16. The perimeter of a triangle is 15x2 − 23x + 9 and two of its sides are
5x2 + 8x − 1 and 6x2 − 9x + 4. Find the third side.
17. The two adjacent sides of a rectangle are 2x2 −5xy+3z 2 and 4xy−x2 −z 2 .
Find its perimeter.
18. The base and the altitude of a triangle are (3x − 4y) and (6x + 5y)
respectively. Find its area.
19. The sides of a rectangle are 2x + 3y and 3x + 2y. From this a square of
side length x+y is removed. What is the area of the remaining region?
Glossary
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20. If a, b, c are rational numbers such that a2 + b2 + c2 − ab − bc − ca = 0,
prove that a = b = c.
No
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Constant: any symbol which has a fixed value.
Variable: a symbol which can be given any value as desired.
Algebraic expression: a combination of constants and variables connected by algebraic operations.
Polynomial: an algebraic expression in which the variables have nonnegative integral powers.
Term: a part of an algebraic expression which does not involve addition
and subtraction, but may be connected by multiplication and division.
Coefficient: the companion term of a variable.
Monomial: an algebraic expression containing only one term.
Binomial: an algebraic expression containing two terms.
Trinomial: an algebraic expression containing three terms.
Algebraic Expressions
147
Degree: the largest power of a variable in a polynomial(if it has more than
one variable, then one has to take the sum of the powers of variables in
each term and take the maximum of all these sums).
Identity: equality of two algebraic expressions which is valid for all the
values of the variables in it.
d
Points to remember
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• An expression involving variables and constants combined using the
algebraic operations, namely addition, multiplication, subtraction and
division is an algebraic expression.
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• While adding two algebraic expressions, we add only like terms.
• While multiplying two expressions, we multiply term-by-term and use
the laws of exponents to simplify it.
• In a polynomial, the powers of variable(s) in it are non-negative integers.
• An identity is an equality of two algebraic expressions which is valid
for all values of the variable(s) in it.
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All the religions of the world, while they may differ in other respects,
unitedly proclaim that nothing lives in this world but Truth.
——Gandhiji
CHAPTER 2
UNIT 2
FACTORISATION
After studying this unit, you learn to:
• factorise an expression by grouping appropriate terms;
d
• factorise an algebraic expression by taking out common factors;
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• factorise a trinomial expression;
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• factorise an expression which is a difference of two square expressions;
• factorise a square trinomial expression by using known identities.
2.2.1 Introduction
The process of writing a given algebraic expression as a product of
two or more expressions is called factorisation. Each of the expression(constant or variable) which form the product is called a factor of
the given algebraic expression.
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For example: (i) Consider 7xy. It has factors: 7, x, y, 7x, 7y, xy and 7xy;
(ii) (x + 3) and (x + 2) are factors of x2 + 5x + 6.
Recall the process of factorisation of a number. Given an integer, we
write this as a product of other integers. This ultimately leads to prime
factorisation of a number. In some sense, algebraic expressions also follow
similar way. Given an expression, you will be able to write the expression
as a product of its factors.
Note: (1) Factorisation and multiplication are reverse processes. In
multiplication, we multiply different expressions and get a new expression. In factorisation, we split the given expression in to simpler
expressions whose product turns out to be the given expression.
(2) You can always use 1 as a factor: (x + 5) = 1 × (x + 5). But this
does not give any thing new. This is called the trivial factorisation.
Many times one has to consider such factorisations as well.
Factorisation
149
2.2.2 Different methods of factorisation
There are many ways of factorising a given expression. We study them
here.
1. Factorisation taking common factors
Consider the following examples:
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Example 1: 5x2 − 10x
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Here you see that 5x is the HCF of 5x2 and 10x. We can remove them from
both the terms. Thus we get
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5x2 − 10x = (5x)(x) − (5x)2 = (5x)(x − 2).
Example 2: 4a + 12b = 4(a + 3b).
Example 3: 3x2 y − 6xy 2 + 9xy = 3xy(x − 2y + 3).
Example 4: a3 − a2 + a = a(a2 − a + 1).
You see that we look at the HCF of all the terms in the expression and
take out the HCF to get a factorisation.
2. Factorisation by grouping
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We follow several steps in this process:
Step I: Arrange the terms of the given expression in suitable groups
such that each group has a common factor;
Step II: Factorise each group;
Step III: Take out the factor which is common to each group.
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Example 5: Factorise ax − bx + ay − by.
Solution: We group them as
(ax − bx) + (ay − by) = (a − b)x + (a − b)y = (a − b)(x + y).
Do you see that the distributive law is used in a reverse way? We can also
do this as follows:
ax − bx + ay − by = (ax + ay) − (bx + by)
= a(x + y) − b(x + y)
= (a − b)(x + y).
Unit 2
150
Example 6: Get the factors of y 3 − 3y 2 + 2y − 6 − xy + 3x.
Solution: Here again we group the terms in the expression and factorise:
y 3 − 3y 2 + 2y − 6 − xy + 3x = (y 3 − 3y 2) + (2y − 6) − (xy − 3x)
= y 2 (y − 3) + 2(y − 3) − x(y − 3) = (y 2 + 2 − x)(y − 3).
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3. Factorisation of difference of two squares
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We know from earlier unit that (a + b)(a − b) = a2 − b2 for all a, b. This
leads to a nice factorisation when the given expression can be written as
difference of two squares.
Example 7: Factorise 36a2 − 49b2 .
Solution: Observe that 36a2 = (6a)2 and 49b2 = (7b)2 . Thus we get
36a2 − 49b2 = (6a)2 − (7b)2 = (6a + 7b)(6a − 7b).
9
x2
Example 8: Factorise 2 − .
16
y
Solution: Here again, we write
2 2 9
x
x2
3
x 3
x 3
−
=
+
−
−
=
.
y
4
y 4
y 4
y 2 16
Example 9: Compute (4.5)2 − (1.5)2 .
to
Solution: We have
(4.5)2 − (1.5)2 = (4.5 + 1.5)(4.5 − 1.5) = 6 × 3 = 18.
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Exercise 2.2.2
1. Resolve in to factors:
(i) x2 + xy;
2. Factorise:
(ii) 3x2 − 6x;
(iii) (1.6)a2 − (0.8)a;
(i) a2 + ax + ab + bx; (ii) 3ac + 7bc − 3ad − 7bd;
(iv) y 3 − 3y 2 + 2y − 6 − xy + 3x.
3. Factorise:
(i) 4a2 − 25;
9
(ii) x2 − ;
(v) (0.7)2 − (0.3)2 ;
(iii) x4 − y 4 ;
(iv) 5 − 10m − 20n.
(iii) 3xy − 6zy − 3xt + 6zt;
2 2
3
1
(iv) 7 10 − 2 10 ;
16
(vi) (5a − 2b)2 − (2a − b)2 .
Factorisation
151
2.2.3 Factorisation of trinomials
We have seen earlier how to multiply two binomials of the from (x + a)
and (x + b):
(x + a)(x + b) = x2 + (a + b)x + ab.
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We can also proceed in the reverse direction. Given the trinomial of the
form x2 +(a+b)x+ab, we can factorise this to get x2 +(a+b)x+ab = (x+a)(x+b).
But generally, the trinomial is not given in this form. You may be given in
the form x2 + mx + n, where m, n are some numbers. You must be able to
write m = (a+b) and n = a×b to bring the given trinomial in to a factorisable
form. This needs some properties of numbers. We study them here.
The sum and product of two numbers are positive if and
only if both the numbers are positive.
This says that if a + b and ab are positive then so are a, b. The converse
is also true. Thus 6 and 5 are positive; 5 = 3 + 2 and 6 = 3 × 2; both 3 and
2 are positive.
The sum of two numbers is negative and their product positive if and only if both the numbers are negative.
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Thus a+b negative and ab positive if and only if both a and b are negative.
If we are given numbers 21 and −10, we see that −10 = (−7) + (−3) and
21 = (−7)(−3).
We say 7 is the absolute value of both 7 and −7. Thus given an integer
a, we define its absolute value by |a| = a if a > 0; |a| = −a if a < 0; and
|a| = 0 if a = 0. Observe −8 < −6 but | − 8| = 8 > 6 = | − 6|.
The sum of two numbers is positive and their product negative if and only if one of the numbers is positive and the
other negative, and the positive number has larger absolute
value than the negative number.
This means a + b is positive and ab negative only if one of a, b is positive
and the other negative; and if a is positive and b is negative, then |a| > |b|;
if a is negative and b positive, then |a| < |b|. For example, we see that if
a + b = 7 and ab = −18, then a = 9 and b = −2 or a = −2 and b = 9.
Unit 2
152
The sum of two numbers is negative and their product negative if and only if one of the numbers is positive and the
other negative, and the positive number has smaller absolute value than the negative number.
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Thus a + b is negative and ab negative if and only if one of a, b is positive
and the other negative; if a is positive and b is negative, then |a| < |b|; if a
is negative and b positive, then |a| > |b|. For example, if a + b = −12 and
ab = −28, we can write a = 2 and b = −14 or a = −14 and b = 2.
Here we have not mentioned any thing about the nature of numbers.
They can be integers, rational numbers or even real numbers, which you
will study in your next class.
Example 10. Factorise 6x2 + 11x + 3.
Solution: Here you can adopt the standard method of splitting and grouping:
6x2 + 11x + 3 = 6x2 + 9x + 2x + 3 = (6x2 + 9x) + (2x + 3)
= 3x(2x + 3) + 1(2x + 3) = (3x + 1)(2x + 3).
The genuine problem is how to arrive at this splitting. Suppose we are
looking for a factorisation of the form 6x2 + 11x + 3 = (ax + b)(cx + d). This
gives, after expansion,
to
6x2 + 11x + 3 = acx2 + (ad + bc)x + bd.
No
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After comparing terms of different degrees,
ac = 6,
ad + bc = 11,
bd = 3.
Thus acbd = 18 or (ad)(bc) = 18; and ad + bc = 11. You have two numbers
whose product is 18 and their sum is 11. You will immediately conclude
that ad = 9, bc = 2 or ad = 2, bc = 9. Thus you may write
6x2 + 11x + 3 = acx2 + (ad + bc)x + bd = 6x2 + 9x + 2x + 3,
which is the splitting we have used. Observe we can also use the other
pair of values for (ad, bc); ad = 2, bc = 9. We get
6x2 + 11x + 3 = (6x2 + 2x) + (9x + 3) = 2x(3x + 1) + 3(3x + 1) = (2x + 3)(3x + 1),
Factorisation
153
which is the same as the original factorisation.
Rule: If you want to factorise a trinomial of the form x2 + px + q, you
must be able to find numbers a and b such that a · b = q and a + b = p.
Then x2 + px + q = (x + a)(x + b).
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You may notice here that a polynomial of degree 2 is written as
a product of two polynomials, each of degree 1, in the process
of factorisation. This helps us in understanding the structure
of a polynomial of degree 2. You will see in later classes that
this is an easy way of solving a polynomial equation, if you are
able to factorise the polynomial.
Example 11. Factorise x2 − 9x + 20.
Solution: Again you must be able to find two numbers a and b such that
ab = 1 × 20(the product of the coefficients of x2 term and the constant term)
and a+b = −9. Here the product is positive and the sum is negative. Hence
both the numbers must be negative. This can be achieved by taking a = −5
and b = −4. Thus
x2 − 9x + 20 = (x2 − 5x) + (−4x + 20) = x(x − 5) − 4(x − 5) = (x − 4)(x − 5).
Factorising a square trinomial
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Any algebraic expression, which can be written either in the form a2 +
2ab + b2 or in the form a2 − 2ab + b2 , is called a square trinomial. For
example, x2 + 2x + 1 is a square trinomial. We can have an immediate
factorisation for such a trinomial using a2 + 2ab + b2 = (a + b)(a + b) or
a2 − 2ab + b2 = (a − b)(a − b).
Example 12. Factorise 4x2 + 12xy + 9y 2 .
Solution: We observe that
4x2 + 12xy + 9y 2 = (2x)2 + 2(2x)(3y) + (3y)2 = (2x + 3y)2.
Thus its factors are two equal expressions: 2x + 3y and 2x + 3y.
Example 13. Is x2 − 6xy + 36y 2 a square trinomial?
We observe that
x2 − 6xy + 36y 2 = (x)2 − (x)(6y) + (6y)2 ,
Unit 2
154
which is not in the standard form,. Thus x2 − 6xy + 36y 2 is not a square
trinomial.
d
Think it over!
Is it possible to factorise x2 + 1? Alternatively, is it possible to find
two numbers whose sum is zero and whose product is one?
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Exercise 2.2.3
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1. In the following, you are given the product pq and the sum p + q.
Determine p and q:
(i) pq = 18 and p + q = 11; (ii) pq = 32 and p + q = −12; (iii) pq = −24
and p + q = 2; (iv) pq = −12 and p + q = 11; (v) pq = −6 and p + q = −5;
(vi) pq = −44 and p + q = −7.
2. Factorise:
(i) x2 + 6x + 8; (ii) x2 + 4x + 3;
a2 − 3a − 40; (vi) x2 − x − 72.
(iii) a2 + 5a + 6;
(iv) a2 − 5a + 6;
(v)
3. Factorise:
(i) x2 + 14x + 49; (ii) 4x2 + 4x + 1; (iii) a2 − 10a + 25;
(v) p2 − 24p + 144; (vi) x3 − 12x2 + 36x.
(iv) 2x2 − 24x + 72;
Additional problems on “Factorisation”
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1. Choose the correct answer:
(a) 4a + 12b is equal to
B. 12b C. 4(a + 3b)
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A. 4a
D. 3a
(b) The product of two numbers is positive and their sum negative
only when
A. both are positive
B. both are negative
C. one positive the other negative
D. one of them equal to zero
(c) Factorising x2 + 6x + 8, we get
A. (x+1)(x+8) B. (x+6)(x+2) C. (x+10)(x−2) D. (x+4)(x+2)
Factorisation
155
(d) The denominator of an algebraic fraction should not be
A. 1
B. 0 C. 4 D. 7
(e) If the sum of two integers is −2 and their product is −24, the
numbers are
A. 0.4 B. 0.04 C. 0.49 D. 0.56
2. Factorise the following:
d
(f) The difference (0.7)2 − (0.3)2 simplifies to
D. 6 and −4
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A. 6 and 4 B. −6 and 4 C. −6 and −4
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(i) x2 + 6x + 9; (ii) 1 − 8x + 16x2 ; (iii) 4x2 − 81y 2; (iv) 4a2 + 4ab + b2 ; (v)
a2 b2 + c2 d2 − a2 c2 − b2 d2 .
3. Factorise the following:
(i) x2 + 7x + 12; (ii) x2 + x − 12; (iii) x2 − 3x − 18; (iv) x2 + 4x − 21; (v)
x2 − 4x − 192; (vi) x4 − 5x2 + 4; (vii) x4 − 13x2 y 2 + 36y 4.
4. Factorise the following:
(i) 2x2 + 7x + 6; (ii) 3x2 − 17x + 20; (iii) 6x2 − 5x − 14; (iv) 4x2 + 12xy + 5y 2;
(v) 4x4 − 5x2 + 1.
5. Factorise the following:
(i) x8 − y 8 ; (ii) a12 x4 − a4 x12 ; (iii) x4 + x2 + 1; (iv) x4 + 5x2 + 9.
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6. Factorise x4 + 4y 4 . Use this to prove that 20114 + 64 is a composite
number.
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Glossary
Common factor: given two or more expressions, the factor of each expression which is common to all the expressions.
Factorisation: the process of writing an algebraic expression as a product
of more than one algebraic expressions.
Points to remember:
• Factorisation is the reverse process to the formation of the products;
• One can factorise some expressions using proper grouping and splitting of its terms.
CHAPTER 2
UNIT 3
LINEAR EQUATIONS IN ONE VARIABLE
After studying this unit, you learn:
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• to solve a linear equation in one variable;
d
• the meaning of linear equation in one variable;
• to formulate a linear equation from a verbal problem;
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• to verify the solution.
2.3.1 Introduction
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In this chapter, we study linear equations in one variable with rational
co-efficients and solve them in rational number system.
Equality of two different algebraic expression is called an equation. The
standard form of an equation is a statement that an algebraic expression
is equal to 0. The statement need not be true for any value of the variables
in it; or may be true only for certain values of the variables in it. For
example: consider 3x − 5 = 0. If we are looking for integers x for which
this statement is true, you see that no integer satisfies it. On the other
hand, if you look for a rational number, then x = 5/3 makes the statement
3x − 5 = 0 true. The value of the variable which makes the statement true
is called a solution of the equation.
Some times you may come across statements like 2x − 5 = x + 6. This is
also an equation, but this is not in the standard form. However, this can
be brought to standard form; x − 11 = 0.
Given an equation, it may not have any solution in some system and
may have solution in some other system, as you have noticed earlier. The
equation x2 − 2 = 0 has no solution in the set of all rational numbers, but
it can be solved in the set of real numbers. In fact one of the motivations
to introduce new number systems is that we should be able to solve such
equations. You will learn later that the equation x2 + 1 = 0 also cannot be
solved in the set of all real numbers, but it can be solved in an enlarged
number system called the set of all complex numbers. Thus it is very
important to mention where you are seeking the solution.
Linear equations
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An equation in one variable is a statement that an algebraic expression
in one variable is equal to 0. If the degree of the expression is also 1, we
say that the equation is a linear equation. The standard form of a linear
equation in one variable is ax + b = 0, where a 6= 0. Here x is the variable
and a, b are constants(some numbers). For example: x − 9 = 0; 5x − 30 = 0;
3
1
x + = 0.
5
3
What do we mean by solving an equation?. The statement that some
expression equals to 0 may not be valid for every value of the variable in
it. However, some value(s) of the variable may make the statement true.
In this case, that value of the variable which makes the statement true is
called a solution of the given equation.
The process of finding solution or solutions of a given equation is called
solving the equation.
As you have observed, given a linear equation ax+b = 0, the existence of
solution depends on the number system where you are seeking solution.
If a and b are integers, then there may not be any integer solution unless a
divides b. However, if you are in the system of rational numbers, you can
always get a rational number as a solution. This is the advantage we had
from moving from integers to rational numbers. We are ready to exploit it.
With this back-ground we can now make the statement:
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A linear equation with rational coefficients, ax+b = 0, a 6= 0,
has a unique solution in the rational number system.
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In fact the solution can be written as x = −ba−1 , where a−1 is the multiplicative inverse of a. You have seen earlier, while studying rational numbers, that any non-zero rational number has its multiplicative inverse in
the system of rational numbers.
You will see later that this statement is also true in the real number
system, provided you look for solution in the real number system. (In fact
this is true in any number system in which every nonzero number has
multiplicative inverse.)
2.3.2 Solving a linear equation in one variable
Consider the equation 5x − 15 = 0. We want to find the value of x for
Unit 3
158
which 5x − 15 = 0 holds. Here we make use of important axioms. You will
again come across them in Geometry.
1. If equals are added to equals, we again get
equals.
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If we know a = b, then for any c, we get a + c = b + c. For example x − 5 = 0
is given. We add equal number 5 both sides. What the axioms says is that
equality remains intact. Thus (x − 5) + 5 = 0 + 5 which is same as x = 5. The
important point to note is that here a, b, c can be algebraic expressions as
well. Thus 3x + 2 = 5 − x implies that (3x + 2) + (x − 5) = (5 − x) + (x − 5) = 0
or 4x − 3 = 0. This axiom and others are important tools in manipulations
of expressions.
2. If equals are subtracted from equals, we again get equals.
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Thus a = b implies that a − c = b − c. Given x + 5 = 2x − 6, we can subtract
x − 6 from both sides and get (x + 5) − (x − 6) = (2x − 6) − (x − 6) or 11 = x.
Can you see that there is not much difference between 1 and 2 once
you consider subtraction as addition of additive inverse. Just like you
define additive inverse to numbers, you can do this for polynomials as
well and all the properties which are true for integers are true again for
polynomials.
Example 1. Solve the equation x − 15 = 0.
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Solution: We add 15 both sides and invoke axiom 1 to get (x − 15) + 15 =
0 + 15 = 15. This reduces to x = 15.
Example 2. Solve the equation x + 9 = 20.
Solution: Do you see that this is not in the standard form. We can bring
this to standard form by subtracting 20 both sides. Thus 2 gives (x + 9) −
20 = 20 − 20 = 0 or x − 11 = 0. Now this is in standard form. By adding
11 both sides, you will get x − 11 + 11 = 0 + 11 or x = 11. We can combine
both these and reduce the number of steps: subtract 9 both sides. Thus
(x + 9) − 9 = 20 − 9 or x = 11.
Example 3. Solve 2x − 3 = x + 8.
Linear equations
159
Solution: Here you see that both the sides contain algebraic expression.
Suppose you subtract x − 3 from both the sides. You get
(2x − 3) − (x − 3) = (x + 8) − (x − 3) or equivalently x = 11.
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3. If equals are multiplied by equal quantities, we again get
equals.
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x
Thus if a = b, then ac = bc for any c. For example, suppose
= 1 is
2
x
given. You can multiply both the sides from 2 and get × 2 = 1 × 2 or x = 2.
2
4. If equals are divided by non-zero equal quantities, we get
equals.
a
b
This says that if a = b and c 6= 0, then = . This again is the same as 3
c
c
once you think that division is the same as multiplication by multiplicative
inverse.
x
Example 4. Solve = 9.
3
Solution: We multiply both the sides from 3 and get
x
× 3 = 9 × 3 = 27.
3
Thus x = 27.
2x
= 5.
9
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Example 5. Solve
9
. Note this is
2
same as multiplying first by 9 and then dividing by 2. We get
2x 9
9
45
× =5× = .
9
2
2
2
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Solution: We multiply both the sides by the same number
45
.
2
Example 6. Solve 15x = 120.
We obtain x =
Solution: We divide both the sides by 15 and get
15x
120
=
= 8.
15
15
This gives x = 8.
Unit 3
160
Example 7. Solve 13y = 100.
We divide both sides by 13 and get
13y
100
=
.
13
13
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This gives y = 100/13.
Another important component of solving an equation is that of verifying the solution. We have to check whether the so called solution we
obtain makes the given statement true. This is done by substituting the
value of the variable we have got in the given equation and by verifying the
truth of the statement.
Example 8. Is 2 a solution of the equation 3x − 5 = 19 ?
Solution: We substitute x = 2, in the given relation. We obtain 3x − 5 =
3(2) − 5 = 6 − 5 = 1, which is the left hand side or LHS. But the right hand
side or RHS is 19. Since 1 6= 19, we see that LHS is not equal to RHS for
x = 2. Hence x = 2 is not a solution.
Example 9. Is 7 a solution of 2x − 4 = 10 ?
Solution: Substitute x = 7 in the given relation. We get LHS = 2x − 4 =
(2 × 7) − 4 = 14 − 4 = 10 and RHS = 10. Thus we see that LHS = RHS.
Therefore, x = 7 is the solution of 2x − 4 = 10.
Example 10. Solve the equation 2x − 3 = 7.
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Solution: Adding 3 both the sides, we get 2x − 3 + 3 = 7 + 3. This gives
2x
10
2x = 10. Now dividing both the sides by 2, we get
=
. Hence x = 5.
2
2
Let us verify whether we have got a solution. Putting x = 5, we get 2x − 3 =
2(5) − 3 = 10 − 3 = 7. Thus the equation is satisfied.
We can reduce the number of steps in solving an equation. Consider
the equation 2x − 3 = 7. We can simply write this as 2x = 7 + 3. Actually you
are adding 3 both sides. But you drop that as a separate step and do the
addition mentally to write 2x−3+3 = 2x. We say, we have transposed −3 to
the other side. In fact this is a convenient way for doing fast calculations.
Again you can divide both the sides by 2 in mind and write x = 5.
We can also transpose algebraic expressions as well. For example, given
4x − 3 = 3x + 2, we transpose 3x to the left side and −3 to the write side to
get (4x − 3x) = (3 + 2); i.e., x = 5.
Linear equations
161
Example 11. Solve 5x − 12 = 10 − 6x.
Solution: By transposing −6x to the right side and −12 to the left side, we
get 5x + 6x = 10 + 12. Hence 11x = 22 or x = 2.
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Rule for transposing: when you transpose an expression
from one side to the other side of an equality, you have
to change the sign of the expression you are transposing.
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Example 12. Solve 8x − 3 = 9 − 2x.
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Thus −6x in the previous example becomes +6x when transposed from
the left side to the right side.
Solution: We transpose the variables on one side and constants on the
other side and get 8x + 2x = 9 + 3. Thus 10x = 12 or x = 12/10. This reduces
6
to x = .
5
Example 13. Solve 8x + 9 = 3(x − 1) + 7.
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Solution: The equation is 8x + 9 = 3x − 3 + 7 = 3x + 4. By transposing, we
get 8x − 3x = 4 − 9 or 5x = −5. We thus get x = −1. We can verify this:
8x + 9 = 8(−1) + 9 = −8 + 9 = 1 and 3(x − 1) + 7 = 3(−1 − 1) + 7 = 3(−2) + 7 =
−6 + 7 = 1. Thus 8x + 9 = 3(x − 1) + 7 is true for x = −1. This checks that
x = −1 is a solution.
3
7
2
Example 14. Solve x = x + .
3
8
12
Solution: Here we multiply by the LCM of the denominators: LCM of
3,8,12 is 24. (Which axiom are you using here?) Thus
2
3
7
x 24 =
x+
24,
3
8
12
and this simplifies to
16x = 9x + 14.
Hence 7x = 14 or x = 2. Check that x = 2 is indeed the solution.
Do you see the advantage of multiplying both the sides by the LCM of all
the denominators?. The new expression becomes an equation with integer
coefficients. You can easily handle this equation. Thus Axiom 3 helps
in transforming an equation with rational coefficients to an equivalent
equation with integer coefficients.
Unit 3
162
Example 15. Solve the equation
2x + 7 3x + 11
2x + 8
−
=
− 5.
5
2
3
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Solution: Here 2,3,5 appear in the denominators of various fractions.
Their LCM is 30. We multiply through out by 30. Thus
2x + 7
3x + 11
2x + 8
× 30 −
× 30 =
× 30 − (5 × 30).
5
2
3
6(2x + 7) − 15(3x + 11) = 10(2x + 8) − 150.
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This reduces to
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=⇒ 12x + 42 − 45x − 165 = 20x + 80 − 150.
Transposing appropriate quantities, we get
12x − 45x − 20x = −42 + 165 + 80 − 150 =⇒ −53x = 53.
Hence x = −1 is the solution. You may verify this by substituting in the
equation and checking whether LHS and RHS agree.
Example 16. Solve (x + 4)2 − (x − 5)2 = 9.
Solution: This apparently looks like an equation which is not linear. But
expanding this using identities, we get
x2 + 8x + 16 − x2 + 10x − 25 = 9.
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This reduces to 18x−9 = 9. After transposing, we get 18x = 18. Hence x = 1.
It is easy to verify that x = 1 is a solution.
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Think it over!
(1) You have learnt how to solve an equation of the form
ax + b = 0, where a 6= 0. You get x = (−b/a), as the unique
solution. Suppose you have two variables x and y; say an
equation ax+by +c = 0, where a 6= 0 and b 6= 0. This is again a
linear equation, but the number of variables is now 2. Can
you solve such an equation? How many solutions (x, y) are
there for such an equation? If a, b, c are integers, is it always possible to find integers x, y satisfying the equation?
(2) Can you solve (x + 1)2 = x2 + 2x + 1 ?
Linear equations
163
Exercise 2.3.2
1. Solve the following:
3
9
(ii) y − 9 = 21; (iii) 10 = z + 3; (iv)
+x= ;
11
11
3x
x
s
= 10; (viii) 1.6 =
;
(v) 10x = 30; (vi) = 4; (vii)
7
6
1.5
x
7
x
3x
(ix) 8x − 8 = 48; (x) + 1 = ; (xi) = 12; (xii)
= 15;
3
15
5
5
x
(xiii) 3(x + 6) = 24; (xiv) − 8 = 1; (xv) 3(x + 2) − 2(x − 1) = 7.
4
2. Solve the equations:
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(i) x + 3 = 11;
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(i) 5x = 3x + 24; (ii) 8t + 5 = 2t − 31; (iii) 7x − 10 = 4x + 11;
(iv) 4z + 3 = 6 + 2z; (v) 2x − 1 = 14 − x; (vi) 6x + 1 = 3(x − 1) + 7;
2x 3
x
x−3
2x
(vii)
− = + 1; (viii)
−2 =
; (ix) 3(x + 1) = 12 + 4(x − 1);
5
2
2
5
5
(x) 2x − 5 = 3(x − 5); (xi) 6(1 − 4x) + 7(2 + 5x) = 53;
m
3
2m
+ 8 = − 1; (xiv) (x − 1) = x − 3.
(xii) 3(x + 6) + 2(x + 3) = 64; (xiii)
3
2
4
2.3.3 Application of linear equations
We shall take up some practical situations leading to linear equations
in one variable.
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Example 17. Seven times a number, if increased by 11, is 81. Find the
number.
Solution: We do this in several steps.
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Step 1: First we convert the given data to an appropriate equation.
Let the number be x. Apriori, we do not know what this number is.
We formulate a linear equation involving the unknown x, using the
given data. Now seven times the number means 7x. Increasing this
by 11 leads to 7x + 11. The problem says that 7x + 11 = 81. Can you
see now that we have a linear equation in the unknown x?.
Step 2: We now have to solve the equation 7x + 11 = 81. You have
already learnt methods for solving such an equation. Transpose 11
to the other side to get 7x = 81 − 11 = 70. Divide by 7 and you get
x = 10.
Unit 3
164
Step 3: We have to check whether the number 10 satisfies the statement of our problem. Now seven times the number gives 7 × 10 = 70.
Adding 11 to this gives 81. And that is precisely the data says. Thus
x = 10 is indeed the solution.
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Example 18. The present age of Siri’s mother is three times the present
age of Siri. After 5 years, their ages add to 66 years. Find their present
ages.
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Solution: Again we go through several steps in the solution of this problem. Suppose Siri’s present age is x years. Then her mothers age is 3x
years. After 5 years, Siri’s and her mother’s respective ages would be x + 5
and 3x + 5 years. The data says that these two numbers would add up to
66. Thus we get the equation
(x + 5) + (3x + 5) = 66.
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Now it is easy to find the value of x. The equation reduces to 4x + 10 =
66. After transposing 10 to the other side, we obtain 4x = 56 or x = 14.
This means Siri’s present age is 14 years and her mother’s present age is
14 × 3 = 42 years.
Let us now verify whether these numbers match with the statement of
the problem. After 5 years, Siri’s age would be 14 + 5 = 19 years. Her
mother’s age would be 42 + 5 = 47 years. Their sum is 19 + 47 = 66 years
which completely matches with the given statement. We conclude that
Siri’s present age is 14 years and her mother’s age is 42 years.
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Example 19. The sum of three consecutive even numbers is 252. Find
them.
Solution: Let x be the least number among these three consecutive even
numbers. Then the other numbers are x + 2 and x + 4. This is because any
two consecutive even numbers differ by 2. The given condition says that
x + (x + 2) + (x + 4) = 252.
Thus we get 3x + 6 = 252. This reduces to 3x = 246. Solving this, we obtain
x = 82. Hence the numbers are 82, 82 + 2 = 84 and 82 + 4 = 86. We check
that
82 + 84 + 86 = 252,
Linear equations
165
and this verifies the validity of the solution.
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Example 20. If the perimeter of a triangle is 14 cm and the sides are x + 4,
3x + 1 and 4x + 1, find x.
Solution: You know that the
perimeter of a triangle is the sum
C
of its three sides. Since the sides
are given to be x + 4, 3x + 1 and
4x + 1, the perimeter is (x + 4) +
3 x+1
4 x +1
(3x + 1) + (4x + 1) = 8x + 6. The
given condition is 8x + 6 = 14. On
solving this, you get x = 1. Hence
the sides are 1+4 = 5, (3×1)+1 = 4
A
B
x +4
and (4 × 1) + 1 = 5 cm. You get a
triangle with sides 5, 4, 5 cm.
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One of the important things to keep in mind is that a mathematical formulation of a problem may not always be physically feasible.
Suppose you are asked to find the sides of a triangle, given that
they are x, x + 1 and x + 3, and its perimeter is 10 units. You write
x + (x + 1) + (x + 3) = 10 and solve this to get x = 2. You may conclude that the sides are 2, 2 + 1 = 3, 2 + 3 = 5. But, there is no triangle
with sides 2, 3, 5, since the sides of a triangle must satisfy the triangle
inequality; the sum of any two sides is greater than the third side.
Thus one has to check whether the solution one obtains is a physically valid solution. This is a very important part in the solution of a
given problem.
There is nothing strange in this. Mathematics is a game you play
using certain rules. As long as you play it safe confining to rules, you
will always get an end result. Whether that result is right or wrong,
no mathematical law will tell you. You have to go back to the physical
situation and see whether the solution you got is a physically correct
solution.
Example 21. Let P be a point on a line AB such that P lies between A
and B, and AP = 3P B. Given that AB = 10 cm, find the length of AP .
Unit 3
166
d
Solution: Since P lies between A
and B, we have AB = AP + P B.
A
P
B
Thus 10 = 3P B + P B = 4P B. We
5
15
can solve for P B and get P B = 10/4 = 5/2. Hence AP = 3P B = 3 × =
2
2
cm. (Here we are not putting P B = x and get an equation in x. You may
directly treat P B as a variable and get an equation involving P B.)
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Activity 1:
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In the previous problem you have taken that P lies between A and B. It
may happen that, on the line AB, P may lie to the left of A or to the right
of B. Formulate appropriate equations for both these cases and solve
them. One case gives you negative number. This is not a feasible practical
situation, as length is non-negative number.
Example 22. The sum of the digits of a two digit number is 12. If the new
number formed by reversing the digits is greater than the original number
by 54, find the original number.
Solution: Let x be the digit in units place. Then 12 − x is the digit is ten’s
place. Hence the number is
(12 − x) × 10 + x = 120 − 9x.
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The number obtained by reversing the digits is 10 × x + 12 − x = 12 + 9x. The
given condition
120 − 9x + 54 = 12 + 9x.
No
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From this we obtain 18x = 120 + 54 − 12 = 162. Hence x = 9. Therefore the
digit in units place 9. The digit is ten’s place is 12 − 9 = 3. The number is
hence 39.
Alternate solution: Let the digit in the unit’s place be x and the digit in the ten’s place
be y. Thus the number is 10y + x. We know that x + y = 12. The number obtained by
reversing the digits is 10x + y. The second condition tells that 10x + y = 10y + x + 54. This
reduces to 9(x − y) = 54 or x − y = 6.
Observe that, we have two equations x + y = 12 and x − y = 6. Adding these, you obtain
(x + y) + (x − y) = 12 + 6 = 18. Thus 2x = 18 or x = 9. Since y = 12 − x, you will also get
y = 12 − 9 = 3. This shows that the original number is 39.
We verify this. The number obtained by reversing the digits is 93. You
may easily check that 93 = 39 + 54.
Linear equations
167
Example 23. The sum of two numbers is 75 and they are in the ratio 3:2.
Find the numbers.
Solution: The numbers which are in the ratio 3:2 are 3x and 2x. We are
given
3x + 2x = 75.
Thus
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5x = 75.
Solving for x, we get x = 15. Hence the numbers are 3x = 45 and 2x = 30
We verify: 45/30 = 3/2 and 45 + 30 = 75.
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Exercise 2.3.3
1. If 4 is added to a number and the sum is multiplied by 3, the result is
30. Find the number.
2. Find three consecutive odd numbers whose sum is 219.
3. A rectangle has length which is 5 cm less than twice its breadth. If
the length is decreased by 5 cm and breadth is increased by 2 cm,
the perimeter of the resulting rectangle will be 74 cm. Find the length
and breadth of the original rectangle.
4. A number subtracted by 30 gives 14 subtracted by 3 times the number. Find the number.
to
5. Sristi’s salary is same as 4 times Azar’s salary. If together they earn
Rs 3,750 a month, find their individual salaries.
No
t
6. Prakruthi’s age is 6 times Sahil’s age. After 15 years, Prakruthi will
be 3 times as old as Sahil. Find their age.
D
C
x +40
7. In the figure, AB is a straight line. Find
x.
x +20
A
x
B
Unit 3
168
8. If 5 is subtracted from three times a number, the result is 16. Find
the number.
9. Find two numbers such that one of them exceeds the other by 9 and
their sum is 81.
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10. The length of a rectangular field is twice its breadth. If the perimeter
of the field is 288 m, find the dimensions of the field.
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11. Ahmed’s father is thrice as old as Ahmed. After 12 years, his age will
be twice that of his son. Find their present age.
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12. Sanju is 6 years older than his brother Nishu. If the sum of their ages
is 28 years, what are their present age ?
13. Viji is twice as old as his brother Deepu. If the difference of their ages
is 11 years, find their present age.
14. Mrs. Joseph is 27 years older than her daughter Bindu. After 8 years
she will be twice as old as Bindu. Find their present age.
15. After 16 years, Leena will be three times as old as she is now. Find
her present age.
Additional problems on “Linear equations in one variable”
to
1. Choose the correct answer
(a) The value of x in the equation 5x − 35 = 0 is:
B. 7 C. 8 D. 11
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A. 2
(b) If 14 is taken away from one fifth of a number, the result is 20.
The equation expressing this statement is:
A. (x/5) − 14 = 20
B. x − (14/5) = (20/5)
C. x − 14 = (20/5)
D. x + (14/5) = 20
(c) If five times a number increased by 8 is 53, the number is:
A. 12 B. 9 C. 11
D. 2
(d) The value of x in the equation 5(x − 2) = 3(x − 3) is:
A. 2
B. 1/2 C. 3/4 D. 0
Linear equations
169
(e) If the sum of two numbers is 84 and their difference is 30, the
numbers are:
A. −57 and 27
B. 57 and 27 C. 57 and −27 D. −57 and −27
d
(f) If the area of a rectangle whose length is twice its breadth is 800
m2 , then the length and breadth of the rectangle are:
A. 60 m and 20 m
B. 40 m and 20 m
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C. 80 m and 10 m
D. 100 m and 8 m
(g) If the sum of three consecutive odd numbers is 249, the numbers
are
A. 81,83,85 B. 79,81,83 C. 103,105,107 D. 95,97,99
(h) If (x + 0.7x)/2 = 0.85, the value of x is:
A. 2
B. 1 C. −1
D. 0
(i) If 2x − (3x − 4) = 3x − 5, then x equals:
A. 4/9 B. 9/4 C. 3/2 D. 2/3
2. Solve: (i) (3x + 24)/(2x + 7) = 2:
(ii)(1 − 9y)/(11 − 3y) = (5/8).
3. The sum of two numbers is 45 and their ratio is 7:8. Find the numbers.
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4. Shona’s mother is four times as old as Shona. After five years, her
mother will be three times as old as Shona(at that time). What are
their present age?
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5. The sum of three consecutive even numbers is 336. Find them.
6. Two friends A and B start a joint business with a capital 60, 000. If
A’s share is twice that of B, how much have each invested?
7. Which is the number when 40 is subtracted gives one-third of the
original number?
8. Find the number whose sixth part exceeds its eigth part by 3.
9. A house and a garden together cost 8, 40, 000. The price of the garden
5
is times the price of the house. Find the price of the house and the
7
garden.
Unit 3
170
10. Two farmers A and B together own a stock of grocery. They agree to
divide it by its value. Farmer A takes 72 bags while B takes 92 bags
and gives 8, 000 to A. What is the cost of each bag?.
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11. A father’s age is four times that of his son. After 5 years, it will be
three times that of his son. How many more years will take if father’s
age is to be twice that of his son?
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12. Find a number which when multiplied by 7 is as much above 132 as
it was originally below it.
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13. A person buys 25 pens worth 250, each of equal cost. He wants to
keep 5 pens for himself and sell the remaining to recover his money.
What should be the price of each pen?
14. The sum of the digits of a two-digit number is 12. If the new number
formed by reversing the digits is greater than the original number by
18, find the original number. Check your solution.
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15. The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks and cross each
other. The speed of one of the them is greater than that of the other
by 5 km/hr. If the distance between two trains after 2 hours of their
start is 30 km., find the speed of each train.
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16. A steamer goes down stream and covers the distance between two
ports in 4 hours while it covers the same distance up stream in 5
hours. If the speed of the steamer upstream is 2 km/hour, find the
speed of steamer in still water.
17. The numerator of the rational number is less than its denominator
by 3. If the numerator becomes three times and the denominator is
increased by 20, the new number becomes 1/8. Find the original
number.
18. The digit at the tens place of a two digit number is three times the
digit at the units’ place. If the sum of this number and the number
formed by reversing its digits is 88, find the number.
Linear equations
171
19. The altitude of a triangle is five- thirds the length of its corresponding
base. If the altitude is increased by 4 cm and the base decreased by
2 cm, the area of the triangle would remain the same. Find the base
and altitude of the triangle.
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20. One of the angles of a triangle is equal to the sum of the other two
angles. If the ratio of the other two angles of the triangle is 4:5, find
the angles of the triangle.
Glossary
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Equation: a statement that a non constant algebraic expression is equal
to 0.
Solution: given an equation, any value of the variable which makes the
statement true.
Linear equation: an equation of degree one.
Transposition: moving a part of the expression to the other side of the
equality; while moving, the sign of the part which has moved changes.
Verification: to check whether the solution obtained satisfies the equation.
Points to remember
• An equation is valid for a certain set of values of the variable(s) in it;
an identity is valid for all values of the variable(s) in it.
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• Given a problem, setting up an equation conforming to the given data
is an important step in the solution of the given problem.
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• A mathematical solution may not always be valid physically. One
has to check whether the solution obtained by a valid mathematical
procedure is also correct for the given physical situation.
CHAPTER 2
UNIT 4
EXPONENTS
• the concept of an integral power to a non-zero base;
• to write large numbers in exponential form;
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After studying this unit, you learn
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• about the various laws of exponents and their use in simplifying complicated expressions;
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• about the validity of these laws of exponents for algebraic variables.
2.4.1 Introduction
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Suppose somebody asks you: how far is the Sun from the Earth? What
is your answer? Perhaps some search in books will give you an idea how
far is the Sun from us. A ray of light travels approximately at the speed of
2,99,792 km per second. It takes roughly eight and a half minutes for a
ray of light to reach the Earth starting from the Sun. Hence the distance
from the Earth to the Sun is about 15, 29,00,000 km. Apart from the Sun,
do you know how far is the nearest star to us? Proxima Centauri is the
closest star to us and it is at a distance of 4.3 light years from us; that is,
the distance a ray of light travels in 4.3 years at the speed of 2,99,792 km
per second. This is equal to
4.3 × 365 × 24 × 60 × 60 × 299792km,
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which you can see is a very huge number.
At the other end what would be the size of an atom ? As you can expect,
it should be very small. The diameter of an atom is roughly 1/100000000000
meters. The subatomic particles have still smaller sizes. Hence there is
a need to represent either big or small numbers in a compact form. The
exponential notation is a very useful and handy notation used for this
purpose. As we shall see there is a methodical way of handling exponentials.
Consider the number 128. An easy factorisation gives
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2.
Exponents
173
Thus 128 is obtained by multiplying 2 with itself seven times. We write
this as 27 . Here 2 is called the base and 7 is called the exponent. We read
this as 2 raised to the power 7. The number 27 is called the exponential
form of 128. Similarly, we observe that
243 = 3 × 3 × 3 × 3 × 3 = 35 .
Observe!
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Here 3 is the base and 5 is the exponent. We read this as 3 raised to the
power 5.
On the other-hand, consider 72 = 2 × 2 × 2 × 3 × 3. In the case of 128, the
only prime factor is 2 and we are able to write 128 = 27 . Similarly, 343 = 35 ,
as 3 is the only prime factor of 343. However, 72 has two prime factors,
namely, 2 and 3. We see that 2 occurs 3 times and 3 occurs 2 times. We
write this in the form 72 = 23 · 32 . We read this as 2 to the power of 3 times
3 to the power of 2.
Let us consider 81 = 3×3×3×3 = 34 . We also observe that 81 = 9×9 = 92 .
Thus the same number 81 can be written in two ways: 81 = 34 = 92 .
In the first representation, 3 is the base and 4 is the exponent. In the
second representation, 9 is the base and 2 is the exponent. Thus the
same number may be represented using different bases and exponents.
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2
2
9 = 3 × 3 = 32 . Hence 92 = 32 . Thus you obtain 32 = 92 = 81 = 34 .
Can you recognise some thing?
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It is not necessary that we use only numbers. For example, if we have
an algebraic variable a, we write a × a × a × a = a4 and read this as a raised
to the power four. Many times we suppress the word raised, and read a
to the power four or simply a power four. An expression of the form ab4
is read as a times b power four.
Observe:
ab4 = a × b × b × b × b = a(b4 ), but not ab × ab × ab × ab which is
actually a4 b4
To avoid such confusions, it is always better to use braces: instead of
writing ab4 , write this as a(b4 ). If you want 529 = 23 × 23, write this as (23)2 .
Unit 4
174
The notation 232 may some time be confused with 2 × 32 which is 18. Thus
use proper braces: 2(32 ) to represent 18 and (23)2 to represent 529.
Another thing you may notice is that it is not necessary to use only
integers as bases. In fact you may use any real number for base:
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(−1.2)3 = (−1.2) × (−1.2) × (−1.2) = −1.728,
4
1
1 1 1 1
1
=
× × × =
,
5
5 5 5 5
625
√
( 2)2 = 2.
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(0.1)5 = 0.1 × 0.1 × 0.1 × 0.1 × 0.1 = 0.00001,
Think it over!
Is it possible to take for exponent a real number which is not an
integer?
In general, given a number a or an algebraic variable which we again
denote by a, and a natural number n, we define
an = |a × a × a{z× · · · × a}.
n times
This is read as a raised to the power n or simply a power n. Here, we say
a is the base and n is the exponent.
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Observe:
0n = 0 for any natural number n and a1 = a for any number a.
Moreover, am = an if and only if m = n for any number a 6= 0,
a 6= 1 or a 6= −1
Let us look at more examples:
Number
1000
m6
1
1024
625
625
Expanded
Exponential
Base and
form
form
exponent
10 × 10 × 10
103
base 10, exponent 3
6
m×m×m×m×m×m
m
base m, exponent 6
10
1 1
1
1
× × · · · (10 times)
base , exponent 10
2 2
2
2
4
5×5×5×5
5
base 5, exponent 4
4
(−5) × (−5) × (−5) × (−5)
(−5)
base −5, exponent 4.
Exponents
175
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Look at the last two examples. The number 625 is represented as 54
(with base 5 and exponent 4) and also as (−5)4 (with base −5 and exponent
4). Thus the same number has representation in base 5 and base −5.
For a number a 6= 0, and a natural number n, we define
n
1
1
−n
= n.
a =
a
a
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This extends the definition of power to negative integer exponents. Here
are some more examples:
1
1
1
1
−4
3
=
×
×
×
; (read as 3 to the power −4.)
3
3
3
3
5
1
−5
= (−10)5 = −100000; (read as (−0.1) to the power −5.)
(−0.1)
=
(−0.1)
−6 6
4
5
5 5 5 5 5 5
56
=
= × × × × × = 6.
5
4
4 4 4 4 4 4
4
If a = bn for some integer n 6= 0, 1 and number b 6= 0, we say a is in
exponential form. Here b is called the base and n is called the exponent.
Earlier you have observed that 81 = 34 = 92 . Thus the same number may
have different exponential forms.
Caution!
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bn is not defined for b = 0, and n = 0 or
n < 0.
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Example 1. Express 1024 and 0.1024 using base 2.
Solution: Consider the number 1024 = 210 . Here we have base 2 and
exponent 10. However, 1024 is a number in decimal notation: here 4 is in
unit’s place; 2 in 10’s place; 0 in 100’s place; and 1 in 1000’s place. Thus
we have
1024 = (1 × 1000) + (0 × 100) + (2 × 10) + (4 × 1)
= (1 × 103 ) + (2 × 101 ) + 4.
Suppose we define b0 = 1 for any number b 6= 0. Then we see that 1024 =
1 · 103 + 2 · 101 + 4 · 100. Do you see that we have done this in the unit playing
Unit 4
176
with numbers? Similarly, consider 0.1024. You observe that
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0.1024 = (0.1) + (0.002) + (0.0004)
1
2
4
=
+
+
10 1000 10000
1
2
4
=
+ 3+ 4
10 10
10
−1
= 1 · 10 + 2 · 10−3 + 4 · 10−4 .
Example 2. Express 1000 using base 2 and exponents.
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Solution: Consider 1000. How do you express this using base 2 and
exponents? You observe that 512 = 29 , 256 = 28 , 128 = 27 and 26 = 64. Thus
you get the sum
512 + 256 + 128 + 64 = 960.
You are still short by 40 to reach 1000. But 40 = 32 + 8 = 25 + 23 . Thus you
obtain
1000 = 29 + 28 + 27 + 26 + 25 + 23
= 1 · 29 + 1 · 28 + 1 · 27 + 1 · 26 + 1 · 25 + 0 · 24 + 1 · 23 + 0 · 22 + 0 · 21 + 0 · 20 .
You observe that this is precisely the binary representation of 1000. We
write this 1000 = (1111101000)2.
Some uses of exponents in daily life
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Some examples of how exponents do connect with our everyday lives:
when we speak about square feet, square meters or any such area units;
or about cubic feet, cubic meters, cubic centimeters or any other such
volume units.
The unit square cm is actually 1 cm × 1 cm = 1 cm2 . Similarly, a cubic
cm is 1 cm × 1 cm × 1 cm = 1 cm3 .
Another kind of indirect example is if you talk about extremely small or
extremely big quantities. For example, the term ‘nanometer’ means 10−9
m. The prefix ‘nano’ means the number 10−9 - an extremely small number.
Or, within computer world you often see megabytes, gigabytes, terabytes.
Mega means 106 or one million; giga means 109 , and tera means 1012 .
Suppose you have a chemical concentration 0.000442 gm per litre, or
a star having huge mass of 8,290,000,000,000,000,000,000 kg. Those
Exponents
177
Activity 1:
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zeroes take up a lot of space. Scientific notation simplifies this by writing
the last number, for example, as 829 × 1019 kg, which means 829 followed
by 19 zeroes. On the other-hand, the chemical concentration is written
with a negative exponent: 4.42 × 10−4 gm per litre.
The mass of a proton is 0.000000000000000000000001673 gm. You see how
much space you need to write such a tiny number. If you use exponents,
you may write this in the form 1.673 × 10−24 gm.
Another place where exponential notation helps is in calculators. In a
calculator, no display is possible once the number exceeds certain fixed
number of digits. This fixed number varies from calculator to calculator depending on the capacity of the calculator. Scientific calculators, in
which large or small numbers are calculated, use exponential notation.
For example, a calculator may show only eight significant digits in its display. So to display a number of the form 234587643214878, which has
15 digits, calculators round it off to 234587640000000 and write the resulting number in the form 23458764 × 107 ( or some calculators in the form
.23458764 × 1015).
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Collect information about many more life situations where the exponential
notation is useful.
Indian contribution to large numbers
The Indians had a passion for high numbers, which is intimately related
to their religious thought. For example, in texts belonging to the Vedic literature dated from 1200 BC to 500 BC, we find individual Sanskrit names
for each of the powers of 10 up to a trillion and even 1062. One of these
Vedic texts, the Yajur Veda (1200-900 BC), even discusses the concept of
numeric infinity (purna “fullness”), stating that if you subtract purna from
purna, you are still left with purna.
The Lalitavistara Sutra (a Mahayana Buddhist work) recounts a contest
including writing, arithmetic, wrestling and archery, in which the Buddha
was pitted against the great mathematician Arjuna and showed off his
numerical skills by citing the names of the powers of ten up to 1 ‘tallakshana’, which equals 1053 , but then going on to explain that this is just
one of a series of counting systems that can be expanded geometrically.
Unit 4
178
The last number at which he arrived after going through nine successive
counting systems was 10421 , that is, a 1 followed by 421 zeros.
There is also an analogous system of Sanskrit terms for fractional numbers, capable of dealing with both very large and very small numbers.
Larger number in Buddhism seems
122
or 1037218383881977644441306597687849648128,
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107×2
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which appeared as Bodhisattva’s maths in the Avatamsaka Sutra.
Here are a few large numbers used in India by about 5-th century BC
(See Georges Ifrah: A Universal History of Numbers, pp 422-423):
koti - 107 , ayuta - 109 , niyuta - 1011 , kankara - 1013 , pakoti - 1014 , vivara
- 1015, kshobhya - 1017 , vivaha - 1019 , kotippakoti - 1021 , bahula - 1023 ,
nagabala - 1025 , and so on. They also had name for 10421 as dhvajagranishamani. (wikipedia.org/wiki/History_of_large_numbers)
Two Indian Legends about Large numbers
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1. King Shiraham of India was pleased with his grand Vizier (Chief minister) for inventing the game of Chess and wanted to reward him. Vizier’s
request was simple. He said: “ Your Majesty, give me a grain of wheat
to put on the first square of the chess board, two grains to put on the
second square, four grains to put on the third, eight on the fourth, each
time doubling the number of grains, please give me enough grains to cover
all the 64 squares.” King thought that his minister’s request was very
very modest. He ordered his men to bring the wheat grains and fulfill
his Vizier’s desire. But he soon realised his folly. Do you know how many
grains are needed to comply with the minister’s request? It is 264 −1 grains.
This is roughly equal to the world’s wheat production for more than 2000
years.(At the current rate of production.)
2. In the great temple of Beneras(that is Kashi), there is a brass plate
in which three diamond needles are fixed. On one of the needles, at the
creation, Lord Brahma placed sixty four golden discs of decreasing diameters such that each disc sits only on a disc of larger diameter. Day and
night, the priest on duty transfers discs from one diamond needle to the
other according to the immutable laws of Lord Brahma. This requires that
the priest must move only one disc at a time and only in such a way that
Exponents
179
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there is no larger disc on a smaller disc. When all the sixty four discs get
transferred to another needle in accordance with this rule, the world will
come to an end. Can you make a guess of the time required for this? Even
with the assumption that the priests move one disc per second, the total
time needed would be 264 − 1 seconds. If you convert this to years, it will
be more than 58 thousand billion years (58 × 1012 years). This is five times
more than the current estimated age of our Universe.
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Exercise 2.4.1
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1. Express the following numbers in the exponential form:
1
(i) 1728 (ii)
(iii) 0.000169.
512
2. Write the following numbers using base 10 and exponents:
(i) 12345
(ii) 1010.0101
(iii) 0.1020304
3. Use base 5 and exponents to represent 2010.
4. Express (1234)5 in decimal form
5. Find the value (−0.2)−4 .
2.4.2 The first law of exponents
Consider 1024 = 210 . Observe
210 = 1024 = 2 × 512 = 21 × 29 :
1 + 9 = 10 and 210 = 21+9 ;
210 = 1024 = 8 × 128 = 23 × 27 :
3 + 7 = 10 and 210 = 23+7 ;
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210 = 1024 = 4 × 256 = 22 × 28 :
210 = 1024 = 16 × 64 = 24 × 26 :
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210 = 1024 = 32 × 32 = 25 × 25 :
2 + 8 = 10 and 210 = 22+8 ;
4 + 6 = 10 and 210 = 24+6 ;
5 + 5 = 10 and 210 = 25+5 ;
What do you observe? Consider more examples:
54 = 625 = 25 × 25 = 52 × 52 :
2 + 2 = 4 and 54 = 52+2;
a7 = (a × a × a × a) × (a × a × a) = a4 × a3 :
4 + 3 = 7 and a7 = a4+3 .
Looking at these examples, can you formulate a law?
For any number a and positive integers m, n, am × an =
am+n .
Unit 4
180
This is also true for an algebraic variable x: xm xn = xm+n . This is called
the first law of exponents. This is useful in simplifying large expressions.
Examples:
3. 25 × 26 = 25+6 = 211 ;
4. 33 × 36 × 37 = (33 × 36 ) × 37 = 33+6 × 37 = 39 × 37 = 39+7 = 316 .
1. Simplify:
(ii) 22 × 33 × 24 × 35 × 36 .
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(i) 31 × 32 × 33 × 34 × 35 × 36 .
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Exercise 2.4.2
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5. 25 × 52 × 23 × 5 = (25 × 23 ) × (52 × 5) = 28 × 53 .
2. How many zeros are there in 104 × 103 × 102 × 10?
3. Which is larger: 53 × 54 × 55 × 56 or 57 × 58 ?
2.4.3 The second law of exponents
Look at the following example:
29 = 512 =
28 = 256 =
27 = 128 =
64 =
26 =
64 =
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26 =
1024
210
= 1
2
2
210
1024
= 2
4
2
1024
210
= 3
8
2
1024
210
= 4
16
2
8
256
2
= 2 :
4
2
:
9 = 10 − 1 and 29 = 210−1;
:
8 = 10 − 2 and 28 = 210−2;
:
7 = 10 − 3 and 27 = 210−3;
:
6 = 10 − 4 and 26 = 210−4;
6 = 8 − 2 and 26 = 28−2 .
Study one more:
33 = 27 =
243
35
= 2 :
9
3
3 = 5 − 2 and 33 = 35−2 .
Can you see some pattern? Is it apparent that some law is followed here
again?
For any number a and positive integers m, n, with m > n,
am
= am−n .
an
Exponents
181
The natural question is: what happens if m = n or m < n. A few examples will help to clarify the situation.
1
1
2
21
=
=
=
.
32
64
25
26
Similarly,
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1
1
9
32
=
= 6 :
3−4 = 4 =
81
729
3
3
−5 = 1 − 6.
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2−5 = 6 :
2
d
Recall: we have defined a−n = a1n for any a 6= 0 and positive integer n. Thus
−4 = 2 − 6 and 3−4 = 32−6 ;
1
1000
103
=
= 4 ; −1 = 3 − 4 and 10−1 = 103−4 .
10
10000
10
The above observations may be put in the following form:
For any number a 6= 0 and positive integers m, n, with m 6= n,
am
= am−n .
an
10−1 =
Again recall, we have defined a0 = 1 for all a 6= 0. Observe
5 − 5 = 0 and 25−5 = 20 ;
4 − 4 = 0 and 34−4 = 30 .
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to
25
= 1 = 20 :
25
34
= 1 = 30 :
34
This shows that aam = 1 = a0 for all a 6= 0. We can now reformulate our law:
m
For any number a 6= 0 and positive integers m, n, not necessarily distinct,
am
= am−n .
an
The important thing here is that the only condition on m, n is that they are
positive integers. It may happen that m < n, or m = n or m > n. In all these
cases, the above law holds.
Unit 4
182
There is another way of looking at the second law. Suppose a 6= 0, and
m, n are positive integers. Then
am
1
am−n = n = am × n = am × a−n ,
a
a
d
1
where we have used a−n = n . Thus we obtain
a
am−n = am × a−n .
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Observe that this resembles the first law except that we have negative
integer −n. In this sense, the second law simply extends the first law.
Again observe that for natural numbers m, n,
1
1
(using the first law)
a−(m+n) = m+n = m
a
a × an
1
1
= m × n (property of fractions)
a
a
−m
= a
× a−n .
Thus we see that the first law is also valid for negative integers m, n. We
can combine both the first and the second law and state them together:
If a 6= 0 and m, n are integers, then am × an =
am+n .
Example 6. Simplify
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to
35 × 22 × 7−3
.
7−2 × 3−3 × 24
Solution: The expression is equal to
5 2 −3 3
2
7
×
×
= 35−(−3) × 22−4 × 7−3−(−2)
−3
4
−2
3
2
7
= 35+3 × 2−2 × 7−3+2
38
6561
= 38 × 2−2 × 7−1 = 2
=
.
28
2 ×7
Example 7. If 3l × 32 = 35 , find the value of l.
Solution: Using the second law of exponents, we obtain 3l+2 = 35 . But we
know that for any a 6= 0, 1, −1, if am = an , then m = n. Hence l + 2 = 5 or
l = 3.
Exponents
183
Exercise 2.4.3
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23 × 32 × 54
1. Simplify: (i) 10−1 × 102 × 10−3 × 104 × 10−5 × 106 ; (ii) 3
.
2 × 24
3
×
5
2. Which is larger: 34 × 23 or 25 × 32 ?
3m × 2n
3. Suppose m and n are distinct integers. Can m
be an integer?
2 × 3n
Give reasons.
24
4. Suppose b is a positive integer such that 2 is also an integer. What
b
are the possible values of b?
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2.4.4 The third law of exponents
Consider the following examples:
210 = 22+2+2+2+2 = 22 × 22 × 22 × 22 × 22 = 22
210 = 25+5 = 25 × 25 = 2
5 2
:
5
:
2 × 5 = 10 and (22 )5 = 210 = 22×5;
5 × 2 = 10 and
25
312 = 32+2+2+2+2+2 = 32 × 32 × 32 × 32 × 32 × 32 = 32
to
2 × 6 = 12 and
4
12
3+3+3+3
3
3
3
3
3
= 3
= 3 × 3 × 3 × 3 = 33 :
No
t
312 = 36+6 = 36 × 36 = 36
2
:
3
2
6
2 6
3 × 4 = 12 and
33
6 × 2 = 12 and
36
4
2
= 210 = 25×2 ;
:
= 312 = 32×6;
= 312 = 33×4;
= 312 = 36×2.
What do you see? Can you put your observations in to a rule?
n
If a 6= 0 is a number and m, n are positive integers, then am = amn .
Study more examples:
1
1
2 −4
=
=
2
;
28
(22 )4
2
1
1
1
−10
(−5)×2
−5 2 .
3
=3
= 10 = 5 2 =
=
3
3
(3 )
35
22×(−4) = 2−8 =
Do you see that negative integer exponents follow similar rule?
Unit 4
184
d
−n
If a 6= 0 is a number and m, n are positive integers, then am
=
n
a−mn = a−m .
What do you expect in the case when exponents are negative integers?
Consider the example:
−3
1
1
(−4)×(−3)
12
−4
5
= 5 = −12 =
.
3 = 5
5
5−4
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Are you convinced that the negative exponents also follow the same rule:
−n
a−m
= amn for all numbers a 6= 0 and positive integers m, n?
What if one of m, n is equal to 0? You observe that for a 6= 0, a0 = 1 and
hence (a0 )n = 1n = 1 = a0 and 0 × n = 0.
Thus we can formulate the third law of exponents:
If a 6= 0 is a number and m, n are integers , then am
amn .
n
=
(1024)3 × (81)4
.
(243)2 × (128)4
Solution: Observe 1024 = 210 , 81 = 34 , 243 = 35 , and 128 = 27 . Hence the
expression is
Example 8. Simplify
(210 )3 × (34 )4
230 × 316
=
(35 )2 × (27 )4
310 × 228
= 230−28 × 316−10 = 22 × 36 = 2916.
to
Exercise 2.4.4
2
5−3 × 34
1. Simplify: (i)
(ii)
−3
−2 .
3−2
× 53
2. Can you find two integers m, n such that 2m+n = 2mn ?
4
3. If 2m = 46 , find the value of m.
No
t
6
2
25 × 33
5
3 ;
26 × 32
2.4.5 The fourth law of exponents
Study the following examples.
(1)
105 = 10 × 10 × 10 × 10 × 10
= 2×5×2×5×2×5×2×5×2×5
= (2 × 2 × 2 × 2 × 2) × (5 × 5 × 5 × 5 × 5)
= 25 × 55 .
Exponents
185
Here you may observe that 10 = 2 × 5.
34 × 24 = (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2)
(2)
= (3 × 2) × (3 × 2) × (3 × 2) × (3 × 2)
= 6×6×6×6
(3)
he
Here again 3 × 2 = 6. Consider one more example:
d
= 64 .
(−2)4 × (44 ) = (−2) × (−2) × (−2) × (−2) × 4 × 4 × 4 × 4
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= (−2 × 4) × (−2 × 4) × (−2 × 4) × (−2 × 4)
= (−8) × (−8) × (−8) × (−8) = (−8)4 .
Once again you find that (−2) × 4 = (−8). Let us see what happens if both
the numbers are negative.
(4)
(−3)3 × (−5)3 = (−3) × (−3) × (−3) × (−5) × (−5) × (−5)
= (−3) × (−5) × (−3) × (−5) × (−3) × (−5)
= 15 × 15 × 15
= 153 .
No
t
to
Again you see that expected rule is true: (−3) × (−5) = 15. Now you can
formulate new law:
If a and b are two non-zero numbers and m is any positive integer,
then (a × b)m = am × bm .
Observe that this is also true if m = 0. Since a 6= 0 and b 6= 0, you may
conclude that a × b 6= 0. Thus (a × b)0 = 1. But a0 = 1 and b0 = 1 so that
a0 × b0 = 1. It follows that (a × b)0 = a0 × b0 . What happens if m is negative?
If a, b are non-zero numbers and n is a positive integer, you see that
(a × b)−n =
1
1
1
1
=
=
×
= a−n × b−n .
(a × b)n
an × bn
an bn
Thus the negative exponents also obey the same rule. We may now reformulate the law:
For any two numbers a 6= 0 and b 6= 0, and integer m,
(a × b)m = am × bm .
Unit 4
186
Here are more examples .
154 × 142
.
213 × 103
Solution: The numerator is
Example 9. Simplify:
Similarly the denominator is
he
213 × 103 = (7 × 3)3 × (2 × 5)3 = 73 × 33 × 23 × 53 .
d
154 × 142 = (3 × 5)4 × (2 × 7)2 = 34 × 54 × 22 × 72 .
The given expression is therefore
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34 × 54 × 22 × 72 .
15
3×5
= 34−3 × 54−3 × 22−3 × 72−3 =
= .
3
3
3
3
2×7
14
7 ×3 ×2 ×5
Example 10. Which is larger: (0.25)4 or (0.35)3 ?
Solution: We use the simple observation: if a and b are nonzero numbers,
a
then a < b is equivalent to < 1.
b
1
7
35
1
73
Write 0.25 = and 0.35 =
= . We have to compare 4 and 3 . But
4
100
20
4
20
203 = (4 × 5)3 = 43 × 53 . Thus we see that
53
(0.25)4
43 × 53
=
< 1,
=
(0.35)3
44 × 73
4 × 73
to
since 5 < 7. Thus (0.25)4 < (0.35)3 .
No
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Exercise 2.4.5
1. Simplify: (i)
68 × 53
;
103 × 34
(ii)
(15)−3 × (−12)4
;
5−6 × (36)2
(104 )3
an integer? Justify your answer.
513
3. Which is larger : (100)4 or (125)3 ?
(iii)
(0.22)4 × (0.222)3
.
(0.2)5 × (0.2222)2
2. Is
2.4.6 The fifth law of exponents
Consider
5
3
3 3 3 3 3
3×3×3×3×3
35
= × × × × =
= 5.
4
4 4 4 4 4
4×4×4×4×4
4
Exponents
187
Similarly, you observe that
3 −4
−4
−4
(−4) × (−4) × (−4)
−4
(−4)3
=
×
×
=
=
.
5
5
5
5
5×5×5
53
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What do you infer from these two examples? Do you see that there is yet
another law of exponent?
a m am
If a, b are non-zero numbers and m is a positive integer,
= m.
b
b
You may also observe that this is valid if m = 0. In this case
a 0
1
a0
= 1 = = 0.
b
1
b
No
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What do you think if m happens to be a negative integer? If m < 0, then
n = −m is a positive integer. Hence you can use the law for positive integral
exponent:
a n an
= n.
b
b
1
1
1
1
1
However, we know that an = −n = m . Similarly bn = −n = m or n = bm .
a
a
b
b
b
Thus you may obtain
a n an
1
1
bm
= n = an × n = m × bm = m .
b
b
b
a
a
a n
But we know
bm
1
a m =
= m.
b
a
b
This implies that
am a m
=
.
bm
b
We have obtained law for negative exponent. Thus we may write the fifth
law of exponents:
If a 6= 0 and b 6= 0, and m is an integer, then
am
.
bm
a m
b
=
−3
0.2 4
2
Example 11. Simplify:
×
.
5
0.5
Solution: You may simplify the two fractions separately. Thus
Unit 4
188
0.2
5
1
= 2,
5
2
0.5
= 2 × 2 = 22 .
Thus the expression simplifies to
4
1
1
1
2 −3 =
×
2
= 2
= 0.00000004.
2
8
6
5
5 ×2
5 × 106
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d
Example 12. Which is larger: (2.5)6 or (1.25)12 ?
6
5
56
6
Solution: We write (2.5) =
= 6 . Similarly, you may obtain
2
2
12
5
512
512
(1.25)12 =
= 12 = 24 .
4
4
2
512
56
Thus you have to check which of the numbers 6 ,
is larger. Equiva2
224
lently, you must determine the larger number between 56 and 218 (why?).
However 52 = 25 < 32 = 25 . Hence
56 = (52 )3 < (25 )3 = 215 < 218 .
Thus
56
512
56
56
<
.
=
×
224
218 26
26
Hence (1.25)12 < (2.5)6 .
Exercise 2.4.6
to
8 3
2
6
1. Simplify: (i)
×
;
3
4
(ii) (1.8)6 × (4.2)−3 ; (iii)
(0.0006)9
.
(0.015)−4
(30)3
n is also an integer?
(35)2
m m2
4
2
3. Can it happen that for some integer m 6= 0,
=
?
25
5
n
4. Find all positive integers m, n such that 3m = 3m × 3n .
No
t
2. What is the least positive integer n such that
Can you see that the fifth law of exponents can be deduced from the
a
fourth law? You may write = a × b−1 . Thus you obtain
b
a m
am
= (a × b−1 )m = am × (b−1 )m = am × b−m = m .
b
b
Exponents
189
You may notice that you needed (b−1 )m = b−m which is a consequence of
the third law. Thus the fifth law is a consequence of the fourth and third
laws.
Important points
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d
All these laws are also valid for algebraic variables. So far you have
seen the validity of these laws for numbers. If x is a variable, we have
xm · xn = xm+n ,
for all integers m, n.
(1)
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Here again, we define x0 = 1. Unlike for numbers, where we have defined
1
1
a−n = n for a 6= 0, we cannot so easily introduce . Hence we define x−n ,
a
x
for a natural number n, as that expression for which xn · x−n = 1 holds.
1
1
We write x−n in the form n . Thus we have the result: xn · n = 1, for all
x
x
1 also holds for all natural numbers
natural numbers n. Note that xn = x−n
n. With this understanding, the above law (1) holds. Similarly, you may
write laws:
(xm )n = xmn , for all integers m, n;
(2)
(x · y)m = xm · y m ,
for all integers m and variables x, y.
to
Additional problems on “Exponents”
No
t
1. Mark the correct option:
n
(a) The value of 3m , for every pair of integers (m, n), is
A. 3m+n
B. 3mn C. 3m D. 3m + 3n
y
(b) If x, y, 2x + are nonzero real numbers, then
2
y −1 y −1
−1
2x +
(2x) +
2
2
n
equals
A. 1
B. x · y −1
C. x−1 · y
D. x−1 · y −1
(c) If 2x − 2x−2 = 192, the value of x is
A. 5
B. 6 C. 7 D. 8
(3)
Unit 4
190
(d) The number
A. 66
6
66
6
B. 66 −1
1/6
is equal to
5
C. 6(6 )
6
D. 6(5 )
(e) The number of pairs positive integers (m, n) such that mn = 25 is
B. 1 C. 2 D. more than 2
2. Use the laws of exponents and simplify:
3−4 × 10−5 × (625)
(ii)
;
5−3 × 6−4
2
23
(iii) 3 2 .
(2 )
he
(12)6
(i)
;
162
d
A. 0
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3. The diameter of the Sun is 1.4 × 109 meters and that of the Earth is
about 1.2768 × 107 meters. Find the approximate ratio of the diameter
of the Sun to that of the Earth.
(103 )2 × 10−4
?
102
2 !−2
b−3 · b7 · b−1
.
3
(−b)2 · b2
4. What is the value of
5. Simplify:
No
t
to
6. Find the value of each of the following expressions:
2
2
2
(a) 32 − (−2)3 − − (52 ) ;
0
−2
(b) (0.6)2 − (4.5)0
;
3
5
1
(c) (4−1 )4 × 25 ×
× 8−2 × (642 )3 ;
16
−3
3
3
−3
(d) (−0.75) + (0.3) − −
;
2
8 × (42 )4 × 33 × 272 + 9 × 63 × 47 × (32 )3
;
(e) 24 × (62 )4 × (24 )2 + 144 × (23 )4 × (92 )2 × 42
219 × 273 + 15 × 49 × 94
(f)
.
69 × 210 + 1210
7. How many digits are there in the number 23 × 54 × 205 ?
−2 −3 × a3 4 × a−17 −1
a
8. If a7 = 3, find the value of
.
a7
Exponents
191
9. If 2m × a2 = 28 , where a, m are positive integers, find all possible values
of a + m.
(625)6.25 × (25)2.60
.
(625)7.25 × (5)1.20
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11. Find the value of
d
10. Suppose 3k × b2 = 64 for some positive integers k, b. Find all possible
values of k + b.
Glossary
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12. A person had some rupees which is a power of 5. He gave a part of it
to his friend which is also a power of 5. He was left with 500. How
much did money he have?
Light year: this is the distance travelled by a light ray in one year; it is
equal to 4.3 × 365 × 24 × 60 × 60 × 299792 km.
Exponential notation: writing a × a × · · · × a, where the product is taken
n times in the form an
Base: in the abbreviation an , a is the base.
Exponent: in the abbreviation an , n is the exponent.
Laws of exponents: am × an = am+n ; (am )n = amn ; (ab)m = am × bm hold
for non-zero numbers and integers m, n. (These are also true for algebraic
variables a, b and real numbers m, n.)
to
Points to remember
• 0n = 0 for n > 0; 0n is not defined for n ≤ 0.
No
t
• am = an if and only if m = n for any a 6= 0, 1 or −1.
1
• For any a 6= 0, and natural number n, a−n = n .
a
• For any a 6= 0, and integers m, n,
am × an = am+n .
n
• For any a 6= 0, and integers m, n, am = amn .
• For any a 6= 0, b 6= 0 and integer m,
(ab)m = am × bm .
CHAPTER 2
UNIT 5
INTRODUCTION TO GRAPHS
After learning this unit, you learn to:
d
• identify a point in the plane using a rectangular coordinate system;
he
• set up a coordinate system and draw graphs of linear curves in such
a system;
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• get to know the way the coordinates of a point are related in two
different rectangular coordinate systems with axes parallel to each
other;
• construct the equation of a straight line by looking at the graph of the
straight line on a graph paper.
2.5.1 Introduction
In May, when the summer is at its peak, you may find some days are
hot and some days are not so hot. You may measure the maximum temperature on each day (you may use either Fahrenheit or Celsius to measure temperature). You get a data for the whole month. Perhaps you will
tabulate it in a straight forward way:
2
3
4
5
6
7
8
······
30
31
32◦
34◦
35◦
37◦
32◦
35◦
38◦
······
36◦
36◦ .
No
t
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Day
1
Maximum
Temperature 33◦
Similarly, the rainfall for each month of a year can be measured in
inches or centimeters and we obtain a data. Recording these data is very
useful for future plans. For example, you can see the rain pattern over
several years and plan the agricultural crops accordingly. You may come
across several such important data in daily life which need to be recorded
for a better future. One easy way is to tabulate as earlier. However, this
is a very cumbersome procedure. People look for efficient way of recording
such data, which also help them to analyse it in a better way. One of the
effective method is the use of graphs.
Graphs
193
he
d
What is a graph? We can define a graph as the visual representation of numerical data collected during an experiment. By looking at the
graph, one can easily understand the data. Moreover, the graphs also help
us to analyse the data quickly.
There are different types of graphs: Bar graphs; Pie charts; Histograms;
Cartesian graphs. Each of them is useful for describing a particular type
of data. But the principal behind all these graphical representation is the
same; an easy visual comprehension of the collected data.
2.5.2 Bar graphs
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Suppose a firm makes a profit of 36 lacks during 2006, 48 lakhs
during 2007, 70 lakhs during 2008, 85 lakhs during 2009 and 75
lakhs during 2010. There is an easy visualisation of the firm’s performance during this 5-year period, if we represent the profit by a rectangular
vertical bar as shown below.
85
75
70
48
36
2006
2007
2008
2009
2010
No
t
to
Note that each bar in the above picture represents the profit of the firm
during that year which is given below the bar. You see that the firm has a
good growth, but a set-back in the last year.
96
90
87
80
Science
Mathematics
Kannada
Social Studies
English
65
It is not necessary that the vertical bars in a bar graph be separated. The adjacent graph shows
the marks obtained by Shilpa in
her 8-th standard examination.
It is easy to compare her performance in individual subjects.
Unit 5
194
2.5.3 Pie charts
Rahim is an engineer in an industry and his take home salary is
36,000 per month. His expenditure for a month runs as follows:
d
6,000
—–
—–
6,000
—–
3,600
—–
6,000
—–
6,000
—–
8,400.
per month.
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How can this data be repreHouse rent
sented as a graph? Rahim
Food
would like to see what fracChildren’s education
tion of his income is spent
Miscellaneous
on various heads. The graph
Parental care
must immediately reflect the
Savings
fraction of his expenditure against his total salary income
No
t
to
We may think that 36,000 is represented by a circle. Since the expenditure on each item is a part of Rahim’s total earning, such expenditure
must be represented by a part of the circle. The representation must immediately reflect the ratio of the part to the whole For example, Rahim
spends 6,000 on rent and let us find the ratio of his rental expenditure
1
6000
= . Thus his rental expenses is oneto that of his income. It is
36000
6
sixth of his earnings and we have to represent rent by one-sixth part of
the whole circle. We can use the fact that the total angle at a point is 360◦ .
Consider the centre of the circle. Construct a sector of the circle which
measures 60◦ , since one-sixth of of 360◦ is 60◦ . Thus, this sector repre1
sents exactly part of the whole circle. This we can do for all heads and
6
find out the angle of the sector needed to represent each of the individual
expenditures. Thus
House rent
−→
6000
1
=
36000
6
and
1
× 360◦ = 60◦
6
Food
−→
6000
1
=
36000
6
and
1
× 360◦ = 60◦
6
3600
1
=
36000
10
and
1
× 360◦ = 36◦
10
6000
1
=
36000
6
and
1
× 360◦ = 60◦
6
Children’s education −→
Miscellaneous
−→
Graphs
195
Parental care −→
−→
Savings
6000
1
=
36000
6
and
1
× 360◦ = 60◦
6
8400
7
=
36000
30
and
7
× 360◦ = 84◦
30
he
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The graph we have generated is as follows:
ion
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uc
at
1/10
Food
1/6
us
Miscellaneo
Ed
Par
ent
s
1/6
1/6
Observe!
nt
Re
1/6
Sa
vin
gs
7/30
1 1 1 1 1 7
+ + + + +
= 1.
6 6 10 6 6 30
to
2.5.4 Coordinate system
No
t
Another extremely useful and important method of representing data
is the use of Cartesian graphs. These are also called coordinate graphs
as the basic principle depends on the use of a coordinate system. A
coordinate system is a useful device for representing points on the plane.
If you take a line, then the points on the line correspond to real numbers.
In a coordinate system in the plane, two lines, one perpendicular to the
other are used. This will help us to identify the points on the plane in a
definite manner.
To get an idea how a coordinate system works, suppose there is a rectangular grazing yard and you are standing at one corner of the yard. A
cow is grazing at some point in the yard. You have to identify its location
by some numerical data. What do you do?
Unit 5
196
d
An easy way is: walk in the direction of the cow from where you stand
and measure the distance you move. You may measure it using some
convenient measurement. But when you describe the situation to some
one else, you have to tell the direction in which you had moved and the
distance you had moved. Describing the direction is not so easy. You have
to use a different mechanism.
The best way is to use the boundaries of the grazing yard. Suppose
you are at the point A and the cow
E (cow)
is grazing at E.(Look at the adjaG
cent figure.) You may walk along
100
the boundary AB to a point F
(you)
B
A
300
F
where EF is parallel to AD. Then
make a 90◦ turn anti clock-wise and move in the direction of the cow;
remember EF is parallel to AD. If AF = 300 m and F E = 100 m, you may
say the cow is located at 300 m in the direction of AB and 100 m in the
direction of AD.
Briefly we say E has coordinates (300, 100) with respect to the point A and
the system (AB, AD). Note that 300 is the distance in the direction of AB
and 100 is the distance in the direction of AD.
You may also walk along AD first to a point G where AG = 100 m, and
then walk along GE(which is parallel to AB) to the point E where GE = 300
m. Thus if you use the point A and the system (AD, AB), then E has
coordinates (100, 300).
D
No
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to
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C
Observe!
The point E is described by the ordered pair (300, 100) with respect to the point A and the system (AB, AD). The same point E
is described by the ordered pair (100, 300) now with respect to the
point A and the system (AD, AC). Thus the same point E may
have different coordinates with respect to different systems.
Caution!!
With respect to the point A and the system (AB, AD), the points
(300, 100) and (100, 300) are totally different.
Graphs
197
C
E
(cow)
50
P
Q
50
50
(you)
A
100
R
200
F
B
Instead being at the point A, suppose you are at the point P (See
the figure). How do you describe
the location of the cow with respect to new position? Again, you
may walk along P Q to the point Q,
where P Q is parallel to AB and QE
d
D
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is parallel to AD. Then you may move along QE to reach E. Now you
can describe E with respect to the point P and the system (AB, AD). Note
that P Q is parallel to AB and QE is parallel to AD; we are using the same
system (AB, AD) but a different starting point P . If you know P Q and
QE, you can easily describe E. Draw P R parallel to AD with R on AB.
Suppose AR = 100 m. Then P Q = RF = 200 m. Similarly, if RP = 50 m,
then QE = F E − F Q = F E − RP = 100 − 50 = 50 m. Thus with respect to the
point P and the system (AB, AD), the point E has coordinates (200, 50)
Observe!
to
The point P can be described with respect to A and the
system (AB, AD) by (100, 50), since AR = 100 and RP = 50
units. Using the point P and the system (AB, AD), the
point E is described by (200, 50). But E is described by
(300, 100) using the point A and the system (AB, AD). Can
you see that 300 = 100 + 200 and 100 = 50 + 50? What do you
conclude?
No
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Now you imagine a situation of a huge grazing yard which is endless
and having no boundaries. Again a cow is grazing at some point in the
yard and you are standing at some other point. How do you describe
the position of the cow? Earlier, you had a rectangular yard and you
had a rectangular frame given by the boundaries. Now you are on your
own and you do not have any such frame. You may consider two lines
perpendicular to each other and passing through the point where you are
standing. Let O be the point where you are standing and let X ′ OX, Y ′ OY
be two line perpendicular to each other and passing through O. Let the
cow be grazing at E(See the figure).
Unit 5
198
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Again you may move along
−−→
OX to F , where EF is parY
allel to Y ′ OY .
Then you
may move from F to E
P
along F E. With respect to
F
L
the point O and the system
X’
X
Q
O
(X ′ OX, Y ′ OY ), you can deK
G
scribe E using the lengths
E
OF and F E = OG(here G is
Y’
on the line Y ′ OY such that
GE is parallel to X ′ OX).
But there is a difficulty. If the point E is below the line X ′ OX, as shown
−−→
in the figure, then you have to move along the ray OX to the point F , and
−−→
then you have to move from F to E in the direction of the ray OY ′ . If some
other cow is located at the point K, then you have to move from O to L in
−−→
the direction of the ray OX ′ and then you have to move from L to K in the
−−→
direction of the ray OY ′ . Similarly, you can reach the point P from O, by
−−→
−−→
moving first along OX ′ and then from Q to P in the direction of OY .
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Now you may observe that there are four different directions to move
−−→
−−→ −−→ −−→
starting from O; OX; OX ′ ; OY ; and OY ′ . You may think the line X ′ OX
as the real number line and O corresponding to the number zero. Thus
−−→
all points to the left of O (along OX ′ ) correspond to negative numbers and
−−→
all points to the right of O (along OX) correspond to the positive numbers. Similarly, we can view the line Y ′ OY as another copy of number line
and O corresponding to zero; all points above O corresponding to positive
numbers and all points below O corresponding to negative numbers. Note
that O has the position 0 with respect to both the lines. We say O has
coordinates (0, 0).
Y
7
6
5
P
4
3
2
1
L
Q
0
1
X’ −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 −1 O
2
3
4
5
6
F
−2
−3
K
−4
E
−5
−6
−7
−8
Y’
7
8 9
10
X
For reaching E, you have to move 5
−−→
units to F in the direction OX and 4
−−→′
units in the direction OY , from F to E.
−−→
Since OY ′ corresponds to the negative
part of the number line, we describe E
by (5, −4). We say E has coordinates
Graphs
199
(5, −4) with respect to the rectangular coordinate system X ′ OX ←→
Y ′ OY . (Here X ′ OX ←→ Y ′ OY is a symbol used to represent the coordinate system.)
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Here the line X ′ OX is called the x-axis and Y ′ OY is called the y-axis.
−−→
The point O is called the origin of the coordinate system. The ray OX
−−→
is called as the positive x-axis and OX ′ is called as the negative x−−→
−−→
axis. Similarly, OY is the positive y-axis and OY ′ is the negative y-axis.
The point E has coordinates (5, −4) in this system. We say 5 is the xcoordinate of E and −4 is the y-coordinate of E. Note that the first number is always identifier along the horizontal axis and the second number
is always the identifier along the vertical axis. Similarly, the point K has
−6 as its x-coordinate and −3 as its y-coordinate. We say K has coordinates (−6, −4) in this system. Similarly, P has coordinates (−4, 4).
The lines X ′ OX and Y ′ OY diY
vide the plane in to 4 regions.
The region bounded by the rays
−−→
−−→
OX and OY is called the first
X
X’
O
quadrant. Similarly, you have
the second quadrant, the third
quadrant and the fourth quadY’
rant. If you take any point in
to
−−→
−−→
the first quadrant, you have to move in the directions of OX and OY to
reach that point. Thus you use the positive x and y-axes.
No
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You will now recognise that all points in the first quadrant has non−−→
negative x- and y- coordinates; each point on the ray OX has its y-coordinate
−−→
equal to 0; and each point on the ray OY has its x-coordinate 0. In the
−−→
−−
→
second quadrant, you have to use the rays OX ′ and OY ; hence each point
in the second quadrant has non-positive x-coordinate and non-negative
y-coordinate. Similarly, the points in the third quadrant are described by
the coordinates (x, y), where x ≤ 0 and y ≤ 0. Each point in the fourth
quadrant is represented by the ordered pair (x, y) where x ≥ 0 and y ≤ 0.
The x-coordinate of a point is also called its the abscissa and the ycoordinate is called the ordinate of that point.
Unit 5
200
Observe that the lines X ′ OX and Y ′ OY are completely left to
your choice. The only condition is that they must be perpendicular to each other for obtaining a rectangular coordinate system.
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René Descartes was born on
March 31st, 1596 in the town
of La Haye in the south of
France. In 1606, at the age
of 8, René Descartes started
studying literature, grammar,
science, and mathematics. In
1616, he received his baccalaureate and licentiate degrees in Law. Aside from his
Law degrees, Descartes also
spent time studying philosophy, theology, and medicine.
After a short stay in the military, Descartes went on to lead a quiet life,
continuing his intellectual pursuits, writing philosophical essays, and exploring the world of science and mathematics. In 1637, he published “Geometry”, in which his combination of algebra and geometry gave birth to
analytical geometry, better known as Cartesian geometry.
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We use only rectangular coordinate system throughout this chapter. However, it is possible to have a coordinate system in which
the two axes need not be perpendicular to each other. Such a system
called oblique coordinate system is also useful in solving many practical
problems.
The idea of introducing coordinates to identify the points on the plane
is due to René Descartes, a french mathematician and philosopher. This
has revolutionised the thinking as it helps to convert geometrical problems
to equivalent algebraic problems. To commemorate his contribution, the
system we study is also known as Cartesian coordinate system. This
has developed in to a new branch of mathematics known as Analytic geometry.
Graphs
201
Activity 1:
Locating a given point on a graph paper
Suppose you are given a graph paper. You also fix up your coordinate
system in the graph paper, say X ′ OX and Y ′ OY which are perpendicular
to each other. How do you locate the point P with coordinates (−4, 6)?
d
Y
P(−4,6)
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8
7
6
5
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4
3
2
X’
Q
1
−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0
X
1 2
3
4
5
6
7 8
9 10
−1
−2
−3
−4
−5
−6
−7
−8
Y’
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to
−−→
Starting from the origin O, move 4 units in the direction of the ray OX ′ .
−−→
Since OX ′ represents negative x-axis, when −4 is given as x-coordinate,
−−→
you have to move 4 units in the direction of OX ′ . You end up with the point
−−→
−−→
Q on OX ′ . Now move 6 units up in the direction parallel to OY . Remember
−−
→
OY represents positive y-axis. Hence the number 6(with positive sign)tells
−−→
you that you have to move 6 units upwards parallel to OY . You then reach
a point P whose coordinates are (−4, 6).
Activity 2:
Take a sheet of graph paper. Now set up two rectangular coordinate axes:
X ′ OX ←→ Y ′ OY and X1′ O1 X1 ←→ Y1′ O1 Y1 , where X ′ OX k X1′ O1 X1 . Locate
a point P with coordinates (10, 5) with respect to the system X1′ O1 X1 ←→
Y1′ O1 Y1 . What are the coordinates of P with respect to the system X ′ OX ←→
Y ′ OY ? Find the coordinates of O1 with respect to the system X ′ OX −Y ′ OY .
Now can you find the relation among the coordinates of P in the system
Unit 5
202
X ′ OX ←→ Y ′ OY , the coordinates of O1 in the system X ′ OX ←→ Y ′ OY and
the coordinates of P in the system X1′ O1 X1 ←→ Y1′ O1 Y1 ?
Y
Y1
7
17
6
P
16
5
15
4
14
3
13
2
11
−11 −10
−9
−8
−7
−6
−5
−4
−3
−1
−2
−1
9
−2
8
−3
7
−4
6
−5
5
−6
4
−7
3
X’
O1
0
10
1
2
3
4
5
6
8
7
9
10
11
12
13
X1
14
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−13 −12
he
1
X 1’
d
12
O
−4
−3
−2
−1
−8
2
−9
1
−10
0
1
2
3
4
5
6
−1
7
8
9 −11 10
−2
−13
−3
−14
−4
−15
−5
−16
Y’
11
12
13
14
15
16
17
18
19
20
21
22
23
24
X
−12
Y 1’
In the above figure, you see that O1 has coordinates (11, 11) in the system X ′ OX ←→ Y ′ OY . The point P has the coordinates (21, 16) in the system X ′ OX ←→ Y ′ OY and (10, 5) in the system X1′ O1 X1 ←→ Y1′ O1 Y1 . Observe
that
21 = 11 + 10, 16 = 11 + 5.
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We write this in the form (21, 16) = (11, 11) + (10, 5). Repeat this with various
positions of P and different systems X ′ OX ←→ Y ′ OY , X1′ O1 X1 ←→ Y1′ O1 Y1 .
What is the conclusion you draw from these activities?
Observe!
Suppose X ′ OX ←→ Y ′ OY and X1′ O1 X1 ←→ Y1′ O1 Y1 are two system
of coordinates such that X ′ OX k X1′ O1 X1 . Let a point P have
coordinates (x, y) in the system X ′ OX ←→ Y ′ OY and (x′ , y ′ ) in the
second system X1′ O1 X1 ←→ Y1′ O1 Y1 . If O1 has coordinates (a, b)
with respect to the system X ′ OX − Y OY , then x = a + x′ and
y = b + y ′.
Graphs
203
Activity 3:
Take a sheet of graph paper and set up your coordinate system X ′ OX ←→
Y ′ OY . Locate a point P with coordinates (−5, 8). Trace the point with
coordinates (8, −5). Do you get the same point P ?
Example 1.
−3
3 × (−3)
y
−9
(x, y) (−3, −9)
−2
3 × (−2)
−6
(−2, −6)
−1
3 × (−1)
−3
(−1, −3)
0
1
2
3
3×0 3×1 3×2 3×3
0
3
6
9
(0, 0) (1, 3) (2, 6) (3, 9)
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Consider the number 3. Let us tabulate the multiples of 3, some positive
and some negative.
Y
(4,12)
12
11
10
(3,9)
9
8
7
(2,6)
6
5
4
(1,3)
3
2
X’
1
−13 −12 −11 −10 −9
−8 −7
−6 −5 −4 −3 −2
−1
0
(0,0)
O1
2
3
4
5
6
7
8
9 10
11
12
13
X
−1
−2
(−1,−3)
−3
−4
−5
(−2,−6)
−6
−7
−8
to
(−3,−9)
−11
−12
Y’
No
t
(−4,−12)
−9
−10
Take a sheet of graph paper, and set up your own coordinate system,
X ′ OX ←→ Y ′ OY . Locate the points (x, y) tabulated in the last row. Can
you see that all these lie on a straight-line?
Example 2. (Area versus perimeter of a square)
Consider a square of side length 1 unit. What is its area? You know that
the area of a square is l2 , where l is the length of the square. And its
perimeter is l + l + l + l = 4l. Hence the area of a square of side length 1 unit
is 1 square unit and its perimeter is 4 units. What are these for a square
of length 2 units? You see that they are respectively 4 square units and 8
Unit 5
204
units. Let us tabulate the perimeter and area of squares of different side
lengths.
d
1
2
3
4
5
6
4
8
12
16
20
24
1
4
9
16
25
36
(4,1) (8,4) (12,9) (16,16) (20,25) (24,36)
he
l
x = 4l
y = l2
(x, y)
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Y
40
(24,36)
35
30
(20,25)
25
20
(16,16)
15
10
(12,9)
5
1
X’
to
(8,4)
(4,1)
O 0
5
10
15
20
X
25
30
No
t
Y’
Again set up your own coordinate system on a graph paper and plot
these points (x, y) on the graph paper. Do you think that these points lie
on a straight-line?
Example 3. (Simple interest versus number of years)
Suppose Shiva deposits 1,000 in a bank for 5 years for simple interest
at the rate of 8% per annum. You can easily calculate what he earns as
interest over the years and tabulate it.
Graphs
205
Year=x
Interest=y
(x, y)
1
2
3
4
5
80
160
240
320
400
(1, 80) (2, 160) (3, 240) (4, 320) (5, 400)
Again you are required
to plot these points on
a graph paper.
Here
you face a practical problem. The interest Shiva
obtains are big numbers
and it is difficult to loO
cate numbers like 240,
X
X’
320, 400 on a graph paper, unless the graph paper you take is very huge.
Y’
However, this difficulty may be avoided using different scaling. For example, you may use 1 unit=40 along y-axis. Here again you notice that all
the points lie on a straight-line.
Activity 4:
Y
(5,400)
400
d
360
(4,320)
320
280
he
(3,240)
240
200
(2,160)
160
120
(1,80)
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80
40
−5 −4 −3 −2
−1
0
1
2
3
4
5
6
7
8
9 10
11
12
13
to
Suppose a car moves with a constant speed of 40 km per hour. Make a
table of the distance covered by the car at the end of 1-st, 2-nd, 3-rd, 4-th,
5-th, 6-th ,7-th and 8-th hour. Trace the points on a graph paper where
x-axis represents the time and y-axis represents the distance traveled by
the car.
No
t
Exercise 2.5.4
1. Fix up your own coordinate system on a graph paper and locate the
following points on the sheet:
(i) P (−3, 5);
(ii) Q(0, −8);
(iii) R(4, 0);
(iv) S(−4, −9).
2. Suppose you are given a coordinate system. Determine the quadrant
in which the following points lie:
(i) A(4, 5);
(ii)B(−4, −5);
(iii)C(4, −5).
3. Suppose P is a point with coordinates (−8, 3) with respect to a coordinate system X ′ OX ←→ Y ′ OY . Let X1′ O1 X1 ←→ Y1′ O1 Y1 be another
system with X ′ OX k X1′ O1 X1 and suppose O1 has coordinates (9, 5)
Unit 5
206
d
with respect to X ′ OX ←→ Y ′ OY . What are the coordinates of P in the
system X1′ O1 X1 ←→ Y1′ O1 Y1 ?
4. Suppose P has coordinates (10, 2) in a coordinate system X ′ OX −Y ′ OY
and (−3, −6) in another coordinate system X1′ O1 X1 ←→ Y1′ O1 Y1 with
X ′ OX k X1′ O1 X1 . Determine the coordinates of O with respect to the
system X1′ O1 X1 ←→ Y1′ O1 Y1 .
he
2.5.5 Linear graphs
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Take a re-look at Example 1, where you have plotted integral multiples
of 3. Instead of taking integer multiples of 3, suppose you take real multiples of 3: for each real number x, consider y = 3x(obtained by multiplying
x by 3).
x 0
1 2 3
−1
−2
y
3 6 9
−3
−6
0
1
2
3
2
−1
2
−3
2
1
3
1
As you assign different real values for x, you get back different real
numbers y = 3x. Thus for each real number x, you get a point whose
coordinates are (x, 3x). Using a coordinate system, we can trace these
points on a graph sheet.
Y
12
11
(3,9)
10
to
9
8
7
(2,6)
6
5
4
(1,3)
No
t
3
X’
−13 −12 −11 −10 −9
2
(1/2,3/2)
1 (1/3, 1)
−8 −7
−6
−5
−4
−3
−2
0 O1
−1
(−1/2, −3/2)
−1
2
3
4
5
6
7
8
9
10
11
12
13
X
−2
(−1,−3)
−3
−4
−5
(−2,−6)
−6
−7
−8
−9
−10
−11
−12
Y’
If you take more and more values of x, you get more and more points
(x, 3x). As you go on tracing them on graph paper, you see that these points
Graphs
207
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get more and more clustered. Of course, you have to get finer and finer
graph paper to trace these points. However, you see that all these points
lie on a straight-line. Now it is clear that as x exhaust all real numbers,
the points (x, y), where y = 3x, trace a straight-line in the plane. We say
the equation y = 3x represents a straight-line on the coordinate plane.
Activity 5:
1
Use the scaling 1 unit = on the x-axis. Calculate more values of y = 3x
3
as x varies over real numbers and trace the points (x, y) on a graph sheet
using the suggested scaling.
The relation y = 3x is of the first degree. Such a relation always represents a straight-line. The general form of the equation for a straight-line
is y = ax + b.(Here a and b are real numbers.) If a = 0, then y = b is a
straight-line parallel to y-axis.
Example 4. Suppose a person deposits 10,000 in a bank for simple
interest at the rate of 7% per annum. Obtain a relation for accrued interest
in terms of years. Draw the graph of this relation. Use this to read the
interest at the end of 10 years.
Solution: Let y be the interest obtained in x years. The interest for x years
7
× 10000 × x = 700x. Thus you get y = 700x as the required relation.
is
100
Now use the scaling: 1 unit= 700 along y-axis. Note that
(x, y) = (1, 700),
for x = 2, you get y = 1400 :
(x, y) = (2, 1400).
to
for x = 1, you get y = 700 :
No
t
Y
7700
7000
S
R
6300
5600
4900
4200
3500
2800
2100
1400
Q(2,1400)
700
X’
P(1,700)
O 0 1
Y’
2
3
4
5
6
7
8
9
T
10 11 12
13 14 15
X
Take a sheet of graph paper and set up coordinate
axes X ′ OX ←→ Y ′ OY . Locate points P (1, 700) and
Q(2, 1400). Remember you are
using the scale 1 unit =700
along y-axis. Now the equation is a straight-line and hence P , Q determine
this straight-line.
Unit 5
208
Observe!
To draw a straight-line, it is enough to know only two points on
that line; given any two distinct points in the plane, you can
use the scale to draw the unique line passing through those two
points.
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Use a straight-edge to join P Q and extend it to possible extent. This
straight-line represents the curve of the interest versus years for this
problem. Now look at the point 10 on x-axis, say T . Draw a perpendicular to x-axis at T to meet the straight-line y = 700x at R. From R draw
a perpendicular to y-axis. Record the point where it cuts y-axis. That gives
the interest at the end of 10 years, which you read as 7,000.
Example 5. Draw the graph of y = 3x + 5.
Solution: Give different values for x and get values for y. Tabulate them.
x 0 1 2
y 5 8 11
3 −1
14 2
−2
−1
20
15
to
(3,14)
(2,11)
No
t
10
(1,8)
5
(0,5)
0
1
(−1,2)
−2 −1
2
3
4
5
(−2, −1)
Let us trace these on a graph paper. Fix x and y axes and locate the
points
(x, y) = (0, 5), (1, 8), (2, 11), (3, 14), (−1, 2), (−2, −1).
Graphs
209
Do you see that all these points lie on a straight line? Use a straightedge and join all these points by a straight line segment.
Activity 6: Draw the graphs of y = x + 4, y = 2x − 3, y = 3, x = 2y + 1, x = 2.
All these are straight lines. Do you see that y = ax + b is always a straight
line?
Example 6.
graph.
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Determine the equation of the line in the following given
Y
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Solution:
Recall
that
the general equation of a
straight-line is y = ax + b,
where a, b are real constants.
The given graph shows that
Q
a 6= 0. (If a = 0, then y = b
represents a straight-line
P
O
X
X’
parallel to y-axis.) If x = 0,
then y = b. Thus (0, b) is a
point on the straight-line.
Y’
b
b
Similarly, taking y = 0, you get ax + b = 0 or x = − . Thus − , 0 is also
a
a
a point on the line. Looking at the graph, we see that it cuts y-axis at (0, 4)
10
9
8
7
6
5
4
3
2
1
−5 −4 −3 −2
−1
0
1
2
3
4
5
6
7
8
9 10 11
12 13
−1
−2
to
−3
No
t
and the x-axis at (3, 0). We conclude that
(0, b) = (0, 4),
b
− , 0 = (3, 0).
a
b
4
= 3, which gives a = − . Hence the equation of the
a
3
4
given graph is y = − x + 4. This may also be written as 3y + 4x = 12(why?)
3
Thus b = 4 and −
Example 7.
graph.
Determine the equation of the line in the following given
Unit 5
210
Y
8
7
6
5
4
Q
3
2
1
O
−5 −4 −3 −2
−1
0
X’
1
2
3
4
5
6
7
8
9 10
11
12
13
X
−1
−2
P
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−3
d
9
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10
Solution: We may take
the equation in the form
y = ax + b. Observe that
the straight-line passes
through (0, 0).
This
means that y = 0 when
x = 0. But the substitution x = 0 in the equation gives 0 = y = b. Thus
b = 0 and so that y = ax.
Y’
We also observe that the line passes through (1, −3); that is y = −3, whenever x = 1. This gives −3 = a × 1 or a = −3. The equation of the line is
therefore y = −3x.
Exercise 2.5.5
1. Draw the graphs of the following straight-lines:
(i) y=3-x; (ii) y=x-3; (iii) y=3x-2; (iv) y=5-3x ;
(v) 4y = −x + 3; (vi) 3y = 4x + 1; (vii) x = 4; (viii) 3y = 1.
y
y+1
2. Draw the graph of =
.
x
x+2
3. Determine the equation of the line in each of the following graphs:
Y
to
Y
10
9
9
8
8
No
t
10
7
7
6
6
5
5
4
4
3
3
2
2
1
1
O
−5 −4 −3 −2 −1 0
X’
−1
1
2
3
4
5
O
6
7
8
9 10 11
12 13
−5 −4 −3 −2 −1 0
X
X’
−2
−2
−3
−3
Y’
1
2
3
4
5
6
7
8
9 10 11
12 13
−1
Y’
4. A boat is moving in a river, down stream, whose stream has speed 8
km per hour. The speed of the motor of the boat is 22 km per hour.
Draw the graph of the distance covered by the boat versus hour.
X
Graphs
211
5. Find the point of intersection of the straight-lines 3y + 4x = 7 and
4y + 3x = 7, by drawing their graphs and looking for the point where
they meet.
(a) The point (4, 0) lie on the line ——————–
A. y − x = 0 B. y = 0 C. x = 0
D. y + x = 0
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(b) The point (−5, 4) lie in ——————–
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1. Choose the correct option:
d
Additional problems on “Introduction to graphs”
A. the first quadrant B. the second quadrant C. the third
quadrant D. the fourth quadrant
(c) If a straight-line pass through (0, 0) and (1, 5), then its equation is
——————–
A. y = x B. y = 5x C. 5y = x D. y = x + 5
(d) If a point P has coordinates (3, 4) in a coordinate system X ′ OX ←→
Y ′ OY , and if O has coordinates (4, 3) in another system X1′ O1 X1 ←→
Y1′ O1 Y1 with X ′ OX k X1′ O1 X1 , then the coordinates of P in the new
system X1′ O1 X1 ←→ Y1′ O1 Y1 is ——————–
A. (3, 4) B. (1, −1)
C. (7, 7) D. (−1, 1)
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(e) The coordinates of a point P in a system X ′ OX ←→ Y ′ OY are (5, 8).
The coordinates of the same point in the system Y ′ OY ←→ XOX ′
are ——————–
A. (−8, 5) B. (8, 5) C. (8, −5) D. (−8, −5)
(f) The signs of the coordinates of a point in the third quadrant are
——————–
A. (+, −)
B. (−, +)
C. (+, +) D. (−, −)
(g) If a person moves either 1 unit in the direction of positive x-axis or
1 unit in the direction of positive y-axis per step, then the number
of steps he requires to reach (10, 12) starting from the origin (0, 0)
is ——————–
A. 10 B. 12 C. 22 D. 120
Unit 5
212
(h) The y-coordinate of the point of intersection of the line y = 3x + 4
with x = 3 is ——————–
A. 4
B. 7 C. 10 D. 13
(i) The equation of the line which passes through (0, 0) and (1, 1) is
——————–
C. y = 1 D. x = 1
(i) (5, 10); (ii) (−8, 9);
(iii) (−800, −3000);
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2. Find the quadrant in which the following points lie:
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A. y = x B. y = −x
(iv) (8, −100).
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3. Match the following:
(A) On the x-axis
(B) In the second quadrant
(C) The line y = 3x + 4
(i) x coordinate is negative
(ii) cuts the y-axis at (0, 4)
(iii) coordinates of a point are of
the form (a, 0).
4. Fill in the blanks:
(a) The y-coordinate of a point on the x-axis is —————–.
(b) The x- coordinate is called as —————–.
(c) The x-axis and y-axis intersect at —————–.
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(d) If a point (x, y) 6= (0, 0) is in the third quadrant, then x + y has
—————– sign.
(e) If a point (x, y) lies above horizontal axis, then y is always ———
——–.
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(f) The point of intersection of x = y and x = −y is —————–.
(g) The line y = 4x + 5 intersects y-axis at the point —————–.
5. True or false?
(a) The equation of the x-axis is x = 0.
(b) The line x = 4 is parallel to y-axis.
(c) The line y = 8 is perpendicular to x-axis.
(d) The lines x = y and x = −y are perpendicular to each other.
(e) The lines x = 9 and y = 9 are perpendicular to each other.
Graphs
213
(f) The graph of y = x2 is a straight line.
(g) The line y = 3x + 4 does not intersect x-axis.
(h) In a rectangular coordinate system, the coordinate axes are chosen such that they form a pair of perpendicular lines.
d
6. Determine the equation of the line which passes through the points
(0, −8) and (7, 0).
Y
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7. Determine the equation of the line in each of the following graph:
(i)
(ii)
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Y
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
X’
O
O
−5 −4 −3 −2
−1
0
1
2
3
4
5
X
−1
−2
−3
Y’
−5
X’
−4
−3
−2
−1
0
1
2
3
4
5
6
7
8
9 10
11
12
13
−1
X
−2
−3
Y’
to
8. A point P has coordinates (7, 10) in a coordinate system X ′ OX ←→
Y ′ OY . Suppose it has coordinates (10, 7) in another coordinate system
X1′ O1 X1 ←→ Y1′ O1 Y1 with X ′ OX k X1′ O1 X1 . Find the coordinates of O1
in the system X ′ OX ←→ Y ′ OY .
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9. Sketch the region {(x, y) : x ≥ 0, y ≥ 0, x + 2y ≤ 4} in a coordinate
system set up by you.
10. Draw the graphs of lines 3y = 4x − 4 and 2x = 3y + 4 and determine the
point at which these lines meet.
11. If a ⋆ b = ab + a + b, draw the graph of y = 3 ⋆ x + 1 ⋆ 2.
Glossary
Graph: a visual representation of numerical data.
Bar graph: graphs in which data is represented by rectangular bars.
Pie chart: a graph in which data is represented by sectors of a circle.
Unit 5
214
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Rectangular frame: two perpendicular lines in a plane which helps to locate points on the plane.
Rectangular coordinate system: a system which making use of rectangular frames locates points as an ordered pair of real numbers.
x-axis: the horizontal line in a rectangular coordinate system.
y-axis: the vertical line in a rectangular coordinate system.
Quadrant: the four parts that a plane gets divided by a rectangular coordinate system.
Abscissa: the x-coordinate of a point.
Ordinate: the y-coordinate of a point.
Cartesian coordinate system: a coordinate system in which every point
in the plane is determined by a pair of real numbers called the coordinates
of the point.
Analytic geometry: the geometry in which all the geometrical concepts
are studied using the coordinate system.
Points to remember
• You can set up your own coordinate system; there is no sacred coordinate system.
• The coordinates of a point depends on the coordinate system you
choose.
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• A rectangular coordinate system is only a convenient coordinate system; it is not necessary that one should always use a rectangular
system.
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• A graph is visual representation of the numerical data.
Answers to Exercises in Chapter 2.
Exercise 2.1.1
−3 √
−x √
2
5xy
1. Constants: 15,
, 3, 7; Variables: 12 + z,
, x, xy;
, 7 − x,
7
5
3
2
8yz
y 2x
6x + 4y, −7z,
, y + 4, ,
.
4x
4 8yz
8x 8 2 2
2. Monomials: 7xyz,
, x y ; Binomials: 9 − 4y, 4y 2 − xz, 7x + z 2 ;
y 5
Trinomials: x − 2y + 3z, 4 + 5y − 6z.
Answers
215
Exercise 2.1.3
1
x , {7xyz}.
3
3. (i) 9a + 6b; (ii) 2x2 y + 5xy 2 + 2y 3.
1. {4x2 , 3x2 }, {xy, 8xy}, {−8x3 , 6x3 , −74x3 },
2. (i) 4x − 14y + 11;
4. (i) 10x2 y − 3xy 2 ;
(ii) 4x2 − 3xy − 9y 2 .
(ii) 3a + 7b.
Exercise 2.1.4
First →
3x
Second ↓
3x
9x2
−6y
−18xy
2
4x
12x3
−8xy
−24x2 y
9x2 y
27x3 y
−11x3 y 2 −33x4 y 2
−6y
4x2
−8xy
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1.
9x2 y
−11x3 y 2
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12x3
−24x2 y
27x3 y
−33x4 y 2
−24x2 y
48xy 2
−54x2 y 2 66x3 y 3
16x4
−32x3 y
36x4 y
−44x6 y 2
−32x3 y
64x2 y 2 −72x3 y 2 88x4 y 3
36x4 y
−72x3 y 2
8x4 y 2
−99x5 y 3
−44x5 y 2 88x4 y 3 −99x5 y 3 121x6 y 4
6
6
2. (i) 15x2 + 24x; (ii) 45p4 q 3 + 3pq 4 ; (iii) a3 x − b3 x; (iv) −x3 + 15x.
5
5
3. (i) 6x3 y 2 − 10x2 y − 3x2 y 2 + 5xy; (ii) 12x3 y 3 − 18x3 y 4 + 4xy − 6xy 2 ;
(iii) 6x4 + 4x3 + 9x2 + 6x; (iv) 10m3 + 12m2 − 3m.
−18xy
36y 2
−24x2 y
48xy 2
−54x2 y 2
66x3 y 3
Exercise 2.1.5
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1. (i) a2 +8a+15; (ii) 9t2 +15t+4; (iii) a2 −6a−16; (iv) a2 −8a+12. 2. (i) 2915;
(ii) 10812; (iii) 1224; (iv) 9888. 3. x3 + (a + b + c)x2 + (ab + bc + ca)x + abc.
4. (i) a2 + 12a + 36; (ii) 9x2 + 12xy + 4y 2; (iii) 4p2 + 12pq + 9q 2; (iv) x4 + 10x2 + 25.
5. (i) 1156; (ii) 104.04; (iii) 2809; (iv) 1681. 6. x2 − 12x + 36;
(ii) 9x2 − 30xy + 25y 2; (iii) 25a2 − 40ab + 16b2 ; (iv) p4 − 2p2 q 2 + q 4 .
2
2
7. (i) 2401; (ii) 96.04;
2 (iii)3481; (iv) 39204. 8. x − 36; (ii) 9x − 25;
4x
(iii) 4a2 − 16b2 ; (iv)
− 1 . 9. (i) 2475; (ii) 851; (iii) 80.75;
9
1 4
1
4
4
4
(iv) 9996. 10. (i) x − 81; (ii) 16a − 81; (iii) p − 16; (iv)
m −
;
16
81
(v) 16x4 − y 4 ; (vi) 16x4 − 81y 4.
Additional problems on “Algebraic expressions”
1. (a) B. (b) B. (c) C. (d) A. (e) A. (f) C. (g) C. (h) B. (i) C.
Answers
216
17. 2x2 − 2xy + 4z 2 .
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15. (i) 13x − 4y; (ii) −28x − 11y. 16. 4x2 − 22x + 6.
18. (18x2 − 9xy − 20y 2)/2. 19. 5x2 + 11xy + 5y 2.
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2. 2x+ 17y + p −q. 3. (i) 16x2 + 24x+ 9; (ii) x2 + 4xy + 4y 2; (iii) x2 + (1/x2 ) + 2;
(iv) x2 + (1/x2 ) − 2. 4. (i) 4t2 − 25; (ii) x2 y 2 − 25; (iii) 4x2 − 9y 2 .
5. (i) n4 − 1; (ii) n4 − (1/n4 ); (iii) x8 − 1; (iv) 16x4 − y 4 . 6. (i) 10609; (ii)
9216; (iii) 9951; (iv) 999936; (v) 21000.
7. 25.
8. 80.
9. ±8.
2
2
10. 29 11. 7 and 18. 12. 34 and 1154. 13. (i) 2(x + y );
3(x + z) 2
(z − x) 2
4
2
2
4
(ii) x − 2x y + y . 14. (i)
−
; (ii) (x + 2y)2 − (2x + y)2 ;
2
2
(iii) (x + 100)2 − 12 ; (iv) 5002 − 52 .
Exercise 2.2.2
1. (i) x(x + y); (ii) 3x(x − 2); (iii) (0.8)a(2a − 1); (iv) 5(1 − 2m − 4n).
2. (i) (a+x)(a+b); (ii) (3a+7b)(c−d);
(iv) (y−3)(y 2 +2−x).
(iii) (x−2z)(3y−3t);
3
3
3. (i) (2a + 5)(2a − 5); (ii) x +
x−
; (iii) (x2 + y 2 )(x + y)(x − y);
4
4
1222
; (v) 0.4; (vi) (7a − 3b)(3a − b).
(iv)
25
Exercise 2.2.3
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1. (i) p = 9, q = 2; (ii) p = −8, q = −4; (iii) p = 6, q = −4; (iv) p = 12, q = −1;
(v) p = −6, q = 1; (vi) p = −11’ q = 4. 2. (i) (x + 4)(x + 2); (ii) x + 3)(x + 1);
(iii) (a + 3)(a + 2); (iv) (a − 3)(a − 2); (v) (a − 8)(a + 5); (vi) (x − 9)(x + 8).
3. (i) (x + 7)(x + 7); (ii) (2x + 1)(2x + 1); (iii) (a − 5)(a − 5); (iv) 2(x − 6)(x − 6);
(v) (p − 12)(p − 12); (vi) x(x − 6)(x − 6).
Additional problems on “Factorisation”
1. (a) C.
(b) B. (c) D. (d) B. (e) D. (f) A.
2. (i) (x + 3)2 ; (ii) (1 − 4x)2 ; (iii) (2x + 9y)(2x − 9y); (iv) (2a + b)2 ;
(v) (a2 − d2 )(b2 − c2 ).
3. (i) (x + 3)(x + 4); (ii) (x + 4)(x − 3); (iii) (x − 6)(x + 3); (iv) (x + 7)(x − 3);
(v) (x − 16)(x + 12); (vi) (x − 1)(x + 1)(x − 2)(x + 2);
(vii) (x − 2y)(x + 2y)(x − 3y)(x + 3y).
4. (i) (2x−3)(x+2); (ii) (x−4)(3x−5); (iii) (x−2)(6x+7); (iv) (2x+y)(2x+5y);
(v) (2x − 1)(2x + 1)(x − 1)(x + 1). 5. (i) (x − 4)(x + 4)(x2 + y 2 )(x4 + y 4 );
Answers
217
(ii) a4 x4 (a − x)(a + x)(a2 + x2 )(a4 + x4 ); (iii) (x2 + x + 1)(x2 − x + 1);
(iv) (x2 + x + 3)(x2 − x + 3). 6. (x2 + 2xy + 2y 2)(x2 − 2xy + 2y 2).
Exercise 2.3.2
6
; (v) x = 3; (vi) s = 28;
11
−8
(vii) x = 20; (viii) x = 2.4; (ix) x = 7; (x) x =
; (xi) x = 60; (xii) x = 25;
5
(xiii) x = 2; (xiv) x = 36; (xv) x = −1. 2. (i) x = 12; (ii) t = −6; (iii) x = 7;
3
(iv) z = ; (v) x = 5; (vi) x = 1; (vii) x = −25; (viii) x = −13; (ix) x = −5;
2
(x) x = 10; (xi) x = 3; (xii) x = 8; (xiii) x = −54; (xiv) x = 9.
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1. (i) x = 8; (ii) y = 30; (iii) z = 7; (iv) x =
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Exercise 2.3.3
1 6. 2. 71, 73, 75. 3. 25 cm and 15 cm. 4. 11. 5. 3,000 and
750. 6. 60 and 10. 7. x = 40. 8. 7. 9. 45 and 36. 10. 96 m and
48 m. 11. Ahmed’s age 12 and his father’s age 36 years. 12. Nishu’s
age 11 and Sanju’s age 17 years. 13. Deepu’s age 11 and Viji’s age 22
years. 14. Bindu’s age 19 and Mrs. Joseph’s age 46 years. 15. 8 years.
Additional problems on “Linear equations in one variable”
to
1 (a) B. (b) A. (c) B. (d) B. (e) B. (f) B. (g) A. (h) B.
47
2. (i) 10; (ii) − . 3. 21 and 24. 4. 10 and 40. 5. 110, 112, 114.
57
6. A’ share 40,000 and B’s share 20,000. 7. 60. 8. 72. 9. garden
35,000 and house 49,000. 10. 800. 11. 15 years. 12. 33.
13. 12.50 14. 57. 15. 90 km and 95 km. 16. 2.25 km. 17. 1/4.
18. 62. 19. altitude 20 cm and base 12 cm. 20. 40◦ , 50◦ and 90◦ .
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Exercise 2.4.1
1. (i) 123 ; (ii) 2−9 ; (iii) (0.013)2. 2. (i) 104 + 2 · 103 + 3 · 102 + 4 · 10 + 5;
1
1
2
1
3
4
(ii) 103 + 101 + 2 + 4 ; (iii)
+ 2 + 5 + 7 . 3. 3 · 54 + 1 · 53 + 2 · 5.
10 10
10
10
10
10
4. 194. 5. 625.
Exercise 2.4.2
1. (i) 321 ; (ii) 26 · 314 .
Exercise 2.4.3
1. (i) 103 ; (ii)
25
.
6
2. 10 zeros.
3. 53 × 54 × 55 × 56 > 57 × 58 .
2. 34 × 23 > 25 × 32 .
4 b = 1, 2 or 4.
Answers
218
Exercise 2.4.4
1. (i) 1; (ii)
1
.
9
2. m = 2, n = 2.
3. m = 3.
Exercise 2.4.5
1. (i) 2592; (ii)
(1.11)3 × (1.1)2
2000
; (iii)
.
27
(1.01)2
2. No.
3. 1004 > 1253 .
32
;
243
(ii)
2. n = 49.
3. can happen for
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m = 2.
4. (m, n) = (2, 2).
313
(iii) 39
.
2 × 544
19683
;
42875
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Exercise 2.4.6
Additional problems on “Exponents”
1. (i) B. (ii) D. (iii) D. (iv) C. (v) B.
2. (i) 1152; (ii) 25/2; (iii) 8.
3. 109.65. 4. 1.
7. 11.
8. 81.
5. b12 .
6. (a) 642; (b) 0; (c) 2−9 ; (d) 944/27; (e) 1/2.
9. a + m = 8.
10. k + 2 = 8 or 14.
11. 1.
12. 625.
Exercise 2.5.4
2. (i) first quadrant; (ii) third quadrant; (iii) fourth quadrant.
3. (−17, −2). 4. (−13, −8).
Exercise 2.5.5
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3. (i) 3x − 2y = 6; (ii) x + y = 5. 5. (1, 1).
Additional problems on “Introduction to graphs”
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1. (a) B. (b) B. (c) C. (d) C.
(e) C. (f) D. (g) C. (h) B. (i) A.
2. (a) First; (b) second; (c) third; (d) fourth.
3. (A) −→ (iii); (B) −→ (i); (C) −→ (ii).
4. (a) zero;
(g) (0, 5).
5. (a) false;
(h) true.
(b) abscissa;
(b) true;
(c) (0, 0);
(c) false;
6. 7y = 8(x − 7); 7y = 8x.
8. (−3, 3).
(d) negative;
(d) true;
(e) positive;
(e) true;
(f) false;
7. (i) 6y = 11x + 15; (ii) 16y = −5x + 12.
(f) (0, 0);
(g) false;
CHAPTER 3
UNIT 1
AXIOMS, POSTULATES AND THEOREMS
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After studying this unit you learn:
• the meaning of undefined objects, axioms, postulates and hypothesis;
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• that the lines, points, plane, space are undefined objects in Euclidean
geometry;
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• various types of angles and relation among these angles;
• the properties of parallel lines and about Euclid’s fifth postulate.
3.1.1 Introduction
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In earlier classes, you have studied many geometrical objects: straight
lines, triangles, quadrilaterals and circles. You have also studied some
geometrical properties of these objects: angles and different types of angles; triangle inequality(the sum of two sides of a triangle is greater than
the third side); medians; altitudes; area of a triangle and a circle. Most of
these are taught to you through observations. You may wonder that these
were developed by our ancestors more than 2000 years ago.
Indeed, the concept of geometry is very old. Egyptian civilisation developed the early geometrical methods and measurements. In fact, geometry
is derived from two Greek words: Geo to mean Earth and metron meaning
measurement. When the Nile river flooded the whole region, the cultivated
land used to submerge in water erasing all the boundaries. Hence Egyptians developed certain geometrical methods to demarcate the boundaries
afresh. They also introduced area of plane figures and volume of some
three dimensional objects which were used as granaries. Perhaps Pyramids, which still occupy a place among seven wonders of the world and
whose construction is certainly one of the greatest human achievements,
will give you an idea how much Egyptians were advanced in the use of
geometry.
Thus the ancient geometry developed through the practical requirement
of measuring land. However, a systematic treatment of geometry started
Unit 1
220
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with ancient Greek around 2500 years ago. They are the first one to realise
the need to conceptualise the geometrical ideas.
The practical geometry used point, line and plane without bothering
much what they mean. But, the Greek Philosophers and Mathematicians
were more interested in proving statements by deductive reasoning. It
was, perhaps, Thales(640 BC- 546 BC) who first introduced the concept of
proof. He realised the need for proving a statement by logical reasoning.
Many more Greeks, like Appolonius, Plato, Pythagoras, Diophantus and
Ptolemy made enormous contributions to the systematic development of
Geometry and other areas of Mathematics and laid the firm foundation to
make Mathematics a science of logical reasoning.
However, it was Euclid who collected all these contributions to Geometry and other branches of mathematics in to thirteen volumes of a book
called the Elements along with his own original ideas.
Euclid(around 300 BC) was a Greek
mathematician often called the Father
of Geometry. He was a contemporary
of Ptolemy(323 BC-283 BC), another famous Greek mathematician of antiquity. Euclid’s work Elements is one of
the most influential work in the whole
history of mathematics which changed
the face of Mathematics laying the foundation for the future development.
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Euclid deduced his results, what is now known as Euclidean Geometry,
from a small number of principles called Axioms and Geometrical postulates. Euclid has also contributed to other branches of Mathematics. His
proof that there are infinitely many prime numbers is a classic example of
the deductive reasoning that Euclid employed in his works. Nothing much
is known about Euclid’s life. The date and place of his birth are unknown.
Like-wise, the date and circumstances of his death are also unknown. All
we know about him is through the references made by other people in their
work. The picture of Euclid we have today is also through the imagination of
an artist.
In ancient India, Śulva Sūtras are perhaps the first record of Math-
Axioms
221
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ematical lore, especially geometry(600 BC to 300 BC). These are records
of the mathematical principles developed during Vedic period and subsequent time. The Śulva Sūtras contain several geometrical principles.
The Indian geometry developed out of religious needs of constructing
sacrificial altars to propitiate Gods and later for the study of Eclipses.
The Baudháyana Sūtra, the most ancient among the Śulva Sūtras says
the diagonals of a rectangle bisect each other. This also contained the idea
of the celebrated Pythagoras’s theorem, but unfortunately no proof was
given.
The Śulva Sūtras also gave methods of constructing a square whose
area equals the area of a given circle. The construction involved approximating π and the approximation used was 3.088, which is quite close to
the present day approximations to π.
With the passage of time, others also made significant contributions:
Āryabhata I, Bháskara I, Varáhamihara, Brahmagupta, Mahávíráchárya,
Bháskárácárya II, Madhava, Nilakantha Somayaji, and many others made
innumerable contribution to the advancement of Mathematics.
3.1.2 Axioms and Postulates
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You have seen that a straight angle is defined as the angle measuring
180 degrees. You use a protractor to measure angle. However, the calibration of protractor is such that when you place it on a straight line, you
read 180 degrees. Thus the protractor is designed to measure 180 degrees
when a straight angle is given. Can you see that you have to go around
from protractor to straight angle to protractor?
This was the major difficulty faced by Greek mathematicians while developing Geometry as a pure deductive science. They had to depend on
certain primitive notions like points, straight lines and planes and space.
But this was not enough to deduce every thing. They had to set up certain statements, whose validity was accepted unquestionably, applicable
to geometry alone. They had to depend some more statements applicable
to all of Mathematics and science in general and Geometry in particular.
The general statements which are accepted without question and which
are applicable to all branches of science are commonly referred as Axioms.
Unit 1
222
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The statements which are particular to Geometry and accepted without
question are called Geometrical Postulates. Any result you further prove
depends on these axioms and postulates.
Another problem with the deductive method is how to define some geometrical terms. For example, you all have an intuitive idea what a point
is. But can you define a point? When you define some thing, you must
do it so using what you know already. Hence you have to depend on some
undefined terms.
In Euclid’s geometry, the undefined terms are point, line, plane. They
are only certain abstract ideas. Thus you cannot see a point. If you take
a sharp pencil and make a dot on a paper, that approximately resembles
a point. Similarly, you cannot see a line. When a point moves in both the
directions, it produces a straight line. A line is endless.
←→
If A and B are points on a line, we denote the straight line by AB. If a
straight line is cut, you get two pieces: each piece is called a ray. Thus a
ray has an initial point and extends indefinitely in one direction. If A is the
initial point of a ray and B is any other point on a ray, we denote the ray
−→
by AB. Take a line and choose any two points A and B on it. The part of
the line between A and B is called a line segment and is denoted by AB.
A
B
to
Straight Line
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A
A
Ray
Segment
B
B
Likewise, you cannot define a plane. Intuitively, a plane is flat infinite
surface without any thickness. A black board or the surface of still water in
a big tank resemble finite part of a plane. Thus, there are undefined terms
in Euclid’s geometry. With suitable axioms and geometrical postulates,
Euclid’s geometry tells you what edifice can be built. Let us study these
axioms and postulates.
Axioms
223
I. Axioms
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There are certain elementary statements, which are self evident and
which are accepted without any questions. These are called axioms.
These statements are also applicable to other areas of mathematics and
science. Euclid used the following statements which he called Common
Notions.
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Take a sheet of paper. Draw a
Axiom 1: Things which are line segment AB of length 10 cm.
equal to the same thing are Draw a second line CD having
equal to one another.
length equal to that of AB, using
a compass. Measure the length
A
B
of CD. Do you observe that CD
AB = 10 cm
implies CD= 10 cm
has length equal to 10 cm? You
C
D
may write this as: CD = AB and
CD = AB
AB = 10 cm implies CD = 10 cm.
to
Suppose you have three baskets A, B and C having mangoes, oranges
and bananas. Suppose A, B have equal number of fruits and B, C also
have equal number of fruits. Can you conclude that A and C have equal
number of fruits?
No
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Suppose you have two line segments AB and DE of equal length.
Axiom 2: If equals are added to
Add BC to AB and add EF to DE.
equals, the wholes are equal.
If BC = EF , then AC = DF .
A
B
D
E
C
F
AB=DE and BC=EF implies AC=DF
Take a basket A of 10 mangoes
and a basket B of 10 oranges.
Add 5 apples to both of these baskets. Do you see that the number
of fruits in both the baskets are
equal?(equal to 15)
Unit 1
224
Suppose you have two line segAxiom 3: If equals are sub- ments AC and DF of equal length.
tracted from equals, then the Remove BC from AC and EF from
remainders are equal.
DF respectively. If BC = EF , then
AB = DE.
B
C
A
F
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AC=DF and BC=EF implies AB=DE
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D
Take a basket A of 10 mangoes
and B of 10 oranges. Remove 2
fruits from each basket. Then A
and B again have equal number
of fruits.
Axiom 4: Things which coincide with one another must be equal to
one another.
This means that if two geometric figures can fit completely one in to other,
then they are essentially the same.
Axiom 5: The whole is greater than the part.
to
Take a container of water. Remove some water from it. Will the remaining
volume of water the same as the original volume?
No
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Euclid’s common notions are these first five axioms. The first three
concern “equals” or “equal things”; the fourth is interpreted now to
mean that if two figures, such as linesegments, angles, triangles, or
circles, are such that one can be moved to coincide with the other,
the figures are equal. In modern language they are called congruent.
Two ideas run through these common notions: (1) that geometrical
figures can be treated as magnitudes, and (2) that, if one figure is
visibly part of another (perhaps after a motion), then the magnitude
of the part is less than the magnitude of the whole. For adding
and comparing, you should have magnitudes of the same kind. For
example, you cannot add area to length. Or you cannot compare a
triangle with a point. Axiom 5 can be used to define greater than.
Axioms
225
II. Geometrical postulates
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If b is a part of a, then a is greater than b. Again this comparison
is between magnitudes of the same kind. (For example, you cannot
take out a point from a line and say line is greater than a point which
is clearly meaningless.) Since point has no magnitude of any kind,
you cannot compare two points. But you can say some line segment
is larger than some other line segment as you can compare their
lengths.
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Apart from these common notions, Euclid also made the following
geometrical postulates to deduce new propositions.
Postulate 1. A straight line segment can be drawn joining any two
points.
Postulate 2. Any straight line segment can be extended indefinitely
in a straight line.
Postulate 3. Given any straight line segment, a circle can be drawn
having the segment as radius and one endpoint as center.
Postulate 4. All right angles are congruent.
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Postulate 5. If a straight line meets two other lines, so as to make
the two interior angles on one side of it together less than two right
angles, the other straight lines will meet if produced on that side on
which the angles are less than two right angles.
The Postulate 5 is the famous Euclid’s parallel postulate. What it
asserts is that two distinct straight lines in a plane are either parallel
or meet exactly in one point. This postulate cannot be proved as
a theorem, although this was attempted by many people. Euclid
himself used only the first four postulates (“absolute geometry”) for
the first 28 propositions of the Elements, but was forced to invoke
the parallel postulate on the 29-th.
Unit 1
226
d
In 1823, Janos Bolyai and Nicolai Lobachevsky independently realized that entirely self-consistent “non-Euclidean geometries” could
be created in which the parallel postulate did not hold. The parallel postulate is actually equivalent to: any two straight lines in the
plane either do not meet at all or meet in one point.
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While proving his propositions, Euclid made several tacit assumptions.
For example, Postulate 1 says that there is a line passing through any two
given points. But what Euclid had in his mind seems to be: given any two
distinct point in the plane, there is a unique line passing through these two
points. On the other hand, you can draw infinitely many lines through a
given point in the plane.
Postulate 2 says that given a line segment in the plane, this can be
extended to a unique straight line. Postulate 4 is concerned about right
angles. But this is not defined any where by Euclid. What he seems to
have thought was that angle by a straight line is made up of two right
angles.
By present day standards, there are several inconsistencies in Euclid’s
Elements. Nevertheless, it is definitely the first book, based on rigorous
mathematical principles. In recent years, many attempts have been made
to introduce new set of undefined objects, axioms and postulates, so that
the terms hitherto undefined could be defined using these new objects,
axioms and postulates. The advancement of set theory and axioms on
number systems make it possible to define a point, a line or a plane using
coordinate systems.
Exercise 3.1.2
1. What are undefined objects in Euclid’s geometry?
2. What is the difference between an axiom and a postulate?
3. Give an example for the following axioms from your experience:
(a) If equals are added to equals, the wholes are equal.
(b) The whole is greater than the part.
4. What is the need of introducing axioms?
Axioms
227
5. You have seen earlier that the set of all natural numbers is closed
under addition(closure property). Is this an axiom or some thing you
can prove?
3.1.3 Lines and Angles
AOB
O
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B
−→
Suppose you have a ray OA on a plane with
end point O. With the same end point O,
−−→
consider another ray OB in the same plane.
−−→
−→
You observe that OB is obtained from OA
through suitable rotation around the point
−−→
−→
O.We say OB subtends an angle with OA.
The amount of rotation is the measure of
A this angle.
We use a numerical measurement called degree to measure angles. We
−→
−−→
use the notation a◦ to denote a degrees. The rays OA and OB are called
the sides of the angle and O is called the vertex of the angle. The angle
−→
−−→
b
subtended by the rays OA and OB is denoted by ∠AOB or AOB.
−→
−−→
−→
Note: Consider two rays OA and OB. If X is any point on OA, then
−→
−−→
the rays OA and OX are the same. Similarly, for any point Y on the
−−→
−−→
−−→
ray OB, the rays OY and OB are the same. Thus ∠AOB = ∠XOY .
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There is a geometrical instrument called protractor which is used for
measuring angles.
Activity 1:
Construct an angle which measures 40 degrees using protractor.
Activity 2:
−→
−−→
Take two rays OA and OB and measure the angle between them using a
protractor.
Warning! Even with the best protractor, scale and pencil, you may
not be able to produce an angle which measures exactly 40◦ . Your
eyes also play an important role, as parallax error normally creep
in. Nevertheless your construction is good enough for all practical
purposes.
Unit 1
228
Recall what you have studied about different types of angles: straight
angle, right angle, acute angle, obtuse angle, reflex angle, complete
angle, adjacent angles, complimentary angles and supplementary angles.
he
d
Consider a straight line and let O be point on the straight line. Then O
divides the straight line in to two rays. If B is to the left of O on the line
−→
−−→
and A to the right of O, then there are two rays OA and OB. The angle
between these two rays is called a straight angle.
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If you set your protractor such that its centre coincides with O, then you
−→
−−→
see that the angle between OA and OB is 180◦ . However, your protractor
is so calibrated that it measures any straight angle exactly 180◦ . Thus you
cannot define a straight angle using a measuring device.
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We will see that in Euclid’s geometry, this is taken as one of the postulates. But once you know a straight angle, you can define all other types
of angles. For example, a right angle is that angle which measures 90◦
or you need half of a protractor. Similarly, an acute angle is one having
measure less than 90◦ and an obtuse angle is that angle which measures
more than 90◦ but less than a straight angle. A reflex angle is an angle
measuring more than 180◦ but less than 360◦ . Finally, a complete angle
is an angle measuring 360◦ . This corresponds to a complete revolution of
−→
a ray OA around the initial point O.
B
Right angle
Straight angle
A
O
B
O
A
B
B
Acute angle
O
Obtuse angle
A
O
A
Axioms
229
Reflex angle
O
B
Complete angle
B
O
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A
A
B
Right angle
O
B
O
Straight angle
A
A
−→
−−→
We say two rays OA and OB are perpendicular to each other if the angle between them is 90◦ and we write
−→
−−→
−→
OA ⊥ OB. We say two rays OA and
−−→
OB are supplementary rays if the angle between them is 180◦ . Observe that
−→
−−→
in this case OA and OB are in opposite
directions.
to
Suppose you have a straight line and O is a point on this line. Then O
divides the straight line in to two rays: if B is to the left of O and A to the
−→
−−→
right of O, then the straight line is made up of two rays OA and OB, the
angle between them being one straight angle or 180◦ .
No
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Two angles are said to be supplementary angles if their sum is 180◦ .
Similarly, two angles are said to be complementary if they add up to 90◦ .
Two angles are said to be adjacent angles, if both the angles have
←→
←→
a common vertex and a common side. If two straight lines AB and CD
intersect at a point O, then you see that four angles are formed at O: if
−→
−−→
the first line is divided by O in to two rays OA and OB and if the second
−→
−−→
line is divided in to two rays OC and OD, then you get four angles ∠AOC,
∠COB, ∠BOD and ∠DOA. The pair of angles ∠AOC and ∠BOD are called
vertically opposite angles. Observe the other pair ∠COB and ∠DOA is
also a pair of vertically opposite angles.
Unit 1
230
B
B
X
C
A
D
Vertically opposite angles
d
A
Adjacent angles
he
O
O
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While measuring lengths of line segments and angles, we observe the
following rules. These were not stated separately by Euclid, but tacitly
assumed by him in the derivation of new propositions. We take them as
additional postulates.
Rule 1. Every line segment has a positive length. (The length of the
line segment AB is denoted by AB or |AB|.)
Rule 2. If a point C lies on a line segment AB, then the length of AB
is equal to the sum of the lengths of AC and CB; that is AB = AC + CB.
Rule 3. Every angle has a certain magnitude. A straight angle measures 180◦ .
to
−→ −−→
−→
−→
−→
−−→
Rule 4. If OA, OB and OC are such that OC lies between OA and OB,
then ∠AOB = ∠AOC + ∠COB.
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Rule 5. If the angle between two rays is zero then they coincide.
Conversely, if two rays coincide, the angle between them is either
zero or an integral multiple of 360◦ .
Note: While measuring angles, we use the following convention: if
the angle is measured anti-clock-wise, it is positive. If it is measured
clock-wise, then it is negative.
Axioms
231
Activity 3:
←→
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Note: What is the minimum number of axioms needed to develop a
self-consistent geometry is a question in Mathematical Philosophy.
However, here we do not bother about the minimum number required nor about retaining original axioms and postulates made by
Euclid. In fact there were several gaps in the set of axioms used by
Euclid and later mathematicians added some more to those used by
Euclid.
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On a sheet of paper draw a straight line AB. Choose a point O on it.
−→
Draw a ray OC on it. Measure angles ∠BOC and ∠COA using a protractor.
What is the sum ∠BOC + ∠COA? Repeat this with different positions of
−→
OC. What do you find?
You will always find that the sum of these two angles is 180◦ . Can you
prove this using axioms and postulates.?
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Note: Perhaps, you may realise here the need for a logical proof. No
matter which configuration you take with a line and a ray standing
on it, you see that the sum of the two adjacent angles always add
up to 180◦ . However, this does not deny that there is a case of a line
and a ray on it such that the sum of the two adjacent angles is different from 180◦ . This is inherent in the structure itself as there are
infinitely many possibilities of a line and a ray on it, and you cannot
verify your finding with all of them. This is the reason why Mathematicians look for logical proofs based on axioms and postulates or
on the propositions which have already been proved.
Let us see what is the statement we need to prove. We have to take
an arbitrary line and an arbitrary ray standing on it. Then there are two
adjacent angles formed by the line and the ray. We have to show that the
sum of these adjacent angles is 180◦ . We put this as a proposition. A
Proposition is a statement which is to be proved using the axioms and
postulates. It also depends on certain assumptions called hypotheses
which are used in the proof of the statement.
Unit 1
232
←→
−→
Proposition 1. Let AB be a straight line and OC be a ray standing
←→
on the line AB. Then ∠BOC + ∠COA = 180◦ .
Before getting on to the proof, let us
see what are given and what we need
to prove.
−→
Given (or hypothesis): A ray OC
d
←→
A
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C
stands on a straight-line AB forming
two adjacent angles ∠BOC and ∠COA.
To prove: ∠BOC + ∠COA = 180◦ .
Proof: We have ∠BOC + ∠COA = ∠BOA
by rule 4. But ∠BOA is a straight an-
O
B
Fig . 1
←→
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gle determined by the line AB. By rule
3, ∠BOA = 180◦ . Now we can invoke
Axiom 1: ∠BOC + ∠COA and 180◦ are
both equal to the same thing, ∠BOA.
We conclude that ∠BOC + ∠COA = 180◦ .
Look at the proposition once again. It says: if a ray stands on a straight
line, then the sum of two adjacent angles formed is 180◦. That is, the two
adjacent angles are supplementary. This is the general nature of all our
new propositions: given that a certain statement S is true, some other
statement R is true. We say S is the hypothesis and R is the conclusion.(Here a ray stands on a straight line is the statement S which is the
hypothesis and the sum of adjacent angles is 180◦ is the statement R, which
is the conclusion.) Hence if R is not true, then S cannot be true. Formally
we say “S implies R” is equivalent to “not R implies not S.”
This was one of the methods frequently employed by Euclid and also
by later mathematicians to prove new propositions. If you want to prove
that “S implies R”, it is enough to prove “not R implies not F .” This is
called method of reductio ad absurdum. This is a latin word with meaning
“reduction to the absurdity.”
What is the converse of a proposition? Naturally, hypothesis and conclusion must change their places. If our original proposition has S as
hypothesis and R as conclusion, the converse proposition must have R
as hypothesis and S as conclusion. In the present context the converse is:
Axioms
233
he
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−→ −−→
−→
If there are three rays OA, OB and OC such that ∠BOC and ∠COA are adja−→
−→
−−→
cent angles(that is OC is between OA and OB) and if ∠BOC + ∠COA = 180◦ ,
then A, O, B all lie on the same straight line. We say A, O, B are collinear
if all of them lie on the same straight line. We put forth this as another
proposition.
−→ −−→
−→
−→
Proposition 2. Let OA, OB and OC be three rays such that OC is
−→
−−→
between OA and OB. Suppose ∠BOC + ∠COA = 180◦ . Then A, O, B are
collinear,that is, they lie on the same straight-line.
C
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C
B
A
O
D
A
Fig . 2
D
O
B
Fig . 3
−→ −−→
−→
Given: Three rays OA, OB and OC are such that ∠BOC and ∠COA are
adjacent angles which add up to 180◦ .
To prove: A, O, B all lie on the same line..
Proof: Here we do some construction. Extend AO to D such that A, O, D
←→
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all lie on the same line, AD. Using proposition 1, we have ∠DOC + ∠COA =
180◦ . But we are given that ∠BOC +∠COA = 180◦ . Using Axiom 1, it follows
that
∠DOC + ∠COA = ∠BOC + ∠COA.
−−→
Now use Axiom 3 to get ∠DOC = ∠BOC. There are two possibilities: OB
−−→
−→
−−→
−−→
−→
lies between OD and OC(see Fig.2) or OD lies between OB and OC(see Fig.
3). In the first case, using rule 4, we get
∠BOC = ∠DOC = ∠DOB + ∠BOC,
and Axiom 3 implies that ∠DOB = 0. In the second case, rule 4 gives
∠DOC = ∠BOC = ∠BOD + ∠DOC,
and Axiom 3 implies that ∠BOD = 0. Thus the angle between the rays
−−→
−−→
−−→
OB and OD is zero. Using rule 5, we conclude that the rays OB and
←→
−−→
OD coincide. This means B and O are on the line AD. Thus A, O, B are
collinear.
Unit 1
234
d
Note: Proposition 1 and proposition 2 are two geometrical statements which are converses of each other. In geometry, if some
statement is true, then many times its converse is also true. However, this is not universally valid. There may be statements which
are true, but whose converses may fail. Later you will see that an
equilateral triangle is isosceles, but an isosceles triangle need not be
equilateral.
he
Example 1. In the adjoining figure, if
∠COA − ∠BOC = 50◦ , find these angles.
C
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Solution: We know by proposition 1,
∠BOC + ∠COA = 180◦ .
A
O
B
Adding, two relations, we get
Fig . 4
2∠COA = 230◦ .
(Which axiom is used here?) This implies that ∠COA = 115◦ . (Which axiom
is needed here?) Now
∠BOC = 180◦ − ∠COA = 180◦ − 115◦ = 65◦ .
Example 2. In the adjoining figure, if the angles ∠AOB, ∠BOC, ∠COD are
←→
to
in the ratio 1:2:3 and AD is a straight line, find the measures of all the
angles.
Solution: Using the proposition
1, we see that
No
t
C
∠AOB + ∠BOC + ∠COD = 180◦ .
But the given hypothesis is
3
∠BOC = 2∠AOB and ∠COD =
1
3∠AOB. Thus we get 6∠AOB =
D
O
A
180◦ . It follows that ∠AOB = 30◦ .
Fig. 5
This gives ∠BOC = 2 × 30◦ = 60◦
and ∠COD = 3 × 30◦ = 90◦ .
−→
−−→
Definition: Suppose ∠AOB is an angle formed by two rays OA and OB. If
−→
−→
−−→
OP is another ray between OA and OB such that ∠AOP = ∠P OB, we say
2
B
Axioms
235
−→
−→
OP bisects ∠AOB or OP is the angle bisector of ∠AOB. We observe that
in this case ∠AOP = ∠P OB = 21 ∠AOB.
Activity 4:
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←→
←→
−→
Draw a straight line AB and a ray OC on AB. Measure ∠BOC and ∠COA.
−→
−→
Construct ray OP such that it bisects ∠BOC. Similarly, construct ray OQ
such that, it bisects ∠COA. Measure ∠P OQ. Do you see that ∠P OQ = 90◦ ?.
←→
−→
Repeat this taking different rays OC on AB. Do you always find that
∠P OQ = 90◦ ? Can you formulate this as a proposition?
←→
−→
Proposition 3. Let AB be a straight line and let OC be a ray standing
−→
−→
on it. Let OP be the bisector of ∠BOC, and let OQ be the bisector of
∠COA. Then ∠P OQ = 90◦ .
Q
C
P
A
O
B
−→
−→
Given: OP bisects ∠BOC and OQ bisects ∠COA.
To prove: ∠P OQ = 90◦ .
Fig. 6
Proof:
−→
Since OP bisects ∠BOC, we have
1
∠P OC = ∠BOC.
2
(1)
to
−→
Since OQ bisects ∠COA, we also have
(2).
No
t
1
∠COQ = ∠COA.
2
Adding these two and using rule 4, we obtain
1
∠P OQ = (∠BOC + ∠COA).
2
By proposition 1, ∠BOC + ∠COA = 180◦ . Thus we obtain
∠P OQ =
←→
1
× 180◦ = 90◦ .
2
←→
Activity 5: Take two lines AB and CD intersecting at a point O. Measure
∠BOD, ∠DOA, ∠AOC, ∠COB. Compare ∠BOD and ∠AOC. Similarly compare ∠DOA and ∠COB. Do you observe some thing pertinent? Repeat this
Unit 1
236
←→
←→
with different positions of CD with respect to AB. Can you formulate a
new geometrical proposition?
Proposition 4. If two straight lines intersect at a point, then the
vertically opposite angles are equal.
←→
←→
d
Given: AB and CD intersecting at a
point O.
D
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O
C
B
Fig. 7
←→
Proof: Consider the straight line AB
−−→
and the ray OD standing on it. Then
∠BOD and ∠DOA are adjacent angles.
By proposition 1, we have
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A
To prove: ∠BOD = ∠AOC and ∠DOA =
∠COB.
∠BOD + ∠DOA = 180◦ .
(1)
←→
−→
Similarly, considering the straight line CD and the ray OA, we see that
∠DOA and ∠AOC are adjacent angles. Thus proposition 1 again gives
∠DOA + ∠AOC = 180◦ .
(2)
to
Using Axiom 1, we can compare (1) and (2) and get
No
t
∠BOD + ∠DOA = ∠DOA + ∠AOC.
(3)
Since we can remove ∠DOA using Axiom 3, we obtain
∠BOD = ∠AOC.
A similar argument gives ∠DOA = ∠COB.
←→
←→
−→
Example 3. Let AB and CD be straight lines intersecting at O. Let OP be
−→
the bisector of ∠BOD and OQ be the bisector of ∠AOC. Prove that Q, O, P
are collinear.
Axioms
237
Solution:
We have to show that
−→
◦
∠P OQ = 180 . Since OP bisects ∠BOD,
we have
D
P
A
O
B
Similarly, using the given hypothesis
−→
that OQ bisects ∠AOC, we also obtain
Fig. 8
1
∠AOQ = ∠AOC.
2
he
C
(1)
d
Q
1
∠P OD = ∠BOD.
2
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However, we have
(2)
∠P OQ = ∠P OD + ∠DOA + ∠AOQ (using rule 4)
1
= ∠DOA + (∠BOD + ∠AOC) (from (1) and (2))
2
1
= ∠DOA + × 2∠AOC (∠BOD = ∠AOC as vertical opposite angles)
2
= ∠DOA + ∠AOC
= 180◦
(using proposition 1).
It follows that P, O, Q are collinear.
Exercise 3.1.3
1. Draw diagrams illustrating each of the following situation:
No
t
to
(a) Three straight lines which do not pass through a fixed point.
(b) A point and rays emanating from that point such that the angle
between any two adjacent rays is an acute angle.
(c) Two angles which are not adjacent angles, but still supplementary.
(d) Three points in the plane which are equidistant from each other.
2. Recognise the type of angles in the following figures:
(i)
(ii)
(iii)
X
Y
Q
B
B
O
O
R
P
O
B
A
A
A
X
Unit 1
238
3. Find the value of x in each of the following diagrams:
(ii)
(i)
C
90
4x
x
D
x
B
O
A
(iii)
B
he
O
d
2x
A
C
(iv)
C
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C
x
90
A
B
x
x−y
A
x+y
O
B
D
(v)
(vi)
F
C
D
3x
x
A
O
F
x + 30
x
y
E
y
B
A
65
B
E
D
to
C
No
t
4. Which pair of angles are supplementary in the following diagram? Are
there supplementary rays?
C
B
120
50
A
O
130
D
Axioms
239
5. Suppose two adjacent angles are supplementary. Show that if one of
them is an obtuse angle, then the other angle must be acute.
3.1.4 Parallel lines and Euclid’s fifth postulate
←→
←→
d
Take any two distinct straight lines AB and CD. We first show that
there is at most one point common to them. Suppose the contrary and
assume that P and Q are two distinct points which lie on both the lines.
he
←→
But we know that by postulate 1 that there is a unique line P Q passing
←→
←→ ←→
through P and Q. Since P and Q are on AB, we must have AB=P Q. A
←→ ←→
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←→ ←→
similar argument shows that CD=P Q. Using Axiom 1, we obtain AB=CD,
←→
←→
which contradicts the assumption that AB and CD are distinct. Thus
given two distinct lines, either they do not have any point in common or
there is one point common to them. In the latter case, the two straight
line intersect at this common point.
We say two straight lines are parallel to each other, if either they are
←→
←→
to
identical or they do not intersect. Thus two distinct lines AB and CD are
parallel to each other if and only if they do not share any common point.
Let us get back to Euclid’s Postulate 5:
Postulate 5. If a straight line meets two other lines, so as to make
the two interior angles on one side of it together less than two right
angles, the other straight lines will meet if produced on that side on
which the angles are less than two right angles.
No
t
This is one of the most complicated postulate made by Euclid. In later
years, many attempts have been made to arrive at a simpler equivalent
versions of Euclid’s fifth postulate.
←→
←→
←→
Suppose AB and CD are two straight lines and let P Q be a line which
←→
←→
meets AB in L and CD in M.(see Fig. 9) If a line intersects two or more
←→
lines, it is called a transversal to those set of lines. Here P Q is a transversal
←→
←→
to AB and CD. There are eight angles formed by these three lines: ∠1, ∠2,
∠3, ∠4, ∠5, ∠6, ∠7 and ∠8 as shown in figure. In this ∠1, ∠4, ∠7 and ∠6 are
called exterior angles; ∠3, ∠2, ∠5 and ∠8 are called interior angles. The
angles ∠3 and ∠5 are called a pair interior alternate angles. Observe that
Unit 1
240
∠2 and ∠8 are also a pair of interior alternate angles. Similarly, the pairs
∠1, ∠7 and ∠4, ∠6 are called pairs of exterior alternate angles. Angles
∠1 and ∠5 are called a pair corresponding angles. There are three more
pairs of corresponding angles: ∠2, ∠6; ∠3, ∠7; and ∠4, ∠8.
1
3
2
B
he
4
L
A
8
d
Q
5
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C
7
M
6
P
D
Fig. 9
Look at ∠3 and ∠8. They are the interior angles on the same side of
←→
←→
←→
the line P Q. According Postulate 5, if ∠3 + ∠8 < 180◦ , then AB and CD
←→
meet on the left side of P Q. (If it happens that ∠2 + ∠5 < 180◦ , then they
←→
must meet on the right side of P Q.) Let us explore more on the condition
∠3 + ∠8 6= 180◦ . We have the following proposition.
to
Proposition 5. If a transversal cuts two parallel lines, then the sum
of two internal angles on the same side of the transversal is equal to
180◦ .
Given: a transversal intersecting two parallel lines.
No
t
To prove: the sum of internal angles on the same side of trnasversal is
equal to 180◦ .
Proof: Suppose the result is not true. (see fig 9.) If ∠3 + ∠8 6= 180◦ , then
←→
either you must have ∠3 + ∠8 < 180◦ or ∠3 + ∠8 > 180◦ . In the first case AB
←→
←→
and CD meet on the left side of P Q. Suppose ∠3 + ∠8 > 180◦ . We observe
that
∠3 + ∠8 + ∠2 + ∠5 = (∠3 + ∠2) + (∠8 + ∠5)
= ∠ALB + ∠CMD
= 180◦ + 180◦ = 360◦ .
Axioms
241
Thus
∠2 + ∠5 = 360◦ − (∠3 + ∠8) < 360◦ − 180◦ = 180◦ .
←→
←→
=⇒ ∠2 + ∠5 < 180◦ . Hence Postulate 5 tells us that AB and CD meet on the
←→
right side of P Q.
We conclude that: if the sum of the interior angles on the same side of
←→
←→
←→
←→
←→
←→
d
P Q is not equal to 180◦ , then AB and CD meet at some point(either to the
left or to the right of P Q). Thus if AB and CD are parallel, then the sum of
he
←→
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the interior angles on the same side of any transversal P Q is equal to 180◦ .
This completes the proof of the proposition.
Thus Postulate 5 implies that given a pair of parallel lines and a transversal, the sum of the internal angles on the same side of the transversal is
equal to 180◦ .
Is the converse true? Given two straight lines and a transversal such
that the sum of two internal angles on the same side of the transversal is
equal to 180◦ , does it follow that the two lines are parallel?. Here we make
use of a fairly simple equivalent version of parallel postulate of Euclid.
This was first given by a Scottish mathematician called Playfair.
Playfair’s postulate: Given a line in a plane and a point outside the
line in the same plane, there is a unique line passing through the
given point and parallel to the given line.
to
We have the following statement.
No
t
Proposition 6. If a transversal cuts two distinct straight lines in such
a way that the sum of two internal angles on the same side of the
transversal is equal to 180◦ , then the two lines are parallel to each
other.
Q
Y
L
A
B
S
M
C
X
P
Fig. 10
D
Unit 1
242
←→
←→
←→
Given: two straight lines AB and CD and a transversal P Q intersecting
←→
←→
AB in L and CD in M respectively; and ∠ALM + ∠LMC = 180◦ .
←→ ←→
To prove: ABkCD.
←→
←→
Proof: Assume that AB and CD are not parallel. Then they must meet at
some point, say S. (see Fig.10) By Playfair’s postulate, there is a unique
←→
←→
←→
d
line XY passing through S and parallel to P Q.
←→
he
Since XY k P Q, we have ∠QLS + ∠LSY = 180◦ (they are internal angles
←→
←→
←→
on the same side of the transversal SB to the parallel lines XY and P Q).
−→
But ∠QLS + ∠ALM = 180◦ ( they are adjacent angles formed by the ray LA
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←→
standing on the line P Q). Hence it follows that ∠LSY = ∠ALM. (Which
axioms are needed here?) But ∠ALM + ∠LMC = 180◦ (given data). We also
have ∠LMC + ∠MSY = 180◦ (since they are the sum of the internal angles
←→
−→
on the same side of the transversal SD cutting the parallel lines XY and
←→
P Q). Thus we get ∠ALM = ∠MSY . It now follows that ∠LSY = ∠MSY . But
←→
←→
∠MSY = ∠MSL+∠LSY . We obtain ∠MSL = 0. Hence SB and SD coincide.
←→
←→
This forces that the straight lines AB and CD are the same, contradicting
←→
←→
that they are distinct lines. We conclude that AB k CD.
Activity 6:
Draw two parallel lines. Draw a transversal and measure different angles
formed by intersections. You will see that:
to
1. Any pair of alternate angles are equal.
2. Any pair of corresponding angles are equal.
No
t
Repeat the same with different positions of the transversal. You will see
that the same results repeat. We formulate this as a theorem.
Theorem 1. If two parallel lines are cut by a transversal, then
(i) each pair of alternate angles are equal;
(ii) each pair of corresponding angles are equal.
←→
←→
←→
Given: AB and CD two distinct parallel lines and P Q a transversal inter←→
←→
secting AB in L and CD in M. (see Fig. 11)
To prove: ∠3 = ∠5 and ∠1 = ∠5.
Axioms
243
Proof: Since ∠3 and ∠8 are two
internal angles on the same side
Q
4
A
3
8
C
7
1
2
←→
of the transversal P Q cutting the
B
←→
L
5
6
←→
parallel lines AB and CD, we
know that
M
D
∠3 + ∠8 = 180◦ .
−−→
But ∠8 and ∠5 are the adjacent angles formed by the ray MP standing
d
Fig. 11
←→
on the line CD. Hence we also know that
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∠8 + ∠5 = 180◦ .
he
P
Comparing, we see that ∠3 = ∠5.
Again observe that ∠2 + ∠3 = 180◦ = ∠8 + ∠5. Using ∠3 = ∠5, we get
∠2 = ∠8.
We also observe that ∠1 = ∠3, since they are vertically opposite angles.
Using this with ∠3 = ∠5, we conclude that ∠1 = ∠5.
Thus the pair ∠1, ∠5 of corresponding angles are equal.
Similarly we can prove ∠2 = ∠6, ∠4 = ∠8 and ∠3 = ∠7, ∠1 = ∠7, ∠4 = ∠6.
No
t
to
Think it over!
There are two statements about parallel lines and a transversal:
(i) if a transversal cuts two distinct parallel lines, then any pair
of alternate angles are equal; (ii) if a transversal cuts two distinct parallel lines, then any pair of corresponding angles are
equal. But these two are not independent statements. You can
easily prove any of them assuming the other and using propositions 1 and 4.
What is the converse of theorem 1. If there are two distinct straight
lines and a transversal such that any pair of alternate angles are equal,
can we prove that the two lines are parallel to each other. Note that if we
are able to prove this result, you can also prove that: given two distinct
lines and a transversal such that any pair of corresponding angles are
equal, then the lines are parallel(use the previous observation). Thus we
have the following theorem.
Unit 1
244
Theorem 2. Suppose a transversal cuts two distinct straight lines
such that a pair of alternate angles are equal. Then the two lines are
parallel to each other.
←→
←→
←→
Given: two straight lines AB and CD and a transversal P Q (see Fig. 11)
and ∠3 = ∠5.
←→
←→
d
To prove: AB k CD.
←→
he
Proof: We know that ∠8 + ∠5 = 180◦ (since they are supplementary angles).
By the given hypothesis, we know that ∠3 = ∠5. We thus obtain ∠3 + ∠8 =
180◦ . However, ∠3 and ∠8 are the internal angles on the same side of the
←→
←→
←→
AB k CD.
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transversal to the lines AB and CD. By proposition 6, we conclude that
Corollary: If a transversal cuts a pair of straight lines in such a way
that a pair of corresponding angles are equal, then the two lines are
parallel to each other.
Proof: Referring to Fig. 11, we are given, say, ∠1 = ∠5. But ∠1 = ∠3, since
they are vertically opposite angles. We obtain ∠3 = ∠5. Hence by theorem
←→
←→
2, AB k CD.
Activity 7:
←→
←→
←→
←→
Draw two parallel lines AB and CD. Draw a line XY parallel to CD. Draw
←→
←→
←→
←→
to
a transversal P Q. Let it cut AB in L, CD in M and XY in N. Measure
∠BLQ and ∠Y NQ. Do you see that they measure the same. Repeat this
←→
with different positions of P Q.
No
t
Example 4. Two lines which are parallel to a common line are parallel to
each other.
←→
Q
L
A
M
C
B
D
Y
P
←→
to a common line CD. We show
←→ ←→
that AB k XY . Draw a transversal
←→
Fig. 12
←→
←→
P Q, cutting AB at L, CD at M and
←→
N
X
←→
Solution: Suppose AB and XY
are two lines which are parallel
XY at N, respectively. We observe
that
∠BLP = ∠DMP,
Axioms
245
←→
as they are corresponding angles made by transversal P Q with the parallel
←→
←→
lines AB and CD. Similarly,
∠DMP = ∠Y NP
(Why?).
Using Axiom 1, we obtain ∠BLP = ∠Y NP . Now we can use corollary to
←→
←→
←→
d
←→
theorem 2 and conclude that AB k XY .
←→
Example 5. Let AB be a straight line. Let CD and EF be two straight lines
←→
←→
he
←→
such that each of them is perpendicular to AB. Prove that CD k EF .
←→
←→
Solution: Let AB intersect CD
←→
and EF at L and M respectively.
F
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D
←→ ←→
L
A
M
C
B
E
Fig. 13
←→
←→
Since CD⊥AB, we have ∠DLA =
←→ ←→
90◦ . Using EF ⊥AB, we also get
∠F MA = 90◦ . Thus ∠DLA =
∠F MA.
But these are corresponding angles made by the
←→
transversal AB with the lines CD and EF . Hence by corollary to theorem
←→
←→
2, we conclude that CD k EF .
Example 6. Show that the angle bisectors of a pair of alternate angles
made by the transversal with two parallel lines are parallel to each other.
Solution:
We are given two
to
←→
←→
No
t
Q
L
R
A
X
C
P
B
Y
D
M
S
Fig. 14
←→
parallel lines AB and CD, and
transversal P Q. Consider the pair
of alternate angles ∠ALM and
−−→
∠LMD. Let LX be the bisector of
−−→
∠ALM; let MY be the bisector of
−−→
∠LMD. Extend the ray LX to the
←→
−−→
straight line XR and the ray MY
←→
to the straight line SY as shown
in the figure(see Fig. 14). We
−−→ −→
have to show that XRkSY . Con←→
←→
sider the lines XR and SY with
←→
transversal P Q.
Unit 1
246
We have
1
1
∠XLM = ∠ALM, ∠LMY = ∠LMD.
2
2
However, ∠ALM = ∠LMD(why?). We hence obtain ∠XLM = ∠LMY . But
∠XLM and ∠LMY are a pair of alternate angles made by the transversal
←→
←→
←→
←→
←→
P Q with the lines XR and SY . It follows, by theorem 2, that XR k SY .
d
Exercise 3.1.4
he
1. Find all the angles in the follow2. Find the value of x in the diagram
ing figure.
below.
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Q
130
L
A
B
x
M
C
D
90
135
P
3. Show that if a straight line is perpendicular to one of the two or more
parallel lines, then it is also perpendicular to the remaining lines.
←→
←→
←→
to
4. Let AB and CD be two parallel lines and P Q be a transversal. Show
that the angle bisectors of a pair of two internal angles on the same
side of the transversal are perpendicular to each other.
Additional problems on “Axioms, postulates and theorems
No
t
1. Choose the correct option:
(i) If a = 60 and b = a, then b = 60 by ————
A. Axiom 1 B. Axiom 2 C. Axiom 3 D. Axiom 4
(ii) Given a point on the plane, one can draw ————
A. unique B. two C. finite number of D. infinitely many
lines through that point.
(iii) Given two points in a plane, the number of lines which can be
drawn to pass through these two points is ————
A. zero B. exactly one C. at most one D. more than one
(iv) If two angles are supplementary, then their sum is ————
A. 90◦
B. 180◦
C. 270◦
D. 360◦
Axioms
247
(v) The measure of an angle which is 5 times its supplement is
————
A. 30◦
B. 60◦
C. 120◦
D. 150◦
2. What is the difference between a pair of supplementary angles and a
pair of complimentary angles?
he
d
3. What is the least number of non-collinear points required to determine a plane?
4. When do you say two angles are adjacent?
←→
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5. Let AB be a segment with C and D between them such that the order
of points on the segment is A, C, D, B. Suppose AD = BC. Prove that
AC = DB.
←→
←→
−−→
6. Let AB and CD be two straight lines intersecting at O. Let OX be the
−−→
−−→
−→
−−→ −−→
bisector of ∠BOD. Draw OY between OD and OA such that OY ⊥ OX.
−−→
Prove that OY bisects ∠DOA.
←→
←→
←→
7. Let AB and CD be two parallel lines and P Q be a transversal. Let P Q
←→
←→
intersect AB in L. Suppose the bisector of ∠ALP intersect CD in R
←→
and the bisector of ∠P LB intersect CD in S. Prove that
∠LRS + ∠RSL = 90◦ .
←→
←→
S
←→
to
8. In the adjoining figure, AB and CD
are parallel lines. The transversals
←→
P Q and RS intersect at U on the line
No
t
←→
AB. Given that ∠DW U = 110◦ and
∠CV P = 70◦ , find the measure of
∠QUS.
Q
x
U
A
B
110
C
V
70
W
D
P
R
9. What is the angle between the hour’s hand and minute’s hand of a
clock at (i) 1.40 hours, (ii) 2.15 hours? (Use 1◦ = 60 minutes.)
10. How much would hour’s hand have moved from its position at 12
noon when the time is 4.24 p.m.?
11. Let AB be a line segment and let C be the midpoint of AB. Extend AB
to D such that B lies between A and D. Prove that AD + BD = 2CD.
Unit 1
248
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←→
←→
−−→
12. Let AB and CD be two lines intersecting at a point O. Let OX be a
−−→
ray bisecting ∠BOD. Prove that the extension of OX to the left of O
bisects ∠AOC.
−−→
−→
−−→
13. Let OX be a ray and let OA and OB be two rays on the same side of
−−→
−→
−−→
−−→
−→
OX, with OA between OX and OB. Let OC be the bisector of ∠AOB.
Prove that
∠XOA + ∠XOB = 2∠XOC.
−→
−−→
−−→
−→
−−→
14. Let OA and OB be two rays and let OX be a ray between OA and OB
−→
such that ∠AOX > ∠XOB. Let OC be the bisector of ∠AOB. Prove
that
∠AOX − ∠XOB = 2∠COX.
−→ −−→ −→
−→
−→
−−→
15. Let OA, OB, OC be three rays such that OC lies between OA and OB.
Suppose the bisectors of ∠AOC and ∠COB are perpendicular to each
other. Prove that B, O, A are collinear.
←→
←→
16. In the adjoining figure, AB k DE.
Prove that
A
B
D
E
Glossary
to
∠ABC − ∠DCB + ∠CDE = 180◦ .
C
17. Consider two parallel lines and a transversal. Among the measures of
8 angles formed, how many distinct numbers are there?
No
t
Undefined objects: those objects in mathematics which cannot be defined using the terms already known.
Axioms: certain statements which are valid in all branches of mathematics whose validity is taken for granted without seeking mathematical
proofs.
Postulates: some statements which are taken for granted in a particular
branch of mathematics.
Hypothesis: certain conditions assumed while proving a proposition.
Adjacent angles: a pair of angles made by a ray standing on a line.
Complementary angles: a pair of angles which add up to 90◦ .
Supplementary angles: a pair of angles which add up to 180◦.
Axioms
249
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Straight angle: an angle formed by a straight line; equal to 180◦ .
Complete angle: an angle which measures 360◦ .
Reflex angle: an angle which measures more than 180◦ , but less than
360◦ .
Linear pair: a pair of angles which make a starlight line.
Vertically opposite angles: when two straight lines intersect each other,
a pair of angles which do not form a linear pair are vertically opposite angles.
Collinear: points all lying on the same straight line.
Parallel lines: a collection of lines which do not intersect pairwise.
Alternate angles: when a transversal cuts a pair of lines, the angles
formed by the transversal which is not a linear pair and lying on both
sides of the transversal.
Corresponding angles: when a transversal cuts a pair of lines, the angles
formed by the transversal which lie on the same side of the transversal
and also on the similar side of the two lines.
Points to remember
• Mathematics is a game in which certain objects are given and you
have to play the game according to certain pre-laid rules.
• In geometry, the objects are points, lines plane; the rules are axioms
and postulates of geometry.
No
t
to
• Euclid’s fifth postulate– equivalent formulation is that given any line
and a point out side that line, there is a unique line passing through
the given point and parallel to the given line– is the one which gives
Euclidean geometry. Changing this postulate will lead to different
geometry.
Purity, patience, and perseverance are three essentials to sucess and
above all, love.
—–Swamy Vivekananda
CHAPTER 3
UNIT 2
THEOREMS ON TRIANGLES
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After studying this unit, you learn to:
• identify a triangle in a collection of figures;
• classify different types of triangles based on sides and angles;
• recognise the angle sum property of a triangle;
• identify the interior and exterior angles of a triangle;
• establish the relationship between the exterior angle and interior opposite angles;
• prove logically angle sum property of the triangle;
• solve problems based on the angles of a triangle.
3.2.1 Introduction
No
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In the previous chapter you have studied th properties lines and angles.
You have seen how the axioms of Euclidean geometry helps you to build
some nice relations between angles and lines. In this chapter, you shall
study about a closed plane figure formed by three non parallel lines, a
triangle.
A plane figure bound by three non concurrent line segments in plane
is called a triangle.
This needs an explanation. When we say a plane figure, we actually
mean the linear figure, not the two-dimensional figure. Let A,B,C be three
points such that they are not on the same line; we say A, B, C are noncollinear. Join AB, BC and CA. You get a linear figure which consists
of three line segments which meet only at their end points. Such a linear
figure is called a triangle. We say A, B and C are the vertices of the
triangle ABC. The segments AB, BC, and CA are called the sides of the
triangle; and the angles ∠BAC, ∠ABC and ∠ACB are called the angles of
the triangle ABC (or the interior angles of ABC).
A triangle consists 9 elements:
A
Vertices Sides
Angles
A
AB
∠BAC or ∠A
B
BC
∠ABC or ∠B
C
AC
∠ACB or ∠C
B
C
Triangles
251
Note: When there is no confusion, the sides of a triangle ABC are
also denoted by AB, BC and AC. These symbols are also used to
denote their respective lengths. The context tells whether the side or
the length to be taken.
Points to ponder:
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(i) A triangle cut in a plane sheet of paper is a triangular sheet,
but not a triangle. Only the three line segments constitute a
triangle. (ii) Triangular sheet as a plane figure has only area,
but has no thickness.
Triangles are classified based on the measure of sides and angles.
classification based on sides:
(i) Equilateral triangle;
(ii) Isosceles triangle;
A
B
(iii) Scalene triangle.
A
A
B
C
Isosceles triangle:
A triangle in which
two sides are of equal
length is called an
isosceles triangle. In
the above triangle
ABC,
No
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Equilateral triangle:
A triangle in which
all sides are of equal
length is called an
equilateral triangle.
In triangle ABC,
AB = BC = CA.
C
B
C
Scalene triangle:
A triangle in which
all sides are of different lengths is called
a scalene triangle In
the above triangle
ABC,
AB = BC.
AB 6= BC 6= CA 6= AB.
Do you see that an equilateral triangle is also isosceles? But an
isosceles triangle need not be equilateral.
Unit 2
252
classification based on angles :
(i) Acute angled triangle;
triangle.
(ii) Right angled triangle;
(iii) Obtuse angled
A
A
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A
B
C
Acute angled triangle:
A triangle in which all
the angles are smaller
than 90◦ is called an
acute angled triangle. In
the above triangle ABC,
∠ABC < 90◦ , ∠BCA < 90◦ ,
∠CAB < 90◦ .
Activity 1:
B
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B
C
Right angled triangle: A triangle having an angle equal
to 90◦ is called a
right angled triangle.
In the above triangle
ABC,
Obtuse angled triangle: If a triangle
has an angle greater
than 90◦ , it is called
an obtuse angled triangle. In the above
triangle ABC,
∠ABC = 90◦ .
∠ABC > 90◦ .
to
Name the types of triangles given below :
100
(ii)
(iii)
No
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(i)
(iv)
C
(vi)
(v)
(ix)
(vii)
(viii)
130
Triangles
253
(x)
(xi)
(xii)
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Exercise 3.2.1
1. Match the following:
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(1)
(a) Equilateral triangle
(2)
(b) Acute angled triangle
(3)
(c) Right angled triangle
(d) Obtuse angled triangle
to
(4)
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2. Based on the sides, classify the following triangles (figures not drawn
to the scales):
5 cm
5 cm
7 cm
3 cm
3.5cm
4.5 cm
4 cm
4 cm
(i)
(ii)
6 cm
(iii)
Unit 2
254
6.5 cm
6.5 cm
5.6 cm
4.1 cm
2.5 cm
4.3 cm
(iv)
(v)
3.2 cm
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4 cm
d
3 cm
(vi)
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3 cm
5 cm
5 cm
9 cm
3 cm
3 cm
6 cm
6 cm
5 cm
6 cm
3.5 cm
(viii)
(vii)
8 cm
(x)
(ix)
3.2.2 Sum of interior angles
Let us do some paper activity before guessing a geometrical result.
Activity 2:
No
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Take a sheet of paper, fold it into four folds. On one of its folds, draw a
triangle using a scale and a pencil. Then cut the triangle from a pair of
scissors. Now you have four identical triangular sheets. Select three of
them and mark identical angles as 1, 2, and 3 on each sheet of the paper
as shown below.
2
1
3
2
2
1
3
1
3
Draw a straight line on a sheet of your note book. Arrange the triangles
such that angle 1 of the first triangle, 2 of the second and 3 of the third as
shown in the figure.
Triangles
255
2
3
12
3
1
2
1
3
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Activity 3:
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You can find that all the three angles together form a straight angle. But
you see that these three angles are precisely the angles of a triangle. Thus
you can guess that the sum of three angles of a triangle is 180◦ .
Draw a triangle ABC on a sheet as shown in the figure. Cut the remaining
part of the paper. Fold the triangular sheet such that the vertex A touches
the base line of the paper at M. Now fold the vertex B and C to meet the
point M. You will find that they make a straight angle. (See the figure
below.)
A
A
B
to
A
B
C
B
C
M
C
No
t
Draw right angled triangle such that angle ∠2 = 90◦ . (See the figure below.)
1
2
1
3
2
4
3
We want to find the sum of the three angles: ∠1+∠2+∠3. Draw the parallel
line to base of the triangle and passing through the vertex at the top.
Unit 2
256
Now we find another angle ∠4 equal to ∠3, because they are alternate
interior angles between two parallel lines. Therefore ∠1+∠3 = ∠1+∠4 = 90◦ .
This implies that
∠1 + ∠2 + ∠3 = ∠2 + (∠1 + ∠3) = 90◦ + 90◦ = 180◦ .
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Thus the sum of interior angles of a right triangle is 180◦ . Now we make
use of this to show that the sum of three interior angles of a triangle is
180◦ .
Take an arbitrary triangle. This
can be split into two right triangles, by drawing a perpendicular
to the base. We know that,
1
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2 4
3 5
6
∠1 + ∠2 + ∠3 = 180◦ ,
∠4 + ∠5 + ∠6 = 180◦ .
Adding these we obtain
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360◦ .
But angles ∠3 and ∠5 are supplementary angles and they make a straight
line. Therefore ∠3 + ∠5 = 180◦ . Thus we get
to
∠1 + ∠2 + ∠4 + ∠6 = 360◦ − (∠3 + ∠5) = 360◦ − 180◦ = 180◦ .
No
t
But can you see that ∠2+∠4 is the angle at one of the vertex of the triangle?
Hence the sum of the three angles of triangle is 180◦
If you observe the proof given above, it consists of two parts. In the first
part you prove that the sum of three interior angles of a right triangle is
180◦ , using a construction; drawing a line parallel to the base line through
the top vertex. A general triangle is split in to two right angled triangles
and we use the result for right triangle for getting on to a general triangle.
Can’t we construct a parallel to the base through the top vertex for a
general triangle and proceed with the proof? We take up this approach
below.
Theorem 1. In any triangle, the sum of the three interior angles is
180◦ . (Interior angle theorem)
Triangles
257
A
F
B
Given: ABC is a triangle.
To prove:
∠ABC + ∠BCA +
◦
∠CAB = 180 .
Construction: Through point A,
draw the line EF k BC.
C
d
E
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Proof: Below we give several statements and the reason for the truth of
each statement. Finally we arrive at the desired conclusion.
Statement
∠ABC = ∠EAB
∠BCA = ∠F AC
∠EAB + ∠BAC + ∠F AC = 180◦
Reason
alternate angles by the transversal AB
with the parallel lines BC and EF
alternate angles by the transversal AC
with the parallel lines
sum of the linear angles at A.
By substituting ∠EAB = ∠ABC and ∠BCA = ∠F AC, we finally get
∠ABC + ∠BAC + ∠BCA = 180◦ .
to
This completes the proof.
Example 1 In a triangle ABC, it is given that ∠B = 105◦ and ∠C = 50◦ .
Find ∠A.
No
t
A
105
B
Solution: We have, in triangle
ABC, (by Theorem 1),
∠A + ∠B + ∠C = 180◦
=⇒ ∠A + 105◦ + 50◦ = 180◦
=⇒ ∠A + 155◦ = 180◦
=⇒ ∠A = 180◦ − 155◦
=⇒ ∠A = 25◦ .
50
C
Thus ∠A measures 25◦ .
Example 2 In the given figure, find all the angles.
Unit 2
258
Solution: In triangle ABC, if we
make use of theorem 1, we get
∠A + ∠B + ∠C = 180◦ . Hence
A
5x + 3x + 2x = 180◦ =⇒ 10x = 180◦
5x
3x
Hence,
2x
∠A = 5x = 90◦ ;
∠B = 3x = 54◦ ;
∠C = 2x = 36◦ .
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=⇒ x = 180◦ /10 =⇒ x = 18◦ .
Example 3 If the bisectors of the angles ∠ABC and ∠ACB of a triangle
1
ABC meet at a point O, then Prove that ∠BOC = 90◦ + ∠BAC.
2
Solution:
Given: A triangle ABC and the bisectors of ∠ABC and ∠ACB meeting at
point O.
1
To prove: ∠BOC = 90◦ + ∠BAC.
2
Proof: In triangle BOC we have
A
∠1 + ∠2 + ∠BOC = 180◦
to
In triangle ABC, we have ∠A +
∠B + ∠C = 180◦ . Since BO and CO
are bisectors of ∠ABC and ∠ACB
respectively, we have
O
2
No
t
1
B
(1)
C
∠B = 2∠1 and ∠C = 2∠2.
We therefore get ∠A + 2(∠1) + 2(∠2) = 180◦ . Dividing by 2, we get
∠2 = 90◦ . This gives
∠1 + ∠2 = 90◦ −
∠A
2
From (1) and (2), we get
90◦ −
∠A
+ ∠BOC = 180◦ .
2
∠A
+ ∠1 +
2
(2)
Triangles
259
Hence
1
∠BOC = 90◦ + ∠BAC.
2
Exercise 3.2.2
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1. In a triangle ABC, if ∠A = 55◦ and ∠B = 40◦ , find ∠C.
2. In a right angled triangle, if one of the other two angles is 35◦ , find the
remaining angle.
3. If the vertex angle of an isosceles triangle is 50◦ , find the other angles.
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4. The angles of a triangle are in the ratio 1:2:3. Determine the three
angles.
A
5. In the adjacent triangle ABC, find
the value of x and calculate the measure of all the angles of the triangle.
x +15
x −15
x +30
B
C
6. The angles of a triangle are arranged in ascending order of their magnitude. If the difference between two consecutive angles is 10◦ , find
the three angles.
No
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A
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3.2.3 Exterior angles
B
C
D
Consider a triangle ABC. If the
side BC is produced externally
−−→
to form a ray BD, then ∠ACD is
called an exterior angle of triangle ABC at C and is denoted by
Ext∠C.
Note: If you produce AC to a point E (instead of BC), you get an
angle ∠BCE. But ∠ACD = ∠BCE as they are vertically opposite
angles. Thus Ext∠C is the same whether you use the side BC
or the side AC; it depends only on ∠C of the triangle ABC.
Unit 2
260
With respect to Ext∠C of triangle ABC, ∠A and ∠B are called interior
opposite angles.
E
Now in triangle ABC side CA,
BC, and AB are produced to form
−
−→ −−→
−→
rays CE, BD and AF .
Then
∠BAE, ∠ACD and ∠CBF are exterior angles of the triangle ABC.
D
C
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B
d
A
F
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Activity 4: Take three sheets of
paper of size 8 × 10 cm. Place
3
one above the other and cut three
right angled triangles such that
one of the corners of each sheet
becomes the right angle of the tri1
2
B
C
angle. Now you will have three
identical triangles. Mark the angles of each triangle as 1,2,3 as shown in
the figure.
A
A
1
2
B
C
Q
No
t
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to
3
A
2
1
3
3
P
1
B
2
3
1
C
2
Q
Draw straight line P Q on another sheet of paper and place
one of the triangular sheet on the
line as shown in figure such that
∠ACQ forms an exterior angle of
the triangle ABC.
Now place the remaining two triangular sheets with angles 3 and
1 as shown in the adjacent figure. Can you see that ∠ACQ =
∠3 + ∠1? Thus, the measure of
an exterior angle of a triangle
is equal to the sum of the corresponding two interior opposite
angles.
Triangles
261
Theorem 2. If a side of triangle is produced, the exterior angle so
formed is equal to the sum of the corresponding interior opposite
angles. (Exterior angle theorem.)
Given: In triangle P QR, produce
QR to S. Then ∠P RS is an exterior angle and the corresponding interior opposite angles are
∠P QR and ∠QP R.
To prove:∠P RS = ∠QP R+∠P QR.
S
Q
Proof:
R
Statement
Reason
◦
∠QP R + ∠P QR + ∠P RQ = 180 ;
interior angle theorem;
◦
∠P RQ + ∠P RS = 180 ;
linear pair
∠QP R + ∠P QR + ∠P RQ = ∠P RQ + ∠P RS; Axiom 1(see Chap 3, unit 1);
∠QP R + ∠P QR = ∠P RS;
Axiom 3(see Chap 3, unit 1).
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P
This completes the proof.
Example 4 An exterior angle of a triangle is 100◦ and one of the interior
opposite angles is 45◦ . Find the other two angles of the triangle.
No
t
A
to
Solution: Let ABC be a triangle whose side BC is produced to form an
exterior angle ∠ACD such that Ext∠C = 100◦ . Let ∠B = 45◦ . By exterior
angle theorem we have
100
B
C
D
∠ACD = ∠B + ∠A
=⇒ 100◦ = 45◦ + ∠A
=⇒ ∠A = 100◦ − 45◦
Hence ∠A = 55◦ . Thus we get
∠C = 180◦ − (∠A + ∠B)
= 180◦ − (55◦ + 45◦ ) = 80◦ . Hence
∠C = 80◦ .
Example 5 In the given figure, sides QP and RQ of a triangle P QR are
produced to the points S and T respectively. If ∠SP R = 135◦ and ∠P QT =
110◦ , find ∠P RQ.
Unit 2
262
Solution: Since Q, P and S all lie
on the same line,
∠QP R + ∠SP R = 180◦ .
Hence
∠QP R + 135◦ = 180◦
or
∠QP R = 180◦ − 135◦ = 45◦ .
Using exterior angle property in
triangle P QR, we have
S
135
T
d
110
R
Q
∠P QT = ∠QP R + ∠P RQ.
110◦ = 45◦ + ∠P RQ.
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This gives
Solving for ∠P RQ, we get ∠P RQ = 110◦ − 45◦ = 65◦ .
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Example 6 The side BC of a triangle ABC is produced on both sides. Show
that the sum of the exterior angles so formed is greater than ∠A by two
right angles.
A
1
4 2
B
3
5
C
to
E
F
Solution: Draw a triangle ABC
and produce BC on both sides to
points D and F . Denote the angles as shown in the figure. We
have to show that
∠4 + ∠5 = ∠1 + 180◦ .
By exterior angle theorem, we
have
∠4 = ∠1 + ∠3 and ∠5 = ∠1 + ∠2.
No
t
Adding these two we get
∠4 + ∠5 = (∠1 + ∠3) + (∠1 + ∠2) = ∠1 + (∠1 + ∠2 + ∠3) = ∠1 + 180◦ ,
since the sum of all the interior angles of a triangle is 180◦ .
Exercise 3.2.3
1. The exterior angles obtained on producing the base of a triangle both
ways are 104◦ and 136◦ . Find the angles of the triangle.
2. Sides BC, CA and AB of a triangle ABC are produced in an order,
forming exterior angles ∠ACD, ∠BAE and ∠CBF . Show that ∠ACD +
∠BAE + ∠CBF = 360◦ .
Triangles
263
3. Compute the value of x in each of the following figures:
(i)
(ii)
(iii)
65
130
(iv)
106
x
(v)
120
100
x
d
x
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50
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20
x
112
x
P
4. In the figure, QT ⊥ P R, ∠T QR = 40◦
and ∠SP R = 30◦ . Find ∠T RS and
∠P SQ.
30
T
40
Q
S
R
5. An exterior angle of a triangle is 120◦ and one of the interior opposite
angles is 30◦ . Find the other angles of the triangle.
No
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6. Find the sum of all the angles at the
five vertices of the adjoining star.
Additional problems on “Theorems on triangles”
1. Fill up the blanks to make the following statements true:
(a) Sum of the angles of a triangle is————
(b) An exterior angle of a triangle is equal to the sum of ——————
——— opposite angles.
(c) An exterior angle of a triangle is always ————— than either of
the interior opposite angles.
Unit 2
264
(d) A triangle cannot have more than ————-right angle.
(e) A triangle cannot have more than———————– obtuse angle.
2. Choose the correct answer from the given alternatives:
(a) In a triangle ABC, ∠A = 80◦ and AB = AC, then ∠B is ———
B. 60◦
C. 40◦
D. 70◦
d
A. 50◦
A. 65◦
B. 55◦
C. 75◦
D. 45◦
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(b) In right angled triangle, ∠A is right angle and ∠B = 35◦ , then ∠C
is ———
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(c) In a triangle ABC, ∠B = ∠C = 45◦ , then the triangle is ———
A. right triangle B. acute angled triangle C. obtuse angle
triangle D. equilateral triangle
(d) In an equilateral triangle, each exterior angle is ———
A. 60◦
B. 90◦
C. 120◦
D. 150◦
(e) Sum of the three exterior angles of a triangle is ———
A. two right angles B. three right angles C. one right angle
D. four right angles
3. In a triangle ABC, ∠B = 70◦ . Find ∠A + ∠C.
to
4. In a triangle ABC, ∠A = 110◦ and AB = AC. Find ∠B and ∠C.
5. If three angles of a triangle are in the ratio 2: 3: 5, determine three
angles.
No
t
6. The angles of triangle are arranged in ascending order of magnitude.
If the difference between two consecutive angles is 15◦ , find the three
angles.
7. The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.
8. In a triangle ABC, if 2∠A = 3∠B = 6∠C, determine ∠A, ∠B and ∠C.
1
9. The angles of triangle are x − 40◦ , x − 20◦ and x + 15◦ . Find the value
2
of x.
Triangles
265
10. In triangle ABC, ∠A − ∠B = 15◦ and ∠B − ∠C = 30◦ , find ∠A, ∠B and
∠C.
11. The sum of two angles of a triangle is 80◦ and their difference is 20◦ .
Find the angles of the triangle.
d
12. In a triangle ABC, ∠B = 60◦ and ∠C = 80◦ . Suppose the bisector of
∠B and ∠C meet at I. Find ∠BIC.
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13. In a triangle, each of the smaller angles is half the largest angle. Find
the angles.
14. In a triangle, each of the bigger angles is twice the third angle. Find
the angles.
15. In a triangle ABC, ∠B = 50◦ and ∠A = 60◦ . Suppose BC is extended
to D. Find ∠ACD.
16. In an isosceles triangle, the vertex angle is twice the sum of the base
angles. Find the angles of the triangle.
Glossary
to
Exterior angle: the angle subtended externally when a side of a triangle
is extended.
Interior opposite angles: angles opposite to the interior angle which is
supplementary to the exterior angle.
Points to remember
No
t
• Triangles are classified according to their angles and their sides.
• The sum of the three interior angles of a triangle is 180◦ .
• An exterior angle of a triangle is the sum of two interior opposite
angles.
CHAPTER 3
UNIT 3
CONGRUENCY OF TRIANGLES
After studying this unit, you learn to:
d
• identify the congruent figures;
• identify the congruent triangles;
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• state the postulates for congruency of triangles;
he
• identify the corresponding sides and corresponding angles of congruent triangles;
• understand that particular triples of elements determine the congruency of triangles;
• deduce logical methods for proving theorems;
• solve problems based on different postulates of congruency;
• appreciate the use of congruency of triangles in solving practical day
to day problems.
3.3.1 Introduction
No
t
to
Suppose you have two equilateral triangles, each of side length, say 1
cm. Can you place one on the other so that all the three sides collace?
That is all the three sides of one triangle sit exactly on three sides of the
other triangle. Take equilateral triangles of side lengths 1 cm and 2 cm.
Can you put one on the other so that one exactly fits the other? No matter
how you adjust them, you see that it is impossible to superpose one on
the other. The exact geometrical idea which needed to understand these
is congruency.
Congruency is one of the fundamental concepts in geometry. This concept is used to classify the geometrical figures on the basis of their shapes.
Two geometrical figures are said to be congruent, if they have same shape
and size. For example:
1. Two line segments are congruent
if they have same length.
A
6 cm
B
C
6 cm
D
Congruency
267
P
2. Two angles are congruent if they
have same measure.
X
28 o
28
R
3. Two circles are congruent if they
have same radii.
Y
S
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Q
2 cm
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2 cm
M
Z
d
Q
o
T
4. Two squares are congruent if they
have sides of same length.
N 2.2 cm P
V 2.2 cm U
Note: 1. Two geometrical congruent figures can be made to superimpose
one on the other so that one exactly covers the other.
2. Two congruent geometrical figures have same parameters.
Congruency of triangles
No
t
A
to
Two triangles are said to be congruent if all the sides and angles of one
triangle are equal to the corresponding sides and angles of the other
triangle.
B
C
E
In triangles ABC and DEF , you
observe AB = DE, AC = DF and
BC = EF ; ∠A = ∠D, ∠B = ∠E,
and ∠C = ∠F .
D
F
Therefore, triangles
DEF are congruent.
ABC
and
We write this as
∆ABC ∼
= ∆DEF.
A few words about the use of this notation. When we write
∆ABC ∼
= ∆DEF , the important thing is to observe that the vertices A, B, C
correspond to the vertices D, E, F in that order. If we write ∆ABC ∼
= ∆EF D,
Unit 3
268
this gives a different meaning. This means ∠A = ∠E, ∠B = ∠F and
∠C = ∠D; and AB = EF , BC = F D and CA = DE. Can you see the
difference? So while using the notation for congruency, keep the order of
the vertices in proper way.
Corresponding sides and angles
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Recall that there are six elements associated with a triangle. Two triangles are congruent if and only if these six elements match in a suitable
sense. This will be made clear by understanding corresponding sides and
angles.
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Let us say that, on superposition, triangle ABC covers triangle DEF
exactly in such way that
1. ∠A = ∠D, ∠B = ∠E and ∠C = ∠F ;
2. AB = DE, BC = EF , AC = DF .
to
Then ∆ABC ∼
= ∆DEF . Angles which coincide on superposition are called
corresponding angles. Sides which coincide on superposition are called
corresponding sides. Generally it is not always possible to superimpose
one triangle over other triangle to know which angles are the corresponding angles and which are the corresponding sides. Can you see that if two
triangles are congruent, then there are corresponding vertices as well.
However, vertices are points and do not have any numerical quantities associated with them. Hence we use only sides and angles for determining
congruency.
No
t
Geometrically, in two congruent triangles ABC and DEF , angles opposite to equal sides are corresponding angles and so they are equal. Therefore, BC = EF implies that angle opposite BC= angle opposite to EF .
Symbolically BC = EF =⇒ ∠A = ∠D. Similarly, AC = DF =⇒ ∠B = ∠E
and AB = DE =⇒ ∠C = ∠F .
Exercise 3.3.1
1. Identify the corresponding sides and corresponding angles in the following congruent triangles:
Congruency
269
(b)
A
R
B
C
d
(a)
P
Q
he
2. Pair of congruent triangles and incomplete statements related to them
are given below. Observe the figures carefully and fill up the blanks:
C
D
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(a) In the adjoining figure
if ∠C = ∠F , then AB = − −
− − − and BC = − − − − −.
E
F
A
B
A
D
(b) In the adjoining figure
if BC = EF , then ∠C = − −
− − − and ∠A = − − − − −.
B
No
t
E
F
A
D
to
(c) In the adjoining figure,
if AC = CE and ∆ABC ∼
= ∆DEC,
then ∠D = − − − − − and ∠A =
− − − − −.
C
C
B
E
3.3.2 SAS postulate for the congruency of triangles
Least number of conditions for congruency
Now we know that two triangles are congruent if and only if the corresponding sides and corresponding angles of two triangles are equal (six
elements), i.e., three for corresponding angles and three for corresponding
sides. Now the natural question is; what is the least number of conditions
required to ensure the congruency of two triangles? Do we need the corresponding equality of all the six elements or a lesser number of conditions
Unit 3
270
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Side-angle-side postulate [SAS postulate]
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will suffice to ensure the congruency of two triangles? In this and subsequent sections, we shall see that if three properly chosen elements out of
six of a triangle are equal to the corresponding there elements of another
triangle, then other three elements of two triangles coincide in order and
we have the congruency of two triangles. Let us discuss those three conditions which will ensure the congruency of two triangles. One has to be
careful in choosing three conditions. For example three angles will not do.
(Draw two non-congruent triangles which have same set of angles.)
If the two sides and included angle of one triangle are equal to the
corresponding two sides and the included angle of the other triangle,
then the two triangles are congruent.
A
In triangles ABC and DEF , you
D
observe that AB = DE, AC = DF
and ∠A = ∠D. Hence SAS postulate tells
B
C
E
F
∆ABC ∼
= ∆DEF.
Look at the following triangles P QR and XY Z.
You
60
observe that P Q = XY and
QR = Y Z. Moreover ∠P =
60
60◦ = ∠Y . Still the triangles
Q
R Y
Z
P QR and XY Z need not be
congruent because the included angles ∠Q and ∠Y , which are corresponding angles, need not be equal.
X
No
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P
Example 1. In the figure O is the midpoint of AB and CD. Prove that
(i) ∆AOC ∼
= ∆BOD; (ii) AC = BD.
Congruency
271
Solution: In triangles AOC and BOD, we
have
A
AO = BO,
(O, the midpoint of AB);
∠AOC = ∠BOD, (vertically opposite angles);
O
CO = OD,
(O, the midpoint of CD).
So by SAS postulate we have
B
C
∆AOC ∼
= ∆BOD.
Hence AC = BD, as they are corresponding parts of congruent triangles.
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Example 2. In the figure, it is given that AE = AD and BD = CE. Prove
that ∆AEB is congruent ∆ADC.
A
E
C
D
B
By SAS postulate ∆AEB ∼
= ∆ADC.
Solution: We have AE = AD
and CE = BD.
Adding, we
get AE + CE = AD + BD
=⇒ AC = AB.
In triangles AEB and ADC, we have
AE = AD,
(given);
AB = AC,
(proved);
∠EAB = ∠DAC, (common angle).
Example 3. In a quadrilateral ACBD, AC = AD and AB bisect ∠A . Show
that ∆ABC is congruent to ∆ABD.
to
C
No
t
A
B
D
(Can you see that BA bisects ∠CBD?)
Solution: In triangles ABC and
ABD, we have
AC = AD,
(given);
∠CAB = ∠DAB, (AB bisects ∠A);
AB = AB,
(common side).
Hence
∆ABC ∼
= ∆ABD.
Exercise 3.3.2
P
1. In the adjoining figure P QRS is a
rectangle. Identify the congruent triangles formed by the diagonals.
Q
O
S
R
Unit 3
272
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2. In the figure ABCD is a square,
M, N, O and P are the midpoints of
sides AB, BC, CD and DA respectively. Identify the congruent triangles.
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3. In a triangle ABC, AB = AC. Points E on AB and D on AC are such
that AE = AD. Prove that triangles BCD and CBE are congruent.
E
4. In the adjoining figure, the sides BA
D
and CA have been produced such
A
that BA = AD and CA = AE. Prove
that DE k BC. [ hint:-use the concept of alternate angles.]
B
C
3.3.3 Consequences of SAS postulate
Now you have learnt how to compare two triangles using SAS condition.
This comparison will lead to some very interesting consequences about the
properties of triangles. We study a few of them here.
to
Theorem 1. In a triangle, the angles opposite to equal sides are equal.
No
t
A
B
D
Given: a triangle ABC in which
AB = AC.
To prove: ∠C = ∠B.
C
Proof:
Statement
AB = AC;
AD = AD;
∠BAD = ∠CAD;
Construction: Draw the angle bisector of ∠A. Let it cut BC at D.
Let us compare triangles ABD and
ACD:
Reasons
given;
common side;
by construction.
Congruency
273
We can use SAS postulate to conclude that ∆ADB ∼
= ∆ADC. Hence
∠ABC = ∠ACB, since these are corresponding angles of congruent triangles. Thus the theorem is proved.
Corollary 1: In an isosceles triangle, the angle bisector of the apex angle
is the perpendicular bisector of the base.
Proof:
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Given: ABC is a triangle in which AB = AC and apex angle ∠A.
Construction: draw the angle bisector AD from A on BC. (Refer to the
figure of the previous theorem.)
To prove: AD is the perpendicular bisector of BC. Equivalently, we have
to show that BD = DC and ∠ADB = ∠ADC = 90◦ .
∆ADB ∼
= ∆ADC, by (theorem 1)
=⇒ DB = DC and ∠ADB = ∠ADC.
But ∠ADB + ∠ADC = 180◦ (linear pair)
=⇒ ∠ADB + ∠ADC = 180◦
=⇒ 2∠ADB = 180◦
=⇒ ∠ADB = 180◦ /2 = 90◦ .
to
=⇒ ∠ADC = 180◦ − ∠ADB = 90◦ .
No
t
We also observe that ∆ABD ∼
= ∆ACD by SAS postulate. Hence
BD = CD.
This completes the proof.
Now you may wonder whether the converse of the theorem is true: in
any triangle, the sides opposite to the equal angles are equal. It is also
true, but its proof needs a different condition of congruency, which you
will study later.
Example 4. In the figure AB = AC and DB = DC. Prove that
∠ABD = ∠ACD.
Unit 3
274
A
Solution: In ∆ABC, we have
AB = AC =⇒ ∠ABC = ∠ACB
(angles opposite to equal sides).
D
B
C
d
Again in ∆DBC, we have
he
DB = DC (given) =⇒ ∠DBC = ∠DCB (angles opposite to equal sides).
Hence we obtain
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This gives
∠ABC − ∠DBC = ∠ACB − ∠DCB.
∠ABD = ∠ACD.
Exercise 3.3.3
1. In a ∆ABC, AB = AC and ∠A = 50◦ . Find ∠B and ∠C.
2. In ∆ABC, AB = BC and ∠B = 64◦ . Find ∠C.
3. In each of the following figure, find the value of x:
A
A
to
40
65
x
B
C
30
x
D
C
B
No
t
AB=AC
55
D
AC=CD
A
A
x
x
75
B
50
C
AB=AC
B
D
C
BD=DC=AD
4. Suppose ABC is an equilateral triangle. Its base BC is produced to D
such that BC = CD. Calculate (i) ∠ACD and (ii) ∠ADC.
Congruency
275
5. Show that the perpendiculars drawn from the vertices of the base of
an isosceles triangle to the opposite sides are equal.
6. Prove that a ∆ABC is an isosceles triangle if the altitude AD from A
on BC bisects BC.
he
3.3.4 ASA postulate for congruency
d
7. Suppose a triangle is equilateral. Prove that it is equiangular.
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If two angles and the common side of one triangle are equal to the
corresponding two angles and the common side of the other triangle,
then the two triangles are congruent.
Given two triangles ABC
and DEF such that
∠B = ∠E, ∠C = ∠F and
BC = EF , the ASA postulates tells
B
C
E
F
∆ABC ∼
= ∆DEF .
Earlier you have seen that the angles opposite to equal sides of a triangle are equal. Now we are ready to prove the converse of this result.
A
D
Theorem 2. If in a triangle two angles are equal, then the sides opposite to them are equal. (Converse of Theorem 1.)
No
t
to
A
B
D
Given: triangle ABC in which
∠B = ∠C.
To prove: AC = AB.
Construction: draw AD ⊥ BC.
C
Proof: Then ∠ADB = ∠ADC = 90◦ . We are given ∠DBA = ∠DCA. Consider
triangles ADB and ADC. We have
∠ADB + ∠DBA + ∠BAD = 180◦ = ∠ADC + ∠DCA + ∠CAD.
It follows that ∠BAD = ∠CAD(why?). Consider triangles ADB and ADC.
We have
Unit 3
276
∠BAD = ∠CAD
∠ADB = ∠ADC
AD = AD
(proved);
(both are right angles);
(common side.)
d
=⇒ ∆ADB ∼
= ∆ADC, by ASA condition. We conclude that AB = AC, by
property of congruency. This completes the proof of the converse of theorem 1.
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Example 5. In a triangle ABC, AB = AC and the bisectors of angles B
and C intersect at O. Prove that BO = CO and AO is the bisector of angle
∠BAC.
A
Solution: Since the angles opposite to equal sides are equal,
AB = AC
=⇒ ∠C = ∠B
∠B
∠C
=⇒
=
.
2
2
B
C
Since BO and CO are bisectors of ∠B and ∠C, we also have
∠ABO = ∠B/2 and ∠ACO = ∠C/2.
Hence
∠C
∠B
∠ABO =
=
= ∠ACO.
2
2
Consider ∆BCO:
to
O
No
t
∠OBC = ∠OCB
=⇒ BO = CO,
(Why?)
(Theorem 2).
Finally, consider triangles ABO and ACO.
BA = CA
BO = CO
∠ABO = ∠ACO
(given);
(proved);
(proved).
Hence by SAS postulate
∆ABO ∼
= ∆ACO
=⇒ ∠BAO = ∠CAO =⇒ AO bisects ∠A.
Congruency
277
Example 6. Diagonal AC of a quadrilateral ABCD bisects the angles ∠A
and ∠C. Prove that AB = AD and CB = CD.
Solution:
Since diagonal AC
bisects the angles ∠A and ∠C,
we have ∠BAC = ∠DAC and
∠BCA = ∠DCA. In triangles ABC
and ADC, we have
∠BAC = ∠DAC (given);
∠BCA = ∠DCA (given);
AC = AC
(common side).
B
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C
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D
So, by ASA postulate, we have
∆BAC ∼
= ∆DAC
=⇒ BA = AD and CB = CD (Corresponding parts of congruent triangle).
Example 7. Two parallel lines l and m are intersected by another pair of
parallel lines p and q as in the figure. Show that triangles ABC and CDA
are congruent.
p
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q
A
D
l
4
No
t
3
2
1
B
C
m
Solution: Since l and m are parallel lines and AC is a transversal, ∠1 = ∠4. Similarly transversal
AC cuts parallel lines p and q, so
that ∠2 = ∠3. In triangles ABC
and CDA, we have
∠1 = ∠4 (proved);
∠2 = ∠3 (proved);
AC = AC (common side).
By ASA postulate,
∆ABC ∼
= ∆CDA.
Example 8. In the figure, ∠BCD = ∠ADC and ∠ACB = ∠BDA. Prove that
AD = BC and ∠A = ∠B.
Unit 3
278
1
2
3
C
4
D
d
B
he
A
Solution: We have ∠1 = ∠2 and
∠3 = ∠4
=⇒ ∠1 + ∠3 = ∠2 + ∠4
=⇒ ∠ACD = ∠BDC.
Thus in triangles ACD and BDC,
we have,
∠ADC = ∠BCD (given);
CD = CD
(common);
∠ACD = ∠BDC (proved.)
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By ASA condition ∆ACD ∼
= ∆BDC. Therefore
AD = BC and ∠A = ∠B.
Think it over!
We have taken ASA condition as a postulate. But actually, it
can be proved as a theorem based on SAS postulate. Try to construct a proof. In deductive geometry, one takes only a minimum number of postulates and try to construct as many results
as possible using these minimum number of postulates. One
can also take ASA as a postulate and obtain SAS condition as a
theorem.
to
Exercise 3.3.4
D
C
No
t
1. In the given figure, If AB k DC and P
is the midpoint of BD, prove that P
is also the midpoint of AC.
P
A
B
A
2. In the adjacent figure, CD and BE
are altitudes of an isosceles triangle
ABC with AC = AB. Prove that AE =
AD.
D
B
E
C
Congruency
279
Q
3. In figure, AP and BQ are perpendiculars to the line segment AB and
AP = BQ. Prove that O is the midpoint of line segment AB as well as
P Q.
A
B
O
d
P
he
4. Suppose ABC is an isosceles triangle with AB = AC; BD and CE are
bisectors of ∠B and ∠C. Prove that BD = CE.
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5. Suppose ABC is an equiangular triangle. Prove that it is equilateral. (You have seen earlier that an equilateral triangle is equiangular.
Thus for triangles equiangularity is equivalent to equilaterality.)
3.3.5 SSS postulate for congruency
We will study one more condition for congruency of two triangles. It
depends only on the sides.
If three sides of one triangle are equal to the three corresponding
sides of the other triangle, then the triangles are congruent.
Activity 1:
No
t
to
Take three sheets of papers; one in the shape of a square, another rectangular and a third-one in the shape of a parallelogram (for this you may
have to draw a parallelogram on a sheet of paper and cut it along the
boundary). Draw diagonals as shown in the figure. Cut the sheets along
the diagonals.
A
D
B
C
1
P
S
W
Q
2
R
Z
X
3
Y
You will get two triangles from each sheet. Now you place one of the
triangular sheet obtained from each figure on the other triangular sheet
Unit 3
280
Think it over!
he
d
of the same figure such that it exactly covers the other triangular sheet.
You will notice that each pair of triangles are congruent and in each case
three sides of one triangle are equal to the corresponding sides of the other
triangle.
Now we look for the converse of this. If three sides of one triangle are
equal to the three corresponding sides of another triangle, can we put one
on the other such that each covers the other exactly? SSS congruency
condition says that it is indeed the case.
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Since there is SSS congruence postulate, can we have AAA
postulate and SSA postulate?
Example 9. In the figure, it is given that AB = CD and AD = BC. Prove
that triangles ADC and CBA are congruent.
Solution: In triangles ADC and
D
B
CBA, we have
AB = CD
AD = BC
AC = AC
C
A
∼
By SSS congruency condition, ∆ADC = ∆CBA.
(given);
(given);
(common side.)
No
t
to
Example 10. In the figure AD = BC and BD = CA. Prove that
∠ADB = ∠BCA and ∠DAB = ∠CBA.
Solution: In triangles ABD and
A
B
BAC, we have
AD = BC given;
AB = AB (common);
AC = BD (given.)
C
D
We can use SSS condition to conclude that ∆ABD ∼
= ∆BAC. From this we
conclude that
∠ADB = ∠BCA and ∠DAB = ∠CBA.
Example 11. In the adjoining figure, AB = CD and AD = BC. Show that
∠1 = ∠2.
Congruency
281
A
B
2
C
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D
d
1
Solution: In triangles ABD and
CDB, we have
AB = CD (given);
AD = BC (given);
BD = DB (common side.)
Hence ∆ABD ∼
= ∆CDB, by SSS
postulate. Comparing the angles,
we get
∠1 = ∠2.
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(Later you will see that, under the given conditions, ABCD is a parallelogram so that AD k BC, and ∠1 and ∠2 are alternate angles formed by the
transversal BD.)
Think it over!
We have taken SSS condition as a postulate. But this is
also a consequence of SAS postulate. You can get SSS as a
theorem from SAS postulate.
Exercise 3.3.5
to
1. In a triangle ABC, AC = AB and the altitude AD bisects BC. Prove
that ∆ADC ∼
= ∆ADB.
2. In a square P QRS, diagonals bisect each other at O. Prove that
∆P OQ ∼
= ∆QOR ∼
= ∆ROS ∼
= ∆SOP .
No
t
A
3. In the figure, two sides AB, BC and
the median AD of ∆ABC are respectively equal to two sides P Q, QR and
median P S of ∆P QR. Prove that
(i) ∆ADB ∼
= ∆P SQ;
(ii) ∆ADC ∼
= ∆P SR.
Does it follow that triangles ABC and
P QR are congruent?
B
Q
P
D
C
S
R
Unit 3
282
Q
4. In △P QR, P Q = QR; L, M and N
are the midpoints of the sides of P Q,
QR and RP respectively. Prove that
LN = MN.
L
N
R
d
P
M
3.3.6 RHS theorem
he
A
D
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A
B
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A
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Activity 2:
Draw an equilateral triangle ABC
on a sheet of paper. From the vertex A, draw the perpendicular AD
to the base the BC. Cut the sheet
of paper along the triangle. Now
B
D
C
fold it along the perpendicular line. You will notice that two right angled
triangles superpose one on the other. So the two triangles are congruent.
C
B
Activity 3:
Take a square sheet of paper. Fold
the sheet of the paper along one of
its diagonal. You will notice that
two triangles so formed by folding the sheet are right angled triangles and they superpose one on
the other.
Activity 4:
Take a rectangular sheet of paper,
such that one of its side length is
equal to the length of the square
you had taken earlier and the
D
C
other is different from the length
of the square . Draw one of its diagonals. Cut the sheet along the diagonal,
to obtain two right triangles. Can you see that here also you can superpose
one right triangle on the other?
Now place one of the triangular sheets from the square and another
from the rectangular sheet. You will notice that even though the two tri-
Congruency
283
d
angles are right angled and one set of corresponding sides are equal, they
do not superpose one on the other.
In SAS postulate, You have noticed earlier that you need the equality of
two corresponding sides and the included angle for congruency. However
for a right angled triangle, you can relax this a bit to get what is known as
RHS condition. We have the following theorem on right angled triangles.
D
E
F
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A
he
Theorem 3. Two right-angled triangles are congruent if the hypotenuse
and a side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle. (RHS theorem.)
B
C
G
to
Given: two right-angled triangles ABC and DEF such that
(i) ∠B = ∠E = 90◦ ;
(ii) Hypotenuse AC = Hypotenuse DF ; and
(iii) AB = DE.
To prove: ∆ABC ∼
= ∆DEF .
Construction: Produce F E to G so that EG = BC. Join DG.
No
t
Proof: In triangles ABC and DEG, observe that
AB = DE
(given);
BC = EG
(by construction);
∠ABC = ∠DEG (each equal to 90◦ .)
Hence by SAS, ∆ABC ∼
= ∆DEG =⇒ ∠ACB = ∠DGE and AC = DG. But
AC = DF , by the given hypothesis. We thus get
DG = AC = DF .
In triangle DGF , we have got DG = DF (just proved). This implies that
∠G = ∠F (angles opposite to equal sides are equal).
In triangles DEF and DEG,
Unit 3
284
∠G = ∠F
∠DEG = ∠DEF
(proved);
(both equal to 90◦ .)
Hence
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DG = DF
(proved);
DE = DE
(common);
∠GDE = ∠F DE (proved).
Hence by SAS condition,
∆DEG ∼
= ∆DEF .
But we have already proved that
∆ABC ∼
= ∆DEG.
It follows that
∆ABC ∼
= ∆DEF .
d
∠GDE = 180◦ − (∠G + ∠DEG) = 180◦ − (∠F + ∠DEF ) = ∠F DE.
Consider triangles DEG and DEF . We have
Note: We have used an important result: If there are three triangles ABC,
DEF and JKL such that ABC is congruent to DEF and DEF is congruent
to JKL, then ABC is congruent to JKL. This is precisely Axiom 3 in the
unit 1 of chapter 3.
No
t
to
Think it over!
If two sides of a right triangle are respectively equal to the corresponding sides of another right triangle, can you conclude that
the two triangles are congruent? (We are not demanding that
hypotenuse of two triangles be equal.)
Example 12. Suppose ABC is an isosceles triangle such that AB = AC
and AD is the altitude from A on BC. Prove that (i) AD bisects ∠A, (ii) AD
bisects BC.
A
B
D
C
Solution: We have to show that
∠BAD = ∠CAD and BD = DC.
In right triangles ADB and ADC, we
have
AB = AC (given);
AD = AD (common side).
Congruency
285
So by RHS congruency of triangles, we have ∆ABD ∼
= ∆ACD. Hence
∠BAD = ∠CAD and BD = DC.
Example 13. Suppose the altitudes AD, BE and CF of triangle ABC are
equal. Prove that ABC is an equilateral triangle.
B
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F
Solution: In right triangles BCE
and CBF , we have,
BC = BC (common hypotenuse);
BE = CF (given).
Hence BCE and CBF are congruent,
by RHS theorem. Comparing the triangles, we get ∠B = ∠C.
d
A
D
C
This implies that
AC = AB (sides opposite to equal angles).
Similarly,
AD = BE =⇒ ∠B = ∠A
=⇒ AC = BC.
Together, we get AB = BC = AC or ABC is equilateral.
Exercise 3.3.6
to
1. Suppose ABCD is rectangle. Using RHS theorem, prove that triangles
ABC and ADC are congruent.
No
t
2. Suppose ABC is a triangle and D is the midpoint of BC. Assume that
the perpendiculars from D to AB and AC are of equal length. Prove
that ABC is isosceles.
3. Suppose ABC is a triangle in which BE and CF are respectively the
perpendiculars to the sides AC and AB. If BE = CF , prove that
triangle ABC is isosceles.
3.3.7 Some consequences
You have seen earlier that in a triangle the angles opposite to equal
sides are equal. And conversely, sides opposite to equal angles are equal.
So the natural question is: if angles are of different measures, can we
Unit 3
286
compare the sides opposite to them? Can we say some thing about angles
if sides are of different lengths?
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Proposition 1. Suppose two sides of a triangle are not equal. Then
the angle opposite to a larger side is greater than the angle opposite
to the smaller side.
Given: a triangle ABC in which
AC > AB.
A
To prove: ∠B > ∠C.
Construction: take a point D on
AC such that AB = AD. (This is
D
possible since AC > AB.)
B
C
Join BD.
Proof: In triangle ABD, we have
AB = AD(by construction) =⇒ ∠ABD = ∠ADB (angles opposite to equal sides).
Now ∠BDC is an exterior angle for triangle BCD. Hence it is larger than
interior opposite angle ∠BCD. We thus get
∠C < ∠BDA = ∠ABD < ∠ABC = ∠B.
This completes the proof.
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Proposition 2. In a triangle, if two angles are unequal, then the side
opposite to the larger angle is greater than the side opposite to the
smaller angle.
Given: a triangle ABC in which
∠B > ∠C.
A
To prove: AC > AB.
Proof: Observe that
∠B > ∠C =⇒ AC 6= AB.
For, AC = AB implies that ∠B =
B
C
∠C(angles opposite to equal sides
are equal).
Thus either
AC < AB or AC > AB.
If AC < AB, then by previous proposition ∠B < ∠C; but this contradicts
the given hypothesis. The only possibility left out is
AC > AB.
Congruency
287
Note: Here we are using law of trichotomy. Given any two real numbers a
and b, you have only one of the possibilities: a < b; a = b; or a > b.
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Proposition 3. In a triangle, the sum of any two sides is greater than
the third side.
Given: a triangle ABC.
D
To prove: AB + AC > BC.
Construction: Extend BA to D such that
AD = AC and join DC.
A
Proof: Then
BD = BA + AD = BA + AC.
Since AD = AC, we have
C
B
∠ADC = ∠ACD (angle opposite to equal sides).
Hence we obtain
∠BCD > ∠ACD = ∠ADC = ∠BDC.
In triangle BCD, we have
∠BCD > ∠BDC =⇒ BD > BC(by proposition 2).
But BD = BA + AC, as we have observed earlier. We thus get
BA + AC > BC.
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You can similarly prove CA < AB + BC and AB < BC + CA.
Note: The inequalities BC < CA+AB, CA < AB+BC and AB < BC +CA are
called triangle inequalities. They are necessary conditions for the existence
of a triangle with sides AB, BC and CA. This tells that the straight line
is the shortest path between any two point. Given three numbers a, b, c,
necessary conditions for the existence of a triangle with sides a, b, c are that
a < b + c, b < c + a and c < a + b. These conditions are also sufficient which
you will see while constructing a triangle with three given sides.
Example 14. Show that in a right angled triangle, hypotenuse is larger
than any of the remaining sides.
Solution: Suppose ABC is a right angled triangle with ∠B = 90◦ . Then
AC is the hypotenuse. Observe that ∠BAC < ∠B and ∠BCA < ∠B. Now
the side opposite to ∠BAC is BC and side opposite ∠BCA is AB. Hence by
proposition 2, BC < AC and AB < AC.
Example 15. In the adjoining figure, ∠B = 70◦ , ∠C = 50◦ and AD is the
bisector of ∠A. Prove that AB > AD > CD.
Unit 3
288
A
Solution: Observe that
∠A = 180◦ − (∠B + ∠C)
= 180◦ − (70◦ + 50◦ ) = 60◦ .
50
D
C
Hence ∠BAD = ∠DAC = 30◦ .
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Consider triangle BAD. We can compute ∠ADB:
∠ADB = 180◦ − (70◦ + 30◦ ) = 80◦ > ∠ABD
Hence AB > AD, by proposition 2. In triangle ADC, we have
∠DAC = 30◦ < 50◦ = ∠ACD.
Again proposition 2 gives CD < AD.
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B
Exercise 3.3.7
1. In a triangle ABC, ∠B = 28◦ and ∠C = 56◦ . Find the largest and the
smallest sides.
2. In a triangle ABC, we have AB = 4cm, BC = 5.6cm and CA = 7.6cm.
Write the angles of the triangle in ascending order of measures.
3. Let ABC be a triangle such that ∠B = 70◦ and ∠C = 40◦ . Suppose D
is a point on BC such that AB = AD. Prove that AB > CD.
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4. Let ABCD be a quadrilateral in which AD is the largest side and BC
is the smallest side. Prove that ∠A < ∠C. (Hint: Join AC)
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5. Let ABC be a triangle and P be an interior point. Prove that AB +
BC + CA < 2(P A + P B + P C).
Additinal problems on “Congruency of triangles”
1. Fill in the blanks to make the statements true .
(a) In right triangle the hypotenuse is the ————— side.
(b) The sum of three altitudes of a triangle is ————— than its
perimeter.
(c) The sum of any two sides of a triangle is —————- than the
third side.
Congruency
289
(d) If two angles of a triangle are unequal , then the smaller angle
has the ————– side opposite to it.
(e) Difference of any two sides of a triangle is —————- than the
third side.
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2. Justify the following statements with reasons:
d
(f) If two sides of a triangle are unequal, then the larger side has
—————- angle opposite to it.
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(a) The sum of three sides of a triangle is more than the sum of its
altitudes.
(b) The sum of any two sides of a triangle is greater than twice the
median drawn to the third side.
(c) Difference of any two sides of triangle is less than the third side.
3. Two triangles ABC and DBC have common base BC. Suppose AB =
DC and ∠ABC = ∠BCD. Prove that AC = BD.
4. Let AB and CD be two line segments such that AD and BC intersect
at O. Suppose AO = OC and BO = OD. Prove that AB = CD.
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5. Let ABC be a triangle. Draw a triangle BDC externally on BC such
that AB = BD and AC = CD. Prove that ∆ABC ∼
= ∆DBC.
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6. Let ABCD be a square and let points P on AB and Q on DC be such
that DP = AQ. Prove that BP = CQ.
7. In a triangle ABC, AB = AC. Suppose P is point on AB and Q is a
point on AC such that AP = AQ. Prove that ∆AP C ∼
= ∆AQB.
8. In an isosceles triangle, if the vertex angle is twice the sum of the base
angles, calculate the angles of the triangle.
9. If the bisector of the vertical angle of a triangle bisects the base, show
that the triangle is isosceles.
10. Suppose ABC is an isosceles triangle with AB = AC. Side BA is
produced to D such that BA = AD. Prove that ∠BCD is a right angle.
Unit 3
290
11. Let AB, CD be two line segments such that AB k CD and AD k BC.
Let E be the midpoint of BC and let DE extended meet AB in F . Prove
that AB = BF .
Glossary
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Congruency: two geometrical figures are identical in both shape and size.
Superpose: one figure sitting exactly on the other.
Corresponding sides: while comparing two triangles, we index the sides
of the triangles in an ordered way.
Corresponding angles: indexing of the angles in an ordered way.
SAS postulate: Side-Angle-Side postulate.
ASA postulate: Angle-Side-Angle postulate.
SSS postulate: Side-Side-Side postulate.
RHS theorem: Right angle-Hypotenuse-Side theorem.
Triangle inequality: any side of a triangle is smaller than the sum of the
remaining two.
Points to remember
• Two triangles are congruent if we can superpose one on the other.
• Two triangles are congruent if two sides and the included angle of
one triangle are respectively equal to the corresponding sides and the
corresponding angle of the other.(SAS)
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• Two triangles are congruent if two angles and the common side to
these angles of one triangle are respectively equal to the corresponding angles and the corresponding side of the other.(ASA)
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• Two triangles are congruent if three sides of one triangle are respectively equal to three corresponding sides of the other. (SSS)
• Two right angled triangles are congruent if they have equal hypotenuse
and, apart from hypotenuse, a side of one triangle is equal to a side
of of the other.(RHS)
• Any side of a triangle is smaller than the sum of the other two.(Triangle
inequality)
CHAPTER 3
UNIT 4
Note: The sections 3.4.5, 3.4.6, 3.4.12 and 3.4.13 are optional in
this unit. This need not be used for examination.
–
–
–
–
–
–
–
–
–
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After studying this unit, you learn:
• to construct a triangle:
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CONSTRUCTION OF TRIANGLES
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when three sides are given;
when two sides and included angle are given;
when two angles and included side are given;
when two sides and altitude on third side are given;
one side and hypotenuse of a right angled triangle are given;
an isosceles triangle, whose base and height are given;
perimeter and ratio of the sides of a right triangle are given;
whose perimeter and base angles are given;
when the base, sum of the other two sides and one base angle of
a right triangle are given;
– when the base, difference of the other two sides and one base
angle of a right triangle are given.
3.4.1 Introduction
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In earlier classes, you have learnt that given a triangle, there are six
elements associated with it, namely, three sides and three angles. You may
be wondering whether you need all these to be known for constructing a
triangle. If all are known, it is well and good. But in a variety of practical
situations, you may not know all these. If only two are known, you cannot
hope to construct a triangle. Even if three of these are known, you may not
be able to construct. For example, If two sides and an angle ( not included
angle ) are given, then it is not possible to construct such triangle.
We will take up different situations when we will be able to construct
a triangle. Of course along with these basic six elements, you can also
associate many other elements like medians, angle bisectors altitudes. You
get several other combinations. And constructing triangles with minimum
number of data which includes these additional elements is interesting
and challenging. We will not go beyond the traditional constructions.
Unit 4
292
3.4.2 When three sides are given
Example 1 Construct a triangle ABC in which AB = 5 cm, BC= 4.3 cm
and AC= 4 cm.
Solution: We follow several steps in the construction:
2. Locate points A and B on it such that AB = 5 cm.
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1. Draw a line segment which is sufficiently long using ruler.
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3. With A as centre and radius 4 cm, draw an arc(see figure).
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4. With B as centre and radius 4.3 cm, draw another arc cutting the
previous arc at C.
5. Join AC and BC.
Then ABC is the required triangle.
C
4.3 cm
4 cm
A
5 cm
B
Think it over!
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The arc with B as centre and radius 4.3 cm always cuts the arc
with A as center and radius 4 cm, whenever AB=5 cm. Can you
relate this to triangle inequality?
Exercise 3.4.2
1. Construct a triangle ABC in which AB=5 cm and BC= 4.6 cm and
AC= 3.7 cm.
2. Construct an equilateral triangle of side 4.8 cm.
3. Construct a triangle P QR, given that P Q =5.6 cm, P R=7 cm and
QR=4.5 cm.
4. Construct a triangle XY Z in which XY =7.8 cm, Y Z=4.5 cm and
XZ=9.5 cm.
Constructions
293
5. Construct a triangle whose perimeter is 12 cm and the ratio of their
sides is 3:4:5.
3.4.3 When two sides and their included angle are given
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Example 2. Construct a triangle P QR, given that P Q=4 cm, QR=5.2 cm
and ∠Q = 60◦ .
Solution: Steps of construction:
1. Draw a line segment which is sufficiently long using ruler.
2. Locate points Q and R on it such that QR = 5.2 cm.
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3. At Q, construct a line segment QM, sufficiently large, such that ∠MQR =
60◦ ; use protractor to measure 60◦ .
4. With Q as center and radius 4 cm., draw an arc cutting QM at P ; join
P R.
Then, P QR is required the triangle.
M
4c
m
P
60
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Q
5.2 cm
R
Think it over!
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Without using protractor, can you construct a line segment QM
at Q such that ∠MQR = 60◦ ?
Exercise 3.4.3
1. Construct a triangle ABC, in which AB=4.5 cm, AC=5.5 cm and
∠BAC = 75◦ .
2. Construct a triangle P QR in which P Q=5.4 cm, QR=5.5 cm and ∠P QR =
55◦ .
3. Construct a triangle XY Z in which XY =5 cm, Y Z=5.5 cm and ∠XY Z =
100◦ .
Unit 4
294
4. Construct a triangle LMN in which LM=7.8 cm, MN=6.3 cm and
∠LMN = 45◦ .
3.4.4 When two angles and included side are given
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Example 3. Construct a triangle XY Z in which XY = 4.5 cm and ∠X = 100◦
and ∠Y = 50◦ .
Solution: Steps of construction
1. Draw a line segment which is sufficiently long using ruler.
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2. Locate points X and Y on it such that XY = 4.5 cm.
3. Construct a line segment XP such that ∠P XY = 100◦ ; construct a
line segment Y Q such that ∠XY Q = 50◦ .
4. Extend XP and Y Q to intersect at Z.
Then XY Z is the required triangle.
P
Q
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Z
X
100
50
4.5 cm
Y
Think it over!
Suppose it is given that XY = 4.5 cm, ∠X = 100◦ and ∠Y = 80◦ .
Can you construct a triangle now? Where does the construction
break down and why?
Exercise 3.4.4
1. Construct a triangle ABC in which AB=6.5 cm, ∠A = 45◦ and ∠B =
60◦ .
Constructions
295
2. Construct a triangle P QR in which QR=4.8 cm, ∠Q = 45◦ and ∠R =
55◦ .
3. Construct a triangle ABC in which BC=5.2 cm, ∠B = 35◦ and ∠C =
80◦ .
4. Construct a triangle ABC in which BC=6 cm, ∠B = 30◦ and ∠C = 125◦ .
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3.4.5 To construct an equilateral triangle of given height
(Optional)
Example 4. Construct an equilateral triangle of height 3.2 cm.
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Solution: First observe that the altitudes from any vertex to the opposite
sides of an equilateral triangle are all of equal length (prove this statement). Hence we can define the height of an equilateral triangle as this
common value of three altitudes.
Steps of construction
1. Draw any line segment XY .
2. Take any point M on XY . Draw ZM ⊥ XY .
3. With M as center and radius 3.2 cm, draw an arc, cutting MZ at A.
4. Construct ∠MAB = 30◦ and ∠MAC = 30◦ , with B and C on XY .
Then ABC is the required triangle.
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Z
X
30
A
3.2 cm
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30
B
M
C
Y
Think it over!
Why is that ABC so constructed is equilateral? Can you think
of a proof?
Unit 4
296
Exercise 3.4.5
1. Construct an equilateral triangle of height 4.5 cm. Measure approximate length of the its side.
2. Construct an equilateral triangle of height 5.2 cm. Measure approximate length of the its side.
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3. Construct an equilateral triangle of height 6. cm. Measure approximate length of the its side.
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3.4.6 When two sides and altitude on the third side are
given (Optional)
Example 5. Construct a triangle ABC in which AB =4.5 cm, AC= 5 cm
and length of perpendicular from A on BC is 3.6 cm.
Solution: Steps of construction
1. Draw a line segment XY .
2. Take a point M on XY .
3. Draw ZM ⊥ XY , with MZ sufficiently large.
4. With M as center and radius 3.6 cm, draw an arc, cutting MZ at A.
Z
5. With A as center and
radii 4.5 cm and 5 cm,
draw arcs cutting XY
at B and C respectively;
join AB and AC.
cm
5
3.6 cm
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X
C
M
cm
Then ABC is the required triangle.
4.5
to
A
Y
B
Think it over!
(1) Why is that the arcs with centre A and radii 4.4 cm and 5 cm
cut the line segment XY ? Which part of the data ensure it?
(2) There are two more triangles possible other than the one
given. Construct them.
Constructions
297
Exercise 3.4.6
1. Construct a triangle P QR in which P Q =5.5 cm, P R=6.2 cm and
length of the perpendicular from P on QR is 4 cm.
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2. Construct a triangle MNP in which MN =4.5 cm, MP = 5.2 cm and
length of perpendicular from M on NP is 3.8 cm.
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3.4.7 To construct a right triangle whose one side and
hypotenuse are given
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Example 6. Construct a right triangle LMN in which ∠M = 90◦ , MN = 4
cm and LN=6.2 cm.
Solution:
Steps of construction
1. Draw a line segment
XY .
P
2. Locate M, N on XY such
that MN =4 cm.
L
2
6.
cm
M
4 cm
Y
N
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X
Then, LMN is the required triangle.
3. Construct
a
line
segment MP ,
sufficiently large, such that
∠NMP = 90◦ .
4. With N as centre and radius 6.2 cm draw an arc,
cutting MP at L; join
NL.
Think it over!
Why is that the arc with centre N and radius 6.2 cm cuts the
line segment MP ? Which part of the data ensures it?
Exercise 3.4.7
1. Construct a right angle triangle ABC in which ∠B = 90◦ , AB = 5 cm
and AC=7 cm.
Unit 4
298
2. Construct a right angle triangle P QR in which ∠R = 90◦ , P Q = 4 cm
and QR=3 cm.
3. Construct a right angle triangle ABC in which ∠B = 90◦ , BC = 4 cm
and AC=5 cm.
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3.4.8 To construct an isosceles triangle whose base and
corresponding altitude are given
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Example 7. Construct an isosceles triangle ABC in which base BC=5.8
cm and altitude from A on BC is 4.8 cm.
Solution:
Steps of construction
P
1. Draw a line segment BC
whose length is 5.8 cm.
A
4.8 cm
2. Draw the perpendicular
bisector of BC; call it MP ,
with M on BC.
B
Y
M 5.8 cm
C
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X
3. With M as centre and radius 4.8 cm, draw an arc
cutting MP at A; join AB
and AC.
Then ABC is the required triangle.
Think it over!
No
t
Which results on triangle is used to conclude that ABC is the
required triangle?
Exercise 3.4.8
1. Construct an isosceles triangle ABC in which base BC=6.5 cm and
altitude from A on BC is 4 cm.
2. Construct an isosceles triangle XY Z in which base Y Z=5.8 cm and
altitude from X on Y Z is 3.8 cm.
3. Construct an isosceles triangle P QR in which base P Q=7.2 cm and
altitude from R on P Q is 5 cm.
Constructions
299
3.4.9 To construct an isosceles triangle when its altitude
and vertex angle are given
Example 8. Construct an isosceles triangle whose altitude is 4 cm and
vertex angle is 80◦ .
P
1. Draw a line segment XY .
A
40
2. Take a point M on XY and
draw a line MP ⊥ XY .
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40
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Solution:
Steps of construction
4 cm
3. With M as centre and radius 4 cm, draw an arc cutting
Y
X
MP at A.
B
M
C
4. Construct B and C on XY such that ∠MAB = 80◦ /2 = 40◦ and ∠MAC =
80◦ /2 = 40◦ .
Then ABC is the required triangle.
Think it over!
Why is triangle ABC isosceles? Which results on triangles are
used to conclude that ABC is the required triangle?
Exercise 3.4.9
1. Construct an isosceles triangle whose altitude is 4.5 cm and vertex
angle is 70◦ .
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2. Construct an isosceles triangle whose altitude is 6.6 cm and vertex
angle is 60◦ .
3. Construct an isosceles triangle whose altitude is 5 cm and vertex angle is 90◦ .
3.4.10 To construct a triangle whose perimeter and ratio
of the sides are given
Example 9. Construct a triangle ABC, whose perimeter is 12 cm and
whose sides are in the ratio 2:3:4.
Solution: Steps of construction
Unit 4
300
1. Draw a line segment and locate points X, Y such that XY = 12 cm.
2. Draw a ray XZ, making an acute angle with XY and drawn in the
down-ward direction.
3. From X, locate (2+3+4)= 9 points at equal distances along XZ.
4. Mark point L, M, N on XZ such that XL=2 parts, LM=3 parts and
MN= 4 parts.
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5. Join NY . Through L and M, draw LB k NY and MC k NY , intersecting XY in B and C respectively.
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6. With B as center and BX as radius, draw an arc; with C as a centre
and CY as radius, draw an arc cutting the previous arc at A.
7. Join AB and AC.
A
X
Y
B
C 12 cm
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M
N
Z
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Then ABC is the required triangle.
Think it over!
Why is that the sides of ABC are in the required ratio? Is it possible to construct a triangle if the sides are in the ratio 2:3:5?
Exercise 3.4.10
1. Construct a triangle ABC, whose perimeter is 13 cm and whose sides
are in the ratio 3:4:5.
2. Construct a triangle P QR, whose perimeter is 14 cm and whose sides
are in the ratio 2:4:5.
Constructions
301
3. Construct a triangle MNP , whose perimeter is 15 cm and whose sides
are in the ratio 2:3:4.
3.4.11 To construct a triangle whose perimeter and base
angles are given
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Example 10. Construct a triangle ABC whose perimeter is 12.5 cm and
whose base angles are 60◦ and 75◦ .
A
L
75
60
P
B
C
Q
12.5 cm
to
Solution: Steps of construction
1. Draw a line segment and locate points P, Q such that P Q=12.5 cm.
−→
−→
2. Construct rays P R such that ∠QP R = 60◦ and QS such that ∠P QS =
75◦ .
No
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3. Draw the bisectors P L and QM of ∠QP R and ∠P QS respectively. Let
these intersect at A.
4. Draw the perpendicular bisector of AP and AQ and let these intersect
P Q at B and C respectively.
5. Join AB and AC.
Then ABC is the required triangle.
Think it over!
What properties of triangles ensure that we get required base
angles? And the perimeter is also as required?
Unit 4
302
Exercise 3.4.11
1. Construct a triangle ABC whose perimeter 12 cm and whose base
angles are 50◦ and 80◦ .
2. Construct a triangle XY Z whose perimeter 15 cm and whose base
angles are 60◦ and 70◦ .
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3. Construct a triangle ABC whose perimeter 12 cm and whose base
angles are 65◦ and 85◦ .
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3.4.12 To construct triangle when its base, sum of the
other two sides and one base angle are given (Optional)
Example 11. Construct a triangle ABC in which AB=5.8 cm, BC + CA=8.4
cm and ∠B = 60◦ .
Solution:
X
Steps of construction
1. Draw a line segment AB of
length 5.8 cm.
D
8.4
cm
2. Draw a line segment BX,
sufficiently large such that
∠ABX = 60◦ .
C
4. Join AD.
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B
5.8 cm
A
3. From the segment BX, cut
off line segment BD of length
8.4 cm.
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5. Draw the perpendicular bisector of AD and let it meet BD at C.
6. Join AC.
Then ABC is the required triangle.
Think it over!
How is that the sum of CA and CB is equal to the given sum?
Exercise 3.4.12
1. Construct a triangle ABC in which BC=3.6 cm, AB + AC=4.8 cm and
∠B = 60◦ .
Constructions
303
2. Construct a triangle ABC in which AB + AC=5.6 cm, BC=4.5 cm and
∠B = 45◦ .
3. Construct a triangle P QR in which P Q + P R=6.5 cm, QR=5.4 cm and
∠Q = 40◦ .
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3.4.13 To construct a triangle when its base, difference of
the other two sides and one base angle are given (Optional)
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Example 12. Construct a triangle ABC in which base AB= 5 cm, ∠A = 30◦
and AC − BC=2.5 cm.
Solution:
X
Steps of construction
1.
Draw
a line segment AB of
C
length 5 cm.
2. Draw another line segment
D
AX, sufficiently large, such
m
c
that ∠BAX = 30◦ .
2.5
3. From the segment AX, cut
5 cm
off line segment AD=2.5 cm,
B
A
which is equal to (AC − BC).
4. Join BD.
6. Join BC.
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5. Draw the perpendicular bisector of BD and let it cut AX at C.
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Then ABC is the required triangle.
Think it over!
Can you see why AD is equal to the difference AC − BC? Can
you take AC − BC > AB and still construct a triangle?
Exercise 3.4.13
1. Construct triangle ABC in which BC=3.4 cm, AB − AC=1.5 cm and
∠B = 45◦ .
2. Construct triangle ABC in which BC=5 cm, AB − AC=2.8 cm and
∠B = 40◦ .
Unit 4
304
3. Construct triangle ABC in which BC=6 cm, AB − AC=3.1 cm and
∠B = 30◦ .
Additional problems on “Constructions of triangles”
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1. Construct a triangle ABC in which AB=5cm, BC=4.7cm and AC=4.3cm.
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2. Construct a triangle ABC in which AB=5cm, BC=5cm and AC=4.3cm.
3. Construct a triangle P QR in which P Q=4cm, QR=4.5cm and ∠Q = 60◦ .
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4. Construct a triangle P QR in which P Q=4cm, ∠P = 60◦ and ∠Q=60◦ .
5. Construct a triangle ABC in which AB=3.5cm, AC=4cm and length of
the perpendicular from A to BC is 3cm.
6. Construct an isosceles triangle ABC in which base BC=4.5cm and
altitude from A on BC is 3.8cm.
7. Construct an isosceles triangle whose altitude is 5 cm and whose
vertex angle is 70◦ .
8. Construct an isosceles triangle whose altitude is 5 cm and whose
vertex angle is 80◦ .
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9. Construct an equilateral triangle of height 3.5 cm. (Optional)
10. Construct an equilateral triangle of height 4.3 cm. (Optional)
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11. Construct right angle triangle LMN in which ∠M = 90◦ , MN=4.5 cm
and LN=5.6 cm.
12. Construct right angle triangle P QR in which ∠Q = 90◦ , QR=4.5 cm
and ∠R = 50◦ .
13. Construct a triangle P QR, whose perimeter is 13 cm and whose sides
are in the ratio 2:3:4.
14. Construct a triangle P QR, whose perimeter is 15 cm and whose sides
are in the ratio 3:4:6.
Constructions
305
15. Construct a triangle ABC, whose perimeter is 13.5 cm and whose
base angles are 60◦ and 75◦ .
16. Construct a triangle ABC, whose perimeter is 12.5 cm and whose
base angles are 50◦ and 80◦ .
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17. Construct a triangle XY Z in which Y Z=4.5 cm, ∠Y = 60◦ and sum of
other two sides is 7.5 cm. (Optional)
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18. Construct a triangle ABC whose perimeter is 9 cm and the angles are
in the ratio 3:4:5.
19. Construct a triangle ABC whose perimeter is 12 cm and the angles
are in the ratio 2:3:5.
20. Construct a triangle ABC in which BC=4.5 cm, ∠B = 35◦ and difference between the other two sides is 2.8 cm. (Optional)
Glossary
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Perpendicular bisector: the line which is perpendicular to the given line
segment and also bisects the line segment.
Angle bisector: the line which bisects the given angle.
Perimeter: the length of the boundary of any plane figure.
Altitude: the perpendicular from a point to a line; this is also used for the
length of such a perpendicular.
Arc: part of the circumference of a circle.
Base angle: any of the angles formed by the base of a triangle, with the
other sides.
Vertex angle: the angle at the top of an isosceles triangle.
Points to remember
• At least three parameters are needed to construct a triangle.
• Not all combination of three parameters will enable one to construct
a triangle.
CHAPTER 3
UNIT 5
QUADRILATERALS
After studying this unit, you learn to:
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• recognise quadrilaterals from a given list of figures;
• list out common properties of a quadrilateral;
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• solve the sums related to quadrilaterals;
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• classify different types of quadrilaterals and recognise their distinct
properties;
• transform the problems which occur in daily life related to quadrilaterals to numerical sums and solve them.
3.5.1 Introduction
You have learnt earlier that a triangle is a plane figure bounded by three
sides. Triangles are classified based on the measures of sides and angles.
Activity 1: Name the triangles based on their sides. Name the triangles
based on their angles.
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Now let us take four points on the plane and see what we obtain on
joining them in pairs in some order.
A
B
C
Fig 1
D
B
C
C
D
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A
A
B
A
C
Fig 2
B
D
Fig 3
D
Fig 4
If all the points are collinear, that is they all lie on the same straight line,
we obtain a line segment (Fig 1). If three points out of the four are collinear,
we get a triangle (Fig 2). If no three points out of four points are collinear,
we get a closed figure with four sides (Fig 3 and Fig 4). Such a closed
figure having four sides formed by joining four points, no three of which
are collinear, in an order, is called quadrilateral.
Look at the following figures:-
Quadrilaterals
307
A
W
P
B
D
Q
M
X
N
S
Z
P
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C
Y
O
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They are closed figures. Their boundary consists of four line segments.
The common name to all the above plane figures is quadrilateral.
to
A quadrilateral is named by referring to its vertices in a particD
ular order. In the adjoining figC
ure, we can read it as ABCD or
ADCB , we cannot read it ADBC
and such names. You may keep
A
in mind that the vertices should
be read in such a way that if
you join adjacent letters in the
B
name, then there should not be
any crossing of line segments.
For
example in the name ADBC , we see that AC and BD cross each other.
Observe the following figures.
A
A
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B
B
D
C
D
C
Are these two figures quadrilaterals? No they are not
quadrilaterals; in the first,
there is crossing of sides; in
the second, the sides are not
all line segments.
So we may redefine the quadrilateral as: A quadrilateral is the union of
four line segments that join four coplanar points, no three of which
are collinear and each segment meet exactly two other lines, each at
their end point.
Unit 5
308
Here again we do not distinguish between the closed curve which
is the union of four line segments and the plane figure which is
bounded by these four line segments. The context makes it clear
which form is taken.
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Observe that a quadrilateral consists of four sides and four internal
angles. Based on these internal angles, we can classify quadrilaterals in
to two types: convex and concave quadrilaterals.
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A quadrilateral is convex if each of the internal angle of the quadrilateral is less than 180◦ . Otherwise it is called a concave quadrilateral.
Concave quadrilateral
Convex quadrilateral
N
D
C
A
B
L
K
M
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Think it over:
Suppose you are given four sticks of different lengths. Can you
put them together and make a quadrilateral?
Like triangles, quadrilaterals also enjoy nice properties. We will study
some of them in the coming sections.
3.5.2 Properties of quadrilaterals
Let ABCD be a quadrilateral.
1. The points A, B, C and D are called the vertices of the quadrilateral.
2. The segments AB, BC, CD and DA are the four sides of the quadrilateral.
Quadrilaterals
309
D
3. The angles ∠DAB , ∠ABC, ∠BCD
and ∠CDA are the four angles of
the quadrilateral.
C
A
B
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4. The segments AC and BD are
called as the diagonals of the
quadrilateral.
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Adjacent sides and opposite sides
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Note: A quadrilateral has four sides, four angles and two diagonals.
In all it has ten elements.
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1. Two sides of a quadrilateral are said
to be adjacent sides or consecutive sides if they have a common
C
end point. In the adjoining figure, AB
and AD are adjacent or consecutive
sides. Identify the other pair of adjaA
B
cent sides.
2. Two sides are said to be opposite sides, if they do not have a common
end point. In the above figure AB and DC are opposite sides. Identify
the other pair of opposite sides.
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Adjacent angles and opposite angles.
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3. Two angles of a quadrilateral are adjacent angles or consecutive
angles, if they have a side common to them. Thus ∠DAB and ∠ABC
are adjacent or consecutive angles. Identify the other pair of adjacent
angles.
4. Two angles of a quadrilateral are said to be opposite angles, if they
do not contain a common side. Here ∠DAB and ∠BCD are opposite
angles. Identify other pair of opposite angles.
Diagonal property
The diagonal AC divides the quadrilateral in to two triangles, namely, triangle ABC and triangle ADC. Name the two triangles formed when the
diagonal BD is drawn.
Unit 5
310
Angle sum property
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Activity 2: Take a cut-out of a quadrilateral drawn on a card board. Cut
it along the arms making each angle of the quadrilateral as in the figure
below. The four pieces so obtained are numbered as 1,2,3 and 4. Arrange
the cut outs as shown in next figure. Are they meeting at a point? What
can you say about the sum of these four angles? The sum of the measures
of the four angles is 360◦ .
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2
3
1
1
2
3
4
4
Note: Recall the sum of the angles at any point is 360◦ .
Theorem 1. The sum of the angles of quadrilateral is 360◦ .
Given: ABCD is a quadrilateral.
To prove: ∠A + ∠B + ∠C + ∠D = 360◦ .
Construction: Draw the diagonal AC.
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D
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2
A
1
4
5
B
3
6
C
Proof: In triangle ADC,
∠1 + ∠2 + ∠3 = 180◦ ;
(angle sum property.) In triangle
ABC,
∠4 + ∠5 + ∠6 = 180◦ .
( again angle sum property.)
Adding these,
∠1 + ∠4 + ∠2 + ∠5 + ∠3 + ∠6 = 360◦ .
But ∠1 + ∠4 = ∠A and ∠3 + ∠6 = ∠C. Therefore ∠A + ∠D + ∠B + ∠C = 360◦ .
Thus the sum of the angles of the quadrilateral is 360◦ .
Example 1. The four angles of a quadrilateral are in the ratio 2:3:4:6.
Find the measures of each angle.
Quadrilaterals
311
Solution:
Given: The ratio of the angles as 2:3:4:6.
To find: The measure of each angle.
Observe that 2+3+4+6 = 15 (sum of the terms of the ratio). Thus 15 parts
accounts for 360◦ . Hence
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15 parts −→ 360◦ ;
360◦
× 2 = 48◦ ;
2 parts −→
15
360◦
3 parts −→
× 3 = 72◦ ;
15
360◦
4 parts −→
× 4 = 96◦ ;
15
360◦
6 parts −→
× 6 = 144◦ .
15
Thus the angles are 48◦ , 72◦ , 96◦ and 144◦ . Can you see that their sum is
360◦ ?
Example 2. In a quadrilateral ABCD, ∠A and ∠C are of of equal measure;
∠B is supplementary to ∠D. Find the measure of ∠A and ∠C.
Solution: We are given ∠B + ∠D = 180◦ . Using angle-sum property of a
quadrilateral, we get
∠A + ∠C = 360◦ − 180◦ = 180◦ .
180◦
= 90◦ .
2
Can you draw a quadrilateral in which ∠A = ∠C = 90◦ and ∠B, ∠D are
complementary ?
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Since ∠A and ∠C are of equal measure, we obtain ∠A = ∠C =
Example 3. Find all the angles in the given quadrilateral below.
S
Solution: We know that ∠P +∠Q+∠R+
∠S = 360◦ (angle sum property). Hence
x + 2x + 3 + x + 3x − 7 = 360◦ .
This gives 7x − 4 = 360◦ or 7x = 364◦ .
Therefore x = 364/7 = 52◦ .
3 x −7
P
x
x
2x+3
Q
R
Unit 5
312
We obtain, ∠P = 52◦ ; ∠R = 52◦ ; ∠Q = 2x + 3 = 2 × 52 + 3 = (104 + 3) = 107◦ ;
∠S = 3 × 52◦ − 7◦ = 156◦ − 7◦ = 149◦ . Check that ∠P + ∠Q + ∠R + ∠S = 360◦ .
Exercise 3.5.2
1. Two angles of a quadrilateral are 70◦ and 130◦ and the other two angles
are equal. Find the measure of these two angles.
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2. In the fig, suppose ∠P and ∠Q
are supplementary angles and
∠R = 125◦ . Find the measures
of ∠S.
125
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R
P
Q
3. Three angles of a quadrilateral are in the ratio 2:3:5 and the fourth
angle is 90◦ . Find the measures of the other three angles.
4. In the adjoining figure, ABCD
C
D
is a quadrilateral such that
P
∠D + ∠C = 100◦ . The bisectors
of ∠A and ∠B meet at ∠P . Determine ∠AP B.
A
B
3.5.3 Trapezium
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Based on the nature of the sides or angles, a quadrilateral gets special
name.
P
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S
A
D
E
Q
R
K
F
H
L
G
B
W
N
C
Set I
M
X
Y
Z
Set II
Observe the above figures given in two sets. Discuss with your friends
what is the difference you observe in the first set and the second set of
quadrilaterals. [Note: the arrow mark indicates parallel lines].
Quadrilaterals
313
The first set of quadrilaterals have a pair of opposite sides which are
parallel. Such a quadrilateral is known as trapezium. In the first set
each quadrilateral is a trapezium. In the second set no quadrilateral is a
trapezium.
Activity 3:
Take identical cut-outs of congruent triangles of sides 3 cm, 4 cm
5
and 5 cm. Arrange them as shown
3
3
in the figure. Which figure do you
5
3
5
get? It is a trapezium. Which are
4
4
the parallel sides?
Can you get a trapezium in which non parallel sides are equal?
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4
D
C
Activity 4:
Cut three identical equilateral
triangles and place them as
shown in the figure. Measure AD
3
3
3
3
and BC. Are they equal? Mea3
3
A
B
sure angles A and B.
Are they equal? Measure angles D and C. Are they equal? Measure AC
and BD. Are they equal?
The special name given to the above trapezium is isosceles trapezium.
You see that in an isosceles trapezium;
3
3
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3
1. the non parallel sides are equal;
2. the base angles are equal;
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3. the adjacent angles corresponding to parallel sides are supplementary;
4. the diagonals are equal.
Think it over!
Is there a trapezium whose all angles are
equal?
Example 4. In the figure ABCD, suppose AB k CD; ∠A = 65◦ and
∠B = 75◦ . What is the measure of ∠D and ∠C?
Unit 5
314
D
65
C
75
B
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A
Solution: Observe that, ∠A +
∠D = 180◦ (a pair of adjacent angles of a trapezium is supplementary). Thus 65◦ + ∠D = 180◦ . This
gives ∠D = 180◦ − 65◦ = 115◦ . Similarly ∠B + ∠C = 180◦ , which gives
75◦ + ∠C = 180◦ . Hence
∠C = 180◦ − 75◦ = 105◦ .
P
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Example 5. In an isosceles trapezium P QRS, ∠P and ∠S are in the ratio
1:2. Find the measure of all the angles.
Solution: In an isosceles trapezium
base angles are equal; ∠P = ∠Q.
R
S
Let ∠P = x◦ and ∠S = 2x◦ . Since
∠P + ∠S = 180◦ (one pair of adjacent
angles of a trapezium is supplementary), we get
Q
x◦ + 2x◦ = 180◦ .
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Thus 3x◦ = 180◦ or x◦ = 180◦ /3 = 60◦ . Hence ∠P = 60◦ and
∠S = 2 × 60◦ = 120◦ . Since, ∠P = ∠Q, we get ∠Q = 60◦ . But, ∠Q + ∠R = 180◦
(one pair of adjacent angles are supplementary). Hence we also get
∠R = 180◦ − 60◦ = 120◦ .
Exercise 3.5.3
No
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1. In a trapezium P QRS, P Q k RS; and ∠P = 70◦ and ∠Q = 80◦ . Calculate
the measure of ∠S and ∠R.
2. In a trapezium ABCD with AB k CD, it is given that AD is not parallel
to BC. Is ∆ABC ∼
= ∆ADC? Give reasons.
R
S
3. In the figure, P QRS is an isosceles trapezium; ∠SRP = 30◦ , and
∠P QS = 40◦ . Calculate the angles
∠RP Q and ∠RSQ.
30
40
P
Q
4. Prove that the base angles of an isosceles trapezium are equal.
Quadrilaterals
315
5. Suppose in a quadrilateral ABCD, AC = BD and AD = BC. Prove
that ABCD is a trapezium.
3.5.4 Parallelograms
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Look at the following sets of quadrilaterals.
Set I
Set II
How many pairs of parallel sides do you see in each of the quadrilaterals in set I ? How many pairs of parallel sides do you see in each of the
quadrilaterals in set II ? What is your conclusion?
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A quadrilateral in which both the pairs of opposite sides are parallel is
called a parallelogram.
The quadrilaterals in set II are not parallelogram, while all the quadrilaterals in set I are parallelograms. Can you see that a parallelogram is
a particular type of trapezium? Hence, whatever properties are true for
trapezium also hold good for parallelograms. You will see that additional
properties are also true because the parallelness of one more pair of sides.
Activity 5:
D
E
A
C
E
B
Take a rectangular card board,
ABCD and mark a point E on AB
(as shown in the fig). Join CE
by dotted line and cut the card
board along CE. Place the triangular part EBC to the left of the
rectangle such that BC coincides
with AD to get a quadrilateral.
Unit 5
316
Which type of quadrilateral is this? It is a parallelogram.
Trace the above cut-out card board in your book and measure the opposite sides, opposite angles, Repeat the same activity with two more parallelograms and mark them as P QRS and KLMN and tabulate the results
as follows.
(vii)
AB =
PQ =
KL =
BC =
QR =
LM =
(viii)
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(vi)
CD = DA =
RS = SP =
MN = NK =
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(v)
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Parallelo(i)
(ii)
(iii)
(iv)
gram
ABCD
∠A = ∠B = ∠C = ∠D =
P QRS
∠P = ∠Q = ∠R = ∠S =
KLMN
∠K = ∠L = ∠M = ∠N =
What relations are there among angles? What relations are there among
sides? Do you observe that opposite angles are equal and opposite sides
are also equal? These observations can be proved logically.
Proposition 1. In a parallelogram, opposite sides are equal and opposite
angles are equal.
D
C
4 2
1
B
to
A
3
Proof: Let ABCD is a parallelogram.
Join BD. Then ∠1 = ∠2, and
∠3 = ∠4. (Why ? See figure.) In triangles ABD and CBD, we observe
∠1 = ∠2,
∠4 = ∠3,
BD (common).
Hence ∆ABD ∼
= ∆CDB (ASA postulate). It follows that
AD = BC and ∠A = ∠C.
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AB = DC,
Similarly join AC, and you can prove that ∆ADC ∼
= ∆CBA. Hence ∠D =
∠B.
Activity 6:
Trace a parallelogram as in the previous case on a card board; draw the diagonals and mark the point of intersection. Measure the intercepts formed
by the point of intersection. What is the inference drawn from this activity?
The diagonals bisect each other. Thus you may describe the properties of
parallelogram as:
Quadrilaterals
317
1. The opposite sides are equal and parallel.
2. The opposite angles are equal.
3. The adjacent angles are supplementary
4. The diagonals bisect each other.
5. Each diagonal bisects the parallelogram into two congruent triangles.
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Example 6. The ratio of two sides of parallelogram is 3:4 and its perimeter
is 42 cm. Find the measures of all sides of the parallelogram.
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Solution: Let the sides be 3x and 4x. Then the perimeter of the parallelogram is 2(3x + 4x) = 2 × 7x = 14x. The given data implies that 42 = 14x, so
that x = 42/14 = 3. Hence the sides of the parallelogram are 3 × 3 = 9 cm
and 3 × 4 = 12 cm.
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Example 7. In the adjoining figure, P QRS is a parallelogram. Find x and
y in cm.
Solution: In a parallelogram, we know
that the diagonals bisect each other.
S
R
Therefore SO = OQ. This gives 16 =
16
y +7
x + y. Similarly, P O = OR, so that
20
O x+y
20 = y + 7. We obtain y = 20 − 7 = 13cm.
Substituting the value of y in the first
P
Q
relation, we get 16 = x + 13. Hence
x = 3cm.
Exercise 3.5.4
No
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1. The adjacent angles of a parallelogram are in the ratio 2:1. Find the
measures of all the angles.
2. A field is in the form of a parallelogram, whose perimeter is 450 m
and one of its sides is larger than the other by 75 m. Find the lengths
of all sides.
3. In the figure, ABCD is a parallelogram. The diagonals AC and BD intersect at O; and ∠DAC = 40◦ , ∠CAB =
35◦ ; and ∠DOC = 110◦ . Calculate the
∠ABO, ∠ADC, ∠ACB, and ∠CBD.
D
C
110
O
40
35
A
B
Unit 5
318
4. In a parallelogram ABCD, the side DC is produced to E and ∠BCE =
105◦ . Calculate ∠A, ∠B, ∠C, and ∠D.
5. Prove logically the diagonals of a parallelogram bisect each other.
Show conversely that a quadrilateral in which diagonals bisect each
other is a parallelogram.
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6. In a parallelogram KLMN, ∠K = 60◦ . Find the measures of all the
angles.
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7. Let ABCD be a quadrilateral in which ∠A = ∠C, and ∠B = ∠D. Prove
that ABCD is a parallelogram.
8. In a quadrilateral ABCD, suppose AB = CD and AD = BC. Prove that
ABCD is a parallelogram.
3.5.5 Special kinds of parallelograms
There are special kind of parallelograms which enjoy different types of
properties. We study them here.
Rectangle
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Activity 7:
Take a sheet from your note book and paste it on a card board; cut the
cardboard along the boundary, measure all the sides, all the angles. Write
down the observations in the following chart. Repeat the activity with
sheets of different sizes.
Parallelogram
(i)
ABCD
∠A =
P QRS
∠P =
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
∠B = ∠C = ∠D = AB = BC = CD = DA =
∠Q = ∠R = ∠S = P Q = QR = RS = SP =
From the above activity, you can infer that:
• all angles are equal to 90◦ ;
• opposite sides are equal;
• opposite sides are parallel;
Quadrilaterals
319
The special name given to such a parallelogram is rectangle. A rectangle is a parallelogram whose all angles are right angles.
Activity 8:
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Take a rectangular sheet of paper and name it as ABCD. Fold it along
the diagonals. Mark the point of intersection as O. Answer the following
questions:
When you fold along any of the diagonals, are the two triangles formed
congruent? Measure the length of the diagonals. Are they equal? Measure
the line segments OA, OC, OB and OD. Do you find any relation among
the length of these segments?
Diagonal properties of a rectangle
(i) The diagonals of a rectangle are equal.
(ii) The diagonals of a rectangle bisect each other.
Example 8. In a rectangle XY W Z, suppose O is the point of intersection
of its diagonals. If ∠ZOW = 110◦ , calculate the measure of ∠OY W .
No
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to
Solution:- We know ∠ZOW = 110◦ .
Z
W
Hence, ∠W OY = 180◦ − 110◦ = 70◦ (sup110
plementary angles). Now OY W is an
isosceles triangle, as OY = OW . Hence
O
∠OY W = ∠OW Y = (180◦ − 70◦ )/2 =
110◦ /2 = 55◦ . (Can you suggest an alX
Y
ternate method?)
Example 9. In a rectangle RENT , the diagonals meet at O. If OR = 2x + 4
and OT = 3x + 1, find x.
Solution: Observe that OR = OT (diT
N
agonal bisect each other and they are
3x+
1
equal in a rectangle). Hence
2x + 4 = 3x + 1.
+4
O
2x
This implies that 4 − 1 = 3x − 2x = 3 = x.
R
E
Hence x = 3.
Unit 5
320
Rhombus
Activity 9:
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Construct four identical right anD
gle triangles with measures 3 cm,
4 cm and 5 cm using card board.
5cm
5cm
Arrange them as shown in the fig1 2
ure, on a plane sheet. Draw the
A
C
4 3
boundary of the figure. Measure
5cm
5cm
the sides and angles of the figure.
Tabulate them. Repeat this with
B
right triangles of different dimensions. What do you observe?
Can you conclude that ∠A = ∠C, ∠B = ∠D; AB = BC = CD = DA?
Such a figure is called a rhombus. A rhombus is a parallelogram in
which all the four sides are equal. Being a parallelogram, a rhombus
has all the properties of a parallelogram and more:
(i) all the sides of a rhombus are equal;
(ii) opposite sides are parallel;
(iii) the diagonals bisect each other at right angles;
(iv) the two diagonals divide the rhombus into four congruent right angled
triangles;
(v) angles are also bisected by the diagonals.
12 cm
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Example 10. The diagonals of a rhombus are 24 cm and 10 cm. Calculate
the area of the rhombus.
Solution: We are given that AC = 24
cm; BD = 10 cm. We know that the diagonals of a rhombus bisect each other
C
at right angles. Let O be the point of intersection of these diagonals. Then we
have AO = CO = 12 cm and BO = DO =
D
B
5 cm
5 cm. We also know that AOD is a right
angled triangle. Hence the area of AOD
is
A
1
1
× OA × OD = × 12 × 5 = 30cm2 .
2
2
Quadrilaterals
321
Since a rhombus has four congruent right triangles, is area is 4 × 30 = 120
cm2 .
Example 11. In a rhombus ABCD, ∠BAC = 38◦ . Find (i) ∠ACD, (ii) ∠DAC
and (iii) ∠ADC.
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Solution: We know that ∠BAC = 38◦ . But in a rhombus ABCD, since
ABC is an isosceles triangle, we see that ∠BAC = ∠ACB = 38◦ . Moreover
∠DAC = 38◦ , since the diagonal AC bisect ∠A. Since ADC is also an
isosceles triangle, we get ∠ACD = ∠DAC = 38◦ . Finally,
C
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∠ADC = 180◦ − (∠DAC + ∠DCA)
= 180◦ − (38◦ + 38◦ )
◦
= 180 − 76
= 104◦ .
Square
D
B
◦
A
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There is a type of parallelogram which is simultaneously a rectangle and
a rhombus. All its angles are equal and all its sides are equal. Recall what
you have for triangles: triangle which is at the same time have all angles
equal and all sides equal. These are equilateral triangles. In the case
of triangles, you have seen that whenever all the angles are equal, all the
sides are also equal. Conversely, if all the sides of a triangle are equal, then
all the angles are also equal. But when you move to quadrilaterals, you
do not have such a nice property. A rectangle is a quadrilateral in which
all the angles are equal, but the sides need not be equal. On the other
hand a rhombus is a quadrilateral in which all the sides are equal, but all
the angles need not be equal. A quadrilateral in which all the angles are
equal and all the sides are equal is given a special name; a square. Thus a
square is an a quadrilateral in which all the nice properties come together.
A square is a parallelogram in which
D
C
((i) all the sides are equal;
(ii) each angle is a right angle;
(iii) diagonals are equal;
A
B
(iv) diagonals bisect at right angles.
Unit 5
322
Think it over!
(iii) and (iv) are consequences of (i) and (ii). Can you prove them?
A square can also be defined as:
(a) a rectangle in which adjacent sides are equal;
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(b) a rhombus in which each angle is 90◦ .
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Think it over!
Suppose the perimeters of a square and a rhombus are equal. Do
they have equal area?
Example 12. A field is in the shape of a square with side 20 m. A pathway
of 2 m width is surrounding it. Find the outer perimeter of the pathway.
D
Solution: Width of the pathway
is 2 m. Length of the side of the
outer square= (20+2+2)= 24 m.
Hence perimeter = 4 × 24 = 96 m..
C
20 m
2m
A
B
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Example 13. The square field has area 196 m2 . Find the length of the
wire required to fence it around 3 times.
No
t
Solution: Suppose s is the side-length of a square. Then its area is s2 sq.
units. We are given that s2 = 196m2 . Therefore S = 14m. Thus perimeter =
4 × s = 4 × 14 = 56 m.
The length of the wire required to fence around it 3 times is 56 × 3 = 168 m.
Kite
You have seen that in a rhombus, a diagonal divides the rhombus in to
two congruent isosceles triangles. Suppose we take two isosceles triangles
whose bases are of equal length and glue them together to get a quadrilateral. You get a special type of quadrilateral. Such a quadrilateral is called
a kite.
Quadrilaterals
323
A
O
D
C
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B
Kite is a quadrilateral in which two isosceles triangles are joined along the common
base. In the adjoining figure AB = AD,
BC = CD, and BD is the common base. It is
important to observe that triangles ABC and
ADC are congruent, but the triangles ABD
and CBD need not be congruent. Can you see
that if ABD and CBD are also congruent, the
quadrilateral ABCD reduces to a rombhus?
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Properties of kite
Like rhombus, kite has some properties which we record here:
1. There are two pairs of equal sides; AB = AD and CB = CD in the
previous diagram.
2. One of the diagonals bisect the other diagonal perpendicularly; the
diagonal BD is bisected perpendicularly by AC in the previous figure.
3. One of the diagonals bisect the apex angles; the diagonal AC bisects
the apex angles ∠A and ∠C.
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Example 14. In the figure P QRS is a kite; P Q = 3 cm and QR = 6 cm.
Find the perimeter of P QRS.
S
Solution: We have P Q = P S = 3
cm, QR = SR = 6 cm. Hence the
P
R
O
perimeter = P Q+QR+RS +P S =
3 + 6 + 6 + 3 = 18 cm.
Q
Exercise 3.5.5
1. The sides of the rectangle are in the ratio 2:1. The perimeter is 30 cm.
Calculate the measure of all the sides.
D
C
2. In the adjacent rectangle ABCD,
30
∠OCD = 30◦ . Calculate ∠BOC. What
O
type of triangle is BOC?
A
B
3. All rectangles are parallelograms, but all parallelograms are not rectangles. Justify this statements.
Unit 5
324
4. Prove logically that the diagonals of a rectangle are equal.
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5. The sides of a rectangular park are in the ratio 4:3. If the area is 1728
m2 , find the cost of fencing it at the rate of Rs. 2.50/m.
6. A rectangular yard contains two
D
F
E
C
flower beds in the shape of congruent isosceles right triangle.
The remaining portion is a yard
A
B
of trapezoidal
shape (see fig). whose parallel sides have lengths 15 m and 25 m.
What fraction of the yard is occupied by the flower bed?
7. In a rhombus ABCD, ∠C = 70◦ . Find the other angles of the rhombus.
8. In a rhombus P QRS, ∠SQR = 40◦ and P Q = 3 cm. Find ∠SP Q, ∠QSR
and the perimeter of the rhombus.
9. In a rhombus P QRS, if P Q = 3x − 7 and QR = x + 3, find P S.
10. Let ABCD be a rhombus and ∠ABC = 124◦ . Calculate ∠A, ∠D and
∠C.
11. Rhombus is a parallelogram: justify.
12. In a given square ABCD, if the area of triangle ABD is 36 cm2 , find
(i) the area of triangle BCD;
(ii) the area of the square ABCD.
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13. The side of a square ABCD is 5 cm and another square P QRS has
perimeter equal to 40 cm. Find the ratio of the perimeter of ABCD to
perimeter of P QRS. Find the ratio of the area ABCD to the area of
P QRS.
No
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14. A square field has side 20 m. Find the length of the wire required to
fence it four times.
15. List out the differences between square and rhombus.
16. Four congruent rectangles are
D
placed as shown in the figure.
S
Area of the outer square is 4
R
times that of the inner square.
What is the ratio of length to
P
Q
breadth of the congruent rectA
angles?
C
B
Quadrilaterals
325
Additional problems on “Quadrilaterals”
1. Complete the following:
(a) A quadrilateral has ————————— sides.
(b) A quadrilateral has ————————— diagonals.
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(c) A quadrilateral in which one pair of sides are parallel to each
other is —————————.
(d) In an isosceles trapezium, the base angles are —————————.
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(e) In a rhombus, the diagonals bisect each other in ———————
—— angles.
(f) In a square, all the sides are —————————.
2. Let ABCD be a parallelogram. What special name will you give it:
(a) if AB = BC?
(b) if ∠BAD = 90◦ ?
(c) if AB = AD and ∠BAD = 90◦ ?
3. A quadrilateral has three acute angles each measuring 70◦ . What is
the measure of the fourth angle?
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4. The difference between the two adjacent angles of a parallelogram is
20◦ . Find measures of all the angles of the parallelogram.
No
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5. The angles of a quadrilateral are in the ratio 1:2:3:4. Find all the
angles of the quadrilateral.
6. Let P QRS be a parallelogram with P Q= 10 cm and QR= 6 cm. Calculate the measures of the other two sides and the perimeter of P QRS.
7. The perimeter of a square is 60 cm. Find its side-length.
8. Let ABCD be a square and let AC = BD = 10 cm. Let AC and BD
intersect in O. Find OC and OD.
9. Let P QRS be a rhombus, with P R = 15 cm and QS = 8 cm. Find the
area of the rhombus.
Unit 5
326
10. Let ABCD be a parallelogram and suppose the bisectors of ∠A and
∠B meet at P . Prove that ∠AP B = 90◦ .
11. Let ABCD be a square. Locate points P , Q, R, S on the sides AB,
BC, CD, DA respectively such that AP = BQ = CR = DS. Prove that
P QRS is a square.
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12. Let ABCD be a rectangle and let P, Q, R, S be the mid-points of AB,
BC, CD, DA respectively. Prove that P QRS is a rhombus.
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13. Let ABCD be a quadrilateral in which the diagonals intersect at O
perpendicularly. Prove that AB + BC + CD + DA > AC + BD.
14. Let ABCD be a quadrilateral with diagonals AC and BD. Prove the
following statements ((Compare these with the previous problem);
(a) AB + BC + CD > AD;
(b) AB + BC + CD + DA > 2AC;
(c) AB + BC + CD + DA > 2BD;
(d) AB + BC + CD + DA > AC + BD.
15. Let P QRS be a kite such that P Q > P S. Prove that ∠P QR > ∠P SR.
(Hint: Join QS.)
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16. Let ABCD be a quadrilateral in which AB is the smallest side and CD
is the largest side. Prove that ∠A > ∠C and ∠B > ∠D.(Hint: Join AC
and BD.)
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17. In a triangle ABC, let D be the mid-point of BC. Prove that AB + AC >
2AD.(What property of quadrilateral is needed here?)
18. Let ABCD be a quadrilateral and let P, Q, R, S be the mid-points of AB,
BC, CD, DA respectively. Prove that P QRS is a parallelogram.(What
extra result you need to prove this ?)
Glossary
Quadrilateral: a linear figure on a plane consisting of four line segments,
which are placed in an ordered way such that the adjacent segments meet
only at their end points.
Convex quadrilaterals: a quadrilateral in which each interior angle is less
Answers
327
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than 180◦ .
Concave quadrilateral: a quadrilateral in which some angle exceeds 180◦ .
Diagonal: the line segment joining opposite vertices of a quadrilateral.
Trapezium: a quadrilateral in which a pair of sides are parallel.
Parallelogram: a quadrilateral in which two pairs of sides are parallel.
Rhombus: a quadrilateral in which all the sides are equal.
Square: a quadrilateral in which all the sides are equal and all the angles
are equal to 90◦ .
Rectangle: a quadrilateral in which all the angles are equal to 90◦ .
Kite: a quadrilateral formed by a pair of isosceles triangle glued along a
common side.
Points to remember
• The sum of all the four angles of a quadrilateral is 360◦.
• For a quadrilateral, equiangularity is not the same as equilaterality,
unlike for triangles.
• The diagonals intersect perpendicularly in a rhombus and a kite.
• Any quadrilateral in which diagonals bisect each other is a parallelogram.
Answers to Exercises in Chapter 3.
Exercise 3.1.3
to
3. (i) 60◦ ; (ii) 18◦ ; (iii) 135◦ ; (iv) 90◦ ; (v) 30◦ ; (vi) 65◦ .
Exercise 3.1.4
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1. ∠DML = 45◦ ; ∠BLQ = 45◦ ; ∠MLB = 135◦ ; ∠CMP = 45◦ ; ∠CML = 135◦ ;
MLA = 45◦ ; ∠QLA = 135◦ . 2. 40◦ .
Additional problems on “Axioms, postulates and theorems”
1. (i) A. (ii) D. (iii) B. (iv) B. (e) D.
3. 3.
8. 40◦ .
9. (i) 190◦ ; (ii) 22◦ 30′ .
10. 144◦ .
Exercise 3.2.1
1. (i) → (C); (ii) → (D); (iii) → (A); (iv) → (B). 2. (i) scalene; (ii) scalene;
(iii) scalene; (iv) isosceles; (v) scalene; (vi) scalene; (vii) scalene; (viii)
equilateral; (ix) isosceles; (x) isosceles.
Answers
328
Exercise 3.2.2
1. 85◦ . 2. 55◦ . 3. 65◦ each. 4 30◦ , 60◦ and 90◦ .
∠B = 35◦ , ∠C = 80◦ . 6. 50◦ , 60◦ and 70◦ .
5 x = 50◦ , ∠A = 65◦ ,
Exercise 3.2.3
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1. If Ext∠B = 136◦ and Ext∠C = 104◦ , then ∠A = 60◦ , ∠B = 44◦ and
∠C = 76◦ . 3. (i) 130◦ ; (ii) 56◦ ; (iii) 35◦ ; (iv) 52◦ ; (v) 40◦ . 4. ∠T RS = 50◦ ,
∠P SQ = 80◦ .
5. The other interior opposite angle is 90◦ and the third
angle is 60◦ . 6. 180◦ .
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Additional problems on “Theorems on triangles”
1. (a) 180◦ ; (b) the interior; (c) larger; (d) one; (e) one.
2. (a) A. (b) B. (c) A. (d) C. (e) D.
3. 110◦ . 4. 35◦ each. 5. 36◦ , 54◦ , 90◦ . 6. 45◦ , 60◦ , 75◦ . 7. 90◦ .
8. ∠C = 30◦ , ∠B = 60◦ , ∠A = 90◦ . 9. x = 90◦ . 10. ∠A = 80◦ , ∠B = 65◦ ,
∠C = 35◦ . 11. 30◦ , 50◦ , 100◦ . 12. 110◦ . 13. 45◦ , 45◦ , 90◦ . 14. 36◦ , 72◦ ,
72◦ . 15. 110◦ . 16. 30◦ , 30◦ , 120◦ .
Exercise 3.3.3
1. ∠B = ∠C = 65◦ . 2. 58◦ . 3. (i) 110◦ ; (ii) 55◦ ; (iii) 20◦ ; (iv) 40◦ .
4. ∠ACD = 120◦ , ∠ADC = 30◦ .
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Exercise 3.3.7
1. BC is the largest and CA is the smallest. 2. ∠C < ∠A < ∠B.
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Additional problems on “Congruency of triangles”
1. (a) largest (b) less (c) larger (d) smaller (e) less (f) larger
9. 30◦ , 30◦ , 120◦ .
Exercise 3.5.2
1. 80◦ each.
2. 55◦ .
3. 54◦ , 81◦ and 108◦ .
4. 50◦ .
Exercise 3.5.3
1. ∠S = 110◦ and ∠R = 100◦ .
Exercise 3.5.4
3. ∠RP Q = 30◦ and ∠RSQ = 40◦ .
Answers
329
1. 120◦ , 60◦ , 120◦ , 60◦ . 2. 150 m and 75 m. 3. ∠ABO = 35◦ , ∠ADC = 105◦ ,
∠ACB = 40◦ , ∠CBD = 70◦ . 4. ∠A = ∠C = 75◦ , ∠B = ∠D = 105◦ .
6. 60◦ , 120◦ , 60◦ , 120◦ .
Exercise 3.5.5
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Additional problems on “Quadrilaterals”
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1. 10 cm, 5 cm, 10 cm, 5 cm. 2. ∠BOC = 60◦ ; ∆BOC is equilateral.
1
5. . 420. 6. . 7. 110◦ , 70◦ , 110◦ . 8. ∠SP Q = 100◦ , ∠QSR = 40◦ and
5
perimeter= 12 cm. 9. 8. 10. 56◦ , 124◦ , 56◦ . 12. (i) 36 cm2 ; (ii) 72 cm2 .
13. (i) 1:2 (ii) 1:4. 14 320 m. 16. 3:1.
1. (a) four; (b) two; (c) a trapezium; (d) equal; (e) right; (f) equal.
2. (a) rhombus; (b) rectangle; (c) square.
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3. 150◦ . 4. 80◦ , 100◦ , 80◦ , 100◦ . 5. 36◦ , 72◦ ; 108◦ , 144◦ . 6. RS = 10 cm,
SP = 6 cm; perimeter is 32 cm. 7. 15 cm. 8. OC = OD = 5 cm. 9. 120
cm2 .
CHAPTER 4
UNIT 1
MENSURATION
After studying this unit you learn to:
• relate formulae to given problems;
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• list out the properties of cubes and cuboids;
d
• recognise the cubes and cuboidal objects used in our day to day life;
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• substitute the data in the given formula and solve the problems.
4.1.1 Introduction
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When we look at an empty box, an empty bowl and an empty container,
it has some space and we can keep things in that empty space. A class
room has space for the students to sit in.
A solid occupies fixed amount of space. Solids occur in different shapes.
Observe the following diagrams. These shapes (cuboid, cube, cylinder,
sphere, cone, triangular prism etc) are known as three dimensional objects.
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Cuboid
Cube
Cylinder
Cone
Triangular
Prism
Do you see that each solid occupies some space. Each solid also has
some surface and hence has associated surface area. Since each occupies
Mensuration
331
book case
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4.1.2 Surface area of a cuboid
Almirah
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wooden box
match box
These are in the shape of cuboids.
d
space, it has some volume. We shall study these aspects corresponding to
some simple figures.
Observe the following figures
Let us understand the cuboids by doing an activity.
Activity 1:
Take a box in the shape of cuboid and cut it open along one edge, open
up the lids and spread it over a sheet of paper and fasten it with pins.(See
the figure)
D
C
6
A
B
1
H
2
E
to
G
F
5
3
4
How many faces has a
cuboid? Find the number of edges and vertices.
A cuboid has 6 faces, 12
edges and 8 vertices. Do
you see that a cuboid has
rectangular faces?.
No
t
Any face of a cuboid may be called its base (can you give the reason?).
The four faces which meet the base are called the lateral faces of a cuboid.
In the given figure above, the cuboid has 6 faces. They are ABCD,
EF GH, EF BA, HGCD, EHDA and F GCB. Any two adjacent faces of a
cuboid meet in a line segment, called an edge of the cuboid. The 12 edges
are AB, BC, CD, DA, EF , F G, GH, EH, AE, DH, GC and BF . The point of
intersection of three edges of a cuboid is a vertex of the cuboid. The eight
vertices are A, B, C, D, E, F , G and H.
Activity 2:
Take any cuboidal box. Fix a base (observe any face can be taken as
Unit 1
332
base). Place it vertically and wrap a thick sheet of paper such that it just
fits around the surface. Remove the paper and measure the area of the
paper. It is the lateral surface area (L.S.A) of the cuboid. Do this with
different cuboids.
l
V
d
b
I
b
l
he
h
h
II
b
III
l
IV
l
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Cut open the cuboidal box
and lay it flat on a sheet of
paper. Let the length of the
base be l; breadth of the
base be b; and the height
of the cuboid corresponding to this base be h units.
VI
b
b
l
Compute 2(lh + bh) and compare this with the area of the lateral surface
you have measured. Do they match? What is 2(lh + hb + bl)?
to
If you change the base, you may observe that the units l, b, h do
not change; only their representation as length, breadth and height may
change.
There are 6 faces in a cuboid. All the faces are rectangular in shape.
The sum of the areas of all the six surfaces is called the total surface
area (T.S.A) of the cuboid. Let us find a formula for total surface area and
lateral surface area of a cuboid. The total surface area of cuboid is
No
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Area of I + area of II + area of III + area of IV + area of V + area of VI (see
fig)
Hence
A = (l × h) + (l × b) + (b × h) + (l × h) + (b × h) + (l × b) sq. units.
Thus
A = 2lh + 2lb + 2bh = 2(lh + lb + bh) sq. units.
The lateral surface area of a cuboid is
area I + area II + area III + area IV.
Hence
L.S.A = (l × h) + (b × h) + (l × h) + b × h = 2h(l + b)
Mensuration
333
sq.units.
Here you may observe that the lateral surface area of a cuboid
depends on the base you choose for the cuboid. If you change the
base the lateral surface area changes. However, the total surface area
of the cuboid remains the same.
d
Cube
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Activity 3:
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A cuboid whose length, breadth and height are all equal is called a cube.
Ice cubes, sugar cubes, dice are some examples of cubes.
R
4 cm
G
G
G
G
4 cm
R
4 cm
Construct a cube of side 4 cm by
using card board. Paint any two
opposite sides with red paint and
remaining with green. Place the
cube on table with one of the red
faces as base. Identify the number of green faces bounding the
red faces.
to
The green faces are called lateral faces of the cube. What is the area of
the lateral faces? Do you see that: L.S.A of this cube is 4 × 16 = 64 cm2 ;
T.S.A of this cube is 6 × 16 = 96 cm2 .
No
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l
l
l
Observe the adjoining figure of a
cube of side l units. It has six
square faces. Area of each face
of the cube is l2 units. Area of 6
faces is therefore 6l2 units. And
the area of 4 lateral faces is 4l2
units.
Think it over!
What should be the maximum length of the ladder which
can be placed from the bottom of the floor to reach the
opposite corner of the roof of a room which is in the shape
of a cube?
Unit 1
334
Example 1. Find the lateral surface area and total surface area of a cuboid
which is 8 m long, 5 m broad and 3.5 m high.
Solution: We are given l = 8 m, b = 6 m, h = 3.5 m. We know that
L.S.A = 2h(l + b) = 2 × 3.5(8 + 6) = 7 × 14 = 98 m2 .
d
Similarly,
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T.S.A = 2(lb+bh+lh) = 2(8×6+6×3.5+8×3.5) = 2(48+21+28) = 2×(97) = 194 m2 .
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Example 2. How many tiles each of 30 cm × 20 cm are required to cover
the floor of hall of dimension 15 m by 12 m?
Solution: Since the tiles are in cm2 , we have to convert the dimensions of
the hall to cm first. It will be 1500 cm by 1200 cm. Hence the area of the
floor is
1500 × 1200 = 1800000 cm2 .
Area of each tile is 30 × 20 = 600 cm2 . Hence the number of tiles required
to cover the floor is:
1800000
= 3000.
600
Example 3. Find the length of each side of a cube having the total surface
area is 294 cm2 .
to
Solution: Given T.S.A of a cube as 294 cm2 , we have to find its length l.
We know T.S.A of a cube is equal to 6l2 . Thus 6l2 = 294 or l2 = 294/6 = 49.
Hence l = 7 cm.
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Example 4. The total surface area of a cube is 600 cm2 . Find the lateral
surface area of the cube.
Solution: Given T.S.A of a cube as 600 cm2 , we have to find its L.S.A. But
we know T.S.A = 6l2 , where l is the length of the cube. Hence 600 = 6l2 or
l2 = 100 units. Taking square-root, we get l = 10 cm. But we also know
that L.S.A. = 4l2 . Hence
L.S.A. = 4 × 10 × 10 = 400 cm2 .
Example 5. Find the area of a metal sheet required to make a cube of
length 2 m. Find the cost of metal sheet required at the rate of 8/ m2 to
Mensuration
335
make the cube.
Solution: We know that l = 2 m. We have to find T.S.A. of the cube. But
T.S.A. = 6l2 = 6 × 2 × 2 = 24 m2 .
Exercise 4.1.2
d
The cost of the metal sheet required is therefore 24 × 8 = 192 rupees.
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1. Find the total surface area of the cuboid with l = 4 m, b = 3 m and
h = 1.5 m.
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2. Find the area of four walls of a room whose length 3.5 m, breadth 2.5
m and height 3 m.
3. The dimensions of a room are l = 8 m, b = 5 m, h = 4 m. Find the cost
of distempering its four walls at the rate of 40/ m2 .
4. A room is 4.8 m long, 3.6 m broad and 2 m high. Find the cost of
laying tiles on its floor and its four walls at the rate of 100/ m2 .
5. A closed box is 40 cm long, 50 cm wide and 60 cm deep. Find the
area of the foil needed for covering it.
6. The total surface area of a cube is 384 cm2 . Calculate the side of the
cube.
7. The L.S.A of a cube is 64 cm2 . Calculate the side of the cube.
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8. Find the cost of white washing the four walls of a cubical room of side
4 m at the rate of 20/ m2 .
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9. A cubical box has edge 10 cm and another cuboidal box is 12.5 cm
long, 10 cm wide and 8 cm high.
(i) Which box has smaller total surface area?
(ii) If each edge of the cube is doubled, how many times will its T.S.A
increase?
4.1.3 Volume of cubes and cuboids
Amount of space occupied by a three dimensional object is called its
volume. Volume of a room is bigger than the volume of a brick or shoe
box. Remember we use square units to measure the area of a region or a
surface. Similarly, we use cubic units to measure the volume of a solid,
as solids are three dimensional objects.
Unit 1
336
l
h
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b
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The figures above are cube and cuboid which occupy space in three
dimension and hence posses volume. We are interested in finding their
volume.
Activity 4:
Take four cubes each of
length 1 unit and arrange
as shown in figure. Take
another 4 cubes of same
size and arrange them on
the top of the cubes arranged earlier, as shown
in the next figure.
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Do you get another cube? You have used 8 cubes, i.e, the volume of the
new cube is 8 cubic units. Here l = b = h = 2 units.
Volume of this cube = 2 units × 2 units × 2 units = 8 cubic units.
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In general volume of a cube = side × side × side. Thus
V = l × l × l = l3
cubic units. Cubic units are used to measure volume:
1 cm3 = 1 cm × 1 cm × 1 cm;
1 m3 =1 m × 1 m × 1 m = 100 cm × 100 cm × 100 cm = 106 cm3 .
Volume of a cuboid
Activity 5:
Take 24 cubes of equal size. Arrange them to form a cuboid. You can
arrange them in many ways to get a cuboid. Observe the following table:
Mensuration
337
2
6
2
3
4
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12
4
1
d
1
h
l×b×h
1 24 cubic units
1 24 cubic units
2 24 cubic units
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l
b
12 2
6 4
4 3
Since we have used 24 cubes in making these cuboids, volume of each
cuboid is 24 cubic units. Thus we may arrive at the conclusion that the
volume of each cuboid is equal to the product of its length, breadth and
height: volume of the cuboid = l × b × h. Since l × b = area of the base, we
can also write that
volume of the cuboid = area of the base × height.
Think it over!
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There are 36 cubes having of side-length 1cm. How many
cuboids of different dimensions can be prepared by using
all of them.
Example 6 A match box measure 4 cm, 2.5 cm and 1.5 cm. What will be
the volume of the packet containing 12 such boxes.
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Solution: The volume of each box is 4 × 2.5 × 1.4 = 15 cm3 . Volume of a
packet containing 12 such boxes is 15 × 12 = 180 cm3 .
Example 7 How many 3 meter cubes can be cut from a cuboid measuring
18 m × 12 m × 9 m.
Solution: Volume of the cuboid = 18 × 12 × 9 m3 . Volume of each cube to
be cut is = 3 × 3 × 3 m3 . Hence number of cubes that can be cut out from
the cuboid is
18 × 12 × 9
= 72.
3×3×3
Unit 1
338
Exercise 4.1.3
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1. Three metal cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find (i) side-length (ii)
total surface area of the new cube. What is the diifernece between the
total surface area of the new cube and the sum of total surface areas
of the original three cubes?
2. Two cubes, each of volume 512 cm3 are joined end to end. Find the
lateral and total surface areas of the resulting cuboid.
3. The length, breadth and height of a cuboid are in the ratio 6:5:3. If
the total surface area is 504 cm2 , find its dimension. Also find the
volume of the cuboid.
4. How many m3 of soil has to be excavated from a rectangular well 28
m deep and whose base dimensions are 10 m and 8 m. Also find the
cost of plastering its vertical walls at the rate of 15/m2 .
5. A solid cubical box of fine wood costs 256 at the rate 500/m3.
Find its volume and length of each side.
Additional problems on “Mensuration”
1. Find the total surface area and volume of a cube whose length is 12
cm.
2. Find the volume of a cube whose surface area is 486 cm2 .
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3. A tank, which is cuboidal in shape, has volume 6.4 m3 . The length
and breadth of the base are 2 m and 1.6 m respectively. Find the
depth of the tank.
4. Find the area of four walls of a room having length, breadth and
height as 8 m, 5 m and 3 m respectively. Find the cost of whitewashing the walls at the rate of Rs. 15/m2.
5. A room is 6 m long, 4 m broad and 3 m high. Find the cost of laying
tiles on its floor and four walls at the cost of Rs. 80/m2 .
6. The length, breadth and height of a cuboid are in the ratio 5:3:2. If its
volume is 35.937 m3 , find its dimension. Also find the total surface
area of the cuboid.
Mensuration
339
7. Suppose the perimeter of one face of a cube is 24 cm. What is its
volume?
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8. A wooden box has inner dimensions l = 6 m, b = 8 m and h = 9 m and
it has uniform thickness of 10 cm. The lateral surface of the outer
side has to be painted at the rate of Rs. 50/ m2 . What is the cost of
painting?
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9. Each edge of a cube is increased by 20%. What is the percentage
increase in the volume of the cube?
Glossary
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10. Suppose the length of a cube is increased by 10% and its breadth is
decreased by 10%. Will the volume of the new cuboid be the same as
that of the cube? What about the total surface areas? If they change,
what would be the percentage change in both the cases?
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Mensuration: finding the area of a planar objects or volume of a three
dimensional objects.
Solids: objects which occupy space in three dimensions.
cube: a cuboid having equal length, breadth and height.
Lateral surface: the surface of a cuboid which is neither a base nor the
top surface.
Edges: the line along which surfaces meet.
Surface area: the area of the faces of a cuboid.
Volume: measure of the space occupied by a solid.
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Points to remember
• A solid is three dimensional figure; it occupies space(in three dimensions).
• A solid has two quantities associated with it; its surface area and it
volume.
• Area is measured in square units whereas volume is measured in
cubic units.
Answers
340
Answers to Exercises in Chapter 4.
Exercise 4.1.2
1. 45 m2 . 2. 36 m2 . 3.
cm. 7. 4 cm. 8. 1280.
4160. 4. 5088. 5. 14,800 cm2 .
9. (i) cube. (ii) 4 times.
6. 8
Exercise 4.1.3
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1. (i) 6 cm; (ii) 216 cm2 ; (iii) 84 cm2 . 2. L.S.A= 384 cm2 and T.S.A=
640 cm2 . 3. dimension: length=12 cm, breadth= 10 cm, height=6 cm;
volume= 720 cm3 . 4. 2240 m3 and 15,120. 5. volume = 0.512 m3 ;
side-length= 80 cm.
Additional problems on “Mensuration”
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1. 864 cm2 , 1728 cm3 . 2. 729 cm3 . 3. 2 m. 4. 78 m2 and 1,170.
6,720. 6. 1.65 m, 0.99 m, 0.66 m; T.S.A= 6.7918 m2 .
7. 216
5.
3
cm . 8. 8,931.5 9. 1.728 % 10. Volume decreases by 1% and T.S.A
decreases by 2%.
OPTIONAL PROBLEMS
These problems are included to pose challenge to those students who are looking for it. These are neither for class
room discussion nor for examination.
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1. Find a proper fraction greater than 1/3, given that the fraction does
not change if the numerator is increased by a positive integer and the
denominator is multiplied by the same positive integer.
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2. Find all rational numbers p/q such that
p
p2 + 30
= 2
.
q
q + 30
3. Show that the number of distinct remainders which occur when a
perfect square is divided by an odd prime p is (p + 1)/2.
4. Find the number of positive divisors of 22 , 32 , 42 , 52 , 102 . Do you
see that the number of divisors is odd? Prove the proposition that
the number of positive divisors of a perfect square is always an odd
number.
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5. Find all odd natural numbers n for which there is a unique perfect
square strictly between n2 and 2n2 .
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6. A person was born in 19-th century. His age was x years in the year
x2 . If he passed away in 1975, what was his age at the time of his
demise?
7. There is a unique 4-digit number n = abcd such that n2 also ends in
abcd. Find this number.
8. In the adjoining figure, you are given
a skeleton of a 4 × 4 magic square
which uses numbers from 1 to 16.
Find A and B. Hard: Complete the
magic square.
14
11
5
A
8
12
3
B
Optional Problems
342
9. (Hard) Prove that the magic sum of a 3 × 3 magic square is three times
the central number.
10. The sum of the squares of three natural numbers is divisible by 9.
Prove that one can select two among these three such that their difference is divisible by 9.
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11. Suppose p and p2 + 2 are prime numbers. Prove that p3 + 2 is also a
prime number.
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12. Find all perfect squares which when divided by 11 give a prime as
quotient and 4 as remainder.
13. The number 60 is written on a board. Two players take turn and
play the following game: they can subtract any positive divisor of
the number on the board and replace the number by the result of
this subtraction. The player who writes 0 on the board wins. Which
player wins, the first or the second?
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14. There are three pile of stones: one with 10 stones, another with 15
stones and the third with 20. Two players play the following game: at
each turn, a player can choose one of the pile and divide it in to two
smaller pile. The player who cannot do this is the loser. Who will win,
the first or the second?
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15. The numbers 1 to 20 are written on a black board in a row. Two
players take turns and put either + sign or − sign between these
numbers according to their choice, one at a time. After all the signs
are put, the sum is evaluated. The first player wins if the number
obtained is even and the second wins if the number is odd. Who
wins?
16. Suppose a natural number n is such that m2 < n < (m + 1)2 for some
natural number m. If n − l = m2 and n + k = (m + 1)2 , prove that n − kl
is a perfect square.
17. If x, y, z are integers such that x2 + y 2 = z 2 , prove that one of x, y is
divisible by 3.
Optional Problems
343
18. (Hard) Suppose x, y, z are three natural numbers such that they do not
have any common factor and x2 + y 2 = z 2 . Prove that xyz is divisible
by 60.
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19. For any x, suppose [x] denotes the largest integer not exceeding x; for
example [2.5] = 2 and [−1.6] = −2. Find all positive real numbers a
such that a[a] = 8.
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20. How many positive integers less than 1000 are 6 times the sum of
their digits?
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21. How many palindromes between 1000 and 10000 are there which are
divisible by 7?
22. (Hard) Suppose there are two mirrors inclined at an angle 8◦ . A ray of
light starts at the point A gets reflected in B and then gets reflected
in C and so on n times in successive mirrors until it hits a mirror at
right angle and then traces back the same path. What is the largest
possible value of n?(The following figure is an example of 7 reflections.)
B
O
C
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8
A
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23. Show that the shortest distance from a point to a line is the perpendicular distance.
24. (Hard) Let P be any point inside a triangle ABC. Prove that
BC + CA + AB
< P A + P B + P C < BC + CA + AB.
2
25. Let D be the mid-point of the side BC in a triangle ABC. Prove that
AB + AC > 2AD.
26. Suppose AD and BE are respectively the medians drawn from A and
B on to the opposite sides in a triangle ABC. If BC > CA, prove that
BE > AD.
Optional Problems
344
27. Let ABCD be a quadrilateral in which AB is the smallest side and CD
is the largest side. Prove that ∠A > ∠C and ∠B > ∠D.
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D
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28. In the adjoining figure, a corner of
a shaded star is at the mid-point
S
Q
of each side of the large square.
What fraction of the large square is
B
shaded?
A
P
29. In the adjoining figure, the outer
C
equilateral triangle ABC has area 1
E
and the points D, E, F are such that
DB = EC = F A and each equal to
F
one-fourth the side of the triangle
A
D B
ABC. What is the area of DEF .
30. In the quadrilateral ABCD, AB = 5, BC = 17, CD = 5 and DA = 9. It
is known that BD is an integer. What is BD?
31. In a triangle ABC, AB = 2AC. Let D, E be points respectively on the
segments AB, BC such that ∠BAE = ∠ACD. Let F be the point of
intersection of AE and CD. Suppose CF E is an equilateral triangle.
What is ∠ACB?
32. Let ABC be a triangle and let AD be the bisector of ∠A with D on BC.
Prove that AB/AC = BD/DC.
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33. Prove that the sum of the interior angles of pentagon is 540◦ . What
do you expect for a hexagon? What about an octogon? Can you
generalise this to a general n-gon? Can you prove your guess? What
tools you need?
34. Suppose ABCD is a parallelogram. Equilateral triangles CBX and
DCY are constructed externally respectvely on the sides DC and CB.
Prove that AXY is an equilateral triangle.
35. One dimension of a cube is increased by 1, another is decreased by 1
and the third remains as it is. The volume of the resulting cuboid is 5
less than that of the original cube. What is the volume of the original
cube?
36. A solid cube has side length 3 units. A 2 × 2 square hole is cut into the
centre of each face. The edge of each square is parallel to the sides of
Optional Problems
345
the cube and each cut goes all the way through the cube. What is the
volume of the resulting solid?
37. Let ABCD be a trapezium in which AB k CD and AD ⊥ DC. Suppose
AB > BC and draw CM ⊥ AB. Suppose BC = 5 cm, MB = 3 cm and
DC = 8 cm. Find the perimeter of ABCD. What is the area of ABCD?
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38. The diagonals of a rhombus are 24 cm and 10 cm respectively. Find
its sides.