Download Mathematics - RESONANCE PCCP IDEAL for NTSE, IJSO

Resonance Pre-foundation Career Care Programmes
(PCCP) Division
WORKSHOP TAPASYA
SHEET
MATHEMATICS
COURSE : NTSE (STAGE-I)
CLASS-X
NTSE
Subject : Mathematics
S. No.
Topics
Page No.
1.
Number System
1 - 10
2.
Polynomial
11 - 18
3.
Linear Equation (2 variables)
19 - 24
4.
Quadratic equation
25 - 29
5.
Progression
30 - 33
6.
Lines and Angles, Triangles
34 - 42
7.
Quadrilateral
43 - 47
8.
Circle
48 - 56
© Copyright reserved.
All right reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly
prohibited. This material belongs to only the enrolled student of RESONANCE. Any sale/resale of this material is
punishable under law. Subject to Kota Jurisdiction only.
.13RPCCP
NUMBER
SYSTEM
(viii) Real numbers : Numbers which can represent
INTRODUCTION
actual physical quantities in a meaningful way are
Number System is a method of writing numerals to
represent numbers.


known as real numbers. These can be represented
on the number line. Number line is geometrical straight
Ten symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 are used to
represent any number (however large it may be) in our
number system.
line with arbitrarily defined zero (origin).
(ix) Prime numbers : All natural numbers that have
Each of the symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 is called a
digit or a figure.
one and itself only as their factors are called prime
numbers i.e. prime numbers are exactly divisible by 1
and themselves. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23,...etc.
If P is the set of prime number then P = {2, 3, 5, 7,...}.
(i) Natural numbers :
Counting numbers are known as natural numbers.
N = { 1, 2, 3, 4, ... }.
(x) Co-prime Numbers : If the H.C.F. of the given
numbers (not necessarily prime) is 1 then they are
known as co-prime numbers. e.g. 4, 9 are co-prime
(ii) Whole numbers :
All natural numbers together with 0 form the collection
of all whole numbers.
W = { 0, 1, 2, 3, 4, ... }.
as H.C.F. of (4, 9) = 1.

Any two consecutive numbers will always be
co-prime.
(iii) Integers : All natural numbers, 0 and negative of
natural numbers form the collection of all integers.
I or Z = { ..., – 3, – 2, – 1, 0, 1, 2, 3, ... }.
(xi) Twin primes : Pairs of prime numbers which have
(iv) Even Numbers : All integers which are divisible by
2 are called even numbers. Even numbers are denoted
by the expression 2n, where n is any integer. So, if E is
a set of even numbers, then E = { ..., – 4, –2, 0, 2, 4,...}.
3, 5 ; 5, 7 ; 11,13 ; 17, 19 ; 29, 31 ; 41, 43 ; 59, 61 and
(v) Odd Numbers : All integers which are not divisible
by 2 are called odd numbers. Odd numbers are
denoted by the general expression 2n –1 where n
is any integer. If O is a set of odd numbers, then
O = {..., –5, –3, –1, 1, 3, 5,...}.
are not prime are composite numbers. If C is the set
(vi) Rational numbers : These are real numbers which
p
can be expressed in the form of , where p and q are
q
integers and q  0 . e.g. 2/3, 37/15, –17/19.

All natural numbers, whole numbers and integers are
rational numbers.

Rational numbers include all Integers (without any
decimal part to it), terminating fractions ( fractions in
which the decimal parts are terminating e.g. 0.75, –
0.02 etc.) and also non-terminating but recurring decimals
e.g. 0.666....., – 2.333...., etc.
(vii) Irrational Numbers : All real number which are
not rational are irrational numbers. These are nonrecurring as well as non-terminating type of decimal
numbers.
For Ex. :
2,
3
4 , 2 3 ,
2 3 ,
47
3 etc.
only one composite number between them are called
Twin primes.
71, 73 etc. are twin primes.
(xii) Composite numbers : All natural numbers, which
of composite number then C = {4, 6, 8, 9, 10, 12,...}.

1 is neither prime nor composite number.
(xiii) Imaginary Numbers : All the numbers whose
square is negative are called imaginary numbers.
e.g. 3i, -4i, i, ... ; where i =
- 1.
IDENTIFICATION OF PRIME NUMBER
Step (i) : Find approximate square root of given number.
Step (ii) : Divide the given number by prime numbers less
than approximately square root of number. If given
number is not divisible by any of these prime number
then the number is prime otherwise not.
Ex.1 Is 673 a prime number ?
Sol. Approximate square root = 26
Prime number < 26 are 2, 3, 5, 7, 11, 13, 17, 19, 23. But
673 is not divisible by any of these prime number. So,
673 is a prime number.
PAGE # 11
FRACTIONS
(a) Common fraction : Fractions whose denominator
is not 10.
(b) Decimal fraction : Fractions whose denominator is
10 or any power of 10.
(c) Proper fraction : Numerator < Denominator i.e.
(d) Improper fraction : Numerator > Denominator i.e.
3
5
.
5
.
3
(e) Mixed fraction : Consists of integral as well as
fractional part i.e. 3
Short cut method for mixed recurring decimal : Form
a fraction in which numerator is the difference between
the number formed by all the digits after the decimal
point taking the repeated digits only once and that
formed by the digits which are not repeated and the
denominator is the number formed by as many nines
as there are repeated digits followed by as many zeros
as the number of non-repeated digits.
Ex.2 Change 2.76 45 in the form of p/q.
Sol. 2.7645 =
3041
27645  276
27369
=
=
.
9900
9900
1100
2
.
7
(f) Compound fraction : Fraction whose numerator and
The absolute value of a real number I x I is defined as
2/3
denominator themselves are fractions. i.e.
.
5/7

Improper fraction can be written in the form of mixed
fraction.
x, if x  0
|x|= 
 x, if x  0
For example :
| 2 |= 2 ; 2 > 0 and | – 2 | = – (– 2 ) = 2; – 2 < 0.
COMPARISON OF FRACTIONS
Suppose some fractions are to be arranged in
ascending or descending order of magnitude. Then,
convert each one of the given fractions in the decimal
form, and arrange them accordingly.
Now,
3
6
7
= 0.6,
= 0.857,
= 0.777.....
5
7
9
Since 0.857 > 0.777....> 0.6, so
6
7
3
>
> .
7
9
5
PURE RECURRING DECIMAL
If in a decimal fraction, a figure or a set of figures is
repeated continuously, then such a number is called
recurring decimal.
Thus,
1
= 0.333...... = 0.3 ,
3
22
= 3.142857142857.....
7
= 3. 142857 .
Conversion of decimal numbers into rational
numbers of the form p/q.
Short cut method for pure recurring decimal: Write
the repeated digit or digits only once in the numerator
and take as many nines in the denominator as there
are repeating digits in the given number.
BODMAS RULE
This rule depicts the correct sequence in which the
operations are to be executed so as to find out the
value of a given expression.
Here,‘B’ stands for ‘Bracket’, ‘O’ for ‘Of’, ‘D’ for
‘Division’, ‘M’ for ‘Multiplication’, ‘A’ for ‘Addition’ and
‘S’ for ‘Subtraction’.
Thus, in simplifying an expression, first of all the
brackets must be removed, strictly in the order ( ), { }
and [ ].
After removing the brackets, we must use the following
operations strictly in the order :
(i) Of
(ii) Division
(iii) Multiplication
(iv) Addition (v) Subtraction.
Vinculum (or Bar) : When an expression contains
Vinculum, before applying the ‘BODMAS’ rule, we
simplify the expression under the Vinculum.

MIXED RECURRING DECIMAL
A decimal is said to be a mixed recurring decimal if
there is at least one digit after the decimal point, which
is not repeated.
 
3
5
1

7  6  
Sol. 5 –  4   2   2  42 




e.g., (i) 0. 3 = 3/9 or 1/3
(ii) 0. 387 = 387/999

3  1 
1 1 
Ex.3 Simplify : 5 –  4  2 2   0.5  6  7   :


 


3
5
1
3
5
22 
1  
= 5 –  4   2   2  42 




3
83 
= 5 –  4   2  42  = 5 –  4  42 





= 5–
 420  229 
229
191
23
 =
= 
=2
.
84
84
84
84


PAGE # 22
Properties of Square Roots :
Squares : When a number is multiplied by itself then
the product is called the square of that number.
Perfect Square : A natural number is called a perfect
square if it is the square of any other natural number
e.g. 1, 4, 9,... are the squares of 1, 2, 3,... respectively.
Ex.4 Find the smallest number by which 180 must be
multiplied so that the product is a perfect square.
Sol. Given number is 180, first we resolve it into prime
factors.
(i) If the unit digit of a number is 2, 3, 7 or 8, then it does
not have a square root in N.
(ii) If a number ends in an odd number of zeros, then it
does not have a square root in N.
(iii) The square root of an even number is even and
square root of an odd number is odd.e.g. 81 = 9,
256 = 16, 324 = 18 ...etc.
(iv) Negative numbers have no square root in set of
real numbers.
Ex.6 Find the square root of 5 +
Sol. Let
2 180
2
90
3
45
3
15
5
5

1

5+
5 3 =
p +
3 =p+q+2
180 = 2 × 2 × 3 × 3 × 5
Clearly 5 has no pair. Thus if we multiply it by 5 then
product will be a perfect square.


2 pq = 3
...(ii)
4pq = 3
...(iii)
(p – q)2 = (p + q)2 – 4 pq
(p – q)2 = 25 – 3
(p – q)2 = 22


p=
1
5  22
2

q=
1
5  22
2
 Required smallest number is 5.
Ex.5 Find the smallest number by which 28812 must be
divided so that the quotient becomes a perfect square.
...(i)
p – q = 22
p+q=5

Sol. Given number is 28812, first we write it as the product

q
pq
p+q=5

5
[By squaring both sides]
[By equating the parts]
[By squaring both sides ]
...(iv)
[By eqn (i)]


3 = 
of prime factors.
3.
[On adding (i) & (iv)]

[On subtracting (i) & (iv)]
1


 5  22  5  22 

2 
2 28812
2 14406
Cube : If any number is multiplied by itself three times
then the result is called the cube of that number.
3
7203
7
2401
7
343
7
49
7
7
Ex.7 What is the smallest number by which 392 must be
multiplied so that the product is a perfect cube.
1
Sol. Resolving 392 into prime factor, we get
Perfect cube : A natural number is said to be a perfect
cube if it is the cube of any other natural number.
28812 = 2 × 2 × 3 × 7 × 7 × 7 × 7
Clearly, 3 has no pair, so if we divide it by 3 then quotient
become a perfect square.
Square roots : The square root of a number x is that
number which when multiplied by itself gives x as the
product. As we say square of 3 is 9, then we can also
say that square root of 9 is 3.
The symbol use to indicate the square root of a number
is ‘

’ , i.e.
81 = 9,
225 = 15 ...etc.
We can calculate the square root of positive numbers
only. However the square root of a positive number
may be a positive or a negative number.
e.g.
25 = + 5 or – 5.
392 = {2 × 2 × 2} × 7 × 7
Grouping the factor in triplets of equal factors, we get
392 = {2 × 2 × 2} × 7 × 7
We find that 2 occurs as a prime factor of 392 thrice but
7 occurs as a prime factor only twice. Thus, if we
multiply 392 by 7,7 will also occur as a prime factor
thrice and the product will be 2 × 2 × 2 × 7 × 7 × 7, which
is a perfect cube.
Hence, we must multiply 392 by 7 so that the product
becomes a perfect cube.
PAGE # 33
Ex.8 What is the smallest number by which 3087 must be
divided so that the quotient is a perfect cube ?
Sol. Resolving 3087 into prime factors, we get
3087 = 3 × 3 × 7 × 7 × 7
Grouping the factors in triplets of equal factors, we get
3087 = 3 × 3 × {7 × 7 × 7}
Clearly, if we divide 3087 by 3 × 3 = 9, the quotient
would be 7 × 7 × 7 which is a perfect cube. Therefore,
we must divide 3087 by 9 so that the quotient is a
perfect cube.
Factors : ‘a’ is a factor of ‘b’ if there exists a relation
such that a × n = b, where ‘n’ is any natural number.

1 is a factor of all numbers as 1 × b = b.

Factor of a number cannot be greater than the number
(infact the largest factor will be the number itself). Thus
factors of any number will lie between 1 and the number
itself (both inclusive) and they are limited.
Multiples : ‘a’ is a multiple of ‘b’ if there exists a relation
of the type b × n = a. Thus the multiples of 6 are
6 × 1 =6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, and so on.

The smallest multiple will be the number itself and the
number of multiples would be infinite.

NOTE :
To understand what multiples are, let’s just take an
example of multiples of 3. The multiples are 3, 6, 9,
12,.... so on. We find that every successive multiples
appears as the third number after the previous.
So if one wishes to find the number of multiples of 6
less than 255, we could arrive at the number through
255
= 42 (and the remainder 3). The remainder is of
6
no consequence to us. So in all there are 42 multiples.
255
=7
36
(and the remainder is 3). Hence, there are 7 multiples
of 36.
Factorisation : It is the process of splitting any number
into form where it is expressed only in terms of the
most basic prime factors.
For example, 36 = 22 × 3 2. 36 is expressed in the
factorised form in terms of its basic prime factors.
Number of factors : For any composite number C,
which can be expressed as C = ap × bq × cr ×....., where
a, b, c ..... are all prime factors and p, q, r are positive
integers, then the number of factors is equal to (p + 1)
× (q + 1) × (r + 1).... e.g. 36 = 22 × 32. So total no. of
factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9.
If one wishes to find the multiples of 36, find
Ex.9 Directions : (i to v) Read the following information
carefully and answer the questions given below.
In a big hostel, there are 1,000 rooms. In that hostel
only even numbers are used for room numbers, i.e.
the room numbers are 2, 4, 6, ...., 1998, 2000. All the
rooms have one resident each. One fine morning, the
warden calls all the residents and tells them to go
back to their rooms as well as multiples of their room
numbers. When a guy visits a room and finds the door
open, he closes it, and if the door is closed, he opens
it, All 1,000 guys do this operation. All the doors were
open initially.
(i) The last room that is closed is room number ?
(ii) The 38th room that is open is room number ?
(iii) If only 500 guys, i.e. residents of room number 2
to 1000 do the task, then the last room that is closed is
room number ?
(iv) In the case of the previous question, how many
rooms will be closed in all ?
(v) If you are a lazy person, you would like to stay in a
room whose number is ?
Sol (i) If a room has an odd number of visitors, it will be
closed. Any room number that is twice a perfect square
will have an odd number of visitors. The room with the
largest such number (twice a perfect square) will be
the last room to have an odd number of visitors. Hence
the last number is 2 × 961 = 1922.
(ii) Note that the 38th room with an open door will be
the 38th room whose number is not twice a perfect
square, which happens to be 88.
(iii) Note that if only occupants of 2 to 1000 do the
task, then only 1 person, i.e. the occupant of number
1000 will go to room number 2000, (since 2000 is a
multiple of 1000), and this room will therefore be closed.
(Since it is initially open).
(iv) Number of room closed upto 1000 = 22, since
there are 22 twice a perfect square numbers. Number
of remaining room = 500. Number of twice a perfect
number between 1000 and 2000 = 9.
All rooms with twice a perfect square between 1000
and 2000 will be open.
 Number of room closed = 22 + 500 – 9 = 513.
(v) As regards questions 34, anyone who lives in a
room with a number greater than 1000 will obviously
have to visit only that particular room, as it will not have
any multiple which is not greater than 2000.
Hence, the answer is greater than 1000.
LCM (least Common Multiple) : The LCM of given
numbers, as the name suggests is the smallest
positive number which is a multiple of each of the given
numbers.
PAGE # 44
HCF (Highest Common factor) : The HCF of given
numbers, as the name suggests is the largest factor
of the given set of numbers.
Consider the numbers 12, 20 and 30. The factors and
the multiples are :
Factors
1, 2, 3, 4, 6, 12
1, 2, 4, 5, 10, 20
1, 2, 3, 5, 6, 10, 15, 30

Given
numbers
12
20
30
Multiples
12, 24, 36, 48, 60, 72, 84, 96, 108, 120....
20, 40, 60, 80, 100, 120.....
30, 60, 90, 120....
The common factors are 1 and 2 and the common
multiples are 60, 120...
Thus the highest common factor is 2 and the meaning
of HCF is that, HCF is the largest number that divides
all the given numbers.
Also since a number divides its multiple, the meaning
of LCM is that it is smallest number which can be
divided by the given numbers.
HCF will be lesser than or equal to the least of the
numbers and LCM will be greater than or equal to the
greatest of the numbers.
There are two methods of finding the H.C.F. of a given
set of numbers :
(i) Factorization Method : Express each one of the
given numbers as the product of prime factors. The
product of least powers of common prime factors gives
H.C.F. and product of highest powers of common prime
factors gives L.C.M.
Some important points :

Ex.12 Find the greatest number that will exactly divide 65,
52 and 78.
Sol. Required number = HCF of 65, 52 and 78 = 13.

(ii) Division Method : Suppose we have to find the H.C.F.
of two given numbers. Divide the larger number by the
smaller one. Now, divide the divisor by the remainder.
Repeat the process of dividing the preceding divisor
by the remainder last obtained till zero is obtained as
remainder. The last divisor is the required H.C.F.
Ex.11 Find the HCF of (72, 108 ).
Sol.
So, HCF of (72, 108 ) = 36.
Finding the H.C.F. of more than two numbers :
Suppose we have to find the H.C.F. of three numbers.
Then, H.C.F. of [(H.C.F. of any two) and (the third
number) gives the H.C.F. of three given numbers.
Similarly, the H.C.F. of more than three numbers may
be obtained.
If we have to find the greatest number that will divide p,
q and r leaving remainders a, band c respectively, then
the required number = HCF of (p – a), (q – b) and (r – c).
Ex.13 Find the greatest number that will divide 65, 52 and
78 leaving remainders 5, 2 and 8 respectively.
Sol. Required number = HCF of (65 - 5), (52 - 2) and (78 - 8)
= HCF of 60, 50 and 70 = 10.

If we have to find the greatest number that will divide p,
q and r leaving the same remainder in each case, then
required number = HCF of the absolute values of
(p – q), (q – r) and (r – p).
Ex.14 Find the greatest number that will divide 65, 81 and
145 leaving the same remainder in each case.
Sol. Required number
= HCF of (81 – 65) (145 – 81) and (145 – 65)
= HCF of 16, 64 and 80 = 16.
/
Ex.10 Find the HCF of (72, 108 ).
Sol. HCF of (72, 108 )
= HCF of (23 × 32, 22 × 33 )
= 22 × 32 = 4 × 9 = 36.
If we have to find the greatest number that will exactly
divide p, q and r, then required number = HCF of p, q
and r.
Least Common Multiple (L.C.M.) : The least number
which is exactly divisible by each one of the given
numbers is called their L.C.M.
(i) Factorization Method of Finding L.C.M. : Resolve
each one of the given numbers into a product of prime
factors. Then, L.C.M. is the product of highest powers
of all the factors.
Ex.15 Find the LCM of (72, 108 )
Sol. The LCM of (72, 108 )
The LCM of (23 × 32, 22 × 33)
= 23 × 33 = 8 × 27 = 216.
(ii) Common Division Method (Short-cut Method) of
finding L.C.M. : Arrange the given numbers in a row in
any order. Divide by a number which divides exactly at
least two of the given numbers and carry forward the
numbers which are not divisible. Repeat the above
process till no two of the numbers are divisible by the
same number except 1. The product of the divisors
and the undivided numbers is the required L.C.M. of
the given numbers.
Ex.16 Find the LCM of (72, 108).
Sol.
2 72, 108
2 36,
2 18,
54
27
3
9,
27
3
3,
9
3
1,
3
1,
1
So, LCM of (72, 108) = 2 × 2 × 2 × 3 × 3 × 3 = 216.
PAGE # 55
Some important points :

If we have to find the least number which is exactly
divisible by p, q and r, then the required number
= LCM of p, q and r .
Ex.17 Find the least number that is exactly divisible by 6, 5
and 7.
Sol. Required number = LCM of 6, 5 and 7 = 210.
Ex.21 In a school 437 boys and 342 girls have been divided
into classes, so that each class has the same number
of students and no class has boys and girls mixed.
What is the least number of classes needed?
Sol. We should have the maximum number of students in
a class. So we have to find HCF (437, 342) = 19.
HCF is also the factor of difference of the number.


If we have to find the least number which when divided
by p, q and r leaves the remainders a, b and c
respectively then if it is observed that
(p – a) = (q – b) = (r – c) = K (say), then the required
number = (LCM of p, q and r) – (K).
Ex.18 Find the least number which when divided by 6, 7
and 9 leaves the remainders 1, 2 and 4 respectively.
Sol. Here, (6 – 1) = (7 – 2) = (9 – 4) = 5.
Required number = (LCM of 6, 7 and 9) – 5
= 126 – 5 = 121.

If we have to find the least number which when divided
by p, q and r leaves the same remainder 'a' in each
case, then required number = (LCM of p, q and r) + a.

L.C.M. of Numerators
L.C.M. = H.C.F. of Deno min ators .
HCF of (
HCF of (a, c, e )
a c e
, , )=
LCM of (b, d, f )
b d f
LCM of (
LCM of (a, c, e )
a c e
, , )=
HCF of (b, d, f )
b d f
H.C.F. and L.C.M. of Decimal Fractions : In given
numbers, make the same number of decimal places
by annexing zeros in some numbers, if necessary.
Considering these numbers without decimal point, find
H.C.F. or L.C.M. as the case may be. Now, in the result,
mark off as many decimal places as are there in each
of the given numbers.
 30

 1
Therefore in 30 min they will toll together 
 2

= 16 times
1 is added as all the bells are tolling together at the
start also, i.e. 0th second.
Ex.23 LCM of two distinct natural numbers is 211. What is
their HCF ?
Sol. 211 is a prime number. So there is only one pair of
distinct numbers possible whose LCM is 211,
i.e. 1 and 211. HCF of 1 and 211 is 1.
Ex.24 An orchard has 48 apple trees, 60 mango trees and
96 banana trees. These have to be arranged in rows
such that each row has the same number of trees and
all are of the same type. Find the minimum number of
such rows that can be formed.
Sol. Total number of trees are 204 and each of the trees
are exactly divisible by 12. HCF of (48, 60, 96).

=
500 2060 1025
,
,
)
1000 1000 1000
5
= 0.005
1000
204
= 17 such rows are possible.
12
Division Algorithm : General representation of result is,
Dividend
Re mainder
 Quotient 
Divisor
Divisor
Ex.20 The HCF of (0.5, 2.06, 1.025 )
Sol. HCF of (0.500, 2.060, 1.025 )
= HCF of (
For any two numbers x and y :
x × y = HCF (x, y) × LCM (x, y).
Make sure the fractions are in the most reducible form.
Ex.22 Six bells start tolling together and they toll at intervals
of 2, 4, 6, 8, 10, 12 sec. respectively, find.
(i) After how much time will all six of them toll together ?
(ii) How many times will they toll together in 30 min ?
Sol. The time after which all six bells will toll together must
be multiple of 2, 4, 6, 8, 10, 12.
Therefore, required time = LCM of time intervals.
= LCM (2, 4, 6, 8, 10, 12) = 120 sec.
Therefore after 120 s all six bells will toll together.
After each 120 s, i.e. 2 min, all bells are tolling together.
Ex.19 Find the least number which when divided by 15, 20
and 30 leaves the remainder 5 in each case.
Sol. Required number = (LCM of 15, 20 and 30) + 5
= 60 + 5 = 65
H.C.F and L.C.M. of fractions :
H.C.F. of Numerators
H.C.F = L.C.M. of Deno min ators
437
342
+
= 23 + 18
19
19
= 41 classes.
Number of classes =
Dividend = (Divisor × Quotient ) + Remainder

NOTE :
(i) (xn – an) is divisible by (x – a) for all the values of n.
(ii) (xn – an) is divisible by (x + a) and (x – a) for all the
even values of n.
(iii) (xn + an) is divisible by (x + a) for all the odd values of n.
PAGE # 66
Test of Divisibility :
No.
Divisiblity Test
2
Unit digit should be 0 or even
3
The sum of digits of no. should be divisible by 3
4
The no formed by last 2 digits of given no. should be divisible by 4.
5
Unit digit should be 0 or 5.
6
No should be divisible by 2 & 3 both
8
The number formed by last 3 digits of given no. should be divisible by 8.
9
Sum of digits of given no. should be divisible by 9
The difference between sums of the digits at even & at odd places
11
should be zero or multiple of 11.
25 Last 2 digits of the number should be 00, 25, 50 or 75.
Rule for 7 : Double the last digit of given number and
subtract from remaining number the result should be
zero or divisible by 7.
Ex.25 Check whether 413 is divisible by 7 or not.
Sol. Last digit = 3, remaining number = 41, 41 – (3 x 2) = 35
(divisible by 7). i.e. 413 is divisible by 7.
This rule can also be used for number having more
than 3 digits.
Ex.26 Check whether 6545 is divisible by 7 or not.
Sol. Last digit = 5, remaining number 654, 654 – (5 x 2)
= 644; 64 – (4 x 2) = 56 divisible by 7. i.e. 6545 is
Ex.31 Find the largest four digit number which when
reduced by 54, is perfectly divisible by all even natural
numbers less than 20.
Sol. Even natural numbers less than 20 are 2, 4, 6, 8, 10,
12, 14, 16, 18.
Their LCM = 2 × LCM of first 9 natural numbers
= 2 × 2520 = 5040
This happens to be the largest four-digit number
divisible by all even natural numbers less than 20. 54
was subtracted from our required number to get this
number.
Hence, (required number – 54) = 5040
 Required number = 5094.
Ex.32 Ajay multiplied 484 by a certain number to get the
result 3823a. Find the value of ‘a’.
Sol. 3823a is divisible by 484, and 484 is a factor of 3823a.
4 is a factor of 484 and 11 is also a factor of 484.
Hence, 3823a is divisible by both 4 and 11.
To be divisible by 4, the last two digits have to be
divisible by 4.
‘a’ can take two values 2 and 6.
38232 is not divisible by 11, but 38236 is divisible by
11.
Hence, 6 is the correct choice.
divisible by 7.
Rule for 13 : Four times the last digit and add to
remaining number the result should be divisible by
13.
Ex.27 Check whether 234 is divisible by 13 or not .
Sol. 234, (4 x 4) + 23 = 39 (divisible by 13), i.e. 234 is divisible
by 13.
Ex.33 Which digits should come in place of  and $ if the
number 62684$ is divisible by both 8 and 5 ?
Sol. Since the given number is divisible by 5, so 0 or 5 must
come in place of $. But, a number ending with 5 in
never divisible by 8. So, 0 will replace $.
Now, the number formed by the last three digits is 40,
which becomes divisible by 8, if  is replaced by 4 or 8.
Hence, digits in place of  and $ are 4 or 8 and 0
respectively.
Rule for 17 : Five times the last digit of the number and
subtract from previous number the result obtained
should be either 0 or divisible by 17.
Ex.28 Check whether 357 is divisible by 17 or not.
Sol. 357, (7 x 5) – 35 = 0, i.e. 357 is divisible by 17.
Rule for 19 : Double the last digit of given number and
add to remaining number The result obtained should
be divisible by 19.
Ex.29 Check whether 589 is divisible by 19 or not.
Ex.34 On dividing 15968 by a certain number, the quotient
is 89 and the remainder is 37. Find the divisor.
Dividend  Re mainder 15968  37

Sol. Divisor =
= 179.
Quotient
89
Ex.35 How many numbers between 200 and 600 are
divisible by 4, 5, and 6 ?
Sol. Every such number must be divisible by L.C.M. of 4, 5,
6, i.e.60.
Such numbers are 240, 300, 360, 420, 480, 540
Clearly, there are 6 such numbers.
Sol. 589, (9 x 2) + 58 = 76 (divisible by 19), i.e. the number
is divisible by 19.
The method of finding the remainder without actually
performing the process of division is termed as
Remainder Theorem.
Ex.30 Find the smallest number of six digits which is exactly
divisible by 111.
Sol. Smallest number of 6 digits is which is 100000.
On dividing 100000 by 111, we get 100 as remainder.
 Number to be added = (111 – 100) = 11.
Hence, required number = 100011.

Remainder should always be positive. For example if
we divide –22 by 7, generally we get –3 as quotient and
–1 as remainder. But this is wrong because remainder
is never be negative hence the quotient should be –4
and remainder is +6. We can also get remainder 6 by
adding –1 to divisor 7 ( 7–1 = 6).
PAGE # 77
Ex.36 Two numbers, x and y, are such that when divided by
6, they leave remainders 4 and 5 respectively. Find the
remainder when (x2 + y2) is divided by 6.
Sol. Suppose x = 6k1 + 4 and y = 6k2 + 5
x2 + y2 = (6k1 + 4)2 + (6k2 + 5)2
= 36k12 + 48k1 + 16 + 36k22 + 60k2 + 25
= 36k12 + 48k1 + 36k22 + 60k2 + 41
Obviously when this is divided by 6, the remainder will
be 5.
Ex.37 A number when divided by 259 leaves a remainder
139. What will be the remainder when the same
number is divided by 37 ?
Sol. Let the number be P.
So, P – 139 is divisible by 259.
P  139
Let Q be the quotient then,
=Q
259
P = 259Q + 139
P
259 Q  139

=
37
37
 259 is divisible by 37,
 When 139 divided by 37, leaves a remainder of 28.

Ex.38 A number being successively divided by 3, 5 and 8
leaves remainders 1, 4 and 7 respectively. Find the
respective remainders if the order of divisors be
reversed.
3 x
5 y 1
Sol.
8 z 4
1 7
 z = (8 × 1 + 7) = 15 ; y = (5z + 4) = (5 × 15 + 4) = 79
; x = (3y + 1) = (3 × 79 + 1) = 238.
8 238
5 29 6
Now, 3 5
4
1
2
 Respective remainders are 6, 4, 2.
Ex.39 A number was divided successively in order by 4, 5
and 6. The remainders were respectively 2, 3 and 4.
Then find out the number.
4 x
Sol.
5 y 2
6 z 3
1 4
 z = (6 × 1 + 4) = 10
 y = (5 × z + 3) = (5 × 10 + 53) = 53
 x = (4 × y + 2) = (4 × 53 + 2) = 214
Hence, the required number is 214.
Ex.40 In dividing a number by 585, a student employed the
method of short division. He divided the number
successively by 5, 9 and 13 (factors of 585) and got the
remainders 4, 8 and 12. If he had divided number by
585, then find out the remainder.
Sol.
5 x
9 y 4
13 z 8
1 12
Now, 1169 when divided by 585 gives remainder
= 584.
We are having 10 digits in our number systems and
some of them shows special characterstics like they,
repeat their unit digit after a cycle, for example 1 repeat
its unit digit after every consecutive power. So, its
cyclicity is 1 on the other hand digit 2 repeat its unit
digit after every four power, hence the cyclicity of 2 is
four. The cyclicity of digits are as follows :
Digit
Cyclicity
0, 1, 5 and 6
1
4 and 9
2
2, 3, 7 and 8
4
So, if we want to find the last digit of 245, divide 45 by 4.
The remainder is 1 so the last digit of 245 would be
same as the last digit of 21 which is 2.
Ex.41 Find the last digit of
(i) 357
(ii) 1359
57
gives the remainder
4
57
1. So, the last digit of 3 is same as the last digit of 31,
i.e. 3.
(ii) The number of digits in the base will not make a
difference to the last digit. It is last digit of the base
which decides the last digit of the number itself. For
Sol. (i) The cyclicity of 3 is 4. Hence,
59
which gives a remainder 3. So the
4
last digit of 1359 is same as the last digit of 33, i.e. 7.
1359, we find
Ex.42 Find the last digit of the product 723 x 813.
Sol. Both 7 and 8 exhibit a cyclicity of 4. The last digits are
71 = 7
81 = 8
72 = 9
82 = 4
3
7 =3
83 = 2
4
7 =1
84 = 6
5
7 =7
85 = 8
The cycle would repeat itself for higher powers.
723 ends with the same last digit as 73, i.e. 3.
813 ends with the same last digit as 81, i.e. 8. Hence,
the product of the two numbers would end with the
same last digit as that of 3 × 8, i.e. 4.
Ex.43 Find unit’s digit in y = 717 + 734.
Sol. Unit digit of (7 17 + 7 34 ) is same as unit digit of
(71 + 72 = 56), Hence the unit digit is 6.
Ex.44 What will be the last digit of (73 )75
Sol. Let (73 )75
6476
= (73)x where x = 75 64
6476
76
= (75)even power
 Cyclicity of 3 is 4
 To find the last digit we have to find the remainder
when x is divided by 4.
x = (75)even power = (76 – 1)even power , where n is divided by
4 so remainder will be 1.
Therefore, the last digit of (73 )75
6476
will be 31 = 3.
PAGE # 88
Ex.45 What will be the unit digit of (87 )75
55
75 63
6355
.
55
Sol. Let (87 )
= (87)x where x = 75 63 = (75)odd
 Cyclicity of 7 is 4.
 To find the last digit we have to find the remainder
when x is divided by 4.
x = (75)odd power = (76 – 1)odd power
where x is divided by 4 so remainder will be –1 or 3, but
remainder should be positive always
therefore, the last digit of (87 )75
6355
Hence, the last digit is of (87 )
digits, then A + B + C = ?
Sol. As the unit digit of the product is 5, therefore, the unit
digit of one of the numbers is 5 and the unit digit of the
other number is odd. Therefore, AB5 x 5BA = 65125,
where A = 1, 3, 5, 7 or 9.
As the product of two three-digit numbers is a five-digit
number, and not a six-digit number, A can only be equal
will be 73 = 343.
to 1. 1B5 x 5B1 = 65125.
is 3.
The digit sums of both numbers, 1B5 and 5B1 will be
55
75 63
Ex.49 If ABC x CBA = 65125, where A, B and C are single
same. Therefore, the product would give digit sum of a
ALPHA NUMERIC PUZZLES
perfect square. The digit sum on the R.H.S. is 1.
Therefore, the digit sum of each number can be 1or 8.
Ex.46 If a – b = 2,
Correspondingly B will be 4 or 2 (as digit sum cannot
a a
and
be equal to 1).
b b
_____
then find the value of a, b and c.
cc 0
Sol. These problems involve basic number
(i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers.
Hence, their sum cannot exceed 198. So, c must be 1.
(iii) Hence, cc0 = 110. This implies a + b = 10 or a = 6
and b = 4.
Such problems are part of a category of problems
called alphanumerics.
a 3b
Ex.47 If
 a c
_____
a a 9
Rules on Counting Numbers :
(i) Sum of first n natural numbers =
n(n  1)
2
(ii) Sum of first n odd numbers = n2
(iii) Sum of first n even number = n(n + 1)
(iv) Sum of the squares of first n natural numbers
=
n(n  1)(2n  1)
6
(v) Sum of the cubes of first n natural number
then find a, b and c if each of them is
distinctly different digit.
Sol. (i) since the first digit of (a 3 b) is written as it is after
subtracting ac carry over from a to 3.
(ii) there must be a carry over from 3 to b, because if no
carry over is there, it means 3 – a = a.
 2a = 3
3
 a=
2
which is not possible because a is a digit. For a carry
over 1, 2 – a = a
a=1
(iii) it means b and c are consecutive digit (2, 3),
(3, 4),.... (8, 9).
Ex.48 If
Keeping B = 2, we can see that 125 x 521 = 65125.
1a4
x3b
8c8
s 72
t 5d8
then, find the value of a, b, c, d, s and t, where all of
them are different digits.
Sol. Let us consider 1 a 4 × 3 = s72.
a × 3 results in a number ending in 6.
As 16 and 26 is ruled out, a is 2.
Thus, s = 3, t, 4
Now 1 a 4 × b = 8c8 ; b = 2 or 7
Again 2 is ruled out because in that case, product would
be much less than 800.
 b = 7.
Hence, a = 2, b = 7, c = 6, d = 8, s = 3 and t = 4.
2
 n(n  1) 
= 
 .
 2 
LOGARITHM
If ‘a’ is a positive real number, other than 1 and x is a
rational number such that ax = N, then x is the
logarithm of N to the base a.
 If ax = N then loga N = x.
[ Remember N will be +ve]
Systems of Logarithm :
There are two systems of logarithm which are generally
used.
(i) Common logarithm : In this system base is always
taken as 10.
(ii) Natural logarithm : In this system the base of the
logarithm is taken as ‘e’. Where ‘e’ is an irrational
number lying between 2 and 3. (The approximate value
of e upto two decimal places is e = 2.73)
Some Useful Results :
(i) If a > 1 then
(a) loga x < 0 [for all x satisfying 0 < x < 1]
(b) loga x = 0 for x = 1
(c) loga x > 0 for x > 1
(d) x > y  loga x > loga y i.e. logax is an increasing
function.
 Graph of y = loga x, a > 1
PAGE # 99
Ex.50 If log3a = 4, find value of a.
y
Sol. 
y = logax, a > 1
0
x'
(1,0)
log3a = 4
 a = 34
 a = 81.
x
Ex.51 Find the value of log
Sol. Given :
y'
(ii) If 0 < a < 1, then
(a) loga x < 0 for all x > 1
(b) loga x = 0 for x = 1
(c) logax > 0 for all x satisfying 0 < x < 1
(d) x > y

9
27
3
– log
 log
8
32
4
logax < loga y i.e. loga x is a decreasing
log
3
9
27
3
 9 27 
– log
 log  log 
  log
4
8
32
4
 8 32 
 9 32 3 
 log 
 
 8 27 4 
function.
 Graph of y = loga x, 0 < a < 1.
y = logax, 0 < a < 1.
0
Sol. Given
2log4x = 1 + log4(x – 1)
 log4x2 – log4(x – 1) = 1
(1,0)
x
y'
Fundamental Laws of Logarithm :
Logarithm to any base a (where a > 0 and a  1 ).
(i)
loga a = 1
x2
=1
x –1
41 =
 x2 = 4x – 4

x2 – 4x + 4 = 0
 (x – 2)2 = 0

x = 2.
Ex.53 Evaluate : 3 2 – log3 5 .
Sol. Given 3 2 – log3 5 = 3 2.3 – log3 5
= 9. 3
m
(v) loga   = logam – logan
n
(vi) loga(m)n = n logam
log m
b
=
log3 5 –1
9
.
5
Ex.54 If A = log27625 + 7
log11 13
Sol. A = log27625 + 7
or, A =
log11 13
(ix) If ‘a’ is a positive real number and ‘n’ is a positive
rational number, then
or, B = log 3 2 5 3 + 7
log11 13
3
log 13
log35 + 7 11
2
By (i) and (ii) we have,
or, B =
A–
q
(xi) ploga q  qloga p
...(ii)
3
4
log35 = B – log35
3
2

4
3
log35 <
log35
3
2

A < B.
(xii) logax = logay  x = y
....(i)
log11 7
and,B = log9125 + 13
loga q np  p log n
a
log11 7
4
log 13
= log 3 3 5 + 7 11
4
log 13
log35 + 7 11
3
(viii)logam . logma = 1
(x) If ‘a’ is a positive real number and ‘n’ is a positive
rational number, then
and B = log9125 + 13
then find the relation between A and B.
log a
b
a loga n  n
[ am + n = am.an]
= 9 × 5–1
(iii) loga (–ve no.) = not defined.
[As in loga N, N will always be (+ ve)]
(iv) loga (mn) = loga m + logan
[Where m and n are +ve numbers]
x2
x –1

 log4
(ii) loga 0 = not defined
[As an = 0 is not possible, where n is any number]
(vii) logam 
[ loga1 = 0]
Ex.52 If 2log4x = 1 + log4(x – 1), find the value of x.
y
x'
= log1 = 0.
PAGE # 1010
,
POLYNOMIALS
An
algebraic
expression
f(x)
2
of
the
(B) Linear Polynomial : A polynomial of degree one is
called a linear polynomial. The general form of linear
polynomial is ax + b, where a and b are any real constant
and a  0.
form
n
f(x) = a0 + a1x + a2x +...........+ anx , where a0, a1,
a2,............,an are real numbers and all the index of ‘x’
(C) Quadratic Polynomial : A polynomial of degree two
is called a quadratic polynomial. The general form of a
quadratic polynomial is ax2 + bx + c, where a  0.
are non-negative integers is called a polynomials in x.
Identification of Polynomial :
(D) Cubic Polynomial : A polynomial of degree three is
called a cubic polynomial. The general form of a cubic
polynomial is ax3 + bx2 + cx + d, where a  0.
For this, we have following examples :
(i)
3 x2 + x – 5 is a polynomial in variable x as all the
exponents of x are non negative integers.
(ii)
3 x2 +
(E) Biquadratic (or quartic) Polynomials :
A polynomial of degree four is called a biquadratic
(quartic) polynomial. The general form of a biquadratic
polynomial is ax4 + bx3 + cx2 + dx + e, where a  0.
x – 5x is not a polynomial as the exponent
of second term ( x = x1/2) is not a non–negative

integer.
(iii) 5x3 + 2x2 + 3x –
5
+ 6 is not a polynomial as the
x
NOTE : A polynomial of degree five or more than five does
not have any particular name. Such a polynomial usually
called a polynomial of degree five or six or ....... etc.
(ii) Based on number of terms
There are three types of polynomials based on number
of terms. These are as follows :
 5
1 
exponent of fourth term   5 x  is not
 x

non–negative integer.
(A) Monomial : A polynomial is said to be a monomial
if it has only one term. e.g. x, 9x2, 5x3 all are monomials.
(a) Degree of the Polynomial :
(B) Binomial : A polynomial is said to be a binomial if it
contains two terms. e.g. 2x2 + 3x,
all are binomials.
Highest index of x in algebraic expression is called the
degree of the polynomial, here a0, a1x, a2x2,............., anxn,
it contains three terms. e.g. 3x3 – 8x +
polynomial f(x).
For example:
(ii) q(x) = 5x4 + 2x5 – 6x6 – 5 is a polynomial of degree 6
x + 5x3, – 8x3 + 3,
(C) Trinomials : A polynomial is said to be a trinomial if
are called the terms of the polynomial and a 0, a 1,
a2,.........., a n are called various coefficients of the
(i) p(x) = 3x4 – 5x2 + 2 is a polynomial of degree 4
1
2
5
, 5 – 7x + 8x9,
2
7 x10 + 8x4 – 3x2 are all trinomials.

NOTE :
A polynomial having four or more than four terms does
not have particular name. These are simply called
polynomials.
(iii) f(x) = 2x3 + 7x – 5 is a polynomial of degree 3.
(b) Different Types of Polynomials :
Some important identities are :
Generally, we divide the polynomials in the following
categories.
(i) (a + b)2 = a2 + 2ab + b2
(i) Based on degrees :
There are four types of polynomials based on degrees.
(iii) a2 – b2 = (a + b) (a – b)
These are listed below :
(A) Zero degree polynomial : Any non-zero number
(constant) is regarded as a polynomial of degree zero
or zero degree polynomial. i.e. f(x) = a, where a  0 is
a zero degree polynomial, since we can write f(x) = a
as f(x) = axo.
(ii) (a – b)2 = a2 – 2ab + b2
(iv) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2 bc + 2ca
(v) a3 + b3 = (a + b) (a2 – ab + b2)
(vi) a3 – b3 = (a – b) (a2 + ab + b2)
(vii) (a + b)3 = a3 + b3 + 3ab (a + b)
(viii) (a – b)3 = a3 – b3 – 3ab (a – b)
(ix) a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)
Special case : if a + b + c = 0 then a3 + b3 + c3 = 3abc.
PAGE # 1111
Ex.3 If a + b + c = 15, a2 + b2 + c2 = 83, then find the value of
Value Form :
(i) a2 + b2 = (a + b)2 – 2ab, if a + b and ab are given.
2
2
2
(ii) a + b = (a – b) + 2ab, if a – b and ab are given.
(iii) a + b =
a  b 2  4ab , if a – b and ab are given.
(iv) a – b =
a  b 2  4ab , if a + b and ab are given.
2
1
(v) a2 +
1

1
=  a   – 2, if a +
is given.
a
a

a2
(vi) a2 +
2
1

1
=  a   + 2, if a –
is given.
a

a
1
a
2
a3 + b3 + c3 – 3abc.
Sol. a + b + c = 15
a2 + b2 + c2 = 83
To find the value of : a3 + b3 + c3 – 3abc
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
(a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
(15)2 = 83 + 2 (ab + bc + ca)
2 (ab + bc + ca) = 225 – 83
2 (ab + bc + ca) = 142
(ab + bc + ca) = 71
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
= (15) (83 – 71)
(vii) a3 + b3 = (a + b)3 – 3ab (a + b), if (a + b) and ab are
= (15) (12) = 180.
given.
(viii) a3 – b3 = (a – b)3 + 3ab (a – b), if (a – b) and ab are
Ex.4 Prove that a2 + b2 + c2 – ab – bc – ca is always
non - negative for all values of a, b & c.
given.
(ix) a3 
(x) a 3 
3
1
1
1


=  a   – 3  a   , if a + is given.
a
a
a


1
a3
3
1

1

1
=  a   + 3  a   , if a –
is given.
3
a


a
a


a
Sol. a2 + b2 + c2 – ab – bc – ca
=
1
[ 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca ]
2
=
1
[(a – b)2 + (b – c)2 + (c – a)2]
2
1
(xi) a4 – b4 = (a2 + b2) (a2 – b2) = [(a + b)2 – 2ab](a + b) (a – b).
 We know square of any negative or non negative
Ex.1 Evaluate :
1

(i) (5x + 4y)
(ii) (4x – 5y)
(iii)  2x  
x

2
2
2
Sol. (i) (5x + 4y) = (5x) + (4y) + 2(5x)(4y)
2
2
2
value is positive and sum of then will also positive
thus we can conclude given eqn will be non–negative
for all values of a, b, c.
= 25x2 + 16y2 + 40xy
2
(ii) (4x – 5y) = (4x)2 + (5y)2 – 2(4x)(5y)
To express a given polynomial as the product of
= 16x2 + 25y2 – 40xy
 1
1

 1
(iii)  2x   = (2x)2 +   – 2 (2x)  
x
x

x
1
= 4x2 + 2 – 4.
x
Ex.2 If a + b = 10 and a2 + b2 = 58, find the value of a3 + b3.
polynomials, each of degree less than that of the given
Sol. Given a + b = 10
... (i)
divide each term by this factor and take out as a multiple.
and a2 + b2 = 58
... (ii)
2
2
we know a3 + b3 = (a + b)(a2 + b2 – ab)
 We know (a + b)2 = a2 + b2 + 2ab
... (iii)
polynomial such that no such a factor has a factor of
lower degree, is called factorization.
(a) Factorization by taking out the common factor :
When each term of an expression has a common factor,
Ex.5 Factorize : 6x3 + 8x2 – 10x
Sol. 6x3 + 8x2–10x = 2x (3x2 + 4x – 5)
(b) Factorization by grouping :
Thus 2ab = (a + b)2 – (a2 + b2)
Put values from eqn (i) & eqn (ii)
Ex.6 Factorize : ax + by + ay + bx
2ab = (10)2 – (58)
Sol. ax + by + ay + bx = ax + ay + bx + by = a (x + y) + b(x+y)
= 100 – 58 = 42
ab = 21
a3 + b3 = (a + b)(a2 + b2 – ab)
= 10 × (58 – 21)
= 10 × 37 = 370.
= (x + y) (a + b)
(c) Factorization by making a perfect square :
Ex.7 Factorize : 4x2 + 12x + 9
Sol. 4x2 + 12x + 9 = (2x)2 + 2 (2x) (3) + 32
= (2x + 3)2
PAGE # 1212
(d) Factorization the difference of two squares :

2
Ex.8 Factorize : 4x – 25.
Sol. 4x2 – 25 = (2x)2 – (5)2 = (2x – 5) (2x + 5)
(e) Factorization of a quadratic polynomial by splitting
the middle term :
Ex.9 Factorize : x2 + 4 2 x + 6
Sol. x2 + 4 2 x + 6
= x2 + 3 2 x +
2 x+6
= x(x + 3 2 ) + 2 (x + 3 2 )
= (x + 3 2 )(x + 2 )
(f) Factorization of a algebraic expression as the sum
or difference of two cubes :
Ex.10 Factorize : 16a3b–250b4.
16a3b–250b 4=2b(8a 3–125b 3)
Sol.
3
3
=
2b{(2a) –(5b) }
=
2b{(2a–5b)(4a2+25b2+10ab)}
(g) Factorization of a algebraic expression of the form
a3 + b3 + c3 – 3abc :
L.C.M. (Least Common Multiple ) : A polynomial P(x) is
called the LCM of two or more given polynomials, if it
is a polynomial of smallest degree which is divided by
each one of the given polynomials. For any two
polynomials P(x) and Q(x). We have :
P(x) × Q(x) = [HCF of P(x) and Q(x)]
× [LCM of P(x) and Q (x)]
Ex.12 If p(x) = (x + 2)(x2 – 4x–21), Q(x) = (x– 7) (2x2 + x – 6)
find the HCF and LCM of P(x) and Q(x).
Sol. P(x) = (x + 2) (x2 – 4x – 21)
= (x + 2) (x2 – 7x + 3x – 21)
= (x + 2) (x – 7) (x + 3)
Q(x) = (x – 7) (2x2 + x – 6)
= (x – 7)(2x2 + 4x – 3x – 6)
= (x – 7) [2x (x + 2) – 3 (x + 2)]
= (x – 7) (2x – 3) (x + 2)
HCF = (x + 2)(x– 7)
LCM = (x + 3)(x – 7) (2x – 3) (x + 2).
Ex.13 If HCF & LCM of P(x) and Q(x) are (x + 2) and (x + 3)
(x2 + 9x + 14) respectively if P(x) = x2 + 5x + 6, find Q(x).
Sol. P(x) = (x2 + 5x + 6) = (x + 2) (x + 3)
LCM = (x + 3) (x2 + 9x + 14) = (x + 3)(x + 7)(x + 2)
We know that HCF  LCM = P(x)  Q(x)
Q(x) =
Ex.11 Factorize :
( x  2)( x  3)( x  7)( x  2)
( x  2)( x  3)
= (x + 7) (x+2) = x2 + 9x + 14.
(i) 2 2a 3  8b 3  27c 3  18 2abc
(ii) (x – y)3 + (y – z)3 + (z –x)3
Sol. (i) 2 2a 3  8b 3  27c 3  18 2abc
=
3
3
 2a  (2b)  (3c )
=  2a  2b  3c 
 2a 2  2b 2   3c 2 

 
=
 2a  2b  3c  2a
2
3
(a) Value of a Polynomial :
The value of a polynomial f(x) at x =  is obtained by
 3( 2a)(2b )(3c )
substituting x =  in the given polynomial and is
denoted by f().
Consider the polynomial f(x) = x3 – 6x2 + 11x – 6,
 2a2b  2b 3c    3c  2a
 4b 2  9c 2  2 2ab  6bc  3 2ac
If we replace x by – 2 everywhere in f(x), we get

(ii) (x – y)3 + (y – z)3 + (z –x)3
Let a =x – y, b = y–z and c = z–x
So, (x – y)3 + (y – z)3 + (z –x)3 = a3 + b3 + c3
Now, a + b + c = x – y + y – z + z – x = 0
So, a3 + b3 + c3 = 3abc
3
3
3
 (x – y) + (y – z) + (z –x) = 3(x – y) (y – z)(z –x)
H.C.F. AND L.C.M. OF POLYNOMIALS

A polynomial D(x) is a divisor of the polynomial P(x) if it
is a factor of P(x). Where Q(x) is another polynomial
such that P(x) = D(x) × Q(x)

HCF/GCD (Greatest Common Divisor) : A polynomial
h(x) is called the HCF or GCD of two or more given
polynomials, if h(x) is a polynomial of highest degree
dividing each of one of the given polynomials.
 f(– 2) = (– 2)3 – 6(– 2)2 + 11(– 2) – 6
f(– 2) = – 8 – 24 – 22 – 6
f(– 2) = – 60  0.
So, we can say that value of f(x) at x = – 2 is – 60.
(b) Zero or root of a Polynomial :
The real number  is a root or zero of a polynomial
f(x), if f( = 0.
Consider the polynomial f(x) = 2x3 + x2 – 7x – 6,
If we replace x by 2 everywhere in f(x), we get
 f(2) = 2(2)3 + (2)2 – 7(2) – 6
= 16 + 4 – 14 – 6 = 0
Hence, x = 2 is a root of f(x).
Let ‘p(x)’ be any polynomial of degree greater than or
equal to one and a be any real number and If p(x) is
divided by (x – a), then the remainder is equal to p(a)
PAGE # 1313
Ex.14 If x51 + 51 is divided by (x + 1) then find the remainder.
Sol. Let P(x)=x51 + 51
In algebraic or in set theoretic language the graph
When P(x) is divided by (x+1)
of a polynomial f(x) is the collection (or set) of all
Then remainder = P (–1)
points (x, y), where y = f(x). In geometrical or in
= (–1 )51 + 51
graphical language the graph of a polynomial f(x) is
= – 1 + 51
a smooth free hand curve passing through points
(x 1, y1), (x 2, y2), (x 3, y3), ... etc, where y1, y2, y3, ... are
= 50
3
2
3
2
Ex.15 If the polynomial 2x + ax + 3x – 5 and x + x – 4x + a
the values of the polynomial f(x) at x 1, x 2, x 3, ...
leave the same remainder when divided by x – 2, find
respectively.
the value of a.
In order to draw the graph of a polynomial f(x), follow
Sol. Let p(x) = 2x3 + ax2 + 3x – 5 and q(x) = x3 + x2 – 4x + a
 When divided by x – 2 leave same remainder therefore p(a) = q(a)
the following algorithm.
ALGORITHM :
Step (i) Find the values y1, y2, ...., yn, .... of polynomial
or p(2) = q(2)
2(2)3 + a(2)2 + 3(2) – 5 = (2)3 + (2)2 – 4 × 2 + a
16 + 4a + 6 – 5 = 8 + 4 – 8 + a
f(x) on different points x 1, x 2, ...., x n, .... and prepare
a table that gives values of y or f(x) for various values
of x.
3a + 13 = 0
x:
x1
x2
y = f(x) : y1 = f(x1) y2 = f(x2)
13
a=
.
3
...
...
xn
yn = f(xn)
xn + 1
...
yn + 1 = f(xn + 1) ...
Step (ii) Plot that points (x 1, y1), (x 2, y2), (x 3, y3), ...
(x n , yn), ... on rectangular co-ordinate system.
FACTOR THEOREM
In plotting these points use different scales on the
Let p(x) be a polynomial of degree greater than or equal
X and Y axes.
to 1 and ‘a’ be a real number such that p(a) = 0, then
(x – a) is a factor of p(x). Conversely, if (x – a) is a factor
Step (iii) Draw a free hand smooth curve passing
of p(x), then p(a) = 0.
through points plotted in step 2 to get the graph of
Ex.16 If x2 – 4 is a factor of 2x3 + ax2 + bx + 12, where a and
b are constant. Then the values of a and b are :
the polynomial f(x).
(a) Graph of a Linear Polynomial :
Sol. x2 – 4 is a factor of 2x3 + ax2 + bx + 12
 x2 – 4 is factor of given expression
Consider a linear polynomial f(x)= ax + b, a  0.
Let x – 4 = 0
Graph of y = ax + b is a straight line. That is why
x=±2
f(x) = ax + b is called a linear polynomial. Since two
f (x) = 2x3 + ax2 + bx + 12
points determine a straight line, so only two points
f (2) = 0
need to plotted to draw the line y = ax + b. The line
2
16 + 4a + 2b + 12 = 0
2a + b = – 14
represented by y = ax + b crosses the
... (i)
f (–2) = 0
– 16 + 4a – 2b + 12 = 0
2a – b = 2
... (ii)
Solve eqn (i) & eqn (ii) (add them)
X-axis at
 b 
exactly one point, namely   , 0  .
 a 
Ex.17 Draw the graph of y = x – 4.
Sol. y = x – 4
y=x–4
2a + b = – 14
2a – b = 2
x
0
4
5
4a = – 12
y
–4
0
1
a = – 3 and b = – 14 – 2a = – 14 + 6 = – 8.
(a, b)  (– 3, – 8)
PAGE # 1414
Conversely, if the parabola y = ax2 + bx + c intersects
the X-axis at a point (, 0), then (, 0) satisfies the
equation y = ax2 + bx + c
 a2 + b + c = 0 [ is a real root of ax2 + bx + c = 0]
Thus, the intersection of the parabola y = ax2 + bx + c
with X-axis gives all the real roots of ax2 + bx + c = 0.
Following conclusions may be drawn :
(i) If D > 0, the parabola will intersect the x-axis in two
distinct points and vice-versa.
The parabola meets x-axis at  
(b) Graph of a Quadratic Polynomial :

–b– D
and
2a
–b D
.
2a
Let a, b, c be real numbers and a  0. Then
f(x) = ax2 + bx + c is known as a quadratic polynomial
in x. Graph of the quadratic polynomial i.e. the curve
whose equation is y = ax2 + bx + c, a  0. Graph of a
quadratic polynomial is always a parabola.
Let y = ax2 + bx + c, where a  0.
(ii) If D = 0, the parabola will just touch the x-axis at
one point and vice-versa.
2 2
 4ay = 4a x + 4abx + 4ac
 4ay = 4a2x2 + 4abx + b2 – b2 + 4ac
 4ay = (2ax + b)2 – (b2 – 4ac)
 4ay + (b2 – 4ac) = (2ax + b)2
 4ay + (b2 – 4ac) = 4a2(x + b/2a)2

 4ay 


 y 

b 2 – 4ac 
b 
2

  4a  x 
4a 
2a 

D 
b 

  a x 

4a 
2a


2
2
.
....(i)
(iii) If D < 0, the parabola will not intersect x-axis at all
and vice-versa.
where, D  b2 – 4ac is the discriminant of the quadratic
equation.

REMARKS :
D 
 b
,–
Shifting the origin at  –
 , we have
4a 
 2a
(–D)
 b 
 and Y = y –
.
X = x – –
 2a 
4a
Substituting these values in (i), we obtain Y = aX2 ...(ii)
which is the standard equation of parabola.
Clearly, this is the equation of a parabola having its
D 
 b
,–
vertex at  –
.
2
a
4
a

The parabola opens upwards or downwards
according as a > 0 or a < 0.
Let  be a real root of ax 2 + bx + c = 0. Then,
a2 + b + c = 0. Point (, 0) lies on y = ax2 + bx + c.
Thus, every real root of ax2 + bx + c = 0 represents a
point of intersection of the parabola with the X-axis.

REMARKS :
  x  R, y > 0 only if a > 0 & D  b²  4ac < 0
  x  R, y < 0 only if a < 0 & D  b²  4ac < 0
Ex.18 Draw the graph of the polynomial f(x) = x2 – 2x – 8.
Sol. Let y = x2 – 2x – 8.
The following table gives the values of y or f(x) for
various values of x.
x
–4
–3
–2
–1
0
1
2
3
4
5
6
y = x – 2x – 8
16
7
0
–5
–8
–9
–8
–5
0
7
16
2
PAGE # 1515
Let us plot the points (–4,16), (–3, 7), (–2, 0),
(–1, –5), (0, –8), (1, –9), (2, –8), (3, –5), (4, 0), (5, 7)
and (6, 16) on a graphs paper and draw a smooth
free hand curve passing through these points. The
Ex.19 Draw the graphs of the polynomial f(x) = x3 – 4x.
Sol. Let y = f(x) or, y = x3 – 4x.
The values of y for variable value of x are listed in
the following table :
x
–3
–2
–1
0
1
2
3
y = x3 – 4x
– 15
0
3
0
–3
0
15
curve thus obtained represents the graphs of the
polynomial f(x) = x2 – 2x – 8. This is called a parabola.
Thus, the curve y = x3 – 4x passes through the points
(–3, –15), (–2, 0), (–1, 3), (0, 0), (1, –3), (2, 0), (3, 15),
(4, 48) etc.
Plotting these points on a graph paper and drawing
a free hand smooth curve through these points, we
obtain the graph of the given polynomial as shown
figure.
The lowest point P, called a minimum point, is the
vertex of the parabola. Vertical line passing through
P is called the axis of the parabola. Parabola is
symmetric about the axis. So, it is also called the
line of symmetry.

Observations :
From the graphs of the polynomial f(x) = x2 – 2x – 8,
following observations can be drawn :
(i) The coefficient of x2 in f(x) = x2 – 2x – 8 is 1
(a positive real number) and so the parabola opens
upwards.
(ii) D = b 2 – 4ac = 4 + 32 = 36 > 0. So, the parabola
cuts X-axis at two distinct points.
(iii) On comparing the polynomial x2 – 2x – 8 with
ax2 + bx + c, we get a = 1, b = – 2 and c = – 8. The
vertex of the parabola has coordinates (1, – 9)
 b D
,
 , where D  b 2 – 4ac.
i.e. 
 2a 4a 
(iv) The polynomial f(x) = x2 – 2x – 8 = (x – 4) (x + 2)
is factorizable into two distinct linear factors
(x – 4) and (x + 2). So, the parabola cuts X-axis at
two d is t inc t poi nts (4 , 0 ) a nd (– 2, 0) . The
x-coordinates of these points are zeros of f(x).
Graphs of a cubic polynomial does not have a fixed
standard shape. Cubic polynomial graphs will
always cross X-axis at least once and at most thrice.

Observations :
From the graphs of the polynomial f(x) = x3 – 4x,
following observations are as follows :
(i) The polynomial f(x) = x 3 – 4x = x (x 2 – 4)
= x (x – 2) (x + 2) is factorisable into three distinct
linear factors. The curve y = f(x) also cuts X-axis at
three distinct points.
(ii) We have, f(x) = x (x – 2) (x + 2)
Therefore, 0, 2 and –2 are three zeros of f(x). The
c u rve y = f (x) c u ts X- axi s a t t hre e p oin ts
O (0, 0), P (2, 0) and Q (–2, 0).
Let  and  be the zeros of a quadratic polynomial
f(x) = ax2 + bx + c. By factor theorem (x – ) and (x – )
are the factors of f(x).
 f(x) = k (x – ) (x – ) are the factors of f(x)
 ax2 + bx + c = k{x2 – ( + ) x + }
 ax2 + bx + c = kx2 – k ( + ) x + k
PAGE # 1616
Comparing the coefficients of x2, x and constant terms
on both sides, we get
a = k, b = – k ( + ) and c = k
c
b
 +=–
and  =
a
a

 +  = –
and
 =
Let , ,  be the zeros of a cubic polynomial
Coefficient of x
f(x) = ax3 + bx2 + cx + d, a  0. Then, by factor theorem,
2
x – , x –  and x –  are factors of f(x) . Also, f(x) being a
Coefficient of x
cubic polynomial, cannot have more than three linear
Constant term
Coefficien t of x
2
factors.
Hence,
Coefficient of x
b
Sum of the zeros = –
=–
Coefficient of x 2
a

Constant term
c
Product of the zeros =
=
.
a
Coefficient of x 2
REMARKS :
ax3 + bx2 + cx + d = k(x – ) (x – ) (x – )

ax3 + bx2 + cx + d = k{x3 – ( +  + ) x2 + ( +  + ) x – }

ax3 +bx2 +cx+d =k x3 –k (+ + )x2 +k ( ++)x– k
a = k, b = – k ( +  + ), c = k ( +  + ) and d = – k ()
the zeros and their coefficients.
Sol. p (x) = 4x2 + 24x + 36,
Here a = 4, b = 24, c = 36
p(x) = 4x2 + 12x + 12x + 36
= 4x (x + 3) + 12 (x + 3)
= (x + 3) (4x + 12)
= 4 (x + 3)2
So, the zeros of quadratic polynomial is – 3 and – 3.
Let  = – 3 and = – 3
Sum of zeros = + 
=–3–3 =–6
++ =–

 +  + =



b –24
=
=–6
a
4
Coefficient of x
c
=
a
Coefficient of x 3
Product of the zeros = –
d
Constant term
=–
a
Coefficient of x 3
given by
f(x) = k (x – ) (x – ) (x – )
f(x) = k{x3 – ( +  + ) x2 + ( +  + ) x – },
where k is any non-zero real number.
cons tan t term
So, product of zeros = =
Coefficien t of x 2
b
=–
a
Coefficien t of x 3
Sum of the products of the zeros taken two at a
Cubic polynomial having ,  and  as its zeros is
coefficien t of x
coefficien t of x
c 36
= =
=9
a
4
d
a
REMARKS :
coefficien t of x 2
product of zeros = = (– 3) (– 3) = 9
Also, product of zeros =
c
a
Sum of the zeros = –
time =

coefficien t of x 2
b
a

And,= –
coefficien t of x
So, sum of zeros =  + = –

terms on both sides, we get
Ex.20 Find the zeros of quadratic polynomial
p(x) = 4x2 + 24x + 36 and verify the relationship between
= –
f(x) = k(x – ) (x – ) (x – )
Comparing the coefficients of x3, x2, x and constant
If  and  are the zeros of a quadratic polynomial f(x).
Then, the polynomial f(x) is given by
f(x) = k{x2 – ( + ) x + }
or, f(x) = k{x2 – (Sum of the zeros) x + Product of the
zeros}
Also, sum of zeros = –

1 3
Ex.22 Verify that  5, , are zeros of cubic polynomial
2 4
8x3 + 30x2 – 47x + 15. Also verify the relationship
2
cons tan t term
2
.
coefficien t of x
Ex.21 Find a quadratic polynomial whose zeros are
3  5 and 3  5 .
Sol. Given : = 3 + 5 and = 3 – 5
Sum of zeros = + 
=3+ 5 +3– 5=6
Products of zeros = 
= (3 + 5 ) (3 – 5 )
=9–5=4
So, quadratic polynomial isk {x2 – 6x + 4}, where k is
any constant.
between the zeros and the coefficients.
Sol. Given : f (x) = 8x3 + 30x2 – 47x + 15
Here, a = 8, b = 30, c = – 47, d = 15
f (– 5) = 8 (– 5)3 + 30 (– 5)2 – 47 (– 5) + 15
= – 1000 + 750 + 235 + 15 = 0
f(
1
)=
2
= 8
=
3
2
 1
 1
 1
8   30  – 47   15
2
2
 
 
2
1
1 47
 30  –
 15
2
8
4
8  60 – 188  120
8
=0
PAGE # 1717
3
and f (
= 8
Also, sum of product of zeros taken two at a time
2
3
3
3
3
) = 8   30  – 47   15
4
4
4
4
coefficient of x
–47
c
= =
coefficient of x 3
a
8
So, sum of product of zeros taken two at a time
=
27
9 141
 30 
–
 15
4
64
16
coefficien t of x
27 135 141

–
 15
=
4
8
8
=
= + + =
Product of zeros = 
27  135 – 282  120
=0
8
So, – 5,
= (– 5) (
1
3
and
are the zeros of the polynomial f (x).
2
4
= –
= –
coefficient of x 2
coefficient of x 3
–30
–15
b
=
=
8
4
a
So, sum of zeros =  + + 
= –
coefficient of x 2
coefficient of x 3
Sum of product of zeros taken two at a time
= + + 
= (– 5) (
1
1 3
3
) + ( ) ( ) + ( ) (– 5)
2
2 4
4
=
–5 3 15
 –
2 8 4
=
–20  3 – 30
–47
=
8
8
cons tan t term
coefficient of x 3
15
d
= –
.
8
a
So, product of zeros = 
1 3
–15
= –5   =
2 4
4
Also, sum of zeros = –
1 3
15
)( )= –
2
4
8
Also, product of zeros = –
1
3
Let,  = – 5, =
and =
2
4
Sum of zeros = + + 
coefficien t of x 3
= –
cons tan t term
coefficient of x 3
Ex.23 , ,  are zeros of cubic polynomial x3 – 12x2 + 44x + c.
If , ,  are in A.P., find the value of c.
Sol. f (x) = x3 – 12x2 + 44x + c.
, , are in A.P.
Let  = a – d, = a, = a + d
Sum of zeros = + + 
=a–d+a+a+d
= 3a
So, 3a = – (– 12)
a = 4.
Sum of product of zeros taken two at a time
 + + = 44
 (a – d) a + (a) (a + d) + (a + d) (a – d) = 44
 a2 – ad + a2 + ad + a2 – d2 = 44
 3a2 – d2 = 44
 3 (4)2 – d2 = 44 d2 = 4
 d=±2
product of zeros = = – c
 (a – d) a (a + d) = – c
 (4 – 2) (4) (4 + 2) = – c
 (2) (4) (6) = – c
 c = – 48.

PAGE # 1818
LINEAR EQUATIONS IN TWO VARIABLES
(b) Elimination by Equating the Coefficients :
An equation of the form Ax + By + C = 0 is called a linear
equation.
Where A is called coefficient of x, B is called coefficient
of y and C is the constant term (free from x & y)
A, B, C, R [  belongs to, R  Real No.]
But A and B can not be simultaneously zero.
f A  0, B = 0 equation will be of the form Ax + C = 0.
[Line || to Y-axis]
f A = 0, B  0 , equation will be of the form By + C = 0.
[Line || to X-axis]
f A  0 , B  0 , C = 0 equation will be of the form A x + By = 0.
[Line passing through origin]
f A  0 , B  0 , C  0 equation will be of the form A x + By + C = 0.
t is called a linear equation in two variable because
the two unknowns (x & y) occurs only in the first power,
and the product of two unknown quantities does not
occur.
Since it involves two variables therefore a single
equation will have infinite set of solution i.e.
indeterminate solution. So we require a pair of equation
i.e. simultaneous equations.
Standard form of linear equation :
(Standard form refers to all positive coefficients)
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
....(i)
....(ii)
For solving such equations we have three methods.
(a) Elimination by substitution
(b) Elimination by equating the coefficients
(c) Elimination by cross multiplication.
(a) Elimination By Substitution :
Ex.1 Solve for x and y :
2x + y = 7, 4x – 3y + 1 = 0
Sol. The given system of equations is
2x + y = 7
4x – 3y = – 1
....(i)
....(ii)
From (i), we get : y = (7 – 2x)
substituting y = (7 –2x) in (ii) , we get :

4x – 3 (7 – 2x) = – 1
4x – 21 + 6x = – 1


10x = 20
x=2
Substituting x = 2 in (i), we get
2  2 + y = 7  y = (7 – 4) = 3
Hence, the solution is x = 2, y = 3.
Ex.2 Solve for x and y : 10x + 3y = 75, 6x – 5y = 11.
Sol. The given equations are
10x + 3y = 75
..(i)
6x – 5y = 11
..(ii)
Multiplying (i) by 5 and (ii) by 3, we get :
50x + 15y = 375
..(iii)
18x – 15y = 33
..(iv)
Adding (iii) and (iv), we get
68x = 408  x =
408
x=6
68
Putting x = 6 in (i), we get :
(10  6) + 3y = 75  60 + 3y = 75
 3y = (75 – 60)
 3y = 15  y = 5
 x = 6, y = 5.
Ex.3 Solve for x and y :
ax + by – a + b = 0, bx – ay – a – b = 0
Sol. The given equation may be written as
ax + by = a – b
...(i)
bx – ay = a + b
... (ii)
Multiplying (i) by a and (ii) by b, we get
a2x + aby = a2 – ab
... (iii)
b2x – aby = ab + b2
...(iv)
Adding (iii) and (iv) , we get
(a2 + b2) x = (a2 + b2)  x =
(a 2  b 2 )
(a 2  b 2 )
x=1
Putting x = 1 in (i) we get
(a  1) + by = a – b  a + by = a – b
 by = a – b – a
 by = – b
y=
b
=–1
b
y=–1
x = 1 and y = – 1 is the required solution
Ex.4 Solve for x and y :
15
22
40
55
+
= 5,
+
= 13, x  y and x  –y
xy xy
xy xy
Sol. Putting
1
1
= u and
= v, the given equation
xy
xy
become
15u + 22v = 5
40u + 55v = 13 `
Multiplying (i) by 5 and (ii) by 2, we get
75u + 110v = 25
80u + 110v = 26
..(i)
..(ii)
..(iii)
..(iv)
PAGE # 1919
On subtracting (iii) from (iv), we get
1
5u = 1  u =
5
1
Putting u =
in (iii), we get
5

 x =
15 + 110v = 25  110v = 10
v=

Hence, x = 4 and y = 3 is the required solution.
(a) Unique Solution :
Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, if the
denominator a1b2 – a2 b1  0 then the given system of
equations have unique solution (i.e. only one solution)
and solutions are said to be consistent.
1
1

 u  5 and v  11



x – y = 5 and x + y = 11
2x = 16 and 2y = 6
x = 8 and y = 3
x = 8 and y = 3 is the required solution.
(c) Elimination by Cross Multiplication :
a1x + b1y + c1 = 0
 a1 b1 



 a2 b2 
a2x + b2y + c2= 0
a1b2 – a2 b1  0

a1
b
 1
a2
b2
(b) No Solution :
Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, if the
denominator a1b2 – a2 b1 = 0 then the given system of
equations have no solution and solutions are said to
be inconsistent.
i.e.
a1
b1
c1
=

a2
b2
c2
b1
c1
a1
b1
(c) Many Solution (Infinite Solutions) :
Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,
b2
c2
a2
b2
if
[Write the coefficient in this manner]
x
y
1
=
=
b1c 2 – b 2c 1 a 2c 1 – a1c 2
a1b 2 – a 2b1
x
1
=
b1c 2 – b 2c 1 a1b 2 – a 2b1

b1c 2 – b 2c 1
x= a b –a b
1 2
2 1

a1
b1
c1
=
=
then system of equations has many
a2
b2
c2
solution and solutions are said to be consistent.
Ex.6 Find the value of 'p' does the pair of equations given
below has unique solution?
4x + py + 8 = 0 & 2x + 2y + 2 = 0
Sol. 4x + py + 8 = 0
2x + 2y + 2 = 0
Here a1 = 4, b1 = p, c1 = 8
a2 = 2, b2 = 2, c2 = 2
For unique solution
a1
b1

a2
b2
y
1
Also a c – a c = a b – a b
2 1
1 2
1 2
2 1
a 2c 1 – a1c 2
y= a b –a b
1 2
2 1


Ex.5 Solve the system of equations 2x + 3y = 17,
3x – 2y = 6 by the method of cross multiplication.
Sol. The given equations may be written as
2x + 3y – 17 = 0
..(i)
3x – 2y – 6 = 0
..(ii)
By cross multiplication, we have
y
x 
1 




 
 17
3
3
2
2

52
39
=4,y=
=3
 13
 13
1
11
1
1
1
1
=
and
=
xy
xy
5
11




x
y
1
=
=
 52
 39  13
6
3
2
x
y
=
{3  ( 6)  ( 2)  ( 17)}
{( 17 )  3  ( 6 )  2}
1
=
{2  ( 2)  3  3}

x
y
1
=
=
( 18  34)
( 51  12) ( 4  9)

4
p

2
2
p  4.
Ex 7.If the system of the given equation has no solution
then find the value of k.
2x + ky = 7, 2kx + 3ky = 20
Sol. 2x + ky = 7
2kx + 3ky = 20
For no solution
a1
b1 c 1

=
a2
b2 c 2
2
k
7


2k 3k 20
2
k
2
7
=
and

2k
3k
2k
20
k =3 and k 
20
7
So, the value of k is 3.
PAGE # 2020
Ex.8 Find the values of p and q for which the following
system of equations has infinite number of solution :
2x + 3y = 7
(p + q)x + (2p – q)y = 3(p + q + 1)
Sol. 2x + 3y = 7
(p + q)x + (2p – q)y = 3(p + q + 1)
For infinitely many solution
(iii) x = 0
a1
b1
c1
=
=
a2 b2 c 2
Graphs of the type (ii) ay = b.
2
3
7
=
=
pq
2p – q
3(p  q  1)
2
3
=
pq
2p – q
Ex.10 Draw the graph of following :
(i) y = 0
(ii) y – 2 = 0
(iii) 2y + 4 = 0
Sol. (i) y = 0


4p – 2q = 3p + 3q
p = 5q ... (i)
and
3
7
=
2p – q
3(p  q  1)
4p – 3p = 5q
Put p = 5q
3
7
=
10q – q 3(5q  q  1)

(ii) y – 2 = 0
3
7
=
9q 18q  3
 21q = 18q + 3
 q = 1.
 p = 5 × 1 = 5.
3q = 3
(iii) 2y + 4 = 0  y = – 2
Graphs of the type (i) ax = b
Ex.9 Draw the graph of following :
(i) x = 2
(ii) x + 4 = 0
(iii) x = 0
Sol. (i) x = 2
Graphs of the type (iii) ax + by = 0
(Passing through origin)
(ii) x + 4 = 0 x = – 4
Ex.11 Draw the graph of following :
(i) x = y
(ii) x = – y
Sol. (i) x = y
x
1
4
–3
0
y
1
4
–3
0
PAGE # 2121
(ii) x = –y
x
1
–2
0
y
–1
2
0
Let equations of two lines are a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0.
(i) Lines are consistent (unique solution) i.e. they meet
at one point and condition is
a1
b
 1
a2
b2
(ii) Lines are inconsistent (no solution) i.e. they do not
meet at one point and condition is
a1
b
c
= 1  1 
a2
b2
c2
Graphs of the Type (iv) ax + by + c = 0.
(Making Interception x-axis , y-axis)
Ex.12 Solve the following system of linear equation
graphically : 4x – 3y + 4 = 0, 4x + 3y – 20 = 0.
Shade the Area bounded by the above two equation
and the x – axis, also find its area.
Sol. 4x – 3y + 4 = 0
...... (i)
4x + 3y – 20 = 0
...... (ii)
In (i) 4x + 4 = 3y
y=
4x  4
=
3
x
y
2
4
1
0
4
4
5
0
8
4
In (ii) 3y = 20 × 4x
20  4 x
y=
=
3
x
y
2
4
Drawing the equations, we get :
Since two lines are meeting at (2,4) their solution is
x = 2, y = 4
Area  =
1
1
BC × h =  6  4
2
2
= 12 sq.unit
So the solution of equation is x = 2 and y = 4 and the
area between the two lines and x-axis is 12 sq. unit
(iii) Lines are coincident (infinite solution)
i.e. overlapping lines (or they are on one another) and
condition is
a1 b1 c 1
=
=
.
a2 b2 c 2
Ex.13 Show graphically that the system of equations
2x + 4y = 10 and 3x + 6y = 12 has no solution.
Sol. Graph of 2x + 4y = 10
We have,
2x + 4y = 10
 4y = 10 – 2x  y =
5x
2
When x = 1, we have : y =
5 1
=2
2
PAGE # 2222
Type of Problems :
53
=1
2
Thus, we have the following table :
When x = 3, we have y =
(i) Determining two numbers when the relation
between them is given.
x
1
3
(ii) Problems regarding fractions, digits of a number,
ages of persons.
y
2
1
(iii) Problems regarding current of a river, regarding
time & distance.
Plot the points A (1,2) and B (3,1) on a graph paper.
Join A and B and extend it on both sides to obtain the
graph of 2x + 4y = 10 as shown in figure.
Graph of 3x + 6y = 12
We have,
3x + 6y = 12  6y = 12 – 3x  y =
When x = 2, we have : y =
4x
2
42
=1
2
40
=2
2
Thus, we have the following table :
When x = 0, we have : y =
x
2
0
y
1
2
Plot the point C (2,1) and D (0,2) on the same graph
paper. Join C and D and extend it on both sides to
obtain the graph of 3x + 6y = 12 as shown in figure.
(iv) Problems regarding mensuration and geometry.
(v) Problems regarding time & work.
(vi) Problems regarding mixtures, cost of articles, profit
& loss, discount etc.
Ex.14 3 bags and 4 pens together cost Rs.257, where as 4
bags and 3 pens cost Rs. 324, find the cost of a bag.
Sol.Let cost of a bag & a pen be x & y respectively
3x + 4y = 257
...(i)
4x + 3y = 324
...(ii)
Add (i) &(ii)
7x + 7y = 581
or x + y = 83 ...(iii)
Subtract (i) from (ii)
x – y = 67
...(iv)
Add (iii) & (iv)
2x = 150
x = Rs.75
 y = Rs.8
Hence the cost of 1 bag is Rs.75.
Ex15. The sum of numerator and denominator of a fraction
is 8. If 3 is added to both the numerator and denominator the fraction becomes
3
, find the fraction.
4
Sol. Let numerator be x and denominator be y. So, the
fraction is
x
y.
x+y=8
.... (i)
x3 3

and
y3 4
4x + 12 = 3y + 9
4x – 3y = – 3
Solving equation (i) and (ii)
x = 3 and y = 5
so, the fraction is
We find the lines represented by given equations are
parallel. So, the two lines have no common point.
Hence, the given system of equations has no solution.
For solving daily – life problems with the help of
simultaneous linear equation in two variables or
equations reducible to them proceed as :(i) Represent the unknown quantities by some variable
x and y, which are to be determined.
(ii) Find the conditions given in the problem and
translate the verbal conditions into a pair of
simultaneous linear equation.
(iii) Solve these equations & obtain the required
quantities with appropriate units.
.... (ii)
3
.
5
Ex16. Seven times a positive two digit number is equal to
four times the number obtained by reversing the
digits. If the difference between the digits is 3. Find the
number.
Sol. Let the unit digit and ten’s of two digit number be y, x
respectively
 the two digit positive number is 10x + y.
Reversed two digit number = 10 y + x
 7(10x + y) = 4(10y + x)
 66x – 33y = 0
 y = 2x
...(i)
And y – x = 3
2x – x = 3
from (i)
x=3
And y = 6
 the two digit positive number = 10x + y
= 10 × 3 + 6
= 30 + 6 = 36.
PAGE # 2323
Ex.17A man walks a certain distance with certain speed. If
he walks 1/2 km an hour faster, he takes 1 hour less.
But if he walks 1 km an hour slower, he takes 3 more
hours. Find the distance covered by man and his
original rate of walking.
Sol. Let the speed man = x km/hr
and Time taken = y hr
1

 x   (y – 1) = xy
2


xy – x +

–x+
1
1
y–
= xy
2
2
1
1
y=
2
2
or – 2x + y = 1
....(i)
And (x – 1) (y + 3) = xy
 xy + 3x – y – 3 = xy
 3x – y = 3
....(ii)
Solving equation (i) & (ii)
x = 4 and y = 9
So, total distance covered = xy km
= 4 × 9 = 36 km
and original rate of walking = 4 km/hr
Ex.18 If 4x + 3y = 120, find how many positive integer
solutions are possible ?
Sol. We can write the equation in another form.
4(x – 3) +3 (y + 4) = 120
Now, we make a table
x
30
27
24
21
18
15
12
9
6
3
0
y
0
4
8
12
16
20
24
28
32
36
40
W e observe a pattern that x reduces by 3 &
y increases by 4. But, both x & y cannot be negative
or 0. So, the value of x = 0 or y = 0 is ruled out.
Hence, nine such positive integer solutions are
possible.
Alternately, we can write the given equation as
4(x + 3) + 3(y – 4) = 120. We get the same values of x & y.

PAGE # 2424
QUADRATIC EQUATIONS
If P(x) is quadratic expression in variable x, then
P(x) = 0 is known as a quadratic equation.
(a) General form of a Quadratic Equation :
The general form of a quadratic equation is
ax2 + bx + c = 0, where a, b, c are real numbers and a  0.
Since a  0, quadratic equations, in general, are of
the following types:(i)
b = 0, c  0, i.e., of the type ax2 + c = 0.
Consider the quadratic equation, a x2 + b x + c = 0
having  and  as its roots and b2  4ac is called
discriminant of roots of quadratic equation. It is
denoted by D or .
Roots of the given quadratic equation may be
(i) Real and unequal
(ii) Real and equal
(iii) Imaginary and unequal.
Let the roots of the quadratic equation ax2 + bx + c = 0
(where a  0, b, c  R) be  and  then
(ii) b  0, c = 0, i.e., of the type ax2 + bx = 0.
=
(iii) b = 0, c = 0, i.e., of the type ax2 = 0.
(iv) b  0, c  0, i.e., of the type ax2 + bx + c = 0.
The value of x which satisfies the given quadratic
equation is known as its root. The roots of the given
equation are known as its solution.
General form of a quadratic equation is :
ax2 + bx + c = 0
or
or
or
or
4a2x2 + 4abx + 4ac = 0 [Multiplying by 4a]
4a2x2 + 4abx = – 4ac
[By adding b2 both sides]
2 2
2
2
4a x + 4abx + b = b – 4ac
(2ax + b)2 = b2 – 4ac
Taking square root of both the sides
2ax + b = ±
or
x=

b 2  4ac
 b  b 2  4ac
2a
 b  b 2  4ac
 b  b 2  4ac
and
.
2a
2a
REMARKS :
 b  b 2  4ac
... (ii)
2a
The nature of roots depends upon the value of
expression ‘b2 – 4ac’ with in the square root sign. This
is known as discriminant of the given quadratic
equation.
and
=
Consider the Following Cases :
Case-1 When b2 – 4ac > 0, (D > 0)
In this case roots of the given equation are real and
distinct and are as follows
=
 b  b 2  4ac
 b  b 2  4ac
and  =
2a
2a
(ii) When a( 0), b, c  Q and b2 – 4ac is not a perfect
square
In this case both the roots are irrational and distinct.
[See remarks also]
Case-2 When b2 – 4ac = 0, (D = 0)
In this case both the roots are real and equal to –
A quadratic equation is satisfied by exactly two
values of ' x ' which may be real or imaginary. The
equation, a x2 + b x + c = 0 is:

A quadratic equation if a  0.
[Two roots]

A linear equation if a = 0, b  0.
[One root]

A contradiction if a = b = 0, c  0.
[No root]

An identity if a = b = c = 0.

A quadratic equation cannot have more than two roots.

... (i)
(i) When a( 0), b, c  Q and b2 – 4ac is a perfect
square
In this case both the roots are rational and distinct.
Hence, roots of the quadratic equation ax2 + bx + c = 0
are
 b  b 2  4ac
2a
[Infinite roots]
It follows from the above statement that if a quadratic
equation is satisfied by more than two values of x, then
it is satisfied by every value of x and so it is an identity.
b
.
2a
Case-3 When b2 – 4ac < 0, (D < 0)
In this case b2 – 4ac < 0, then 4ac – b2 > 0

=
and  =
or
=
 b   ( 4ac  b 2 )
2a
 b   ( 4ac  b 2 )
2a
 b  i 4ac  b 2
2a
 b  i 4ac  b 2
[  1 = i ]
2a
i.e. in this case both the roots are imaginary and distinct.
and  =
25
25
PAGE # 25
25

REMARKS :

If a, b, c  Q and b2 – 4ac is positive (D > 0) but not a
perfect square, then the roots are irrational and they
(ii) If only one root is common, then the common
root '  ' will be :
=
always occur in conjugate pairs like 2 + 3
and 2 – 3 . However, if a, b, c are irrational numbers
and b2 – 4ac is positive but not a perfect square, then
the roots may not occur in conjugate pairs.
Hence the condition for one common root is:
 c 1 a 2  c 2 a1 2 = a1 b 2  a 2 b1  b1 c 2  b 2 c 1 


If b2 – 4ac is negative (D < 0), then the roots are complex
conjugate of each other. In fact, complex roots of an
equation with real coefficients always occur in conjugate
pairs like 2 + 3i and 2 – 3i. However, this may not be true
in case of equations with complex coefficients.
For example, x2 – 2ix – 1 = 0 has both roots equal to i.

If a and c are of the same sign and b has a sign opposite
to that of a as well as c, then both the roots are positive,
the sum as well as the product of roots is positive
(D  0).

If a, b, c are of the same sign then both the roots are
negative, the sum of the roots is negative but the product
of roots is positive (D  0).
Relation Between Roots & Coefficients
(i) The solutions of quadratic equation a x2 + b x + c = 0
are given by
x=
 b  b2  4 a c
2a
2
(ii) The expression b  4 a c  D is called discriminant
of the quadratic equation a x2 + b x + c = 0.
If ,  are the roots of the quadratic equation
a x2 + b x + c = 0, then
coefficien t of x
(a) Sum of the roots = –
 +  = 
coefficien t of x 2
b
a
(b) Product of the roots =
NOTE :
If f(x) = 0 & g(x) = 0 are two polynomial equation having
some common root(s) then those common root(s)
is/are also the root(s) of h(x) = a f(x) + bg (x) = 0.
Ex.1 If x2 – ax + b = 0 and x2 – px + q = 0 have a root in
common and the second equation has equal roots,
ap
.
2
Sol. Given equations are : x2 – ax + b= 0
..... (i)
and x2 – px + q = 0
.... (ii)
Let  be the common root. Then roots of equation
(ii) will be  and . Let  be the other root of equation
(i). Thus roots of equation (i) are ,  and those of
equation (ii) are , .
Now
+=a
...(iii)
 = b
...(iv)
2 = p
...(v)
2 = q
...(vi)
L.H.S. = b + q =  + 2 = ( + )
...(vii)
show that b + q =
ap
(   ) 2
=
=  ( + )
2
2
From (vii) and (viii), L.H.S. = R.H.S.
And R.H.S. =


c
a
b
=
=
9
1
2
a:b:c=1:2:9
(a) By Factorisation :
ALGORITHM :
Condition for common Root
2
Consider two quadratic equations, a1 x + b1 x + c1 = 0
& a 2 x2 + b 2 x + c 2 = 0.
(i) If two quadratic equations have both roots
common, then the equations are identical and their
coefficients are in proportion. i.e.
a1
b
c
= 1 = 1 .
a2
b2
c2
... (viii)
Ex.2 If a, b, c  R and equations ax2 + bx + c = 0 and
x2 + 2x + 9 = 0 have a common root, show that
a : b : c = 1 : 2 : 9.
Sol. Given equations are : x2 + 2x + 9 = 0 ...(i)
and ax2 + bx + c = 0
...(ii)
Clearly roots of equation (i) are imaginary since
equation (i) and (ii) have a common root, therefore
common root must be imaginary and hence both
roots will be common.
Therefore equations (i) and (ii) are identical
constant term
coefficien t of x 2
c
 =
a
(iii) A quadratic equation whose roots are  and  is
(x ) (x ) = 0
i.e. x2  (sum of roots) x + (product of roots) = 0.
c 1 a 2  c 2 a1
b1 c 2  b 2 c 1
=
a1 b 2  a 2 b1
c 1 a 2  c 2 a1
Step (i) Factorise the constant term of the given
quadratic equation.
Step (ii) Express the coefficient of middle term as the
sum or difference of the factors obtained in step 1.
Clearly, the product of these two factors will be equal to
the product of the coefficient of x2 and constant term.
Step (iii) Split the middle term in two parts obtained in
step 2.
Step (iv) Factorise the quadratic equation obtained in
step 3.
26
26
PAGE # 26
26
Ex.3 Solve the following quadratic equation by factorisation
method : x2 – 2ax + a2 – b2 = 0.
Sol. Here, factors of constant term (a2 – b2) are (a – b) and (a + b).
Sol. We have,
x  3 x – 3 2x – 3


x –1
x2 x–2
Also,Coefficient of the middle term = – 2a
= – [(a – b) + (a + b)]

x2 – 2ax + a2 – b2 = 0
2

x – {(a – b) + (a + b)} x + (a – b) (a + b) = 0

x2 – (a – b) x – (a + b) x + (a – b) (a + b) = 0

x{x – (a – b)} – (a + b) {x – (a – b)} = 0

{x – (a – b)} {x – (a + b)} = 0

x – (a – b) = 0 or, x – (a + b) = 0

x = a – b or x = a + b
( x  2)  1 ( x – 2) – 1 2( x – 1) – 1


x–2
x –1
x2
1
1
1
1
 1–
 2–
x–2
x –1
x2
1
1
1
–
–
x2 x–2
x –1
x–2–x–2
1
–
2
x –1
x –4
–4
–1

x2 – 4 x – 1




Ex.4 Solve 9x2 – 49 = 0

Sol. We have 9x2 – 49 = 0
 4(x – 1) = x2 – 4
or
(3x)2 – (7)2 = 0
or
(3x + 7)(3x – 7) = 0
 x2 – 4x = 0
i.e., 3x + 7 = 0 or 3x – 7 = 0.
 x(x – 4) = 0
This gives x = –
Thus, x = –
 x = 0, 4.
7
7
or
.
3
3
7
3
or
x  3 x – 3 2x – 3


.
x –1
x2 x–2
Ex.8 Solve the equation :
7
3
Hence, the roots of the given equation are 0, 4.
are solutions of the given
(b) By the Method of Completion of Square :
equation.
ALGORITHM :
2
Ex.5 Solve the quadratic equation : 3x – 15x = 0.
Sol. The given equation may be written as 3x(x – 5) = 0
Step-(i) Obtain the quadratic equation. Let the quadratic
equation be ax2 + bx + c = 0, a  0.
This gives x = 0 or x = 5.
Step-(ii) Make the coefficient of x2 unity, if it is not unity.
x = 0, 5 are the required solutions.
i.e., obtain x2 +
Ex. 6 Solve the quadratic equation : 48y2 – 13y – 1 = 0.
Step-(iii) Shift the constant term
Sol. 48y2 – 13y – 1 = 0

48 y2 – 16 y + 3y – 1 = 0

16 y (3y – 1) + 1(3y – 1) = 0

(16y + 1) (3y – 1) = 0

y=
b
c
x+
= 0.
a
a
c
on R.H.S. to get
a
b
c
x=–
a
a
x2 +
Step-(iv) Add square of half of the coefficient of x.i.e.
2
 b 
  on both sides to obtain
 2a 
–1
1
,y=
16
3
2
2
 b 
 b 
 b 
c
x +2  x+   =   –
 2a 
 2a 
 2a 
a
2
2
Ex.7 Solve the quadratic equation : 6 2 x  5 x – 3 2  0 .
Sol. 6 2 x2 + 5x – 3
2 =0
Step-(v) Write L.H.S. as the perfect square of a binomial
expression and simplify R.H.S. to get
2

6 2 x2 + 9x – 4x – 3 2 = 0

3x (2 2 x + 3) – 2 (2 2 x + 3)= 0

(3x – 2 ) (2 2 x + 3) = 0

x=
2
–3
or x =
3
2 2
b 
b 2  4ac

x 
 =
.
2a 
4a 2

Step-(vi) Take square root of both sides to get
x+
b
=±
2a
b 2  4ac
4a 2
Step (vii) Obtain the values of x by shifting the constant
term
b
on RHS.
2a
27
27
PAGE # 27
27
Ex.9 Solve : x2 + 5x + 6 = 0.
Sol. We have
x2 + 5x + 6 = 0
Add and subtract (
5
2
2
1
coefficient of x)2 in L.H.S. and get
2
5
2
2
by x =
x2 + 5x + 6 +   –   = 0
2
2

5
5 5
x2 + 2   x +   –   + 6 = 0
2
2 2
 

5
1

x   –  0
2
4


5
1

x   
2
4


x
2
2
5
1

2
2
This gives x = –3 or x = – 2
Therefore, x =–3,–2 are the solutions of the given
equation.
Ex.10 By using the method of completing the square, show
that the equation 4x2 + 3x + 5 = 0 has no real roots.
Sol. We have, 4x2 + 3x + 5 = 0
3
5
x+
=0
4
4

x2 +

3 
5
x2 + 2  x  = –
8 
4

3
x2 + 2   x +
8

3

71
x   = –
8
64

Ex11 Solve the quadratic equation x2 – 7x – 5 = 0.
Sol. Comparing the given equation with ax2 + bx + c = 0,
a = 1, b = –7 and c = –5.
Therefore, D = (–7)2 – 4 × 1 × (–5) = 49 + 20 = 69 > 0
Since D is positive, the equation has two roots given

2
2
3
3
5
  =  –
8
8
4
 
 
Clearly, RHS is negative.
2
3

But,  x   cannot be negative for any real value of x.
8

Hence, the given equation has no real roots.
7  69 7 – 69
,
are the required solutions.
2
2
Ex.12 Find the value of P, If the quadratic equation
x2 – 2x (1 + 3P) + 7(3 + 2P) = 0 has real & equal roots.
Sol. x2 – 2x (1 + 3P) + 7 (3 + 2P) = 0
If it has real and equal roots, then D = 0
D = b2 – 4ac
a = 1, b = – 2 (1 + 3P), c = 7 (3 + 2P)
D = 4 (1 + 3P)2 – 4 × 1 × 7 (3 + 2P)
= 4[(1 + 3p)2 – 7(3 + 2P)]
= 4 [1 + 6P + 9P2 – 21 – 14 P]
D = 4 [9P2 – 8P – 20]
Now, D = 0
[according to condition]
4[9P2 – 8P – 20] = 0
9P2 – 8P – 20 = 0
9p2 – 18p + 10p – 20 = 0
9P (P – 2) + 10 (P – 2) = 0
(9P + 10) (P – 2) = 0
So, 9P + 10 = 0 or p – 2 = 0
P = – 10/9 or p = 2.
2 1 1
  .
b a c
Sol. Since the roots of the given equations are equal, so
discriminant will be equal to zero.
 b2(c – a)2 – 4a(b – c)c(a – b) = 0
 b2(c2 + a2 – 2ac) – 4ac(ba – ca – b2 + bc) = 0,
 a2b2 + b2c2 + 4a2c2 + 2b2ac – 4a2bc – 4abc2 = 0
 (ab + bc – 2ac)2 = 0
 ab + bc – 2ac = 0
 ab + bc = 2ac

1 1 2
+ =
c a b

2
1
1
=
+ .
b
a
c
(c) By Using Quadratic Formula :
Solve the quadratic equation in general form
viz. ax2 + bx + c = 0.
Step (i) By comparison with general quadratic equation,
find the value of a, b and c.
Step (ii) Find the discriminant of the quadratic equation.
D = b2 – 4ac
Step (iii) Now find the roots of the equation by given
equation

x=
Ex.13 If the roots of the equation
a(b – c) x2 + b(c – a)x + c (a – b) = 0 are equal, show that
2
x=
7  69 7 – 69
,
2
2
b D b D
,
2a
2a
REMARK :
If b2 – 4ac < 0, i.e., negative, then
b 2 – 4ac is not real
and therefore, the equation does not have any real roots.
Hence Proved.
Ex.14 For what value of K does the equation
(k + 3)x2 – (2k + 5)x + k + 1 = 0 have no real roots.
Sol. (k + 3)x2 – (2k + 5) x + k + 1 = 0
If it has no real roots, then D < 0
D = b2 – 4ac
{– (2k + 5)}2 – 4 × (k + 3) × (k + 1)
= 4k2 + 20 k + 25 – 4 [k2 + k + 3k + 3]
= 4k + 13
SInce D < 0
4k + 13 < 0
Hence , k <
13
.
4
28
28
PAGE # 28
28


4500 = x2 – 15x
x2 – 15x – 4500 = 0
The method of problem solving consists of the
following three steps :

x=
15  225  18000
2
Step (i) Translating the word problem into symbolic
language (mathematical statement) which means
identifying relationships existing in the problem and
then forming the quadratic equation.
Step (ii) Solving the quadratic equation thus formed.
Step (iii) Interpreting the solution of the equation, which
means translating the result of mathematical statement
into verbal language.

x=
15  135
= 75, – 60
2
ALGORITHM :
Ex.15 Sum of a number and its reciprocal is 2
1
. Find the
30
number.
Sol. Let the number be x. Then according to given condition
x+

expenses =
360
x4
Now, according given condition
 30x2 + 30 = 61x
 30x2 – 61x + 30 = 0
By using quadratic formula
360
360
–
=3
x
x4
61 3721 – 3600
60
61 11
 x=
60
61 11 72
6
x=
=
=
60
60
5
x=
 1440 = 3x2 + 12x
 x2 + 4x – 480 = 0
 (x + 24) (x – 20) = 0
 x = 20
So, the original duration of the tour = 20 days.
61 11 50 5
x=
=
=
60
60 6
Hence the required number is
5
6
or
.
6
5
Ex.16 A fast train takes 3 hours less than a slow train for a
journey of 900 km. If the speed of slow train is
15 km/ hr. less than that the fast train. Find the speeds
of the two trains.
Sol. Let the speed of fast train = x km/h
and the speed slow train = (x – 15) km/h
900
hr
x
900
the time taken by slow train =
hr
x – 15
then according the given the condition
300 
 300
3
–
 =3
x
–
15
x 

Total money
360
=
Number of day
x
when he extend his tour by 4 days, then his daily
x 2  1 61
=
30
x
900
900
–
=3
x – 15
x
Ex.17 A person on tour has Rs. 360 for his expenses. If he
extends his tour 4 days, he has to cut down his daily
expenses by Rs. 3, Find the original duration of the
tour.
Sol. let x be the number of days
So his daily expenses =
61
1
=
30
x
the time taken by fast train =
 x – 15 = 60
 x = 75
So speed of fast train is 75 km/hr and the speed of
slow train (75 – 15) km/hr = 60 km/hr.
Ex.18 A two digit number is such that the product of its
digits is 10 when 27 is subtracted from the number,
the digit interchange their places. Find the number.
Sol. Let the tens and unit digit of required number be x and
y respectively. Then according given condition
xy = 10  y =
10
x
and 10x + y – 27 = 10y + x
 10x +
 10 
10
+x
– 27 = 10 
 x 
x

10 x 2  10  27 x 100  x 2
=
x
x







9x2 – 27x – 90 = 0
9(x2 – 3x – 10) = 0
x2 – 3x – 10 = 0
x2– 5x + 2x – 10 = 0
x(x – 5) + 2(x – 5) = 0
(x – 5)(x + 2) = 0
x = – 2 or 5
So, x = 5, y =
10 10
=
=2
x
5
So the required number is 10x + y = 10 × 5 + 2 = 52.
29
29
PAGE # 29
29
PROGRESSIONS
SEQUENCE
A sequence is an arrangement of numbers in a
definite order according to some rule.
e.g. (i) 2, 5, 8, 11, ...
(ii) 4, 1, – 2, – 5, ...
(iii) 3, –9, 27, – 81, ...
Types of Sequence
On the basis of the number of terms there are two
types of sequence :
(i) Finite sequences : A sequence is said to be
finite if it has finite number of terms.
(ii) Infinite sequences : A sequence is said to be
infinite if it has infinite number of terms.
Ex.1 Write down the sequence whose n th term is n 2+1 :
Sol. Let tn = n 2+1
Put n = 1,2,3,4,.............. we get
t1 =2, t2 = 5, t3 = 10,....
So the sequence is 2,5,10,17,.....
PROGRESSIONS
Those sequence whose terms follow certain patterns
are called progressions. Generally there are three types
of progressions.
(i) Arithmetic Progression (A.P.)
(ii) Geometric Progression (G.P.)
(iii) Harmonic Progression (H.P.)
ARITHMETIC PROGRESSION
A sequence is called an A.P., if the difference of a term
and the previous term is always same.
i.e. d = tn + 1 – tn = Constant for all n  N. The constant
difference, generally denoted by ‘d’ is called the
common difference.
Ex.2 Find the common difference of the following
A.P. : 5,9,13,17,21,.....
Sol. 9 – 5 = 13 – 9 = 17 – 13 = 21 – 17 = 4 (constant).
 Common difference (d) = 4.
GENERAL FORM OF AN A.P.
If we denote the starting number i.e. the 1st number by
‘a’ and a fixed number to be added is ‘d’ then
a, a + d, a + 2d, a + 3d, a + 4d,........... forms an A.P.
Ex.3 Find the A.P. whose 1st term is 10 & common difference
is 5.
Sol. Given :First term (a) = 10 & Common difference (d) = 5.
 A.P. is 10, 15, 20, 25, 30,.......
th
n TERM OF AN A.P.
Let A.P. be a, a + d, a + 2d, a + 3d,...........
Then, First term (a1) = a + 0.d
Second term (a2) = a + 1.d
Third term (a3)
= a + 2.d
.
.
.
.
.
.
nth term (an)
= a + (n – 1) d
 an = a + (n – 1) d is called the nth term.
Ex.4. Which term of the A.P. 21, 42, 63, 84, .............. is 420?
Sol. 21, 42, 63, ........ 420
first term (a) = 21, common difference (d) = 21
Let 420 will come at an term.
an=a+(n–1)d
420 = 21 + (n – 1) 21
420 = 21 + 21n – 21
21n = 420
n = 20
Ex.5. The 14th term of an A.P. is twice its 7th term. If its 5th
term is 10. Find its first term and the common difference. Also find its 53 rd term.
Sol.
a14 = 2 a7
a + 13 d = 2 ( a + 6 d)
a + 13 d = 2a + 12 d
a=d
and a5 = 10
a + 4d = 10
a + 4a = 10
[a = d ]
5 a = 10
a = 2.
So, d = 2.
1st term = 2, common difference = 2
a53 = a + 52 d
a53 = 2 + 52 × 2
a53 = 2 + 104
a53 = 106.
Ex.6 Is 51 a term of the A.P. 5,8,11,14,.............?
Sol. 5, 8, 11, 14 .............
first term (a)= 5
common difference (d) = 3
Let 51 be the nth term a term of AP
and we know an=a+(n–1)d
 51 = 5 + (n – 1)3
51 = 5 + 3n – 3
51 = 2 + 3n
49 = 3n
n=
49
3
But n can never be a fraction in an AP
 51 is not a term of AP.
PAGE # 30
Ex.9 Find the sum of 16 terms of the A.P. 3,8,13,18,.......
Sol. a = 3, d = 5
th
m TERM OF AN A.P. FROM THE END
Let ‘a’ be the 1st term and ‘d’ be the common difference
of an A.P. having n terms. Then mth term from the end
is (n – m + 1)th term from beginning or {n – (m – 1)}th
term from beginning.
Ex.7 Find 12th term from the end of an A.P. 5,12,19,... 495.
Sol. 495 = 5 + (n – 1) 7 n = 71
 12h term from end m = 12
a71 – (12 – 1) = a71 – 11 = a60 from the beginning.
a60 = 5 + (60 – 1) 7 = 418.
SELECTION OF TERMS IN AN A.P.
Sometimes we require certain number of terms in A.P.
The following ways of selecting terms are generally
very convenient.
No. of Terms
Terms
Common Difference
For 3 terms
a – d, a, a + d
d
For 4 terms
a – 3d, a – d, a + d, a + 3d
2d
For 5 terms
a – 2d, a – d, a, a + d, a + 2d
d
For 6 terms
a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d
2d
Ex.8 The sum of three numbers in A.P. is –3 and their product
is 8. Find the numbers.
Sol. Let three no.’s in A.P. be a – d, a, a + d
 a – d + a + a + d = –3
 3a = –3 a = –1
& (a – d) a (a + d) = 8
 a (a2 – d2) = 8
 (–1) (1 – d2) = 8
 1 – d2 = – 8
 d2 = 9
 d= 3
If a = – 1 & d = 3 numbers are – 4, –1, 2.
If a = – 1 & d = –3 numbers are 2, –1, – 4.
SUM OF n TERMS OF AN A.P.
S16 =
n
[2a + (n – 1)d]
2
16
[2(3) + (16 – 1)5]
2
= 648.
Ex.10. Find the value of x when 1 + 6 + 11 + 16 + ........ + x = 148.
Sol. Clearly, terms of the given series form an A.P. with first
tern a = 1 and common difference d = 5. Let there be n
terms in this series. Then,
1 + 6 + 11 + 16 + ...+ x = 148.
 Sum of n terms = 148

n
[ 2+ (n – 1) × 5] = 148
2
 5n2 – 3n – 296 = 0
 (n – 8) (5n + 37) = 0
 n=8
Now, x = nth term
 x = a + (n – 1)d
 x = 1 + (8 – 1) × 5 = 36
[ n is not negative]
[ a =1, d = 5, n = 8]
PROPERTIES OF A.P.
(i) For any real numbers a and b, the sequence whose
nth term is an = an + b is always an A.P. with common
difference ‘a’ (i.e. coefficient of term containing n).
(ii) If any nth term of a sequence is a linear expression
in n then the given sequence is an A.P.
(iii) If a constant term is added to or subtracted from
each term of an A.P. then the resulting sequence is
also an A.P. with the same common difference.
(iv) If each term of a given A.P. is multiplied or divided
by a non-zero constant K, then the resulting
sequence is also an A.P. with common difference
d
respectively. Where d is the common
K
difference of the given A.P.
Let A.P. be a, a + d, a + 2d, a + 3d,............., a + (n – 1)d
Then,
Sn = a + (a + d) +...+ {a + (n – 2) d} + {a + (n – 1) d} ..(i)
also
Sn= {a + (n – 1) d} + {a + (n – 2) d} +....+ (a + d) + a ..(ii)
Add (i) & (ii)
Kd or
 2Sn = 2a + (n – 1)d + 2a + (n – 1)d +....+ 2a + (n – 1)d
(vi) If three numbers a, b, c are in A.P. , then 2b = a + c.

2Sn = n [2a + (n – 1) d]

Sn 
n
2a ( n  1) d 
2
Sn =
n
[a + a + (n – 1)d]
2
n
[a +  ]
2
n
 S n  a    where, is the last term.
2
rth term of an A.P. when sum of first r terms is Sr is
given by, tr = Sr – Sr – 1.
=

Sn =
(v) In a finite A.P. the sum of the terms equidistant from
the beginning and end is always same and is equal to
the sum of 1st and last term.
Ex.11 Check whether an = 2n2 + 1 is an A.P. or not.
Sol. an = 2n2 + 1
Then, an + 1 = 2(n + 1)2 + 1
 an + 1 – an = 2(n2 + 2n + 1) + 1 – 2n2 – 1
= 2n2 + 4n + 2 + 1 – 2n2 – 1
= 4n + 2, which is not constant
 The above sequence is not an A.P..
Arithmetic Mean (Mean or Average) (A.M.)
If three terms are in A.P. then the middle term is
called the A.M. between the other two, so if a, b, c
are in A.P., b is A.M. of a & c.
PAGE # 31
(ii) Sum of the first n terms.
A.M. for any n number a 1, a 2,..., a n is;
A=
a1  a 2  a 3  .....  a n
.
n
n - Arithmetic
Numbers :
Means
Between
Two
If a, b are any two given numbers & a, A1, A2,...., An, b are
in A.P. then A1, A2,... An are the n-A.M.’s between a & b.
Total terms are n + 2.
ba
.
n1
2(b  a )
= a +
,. .. ... ... .. .,
n 1
Last term b = a + (n+2–1)d.Now, d =
ba
, A2
n 1
n (b  a)
An = a +
.
n 1
A1 = a +

NOTE :
Sum of all n-A.M.’s inserted between a & b is equal
to n times the single A.M. between a & b.
n
i.e.

r 1
Ar = nA where A is the single A.M. between a & b.
13
, an even
6
number of A.M.s is inserted, the sum of these
means exceeds their number by unity. Find the
number of means.
Sol. Let a and b be two numbers and 2n A.M.s are
inserted between a and b then
2n
(a + b) = 2n + 1.
2
Ex.12 Between two numbers whose sum is



a rn  1

 r 1
Sn = 
a 1 rn

 1  r
13 

 13 
n   = 2n + 1.
Given a  b  6 


6
 
n = 6.
Number of means = 12.
Ex.13 Insert 20 A.M. between 2 and 86.
Sol. Here 2 is the first term and 86 is the 22 nd term of A.P.
So, 86 = 2 + (21)d
 d=4
So, the series is 2, 6, 10, 14,......., 82, 86
 required means are 6, 10, 14,...82.


, r 1


, r 1
(iii) Sum of an infinite G.P. when r < 1. When
n  rn  0 if r < 1 therefore,
a
(| r |  1) .
S =
1– r
Ex.14 Find 10th term of the G.P. 7, 14, 28, .....
Sol. In this G.P. we have a = 7 and r = 2
 10th term = ar(10–1) = ar9 = 7  29 = (7  512) = 3584.
Ex.15 How many terms are there in the G.P. 16,8,4.....,
Sol. In this G.P. we have a = 16, r =
8
1
=
16
2
Let the number of terms be n. Then,
 1
 16   
2
 1
 
2
n 1
=
n 1
=
1
16
1
 1
=  
16  16  2 
8
 n – 1 = 8 or n = 9
Ex.16 A boy saves Re 1 on first day of the month, Rs. 2 on
2nd day of the month, Rs. 4 on third day, Rs. 8 on 4th
day, Rs. 16 on 5th day and so on. How much does he
save in 12 days ?
Sol. His savings are 1,2,4,8,16, .... upto terms.
This is a G.P. with a = 1 and r =
2
=2
1
 Savings = Sum of above G.P. upto 12 terms.
G.P. is a sequence of numbers whose first term is
non zero & each of the succeeding terms is equal
to the preceding terms multiplied by a constant.
Thus in a G.P. the ratio of successive terms is
c o ns t ant . Thi s c ons tan t f ac t or is c a lle d t he
common ratio of the series & is obtained by dividing
any term by that which immediately precedes it.
1 1 1 1
Example : 2, 4, 8, 16 ... & , ,
,
...are in G.P.
.P.
3 9 27 81
(i) Therefore a, ar, ar2, ar3, ar4,...... is a G.P. with ‘a’ as
the first term and ‘r’ as common ratio.
nth term = a rn1
1
?
16
=
a(r n  1) 1 ( 212  1)
=
= 4095.
(r  1)
(2  1)
Ex.17 Find the sum of 10 terms of the sequence 5, 5.5,
5.55, 5.555, ......
Sol. 5, 5.5., 5.55, 5.555, ..... 10 terms
= 5(1 + 1.1 + 1.11 + ..... 10 terms)
=
5
(9 + 9.9 + 9.99 + .... 10 terms)
9
=
5
(10 – 1 + 10 – 0.1 + 10 – 0.01 + ..... 10 terms)
9
=
5
(10 + 10 + 10 +.....10 terms) – (1 + 0.1 + 0.01
9
+.....10 terms)
PAGE # 32
=
5
(100 – sum of G.P. of 10 terms with a = 1 & r = 0.1)
9
Sol. Let G1, G2, G3 be three geometric means between 4
10
5
= 9 100 

 = 5
Ex.18 Find three geometric means between 4 and 64.
1 (1  (0.1) )  5 
10 
1 
  100  1  10 
1  0.1
9
9
 10 


and 64.
Then 4, G1, G2, G3, G4, 64 are in G.P.
 5th term = 64 Let common ratio be r.
5
(890  10  9 ) .
81
Then, 4  r4 = 64 r4 = 16 = 24
r=2
 G1 = (4  2) = 8, G2 = (8  2) = 16 & G3 = (16  2) = 32
(i) If a, b, c are in G.P. then b 2 = ac, in general if
a 1, a 2, a 3, a 4,......... a n – 1 , a n are in G.P.,
then a 1a n = a 2a n – 1 = a 3 a n – 2 = ..............
So, the required geometric means are 8,16, 32.
Important results :
n(n  1)
2
(ii) Any three consecutive terms of a G.P. can be
i.
Sum of the first n natural numbers =
a
, a , ar..
r
(iii) Any four consecutive terms of a G.P. can be taken
a a
as 3 , , ar, ar3.
r
r
ii.
Sum of the squares of first n numbers
taken as
(iv) If each term of a G.P. be multiplied or divided or
raised to power by the some nonzero quantity, the
resulting sequence is also a G.P.
=
n(n  1)(2n  1)
6
 n(n  1) 

iii. Sum of the cubes of first n numbers = 
 2 
iv.
Sum of the first n even numbers = n(n + 1)
v.
Sum of the first n odd numbers = n2
If a, b, c are in G.P., b is the G.M. between a & c.
b² = ac, therefore b = a c ; a > 0, c > 0.
n-Geometric Means Between a, b :
If a, b are two given numbers & a, G1, G2,....., Gn, b
are in G.P.. Then,
G1, G2, G3,...., Gn are n-G.M.s between a & b.
G1 = a(b/a)1/n+1, G2 = a(b/a)2/n+1,......, Gn = a(b/a)n/n+1

NOTE :
The product of n G.M.s between a & b is equal to the
nth power of the single G.M. between a & b
n
i.e.
n
 G =  ab  = G
r
n
where G is the single G.M.
r 1
between a & b.

PAGE # 33
2
LINES AND ANGLES, TRIANGLES
(iii) Obtuse angle : An angle whose measure is more
than 90º but less than 180º is called an obtuse angle.
A line has length but no width and no thickness.
An angle is the union of two non-collinear rays with a
common initial point. The common initial point is called
the ‘vertex’ of the angle and two rays are called the
‘arms’ of the angles.

90º < AOB < 180º.
(iv) Straight angle : An angle whose measure is 180º
is called a straight angle.
(v) Reflex angle : An angle whose measure is more
than 180º is called a reflex angle.
REMARK :
Every angle has a measure and unit of measurement
is degree.
One right angle = 90º
1º = 60’ (minutes)
1’ = 60” (Seconds)
Angle addition axiom : If X is a point in the interior of
BAC, then m BAC = m BAX + m XAC.
(a) Types of Angles :
(i) Right angle : An angle whose measure is 90º is
180º < AOB < 360º.
(vi) Complementary angles : Two angles, the sum of
whose measures is 90º are called complementary
angles.
AOC & BOC are complementary as their sum is
90 º.
called a right angle.
(vii) Supplementary angles : Two angles, the sum of
whose measures is 180 º , are called the
supplementary angles.
(ii) Acute angle : An angle whose measure is less
AOC & BOC are supplementary as their sum is
180 º.
than 90 is called an acute angle.
º
(viii) Angle Bisectors : A ray OX is said to be the bisector
of AOB , if X is a point in the interior of AOB, and
AOX = BOX.
B
O
0
0 < BOA < 90
A
0
PAGE # 3434
(ix) Adjacent angles : Two angles are called adjacent
angles, if
(A) they have the same vertex,
(B) they have a common arm,
(C) non common arms are on either side of the
common arm.
AOX and BOX are adjacent angles, OX is common
arm, OA and OB are non common arms and lies on
either side of OX.
(x) Linear pair of angles : Two adjacent angles are
said to form a linear pair of angles, if their non common
arms are two opposite rays.
AOC + BOC = 180º.
(xi) Vertically opposite angles : Two angles are called
a pair of vertically opposite angles, if their arms form
two pairs of opposite rays.
 AOC &  BOD form a pair of vertically opposite
(i) Corresponding angles : Two angles on the same
side of a transversal are known as the corresponding
angles if both lie either above the two lines or below
the two lines, in figure 1 & 5, 4 & 8, 2 & 6,
3 & 7 are the pairs of corresponding angles.
If a transversal intersects two parallel lines then the
corresponding angles are equal i.e.  1 =  5,
4 = 8, 2 = 6 and 3 = 7.
(ii) Alternate interior angles : 3 & 5, 2 & 8, are
the pairs of alternate interior angles.
If a transversal intersects two parallel lines then the
each pair of alternate interior angles are equal i.e.
3 = 5 and 2 = 8.
(iii) Co- interior angles : The pair of interior angles on
the same side of the transversal are called pairs of
consecutive or co - interior angles. In figure
2 &5, 3 & 8, are the pairs of co-interior angles.
If a transversal intersects two parallel lines then each
pair of consecutive interior angles are supplementary
i.e. 2 + 5 = 180º and 3 + 8 = 180º.
Ex.1. Find the measure of an angle, if six times its
complement is 12º less than twice its supplement.
Sol. Let the measure of the required angle be xº.
Then, measure of its complement = (90 – x)º.
Measure of its supplement = (180 – x)º.
 6 (90 – x) = 2 (180 – x) – 12
 540 – 6x = 360 – 2x – 12
 4x = 192
 x = 48º.
Hence, the required angle is 48º.
Ex.2. In the adjoining figure, ABC = 100º, EDC = 120º
and AB || DE. Then, find BCD .
angles. Also AOD & BOC form a pair of vertically
opposite angles.
D
If two lines intersect, then the vertically opposite angles
are equal i.e. AOC = BOD and BOC = AOD.
A
xº
(b) Angles Made by a Transversal with two
Parallel Lines :
C
Transversal : A line which intersects two or more given
parallel lines at distinct points is called a transversal
of the given lines.
E
B
D
Sol. A
B
E
F
xº
C
Produce AB to meet CD at F.
BFD = EDF = 120°
[alternate interior s]
 BFC = (180° – 120°) = 60°
 CBF = (180° – 100°) = 80°
 BCF = 180°– (60° + 80°) = 40°.
PAGE # 3535
POLYGON
A closed plane figure bounded by line segments is
called a polygon.
A polygon is named according to the number of sides
it has :
No. of sides
Figure
3
4
5
6
7
8
Ex.3 If the sum of interior angles of a polygon is 1620º, find
its number of sides.
Sol. We know that sum of all interior angles of a n sided
polygon = (2n – 4) right angles
= [(2n – 4) 900]
but given sum of interior angles = 1620º
 (2n – 4) 900 = 1620º
10
 2n – 4 =
Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Decagon
 2n – 4 = 18
 2n = 22
 n = 11.
In general, a polygon having n sides is called 'n' sided
polygon.

Diagonal of Polygon :
Line segment joining any two non-consecutive vertices
of a polygon is called its diagonal.

TRIANGLE
A plane figure bounded by three lines in a plane is
called a triangle. Every triangle have three sides and
three angles. If ABC is any triangle then AB, BC & CA
are three sides and A,B and C are three angles.
Convex Polygon :
If all the interior angles of a polygon are less than 1800,
it is called a convex polygon.

1620 º
90 º
Concave Polygon :
If one or more of the interior angles of a polygon is
greater than 1800 i.e. reflex, it is called a concave
polygon.

Types of triangles :
A.
On the basis of sides we have three types of triangle.
1. Scalene triangle – A triangle in which no two sides
are equal is called a scalene triangle.
2. Isosceles triangle – A triangle having two sides equal
is called an isosceles triangle.

3. Equilateral triangle – A triangle in which all sides
are equal is called an equilateral triangle.
Regular Polygon :
A polygon is called a regular polygon if all its sides
have equal length and all the angles have equal
measure.
B.
On the basis of angles we have three types :
1. Right triangle – A triangle in which any one angle is
right angle is called right triangle.
2. Acute triangle – A triangle in which all angles are
acute is called an acute triangle.
3. Obtuse triangle – A triangle in which any one angle
is obtuse is called an obtuse triangle.


REMARKS :
(1) The sum of the interior angles of a convex polygon of n
sides is (2n – 4) right angles or (2n – 4) 90º.
(2) The sum of the exterior angles of a convex polygon is
4 right angles or 360º.
(3) Each interior angle of a n-sided regular polygon is
2n  4 90º
n
(4) Each exterior angle of a regular polygon of n sides
 3600 


=  n 


(5) If a polygon has n sides, then the number of diagonals
nn  3 
of the polygon =
.
2
SOME IMPORTANT THEOREMS :
Theorem : The sum of interior angles of a triangle is
180º.
Theorem : If the bisectors of angles ABC and ACB
of a triangle ABC meet at a point O, then
1
BOC = 90º + A.
2
Exterior Angle of a Triangle :
If the side of the triangle is produced, the exterior angle
so formed is equal to the sum of two interior opposite
angles.
Corollary : An exterior angle of a triangle is greater
than either of the interior opposite angles.
Theorem : The sides AB and AC of a ABC are produced
to P and Q respectively. If the bisectors of PBC and
1
QCB intersect at O, then BOC = 90º – A.
2
PAGE # 3636
Ex.4 In figure, TQ and TR are the bisectors of Q and R
respectively. If QPR = 80º and PRT = 30º, determine
TQR and QTR.
Sol. Since the bisectors of Q and R meet at T.
(b) Sufficient Conditions for Congruence of two
Triangles :
(i) SAS Congruence Criterion :
A
P
P
80º
T
B
Q
C
R
Two triangles are congruent if two sides and the
included angle of one are equal to the corresponding
sides and the included angle of the other triangle.
30
º
Q
R
1
 QTR = 90º + QPR
2
1
 QTR = 90º + (80º)
2
 QTR = 90º + 40º = 130º
In QTR, we have
TQR + QTR + TRQ = 180º
 TQR + 130º + 30º = 180º [ TRQ = PRT = 30º ]
 TQR = 20º
Thus, TQR = 20º and QTR = 130º.
(ii) ASA Congruence Criterion :
A
P
B
Q
C
R
Two triangles are congruent if two angles and the
included side of one triangle are equal to the
corresponding two angles and the included side of
the other triangle.
The figures are called congruent if they have same
shape and same size. In other words, two figures are
called congruent if they are having equal length, width
and height.
Fig.(i)
(iii) AAS Congruence Criterion :
A
P
B
Q
C
R
If any two angles and a non included side of one triangle
are equal to the corresponding angles and side of
another triangle, then the two triangles are congruent.
Fig.(ii)
(iv) SSS Congruence Criterion :
In the above figures {fig.(i) and fig.(ii)} both are equal in
length, width and height, so these are congruent
figures.
A
P
(a) Congruent Triangles :
Two triangles are congruent if and only if one of them
can be made to superimposed on the other, so as to
cover it exactly.
B
C
Q
R
Two triangles are congruent if the three sides of one
triangle are equal to the corresponding three sides of
the other triangle.
( v ) RHS Congruence Criterion :
A
If two triangles ABC and DEF are congruent then
A = D, B = E, C = F and AB = DE, BC = EF,
AC = DF.
If two ABC & DEF are congruent then we write
B
Q
C
R
Two right triangles are congruent if the hypotenuse
and one side of one triangle are respectively equal to
the hypotenuse and one side of the other triangle.
ABC  DEF, we can not write as ABC  DFE or
ABC  EDF.
Hence, we can say that “two triangles are congruent if
and only if there exists a one-one correspondence
between their vertices such that the corresponding
sides and the corresponding angles of the two
triangles are equal.
P

NOTE :
If two triangles are congruent then their corresponding
sides and angles are also congruent by CPCT
(corresponding parts of congruent triangles are also
congruent).
Theorem : If the bisector of the vertical angle bisects
the base of the triangle, then the triangle is isosceles.
PAGE # 3737
Ex.5 In figure, line m is the bisector of an angle A and O is
any point on m. OQ and OP are perpendiculars from O
to the arms of A. Prove that :
In triangle ABC, AB = c, BC = a & CA = b then,
A
P
m
O
c
b
B
A
Q
(i) AOQ AOP
(ii) OQ = OP.
Sol. (i) In AOQ and AOP,
OAQ = OAP [ line m is the bisector of A]
AQO = APO = 90º
[Given]
and
AO = AO
[Common]
By AAS congruency
 AOQ  AOP
(ii) OQ = OP
[C.P.C.T.]
Ex.6 If D is the mid-point of the hypotenuse AC of a right
1
triangle ABC, prove that BD = AC.
2
Sol. Given : ABC is a right triangle such that B = 900 and
D is mid point of AC.
1
To prove : BD = AC.
2
Construction : Produce BD to E such that BD = DE and
join EC.
Proof :
In ADB and CDE
AD = DC
[Given]
BD = DE
[By construction]
And, ADB = CDE [Vertically opposite angles]
 By SAS criterion of congruence we have
ADB  CDE
 EC = AB and CED = ABD ... (i) [By CPCT]]
But CED & ABD are alternate interior angles
 CE || AB
 ABC + ECB = 1800
[Consecutive interior angles]
 90 + ECB = 1800
 ECB = 900.
Now, In ABC & ECB we have
AB = EC
[By (i)]
BC = BC
[Common]
And, ABC = ECB = 900
 By SAS criterion of congruence
ABC  ECB
[By CPCT]
 AC = EB
1
1
AC = EB
2
2
1
 BD = AC.
2

Hence Proved.
a
C
(i) The sum of any two sides of a triangle is greater
than its third side. i.e. in ABC,
(A) a + b > c
(B) b + c > a
(C) a + c > b
(ii) If two sides of a triangle are unequal, then the longer
side has greater angle opposite to it
i.e. in ABC, if AB > AC then C > B.
(iii) Of all the line segments that can be drawn to a
given line, from a point, not lying on it, the perpendicular
line segment is the shortest. i.e. in PMN,
P
M
N
(A) PM < PN.
(iv) The difference between any two sides of a triangle
is less than its third side. i.e. in ABC,
(A) a – b < c
(B) b – c < a
(C) a – c < b
(v) In a right angle triangle the sum of squares of two
smaller sides is equal to the square of its third side.
i.e. in ABC, a2 + b2 = c2.
(vi) If sum of squares of two smaller sides is greater
than the square of its third side then that triangle is
acute angled triangle.
i.e in ABC, a2 + b2 > c2.
(vii) If sum of squares of two smaller sides is lesser
than the square of its third side then that triangle is
obtuse angled triangle.
i.e in ABC, a2 + b2 < c2.
Ex.7 From which given triplet we can make the sides a
triangle.
(i) {15, 7, 8}
(ii) {3.5, 4.5, 5.5}
Sol. As we know sum of two sides is always greater then
third side.
(i) we can not make the triangle because
here the sum of two side is equal to third side
i.e 7+8=15
(ii) We can make the triangle because sum of two
sides is always greater then third side
3.5 +4.5 >5.5
3.5 +5.5 >4.5
4.5 +5.5 >3.5
PAGE # 3838
Ex.8 Prove that any two sides of the triangle are together
greater than twice the median drawn to the third side.
Sol. Given :ABC and AD is the median.
To prove : AB + AC > 2AD
A
(ii) (SSS Similarity) If the corresponding sides of two
triangles are proportional, then they are similar.
(iii) (SAS Similarity) If in two triangles, one pair of
corresponding sides are proportional and the included
angles are equal then the two triangles are similar.
D
B
(b) Results Based Upon Characteristic Properties
C
of Similar Triangles :
(i) If two triangles are equiangular, then the ratio of the
corresponding sides is the same as the ratio of the
corresponding medians.
E
Construction : Produce AD to E such that AD = DE.
Join EC.
(ii) If two triangles are equiangular, then the ratio of the
Proof : In ADB and CDE
corresponding sides is same as the ratio of the
AD = DE
[By construction]
BD = DC
[AD is the median]
 ADB =  CDE
ADB  CDE
[Vertically opposite angles]
[By SAS congruency]
corresponding angle bisector segments.
(iii) If two triangles are equiangular then the ratio of the
corresponding sides is same as the ratio of the
So, by CPCT
corresponding altitudes.
AB = EC
(iv) If one angle of a triangle is equal to one angle of
In AEC
another triangle and the bisectors of these equal
AC + EC > 2AD.
angles divide the opposite side in the same ratio, then
[Sum of two sides of a triangle is always greater than
the third side]
the triangles are similar.
So,
(v) If two sides and a median bisecting the third side of
AC + AB > 2AD.
[As EC = AB].
a triangle are respectively proportional to the
corresponding sides and the median of another
Two triangles ABC and DEF are said to be similar if
their
(i) Corresponding angles are equal.
triangle, then two triangles are similar.
(vi) If two sides and a median bisecting one of these
sides of a triangle are respectively proportional to the
i.e. A = D, B = E, C = F
And,
two sides and the corresponding median of another
triangle, then the triangles are similar.
Statement : If a line is drawn parallel to one side of a
triangle to intersect the other two sides in distinct points,
then the other two sides are divided in the same ratio.
i.e., In ABC in which a line parallel to side BC intersects
(ii) Corresponding sides are proportional.
other two sides AB and AC at D and E respectively.
AB
BC
AC
i.e.
=
=
.
DE
EF
DF
Then,
AE
AD
=
.
EC
DB
A
(a) Characteristic Properties of Similar Triangles :
(i) (AAA Similarity) If two triangles are equiangular,
then they are similar.
D
B
E
C
PAGE # 3939
Ex.9 In a trapezium ABCD, AB || DC and DC = 2AB. EF
drawn parallel to AB cuts AD in F and BC in E such that
Corollary :
If in a ABC, a line DE || BC, intersects AB in D and AC
in E, then
(i)
DB EC

AD AE
BE 3
 . Diagonal DB intersects EF at G. Prove that
EC 4
7FE = 10AB.
Sol. In DFG and DAB,
1 = 2
AB AC

(ii)
AD AE
A
B
2
F
E
D
AB AC

DB EC

(v)
G
1
AD AE

(iii)
AB AC
(iv)
[Corresponding s  AB ||FG]
DB EC

AB AC

Converse of Basic Proportionality Theorem :
Statement : If a line divides any two sides of a triangle

in the same ratio, then the line must be parallel to

the third side.

Some Important Results and Theorems :
(i) The internal bisector of an angle of a triangle divides
the opposite side internally in the ratio of the sides
containing the angle.


(ii) In a triangle ABC, if D is a point on BC such that
C
FDG = ADB [Common]
DFG ~ DAB [By AA rule of similarity]
DF FG

DA AB
Again in trapezium ABCD
EF ||AB ||DC
AF BE

DF EC
… (i)
 BE 3

 EC  4 ( given )


AF 3

DF 4
AF
3
1 1
DF
4
AF  DF 7

DF
4
AD 7

DF 4
DF 4

AD 7
D divides BC in the ratio AB : AC, then AD is the bisector

of A.
From (i) and (ii), we get
FG 4
4

i.e., FG = AB … (iii)
AB 7
7
(iii) The external bisector of an angle of a triangle divides
the opposite sides externally in the ratio of the sides
containing the angle.
(iv) The line drawn from the mid-point of one side of a
triangle parallel to another side bisects the third side.
(v) The line joining the mid-points of two sides of a
In BEG and BCD, we have
BEG = BCD [Corresponding angle  EG || CD]
GBE = DBC
[Common]
 BEG ~ BCD
[By AA rule of similarity]
BE EG


BC CD
triangle is parallel to the third side.
(vi) The diagonals of a trapezium divide each other
proportionally.
(vii) If the diagonals of a quadrilateral divide each other
proportionally, then it is a trapezium.
(viii) Any line parallel to the parallel sides of a
trapezium
divides
the
non-parallel
s ides
proportionally.

3 EG

7 CD
EC 4
EC  BE 4  3
BC 7 
 BE 3



 EG  7 i.e., BE  3  BE
3
BE 3 

3
3
 EG = CD  (2 AB)  CD  2AB (given)
7
7
6
 EG = AB
... (iv)
7
Adding (iii) and (iv), we get
FG + EG =
(ix) If three or more parallel lines are intersected by
two transversals, then the intercepts made by them on
the transversals are proportional.
… (ii)

EF =
4
6
10
AB  AB 
AB
7
7
7
10
AB i.e., 7EF = 10AB.
7
Hence proved.
PAGE # 4040
Ex.10 In the given figure, PA, QB and RC are each
perpendicular to AC. Prove that
1 1 1
  .
x z y
Properties of Areas of Similar Triangles :
(i) The areas of two similar triangles are in the ratio of
the squares of corresponding altitudes.
Sol. In PAC, we have BQ || AP
(ii) The areas of two similar triangles are in the ratio of

BQ CB

AP CA
[ CBQ ~ CAP]

y CB

x CA
… (i)
the squares of the corresponding medians.
(iii) The area of two similar triangles are in the ratio of
the squares of the corresponding angle bisector
segments.
In ACR, we have BQ || CR

BQ AB

CR AC
Ex.11 Prove that the area of the equilateral triangle described
on the side of a square is half the area of the equilateral
triangle described on its diagonal.
Sol. Given : A square ABCD. Equilateral triangles BCE and
ACF have been described on side BC and diagonal
AC respectively.
 ABQ ~ ACR 
y AB

… (ii)
z AC
Adding (i) and (ii), we get

1
. Area (ACF)
2
Proof : Since BCE and ACF are equilateral.
Therefore, they are equiangular ( each angle being
equal to 60º) and hence BCE ~ ACF.
P
To Prove : Area (BCE) =
R
Q
x
z
y
A
C
B
y y CB AB
 

x z AC AC

y y AB  BC
 
x z
AC

y y AC
 
x z AC

y y
 1
x z

1 1 1
  .
x z y

Area( BCE)
BC 2
=
Area( ACF)
AC 2

Area( BCE)
=
Area( ACF)

Area( BCE)
1
= .
Area( ACF)
2
Hence Proved.
Statement :The ratio of the areas of two similar
triangles is equal to the square of the ratio of their
BC 2
2
 2BC
=
1
2
Hence Proved.
corresponding sides. i.e., In 2 triangles ABC and PQR
PYTHAGORA S THEOREM
such that  ABC ~  PQR [Shown in the figure]
Statement : In a right triangle, the square of the
hypotenuse is equal to the sum of the squares of the
P
A
other two sides. i.e., In a right triangle ABC, right angled
at B.
B
M
C Q
2
Then,
R
N
2
2
ar( ABC)  AB 
 BC 
 CA 
 =
 =
 .
=
ar(PQR)  PQ 
 QR 
 RP 
Then, AC2 = AB2 + BC2
PAGE # 4141
Converse of Pythagoras Theorem :
,
Statement : In a triangle, if the square of one side is
equal to the sum of the squares of the other two sides,
Ex.12 In an equilateral triangle ABC, the side BC is trisected
at D. Prove that 9 AD2 = 7AB2.
Sol. ABC be an equilateral triangle and D be point on BC
then the angle opposite to the first side is a right angle.
such that BD =
Some Results Deduced From Pythagoras Theorem :
(i) In the given figure ABC is an obtuse triangle,
obtuse angled at B. If AD  CB,
then AC2 = AB2 + BC2 + 2BC. BD
1
BC (Given)
3
Draw AE  BC, Join AD.
BE = EC ( Altitude drown from any vertex of an equilateral
triangle bisects the opposite side)
A
D
A
B
C
(ii) In the given figure, if B of ABC is an acute angle
and AD  BC, then AC2 = AB2 + BC2 – 2BC . BD
B
So, BE = EC =
D
E
BC
2
In  ABC
AB2 = AE2 + EB2
AD2 = AE2 + ED2
From (i) and (ii)
AB2 = AD2 – ED2 + EB2
(iii) In any triangle, the sum of the squares of any two
sides is equal to twice the square of half of the third
side together with twice the square of the median which
bisects the third side.
(iv) Three times the sum of the squares of the sides of
a triangle is equal to four times the sum of the
squares of the medians of the triangle.
C
......(i)
.....(ii)
BC2
BC2
+
36
4
BC
BC
BC
[  BD + DE =

+ DE =
 DE =
3
2
2
BC 2
BC2
 AB2 +
–
= AD2
[ EB =
36
4




AB2 = AD2 –
AB2 +
AB 2
AB2
–
= AD2
36
4
BC
]
6
BC
]
2
[ AB = BC]
36 AB 2  AB 2  9 AB 2
= AD2
36
28 AB 2
= AD2

7AB2 = 9AD2.
36

PAGE # 4242
QUADRILATERALS
QUADRILATERAL
A quadrilateral is a four sided closed figure.
D
(ii) Rectangle : A rectangle is a parallelogram, in which
each of its angle is a right angle. If ABCD is a rectangle
then A = B = C = D = 90°, AB = CD, BC = AD and
diagonals AC = BD.
C
D
A
C
900
A
B
(iii) Rhombus : A rhombus is a parallelogram in which
all its sides are equal in length. If ABCD is a rhombus
then, AB = BC = CD = DA.
B
Let A, B, C and D be four points in a plane such that :
(i) No three of them are collinear.
(ii) The line segments AB, BC, CD and DA do not
intersect except at their end points, then figure
obtained by joining A, B, C & D is called a quadrilateral.
Convex and Concave Quadrilaterals :
(i) A quadrilateral in which the measure of each interior
angle is less than 180° is called a convex
quadrilateral. In figure, PQRS is convex quadrilateral.
R
The diagonals of a rhombus are perpendicular to each
other.
(iv) Square : A square is a parallelogram having all
sides equal and each angle equal to right angle. If ABCD
is a square then AB = BC = CD = DA, diagonal AC = BD
and A = B = C = D = 90°.
S
Q
P
(ii) A quadrilateral in which the measure of one of the
interior angles is more than 180° is called a concave
quadrilateral. In figure, ABCD is concave quadrilateral.
(v) Trapezium : A trapezium is a quadrilateral with only
one pair of opposite sides parallel. In figure, ABCD is
a trapezium with AB || DC.
B
D
The diagonals of a square are perpendicular to each
other.
C
D
C
A
Special Quadrilaterals :
(i) Parallelogram : A parallelogram is a quadrilateral
in which both pairs of opposite sides are parallel. In
figure, AB || DC, AD || BC therefore, ABCD is a
parallelogram.
D
A
B
(vi) Kite : A kite is a quadrilateral in which two pairs of
adjacent sides are equal. If ABCD is a kite then AB = AD
C
and BC = CD.
C
B
A
B
Properties :
(a) A diagonal of a parallelogram divides it into two
congruent triangles.
D
A
(vii) Isosceles trapezium : A trapezium is said to be an
isosceles trapezium, if its non-parallel sides are equal.
Thus a quadrilateral ABCD is an isosceles trapezium,
if AB || DC and AD = BC.
(b) In a parallelogram, opposite sides are equal.
(c) The opposite angles of a parallelogram are equal.
(d) The diagonals of a parallelogram bisect each other.
In isosceles trapezium A = B and C =D.
PAGE # 4343
Ex.2. In the figure below, P and Q are the mid-points of the
PROPERTIES
sides AB and BC of the rectangle ABCD
Theorem 1 : The sum of the four angles of a
quadrilateral is 360°.
D
C
Theorem 2 : A diagonal of a parallelogram divides it
into two congruent triangles.
Q
Theorem 3 : In a parallelogram, opposite sides are
equal.
A
Theorem 6 : Each of the four angles of a rectangle is a
right angle.
Theorem 7 : Each of the four sides of a rhombus is of
the same length.
Theorem 8 : Each of the angles of a square is a right
angle and each of the four sides is of the same length.
Theorem 9 : The diagonals of a rectangle are of equal
length.
what is the area of the whole rectangle ?
Sol.
D
C
Q
A
P
B
In QAB, QP is median
So, ar QAB = 2 area APQ
= 2 × 1 = 2 cm2
In ABC, AQ is median
Area ABC = 2 area ABQ
Theorem 10 : The diagonals of a rhombus are perpendicular to each other.
Theorem 11 : The diagonals of a square are equal
and perpendicular to each other.
Theorem 12 : Parallelogram and Triangles on the
same base (or equal bases) and between the same
parallels, then area of parallelogram is twice the area
of triangle.
= 2 × 2 = 4 cm2
Area of rectangle ABCD = 2 ar. ABC
= 2 × 4 = 8 cm2.
Ex.3 In the adjoining figure, ABCD is a parallelogram and
the bisector of A bisect BC at X. Prove that AD = 2AB.
A quadrilateral become a parallelogram when :
(i) Opposite angles are equal.
(ii) Both the pair of opposite sides are equal
(iii) A pair of opposite side is equal as well as parallel
(iv) Diagonals of quadrilateral bisect each other.

REMARK :
(i) Square, rectangle and rhombus are all
parallelograms.
(ii) Kite and trapezium are not parallelograms.
(iii) A square is a rectangle.
(iv) A square is a rhombus.
(v) A parallelogram is a trapezium.
Ex.1 In the parallelogram, Find the value of “x”.
80º
xº
D
C
X
2
A
1
B
Sol. ABCD is a parallelogram.

AD || BC and AX cuts them.

BXA = DAX =
1
A [Alternate interior angles]
2
1
A.
2
1
Also, 1 = A
2

2 =

2 = 1
150º
Sol. A = C = 80º
(Opposite angle of a parallelogram)
 FEB = x +  A
(Exterior angle of a triangle is equal to interior
opposite angle)
D
C
 150° = x + 80
80
 x = 70°
F
x
150
A
B
E
B
If the area of the triangle APQ is 1 square centimeter,
Theorem 4 : The opposite angles of a parallelogram
are equal.
Theorem 5 : The diagonals of a parallelogram bisect
each other.
P
 AB = BX
 AB = 1 BC
2
1
 AB = AD
2
 AD = 2 AB.
PAGE # 4444
In  ADC, SR || AC and SR =
In a triangle, the line segment joining the mid-points
of any two sides is parallel to the third side and is
half of it.
i.e., In a triangle ABC in which P is the mid-point of side
AB and Q is the mid-point of side AC.
A
–
P
1
AC
... (ii)
2
[By mid-point theorem]
 PQ = SR and PQ || SR
[From (i) and (ii)]
 PQRS is a parallelogram.
Now, PQRS will be a rectangle if any angle of the
parallelogram PQRS is 90º.
Q
–
C
B
then, PQ is parallel to BC and is half of it i.e., PQ || BC
1
and PQ =
BC.
2
PQ || AC
[By mid-point theorem]
QR || BD
[By mid-point theorem]
But, AC  BD
[Diagonals of a rhombus are perpendicular to each other]
 PQ  QR
Converse of the Mid-Point Theorem :
The line drawn through the mid-point of one side of a
triangle parallel to the another side; bisects the third
side.
i.e., A triangle ABC in which P is the mid-point of side
AB and PQ is parallel to BC.
A
P
[Angle between two lines = angle between their parallels]
 PQRS is a rectangle.
Hence Proved.
Ex.5 In the given figure, E and F are respectively, the
mid-points of non-parallel sides of a trapezium ABCD.
Prove that :
(i) EF || AB
Q
B
(ii) EF =
1
(AB + DC).
2
C
then, PQ bisects the third side AC i.e., AQ = QC.

REMARK :
In quadrilateral ABCD, if side AD is parallel to side BC;
ABCD is a trapezium.
Sol. Join BE and produce it to intersect CD produced at
point P. In  AEB and  DEP, AB || PC and BP is
transversal.
Now, P and Q are the mid-points of the non-parallel
1
sides of the trapezium; then PQ =
(AD + BC).
2
i.e. The length of the line segment joining the
mid-points of the two non-parallel sides of a trapezium
is always equal to half of the sum of the lengths of its
two parallel sides.
Ex.4 ABCD is a rhombus and P, Q, R and S are the
mid-points of the sides AB, BC, CD and DA respectively.
Prove that the quadrilateral PQRS is a rectangle.
Sol.
ABE = DPE
[Alternate interior angles]
AEB = DEP
[Vertically opposite angles]
And AE = DE
[E is mid-point of AD]
So, AEB  DEP
[By ASA congruency]
BE = PE
[By CPCT]
And AB = DP
[By CPCT]
Since, the line joining the mid-points of any two sides
of a triangle is parallel and half of the third side,
Therefore, in  BPC,
E is mid-point of BP
[As, BE = PE]
and F is mid-point of BC
[Given]
1
PC
2
 EF || DC and EF = 1 (PD + DC)
2
1
 EF || AB and EF = (AB + DC)
2
[As, DC || AB and PD = AB]
 EF || PC and EF =
In  ABC, PQ || AC and PQ =
1
AC
... (i)
2
[By mid-point theorem]
Hence Proved.
PAGE # 4545
Ex.6 AD and BE are medians of ABC and BE || DF. Prove
that CF =
A
1
AC.
4
then, ar(||gm ABCD) = ar(||gm ABEF)
Theorem : The area of parallelogram is the product
of its base and the corresponding altitude.
i.e., In a ||gm ABCD in which AB is the base and AL is the
corresponding height.
E
F
Sol.
B
D
C
In BEC, DF is a line through the mid - point D of BC
and parallel to BE intersecting CE at F. Therefore, F is
the midpoint of CE. Because the line drawn through
the mid point of one side of a triangle and parallel to
another sides bisects the third side.
Now, F is the mid point of CE
1
CE
2
1 1
 CF = ( AC)
2 2
 CF = 1 AC.
4
 CF =
Ex.7 ABCD is a trapezium with AB || DC. A line parallel to AC
intersects AB at X and BC at Y. Prove that
ar(ADX) = ar(ACY).
Sol.
D
C
Y
A
(a) Base and Altitude of a Parallelogram :
(i) Base : Any side of a parallelogram can be called its
base.
(ii) Altitude : The length of the line segment which is
perpendicular to the base from the opposite side is
called the altitude or height of the parallelogram
corresponding to the given base.
D
B
X
Join CX, DX and AY.
Clearly, triangles ADX and ACX are on the same base
AX and between the parallels AB and DC.
 ar (ADX) = ar (ACX)
... (i)
Also,  ACX and  ACY are on the same base AC and
between the parallels AC and XY.
 ar (ACX) = ar (ACY)
...(ii)
From (i) and (ii), we get
ar (ADX) = ar (ACY).
C
M
A
then, area (||gm ABCD) = AB × AL.
B
L
gm
(i) DL is the altitude of ||
ABCD, corresponding to the
base AB.
(ii) DM is the altitude of ||gm ABCD, corresponding to the
Theorem : Two triangles on the same base (or equal
bases) and between the same parallels are equal in
area.
i.e., If two triangles ABC and PBC are on the same base
BC and between the same parallel lines BC and AP.
then, ar(  ABC) = ar(  PBC).
base BC.
Theorem : A diagonal of a parallelogram divides it
into two triangles of equal area.
between the same parallels are equal in area.
i.e., If two ||gms ABCD and ABEF are on the same base
Theorem : The area of a trapezium is half the product
of its height and the sum of the parallel sides.
i.e., In a trapezium ABCD in which AB || DC, AL  DC,
AB and between the same parallels AB and FC.
CN  AB and AL = CN = h (say), AB = a, DC = b.
Theorem : Parallelograms on the same base and
b
L
D
C
h
A
h
a
N
B
PAGE # 4646
then, ar(trapezium ABCD) =
1
h × (a + b).
2
Theorem : Median of a triangle divides it into two
triangles of equal area.
i.e., In a ABC in which AD is the median.
A
B
D
then, ar (ABD) = ar(ADC) =
C
1
ar(ABC)
2
Ex.8 ABC is a triangle in which D is the mid-point of BC and
E is the mid-point of AD. Prove that the area of  BED
1
area of  ABC.
4
Sol. Given : A  ABC in which D is the mid-point of BC and
=
E is the mid-point of AD.
To prove : ar(  BED) =
1
ar(  ABC).
4
Proof :  AD is a median of  ABC.
1
ar (  ABC)
... (i)
2
[ Median of a triangle divides it into two triangles of
equal area]
Again,
 ar (  ABD) = ar(  ADC) =
 BE is a median of  ABD.
1
ar (  ABD)
2
[ Median of a triangle divides it into two triangles of
equal area]
1
And ar(  BED) = ar(  ABD)
2
 ar (  BEA) = ar(  BED) =
=
 ar (  BED) =
1 1
× ar (  ABC)
2 2
1
ar(  ABC).
4
[From (i)]
Hence Proved.
PAGE # 4747
CIRCLES

Circumference : The length of the complete circle
is called its circumference.

Circle : The collection of all the points in a plane,
which are at a fixed distance from a fixed point in the
plane, is called a circle.
The fixed point is called the centre of the circle and the
fixed distance is called the radius of the circle.
Segment : The region between a chord and either
of its arcs is called a segment of the circular region or
simply a segment of the circle. There are two types of
segments which are the major segment and the minor
segment (as in figure).
O
Major segment
P
P
In figure, O is the centre and the length OP is the radius
of the circle. So the line segment joining the centre
and any point on the circle is called a radius of the
circle.

Chord : If we take two points P and Q on a circle, then
the line segment PQ is called a chord of the circle.
Sector : The region between an arc and the two radii,
joining the centre to the end points of an arc is called a
sector. Minor arc corresponds to the minor sector and
the major arc corresponds to the major sector. When
two arcs are equal, then both segments and both
Q
P
Major sector
O
Diameter : The chord which passes through the
centre of the circle, is called the diameter of the circle.
P
Q
Q
Theorem : Equal chords of a circle subtend equal
angles at the centre.
O
A
Given : AB and CD are the two equal chords of a circle
with centre O.
A diameter is the longest chord and all diameters of
same circle have the same length, which is equal to
two times the radius. In figure, AOB is a diameter of
circle.
To Prove : AOB = COD.
A
arc PQ is also denoted by PQ and the major arc PQ by
QP . When P and Q are ends of a diameter, then both
arcs are equal and each is called a semi circle.
O
D
`
Arc : A piece of a circle between two points is called
an arc. The longer one is called the major arc PQ and
the shorter one is called the minor arc PQ. The minor
B C
Proof : In  AOB and  COD,
OA = OC
OB = OD
[Radii of a circle]
[Radii of a circle]
AB = CD
[Given]
 AOB  COD [By SSS congruency]
Hence Proved.
 AOB = COD. [By CPCT]
R
Major arc PQ
P
Semicircular
region
O
Semicircular
region
Minor
sector
P
B

Q
sectors become the same and each is known as a
semicircular region.
O

Minor segment
Converse :
Q
P
Minor arc PQ
Q
If the angles subtended by the chords of a circle at the
centre are equal, then the chords are equal.
PAGE # 4848
Theorem : The perpendicular from the centre of a
circle to a chord bisects the chord.
O
A
B
M
Ex.1The radius of a circle is 13 cm and the length of one of
its chords is 10 cm. Find the distance of the chord from
the centre.
Sol. Let O be the center of the circle of radius 13 cm and AB
is the chord of length 10 cm
OC AB
AB
10
=
=5
2
2
Given : A circle with centre O. AB is a chord of this circle.
AC =
OM  AB.
To Prove : MA = MB.
Construction : Join OA and OB.
Proof : In right triangles OMA and OMB,
OA = OB
[Radii of a circle]
OM = OM
[Common]
OMA = OMB
[90º each]
[Line perpendicular from centre to chord bisect the
chord]
In AOC
(OC)2 + (AC)2 = (AO)2
(OC)2 = (13)2 – (5)2
OC = 12 cm.
  OMA   OMB
 MA = MB
[By RHS]
[By cpctc]
Hence Proved.
Converse :
The line drawn through the centre of a circle to bisect
a chord is perpendicular to the chord.
Theorem : There is one and only one circle passing
through three given non-collinear points.
Proof : Take three points A, B and C, which are not in
the same line, or in other words, they are not collinear
[as in figure ]. Draw perpendicular bisectors of AB and
BC say, PQ and RS respectively. Let these
perpendicular bisectors intersect at one point O.(Note
that PQ and RS will intersect because they are not
parallel) [as in figure].
C
P
R
O
S
A
Q
B
 O lies on the perpendicular bisector PQ of AB.
 OA = OB
[ Every point on the perpendicular bisector of a line
segment is equidistant from its end points]
Similarly,
 O lies on the perpendicular bisector RS of BC.
 OB = OC
[ Every point on the perpendicular bisector of a line
segment is equidistant from its end points]
So, OA = OB = OC
i.e., the points A, B and C are at equal distances from
the point O.
So, if we draw a circle with centre O and radius OA it
will also pass through B and C. This shows that there
is a circle passing through the three points A, B and C.
We know that two lines (perpendicular bisectors) can
intersect at only one point, so we can draw only one
circle with radius OA. In other words, there is a unique
Hence Proved.
circle passing through A, B and C.
Ex.2 In figure,
and O is the centre of the circle.

Prove that OA is the perpendicular bisector of BC.
Sol. Given : In figure,
and O is the centre of the

circle.
To Prove : OA is the perpendicular bisector of BC.
Construction : Join OB and OC.
Proof :

[Given]

 chord AB = chord AC.
[  If two arcs of a circle are congruent, then their
corresponding chords are equal]
 AOB = AOC
...(i)
[ Equal chords of a circle subtend equal angles at
the centre]
In  OBD and  OCD,
DOB = DOC
[From (i)]
OB = OC
[Radii of the same circle]
OD = OD
[Common]
 OBD  OCD [By SAS congruency]
[By CPCT]
 ODB = ODC ...(ii)
And, BD = CD
...(iii)
[By CPCT]
But BDC = 180º
 ODB + ODC = 180º
 ODB + ODB = 180º
[From equation(ii)]
 2ODB = 180º

ODB = 90º
 ODB = ODC = 90º
...(iv)
[From (ii)]
So, by (iii) and (iv), OA is the perpendicular bisector of
BC.
Hence Proved.
PAGE # 4949
Theorem : Equal chords of a circle (or of congruent
circles) are equidistant from the centre (or centres).

Angle Subtended by an Arc of a Circle :
In figure, the angle subtended by the minor arc PQ at O
C
A
O
is POQ and the angle subtended by the major arc PQ
N
at O is reflex angle POQ.
M
D
B
O
Given : A circle have two equal chords AB & CD. i.e.
AB = CD and OM  AB, ON  CD.
To Prove : OM = ON
Construction : Join OB & OD
Proof : AB = CD (Given)
[ The perpendicular drawn from the centre of a circle
to a chord bisects the chord]
P
Q
Theorem : The angle subtended by an arc at the
centre is double the angle subtended by it at any
1
1
AB = CD
2
2
 BM = DN
point on the remaining part of the circle.
In  OMB &  OND
OMB = OND = 90º
OB = OD
BM = DN
of the circle.

Given : An arc PQ of a circle subtending angles POQ at
the centre O and PAQ at a point A on the remaining part
  OMB   OND
 OM = ON

REMARK :
To Prove : POQ = 2PAQ.
[Given]
[Radii of same circle]
[Proved above]
Construction : Join AO and extend it to a point B.
[By R.H.S. congruency]
[By CPCT]
Hence Proved.
REMARK :
Chords equidistant from the centre of a circle are
equal in length.
Ex.3 AB and CD are equal chords of a circle whose centre
is O. When produced, these chords meet at E. Prove
that EB = ED.
Sol. Given : AB and CD are equal chords of a circle whose
centre is O. When produced, these chords meet at E.
To Prove : EB = ED.
Construction : From O draw OP  AB and OQ  CD.
Join OE.
Proof :  AB = CD
[ Given]
 OP = OQ
[  Equal chords of a circle are equidistant from the
centre]
A
(A)
(A) arc PQ is minor
(B) arc PQ is a semi-circle
(C) arc PQ is major.
In all the cases,
BOQ = OAQ + AQO
the two interior opposite angles]
In OAQ,
OA = OQ
E

...(ii)
From (i) and (ii)
BOQ = 2OAQ
[By RHS congruency]
[By CPCT]
 EB = ED.
Hence Proved.
[  AB = CD (Given)]
...(iii)
Similarly,
[Each 90º]
[ Common]
[ Proved above]
 OPE   OQE
PE  QE
1
1
 PE – AB = QE – CD
2
2
 PE – PB = QE – QD


[Radii of a circle]
OQA = OAQ
[Angles opposite equal sides of a triangle are equal]
D
Q
Now in OPE and OQE,
OPE = OQE
OE = OE
OP = OQ
...(i)
[  An exterior angle of a triangle is equal to the sum of
B
C
(C)
Proof : There arises three cases:-
P
O
(B)
BOP = 2OAP
...(iv)
Adding (iii) and (iv), we get
BOP + BOQ = 2(OAP + OAQ)
 POQ = 2PAQ.

...(v)
NOTE :
For the case (C), where PQ is the major arc, (v) is
replaced by reflex angles.
Thus, reflex POQ = 2PAQ.
PAGE # 5050
Theorem : Angles in the same segment of a circle
are equal.
Proof : Let P and Q be any two points on a circle to form
a chord PQ, A and C any other points on the remaining
part of the circle and O be the centre of the circle. Then,
Ex.5. In the given figure, the chord ED is parallel to the
diameter AC. Find CED.
B
50º
A
O
P
1
C
2
3
E
POQ = 2PAQ
...(i)
And
POQ = 2PCQ
...(ii)
[Angle subtended at the centre is double than the angle
subtended by it on the remaining part of the circle]
From equation (i) & (ii)
2PAQ = 2PCQ
PAQ = PCQ.

D
Sol. CBE = 1 [Angles in the same segment]
 1 = 50º
…(i)
[  CBE = 50º]
 AEC = 90º
..(ii)
[Angle in a semicircle is a right angle]
Hence Proved.
Theorem : Angle in the semicircle is a right angle.
Now, in AEC,
1 + AEC + 2 = 180º
50º + 90º + 2 = 180º
Proof :  PAQ is an angle in the segment, which is a
semicircle.

1
1
 PAQ = POQ =
× 180º = 90º
2
2
[ POQ is straight line angle or POQ = 180º]
If we take any other point C on the semicircle, then
again we get
Also, ED || AC
 2=3
1
1
POQ =
× 180º = 90º.
2
2
Hence Proved.
Ex.4. A chord of a circle is equal to the radius of the circle,
find the angle subtended by the chord at a point on the
minor arc and also at a point on the major arc.
Sol.
 2 = 180º – 140º = 40º
Thus  2 = 40º
...(iii)
[Given]
[Alternate angles]
  3 = 40º
Hence,  CED = 40º.
PCQ =
A quadrilateral ABCD is called cyclic if all the four
vertices of it lie on a circle.
C
O
B
A
D
Let AB be a chord of the circle with centre at O such that
OA = OB = AB.
OAB is equilateral.
 AOB = 60º
1
 ACB = AOB = 30º
2
Consider arc ACB.
Clearly, it makes 360º – 60º = 300º at the centre O.
Theorem : The sum of either pair of opposite angles
of a cyclic quadrilateral is 180º.
Given : A cyclic quadrilateral ABCD.
D
C
A
B
PAGE # 5151
To Prove : A + C = B + D = 180º.
Construction : Join AC and BD.
Proof : ACB = ADB
...(i)
And BAC = BDC
...(ii)
[Angles of same segment of a circle are equal]
Adding equation (i) & (ii)
 ACB + BAC = ADB + BDC
ACB + BAC = ADC.
Adding ABC to both sides, we get
ACB + BAC + ABC = ADC + ABC.
 ADC + ABC = 180º
i.e., D + B = 180º
 A + C = 360º – (B + D) = 180º
[ A  B  C  D  360 º ]
Hence Proved.
Ex.6 If a side of a cyclic quadrilateral is produced, then the
exterior angle is equal to the interior opposite angle.
Sol. Let ABCD be a cyclic quadrilateral inscribed in a circle
with centre O. The side AB of quadrilateral ABCD is
produced to E. Then, we have to prove that
CBE = ADC.
Ex.8 If the nonparallel side of a trapezium are equal, prove
that it is cyclic.
Sol. Given : ABCD is a trapezium whose two non-parallel
sides AD and BC are equal.
To Prove : Trapezium ABCD is a cyclic.
Construction : Draw BE || AD.
Proof : 
AB || DE [Given]
and
AD || BE
[By construction]
 Quadrilateral ABED is a parallelogram.
[Opp. angles of a ||gm]
 BAD = BED ...(i)
And ,
AD = BE
...(ii)
[Opp. sides of a ||gm]
But
AD = BC
...(iii)
[Given]
From (ii) and (iii),
BE = BC
 BCE = BEC ...(iv)
[Angles opposite to equal sides]
BEC + BED = 180º
[Linear Pair Axiom]
 BCE + BAD = 180º
[From (iv) and (i)]
 Trapezium ABCD is cyclic.
[ If a pair of opposite angles of a quadrilateral is 180º,
then the quadrilateral is cyclic]
Hence Proved.
Ex.9 In figure, O is the centre of the circle. Prove that
x + y = z.
Since the sum of opposite pairs of angles of a cyclic
quadrilateral is 180°
ABC + ADC = 180°

But
ABC + CBE = 180°
[ ABC and CBE form a linear pair]
ABC + ADC = ABC + CBE

ADC = CBE

or
CBE = ADC.
Hence proved
1
1
EOF = z
2
2
[ Angle subtended by an arc of a circle at the centre is
Sol. EBF =
twice the angle subtended by it at any point of the
remaining part of the circle]
Ex.7 Find the value of a & b.
C
D
130º
A
b
1
z
...(i) [Linear Pair Axiom]
2
1
1
EDF = EOF = z
2
2
[ Angle subtended by any arc of a circle at the centre
 ABF = 180º –
a
O
B
Sol. In ADC
AD = DC
DAC = DCA = x (say)
130º + x + x = 180º
2x = 50º

x = 25º
ADC + ABC = 180º
[Opposite angles in a cyclic quadrilateral are supplementary]
130º + ABC = 180º
ABC = 50º
b = 90º
[Angle in a semicircle]
a = 180º – ABC – b
= 180º – 50º – 90º
a = 180º– 140º = 40º.
So, a = 40º and b = 90º.
is twice the angle subtended by it at any point of the
remaining part of the circle]
 ADE = 180º –
1
z
2
...(ii)
[Linear Pair Axiom]
BCD = ECF = y
[Vertically Opp. Angles]
BAD = x
In quadrilateral ABCD
ABC + BCD + CDA + BAD = 2 × 180º
 180º –
1
1
z + y + 180º – z + x = 2 × 180º
2
2
 x + y = z.
Hence Proved.
PAGE # 5252
Ex.10 AB is a diameter of the circle with centre O and chord
CD is equal to radius OC. AC and BD produced meet
at P. Prove that CPD = 60º.
Sol. Given : AB is a diameter of the circle with centre O and
chord CD is equal to radius OC. AC and BD produced
meet at P.
To Prove : CPD = 60º.
O
A
B
Construction : Join AD.
Proof : In  OCD,
C
OC = OD
...(i)
[Radii of the same circle]
OC = CD
...(ii) [Given]
From (i) and (ii),
OC = OD = CD
 OCD is an equilateral triangle.
 COD = 60º
D
1
1
COD = (60º) = 30º
2
2
[ Angle subtended by any arc of a circle at the centre
is twice the angle subtended by it at any point of the
remaining part of the circle]
 PAD = 30º
...(iii)
...(iv)
[Angle in a semi-circle]
 ADB + ADP = 180º
[Linear Pair Axiom]
 90º + ADP = 180º
[From (iv)]
 ADP = 90º
...(v)
In ADP,
APD + PAD + ADP = 180º
Lengths of two tangents drawn from an external point
to a circle are equal.
Given : AP and AQ are two tangents drawn from a point
A to a circle C (O, r).
To prove : AP = AQ
Construction : Join OP, OQ and OA.
Proof : In AOQ and APO
OQA = OPA [Tangent at any point of a circle is perp.
to radius through the point of contact]
AO = AO
[Common]
OQ = OP
[Radius]
So, by R.H.S. criterion of congruency AOQ  AOP
 AQ = AP
[By CPCT]
Hence Proved.
 RESULTS :
 APD + 30º + 90º = 180º [From (iii) and (v)]
 APD + 120º = 180º
 APD = 180º – 120º = 60º
 CPD = 60º.
THEOREM
P
 CAD =
And,ADB = 90º
Clearly OP = OR (Radius)
Now, OQ = OR + RQ
 OQ > OR
 OQ > OP
( OP = OR)
Thus, OP is shorter than any other segment joining O
to any point of AB.
Hence, OP  AB.
Hence Proved.
THEOREM
A tangent to a circle is perpendicular to the radius
(i) If two tangents are drawn to a circle from an external
point, then they subtend equal angles at the centre.
QOA = POA [By CPCT]
(ii) If two tangents are drawn to a circle from an external
point, they are equally inclined to the segment, joining
the centre to that point OAQ = OAP [By CPCT]
Ex.11 In figure, a circle touches all the four sides of a
quadrilateral ABCD with AB = 6 cm, BC = 7 cm and
CD = 4 cm. Find AD.
through the point of contact.
Given : A circle C (O, r) and a tangent AB at a point P.
To prove : OP  AB
Construction : Take any point Q, other than P on the
tangent AB. Join OQ. Suppose OQ meets the circle at R.
Proof : Among all line segments joining the point O to
a point on AB, the shortest one is perpendicular to AB.
So, to prove that OP  AB, it is sufficient to prove that OP
is shorter than any other segment joining O to any
point of AB.
Sol. PB = BQ, CR = CQ, DR = DS, AP= AS
[ Length of tangents drawn from external point are
equal]
Add
PB + CR + DR + AP = BQ + CQ + DS + AS
(PB + AP) + (CR + DR) = (BQ + CQ) + (DS + AS)
AB + CD = BC + AD
AD = AB + CD – BC
AD = 6 + 4 – 7
AD = 3 cm.
PAGE # 5353
Ex.12 Two tangents TP and TQ are drawn to a circle with
centre O from an external point T. Prove that
PTQ = 2OPQ.
Sol. We know that lengths of tangents drawn from an
external point to a circle are equal.
Proof :
Case 1. When AB and CD intersect at P inside the
circle.
In ’s APC and BPD,
APC = BPD
[Vertically opposite angles]

TPQ is an isosceles triangle.
PAC = PDB
[Angles in the same segment]
 APC ~ BPD

TPQ = TQP
Hence,
TP = TQ

PA PC
=
`
PD PB
PA × PB = PC × PD.
Case 2. When AB and CD when produced intersect at
P outside the circle.
In PBD and PAC, PBD = ACP
[ABCD is a cyclic quadrilateral and exterior angle is
In TPQ, we have

TPQ + TQP + PTQ = 180º

2TPQ = 180º – PTQ
1
 TPQ = 90º –
PTQ
2
1
PTQ = 90º – TPQ
2
Since,
equal to the interior opposite angle]
PDB = CAP
[ABCD is a cyclic quadrilateral and exterior angle is
equal to the interior opposite angle]
PBD ~ PCA
PB PD
Hence,
=
`
PC PA

PA × PB = PC × PD.

....(i)
OP  TP
OPT = 90º
THEOREM
OPQ + TPQ = 90º
OPQ = 90º – TPQ
...(ii)
From (i) and (ii), we get
1
PTQ = OPQ
2

PTQ = 2OPQ.
SEGMENTS OF A CHORD
Let AB be a chord of a circle, and let P be a point on AB
If PAB is a secant to a circle intersecting the circle at A
and B and PT is a tangent segment, then
PA × PB = PT2.
Given : A circle with centre O; PAB is a secant
intersecting the circle at A and B and PT is a tangent
segment to the circle.
To prove : PA × PB = PT2.
Construction : Join OA, OP and OT. Draw OD  AB.
T
inside the circle. Then, P is said to divide AB internally
into two segments PA and PB.
O
THEOREM
P
If two chords of a circle intersect inside or outside the
circle when produced, the rectangle formed by two
segments of one chord is equal in area to the rectangle
formed by the two segments of another chord.
D
A
P
B
C
Fig. (i)
Given : Two chords AB and CD of a circle C(O, r)
intersecting at P, inside in fig. (i) and outside in fig.(ii).
To prove : PA . PB = PC. PD
Construction : Join AC and BD.
A
D
B
Proof : Since OD  chord AB

AD = DB.
Now PA × PB = (PD – AD) (PD + DB)
= (PD – AD)(PD + AD)
[ AD = DB]
= PD2 – AD2
= (OP2 – OD2) – AD2
[From right-angled ODP]
= OP2 – (OD2 + AD2)
= OP2 – OA2
[From right-angled ODA]
= OP2 – OT2
= PT2
 OA = OT [Radii]
[From right-angled OTP]
Hence,
PA × PB = PT2.
PAGE # 5454
 BEA = BAT
But BEA = ACB
Ex.13 Two chords AB and CD of a circle intersect each
other at O internally. If AB = 11 cm, CO =6 cm and DO =
[Angles in the same segment]
ACB = BAT
i.e. BAT = ACB.
This proves the first part.
Again, since ADBC is a cyclic quadrilateral.
 ACB + ADB. = 180º
Also BAT + BAP = 180º
[Linear pair]
 ACB + ADB =BAT + BAP
 BAT + ADB = BAT + BAP
[ ACB = BAT]
 ADB = BAP
 BAP = ADB
Hence, BAT = ACB and BAP = ADB.
4 cm, find OA and OB.
Sol.
Let OA = x cm, then OB = (11– x) cm
We have, AO × OB = CO × OD
 (x) × (11–x) = 6 × 4 = 24
 11x – x2 = 24
 x2–11x+24 = 0
x–3x–8
x= 3 or 8
so, if x = 3 then OA = 3cm, OB = 8cm
but if x = 8 then OA = 8cm, OB = 3cm
Ex.14 ABCD is a cyclic quadrilateral and PQ is a tangent to
the circle at C. If BD is a diameter and DCQ = 40º and
ABD = 60º ; find the measure of the following angles :
(i) DBC
(ii) BCP
(iii) ADB.
A
ANGLES IN THE ALTERNATE SEGMENTS
Let PAQ be a tangent to a circle at point A and AB be a
chord. Then, the segment opposite to the angle formed
by the chord of a circle with the tangent at a point is
called the alternate segment for that angle.
THEOREM
A line touches a circle and from the point of contact a
chord is drawn. Prove that the angles which the chord
makes with the given line are equal respectively to
angles formed in the corresponding alternate
segments.
Given : A circle with centre O and PAT is a tangent to the
circle at A. A chord AB is drawn. Let C and D be the
points on the circumference on opposite sides of AB.
To prove : BAT = ACB and BAP = ADB.
Construction : Draw the diameter AOE and Join EB.
Proof : Since AE is the diameter,
 ABE = 90º
[Angle in semi circle]
 BEA + EAB = 90º
...(i)
[Remaining s of BEA]
But since EA  AT
[ Radius is  to the tangent]
 EAB + BAT = 90º
...(ii)
Hence from (i) and (ii)
BEA + EAB = EAB + BAT
...(iii)
60º
B
D
Oº
Sol.
40º
P
C
Q
Since BD is a diameter of the circle.

BAD = 90º
and also
BCD = 90º
(i)DBC = DCQ = 40º
[s is the alternate segments]
(ii) BCP + BCD + DCQ = 180º
 BCP + 90º + 40º = 180º
 BCP = 50º
(iii) Similarly from BAD, 60º + BAD + ADB = 180º
 60º + 90º + ADB = 180º
ADB = 30º.
Ex.15 In a right triangle ABC, the perpendicular BD on the
hypotenuse AC is drawn. Prove that
(i) AC × AD = AB2
(ii) AC × CD = BC2
Sol. We draw a circle with BC as diameter. Since  BDC = 90º.
 The circle on BC as diameter will pass through D.
Again

BC is a diameter and AB  BC.

AB is a tangent to the circle at B.
Since AB is a tangent and ADC is a secant to the circle.
PAGE # 5555

AC × AD = AB2
This proves (i)
Again
AC × CD = AC × (AC – AD) = AC2 – AC × AD
= AC2 – AB2
[Using (i)]
2
= BC
[ABC is a right triangle]
2
Hence, AC × CD = BC . This proves (ii).
Ex.16 Two circles touch externally at P and a common
tangent touches them at A and B. Prove that
(i) the common tangent at P bisects AB.
(ii) AB subtends a right angle at P.
Sol. Let PT be the common tangent at any point P. Since the
tangent to a circle from an external point are equal,

TA = TP, TB = TP
 TA = TB
i.e. PT bisects AB at T
TA = TP gives  TAP =  TPA
(from PAT)
(a) In fig. (i) d > r1 + r2 i.e. two circles do not intersect.
In this case, four common tangents are possible.
The tangent lines l and m are called direct common
tangents and the tangent lines p and q are called
indirect (transverse) common tangents.
TB = TP gives  TBP =  TPB
[from PBT]
(b) In fig. (ii), d = r1 + r2. In this case, two circles touch

 TAP +  TBP = TPA + TPB = APB
externally and there are three common tangents.

 TAP +  TBP + APB = 2  APB
(c) In fig.(iii) d < r1 + r2. In this case two circles intersect

2  APB = 180º

 APB = 90º
[sum of s of a  = 180º]
Hence proved
in two distinct points and there are only two common
tangents.
(d) In fig. (iv), d = r1 – r2 (r1 > r2), in this case, two circles
Definition : A line which touches the two given circles is
touch internally and there is only one common tangent.
called common tangent to the two circles. Let
(e) In fig. (v), the circle C(O2, r2) lies wholly in the circle
C(O1, r1), C(O2, r2) be two given circles. Let the distance
C(O1, r1) and there is no common tangent.
between centres O1 and O2 be d i.e., O1O2 = d.
Ex.17 Two circles of radii R and r touch each other externally
and PQ is the direct common tangent. Then show that
PQ2 =4rR
Sol. Draw O'S|| PQ,
O'SPQ is rectangle
O'S=PQ, PS=QO'=r
OO'=R+r
OS=OP–PS=R–r
In O'OS
(O'S)2 = (OO')2–(OS)2
PQ2=(R+r)2–(R–r)2
PQ2=R2+r2+2Rr–(R2+r2–2Rr)
PQ2=R2+r2+2Rr–R2–r2+2Rr
PQ2=4rR
PAGE # 5656