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Resonance Pre-foundation Programmes
(PCCP) Division
Career Care
COURSE : OLYMPIAD
MATHEMATICS
WORKSHOP TAPASYA
SHEET
CLASS-VIII
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Subject :Mathematics Olympiads_VIII
S. No. Topics Page No.
1. Number system 1 - 3
Algebra 5 11
4. Time, Speed & Distance 12 - 13
5. Work and Time Sheet & Exiesice 14 - 15
6. Lines and angles and Triangle 16 - 17
7. Quadrilateral 18 - 19
8. Circle 20 - 21
9. Mensuration 22 - 27
2. Surds and Exponents 4
3. -
.13RPCCP
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1PAGE # 1
INTRODUCTION
Number System is a method of writing numerals torepresent numbers.
Ten symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 are used torepresent any number (however large it may be) in ournumber system.
Each of the symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 is called adigit or a figure.
NATURAL NUMBER
Counting numbers 1, 2, 3, 4, 5, ....are called naturalnumbers.
The set of natural numbers is denoted by Ni.e., N = {1, 2, 3, 4, 5, .........}.
EVEN NUMBERS
Whole numbers which are exactly divisible by 2 arecalled even numbers.
The set of even numbers is denoted by 'E', such thatE = {0, 2, 4, 6, 8, .....}.
ODD NUMBERS
Natural numbers which are not exactly divisible by 2are called odd numbers. O = {1, 3, 5, 7, 9.....}
PRIME NUMBERS
Natural numbers having exactly two distinct factors i.e.1 and the number itself are called prime numbers.2, 3, 5, 7, 11, 13, 17, 19,... are prime numbers.
2 is the smallest and only even prime number.
IDENTIFICATION OF PRIME NUMBER
Step (i) Find approximate square root of given number.
Step (ii) Divide the given number by prime numbers lessthan approximately square root of number. If givennumber is not divisible by any of these prime numberthen the number is prime otherwise not.
COMPOSITE NUMBERS
Natural numbers having more than two factors arecalled composite numbers.
4, 6, 8, 9, 10, 12, 14, 15, 16, 18... are composite
numbers.
Number 1 is neither prime nor composite number.
All even numbers except 2 are composite numbers.
Every natural number except 1 is either prime or
composite number.
There are infinite prime numbers and infinite composite
numbers.
CO-PRIME NUMBER OR RELATIVELY
PRIME NUMBERS
Two natural numbers are said to be co-primenumbers or relatively prime numbers if they have only
1 as common factor. For ex. 8, 9 ; 15, 16 ; 26, 33 etc. are
co-prime numbers.
Co-prime numbers may not themselves be prime
numbers. As 8 and 9 are co-prime numbers, but neither
8 nor 9 is a prime number.
Every two consecutive natural numbers are co - primes.
IRRATIONAL NUMBERS
All real numbers which are not rational are irrationalnumbers. These are non-recurring as well as non-
terminating type of decimal numbers.
For example : 2 , 3 4 , 32 , 32 , 4 7 3 etc.
PURE RECURRING DECIMAL
A decimal is said to be a pure recurring decimal if the
a digit or set of digits after the decimals are repeated.
Thus, 3
1 = 0.333...... = 3.0 ,
7
22= 3.142857142857..... = 3.142857 .
Conversion of decimal numbers into rationalnumbers of the form plq.
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.
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.
NUMBER SYSTEM
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2PAGE # 2
Short cut method for mixed recurring decimal : Form
a fraction in which numerator is the difference between
the number formed by all the digits (take the digits
once which are repeating after decimal) 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.
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.
SQUARE AND 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.
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 used 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. e.g. 25 = 5.
To find Square root by Division Method :
Square root of a perfect square by the long divisionmethod. When numbers are very large, the method offinding their square roots by factorization becomeslengthy and difficult. So , we use long division methodwhich is explained in the following steps :
(i) Group the digits in pairs, starting with the digit in theunit place. Each pair and the remaining digit (if any) iscalled a period.(ii) Think of the largest number whose square is equalto or just less than the first period.
(iii) Subtract the product of the divisor and the quotientfrom the first period and bring down the nextperiod to the right of the remainder. This becomes thenext dividend.
(iv) Now, the new divisor is obtained by taking two timesthe quotient and annexing with it a suitable digit whichis also taken as the next digit of the quotient, chosen insuch a way that the product of the new divisor and thedigit is equal to or just less than the new dividend.
Repeat steps (ii), (iii) and (iv) till all the period havebeen taken up. Now, the quotient so obtained is therequired square root of the given number.
Properties of Square Roots :
(i) If the unit digit of a number is 2, 3, 7 or 8, then it doesnot have a square root in N.
(ii) If a number ends in an odd number of zeros, then itdoes not have a square root in N.
(iii) The square root of an even number is even andsquare 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 ofreal numbers.(v) A triplet (x, y, z) of three natural numbers x, y, and z iscalled a Pythagorean triplet, if x2 + y2 = z2.For example : (6, 8, 10) is a Pythagorean triplet. Since62 + 82 = 36 + 64 = 100 and 102 = 100.For any number n greater than 1, the Pythagoreantriplet is given by (2n, n2 � 1, n2 + 1)
CUBE AND CUBE ROOTS
Cube : If any number is multiplied by itself three timesthen the result is called the cube of that number.
Perfect cube : A natural number is said to be a perfectcube if it is the cube of any other natural number.
Cube roots : The cube root of a number x is thatnumber whose cube gives x.
The cube root of x is denoted by the symbol 3 x . Thus,
3 8 = 2, 3 27 = 3, 3 64 = 4, 3 125 = 5 and so on.
3PAGE # 3
FACTORS
Factors : �a� is a factor of �b� if there exists a relationsuch that a × n = b, where �n� is any natural number.
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)....
For example : 36 = 22 × 32.So the factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9.
DIVISIBILITY
Division Algorithm : General representation of result is,
DivisormainderRe
QuotientDivisor
Dividend
Dividend = (Divisor × Quotient ) + Remainder
We are having 10 digits in our decimal number system
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 Cyclicity0, 1, 5 and 6 14 and 9 22, 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.
4PAGE # 4
SURDS
Let a be a rational number and n be a positive integer,
then irrational number is of the form n a is given a
special name surd, where �a� is called radicand and itshould always be a rational number. Also the symbol
n is called the radical sign and the index n is called
order of the surd. n a is read as �nth root of a� and can
also be written as n
1
a .
Laws of radicals :
(i) nn a =
n
n1
a
= n
n1
a
= a.
(ii) nnn abb.a
(iii) n
n
b
a = n
ba
(iv) mn )a( = n ma = am / n
(v) m n a = mn a = n m a
Ex. Simplify : 33 4.2 .
Sol. 33 4.2 = 3 42 = 3 32 = 31
3 )2( = 2.
Rationalising Factor :
If the product of two surds is a rational number, theneach surd is called a rationalising factor (RF) of theother.
Rationalisation of surds :The process of converting a surd into rational number
by multiplying it with a suitable RF, is called the
rationalisation of the surd.
Monomial surds and their RF :
The general form of a monomial surd is n a and its
RF is n1
1a
.
Ex. Find rationalisation factor of 3 5 .
Sol. Rationalisation factor of 3 5 is 33 232
31
125555
.
Binomial surds and their RF :
The surds of the types : ba , ba , ba ,
and ba are called binomial surds.
Conjugate Surds : The binomial surds which differ
only in sign between the terms separating them are
known as conjugate surds. In binomial surds, the
conjugate surds are RF of each other.
For example :
(i) RF of ba is ba .
(ii) RF of ba is ba .
Trinomial surds :
A surd which consists of three terms, atleast two of
which are monomial surds, is called a trinomial surd.
Example : 537
In order to rationalize = cba
x
(i) Multiply and divide by a + b � c
(ii) Multiply and divide by (a + b � c) � 2 ab
EXPONENTS
The repeated multiplication of the same factor can be
written in a more compact form, called exponentialform.
Laws of exponents :If a is any non � zero rational number and m, n are
whole numbers, then
(i) On the same base in multiplication, powers are
added. am × an = am + n
For example : 32 × 34 = 32 + 4 = 36.
(ii) On the same base in division, powers are
subtracted. n
m
a
a= am � n
For Example : 2
5
3
3= 35 � 2 = 33.
(iii) n
m
a
a=
mna
1
, n > m.
For Example : 21
2
1
2
2344
3
.
(iv) (am)n = amn
For Example : (22)3 = 22 × 3 = 26.
(v) an × a� n = a0 = 1
(vi) am × bm = (ab)m
For Example : 22 ×32 = (2 ×3)2 = 62 = 36.
SURDS & EXPONENTS
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PAGE # 5
INTRODUCTION
Algebra is that branch of Mathematics in which lettersrepresent any value which we can assign accordingto our requirement. These letters are generally of twotypes : constants and variables (or literal numbers).
(a) Definition :
An algebraic expression f(x) of the formf(x) = a
0 + a
1x + a
2x2 + ..........+ a
nxn, Where a
0 ,a
1, a
2.....a
n
are real numbers, x is variable and all the indices of xare non negative integers, is called a polynomial in xand the highest index n is called the degree of the
polynomial, if an 0. Here a
0 , a
1x, a
2x2 .....,a
nxn are
called the terms of the polynomial and a0, a
1, a
2, ...... a
n
are called various co-efficients of the polynomial f(x).A polynomial in x is said to be in standard form whenthe terms are written either in increasing order ordecreasing order of the indices of x in various terms.
EXAMPLES :
(i) 2x3 + 4x2 + x + 1 is a polynomial of degree 3.
(ii) x7 + x5 + x2 + 1 is a polynomial of degree 7.
(iii) x3/2 + x2 + 1 is not a polynomial as the indices of xare not all non negative integer.
(iv) x2 + 2 x + 1 is a polynomial of degree 2.
(v) x�2 + x + 1 is not a polynomial as �2 is not non
negative.
(a) Polynomial Based on Degree :
There are five types of polynomials based on degree.
(i) Constant polynomial :
A polynomial of degree zero is called a zero degreepolynomial or constant polynomial. e.g. f(x) = 4 = 4x0.
(ii) Linear polynomial :
A polynomial of degree one is called a linearpolynomial. The general form of a linear polynomial is
ax + b, where a and b are any real numbers and a 0.e.g. 4x + 5, 2x + 3, 5x + 3 etc.
(iii) Quadratic polynomial :
A polynomial of degree two is called a quadraticpolynomial. The general form of a quadratic
polynomial is ax2 + bx + c where a 0.
e.g. x2 + x + 1, 2x2 + 1, 3x2 + 2x + 1 etc.
ALGEBRA
(iv) Cubic polynomial :
A polynomial of degree three is called a cubicpolynomial. The general form of a cubic polynomial is
ax3 + bx2 + cx + d, where a 0.e.g. x3 + x2 + x + 1, x3 + 2x + 1, 2x3 + 1 etc.
(v) Biquadratic polynomial :
A polynomial of degree four is called a biquadratic orquartic polynomial. The general form of biquadratic
polynomial is ax4 + bx3 + cx2 + dx + e where a 0.e.g. x4 + x3 + x2 + x + 1 , x4 + x2 + 1 etc.
NOTE :A polynomial of degree five or more than five does nothave any particular name. Such a polynomial is usuallycalled a polynomial of degree five or six or ..... etc.
(b) Polynomial Based on Terms :
There are three types of polynomial based on numberof terms.
(i) Monomial : A polynomial is said to be a monomial ifit has only one term.For example: x, 9x2, �5x2 are all monomials.
(ii) Binomial : A polynomial is said to be a binomial if itcontains two terms.
For example : 2x2 + 3x, 3 x + 5x4, � 8x3 + 3 etc are all
binomials.
(iii) Trinomial : A polynomial is said to be a trinomial ifit contains three terms.
For Example : 3x3 � 8x +25
, 7 x10 + 8x4 � 3x2 etc are
all trinomials.
REMARKS :
(i) A polynomial having four or more than four termsdoes not have any particular name. They are simplycalled polynomials.
(ii) A polynomial whose coefficients are all zero is calleda zero polynomial, degree of a zero polynomialis not defined.
POLYNOMIAL IN TWO OR MORE VARIABLES
An algebraic expression containing two or morevariables with the powers of the variables as positiveintegers (or natural numbers), is called a polynomialin two or more variables.
Degree of a Polynomial in Two or More Variables :Take the sum of the powers (or indices or exponents)of the variables in each term; the greatest sum is thedegree of the polynomial.The sum of the powers (or indices or exponents) ofthe variables in each term is called the degree of thatterm.
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PAGE # 6
For addition of two polynomials, first we arrange boththe polynomials in standard form and then we add thecoefficients of like powers of variable. For examplesuppose we have to find the sum of two polynomialsf(x) = x + x2 + 1 and g(x) = 2x2 � 4x + 3, for addition first we
arrange both the polynomials in standard form asfollows f(x) = x2 + x + 1, g(x) = 2x2 � 4x + 3. Now
f(x) + g(x) = (x2 + x + 1) + (2x2 � 4x + 3) after arranging
both the polynomials as above we find the sum ofcoefficients of like power of x as follows.
f(x) + g(x) = (1 + 2)x2 + (1 � 4)x + (1 + 3)
= 3x2 � 3x + 4.
Thus, we get the required sum.
REMARK :The process of subtraction is same as addition. Insubtraction after arrangement, we find the differenceof coefficients of like powers of variable.e.g. f(x) � g(x) = (x2 + x + 1) � (2x2 � 4x + 3)
= (1 � 2)x2 + (1 + 4)x + (1 � 3)
= � x2 + 5x � 2.
Ex. Find the sum of f(x) & g(x) where :f(x) = 4x5 + 3x3 + 4x2 + x + 1 & g(x) = 5x4 + x5 + x3 + 3.
Sol. Arrange in standard formf(x) = 4x5 + 3x3 + 4x2 + x + 1
f(x) = 4x5 + 0.x4 + 3x3 + 4x2 + x + 1and g(x) = x5 + 5x4 + x3 + 3 g(x) = x5
+ 5x4 + x3 + 0.x2 + 0.x + 3
f(x) + g(x) = (4x5 + 0.x4 + 3x3 + 4x2 + x + 1) + (x5 + 5x4
+ x3 + 0.x2 + 0.x + 3) f(x) + g(x) = (4 + 1)x5 + (0 + 5)x4 + (3 + 1)x3 + (4 + 0)x2
+ (1 + 0)x + (1 + 3) = (5x5 + 5x4 + 4x3 + 4x2 + x + 4)
Ex. Subtract g(x) from f(x) where f(x) = 2 + x2 + 4x3,g(x) = x4 + x2 + 3x + 5
Sol. f(x) = 4x3 + x2 + 0.x + 2 = 0.x4 + 4x3 + x2 + 0.x + 2g(x) = x4 + 0.x3 + x2 + 3x + 5f(x) � g(x) = (0.x4 + 4x3 + x2 + 0.x + 2) � (x4 + 0x3 + x2 + 3x + 5)f(x) � g(x) = ( 0 � 1)x4 + (4 � 0) x3 + (1 � 1) x2 + (0 � 3) x
+ ( 2 � 5)
= � x4 + 4x3 + 0. x2 � 3x � 3
= � x4 + 4x3 � 3x � 3
Ex. Subtract h(x) from the sum of f(x) & g(x) wheref(x) = x3 + x2 + x + 1, g(x) = 2x3 � 3x2 + 1,h(x) = 3x2 � 4x3 + 5x + 7.
Sol. f(x) = x3 + x2 + x + 1 ; g(x) = 2x3 � 3x2 + 0.x + 1h(x) = � 4x3 + 3x2 + 5x + 7f(x) + g(x) �h(x) = (x3 + x2 + x + 1) + (2x3 � 3x2 + 0.x + 1)� (�4x3 + 3x2 + 5x + 7)f(x) + g(x) � h(x) = ( 1 + 2 + 4)x3 + (1 � 3 � 3)x2 + (1 + 0� 5)x + (1 + 1 � 7)
= 7x3 � 5x2 � 4x � 5
To get the product of two polynomials, carry out thefollowing steps.
Step (i) Multiply each term of the first polynomial witheach term of the second polynomial.
Step (ii) Add all the products obtained is step I.
Ex. Find the product of (x + y) & (x2 + xy + y2).Sol. Step (i) (x + y) (x2 + xy + y2)
= x (x2 + xy + y2) + y(x2 + xy + y2)= x3 + x2y + xy2 + yx2 + xy2 + y3
Step (ii) Product = x3 + 2x2y + 2xy2 + y3 (Grouping liketerms)Thus, the product = x3 + 2x2y + 2xy2 + y3.
Ex. Find the product of f(x) & g(x) where f(x) = (x2 + 2x + 2),g(x) = 3x2 � x � 1.
Sol. Step (i)f(x) . g(x) = (x2 + 2x + 2) (3x2 � x �1)
= x2(3x2 � x �1) + 2x(3x2 � x �1) + 2 (3x2 �x �1)
= 3x4� x3 � x2 + 6x3� 2x2 � 2x + 6x2 � 2x � 2
Step (ii)f(x).g(x) = 3x4 + (�1 + 6)x3 + (�1 �2 + 6)x2 + (� 2 �2)x � 2
= 3x4+ 5x3 + 3x2 � 4x � 2.
Division Algorithm :Dividend = quotient × divisor + remainder. Theprocess of division of a polynomial by anotherpolynomial is also same as we use in number system.The process of division may be divided in three casesin polynomials.(i) Division of a monomial by another monomial.
(ii) Division of a polynomial by monomial.
(iii) Division of a polynomial by another polynomial.
We have two methods for division of a polynomial byanother polynomial.
(a) Factor method : Factor method is generally usedwhen remainder is zero.
(b) Long division : This method generally used whenremainder is not zero. This method can also be usedwhen remainder is zero. We shall discuss these threecases one by one.
Case1. Division of a monomial by another monomial
Ex. Divide :(i) 15a5 by 5a3 (ii) 36a3b5 by �12a2b
(iii) 2x3 by 2 x
Sol. (i) Quotient = 3
5
a5
a15 =
515
3
5
a
a = 3a2
(ii) Quotient = ba12
ba362
53
=
1236
2
3
a
a
bb5
= � 3ab4
(iii) Quotient = x2
x2 3
=
2
2
xx3
= 2 x2
PAGE # 7
Case 2. Division of a polynomial by a monomial
Divide each term of the polynomial by the monomialand then write the resulting quotients with their propersigns.
Ex. Divide : � 4x3 � 6x2 + 8x by 2xSol. Dividing each term of the dividend by the divisor, we
get
Quotient = x2
x8x6x4 23
= x2x4 3
�
x2x6 2
+x2
x8
= � 2x2 � 3x + 4.
Case3. Division of a polynomial by another polynomial
It is advisable in this case to rearrange the dividendand the divisor in descending order of powers ofvariable .
Ex. Divide x2 + 5x + 6 by x + 3.Sol.
Quotient = x + 2Remainder = 0
EXPLANATION :
(i) Divide the first term (x2) of the dividend by the firstterm (x) of the divisor.The result x2 x = x is the first term of the quotient.
(ii) Multiply the divisor x + 3 by x, the first term of thequotient.
(iii) Subtract the product (x + 3) x = x2 + 3x from thedividend x2 + 5x + 6. i.e. (x2 + 5x + 6 ) � (x2 + 3x) = 2x + 6.
(iv) Proceed with this remainder 2x + 6 as with the originaldividend i.e., divide 2x by x, The result 2x x = 2 is thesecond term of the quotient.
(v) Multiply the divisor (x + 3) by 2, the second term of thequotient. Now subtract 2(x + 3) from 2x + 6 i.e.,2x + 6 � 2 (x + 3) = 2x + 6 � 2x � 6 = 0. The remainder is 0.
Hence, the required quotient = x + 2.
Ex. Divide 2x3 + x2 � 3x � 3 by 2x � 1 and verify your answer.
Sol.
Quotient = x2 + x � 1,
Remainder = � 4
VERIFICATION :
We know that, Dividend = Divisor × Quotient +
Remainder ....(i)
Divisor Quotient = (2x � 1) (x2 + x �1)
= 2x3 + x2� 3x + 1
Now R.H.S. of (i) = 2x3 + x2 � 3x + 1 + (�4)
= 2x3 + x2 � 3x � 3 = L.H.S.
Hence, the answer is correct.
REMARK :
When the dividend and the divisor are polynomials of
one variable, the degree of the polynomial in the
remainder is always less than the degree of the
polynomial of the divisor.
Ex. Divide 3x4 + 5x3 � x2 + 13x + 9 by 3x + 2 and verify that :
Dividend = Divisor Quotient + Remainder..
Sol. First we divide 3x4 + 5x3 � x2 + 13x + 9 by 3x + 2
3x + 2 3x + 5x x +13x4 3 2 � + 9 x + x � x+
3 2 5
3x x 3 2�
3x + 2x 3 2
� 3x + 13x2
� � 3x 2x2
15x + 9 15x + 10
3x + 2x4 3
� 1Quotient = x3 + x2 � x + 5,
Remainder = �1
Now, Divisor Quotient + Remainder
= (3x + 2) (x2 + x2 � x + 5) � 1
= 3x (x3 + x2 � x + 5) + 2 (x3 + x2� x + 5) � 1
= 3x4 + 3x3 � 3x2 + 15x + 2x3 + 2x2 � 2x + 10 � 1
= 3x4 + 5x3 � x2 + 13x + 9
= Dividend
REMARK :
When the remainder is zero, the divisor is called a
factor of the dividend.
Ex. Find the value of a if 2x � 3 is a factor of 2x4 � x3 � 3x2 � 2x
+ a.
Sol. First we divide 2x4 � x3 � 3x2 � 2x + a by 2x � 3.
2x - 3 is a factor of 2x4 � x3� 3x2 � 2x + a if, a � 3 = 0.
Hence, a = 3.
PAGE # 8
(a) Value of a Polynomial :
The value of a polynomial f(x) at x = is obtained bysubstituting x = in the given polynomial and isdenoted by f().Consider the polynomial f(x) = x3 � 6x2 + 11x � 6,
If we replace x by � 2 everywhere in f(x), we get
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 of a Polynomial :
The real number is a 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 zero of f(x).
Let �p(x)� be any polynomial of degree greater than orequal to one and a be any real number and If p(x) isdivided by (x � a), then the remainder is equal to p(a).
Ex. Find the remainder, when f(x) = x3 � 6x2 + 2x � 4 is
divided by g(x) = 1 � 2x.
Sol. f(x) = x3 � 6x2 + 2x � 4
Let, 1 � 2x = 0
2x = 1
x = 21
Remainder =
21
f
21
f = 4�2
12
2
16�
2
123
= 4�123
�81
= 8
35�
832�812�1
.
FACTOR THEOREM
Let p(x) be a polynomial of degree greater than orequal 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) isa factor of p(x), then p(a) = 0.
Ex. Show that x + 1 and 2x � 3 are factors of
2x3 � 9x2 + x + 12.Sol. To prove that (x + 1) and (2x � 3) are factors of
2x3 � 9x2 + x + 12 it is sufficient to show that p(�1) and
23
p both are equal to zero.
p (� 1) = 2 (� 1)3 � 9 (� 1)2 + (� 1) + 12
= � 2 � 9 � 1 + 12
= � 12 + 12 = 0.
and
23
p = 1223
23
9�23
223
= 1223
481
�4
27
= 4
48681�27
=4
8181�
= 0.Hence, (x + 1) and (2x � 3) are the factors
2x3 � 9x2 + x + 12.
Ex. Find the values of a and b so that the polynomialsx3 � ax2 � 13x + b has (x � 1) and (x + 3) as factors.
Sol. Let f(x) = x3 � ax2 � 13x + b
Because (x � 1) and (x + 3) are the factors of f(x),
f(1) = 0 and f(� 3) = 0
f(1) = 0 (1)3 � a(1)2 � 13(1) + b = 0
1 � a � 13 + b = 0
� a + b = 12 .... (i) f(�3) = 0
(� 3)3 � a(� 3)2 � 13(� 3) + b = 0
� 27 � 9a + 39 + b = 0
� 9a + b = �12 ...(ii)Subtracting equation (ii) from equation (i) (� a + b) � (� 9a + b) = 12 + 12
� a + 9a = 24
8a = 24 a = 3.
Put a = 3 in equation (i)� 3 + b = 12
b = 15.Hence, a = 3 and b = 15.
Some important identities are :
(i) (a + b)2 = a2 + 2ab + b2
(ii) (a � b)2 = a2 � 2ab + b2
(iii) a2 � b2 = (a + b) (a � b)
(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.But converse is not true i.e.a3 + b3 + c3 = 3abc thena + b + c = 0 or a = b = c.
PAGE # 9
Value Form :
(i) a2 + b2 = (a + b)2 � 2ab, if a + b and ab are given.
(ii) a2 + b2 = (a � b)2 + 2ab, if a � b and ab are given.
(iii) a + b = ab4ba 2 , if a � b and ab are given.
(iv) a � b = ab4ba 2 , if a + b and ab are given.
(v) a2 + 2a
1 =
2
a
1a
� 2, if a +
a1
is given.
(vi) a2 + 2a
1 =
2
a
1a
+ 2, if a �
a1
is given.
(vii) a3 + b3 = (a + b)3 � 3ab (a + b), if (a + b) and ab aregiven.
(viii) a3 � b3 = (a � b)3 + 3ab (a � b), if (a � b) and ab aregiven.
(ix) 3
3
a
1a =
3
a1
a
� 3
a1
a , if a +a1
is
given.
(x) 33
a
1a =
3
a1
a
+ 3
a1
a , if a � a1
is given.
(xi) a4 � b4 = (a2 + b2) (a2 � b2) = [(a + b)2 � 2ab](a + b) (a � b).
Ex. Expand :
(i) 2
x3
1x2
(ii) y3x2 y3x2
Sol. (i)2
x3
1x2
= (2x)2 � 2(2x)
x31
+ 2x3
1
= 4x2 � 34
+ 2x9
1.
(ii) ( 2 x � 3y)( 2 x + 3y)
= ( 2 x)2 � (3y)2
= 2x2 � 9y2
Ex. Simplify :
44
22
x
1x
x
1x
x1
xx1
x .
Sol.
x1
x
x1
x
22
x
1x
44
x
1x
=
22
x
1x
22
x
1x
44
x
1x
=
2
222
x
1)x(
44
x
1x
=
44
x
1x
44
x
1x
= (x4)2 � 2
4x
1
= x8 �
8x
1.
Ex. If 4x + 5y = 30 and xy = 12, then evaluate 16x2 + 25y2.
Sol. 4x + 5y = 30
(4x + 5y)2 = 302
16x2 + 25y2 + 2(4x)(5y) = 900
16x2 + 25y2 + 40xy = 900
16x2 + 25y2 + 40(12) =900
16x2 + 25y2 + 480 =900
16x2 + 25y2 = 900 � 480
16x2 + 25y2 = 420.
Ex. If a + a1
= 6 then find the value of a4 + 4a
1 .
Sol. a + a1
= 6
2
a1
a
= 62
a2 + 2a
1 + 2 = 36
a2 + 2a
1 = 34
2
22
a
1a
= 342
a4 + 4a
1 + 2 = 1156
a4 + 4a
1 = 1156 � 2 = 1154.
To express a given polynomial as the product of
polynomials, each of degree less than that of the given
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,
divide each term by this factor and take out as a
multiple.
Ex. Factorize :
(i) 2a (x + y) � 3b (x + y) (ii) 8(3a � 2b)2 � 10(3a � 2b)
Sol. (i) 2a (x + y) � 3b (x + y)
= (x + y) (2a � 3b).
(ii) 8 (3a � 2b)2 � 10 (3a � 2b)
= 2 (3a � 2b) [4 (3a � 2b) � 5]
= 2 (3a � 2b) [12a � 8b � 5].
(b) Factorization by grouping :
Ex. Factorize : ax + by + ay + bx.
Sol. ax + by + ay + bx
= ax + ay + bx + by
= a(x + y) + b (x + y)
= (x + y)(a + b)
PAGE # 10
(c) Factorization by making a perfect square :
Ex. Factorize : 25x2 + 4y2 + 9z2 � 20xy � 12yz + 30xz
Sol. 25x2 + 4y2 + 9z2 � 20xy � 12yz + 30xz
= (5x)2 + (� 2y)2 + (3z)2 + 2(5x)(�2y) + 2(�2y)(3z) + 2 (3z)(5x)
= (5x � 2y + 3z)2.
(d) Factorization by the difference of two squares :
Ex. Factorize : a2 + b2 + 2ab �1.
Sol. a2 + b2 + 2ab �1
(a + b)2 � (1)2
(a + b + 1)(a + b �1)
(e) Factorization of a quadratic polynomial by splittingthe middle term :
Ex. Factorize : 8x2 + 6x � 5.
Sol. 8x2 + 6x - 5
= 8x2 + 10 x � 4x - 5
= 2x (4x + 5) �1 (4x + 5)
= (4x + 5)(2x �1).
Ex. Find the value of 96.5
02.202.2�98.798.7 .
Sol.96.5
02.202.2�98.798.7
=96.5
)02.2�98.7)(02.278.7(
=96.5
96.510 = 10.
Ex. Find the value of
aba
b�a
b5ab6�a
ab5�a2
22
22
2
.
Sol. 22
2
b5ab6�a
ab5�a
×
aba
b�a2
22
= )b5�a)(b�a()b5�a(a
× )ba(a)ba)(b�a(
= 1
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.
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. 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. 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) = )3x)(2x()2x)(7x)(3x)(2x(
= (x + 7) (x+2) = x2 + 9x + 14.
LINEAR EQUATION IN ONE VARIABLE
An equation involving a variable in first degree is calleda linear equation in one variable. It is of the formax + b = 0 , a 0, x is variable.
x = � ab
is the root or solution of this equation.
Rules for solving a linear equation:
Rule 1 : Same quantity (number) can be added to bothsides of an equation without changing the equality.
Rule 2 : Same quantity can be subtracted from bothsides of an equation without changing the equality.
Rule 3 : Both sides of an equation may be multipliedby the same non-zero number without changing theequality.
Rule 4 : Both sides of an equation may be divided bythe same non-zero number without changing theequality.
Rule 5 : (Transposition) Any term of an equation maybe taken to the other side with the sign changed. Thisprocess is called transposition.
Ex. 3x
� 5
3x2
2 = 4
Sol.3x
� 5
3
x22 = 4
3x
� 10 + 3
x10 = 4
3x
+ 3
x10 = 14
3x11
= 14
x = 14 × 113
x = 1142
PAGE # 11
Ex. Ratio of present ages of Ramesh and Mohan is 2 : 3.After 3 years this ratio becomes 3 : 4 Find the presentage of Mohan.
Sol. Let Ramesh present age = 2x yearsMohan present age = 3x yearsAfter 3 yearRamesh age will become 2x + 3.Mohan age will become 3x + 3.According to problem
3x33x2
=
43
8x + 12 = 9x + 9 x = 3. Present age of Mohan 3 × 3 = 9 years.
PAGE # 12
TIME, SPEED & DISTANCE
(i) Speed =
TimecetanDis
, Time =
SpeedcetanDis
,
Distance = (Speed × Time)
(ii) x km/hr =
185
x m/sec
(iii) x m/sec =
518
x km/hr
(iv) If the ratio of the speeds of A and B is a : b, then theratio of the times taken by them to cover the same
distance is a1
: b1
or b : a.
(v) Suppose a man covers a certain distance atx km/hr and an equal distance at y km/hr. Then, average
speed during the whole journey is
yxxy2
km/hr..
Ex. An athlete ran 100 metres in 10 seconds. Then find hisspeed in km/hr.
Sol. Speed =time
cetandis =
10100
m/sec
= 10 m/sec = 10 × 5
18km/hr = 36 km/hr..
Ex. A car travels 120 km from A to B at 30 km per hour butreturns the same distance at 40 km per hour. Find theaverage speed for the round trip.
Sol. Average speed = 4030
40302
km/hr
= 7
240km/hr =
72
34 km/hr..
Ex. If a person has a speed of 60 km/hr he reaches theoffice 5 min late and if he increases his speed to80 km/hr, he reaches the office 3 min early. Calculatethe distance to be covered.
Sol. Let his distance from office be x km and actual time toreach office by y hrs.When his speed is 60 km/hr he reaches, the office5 min late
605
y60x
y =605
60x
... (i)
When he increases his speed to 80 km/hr, he reachesthe office 3 min early.
603
y80x
y =603
80x
.....(ii)
From (i) and (ii) we get
605
60x
=603
80x
605
603
80x
60x
60
53240
x3x4
608
240x
x = 240608 = 32 km.
(i) Time taken by a train of length �a� metres to pass apole or a standing man or a signal post is equal to thetime taken by the train to cover �a� metres.
(ii) Time taken by a train of length �a� metres to pass astationary object of length �b� metres is the time takenby the train to cover (a + b) metres.
(iii) Suppose two trains or two bodies are moving inthe same direction at u m/s and v m/s, where u > v,then their relatives speed = (u � v) m/s.
(iv) Suppose two trains or two bodies are moving inopposite direction at u m/s and v m/s, where u > v,then their relative speed = (u + v) m/s.
(v) If two trains of length �a� metres and �b� metres aremoving in opposite directions at u m/s and v m/s, thentime taken by the trains to cross each other
= v)( u
b)( a
sec.
(vi) It two trains of length �a� metres and �b� metres aremoving in the same direction at u m/s and v m/s thenthe time taken by the faster train to cross the slower
train = v)( u
b)( a
�
sec.
(vii) If two trains (or bodies) start at the same timefrom points A and B towards each other and aftercrossing they take �a� and �b� sec in reaching B and A
respectively, then (A�s speed) : (B�s speed) = )a:b( .
TIME, SPEED AND DISTANCE
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PAGE # 13
Ex. A train passes a station platform in 36 seconds and aman standing on the platform in 20 seconds. If thespeed of the train is 54 km/hr, what is the length of theplatform ?
Sol. Speed =
185
54 m/sec = 15 m/sec.
Length of the train = (15 × 20) m = 300 m.
Let the length of the platform be x metres.
Then, 36300x
= 15
x + 300 = 540
x = 240 m.
So, the length of the platform is 240 metres.
PAGE # 14
DIRECT VARIATION
Two quantities are said to vary directly if the increase
(or decrease) in one quantity cause the increase
(or decrease) in the other quantity.
Some Examples :
(i) The cost of articles varies directly as the number of
articles (More articles, More cost).
(ii) The distance covered by a moving object varies
directly as its speed (More speed, more distance
covered in the same time).
(iii) The work done varies directly as the number of
men at work (More men at work, more is the work
done in the same time).
(iv) The work done varies directly as the working time
(More is the working time, more is the work done).
Ex. If a car covers 75 km in 3 hours, how much it travel in 5
hours.
Sol. In 3 hr, car cover a distance of 75 km.
In 1 hr, car cover a distance of 3
75 km.
[Less time, Less distance]
In 5 hr, car cover a distance of 5 375
km = 125 km.
[More time, More distance]
Ex. If the wages of 15 workers for 6 days are Rs. 9450, find
the wages of 19 workers for 5 days.
Sol. Wages of 15 workers for 6 days = Rs. 9450
Wages of 1 workers for 6 days = Rs. 15
9450
[Less workers, Less wages]
Wages of 1 workers for 1 day = Rs. 1569450
[Less days, Less wages]
Wages of 19 workers for 1 day = Rs. 19 615
9450
[More workers, More wages]
Wages of 19 workers for 5 days = Rs. 519615
9450
[More days, More wages]
= Rs. 9975
INVERSE VARIATION
Two quantities are said to vary inversely if the increase
(or decrease) in one quantity cause the decrease
(or increase) in the other quantity.
WORK AND TIME
Examples :
(i) The time taken to finish a piece of work variesinversely as the number of men at work (More men atwork, less is the time to finish it).
(ii) The speed varies inversely as the time taken tocover a distance (More is the speed, less is the timetaken to cover a distance).
Ex. A fort had provision for food for 300 men for 90 days.
After 20 days, 50 men left the fort. How long would the
food last at the same rate ?
Sol. Remaining number of men = (300 � 50) = 250.
Remaining number of days = (90 � 20) days
= 70 days.
300 men had provision for 70 days.
1 men had provision for 300 × 70 days.
[Less men, More days]
250 men had provision for 250
70300 days.
[More men , Less days]
= 84 days
Hence, the remaining food will last for 84 days.
Ex. 6 oxen or 8 cows can graze the field in 28 days. How
would 9 oxen and 2 cows take to graze the same field ?
Sol. 6 oxen 8 cows
1 ox 68
cows
9 oxen
9
68
cows = 12 cows.
(9 oxen + 2 cows) (12 cows + 2 cows) = 14 cows
Now, 8 cows can graze the field in 28 days.
1 cow can graze the field in 28 × 8 days.
14 cows can graze the field in 14
828 days = 16 days.
Hence, 9 oxen and 2 cows can graze the field in
16 days.
WORK AND TIME
(i) If A can do a piece of work in m days, then :
A�s 1 day work is = m1
.
(ii) If B�s 1 day work = n1
, then :
B can finish the hole work in n days.
(iii) If A is thrice as good a workman as B, then :
Ratio of work done by A and B is = 3 : 1.
Ratio of times taken by A and B to finish a work is = 1 : 3
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PAGE # 15
Ex. A and B can do a piece of work in 10 days, B alone cando it in 15 days. In how many days �A� alone can
complete the same work ?Sol. Let A complete the work in x days
So, work done by A in 1 day = x1
According to problem :
x1
+ 151
= 101
x1
=101
�151
x1
= 30
23
x = 30.So, A complete the work in 30 days.
PIPES AND CISTERNS
Inlet Pipe : A pipe connected with a tank or a cistern ora reservoir, that fills it, is known as an inlet.
Outlet Pipe : A pipe connected with a tank or a cisternor a reservoir, emptying it, is known as an outlet.
(i) If a pipe can fill a tank in x hours, then ;
Part filled in 1 hour = x1
.
(ii) If a pipe can empty a full tank in y hours, then,
Part emptied in 1 hour = y1
.
(iii) If a pipe can fill a tank in x hours and another pipecan empty the full tank in y hours (where y> x), then onopening both the pipes, the net part filled in 1 hour
=
y1
x1
.
(iv) If a pipe can fill a tank in x hours and another pipecan empty the full tank in y hours (where x> y), then onopening both the pipes, the net part emptied in 1 hour
=
x1
y1
.
Ex. Three pipes A, B & C can fill a tank in 8 hrs. After workingat it together for 2 hrs., C is closed and A and B can fillit in 10 hrs. Find the number of hours taken by C aloneto fill the cistern.
Sol. Three pipes A, B & C can fill a tank in 8 hrs.
In 1 hr. they can fill the 81
part of tank
Part of tank filled in 2 hrs. = 241
81
Remaining part = 1 � 41
= 43
(A + B)�s 10 hrs. work = 43
(A + B)�s 1 hr. work = 403
101
43
Let C fill the tank in x hr. alone.So, (A + B)�s 1 hr. work + C�s 1 hr. work = (A + B + C)�s
1 hr. work
81
x1
403
403
81
x1
x1
=201
402
4035
.
x = 20 hours C fill the tank in 20 hours.
16PAGE # 16
Angles Made by a Transversal with two
Parallel Lines :
Transversal : A line which intersects two or more given
parallel lines at distinct points is called a transversal
of the given lines.
(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.
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.
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.
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.
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
BOC = 90º + 21A.
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 QCB intersect at O, then
BOC = 90º � 21A.
Property : The sum of any two sides of a triangle is
greater than the third side.
Property : Angles opposite to equal sides of a triangle
are equal.
Property : Sides opposite to equal angles of a triangle
are equal.Property : In a right triangle, if a, b are the lengths of
LINES & ANGLES
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17PAGE # 17
the sides and c that of the hypotenuse, then c2 = a2 +b2.(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Property : If the sides of a triangle are of lengths a, band c such that c2 = a2 + b2, then the triangle is rightangled and the side of length c is the hypotenuse.
NOTE : Three positive numbers a, b, c in this order aresaid to form a pythagorean triplet, if c2 = a2 + b2. Triplets(3, 4, 5) (5, 12, 13), (8, 15, 17), (7, 24, 25) and (12, 35, 37)are some pythagorean triplets.
Statement : If a line is drawn parallel to one side of atriangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.If in a ABC, a line DE || BC, intersects AB in D and ACin E, then
(i) AE
EC
AD
DB
(ii) AEAC
ADAB
(iii) ACAE
ABAD
(iv) EC
AC
DB
AB
(v) ACEC
ABDB
18PAGE # 18
QUADRILATERAL
A quadrilateral is a four sided closed figure.
A
D
C
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 fig., PQRS is convex quadrilateral.
S
P Q
R
(ii) A quadrilateral in which the measure of one of the
interior angles is more than 180° is called a concave
quadrilateral. In fig., ABCD is concave quadrilateral.
A
B
C
D
Special Quadrilaterals :
(i) Parallelogram : A parallelogram is a quadrilateral
in which both pairs of opposite sides are parallel. In
fig., AB || DC, AD || BC therefore, ABCD is a
parallelogram.
A B
CD
(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°.
A B
CD
900
(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.
(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 and
A = B = C = D = 90°.
(v) Trapezium : A trapezium is a quadrilateral with only
one pair of opposite sides parallel. In fig., ABCD is a
trapezium with AB || DC.
A B
CD
(vi) Kite : A kite is a quadrilateral in which two pairs of
adjacent sides are equal. If ABCD is a kite then
AB = AD and BC = CD.
A
B
C
D
(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.
QUADRILATERALS
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19PAGE # 19
PROPERTIES
Theorem 1 : The sum of the four angles of a
quadrilateral is 360°.
Theorem 2 : A diagonal of a parallelogram divides it
into two congruent triangles.
Theorem 3 : In a parallelogram, opposite sides are
equal.
Theorem 4 : The opposite angles of a parallelogram
are equal.
Theorem 5 : The diagonals of a parallelogram bisect
each other.
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.
Theorem 10 : The diagonals of a rhombus are per-
pendicular to each other.
Theorem 11 : The diagonals of a square are equal
and perpendicular to each other.
Theorem 12 : Parallelograms on the same base and
between the same parallels are equal in area.
Theorem 13 : Two triangles on the same base (or
equal bases) and between the same parallels are
equal in area.
Theorem 14 : 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.
Theorem 15 : Median of a triangle divides it into two
triangles of equal area.
Theorem 16 : A diagonal of a parallelogram divides it
into two triangles of equal area.
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.
20PAGE # 20
DEFINITIONS
Circle :
The collection of all the points in a plane, which are ata fixed distance from a fixed point in the plane, is calleda circle.The fixed point is called the centre of the circle and thefixed distance is called the radius of the circle.
In figure, O is the centre and the length OP is the radiusof the circle. So the line segment joining the centreand any point on the circle is called a radius of thecircle.
Chord :
If we take two points P and Q on a circle, then the linesegment PQ is called a chord of the circle.
QO
P
If the chord which passes through the centre of thecircle, is called a diameter of the circle.
Arc :
A piece of a circle between two points is called an arc.
P Q
R
The longer one is called the major arc and the
shorter one is called the minor arc
Secant :Secant to a circle is a line which intersects the circle intwo distinct points.
Tangent :A tangent to a circle is a line that intersects the circle inexactly one point.
Theorem-1 : Equal chords of a circle subtend equalangles at the centre.If AB = CD then AOB = COD
Theorem-2 : The perpendicular from the centre of acircle to a chord bisects the chord.If OM AB then AM = BM
A M B
O
Theorem-3 : There is one and only one circle passingthrough three given non-collinear points.
Theorem-4 : The angle subtended by an arc at thecentre is double the angle subtended by it at any pointon the remaining part of the circle.POQ = 2PAQ
BO
A
PQ
Theorem-5 : Angles in the same segment of a circleare equal. PAQ = PCQ
O
P
A
Q
C
Theorem-6 : Angle in the semicircle is a right angle.PCQ = PAQ = 90º
Theorem-7 : The sum of either pair of opposite anglesof a cyclic quadrilateral is 180º.BCD + BAD = 180º
ABC + ADC = 180º
Theorem-8 : Equal chords of a circle (or of congruentcircles) are equidistant from the centre (or centres).If AB = CD then ON = OM
ON
C
DB
M
A
CIRCLES
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21PAGE # 21
Theorem-9 : A tangent to a circle is perpendicular to
the radius through the point of contact.
OP AB
Theorem-10 : Lengths of two tangents drawn from an
external point to a circle are equal.
If AP and AQ are two tangents then AP = AQ.
Theorem-11 : If two chords of a circle intersect inside
or outside the circle when produced, then the area
of rectangle formed by the two segments of one chord
is equal in area to the rectangle formed by the two
segments of the other chord. (PA × PB = PC × PD)
A
BC
D
P
Theorem-12 : If PAB is a secant to a circle intersecting
the circle at A and B and PT is tangent segment, then
PA × PB = PT2
P
AB
T
22PAGE # 22
PERIMETER AND AREA
Area : The magnitude of measurement of a planeregion enclosed by a simple closed figure is called itsarea.
Perimeter :
The measurement of the boundary of a plane figure isknown as its perimeter.
Triangle :
D
(i) Scalene triangle :Perimeter = a + b + c
Area = 21
× Base × Height = 21
ah
(ii) sosceles triangle :
Area = 21 base
22 )base(41
)sideequal(
(iii) Right-angled triangle :
For an right-angled triangle, let b be the base, h be theperpendicular and d be the hypotenuse. Then :(A) Perimeter = b + h + d
(B) Area = 21
(Base × Height) = 21
bh
B C
A
h
b
d
(C) Hypotenuse, d = 22 hb [Pythagoras theorem]
(iv) sosceles right-angled triangle :
For an isosceles right-angled triangle, let a be theequal sides, then
(A) Hypotenuse = 22 aa = a2
(B) Perimeter = 2a + a2
B C
A
a
a a2
(C) Area = 21
(Base × Height) = 21
(a × a) = 21
a2.
(v) Equilateral triangle :
Area = 43
(side)2, Perimeter = 3(side).
Rectangle :
Perimeter = 2 (+ b)
Area = × b
Length of diagonal = 22 b.
Square :
Perimeter = 4a
Area = a2
Length of diagonal = a 2 .
Parallelogram :
Perimeter = 2 (a + b)
Area = ah1 = bh2
Rhombus :
Perimeter = 4a = 2 22
21 dd
Area = 21
d1d2
MENSURATION
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23PAGE # 23
Quadrilateral :
Let AC = d
Area = 21
d (h1 + h2)
Trapezium :
h
b
a
D
A B
C
Area = 21
h (a + b).
AREA RELATED TO CIRCLE
Circle : Circle is a path of a moving point, which movesin such a manner that its distance from a fixed point isalways equal. The fixed point is called centre of thecircle and the fixed distance is called radius of thecircle.
Circle
r
C
Area of circle (A) = r2
Circumference (C) = 2 rDiameter (D) = 2r
Results :(i) Distance moved by a rotating wheel in onerevolution is equal to the circumference of the wheel.
(ii) Number of revolutions completed by a rotating wheel
in one minute =nceCircumfere
uteminoneinmovedcenatDis.
Semicircle :
Semi-Circle
rC
r
Perimeter = r + 2r = ( + 2) r
Area (A) = 2r2
SECTOR OF A CIRCLE AND ITS AREA
The region bounded by an arc of a circle and its twobounding radii is called a sector of the circle.
Length of Arc and Area of sector :Let r be the radius of the circle with centre O and AOBbe a sector of the circle such that AOB = .
When an arc subtends angle at the centre, then
length of the arc =
180r
=
180r
.
When an arc subtends an at the centre,then area
of the sector =
360
2r
MENSURATION (SOLID FIGURES)
If any figure such as cuboid, which has threedimensions length, width and height are known asthree dimensional figures.Some of the main solid figures are :
Cuboid :
Total Surface Area (T.S.A.) : The area of surface fromwhich cuboid is formed. There are six faces(rectangular), eight vertices and twelve edges in acuboid.
(i) Total Surface Area (T.S.A.)= 2 [ × b + b × h + h × ]
(ii) Lateral Surface Area (L.S.A.) (or Area of 4 walls)
= 2 [b × h + h × ]= 2 h [ + b]
(iii) Volume of Cuboid= (Area of base) × height
= × b × h
(iv) Length of Diagonal = 222 hb
Cube :Cube has six faces. Each face is a square.
24PAGE # 24
(i) T.S.A. = 2 [x x + x x + x x]= 2 [x2 + x2 + x2] = 2 (3x2) = 6x2
(ii) L.S.A. = 2 [x2 + x2] = 4x2
(iii) Volume = (Area of base) × Height
= (x2) x = x3
(iv) Length of Diagonal = x 3
Cylinder :
(i) C.S.A. of Cylinder = (2 r) × h = 2 rh.(ii) Total Surface Area (T.S.A.) : T.S.A. = C.S.A. + Area of circular top & bottom
= 2 rh + 2 r2
= 2 r (h + r)
(iii) Volume of Cylinder :Volume = Area of base × height
= ( r2) × h
= r2hCone :
r
h
(i) C.S.A. = r
(ii) T.S.A. = C.S.A. + base area= r + r2
= r ( + r)
(iii) Volume = 31
r2h
Where, h = height r = radius of base = slant height
Sphere :
T.S.A. = S.A. = 4 r2
Volume =34
r3
Hemisphere :
C.S.A. = 2 r2
T.S.A. = C.S.A. + base area = 2 r2 + r2
= 3 r2
Volume = 32
r3
Ex. If the area of an equilateral triangle is 34 square cm,then find its perimeter.
Sol. Let the side of equilateral be a.
Area = 2a43
= 4 3
a2 = 16 a = 4 cm Perimeter = 3a = 12 cm.
Ex. Find the area of a triangle whose sides are 8 cm,
13 cm and 15 cm.
Sol. Side of are 8 cm, 13 cm, 15 cm
a = 8 cm, b = 13 cm, c = 15 cm
s = 2
cba =
215138
= 18 cm
Area of = )c�s)(b�s)(a�s(s
= 351018
= 3525332
= 330 cm2
Ex. In the given figure, ABC is an equilateral triangle the
length of whose side is equal to 10 cm, and DBC is
right-angled at D and BD = 8 cm. Find the area of the
shaded region. [Take 3 = 1.732]
25PAGE # 25
Sol. Ar. of ABC = 43
a2
= 10043
= 325 cm2
In DBC
(BC)2 = (BD)2 + (DC)2
(10)2 = (8)2 + (DC)2
DC = 6 cm
Area DBC = 21
× 8 × 6 = 24 cm2
Shaded region = 325 � 24
= 25 x 1.73 �24
= 43.25 � 24
= 19.25 cm2
Ex. Diagonals of rhombus are 15 cm and 20 cm. Find itsarea and perimeter.
Sol. Diagonal D1 = 15 cm
Diagonal D2 = 20 cm
Area of rhombus = 21
(Product of Diagonals)
= 21
× 15cm × 20cm
= 150 cm2
The diagonals bisect each other pendicularly.OD OC
OD = 21
DB = 21
× 2 = 10 cm
OC = 21
AC = 21
× 15 = 2
15
(OD)2 + (OC)2 = (DC)2
(10)2 +
2
215
= (DC)2
100 + 4
225 = (DC)2
4
625 = DC
2
25 = DC
Perimeter = 4 × side
= 4 × 2
25 = 50 cm.
Ex. Two parallel sides of a trapezium are 60 cm and77 cm and other sides are 25 cm and 26 cm. Find thearea of the trapezium.
Sol.
In trapezium we draw a line from vertex C and parallelto AD, which meet AB at E.as AE II DC and AD II ECAECD is a parallelogram EC = AD = 25 cm AE = DC = 60 cm EB = AB � AE = 77 � 60 = 17 cm
Area BEC
s = 2
172625 = 34 cm
Area BEC = 178934
= 17 3 2 2= 204 cm2
Let h cm is the length of perpendicular to EB.
Area BEC = 21
17 h = 204 cm2
h = 17
2204 = 24 cm.
Area trap. ABCD = 21
(60 + 77) 24
= 137 12 = 1644 cm2.
Ex. From a square metal sheet of side 28 cm, a circular
sheet of largest possible radius is cut off then find the
area of the remaining sheet.
Sol.
According to the question
Diameter of the circle = side of the square
2r = 28
r = 14 cm
Area of the remaining portion = area of the square �
area of the circle
= 28 × 28 � 1414722
= 784 � 616 = 168 cm2.
26PAGE # 26
Ex. A lawn is the form of a square of side 30 m. A cow is tied
with a rope of 10 m to a pole standing at one of its
corner. Find the maximum area of the lawn grazed by
this cow.
Sol.D C
30 m
30 m
BA 10 m
Maximum area of the lawn grazed by the cow is the
area of shaded region.
Area of the shaded region = º360
r2
here, r = 10 m (length of rope)
= 90º (each angle of rectangle is 90º)
So, area of the shaded region = 722
× 360
901010
= 7
530= 78.57 m2
Ex. Find the perimeter of figure, where is a
semi-circle and ABCD is a rectangle.
Sol. Perimeter of = r = 7
22× 7 = 22 cm.
So, perimeter of figure = 14 + 20 + 20 + 22 = 76 cm.
Ex. A metallic sheet is of the rectangular shape with
dimension 48 cm 36cm. From each one of corners,
a square of 8 cm is cut off. An open box is made of the
remaining sheet. Find the volume of the box.
Sol. In order to make an open box, a square of side 8 cm is
cut off from each of the four corners and the flaps are
folded up.
Thus, the box will have the following dimensions :
8 cm 8 cm
8 cm 8 cm
48cm
Length = (48 � 8 � 8) cm = 32 cm,
Breadth = (36 � 8 � 8) cm = 20 cm,
Height = 8 cm
Volume of the box formed = (32 20 8) cm3
= 5120 cm3
Ex.10. If v is the volume of a cuboid of dimensions a, b, andc and s is its surface area, then prove that
c1
b1
a1
s2
v1
.
Sol. L.H.S. = v1
= abc
1... (i)
R.H.S. =
c1
b1
a1
s2
= )cabcab(22
abc
abcabc
= abc
1... (ii)
By compairing (1) and (2)
c
1
b
1
a
1
s
2
v
1.
Ex. By melting a solid cylindrical metal, a few conicalmaterials are to be made. If three times the radius ofthe cone is equal to twice the radius of the cylinder andthe ratio of the height of the cylinder and the height ofthe cone is 4 : 3, find the number of cones which canbe made.
Sol. Let R be the radius and H be the height of the cylinderand let r and h be the radius and height of the conerespectively. Then,3r = 2Rand H : h = 4 : 3 ....(i)
34
hH
3H = 4h ....(ii)Let n be the required number of cones which can bemade from the materials of the cylinder. Then, thevolume of the cylinder will be equal to the sum of thevolumes of n cones. Hence, we have
R2H = 3n r2h
3R2H = nr2h
n = hr3
4h
4
9r3
hr
H3R2
2
2
2
[From (i) and (ii), R = 2r3
and H = 3h4
]
n = 43
493
n = 9.Hence, the required number of cones is 9.
27PAGE # 27
Ex. The base diameter of a solid in the form of a cone is 6cm and the height of the cone is 10 cm. It is melted andrecast into spherical balls of diameter 1 cm. Find thenumber of balls.
Sol. Let the number of spherical balls be n. Then, thevolume of the cone will be equal to the sum of thevolumes of the �n� spherical balls. The radius of the
base of the cone = 26
cm = 3 cm
and the radius of the sphere = 21
cm.
Now, the volume of the cone
= 31
× 32 × 10 cm3 = 30 cm3
and, the volume of each sphere
= 34
3
21
cm3 =
6
cm3
Hence, we have
n6
= 30
n = 6 × 30 = 180
Hence, the required number of balls = 180.
Ex. A conical empty vessel is to be filled up completely bypouring water into it successively with the help of acylindrical can of diameter 6 cm and height 12 cm. Theradius of the conical vessel is 9 cm and its height is 72cm. How many times will it require to pour water intothe conical vessel to fill it completely, if, in each time,the cylindrical can is filled with water completely?
Sol. Let n be the required number of times. Then, the volumeof the conical vessel will be equal to n times the volumeof the cylindrical can. Now, the volume of theconical vessel
= 31
× 92 × 72 cm3 = 24 × 81 cm3
And the volume of the cylindrical can= × 32 × 12 cm3 = 9 × 12 cm3
Hence, 24 × 81 = 9 × 12 × n
n = 1298124
= 18
Hence, the required number of times = 18.