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Resonance Pre-foundation Programmes

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COURSE : OLYMPIAD

MATHEMATICS

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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


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


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


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.


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


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


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


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


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


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


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.

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