X CBSE Maths Sem 01.pdf

Table of contents

Table of Contents CHAPTER 01: REAL NUMBERS.............................................................................. 1 CHAPTER MAP: ........................................................................................................................................... 1 Introduction: ....................................................................................................................................... 1 Euclid’s Division Lemma: .................................................................................................................. 1 Solved Examples 1.1: .............................................................................................................. 2 Unsolved Exercise 1.1: ............................................................................................................ 3 The Fundamental Theorem of Arithmetic:......................................................................................... 3 Solved Examples 1.2: .............................................................................................................. 4 Unsolved Exercise 1.2: ............................................................................................................ 5 Irrational Numbers: ............................................................................................................................ 6 Solved Examples 1.3: .............................................................................................................. 7 Unsolved Exercise 1.3: ............................................................................................................ 8 Rational Numbers and their Decimal Expansions: ............................................................................ 8 Unsolved Exercise 1.4: ............................................................................................................ 8 Miscellaneous Exercise: .......................................................................................................... 9 Multiple Choice Questions: .................................................................................................... 11 Column Matching Questions: ................................................................................................ 13 Answers to Unsolved Exercise: ............................................................................................. 14

CHAPTER 02: POLYNOMIALS .............................................................................. 16 CHAPTER MAP: ......................................................................................................................................... 16 INTRODUCTION: ......................................................................................................................................... 16 Geometrical Meaning of the Zeroes of a Polynomial: ..................................................................... 17 Solved Example 2.1: .............................................................................................................. 18 Unsolved Exercise 2.1: .......................................................................................................... 19 Relationship between Zeroes and Coefficients of a Polynomial: .................................................... 19 Solved Examples 2.2: ............................................................................................................ 20 Unsolved Exercise 2.2: .......................................................................................................... 21 Division Algorithm for Polynomials: ............................................................................................ 22 Solved Examples 2.3: ............................................................................................................ 22 Unsolved Exercise 2.3: .......................................................................................................... 23 Miscellaneous Exercise: ........................................................................................................ 24 Multiple Choice Questions: .................................................................................................... 26 Column Matching Questions: ................................................................................................ 27 Answer to Unsolved Exercise: ............................................................................................... 29

CHAPTER 03: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES .................. 31 CHAPTER MAP: ......................................................................................................................................... 31 BASIC FUNDAMENTALS: ............................................................................................................................ 31 GENERAL FORM OF A LINEAR EQUATION IN TWO VARIABLES: ..................................................................... 31 Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

Solved Exercise 3.1: .............................................................................................................. 33 Unsolved Exercise 3.1: .......................................................................................................... 33 Pair of Linear Equations in Two Variables: ..................................................................................... 34 Solution of a Pair of Linear Equation in Two Variables .............................................................. 34 Graphical Method of solving a pair of Linear Equations: ................................................................ 35 Solved Examples 3.2: ............................................................................................................ 36 Unsolved Exercise 3.2: .......................................................................................................... 37 Algebraic Methods of solving a pair of Linear Equations: ............................................................... 39 Solved Exercise 3.3: .............................................................................................................. 39 Unsolved Exercise 3.3: .......................................................................................................... 40 Solved Exercise 3.4: .............................................................................................................. 42 Unsolved Exercise 3.4: .......................................................................................................... 43 Cross Multiplication Method: ...................................................................................................... 43 Solved Examples 3.5: ............................................................................................................ 44 Unsolved Exercise 3.5: .......................................................................................................... 46 Equations Reducible to a Pair of Linear Equations in Two Variables ........................................ 47 Solved Exercise 3.6: .............................................................................................................. 47 Unsolved Exercise 3.6: .......................................................................................................... 48 WORD PROBLEMS:.................................................................................................................................... 49 Solved Examples 3.7: ............................................................................................................ 49 Unsolved Exercise 3.7: .......................................................................................................... 50 Miscellaneous Exercise: ........................................................................................................ 52 Multiple Choice Questions: .................................................................................................... 55 Column Matching Questions: ................................................................................................ 57 Answers to Unsolved Exercises: ........................................................................................... 58

CHAPTER 06: TRIANGLES .................................................................................... 62 CHAPTER MAP: ......................................................................................................................................... 62 Congruent Figures:.......................................................................................................................... 62 Unsolved Exercise 6.1: .......................................................................................................... 63 SIMILARITY OF TRIANGLES: ....................................................................................................................... 63 Congruence of Triangles: ................................................................................................................ 63 Theorem 6.1 Basic Proportionality Theorem (Thales Theorem):............................................... 64 Solved Example 6.2: .............................................................................................................. 65 Unsolved Exercise 6.2: .......................................................................................................... 66 CRITERIA FOR SIMILARITY OF TRIANGLES: ................................................................................................. 68 Tests for Similarity of Triangles: ...................................................................................................... 68 Theorem 6.3: A–A–A Criterion of Similarity ............................................................................... 68 Theorem 6.4: S–S–S Criterion of Similarity ............................................................................... 68 Theorem 6.5: S–A–S Criterion of Similarity ............................................................................... 68 Solved Example 6.3: .............................................................................................................. 69 Unsolved Exercise 6.3: .......................................................................................................... 70 Areas of Similar Triangles: .............................................................................................................. 72 Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Table of contents

Theorem 6.6: .............................................................................................................................. 72 Solved Example 6.4: .............................................................................................................. 73 Unsolved Exercise 6.4: .......................................................................................................... 73 PYTHAGORAS THEOREM: .......................................................................................................................... 75 Similarity in Right angled Triangles: ........................................................................................... 75 Theorem 6.7: .............................................................................................................................. 75 Theorem (6.8) Pythagoras Theorem: ......................................................................................... 75 Theorem 6.9: Converse of Pythagoras Theorem:...................................................................... 75 Solved Example 6.5: .............................................................................................................. 76 Unsolved Exercise 6.5: .......................................................................................................... 77 Proof of Theorems:.......................................................................................................................... 78 Theorem 6.7: Similarity in Right Angled Triangles ..................................................................... 81 Angle Bisector Property: ............................................................................................................ 81 Applications of Pythagoras Theorem: ............................................................................................. 82 Acute Angled Triangle: ............................................................................................................... 82 Obtuse Angled Triangle: ............................................................................................................ 83 Appollonius Principle: ................................................................................................................. 83 Miscellaneous Exercise: ........................................................................................................ 83 Multiple Choice Questions: .................................................................................................... 89 Column Matching Questions: ................................................................................................ 91 Answers to Unsolved Exercise: ............................................................................................. 93

CHAPTER 08: INTRODUCTION TO TRIGONOMETRY ......................................... 94 CHAPTER MAP: ......................................................................................................................................... 94 INTRODUCTION: ......................................................................................................................................... 94 What is Trigonometry? ............................................................................................................... 94 Use of learning trigonometry: ..................................................................................................... 94 Trigonometric Ratios: ...................................................................................................................... 94 Solved Examples 8.1: ............................................................................................................ 96 Unsolved Exercise 8.1: .......................................................................................................... 97 Trigonometric Ratio of some specific angles: ................................................................................. 99 Triangle Method: ...................................................................................................................... 102 Solved Examples 8.2: .......................................................................................................... 102 Unsolved Exercise 8.2: ........................................................................................................ 103 COMPLEMENTARY ANGLES: .................................................................................................................... 105 Trigonometric Inter-relationships:............................................................................................. 105 Solved Examples 8.3: .......................................................................................................... 106 Unsolved Exercise 8.3: ........................................................................................................ 106 TRIGONOMETRIC IDENTITIES: ................................................................................................................... 108 Solved Examples 8.4: .......................................................................................................... 109 Unsolved Exercise 8.4: ........................................................................................................ 111 Miscellaneous Exercise: ...................................................................................................... 112 Multiple Choice Questions: .................................................................................................. 116 Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

Column Matching Questions: .............................................................................................. 119 Answers to Unsolved Exercises: ......................................................................................... 120

CHAPTER 14: STATISTICS.................................................................................. 123 CHAPTER MAP: ....................................................................................................................................... 123 Calculation of Central tendencies for grouped data ...................................................................... 123 14.1 Mean of grouped data: ..................................................................................................... 123 Solved Examples 14.1: ........................................................................................................ 124 Unsolved Exercise 14.1: ...................................................................................................... 126 To find mean by Assumed mean method: ............................................................................... 127 To determine mean by step deviation method: ........................................................................ 127 Solved Examples 14.2: ........................................................................................................ 128 Unsolved Exercise 14.2: ...................................................................................................... 130 14.2 Mode of grouped data: .......................................................................................................... 132 Solved Examples 14.3: ........................................................................................................ 132 Unsolved Exercise 14.3: ...................................................................................................... 133 14.3 Median of Grouped Data: ...................................................................................................... 134 Solved Examples 14.4: ........................................................................................................ 135 Unsolved Exercise 14.4: ...................................................................................................... 137 Comparative Study: .................................................................................................................. 139 Graphical Representation of Cumulative frequency Distribution: ................................................. 139 Ogive of Less than type: .......................................................................................................... 139 Ogive of more than type: .......................................................................................................... 140 Solved Examples 14.5: ........................................................................................................ 141 Unsolved Exercise 14.5: ...................................................................................................... 142 Miscellaneous: ..................................................................................................................... 143 Multiple Choice Questions: .................................................................................................. 146 Answer to the Unsolved Exercise: ....................................................................................... 147

ANSWER TO THE MCQS: .................................................................................... 149

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 01: Real Numbers

1

Chapter 01: Real Numbers Chapter Map: → Euclid’s Division Lemma → The Fundamental Theorem of Arithmetic → Irrational Numbers → Rational Numbers and their Decimal Expansions

Introduction: ¾ An algorithm is a series of well defined steps which gives a procedure for solving a type of problem. ¾ A lemma is a proven statement used for proving another statement.

Euclid’s Division Lemma: Theorem 1.1 (Euclid’s Division Lemma): z

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.

z

Consider the numbers 455 and 42; 455 can be uniquely expressed as 455 = 42 × 10 + 35.

z

If we consider 24 and 6 it can be uniquely expressed as 24 = 6 × 4 + 0.

Note: Euclid’s division lemma is a technique to compute the Highest Common Factor (HCF) of two given positive integers. Recall that the HCF of two positive integers a and b is the largest positive integer d that divides both a and b.

Euclid’s division algorithm: z

Euclid’s Division lemma can be expressed in words as: Dividend = Divisor × Quotient + Remainder. To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below: Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d. Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r. Step 3: Continue the process till the remainder is zero. The divisor at the last stage will be the required HCF. „

z z z z

This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes the HCF of c and d etc. z

Eg: If we consider the numbers 420 and 272; 420 and 272 can be expressed as 420 = 272 × 1 + 148 Again consider 272 = 148 it can be expressed as 272 = 148 × 1 + 124 Again 148 = 124 × 1 + 24

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

1

2

Similarly 124 = 24 × 5 + 4 Finally 24 = 4 × 6 + 0 So HCF is 6 i.e. 6 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) = HCF (420, 272)

SOLVED EXAMPLES 1.1: 1) Use Euclid’s algorithm to find the HCF of 4052 and 12576. Sol: Step 1: Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 = 4052 × 3 + 420

2) Sol:

3) Sol:

Step 2: Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get 4052 = 420 × 9 + 272 Step 3: We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get 420 = 272 × 1 + 148 We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get, 272 = 148 × 1 + 124 We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get, 148 = 124 × 1 + 24 We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get, 124 = 24 × 5 + 4 We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get, 24 = 4 × 6 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4. Notice that 4 = HCF(24, 4) = HCF(124, 24) = HCF(148, 124) = HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052). Euclid’s division algorithm is not only useful for calculating the HCF of very large numbers, but also because it is one of the earliest examples of an algorithm that a computer had been programmed to carry out. Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer. Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r, for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1. If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1. What is the greatest number which divides each of the numbers 2261 and 2527 exactly? The greatest no. which divides each of the no. 2261 & 2527 exactly is the HCF of 2261 & 2527 By Euclid’s algorithm, 2527 = 2261 × 1 + 266 ⇒ 2261 = 266 × 8 + 133 ⇒ 266 = 133 × 2 + 0 ∴ 133 is the HCF of 2527 and 2261 HCF (2527, 2261) = 133

2

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 01: Real Numbers

3

UNSOLVED EXERCISE 1.1: CW Exercise: 1) Use Euclid’s division algorithm to find the HCF of: i) 135 and 225 ii) 196 and 38220 iii) 81 and 127 2) Show that any positive integer is of the form 3q, 3q + 1 or 3q + 2 where q is some integer. 3) Show that any odd positive integer p can be expressed in the form i) 6q + 1, or 6q + 3, or 6q + 5 ii) 4q + 1 or 4q + 3 4) Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [CBSE–08] 5) What is the greatest number which divides 209 and 1195 leaving remainder 5 in each case? 6) What is the largest number which when divides 63, 77 and 112 leaves 3, 5 and 4 as remainders respectively? 7) Find the HCF of 65 and 117 and express it in the form 65m + 117n. 8) An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? 9) Find the greatest number of 6 digits exactly divisible by 24, 15 and 36. HW Exercise: 1) Use Euclid’s division algorithm to find the HCF of: i) 867 and 255 ii) 3638 and 3587 2) Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. 3) A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose? 4) The length, breadth and height of a room are 8m 25 cm, 6m 75 cm and 4 m 50 cm, respectively. Determine the longest tape which can measure the three dimensions of the room exactly. 5) Show that any positive odd integer is of the form 6q + 1 or 6q + 5 or 6q + 3, where q is some integer. 6) Show that the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m. 7) If the HCF of 210 and 55 is expressible in the form 210 × 5 + 55y, find y. 8) If d is the HCF of 56 and 72, find x, y satisfying d = 56x + 72y. Also, show that x and y are not unique. 9) Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively. 10) Find the largest number that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively.

The Fundamental Theorem of Arithmetic: Theorem 1.2 (Fundamental Theorem of Arithmetic): z

Volume

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. Universal Tutorials – X CBSE (2012–13) – Mathematics

3

4 z

Eg: Consider the number 32760, it can be expressed as

z

32760 = 2 ×2 ×2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13 This prime factorization is unique apart from the order in which the prime factor occurs.

Note: In general, given a composite number x, we factorise it as x = p1p2 … pn, where p1,p2, …, pn are primes and written in ascending order, i.e. p1 ≤ p2 ≤ … ≤ pn. If we combine the same primes, we will get powers of primes. z z z

HCF of numbers is the Product of the smallest power of each common prime factor in the numbers. LCM of numbers is the Product of the greatest power of each prime factor, involved in the numbers. For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.

Note: For any positive integers a, b and c. a × b × c ≠ HCF(a, b, c) × LCM (a, b, c) LCM (p, q, r) =

p ⋅ q ⋅ r ⋅ HCF ( p, q, r ) p ⋅ q ⋅ r ⋅ LCM ( p, q, r ) ; HCF (p, q, r) = HCF ( p, q ) ⋅ HCF (q, r ) ⋅ HCF ( p, r ) LCM ( p, q ) ⋅ LCM (q, r ) ⋅ LCM ( p, r )

SOLVED EXAMPLES 1.2: 1) Consider the numbers 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. Sol: If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4n would contain the prime 5. This is not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n. So, there is no natural number n for which 4n ends with the digit zero. 2) Find the LCM and HCF of 6 and 20 by the prime factorisation method. Sol: We have: 6 = 21 × 31 and 20 = 2 × 2 × 5 = 22 × 51. You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in your earlier classes Note that HCF(6, 20) = 21 = Product of the smallest power of each common prime factor in the numbers = 2. LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers = 60. 3) Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM. Sol: The prime factorisation of 96 and 404 gives: 96 = 25 × 3, 404 = 22 × 101 Therefore, the HCF of these two integers is 22 = 4. 96 × 404 96 × 404 = = 9696. Also, LCM (96, 404) = HCF (96, 404) 4 4) Find the HCF and LCM of 6, 72 and 120, using the prime factorization method. Sol: We have: 6 = 2 × 3, 72 = 23 × 32, 120 = 23 × 3 × 5 Here, 21 and 31 are the smallest powers of the common factors 2 and 3 respectively. So, HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6 23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers. So, LCM (6, 72, 120) = 23 × 32 × 51 = 360 4

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 01: Real Numbers

5

UNSOLVED EXERCISE 1.2: CW Exercise: 1) Express each number as a product of its prime factors: i) 140 ii) 3825 iii) 372 iv) 9072 v) 462 2) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. i) 26 and 91 ii) 510 and 92 iii) 336 and 54 iv) 84,144 3) Find the LCM and HCF of the integers given below by applying the prime factorization method. Also verify that for 3 numbers LCM × HCF = product is true or not. i) 12, 15 and 21 ii) 7, 13, and 19 iii) 20, 18, and 75 iv) 52, 75 and 77 4) Check whether 6n can end with the digit 0 for any natural number n. Also count the number of zeros in the number given by 24n × 25n × 26n 5) Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. 6) Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively. 7) Answer the questions given below based on the information given about two numbers x, y: i) HCF(x, y) = 16 and x. y = 3072. Find LCM(x, y). ii) LCM(x, y) = 6, HCF(x, y) = 180 and x = 30 then find the value of y. 8) In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room the same number of participants is to be seated and all of them being in the same subject. 9) On CST–Thane route one cycle of journey for a local train is considered from CST to Thane and back to CST. A fast local takes 90 min. and slow local takes 120 min. to complete a cycle on CST–Thane route. A fast local and a slow local start together from CST at 10:00 am, at what time they will meet again at CST considering there is no time gap between the cycles of journey? 10) Four bells strike at intervals of 6, 8, 9 and 12 minutes. An alarm is set in a mobile phone such that it alarms after every hour. It was noticed that all the bells struck simultaneously when mobile phone alarmed at 10:00 am. At what time all the bells will strike together with the alarm? HW Exercise: 1) Express each number as a product of its prime factors: i) 156 ii) 5005 iii) 7429 iv) 19530 v) 6006 2) Find the LCM and HCF of the 336 and 54 and verify that LCM × HCF = product of the two numbers. 3) Find the LCM and HCF of the following integers by applying the prime factorization method. i) 17, 23 and 29 ii) 8, 9 and 25 iii) 24, 36 and 60 4) Given that HCF (306, 657) = 9, find LCM (306, 657). 5) Explain why 7 × 19 × 23 + 23 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 7 are composite numbers. 6) There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? 7) Consider the number 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. 8) Three sets of English, Hindi and Mathematics book have to be stacked in such a way that all the books are stored topic wise and the highest of each stack is the same. The number of English book is 96, the number of Hindi book is 240 and the number of Mathematics book is 336. Assuming that the books are of the same thickness, determine the number of stacks of English, Hindi and Mathematics books.

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

5

6

Irrational Numbers: ¾ Recall, a number ‘s’ is called irrational if it cannot be written in the form

p , where p and q are q

integers and q ≠ 0. Some examples of irrational numbers, with which you are already familiar, are: 2,

3,

15 , π, –

2 3

, 0, 10110111011110 …, etc.

Theorem 1.3: z z

Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. Proof : Let the prime factorisation of a be as follows: a = p1p2 . . . pn, where p1, p2, . . ., pn are primes, not necessarily distinct. Therefore, a2 = (p1p2 . . . pn)(p1p2 . . . pn) = p12 … pn2 .

z z z z z z

Now, we are given that p divides a2. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a2 are p1, p2, . . ., pn. So p is one of p1, p2, . . ., pn. Now, since a = p1 p2 . . . pn. Therefore, p divides a.

Theorem 1.4: 2 is irrational: z

Proof : Let us assume, to the contrary, that

2 is rational.

So, we can find integers r and s (≠ 0) such that

2 =

r s

Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get,

2 =

a , where a and b are co prime. b

So, b 2 = a. Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2. Now, by Theorem 1.3, it follows that 2 divides a. So, we can write a = 2c for some integer c. Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2. This means that 2 divides b2, and so 2 divides b(again using Theorem 1.3 with p = 2). Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that So, we conclude that

2 is rational

2 is irrational.

Note: The sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational. 6

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 01: Real Numbers

7

SOLVED EXAMPLES 1.3: 1) Prove that

3 is irrational.

[CBSE–08]

3 is rational.

Sol: Let us assume, to the contrary, that

a . b Suppose a and b have a common factor other than 1, then we can divide by the common factor,

That is, we can find integers a and b (≠ 0) such that

3 =

3b=a Squaring on both sides, and rearranging, we get 3b2 = a2. Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3. So, we can write a = 3c for some integer c. Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2. This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3) Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are co prime. This contradiction has arisen because of our incorrect assumption that 3 are rational.

and assume that a and b are co prime. So,

3 is irrational.

So, we conclude that 2) Show that 5 –

3 is irrational. 3 is rational.

Sol: Let us assume, to the contrary, that 5 –

That is, we can find co prime a and b (b ≠ 0) such that 5 – Therefore, 5 –

a = b

3 =

a b

3

Rearranging this equation, we get

3 =5–

Since a and b are integers, we get 5 – But this contradicts the fact that

a 5b − a = b b

a is rational, and so b

3 is rational.

3 is irrational.

This contradiction has arisen because of our incorrect assumption that 5 – So, we conclude that 5 –

3 is rational.

3 is irrational.

3) Show that 3 2 is irrational. Sol: Let us assume, to the contrary, that 3 2 is rational. That is, we can find co prime a and b (b ≠ 0) such that 3 2 = a . 3b a Since 3, a and b are integers, is rational, and so 3b

Rearranging, we get

a . b

2 =

But this contradicts the fact that

2 is rational.

2 is irrational.

So, we conclude that 3 2 is irrational. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

7

8

UNSOLVED EXERCISE 1.3: CW Exercise: 1) Prove that the following are irrationals: 1 ii) 7 5 i) 2 v) 3 + 2 5

vi) 5 3

vii)

HW Exercise: 1) Prove that the following are irrationals: 1 i) ii) 7 10 3 v) 5 – 2 3 [CBSE–08] vi) (3 –

iii) 3 +

5 )2

2 +

iii) 6 + vii)

2 [CBSE 08] iv)

5 [CBSE 08–09]

3

5

6 2 3

iv) viii)

6 45

Rational Numbers and their Decimal Expansions: Theorem 1.5: z

Let x be a rational number whose decimal expansion terminates then x can be expressed in p the form , where p and q are co prime and the prime factorisation of q is of the form 2n5m, q where n, m are non-negative integers.

Theorem 1.6: z

Let x =

p be a rational number, such that the prime factorization of q is of the form 2n5m, q

where n, m are non-negative integers then x has a decimal expansion which terminates.

Theorem 1.7: z

Let x =

p be a rational number, such that the prime factorization of q is not of the form q

2n5m, where n, m are non-negative integers then, x has a decimal expansion which is nonterminating repeating (recurring). Note: From theorem 1.5, 1.6 and 1.7, we can conclude that the decimal expansion of every rational number is either terminating or non-terminating repeating.

UNSOLVED EXERCISE 1.4: CW Exercise: 1) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 13 64 15 35 11 23 ii) iii) iv) 3 2 v) vi) i) 3125 455 1600 50 1000 2 5 2) Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

8

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 01: Real Numbers

9

3) The following real numbers have decimal expansions as given below. In each case, decide p whether they are rational or not. If they are rational, and of the form , what can you say about q the prime factors of q? i) 43.123456789 ii) 0.120120012000120000 . . . iii) 43.123456789 HW Exercise: 1) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 17 29 6 77 13 129 i) ii) iii) 2 7 5 iv) v) vi) 8 343 15 210 8000 2 5 7 2) Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. 3) The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p/q, what can you say about the prime factors of q? i) 2.12342543

ii) 3.01001000100001 . . .

iii) 25.912345678

MISCELLANEOUS EXERCISE: 1) Use Euclid’s algorithm to find the HCF of: i) 2527 and 1653 ii) 1261 and 442 iii) 576 and 252 2) Use Euclid’s algorithm to find the HCF of: i) 1320 and 935 ii) 1624 and 1276 3) Use Euclid’s algorithm to find the HCF of: i) 963 and 657 ii) 3638 and 3587 iii) 468 and 222 iv) 495 and 657 4) Show that any positive odd integer is of the form 8q + 1, 8q + 3, 8q + 5 or 8q + 7, where q is some integer. 5) Show that the square of any positive odd integer is of the form 8m + 1, for some integer m. [CBSE–09] 6) Find the greater number which will divides 3457 and 9375 leaving 6 and 8 as remainder respectively. 7) Find the greater number which will divide 410, 751 and 1030 so as to leave remainder 7 in each case. 8) Two masses of gold weighing 3318 and 3054 gram respectively are each to be made into medals of the same size. What is the weight of the largest possible medal? 9) Express each number as a product of its prime factor. i) 1560 ii) 3990 10) Find the HCF by prime factorisation method of: i) 81 and 17

ii) 225 and 450

11) Find the HCF by prime factorisation method of, 106, 159 and 265 12) Find the LCM of 45, 105 and 165 by finding the prime factors. 13) If HCF (12, 15) = 3. Find LCM of (12, 15). 14) Find the HCF of 96 and 404 by prime factorisation method. Hence, find their LCM. 15) Find the HCF and LCM of 6, 72 and 120, using prime factorisation method. 16) Find the HCF of 16 and 40 by prime factorisation method. Hence, find their LCM. 17) Find the smallest number which when divided by 25, 40 and 60 leaves remainder 7 is each case. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

9

10

18) Can two numbers have 14 as their HCF and 204 as their LCM. Give reasons in support of your answer. 19) Three horses run round a circular path, 1760 metres in circumference, at the rate of 440 m, 352 m and 264 m per minute. When will they again be together at the starting point? 20) In a morning walk three persons step off together. Their steps measures 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps? 21) Find the smallest length of a rope which can be measured exact number of times by three taps measuring 1 m, 20 cm, 75 cm and 1 m. 22) Telegraph poles occur at equal distances of 220 m along and heaps of stones are put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole? 23) Determine the number nearest to 110000 which is exactly divisible by each of 8, 15 and 21. 24) Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468. 25) A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48, 60 and 72 km a day, round the field. When will they meet again? 26) The LCM and HCF of two numbers are 180 and 6 respectively. If one of the numbers is 30, Find the other number. 27) Write down the decimal expansions of the following rational numbers by writing their denominators in the form 2m × 5n, where m, n are non–negative integers. 3 8

i)

ii)

13 125

iii)

7 80

iv)

14588 625

v)

129 2

2 × 57

28) Examine each of the following as rational or irrational i) 5 +

5

⎛ 1 ⎞ ⎟⎟ ii) ⎜⎜ 3 + 3⎠ ⎝

2

⎛ 1 ⎞ ⎟⎟ iii) ⎜⎜ 3 + 2⎠ ⎝

2

iv)

15 6 5

29) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion. i)

7 1250

ii)

v)

24 729

vi)

15 24 13 25 5 3

iii)

32 455

iv)

vii)

3 8

viii)

18 4000 131 2 5 4 75 3

71 17 x) 630 1500 30) The following real numbers have decimal expansions as given below. In each case decide, p what can you say abut whether they are rational or not. If they are rational, and of the form q the prime factor of q?

ix)

i) 0.0875

ii) 0.130130013000130000

iii) 0. 142857

31) Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q. 32) Prove that the product of three consecutive positive integers is divisible by 6. 33) If the HCF of 408 and 1032 is expressible in the form 1032m – 408 × 5, find m. 34) If the HCF of 657 and 963 is expressible in the form 657x + 963 × – 15, find x. 10

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 01: Real Numbers

11

35) Find the largest number which divides 280 and 1245 leaving remainders 4 and 3, respectively. 36) What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively. 37) 105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip? 38) 15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets and how many pastries will each box contain? 39) A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10 ft. by 8 ft. What would be the size in inches of the tile required that has to be cut and how many such tiles are required? 40) Two brands of chocolates are available in pack of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy? 41) 144 cartons of Coke Cans and 90 cartons of Pepsi Cans are to be stacked in a Canteen. If each stack is of the same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have? 42) During a sale, colour pencils were being sold in pack of 24 each and crayons in packs of 32 each. If you want full pack of both and the same number of pencils and crayons, how many of each would you need to buy? 43) Find the greatest number of 6 digits exactly divisible by 24, 15 and 36. 44) A rectangular courtyard is 18 m 72 cm long and 13m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles. 45) What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case? 46) Prove that the following are irrational: i)

6

v) 2 +

ii)

3

1

iii)

5

vi) 3 –

5

iv)

11

vii) 7 + 3 2

3 –

2

[CBSE–09]

MULTIPLE CHOICE QUESTIONS: CW Exercise: 1) 21 .234 78 is a) an integer

b) a rational

c) an irrational

d) none of these.

8 is a) rational b) irrational c) none of these. 3) Which one of the following is an irrational number? a) x2 = 9 b) y2 = 64 c) z2 = 8 d) none of these. p 4) A rational number is terminating decimals only when prime factors of q are only: q 2)

a) 2 or 3

Volume

b) 3 or 5

c) 3 and 4

Universal Tutorials – X CBSE (2012–13) – Mathematics

d) 2 or 5.

11

12

13 is 8 a) 0.175 b) 0.625 c) 1.625 Which rational number is represented by 11.125 89 98 93 a) b) c) 8 4 8 The LCM of two numbers 26 and 91 is: a) 91 b) 182 c) 1183 The sum of a rational number and irrational number is always a) a rational number b) an irrational number c) an integer 2 If p is an even integer then p is an a) odd integer b) even integer c) multiple of 3

5) The decimal expansion of 6)

7) 8) 9)

d) 1.525. d)

88 9

d) 637. d) none of these. d) none of these.

10) 4 5 is

a) rational b) irrational c) not real d) none of these. 11) Every composite number can be expressed as a product of a) coprimes b) primes c) none of these. HW Exercise: 1) 0.101001000100001000001 is a) an irrational b) a rational c) an integer d) none of these. 2) The decimal representation of an irrational number is always a) terminating b) terminating, repeating c) non terminating, repeating d) non-terminating, non-repeating. 3) HCF (a, b) × LCM (a, b) = a) a + b b) a – b c) a × b d) none of these. 4) Every terminating decimal is a) an integer b) a rational c) an irrational d) none of these. 5) Which rational number is represented by 3.41 .

307 338 341 34 b) c) d) 90 99 900 990 The HCF of two numbers 867 and 255 is a) 17 b) 34 c) 51 d) 68. The LCM of three numbers is 28, 44, 132 is a) 528 b) 231 c) 462 d) 924. Circumference of the circle π= is Diameter of the circle a) an irrational b) a rational c) none of these. The decimal expansion of a rational number is always: a) non-terminating b) non-terminating and non repeated c) terminating or non-terminating repeated d) none of these. a)

6) 7) 8) 9)

10) The given number: 3 + 2 2 is an a) rational number b) irrational number c) not real. 11 11) The rational number is a 24 a) terminating decimal b) non-terminating repeating 12

Universal Tutorials – X CBSE (2012–13) – Mathematics

c) none of these. Volume

Chapter 01: Real Numbers

13

COLUMN MATCHING QUESTIONS: 1) Listed in column I are some statements. For each statement in column I. Applying Euclid’s division lemma choose all the correct options in column II. Column I

Column II

i)

A) 9m or 9m+1 for some integer m

The square of any positive integer is of the form

ii) Any positive odd integer is of the form

B) 6m+1 or 6m+3 or 6m+5 for some m

iii) The cube of any positive integer is of the form

C) 5m, 5m+1, 5m+4 for some integer m D) 3m or 3m+1 for some m E) 4m+1 or 4m+3 for some m

2) Given in column I are some types of real numbers. For each item in column I. Choose all the correct options in column II. Column I

Column II

i)

Irrational Number

A)

(5+ 3 ) (5– 3 )

ii)

Integer

B)

43 .12345

iii)

Rational Number

C) 43.1234567 D) (3– 5 ) (3+ 7 ) E)

18 2

3) Given in column I are some real numbers. For each item in column I. Choose all correct options in column II. Column I Column II i) ii) iii)

π

A) Terminating decimal representation B) Irrational

135 3

2 ×5 5

3

C) Rational number D) Non–terminating and non–repeating decimal representation E) Non–terminating and repeating decimal

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

13

14

4) Given in col. I are values of missing entries named x & y.Choose all correct options in column II Column I Column II i) x = 7, y = 2 A) 156 x

78 2

39 y

ii)

x = 2, y = 3

B)

13

240 2

120 2

60 2

30 y

15 x

iii)

x = 3, y = 2

C)

5

210 x

105 y

35 5

D)

7

462 3

154 y

77 x

E)

11

372 x

186 2

93 y

31

ANSWERS TO UNSOLVED EXERCISE: CW Exercise 1.1: 1) (i) 45 (ii) 196 (iii) 1 5) 34 6) 12 7) HCF=13, m=2, n –1 8) 8 columns HW Exercise 1.1: 1) (i) 51 ii) 17 3) 10 4) 75 7) y = –19 8) x = –68, y = 53 9) 64 10) 17 CW Exercise 1.2: 1) (i) 22 × 5 × 7 (ii) 32 × 52 × 17 (iii) 22 × 3 × 31 (iv) 24 × 34 × 7 (v) 3 × 2 × 7 × 11 14

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 01: Real Numbers

15

2) (i) 182, 13 (ii) 23460, 2 (iii) 3024, 6 (iv) 1008, 2 3) (i) 420, 3 (ii) 1729, 1 (iii) 900, 1 (iv) 300300, 1 4) 2n 6) 204 7) (i)192 (ii) 36 8) 21

9) 4:00 pm

10) 4:00 pm

HW Exercise 1.2: 1) (i) 22 × 3 × 13 (ii) 5 × 7 × 11 × 13 (iii) 17 × 19 × 23 (iv) 2 × 32 × 5 × 7 × 31 (v) 2 × 3 × 7 × 11 × 13 2) 3024, 6 3) (i) 11339, 1 (ii) 1800, 1 (iii) 360, 12 (iv) 4) 22338 6) 36 minutes 8) 2, 5, 7 CW Exercise 1.4: 1) (i, iii, iv, v, vi) Terminating (ii) Non–terminating 2) (i) 0.00416 (iii) 0.009375 (iv) 0.115 (v) 0.7 3) (i) Rational (ii) Non–rational (iii) Rational HW Exercise 1.4: 1) (i, iv, vi) Terminating (ii, iii, v) Non–terminating 2) (i) 2.125 (iv) 0.4 3) (i) Rational (ii) Non–rational (iii) Rational Miscellaneous: 1) (i) 19 (ii) 13 (iii) 36 2) (i) 55 (ii) 116 3) (i) 9 (ii) 17 (iii) 6 (iv) 45 6) 493 7) 31 8) 6 gm 17) 607 18) (i) Two numbers cannot have 14 as HCF and 204 as LCM (ii) Can never end with 0 19) After 20 minutes 20) After covering a distance of 12240 cm. from the staring point 21) 6 metres 22) 3300 m from first telegraph pole 23) 110040 24) 4663 25) 60 days 26) 36 28) i, iii, iv) Irrational (ii) Rational 29) i, ii, iv, vi, vii) Terminating iii, v, viii, ix, x) Non-terminating repeating 30) i) Rational, prime factors of q will be either 2 or 5 or both only (ii) Not rational (iii) Rational prime factors of q will also have a factor other than 2 or 5. 33) 2 34) 22 35) 138 36) 625 37) 35 38) 4 Biscuit packets, 5 Pastries 39) 24 inches, 20 tiles 40) 8 of second kind, 5 of 1st kind 41) 18 42) 4 packets of Pencils, 3 packets of Crayons 43) 999720 44) 4290 45) 3647 Column Matching Question: 1) i–CD; ii–BE; iii–A 3) i–BD; ii–AC; iii–BD

Volume

2) i–D; ii–AE; iii–ABEC 4) i–D; ii–ACE; iii–B

Universal Tutorials – X CBSE (2012–13) – Mathematics

15

16

Chapter 02: Polynomials Chapter Map: → Introduction → Geometrical meaning of Zeroes of a Polynomial → Relationship between zeroes and coefficients of a Polynomial → Division Algorithm for Polynomials

Introduction: Degree of a Polynomial: z

If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x).

Linear polynomial: z

A polynomial of degree 1 is called a linear polynomial.

z

For example, 2x – 3,

z

General form of linear polynomial is ax + b, where a, b are real numbers and a ≠ 0.

3 x + 5, y +

2,x–

2 2 , 3z + 4, u + 1 etc. are linear polynomials. 11 3

Quadratic polynomial: z z z

z

A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’. 2 2 u 2 1 , y – 2, 2 – x2 + 3 x, − 2u2 + 5, 5 v2 – v, 4z2 + are some examples 5 3 3 7 of quadratic polynomials (whose coefficients are real numbers). Generally, any quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers and a ≠ 0.

2x2 + 3x –

Cubic polynomial: z

A polynomial of degree 3 is called a cubic polynomial.

z

Some examples of a cubic polynomial are 2 – x3, x3,

z

In fact, the most general form of a cubic polynomial is ax3 + bx2 + cx + d, where, a, b, c, d are real numbers and a ≠ 0.

3

2

3

3

2

2 x , 3 – x + x , 3x – 2x + x – 1.

Value of a polynomial: z z z

16

If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k). Consider the polynomial P(x) = x2 – 3x – 4 value of the polynomial at x = 3 is denoted by P(3) and P(3) = 32 – 3(3) – 4 = –4. Value of the above polynomial at x = –1 is given by P(–1) = (–1)2 – 3(–1) – 4 = 0.

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 02: Polynomials

17

Zero of a polynomial: z z z z

z

A real number k is said to be a zero of a polynomial p(x), if p(k) = 0. If x – a is a factor of the polynomial p(x), then p(a) = 0 If x + a is a factor of the polynomial p(x), then p(–a) = 0 Consider the polynomial P(x) = x2 – 5x + 6 P(2) = 22 – 5(2) + 6 = 0 and P(3) = 32 – 5(3) + 6 = 0 As P(2) = 0 and P(3) = 0, 2 and 3 are called the zeros of the polynomial x2 – 5x + 6.

Geometrical Meaning of the Zeroes of a Polynomial: ¾ ¾ ¾ ¾

Now, let us look for the geometrical meaning of a zero of a quadratic polynomial. Consider the quadratic polynomial x2 – 3x – 4. Let us see what the graph of y = x2 – 3x – 4 looks like. Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x as given in Table. –2 –1 0 1 2 3 4 5 x 2 6 0 –4 –6 –6 –4 0 6 y = x – 3x – 4 (–2,6)

(5, 6)

6 5 4 3 2

(–1,0)

Note from the figure that the curve intersect x–axis at the points –1 and 4. Thus the zeros of the Polynomial x2 – 3x – 4 are –1 and 4.

1

–3 –2 –1 0 –1

(4,0) 1

2

3

4

5

–2 –3 –4

(0,–4)

(3,–4)

–5 –6 (1,–6)

(2,–6)

¾ From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen: ¾ Case (i) : Here, the graph cuts x-axis at two distinct points A and A′. ¾ The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see given below). Y

A X′

0

Volume

A′

A′ X

Y′ (i)

Y

X′

A 0

X

Y′ (ii)

Universal Tutorials – X CBSE (2012–13) – Mathematics

17

18 z

z

Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e. at two coincident points. So, the two points A and A′ of Case (i) coincide here to become one point A (see Fig. given below). The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case. Y

Y

A 0

X′

Y′

z

X

Y′ (ii)

(i) z

A

0

X′

X

Case (iii) : Here, the graph is either completely above the x-axis or completely below the xaxis. So, it does not cut the x-axis at any point (see Fig. given below). So, the quadratic polynomial ax2 + bx + c have no zero in this case. Y

Y

0

X′

0

X′

X

Y′

X

Y′ (ii)

(i)

Number of Zeros of a Polynomial: z z

Given a polynomial P(x) of degree n, the graph of y = P(x) intersects the x–axis at most n points. Therefore a polynomial P(x) of a degree n has at most n zeros. That is, a quadratic polynomial can have at most two zeros and so on.

SOLVED EXAMPLE 2.1: 1) Look at the graphs in Fig. given below. Each is the graph of y = p(x), where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x). Y

X′

0

X

Y′ (i)

18

Y

X′

0

Y

X

X′

Y′ (ii)

Universal Tutorials – X CBSE (2012–13) – Mathematics

0

X

Y′ (ii)

Volume

Chapter 02: Polynomials

19 Y

Y

0

X′

0

X′

X

Y

Y′ (iv)

0

X′

X

Y′ (v)

X

Y′ (vi)

Sol: i) The number of zeroes is 1 as the graph intersects the x-axis at one point only. ii) The number of zeroes is 2 as the graph intersects the x-axis at two points. iii) The number of zeroes is 3. iv) The number of zeroes is 1. v) The number of zeroes is 1. vi) The number of zeroes is 4.

UNSOLVED EXERCISE 2.1: 1) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. Y

X′

0

Y

X

X′

Y′

0

Y

0

Y′ (iv)

X

X′

Y′ (ii)

(i)

X′

Y

0

Y′ (ii)

Y

X

X′

0

X

Y

X

X′

Y′ (v)

0

X

Y′ (vi)

Relationship between Zeroes and Coefficients of a Polynomial: ¾ If α and β are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c, a ≠ 0, then you know that x – α and x – β are the factors of p(x). ¾ Therefore, ax2 + bx + c = k(x – α) (x – β), where k is a constant

= k[x2 – (α + β)x + αβ] = kx2 – k(α + β)x + kαβ = k[x2 – (sum of zero)x + (Product of zero)] Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

19

20

¾ Comparing the coefficients of x2, x and constant terms on both the sides, we get a = k, b = – k(α + β) and c = kαβ. b c − ¾ This gives α + β = a , αβ = a ¾ i.e., sum of zeroes = α + β = − ¾ Product of zeroes = αβ =

b −(Coefficien t of x) = , a Coefficien t of x 2

c Consta nt term = a Coefficien t of x 2

In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial b c d ax3 + bx2 + cx + d, then α + β + γ = − , αβ + βγ + γα = , α β γ = − a a a

SOLVED EXAMPLES 2.2: 1) Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.

Sol: We have, x2 + 7x + 10 = (x + 2)(x + 5) So, the value of x2 + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0 i.e., when x = – 2

or

x = –5

2

∴ the zeroes of x + 7x + 10 are – 2 and – 5. −(7) −(Coefficien t of x) = 1 Coefficien t of x 2

Now, sum of zeroes = –2 + (–5) = – (7) = Product of zeroes = (–2) × (–5) = 10 =

10 Consta nt term = 1 Coefficien t of x 2

2) Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients.

Sol: Recall the identity a2 – b2 = (a – b)(a + b) Using it, we can write: x2 – 3 = (x –

3 )(x +

So, the value of x2 – 3 is zero when x = Therefore, the zeroes of x2 – 3 are Now, sum of zeroes =

3 –

3)

3 or x = – 3

3 and − 3

3 =0=

−(Coefficien t of x) Coefficien t of x 2

Product of zeroes = ( 3 )(– 3 ) = –3 = −

3 Consta nt term = 1 Coefficien t of x 2

3) Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2, respectively.

Sol: Let the quadratic polynomial be ax2 + bx + c, and its zeroes be α and β. We have, α + β = – 3 = −

b a

and

αβ = 2 =

c . a

If a = 1, then b = 3 and c = 2. So, one quadratic polynomial which fits the given conditions is x2 + 3x + 2. 20

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 02: Polynomials

21 2

4) If α and β are zero of the Quadratic Polynomial f(x) = ax + bx + c, then evaluate (i) α + β2 α β (ii) + . β α

Sol: α and β are zero of ax2 + bx + c

∴α+β=– 2

⎛ b⎞ ⎛c ⎞ i) α2 + β2 = (α + β)2 – 2αβ = ⎜ − ⎟ – 2 ⎜ ⎟ a ⎝ ⎠ ⎝a⎠

∴ α2 + β2 =

2

b c , αβ = a a

b 2 − 2ac a2

b 2 − 2ac 2

ii)

α α +β β + = β α αβ

2

a2 c a

=

=

b 2 − 2ac ac

UNSOLVED EXERCISE 2.2: CW Exercise 1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. i) x2 – 2x – 8 ii) 4s2 – 4s + 1 iii) 3x2 – x – 4 iv) 7z2 – 343 2 v) 5p + 20p 2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 1 1 1 ii) – , iii) 4, 1 i) 2 , 3 4 4

3) Find the zeros of the polynomial f(x) = 4 3 x 2 + 5 x − 2 3 , and verify the relationship between the zeros and its coefficients. 4) Find the zeros of the quadratic polynomial f(x) = abx2 + (b2 – ac) x – bc, and verify the relationship between the zeros and its coefficients. 5) If α and β are the zeros of the quadratic polynomial f(x) = x2 – px + q, then find the values of 1 1 ii) + i) α2 + β2 α β 6) If α and β are the zeros of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate: i) α2 + β2

ii)

β α + α β

iii)

α2 β2 + β α

7) If α and β are the zeros of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate: i) α4 + β4

ii)

α2 β

2

+

β2 α2

8) If α and β are the zeros of the quadratic polynomial f(x) = kx2 + 4x + 4 such that α2 + β2 = 24, find the value of k. 9) If α and β are the zero of the quadratic polynomial f(x) = x2 – x – 2, find a polynomial whose zero are 2α + 1 and 2β + 1. HW Exercise: 1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. i) 6x2 – 3 – 7x [CBSE 08] ii) 4u2 + 8u iii) t2 – 15 Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

21

22

2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 1 ii) 0, 5 iii) 1, 1 i) , –1 4 1 1 3) If α and β are the zeros of the quadratic polynomial f(x) = x2 – x – 4, find the value of + − αβ . α β 4) If α & β are the zeros of the quadratic polynomial P(x) = 4x2 –5x – 1, find the value of α2β + αβ2. 5) If α & β are the zeros of the polynomial f(x) = x2 – 5x + k such that α – β = 1, find the value of k. 21 6) If α, β are the zeros of the polynomial f(x) = 2x2 + 5x +k satisfying the relation α2 + β2 + αβ = , 4 then find the value of k for this to be possible. 7) If sum of the square of the zeros of the quadratic polynomial f(x) = x2 – 8x + k is 40, find the value of k. 8) If α and β are the zeros of the quadratic polynomial f(x) = 2x2 – 5x + 7, find a polynomial whose zeros are 2α + 3β and 3α + 2β.

Division Algorithm for Polynomials: If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x). This result is known as the Division Algorithm for polynomials. i.e. Dividend = Divisor × Quotient + Reminder

SOLVED EXAMPLES 2.3: 3

1) Divide 3x + x + 2x + 5 by 1 + 2x + x2. Sol: We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Recall that arranging the terms in this order is called writing the polynomials in standard form. In this example, the dividend is already in standard form, and the divisor, in standard form, is x2 + 2x + 1. Step 1: To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e. 3x3) by the highest degree term of the divisor (i.e. x2). This is 3x. Then carry out the division process. What remains is – 5x2 – x + 5. Step 2: Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend (i.e. –5x2) by the highest degree term of the divisor (i.e. x2). This gives –5. Again carry out the division process with –5x2 – x + 5. Step 3: What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the divisor x2 + 2x + 1. So, we cannot continue the division any further. So, the quotient is 3x – 5 and the remainder is 9x + 10. Also, (x2 + 2x + 1) × (3x – 5) + (9x + 10) = 3x3 + 6x2 + 3x – 5x2 – 10x – 5 + 9x + 10 = 3x3 + x2 + 2x + 5 Here again, we see that, Dividend = Divisor × Quotient + Remainder 2) Divide 3x2 – x3 – 3x + 5 by x – 1 – x2, and verify the division algorithm. Sol: Note that the given polynomials are not in standard form. To carry out division, we first write both the dividend and divisor in decreasing orders of their degrees. So, dividend = –x3 + 3x2 – 3x + 5 and divisor = –x2 + x – 1. Division process is shown on the right side. We stop here since degree (3) = 0 < 2 = degree (–x2 + x – 1). So, quotient = x – 2, remainder = 3. 22

2

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 02: Polynomials

23 2

Now, Divisor × Quotient + Remainder = (–x + x – 1) (x – 2) + 3 = –x3 + x2 – x + 2x2 – 2x + 2 + 3 = –x3 + 3x2 – 3x + 5 = Dividend In this way, the division algorithm is verified. 3) Find all the zeroes of 2x4 – 3x3 – 3x2 + 6x – 2, if you know that two of its zeroes are

2

[CBSE–08]

and – 2 . 2

Sol: Since two zeroes are 2 and − 2 , (x – 2 )(x + 2 ) = x – 2 is a factor of the given polynomial. Now, we divide the given polynomial by x2 – 2. First term of quotient is

2x 4 = 2x2 x2

Second term of quotient is

− 3x 3 = –3x x2

x2 =1 x2 So, 2x4 – 3x3 – 3x2 + 6x – 2 = (x2 – 2)(2x2 – 3x + 1). Now, by splitting –3x, we factorise 2x2 – 3x + 1 as (2x – 1)(x – 1).

Third term of quotient is

So, its zeroes are given by x =

1 2

and

x = 1.

Therefore, the zeroes of the given polynomial are

2,– 2,

1 and 1. 2

UNSOLVED EXERCISE 2.3: CW Exercise: 1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2 2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2 i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12 4 3 3) The graph of a polynomial f(x) = 3x + 6x – 2x2 – 10x – 5, intersects x axis at four different ⎛ 5 ⎞ ⎛ 5 ⎞⎟ , 0 ⎟ and ⎜ − ,0 points P, Q, R and S. If the co-ordinates of points P and Q are ⎜ ⎜ 3 ⎟ ⎜ 3 ⎟⎠ ⎝ ⎠ ⎝ respectively then find the co-ordinates of R and S. 4) Polynomial f(x) = x3 – 2x – 4 on dividing by another polynomial g(x) gives equal quotient and remainder. Find out the value of g(x) if the remainder of the division is x-2. 5) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and i) deg p(x) = deg q(x) ii) deg q(x) = deg r(x) iii) deg r(x) = 0 6) Apply the division algorithm to find the quotient remainder on dividing f(x) by g(x) as given below. ii) f(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x i) f(x) = x3 – 6x2 + 11x – 6, g(x) = x + 2 7) By applying division algorithm prove that the polynomial g(x) = x2 + 3x + 1 is a factor of the polynomial f(x) = 3x4 + 5x3 – 7x2 + 2x + 2. 8) What must be subtracted from 8x4 + 14x3 – 2x2 + 7x – 8 so that the resulting polynomial is exactly divisible by 4x2 + 3x – 2. 9) Find the values of a and b so that x4 + x3 + 8x2 + ax + b is divisible by x2 + 1. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

23

24

10) A polynomial f(x) = x4 – 3x3 + 6x – 4 is factorized into three polynomials such that f(x) = p(x).q(x).r(x). If p(x) = x2 – 3x + 2 and q(x) = x – 2 , then find r(x). 11) For which value of a, (x + a) is a factor of 2x2 + 2ax + 5x + 10? HW Exercise: 1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in p(x) = x4 – 5x + 6, g(x) = 2 – x2. 2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1. [CBSE–08] 3) Obtain all other zeroes of x4 + x3 – 34x2 – 4x + 120, if two of its zeroes 2 and – 2. 4) On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x). 5) If (x – 2) is a factor of polynomial x3 + ax2 + bx + 16 and a – b = 6 then find the value of a and b. 6) Apply the division algorithm to find the quotient remainder on dividing f(x) by g(x) as given below. ii) f(x) = x4 – 5x + 6, g(x) = 2 – x2 i) f(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2 4 3 2 7) What must be added to f(x) = 4x + 2x – 2x + x – 1 so that the resulting polynomial is divisible by g(x) = x2 + 2x – 3. 8) If the polynomial f(x) = x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.

MISCELLANEOUS EXERCISE: 1) Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: 1 ii) x3 – 4x2 + 5x – 2; 2, 1, 1 i) 2x3 + x2 – 5x + 2; , 1, – 2 2 2) Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively. 3) If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b. 4) If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± 3 find other zeroes. 5) If the polynomial 6x4 + 8x3 + 17x2 + 21x + 7 is divided by another polynomial 3x2 + 4x + 1, the remainder comes out to be (ax + b), find a and b. [CBSE 09] 6) Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients: i) f(x) = x2 – 2x – 8 ii) g(s) = 4s2 – 4s + 1 iii) h(t) = t2 – 15

iv) p(x) = x2 + 2 2 x – 6

v) q(x) = 3 x2 + 10x + 7 3 vi) f(x) = x2 – ( 3 + 1)x + 3 vii) g(x) = a(x2 + 1) – x(a2 + 1) 7) If α and β are the zeros of the quadratic polynomials f(x) = ax2 + bx + c, then evaluate: i) α – β

ii)

1 1 − α β

iii)

1 1 + − 2αβ α β

iv) α2β + αβ2

8) If α and β are the zeros of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of 9) If α and β are the zeros of the quadratic polynomial f(x) = x2 + x – 2, find the value of

α β + . β α

1 1 − . α β

10) If α and β are the zeros of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of 1 1 + – 2αβ. α β 24

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 02: Polynomials

25 2

4 3

3 4

11) If α and β are the zeros of the quadratic polynomial f(t) = t – 4t + 3, find the value of α β + α β . 12) If α and β are the zeros of the quadratic polynomial p(y) = 5y2 – 7y + 1, find the value of

1 1 + . α β

13) If the sum of the zeros of the quadratic polynomial f(t) = kt2 + 2t + 3k is equal to their product, find the value of k. 14) If one zero of the quadratic polynomial f(x) = 4x2 – 5kx – 9 is negative of the other, find the value of k. 15) If α and β are the zeros of the quadratic polynomial f(x) = x2 – 1, find a quadratic polynomial 2β 2α whose zeros are and . α β 16) Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case: i) f(x) = 2x3 + x2 – 5x + 2; ½, 1, –2 ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1 17) Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and product of its zeros as 3, –1 and –3 respectively. 18) Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following: i) f(x) = x3 – 6x2 + 11x – 6, g(x) = x2 + x + 1 ii) f(x) = 10x4 + 17x3 – 62x2 + 30x – 3, g(x) = 2x2 + 7x + 1 iii) f(x) = 4x4 + 8x + 8x2 + 7, g(x) = 2x2 – x + 1 iv) f(x) = 15x3 – 20x2 + 13x – 12, g(x) = 2 – 2x + x2 19) Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm: i) g(t) = t2 – 3, f(t) = 2t4 + 3t3 – 2t2 – 9t – 12 ii) g(x) = x3 – 3x + 1, f(x) = x5 – 4x3 + x2 + 3x + 1 iii) g(x) = 2x2 – x + 3, f(x) = 6x5 – x4 + 4x3 – 5x2 – x – 15 20) Obtain all zeros of the polynomial f(x) = 2x4 + x3 – 14x2 – 19x – 6, if two of its zeros are –2 & –1. 21) Obtain all zeros of f(x) = x3 + 13x2 + 32x + 20, if one of its zeros is –2. 22) Obtain all zeros of the polynomial f(x) = x4 – 3x3 – x2 + 9x – 6, if two of its zeros are – 3 &

3.

3 3 & . 2 2 24) What must be added to the polynomial f(x) = x4 + 2x3 – 2x2 + x – 1 so that the resulting polynomial is exactly divisible by x2 + 2x – 3? 25) What must be subtracted from the polynomial f(x) = x4 + 2x3 – 13x2 – 12x + 21 so that the resulting polynomial is exactly divisible by x2 – 4x + 3?

23) Find all zeros of the polynomial f(x) = 2x4 –2x3 – 7x2 + 3x + 6, if two of its zeros are –

26) Find all the zeroes of the polynomial x3 + 3x2 – 2x – 6, if two of its zeroes are – 2 and 2 . [CBSE 09] 27) If the polynomial 6x4 + 8x3 – 5x2 + ax + b is exactly divisible by the polynomial 2x2 – 5, then find the values of a and b. [CBSE 09] 28) If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is reciprocal of the other, find the value of a. [CBSE 08] 2 [CBSE 08] 29) If the product of zeroes of the polynomial ax – 6x – 6 is 4, find the value of ‘a’. 30) Find the condition which must be satisfied by the coefficient of the polynomial f(x) = x5 – px2 + qx – r when the sum of its two zeros is zero. 31) Find the value of constant k if two of the zeros of the polynomial f(x) = x3 – 3x2 – 4x + k, are equal in magnitude but opposite in sign. 32) Find the zeros of the polynomial f(x) = x3 – 12x2 + 39x – 28, if it is given that the zeros are in A.P. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

25

26

33) Find the value of p if three consecutive odd integers a, b and c are zeros of the polynomial f(x) = x3 – 15x2 + 71x + p. 34) Find the condition that the zeros of the polynomial f(x) = x3 – px2 + qx – r may be in arithmetic progression. 35) Find the zeros of the polynomial f(x) = x3 – 5x2 – 2x + 24, if it is given that the product of its two zeros is 12. 36) If the zeros of the polynomial f(x) = x3 – 3x2 + x + 1 are a– b, a, a + b, find a and b.

MULTIPLE CHOICE QUESTIONS: CW Exercise: 1) If α, β are the zeros of the polynomial f(x) = x2 + x + 1, then

1 1 + = α β

a) 1 b) –1 c) 0 d) None of these 2) If one zero of the polynomial f(x) = (k2 + 4)x2 + 13x + 4k is reciprocal of the other, then k = a) 2 b) –2 c) 1 d) –1 1 3) If α and β are the zeros of the polynomial f(x) = x2 + px + q, then a polynomial having α 1 is its zeros is and β 4)

5)

6)

7)

8)

9)

10)

a) x2 + qx + p b) x2 – px + q c) qx2 + px + 1 d) px2 + qx + 1 If α, β are the zeros of the polynomial f(x) = x2 – p (x + 1) – c such that (α + 1) (β + 1) = 0, then c = a) 1 b) 0 c) –1 d) 2 3 2 If the product of zeros of the polynomial f(x) = ax – 6x + 11x – 6 is 4, then a = 3 3 2 2 a) b) – c) d) – 2 2 3 3 If one root of the polynomial f(x) = 5x2 + 13x + k is reciprocal of the other, then the value of k is 1 a) 0 b) 5 c) d) 6 6 If x = 2 and x = 3 are zeros of the quadratic polynomial x2 + ax + b, the values of a and b respectively are a) 5, 6 b) –5, –6 c) –5, 6 d) 5, 6. 3 2 If f(x) = 4x – 6x + 5x – 1 and α, β and γ are its zeros then αβγ is equal to: 3 3 5 1 a) b) c) – d) 4 4 2 2 On dividing x3 – 3x2 + x + 2 by polynomial g(x), the quotient and remainder were x – 2 and 4 – 2x respectively then g(x) a) x2 + x + 1 b) x2 + x – 1 c) x2 – x – 1 d) x2 – x + 1. 1 If sum of zeros = 2 , product of its zeros = . The quadratic polynomial is 3 a) 3x2 – 3 2 x + 1

b)

2

2 x + 3x + 1

c) 3x2 – 2 3 x + 1

d)

2

2x +x+3

11) Let p(x) = ax2 + bx + c be a quadratic polynomial can have at most a) one zero b) two zeros c) three zeros d) none of these.

26

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 02: Polynomials

27

HW Exercise: 1) If α, β are the zeros of the polynomial p(x) = 4x2 + 3x + 7, then

1 1 + is equal to α β

7 7 3 3 b) – c) d) – 3 3 7 7 If the sum of the zeros of the polynomial f(x) = 2x3 – 3kx2 + 4x – 5 is 6, then the value of k is a) 2 b) 4 c) –2 d) –4 If α, β are the zeros of polynomial f(x) = x2 – p (x + 1) – c, then (α + 1) (β + 1) = a) c – 1 b) 1 – c c) c d) 1 + c If the product of two zeros of the polynomial f(x) = 2x3 + 6x2 – 4x + 9 is 3, then its third zero is 3 3 9 9 a) b) – c) d) – 2 2 2 2 If the polynomial f(x) = ax3 + bx – c is divisible by the polynomial g(x) = x2 + bx + c, then ab = 1 1 a) 1 b) c) –1 d) – c c If 3 is a zero of the polynomial f(x) = x4 – x3 – 8x2 + kx + 12, then the value of k is: 3 a) –2 b) 2 c) –3 d) 2 The sum and product zeros of the quadratic polynomial are –5 and 3 respectively the quadratic polynomial is equal to: a) x2 + 2x + 3 b) x2 – 5x + 3 c) x2 + 5x + 3 d) x2 + 3x – 5. 2 3 If the polynomial 3x – x – 3x + 5 is divided by another polynomial x – 1 – x2, the remainder comes out to be 3, then quotient polynomials is a) 2 – x b) 2x – 1 c) 3x + 4 d) x – 2. If α, β and γ are the zeros of the cubic polynomial such that α + β + γ = 2, αβ + βγ + γα = –7, αβγ = –14 then cubic polynomial is: a) x3 – 7x2 – 2x + 14 b) x3 + 2x2 + 7x – 14 c) x3 – 2x2 + 7x + 14 d) x3 – 2x2 – 7x + 14. 3 2 If the sum of zeros of the polynomial p(x) = kx – 5x – 11x – 3 is 2 then k is equal to: 5 2 5 a) k = – b) k = c) k = 10 d) 2 5 2 1 If 2 and – as the sum and product of its zeros respectively then the quadratic 2 polynomial f(x) is: a) x2 – 2x – 4 b) 4x2 – 2x + 1 c) 2x2 + 4x – 1 d) 2x2 – 4x – 1.

a)

2)

3) 4)

5)

6)

7)

8)

9)

10)

11)

COLUMN MATCHING QUESTIONS: 1) Listed in column I are some types of Polynomials. Choose all correct options for each item in column I in column II. Column I Column II i) Linear A) 3x2 + 1 ii) Binomial B) x3 – 2 iii) Quadratic C) x3 – 5x + 1 iv) Trinomial D) 2x –1 E) x2 – 5x + 6 Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

27

28

2) Given in column I are the number of zeros of the Polynomials and column II shows the graphs of the Polynomial. Choose all correct options for each item in column I in column II. Column I Column II i) 2 A) Y 0

X

X’ Y’

ii)

1

B)

Y 0

X’

X

Y’

iii)

0

C)

Y 0

X’

X

Y’

iv) 3

D)

Y 0

X’

X

Y’

E)

Y 0

X’

X

Y’

3) Listed in column I are the zeros of the polynomials and in column II some polynomials. Choose all correct options for each item in column I in column II. Column I Column II i) –1 and 2 A) x2 – x + 2 ii) –2 and 1 B) x2 + 3x + 2 iii) –2 and –1 C) x2 + x + 2 iv) No zeros D) x2 – x – 2 E) x2 + x – 2 4) In column I are the sum and product of the zeroes of the polynomials. For each item in column I. Choose all correct options in column II. Column I Column II i) S = 3/2, P = –9/2 A) x2 + x – 6 ii) S = –1, P = –6 B) x2 + 3x + 2 iii) S = –3, P = 2 C) 3x2 + 2x + 2 D) 2x2 + 2x – 12 E) 2x2 – 9 – 3x 28

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 02: Polynomials

29

5) Given in column I are some polynomials with one of the zeroes given in brackets. Choose one correct option from column II for each item in column I. Column I Column II i) x2 – x – (2k + 2), (–4) A) 1 The value of k is ii) x2 – 2x – (7p + 3), (–4) B) –5 The value of k is iii) kx2 – 3(k – 1)x – 1, (1) C) 9 The value of k is iv) x3 – 4x2 + kx – 2, (2) D) 3 The value of k is E) 5 6) Given in column I are some polynomials and column II shows the zeroes of the polynomial. Choose one correct option from column II for each item in column I. Column I Column II i) 3x2 + 11x – 4 A) 4, –1 ii) x2 – 16 B) –4, 2 iii) x2 – 3x – 4 C) 4, –4 2 iv) x + 5x + 4 D) –4, 1/3 E) –1, –4

ANSWER TO UNSOLVED EXERCISE: CW Exercise 2.1: 1) (i) No (ii) 1 (iii) 3 (iv) 2 (v) 4 (vi) 3 CW Exercise 2.2: 2) (i) 3x2 –3 2 x + 1 (ii) 4x2 +x + 1 (iii) x2 –4x + 1

1) (i) –2, 4 (ii) ½, ½ (iii) –1, 4/3 iv) 7, –7 3) α = 6) (i) 7) (i)

3 −2 ,β= 4 3

5) (i) p2 – 2q, (ii) p/q

4) x = –b/a, c/b

3abc − b3 3abc − b3 3abc − b3 b 2 − 2ac b 2 − 2ac (ii) (iii) (iv) (v) ac a2 a3 c3 a 2c (b 2 − 2ac )2 − 2a 2c 2 a4

(ii)

(b 2 − 2ac )2 − 2a 2c 2 8) k = –1 or 2/3 a 2c 2

9) x2 – 4x – 5

HW Exercise 2.2: 1) (i) –1/3, 3/2 (ii) –2, 0 (iii) – 15 ,

2) (i) 4x2 – x – 4 (ii) x2 +

15

3) 15/4

4) –5/16

7) 12

8) x2 –

5) 6

5 (iii) x2 – x + 1 6) 2

25 x + 41 2

CW Exercise 2.3: 1) (i) Q = x – 3; R = 7x – 9 (ii) Q = x2 + x – 3; R = 8 2) (i and ii) Yes 2

4) x2 + 2x +1

3) –1, 1 2

5) i) P(x) = 2x – 2x + 14; g(x) = 2; q(x) = x – x + 7; r(x) = 0 Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

29

30

ii) P(x) = x3 + x2 + x + 1; g(x) = x2 – 1; q(x) = x + 1; r(x) = 2x + 2 iii) P(x) = x3 + 2x2 – x + 2; g(x) = x2 – 1; q(x) = x + 2; r(x) = 4 6) i) Q = x2 – 8x + 27, R = –60

ii) Q = x2 + x – 3, R = 8

8) 14x – 10

10) (x +

9) a = 1, b = 7

2)

11) 2

HW Exercise 2.3: 1) i) Q = –x2 – 2; R = –5x + 10

2) No

4) x2 – x + 1

5) a = –2, b = –8

6) i) Q = x–3, R = 7x–9 (ii) Q = –x2–2, R=–5x+10

7) 61x – 65

8) k = 5, a = –5

3) 5, –6

Miscellaneous: 2) x3 – 2x2 – 7x + 14 7) i) 8) −

3) a = 1; b = ± 2

b 2 − 4ac a 25 12

12) 7

4) –5, 7

5) k = 5 and a = –5 iv) −

ii)

b 2 − 4ac c

⎛ b 2c ⎞ iii) − ⎜ + ⎟ ⎝c a ⎠

9) −

3 2

10) −

27 4

bc a2

11) 108 15) f(x) = k(x2 + 4x + 4)

13) –2/3

14) 0

21) –10, –1, –2

22) – 3 , 3 , 1, 2

24) x – 2

25) 2x – 3

26) –3

30) pq = r

31) 12

17) f(x) = k(x3 – 3x2 – x + 3) 20) –

1 , 3, –2, –1 2

23) 2, –1,

3 3 ,– 2 2

3 2

28) 3

29) –

32) 1, 4, 7

33) –105

34) 2p3 – 9pq + 27r = 0 35) 3, 4, –2

36) a = 1, b = ± 2 Column Matching Question:

30

1) i–D; ii–ABD; iii–AE; iv–CE

2) i–C; ii–DE; iii–B; iv–A

3) i–D; ii–E; iii–B; iv–AC

4) i–E; ii–AD; iii–B

5) i–C; ii–D; iii–A; iv–E

6) i–D; ii–C; iii–A; iv–E

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

31

Chapter 03: Pair of Linear Equations in Two Variables Chapter Map: Basic Fundamentals Equation

Variable (Dimension)

Coefficient/ Constants

Linear (Degree)

General Form of a Linear Equation in two variables Solution by Graphical Method Solving Simultaneous Linear Equations by Algebraic Methods Substitution Method

Elimination Method

Cross Multiplication Methods

Equations Reducible to a pair of Linear Equations in Two varibales Word Problems (Tips for solving word problems)

Basic Fundamentals: ™ For a better understanding of linear equations, let us first understand the basic terminology used to describe any equation. There are basically four basic terms, which are described as below:

Equation: z

When a relation is such that the L.H.S. = R.H.S., the relation is said to be an equation E.g. x = 8, x + 2y = 13

Variable: z

If the equation is of the form ax + by + c = 0, x and y are said to be variables.

Coefficient/ Constants: z

If the equation is of the form ax + by + c = 0 then a is the coefficient of x, b is the coefficient of y and c is called the constant term.

Linear (Degree): z

An equation in two variables x and y is said to be linear if it is equivalent to an equation of the form ax + by + c = 0 where a, b and c are constants. The degree of the equation is 1.

General Form of a Linear Equation in two variables: ™ The general form of the linear equation in two variables is ax + by + c = 0, where a, b and c are real numbers, a and b are not both zero. We often denote the condition a and b are not both zero by a2 + b2 ≠ 0. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

31

32

Solution of Linear Equation in two Variables: A solution the equation ax + by + c = 0 is a pair values, one for x and the other for y which makes the two sides of the equation equal. Example: Consider the equation 2x + 3y = 5. Let us substitute x = 1 and y = 1 in the LHS of the equation LHS = 2(1) + 3(1) = 5 = RHS. ∴ x = 1 and y = 1 is a solution of the equation 2x + 3y = 5. If we consider x = 1 and y = 3 in the equation 2x + 3y = 5. Then LHS = 2(1) + 3(3) = 2 + 9 = 11 which is not equal to RHS. ∴ x = 1 and y = 3 is not a solution of the equation.  Remember that a linear equation has infinite solution. z

Note: 1) Graphically, a linear equation represents a line 2) Every solution of the equation is a point on the line representing it. 3) Each solution (x, y) of a linear equation in two variables corresponds to a point on the line representing the equation and vice versa.

Graphing a Linear Equation in two variables:  We know that the solution of linear equation ax + by + c = 0 is the value of an ordered pair (x, y) which represents a point on the Cartesian co–ordinates. If all the values of ordered pair (x, y) are plotted on the Cartesian plane and joined together, they form a straight line passing through all such points.  The graph of a linear equation can be drawn using any of the following methods:

Three Point Method: z z z

Step 1: Find three solutions (points) of the line represented in ordered pair(x, y) form Step 2: Plot the three points on Cartesian plane Step 3: Draw a line joining these three points which represents the given equation

Note: Although only two points are sufficient to draw a line but third point is taken for confirmation

X, Y Intercept Method: z z z z

32

Step 1: To get x–intercept, put y = 0 in ax + by + c = 0 which gives x = –c/a Step 2: To get y–intercept, put x = 0 in ax + by + c = 0 which gives y = –c/b Step 3: Locate the point (–c/a, 0) on x– axis and (0, –c/b) on y–axis Step 4: Join the two points located on the Cartesian plane to get the graph of the line.

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: 0 Pair of Lin near Equations in Two Va ariables

33

SOLVE ED EXERCISE 3.1: 1)

Drraw the grap ph of

2x + 3y + 5 = 0.

Sol:

Step I: 2x = – 3y 3 –5

⇒x=

−3 y − 5 2

Step II: Substittution When y = 1, x =

−3(1) − 5 = –4 2

When y = 0, x =

−3(0) − 5 = –2.5 5 2

When y = –1, x =

−3( −1) − 5 = –1 – 2 –2.5

–4 –1 x 1 0 –1 y (–4, 1) (–2.5, 0) (–1, –1) (x, y) 2) Ro omila went to a station nery shop and a purchas sed 2 pencils and 3 errasers for Rs.9. R Her friend Sonali saw the new w variety of pencils and d erasers wiith Romila, a and she also o bought 4 pencils p and d 6 erasers of o the same e kind for Rs s.18. Repres sent this sittuation alge ebraically an nd graphicallly. Sol: Le et us denote e the cost of o 1 pencil by Rs x an nd one eraser by Rs yy. Then the algebraic rep presentation is given by the t following g equations: –– (2) 4x + 6y = 18 2xx + 3y = 9 –– –– (1) To o obtain the equivalent geometric g representation n, we find tw wo points on the line rep presenting ea ach equation. That is, we find two solu utions of eac ch equation. Y 5 Th hese solution ns are given below in Tab ble 3.2. 0 x 4.5 4 9 − 2x 3

3

x

0

3

18 − 4 x y= 6

3

1

y=

0

3

P(0, 3)

2 (3, 1)

1

Q(4.5, Q 0) X

(0, 0)

X′ – –1

–1

1

2

3

4

5

Y′

We plot these points in a graph g paper and a draw the e lines. We find that both the lines coiincide. Th his is so, beccause, both th he equationss are equivalent, i.e., one e can be derived from the e other.

UNSOLV VED EXER RCISE 3.1: CW Exerrcise: 1) Re epresent the equation 3xx + 4y = 20 grraphically. 2) Drraw the graph of the following c) y = 2 b) x = 3 a) 4x + 2y = 5 d) 3 3y = 4x 3) Th he coach of a cricket team m buys 3 batts and 6 balls for Rs.390 00. Later, he buys anothe er bat and 2 more balls of the same kind fo or Rs.1300. Represent this situatio on algebraic cally and ge eometrically. Volume

Universal Tutorials – X CBSE (2012–13) – Math hematics

33

34

4) Astha’s mother tells all the kids on Astha’s birthday party “Seven years ago, I was seven times as old as Astha was then. Also, three years from now, I shall be three times as old as Astha will be.” Represent this situation algebraically and graphically. 5) Draw the graph of the equation 2y – x = 7 and determine from the graph whether x = 3, y = 2 is a solution or not. HW Exercise: 1) Represent the following graphically b) 3x + 4y – 10 = 0 a) x – 2y = 0 2) The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically. 3) Akhila goes to a fair with Rs.20 and wants to have rides on the Giant Wheel and play Hoopla. The number of times she played hoopla is ½ the no of rides on the Giant wheel. Each rides costs Rs.3 and of Hoopla costs Rs.4. Represent this situation algebraically and graphically. 4) The motion of two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically. 5) If 3x + 7y = 14, express y in terms of x. Check whether (3, –2) is a point on the given line.

Pair of Linear Equations in Two Variables: ¾ The general form for a pair of linear equations in two variables x and y is a1x + b1y + c1 = 0 and a2x + b2 y + c2 = 0, where a1, b1, c1, a2, b2, c2 are all real numbers and a12 + b12 ≠ 0, a22 + b22 ≠ 0. ¾ Since the linear equation represents a line, a pair of linear equations represents a pair of lines. ¾ If we consider two lines in a plane, only one of the following three possibilities can happen.  The two lines will be intersect at one point  The two lines will not intersect, i.e. they are parallel.  The two lines will be coincident.

Solution of a Pair of Linear Equation in Two Variables  The common pair of values of the variables which satisfy the given system of simultaneous equations is called the solution of a pair of linear be equation in two variables. Example: Consider the equations x – 2y = 0 and 3x + 4y = 20 Let us take x = 4, y = 2. Substituting the values of x and y be equation, we get 4 – 2 × 2 = 0, and 3 (4) + 4 (20) = 20 So x = 4, y = 2 satisfy both the equations ∴ x = 4, y = 2 is the solution of the given pair of equations.

Solving Simultaneous Linear Equations: z

There are two methods available to solve a system of two linear equations in two variables. These are: Solution I. Graphical method

34

II. Algebraic method

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

35

Consistent and Inconsistent Systems of Simultaneous Linear Equation: z

A system consisting of two simultaneous linear equations is said to be consistent if it has at least one solution, otherwise it is called an inconsistent.

Graphical Method of solving a pair of Linear Equations: ¾ As we have studied a pair of linear equations represents a pair of lines. If we consider two lines in a plane, only one of the following there possibilities can happen. Three Possibilities

Intersecting Lines (Consistent)

Coincident lines (Dependent – consistent)

Parallel Lines (Inconsistent)

Consider the system of simultaneous linear equations in x and y. ––– (i) Let the equations be a1x + b1y + c1 = 0 ––– (ii) And a2x + b2y + c2 = 0 Draw the graphs of the equations (i) and (ii) Let the line l1 and l2 represents the graphs of (i) and (ii) respectively. Case (i): If the lines l1 and l2 intersect at the point and (a, b) are the co–ordinates of this point, then the equations has a unique solution and the solution is given by x = a, y = b. l1 Y (a,b)

l2

0

X Case (ii): If the lines l1 and l2 are parallel, then the given pair of equation has no solution l1 Y l2

0

X Case (iii): If the lines l1 and l2 are coincident, then the pair of equations has infinitely many solutions. In this case every point on the line is a solution. Y

0

l1 l2

X

¾ A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. Graphically they represent a pair of intersecting lines. ¾ A pair of linear equations which has no solution is called an inconsistent pair of linear equations. Graphically they represent a pair of parallel lines. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

35

36

¾ A pair of linearr equations which w are equivalent ha as infinitely many m distincct common solutions. Succh a pair is ca alled a depe endent pair of o linear equa ations in two variables. N Note that a dependent pairr of linear equ uations is alw ways consisttent. Graphic cally they rep present coinccident lines. ¾ if the e lines repre esented by th he equation, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are  Intersecting, then  Parallel, P then n

b a1 ≠ 1 a2 b2

a1 b c = 1 ≠ 1 a2 b2 c2

 Coincident, C th hen

Pair of Liness Coefficient Ratios Algebraic Interpretation Graphical Interpretation Consistencyy

b a1 c = 1 = 1 a2 b2 c2

Solution of a pair of Line ear Equation ns x + 2y – 2 = 0 x+y–4=0 x + 2y – 4 = 0 x – 2y + 2 = 0 b b c a1 a1 ≠ 1 = 1 ≠ 1 a2 a2 b2 b2 c2

2x + 3yy – 6 = 0 6x + 9yy –18 = 0 b c a1 = 1 = 1 a2 b2 c2

One solution

No so olution

Infinite solutions

Interrsecting

Parrallel

Coincident

Con nsistent

Inconsistent

Dependentt (consistent))

Graphs

SOLVED D EXAMPL LES 3.2: 1) Solv ve graphica ally the syste em of equattion, 5x – y – 7 = 0 –– –– (1) and x – y + 1 = 0 Sol: Let us draw the graphs of eq quation (1) an nd (2). To do d this, we find two solutiions of each of the eq. (1) and (2). 5x – y – 7 = 0 x

0

1

y

– –7

–2

x

– –1

0

y

0

1

––– (2).

x–y+1=0

From the graph, we get the e solution (2, 3)

36

Universal Tutorials – X CBSE (2012–13) – Math hematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

37

2)

Determine graphically the coordinates of the vertices of triangle, the equations of whose sides are y = x, 3y = x, x + y = 8. Also shade the region bounded by these lines. Y Sol: Step – I: y=x

3y = x

x+y=8

When x = 0, y = 0

When x = 0, y = 0

When x=3, y=5

When x=–1, y =–1

When x = 3, y = 1

When x=6, y=2

When x = 2, y = 2

When x = –3,y=–1

When x=0, y=8

After plotting the points on the graph we get O (0, 0), B (6, 2), C (4, 4)

8 7

6 5

C (4, 4)

4 3

(6, 2)

2

B

1 X′ O (0,0)

2

3

4

5

6

X

Y′

3) Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa bought. y = 2x – 2 Y Sol: Let us denote the number of pants by x and the number of y = 4x – 4 3 skirts by y. Then the equations formed are: A(2, 2) y = 2x – 2 ––– (1) and y = 4x – 4 ––– (2) 2 Let us draw the graphs of Equations (1) and (2) by 1 finding two solutions for each of the equations. (0, 0) X X′ They are given in Table 3.6. –1 1 2 3 4 5 x 2 0 –1 Q(1, 0) y = 2x – 2 2 –2 B(0, –2) –2 –3

x 0 1 P(0, –4) –4 y = 4x – 4 –4 0 –5 Plot the points and draw the lines passing through Y′ them to represent the equations, as shown in Fig. The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

UNSOLVED EXERCISE 3.2: CW Exercise: Graphical Solution a b c 1) On comparing the ratios 1 , 1 and 1 , find out whether the lines representing the following a2 b2 c2 pairs of linear equations intersect at a point, are parallel or coincident: i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0 ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0 iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0 Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

37

38

2) Solve the following system of simultaneous linear equations graphically: 5y b) 2x + 3y = 6, 3x – =2 a) 2x + y = 4, 3x + 2y = 5 2 d) 5x + 2y = 9, 15x + 6y = 27 c) 4x + 2y = 5, 4y – 2x = 5 e) x – 3y = 3, 3x – 9y = 2 3) Ten students of Class X took part in a Mathematics quiz. The number of girls who participated in the quiz is 4 more than the number of boys. Obtain the linear equations satisfying the conditions and solve them graphically. State the number of boys and girls who took part in the quiz from the result so obtained. 4) Kevin has only 2 rupees and 5 rupees coins in his purse. He has 8 coins in all totaling Rs.25. How many coins of each does he have? 5) Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. 6) For each linear equation given in the table write down another equation which forms intersecting line, parallel line and coincident line with the given line. Linear Equation Intersecting Line Parallel Line Coincident Line 2x + 3y – 8 = 0 3x – 4y – 12 = 0 7) Shade the area of the triangles formed by y = x, y = 1, y = –1 and x = 0. 8) Draw the graph of the equation, 4x – y = 4 and 4x + y = 12. Determine the vertices of the triangle formed by the lines representing these equations and the x–axis. Shade the triangle region so formed. 9) Draw the graphs of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and x–axis. Find the area of the shaded region. 10) Solve the system of linear equations graphically x – y = 1 and 2x + y = 8. Shade the area bounded by these two lines and y–axis. Also, determine this area. 11) Draw the graphs of the equations, 2x – y – 2 = 0, 4x + 3y – 24 = 0 and y + 4 = 0. Obtain the vertices of the triangle so obtained. Also, determine its area. HW Exercise: Graphical Solution 1) Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically: ii) x – y = 8, 3x – 3y = 16 i) x + y = 5, 2x + 2y = 10 iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0 iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 a b c 2) On comparing the ratios 1 , 1 and 1 , find out whether the following pair of linear equations a2 b2 c2 are consistent, or inconsistent. ii) 2x – 3y = 8; 4x – 6y = 9 i) 3x + 2y = 5; 2x – 3y = 7 3 5 iii) x + y = 7; 9x – 10y = 14 iv) 5x – 3y = 11; – 10x + 6y = –22 2 3 4 v) x + 2y = 8; 2x + 3y = 12 3 3) Draw the graph of 2y = 4x – 6, 2x = y + 3 and determine whether this system of linear equations has a unique solution or not. 4) Determine by drawing graphs whether the following system of linear equations has a unique solution or not: x – 3y = 3 and 3x – 9y = 2. 5) Form the pair of linear equations in the scenario given and find the solutions graphically.5 pencils and 7 pens together cost Rs.50, whereas 7 pencils and 5 pens together cost Rs.46. Find the cost of one pencil and that of one pen. 38

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

39

6) Determine graphically the vertices of the triangle the equation of whose sides are given below 2y – x = 8, 5y – x = 14, y – 2x = 1 7) Find graphically the vertices of the triangle whose sides have the equations, y = x, y = 0 and 2x + 3y = 10. Find the area of the triangle formed by these lines. 8) Determine graphically whether the following equations 3x – 4y = 1, 8y – 6x = 4 is consistent or inconsistent. 9) Determine the vertices of the triangle formed by the lines representing the equations, x + y = 5; x – y = 5 and x = 0. Find the area of the triangle formed by these lines. 10) Determine graphically the co–ordinates of the vertices of a triangle, the equations of whose sides are y = x, y = 2x and x + y = 6. 11) Solve graphically: 3x – 5y = 19; 3y – 7x + 1 = 0. Does the point (9, 4) lie on any of lines?

Algebraic Methods of solving a pair of Linear Equations: ¾ The simultaneous linear equation can also be solved algebraically, without drawing the graph. It is possible to solve such a system of equations, because there are two distinct equations for two variables. The following are the three methods: Algebraic Methods

Substitution Method

Elimination Method

Substitution Method: z z z

z

Cross Multiplication Method

To understand the substitution method more clearly, let us consider it stepwise: Step 1 : Find the value of one variable, say y in terms of the other variable, i.e. x from either equation, whichever is convenient. Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e. in terms of x, which can be solved. Sometimes, you can get equations with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent. Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

Remark: We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method.

SOLVED EXERCISE 3.3: 1) Solve: 5x – 3y = 1; 2x + 5y = 19 by Substitution Method. Sol: 5x – 3y = 1 –– (1) 2x + 5y = 19 –––– (2) 1 + 3y From (1) we get, 5x = 1 + 3y ⇒ x = –– (A) 5 2 + 6 y + 25 y 1 + 3y 2(1 + 3 y ) in equation (2), + 5y = 19 ⇒ = 19 Substitute x = 5 5 5 2 + 31y = 95 ⇒ 31y = 93, y = 3 1+ 3 × 3 = 2 Solution set = {(2, 3)} Substitute y = 3 in equation (A) x= 5

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

39

40

2) Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Solve by the method of substitution. Sol: Let s and t be the ages (in years) of Aftab and his daughter, respectively. Then, the pair of linear equations that represent the situation is s – 7 = 7 (t – 7), i.e. s – 7t + 42 = 0 ––– (1) i.e. s – 3t = 6 ––– (2) and s + 3 = 3 (t + 3), Using Equation (2), we get s = 3t + 6. Putting this value of s in Equation (1), we get (3t + 6) – 7t + 42 = 0, i.e. 4t = 48, which gives t = 12. Putting this value of t in Equation (2), we get s = 3(12) + 6 = 42 So, Aftab and his daughter are 42 and 12 years old, respectively. Verify this answer by checking if it satisfies the conditions of the given problems. 3) The cost of 2 pencils and 3 erasers is Rs.9 and the cost of 4 pencils and 6 erasers is Rs.18. Find the cost of each pencil and each eraser. Sol: The pair of linear equations formed were: ––– (1) 2x + 3y = 9 ––– (2) 4x + 6y = 18 We first express the value of x in terms of y from the equation 2x + 3y = 9, to get 9 − 3y x= ––– (3) 2 Now we substitute this value of x in Equation (2), to get 4(9 − 3 y ) + 6y = 18 i.e., 18 – 6y + 6y = 18 i.e., 18 = 18 2 This statement is true for all values of y. However, we do not get a specific value of y as a solution. Therefore, we cannot obtain a specific value of x. This situation has arisen because both the given equations are the same. Therefore, Equations (1) and (2) have infinitely many solutions. Observe that we have obtained the same solution graphically also. We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.

UNSOLVED EXERCISE 3.3: CW Exercise: 1) Solve the following pair of linear equations by the substitution method. i) x + y = 14; x – y = 4 s t + =6 ii) s – t = 3; 3 2 iii) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3 2) Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3. 2x y x y 3) Solve the systems of equation by using the method of substitution, = 2; − = 4. + a b a b 4) Solve: ax + by = a – b; bx – ay = a + b. 5) Solve by substitution method, 2x + 7y = 21, 6x + 21y = 63. 6) Solve: 3x – 8y = 2, 9x – 24y = 21. Form the pair of linear equations for the following problems and find their solution by substitution method. 7) The difference between two numbers is 26 and one number is three times the other. Find them. 40

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

41

8) Vinayak, Amar and Sherwin after attending the prize distribution function of UT at Vashi decided to go to their respective destinations by Auto. The auto charges in Navi Mumbai consist of a fixed charge and a variable charge at the rate of per km. Vinayak paid Rs.38 for a 5 km travel to his house in Nerul, and Amar paid Rs.63 for a 10 km travel to his house at Seawoods from Vashi. What are the fixed and variable charges? How much did Sherwin pay if his house is 3 km away from Vashi? 9) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. HW Exercise: 1) Solve the following pair of linear equations by the substitution method. i) 3x – y = 3; 9x – 3y = 9

ii)

2x+

3 y = 0;

3x+

8y=0

3x 5y x y 13 – = –2; + = 2 3 3 2 6 2) Find the value of k in the equation 2y = x + k if x and y are related by the pair of linear equations given by 2x – 3y = 16 and 4x + 3y = 14.

iii)

3) Solve:

x y x y = a + b; 2 + 2 = 2. + a b a b

x y = 2; ax – by = a2 – b2. + a b Solve by substitution method, ii) 3x – 4y = 2, 9x – 12y = 6. i) 5x + 2y = 11, 20x + 8y = 10 Form the pair of linear equations for the following problems and find their solution by substitution method. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. The coach of a cricket team buys 7 bats and 6 balls for Rs.3,800. Later, she buys 3 bats and 5 balls for Rs.1, 750. Find the cost of each bat and each ball. Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages? Four times a number is equal to seven times the number formed by reversing its digits. Sum of the digits is equal to three. Find out the number.

4) Solve: 5)

6) 7) 8) 9)

Elimination by equating the coefficients: z z z

z z

Volume

Let us note down the steps in the elimination method: Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal. Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3. If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions. If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent. Step 3: Solve the equation in one variable (x or y) so obtained to get its value. Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable.

Universal Tutorials – X CBSE (2012–13) – Mathematics

41

42

SOLVED EXERCISE 3.4: 1) Solve: 5x + 3y = 70; 3x – 7y = 130 by elimination method. Sol: 5x + 3y = 70 –– (1) 3x – 7y = 130 ––– (2) –––––– (3) Equation (1) × 3; 15x + 9y = 210 –––––– (4) Equation (2) × 5; 15x – 35y = 650 Equations (3) – (4), 44y = – 440 ⇒ y = – 10 Substitute y = – 10 in equation (1) 5x + 3 × (–10) = 70 5x = 100 x = 20 Solution set {(20, –10)} y 11 = 10 2) Solve algebraically the following system of equations: x + = 0.9, y 3 x+ 2 Sol:

x+

y = 0 .9 3

11 = 10 y x+ 2

––– (1)

Equation (1) × 3; 3x + y = 2.7

–––– (3)

Equation (2) × x +

y 2

y⎞ ⎛ 11 = 10⎜ x + ⎟ 2⎠ ⎝

11 = 10x + 5y

3x + y = 2.7

–––– (3)

Eqn (3) × 5; 15x + 5y = 13.5

––– (2)

–––– (4) 10x + 5y = 11

–– (5) Eqn (4) × 1; 10x + 5y = 11.0

–––– (4) –––– (6)

Eqn. (5) – (6); 5x = 2.5 ⇒ x = 0.5 Substituting x = 0.5 in eqn. (1), 0.5 +

y = 0.9, 3

y y = 0.9 – 0.5, = 0.4, y = 1.2 Solution Set {(0.5, 1.2)} 3 3 3) The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? Sol: Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number may be written as 10x + y in the expanded form (for eg. 56 = 10(5) + 6) When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x (for eg. when 56 is reversed, we get 65 = 10(6) + 5) i.e. x + y = 6 (1) According to the given condition, (10x + y) + (10y + x) = 66 i.e. 11(x + y) = 66 We are also given that the digits differ by 2, ––– (2) or y – x = 2 –– (3) therefore, either x – y = 2 If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. In this case, we get the number 42. If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. In this case, we get the number 24. Thus, there are two such numbers 42 and 24. Verification: Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2.

42

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

43

UNSOLVED EXERCISE 3.4: CW Exercise: 1) Solve by elimination method:

iii) 3x – y = 12, 5x – 3y = 16

11 31 , –7x + 5y = 3 3 iv) x – 5y = 11, 2x + 3y = 0

v) 3x + 4y = 25, 5x – 6y = –9

vi)

i) x – 5y = 11, 2x + 3y = –4

ii) 3x + 2y =

2x − 3y = 0 ,

5 x + 2y = 0

2) Solve, 217x + 131y = 913, 131x + 217y = 827. 3) Solve, 37x + 41y = 70, 41x + 37y = 86. x y 4) Solve: = ; ax + by = a2 + b2. a b x y x y = a + b; 2 + 2 = 2. 5) Solve: + a b a b

[CBSE–08]

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: 6) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction? 7) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? 8) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. HW Exercise: 1) Solve by elimination method: i) x + y = 7, 3x – 2y = 11 ii) x – y = – 1, 3x – 5y = – 1 iv) 2x – 3y = 1.3, y – x = 0.5 iii) 3x + 5y = –7, 11x – 8y = 27 2) Solve, 99x + 101y = 499, 101x + 99y = 501. 3) Solve, 23x – 29y = 98, 29x – 23y = 110. 5 2 15 7 – = –1; + = 10. 4) Solve, x+y x−y x+y x−y Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: 5) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. 6) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.

Cross Multiplication Method:  The two general linear equations in two variables are given as below: a1x + b1y + c1 = 0 ––– (1) a2x + b2y + c2 = 0 ––– (2)

Volume

i) When

a1 b ≠ 1 , we get a unique solution a2 b2

ii) When

a1 b c = 1 = 1 , there are infinitely many solutions. a2 b2 c2 Universal Tutorials – X CBSE (2012–13) – Mathematics

43

44

iii) When

a1 b c = 1 ≠ 1 , there is no solution. a2 b2 c2

General solution: „

x=

b1c2 − b2c1 c a − c2a1 and y = 1 2 a1b2 − a2 b1 a1b2 − a2b1

––– (3)

x y 1 = = b1c 2 − b2c1 c1a2 − a1c 2 a1b2 − a2 b1

„

Also written as,

„

Diagrammatically represented as: x y b1 c1

––– (4)

1 a1

b1

b2 c2 a2 b2  The arrows between the two numbers indicate that they are to be multiplied and the second product is to be subtracted from the first.  For solving a pair of linear equations by this method, we will follow the following steps : z Step 1: Write the given equations in the form (1) and (2). z Step 2: Taking the help of the diagram above, write Equations as given in (4). z Step 3: Find x and y, provided a1b2 – a2b1 ≠ 0 Step 2 above gives you an indication of why this method is called the cross-multiplication method.

SOLVED EXAMPLES 3.5: 1) In case of the following system of equations determine whether the system has a unique solution, no solution or infinite number of solutions. In case there is a unique solution, find it. a) 5x + 7y + 2 = 0, 4x – 3y – 7 = 0 Sol: 5x + 7y + 2 = 0 4x – 3y – 7 = 0 Comparing these equations with the general equations. a1x + b1y + c1 = 0; a2x + b2y + c2 = 0 a2 = 4, b2 = – 3, c2 = – 7 Here a1 = 5, b1 = 7, c1 = 2 a1 5 b1 2 a b 7 c1 = , = ⇒ 1 ≠, 1 , = a2 4 b2 − 3 c 2 − 7 a2 b2 ∴ The system has a unique solution. b c − b2c1 −49 + 6 −43 7 × ( − 7 ) − ( − 3 )( 2 ) x= 1 2 = = = =1 − 15 − 28 − 43 5( − 3 ) − 4( 7 ) a1b2 − a2b1 y=

c1a2 − a1c 2 2( 4) − 5( −7) 43 = = = −1 − 43 − 43 a1b2 − a2b1

2) Solve for x and y: ax + by = a – b, bx – ay = a + b. Sol: ax + by = a – b ––– (1) bx – ay = a + b ––– (2) ––– (3) Multiply eq (1) × b; abx + b2y = ab – b2 ––– (4) Multiply eq (2) × a; abx – a2y = a2 + ab y (b2 + a2) = a (b – a) – b (b + a) Eq (3) – Eq (4)

y= 44

ab − a 2 − b 2 − ab − (a 2 + b 2 ) = −1 = (a 2 + b 2 ) b2 + a2

∴y=–1

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

45

Substituting y = – 1 in equation (1) ax – b = a – b ax = a – b + b ax = a ∴ x = 1, y = –1 3) For what value of k the following system of equations have, a) Unique solution b) No solution. 3x + ky = 1; 3x – 5y = 7 Sol: 3x + ky = 1 ––– (1) 3x – 5y = 7 ––– (2) 3x – 5y – 7 = 0 3x + ky – 1 = 0 Comparing with general equation a1x + b1y + c1 = 0 a1 = 3, b1 = k a2x + b2y + c2 = 0 a2 = 3, b2 = –5 a1 b1 3 k [The system has a unique solution] k≠–5 ≠ ≠ 3 −5 a2 b2 If the system has no solution then,

a1 b1 c1 3 k ; = = ≠ a2 b2 c2 3 − 5

k = –5

4) Find the value of k for which the following system of equations 5x – 3y = 0 and 2x + ky = 0 has a non–zero solution. b1 -3 a1 5 c1 0 = Sol: = = b2 k a2 2 c2 0

For unique solution:

a1 b1 5 −3 6 i.e. ≠ or k ≠ – ≠ 5 a2 b2 2 k

But if we take this value of k, the unique solution is (0, 0), since constants are zero. Now a non–zero solution can only be found if the two lines are coincident. a b 6 ∴ 1 = 1 i.e. k = – a2 b2 5 5) After covering a distance of 30km with a uniform speed there is some defect in a rail engine and therefore, its speed is reduced to 4/5 of its original speed. Consequently, the train reaches its destination late by 45 minutes. Had it happened after covering 18 kms more, the train would be late by only 36 minutes. Find the speed of the train and the distance of journey. Sol: Distance covered after the speed reduced be = D km Let the speed be = x km/hr D 5D ∴Time = ∴ New Time = 4x x 5D D 45 3 – = According to the given condition, = 60 4 4x x 5D − 4D 3 D 3 ⇒ D = 3x ––– (I) = = 4x 4 4x 4 According to the 2nd condition, D − 18 5(D − 18 ) New time taken = Change time = x 4x 5(D − 18 ) D − 18 36 3 – = = 4x x 60 5 D − 18 3 5D – 90 = 12x 5 × 3x – 90 = 12x = 4x 5 15x – 12x = 90 ⇒ 3x = 90 ⇒ x = 30 Total distance = 30 + D = 30 + 30 × 3 = 120 km Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

45

46

UNSOLVED EXERCISE 3.5: CW Exercise: General Solutions 1) In case of the following system of linear equations determine whether the system has a unique solution, no solution or infinite no. of solutions. In case there is a unique solution find it by cross multiplication method. a) 2x + 3y = 5, 10x + 15y = 16 b) 3x – 5y = 4, 6x – 10y = 8 2) Find the values of k for which the system of equations x – 2y = 3, 3x + ky = 1 has a unique solution. 3) For what value of k will the equations x + 2y + 7 = 0, 2x + ky + 14 = 0 represent coincident lines. 4) Find the value of k, if it exists such that the system of equations kx + 3y = k – 3, 12x + ky = k has infinitely many solutions. 5) Express y in terms of x in the equation 2x + 3y = 11. Find the coordinates of the point where the line of the equation 2x + 3y = 11 cuts the y axis without drawing the graph. 6) Determine the value of k for which the given system of equations becomes consistent 7x – y = 5, 21x – 3y = k. 7) Find the value or values of k for which the system of equations kx – y = 2, 6x –2y = 3 has i) a unique solution ii) No solution iii) Is there a value of k for which the system has infinitely many solutions. 8) Find the relation between p and q for which the following system of equations has finite number of solutions. 2x + 3y = 7; (p + q)x + (2p – q)y = 21. 9) For which values of a and b does the following pair of linear equations have an infinite number of solutions? (1) 2x + 3y = 7 (2) (a – b)x + (a + b)y = 3a + b – 2. 10) Solve for x and y: (a – b)x + (a + b)y = a2 – 2ab – b2, (a + b)(x + y) = a2 + b2. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method: 11) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1,000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1,180 as hostel charges. Find the fixed charges and the cost of food per day. 1 1 when 1 is subtracted from the numerator and it becomes when 8 is 3 4 added to its denominator. Find the fraction. HW Exercise: General Solution 1) In case of the following system of equations determine whether the system has a unique solution, no solution or infinite number of solutions. In case there is a unique solution find it.

12) A fraction becomes

a) x + 6y = 5, 3x – 5y = – 8 c) 5x + 3y = 1, 2x +

6 2 y= 5 5

b) 6x – 7y = 13, 12x + 14y = – 26 d) 5x – 6y – 10 = 0, 10x – 12y + 20 = 0

6 9 15 y= 2 f) – 3x + 4y = 5, x − 6 y = 5 2 2 2) Determine the values of k for which the following system of equations has a unique solution 2x – 3y = 1, kx + 5y = 7.

e) 5x + 2y = 16, 3x +

3) Find the value of k for which the following system of equations has no solution 3x – 4y + 7 = 0, kx + 3y – 5 = 0. 4) Determine the value of k for which the following system of equations has an infinite no. of solutions (k – 3)x + 3y = k, kx + ky = 12. 5) Find the value of k for which the equations 3x – y = 6, 6x + ky = 12 represents coincident lines. 46

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

47

6) Solve for x and y i) 4x +

7) 8) 9) 10)

6 8 = 15 , 6x – = 14 and hence find ‘p’ for which y = px – 2 y y

ii) 43x + 67y = 2870, 67x + 43y = 2630 Obtain the condition for the following system of linear equation to have a unique solution ax + by = c, lx + my = n. Determine whether the system of equations 3x – y = 7, 9x – 3y + 25 = 0 has a unique solution, no solution or infinitely many solutions. Justify your answer. For what value of k will the following system of linear equations have no solutions 3x + y = 1; (2k – 1)x + (k – 1)y = 2k + 1. Solve, x + y = a + b, ax – by = a2 – b2.

11) Solve,

a 2b a ab 2 b – = 0, + = a2 + b2, where x, y ≠ 0. y x y x

12) Solve, a(x + y) + b(x – y) = a2 – ab + b2, a(x + y) – b(x – y) = a2 + ab + b2. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method: 13) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? [CBSE–09]

Equations Reducible to a Pair of Linear Equations in Two Variables  In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. We now explain this process through some examples. 1 7 + = 10. Which is not a linear equation. But by making x y some substitution we can reduce it to a linear equation, which we know how to solve. Then using that solution, we can solve the unique equation.

 Sometimes we find equations like

SOLVED EXERCISE 3.6: 1) Solve the following pair of equations by reducing them to a pair of linear equations : 5 1 + =2 x −1 y −2

Sol: Let us put

6 3 – =1 x −1 y −2

[CBSE–09]

1 1 = p and = q. Then the given equations x −1 y −2

⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 1 ⎟⎟ = 1 ––– (2) 5⎜ 6⎜ = 2 ––– (1) ⎟ + ⎟ – 3⎜⎜ − 1 −1 x x y − 2 ⎝ ⎠ ⎝ ⎠ ⎝ y −2⎠ can be written as : 5p + q = 2 ––– (3) 6p – 3q = 1 ––– (4) Equations (3) and (4) form a pair of linear equations in the general form. Now, you can use any method to solve these equations. 1 1 We get p = and q = . 3 3

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

47

48

1 1 1 for p, we have = i.e., x – 1 = 3, i.e., x = 4. x −1 x −1 3 1 1 1 for q, we get = i.e., 3 = y – 2, i.e., y = 5 Similarly, substituting 3 y −2 y −2

Now, substituting

Hence, x = 4, y = 5 is the required solution of the given pair of equations. 2 3 5 4 2) Solve for (x, y) using cross multiplication, + = 13, − = −2; where ; x ≠ 0, y ≠ 0 x y x y Sol: The equation can be written as 2 3 –– (1) + − 13 = 0 x y

5 4 − +2 = 0 x y

–– (2)

1 1 = A, = B The given expression becomes x y

Let

–– (3) 5A – 4B + 2 = 0 2A + 3B – 13 = 0 Comparing with the general equations. a1A + b1B + c1 = 0 a2A + b2B + c2 = 0 b c − b2c1 2 × 3 + 4( −13 ) 6 − 52 −46 A= 1 2 = = = = +2 − 8 − 15 − 23 − 23 a1b2 − a2 b1 B=

c1a2 − c 2a1 ( −13)5 − 2(2) −65 − 4 −69 = = = =3 − 23 − 23 − 23 a1b2 − a2 b1

A=

1 x

2x =1;

x=½

B=

1 y

3y =1;

–– (4)

y=

1 3

UNSOLVED EXERCISE 3.6: CW Exercise: 1) Solve for (x, y) by reducing them to a pair of linear equation, 5 2 15 7 1 1 1 1 a) − + 1 = 0, + − 10 = 0 + = 3, − = 5 where (x≠0, y≠0) b) 2x 3y 7x 6y x+y x−y x+y x−y

c)

x+y x−y = 2, = 6; where ; x ≠ 0, y ≠ 0 xy xy

e)

xy 6 xy = , = 6;where: x + y ≠ 0, y − x ≠ 0 x+y 5 y −x

d)

2 3 9 4 9 21 + = , + = ; where : x ≠ 0, y ≠ 0 x y xy x y xy

2) Solve the following pairs of equations by reducing them to a pair of linear equations: 13 1 1 2 4 3 9 1 1 i) + = 2; + = ii) + = 2; + =5 2x 3x 3y 2y 6 x x y y

iii)

3 1 1 1 1 1 + = ; – =– 4 2(3 x + y ) 8 3x + y 3x − y 2(3 x − y )

3) Solve, 8v – 3u = 5uv; 6v – 5u = –2uv. 4) Solve, 3(2u + v) = 7uv; 3(u + 3v) = 11uv. 1 1 12 7 4 5) Solve, + = ; + = 2, where 2x + 3y ≠ 0 and 3x – 2y ≠ 0. 2 2 x + 3 y 3 x − 2y 2( 2 x + 3 y ) 7(3 x − 2 y ) Formulate the following problems as a pair of equations, and hence find their solutions: 6) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. 48

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

49

7) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by one man and one woman alone to finish the work. HW Exercise: 1) Solve the following pairs of equations by reducing them to a pair of linear equations: 4 3 7 x − 2y 8x + 7y i) + 3y = 14; – 4y = 23 ii) = 5; = 15 x x xy xy iii) 6x + 3y = 6xy; 2x + 4y = 5xy

iv)

10 2 15 5 + = 4; + = –2 x+y x−y x+y x−y

2) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately. 3) Solve for (x, y) by reducing them to a pair of linear equation, x + y − 8 x + 2y − 14 3 x + y − 12 3 a) x + 2y = 1.3, b) = = =1 2 3 11 (2 x + 5 y )

c) 4 x + 4) Solve,

6 8 = 15, 6 x − = 14; ( y ≠ 0 ) y y

5 1 10 5 2 2 – = ; + = , where x ≠ –1 and y ≠ 1. x +1 2 x +1 2 y −1 y −1

5) Solve, x + y = 5xy; 3x + 2y = 13xy, where x ≠ 0 and y ≠ 0. 6) Solve, 2(3u – v) = 5uv; 2(u + 3v) = 5uv. 17 2 3 5 1 7) Solve, + = ; + = 2. 5 3 x + 2y 3 x + 2y 3 x − 2y 3 x − 2y 8) Find the real values of x and y which will make (2x – 3y –13)2 + (3x + 5y + 9)2 = 0.

Word Problems: Tips on solving word problems: z z z z

Assume the unknown value (value to find) as variable. Identify the conditions given. Convert these conditions in an equation. Solve the simultaneous equation for finding the desired value.

SOLVED EXAMPLES 3.7: 1) A fraction becomes 4/5 if 1 is added to both numerator and denominator. If however 5 is subtracted from both numerator and denominator, the fraction becomes 1/2. What is the fraction? Sol: Let the numerator of the fraction be x and the denominator of the fraction y x ∴ Fraction = y

According to the 1st condition given According to 2nd condition given

Volume

x +1 4 = y +1 5

x −5 1 = y −5 2

Universal Tutorials – X CBSE (2012–13) – Mathematics

–––– (1) –––– (2)

49

50

2)

Sol:

3)

Sol:

∴ 5(x + 1) = 4(y + 1); 5x + 5 = 4y + 4; 5x – 4y = – 1 –––– (1) 2(x – 5) = 1(y – 5) 2x – 10 = y – 5 2x – y = 5 –––– (2) Solve for x and y 5x – 4y = –1 2x – y = 5 ∴ x = 7 and y = 9 fraction = 7/9 Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B at the same time. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other they meet in one hour. What are their speeds? Let the speed of car from A = x km/hr Distance travelled in 7 hrs = 7x km Let the speed of car from B = y km/hr Distance travelled in 7 hrs = 7y km ––– (1) According to 1st condition, 7x = 70 + 7y, x – y = 10 nd According to 2 condition, x + y = 70 ∴ x = 40 km/hr y = 30 km/hr ∴ Speed of car from A = 40 km/hr Speed of car from B = 30 km/hr A bag contains 94 coins of 50 paise and 25 paise denominations. If the total worth of these coins were Rs.29.75, find the number of coins of each kind. Let the number of 50 paise coins be x; Let the number of 25 paise coins be y Their values = 50x + 25y = 2975 paise ∴ x = 25, y = 69 x + y = 94 ∴ Number of 50 paise coins = 25 Number of 25 paise coins = 69

UNSOLVED EXERCISE 3.7: CW Exercise: 1) I am three times as old as my son. Five years later, I shall be two–and–a–half times as old as my son. How old am I and how old is my son? 2) A is two years older than B. A’s father D is twice as old as A and B is twice as old as his sister C. The ages of D and C differ by 40 years. Find the ages of A and B. 3) A fraction is such that if the numerator is multiplied by 3 and the denominator reduced by 3 we get 18/11; but if the numerator is increased by 8 and the denominator is doubled we get 2/5. Find the fraction. 4) The sum of the digits of a two–digit number is 9. The number obtained by reversing the order of digits of the given number exceeds the given number by 27. Find the given number. 5) Seven times a two–digit number is equal to four times the number obtained by interchanging the digits, and the digit in units place exceeds digit at 10’s place by 3. Find the number. 6) A two–digit number is obtained by either multiplying sum of the digits by 8 and adding 1 or by multiplying the difference of the digits by 13 and adding 2. Find the number. 7) If three times the larger of the two numbers is divided by the smaller one, we get 4 as quotient and 3 as remainder. Also, if seven times the smaller number is divided by the larger one, we get 5 as quotient and 1 as remainder. Find the numbers. 8) In a cyclic quadrilateral ABCD, find the four angles: ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x – 5)° 9) In a ΔABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles. 10) 3 chairs and 2 tables cost Rs.700 and 5 chairs and 3 tables cost Rs.1100. What is the cost of 2 chairs and 2 tables? 11) On selling a tea–set at 5% loss and a lemon–set at 15% gain, a crockery seller gains Rs.7. If he sells the tea–set at 5% gain and the lemon–set at 10% gain, he gains Rs.13. Find the actual prices of the tea–set and the lemon–set. 12) A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Bombay to Ahmedabad costs Rs.216 and 50

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

13)

14)

15)

16)

17) 18) 19) 20) 21)

22)

51

one full and one half–reserved first class tickets cost Rs.327. What is the basic first–class full fare and what is the reservation charge? From Delhi station if we buy 2 tickets to station A and 3 tickets to station B, the total cost is Rs.77, but if we buy 3 tickets to station A and 5 tickets to station B, the total cost is Rs.124. What are the fares from Delhi to station A and to station B? A train covered a certain distance at a uniform speed. If the train had been 6km/h faster, it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6km/h, it would have taken 6 hours more than the scheduled time. Find the length of the journey. A and B each has some money. If A gives Rs.30 to B, then B will have twice the money left with A. But, if B gives Rs.10 to A, then A will have thrice as much as is left with B. How much money does each have? The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and the breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, the area is increased by 67 square units. Find the length and breadth of the rectangle. A takes 3 hours more than B to walk 30km. But, if A doubles his pace, he is ahead of B by 1½ hour. Find their speeds of walking. One number is greater than thrice the other number by 2 and 4 times the smaller number exceeds the greater by 5. Find the numbers. The income of A and B are in the ratio of 8:7 and their expenditures are in the ratio of 19:16. If each saves Rs.1,250. Find their incomes. 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by 1 man alone and that by 1 boy alone to finish the work. A boat covers 32km upstream and 36km downstream in 7 hours. Also, it covers 40km upstream and 48km downstream in 9 hours. Find the speed of the boat in the still water and that of the stream. ABCD is a cyclic quadrilateral. C Find the angles of the cyclic quadrilateral. –4x B 3y–5

4y+20

–7x+5

A D HW Exercise: 1) The sum of the prices of an Almirah and a table is Rs.2, 340 and the difference of their price is Rs.140. Find the price of each. 2) 3 chairs and 4 tables cost Rs.2250 and 4 chairs and 3 tables cost Rs.1950. Find the cost for 2 chairs and 1 table. 3) In a quadrilateral the measures of its angles are x + 6, 2x + 3, 4x, 5x – 9. Find x and find the measures of each angle. 4) In ΔPQR, QR is smaller than twice the length of side PQ by 2cm. The length of side PR exceeds PQ by 10cm. The perimeter of the triangle is 40cm. Find the length of each side. 5) A and B each have certain number of oranges. A says to B, ‘If you give me 10 of your oranges, I will have twice the number of oranges left with you’. B replies, ‘If you give me 10 of your oranges, I will have the same number of oranges as left with you’. Find the number of oranges with A and B respectively. 6) A father is four times the age of his son. 5 years hence father will be three times the age of his son. Find their present ages. 7) The perimeter of a rectangular plot is 32 meters. If the length increased by 2 meters and the breadth is decreased by 1 meter, the area of the plot remains unchanged. Find the dimensions of the plot. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

51

52

8) Ramesh travels 760 km to his home, partly by train and partly by car. He takes 8 hours if he travels 160 km by train and the rest by car. He takes 12 minutes more if he travels 240km by train and the rest by car. Find the speed of the train and the car separately. 9) A man has only 20 paise coins and 25 paise coins in his purse. If he has 50 coins in totaling Rs.11.25, how many coins of each type does he have? 10) A person invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. He received yearly interest of Rs.130. But if he had interchanged the amounts invested, he would have received Rs.4 more as interest. How much amount did he invest at different rates? 11) The ratio of incomes of two persons is 9:7 and the ratio of their expenditure is 4:3. If each of them saves of Rs.200 per month. Find their monthly income. 12) A dealer sold a VCR and a TV for Rs.25820, making a profit of 10% on VCR and 15% on TV. By selling them for Rs.25390, he would have realized a profit of 15% on VCR and 10% on TV. Find the cost price of each. 13) A man when asked how many hens and buffaloes he has told that his animals have 120 eyes and 180 legs. How many hens and buffaloes does he have? 14) A fraction is such that if the numerator is multiplied by 2 and the denominator is increased by 2, we get 5/4. But if the numerator is increased by 1 and the denominator is doubled, we get 1/2. Find the fraction. 15) A 90% acid solution is mixed with a 97% acid solution to obtain 21 litres of a 95% solution. Find the quantity of each of the solutions to get the resultant mixture. 16) There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is the same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room. 17) 2 men and 5 boys together can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work. 18) A man sold a chair and a table together for Rs.760 thereby making a profit of 25% on the chair and 10% on the table. By selling them together for Rs.767.50, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each. 19) A sailor goes 8km downstream in 40 minutes and returns back to the starting point in 1 hour. Find the speed of the sailor in still water and the speed of the current. 20) A motorboat takes 6 hours to cover 100km downstream and 30km upstream. If the motorboat goes 75km downstream and returns back to its starting point in 8 hours, find the speed of the motorboat in still water and the rate of the stream. 21) In covering a distance of 30km. Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would have taken 1 hour less than Amit. Find their rates of walking.

MISCELLANEOUS EXERCISE: 1) Solve, 19x – 17y = 55, 17x – 19y = 53 2) In case of the following system of equations determine whether the system has a unique solution, no solution or infinite no. of solutions. In case there is a unique solution find it. 3 15 ii) 6x + 5y = 11, 9x + y = 21 i) 7x – 2y = 3, 11x – y = 8 2 2 iii) 2x + 3y = 7, 6x + 5y = 11 3) For what value of k will the equations x + 5y – 7 = 0 & 4x + 20y + k = 0 represent coincident lines 4) Find the value of k for which the system of equations 3x+5y =0, kx+10y=0 has an infinite solution 5) Find the value of k for which the system of equations is inconsistent, – 2x + 3y = 2, 3x + ky = 1. 6) Solve for x and y i) 147x – 231y = 525, 77x – 49y = 203 ii) 35x + 23y = 209, 23x + 35y = 197 52

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

53

7) For what value of k, will the equations 2x + 3y – 5 = 0, 6x + ky – 15 = 0 have an infinite number of solutions. 8) For what value of k, the system of equations is kx + 2y = 5, 3x + y = 1 will be coincident 9) For what value of k, the system of equation is kx – 5y = 2, 6x + 2y = 7 will be inconsistent? 10) Show that each one of the following systems of equations has a unique solution. Find the solution. Find the solution by using the method of cross multiplication: i) 2x + 3y = 19, 3x + 5y = 31 ii) x – 8y = 70, 3x – 7y = 57 11) Solve each of the following systems of equations by eliminating x (by substitution): i) x + y = 7; 12x + 5y = 7 ii) 2x – 7y = –1; 4x + 3y = 15 12) Solve each of the following systems of equations by eliminating y (by substitution): i) 7x + 11y – 3 = 0; 8x + y – 15 = 0 ii) 2x + y – 17 = 0; 17x – 11y – 8 = 0 13) Solve the following systems of equations by eliminating method (by equ. coefficient of x or of y) 29 =0 ii) 7x – 8y – 11 = 0; 8x – 7y – 7 = 0 i) 4x – 3y – 8 = 0; 6x – y – 3 14) Solve the following systems of equations: 15 2 1 1 36 + = 17 ; + = i) 23x – 17y + 11 = 0; 31x + 13y – 57 = 0 ii) u v u v 5 7 11 uv . 15) Find the solution such that u ≠ 0, v ≠ 0 of, 2u + v = uv ; u + 3v = 3 3 16) Solve the following systems of equations by using cross multiplication method: 35 2 y = 25 ii) 4x + y – 1 = 0; 6x – y + 2 = 0 i) 4x + 7y = 10; 10x – 2 3 17) Determine, graphically whether the following system of linear equations has a unique solution or not: 2x – 3y = 6, x + y = 8. 18) Solve graphically 2 (x – 1) = y, x + 3y = 15. Find the coordinates of the points where the lines meet y–axis. 19) Solve graphically: x – 2y = 6, 3x – 6y = 0 20) Solve the equation graphically 2x + 3y = 12, x – y = 1, shade the region between the two lines and the x–axis. 21) Solve

2 2 1 3 2 + = , + = 0 and hence find the value of a for which y = ax + 4, (x ≠ 0) x 3y 6 x y

22) Solve

1 1 1 5 2 1 + = , − = , where 3x + 4y ≠ 0, 2x – 3y ≠ 0 2(3 x + 4 y ) 5( 2 x − 3 y ) 4 (3 x + 4 y ) 5( 2 x − 3 y ) 10

23) 3 bags and 4 pens together cost Rs.257 whereas 4 bags and 3 pens together cost Rs.324. Find the total cost of 1 bag and 10 pens. 24) The average age of Mahesh, Prakash and Akash is 20 years. Prakash present age is 15 years. 5 years ago the sum of ages of Prakash and Mahesh was 5/4 times than age of Akash. Find their present ages. 25) Five years ago, A was thrice as old as B and ten years later A shall be twice as old as B. What are the present ages of A and B? 26) In a two digit number the unit’s digit is twice the tens digit. If 27 is added to the number the digits interchange their place. Find the number. 27) The sum of the numerator and denominator of a fraction is 8. If the numerator and denominator are increased by 3, the fraction become 3/4. Find the fraction. 28) Point A and B are 90 km apart from each other on a highway. A car starts from A and another from B at the same time. If they go in the same direction, they meet in 9 hours and if they go towards opposite directions, they meet in 9/7 hours. Find their speeds. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

53

54

29) A man rowing at the rate of 5 km an hour in still water takes thrice as much time in going 40 km up the river as in going 40 km down. Find the rate at which the river flows. 30) A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs.1,500/– after 4 years of service and Rs.1,800/– after 10 years of service, what was his starting salary and what is the annual income? 31) The taxi charges in a city comprise of a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is Rs.75 and for a journey of 15 km, the charge paid is Rs.110. What will a person have to pay for traveling a distance of 25 km? 32) In an examination paper, one mark is awarded for every correct answer while ¼ mark is deducted for every wrong answer. A student answered 120 questions and got 90 marks. How many questions did he answer correctly? 33) The largest angle of a triangle is equal to the sum of the other two angles. The smallest angle is 1/4 of the largest angle. Find the angles of the triangle. 34) Use a single graph paper and draw the graph of the equations, 2y – x = 8; 5y – x = 14; y –2x = 1. 35) If 2x + y = 35 and 3x + 3y = 65 find the value of x/y. 36) Find the values of α and β for which the following system of linear equations has infinite number of solutions, 2x + 3y = 7; 2αx + (α + β)y = 28. 37) In a bag containing only white and black balls, half the number of white balls is equal to one– third of the number of black balls. Twice the total number of balls exceeds three times the number of black balls by 4. How many balls of each type does the bag contain? 38) Solve the following system of linear equations graphically, x – y = 1; 2x + y = 8. Shade the area bounded by these two lines and y–axis. Also, determine this area. 39) Draw the graph of the following equations, 2x – y – 2 = 0; 4x + 3y – 24 = 0; y + 4 =0. Obtain the vertices of the triangle so obtained. Also, determine its area. 40) After covering a distance of 30 km with a uniform speed there is some defect in a train engine and therefore, its speed is reduced to 4/5 of its original speed. Consequently, the train reaches its destination late by 45 minutes. Had it happened after covering 18 kilometers more, the train would have reached 9 minutes earlier. Find the speed of the train and the distance of journey. 41) By selling a table and a chair for Rs.1896, a trader gains 25% on the table and 10% on the chair. If he sells them for Rs.1770, he makes a profit of 10% on the table and 25% on the chair. Find the cost price of each. 42) Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. 43) The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju. 44) One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)]. 45) A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 46) The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class. 47) Draw the graphs of the equations 3x + y = 5 and 2x – y = 5. Determine the points where the intersect y–axis [CBSE–08] 48) Solve the following pair of linear equations using cross multiplication method: [CBSE–08] i) px + qy = p – q; qx – py = p + q ii) ax + by = c; bx + ay = 1 + c 54

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

55

x y – = 0; ax + by = a2 + b2 iv) 152x – 378y = – 74; –378x + 152y = – 604 a b 49) A traffic police at a Panvel Check Naka collected Rs.1400 for Challans paid by drivers who did not follow traffic rules. He fined Rs.50 for not wearing helmet and Rs.100 for not carrying proper documents. The total number of defaulters he caught was 20.How many people did he caught not wearing helmet? 50) Find the value(s) of k for which the pair of linear equations kx + 3y = k – 2 and 12x + ky = k has no solution. [CBSE–09]

iii)

ax by – = a + b, ax – by = 2ab [CBSE–09] b a 52) Without drawing the graph, find out whether the lines representing the following pair of linear 3 5 7 equations intersect at a point, are parallel or coincident: 9x – 10y = 21; x – y = [CBSE–09] 2 3 2

51) Solve for x and y:

MULTIPLE CHOICE QUESTIONS: CW Exercise: 1) The value of k for which the system of equations kx – y = 2; 6x – 2y = 3 has a unique solution, is a) = 3 b) ≠ 3 c) ≠ 0 d) = 0 2) The value of k for which the system of equations x + 2y – 3 = 0 and 5x + ky + 7 = 0 has no solution, is a) 10 b) 6 c) 3 d) 1 3) If the system of equations 2x + 3y = 7; (a + b)x + (2a – b)y = 21 has infinitely many solutions, then a) a = 1, b = 5 b) a = 5, b = 1 c) a = –1, b = 5 d) a = 5, b = –1 4) If am ≠ bl, then the system of equations ax + by = c; lx + my = n a) has a unique solution b) has no solution c) has infinitely many solutions d)may or may not have a solution. 5) The value of k for which the system of equations x + 2y = 5; 3x + ky + 15 = 0 has no solution is a) 6 b) –6 c) 3/2 d) None of these 6) A system of two linear equations in two variables is consistent, if their graphs a) are parallel b) are perpendicular c) intersect in a point d) none of these. 7) A dependent system of linear simultaneous equations will have a) one solution b) infinite solutions c) two solutions d) none of these. 8) The graph of x = 5 is a line: a) ⊥ to y-axis b) || to x-axis c) || to y-axis d) ⊥ to x-axis. 9) The equations x + 2y = 4 and 2x + y = 5 are: a) consistent and have a unique solution b) consistent and have infinite solutions c) inconsistent d) homogeneous linear equations. 10) If the system 2x + 3y – 5 = 0, 4x + ky – 10 = 0 has an infinite solution then: 3 3 a) k = b) k ≠ c) k ≠ 6 d) k = 6. 2 2 11) The equations 4x – 3y = 12, 12x – 9y = 21 have: a) two solution b) one solution c) no solutions d) infinite solutions. x y 12) The value of k for which kx + (k + 1)y = 12, + = 1 will have infinite solutions will be: 4 3

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

55

56

8 d) –2. 3 13) If the system of equation 2x + 3y = 11 and 2x – 4y = – 24 has a solution: a) x = 5, y = –2 b) x = 2, y = – 5 c) x = –2, y = 5 d) x = –5, y = 2. HW Exercise: 1) The value of k for which the system of equations 2x + 3y = 5; 4x + ky = 10 has infinite number of solutions, is a)1 b) 3 c) 6 d) 0 2) The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has a nonzero solution, is a) 0 b) 2 c) 6 d) 8 3) If the system of equations 3x + y = 1; (2k – 1)x + (k – 1)y = 2k + 1 is inconsistent, then k = a) 1 b) 0 c) –1 d) 2 4) If the system of equations 2x + 3y = 7; 2ax + (a + b) y = 28 has infinitely many solutions, then a) a = 2b b) b = 2a c) a + 2b = 0 d) 2a + b = 0 5) If 2x – 3y = 7 and (a + b)x – (a + b – 3)y = 4a + b represent coincident lines, then a and b satisfy the equation a) a + 5b = 0 b) 5a + b = 0 c) a – 5b = 0 d) 5a – b = 0 6) A inconsistent system of linear simultaneous equations will have a) no solution b) infinite solutions c) one solution d) none of these. 7) The graph of y = k (constant) is a line: a) parallel to x-axis b) parallel to y-axis c) perpendicular to x-axis d) none of these. 8) The graph of y = x is a line which is: a) parallel to x-axis b) parallel to y-axis c) ⊥ to x-axis d) none of these. 9) For what value of k will the equations 2x + 32y + 3 = 0 and 3x + 48y + k = 0 represent coincident lines? 3 2 9 a) b) c) d) 1. 2 3 2 10) The equations 2x – 3y + 6 = 0 and 6x – 9y + 7 = 0 have: a) unique solution b) no solution c) infinite solution d) none of these. 11) The value(s) of k for which the simultaneous equations 2kx + 6y + 9 = 0 and 4x + (2k + 2) y + 5 = 0 will have no solution, will be: a) 1, – 2 b) – 2, – 3 c) – 2, 3 d) – 3, 2. 12) The value of k for which kx + y = k2, x + ky = 1 will have no solution, will be: a) 1 b) ± 1 c) 2 d) – 1. 13) If the system 3x + 5y = 0, kx + l5y = 0 has a non-zero solution, then: a) k = 0 b) k = 9 c) k ≠ 0 d) k ≠ 3. 14) If the system 5x + ky = – 7 and x + 2y = 3 is inconsistent then: 2 2 a) k = 10 b) k= – c) k ≠ 10 d) k ≠ 5 5

a) 3

56

b) –4

c)

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

57

COLUMN MATCHING QUESTIONS: 1) Given in column I are types of graphs shown by some pair of linear equations in two variables. From column II choose correct option for each item in column I. Column I

Column II

i)

A) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0

Intersect at a point

ii) Lines are parallel

B) 2x + 3y – 8 = 0; 5x + 4y + 2 = 0

iii) Coincident lines

C) 6x – 3y + 10 = 0; 2x – y + 9 = 0 D) 5x – 3y = 11; –10x + 6y = –22

E) 3x – 2y = 4; 6x + 2y = 4 2) Consider the equation, 3(a1x + b1y – c1)2 + 2(a2x + b2y – c2)2 = 0 in real variables x and y, where a1, a2, b1, b2, c1, c2 are non–zero real numbers. For each item in column I, choose all the correct options in column II. Column I

Column II

i)

A)

a1 b ≠ 1 a2 b2

ii) No solution

B)

a1 b c = 1 = 1 a2 b2 c2

iii) Infinitely many solutions

C)

a1 b c = 1 ≠ 1 a2 b2 c2

D)

a1 b = 1 a2 b2

Unique solution

E) a22 b12 + a12b22 > 2a1a2b1b2 3) On comparing the ratios

a1 b1 c1 , , in column II choose correct option for each item in column I a2 b2 c 2

Column I

Column II

i)

A) 5x – 3y = 11; –10x + 6y = –22

No solution

ii) Unique solution

B) 3x – 2y = 4; 9x – 6y = 12

iii) Infinitely many solutions

C) 3x – 2y = 4; 6x + 2y = 4 D) 2x – 3y = 8; 4x – 6y = 9

E) 3x + 2y = 5; 2x – 3y = 7 4) Given in column I are linear equations having infinitely many solutions for some value of k. Choose correct option in column II for each item in column I. Column I

Column II

i)

A) 4

2x + 3y = 4; (k + 2)x + 6y = 3k + 2

ii) 3x + 2y = 1; (2k + 1)x + (k + 2)y = k – 1

B) 3

iii) x + (k + 1)y = 5; (k + 1)x + 9y = 8k – 1

C) –2

iv) (k – 1)x – y = 5; (k + 1)x + (1 – k)y = 3k + 1

D) 2 E) –3

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

57

58

5) Given in column I are the graphs of pair of linear equations. From Column II choose correct option for each item in column I. Column I Column II i)

A) 2y – x = 9; 6y – 3x = 21

ii)

B) 4y – 2x = 18; 12y – 6x = 42

iii)

C) x + 3y = 6; 2x – 3y = 12

D) 5x – 8y + 1= 0; 3x –

24 3 y+ =0 5 5

ANSWERS TO UNSOLVED EXERCISES: CW Exercise 3.1: 3) x + 2y = 1300, x + 2y = 1300 HW Exercise 3.1: 2) 2x + y = 160, 4x + 2y = 300 14 − 3 x 5) y = , No 7 58

4) x – 7y + 42 = 0, x – 3y – 6 = 0

5) No

3) x – 2y = 0, 3x + 4y = 20

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

59

CW Exercise 3.2: 1) (i) intersect at a point (ii) Coincident (iii) parallel 2) a) (3,–2) b) 1.5, 1 c) 0.5, 1.5 d, e) No Sol 3) x + y = 10, x – y = 4 , boys = 3, girls = 7 4) Rs 2 coins = 5, Rs 5 coins = 3 5) Length 20 m, breadth 60 m 6) i) 3x + 2y – 7 = 0, 2x + 3y – 12 = 0, 4x + 6y – 16 = 0 ii) 4x + 3y + 12 = 0, 3x – 4y – 2 = 0, 9x – 12y – 36 = 0 9) 8 Sq. Unit 10) 13.5 Sq. Unit 11) 40 Sq. Unit HW Exercise 3.2: iii) consistent, x = 2, y = 2 iv) inconsistent 1) i and ii) inconsistent 2) (i) consistent (ii) inconsistent (iii to v) consistent 3) No 4) No 5) 5x + 7y = 50, 7x + 5y = 46, (Rs 3, Rs 5) 6) (–4, 2), (2, 5), (1, 3) 7) (0, 0), (2, 2), (5, 0) Area = 5 8) Inconsistent 9) (5, 0), (0, 5), (0, –5) Area = 25 10) (0, 0), (2, 4), (3, 3) 11) 3x – 5y = 19 – No, 3y – 7x + 1 = 0 – Yes CW Exercise 3.3: 2) x = –2, y = 5, m = –1 1) (i) (9, 5), (ii) s = 9, t = 6 (iii) x = 2, y = 3 3) x = 2a, y = –2b 4) x = 1, y = –1 5) Infinite Sol – Coincident Lines 6) No Sol – Parallel Lines 7) x – y = 26, x = 3y, x = 39, y = 13 8) x + 5y = 38, x +10y = 63, x = 13, y = 5, 28 7 x 9) 11x – 9y + 4 = 0, 6x – 5y + 3 = 0, = 9 y HW Exercise 3.3: ii) x = 0, y = 0 iii) x = 2, y = 3 1) i) x can take infinite values 2) - 9 3) x=a2, y=b2 4) x = a, y = b 5) i)Parallel ii)Coincident 6) x – y = 18, x + y = 180, x = 99, y = 81 7) 7x+6y = 3800, 3x +5y = 1750, x = 500, y = 50 8) x – 3y – 10 = 0, x – 7y + 30 = 0, Jacob = 40, Son = 10 9) x = 2y, x + y = 3, Number = 21 CW Exercise 3.4: ⎛ 7 170 ⎞ ⎟⎟ iii) (5, 3) , ii) ⎜⎜ − ⎝ 87 87 ⎠ 2) x = 3, y = 2 3) x = 3, y = –1 3 6) 7) Nuri=50 & Sonu=20 5 HW Exercise 3.4: 1) i) (5, 2) (ii) (–2, –1) (iii) (1, –2) (iv) (–2.8, –2.3) 4) x = 3, y = 2 5) 18 CW Exercise 3.5: 2) k ≠ –6 1) (a) No (b) infinite

1) i) (1, –2)

22 ⎞ ⎛ 33 iv) ⎜ , − ⎟ 13 13 ⎠ ⎝

v) (3, 4) (vi) (0, 0)

4) x = a, y = b

5) x = a2, y = b2

8) x = 15, y = 3 2) x = 3, y = 2 3) x = 3, y = –1 6) Rs.50 = 10 & Rs.100 = 15 3) k = 4

4) k = 6

⎛ 11⎞ 5) ⎜ 0, ⎟ ⎝ 3⎠

6) k = 15

7) (i) k ≠ 3 (ii) k = 3

8) p ≠ 5q

9) a = 5, b = 1

10) x = a+b, y = –

11) 400, 130 Volume

12)

(iii) No value

2ab a+b

5 12

Universal Tutorials – X CBSE (2012–13) – Mathematics

59

60

HW Exercise 3.5: 1) a) Unique (–1, 1) 2) k ≠ −

10 3

6) (i) (3, 2) p =

13 ⎞ ⎛ b) Unique ⎜ 0, − ⎟ 7 ⎠ ⎝ 3) −

9 4

4 (ii) (20, 30) 3

9) k = 2

r 10) x = a, y = b

c to f) infinite 4) k = 6

5) k = –2

7) am ≠ bl

8) No solution

11) x = a, y = b

12) x=

b2 2a 2 + b 2 , y= 2a 2a

13) 60, 40 CW Exercise 3.6:

⎛ 1 1⎞ ⎛ 1 1⎞ 1) (a) ⎜ , ⎟ (b) (3, 2) (c) ⎜ − , ⎟ (d) (1, 3) (e) (2, 3) 6 14 ⎝ ⎠ ⎝ 2 4⎠ 1 1 ,y= (ii) x = 4, y = 9 (iii) x = 1, y = 1 2 3 11 22 v = 0 or , u = 0 or 4) v = 0 or 3/2, u = 0 or 1 23 31 x = 2, y = 1 speed in still water (u) + speed of the current (v) = 10, u – v = 2, u = 6, v = 4 2 5 6 1 3 1 + = ; + = , m = 36, w = 18 w m 4 w m 3

2) (i) x = 3) 5) 6) 7)

HW Exercise 3.6: 1 1) i) x = , y = –2 5 2) u = 60, v = 80 4) x = 4, y = 5 8) (2, –3)

ii) x = 1, y = 1

iii) x = 1, y = 2

iv) (x=3, y=2)

3) (a) (0.5, 0.4) (b) (2, 6) (c) (3, 2) 5) x = 1/2, y = 1/3 6) u = 2, v = 1

7) x = 1, y = 1

CW Exercise 3.7: 1) 45, 15 5) 36 9) 20, 40, 120 13) 13, 17 10 17) ,5 3 21) 10 km/hr, 2 km/hr HW Exercise 3.7: 1) 1240, 1120 4) 8, 14, 18 8) 80, 100 60

6) 41 10) 600 14) 720 km

12 25 7) 25, 18 11) 100, 80 15) 62, 34

18) 20, 7

19) 6000, 5250

2) 26, 24

3)

4) 36 8) 70, 53, 110, 127 12) 210, 6 16) 17, 9 20) 140, 280

22) ∠A = 120, ∠B = 70, ∠C = 60, ∠D = 110

2) 750 5) A = 70, B = 50 9) 25, 25

3) x = 30, Angle 36, 63, 120, 141 6) 40, 7 7) 10, 6 10) 500, 700 11) 1800, 1400

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 03: Pair of Linear Equations in Two Variables

12) 12600, 10400

13) 30, 30

16) 100, 80 17) 18, 36 21) 7.5, 5 Miscellaneous Exercise: 1) (2, –1) 3) k = –28 7) k = 9

18) 350, 300

61

5 6 19) 10, 2

14)

15) 6, 15 20) 20, 5

2) i) unique solution (ii) No solution (iii) unique solution 9 4) k = 6 5) k = − 6) i) (2, –1) ii) (4, 3) 2 8) k = 6 9) k = –15 10) i) (2, 5) ii) (–2, –9)

11) i) (–4, 11) ii) (3, 1)

12) i) (2, –1) ii) (5, 7)

⎛3 −2⎞ ⎛ 7 13 ⎞ 13) i) ⎜ , ⎟ ii) ⎜ − ,− ⎟ 3 2 ⎝ 5 5 ⎠ ⎝ ⎠

⎛ 1⎞ 14) i) (1, 2) ii) ⎜ 5, ⎟ ⎝ 7⎠

⎛ 1 1⎞ 15) ⎜ , ⎟ ⎝2 3⎠

⎛ 5 ⎞ ⎛ −1 7 ⎞ 16) i) ⎜ ,0 ⎟ ii) ⎜ , ⎟ 17) (6, 2) ⎝ 2 ⎠ ⎝ 24 4 ⎠

18) (3, 4) (0, –2) (0, 5)

19) No solution

20) (3, 2)

21) (6, –4), a = −

22) (2, 1)

23) Rs.155 3 27) 5 31) Rs.5; Rs.7

24) 15, 20, 25 years

25) 50, 20 years

28) 40, 30 km/hr.

29) 2.5 km/hr.

26) 36

4 3

30) Rs.1,300, Rs.50 32) 96 33) 22.5°, 67.5°, 90° 8 35) 36) α = 4, β = 8 37) White = 8, black = 12 5 38) (3, 2) Area 13.5 sq.units 39) (3, 4); (–1, –4); (9, –4) Area 40 sq.units 40) speed 30 km/hr.; distance 120 km 41) Rs.1,200, Rs.360 42) (–1, 0), (4, 0), (2, 3) 43) A= 19 or 21, B = 16 or 24 44) 40, 170 46) 36 45) 600 km 47) (0, –5), (0, 5) c (a − b ) − b c (a − b ) + a 48) i) (1, –1) (ii) (iii) (a, b) (iv) x = 2, y = 1 , a2 − b2 a2 + b2 49) 12 Column Matching Question: 1) i–BE; ii–C; iii–AD

2) i–AE; ii–C; iii–B

3) i–D; ii–CE; iii–AB

4) i–D; ii–A; iii–D; iv–B

5) i–C; ii–D; iii–AB

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

61

62

Chapter 06: Triangles Chapter Map: Similar Figures Similarity of Triangles

Basic Proportionality Theorem (Thales Theorem)

Converse of Basic Proportionality Theorem

Criteria for Similarity of Triangles

Congruence of Triangles

Similarity of Triangles

A–A–A Similarity (A–A Similarity)

S–S–S Similarity

Tests for Similarity of Triangles

S–A–S Similarity

Areas of Similar Triangles Pythagoras Theorem

Pythagoras Theorem

Similarity in Right Angled Triangles

Converse of Pythagoras Theorem

Applications of Pythagoras Theorem

Congruent Figures: ¾ ¾ ¾ ¾

Two line segments are said to be congruent if they are of equal lengths. Circles with equal radii are congruent. Squares with equal sides are congruent. Equilateral triangles with equal sides are congruent.

Similar Figures: z z z z z

Two similar figures have the same shape but not necessarily the same size. All circles are similar. All equilateral triangles are similar. All squares are similar. All congruent figures are similar but the similar figures need not be congruent

Similar Polygons: z

62

Two polygons of the same number of sides are similar, if „ their corresponding angles are equal (i.e. they are equiangular) and Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles „

63

their corresponding sides are in the same ratio (or proportional)

Note: Two quadrilaterals (a square and a rectangle) of corresponding angles are equal, but their corresponding sides are not in the same ratio then they are not similar.

Two quadrilaterals (a square and a rhombus) of corresponding sides are in the same ratio, but their corresponding angles are not equal, then they are not similar. Thus, either of the above two conditions of similarity of two polygons in not sufficient for them to be similar.

UNSOLVED EXERCISE 6.1: CW Exercise: 1) Fill in the blanks using the correct word given in brackets: i) All circles are _____ (congruent, similar) ii) All squares are ____ (similar, congruent) iii) All ____ triangles are similar. (isosceles, equilateral) iv) Two polygons of the same number of sides are similar, if a) their corresponding angles are ______ and b) their corresponding sides are ____ (equal, proportional) 2) Give two different examples of pair of i) Similar figures ii) non–similar figures. 3) State whether the following quadrilateral are similar or not: D

3 cm

C S

3 cm

A

3 cm

3 cm

1.5 cm

1.5 cm

B

P

R 1.5 cm

1.5 cm

Q

Similarity of Triangles: Congruence of Triangles: ¾ Two triangles are said to be congruent if:  Their corresponding angles are congruent, and  Their corresponding sides are equal/ congruent

Similar Triangles: z

Two triangles are said to be similar if: „ Their corresponding angles are congruent, and „ Their corresponding sides are proportional

Note: In congruent triangles the corresponding sides are equal whereas in similar triangles the corresponding sides are proportional.

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

63

64

œ Theorem 6.1 Basic Proportionality Theorem (Thales Theorem):  If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then it divides the two sides in the same ratio. [CBSE–09]

Given:

A

In ΔABC, DE || BC, and intersects AB in D and AC in E

N

F

To Prove:

D

AD AE = DB EC

E

B

Construction:

C

Join BE, CD, and draw EF ⊥ AB, DN ⊥ AC

Proof: 1) A(ΔADE) = ½AD × EF

[Area of a triangle = ½bh]

2) A(ΔBDE) = ½BD × EF

[Area of a triangle = ½bh]

3)

A( ΔADE ) AD × EF = A( ΔBDE ) BD × EF

4)

A( ΔADE ) AD = A( ΔBDE ) BD

5) Similarly,

[Statement (1) and (2)]

A( ΔADE ) = A( ΔCDE )

1 × AE × DN 2 1 × EC × DN 2

AE EC

=

6) But A(ΔBDE) = A (ΔCDE) [Δ on the same base DE & between the same || DE & BC] 7) ∴

AD AE = BD EC

[Statement 4, 5, 6]

Corollary: z

If in a ΔABC a line parallel to BC intersects other two sides AB & AC at D and E.

z

Then

AD AE = AB AC

A

AD AE = AB AC

Data: DE || BC

TPT:

Proof: DE || BC

AD AE ∴ = BD EC

According to invertendo

E

[BPT]

BD EC = AD AE

B

BD EC +1 = +1 AD AE

[Adding 1 to both the sides]

AB AC = AD AE

[Taking LCM]

Taking their reciprocals, 64

D

C

AD AE = AB AC

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

65

Converse of Basic Proportionality (Thales) Theorem: Theorem 6.2: z

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. A

Interpretation:

D

In ΔABC, line l intersects AB in D and AC in E, such that AD AE = DB EC Then, line l parallel to BC.

B

E

l C

SOLVED EXAMPLE 6.2: 1)

Using Basic Proportionality Theorem, prove that the line, drawn from the mid point of one side of a triangle, parallel to another side bisects the third side. Sol: Data: In ΔABC, D is the midpoint of seg AB. DE || BC. TPT: AE = CE A Proof: DE || BC [Given] AD AE D E = [B.P.T] BD CE C B AD = BD [Given] AD AE ∴ = 1= ∴ AE = EC BD CE 2) If the diagonals of a quadrilateral divide each other proportionally prove that it is a [CBSE–08] trapezium. AO OB D Sol: Data: In quadrilateral ABCD, = C OC OD O RTP: Quadrilateral ABCD is a trapezium E Construction: Draw OE || CD. Proof: OE || CD [By construction] A B BO BE = [B.P.T] OD CE AO OB = [Given] But OC OD AO BE ∴ = ∴ OE || AB [Converse of B.P.T.] OC CE ∴ OE || CD || AB ∴ AB || CD ∴ ABCD is a trapezium. 3) If three or more parallel lines are intersected by two transversals, prove that the intercepts made by them on the transversals are proportional. Sol: Data: l || m || n, p and q are two transversals making intercepts AB, BC and DE and EF AB DE = BC EF Construction: Join AF intersecting BE at O.

RTP:

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

65

66

Proof: In ΔACF, BO || CF [Part of the parallel lines] ∴

AB AO = BC OF

[B.P.T]

In ΔAFD, OE || AD

(Part of the parallel lines)

OF EF = OA DE

[BPT]

∴

AB AO DE = = BC OF EF

[II]

AB DE = [from I and II] BC EF 4) M & N are points on the sides PQ & PR respectively of a ΔPQR. State whether MN || QR. Given: PM = 4, QM = 4.5, PN = 4, NR = 4.5 P Sol: Proof:

i.e.

PM 4 PN = = MQ 4 .5 NR

∴

MN is parallel to QR

[Converse of B.P.T]

M

PM PN = MQ NR

N R

Q

UNSOLVED EXERCISE 6.2: A

CW Exercise: 1) If in figure PQ || BC, AP = 3 cm; BP = 6 cm; CQ = 5.3 cm. Find AQ.

P

Q C

B

2) M and N are points on the sides PQ and PR respectively of a ΔPQR. For the following case state whether MN || QR. PQ = 1.28, PR = 2.56, PM = 0.16, PN = 0.32. C 3) In figure, if

AD BE = and ∠CDE = ∠CED, prove that ΔCAB is isosceles. DC EC

A

E B

D A

D 4)

In figure, DE || AC and DC || AP. Prove that B

E

C

BE BC = . EC CP

P D

P E

5) In figure, DE || AQ and DF || AR. Prove that EF || QR.

F A

Q

R

6) Using Basic Proportionality Theorem, prove that the line, drawn from the mid point of one side of a triangle, parallel to another side bisects the third side. 7) ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. AO CO = . Show that BO DO 66

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

67

C 8)

L

In fig. LM || AB. If AL = x – 3, AC = 2x, BM = x – 2 and BC = 2x + 3, find the value of x.

M

A B 9) Let X by any point on the side BC of a ΔABC. If XM, XN are drawn parallel to BA and CA meeting CA, BA in M, N respectively; MN meets BC produced in T, prove that: TX2 = TB × TC. A 1 5cm D

HW Exercise:

1) In fig, i) and ii) DE || BE.

1 cm E

A 1.8 cm

D

7.2 cm

E

B

3 cm

5.4 cm

Find EC in (i) and AD in (ii). B

(i)

C

C

(ii)

2) E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9cm iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm 3) The diagonals of a quadrilateral ABCD intersect each other at the point O such that B

Show that ABCD is a trapezium 4) In figure, if PQ || BC and PR || CD, prove that A

Q

AR AQ = . AD AB

A

P

C

R

P

5) In figure, PQ || AB and PR || AC.

AO CO = BO DO

D

Prove that QR || BC. Q B

O

R C

6) Using Converse of Basic Proportionality Theorem, prove that the line joining the mid points of two sides of the triangle is parallel to the third side. 7) Prove that any line parallel to the parallel sides of a trapezium divides the non–parallel sides proportionately. 8) ABCD is a parallelogram, P is a point on side BC and DP when produced meets AB produced at L. Prove that i)

DP DC = PL BL

ii)

DL AL = DP DC

9) Two ΔABC and DBC lie on the same side of the base BC. From a point P on BC, PQ || AB and PR || BD are drawn. They meet AC in Q & DC in R respectively. Prove that QR || AD. [CBSE-09] 10) In a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. i) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x. ii) If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = (3x – 1), find the value of x.

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

67

68

Criteria for Similarity of Triangles: ™ It must be noted that as done in the case of congruency of two triangles, the similarity of two triangles should also be expressed symbolically, using correct correspondence of their vertices. ™ For example, for the similarity of triangle ABC and DEF. ™ We cannot write ΔABC ~ ΔEDF or ΔABC ~ ΔFED. However, we can write ΔBAC ~ ΔEDF.

Tests for Similarity of Triangles: ¾ A group of sufficient conditions (particular conditions that are sufficient to ensure similarity) are called as ‘Tests for Similarity of Triangles’. ¾ There are primarily 3 tests available: 1) A–A–A test (A–A test) 2) S–S–S test 3) S–A–S test

Theorem 6.3: A–A–A Criterion of Similarity  If in two triangles, corresponding angles are equal, then their corresponding sides are proportional (i.e. in the same ratio) and hence the two triangles are similar.

A

Interpretation:

P

In ΔABC and ΔPQR If ∠A = ∠P; ∠B = ∠Q;

C

B

R

Q

∠C = ∠R then ΔABC ∼ ΔPQR

Corollary: (A – A Similarity) If two angles of a triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This is referred to as the AA Similarity criterion for two triangles A P Interpretation: z

In Δ ABC and ΔPQR If ∠A = ∠P ∠B = ∠Q then ΔABC ∼ ΔPQR

C

B

R

Q

Theorem 6.4: S–S–S Criterion of Similarity  If in two triangles, sides of one triangle are proportional to (i.e. in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. A P Interpretation:

In ΔABC and ΔPQR If

AB BC AC then ΔABC ∼ ΔPQR = = PQ QR PR

B

C

Q

R

Theorem 6.5: S–A–S Criterion of Similarity  If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. 68

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

69

Interpretation:

A

P

In ΔABC and ΔPQR If

AB BC and ∠B = ∠Q = PQ QR

Then ΔABC ∼ ΔPQR

B

C

R

Q

SOLVED EXAMPLE 6.3: 1)

Sol:

2)

Sol:

3)

Sol:

If two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle, prove that the triangles are similar. A P Data: In ΔABC and ΔPQR, BX and CY are the medians. AB AC BX = = X Y PQ PR QY To prove: ΔABC ∼ ΔPQR. B C Q R AB AC Proof: [Given] = PQ PR 1 AC AB = 2 [Given] 1 PQ PR 2 AB AX ⇒ [AX = XC; PY = YR] = PQ PY BX [given] = QY ΔABX ∼ ΔPQY [SSS ∼ test] ∠BAX ≅ ∠QPY [Corresponding angles of similar Δs] AB AC In ΔABC and ΔPQR, [Given] = PQ PR ∠A = ∠P proved above ∴ ΔABC and ΔPQR [SAS Test] P and Q are points on the sides AB and AC respectively of a triangle ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm, QC = 6 cm, prove that BC = 3 PQ. AP 2 1 AQ A = = = PB 4 2 QC P Q ∴ PQ || BC [Converse of B.P.T] ∴ ΔAPQ ∼ ΔABC [AA Corollary] C B AP PQ = [Corresponding sides of similar triangles are proportional] AB BC 2 PQ = ⇒ 2BC = 6PQ ⇒ BC = 3PQ 6 BC A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time a tower casts the shadow 40 m long on the ground. Determine the height of the tower. E Let AB be the vertical stick and DE be the tower ∠C = ∠C [Common angle] A ∠ABC = ∠EDC = 90° ∴ ΔCDE ∼ ΔCBA [AA Corollary] C

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

B

D

69

70

DE DC = AB BC 40 h = ⇒ h = 60cm 12 8

[Corresponding sides of similar triangles] ∴ Height of the tower = 60cm

UNSOLVED EXERCISE 6.3: CW Exercise: 1) State which pairs of triangle in the fig. are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form: P

L A

i)

6

5

3

2 B

2.5

C

Q

4

ii)

R

D 3

2.7 2

M

6

4 E

P

5

F

P

M

iii) 2.5

6

70° 5

70° L

N

Q

10

R

2) In fig., ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB. C D 70° 125° O A

B

QT QR and ∠1 = ∠2. Prove that ΔPQS ∼ ΔTQR. 3) In figure = PR QS

4) If CD and GH are respectively bisectors of ∠ACB and ∠EGF such that D and H lie on AB and FE of ΔABC and ΔFEG respectively. If ΔABC ∼ ΔFEG, prove that CD AC ii) ΔDCA ∼ ΔHGF iii) = i) ΔDCB ∼ ΔHGE GH FG 5) In Figure, ΔFCE ≅ ΔGBD and ∠1 = ∠2. Prove that ΔADE ∼ ΔABC

6) Sides AB and AC and median AD of a triangle ABC are proportional to sides PQ and PR and median PM of another triangle PQR. Prove that ΔABC ~ ΔPQR. 7) A girl of height 90 cm is walking away from the base of a lamp–post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds. CA CB 8) D is a point on the side BC of ΔABC such that ∠ADC = ∠BAC. Prove that = CD CA A 9) In fig, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

D

B

70

E C

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

71

10) If AD and PM are medians of triangles ABC and PQR respectively where ΔABC ~ ΔPQR prove AB AD that = PQ PM 11) In a triangle ABC, P, Q are points on AB, AC respectively and PQ || BC. Prove that the median AD bisects PQ. 12) Through the vertex D of a parallelogram ABCD, a line is drawn to intersect the sides CB in F and DA FB FC = = . AB produced at E. Prove that A AE BE CD 13) Trapezium PQRS is carved out of a triangle ∆ABC such that its parallel S P sides are parallel to BC as shown in the figure. Find the dimensions of ∆ABC, given that RC = 4.8, PQ = SR = 6, PS = 10 and QR = 15. Q R (All the lengths are in centimeter) C B HW Exercise: 1) A vertical pole of length 6m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower 2) Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. OA OB Using similarity criterion for two triangles show that = OC OD 3) S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS. 4) E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show [CBSE–08] that ΔABE ~ ΔCFB. 5) The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle. 6) Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side [CBSE–08] of BC. If AC and DB intersect at P, prove that AP × PC = BP × PD. 7) Examine each pair of triangle in figure and state which pair of triangle are similar. Also state the similarity criterion used by you for answering the question and write the similarity relation in E symbolic form. P A A 60°

40°

B

80°

60°

40°

C Q Q.7 (i)

80°

5 cm

80°

R

B

3 cm

80°

C F Q.7 (ii)

6 cm

D

R

S

In figure, if PS || QR, prove that ΔPOS ~ ΔROQ.

8) O P

Q.8

9) In figure, AD and CE are two altitudes of ΔABC intersect each other at the point F Prove that ii) ΔABD ~ ΔCBE i) ΔAEF ~ ΔCDF iv) ΔFDC ~ ΔBEC iii) ΔAEF ~ ΔADB Volume

C

Q

D F A

Universal Tutorials – X CBSE (2012–13) – Mathematics

E

B 71

72

10) In ΔABC, AB = 5, BC = 12, and AC = 13. Which of the following triangles is similar to ΔABC? i) A triangle with sides 3, 5, and 12 ii) A triangle with sides 15, 36, and 39 iii) A triangle with sides 10, 22, and 24 iv) A triangle with sides 3, 4, and 5 11) If the angles of one triangle are respectively equal to the angles of another triangle, prove that the ratio of the corresponding sides is the same as the ratio of the corresponding. i) Medians ii) Bisectors of angles (angle bisector segments) iii) Altitudes 12) In a triangle ABC, a point D is located on side BC such that ∠ CAB = ∠CDA. Show that CA2 – CD2 = DB.CD.

Areas of Similar Triangles: œ Theorem 6.6:  The ratio of the areas of two similar triangles is equal to the square ratio of their corresponding sides. [CBSE–09]

Given: ΔABC ~ ΔPQR

P

Area ( ΔABC ) AB 2 BC 2 AC 2 To Prove: = = = Area ( ΔPQR) PQ 2 QR 2 PR 2

Construction: Draw AD ⊥ BC, PS ⊥ QR. 1 Proof: ar(ΔABC) = BC × AD 2

A

B D

C

Q

R S

1 1 QR × PS (Area of triangle = base × height) 2 2 1 BC × AD ar ( ΔABC ) 2 BC × AD ––– (I) = = 1 ar ( ΔPQR ) QR × PS QR × PS 2 [Given ΔABC ~ ΔPQR] In ΔADB ∼ ΔPSQ, ∠B = ∠Q ∴ ∠ADB = ∠PSQ [By construction, each 90°] ΔADB ∼ ΔPSQ [AA corollary] AD AB [Corresponding sides of similar triangles are proportional] = PS PQ AB BC But [Corresponding sides of ΔABC ∼ ΔPQR] = PQ QR AD BC ∴ [II] = PS QR

ar(ΔPQR) =

ar ( ΔABC ) BC × BC BC 2 [I and II] = = ar ( ΔPQR ) QR × QR QR 2 AB BC AC (As ΔABC ∼ ΔPQR) = = PQ QR PR

∴

Hence,

72

AB 2 BC 2 AC 2 ar (Δ ABC ) = = = ar (ΔPQR ) PQ 2 QR 2 PR 2

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

73

SOLVED EXAMPLE 6.4: 1)

Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal. [CBSE–09]

Sol:

ΔADE and ΔACP are equilateral triangles ∴ ΔADE ~ ΔACP Let the side of square ABCD = a ∴ Its diagonal = a 2 2

AD ar ( Δ ADE ) = ar ( Δ ACP ) AC 2 =

[Given]

C

D

E

a 2

[Area of similar Δs]

a2 (a 2 )

2

=

1 2

A

B

a

1 ar(ΔACP) 2 If areas of two similar triangles are equal prove that the triangles are congruent.

∴ ar(ΔADE) = 2)

Sol.: Data: ΔABC ~ ΔDEF

A(ΔABC) = A(ΔDEF)

A

To prove that: ΔABC ≅ ΔDEF Proof: ΔABC ~ ΔDEF ∴

A( ΔABC ) BC = A( ΔDEF ) EF 2

P

D

(Given)

2

(Area of similar triangle)

B P

Put A(ΔABC) = A(ΔDEF)

C

E

F

Q

BC 2

=1 i.e. BC2 = EF2 or BC = EF EF 2 Similarly we can prove that AB = DE, AC = DF.

∴

∴ ΔABC ≅ ΔDEF

(SSS ≅ test)

UNSOLVED EXERCISE 6.4: CW Exercise:

1) The areas of two similar triangles ABC and PQR are 64 cm2 and 121 cm2, respectively. A If QR = 15.4 cm, find BC. X 2) In figure, XY || AC and XY divides triangular region ABC into two parts of equal areas. Determine

AX . AB

[CBSE–08]

Y C Q.2 3) D, E, F are the mid points of the sides BC, CA and AB respectively of a triangle ABC. Determine the ratio of the areas of triangles DEF and ABC. B

A

4) In the figure, ABC and DBC are two triangle on the same base BC. If AD intersect BC at O ar (ΔABC ) AO Prove that = ar (ΔDBC ) DO

Volume

C O

B

Universal Tutorials – X CBSE (2012–13) – Mathematics

D

73

74

5) Prove that the ratio of areas of two similar triangles is the same as the ratio of the squares of their corresponding medians. 6) ABC and BDE are two equilateral triangles such that D is the mid–point of BC. Ratio of the areas of triangles ABC and BDE is a) 2:1

b) 1:2

c) 4:1

7) In fig., DE || BC and AD : DB = 5 : 4. Find

d) 1:4 A

Area ( ΔDEF ) . Area ( ΔCFB )

D

E F

B C 8) ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that i) ΔAOB ~ ΔCOD ii) If OA = 6 cm, OC = 8 cm. Find: a)

Area ( ΔAOB ) Area ( ΔAOD ) (b) Area ( ΔCOD ) Area ( ΔCOD ) D A

9) ΔABC and ΔDEF are similar. The area of ΔABC is 9 sq. cm and area of ΔDEF is 16 sq. cm. If BC = 2.1 cm. find the length of EF.

C E

B

HW Exercise:

D

1) In figure, ABCD is a trapezium in which AB || CD and

F

C O

AB = 2CD. Find the ratio of the areas of triangles AOB and COD.

B

A

2) ΔABC ~ ΔPQR. A(ΔABC) = 4 A(ΔPQR). If BC = 12cm. Find QR. 3) ΔABC is a right Angled triangle right angled at A and AD ⊥ BC. If BC = 13 cm and AC = 5cm. Find the ratio of the areas of ΔABC and ΔADC. 4) The areas of two similar triangles are 121 cm² and 64cm². If the median of one triangle is 12.1 cm. Find the corresponding median of the other. 5) In a ΔABC, P divides the side AB such that AP:PB = 1:2, Q is a point on AC such that PQ || BC. Find the ratio of the areas of ΔAPQ and trapezium BPQC. 6) Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal. [CBSE–09] 7) Two isosceles triangles have equal vertical angles and their areas are in the ratio 9:16. Find the ratio of their corresponding heights (altitudes). 8) Sides of two similar triangles are in the ratio 4:9 Areas of these triangles are in the ratio a) 2:3

b) 4:9

c) 81:16

d) 16:81 P

9) In fig. ΔACB ~ ΔAPQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.

B

A

Q

Also, find the area (ΔACB) : area (ΔAPQ). C 10) If areas of two similar triangles are equal prove that the triangles are congruent. 74

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

75

Pythagoras Theorem: Similarity in Right angled Triangles: Theorem 6.7:  If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

œ Theorem (6.8) Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. B Given: A right triangle ABC, right angled at B To Prove: AC2 = AB2 + BC2 Construction: Draw BD ⊥ AC Proof: In ΔABC

A

D

C

∠ABC = 90°

[Given]

BD ⊥ AC

[Construction]

∴ ΔADB ∼ ΔABC

[If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other.]

AD AB = AB AC

[Corresponding sides of similar triangles]

AB2 = AD × AC

––– (1)

Also, ΔBDC ∼ ΔABC

[Above theorem]

CD CB = CB CA

BC2 = DC × AC Adding (1) and (2) we get,

––– (2)

AB2 + BC2 = AD × AC + DC × AC = AC [AD + DC] = AC × AC = AC2 This theorem is also known as Baudhayan Theorem.

œ Theorem 6.9: Converse of Pythagoras Theorem:  In a triangle, if the square of one side is equal to sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. Given: In triangle ABC, AC2 = AB2 + BC2 A

To Prove: ∠B = 90°

P

Construction: Construct a right triangle PQR,

right angled at Q, such that PQ = AB and QR = BC Proof: In ΔPQR, ∠Q = 90° Volume

B

C

Q

R

[By construction]

Universal Tutorials – X CBSE (2012–13) – Mathematics

75

76

PQ2 + QR2 + = PR2 [Pythagoras theorem] 2 2 2 ∴ AB + BC = PR –– (1) [By construction PQ = AB, QR = BC] But, AB2 + BC2 = AC2 –– (2) [Given] ∴ PR2 = AC2 [from (1) and (2)] i.e. PR = AC In ΔABC and ΔPQR, AB = PQ [construction] BC = QR [construction] AC = PR [Proved above] ΔABC ≅ ΔPQR [SSS test] ∴ ∠B = ∠Q [CPCT] ∠Q = 90° [construction] ∴ ∠B = 90°

SOLVED EXAMPLE 6.5: 1)

ABC is a triangle in which AB = AC & D is any point in BC. Prove that AB2 – AD2 = BD.CD.

Sol:

Construction: Draw AE ⊥ BC Proof: AB = AC [Given] 2 2 2 [Pythagoras theorem] AB = AE + BE 2 2 2 AD = AE + DE [Pythagoras theorem] AB2 – AD2 = AE2 + BE2 – AE2 – DE2 = BE2 – DE2 = (BE + DE) (BE – DE) = (BE + DE) BD = (CE + DE)BD [BE = CE] 2

A

B

D

E

C

2

AB – AD = BD × CD 2)

Sol:

The perpendicular AD on the base BC of a ΔABC intersects BC at D so that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2 [CBSE 2003] BD = 3CD A [Pythagoras Theorem] – I AB2 = AD2 + BD2 [Pythagoras Theorem] – II AC2 = AD2 + CD2 From statement (1) and (2) AB2 = AC2 – CD2 + BD2 = AC2 + (BD2 – CD2) C B D = AC2 + (BD + CD) (BD – DC) = AC2 + BC [3CD – CD]

= AC2 + BC(2CD) = AC2 + BC × 3)

Sol:

∴ 2AB2 = 2AC2 + BC2 ABC is a right triangle, right angled at C. If p is the length of the perpendicular from C to 1 1 1 AB and AB = c, BC = a and CA = b, then prove that, (i) pc = ab (ii) 2 = 2 + 2 p a b Data: ΔABC is a right angled triangle right angled at C. C CP ⊥ AB, CP = p, BC = a, AB = c, AC = b 1 1 1 TPT: 1) pc = ab 2) 2 = 2 + 2 p a b Proof: A(ΔABC) = ½bh = ½pc A B A(ΔABC) = ½ba [∠C = 90°] P ∴ ab = pc ½ab = ½pc According to Pythagoras theorem, C2 = a2 + b2

pc = ab ⇒ p = 76

1 BC 2 BC [ 4CD = BC ] = AC2 + 2 2

ab 1 C , = c p ab

1 C2 a2 + b2 1 1 = 2 2 = 2 2 = 2+ 2 2 p a b a b b a

Universal Tutorials – X CBSE (2012–13) – Mathematics

i.e.

1 1 1 = 2 + 2 2 p b a Volume

Chapter 06: Triangles

77

UNSOLVED EXERCISE 6.5: CW Exercise: 1) PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM. MR 2) The sides of certain triangles are given below. Determine which of them are right triangles. i) 6 cm, 8 cm, 10 cm ii) 5 cm, 8 cm, 11 cm 3) A man goes 150 m due east and then 200 m due north. How far is he from the starting point? 4) A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the foot of the ladder from the building. 5) A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its tip reaches a window 12 m above the ground. Determine the length of the ladder. 6) From a point O in the interior of a ΔABC, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that i) AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2 ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2 7) P and Q are the mid points of the sides CA and CB respectively of a ΔABC right angled at C. Prove that i) 4AQ2 = 4AC2 + BC2 ii) 4BP2 = 4BC2 + AC2 iii) 4(AQ2 + BP2) = 5 AB2 8) In an isosceles triangle ABC, with AB = AC, BD is perpendicular from B to the side AC. Prove that BD2 – CD2 = 2CD AD 9) ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2 C 10) If ABC is an equilateral triangle of side 2a, prove that altitude AD = a 3 .

11) In figure, ∠ACB = 90° and CD ⊥ AB.

CB 2

BD A B AD CA D 2 2 12) ABC is an isosceles triangle with AC = BC. If AB = 2AC , prove that ABC is a right triangle. 13) An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per 1 hour. How far apart will be the two planes after 1 hours? 2 14) The perpendicular AD on the base BC of a ΔABC intersects BC at D so that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2. 15) A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. [Hint: Through O draw a line parallel to BC] Prove that OB2 + OD2 = OC2 + OA2. 16) Two poles of heights 6m and 11m stand vertically on a plane ground. If the distance between their feet is 12 m, determine the distance between their tops. 17) A ladder reaches a window, which is 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9m high. Find the width of the street if the length of the ladder is 15m. 18) In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9AD2 = 7AB2 19) P and Q are points on the sides CA and CB respectively of a ΔABC right angled at C. Prove that AQ2 + BP2 = AB2 + PQ2. 20) In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. HW Exercise: 1) ABC is an isosceles triangle right angled at C Prove that AB2 = 2 AC2 Prove that,

Volume

2

=

Universal Tutorials – X CBSE (2012–13) – Mathematics

77

78

2) A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut? 3) In a triangle ABC, AD is drawn perpendicular to BC. Prove that AB2 – BD2 = AC2 – CD2 5) In a triangle ABC, ∠B > ∠C, D is the mid point of BC and AE ⊥ BC. Prove that 1 1 1 i) AC2 = AD2 + BC.DE + BC 2 (ii) AB2 = AD2 – BC.DE + BC 2 (iii) AB2 + AC2 = 2AD2 + BC 2 4 4 2 D 6) ABD is a triangle in which ∠DAB = 90° and AC ⊥ BD. Prove that i) AB2 = BC × BD ii) AC2 = BC × DC iii) AD2 = BD × CD

C

[CBSE–09] A B 7) In ΔABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5 cm and AD =

3 5 cm, find the length of CE. 2

8) BL and CM are medians of a ΔABC right angled at A. Prove that, 4(BL2 + CM2) = 5 BC2

Proof of Theorems: Theorem 6.2: (Converse of Thales Theorem) z

If a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side.

Given: z

A

In ΔABC, line l intersects AB in D and AC in E, such that

F

AD AE = DB EC

D E

To Prove: z

B

Line l || BC

l C

Construction: z

Let line l not parallel to BC. Through D draw DF || BC

Proof: z z

z

78

DF ||BC

[BPT]

AD AF = DB FC

But

AD AE = DB EC

[Given]

z

AF AE = FC EC

z

AF + FC AE + EC = FC EC

[Componendo]

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles z z z

79

AC AC ⇒ FC = EC = FC EC But this is impossible unless F and E coincide i.e. DE is l itself Hence l || BC.

Theorem 6.3: If in two triangle, corresponding angles are equal, i.e., the two triangles are equiangular, then the triangles are similar.

Given: z

Two ΔABC and ΔDEF such that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

To prove: z

ΔABC ~ ΔDEF

Construction: z z

We mark point P on the line DE and Q on the line DF such that AB = DP and AC = DQ. We join PQ. D

Proof:

A

z

AB < DE. Thus, P lies in DE.

z

In ΔABC and ΔDPQ,

z

AB = DP, AC = DQ and ∠A = ∠D

z

∴ ΔABC ≅ ΔDPQ

z

So, ∠B = ∠DPQ

z

But, ∠B = ∠E [Given]

z z

∴ ∠E = ∠DPQ Consequently, PQ || EF

z

∴

z

i.e.,

z

Similarly,

z

From (1) and (2) we get,

z

Since, corresponding angles are given equal, we conclude that, ΔABC ~ ΔDEF.

P B

C

E

Q F

[SAS criterion of congruence]

DP DQ = DE DF

[Corollary to Basic Proportionally Theorem]

AB AC = DE DF AB BC = DE EF

–– (1) [Construction] –– (2) AB BC AC = = DE EF DF

Theorem 6.4: If the corresponding sides of two triangles are proportional, then they are similar.

Given: z

Volume

Two ΔABC and ΔDEF such that

AB BC AC = = DE EF DF

Universal Tutorials – X CBSE (2012–13) – Mathematics

79

80

To prove: z

ΔABC ~ ΔDEF

Construction: z

We mark point P on DE and Q on DF such that AB = DP and AC = DQ. We join PQ. D

Proof: z

Since

A

AB AC = ,we get DE DF

P

Q

z

DP DQ = DE DF

z

∴ PQ || EF

[Converse of Basic Proportionally Theorem]

z

So, ∠DPQ = ∠E and ∠DQP = ∠F

[Corresponding angles]

z

By AAA similarity, ΔDPQ ~ ΔDEF

z

This gives,

z

But,

z

From (1) and (2) we get,

z

So, by SSS congruence criterion, ΔABC ≅ ΔDPQ

z

Since, ΔDPQ ~ ΔDEF

B

C

E

F

DP PQ AB PQ = or = –– (1) DE EF DE EF

AB BC = DE EF

–– (2) PQ BC = EF EF

or

[Given]

PQ = BC

We get, ΔABC ~ ΔDEF

Theorem 6.5: If in two triangles, one pair of corresponding sides is proportional and the included angles are equal, then the two triangles are similar.

Given: z

Two ΔABC and ΔDEF such that

AB AC = DE DF

and ∠A = ∠D

To prove: z

ΔABC ~ ΔDEF

Construction: z

We mark point P in DE and Q in DF such that AB = DP and AC = DQ. We join PQ.

Proof: z

In ΔABC and ΔDPQ, we get

z

AB = DP; AC = DQ; ∠A = ∠D

z

∴ ΔABC ≅ ΔDPQ

[SAS criterion of congruence]

z

AB AC Now, = DE DF

[Given]

DP DQ = DE DF

[By construction]

z

80

–– (1)

D

A P B

C

Universal Tutorials – X CBSE (2012–13) – Mathematics

E

Q F Volume

Chapter 06: Triangles

81

z

So, PQ || EF

[Converse of Basic Proportionally Theorem]

z

∴ ∠DPQ = ∠E and ∠DQP = ∠F

z

Consequently, by AA similarity, ΔDPQ ~ ΔDEF

z

Hence, ΔABC ~ ΔDEF

[By (1)]

Theorem 6.7: Similarity in Right Angled Triangles  If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other. z The triangles on each side of the hypotenuse are similar to each other and to the original triangle z The square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse A Given: D A right triangle ABC, right angled at B

BD ⊥ AC

B

To Prove:

1) ΔADB ~ ΔBDC

ΔADB ~ ΔABC

C

ΔBDC ~ ΔABC, and

2) BD2 = AD × DC Proof: ∠ABD + ∠DBC = 90° ∠C + ∠DBC = 90° ∠ABD = ∠C ΔADB ∼ ΔBDC In ΔADB and ΔABC ∠A = ∠A ∠ADB = 90° – ∠ABC ∴ ΔADB ∼ ΔABC Similarly, ΔBDC ∼ ΔABC ∴ ΔADB ∼ ΔBDC

∴

[Angle addition property] [BD ⊥ AC] [I] ∠ADB = 90° = ∠BDC [AA Corollary] [common] [AA Corollary] [AA Corollary] AD BD = i.e. BD2 = AD × DC BD DC

Note: BD is called the Geometric mean of AD and DC.

Angle Bisector Property:  The bisector of an angle of a triangle, divides the opposite side in the ratio of the sides containing the angle. E

Given:  In < ABC, bisector of ∠BAC intersects BC in D

A

To Prove: z

BD AB = DC AC B

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

D

C

81

82

Construction: z

Through C draw CE || AD meeting BA produced at E

Proof: z

Since CE || AD, ∠CAD = ∠ACE –– (I) [Alternate angles of parallel lines]

z z

∠BAD = ∠AEC From (I) and (II)

z

∠ACE = ∠AEC

–– (II) [Corresponding angles of parallel lines]

∴ AC = AE z

z

[Properties of isosceles triangle]

Using the property of BPT

BD BA = DC AE

BD BA = DC AE

[AC =AE]

The bisector of the exterior ∠A of a triangle ABC intersects the side BC produced in D. AB BD = . Prove that AC CD F [Note: We say that the point D divides BC externally in the ratio AB:AC] Sol: Data: In ΔABC, ray AD bisects the exterior ∠A intersecting BC at D. A AB BD = RTP: E AC CD Construction: Through C draw CE || AD. Proof: CE || AD [By construction] BE BC B C D In ΔABD, = [B.P.T] EA CD ∠FAD = ∠CAD [Ray AD is the bisector] ∠DAC = ∠ACE [Alternate angles] ∠AEC = ∠FAD [Corresponding angles] ∠AEC = ∠ACE Statement (3), (4), (5) ∴ AE = AC [Prop of isosceles triangles] BE + EA BC + CD = [Componendo from statement (2)] EA CD AB BD = Statement (8) EA CD AB BD = Statement (7) and (8) i.e. AC CD

Applications of Pythagoras Theorem: Acute Angled Triangle:

A

Given: A triangle ABC, such that AD ⊥ BC ∠B < 90 To Prove: AC2 = AB2 + BC2 – 2 BC.BD Construction: Draw AD ⊥ BC 82

B

Universal Tutorials – X CBSE (2012–13) – Mathematics

D

C

Volume

Chapter 06: Triangles

83

Proof: In ΔABD, ∠ADB = 90° 2

2

AB = BD + AD

[AD ⊥ BC]

2

[Pythagoras theorem]

AC2 = AD2 + CD2

[Pythagoras theorem]

2

2

2

= AD + [BC – BD] = AD + BC2 – 2BC.BD + BD2 = AB2 + BC2 – 2BC.BD

[AB2 = AD2 + BD2]

Obtuse Angled Triangle:

A

In an obtuse angled ΔABC, obtuse angled at B, AD ⊥ BC Given: A triangle ABC, such that AD ⊥ DC; ∠B > 90° To Prove: AC2 = AB2 + BC2 + 2 BC BD

D

Construction: Draw AD ⊥ CB

C

B

Proof: ΔADC is a right–angled triangle right angled at D (AD ⊥ BC) ∴ AC2 = AD2 + CD2

[Pythagoras theorem]

2

= AD + (BD + BC) 2

2

[C – B – D]

2

2

= AD + BD + 2BD.BC + BC = AB2 + BC2 + 2BD.BC

[AD2 + BD2 = AB2]

Hence proved

Appollonius Principle:

A

In an acute angled ΔABC, D is the midpoint to side BC. Prove: AB² + AC² = 2AD² + 2DC² Given: A triangle ABC, such that BD = DC 2

2

2

To Prove: AB + AC = 2AD + 2DC

B

D

2

C

E

Construction: Construct a perpendicular AE on side BC Proof: AB2 = BE2 + AE2

[Pythagoras theorem] 2

2

= [BD + DE] + AE 2

2

2

2

2

2

[B–D–E]

2

AC = AE + EC = AE + (CD – DE)

2

2

[D–E–C] 2

AB + AC = BD + 2BD.DE + DE + 2AE + CD2 – 2CD.DE + DE2 = BD2 + 2AE2 + CD2 + 2DE2 = 2CD2 + 2[AE2 + DE2] 2

2

= 2CD + 2AD

[Since BD = CD] [Since AE2 + DE2 = AD2]

MISCELLANEOUS EXERCISE: 1) In fig. D is a point on side BC of ΔABC such that

Prove that AD is the bisector of ∠BAC. 2) In Fig ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: CA BC i) ΔABC ~ ΔAMP = ii) PA MP

C

B M

A

Volume

A

BD AB = . CD AC

B

Universal Tutorials – X CBSE (2012–13) – Mathematics

D

C

P

83

84

3) If the angles of one triangle are respectively equal to the angles of another triangle, prove that the ratio of the corresponding sides is the same as the ratio of the corresponding. i) Medians ii) Bisectors of angles (angle bisector segments) iii) Altitudes 4) ABCD is a trapezium with AB || DC. If AC and BD intersect at E and ΔAED is similar to ΔBEC, prove that AD = BC. 5) Through the vertex D of a parallelogram ABCD, a line is drawn to intersect the sides AB and CB DA FB FC = = . produced at E and F respectively. Prove that AE BE CD BD DA = . Prove that ΔABC is a right triangle. 6) In ΔABC, AD ⊥ BC. If DA DC 7) Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle. 8) In adjoining figure, DE is parallel to BC. If find AE.

AD 2 = and AC = 18 cm, DB 3

A D

K

B

E C

KP 4 P = and 9) In the given figure, PQ is parallel to MN. If Q PM 13 M N KN = 20.4 cm, find KQ. 10) Prove that the area of a semicircle on the hypotenuse of a right angled triangle is equal to the sum of the areas of semicircle on the other two sides. A

11) ∠A = ∠B; AD = BE. Show that DE || AB.

C B

A

12) B

C

D E

If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is a right–angled triangle. a) In right triangle in the figure 1 if AB = BC = x cm, then find AC A

b) In given figure 2, ABC is a right angled triangle in which AC is 8 m more than AB. If BC is 7 m more than AB. Find the dimensions of ΔABC if ∠B = 90°.

B

C

13) ABCD is a square. F is the mid–point of AB. BE is one third of BC. If the area of the ΔFBE is 108 sq. cm, find the length of AC. 14) A man goes 10 m due east and then 24 m due north. Find his distance from the starting point. 15) M and N are points on sides AB and AC of ΔABC such that AM = 4 cm, MB = 8 cm, AN = 6 cm, NC = 12 cm. Prove that BC = 3 MN. 16) A man goes 150 m due East and then 200 m due North. How far is he from the starting point? A

B x

17)

D

84

4x – 4

2x – 1

O

2x + 4

In trapezium ABCD, AB || CD. Find the value of x (Refer the diagram). C

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

85

P 18) PA, QB and RC are each perpendicular to AC. 1 1 1 Prove that + = (Refer the diagram) x y z D

B

A

x

1.5 cm

20) In figure, PQ || BC. Find QC.

C

B

A

P

1.3 cm

Q.20

C

3 cm

y

z

A In a quadrilateral ABCD, CA = CD, ∠B = 90°, AD2 = AB2 + BC2 + CA2. Prove that ∠ACD= 90°

C

19)

R

Q

Q

B D

21) In figure, ΔEDC ~ ΔEBA, ∠BEC = 115° and ∠EDC = 70°. Find ∠DEC, ∠DCE, ∠EAB, ∠AEB and ∠EBA. R

E

115°

A 22) P

B

In figure, if ∠P = ∠RTS, Prove that ΔRPQ ~ ΔRTS.

T

S

C

70°

Q

Q.22

A

C

O

OA OD . Prove that ∠A = ∠C and ∠B = ∠D. = 23) In figure, OC OB

D

B

Q.23

24) Any point X inside ΔDEF is joined to its vertices. From a point P in DX, PQ is drawn parallel to DE meeting XE at Q and QR is drawn parallel to EF meeting XF in R. Prove that PR || DF. P

25) In the given figure, ΔACB ~ ΔAPQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, find CA and AQ.

A

B

Q

C

26) The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB. 27) A vertical stick 15 cm. long casts its shadow 10 cm long on the ground. At the same time, the flag–pole casts a shadow 60 cm. long. Find the height of the flag–pole. P

28) In the given figure, QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Find AQ. D

A

29) 1.2 cm B

Volume

E

O B

Q

In the given figure, ΔABC ~ ΔDEF, AB = 1.2 cm and DE = 1.4 cm. Find the ratio of areas of ΔABC and ΔDEF.

1.4 cm C

A

F

Universal Tutorials – X CBSE (2012–13) – Mathematics

85

86 A

AO BO 1 = = and AB = 5 cm. OC OD 2 Find the value of DC.

5 cm

B

30) In the given figure,

O D

P

31)

O

A

B

C

In the given figure, PB and QA are perpendiculars to segment AB. If PO = 5 cm, QO = 7 cm and area (ΔPOB) = 150 cm2, C find the area (ΔQOA). F

Q

D

32) AB || DE, BD || EF. Prove that DC2 = CF × AC. C D 33)

E

B

A

In figure below, ABC is a triangle, right angled at B. FG and DE are each parallel to CB. Also AG = GE = EB. Find the combined lengths of DE and FG if BC = 12cm.

F

A G E B 34) ABC is a right triangle, right angled at C. If p is the length of the perpendicular from C to AB and AB = c, BC = a and CA = b, then prove that, (i) pc = ab (ii)

1 1 1 = + p2 a2 b2

35) In trapezium ABCD, AB || DC and DC = 2AB.EF is drawn parallel to BE 3 F = . Diagonal DB intersects AB cuts AD in F & BC in E such that EC 4 D FE at G. Prove 7FE = 10 AB. 36) L and M are the mid points of AB and BC resp. of ΔABC, right angled at B. Prove that 4LC² = AB² + 4BC² A E

37) In fig ΔABC is right angled at C and DE ⊥ AB. Prove that ΔABC ∼ ΔADE and hence find the lengths of AE and DE.

38) C

39) The areas of two similar triangles are 81 cm2, 49 cm2 respectively. If the attitude of bigger triangle is 4.5 cm. Find the corresponding attitude of the smaller triangle. B

A D

40) B 86

3 cm

[CBSE–09]

2 cm 12 cm

C

In figure, M is mid–point of side CD of a parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced at E. Prove that EL = 2 BL. [CBSE–09]

M

L

C

E

D

B

E

G

D

B A

B

A

E

A

D

81

49

X

C

E

Y

A

In figure, if ΔABE ≅ ΔACD, prove that ΔADE ~ ΔABC. C Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

87

C 41) In figure, BD ⊥ AC and CE ⊥ AB. Prove that i) ΔAEC ~ ΔADB CA CE ii) = AB DB

D A

F B

E Q.41

42) E is a point on side AD produced of a parallelogram ABCD and BE intersects CD at F. Prove that ΔABE ~ ΔCFB. A F

43)

In figure, E is a point on side CB produced of an isosceles ΔABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

C D Q.43 44) ABC is an isosceles triangle with AB = AC and D is a point on AC such that BC2 = AC × CD. Prove that BD = BC. A E

B

45) ABC is a right triangle with ∠ABC = 90°, BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that ii) DN2 = DM × AN i) DM2 = DN × MC

N

D

C M Q.45 46) ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle. 47) In ΔPQR, QM ⊥ PR and PR2 – PQ2 = QR2. Prove that QM2 = PM × MR. C B

48) In ΔABC, ∠A = 90°. If AD ⊥ BC, prove that AB2 + CD2 = BD2 + AC2.

D B

Q.48

A

49) Let ABC be a triangle and D and E be two points on side AB such that AD = BE. If DP || BC and EQ || AC, then prove that PQ || AB. A

50) In figure, P is the mid–point of BC and Q is the mid–point of AP. 1 If BQ when produced meets AC at R, prove that RA = CA. 3

Q

R S

C P Q.50 51) D is the mid point of side BC at a ΔABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE: EX = 3: 1. [CBSE, 1987] 52) The diagonal BD of a parallelogram ABCD intersects the segment AE at point F, where E is any [CBSE, 94] point on the side BC. Prove that DF × EF = FB × FA. Volume

B

Universal Tutorials – X CBSE (2012–13) – Mathematics

87

88

53) Two poles of height a and b metres (b > a) are c metres apart. Prove that the height in metres of ab . the intersection of the lines joining the top of each pole to the foot of the opposite pole is a+b A 54) A triangle has sides 5 cm, 12 cm and 13 cm. Find the length, 12cm 5cm to 1 decimal place of the ⊥ from the opposite vertex to the side whose length is 13 cm. B

D

C 13cm

55) D & E are points on the sides AB and AC respective of ΔABC. Such that DE || BC and AD:DB = 4:5. CD & BE intersect at F. Find the ratio of the areas of ΔDEF and ΔBCF. A

56) In the figure, ΔFEC ≅ ΔGDB and ∠ADE = ∠AED. Prove that ΔADE ~ ΔABC.

D

E

F

G B

57) In ΔABC, ∠A is acute. BD and CE are ⊥ on AC and AB respectively. Prove that AB × AE = AC × AD.

C A

D

58) BD ⊥ BC, DE ⊥ BA, AC ⊥ BC. Prove that ΔBDE ~ ΔABC.

[CBSE–08]

A

E

P

B

Q

59) B

S

C

R

D

C

P, Q, R, S are the mid points of AB, BC, CD and AD respectively. Prove that quadrilateral PQRS is a parallelogram. A 90° G

60) In the following figure, DEFG is a square and ∠BAC = 90°. Prove that DE2 = BD × EC.

F [CBSE–09]

B

C D E 61) Equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides. 62) In Δ ABC, ∠B = 90°. Points D and E divides the side BC in to three equal parts. Prove that: 8 AE2 = 3 AC2 + 5 AD2 [CBSE–09] C 63) In figure, AD and BE are respectively perpendiculars to BC and AC. E D Show that i) ΔADC ~ ΔBEC ii) CA × CE = CB × CD B A iii) ΔABC ~ ΔDEC Q.63 iv) CD × AB = CA × DE 64) Prove that sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of its sides. 65) In a triangle ∆ABC, a line PQ is drawn parallel to AB. If AP = x – 3, AC = 2y, BQ = x – 2 and BC = 2y + 3, then prove that 3x - 2y = 9. 88

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

89

66) The side BC of a ΔABC is bisected at D; O is any point in AD. BO and CO produced meet AC and AB in E and F respectively and AD is produced to X so that D is the mid–point of OX. Prove F that AO : AX = AF : AB and show that FE || BC. B 67) In figure, we have AB || CD || EF. If AB = 6 cm, CD = x cm, 10 cm 4 cm D 6 cm EF = 10 cm, BD = 4 cm and DE = y cm, y cm E x cm calculate the values of x and y. A E C Q.68 68) In each of the fig. given below, an altitude is drawn to the hypotenuse by a right–angled triangle. The length of different line–segments are marked in each fig. Determine x, y, z in each case. A

4

a) x

P D

y B

b) 6

5 z

C

Q

4

S x

y z

R

MULTIPLE CHOICE QUESTIONS: CW Exercise: 1) A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of the tower is a) 100 m b) 120 m c) 25 m d) 200 m. 2) The areas of two similar triangles are in respectively 9 cm2 and 16 cm2. The ratio of their corresponding sides is a) 3:4 b) 4:3 c) 2:3 d) 4:5 3) ΔABC and ΔBDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is a) 2: 1 b) 1 : 2 c) 4: 1 d) 1: 4 4) If ΔABC and ΔDEF are similar such that 2AB = DE and BC = 8 cm, then EF = a) 16 cm b) 12 cm c) 8 cm d) 4 cm. 5) ΔABC is such that AB = 3cm, BC = 2 cm and CA = 2.5 cm. If ΔDEF ~ ΔABC and EF = 4 cm, then perimeter of ΔDEE is a) 7.5 cm b) 15 cm c) 22.5cm d) 30cm. 6) Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is a) 12m b) 14m c) 13m d) 11 m 7) In ΔABC, D and E are points on side AB and AC respectively such that DE || BC and AD : DB = 3:1. If EA = 3.3 cm, then AC = a) 1.1 cm b) 4cm c) 4.4cm d) 5.5cm 8) If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 83°, then ∠C = a) 50° b) 60° c) 70° d) 80° 9) In a ΔABC, ∠A = 90°, AB = 5 cm and AC = 12 cm. If AD ⊥ BC, then AD = 13 60 13 cm b) cm c) cm 2 13 60 10) If ΔABC is an equilateral triangle such that AD ⊥ BC, then AD2 = 3 a) DC2 b) 2DC2 c) 3CD2 2

a)

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

d)

2 15 cm 13

d) 4DC2 89

90

11) In a ΔABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium. If DE:BC = 3:5, then Area(ΔADE) : Area (†BCED) = a) 3:4 b) 9:16 c) 3:5 d) 9:25 12) In a ΔABC, AD is the bisector of ∠BAC. If AB = 8cm, BD = 6cm and DC = 3cm. Find AC a) 4cm b) 6cm c) 3cm d) 8cm 13) If ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC, then a) AB2 – AD2 = BD.DC b) AB2 – AD2 = BD2 – DC2 2 2 d) AB2 + AD2 = BD2 – DC2 c) AB + AD = BD.DC 14) If ABC is a right triangle right-angled at B and M, N are the mid-points of AB and BC respectively, then 4(AN2 + CM2) = 5 a) 4 AC2 b) 5 AC2 c) AC2 d) 6 AC2 4 15) In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and BC respectively, then a) AQ2 + CP2 = 2(AC2 + PQ2) b) 2(AQ2 + CP2) = AC2 + PQ2 1 c) AQ2 + CP2 = AC2 + PQ2 d) AQ + CP = (AC + PQ). 2 HW Exercise: 1) Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio. a) 2 : 3 b) 4: 9 c) 81 : 16 d) 16 : 81 2) The areas of two similar triangles ΔABC and ΔDEF are 144 cm2 and 81 cm2 respectively. If the longest side of larger ΔABC be 36 cm, then the longest side of the smaller triangle ΔDEF is a) 20 cm b) 26 cm c) 27cm d) 30cm 3) Two isosceles triangles have equal angles and their areas are in the ratio 16: 25. The ratio of their corresponding heights is a) 4:5 b) 5:4 c) 3:2 d) 5:7 AB BC CA 2 4) If ΔABC & ΔDEF are two triangles such that = = = , Area(ΔABC):Area(ΔDEF) = DE EF FD 5 a) 2 : 5 b) 4 : 25 c) 4: 15 d) 8: 125 5) XY is drawn parallel to the base BC of a ΔABC cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY = a) 2 cm b) 4 cm c) 6cm d) 8cm. 6) In ΔABC, a line XY parallel to BC cuts AB at X and AC at Y. If BY bisects ∠XYC, then a) BC = CY b) BC = BY c) BC ≠ CY d) BC ≠ BY 7) In triangles ABC and DEF, ∠A = ∠E = 40°, AB : ED = AC : EF and ∠F = 65°, then ∠B = a) 35° b) 65° c) 75° d) 85° 8) If D, E, F are the mid-points of sides BC, CA and AB respectively of ΔABC, then the ratio of the areas of triangles DEF and ABC is a) 1: 4 b) 1 : 2 c) 2: 3 d) 4 : 5 9) In an equilateral triangle ABC, if AD ⊥ BC, then a) 2AB2 = 3AD2 b) 4AB2 = 3AD2 c) 3AB2 = 4AD2 d) 3AB2 = 2AD2 10) In a ΔABC, perpendicular AD from A on BC meets BC at D. If BD = 8cm, DC = 2 cm and AD = 4cm, then a) ΔABC is isosceles b) ΔABC is equilateral c) AC = 2 AB d) ΔABC is right-angled at A. 90

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

91

11) In a ΔABC, AD is the bisector of ∠BAC. If AB = 6cm, AC = 5cm and BD = 3cm, then DC = a) 11.3 cm b) 2.5 cm c) 3:5cm d) None of these. 12) ABCD is a trapezium such that BC || AD and AD = 4 cm. If the diagonals AC and BD AO DO 1 = = , then BC = intersect at O such that OC OB 2 a) 7cm b) 8cm c) 9cm d) 6cm BD 13) ΔABC is a right triangle right-angled at A and AD ⊥ BC. Then, = DC 2

2

AB AB ⎛ AB ⎞ ⎛ AB ⎞ a) ⎜ b) c) ⎜ d) ⎟ ⎟ AC AD ⎝ AC ⎠ ⎝ AD ⎠ 14) If E is a point on side CA of an equilateral triangle ABC such that BE ⊥ CA, then AB2 + BC2 + CA2 = a) 2 BE2 b) 3 BE2 c) 4 BE2 d) 6 BE2

COLUMN MATCHING QUESTIONS: 3) In triangle ABC, DE || BC and column I. Column I

2 AD = . From column II choose correct option for each item in DB 3

Column II

i)

ar ( ΔADE ) ar ( ΔABC )

A)

21 25

ii)

ar (Trap DECB ) ar ( ΔABC )

B)

4 21

iii)

ar ( ΔADE ) ar (Trap DECB )

C)

21 4

4 25 2) Given in column I are triangles with DE || BC in each. Using Basic Proportionality Theorem. Choose correct option in column II for each item in column I. Column I Column II i) AB = 2x, A AC = 2x + 3 DB = x – 3 and A) 11 E D EC = x – 2 B C The value of x is ii) AD = x, A DB = x – 2 AE = x + 2 and B) 3 D E EC = x – 1 C The value of x is B iii) AD = x – 2 A AB = x C) –4 AE = x – 1 and D E AC = x + 2 B C The value of x is D)

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

91

92

iv) AD = 4 cm BD = (x–4) cm AE = 8 cm and D EC = (3x – 19)cm The value of x is B

A D) 9

E C

E) 4 3) Given in column I are pairs of triangles. From column II choose correct option(s) for each item in column I. Column I Column II i) A P 60° 40°

80°

B ii)

60°

C

40°

80°

Q

A 2.7 cm

B

Q

P 6 cm

3 cm

C

2.5 cm

Q

5 cm

4 cm

A

C) ΔPQR ~ ΔABC R

P

3.1 cm

3.1 cm

60°

B

R

5 cm

A

B iv)

B) ΔPQR ? ΔABC

6 cm

4 cm

C

2 cm

2 cm

R

P 3 cm

iii)

A) ΔPQR ≅ ΔABC

C

4 cm

D) ΔABC ~ ΔQRP

60°

Q

4 cm

R

4) Given in column I are formations of some figures. From column II choose correct option for each item in column I. Column I Column II i) A A) AB2 = AC2 + BC2 – 2BC . CD B

C

D

ii)

A B) AB2 + CD2 = AC2 + BD2 B

iii)

D

C A

C) AD2 = BD . CD B

92

D

C D) AB2 = AC2 + BC2 + 2BC . CD E) AC2 = AB2 + BC2 – 2BC . BD Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 06: Triangles

93

ANSWERS TO UNSOLVED EXERCISE: CW Exercise 6.1: 1) (i and ii) Similar

iii) Equilateral

iv) (a) Equal (b) proportional)

2) Yes

8) 9

CW Exercise 6.2: 1) 2.65 cm

9) AB=AC=12, BC = 40

HW Exercise 6.2: 1) (i) 2 cm (ii) 2.4 cm

2) (i) No (ii and iii) Yes 10) (i) x = 4 (ii) x = 1

CW Exercise 6.3: 1) (i) ΔABC ~ ΔQRP SSS (ii and iii) No

2) 55°, 55°, 55°

7) 1.6 m

HW Exercise 6.3: 1) 42 m

5) 8 cm

7) (i) ΔABC ~ ΔQPR by A–A–A similarity (ii) No

⎛2− 2 ⎞ ⎟ 2) ⎜ ⎜ 2 ⎟ ⎠ ⎝

3) 1:4

10) (ii) CW Exercise 6.4: 1) 11.2 cm 7)

25 81

8) a)

9 16

b)

3 4

6) c 9) 2.8 cm

HW Exercise 6.4: 1) 4:1

2) 6 cm

3) 169:25

4) 8.8 cm

5) 1:8

6) 3:4

9) d

4) 15 m

5) 13 m

13) 300 61 km

6) 13 m

7) 21 m

7) 2 5 cm

9) 4.8 m

13) 50.904 cm

14) 26 m

17) 2/5

20) 2.6 cm

21) 65°,45°,45°,65°,70°

25) CA = 5.6 cm, AQ = 3.25 cm

26) 15 cm

27) 90 cm

28) 15 cm

29) 36/49

30) 10 cm

31) 294 cm2

33) 12 cm

37) CE = 2 5 cm

38) 3:4

39) 3.5 cm

54) 4.6 cm

55) 16:81

CW Exercise 6.5: 3) 250 m HW Exercise 6.5: 2) 6 7 m Miscellaneous: 8) 7.2 cm 16) 250 m

67) x = 15/4 cm or 3.75 cm, y = 20/3 cm or 6.67 cm Column Matching Question: 1) i–D; ii–A; iii–B

2) i–D; ii–E; iii–E; iv–A

3) i–C; ii–B; iii–D

4) i–BC; ii–D; iii–AEB

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

93

94

Chapter 08: Introduction to Trigonometry Chapter Map: Introduction

Basic Ratios

Range of θ

Standard Angles Table Method

Triangles Method

Basic Identities

Prove Identities

Verify if Identity

Solve for θ

Complementary Angles

Direct Applications

Prove Identities

Find Value

Introduction: What is Trigonometry?  The word trigonometry is derived from the Greek words ‘tri’ means 3, ‘gon’ means sides and metron means measure. Trigonometry is the study of relationships between the sides and angles of a triangle.

Use of learning trigonometry:  Trigonometry is used in astronomy, surveying geography, physics and navigation. The captains of ships use trigonometry to calculate the distances from far off islands, sea shores, cliffs and other ships in the ocean.  For learning trigonometry we should know the trigonometrical ratios.

Trigonometric Ratios: ¾ Let us take a right triangle ABC as shown in Fig.

C

¾ Note the position of the side BC with respect to angle A. It faces ∠A. ¾ We call it the side opposite to angle A. AC is the hypotenuse of the right triangle and the side AB is a part of ∠A. ¾ So, we call it the side adjacent to angle A.

Hypotenuse

A

94

Universal Tutorials – X CBSE (2012–13) – Mathematics

Side opposite to ∠A

¾ Here, ∠CAB (or, in brief, angle A) is an acute angle.

Side adjacent to ∠A

B

Volume

Chapter 08: Introduction of Trigonometry

95

Side adjacent to ∠C

C Hypotenuse

Note: If we consider ∠C then side AB faces ∠C. Hence AB is the side opposite to ∠C and BC is the side adjacent to ∠C. Remember the hypotenuse remains unchanged.

A

Side Opposite to ∠C

B

¾ You have studied the concept of ‘ratio’ in your earlier classes. ¾ We now define certain ratios involving the sides of a right triangle and call them trigonometric ratios. ¾ The trigonometric ratios of the angle A in right triangle ABC are defined as follows: BC side opp. to ∠ A = AC hypotenuse

 cosine of ∠A =  tangent of ∠A =

AB side adj. to ∠A = AC hypotenuse

Hypotenuse

BC side opp. to ∠ A = AB side adj. to ∠ A 1 AC hypotenuse = = sin of ∠A BC side opp. to ∠ A

 cosecant of ∠A =  secant of ∠A =

C Side opposite to ∠A

 sine of ∠A =

A

Side adjacent to ∠A

B

AC 1 hypotenuse = = BC co sin of ∠A side adj. to ∠A

 cotangent of ∠A =

AB 1 side adj. to ∠ A = = BC tangent of ∠ A side opp. to ∠ A

¾ The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A. BC ¾ Also, observe that tan A = AB =

BC AC AB AC

sin A cos A cos A = and cot A = sin A .

¾ So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.

Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’. sin A is not the product of ‘sin’ and A. ‘sin’ separated from A has no meaning.

Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also. ¾ Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to Q AB extended C ¾ By AA similarity criterion. you will see that the triangles PAM and CAB are similar. P ¾ Therefore, by the property of similar triangles, the corresponding Hypotenuse sides of the triangles are proportional. ¾ So, we have Volume

AM AP MP = = AB AC BC Universal Tutorials – X CBSE (2012–13) – Mathematics

A

M

B

N

95

96

¾ From this, we find

MP BC = = sin A AP AC

AM AB MP BC = = cos A, = = tan A and so on. AP AC AM AB ¾ This shows that the trigonometric ratios of angle A in ΔPAM do not differ from those of angle A in ΔCAB. ¾ In the same way, we conclude that the value of sin A (and also of other trigonometric ratios) remains the same in ΔQAN also. ¾ From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.

¾ Similarly,

Note: Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1).

SOLVED EXAMPLES 8.1: 4 , find the other trigonometric ratios of the angle A. 3 Sol: Let us first draw a right ΔABC BC 4 Now, we know that tan A = = AB 3 ∴ if BC = 4k, then AB = 3k, where k is a positive number. Now, by using the Pythagoras Theorem, A We have AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 So, AC = 5k Now, we can write all the trigonometric ratios using their definitions. BC 4k 4 = = sin A = AC 5k 5 BC 4k 4 AB 3k 3 = = and tan A = = = cos A = AB 3k 3 AC 5k 5 3 1 1 5 1 5 ∴ cot A = = cosec A = = and sec A = = . 4 tan A sin A 4 cos A 3

1) Given tan A =

2)

If tan θ =

Sol: tan θ = sin θ =

7 and 0° < θ < 90° find all other trigonometric ratios of θ. 24

Thus sin θ = cosec θ =

96

B

3k

7x

7 24 ; cos θ = 25 25 24 1 25 , cot = = 7 sin θ 7

4k

A

7 , AC = 25 24

cosec θ =

C

C

θ

24x

B

24 7 24 ; cos θ = ; cot θ = ; 25 25 7

25 25 ; sec θ = 7 24

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry

3)

If sec θ =

Sol: sec θ = ∴

97

⎡ 13 sin θ − 5 tan θ + 1⎤ 13 find the value of ⎢ ⎥. 5 ⎣ 12(cot θ + cos ecθ) ⎦

13 =1 5

⇒ sec θ =

tan θ 12 = sin θ ∴ sin θ = sec θ 13

1 13 5 12 = , cos θ = , tan θ = cos θ 5 13 5

⇒ cosec θ =

13 5 , cot θ = 12 12

⎛ 12 ⎞ ⎛ 12 ⎞ 13⎜ ⎟ − 5⎜ ⎟ + 1 12 − 12 + 1 1 ⎝ 13 ⎠ ⎝ 5 ⎠ = = 5 13 5 + 13 18 ⎛ ⎞ ⎛ ⎞ 12⎜ ⎟ + 12⎜ ⎟ ⎝ 12 ⎠ ⎝ 12 ⎠

13 sin θ − 5 tan θ + 1 ∴ = 12 cot θ + 12 cos ecθ

UNSOLVED EXERCISE 8.1: CW Exercise: 1) Complete the table given below with other trigonometric ratios of Angle θ from the given trigonometric ratio: sin θ cos θ tan θ cot θ sec θ cosec θ a) 4/5 b) 8/15 c) √10 2) Triangle Δ PQR is a right angle triangle in which PQ = 3 cm and PR = 6 cm P If ∠ PRQ = θ then determine the values of 6 cm 3 cm i) sin2 θ + cos2 θ 2 2 ii) cot θ – tan θ Q R 3) In a ∆ABC, ∠ A + ∠ C = 90°, AB = 5 cm and BC = 12 cm. Find out i) sin A, sin C ii) cos A, cos C iii) tan A, tan C 4) If 9 cos θ – 5 sin θ = 4 cos θ + 7 sin θ, find the value of tan θ.

5) If cos θ =

1 2 sec θ Find the value of 2 1 + tan 2 θ

6) If tan θ =

4 . Show that 3

1 − sin θ 1 = 1 + sin θ 3

7) If 4 cot θ = 3 find the value of 4 cos ecθ + 5 sin θ + 1 2 sin θ − 3 cos θ ii) 5 cos θ − 3 sec θ + 3 tan θ 2 sin θ + 3 cos θ 8) In a right angle triangle ∆PQR, ∠ Q = 900 and PQ < QR. If PQ + QR = 17 and PR = 13 then find out the values of sin P, cos P and tan P. 9) If tan θ = x/y, prove that:

i)

i)

x sin θ − y cos θ = (x2 – y2) / (x2 + y2) x sin θ + y cos θ 2

2

2

A 2

ii) (1 – tan θ) / (1 + tan θ) = cos θ – sin θ 10) In fig., AD = DB and ∠B is a right angle. Determine: i) sin θ

ii) cos θ

iii) tan θ

iv) sin2 θ + cos2 θ

Volume

b C

Universal Tutorials – X CBSE (2012–13) – Mathematics

θ

D

a B 97

98

HW Exercise: 3 calculate cos A and tan A. 4 2) Given 15 cot A = 8, find sin A and sec A.

1) If sin A =

13 , calculate all other trigonometric ratios. 12 4) If cos A = 1 / 3, determine the value of sin A sec A + tan A cosec A. 5) The adjacent figure shows a right angle triangle ∆PQR where

3) Given sec θ =

P 10 cm 6 cm

PQ = 6 cm and PR = 10 cm. If ∠PRQ = θ, determine the value of i) cot θ – tan θ R ii) sin θ + cos2 θ 6) If sin θ = x/y, prove that (cos θ + cot θ) xy = x √ (y2 – x2) + y √ (y2 – x2) 7) If cos θ = 3 / 5, find the value of 8) If cot θ = i)

Q

4 tan θ − 5 cos θ . sec θ + 4 cot θ

7 , evaluate : 8

(1 + sin θ )(1 − sin θ ) (1 + cos θ )(1 − cos θ )

9) If 3 cot A = 4, check whether

ii) cot2 θ

1 − tan 2 A 1 + tan 2 A

= cos2 A – sin2 A or not.

10) In triangle ABC, right-angled at B, if tan A = 1/ 3 and tan c =

3 find the value of:

i) sin A cos C + cos A sin C ii) cos A cos C – sin A sin C 11) If cosec θ = 2, find the value of

1 sin θ + . tan θ 1 + cos θ

12) In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. 13) If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B. 14) State whether the following are true or false. Justify your answer. i) The value of tan A is always less than 1. 12 for some value of angle A. 5 iii) cos A is the abbreviation used for the cosecant of angle A. iv) cot A is the product of cot and A.

ii) sec A =

v) sin θ =

4 for some angle θ. 3

A

θ φ

15) ΔABC is a right angled D is the midpoint of BC. Show that

98

1 tan θ = . 2 tan φ

B

D

Universal Tutorials – X CBSE (2012–13) – Mathematics

C

Volume

Chapter 08: Introduction of Trigonometry

99

Trigonometric Ratio of some specific angles: Trigonometric Ratio of 0° and 90°: z z z

Let ∠ XAY = θ be an acute angle and let P be a point on its terminal side AY. P Draw perpendicular PM from P on AX. In Δ AMP, we have

Y

PM AM PM θ cos θ = and tan θ = X A AP AP AM M It is evident from Δ AMP that as θ becomes smaller and smaller, line segment PM also becomes smaller and smaller; and finally when θ become 0°; the point P will coincide with M. Consequently, we have PM = 0 and AP = AM sin θ =

z

z

∴ sin 0° =

PM 0 AM AP PM 0 = = 0, cos 0° = = = 1 and, tan 0° = = =0 AP AP AP AP AP AP

z

Thus, we have sin 0° = 0, cos 0° = 1 and tan 0° = 0

z

From Δ AMP, it is evident that as θ increase, line segment AM becomes smaller and smaller and finally when θ becomes 90° the point M will coincide with A. Consequently, we have AM = 0, AP = PM

z

∴ sin 90° = z

PM PM AM 0 = = 1 and cos 90° = = =0 AP PM AP AP

Thus, we have sin 90° = 1 and cos 90° = 0

Remark: It is evident from the above discussion that tan 90° =

PM PM = is not defined. AM 0

Similarly, sec 90°, cosec 0°, cot 0° are not defined.

Trigonometric Rations of 30° and 60°: z

Consider an equilateral triangle ABC with each side of length 2a. Since each angle of an equilateral triangle is of 60°.

z z

Therefore, each angle of Δ ABC is of 60°. Let AD be perpendicular from A on BC. A Since the triangle is equilateral.

z

Therefore, AD is the bisector of ∠A and D is the mid–point of BC.

30°

∴ BD = DC = a and ∠BAD = 30° z z

Thus, in Δ ABD, ∠D is a right angle, hypotenuse AB = 2a and BD = a. So by Pythagoras theorem, we have AB2 = AD2 + BD2 2

2

2

⇒ (2a) = AD + a

2

2

2

⇒ AD = 4a – a

⇒ AD =

60° B

3a

Trigonometric ratios of 30°: z

A

In right triangle ADB, we have Base = AD =

30° 30°

3a , Perpendicular = BD = a,

2a

Hypotenuse = AB = 2a and ∠DAB = 30° B

Volume

C

D

Universal Tutorials – X CBSE (2012–13) – Mathematics

2a

60 a

D

60 a

C

99

100 z

∴ sin 30° =

AD 3a 3 BD a 1 = = = = , cos 30° = AB 2a 2 AB 2a 2

tan 30° =

1 BD a 1 , cosec 30° = =2 = = sin 30° AD 3a 3

sec 30° =

1 2 = cos 30° 3

and cot 30° =

1 = 3 tan 30°

Trigonometric ratios of 60°: z

In right triangle ADB, we have Base = BD = a, Perpendicular = AD =

3a

Hypotenuse = AB = 2a and ∠ ABD = 60° z

BD a 1 AD 3a 3 = = , cos 60° = = = AB 2a 2 AB 2a 2

∴ sin 60° =

3a = a

1 2 = sin 60° 3

tan 60° =

AD = BD

sec 60° =

1 1 1 = 2 and cot 60° = = cos 60° tan 60° 3

3 , cosec 60° =

Trigonometric Ratios of 45°: z

Consider a right triangle ABC with right angle at B such that ∠A = 45°.

z

Then, ∠A + ∠B + ∠C = 180°

C

⇒ 45° + 90° + ∠C = 180°

2a

⇒ ∠C = 45° ∴ ∠A = ∠C z

⇒ AB = BC Let AB = BC = a. Then, by Pythagoras theorem, we have AC2 = AB2 + BC2

A

45 a

45 a B

⇒ AC2 = a2 + a2 ⇒ AC2 = 2a2 ⇒ AC = z

2a

Thus, in Δ ABC, we have ∠A = 45°, Base = AB = a, Perpendicular = BC = a

and Hypotenuse = AC = ∴ sin 45° =

100

2a

BC a 1 AB a 1 , cos 45° = = = = = AC AC 2a 2 2a 2

tan 45° =

BC a 1 = = 1 , cosec 45° = = 2 AB a sin 45°

sec 45° =

1 1 1 = 2 and cot 45° = = =1 cos 45° tan 45° 1

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry z

101

Following table gives the value of various trigonometric rations of 0°, 30°, 45°, 60° and 90° for ready reference. T. ratios ↓

0°

30°

45°

60°

90°

sin θ

0

1 2

1

3 2

1

cos θ

1

3 2

1 2

1 2

0

tan θ

0

1

3

Not Defined

cosec θ

Not Defined

sec θ

1

cot θ

Not Defined

2

1 3

2

2

2

2

3

1

2

1

3

2

Not Defined

1

0 3 ¾ Certain angles like 0°, 30°, 45°, 60° and 90° are considered standard angles and ratios of these are expected to be memorised.

3

Short–cut way to remember the table: No 1

Ratio / Operation Write numbers from 0 to 4 in each column

0° 0

2

Divide each number by 4

0

3

Take Square Root for each number (these are the values for sin θ)

0

4

sin θ (copy the values from row 3)

0

1 2

5

cos θ (invert values from row 4)

1

3 2

6

tan θ (row 3 value ÷ row 4 value)

0

7

cot θ (invert values from row 6)

Not defined

3

1

8

cosec θ (1 ÷ row 4 values)

Not defined

2

2

9

sec θ (invert values from row 8)

θ

sin θ

Volume

0° 0 4

1

30°

45°

60°

90°

1 4

2 4

3 4

4 4

30° 1 1 4 1 2

1 3

2 3

Universal Tutorials – X CBSE (2012–13) – Mathematics

45° 2 1 2 1

60° 3 3 4

90° 4

1

3 2

1

3 2

1

2

1 2

0

1

3

Not defined

2 1 2 1

2

1 3 2 3

2

0 1 Not defined

101

102

4 4

cos θ

3 4

2 4

1 4

0 4

Triangle Method:  Draw two triangles as shown below and with these values, all trigonometric ratios for all standard angles can be easily computed.

30° – 60° – 90° Triangle

45° – 45° – 90° Triangle

A

A 60°

45°

2

1

√2

1 30°

B

45°

C

√3

B

1

C

SOLVED EXAMPLES 8.2: 1) Find the value of the following expressions. i) cos 60° cos 45° – sin 60° sin 45° tan 45° sec 60° 2 sin 90° + − ii) cos ec 30° cot 45° cos 0°

Sol: i)

1 1 1 1− 3 3 × – × = 2 2 2 2 2 2

1 2 2 ×1 5 2 1 + – = – = 2 1 1 2 1 2 2) Find the value of:

ii)

i) sin 30° Using the formula sin θ =

1 1 − cos 2θ . Given that cos 60° = 2 2

ii) sin 15° Using the formula sin(A – B) = sin A cos B – cos A sin B. 1− cos 60 = 2

Sol: i) sin 30° =

1− 2

1 2

=

1 1 = 4 2

ii) sin 15° sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° = 3)

1 2

×

1 1 1( 3 − 1) 3 – × = 2 2 2 2 2

Verify sin (A + B) = sin A cos B + cos A sin B for A = 60° and B = 30°.

Sol: L.H.S. = sin (A + B) = sin (60° + 30°) = sin 90° = 1

–– (1)

R.H.S. = sin A cos B + cos A sin B = sin 60° cos 30° + cos 60° sin 30° =

3 3 1 1 3 1 × + × = + =1 2 2 2 2 4 4

From 1 and 2 4)

–– (2)

L.H.S. = R.H.S., hence verified

Find the value of sin 60° cos 30° – cos 60° sin 30°

Sol: Given expression = sin 60° cos 30° – cos 60° sin 30°

102

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry

=

103

3 3 1 1 3 1 1 × − × = − = . 2 2 2 2 4 4 2

UNSOLVED EXERCISE 8.2: CW Exercise: 1) Find the value of the following expressions.

i) sin 45° cos 30° – cos 45° sin 30°

ii)

iii) 4tan230° + cosec260° – cos245° – cos20°

iv)

v)

tan 45° sin 30° + cos 60°

vi)

sin 30° tan 45° sin 60° cos 30° + − − cos 45° sec 60° cot 45° sin 90° tan 2 60 ° + 4 cos 2 45 ° + 3 sec 2 30 ° + 5 cos 2 90 ° cos ec 30 ° + sec 60 ° − cot 2 30 °

sin 60 ° cos 2 45 °

– cot 30° + 15cos 90°

2) Solve the following for θ when 00 < θ < 900 i) 2 sin 3θ = 1 ii) √3 cot 2θ – 3 = 0 iii) 3 cos2θ - sin2θ = 0 4 4 3) What must be subtracted from 4[sin 30 + cos 60] – 2[cos2 45° – sin2 60] to get zero? 4) If 0° < θ < 90°, find the value of θ when

sec θ − tan θ 2 − 3 = . sec θ + tan θ 2 + 3

5) Verify the following. i) cos(A + B) = cos A cos B – sin A sin B for A = 60° and B = 30° ii) sin(A – B) = sin A cos B – cos A sin B for A = 60° and B = 30° iii) tan(A – B) (1 + tan A tan B) = tan A – tan B for A = 60° and B = 30° iv) (sin 2A) (1 + tan2 A) = 2tan A for A = 45° v) (cos 2A) (1 + tan2 A) = 1 – tan2 A for A = 30° 6) Determine the value of required trigonometric ratio from the relation given below i) sin 15° from sin(A – B) = sin A cos B – cos A sin B ii) cos 75° from cos(A + B) = cos A cos B – sin A sin B 7) If θ is an acute angle and tan θ + cot θ = 2, find the value of tan7 θ + cot7 θ. 8) For x > y, find the acute angles x and y if sin (x + 2y) = √3/2 and cos (x + 4y) = 0. 9) In Δ ABC, right-angled at B, AB = 5 cm and ∠ ACB = 30° (Fig.). Determine the lengths of the sides BC and AC. 10) In Δ PQR, right-angled at Q (Fig.), PQ = 3 cm and PR = 6 cm. Determine ∠ QPR and ∠ PRQ. 11) 12)

13) 14)

A 5 cm

30°

C 6 cm

P

B

3 cm

Q R 1 1 If sin(A–B) = , cos(A+B) = , 0° < A + B ≤ 90°, A > B, find A and B. 2 2 Find the value of x in each of the following: ii) cos x = cos 60° cos 30° + sin 60° sin 30° i) tan 3x = sin 45° cos 45° + sin 30° iii) sin 2x = sin 60° cos 30° – cos 60° sin 30° In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°.Calculate side BC. A rhombus of side 20 cm has two angles of 60° each. Find the length of the diagonals.

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

103

104

15) If each of α, β and γ is a positive acute angle such that sin (α + β – γ) =

1 1 , cos (β + γ – α) = 2 2

and tan (γ + α – β) =1, find the values of α, β and γ. HW Exercise: 1) Find the value of the following expressions. i) cos 45° cos 30° + sin 45° sin 30° ii) cos 45° cos 60° – sin 45° sin 60° iii) sec2 30° + cosec2 30° – cos2 45°+3cos2 60° iv) sin 60° cos 30° + sin 30° cos 60° cos 45° vi) v) 2 tan2 45° + cos2 30° – sin2 60° sec 30° + cosec 30°

vii)

sin 30° + tan 45° − cosec 60° sec 30° + cos 60° + cot 45°

2) Evaluate:

5 cos2 60° + 4 sec 2 30° − tan2 45°

sin2 30° + cos 2 30° 3) Verify the following. i) For A = 30°, verify that: cos 2A = 2cos2A – 1 = 1 – 2sin2A ii) If ∠A = 30°, verify that sin2A =

2 tan A

1 + tan 2 A 4) Find the value of unknown k in the following expressions. i) cos2 45° + tan2 60° = 3(sin2 45° - tan2 30°) + k ii) cos 45° sin 45° + sin 300 = tan 3k

5) If sin θ + cosec θ = 2, prove that (sin θ)m + (cosec θ)m = 2 for all integral values of m. 6) If sin θ = cos θ, solve the equations for x and y: x cot2 θ + y = 5 and x sin2 θ – y cos2 θ = –1/2. 7) Evaluate tan 15° from the relation tan(A – B) (1 + tan A tan B) = tan A – tan B. 1 ; 0° < A + B ≤ 90°; A > B, find A and B. 8) If tan (A + B) = 3 and tan (A – B) = 3 9) Choose the correct option and justify your choice: i)

2 tan 30 ° 1 + tan 2 30 °

=

a) sin 60° b) cos 60° 2 1 − tan 45° ii) = 1 + tan2 45° a) tan 90° b) 1 iii) sin 2A = 2 sin A is true when A = a) 0° b) 30° iv)

2 tan 30 ° 1 − tan 2 30 °

c) tan 60°

d) sin 30°

c) sin 45°

d) 0

c) 45°

d) 60°

=

a) cos 60° b) sin 60° c) tan 60° 10) Solve each of the following equations when 0° < θ < 90°. 1 1 i) 2 cos θ = 1 ii) 2 cos2 θ = iii) 2 sin2 θ = 2 2 cos θ − sin θ 1− 3 = . 11) Find an acute angle θ, when cos θ + sin θ 1+ 3

d) sin 30° iv) 3 tan2 θ – 1 = 0

12) The altitude AD of a ΔABC, in which ∠A obtuse and AD = 10 cm. If BD = 10 cm and CD = 10 3 cm, determine ∠A. 13) An equilateral triangle is inscribed in a circle of radius 6 cm. Find its side. 104

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry

105

14) In an acute angled triangle ABC, if tan (A + B – C) = 1 and sec (B + C – A) = 2, find the value of A, B and C. 15) Given that tan(θ + φ) =

tan θ + tan φ where θ and φ are acute angles. 1 − tan θ tan φ

Calculate θ + φ when tan θ = ½ and tan φ = 1/3.

Complementary Angles: Definition: z z

If the sum of two angles is equal to 90°, the angles are said to be complementary angles. In a right triangle, the two acute angles are complementary angles.

Theorem: z

If θ is an acute angle, then prove that sin (90° – θ) = cos θ,

cos(90° – θ) = sin θ,

tan (90° – θ) = cot θ,

cot(90° – θ) = tan θ,

sec (90° – θ) = cosec θ and cosec (90° – θ) = sec θ

Proof: z

Consider a right triangle OPM, right angled at M as shown in fig. Let MOP = θ, then ∠OPM = (90° – θ) For the reference angle θ, we have sin θ =

OP OM OM PM OP PM , cos θ = , tan θ = , cosec θ = , cot θ = and sec θ = –– (i) OP OP OM PM PM OM

For the reference angle (90° – θ), we have Base = PM, Perpendicular = OM and Hypotenuse = OP

y P

PM OM ∴ sin(90° – θ) = , cos(90° – θ) = , OP OP

tan (90° – θ) =

OM OP , cosec (90° – θ) = , PM OM

OP PM and cot(90° – θ) = sec(90° – θ) = PM OM

x′

Q O

M

x

––– (ii) y′

From (i) and (ii), we obtain, sin (90° – θ) = cos θ,

cos(90° – θ) = sin θ,

tan (90° – θ) = cot θ,

cot(90° – θ) = tan θ,

sec(90° – θ) = cosec θ and cosec (90° – θ) = sec θ

Trigonometric Inter-relationships:  The complementary angles have a special property that inter - relates their trigonometry ratios as follows:

Volume

z

sin θ = cos (90° – θ)

z

cos θ = sin (90° – θ)

z

tan θ = cot (90° – θ) Universal Tutorials – X CBSE (2012–13) – Mathematics

105

106 z

cot θ = tan (90° – θ)

z

cosec θ = sec (90° – θ)

z

sec θ = cosec (90° – θ)

SOLVED EXAMPLES 8.3: 1)

Sol: 2)

Without using trigonometric tables evaluate,

cos 53° . sin 37°

cos 53° cos(90° − 37°) sin 37° = =1 = sin 37° sin 37° sin 37° Prove the following identities without using the tables.

i) sin 43° cos 47° + cos 43° sin 47° = 1

Sol: L.H.S. = sin 43°cos 47° + cos 43°sin 47° = sin 43°cos(90° – 43°) + cos 43°sin(90° – 43°) = sin243°+ cos243° = 1 = R H S 2

ii)

2 tan 53° ⎛ cot 80° ⎞ 2 −⎜ ⎟ − 4 sin 30° = 0 cot 37° ⎝ tan 10° ⎠ 2

Sol: L.H.S. =

2 tan 53° ⎛ cot 80° ⎞ −⎜ ⎟ − 4 sin2 30° cot 37° ⎝ tan 10° ⎠ 2

= iii)

2

2 cot 37° ⎛ tan 10° ⎞ ⎛ 1⎞ −⎜ ⎟ − 4⎜ ⎟ = 2 − 1 − 1 = 0 = R.H.S cot 37° ⎝ tan 10° ⎠ ⎝2⎠

sin θ cos θ = sec( 90 ° − θ ) cos ec (90 ° − θ ) + sin( 90 ° − θ ) cos( 90 ° − θ )

Sol: L.H.S. =

sin θ cos θ sin θ cos θ sin2 θ + cos2 θ + = + = sin(90° − θ) cos(90° − θ) cos θ sin θ sin θ cos θ

=

1 1 1 = × = cos ecθ sec θ = sec(90° − θ) cos ec (90° − θ) = R.H.S. sin θ cos θ sin θ cos θ

UNSOLVED EXERCISE 8.3: CW Exercise: 1) Express the following in terms of trigonometric ratios of angles between 0° and 45° ii) cot 85° + cos 75° i) tan 68° + sec 69° 2) Evaluate the following:

i) 2 + iii)

1 sin 49° 4 cos 41°

ii)

sin 30° sin 40° sin 50° sin 60° cos 30° cos 40° cos 50° cos 60°

sec 70° sin59° + cos ec 20° cos31° tan 53° cot80° vii) 2 cot 37° tan10° 2 cos 67° tan40° – sin 90° ix) sin 23° cot50°

v)

106

sin 40° 1 ⎛ tan 35° ⎞ − ⎜ ⎟ cos 50° 2 ⎝ cot 55° ⎠

iv) sec 50° sin 40° + cos 40° cosec 50° cot 54° tan20° + -2 tan 36° cot70° cos 80° viii) + cos59°cosec31° sin10°

vi)

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry

⎛ cos 58° ⎞ ⎟⎟ – 3) Evaluate the following: 2⎜⎜ ⎝ sin 32° ⎠

107

⎛ cos 38° cos ec 52° ⎞ ⎟⎟ 3 ⎜⎜ ⎝ tan 15° tan 60° tan 75° ⎠

2 cos 67° tan 40° − – cos 0° + tan 15°.tan25°.tan60°.tan65°.tan75°. sin 23° cot 50° 5) What must be subtracted from the expression given below such that its value becomes zero cot 54 ° + sin 20° × sec 70° tan 7° × tan 23° × tan 60° × tan 67° × tan 83° + tan 36 ° 6) Evaluate the following: i) cosec (65° + θ) – sec (25° – θ) – tan (55° – θ) + cot (35° + θ) tan( 90° − θ) sec( 90 ° − θ ) cos( 90 ° − θ ) tan θ + ii) cot θ cos ec ( 90 ° − θ ) cot( 90 ° − θ ) sin( 90 ° − θ )

4) Evaluate

7) Prove the following identities without using the tables. i) sec 70°sin 20°– cos 20°cosec 70° = 0 ii) sin(60° + θ) = cos (30° – θ) iii) tan1° tan2° tan3° tan87° tan88° tan89° = 1 iv) tan 10° tan 15° tan 75° tan 80° = 1 v) tan 7° tan 23° tan 60° tan 67° tan 83° =

3

8) Prove that: i) cot 1° cot 2° cot 3° …… cot 89° = 1 ii) sin 1° cos 1° sin 2° cos 2° sin 3° cos 3° …… sin 180° cos 180° = 0 9) Prove that

1 1 = –2 tan θ cosec(90° – θ) − 1 + cos( 90 ° − θ ) 1 − cos( 90 ° − θ )

10) If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A. 11) If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A. A C ⎛B +C ⎞ ⎛A+B⎞ 12) Prove that tan ⎜ tan where A, B, C are interior angles of a ΔABC. ⎟ cot ⎜ ⎟ = cot 2 2 2 2 ⎝ ⎠ ⎝ ⎠

HW Exercise: 1) Without using trigonometric tables evaluate the following.

i)

sin10° cos ec19° + cos 80° sec 71°

ii) 2 –

iii) tan(55° – θ) – cot(35° + θ) ⎛ sin 49° ⎞ ⎛ cos 41° ⎞ v) ⎜ ⎟+⎜ ⎟ ⎝ cos 41° ⎠ ⎝ sin 49° ⎠

cot 40° 1 ⎛ cos 35° ⎞ − ⎜ ⎟ tan 50° 2 ⎝ sin 55° ⎠

iv)

2

vi) sec 70°sin 20° – cos 20°cosec 70°

vii) cos(70 + θ) – sin(20 – θ) 2

1 ⎛ tan 79° ⎞ ⎜ ⎟ 5 ⎝ cot 11° ⎠

viii)

2

⎛ sin 47° ⎞ ⎛ cos43° ⎞ ix) ⎜ ⎟ +⎜ ⎟ – 2cos²45° ⎝ cos 43° ⎠ ⎝ sin47° ⎠

x)

2 sin 43° cot30° - 2 sin 45° cos 47° tan60°

sin 50° ⎛ cos ec 40° ⎞ +⎜ ⎟ − 4cos 50°cosec 40° cos 40° ⎝ sec 50° ⎠

tan 50 ° sec 50° + + cos 40° cosec 50° cot 40 ° cos ec 40° 2) Prove the following identities. i) sin(90° – θ) cos θ + cos(90° – θ) sin θ = 1

xi)

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

107

108

ii) sec(90° – θ) cosec θ – tan (90° – θ) cot θ = 1 sin( 90 ° − θ ) cos( 90 ° − θ ) iii) + =1 cos ec ( 90 ° − θ ) sec( 90 ° − θ ) iv)

sin θ cos θ = sec(90° – θ) cosec(90° – θ) + sin( 90 ° − θ ) cos( 90 ° − θ )

v) sec(90° – θ) cosec(90° – θ) = sec2 θ cot θ = sec2(90° – θ) cot(90° – θ) vi)

1 + cos 60° 4 tan2 30° = 1 − cos 60° (1 − tan2 30°)2

vii)

10 3 4 cot 2 30° + 3 sin2 60° − 2 cos ec 2 60° − tan 2 30° = 3 4 3 2

2

⎛ sin 47° ⎞ ⎛ cos 43° ⎞ viii) ⎜ ⎟ +⎜ ⎟ − 4 cos 2 45° = 0 ⎝ cos 43° ⎠ ⎝ sin 47° ⎠ 3) Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. 4) i) If sin θ + cos θ =

2 cos(90° – θ). Find cot θ.

ii) Determine value of x such that 2cosec² 30° + x sin² 60° –

3 tan² 30° = 10 4

iii) Verify sin 60° = 2cos 30°sin 30°. iv) If tan 11A = cot 7A. Find the value of sin2 6A + sec2 9A v) If sin θ + cos θ =

2 sin (90° – θ), determine cot θ

2 sin2 A + 3 cot 2 A 4 tan2 A − cos 2 A 5) If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. 6) If tan A = cot B, prove that A + B = 90°. 11 sin 70 ° 4 cos 53 ° cos ec 37 ° 7) Evaluate the value of the expression: − 7 cos 20 ° 7 tan 15 ° tan 35 ° tan 55 ° tan 75 °

vi) If cosec A =

2 , find the value of,

8) Find the value of x satisfying the equations given below: i) x = sin 1° cos 1° sin 2° cos 2° sin 3° cos 3° ………… sin 180° cos 180° 3 ii) 2cosec2 30° + x sin2 60° – tan2 30° = 10 4 A ⎛B +C ⎞ 9) If A, B and C are interior angles of a triangle ABC, then show that sin ⎜ ⎟ = cos . 2 ⎝ 2 ⎠

Trigonometric Identities: ™ An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved. Following are the identifies: A  sin2 A + cos2 A = 1 2 2  1 + tan A = sec A  1 + cot2 A = cosec2 A for all values of A ™ In ΔABC, right-angled at B (Fig.)

we have: sin A =

108

AB AB AC ; cos A = ; sec A = ; AC AB AC

C

Universal Tutorials – X CBSE (2012–13) – Mathematics

B

Volume

Chapter 08: Introduction of Trigonometry

109

BC AC AB ; cot A = ; cosec A = AB BC BC

tan A =

By Pythagoras theorem we have, AB2 + BC2 = AC2 –––– (1) Dividing each term of (1) by AC2, we get 2

2

⎛ AB ⎞ ⎛ BC ⎞ ⎛ AC ⎞ i.e. ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ ⎝ AC ⎠ ⎝ AC ⎠ ⎝ AC ⎠

AB

2

+

BC 2 AB

2

=

AC 2 AB

2

AC

2

+

BC 2 AC

2

=

AC 2 AC 2

2

i.e. (cos A)2 + (sin A)2 = 1

i.e. cos2 A + sin2 A = 1 ™ Let us now divide (1) by AB2. We get AB 2

AB2

2

sin2 A + cos2 A + = 1 2

⎛ AB ⎞ ⎛ BC ⎞ ⎛ AC ⎞ or ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ AB AB ⎝ ⎝ ⎝ AB ⎠ ⎠ ⎠

––– (2)

2

i.e. 1 + tan2 A = sec2 A ––– (3)

™ Well, tan A and sec A are not defined for A = 90°. So, (3) is true for all A such that 0° ≤ A < 90°. ™ Let us see what we get on dividing (1) by BC2. We get

AB2 BC

2

+

BC 2 BC

2

=

AC 2 BC

2

2

2

⎛ AB ⎞ ⎛ BC ⎞ ⎛ AC ⎞ i.e. ⎜ ⎟ +⎜ ⎟ =⎜ ⎟ ⎝ BC ⎠ ⎝ BC ⎠ ⎝ BC ⎠

2

i.e. cot2 A + 1 = cosec2 A

2

1 + cot A = cosec2 A ––– (4) ™ Note that cosec A and cot A are not defined for A = 0°. ™ Therefore (4) is true for all A such that 0° < A ≤ 90°.

Identical expressions of: sin2 A + cos2 A = 1 are 2 2 z 1 + tan A = sec A are 2 2 z 1 + cot A = cosec A are ™ Using these identities, we can ratios. i.e., if any one of the trigonometric ratios. z

1 – sin2 A = cos2 A and 1 –cos2 A = sin2 A sec2 A – tan2 A = 1 and sec2 A – 1 = tan2 A cosec2 A – cot2 A = 1 and cosec2 A – 1 = cot2 A express each trigonometric ratio in terms of other trigonometric ratios is known, we can also determine the values of other

SOLVED EXAMPLES 8.4: 1)

Prove the following identities, 1 − cos θ sin θ = i) sin θ 1 + cos θ

Sol: L.H.S. = =

1 − cos θ 1 − cos θ 1 + cos θ = [Multiplying numerator and denominator by (1+cos θ)] × sin θ sin θ 1 + cos θ

1 − cos2 θ sin2 θ = sin θ(1 + cos θ) sin θ(1 + cos θ)

sin θ =RHS [By the identity sin2θ = 1 – cos2θ] 1 + cos θ ii) cos4 θ – cos2 θ = sin4 θ – sin2θ Sol: L.H.S. = cos4θ – cos2θ = cos2θ (cos2θ – 1) [Using cos2θ = 1 – sin2θ] = (1 – sin2θ) (1 – sin2θ – 1) = (1 – sin2θ) (–sin2θ) = –sin2θ + sin4θ = sin4θ – sin2θ = R.H.S.

=

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

109

110

iii)

1 = sec θ + tan θ sec θ − tan θ

(sec θ − tan θ ) (sec θ − tan θ )

Sol: R.H.S. = sec θ + tan θ = (sec θ + tan θ)

[Multiplying numerator and denominator by (sec θ – tan θ)] = 2)

2

sec θ − tan 2 θ = sec θ + tan θ = L.H.S. sec θ − tan θ

Determine whether the following equations are identities. i)

sin θ tan θ + = sec θ cosec θ + cot θ 1 − cos θ 1 + cos θ

Sol: Consider

sin θ tan θ sin θ + sin θ cos θ + tan θ − tan θ cos θ + = 1 − cos θ 1 + cos θ 1 − cos 2 θ

sin θ + sin θ cos θ + tan θ − sin θ = = sin2 θ =

sin θ cos θ + sin2 θ

sin θ cos θ

sin θ cos 2 θ + sin θ sin θ cos 2 θ sin θ = + sin 2 θ cos θ sin 2 θ cos θ sin 2 θ cos θ

cos θ 1 + = cosec θ sec θ + cot θ sin θ sin θ cos θ LHS = RHS So it is an identify

=

ii)

1 − tan θ 1 − cot θ = 1 + tan θ 1 + cot θ

1 1− 1 − cot θ tan θ = tan θ − 1 = − (1 − tan θ) = Sol: Consider 1 + cot θ 1 + 1 tan θ + 1 1 + tan θ tan θ

⇒

1 − tan θ 1 − tan θ 1 − cot θ if = =0 1 + tan θ 1 + cot θ 1 + tan θ

i.e. if and only if tan θ = 1 2

θ = 45°

Hence this is not an identity.

2

iii) cot θ + cos θ = sin θ Sol: The variable θ can take any value in the range 0° < θ ≤ 90° as cot θ is not defined for θ = 0° Now

cos 2 θ + cos θ = sin 2 θ sin 2 θ

⇒

cos 2 θ + sin 2 θ cos θ = sin 2 θ sin 2 θ

For θ = 90° LHS of (II) = 0 and RHS = 1

→ (II)

i.e. RHS ≠ LHS. Hence it is not an identity.

cos θ cos θ + = 4. 1 − sin θ 1 + sin θ

3)

Solve the equation,

Sol:

cos θ cos θ =4 + 1 − sin θ 1 + sin θ

⇒ cos θ (1 + sin θ) + cos θ (1 – sin θ) = 4 (1 – sin2 θ) ⇒ cos θ + cos θ sin θ + cos θ – cos θ sin θ = 4cos2 θ ⇒ 2 cos θ = 4cos2 θ ⇒ 2 cos2θ – cos θ = 0 or cos θ (2 cosθ – 1) = 0 110

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry

⇒ cos θ = 0 or cos θ = ½

111

θ = 90° or 60°

UNSOLVED EXERCISE 8.4: CW Exercise: 1) Write all the other trigonometric ratios of ∠A in terms of sec A 2) Prove the following identities i) (1 + tan2θ) (1 – sin θ) (1 + sin θ) = 1

ii)

iii) (1 + cotθ – cosec θ) (1 + tanθ + secθ) = 2

iv) (cosec θ – cot θ)2 =

v) sin6 θ + cos6 θ = 1 – 3 sin2 θ cos2θ tan θ + sec θ − 1 1 + sin θ = tan θ − sec θ + 1 cos θ 1 1 + =1 ix) 2 1 + tan θ 1 + cot 2 θ 3) Find the value of

vii)

i) cot θ + cosec θ; if cos θ = iii) sin2θ – cos2 θ, if

3 5

sec 2 θ − 1cos ec 2 θ = tanθ + cotθ

1 + sin θ cos θ + = 2secθ cos θ 1 + sin θ sin θ 1 + cos θ viii) + = 2cosec θ 1 + cos θ sin θ

vi)

ii)

cos ec 2 θ − sec 2 θ 2 when tanθ = 2 2 cos ec θ + sec θ 5

3 tan θ = 1

iv) cos(θ + φ); θ and φ are acute angles if tan θ = 1 and sin φ = 2

4) Evaluate,

1 − cos θ 1 + cos θ

1 2

2

sin 63° + sin 27° cos 2 17° + cos2 73°

5) If sec θ + tan θ = m, sec θ – tan θ = n than P:T

mn = 1

1 1 6) If secθ = x + P:T sec θ + tan θ = 2x or 4x 2x 7) Prove the following identities. i) (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ 1 1 1 1 ii) − = − sec x − tan x cos x cos x sec x + tan x cos α cos α = n then (m2 + n2) cos2β = n2 iii) If = m and cos β sin β

iv) x = r cos α sin β; y = r cos α cos β and z = r sin α then x2 + y2 + z2 = r2 v) If tan θ + sin θ = m and tan θ – sin θ = n. Show that m2 – n2 = 4 mn tan θ cot θ + = 1 + sec θ cosec θ 1 − cot θ 1 − tan θ 8) If x = a cos θ – b sin θ, y = a sinθ + b cosθ then prove that x2 + y2 = a2 + b2 9) If sinθ + cosθ = m, secθ + cosecθ = n, prove that n(m2 – 1) = 2m HW Exercise: 1) Express the trigonometric ratios sin A sec A and tan A in terms of cot A 2) Prove the identities

vi)

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

111

112

sin θ + cos θ sin θ − cos θ 2 2 + = = 2 2 sin θ − cos θ sin θ + cos θ sin θ − cos θ 2 sin2 θ − 1 ii) sin4θ + cos4θ = 1 – 2 sin2θ cos2θ cos ecθ + cot θ iii) = 1 + 2 cosec θ cot θ + 2cot2 θ cos ecθ − cot θ iv) sec2θ + cosec2θ = sec2θ cosec2θ

i)

(1+ tan θ)cot θ 2

v)

cos ec 2 θ

= tan θ

⎛ cos 2 A ⎞ 1 + 1⎟ tan2 A = vi) ⎜⎜ 2 2 ⎟ sin A cos A ⎠ ⎝ vii)

1 + sin θ = sec θ + tan θ 1 − sin θ

viii) tan4 θ + tan2 θ = sec4 θ – sec2 θ 3) Find the value of 4 sin θ − 2 cos θ 3 i) if tanθ = 4 sin θ + 3 cos θ 4 b iii) sin2 A, a, b ∈ R if tan A = a

ii) sin2θ – cos2θ if

3 tan θ = 3 sinθ

MISCELLANEOUS EXERCISE: 1) If the ratio of the cosine of an angle to its sine is 8:15 find all the trigonometric ratios of that angle 2) Find the value of the following expressions. i) 4(sin4 30° + cos4 60°) – 3(cos2 45° – sin2 90°) ii) (cosec2 45° sec2 30°) (sin2 30° + 4cot2 45° – sec2 60°) 3 iii) cot2 30° – 2cos2 60° – sec2 45° – 4sec2 30° 4 1 iv) cosec2 30° + sec2 30° – tan2 45° – sin2 90° 2 1 v) cos2 30°cos2 45° + 4cos2 60° + cos2 90° – 2tan2 60° 2 3) Find the value of: 4 sin θ − 2 cos θ 3 i) , if tan θ = ii) sin² θ – cos² θ, if 3 tan θ = 3 sin θ 4 sin θ + 3 cos θ 4 b iii) sin² A, given that tan A = a 4) a) If (tan θ + cot θ) = 2 find the value of sec2 θ + cosec2 θ b) If x = a sin A cos B, y = a sin A sin B, z = a cos A prove that x2 + y2 + z2 = a2 5) Verify the following: i) cos (A – B) = cos A cos B + sin A sin B for A = 60° B = 30° ii) sin(A + B) = sin A cos B + cos A sin B for A = B = 45° iii) sin(A + B) = sin A cos B + cos A sin B for A = B = 30° 6) Prove the following identities: 112

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry

i) (1 + sin θ) (1 – sin θ) =

1 2

sec θ

iii) tan2 φ + cot2 φ + 2 = sec2 φ ⋅ cosec2 φ 1 1 + = 2 sec2 θ 1 − sin θ 1 + sin θ cos θ + sin θ = 1 + cot θ vii) sin θ

v)

tan 2 θ ix) sec θ + 1 = sec θ –1

xi)

sin4 θ − cos 4 θ 2

2

113

ii)

tan A tan A + = 2 cosec A sec A − 1 sec A + 1

iv)

1+ cos 2 A = 2 cosec2 A – 1 2 sin A

vi)

tan θ cot θ + = 1 + tan θ + cot θ 1 − cot θ 1 − tan θ

viii) (cosec θ + cot θ) (1 – cos θ) = sin θ

x)

cos 2 θ + tan2 θ − 1 = tan2 θ sin2 θ

xii) tan2θ + cot2θ = cosec2 θ . sec2θ – 2

=1

sin θ − cos θ xiii) (sin8 θ – cos8 θ) = (sin2 θ – cos2 θ) (1 – 2sin2 θ cos2 θ) 1 + cos θ + sin θ 1 + sin θ xiv) = xv) (sin θ – cos θ)2 = 1 – 2sin θ cos θ 1 + cos θ − sin θ cos θ xvi) sin4θ – cos4θ = (sin2θ – cos2θ) = 2sin2θ – 1 = 1 – 2cos2θ 2

1 + cos A ⎛ 1 + cos A ⎞ xvii) ⎜ ⎟ = 1 − cos A ⎝ sin A ⎠

xix)

(tan θ + sin θ )

(tan θ − sin θ )

=

xviii)

sin θ − 2 sin3 θ = tan θ 2 cos3 θ − cos θ

(sec θ + 1) (sec θ − 1)

xx) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1 1 + sin θ 2 tan θ xxi) = 1+ + 2 tan 2 θ 1 − sin θ cos θ xxii) xxiii)

tan 3 θ cot 3 θ + = cosec θ sec θ – 2 sin θ cos θ 2 1 + tan θ 1 + cot 2 θ

tan2 θ cos ec 2θ 1 + = tan2 θ − 1 sec 2 θ − cos ec 2 θ sin2 θ − cos 2 θ

sec 2 θ − sin2 θ tan 3 θ − 1 = cosec2 θ – cos2 θ xxv) = sec2 θ + tan θ 2 tan θ − 1 tan θ (cos ec θ − cot θ )(cos ec θ + cot θ ) xxvi) (tan2 θ) (1 – sin θ) (1 + sin θ) = 1 + cot 2 θ xxvii) If cosec θ – sin θ = m and sec θ – cos θ = n then show that (m2n)2/3 + (mn2)2/3 = 1

xxiv)

xxviii) If sin A + cos A = a and tan A + cot A = b. Prove that xxix) If cos θ + sec θ =

1 a2 − 1 = b 2

3 , prove cos3 θ + sec3 θ = 0

xxx) If sin θ + cos θ = p and sec θ + cosec θ = q then q (p2 – 1) = 2p. 7) Find the value of: 1 2 tan θ i) tan 60° using the formula tan 2θ = ; given that tan 30° = . 2 3 1 − tan θ ii) sin 75° using the formula sin(A + B) = sin A cos B + cos A sin B] 8) Solve the following equations Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

113

114

sin θ sin θ + =4 ii) 3cos θ = 2sin2 θ iii) sec2 θ – 2tan θ = 0 1 − cos θ 1 + cos θ 9) Without using trigonometric tables evaluate the following. tan 20° 2 tan 53° cot 80° cos 70° cos 59° i) − + 1− ii) + − 8 sin2 30° cot 70° cot 37° tan 10° sin 20° sin 31° 2 3 tan 25. tan 40. tan 50. tan 65 − 1 tan2 60 sin 35° ⎛ cos 55° ⎞ 2 +⎜ iv) iii) ⎟ − 2 cos 60° cos 55° ⎝ sin 35° ⎠ 4(cos2 29 + cos2 61)

i)

2

⎛ tan 20 ⎞ ⎛ cot 20 ⎞ ⎟⎟ + ⎜ v) ⎜⎜ ⎟ ec cos 70 ⎝ sec 70 ⎠ ⎝ ⎠

2

sin2 20° + sin2 70° cos(90° − θ) cos θ sin(90° − θ) sin θ = + 2 2 cot θ tan θ cos 20° + cos 70° cos 70° cos 55° cos ec 35° + vii) sin 20° tan 5° tan 25° tan 45° tan 65° tan 85°

vi)

tan 20 ° ⎛ sin11° ⎞ viii) ⎜ ⎟ + cos 79 ° cot 70° ⎝ ⎠

ix)

cos 81° cos 14° cot 54° + −2 sin 9° sin 76° tan 36° 1 cosec2 30° xii) sin2 82° + sin2 8° – 2

x)

xiv)

cot 40° 1 ⎛ cos 35° ⎞ − ⎜ ⎟ tan 50° 2 ⎝ sin 55° ⎠

cos 75° sin12° cos 18° + − sin15° cos 78° sin 72° tan 35° cot 78° xiii) + −1 cot 55° tan 12°

xi)

xv) tan 5 $ tan 25 $ tan 30 $ tan 65 $ tan 85.

xvi) cos(40 + θ) – sin(50 – θ) + 2

sin 2 49 ° cos 2 41 ° + 2 cos 41 ° sin 2 49 °

cos2 40 + cos2 50 sin2 40 + sin2 50

2

3 sin 62° sec 42° ⎛ sin 27° ⎞ ⎛ cos 63° ⎞ − xvii) ⎜ xviii) ⎟ +⎜ ⎟ cos 28° cos ec 48° ⎝ cos 63° ⎠ ⎝ sin 27° ⎠ 10) Solve the following: i) If tan θ = 1, prove that 2 sin θ cos θ = 1. 10 cos θ − 7 sin θ ii) If 5 cos θ = 7 sin θ, find the value of 10 cos θ + 7 sin θ 3 iii) Find the value of 2 cosec2 30 + sin2 60 – tan2 30 4 iv) If A and B are acute angles and sin A = cos B, prove that A + B = 90°. 3 v) cot2 30° + tan2 60° – sin2 45° + cosec2 30° 4 2 2 vi) cosec 60° + sec 60° – cot2 60° + tan2 30° 11) i) Verify: For A = 30°, verify that: cos 3A = 4cos3 A – 3cos A ii) Prove the identities, (sin θ + sec(90°–θ))2 + (cos θ + cosec(90°–θ))2 = 7 + tan2 θ + tan2(90°–θ) 12) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

i) 114

cos A 1 + sin A + = 2 sec A 1 + sin A cos A

iii)

1+ sec A sin 2 A = sec A 1 − cos A

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry

iv)

115

cos A − sin A + 1 = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A. cos A + sin A − 1

sin θ − 2 sin3 θ

= tan θ 2 cos3 θ − cos θ 13) a) Solve the following: v)

i) Solve for θ cosec2θ – cot θ (1 + ii) If

3) + 3 – 1 = 0

cos 2 θ = 3. Find the value of θ. cot θ − cos 2 θ 2

b) Prove the following identities i) (sin2 63° + sin2 27°) + (cos2 73° – sin2 17°) = 1 ii) (sec(90° – θ) – sin θ) (cosec(90° – θ) – cos θ) (cot(90° – θ) + cot θ) = 1 iii) cos (81° – θ) = sin (9° + θ) v)

tan (90 ° − A )cot A 2

cos ec A

iv) sin θ sin (90° – θ) – cos θ cos (90° – θ) = 0

– cos2 A = 0

vi) sin (50°+θ) – cos (40°–θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1 vii)

cos (90 ° − θ ) cos θ =2 + sin θ sin (90 ° − θ )

viii)

cos 20 ° cos θ =2 + sin 70 ° sin (90 ° − θ )

14) Determine whether the following equations are identities. 1 1 = 2cot θ cosec θ ii) sin2θ + sin θ = 1 i) + sec θ − 1 sec θ + 1 iii) cot4 θ + cot6 θ = cot3 θ cosec2 θ 15) Solve the following equation 1 i) sin θ – cos θ = 0 ii) 2sin2 θ = 2 iv) 3tan θ – 5cosec θ + cot θ = 0 iii) tan2 θ + cot2 θ = 2 tan A + sec A − 1 1 + sin A 16) Prove that, = tan A − sec A + 1 cos A 17) Find the value of the following expressions. b cos θ + sin θ , find the value of a cos θ − sin θ 18) If (sec A + tan A)(sec B + tan B)(sec B + tan C) = (sec A – tan A)(sec B – tan B) (sec C – tan C). Prove that each side equal to 1.

i) If 7 sin2 θ+3 cos2 θ =4, find the value of tanθ

19) Simplify:

ii) If cot θ =

sin 3 θ + cos 3 θ + sin θ cos θ sin θ + cos θ

[CBSE 09]

20) Find the value of sin 30° geometrically. 21) Without using trigonometric tables, evaluate: cos 58° sin 22° cos 38° cos ec 52° + – sin 32° cos 68° tan 18°. tan 35° tan 60° tan 72° tan 55° 15 (2 + 2 sin θ )(1 − sin θ ) . , then evaluate 22) If cot θ = 8 (1 + cos θ )(2 − 2 cos θ )

[CBSE 09]

23) Find the value of tan 60°, geometrically.

[CBSE 09]

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

[CBSE 09] [CBSE 09]

115

116

24) Evaluate:

2 2 5 cosec2 58° – cot 58° tan 32° – tan 13° tan 37° tan 45° tan 53° tan77°.[CBSE 09] 3 3 3

25) Prove that sec2θ – 26)

27) 28)

29) 30) 31) 32)

sin2 θ − 2 sin4 θ

=1 2 cos4 θ − cos2 θ Without using the trigonometric tables, evaluate the following: 11 sin 70° 4 cos 53° cos ec 37° – 7 cos 20° 7 tan 15° tan 35° tan 55° tan 75° cot A − cos A cos ecA − 1 = Prove that: cot A + cos A cos ecA + 1 Without using trigonometric tables, evaluate the following: sin18° + 3 [tan 10° tan 30° tan 40° tan 50° tan 80°] cos 72° Without using trigonometric tables, evaluate the following: (cos2 25° + cos2 65°) + cosec θ sec(90° – θ) – cot θ. tan (90° – θ) 1 If 7 sin2 θ + 3 cos2 θ = 4, show that tan θ = 3 Prove that: (1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A. Without using trigonometric tables, evaluate the following:

⎛ cos 58° ⎞ 2⎜ ⎟ – ⎝ sin 32° ⎠

[CBSE 09]

[CBSE 08] [CBSE 08]

[CBSE 08] [CBSE 08] [CBSE 08] [CBSE 08]

⎛ cos 38° cos ec 52° ⎞ 3 ⎜ ⎟ ⎝ tan 15° tan 60° tan 75° ⎠

[CBSE 08]

33) If sec 2A = cosec (A – 42)°, where 2A is an acute angle, find the value of A. 34) If cos θ + sin θ =

2 cos θ S.T cos θ – sin θ =

[CBSE 08]

2 sin θ

MULTIPLE CHOICE QUESTIONS: CW Exercise: 3 sin θ tan θ − 1 , then = 5 2 tan2 θ 16 1 3 160 a) b) c) d) 625 36 160 3 5 sin θ − 4 cos θ is 2) If 5 tan θ – 4 = 0, then the value of 5 sin θ + 4 cos θ 5 5 1 a) b) c) 0 d) 3 6 6 (1 + sin θ)(1 − sin θ) 8 is 3) If θ is an acute angle such that tan2 0 = , then the value of 7 (1 + cos θ)(1 − cos θ)

1) If θ is an acute angle such that cos θ =

a)

7 8

4) The value of a)

1 2

b)

8 7

cos3 20° − cos3 70° sin3 70° − sin3 20° 1 b) 2

c)

7 4

d)

64 49

is c) 1

d) 2

5) If angles A, B, C of a ΔABC form an increasing AP, then sin B = 116

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 08: Introduction of Trigonometry

a)

1 2

117

3 2

b)

c) 1

1

d)

2

6) The value of tan 1° tan 2° tan 3° … tan 89° is a) 1 b) – 1 c) 0 d) None of these 7) If θ and 2θ–45° are acute angles such that sin θ = cos (2θ – 45°), then tan θ is equal to 1 d) a) 1 b) –1 c) 3 3 8)

1 − tan2 45° 1 + tan2 45° a) tan 90°

is equal to c) sin 45°

b) 1

d) sin 0°

⎛B+C⎞ 9) If A, B and C are interior angles of a triangle ABC, then sin ⎜ ⎟ = ⎝ 2 ⎠ A A A b) cos c) –sin 2 2 2 sec4 A – sec2 A is equal to b) tan4 A – tan2 A c) tan4 A + tan2 A a) tan2 A – tan4 A If x = a cos θ and y = b sin θ, then b2x2 + a2y2 = b) ab c) a4 b4 a) a2 b2 2 2 The value of sin 29° + sin 61° is a) 1 b) 0 c) 2 sin2 29° 2 2 4 If sin θ + sin θ = 1, then cos θ + cos θ = a) –1 b) 1 c) 0 (1 + tan θ + sec θ) (1 + cot θ – cosec θ) = a) 0 b) 2 c) 1

d) –cos

a) sin

10) 11) 12) 13) 14)

A 2

d) tan2 A + tan4 A d) a2 + b2 d) 2cos2 61 d) None of these d) –1

2

15)

1 + tan A is equal to 1 + cot 2 A a) sec2 A b) –1

c) cot2 A

d) tan2 A

HW Exercise: 1) If tan θ = a)

a a sin θ + b cos θ , then is equal to a sin θ − b cos θ b

a2 + b2

b)

a2 − b2

2) If 16 cot x = 12, then a)

1 7

a2 − b2 a2 + b2

sin x − cos x equals sin x + cos x 3 b) 7

3) If 3 cos θ = 5 sin θ, then the value of a)

Volume

271 979

b)

316 2937

c)

a+b a−b

d)

c)

2 7

d) 0

5 sin θ − 2 sec 3 θ + 2 cos θ 5 sin θ + 2 sec 3 θ − 2 cos θ c)

a−b a+b

is

542 2937

Universal Tutorials – X CBSE (2012–13) – Mathematics

d) None of these

117

118

4) If

x cos ec 2 30° sec2 45° 8 cos2 45° sin2 60°

a) 1

= tan2 60° – tan2 30°, then x =

b) –1

c) 2

d) 0 2

tan θ − cos ec 2 θ

5) If θ is an acute angle such that sec2 θ = 3, then the value of

tan2 θ + cos ec 2 θ

4 3 2 b) c) 7 7 7 6) The value of cos 1° cos 2° cos 3° … cos 180° is a) 1 b) 0 c) – 1

d)

a)

1 7

d) None of these

7) If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ – to a) 1

b) 0

8) sin 2A = 2 sin A is true when A = a) 0° b) 30°

3 tan 3θ is equal

c) –1

d) 1 +

c) 45°

d) 60°

c) sec2 θ + tan2 θ

d) sec2 θ – tan2 θ

3

1 + sin θ is equal to 1 − sin θ

9)

a) sec θ + tan θ sin θ 10) is equal to 1 + cos θ 1 + cos θ a) sin θ

b) sec θ – tan θ

b)

1 − cos θ cos θ

c)

1 − cos θ sin θ

d)

11) 2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) is equal to a) 0 b) b c) –1 12) If x = r sin θ cos φ, y = r sin θ sin φ and z = r cos θ, then a) x2 + y2 + z2 = r2 b) x2 + y2 – z2 = r2 c) x2 – y2 + z2 = r2 13) If x = a sec θ cos φ, y = b sec θ sin φ and z = c tan θ, then a)

z2

b) 1 –

z2

c2 c2 14) (sec A + tan A) (1 – sin A) = a) sec A b) sin A

118

is

c)

z2 c2

x2 a

2

+

1 − sin θ cos θ

d) None of these d) z2 + y2 – x2 = r2 y2 b2

–1

c) cosec A

Universal Tutorials – X CBSE (2012–13) – Mathematics

= d) 1 +

z2 c2

d) cos A

Volume

Chapter 08: Introduction of Trigonometry

119

COLUMN MATCHING QUESTIONS: 1) Given in column I are some right angled triangles. For each item in Column I, choose correct option(s) in column II. Column I Column II i) C AB A) sin C = AC B A ii) A BC B) cos A = AC B C B iii) CA C) tan B = AB A C iv) B BC D) cot C = AB C A AB E) BC 2) From column II choose correct option(s) for each item in column I. Column I Column II i)

2sin 2θ =

ii) 2cos 3θ = iii)

3 then θ =

A) 12°

3 then θ =

B) 30°

3 tan 5θ – 3 = 0 then θ =

C) 45° D) 20°

3) It is given that θ < (A + B) ≤ 90° and A > B. For the value of A and B, choose correct option from column II for each item in column I. Column I Column II i)

sin (A – B) =

ii) tan (A + B) =

1 1 , cos (A + B) = 2 2 3 , tan (A – B) =

iii) sin (A + B) = 1, cos (A – B) =

3 2

A) A = 45°, B = 30° 1 3

B) A = 60°, B = 30° C) A = 45°, B = 15° D) A = 30°, B = 15°

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

119

120

4) Choose correct option in column II for each item in column I. Column I Column II i) sin x = cos 60° cos 30° – sin 60° sin 30°, then x = A) 45° ii) cot 3x = sin 45° cos 45° + sin 30°, then x = B) 60° iii) cos 2x = sin 60° cos 30° – cos 60° sin 30°, then x = C) 0° iv) tan 2x =

2 sin 30 ° cos 30 ° cos 2 30 ° − cos 2 60 °

, then x =

D) 15°

E) 30° 5) For each item in column I choose correct option in column II. Column I i) sin 3A = cos (A – 26°) where 3A is acute angle, then A = ii) tan 2A = cot (A – 18°) where 2A is acute angle, then A = 0 iii) sec 4A = cosec (A – 20°) where 4A is acute angle, then A = 0 iv) cot 4A = tan (A – 20°) where 4A is acute angle, then A = 0 6) For each item in column I choose correct option in column II. Column I Column II i) If sin A + sin B + sin C = 3, then A) m∠A = 90° B) sin B > sin C ii) In ΔABC, ∠B = 90°, then C) sin2 A + sin2 C = 1 iii) In ΔABC, ∠B > ∠C > ∠A

Column II A) 22° B) 29° C) 27° D) 36°

D) ∠B = ∠C E) cosec B + cosec C > 2 7) For each item in column I choose correct option in column II. Column I Column II i) cos 3° × cos 6° × cos 9° × … × cos 90° = A) 45° 2 2 B) 0 ii) If sec θ = cosec θ, θ = iii) cot 3° × cot 6° × … × cot 90° =

C) 180° – 3θ D) sin2 20° – cosec2 20° + sin2 70° + tan2 70 E) 90° 8) For each item in column I choose correct option in column II. Column I Column II i) Maximum value of sin A + cos A is A) 2 ii) Minimum value of cosec A + sec B is

B) 2 C) 0 D) 3 E) 1

iii) In a right triangle, sin2 A + sin2 B + sin2 C =

ANSWERS TO UNSOLVED EXERCISES: CW Exercise 8.1: 1) a)

15 8 15 17 17 3 3 4 5 5 , , , , (b) , , , , (c) 5 4 3 4 3 8 15 17 17 8

2) i) 1 (ii) 4) 5/12 120

544 225

1 10

,

3 10

,

1 , 3, 3

10 3

3) i) 12/13, 5/13 (ii) 5/13, 12/13 (iii) 12/5, 5/12 5) 1

7) i) 5 (ii) –1/17

Universal Tutorials – X CBSE (2012–13) – Mathematics

8) 12/13, 5/13, 12/5 Volume

Chapter 08: Introduction of Trigonometry

a

10) i)

2

4 b − 3a

2

(ii)

2 b2 − a2 2

4b − 3a

2

(iii)

121

a 2

2 b − a2

(iv) 1

HW Exercise 8.1: 1)

7 3 , 4 7

2)

5) i) 7/12 (ii) 31/25

15 17 , 17 8

7) 1/2

10) i) 1 (ii) 0

11) 2

3)

5 12 5 12 13 , , , , 13 13 12 5 5

49 49 ii) 64 64 12 5 12 , , 12) 13 13 5

8) i)

4) 3 + 2√2 9) yes 14) i)F ii)T iii)F iv)F v) F

CW Exercise 8.2: 1) i)

3 −1 2 2

(ii)

7 2 + 1− 2 3 (iii) (iv) 9 (v) 1 (vi) 0 6 2

2) i) 10° (ii) 15° (iii) 60° 3 −1

3 −1

3) 1

4) 60°

6) i)

8) X = 30°, Y = 15°

9) 5 3 cm, 10 cm

10) 60°, 30°

13) 20 3

14) 20 cm and 20 3 cm

12) (i) 15° (ii) 30° (iii) 15°

2 2

(ii)

2 2

7) 2 11) 45°, 15°

15) α = 37½°, β = 45°, γ = 52½° HW Exercise 8.2: 1) (i)

3 +1 2 2

(ii)

1− 3 2 2

(iii)

73 67 3 2− 6 43 − 24 3 (iv) 1 (v) 2 (vi) (vii) 2) 12 12 8 11 − 1+ 3

4) i) 3 (ii) 15°

6) x = 2, y = 3

9) (i) a (ii) d (iii) a (iv) c

10) (i) 60° (ii) 60° (iii) 30° (iv) 30°

12) 105°

13) 6 3 cm

7)

1+ 3

8) 45°, 15° 11) 60°

14) A = 60°, B = 52½°, C = 67½°

15) 45° CW Exercise 8.3: 1) i) cot 22° + cosec 21° (ii) tan 5° + sin 15° 2) (i)

9 1 (ii) 0 (iii) (iv) 2 (v) 2 (vi) 2 (vii) 0 (viii) 1 (ix) 2 (x) 0 4 2

3) 1

4)

3

5) 2 +

3

10) 29° 11) 22° HW Exercise 8.3: 9 1 1) (i) 2 (ii) (iii) 0 (iv) (v) 2 (vi) 0 (vii) 0 (viii) 0 (ix) 1 (x) –2 (xi) 2 5 2 9 8 4) (i) 2 – 1 (ii) 3 (iv) (v) 2 + 1 (vi) 5) 36° 4 7 8) i) x = 0 (ii) x = 3 9) 2 CW Exercise 8.4: 1) sin A= Volume

6) i) 0 (ii) 2

3) cos 23° + sin 15° 7) 1

1 1 sec A sec A − 1 cos A = tan A= sec 2 A − 1 cot A = cosec A= 2 sec A sec A sec A − 1 sec 2 A − 1 Universal Tutorials – X CBSE (2012–13) – Mathematics

121

122

2) i) 1 (ii) 1 3) c 7) (i, ii) No (iii) yes HW Exercise 8.4: 1) 1 2) 1 6) (i, ii) Yes (iii) No Miscellaneous Exercise: 8 15 15 8 17 17 , tan θ = , sin θ = , cos θ = , cosec θ = , sec θ = 1) cot θ = 15 8 17 17 15 8 2) (i) 2 (ii) 7) (i)

2 −13 23 −37 (iii) (iv) (v) 3 3 6 8 3 +1

3 (ii)

3) (i)

b2 1 1 (ii) (iii) 2 6 3 a + b2

4) 4

8) (i) 30° (ii) 60° (iii) 45°

2 2

3 1 (v) 1 (vi) 2 (vii) 2 (viii) 2 (ix) 2 (x) 0 (xi) 1 (xii) –1 (xiii) 1 (xiv) ½ (xv) 8 3 1 17 77 16 (iv) (v) (xvi) 1 (xvii) 2 (xviii) 2 10) (ii) (iii) 3 2 8 3 14) (i) Yes (ii, iii) No 13) (a) (i) θ = 45° or θ = 30° (ii) θ = 60°

9) (i) –1 (ii) 0 (iii) 1 (iv)

15) (i) 45° (ii) 30° (iii) 45° (iv) 60° 26) 1 34)

28) 2 1 1 + cot 2 A

29) 2

32) 1

1 3

(ii)

b+a b−a

33) 44°

2

,

1 + cot A 1 , cot A cot A

Column Matching Question: 1) i–AE, ii–BAC, iii–C, iv–DE 3) i–C, ii–C, iii–B 5) i–B, ii–D, iii–A, iv–A 7) i–BD, ii–AC, iii–BD

122

17) (i)

2) i–B, ii–D, iii–A 4) i–C, ii–D, iii–E, iv–E 6) i–AD, ii–CE, iii–BE 8) i–B, ii–A, iii–A

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

123

Chapter 14: Statistics Chapter Map: → Data → Ungrouped → Grouped → Mean → Direct → Assumed → Step Deviation → Mode → Median → Graphical Representation of cumulative frequency distribution

Calculation of Central tendencies for grouped data 14.1 Mean of grouped data:  Mean of observations is the sum of observations divided by the total number of observations.

14.1A To find mean by Direct Method: z z

If x1, x2,…….., xn are observations with respective frequencies f1, f2, fn, then this means observation x1 occurs f1 times, x2 occurs f2 times, and so on. Now, the sum of the values of all the observations = f1 x1 + f2 x2 + ……… + fnxn, and the number of observations = f1 + f2 + ……. + fn. f1x1 + f2 x2 + .... + fn xn f1 + f2 + ... + fn

z

So, the mean = x =

z

Recall that we can write this in short form by using the Greek letter ∑ (capital sigma) which n

means summation. That is, x =

∑ fi xi i =1 n

∑ fi i =1

, it is understood that i varies from 1 to n.

Which, more briefly, is written as x =

z

In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean. Now, for each class–interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each class–interval is centred around its mid–point. So, the mid–point (or class mark) of each class can be chosen to represent

z

Volume

∑ fi xi ∑ fi

z

Universal Tutorials – X CBSE (2012–13) – Mathematics

123

124

the observations falling in the class. Recall that we find the mid–point of the class (or its class mark) by finding the average of its upper and lower limits. That is, Upper class lim it + Lower Class lim it 2 These class marks serve as our xi’s. Now, in general, for the ith class interval, we have the frequency fi corresponding to the class mark xi.

Class mark =

z

SOLVED EXAMPLES 14.1: 1) The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students. Marks obtained(xi) Number of students (fi)

10 1

20 1

36 3

40 4

50 3

56 2

60 4

70 4

72 1

80 1

88 2

92 3

95 1

Sol: Recall that to find the mean marks, we require the product of each xi with The corresponding frequency fi. So, let us put them in a column as shown in Table. Number of Marks fixi students (fi) obtained (xi) 10 1 10 20 1 20 36 3 108 40 4 160 50 3 150 56 2 112 60 4 240 70 4 280 72 1 72 80 1 80 88 2 176 92 3 276 95 3 276 95 1 95 Total Σfi = 30 Σfixi = 1779 ∑ fi xi 1779 = = 59.3 Therefore, the mean marks obtained is 59.3. x = ∑ fi 30 2) Calculate the mean daily earning of a drug store for the above data and find the value of Σfi(xi – x ). Daily earnings of 30 drug stores are given below Daily earnings No. of stores 0 – 50 3 50 –100 7 100 – 150 4 150 – 200 5 200 – 250 4 250 – 300 3 300 – 350 2 350 – 400 2 Total 30

124

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

125

Sol: Sr. 1

Daily Earnings 0 – 50

Mid value (xI) 25

Frequency (fi) 3

25 × 3 = 75

2

50 – 100

75

7

75 × 7 = 525

3

100 – 150

125

4

125 × 4 = 500

4

150 – 200

175

5

175 × 5 = 875

5

200 – 250

225

4

225 × 4 = 900

6

250 – 300

275

3

275 × 3 = 825

7

300 – 350

325

2

325 × 2 = 650

8

350 – 400

375

2

375 × 2 = 750

Σfi = 30

Σfixi = 5100

xi f i

n

_

∑ fi x i

Mean = x = i = 1 ∑ fi

=

5100 = 170 30

∴ Mean daily earning of a drug store = Rs.170. fi (xi – x ) = {3 (25 –170) + 7(75 – 170) + 4(125 – 170) + 5(175 – 170) + 4(225 – 170) + 3 (275 – 170) + 2(325 – 170) + 2(375 – 170)} = – 435 – 665 – 180 + 25 + 220 + 315 + 310 + 410 = – 1280 + 1280 = 0 ∴ Σfi(xi – x ) = 0 3) If the mean of the following data is 13.5 find p. 5 10 20 25 x p 10 10 10 2 8 f Sol: Sr. No. fi xi 1 5 10 2 10 10 3 10 p 4 20 2 5 25 8 Total ∑fi = 40

[CBSE 1992]

xifi 5 × 10 = 50 10 × 10= 100 p × 10= 10p 20 × 2 = 40 25 × 8 = 200 ∑fixi = 10p + 390

n

Mean x =

∑ fi x i

i =1

∑ fi

⇒ 13.5 =

10 p + 390 40

13.5 × 40 = 10p + 390 540 – 390 = 10p ∴150 = 10p 150 = 15.0 = 15 p= 10

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

125

126

UNSOLVED EXERCISE 14.1: CW Exercise: 1) The following table gives the literacy rate in percentage) of 35 cities. Find the mean literacy rate. 45 – 55 55– 65 65–75 75–85 85–95 Literacy rate (in %) 3 10 11 8 3 Number of cities 2) Find the mean of the following distribution.

3)

4)

5)

6)

Weekly wage in (Rs.) 12.50–17.50 17.50–22.50 22.50–27.50 27.50–32.50 32.50–37.50 2 22 19 14 3 Number of workers The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f. 11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25 Daily pocket allowance (in Rs.) 7 6 9 13 5 4 Number of children f Find the mean of the following distributions: [CBSE 1992] 0–30 30–60 60–90 90–120 120–150 150–180 Class interval 7 6 9 13 10 5 Number of children If the mean of the following data is 21, find the value p. 10 15 20 25 35 x 6 10 10 8 p f Find the missing frequencies in the following frequency distribution if it is known that the mean of the distributions is 1.46. 0 1 2 3 4 5 Total No. of accident (x) 46 ? ? 25 10 5 200 Frequency (f)

HW Exercise: 1) A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. 0–2 2–4 4–6 6–8 8 – 10 10 –12 12 – 14 Number of plants 1 2 1 5 6 2 3 Number of houses 2) If the mean of the following data is 25, find the value k. [CBSE 2001 OD] 5 15 25 35 45 x 3 k 3 6 2 f 3) If the mean of the following data is 21.5, find the value k. [CBSE 2001] 5 15 25 35 45 x 6 4 3 2 k f 4) The table below gives the distribution of villages under different heights from sea level in a certain region. Compute the mean height of the region. 200 600 1000 1400 1800 2200 Heights (in metres) 142 265 560 271 89 16 No. of villages 5) Determine the mean of the following distribution. 10–16 16–22 22–28 28–34 34–60 Group 1 10 5 3 6 Frequency 126

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

127

6) Calculate the mean of the following distribution. 0–5 5–10 10–15 15–20 20–25 25–30 30–35 35–40 Total Cl. interval 5 8 15 21 25 15 8 6 103 Frequency 7) Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50. 10 30 50 70 90 Total x 17 32 f2 19 120 f1 f

To find mean by Assumed mean method:  Sometimes when the numerical values of xi and fi are large, finding the product of xi and fi becomes tedious and time consuming. So, for such situations, let us think of a method of reducing these calculations.  The first step is to choose one among the xi’s as the assumed mean, and denote it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that xi which lies in the centre of x1, x2, … xn.  The next step is to find the difference di between a and each of the xi’s, that is, the deviation of ‘a’ from each of the xi’s. i.e. di = xi – a.  The third step is to find the product of di with the corresponding fi and take the sum of all the Σf d fidi’s the mean of the deviations, d = i i . Σfi  ∴ Mean of deviation, d =

Note: d =

Σfi ( x i − a ) Σfi

So x = a + d

Σfi d i . Σfi

Σfi x i Σfi a − Σfi Σfi

i.e. x = a +

x –a

Σfi = x–a Σfi

Σfi d i Σfi

To determine mean by step deviation method:  Determine the class size (h the common multiple of di = xi – a, where a is the assumed mean) x −a  Calculate ui for each class interval i.e. ui = i h Σfi ui  Calculate fiui, then u = Σfi

⎛ Σf u  Now x = a + h u . i.e. x = a + h ⎜⎜ i i ⎝ Σfi

⎞ ⎟⎟ ⎠

We note that:

1) The step-deviation method will be convenient to apply if all the di’s have a common factor. 2) The mean obtained by all the three methods is the same. 3) The assumed mean method and step-deviation method are just simplified forms of the direct method. 4) The formula x = a + h u still holds if a and h are not as given above, but are any non-zero x −a numbers such that ui = i . h Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

127

128

SOLVED EXAMPLES 14.2: 1) Find the mean by assumed mean method 10-25 25-40 40-55 55-70 70-85 85-100 Class interval 2 3 7 6 6 6 No. of students Sol: Class interval No. of students (fi) Class mark (xi) di = xi – 47.5 fidi 10-25 2 17.5 –30 –60 25-40 3 32.5 –15 –45 40-55 7 47.5 0 0 55-70 6 62.5 15 90 70-85 6 77.5 30 180 85-100 6 92.5 45 270 Total ∑fi = 30 ∑fidi = 435 Substituting the values of a, Σfidi and Σfi from Table we get 435 = 47.5 + 14.5 = 62. x = 47.5 + 30 Therefore, the mean of the marks obtained by the students is 62. 2) The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section. 15–25 25–35 35–45 45–55 55–65 65–75 75–85 % of female teacher 6 11 7 4 4 2 1 Number of States/ U.T. Sol: Find the class marks, xi, of each class, and put them in a column x − 50 Take a = 50, h = 10, then di = xi – 50 and ui = i 10 % of female teacher

No. of states/ U.T (fi)

xi

di = xi – 50

ui =

x i − 50 10 –3 –2 –1 0 1 2 3

fixi

15 – 256 20 –30 120 25 – 33 11 30 –20 330 35 – 45 7 40 –10 280 45 – 55 4 50 0 200 55 – 65 4 60 10 240 65 – 75 2 70 20 140 75 – 85 1 80 30 80 Total 35 1390 From the table above, we obtain ∑fi = 35, ∑fixi = 1390. ∑fidi = – 360, ∑fiui = – 36 ∑ fi x i 1390 = = 39.71 Using the direct method, x = ∑ fi 35 Using the assumed mean method, x = a +

fidi

fiui

–180 –220 –70 0 40 40 30 –360

–18 –22 –7 0 4 4 3 –36

(− 360 ) = 39.71 ∑ fi d i = 50 + ∑ fi 35

⎛ ∑f u Using the step–deviation method, x = a + ⎜⎜ i i ⎝ ∑ fi

⎞ ⎛ − 36 ⎞ ⎟⎟ × h = 50 + ⎜ ⎟ × 10 = 39.71 ⎝ 35 ⎠ ⎠

Therefore, mean percentage of female teachers in the primary schools of rural areas is 39.71. 128

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

129

3) The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify? 20–60 60–100 100–150 150–250 250–350 350–450 No. of wickets 7 5 16 12 2 3 No. of bowlers Sol: The class size varies and the xi’s are large apply the step deviation method with a =200 & h = 20 No. of wickets taken

No. of bowlers (fi)

xi

di = xi – 200

7 5 16 12 2 3 45

40 80 125 200 300 400

–160 –120 –75 0 100 200

20 – 60 60 – 100 100 – 150 150 – 250 250 – 350 350 – 450 Total

di 20 –8 –6 –3.75 0 5 10

ui =

fi ui

–56 –30 –60 0 10 30 –106

Σfi u i −106 ⎛ − 106 ⎞ = . Therefore, x = a + h u = x = 200 + 20 ⎜ ⎟ = 200 – 47.11 = 152.89. Σf i 45 ⎝ 45 ⎠ i.e. on an average the number of wickets taken by these 45 bowlers in one–day cricket is 152.89 4) The frequency distribution of the lifetimes of 400 T.V. picture tubes tested in a Tube Company is given below. Find the average life time of a tube. Lifetime (in hours) No. of tubes 300–399 14 400–499 46 500–599 58 600–699 76 700–799 68 800–899 62 900–999 48 1000–1099 22 1100–1199 6 Total 400 Sol:

So, u =

C.I.

xi

fi

300–399 349.5 14 400 – 499 449.5 46 500 – 599 549.5 58 600–699 649.5 76 700–799 a = 749.5 68 800–899 849.5 62 900–999 949.5 48 1000–1099 1049.5 22 1100–1199 1149.5 6 Total ∑fi = 400 Let the assumed mean ‘a’ = 749.5 Volume

di = xi – a

ui =

– 400 – 300 –200 –100 0 100 200 300 400

–4 –3 –2 –1 0 1 2 3 4

xi − a h

fiui

– 56 – 138 –116 –76 0 62 96 66 24 ∑fiui = –138

Here h = 100

Universal Tutorials – X CBSE (2012–13) – Mathematics

129

130

We know that, u =

Σf i u i = Σf i

138 400

⎛ − 138 ⎞ ⎟ × 100 = 749.5 – 34.5 = 715. ⎝ 400 ⎠

x = a + h = 749.5 + ⎜

Hence, mean = 715 hours.

UNSOLVED EXERCISE 14.2: CW Exercise: 1) The following is the frequency distribution of the number of teachers in Higher Secondary School, in 1978 in India. Find the average number of teachers per Higher Secondary School in India for 1978 by using step deviation method. No. of teachers No. of H.S. School 6 – 10 955 11 – 15 1067 16 – 20 1663 21 – 25 1492 26 – 30 1220 31 – 35 1129 36 – 40 745 41 – 45 637 46 – 50 442 2) Consider the following distribution of daily wages of 50 workers of a factory. 100–120 120–140 140–160 160–180 180–200 Daily wages (in Rs.) 12 14 8 6 10 Number of workers Find the mean daily wages of the workers of the factory by using assumed mean method. 3) Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, by step deviation method. 65–68 68–71 71–74 74–77 77–80 80–83 83–86 No. of heart beats per minute Number of women

2

4

3

8

7

4

2

4) In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. No. of mangoes No. of boxes

50–52

53–55

56–58

59–61

62–64

15

110

135

115

25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? 5) The table below shows the daily expenditure of food of 25 households in a locality. Daily expenditure (in Rs.) No. of households

100–150

150–200

200–250

250–300

300–350

4

5

12

2

2

Find the mean daily expenditure on food by a suitable method.

130

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

131

6) To find out the concentration of SO2 in the air (in parts per million i.e. ppm), the data was collected for 30 localities in a certain city and is presented below: Concentration of SO2(in ppm) Frequency 0.00 – 0.04 4 0.04 – 0.08 9 0.08 – 0.12 9 0.12 – 0.16 2 0.16 – 0.20 4 0.20 – 0.24 2 Find the mean concentration of SO2 in the air 7) A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. 0–6 6–10 10–14 14–20 20–28 28–38 38–40 No. of days 11 10 7 4 4 3 1 No. of students 8) If the mean of the following data is 12, find the value p. 4 8 P 16 20 x 5 3 12 5 4 f HW Exercise: 1) Find the missing frequency k from the following data if the AM is 16. 5 10 15 20 25 x 2 8 10 5 k y 2) Find the mean by step deviation method 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Variety 12 15 16 19 12 6 Frequency 3) The frequency distribution of the life times of 400 TV picture tubes tested in a Tube company is given below. Find the average lifetime of a tube. Life time Tubes

300-399 14

400-499 46

500-599 58

600-699 76

700-799 68

800-899 62

900-999 48

1000-1099 22

1100-1199 6

4) The following table shows the marks of 120 students at ICSE examination in Maths. Calculate the mean mark by assumed mean method. 30 – 39 40 – 49 50 – 59 60 – 69 70 –79 80 – 89 90 – 99 Marks % 1 3 11 21 43 32 9 Students 5) Use the short cut method to calculate the mean of the following data. (step deviation method) 0-20 20-40 40-60 60-80 80-100 100-120 Class interval 12 18 23 17 9 11 Frequency 6) The ages of workers in a company are as follows. Calculate the average age of the group. Age (in years) No of workers 0-10 2 10-20 6 20-30 9 30-40 7 40-50 4 50-60 2

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

131

132

7) Find the mean marks of students from the following cumulative frequency table: Marks No. of students 0 and above 80 10 and above 77 20 and above 72 30 and above 65 40 and above 55 50 and above 43 60 and above 28 70 and above 16 80 and above 10 90 and above 8 100 and above 0 8) The following table shows marks secured by 140 students in an examination: 0–10 10–20 20–30 30–40 40–50 Marks 20 24 40 36 20 No. of students Calculate mean marks by using all the three methods, i.e. direct method, assumed mean method and step deviation method.

14.2 Mode of grouped data: Definition:

z

Mode is that value among the observations which occurs most often i.e. the value of the observation having the maximum frequency. In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is the value inside the modal class.

z

⎛ f1 − f0 In a grouped frequency mode = l + ⎜⎜ ⎝ 2f1 − f0 − f2

z

⎞ ⎟⎟ × h ⎠

Where l = lower limit of the modal class; h = Size of the class interval (assuming all class sizes to be equal) f1 = frequency of the modal class f0 = frequency of the class preceding the modal class f2 = frequency of the class succeeding the modal class.

SOLVED EXAMPLES 14.3: 1) A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household: 1–3 3–5 5–7 7–9 9 – 11 Family size 7 8 2 2 1 No. of families Find the mode of this data. Sol: Here the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5. So, the modal class is 3 – 5. Now, Modal class = 3 – 5, lower limit (l) of modal class = 3, class size (h) = 2 Frequency (f1) of the modal class = 8, Frequency (f0) of class preceding the modal class = 2 132

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

133

Now, let us substituted these values in the formula: ⎛ f1 − f0 ⎞ 2 8−7 ⎞ ⎛ ⎟⎟ × h = 3 + ⎜ Mode = l + ⎜⎜ ⎟ × 2 = 3 + = 3.286 × − − 7 2 8 7 2 − − 2 f f f ⎠ ⎝ ⎝ 1 0 2⎠ Therefore, the mode of the data above is 3.286 2) The marks distribution of 30 students in a mathematics examination is given in solved example 14.1 example 1. Find the mode of this data. Also compare and interpret the mode and the mean. Sol: Refer to table of example 1 since the maximum number of students i.e. 7 have got marks in the interval 40 – 55, the modal class is 40 – 55. Therefore, the lower limit (l) of the modal class = 40, the class size (h) = 15. The frequency (f1) of modal class = 7. The frequency (fu) of the class preceding the modal class = 3, the frequency (f2) of the class succeeding the modal class = 6. ⎛ f1 − f0 ⎞ ⎛ 7−3 ⎞ ⎟⎟ × h , ⇒ 40 + ⎜ Now, using the formula: Mode = l + ⎜⎜ ⎟ × 15 = 52 2 f f f − − ⎝ 14 − 6 − 3 ⎠ ⎝ 1 0 2⎠

So, the mode marks is 52. Now, from example 1, you know that the mean marks is 62. So, the maximum number of students obtained 52 marks, while on an average a student obtained 62 marks.

UNSOLVED EXERCISE 14.3: CW Exercise: 1) The following table shows the ages of the patients admitted in a hospital during a year: 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65 Age (in ears) 6 11 21 23 14 5 No. of patients Find the mode and mean of the data given above. Compare and interpret the two measures of central tendency. 2) The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure: Expenditure (in Rs.) No. of families 1000 – 1500 24 1500 – 2000 40 2000 – 2500 33 2500 – 3000 28 3000 – 3500 30 3500 – 4000 22 4000 – 4500 16 4500 – 5000 7 3) The given distribution shows the number of runs scored by some top batsmen of the world in one–day international cricket matches. Runs scored No. of batsmen 3000 – 4000 4 4000 – 5000 18 5000 – 6000 9 6000 – 7000 7 7000 – 8000 6 8000 – 9000 3 9000 – 10000 1 10000 – 11000 1 Find the mode of the data.

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

133

134

HW Exercise: 1) The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 Lifetimes (in hours) 10 35 52 61 38 29 Frequency Determine the modal lifetimes of the components. 2) The following distribution gives the state–wise teacher–student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. No. of students per teacher No. of states/U.T. 15 – 20 3 20 – 25 8 25 – 30 9 30 – 35 10 35 – 40 3 40 – 45 0 45 – 50 0 50 – 55 2 3) A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data: 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 No. of cars 7 14 13 12 20 11 15 8 Frequency

14.3 Median of Grouped Data: ¾ Definition: Median is a measure of central tendency which gives the value of the middle–most observation in the data. ¾ First we arrange the data values of the observation in ascending order. If n is odd, the median is th

th

n ⎛ n + 1⎞ ⎛ n + 1⎞ the ⎜ ⎟ observation. If n is even then the median will be average of the th and ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ 2 ⎠ observation. ¾ Now in a grouped data, we may not be able to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves.

¾ To find this class, we find the cumulative frequencies of all the classes and

class whose cumulative frequency is greater than (and nearest to)

n . We now locate the 2

n . This is called the median 2

class.  After finding the median class, we use the following formula for calculating the median.

⎛ n − cf ⎞ ⎟⎟ × h  Median = l + ⎜⎜ 2 ⎝ f ⎠ Where, l = Lower Limit of median Class; n = number of observation; cf = cumulative frequency of class preceding the median class; f = frequency of median class; h = class size (assuming class size to be equal) 134

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

135

SOLVED EXAMPLES 14.4: 1) The marks obtained out of 50 by 102 students in a test were according to the frequency table given below: 20 22 23 24 26 31 38 43 Marks 8 15 28 27 20 2 1 1 Frequency Obtain the median and describe what information it conveys Sol: Arranging the terms in ascending order and preparing the cumulative frequency table, we have: Marks obtained Frequency Cumulative Frequency 20 8 8 22 15 23 (23) 28 (51) (24) 27 (78) 26 20 98 31 2 100 38 1 101 43 1 102

∴ Total number of terms, n = 102. So,

n ⎛n ⎞ = 51 ⎜ + 1⎟ = 52 2 ⎝2 ⎠

∴ Median =

Value of 51st term + value of 52in term 2

⎛ 23 + 24 ⎞ 47 = 23.5 = ⎜ ⎟= 2 ⎝ 2 ⎠ [∴ 51st term lies in the row in which c.f. is 51 and 52nd term lies in the row in which c.f. is 78] The conclusion that median is 23.5 conveys that about 50% of the students obtained less than 23.5 marks out of 50 in the test 2) The marks obtained by 60 students in a certain paper out of 75 are given below: Marks No. of Students 15 – 20 4 20 – 25 5 25 – 30 11 30 – 35 6 35 – 40 5 40 – 45 8 45 – 50 9 50 – 55 6 55 – 60 4 60 – 65 2 Total 60 Calculate the median Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

135

136

Sol: Marks 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40(M.C) 40 – 45 45 – 50 50 – 55 55 – 60 60 – 65 Total

No. of. students 4 5 11 6 5 8 9 6 4 2 60

Cumulative Frequency 4 9 20 26 31 39 48 48 54 60

n = 30 2 Now, 30 lies in the row with cumulative frequency 31. [∴ 26, 30, 31] So, the median class is 35 – 40 ∴ l = 35, h = (40 – 35) = 5, n = 60, f1 = 5 and f0 = 26

Total number of observation, n = 60

∴

h⎛n 5 ⎛ 60 ⎞ ⎞ − 26 ⎟ = 39 ⎜ − f0 ⎟ = 35 + ⎜ f1 ⎝ 2 5 ⎝ 2 ⎠ ⎠ 3) The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Class interval Frequency 0 – 100 2 100 – 200 5 200 – 300 x 300 – 400 12 400 – 500 17 500 – 600 20 600 – 700 y 700 – 800 9 800 – 900 7 900 – 1000 4 Sol: Class intervals Frequency Cumulative frequency 0 – 100 2 2 100 – 200 5 5 7 200 – 300 x 7+x 300 – 400 12 19 + x 400 – 500 17 36 + x 500 – 600 20 56 + x 600 – 700 56 + x + y y 700 – 800 9 65 + x + y 800 – 900 7 72 + x + y 900 – 1000 4 76 + x + y Hence, Median = l +

136

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

137

It is given that n = 100 So, 76 + x + y = 100, i.e. x + y = 24 The median is 525, which lies in the class 500 – 600 So, l = 500, f = 20, cf = 36 + x, h = 100

⎛ n / 2 − cf ⎞ Using the formula: Median = l + ⎜ ⎟ h, we get f ⎝ ⎠ ⎛ 50 − 36 − x ⎞ 525 = 500 + ⎜ ⎟ × 100 20 ⎝ ⎠ i.e. 525 – 500 = (14 – x) × 5 i.e. 25 = 70 – 5x ⇒ 5x = 70 – 25 = 45 ⇒ x = 9 Therefore, from (1) we get 9 + y = 24 y = 15

UNSOLVED EXERCISE 14.4: CW Exercise: 1) The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table: Length (in mm) No. of leaves 118 – 126 3 127 – 135 5 136 – 144 9 145 – 153 12 154 – 162 5 163 – 171 4 172 – 180 2 Find the median length of the leaves 2) The following table gives the frequency distribution of married women by age at marriage: Age (in years) Frequency 15 – 19 53 20 – 24 140 25 – 29 98 30 – 34 32 35 – 39 12 40 – 44 9 45 – 49 5 50 – 54 3 55 – 59 3 60 and above 2 Calculate the median and interpret the result. 3) If the median of the distribution given below is 28.5, find the values of x and y. 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Class interval Total 5 x 20 15 y 5 Frequency 60 4) A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

137

138

Age(in years)

No. of policy holders

Below 20

2

Below 25

6

Below 30

24

Below 35

45

Below 40

78

Below 45

89

Below 50

92

Below 55

98

Below 60

100

HW Exercise: 1) The following table gives the distribution of the life time of 400 neon lamps: Life time (in hours) No. of lamps 1500 – 2500 14 2000 – 2500 56 2500 – 3000 60 3000 – 3500 86 3500 – 4000 74 4000 – 4500 62 4500 – 5000 48 Find the median life time of a lamp 2) 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: 1–4 4–7 7 – 10 10 – 13 13 – 16 16 – 19 No. of letters 6 30 40 16 4 4 No. of surnames Determine the median number of letters in the surname. Find the mean number of letters in the surnames? Also, find the modal size of the surnames. 3) The distribution below gives the weights of 30 students of a class. Find the median weight of the students 40 – 45 45 – 50 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 Weight (in kg) 2 3 8 6 6 3 2 No. of students 4) Following is the distribution of rents in a certain city for a one room set: Rent (in rupees) No. of one room sets 150–175 10 175–200 13 200–225 17 225–250 15 250–275 16 275–300 10 300–325 7 325–350 5 350–375 4 375–400 3 Total 100 Compute the median rent 138

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

139

5) Calculate the mean and median of: Variable Frequency Variable Frequency 0–5 6 25 – 30 250 5 – 10 12 30 – 35 185 10 – 15 50 35 – 40 110 15 – 20 220 40 – 45 32 20 – 25 125 45 – 50 10 6) Find the missing frequencies in the following distribution, if n = 100 and median is 30: 0–10 10–20 20–30 30–40 40–50 50–60 Marks 10 ? 25 30 ? 10 No. of Students

Comparative Study:  Now, that you have studied about all the three measures of central tendency, let us discuss which measure would be best suited for a particular requirement.  The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes, i.e. the largest and the smallest observations of the entire data. It also enables us to compare two or more distributions.  For example, by comparing the average (mean) results of students of different schools of a particular examination, we can conclude which school has a better performance.  However, extreme values in the data affect the mean. For example, the mean of classes having frequencies more or less the same is a good representative of the data.  But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20, 21, 18, then the mean will certainly not reflect the way the data behaves. So, in such cases, the mean is not a good representative of the data.  In problems where individual observations are not important, and we wish to find out a ‘typical’ observation, the median is more appropriate, e.g., finding the typical productivity rate of workers, average wage in a country, etc. These are situations where extreme values may be there. So, rather than the mean, we take the median as a better measure of central tendency.  In situations which require establishing the most frequent value or most popular item, the mode is the best choice, e.g., to find the most popular T.V. programme being watched, the consumer item in greatest demand, the colour of the vehicle used by most of the people, etc.

Remarks: There is a empirical relationship between the three measures of central tendency: 3 Median = Mode + 2 Mean

Graphical Representation of Cumulative frequency Distribution: ¾ Graphical representation of cumulative frequency distribution is known as ogive. Ogives are: i) Less than type ii) More than type

Ogive of Less than type:  First prepare a cumulative frequency table less than the upper limit.  To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis (y-axis), choosing a convenient scale.  The scale may not be the same on both the axis.  Let us now plot the points corresponding to the ordered pairs given by (upper limit, corresponding cumulative frequency), on a graph paper and join them by a free hand smooth curve. Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

139

140

 The curve we get is called a cumulative frequency curve, or an ogive of less than type.

Marks 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100

Number of students 5 3 4 3 3 4 7 9 7 8

Marks obtained

No. of students (Cumulative frequency) 5 5+3=8 8+4=12 12+3=15 15+3=18 18+4=22 22+7=29 29+9=38 38+7=45 45+8=53

Cumulative frequency →

Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 Less than 100 60 50

‘Less than’ ogive

40 30 20 10 10

20

30

40

50

60

70

80

90

100

Upper limits →

Ogive of more than type:  First prepare a cumulative frequency table more than the lower limit.  To represent ‘the more than type’ graphically, we plot the lower limits on the x-axis and the corresponding cumulative frequencies on the y-axis.  Then we plot the points (lower limit, corresponding cumulative frequency), on a graph paper, and join them by a free hand smooth curve. 140

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

141

 The curve we get is a cumulative frequency curve, or an ogive (of the more than Type)

Marks obtained

Number of students (Cumulative frequency) 53 53–5=48 48–3=45 45–4=41 41–3=38 38–3=35 35–4=31 31–7=24 24–9=15 15–7=8

Cumulative frequency →

More than or equal to 0 More than or equal to 10 More than or equal to 20 More than or equal to 30 More than or equal to 40 More than or equal to 50 More than or equal to 60 More than or equal to 70 More than or equal to 80 More than or equal to 90 60 50

‘More than’ ogive

40 30 20 10 10

20

30

40

50

60

70

80

90

100

Lower limits →

Note: The median of grouped data can be obtained graphically as the x–co–ordinate of the point of intersection of two ogives for this data.

SOLVED EXAMPLES 14.5: 1) The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution: Profit (in lakhs Rs.) No. of shops (frequency) More than or equal to 5 30 More than or equal to 10 28 More than or equal to 15 16 More than or equal to 20 14 More than or equal to 25 10 More than or equal to 30 7 More than or equal to 35 3 Draw both ogives for the data above. Hence obtain the median profit.

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

141

142

Cumulative frequency →

Sol Draw the coordinate axes, with lower limits of the profit along the horizontal axis, and the cumulative frequency along the vertical axes. We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3), we join these points with a smooth curve to get the ‘more Y than’ ogive as shown 50

40 30 20 10 X 10

20

30

40

50

Lower limits of profit (in lakhs Rs.) →

5 – 10 10–15 15–20 20–25 25–30 30–35 35–40 Classes 2 12 2 4 3 4 3 No of shops 2 14 16 20 23 27 30 Cumulative frequency Using these values, we plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27), (40, 30) on the same axes, as in fig to get the ‘less than’ ogive, Cumulative frequency →

Y 50 40

More than ogive

30 Less than ogive

20 10

X 10

20

30

40

50

Median (17.5) Profit (in lakhs Rs.) →

The abscissa their point of intersection is nearly 17.5, which is the median. This can also be verified by using the formula. Hence, the median profit (in lakhs) is Rs.17.5

UNSOLVED EXERCISE 14.5: CW Exercise: 1) During the medical check–up of 35 students of a class, their weights were recorded as follows: Weight (in kg) Number of students Less than 38 0 Less than 40 3 Less than 42 5 Less than 44 9 Less than 46 14 Less than 48 28 Less than 50 32 Less than 52 35 Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula. 142

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

143

2) Suppose we are given the following discrete series of marks Marks obtained Number of students 20–30 6 30–40 18 40–50 25 50–60 22 60–70 17 70–80 12 Total 100 From this table form the (i) ‘Less than’ and (ii) ‘More than’ series 3) From the following continuous series prepare the: i) Less than ii) More than cumulative series Marks obtained Number of students 0 – 10 7 10 – 20 11 20 – 30 9 30 – 40 25 40 – 50 8 4) Draw an ogive for the following Class Frequency 0–4 4 4–8 6 8 – 12 10 12 –16 8 16 – 20 4 HW Exercise: 1) The following distribution gives the daily income of 50 workers of a factory. 100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 Daily income (in Rs.) 12 14 8 6 10 No. of workers Convert the distribution above to a less than type cumulative frequency distribution and draw ogive. 2) The following table gives production yield per hectare of wheat of 100 farms of a village. 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 75 – 80 Production yield (in kg/ha) 2 8 12 24 38 16 No. of farms Change the distribution to a more than type distribution, and draw its ogive.

MISCELLANEOUS: 1) Find the mean of the following frequency distribution, using ‘direct method’. Marks 10-20 20-30 30-40 40-50 Students 5 8 12 10 5 2) Find the mean using ‘direct method’: Weight in kg 30–34 34-38 38–42 42–46 46-50 Students 6 8 12 9 5 3) Find the mean by ‘short-cut method’: Weight in kg 11–15 16-20 21-25 26-30 31-35 36-40 41-45 46-50 Students 8 12 13 16 12 9 8 2 Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

143

144

4) Find the mean, by ‘step-deviation method’: Marks

10-20

20-30

30-40

40-50

50-60

60-70

6

8

12

15

10

9

Students

5) Find the mean, by ‘step-deviation method’: Variate

10-20

20-30

30-40

40-50

50-60

60-70

12

15

16

19

12

6

Frequency

6) Compute the mean of the following frequency table by i) a ‘direct method’ and ii) a ‘short-cut method’ Class

5-10

10-15

15-20

20-25

25-30

30-35

35-40

40-45

45-50

10

6

4

12

8

4

2

1

3

Frequency

7) Find the mean for the following frequency distribution: Class

0-10

10-20

20-30

30-40

40-50

6

8

12

8

6

Frequency

8) Find the mean for the following distribution: Class

1-10

11-20

21-30

31-40

41-50

3

5

8

6

3

Frequency

9) Find the mean for the distribution: Class

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

6

12

18

20

25

32

24

18

6

1

Frequency

∑f = 162 10) Find the mean for the following distribution: Class

1-10

11-20

21-30

31-40

41-50

51-60

61-70

71-80

81-90

91-100

8

15

16

17

20

18

11

8

2

1

Frequency

11) The frequency distribution of marks obtained by 40 students is a under. Calculate the Arithmetic mean: Marks Students

0-8

8-16

16-24

24-32

32-40

40-48

5

3

10

16

4

2

12) Find the missing frequency for the following distribution, if the mean is 12.9. Class

0-5

5-10

10-15

15-20

20-25

3

F

8

5

4

Frequency

13) The mean of the following frequency table is 50. But f1 and f2 in classes 20–40 and 60–80 are missing. Find the missing frequencies. Class Frequency

0–20

20–40

40–60

60–80

80–100

Total

17

f1

32

f2

19

120

14) Find the value of P if the mean of the following distribution is 20. x

15

17

19

20+p

23

f

2

3

4

5p

6

15) Find the mode of the following data: Class interval

0–6

6–12

12–18

18–24

24–30

Frequency

6

8

10

9

7

Also compare and interpret the mode and the mean. 144

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Chapter 14: Statistics

145

16) The following table gives the distribution of total expenditure (in rupees) of 100 workers in a city: Expenditure (in Rupees) No. of Workers 100–150 12 150–200 20 200–250 16 250–300 14 300–350 15 350–400 11 400–450 8 450–500 4 Calculate the mode of the above data. 17) Calculate the mean, mode of the following data and interpret. Class Interval 0–20 20–40 40–60 60–80 80–100 100–120 Frequency 16 11 25 16 12 10 18) The data below gives the monthly earnings of 38 workers in a flour mill. Monthly Earnings (Rs) 500–520 520–540 540–560 560–580 580–600 600–620 No of workers 4 6 12 8 6 2 Calculate the mean monthly earnings of the group and the mode value. 19) Find the mode of the following: Class Interval 40–50 50–60 60–70 70–80 80–90 90–100 Frequency 10 25 28 12 10 15 20) The frequency distribution of scores obtained by 230 candidates in a medical entrance test is as follows: Scores No. of Candidates

400–450

450–500

500–550

550–600

600–650

650–700

700–750

750–800

Total

20

35

40

32

24

27

18

34

230

Draw a cumulative frequency curve (O give) to represent the above data. 21) The frequency distribution of marks obtained by 100 students. Marks 0–10 10–20 20–30 30–40 40–50 No.of students 7 10 23 51 6 Draw a cumulative frequency curve (o give) to represent the above data. 22) Find the mode for the following frequency distribution: Class 0–10 10–20 20–30 30–40 40–50 Frequency 6 8 12 8 6 23) Find the mode for the following frequency distribution: Class Frequency 0–10 6 10–20 12 20–30 18 30–40 20 40–50 25 50–60 32 60–70 24 70–80 18 80–90 6 90–100 1 Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

50–60 3

Total 100

145

146

24) Find the median for the following distribution by drawing an ogive. Class 1–10 11–20 21–30 31–40 41–50 Frequency 3 5 8 6 3 25) Find the median for the distribution and draw an ogive for the same. Calculate the value of median from the graph. Class 0–5 5–10 10–15 15–20 20–25 Frequency 1 3 5 4 2 26) Following table shows the marks of 120 students at ICSE examination in Maths. Draw an ogive for the table. Marks 30–39 40–49 50–59 60–69 70–79 80–89 90–99 No. of students 1 3 11 21 43 32 9 Estimate: (i) Median (ii) Lowest mark scored by top 25% of the class (iii) The highest mark obtained by the 20% of the class. 27) Find the median for the following distribution by drawing an ogive. Class 0–5 5–10 10–15 15–20 20–25 Frequency 3 5 8 5 4 28) For the following distribution construct the “less than” frequency table and draw the ogive. Class 0–2 3–5 6–8 9–11 12–14 Frequency 17 22 29 18 9 From the ogive estimate the median. 29) The following table gives the daily income of 50 workers of a factory: 100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 Daily income (in Rs.) 12 14 8 6 10 No. of workers Find the Mean, Mode and Median of the above data. [CBSE-09] 30) During the medical check-up of 35 students of a class their weights were recorded as follows: Weight (in kg) Number of students 38 - 40 3 40 - 42 2 42 - 44 4 44 - 46 5 46 – 48 14 48 – 50 4 50 - 52 3 Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph. [CBSE-09]

MULTIPLE CHOICE QUESTIONS: CW Exercise: 1) Which of the following is not a measure of central tendency? a) Mean b) Median c) Mode 2) The arithmetic mean of 1, 2, 3, ..., n is n +1 n −1 n a) b) c) 2 2 2 3) Which of the following cannot be determined graphically? a) Mean b) Median c) Mode 146

Universal Tutorials – X CBSE (2012–13) – Mathematics

d) Standard deviation d)

n +1 2

d) None of these Volume

Chapter 14: Statistics

147

4) The mode of a frequency distribution can be determined graphically from a) Histogram b) Frequency polygon c) Ogive d) Frequency curve 5) The mean of n observations is X . If the first item is increased by 1, second by 2 and so on, then the new mean is n n +1 a) X + n b) X + c) X + d) None of these 2 2 6) If mean = 75 and median = 60 then mode is: a) 105 b) 345 c) 30 d) 330 7) The mode for the following distribution is: 0–10 10–20 20–30 30–40 40–50 Class 6 8 12 8 6 Frequency a) 25 b) 24 c) 22 d) 28 HW Exercise: 1) The algebraic sum of the deviations of a frequency distribution from its mean is a) always positive b) always negative c) 0 d) a non-zero number 2) For a frequency distribution, mean, median and mode are connected by the relation a) Mode = 3 Mean – 2 Median b) Mode = 2 Median – 3 Mean c) Mode = 3 Median – 2 Mean d) Mode = 3 Median + 2 Mean 3) The median of a given frequency distribution is found graphically with the help of a) Histogram b) Frequency curve c) Frequency polygon d) Ogive 4) Mode is a) least frequent value b) middle most value c) most frequent value d) None of these 5) If the mean of the following distribution is 2.6, then the value of y is Variable (x): 1 2 3 4 5 Frequency: 4 5 y 1 2 a) 3 b) 8 c) 13 d) 24 6) Which measure of central tendency is given by the x–coordinate of the point of intersection of the ‘more than ogive’ and ‘less than ogive’? a) Mean b) Median c) Mode d) None of these 7) The median of the following data is: 0–5 5–10 10–15 15–20 20–25 Class 1 3 5 4 2 Frequency a) 15.5 b) 14.5 c) 16.5 d) 13.5

ANSWER TO THE UNSOLVED EXERCISE: CW Exercise 14.1: 1) 69.43 5) p = 26

2) 24.7 6) f1 = 76, f2 = 38

3) f = 20

4) 91.8

HW Exercise 14.1: 1) 8.1 plants 5) 25.72

2) K = 14 6) 20.27

3) K = 5 7) f1 = 28, f2 = 24

4) 984.5

CW Exercise 14.2: 1) 25.04 5) Rs.211

2) 145.20 6) 0.099 ppm

3) 75.9 7) 12.38 days

4) 57.19 8) 12

Volume

Universal Tutorials – X CBSE (2012–13) – Mathematics

147

148

HW Exercise 14.2: 1) K = 15 5) 73.67

2) 37.75 6) 28.67

3) 715 hrs 7) 51.7 marks

4) 74.5 8) 25.857

CW Exercise 14.3: 1) Mean = 35.37; Mode = 36.8. Both the measures are approx. same in this case 2) Modal monthly expenditure = Rs.1847.83, Mean monthly expenditure = Rs.2662.5 3) Mode = 4608.7 runs HW Exercise 14.3: 1) 62.625 hrs. 2) Mode = 30.6, Mean = 29.2. Most states / UT have a student teacher ratio of 30.6 and on an average this ratio is 29.2 3) Mode = 44.7 cars CW Exercise 14.4: 1) Median length = 146.75 mm 2) Median = 23.58 yrs. This means that the age of about 50% of the girls is less than this age and 50% are taller than this height 3) x = 8, y = 7 4) Median age = 35.76 yrs HW Exercise 14.4: 1) Median life = 3406.98 hrs. 3) Median weight = 56.67 kgs 5) Mean = 26.24, Median = 26.74

2) Median=8.05, Mean =8.32, Modal size =7.88 4) Rs.241.67 6) (i) 15 (ii) 10

Miscellaneous: 1) 25.5 2) 39.9 kg 5) 37.75 6) 22 9) 47.9 10) 41.28 13) 28 and 24 14) p = 1 17) Mean = 56, Mode = 52.1 19) Mode: 61.5 22) Mode : 25 24) Median = 26.6 25) Median = 14

3) 28.06 kg 4) 42 7) 25 8) 25.9 11) 23.4 12) 5 15) 16 16) 183.3 18) Mean: Rs.556.32; Mode: Rs.552 23) Mode = 54.7 26) i) 77.8 ii) 83 iii) 64 27) Median = 12.8

148

Universal Tutorials – X CBSE (2012–13) – Mathematics

Volume

Answers

149

Answer to the MCQs: Chapter 1: Real Numbers CW Exercise: 1) b 2) b 11) b HW Exercise: 1) a 2) d 11) b

3) c

4) d

5) c

6) a

7) b

8) b

9) b

10) b

3) c

4) b

5) b

6) c

7) d

8) a

9) c

10) b

3) c

4) a

5) a

6) b

7) c

8) d

9) d

10) a

3) b

4) b

5) a

6) b

7) c

8) d

9) d

10) d

Chapter 2: Polynomials CW Exercise: 1) b 2) a 11) b HW Exercise: 1) d 2) b 11) d

Chapter 3: Pair of Linear Equations CW Exercise: 1) b 2) 11) c 12) HW Exercise: 1) c 2) 11) d 12)

a a

3) b 13) c

4) a

5) a

6) c

7) b

8) c

9) a

10) d

c d

3) d 13) b

4) b 14) a

5) c

6) a

7) a

8) b

9) c

10) b

Chapter 6: Triangles CW Exercise: 1) a 2) 11) b 12) HW Exercise: 1) d 2) 11) b 12)

a a

3) c 13) a

4) a 14) b

5) b 15) c

6) c

7) c

8) a

9) b

10) c

c b

3) a 13) a

4) b 14) c

5) c

6) a

7) c

8) a

9) c

10) d

Chapter 8: Introduction to Trigonometry CW Exercise: 1) c 2) 11) a 12) HW Exercise: 1) a 2) 11) c 12)

c a

3) a 13) b

4) c 14) b

5) b 15) d

6) a

7) a

8) d

9) b

10) c

a a

3) a 13) d

4) a 14) d

5) d

6) b

7) b

8) a

9) a

10) c

Chapter 14: Statistics CW Exercise: 1) d 2) a HW Exercise: 1) c 2) c Volume

3) a

4) a

5) c

6) c

7) a

3) d

4) c

5) b

6) b

7) d

Universal Tutorials – X CBSE (2012–13) – Mathematics

149