Untouchability is a crime. Untouchability is inhuman. TAMILNADU. TEXTBOOK
CORPORATION. COLLEGE ROAD, CHENNAI – 600 006. www.kalvisolai.com ...
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MATHEMATICS STANDARD − IX
Untouchability is a sin Untouchability is a crime Untouchability is inhuman
TAMILNADU TEXTBOOK CORPORATION COLLEGE ROAD, CHENNAI – 600 006.
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© Government of Tamilnadu First Edition 2003 Revised Edition 2004 Reprint 2006 CHAIRPERSON S. UDAYABASKARAN, Reader in Mathematics, Presidency College (Autonomous), Chennai – 600 005.
REVIEWERS Thiru. E. ARJUNAN,
Thiru K. THANGAVELU,
Lecturer (S.G.) in Mathematics, L. N. Govt. Arts College, Ponneri 601 204.
Lecturer (S.S.) in Mathematics, Pachaiyappa’s College, Chennai − 600 030.
AUTHORS Thiru. V. SRIRAM,
Thiru K. ARIVAZHAGAN,
School Assistant (Mathematics), P.C.K.G. Govt. Hr. Sec. School, Kodambakkam, Chennai − 600 024.
School Assistant (Mathematics), Govt. Boys Hr. Sec. School, Ulundurpet − 606 107.
Tmt. R. NAMBIKAI JAYARAJ,
Tmt. S. VIJAYA,
P.G. Assistant (Mathematics), St. Anne’s Girls Hr. Sec. School, Royapuram, Chennai − 600 013.
P.G. Assistant (Mathematics), B.P.G. Hr. Sec. School, Kailasapuram, Trichy − 620 014.
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CONTENTS PAGES
1.
NUMBER SYSTEMS 1.1 1.2
2.
MEASUREMENTS 2.1 2.2
3.
Conjectures and Proofs Mathematical Models
THEORETICAL GEOMETRY 6.1 6.2
7.
Polynomials Algebraic Identities Factorization Division of a Polynomial by a Polynomial
PROBLEM SOLVING TECHNIQUES 5.1 5.2
6.
Scientific Notation Notation of Logarithms Set Notation
ALGEBRA 4.1 4.2 4.3 4.4
5.
Area and Perimeter Combined Figures
SOME USEFUL NOTATION 3.1 3.2 3.3
4.
Number Systems The Real Number Line
Theorems for Verification Theorems with logical Proofs
ALGEBRAIC GEOMETRY 7.1 7.2 7.3
The Cartesian Coordinates system Slope of a Line The Distance between any two points (x1, y1) and (x2, y2)
1 - 30 1 17 31 - 47 31 36 48 - 87 48 51 71 88 - 117 89 93 103 112 118 - 130 119 127 131 - 163 131 153 164 - 182 164 168 174
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8.
TRIGONOMETRY 8.1 8.2 8.3
9.
PRACTICAL GEOMETRY 9.1 9.2
10.
Trigonometric ratios Trigonometric Identities Trigonometric Ratios for Complementary Angles
Concurrency in a triangle Geometric interpretation of averages
HANDLING DATA 10.1 Measures of Central Tendency
11.
GRAPHS 11.1 Linear Graphs 11.2 Application of Linear Graphs
183 - 201 185 194 199 202 - 212 203 209 213 - 223 215 224 - 235 224 228
Logarithms
236 - 237
Anti Logarithms
238 - 239
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1. NUMBER SYSTEMS Numbers occur everywhere in our day-to-day life. “Numbers are in every thing”, said Pythagoras, an ancient Greek mathematician. If we understand more about numbers, then we know more about mathematics. “Mathematics is the Queen of the Sciences, and the Theory of Numbers is the Queen of Mathematics”, said Carl Friedrich Gauss, a German mathematician. Numbers possess very nice properties and the properties will help us to solve problems of other Sciences. “Numbers are my Friends”, said Srinivasa Ramanujan, an Indian mathematician of our modern times. “God created the natural numbers and all the rest is the work of man”, exclaimed Kronecker, a German mathematician. In our earlier classes, we introduced natural numbers, integers, rational and irrational numbers and real numbers. Natural numbers were introduced as counting numbers and other numbers were developed from them to fulfill our requirements. In this chapter, we shall study some of the properties of the numbers.
1.1
Number Systems
1.1.1
The Natural Numbers
The numbers 1, 2, 3,…. are called natural numbers. They are also called counting numbers since they are used for counting objects. The collection of all natural numbers is denoted by the letter N. Even though it is not possible to list all the elements of N, we write N = {1, 2, 3, …}. In the above representation, we write " … " after the element 3 to indicate that the other elements of N are listed following the pattern of 1, 2, 3. The numbers 1, 3, 5,…. are called odd numbers. The numbers 2, 4, 6, …. are called even numbers. In the collection N, we can solve equations such as x − 9 = 0, x−16 = 0, x−54 = 0. However, we note that the equations such as x + 5 =5, x + 9 = 9 have no solution in the collection N, since they are satisfied by the number 0. 1.1.2
The Whole Numbers
The numbers 0, 1, 2, … are called whole numbers. The collection of all whole numbers is denoted by the letter W. We observe that W = {0, 1, 2,…}. Now the equations such as x + 5 = 5 and x + 9 = 9 have solutions in W. We note that all natural numbers are whole numbers but there is the whole number 0 which is not a natural number. However, we note that the equations such as x + 25 = 15, x + 12 = 9 have no solution in the collection W, since they are satisfied respectively by the numbers −10 and −3 which are not whole numbers.
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1.1.3
The Integers
The numbers 0, 1, −1, 2, −2, … are called integers of which 1, 2, 3, … are called positive integers and −1, −2, −3,… are called negative integers. The collection of all integers is denoted by the letter Z. Thus Z = {…, −3, −2, −1, 0, 1, 2, 3,…}. We observe that all whole numbers are integers but the negative integers −1, −2, −3,… are not whole numbers. Now, the equations such as x + 25 = 15, x + 12 = 9 have solution in the collection Z, since they are satisfied respectively by the numbers −10 and −3 which are in Z. However, the equations such as 2x + 5 = 12, 3x + 9 = 4 have no solution in the collection Z, since they are 7 −5 satisfied respectively by the numbers and which are not in Z . 2 3 1.1.4 The Rational Numbers p where p and q are integers and q ≠ 0 is called a rational q p number. The collection of all rational numbers is denoted by Q. A rational number is said q to be in the proper form if q is a positive integer and p and q have no common factor other A number of the form
than 1. For example, the rational numbers
7
,
−2 5
11
numbers
are in the proper form where as the rational
12 24 and are not in the proper form. But every rational number has an 15 − 22
equivalent proper form. For example, we can write
12 15
=
4 5
and
24 − 22
=
− 12
which are in
11
proper form. Every integer is a rational number. For example, the integer −19 can be written as
− 19 1
, where −19, 1 are in Z and the denominator 1 ≠ 0. Thus, we note that all integers are
rational numbers. But there are rational numbers which are not integers. For example,
4
is a
5
rational number which is not an integer. 1.1.5
Some properties of +, −, ×, ÷ in N, W, Z and Q
From our experience with numbers, we observe that the addition '+' and the multiplication '×' have the following properties in Q: 1. If x, y are rational numbers, then x + y is also a rational number. For example, 11 and −2
are in Q and their sum 11 + ⎛⎜ − 2 ⎞⎟ = 11 + ⎛⎜ − 2 ⎞⎟ = 33 + (−2) = 31 is also in Q. This property is 3 3 3 ⎝ 3 ⎠ 1 ⎝ 3 ⎠ called the closure property with respect to addition in Q. 2.
If x, y are rational numbers, then x + y = y + x.
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For example, ⎛⎜ − 4 ⎞⎟ + ⎛⎜ − 2 ⎞⎟ = (−12) + (−22) = − 34 and ⎛⎜ − 2 ⎞⎟ + ⎛⎜ − 4 ⎞⎟ = (−22) + (−12) = − 34 . 3 So
⎝ 11 ⎠ ⎝ ⎠ ⎛−4⎞ ⎛−2⎞ ⎛−2⎞ ⎛−4⎞. ⎜ ⎟+⎜ ⎟ = ⎜ ⎟+⎜ ⎟ ⎝ 11 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 11 ⎠
33
⎝ 3 ⎠ ⎝ 11 ⎠
33
33
33
This property is called the commutative property with respect to
addition in Q. 3. If x, y, z are rational numbers, then x + (y + z) = (x + y) + z. For example, 2 , − 4 , − 2 are in Q and
3 5 7 ⎛ 2 ⎞ ⎡⎛ − 4 ⎞ ⎛ − 2 ⎞⎤ ⎛ 2 ⎞ ⎡ (−28) + (−10) ⎤ = ⎛ 2 ⎞ ⎛ − 38 ⎞ = 70 + (−114) = − 44 . ⎜ ⎟+⎜ ⎟ ⎜ ⎟ + ⎢⎜ ⎟+⎜ ⎟⎥ = ⎜ ⎟ + ⎢ ⎥ 105 105 35 ⎝ 3 ⎠ ⎝ 35 ⎠ ⎝ 3 ⎠ ⎣⎝ 5 ⎠ ⎝ 7 ⎠⎦ ⎝ 3 ⎠ ⎣ ⎦ ⎡ 2 ⎛ − 4 ⎞⎤ ⎛ − 2 ⎞ ⎡10 + (−12) ⎤ ⎛ − 2 ⎞ = ⎛ − 2 ⎞ ⎛ − 2 ⎞ = (−14) + (−30) = − 44 . ⎜ ⎟+⎜ ⎟ ⎢ 3 + ⎜ 5 ⎟⎥ + ⎜ 7 ⎟ = ⎢ ⎥+⎜ 7 ⎟ 105 105 15 ⎝ 15 ⎠ ⎝ 7 ⎠ ⎝ ⎠⎦ ⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎣ ⎛ 2 ⎞ ⎡⎛ − 4 ⎞ ⎛ − 2 ⎞⎤ ⎡ 2 ⎛ − 4 ⎞⎤ ⎛ − 2 ⎞ ∴ ⎜ ⎟ + ⎢⎜ ⎟+⎜ ⎟⎥ = ⎢ + ⎜ ⎟⎥ + ⎜ ⎟. ⎝ 3 ⎠ ⎣⎝ 5 ⎠ ⎝ 7 ⎠⎦ ⎣ 3 ⎝ 5 ⎠⎦ ⎝ 7 ⎠
This property is called the associative property of addition in Q. 4. The number 0 is a rational number and 0 + x = x + 0 = x for all rational number x. For example, 0 + 11 = 0 + 11 = 0 + 11 = 11 and 11 + 0 = 11 + 0 = 11 + 0 = 11 . 3
1
3
3
3
3
3
1
3
3
The rational number 0 is called the additive identity in Q. 5. For every rational number x, there is a rational number −x such that x + (−x) = (−x) + x = 0. The rational number –x is called the negative of x or the additive − 11 inverse of x in Q . For example, for the rational number , we have the rational number 3 11 3
− 11 ⎞ ⎛ 11 ⎞ ( −11) + 11 0 such that ⎛⎜ = = 0. ⎟+⎜ ⎟ = ⎝ 3 ⎠ ⎝3⎠
3
3
If x, y are rational numbers, then x × y is also a rational number. We write x × y simply 2 as xy. For example, (−5), are rational numbers and 3 ⎛ 2 ⎞ ⎛ − 5 ⎞⎛ 2 ⎞ (−5) × 2 − 10 (−5) ⎜ ⎟ = ⎜ = is a rational number. ⎟⎜ ⎟ = 1× 3 3 ⎝ 3 ⎠ ⎝ 1 ⎠⎝ 3 ⎠ This property is called the closure property of multiplication in Q. 6.
7.
If x, y are rational numbers, then xy = yx. ⎛ −5⎞ For example, (−3), ⎜ ⎟ are rational numbers and ⎝ 7 ⎠ 15 ⎛ − 5 ⎞ 15 ⎛ − 5 ⎞ ⎛ −5⎞ ⎛ −5⎞ (−3)⎜ ⎟= , ⎜ ⎟(−3) = . So (−3) ⎜ ⎟ = ⎜ ⎟(−3). 7 ⎝ 7 ⎠ 7 ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ This property is called the commutative property of multiplication in Q.
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8. If x, y, z are rational numbers, then x(yz) = (xy)z. We observe that (−xy) = [(−1)(x)]y = (−1)xy, −(−x) = (−1)[(−1)x] =[(−1)(−1)x] = 1x = x. For example, 2, −3, − 7 are rational numbers, and (2) ⎡⎢(−3)⎛⎜ − 7 ⎞⎟⎤⎥ = (2)⎛⎜ 21 ⎞⎟ = 42 , 5 5 5 5 ⎝
⎣
⎠⎦
⎝
⎠
⎡ −7⎞ ⎛−7⎞ ⎛ − 7 ⎞⎤ ⎛ − 7 ⎞ 42 So we have [(2) (−3)] ⎛⎜ (2) ⎢(−3)⎜ ⎟. ⎟⎥ = [(2)(−3)] ⎜ ⎟ = (−6)⎜ ⎟= . 5 5 5 5 5 ⎝
⎠
⎝
⎠
⎣
⎝
⎠⎦
⎝
⎠
This property is called the associative property of multiplication in Q. 9.
The number 1 is a rational number and 1(x) = (x)1 = x for all rational numbers x. ⎛ 5 ⎞ 1× 5 5 For example, (1) ⎜ ⎟ = = . 3 3 ⎝ 3⎠ 10. For every non-zero rational number x, 1 is a rational number and x
⎛1⎞ ⎛1⎞ x⎜ ⎟ = ⎜ ⎟ x = 1. For example, if x is the rational number ⎝ x⎠ ⎝ x⎠
− 21 we , 4
observe that x ≠ 0 and
4 1 1 − 4 is a rational number and x ⎛ 1 ⎞ = ⎛ − 21 ⎞ ⎛ − 4 ⎞ = 84 = 1 . The number ⎜ ⎟ ⎜ ⎟⎜ ⎟ = = = ⎝ x ⎠ ⎝ 4 ⎠ ⎝ 21 ⎠ 84 x − 21 − 21 21 4 1 x
is called the reciprocal or multiplicative inverse of x in Q.
11.
If x, y, z are rational numbers, then x(y + z) = xy + xz, (x+y) z = xz+ yz. 2
For example, if x = , y = 3
−1 , z = 5, we 2
have
⎛ 2 ⎞ ⎡⎛ − 1 ⎞ ⎤ ⎛ 2 ⎞ ⎡ − 1 + 10 ⎤ ⎛ 2 ⎞⎛ 9 ⎞ x (y + z) = ⎜ ⎟ ⎢⎜ ⎟ + 5⎥ = ⎜ ⎟ ⎢ = ⎜ ⎟⎜ ⎟ = 3, ⎝ 3 ⎠ ⎣⎝ 2 ⎠ ⎦ ⎝ 3 ⎠ ⎣ 2 ⎥⎦ ⎝ 3 ⎠⎝ 2 ⎠ ⎛ − 1 ⎞ ⎛ 10 ⎞ (−1) + 10 9 ⎛ 2 ⎞⎛ − 1 ⎞ ⎛ 2 ⎞ xy + xz = ⎜ ⎟⎜ ⎟ + ⎜ ⎟(5) = ⎜ ⎟ + ⎜ ⎟ = = = 3. 3 3 ⎝ 3 ⎠ ⎝ 3⎠ ⎝ 3 ⎠⎝ 2 ⎠ ⎝ 3 ⎠ So, x(y+z) = xy + xz. Similarly, we have ⎡⎛ 2 ⎞ ⎛ − 1 ⎞⎤ 5 ⎡ 4 + ( −3) ⎤ ⎛1⎞ (x+y)z = ⎢⎜ ⎟ + ⎜ ⎟⎥5 = ⎢ 5 = ⎜ ⎟5 = , ⎥ 6 ⎣ 6 ⎦ ⎝6⎠ ⎣⎝ 3 ⎠ ⎝ 2 ⎠⎦ 10 ⎛ − 5 ⎞ 20 + ( −15) 5 ⎛ 2 ⎞ ⎛ −1⎞ xz + yz = ⎜ ⎟5 + ⎜ ⎟5 = = . +⎜ ⎟= 6 6 3 ⎝ 2 ⎠ ⎝3⎠ ⎝ 2 ⎠ So, (x+y) z = xz + yz. These properties are called the distributive properties of multiplication over addition in Q. The properties 1, 2, 3, 6, 7, 8 and 11 do not depend on any particular element of Q and they are also valid for the elements of N, W and Z. The property 4 depends on the number 0. Since 0 is not in N, the property is not valid in N. Since 0 is in W and 0 is also in Z, the property 4 is valid in W and Z.
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The property 5 depends on negative numbers. So it is not valid in N and W. But it is valid in Z. The property 9 depends on the number 1. Since 1 is in N and in W and also in Z, the property 9 is valid in N, W and Z. The property 10 depends on the reciprocals of non-zero numbers. But the reciprocals of non zero integers are not in N, W and Z. So the property 10 does not hold in N, W and Z. The operation ‘−’, that is, the subtraction is defined in terms of addition ‘+’ as follows: If x, y are in Q, then x − y = x + (−y). The operation ‘−’ does not satisfy the commutative property in Q. For example, 4 − 5 = −1, 5 − 4 = 1 and so 4 −5 ≠ 5 − 4. The operation ‘÷’, that is, the division is defined in terms of multiplication ‘×’ as ⎛1⎞ follows: If x, y are in Q and y ≠ 0, then x ÷ y = x × ⎜⎜ ⎟⎟ . The operation ‘÷’ does not satisfy the ⎝ y⎠ 4 5 commutative property in Q. For example, 4 ÷ 5 = , 5 ÷ 4 = and so 4 ÷ 5 ≠ 5 ÷ 4. 5 4 The operation ‘−’ does not satisfy the associative property in Q. 2 For example, let us consider − 3, 11, in Q. Then 3 ⎡ 31 − 9 − 31 − 40 ⎛ 2 ⎞⎤ ⎡ 33 − 2 ⎤ (−3) − ⎢11 − ⎜ ⎟⎥ = ( −3) − ⎢ = ( − 3 ) − = = , 3 3 3 ⎝ 3 ⎠⎦ ⎣ 3 ⎥⎦ ⎣ [(−3) − 11] − 2 = (−14) − 2 = (−42) − 2 = − 44 . 3 3 3 3 ⎡ ⎛ 2 ⎞⎤ ⎛2⎞ So, (−3) − ⎢11 − ⎜ ⎟⎥ ≠ [(−3) − 11] − ⎜ ⎟. ⎝ 3 ⎠⎦ ⎝3⎠ ⎣ Similarly, we have ⎡⎛ 11 ⎞⎛ 3 ⎞⎤ ⎡ ⎛ 2 ⎞⎤ ⎛ 33 ⎞ (−3) ÷ ⎢11 ÷ ⎜ ⎟⎥ = (−3) ÷ ⎢⎜ ⎟⎜ ⎟⎥ = (−3) ÷ ⎜ ⎟ ⎝ 2⎠ ⎝ 3 ⎠⎦ ⎣⎝ 1 ⎠⎝ 2 ⎠⎦ ⎣ ⎛ − 3 ⎞ ⎛ 33 ⎞ ⎛ − 3 ⎞⎛ 2 ⎞ − 2 =⎜ , ⎟÷⎜ ⎟ = ⎜ ⎟⎜ ⎟ = ⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 1 ⎠⎝ 33 ⎠ 11 [(−3) ÷ 11] ÷ ⎛⎜ 2 ⎞⎟ = ⎡⎢⎛⎜ − 3 ⎞⎟ ÷ ⎛⎜ 11 ⎞⎟⎤⎥ ÷ ⎛⎜ 2 ⎞⎟ = ⎡⎢⎛⎜ − 3 ⎞⎟⎛⎜ 1 ⎞⎟⎤⎥ ÷ ⎛⎜ 2 ⎞⎟ ⎝ 3 ⎠ ⎣⎝ 1 ⎠ ⎝ 1 ⎠⎦ ⎝ 3 ⎠ ⎣⎝ 1 ⎠⎝ 11 ⎠⎦ ⎝ 3 ⎠ ⎛ − 3 ⎞ ⎛ 2 ⎞ ⎛ − 3 ⎞⎛ 3 ⎞ − 9 =⎜ . ⎟⎜ ⎟ = ⎟ ÷⎜ ⎟= ⎜ ⎝ 11 ⎠ ⎝ 3 ⎠ ⎝ 11 ⎠⎝ 2 ⎠ 22
⎡ ⎛ 2 ⎞⎤ ⎛2⎞ So (−3) ÷ ⎢11 ÷ ⎜ ⎟⎥ ≠ [(−3) ÷ 11] ÷ ⎜ ⎟ . ⎝ 3 ⎠⎦ ⎝3⎠ ⎣
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The subtraction operation ‘−’ does not satisfy the closure property in N, since if we consider the members 5 and 7 in N, we get 5 −7 = −2 which is not in N. Similarly the division 5 operation ÷ does not satisfy the closure property in N, since 5, 7 are in N but 5 ÷ 7 = which 7 is not in N. Using the properties of N, W, Z and Q, we can ascertain whether an equation has a solution in a particular number system or not. For example, we consider the equation 5x −10 = 0. Solving the equation, we get x = 2. Since 2 is in N, we say that equation 5x − 10 = 0 has a solution in N. Next, we consider the equation 5x = 0. Solving the equation, we get x = 0. Since 0 is not in N but 0 is in W, we say that the equation 5x = 0 has no solution in N but has solution in W. As another example, we consider the equation 5x + 10 = 0. Solving the equation, we get x = −2. Since −2 is not in N, −2 is not in W and −2 is in Z, we say that equation 5x + 10 = 0 has no solution in N or in W, but has solution in Z. −5 −5 Let us consider another equation 3x + 5 = 0. Solving the equation, we get x = . Since 3 3 −5 −5 is not in W and is not in Z, we say that that the equation 3x + 5 = 0 does is not in N, 3 3 −5 not possess a solution in N, W and Z. Since is in Q, we say that 3x + 5 = 0 possesses a 3 solution in Q. However, since we know that 2 , 3 , π, … are not rational numbers, the equations such as x −
2 = 0, x −
3 = 0, x − π = 0 have no solutions in Q.
Now we proceed to know about numbers which are not rational numbers. For this, we review what we have learnt about the decimal representation of rational numbers. 1.1.6
Decimal Representation of rational numbers
We have already learnt how to obtain the decimal representation of a rational number − 15 51 by the long division process. For example, the decimal representations of and are 32 7 obtained as follows: − 15 = −0.46875. 32
51 =7.28571428 7
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Let us understand the rule followed in the above long division process. For this, let us recall the representation of integers. For example, when we consider the integer 324, we mean that the integer 324 is the sum (addition) of 3 hundreds, 2 tens and 4 ones. That is, 324 = 3 × 102 + 2 × 101 + 4 × 100. Similarly 2003 = 2 × 103 + 0 × 102 + 0 × 101 + 3 × 100. Thus, when we write integers, we use the numerals 0, 1, 2, …, 9 and fix their face values as multiples of 100, 101, 102, …. In the same way, we can use numerals 0, 1, 2, …, 9 and fix their face values as multiples of 10-1, 10-2, 10-3, …. to get fractions. For example, 6× 10-1 + 2 × 10-2 + 5× 10-3 =
6 2 5 600 + 20 + 5 625 5 × 125 5 + 2 + 3 = = = = 10 10 1000 1000 8 × 125 8 10
5 as 0.625. Here the first numeral 6 from the right side of dot called decimal 8 6 2 point has the face value , the second numeral 2 has the face value and so on. The first 10 100 numeral 0 from the left side of dot has the face value 0 × 100 = 0. That is, the dot in the above representation is used to separate the integral part and the fractional part or decimal part 5 of the rational number . 8 With this notation, 3.025 means 25 1 121 3 × 100 + 0 × 10-1 + 2 × 10-2 + 5 × 10-3 = 3 + =3+ = . 1000 40 40 − 15 Now, we consider the fraction . Here 32 15 1 ⎛ 150 ⎞ = ⎜ ⎟ 32 10 ⎝ 32 ⎠ 1 ⎛ 22 ⎞ = ⎜4 + ⎟ 10 ⎝ 32 ⎠ 4 1 ⎛ 22 ⎞ = + ⎜ ⎟ 10 10 ⎝ 32 ⎠ 4 1 ⎛ 220 ⎞ = + 2⎜ ⎟ 10 10 ⎝ 32 ⎠ 4 1 ⎛ 28 ⎞ = + 2 ⎜6 + ⎟ 10 10 ⎝ 32 ⎠ 4 6 1 ⎛ 28 ⎞ = + 2 + 2⎜ ⎟ 10 10 10 ⎝ 32 ⎠ 4 6 1 ⎛ 280 ⎞ = + 2 + 3⎜ ⎟ 10 10 10 ⎝ 32 ⎠
and we denote
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4 6 + 2 10 10 4 6 = + 2 10 10 4 6 = + 2 10 10 4 6 = + 2 10 10 4 6 = + 2 10 10 4 6 = + 2 10 10 4 6 = + 2 10 10 = 0.46875
=
1 10 3 8 + 3 10 8 + 3 10 8 + 3 10 8 + 3 10 8 + 3 10 8 + 3 10 +
24 ⎞ ⎛ ⎜8 + ⎟ 32 ⎠ ⎝ 1 ⎛ 24 ⎞ + 3⎜ ⎟ 10 ⎝ 32 ⎠ 1 ⎛ 240 ⎞ + 4⎜ ⎟ 10 ⎝ 32 ⎠ 1 ⎛ 16 ⎞ + 4 ⎜7 + ⎟ 32 ⎠ 10 ⎝ 7 1 ⎛ 16 ⎞ + 4 + 4⎜ ⎟ 10 10 ⎝ 32 ⎠ 7 1 ⎛ 160 ⎞ + 4 + 5⎜ ⎟ 10 10 ⎝ 32 ⎠ 7 5 + 4 + 5 (The process terminates.) 10 10
− 15 = −0.46875. 32 Next consider the following process: 51 ⎛2⎞ = 7+⎜ ⎟ 7 ⎝7⎠ ⎡ 1 ⎛ 20 ⎞⎤ = 7 + ⎢ ⎜ ⎟⎥ ⎣10 ⎝ 7 ⎠⎦
∴
(A)
⎡1 ⎛ 6 ⎞⎤ = 7 + ⎢ ⎜ 2 + ⎟⎥ 7 ⎠⎦ ⎣10 ⎝ ⎡2 1 ⎛ 6 ⎞⎤ = 7 + ⎢ + ⎜ ⎟⎥ ⎣10 10 ⎝ 7 ⎠⎦ ⎡2 1 ⎛ 60 ⎞⎤ = 7 + ⎢ + 2 ⎜ ⎟⎥ ⎣10 10 ⎝ 7 ⎠⎦ ⎡2 1 ⎛ 4 ⎞⎤ = 7 + ⎢ + 2 ⎜ 8 + ⎟⎥ 7 ⎠⎦ ⎣10 10 ⎝ ⎡2 8 1 ⎛ 4 ⎞⎤ = 7 + ⎢ + 2 + 2 ⎜ ⎟⎥ 10 ⎝ 7 ⎠⎦ ⎣10 10 ⎡2 8 1 ⎛ 40 ⎞⎤ = 7 + ⎢ + 2 + 3 ⎜ ⎟⎥ 10 ⎝ 7 ⎠⎦ ⎣10 10 ⎡2 8 1 ⎛ 5 ⎞⎤ = 7 + ⎢ + 2 + 3 ⎜ 5 + ⎟⎥ 7 ⎠⎦ 10 ⎝ ⎣10 10 ⎡2 8 5 1 = 7+⎢ + 2 + 3 + 3 10 10 ⎣10 10
⎛ 5 ⎞⎤ ⎜ ⎟⎥ ⎝ 7 ⎠⎦ 8
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⎡2 8 5 1 ⎛ 50 ⎞⎤ = 7 + ⎢ + 2 + 3 + 4 ⎜ ⎟⎥ 10 10 ⎝ 7 ⎠⎦ ⎣10 10 ⎡2 8 5 1 ⎛ 1 ⎞⎤ = 7 + ⎢ + 2 + 3 + 4 ⎜ 7 + ⎟⎥ 7 ⎠⎦ 10 10 ⎝ ⎣10 10 ⎡2 8 5 7 1 ⎛ 1 ⎞⎤ = 7 + ⎢ + 2 + 3 + 4 + 4 ⎜ ⎟⎥ 10 10 10 ⎝ 7 ⎠⎦ ⎣10 10 ⎡2 8 5 7 1 ⎛ 10 ⎞⎤ = 7 + ⎢ + 2 + 3 + 4 + 5 ⎜ ⎟⎥ 10 10 10 ⎝ 7 ⎠⎦ ⎣10 10 ⎡2 8 5 7 1 ⎛ 3 ⎞⎤ = 7 + ⎢ + 2 + 3 + 4 + 5 ⎜1 + ⎟ ⎥ 10 10 10 ⎝ 7 ⎠⎦ ⎣10 10 ⎡2 8 5 7 1 1 = 7+⎢ + 2 + 3 + 4 + 5 + 5 10 10 10 10 ⎣10 10 ⎡2 8 5 7 1 1 = 7+⎢ + 2 + 3 + 4 + 5 + 6 10 10 10 10 ⎣10 10
⎛ 3 ⎞⎤ ⎜ ⎟⎥ ⎝ 7 ⎠⎦ ⎛ 30 ⎞⎤ ⎜ ⎟⎥ ⎝ 7 ⎠⎦
⎡2 8 5 7 1 1 ⎛ 2 ⎞⎤ = 7 + ⎢ + 2 + 3 + 4 + 5 + 6 ⎜ 4 + ⎟⎥ 7 ⎠⎦ 10 10 10 10 ⎝ ⎣10 10 8 5 7 1 4 ⎤ 1 ⎛2⎞ ⎡2 = 7×100+ ⎢ + 2 + 3 + 4 + 5 + 6 ⎥ + 6 ⎜ ⎟ 10 10 10 10 ⎦ 10 ⎝ 7 ⎠ ⎣10 10 = 7.285714 +
1 10 6
⎛2⎞ ⎜ ⎟. ⎝7⎠
The process repeats from (A)
We observe that in the decimal representation of remainder) and we say that
(B)
− 15 , the process terminates (zero 32
− 15 has a terminating decimal expansion. But in the decimal 32
51 , the process does not terminate (non zero remainder) at any stage. 7 However, we notice that the remainder that we get at the stage (B) is the same as the remainder at the stage (A). So the numerals 2, 8, 5, 7, 1, 4 between the two stages (A) and (B) repeat in the same order in the long division process. In this case, we say that the decimal representation is non terminating and recurring. We write
representation of
51 = 7 .285714285714285714… = 7.285714 , 7
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where the bar over 285714 indicates that the numerals under the bar repeat endlessly in the 15 same order in the long division process. The terminating decimal expansion 0.46875 for 32 can be considered to be a non-terminating and repeating, since 4 6 8 7 5 0.46875 = + 2 + 3+ 4 + 5 10 10 10 10 10 4 6 8 7 5 0 0 = + 2 + 3 + 4 + 5 + 6 + 7 + ..... = 0.468750 . 10 10 10 10 10 10 10 We also have 4 6 8 7 5 0.46875 = + 2 + 3 + 4 + 5 10 10 10 10 10 4 6 8 7 1 = + 2 + 3 + 4 + 5 (4 + 1) 10 10 10 10 10 4 6 8 7 4 ⎛ 1 ⎞ = + 2 + 3 + 4 + 5 +⎜ 5 ⎟ 10 10 10 10 10 ⎝ 10 ⎠ 4 6 8 7 4 1 = + 2 + 3 + 4 + 5 + 6 (10 ) 10 10 10 10 10 10 4 6 8 7 4 1 = + 2 + 3 + 4 + 5 + 6 (9 + 1) 10 10 10 10 10 10 4 6 8 7 4 9 ⎛ 1 ⎞ = + 2 + 3 + 4 + 5 + 6 +⎜ 6 ⎟ 10 10 10 10 10 10 ⎝ 10 ⎠ 4 6 8 7 4 9 1 = + 2 + 3 + 4 + 5 + 6 + 7 (10 ) 10 10 10 10 10 10 10 4 6 8 7 4 9 1 = + 2 + 3 + 4 + 5 + 6 + 7 (9 + 1) 10 10 10 10 10 10 10 4 6 8 7 4 9 9 ⎛ 1 ⎞ = + 2 + 3 + 4 + 5 + 6 + 7 + ⎜ 7 ⎟ = 0.468749. 10 10 10 10 10 10 10 ⎝ 10 ⎠ Thus, every rational number has a decimal representation which is either terminating or non terminating with repetition. This is the characteristic property of rational numbers which is due to the fact that the remainders that we get in the long division process are non negative integers less than the divisors. At one stage, one remainder in a previous stage starts repeating. So the digits in the quotient begin to repeat. Now we ask the following converse question: What does a terminating or non terminating recurring decimal expansion represent? We investigate the question through examples. (i) Consider the decimal expansion 0.45. 4 5 40 + 5 45 9 0.45 = + 2 = = = 10 10 100 100 20 (ii) Consider the decimal expansion 0.45 Let x = 0.45 = 0.454545…. Then 100x = 45.454545…
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∴ 100 x − x = (45.4545…) − (0.4545…) or 99x = 45 or x =
45 5 = . 99 11
(iii) Consider the decimal expansion − 0.349 Let x = 0.349 = 0.349999…. Then 100x = 34.9999… ∴ 1000x = 349.9999… ∴ 1000x−100x = (349.9999…) − (34.9999…) 315 35 7 ∴ 900x= 315 or x = = = . 900 100 20 7 . ∴ 0.349 = 20 ⎛ 7 ⎞ −7 ∴ − 0.349 = − ⎜ ⎟ = . ⎝ 20 ⎠ 20 Thus, every terminating or non- terminating recurring decimal expansion represents a rational number. That is, a terminating or non-terminating but repeating decimal representation can be put in the “Integer-by-Integer” form. 1.1.7
Irrational numbers
Now let us consider decimal expansions which are non-terminating and non-recurring. As an example, consider the non-terminating and non-recurring decimal representation 0.101001000100001000001… . We observe that the above decimal expansion has the numerals 0’s and 1’s. As we proceed from the right of the dot, the 1’s are separated by 1 zero, 2 zeros, 3 zeros, 4 zeros, … endlessly. So we find no repeating block in the representation. Hence the decimal representation can not represent a rational number. Such decimal expansions are said to represent irrational numbers. Rational numbers and irrational numbers are called real numbers. A decimal representation can be in exactly one of the following forms: (i) Terminating. (ii) Non-terminating but repeating. (iii) Non-terminating and non-repeating. Hence, every decimal representation is a real number. We state that every real number has a decimal representation. A real number x is said to be positive if it has a decimal representation in which at least one of the coefficients of 10n is a positive integer. Similarly, x is said to be negative if it is not positive or zero. For example, 0 0 2 5 2 5 7.00252525… = 7 + + 2 + 3 + 4 + 5 + 6 + .... is positive. 10 10 10 10 10 10 0 2 0 2 2 0 ⎡ ⎤ − 3.0202202220… = − ⎢3 + + 2 + 3 + 4 + 5 + 6 + ...⎥ is negative. 10 10 10 10 ⎣ 10 10 ⎦ The collection of all real numbers is denoted by the letter R. Thus R is the collection formed by the rational and irrational numbers. We observe that all natural numbers, all whole
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numbers, all integers, all rational numbers and all irrational numbers are real numbers. We also observe that no rational number is irrational and no irrational number is rational. The usual laws of addition, subtraction, multiplication and division are satisfied in R. In particular, the commutative and the distributive properties in R are (i) x + y = y + x, xy = y x (ii) x(y + z) = xy + xz (iii) (x + y)z = xz + yz where x, y and z are any three real numbers.
(i) (ii)
(iii)
(iv)
From the above two properties we have x (y − z)=x [y + (− z)]=xy + x(− z) = xy − xz. (x + y)2 = (x + y)(x + y) = (x + y) z, where we have put z = x + y for simplification = xz + yz = x (x + y) + y (x + y) = xx + xy + yx +yy = x2 + xy + xy + y2 = x2 + 2xy + y2 . (x − y)2 = (x − y)(x − y) = (x − y) z, where we have put z =x − y for simplification = xz − yz = x (x − y) − y (x − y) = xx − xy − (yx − yy) = x2 − xy − xy + y2 = x2 − 2xy + y2 . (x+ y) (x − y) = (x + y) z where we have put z =x − y for simplification = xz + yz = x (x − y) + y (x − y) = xx − xy + yx − yy = x2 − y2 .
Example 1: Determine the rational number represented by 0.75 . Solution: Let x = 0.75 . Then x = 0.757575… ∴ 100x = 75.757575… ∴ 100x − x = (75.757575…) − (0.757575…) = 75.0000… 75 25 ∴ 99x = 75 or x = = . 99 33 Example 2: Justify 1.0 = 0.9. Solution: 1 1 1 9 ⎛1⎞ 9 1.0 = 1 = (10) = 9 + 1 (9 + 1) (10) = (9 + 1) = + +⎜ ⎟= 10 10 10 100 10 ⎝ 10 ⎠ 10 100 9 1 9 9 ⎛ 1 ⎞ 9 = (10) = 9 + 9 + 1 (9 + 1) + + + +⎜ ⎟= 10 100 ⎝ 100 ⎠ 10 100 1000 10 100 1000 9 9 9 ⎛ 1 ⎞ = + + +⎜ ⎟. 10 100 1000 ⎝ 1000 ⎠
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The above process continues endlessly and we get 1.0 = 0.9999… or 1.0 = 0.9. Example 3: Find 2.52 + 2.52 in the integer-by-integer form. Solution: Let x = 2.52 . Then x = 2.525252…. ∴ 100x = 252.5252… 250 ∴ 100x − x = 250.000… or 99x = 250 or x = . 99 Let y = 2.52 . Then y = 2.52222….. ∴ 10y = 25.2222… ∴ 100y = 252.2222… 227 ∴ 100y − 10y = 227.0000… or 90y = 227 or y = . 90 250 227 2500 + 2497 4997 Hence 2.52 + 2.52 = + = = . 99 90 990 990 Example 4: Find the sum of the irrational numbers 0.101001000100001… and 0.010110111011110… . If it is a rational number, find it in the integer- by-integer form. Solution: Let x = 0.1010010001… and y = 0.0101101110… . Then x + y = 0.1010010001… + 0.0101101110…= 0.111111… = 0. 1 . Since x + y has a non-terminating but repeating decimal expansion, x + y is a rational number. Let the rational number be a. Then a = 0.11111… Multiplying by 10, we get 10a = 1.1111… 1 ∴ 10a − a = 1.1111… − 0.1111 = 1.0000…. or 9a = 1 or a = . 9 1 ∴x+y= . 9 Note: From the above example, we observe that the sum of two irrational numbers need not be an irrational number. Similarly the product of two irrational numbers need not be an irrational number ( 2 is irrational but rational).
2 ×
2 = 2 is
Now let us consider the irrational number
2 . We know already much about this number. We recall here the method of finding the decimal expansion of and
2 to any number of decimal places.
2 = 1.414213562…
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We observe that the above process neither terminates nor repeats. That is, 2 has a non-terminating and non-repeating decimal expansion. Therefore, we conclude that 2 is an irrational number. Similarly, we can show that 3 , 5 , 7 , … are all irrational numbers. We come across two special irrational numbers in mathematics. They are π and e. When we calculate the ratio of the perimeter of any circle to the length of the diameter of the circle, we observe that the ratio is a fixed real number and it is denoted by the Greek letter π. The decimal expansion of π is 3.1415926… which is non-terminating and non-repeating. We 22 recall that we used the rational number as an approximation for the irrational number π in 7 calculations. Ramanujan, the celebrated Indian mathematician has obtained several formulas involving π. Around the year 1973, 1,000,000 decimal digits of π were computed. Getting the decimal expansion of π to several billion thousands of decimal places is even today a fascinating and challenging task. When we calculate the values of the numbers 2
4
5
⎛3⎞ ⎛4⎞ ⎛5⎞ ⎛6⎞ ⎜ ⎟ , ⎜ ⎟ , ⎜ ⎟ , ⎜ ⎟ , …, ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ we observe that they become closer and closer to a particular real number and this real number is denoted by the letter e. The decimal expansion of e is e = 2.7182818284…. which is non-terminating and non-repeating. We will study more about this irrational number e in higher standards.
1.1.8
3
Order Relation in R
When we arrange certain objects according to some property, we say that the objects are ordered. For example, when students are arranged standing according to their heights from the shortest to the highest, we say that they stand ordered. Likewise, we can arrange real numbers in order. Let x and y be any two real numbers. If y − x is a positive real number, then the real number x is said to be less than the real number y or y is said to be greater than x. The symbol x < y is used to mean that x is less than y and y > x to mean that y is greater than x. Thus, x < y and y > x both mean the same fact that y − x is positive. If y − x is negative, then −(y − x) = x − y is positive and so y < x which is equivalent to x > y. For example, consider 19 and 17. Since 19 − 17 = 2 is a positive number, we get 17 < 19 which is equivalent to 19 > 17. Next, consider −19 and −17. Since (−19) − (−17) = −19 + 17 = −2 is a negative number, we get −19 < −17 which is equivalent to −17 > −19. Let x and y be real numbers. Then three cases arise. (i) x − y is negative (ii) x − y is positive and (iii) x − y = 0. In the first case, x < y. In the second case, x > y. In the third case, x = y. If either x < y or x = y, then we say that x is less than or equal to y and write x ≤ y. Similarly, if either x > y or x = y, then we say that x is greater than or equal to y and write 14
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x ≥ y. If x is positive, then x > 0. If x is non negative, then x ≥ 0. If x is negative, then x < 0. If x is non positive, then x ≤ 0. Thus, given any two real numbers x and y, exactly one of the following is true: x < y, x = y, x > y. This fact is called the law of trichotomy of the order in real numbers. If x, y and z are three real numbers such that x < y and y < z, then y − x is positive and z − y is positive. So, (y − x) + (z − y) is positive; that is, z – x is positive. Hence x < z. Thus, we get that, if x < y and y < z, then x < z. This property is called the transitive property of the order in R. When x < y and y < z, we write x < y < z and we say that y is inserted between x and z or we say that y lies in between x and y. Consider a real number x along with 0. Then, precisely one of the following is true: x < 0 , x = 0 , x > 0. If x < 0, then x2 = x × x = negative number × negative number = positive number. If x = 0, then x2 = x × x = 0 × 0 = 0. If x > 0, then x2 = x × x = positive number × positive number = positive number. Thus, we observe that, for every real number x, we have x2 ≥ 0. In particular, we observe that, for every non-zero real number x (> 0 or < 0), we always have x2 > 0. When x ≥ 0, the absolute value of x is defined to be x; when x < 0, the absolute value of x is defined to be −x .The absolute value of x is denoted by |x|. Thus, |x| = x if x ≥ 0; |x| = −x if x < 0. We observe that |x| ≥ 0, |−x| = |x| and |x|2 = x2. 22 in order relation. 7 22 Solution: π = 3.141528…, = 3.142857… 7 22 − π = 3.142857… − 3.141528… = 0.0013… ∴ 7 22 ∴ − π is positive. 7 22 ∴ π< . 7
Example 5: Put π and
Example 6: Insert any four rational numbers in between the rational numbers 1.201 and 1.202. Solution: Here 1.202 − 1.201 = 0.001 > 0 and so 1.201 < 1.202. Consider the numbers 1.2011, 1.2012, 1.2013, 1.2014. Since these have terminating decimal expansions, they are rational numbers. We find 1.2011 − 1.201 = 0.0001 > 0 and so 1.201 < 1.2011. Since 1.2012 − 1.2011 = 0.0001 > 0, we get 1.2011 < 1.2012. Since 1.2013 − 1.2012 = 0.0001 > 0, we get 1.2012 < 1.2013. Since 1.2014 − 1.2013 = 0.0001 > 0, we get 1.2013 < 1.2014. Since 1.202 − 1.2014 = 0.0006 > 0, we get 1.2014 < 1.202. ∴ 1.201 < 1.2011 < 1.2012 < 1.2013 < 1.2014 < 1.202.
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Example 7: Insert any four irrational numbers between 1.201 and 1.202. Solution: Since 1.202 − 1.201 = 0.001 > 0, we get 1.201 < 1.202. Consider the real numbers a = 1.2011010010001…, b = 1.2012020020002…, c = 1.2013030030003…, d = 1.2014040040004…. Since these four real numbers have distinct non-terminating and non repeating decimal representations, they are distinct irrational numbers. We find a − 1.201 = 1.2011010010001… − 1.2010000000000… = 0.0001010010001… > 0, b − a = 1.2012020020002… − 1.2011010010001… = 0.0001010010001… > 0, c − b = 1.2013030030003… − 1.2012020020002… = 0.0001010010001… > 0, d − c = 1.2014040040004… − 1.2013030030003….= 0.0001010010001… > 0, 1.202 − d = 1.2020000000000… − 1.2014040040004… = 0.0005959959995… > 0. ∴ 1.201 < a < b < c < d < 1.202. Example 8: Insert any two rational numbers in between the irrational numbers 2.003 . Solution:
∴
2.001 and
2.001 = 1.41456… and 2.003 = 1.41527… ∴ 2.003 − 2.001 = 0.0007… > 0 ∴ 2.001 < 2.003 . Consider the numbers 1.41461 and 1.41462. Since they have terminating decimal expansions, they are rational numbers and we have 1.41461 − 2.001 = 1.414610000… − 1.41456…= 0.00004… > 0 ∴ 2.001 < 1.41461. 1.41462 − 1.41461 = 0.00001 > 0 ∴ 1.41461 < 1.41462. 2.003 − 1.41462 = 1.41523… − 1.41462 = 0.00061…> 0 ∴ 1.41462 < 2.003 . ∴ 2.001 < 1.41461 < 1.41462 < 2.003. 16
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From the above examples, we are able to observe that, if a and b are two distinct real numbers such that a < b, then there is a rational number r such that a < r < b. We note that the above property holds even if a and b are very close to each other. This property is usually called the denseness property of Q in R. We also observe as a comparison that, between any two distinct points on a straight line, there is another point distinct from them. This similarity between points on a straight line and real numbers enables us to get a line picture representation of real numbers. We also observe that all rational numbers are real numbers. But there are real numbers
2,
which are not rational numbers. For example,
3 , π, … are real numbers which are not
rational numbers. Now, the equations such as x −
2 = 0, x − 3 = 0, x − π = 0 have solutions in R. However, since x is positive for any non-zero real number x, there is no real number x such that x2 = −1 and so the equation x2 +1 = 0 has no solution in R. We do not study such equations in our standard. 2
Exercise 1.1 1.
Obtain the decimal expansions of 21 11 −7 (i) (ii) − (iii) 64 6 20
(iv)
11 24
2.
Determine the integer-by-integer form of each of the following rational numbers: (i) 0.34 9 (ii) −0. 7 (iii) 0. 125 (iv) 0.5 21 (v) 8. 9
3.
Answer true or false. 2 (i) is a whole number. 7 (ii) − 11 is an integer. 1 (iii) is a rational number. 3 (iv) 0.16 is a rational number. (v) Every decimal expansion is a real number. (vi) Every real number is a rational number. (vii) 0.1212212221… is a rational number.
1.2
The Real Number Line
We have already learnt about real number line in our earlier classes. We know how the real numbers are represented as points on the line. We review the idea of the real number line. Let us consider a straight line and fix arbitrarily a point on it. We name this point as O and say that it represents the number 0. We fix arbitrarily another point A on the line on the right of O 17
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and say that this point A represents the number 1. Now we say that line segment OA is of length 1 unit. We observe that the length of OA will be different for different choices of the point A. But once we have chosen O and fixed the point A, then, for us, OA is of length 1 unit. Using the segment OA as a scale for measuring unit distances, we can represent any real number as a point on the straight line. We call this straight line, the real number line or simply number line. First, we recall how positive integers are represented as points on the number line. Already the point O is there to represents the number 0 and A to represents the number 1. Now we locate points on the number line to the right of O at a distance 2 units (2 times the lengths of OA), 3 units, … . These points correspond to the numbers 2, 3, … respectively (see Figure 1.1).
Figure 1.1 Similarly, we can locate the points on the number line to the left of O at distances 1 unit, 2 units, 3 units, … . These points correspond to the negative integers − 1, − 2, −3,… respectively. Now, we review with an example the method of locating points for rational numbers on the number line. Consider the rational number 2 . To locate the point on the number 3
line corresponding to 2 , we proceed as 3
follows: Locate the point P corresponding to the 2 positive integer 3 (denominator of ). Draw 3
Figure 1.2
2 ) perpendicular to OP. Join OQ. 3 Consider the line through A parallel to PQ. This line meets OQ at the point R. Then the length 2 of AR is times that of OA. This is so because ∆OAR and ∆OPQ are similar and so 3 PQ OP 2 3 2 = or = or AR = . Now, draw a circle with centre at O and radius equal to the AR OA AR 1 3 length of AR. This circle cuts the real number line at a point on the right side of O. This point
the line segment PQ of length 2 (numerator of
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2 (see Figure 1.2). The same circle cuts the real number 3 −2 line at a point on the left side of O and this point corresponds to the rational number .In 3 the same way, we can represent any rational number on the real number line. The real number line is a straight line. Given any two distinct points P and Q on the line, however close they may be, we can find a point between P and Q different from P and Q. That is, there is no gap between any two points on the real number line. We say that the real number line is a continuum of points. We have shown that every rational number corresponds to a unique point on the real number line. Let us now ask the question whether all points of the real number line correspond to rational numbers only. The answer is ‘No’. For example, let us consider the irrational number 2.
corresponds to the rational number
Draw a square OABC with side OA = 1. Then, by Pythagoras theorem OB2 = OA2 + AB2 = 1+1 = 2. So OB = 2. With O as centre and OB as radius, draw a circle. We observe (see Figure 1.3) that this circle intersects the real number line at P on the right side of O and at Q
Figure 1.3
on the left side of O. The point P represents the irrational number 2 , since OP = OB = The point Q represents the irrational number − 2 , since OQ = OB = left of O.
2.
2 and Q lies to the
3 , 5, − 3 ,− 5 on the real number line.
Example 9: Represent the irrational numbers
Solution: Having plotted the point P for 2 , we can now locate the points for as given below: Construct a rectangle OPDC (see Figure 1.4)
3 and − 3
with length OP (= 2 ) and breadth PD = 1. Then, by Pythagoras theorem, OD2 = OP2 + PD2 = ( 2 ) 2 + 12 =2+1=3
Figure 1.4
and so OD = 3 . Now, with O as centre and OD as radius, draw a circle. This circle intersects the real number line at the points say Q and Q′ (Note that Q is just a notation and not representing the collection of rational numbers). Now OQ = OQ′ = OD = Q represents
3 . Since Q and Q′ are on the right and the left of O respectively,
3 and Q′ represents − 3 .
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To plot the points for 5 and − 5 , the above technique is followed. Locate the point R on the real number line to right of O at a distance of 2 units from O (we have already provided a method to locate the points on the number line corresponding to rational numbers). Construct the rectangle ORFC (see Figure 1.5) with length OR (= 2) and breadth RF = 1.
Figure 1.5 Then, by Pythagoras theorem, OF2 = OR2 + RF2 = 22 + 12 = 4 + 1 = 5 and so OF = 5 . With O as centre and OF as radius, draw a circle. This circle intersects the real number line at say S and S′ on the right and left of O respectively. Since OS = OS′ = OF = 5 , S represents 5 and S′ represents − 5 . From the above discussion, we are able to know that any real number corresponds to a unique point on the real number line and that any point on the real number line represents a real number. Points on the right side of O correspond to positive real numbers and those on the left side of O represent negative real numbers. Let P and Q be the points corresponding to two real numbers a and b respectively. If a < b, then P lies to the left of Q which is same as saying that Q lies to the right of P on the real number line (see Figure 1.6). Further, Figure 1.6 if x is a real number such that a < x < b, then the point on the real number line corresponding to x lies between P and Q. If P1 and P2 are any two points on the real line and if x1 and x2 are the real numbers corresponding to P1 and P2, then the distance between P1 and P2 is |x1 − x2|. 1.2.1
Manipulation of irrational numbers Let us recall how the irrational numbers
irrational number x−
2 , 3 , 5 , … originated. For example, the
2 was needed when we wanted to find a solution x for the equation
2 = 0 or x − 2 = 0 or x2 = 2 . Similarly, we may need to obtain a real number x such that 2
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xn = r, where r is a rational number and n is a positive integer. If n is an even positive integer such as 2, 4, 6, … and r < 0, then we can not find a real number x such that xn = r. This is so because x2 = x × x > 0, x4 = x2 × x2 > 0, …, xn > 0 and r < 0 . If n is an even positive integer and r > 0, then it is possible to find a positive real number x such that xn = r. For example, we find 5 for x when x4 = 625. If n is an odd positive integer such as 1, 3, 5, … and r > 0, then it is possible to find a positive real number x such that xn = r. For example, we find 4 for x when x3 = 64. If n is an odd positive integer and r < 0, then it is not possible to find a positive real number x such that xn = r. This is so because if x > 0, then x2 = x × x > 0, x3 = x2 × x > 0, … xn > 0. Thus, if n is a positive integer and r is a positive rational number, then it is possible to find a positive real number x such that xn = r. In this situation, we write the positive real number x =
n
r and say that x is the nth root of r. We call
n
r a radical and
the radical sign. The positive integer n is called the index of the radical number r is called the radicand of irrational number. For example, rational number
p , then q
n
3
r =
n
r . The real number
64 = 4,
4
n
n
is known as
n
r and the rational
r may be a rational number or an
4 = 2. We observe that if r is the nth power of a
p is a rational number because in this case we have q
n
⎛ p⎞ ⎜⎜ ⎟⎟ = r. On the other hand, if the rational number r is not the nth power of some rational ⎝q⎠ number, then
n
r is not a rational number; that is,
n
r is an irrational number. If
n
r is an
irrational number, then
n
r is called a surd. We observe that a surd is an irrational number in
a particular form. Thus
n
r is a surd when the radicand is a rational number which is not the
nth power of a rational number. If the index n is 2, then we call
2
r the square root of r and
simply write it as r with out the index 2. If the index n is 3, then we call of r.
3
r the cube root
In what follows, n r always represents a positive real number and r is a positive rational number. Since radicals are real numbers, we can perform the four fundamental operations +, −, ×, ÷ with them. If the addition of two surds and multiplication of a surd by a rational number yield irrational numbers, then the resulting irrational numbers are also called surds. For example, 3 2 + 5 7 ,−3 2 are surds. Let a and b be two distinct positive rational numbers such that a and b are surds. Then a + b , a − b , a + b , a − b are surds. Here a − b is called the conjugate of a + b . Similarly, a + b is the conjugate of a − b , a + b is the conjugate of a − b and a − b is the conjugate of a + b . The product of a surd and its conjugate is a rational number. For example,
(2 + 3 )(2 − 3 ) = 2 − ( 3 ) = 4 − 3 = 1, ( 2
2
3+ 5
21
)(
) ( 3 ) − ( 5 ) = 3 − 5= −2.
3− 5 =
2
2
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The following laws are used in the manipulation of radicals:
Let a, b be positive rational numbers and m, n be positive integers. Then (i) (ii) (iii)
( a ) = a. ( a) ( b)= n
n
n
a
n
b
(iv) p (v)
n
n
n
n m
=n
n
ab .
a . b
r +q
n
r = ( p + q)
n
r where p and q are real numbers.
a = mn a.
(vi) n a = mn a m . (vii) If a < b, then
n
a < n b.
We observe that in the laws (ii), (iii) and (iv), the surds are of same index. So, when surds of different indices are given and if we are asked to perform any of the four fundamental operations, we first convert the surds into surds of the same index by applying the law (vi) and then proceed to carry out the operations. Law (vi) is actually the combination of law (i) and law (v): By (i), a =
m
am . ∴
n
a =n
m
am =
mn
am .
Using law (vii), we can compare any surds of same order. 1
Note:
n
a is denoted as a n .
Example 10: Answer with reasons whether the following are surds or not: (a)
121 225
(b)
(d)
144 72
(e)
49 25 32 3
7 5
(c) (f)
216
3
4 × 3 16
Solution: 2
11 ⎛ 11 ⎞ = a rational number. ∴ ⎜ ⎟ = 15 ⎝ 15 ⎠
(a)
121 = 225
(b
49 7 ⎛7⎞ = ⎜ ⎟ = = a rational number. ∴ 25 5 ⎝5⎠
(c)
7 7 is a surd since is not a square of a rational number. 5 5
(d)
144 = 2 = a surd. 72
2
22
121 is not a surd. 225
49 is not a surd. 25
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(e)
32 3
216
Since (f)
3
=
16 × 2 3
6×6×6
=
16 × 2 3
6
=
3
4× 2 2 2 = . 6 3
2 2 is an irrational number, 3
32 3
is a surd.
216
4 × 3 16 = 3 4 × 16 = 3 4 × 4 × 4 = 3 4 3 = 4 = a rational number.
∴ 3 4 × 3 16 is not a surd. When we manipulate with surds using the four fundamental operation, some times the result may be a rational number. Example 11: Answer with reasons whether the following are surds or not: (i) (iii) Solution: (i)
(5 + 54 ) + (4 − 3 2 ) (3 + 4 2 )(3 − 32 ) 3
3
(5 +
3
) (
(2 + 3 ) − (3 + 3 )
(ii)
4
(iv)
)
54 + 4 − 3 3 2 = (5 + 4) +
=9+
(3 3
(
3
(
3
432 ÷ 4 2187
27 × 2 − 3 3 2
×3 2 −3 2 3
)
)
)
= 9 + 3 × 2 − 3 × 2 = 9. 3
3
This is a rational number and not a surd although 5 + 3 54 and 4 – 3 3 2 are surds. So the addition of two surds need not be a surd. This means that the surds do not satisfy the closure property with respect to addition. (ii) (2 +
3 ) − (3 +
3 ) = (2 − 3) +
(
)
3 − 3 = −1 + 0 = −1 = a rational number.
∴ the given expression is not a surd. (iii)
(3 + 4 2 )(3 −
)
32 = (3 + 4 2 ) (3 − 16 × 2 )
= (3 + 4 2 ) (3 − 4 2 ) = 3 × 3 − 3 × 4 2 + 4 2 × 3 − 16 × 2 × 2 = 9 − 12 2 + 12 2 − 32 = 9 − 32 = −23. This is a rational number and so the given expression is not a surd. (iv) 432 = 2 × 2 × 2 × 2 × 27 = 24 × 27. ∴ 4 432 = 2 4 27. 2187 = 3 × 3 × 3 × 3 × 27 =34 × 27. ∴ 4 2187 = 3 4 27.
2 4 27
2 . 3 27 3 This is a rational number and so the given expression is not a surd.
∴ 4 432 ÷ 4 2187 =
4
=
23
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Example 12: Simplify each of the following: 20 − 225 + 80
(i)
(iii) 2 6 27 + 50 Solution: 20 = 4 × 5 = 4 × 5 = 2 5 . (i) 9 × 25 =
225 =
40 − 2 3 135
3
(ii)
(iv) 3 3 128 − 3 2000
9 × 25 = 3 × 5 = 15.
80 = 16 × 5 = 16 × 5 = 4 5 .
∴ 20 − 225 + 80 = 2 5 − 15 + 4 5 = 6 5 − 15. 3
(ii)
∴
40 = 3 8 × 5 = 3 8 × 3 5 = 2 3 5 ,
3
135 = 3 27 × 5 = 3 27 × 3 5 = 3 3 5 .
40 − 23 135 = 2 3 5 − 2 × 3 3 5 = ( 2 − 6) 3 5 = −43 5.
3
6
(iii)
27 = 2×3 33 = 2
33 = 2 3 = 3 and
3
50 = 2 × 25 = 5 2 .
∴ 2 6 27 + 50 = 2 3 + 5 2 . 3
(iv) 3
128 =
3
43 × 2 = 4
3
2.
2000 = 3 10 3 × 2 = 10 2 . 3
∴ 3 3 128 − 3 2000 = 3(43 2 ) − 10 3 2 = 12 3 2 − 10 3 2 . = (12 − 10) 3 2 = 2 3 2 . Example 13: Write each of the following into a single surd: 3 (i) 7 ×36 (ii) 5 × (iii) 8 Solution: (i) write 3
7 and 3 2
6 =
(ii)
(iii) 8 (iv)
3
3
Here 3
4
2
4
4 ÷6
81 ÷
(
3
6 as surds of index 6. Here
3 = 4
12
625 ×
2 =
7 × 12
83 4 642
)
5 + 20 =
3
(iv)
7 is of index 2 and
6 2 = 6 36. So,
5 ×
3
4 ÷6
3
3
6 =
27 =
= 3
12
6
7 =
5 + 20
(
5 + 20
)
2
7 =
2 3
7 3 = 6 343, and
343 × 6 36 = 6 343 × 36 = 6 12348 .
625 × 27 =
=
81 ÷
3.
6 is of index 3. The l.c.m. of 2 and 3 is 6. So we
12
16875 .
4 12 256 4 12 256 4 × = × = 3 12 8 3 8 3 81
4
3
27 × 3
5 + 4×5
24
=
12
3× 3 3 3 5
32.
=
3
3 5
=
6 6
9 125
=
6
9 . 125
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Example 14: Arrange the following in the ascending order of magnitudes: 4 3 , 6 10 , 12 25 Solution: We shall first find the common index. For this, we find the l.c.m. of the indices 4, 6, 12. The l.c.m. of 4, 6, 12, is 12. Then we convert the surds into surds with index 12. 4
3=4
We observe
12
3
3 × 3 × 3 = 12 27 and
10 = 6 2 10 × 10 = 12 100 .
6
25 < 12 27 < 12 100.
∴ 12 25 < 4 3 < 6 10. The product of a surd and its conjugate is always a rational number. For example,
(2 + 3 ) (2 − 3 ) = (2) − 2 3 + 2 3 − ( 3 ) = 4 − 3 = 1 and ( 3 + 5 ) ( 3 − 5 ) = ( 3 ) − 15 + 15 − ( 5 ) = 3 − 5 = −2. 2
2
2
2
x may be a surd. In such ratios, the y denominator can be made as a rational number by a suitable procedure. This procedure is called the rationalization of the denominator. We give the procedure as below: (i) If the denominator y is in the form a where a is a rational number, then multiply both the numerator and denominator by a . For example, Some times, the denominator y of a ratio
3
=
3× 5
=
15 5
5 5× 5 (ii) If the denominator y is in the form a +
b where a and b are rational numbers, then multiply the numerator and denominator by a − b . Then the denominator becomes (a + b ) (a − example, 2 3+ 2
=
b ) = a2 − a b + a b − 2 × (3 − 2) (3 + 2 )(3 − 2 )
=
( b)
2
2×3− 2× 2 3 − ( 2) 2
2
= a 2 − b which is a rational number. For
=
6−2 2 6−2 2 = . 9−2 7
(iii) If the denominator y is in the form a −
b where a and b are rational numbers, x then multiply both the numerator and denominator of by a + b . For example, y 3+ 2 3− 2
=
(3 + 2) × (3 + 2) (3 − 2 )(3 + 2 )
(3 + 2 ) 2
=
( )
32 − ( 2 ) 2
2
32 + 2 × 3 × 2 + 2 9 + 6 2 + 2 11 + 6 2 = = = . 9−2 7 7 (iv) If y is in the form a + b , where a and b are rational numbers, then multiply x the numerator and denominator of by a − b . For example, y
25
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) 3 + 5 ( 3 + 5 )( 3 − 5 ) ( 3 ) − ( 5 ) 2( 3 − 5 ) 2( 3 − 5 ) = = = −1( 3 − 5 ) = 5 − −2 3−5 2
2×
=
(
3− 5
)
=
(
2 3− 5 2
2
3.
a − b , where a and b are rational numbers, then multiply x the numerator and denominator of by a + b . For example, y (v) If y is in the form
5
=
5− 7
5×
(
(
5+ 7
5− 7
)(
5+
)
( 5 ) + 5 7 = 5 + 35 = − 1 (5 + = 5−7 2 7) ( 5) − ( 7) 2
2
1
Example 15: Rationalize the denominator of Solution: 1 5 + 14 =
=
1×
(
(
5 − 14
)
35 .
.
5 + 14
5 − 14
=
) ( 5 ) − ( 14 ) 5 − 14 1 = ( 14 − 5 ). −9 9
5 + 14
5 − 14 = 5 − 14
)(
)
2
5 − 14
2
Note: If we want to rationalize the numerator of
2
x , where x is of the form y
a or a + b or
a − b or a + b or a − b , then we multiply the numerator and denominator by the conjugate of the numerator. 11 − 3 . 4
Example 16: Rationalize the numerator in Solution:
2
(
=
Example 17: If Solution:
( 11 − 3 )× ( 11 + 3 ) = ( 11) − ( 3 ) 4 × ( 11 + 3 ) 4 × ( 11 + 3 )
11 − 3 = 4
11 − 3
4 11 + 3
1 1− 2 + 3
1 1− 2 + 3
= =
(1 −
)
=
(
8
4 11 + 3
)
=
2
2 11 + 3
= a + b 2 + c 6 , find a + b +c.
(
1× 1 − 2 − 3
)(
)
2 + 3 1− 2 − 3
1− 2 − 3
(1 − 2 ) − ( 3 ) 2
2
26
)
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1− 2 − 3
(1 − 2
= =
∴ ∴ ∴
)
2 +2 −3
1− 2 − 3
=
(1 −
2− 3
)×
2
−2 2 −2 2 2 2− 2 2− 3 2 2 −2− 6 = = −4 −4 2− 2 + 6 1 1 1 = = − 2+ 6. 4 2 4 4 1 1 1 2+ a+b 2 +c 6 = − 6 2 4 4 1 −1 1 a= ,b= ,c= . 2 4 4 1 1 1 1 a+b+c = − + = . 2 4 4 2 In the surd p + q n a , where p and q are rational numbers and
n
a is a surd, p is called
the rational part and q n a is called the irrational part. Two surds are said to be equal if their rational parts are equal and their irrational parts are equal. 3 −1
Example 18: If
3 +1
( = 3 +1 (
3 +1
+
3 −1
= x + y 3, find x2 + y2.
)( 3 − 1) = ( 3 ) − 2 3 + 1 = 3 − 2 3 + 1 = 4 − 2 Solution: 3 −1 2 3 + 1)( 3 − 1) ( 3) −1 3 +1 1 1 × (2 + 3 ) 2+ 3 2+ 3 ∴ = = = = = 2 + 3. 4−3 3 − 1 2 − 3 (2 − 3 )(2 + 3 ) 2 − ( 3 ) 3 −1
2
3 −1
2
2
2
3 −1 3 +1
+
3 +1 3 −1
(
) (
2
)
= 2− 3 + 2+ 3 = 4= 4+0 3 .
∴ x+ y 3 = 4+0 3 ∴ x = 4, y =0 ∴ x2 + y2 = 16 + 0 = 16.
27
3
= 2− 3.
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3+ 2
1 , find x + . x 3− 2
Example 19: If x = Solution: x =
3+ 2 3− 2
( (
)( 3 + 2 ) 3 − 2 )( 3 + 2 ) ( 3 + 2) ( 3) − ( 2 )
=
3+ 2
2
=
2
3+ 2
=
∴
1 1 = = x 3+ 2 =
∴ x+
1 = x
(
3+
(
2
=
3− 2 1 1 3− 2 × 3+ 2 3− 2
( ) (
3− 2
2
3+ 2 2 3−2 2
= 3 + 2.
) )
=
( 3) − ( 2 ) 2 )+ ( 3 − 2 ) = 2 2
Example 20: If a =
3+ 2
3− 2 = 3−2
3− 2 = 3− 2. 1
3.
, find the value a2 (a− 6)2.
Solution:
(3 + 2 2 ) × (3 + 2 2 ) = (3 + 2 2 ) = 3 + 2 2 = 3 + 2 2 . a= (3 − 2 2 ) (3 + 2 2 ) 3 − 4( 2 ) 9 − 8 ∴ a − 6 = (3 + 2 2 ) − 6 = −3 + 2 2. ∴ a (a − 6) = (3 + 2 2 ) (− 3 + 2 2 ) = [(2 2 + 3)(2 2 − 3)] = ⎡(2 2 ) − 3 ⎤ = (8 − 9 ) = 1. ⎢⎣ ⎥⎦ 2
2
2
2
2
2
2
2
2
2 = 1.414…., find an approximate value of
Example 21: If Solution:
2
2
2 +1 2 −1
=
2 +1 2 −1
2 +1
×
2 +1
( 2 + 1) = ( 2 + 1) = ( 2 + 1) . = ( 2) −1 2 −1 ( 2 + 1) = 2 + 1 = 1.414… + 1 = 2.414… 2
2
2
2
∴
2 +1 2 −1
=
2
2
28
2
2 +1 2 −1
.
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Exercise 1.2 1.
Answer with reasons whether the following are surds or not: 3
(i) 2.
16
(ii)
(5
(iii)
32
)(
2 −2 3 4 3+3 2
(v)
3
5 5
(vii)
5
4 ÷ 15 12
729
(iv)
4
2
4
2
(v)
50
2
(ii)
)
(iv) (vi)
(viii)
4
3
(
320 − 3 40 20 − 12
3 ÷
3 ÷3 5
3
)(
75 + 45
)
6 25 ÷ 5
(ix)
3
2,
5,
Arrange in ascending order: (i)
4.
5
(iii)
Simplify each of the following to the simplest form: (i) 5 2 +
3.
2 4
3
3,
5 , 6 11
(ii)
5,
3
7,
4
9
(iii)
3
4
3
Rationalize the denominator in each of the following: (i)
18
(ii)
6 1
(iv)
(v)
3+ 5
4
(iii)
1+ 2 3 2− 3
(vi)
2+ 3
5 +1 5 −1 3 2 −2 3 3 3+2 2
5.
Find x and y in each of the following: 2+ 3 5+4 3 (i) =x+y 3 (ii) = x+ y 3 2− 3 4+5 3 3+ 5 3− 5 5− 7 5+ 7 (iii) + = x+ y 5 (iv) − = x + y 35 3− 5 3+ 5 5+ 7 5− 7
6.
If a =
7.
If a =
8.
If
3 = 1.732…, find an approximate value of
9.
If
2 = 1.414… and
2+ 3 2− 3 2 +1 2 −1
, find the value of a2 (a − 4)2.
, find the value of a2 +
1 . a2
3 +1 3 −1
.
3 = 1.732…, find an approximate value of
29
1 2+ 3
.
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Answers Exercise 1.1 1. 2. 3.
(i) 0.328125 (ii) − 1.83 (iii) − 0.35 (iv) 0.4583 7 7 125 86 9 (ii) − (iii) (iv) (v) (i) 20 9 999 165 1 (i) F (ii) T (iii) T (iv) T (v) T (vi) F
(vii) F
Exercise 1.2 1. 2.
(i) surd
4.
(i)
6
3 4
(vii) 3
11 ,
5, 3
(i) 3 6 (iv)
5.
6
1 2
(
5− 3
(i) x = 7, y = 4 (iii) x = 7, y = 0
6. a2 (a − 4)2 = 1
)
15
(iv) surd
(iii) 14 6 + 6
(ii) 23 5
(i) 9 2 (vi)
3.
(ii) not a surd (iii) surd
16 3
(viii) 3
(ii)
4
(ii)
4 2 3 −1 11
9,
7,
(
5
)
(v) 2 6 − 5
(iii)
3
2,
(
4
9. 0.318
(v)
5
3,
5.
)
1 3+ 5 2 1 (vi) 13 6 − 30 19
(iii)
9 40 , y= 59 59 (iv) x = 0, y = 2
30
6
(ix)
(ii) x =
7. 34 8. 3.732
(iv) 4 15
27 625
12
(v) not a surd
(
)
6
3125
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2. MEASUREMENTS We do measurements in our routine life in several situations. For example, we measure the length of a cloth for stitching, the area of a wall for white washing, the perimeter of a land for fencing and the volume of a container for filling. Based upon the measurements, we do further calculations according to our needs. The branch of mathematics which deals with the measurement of lengths, angles, areas, perimeters and volumes of plane and solid figures is called mensuration. In our earlier classes, we have learnt about the areas and perimeters of some plane geometrical figures such as triangles, quadrilaterals and circles. (All geometrical figures are drawn in a plane). In this chapter, we shall study about some combinations of plane figures which are obtained by placing two or more triangles, quadrilaterals or circles in juxta position. As all figures we consider lie in a plane, we shall call a plane figure, simply a figure.
2.1
Area and Perimeter We recall the formulae for the perimeters and areas of plane geometrical figures.
2.1.1
Rectangle Area = l × b sq.units Perimeter = 2 (l + b) units d=
2.1.2
Figure 2.1
l 2 + b 2 units.
Parallelogram: Area = b × h sq.units Perimeter = 2(a + b) units.
2.1.3
Figure 2.2
Triangle with a given base and height: Area =
1 b × h sq.units 2
Figure 2.3
31
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2.1.4
Right triangle: 1 Area = b × h sq.units 2 Perimeter = b + h + d units d=
2.1.5
2.1.6
Figure 2.4
b 2 + h 2 units.
Equilateral triangle: 3 a units altitude = h = 2 3 2 Area = a sq. units 4 Perimeter = 3a units.
Figure 2.5
Isosceles triangle: Area = h a 2 − h 2 sq. units
(
Figure 2.6
)
Perimeter = 2 a + a 2 − h 2 units. 2.1.7
Scalene triangle:
s( s − a)(s − b)(s − c) sq. units a+b+c where s = units 2 Perimeter = a + b +c units.
Area =
2.1.8
Trapezium: Area =
2.1.9
Figure 2.7
1 (a + b) × h sq. units. 2
Figure 2.8
Quadrilateral: 1 Area = d × (h1 + h2) 2 sq.units.
Figure 2.9
2.1.10 Rhombus: 1 Area = d1 × d2 sq. units 2 Perimeter = 2 d1 + d 2 2
2
Figure 2.10
= 4a units.
32
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2.1.11 Circle: Area of the circle = πr2 sq. units Perimeter of the circle = 2πr units 1 Area of a semicircle = πr2 sq. units 2 Arc length of the semicircle = πr units 1 Area of a quadrant circle = πr2 sq. units 4 1 Arc length of a quadrant circle = πr units. 2
Figure 2.11
Note: A line segment joining the points A and B is denoted by AB or AB. We shall also use AB to denote the length of AB . Example 1: A wall in the form of a rectangle has base 15m and height 10m. If the cost of painting the wall is Rs. 16 per square metre, find the cost for painting the entire wall. Solution: Let b = 15 and h = 10. Then the area of the rectangle = b × h = 15 × 10 = 150 sq. metres. Since the cost of painting 1sq. metre is Rs. 16, the cost for painting the entire wall = 16 × 150 = Rs. 2400. Figure 2.12 Example 2: The dimensions of a rectangular metal sheet are 4m × 3m. The sheet is to be cut into square sheets each of side 4 cm. Find the number of square sheets. Solution: Area of the metal sheet = 400 × 300 = 12,0000 cm2. Area of a square sheet = 4 × 4 = 16 cm2. 12,0000 ∴ No. of square sheets = = 7500. 16 Figure 2.13 Example 3: Find the base of a parallelogram if its area is 40 cm2 and altitude is 15 cm. Solution: Area = b × h. ∴ 40 = b × 15. 40 8 ∴ b= = . 15 3 8 ∴ Base = cm. 3 Figure 2.14
33
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Example 4: If the lengths of the sides of a triangle are 11 cm, 60 cm and 61 cm, find the area and perimeter of the triangle. Solution: Area = s ( s − a)(s − b)(s − c) . Here 2s = a + b + c = 11 + 60 + 61 = 132. ∴ s = 66, s − a = 66 − 11 = 55, s − b = 66 − 60 = 6, s − c = 66 − 61 = 5. ∴ Area = 66 × 55 × 6 × 5 = 330 sq.cm. Perimeter = a + b + c = 11 + 60 + 61 =132 cm.
Figure 2.15
Example 5: Find the area of the quadrilateral ABCD given in Figure 2.16. 1 1 Solution: Area = d (h1 + h2 ) = × 50 × (10 + 20) 2 2 = 25 × 30 = 750 m2 Figure 2.16 Example 6: Find the area of the trapezium ABCD given in Figure 2.17 1 Solution: Area = (a + b) × h 2 1 = (12 + 5) × 4 2 = 34 sq. units. Figure 2.17 Example 7: Cost of levelling a land is Rs. 12 per square metre. A land is in the form of a trapezium whose parallel sides are of lengths 18m and 12 m. If its other two sides are each of length 5m, find the total cost incurred in levelling the land. Solution: ABCD is the given trapezium where AB = 18m, CD = 12 m, AD = BC = 5 m. Draw CE parallel to DA (see Figure 2.18). ∆EBC is isosceles whose height h = 5 2 − 3 2 = 16 = 4cm. Now, the area of the trapezium ABCD 1 1 (a + b) × h = (18 + 12) × 4 = 2 2 = 2 × 30= 60 sq. metre. The cost of leveling 1 sq. metre is Rs. 12. So the cost of levelling the entire land = 60 × 12 = Rs. 720.
34
Figure 2.18
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Example 8: The perimeter of a rhombus is 20 cm. One of the diagonals is of length 8 cm. Find the length of the other diagonal and the area of the rhombus. Solution: Let d1 and d2 be the lengths of the diagonals. Then perimeter = 2 d1 + d 2 . But the perimeter is 2
20 cm. ∴ 2 d1 + d 2 2
2
2
= 20 cm or d 1 + d 2 = 100. Here 2
2
one of the diagonals is of length 8 cm. Take d1 = 8. Then 64 + d22 = 100 or d22 = 36. ∴ d2 = 6 cm. The area of the 1 1 × 8 × 6 = 24 cm2. rhombus is d1 × d2 = 2 2
Figure 2.19
Example 9: A wire of length 264 cm is cut into two equal portions. One portion is bent in the form of a circle and the other in the form of an equilateral triangle. Find the ratio of the areas 22 ) enclosed by them.(use π ≈ 7 264 Solution: Perimeter of the circle = =132 cm. 2 But perimeter of the circle = 2πr. 22 ∴2× × r = 132 or r = 21 cm. 7 22 × 21 × 21 = 1386 cm2. ∴ Area of the circle = πr2 = 7 Perimeter of the equilateral triangle = 3a But perimeter = 132 cm.∴ 3a = 132 or a = 44 cm. ∴ Area of the equilateral triangle =
Figure 2.20
3 × a2 4
3 × 44 2 =484 3 cm2 4 ∴ The ratio of the area of circle to that of the equilateral triangle
=
= 1386 : 484 3 = 21 3 : 22
Figure 2.21
35
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Exercise 2.1 1.
2.
3. 4. 5. 6. 7.
8.
2.2
Find the area of a triangle when (i) base length = 18 cm, height = 3 cm. (ii) the three sides are of lengths 20 cm, 48 cm and 52 cm. (iii) the triangle is equilateral with side length = 8 cm. Find the area of the quadrilateral ABCD where the diagonal AC is of length 44 cm and the lengths of the perpendicular from B and D to AC are 20 cm and 12 cm respectively. Find the area of the quadrilateral one of whose diagonals is of length 15 cm and the lengths of the altitudes to this diagonal are 3 cm and 5 cm. The perimeter of a rhombus is 260m and one of its diagonals is of length 66m. Find the length of the other diagonal and also find the area of the rhombus. If the area of a parallelogram is 612 cm2 and the height is 18 cm, find the base length. The area of a trapezium whose parallel sides have lengths 7 cm and 8 cm is 30 cm2. Find the distance between the parallel sides. The distance between the parallel sides of a trapezium is 5 cm and the length of one of the parallel side is 8 cm. If the area of the trapezium is 45 cm2, find the length of the other parallel side. Find the perimeter of the circular land whose area is that of a rectangular land of dimensions 22 cm × 14 cm.
Combined Figures Consider a quadrilateral ABCD (see Figure
2.22). Join the line segment BD . Now, the quadrilateral is divided into two triangles ABD and BCD. The two triangles have the side BD in Figure 2.22 common. Looking in the reverse order, the two triangles ∆ABD and ∆BCD are put in juxta position with BD as the common side and the quadrilateral ABCD is obtained. Thus ABCD is the combination of two triangles or ABCD is a combined figure. Similarly, a trapezium is a combined figure obtained by placing a rectangle Figure 2.23 and two right triangles in juxta position (see Figure 2.23). We observe that two figures can be placed in juxta position if one has a side equal in length to a side of the other.
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Some combined figures are given in Figure 2.24 to Figure 2.35.
Figure 2.25
Figure 2.24
Figure 2.26
Figure 2.27 Figure 2.28
Figure 2.29
Figure 2.30
Figure 2.31 Figure 2.32 37
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Figure 2.33 Figure 2.34
Figure 2.35
The sides that are in juxta position are shown by dotted lines. We can easily identify the figures that are combined in Figure 2.24 to Figure 2.35. For example, Figure 2.25 is the combination of a triangle and a semicircle. It looks like the vertical cross section of a top. Figure 2.27 is the combined form of a rectangle and a semicircle. It can be viewed as a rectangular window surmounted by a semicircle. In Figure 2.29, we have the combination of a rectangle and two quadrant circles. Figure 2.33 is the combined figure of a rectangle, a triangle and a trapezium. It looks like a rocket. Since the combined figures are the combination of triangles, quadrilaterals and circles, their perimeters and areas can be calculated by applying the formulae that we have already learnt in our previous classes. We now consider two important combined plane figures namely trapeziums and polygons. 2.2.1
Trapezium
A trapezium is a four-sided plane figure in which two sides are parallel (see Figure 2.36) Consider the trapezium ABCD where AB and DC are parallel. Let AB = a and CD = b. Let h be the distance between the parallel sides. We can consider the trapezium ABCD as the combined figure of the triangles ABC and ACD. For the triangle ABC, the
Figure 2.36
side, AB is the base and h is the height. For the triangle ACD, CD is the base and h is the height. So, 1 1 Area of ∆ABC = × a × h and Area of ∆ACD = ×b×h 2 2 1 1 1 ∴Area of the trapezium = × a × h + × b × h = (a + b)h sq. units. 2 2 2
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2.2.2
Polygon
A polygon is a plane figure formed by n line segments. We observe that combined figure of several triangles is a polygon. If the sides and angles of a polygon are equal, then the polygon is known as a regular polygon. A regular polygon of six sides is called regular hexagon. In a regular hexagon, all the six sides are equal and all the included angles are equal to 120° (see Figure 2.37). As this particular regular polygon is quite often used we shall derive its Figure 2.37 perimeter and area. Let ABCDEF be a regular hexagon. Then the sides AB, BC, CD, DE, EF and FA are of equal length. Let each side be of length a units. Then the perimeter of the regular hexagon is a + a + a + a + a + a = 6a units. We shall now derive a formula for the area of the regular hexagon. The diagonals AD , BE , CF meet at a point, say O. Then the triangles OAB, OBC, OCD, ODE, OEF, OFA are equilateral triangles. So, the area of each triangle is area of the regular hexagon = 6 ×
3 2 a . Hence the 4
3 2 3 3 2 a = a sq. units. 4 2
Example 10: Find the area of Figure 2.38 Solution: The figure ABCDE is the combination of ABDE and BCD with the side BD in juxta position. ABDE is a trapezium where the parallel sides AE and BD have Figure 2.38 lengths 10 cm and 16 cm respectively. The distance between the parallel sides is 9 cm. So, the area of the trapezium ABDE is 1 1 ( a + b) × h = (10 + 16) × 9 = 117cm 2 . 2 2 BCD is a triangle whose base BD is of length 16 cm and height 8 cm. So its area is 1 1 b × h = × 16 × 8 = 64cm 2 . 2 2 ∴ The area of the combined figure ABCDE is Area of ABDE + Area of BCD= 117+64 = 181 cm2.
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Example 11: A surveyor has sketched the measurements of a land as below.
Figure 2.39
Find the area of the land. Solution: Let P, Q, R, S, be the surveyors marks from A to D. Then AP = 5m, AQ = 7m, AR = 12m, AS = 15m, AD = 17m, BP = 10m, FQ = 8m, CR = 8m, ES = 9m. The given land is the combination of the trapeziums. PRCB, FESQ and triangles AQF, APB, DSE and CRD (see Figure 2.40). Area of the trapezium PRCB: The parallel sides are BP and CR and height is PR. We have BP = 10m, CR = 8m, and PR = AR − AP = 12 − 5 = 7m. So, the area of PRCB is 1 2
(BP + CR) × PR = 1 (10 + 8) × 7 = 63 m2 2
Figure 2.40 Area of the trapezium QFES: The parallel sides are ES and FQ and height is QS. We have ES = 9m, FQ = 8m and QS = AS − AQ = 15 − 7 = 8m. So, the area of QFES is 1 (ES + FQ) × QS = 1 (9 + 8) × 8 = 17 × 4 = 68 m2 2
2
Area of the triangle AQF = 1 × AQ × FQ = 1 × 7 × 8 = 28 m2 Area of the triangle DSE =
Area of the triangle CRD =
2 1 2
2 × DS × ES = 1 ×(AD − AS) × 9 2 = 1 (17 − 15) × 9 = 1 2 2 1 × RD × CR = 1 × (AD − AR) × 8 2 2
× 2 × 9 = 9 m2.
= 4 × (17 − 12)= 4 × 5 = 20 m2. Area of the triangle APB = 1 × AP × BP = 1 × 5 × 10 = 25m2 2
2
∴ Area of the land = 63 + 68 + 28 + 9 + 20 + 25 = 213 m2.
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Example 12: Find the area of the design as shown in Figure 2.41. (π ≈ 22 ) 7
Figure 2. 41 Solution: We observe that the plot is a combination of the rectangle ABDE, the semi-circle AFE and the equilateral triangle BCD. The area of the rectangle ABDE = 20 × 14 = 280 cm2. The area of the semicircle AFE = 1 π × r2 = 1 × 22 × 7 × 7 = 77 cm2 2
The area of the equilateral triangle BCD =
2 3 4
7
a = 3 × 14 × 14 = 49 3 cm2. 2
4
∴ The area of the plot = 280 + 77 + 49 3 = 357 + 49 × 1.732 = 357 + 84.868 = 441.868 cm2 Example 13: Find the area of the design as in Figure 2.42. (Take π≈ 22 ) 7
Figure 2.42 Solution: We observe that the plot is the combination of the rectangle ABCD, the semi-circle CDE and the quadrant circles AFD, BCG. The area of the rectangle ABCD = 12 × 4 = 48 cm2. The area of the semi-circle CDE = 1 π × 6 × 6 = 22 × 3 × 6 = 396 = 56 4 cm2. 2
7
7
7
4 The area of the quadrant circle AFD = 1 π × 4 × 4 = 22 × 4 = 88 = 12 cm2. 7 4 7 7 4 The area of the quadrant circle BCG = 12 cm2. 7 4 4 4 12 5 ∴ The area of the given plot = 48 + 56 + 12 + 12 = 128 = 129 cm2. 7 7 7 7 7
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Example 14: Find the area enclosed by Figure 2.43
Figure 2.43 Solution: The figure is the combination of the rectangle CDFG, the semi circle DEF and the trapezium ABCG. The area of the rectangle CDFG = 28 × 13 = 364 cm2. The area of the trapezium ABCG = 1 (36 + 28) × 14 = 64 × 7 = 448 cm2. The area of the semi-circle DEF =
2 1 × 22 2 7
× 14 × 14 = 22 × 14 = 308 cm2
∴ The area of the given design = 364 + 448 + 308 = 1120 cm2. Sometimes, we come across plane figures which are obtained by cutting out and removing plane figures from a bigger one. Their areas can be found as before but instead of summing up the smaller areas, we subtract the areas of the removed parts from the area of the bigger figure. Area of a circular ring A circular ring is the region in between two concentric circles (see Figure 2.44). The area of the ring is equal to the area of the outer circle minus the area of the inner circle; that is, π R2 − π r2 or π (R2 − r2).
Figure 2.44
∴ The area of the semi-circular ring is 1 π (R2 − r2) sq. units. 2
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Example 15: Find the area of the shaded region in Figure 2.45, where the boundaries of the region are quadrants of a circle. (Take π ≈ 22 ). 7
Figure 2.45. Solution: The given region is that which remains after cutting away four equal quadrants each of radius 14 cm from a square of side 28 cm. The area of the square = 28 × 28 = 784 cm2. The area of one quadrant circle = 1 π × 14 × 14 = 1 × 22 × 14 × 14 = 154 cm2. 4
4
7
∴ Required area = 784 − 4(154) = 784 − 616 = 168 sq. cm. Example 16: A running track of 7m wide is as shown in Figure 2.46. The inside perimeter is 720m and the length of each straight portion is 140m. The curved portions are in the form of semi-circles. Find the area of the track. (use π ≈ 22 ) 7
Figure 2.46 Solution: Let r be the radius of the inner semicircles. Then the inside perimeter is 2 × 140 + 2 × (π × r) or 280 + 2πr. But this is given as 720m. ∴ 280 + 2πr = 720 or 2πr = 440 or r = 440 = 440 × 7 = 70 m. 2π
2 × 22
So the radius of the inner semicircle r = 70m. ∴The radius of the outer semicircle R = 70 + 7 = 77m. Now the area of the running track is equal to the sum of the areas of the semicircular tracks and the areas of the rectangular tracks. But, the area of one semi-circular track = 1 π (R2 − r2) = 1 × 22 (772 − 702) = 11 × 147 × 7 = 1617 sq.m 2
2
7
7
The area of one rectangular track = 140 × 7 = 980 sq. m. ∴ Area of the track = 2 × 1617 + 2 × 980 = 3234 + 1960 = 5194 sq.m. 43
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Example 17: A cow is tied up for grazing at one outside corner of a square building of length 4.2m. If the length of the rope is 4.9 m, find the area the cow can graze. Solution: The cow is tied up at the corner point A of the square (see Figure 2.47). The rope is of length 4.9m and the side wall is of length 4.2m. As the cow cannot cross the wall, its rope can go upto D and G at the corners B and E of the square. Thus, the cow can graze the 3 th of a 4
circular region of radius 4.9m and two quadrant circular regions of radius 4.9 − 4.2 = 0.7m.
∴ Area, the cow can graze = 3 × π × 4.9 × 4.9 + 2 × 1 × π × 0.7 × 0.7 = =
4 4 3 × 22 × 4.9 × 4.9 + 1 × 22 × 0.7 × 0.7 4 7 2 7 33 × 0.7 × 4.9 + 11 × 0.1 × 0.7= 56. 595 + 2
Figure 2.47
0.77 = 57.365m2.
Example 18: Find the area of the shaded portion in Figure 2.48 (Take π≈ 22 ) 7
Figure 2.48 Solution: The area of the shaded portion is equal to the area of the semicircle of radius 14 cm minus the area of the semicircle of radius 7 cm. That is, 1 × π × (14)2 − 1 × π × (7)2 or
1 2
×
2 22 7
2
× 14 × 14
−1 2
× 22 × 7 × 7 = 11 × 2 × 14 − 11 × 7 7
= 308 − 77 = 231 cm2.
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Exercise 2.2 1.
From each of the following notes in the field book of a surveyor, make a rough plan of the field and find its area (i)
(ii)
Figure 2.49
Figure 2.50
(iii)
Figure 2.51
2.
A play ground is to be constructed with two straight segments and two semicircular segments as shown in Figure 2.52. The radius of each semicircular segment is 21m. The length of each of the straight segment is 85m. Find the area of the playground. (Take π ≈
3.
Figure 2.52
22 ) 7
A trapezium has parallel sides of lengths 22 cm and 12 cm. Find its area, if the other two sides are each of length 10 cm.
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If the area of a regular hexagon is 150 3 cm2, find the side.
4.
5. Find the area of the shaded region is Figure 2.53.
Figure 2.53 6. Find the area of the shaded region in the following figures: (i)
(ii)
Figure 2.55
Figure 2.54 (iii)
(iv)
Figure 2.57
Figure 2.56 7.
A circle has diameter 54 cm. One of its diameter is AB . C is a point on the line segment AB such that BC = 10 cm. A circle is drawn on AC as diameter. Find the area included between them. Take π ≈ 22 . 7
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8.
Find the area and perimeter of the shaded portion in Figure 2.58.
Figure 2.58 9. Four cows are tied to the four corners of a square plot of side measuring 14 m. so that each can reach just two of the other cows. These cows eat the grass inside the plot within their range. Find what area of the plot is left ungrazed. Take π ≈ 22 . 7
10.
ABCD is a rectangular plot of dimensions 36m × 24m. 4 horses are tied to the four corners of the plot, each with a rope of length 10m. Each horse reaches as far as possible for grazing. Find the area of the portion of the plot which is left ungrazed. Take π ≈ 22 . 7
Answers Exercise 2.1 1.
(i) 27 cm2
(ii) 480 cm2
(iii) 27.71 cm2
2.
704 cm2
3. 60cm2
4. 112 m, 3696 m2
5.
34 cm
6. 4 cm
7. 10 cm
8.
62.22cm Exercise 2.2
1.
(i). 27,200 sq.m
2.
4,956 sq.m
5.
140 m2
6.
(i) 36.33 sq.cm
7.
770 cm2
(ii). 15,100 sq.m
3. 147.22 cm2 (ii) 25m2
(iii). 7,525 sq.m
4. 10cm (iii) 37.71cm2
8. 354.37cm2, 94cm
9. 42m2
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(iv) 240.28 cm2 10. 549.71m2
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3. SOME USEFUL NOTATION 3.1
Scientific Notation
In subjects, such as astronomy, physics, chemistry, biology and engineering, we come across very large numbers and very small numbers. For example, we have, (i) the distance of sun from earth is about 92,900,000 miles. (ii) the average cell contains about 200, 000,000,000,000 molecules. (iii) the life-time of an elementary particle is 0.000000000251 seconds. (iv) the diameter of an electron is about 0.000000000004 centimeter. Such numbers are not so easy to write and manipulate in the decimal form. However, they can be written and manipulated easily using the laws of indices. We recall the laws of indices which we have already learnt in our earlier classes. If m is a natural number and a is a real number, then am means the product of m numbers each equal to a; that is, am = a × a ×…. m factors. Here a is called the base and m, the power or exponent or index. The notation am is read as a to the power m or a raised to m. For example, a5 = a × a × a × a × a. The laws of indices are given below: (i) am × an = a m+ n (Product law) am (ii) = a m − n , a ≠ 0, m > n (Quotient law) an (iii) (am)n = amn (Power law) (iv) am × bm = (a × b)m (Combination law) 1 When a ≠ 0, we denote m as a− m and define a0 = 1. a Now, using the laws of indices, any positive real number can be written in the form a × 10n, where 1 ≤ a < 10 and n is an integer. For example, (i) 7.32 = 7.32 × 100 (ii) 11.2 = 1.12 × 10 =1.12×101 (iii) 226 = 2.26 × 100 = 2.26 × 102 (iv) 6435.7 = 6.4357 × 1000 = 6.4357 × 103 (v) 92900000 = 9.29 × 10000000 = 9.29 × 107 2.56 (vi) 0.256 = = 2.56 × 10−1 10 7.86 (vii) 0.00786 = = 7.86 × 10−3 1000 (viii) 0.000000537 = 5.37 × 10−7 (ix) 0.0000000279 = 2.79 × 10−8
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Here after, by a number, we shall mean a positive number only. We again mention that, when a number is written in scientific notation a × 10n, the integral part of the number, a is a digit from 1 to 9 and the power of 10 is an integer (positive, negative or zero). We also observe that while converting a given number into the scientific notation, if the decimal point is moved r places to the left, then this movement is compensated by the factor 10r; and if the decimal point is moved r places to the right, then this movement is compensated by the factor 10− r .
When very large or very small numbers are put in the scientific notation, they can be multiplied or divided easily in this form. Example 1: Write the following numbers in scientific notation: (i) 7493 (ii) 105001 (iii) 3449099.93 (iv) 0.00567 (v) 0.0002079 (vi) 0.000001024 Solution:
Example 2: Write the following numbers in decimal form: (i) 3.25 × 105 (ii) 1.86 × 107 (iii) 9.87 × 109 (iv) 4.02 × 10−4 (v) 1.423 × 10−6 (vi) 3.25 × 10−9.
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325 × 105 = 325 × 105 − 2 = 325 × 103 = 325000 2 10 186 7 (ii) 1.86 × 10 = 2 × 107 = 186 × 107−2 = 186 × 105 = 18600000. 10 987 (iii) 9.87 × 109 = 2 × 109 = 987 × 109 − 2 = 987 × 107 = 9870000000. 10 402 (iv) 4.02 × 10−4 = 2 × 10−4 = 402 × 10−4− 2 = 402 × 10−6 = 0.000402. 10 1423 (v) 1.423 × 10−6 = × 10−6 = 1423 × 10−6 − 3 = 1423 × 10−9 =0.000001423. 10 3 325 (vi) 3.25 × 10−9 = 2 × 10−9 = 325 × 10−9 − 2 = 325 × 10−11 =0.00000000325. 10
Solution: (i) 3.25 × 105 =
Example 3: Perform the calculation and write the answer of the following in scientific notation. (i) (3000000)3 (ii) (4000)5 × (200)3 (iii) (0.00005)4 (iv) (2000)2 ÷ (0.0001)4 Solution : (i) 3000000 = 3.0 × 106. ∴ (3000000)3 = (3.0 × 106)3 = (3.0)3 × (106)3 = 3 × 3 × 3 × 106×3 = 27 × 1018 = 2.7 × 10 × 1018 = 2.7 × 1019. (ii) 4000 = 4.0 × 103, 200 = 2.0 × 102 ∴ (4000)5 × (200)3 = (4.0 × 103)5 × (2.0 × 102)3 = (4.0)5 × (103)5 × (2.0)3 × (102)3 = 1024 × 103×5 × 8 × 102×3 = 1024 × 1015 × 8 × 106 = 8192 × 1021= 8.192 × 103 × 1021= 8.192 × 1024. (iii) 0.00005 = 5.0 × 10−5 ∴ (0.00005)4 = (5.0 × 10−5)4 = (5.0)4 × (10−5)4 = 625 × 10−5×4 = 625 × 10−20 = 6.25 × 102 × 10−20 = 6.25 × 10−18.
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(iv) 2000 = 2.0 × 103, 0.0001 = 1.0 × 10−4 ∴ (2000)2 ÷ (0.0001)4 = =
(2.0 × 10 3 ) 2 (2.0) 2 × (10 3 ) 2 4 × 10 3×2 = = (1.0 × 10 − 4 ) 4 (1.0) 4 × (10 − 4 ) 4 1 × 10 − 4×4 4 × 10 6 = 4 × 10 6−( −16 ) = 4 × 10 22 . 10 −16 Exercise 3.1
1.
Represent the following numbers in the scientific notation: (i) 29980000000 (ii) 1300000000 (iii) 1083000000000 (iv) 4300000000 (v) 9463000000000000 (vi) 534900000000000000 (vii) 0.0037 (viii) 0.000107 (ix) 0.00008035 (x) 0.0000013307 (xi) 0.00000000011 (xii) 0.0000000000009
2.
Write the following numbers in decimal form: (i) 3.25 × 10−6 (ii) (iii) 4.132 × 10−4 (iv) (v) 3.25 × 106 (vi) 4 (vii) 4.132 × 10 (viii)
3.
Find the value of the following in scientific notation: (i) (100)3 × (40)5 (ii) (21000)2 × (0.001)4 (iii) (18000)4 ÷ (30000)2 (iv) (0.002)8 × (0.0001)3 ÷ (0.01)4 2 (v) (120000) × (0.0005) ÷ (400000)
3.2
Notation of logarithm
4.02 × 10−5 1.432 × 10−3 4.02 × 105 1.432 × 103
John Napier, an English mathematician introduced the notation of logarithm as a mathematical device to do calculations easily and quickly. The word logarithm is derived from two Greek words ‘logos’ and ‘arithmos’. The word logos means reckoning and arithmos means number. Thus logarithm means reckoning number. To introduce the notation of logarithm, we shall first know about exponential notation. 3.2.1 Exponential notation Let a be a positive number. We have already introduced the notation ax ,where x is an integer. p When x is a rational number, say with p, an integer and q, a positive integer, we define ax q p ⎛q ⎞ x by a = ⎜ a ⎟ ⎝ ⎠ 8 3
For example, 5 =
( 5) 3
8
,7
−4 11
=
( 7) 11
−4
.
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When x is an irrational number, ax can be defined to represent a real number. But the definition requires some advanced topics in mathematics. Although it is not required in our standard, we accept now that, for any a > 0, ax can be defined and that it represents an unique real number u and write u = ax. In this situation, we say that the real number u is written in the exponential form or in the exponential notation. Here the positive number a is called the base and x, the index or the power or the exponent. The laws of indices which we have stated for integer exponents can be obtained for all real exponents. We state them here and call them, the laws of exponents: (i) a x × a
y
=a
x+ y
ax x− y =a y a y xy (iii) ⎜⎛ a x ⎞⎟ = a ⎝ ⎠ (ii)
(iv) a − x =
1 ax
(v) a x × b x = (a × b ) x (vi) a 0 = 1
Now we are in a position to introduce the notation of logarithm. 3.2.2
Logarithmic notation
Let b be a positive number and b ≠ 1. We have already understood that, for any real number x, bx represents a unique real number, say a. If we write a = bx, then the exponent x is called the logarithm of a to the base b. We also call x, the value of log a. Thus, x = log a b b x is an equivalent form of a = b . We say that x = log a is the logarithmic form of the b x exponential form a = b . In both the forms, the base is same. We observe that x = log a is an b x equivalent way of writing a = b . The notation x = log a is called the logarithmic notation b x and it means the equation a = b . For example, (i) 3 = log 9 729 is equivalent to 93 = 729; 1
1 (ii) = log 8 2 is equivalent to 8 3 = 2; 3 (iii) −3 = log 10 0.001 is equivalent to 10−3 = 0.001;
(iv) 2 = log 7 49 is equivalent to 72 = 49; 1
1 = log 9 3 is equivalent to 9 2 = 3 or 9 = 3 ; 2 −3 3 1 ⎛1⎞ (vi) − = log 4 ⎜ ⎟ is equivalent to 4 2 = . 2 8 ⎝8⎠
(v)
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Note: The base must be specified in logarithmic notation. If we write y = log x, then it is meaningless since its equivalent can not be written unless the base is given. However, in some situations, we write logarithms, omitting their bases. In such cases, it is understood that all logarithms have the same base. Example 4: Change the following from logarithmic form to exponential form: 1 1 ⎛1⎞ ⎛1⎞ (ii) log 2 ⎜ ⎟ = −2 (iii) log 216 6 = (iv) log 3 ⎜ ⎟ = −2 (i) log 25 5 = 2 3 ⎝4⎠ ⎝9⎠ Solution: As base is same in both forms, 1 1 (i) log 25 5 = is equivalent to (25) 2 = 5. 2 1 ⎛1⎞ −2 (ii) log 2 ⎜ ⎟ = −2 is equivalent to (2 ) = . 4 ⎝4⎠ 1 1 (iii) log 216 6 = is equivalent to (216 ) 3 = 6. 3 1 ⎛1⎞ (iv) log 3 ⎜ ⎟ = −2 is equivalent to (3)−2= . 9 ⎝9⎠ Example 5: Change the following from exponential form to logarithmic form: −2
1 ⎛1⎞ 3 1 (i) 2 = 64 (ii) 9 = (iii) ⎜ ⎟ = 4 729 ⎝8⎠ Solution: As base is same in both forms, we have 1 1 (i) 2 = 64 6 is equivalent to = log 64 2. 6 1 ⎛ 1 ⎞ (ii) 9−3 = is equivalent to −3 = log 9 ⎜ ⎟. 729 ⎝ 729 ⎠ 1 6
−3
(iv)
1 = 7 −1 7
2
2 ⎛1⎞3 1 ⎛1⎞ (iii) ⎜ ⎟ = is equivalent to = log 1 ⎜ ⎟. 4 3 4 ⎝8⎠ 8⎝ ⎠ 1 ⎛1⎞ (iv) = 7−1 is equivalent to −1 = log 7 ⎜ ⎟. 7 ⎝7⎠ Example 6: Evaluate ⎛ 1 ⎞ (iii) log 9 ⎜ ⎟ (iv) log 3 ( 243) −1 . ⎝ 27 ⎠ x ⎛ 1 ⎞ Solution : (i) Let x = log 9 729. Then 9 = 729 = 93. (iii) Let x = log 9 ⎜ ⎟ . ⎝ 27 ⎠ ∴ x = 3. 1 1 (ii) Let x = log 4 8. Then 4x = 8 = 23 Then 9x= = 3 = 3 −3 27 3 But 4x = (22)x = 22x. x 2 x But 9 = (3 ) = 32x 3 2x 3 ∴ 2 = 2 . ∴ 2x = 3 or x = −3 . ∴ 32x = 3−3, 2x= −3 or x = 2 2
(i) log 9 729
(ii) log 4 8
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(iv) Let x = log 3 ( 243) −1 Then 3x = (243)−1 = or 3x =3−5
or
x = −5.
1 1 = 5 = 3−5 243 3
Example 7: Solve the equations: (i) log 3 x = −2 (iii) x = log ⎛ 1 ⎞ 512
(ii) log b 100 = 2 (iv) x + 2 log 27 9 = 0.
⎜ ⎟ ⎝8⎠
Solution: (i) log 3 x = −2
∴ 3−2 = x or x =
(ii) log b 100 = 2 ∴ b2 = 100 = 102. ∴ b = 10.
1 1 = . 2 3 9
x
⎛1⎞ (iii) x = log ⎛ 1 ⎞ 512. ∴ ⎜ ⎟ = 512 = 8 3 . ⎜ ⎟ ⎝8⎠ ⎝8⎠
(iv) x + 2 log 27 9 = 0 −x ∴ x = −2 log 27 9 or = log 27 9 2 ∴ (27) ∴
−x 2
= 9 or (33 )
−x 2
= 3 2 or (3)
or (8−1)x = 83 or 8−x = 83. ∴ −x = 3 or x = −3.
−3 x 2
= 32
− 3x −4 . = 2 or x = 2 3
Now we proceed to state and prove some properties of logarithms of positive numbers. All positive numbers other than 1 are considered. (i) Product rule: If a, m and n are positive numbers and a ≠1, then log ( mn) = log m + log n a a a Proof: Let log m = x and log n = y . a a Then, m = ax and n = ay. m × n = ax × ay or mn = ax+y. ∴ This is in exponential form. Writing this in the logarithmic form, we get log ( mn) = x + y a
or
log ( mn) = log m + log n . a a a
In words, the above rule states that the logarithm of the product of two positive numbers is equal to sum of the logarithms of the numbers. (ii) Quotient rule: If m, n and a are positive numbers and a ≠ 1, then, ⎛m⎞ log ⎜ ⎟ = log m − log n . a⎝ n ⎠ a a
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∴
Proof: Let log m = x and log n = y . a a x Then m = a and n = ay. m ax = y = ax−y. n a
This is in exponential form. Writing this in logarithmic form, we get ⎛m⎞ log ⎜ ⎟ = x − y or a⎝ n ⎠
⎛m⎞ log ⎜ ⎟ = log m − log n. a⎝ n ⎠ a a
In words, the quotient rule states that the logarithm of the quotient difference log m − log n . a a
m is equal to the n
(iii) Power rule: If a and m are positive numbers, a ≠ 1 and n is a real number, then log m n = n log m . a a x Proof: Let log m = x . Then m = a . Raising to the power n on both sides, we get a mn = (ax)n =ax n . This is in exponential form. Writing this in the notation of logarithm, we get log m n = nx or a log m n = n log m. a a (iv) If a is a positive number, then log 1 = 0 a Proof: Let x = log 1 . Then ax = 1 = a0. a ∴ x = 0 or log 1 = 0. a (v) If a is a positive number, then log a a = 1. Proof: Let x = log a . Then ax = a = a1. a log a = 1. ∴ x = 1 or a (vi) Change of base rule: If m, n and p are positive numbers and n ≠ 1, p ≠ 1, then log m = ⎛⎜ log m ⎞⎟ × log p . n p ⎠ n ⎝ Proof: Let x = log m and y = log p. p n x y Then p = m and n = p. Eliminating p with these equations, we get
(
(n )
y x
= m or n xy = m.
55
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This is in exponential form. Writing this in the logarithmic form, we get log n m = xy or
log n m = (log p m )× (log n p ). (vii) Reciprocal rule: If m and n are positive numbers other than 1, then 1 . log n m = log m n x 1 ×x ⎛ 1x ⎞ x x n = x . Then m = n = n = ⎜⎜ n ⎟⎟ . Proof: Let log m ⎝ ⎠ 1 x
∴ m = n . This is in exponential form. Writing this in the logarithmic form, we get 1 or log m = n x
log n m =
1 . log m n
log a (viii) If a and b are any two positive numbers and b ≠ 1, then b b = a . Proof: Let x = log a . Then bx = a. Substituting for x in this equation, we get b
b
log a b
= a.
(ix) Let m, n and a be positive numbers and a ≠ 1. If log m = log n, then m = n. a a Proof: Let x = log m .Then x = log n . a a log m ∴ ax = n or (by property (viii)). a a = n or m = n Note: We are avoiding 1 in the base of all logarithms because if we consider one such logarithm, say log1 9 with 1 in the base, then x = log1 9 would give 1x = 9. We know that there is no real number x such that 1x = 9. Caution: Some errors which are commonly committed are ⎛ m ⎞ log a m (1) log ⎜ ⎟ = , a ⎝ n ⎠ log n a (2) log (m + n) = log m + log n. a a a log m a . (1) is wrong since L.H.S. = log m − log n ≠ a a log n a (2) is incorrect since R.H.S. = log (mn) ≠ log (m + n). a a
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1 1000 Solution: (i) Since the expression is the sum of two logarithms and the bases are equal, we can apply the product rule. log 3 27 + log 3 729 = log (27 × 729)
Example 8: Simplify: (i) log 3 27 + log 3 729
(ii) log 5 8 + log 5
3
= log 3 (33 × 3 6 ) = log 3 39 = 9 × log 3 3 = 9 × 1 = 9. (ii) log 5 8 + log 5
1 1 ⎞ ⎛ = log 5 ⎜ 8 × ⎟ 1000 ⎝ 1000 ⎠ ⎛ 1 ⎞ = log 5 ⎜ ⎟ ⎝ 125 ⎠ ⎛ 1⎞ = log 5 ⎜ 3 ⎟ = log 5 (5 −3 ) ⎝5 ⎠ = (−3) × log 5 5 = (−3) × 1 = −3.
Example 9: Simplify:
(i) log 7 98 − log 7 14
(ii)
1 log 9 36 + 2 log 9 4 − 3 log 9 4 2
⎛ 98 ⎞ Solution: (i) log 7 98 − log 7 14 = log 7 ⎜ ⎟ = log 7 7 = 1. ⎝ 14 ⎠ ⎛ 12 ⎞ 1 (ii) log 9 36 + 2 log 9 4 − 3 log 9 4 = log 9 ⎜⎜ 36 ⎟⎟ + log 9 4 2 − log 9 4 3 2 ⎠ ⎝ = log 9 6 + log 9 16 − log 9 64
= log 9 (6 × 16) − log 9 64 = log 9 96 − log 9 64 ⎛ 96 ⎞ ⎛3⎞ = log 9 ⎜ ⎟ = log 9 ⎜ ⎟. ⎝ 64 ⎠ ⎝2⎠
Example 10: Prove
(i) log10 1250 = 4 − 3 log10 2
(ii) log 5 1875 =
1 1 log 5 36 − log 5 8 + 20 log 32 2. 2 3
Solution: (i) R.H.S. = 4 − 3 log10 2 = 4 − log 10 2 3 = 4 × log10 10 − log10 8
= log 10 10 4 − log 10 8 = log10 10000 − log10 8 ⎛ 10000 ⎞ = log 10 ⎜ ⎟ = log 10 1250 = L.H.S. ⎝ 8 ⎠
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1
(ii)
1
1 1 R.H.S. = log 5 36 − log 5 8 + 20 log 32 2 = log 5 (36) 2 − log 5 (8) 3 + 4 × 5 log 32 2 2 3 = log 5 6 − log 5 2 + 4 log 32 2 5 ⎛6⎞ ⎛6⎞ = log 5 ⎜ ⎟ + 4 log 32 32 = log 5 ⎜ ⎟ + 4 ⎝2⎠ ⎝2⎠ = log 5 3 + 4 log 5 5 = log 5 3 + log 5 5 4 = log 5 3 + log 5 625 = log 5 (3 × 625) = log 5 1875 = L.H .S .
Example 11: Prove that log 3 4 × log 4 5 × log 5 6 × log 6 7 × log 7 8 × log 8 9 = 2 . Solution: L.H.S. = (log 3 4 × log 4 5) × (log 5 6 × log 6 7 ) × (log 7 8 × log 8 9) = log 3 5 × log 5 7 × log 7 9 = log 3 5 × (log 5 7 × log 7 9)
= log 3 5 × log 5 9 = log 3 9 = log 3 3 2 = 2 log 3 3 = 2 × 1 = 2 = R.H.S. Example 12: Solve log10 (2 x + 50) = 3. Solution: Writing the equation in the exponential form, we get 2x + 50 = 103 = 1000 or 2x = 1000 − 50 = 950 or x = 475. − 3 log 9 2 Example 13: Find the value of 81 . − 3 log 2 9 − 3 log 9 2 ⎛ 2 ⎞ Solution: 81 = ⎜9 ⎟ ⎝ ⎠
log 2 − 6 − 6 log 9 2 log a = 9 9 =9 = 2 − 6 . (Since b b = a ) 1 1 = 6 = . 64 2 Example 14: Solve log 6 2 x − log 6 ( x + 1) = 0. ⎛ 2x ⎞ Solution: Using the quotient law, we can write the equation as log 6 ⎜ ⎟ = 0. Changing ⎝ x +1⎠ into exponential form, we get 2x = 6 0 = 1 or 2x = x + 1 or x = 1. x +1
Example 15: Solve log 3 (7 − x) − log 3 (1 − x) = 1. ⎛7− x⎞ Solution: Using the quotient law, the equation can be written as log 3 ⎜ ⎟ = 1. ⎝ 1− x ⎠ Writing in the exponential form, we get 7−x = 31 = 3 or 7 − x = 3(1 − x ) or 7 − x = 3 − 3x or 2x = −4 or x = −2. 1− x
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Example 16: Solve log 2 (log 3 x ) = 2 . Solution: Put y = log 3 x. Then the equation becomes log 2 y = 2. Writing the equation in the
exponential form, we get y = 2 2 = 4 or log 3 x = 4. Again, writing in the exponential form, we get x = 3 4 or x = 81. Example 17: Solve 2 log 5 3 × log 9 x + 1 = log 5 3 . Solution: Rewriting the equation, we get log 5 3 2 × log 9 x = log 5 3 − 1
or
⎛3⎞ log 5 9 × log 9 x = log 5 3 − log 5 5 = log 5 ⎜ ⎟. ⎝5⎠
3 ⎛3⎞ Using the change of base rule, we get log 5 x = log 5 ⎜ ⎟ . ∴ x = . 5 ⎝5⎠
Exercise 3.2.1 1. Write true or false in the following: (i) log 3 243 = 5
(ii) log 1 27 = 3 3
16 ⎛ 16 ⎞ (iii) log 2 ⎜ − 4 ⎟ = log − log 2 4. 3 ⎝ 3 ⎠ (v) log 1 a = −1
(iv) log 2 (8 − 4) = log 2 8 − log 2 4. (vi) log a (m + n) = log a m + log a n.
a
2. Obtain the equivalent logarithm form of the following:
(i) 5 −2 = 0.04. (iv) 36= 729.
⎛1⎞ (ii) ⎜ ⎟ ⎝8⎠ (v) 36
−
−
3 2
2 3
=
= 4.
(iii) 4 4 = 256.
1 216
(vi) 10 −3 = 0.001.
3. Find the value in the following: (i) log 5 625 (ii) log 6 216
(iv) log 9 (vii) 10
1 3
2 log 6 10
(v) log 1 81 (viii) 25
4. Solve for the unknown: (vi) (i) logx 0.001 = −3
(ii) log 1 x = 7
(iii) log
3
9
(vi) log 2 4 2
3 log 8 5
(ix) 9
log 2 x = 3
(vii) log 5 25 c = 4
2
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(iii) log x 100 = −2
⎛N⎞ (viii) log 3 ⎜ ⎟ = 4 ⎝2⎠
(iv) log b 125 = 3
⎛ 1 ⎞ (ix) log 10 ⎜ ⎟ =1 ⎝ 1000 ⎠
⎛1⎞ (v) log 2 ⎜ ⎟ = 5 ⎝ x⎠
(x) 2 log 9 N = 1
a
5. Choose the correct answer from the alternatives given for each of the following: (i) If 3 log x 5 = 1, then x is (A) 5 (B) 25 (C) 125 (D) 625 1 (ii) log14 + log12 144 = 196 (A) 0 (B) 1 (C) 2 (D) 3 (iii) The value of log 4 1024 = (A) 10 (B) 8 (C) 7 (D) 5 (iv) If log a 4 = 2 log a x, then x = (A) 0 (B) 1 (C) 2 (D) 3 (v) If 2 log16 x = 1, then x = (A) 4 (B) 8 (C) 16 (D) 32 (vi) If log 5 x = 2 then x =
(A) 5
(B) 25
(C) 125
(D) 625
6. Simplify the expression into a single logarithm in each of the following: (i) log10 2 + log10 9. (ii) 3 log 3 2 + 4 log 3 3 − 8.
(iii) 5 − 2 log 2 3 + 3 log 2 4 + 2 log 2 6. (iv) 2 log 4 2 + 3 log 2 5 − log 2 15 + log 2 7 . (v) 5 log10 2 + 2 log10 3 − 6 log 64 4 . (vi) log10 5 + log10 20 − log10 24 + log10 25 − 3. 7. Given log a 2 = x, log a 3 = y, log a 5 = z and log a 7 = t , find value in each of the following in terms of x, y, z and t. (i) log a 6 (ii) log a 4 (iii) log a 1.5 (iv) log a 27 1 8 (vi) log a 600 (vii) log a (viii) log a 15 (v) log a 2 3 27 14 (ix) log a 35 (x) log a 12 (xi) log a 4.9 (xii) log a 15
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8. Solve the equation in each of the following:
(i) 2 log 5 x = 3 log 5 2
(ii)
log 3 x + log 3 7 = log 3 11
(iii) x log16 8 + 1 = 0
(iv) log 4 x × log 4 16 = log 4 256
(v) log 4 ( x + 2) + log 4 3 = 2
(vi) log 3 (2 x + 1) − log 3 (2 x − 1) = log 3 4
(vii) log 3 10 x − log 3 ( x + 1) = 2 (ix) log 5 (10 + x) = log 5 (3 + 4 x) 1 (xi) log 3 10 x + 5 − = log 3 x + 1 2
(viii) log 2 (7 x + 3) − log 2 (5 x − 1) = log 2 3 − 1 (x) log 5 (5 log 3 x) = 2.
9. Prove the equation in the following:
(i) (iii) (v) (vii)
log 3 135 = 3 + log 3 5 log10 2500 = 2 + 2 log10 5 log10 25 = 2 − 2 log10 2 log16 0.000256 = 2 − 6 log16 10
(ii) log10 1600 = 2 + 4 log10 2 (iv) log10 125 = 3 − 3 log10 2 (vi) log 3 0.0027 = 3 − 4 log 3 10
10. If a, b and c are positive numbers other than one, prove that log b a × log c b × log a c = 1. 3.2.3
Common logarithms
While defining logarithms, we stressed that the logarithm of a positive number is defined only if the base is specified and the base can be any positive real number other than 1. If we choose the base as the irrational number ‘e’, then such logarithms are called Natural logarithms. If we choose the base as 10, then such logarithms are called Common logarithms. Natural logarithms were introduced by John Napier and common logarithms by his friend Henry Briggs, an English mathematician. To honour John Napier, the founder of logarithms, we denote the natural logarithm log e x simply as ln x. We will study more about ln x in higher classes. Now, we proceed to bring out the use of common logarithms in computations. We denote the common logarithm log10 x as log x, omitting the base 10. Thus, log x = y means log10 x = y and is equivalent to x = 10y. If we substitute = log x, we get y = log
1 = log 10 10 −3 = −3 . 1000
1 1 , , 1, 10, 100, 1000, …, we correspondingly get 100 10 −2, −1, 0, 1, 2, 3, ….
Similarly, for x =
61
1 for x in 1000
y
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x log10 x
10−3 −3
10−2 −2
10−1 −1
100 0
101 1
102 2
103 3
We observe that as x increases along the positive real axis, log10 x also increases. We also observe that log10 x is positive for all values of x > 1 and is negative whenever x < 1. We observe that the values of log10 x is as given in the table below: Location of log10 x
The value of log10 x
−5
−4
−5 < log10 x AB (see Figure 6.56). Proof: The lengths of AB and AC are positive numbers. So three cases arise (i) AC < AB (ii) AC = AB (iii) AC > AB Figure 6.56 Case (i) Suppose that AC < AB. Then the side AB has longer length than the side AC. So the angle ∠C which is opposite to AB is larger measure than that of ∠B which is opposite to the shorter side AC. That is, m∠C > m∠B. This contradicts the given fact that m∠B > m∠C. Hence the assumption that AC < AB is wrong. ∴AC < AB. Case (ii) Suppose that AC = AB. Then the two sides AB and AC are equal. So the angles opposite to these sides are equal. That is ∠B = ∠C. This is again a contradiction to the given fact that ∠B > ∠C. Hence AC = AB is impossible. Now Case (iii) remains alone to be true. Hence the theorem is proved.
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Theorem 24: A parallelogram is a rhombus if its diagonals are perpendicular. Given: ABCD is a parallelogram where the diagonals AC and BD are perpendicular. To prove: ABCD is a rhombus. Construction: Draw the diagonals AC and BD. Let M be the point of intersection of AC and BD (see Figure 6.57). Proof: In triangles AMB and BMC, (i) ∠AMB = ∠BMC = 90° (ii) AM = MC (iii) BM is common. ∴ By SSA criterion, ∆AMB ≡ ∆BMC. ∴Corresponding sides are equal. In particular, AB = BC. Since ABCD is a parallelogram, AB = CD, BC = AD. ∴ AB = BC = CD = AD. Figure 6.57 Hence ABCD is a rhombus. The theorem is proved. Example 3: Find the complement of the following angles: (i) 30° (ii) 45° (iii) 55° (iv) 81° Solution: Since the sum of complementary angles is 90° (i) the complement of 30° is 90° − 30° = 60°. (ii) the complement of 45° is 90° − 45° = 45°. (iii) the complement of 55° is 90° − 55° = 35°. (iv) the complement of 81° is 90° − 81° = 9°. Example 4: Find the supplement of the following angles: (i) 70° (ii) 45° (iii) 120° (iv) 155° Solution: Since the sum of supplementary angles is 180°, (i) the supplement of 70° is 180° − 70° = 110°. (ii) the supplement of 45° is 180° − 45° = 135°. (iii) the supplement of 120° is 180° − 120° = 60°. (iv) the supplement of 155° is 180° − 155° = 25°. Example 5: Find the angles in each of the following: (i) The angles are supplementary and the larger is twice the small. (ii) The angles are complementary and the larger is 20° more than the other (iii) The angles are adjacent and form an angle of 120°. The larger is 20° less than three times the smaller. (iv) The angles are vertically opposite and complementary. Solution: (i) Let the smaller angle be x°. Then the larger angle = 2x°. Since the two angles are supplementary x° + 2x° = 180° or 3x° = 180° or x° = 60°. ∴ smaller angle = 60°, larger angle = 120°. Figure 6.58
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(ii) Let x° be the smaller angle. Then the larger angle = x° + 20°. Since the angles are complementary, x° + (x° + 20°) = 90° or 2x° = 70° or x° = 35°. ∴ smaller angle = 35°, larger angle = 35° + 20° = 55°. Figure 6.59 (iii) Let x° be the smaller angle. Then the larger angle = 3x° − 20° The angles are adjacent and form an angle of 120°. ∴ x° + (3x° − 20°) = 120°. ∴ 4x° = 140° or x° = 35°. ∴ smaller angle = 35°, larger angle = 3 × 35° − 20° = 105° − 20° = 85°. (iv) Let the vertically opposite angles be x° each. Since they are complementary, x° + x° = 90° or 2x° = 90° or x° = 45°. ∴ The angles are 45° and 45°.
Figure 6.60
Figure 6.61
Example 6: In Figure 6.62 the line l3 is a transversal to the parallel lines l1 and l2. Find the angles x and y. Solution: Alternate angles are equal. ∴ x = 130°. Interior angles on the same side of the transversal are supplementary. Figure 6.62 ∴ y + 130° = 180° or y = 180° − 130° = 50°. Example 7: Find x and y in Figure 6.63 where the line l4 is a transversal to the parallel lines l1, l2 and l3. Solution: Corresponding angles are equal. ∴ x = 75°. Interior angles on the same side of the transversal are supplementary. ∴ y + 75° = 180° or y = 180° − 75° = 105°.
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Figure 6.63
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Example 8: Find the angles x and y in Figure 6.64 where the lines l1 and l2 are parallel and l3 is a transversal to l1 and l2. Solution: 4y + 92° = 180° (since the interior angles on the same side of the transversal are supplementary). ∴ 4y = 180° − 92° = 88° or y = 22°. Now, since the corresponding angles are equal, x + 2y = 92° ⇒ x + 44° = 92° ⇒ x = 92° − 44° = 48°. Figure 6.64 Example 9: If the angles of a triangle are in the ratio 3 : 4 : 5, find them. Solution: Let the angles be 3x, 4x, 5x. Then 3x + 4x + 5x = 180° or 12x = 180° or x = 15°. ∴ The angles are 3 × 15°, 4 × 15°, 5 × 15°, or 45°, 60°, 75°.
Example 10: Find the angles x and y marked in Figure 6.65. Solution: In ∆ABC, x + 65° + 90° = 180° or x + 155° = 180° or x = 180° − 155° = 25° In ∆BDC, x + y + 90° = 180° or 25° + y + 90° = 180° or y + 115° = 180° or y = 180° − 115° = 65°.
Example 11: Find x and y from the following figures: (i)
Figure 6.65
(ii)
Figure 6.66
Figure 6.67
Solution: (i) AD = BC, AB = CD. ∴ ABCD is a parallelogram. ∴ 2x = 24°, 3y = 60° (alternate angles are equal) ∴ x = 12°, y = 20°. (ii) In triangles ACD and ACB, AD = AB, CD = BC and AC is common. ∴ ∆ADC ≡ ∆ABC. ∴ Corresponding angles are equal. ∴ x + 20° = 26°, y − 5° = 42° or x = 26° − 20°, y = 42 + 5 or x = 6°, y = 47°.
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Example 12: Prove that the bisector of the vertex angle of an isosceles triangle is a median to the base. Solution: Let ABC be an isosceles triangle where AB = AC. Let AD be the bisector of the vertex angle ∠A. We have to prove that AD is the median of the base BC. That is, we have to prove that D is the mid point of BC. In the triangles ADB and ADC, Figure 6.68 we have AB = AC, m∠BAD = m∠DAC AD is an angle (bisector), AD is common. ∴ By SAS criterion, ∠ABD ≡ ∆ACD. ∴ The corresponding sides are equal. ∴ BD = DC. i.e., D is the mid point of BC. Example 13: ABC is a triangle and D is the mid point of BC. DA is drawn. If DA = DC, prove that ∠BAC is a right angle. Solution: Given DA = DC. Since D is the mid point of BC, BD = DC. The triangles ABD and ACD are isosceles. ∴∠DAB = ∠DBA (1) ∠DAC = ∠DCA (2) (1) + (2) ⇒ ∠DAB + ∠DAC = ∠DBA + ∠DCA ⇒ ∠BAC = ∠DBA + ∠DCA ⇒ ∠BAC = ∠CBA + ∠BCA (3) But ∠BAC + ∠CBA + ∠BCA = 180° (4) Figure 6.69 (4) ⇒ ∠BAC + (∠CBA + ∠BCA) = 180° ⇒ ∠BAC + ∠BAC = 180° (using (3)) ⇒ 2∠BAC = 180° ⇒∠BAC = 90°. Example 14: Prove that the sum of the four angles of a quadrilateral is 360°. Solution: Let ABCD be the given quadrilateral. We have to prove that ∠A + ∠B + ∠C + ∠D = 360°. Draw the diagonal AC. From the triangles ACD and ABC, we get ∠DAC + ∠D + ∠ACD = 180° (1) ∠CAB + ∠B + ∠ACB = 180° (2) (1) + (2) ⇒ ∠DAC + ∠D + ∠ACD + ∠CAB Figure 6.70 + ∠B + ∠ACB = 360° ⇒ (∠DAC + ∠CAB) + ∠B + (∠ACD + ∠ACB)+ ∠D = 360° ⇒ ∠A + ∠B + ∠C + ∠D = 360°.
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Example 15: ABC is an isosceles triangle with AB = AC. D is a point inside the triangle ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠A. Solution: Since ∠DBC = ∠DCB, the triangle DBC is an isosceles triangle. ∴ BD = DC. Already we have AB = AC and AD is common. So by the SSS criterion, ∆ADB ≡∆ADC. In particular, ∠BAD = ∠CAD. ∴ AD bisects ∠A.
Figure 6.71
Example 16: AD and BE are two altitudes of a triangle ABC such that AE = BD. Prove that AD = BE. Solution: In triangles ADB and AEB, we have (i) ∠ADB = ∠AEB = 90° (ii) AB is common. (iii) BD = AE. ∴ By RHS criterion, ∆ADB ≡ ∆AEB. ∴ AD = BE. Figure 6.72 Example 17: In a rectangle ABCD, E is the mid point of BC. Prove that AE = ED (see Figure 6.73) Solution: In triangles ABE and DCE, we have (i) ∠ABE = ∠DCE = 90° (ii) BE = CE (since E is the mid point of BC) (iii) AB = CD (ABCD is a rectangle) ∴By SAS criterion, ∆ABE ≡ ∆DCE. ∴AE = ED.
Figure 6.73
Example 18: In a rhombus, prove that the diagonals bisect each other at right angles. Solution: Let ABCD be a rhombus, Draw the diagonals AC and BD. Let them meet at O. We have to prove that O is the mid point of both AC and BD and that AC is perpendicular (⊥) to BD. Since a rhombus is a parallelogram, the diagonals AC and BD bisect each other. ∴ OA = OC, OB = OD. In triangles AOB and BOC, we have (i) AB = BC (ii) OB is common (iii) OA = OC Figure 6.74 ∴ ∆AOB = ∆BOC, by SSS criterion. ∴ ∠AOB = ∠BOC. Similarly, we can get ∠BOC = ∠COD, ∠COD = ∠DOA. ∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = x (say) But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°.
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∴ x + x + x +x = 360° 360° ∴ 4x = 360° or x = = 90°. 4 ∴ The diagonals bisect each other at right angles. Example 19: Prove that a diagonal of a rhombus bisects each vertex angles through which it passes. Solution: Let ABCD be the given rhombus. Draw the diagonals AC and BD. Since AB || CD and AC is a transversal to AB and CD. We get ∠BAC = ∠ACD (alternate angles are equal) (1) But AD = CD (since ABCD is a rhombus) ∴∆ADC is isosceles. ∴∠ACD = ∠DAC (angles opposite to the equal sides are equal) (2) From (1) and (2), we get Figure 6.75 ∠BAC = ∠DAC i.e., AC bisects the angle ∠A. Similarly we can prove that AC bisects ∠C, BD bisects ∠B and BD bisects ∠D. Example 20: AB and CD are parallel lines. A point O lies in between AB and CD (see Figure 6.76) such that ∠APO = 45° and ∠OQC = 35°. Find ∠POQ Solution: Produce PO to meet CD at X. Produce QO to meet AB at Y. Since AB || CD and PX is a transversal to AB and CD, ∠OXQ = ∠OPY = 45° (alternate angles). Figure 6.76 In the triangle OXQ, ∠POQ is the exterior angle and it is equal to the sum of the interior opposite angles ∠OXQ and ∠OQX. So ∠POQ = ∠OXQ + ∠OQX = 45° + 35° = 80°. Example 21: In the ∆ABC the angle B is bisected and the bisector meets AC in D. If ∠ABC = 80° and ∠BDC = 95°, find ∠A and ∠C. Solution: See the Figure From ∆BDC, 40° + 95° + ∠C = 180° ⇒ ∠C = 180° − 135° = 45° From ∆ABC, ∠A + ∠B +∠C = 180° ∴∠A + 80° + 45° = 180° or ∠A =180° − 125° = 55°.
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Figure 6.77
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Example 22: ABCD is a trapezium is which AB is parallel to CD. If AD = BC, prove that ∠ADC = ∠BCD. Solution: Draw BE parallel to AD. Since ABED is a parallelogram, BE = AD. But AD = BC. ∴ BC = BE. So the triangle BEC is isosceles. ∴ ∠BCE = ∠BEC. But AD || BE and DEC is a transversal to AD and BE. ∴∠ADC = ∠BEC (corresponding angles) From (1) and (2), we get ∠BCE = ∠ADC or ∠BCD = ∠ADC.
Figure 6.78 (1) (2)
Exercise 6.2 1. (i) (ii) (iii) (iv) (v) (vi) (vii)
Which of the following statements are true and which are false: If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° If two lines intersect, then vertically opposite angles are equal. A triangle can have two obtuse angles. The sum of the angles of a quadrilateral is 180° If ∆ABC ≡ ∆PQR, then ∠A = ∠Q If ∆DEF ≡ ∆XYZ, then DE = XY In a parallelogram, the diagonals bisect each other.
2. Find the complement of the following: (i) 20° (ii) 65° (iii) 70° (iv) 78° 3. Find the supplement of the following? (i) 50° (ii) 130° (iii) 80° (iv) 152°. 4. Find the angles in each of the following: (i) The angles are complementary and the smaller is 40° less than the other. (ii) The angles are complementary and the larger is 4 times the smaller. (iii) The angles are supplementary and the larger is 58° more than the smaller. (iv) The angles are supplementary and the larger is 20° less than three times the smaller. (v) The angles are adjacent and form an angle of 140°. The smaller is 28° less than the larger. (vi) The angles are vertically opposite and supplementary.
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5.
Find x and y from the following figures: (i)
(ii)
Figure 6.80 Figure 6.79
(iii)
(iv)
Figure 6.82
Figure 6.81 6.
In each of the following, find x and y: (i)
(ii)
Figure 6.83
(iii)
Figure 6.84
Figure 6.85 162
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Answers Exercise 6.1 1. F 2. T 3. F Exercise 6.2 1. 2. 3. 4. 5.
6.
(i) T (ii) T (iii) F (iv) F (vi) F (vii) T (i) 70° (ii) 25° (iii) 20° (i) 130° (i) (v) (i) (iii)
(ii) 50°
(iii) 100°
(v) F (iv) 12° (iv) 28°
25°, 65° (ii) 18° (iii) 61° , 119° (iv) 50°, 130° 56°, 84° (vi) 90°, 90° x = 130°, y = 50° (ii) x = 80°, y = 70° x = 20°, y = 30° (iv) x = 50°, y = 130°
(i) x = 19, y = 8
(ii) x = 48°, y = 12° (iii) x = 6, y = 3.
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7. ALGEBRAIC GEOMETRY In Chapter 1, we saw that every point on a straight line is associated with exactly one real number. In this chapter, we shall examine how a point in a plane can be represented by real numbers. Rene Descartes, a renowned French mathematician first introduced an algebraic method (method of using numbers and the four fundamental operations) to analyse geometry and hence the subject of analysing geometry using algebraic method is known as algebraic geometry or analytical geometry. As the formulation of this subject was first made by Rene Descartes, he is known as the father of analytical geometry.
7.1
The Cartesian Coordinate System
We want to study the properties of some figures drawn in a plane. A figure in a plane is a collection of points of the plane. So a point is a fundamental concept in geometry. We now proceed to associate a pair of real numbers to every point in the plane. Consider the plane of the paper as the plane and in the plane, draw two fixed perpendicular straight lines. We usually draw one straight line horizontally and the other line vertically as in Figure 7.1. However, they can be drawn in any way as indicated in Figure 7.2. These two lines intersect at the point named as O and called the origin. The point O is fixed since the perpendicular lines are fixed. Now, let us scale the lines with the point O
Figure 7.1 Figure 7.2 representing the number 0 for both the lines. We use the same scaling on the two lines. Now the two perpendicular lines become two perpendicular number lines. The positive numbers for
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the horizontal line are to the right of O and the positive numbers for the vertical line are above O. Similarly the left of O and below of O are for negative numbers. This procedure is indicated by placing arrow heads on the lines as shown in the Figure 7.1. The arrow heads indicate the ordering of the numbers on the lines. We call the horizontal number line, x–axis and the vertical number line, the y–axis. The two lines divide the plane into four regions, called quadrants. These quadrants are named I quadrant, II quadrant, III quadrant and IV quadrant as shown in Figure 7.1. The point O is common to all the four quadrants. Consider any point P in the plane. This point P lies in a quadrant. From P, draw a straight line parallel to the y–axis to meet the x–axis at the point L, and draw a straight line parallel to the x–axis to meet the y–axis at the point M. Let a be the real number representing the point L on the x–axis and b be the real number representing the point M on the y–axis. If P lies on the x– axis, then we observe that b = 0. If P lies on the y–axis, then we observe that a = 0. If P is not on the x and y axes, but lies within the I quadrant, then a > 0 and b > 0. If a < 0 and b > 0, then P lies within the II quadrant. If P lies within the III quadrant, then a < 0 and b < 0. If a > 0 and b < 0, then P lies within the IV quadrant. If P is the point O, then a = 0 and b = 0. The number a is called the abscissa or x–coordinate of the point P and the number b the ordinate or y–coordinate of P (see Figure 7.3). We write the numbers a and b within the parentheses ( , ) separated by a comma as (a, b) and call it the ordered pair of a and b. It is called an ordered pair because the number to the left of the comma is the x–coordinate and the number to the right of the comma is the y–coordinate of the point P. The ordered pair (a, b) is unique for the point P. That is, there is no other ordered pair of numbers for the same point P. The point P is represented as P(a, b) or simply (a, b). We say that P has coordinates (a, b). Thus, every point in the plane is Figure 7.3 represented as an ordered pair of real numbers. The plane now is called the Cartesian plane to honour the great work of Rene Descartes. It is also called the rectangular coordinate plane or the xy–plane. The system of representation of points in the plane by ordered pairs of numbers is called the Cartesian or rectangular or xy- coordinate system. The two axes are called rectangular or coordinate axes.
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We observe that, (i) The origin O has coordinates (0, 0). (ii) Any point on the x–axis has its y–coordinate 0. (iii) Any point on the y–axis has its x–coordinate 0. (iv) Whenever an ordered pair of real numbers is given, we can locate a unique point in the Cartesian plane and plot it by a dot at an appropriate place in the plane. (v) If a point lies within the I quadrant, then both of its coordinates are positive. If the point lies within the II quadrant, then its x–coordinate is negative and y–coordinate is positive. If the point lies within the III quadrant, then both of its coordinates are negative. If the point lies within the IV quadrant, then its x–coordinate is positive and the y–coordinate is negative. The algebraic signs of the coordinates of any point are as shown in Figure 7.4.
Figure 7.4 (vi) All points on a line parallel to x-axis have the same y-coordinate(see Figure 7.5)
Figure 7.5 (vii) All points on a line parallel to y-axis have the same x-coordinate (see Figure 7.6).
Figure 7.6 166
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If A(x1, ,y1) and B(x2, y2) are any two points in the Cartesian plane, then the horizontal distance between A and B is x2 − x1 if x2 > x1, x1 − x2 if x1 > x2, 0 if x1 = x2 and the vertical distance between A and B is y2 − y1 if y2 > y1, y1 − y2 if y1 > y2, 0 if y1 = y2. Figure 7.7 They are respectively denoted by x2 − x1 and
y 2 − y1 . For example, in the Figure 7.7, BN is the horizontal distance and AN is the vertical distance between A and B. We observe that BN = OL + OM = (−x1) + (x2) = x2 − x1, AN = AL + LN = y1 + MB = y1+(−y2) = y1−y2. Similarly, in the Figure 7.8, the horizontal and vertical distances between A(x1, y1) and B(x2, y2) are respectively BN = ML = OM – OL = −x2 – (−x1) = x1 – x2 , AN = LN – AL = BM – (−y1) = (−y2) + y1 = y1−y2.
Figure 7.8
Example 1: Plot the points A (3, 0), B (0, 2 ), C (4, − 4), D (3, 3), E (−2.5, 1), F (−1, −3), G (−1, 0) and H (0, −4). Also specify the quadrant in which each point lies. Solution: The points are plotted in the coordinate plane as in Figure 7.9. The point A lies on the positive side of the x–axis, the point B lies on the positive side of the y–axis, the point C lies within the IV quadrant, the point D lies in the I quadrant, the point E lies in the II quadrant, the point F lies in the III quadrant, the point G lies on the negative side of the x–axis and the point H lies on the negative side of the y–axis.
Figure 7.9
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Example 2: Find the horizontal and the vertical distances between the points (−3, −4) and (−9, 11). Solution: The horizontal distance between the points (−3, −4) and (−9, 11) is a distance between the point corresponding to x coordinates−3 and −9 on the number line x- axis; i.e., (−3) − (−9) = 9 − 3 = 6. and the vertical distances between (−3, −4) and (−9, 11) is the distance between the points corresponding to y co-ordinates −4 and 11 on the number line y axis; i.e., (11) − (−4) = 15. Exercise 7.1 1. Plot the following points and specify in which quadrant each point lies. (i) (2, 3) (ii) (7, 6) (iii) (−2, −3) (iv) (6, −2) (v) (−9, 0) (vi) (5, 0) (vii) (0,11) (viii) (−3, 2) 2. Answer true or false (i) (9, −1) lies in the II quadrant. (ii) (1, 0) lies on the y−axis.. (iii) (−3,1) lies to the right of y–axis. (iv) (1, −1) lies below the x–axis. (v) (0, 0) is the point of intersection of the coordinate axes. (vi) (− 2, 2 ) lies in the II quadrant. (vii) (−π,− 3 ) lies in the III quadrant. (viii) (ix) (x) (xi) (xii)
( 2 − 3 , −1) lies in the IV quadrant. (0,−3) lies to the left of x−axis. (5, 0) lies below the x–axis. Any two points on a line parallel to x-axis have equal x-coordinates. If (a, b) and (c, d) are two points on a line parallel to y-axis, then a = c.
3. Find the horizontal and vertical distances between (i) (1, 4) and (3, 5). (ii) (−2, 3) and (4, −6). (iii) (−3, −5) and (7,2). (iv) (−2, −1) and (−4, −3).
7.2
Slope of a line
First let us proceed to define the slope of a straight line LL′ which is not parallel to x-axis or parallel to y-axis. For this, we think of a man as a point P(x, y) running along the line. We observe that the point P can run in one of the two directions (see Figure 7.10 or Figure 7.11).
Figure 7.10
Figure 7.11
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As P moves in one particular direction along the line, the x-coordinate of P increases (see Figures 7.12 and 7.13). We call this particular direction, the positive direction of the line.
Figure 7.12
Figure 7.13
The other direction is called the negative direction of the line. We observe that if P moves in the negative direction, then its x-coordinate decreases. Let a point P move along the line in the positive direction from the point P1(x1, y1) to the point P2(x2, y2).Then x2 > x1. We observe that the x-coordinate of P changes from the value x1 to x2 and the y-coordinate of P correspondingly changes from the value y1 to the value y2. The change in the x-coordinate value is x2−x1 and is called the run of the moving point P. The corresponding change in the y-coordinate values is y2 − y1 and is called the rise of the moving point P. We observe that the run x2 −x1 is positive (see Figures 7.12 and 7.13) In Figure 7.12, the point P is moving up the line, that is it is rising up the line from P1 to P2 and the point P2 is at a higher position than the point P1. So the rise y2−y1 is positive and y − y1 is positive. In Figure 7.13, the point P is moving down the line; that hence the ratio 2 x 2 − x1 is, it is falling down the line from P1 to P2 and the point P2 is at a lower position than the y − y1 is negative. Thus, the ratio point P1. So the rise y2 −y1 is negative and hence the ratio 2 x 2 − x1 y 2 − y1 y − y1 is positive for a rising line and it is negative for a falling line. The ratio 2 is x 2 − x1 x 2 − x1 called the slope of the line. Next, we shall examine the slope of a line parallel to the x-axis. In Figure 7.14, the line LL′ is parallel to the x-axis and we observe that all points on the line have the same ycoordinate. If P1(x1, y1) and P2(x2, y2) are two points on the parallel line, then y1 = y2 and so the rise y2 − y1 = 0. y − y1 Hence the slope 2 = 0. x 2 − x1
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Figure 7.14
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Next, let us examine the slope of a line parallel to the y-axis. Let P1(x1, y1) and P2(x2, y2) be any two points on the parallel line (see Figure 7.15). Then x1 = x2. So, the run = x2 − x1 = 0. Since P1 and P2 are distinct, y1 ≠ y2. Hence, the slope y − y1 y − y1 rise m= = 2 = 2 . run x 2 − x1 0 This is undefined. ∴ The slope of a line perpendicular to x-axis is undefined. y2 x2 y2 x2 y2 x2 y2 x2
− y1 − x1 − y1 − x1 − y1 − x1 − y1 − x1
Figure 7.15
> 0 for rising line. < 0 for falling line = 0 for the line parallel to x-axis. is undefined for the line parallel to the y-axis.
y 2 − y1 − ( y1 − y 2 ) y − y2 = = 1 .Thus, the slope of the line joining the two x 2 − x1 − (x1 − x 2 ) x1 − x 2 y − y1 y − y2 = 1 . points (x1, y1) and (x2, y2) is 2 x 2 − x1 x1 − x 2 From this, we observe that the slope is independent of the direction of the line. The y − y1 slope 2 of the line is also independent of x 2 − x1 the particular choice of the points P1 and P2. To understand this fact, consider any other two points P3 (x3, y3) and P4 (x4, y4) on the line Figure 7.16 (see Figure 7.16). Then the slope of the line y − y3 from P3 to P4 is 4 . x4 − x 3 y − y1 The slope of the line from P1 and P2 is 2 . x 2 − x1 We note that
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But the triangles ∆P1AP2 and ∆P3BP4 are similar. P A AP2 BP4 AP2 ∴ 1 = or = . P3 B BP 4 P1 A P3B y − y3 y − y1 ∴ 2 = 4 . x 4 − x3 x 2 − x1 That is, the slope is independent of the positions of two points on the line. Note: Through two given points (x1, y1) and (x2, y2) one and only one straight line can be y − y1 drawn. The slope of the line is 2 . x 2 − x1 Example 3: Find the slope of the line passing through (5,6) and (15,9) and state whether the line is rising up or falling down. Solution: Take (5,6) as (x1, y1) and (15, 9) as (x2, y2). Then the slope of the line is y − y1 9−6 3 .= m= 2 = . x 2 − x1 15 − 5 10 The slope is a positive number and so the line is rising up as shown in Figure 7.17. Figure 7.17 Example 4: Find the slope of the line passing through (−16, 29) and (40, −6) and state whether the line is rising up or falling down. Solution: The slope of the line is − 6 − 29 − 35 − 5 m= = . = 40 − (−16) 56 8 Since m is a negative number, the line is falling down as indicated in Figure 7.18. Figure 7.18 Example 5: Interpret the slopes of the following lines joining (i) (6,4) and (−7, 4) (ii) (−2,8) and (−2, 7). Solution: y − y1 4−4 0 (i) Slope of the line = 2 = = = 0, x 2 − x1 − 7 − 6 − 13 ∴ The line is parallel to the x-axis.
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y 2 − y1 −1 7−8 = = not defined. = x 2 − x1 − 2 + 2 0 ∴ The line is perpendicular to the x-axis. (ii) Slope of the line =
Example 6: Find another point on the line with slope −
3 which passes through the point 5
(−2, 3).
−3 (i.e., with the denominator as a positive 5 number). Designate the given point (−2, 3) as P (see Figure 7.19). From P, move 5 units to the right (since the run = 5) to reach the point Q (−2 + 5, 3); i.e., Q (3,3). From Q, move 3 units down (since the rise = −3) to reach the point R (3, 3 + (−3)); i.e., R (3,0). The point R (3,0) is another point on the line. We can verify that the slope of the line joining P and R is Figure 7.19 0−3 −3 or . 3 − (−2) 5 7.2.1 The equation of a straight line
Solution: First we shall write the slope as
Let P (x, y) be a variable point on a given straight line. Then an algebraic equation connecting the variables x and y is called the equation of the straight line. The co-ordinates x and y of any point on the straight line satisfy the equation of the line. By plotting the ordered pairs (x, y) as points in the Cartesian plane, we get the graph of the straight line. A straight line is hereafter called simply a line. The graph crosses the x-axis at a unique point A and it crosses the y-axis at a unique point B. Since A lies on the x-axis its y-coordinate is 0. If a is the x-coordinate of A, then (a, 0) should satisfy the equation of the line. Substituting a for x and 0 for y in the equation of the line, we can solve for the value of a. This value of a is called the x-intercept of the line. That is, the x-intercept of the line is the x-coordinate of the point where the line crosses the x-axis. Similarly, since B lies on the y-axis, its x-coordinate is 0. So, if b is the y-coordinate of B, then (0, b) should satisfy the equation of the line. Replacing x by 0 and y by b in the equation of the line, we can solve for b. This value b is called the y-intercept of the line. Thus, the y-intercept of the line is the y-coordinate of the point where the line crosses the y-axis.
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Now, let us derive the equation of the line whose slope is m and y-intercept is c. Since the y-intercept of the line is c, the point P1 (0, c) is the point at which the line crosses the y-axis (see Figure 7.20). Let P (x, y) be any point on the line. Then the slope of the line is y−c y−c or . x−0 x But the slope of the line is given to be m. y−c = m or y − c = mx or y = mx + c. ∴ x The above equation is called the slope – intercept formula for the equation of a line.
Figure 7.20
Note: If the line passes through the origin (0,0), then its y-intercept is c =0. So the equation of the line is y = mx + 0 or y = mx. 1 and y-intercept −3. 2 Solution: Applying the slope-intercept formula, the equation of the line is 1 1 y= x + (−3) m= 2 2 or 2y = x − 6 c = −3 or x− 2y − 6 = 0. y = mx + c
Example 7: Find the equation of the line having slope
Example 8: Find the slope and the y-intercept of the line whose equation is 3x + 4y + 5 = 0. −3 ⎛ −5⎞ Solution: Rewriting the equation, we get 4y = −3x − 5 or y = x+⎜ ⎟. 4 ⎝ 4 ⎠ −3 −5 Comparing this equation with y = mx + c, we get slope m = and y-intercept c = . 4 4
Exercise 7.2 1. Find the slope of the line joining the two given points (i) (−4,1) and (−5, 2). (ii) (4,−8) and (5,−2). (iii) (−5,0) and (0, −8). (iv) (0,0) and ( 3 , 3). (v) (2a, 3b) and (a, −b). (vi) (a, 0) and (0, b). 2. Find another point on the line (i) through (5, 6) with slope 1.
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(ii) through (0, 4) with slope 1 . 4
(iii) through (2, −2) with slope −1. (iv) through (1, −3) with slope 4. (v) through (−1, −4) with slope 7 . 3
3. Find the equation of the line whose slope and y-intercept are (i) −3 and −7. (ii) 5 and 9. (iii) −2 and 15. (iv) 6 and −11. (v) − 3 and 1.
(vi)
5 − 2 and 8 . 5 5
4. Find the slope and y-intercept of the line whose equation is (i) 3x + 2y = 4. (ii) 2x = y. (iii) x − y − 3 = 0. (iv) 5x − 4y = 8.
7.3
The distance between any two points (x1, y1) and (x2, y2)
The distance between two points is a basic concept in geometry. We now give an algebraic expression for the same. Let P1 (x1, y1) and P2 (x2, y2) be two distinct points in the Cartesian plane and denote the distance between P1 and P2 by d(P1, P2) or by P1P2. Draw the line segment P1 P2 . Three cases arise. Case (i): The segment P1 P2 is parallel to the x-axis (see Figure 7.21). Then y1 = y2. Draw P1L and P2M, perpendicular to the x-axis. Then d(P1,P2) is equal to the distance between L and M. But L is (x1, 0) and M is (x2, 0). So the length LM = x1 − x2 . Hence d (P1, P2) = x1 − x2 .
Figure 7.21
Case (ii): The segment P1 P2 is parallel to the y-axis
(see Figure 7.22). Then x1 = x2 .Draw P1L and P2M, perpendicular to the y-axis. Then d(P1, P2) is equal to the distance between L and M. But L is (0, y1) and M is (0, y2). So the length LM =
y1 − y 2 . Hence
d(P1, P2) = y1 − y 2 .
Figure 7.22 174
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Case (iii): The line segment P1 P2 is neither parallel to the x-axis nor parallel to the y-axis (see Figure 7.23). Draw a line through P1 parallel to x-axis and a line through P2 parallel to y-axis. Let these lines intersect at the point P3. Then P3 (x2, y1). The length
of the line segment P1P3 is x1 − x2 and the length of the segment P3P2 is y1 − y 2 . We observe that
Figure 7.23
the triangle ∆P1P3P2 is a right triangle. ∴ [d (P1 , P2 )] = [d (P1 , P3 )] + [d (P3 , P2 )] = x1 − x 2 + y1 − y 2 2
2
2
2
2
= (x1 − x2)2 + (y1 − y2)2 = (x2 − x1)2 + (y2 − y1)2. ∴ d(P1, P2) =
( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 .
This is called the distance formula which gives the distance d between the two given points (x1, y1) and (x2, y2). We observe that d(P1, P2) = d(P2, P1). The formula has been derived for two points which are not on a horizontal line or vertical line. But the formula holds for these cases as well. When P1 and P2 lie on the same horizontal line, then y1 = y2 and so d(P1, P2) =
( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 =
x 2 − x1 + 0 2 = x 2 − x1 . 2
When P1 and P2 lie on the same vertical line, then x1 = x2 and so d(P1, P2) =
( x1 − x 2 ) 2 + ( y1 − y 2 ) 2 =
0 2 + y1 − y 2
2
= y1 − y 2 .
Note: Since the origin O is (0,0), then, for any point P (x, y), we have OP =
( x − 0) 2 + ( y − 0) 2 =
This distance
(i) (ii)
x2 + y2 .
x 2 + y 2 is called the radius vector of the point (x, y) from the origin.
Using the distance formula, we can show whether three given points are collinear or form a right triangle, isosceles triangle or equilateral triangle. four given points form a parallelogram, rectangle, square or rhombus.
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Example 9: Find the distance between the points A(−15, −3) and B (7, 1). Solution: Let d be the distance between A and B. ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2
Then d (A, B) = =
(7 + 15) 2 + (1 + 3) 2
= = =
22 2 + 4 2 484 + 16
(x1, y1)
(−15, −3)
(x2, y2)
(7, 1)
500 = 10 5 .
Example 10: Show that the points (−4, −9), (2, 0) and (4, 3) are collinear. Solution: Let A, B and C be the given points respectively. Then
AB =
A
(−4,−9)
B
(2, 0) ( 2 + 4) 2 + ( 0 + 9) 2
Rough Sketch
= 6 2 + 9 2 = 36 + 81 = 117 = 9 × 13 = 3 13 .
Figure 7.24 B
(2, 0)
C
(4, 3)
BC =
= AC =
( 4 − 2 ) 2 + (3 − 0 ) 2
2 2 + 32 =
4 + 9 = 13
( 4 + 4 ) 2 + (3 + 9 ) 2
= 8 2 + 12 2 = 64 + 144 = 208 = 16 × 13 = 4 13 We observe that AB + BC = AC. 3 13 + 13 = 4 13 . ∴ A, B and C are collinear. Example 11: Show that the points (3, −2), (2, 5) and (8, −7) form an isosceles triangle. Solution: Let the given points be P, Q and R respectively. One way of proving that ∆PQR is an isosceles triangle is to show that two of its sides are of equal length. Here we have
Figure 7.25
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( 2 − 3) 2 + (5 + 2) 2 = 12 + 7 2 = 1 + 49 = 50 = 5 2.
d(P,Q) = d (Q, R) =
(8 − 2) 2 + ( −7 − 5) 2 = 6 2 + 12 2 = 36 + 144 = 180 = 6 5.
d (R, P) = (8 − 3) 2 + ( −7 + 2) 2 = 5 2 + ( −5) 2 = 25 + 25 = 50 = 5 2. ∴ d (P, Q) = d (R, P) ≠ d (Q, R). ∴ ∆PQR is an isosceles triangle but not an equilateral triangle.
(
)
Example 12: Show that the points (0, 3), (0,1) and 3 , 2 are the vertices of an equilateral triangle. Solution: Let the points be A, B and C respectively. One way of showing that ∆ABC is an equilateral triangle is to show that all its sides are of equal length. Here we find that d (A, B) =
(0 − 0) 2 + (1 − 3) 2 = 0 2 + ( −2) 2 = 4 = 2.
d (B, C) =
( 3 − 0) 2 + (2 − 1) 2 = 3 + 1 = 4 = 2.
d (C, A) = (0 − 3 ) 2 + (3 − 2) 2 = 3 + 1 = 4 = 2. ∴ d (A, B) = d (B, C) = d (C, A). ∴∆ABC is an equilateral triangle.
Figure 7.26
Example 13: Examine whether the points P (7, 1), Q (−4,−1) and R (4,5) are the vertices of a right triangle. Solution: The points P ,Q, R form a triangle. To show that ∆PQR is a right triangle, we have to show that one vertex angle is 90°. This is done by showing that the lengths of the sides of the triangle satisfy Pythagoras theorem. Here PQ =
(−4 − 7) 2 + (−1 − 1) 2 = 121 + 4 = 125 = 5 5.
QR =
( 4 + 4) 2 + (5 + 1) 2 = 64 + 36 = 100 = 10.
PR = (4 − 7) 2 + (5 − 1) 2 = 9 + 16 = 25 = 5. ∴ PQ2 = 125, QR2 = 100 and PR2 = 25. We observe that QR2 + PR2 = PQ2. ∴ The Pythagoras formula is satisfied. ∴ ∆PQR is a right triangle with right angle at R.
Figure 7.27
Example 14: Show that the points (1, 2), (2, −1), (5, 3) and (4, 6) taken in order form a parallelogram. Is it a rectangle ? Justify. Solution: Let the points be P1, P2 ,P3 and P4 respectively. One way of showing that P1 P2 P3 P4 is a parallelogram is to show that the opposite sides are of equal length. Here we find P1P2 =
(2 − 1) 2 + ( −1 − 2) 2 = 1 + 9 = 10.
P2P3 =
(5 − 2) 2 + (3 + 1) 2 = 9 + 16 = 25.
Figure 7.28 177
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P3P4 =
( 4 − 5) 2 + (6 − 3) 2 = 1 + 9 = 10.
P4P1 =
(4 − 1) 2 + (6 − 2) 2 = 9 + 16 = 25.
∴ P1P2 = P3P4 = 10 and P2P3 = P4P1 = ∴ P1P2 P3P4 is a parallelogram. Since
25 .
P1P3 = (5 − 1) 2 + (3 − 2) 2 = 16 + 1 = 17 and (P1P2 )2 + (P2P3)2 = 10 + 25 = 35, (P1P3)2 = 17, (P1P2 )2 + (P2P3)2 ≠ (P1P3)2. ∴ ∆P1P2 P3 is not a right triangle. ∴ ∠P1P2 P3 is not a right angle. ∴ P1P2 P3P4 is not a rectangle. Example 15: Show that the points (0, −1), (−2, 3), (6, 7) and (8, 3), taken in order form the vertices of a rectangle. Solution: Let the points be A, B, C and D respectively. One way of showing that ABCD is rectangle is to show that the opposite sides are of equal length and one corner angle is 90°. One way of showing that one corner angle is 90° is to show that the lengths of the sides of ∆ABC satisfy the Pythagoras theorem. Here we find. AB =
( −2 − 0) 2 + (3 + 1) 2 = 4 + 16 = 20 = 2 5.
BC =
(6 + 2) 2 + (7 − 3) 2 = 64 + 16 = 80 = 4 5.
CD =
(8 − 6) 2 + (3 − 7) 2 = 4 + 16 = 20 = 2 5.
AD =
(8 − 0) 2 + (3 + 1) 2 = 64 + 16 = 80 = 4 5.
AC =
(6 − 0) 2 + (7 + 1) 2 = 36 + 64 = 100 = 10
We observe that AB = CD = 2 5 , BC = AD = 4 5 and AB2 + BC2 = 20 + 80 = 100 = AC2 ∴ ABCD is a rectangle but not a square.
Figure 7.29
Example 16: Show that the points (0, −1), (2, 1) (0, 3) and (−2, 1) taken in order form the vertices of a square. Solution: Let A, B, C, D be the given points respectively.
One way of showing that ABCD is a square is to show that all its sides are of equal length and the diagonals are of equal length. AB =
(2 − 0) 2 + (1 + 1) 2 = 4 + 4 = 8 = 2 2,
BC =
(0 − 2) 2 + (3 − 1) 2 = 4 + 4 = 8 = 2 2,
CD =
( −2 − 0) 2 + (1 − 3) 2 = 4 + 4 = 8 = 2 2,
AD =
(−2 − 0) 2 + (1 + 1) 2 = 4 + 4 = 8 = 2 2,
178
Figure 7.30
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BD =
( −2 − 2) 2 + (1 − 1) 2 = 16 + 0 = 16 = 4,
AC = (0 − 0) 2 + (3 + 1) 2 = 0 + 16 = 16 = 4. We observe here that AB = BC = CD = AD = 2 2 and BD = AC = 4. ∴ ABCD is a square. Example 17: Prove that the points A(2, −3), B(6, 5), C(−2, 1) and D(−6, −7), taken in order form a rhombus but not a square. Solution: One way of showing that ABCD is a rhombus is to show that all its sides are of equal length. One way is showing that a rhombus is not a square is to show that the diagonals are of unequal length. Here we find AB =
(6 − 2) 2 + (5 + 3) 2
= 16 + 64 = 80 BC =
( −2 − 6) 2 + (1 − 5) 2
= 64 + 16 = 80 AC =
( −2 − 2) 2 + (1 + 3) 2
Figure 7.31
= 16 + 16 = 32 BD =
( −6 − 6) 2 + (−7 − 5) 2
= 144 + 144 = CD =
288
(−6 + 2) + ( −7 − 1) 2 2
= 16 + 64 = 80 AD =
=
( −6 − 2) 2 + ( −7 + 3) 2
64 + 16 = 80 .
∴ AB = BC = CD = AD, AC ≠ BD. ∴ ABCD is a rhombus but not a square.
Exercise 7.3 1. Find the distance between the following pair of points:
(i) (1, 2) and (4, 3) (ii) (3, 4) and (−7, 2) (iii) (−7, 2) and (3, 2)
(vi) (a, −b) and (−b, a) (vii) ( 2 + 1, 1) and (1, 3) ⎛2 5⎞ (viii) ⎜ , ⎟ and (− 1, 2) ⎝3 4⎠
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(ix) (2, 0) and (5, −4) (iv) (4, −5) and (−4, 5) (v) (a, b) and (b, a) (x) (−2, 3) and (−1, −5) 2. Examine whether the following points are collinear: (i) (5, 2), (3, −2) and (8, 8) 1 2
(ii) ( , 1), (1, 2) and (0,
3 ) 2
(iii) (1, 4), (3, −2) and (−3, 16) (iv) (−4, 8), (2, −4) and (3, 16) (v) (8, 4), (5, 2) and (9, 6). 3. Examine whether the following points form an isosceles triangle: (i) (5, 4), (2, 0) and (−2, 3). (ii) (6, − 4), (−2, − 4) and (2, 10). (iii) (2, −1), (− 4, 2) and (2, 5). 4. Examine whether the following points form an equilateral triangle: (i) (− 3 , 1), (2 3 , −2) and (2 3 , 4). (ii) ( 3 , 2) , (0, 1) and (0, 3). 5. 6. 7.
8.
9.
(iii) (0, 3) (0, 5) and ( 3 , 4). Examine whether the following points are the vertices of a right triangle: (i) (4, 4), (3, 5) and (−1, −1). (ii) (2, 0), (−2, 3) and (−2, −5). Find the type of the triangle whose vertices are given below: (i) (−3, 7), (−4, 0) and (−10, 8). (ii) (−5, −2), (0, 6) and (8, 1). Examine whether the following points taken in order form a parallelogram: (i) (3, −5), (−5, −4), (7, 10) and (15, 9). (ii) (5, 8), (6, 3), (3, 1) and (2, 6). (iii) (6, 1), (5, 6), (−4, 3) and (−3, −2). (iv) (0, 3), (4, 4), (6, 2) and (2, 1). Examine whether the following points taken in order form a rectangle: (i) (8, 3), (0, −1), (−2, 3) and (6, 7) (ii) (−2, 7), (5, 4) (−1, −10) and (−8, −7). (iii) (−3, 0), (1,−2), (5, 6) and (1, 8). (iv) (−1, 1), (0, 0) (3, 3) and (2, 4) Examine whether the following points taken in order form a square: (i) (1, 2), (2, 2), (2, 3) and (1, 3). (ii) (−1, −8), (4, −6), (2, −1) and (−3, −3). (iii) (1, −1), (0, −4), (7, −3) and (8, −10). (iv) (12, 9), (20, −6), (5, −14) and (−3, 1). (v) (−1, 2), (1, 0), (1, 4) and (3, 2).
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10. Examine whether the following points taken in order form the vertices of a rhombus: (i) (0, 0), (3, 4), (0, 8) and (−3, 4). (ii) (2, −3), (6, 5), (−2, 1) and (−6, −7). (iii) (1, 4), (5, 1), (1, −2) and (−3, 1)
Answers Exercise 7.1 1.
(i) I (ii) I (vi) No quadrant
2.
(i) F (viii) F
3.
(i) (2, 1)
(ii) F (ix) F
(iii) III (iv) IV (vii) No quadrant (viii) II (iii) F (x) F
(ii) (6, 9)
(v) No quadrant
(iv) T (xi) F
(v) T (xii) T
(iii) (10, 7)
(iv) (2, 2)
(vi) T
(vii) T
Exercise 7.2 −8 5 (ii) (4, 5)
⎛b⎞ (v) 4⎜ ⎟ ⎝a⎠ (iv) (2, 1)
1.
(i) −1 (ii) 6 (iii)
(iv)
2.
(i) (6, 7) (v) (2, 3)
(iii) (3, −3)
3.
(i) 3x + y + 7 = 0 (ii) 5x − y + 9 = 0 (iv) 6x − y − 11 = 0 (v) 3x + 5y − 5 = 0
(iii) 2x + y − 15 = 0 (vi) 2x + 5y − 8 = 0
4.
⎛−3 ⎞ (i) ⎜ , 2 ⎟ (ii) (2, 0) ⎝ 2 ⎠
⎛5 ⎞ (iv) ⎜ , − 2 ⎟ ⎝4 ⎠
3
(iii) (1, −3)
(vi)
−b a
Exercise 7.3 1.
(i)
10
(vi) (a + b) 2 2.
(ii) 2 26 (vii)
481 (viii)
(i) Collinear (ii) Non-collinear (iv) Non-Collinear
3.
6
(iv) 2 41
(iii) 10
12
(iii) Collinear
(v) Non-Collinear
(i) Isosceles (ii) Isosceles (iii) Isosceles
181
(ix) 5
(v) (a − b) (x)
65.
2
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4.
(i) Equilateral
5. 6. 7.
(i) Right triangle (ii) Not a right triangle (i) right angled isosceles triangle (ii) right angled isosceles triangle (i) parallelogram (ii) parallelogram (iii) parallelogram (iv) parallelogram (i) Rectangle (ii) Rectangle (iii) Rectangle (iv) Rectangle (i) square (ii) square (iii) not a square (iv) square (v) Not a square (i) Rhombus (ii) Rhombus (iii) Rhombus
8. 9. 10.
(ii) Equilateral
182
(iii) Equilateral
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8. TRIGONOMETRY This branch of mathematics originated several centuries ago in the study of astronomy. Hipparchus, a Greek astronomer and mathematician developed the subject trigonometry and used its principles to a large extent in predicting the paths and positions of the heavenly bodies. The word ‘trigonometry’ is derived from two Greek words ‘trigon’ and ‘metra’. The word ‘trigon’ means triangle and ‘metra’ means measurement. Thus, the name trigonometry deals with the subject which provides the relationships between the measurements of sides and the angles of a triangle. To begin our study of trigonometry, we have to refresh our ideas about angles and their measures. Angles and their measures We say that an angle is formed when two rays originate from a common point. One of
Figure 8.1
Figure. 8.2
the rays is called the initial arm(side) and the other ray the terminal arm (side) of the angle. The common point is called the vertex. We note that when a ray originating from the vertex rotates from the position of the initial arm to the position of the terminal arm, the angle is formed. The rotation of the ray can be performed either in the anti-clockwise direction (see Figure 8.1) or in the clockwise direction (see Figure 8.2). If OA and OB are the initial and terminal sides of an angle, then the angle is
183
Figure 8.3
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denoted by the symbol ∠ AOB. Sometimes it is convenient to position an angle in a Cartesian coordinate plane by taking the vertex as the origin and the initial arm as the positive x-axis (see Figure 8.3) when an angle is positioned in the above way, we say that it is in the standard position. To measure an angle we use an unit called degree. Degree measure When a ray makes one complete rotation in the anticlockwise direction, we say that an angle of measure 360 degrees (written as 360°) is formed. Measurement of all other angles are based on a 360° angle. When a ray makes no rotation, we say that an angle of measure 0° is formed. For example, when a ray makes 1 th of one complete rotation in the anticlockwise 4
direction, an angle of measure 1 (360°) = 90° is formed. When a ray makes 1 th of one 4 4 complete rotation in the clockwise direction, an angle of measure − 1 (360°) = −90°.is formed 4 Thus, rotations in the anticlockwise direction yield positive angles and rotations in the clockwise direction yield negative angles. An angle whose measure lies between 0° and 90° is called an acute angle. A 90° angle is called a right angle and a 180°angle is called a straight angle. If the sum of two acute angles is 90°, then the two angles are said to be complementary. When the sum of two positive angles is 180°, the two angles are said to be supplementary. Right triangle and Pythagoras Theorem If an angle of a triangle is of measure 90°, then the triangle is called a right angle. Let ABC be a right triangle in which the measure of ABC = 90° (see Figure 8.4). The side AC is called the hypotenuse of the right triangle. It is the longest side and is opposite to the right angle. Greek mathematician Pythagoras found that the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. That
Figure 8.4
is, AC = AB + BC . This is known as Pythagoras Theorem. 2
2
2
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8.1 Trigonometric Ratios Let us consider any acute angle ∠AOB and denote it by the Greek letter θ. Let P be a point on the ray OB and PQ be drawn perpendicular to the ray OA . Then the triangle OQP is a right triangle having right angle at the vertex Q. The side OP is the hypotenuse side of ∆OQP. The side PQ is opposite to the angle θ and the side OQ is the adjacent to the angle θ. We shall denote the lengths of these sides by OP, PQ, OQ respectively. Using these lengths, we define the six trigonometric ratios as follows:
Figure 8.5
length of opposite side PQ , = length of hypotenuse side OP length of adjacent side OQ cosine θ = , = length of hypotenuse side OP length of opposite side PQ tangent θ = , = length of adjacent side OQ length of hypotenuse side OP cosecant θ = , = length of opposite side PQ length of hypotenuse side OP secant θ = , = length of adjacent side OQ length of adjacent side OQ cotangent θ = . = length of opposite side PQ
sine θ =
We abbreviate the names of the above ratios as sinθ, cosθ, tanθ, cosecθ, secθ, cotθ respectively. The values of the above ratios do not depend on the size of the right triangle OQP. To know this let P′ be any other point on the ray OB and P′Q′ be drawn perpendicular to the ray OA (see Figure 8.5). Since the right triangles OQP and OQ′P′ are similar, we get
PQ OQ OP . = = P ′Q ′ OQ ′ OP ′
From this, we get
or
PQ P ′Q ′ = , OP OP ′ OP OP ′ = , PQ P ′Q ′
OQ OQ ′ PQ P ′Q ′ = = , OP OP ′ OQ OQ ′ OP OP ′ OQ OQ ′ = , = OQ OQ ′ PQ P ′Q ′
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Thus the six ratios have the same value regardless of the position of the point P on the ray OB. From the above six ratios, we find that PQ OP 1 1 sin θ × cosecθ = × =1 , . cos ec θ = , sin θ = OP PQ sin θ cos ec θ OQ OP 1 1 cosθ × secθ = × =1 , . sec θ = , cos θ = OP OQ cos θ sec θ PQ OQ 1 1 tan θ × cot θ = × =1 . . cot θ = , tan θ = OQ PQ tan θ cot θ
⎛ PQ ⎞ ⎜ ⎟ sin θ OP ⎠ PQ OP PQ ⎝ = = × = = tan θ . We also note that, cosθ ⎛ OQ ⎞ OP OQ OQ ⎜ ⎟ ⎝ OP ⎠ cos θ 1 Taking reciprocals, we get = = cot θ . sin θ tan θ cos θ sin θ Thus, we have = tan θ , = cot θ . sin θ cos θ Note: When θ is acute and one of the six trigonometric ratios is known, we can find the other trigonometric ratios by applying the above formulae. Example 1: Find the six trigonometric ratios sinθ, cosθ, tanθ, cosecθ, secθ and cotθ from the given right triangle. Solution: We note that for the angle θ, length of opposite side = 6; length of adjacent side = 8. (see Figure 8.6) By Pythagoras theorem, (length of hypotenuse side)2 = 82 + 62 = 64 + 36 = 100. ∴ length of hypotenuse side = 100 = 10.
Figure 8.6
length of opposite side 6 3 = = , length of hypotenuse side 10 5 length of adjacent side 8 4 cos θ = = = , length of hypotenuse side 10 5 length of opposite side 6 3 tan θ = = = , length of adjacent side 8 4
sin θ =
length of hypotenuse side 10 5 = = , length of opposite side 6 3 length of hypotenuse side 10 5 sec θ = = = , length of adjacent side 8 4 length of adjacent side 8 4 cot θ = = = . length of opposite side 6 3
cosec θ =
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Example 2: In ∆ABC, m∠B= 90°, AB = 8cm, AC = 17cm. Find all the trigonometrical ratios for the angles A and C. Solution: Here A = m∠BAC and C = m∠BCA (see Figure 8.7) By Pythagoras formula, AC2 = AB2 + BC2 ∴ BC2 = AC2 – AB2 = 172 – 82 = 289 – 64 = 225. ∴ BC = 225 = 15. Hence we have BC 15 = , AC 17 1 8 = , cot A = tan A 15 AB 8 sin C = = , AC 17 1 15 cot C = = , tan C 8
sin A =
AB 8 BC 15 = , tan A = = , AC 17 AB 8 1 17 1 17 sec A = = , cos ec A = = , cos A 8 sin A 15 BC 15 AB 8 cos C = = , tan C = = , AC 17 BC 15 1 17 1 17 sec C = = , cos ec C = = , cos C 15 sin C 8
cos A =
Figure 8.7
Note: In the above problem, we observe that sin C = cos A, cos C = sin A, tan C =cot A,…. This is so because the angles A and C are complementary. 7 Example 3: If sin θ = , find the other trigonometric ratios. 25 7 Solution: Since sinθ = , let us consider a right triangle ABC in which m∠ABC = 90°, 25 m∠ACB=θ, AB = 7 and AC = 25 (see Figure 8.8). By Pythagoras formula, AC2 = AB2 + BC2 ∴ 252 = 72 + BC2 or 625 = 49 + BC2 ∴ BC2 = 625−49 = 576.
∴
BC= 576 = 24. BC 24 = , Hence, cos θ = AC 25 AB 7 tan θ = = , BC 24 1 25 cos ec θ = = , sin θ 7 1 25 sec θ = = , cos θ 24 1 24 Figure 8.8 cot θ = = . tan θ 7 Example 4: If cosec A = 2 , find (i) sin A + cos A (ii) tan A + cot A. Solution: Since cosec A =
2=
2 length of hypotenuse side , = 1 length of opposite side
we consider a right triangle PQR where m∠QRP = A,
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PR =
2 and PQ = 1 (see Figure 8.9). By Pythagoras theorem,
PR2 = PQ2 + QR2 ∴ ( 2 )2 = (1)2 + QR2 ∴ 2 = 1 + QR2
∴ QR2 = 2 – 1 = 1 ∴ QR = 1. Hence we get PQ 1 QR 1 sin A = = , cos A = = , PR PR 2 2 PQ 1 QR 1 tan A = = =1, cot A = = = 1. QR 1 PQ 1 ⎛ 1 ⎞ = 2⎜ ⎟ = 2 2 ⎝ 2⎠ (ii) tan A + cot A = 1 + 1 = 2.
∴(i) sin A + cos A =
1
+
1
Figure 8.9
2,
Note : Whenever we are asked to prove an equation, we adopt any one of the following methods: Method 1 : Simplify the expression in the L.H.S. or R.H.S and obtain the expression on the other side. Method 2 : Simplify the expression on the L.H.S to a form (1). Next simplify the R.H.S to a form (2). Show (1) = (2). Example 5: Prove that
tan A + tan B sin A cos B + cos A sin B . = 1 − tan A tan B cos A cos B − sin A sin B
Solution:
sin A sin B sin A cos B + cos A sin B + tan A + tan B cos A cos B L.H.S = = cos A cos B = sin A sin B cos A cos B − sin A sin B 1 − tan A tan B 1− × cos A cos B cos A cos B (sin A cos B + cos A sin B ) cos A cos B = × cos A cos B cos A cos B − sin A sin B sin A cos B + cos A sin B = = R.H.S. cos A cos B − sin A sin B tan A + cot B tan A . = cot A + tan B tan B tan A tan B + 1 tan A 1 + tan A + cot B tan B tan B = Solution: L.H.S = = 1 1 + tan A tan B 1 tan B cot A + tan B + tan A tan A 1
Example 6: Prove that
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(tan A tan B + 1) tan A . × tan B (1 + tan A tan B ) tan A = = R.H.S. tan B
=
1 + tan θ sin θ + tan θ . = 1 + cot θ 1 + cos θ 1 + tan θ 1 + tan θ Solution: L.H.S. = = 1 1 + cot θ 1+ tan θ (1 + tan θ ) tan θ ( 1 + tan θ) = = tan θ = × 1 ( tan θ + 1 ) ⎛ tan θ + 1 ⎞ ⎜ ⎟ ⎝ tan θ ⎠ sin θ sin θ + sin θ + tan θ cos θ R.H.S. = = 1 + cos θ 1 + cos θ (sin θ cos θ + sin θ ) sin θ (cos θ + 1) 1 1 = = × × cos θ (1 + cos θ ) cos θ (1 + cos θ ) sin θ = tanθ` = cos θ From (1) and (2), L.H.S = R.H.S.
Example 7: Show that
(1)
(2)
Trigonometric ratios of certain angles We shall find the values of the six trigonometric ratios of the angles whose measures are 30°, 45° and 60°. (i) Trigonometric ratios of 30° and 60° angles We consider an equilateral triangle ABC with sides of length 2 (see Figure 8.10) and draw CD perpendicular to AB . Then D bisects the side AB . Now AD = 1, AC = 2. m∠DAC = 60°, m∠ACD = 30° and ∆ADC is a right triangle. By Pythagoras theorem,
AC2 = AD2 + DC2 22 = 12 + DC2 DC2 = 3 DC = 3 . Figure 8.10
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Here from the right triangle ADC, (see Figure 8.11) we get 3 DC AD 1 sin 30° = sin 60° = = = AC 2 AC 2 3 DA 1 DC cos 30° = cos 60° = = = AC 2 AC 2 AD 1 3 DC tan 30° = = tan 60° = = = 3 DC DA 1 3 cot 60° =
DA 1 = DC 3
cot 30° =
3 DC = = 3 AD 1
sec 60° =
AC 2 = =2 DA 1
sec 30° =
AC 2 2 3 = = DC 3 3
cosec 60° =
AC 2 2 3 = = DC 3 3
cosec 30° =
Figure 8.11
AC 2 = =2 AD 1
(ii) Trigonometric ratios of a 45° angle. We consider the isosceles right angle ABC where m∠B = 90°, AB = BC = 1(see Figure 8.12). Then m∠CAB = 45° and m∠BCA = 45° . Now from the right triangle ABC, we get, by Pythagoras theorem, AC2 = AB2 + BC2 = 1 + 1 = 2 and so AC = 2 , AB 1 = sin 45° = AC 2 AB 1 cos 45° = = AC 2 BC 1 tan 45° = = =1 AB 1 AB 1 cot 45° = = =1 BC 1 2 AC sec 45° = = = 2 AB 1 Figure 8.12 2 AC cosec 45° = = = 2 BC 1 (iii) Trigonometric ratios of a 0° angle and a 90° angle To get the trigonometric ratios of these two angles, we consider a circle of radius r with center at the origin in the Cartesian coordinate plane. Let P be any point on the arc of the circle in the positive quadrant of the coordinate plane (see Figure 8.13). Let PM be drawn perpendicular to the x – axis. Let the coordinates of P be x and y. Figure 8.13
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Then OM = x and PM = y. By applying Pythagoras theorem in the right triangle OMP, we get x2 + y2 = r2; ∴ r = x 2 + y 2 . Let ∠MOP = θ. Then θ is an acute angle and y x sinθ = , cos θ = . r r Choosing P at different positions on the arc , we note that as the ray OP turns from the position OA to the position OB , the angle θ increases from 0° to 90° , x decreases from r to x 0 and y increases from 0 to r. So as θ increases from 0° to 90° , decreases from 1 to 0 and r y increases from 0 to 1. That is, as θ increases from 0° to 90°, cos θ decreases from 1 to 0 r and sin θ increases from 0 to 1. Further, we also observe that for each acute angle θ, x and y are unique and so the trigonometric ratios are unique. When OP is in the position OA , θ = 0 r 0°, x = r and y = 0. So we have sin 0 o = = 0 , cos 0 o = = 1 . When OP is in the position r r 0 r OB , θ = 90°, x = 0 and y = r. So, we have sin 90 o = = 1, cos 90 o = = 0 . r r o o cos 0 1 sin 0 0 cot 0 o = = = = 0, not defined, Now, we have tan 0 o = o sin 0 o 0 cos 0 1 1 1 1 1 cos ec 0 o = sec 0 o = = =1 ; = not defined, o o sin 0 0 cos 0 1 [
sin 90 o 1 cos 90 o 0 o = not defined, cot 90 = = =0, cos 90 o 0 sin 90 o 1 1 1 1 1 sec 90 o = = not defined, cos ec 90 o = = =1 . o o cos 90 0 sin 90 1 All the trigonometric ratios for angle of measures 0° , 30°, 45°, 60° and 90° are provided in the following table: tan 90 o =
θ
0°
sin θ
0
cos θ
1
tan θ
0
cot θ
Not defined
sec θ
1
cosec θ
Not defined
30°
45°
60°
90°
1 2 3 2 1 3
1
3 2
1
1 2
0
2 1 2
1 1
3 2 3
2
2
2 Table - 1
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3 1 3
2 2 3
Not defined 0 Not defined 1
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Notation. We write (sinθ)2 as sin2θ but not as sin θ 2. Similarly (tan θ)3 is written as tan3θ. Example 8: Evaluate 2cos2 30° tan260° − sec245° sin260°. 3 3 Solution: cos 30° = , tan 60° = 3 , sec 45° = 2 , sin 60° = . 2 2 ∴2cos230° tan260° − sec2 45° sin2 60° 2
2
( ) ( )
⎛ 3⎞ 2 2 ⎛ 3⎞ ⎟ ⎜ ⎟ 3 2 = 2 ⎜⎜ − ⎜ 2 ⎟ ⎟ 2 ⎠ ⎝ ⎠ ⎝ 3 3 9 3 = 2 × × 3 – 2 × = − = 3. 4 4 2 2 Example 9: Find the acute angle A if tan A =
sin 60 o . 1 + cos 60 o
3 1 , cos 60° = 2 2 3 3 3 1 1 . ∴ A = 30°. = . But tan 30° = ∴ tan A = 2 = 2 = 1 3 3 3 3 1+ 2 2
Solution: sin 60° =
2 cos B = 1 , find A and B.
Example 10: If 2 sin (A + B) =
3 and
Solution: Since 2 sin (A + B) =
3 , we get sin (A + B) =
But sin 60° = Since
3 . So A + B = 60°. 2
2 cos B = 1, we get cos B =
(1)
1 2
1
. So B = 45°. 2 A = 15°. Solving (1) and (2),
But cos 45° =
3 . 2
. (2)
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Exercise 8.1
1.
In problems 1 to 4 find the indicated trigonometric ratios in the given right triangle. 2.
sin B, cos C, tan B
sec X, cot Z, cosec Z
Figure 8.14
Figure 8.15
3.
4.
tan M, sec N, cosec N cos Q , tan R, cot Q Figure 8.16
Figure 8.17
In problems 5 to 10 find the other trigonometric ratios. 12 13 1 8. cosec θ = 10 9. cot θ = 7 sec A + tan A 35 11. If cos A = , find 37 sec A − tan A
5. cos θ =
3 5
6. sin θ =
13. If cosec θ = 2, find the value of cot θ +
2
7. sec θ = 10. tan θ =
3 2 5
12. If sin θ = sin θ . 1 + cos θ
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3 cos ecθ , find 5 cot θ − cos θ
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1 − cos 2 θ 3 14. If cot θ = = . , show that 2 − sin 2 θ 5 3 3 sin θ + 2 cos θ 15. If 3 cot θ = 4, find the value of . 2 sec θ + 3 cos ec θ 16. Evaluate (i) cosec2 45° cot230° + sin2 60° sec2 30° (ii) cos2 30° − sin2 30° − cos 60° tan 45 o (iii) 8 sin2 60° cos 60° (iv) tan 30 o + tan 60 o 17. Verify the following: (i) sin230° + cos230° = 1 (ii) sec260° − 1 = tan2 60° (iii) 1 + cot2 30° = cosec2 30° 1
18. If sin (A+B) = 2 sin (A – B) = 1, find A and B.
8.2 Trigonometric Identities We shall derive three fundamental trigonometric identities. Although we derive these identities for acute angles, they also hold for general angles. Let θ be an acute angle. The vertex of θ is taken as the origin and the initial arm of θ is taken as the positive x-axis. Let P (x, y) be on the terminal arm of θ (see Figure 8.18). Let PQ be drawn perpendicular to x – axis. Then OQ = x, PQ = y. Let OP = r. Applying Pythagoras identity in the right triangle OQP, we get x2+ y2 = r2 . 2
Dividing both sides by r ,
x2 + y2 r2
=
r2 r2
(or) Figure 8.18
x2 y2 + =1 . r2 r2 But sin θ =
y x , cos θ = . ∴ (cos θ)2+ (sin θ)2 = 1. r r or cos2θ + sin2θ = 1
(1)
Dividing both sides of (1) by cos2θ, we get cos 2 θ + sin 2 θ 1 = 2 cos θ cos 2 θ 2
⎛ sin θ ⎞ ⎛ 1 ⎞ ⎟⎟ = ⎜⎜ ⎟⎟ 1 + ⎜⎜ ⎝ cos θ ⎠ ⎝ cos θ ⎠
(or)
cos 2 θ sin 2 θ ⎛ 1 ⎞ ⎟ + =⎜ cos 2 θ cos 2 θ ⎜⎝ cos θ ⎟⎠
2
(or) 1 +(tan θ)2 = (sec θ)2
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1 + tan2θ = sec2 θ
(or)
(2)
Dividing both sides of (1) by sin2θ, we get PQ OQ OP = = ′ ′ ′ P Q OQ OP ′ 2
(or)
cos 2 θ sin 2 θ ⎛ 1 ⎞ ⎟ = (cosec θ ) 2 (or) cot2θ + 1 = cosec2θ + =⎜ sin 2 θ sin 2 θ ⎜⎝ sin θ ⎟⎠
(or)
1 + cot2θ = cosec2θ
(3)
The three identities (1), (2) and (3) are based on the Pythagoras identity. We deduce some more identities from them. Using the fundamental identity (1), we have (i) sin2θ = (sin2θ + cos2θ)− cos2θ = 1 – cos2θ. (ii) cos2θ = (cos2θ + sin2θ) – sin2θ = 1 – sin2θ. Using the fundamental identity (2), we get (i) tan2θ = (1 + tan2θ) – 1= sec2θ − 1 (ii) sec2θ − tan2θ = (1 + tan2θ) – tan2θ = 1. Using fundamental identity (3), we get (i) cot2θ = (1 + cot2θ) – 1 = cosec2θ − 1 (ii) cosec2θ − cot2θ = (1 + cot2θ) – cot2θ = 1. We list the identities in the following table sin2 θ + cos2θ ≡ 1 1 + tan2θ ≡ sec2θ 1 + cot2θ ≡ cosec2θ sin2θ ≡ 1 – cos2θ cos2θ ≡ 1 – sin2θ tan2θ ≡ sec2θ − 1 sec2θ − tan2θ ≡ 1 cot2θ ≡ cosec2θ − 1 cosec2θ − cot2θ ≡ 1
We again mention here that an identity is applied in both ways, left to right or right to left. Example 11: Prove that sin4θ + cos4θ = 1 – 2sin2θ cos2θ . Solution: L.H.S. = sin4θ + cos4θ = (sin2θ)2 + (cos2θ)2 = [sin2θ + cos2θ]2 – 2 (sin2θ)(cos2θ) = (1)2 – 2sin2θ cos2θ = 1 – 2sin2θ cos2θ = R.H.S.
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(
a2 + b2 = (a + b)2 – 2ab)
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Example 12: Prove that Solution:
cos θ = secθ − tan θ. 1 + sin θ
cos θ cos θ 1 − sin θ = × 1 + sin θ 1 + sin θ 1 − sin θ cos θ ( 1 − sin θ) cos θ ( 1 − sin θ) = = 2 cos 2 θ 1 − sin θ 1− sin θ sin θ 1 = = − cos θ cos θ cos θ = sec θ − tan θ = R.H.S.
L.H.S. =
Example 13: Prove that Solution :
L.H.S =
1 + cos A = (cos ec A + cot A) 2 . 1 − cos A
1 + cos A 1 + cos A × 1 − cos A 1 + cos A
(1 + cos A) 2 ⎡1 + cos A ⎤ cos A ⎞ (1 + cos A) 2 ⎛ 1 =⎜ + =⎢ = = ⎟ 2 2 ⎥ 1 − cos A sin A ⎣ sin A ⎦ ⎝ sin A sin A ⎠ = (cosec A + cot A)2 = R.H.S. 2
Alternately, 2
⎛ 1 ⎛ 1 + cos A ⎞ cos A ⎞ ⎟⎟ = ⎜⎜ ⎟⎟ R.H.S. = (cosec A + cot A) = ⎜⎜ + A A A sin sin sin ⎝ ⎝ ⎠ ⎠ 2 AC 2 (1+ cos A) = =2 = = AD 1 1− cos 2 A
2
2
=
(1 + cos A) 2 1 + cos A = = L.H.S. (1 + cos A) (1 − cos A) 1 − cos A
Example 14: Prove that sin4θ − cos4θ = sin2θ − cos2θ. Solution: L.H.S. = sin4θ − cos4θ = (sin2θ)2 – (cos2θ)2 = (sin2θ + cos2θ) (sin2 θ − cos2θ ) =(1) (sin2θ − cos2θ) = sin2θ − cos2θ = R.H.S. 1 Example 15: Prove that sec A – tan A = . sec A + tan A Solution : sec A − tan A 1 1 R.H.S. = = × sec A + tan A sec A + tan A sec A − tan A sec A − tan A sec A − tan A = = sec A – tan A = L.H.S. = 2 2 1 sec A − tan A
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Example 16: Prove that (sec θ + cosθ) (secθ − cosθ) = tan2θ + sin2θ. Solution : L.H.S. = (sec θ + cosθ) (secθ − cosθ) = sec2θ − cos2θ = (1 + tan2θ) – cos2θ = tan2θ + (1 – cos2θ) = tan2θ + sin2θ = R.H.S. 1 1 = 2cosec2θ. + 1 + cos θ 1 − cos θ
Example 17: Prove that Solution :
L.H.S. =
=
1 1 1(1 − cos θ ) + 1(1 + cos θ ) = + 1 + cos θ 1 − cos θ (1 + cosθ ) (1 − cos θ ) 1 − cos θ + 1 + cos θ 2 = = 2cosec2θ = R.H.S. 2 2 sin θ 1 − cos θ
Example 18: Prove that sin2A sin2B + cos2A cos2B + sin2A cos2B + cos2A sin2B = 1. Solution : L.H.S. = (sin2A sin2B + sin2A cos2B) + (cos2A cos2B + cos2A sin2B) = sin2A (sin2B + cos2B) + cos2A (cos2B + sin2B) = sin2A(1) + cos2A (1) = sin2A + cos2A = 1 = R.H.S Example 19: If m = tan A + sin A, n = tan A – sin A. Prove that m2 – n2 = 4 mn . Solution : L.H.S. = m2 – n2 = (tan A + sin A)2 – (tan A – sin A)2 = tan2 A + sin2 A + 2 tan A sin A – (tan2A + sin2A – 2 tan A sin A) = 4 tan A sin A y R.H.S = 4 mn = r
sin 2 A − sin 2 A = 4 tan A − sin A = 4 2 cos A 2
2
sin 2 A − sin 2 A cos 2 A sin 2 A (1 − cos 2 A) =4 = 4 cos 2 A cos 2 A = 4 sin 2 A tan 2 A = 4 sin A tan A = L.H.S.
Example 20: Prove that cos6θ + sin6θ = 1 – 3 cos2θ sin2θ Solution : L.H.S = cos6θ + sin6θ = (cos2θ)3 + (sin2θ)3 = (cos2θ + sin2θ) (cos4θ − cos2θ sin2θ + sin4θ) = (1) (cos4θ + sin4θ − cos2θ sin2θ) = [(cos2θ)2 + (sin2θ)2] – cos2θ sin2θ = [(cos2θ + sin2θ)2 – 2 cos2θ sin2θ] – cos2θ sin2θ = (1)2 – 3 cos2θ sin2θ = 1 – 3 cos2θ sin2θ = R.H.S.
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Example 21: Prove that 1 + 2 sin θ cos θ = sin θ + cos θ Solution:
L.H.S =
sin 2 θ + cos 2 θ + 2 sin θ cos θ
1 = sin 2 θ + cos 2 θ
= ( sin θ + cos θ) 2 = sinθ + cosθ = R.H.S
Exercise 8.2 1. Prove that sec A − 1 = sin A sec A . 2. Prove that (sin A + cos A)2 + (sin A – cos A)2 = 2. cot 2 θ − cos ec 2 θ . 3. Simplify : sec 2 θ − tan 2 θ sec A + tan A 1 + sin A 4. Prove that = . sec A − tan A 1 − sin A 1 1 5. Prove that + = 2 sec 2 θ . 1 + sin θ 1 − sin θ 6. If x = r sin A sin B, y = r sin A cos B, z = r cos A, find the value of x2 + y2 +z2. 7. Show that tan A + cot A = cosec A sec A. 1 − tan 2 A 8. Prove that = 2 cos 2 A −1 . 2 1 + tan A 1 9. Prove that = cosec θ + cot θ. cos ec θ − cot θ 10. Prove that (tan A + cot A)2 = sec2 A+ cosec2A. cos A sin A 11. Prove that + = sin A + cos A . 1 − tan A 1 − cot A tan A + sec A − 1 1 + sin A 12. Prove that . = tan A − sec A + 1 cos A 13. Prove that (tan A – tan B)2 + (1 + tan A tan B)2 = sec2A sec2B. tan θ cot θ 14. Prove that + = sec θ cosec θ +1. 1 − cot θ 1 − tan θ 2
2
⎛ 1 + sin θ − cos θ ⎞ 1 − cos θ ⎟⎟ = 15. Prove that ⎜⎜ . 1 + cos θ ⎝ 1 + sin θ + cos θ ⎠ 16. Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2θ + cot2θ. cos 3 θ + sin 3 θ cos 3 θ − sin 3 θ 17. Prove that + = 2. cos θ + sin θ cos θ − sin θ
1 − cos 4 θ + sin 4 θ = tan 2 θ . 18. Prove that 4 4 1 − sin θ + cos θ
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8.3 Trigonometric Ratios For Complementary Angles We have already learnt about complementary angles in a right triangle. In the right triangle OQP (see figure 8.19), right angled at Q, the angles ∠QOP and ∠OPQ are called complementary angles. Since the sum of their measures is 90°. Let ∠QOP = θ. Then ∠OPQ = 90° − θ. Using the definition of trigonometric ratios, for the angle θ, we get sin θ =
PQ OQ PQ , cosθ = , tan θ = , OP OP OQ
Figure 8.19 ………..
OP OP OQ , secθ = , cot θ = PQ OQ PQ Similarly, for the angle 90°− θ, we get PQ OQ 3 cos (90 o −θ ) = , tan(90 o −θ ) = , sin (90° - θ) = OP PQ 4 OP OP PQ cos ec (90 o −θ ) = , sec (90 o −θ ) = , cot (90 o −θ ) = OQ PQ OQ Comparing (1) and (2), we get PQ = sin θ = cos (90 o − θ ) OP OQ = cos θ = sin (90° − θ) OP PQ = tan θ = cot (90°− θ) OQ OP = cosec θ = sec (90° − θ) PQ OP = sec θ = cosec (90° − θ) OQ OQ = cot θ = tan (90° − θ). PQ
(1)
cos ecθ =
Hence, we obtain the following table: sin (90° − θ) = cosθ cos (90° − θ) = sin θ tan (90° − θ) = cot θ cot (90° − θ) = tan θ sec (90° − θ) = cosec θ cosec (90° − θ) = sec θ
199
………..
(2)
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tan 65 o . Example 22: Evaluate cot 25 o Solution : tan 65° = tan(90° − 25°) = cot 25° tan 65 o cot 25 o = = 1. ∴ cot 25 o cot 25 o Example 23: Evaluate sin 20° tan 60° sec 70° 1 sin 20 o ∴sin 20° tan 60° sec 70° = sin 20° tan 60° cosec 20° 1 =sin 20° × 3 × = 3. sin 20 o
Solution : sec 70° = sec(90°−20°) = cosec 20°=
Example 24: Find x° if cosec x° = sec 25°. Solution: Since cosec x° =sec(90°− x°),we have sec(90°− x°)= sec 25°. ∴ 90°− x°=25°. ∴ x° = 90°− 25°= 65°. Note. The above value of x is obtained not by canceling sec on both sides but by using the property of uniqueness of trigonometric ratios for each acute angle.
o
Exercise 8.3 tan 35o (ii) cot 55o
1. Evaluate (i)
sin 36 cos 54 o
2. Simplify: (i)
tan 33o 1 sin 42 o 3 sec 51o + + . 2 cos 48 o 2 cos ec 39 o cot 57 o
(ii) 3
(iii) sin θ sec (90° − θ)
sin 23o sec 47 o + . 4 cos 67 o cos ec 43 o
3. Find x if (i) sin 60° = cos x°
(ii) cosec x° cos54° = 1
(iii) sec x° = cosec 25°
(iv) tan x° tan 35° = 1
Answers Exercise 8.1 5 5 5 5 1 5 8 8 8 2. 3. , , , , , , 13 13 12 2 2 2 17 15 15 4 4 5 5 3 5. sin θ = , tan θ = , cos ec θ = ,sec θ = , cot θ = . 5 3 4 3 4
1.
200
4. 1,
2,
2
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5 12 13 13 5 , tan θ = ,cos ec θ = , sec θ = , cot θ = . 13 5 12 5 12 1 3 1 7. sin θ = ,cos θ = , tan θ = , cos ec θ = 2, cot θ = 3 . 2 2 3
6. cos θ =
10 1 , tan θ = ,sec θ = ,cot θ = 3 . 3 3 10 10 5 2 7 1 sin θ = ,cos θ = , tan θ = 7 , cos ec θ = , sec θ = 5 2 . 7 5 2 5 2 2 5 29 29 5 sin θ = , cos θ = , cos ec θ = ,sec θ = ,cot θ = . 2 5 2 29 29 49 25 34 12. 13. 2 15. 25 8 75 3 (i) 7 (ii) 0 (iii) 3 (iv) 4 A = 60°, B = 30°
8. sin θ = 9. 10. 11. 16. 18.
1
, cos θ =
3
Exercise 8.2 3. −1
6. r2 Exercise 8.3
1.
(i) 1
(ii) 1
2.
(i) 3
(ii) 7
3.
(i) 30°
(ii) 36°
(iii) 1
(iii) 65°
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(iv) 55°
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9. PRACTICAL GEOMETRY In Theoretical geometry or pure geometry, we give proofs for theorems on the properties of geometrical figures by applying axioms and reasoning. Here, we do not construct exactly the geometrical figures but draw rough sketches of the figures to give support to our logical reasoning. No geometrical instrument is needed in studying theoretical geometry. For example in theoretical geometry, when we say that the line segment PQ is the perpendicular bisector of the line segment AB , we do not actually construct PQ but roughly draw PQ perpendicular to AB . However, for constructing such geometrical figures, much ingenuity and skill are needed. To draw geometrical figures, several geometrical instruments are available. Drawing geometrical figures using geometrical instruments is the subject matter of practical geometry. However, a challenge is always made to use only two geometrical instruments namely an ungraduated ruler (also called a straight edge) and a pair of compasses in construction problems. Great many facts and theorems (for example, the Pythagoras theorem) have been formulated by considering construction problems. In our earlier classes, We have learnt the following: (i) Construction of perpendicular bisector of a line segment. (ii) Construction of the bisector of an angle. (iii) Construction of an equilateral triangle. (iv) Division of a line segment in a given ratio. (v) Construction of a right triangle. (vi) Construction of a parallelogram. (vii) Construction of a Rhombus. (viii) Construction of concentric circles. (ix) Construction of a trapezium. (x) Construction of a triangle when the side lengths are given. In the present chapter, we shall know the following methods: (i) To locate the centroid, orthocentre, circumcentre and incentre of a triangle. (ii) To construct the arithmetic and geometrical means. (iii) To construct the mean proportional of two numbers.
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9.1 Concurrency in a Triangle In theoretical geometry, we have learnt the following: (i) the medians of a triangle are concurrent and the point of concurrence called the centroid of the triangle, divides each median in the ratio 2 : 1. (ii) the perpendicular bisectors of the sides of a triangle are concurrent and the point of concurrence is called the circumcentre of the triangle . (iii) the angular bisectors or the internal bisectors of the angles of a triangle are concurrent and the point of concurrence is called the incentre of the triangle. (iv) the altitudes of a triangle are concurrent, and the point of concurrence is called the orthocentre of the triangle Now we proceed to know how to locate these points of concurrency in practice. 9.1.1 Centroid The line segment joining a vertex of a triangle and the midpoint of the side opposite to the vertex is called a median. As there are three vertices, there are three medians of a triangle. The medians of a triangle are concurrent. The point of concurrency is called the centroid of the triangle and is usually denoted by the letter G. The point G divides each median in the ratio 2 : 1. G is more near to the side than to the vertex. Based upon the properties of the centroid G, we give below the procedure to locate the point G. Step 1: Draw the given triangle ABC. Step 2: Locate the mid point D of the side BC and draw the median AD. Step 3: Locate the mid point E of the side CA and draw the median BE. Step 4: AD & BE meet at G. G is the centroid of ∆ABC. Note: In the above procedure, we did not find the third median to locate G since two medians are sufficient to locate the point of intersection, namely the centroid. If we draw the third median, we observe that it passes through G. In any construction or drawing problem, we first know what are the given measurements and what is required. Then we draw a rough sketch where the steps for the construction or drawing are indicated. Example 1: Draw ∆ABC if AB = 7 cm, AC = 7.5 cm and BC = 5.5 cm and find its centroid G. Write down the ratio in which G divides AD. Solution: We draw the rough figure of ∆ABC and mark the given measurements (see Figure 9.1). Now we proceed to locate the centroid. The steps are given below:
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Step 1: Draw the line segment BC with BC = 5.5 cm. With B as centre, cut an arc with radius 7 cm. Similarly with C as centre, cut an arc of radius 7.5 cm. These two arcs intersect at A. Now draw AB and AC . The triangle ABC is drawn. Step 2: We locate the mid point D of BC and the mid point E of AC by the perpendicular bisector method. Step 3: Draw AD and BE. They meet at the point G. G is the centroid of ∆ABC.
Figure 9.1
Step 4: Locate the mid point F of AB. Draw CF . We observe that it passes through G. Step 5: Measure the length AG and GD. We find AG = 4.5 cm, GD = 2.25cm. We observe AG 4 .5 2 that = = . That is G divides AD in the ratio 2:1. 2.25 1 GD
Figure 9.2
9.1.2 Circumcentre The perpendicular bisectors of the sides of a triangle are concurrent at a point. This point is called the circumcentre of the triangle and is usually denoted by the letter S. It is at an equal distance R from the vertices of the triangle. The circle drawn with S as the centre and the equidistance R as radius passes through the vertices of the triangle. This circle is called the circumcircle of the triangle and R is called its circumradius. To locate the circumcentre and the circumcircle, we adopt the following procedure: Step 1: Draw the triangle ABC. Step 2: Draw the perpendicular bisectors of BC and AC. Step 3: Mark the point of intersection of the perpendicular bisectors of BC and AC as S. This point S is the circumcentre of ∆ABC.
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Step 4: Draw the perpendicular bisector of AB. Observe that this bisector passes through S. Measure the lengths SA, SB and SC and observe that SA = SB = SC. Draw the circumcircle. Example 2: Draw the triangle ∆ABC if AB = 7cm, m ∠B = 45º, BC = 6cm. Construct the circumcircle. Solution: Draw the triangle ABC with the given measurements (SAS construction). Then, the following steps are followed: Step 1: Draw the perpendicular bisectors of BC and AB. Step 2: Mark the meeting point S of the perpendicular bisectors. S is the circumcentre. Step 3: Measure the lengths SA, SB and SC. We find SA = SB = SC = 3.6 cm Step 4: Draw a circle with S as the centre and SA as the radius. This circle passes through A, B and C and it is the required circumcircle.
Figure 9.3
Figure 9.4 Note: When the circumcircle of a triangle is drawn, we say that the triangle is circumscribed. 9.1.3
Incentre The internal bisectors of angles of a triangle are concurrent at a point. This point is called the incentre of the triangle and is denoted by the letter I. An important property of the incentre is that the perpendicular segments IL , IM , IN from I to the sides are equal in length. The Figure 9.5 equal distance is called the inradius of the circle and it is denoted by r. The circle drawn with I as centre and r as radius touches all the sides of the triangle internally. The circle is said to be inscribed in the triangle and it is called the incircle of the triangle (see Figure 9.5) 205
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To locate the incentre, to measure the inradius and to draw the incircle of a triangle, the steps are given below: Step 1: Draw the triangle ∆ABC. Step 2: Draw the internal bisectors of the angles ∠B and ∠C. Step 3: The point of intersection of the internal bisectors is located as I. The point I is the required incentre. Step 4: Verify that the internal bisector of ∠A also passes through I. Step 5: Draw the perpendicular line segment from I to the side BC. Measure its length. This gives the inradius r. Step 6: Draw the circle with I as centre and r as radius. We get the incircle. Example 3: Draw the incircle of ∆PQR if PQ = 8 cm, m∠P = 50º, m∠Q = 60º. Also measure the inradius and draw the incircle. Solution: In the rough figure of ∆PQR, we have marked the given measurements. Step 1: Draw the ∆PQR using ASA method. Step 2: Draw the angle bisector of ∠P. Step 3: Draw the angle bisector of ∠Q. Step 4: Mark the point of intersection of the bisectors as I. The point I is the incentre Step 5: Draw the perpendicular segment ID to the side Figure 9.6 PQ. Step 6: Measure the length of ID . This length is the inradius of the triangle. We find ID = 2 cm. Step 7: With I as centre and ID as radius, draw a circle. This is the incircle of the triangle.
Figure 9.7
9.1.4
Orthocentre We recall that an altitude of a triangle is a perpendicular line segment drawn from a
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vertex of the triangle to the side of the triangle opposite to the vertex. We observe that there are three altitudes in a triangle. We have already learnt that the altitudes of a triangle are concurrent at a point. The point of concurrence is called the orthocentre of the triangle and it is denoted by the letter H. In figure 9.5, ABC is a triangle. AL, BM and CN are altitudes. They meet at the point
Figure 9.8
H, the orthocentre of ∆ABC. The steps for locating the orthocentre H are given below: Step 1: Draw the triangle ABC with the given measurements. Step 2: Draw the altitudes AL, and BM . Step 3: Mark the meeting point of AL and BM as H. H is the orthocentre of the triangle ABC. Example 4: Locate the orthocentre of ∆XYZ if XY = 9cm, YZ = 8 cm and ZX = 7cm. Solution: Draw a rough figure of ∆XYZ and mark the given measurements. Step 1: Draw the triangle XYZ using SSS Rough Figure procedure. Step 2: Draw the altitudes XL, YM . Step 4: Mark the point of intersection of XL and YM as H. H is the orthocentre of ∆XYZ. Step 5: Join ZH and produce it to meet XY at N. We observe that ZN is the altitude through Z. Figure 9.9
Figure 9.10 If we locate the centroid, circumcentre, incentre and orthocentre of various types of triangles, we observe that their positions are as given below:
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Table Point of concurrence Centroid Circumcentre Incentre Orthocentre
Type of triangle Any type of triangle Acute angled triangle Right triangle Obtuse angled triangle Any type of triangle Acute angled triangle Right triangle Obtuse angled triangle
Location of the point of concurrence Inside the triangle Inside the triangle Mid point of the hypotenuse Outside the triangle Inside the triangle Inside the triangle Vertex of the right angle Outside the triangle
Exercise 9.1 In problems 1 to 4, locate the centroid G of the given triangle. 1. 2. 3. 4.
∆ABC, where BC = 6cm, m∠B = 40º, m∠C = 60º. ∆ABC, where all sides are of 6.5 cm long. ∆PQR, where m∠R = 90º, PQ = 7 cm, PR = 6 cm. ∆LMN, where LM = 6cm, m∠L = 95º, MN = 8cm.
In problems 5 to 8, draw the circumcircle of the given triangle. Find also the circumradius. 5. 6. 7. 8.
∆ABC, where AB = 8 cm, BC = 5cm, AC = 7 cm. ∆PQR, where PQ = 5 cm, PR = 4.5 cm, m∠P = 100º. ∆XYZ, where XY = 7 cm, m∠X = 70º, m∠Y = 60º. ∆PQR, where each side is of length 5.5 cm.
In problems 9 to 12, draw the incircle and measure its inradius. 9. 10. 11. 12.
∆ABC, where AB = 9cm, BC = 7cm, CA = 5cm. ∆XYZ, where XY = YZ = ZX = 8cm. ∆PQR, where PQ = 10cm, m∠P = 90º, m∠Q = 60º. ∆ABC, where AB = 5.4 cm, m∠A = 50º, AC = 5cm.
In problems 13 to 16, locate the orthocentre of the triangle. 13. 14. 15. 16.
∆ABC, where BC = 5.6cm, m∠B = 55º, m∠C = 65º. ∆PQR, where m∠P = 90º, m∠Q = 30º, PQ = 4.5 cm. ∆LMN, where, LM = 7cm, m∠M = 130º, MN = 6 cm. ∆XYZ, where XY = 7cm, YZ = 5 cm, ZX = 6 cm.
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9.2
Geometrical Interpretation of Averages
We shall now study geometrical construction methods to find the arithmetic mean and geometric mean between two positive numbers. 9.2.1. Arithmetic mean The Arithmetic mean between two numbers a and b is a+b geometrically, we proceed as follows: 2 Step 1: Take a line segment PQ whose length is a + b.
a+b . To find the number 2
Step 2: Draw the perpendicular bisector of PQ . Step 3: The meeting point of PQ and its perpendicular bisector is marked and its distance a+b . The reason is that if a line segment XY from P or Q is measured. This distance gives 2 l is of length l, then the midpoint of XY is at a distance from X or Y. 2 Example 5: Find the arithmetic mean between 6 and 9. Solution: The procedure is given below: Step 1: Draw a line segment sufficiently long. Cut off a line segment AB on it such that AB is of length 6 cm. Step 2: Cut off a line segment BC on AX to the right of B such that BC is of length 9 cm. Step 3: Draw the perpendicular bisector AC . Mark the meeting point of this bisector with AC . Name the meeting point as M. Step 4: Measure the distance of M from A or C. We observe that AM is of length 7.5 cm. This gives the arithmetic mean between 6 and 9.
Figure 9.11 9.2.2
The geometric mean or mean proportional between two numbers a and b Let a and b be two positive numbers. Then the geometric mean of a and b is ab. If a x x = ab, then x2 = ab or x × x = ab or = or a : x = x : b. So x is also called the mean x b proportional between a and b. (If a number x is such that a : x = x : b, then x is called the mean proportional between a and b). To find ab through geometrical constructions, we proceed as follows:
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Step 1: Draw a line segment sufficiently long. Step 2: Along it, cut off the line segments AB and BC of lengths a and b respectively.
Step 3: Draw a circle with AC as diameter. Step 4: Draw a chord DE through B perpendicular to AC . Step 5: Measure the length of BD or BE. This length gives the mean proportion between a and b. Let us now understand how the length of BD gives the mean proportional between the lengths of AB and BC.
Figure 9.12
AC is a diameter and D is a point on the circle. So ∠ADC = 90º. BD is perpendicular to AC .∠B and the side BD are common to the right triangle ADB and DCB. So these two triangles are similar. Hence the sides are proportional. AB BD Then = or BD2 = AB × BC or BC BD x2 = ab or x =
ab.
Example 6: Find the geometric mean between two segments of lengths 9 cm and 3 cm. Solution: Step 1: Draw a line segment AX sufficiently long. Step 2: Cut off from it line segments AB and BC of lengths 9 cm and 3 cm respectively. Step 3: Draw the perpendicular bisector of AC and make the meeting point of AC and the bisector as O. Step 4: With O as centre and OA as radius, draw a circle. Step 5: Draw the perpendicular chord DE through B. Step 6: BD or BE represents the geometric mean between AB and BC . Measure the length of BD or BE . This length is the geometric mean between 9 and 3. We measure BD and find BD = 5.2 cm.
Figure 9.13
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Example 7: Find the mean proportional between 4 and 9. Solution: We know that the mean proportional between 4 and 9 is 4 × 9 = 6. We find this value 6 through geometrical construction. Step 1: First draw a line segment AX . Step 2: On AX cut off line segment AB of length 4 cm. Step 3: Starting with B, cut off BC of length 9 cm. to the right of B. Step 4: Mark the mid point of AC as O. Step 5: Draw a circle with O as centre and OA as radius. Step 6: Draw the chord DE through B, perpendicular to AC . Step 7: BD represents the mean proportional between AB and BC. Step 8: Measure the length of BD . We find BD = 6 cm. The number 6 is the mean proportional between the numbers 4 and 9.
Figure 9.14 Example 8: Find geometrically the value of 12. Solution: We observe that 12 = 4 × 3. So 12 can be considered as the mean proportional between 4 and 3. Applying the geometrical construction for getting the mean proportional between 4 and 3, we get Step 1: Cut off line segments AB and BC on a line AX Such that AB = 4 and BC = 3. Step 2: Draw the circle with AC as one of its diameter. Step 3: Draw the chord DE through B perpendicular to AC . Step 4: Measure the length of BD . We find that it is 3.4 cm.
Figure 9.15
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Exercise 9.2 In problems 1 to 4, find the arithmetic mean between the given numbers. 1. 6 and 4
2. 10 and 5
3. 9 and 3
4. 2.5 and 6.5
In problems 5 to 8, find the geometric mean between the given numbers. 5. 3.2 and 1.8
6. 2.2 and 5
7. 4 and 1.6
8. 6 and 4
In problems 9 to 12, find the square root of the given number. 9. 15
10. 18
11. 21
12. 24
Answers Exercise 9.1 5. 4 cm 9. 1.65 cm
6. 3.7 cm 10. 2.3 cm
7. 4.7 cm 11. 3.7 cm
Exercise 9.2
1.
5
2. 7.5
3. 6
4. 4.5
5. 2.4
6. 3.3
7. 2.5
8. 4.9
9. 3.9
10. 4.2
11. 4.6
12. 4.9
212
8. 3.2 cm 12. 1.5 cm
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10. HANDLING DATA We come across innumerable numerical figures, called data, in our day-to-day life. For example, when we read a newspaper, we find information about the storage level of water in a dam or the quantity of inflow of water into the dam. These numerical facts are recorded at regular time intervals in order to know about their future trend. The data collected may be huge in size and so a scientific method is needed to handle them in order to derive purposeful information. Statistics is that branch of applied mathematics which deals with the scientific analysis of data. The subject had been started in the early days as an arithmetic to assist a ruler(a political state) who needed to know the wealth of his subjects to levy new taxes. Now, it has developed to a great extent and plays a vital role in almost all organizations in their decision making and planning. The word ‘Statistics’ is derived from the Latin word ‘Status’ which means ‘political state’. We shall review what we have already learnt in our earlier classes about collection of data and their presentation. Data are of two kinds, primary data and secondary data. The data collected by the investigator himself is known as a primary data. Sometimes an investigator utilizes the primary data of another investigator collected for a different purpose. Such data are called secondary data. The data collected by an investigator is called ungrouped data or raw data. This raw data can be condensed in a proper way by grouping and presenting it in the form of a table, called a frequency table. Data presented in the form of a frequency table is said to form a grouped data. For example, consider the raw data of marks of 30 students in mathematics given below. 31 39 37 46 39 49 42 31 31 40 43 46 48 42 30 43 42 42 46 48 40 56 56 50 37 50 45 37 45 48 Let us arrange the given marks in the ascending order. Then, we get 30, 31, 31, 31, 37, 37, 37, 39, 39, 40, 40, 42, 42, 42, 42, 43, 43, 45, 45, 46, 46, 46, 48, 48, 48, 49, 50, 50, 56, 56 In the above list, the mark 30 appears once, 31 thrice, 37 thrice and so on. Counting the data in this way we get a following table called a frequency table for ungrouped data. Let x denote the mark and f denote the number of students or frequency of mark x. x 30 31 37 39 40 42 43 45 46 48 49 50 56 f 1 3 3 2 2 4 2 2 3 3 1 2 2
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Here, x is called the variable or variate of measurement (here mark) and f, the frequency or the number. of times of the occurrence of a particular value of the variable. Here, the largest value = 56 and the smallest value = 30. So, The range = The largest value − The smallest value = 56 − 30 = 26. We shall form intervals called class intervals to include the given marks. The first class interval is 30-34. This interval includes the marks 30, 31, 32 ,33 and 34. The next class interval is 35-39. Proceeding in this way the last class interval is 55-59 which includes the marks 55, 56, 57, 58 and 59. The class intervals are inclusive since the lower and upper limits of each interval are included in that interval. Now, we shall form the frequency table. Class Interval 30-34 35-39 40-44 45-49 50-54 55-59
Tally bars Frequency |||| 4 |||| 5 |||| ||| 8 |||| |||| 9 || 2 || 2 Total 30
Read the observation in the data one by one. For each observation, locate the class interval in which the observation lies and to account this, put a vertical bar like ‘|’ (called tally bar) in the box against the class interval. For every 5th observation that occurs in a class interval, put a cross tally bar like ‘ \ ’ across the four tally bars already there. This process is carried out till all observations are exhausted. In the above table, the number of tally bars marked for a particular class is called the frequency of the class. The table is called a frequency table for grouped data. In this table, the intervals do not cover marks such as 34.5, 39.5. To cover such situations, we can alter the intervals as 29.5-34.5, 34.5-39.5, …, 54.5-59.5 with the convention that each interval does not include its upper limit. The modified frequency table is presented below. Class Interval 29.5-34.5 34.5-39.5 39.5-44.5 44.5-49.5 49.5-54.5 54.5-59.5
Tally bars Frequency |||| 4 |||| 5 |||| ||| 8 |||| |||| 9 || 2 || 2 Total 30
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Here, we say that the frequency table represents continuous variation of the variate x. In this representation, the difference between the upper limit and the lower limit of a class interval is called the size of the class interval and the average of the upper limit and the lower limit is called the class mark of that interval. Here, for the class interval 34.5-39.5, the size is 5 and the class mark is 37, the mid value of the interval. From the above table we observe that the number of students who have obtained marks below 34.5 is 4, the number of students who have obtained marks below 39.5 is 4 + 5 = 9, the number of students who have obtained marks below 44.5 is 4 + 5 + 8 = 17, the number of students who have obtained marks below 49.5 is 4 + 5 + 8 + 9 = 26, the number of students who have obtained marks below 54.5 is 4 + 5 + 8 + 9 + 2 = 28 and the number of students who have obtained marks below 59.5 is 4 + 5 + 8 + 9 + 2 + 2 = 30. These frequencies are called cumulative frequencies (c.f) corresponding to the frequency table. The table of c.f’s is given below. Class Interval Mid-value x f 29.5-34.5 32 4 34.5-39.5 37 5 39.5-44.5 42 8 44.5-49.5 47 9 49.5-54.5 52 2 54.5-59.5 57 2 Total 30
c.f 4 9 17 26 28 30
10.1 Measures of Central Tendency Having presented the raw data in the form of a frequency table, we are able to get a satisfactory picture of the data. To get more information about the tendency of the data to deviate about a particular value, there are certain measures which characterize the entire data. These measures are called the Measures of Central Tendency. They are also called the Measures of Location. Some such measures are 1. Arithmetic mean 2. Median 3. Mode. 10.1.1 Arithmetic mean Consider the observations 11, 22, 7, 33, 27. If we subtract 20 from each of the observations, we get −9, 2, −13, 13, 7. Adding all these differences, we get 0. This means that the number 20 is centrally located to the given 5 observations. It is the mean or average or arithmetic mean of the observations. In general, the arithmetic mean (A.M) or simply the mean or average of n observations x1, x2, …, xn is defined to be the number x such that the sum of the deviations of the observations from x is 0. That is, the arithmetic mean x of n observations x1, x2, …, xn is given by the equation (x1 − x ) +(x2 − x ) + ... +(xn − x ) = 0 or ( x1 + x 2 + ... + x n ) − n × x = 0.
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Hence
x=
x 1 + x 2 + ... + x n . n
In mathematics, the symbol ∑, called sigma notation is used to represent summation. With n
this symbol, the sum x1 + x 2 + ... + x n is denoted as x=
∑x n
i
∑x i =1
i
∑x
or simply as
i
. Then, we have
.
If the observations are represented in the form of a frequency table, the mean x is given by f x + f 2 x 2 + ... + f n x n x= 1 1 , f1 + f 2 + .... + f n where x1, x2, …, xn are the individual values or the mid-values of the class intervals whose frequencies are f1, f2,…,fn. In this case, with sigma notation, we have ∑ f i xi , where N = f + f + .... + f . x = 1 2 n N If the observed values x1, x2, …, xn are numerically large, the mean can be calculated by a short-cut method. Let A be a suitably chosen number. We form the deviations x1− A, x2 − A, …, xn − A. If these deviations have a common factor c, then form the ratios x −A x1 − A x 2 − A , ,..., n . c c c Let them be d1, d2, …,dn.. Then x −A x −A x −A ∑ f i d i = f1 × d1 + f2 × d2 + …..+ fn × dn = f1 × 1 c + f 2 × 2 c + ... + f n × n c . 1 = [( f 1 x1 − f 1 A) + ( f 2 x 2 − f 2 A) + ... + ( f n x n − f n A)] c 1 1 = [( f 1 x1 + f 2 x 2 + ... + f n x n ) − A( f 1 + f 2 + ... + f n )] = [∑ f i xi − A × N ]. c c ∴ ∑ f i xi − A × N = c × ∑ f i d i or
∑fx i
i
= A × N + c ∑ f i d i or x =
∑fx i
i
N
= A + c×
∑fd i
N
Example 1: Calculate the mean of the data 9, 11, 13, 15, 17, 19. ∑ xi = 9 + 11 + 13 + 15 + 17 + 19 = 14. Solution : x = 6 N
Example 2: Compute the A.M. of the following data: 10 11 13 15 x 4 5 8 6 f
216
16 4
19 3
i
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Solution: Direct Method:
Short-cut Method: Take A = 14, c = 1, d = x − A
f×x x f 10 4 40 11 5 55 13 8 104 15 6 90 16 4 64 19 3 57 Total N = 30 ∑ fx = 410
∴
x=
∑fx i
N
i
=
x f 10 4 11 5 13 8 15 6 16 4 19 3 Total N=30
410 = 13.67. 30
x = A+ c ×
d f×d −4 −16 −3 −15 −1 − 8 1 6 2 8 5 15 ∑fd =29−39 = −10
∑ fd =14 + 1
N ≈14 − 0.33 = 13.67.
×
− 10 30
Example 3: Calculate the A.M. for the following data: 80 85 5 6 x− A Solution: Take A = 90 , c = 5 and d = . c X f
90 6
Marks No. of students
100 1
f×d
d
80
5
−2
−10
85
6
−1
−6
90
6
0
0
95
2
1
2
100
1
2
2 ∑ fd = −12
N = 20 ∴
95 2
A.M. = x = A+ c ×
∑ fd
N − 12 = 90 + 5 × = 90 − 3 = 87. 20
Example 4: Calculate A .M for the following data: Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 12 18 27 20 17 6 Marks
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Solution: Direct Method: f×x
Class Mid-value x Frequency f
0-10
5
12
60
10-20
15
18
270
20-30
25
27
675
30-40
35
20
700
40-50
45
17
765
50-60
55
6
330 ∑ fx = 2800
N = 100
From the table, we get N = the total frequency = 100, ∴
x=
Short-cut Method:
Let A = 30, c = 5 and d =
∑ fx N
=
∑ fx = 2800.
2800 = 28. 100
x− A . c
Class Mid-value x d Frequency f 0-10 5 −5 12 10-20 15 −3 18 20-30 25 −1 27 30-40 35 1 20 40-50 45 3 17 50-60 55 5 6 N = 100
∴
A.M. = x = A+ c ×
f×d
−60 −54 −27 20 51 30 ∑ fd = 101−141= −40
∑ fd
N − 40 = 30 + 5 × = 30 − 2 = 28. 100
Exercise 10.1.1 1. Calculate the mean of 7, 12, 18, 14, 19, 20. 2. If in a class of 15 students, 4 students have scored 66 marks, 5 students have scored 67 marks and 6 students have scored 68 marks, then compute the mean of the class. 3. Calculate the Arithmetic mean of the following data:
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5 4
X f
10 5
15 7
20 4
25 3
30 2
70 6
75 3
80 6
85 4
90 10
100 4
45 14
55 16
65 11
75 7
85 8
4. Obtain the A.M. of the following: Marks No. of students
65 11
5. Find the A. M. of the following data: Variable Frequency
15 12
25 20
35 15
6. Find the Arithmetic mean from the following table: Class interval Frequency
10-20
20-30
30-40
40-50
50-60
8
14
7
10
11
10.1.2 Median
When the given raw data are arranged in ascending or descending order, we can find a value which is centrally located in the arranged order. This central value or the middle most value is called the median of the data. For example, consider the data 14, 28, 20, 29, 18, 25, 26, 17, 36. Arranging them in the ascending order, we get 14,17,18,20,25,26,28,29,36. We observe that 25 is centrally located in the series. Hence, it is the median of the data. We note that there are odd number of observations and so we are able to locate the median as an observed value in the series. Consider the data 85, 79, 57, 59, 66, 26, 40, 33, 48, 53. Arranging them in the ascending order, we get 26, 33, 40, 48, 53, 57, 59, 66, 79, 85. Since there are even number of observations, we get 53 and 57 centrally located in the series. Hence, we take their average, 53 + 57 110 namely = = 55 as the middle most value for the series. This is the median of the 2 2 given data. Thus, to get the median for a raw data, we proceed as follows: First we arrange the entire data in the ascending order. Let N be the number of ⎛ N + 1 ⎞ th observations. If N is an odd integer, then there is only one middle term and it is the ⎜ ⎟ ⎝ 2 ⎠ term of the given set of observations. This is the median. If N is an even integer, there are two
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middle terms. They are the
N 2
th
⎛N ⎞ term and ⎜ + 1⎟ ⎝2 ⎠
th
term. Hence the median is the average of
these two terms. When the data are arranged in the form of a frequency table, we proceed to find the median as follows: N First, form the cumulative frequency column and then find the value of where N is 2 the total frequency. Then we locate the class interval or the value of the variate in the table for N N or just greater than . This is the which the cumulative frequency is either equal to 2 2 median class of this distribution. The value of the variable in the table corresponding to the median class is the median of the given data. Example 5: Find the median of 23, 25, 29, 30, 39. Solution: The given values are already in the ascending order. No. of observations N = 5. ⎛ N + 1 ⎞ th ⎛ 5 + 1 ⎞ th This is an odd number. So the median = ⎜ ⎟ term = ⎜ ⎟ term ⎝ 2 ⎠ ⎝ 2 ⎠ = 3 rd term =29. ∴ Median = 29. Example 6: Find the median of 3, 4, 10, 12, 27, 32, 41, 49, 50, 55, 60, 63, 71, 75, 80. Solution: The given values are already in the ascending order. N = No. of observations = 15, odd number. ⎛ N + 1 ⎞ th ⎛ 15 + 1 ⎞ th th ∴ Median = ⎜ ⎟ term = ⎜ ⎟ term = 8 term = 49. ⎝ 2 ⎠ ⎝ 2 ⎠ Example 7: Find the median of 29, 23, 25, 29, 30, 25, 28. Solution: Arranging the observations in the ascending order, we get 23, 25, 25, 28, 29, 29, 30. N = Number of observations = 7, odd integer. ⎛ N + 1 ⎞ th ⎛ 7 + 1 ⎞ th th ∴ Median = ⎜ ⎟ term = ⎜ ⎟ term = 4 term = 28. ⎝ 2 ⎠ ⎝ 2 ⎠ Example 8: Find the median of 26, 25, 29, 23, 25, 29, 30, 25, 28, 30. Solution: Arranging the observations in the ascending order, we get 23, 25, 25, 25, 26, 28, 29, 29, 30, 30. N = No. of observations = 10, an even integer. N th ⎛N ⎞ and ⎜ + 1⎟ th terms = average of 5th and 6th terms ∴ Median = average of 2 ⎝2 ⎠ 26 + 28 = average of 26 and 28 = = 27. 2
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Example 9: Calculate the median of the following table: Variable ( x) Frequency( f )
5 3
10 6
15 10
20 8
25 2
30 3
Solution: x
f
5 10 15 20 25 30
3 6 10 8 2 3
Total frequency = N = ∑f = 32 and so
Cumulative frequency 3 9 19 27 29 32
N = 16. 2
⎛N⎞ The median is the ⎜ ⎟ th value = 16th value. But the 16th value occurs in the class whose ⎝2⎠ cumulative frequency is 19. The corresponding value of the variate is 15. Hence, the median = 15.
Exercise 10.1.2 1. Find the median of the following set of variables: (i) 66, 63, 55, 60, 46, 10 (ii) 35, 39, 36, 34, 28, 27, 45, 41 (iii) 60, 61, 60, 58, 57, 59, 70 (iv) 41, 45, 36, 37, 43, 45, 41, 36 2. Find the median for the marks of 40 students 24 20 Marks 4 7 No. of students 3. Find the median for the following data 1 2 x 4 6 f
3 5
35 3
52 9
4 3
50 5
48 12
5 2
6 5
55 10
65 5
4. The wages of 43 employees are given. Find the median. Wage No. of Employees
25 3
35 5
221
45 20
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10.1.3 Mode
Mode is also a measure of central tendency. (i) In a set of individual observations, mode is defined as the value which occurs most often. (ii) If the data are arranged in the form of a frequency table, the class corresponding to the maximum frequency is called the modal class. The value of the variate of the modal class is the mode. Example 10: Find the mode of 7, 4, 5, 1, 7, 3, 4, 6,7. Solution: Arranging the data in the ascending order, we get 1, 3, 4, 4, 5, 6, 7, 7, 7. In the above data 7 occurs maximum number of times. Hence mode = 7. Example 11: Find the mode of 19, 20, 21, 24, 27, 30. Solution: The data are already in the ascending order. Each value occurs exactly one time in the series. Hence there is no mode for this data. Example 12: Find the mode for 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, 15. Solution: Arranging the data in the ascending order, we get 11, 12, 12, 12, 15, 15, 15, 19, 19, 20, 24, 27. Here 12 occurs 3 times and 15 also occurs 3 times. ∴ both 12 and 15 are the modes for this data. We observe that there are two modes for the given data. Example 13: Find the mode from the following frequency table: Wage No. of Employees
45 12
50 11
55 14
60 13
65 12
70 10
75 9
Solution: We observe from the table that the maximum frequency is 14. The value of the variate (wage) corresponding to the maximum frequency 14 is 55. This is the mode of the data.
Exercise 10.1.3 1. Find the mode of the following data:
(i) (ii) (iii) (iv)
84, 91, 72, 68, 87, 84 65, 61, 72, 81, 51, 31 38, 31, 22, 20, 31, 61, 15, 20 15, 11, 18, 23, 11, 19, 11
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2. Find the mode for the following distribution x f
10 8
20 15
30 12
40 10
50 9
60 6
62 14
63 16
64 10
65 7
3. Find the mode for the following table. x f
60 5
61 8
Answers Exercise 10.1.1 1. 15
2. 67.1
3. 15.6
4. 79.1
5. 45
Exercise 10.1.2 1. (i) 57.5
(ii) 35.5
(iii) 60
3. 3
4. 45
2. 48
(iv) 41
Exercise 10.1.3 1. (i) 84 2. 20
(ii) No mode
(iii) 20, 31
3. 63
223
(iv) 11
6. 35.4
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11. GRAPHS In fields such as science, engineering and business, we come across several variables which take real values. For example, in business, the supply (s) and price (p) of a commodity are real variables. These variables may be connected by an equation. Using this equation, we can get a value for p for each value of s and obtain a set of ordered pairs (s, p) of real numbers. All these ordered pairs (s, p) can be plotted as points in the Cartesian plane where its horizontal axis is the s-axis and the vertical axis is the p-axis. These points now define what is called the graph of the relation. The graph displays the nature of relationship between the variables. One of the most useful graph that we obtain quite often is the linear graph. We now proceed to know about linear graphs, how they are drawn and applied to solve some equations.
11.1 Linear Graphs Let x and y be two variables. If they are connected by an equation of the form y = mx + c, then we say that x and y are linearly related. We have already seen in the chapter on algebraic geometry that the equation y = mx + c represents a straight line in the Cartesian plane. This is the reason why the relationship between x and y is called linear. For each value of x, the equation y = mx + c gives a value of y and we obtain an ordered pair (x, y) of numbers. The set of all such ordered pairs defines the graph of y = mx + c, called a linear graph. We recall that in the equation y = mx + c, the number m is called the slope of the line and c is known as the y-intercept. The y-intercept is the value of y when x = 0. Sometimes the y-intercept c is 0. In this situation, the equation of the line is y = mx and we say that the line passes through the origin. We now proceed to draw linear graphs under several situations. The basic principle behind drawing a linear graph is that we need only two points to graph a straight line. The following procedure is followed in drawing linear graphs: Step 1:By substituting two different values for x in the equation y = mx + c, we get two values for y. Thus we get two points (x1, y1) and (x2, y2) on the line. Step 2:Draw the x-axis and y-axis on the graph paper and choose a suitable scale on the coordinate axes. The scale for both the axes is chosen based on the values of the coordinates obtained in step 1. If the coordinate values are large, then 1 cm along the axes may be taken to represents large number of units. Step 3:Plot the two points (x1, y1) and (x2, y2) in the Cartesian plane of the paper. Step 4:Join the two points by a line segment and extend it in both directions of the segment. This is the required linear graph.
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Example 1: Draw the graph of the line joining the points (2, 3) and (−4, 1). Solution: Draw the x-axis and y-axis on a graph paper and take 1 cm = 1 unit on both the axes. Let A and B be the points (2, 3) and (−4, 1). We mark these points on the graph paper. We join the points A and B by a line segment and extend it along the two directions. The required graph is now obtained (see Figure 11.1)
Figure 11.1 Example 2: Draw the graph of y = 2x. Solution: Since the equation of the line is y = 2x, the line passes through the origin. Substituting x = −1, 0, 1 in the equation of the line , we get correspondingly y = −2, 0, 2. We form the table as given below:
x
−1
0
1
y
−2
0
2
We plot the points (−1, −2), (0, 0) and (1, 2) in Figure 11.2 the graph sheet by taking 1 cm = 1 unit for both the axes. We join the points by a line segment and extend it in both directions. We get the required linear graph (see Figure 11.2). Example 3: Draw the graph y = 3x −1. Solution: Substituting x = −1, 0, 1 in the equation of the line, we get y = −4, −1, 2 correspondingly. In a graph, plot the points (−1, −4), (0, −1) and (1, 2).
x
−1
0
1
y
−4
−1
2
Join the points by a line segment and extend it in both directions. Thus we get the required linear graph ( see Figure 11.3).
Figure 11.3
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Example 4: Draw the graph of the line whose slope is − 3 and y-intercept is −3. 2
Solution: The equation of the line is y = mx + c or −3 −3 y= x + (−3) or y = x −3 2 2 Substituting x = −2, 0, 2, we get y = 0, −3, −6 respectively. Plot the points (−2, 0), (0, −3) and (2,−6) in the graph paper. x y
−2 0
0
2
−3
−6
Join the points by a line segment and extend it in both directions. We now get the required linear graph (see Figure 11.4).
Figure 11.4
Example 5: Draw the graph of the line 2x + 3y = 12. Solution: The given equation is rewritten as ⎛−2⎞ 3y = −2x + 12 or y = ⎜ ⎟ x + 4. ⎝ 3 ⎠ Substituting x = −3, 0, 3, we get y = 6, 4, 2 respectively. Plot (−3, 6), (0, 4) and (3, 2) in the graph sheet. x y
−3 6
0 4
3 2 Figure 11.5
Join the points by a line segment and extend it in both the directions. This is the required linear graph.(see Figure 11.5). Example 6: Draw the graph x = 3. Solution: We observe that y is not specified in the equation x = 3. So, any value of y gives x = 3. Choose any two values say 1 and 2 for y and get two points (3, 1) and (3, 2)on the line x = 3. x y
3 1
3 2 Figure 11.6 226
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Plot these points on the graph paper. Join these two points by a line segment and extend it in both the directions. We get the required linear graph (see Figure 11.6). Note that the line is parallel to the y-axis. Example 7: Draw the graph of y = −4. Solution: We observe that the value of y is fixed as −4 and the value of x is not specified in the equation. So, we choose any two values for x, say x = −2, 2. Then we get two points (−2, −4) and (2, −4) on the line y = −4. Plot these two points in the graph paper. x y
−2 −4
2 −4 Figure 11.7
Join the two points and extend it in the both directions. The required graph is the resulting graph(see Figure 11.7) The line is parallel to the x–axis. Exercise 11.1 1. Draw the linear graph through the points (i) (2, 3) and (4, −6) (ii) (−1, 0) and (−2, −5) (iii) (−3, 2) and (5, −1) (iv) (−2, −3) and (5, −4) 2. Draw the graph of the following: (i) y = −2x (ii) y = 3x (iii) x = 5y (iv) x = −4y 3. Draw the graphs of the following: (i) x = −3 (ii) y = 5 (iii) x = 5 (iv) y = −4 (v) 2x + 3 = 0. (vi) 1 + 2y =0. 4. Draw the graph for y = mx + c when −2 and c = 3. (i) m = 3 and c = 4. (ii) m = 3 (iii) m = −3 and c = −4 (iv) m = 2 and c = −5 5. Draw the graph of the following equations: (i) 2x + 3y = 12. (ii) x −5y = 10. (iii) y + 2x −5 = 0. (iv) x −2y + 1 = 0.
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11.2 Application of Linear Graphs Without doing any algebraic manipulations, we can solve two simultaneous linear equations in x and y by drawing the graphs corresponding to the equations together. A linear equation in x and y is of the form ax + by + c = 0. The equation represents a straight line, So, the problem of solving two simultaneous linear equations in x and y reduces to the problem of finding the common point between the two corresponding lines. Here, three cases arise. In the first case, the two linear graphs i.e., lines are coincident; that is, the graphs are one and the same. In this situation, there are infinitely many points common to both the graphs. That is, there are infinitely many solutions to the given equations. In the second case, the linear graphs are not coincident but are parallel. In this situation, the two linear graphs do not meet at all. So there is no point common to both the lines. Hence the simultaneous equations have no solution. In the third case, the two linear graphs intersect exactly at one point. In this situation, the given simultaneous equations possess a unique solution namely the coordinates of the intersecting point. Example 8: Solve graphically the simultaneous equations 2x + y = 1 and 4x + 2y = 2. Solution: We graph the two equations together. Line 1: y = −2x + 1 x y
−1 3
Line 2: 2y = −4x + 2 i.e., y = −2x + 1 x y
1 −1
−1 3
Plot the points corresponding to the two equations in the same graph paper. Join the corresponding points by line segment and extend them in both the directions. Then we get two coincident lines (see Figure 11.8). Any point on one line is also a point on the other. That is, there are infinitely many points common to both the equation. So there are infinitely many solutions for the given simultaneous equations. Figure 11.8
228
1 −1
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Example 9: Draw the graphs x − 2y = 4 and x − 2y = −6 and hence solve the simultaneous equations. Solution: We find points for plotting the two lines. Line 1:
Line 2: x −2y = −6 or 2y = x + 6
x − 2y = 4
or
2y = x −4
or
1 y = ⎛⎜ ⎞⎟ x −2 ⎝2⎠
X Y
1 y = ⎛⎜ ⎞⎟ x + 3
or
0 −2
⎝2⎠
0 3
x y
2 −1
2 4
We plot the points (0, −2) and (2 ,−1) in the graph paper and draw the line through them. Next, we plot the points (0, 3) and (2, 4) in the same graph paper and draw the line through them. We find that the two linear graphs are parallel. So they do not intersect. Hence the simultaneous equations have no solution (see Figure 11.9).
Figure 11.9 Example 10: Solve graphically the simultaneous equations x + y = 5, x − y = 3. Solution: We draw the graphs for the two equations in the same graph sheet. Line (1): x + y = 5 Line (2) : x −y = 3 or y = −x + 5 or y = x −3 X Y
−2 7
−1 6
x y
3 2
Plot the points (−2, 7) , (−1, 6) and (3, 2) on the graph paper. Draw the line passing through them. This is the linear graph for line (1). Next, plot the points (1, −2) (0, −3) and (3, 0) in the same graph paper. Draw the line passing through these points. This is the linear graph for line (2). The two linear graphs intersect at the point P (4, 1) (see Figure 11.10). Since this point lies on both the lines, the solution of the simultaneous equation is x = 4, y = 1.
229
1 −2
0 −3
Figure 11.10
3 0
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Exercise 11.2 In problems 1 to 10, solve graphically the following system of equations: 6. 2x + y = 1, 4x + 2y = 2. 1. x + y = 0, x = 4. 7. x + 2y = 4, x + 2y = 6. 2. x − y = 0, y = −3. 8. x − 3y = 4, x + 2y = −1. 3. x + y = 2, x − y = 2. 9. 3x + y = 2, 6x − y = 7. 4. x −y = 6, 2x + y = 9. 10. 2x + 3 = 0, 4x + y + 4 = 0. 5. x + y = 5, x − y = 1. Answers Exercise 11.1 1. (ii)
1. (i)
1. (iii)
1. (iv) 2 (i) x y
−1 2
0 0
1 −2
2. (ii) x y
−1 −3
0 0
1 3
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2. (iii)
2. (iv) 0 0
x y
3.
−5 −1
5 1
(i)
x y
0 0
4 −1
−4 1
x y
1 5
2 5
−1 5
x y
0 −4
1 −4
−1 −4
(ii) x y
−3 1
−3 2
−3 −1
(iii)
(iv) x y
5 0
5 1
5 2
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(v)
(vi) x y
4
−1.5 0
−1.5 1
−1.5 −1
(i)
x y
0 −0.5
1 −0.5
−1 −0.5
−3 5
0 3
3 1
0 −5
1 −3
2 −1
(ii) x y
−1 1
0 4
1 7
(iii)
x y
(iv) x y
0 −4
1 −7
−1 −1
x y
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5
(i)
(ii) x y
−3 6
0 4
x y
3 2
(iii)
5 −1
0 −2
−5 −3
1 1
3 2
5 3
(iv) x y
−1 7
0 5
x y
1 3
Exercise 11.2 1. x
−1
0
1
x
4
4
4
y
1
0
−1
y
1
0
−1
Solution is x = 4; y = −4. 2. x
−1
0
1
x
0
1
2
y
−1
0
1
y
−3
−3
−3
Solution is x = −3; y = −3.
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3. x
0
1
2
x
3
4
5
y
2
1
0
y
1
2
3
Solution is x = 2; y = 0. 4. x
3
4
5
x
2
3
4
y
−3
−2
−1
y
5
3
1
Solution is x = 5; y = −1. 5. x
2
3
4
x
2
3
4
y
3
2
1
y
1
2
3
x y
−1 0 3 1
Solution is x = 3; y =2. 6.
x y
-1 3
0 1
1 −1
1 −1
Infinitely many solutions.
7. x
0
2
4
x
0
2
4
y
2
1
0
y
3
2
1
No solution.
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8. x
−2
1
4
x
−1
1
3
y
−2
−1
0
y
0
−1
−2
Solution is x = 1; y = −1.
9. x
0
1
2
x
0
1
2
y
2
−1
−4
y
−7
−1
5
x
−2
−1
0
y
4
0
−4
Solution is x = 1; y = −1.
10.
x
−3 2
−3 2
−3 2
y
−2
0
2
Solution is x =
−3; 2
y = 2.
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LOGARITHMS Mean Differences 0
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1.0
.0000
.0043
.0086
.0128
.0170
.0212
.0253
.0294
.0334
.0374
4
8
12
17
21
25
29
33
37
1.1 1.2 1.3 1.4 1.5
.0414 .0792 .1139 .1461 .1761
.0453 .0828 .1173 .1492 .1790
.0492 .0864 .1206 .1523 .1818
.0531 .0899 .1239 .1553 .1847
.0569 .0934 .1271 .1584 .1875
.0607 .0969 .1303 .1614 .1903
.0645 .1004 .1335 .1644 .1931
.0682 .1038 .1367 .1673 .1959
.0719 .1072 .1399 .1703 .1987
.0755 .1106 .1430 .1732 .2014
4 3 3 3 3
8 7 6 6 6
11 10 10 9 8
15 14 13 12 11
19 17 16 15 14
23 21 19 18 17
26 24 23 21 20
30 28 26 24 22
34 31 29 27 25
1.6 1.7 1.8 1.9 2.0
.2041 .2304 .2553 .2788 .3010
.2068 .2330 .2577 .2810 .3032
.2095 .2355 .2601 .2833 .3054
.2122 .2380 .2625 .2856 .3075
.2148 .2405 .2648 .2878 .3096
.2175 .2430 .2672 .2900 .3118
.2201 .2455 .2695 .2923 .3139
.2227 .2480 .2718 .2945 .3160
.2253 .2504 .2742 .2967 .3181
.2279 .2529 .2765 .2989 .3201
3 2 2 2 2
5 5 5 4 4
8 7 7 7 6
11 10 9 9 8
13 12 12 11 11
16 15 14 13 13
18 17 16 16 15
21 20 19 18 17
24 22 21 20 19
2.1 2.2 2.3 2.4 2.5
.3222 .3424 .3617 .3802 .3979
.3243 .3444 .3636 .3820 .3997
.3263 .3464 .3655 .3838 .4014
.3284 .3483 .3674 .3856 .4031
.3304 .3502 .3692 .3874 .4048
.3324 .3522 .3711 .3892 .4065
.3345 .3541 .3729 .3909 .4082
.3365 .3560 .3747 .3927 .4099
.3385 .3579 .3766 .3945 .4116
.3404 .3598 .3784 .3962 .4133
2 2 2 2 2
4 4 4 4 3
6 6 6 5 5
8 8 7 7 7
10 10 9 9 9
12 12 11 11 10
14 14 13 12 12
16 15 15 14 14
18 17 17 16 15
2.6 2.7 2.8 2.9 3.0
.4150 .4314 .4472 .4624 .4771
.4166 .4330 .4487 .4639 .4786
.4183 .4346 .4502 .4654 .4800
.4200 .4362 .4518 .4669 .4814
.4216 .4378 .4533 .4683 .4829
.4232 .4393 .4548 .4698 .4843
.4249 .4409 .4564 .4713 .4857
.4265 .4425 .4579 .4728 .4871
.4281 .4440 .4594 .4742 .4886
.4298 .4456 .4609 .4757 .4900
2 2 2 1 1
3 3 3 3 3
5 5 5 4 4
7 6 6 6 6
8 8 8 7 7
10 9 9 9 9
11 11 11 10 10
13 13 12 12 11
15 14 14 13 13
3.1 3.2 3.3 3.4 3.5
.4914 .5051 .5185 .5315 .5441
.4928 .5065 .5198 .5328 .5453
.4942 .5079 .5211 .5340 .5465
.4955 .5092 .5224 .5353 .5478
.4969 .5105 .5237 .5366 .5490
.4983 .5119 .5250 .5378 .5502
.4997 .5132 .5263 .5391 .5514
.5011 .5145 .5276 .5403 .5527
.5024 .5159 .5289 .5416 .5539
.5038 .5172 .5302 .5428 .5551
1 1 1 1 1
3 3 3 3 2
4 4 4 4 4
6 5 5 5 5
7 7 6 6 6
8 8 8 8 7
10 9 9 9 9
11 11 10 10 10
12 12 12 11 11
3.6 3.7 3.8 3.9 4.0
.5563 .5682 .5798 .5911 .6021
.5575 .5694 .5809 .5922 .6031
.5587 .5705 .5821 .5933 .6042
.5599 .5717 .5832 .5944 .6053
.5611 .5729 .5843 .5955 .6064
.5623 .5740 .5855 .5966 .6075
.5365 .5752 .5866 .5977 .6085
.5647 .5763 .5877 .5988 .6096
.5658 .5775 .5888 .5999 .6107
.5670 .5786 .5899 .6010 .6117
1 1 1 1 1
2 2 2 2 2
4 3 3 3 3
5 5 5 4 4
6 6 6 5 5
7 7 7 7 6
8 8 8 8 8
10 9 9 9 9
11 10 10 10 10
4.1 4.2 4.3 4.4 4.5
.6128 .6232 .6335 .6435 .6532
.6138 .6243 .6345 .6444 .6542
.6149 .6253 .6355 .6454 .6551
.6160 .6263 .6365 .6464 .6561
.6170 .6274 .6375 .6474 .6571
.6180 .6284 .6385 .6484 .6580
.6191 .6294 .6395 .6493 .6590
.6201 .6304 .6405 .6503 .6599
.6212 .6314 .6415 .6513 .6609
.6222 .6325 .6425 .6522 .6618
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
4.6 4.7 4.8 4.9 5.0
.6628 .6721 .6812 .6902 .6990
.6637 .6730 .6821 .6911 .6998
.6646 .6739 .6830 .6920 .7007
.6656 .6749 .6839 .6928 .7016
.6665 .6758 .6848 .6937 .7024
.6675 .6767 .6857 .6946 .7033
.6684 .6776 .6866 .6955 .7042
.6693 .6785 .6875 .6964 .7050
.6702 .6794 .6884 .6972 .7059
.6712 .6803 .6893 .6981 .7067
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 3
5 5 4 4 4
6 5 5 5 5
7 6 6 6 6
7 7 7 7 7
8 8 8 8 8
5.1 5.2 5.3 5.4
.7076 .7160 .7243 .7324
.7084 .7168 .7251 .7332
.7093 .7177 .7259 .7340
.7101 .7185 .7267 .7348
.7110 .7193 .7275 .7356
.7118 .7202 .7284 .7364
.7126 .7210 .7292 .7372
.7135 .7218 .7300 .7380
.7143 .7226 .7308 .7388
.7152 .7235 .7316 .7396
1 1 1 1
2 2 2 2
3 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 6 6
8 7 7 7
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LOGARITHMS Mean Differences 0
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
5.5
.7404
.7412
.7419
.7427
.7435
.7443
.7451
.7459
.7466
.7474
1
2
2
3
4
5
5
6
7
5.6 5.7 5.8 5.9 6.0
.7482 .7559 .7634 .7709 .7782
.7490 .7566 .7642 .7716 .7789
.7497 .7574 .7649 .7723 .7796
.7505 .7582 .7657 .7731 .7803
.7513 .7589 .7664 .7738 .7810
.7520 .7597 .7672 .7745 .7818
.7528 .7604 .7679 .7752 .7825
.7536 .7612 .7686 .7760 .7832
.7543 .7619 .7694 .7767 .7839
.7551 .7627 .7701 .7774 .7846
1 1 1 1 1
2 2 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 4 4 4
5 5 5 5 5
6 6 6 6 6
7 7 7 7 6
6.1 6.2 6.3 6.4 6.5
.7853 .7924 .7993 .8062 .8129
.7860 .7931 .8000 .8069 .8136
.7868 .7938 .8007 .8075 .8142
.7875 .7945 .8014 .8082 .8149
.7882 .7952 .8021 .8089 .8156
.7889 .7959 .8028 .8096 .8162
.7896 .7966 .8035 .8102 .8169
.7903 .7973 .8041 .8109 .8176
.7910 .7980 .8048 .8116 .8182
.7917 .7987 .8055 .8122 .8189
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 3 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 5 5 5
6 6 6 6 6
6.6 6.7 6.8 6.9 7.0
.8195 .8261 .8325 .8388 .8451
.8202 .8267 .8331 .8395 .8457
.8209 .8274 .8338 .8401 .8463
.8215 .8280 .8344 .8407 .8470
.8222 .8287 .8351 .8414 .8476
.8228 .8293 .8357 .8420 .8482
.8235 .8299 .8363 .8426 .8488
.8241 .8306 .8370 .8432 .8494
.8248 .8312 .8376 .8439 .8500
.8254 .8319 .8382 .8445 .8506
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
3 3 3 2 2
3 3 3 3 3
4 4 4 4 4
5 5 4 4 4
5 5 5 5 5
6 6 6 6 6
7.1 7.2 7.3 7.4 7.5
.8513 .8573 .8633 .8692 .8751
.8519 .8579 .8639 .8698 .8756
.8525 .8585 .8645 .8704 .8762
.8531 .8591 .8651 .8710 .8768
.8537 .8597 .8657 .8716 .8774
.8543 .8603 .8663 .8722 .8779
.8549 .8609 .8669 .8727 .8785
.8555 .8615 .8675 .8733 .8791
.8561 .8621 .8681 .8739 .8797
.8567 .8627 .8686 .8745 .8802
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
3 3 3 3 3
4 4 4 4 3
4 4 4 4 4
5 5 5 5 5
5 5 5 5 5
7.6 7.7 7.8 7.9 8.0
.8808 .8865 .8921 .8976 .9031
.8814 .8871 .8927 .8982 .9036
.8820 .8876 .8932 .8987 .9042
.8825 .8882 .8938 .8993 .9047
.8831 .8887 .8943 .8998 .9053
.8837 .8893 .8949 .9004 .9058
.8842 .8899 .8954 .9009 .9063
.8848 .8904 .8960 .9015 .9069
.8854 .8910 .8965 .9020 .9074
.8859 .8915 .8971 .9025 .9079
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
3 3 3 3 3
3 3 3 3 3
4 4 4 4 4
5 4 4 4 4
5 5 5 5 5
8.1 8.2 8.3 8.4 8.5
.9085 .9138 .9191 .9243 .9294
.9090 .9143 .9196 .9248 .9299
.9096 .9149 .9201 .9253 .9304
.9101 .9154 .9206 .9258 .9309
.9106 .9159 .9212 .9263 .9315
.9112 .9165 .9217 .9269 .9320
.9117 .9170 .9222 .9274 .9325
.9122 .9175 .9227 .9279 .9330
.9128 .9180 .9232 .9284 .9335
.9133 .9186 .9238 .9289 .9340
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
3 3 3 3 3
3 3 3 3 3
4 4 4 4 4
4 4 4 4 4
5 5 5 5 5
8.6 8.7 8.8 8.9 9.0
.9345 .9395 .9445 .9494 .9542
.9350 .9400 .9450 .9499 .9547
.9355 .9405 .9455 .9504 .9552
.9360 .9410 .9460 .9509 .9557
.9365 .9415 .9465 .9513 .9562
.9370 .9420 .9469 .9518 .9566
.9375 .9425 .9474 .9523 .9571
.9380 .9430 .9479 .9528 .9576
.9385 .9435 .9484 .9533 .9581
.9390 .9440 .9489 .9538 .9586
1 0 0 0 0
1 1 1 1 1
2 1 1 1 1
2 2 2 2 2
3 2 2 2 2
3 3 3 3 3
4 3 3 3 3
4 4 4 4 4
5 4 4 4 4
9.1 9.2 9.3 9.4 9.5
.9590 .9638 .9685 .9731 .9777
.9595 .9643 .9689 .9736 .9782
.9600 .9647 .9694 .9741 .9786
.9605 .9653 .9699 .9745 .9791
.9609 .9657 .9703 .9750 .9795
.9614 .9661 .9708 .9754 .9800
.9619 .9666 .9713 .9759 .9805
.9624 .9671 .9717 .9763 .9809
.9628 .9675 .9722 .9768 .9814
.9633 .9680 .9727 .9773 .9818
0 0 0 0 0
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
3 3 3 3 3
3 3 3 3 3
4 4 4 4 4
4 4 4 4 4
9.6 9.7 9.8 9.9
.9823 .9868 .9912 .9956
.9827 .9872 .9917 .9961
.9832 .9877 .9921 .9965
.9836 .9881 .9926 .9969
.9841 .9886 .9930 .9974
.9845 .9890 .9934 .9978
.9850 .9894 .9939 .9983
.9854 .9899 .9943 .9987
.9859 .9903 .9948 .9991
.9863 .9908 .9952 .9996
0 0 0 0
1 1 1 1
1 1 1 1
2 2 2 2
2 2 2 2
3 3 3 3
3 3 3 3
4 4 4 3
4 4 4 4
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ANTILOGARITHMS Mean Differences 0
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
.00
1.000
1.002
1.005
1.007
1.009
1.012
1.014
1.016
1.019
1.021
0
0
1
1
1
1
2
2
2
.01 .02 .03 .04 .05
1.023 1.047 1.072 1.096 1.122
1.026 1.050 1.074 1.099 1.125
1.028 1.052 1.076 1.102 1.127
1.030 1.054 1.079 1.104 1.130
1.033 1.057 1.081 1.107 1.132
1.035 1.059 1.084 1.109 1.135
1.038 1.062 1.086 1.112 1.138
1.040 1.064 1.089 1.114 1.140
1.042 1.067 1.091 1.117 1.143
1.045 1.069 1.094 1.119 1.146
0 0 0 0 0
0 0 0 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 2 2
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
.06 .07 .08 .09 .10
1.148 1.175 1.202 1.230 1.259
1.151 1.178 1.205 1.233 1.262
1.153 1.180 1.208 1.236 1.265
1.156 1.183 1.211 1.239 1.268
1.159 1.186 1.213 1.242 1.271
1.161 1.189 1.216 1.245 1.274
1.164 1.191 1.219 1.247 1.276
1.167 1.194 1.222 1.250 1.279
1.169 1.197 1.225 1.253 1.282
1.172 1.199 1.227 1.256 1.285
0 0 0 0 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
2 2 3 3 3
.11 .12 .13 .14 .15
1.288 1.318 1.349 1.380 1.413
1.291 1.321 1.352 1.384 1.416
1.294 1.324 1.355 1.387 1.419
1.297 1.327 1.358 1.390 1.422
1.300 1.330 1.361 1.393 1.426
1.303 1.334 1.365 1.396 1.429
1.306 1.337 1.368 1.400 1.432
1.309 1.340 1.371 1.403 1.435
1.312 1.343 1.374 1.406 1.439
1.315 1.346 1.377 1.409 1.442
0 0 0 0 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
2 2 3 3 3
3 3 3 3 3
.16 .17 .18 .19 .20
1.445 1.479 1.514 1.549 1.585
1.449 1.483 1.517 1.552 1.589
1.452 1.486 1.521 1.556 1.592
1.455 1.489 1.524 1.560 1.596
1.459 1.493 1.528 1.563 1.600
1.462 1.496 1.531 1.567 1.603
1.466 1.500 1.535 1.570 1.607
1.469 1.503 1.538 1.574 1.611
1.472 1.507 1.542 1.578 1.614
1.476 1.510 1.545 1.581 1.618
0 0 0 0 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
2 2 2 3 3
3 3 3 3 3
3 3 3 3 3
.21 .22 .23 .24 .25
1.622 1.660 1.698 1.738 1.778
1.626 1.663 1.702 1.742 1.782
1.629 1.667 1.706 1.746 1.786
1.633 1.671 1.710 1.750 1.791
1.637 1.675 1.714 1.754 1.795
1.641 1.679 1.718 1.758 1.799
1.644 1.683 1.722 1.762 1.803
1.648 1.687 1.726 1.766 1.807
1.652 1.690 1.730 1.770 1.811
1.656 1.694 1.734 1.774 1.816
0 0 0 0 0
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
3 3 3 3 3
3 3 3 3 3
3 3 4 4 4
.26 .27 .28 .29 .30
1.820 1.862 1.905 1.950 1.995
1.824 1.866 1.910 1.954 2.000
1.828 1.871 1.914 1.959 2.004
1.832 1.875 1.919 1.963 2.009
1.837 1.879 1.923 1.968 2.014
1.841 1.884 1.928 1.972 2.018
1.845 1.888 1.932 1.977 2.023
1.849 1.892 1.936 1.982 2.028
1.854 1.897 1.941 1.986 2.032
1.858 1.901 1.945 1.991 2.037
0 0 0 0 0
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
3 3 3 3 3
3 3 3 3 3
3 3 4 4 4
4 4 4 4 4
.31 .32 .33 .34 .35
2.042 2.089 2.138 2.188 2.239
2.046 2.094 2.143 2.193 2.244
2.051 2.099 2.148 2.198 2.249
2.056 2.104 2.153 2.203 2.254
2.061 2.109 2.158 2.208 2.259
2.065 2.113 2.163 2.213 2.265
2.070 2.118 2.168 2.218 2.270
2.075 2.123 2.173 2.223 2.275
2.080 2.128 2.178 2.228 2.280
2.084 2.133 2.183 2.234 2.286
0 0 0 1 1
1 1 1 1 1
1 1 1 2 2
2 2 2 2 2
2 2 2 3 3
3 3 3 3 3
3 3 3 4 4
4 4 4 4 4
4 4 4 5 5
.36 .37 .38 .39 .40
2.291 2.344 2.399 2.455 2.512
2.296 2.350 2.404 2.460 2.518
2.301 2.355 2.410 2.466 2.523
2.307 2.360 2.415 2.472 2.529
2.312 2.366 2.421 2.477 2.535
2.317 2.371 2.427 2.483 2.541
2.323 2.377 2.432 2.489 2.547
2.328 2.382 2.438 2.495 2.553
2.333 2.388 2.443 2.500 2.559
2.339 2.393 2.449 2.506 2.564
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
3 3 3 3 3
3 3 3 3 4
4 4 4 4 4
4 4 4 5 5
5 5 5 5 5
.41 .42 .43 .44 .45
2.570 2.630 2.692 2.754 2.818
2.576 2.636 2.698 2.761 2.825
2.582 2.642 2.704 2.767 2.831
2.588 2.649 2.710 2.773 2.838
2.594 2.655 2.716 2.780 2.844
2.600 2.661 2.723 2.786 2.851
2.606 2.667 2.729 2.793 2.858
2.612 2.673 2.735 2.799 2.864
2.618 2.679 2.742 2.805 2.871
2.624 2.685 2.748 2.812 2.877
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
2 2 3 3 3
3 3 3 3 3
4 4 4 4 4
4 4 4 4 5
5 5 5 5 5
5 6 6 6 6
.46 .47 .48 .49
2.884 2.951 3.020 3.090
2.891 2.958 3.027 3.097
2.897 2.965 3.034 3.105
2.904 2.972 3.041 3.112
2.911 2.979 3.048 3.119
2.917 2.985 3.055 3.126
2.924 2.992 3.062 3.133
2.931 2.999 3.069 3.141
2.938 3.006 3.076 3.148
2.944 3.013 3.083 3.155
1 1 1 1
1 1 1 1
2 2 2 2
3 3 3 3
3 3 4 4
4 4 4 4
5 5 5 5
5 5 6 6
6 6 6 6
238
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ANTILOGARITHMS Mean Differences 0
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
0.50
3.162
3.170
3.177
3.184
3.192
3.199
3.206
3.214
3.221
3.228
1
1
2
3
4
4
5
6
7
0.51 0.52 0.53 0.54 0.55
3.236 3.311 3.388 3.467 3.548
3.243 3.319 3.396 3.475 3.556
3.251 3.327 3.404 3.483 3.565
3.258 3.334 3.412 3.491 3.573
3.266 3.342 3.420 3.499 3.581
3.273 3.350 3.428 3.508 3.589
3.281 3.357 3.436 3.516 3.597
3.289 3.365 3.443 3.524 3.606
3.296 3.373 3.451 3.532 3.614
3.304 3.381 3.459 3.540 3.622
1 1 1 1 1
2 2 2 2 2
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
5 5 6 6 6
6 6 6 6 7
7 7 7 7 7
0.56 0.57 0.58 0.59 0.60
3.631 3.715 3.802 3.890 3.981
3.639 3.724 3.811 3.899 3.990
3.648 3.733 3.819 3.908 3.999
3.656 3.741 3.828 3.917 4.009
3.664 3.750 3.837 3.926 4.018
3.673 3.758 3.846 3.936 4.027
3.681 3.767 3.855 3.945 4.036
3.690 3.776 3.864 3.954 4.046
3.698 3.784 3.873 3.963 4.055
3.707 3.793 3.882 3.972 4.064
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
3 3 4 4 4
4 4 4 5 5
5 5 5 5 6
6 6 6 6 6
7 7 7 7 7
8 8 8 8 8
0.61 0.62 0.63 0.64 0.65
4.074 4.169 4.266 4.365 4.467
4.083 4.178 4.276 4.375 4.477
4.093 4.188 4.285 4.385 4.487
4.102 4.198 4.295 4.395 4.498
4.111 4.207 4.305 4.406 4.508
4.121 4.217 4.315 4.416 4.519
4.130 4.227 4.325 4.426 4.529
4.140 4.236 4.335 4.436 4.539
4.150 4.246 4.345 4.446 4.550
4.159 4.256 4.355 4.457 4.560
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
0.66 0.67 0.68 0.69 0.70
4.571 4.677 4.786 4.898 5.012
4.581 4.688 4.797 4.909 5.023
4.592 4.699 4.808 4.920 5.035
4.603 4.710 4.819 4.932 5.047
4.613 4.721 4.831 4.943 5.058
4.624 4.732 4.842 4.955 5.070
4.634 4.742 4.853 4.966 5.082
4.645 4.753 4.864 4.977 5.093
4.656 4.764 4.875 4.989 5.105
4.667 4.775 4.887 5.000 5.117
1 1 1 1 1
2 2 2 2 2
3 3 3 3 4
4 4 4 5 5
5 5 6 6 6
6 7 7 7 7
7 8 8 8 8
9 9 9 9 9
10 10 10 10 11
0.71 0.72 0.73 0.74 0.75
5.129 5.248 5.370 5.495 5.623
5.140 5.260 5.383 5.508 5.636
5.152 5.272 5.395 5.521 5.649
5.164 5.284 5.408 5.534 5.662
5.176 5.297 5.420 5.546 5.675
5.188 5.309 5.433 5.559 5.689
5.200 5.321 5.445 5.572 5.702
5.212 5.333 5.458 5.585 5.715
5.224 5.346 5.470 5.598 5.728
5.236 5.358 5.483 5.610 5.741
1 1 1 1 1
2 2 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 6 6 7
7 7 8 8 8
8 9 9 9 9
10 10 10 10 10
11 11 11 12 12
0.76 0.77 0.78 0.79 0.80
5.754 5.888 6.026 6.166 6.310
5.768 5.902 6.039 6.180 6.324
5.781 5.916 6.053 6.194 6.339
5.794 5.929 6.067 6.209 6.353
5.808 5.943 6.081 6.223 6.368
5.821 5.957 6.095 6.237 6.383
5.834 5.970 6.109 6.252 6.397
5.848 5.984 6.124 6.266 6.412
5.861 5.998 6.138 6.281 6.427
5.875 6.012 6.152 6.295 6.442
1 1 1 1 1
3 3 3 3 3
4 4 4 4 4
5 5 6 6 6
7 7 7 7 7
8 8 8 9 9
9 10 10 10 10
11 11 11 11 12
12 12 13 13 13
0.81 0.82 0.83 0.84 0.85
6.457 6.607 6.761 6.918 7.079
6.471 6.622 6.776 6.934 7.096
6.486 6.637 6.792 6.950 7.112
6.501 6.653 6.808 6.966 7.129
6.516 6.668 6.823 6.982 7.145
6.531 6.683 6.839 6.998 7.161
6.546 6.699 6.855 7.015 7.178
6.561 6.714 6.871 7.031 7.194
6.577 6.730 6.887 7.047 7.211
6.592 6.745 6.902 7.063 7.228
2 2 2 2 2
3 3 3 3 3
5 5 5 5 5
6 6 6 6 7
8 8 8 8 8
9 9 9 10 10
11 11 11 11 12
12 12 13 13 13
14 14 14 15 15
0.86 0.87 0.88 0.89 0.90
7.244 7.413 7.586 7.762 7.943
7.261 7.430 7.603 7.780 7.962
7.278 7.447 7.621 7.798 7.980
7.295 7.464 7.638 7.816 7.998
7.311 7.482 7.656 7.834 8.017
7.328 7.499 7.674 7.852 8.035
7.345 7.516 7.691 7.870 8.054
7.362 7.534 7.709 7.889 8.072
7.379 7.551 7.727 7.907 8.091
7.396 7.568 7.745 7.925 8.110
2 2 2 2 2
3 3 4 4 4
5 5 5 5 6
7 7 7 7 7
8 9 9 9 9
10 10 11 11 11
12 12 12 13 13
13 14 14 14 15
15 16 16 16 17
0.91 0.92 0.93 0.94 0.95
8.128 8.318 8.511 8.710 8.913
8.147 8.337 8.531 8.730 8.933
8.166 8.356 8.551 8.750 8.954
8.185 8.375 8.570 8.770 8.974
8.204 8.395 8.590 8.790 8.995
8.222 8.414 8.610 8.810 9.016
8.241 8.433 8.630 8.831 9.036
8.260 8.453 8.650 8.851 9.057
8.279 8.472 8.670 8.872 9.078
8.299 8.492 8.690 8.892 9.099
2 2 2 2 2
4 4 4 4 4
6 6 6 6 6
8 8 8 8 8
9 10 10 10 10
11 12 12 12 12
13 14 14 14 15
15 15 16 16 17
17 17 18 18 19
0.96 0.97 0.98 0.99
9.120 9.333 9.550 9.772
9.141 9.354 9.572 9.795
9.162 9.376 9.594 9.817
9.183 9.397 9.616 9.840
9.204 9.419 9.638 9.863
9.226 9.441 9.661 9.886
9.247 9.462 9.683 9.908
9.268 9.484 9.705 9.931
9.290 9.506 9.727 9.954
9.311 9.528 9.750 9.977
2 2 2 2
4 4 4 5
6 7 7 7
8 9 9 9
11 11 11 11
13 13 13 14
15 15 16 16
17 17 18 18
19 20 20 20
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