std09 maths em
TRANSCRIPT
MATHEMATICS
STANDARD − IX
Untouchability is a sin
Untouchability is a crime
Untouchability is inhuman
TAMILNADU TEXTBOOK CORPORATION COLLEGE ROAD, CHENNAI – 600 006.
© 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, Lecturer (S.S.) in Mathematics, L. N. Govt. Arts College, Pachaiyappa’s College, Ponneri 601 204. Chennai − 600 030.
AUTHORS Thiru. V. SRIRAM, Thiru K. ARIVAZHAGAN, School Assistant (Mathematics), School Assistant (Mathematics), P.C.K.G. Govt. Hr. Sec. School, Govt. Boys Hr. Sec. School, Kodambakkam, Chennai − 600 024. Ulundurpet − 606 107. Tmt. R. NAMBIKAI JAYARAJ, Tmt. S. VIJAYA, P.G. Assistant (Mathematics), P.G. Assistant (Mathematics), St. Anne’s Girls Hr. Sec. School, B.P.G. Hr. Sec. School, Royapuram, Chennai − 600 013. Kailasapuram, Trichy − 620 014.
CONTENTS PAGES 1. NUMBER SYSTEMS 1 - 30 1.1 Number Systems 1 1.2 The Real Number Line 17 2. MEASUREMENTS 31 - 47 2.1 Area and Perimeter 31 2.2 Combined Figures 36 3. SOME USEFUL NOTATION 48 - 87 3.1 Scientific Notation 48 3.2 Notation of Logarithms 51 3.3 Set Notation 71 4. ALGEBRA 88 - 117 4.1 Polynomials 89 4.2 Algebraic Identities 93 4.3 Factorization 103
4.4 Division of a Polynomial by a Polynomial 112 5. PROBLEM SOLVING TECHNIQUES 118 - 130 5.1 Conjectures and Proofs 119 5.2 Mathematical Models 127 6. THEORETICAL GEOMETRY 131 - 163 6.1 Theorems for Verification 131 6.2 Theorems with logical Proofs 153 7. ALGEBRAIC GEOMETRY 164 - 182 7.1 The Cartesian Coordinates system 164 7.2 Slope of a Line 168 7.3 The Distance between any two points (x1, y1) and (x2, y2) 174
8. TRIGONOMETRY 183 - 201 8.1 Trigonometric ratios 185 8.2 Trigonometric Identities 194
8.3 Trigonometric Ratios for Complementary Angles 199 9. PRACTICAL GEOMETRY 202 - 212 9.1 Concurrency in a triangle 203 9.2 Geometric interpretation of averages 209 10. HANDLING DATA 213 - 223 10.1 Measures of Central Tendency 215 11. GRAPHS 224 - 235 11.1 Linear Graphs 224 11.2 Application of Linear Graphs 228 Logarithms 236 - 237 Anti Logarithms 238 - 239
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.
1
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
satisfied respectively by the numbers 27 and
35− which are not in Z .
1.1.4 The Rational Numbers
A number of the form qp where p and q are integers and q ≠ 0 is called a rational
number. The collection of all rational numbers is denoted by Q. A rational number qp is said
to be in the proper form if q is a positive integer and p and q have no common factor other
than 1. For example, the rational numbers 11
7 , 5
2− are in the proper form where as the rational
numbers 1512 and
2224−
are not in the proper form. But every rational number has an
equivalent proper form. For example, we can write 15
12 =5
4 and 22
24
−=
11
12− which are in
proper form. Every integer is a rational number. For example, the integer −19 can be written
as 1
19− , 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, 5
4 is a
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
3
2− are in Q and their sum 11 +3
31
3)2(33
32
111
32
=−+
=⎟⎠⎞
⎜⎝⎛ −+=⎟
⎠⎞
⎜⎝⎛ − is also in Q. This property is
called the closure property with respect to addition in Q. 2. If x, y are rational numbers, then x + y = y + x.
2
For example,33
34
33
)22()12(32
11
4 −=
−+−=
−+
−⎟⎠⎞
⎜⎝⎛⎟
⎠⎞
⎜⎝⎛ and
33
34
33
)12()22(
11
4
3
2 −=
−+−=
−+
−⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ .
So ⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ −
+−
=−
+−
11
4
3
2
3
2
11
4 . 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,
72 ,
54 ,
32 −− are in Q and
⎥⎦⎤
⎢⎣⎡ −+−
+⎟⎠⎞
⎜⎝⎛=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛
35)10()28(
32
72
54
32 = ⎟
⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛
3538
32 =
105)114(70 −+ =
10544− .
⎟⎠⎞
⎜⎝⎛ −+⎥⎦
⎤⎢⎣⎡ −+
=⎟⎠⎞
⎜⎝⎛ −+⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+
72
15)12(10
72
54
32 = ⎟
⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ −
72
152 =
105)30()14( −+− = .
10544−
∴ .72
54
32
72
54
32 ⎟
⎠⎞
⎜⎝⎛ −+⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛
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,
311
3110
311
10
3110 =
+=+=+ and .
311
3011
10
3110
311
=+
=+=+
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
inverse of x in Q . For example, for the rational number 311− , we have the rational number
311 such that 0
30
311)11(
311
311
==+−
=⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ − .
6. If x, y are rational numbers, then x × y is also a rational number. We write x × y simply
as xy. For example, (−5), 32 are rational numbers and
(−5) ⎟⎠⎞
⎜⎝⎛
32
= ⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛ −
32
15 =
312)5(
××− =
310− is a rational number.
This property is called the closure property of multiplication in Q. 7. If x, y are rational numbers, then xy = yx.
For example, (−3), ⎟⎠⎞
⎜⎝⎛ −
75 are rational numbers and
.7
15)3(75 ,
715
75)3( =−⎟
⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −− So (−3) ).3(
75
75
−⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −
This property is called the commutative property of multiplication in Q.
3
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, 57− are rational numbers, and ,
542
521)2(
57)3()2( =⎟
⎠⎞
⎜⎝⎛=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −−
[(2) (−3)] .542
57)6(
57
=⎟⎠⎞
⎜⎝⎛ −−=⎟
⎠⎞
⎜⎝⎛ − So we have [ ] .
57 )3)(2(
57)3()2( ⎟
⎠⎞
⎜⎝⎛ −−=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −−
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.
For example, (1) .35
351
35
=×
=⎟⎠⎞
⎜⎝⎛
10. For every non-zero rational number x, x1 is a rational number and
.111=⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛ x
xxx For example, if x is the rational number ,
421− we observe that x ≠ 0 and
214
214
42111 −
=−
=−
=x
is a rational number and x .18484
214
4211
==⎟⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛
xThe number
x1 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.
For example, if ,5 ,21
,3
2=
−== zyx we have
x (y + z) = ⎥⎦
⎤⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛ −
⎟⎠⎞
⎜⎝⎛ 5
21
32 = ⎥⎦
⎤⎢⎣⎡ +−⎟⎠⎞
⎜⎝⎛
2101
32 = ,3
29
32
=⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛
xy + xz = )5(32
21
32
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ = ⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ −
310
31 = .3
39
310)1(
==+−
So, x(y+z) = xy + xz. Similarly, we have
(x+y)z = 521
32
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ = 5
6)3(4⎥⎦⎤
⎢⎣⎡ −+ = ,
655
61
=⎟⎠⎞
⎜⎝⎛
xz + yz = 5215
32
⎟⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ = ⎟
⎠⎞
⎜⎝⎛ −+
25
310 = .
65
6)15(20=
−+
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.
4
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
follows: If x, y are in Q and y ≠ 0, then x ÷ y = ⎟⎟⎠
⎞⎜⎜⎝
⎛×
yx 1 . The operation ‘÷’ does not satisfy the
commutative property in Q. For example, 4 ÷ 5 =54 , 5 ÷ 4 =
45 and so 4 ÷ 5 ≠ 5 ÷ 4.
The operation ‘−’ does not satisfy the associative property in Q.
For example, let us consider − 3, 11, 32 in Q. Then
(−3) − ⎥⎦⎤
⎢⎣⎡ −
−−=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−
3233)3(
3211 = ,
340
3319
331)3( −
=−−
=−−
[ ]32)14(
3211)3( −−=−−− = .
344
32)42( −=
−−
So, (−3) − [ ] .3211)3(
3211 ⎟
⎠⎞
⎜⎝⎛−−−≠⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−
Similarly, we have
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛÷−=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛÷÷−
23
111)3(
3211)3( = (−3) ÷ ⎟
⎠⎞
⎜⎝⎛
233
= ,11
2332
13
233
13 −
=⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛÷⎟
⎠⎞
⎜⎝⎛ −
[ ] ⎟⎠⎞
⎜⎝⎛÷⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛÷⎟
⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛÷÷−
32
111
13
3211)3( = ⎟
⎠⎞
⎜⎝⎛÷⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛ −
32
111
13
= ⎟⎠⎞
⎜⎝⎛÷⎟
⎠⎞
⎜⎝⎛ −
32
113 = .
229
23
113 −
=⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛ −
So (−3) ÷ [ ] ⎟⎠⎞
⎜⎝⎛÷÷−≠⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛÷
3211)3(
3211 .
5
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
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
operation ÷ does not satisfy the closure property in N, since 5, 7 are in N but 5 ÷ 7 = operation ÷ does not satisfy the closure property in N, since 5, 7 are in N but 5 ÷ 7 = 75 which
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.
Let us consider another equation 3x + 5 = 0. Solving the equation, we get x = 35− . Since
35−
is not in N, 35− is not in W and
35− is not in Z, we say that that the equation 3x + 5 = 0 does
not possess a solution in N, W and Z. Since 35− is in Q, we say that 3x + 5 = 0 possesses a
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
by the long division process. For example, the decimal representations of 751and
3215− are
obtained as follows:
15− = −0.46875.
32751=7.28571428
6
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 = 85
12581255
1000625
1000520600
105
102
106
32 =××
==++
=++
and we denote 85 as 0.625. Here the first numeral 6 from the right side of dot called decimal
point has the face value 106 , the second numeral 2 has the face value
1002 and so on. The first
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
of the rational number 85 .
With this notation, 3.025 means
3 × 100 + 0 × 10-1 + 2 × 10-2 + 5 × 10-3 = 3 + 1000
25 = 3 + 401 =
40121 .
Now, we consider the fraction .3215− Here
⎟⎠⎞
⎜⎝⎛=
32150
101
3215
= ⎟⎠⎞
⎜⎝⎛ +
32224
101
= ⎟⎠⎞
⎜⎝⎛+
3222
101
104
= ⎟⎠⎞
⎜⎝⎛+
32220
101
104
2
= ⎟⎠⎞
⎜⎝⎛ ++
32286
101
104
2
= ⎟⎠⎞
⎜⎝⎛++
3228
101
106
104
22
= ⎟⎠⎞
⎜⎝⎛++
32280
101
106
104
32
7
= ⎟⎠⎞
⎜⎝⎛ +++
32248
101
106
104
32
= ⎟⎠⎞
⎜⎝⎛+++
3224
101
108
106
104
332
= ⎟⎠⎞
⎜⎝⎛+++
32240
101
108
106
104
432
= ⎟⎠⎞
⎜⎝⎛ ++++
32167
101
108
106
104
432
= ⎟⎠⎞
⎜⎝⎛++++
3216
101
107
108
106
104
4432
= ⎟⎠⎞
⎜⎝⎛++++
32160
101
107
108
106
104
5432
= 5432 105
107
108
106
104
++++ (The process terminates.)
= 0.46875
∴ 3215− = −0.46875.
Next consider the following process:
⎟⎠⎞
⎜⎝⎛+=
727
751 (A)
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+
720
1017
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ ++
762
1017
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛++
76
101
1027
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛++
760
101
1027 2
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +++
748
101
1027 2
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+++
74
101
108
1027 22
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+++
740
101
108
1027 32
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ ++++
755
101
108
1027 32
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛++++
75
101
105
108
1027 332
8
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛++++
750
101
105
108
1027 432
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +++++
717
101
105
108
1027 432
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+++++
71
101
107
105
108
1027 4432
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+++++
710
101
107
105
108
1027 5432
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ ++++++
731
101
107
105
108
1027 5432
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛++++++
73
101
101
107
105
108
1027 55432
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛++++++
730
101
101
107
105
108
1027 65432
= ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +++++++
724
101
101
107
105
108
1027 65432
= 7×100+ ⎟⎠⎞
⎜⎝⎛+⎥⎦
⎤⎢⎣⎡ +++++
72
101
104
101
107
105
108
102
665432 (B)
= 7.285714 + ⎟⎠⎞
⎜⎝⎛
72
101
6 . The process repeats from (A)
We observe that in the decimal representation of ,3215− the process terminates (zero
remainder) and we say that 3215− has a terminating decimal expansion. But in the decimal
representation of ,751 the process does not terminate (non zero remainder) at any stage.
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
7751
= .285714285714285714… = 285714.7 ,
9
where the bar over 285714 indicates that the numerals under the bar repeat endlessly in the
same order in the long division process. The terminating decimal expansion 0.46875 for 3215
can be considered to be a non-terminating and repeating, since
0.46875 = 5432 105
107
108
106
104
++++
= .....10
010
010
510
710
810
6104
765432 +++++++ = 046875.0 .
We also have
0.46875 = 5432 105
107
108
106
104
++++
= ( )1410
110
710
810
6104
5432 +++++
= ⎟⎠⎞
⎜⎝⎛+++++ 55432 10
110
410
710
810
6104
= ( )1010
110
410
710
810
6104
65432 +++++
= ( )1910
110
410
710
810
6104
65432 ++++++
= ⎟⎠⎞
⎜⎝⎛++++++ 665432 10
110
910
410
710
810
6104
= ( )1010
110
910
410
710
810
6104
765432 ++++++
= ( )1910
110
910
410
710
810
6104
765432 +++++++
= ⎟⎠⎞
⎜⎝⎛+++++++ 7765432 10
110
910
910
410
710
810
6104 = .946874.0
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.
0.45 = 209
10045
100540
105
104
2 ==+
=+
(ii) Consider the decimal expansion 45.0 Let x = 45.0 = 0.454545…. Then 100x = 45.454545…
10
∴ 100 x − x = (45.4545…) − (0.4545…) or 99x = 45 or x = 115
9945
= .
(iii) Consider the decimal expansion 934.0− Let x = 934.0 = 0.349999…. Then 100x = 34.9999… ∴ 1000x = 349.9999… ∴ 1000x−100x = (349.9999…) − (34.9999…)
∴ 900x= 315 or x = .207
10035
900315
==
∴ 934.0 = 207 .
∴ 934.0− = − 20
7207 −
=⎟⎠⎞
⎜⎝⎛ .
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,
7.00252525… = ....10
510
210
510
210
01007 65432 +++++++ is positive.
− 3.0202202220… = − ⎥⎦⎤
⎢⎣⎡ +++++++ ...
100
102
102
100
102
1003 65432 is negative.
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
11
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.
From the above two properties we have (i) x (y − z)=x [y + (− z)]=xy + x(− z) = xy − xz. (ii) (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 . (iii) (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 . (iv) (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 75.0 . Solution: Let x = 75.0 . Then x = 0.757575… ∴ 100x = 75.757575… ∴ 100x − x = (75.757575…) − (0.757575…) = 75.0000…
∴ 99x = 75 or x = .3325
9975
=
Example 2: Justify 0.1 = .9.0 Solution:
0.1 = 1 = )10(101 = )19(
101
+ = ⎟⎠⎞
⎜⎝⎛+101
109 = ( )10
1001
109+ = ( )19
1001
109
++
= ⎟⎠⎞
⎜⎝⎛++100
1100
9109 = ( )10
10001
1009
109
++ = ( )191000
1100
9109
+++
= ⎟⎠⎞
⎜⎝⎛+++1000
11000
9100
9109 .
12
The above process continues endlessly and we get 0.1 = 0.9999… or 0.1 = .9.0
Example 3: Find 52.2 + 25.2 in the integer-by-integer form. Solution: Let x = 52.2 . Then x = 2.525252…. ∴ 100x = 252.5252…
∴ 100x − x = 250.000… or 99x = 250 or x = .99250
Let y = 25.2 . Then y = 2.52222….. ∴ 10y = 25.2222… ∴ 100y = 252.2222…
∴ 100y − 10y = 227.0000… or 90y = 227 or y = .90227
Hence 52.2 + 25.2 = 99250 +
90227 =
99024972500 + =
9904997 .
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…
∴ 10a − a = 1.1111… − 0.1111 = 1.0000…. or 9a = 1 or a = 91 .
∴ x + y = 91 .
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 2 × 2 = 2 is rational).
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 2 to any number of decimal places.
and 2 = 1.414213562…
13
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
recall that we used the rational number 722 as an approximation for the irrational number π in
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
23⎟⎠⎞
⎜⎝⎛ ,
3
34⎟⎠⎞
⎜⎝⎛ ,
4
45⎟⎠⎞
⎜⎝⎛ ,
5
56⎟⎠⎞
⎜⎝⎛ , …,
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 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
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.
Example 5: Put π and 722 in order relation.
Solution: π = 3.141528…, 722 = 3.142857…
∴ 722 − π = 3.142857… − 3.141528… = 0.0013…
∴ 722 − π is positive.
∴ π < 722 .
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.
15
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 001.2 and
003.2 . Solution: ∴ 001.2 ∴ 003.2 −∴ 001.2 Consider ththey are rati1.41461 − ∴ 001.2 1.41462 − 1∴ 1.41461
003.2 − 1∴ 1.41462 ∴ 001.2
= 1.41456… and 003.2 = 1.41527… 001.2 = 0.0007… > 0
< 003.2 . e numbers 1.41461 and 1.41462. Since they onal numbers and we have
001.2 = 1.414610000… − 1.41456…= 0.00< 1.41461. .41461 = 0.00001 > 0 < 1.41462. .41462 = 1.41523… − 1.41462 = 0.00061…>< 003.2 . < 1.41461 < 1.41462 < .003.2
16
have terminating decimal expansions,
004… > 0
0
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
which are not rational numbers. For example, 2 , 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 x2 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.
Exercise 1.1 1. Obtain the decimal expansions of
(i) 6421 (ii) −
611 (iii)
207− (iv)
2411
2. Determine the integer-by-integer form of each of the following rational numbers:
(i) 934.0 (ii) −0. 7 (iii) 0.125 (iv) 0.5 21 (v) 8. 9 3. Answer true or false.
(i) 72 is a whole number.
(ii) − 11 is an integer.
(iii) 31 is a rational number.
(iv) 61.0 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
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 .
32 To locate the point on the number
line corresponding to 32 , we proceed as
follows: Locate the point P corresponding to the
positive integer 3 (denominator of 32 ). Draw
the line segment PQ of length 2 (numerator of
Consider the line through A parallel to PQ. This
of AR is 32 times that of OA. This is so be
or 132or ==
AROAOP
ARPQ
AR =32 . Now, draw a
length of AR. This circle cuts the real number l
18
Figure 1.2
32 ) perpendicular to OP. Join OQ.
line meets OQ at the point R. Then the length
cause ∆OAR and ∆OPQ are similar and so
circle with centre at O and radius equal to the
ine at a point on the right side of O. This point
corresponds to the rational number 32 (see Figure 1.2). The same circle cuts the real number
line at a point on the left side of O and this point corresponds to the rational number 32− .In
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
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
on the left side of O. The point P represents the i
The point Q represents the irrational number −left of O. Example 9: Represent the irrational numbers 3
Solution: Having plotted the point P for 2 , weas given below: Construct a rectangle OPDC (see Figure 1.4)
with length OP (= 2 ) and breadth PD = 1. Then, by Pythagoras theorem, OD2 = OP2 + PD2
= ( 2 )2 + 12 = 2 + 1 = 3 and so OD = 3 . Now, with O as centre and OD as radius, draw a circle. This circle intersectsQ′ (Note that Q is just a notation and not repreNow OQ = OQ′ = OD = 3 . Since Q and Q′ ar
Q represents 3 and Q′ represents − 3 .
19
Figure 1.3
rrational number 2 , since OP = OB = .2
2 , since OQ = OB = 2 and Q lies to the
, ,5 − 3 ,− 5 on the real number line.
can now locate the points for 3 and − 3
Figure 1.4
the real number line at the points say Q and senting the collection of rational numbers). e on the right and the left of O respectively,
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. Then, by Pythagoras t
and so OF = 5 . Witreal number line at sa
S represents 5 and S
From the abovunique point on the rreal number. Points othe left side of O reprtwo real numbers a anIf a < b, then P lies same as saying that Qthe real number line (if x is a real number sthe point on the real nany two points on theP2, then the distance b 1.2.1 Manipulation
Let us recall h
irrational number 2
x − 2 = 0 or x2 − 2 =
Figure 1.5
heorem, OF2 = OR2 + RF2 = 22 + 12 = 4 + 1 = 5
h O as centre and OF as radius, draw a circle. This circle intersects the y S and S′ on the right and left of O respectively. Since
OS = OS′ = OF = 5 , ′ represents − 5 .
e discussion, we are able to know that any real number corresponds to a eal number line and that any point on the real number line represents a n the right side of O correspond to positive real numbers and those on esent negative real numbers. Let P and Q be the points corresponding to d b respectively. to the left of Q which is lies to the right of P on see Figure 1.6). Further,
uch that a < x < b, then umber line corresponding to
real line and if x1 and x2 are tetween P1 and P2 is |x1 − x2|.
of irrational numbers
ow the irrational numbers
was needed when we wan
0 or x2 = 2 . Similarly, we m
20
Figure 1.6
x lies between P and Q. If P1 and P2 are he real numbers corresponding to P1 and
2 , 3 , 5 , … originated. For example, the
ted to find a solution x for the equation
ay need to obtain a real number x such that
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 n is known as
the radical sign. The positive integer n is called the index of the radical n r and the rational
number r is called the radicand of n r . The real number n r may be a rational number or an
irrational number. For example, ,4643 = .244 = We observe that if r is the nth power of a
rational number qp , then n r =
qp is a rational number because in this case we have
.rqp
n
=⎟⎟⎠
⎞⎜⎜⎝
⎛On the other hand, if the rational number r is not the nth power of some rational
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 3 r the cube root of r.
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, 23,72 53 −+ 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,
( )( )32 32 −+ = ( )22 32 − = 4 − 3 = 1, ( )( )53 53 −+ = ( ) ( )2253 − = 3 − 5= −2.
21
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) ( ) .aann =
(ii) ( ) ( ) .nnn abba =
(iii) .nn
n
ba
ba=
(iv) p n r + q n r = ( p + q) n r where p and q are real numbers.
(v) mnn m aa .=
(vi) =n a mn ma . (vii) If a < b, then nn ba .<
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 ma . ∴ n m mn aa = = mn ma . Using law (vii), we can compare any surds of same order.
Note: n a is denoted as na1
. Example 10: Answer with reasons whether the following are surds or not:
(a) 225121 (b)
2549 (c)
57
(d) 72
144 (e) 3 216
32 (f) 33 164 ×
Solution:
(a) 225121 =
1511
1511 2
=⎟⎠⎞
⎜⎝⎛ = a rational number. ∴
225121 is not a surd.
(b =2549 2
57⎟⎠⎞
⎜⎝⎛ =
57 = a rational number. ∴
2549 is not a surd.
(c) 57 is a surd since
57 is not a square of a rational number.
(d) 272
144= = a surd.
22
(e) 3 216
32 = =×××
3 666216 .
322
624
6216
3 3=
×=
×
Since 3
22 is an irrational number, 3 216
32 is a surd.
(f) 44444164164 3 33333 ==××=×=× = 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) ( ) ( )33 234545 −++ (ii) ( ) ( )3332 +−+ (iii) ( )( )323 243 −+ (iv) 4 432 ÷ 4 2187
Solution: (i) ( ) ( )33 234545 −++ = (5 + 4) + ( )33 23227 −×
= 9 + ( )333 3 2323 −× = 9 + ( )33 2323 ×−× = 9.
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) + ( )33 − = −1 + 0 = −1 = a rational number. ∴ the given expression is not a surd.
(iii) ( )( )323 243 −+ = (3 + 4 )2 (3 − )216× = (3 + 4 )2 (3 − 4 )2 = 221632424333 ××−×+×−× = 322122129 −+− = 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. ∴ 44 .272432 = 2187 = 3 × 3 × 3 × 3 × 27 =34 × 27. ∴ 44 .2732187 =
∴ .32
2732722187432
4
444 ==÷
This is a rational number and so the given expression is not a surd.
23
Example 12: Simplify each of the following:
(i) 8022520 +− (ii) 33 135240 − (iii) 50272 6 + (iv) 33 20001283 −
Solution: (i) 52545420 =×=×= .
225 = 259× = 259 × = 3 × 5 = 15.
80 = 516× = 54516 =× .
∴ 5415528022520 +−=+− = 6 .155 −
(ii) 33333 52585840 =×=×= , 33333 53527527135 =×=×= .
∴ .545)62(53252135240 333333 −=−=×−=−
(iii) 333327 22 3 332 36 ==== × and 2525250 =×= .
∴ 50272 6 + = 2532 + .
(iv) 3 128 = 3 3 24 × = 3 24 .
33 33 2102102000 =×= .
∴ 3 ( ) 3333 2102432000128 −=− = 12 33 2102 − . = (12 − 10) 3 2 = 2 3 2 .
Example 13: Write each of the following into a single surd:
(i) 7 × 3 6 (ii) 3 5 × 4 3 . (iii) 8 3 4 ÷6 4 2 (iv) 3 81 ÷ ( )205 +
Solution: (i) Here 7 is of index 2 and 3 6 is of index 3. The l.c.m. of 2 and 3 is 6. So we
write 7 and 3 6 as surds of index 6. Here 7 = 2 7 = ,3437 62 3 3 = and 3 6 = 63 2 2 .366 = So, 7 × 3 6 = 666 3634336343 ×=× = 6 12348 .
(ii) 3 5 × 4 3 = 12 625 × 12 27 = 12 27625× = 12 16875 .
(iii) 8 3 4 ÷ 6 4 2 = 4
3
2648 = 1212
12
12
.3234
8256
34
8256
34
=×=×
(iv) 3 81 ÷ ( )545
327205
8120533
×+×
=+
=+ = ===×
6
633
1259
53
5333 6
1259 .
24
Example 14: Arrange the following in the ascending order of magnitudes: 1264 25,10,3
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.
124 34 273333 =××= and 126 26 100101010 =×= . We observe 121212 .1002725 << ∴ 6412 .10325 << The product of a surd and its conjugate is always a rational number. For example, ( ) ( ) ( ) ( ) 1343323223232
22 =−=−+−=−+ and
( ) ( ) ( ) ( ) .253515153535322
−=−=−+−=−+
Some times, the denominator y of a ratio yx may be a surd. In such ratios, the
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,
53 = =
××
5553
515
(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 − b ) = ( ) babbabaa −=−+− 222 which is a rational number. For example,
22 )2(32232
)23)(23()23(2
232
−×−×
=−+
−×=
+ = .
7226
29226 −=
−−
(iii) If the denominator y is in the form a − b where a and b are rational numbers,
then multiply both the numerator and denominator of yx by a + b . For example,
22
2
)2(3)23(
)23)(23()23()23(
2323
−+
=+−
+×+=
−+
= ( )7
26117
226929
2232322 +
=++
=−
+××+ .
(iv) If y is in the form ,ba + where a and b are rational numbers, then multiply
the numerator and denominator of yx by .ba − For example,
25
( )( )( )
( )( ) ( )22
53
5325353
53253
2
−
−=
−+−×
=+
= ( ) ( ) ( ) .355312
53253
532−=−−=
−−
=−−
(v) If y is in the form ba − , where a and b are rational numbers, then multiply
the numerator and denominator of yx by ba + . For example,
( )( )( )
( )( ) ( )22
2
75
7557575
75575
5
−
+=
+−+×
=−
= ( )35521
75355
+−=−
+ .
Example 15: Rationalize the denominator of 145
1+
.
Solution:
1451+
= ( )( )( ) ( ) ( )22
145
145145145
1451
−
−=
−+−×
= ( ).51491
9145
145145
−=−−
=−−
Note: If we want to rationalize the numerator of yx , where x is of the form a or ba + or
ba − or ba + or ba − , then we multiply the numerator and denominator by the conjugate of the numerator.
Example 16: Rationalize the numerator in 4
311 − .
Solution: 4
311 − = ( ) ( )( )3114
311311+×
+×− = ( ) ( )( )3114
31122
+×−
= ( )3114311+− = ( )3114
8+
= 311
2+
Example 17: If 321
1+−
= a + b 2 + c 6 , find a + b +c.
Solution:321
1+−
= ( )( )( )321321
3211−−+−
−−×
= ( ) ( )22321
321
−−
−−
26
= ( ) 32221321−+−
−−
= ( )22
22321
22321
×−
−−=
−−−
= 4
6224
23222−−−
=−
−−
= .6412
41
21
4622
+−=+−
∴ 6412
41
2162 +−=++ cba
∴ .41,
41,
21
=−
== cba
∴ .21
41
41
21
=+−=++ cba
In the surd p + n aq , where p and q are rational numbers and n a is a surd, p is called
the rational part and n aq is called the irrational part. Two surds are said to be equal if their rational parts are equal and their irrational parts are equal.
Example 18: If ,31313
1313 yx +=
−+
++− find x2 + y2.
Solution: ( ) ( )( ) ( )
( )( ) 13
1323
13
132313131313
1313
22
2
−+−
=−
+−=
−+−−
=+− = 32
2324
−=− .
∴ ( )( )( ) ( ) .32
3432
32
323232
32132
11313
22+=
−+
=−
+=
+−+×
=−
=−+
( ) ( ) 432321313
1313
=++−=−+
++− = 304 + .
∴ 3043 +=+ yx
∴ x = 4, y =0
∴ x2 + y2 = 16 + 0 = 16.
27
Example 19: If x = ,2323
−+ find .1
xx +
Solution: x =2323
−+ =
( )( )( )( )2323
2323+−++
= ( )( ) ( )22
2
23
23
−
+
= .231
232323
+=+
=−+
∴ ( )( )( )23
2323
123
11−−
×+
=+
=x
= ( ) ( ) 231
2323
23
23
2322 −=
−=
−−
=−
− .
∴ ( ) ( ) .3223231=−++=+
xx
Example 20: If a = ,223
223
−
+ find the value a2 (a− 6)2.
Solution:
a= ( )( )
( )( )223
223223223
++
×−+ = ( )
( ) 22389223
243
22322
2
+=−
+=
−
+ .
∴ a − ( ) .22362236 +−=−+=
∴ a2(a − ( ) ( )222 223223)6 +−+= = ( )( )[ ]2322 322 −+
= ( ) 222
322 ⎥⎦⎤
⎢⎣⎡ − = ( ) .198 2 =−
Example 21: If 2 = 1.414…., find an approximate value of 1212
−+ .
Solution: 1212
1212
1212
++
×−+
=−+
= ( )( ) 22
2
12
12
−
+ = ( ) ( )22
121212
+=−+ .
∴ 1212
−+ = ( ) 1212
2+=+ = 1.414… + 1 = 2.414…
28
Exercise 1.2 1. Answer with reasons whether the following are surds or not:
(i) 3 16 (ii) 42 (iii) 5 729 (iv) 4 2 4 2 (v) 50 2 2. Simplify each of the following to the simplest form:
(i) 25 + 32 (ii) 33 40320 −
(iii) ( ) ( )23343225 +− (iv) ( ) ( )45751220 +−
(v) 553 (vi) 3 ÷ 3 6
(vii) 155 124 ÷ (viii) 34 53 ÷ (ix) 5253 ÷ 3. Arrange in ascending order:
(i) ,3 3 5 , 6 11 (ii) ,5 3 7 , 4 9 (iii) 3 2 , 5 , 4 3 4. Rationalize the denominator in each of the following:
(i) 6
18 (ii) 321
4+
(iii) 1515
−+
(iv) 53
1+
(v) 3232
+− (vi)
22333223
+−
5. Find x and y in each of the following:
(i) 3232
−+ = x + y 3 (ii) 3
354345 yx +=
++
(iii) 55353
5353 yx +=
+−
+−+ (iv) 35
7575
7575 yx +=
−+
−+−
6. If a = 3232
−+ , find the value of a2 (a − 4)2.
7. If a = ,1212
−+ find the value of a2 + 2
1a
.
8. If 3 = 1.732…, find an approximate value of .1313
−+
9. If 2 = 1.414… and 3 = 1.732…, find an approximate value of 32
1+
.
29
Answers
Exercise 1.1 1. (i) 0.328125 (ii) − 38.1 (iii) − 0.35 (iv) 3458.0
2. (i) 207 (ii)
97
− (iii) 999125 (iv)
16586 (v)
19
3. (i) F (ii) T (iii) T (iv) T (v) T (vi) F (vii) F
Exercise 1.2 1. (i) surd (ii) not a surd (iii) surd (iv) surd (v) not a surd 2. (i) 29 (ii) 3 52 (iii) 6614 + (iv) 154 (v) 6 3125
(vi) 6
43 (vii) 15
316 (viii) 12
62527 (ix) 6 5
3. (i) 6 11 , 3 5 , 3 (ii) 4 9 , 3 7 , 5 (iii) 3 2 , 4 3 , 5 .
4. (i) 63 (ii) ( )132114
− (iii) ( )5321
+
(iv) ( 3521
− ) (v) 562 − (vi) ( )30613191
−
5. (i) x = 7, y = 4 (ii) x = ,5940
599
=y
(iii) x = 7, y = 0 (iv) x = 0, y = 2 6. a2 (a − 4)2 = 1 7. 34 8. 3.732 9. 0.318
30
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 = 22 bl + units.
2.1.2 Parallelogram:
Area = b × h sq.units Perimeter = 2(a + b) units.
2.1.3 Triangle with a given base and hei
Area = 21 b × h sq.units
Figure 2.1
ght:
Figure 2.2
3
Figure 2.3
1
2.1.4 Right triangle:
Figure 2.4 Area = 21 b × h sq.units
Perimeter = b + h + d units d = 22 hb + units.
2.1.5 Equilateral triangle:
altitude = h = 23 a units
Figure 2.5 Area =
43 a2 sq. units
Perimeter = 3a units. 2.1.6 Isosceles triangle:
Figure 2.6 Area = h 22 ha − sq. units
Perimeter = 2 ( )22 haa −+ units. 2.1.7 Scalene triangle:
Area = ))()(( csbsass −−− sq. units
where 2
cbas ++= units Figure 2.7
Perimeter = a + b +c units. 2.1.8 Trapezium:
Area = 21 (a + b) × h sq. units. Figure 2.8
2.1.9 Quadrilateral:
Area = 21 d × (h1 + h2)
sq.units. 2.1.10 Rhombus:
Area = 21 d1 × d2 sq. units
Perimeter = 2 22
21 dd + = 4
a units.
3
Figure 2.9
2
Figure 2.10
2.1.11 Circle:
Area of the circle = πr2 sq. units Perimeter of the circle = 2πr units
Area of a semicircle = 21πr2 sq. units
Arc length of the semicircle = πr units
Area of a quadrant circle = 41πr2 sq. units Figure 2.11
Arc length of a quadrant circle = 21 πr units.
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.
Figure 2.12
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. 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.
∴ No. of square sheets = 160000,12 = 7500.
Example 3: Find the base of a parallelogram if its area is 40 cm2 aSolution: Area = b × h. ∴ 40 = b × 15.
Figure
∴ b = 38
1540
= .
∴ Base = 38 cm.
33
Figure 2.13
nd altitude is 15 cm.
2.14
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 = ))()(( csbsass −−− . 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 = 565566 ××× = 330 sq.cm. Perimeter = a + b + c = 11 + 60 + 61 =132 cm. Example 5: Find the area of the quadrilateral ABCD g
Solution: Area = )(21
21 hhd + = )2010(5021
+××
= 25 × 30 = 750 m2
Example 6: Find the area of the trapezium ABCD give
Solution: Area = hba ×+ )(21
= 4)512(21
×+
= 34 sq. units. Example 7: Cost of levelling a land is Rs. 12 per sqtrapezium whose parallel sides are of lengths 18m andlength 5m, find the total cost incurred in levelling the lSolution: ABCD is the given trapezium where AB = 1CE parallel to DA (see Figure 2.18). ∆EBC is isosceles
h = 22 35 − = 16 = 4cm. Now, the area of the trapezium ABCD
= 21 (a + b) × h =
21 (18 + 12) × 4
= 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.
34
Figure 2.15
iven in Figure 2.16.
Figure 2.16
n in Figure 2.17
Figure 2.17
uare metre. A land is in the form of a 12 m. If its other two sides are each of and. 8m, CD = 12 m, AD = BC = 5 m. Draw whose height
720.
Figure 2.18
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 = 22
212 dd + . But the perimeter is
20 cm. ∴ 22
212 dd + = 20 cm or = 100. Here
one of the diagonals is of length 8 cm. Take d
22
21 dd +
1 = 8. Then 64 + d2
2 = 100 or d22 = 36. ∴ d2 = 6 cm. The area of the
rhombus is21 d1 × d2 =
21 × 8 × 6 = 24 cm2.
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
enclosed by them.(use π ≈ 722 )
Solution: Perimeter of the circle = 2
264 =132 cm.
But perimeter of the circle = 2πr.
∴ 2 × 722 × r = 132 or r = 21 cm.
∴ Area of the circle = πr2 = 722 × 21 × 21 = 1386 cm2.
Perimeter of the equilateral triangle = 3a But perimeter = 132 cm.∴ 3a = 132 or a = 44 cm.
∴ Area of the equilateral triangle = 2
43 a×
= 24443× =484 3 cm2
∴ The ratio of the area of circle to that of the equilateral triangle
= 1386 : 484 3 = 21 3 : 22
35
Figure 2.21
Figure 2.19
Figure 2.20
Exercise 2.1 1. 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.
2. 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.
3. 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.
4. 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.
5. If the area of a parallelogram is 612 cm2 and the height is 18 cm, find the base length.
6. 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.
7. 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.
8. Find the perimeter of the circular land whose area is that of a rectangular land of dimensions 22 cm × 14 cm.
2.2 Combined Figures
Figure 2.22
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 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 and two right triangles in juxta position (see Figure 2.23). We observe that two figures can be placeequal in length to a side of the other.
36
Figure 2.23
d in juxta position if one has a side
Some combined figures are given in Figure 2.24 to Figure 2.35.
Figure 2.25
Figure 2.24
Figure 2.29
F
37
Figure 2.26
Figure 2.27
Figure 2.28Figure 2.30
Figure 2.31
igure 2.32
thcoFireretricoca W 2.2
wh
Coardithtri
sidhe
Ar
∴
Figure 2.33
Figure 2.34
The sides that are in juxta position are shown by dotted linese figures that are combined in Figure 2.24 to Figure 2.35. For exambination of a triangle and a semicircle. It looks like the verticagure 2.27 is the combined form of a rectangle and a semicirclectangular window surmounted by a semicircle. In Figure 2.29, we hctangle and two quadrant circles. Figure 2.33 is the combined angle and a trapezium. It looks like a rocket. Since the commbination of triangles, quadrilaterals and circles, their perimelculated by applying the formulae that we have already learnt in our
e now consider two important combined plane figures namely trape
.1 Trapezium
A trapezium is a four-sided plane figure in ich two sides are parallel (see Figure 2.36)
nsider the trapezium ABCD where AB and DC e parallel. Let AB = a and CD = b. Let h be the stance between the parallel sides. We can consider e trapezium ABCD as the combined figure of the angles ABC and ACD. For the triangle ABC, the
e, AB is the base and h is the height. For the triangle ACD, CDight. So,
ea of ∆ABC = 21 × a × h and Area of ∆ACD =
21 × b × h
Area of the trapezium = 21 × a × h +
21 × b × h =
21 (a + b)h sq. un
38
Figure 2.35
. We can easily identify mple, Figure 2.25 is the l cross section of a top. . It can be viewed as a ave the combination of a figure of a rectangle, a
bined figures are the ters and areas can be previous classes.
ziums and polygons.
Figure 2.36
is the base and h is the
its.
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
aht
T
O
a
EScBwldS
B
∴
Figure 2.37
perimeter and area. Let ABCDEF be a regular hexagon. Then the sides AB, BC, CD, DE, EF and FA
re of equal length. Let each side be of length a units. Then the perimeter of the regular exagon is a + a + a + a + a + a = 6a units. We shall now derive a formula for the area of he regular hexagon.
he diagonals AD , BE , CF meet at a point, say O. Then the triangles OAB, OBC, OCD,
DE, OEF, OFA are equilateral triangles. So, the area of each triangle is 2
43 a . Hence the
rea of the regular hexagon = 6 × 2
43 a = 2
233 a sq. units.
Figure 2.38
xample 10: Find the area of Figure 2.38 olution: The figure ABCDE is the ombination of ABDE and BCD with the side D in juxta position. ABDE is a trapezium here the parallel sides AE and BD have
engths 10 cm and 16 cm respectively. The istance between the parallel sides is 9 cm. o, the area of the trapezium ABDE is
.cm1179)1610(21)(
21 2=×+=×+ hba
CD is a triangle whose base BD is of length 16 cm and height 8 cm. So its area is
.cm6481621
21 2=××=× hb
The area of the combined figure ABCDE is Area of ABDE + Area of BCD= 117+64 = 181 cm2.
39
Example 11: A surveyor has sketched the measurements of a land as below. Find the area of the land.Solution: Let P, Q, R, SThen AP = 5m, AQ = 7mES = 9m. The given land is the coDSE and CRD (see Figur Area of the trapezium
sides are BP and CR ahave BP = 10m, CR = 8m
PR = AR − AP =So, the area of PRCB is
21 (BP + CR) × PR =
21 (
Area of the trapezium QES = 9m, FQ = 8m and Q
21 (ES + FQ) × QS =
21 (
Area of the triangle AQF
Area of the triangle DSE
Area of the triangle CRD
Area of the triangle APB
∴ Area of the land = 63
Figure 2.39
, be the surveyors marks from A to D. , AR = 12m, AS = 15m, AD = 17m, BP = 10m, FQ = 8m, CR = 8m,
mbination of the trapeziums. PRCB, FESQ and triangles AQF, APB, e 2.40).
PRCB: The parallel
nd height is PR. We , and
Figure 2.40
12 − 5 = 7m.
10 + 8) × 7 = 63 m2
FES: The parallel sides are ES and FQ and height is QS. We have S = AS − AQ = 15 − 7 = 8m. So, the area of QFES is 9 + 8) × 8 = 17 × 4 = 68 m2
= 21 × AQ × FQ =
21 × 7 × 8 = 28 m2
= 21 × DS × ES =
21 ×(AD − AS) × 9
= 21 (17 − 15) × 9 =
21 × 2 × 9 = 9 m2.
= 21 × RD × CR =
21 × (AD − AR) × 8
= 4 × (17 − 12)= 4 × 5 = 20 m2. =
21 × AP × BP =
21 × 5 × 10 = 25m2
+ 68 + 28 + 9 + 20 + 25 = 213 m2.
40
Example 12: Find the area of the design as shown in Figure 2.41. (π ≈ 722 )
Solution: We observe thAFE and the equilateral tThe area of the rectangleThe area of the semicircl
The area of the equilater
∴ The area of the plot =
Example 13: Find the ar
Solution: We observe thCDE and the quadrant ciThe area of the rectangleThe area of the semi-circ
The area of the quadrant
The area of the quadrant
∴ The area of the given
Figure 2. 41
at the plot is a combination of the rectangle ABDE, the semi-circle riangle BCD. ABDE = 20 × 14 = 280 cm2. e AFE =
21 π × r2 =
21 ×
722 × 7 × 7 = 77 cm2
al triangle BCD = 43 a2 =
43 × 14 × 14 = 49 3 cm2.
280 + 77 + 49 3 = 357 + 49 × 1.732 = 357 + 84.868 = 441.868 cm2
ea of the design as in Figure 2.42. (Take π≈722 )
Figure 2.42
at the plot is the combination of the rectangle ABCD, the semi-circle rcles AFD, BCG. ABCD = 12 × 4 = 48 cm2. le CDE =
21 π × 6 × 6 =
722 × 3 × 6 =
7396 = 56
74 cm2.
circle AFD = 41 π × 4 × 4 =
722 × 4 =
788 = 12
74 cm2.
circle BCG = 1274 cm2.
plot = 48 + 56 74 + 12
74 + 12
74 = 128
712 = 129
75 cm2.
41
Example 14: Find the area enclosed by Figure 2.43 Solution: The figure is ttrapezium ABCG. The area of the rectangleThe area of the trapezium
The area of the semi-circ
∴ The area of the given
Sometimes, we removing plane figures fsumming up the smaller bigger figure. Area of a circular ring
A circular ring isarea of the ring is equal is, π R2 − π r2 or π (R2 − ∴ The area of the semi-c
Figure 2.43
he combination of the rectangle CDFG, the semi circle DEF and the
CDFG = 28 × 13 = 364 cm2. ABCG =
21 (36 + 28) × 14 = 64 × 7 = 448 cm2.
le DEF = 21 ×
722 × 14 × 14 = 22 × 14 = 308 cm2
design = 364 + 448 + 308 = 1120 cm2.
come across plane figures which are obtained by cutting out and rom a bigger one. Their areas can be found as before but instead of areas, we subtract the areas of the removed parts from the area of the
the region in between two concentric circles (see Figure 2.44). The to the area of the outer circle minus the area of the inner circle; that r2).
Figure 2.44
ircular ring is 21 π (R2 − r2) sq. units.
42
Figure 2.45.
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 ≈
722 ).
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 =
41 π × 14 × 14 =
41 ×
722 × 14 × 14 = 154 cm2.
∴ 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 ≈
722 )
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 =
π2440 =
2227440
×× = 70 m.
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
= 21 π (R2 − r2) =
21 ×
722 (772 − 702) =
711 × 147 × 7 = 1617 sq.m
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
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
43 th of a
circular region of radius 4.9m and two quadrant circular regions of radius 4.9 − 4.2 = 0.7m.
Figure 2.47
∴ Area, the cow can graze
= 43 × π × 4.9 × 4.9 + 2 ×
41 × π × 0.7 × 0.7
= 43 ×
722 × 4.9 × 4.9 +
21 ×
722 × 0.7 × 0.7
= 2
33 × 0.7 × 4.9 + 11 × 0.1 × 0.7= 56. 595 + 0.77 = 57.365m2.
Example 18: Find the area of the shaded portion in Figure 2.48 (Take π≈
722 )
Solution: The area of the shadedthe area of the semicircle of radThat is,
21 × π × (14)2 −
21 × π ×
or 21 ×
722 × 14 × 14 −
21 ×
722 ×
Figure 2.48
portion is equal to ius 14 cm minus the area of the semicircle of radius 7 cm. (7)2
7 × 7 = 11 × 2 × 14 − 11 × 7
= 308 − 77 = 231 cm2.
44
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 ≈
722 )
3. A trapezium has parallel sides of lengths 22 cm atwo sides are each of length 10 cm.
45
Figure 2.52
n
d 12 cm. Find its area, if the other
4. If the area of a regular hexagon is 150 3 cm2, find the side. 5. Find the area of the shaded region is Figure 2.53.
6. Find the area of the shaded region in the following figures:
Figure 2.55
(i) (ii) (iii) 7.
Figure 2.54
(iv)
Figure 2.56
A circle has diameter 54 cm. One of its diametsegment AB such that BC = 10 cm. A circle is darea included between them. Take π ≈
722 .
46
Figure 2.57
Figure 2.53
er is AB . C is a point on the line rawn on AC as diameter. Find the
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 ≈
722 .
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 ≈
722 .
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 (ii). 15,100 sq.m (iii). 7,525 sq.m
2. 4,956 sq.m 3. 147.22 cm2 4. 10cm
5. 140 m2
6. (i) 36.33 sq.cm (ii) 25m2 (iii) 37.71cm2 (iv) 240.28 cm2
7. 770 cm2 8. 354.37cm2, 94cm 9. 42m2 10. 549.71m2
47
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)
(ii) n
m
aa = a m − n , a ≠ 0, m > n (Quotient law)
(iii) (am)n = amn (Power law) (iv) am × bm = (a × b)m (Combination law)
When a ≠ 0, we denote ma1 as a− m and define a0 = 1.
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
(vi) 0.256 = 1056.2 = 2.56 × 10−1
(vii) 0.00786 = 1000
86.7 = 7.86 × 10−3
(viii) 0.000000537 = 5.37 × 10−7
(ix) 0.0000000279 = 2.79 × 10−8
48
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 largemultiplied or divid Example 1: Writ ( (ivSolution: Example 2: Write (i) 3 (iv) 4
or very small numbers are put in the scientific notation, they can be ed easily in this form.
e the following numbers in scientific notation: i) 7493 (ii) 105001 (iii) 3449099.93 ) 0.00567 (v) 0.0002079 (vi) 0.000001024
the following numbers in decimal form: .25 × 105 (ii) 1.86 × 107 (iii) 9.87 × 109
.02 × 10−4 (v) 1.423 × 10−6 (vi) 3.25 × 10−9.
49
Solution: (i) 3.25 × 105 = 210325 × 105 = 325 × 105 − 2 = 325 × 103 = 325000
(ii) 1.86 × 107 = 210186 × 107 = 186 × 107−2 = 186 × 105 = 18600000.
(iii) 9.87 × 109 = 210987 × 109 = 987 × 109 − 2 = 987 × 107 = 9870000000.
(iv) 4.02 × 10−4 = 210402 × 10−4 = 402 × 10−4− 2 = 402 × 10−6 = 0.000402.
(v) 1.423 × 10−6 = 3101423 × 10−6 = 1423 × 10−6 − 3 = 1423 × 10−9 =0.000001423.
(vi) 3.25 × 10−9 = 210325 × 10−9 = 325 × 10−9 − 2 = 325 × 10−11 =0.00000000325.
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) 000000 = 3.0 × 106. ∴ (3 (ii) ∴ (4
(iii) ∴ (0
3
000000)3 = (3.0 × 106)3 = (3.0)3 × (106)3
= 3 × 3 × 3 × 106×3
= 27 × 1018 = 2.7 × 10 × 1018 = 2.7 × 1019.
000 = 4.0 × 103, 00 = 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
= 92 × 1021= 8.192 × 103 × 1021= 8.192 × 1024.
00005 = 5.0 × 10−5
0.
.00005)4 = (5.0 × 10 = (5.0)4 × ( = 25 × 10
81
−5)4
10−5)4 = 625 × 10−5×4
−20 = 6.25 × 102 × 10−20 = 6.25 × 10−18.
62
50
(iv) 000 = 2.0 × 103, 0001 = 1.0 × 10−4
∴ (2
1.
2.
3.
3.2
mathfromarithof log 3.2.1Let a
When
by a
For e
2
000)2 ÷ (0.0001)4 =
=
Represent the fo (i) 29980000 (iii) 10830000 (v) 94630000(vii) 0.0037 (ix) 0.0000803 (xi) 0.0000000
Write the follow (i) 3.25 × 10−
(iii) 4.132 × 10 (v) 3.25 × 106
(vii) 4.132 × 10
Find the value o (i) (100)3 × (4(iii) (18000)4 ÷ (v) (120000) ×
Notation of lo
John Napier, anematical device to two Greek wordmos means numbearithm, we shall f
Exponential no be a positive num
x is a rational nu
pq ax ⎟⎠⎞
⎜⎝⎛=
xample, ( )338
55 =
0.
44
23
)1001()1002(
−××
. . = 444
232
)10()01()10()02(
−××
. . = 44
23
101104
×−
×
××
22)16(616
6
10410410
104×=×=
× −−−
.
Exercise 3.1
llowing numbers in the scientific notation: 000 (ii) 1300000000 00000 (iv) 4300000000 00000000 (vi) 534900000000000000
(viii) 0.000107 5 (x) 0.0000013307 0011 (xii) 0.0000000000009
ing numbers in decimal form: 6 (ii) 4.02 × 10−5
−4 (iv) 1.432 × 10−3
(vi) 4.02 × 105
4 (viii) 1.432 × 103
f the following in scientific notation: 0)5 (ii) (21000)2 × (0.001)4
(30000)2 (iv) (0.002)8 × (0.0001)3 ÷ (0.01)4
(0.0005)2 ÷ (400000)
garithm
English mathematician introduced the notation of logarithm as a do calculations easily and quickly. The word logarithm is derived s ‘logos’ and ‘arithmos’. The word logos means reckoning and r. Thus logarithm means reckoning number. To introduce the notation irst know about exponential notation.
tation ber. We have already introduced the notation ax ,where x is an integer.
mber, say qp with p, an integer and q, a positive integer, we define ax
8 , ( ) 411114
77−
−
= .
51
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:
Now we 3.2.2 Logarit
Let b benumber x, bx recalled the logar
is an equivalen
exponential for
equivalent way
and it means thFor example,
(i) (iv) yxayaxa +=×xa
xa 1=−
(ii) yxaya
xa −= (v) ( )xbaxbxa ×=×
y
(iii) xyaxa =⎟⎠⎞⎜
⎝⎛ (vi) 10 =a
are in a position to introduce the notation of logarithm.
hmic notation
a positive number and b ≠ 1. We have already understood that, for any real presents a unique real number, say a. If we write a = bx, then the exponent x is ithm of a to the base b. We also call x, the value of a. Thus, x = a
t form of a = bblog blog
x. We say that x = a is the logarithmic form of the
m a = bblog
x. In both the forms, the base is same. We observe that x = a is an
of writing a = bblog
x. The notation x = a is called the logarithmic notation
e equation a = bblog
x.
(i) 3 = 729 is equivalent to 99log 3 = 729;
(ii) 2log31
8= is equivalent to 31
8 = 2;
(iii) −3 = 0.001 is equivalent to 1010log −3 = 0.001; (iv) 2 = is equivalent to 749log7
2 = 49;
(v) 3log21
9= is equivalent to 39or 39 21
== ;
(vi) ⎟⎠⎞
⎜⎝⎛=−
81log
23
4 is equivalent to 814 2
3
=−
.
52
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:
(i) 5 = 25log21 (ii) ⎟
⎠⎞
⎜⎝⎛
41log 2 = −2 (iii)
316log 216 = (iv) ⎟
⎠⎞
⎜⎝⎛
91log3 = −2
Solution: As base is same in both forms,
(i) 5 = 25log21 is equivalent to ( ) .525 2
1=
(ii) ⎟⎠⎞
⎜⎝⎛
41log 2 = −2 is equivalent to ( ) .
412 2 =−
(iii) 316log 216 = is equivalent to ( ) .6216 3
1=
(iv) ⎟⎠⎞
⎜⎝⎛
91log3 = −2 is equivalent to (3)−2= .
91
Example 5: Change the following from exponential form to logarithmic form:
(i) 2 = 61
64 (ii) 9−3 = 7291 (iii)
41
81 3
2
=⎟⎠⎞
⎜⎝⎛
−
(iv) 1771 −=
Solution: As base is same in both forms, we have
(i) 2 = 61
64 is equivalent to 61 = .2log64
(ii) 9−3 =7291 is equivalent to −3 = .
7291log9 ⎟
⎠⎞
⎜⎝⎛
(iii) 41
81 3
2
=⎟⎠⎞
⎜⎝⎛ is equivalent to .
41log
32
81 ⎟
⎠⎞
⎜⎝⎛=
(iv) 71 = 7−1 is equivalent to −1 = .
71log7 ⎟
⎠⎞
⎜⎝⎛
Example 6: Evaluate
(i) (ii) (iii) 729log9 8log4 ⎟⎠⎞
⎜⎝⎛
271log9 (iv) .)243(log 1
3−
(iii) Let x = ⎟⎠⎞
⎜⎝⎛
271log . 9
Then 9x= 33 3
31
271 −==
But 9x = (32)x = 32x
∴ 32x = 3−3, 2x= −3 or x = 23− .
Solution : (i) Let x = Then 9.729log9x = 729 = 93.
∴ x = 3. (ii) Let x = Then 4.8log4
x = 8 = 23
But 4x = (22)x = 22x.
∴ 22x = 23. ∴ 2x = 3 or x = 23
53
(iv) Let x = 13 )243(log −
Then 3x = (243)−1 =2431 = 53
1 = 3−5
or 3x =3−5 or x = −5. Example 7: Solve the equations: (i) (ii) 2log3 −=x 2100log =b (iii) x = 512log
81
⎟⎠⎞
⎜⎝⎛ (iv) x + 2 .09log27 =
Solution: (ii) 2100log =b ∴ b2 = 100 = 102. ∴ b = 10.
(iii) x = .512log ∴ 81
⎟⎠⎞
⎜⎝⎛
3851281
==⎟⎠⎞
⎜⎛ . ⎝
x
or (8−1)x = 83 or 8−x = 83. ∴ −x = 3 or x = −3.
(i) 2log3 −=x
∴ 3−2 = x or x = 231 =
91 .
(iv) x + 2 09log27 =
∴ x = −2 or 9log27 9log2 27=
− x
∴ 9)27( 2 =− x
or 223 3)3( =− x
or 223
3)3( =− x
∴ 223
=− x or x =
34− .
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 mna = nama loglog +
Proof: Let xma =log and yna =log .
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 mna = x + y or = )(log mna nama loglog + .
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,
namanm
a logloglog −=⎟⎠⎞
⎜⎝⎛ .
54
Proof: Let xma =log and yna =log .
Then m = ax and n = ay.
∴ nm = y
x
aa = ax−y.
This is in exponential form. Writing this in logarithmic form, we get
yxnm
a −=⎟⎠⎞
⎜⎝⎛log or m⎛ ⎞
In words, the quotient r
difference ama loglog −
(iii) Power rule: If
Proof: Let . Txma =log
mn = (ax)n =ax n . This is in
or nxnma =log
(iv) If a is a positive num
Proof: Let . Th1logax =
∴ x = 0 or (v) If a is a positive numProof: Let . Thaax log=
∴ x = 1 or (vi) Change of base rule
Proof: Let x = anlog mpThen px = m and ny = p. E
.logna ⎜
⎝loglog nama −=⎟
⎠
ule states that the logarithm of the quotient nm is equal to the
. n
a and m are positive numbers, a ≠ 1 and n is a real number, then . mannma loglog =
hen m = ax. Raising to the power n on both sides, we get
exponential form. Writing this in the notation of logarithm, we get
.loglog mannma =
ber, then 01log =a
en ax = 1 = a0.
.01log =
ber, then log a
en ax = a = a
: If m, n and
mnlog
.logd pny =
liminating p w( )n xy
a
.1=a 1.
.1log =a
p are
lo⎜⎝⎛=
ith thm=
a
positive numbers and n ≠ 1, p ≠ 1, then
( )pnmp logg ×⎟⎠⎞ .
ese equations, we get .or mn xy =
55
This is in exponential form. Writing this in the logarithmic form, we get xymn =log or
( ) ( ).logloglog pmm npn ×= (vii) Reciprocal rule: If m a
Proof: Let . Thexnm =log
∴ xnm1
= . This is in exponen
xmn
1log = or
(viii) If a and b are any two Proof: Let x = . Then ablog
(ix) Let m, n and a be positiv
Proof: Let .Then max log=
∴ ax = n or maa
log
Note: We are avoiding 1 inlogarithm, say with 1 9log1
there is no real number x such Caution: Some errors which
(1) ,log
loglog
na
manm
a =⎟⎠⎞
⎜⎝⎛
(2) llog)(log manma +=+
(1) is wrong since L.H
(2) is incorrect since R
nd n are positive numbers other than 1, then
nm
mn log
1log = .
n x
nnnxm xx
x⎟⎟⎠
⎞⎜⎜⎝
⎛===
×11
.
tial form. Writing this in the logarithmic form, we get
positive bx = a. S
e numb
ax log=
on=
the bain the b that 1x
are com
.og na
.S. = lo
.H.S. =
1
nmm
n loglog = .
numbers and b ≠ 1, then . aabb =
log
ubstituting for x in this equation, we get
alog
ab b = .ers and a ≠ 1. If ,loglog nama = then m = n.
. n
r m = n (by property (viii)).
se of all logarithms because if we consider one such ase, then x = would give 19log1
x = 9. We know that = 9.
monly committed are
.log
loglogg
na
manama ≠−
).(log)(log nmamna +≠
56
Example 8: Simplify: (i) (ii) 729log27log 33 +1000
1log8log 55 +
Solution: (i) Since the expression is the sum of two logarithms and the bases are equal, we can apply the product rule. )72927(log729log27log
333 ×=+
= )33(log 633 ×
= = 9 × 1 = 9. 3log93log 39
3 ×=
(ii) ⎟⎠⎞
⎜⎝⎛ ×=+
100018log
10001log8log 555
= ⎟⎠⎞
⎜⎝⎛125
1log5
= ( )3535 5log
51log −=⎟
⎠⎞
⎜⎝⎛
= (−3) × 5log = (−3) × 1 = −3. 5
Example 9: Simplify:
(i) (ii) 14log98log 77 − 4log34log236log21
999 −+
Solution: (i) .17log1498log14log98log 7777 ==⎟
⎠⎞
⎜⎝⎛=−
(ii) 39
29
21
9999 4log4log36log4log34log236log21
−+⎟⎟⎠
⎞⎜⎜⎝
⎛=−+
= 64log16log6log 999 −+ = 64log)166(log 99 −× = 64log96log 99 −
= .23log
6496log 99 ⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
Example 10: Prove
(i) (ii) 2log341250log 1010 −= .2log208log3136log
211875log 32555 +−=
Solution: (i) R.H.S. 2log34 10−= = 4 − 310 2log
= 4 × 8log10log 1010 −
= 8log10log 104
10 − = 8log10000log 1010 −
= 1250log8
10000log 1010 =⎟⎠⎞
⎜⎝⎛ = L.H.S.
57
(ii) R.H.S. 2log208log3136log
21
3255 +−= = 2log54)8(log)36(log 3231
521
5 ×+−
= 53255 2log42log6log +−
= 426log32log4
26log 5325 +⎟
⎠⎞
⎜⎝⎛=+⎟
⎠⎞
⎜⎝⎛
= 45555 5log3log5log43log +=+
= 625log3log 55 + = ( )6253log5 × = ...1875log5 SHL= Example 11: Prove that 29log8log7log6log5log4log 876543 =××××× . Solution: L.H.S. = ( ) ( ) ( )9log8log7log6log5log4log 876543 ××××× = = 9log7log5log 753 ×× ( )9log7log5log 753 ×× = R.H.S.2123log23log9log9log5log 3
23353 ==×====×
Example 12: Solve .3)502(log10 =+xSolution: Writing the equation in the exponential form, we get 2x + 50 = 103 = 1000 or 2x = 1000 − 50 = 950 or x = 475.
Example 13: Find the value of . 2log381 9−
Solution: = 2log381 9− 2log3299−
⎟⎠⎞⎜
⎝⎛
= = 2log69 9− 6262log9 9 −=
−. (Since ) aab b =log
= .641
21
6 =
Example 14: Solve .0)1(log2log 66 =+− xx
Solution: Using the quotient law, we can write the equation as .01
2log6 =⎟⎠⎞
⎜⎝⎛
+xx Changing
into exponential form, we get
161
2 0 ==+xx or 2x = x + 1 or x = 1.
Example 15: Solve .1)1(log)7(log 33 =−−− xx
Solution: Using the quotient law, the equation can be written as .117log3 =⎟
⎠⎞
⎜⎝⎛
−−
xx
Writing in the exponential form, we get
3317 1 ==
−−
xx or )1(37 xx −=− or xx 337 −=− or 2x = −4 or x = −2.
58
Example 16: Solve . ( ) 2loglog 32 =xSolution: Put Then the equation becomes .log3 xy = .2log2 =y Writing the equation in the exponential form, we get or 422 ==y .4log3 =x Again, writing in the exponential form, we get .81or 34 == xx Example 17: Solve 3log1log3log2 595 =+× x . Solution: Rewriting the equation, we get 13loglog3log 59
25 −=× x
or .53log5log3loglog9log 55595 ⎟
⎠⎞
⎜⎝⎛=−=× x
Using the change of base rule, we get ⎟⎠⎞
⎜⎝⎛=
53loglog 55 x . ∴ x = .
53
Exercise 3.2.1
1. Write true or false in the following: (i) (ii) 5243log3 = 327log
31 =
(iii) .4log3
16log43
16log 22 −=⎟⎠⎞
⎜⎝⎛ − (iv) .4log8log)48(log 222 −=−
(v) 1log 1 −=aa
(vi) .loglog)(log nmnm aaa +=+
2. Obtain the equivalent logarithm form of the following:
(i) (ii) .04.05 2 =− .481 3
2
=⎟⎠⎞
⎜⎝⎛
−
(iii) .25644 =
(iv) 36= 729. (v) 216136 2
3
=−
(vi) .001.010 3 =−
3. Find the value in the following: (i) (ii) (iii) 625log5 216log6 9log 3
(iv) 31log9 (v) 81log
31 (vi) 24log 2
(vii) (viii) (ix) 610log2
1085log
2529log2
9−
4. Solve for the unknown: (i) logx 0.001 = −3 (vi) 3log2 =x (ii) 7log
21 =x (vii) = 4 c25log5
59
(iii) (viii) 2100log −=x 42
log3 =⎟⎠⎞
⎜⎝⎛ N
(iv) (ix) 3125log =b 11000
1log10 =⎟⎠⎞
⎜⎝⎛
a
(v) 51log2 =⎟⎠⎞
⎜⎝⎛
x (x) 2 1log9 =N
5. Choose the correct answer from the alternatives given for each of the following: (i) If 3 then x is ,15log =x
(A) 5 (B) 25 (C) 125 (D) 625
(ii) =+ 144log196
1log 1214
(A) 0 (B) 1 (C) 2 (D) 3 (iii) The value of =1024log4
(A) 10 (B) 8 (C) 7 (D) 5 (iv) If then x = ,log24log xaa = (A) 0 (B) 1 (C) 2 (D) 3 (v) If ,1log2 16 =x then x = (A) 4 (B) 8 (C) 16 (D) 32 (vi) If 2log5 =x then x = (A) 5 (B) 25 (C) 125 (D) 625 6. Simplify the expression into a single logarithm in each of the following: (i) .9log2log 1010 + (ii) .83log42log3 33 −+ (iii) .6log24log33log25 222 ++− (iv) 7log15log5log32log2 2224 +−+ . (v) . 4log63log22log5 641010 −+ (vi) .325log24log20log5log 10101010 −+−+ 7. Given zyx aaa === 5log,3log,2log and ,7log ta = find value in each of the following in terms of x, y, z and t. (i) (ii) (iii) (iv) 6log a 4loga 5.1loga 27log a
(v) 312log a (vi) (vii) 600log a 27
8log a (viii) 15loga
(ix) (x) (xi) (xii) 35log a 12log a 9.4log a 1514log a
60
8. Solve the equation in each of the following: (i) (ii) 2log3log2 55 =x 11log7loglog 333 =+x (iii) 018log16 =+x (iv) 256log16loglog 444 =×x (v) 23log)2(log 44 =++x (vi) 4log)12(log)12(log 333 =−−+ xx (vii) (viii) 2)1(log10log 33 =+− xx 13log)15(log)37(log 222 −=−−+ xx (ix) )43(log)10(log 55 xx +=+ (x) .2)log5(log 35 =x
(xi) 1log21510log 33 +=−+ xx
9. Prove the equation in the following: (i) 5log3135log 33 += (ii) 2log421600log 1010 += (iii) 5log222500log 1010 += (iv) 2log33125log 1010 −= (v) 2log2225log 1010 −= (vi) 10log430027.0log 33 −= (vii) 10log62000256.0log 1616 −= 10. If a, b and c are positive numbers other than one, prove that .1logloglog =×× cba acb 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 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 as log x, omitting the base 10. Thus,
log x = y means
xelog
x10log
yx =10log and is equivalent to x = 10y. If we substitute 1000
1 for x in y
= log x, we get
y = log 1000
1 = . 310log 310 −=−
Similarly, for x = 100
1 , 101 , 1, 10, 100, 1000, …, we correspondingly get
−2, −1, 0, 1, 2, 3, ….
61
x 10−3 10−2 10−1 100 101 102 103
x10log −3 −2 −1 0 1 2 3 We observe that as x increases along the positive real axis, also increases. We also observe that is positive for all values of x > 1 and is negative whenever
x10logx10log
x < 1. We observe that the values of is as given in the table below: x10log
Range of x Location of x10log The value of x10log
10−5 < x < 10−4 −5 < <−4 x10log −5 + 0.a1a2… 10−4 < x < 10−3 −4 < <−3 x10log −4 + 0.b1b2… 10−3 < x < 10−2 −3 < <−2 x10log −3 + 0.c1c2… 10−2 < x < 10−1 −2 < <−1 x10log −2 + 0.d1d2… 10−1 < x < 100 −1 < <0 x10log −1 + 0.e1e2… 100 < x < 101 0 < <1 x10log 0 + 0.f1f2… 101 < x < 102 1 < <2 x10log 1 + 0.g1g2… 102 < x < 103 2 < <3 x10log 2 + 0.h1h2… 103 < x < 104 3 < <4 x10log 3 + 0.i1i2… 104 < x < 105 4 < <5 x10log 4 + 0.j1j2…
From the above table, we note that the value of the common logarithm can be expressed as
x10log
(an integer) + (a decimal expansion of the form 0.r1r2r3r4)
In this form, the integer part is called the characteristic of and the decimal part is called mantissa of . We usually write the mantissa in 4 decimal places. We observe that the mantissa of always represents a positive number between 0 and 1, and the characteristic of is a positive integer or negative integer or zero depending on the value of x. If x < 1, then the characteristic is a negative integer. If
x10logx10log
x10logx10log
x >10, then the characteristic of is a positive integer. If x lies between 1 and 10, then the characteristic of is 0.
x10logx10log
Example 18: Find the characteristics of the common logarithms of (i) 2003 (ii) 200.3 (iii) 20.03 (iv) 2.003 (v) 0.2003 (vi) 0.02003 (vii) 0.002003 (viii) 0.0002003 (ix) 0.00002003 (x) 0.000002003 Solution: (i) As 2003 lies between 103 and 104, is 3 + 0. d2003log10 1d2d3d4. Then the characteristic of is 3 and the mantissa is Alternatively, writing 2003 in the scientific notation,
2003log100.d1d2d3d4.
62
2.003 × 103. So we have )10003.2(log2003log 31010 ×=
= 31010 10log003.2log +
= 10log3003.2log 1010 + = 432110 .03003.2log3 dddd+=+ . Thus, when we write the number x in the scientific notation a × 10n , the integer n is the characteristic of and the value is the mantissa of . x10log a10log x10log(ii) 200.3 = 2.003 × 102
The characteristic of 200.3 is 2 logThe mantissa of log 200.3 in 2.003 which is same as the mantissa of log 2003. 10log
(iii) 20.03 = 2.003 × 101
∴ The characteristic of log 20.03 is 1 and the mantissa of log 20.03 is log 2.003.
(iv) 2.003 = 2.003 × 100
∴ The characteristic of log 2.003 is 0 and the mantissa of log 2.003 is log 2.003. (v) 0.2003 = 2.003 × 10−1
∴ The characteristic of log 0.2003 is −1 and the mantissa of log 0.2003 is log 2.003. (vi) 0.02003 = 2.003 × 10−2
∴ The characteristic of log 0.02003 is −2 and the mantissa of log 0.02003 is log 2.003. (vii) 0.002003 = 2.003 × 10−3
∴ The characteristic of log 0.002003 is −3 and the mantissa of log 0.002003 is log 2.003. (viii) 0.0002003 = 2.003 × 10−4
∴ The characteristic of log 0.0002003 is −4 and the mantissa of log 0.0002003 is log 2.003. (ix) 0.00002003 = 2.03 × 10−5
∴ The characteristic of log 0.00002003 is −5 and the mantissa of log 0.00002003 is log 2.003 (x) 0.000002003 = 2.003 × 10−6
∴ The characteristic of log 0.000002003 is −6 and the mantissa of log 0.000002003 is log 2.003. From the above examples, we observe the following: (i) If the integral part of x is a nonzero n digit number, then the characteristic of is n − 1.
x10log
63
(ii) If the integral part of x is zero and the decimal (fractional) part has n zeros before the first non zero digit on the right side of the decimal point, then the characteristic of is − (n + 1).
x10log
(iii) The numbers 2003, 200.3, …, 0.000002003 have the same mantissa irrespective of the position of the decimal point. Thus, the mantissa is the same for all numbers with identical significant digits in a given order. Example 19: Given that log 162 = 2.2095, find (i) log 1620 (ii) log 16.2 (iii) log 1.62 (iv) log 0.162 (v) log 0.0162 (vi) log 0.00162. Solution: The mantissa of log 162 is 0.2095. (i) log 1620 = 3.2095 (ii) log 16.2 = 1.2095 (iii) log 1.62 = 0.2095 (iv) log 0.162 = −1 + 0.2095. (v) log 0.0162 = −2 + 0.2095 (vi) log 0.00162 = −3 + 0.2095 Note: −3 +0.2095 is written simply as 3 .2095. Likewise −5 + 0.1023 is written as 5 .1023. While doing calculations, we may get a negative number for a common logarithm. Adding and subtracting a suitable positive integer, the negative number can be written in the form (a negative integer) + 0.d1d2d3d4. Example 20: Given log 0.25 = 1 .3979 and log 2003 = 3.3016, find
(i) log 0.025 (ii) log 0.0025 (iii) log ⎟⎠⎞
⎜⎝⎛
200325 (iv) log ⎟
⎠⎞
⎜⎝⎛
2500003.2
Solution: log 0.25 = 1 .3979 = −1 + 0.3979. So the mantissa of log 0.25 is 0.3979. log 2003 = 3.3016 = 3 + 0.3016. So the mantissa of log 2003 is 0.3016. (i) log 0.025 = 2 .3979 (ii) log 0.0025 = 3 .3979
(iii) log ⎟⎠⎞
⎜⎝⎛
200325 = log 25 − log 2003 3.3016
1.3979
1.9037 = 1.3979 − 3.3016 = − 1.9037. = − 2 + 2 −1.9037 = 2 .0963.
(iv) log ⎟⎠⎞
⎜⎝⎛
2500003.2 = log 2.003 − log 2500
= 0.3016 − 3.3979 = −4 + 4. 3016 − 3.3979 = 4 .9037.
64
2.0000 1.9037
0.0963
3.2.4 Table of Logarithms
Common logarithms of positive numbers from 1.000 to 9.999 (3 digit decimal part) have been calculated and listed in the form of a ready-made table. This table is called “Table of logarithms”. Using this table, we can find the common logarithm of any positive number. Before we proceed to know the use of common logarithms, we shall familiarize our self with the method of reading the common logarithms from the logarithms table. Example 21: Find log 36.78. Solution: 36.78 = 3.678 × 101.
The characteristic is 1. To get the mantissa, we consider 3.678 and locate the number 3.6 in the extreme left column of the table. Read along the row corresponding to 3.6 and down the column under 7. We find 0.5647. We go further along the row and reach the column under 8 in the mean difference. Here we find 10. We add this 10 to 0.5647 and get 0.5657 as the common logarithm of 3.678. This is the required mantissa of the given number and hence log 36.78 = 1.5657. Example 22: Find log 0.00200316. Solution: 0.00200316 = 2.00316 × 10−3.
The characteristic = −3. To get the mantissa, consider 2.00316. Common logarithms have been tabulated for numbers with 3 digits in the decimal part. So we approximate 2.00316 as 2.003(since 1 in the 4th decimal place is less than 5). Now, we get as before from the tables, log 2.003 = 0.3016. Hence log 0.00200316 ≈ −3 + 0.3016 = 3 .3016. Example 23: Find log 730.391. Solution: 730.391 = 7.30391 × 102.
The characteristic = 2. To get the mantissa, consider 7.30391 and approximate it as 7.304 (since 9 in the 4th decimal place is not less than 5). Now, we get as before from the tables, log 7.304 = 0.8635. Hence log 730.391 ≈ 2 + 0.8635 = 2.8635. 3.2.5 Table of antilogarithms
If log x = y, then x is called the antilogarithm of y. That is, if = y, then x10log
x = 10y is the antilogarithm of y. Thus, antilog of y = 10y. We shall abbreviate antilogarithm of y as antilog of y. We observe that getting antilogarithm is the reverse (opposite) process of getting logarithm. For example, log 20 = 1.3010 and so antilog of 1.3010 = 20.
A table of antilogarithms of numbers ranging from 0.0000 to 0.9999 (4 digits in the decimal part) is provided at the end of this text book. Actually, this table gives us the values of 10y where y ranges from 0.0000 to 0.9999 (4 digits in the decimal part). Using this table, we can calculate the antilogarithm of any given number.
65
Example 24: Find the antilogarithm of (i) 1 .2305 (ii) 3 .4629 (iii) 1.8658 (iv) 2.0578 To get the antilog of y, consider first the mantissa only, locate the antilogarithm corresponding to the first three digits of the mantissa and to this add the mean difference corresponding to the fourth digit of the mantissa. Solution:
from antilogarithm tables 0.230 1.698 mean difference corresponding to 5 2
1.700
2.897 6 2.903
7.328 13 7.341
1.140 2 1.142
(i) antilog of 1 .2305 = 2305.110 = 10−1 + 0.2305
= 10 −1 × 100.2305
= 700.1101
×
= 0.1700 (ii) antilog of 3 .4629 =10−3 + 0.4629
= 10− 3× 100.4629
= 903.210
13 ×
= 0.002903. (iii) antilog 1.8658 = 101.8658
= 101 + 0.8658
= 101 × 100.8658
= 10 × 7.341 = 73.41
(iv) antilog 2.0578 = 102.0578
= 102 + 0.0578
= 102 × 100.0578
= 102 × 1.142 = 114.2
Note: From the above example, we observe that (i) If the characteristic of log x is a non negative integer n, then the decimal point in the antilog of log x appears after the ( n + 1)th digit.
Characteristic Position of the decimal point 0 after the 1st digit 1 after the 2nd digit 2 after the 3rd digit 3 after the 4th digit 4 after the 5th digit
66
(ii) If the characteristic of log x is a negative integer, say − n, then the decimal point in the antilog (log x ) is inserted such that the first significant digit occurs at the nth place; that is, the first n − 1 digits are zero.
Characteristic Position of the decimal point −1 0.d1d2d3d4
−2 0.0d1d2d3d4
−3 0.00d1d2d3d4
−4 0.000d1d2d3d4
−5 0.0000d1d2d3d4
In the above table, d1. d2d3d4 is the antilogarithm corresponding to the mantissa of the given logarithm. 3.2.6 Calculations using logarithms
In the present day world, we have electronic calculators and computers to do computations quickly and more accurately. But, before these instruments came into existence, computations were done by hand. Logarithms were introduced in order to make computations quickly. Table of logarithms and antilogarithms were prepared as a ready-reckoner. Now we proceed to do some examples to show the usage of logarithms in calculations. We need mainly the following formulae:
(i) namamna loglog)(log +=
(ii) namanm
a logloglog −=⎟⎠⎞
⎜⎝⎛
(iii) .loglog mannma =
Example 25: Find (i) 27.91 × 5.49 (ii) 0.02871 × 0.00099 × 482.49 Solution: (i) Let x = 27.91 × 5.49
1.4456 2 1.4458
1.531 1 1.532
Then log x = log (27.91 × 5.49) = log 27.91 + log 5.49 = 1.4458 + 0.7396 = 2.1854
∴ x = antilog of 2.1854
= 102.1854
= 102 + .1854
= 102 ×10.1854
= 102 × 1.532= 153.2
67
(ii) Let x =0.02871 × 0.00099 × 482.49 Then log x = log 0.02871 + log 0.00099 + log 482.49
= 6835.29956.44581.2 ++ = 1372.2 .
x = antilog( 1372.2 )=0.01372. Example 26: Evaluate
(i) 23.16
2003 (ii) 09782.03421.0
Solution: (i) Let x = 23.16
2003 3.3016 1.2103 2.0913
Then log x = log 2003 − log 16.23 = 3.3016 − 1.2103 = 2.0913
∴ x = antilog of 2.0913 = 123.4
(ii) Let x = 09782.03421.0 .
Then log x = log 0.3421 − log 0.09782 = 1 .5341 − 2 .9904 = (−1+0.5341) − (−2+0.9904) = 1 + 0.5341 − 0.9904 = 1.5341 − 0.9904 = 0.5437
∴ x = antilog of 0.5437 = 3.497
Example 27: Compute the value of (i) (29.76)5 (ii) (0.3749)7
Solution: Let x = (29.76)5
Then log x = log (29.76)5 = 5 × log (29.76) = 5 × 1.4737 = 7.3685. ∴ x = antilog of 7.3685 = 23360000. (ii) Let x = (0.3749)7
Then log x =7× log( 0.3749) = 7 × 1 .5739 = (−1 + 0.5739) × 7 = −7 + 4.0173 = −7 + 4 + 0.0173 = −3 + 0.0173 = 3 .0173
∴ x = antilog of 3 .0173 = 0.001041. Example 28: Find the value of 5 2713.0
Solution: Let x = 5 2713.0 = (0.2713) 51
.
Then log x = log (0.2713) 51
.
68
= 51 . × log 0.2713 =
51 . × 1 .4335 =
54335.01+− =
54335.45 +−
= −1 + 0.8867 = 1 .8867. ∴ x = antilog of ( 1 .8867) =0.7702
Example 29: Find the value of 46.1828
159.2223.175 ×
Solution: Let x = 46.1828
159.2223.175 × . Then we have
log x = log 175.23 + log 22.159 − log 1828.46 We have to make the following approximations since the table of logarithms has been prepared for numbers ranging from 1.000 to 9.999 (with three digits in the fractional part).
175.23 = 1.7523 × 102
≈ 1.752 × 102. 22.159 = 2.2159 × 101
≈ 2.216 × 101. 1828.46 = 1.82846 × 103
≈ 1.829 × 103. ∴ log x = 2.2435 + 1.3456 − 3.2622
= 3.5891 − 3.2622 = 0.3269. ∴ x = antilog of (0.3269) = 2.122.
Example 30: Find the value of 04623.0)75.42(
928.1)25.76(5
33
××
Solution: Let x be the given expression. Then log x = log (76.25)3 + log 3 928.1 −[log (42.75)5 + log 0.04623]
= 3 log 76.25 + 31 log 1.928−[5 log 42.75 + log 0.04623]
= 3 × 1.8823 + 31 × 0.2851 − [5 × 1.6309 + 2 .6649]
= 5.6469 + 0.0950 − [8.1545 + 2 .6649] = 5.7419 − 6.8194 = − 2 + (7.7419 − 6.8194) = − 2 + 0.9225 = 2 .9925
x = anti log 2 .9925 =0.08366. Example 31: Find the value of 4.3562 12logSolution: Here the base is 12. To use the table of logarithms, we have to get 10 in the base. Using the change of base rule,
12log 4.3562 = 4.3562 × 10 10log 12log
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= 4.3562 × 10log12log
1
10
= .say0792.16391.0 x= Then log x = log ⎟
⎠⎞
⎜⎝⎛
0792.16391.0
= 1 .8056 − (0.0331) = −1 + (0.8056 − 0.0331) = −1 + (0.7725) = 1 .7725
∴ x = antilog ( 1 .7725) = 0.5923
Exercise 3.2.2 1. Find the characteristics of the common logarithms of the numbers. (i) 1234 (ii) 27.36 (iii) 3.65 (iv) 0.7851 (v) 0.084 (vi) 0.00532 (vii) 0.00003 (viii) 0.032 × 104
2. Find the number of digits in the integral part of the numbers whose common
logarithms are (i) 2.345 (ii) 1.456 (iii) 3.4567 (iv) 0.1234 (v) 0.9876 (vi) 3 × 0.982 3. The common logarithms of numbers are given below. Find the number of zeros after
the decimal point of the numbers. (i) 3456.1 (ii) 2 .2345 (iii) 123.3 (iv) 4 .7877 (v) 5 .7245 (vi) 4 .102 4. The mantissa of the common logarithm of 32740 is 0.5151. Write down the common
logarithms of the following. (i) 32740 (ii) 3274 (iii) 327.4 (iv) 32.74 (v) 3.274 (vi) 0.3274 (vii) 0.0003274 (viii) 0.03274 × 10−5
5. Find the common logarithms of the following numbers using the table of logarithms. (i) 8273 (ii) 843250 (iii) 0.001439 (iv) 0.0000324 (v) 0.00468 (vi) 0.2356 6. Find the antilogarithms of the following common logarithms. (i) 2.8903 (ii) 0.4321 (iii) 1 .4583 (iv) 4261.3 (v) 5 .5201 (vi) .0930.3 7. Find x, if the common logarithm of x is (i) 5.3027 (ii) 1.9168 (iii) − 2.0411 (iv) − 3.1773 (v) − 0.3916 (vi) − 4.1083 (vii) 2.12.3 + (viii) 1.24.5 − (ix) 2.1 − 5.4 (x) 23.1 × (xi) 341.3 × (xii) 5 .5 ÷ 3
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8. Evaluate the following: (i) 25.46 × 80.17 (ii) 37.42 × 816.3 (iii) 1.231 × 0.0084 (iv) 86.3 ÷ 0.0625 (v) (0.0275)3 (vi) (50.49)5
(vii) (525.9)8 (viii) 3 3.452
(ix) 5 08745.0 (x) 0543.0
37.454935.0 ×
(xi) 2
3
)4.132(5.27)23.4( × (xii)
23.394.163.8471.24
××
(xiii) (xiv) 326.1log5 28.63log9
3.3 Set Notation
Set notation is an achievement of our modern mathematics. It appears in all branches of mathematics. It originated when mathematicians attempted to axiomatize mathematics with in the frame work of logic. George Cantor (1845 − 1918), a German mathematician, developed the theory of sets which has become a milestone in the growth of mathematics. We now proceed to introduce the concept of a set and some elementary aspects of set theory. 3.3.1 The Concept of a Set
A collection of well defined objects is called a set. For example, the collection of all natural numbers, the collection of all equilateral triangles in a plane, the collection of all IX Standard students of Government Boys Higher Secondary School, Tiruthani, the collection of all real numbers, the collection of all vowels in English alphabet are some examples of sets since we can definitely say what objects are there in each of the collections. Consider the following statements: (i) The set of all tall students in your class. (ii) The set of good books you have studied. Both the above statements are not well defined, since there is no criteria for ‘tall’ and ‘good’ and so we cannot list the elements described by the above statements. Note: In a set, the objects are all distinct.
An object of a set is called an element or a member or an individual of the set. We usually denote a set by an upper-case letter like A or B and an element of a set by a lower-case letter such as x or y. If x is an element of a set A, we indicate this fact symbolically by writing x ∈ A. The symbol ∈ stands for ‘ is an element of ’ or ‘is a member of ’or ‘belongs to’. When an object x is not an element of the set A, we denote this fact by writing x ∉ A. Here, the
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symbol ∉ stands for ‘is not an element of ’ or ‘is not a member of ’ or ‘ does not belongs to’. For example, if A is the set {1, 3, 4, 5}, then the elements of A are 1, 3, 4 and 5; that is, we have 1 ∈ A, 3 ∈ A, 4 ∈ A and 5 ∈ A. We note that 6 ∉ A, −11 ∉ A, 9 ∉ A, Kumar ∉ A.
To represent a set, we adopt two methods: (i) the tabular or roaster method (ii) the set-builder or rule method.
3.3.2 Tabular or Roaster representation of a set
A set is usually written by listing all its objects, separating them by commas between the braces { }. For example, the set of vowels in the word mathematics is {a, e, i}. We note here that although the object a appears twice in the word mathematics, it is listed within the braces once only, since the objects of a set are distinct. When a set is written by listing all its objects separated by commas within braces { }, we say that the set is represented in the tabular or roaster form. Some sets in their tabular representation are given below:
(i) The set of prime numbers less than 13 is {2, 3, 5, 7, 11}. (ii) The set of letters in the word FOOTBALL is{F, O, T, B, A, L,} (iii) The set of all natural numbers 1, 2, 3,… is {1, 2, 3,…}. Here by writing …, we mean that the elements occur one after other in the representation without any omission. This set is specially denoted by the letter N. (iv) The set of all whole numbers 0, 1, 2, 3, … is {0, 1, 2, 3,…}.This set is specially denoted by the letter W. (v) The set of all integers 0, 1, −1, 2, −2, 3, −3,… is {0, 1, −1, 2, −2, 3, −3, …}. This set is specially denoted by the letter Z.
Example 32: Write the following set in the tabular form. Even natural numbers which are multiples of 5 and less than 50. Solution: The even natural numbers divisible by 5 are 10, 20, 30, 40, 50, … ∴ The set of even natural numbers divisible by 5 and less than 50 = {10, 20, 30, 40}. 3.3.3 Set-builder or Rule representation of a set
A set can also be written by another form known as set-builder form. To write a set in this form, we first find a common property of the elements of the set. This common property should be such that it should specify the objects of the set only. For example, let us consider the set {6, 36, 216}. The elements of the set are 6, 36 and 216. These numbers have a common property that they are powers of 6. So the condition x = 6n, where n = 1, 2 and 3 yields the numbers 6, 36 and 216. No other number can be obtained from the condition. Thus we observe that the set {6, 36, 216} is the collection of all numbers x such that x = 6n, where n = 1, 2, 3. This fact is written in the following form { x | x = 6n, n = 1, 2, 3}. In words, we read it as the set consisting of all x such that x = 6n, where n = 1, 2, 3. Here also, the braces
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{ } are used to mean ‘the set consisting of ’. The vertical bar ‘ | ’ within the braces is used to mean ‘such that ’. The common property ‘x = 6n, where n = 1, 2 and 3 ’ acts as a builder for the set and hence this representation is called the set–builder or rule form. If P is the common property possessed by each object of a given set A and no object other than these objects possesses the property P, then the set A is represented by { x | x has the property P} and we say that A is the set of all elements x such that x has property P. Example 33: Represent the following sets in Rule Form: (i) The set of all natural numbers less than 6. (ii) The set of vowels in English alphabet.
(iii) The set of the numbers 2, 4, 6, … . Solution: (i) A natural number less than 6 can be described by the statement: x ∈ N, x < 6. ∴ the set is { x | x ∈ N, x < 6}. (ii) A vowel in English alphabets can be described by the statement: x is a vowel in English alphabet. ∴ the set is {x | x is a vowel in English alphabet}.
(iii) A number x of the form 2, 4, 6, … can be described by the statement: x = 2n, n ∈ N. ∴ the set is { x | x = 2n, n ∈ N}.
Note: The set-builder method is also called descriptive method since here the elements are not listed but are indicated by the description of their characteristics. Example 34: Write the following set in the tabular form A = {x | x + 5 = 7, x ∈ N}. Solution: x + 5 = 7 ⇒ x = 7 − 5 = 2. Here 2 ∈ N. ∴ A = {2}.
Example 35: Obtain the set builder representation of the set A = ⎭⎬⎫
⎩⎨⎧
71 ,
61 ,
51 ,
41 ,
31 ,
21 1, .
Solution: Since the given elements are the reciprocals of the first seven natural numbers 1, 2, 3, 4, 5, 6, 7, we have
A = ⎭⎬⎫
⎩⎨⎧ ≤∈= 7 and 1| nn,
nxx N .
Note: ‘x, y’ means ‘x and y’.
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3.3.4 Finite and Infinite Sets
Suppose that we count the number of elements one by one in a set A. If we come to an end in the process, then the set A is said to be a finite set. If we never come to an end in counting the elements, then the set A is said to be an infinite set. The number of elements in a set A is called the cardinal number of the set A and is denoted by the symbol n(A) which is read as the number of elements in the set A. We observe that if A is a finite set, then n(A) is a whole number. Example 36: Identify finite and infinite sets from the following:
(i) A = {x | x ∈W, x < 5}. (ii) {All schools in Tamil Nadu}. (iii) {Students in IX standard in your school}. (iv) N. (v) W. (vi) Z. (vii) The set of all prime numbers.
Solution: (i) x ∈W, x < 5 ⇒ x = 0, 1, 2, 3, 4. ∴ A = {0, 1, 2, 3, 4}. When we count elements of A one by one as 1 for 0, 2 for 1, 3 for 2, 4 for 3, 5 for 4, we come to an end . ∴ n(A) = 5 and so A is a finite set. (ii) All schools in Tamil Nadu can be counted one by one and we come to an end in the counting process. So the set {All schools in Tamil Nadu} is a finite set. (iii) When we count the students of IX standard in your school, we come to an end in the counting process. So the set {Students in IX standard in your school} is a finite set. (iv) N = {1, 2, 3,…}. When we count elements of N one by one as 1 for 1, 2 for 2, 3 for 3, 4 for 4, we are not able to come to an end in the counting process. ∴ the set N is an infinite set. (v) W = {0, 1, 2, 3,…}. When we count elements of W one by one as 1 for 0, 2 for 1, 3 for 2, 4 for 3, we are unable to come to an end in the counting process. ∴ the set W is an infinite set. (vi) Z = {0, 1, −1, 2, −2,…}. When we count elements of Z one by one as 1 for 0, 2 for 1, 3 for −1, 4 for 2, 5 for −2 and so on, we are not able to come to an end in the counting process. ∴ the set Z is an infinite set. (vii) When we write the prime numbers one by one as 2, 3, 5, 7, 11, 13, 17 and so on, we are unable to come to an end in the counting process. ∴ the set of all prime numbers is an infinite set. 3.3.5 Empty set or Null Set or Void set
A set which contains no elements is called the empty or null or void set. It is denoted by the symbol Ø. Thus Ø = { }. We observe that n(Ø) = 0.
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Example 37: Which of the following sets are empty? (i) {Odd natural numbers which are divisible by 2}. (ii) {Prime numbers which have 4 as a factor}. (iii) {x | x ∈W, x ∉ N}. (iv) {Ø}. Solution: (i) There is no odd number which is divisible by 2.
∴ the set {x | x is an odd natural number and x is divisible by 2} = Ø. (ii) A prime number is a natural number which is not divisible by any other natural number except 1 and itself. So a prime number cannot have a factor other than 1 and itself. Hence there is no prime number which has 4 as a factor. ∴ {Prime numbers which have 4 as a factor} = Ø. (iii) We observe that 0 is the only element which is in W but not in N. i.e., x ∈W, x ∉ N ⇒ x = 0. ∴ {x | x ∈W, x ∉ N} = {0}. This set contain one element. So it is not an empty set. (iv) Empty set is an object. So {Ø} is a set containing one object. ∴ {Ø} is not an empty set. i.e., {Ø} ≠ Ø. 3.3.6 Equivalent sets
Two sets A and B are said to be equivalent if they contain the same number of elements. If the set A is equivalent to the set B, then n(A) = n(B) and we write A ≈ B. For example, if A = {1, 2, 3}, B = {11, 9, 23}, n(A) = 3, n(B) = 3. So A ≈ B. 3.3.7 Equal Sets
If two sets A and B contain the same elements, then they are said to be equal and we write A = B. For example, if A = {1, 2, 3, 4} and B = {x | x ∈ N, x < 5}, then A and B are equal sets since if we list the elements of B, we get the tabular form of B as {1, 2, 3, 4}. We observe that the equality of two sets is ensured by the presence of the same elements in the two sets. Hence, the elements of a set can be listed in the set in any order we like. For example, the sets {1, 2 ,3, 4} and {4, 3, 1, 2} are equal. Note: If two sets are equal, then they must be equivalent. However two equivalent sets need not be equal. For example, {1, −1, 2, −2} and {1, 2, −1, −2} are equal as well as equivalent sets, but , {1, −1, 2, −2} and {1, 2, 3, 4} are equivalent sets. But they are not equal sets, since the elements −1, −2 ∈ {1, −1, 2, −2} and −1, −2 ∉ {1, 2, 3, 4}. 3.3.8 Singleton set
A set which has only one element is called a singleton set. For example, the set of all even prime numbers. We know that the only even prime number is 2. So {All even prime numbers} = {2} and n(A) = 1. Hence it is a singleton set.
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3.3.9 Universal Set
The elements of all sets which occur in a mathematical investigation belong to a set. This set is called the universal set. The universal set is denoted by the symbol U or by the symbol ∈. For example, suppose the sets of a mathematical investigation are A = {2, 3, 4, 5}, and B = {1, 3, 7, 11}. Then we can take the universal set U as
U = {1, 2, 3, 4, 5, 7, 11,} or U = N or U = W or U = Z. 3.3.10 Subset
Let A and B be two sets. If every element of A is also an element of B, then A is called
a subset of B or B is a super set of A. We write this fact by the symbol A ⊆ B or B ⊇ A. Here the symbol ⊆ stands for is a subset of or is contained in and ⊇ means is a super set of or contains. For example, we consider two sets A = {1, −1, 2, −2, 3} and B = {1, 2, −1, −2, 3, −3}. We observe that 1, 2, −1, −2, 3 ∈ A and 1, 2, −1, −2, 3 ∈ B. That is, every element of the set A is also an element of the set B. So A is a subset of B, that is, A ⊆ B. We also observe here that −3 ∈ B and −3 ∉ A. So B is not a subset of A and this fact is written as B ⊄ A. Here the symbol ⊄ stands for is not a subset of or is not contained in. Note: If a set X is a subset of a set Y and the set Y is a subset of the set X, then the two sets X and Y have the same elements and so they are equal. If X = Y, then every element of X is also an element of Y and every element of Y is also an element of X. We should note that the sets X and Y are not numbers, still we use the equality sign ‘=’ between them whenever they are equal. The equality symbol is used with the understanding that all elements of X are in Y and all elements of Y are in X; that is X and Y have precisely the same elements. Note: Every set A is a subset of itself since every element of A is in A itself. The empty set Ø is a subset of every set A because if Ø cannot be a subset of A, then there would be an element in the empty set Ø which would not be in A. Thus, A ⊆ A and Ø ⊆ A. 3.3.11 Proper Subset
Let X and Y be two sets. If X is a subset of Y and there is an element u ∈ Y and u ∉ X,
then the set X is called a proper subset of Y. It is denoted by X ⊂ Y. In this situation, we also write Y ⊃ X. For example, let A = {1, 2, 3, 4} B = {0, 1, 2, 3, 4, 5}. We observe that A is a subset of B and 5 ∈ B but 5 ∉ A. ∴ A is a proper subset of B; that is, A ⊂ B . Note: Although A is a subset of A, it is not a proper subset of A. Hence it is called an improper subset of A.
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3.3.12 Power Set
The collection of all subsets of a given set A is called the power set of A. It is denoted by the symbol p(A). For example if A = {a, b}, then p(A) is{ }}{ },{ },{ {}, a,bba and the cardinal number of p(A) is n[p(A)] = 4 = 22. Similarly, if A = {a, b, c}, then p(A) is{ and the cardinal number of p(A) is n[p(A)] = 8 = 2
}},{},{ },{ },{},{ },{ },{ {}, b,cac,ab,ca,bcba3. Thus, we observe that if n(A) = m, then n[p(A)] = 2m.
Exercise 3.3.1 1. Rewrite the following sets using tabular form: (i) A = {The Vowels in the word SUNDAY}. (ii) B = {The Seasons of the year}. (iii) C = {The set of all letters in the word MATHEMATICS}. (iv) D = {The set of all letters in the word TAMILNADU}. 2. Write the following sets in the roaster form: (i) P = {x | x is a letter of the word TAMILNADU}. (ii) Q = {x | x is a whole number and −3 ≤ x < 7}. (iii) R = {x | x is a two digit natural number such that sum of its digits is 9}. 3. Write the following sets in the builder form: (i) {3, 6, 9, 12} (ii) {5, 25, 125, 625} (iii) {1, 3, 5,…} (iv) {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}. 4. Match each of the sets on the left described in the roaster form with the same set on the right described in set builder form. (i) {1, 2, 3, 6} (a) {x | x is a prime number and a divisor of 6}. (ii) {2, 3} (b) {x | x is an odd natural number less than 10}. (iii) {2, 4, 6, 8} (c) {x | x is a positive integer and a divisor of 6}. (iv) {1, 3, 5, 7, 9} (d) {x | x is an even natural number less than 10}. 5. Find the cardinal number of the following sets: (i) The set of vowels in English alphabet. (ii) The set of all perfect square numbers less than 100. (iii) A = { x | x is a letter in the word “mathematics”}. (iv) B = {x | x < 0, x ∈ W}. (v) C = {x | − 3 ≤ x < 4, x ∈ Z}. (vi) The set of all prime numbers between 10 and 20. 6. Write down the elements that have been left out in the following finite sets: (i) A = {1, 10, 100, ______, _______, 1,00,000}. (ii) B = {2, 5, 8, 11, ______, _______, 20, 23}.
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7. Write down the next three elements of the following infinite sets: (i) C = {3, 6, 12, 24, ______, ______, ______, ….}. (ii) D = {−4, −3, −2, −1, ______, ______, _____, …}. 8. State whether the following sets are empty sets: (i) A = {Even natural numbers divisible by 3}. (ii) B = { x | x ∈ R , x2 + 1 = 0}. (iii) C = {Polygons with four sides}. (iv) D = {Quadrilateral with five sides}. 9. Let A = {p, q, r, s}, B = {1, 3, 5, 7}, C = {q, r}, D = {8, 4, 6, 2}, E = {r, q, s, p} F = {2, 6, 4, 8}. Write true or false: (i) A and C are equivalent sets. (ii) A and E are equal sets. (iii) F and B are equivalent sets. (iv) A and B are equal sets. (v) F and D are equal sets. 10. Write down the power set for each of the following sets: (i) A = {1, 2} (ii) B = { x, y, z} (iii) C = {a, b, c, d} 11. (i) If n(A) = 5, find n[p(A)] (ii) If n[p(A)] = 128, find n(A). 3.3.13 Set Operations We shall now study about the set operations
(i) union of two sets (ii) intersection of two sets (iii) complement of a set
(i) Union of two sets
Let A and B be two given sets. The set of all elements that belong either to A or to B or to both is called the union of A and B. We denote the union of A and B by A U B. Thus,
A B = { x | x ∈ A or x ∈ B or x ∈ A and B}. U
We write A B = {x| x ∈ A or x ∈ B} where it is understood that the word or is used in the inclusive sense; that is, x ∈ A or x ∈ B stands for x ∈ A or x ∈ B or x ∈ A and B.
U
Example 38: If A = {1, 2, 3, 4} and B = {2, 4, 6}, find A U B. Solution: Listing all the elements A and B together and omitting the repetition, we get Thus, A U B = {1, 2, 3, 4, 6}. 1, 2, 3, 4, 2, 4, 6.
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(ii) Intersection of two sets
Let A and B be two given sets. The set formed by the elements that are common to both A and B is called the intersection of A and B. We denote the intersection of A and B by A I B. Thus A B = {x | x ∈ A and x ∈ B}. I Example 39: If A = {1, 2, 3} and B = {2, 3, 4}, find A I B. Solution: All elements in A and B: 1, 2, 3, 2, 3, 4
2, 3, 2, 3 Common elements in A and B: ∴ A B = {2, 3}. I 3.3.14 Disjoint Sets
If two sets A and B have no elements in common, then they are called disjoint sets; i.e., if A B = Ø or { }, then A and B are disjoint sets. For example, if A = {1, 2, 3, 7} and B = {4, 5, 6}, then A I B = { } and so A and B are disjoint sets.
I
3.3.15 Difference Set
Let A and B be two given sets. The set of all elements of A which are not in B is called the difference set. It is denoted by A – B. Thus, A − B = {x | x ∈ A, x ∉ B}. Note: B − A = {x | x ∈ B, x ∉ A} Example 40: If A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 7}, find A − B and B − A. Solution: A−B = {2, 4, 5, 6}. B −A = {7}. Note: A − B ≠ B − A. 3.3.16 Complement of a set
Let A be a given set and U be the universal set. The set of all elements of U which are not in A is called the complement of the set A and is denoted by A′ or Ac or A . Note: Ac = U − A. Example 41: If U = {1, 2, 3, 4, 5} and A = {3, 4}, find Ac. Solution: Ac = {1, 2, 5}. 3.3.17 An Identity in set theory
There is an useful identity in set theory which provides the number in the union of the sets. It is stated as follows: If A and B are two sets, then n(A B) ≡ n(A) + n(B) − n(A B). U I
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Example 42: If A = {1, 3, 4, 5, 6, 7, 8, 9} and B = {1, 2, 3, 5, 7}, find n(A), n(B), n(A B) and n(A B) and verify the identity n(A B) ≡ n(A) + n(B) − n(A B).
U
I U I
Solution: We observe that A B = {1, 2, 3, 4, 5, 6, 7, 8, 9} A B ={1, 3, 5, 7}. U I
n(A) = 8, n(B) = 5, n(A U B) = 9 and n(A I B) = 4. We find n(A) + n(B) − n(A I B) = 8 + 5 − 4 = 9. Here, n(A B) = 9. So n(A B) = n(A) + n(B) − n(A B). U U I
In fact, this result is true for any two finite sets.
Exercise 3.3.2 1. Find A B and A B for the following sets: U I (i) A = {a, e, i, o, u} and B = {a, b}. (ii) A = {1, 3, 5} and B ={1, 2, 3}. (iii) A = {x | x is a natural number and 1 < x ≤ 6)} and B = {x | x is a natural number 6 < x < 10}. (iv) A = {p, q, r} and B = Ø. 2. Find A − B, A − C and B − A for the following sets: (i) A = {a, b, c, d, e, f, i, o, u}, B = {a, b, c, d} and C = {a, e, i, o, u}. (ii) A = {3, 4, 5}, B = {5, 6, 7, 8} and C = {7, 8, 9}. 3. If U = {a, b, c, d, e, f, g, h}, A = {a, c, g} and B = {a, b, c, d, e, f}, find (i) Ac (ii) Bc (iii) (A U B) c
(iv) (AcI B) c (v) Ac BI c (vi) Ac BU c. 3.3.18 Venn Diagram
As an aid to visualize the operations (forming union and intersection and taking complement) on sets, John Venn, an English mathematician introduced a diagrammatic way of representing sets. The diagrams representing sets are called Venn Diagrams. Usually the Universal set is represented by a rectangle and its proper subsets are by circles drawn within the rectangle. We now give the representation of different sets in Venn diagram:
Universal set Figure 3.1
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A I B
A B UFigure 3.2
Figure 3.3Ac
.Figure 3.4 Bc
Figure 3.5
(A U B) c
Figure 3.6 (A I B) c
Figure 3.7
A − B Figure 3.8
B − A Figure 3.9
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Example 43: From the diagram given below(see Figure 3.10), find the sets (i) A B (ii) A B (iii) (A B)U I U c
Solution: Here U = {1, 2, 3, 4, 5, 6, 7}.
Figure 3.10
(i) A B = {1, 2, 3, 4, 5, 6}. U
(ii) A B = {2, 5}. I (iii) (A U B) c = {7}.
Example 44: Using Venn Diagram, exhibit the sets A = {−2, 3, 5, 7} and B = {3, 9, 11}. Verify the formula: n(A B) = n(A) + n(B) − n(A B). U I
Solution:
Figure 3.11
A U B = {−2, 3, 5, 7} U {3, 9, 11} = {−2, 3, 5, 7, 9, 11} ∴ n(A B) = 6. (1) U
A I B = {−2, 3, 5, 7} I {3, 9, 11} = {3} ∴ n(A B) = 1. I
We have n(A) = 4, n(B) = 3. ∴ n(A) + n(B) = 4 + 3 = 7. ∴ n(A) + n(B) − n(A B) = 7 − 1 = 6. (2) I
From (1) and (2) we get (A U B) = n(A) + n(B) − n(A B) I Note: If A and B are disjoint then A B = Ø and n(A B) = 0. I IThis gives n(A B) = n(A) + n(B). U Example 45 : The number of girls in a village who attended tailoring classes was 45, the number of girls who attended classes on gardening was 70. If 30 of these attended both the classes, using Venn Diagram, find
(i) how many attended only one type of class (ii) how many totally attended either of these classes.
Figure 3.12
Solution: If A and B represent respectively, the set of girls who attended the tailoring classes and the set of girls who attended classes in gardening we have n(A) = 45, n(B) = 70 By data n(A B) = 30. From the Venn diagram we see that 45 − 30 = 15 girls attended only tailoring classes and 70 − 30 = 40 girls attended only classes on gardening. Thus 15 + 30 + 40 = 85 attended either of these classes.
I
The number of girls who attended either of these classes is given by n(A U B) = n(A) + n(B) − n(A B) = 45 + 70 − 30 = 85. I
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Example 46: Out of 45 houses in a village 25 houses have T.V and 30 houses have radio. Find how many of them have both. Solution: Let A = {the number of houses having T.V} B = {the number of houses having Radio} A I B = {The number of houses having both T.V and Radio}. Let n(AI B) = x From the figure 3.13, 25 − x + x + 30 − x = 45
55 − x = 45 −x = 45 − 55 −x = −10 x = 10
∴ The number of houses having both T.V and Radio = 10.
Figure 3.13
Example 47: In a class of 35 students, 28 students passed in History, 22 in Mathematics and 18 in both Subjects. How many failed in both subjects? Using Venn Diagram, find the number of students who passed in (i) History alone and (ii) Mathematics alone. Solution: If H and M denote respectively the set of students who passed in History and those who passed in Mathematics, we find from the given data
Figure 3.14
n(H) = 28, n(M) = 22 and n(H M) = 18. I
The number of students who have passed in History or in Mathematics or on both n(H M) = n(H) + n(M) − n(H ∩ M) U
i.e., n(H U M) = 28 + 22 − 18 = 32. Since there are 35 students in the class, the number of students who have failed in both subjects = 35 − 32 = 3. Using Venn diagram we find that the (i) Number of students passed in History alone = n(H ) − n(H M) = 25 − 18 = 10 I (ii) The number of students who passed in Mathematics alone = n(M) − n(H M) I
= 22 − 18 = 4.
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Exercise 3.3.3 1. Given A = {2, 3, 5, 8, 10},B = {3, 7, 8, 9} and C = {1, 2, 5, 8, 12}. Find (i) n(A B) (ii) n(B C) (iii) n(A − B) (iv) n(C − B) U I
2. If n(A) = 30, n(B) = 43 and n(A B) = 11. Find n(A B) I U
3. A student in IX standard has to attend at least one of two tests A and B. If 40 students
attend test A, 30 students attend test B and 20 students attend both the tests. Find the number of students in the class.
4. In a survey taken among 400 residents of a colony, 250 bought English news papers,
170 bought Tamil news papers and 65 bought both. How many residents did not buy any paper at all.
5. In a village there are 200 families. Two brands of soaps A and B were popular there
160 families used brand A and 140 families used brand B. If all the families used at least one of these brands find how many families used both the brands.
6. In a party attended by 250 persons, 210 took coffee, 50 took tea and some persons
took both coffee and tea. If 20 persons did not drink coffee or tea. Find the number of persons who took both coffee and tea.
7. A college magazine reported that 150 students had combined membership in physics
club and mathematics club. Find the membership in physics club if 70 students were members of the mathematics club and 50 students were members of both the clubs.
8. In a class of 30 girls, 20 girls took part in singing competition and 10 took part in
singing and dancing. If 5 girls did not take part in any one of them. Find how many took part in dance competition only.
9. During Summer Vacation, 35 students of class XII attended computer classes. 25
students coaching classes for entrance examinations and 15 attended both. Find how many students attended neither if the class strength was 50.
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Answers
Exercise 3.1 1. (i) (ii) 1.3 × 101010998.2 × 9 (iii) 1.083 × 1012 (iv) 4.3 × 109
(v) 9.463 × 1015 (vi) 5.349 × 1017 (vii) 3.7 × 10−3 (viii) 1.07 × 10−4
(ix) 8.035 × 10−5 (x) 1.3307 × 10−6 (xi) 1.1 × 10−10 (xii) 9 × 10−13
2. (i) 0.00000325 (ii) 0.0000402 (iii) 0.0004132 (iv) 0.001432
(v) 3250000 (vi) 402000 (vii) 41320 (viii) 1432 3. (i) 1.024 × 1014 (ii) 4.41 × 10−4 (iii) 1.1664 × 108
(iv) 2.56 × 10−26 (v) 7.5 × 10−8
Exercise 3.2.1
1. (i) T (ii) F (iii) F (iv) F (v) T (vi) F
2. (i) 204.0log5 −= (ii) 324log
81
−= (iii) 4256log4 =
(iv) 6729log3 = (v) 23216log36
−= (vi) 3001.0log10 −=
3. (i) 4 (ii) 3 (iii) 4 (iv) 21− (v) −4
(vi) 25 (vii) 36 (viii) 64 (ix)
41
4. (i) 10 (ii) 128
1 (iii) 101 (iv) 5 (v)
321
(vi) 512 (vii) 2 (viii) 162 (ix) 31− (x) 3
5. (i) C (ii) A (iii) D (iv) C (v) A (vi) D
6. (i) (ii) 18log10 ⎟⎠⎞
⎜⎝⎛
818log3 (iii) 13 (iv) ⎟
⎠⎞
⎜⎝⎛
3350log 2
(v) ⎟⎠⎞
⎜⎝⎛
2572log10 (vi) ⎟
⎠⎞
⎜⎝⎛
485log10
7. (i) x + y (ii) 2x (iii) y − x (iv) 3y (v) t − y (vi) 3x + y + 2 z (vii) 3(x − y) (viii) y + z (ix) z + t (x) 2x + y (xi) 2t − x − z (xii) x − y − z + t
8. (i) 22 (ii) 7
11 (iii) 34− (iv) 16 (v)
310 (vi)
65
(vii) 9 (viii) 9 (ix) 37 (x) 243 (xi)
72−
Exercise 3.2.2
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1. (i) 3 (ii) 1 (iii) 0 (iv) −1 (v) −2 (vi) −3 (vii) −5 (viii) 2 2. (i) 3 (ii) 2 (iii) 4 (iv) 1 (v) 1 (vi) 3 3. (i) 0 (ii) 1 (iii) 2 (iv) 3 (v) 4 (vi) 3 4. (i) 4.5151 (ii) 3.5151 (iii) 2.5151 (iv) 1.5151 (v) 0.5151 (vi) .51511 (vii) .51514 (viii) .51517 5. (i) 4.9177 (ii) 5.926 (iii) .1583 (iv). .51055 (v). .67023 (vi) .37221 6. (i) 776.7 (ii) 2.705 (iii) 0.2873 (iv) 0.002668 (v) 0.00003312 (vi) 0.001239 7. (i) 200700 (ii) 82.56 (iii) 0.01099 (iv) 0.0006
(v) 0.4059 (vi) 0.00007789 (vii) 0.00003981 (viii) 0.0005012 (ix) 0.0005012 (x) 0.03981 (xi) 0.00000001698 (xii) 0.03162.
8. (i) 2041 (ii) 30550 (iii) 0.01034 (iv) 1380
(v) 0.00002079 (vi) 3285000 (vii) 5851 × 1018
(viii) 7.670 (ix) 0.6142 (x) 412.3 (xi) 0.02263 (xii) 0.8339 (xiii) 0.1752 (xiv) 1.893.
Exercise 3.3.1
1. (i) A = {U, A} (ii) B = { Summer, Winter, Spring, Autumn} (iii) C = { M, A, T, H, I, C, S, E.} (iv) D = { T, A, M, I, L, N, D, U} 2. (i) P = { T, A, M, I, L, N, D, U} (ii) Q = {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6} (iii) R = { 18, 27, 36, 45, 54, 63, 72, 81, 90} 3. (i) A = {x | x = 3n, n = 1, 2, 3, 4}
(ii) B = {x | x = 5n, n = 1, 2, 3, 4} (iii) C = {x | x ∈ N, x is odd} (iv) D = {x | x ∈ N, x = n2, 0 < n ≤ 10}
4. (i) → (c) ; (ii) → (a); (iii) → (d); (iv) → (b) 5. (i) 5 (ii) 9 (iii) 8 (iv) 0 (v) 7 (vi) 4
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6. (i) 1000, 10000 (ii) 14, 17 7. (i) 48, 96, 192 (ii) 0, 1, 2 8. (i) No (ii) Empty set (iii) No (iv) Empty set 9. (i) F (ii) T (iii) T (iv) F (v) T 10. (i) p(A)= { } { } { } { }{ },2,1,2,1
(ii) p(B)= { } { } { } { } { } { } { } { }{ } ,,, ,, ,, ,, , , , zyxzyzxyxzyx
(iii) p(C)= {{a},{b},{c},{d},{a, b}, {a, c}, {a, d}, {b, c},{b, d},{c, d}, {a, b, d}, {b, c, d},{c, d, a}, {a, b, c},{a, b, c, d},{}}
11. (i) 32 (ii) 7
Exercise 3.3.2 1. (i) A B = {a, e, i, o, u, b}, A I B = {a} U
(ii) A B = {1, 2, 3, 5}, A I B = {1, 3} U
(iii) A U B = {2, 3, 4, 5, 6, 7, 8, 9}, A I B = Ø (iv) A B = {p, q, r}, A B = Ø U I
2. (i) A − B = {e, f, i, o, u }, A − C = {b, c, d, f }, B − A = Ø
(ii) A − B = {3, 4}, A − C = {3, 4, 5}, B − A = {6, 7, 8} 3. (i) Ac = {b, d, e, f, h}
(ii) Bc = {g, h} (iii) (A U B) c = {h} (iv) (Ac I B) c= {a, c, g, h} (v) Ac I B c= {h} (vi) AcU B c = {b, d, e, f, g, h }
Exercise 3.3.3
1. (i) 7 (ii) 1 (iii) 3 (iv) 4 2. 62 3. 50 4. 45 5. 100 6. 30 7. 130 8. 5 9. 5
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4. ALGEBRA
The word algebra is derived from the Arabic word al–jabr. In Arabic language, ‘al’ means ‘the’ and ‘jabr’ means ‘reunion of broken parts’. The usage of the word can be understood by a simple example. In the equation x + 5 = 9, the left hand side is the addition (sum) of two parts x and 5. If we add (unite) (–5) to each side of the equation, we get (x + 5) + (–5) = 9 + (–5) or x + [5 + (–5)] = 9 – 5 or x + 0 = 4 or x = 4. Here 9 and −5 are reunited to get 4. This type of mathematics is called algebra. Indian mathematicians like Aryabhatta, Brahmagupta, Mahavir, Sridhara, Bhaskara II have developed this subject very much. The Greek mathematician Diophantus has developed this subject to a great extent and hence we call him the father of Algebra.
In this branch of mathematics, we use letters like a, b, x and y to denote numbers. Performing addition, subtraction, multiplication, division or extraction of roots on these symbols and real numbers, we obtain what are called algebraic expressions. The following are some examples of algebraic expressions:
2x + 3, (3a + b) (2x – y), yxxx 1914,
7191225
+++ .
If two algebraic expressions are equated, we get what are called algebraic equations. The following are some examples of algebraic equations:
2x + 3 = x + 6, 2532
7 2 +=+− x
xx , 2x + 11 = 0.
Symbols in an algebraic expression are called variables of the expression. For example, in ax + b, if a and b are specific numbers and x is not specified, then x is the variable of ax + b. In 2x2 + 3xy + y2, x and y are variables. If the variables in an algebraic expression are replaced with specific numbers, then the expression yields a number and this number is called a value of the expression. For example, 2x2 + y is an algebraic expression and x and y are variables of the expression. When we substitute 2 for x and 1 for y, the value of 2x2 + y is 2(2)2 + 1 = 9. If we substitute (–1) for x and 2 for y, then the value of 2x2 + y is 2(–1)2 + 2 = 4. An algebraic expression may not have a real number value for some real number values of the variables of the expression. For example, the expression 3−x has no real number value when x = 1 because the square of a real number can not be a negative real number and hence the square root of a negative real number cannot be a real number. In this chapter, we shall study about algebraic expressions in one or several variables, the variables and the algebraic expressions represent real numbers only. Here after, by a number we shall mean real number only.
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4.1 Polynomials
An Algebraic expression of the form axn is called a monomial in x, where a is a known number, x is a variable and n is a non-negative integer. The number a is called the coefficient of xn and n, the degree of the monomial. For example, 7x3 is a monomial in x of degree 3 and 7 is the coefficient of x3. The sum of two monomials is called a binomial and the sum of three monomials is called a trinomial. For example, 2x3 + 3x is a binomial and 2x5 – 3x2 + 3 is a trinomial. The sum of a finite number of monomials in x is called a polynomial in x. The coefficients of the monomials in a polynomial are called the coefficients of the polynomial. If all the coefficients of a polynomial are zero, then the polynomial is called the zero polynomial. Hereafter, by a polynomial, we shall mean non–zero polynomial in x only, that is, a polynomial with at least one non–zero coefficient. The monomials in a polynomial are called the terms and the highest degree of the terms, the degree of the polynomial. The coefficient of the highest power of x in a polynomial is called the leading coefficient of the polynomial. For example, 3x – 1 + 3x2 + 6x5 – 4x3 is a polynomial of degree 5 with terms 3x, –1, 3x2, 6x5 and –4x3 and leading coefficient 6. Since the terms of a polynomial are real numbers, they can be rearranged using the properties of real numbers such that the terms of the polynomial occur in the descending powers of x.
For example, the polynomial 3x2 – 1 – 5x – 3x3 is rewritten as – 3x3 + 3x2 – 5x – 1. When we write a polynomial in descending powers of x, we say that the polynomial is in the standard form. Special names are given to lower degree polynomials. A first degree polynomial is called a linear polynomial. A second degree polynomial is called a quadratic polynomial. A third degree polynomial is called a cubic polynomial and a fourth degree polynomial is called a bi-quadratic polynomial. We shall denote polynomials in x by symbols such as f(x), g(x), p(x), q(x), r(x). The degree of a polynomial f(x) is denoted by the symbol deg(f(x)). For example, if f(x) is the polynomial 1 – 3x2 + 5x3 – 2x, then deg(f(x)) = 3. Since the terms of a polynomial are numbers, all polynomials represent numbers. So we can add, subtract, multiply and divide polynomials. We shall know about the division of one polynomial by another polynomial later. 4.1.1 Addition of Polynomials
We add two polynomials by adding the coefficients of the like powers. Example 1: Find the sum of 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1. Solution: Using the associative and distributive properties of real numbers, we obtain (2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1) = 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1 = 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2 = 2x4 + 6x3 – 9x2 + 9x + 2.
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The following scheme is helpful in adding two polynomials
2x4 + 0x3 – 3x2 + 5x + 3 0x4 + 6x3 – 6x2 + 4x – 1
2x4 + 6x3– 9x2 + 9x + 2
4.1.2 Subtraction of Polynomials
We subtract polynomials like addition of polynomials. Example 2: Subtract 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6. Solution: Using associative and distributive properties, we have (x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1) = x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1 = x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1 = (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1) = –x3 + 8x2 – 4x – 5. The subtraction can also be performed in the following way: Line (1): x3 + 5x2 – 4x – 6. Line (2): 2x3 – 3x2 – 1. Changing the signs of the polynomial in Line (2), we get Line (3): –2x3 + 3x2 + 1. Adding the polynomials in Line (1) and Line (3), we get –x3 + 8x2 – 4x – 5 The above procedure is written as follows : x3 + 5x2 – 4x – 6 2x3 – 3x2 – 1 – + + – x3 + 8x2 – 4x – 5 4.1.3 Multiplication of two polynomials
To find the multiplication or product of two polynomials, we use the distributive properties and the law of exponents. Example 3: Find the product of x3 – 2x2 – 4 and 2x2 + 3x – 1 . Solution: (x3 – 2x2 – 4) (2x2 + 3x – 1)
= x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–4) (2x2 + 3x – 1) = (2x5 + 3x4 – x3) + (–4x4 – 6x3 + 2x2) + (–8x2 – 12x + 4) = 2x5 + 3x4 – x3 – 4x4 – 6x3 + 2x2 – 8x2 – 12x + 4 = 2x5 + (3x4 – 4x4) + (–x3 – 6x3) + (2x2 – 8x2) + (–12x) +4 = 2x5 – x4 – 7x3 – 6x2 – 12x + 4.
When finding the product of two polynomials, we multiply each term of one
polynomial by each term of the other polynomial and then the products are added (summed). The following scheme may be helpful for the beginners.
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x3 – 2x2 – 4 × 2x2 + 3x – 1
x3 (2x2 + 3x – 1) : 2x5 + 3x4 – x3
–2x2 (2x2 + 3x – 1) : – 4x4 – 6x3 + 2x2
–4(2x2 + 3x – 1) : – 8x2 – 12x + 4
2x5 – x4 – 7x3 – 6x2 – 12x + 4
Sometimes we require the coefficient of a particular term in a product of polynomials.
To save time and space, we can find the coefficient without actually multiplying the polynomials. For example, if we want to obtain the coefficient of x3 in a product of two polynomials A and B, then the following scheme may be useful for the beginners: Step 1: Get the products given below: Coefficient of x3 term in A × Constant term in B Coefficient of x2 term in A × Coefficient of x term in B Coefficient of x term in A × Coefficient of x2 term in B Constant term in A × Coefficient of x3 term in B Step 2: Add the above products in step 1 and the resulting value is the coefficient of x3 in the product of the polynomials A and B Example 4: Find the coefficients of x4, x3, x2 and x terms in the product of 7x3 – 6x2 – 9x + 8 and 5x2 – 3x + 5 without doing actual multiplication. Solution: To get the coefficient of x4: Coefficient of x4 term in A × Constant term in B = 0 × 5 = 0. Coefficient of x3 term in A × Coefficient of x term in B = 7 × – 3 = –21. Coefficient of x2 term in A × Coefficient of x2 term in B = – 6 × 5 = –30. Coefficient of x term in A × Coefficient of x3 term in B = – 9 × 0 = 0. Constant term in A × Coefficient of x4 term in B = 8 × 0 = 0. So the coefficient of x4 in the product of A × B is 0 + (–21) + (–30) + 0 + 0 = –51. To get the coefficient of x3: Coefficient of x3 term in A × Constant term in B = 7 × 5 = 35. Coefficient of x2 term in A × Coefficient of x term in B = – 6 × – 3 = 18. Coefficient of x term in A × Coefficient of x2 term in B = – 9 × 5 = –45. Constant term in A × Coefficient of x3 term in B = 8 × 0 = 0. So the coefficient of x3 in A × B is 35 + 18 + (– 45) + 0 = 8.
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To get the coefficient of x2: Coefficient of x2 term in A × Constant term in B = – 6 × 5 = –30. Coefficient of x term in A × Coefficient of x term in B = –9 × –3 = 27. Constant term in A × Coefficient of x2 term in B = 8 × 5 = 40. So the coefficient of x2 in A × B is (–30) + 27 + 40 = 37. To get the coefficient of x term: Coefficient of x term in A × Constant term in B = –9 ×5 = –45. Constant term in A × Coefficient of x term in B = 8 × –3 = –24. So the coefficient of x term in A × B is (–45) + (–24) = – 69. 4.1.4 Polynomials in several variables
A monomial in x and y is of the form axnym where a is a real number, x and y are variables and n and m are non-negative integers. For example, 5x3y2 is a monomial in x and y. The sum of a finite number of monomials in x and y is called a polynomial in x and y. For example, 5x2y + 3x + y2, 3x – 8y, 2x2 + 3xy + 2y2 are polynomials in x and y. Similarly we have polynomials in more variables. As we did for polynomials of one variable, we can add, subtract and multiply polynomials of several variables. Example 5: Find the sum of x3y + x2y2 – 3xy3 and x3 – 3x3y + y3 + 4xy3. Solution: (x3y + x2y2 – 3xy3) + (x3 – 3x3y + y3 + 4xy3)
= x3y + x2y2 – 3xy3 + x3 – 3x3y + y3 + 4xy3
= (x3y – 3x3y) + (x2y2) + (–3xy3 + 4xy3) + (x3) + (y3) = –2x3y + x2y2 + xy3 + x3 + y3 .
Example 6: Find the product of 2x + 3y and x2 – xy + y2. Solution:
2x + 3y × x2 – xy + y2
2x(x2 – xy + y2) : 2x3 – 2x2y + 2xy2
3y(x2–xy + y2) : 3x2y – 3xy2 + 3y3
2x3 + x2y – xy2 + 3y3
The process of division of one polynomial b
Ex1. Answer true or false:
(i) 2x + x3 is a polynomial.
(ii) 22x + 3x3 + 1 is a second de
Or (2x + 3y) (x2 – xy + y2)
= 2x (x2 – xy + y2) + 3y (x2 – xy + y2)
= 2x3 – 2x2y + 2xy2 + 3x2y – 3xy2 + 3y3
=2x3 + (–2x2y + 3x2y) + (2xy2 – 3xy2) + 3y3
= 2x3 + x2y – xy2 + 3y3.
y another polynomial is discussed in section 4.4.
ercise 4.1
gree polynomial.
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(iii) The coefficient of x2 in 5 – 2x – 3x2 + x3 is 3. (iv) 5xy is a binomial. (v) 2x + 3y + 5z is a trinomial. In each of the problems 2−4, find the sum and write it in the standard form :
2. (x3 + 3x – 1) + (2x2 – 4x + 5) 3. (2x4 + x2 + 3x) + (x4 – 3x2 + 7x – 8) 4. (6–10x + 5x2 + x3) + (2x3 – 3x – 4)
In each of the problems 5 to 7, find the subtraction and write it in the standard form :
5. (x3 + 5x2 – 10x + 6) – (2x3 – 3x – 4) 6. (x4 – 3x2 + 7x – 8) – (2x4 + x2 + 3x) 7. (3x5 – 5x2 + 4x – 7) – (1– 2x + 3x2 – x3)
In each of the problems 8 to 10, find the multiplication and write it in the standard form:
8. (2x2 – 6x + 3) (3x2 – 4x + 9) 9. (3x2 – 4x + 5x3 – 7) (2x2
– x + 4) 10. (7 – x – x2 ) (x3 – 5x2 + 3x)
Without doing actual multiplication, find the coefficients of x3, x2 and x in each of the products in problems 11 to 13:
11. (x2 – 4x + 4) (x2 + 2x –3) 12. (3x – 2 – x2) (1 + 3x – x2) 13. (7x3 – 6x2 – 9x – 1) (2x3 – 3x2 – 1)
Find the products in problems 14 to 16 in the standard form:
14. (ax + by) (cx + dy) 15. (x+y) (2x2 – 3xy – 2y2) 16. (x2 – xy + y2) (x2 + xy + y2) 17. If the coefficient of x2 in the product (x3 – px2 + 9x – 1) (2x3 – 3x2 – x + 2) is 12, find the value of p. 18. If the coefficient of x in the product (x3 – 2x + 5) (a – 3x – x2) is equal to the coefficient of x2 in the product (2x2 + x – 1) (x2 – 3x – 2), find a. 19. If the sum of the coefficients of x2 and x in the product (1 – 2x – x2) (2x2 – mx + 3) is 5, find the value of m. 4.2. Algebraic Identities
We study about certain algebraic equations called algebraic identities. We have already learnt about algebraic equations. An algebraic equation may involve one or more variables. For example, in the algebraic equation 2x + 3 = 6 – x, x is the variable. When we substitute 1 for x, the equation becomes 5 = 5, a true statement. When we substitute any other number say 2 for x, the equation becomes 7 = 4, a false statement. If a number substituted for the variable of the equation makes the equation a true statement, then the number is called a solution or root of the equation. If a number is a solution of an equation, then it is said to satisfy the equation. For example, 1 satisfies the equation 2x + 3 = 6 – x but 2 does not satisfy
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the equation. We observe that the equation x2 –1 = (x + 1) (x – 1) is satisfied by any number. Such an equation is called an algebraic identity. Thus, an algebraic identity is an algebraic equation which is satisfied by all numbers. If an algebraic equation A = B is an algebraic identity, we write A ≡ B. Reversing the sides, the algebraic identity is also written as B ≡ A. Now we proceed to derive some algebraic identities. 4.2.1 Algebraic identity for (x + a)(x + b)
By using the distributive properties of numbers, (x + a )(x + b ) = x(x + b) + a(x + b) = x2 + xb + ax + ab = x2 + ax + bx + ab= x2 + (a + b)x + ab.
Thus we get (x + a)(x + b) ≡ x2 + (a + b) x + ab
We give a geometrical explanation for the above identity.
The area of the rectangle ABCD is equal to the sum of the area of the square AHFE and the areas of the rectangles HBGF, FGCI and EFID (see Figure 4.2.1). So we have (x + a)(x + b) = x2 + ax + ab + xb
= x2+ (a+b)x + ab.
We shall derive some important identities using the above identity.
(i) (x – a)(x + b) = [x + (–a)] (x + b) = x2 + [(–a) + b]x + (–a)b
= x2 + (b – a)x – ab. (ii) (x + a)(x – b) = (x + a) [x + (–b)] = x2 + [a + (–b)]x + a(–b) = x2 + (a – b)x – ab. (iii) (x – a)(x – b) = [x+(–a)] [x + (–b)] = x2 + [(–a) + (–b)] x + (–a) (–b)
= x2 – (a + b)x + ab. (iv) (a + b)2 = (a + b)(a + b) = a2 + (b + b)a + b2 = a2 + 2ab + b2. (v) (a – b)2 = (a – b)(a – b) = [a+(–b)] [a+(–b)]
= a2 + [(–b) + (–b)]a + (–b) (–b) = a2 –2ab + b2. (vi) (a + b)(a–b) = a2 + [b + (–b)]a + (b)(–b) = a2 + 0 × a – b2 = a2 – b2.
Thus, we have (x + a)(x + b) ≡ x2 + (a + b)x + ab (x – a)(x + b) ≡ x2 + (b – a)x – ab (x + a)(x – b) ≡ x2 + (a – b)x – ab (x – a)(x – b) ≡ x2 – (a + b)x + ab (a + b)2 ≡ a2 + 2ab + b2
(a – b)2 ≡ a2 – 2ab + b2
(a + b)(a – b) ≡ a2 – b2
Figure 4.2.1
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The identity for (x – a)(x – b) is x2 – (a + b)x + ab. It is a polynomial of 2nd degree. The coefficient of x and the constant term in it are –(a + b) and ab respectively. The above identities are also called product formulae since they are based on the expansion of the product (x + a)(x + b). These formulae can be used to evaluate product of two binomials. Example 7: Find the following products:
(a) (x + 3) (x + 5) (b) (p + 9) (p – 2) (c) (z – 7) (z – 5) (d) (x – 8) (x + 2) Solution: (a) (x + 3) (x + 5) = x2 + (3 + 5) x + 3 × 5 = x2 + 8x + 15. (b) (p + 9) (p – 2) = p2 + (9 – 2) p – 9 × 2 = p2 + 7p – 18. (c) (z – 7) (z – 5) = z2 – (7 + 5) z + 7 × 5 = z2 – 12z + 35. (d) (x – 8) (x + 2) = x2 + (2 – 8)x – 8 × 2 = x2 – 6x – 16. Example 8: Evaluate the following products using the product formulae: (a) 107 × 103 (b) 56 × 48 Solution: (a) 107 × 103 = (100 + 7) (100+3) (b) 56 × 48
= (50 + 6) (50 –2) = 502 + (6 – 2) × 50 – 6 × 2 = 2500 + 200 – 12 =2688.
= 1002 + (7 + 3) × 100 + 7 ×3 = 10000 + 10 × 100 + 21 = 10000 + 1000 + 21 = 11021. Example 9: Expand the following: (i) (3x + 7y)2 (ii) (11a – 7b)2 (iii) (2p + 5q)(2p – 5q) Solution: (i) (3x + 7y)2 = (3x)2 + 2(3x)(7y) + (7y)2 =9x2+ 42xy + 49y2. (ii) (11a – 7b)2 = (11a)2 – 2(11a) (7b) + (7b)2 = 121a2 – 154ab + 49b2. (iii) (2p + 5q)(2p – 5q) = (2p)2 – (5q)2 = 4p2 – 25q2. Example 10: Find the value of the following using the product formulae: (i) 1032 (ii) 982 (iii) 104 × 96 Solution: (i) 1032 = (100+3)2 = 1002 + 2(100)(3) + 32 = 10000 + 600 +9 = 10609. (ii) 982 = (100–2)2 = 1002 – 2 (100)(2) + 22 = 10000 – 400 + 4 = 9604. (iii) 104 × 96 = (100 + 4) (100 – 4) = 1002 – 42 = 10000 – 16 = 9984. We now deduce some useful identities from the product formulae: (i) (a + b)2 + (a – b)2 = (a2 + 2ab + b2) + (a2 – 2ab + b2) = (a2 + a2) + (2ab – 2ab) + (b2 + b2) = 2a2 + 2b2.
So 21 [(a + b)2 + (a – b)2] =
21 [2(a2 + b2)] = a2 + b2.
(ii) (a + b)2 – (a – b)2 = (a2 + 2ab + b2) – (a2 – 2ab + b2)
= a2 + 2ab + b2 – a2 + 2ab – b2 = 4ab.
95
So 41 [(a + b)2 – (a – b)2] =
41 [4ab] = ab.
(iii) (a + b)2 – 2ab = (a2 + 2ab + b2) – 2ab = a2 + 2ab + b2 – 2ab = a2 + b2.
(iv) (a + b)2 – 4ab = (a2 + 2ab + b2) – 4ab = a2 + 2ab + b2 – 4ab
= a2 – 2ab + b2 = (a–b)2.
(v) (a – b)2 + 2ab = (a2 – 2ab + b2) + 2ab = a2 – 2ab + b2 + 2ab = a2 + b2.
(vi) (a – b)2 + 4ab = (a2 – 2ab + b2) + 4ab = a2 – 2ab + b2 + 4ab
= a2 + 2ab + b2 = (a + b)2.
Thus we have the following useful identities: 2
1 [(a + b)2 + (a – b)2] = a2 + b2
41 [(a + b)2 – (a – b)2] = ab
(a + b)2 – 2ab = a2 + b2
(a + b)2 – 4ab = (a – b)2
(a – b)2 + 2ab = a2 + b2
(a – b)2 + 4ab = (a + b)2
Reversing the sides of the above identities, we get a2 + b2 = 2
1 [(a + b)2 + (a – b)2] ab = 4
1 [(a + b)2 – (a – b)2] a2 + b2 = (a + b)2 – 2ab (a – b)2 = (a + b)2 – 4ab a2 + b2 = (a – b)2 + 2ab (a + b)2 = (a – b)2 + 4ab Example 11: If the values of a + b and a – b are 7 and 4 respectively, find the values of a2 + b2 and ab.
ab = 41 [(a + b)2 – (a – b)2]
= 41 [(7)2 – (4)2]
= 41 (49 – 16) =
433 .
Solution: a2 + b2 = 21 [(a+b)2 + (a–b)2]
= 21 [(7)2 + (4)2]
= 21 (49 + 16) =
265 .
Example 12: If the values of a + b and ab are 12 and 32 respectively, find the values of a2 + b2 and (a–b)2.
96
Solution: a2 + b2 = (a + b)2 –2ab = (12)2 – 2(32) = 144 – 64 = 80.
(a – b)2 = (a + b)2 – 4 ab = (12)2 – 4(32) =144 – 128 = 16.
Example 13: If the values of a – b and ab are 6 and 40 respectively, find the values of a2 + b2 and (a + b)2. Solution: a2 + b2 = (a – b)2 + 2ab (a + b)2 = (a – b)2 + 4ab
= 62 + 4(40)= 36 + 160 = 196. = 62 + 2(40) = 36 + 80 = 116. Example 14: If (x + p)(x + q) = x2 – 5x – 300, find the value of p2 + q2. Solution: By product formula, we have (x + p) (x + q) = x2 + (p + q)x + pq. So, by comparison, we get p + q = –5, pq = –300. Now, we have p2 + q2 = (p + q)2 – 2 pq = (–5)2 –2(–300) = 25 + 600 = 625. We now derive the algebraic identity for (a + b + c)2
(a + b + c)2 = [(a + b) + c]2 = (a + b)2 + 2(a + b) c+ c2
= (a2 + 2ab + b2) + 2ac + 2bc + c2 = a2 + b2 + c2 + 2ab + 2bc + 2ca = a2 + b2 + c2 + 2(ab + bc + ca). So we have the identity
(a + b + c)2 ≡ a2 + b2 + c2 + 2(ab + bc + ca) Reversing the sides, we get
a2 + b2 + c2 + 2(ab + bc + ca) ≡ (a + b + c)2
Also we have (a + b + c)2 – 2 (ab + bc + ca) = a2 + b2 + c2 + 2(ab + bc + ca) – 2(ab + bc + ca) = a2 + b2 + c2. Thus, we get another algebraic identity
(a + b + c)2 – 2(ab + bc + ca) ≡ a2 + b2 + c2
Reversing the sides, we get
a2 + b2 + c2 ≡ (a + b + c)2 – 2(ab + bc + ca) Example 15: Expand the following: (i) (2x + y + 2z)2 (ii) (x – 2y + z)2 (iii) (2p – 3q – r)2 (iv) (2a + 3b − 2c)2
Solution: (i) (2x + y + 2z)2 = [(2x) + y + (2z)]2 = (2x)2 + y2 + (2z)2 + 2(2x)y + 2y(2z) + 2(2z)(2x) = 4x2 + y2 + 4z2 + 4xy + 4yz + 8zx. (ii) (x – 2y + z)2 = [x + (–2y) + z]2= x2 + (–2y)2 + z2 + 2x(–2y) +2 (–2y)z + 2zx = x2 + 4y2 + z2 – 4xy – 4yz + 2zx. (iii) (2p – 3q – r)2 = [(2p) + (–3q) + (–r)]2
= (2p)2 + (–3q)2 + (–r)2 + 2(2p) (–3q) + 2(–3q) (–r) + 2(–r)(2p). = 4p2 + 9q2 + r2 – 12pq + 6qr – 4rp. (iv) (2a + 3b – 2c)2 = [(2a) + (3b) + (–2c)]2
= (2a)2 + (3b)2 + (–2c)2 + 2(2a)(3b) + 2(3b)(–2c) + 2(–2c)(2a) = 4a2 + 9b2 + 4c2 + 12ab – 12bc – 8ca.
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4.2.2 Algebraic Identity for (x + a) (x + b) (x + c) (x + a)(x + b)(x + c) = (x + a)[(x + b)(x + c)] = (x + a)[x2 + (b + c)x + bc] = (x + a)(x2 + bx + cx + bc) = x(x2 + bx + cx + bc) + a(x2 + bx + cx + bc) = x3 + bx2 + cx2 + bcx + ax2 + abx + acx + abc = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc. Hence, (x + a)(x + b)(x + c) ≡ x3 + (a + b + c)x2 + (ab + bc + ca)x + abc In the above identity, replacing a, b and c by –a, –b and –c, we get (x – a)(x – b)(x – c) = [x+(–a)][x+(–b)][x+(–c)] = x3 + [(–a) + (–b) + (–c)]x2 + [(–a) (–b) + (–b) (–c) + (–c) (–a)]x + (–a) (–b) (–c) = x3 – (a + b + c)x2 + (ab + bc + ca)x – abc. Thus, (x – a)(x – b)(x – c) ≡ x3 – (a + b + c)x2 + (ab + bc + ca)x – abc. In the above identity, coefficient of x2 = – (a + b + c), coefficient of x = ab + bc + ca, constant term = –abc. Example 16 : Find the expansions of (i) (x + 4)(x + 3)(x + 5) (ii) (2x + 1)(2x – 3)(2x + 5) (iii) (3 – 2x)(2x + 7)(2x + 1) (iv) (x – a)(x – 2a)(x – 3a) Solution : (i) (x + 4)(x + 3)(x + 5) = x3 + (4 + 3 + 5)x2 + [4 × 3 + 3 × 5 + 5 × 4]x + 4 × 3 × 5 = x3 + 12x2 + [12 + 15 + 20]x + 60 = x3 + 12x2
+ 47x + 60. (ii) (2x + 1)(2x – 3)(2x + 5) = (2x + 1)[2x + (–3)](2x + 5) = (2x)3 + [1 + (–3) + 5](2x)2 + [1 × (–3) + (–3) ×5 + 5 × 1](2x) + 1 × (–3) × 5 = 8x3 + 3(4x2) + [–3 – 15 + 5](2x) – 15 = 8x3 + 12x2 – 26x – 15. (iii) (3 – 2x)(2x + 7)(2x + 1) = [–(2x – 3)](2x +7 )(2x + 1) = – [2x + (–3)](2x + 7)(2x + 1) = – [(2x)3 + {(–3) + 7 + 1}(2x)2 + {(–3) × 7 + 7×1+ 1×(–3)}(2x) + (–3) × 7 × 1] = – [8x3 + (5)(4x2) + (–21 + 7 − 3)(2x) –21] = – [8x3 + 20x2 – 34x – 21] = 21 + 34x – 20x2 – 8x3. (iv) (x – a)(x – 2a)(x – 3a) = [x+(–a)][x+(–2a)][x+(–3a)] = x3 + {(–a) + (–2a)+( –3a)}x2+{(–a) (–2a) + (–2a) (–3a) + (–3a) (–a)}x + (–a) (–2a) (–3a) = x3 + {–6a}x2 + {2a2 + 6a2 + 3a2}x – 6a3 = x3 – 6ax2 + 11a2x – 6a3. Example 17 : Using the product formulae, find the coefficients of x2 term, x term and constant term in (i) (x + 3)(x + 5)(x + 6) (ii) (x – 7)(x + 2)(x + 4) (iii) (x – 5)(x – 2)(x + 4) (iv) (2x – 3 )(2x – 5)(7 – 2x)
98
Solution: (i) Comparing with (x + a)(x + b)(x +c ), we get a = 3, b = 5, c = 6. ∴ Coefficient of x2 = a + b + c = 3 + 5 + 6 = 14, Coefficient of x = ab + bc + ca = (3 × 5) + (5 × 6) + (6 × 3) = 15 + 30 + 18 = 63, Constant term = abc = 3 × 5 × 6 = 90. (ii) (x – 7)(x + 2)(x + 4) = [x + (–7)](x + 2)(x + 4) Comparing with (x + a)(x + b)(x + c), we get a = –7, b = 2, c = 4. ∴ Coefficient of x2 = a + b + c = (–7) + 2 + 4 = –1, Coefficient of x = ab + bc + ca = (–7) × 2 + 2 × 4 + 4 × (–7) = (–14) + 8 + (–28)= (–42) + 8 = –34, Constant term = abc = (–7) × 2 × 4 = –56. (iii) (x – 5)(x – 2)(x + 4) = [x + (–5)][x + (–2)](x + 4) Comparing with (x + a)(x + b)(x + c), we get a = –5, b = –2, c = 4. ∴ Coefficient of x2 = a + b + c = (–5) + (–2) + 4 = (–7) + 4 = –3, Coefficient of x = ab + bc + ca = (–5) × (–2) +(–2) × 4 + 4 × (–5)= 10 – 8 – 20 = –18, Constant term = abc = (–5) (–2)4 = 40. (iv) Let A be algebraic expression (2x – 3)(2x – 5)(7 – 2x) and put y=2x. Then, A = (y – 3)(y – 5)(7 – y) = (y – 3)(y – 5) [–(y – 7)] = –[(y – 3)(y – 5 )(y – 7)] = –[y3 + {(–3) + (–5) + (–7)}y2 + {(–3)(–5) + (–5)(–7) + (–7)(–3)}y + (–3)(–5)(–7)] = –[y3 – 15y2 + (15 + 35 + 21)y – 105] = –y3 + 15y2 – 71y + 105 = –(2x)3 + 15(2x)2 – 71(2x) + 105 = –8x3 + 60x2 – 142x + 105 ∴ Coefficient of x2 = 60, Coefficient of x = –142, Constant term = 105 Alternately, (2x – 3)(2x – 5)(7 – 2x) =
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −−⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −
272
252
232 xxx = ⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+−
27
25
238 xxx
∴ Coefficient of x2 = –8(a + b + c) = ⎥⎦⎤
⎢⎣⎡ −−−−
27
25
238 = ⎟
⎠⎞
⎜⎝⎛−−
2158 = 60,
Coefficient of x = –8( ab + bc + ca) = ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −−
23
27
27
25
25
238
= –8 ⎥⎦⎤
⎢⎣⎡ ++
421
435
415 = –8 ⎥⎦
⎤⎢⎣
⎡ ++4
213515 = (–2) (71)= –142,
Constant term = –8(abc) = ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −−
27
25
238 = –8 ⎟
⎠⎞
⎜⎝⎛ −
8105
= 105.
Example 18 : If (x + a)(x + b)(x + c) ≡ x3 – 6x2 + 11x – 6, find the value of a2 + b2 + c2. Solution: From the product formula, we have (x+a)(x+b)(x+c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc. Comparing, we get a + b + c = –6, ab + bc + ca = 11, abc = –6. ∴ a2 + b2 + c2 = (a + b + c)2 –2 (ab + bc + ca) = (– 6)2 – 2(11) = 36 – 22 = 14.
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We shall obtain some identities from the identity (x + a)(x + b)(x + c) (i) Identity for (a + b)3
(a + b)3 = (a + b)(a + b)(a + b) = a3 + (b + b + b)a2 + (b × b + b × b + b × b)a + b × b × b = a3 + 3a2b + 3ab2 + b3
(a + b)3 ≡ a3 + 3a2b + 3ab2 + b3
∴ Reversing the sides, (ii) Identity for (a – b)3
(a – b)3 = [a + (–b)]3
= a3 + 3a2(–b) + 3a(–b)2 + (–b)3= a3 – 3a2b + 3ab2 – b3
∴ Reversing the sides, Based on the identities for (a+b)3 and (a–b)3, we deduce the following identities:
(i) (a+b)3–3ab(a+b) = a3 + 3a2b + 3ab2 + b3 – 3a2b – 3ab2 = a3+b3.
∴
Reversing the sides, (ii) (a – b)3+3ab(a – b) = a3 – 3a2b + 3ab2 – b3 + 3a2b – 3ab2 = a3–b3.
(a − b)3 + 3ab(a − b) ≡ a3 − b3 ∴
Reversing the sides, (iii) a3 + b3 = (a + b)3 – 3ab(a + b) = (a + b)(a + b)2 – 3ab(a + b) = (a + b)[(a + b)2 – 3ab] = (a + b)[(a2 + 2ab + b2) – 3ab] = (a + b)(a2 – ab + b2)
a3 + b3 ≡ (a + b)(a2 – ab + b2) ∴
Reversing the sides, (iv) a3 – b3 = (a – = (a – = (a –
∴ Reversing the sides,
(a + b)(a2 – ab + b2) ≡ a3 + b3
bbb
a
(
a3 + 3a2b + 3ab2 + b3 ≡ (a + b)3
a
(a − b)3 ≡ a3 − 3a2b + 3ab2 − b3
a3 − 3a2b + 3ab2 − b3 ≡ (a − b)3
(a + b)3 − 3ab(a + b) ≡ a3 + b3
a3 + b3 ≡ (a + b)3 − 3ab(a + b)
a3 − b3 ≡ (a − b)3 + 3ab(a− b)
)3 + 3ab(a – b) = (a – b)(a – b)2 + 3ab(a – b) )[(a – b)2 + 3ab] = (a – b)[(a2 – 2ab + b2) + 3ab] )(a2 + ab + b2) 3 − b3 ≡ (a−b)(a2 + ab + b2)
− b)(a2 + ab + b2) ≡ a3 − b3
100
Identity for a2 + b2 + c2 − ab − bc − ca We have
a2 + b2 + c2 − ab − bc − ca = )222222(21 222 cabcabcba −−−++
= [ ])2()2()2(21 222222 acaccbcbbaba +−++−++−
= [ ].)()()(21 222 accbba −+−+−
Thus, we get the following identity:
2221 Reversing Identity f
Us (a + b + c = = = Thus, we Reversing
2 2 2
a + b + c −ab − bc − ca ≡ [ ])()()(2accbba −+−+−
the sides
2221 2 2 2
[ )()()( ]2accbba −+−+− ≡ a + b + c −ab − bc − ca
or (a + b + c) (a2 + b2 + c2 − ab − bc − ca)
ing the distributive properties, we get
) (a2 + b2 + c2 − ab − bc − ca) a (a2 + b2 + c2 − ab − bc − ca) + b (a2 + b2 + c2 − ab − bc − ca)
+ c (a2 + b2 + c2 − ab − bc − ca)
a3 + ab2 + c2a − a2b − abc − ca2 + a2b + b3 + bc2 − ab2 − b2c − abc +ca2 + b2c + c3 − abc − bc2 − c2a
a3 + b3 + c3 − 3abc.
have
(a + b + c) (a2 + b2 + c2 − ab − bc − ca) ≡ a3 + b3 + c3 − 3abc
the sides, we get
a3 + b3 + c3 − 3abc ≡ (a + b + c) (a2 + b2 + c2 − ab − bc − ca).
101
Example 19 : Expand the following : (i) (3x+2y)3 (ii) (2x2–3y)3
Solution: (i) (3x+2y)3 = (3x)3 + 3(3x)2(2y) + 3(3x)(2y)2 + (2y)3
= 27x3 + 3(9x2)(2y) + 3(3x)(4y2) + 8y3
= 27x3 + 54x2y + 36xy2 + 8y3. (ii) (2x2– 3y)3 = (2x2)3 – 3(2x2)2 (3y) + 3(2x2) (3y)2 – (3y)3
= 8x6 – 3(4x4)(3y) + 3(2x2)(9y2) – 27y3
= 8x6– 36x4y + 54x2y2 – 27y3
Example 20 : If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3. Solution: a3+ b3 = (a+b)3 – 3ab(a+b) = (4)3 – 3(1)(4) = 64 – 12 = 52. Example 21: If a – b = 4 and ab = 2, find a3 – b3. Solution: a3 – b3 = (a–b)3+3ab(a–b)= (4)3+3(2)(4) =64+24=88. Example 22: If a+b=2 and a2+b2=8,find a3+b3 and a4+b4. Solution: a2 + 2ab + b2= (a+b)2
2ab = (a+b)2 – (a2 + b2) = (2)2 – (8) = 4 – 8 = – 4. ∴ ab = 2
1 (2ab) = 21 (–4) = – 2.
a3+ b3 = (a+b)3 – 3ab(a+b) = (2)3 – 3(–2)(2) = 8 – 3 (– 4) = 8 + 12 = 20. Alternately, a3 + b3 = (a+b)(a2 – ab + b2) = (a+b)(a2 + b2 – ab) = (2) [8 – (–2)] = 2(10) = 20. a4 + b4 = (a2)2 + (b2)2 = [(a2) + (b2)]2 – 2 (a2)(b2) = (a2 + b2)2 – 2a2b2 = (a2 + b2)2 – 2(ab)2
= (8)2 – 2(–2)2 = 64 – 2(4)= 64 – 8 = 56.
Exercise 4.2 1. Using the product formula, find
(i) (x + 9) (x + 2) (ii) (x + 8) (x – 2) (iii) (t – 2)(t + 6) (iv) (p – 4)(p – 3) (v) 102 × 106 (vi) 59 × 62 (vii) 34 × 36 (viii) 53 × 55
2. Using product formulae, find (i) (5x + 8y)2 (ii) (3s – 4t)2 (iii) (4p + 7q)(4p – 7q) (iv) (101)2 (v) (98)2 (vi) 101 × 98 3. If a + b = 5 and a – b = 4, find a2 + b2 and ab. 4. If a + b = 10 and ab = 20, find a2 + b2 and (a–b)2.
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5. If (x + l)(x + m) = x2 + 4x + 2, find l2 + m2 and (l – m)2. 6. Expand the following :
(i) (3x + y + 2z)2 (ii) (4x – 2y + 3z)2
(iii) (2p + 3q – 2r)2 (iv) (3a – 2b – 2c)2
7. If a + b + c = 11 and ab + bc + ca = 38, find a2 + b2 + c2. 8. Using the product formula, find (i) (x + 2 )(x + 3)(x + 4) (ii) (x + 2)(x + 3)(x – 4) (iii) (x + 2)(x – 3)(x + 4) (iv) (x + 2)(x – 3)(x – 4) (v) (x – 2)(x – 3)(x – 4) 9. Using the product formula, find the coefficients of x2 term, x term and the constant
term in each of the following: (i) (x + 10)(x – 3)(x + 2) (ii) (2x – 3)(2x + 4)(2x – 1) (iii) (6x + 1)(6x – 5)(7 – 6x)
10. If (x + a)(x + b)(x + c) ≡ x3 – 9x2 + 23x – 15, find a + b + c, cba111
++ and a2 + b2 + c2.
11. Expand using the product formulae:
(i) (2x + y2)3 (ii) (2u – 7v)3 (iii) 31⎟⎠⎞
⎜⎝⎛ −
xx (iv) (x2y3 + 2)3
12. If 2a – 3b = 2 and ab = 6, find 8a3 – 27b3.
13. If ,31=+
xx find x2 + 2
1x
and x3+ 3
1x
.
14. If x + y = 6 and xy = 8, find x2 + y2 and x3 + y3. 15. If p + q = 6 and p2 + q2 = 32, find pq, p3 + q3 and p4 + q4. 4.3. Factorization In the previous section, we have learnt how to multiply two or more polynomials to get another polynomial. Now, we proceed to learn how to write a polynomial as a product of two or more polynomials. The process of writing a polynomial as a product of two or more simpler polynomials is called factorization. Each simpler polynomial in the product is called a factor of the given polynomial. For example, x + 3 and x – 3 are factors of x2 – 9 because x2 – 9 = (x + 3)(x – 3). Here x2 – 9 is a second degree polynomial whereas x + 3 and x – 3 are first degree polynomials. Thus factorization is useful in simplifying expressions. The process of factorization is also known as the resolution into factors.
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4.3.1. The Process of Factorization Step 1: (Finding a common factor) When the terms of an algebraic expression A have a common factor B, we divide each term of A by B and get an expression C. Now, A is factored as B × C. Example 23 : Factorize 6x4y3 – 4x2y2 + 10xy3. Solution : We observe that 2xy2 is a common factor .
∴ 6x4y3 – 4x2y2 + 10xy3 = 2xy2 ⎟⎟⎠
⎞⎜⎜⎝
⎛+− 2
3
2
22
2
34
210
24
26
xyxy
xyyx
xyyx = 2xy2(3x3y – 2x + 5y).
Step 2: (Grouping the terms) When the terms of an algebraic expression do not have a common factor, the terms may be grouped in an appropriate manner and a common factor is determined. Example 24 : Factorize x2 – 2xy – x + 2y. Solution: The terms of the expression do not have a common factor. However, we observe that the terms can be grouped as follows: x2 – 2xy – x + 2y = (x2 – 2xy) – (x – 2y) = x(x – 2y) + (–1) (x – 2y) = (x – 2y) [x + (–1)] = (x – 2y) (x – 1). Example 25 : Factorize 6x5y2 + 6x4y3 + 9x2y4 + 9xy5. Solution : Applying both step 1 and step 2, we have 6x5y2 + 6x4y3 + 9x2y4 + 9xy5 = 3xy2(2x4 + 2x3y + 3xy2 + 3y3) = 3xy2 [(2x4 + 3xy2) + (2x3y + 3y3)] = 3xy2 [x(2x3 + 3y2) + y(2x3 + 3y2)] = 3xy2 (2x3 + 3y2) (x + y).
Exercise 4.3.1 Resolve into factors by finding a common factor in the following : 1. 9m – 3n 2. 4a3 – 8a2 + 16a 3. x5 + 4x 4. 6x5y5 + 3x2y3 + 14xy3 5. 7pq – 21p2q2
Resolve into factors by finding a common factor or by grouping method: 6. mn – 2p – pn + 2m 7. x3 – 2x2 – 2x + 4 8. x3 – x2 – ax + a 9. 2p3 – p2 + 2p – 1 10. 8x3 + 4x2 + 4x + 2 4.3.2. Factorization using factorization formulae: Sometimes, in resolving a polynomial into factors, we use factorization formulae. These factorization formulae are obtained from the product formulae. We have already learnt that the product formulae are (i) (X + Y)2 = X 2 + 2XY + Y 2 (ii) (X – Y)2 = X 2 – 2XY + Y 2 (iii) (X + Y)(X–Y) = X2 – Y 2 (iv) ( X + Y) (X 2 – XY + Y 2) = X 3 + Y 3 (v) (X – Y)(X 2 + XY + Y 2) = X 3 – Y 3 (vi) (X + Y)3 = X 3 + Y 3 + 3X 2Y + 3XY 2 = X 3 + Y 3 + 3X Y (X +Y )
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(vii) (X – Y)3 = X 3 – Y 3 – 3X 2Y + 3XY 2 = X 3 – Y 3 – 3X Y (X –Y ) (viii) (X + Y + Z)2 = X 2 + Y 2 + Z 2 + 2XY + 2YZ + 2ZX (ix) (X + Y + Z) (X 2 + Y 2 +Z 2 – XY – YZ – ZX) = X 3 + Y 3 + Z 3 – 3XYZ Reading the above formulae from the Right Hand Side (R.H.S.) to Left Hand Side (L.H.S.), we get the following factorization formulae:
(i) X 2 + 2XY + Y 2 = (X +Y)2
(ii) X 2 – 2XY + Y 2 = (X – Y)2
(iii) X 2 – Y 2 = (X + Y) (X – Y) (iv) X 3 + Y 3 = ( X + Y) (X 2 – XY + Y 2) (v) X 3 – Y 3 = (X – Y) (X 2 + XY + Y 2) (vi) X 3 + Y 3 + 3X 2Y+ 3XY 2 = (X + Y)3
(vii) X 3 – Y 3 – 3X 2Y + 3XY 2 = (X – Y)3
(viii) X 2 + Y 2 + Z 2 + 2XY + 2YZ + 2ZX = ( X + Y + Z)2
(ix) X 3+Y 3+Z 3–3XYZ = (X + Y + Z) (X 2 + Y 2 + Z 2 – XY – YZ – ZX) Factorization using X 2 + 2XY + Y 2 ≡ (X + Y)2 Example 26: Resolve into factors 4x2 + 12xy + 9y2. Solution: The given expression can be written as follows: 4x2+ 12xy + 9y2 = (2x)2 + 2(2x)(3y) + (3y)2
Setting X = 2x, Y = 3y, the R.H.S. is X 2 + 2XY + Y 2 and so it is factored as (X + Y)2. Hence we get 4x2 + 12xy + 9y2 = (2x + 3y)2. Factorization using X 2 – 2XY + Y 2 ≡ (X – Y)2 Example 27 : Factorize p2 – 18pq + 81q2. Solution : The given polynomial can be written as follows: p2 – 18pq + 81q2 = p2 – 2(p)(9q) + (9q)2
Setting X = p and Y = 9q, the R.H.S. is X 2 – 2XY + Y 2 and so it is factorised as (X – Y)2. Hence we get p2 – 18pq + 81q2 = (p – 9q)2. Factorization using X 2 – Y 2 ≡ (X + Y) (X – Y) Example 28: Factorize 16x4y2 – 25. Solution : Since 16x4y2 = (4x2y)2 and 25 = (5)2, we have 16x4y2 – 25 = (4x2y)2 – (5)2 = (4x2y + 5) (4x2y – 5) Factorization using X 3 + Y 3 ≡ (X + Y) (X 2 – XY + Y 2) Example 29: Resolve into factors 125a3 + 64b3. Solution : Since 125a3 = (5a)3 and 64b3 = (4b)3, we take X = 5a and Y = 4b . Then 125a3 + 64b3 = (5a)3 + (4b)3 X 3 + Y 3 = (X + Y) (X 2 – XY + Y 2) = (5a + 4b) [(5a)2 – (5a)(4b) + (4b)2] = (5a+ 4b) (25a2 – 20ab + 16b2)
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Factorization using X 3 – Y 3 ≡ (X – Y) (X 2 + XY + Y 2) Example 30 : Factorize 216p3–8q3. Solution: We can write 216p3 = (6p)3 and 8q3 = (2q)3. So, taking X = 6p and Y = 2q, we get 216p3 – 8q3 = (6p)3 – (2q)3
X 3 – Y 3 = (X – Y) (X 2 + XY + Y 2) = (6p – 2q) [(6p)2 +(6p)(2q) +(2q)2] = (6p –2q)(36p2+12pq+4q2). Factorization using X 3 + Y 3 + 3X 2Y + 3XY 2 ≡ (X + Y)3
Example 31: Factorize 8x3 + y 3 + 12x2y + 6xy2. Solution: 8x3 + y3 + 12x2y + 6xy2 = (2x)3 + y3 + 3(2x)2y + 3(2x)y2= [(2x) + y]3 = (2x + y)3. Factorization using X 3 – Y 3 – 3X 2Y + 3XY 2 ≡ (X – Y)3 Example 32: Resolve into factors 8x3 – 27y3 – 36x2y + 54xy2. Solution : 8x3 – 27y3 – 36x2y + 54xy2 = (2x)3 – (3y)3 – 3(2x)2(3y) + 3(2x)(3y)2= (2x – 3y)3. Factorization using X 2+ Y 2 + Z 2 + 2XY + 2YZ + 2ZX ≡ (X + Y + Z)2
Example 33 : Factorize x2+ 9y2 – 6xy + 4x – 12y + 4. Solution : The expression contains a sum x2 + 9y2 + 4. This is a sum of three squares. So, we write x2 + 9y2 – 6xy + 4x – 12y + 4 = x2 + (3y)2 + 22 + 2x (–3y) + 2x(2) + 2(2)(–3y) = [x + (–3y) + 2]2 = (x – 3y + 2)2. Factorization using X 3 + Y 3 + Z 3 – 3XYZ ≡ (X + Y + Z) (X 2 + Y 2 + Z 2 – XY – YZ – ZX) Example 34: Factorize x3 – 8y3 + 27z3 + 18xyz. Solution: Since –8y3 = (–2y)3 and 27z3 = (3z)3, the given expression P contains a sum of three cubes. So, we write P = (x)3 + (–2y)3 + (3z)3 – 3(x)(–2y)(3z) = [(x) + (–2y) + (3z)] [(x)2 + (–2y)2 + (3z)2 – (x) (–2y) – (–2y)(3z) – (3z)(x)] = (x – 2y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3zx) Factorization of X 3 + Y 3 + Z 3 when X + Y + Z =0 X 3 + Y 3 + Z 3 = (X 3 + Y 3 + Z 3 –3XYZ) + 3XYZ = (X + Y + Z) (X 2 + Y 2 + Z 2 – XY – YZ – ZX) + 3XYZ = (0) (X 2 + Y 2 + Z 2 – XY – YZ – ZX) + 3XYZ = 3XYZ. Note: When X + Y + Z = 0 , X + Y = – Z. ∴ (X + Y)3 = (– Z)3 = – Z3 . That is X 3 + Y 3 + 3XY(X +Y ) = – Z3 . That is, X 3 + Y 3 + Z 3 + 3XY(– Z) = 0. ∴ X 3 + Y 3 + Z 3 = 3XY Z. Example 35: Factorize (x – y)3 + (y – z)3 + (z – x)3. Solution: Put X = x – y, Y = y – z, Z = z – x. Then X + Y + Z = (x – y) + (y – z) + (z – x) = x – y + y – z + z – x = 0. ∴ X 3 + Y 3 + Z 3 = 3XYZ
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Substituting for X, Y and Z (x – y)3 + (y – z)3 + (z –x)3 = 3(x – y) (y – z) (z – x). Factorization using grouping technique Example 36: Factorize 4x2 + 20xy + 25y2 – 10x – 25y. Solution: The given expression is 4x2 + 20xy + 25y2 – 10x – 25y = (2x)2 + 2(2x)(5y) + (5y)2 – 5(2x) – 5(5y) = [(2x) + (5y)]2 – 5[(2x) + (5y)] = (2x + 5y)2 – 5(2x + 5y) = (2x + 5y) (2x + 5y –5). Example 37: Factorize 4a2 – 4ab + b2 – 2a + b. Solution: The given polynomial is 4a2 – 4ab + b2 – 2a + b = (2a)2 – 2(2a)(b) + (b)2 – (2a – b) = (2a – b)2 – 1 (2a – b) = (2a – b) (2a – b – 1). Example 38: Factorize 81x2 – 18x + 1 – 25y2. Solution : Grouping the terms, we get 81x2 – 18x + 1 – 25y2 = [(9x)2 – 2(9x)(1) + (1)2] – (5y)2
= [(9x) – (1)]2 – (5y)2
= (9x – 1) 2 – (5y)2
= [(9x –1) + (5y)][(9x – 1) – (5y)] = (9x + 5y – 1) (9x – 5y –1). Example 39: Factorize x4 +1. Solution: Adding and subtracting 2x2, the given expression is x4 + 1 = (x4 + 2x2 + 1) – 2x2
= [(x2)2 + 2(x2)(1) + (1)2] – 2x2
= (x2 + 1)2 – ( 2 x)2
= [(x2 + 1) + ( 2 x)][(x2 + 1) – ( 2 x)] = (x2 + 2 x + 1) (x2 – 2 x + 1). Example 40: Factorize x4 + x2y2 + y4. Solution: Adding and subtracting x2y2, we get x4 + x2y2 + y4 = (x4 + 2x2y2 + y4) – x2y2
= [(x2)2 + 2(x2)(y2) + (y2)2] – (xy)2
= (x2 + y2)2 – (xy)2
= [x2 + y2 + (xy)] [x2+y2 – (xy)] = (x2 + xy + y2) (x2 – xy + y2). Example 41: Factorize x4 + 5x2 + 9. Solution : Adding and subtracting x2, we get x4 + 5x2 + 9 = (x4 + 6x2 + 9) – x2
= (x2 + 3)2 – x2 = [(x2 + 3) + x][(x2+3) – x] = (x2 + x + 3) (x2 – x + 3).
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Example 42: Resolve into factors x8 – x2y6. Solution: Taking x2 as a common factor, x8 – x2y6 = x2 (x6 – y6) = x2 [(x3)2 – (y3)2] = x2 (x3 + y3) (x3 – y3) = x2[(x + y)(x2 – xy + y2)] [(x–y)(x2 + xy + y2)] = x2 (x +y)(x–y)(x2 + xy + y2) (x2 – xy + y2).
Exercise 4.3.2. Answer True or False in the following:
1. 2x2 + 2 2 x + 1 = (1 + x2 )2
2. x6 – 4x3 + 4 = (x3 + 1)2
3. a2 – b2 = (a – b)2
4. a3 + b3 = (a + b)(a2 + ab + b2) 5. a3 – b3 = (a – b)(a2 + ab + b2)
Factorize the polynomial using the factorization formulae: 6. 1 + 6x + 9x2 7. 144x2 – 72x + 9 8. 4a2b2 + 20abcd + 25c2d2 9. x2 + y2 – a2 – b2 + 2xy + 2ab 10. 3 3 x3y3 + 27z3 11. (x + y)3 + 8y3
12. (x2 + 1)3 + (x2 – 1)3 13. x6 – y6 14. (x+y)3 − (x–y)3 15. (p+q)3 + (p–q)3 + 6p (p2 – q2) 16. 27x3 + y3 + 27x2y + 9xy2 17. x3 – 12x2 +48x – 64 18. 8x3 – 27y3 – 36x2y + 54xy2 19. 4x2 + 9y2 + z2 + 12xy + 4xz + 6yz 20. a2 + b2 + 9c2 + 2ab – 6ac – 6bc 21. x3 – y3 + 1 + 3xy 22. 8x3 – 125y3 + 180xy + 216 23. 8x3 – 27y3 +z3+18xyz 24. 3 3 a3 – 8b3 – 125c3 − 30 3 abc 25. (a –2b)3 + (2b – 3c)3 + (3c – a)3
26. (x + y – 2z)3 + (y + z – 2x)3 + (z + x – 2y)3
27. (a2 – b2)3 + (b2 – c2) 3 + (c2 – a2)3 28. a3(b – c)3 + b3(c – a)3 + c3(a – b)3
29. x(x + z) – y(y + z) 30. 1–2xy–x2 – y2 31. x4 +4 4.3.3. Factorization of the quadratic expression ax2 + bx + c We assume that the coefficients a, b and c are all integers and a ≠ 0. When the coefficients a, b and c satisfy certain conditions, the algebraic expression ax2 + bx +c can be factorized. We want to find these conditions and the factors of the expression. First, we consider a simpler case with a = 1 and b and c as integers. Now, we have to factorize x2 + bx + c. We try to write the integer constant term c as a product of two integers p and q such that p + q = b. If we succeed in our attempt, then x2 + bx + c = x2 + (p + q)x + pq = (x2 + px) + (qx + pq) = x(x + p) + q(x + p) = (x + p) (x + q) Thus, we have achieved what we wanted. Rule : If the constant term c of x2 + bx + c can be expressed as a product of two integers p and q such that the sum p + q is the coefficient b of x, then x2 + bx + c = (x + p) (x + q).
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Example 43: Factorize x2 + 9x + 18. Solution : The given expression cannot be written in the form X2 + 2XY + Y2 and so the factorization formula X2 + 2XY + Y2 = (X + Y)2 cannot be used directly. Then we try to factorize the constant term 18. The list of all possible factorization of 18 is 18 = 1 × 18 = 18 × 1 = –1 × –18 = –18 × –1 18 = 2 × 9 = 9 × 2 = –2 × –9 = –9 × –2 18 = 3 × 6 = 6 × 3 = –3 × –6 = –6 × –3 We list below the sum of the factors: 18 + 1 = 1 + 18 = 19 (–18) + (–1) = (–1) + (–18) = –19 2 + 9 = 9 + 2 = 11 (–2) + (–9) = (–9) + (–2) = –11 3 + 6 = 6 + 3 = 9 (–3) + (–6) = (–6) + (–3) = –9. We compare the coefficient of x and the sum of the factors. We find that the sum of the factors 3 and 6 is the coefficient of x. Hence the factorization is
x2 + 9x + 18 = (x + 3) (x + 6). Example 44: Factorize x2 – 15x + 54. Solution: The constant term = 54 and the coefficient of x = –15. We list below the factors of 54 and their sum.
Factors Sum {1, 54} 55 {–1, –54} –55 {2, 27} 29 {–2, –27} –29 {3, 18} 21 {–3, –18} –21 {6, 9} 15 {–6, –9} –15
Hence x2 – 15x + 54 = (x – 6) (x – 9). Example 45: Factorize 15 – 2x – x2. Solution: Writing in the standard form, 15 – 2x – x2 = –x2 – 2x + 15 = (–1) (x2 + 2x – 15). Here, we find –15 = 5 × –3, 5 + (–3) = 2 Hence, we get 15 – 2x – x2 = (–1) [(x+5) {x + (–3)}] = (–1) (x +5)(x – 3) = (x + 5) ( 3 – x). Example 46: Factorize x2 – x – 132. Solution: We find –132 = (–12) × (11), (–12) + 11 = –1. Hence we get x2 – x – 132 = [x + (–12)] (x + 11) = (x – 12) (x + 11).
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Next, we consider the quadratic polynomial ax2 + bx + c where a, b, c are integers and a ≠0, 1. If we are able to find two integers p and q such that pq = ac and p + q = b. Then
ax2 + bx + c =a1 (a2x2 + abx + ac)
=a1 [a2x2 + a(p+q)x + pq] =
a1 [a2x2 + apx + aqx + pq] =
a1 [ax (ax + p) + q(ax + p)]
= a1 (ax + p) (ax + q)
Thus, we are able to factorize the expression. Rule : If the product a × c can be expressed as a product p × q such that b = p + q, then
ax2 + bx + c = a1 (ax + p) (ax + q).
Example 47: Factorize 2x2+ 7x + 3. Solution: Here a = coefficient of x2 = 2 b = coefficient of x = 7 c = constant term = 3 We find a × c = 2 × 3 = 6 = 6 × 1, 6 + 1 = 7 = b. Hence
2x2 + 7x + 3 = 21 (2x + 6) (2x + 1) =(x+3)(2x+1).
Instead of applying the final result of the rule, we can also do the factorization by splitting the middle term and grouping as follows: 2x2 + 7x + 3 = 2x2 + (6 + 1)x + 3 = 2x2 + 6x + x + 3 = 2x(x + 3) + (1)(x+3) = (2x+1) (x+3). Example 48: Factorize 8a2 + 2a – 3. Solution : Here, we find 8 ×– 3 = –24 = 6 × – 4, 6+(–4) = 2 By splitting and grouping, we get 8a2 + 2a – 3 = 8a2 + 6a – 4a – 3 = 2a(4a + 3) – (1)(4a+3) = (4a + 3) (2a – 1) Example 49: Factorize 6 +
211 x + x2.
Solution: Rewriting the expression, 6 + 2
11 x + x2 = 21 (2x2 + 11x + 12).
Here 2 × 12 = 24 = 8 × 3 and 8 + 3 = 11 Hence, by splitting the middle term and grouping,
6 + 2
11 x + x2 = 21 (2x2 + 8x + 3x + 12) =
21 [2x(x + 4) + 3(x + 4)] =
21 (x + 4) (2x + 3).
The method of splitting the middle term and grouping can also be tried to factorize ax2 + bx + c when a, b, c are real numbers.
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Example 50: Factorize 241027 2 −− xx . Solution : Here 27 × 24− = –56 = (–14) × 4 and (–14) + 4 = –10. Hence, by splitting the middle and grouping ,we get 7 2 x2 – 10x – 4 2 = 7 2 x2 – 14x + 4x – 4 2 = 7x ( 2 x– 2)+ 2 2 ( 2 x – 2)= ( 2 x – 2) (7x + 2 2 ). Example 51: Factorize 3x2 + 5 3 x+ 6 Solution: Here 3 × 6 = 18 = (3 3 ) (2 3 ) and 3 3 + 2 3 = 5 3 . Hence by splitting the middle term and grouping, 3x2 + 5 3 x + 6 = 3x2 + 3 3 x + 2 3 x + 6 = 3x (x + 3 ) + 2 3 (x + 3 ) = (x + 3 ) (3x+ 2 3 ). The method of splitting the middle term and then grouping the terms can be followed to factorize algebraic expressions of the form aX 2 + bXY + cY 2. Example 52: Factorize 15x2 + 17xy + 4y2. Solution: Here we find 15 × 4 = 60 = 12 × 5 and 17 = 12 + 5. Hence by splitting the middle term and grouping, 15x2 + 17xy + 4y2 = (15x2 + 12xy) + (5xy + 4y2) = 3x(5x + 4y) + y(5x + 4y)= (5x + 4y) (3x +y). Example 53: Factorize 6(a – 1)2b – 5(a –1)b2 – 6b3
Solution: Putting x = a – 1, we get 6(a – 1)2b – 5(a–1)b2 – 6b3 = 6x2b – 5xb2 – 6b3
= b(6x2 – 5bx –6b2) Here we find 6 × – 6 = –36 = (–9) × 4 and (–9) + 4 = –5. Hence we get = b(6x2 – 9bx + 4bx – 6b2) = b[3x(2x – 3b) + 2b(2x – 3b)] = b(2x – 3b) (3x + 2b) = b[2(a – 1) – 3b] [3(a – 1) + 2b] = b[(2a –3b –2) (3a + 2b –3)]. There are quadratic polynomials with integer coefficients which cannot be factored with integer coefficients. Example 54: Factorize x2 + 3x – 1. Solution: Comparing with ax2 + bx + c, we get a = 1, b = 3, c = –1. Here ac = 1 × –1 = –1 = –1 × 1 and (–1) + 1 = 0 ≠ b Hence x2 + 3x – 1 cannot be factored with integer coefficients.
Note: x2 + 3x – 1 = x2 + ⎟⎠⎞
⎜⎝⎛
232 x +
2
23⎟⎠⎞
⎜⎝⎛ –
2
23⎟⎠⎞
⎜⎝⎛ –1 =
2
23⎟⎠⎞
⎜⎝⎛ +x –
413 =
2
23⎟⎠⎞
⎜⎝⎛ +x –
2
213
⎟⎟⎠
⎞⎜⎜⎝
⎛
= ⎟⎟⎠
⎞⎜⎜⎝
⎛++
213
23x ⎟⎟
⎠
⎞⎜⎜⎝
⎛−+
213
23x .
111
Exercise 4.3.3. Resolve into factors each of the following: 1. x2 + 7x + 12 2. x2 + 9x + 20 3. d2 + 10d + 21 4. z2 – 7z – 98 5. a2 – a – 72 6. x2 + x – 90 7. p2 – 8p + 15 8. y2 – 13y + 42 9. y2 – 20y + 99 10. t2 – 28t + 195 Factorize each of the following: 11. 2a2 + 13a + 15 12. 4x2 + 8x + 3 13. 4x2 + 12x + 9 14. 6x2 + x – 1 15. 6p2 + 17p + 10 16. 4a2 – 11a – 15 17. 7m2 + 16m – 15 18. 8p2 + 29p – 12 19. 6x2 + 5x – 6 20. 15y2 – 13y – 6 21. 14x2 – x – 3 22. 9a2 – 9a + 2 23. 2a2 – 13a + 18 24. 12x2 – 7x + 1 25. 16x2 – 32x + 7 Resolve into factors each of the following : 26. 9x2 + 24xy + 15y2 27. 4x2 – 16xy – 9y2
28. 6c2 + 11cd – 10d2 29. 5x2 – 11xy + 6y2 30. 2a2 – 15ab + 28b2
Factorize the following:
31. 103
10132 −− xx 32.
38
3102 +− uu 33.
161
212 +− xx
34. 4784 2 +− xx 35.
61
354 2 +− xx 36. 232 2 ++ xx
37. 36113 2 ++ xx 38. 532055 2 ++ xx 39. 5532 2 ++ xx 40. 21427 2 ++ xx 4.4 Division of a Polynomial by a Polynomial The process of division of a polynomial f(x) by a polynomial g(x) is to find two polynomials q(x) and r(x) such that f(x) ≡ q(x) g(x) + r(x) where either r(x) = 0 or deg(r(x)) is less than deg(g(x)). Here f(x) is called the dividend, g(x) the divisor, q(x) the quotient and r(x) the remainder. In the process of division of f(x) by g(x), we assume that the degree of the dividend f(x) is always greater than or equal to that of the divisor g(x) because if deg(f(x)) is less than deg(g(x)), then q(x) = 0 and r(x) = f(x). There are two methods of finding the quotient and the remainder. Method 1 : The Long Division method: We explain the method through an example. Example 55: Divide 2 + 5x + 3x2 + 2x3 by 1 + x. Solution: Dividend f(x) = 2 + 5x + 3x2 + 2x3
Divisor g(x) = 1 + x, deg(f(x)) = 3, deg(g(x)) = 1. Our aim is to get the quotient q(x) and the remainder r(x).
112
Step 1: We write f(x) and g(x) in the standard form. f(x) = 2x3 + 3x2 + 5x + 2 g(x) = x + 1 Step 2: We divide the first term (2x3) of the dividend by the first term (x) of the divisor and
obtain the first term ⎟⎟⎠
⎞⎜⎜⎝
⎛= 2
3
22 xxx of the quotient.
Step 3: We multiply the divisor (x + 1) by the first term (2x2) of the quotient and subtract the product (2x3 + 2x2) from the dividend. We obtain the remainder (x2 + 5x + 2). The degree of this remainder is greater than that of the divisor. Step 4: We take the above remainder (x2 + 5x + 2) as the new dividend and repeat step 2 to
obtain the second term ⎟⎟⎠
⎞⎜⎜⎝
⎛= x
xx2
of the quotient.
Step 5 : We multiply the divisor (x + 1) by the second term (x) of the quotient and subtract this product (x2 + x) from the new dividend. We obtain the remainder (4x + 2). The degree of this remainder is equal to that of the divisor. Step 6 : We take the remainder (4x + 2) as the new dividend and repeat step 2 to obtain the
third term ⎟⎠⎞
⎜⎝⎛ = 44
xx of the quotient.
Step 7 : We multiply the divisor (x + 1) by the third term (4) of the quotient and subtract this product (4x + 4) from the new dividend. We obtain the remainder (–2). The degree of this remainder is 0 which is less than that of the divisor. So we stop the process and write the quotient and the remainder. The above steps are presented in the following form: 2x2 + x + 4
2x3 + 3x2 + 5x + 2 2x3 + 2x2
– – x2 + 5x + 2 x2 + x – – 4x + 2 4x + 4 – – – 2
x + 1 ⎟⎟⎠
⎞⎜⎜⎝
⎛= 2
3
22 xxx
⎟⎟⎠
⎞⎜⎜⎝
⎛= x
xx 2
⎟⎠⎞
⎜⎝⎛ = 44
xx
Thus, we have Quotient q(x) = 2x2 + x + 4, Remainder r(x) = –2.
f(x)=q(x) g(x) + r(x)
∴ 2x3 + 3x2 + 5x + 2 = (2x2 + x + 4) (x + 1) + (–2)
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Method 2 : Divisor Factorization Method: In this method, the aim is to get f(x) as a sum of two terms. The first term has g(x) as a factor and the next term is a polynomial whose degree is less than that of g(x). In order to achieve this, we add and subtract suitable terms to f(x). Example 56: Divide 9x3 + 3x2 – 5x + 7 by 3x2 + 2x – 1. Solution: 9x3 + 3x2 – 5x + 7 = [(3x2 + 2x – 1) (3x) – 6x2 + 3x] + 3x2 – 5x + 7 = (3x2 + 2x – 1)(3x) + [–6x2 + 3x2] + [3x – 5x] + 7 = (3x2 + 2x – 1)(3x) + (–3x2) + (–2x) + 7 = (3x2 + 2x – 1) (3x) + [(3x2 + 2x – 1) (–1) + 2x – 1] + (–2x) + 7 = (3x2 + 2x – 1) (3x) + (3x2 + 2x – 1) (–1) + [(2x) + (–2x)] + [(–1) + 7] = (3x2 + 2x – 1)(3x) + (3x2 + 2x – 1) (–1) + 6 = (3x2 + 2x – 1) (3x – 1) + 6 So, the quotient is 3x – 1 and the remainder is 6. Example 57: Divide 4 – 17x – 22x2 – 12x3 – 2x4 by x2 – 3x + 4. Solution: We find the solution by both methods. Divisor Factorization method: 4 – 17x – 22x2 – 12x3 – 2x4 = – 2x4 – 12x3 – 22x2 – 17x + 4
= [(x2 – 3x + 4) (–2x2) – 6x3 + 8x2] – 12x3 – 22x2 – 17x + 4. = (x2 – 3x + 4) (–2x2) – 6x3 + 8x2 – 12x3 – 22x2 – 17x + 4 = (x2 – 3x + 4) (–2x2) – 18x3 – 14x2 – 17x + 4 = (x2 – 3x + 4) (–2x2) + [(x2 – 3x + 4) (–18x) – 54x2 + 72x] – 14x2 –17x + 4 =(x2 – 3x + 4) (–2x2) + (x2 – 3x + 4) (–18x) – 54x2 + 72x – 14x2 –17x + 4 = (x2 – 3x + 4) (–2x2) + (x2 – 3x + 4) (–18x) – 68x2 +55x+ 4 = (x2 – 3x + 4) (–2x2) + (x2 – 3x + 4) (–18x) + [(x2 – 3x + 4) (–68) – 204x + 272] +55x + 4 = (x2 – 3x + 4) (–2x2) + (x2 – 3x + 4) (–18x) + (x2– 3x + 4) (–68) –204x + 272 + 55x + 4 = (x2 – 3x + 4) (–2x2 – 18x – 68) + (–149x + 276)
So, the quotient is − 2x2 – 18x – 68 and the remainder is –149x + 276. Long Division method: – 2x2 – 18x – 68 –2x4 – 12x3 – 22x2 – 17x + 4
–2x4 + 6x3 – 8x2
+ – + –18x3 – 14x2 –17x + 4 –18x3 + 54x2 – 72x + – + –68x2 + 55x + 4 –68x2 + 204x – 272 + – + – 149x + 276
x2 – 3x + 4⎟⎟⎠
⎞⎜⎜⎝
⎛−=
− 22
4
22 xx
x
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
− xx
x 18182
3
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
− 68682
2
xx
So the quotient is –2x2 – 18x – 68 and the remainder is –149x + 276.
114
Exercise 4.4 1. Find the quotient and the remainder when 4x3 – 3x2 + x – 7 is divided by (i) 2x + 1 (ii) x – 4 (iii) 1– x 2. Find the quotient and the remainder when 15 + x4 – 8x2 is divided by (i) (x+1)(x+2) (ii) (x – 2)2 (iii) x3+2x.
Answers
Exercise 4.1 1. (i) F (ii) F (iii) F (iv) F (v) T 2. x3 + 2x2 – x + 4 3. 3x4 – 2x2 + 10x – 8 4. 3x3 + 5x2 – 13x + 2 5. –x3 + 5x2
– 7x + 10 6. –x4 – 4x2 + 4x – 8 7. 3x5 + x3 – 8x2 + 6x – 8 8. 6x4 – 26x3 + 51x2 – 66x + 27 9. 10x5 + x4 + 9x3 + 2x2 – 9x – 28 10. – x5 + 4x4 + 9x3 – 38x2 + 21x
Coefficient of x3 Coefficient of x2 Coefficient of x
11. –2 –7 20 12. –6 10 –3 13. 18 9 9 14. acx2 + bdy2 + (ad + bc)xy 15. 2x3 – x2y – 5xy2 – 2y3
16. x4 + x2y2 + y4 17. p = –9 18. a=27
− 19. m = 12
Exercise 4.2 1. (i) x2 + 11x + 18 (ii) x2 + 6x – 16 (iii) t2 + 4t – 12 (iv) p2 – 7p + 12 (v) 10812 (vi) 3658 (vii) 1224 (viii) 2915 2. (i) 25x2 + 80xy + 64y2 (ii) 9s2 – 24st + 16t2 (iii) 16p2 – 49q2
(iv) 10201 (v) 9604 (vi) 9898
3. 241 ,9/4 4. 60, 20 5. 12, 8
6. (i) 9x2 + y2 + 4z2 + 6xy + 4yz + 12xz (ii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz (iii) 4p2 + 9q2 + 4r2 + 12pq – 12qr – 8pr (iv) 9a2 + 4b2 + 4c2 – 12ab + 8bc – 12ac 7. 45 8. (i) x3 + 9x2 + 26x + 24 (ii) x3 + x2 – 14x – 24 (iii) x3 + 3x2 – 10x –24 (iv) x3 – 5x2 − 2x + 24 (v) x3 – 9x2 + 26x – 24 Coefficient of x2 Coefficient of x Constant 9. (i) 9 –16 –60 (ii) 0 –26 12 (iii) 396 –138 35
115
10. –9, 1523
− , 35
11. (i) 8x3 + 12x2y2 + 6xy4 + y6 (ii) 8u3 – 84u2v + 294uv2 – 343v3
(iii) 33 133
xxxx −+− (iv) x6y9 + 6x4y6 + 12x2y3 + 8
12. 304 13. 7, 18 14. 20,72 15. 2, 180,1016
Exercise 4.3.1 1. 3(3m – n) 2. 4a(a2 – 2a + 4) 3. x(x4 + 4) 4. xy3 (6x4y2 + 3x + 14) 5. 7pq (1–3pq) 6. (m – p) (n + 2) 7. (x + 2 ) (x – 2 ) (x – 2) 8. (x + a ) (x – a ) (x – 1) 9. (p2 + 1) (2p – 1) 10. 2(2x + 1) (2x2 + 1)
Exercise 4.3.2 1. T 2. F 3. F 4. F 5. T 6. (1 + 3x)2 7. (12x – 3)2 8. (2ab + 5cd)2 9. (x + y + a − b) (x + y − a + b) 10. )33( zxy + (3x2y2 – xyz33 + 9z2) 11. (x + 3y) (x2 + 3y2) 12. 2x2 (x4 + 3) 13. (x + y) (x − y) (x2 − xy + y2) (x2 + xy + y2) 14. 2y(3x2 + y2) 15. 8p3 16. (3x + y)3
17. (x – 4)3 18. (2x – 3y)3
19. (2x + 3y + z)2 20. (a + b – 3c)2
21. (x – y + 1) (x2 + y2 + 1 + xy + y – x) 22. (2x – 5y + 6) (4x2 + 25y2 + 36 + 10xy + 30y – 12x) 23. (2x – 3y + z) (4x2 + 9y2 + z2 + 6xy + 3yz – 2zx) 24. ( a3 – 2b – 5c) (3a2
+ 4b2 + 25c2 + ab32 – 10bc + ca35 ) 25. 3 (a – 2b) (2b – 3c) (3c – a) 26. 3(x + y – 2z) (y + z – 2x) (z + x – 2y) 27. 3(a + b) (a – b) (b + c) (b − c) (c + a) (c − a) 28. 3abc (a – b) (b – c) (c – a) 29. (x – y) (x + y + z) 30. (1 + x + y) (1 – x – y) 31. (x2 + 2x + 2) (x2 – 2x + 2)
Exercise 4.3.3 1. (x + 3) (x + 4) 2. (x + 4) (x + 5) 3. (d + 3) (d + 7) 4. (z + 7) (z − 14) 5. (a + 8) (a – 9) 6. (x + 10) (x – 9) 7. (p – 3) (p – 5 ) 8. (y – 6) (y – 7) 9. (y – 9) (y – 11) 10. (t – 13) ( t – 15) 11. (a + 5) (2a + 3) 12. (2x + 1) (2x + 3)
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13. (2x + 3) (2x + 3) 14. (2x + 1)(3x − 1) 15. (p + 2) (6p + 5) 16. (a + 1) (4a – 15) 17. (m + 3) (7m – 5) 18. (p + 4) (8p – 3) 19. (2x + 3) (3x – 2) 20. (3y + 1) (5y – 6) 21. (2x – 1) (7x + 3) 22. (3a – 1)(3a – 2) 23. (a – 2) (2a – 9) 24. (3x – 1) (4x – 1) 25. (4x – 1) (4x – 7) 26. (x + y) (9x + 15y) 27. (2x + y) (2x – 9y) 28. (2c + 5d) (3c – 2d) 29. (x – y) (5x – 6y) 30. (a – 4b) (2a – 7b)
31. 101 (2x – 3) (5x +1) 32.
31 (u – 2) (3u – 4) 33.
161 (4x – 1)(4x – 1)
34. 41 (4x – 1) (4x – 7) 35.
61 (4x – 1) (6x – 1) 36. ( x2 + 1) ( x + 2 )
37. ( x3 + 2) (x + 33 ) 38. ( x5 + 3) ( 5x + 5 ) 39. (x + 5 ) (2x + 5 )
40. ( )( )2727 ++ xx
Exercise 4.4 Quotient Remainder
1. (i) 2x2 –47
25
+x 4
35−
(ii) 4x2 + 13x + 53 205 (iii) –4x2 –x – 2 −5
2. (i) x2 – 3x – 1 9x + 17 (ii) x2 + 4x + 4 –1 (iii) x − 10x2 + 15
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5. PROBLEM SOLVING TECHNIQUES
Whenever we are asked to solve a mathematical problem, we first read and understand it thoroughly and make clear what are given in the problem and what are to be found or proved. Then we proceed to decide what mathematical facts are to be used and how these facts may be used to get a solution to the given problem. The process of deciding about the facts to be used and developing a method to use the facts to solve the problem require skill and experience. Through our skill, we learn more and more methods or techniques to solve problems. Quite often, a given problem may be solved by several ways or techniques. For example, consider the following problem.
Problem: If a − b = 4 and a + b = 5, then find the value of .ba
To solve the above problem, our thinking process works as follows. Question: What are given? Answer: a − b = 4 and a + b = 5 Question: What is to be found?
Answer: The value of .ba
Question: How would you proceed? Answer: Technique 1: a − b = 4. (1) a + b = 5. (2)
(1) + (2) ⇒ 2a = 9 ⇒ a = .29
(1) − (2) ⇒ −2b = −1 ⇒ b = .21
21=
−−
∴ ba =
29 ×
12 = 9.
Technique 2:
ba =
)()()()(
22
babababa
ba
−−+−++
= = .919
4545
==−+
118
Technique 3: a − b = 4 (1) a + b = 5 (2)
Let ba = x. Then
1x
ba= or a = bx.
(1) ⇒ bx − b = 4 ⇒ b (x − 1) = 4. (2) ⇒ bx + b = 5 ⇒ b (x + 1) = 5.
∴ =+−
)1()1(
xbxb
54 or
54
11=
+−
xx or 5x − 5 = 4x + 4 or 5x − 4x = 4 +5 or x = 9.
Thus, we observe that our insight, skill, ability and experience are required in
developing strategies and techniques to solve problems. The techniques presented above may not be exhaustive. New techniques may be developed whenever we solve the same problem again and again. Further, our mind should be more flexible in thinking and kept very well open but not blank to receive the flow of ideas. If a technique is not working in solving a problem, then another method may be tried to solve the problem until we exhaust all techniques known to us. In this chapter, we shall discuss some of the techniques that are quite often used in solving problems. 5.1 Conjectures and Proofs Consider the following pattern of numbers
1, 3, 7, 13, 21, …… We want to know what would be the general term of the pattern. Let us analyse the terms of the pattern as follows:
1=1 3=1 + 2 7=1 + 2 + 4 13=1 + 2 + 4 + 6 21=1 + 2 + 4 + 6 + 8
So we are certain that the next number in the pattern is 1 + 2 + 4 + 6 + 8 + 10 = 21 + 10 = 31. That is, the 6 th term is 31 and 31 = 1 + 2(1 + 2 + 3 + 4 + 5). Similarly,
the 7 th term = 1 + 2 + 4 + 6 + 8 + 10 +12 = 31 + 12 = 43 = 1 + 2(1 + 2 + 3 + 4 + 5 + 6), the 8 th term = 1 + 2 + 4 + 6 + 8 + 10 +12 +14 = 43 + 14 = 57
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= 1 + 2(1 + 2 + 3 + 4 + 5 + 6 + 7), the 9 th term = 1 + 2 + 4 + 6 + 8 + 10 +12 +14 + 16 = 57 + 16 = 73 = 1 + 2(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8).
Looking at the way in which the terms occur, we make the statement that the general term (i.e., the n th term, where n = 1, 2, 3, ….) of the pattern is 1 + 2[1 + 2 + 3 + … + (n−1)]. But, in this statement, we have an unknown sum, namely, 1 + 2 + 3 + ….+ (n − 1). Unless, this sum is found, we can not say that we have found the general term of the pattern. Consider for example the sum 1 + 2 + 3 + 4 + 5 + 6 + ….+ 100. This is a sum of 100 consecutive natural numbers starting with 1. If S is its sum, then S = 1 + 2 + 3 + ….+ 98 + 99 + 100. We can get S in the reverse way also as S = 100 + 99 + 98 + ….+ 2 + 1. That is,
S = 1 + 2 + 3 + …98 + 99 + 100 S =100 + 99 + 98 +…+ 3 + 2 + 1
∴2S = 101 + 101 + 101 + … + 101 + 101 + 101. Remember that there are 100 such 101’s.
So 2S=100 ×101 or S = .50502
101100=
×
Looking at the pattern through which the sum was obtained, we can guess and make the statement
1 + 2 + 3 + …..+ (n − 1) = .2
)1( nn −
We can put n = 1, 2, 3, … and verify that the statement is true. But we have only made a statement. We have not given a proof of it. That is, we have not proved it for a general n. (Gauss, the famous German mathematician while he was 10 years old studying in his 4 th Standard gave the answer immediately as 5050, when his teacher asked the whole class to compute the sum 1 + 2 + 3 + …. + 100. When the teacher asked how he could immediately arrive at the answer, Gauss gave the arguments as above.) So, the general term in our pattern is
1 + 2 × 2
)1( −nn = 1 + n(n−1) = n2 – n +1.
Thus, the general term in the pattern 1, 3, 7, 13, …. is n2 – n +1.
In the above discussion, we have observed a pattern and made a statement. In fact, many theorems in every branch of mathematics such as algebra and geometry have been developed by observing and making statements on patterns of numbers and figures. We shall now discuss various types of statements made in mathematics and also provide various techniques followed in analysing statements.
Statements are simply assertions made. Some examples of statements are given below:
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(i) “x + 7 = 5 where x ∈ N.” (ii) “The diagonals of a square are perpendicular to each other.” (iii) “(a + b) (a − b) = a2 − b2 where a, b ∈ R.” (iv) “ 2 is a rational number.” (v) “The diagonals of a rhombus are not perpendicular to each other.”
(vi) “1 + 2 + ….+ n = 2
)1( +nn .”
A statement made may be true or false. For example, the statements (i), (iv) and (v) are false while (ii) (iii) and (vi) are true. 5.1.1 What is called a definition?
There are certain statements which are made to create new concepts from the already existing ones without leading to erroneous results. They are called definitions. Following are some definitions which we have already encountered :
(i) A triangle is called equilateral if all its sides are of same length (ii) If ax=b, then x is called the logarithm of b to the base a. (iii) Two angles are called supplementary if their sum is 180°.
5.1.2 Axiom
There are certain statements which are assumed to be true. These statements are called axioms. Following are some axioms which we come across in Geometry and Algebra.
(i) There is exactly one and only one straight line passing through two given points. (ii) For any two real numbers, x + y and xy are real numbers. (iii) If n is a natural number, then n + 1 is also a natural number. (iv) A straight line segment has one and only one mid point. (v) An angle has one and only one bisector.
5.1.3 The symbols ⇒, ⇒ and ⇔
Mathematics is a subject which is the outcome of logical thinking. The solution or derivation or proof of a problem is a step by step structure. Each step in a derivation follows logically from the previous steps. To indicate the flow of logic, we make use of the symbol ‘⇒’. We shall explain the usage of this symbol ‘⇒’ through an example. Consider the two statements:
P: x = 2, Q: x2 = 4 Suppose that P is true. Then x = 2. ∴ x2 = x × x = 2 × 2 = 4. ∴ Q is true
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Thus, if P is true, then Q is true. We write this symbolically as “P is true ⇒ Q is true” or simply as “P ⇒ Q”. The symbol ⇒ stands for “implies”.
Next, let us explain the symbol ⇒. Consider the same two statements P and Q:
P: x = 2, Q: x2 = 4 Suppose that Q is true. Then x2 = 4. This does not mean that x = 2, since x = − 2 also satisfies the equation. Hence P need not be true. We write this fact by the symbol Q ⇒ P.
We shall now explain the symbol “⇔” Consider the two statements Q1: 2x + 3y = 12 and 5x − 6y = 3. Q2: x = 3 and y = 2.
Suppose that Q1 is true. Then
2x + 3y = 12, (1) 5x − 6y = 3. (2)
∴ (1) × 2 ⇒ 4x + 6y = 24 (3) and (2) ⇒ 5x − 6y = 3. (4) ∴ (3) + (4) ⇒ 9x = 27 ⇒ x = 3. Putting x = 3 in (1), 6 + 3y = 12 or 3y = 6 or y = 2. i.e., x = 3, y = 2 i.e., Q2 is true Suppose that Q2 is true. Then x = 3, y = 2. ∴ 2x + 3y = 2 × 3 + 3 × 2 = 6 + 6 = 12 and 5x −6y = 5 × 3 − 6 × 2 = 15 − 12 = 3. i.e., Q1 is true Thus Q2 ⇒ Q1. Whenever Q1 ⇒ Q2 and Q2 ⇒ Q1, the two results Q1 ⇒ Q2 and Q2 ⇒ Q1 are put together and written as Q1⇔ Q2. In this situation, we say that , Q1 is equivalent to Q2. This means that, if Q1 is true, then Q2 is true and, if Q2 is true, then Q1 is true. That is, Q1 is true if and only if Q2 is true. Thus the symbol “⇔” stands for ‘if and only if ’. 5.1.4 What is called a theorem?
A statement until it is proved or disproved is called a conjecture. A conjecture, if it is proved, becomes a theorem. A conjecture, if it is disproved, becomes a false statement. Thus, a statement which has been already proved to be true is called a theorem. If a statement holds true in a particular case, then we say that the verification of the statement is made. Following are some theorems which we know already in Algebra and Geometry:
(i) If two sides of a triangle are equal, then the angles opposite to them are equal. (ii) The sum of the three angles of a triangle is equal to two right angles.
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(iii) The diagonals of a parallelogram bisect each other. (iv) The square of an odd integer is odd. (v) The square of an even integer is even.
(vi) 2 is an irrational number. (vii) (a + b)2=a2 + 2ab + b2. (viii) log a(mn)=log am + log an.
5.1.5 What is called a proof of a theorem?
A proof of a theorem is an argument that establishes the truth of the theorem. For example, consider the following statement: “The square of an odd integer is odd.” If the above statement has to be a theorem, then it should have a logical proof. Consider the following argument:
“Let n be an odd integer. Then n = 2m + 1 where m is an integer. Now, we have n2 = (2m + 1)2 = 4m2 + 4m + 1 = 2(2m2 + 2m)+1. Since m is an integer, 2m2 + 2m is an integer. So 2(2m2 + 2m) is an even integer. Hence 2(2m2+2m)+1 is an odd integer. Thus n2 is odd.”
There are several techniques by which a proof may be given. The techniques are
broadly classified as follows:
(i) Direct Proof. (ii) Indirect Proof or Proof by contradiction. (iii) Proof by counter examples. (iv) Geometrical Proof Technique (v) Proof by construction
(i) Direct Proof Technique: Suppose that we want to prove that P ⇒ Q. First, take P to be true. Applying the step by step reasoning, we get that Q is true. This method of proving that Q is true from that P is true is called the direct proof method.
Example 1: If ba = 5, show that =
+−
baba
32 .
Solution: Let ba = 5.
Then a= 5b.
∴ .32
64
55
==+−
=+−
bb
bbbb
baba
123
Example 2: Prove, by direct method, that, if ba =
dc , =
+−
baba .
dcdc
+−
Solution: Let .dc
ba= Put .u
ba= Then a = ub. Since ,
dc
ba= we have .u
dc=
∴ c = ud.
∴ =+−
baba .
11
)1()1(
+−
=+−
=+−
uu
bubu
bubbub
Similarly 11
)1()1(
+−
=+−
=+−
=+−
uu
dudu
duddud
dcdc .
∴ dcdc
baba
+−
=+− .
(ii) Indirect Proof Technique: Suppose that we want to prove that a statement P implies the statement Q. In order to do this, first assume that P is true and Q is not true. Then, applying a step by step reasoning, we arrive at a contradiction (a statement opposite to the assumption). The above method of proof is called the indirect method of proof or contradiction method of proof. Example 3: Prove, by contradiction method, that 2 is an irrational number. Solution: Assume that 2 is a rational number. We know that every rational number has a
simplest form. Let qp be the proper (simplest) form of 2 . Then 2 =
qp ,where p and q are
positive integers having no common factor other than 1. But
2 = qp ⇒ p = 2 q ⇒ p2 = 2q2 ⇒ p2 is an even integer
⇒ p is an even integer ⇒ p = 2m, where m is an integer ⇒ 4m2 = 2q2
⇒ q2 = 2m2
⇒ q2 is an even integer ⇒ q is an even integer ⇒ q = 2n, where n is an integer ⇒ p and q have 2 as a common factor ⇒ a contradiction.
Hence 2 is an irrational number.
124
Example 4: Prove, by indirect technique, that if 100 balls are placed in 9 boxes, then there is at least one box which contains 12 or more balls. Solution: If possible, let us assume that 100 balls are placed in 9 boxes and that no box has 12 or more balls. Then each box has at most 11 balls. Then the total number of balls placed in 9 boxes is 9 × 11 = 99. But this contradicts the assumption that 100 balls have been placed in the 9 boxes. Hence there is some box which contains 12 or more balls. Example 5: Prove, by indirect technique, that, if P is a point which divides the line segment AB in the ratio m: n internally, then P is unique. Solution: Assume that P divides AB in the ratio m: n internally. Assume that P is not unique. Then there is another point P1 on AB which divides AB in the ratio m: n internally (see Figure 5.1). Then
nm
PBAP
= and .1
1
nm
BPAP
=
∴ n AP = m PB and n A∴ n AP =m (AB − AP) a∴ n AP =m AB − m AP a∴ (m + n) AP = m AB an
∴ AP = ,nm
mAB+
AP1 = mm
∴ AP = AP1
∴ P and P1 are the sameThis is a contradiction. H (iii) Proof by Counteto know whether P1 ⇒ Pconclude that P1 ⇒ P2. T Example 6: If x is a realSolution: P1: x is a real nP2: x2
≥ x.
For x = ,21 P1 is true but
Here 21 is a counter exam
Figure 5.1
P1 = m P1B nd n AP1 = m (AB − AP1) nd n AP1 = m AB − m AP1
d (m + n) AP1 = m AB
.n
AB+
point. ence P is unique.
r-example: Let P1 and P2 be two statements. Suppose that we want 2. If we can find an example where P1 is true but P2 is false, then we he example which we have found is called a counter-example.
number, does x2 ≥ x? umber.
x2 = 41 <
21 = x and so P2 is not true. Hence P1 ⇒ P2.
ple and we conclude that “x2 ≥ x for all real x” is false.
125
Example 7: If n is a natural number, then prove that n2 ≥ n. Solution: Let n be a natural number. Since n is a natural number, n > 0 and n − 1 ≥ 0. ∴ n (n − 1) ≥ 0. or n2 − n ≥ 0 or n2 ≥ n. Example 8: For every real number x, does x2 − 1 > 0? Solution: P1: x is a real number. P2: x2 − 1 > 0.
Take x = 21 .Then P1 is true. But we get x2 − 1=
41 − 1 = −
43 < 0. So P2 is not true.
Thus, with x = 21 we get that P1 is true and P2 is not true. Hence P1 ⇒ P2.. Hence
21 is a
counter example and the statement “x2 − 1 > 0 for every real number x” is false. Note: In the above example, by choosing a numerical value for x, we are able to establish the falseness of a statement. (iv) Geometrical Proof Technique: Some algebraic problems can be solved by geometric methods. Example 9: Prove that (a + b) (a − b) = a2 − b2. Solution: Consider the following rectangle. ABCD where AB = AI = a, BE = DI = b (see Figure 5.2) Then AD = AI + DI = a + b AE = AB − BE = a − b. ∴ Area of the rectangle AEFD = (a + b) (a − b) But, the area of the rectangle AEFD
= area of the square ABGI − area of the rectangle BEHG + area of the rectangle DFHI = a2 − ab + (area of the rectangle CDIG − area of the square C= a2 − ab + (ab − b2) = a2 − b2. ∴ (a + b) (a − b) = a2 − b2.
(v) Proof by construction:
When we solve some problems in geometry, we do slike adding lines to the figure in order to get the solution. Conmany geometrical problems. But construction should be maproof by construction requires fore-sightedness on the part of t
126
Figure 5.2
FHG)
ome intermediate constructions struction is a good technique in de at appropriate places. Thus he problem solver.
Example 10: Prove that if two sides of a triangle are equal, then the angles opposite to them are equal. Solution: Let ABC be a triangle. Assume that AB = AC. We want to prove that ∠B = ∠C. Let D be the mid point of BC. Join AD. Consider the triangles ADB and ADC. In these triangles, BD = DC, AD is common and AB = AC. So the triangles are congruent. Then the corresponding angles are equal. ∴∠B = ∠C.
Exercise 5.1 1. What is called a conjecture? 2. What is called a theorem? 3. What is a proof of a theorem? 4. Why should verification precede proof? 5. Explain ‘geometric proof’ with an example. 6. Explain ‘proof’ by counter-example. 7. What is called a construction method? 8. Prove, by indirect method, that, if x, y are any two real numb then either x ≥ 1 or y ≥ 1. 5.2 Mathematical Models
Mathematical models are created to understand the real wo
are framed by using subjects such as Algebra and Geometry. formulated based upon geometric ideas, then such models are calgebraic ideas are used in framing mathematical models, thalgebraic models. Some times mathematical models are aexperiments. Conducting an experiment may be very expensive,Mathematical models give proof for the experimental works. Mwere developed when attempts were made to study practical prostudy some algebraic models and geometric models.
We first consider an example. If the cost of 1 note book i
note books = 12 × 10 = Rs. 120 and the cost of 20 note books = of 5 note books = 12 × 5 = 60 and the cost of 3 note books is below provides the variation of cost of note books with respect tbought.
127
Figure 5.3
ers such that x + y ≥ 2,
rld problems. These models If mathematical models are alled geometric models. If en such models are called nalysed before conducting time consuming and risky. any of mathematical ideas blems. Now we proceed to
s Rs. 12, then the cost of 10 12 × 20 = Rs. 240. The cost Rs. 12 × 3 = 36. The table o the number of note books
Number of Note books (x) 2 3 5 6 8 10 Cost of note books (y) 24 36 60 72 96 120
We observe that when the number of note books increases, the cost also increases and when the number of note books decreases, the cost also decreases. Hence we come across two variables which decrease or increase together. Such variables are said to be in direct variation. In the above example, the number of note books and their cost are directly proportional or they are in direct variation.
Let us now take a real life problem. We have recorded the number x of people in the age group 10 through 15 and the number y among them attending high schools in a town for the years 1999 to 2002. The recordings are tabulated as given below:
x 9200 10200 11600 12400 y 4140 4590 5220 5580
Here x stands for the number of people of age group 10 through 15 years and y stands for the number of them enrolled in high schools in a year. We observe that, for each year, the ratio
=xy 0.45.
Then, we get y = 0.45 x as an equation connecting the two variables x and y. Based upon the previous years observations, we make the assumption that this ratio remains a constant in future years also. Then we find that if 15,000 people of the age group 10 through 15 years are expected in the town for the next year, the number enrolled in high schools will be 6,750 among them. Thus, the equation y = 0.45x is said to be a mathematical model representing the relationship between number of persons enrolled in high schools of the town and the number in the age group 10 through 15 years. It is an algebraic model since it is represented by an algebraic equation. In this example, the ratio of the number y enrolled in high schools to the number x of people in the age group 10 through 15 is a positive constant. In this situation, we say that y varies directly as x or that y has a direct variation with x. The equation y = 0.45 x is a direct variation mathematical model. In general, a direct variation mathematical model is of the form y = k x, where k is a constant. Although our problem is not a problem of geometry, we may try to draw a figure and the figure may be an important step towards the solution of the problem. For example, the algebraic model formulated above may be provided with a geometric description as follows. Let us draw the graph of the equation y = 0.45 x. This graph is a straight line (see Figure 5.4) and it provides a geometric model of the direct variation.
Figure 5.4
128
Consider next the following example. Suppose that a train runs with a uniform speed. If the train takes 4 hours to cover the distance of 160 km, then the speed of the train is
4160 = 40 k.m.p.h.
If the train takes 2 hours to cover the distance of 160 k.m., then the speed of the train is
2160 = 80 k.m.p.h.
We observe that, when t, the number of hours is halved, the speed v is doubled. If the train takes 8 hours to cover the distance of 160 k.m., the speed v is
208
160= k.m.p.h.
We observe that, when t, the number of hours is doubled the speed v is halved. From the above discussion, we note that faster the train goes, the lesser will be the time taken, and slower the train goes, more will be the time taken. That is, if v increases, t decreases and vice versa. That is, v is inversely proportional to t. Two such quantities are said to be in inverse variation. Thus, two quantities are said to in inverse variation if an increase in one quantity makes the other quantity decrease and a decrease in one quantity makes the other quantity increase. The following table gives the speed (v) in k.m. per hour and time (t) taken in hours at that speed to cover a specific distance.
Times (t) hour 2 4 5 8 Speed (v) k.m. 80 40 32 20
From the above table, we observe that,
t × v = 2 × 80 = 160 , t × v = 4 × 40 = 160, t × v = 5 × 32 = 160, t × v = 8 × 20 = 160 which means
tv = 160 or .160t
v =
That is, v and t are in inverse variation. This is an algebraic model representing the indirect variation of speed with respect to time. If we draw a graph of v t = 160, we observe that the graph is not a straight line, but is as shown in Figure 5.5. The graph shows a falling trend of v with respect to t. It is a geometric model representing speed-time relation.
129
In physics we come across thvariation. Under constant temand if volume increases, the by considering a rectangle orectangle whose area is A. T
∴ y = xA or x =
yA . As the b
increases, the breadth y decr
1. What is a mathematical 2. Explain direct variation.3. Explain inverse variation
Figure 5.5
e equation PV = constant. This is an example of inverse perature, as the pressure (P) increases, the volume (V) decreases
pressure decreases. The inverse variation can also be understood f given area. Suppose x and y are the length and breadth of the hen xy = A.
readth y increases, the length x decreases; and as the length x
eases.
Exercise 5.2
model? .
130
6. THEORETICAL GEOMETRY
The word “geometry” is derived from the combination of two Greek words “geo” and “metron”. The word “geo” means “earth” and “metron” means “measurement”. Thus the subject of “earth measurement” was originally named as “geometry”. In the early development of this subject, Egyptians applied geometrical principles in surveying and construction of temples, tombs and pyramids. Later, Greeks emphasized the logical reasoning of geometrical facts and they dedicated their knowledge on geometry to the world of mathematics through the works of Pythagoras and Euclid. This culminated into the birth of the subject theoretical geometry. In this subject, the emphasis is on giving proofs of geometrical facts through deductive reasoning without the need of any geometrical instruments. Euclid, a distinguished Greek mathematician, called the father of geometry who lived about 330 B.C. contributed geometry of his times in thirteen volumes called “The Elements”. Studying theoretical geometry develops creative thinking and generates skills in other subjects also. 6.1 Theorems for Verification
In our earlier classes, we have learnt about certain basic undefined terms in geometry such as point, line, plane and angle. We have also studied some principles of triangles, parallel lines and some special quadrilaterals. However, we recall them here before we proceed to know more about them. 6.1.1 Basic Geometrical terms
A point is used to represent a position in space. In practice, we put a small dot on a paper or on a black board to indicate a point. But theoretically, a point has no size or shape. A point can also be understood as the position where two lines intersect each other. But, here we have made use of the concept of a line to understand a point. We understand a line to be the set of points lying at the intersection of two planes. But, here we have made use of the concept of a plane to understand a line. We understand a plane to be a surface extending infinitely in all directions such that all points lying on the line joining any two points on the surface lie on the surface itself. Thus, we observe that the three concepts a point, a line and a plane are basic terms which are to be understood and cannot be defined.
131
Points are denoted by capital letters such as A, B, C and D. If A and B are two points on a line,
then the line is denoted by writing↔AB and is read as ‘the line AB’. The double headed arrow
‘↔’ indicates the fact that the line extends infinitely in two directions. For brevity, we shall refer to a line by a single letter l. If A and B are two points on a line, then the portion of the line between A and B, including the points A and B, is called the line segment between A and B and is denoted by the symbol AB . The length of AB is denoted simply by writing AB. A ray is the portion of a line starting from a point on the line extending in one of the two directions of the line. The starting point is called the initial point of the ray.
If A is the initial point and B is any point on a ray, then the ray is denoted simply by →AB and is
read as the ray AB. The single headed arrow over AB represents the direction of the ray. A ray →AB is simply written as the ray AB without the arrow mark ‘→’ over AB.
Note: A line ↔AB is simply written as the line AB where the double headed arrow ‘↔’ is
omitted. Similarly, the line segment AB is simply written as AB without the bar ‘⎯’ over AB.
Two rays →AB and having the
common initial point A is said to form an angle at A. The point A is called the vertex of the angle. We shall denote the angle as
→AC
BAC∠ or . Here the segments CAB∠ AB and AC are
called a pair of arms of the angle. The formation of the angle at A is denoted by a small arc starting from one arm to the other arm (see Figure 6.1).
Figure 6.1
If the arms are understood, we simply
write the angle simply as A∠ . Just like a line segment, an angle also has a measure. We shall denote the measure of the angle A∠ by m∠A. The measure does not depend on the lengths of the arms. Angles are measured by a unit called degree. When a ray makes one complete rotation on the plane about its initial point, we say that an angle of measure 360 degrees (written as 360º) is formed. (see Figure 6.2) Here the two arms of the angle coincide. Measurement of al
360º angle. When a ray makes th41 part of one complete rot
132
Figure 6.2
l other angles are made based on a
ation, an angle of measure
41 (360º) = 90º is formed. When a ray makes
61 th part of one complete rotation, an angle of
measure 61 (360º) = 60º is formed. When a ray makes
3601 th part of one complete rotation,
an angle of measure 360
1 (360) = 1º is formed.
Figure 6.5
Figure 6.4
If mthanthen
callesegm
theyB∠
are than
Figure 6.3
= 90º then BAC∠ BAC∠ is called a right angle (see Figure 6.3). If m is greater 90º, then is called an obtuse angle (see Figure 6.4). If m
BAC∠BAC∠ BAC∠ is less than 90º,
is called an acute angle (see Figure 6.5). If mBAC∠ BAC∠ = 180º, then is
d a straight angle (see Figure 6.6). If
BAC∠
BAC∠ is a straight angle, then BC is a line ent and the points A, B and C are collinear; that is,
lie on anAC
also v 360º
Figure 6.6
a straight line. When two lines ↔BD and intersect at the point A, the two angles
d
↔CE
DAE∠ are called vertically opposite angles. We observe that and CAD∠ BAE∠ ertically opposite angles (see Figure 6.7). If m BAC∠ is greater than 180º but less , then the angle is called a reflex angle (see Figure 6.8).
Figure 6.9
133
Figure 6.10
Figure 6.7
Figure 6.8
If two angles have a common vertex and lie on the opposite sides of a common arm, then the angles are called adjacent angles. (see Figure 6.9). In the Figure 6.9, BAD∠ and are adjacent angles since they have the common vertex A and lie on the opposite sides of the common arm
DAC∠
AD . We observe that ∠BAC and ∠BAD are not adjacent angles since they are on the same side of the common arm AB .Two adjacent angles are said to be complementary angles if the sum of their measures is 90º (see Figure 6.10). In Figure 6.10, m BAD∠ + m = 90º and so DAC∠ BAD∠ and DAC∠ are complementary angles. Here m BAD∠ = 90º − m and we say that one angle is the complement of the other. DAC∠ If the sum of the measures of two adjacent angles is 180º, then the two angles are called supplementary angles and we say that one angle is the supplement of the other (see Figure 6.11). In figure BAD∠ and
are supplementary angles and we observe that m = 180º − m
DAC∠DAC∠ BAD∠ .
6.1.2 Axioms and Theorems on lines We present below some axioms and theorems on lines for understanding. Activity: Plot two points A and B on a plane. Through A, we can draw severa
there is one and only one linethrough B, we can draw infinite
namely the line ↔AB which pass
From the above activity, we posProperty 1: Given any two poin
Figure 6.12
l (infinite number of ) lines
, namely ↔AB which passe
number of lines. Of these lin
es through A.
tulate the following propertyts on a plane, there is one an
134
Figure 6.11
(see Figure 6.12). Of these lines,
s through the point B. Similarly, es, there is one and only one line,
: d only one line containing them.
Note: From the above property, we observe that (i) Two distinct points in a plane determine a unique line. If X and Y are any two points on a
line, the line ↔XY is denoted simply as the line XY, omitting ‘↔’.
(ii) Three or more points are called collinear points if they all lie on the same line. (iii) Three or more points are called non-collinear points if at least one of them does not lie on the line passing through two of the points. Activity: Draw two distinct lines AB and CD in a plane. We observe that these lines can have either one point in common or no point in common (see Figure 6.13). From the above actProperty 2: Two d Note: If two distinctwo distinct lines non-intersecting lin
Activity: Draw a
observe that there From this activity, Property 3: Giventhrough the given p Note: If three or mlines.
Figure 6.13
ivity, we enunciate the following property. istinct lines cannot have more than one point in common.
t lines have a common point, then the lines are called intersecting lines. If in a plane have no point in common, then the two lines are called es. Two non-intersecting lines are called parallel lines.
line ↔AB and mark a point P not on the line. Draw lines through P. We
is a unique line passing through P and parallel to ↔AB (see Figure 6.14).
we have the following property.
a line and oint and is pa
ore lines pas
Figure 6.14
a point not on it, there is one and only one line that passes rallel to the given line.
s through the same point, then the lines are called concurrent
135
Activity: Draw two intersecting lines ↔AB and Make the point of intersection as O.
Measure the angles , ,
↔.CD
AOC∠ BOD∠ AOD∠ and BOC∠ (see Figure 6.15 We observe that
m = m AOD∠ ,BOC∠m = m AOC∠ .BOD∠
The angles and are said to form a pair of vertically opposite angles. Similarly
AOC∠ BOD∠BOC∠
and are vertically opposite angles. AOD∠ From the above activity, we get the following: Property 4: If two lines intersect, then the vertically opposite angles are equal. Note: Let m = xº and m = yº. Then, usingm = yº. But m + m + m
AOC∠ AOD∠BOC∠ AOC∠ BOC∠ BOD∠ + m
or xº + yº + xº + yº = 360º or 2xº + 2yº = 360º or xº + yº =angles. Thus, we observe that m + mAOC∠ BOC∠ = 1m∠AOC + m∠AOD = 180º and m + mBOC∠ BOD∠∠BOD and ∠AOD, ∠AOC and ∠AOD and ∠BOC andangles.
Activity: Draw two parallel lines ↔AB and . Draw a
↔CD
(see Figure 6.16).
We observe that intersects ↔PQ
↔AB and . Mark the
↔CD
and the common point of and as M. M and
↔CD
↔PQ
,BLM∠ ,ALM∠ ,LMD∠ ,LMC∠ CMQ∠ QMD∠m = mPLB∠ LMD∠ , m BLM∠ = m ,DMQ∠m = m PLA∠ ,LMC∠m ALM∠ = m From the above activity, we get .CMQ∠
136
Figure 6.15
the above axiom, m = xº and BOD∠AOD∠ = 360º
180º. So xº and yº are supplementary 80º and m∠BOD + m∠AOD = 180º, = 180º. That is, ∠AOC and ∠BOC, ∠BOD form pairs of supplementary
line not parallel to ↔PQ
↔AB
Figure 6.16
common point of ↔AB and as L
easure the angles
↔PQ
,PLB∠ ,PLA∠. We observe that
the following property.
Property 5: If a transversal intersects two parallel lines, then any pair of corresponding angles are equal. Note: (i) The angles ALM∠ and LMD∠ are said to form a pair of alternate interior angles. The angles BLM∠ and are alternate interior angles. In the above activity, we observe that
LMC∠BLM∠ = LMC∠ and ALM∠ = LMD∠ . We state this fact as a theorem.
Theorem 1: If a transversal intersects two parallel lines, then any pair of alternate interior angles are equal. (ii) The angles and are said to form a pair of alternate exterior angles. The angles
PLA∠ DMQ∠BLP∠ and are alternate exterior angles. We observe from the activity that CMQ∠
m = m mPLA∠ ,DMQ∠ BLP∠ = m .CMQ∠We state this fact as a theorem. Theorem 2: If a transversal intersects two parallel lines, then any pair of alternate (interior or exterior) angles are equal. (iii) The angles BLM∠ and LMD∠ are said to form a pair of interior angles on the same side of the transversal. Similarly, the angles ALM∠ and LMC∠ are interior angles on the same side of the transversal. We observe from the activity that m BLM∠ + m LMD∠ = 180°,m ALM∠ + m LMC∠ = 180°. We state this fact as a theorem. Theorem 3: If a transversal intersects two parallel lines, then any pair of interior angles are supplementary.
Activity: Draw a line ↔AB . Draw another line
which is not parallel to
↔PQ
↔AB . Mark the intersecting
point of ↔AB and as L. Take a point M on
other than L.. Measure the angle
↔PQ
↔PQ
ALM∠ . Draw a
line through M such that m↔
CD ALM∠ = m LMD∠
(see Figure 6.17). We observe that is a
transversal of the lines
↔PQ
↔AB and , and the two
lines
↔CD
↔AB and do not meet at all. From this
activity, we have the following property.
↔CD
Property 6: If a transversal cuts two lines such that a paithe lines are parallel. As a consequence of the above property, the following the Theorem 4: If a transversal intersects two lines such thaequal, then the lines are parallel. Theorem 5: If a transversal cuts two lines such that a paiof the transversal are supplementary, then the lines are par
137
Figure 6.17
r of alternate angles are equal, then
orems can be proved.
t a pair of corresponding angles are
r of interior angles on the same side allel.
Example 1: In Figure 6.18, the line L1 is parallel to the line L2 and the line L3 is the transversal of the lines L1 and L2 . If the measures of the angles 1∠ and 2∠ are in the ratio 4:5, find the measures of the angles , 1∠ 2∠ , 3∠ , 4∠ , 5∠ , 6∠ , 7∠ , and . 8∠
Solution: m : m = 1∠ 2∠
But m + m = 180° 1∠ 2∠
∴ 54
× m + m = 182∠ 2∠
or 9× m = 180 × 5 or m2∠∴ m = 180° − m = 1∠ 2∠ Now m = m (vertica3∠ 1∠ m = m (vertically o4∠ 2∠m = m (correspond5∠ 1∠m = m (correspond6∠ 2∠m = m (vertically o7∠ 5∠m = m (vertically o8∠ 6∠
Example 2: In Figure 6.19 Solution: ↔
CD is a transversal to theSo the two angles ACD∠
Figure 6.18
4 : 5 ⇒ 54
2m1m
=∠∠ ⇒ m 1∠ =
54
× m 2∠ .
since they are supplementary angles.
0° or 5
2m52m 4 ∠×+∠× = 180°
= 100°. 2∠180° − 100° = 80°.
lly opposite angles) = 80°,
pposite angles) = 100°, ing angles) = 80°, ing angles) = 100°, pposite angles) = 80°, pposite angles) = 100°.
prove that ↔AB is parallel to and is parallel to
↔CD
↔AC
↔BD .
lines ↔A
and ∠
Figure 6.19
and C↔BD and CDBACD ∠+∠ mm = 50° + 130° = 180°.
which are on the same side of the transversal are CDB
138
supplementary. Hence and ↔AC
↔BD are parallel. Then the corresponding angles and
are equal. So m
XCA∠
CDB∠ XCA∠ = 130°. Now, we observe that is a transversal to the lines ↔AC
↔AB and and m
↔CD XCA∠ = m = 130°. That is, the alternate angles and
are equal. Hence the lines
CAB∠ XCA∠
CAB∠↔AB and are parallel.
↔CD
6.1.3 Axioms and Theorems on a triangle We know already about a triangle. A plane triangle or simply a triangle is a geometrical figure formed
by three lines in a plane. In Figure 6.20 ↔DE ,
↔XY
and are three lines. The lines and ↔PQ
↔PQ
↔XY
intersect at A, the lines ↔DE and
↔XY intersect at B,
and the lines and ↔PQ
↔DE intersect at C. We name
the shape bounded by the line segments ,AB BC , CA as the triangle ABC. The line segments ,AB Btriangle ABC. The points A, B and C are called the v
and are called the interior anABC.
,BAC∠ ABC∠ BCA∠
They are also simply denoted by A∠ , B∠ and C∠ ran exterior angle of the triangle ABC. We
and are exterior angles,YBC∠ ,BCQ∠ ECA∠ CAX∠ From what we have studied in our earlier classes, we re (i) If no two sides of a triangle are of equal lengttriangle. (ii) If two sides of a triangle are of equal length, triangle. (iii) If all the sides of a triangle are of equal length, ttriangle. (iv) If each of the three angles of a triangle is an acuacute angled triangle or simply acute triangle. (v) If one angle of a triangle is an obtuse angle, thentriangle or simply obtuse triangle.
139
Figure 6.20
C and CA are called the sides of the ertices of the triangle ABC. The angles gles or simply the angles of the triangle
espectively. The angle is called observe that the angles ,
of the triangle ABC.
PAB∠DBA∠
call the following definitions:
h, then the triangle is called a scalene
then the triangle is called an isosceles
hen the triangle is called an equilateral
te angle, then the triangle is termed an
the triangle is called an obtuse angled
(vi) If one angle of a triangle is a right angle (i.e., of measure 90°), then the triangle is called a right angled triangle or simply right triangle. We now state certain theorems on triangles. Theorem 6: The sum of the measures of the three angles of a triangle is 180°. Theorem 7: If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the interior opposite angles. Proof: ABC is a given triangle. Produce BC and take a point X on the extension of BC as in Figure 6.21. Now ACX∠m is an external angle and A∠ and B∠ are interior opposite angles. We have to show that ACX∠m = A∠ + B∠ . Now
and are supplementary. ACX∠ C∠∴ + = 180°. ACX∠ C∠But A∠ + B∠ + = 180°. C∠∴ + = ACX∠ C∠ A∠ + B∠ + . C∠Cancelling on both sides, we get C∠
ACX∠ = A∠ + B∠ . Activity: Draw a triangle PQR as in Figure 6.22. MQR and RP .Compute the sums PQ + QR, QR + PR an
(i) PQ + QR > PR, (ii) QR + PR > PQ, (iii) PR + PQ > QR.
From the above activity, we have the following proper Property 7: The sum of any two sides of a triangle is gthe third side. Note: We observe that in any triangle ABC, (i) AB < BC + CA (ii) BC < CA + AB (iii) CA These inequalities are called triangle inequalities. Activity: Draw a triangle ABC (see Figure 6.23). Mthe angles A∠ , B∠ and . Measure also the lengthe sides AB, BC and CA. Compare the measures angles and find out which angle has the greater meCompare also the lengths of the sides and find out side has the larger length. We observe that the greaterhas the larger side opposite to it, where we have namangle of greater measure, the greater angle and the sgreater length, the greater side. The above fact is put following property.
C∠
140
Figure 6.21
easure the lengths of the sides PQ , d PR + PQ. We observe that
Figure 6.22
ty.
reater than
< AB + BC.
easure ths of of the asure. which angle ed the ide of as the
Figure 6.23
Property 8: In any triangle, the largest side has the greateangle opposite to it. Note: Consider a line l and a point P not lying on l see Figure 6.24. Draw the perpendicular line-segment PL to the line l. The point L is called the foot of the perpendicular from P to l. Take any point M other than L on l. Now the triangle PLM is a right triangle. Since PL is perpendicular to l, m = 90°. Since m + mPLM∠ PLM∠ LMP∠ + m LPM∠ = 180°, we get 90° + m LMP∠ + m LPM∠ = 180° or m LMP∠ + m LPM∠ = 90° ∴ m LMP∠ < 90°, m LPM∠ < 90° and m LMP∠ = 90° − m LPM∠ . That is LMP∠ and LPM∠ are acute angles and are complementary angles. So is the greater angle in the triangle PLM. Hence the side PM is the greater side. This means that PL < PM, where M is any point on the line. Hence the perpendicular segment PL has length smaller than the length of any other line-segment drawn from P to the line. Thus, the perpendicular line segment PL is the shortest of all line-segments drawn from P to the line l.
PLM∠
6.1.4 Congruent triangles
In the study of geometry, congruent triangles occur frequently. We shall list some important facts, about congruent triangles. Activity: Take two blank papers and put a carbon sheet between them. Now draw a triangle on one paper. We observe that another triangle of identical nature (mirror image or carbon copy) is produced on the other paper. We say that the two triangles are congruent triangles. In the two triangles, we can match the equal sides and the equal angles. The matched sides and matched angles are called corresponding sides and corresponding angles. Thus, we say that two triangles are congruent, if all the sides, and angles of one triangle are equal to the corresponding sides and angles of the other triangle. For example, consider the two triangles ABC and DPX (see Figure 6.25).
Figure 6.25
141
st
Figure 6.24
where AB = DX, BC = PX, CA = PD; .,, DAPCXB ∠=∠∠=∠∠=∠ Observing the corresponding angles, we write that ABC is congruent to DXP. We write this fact as ∆ABC ≡ ∆DXP. There are six corresponding equations when we have congruency of triangles. Conversely, if the six corresponding equations are given, then the triangles are congruent. We have learnt in our earlier classes that a triangle can be constructed if one of the following three sets is given:
(i) The length of two sides and the measure of the included angle. (ii) The length of all the three sides. (iii) Measures of two angles and the length of one side.
The congruency of two triangles can be ensured by establishing any one of the following correspondences:
(i) Side-Angle-Side. (ii) Angle-Side-Angle. (iii) Side-Side-Side
Activity: Draw two triangles ABC and PQR such that AB = PQ, PA ∠=∠ and AC = PR. Now, cut out the two triangles ABC and PQR. Place the triangle PQR on ABC and try to fit one triangle with the other (see Figure 6.26). We are able to adjust tcoincides with A, Q coangle between the sidePQ and PR. Thus, we h Property 9: If any twoand the included angle
Figure 6.26
hem and we observe that the triangles fit exactly such that the vertex P incides with B and R coincides with C. The ∠A is called the included s AB and AC. Similarly, P∠ is the included angle between the sides ave the following property.
sides and the included angle of one triangle are equal to any two sides of another triangle, then the two triangles are congruent.
142
Note: The above property is known as Side-Angle-Side criterion or simply SAS criterion for congruence of triangles. Activity: Draw two triangles ABC and PQR such that
RCQRBCQB ∠=∠=∠=∠ and, (see Figure 6.27) Now cut out the twomanner that they fit ewhich can be provedcriterion as a theorem Theorem 8: Two trtriangle are equal to tNote: The above ccriterion. We observshould correspond toActivity: Draw two tFigure 6.28). Since ABC is a triangSince DEF is a triangBut , EAD =∠∠=∠From (1) and (2), we∴ . FC ∠=∠
Figure 6.27
triangles and place them one over the other and adjust them in such a xactly as one triangle. From this activity, we state the following criterion . However, we do not give the proof here and simply state the important .
iangles are congruent if any two angles and the included side of one he two angles and the included side of the other triangle. riterion is known as Angle-Side-Angle criterion or simply as ASA e that in this criterion, a side and the angles on this side of one triangle a side and the angles on it of another triangle for congruency. riangles ABC and DEF such that EBDA ∠=∠∠=∠ , and BC = EF (see
Figure 6.28
le, =∠+∠+∠ CBA 180° (1) le, =∠+∠+∠ FED 180°
∴.B∠ =∠+∠+∠ FBA 180° (2) get =∠+∠+∠ CBA FBA ∠+∠+∠ .
143
Now, in triangles ABC and DEF, we observe that the side BC and the angles B∠ and on it correspond to the side EF and the angles
C∠E∠ and F∠ on it. Hence by ASA criterion,
∆ABC ≡ ∆DEF. Thus, we have the following theorem. Theorem 9: Two triangles are congruent if any two angles and a side of one triangle are equal to the two angles and the corresponding side of the other triangle. Note: The above criterion for congruency is known as the Angle-Angle-Side or AAS Criterion. Activity: Draw the triangles ABC and DEF such that BC = EF, CA = FD, AB = DE (see Figure 6.29). Cut the triathis positionof the activiTheorem 1three sides o Note: Thetwo triangle
Nowcongruency Activity: Dthe point ofline XY andline segmen
Figure 6.29
ngle DEF and place it over ABC and adjust such that they fit exactly as one. In D stands on A, E stands on B and F stands on C and ∆ABC ≡ ∆DEF. The result ty is stated as the theorem given below. 0: Two triangles are congruent if the three sides of one triangle are equal to the f the other triangle.above criterion is known as Side-Side-Side or SSS criterion for congruence of s.
, we shall examine whether we can have SSA criterion or AAA criterion for of two triangles.
raw a triangle ABC and a line XY parallel to the side BC (see Figure 6.30). Mark intersection of line XY and the side AB as D, and the point of intersection of the the side AC as E. Since the side AB and the side AC are transversal of the parallel ts XY and BC, CEBD ∠=∠∠=∠ , (corresponding angles).
Figure 6.30
144
∴ The two triangles ABC and ADE have AAA property. However, they are not congruent since the corresponding sides are not equal. Hence, we conclude that AAA correspondence cannot be a criterion for congruency of triangles. Activity: Draw a line AX sufficiently long. Draw a line segment AB of length a such that the angle BAX∠ has some specified measure. Draw a circle with B as its centre and some radius b (< a). We observe that the circle crosses the line AX at two points C and D (see Figure 6.31). In triangles ABC andHence, we conclude two triangles. Activity: Draw line
of BC and QR as X
units. Now cut arcsintersection of these AC, PQ and PR. Wewe conclude the follo Theorem 11: Two rare respectively equa Note: The above crcongruence of right t
Figure 6.31
ABD, we have SSA correspondence. But AC ≠ AD. So ∆ABC ≡ ∆ABD. that SSA correspondence cannot be taken as a criterion for congruency of
segments BC and QR such that BC = QR = a units. Mark the mid points
and Y respectively. Draw circles with X and Y as centres and radius 2a
of radius b(< a) units with centres at B and Q. Mark the points of arcs with the circles as A and P respectively (see Figure 6.32). Join AB, observe that AC = PR. Hence ∆ABC ≡ ∆PQR. From the above activity, wing theorem.
Figure 6.32
ight triangles are congruent if the hypotenuse and a side of one triangle l to the hypotenuse and a side of the other triangle.
iterion is known as the Right-Hypotenuse-Side or RHS criterion for riangles.
145
6.1.5 Properties of parallelogram We know that a quadrilateral is a closed figure formed by four line segments and a parallelogram is a uadrilateral in which the opposite sides are parallel to each other. In Figure 6.33, ABPQ || SR (i.e., PQ is Property 1: In a paProof: Let ABCD band BDC. Since ABAD|| BC and BD imboth ABD and BD∆ABD ≡ ∆BDC.
m DBCADB ∠=∠
∴ The correspondiAB = CD and AD =Property 2: In a paof equal measure. Proof: Let ABCD ba transversal to AB m m BDCABD ∠=∠Since AD || BC andto AD and BC, m m CBDADB ∠=∠∴ m AABC ∠=∠ m
= m B∠ = m A∠
Similarly, m BAD∠ Property 3: The diProof: ABCD is a pBy ASA criterion, ∆ ∴AM = CM, BM =∴The diagonals bis
q
Figure 6.33 CD is a quadrilateral. In Figure 6.34 parallel to SR) and PS || QR.
rallelogram, the opposite sides are ofe a parallelogram. Join BD (see Figu || CD and BD is a transversal of AB as a transversal of AD and BC,
The side BD is common to C. Hence, by AAS property, .
ng sides are equal. Hence BC. rallelogram, the opposite angles are
e the parallelogram. Join BD (see Fiand DC,
. BD is a transversal
.DBCBD ∠+ m ADBDC ∠+ m
.DC.m BCD∠=
agonals of a parallelogram bisect eacarallelogram. AC and BD are diagonaAMB ≡ CMD (see Figure 6.37).
DM. ect each other.
146
Figure 6.34
. PQRS is a parallelogram in which
equal length. re 6.35). Consider the triangles ABD nd CD, m .m BDCABD ∠=∠ Since
Figure 6.35
gure 6.36). Since AB || DC and BD is
Figure 6.36
Figure 6.37
h other. ls.
Property 4: If the opposite sides of a quadrilateral are rilateral is a parallelogram. Proof: Let ABCD be a quadrilateral where AB = CD, AD= BC. Join AC. Consider the triangles, ACB and ADC. By SSS criterion, ∆ABC ≡ ∆CDA (see Figure 6.38). Then m m,m ACDBAC ∠=∠ .m ACBCAD ∠=∠ ∴ AB || CD and AD || BC. Hence ABCD is a parallelogram. Property 5: If the opposite angles in a quadrilaterquadrilateral is a parallelogram. Proof: ABCD is a quadrilateral (see Figure 6.39). m and m.m BCDBAD ∠=∠ ADCABC ∠=∠ m . Join BD. Consider the triangles ABD and CDB. Now, m∠1 + m∠2 + m∠BCD = 180° and m∠3 + m∠4 + m∠BAD = 180° ∴ m∠1 + m∠2 + m∠BCD = m∠3 + m∠4 + m∠BAD m∠BCD =∴ m∠1 + m∠2 = m∠3 + m∠4 But ADCABC ∠=∠+∠∠=∠+∠ m4m2m,m3m1m ∴ 4m2m3m1m ∠+∠=∠+∠ i.e., 3m4m2m1m ∠−∠=∠−∠ (1) + (2) ⇒ 2m∠1 = 2m∠4 ⇒ m∠1 = m∠4. ∴ AD || BC. (1) − (2) ⇒ 2m∠2 = 2m∠3 ⇒ m∠2 = m∠3 ∴ AB || CD. Hence ABCD is a parallelogram. Property 6: If the diagonals of a quadrilateral bisect eparallelogram. Proof: ABCD is a quadrilateral. AC and BD are diagonals. AC and BD bisect each other at M (see Figure 6.40). ∴AM = CM, BM = DM. ∴ .mm,mm BMCAMDCMDAMB ∠=∠∠=∠ ∴ By SAS criterion, ∆AMB ≡ ∆CMD, ∆AMD ≡ ∆CMB ∴ AB || CD, AD || BC. Hence ABCD is a parallelogram.
147
of equal length, then the quad
Figure 6.38
al are of equal measure, then the
m∠
ach
Figure 6.39
BAD(1)
2)
other, then the quadrilateral is a
Figure 6.40
Theorem 12: A quadrilateral is a parallelogram if one pair of opposite sides are parallel and equal. Given: ABCD is a quadrilateral where AB || CD and AB = CD.
Figure 6.41
To prove: ABCD is a parallelogram. Construction: Draw the diagonal AC (see Figure 6.41). Proof: In triangles ABC and ADC, (i) AB = CD (given) (ii) AC is common (iii) m∠BAC = m∠ACD ∴ By SAS criterion, ∆ABC ≡ ∆ADC. ∴ Corresponding sides are equal; and corresponding angles are equal ∴ AD = BC, m∠DAC = m∠ACB. ∴ AD || BC. Hence ABCD is a parallelogram. The theorem is proved.
AB || CD, AC is a transversal to AB and CD, ⇒ alternate angles are equal.
Property 7: If there are three or more parallel lines and the intercepts made by them on a transversal are equal, then the corresponding intercepts on any other transversal are also equal. Proof: The lines l1, l2, l3 are parallel to each other. PQ and XY are transversals to l1, l2, l3, AC = CE. We have to prove that BD = DF. Draw AG || BD and CH || DF (see Figure 6.42). We observe that AGDB and CHFD are parallelograms. ∴ AG = BD, CH = DF (1) In triangles ACG and CEH, CE = AC, m∠GAC = m∠HCE (corresponding angles) m∠ACG = m∠CEH (corresponding angles). ∴ by ASA criterion ∆ACG ≡ ∆CEH. ∴ AG = CH (2) From (1) and (2) BD = DF. Property 8: In a triangle, the line joining the mid poinside and is equal to one half of it. Proof: ABC is a triangle. D and E are mid points of AB and AC
To prove DE || BC and DE = ).(21 BC
Draw CF || BD to meet DE extended at F (see Figure 6CFE. AD || CF and AC is transversal to them ∴ m∠DAE = m∠ECF AD || CF and DF is a transversal to them ∴ m∠ADE = m∠CFE Also AE = EC, since E is the mid point of AC.
148
Figure 6.42
ts of two sides is parallel to the third
respectively.
.43). Consider the triangles ADE and
∴ By AAS criterion, ∆ADE ≡ ∆CFE. ∴ AD = CF and DE = EF. But BD = AD, since D is the mid point of AB ∴ BD = CF Already BD || CF. ∴ BCFD is a parallelogram. ∴ DF = BC and DF || BC or DE || BC. ie., DE + EF = BC or DE + DE = BC or 2DE = BC.
Thus DE || BC and DE = ).(21 BC
Property 9: In a triangle, the line drawn through the mid-poside, bisects the third side. Proof: Let ABC be the given triangle and D be the mid point of ABAC at E (see Figure 6.44). We have to prove that E is the miBC || DE and AB is a transversal to BC and DE. ∴ m∠ADE = m∠ABC, AD || CF and DF is a transversal to AD and CF. ∴ m∠ADE = m∠EFC and ∴ m∠AED = m∠BCE, From (1) and (2), m∠ABC = m∠EFC, AD || CF and AC is a transversal to AD and CF. ∴ m∠BAC = m∠ACF. Now m∠BCF = m∠BCE + m∠ECF = m∠AED + ∠ACF by (3) = m∠AED + m∠DAE by (5) = m∠BDE (since ext. angle = sum of int. opp. angles) i.e., m∠BCF = m∠BDE From (4) and (6), BCFD is a parallelogram. ∴ BD = CF, BC = DF Consider the triangles ADE and CFE. Here CF = AD, since CF = BD = AD. ∴ by ASA criterion, ∆ADE ≡ ∆CFE. ∴ DE = EF, AE = EC. ∴ E is the mid point of AC.
149
Figure 6.43
int of one side, parallel to another
Figure 6.44
. Draw DE parallel to BC to meet d point of CA.(1)
(2) (3) (4)
(5)
(6)
6.1.6 Concurrency of lines
If three or more lines pass through the same point P, the lines are said to be concurrent and the point P is called the point of concurrence. Activity : Draw a triangle ABC. Draw the perpendicular bisectors DX and EY of the sides BC and CA (see Figure 6.45). Mark their point of intersection as S. Now draw the perpendicular bisector FZ of the side AB. We observe that FZ passes through S. From the above activity, we understand the following theorem.
Figure 6.45
Theorem 13: The perpendicular bisectors of the sides of a triangle are concurrent. Note: The point of concurrence of the perpendicular bisectors of the sides of a triangle is called the circumcentre of the triangle and denoted by the letter S. Activity: Consider a triangle ABC. Find out the circumcentre S of ABC. Join SA, SB and SC. Measure the lengths of SA, SB and SC. We observe that SA = SB = SC. Draw a circle with centre at S and radius equal to SA. We observe that this circle passes through the three vertices of the triangle. This circle is called the circumcircle of the triangle and its radius namely SA(= SB = SC) is called the circumradius of the triangle. Activity: Draw a triangle ABC. Draw AD perpendicular to BC to meet it at D. Draw BE perpendicular to CA to meet it at E. Mark the point of intersection of AD and BE as H. Now, draw CF perpendicular to AB to meet it at F (see Figure 6.46). We observe that CF passes through the point H. The perpendiculars AD, BE and CF are called the altitudes of the triangle. From the activity, we understand the following theorem. Theorem 14: The altitudes of a triangle are concurrent. Note: The point of concurrence of the altitudes of a triantriangle and is denoted by the letter H.
150
Figure 6.46
gle is called the orthocenter of the
Activity: Draw a triangle ABC. Draw the angle bisector of the angle ∠A. Draw the angle bisector of the angle ∠B. Mark the point of intersection of these two bisectors as I. Now draw the angle bisector of the angle ∠C (see Figure 6.47). We observe that this bisector passes through the point I. From the above activity, we are able to understand the following theorem.
Figure 6.47
Theorem 15: The bisectors of the angles of a triangle are concurrent.. Note: The point of concurrence of the angle bisectors of a triangle is called the incentre of the triangle and is denoted by the letter I. Activity: Draw perpendiculars from I to the sides of the triangle ABC and measure their lengths. We observe that all of them are equal. Now draw a circle with centre at I and radius as the common length of the three perpendiculars (see Figure 6.48). We observe that this circle touches all the sides of the triangle. This circle is called the incircle and its radius is known as the inradius of the triangle ABC. Activity: Draw a triangle ABC. Mark the mid points
of the sides BC, CA and AB as D, E and F
respectively. Join AD and BE. Mark the meeting
point of AD and BE as G. Now, join CF (see Figure
6.49). We observe that CF passes through the point
G. The lines AD, BE and CF are called the medians
of the triangle ABC. Measure the length AG, GD,
BG, GE, CG, GF. We observe that
.12
===GFCG
GEBG
GDAG
From the above activity, we understand the following theore
Theorem 16: The medians of a triangle are concurrent andeach median in the radio 2:1.
151
Figure 6.48
Figure 6.49
m.
the point of concurrency divides
Note: The point of concurrency of the medians of a triangle is called the centroid of the triangle and is denoted by the letter G. Activity: Draw a line segment AB and a line l parallel to AB. Mark two points C and D on l such that AB = CD. We observe that ABCD is a parallelogram. Draw AL perpendicular to l. Measure the length of AL. Now, calculate the area of the parallelogram ABCD. It is equal to base × height = AB × AL. Now, mark two other points P and Q, on l such that PQ = AB (see Figure 6.50). We observe that ABPQ is also a parallelogram. Its area is equal to base × height = AB × AL. Thus, we understand the following theorem. Theorem 17: Parallelograms on the same base and between thearea. Activity: Draw a line segment AB. Draw a line l parallel to AB. Mark a point C on l. Draw AL perpendicular to l. Measure the length of AL. (see Figure 6.51).
We find the area of the triangle ABC as 21 × base × height
= .21 ALAB ×× Mark another point P on l. We find the
area of ∆ABP as 21 × base × height = .
21 ALAB ×× thus,
we observe that the area of the triangle remains the same for all positions of the vertex C on the line l. From this activity, we are able to understand the following theorem. Theorem 18: Triangles on the same base and between the same p
Exercise 6.1 Which of the following statements are true and which are false: 1. If two lines are intersected by a transversal, then the alternate2. If two parallel lines are intersected by a transversal, then thequal. 3. If two parallel lines are intersected by a transversal, then theside of the transversal are equal.
152
Figure 6.50
same parallels are equal in
Figure 6.51
arallels are equal in area.
angles are equal. e corresponding angles are
interior angles on the same
6.2 Theorems with logical proofs So far we have performed certain activities and verified certain axioms and theorems. Now, we proceed to give logical proofs for certain theorem on angle and triangles Theorem 19: If a ray stands on a line, then the sum of the adjacent angles so formed is 180°. Given: The ray PQ stands on the line XY. To Prove: m∠QPX + m∠YPQ = 180°. Construction: Draw PE perpendicular to XY. Proof: m∠QPX = m∠QPE + m∠EPX = m∠QPE + 90° (1) m∠YPQ = m∠YPE − m∠QPE = 90° − m∠QPE (2)
(1) + (2) ⇒ m∠QPX + m∠YPQ = (m∠QPE + 9Thus the theorem is proved. Theorem 20: If two lines intersect, then the vertiGiven: Two lines AB and CD intersect at the poiTo prove: m∠AOC = m∠BOD, m∠BOC = m∠Proof: The ray OB stands on the line CD. ∴ m∠BOD + m∠BOC = 180° (1The ray OC stands on the line AB. ∴ m∠BOC + m∠AOC = 180° (2From (1) and (2), m∠BOD + m∠BOC = m∠BOC + m∠AOC ∴ m∠BOD = m∠AOC. Since the ray OA stands on the line CD, m∠AOC + m∠AOD = 180° (3From (2) and (3), we get m∠BOC + m∠AOC = m∠AOC + m∠AOD ∴ m∠BOC = m∠AOD. Hence the theorem is proved. Theorem 21: The sum of the three angles of a triGiven: ABC is a triangle (see Figure 6.54). To prove: ∠A + ∠B + ∠C = 180°. Construction: Through the vertex A, draw the linXY parallel to the side BC. Proof: XY || BC.
153
Figure 6.52
0°) + (90° − m ∠QPE) = 180°.
cally opposite angles are of equal measure. nt O (see Figure 6.53). AOD.
)
)
)
angle
e
Figure 6.53
is 180°.
Figure 6.54
Now, AB is a transversal to the lines XY and BC. ∴ m∠XAB = m∠ABC (alternate angles).
= ∠B. (1) Next, AC is a transversal to the parallel lines XY and BC. ∴ m∠YAC = m∠ACB (alternate angles) = ∠C. (2) We also have m∠BAC = m∠A. (3) (1) + (2) + (3) ⇒ m∠XAB + m∠YAC + m∠BAC = m∠B + m∠C + m∠A ⇒ (m∠XAB + m∠BAC) + m∠CAY = m∠A + m∠B + m∠C ⇒ m∠XAC + m∠CAY = m∠A + m∠B + m∠C ⇒ 180° = m∠A + m∠B + m∠C. Hence the theorem is proved. Theorem 22: The angles opposite to equal sides of a triangle are equaGiven: ABC is a triangle where AB = AC (see Figure 6.55). To prove: ∠B = ∠C. Construction: Mark the mid point of BC as M and join AM. Proof: In the triangles AMB and AMC (i) BM = CM (ii) AB = AC (iii) AM is common. ∴ By the SSS criterion, ∆AMB ≡ ∆AMC. ∴ Corresponding angles are equal. In particular, ∠B = ∠C. Hence the theorem is proved. Theorem 23: The side opposite to the larger of two angles in a triangopposite to the smaller angle. Given: ABC is a triangle, where ∠B is larger than ∠C, th t is m∠B >To prove: The length of the side AC is longer than the length of the side AB.
Fig
i.e., AC > 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
Case (i) Suppose that AC < AB. Then the side AB has longer length angle ∠C which is opposite to AB is larger measure than that of ∠Bshorter side AC. That is, m∠C > m∠B. This contradicts the given Hence the assumption that AC < AB is wrong. ∴AC < AB. Case (ii) Suppose that AC = AB. Then the two sides AB and AC aopposite to these sides are equal. That is ∠B = ∠C. This is again a cfact that ∠B > ∠C. Hence AC = AB is impossible. Now Case (iii) remHence the theorem is proved.
154
Figure 6.55
l.
le is longer than the side
m∠C.
aure 6.56
than the side AC. So the which is opposite to the fact that m∠B > m∠C.
re equal. So the angles ontradiction to the given ains alone to be true.
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°.
155
Figure 6.58
(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°. (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°. Example 6: In Figure 6.62 the line l3 is a transversal to the paFind the angles x and y. Solution: Alternate angles are equal. ∴ x = 130°. Interior angles on the same side of the transversal are supplementary. ∴ 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°.
156
Figure 6.59
Figure 6.60
Figure 6.61
rallel lines l1 and l2.
Figure 6.62
Figure 6.63
Example 8: Find the angles x and y in Figure 6.64 where e lines l1 and l2 are parallel and l3 is a transversal to l1 and l2.
Figure 6.64
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°. 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) (ii) Solution: (i) A∴ ABCD is a∴ 2x = 24°, 3∴ x = 12°, y = (ii) I∴ ∆ADC ≡ ∆∴ Corresponx = 26° − 20°
Figure 6.66
D = BC, AB = CD. parallelogram. y = 60° (alternate angles are equal) 20°.
n triangles ACD and ACB, AD = AB, CD = BCABC. ding angles are equal. ∴ x + 20° = 26°, y − 5°, y = 42 + 5 or x = 6°, y = 47°.
157
Figure 6.67
th
: Figure 6.65
and AC is common.
= 42° or
Example 12: Prove that the bisector of the vertex angle of edian 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, 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 Bthat ∠BAC is a right angle. Solution: Given DA = DC. Since D is the mid point of BC, BD = Disosceles. ∴∠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) (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 quSolution: 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 + ∠B + ∠ACB = 36
⇒ (∠DAC + ∠CAB) + ∠B + (∠ACD + ∠ACB)+ ⇒ ∠A + ∠B + ∠C + ∠D = 360°.
158
an isosceles triangle is a m
Figure 6.68
C. DA is drawn. If DA = DC, prove
C. The triangles ABD and ACD are
°
adrilateral is 360°.
0° ∠D = 360°
Figure 6.70
Figure 6.69
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. Example 16: AD and BE are two altitudes of a triangleProve 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. Example 17: In a rectangle ABCD, E is the mid point 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. Example 18: In a rhombus, prove that the diagonals biSolution: Let ABCD be a rhombus, Draw the diagonaBD. Let them meet at O. We have to prove that O ipoint of both AC and BD and that AC is perpendicuBD. 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
∴ ∆AOB = ∆BOC, by SSS criterion. ∴ ∠AOB = ∠BOC. Similarly, we can get ∠BOC = ∠COD, ∠COD = ∠DO∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = x (say) But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°.
159
Figure 6.71
ABC
of BC
sect els ACs the lar (⊥
A.
such that AE = BD.
Figure 6.72
.Figure 6.73
ach other at right angles. and mid ) to
Figure 6.74
∴ x + x + x +x = 360°
∴ 4x = 360° or x = 4
360° = 90°.
∴ 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 ∠BAC = ∠DAC i.e., AC bisects the angle ∠A. Similarly we can prove that AC bisects ∠C, BD bisects ∠B a 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). 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° + 3 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°.
160
Figure 6.75
nd BD bisects ∠D.
Figure 6.76
5° = 80°.
Figure 6.77
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. (1But AD || BE and DEC is a transversal to AD and BE. ∴∠ADC = ∠BEC (corresponding angles) (2From (1) and (2), we get ∠BCE = ∠ADC or ∠BCD = ∠ADC.
Exercise 6.2 1. Which of the following statements are true and wh (i) If a ray stands on a line, then the sum of the two adj (ii) If two lines intersect, then vertically opposite angles (iii) A triangle can have two obtuse angles. (iv) The sum of the angles of a quadrilateral is 180° (v) If ∆ABC ≡ ∆PQR, then ∠A = ∠Q (vi) If ∆DEF ≡ ∆XYZ, then DE = XY (vii) 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 (ii) The angles are complementary and the larger is 4 ti(iii) The angles are supplementary and the larger is 58° (iv) The angles are supplementary and the larger is 20° (v) The angles are adjacent and form an angle of 140°. larger. (vi) The angles are vertically opposite and supplementar
161
Figure 6.78
))
ich are false: acent angles so formed is 180° are equal.
.
° less than the other. mes the smaller. more than the smaller. less than three times the smaller. The smaller is 28° less than the
y.
5. Find x and y from the following figures: (i) (ii)
Figure 6.80
(iii) 6. In each (i) (iii)
Figure 6.79
(iv)
Figure 6.81
of the following, find x and y:
(ii)
Figure 6.85
162
Figure 6.82
Figure 6.83
Figure 6.84
Answers
Exercise 6.1 1. F 2. T 3. F
Exercise 6.2 1. (i) T (ii) T (iii) F (iv) F (v) F (vi) F (vii) T 2. (i) 70° (ii) 25° (iii) 20° (iv) 12° 3. (i) 130° (ii) 50° (iii) 100° (iv) 28° 4. (i) 25°, 65° (ii) 18° (iii) 61° , 119° (iv) 50°, 130°
(v) 56°, 84° (vi) 90°, 90° 5. (i) x = 130°, y = 50° (ii) x = 80°, y = 70° (iii) x = 20°, y = 30° (iv) x = 50°, y = 130° 6. (i) x = 19, y = 8 (ii) x = 48°, y = 12° (iii) x = 6, y = 3.
163
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 reprthe t
Figure 7.1
esenting the number 0 for both the lines. We use thwo perpendicular lines become two perpendicular n
164
Figure 7.2
e same scaling on the two lines. Now umber lines. The positive numbers for
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
Drra
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 escartes. It is also called the rectangular coordinate plane or the xy–plane. The system of
epresentation of points in the plane by ordered pairs of numbers is called the Cartesian or ectangular or xy- coordinate system. The two axes are called rectangular or coordinate xes.
165
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.
(vi) All points on a line pathe same y-coordinate(see Fi (vii) All points on a line pa
Figure 7.4
rallel to x-axis have gure 7.5)
rallel to y-axis have the s
166
Figure 7.5
ame x-coordinate (see Figure 7.6).
Figure 7.6
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
Figure 7.7
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.
They are respectively denoted by 12 xx − and
12 yy − . 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. Example 1: Plot the points A (3, 0), B (0, 2 ),G (−1, 0) and H (0, −4). Also specify the quadranSolution: The points are plotted in the coordinate plane as ipositive 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.
167
Figure 7.8
C (4, − 4), D (3, 3), E (−2.5, 1), F (−1, −3), t in which each point lies.
n Figure 7.9. The point A lies on the
Figure 7.9
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) ( 2 − 3 , −1) lies in the IV quadrant. (ix) (0,−3) lies to the left of x−axis. (x) (5, 0) lies below the x–axis. (xi) Any two points on a line parallel to x-axis have equal x-coordinates. (xii) 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
168
Figure 7.11
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. Thth
thvavapoofInan
he
is,
po
xy
ca
linlinthcopoth
He
Figure 7.12
e other direction is called the negative directe negative direction, then its x-coordinate dec
Let a point P move along the line in the point P2(x2, y2).Then x2 > x1. We observelue x1 to x2 and the y-coordinate of P corrlue y2. The change in the x-coordinate valueint P. The corresponding change in the y-co the moving point P. We observe that the ru Figure 7.12, the point P is moving up the lid the point P2 is at a higher position than the
nce the ratio 12
12
xxyy
−−
is positive. In Figure 7
it is falling down the line from P1 to P2 a
int P1. So the rise y2 −y1 is negative and henc
12
12
xy
−−
is positive for a rising line and it is n
lled the slope of the line.
Next, we shall examine the slope of e parallel to the x-axis. In Figure 7.14, the LL′ is parallel to the x-axis and we observ
at all points on the line have the same yordinate. If P1(x1, y1) and P2(x2, y2) are twints on the parallel line, then y1 = y2 and se rise y2 − y1 = 0.
nce the slope 12
12
xxyy
−−
= 0.
16
Figure 7.13
ion of the line. We observe that if P moves in reases. e positive direction from the point P1(x1, y1) to that the x-coordinate of P changes from the
espondingly changes from the value y1 to the is x2−x1 and is called the run of the moving
ordinate values is y2 − y1 and is called the rise n x2 −x1 is positive (see Figures 7.12 and 7.13) ne, that is it is rising up the line from P1 to P2 point P1. So the rise y2−y1 is positive and
.13, the point P is moving down the line; that
nd the point P2 is at a lower position than the
e the ratio 12
12
xxyy
−−
is negative. Thus, the ratio
egative for a falling line. The ratio 12
12
xxyy
−−
is
a e e -o o
9
Figure 7.14
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
m = runrise =
12
12
xxyy
−−
= 0
12 yy −.
This is undefined. ∴ The slope of a line perpendicular to x-axis is undefined.
12
12
xxyy
−− > 0 for rising line.
12
12
xxyy
−− < 0 for falling line
12
12
xxyy
−− = 0 for the line parallel to x
12
12
xxyy
−− is undefined for the line par
We note that 12
12
xxyy
−−
= ( )( )21
21
xxyy
−−−−
= 21
21
xxyy
−−
.Thus,
points (x1, y1) and (x2, y2) is 12
12
xxyy
−−
= 21
21
xxyy
−−
.
From this, we observe that the slope is independent of the direction of the line. The
slope 12
12
xxyy
−−
of the line is also independent of
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 (see Figure 7.16). Then the slope of the line
from P3 to P4 is34
34
xxyy
−−
.
The slope of the line from P1 and P2 is12
12
xxyy
−−
.
170
Figure 7.15
-axis.
allel to the y-axis.
the slope of the line joining the two
Figure 7.16
But the triangles ∆P1AP2 and ∆P3BP4 are similar.
∴ BPAP
3
1 = 4
2
BPAP or
APAP
1
2 = BP
BP
3
4 .
∴ 12
12
xxyy
−− =
34
34
xxyy
−−
.
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
drawn. The slope of the line is 12
12
xxyy
−−
.
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
m = 12
12
xxyy
−−
.= 103
51569=
−− .
The slope is a positive number and so the line is rising up as shown in Figure 7.17. Example 4: Find the slope of the line passing thwhether the line is rising up or falling down. Solution: The slope of the line is
m = )16(40
296−−−− =
85
5635 −
=− .
Since m is a negative number, the line is falling down as indicated in Figure 7.18. Example 5: Interpret the slopes of the following line(i) (6,4) and (−7, 4) (ii) (−2,8) and (−2, 7). Solution:
(i) Slope of the line = 12
12
xxyy
−−
= ,0130
6744
=−
=−−−
∴ The line is parallel to the x-axis.
171
Figure 7.17
rough (−16, 29) and (40, −6) and state
s join
Figure 7.18
ing
(ii) Slope of the line = 12
12
xxyy
−−
=01
2287 −
=+−− = not defined.
∴ The line is perpendicular to the x-axis.
Example 6: Find another point on the line with slope 53
− which passes through the point
(−2, 3).
Solution: First we shall write the slope as 53− (i.e., with the denominator as a positive
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
)2(330−−− or
53− .
7.2.1 The equation of a straight line Let P (x, y) be a variable point on a giv
connecting the variables x and y is called the equand y of any point on the straight line satisfy thepairs (x, y) as points in the Cartesian plane, we gline is hereafter called simply a line. The graph crosses the y-axis at a unique point B. Since A liex-coordinate of A, then (a, 0) should satisfy the efor y in the equation of the line, we can solve fox-intercept of the line. That is, the x-intercept where the line crosses the x-axis. Similarly, sinSo, if b is the y-coordinate of B, then (0, b) shoulby 0 and y by b in the equation of the line, wey-intercept of the line. Thus, the y-intercept owhere the line crosses the y-axis.
172
Figure 7.19
en straight line. Then an algebraic equation ation of the straight line. The co-ordinates x equation of the line. By plotting the ordered et the graph of the straight line. A straight crosses the x-axis at a unique point A and it s on the x-axis its y-coordinate is 0. If a is the quation of the line. Substituting a for x and 0 r the value of a. This value of a is called the of the line is the x-coordinate of the point ce B lies on the y-axis, its x-coordinate is 0.
d satisfy the equation of the line. Replacing x can solve for b. This value b is called the f the line is the y-coordinate of the point
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
Figure 7.20
0−−
xcy or
xcy − .
But the slope of the line is given to be m.
∴ x
cy − = m or y − c = mx or y = mx + c.
The above equation is called the slope –intercept formula for the equation of a line. 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.
Example 7: Find the equation of the line having slope 21 and y-intercept −3.
Solution: Applying the slope-intercept formula, the equation of the line is
y = 21 x + (−3) m =
21
c = −3 y = mx + c
or 2y = x − 6 or x− 2y − 6 = 0.
Example 8: Find the slope and the y-intercept of the line whose equation is 3x + 4y + 5 = 0.
Solution: Rewriting the equation, we get 4y = −3x − 5 or y = .45
43
⎟⎠⎞
⎜⎝⎛ −+
− x
Comparing this equation with y = mx + c, we get slope m = 43− and y-intercept c =
45− .
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.
173
(ii) through (0, 4) with slope 41 .
(iii) through (2, −2) with slope −1. (iv) through (1, −3) with slope 4. (v) through (−1, −4) with slope
37 .
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)
53− and 1.
(vi) 52− and
58 .
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 21PP . Three cases arise. Case (i): The segment 21PP 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 = .21 xx − Hence d (P1, P2) = 21 xx − .
Case (ii): The segment 21PP 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 = 21 yy − . Hence
d(P1, P2) = 21 yy − .
174
Figure 7.21
Figure 7.22
Case (iii): The line segment 21PP 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 21 xx − and the length
of the segment P3P2 is 21 yy − . We observe that
the triangle ∆P1P3P2 is a right triangle.
Figure 7.23
∴ ( )[ ] ( )[ ] ( )[ ] 2
212
212
232
312
21 ,,, yyxxPPdPPdPPd −+−=+= = (x1 − x2)2 + (y1 − y2)2
= (x2 − x1)2 + (y2 − y1)2.
∴ d(P1, P2) = 212
212 )()( yyxx −+− .
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) = 212
212 )()( yyxx −+− = 22
12 0+− xx = 12 xx − . When P1 and P2 lie on the same vertical line, then x1 = x2 and so
d(P1, P2) = 221
221 )()( yyxx −+− = 2
2120 yy −+ = .21 yy −
Note: Since the origin O is (0,0), then, for any point P (x, y), we have OP = 22 )0()0( −+− yx = .22 yx + This distance 22 yx + is called the radius vector of the point (x, y) from the origin. Using the distance formula, we can show whether (i) three given points are collinear or form a right triangle, isosceles triangle or equilateral
triangle. (ii) four given points form a parallelogram, rectangle, square or rhombus.
175
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. Then d (A, B) = 2
122
12 )()( yyxx −+−
= 22 )31()157( +++ (x1, y1) (−15, −3)
(x2, y2) (7, 1) = 22 422 + = 16484 + = 500 = 510 .
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
A (−4,−9)
B (2, 0) AB = 22 )90()42( +++
Rough Sketch
Figure 7.24
= 22 96 + = 8136 + = 117 = 133139 =× .
B (2, 0)
C (4, 3) BC = 22 )03()24( −+−
= 22 32 + = 94 + = 13 AC = 22 )93()44( +++
= 22 128 + = 14464 + = 208 = 1341316 =× We observe that AB + BC = AC. 13413133 =+ . ∴ 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
176
Figure 7.25
d(P,Q) = 22 )25()32( ++− = .255049171 22 ==+=+
d (Q, R) = .5618014436126)57()28( 2222 ==+=+=−−+−
d (R, P) = .25502525)5(5)27()38( 2222 ==+=−+=+−+− ∴ 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 ( )2,3 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) = .24)2(0)31()00( 2222 ==−+=−+−
d (B, C) = .2413)12()03( 22 ==+=−+− Figure 7.26
d (C, A) = .2413)23()30( 22 ==+=−+− ∴ d (A, B) = d (B, C) = d (C, A). ∴∆ABC is an equilateral triangle. 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 = .551254121)11()74( 22 ==+=−−+−−
QR = .101003664)15()44( 22 ==+=+++
PR = .525169)15()74( 22 ==+=−+− ∴ 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. Example 14: Show that the points (1, 2), (2, −1), (5, 3) parallelogram. Is it a rectangle ? Justify. Solution: Let the points be P1, P2 ,P3 and P4 respectively. OP1 P2 P3 P4 is a parallelogram is to show that the opposite sides are of equal length. Here we find P1P2 = .1091)21()12( 22 =+=−−+−
P2P3 = .25169)13()25( 22 =+=++−
177
and
ne w
Figure 7.27
(4, 6) taken in order form a
ay of showing that
Figure 7.28
P3P4 = .1091)36()54( 22 =+=−+−
P4P1 = .25169)26()14( 22 =+=−+−
∴ P1P2 = P3P4 = 10 and P2P3 = P4P1 = 25 . ∴ P1P2 P3P4 is a parallelogram. Since P1P3 = 17116)23()15( 22 =+=−+− 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 = .5220164)13()02( 22 ==+=++−−
BC = .54801664)37()26( 22 ==+=−++
CD = .5220164)73()68( 22 ==+=−+−
AD = .54801664)13()08( 22 ==+=++−
AC = 101006436)17()06( 22 ==+=++−
We observe that AB = CD = 52 , BC = AD = 54 and AB2 + BC2 = 20 + 80 = 100 = AC2
∴ ABCD is a rectangle but not a square. Example 16: Show that the points (0, −1), (2, 1) (0, 3vertices 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 = ,22844)11()02( 22 ==+=++−
BC = ,22844)13()20( 22 ==+=−+−
CD = ,22844)31()02( 22 ==+=−+−−
AD = ,22844)11()02( 22 ==+=++−−
178
Figure 7.29
) and (−2, 1) taken in order form the
.
Figure 7.30
BD = ,416016)11()22( 22 ==+=−+−−
AC = .416160)13()00( 22 ==+=++− We observe here that AB = BC = CD = AD = 22 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 = 22 )35()26( ++−
= 6416 + = 80
BC = 22 )51()62( −+−−
= 801664 =+
AC = 22 )31()22( ++−−
= 321616 =+
BD = 22 )57()66( −−+−−
= 144144+ = 288
CD = 22 )17()26( −−++−
= 806416 =+
AD = 22 )37()26( +−+−−
= 801664 =+ . ∴ 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 point
(i) (1, 2) and (4, 3) (vi) (a, −b) an (ii) (3, 4) and (−7, 2) (vii) 1 ,12( +
(iii) (−7, 2) and (3, 2) (viii) ⎟⎠⎞
⎜⎝⎛
45,
32
179
Figure 7.31
s:
d (−b, a) ) and (1, 3)
and ( )2,1−
(iv) (4, −5) and (−4, 5) (ix) (2, 0) and (5, −4) (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)
(ii) (21 , 1), (1, 2) and (0,
23 )
(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).
(iii) (0, 3) (0, 5) and ( 3 , 4). 5. 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). 6. 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). 7. 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).
8. 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)
9. 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).
180
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 (iii) III (iv) IV (v) No quadrant (vi) No quadrant (vii) No quadrant (viii) II 2. (i) F (ii) F (iii) F (iv) T (v) T (vi) T (vii) T (viii) F (ix) F (x) F (xi) F (xii) T 3. (i) (2, 1) (ii) (6, 9) (iii) (10, 7) (iv) (2, 2)
Exercise 7.2
1. (i) −1 (ii) 6 (iii) 58− (iv) 3 (v) ⎟
⎠⎞
⎜⎝⎛
ab4 (vi)
ab−
2. (i) (6, 7) (ii) (4, 5) (iii) (3, −3) (iv) (2, 1) (v) (2, 3) 3. (i) 3x + y + 7 = 0 (ii) 5x − y + 9 = 0 (iii) 2x + y − 15 = 0 (iv) 6x − y − 11 = 0 (v) 3x + 5y − 5 = 0 (vi) 2x + 5y − 8 = 0
4. (i) ⎟⎠⎞
⎜⎝⎛ − 2 ,
23 (ii) (2, 0) (iii) (1, −3) (iv) ⎟
⎠⎞
⎜⎝⎛ − 2 ,
45
Exercise 7.3
1. (i) 10 (ii) 262 (iii) 10 (iv) 412 (v) (a − b) 2
(vi) (a + b) 2 (vii) 6 (viii) 12481
(ix) 5 (x) .65
2. (i) Collinear (ii) Non-collinear (iii) Collinear
(iv) Non-Collinear (v) Non-Collinear
3. (i) Isosceles (ii) Isosceles (iii) Isosceles
181
4. (i) Equilateral (ii) Equilateral (iii) Equilateral
5. (i) Right triangle (ii) Not a right triangle 6. (i) right angled isosceles triangle (ii) right angled isosceles triangle 7. (i) parallelogram (ii) parallelogram (iii) parallelogram (iv) parallelogram 8. (i) Rectangle (ii) Rectangle (iii) Rectangle (iv) Rectangle 9. (i) square (ii) square (iii) not a square (iv) square (v) Not a square 10. (i) Rhombus (ii) Rhombus (iii) Rhombus
182
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. 8.2
Figure 8.1
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 OAand OB are the initial and terminal sides of an angle, then the angle is
Figure 8.3
183
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
41 th of one complete rotation in the anticlockwise
direction, an angle of measure 41 (360°) = 90° is formed. When a ray makes
41 th of one
complete rotation in the clockwise direction, an angle of measure −41 (360°) = −90°.is formed
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 is, . This is known as Pythagoras Theorem.
222 BCABAC +=
Figure 8.4
184
8.1 Trigonometric Ratios
Figure 8.5
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:
sine θ = OPPQ
=side hypotenuseoflength
sideoppositeoflength ,
cosine θ = OPOQ
side hypotenuseoflengthsideadjacentoflength
= ,
tangent θ = OQPQ
=sideadjacentoflengthsideoppositeoflength ,
cosecant θ = PQOP
=sideoppositeoflength
sidehypotenuseoflength ,
secant θ = OQOP
=sideadjacentoflength
sidehypotenuseoflength ,
cotangent θ = PQOQ
=sideoppositeoflengthsideadjacentoflength .
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 PO
OPQO
OQQP
PQ′
=′
=′′
.
From this, we get
QOQP
OQPQ,
POQO
OPOQ,
POQP
OPPQ
′′′
=′′
=′′′
=
or QPQO
PQOQ
QOPO
OQOP
QPPO
PQOP
′′′
=′′
=′′′
= ,,
185
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
1cosecsin =×=×PQOP
OPPQθθ ,
θecθ,
θθec
cos1sin
sin1cos == .
1seccos =×=×OQOP
OPOQθθ ,
θθ,
θθ
sec1cos
cos1sec == .
1cottan =×=×PQOQ
OQPQθθ .
θθ,
θθ
cot1tan
tan1cot == .
We also note that, θθθ tan
cossin
==×=⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=OQPQ
OQOP
OPPQ
OPOQOPPQ
.
Taking reciprocals, we get θθθ
θ cottan
1sincos
== .
Thus, we have θθθθ
θθ cot
sincos,tan
cossin
== .
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
,53
106
side hypotenuseoflengthsideoppositeoflengthsin ===θ
35
610
sideoppositeoflengthsidehypotenuseoflengthcosec ===θ ,
54
108
side hypotenuseoflengthsideadjacentoflengthcos ===θ ,
45
810
sideadjacentoflengthsidehypotenuseoflengthsec ===θ ,
43
86
sideadjacentoflengthsideoppositeoflengthtan ===θ ,
34
68
sideoppositeoflengthsideadjacentoflengthcot ===θ .
186
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
, A
A, A
A, A
A
,ABBC A,
ACAB A,
ACBC A
1517
sin1eccos
817
cos1sec
158
tan1cot
815tan
178cos
1715sin
======
======
C C,
C C,
C C
,BCAB C,
ACBC C,
ACAB C
81
sin1eccos
1517
cos1sec
815
tan1cot
158tan
1715cos
178sin
======
======
Note: In the above problem, we observe that sin C = cos Atan C =cot A,…. This is so because the angles A and C are
Example 3: If 257sin =θ , find the other trigonometric rati
Solution: Since 257sin =θ , let us consider a right triang
m∠ACB=θ, AB = 7 and AC = 25 (see Figure 8.8). By PythAC2 = AB2 + BC2 ∴ 252 = 72 + BC2 or 625 = 49 +
∴ BC2 = 625−49 = 576. ∴ BC= 576 = 24.
.θ
θ
,θ
θ
,θ
θec
,BCABθ
,ACBCθ
724
tan1cot
2425
cos1sec
725
sin1cos
247tan
2524cos
==
==
==
==
==
Hence,
Example 4: If 2cosec =A , find (i) sin A + cos A (ii) ta
Solution: Since cosec A = 2 = oppositeoflength
hypotenuoflength12=
we consider a right triangle PQR where m∠QRP = A,
187
,7
Figure 8.7
, cos C = sin A, complementary.
os.
le ABC in which m∠ABC = 90°,
agoras formula, BC2
Figure 8.8
n A + cot A.
sidesidese ,
PR = 2 and PQ = 1 (see Figure 8.9). By Pythagoras theorem,
PR2 = PQ2 + QR2
Figure 8.9
∴ ( 2 )2 = (1)2 + QR2 ∴ 2 = 1 + QR2
∴ QR2 = 2 – 1 = 1 ∴ QR = 1.
Hence we get
.111cot,1
11tan
,2
1cos,2
1sin
======
====
PQQRA
QRPQA
PRQRA
PRPQA
∴(i) sin A + cos A = 2
12
1+ = 2 ⎟
⎠
⎞⎜⎝
⎛2
1 = 2 ,
(ii) tan A + cot A = 1 + 1 = 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 BABABABA
BABA
sinsincoscossincoscossin
tantan1tantan
−+
=−
+ .
Solution:
L.H.S = BABA
tantan1tantan
−+ =
BB
AA
BB
AA
cossin
cossin1
cossin
cossin
×−
+ =
BABABA
BABABA
coscossinsincoscos
coscossincoscossin
−
+
=BABA
BABA
BABAsinsincoscos
coscoscoscos
)sincoscos(sin−
×+
= BABABABA
sinsincoscossincoscossin
−+ = R.H.S.
Example 6: Prove that BA
BABA
tantan
tancotcottan
=++ .
Solution: L.H.S = =++
BABA
tancotcottan
1tan
tan1
tan1
1tan
BA
BA
+
+=
ABA
BBA
tantantan1
tan1tantan
+
+
188
= )tantan1(
tantan
)1tan(tanBA
AB
BA+
×+ .
= BA
tantan = R.H.S.
Example 7: Show that θθθ
θθ
cos1tansin
cot1tan1
++
=++ .
Solution: L.H.S. = =++
θθ
cot1tan1
θ
θ
tan11
tan1
+
+
= ⎟⎠⎞
⎜⎝⎛ ++
θθθ
tan1tan)tan1( =
)θ(θθ)(
1tantan
1tan1
+×
+ = tan θ (1)
R.H.S. = θθθ
cos1tansin
++ =
θθθθ
cos1cossinsin
+
+
= )cos1(
1cos
)sincos(sinθθ
θθθ+
×+ =
)cos1(1
cos)1(cossin
θθθθ
+×
+
= θθ
cossin = tanθ` (2)
From (1) and (2), L.H.S = R.H.S. 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
Figure 8.10
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 .
189
Here from the right triangle ADC, (see Figure 8.11) we get
sin 60° = 23
=ACDC sin 30° =
21
=ACAD
cos 60° = 21
=ACDA cos 30° =
23
=ACDC
tan 60° = 313==
DADC tan 30° =
31
=DCAD
cot 60° = 3
1=
DCDA cot 30° = 3
13==
ADDC
sec 60° = 212==
DAAC sec 30° =
332
32
==DCAC
cosec 60° = 3
323
2==
DCAC cosec 30° = 2
12==
ADAC
(ii) Trigonometric ratios of a 45° angle. We consider the isosceles right angle ABC where m∠
Figure 8.12). Then m∠CAB = 45° and m∠BCA = 45° . Now froget, by Pythagoras theorem, AC2 = AB2 + BC2 = 1 + 1 = 2 and s
sin 45° = 2
1=
ACAB
cos 45° = 2
1=
ACAB
tan 45° = 111==
ABBC
cot 45° = 111==
BCAB
sec 45° = 212==
ABAC
Figure 8.12 cosec 45° = 2
12==
BCAC
(iii) Trigonometric ratios of a 0° angle and
Fig
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.
190
Figure 8.11
B = 90°, AB = BC = 1(see m the right triangle ABC, we o = AC 2 ,
ure 8.13
Then OM = x and PM = y. By applying Pythagoras theorem in the right triangle OMP, we get x2 + y2 = r2; ∴ r = 22 yx + . Let ∠MOP = θ. Then θ is an acute angle and
sin[θ = rxθ,
ry
=cos .
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
0 and y increases from 0 to r. So as θ increases from 0° to 90° , rx decreases from 1 to 0 and
ry increases from 0 to 1. That is, as θ increases from 0° to 90°, cos θ decreases from 1 to 0
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°, x = r and y = 0. So we have 10cos000sin ====rr,
roo . When OP is in the position
OB , θ = 90°, x = 0 and y = r. So, we have 0090cos190sin ====r
,rr oo .
Now, we have ,010
0cos0sin0tan ===o
oo
01
0sin0cos0cot ==o
oo not defined,
111
0cos10sec ===
o
o ; 01
0sin10 eccos ==
o
o not defined,
01
90cos90sin90tan ==
o
oo not defined, 0
10
90sin90cos90cot ===
o
oo ,
01
90cos190sec ==
o
o not defined, 111
90sin190eccos ===
o
o .
All the trigonometric ratios for angle of measures 0° , 30°, 45°, 60° and 90° are provided in the following table:
θ 0° 30° 45° 60° 90°
sin θ 0 21
21
23 1
cos θ 1 23 2
1 21 0
tan θ 0 3
1 1 3 Not defined
cot θ Not defined 3 1 3
1 0
sec θ 1 32 2 2 Not
defined
cosec θ Not defined 2 2 3
2 1
Table - 1
191
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°.
Solution: cos 30° = 23 , tan 60° = 3 , sec 45° = 2 , sin 60° =
23 .
∴2cos230° tan260° − sec2 45° sin2 60°
= 2 ( ) ( )2
222
2323
23
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛
= 2 × 43 × 3 – 2 ×
43 =
29−
23 = 3.
Example 9: Find the acute angle A if tan A = o
o
60cos160sin
+.
Solution: sin 60° = 23 , cos 60° =
21
∴ tan A = 3
133
2323
211
23
===+
. But tan 30° = 3
1 . ∴ A = 30°.
Example 10: If 2 sin (A + B) = 3 and 1cos2 =B , find A and B.
Solution: Since 2 sin (A + B) = 3 , we get sin (A + B) = 23 .
But sin 60° = 23 . So A + B = 60°. (1)
Since 2 cos B = 1, we get cos B = 2
1 .
But cos 45° = 2
1 . So B = 45°. (2)
Solving (1) and (2), A = 15°.
192
Exercise 8.1
In problems 1 to 4 find the indicated trigonometric ratios in the given right triangle. 1. 2.
sin B, cos C, tan B sec X, cot Z, cosec Z
Figure 8.14 Figure 8.15
tan M, sec N, cosec N
3. 4.
cos Q , tan R, cot Q
Figure 8.16
Figure 8.17
In problems 5 to 10 find the other trigonometric ratios.
5. cos θ = 53 6. sin θ =
1312 7. sec θ =
32
8. cosec θ = 10 9. cot θ = 71 10. tan θ =
52
11. If 3735cos =A , find
AAAA
tansectansec
−+ 12. If sin θ =
53 , find
θθθ
coscoteccos−
13. If cosec θ = 2, find the value of cot θ + θθ
cos1sin+
.
193
14. If cot θ = 3
1 , show that 53
sin2cos1
2
2
=−−
θθ .
15. If 3 cot θ = 4, find the value of θθ
θθeccos3sec2
cos2sin3++ .
16. Evaluate (i) cosec2 45° cot230° + sin2 60° sec2 30° (ii) cos2 30° − sin2 30° − cos 60°
(iii) 8 sin2 60° cos 60° (iv) oo
o
60tan30tan45tan+
17. Verify the following: (i) sin230° + cos230° = 1 (ii) sec260° − 1 = tan2 60° (iii) 1 + cot2 30° = cosec2 30°
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.
Figure 8.18
Then OQ = x, PQ = y. Let OP = r.
Applying Pythagoras identity in the right triangle OQP, we get x2+ y2 = r2 .
Dividing both sides by r2, 2
2
2
22
rr
ryx
=+ (or)
12
2
2
2
=+ry
rx .
But sin θ = ry , cos θ =
rx . ∴ (cos θ)2+ (sin θ)2 = 1.
or cos2θ + sin2θ = 1 (1) Dividing both sides of (1) by cos2θ, we get
θθ
θθ22
22
cos1
cossincos
=+ (or)
2
2
2
2
2
cos1
cossin
coscos
⎟⎟⎠
⎞⎜⎜⎝
⎛=+
θθθ
θθ
22
cos1
cossin1 ⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛+
θθθ (or) 1 +(tan θ)2 = (sec θ)2
194
(or) 1 + tan2θ = sec2 θ (2)
Dividing both sides of (1) by sin2θ, we get
POOP
QOOQ
QPPQ
′=
′=
′′
(or) 22
2
2
2
2
)(cosecsin
1sinsin
sincos θ
θθθ
θθ
=⎟⎟⎠
⎞⎜⎜⎝
⎛=+ (or) cot2θ + 1 = cosec2θ
(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) ( a2 + b2 = (a + b)2 – 2ab) = (1)2 – 2sin2θ cos2θ = 1 – 2sin2θ cos2θ = R.H.S.
195
Example 12: Prove that θθ
sin1cos+
= secθ − tan θ.
Solution:
L.H.S. = θθ
sin1cos+
=θθ
θθ
sin1sin1
sin1cos
−−
×+
= θθ)(θ
2sin1sin1cos
−− =
θθ)(θ
2cossin1cos −
= θθ
cossin1− =
θθ
θ cossin
cos1
−
= sec θ − tan θ = R.H.S.
Example 13: Prove that 2)cotec(coscos1cos1 AA
AA
+=−+ .
Solution :
L.H.S = AA
AA
cos1cos1
cos1cos1
++
×−+
= A
A2
2
cos1)cos1(
−+ =
22
2
2
sincos
sin1
sincos1
sin)cos1(
⎟⎠⎞
⎜⎝⎛ +=⎥⎦
⎤⎢⎣⎡ +
=+
AA
AAA
AA
= (cosec A + cot A)2 = R.H.S. Alternately,
R.H.S. = (cosec A + cot A)2 =2
sincos
sin1
⎟⎟⎠
⎞⎜⎜⎝
⎛+
AA
A =
2
sincos1
⎟⎟⎠
⎞⎜⎜⎝
⎛ +A
A
= 212==
ADAC
= A
A2
2
cos1)cos1(
−+
= )cos1()cos1(
)cos1( 2
AAA−+
+ = AA
cos1cos1
−+ = L.H.S.
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.
Example 15: Prove that sec A – tan A = AA tansec
1+
.
Solution :
R.H.S. = AA tansec
1+
= AAAA
AA tansectansec
tansec1
−−
×+
= AA
AA22 tansec
tansec−− =
1tansec AA− = sec A – tan A = L.H.S.
196
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.
Example 17: Prove that θθ cos1
1cos11
−+
+ = 2cosec2θ.
Solution :
L.H.S. = θθ cos1
1cos11
−+
+ =
)cos1()cos1()cos1(1)cos1(1
θθθθ
−+++−
= θ
θθ2cos1
cos1cos1−
++− = θ2sin
2 = 2cosec2θ = R.H.S.
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 R.H.S = 4 mn = r
y
= AA 22 sintan4 − = AAA 2
2
2
sincossin4 −
= A
AAA2
222
coscossinsin4 − =
AAA
2
22
cos)cos1(sin4 −
= AA 22 tansin4 = 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.
197
Example 21: Prove that θθ cossin21+ = sinθ + cosθ
Solution: L.H.S = θθθθ cossin2cossin 22 ++ 1 = sin + cos θ2 θ2
= 2cossin θ)θ( + = sinθ + cosθ = R.H.S
Exercise 8.2 1. Prove that AAA secsin1sec2 =− . 2. Prove that (sin A + cos A)2 + (sin A – cos A)2 = 2.
3. Simplify : θθθecθ
22
22
tanseccoscot−− .
4. Prove that AAAA
tansectansec
−+ =
AA
sin1sin1
−+ .
5. Prove that θθθ
2sec2sin11
sin11
=−
++
.
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.
8. Prove that 1cos2tan1tan1 2
2
2
−=+− A
AA .
9. Prove that θθec cotcos
1−
= cosec θ + cot θ.
10. Prove that (tan A + cot A)2 = sec2 A+ cosec2A.
11. Prove that AAA
AA
A cossincot1
sintan1
cos+=
−+
−.
12. Prove that A
AAAAA
cossin1
1sectan1sectan +=
+−−+ .
13. Prove that (tan A – tan B)2 + (1 + tan A tan B)2 = sec2A sec2B.
14. Prove that θθ
cot1tan−
+ θθ
tan1cot−
= sec θ cosec θ +1.
15. Prove that θθ
θθθθ
cos1cos1
cossin1cossin1
2
+−
=⎟⎟⎠
⎞⎜⎜⎝
⎛++−+ .
16. Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2θ + cot2θ.
17. Prove that 2sincossincos
sincossincos 3333
=−−
+++
θθθθ
θθθθ .
18. Prove that θθθθθ 2
44
44
tancossin1sincos1
=+−+− .
198
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
OQPQ
OPOQ
OPPQ
=== θθθ tan,cos,sin ,
PQOQ
OQOP
PQOP
=== θθθ cot,sec,eccos
Similarly, for the angle 90°− θ, we get
sin (90° - θ) = 43 )90tan(,)90(cos
OPPQ
=−=− θθ oo
PQOP
OQOP
−=−=− 90(cot,)90(sec,)90(eccos θθθ ooo
………
Comparing (1) and (2), we get
)90(cossin θOPPQ
−== oθ
=OPOQ cos θ = sin (90° − θ)
OQPQ = tan θ = cot (90°− θ)
=PQOP cosec θ = sec (90° − θ)
OQOP = sec θ = cosec (90° − θ)
PQOQ = cot θ = tan (90° − θ).
Hence, we obtain the following table:
sin (90° − θ) = cocos (90° − θ) = stan (90° − θ) = ccot (90° − θ) = tasec (90° − θ) = ccosec (90° − θ) =
199
Figure 8.19
,PQOQ
OQPQ
=) ……….. (2)
.. (1)
sθ in θ ot θ n θ
osec θ sec θ
Example 22: Evaluate o
o
25cot65tan .
Solution : tan 65° = tan(90° − 25°) = cot 25°
∴ o
o
25cot65tan =
o
o
25cot25cot = 1.
Example 23: Evaluate sin 20° tan 60° sec 70°
Solution : sec 70° = sec(90°−20°) = cosec 20°=o20sin
1
∴sin 20° tan 60° sec 70° = sin 20° tan 60° cosec 20°
=sin 20° × 3 × o20sin
1 = 3 .
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.
Exercise 8.3
1. Evaluate (i) o
o
54cos36sin (ii)
o
o
55cot35tan (iii) sin θ sec (90° − θ)
2. Simplify: (i) o
o
o
o
o
o
39eccos51sec
23
48cos42sin
21
57cot33tan
++ .
(ii) o
o
o
o
43eccos47sec4
67cos23sin3 + .
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
1. 125,
135,
135 2.
25,
21,
25 3.
158,
158,
178 4. 1, 2 , 2
5. 43cot
35sec
45eccos
34tan
54sin ===== θ,θ,θ,θ,θ .
200
6. 125cot
513sec
1213eccos
512tan
135cos ===== θ,θ,θ,θ,θ .
7. 3cot2eccos3
1tan23cos
21sin ===== θ,θ,θ,θ,θ .
8. 3cot310sec
31tan
103cos
101sin ===== θ,θ,θ,θ,θ .
9. 25sec7
25eccos7tan25
1cos25
7sin ===== θ,θ,θ,θ,θ .
10. 25cot
529sec
229eccos
295cos
292sin ===== θ,θ,θ,θ,θ .
11. 2549 12.
825 13. 15. 2
7534
16. (i) 7 (ii) 0 (iii) 3 (iv) 43
18. A = 60°, B = 30°
Exercise 8.2 3. −1 6. r2
Exercise 8.3
1. (i) 1 (ii) 1 (iii) 1
2. (i) 3 (ii) 7
3. (i) 30° (ii) 36° (iii) 65° (iv) 55°
201
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.
Figure 9.1
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.
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
that GDAG = .
12
25.25.4= That is G divides AD in the ratio 2:1.
9.1.2 Circumcentre
The perpendicular bisepoint is called the circumcentan equal distance R from the vthe equidistance R as radius pacircumcircle of the triangle an
To locate the circumceStep 1: Draw the triangle ABCStep 2: Draw the perpendiculaStep 3: Mark the point of intThis point S is the circumcentr
Figure 9.2
ctors of the sides of a triangle are concurrent at a point. This re of the triangle and is usually denoted by the letter S. It is at ertices of the triangle. The circle drawn with S as the centre and sses through the vertices of the triangle. This circle is called the d R is called its circumradius.
ntre and the circumcircle, we adopt the following procedure: . r bisectors of BC and .AC ersection of the perpendicular bisectors of BC and AC as S. e 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 9.4 Note: When the circumcircle of a triangle is drawncircumscribed. 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 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 touinternally. The circle is said to be inscribed in the triangle atriangle (see Figure 9.5)
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Figure 9.3
Figure 9.5
, we say that the triangle is
ches all the sides of the triangle nd it is called the incircle of the
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 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 i Figure 9.7 9.1.4 Orthocentre
We recall that an altitude of a triangle is a perpendicular lin
206
Figure 9.6
ncircle of the triangle.
e segment drawn from a
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 H, the orthocentre of ∆ABC.
Figure 9.8
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 procedure.
Rough Figure
Figure 9.9
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. If we locate the centrowe observe that their p
Figure 9.10
id, circumcentre, incentre and orthocentre of various types of triangles, ositions are as given below:207
Table
Point of concurrence Type of triangle Location of the point of
concurrence Centroid Any type of triangle Inside the triangle
Acute angled triangle Inside the triangle Right triangle Mid point of the hypotenuse Circumcentre Obtuse angled triangle Outside the triangle
Incentre Any type of triangle Inside the triangle Acute angled triangle Inside the triangle Right triangle Vertex of the right angle Orthocentre Obtuse angled triangle Outside the triangle
Exercise 9.1
In problems 1 to 4, locate the centroid G of the given triangle.
1. ∆ABC, where BC = 6cm, m∠B = 40º, m∠C = 60º. 2. ∆ABC, where all sides are of 6.5 cm long. 3. ∆PQR, where m∠R = 90º, PQ = 7 cm, PR = 6 cm. 4. ∆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. ∆ABC, where AB = 8 cm, BC = 5cm, AC = 7 cm. 6. ∆PQR, where PQ = 5 cm, PR = 4.5 cm, m∠P = 100º. 7. ∆XYZ, where XY = 7 cm, m∠X = 70º, m∠Y = 60º. 8. ∆PQR, where each side is of length 5.5 cm.
In problems 9 to 12, draw the incircle and measure its inradius.
9. ∆ABC, where AB = 9cm, BC = 7cm, CA = 5cm. 10. ∆XYZ, where XY = YZ = ZX = 8cm. 11. ∆PQR, where PQ = 10cm, m∠P = 90º, m∠Q = 60º. 12. ∆ABC, where AB = 5.4 cm, m∠A = 50º, AC = 5cm.
In problems 13 to 16, locate the orthocentre of the triangle.
13. ∆ABC, where BC = 5.6cm, m∠B = 55º, m∠C = 65º. 14. ∆PQR, where m∠P = 90º, m∠Q = 30º, PQ = 4.5 cm. 15. ∆LMN, where, LM = 7cm, m∠M = 130º, MN = 6 cm. 16. ∆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 2
ba + . To find the number
2ba + geometrically, we proceed as follows:
Step 1: Take a line segment PQ whose length is a + b. Step 2: Draw the perpendicular bisector of PQ . Step 3: The meeting point of PQ and its perpendicular bisector is marked and its distance
from P or Q is measured. This distance gives 2
ba + . The reason is that if a line segment XY
is of length l, then the midpoint of XY is at a distance 2l from X or Y.
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. 9.2.2 The g
Let a
x = ,ab the
proportionalmean proportproceed as fo
Figure 9.11 eometric mean or mean proportional between two numbers a and b and b be two positive numbers. Then the geometric mean of a and b is .ab If
n x2 = ab or x × x = ab or bx
xa= or a : x = x : b. So x is also called the mean
between a and b. (If a number x is such that a : x = x : b, then x is called the ional between a and b). To find ab through geometrical constructions, we llows:
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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.
Figure 9.12
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
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.
Then BDAB =
BCBD or BD2 = AB × BC or
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 .694 =× 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. Example 8: Find geoSolution: We observ
12 = .34× So considered as the mebetween 4 and 3. Appconstruction for gettiproportional between Step 1: Cut off line sBC on a line AX SuBC = 3. Step 2: Draw the circof its diameter. Step 3: Draw the choperpendicular to ACStep 4: Measure the find that it is 3.4 cm.
Figure 9.14
metrically the value of .12 e that 12 can be
an proportional lying the geometrical
ng the mean 4 and 3, we get
egments AB and ch that AB = 4 and
le with AC as one
rd DE through B . length of BD . We
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 6. 3.7 cm 7. 4.7 cm 8. 3.2 cm 9. 1.65 cm 10. 2.3 cm 11. 3.7 cm 12. 1.5 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
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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 Tally bars Frequency30-34 |||| 4 35-39 |||| 5 40-44 |||| ||| 8 45-49 |||| |||| 9 50-54 || 2 55-59 || 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 Tally bars Frequency29.5-34.5 |||| 4 34.5-39.5 |||| 5 39.5-44.5 |||| ||| 8 44.5-49.5 |||| |||| 9 49.5-54.5 || 2 54.5-59.5 || 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 c.f29.5-34.5 32 4 4 34.5-39.5 37 5 9 39.5-44.5 42 8 1744.5-49.5 47 9 2649.5-54.5 52 2 2854.5-59.5 57 2 30 Total 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 ( nxxx +++ ...21 ) − n × x = 0.
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Hence
.n
x...xxx n21 +++=
In mathematics, the symbol ∑, called sigma notation is used to represent summation. With
this symbol, the sum nxxx +++ ...21 is denoted as or simply as ∑=
n
iix
1∑ ix . Then, we have
.nx
x i∑=
If the observations are represented in the form of a frequency table, the mean x is given by
n
nn
fffxfxfxf
x++++++
=....
...
21
2211 ,
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
x = N
xf ii∑ , where N = nfff +++ ....21 .
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
.,...,, 21
cAx
cAx
cAx n −−−
Let them be d1, d2, …,dn.. Then
∑ ii df = f1 × d1 + f2 × d2 + …..+ fn × dn = ....22
11 c
Axfc
Axfc
Axf nn
−×++
−×+
−×
= ( ) ( ) ([ ]AfxfAfxfAfxfc nnn −++−+− ...1
222111 )
= ( ) ( )[ ]nnn fffAxfxfxfc
+++−+++ ......1212211 = [ ]NAxf
c ii ×−∑1.
∴ NAxf ii ×−∑ = c × ∑ ii df
or or iiii dfcNAxf ∑∑ +×= x =
Nxf ii∑ = A + c×
Ndf ii∑ .
Example 1: Calculate the mean of the data 9, 11, 13, 15, 17, 19.
Solution : x = N
xi∑ = 6
19 17 15 13 11 9 +++++ = 14.
Example 2: Compute the A.M. of the following data:
x 10 11 13 15 16 19 f 4 5 8 6 4 3
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Solution:
Short-cut Method: Take A = 14, c = 1, d = x − A
x f d f × d 10 4 −4 −16 11 5 −3 −15 13 8 −1 − 8 15 6 1 6 16 4 2 8 19 3 5 15
Total N=30 ∑fd =29−39 = −10
x = A+ c × N
fd∑ =14 + 1 × 3010−
≈14 − 0.33 = 13.67.
Direct Method:
x f f × x 10 4 40 11 5 55 13 8 104 15 6 90 16 4 64 19 3 57
Total N = 30 ∑ fx = 410
∴ x = N
xf ii∑ = 30410 = 13.67.
Example 3: Calculate the A.M. for the following data:
Marks 80 85 90 95 100 No. of students 5 6 6 2 1
Solution: Take A = 90 , c = 5 and c
Axd −= .
X f d f × d
80 5 −2 −10
85 6 −1 −6
90 6 0 0
95 2 1 2
100 1 2 2
N = 20 ∑ fd = −12
∴ A.M. = x = A+ c × N
fd∑
= 90 + 5 ×2012− = 90 − 3 = 87.
Example 4: Calculate A .M for the following data:
Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Marks 12 18 27 20 17 6
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Solution: Direct Method:
Class Mid-value x Frequency f f × x
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
N = 100 ∑ fx = 2800
From the table, we get N = the total frequency = 100, ∑ fx = 2800.
∴ x =N
fx∑ = 1002800 = 28.
Short-cut Method:
Let A = 30, c = 5 and c
Axd −= .
Class Mid-value x d Frequency f f × d 0-10 5 −5 12 −60 10-20 15 −3 18 −54 20-30 25 −1 27 −27 30-40 35 1 20 20 40-50 45 3 17 51 50-60 55 5 6 30
N = 100 ∑ fd = 101−141= −40
∴ A.M. = x = A+ c × N
fd∑
= 30 + 5 ×100
40− = 30 − 2 = 28.
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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X 5 10 15 20 25 30 f 4 5 7 4 3 2
4. Obtain the A.M. of the following:
Marks 65 70 75 80 85 90 100 No. of students 11 6 3 6 4 10 4
5. Find the A. M. of the following data:
Variable 15 25 35 45 55 65 75 85 Frequency 12 20 15 14 16 11 7 8
6. Find the Arithmetic mean from the following table:
Class interval 10-20 20-30 30-40 40-50 50-60
Frequency 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,
namely 552
1102
5753==
+ as the middle most value for the series. This is the median of the
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
observations. If N is an odd integer, then there is only one middle term and it is the ⎟⎠⎞
⎜⎝⎛ +
21N th
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 2N th term and ⎟
⎠⎞
⎜⎝⎛ +1
2N 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:
First, form the cumulative frequency column and then find the value of 2N where N is
the total frequency. Then we locate the class interval or the value of the variate in the table for
which the cumulative frequency is either equal to2N or just greater than
2N . This is the
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.
This is an odd number. So the median = ⎟⎠⎞
⎜⎝⎛ +
21N th term = ⎟
⎠⎞
⎜⎝⎛ +
215 th term
= 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.
∴ Median = ⎟⎠⎞
⎜⎝⎛ +
21N th term = ⎟
⎠⎞
⎜⎝⎛ +
2115 th term = 8th term = 49.
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.
∴ Median = ⎟⎠⎞
⎜⎝⎛ +
21N th term = ⎟
⎠⎞
⎜⎝⎛ +
217 th term = 4th term = 28.
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.
∴ Median = average of 2N th and ⎟
⎠⎞
⎜⎝⎛ +1
2N th terms = average of 5th and 6th terms
= average of 26 and 28 = 2
2826 + = 27.
220
Example 9: Calculate the median of the following table:
Variable ( x) 5 10 15 20 25 30 Frequency( f ) 3 6 10 8 2 3
Solution:
x f Cumulative frequency
5 3 3 10 6 9 15 10 19 20 8 27 25 2 29 30 3 32
Total frequency = N = ∑f = 32 and so 2N = 16.
The median is the ⎟⎠⎞
⎜⎝⎛
2N th value = 16th value. But the 16th value occurs in the class whose
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
Marks 24 20 35 52 50 48 No. of students 4 7 3 9 5 12
3. Find the median for the following data
x 1 2 3 4 5 6 f 4 6 5 3 2 5
4. The wages of 43 employees are given. Find the median.
Wage 25 35 45 55 65 No. of Employees 3 5 20 10 5
221
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 45 50 55 60 65 70 75 No. of Employees 12 11 14 13 12 10 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) 84, 91, 72, 68, 87, 84 (ii) 65, 61, 72, 81, 51, 31 (iii) 38, 31, 22, 20, 31, 61, 15, 20 (iv) 15, 11, 18, 23, 11, 19, 11
222
2. Find the mode for the following distribution
x 10 20 30 40 50 60 f 8 15 12 10 9 6
3. Find the mode for the following table.
x 60 61 62 63 64 65 f 5 8 14 16 10 7
Answers
Exercise 10.1.1
1. 15 2. 67.1 3. 15.6 4. 79.1 5. 45 6. 35.4
Exercise 10.1.2 1. (i) 57.5 (ii) 35.5 (iii) 60 (iv) 41 2. 48 3. 3 4. 45
Exercise 10.1.3 1. (i) 84 (ii) No mode (iii) 20, 31 (iv) 11 2. 20 3. 63
223
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.
224
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
Figure 11.1
(see 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 the graph sheet by taking 1 cm = 1 unit for both the axes. We join the points by a line segment and 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).
225
Figure 11.2
extend it in both directions. We get the
Figure 11.3
Example 4: Draw the graph of the line whose slope is 23− and y-intercept is −3.
Solution: The equation of the line is y = mx + c or
y = 23− x + (−3) or y =
23− x −3
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 −2 0 2 y 0 −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). Example 5: Draw the graph of the line 2x + 3y = Solution: The given equation is rewritten as
3y = −2x + 12 or y = ⎟⎠⎞
⎜⎝⎛ −
32 x + 4.
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 −3 0 3 y 6 4 2
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 thequation x = 3. So, any value of y gives x = 3Choose any two values say 1 and 2 for y and getwo points (3, 1) and (3, 2)on the line x = 3.
x 3 3 y 1 2
226
Figure 11.4
12.
e . t
Figure 11.5
Figure 11.6
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 −2 2 y −4 −4
Join the two points and extend it in the both directionsgraph(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) ( (iii) (−3, 2) and (5, −1) (iv) (2. Draw the graph of the following: (i) y = −2x (ii) y = 3x (iii) x = 53. Draw the graphs of the following: (i) x = −3 (ii) y = 5 (iii) (iv) y = −4 (v) 2x + 3 = 0. (vi) 4. Draw the graph for y = mx + c when
(i) m = 3 and c = 4. (ii) m
(iii) m = −3 and c = −4 (iv) m5. Draw the graph of the following equations: (i) 2x + 3y = 12. (ii) (iii) y + 2x −5 = 0. (iv)
227
Figure 11.7
. The required graph is the resulting
−1, 0) and (−2, −5) −2, −3) and (5, −4)
y (iv) x = −4y
x = 5 1 + 2y =0.
= 32− and c = 3.
= 2 and c = −5
x −5y = 10. x −2y + 1 = 0.
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
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.
22
Line 2: 2y = −4x + 2 i.e., y = −2x + 1
x −1 1 y 3 −1
x −1 1 y 3 −1
Figure 11.8
8
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 2: x −2y = −6 or 2y = x + 6
or y = ⎟⎠⎞
⎜⎛ x + 3 ⎝ 2
1
x 0 2 y 3 4
Line 1: x − 2y = 4 or 2y = x −4
or y = ⎟⎠⎞
⎜⎝⎛
21 x −2
X 0 2 Y −2 −1
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). Example 10: Solve graphically the simultaneous equSolution: We draw the graphs for the two equations Line (1): x + y = 5
or y = −x + 5
X −2 −1 3 Y 7 6 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
Figure 11.9
ations x + y = 5, x − y = 3. in the same graph sheet.
Line (2) : x −y = 3 or y = x −3
x 1 0 3 y −2 −3 0
Figure 11.10
Exercise 11.2 In problems 1 to 10, solve graphically the fo
1. x + y = 0, x = 4. 2. x − y = 0, y = −3. 3. x + y = 2, x − y = 2. 4. x −y = 6, 2x + y = 9. 5. x + y = 5, x − y = 1.
Answers
Exercise 11.1
1. (i)
1. (ii)
1. (iii)
1. (iv)
2 (i)
x −1 0 1 y 2 0 −2
2. (ii)
230
x −1 0 1 y −3 0 3
llowing system of equations: 6. 2x + y = 1, 4x + 2y = 2. 7. x + 2y = 4, x + 2y = 6. 8. x − 3y = 4, x + 2y = −1. 9. 3x + y = 2, 6x − y = 7. 10. 2x + 3 = 0, 4x + y + 4 = 0.
2. (iii) 2. (iv) x 0 5 −5
y 0 1 −1
3. (i) (ii) x −3 −3 −3
y 1 2 −1
(iii) (iv) x 5 5 5
y 0 1 2
x 0 4 −4 y 0 −1 1
x 1 2 −1 y 5 5 5 x 0 1 −1 y −4 −4 −4231
(v) (vi) x −1.5 −1.5 −1.5
y 0 1 −1
4 (i) x −1 0 1
y 1 4 7
(iii) x 0 1 −1
y −4 −7 −1
x 0 1 −1 y −0.5 −0.5 −0.5
(ii)
x −3 0 3 y 5 3 1(iv)
x 0 1 2 y −5 −3 −1232
5 (i) (ii) x −3 0 3
y 6 4 2
(iii) (iv) x −1 0 1
y 7 5 3
Exercise 11.2
1.
x −1 0 1
y 1 0 −1
Solution is x = 4; y = −4. 2.
x −1 0 1
y −1 0 1
Solution is x = −3; y = −3.
233
x 5 0 −5 y −1 −2 −3
x 1 3 5 y 1 2 3
x 4 4 4
y 1 0 −1
x 0 1 2
y −3 −3 −3
3.
x 0 1 2
y 2 1 0
Solution is x = 2; y = 0.
4.
x 3 4 5
y −3 −2 −1
Solution is x = 5; y = −1. 5.
x 2 3 4
y 3 2 1
Solution is x = 3; y =2. 6. Infinitely many solutions. 7.
x 0 2 4
y 2 1 0
No solution.
234
x 3 4 5
y 1 2 3
x 2 3 4
y 5 3 1
x 2 3 4
y 1 2 3
x 0 2 4
y 3 2 1
x -1 0 1 y 3 1 −1
x −1 0 1 y 3 1 −1
8.
x −2 1 4
y −2 −1 0
Solution is x = 1; y = −1. 9.
x 0 1 2
y 2 −1 −4
Solution is x = 1; y = −1. 10.
x 23−
23−
23−
y −2 0 2
Solution is x =
23− ; y = 2.
235
x −1 1 3
y 0 −1 −2
x 0 1 2
y −7 −1 5
x −2 −1 0
y 4 0 −4
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 .0414 .0453 .0492 .0531 .0569 .0607 .0645 .0682 .0719 .0755 4 8 11 15 19 23 26 30 34 1.2 .0792 .0828 .0864 .0899 .0934 .0969 .1004 .1038 .1072 .1106 3 7 10 14 17 21 24 28 31 1.3 .1139 .1173 .1206 .1239 .1271 .1303 .1335 .1367 .1399 .1430 3 6 10 13 16 19 23 26 29 1.4 .1461 .1492 .1523 .1553 .1584 .1614 .1644 .1673 .1703 .1732 3 6 9 12 15 18 21 24 27 1.5 .1761 .1790 .1818 .1847 .1875 .1903 .1931 .1959 .1987 .2014 3 6 8 11 14 17 20 22 25
1.6 .2041 .2068 .2095 .2122 .2148 .2175 .2201 .2227 .2253 .2279 3 5 8 11 13 16 18 21 24 1.7 .2304 .2330 .2355 .2380 .2405 .2430 .2455 .2480 .2504 .2529 2 5 7 10 12 15 17 20 22 1.8 .2553 .2577 .2601 .2625 .2648 .2672 .2695 .2718 .2742 .2765 2 5 7 9 12 14 16 19 21 1.9 .2788 .2810 .2833 .2856 .2878 .2900 .2923 .2945 .2967 .2989 2 4 7 9 11 13 16 18 20 2.0 .3010 .3032 .3054 .3075 .3096 .3118 .3139 .3160 .3181 .3201 2 4 6 8 11 13 15 17 19
2.1 .3222 .3243 .3263 .3284 .3304 .3324 .3345 .3365 .3385 .3404 2 4 6 8 10 12 14 16 18 2.2 .3424 .3444 .3464 .3483 .3502 .3522 .3541 .3560 .3579 .3598 2 4 6 8 10 12 14 15 17 2.3 .3617 .3636 .3655 .3674 .3692 .3711 .3729 .3747 .3766 .3784 2 4 6 7 9 11 13 15 17 2.4 .3802 .3820 .3838 .3856 .3874 .3892 .3909 .3927 .3945 .3962 2 4 5 7 9 11 12 14 16 2.5 .3979 .3997 .4014 .4031 .4048 .4065 .4082 .4099 .4116 .4133 2 3 5 7 9 10 12 14 15
2.6 .4150 .4166 .4183 .4200 .4216 .4232 .4249 .4265 .4281 .4298 2 3 5 7 8 10 11 13 15 2.7 .4314 .4330 .4346 .4362 .4378 .4393 .4409 .4425 .4440 .4456 2 3 5 6 8 9 11 13 14 2.8 .4472 .4487 .4502 .4518 .4533 .4548 .4564 .4579 .4594 .4609 2 3 5 6 8 9 11 12 14 2.9 .4624 .4639 .4654 .4669 .4683 .4698 .4713 .4728 .4742 .4757 1 3 4 6 7 9 10 12 13 3.0 .4771 .4786 .4800 .4814 .4829 .4843 .4857 .4871 .4886 .4900 1 3 4 6 7 9 10 11 13
3.1 .4914 .4928 .4942 .4955 .4969 .4983 .4997 .5011 .5024 .5038 1 3 4 6 7 8 10 11 12 3.2 .5051 .5065 .5079 .5092 .5105 .5119 .5132 .5145 .5159 .5172 1 3 4 5 7 8 9 11 12 3.3 .5185 .5198 .5211 .5224 .5237 .5250 .5263 .5276 .5289 .5302 1 3 4 5 6 8 9 10 12 3.4 .5315 .5328 .5340 .5353 .5366 .5378 .5391 .5403 .5416 .5428 1 3 4 5 6 8 9 10 11 3.5 .5441 .5453 .5465 .5478 .5490 .5502 .5514 .5527 .5539 .5551 1 2 4 5 6 7 9 10 11
3.6 .5563 .5575 .5587 .5599 .5611 .5623 .5365 .5647 .5658 .5670 1 2 4 5 6 7 8 10 11 3.7 .5682 .5694 .5705 .5717 .5729 .5740 .5752 .5763 .5775 .5786 1 2 3 5 6 7 8 9 10 3.8 .5798 .5809 .5821 .5832 .5843 .5855 .5866 .5877 .5888 .5899 1 2 3 5 6 7 8 9 10 3.9 .5911 .5922 .5933 .5944 .5955 .5966 .5977 .5988 .5999 .6010 1 2 3 4 5 7 8 9 10 4.0 .6021 .6031 .6042 .6053 .6064 .6075 .6085 .6096 .6107 .6117 1 2 3 4 5 6 8 9 10
4.1 .6128 .6138 .6149 .6160 .6170 .6180 .6191 .6201 .6212 .6222 1 2 3 4 5 6 7 8 9 4.2 .6232 .6243 .6253 .6263 .6274 .6284 .6294 .6304 .6314 .6325 1 2 3 4 5 6 7 8 9 4.3 .6335 .6345 .6355 .6365 .6375 .6385 .6395 .6405 .6415 .6425 1 2 3 4 5 6 7 8 9 4.4 .6435 .6444 .6454 .6464 .6474 .6484 .6493 .6503 .6513 .6522 1 2 3 4 5 6 7 8 9 4.5 .6532 .6542 .6551 .6561 .6571 .6580 .6590 .6599 .6609 .6618 1 2 3 4 5 6 7 8 9
4.6 .6628 .6637 .6646 .6656 .6665 .6675 .6684 .6693 .6702 .6712 1 2 3 4 5 6 7 7 8 4.7 .6721 .6730 .6739 .6749 .6758 .6767 .6776 .6785 .6794 .6803 1 2 3 4 5 5 6 7 8 4.8 .6812 .6821 .6830 .6839 .6848 .6857 .6866 .6875 .6884 .6893 1 2 3 4 4 5 6 7 8 4.9 .6902 .6911 .6920 .6928 .6937 .6946 .6955 .6964 .6972 .6981 1 2 3 4 4 5 6 7 8 5.0 .6990 .6998 .7007 .7016 .7024 .7033 .7042 .7050 .7059 .7067 1 2 3 3 4 5 6 7 8
5.1 .7076 .7084 .7093 .7101 .7110 .7118 .7126 .7135 .7143 .7152 1 2 3 3 4 5 6 7 8 5.2 .7160 .7168 .7177 .7185 .7193 .7202 .7210 .7218 .7226 .7235 1 2 2 3 4 5 6 7 7 5.3 .7243 .7251 .7259 .7267 .7275 .7284 .7292 .7300 .7308 .7316 1 2 2 3 4 5 6 6 7 5.4 .7324 .7332 .7340 .7348 .7356 .7364 .7372 .7380 .7388 .7396 1 2 2 3 4 5 6 6 7
236
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 .7482 .7490 .7497 .7505 .7513 .7520 .7528 .7536 .7543 .7551 1 2 2 3 4 5 5 6 7 5.7 .7559 .7566 .7574 .7582 .7589 .7597 .7604 .7612 .7619 .7627 1 2 2 3 4 5 5 6 7 5.8 .7634 .7642 .7649 .7657 .7664 .7672 .7679 .7686 .7694 .7701 1 1 2 3 4 4 5 6 7 5.9 .7709 .7716 .7723 .7731 .7738 .7745 .7752 .7760 .7767 .7774 1 1 2 3 4 4 5 6 7 6.0 .7782 .7789 .7796 .7803 .7810 .7818 .7825 .7832 .7839 .7846 1 1 2 3 4 4 5 6 6
6.1 .7853 .7860 .7868 .7875 .7882 .7889 .7896 .7903 .7910 .7917 1 1 2 3 4 4 5 6 6 6.2 .7924 .7931 .7938 .7945 .7952 .7959 .7966 .7973 .7980 .7987 1 1 2 3 3 4 5 6 6 6.3 .7993 .8000 .8007 .8014 .8021 .8028 .8035 .8041 .8048 .8055 1 1 2 3 3 4 5 5 6 6.4 .8062 .8069 .8075 .8082 .8089 .8096 .8102 .8109 .8116 .8122 1 1 2 3 3 4 5 5 6 6.5 .8129 .8136 .8142 .8149 .8156 .8162 .8169 .8176 .8182 .8189 1 1 2 3 3 4 5 5 6
6.6 .8195 .8202 .8209 .8215 .8222 .8228 .8235 .8241 .8248 .8254 1 1 2 3 3 4 5 5 6 6.7 .8261 .8267 .8274 .8280 .8287 .8293 .8299 .8306 .8312 .8319 1 1 2 3 3 4 5 5 6 6.8 .8325 .8331 .8338 .8344 .8351 .8357 .8363 .8370 .8376 .8382 1 1 2 3 3 4 4 5 6 6.9 .8388 .8395 .8401 .8407 .8414 .8420 .8426 .8432 .8439 .8445 1 1 2 2 3 4 4 5 6 7.0 .8451 .8457 .8463 .8470 .8476 .8482 .8488 .8494 .8500 .8506 1 1 2 2 3 4 4 5 6
7.1 .8513 .8519 .8525 .8531 .8537 .8543 .8549 .8555 .8561 .8567 1 1 2 2 3 4 4 5 5 7.2 .8573 .8579 .8585 .8591 .8597 .8603 .8609 .8615 .8621 .8627 1 1 2 2 3 4 4 5 5 7.3 .8633 .8639 .8645 .8651 .8657 .8663 .8669 .8675 .8681 .8686 1 1 2 2 3 4 4 5 5 7.4 .8692 .8698 .8704 .8710 .8716 .8722 .8727 .8733 .8739 .8745 1 1 2 2 3 4 4 5 5 7.5 .8751 .8756 .8762 .8768 .8774 .8779 .8785 .8791 .8797 .8802 1 1 2 2 3 3 4 5 5
7.6 .8808 .8814 .8820 .8825 .8831 .8837 .8842 .8848 .8854 .8859 1 1 2 2 3 3 4 5 5 7.7 .8865 .8871 .8876 .8882 .8887 .8893 .8899 .8904 .8910 .8915 1 1 2 2 3 3 4 4 5 7.8 .8921 .8927 .8932 .8938 .8943 .8949 .8954 .8960 .8965 .8971 1 1 2 2 3 3 4 4 5 7.9 .8976 .8982 .8987 .8993 .8998 .9004 .9009 .9015 .9020 .9025 1 1 2 2 3 3 4 4 5 8.0 .9031 .9036 .9042 .9047 .9053 .9058 .9063 .9069 .9074 .9079 1 1 2 2 3 3 4 4 5
8.1 .9085 .9090 .9096 .9101 .9106 .9112 .9117 .9122 .9128 .9133 1 1 2 2 3 3 4 4 5 8.2 .9138 .9143 .9149 .9154 .9159 .9165 .9170 .9175 .9180 .9186 1 1 2 2 3 3 4 4 5 8.3 .9191 .9196 .9201 .9206 .9212 .9217 .9222 .9227 .9232 .9238 1 1 2 2 3 3 4 4 5 8.4 .9243 .9248 .9253 .9258 .9263 .9269 .9274 .9279 .9284 .9289 1 1 2 2 3 3 4 4 5 8.5 .9294 .9299 .9304 .9309 .9315 .9320 .9325 .9330 .9335 .9340 1 1 2 2 3 3 4 4 5
8.6 .9345 .9350 .9355 .9360 .9365 .9370 .9375 .9380 .9385 .9390 1 1 2 2 3 3 4 4 5 8.7 .9395 .9400 .9405 .9410 .9415 .9420 .9425 .9430 .9435 .9440 0 1 1 2 2 3 3 4 4 8.8 .9445 .9450 .9455 .9460 .9465 .9469 .9474 .9479 .9484 .9489 0 1 1 2 2 3 3 4 4 8.9 .9494 .9499 .9504 .9509 .9513 .9518 .9523 .9528 .9533 .9538 0 1 1 2 2 3 3 4 4 9.0 .9542 .9547 .9552 .9557 .9562 .9566 .9571 .9576 .9581 .9586 0 1 1 2 2 3 3 4 4
9.1 .9590 .9595 .9600 .9605 .9609 .9614 .9619 .9624 .9628 .9633 0 1 1 2 2 3 3 4 4 9.2 .9638 .9643 .9647 .9653 .9657 .9661 .9666 .9671 .9675 .9680 0 1 1 2 2 3 3 4 4 9.3 .9685 .9689 .9694 .9699 .9703 .9708 .9713 .9717 .9722 .9727 0 1 1 2 2 3 3 4 4 9.4 .9731 .9736 .9741 .9745 .9750 .9754 .9759 .9763 .9768 .9773 0 1 1 2 2 3 3 4 4 9.5 .9777 .9782 .9786 .9791 .9795 .9800 .9805 .9809 .9814 .9818 0 1 1 2 2 3 3 4 4
9.6 .9823 .9827 .9832 .9836 .9841 .9845 .9850 .9854 .9859 .9863 0 1 1 2 2 3 3 4 4 9.7 .9868 .9872 .9877 .9881 .9886 .9890 .9894 .9899 .9903 .9908 0 1 1 2 2 3 3 4 4 9.8 .9912 .9917 .9921 .9926 .9930 .9934 .9939 .9943 .9948 .9952 0 1 1 2 2 3 3 4 4 9.9 .9956 .9961 .9965 .9969 .9974 .9978 .9983 .9987 .9991 .9996 0 1 1 2 2 3 3 3 4
237
238
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 1.023 1.026 1.028 1.030 1.033 1.035 1.038 1.040 1.042 1.045 0 0 1 1 1 1 2 2 2
.02 1.047 1.050 1.052 1.054 1.057 1.059 1.062 1.064 1.067 1.069 0 0 1 1 1 1 2 2 2
.03 1.072 1.074 1.076 1.079 1.081 1.084 1.086 1.089 1.091 1.094 0 0 1 1 1 1 2 2 2
.04 1.096 1.099 1.102 1.104 1.107 1.109 1.112 1.114 1.117 1.119 0 1 1 1 1 2 2 2 2
.05 1.122 1.125 1.127 1.130 1.132 1.135 1.138 1.140 1.143 1.146 0 1 1 1 1 2 2 2 2
.06 1.148 1.151 1.153 1.156 1.159 1.161 1.164 1.167 1.169 1.172 0 1 1 1 1 2 2 2 2
.07 1.175 1.178 1.180 1.183 1.186 1.189 1.191 1.194 1.197 1.199 0 1 1 1 1 2 2 2 2
.08 1.202 1.205 1.208 1.211 1.213 1.216 1.219 1.222 1.225 1.227 0 1 1 1 1 2 2 2 3
.09 1.230 1.233 1.236 1.239 1.242 1.245 1.247 1.250 1.253 1.256 0 1 1 1 1 2 2 2 3
.10 1.259 1.262 1.265 1.268 1.271 1.274 1.276 1.279 1.282 1.285 0 1 1 1 1 2 2 2 3
.11 1.288 1.291 1.294 1.297 1.300 1.303 1.306 1.309 1.312 1.315 0 1 1 1 2 2 2 2 3
.12 1.318 1.321 1.324 1.327 1.330 1.334 1.337 1.340 1.343 1.346 0 1 1 1 2 2 2 2 3
.13 1.349 1.352 1.355 1.358 1.361 1.365 1.368 1.371 1.374 1.377 0 1 1 1 2 2 2 3 3
.14 1.380 1.384 1.387 1.390 1.393 1.396 1.400 1.403 1.406 1.409 0 1 1 1 2 2 2 3 3
.15 1.413 1.416 1.419 1.422 1.426 1.429 1.432 1.435 1.439 1.442 0 1 1 1 2 2 2 3 3
.16 1.445 1.449 1.452 1.455 1.459 1.462 1.466 1.469 1.472 1.476 0 1 1 1 2 2 2 3 3
.17 1.479 1.483 1.486 1.489 1.493 1.496 1.500 1.503 1.507 1.510 0 1 1 1 2 2 2 3 3
.18 1.514 1.517 1.521 1.524 1.528 1.531 1.535 1.538 1.542 1.545 0 1 1 1 2 2 2 3 3
.19 1.549 1.552 1.556 1.560 1.563 1.567 1.570 1.574 1.578 1.581 0 1 1 1 2 2 3 3 3
.20 1.585 1.589 1.592 1.596 1.600 1.603 1.607 1.611 1.614 1.618 0 1 1 1 2 2 3 3 3
.21 1.622 1.626 1.629 1.633 1.637 1.641 1.644 1.648 1.652 1.656 0 1 1 2 2 2 3 3 3
.22 1.660 1.663 1.667 1.671 1.675 1.679 1.683 1.687 1.690 1.694 0 1 1 2 2 2 3 3 3
.23 1.698 1.702 1.706 1.710 1.714 1.718 1.722 1.726 1.730 1.734 0 1 1 2 2 2 3 3 4
.24 1.738 1.742 1.746 1.750 1.754 1.758 1.762 1.766 1.770 1.774 0 1 1 2 2 2 3 3 4
.25 1.778 1.782 1.786 1.791 1.795 1.799 1.803 1.807 1.811 1.816 0 1 1 2 2 2 3 3 4
.26 1.820 1.824 1.828 1.832 1.837 1.841 1.845 1.849 1.854 1.858 0 1 1 2 2 3 3 3 4
.27 1.862 1.866 1.871 1.875 1.879 1.884 1.888 1.892 1.897 1.901 0 1 1 2 2 3 3 3 4
.28 1.905 1.910 1.914 1.919 1.923 1.928 1.932 1.936 1.941 1.945 0 1 1 2 2 3 3 4 4
.29 1.950 1.954 1.959 1.963 1.968 1.972 1.977 1.982 1.986 1.991 0 1 1 2 2 3 3 4 4
.30 1.995 2.000 2.004 2.009 2.014 2.018 2.023 2.028 2.032 2.037 0 1 1 2 2 3 3 4 4
.31 2.042 2.046 2.051 2.056 2.061 2.065 2.070 2.075 2.080 2.084 0 1 1 2 2 3 3 4 4
.32 2.089 2.094 2.099 2.104 2.109 2.113 2.118 2.123 2.128 2.133 0 1 1 2 2 3 3 4 4
.33 2.138 2.143 2.148 2.153 2.158 2.163 2.168 2.173 2.178 2.183 0 1 1 2 2 3 3 4 4
.34 2.188 2.193 2.198 2.203 2.208 2.213 2.218 2.223 2.228 2.234 1 1 2 2 3 3 4 4 5
.35 2.239 2.244 2.249 2.254 2.259 2.265 2.270 2.275 2.280 2.286 1 1 2 2 3 3 4 4 5
.36 2.291 2.296 2.301 2.307 2.312 2.317 2.323 2.328 2.333 2.339 1 1 2 2 3 3 4 4 5
.37 2.344 2.350 2.355 2.360 2.366 2.371 2.377 2.382 2.388 2.393 1 1 2 2 3 3 4 4 5
.38 2.399 2.404 2.410 2.415 2.421 2.427 2.432 2.438 2.443 2.449 1 1 2 2 3 3 4 4 5
.39 2.455 2.460 2.466 2.472 2.477 2.483 2.489 2.495 2.500 2.506 1 1 2 2 3 3 4 5 5
.40 2.512 2.518 2.523 2.529 2.535 2.541 2.547 2.553 2.559 2.564 1 1 2 2 3 4 4 5 5
.41 2.570 2.576 2.582 2.588 2.594 2.600 2.606 2.612 2.618 2.624 1 1 2 2 3 4 4 5 5
.42 2.630 2.636 2.642 2.649 2.655 2.661 2.667 2.673 2.679 2.685 1 1 2 2 3 4 4 5 6
.43 2.692 2.698 2.704 2.710 2.716 2.723 2.729 2.735 2.742 2.748 1 1 2 3 3 4 4 5 6
.44 2.754 2.761 2.767 2.773 2.780 2.786 2.793 2.799 2.805 2.812 1 1 2 3 3 4 4 5 6
.45 2.818 2.825 2.831 2.838 2.844 2.851 2.858 2.864 2.871 2.877 1 1 2 3 3 4 5 5 6
.46 2.884 2.891 2.897 2.904 2.911 2.917 2.924 2.931 2.938 2.944 1 1 2 3 3 4 5 5 6
.47 2.951 2.958 2.965 2.972 2.979 2.985 2.992 2.999 3.006 3.013 1 1 2 3 3 4 5 5 6
.48 3.020 3.027 3.034 3.041 3.048 3.055 3.062 3.069 3.076 3.083 1 1 2 3 4 4 5 6 6
.49 3.090 3.097 3.105 3.112 3.119 3.126 3.133 3.141 3.148 3.155 1 1 2 3 4 4 5 6 6
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 3.236 3.243 3.251 3.258 3.266 3.273 3.281 3.289 3.296 3.304 1 2 2 3 4 5 5 6 7 0.52 3.311 3.319 3.327 3.334 3.342 3.350 3.357 3.365 3.373 3.381 1 2 2 3 4 5 5 6 7 0.53 3.388 3.396 3.404 3.412 3.420 3.428 3.436 3.443 3.451 3.459 1 2 2 3 4 5 6 6 7 0.54 3.467 3.475 3.483 3.491 3.499 3.508 3.516 3.524 3.532 3.540 1 2 2 3 4 5 6 6 7 0.55 3.548 3.556 3.565 3.573 3.581 3.589 3.597 3.606 3.614 3.622 1 2 2 3 4 5 6 7 7
0.56 3.631 3.639 3.648 3.656 3.664 3.673 3.681 3.690 3.698 3.707 1 2 3 3 4 5 6 7 8 0.57 3.715 3.724 3.733 3.741 3.750 3.758 3.767 3.776 3.784 3.793 1 2 3 3 4 5 6 7 8 0.58 3.802 3.811 3.819 3.828 3.837 3.846 3.855 3.864 3.873 3.882 1 2 3 4 4 5 6 7 8 0.59 3.890 3.899 3.908 3.917 3.926 3.936 3.945 3.954 3.963 3.972 1 2 3 4 5 5 6 7 8 0.60 3.981 3.990 3.999 4.009 4.018 4.027 4.036 4.046 4.055 4.064 1 2 3 4 5 6 6 7 8
0.61 4.074 4.083 4.093 4.102 4.111 4.121 4.130 4.140 4.150 4.159 1 2 3 4 5 6 7 8 9 0.62 4.169 4.178 4.188 4.198 4.207 4.217 4.227 4.236 4.246 4.256 1 2 3 4 5 6 7 8 9 0.63 4.266 4.276 4.285 4.295 4.305 4.315 4.325 4.335 4.345 4.355 1 2 3 4 5 6 7 8 9 0.64 4.365 4.375 4.385 4.395 4.406 4.416 4.426 4.436 4.446 4.457 1 2 3 4 5 6 7 8 9 0.65 4.467 4.477 4.487 4.498 4.508 4.519 4.529 4.539 4.550 4.560 1 2 3 4 5 6 7 8 9
0.66 4.571 4.581 4.592 4.603 4.613 4.624 4.634 4.645 4.656 4.667 1 2 3 4 5 6 7 9 10 0.67 4.677 4.688 4.699 4.710 4.721 4.732 4.742 4.753 4.764 4.775 1 2 3 4 5 7 8 9 10 0.68 4.786 4.797 4.808 4.819 4.831 4.842 4.853 4.864 4.875 4.887 1 2 3 4 6 7 8 9 10 0.69 4.898 4.909 4.920 4.932 4.943 4.955 4.966 4.977 4.989 5.000 1 2 3 5 6 7 8 9 10 0.70 5.012 5.023 5.035 5.047 5.058 5.070 5.082 5.093 5.105 5.117 1 2 4 5 6 7 8 9 11
0.71 5.129 5.140 5.152 5.164 5.176 5.188 5.200 5.212 5.224 5.236 1 2 4 5 6 7 8 10 11 0.72 5.248 5.260 5.272 5.284 5.297 5.309 5.321 5.333 5.346 5.358 1 2 4 5 6 7 9 10 11 0.73 5.370 5.383 5.395 5.408 5.420 5.433 5.445 5.458 5.470 5.483 1 3 4 5 6 8 9 10 11 0.74 5.495 5.508 5.521 5.534 5.546 5.559 5.572 5.585 5.598 5.610 1 3 4 5 6 8 9 10 12 0.75 5.623 5.636 5.649 5.662 5.675 5.689 5.702 5.715 5.728 5.741 1 3 4 5 7 8 9 10 12
0.76 5.754 5.768 5.781 5.794 5.808 5.821 5.834 5.848 5.861 5.875 1 3 4 5 7 8 9 11 12 0.77 5.888 5.902 5.916 5.929 5.943 5.957 5.970 5.984 5.998 6.012 1 3 4 5 7 8 10 11 12 0.78 6.026 6.039 6.053 6.067 6.081 6.095 6.109 6.124 6.138 6.152 1 3 4 6 7 8 10 11 13 0.79 6.166 6.180 6.194 6.209 6.223 6.237 6.252 6.266 6.281 6.295 1 3 4 6 7 9 10 11 13 0.80 6.310 6.324 6.339 6.353 6.368 6.383 6.397 6.412 6.427 6.442 1 3 4 6 7 9 10 12 13
0.81 6.457 6.471 6.486 6.501 6.516 6.531 6.546 6.561 6.577 6.592 2 3 5 6 8 9 11 12 14 0.82 6.607 6.622 6.637 6.653 6.668 6.683 6.699 6.714 6.730 6.745 2 3 5 6 8 9 11 12 14 0.83 6.761 6.776 6.792 6.808 6.823 6.839 6.855 6.871 6.887 6.902 2 3 5 6 8 9 11 13 14 0.84 6.918 6.934 6.950 6.966 6.982 6.998 7.015 7.031 7.047 7.063 2 3 5 6 8 10 11 13 15 0.85 7.079 7.096 7.112 7.129 7.145 7.161 7.178 7.194 7.211 7.228 2 3 5 7 8 10 12 13 15
0.86 7.244 7.261 7.278 7.295 7.311 7.328 7.345 7.362 7.379 7.396 2 3 5 7 8 10 12 13 15 0.87 7.413 7.430 7.447 7.464 7.482 7.499 7.516 7.534 7.551 7.568 2 3 5 7 9 10 12 14 16 0.88 7.586 7.603 7.621 7.638 7.656 7.674 7.691 7.709 7.727 7.745 2 4 5 7 9 11 12 14 16 0.89 7.762 7.780 7.798 7.816 7.834 7.852 7.870 7.889 7.907 7.925 2 4 5 7 9 11 13 14 16 0.90 7.943 7.962 7.980 7.998 8.017 8.035 8.054 8.072 8.091 8.110 2 4 6 7 9 11 13 15 17
0.91 8.128 8.147 8.166 8.185 8.204 8.222 8.241 8.260 8.279 8.299 2 4 6 8 9 11 13 15 17 0.92 8.318 8.337 8.356 8.375 8.395 8.414 8.433 8.453 8.472 8.492 2 4 6 8 10 12 14 15 17 0.93 8.511 8.531 8.551 8.570 8.590 8.610 8.630 8.650 8.670 8.690 2 4 6 8 10 12 14 16 18 0.94 8.710 8.730 8.750 8.770 8.790 8.810 8.831 8.851 8.872 8.892 2 4 6 8 10 12 14 16 18 0.95 8.913 8.933 8.954 8.974 8.995 9.016 9.036 9.057 9.078 9.099 2 4 6 8 10 12 15 17 19
0.96 9.120 9.141 9.162 9.183 9.204 9.226 9.247 9.268 9.290 9.311 2 4 6 8 11 13 15 17 19 0.97 9.333 9.354 9.376 9.397 9.419 9.441 9.462 9.484 9.506 9.528 2 4 7 9 11 13 15 17 20 0.98 9.550 9.572 9.594 9.616 9.638 9.661 9.683 9.705 9.727 9.750 2 4 7 9 11 13 16 18 20 0.99 9.772 9.795 9.817 9.840 9.863 9.886 9.908 9.931 9.954 9.977 2 5 7 9 11 14 16 18 20
239