MATHEMATICS STANDARD IX TAMILNADU TEXTBOOK CORPORATION COLLEGE ROAD, CHENNAI

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1 MATHEMATICS STANDARD IX Untouchability is a sin Untouchability is a crime Untouchability is inhuman TAMILNADU TEXTBOOK CORPORATION COLLEGE ROAD, CHENNAI

2 Government of Tamilnadu First Edition 00 Revised Edition 004 Reprint 006 CHAIRPERSON S. UDAYABASKARAN, Reader in Mathematics, Presidency College (Autonomous), Chennai 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 Chennai 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 Ulundurpet 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 Kailasapuram, Trichy

3 CONTENTS PAGES. NUMBER SYSTEMS - 0. Number Systems. The Real Number Line 7. MEASUREMENTS Area and Perimeter. Combined Figures 6. SOME USEFUL NOTATION Scientific Notation 48. Notation of Logarithms 5. Set Notation 7 4. ALGEBRA Polynomials Algebraic Identities 9 4. Factorization Division of a Polynomial by a Polynomial 5. PROBLEM SOLVING TECHNIQUES Conjectures and Proofs 9 5. Mathematical Models 7 6. THEORETICAL GEOMETRY Theorems for Verification 6. Theorems with logical Proofs 5 7. ALGEBRAIC GEOMETRY The Cartesian Coordinates system Slope of a Line The Distance between any two points (x, y ) and (x, y ) 74

4 8. TRIGONOMETRY Trigonometric ratios Trigonometric Identities Trigonometric Ratios for Complementary Angles PRACTICAL GEOMETRY 0-9. Concurrency in a triangle 0 9. Geometric interpretation of averages HANDLING DATA - 0. Measures of Central Tendency 5. GRAPHS 4-5. Linear Graphs 4. Application of Linear Graphs 8 Logarithms 6-7 Anti Logarithms 8-9

5 . 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.. Number Systems.. The Natural Numbers The numbers,,,. 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 {,,, }. In the above representation, we write " " after the element to indicate that the other elements of N are listed following the pattern of,,. The numbers,, 5,. are called odd numbers. The numbers, 4, 6,. are called even numbers. In the collection N, we can solve equations such as x 9 0, x 6 0, x However, we note that the equations such as x + 5 5, x have no solution in the collection N, since they are satisfied by the number 0... The Whole Numbers The numbers 0,,, are called whole numbers. The collection of all whole numbers is denoted by the letter W. We observe that W {0,,, }. Now the equations such as x and x 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 + 5 5, x + 9 have no solution in the collection W, since they are satisfied respectively by the numbers 0 and which are not whole numbers.

6 .. The Integers The numbers 0,,,,, are called integers of which,,, are called positive integers and,,, are called negative integers. The collection of all integers is denoted by the letter Z. Thus Z {,,,, 0,,,, }. We observe that all whole numbers are integers but the negative integers,,, are not whole numbers. Now, the equations such as x + 5 5, x + 9 have solution in the collection Z, since they are satisfied respectively by the numbers 0 and which are in Z. However, the equations such as x + 5, x have no solution in the collection Z, since they are 7 5 satisfied respectively by the numbers and which are not in Z...4 The Rational Numbers A number of the form q p 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 q p is said to be in the proper form if q is a positive integer and p and q have no common factor other 7 than. For example, the rational numbers, are in the proper form where as the rational 5 4 numbers and are not in the proper form. But every rational number has an equivalent proper form. For example, we can write and which are in 5 5 proper form. Every integer is a rational number. For example, the integer 9 can be written as 9, where 9, are in Z and the denominator 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...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:. If x, y are rational numbers, then x + y is also a rational number. For example, and are in Q and their sum + + ( ) + called the closure property with respect to addition in Q.. If x, y are rational numbers, then x + y y + x. is also in Q. This property is

7 For example, 4 ( ) + ( ) 4 and So ( ) + ( ) This property is called the commutative property with respect to addition in Q.. If x, y, z are rational numbers, then x + (y + z) (x + y) + z. For example, 4 are in Q and,, ( 8) + ( 0) ( 4) ( ) ( 4) + ( 0) 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, and 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, we have the rational number ( ) + 0 such that If x, y are rational numbers, then x y is also a rational number. We write x y simply as xy. For example, ( 5), are rational numbers and ( 5) 5 ( 5) 0 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. 5 For example, ( ), are rational numbers and ( ), ( ). So ( ) ( ) This property is called the commutative property of multiplication in Q.

8 8. If x, y, z are rational numbers, then x(yz) (xy)z. We observe that ( xy) [( )(x)]y ( )xy, ( x) ( )[( )x] [( )( )x] x x. For example,,, 7 are rational numbers, and 7 4 ( ) ( ) (), [() ( )] ( 6). So we have 7 7 ( ) ( ) [()( ) ] This property is called the associative property of multiplication in Q. 9. The number is a rational number and (x) (x) x for all rational numbers x For example, (). 0. For every non-zero rational number x, is a rational number and x x x. x x For example, if x is the rational number 4, we observe that x 0 and 4 4 is a rational number and x The number x x is called the reciprocal or multiplicative inverse of x in Q. x. If x, y, z are rational numbers, then x(y + z) xy + xz, (x+y) z xz+ yz. For example, if x, y, z 5, we have x (y + z) + 5, 0 ( ) xy + xz + (5) +. So, x(y+z) xy + xz. Similarly, we have 4 + ( ) 5 (x+y)z , ( 5) 5 xz + yz So, (x+y) z xz + yz. These properties are called the distributive properties of multiplication over addition in Q. The properties,,, 6, 7, 8 and 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

9 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. Since is in N and in W and also in Z, the property 9 is valid in N, W and Z. The property 0 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 0 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, 5 4 and so 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 x. The operation does not satisfy the y commutative property in Q. For example, , and so The operation does not satisfy the associative property in Q. For example, let us consider,, in Q. Then 9 40 ( ) ( ) ( ), ( 4) 44 [( ) ] ( 4). So, ( ) [( ) ]. Similarly, we have ( ) ( ) ( ), [( ) ] 9. So ( ) [( ) ]. 5

10 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 which is not in N. Similarly the division 5 operation does not satisfy the closure property in N, since 5, 7 are in N but 5 7 which 7 is not in N. Using the properties of N, W, Z and Q, we can ascertain whether an equation has a solution in a particular number system or not. For example, we consider the equation 5x 0 0. Solving the equation, we get x. Since is in N, we say that equation 5x 0 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 Solving the equation, we get x. Since is not in N, is not in W and is in Z, we say that equation 5x has no solution in N or in W, but has solution in Z. Let us consider another equation x Solving the equation, we get x 5 5 is not in N, is not in W and 5 not possess a solution in N, W and Z. Since 5. Since 5 is not in Z, we say that that the equation x does is in Q, we say that x possesses a solution in Q. However, since we know that,, π, are not rational numbers, the equations such as x 0, x 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...6 Decimal Representation of rational numbers We have already learnt how to obtain the decimal representation of a rational number 5 5 by the long division process. For example, the decimal representations of and are 7 obtained as follows:

11 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 4, we mean that the integer 4 is the sum (addition) of hundreds, tens and 4 ones. That is, Similarly Thus, when we write integers, we use the numerals 0,,,, 9 and fix their face values as multiples of 0 0, 0, 0,. In the same way, we can use numerals 0,,,, 9 and fix their face values as multiples of 0 -, 0 -, 0 -,. to get fractions. For example, and we denote 8 5 as Here the first numeral 6 from the right side of dot called decimal 6 point has the face value, the second numeral has the face value and so on. The first 0 00 numeral 0 from the left side of dot has the face value 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 8 5. With this notation,.05 means Now, we consider the fraction Here 7

12 (The process terminates.) Next consider the following process: (A)

13 (B) The process repeats from (A) We observe that in the decimal representation of, the process terminates (zero 5 remainder) and we say that has a terminating decimal expansion. But in the decimal 5 representation of, the process does not terminate (non zero remainder) at any stage. 7 However, we notice that the remainder that we get at the stage (B) is the same as the remainder at the stage (A). So the numerals, 8, 5, 7,, 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 , 7 9

14 where the bar over 8574 indicates that the numerals under the bar repeat endlessly in the 5 same order in the long division process. The terminating decimal expansion for can be considered to be a non-terminating and repeating, since We also have ( 4 + ) ( 0) ( 9 + ) ( 0) ( 9 + ) 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 (ii) Consider the decimal expansion Let x Then 00x

15 x x ( ) ( ) or 99x 45 or x. 99 (iii) Consider the decimal expansion 0.49 Let x Then 00x x x 00x ( ) ( ) x 5 or x 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...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 We observe that the above decimal expansion has the numerals 0 s and s. As we proceed from the right of the dot, the s are separated by zero, zeros, 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 0 n is a positive integer. Similarly, x is said to be negative if it is not positive or zero. For example, is positive 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

16 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) (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 x + xy + xy + y x + xy + y. (iii) (x y) (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) x xy xy + y x xy + y. (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 x y. Example : Determine the rational number represented by Solution: Let x Then x x x x ( ) ( ) x 75 or x. 99 Example : Justify Solution: (0) (9 + ) + ( 0) ( 0) + + ( 9 + ) ( 9 + )

17 The above process continues endlessly and we get or Example : Find in the integer-by-integer form. Solution: Let x. 5. Then x x x x or 99x 50 or x. 99 Let y.5. Then y.5.. 0y 5. 00y y 0y or 90y 7 or y. 90 Hence Example 4: Find the sum of the irrational numbers and If it is a rational number, find it in the integer- by-integer form. Solution: Let x and y Then x + y 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. Multiplying by 0, we get 0a. 0a a or 9a or a 9. x + y 9. Note: From the above example, we observe that the sum of two irrational numbers need not be an irrational number. Similarly the product of two irrational numbers need not be an irrational number ( is irrational but is rational). Now let us consider the irrational number. We know already much about this number. We recall here the method of finding the decimal expansion of to any number of decimal places. and.4456

18 We observe that the above process neither terminates nor repeats. That is, has a non-terminating and non-repeating decimal expansion. Therefore, we conclude that is an irrational number. Similarly, we can show that, 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.4596 which is non-terminating and non-repeating. We recall that we used the rational number 7 as an approximation for the irrational number π in calculations. Ramanujan, the celebrated Indian mathematician has obtained several formulas involving π. Around the year 97,,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 4 5 6,,,,, 4 5 we observe that they become closer and closer to a particular real number and this real number is denoted by the letter e. The decimal expansion of e is e which is non-terminating and non-repeating. We will study more about this irrational number e in higher standards...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 9 and 7. Since 9 7 is a positive number, we get 7 < 9 which is equivalent to 9 > 7. Next, consider 9 and 7. Since ( 9) ( 7) is a negative number, we get 9 < 7 which is equivalent to 7 > 9. 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 4 5 4

19 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 x x x negative number negative number positive number. If x 0, then x x x If x > 0, then x x x positive number positive number positive number. Thus, we observe that, for every real number x, we have x 0. In particular, we observe that, for every non-zero real number x (> 0 or < 0), we always have x > 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 x. Example 5: Put π and 7 in order relation. Solution: π.458, π π is positive. π < 7. Example 6: Insert any four rational numbers in between the rational numbers.0 and.0. Solution: Here > 0 and so.0 <.0. Consider the numbers.0,.0,.0,.04. Since these have terminating decimal expansions, they are rational numbers. We find > 0 and so.0 <.0. Since > 0, we get.0 <.0. Since > 0, we get.0 <.0. Since > 0, we get.0 <.04. Since > 0, we get.04 <.0..0 <.0 <.0 <.0 <.04 <.0. 5

20 Example 7: Insert any four irrational numbers between.0 and.0. Solution: Since > 0, we get.0 <.0. Consider the real numbers a , b , c , d Since these four real numbers have distinct non-terminating and non repeating decimal representations, they are distinct irrational numbers. We find a > 0, b a > 0, c b > 0, d c > 0,.0 d > 0..0 < a < b < c < d <.0. Example 8: Insert any two rational numbers in between the irrational numbers. 00 and.00. Solution: and > <. 00. Consider the numbers.446 and.446. Since they have terminating decimal expansions, they are rational numbers and we have > < > < > < <.446 <.446 <.00. 6

21 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,,, π, are real numbers which are not rational numbers. Now, the equations such as x 0, x 0, x π 0 have solutions in R. However, since x is positive for any non-zero real number x, there is no real number x such that x and so the equation x + 0 has no solution in R. We do not study such equations in our standard. Exercise.. Obtain the decimal expansions of (i) (ii) (iii) (iv) 4. Determine the integer-by-integer form of each of the following rational numbers: (i) 0.49 (ii) 0. 7 (iii) 0.5 (iv) 0.5 (v) Answer true or false. (i) is a whole number. 7 (ii) is an integer. (iii) is a rational number. (iv) 0.6 is a rational number. (v) Every decimal expansion is a real number. (vi) Every real number is a rational number. (vii) 0. is a rational number.. 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 7

22 and say that this point A represents the number. Now we say that line segment OA is of length 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 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. Now we locate points on the number line to the right of O at a distance units ( times the lengths of OA), units,. These points correspond to the numbers,, respectively (see Figure.). Figure. Similarly, we can locate the points on the number line to the left of O at distances unit, units, units,. These points correspond to the negative integers,,, respectively. Now, we review with an example the method of locating points for rational numbers on the number line. Consider the rational number. To locate the point on the number line corresponding to, we proceed as follows: Locate the point P corresponding to the positive integer (denominator of ). Draw Figure. the line segment PQ of length (numerator of ) perpendicular to OP. Join OQ. Consider the line through A parallel to PQ. This line meets OQ at the point R. Then the length of AR is times that of OA. This is so because OAR and OPQ are similar and so PQ OP or or AR. Now, draw a circle with centre at O and radius equal to the AR OA AR length of AR. This circle cuts the real number line at a point on the right side of O. This point 8

23 corresponds to the rational number (see Figure.). 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 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. Draw a square OABC with side OA. Then, by Pythagoras theorem OB OA + AB +. So OB. With O as centre and OB as radius, draw a circle. We observe (see Figure.) that this circle intersects the real number line at P on the right side of O and at Q Figure. on the left side of O. The point P represents the irrational number, since OP OB. The point Q represents the irrational number, since OQ OB and Q lies to the left of O. Example 9: Represent the irrational numbers, 5,, 5 on the real number line. Solution: Having plotted the point P for, we can now locate the points for and as given below: Construct a rectangle OPDC (see Figure.4) with length OP ( ) and breadth PD. Then, by Pythagoras theorem, OD OP + PD ( ) + + Figure.4 and so OD. Now, with O as centre and OD as radius, draw a circle. This circle intersects the real number line at the points say Q and Q (Note that Q is just a notation and not representing the collection of rational numbers). Now OQ OQ OD. Since Q and Q are on the right and the left of O respectively, Q represents and Q represents..in 9

24 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 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.5) with length OR ( ) and breadth RF. Figure.5 Then, by Pythagoras theorem, OF OR + RF and so OF 5. With O as centre and OF as radius, draw a circle. This circle intersects the real number line at say S and S on the right and left of O respectively. Since OS OS OF 5, S represents 5 and S represents 5. From the above discussion, we are able to know that any real number corresponds to a unique point on the real number line and that any point on the real number line represents a real number. Points on the right side of O correspond to positive real numbers and those on the left side of O represent negative real numbers. Let P and Q be the points corresponding to two real numbers a and b respectively. If a < b, then P lies to the left of Q which is same as saying that Q lies to the right of P on the real number line (see Figure.6). Further, Figure.6 if x is a real number such that a < x < b, then the point on the real number line corresponding to x lies between P and Q. If P and P are any two points on the real line and if x and x are the real numbers corresponding to P and P, then the distance between P and P is x x... Manipulation of irrational numbers Let us recall how the irrational numbers,, 5, originated. For example, the irrational number x was needed when we wanted to find a solution x for the equation 0 or x 0 or x. Similarly, we may need to obtain a real number x such that 0

25 x n r, where r is a rational number and n is a positive integer. If n is an even positive integer such as, 4, 6, and r < 0, then we can not find a real number x such that x n r. This is so because x x x > 0, x 4 x x > 0,, x n > 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 x n r. For example, we find 5 for x when x If n is an odd positive integer such as,, 5, and r > 0, then it is possible to find a positive real number x such that x n r. For example, we find 4 for x when x 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 x n r. This is so because if x > 0, then x x x > 0, x x x > 0, x n > 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 x n r. In this situation, we write the positive real number x n r and say that x is the n th 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, 64 4, 4 4. We observe that if r is the n th power of a rational number q p, then n r q p is a rational number because in this case we have p n r. On the other hand, if the rational number r is not the n th power of some rational q number, then n r is not a rational number; that is, n r is an irrational number. If n r is an irrational number, then n r is called a surd. We observe that a surd is an irrational number in a particular form. Thus n r is a surd when the radicand is a rational number which is not the n th power of a rational number. If the index n is, then we call r the square root of r and simply write it as of r. r with out the index. If the index n is, then we call r the cube root In what follows, n r always represents a positive real number and r is a positive rational number. Since radicals are real numbers, we can perform the four fundamental operations +,,, with them. If the addition of two surds and multiplication of a surd by a rational number yield irrational numbers, then the resulting irrational numbers are also called 5 surds. For example, + 7, 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, ( + )( ) ( ) 4, ( + 5)( 5) ( ) ( 5) 5.

26 The following laws are used in the manipulation of radicals: Let a, b be positive rational numbers and m, n be positive integers. Then n (i) ( ) a. a n n n n (ii) ( a ) ( b ) ab. n a a (iii) n. n b b (iv) p n r + q n r ( p + q) n r where p and q are real numbers. (v) n m mn a a. (vi) n a mn m a. (vii) If a < b, then n n a < b. We observe that in the laws (ii), (iii) and (iv), the surds are of same index. So, when surds of different indices are given and if we are asked to perform any of the four fundamental operations, we first convert the surds into surds of the same index by applying the law (vi) and then proceed to carry out the operations. Law (vi) is actually the combination of law (i) and law (v): By (i), a m a m. n n m m a a mn m a. Using law (vii), we can compare any surds of same order. Note: n a is denoted as n a. Example 0: Answer with reasons whether the following are surds or not: (a) (d) Solution: (a) (b 5 (b) (e) (c) 7 5 (f) 4 6 a rational number is not a surd a rational number. is not a surd. 5 5 (c) 7 7 is a surd since is not a square of a rational number (d) a surd. 7

27 (e) Since is an irrational number, is a surd. 6 (f) a rational number. 4 6 is not a surd. When we manipulate with surds using the four fundamental operation, some times the result may be a rational number. Example : Answer with reasons whether the following are surds or not: Solution: (i) ( 5 + ) ( ) (ii) ( + ) ( + ) (iii) ( + 4 )( ) (iv) (i) ( 5 + ) ( ) (5 + 4) + ( ) ( ) 9 + ( ) 9. This is a rational number and not a surd although and 4 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) ( + ) ( + ) ( ) + ( ) + 0 a rational number. the given expression is not a surd. (iii) ( + 4 )( ) ( + 4 ) ( 6 ) ( + 4 ) ( 4 ) This is a rational number and so the given expression is not a surd. (iv) This is a rational number and so the given expression is not a surd..

28 Example : Simplify each of the following: (i) (ii) 40 5 (iii) (iv) Solution: (i) (ii) , ( 6) (iii) 6 7 and (iv) ( 4 ) 0 0. ( 0). Example : Write each of the following into a single surd: (i) 7 6 (ii) 5 4. (iii) (iv) 8 ( ) Solution: (i) Here 7 is of index and 6 is of index. The l.c.m. of and is 6. So we write 7 and 6 as surds of index 6. Here , and So, (ii) (iii) (iv) 8 ( 5 0)

29 Example 4: Arrange the following in the ascending order of magnitudes: 4 6, 0, 5 Solution: We shall first find the common index. For this, we find the l.c.m. of the indices 4, 6,. The l.c.m. of 4, 6,, is. Then we convert the surds into surds with index and We observe 5 < 7 < < < 0.. The product of a surd and its conjugate is always a rational number. For example, ( + )( ) ( ) + ( ) 4 and ( + 5)( 5) ( ) ( 5) 5. Some times, the denominator y of a ratio y x 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, 5 5 (ii) If the denominator y is in the form a + then multiply the numerator and denominator by a (a + b ) (a b ) a a b a b ( b ) a b b where a and b are rational numbers, b. Then the denominator becomes + which is a rational number. For example, ( ) ( + )( ) ( ) 9 7 (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 y x by a + b. For example, + ( + ) ( + ( )( ( ) ) ( + ) ) ( ) (iv) If y is in the form a + b, where a and b are rational numbers, then multiply x the numerator and denominator of by a b. For example, y 5

30 ( 5) ( + 5)( 5) 5 ( 5) ( ) ( ( ) ( ) ( 5) 5. (v) If y is in the form a b, where a and b are rational numbers, then multiply the numerator and denominator of y x by ( ) ( 5 7 )( ) a + b. For example, ( 5) ( 5 + 5) ( 5) ( 7 ) 5 7. Example 5: Rationalize the denominator of Solution: ( 5 4) 5 4 ( 5 + 4)( 5 4) ( ) ( ) ( 4 5) x Note: If we want to rationalize the numerator of, where x is of the form a or a + b or y a b or a + b or a b, then we multiply the numerator and denominator by the conjugate of the numerator. Example 6: Rationalize the numerator in. 4 ( ) ( + ) Solution: 4 4 ( + ) ( ) ( ) ( ) ( ) ( + ) Example 7: If Solution: + + a + b + c 6, find a + b +c. ( ) ( + )( ) ( ) ( ) 6

31 ( ) + ( ) c b a. 4, 4, c b a c b a In the surd p + n a q, where p and q are rational numbers and n a is a surd, p is called the rational part and n a q 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 8: If, y x find x + y. Solution: ( )( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) y x x 4, y 0 x + y

32 Example 9: If x, + find. x x + Solution: x + ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) + + x ( ) ( ). ( ) ( ) x x Example 0: If a, + find the value a (a 6). Solution: a ( ) ( ) ( ) ( ) ( ) ( ) a ( ) a (a ( ) ( ) ) ( )( ) [ ] + ( ) ( ). 9 8 Example : If.44., find an approximate value of +. Solution: ( ) ( ) + ( ) ( ) ( )

33 Exercise.. Answer with reasons whether the following are surds or not: (i) 6 (ii) 4 (iii) 5 79 (iv). Simplify each of the following to the simplest form: 4 4 (v) 50 (i) 5 + (ii) 0 40 (iii) ( 5 ) ( 4 + ) (iv) ( 0 ) ( ) (v) 5 5 (vi) 6 (vii) (viii) 5. Arrange in ascending order: (ix) 5 5 (i), 5, 6 (ii) 5, 7, 4 9 (iii), 5, 4 4. Rationalize the denominator in each of the following: (i) 8 4 (ii) 6 + (iii) (iv) (v) (vi) Find x and y in each of the following: (i) x + y (ii) x + y (iii) + x + y 5 (iv) x + y If a +, find the value of a (a 4) If a, find the value of a + a If.7, find an approximate value of. 9. If.44 and.7, find an approximate value of +. 9

34 Answers Exercise.. (i) 0.85 (ii).8 (iii) 0.5 (iv) (i) (ii) (iii) (iv) (v) (i) F (ii) T (iii) T (iv) T (v) T (vi) F (vii) F Exercise.. (i) surd (ii) not a surd (iii) surd (iv) surd (v) not a surd. (i) 9 (ii) 5 (iii) (iv) 4 5 (v) 6 5 (vi) (vii) 5 7 (viii) 65 (ix) 6 5. (i) 6, 5, (ii) 4 9, 7, 5 (iii), 4, (i) 6 4 (ii) ( ) (iii) ( + 5) 5 ) (v) 6 5 (vi) ( 6 0) (iv) ( (i) x 7, y 4 (ii) x, y 59 (iii) x 7, y 0 (iv) x 0, y 6. a (a 4)

35 . 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.. Area and Perimeter We recall the formulae for the perimeters and areas of plane geometrical figures... Rectangle Area l b sq.units Perimeter (l + b) units d l + b units. Figure... Parallelogram: Area b h sq.units Perimeter (a + b) units. Figure... Triangle with a given base and height: Area b h sq.units Figure.

36 ..4 Right triangle: Area b h sq.units Perimeter b + h + d units d b + h units. Figure.4..5 Equilateral triangle: altitude h a units Area a sq. units 4 Perimeter a units. Figure.5..6 Isosceles triangle: Area h a h sq. units Perimeter ( a a h ) + units. Figure.6..7 Scalene triangle: Area s( s a)( s b)( s c) sq. units a + b + c where s units Figure.7 Perimeter a + b +c units...8 Trapezium: Figure.8 Area (a + b) h sq. units...9 Quadrilateral: Area d (h + h ) sq.units. Figure.9..0 Rhombus: Area d d sq. units Figure.0 Perimeter d d + 4a units.

37 .. Circle: Area of the circle πr sq. units Perimeter of the circle πr units Area of a semicircle πr sq. units Arc length of the semicircle πr units Area of a quadrant circle 4 πr sq. units Figure. Arc length of a quadrant circle 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 : A wall in the form of a rectangle has base 5m and height 0m. If the cost of painting the wall is Rs. 6 per square metre, find the cost for painting the entire wall. Solution: Let b 5 and h 0. Then the area of the rectangle b h sq. metres. Since the cost of painting sq. metre is Rs. 6, the cost for painting the entire wall 6 50 Rs Figure. Example : The dimensions of a rectangular metal sheet are 4m m. 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 ,0000 cm. Area of a square sheet cm.,0000 No. of square sheets Figure. Example : Find the base of a parallelogram if its area is 40 cm and altitude is 5 cm. Solution: Area b h. 40 b b. 5 8 Base cm. Figure.4

38 Example 4: If the lengths of the sides of a triangle are cm, 60 cm and 6 cm, find the area and perimeter of the triangle. Solution: Area s( s a)( s b)( s c). Here s a + b + c s 66, s a 66 55, s b , s c Area sq.cm. Perimeter a + b + c cm. Figure.5 Example 5: Find the area of the quadrilateral ABCD given in Figure.6. Solution: Area d ( h + h ) 50 (0 + 0) m Figure.6 Example 6: Find the area of the trapezium ABCD given in Figure.7 Solution: Area ( a + b) h ( + 5) 4 4 sq. units. Figure.7 Example 7: Cost of levelling a land is Rs. per square metre. A land is in the form of a trapezium whose parallel sides are of lengths 8m and m. If its other two sides are each of length 5m, find the total cost incurred in levelling the land. Solution: ABCD is the given trapezium where AB 8m, CD m, AD BC 5 m. Draw CE parallel to DA (see Figure.8). EBC is isosceles whose height h 5 6 4cm. Now, the area of the trapezium ABCD (a + b) h (8 + ) sq. metre. The cost of leveling sq. metre is Rs.. So the cost of levelling the entire land 60 Rs. 70. Figure.8 4

39 Example 8: The perimeter of a rhombus is 0 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 d and d be the lengths of the diagonals. Then perimeter d + d. But the perimeter is 0 cm. d + d 0 cm or d Here d one of the diagonals is of length 8 cm. Take d 8. Then 64 + d 00 or d 6. d 6 cm. The area of the Figure.9 rhombus is d d cm. Example 9: A wire of length 64 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 7 ) Solution: Perimeter of the circle But perimeter of the circle πr. 7 r or r cm. 64 cm. Area of the circle πr 7 86 cm. Perimeter of the equilateral triangle a But perimeter cm. a or a 44 cm. Area of the equilateral triangle a cm 4 The ratio of the area of circle to that of the equilateral triangle 86 : 484 : Figure.0 Figure. 5

40 Exercise.. Find the area of a triangle when (i) base length 8 cm, height cm. (ii) the three sides are of lengths 0 cm, 48 cm and 5 cm. (iii) the triangle is equilateral with side length 8 cm.. Find the area of the quadrilateral ABCD where the diagonal AC is of length 44 cm and the lengths of the perpendicular from B and D to AC are 0 cm and cm respectively.. Find the area of the quadrilateral one of whose diagonals is of length 5 cm and the lengths of the altitudes to this diagonal are cm and 5 cm. 4. The perimeter of a rhombus is 60m 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 6 cm and the height is 8 cm, find the base length. 6. The area of a trapezium whose parallel sides have lengths 7 cm and 8 cm is 0 cm. 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 cm, 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 cm 4 cm.. Combined Figures Consider a quadrilateral ABCD (see Figure.). 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 Figure. 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. Figure.). We observe that two figures can be placed in juxta position if one has a side equal in length to a side of the other. 6

41 Some combined figures are given in Figure.4 to Figure.5. Figure.4 Figure.5 Figure.6 Figure.7 Figure.8 Figure.9 Figure.0 Figure. Figure. 7

42 Figure. Figure.4 Figure.5 The sides that are in juxta position are shown by dotted lines. We can easily identify the figures that are combined in Figure.4 to Figure.5. For example, Figure.5 is the combination of a triangle and a semicircle. It looks like the vertical cross section of a top. Figure.7 is the combined form of a rectangle and a semicircle. It can be viewed as a rectangular window surmounted by a semicircle. In Figure.9, we have the combination of a rectangle and two quadrant circles. Figure. is the combined figure of a rectangle, a triangle and a trapezium. It looks like a rocket. Since the combined figures are the combination of triangles, quadrilaterals and circles, their perimeters and areas can be calculated by applying the formulae that we have already learnt in our previous classes. We now consider two important combined plane figures namely trapeziums and polygons... Trapezium A trapezium is a four-sided plane figure in which two sides are parallel (see Figure.6) Consider the trapezium ABCD where AB and DC are parallel. Let AB a and CD b. Let h be the distance between the parallel sides. We can consider the trapezium ABCD as the combined figure of the Figure.6 triangles ABC and ACD. For the triangle ABC, the side, AB is the base and h is the height. For the triangle ACD, CD is the base and h is the height. So, Area of ABC a h and Area of ACD b h Area of the trapezium a h + b h (a + b)h sq. units. 8

43 .. 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 0 (see Figure.7). As this particular Figure.7 regular polygon is quite often used we shall derive its perimeter and area. Let ABCDEF be a regular hexagon. Then the sides AB, BC, CD, DE, EF and FA are of equal length. Let each side be of length a units. Then the perimeter of the regular hexagon is a + a + a + a + a + a 6a units. We shall now derive a formula for the area of the regular hexagon. The diagonals AD, BE, CF meet at a point, say O. Then the triangles OAB, OBC, OCD, ODE, OEF, OFA are equilateral triangles. So, the area of each triangle is area of the regular hexagon 6 a 4 a sq. units. a. Hence the 4 Example 0: Find the area of Figure.8 Solution: The figure ABCDE is the combination of ABDE and BCD with the side BD in juxta position. ABDE is a trapezium where the parallel sides AE and BD have lengths 0 cm and 6 cm respectively. The Figure.8 distance between the parallel sides is 9 cm. So, the area of the trapezium ABDE is ( a + b) h (0 + 6) 9 7cm. BCD is a triangle whose base BD is of length 6 cm and height 8 cm. So its area is b h cm. The area of the combined figure ABCDE is Area of ABDE + Area of BCD cm. 9

44 Example : A surveyor has sketched the measurements of a land as below. Figure.9 Find the area of the land. Solution: Let P, Q, R, S, be the surveyors marks from A to D. Then AP 5m, AQ 7m, AR m, AS 5m, AD 7m, BP 0m, FQ 8m, CR 8m, ES 9m. The given land is the combination of the trapeziums. PRCB, FESQ and triangles AQF, APB, DSE and CRD (see Figure.40). Area of the trapezium PRCB: The parallel sides are BP and CR and height is PR. We have BP 0m, CR 8m, and PR AR AP 5 7m. So, the area of PRCB is (BP + CR) PR (0 + 8) 7 6 m Figure.40 Area of the trapezium QFES: The parallel sides are ES and FQ and height is QS. We have ES 9m, FQ 8m and QS AS AQ 5 7 8m. So, the area of QFES is (ES + FQ) QS (9 + 8) m Area of the triangle AQF AQ FQ m Area of the triangle DSE DS ES (AD AS) 9 (7 5) m. Area of the triangle CRD RD CR (AD AR) 8 4 (7 ) m. Area of the triangle APB AP BP 5 0 5m Area of the land m. 40

45 Example : Find the area of the design as shown in Figure.4. (π ) 7 Figure. 4 Solution: We observe that the plot is a combination of the rectangle ABDE, the semi-circle AFE and the equilateral triangle BCD. The area of the rectangle ABDE cm. The area of the semicircle AFE π r cm The area of the equilateral triangle BCD 7 a cm. 4 The area of the plot cm Example : Find the area of the design as in Figure.4. (Take π ) 7 Figure.4 Solution: We observe that the plot is the combination of the rectangle ABCD, the semi-circle CDE and the quadrant circles AFD, BCG. The area of the rectangle ABCD 4 48 cm. The area of the semi-circle CDE π cm The area of the quadrant circle AFD π The area of the quadrant circle BCG 7 4 cm cm The area of the given plot cm

46 Example 4: Find the area enclosed by Figure.4 Figure.4 Solution: The figure is the combination of the rectangle CDFG, the semi circle DEF and the trapezium ABCG. The area of the rectangle CDFG 8 64 cm. The area of the trapezium ABCG (6 + 8) cm. The area of the semi-circle DEF cm 7 The area of the given design cm. Sometimes, we come across plane figures which are obtained by cutting out and removing plane figures from a bigger one. Their areas can be found as before but instead of summing up the smaller areas, we subtract the areas of the removed parts from the area of the bigger figure. Area of a circular ring A circular ring is the region in between two concentric circles (see Figure.44). The area of the ring is equal to the area of the outer circle minus the area of the inner circle; that is, π R π r or π (R r ). Figure.44 The area of the semi-circular ring is π (R r ) sq. units. 4

47 Example 5: Find the area of the shaded region in Figure.45, where the boundaries of the region are quadrants of a circle. (Take π ). 7 Figure.45. Solution: The given region is that which remains after cutting away four equal quadrants each of radius 4 cm from a square of side 8 cm. The area of the square cm. The area of one quadrant circle π cm Required area 784 4(54) sq. cm. Example 6: A running track of 7m wide is as shown in Figure.46. The inside perimeter is 70m and the length of each straight portion is 40m. The curved portions are in the form of semi-circles. Find the area of the track. (use π ) 7 Figure.46 Solution: Let r be the radius of the inner semicircles. Then the inside perimeter is 40 + (π r) or 80 + πr. But this is given as 70m πr 70 or πr 440 or r m. π So the radius of the inner semicircle r 70m. The radius of the outer semicircle R m. 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 π (R r ) (77 70 ) sq.m 7 The area of one rectangular track sq. m. Area of the track sq.m. 7 4

48 Example 7: A cow is tied up for grazing at one outside corner of a square building of length 4.m. 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.47). The rope is of length 4.9m and the side wall is of length 4.m. 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 th of a 4 circular region of radius 4.9m and two quadrant circular regions of radius m. Area, the cow can graze π π m. Figure.47 Example 8: Find the area of the shaded portion in Figure.48 (Take π ) 7 Figure.48 Solution: The area of the shaded portion is equal to the area of the semicircle of radius 4 cm minus the area of the semicircle of radius 7 cm. That is, or π (4) π (7) cm. 44

49 Exercise.. 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) (iii) Figure.49 Figure.50 Figure.5. A play ground is to be constructed with two straight segments and two semicircular segments as shown in Figure.5. The radius of each semicircular segment is m. The length of each of the straight segment is 85m. Find the area of the playground. (Take π ) 7 Figure.5. A trapezium has parallel sides of lengths cm and cm. Find its area, if the other two sides are each of length 0 cm. 45

50 4. If the area of a regular hexagon is 50 cm, find the side. 5. Find the area of the shaded region is Figure.5. Figure.5 6. Find the area of the shaded region in the following figures: (i) (ii) Figure.54 Figure.55 (iii) (iv) Figure.56 Figure A circle has diameter 54 cm. One of its diameter is AB. C is a point on the line segment AB such that BC 0 cm. A circle is drawn on AC as diameter. Find the area included between them. Take π. 7 46

51 8. Find the area and perimeter of the shaded portion in Figure.58. Figure Four cows are tied to the four corners of a square plot of side measuring 4 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. 0. ABCD is a rectangular plot of dimensions 6m 4m. 4 horses are tied to the four corners of the plot, each with a rope of length 0m. Each horse reaches as far as possible for grazing. Find the area of the portion of the plot which is left ungrazed. Take π. 7 7 Answers Exercise.. (i) 7 cm (ii) 480 cm (iii) 7.7 cm. 704 cm. 60cm 4. m, 696 m 5. 4 cm 6. 4 cm 7. 0 cm 8. 6.cm Exercise.. (i). 7,00 sq.m (ii). 5,00 sq.m (iii). 7,55 sq.m. 4,956 sq.m. 47. cm 4. 0cm m 6. (i) 6. sq.cm (ii) 5m (iii) 7.7cm (iv) 40.8 cm cm cm, 94cm 9. 4m m 47

52 . SOME USEFUL NOTATION. 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 9,900,000 miles. (ii) the average cell contains about 00, 000,000,000,000 molecules. (iii) the life-time of an elementary particle is seconds. (iv) the diameter of an electron is about 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 a m means the product of m numbers each equal to a; that is, a m a a. m factors. Here a is called the base and m, the power or exponent or index. The notation a m is read as a to the power m or a raised to m. For example, a 5 a a a a a. The laws of indices are given below: (i) a m a n a m+ n (Product law) m a (ii) a m n, a 0, m > n (Quotient law) n a (iii) (a m ) n a mn (Power law) (iv) a m b m (a b) m (Combination law) When a 0, we denote m as a m and define a 0. a Now, using the laws of indices, any positive real number can be written in the form a 0 n, where a < 0 and n is an integer. For example, (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)

53 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 0 n, the integral part of the number, a is a digit from to 9 and the power of 0 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 0 r ; and if the decimal point is moved r places to the right, then this movement is compensated by the factor 0 r. When very large or very small numbers are put in the scientific notation, they can be multiplied or divided easily in this form. Example : Write the following numbers in scientific notation: (i) 749 (ii) 0500 (iii) (iv) (v) (vi) Solution: Example : Write the following numbers in decimal form: (i) (ii) (iii) (iv) (v) (vi)

54 Solution: (i) (ii) (iii) (iv) (v) (vi) Example : Perform the calculation and write the answer of the following in scientific notation. (i) (000000) (ii) (4000) 5 (00) (iii) ( ) 4 (iv) (000) (0.000) 4 Solution : (i) (000000) ( ) (.0) (0 6 ) (ii) , (4000) 5 (00) (4.0 0 ) 5 (.0 0 ) (4.0) 5 (0 ) 5 (.0) (0 ) (iii) ( ) 4 ( ) 4 (5.0) 4 (0 5 )

55 (iv) , (000) (0.000) 4 (. 0 0 ( ) ) (. 0) (0). (0 (0 ) ) ( 6) Exercise.. Represent the following numbers in the scientific notation: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) Write the following numbers in decimal form: (i) (ii) (iii) (iv).4 0 (v) (vi) (vii) (viii).4 0. Find the value of the following in scientific notation: (i) (00) (40) 5 (ii) (000) (0.00) 4 (iii) (8000) 4 (0000) (iv) (0.00) 8 (0.000) (0.0) 4 (v) (0000) (0.0005) (400000). Notation of logarithm John Napier, an English mathematician introduced the notation of logarithm as a mathematical device to do calculations easily and quickly. The word logarithm is derived from two Greek words logos and arithmos. The word logos means reckoning and arithmos means number. Thus logarithm means reckoning number. To introduce the notation of logarithm, we shall first know about exponential notation... Exponential notation Let a be a positive number. We have already introduced the notation a x,where x is an integer. When x is a rational number, say q p with p, an integer and q, a positive integer, we define a x by x a q p a For example, 5 ( ) 5 8, 7 ( 7 )

56 When x is an irrational number, a x 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, a x can be defined and that it represents an unique real number u and write u a x. In this situation, we say that the real number u is written in the exponential form or in the exponential notation. Here the positive number a is called the base and x, the index or the power or the exponent. The laws of indices which we have stated for integer exponents can be obtained for all real exponents. We state them here and call them, the laws of exponents: (i) (ii) (iii) x a a x y a y a x a x y a + y (iv) a x a x a x b x a b (v) ( ) x y x xy a a (vi) a 0 Now we are in a position to introduce the notation of logarithm... Logarithmic notation Let b be a positive number and b. We have already understood that, for any real number x, b x represents a unique real number, say a. If we write a b x, then the exponent x is called the logarithm of a to the base b. We also call x, the value of log a. Thus, x log a b b is an equivalent form of a b x. We say that x log a is the logarithmic form of the b exponential form a b x. In both the forms, the base is same. We observe that x log a is an b equivalent way of writing a b x. The notation x log a is called the logarithmic notation b and it means the equation a b x. For example, (i) log 9 79 is equivalent to 9 79; (ii) log8 is equivalent to 8 ; (iii) log is equivalent to ; (iv) log 7 49 is equivalent to 7 49; (v) log 9 is equivalent to or 9 ; (vi) log 4 is equivalent to

57 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) log 5 5 (ii) log (iii) log 6 6 (iv) log 4 9 Solution: As base is same in both forms, (i) log 5 5 (ii) (iii) (iv) is equivalent to ( ) log 6 6 is equivalent to ( 6) 6. log is equivalent to (). 9 9 log is equivalent to ( ). Example 5: Change the following from exponential form to logarithmic form: 6 (i) 64 (ii) 9 (iii) 79 8 Solution: As base is same in both forms, we have 6 (i) 64 is equivalent to log (ii) 9 79 (iii) (iv) 8 4 is equivalent to log is equivalent to log. 4 7 is equivalent to log (iv) 7 7 Example 6: Evaluate (i) log 9 79 (ii) log 4 8 (iii) log 9 (iv) log (4). 7 Solution : (i) Let x log 79. Then 9 x (iii) Let x log 9. 7 x. Then 9 x (ii) Let x log 8. Then 4 x 4 8 But 4 x ( ) x x. 7 But 9 x ( ) x x x. x or x x, x or x. 5

58 (iv) Let x log (4) Then x (4) or x 5 or x 5. Example 7: Solve the equations: (i) log x (ii) log b 00 (iii) x log 5 (iv) x + log Solution: (i) log x x or x. 9 8 (iv) x + log x x log 7 9 or log 7 9 x ( 7) 9 or x ( ) or x ( ) (ii) log 00 b b b 0. (iii) x log or (8 ) x 8 or 8 x 8. x or x. x 5 8. x or x 4. Now we proceed to state and prove some properties of logarithms of positive numbers. All positive numbers other than are considered. (i) Product rule: If a, m and n are positive numbers and a, then log ( mn) log m + log n a a a Proof: Let log m x and log n y. a a Then, m a x and n a y. m n a x a y or mn a x+y. This is in exponential form. Writing this in the logarithmic form, we get log ( mn) x + y or log ( mn) a a log m + log n. a a In words, the above rule states that the logarithm of the product of two positive numbers is equal to sum of the logarithms of the numbers. (ii) Quotient rule: If m, n and a are positive numbers and a, then, m log log m log n a. n a a 54

59 Proof: Let log m x and log n y. a a Then m a x and n a y. m a a x y. n y x a This is in exponential form. Writing this in logarithmic form, we get m m log x y a or log. n a log m log n n a a In words, the quotient rule states that the logarithm of the quotient n m is equal to the difference log m log n. a a (iii) Power rule: If a and m are positive numbers, a and n is a real number, then log m n n log m. a a Proof: Let log m x. Then m a x. Raising to the power n on both sides, we get a m n (a x ) n a x n. This is in exponential form. Writing this in the notation of logarithm, we get log m n nx or a log m n n log m. a a (iv) If a is a positive number, then log 0 a Proof: Let x log. Then a x a 0. a x 0 or log a (v) If a is a positive number, then log a a. Proof: Let x log a. Then a x a a. a x or log a. a 0. (vi) Change of base rule: If m, n and p are positive numbers and n, p, then log m log m ( log p) n p. n Proof: Let x log m and y log p. p n Then p x m and n y p. Eliminating p with these equations, we get y x xy ( n ) m or n m. 55

60 This is in exponential form. Writing this in the logarithmic form, we get log m xy or n ( log m) ( log p). log n m p (vii) Reciprocal rule: If m and n are positive numbers other than, then log n m. log n Proof: Let log n x m. Then x m n n x x n m x n x. x m n. This is in exponential form. Writing this in the logarithmic form, we get log m or n x log n m. log n m log a (viii) If a and b are any two positive numbers and b, then b b a. Proof: Let x log a. Then b x a. Substituting for x in this equation, we get b log a b b a. (ix) Let m, n and a be positive numbers and a. If log m log n, then m n. a a Proof: Let x log m.then x log n. a a log m a x n or a a n or m n (by property (viii)). Note: We are avoiding in the base of all logarithms because if we consider one such logarithm, say log 9 with in the base, then x log 9 would give x 9. We know that there is no real number x such that x 9. Caution: Some errors which are commonly committed are m log m () log a, a n log n a () log ( m + n) log m + log n. a a a log m () is wrong since L.H.S. log m log n a. a a log n a () is incorrect since R.H.S. log ( mn) log ( m + n). a a 56

61 Example 8: Simplify: (i) log 7 + log 79 (ii) log log Solution: (i) Since the expression is the sum of two logarithms and the bases are equal, we can apply the product rule. log 7 + log 79 log (7 79) 6 log ( ) 9 9 log log 9 9. (ii) log log 5 log log 5 5 log 5 log 5 ( 5 ) 5 ( ) log 5 ( ). 5 Example 9: Simplify: (i) log 7 98 log 7 4 (ii) log log 9 4 log Solution: (i) log 7 98 log 7 4 log 7 log (ii) log log 9 4 log 9 4 log log 9 4 log 9 4 log9 6 + log9 6 log9 64 log9 (6 6) log9 64 log9 96 log log 9 log Example 0: Prove (i) log log0 (ii) log log 5 6 log log. Solution: (i) R.H.S. 4 log0 4 log 0 4 log0 0 log0 8 4 log0 0 log0 8 log log log0 log050 L.H.S. 8 57

62 (ii) R.H.S. log 5 6 log log log 5 (6) log 5 (8) + 4 5log 5 log 6 log + 4log log 5 + 4log log log + 4log 5 log + log log log 5 65 log 5 ( 65) log L. H. S. Example : Prove that log 4 log 4 5 log5 6 log 6 7 log 7 8 log8 9. Solution: L.H.S. ( log 4 log 4 5) ( log5 6 log 6 7) ( log 7 8 log8 9) log 5 log5 7 log7 9 log 5 ( log 5 7 log 7 9) log 5 log 5 9 log 9 log log R.H.S. Example : Solve log 0 (x + 50). Solution: Writing the equation in the exponential form, we get x or x or x 475. log 9 Example : Find the value of 8. log log 9 Solution: log9 log log b a. (Since b a ) Example 4: Solve log 6 x log 6 ( x + ) 0. x Solution: Using the quotient law, we can write the equation as log 6 0. x + into exponential form, we get x 6 0 x + or x x + or x. Changing Example 5: Solve log (7 x ) log( x). 7 x Solution: Using the quotient law, the equation can be written as log. x Writing in the exponential form, we get 7 x x or 7 x ( x) or 7 x x or x 4 or x. 58

63 ( ) Example 6: Solve log log x. Solution: Put y log x. Then the equation becomes log y. Writing the equation in the exponential form, we get we get x 4 or x 8. y 4 or log x 4. Again, writing in the exponential form, Example 7: Solve log5 log9 x + log5. Solution: Rewriting the equation, we get log 5 log 9 x log 5 or log 5 9 log 9 x log 5 log 5 5 log 5. 5 Using the change of base rule, we get log5 x log5. x. 5 5 Exercise... Write true or false in the following: (i) log 4 5 (ii) log 7 (iii) 6 6 log 4 log log 4. (iv) log (8 4) log 8 log 4. (v) log a (vi) log a ( m + n) log a m + log a n. a. Obtain the equivalent logarithm form of the following: (i) (iv) (ii) (v) (iii) (vi) Find the value in the following: (i) log 5 65 (ii) log 6 6 (iii) log 9 (iv) log 9 (v) log 8 (vi) log 4 log 6 log (vii) 0 (viii) 5 (ix) 4. Solve for the unknown: (i) log x 0.00 (vi) log x (ii) log x 7 (vii) c log log 9 59

64 (iii) (iv) log 00 (viii) N x log 4 a b (ix) log 0 log 5 (v) log 5 x 000 (x) log 9 N 5. Choose the correct answer from the alternatives given for each of the following: (i) If log x 5, then x is (A) 5 (B) 5 (C) 5 (D) 65 (ii) log4 + log (A) 0 (B) (C) (D) (iii) The value of log 4 04 (A) 0 (B) 8 (C) 7 (D) 5 (iv) If log a 4 log a x, then x (A) 0 (B) (C) (D) (v) If log6 x, then x (A) 4 (B) 8 (C) 6 (D) (vi) If log 5 x then x (A) 5 (B) 5 (C) 5 (D) Simplify the expression into a single logarithm in each of the following: (i) log0 + log0 9. (ii) log + 4log 8. (iii) 5 log + log 4 + log 6. (iv) log4 + log 5 log 5 + log 7. (v) 5log0 + log0 6log (vi) log0 5 + log0 0 log0 4 + log Given log a x,log a y,log a 5 z and log a 7 t, find value in each of the following in terms of x, y, z and t. (i) log a 6 (ii) log a 4 (iii) log a. 5 (iv) log a 7 8 (v) log a (vi) log a 600 (vii) log a (viii) log a (ix) log a 5 (x) log a (xi) log a 4. 9 (xii) log a 5 60

65 8. Solve the equation in each of the following: (i) log5 x log5 (ii) log x + log 7 log (iii) x log (iv) log4 x log4 6 log4 56 (v) log4 ( x + ) + log4 (vi) log (x + ) log (x ) log 4 (vii) log 0x log ( x + ) (viii) log (7x + ) log (5x ) log (ix) log5 (0 + x ) log5 ( + 4x) (x) log5 (5log x ). (xi) log 0x + 5 log x + 9. Prove the equation in the following: (i) log 5 + log 5 (ii) log log0 (iii) log log0 5 (iv) log0 5 log0 (v) log0 5 log0 (vi) log log 0 (vii) log log If a, b and c are positive numbers other than one, prove that log a log b log c... 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. If we choose the base as the irrational number e, then such logarithms are called Natural logarithms. If we choose the base as 0, then such logarithms are called Common logarithms. Natural logarithms were introduced by John Napier and common logarithms by his friend Henry Briggs, an English mathematician. To honour John Napier, the founder of logarithms, we denote the natural logarithm log e x simply as ln x. We will study more about ln x in higher classes. Now, we proceed to bring out the use of common logarithms in computations. We denote the common logarithm log 0 x as log x, omitting the base 0. Thus, log x y means log 0 x y and is equivalent to x 0 y. If we substitute for x in y 000 b c a log x, we get y log 000 log 0. 0 Similarly, for x,,, 0, 00, 000,, we correspondingly get 00 0,, 0,,,,. 6

66 x log x 0 0 We observe that as x increases along the positive real axis, log 0 x also increases. We also observe that log 0 x is positive for all values of x > and is negative whenever x <. We observe that the values of log 0 x is as given in the table below: Range of x Location of log0 x The value of log 0 x 0 5 < x < < log0 x < a a 0 4 < x < 0 4 < log0 x < b b 0 < x < 0 < log 0 x < + 0.c c 0 < x < 0 < log0 x < + 0.d d 0 < x < 0 0 < log0 x <0 + 0.e e 0 0 < x < 0 0 < log0 x < f f 0 < x < 0 < log 0 x < + 0.g g 0 < x < 0 < log 0 x < + 0.h h 0 < x < 0 4 < log 0 x <4 + 0.i i 0 4 < x < < log x < j j 0 From the above table, we note that the value of the common logarithm expressed as (an integer) + (a decimal expansion of the form 0.r r r r 4 ) log 0 x can be In this form, the integer part is called the characteristic of log 0 x and the decimal part is called mantissa of log 0 x. We usually write the mantissa in 4 decimal places. We observe that the mantissa of log 0 x always represents a positive number between 0 and, and the characteristic of log 0 x is a positive integer or negative integer or zero depending on the value of x. If x <, then the characteristic is a negative integer. If x >0, then the characteristic of log 0 x is a positive integer. If x lies between and 0, then the characteristic of log 0 x is 0. Example 8: Find the characteristics of the common logarithms of (i) 00 (ii) 00. (iii) 0.0 (iv).00 (v) 0.00 (vi) (vii) (viii) (ix) (x) Solution: (i) As 00 lies between 0 and 0 4, log 0 00 is + 0. dd d d 4. Then the characteristic of log 0 00 is and the mantissa is 0.d d d d 4. Alternatively, writing 00 in the scientific notation, 6

67 So we have log 00 log (.00 0 ) 0 0 log log0 0 log log0 0 log dd dd 4 +. Thus, when we write the number x in the scientific notation a 0 n, the integer n is the characteristic of log0 x and the value log0 a is the mantissa of log 0 x. (ii) The characteristic of log 00. is The mantissa of log 00. in log 0.00 which is same as the mantissa of log 00. (iii) The characteristic of log 0.0 is and the mantissa of log 0.0 is log.00. (iv) The characteristic of log.00 is 0 and the mantissa of log.00 is log.00. (v) The characteristic of log 0.00 is and the mantissa of log 0.00 is log.00. (vi) The characteristic of log is and the mantissa of log is log.00. (vii) The characteristic of log is and the mantissa of log is log.00. (viii) The characteristic of log is 4 and the mantissa of log is log.00. (ix) The characteristic of log is 5 and the mantissa of log is log.00 (x) The characteristic of log is 6 and the mantissa of log is log.00. 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 n. log 0 x is 6

68 (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 log 0 x is (n + ). (iii) The numbers 00, 00.,, 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 9: Given that log 6.095, find (i) log 60 (ii) log 6. (iii) log.6 (iv) log 0.6 (v) log 0.06 (vi) log Solution: The mantissa of log 6 is (i) log (ii) log (iii) log (iv) log (v) log (vi) log Note: is written simply as.095. Likewise is written as 5.0. 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.d d d d 4. Example 0: Given log and log 00.06, find 5.00 (i) log 0.05 (ii) log (iii) log (iv) log Solution: log So the mantissa of log 0.5 is log So the mantissa of log 00 is (i) log (ii) log (iii) log log 5 log (iv) log log.00 log

69 ..4 Table of Logarithms Common logarithms of positive numbers from.000 to ( 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 : Find log Solution: The characteristic is. To get the mantissa, we consider.678 and locate the number.6 in the extreme left column of the table. Read along the row corresponding to.6 and down the column under 7. We find We go further along the row and reach the column under 8 in the mean difference. Here we find 0. We add this 0 to and get as the common logarithm of.678. This is the required mantissa of the given number and hence log Example : Find log Solution: The characteristic. To get the mantissa, consider.006. Common logarithms have been tabulated for numbers with digits in the decimal part. So we approximate.006 as.00(since in the 4 th decimal place is less than 5). Now, we get as before from the tables, log Hence log Example : Find log Solution: The characteristic. To get the mantissa, consider 7.09 and approximate it as 7.04 (since 9 in the 4 th decimal place is not less than 5). Now, we get as before from the tables, log Hence log Table of antilogarithms If log x y, then x is called the antilogarithm of y. That is, if log 0 x y, then x 0 y is the antilogarithm of y. Thus, antilog of y 0 y. 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 0.00 and so antilog of A table of antilogarithms of numbers ranging from to (4 digits in the decimal part) is provided at the end of this text book. Actually, this table gives us the values of 0 y where y ranges from to (4 digits in the decimal part). Using this table, we can calculate the antilogarithm of any given number. 65

70 Example 4: Find the antilogarithm of (i).05 (ii).469 (iii).8658 (iv).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:.05 (i) antilog of from antilogarithm tables mean difference corresponding to (ii) antilog of (iii) antilog (iv) antilog 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 + ) th digit. Characteristic Position of the decimal point 0 after the st digit after the nd digit after the rd digit after the 4 th digit 4 after the 5 th digit 66

71 (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 n th place; that is, the first n digits are zero. Characteristic Position of the decimal point 0.d d d d 4 0.0d d d d d d d d d d d d d d d d 4 In the above table, d. d d d 4 is the antilogarithm corresponding to the mantissa of the given logarithm...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) log ( mn) log m + log n a a a (ii) m log log m log n a n a a (iii) log m n nlog m. a a Example 5: Find (i) (ii) Solution: (i) Let x Then log x log ( ).4456 log log x antilog of

72 (ii) Let x Then log x log log log x antilog(. 7 )0.07. Example 6: Evaluate (i) (ii) Solution: (i) Let x 6. Then log x log 00 log x antilog of (ii) Let x Then log x log 0.4 log ( +0.54) ( ) x antilog of Example 7: Compute the value of (i) (9.76) 5 (ii) (0.749) 7 Solution: Let x (9.76) 5 Then log x log (9.76) 5 5 log (9.76) x antilog of (ii) Let x (0.749) 7 Then log x 7 log( 0.749) ( ) x antilog of Example 8: Find the value of Solution: Let x (0.7) 5. Then log x log (0.7) 5. 68

73 log x antilog of (.8867) Example 9: Find the value of Solution: Let x. Then we have log x log log.59 log We have to make the following approximations since the table of logarithms has been prepared for numbers ranging from.000 to (with three digits in the fractional part) log x x antilog of (0.69).. (76.5).98 Example 0: Find the value of 5 (4.75) Solution: Let x be the given expression. Then log x log (76.5) + log. 98 [log (4.75) 5 + log 0.046] log log.98 [5 log log 0.046] [ ] [ ] ( ) x anti log Example : Find the value of log 4.56 Solution: Here the base is. To use the table of logarithms, we have to get 0 in the base. Using the change of base rule, log 4.56 log 4.56 log

74 log log x say. Then log x log (0.0) + ( ) + (0.775).775 x antilog (.775) Exercise... Find the characteristics of the common logarithms of the numbers. (i) 4 (ii) 7.6 (iii).65 (iv) (v) (vi) (vii) (viii) Find the number of digits in the integral part of the numbers whose common logarithms are (i).45 (ii).456 (iii).4567 (iv) 0.4 (v) (vi) The common logarithms of numbers are given below. Find the number of zeros after the decimal point of the numbers. (i). 456 (ii).45 (iii). (iv) (v) (vi) The mantissa of the common logarithm of 740 is Write down the common logarithms of the following. (i) 740 (ii) 74 (iii) 7.4 (iv).74 (v).74 (vi) 0.74 (vii) (viii) Find the common logarithms of the following numbers using the table of logarithms. (i) 87 (ii) 8450 (iii) (iv) (v) (vi) Find the antilogarithms of the following common logarithms. (i).890 (ii) 0.4 (iii).458 (iv). 46 (v) 5.50 (vi) Find x, if the common logarithm of x is (i) 5.07 (ii).968 (iii).04 (iv).77 (v) 0.96 (vi) 4.08 (vii). +. (viii) 5.4. (ix). 5.4 (x). (xi).4 (xii)

75 8. Evaluate the following: (i) (ii) (iii) (iv) (v) (0.075) (vi) (50.49) 5 (vii) (55.9) 8 (viii) (ix) (x) (xi) ( 4.) (xii) (.4) (xiii) log 5.6 (xiv) log 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 (845 98), 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... 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 7

76 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 {,, 4, 5}, then the elements of A are,, 4 and 5; that is, we have A, A, 4 A and 5 A. We note that 6 A, 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... 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 is {,, 5, 7, }. (ii) The set of letters in the word FOOTBALL is{f, O, T, B, A, L,} (iii) The set of all natural numbers,,, is {,,, }. 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,,,, is {0,,,, }.This set is specially denoted by the letter W. (v) The set of all integers 0,,,,,,, is {0,,,,,,, }. This set is specially denoted by the letter Z. Example : 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 0, 0, 0, 40, 50, The set of even natural numbers divisible by 5 and less than 50 {0, 0, 0, 40}... 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, 6, 6}. The elements of the set are 6, 6 and 6. These numbers have a common property that they are powers of 6. So the condition x 6 n, where n, and yields the numbers 6, 6 and 6. No other number can be obtained from the condition. Thus we observe that the set {6, 6, 6} is the collection of all numbers x such that x 6 n, where n,,. This fact is written in the following form { x x 6 n, n,, }. In words, we read it as the set consisting of all x such that x 6 n, where n,,. Here also, the braces 7

77 { } are used to mean the set consisting of. The vertical bar within the braces is used to mean such that. The common property x 6 n, where n, and 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 : 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, 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, 4, 6, can be described by the statement: x n, n N. the set is { x x n, 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 4: Write the following set in the tabular form A {x x + 5 7, x N}. Solution: x x 7 5. Here N. A {}. Example 5: Obtain the set builder representation of the set A,,,,,, Solution: Since the given elements are the reciprocals of the first seven natural numbers,,, 4, 5, 6, 7, we have Note: x, y means x and y. A x x, n N and n 7. n 7

78 ..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 6: 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,,,, 4. A {0,,,, 4}. When we count elements of A one by one as for 0, for, for, 4 for, 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 {,,, }. When we count elements of N one by one as for, for, for, 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,,,, }. When we count elements of W one by one as for 0, for, for, 4 for, we are unable to come to an end in the counting process. the set W is an infinite set. (vi) Z {0,,,,, }. When we count elements of Z one by one as for 0, for, for, 4 for, 5 for 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,, 5, 7,,, 7 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...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. 74

79 Example 7: Which of the following sets are empty? (i) {Odd natural numbers which are divisible by }. (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. the set {x x is an odd natural number and x is divisible by } Ø. (ii) A prime number is a natural number which is not divisible by any other natural number except and itself. So a prime number cannot have a factor other than 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...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 {,, }, B {, 9, }, n(a), n(b). So A B...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 {,,, 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 {,,, 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 {,,, 4} and {4,,, } are equal. Note: If two sets are equal, then they must be equivalent. However two equivalent sets need not be equal. For example, {,,, } and {,,, } are equal as well as equivalent sets, but, {,,, } and {,,, 4} are equivalent sets. But they are not equal sets, since the elements, {,,, } and, {,,, 4}...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. So {All even prime numbers} {} and n(a). Hence it is a singleton set. 75

80 ..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 {,, 4, 5}, and B {,, 7, }. Then we can take the universal set U as U {,,, 4, 5, 7,,} or U N or U W or U Z...0 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 {,,,, } and B {,,,,, }. We observe that,,,, A and,,,, 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 B and 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... 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 {,,, 4} B {0,,,, 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. 76

81 .. 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 },{ b},{ a,b} and the cardinal number of p(a) is n[p(a)] 4. Similarly, if A {a, b, c}, then p(a) is{ {},{ a},{ b},{ c},{ a,b},{ b,c},{ c,a},{ a, b,c and the cardinal number of p(a) is n[p(a)] 8. Thus, we observe that if n(a) m, then n[p(a)] m. }} Exercise... 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}.. 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 x < 7}. (iii) R {x x is a two digit natural number such that sum of its digits is 9}.. Write the following sets in the builder form: (i) {, 6, 9, } (ii) {5, 5, 5, 65} (iii) {,, 5, } (iv) {, 4, 9, 6, 5, 6, 49, 64, 8, 00}. 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) {,,, 6} (a) {x x is a prime number and a divisor of 6}. (ii) {, } (b) {x x is an odd natural number less than 0}. (iii) {, 4, 6, 8} (c) {x x is a positive integer and a divisor of 6}. (iv) {,, 5, 7, 9} (d) {x x is an even natural number less than 0}. 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 00. (iii) A { x x is a letter in the word mathematics }. (iv) B {x x < 0, x W}. (v) C {x x < 4, x Z}. (vi) The set of all prime numbers between 0 and Write down the elements that have been left out in the following finite sets: (i) A {, 0, 00,,,,00,000}. (ii) B {, 5, 8,,,, 0, }. 77

82 7. Write down the next three elements of the following infinite sets: (i) C {, 6,, 4,,,,.}. (ii) D { 4,,,,,,, }. 8. State whether the following sets are empty sets: (i) A {Even natural numbers divisible by }. (ii) B { x x R, x + 0}. (iii) C {Polygons with four sides}. (iv) D {Quadrilateral with five sides}. 9. Let A {p, q, r, s}, B {,, 5, 7}, C {q, r}, D {8, 4, 6, }, E {r, q, s, p} F {, 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. 0. Write down the power set for each of the following sets: (i) A {, } (ii) B { x, y, z} (iii) C {a, b, c, d}. (i) If n(a) 5, find n[p(a)] (ii) If n[p(a)] 8, find n(a)... 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 U B { x x A or x B or x A and B}. We write A U 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. Example 8: If A {,,, 4} and B {, 4, 6}, find A U B. Solution: Listing all the elements A and B together and omitting the repetition, we get,,, 4,, 4, 6. Thus, A U B {,,, 4, 6}. 78

83 (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 I B {x x A and x B}. Example 9: If A {,, } and B {,, 4}, find A I B. Solution: All elements in A and B:,,,,, 4 Common elements in A and B:,,, A I B {, }...4 Disjoint Sets If two sets A and B have no elements in common, then they are called disjoint sets; i.e., if A I B Ø or { }, then A and B are disjoint sets. For example, if A {,,, 7} and B {4, 5, 6}, then A I B { } and so A and B are disjoint sets...5 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 {,,, 4, 5, 6} and B {,, 7}, find A B and B A. Solution: A B {, 4, 5, 6}. B A {7}. Note: A B B A...6 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 A c or A. Note: A c U A. Example 4: If U {,,, 4, 5} and A {, 4}, find A c. Solution: A c {,, 5}...7 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 U B) n(a) + n(b) n(a I B). 79

84 Example 4: If A {,, 4, 5, 6, 7, 8, 9} and B {,,, 5, 7}, find n(a), n(b), n(a U B) and n(a I B) and verify the identity n(a U B) n(a) + n(b) n(a I B). Solution: We observe that A U B {,,, 4, 5, 6, 7, 8, 9} A I B {,, 5, 7}. 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) Here, n(a U B) 9. So n(a U B) n(a) + n(b) n(a I B). In fact, this result is true for any two finite sets. Exercise... Find A U B and A I B for the following sets: (i) A {a, e, i, o, u} and B {a, b}. (ii) A {,, 5} and B {,, }. (iii) A {x x is a natural number and < x 6)} and B {x x is a natural number 6 < x < 0}. (iv) A {p, q, r} and B Ø.. 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 {, 4, 5}, B {5, 6, 7, 8} and C {7, 8, 9}.. If U {a, b, c, d, e, f, g, h}, A {a, c, g} and B {a, b, c, d, e, f}, find (i) A c (ii) B c (iii) (A U B) c (iv) (A c I B) c (v) A c I B c (vi) A c U B c...8 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. 80

85 A U B Figure. A I B Figure. A c. Figure.4 B c Figure.5 (A U B) c Figure.6 (A I B) c Figure.7 A B Figure.8 B A Figure.9 8

86 Example 4: From the diagram given below(see Figure.0), find the sets (i) A U B (ii) A I B (iii) (A U B) c Solution: Here U {,,, 4, 5, 6, 7}. (i) A U B {,,, 4, 5, 6}. (ii) A I B {, 5}. (iii) (A U B) c {7}. Figure.0 Example 44: Using Venn Diagram, exhibit the sets A {,, 5, 7} and B {, 9, }. Verify the formula: n(a U B) n(a) + n(b) n(a I B). Solution: A U B {,, 5, 7} U {, 9, } {,, 5, 7, 9, } n(a U B) 6. () A I B {,, 5, 7} I {, 9, } {} n(a I B). We have n(a) 4, n(b). n(a) + n(b) Figure. n(a) + n(b) n(a I B) 7 6. () From () and () we get (A U B) n(a) + n(b) n(a I B) Note: If A and B are disjoint then A I B Ø and n(a I B) 0. This gives n(a U B) n(a) + n(b). 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 0 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. 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 I B) 0. From the Venn diagram we see that girls attended only tailoring classes and girls attended only classes on gardening. Thus attended either of these classes. The number of girls who attended either of these classes is given by n(a U B) n(a) + n(b) n(a I B) Figure. 8

87 Example 46: Out of 45 houses in a village 5 houses have T.V and 0 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(a I B) x From the figure., 5 x + x + 0 x x 45 x x 0 x 0 The number of houses having both T.V and Radio 0. Figure. Example 47: In a class of 5 students, 8 students passed in History, in Mathematics and 8 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 n(h) 8, n(m) and n(h I M) 8. The number of students who have passed in History or in Mathematics or on both n(h U M) n(h) + n(m) n(h M) Figure.4 i.e., n(h U M) Since there are 5 students in the class, the number of students who have failed in both subjects 5. Using Venn diagram we find that the (i) Number of students passed in History alone n(h ) n(h I M) (ii) The number of students who passed in Mathematics alone n(m) n(h I M)

88 Exercise... Given A {,, 5, 8, 0},B {, 7, 8, 9} and C {,, 5, 8, }. Find (i) n(a U B) (ii) n(b I C) (iii) n(a B) (iv) n(c B). If n(a) 0, n(b) 4 and n(a I B). Find n(a U B). A student in IX standard has to attend at least one of two tests A and B. If 40 students attend test A, 0 students attend test B and 0 students attend both the tests. Find the number of students in the class. 4. In a survey taken among 400 residents of a colony, 50 bought English news papers, 70 bought Tamil news papers and 65 bought both. How many residents did not buy any paper at all. 5. In a village there are 00 families. Two brands of soaps A and B were popular there 60 families used brand A and 40 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 50 persons, 0 took coffee, 50 took tea and some persons took both coffee and tea. If 0 persons did not drink coffee or tea. Find the number of persons who took both coffee and tea. 7. A college magazine reported that 50 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 0 girls, 0 girls took part in singing competition and 0 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, 5 students of class XII attended computer classes. 5 students coaching classes for entrance examinations and 5 attended both. Find how many students attended neither if the class strength was

89 0. (i) (ii). 0 Answers Exercise. 9 (iii).08 0 (iv) (v) (vi) (vii).7 0 (viii) (ix) (x) (xi). 0 0 (xii) 9 0. (i) (ii) (iii) (iv) (v) (vi) (vii) 40 (viii) 4. (i) (ii) (iii) (iv) (v) Exercise... (i) T (ii) F (iii) F (iv) F (v) T (vi) F. (i) log (ii) log 4 (iii) log (iv) log 79 6 (v) log 6 6 (vi) log (i) 4 (ii) (iii) 4 (iv) (v) 4 (vi) 5 (vii) 6 (viii) 64 (ix) 4 4. (i) 0 (ii) (iii) (iv) 5 (v) 8 0 (vi) 5 (vii) (viii) 6 (ix) (x) 5. (i) C (ii) A (iii) D (iv) C (v) A (vi) D 6. (i) log 0 8 (ii) 8 50 log (iii) (iv) log 8 (v) 7 5 log 0 (vi) log (i) x + y (ii) x (iii) y x (iv) y (v) t y (vi) x + y + z (vii) (x y) (viii) y + z (ix) z + t (x) x + y (xi) t x z (xii) x y z + t 4 (iii) 7 (iv) 6 (vii) 9 (viii) 9 (ix) 7 (x) 4 Exercise.. 8. (i) (ii) (v) (xi) 0 7 (vi)

90 . (i) (ii) (iii) 0 (iv) (v) (vi) (vii) 5 (viii). (i) (ii) (iii) 4 (iv) (v) (vi). (i) 0 (ii) (iii) (iv) (v) 4 (vi) 4. (i) 4.55 (ii).55 (iii).55 (iv).55 (v) 0.55 (vi).55 (vii) 4.55 (viii) (i) (ii) 5.96 (iii).58 (iv) (v)..670 (vi).7 6. (i) (ii).705 (iii) 0.87 (iv) (v) (vi) (i) (ii) 8.56 (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (i) 04 (ii) 0550 (iii) (iv) 80 (v) (vi) (vii) (viii) (ix) 0.64 (x) 4. (xi) 0.06 (xii) 0.89 (xiii) 0.75 (xiv).89. Exercise... (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}. (i) P { T, A, M, I, L, N, D, U} (ii) Q {,,, 0,,,, 4, 5, 6} (iii) R { 8, 7, 6, 45, 54, 6, 7, 8, 90}. (i) A {x x n, n,,, 4} (ii) B {x x 5 n, n,,, 4} (iii) C {x x N, x is odd} (iv) D {x x N, x n, 0 < n 0} 4. (i) (c) ; (ii) (a); (iii) (d); (iv) (b) 5. (i) 5 (ii) 9 (iii) 8 (iv) 0 (v) 7 (vi) 4 86

91 6. (i) 000, 0000 (ii) 4, 7 7. (i) 48, 96, 9 (ii) 0,, 8. (i) No (ii) Empty set (iii) No (iv) Empty set 9. (i) F (ii) T (iii) T (iv) F (v) T 0. (i) p(a) {{}{, }{,,}{, } (ii) p(b) {{}{ x, y}{}{, z, x, y}{, x, z}{, y, z}, { x, y, z}, { }} (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},{}}. (i) (ii) 7 Exercise... (i) A U B {a, e, i, o, u, b}, A I B {a} (ii) A U B {,,, 5}, A I B {, } (iii) A U B {,, 4, 5, 6, 7, 8, 9}, A I B Ø (iv) A U B {p, q, r}, A I B Ø. (i) A B {e, f, i, o, u }, A C {b, c, d, f }, B A Ø (ii) A B {, 4}, A C {, 4, 5}, B A {6, 7, 8}. (i) A c {b, d, e, f, h} (ii) B c {g, h} (iii) (A U B) c {h} (iv) (A c I B) c {a, c, g, h} (v) A c I B c {h} (vi) A c U B c {b, d, e, f, g, h } Exercise... (i) 7 (ii) (iii) (iv)

92 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 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: 5 x + x +, (a + b) (x y),, 4 x + 9 y. 9 x + 7 If two algebraic expressions are equated, we get what are called algebraic equations. The following are some examples of algebraic equations: 7 x x + x + 6, 5 x +, x x 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 x + xy + y, 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, x + y is an algebraic expression and x and y are variables of the expression. When we substitute for x and for y, the value of x + y is () + 9. If we substitute ( ) for x and for y, then the value of x + y is ( ) + 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 x has no real number value when x 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. 88

93 4. Polynomials An Algebraic expression of the form ax n 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 x n and n, the degree of the monomial. For example, 7x is a monomial in x of degree and 7 is the coefficient of x. The sum of two monomials is called a binomial and the sum of three monomials is called a trinomial. For example, x + x is a binomial and x 5 x + 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, x + x + 6x 5 4x is a polynomial of degree 5 with terms x,, x, 6x 5 and 4x 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 x 5x x is rewritten as x + x 5x. 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 x + 5x x, then deg(f(x)). 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.. Addition of Polynomials We add two polynomials by adding the coefficients of the like powers. Example : Find the sum of x 4 x + 5x + and 4x + 6x 6x. Solution: Using the associative and distributive properties of real numbers, we obtain (x 4 x + 5x + ) + (6x 6x + 4x ) x 4 + 6x x 6x + 5x + 4x + x 4 + 6x (+6)x + (5+4)x + x 4 + 6x 9x + 9x +. 89

94 The following scheme is helpful in adding two polynomials x 4 + 0x x + 5x + 0x 4 + 6x 6x + 4x x 4 + 6x 9x + 9x Subtraction of Polynomials We subtract polynomials like addition of polynomials. Example : Subtract x x from x + 5x 4x 6. Solution: Using associative and distributive properties, we have (x + 5x 4x 6) (x x ) x + 5x 4x 6 x + x + x x + 5x + x 4x 6 + (x x ) + (5x + x ) + ( 4x) + ( 6+) x + 8x 4x 5. The subtraction can also be performed in the following way: Line (): x + 5x 4x 6. Line (): x x. Changing the signs of the polynomial in Line (), we get Line (): x + x +. Adding the polynomials in Line () and Line (), we get x + 8x 4x 5 The above procedure is written as follows : x + 5x 4x 6 x x + + x + 8x 4x Multiplication of two polynomials To find the multiplication or product of two polynomials, we use the distributive properties and the law of exponents. Example : Find the product of x x 4 and x + x. Solution: (x x 4) (x + x ) x (x + x ) + ( x ) (x + x ) + ( 4) (x + x ) (x 5 + x 4 x ) + ( 4x 4 6x + x ) + ( 8x x + 4) x 5 + x 4 x 4x 4 6x + x 8x x + 4 x 5 + (x 4 4x 4 ) + ( x 6x ) + (x 8x ) + ( x) +4 x 5 x 4 7x 6x x + 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. 90

95 x x 4 x + x x (x + x ) : x 5 + x 4 x x (x + x ) : 4x 4 6x + x 4(x + x ) : 8x x + 4 x 5 x 4 7x 6x x + 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 x in a product of two polynomials A and B, then the following scheme may be useful for the beginners: Step : Get the products given below: Coefficient of x term in A Constant term in B Coefficient of x term in A Coefficient of x term in B Coefficient of x term in A Coefficient of x term in B Constant term in A Coefficient of x term in B Step : Add the above products in step and the resulting value is the coefficient of x in the product of the polynomials A and B Example 4: Find the coefficients of x 4, x, x and x terms in the product of 7x 6x 9x + 8 and 5x x + 5 without doing actual multiplication. Solution: To get the coefficient of x 4 : Coefficient of x 4 term in A Constant term in B Coefficient of x term in A Coefficient of x term in B 7. Coefficient of x term in A Coefficient of x term in B Coefficient of x term in A Coefficient of x term in B Constant term in A Coefficient of x 4 term in B So the coefficient of x 4 in the product of A B is 0 + ( ) + ( 0) To get the coefficient of x : Coefficient of x term in A Constant term in B Coefficient of x term in A Coefficient of x term in B 6 8. Coefficient of x term in A Coefficient of x term in B Constant term in A Coefficient of x term in B So the coefficient of x in A B is ( 45)

96 To get the coefficient of x : Coefficient of x term in A Constant term in B Coefficient of x term in A Coefficient of x term in B 9 7. Constant term in A Coefficient of x term in B So the coefficient of x in A B is ( 0) To get the coefficient of x term: Coefficient of x term in A Constant term in B Constant term in A Coefficient of x term in B 8 4. So the coefficient of x term in A B is ( 45) + ( 4) Polynomials in several variables A monomial in x and y is of the form ax n y m where a is a real number, x and y are variables and n and m are non-negative integers. For example, 5x y 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, 5x y + x + y, x 8y, x + xy + y 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 x y + x y xy and x x y + y + 4xy. Solution: (x y + x y xy ) + (x x y + y + 4xy ) x y + x y xy + x x y + y + 4xy (x y x y) + (x y ) + ( xy + 4xy ) + (x ) + (y ) x y + x y + xy + x + y. Example 6: Find the product of x + y and x xy + y. Solution: x + y x xy + y Or (x + y) (x xy + y ) x(x xy + y ) : x x y + xy x (x xy + y ) + y (x xy + y ) y(x xy + y ) : x y xy + y x + x y xy + y x x y + xy + x y xy + y x + ( x y + x y) + (xy xy ) + y x + x y xy + y. The process of division of one polynomial by another polynomial is discussed in section Answer true or false: (i) x + x is a polynomial. Exercise 4. (ii) x + x + is a second degree polynomial. 9

97 (iii) The coefficient of x in 5 x x + x is. (iv) 5xy is a binomial. (v) x + y + 5z is a trinomial. In each of the problems 4, find the sum and write it in the standard form :. (x + x ) + (x 4x + 5). (x 4 + x + x) + (x 4 x + 7x 8) 4. (6 0x + 5x + x ) + (x x 4) In each of the problems 5 to 7, find the subtraction and write it in the standard form : 5. (x + 5x 0x + 6) (x x 4) 6. (x 4 x + 7x 8) (x 4 + x + x) 7. (x 5 5x + 4x 7) ( x + x x ) In each of the problems 8 to 0, find the multiplication and write it in the standard form: 8. (x 6x + ) (x 4x + 9) 9. (x 4x + 5x 7) (x x + 4) 0. (7 x x ) (x 5x + x) Without doing actual multiplication, find the coefficients of x, x and x in each of the products in problems to :. (x 4x + 4) (x + x ). (x x ) ( + x x ). (7x 6x 9x ) (x x ) Find the products in problems 4 to 6 in the standard form: 4. (ax + by) (cx + dy) 5. (x+y) (x xy y ) 6. (x xy + y ) (x + xy + y ) 7. If the coefficient of x in the product (x px + 9x ) (x x x + ) is, find the value of p. 8. If the coefficient of x in the product (x x + 5) (a x x ) is equal to the coefficient of x in the product (x + x ) (x x ), find a. 9. If the sum of the coefficients of x and x in the product ( x x ) (x mx + ) is 5, find the value of m. 4.. 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 x + 6 x, x is the variable. When we substitute for x, the equation becomes 5 5, a true statement. When we substitute any other number say 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, satisfies the equation x + 6 x but does not satisfy 9

98 the equation. We observe that the equation x (x + ) (x ) 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.. Algebraic identity for (x + a)(x + b) By using the distributive properties of numbers, (x + a )(x + b ) x(x + b) + a(x + b) x + xb + ax + ab x + ax + bx + ab x + (a + b)x + ab. Thus we get (x + a)(x + b) x + (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..). So we have (x + a)(x + b) x + ax + ab + xb x + (a+b)x + ab. We shall derive some important identities using the above identity. Figure 4.. (i) (x a)(x + b) [x + ( a)] (x + b) x + [( a) + b]x + ( a)b x + (b a)x ab. (ii) (x + a)(x b) (x + a) [x + ( b)] x + [a + ( b)]x + a( b) x + (a b)x ab. (iii) (x a)(x b) [x+( a)] [x + ( b)] x + [( a) + ( b)] x + ( a) ( b) x (a + b)x + ab. (iv) (a + b) (a + b)(a + b) a + (b + b)a + b a + ab + b. (v) (a b) (a b)(a b) [a+( b)] [a+( b)] a + [( b) + ( b)]a + ( b) ( b) a ab + b. (vi) (a + b)(a b) a + [b + ( b)]a + (b)( b) a + 0 a b a b. Thus, we have (x + a)(x + b) x + (a + b)x + ab (x a)(x + b) x + (b a)x ab (x + a)(x b) x + (a b)x ab (x a)(x b) x (a + b)x + ab (a + b) a + ab + b (a b) a ab + b (a + b)(a b) a b 94

99 The identity for (x a)(x b) is x (a + b)x + ab. It is a polynomial of nd 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 + ) (x + 5) (b) (p + 9) (p ) (c) (z 7) (z 5) (d) (x 8) (x + ) Solution: (a) (x + ) (x + 5) x + ( + 5) x + 5 x + 8x + 5. (b) (p + 9) (p ) p + (9 ) p 9 p + 7p 8. (c) (z 7) (z 5) z (7 + 5) z z z + 5. (d) (x 8) (x + ) x + ( 8)x 8 x 6x 6. Example 8: Evaluate the following products using the product formulae: (a) 07 0 (b) Solution: (a) 07 0 (00 + 7) (00+) (b) (7 + ) (50 + 6) (50 ) 50 + (6 ) Example 9: Expand the following: (i) (x + 7y) (ii) (a 7b) (iii) (p + 5q)(p 5q) Solution: (i) (x + 7y) (x) + (x)(7y) + (7y) 9x + 4xy + 49y. (ii) (a 7b) (a) (a) (7b) + (7b) a 54ab + 49b. (iii) (p + 5q)(p 5q) (p) (5q) 4p 5q. Example 0: Find the value of the following using the product formulae: (i) 0 (ii) 98 (iii) Solution: (i) 0 (00+) 00 + (00)() (ii) 98 (00 ) 00 (00)() (iii) (00 + 4) (00 4) We now deduce some useful identities from the product formulae: (i) (a + b) + (a b) (a + ab + b ) + (a ab + b ) (a + a ) + (ab ab) + (b + b ) a + b. So [(a + b) + (a b) ] [(a + b )] a + b. (ii) (a + b) (a b) (a + ab + b ) (a ab + b ) a + ab + b a + ab b 4ab. 95

100 So [(a + b) (a b) ] [4ab] ab. 4 4 (iii) (a + b) ab (a + ab + b ) ab a + ab + b ab a + b. (iv) (a + b) 4ab (a + ab + b ) 4ab a + ab + b 4ab a ab + b (a b). (v) (a b) + ab (a ab + b ) + ab a ab + b + ab a + b. (vi) (a b) + 4ab (a ab + b ) + 4ab a ab + b + 4ab a + ab + b (a + b). Thus we have the following useful identities: Reversing the sides of the above identities, we get [(a + b) + (a b) ] a + b 4 [(a + b) (a b) ] ab (a + b) ab a + b (a + b) 4ab (a b) (a b) + ab a + b (a b) + 4ab (a + b) a + b [(a + b) + (a b) ] ab 4 [(a + b) (a b) ] a + b (a + b) ab (a b) (a + b) 4ab a + b (a b) + ab (a + b) (a b) + 4ab Example : If the values of a + b and a b are 7 and 4 respectively, find the values of a + b and ab. Solution: a + b [(a+b) + (a b) ] ab 4 [(a + b) (a b) ] [(7) + (4) ] (49 + 6) [(7) (4) ] (49 6). 4 4 Example : If the values of a + b and ab are and respectively, find the values of a + b and (a b). 96

101 Solution: a + b (a + b) ab () () (a b) (a + b) 4 ab () 4() Example : If the values of a b and ab are 6 and 40 respectively, find the values of a + b and (a + b). Solution: a + b (a b) + ab (a + b) (a b) + 4ab 6 + (40) (40) Example 4: If (x + p)(x + q) x 5x 00, find the value of p + q. Solution: By product formula, we have (x + p) (x + q) x + (p + q)x + pq. So, by comparison, we get p + q 5, pq 00. Now, we have p + q (p + q) pq ( 5) ( 00) We now derive the algebraic identity for (a + b + c) (a + b + c) [(a + b) + c] (a + b) + (a + b) c+ c (a + ab + b ) + ac + bc + c a + b + c + ab + bc + ca a + b + c + (ab + bc + ca). So we have the identity Reversing the sides, we get (a + b + c) a + b + c + (ab + bc + ca) a + b + c + (ab + bc + ca) (a + b + c) Also we have (a + b + c) (ab + bc + ca) a + b + c + (ab + bc + ca) (ab + bc + ca) a + b + c. Thus, we get another algebraic identity Reversing the sides, we get (a + b + c) (ab + bc + ca) a + b + c a + b + c (a + b + c) (ab + bc + ca) Example 5: Expand the following: (i) (x + y + z) (ii) (x y + z) (iii) (p q r) (iv) (a + b c) Solution: (i) (x + y + z) [(x) + y + (z)] (x) + y + (z) + (x)y + y(z) + (z)(x) 4x + y + 4z + 4xy + 4yz + 8zx. (ii) (x y + z) [x + ( y) + z] x + ( y) + z + x( y) + ( y)z + zx x + 4y + z 4xy 4yz + zx. (iii) (p q r) [(p) + ( q) + ( r)] (p) + ( q) + ( r) + (p) ( q) + ( q) ( r) + ( r)(p). 4p + 9q + r pq + 6qr 4rp. (iv) (a + b c) [(a) + (b) + ( c)] (a) + (b) + ( c) + (a)(b) + (b)( c) + ( c)(a) 4a + 9b + 4c + ab bc 8ca. 97

102 4.. Algebraic Identity for (x + a) (x + b) (x + c) (x + a)(x + b)(x + c) (x + a)[(x + b)(x + c)] (x + a)[x + (b + c)x + bc] (x + a)(x + bx + cx + bc) x(x + bx + cx + bc) + a(x + bx + cx + bc) x + bx + cx + bcx + ax + abx + acx + abc x + (a + b + c)x + (ab + bc + ca)x + abc. Hence, (x + a)(x + b)(x + c) x + (a + b + c)x + (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)] x + [( a) + ( b) + ( c)]x + [( a) ( b) + ( b) ( c) + ( c) ( a)]x + ( a) ( b) ( c) x (a + b + c)x + (ab + bc + ca)x abc. Thus, (x a)(x b)(x c) x (a + b + c)x + (ab + bc + ca)x abc. In the above identity, coefficient of x (a + b + c), coefficient of x ab + bc + ca, constant term abc. Example 6 : Find the expansions of (i) (x + 4)(x + )(x + 5) (ii) (x + )(x )(x + 5) (iii) ( x)(x + 7)(x + ) (iv) (x a)(x a)(x a) Solution : (i) (x + 4)(x + )(x + 5) x + ( )x + [ ]x x + x + [ ]x + 60 x + x + 47x (ii) (x + )(x )(x + 5) (x + )[x + ( )](x + 5) (x) + [ + ( ) + 5](x) + [ ( ) + ( ) ](x) + ( ) 5 8x + (4x ) + [ 5 + 5](x) 5 8x + x 6x 5. (iii) ( x)(x + 7)(x + ) [ (x )](x +7 )(x + ) [x + ( )](x + 7)(x + ) [(x) + {( ) }(x) + {( ) ( )}(x) + ( ) 7 ] [8x + (5)(4x ) + ( + 7 )(x) ] [8x + 0x 4x ] + 4x 0x 8x. (iv) (x a)(x a)(x a) [x+( a)][x+( a)][x+( a)] x + {( a) + ( a)+( a)}x +{( a) ( a) + ( a) ( a) + ( a) ( a)}x + ( a) ( a) ( a) x + { 6a}x + {a + 6a + a }x 6a x 6ax + a x 6a. Example 7 : Using the product formulae, find the coefficients of x term, x term and constant term in (i) (x + )(x + 5)(x + 6) (ii) (x 7)(x + )(x + 4) (iii) (x 5)(x )(x + 4) (iv) (x )(x 5)(7 x) 98

103 Solution: (i) Comparing with (x + a)(x + b)(x +c ), we get a, b 5, c 6. Coefficient of x a + b + c , Coefficient of x ab + bc + ca ( 5) + (5 6) + (6 ) , Constant term abc (ii) (x 7)(x + )(x + 4) [x + ( 7)](x + )(x + 4) Comparing with (x + a)(x + b)(x + c), we get a 7, b, c 4. Coefficient of x a + b + c ( 7) + + 4, Coefficient of x ab + bc + ca ( 7) ( 7) ( 4) ( 8) ( 4) + 8 4, Constant term abc ( 7) (iii) (x 5)(x )(x + 4) [x + ( 5)][x + ( )](x + 4) Comparing with (x + a)(x + b)(x + c), we get a 5, b, c 4. Coefficient of x a + b + c ( 5) + ( ) + 4 ( 7) + 4, Coefficient of x ab + bc + ca ( 5) ( ) +( ) ( 5) , Constant term abc ( 5) ( )4 40. (iv) Let A be algebraic expression (x )(x 5)(7 x) and put yx. Then, A (y )(y 5)(7 y) (y )(y 5) [ (y 7)] [(y )(y 5 )(y 7)] [y + {( ) + ( 5) + ( 7)}y + {( )( 5) + ( 5)( 7) + ( 7)( )}y + ( )( 5)( 7)] [y 5y + ( )y 05] y + 5y 7y + 05 (x) + 5(x) 7(x) x + 60x 4x + 05 Coefficient of x 60, Coefficient of x 4, Constant term 05 Alternately, (x )(x 5)(7 x) 5 7 x x x x + x + x + Coefficient of x (a + b + c) , Coefficient of x 8( ab + bc + ca) ( ) (7) 4, 4 Constant term 8(abc) Example 8 : If (x + a)(x + b)(x + c) x 6x + x 6, find the value of a + b + c. Solution: From the product formula, we have (x+a)(x+b)(x+c) x + (a + b + c)x + (ab + bc + ca)x + abc. Comparing, we get a + b + c 6, ab + bc + ca, abc 6. a + b + c (a + b + c) (ab + bc + ca) ( 6) ()

104 We shall obtain some identities from the identity (x + a)(x + b)(x + c) (i) Identity for (a + b) (a + b) (a + b)(a + b)(a + b) a + (b + b + b)a + (b b + b b + b b)a + b b b a + a b + ab + b (a + b) a + a b + ab + b Reversing the sides, a + a b + ab + b (a + b) (ii) Identity for (a b) (a b) [a + ( b)] a + a ( b) + a( b) + ( b) a a b + ab b Reversing the sides, (a b) a a b + ab b a a b + ab b (a b) Based on the identities for (a+b) and (a b), we deduce the following identities: (i) (a+b) ab(a+b) a + a b + ab + b a b ab a +b. (a + b) ab(a + b) a + b Reversing the sides, a + b (a + b) ab(a + b) (ii) (a b) +ab(a b) a a b + ab b + a b ab a b. Reversing the sides, (a b) + ab(a b) a b a b (a b) + ab(a b) (iii) a + b (a + b) ab(a + b) (a + b)(a + b) ab(a + b) (a + b)[(a + b) ab] (a + b)[(a + ab + b ) ab] (a + b)(a ab + b ) Reversing the sides, a + b (a + b)(a ab + b ) (a + b)(a ab + b ) a + b (iv) a b (a b) + ab(a b) (a b)(a b) + ab(a b) (a b)[(a b) + ab] (a b)[(a ab + b ) + ab] (a b)(a + ab + b ) Reversing the sides, a b (a b)(a + ab + b ) (a b)(a + ab + b ) a b 00

105 Identity for a + b + c ab bc ca We have a + b + c ab bc ca (a + b + c ab bc ca) Thus, we get the following identity: [( a ab + b ) + ( b bc + c ) + ( c ca + a )] [( a b) + ( b c) + ( c a) ]. a + b + c ab bc ca [( a b) + ( b c) + ( c a) ] Reversing the sides [( a b) + ( b c) + ( c a) ] a + b + c ab bc ca Identity for (a + b + c) (a + b + c ab bc ca) Using the distributive properties, we get (a + b + c) (a + b + c ab bc ca) a (a + b + c ab bc ca) + b (a + b + c ab bc ca) + c (a + b + c ab bc ca) Thus, we have a + ab + c a a b abc ca + a b + b + bc ab b c abc +ca + b c + c abc bc c a a + b + c abc. (a + b + c) (a + b + c ab bc ca) a + b + c abc Reversing the sides, we get a + b + c abc (a + b + c) (a + b + c ab bc ca). 0

106 Example 9 : Expand the following : (i) (x+y) (ii) (x y) Solution: (i) (x+y) (x) + (x) (y) + (x)(y) + (y) 7x + (9x )(y) + (x)(4y ) + 8y 7x + 54x y + 6xy + 8y. (ii) (x y) (x ) (x ) (y) + (x ) (y) (y) 8x 6 (4x 4 )(y) + (x )(9y ) 7y 8x 6 6x 4 y + 54x y 7y Example 0 : If the values of a+b and ab are 4 and respectively, find the value of a + b. Solution: a + b (a+b) ab(a+b) (4) ()(4) Example : If a b 4 and ab, find a b. Solution: a b (a b) +ab(a b) (4) +()(4) Example : If a+b and a +b 8,find a +b and a 4 +b 4. Solution: a + ab + b (a+b) ab (a+b) (a + b ) () (8) ab (ab) ( 4). a + b (a+b) ab(a+b) () ( )() 8 ( 4) Alternately, a + b (a+b)(a ab + b ) (a+b)(a + b ab) () [8 ( )] (0) 0. a 4 + b 4 (a ) + (b ) [(a ) + (b )] (a )(b ) (a + b ) a b (a + b ) (ab) (8) ( ) 64 (4) Exercise 4.. Using the product formula, find (i) (x + 9) (x + ) (ii) (x + 8) (x ) (iii) (t )(t + 6) (iv) (p 4)(p ) (v) 0 06 (vi) 59 6 (vii) 4 6 (viii) Using product formulae, find (i) (5x + 8y) (ii) (s 4t) (iii) (4p + 7q)(4p 7q) (iv) (0) (v) (98) (vi) If a + b 5 and a b 4, find a + b and ab. 4. If a + b 0 and ab 0, find a + b and (a b). 0

107 5. If (x + l)(x + m) x + 4x +, find l + m and (l m). 6. Expand the following : (i) (x + y + z) (ii) (4x y + z) (iii) (p + q r) (iv) (a b c) 7. If a + b + c and ab + bc + ca 8, find a + b + c. 8. Using the product formula, find (i) (x + )(x + )(x + 4) (ii) (x + )(x + )(x 4) (iii) (x + )(x )(x + 4) (iv) (x + )(x )(x 4) (v) (x )(x )(x 4) 9. Using the product formula, find the coefficients of x term, x term and the constant term in each of the following: (i) (x + 0)(x )(x + ) (ii) (x )(x + 4)(x ) (iii) (6x + )(6x 5)(7 6x) 0. If (x + a)(x + b)(x + c) x 9x + x 5, find a + b + c, + + and a + b + c.. Expand using the product formulae: (i) (x + y ) (ii) (u 7v) (iii). If a b and ab 6, find 8a 7b. a b c x (iv) (x y + ) x. If x +, find x + x x and x + x. 4. If x + y 6 and xy 8, find x + y and x + y. 5. If p + q 6 and p + q, find pq, p + q and p 4 + q 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 + and x are factors of x 9 because x 9 (x + )(x ). Here x 9 is a second degree polynomial whereas x + and x are first degree polynomials. Thus factorization is useful in simplifying expressions. The process of factorization is also known as the resolution into factors. 0

108 4... The Process of Factorization Step : (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 : Factorize 6x 4 y 4x y + 0xy. Solution : We observe that xy is a common factor. 6x 4 y 4x y + 0xy xy 4 6x y 4x y 0xy + xy xy xy xy (x y x + 5y). Step : (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 4 : Factorize x xy x + y. Solution: The terms of the expression do not have a common factor. However, we observe that the terms can be grouped as follows: x xy x + y (x xy) (x y) x(x y) + ( ) (x y) (x y) [x + ( )] (x y) (x ). Example 5 : Factorize 6x 5 y + 6x 4 y + 9x y 4 + 9xy 5. Solution : Applying both step and step, we have 6x 5 y + 6x 4 y + 9x y 4 + 9xy 5 xy (x 4 + x y + xy + y ) xy [(x 4 + xy ) + (x y + y )] xy [x(x + y ) + y(x + y )] xy (x + y ) (x + y). Exercise 4.. Resolve into factors by finding a common factor in the following :. 9m n. 4a 8a + 6a. x 5 + 4x 4. 6x 5 y 5 + x y + 4xy 5. 7pq p q Resolve into factors by finding a common factor or by grouping method: 6. mn p pn + m 7. x x x x x ax + a 9. p p + p 0. 8x + 4x + 4x 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) X + XY + Y (ii) (X Y) X XY + Y (iii) (X + Y)(X Y) X Y (iv) ( X + Y) (X XY + Y ) X + Y (v) (X Y)(X + XY + Y ) X Y (vi) (X + Y) X + Y + X Y + XY X + Y + X Y (X +Y ) 04

109 (vii) (X Y) X Y X Y + XY X Y X Y (X Y ) (viii) (X + Y + Z) X + Y + Z + XY + YZ + ZX (ix) (X + Y + Z) (X + Y +Z XY YZ ZX) X + Y + Z XYZ 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 + XY + Y (X +Y) (ii) X XY + Y (X Y) (iii) X Y (X + Y) (X Y) (iv) X + Y ( X + Y) (X XY + Y ) (v) X Y (X Y) (X + XY + Y ) (vi) X + Y + X Y+ XY (X + Y) (vii) X Y X Y + XY (X Y) (viii) X + Y + Z + XY + YZ + ZX ( X + Y + Z) (ix) X +Y +Z XYZ (X + Y + Z) (X + Y + Z XY YZ ZX) Factorization using X + XY + Y (X + Y) Example 6: Resolve into factors 4x + xy + 9y. Solution: The given expression can be written as follows: 4x + xy + 9y (x) + (x)(y) + (y) Setting X x, Y y, the R.H.S. is X + XY + Y and so it is factored as (X + Y). Hence we get 4x + xy + 9y (x + y). Factorization using X XY + Y (X Y) Example 7 : Factorize p 8pq + 8q. Solution : The given polynomial can be written as follows: p 8pq + 8q p (p)(9q) + (9q) Setting X p and Y 9q, the R.H.S. is X XY + Y and so it is factorised as (X Y). Hence we get p 8pq + 8q (p 9q). Factorization using X Y (X + Y) (X Y) Example 8: Factorize 6x 4 y 5. Solution : Since 6x 4 y (4x y) and 5 (5), we have 6x 4 y 5 (4x y) (5) (4x y + 5) (4x y 5) Factorization using X + Y (X + Y) (X XY + Y ) Example 9: Resolve into factors 5a + 64b. Solution : Since 5a (5a) and 64b (4b), we take X 5a and Y 4b. Then 5a + 64b (5a) + (4b) X + Y (X + Y) (X XY + Y ) (5a + 4b) [(5a) (5a)(4b) + (4b) ] (5a+ 4b) (5a 0ab + 6b ) 05

110 Factorization using X Y (X Y) (X + XY + Y ) Example 0 : Factorize 6p 8q. Solution: We can write 6p (6p) and 8q (q). So, taking X 6p and Y q, we get 6p 8q (6p) (q) X Y (X Y) (X + XY + Y ) (6p q) [(6p) +(6p)(q) +(q) ] (6p q)(6p +pq+4q ). Factorization using X + Y + X Y + XY (X + Y) Example : Factorize 8x + y + x y + 6xy. Solution: 8x + y + x y + 6xy (x) + y + (x) y + (x)y [(x) + y] (x + y). Factorization using X Y X Y + XY (X Y) Example : Resolve into factors 8x 7y 6x y + 54xy. Solution : 8x 7y 6x y + 54xy (x) (y) (x) (y) + (x)(y) (x y). Factorization using X + Y + Z + XY + YZ + ZX (X + Y + Z) Example : Factorize x + 9y 6xy + 4x y + 4. Solution : The expression contains a sum x + 9y + 4. This is a sum of three squares. So, we write x + 9y 6xy + 4x y + 4 x + (y) + + x ( y) + x() + ()( y) [x + ( y) + ] (x y + ). Factorization using X + Y + Z XYZ (X + Y + Z) (X + Y + Z XY YZ ZX) Example 4: Factorize x 8y + 7z + 8xyz. Solution: Since 8y ( y) and 7z (z), the given expression P contains a sum of three cubes. So, we write P (x) + ( y) + (z) (x)( y)(z) [(x) + ( y) + (z)] [(x) + ( y) + (z) (x) ( y) ( y)(z) (z)(x)] (x y + z) (x + 4y + 9z + xy + 6yz zx) Factorization of X + Y + Z when X + Y + Z 0 X + Y + Z (X + Y + Z XYZ) + XYZ (X + Y + Z) (X + Y + Z XY YZ ZX) + XYZ (0) (X + Y + Z XY YZ ZX) + XYZ XYZ. Note: When X + Y + Z 0, X + Y Z. (X + Y) ( Z) Z. That is X + Y + XY(X +Y ) Z. That is, X + Y + Z + XY( Z) 0. X + Y + Z XY Z. Example 5: Factorize (x y) + (y z) + (z x). 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 + Y + Z XYZ 06

111 Substituting for X, Y and Z (x y) + (y z) + (z x) (x y) (y z) (z x). Factorization using grouping technique Example 6: Factorize 4x + 0xy + 5y 0x 5y. Solution: The given expression is 4x + 0xy + 5y 0x 5y (x) + (x)(5y) + (5y) 5(x) 5(5y) [(x) + (5y)] 5[(x) + (5y)] (x + 5y) 5(x + 5y) (x + 5y) (x + 5y 5). Example 7: Factorize 4a 4ab + b a + b. Solution: The given polynomial is 4a 4ab + b a + b (a) (a)(b) + (b) (a b) (a b) (a b) (a b) (a b ). Example 8: Factorize 8x 8x + 5y. Solution : Grouping the terms, we get 8x 8x + 5y [(9x) (9x)() + () ] (5y) [(9x) ()] (5y) (9x ) (5y) [(9x ) + (5y)][(9x ) (5y)] (9x + 5y ) (9x 5y ). Example 9: Factorize x 4 +. Solution: Adding and subtracting x, the given expression is x 4 + (x 4 + x + ) x [(x ) + (x )() + () ] x (x + ) ( x) [(x + ) + ( x)][(x + ) ( x)] (x + x + ) (x x + ). Example 40: Factorize x 4 + x y + y 4. Solution: Adding and subtracting x y, we get x 4 + x y + y 4 (x 4 + x y + y 4 ) x y [(x ) + (x )(y ) + (y ) ] (xy) (x + y ) (xy) [x + y + (xy)] [x +y (xy)] (x + xy + y ) (x xy + y ). Example 4: Factorize x 4 + 5x + 9. Solution : Adding and subtracting x, we get x 4 + 5x + 9 (x 4 + 6x + 9) x (x + ) x [(x + ) + x][(x +) x] (x + x + ) (x x + ). 07

112 Example 4: Resolve into factors x 8 x y 6. Solution: Taking x as a common factor, x 8 x y 6 x (x 6 y 6 ) x [(x ) (y ) ] x (x + y ) (x y ) x [(x + y)(x xy + y )] [(x y)(x + xy + y )] x (x +y)(x y)(x + xy + y ) (x xy + y ). Exercise 4... Answer True or False in the following:. x + x + ( + x ). x 6 4x + 4 (x + ). a b (a b) 4. a + b (a + b)(a + ab + b ) 5. a b (a b)(a + ab + b ) Factorize the polynomial using the factorization formulae: x + 9x 7. 44x 7x a b + 0abcd + 5c d 9. x + y a b + xy + ab 0. x y + 7z. (x + y) + 8y. (x + ) + (x ). x 6 y 6 4. (x+y) (x y) 5. (p+q) + (p q) + 6p (p q ) 6. 7x + y + 7x y + 9xy 7. x x +48x x 7y 6x y + 54xy 9. 4x + 9y + z + xy + 4xz + 6yz 0. a + b + 9c + ab 6ac 6bc. x y + + xy. 8x 5y + 80xy x 7y +z +8xyz 4. a 8b 5c 0 abc 5. (a b) + (b c) + (c a) 6. (x + y z) + (y + z x) + (z + x y) 7. (a b ) + (b c ) + (c a ) 8. a (b c) + b (c a) + c (a b) 9. x(x + z) y(y + z) 0. xy x y. x Factorization of the quadratic expression ax + 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 ax + 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 and b and c as integers. Now, we have to factorize x + 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 x + bx + c x + (p + q)x + pq (x + 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 x + 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 x + bx + c (x + p) (x + q). 08

113 Example 4: Factorize x + 9x + 8. Solution : The given expression cannot be written in the form X + XY + Y and so the factorization formula X + XY + Y (X + Y) cannot be used directly. Then we try to factorize the constant term 8. The list of all possible factorization of 8 is We list below the sum of the factors: ( 8) + ( ) ( ) + ( 8) ( ) + ( 9) ( 9) + ( ) ( ) + ( 6) ( 6) + ( ) 9. We compare the coefficient of x and the sum of the factors. We find that the sum of the factors and 6 is the coefficient of x. Hence the factorization is x + 9x + 8 (x + ) (x + 6). Example 44: Factorize x 5x Solution: The constant term 54 and the coefficient of x 5. We list below the factors of 54 and their sum. Factors Sum {, 54} 55 {, 54} 55 {, 7} 9 {, 7} 9 {, 8} {, 8} {6, 9} 5 { 6, 9} 5 Hence x 5x + 54 (x 6) (x 9). Example 45: Factorize 5 x x. Solution: Writing in the standard form, 5 x x x x + 5 ( ) (x + x 5). Here, we find 5 5, 5 + ( ) Hence, we get 5 x x ( ) [(x+5) {x + ( )}] ( ) (x +5)(x ) (x + 5) ( x). Example 46: Factorize x x. Solution: We find ( ) (), ( ) +. Hence we get x x [x + ( )] (x + ) (x ) (x + ). 09

114 Next, we consider the quadratic polynomial ax + bx + c where a, b, c are integers and a 0,. If we are able to find two integers p and q such that pq ac and p + q b. Then ax + bx + c a (a x + abx + ac) a [a x + a(p+q)x + pq] a [a x + apx + aqx + pq] a [ax (ax + p) + q(ax + p)] a (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 ax + bx + c a (ax + p) (ax + q). Example 47: Factorize x + 7x +. Solution: Here a coefficient of x b coefficient of x 7 c constant term We find a c 6 6, b. Hence x + 7x + (x + 6) (x + ) (x+)(x+). Instead of applying the final result of the rule, we can also do the factorization by splitting the middle term and grouping as follows: x + 7x + x + (6 + )x + x + 6x + x + x(x + ) + ()(x+) (x+) (x+). Example 48: Factorize 8a + a. Solution : Here, we find , 6+( 4) By splitting and grouping, we get 8a + a 8a + 6a 4a a(4a + ) ()(4a+) (4a + ) (a ) Example 49: Factorize 6 + x + x. Solution: Rewriting the expression, 6 + x + x (x + x + ). Here 4 8 and 8 + Hence, by splitting the middle term and grouping, 6 + x + x (x + 8x + x + ) [x(x + 4) + (x + 4)] (x + 4) (x + ). The method of splitting the middle term and grouping can also be tried to factorize ax + bx + c when a, b, c are real numbers. 0

115 Example 50: Factorize 7 x 0x 4. Solution : Here ( 4) 4 and ( 4) Hence, by splitting the middle and grouping,we get 7 x 0x 4 7 x 4x + 4x 4 7x ( x )+ ( x ) ( x ) (7x + ). Example 5: Factorize x + 5 x+ 6 Solution: Here 6 8 ( ) ( ) and + 5. Hence by splitting the middle term and grouping, x + 5 x + 6 x + x + x + 6 x (x + ) + (x + ) (x + ) (x+ ). The method of splitting the middle term and then grouping the terms can be followed to factorize algebraic expressions of the form ax + bxy + cy. Example 5: Factorize 5x + 7xy + 4y. Solution: Here we find and Hence by splitting the middle term and grouping, 5x + 7xy + 4y (5x + xy) + (5xy + 4y ) x(5x + 4y) + y(5x + 4y) (5x + 4y) (x +y). Example 5: Factorize 6(a ) b 5(a )b 6b Solution: Putting x a, we get 6(a ) b 5(a )b 6b 6x b 5xb 6b b(6x 5bx 6b ) Here we find ( 9) 4 and ( 9) Hence we get b(6x 9bx + 4bx 6b ) b[x(x b) + b(x b)] b(x b) (x + b) b[(a ) b] [(a ) + b] b[(a b ) (a + b )]. There are quadratic polynomials with integer coefficients which cannot be factored with integer coefficients. Example 54: Factorize x + x. Solution: Comparing with ax + bx + c, we get a, b, c. Here ac and ( ) + 0 b Hence x + x cannot be factored with integer coefficients. Note: x + x x + x + x + + x +. x + x + 4

116 Resolve into factors each of the following: Exercise x + 7x +. x + 9x + 0. d + 0d + 4. z 7z a a 7 6. x + x p 8p y y y 0y t 8t + 95 Factorize each of the following:. a + a x + 8x +. 4x + x x + x 5. 6p + 7p a a m + 6m p + 9p 9. 6x + 5x y y 6. 4x x. 9a 9a +. a a x 7x x x + 7 Resolve into factors each of the following : 6. 9x + 4xy + 5y 7. 4x 6xy 9y 8. 6c + cd 0d 9. 5x xy + 6y 0. a 5ab + 8b Factorize the following:. 0 8 x x. u u +. x x x 8x x x x + x + 7. x + x x + 0x x + 5x x + 4x 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 : The Long Division method: We explain the method through an example. Example 55: Divide + 5x + x + x by + x. Solution: Dividend f(x) + 5x + x + x Divisor g(x) + x, deg(f(x)), deg(g(x)). Our aim is to get the quotient q(x) and the remainder r(x).

117 Step : We write f(x) and g(x) in the standard form. f(x) x + x + 5x + g(x) x + Step : We divide the first term (x ) of the dividend by the first term (x) of the divisor and x obtain the first term x of the quotient. x Step : We multiply the divisor (x + ) by the first term (x ) of the quotient and subtract the product (x + x ) from the dividend. We obtain the remainder (x + 5x + ). The degree of this remainder is greater than that of the divisor. Step 4: We take the above remainder (x + 5x + ) as the new dividend and repeat step to x obtain the second term x of the quotient. x Step 5 : We multiply the divisor (x + ) by the second term (x) of the quotient and subtract this product (x + x) from the new dividend. We obtain the remainder (4x + ). The degree of this remainder is equal to that of the divisor. Step 6 : We take the remainder (4x + ) as the new dividend and repeat step to obtain the third term 4 x x 4 of the quotient. Step 7 : We multiply the divisor (x + ) by the third term (4) of the quotient and subtract this product (4x + 4) from the new dividend. We obtain the remainder ( ). 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: x + x + 4 x + x + x + 5x + x + x x + 5x + x + x 4x + 4x + 4 x x x x x x 4x 4 x Thus, we have Quotient q(x) x + x + 4, Remainder r(x). f(x)q(x) g(x) + r(x) x + x + 5x + (x + x + 4) (x + ) + ( )

118 Method : 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 9x + x 5x + 7 by x + x. Solution: 9x + x 5x + 7 [(x + x ) (x) 6x + x] + x 5x + 7 (x + x )(x) + [ 6x + x ] + [x 5x] + 7 (x + x )(x) + ( x ) + ( x) + 7 (x + x ) (x) + [(x + x ) ( ) + x ] + ( x) + 7 (x + x ) (x) + (x + x ) ( ) + [(x) + ( x)] + [( ) + 7] (x + x )(x) + (x + x ) ( ) + 6 (x + x ) (x ) + 6 So, the quotient is x and the remainder is 6. Example 57: Divide 4 7x x x x 4 by x x + 4. Solution: We find the solution by both methods. Divisor Factorization method: 4 7x x x x 4 x 4 x x 7x + 4 [(x x + 4) ( x ) 6x + 8x ] x x 7x + 4. (x x + 4) ( x ) 6x + 8x x x 7x + 4 (x x + 4) ( x ) 8x 4x 7x + 4 (x x + 4) ( x ) + [(x x + 4) ( 8x) 54x + 7x] 4x 7x + 4 (x x + 4) ( x ) + (x x + 4) ( 8x) 54x + 7x 4x 7x + 4 (x x + 4) ( x ) + (x x + 4) ( 8x) 68x +55x+ 4 (x x + 4) ( x ) + (x x + 4) ( 8x) + [(x x + 4) ( 68) 04x + 7] +55x + 4 (x x + 4) ( x ) + (x x + 4) ( 8x) + (x x + 4) ( 68) 04x x + 4 (x x + 4) ( x 8x 68) + ( 49x + 76) So, the quotient is x 8x 68 and the remainder is 49x Long Division method: x x + 4 x 8x 68 x 4 x x 7x + 4 x 4 + 6x 8x + + 8x 4x 7x + 4 8x + 54x 7x x + 55x x + 04x x x x 8x x 68x x x 8x 68 So the quotient is x 8x 68 and the remainder is 49x

119 Exercise 4.4. Find the quotient and the remainder when 4x x + x 7 is divided by (i) x + (ii) x 4 (iii) x. Find the quotient and the remainder when 5 + x 4 8x is divided by (i) (x+)(x+) (ii) (x ) (iii) x +x. Answers Exercise 4.. (i) F (ii) F (iii) F (iv) F (v) T. x + x x + 4. x 4 x + 0x 8 4. x + 5x x + 5. x + 5x 7x x 4 4x + 4x 8 7. x 5 + x 8x + 6x x 4 6x + 5x 66x x 5 + x 4 + 9x + x 9x 8 0. x 5 + 4x 4 + 9x 8x + x Coefficient of x Coefficient of x Coefficient of x acx + bdy + (ad + bc)xy 5. x x y 5xy y 6. x 4 + x y + y 4 7. p 9 8. a 7 9. m Exercise 4.. (i) x + x + 8 (ii) x + 6x 6 (iii) t + 4t (iv) p 7p + (v) 08 (vi) 658 (vii) 4 (viii) 95. (i) 5x + 80xy + 64y (ii) 9s 4st + 6t (iii) 6p 49q (iv) 00 (v) 9604 (vi) ,9/ , 0 5., 8 6. (i) 9x + y + 4z + 6xy + 4yz + xz (ii) 6x + 4y + 9z 6xy yz + 4xz (iii) 4p + 9q + 4r + pq qr 8pr (iv) 9a + 4b + 4c ab + 8bc ac (i) x + 9x + 6x + 4 (ii) x + x 4x 4 (iii) x + x 0x 4 (iv) x 5x x + 4 (v) x 9x + 6x 4 Coefficient of x Coefficient of x Constant 9. (i) (ii) 0 6 (iii)

120 0. 9,, 5 5. (i) 8x + x y + 6xy 4 + y 6 (ii) 8u 84u v + 94uv 4v (iii) x x + (iv) x 6 y 9 + 6x 4 y 6 + x y + 8 x x , ,7 5., 80,06 Exercise 4... (m n). 4a(a a + 4). x(x 4 + 4) 4. xy (6x 4 y + x + 4) 5. 7pq ( pq) 6. (m p) (n + ) 7. (x + ) (x ) (x ) 8. (x + a ) (x a ) (x ) 9. (p + ) (p ) 0. (x + ) (x + ) Exercise 4... T. F. F 4. F 5. T 6. ( + x) 7. (x ) 8. (ab + 5cd) 9. (x + y + a b) (x + y a + b) 0. ( xy + z) (x y xyz + 9z ). (x + y) (x + y ). x (x 4 + ). (x + y) (x y) (x xy + y ) (x + xy + y ) 4. y(x + y ) 5. 8p 6. (x + y) 7. (x 4) 8. (x y) 9. (x + y + z) 0. (a + b c). (x y + ) (x + y + + xy + y x). (x 5y + 6) (4x + 5y xy + 0y x). (x y + z) (4x + 9y + z + 6xy + yz zx) 4. ( a b 5c) (a + 4b + 5c + ab 0bc + 5 ca ) 5. (a b) (b c) (c a) 6. (x + y z) (y + z x) (z + x y) 7. (a + b) (a b) (b + c) (b c) (c + a) (c a) 8. abc (a b) (b c) (c a) 9. (x y) (x + y + z) 0. ( + x + y) ( x y). (x + x + ) (x x + ) Exercise 4... (x + ) (x + 4). (x + 4) (x + 5). (d + ) (d + 7) 4. (z + 7) (z 4) 5. (a + 8) (a 9) 6. (x + 0) (x 9) 7. (p ) (p 5 ) 8. (y 6) (y 7) 9. (y 9) (y ) 0. (t ) ( t 5). (a + 5) (a + ). (x + ) (x + ) 6

121 . (x + ) (x + ) 4. (x + )(x ) 5. (p + ) (6p + 5) 6. (a + ) (4a 5) 7. (m + ) (7m 5) 8. (p + 4) (8p ) 9. (x + ) (x ) 0. (y + ) (5y 6). (x ) (7x + ). (a )(a ). (a ) (a 9) 4. (x ) (4x ) 5. (4x ) (4x 7) 6. (x + y) (9x + 5y) 7. (x + y) (x 9y) 8. (c + 5d) (c d) 9. (x y) (5x 6y) 0. (a 4b) (a 7b). (x ) (5x +). (u ) (u 4). (4x )(4x ) (4x ) (4x 7) 5. (4x ) (6x ) 6. ( x + ) ( x + ) ( x + ) (x + ) 8. ( 5 x + ) ( 5x + 5) 9. (x + 5 ) (x + 5) 40. ( 7 x + )( 7x + ) Exercise 4.4 Quotient Remainder. (i) x x (ii) 4x + x (iii) 4x x 5. (i) x x 9x + 7 (ii) x + 4x + 4 (iii) x 0x + 5 7

122 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. a Problem: If a b 4 and a + b 5, then find the value of. b 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? a Answer: The value of. b Question: How would you proceed? Answer: Technique : a b 4. () a + b 5. () 9 () + () a 9 a. () () b b. b a 9 9. Technique : a a ( a + b) + ( a b) b b ( a + b) ( a b) 5 4 8

123 Technique : a b 4 () a + b 5 () a a x Let x. Then or a bx. b b () bx b 4 b (x ) 4. () bx + b 5 b (x + ) 5. b( x ) 4 x 4 or or 5x 5 4x + 4 or 5x 4x 4 +5 or x 9. b( x + ) 5 x + 5 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. Conjectures and Proofs Consider the following pattern of numbers,, 7,,, We want to know what would be the general term of the pattern. Let us analyse the terms of the pattern as follows: So we are certain that the next number in the pattern is That is, the 6 th term is and + ( ). Similarly, the 7 th term ( ), the 8 th term

124 + ( ), the 9 th term ( ). 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,,,.) of the pattern is + [ (n )]. But, in this statement, we have an unknown sum, namely, (n ). Unless, this sum is found, we can not say that we have found the general term of the pattern. Consider for example the sum This is a sum of 00 consecutive natural numbers starting with. If S is its sum, then S We can get S in the reverse way also as S That is, S S S Remember that there are 00 such 0 s So S00 0 or S Looking at the pattern through which the sum was obtained, we can guess and make the statement ( n ) n (n ). We can put n,,, 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 0 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 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 n( n ) + + n(n ) n n +. Thus, the general term in the pattern,, 7,,. is n n +. 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: 0

125 (i) x where x N. (ii) The diagonals of a square are perpendicular to each other. (iii) (a + b) (a b) a b where a, b R. (iv) is a rational number. (v) The diagonals of a rhombus are not perpendicular to each other. (vi) n( n +) n. 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.. 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 a x b, then x is called the logarithm of b to the base a. (iii) Two angles are called supplementary if their sum is 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 + 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.. 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, Q: x 4 Suppose that P is true. Then x. x x x 4. Q is true

126 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, Q: x 4 Suppose that Q is true. Then x 4. This does not mean that x, since x 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 Q : x + y and 5x 6y. Q : x and y. Suppose that Q is true. Then x + y, () 5x 6y. () () 4x + 6y 4 () and () 5x 6y. (4) () + (4) 9x 7 x. Putting x in (), 6 + y or y 6 or y. i.e., x, y i.e., Q is true Suppose that Q is true. Then x, y. x + y and 5x 6y i.e., Q is true Thus Q Q. Whenever Q Q and Q Q, the two results Q Q and Q Q are put together and written as Q Q. In this situation, we say that, Q is equivalent to Q. This means that, if Q is true, then Q is true and, if Q is true, then Q is true. That is, Q is true if and only if Q is true. Thus the symbol stands for if and only if 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.

127 (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) is an irrational number. (vii) (a + b) a + ab + b. (viii) log a (mn)log a m + log a n 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 m + where m is an integer. Now, we have n (m + ) 4m + 4m + (m + m)+. Since m is an integer, m + m is an integer. So (m + m) is an even integer. Hence (m +m)+ is an odd integer. Thus n is odd. There are several techniques by which a proof may be given. The techniques are broadly classified as follows: (i) (ii) (iii) (iv) (v) Direct Proof. Indirect Proof or Proof by contradiction. Proof by counter examples. Geometrical Proof Technique 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. a a b Example : If 5, show that b a + b Solution: Let b a 5. Then a 5b. a b 5b b 4b. a + b 5b + b 6b.

128 a c a b c d Example : Prove, by direct method, that, if,. b d a + b c + d a c Solution: Let. a a c c Put u. Then a ub. Since, we have u. b d b b d d c ud. a b ub b ( u ) b u. a + b ub + b ( u + ) b u + Similarly c d c + d ud d ud + d ( u ) d ( u + ) d u. u + a b a + b c d c + d. (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 : Prove, by contradiction method, that is an irrational number. Solution: Assume that is a rational number. We know that every rational number has a simplest form. Let q p be the proper (simplest) form of. Then q p,where p and q are positive integers having no common factor other than. But q p p q p q p is an even integer Hence is an irrational number. p is an even integer p m, where m is an integer 4m q q m q is an even integer q is an even integer q n, where n is an integer p and q have as a common factor a contradiction. 4

129 Example 4: Prove, by indirect technique, that if 00 balls are placed in 9 boxes, then there is at least one box which contains or more balls. Solution: If possible, let us assume that 00 balls are placed in 9 boxes and that no box has or more balls. Then each box has at most balls. Then the total number of balls placed in 9 boxes is But this contradicts the assumption that 00 balls have been placed in the 9 boxes. Hence there is some box which contains 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 P on AB which divides AB in the ratio m: n internally (see Figure 5.). Then AP m AP and m. PB n P B n Figure 5. n AP m PB and n AP m P B n AP m (AB AP) and n AP m (AB AP ) n AP m AB m AP and n AP m AB m AP (m + n) AP m AB and (m + n) AP m AB mab mab AP, AP. m + n m + n AP AP P and P are the same point. This is a contradiction. Hence P is unique. (iii) Proof by Counter-example: Let P and P be two statements. Suppose that we want to know whether P P. If we can find an example where P is true but P is false, then we conclude that P P. The example which we have found is called a counter-example. Example 6: If x is a real number, does x x? Solution: P : x is a real number. P : x x. For x, P is true but x 4 < x and so P is not true. Hence P P. Here is a counter example and we conclude that x x for all real x is false. 5

130 Example 7: If n is a natural number, then prove that n n. Solution: Let n be a natural number. Since n is a natural number, n > 0 and n 0. n (n ) 0. or n n 0 or n n. Example 8: For every real number x, does x > 0? Solution: P : x is a real number. P : x > 0. Take x.then P is true. But we get x 4 4 < 0. So P is not true. Thus, with x we get that P is true and P is not true. Hence P P.. Hence is a counter example and the statement x > 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) a b. Solution: Consider the following rectangle. ABCD where AB AI a, BE DI b (see Figure 5.) 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 Figure 5. + area of the rectangle DFHI a ab + (area of the rectangle CDIG area of the square CFHG) a ab + (ab b ) a b. (a + b) (a b) a b. (v) Proof by construction: When we solve some problems in geometry, we do some intermediate constructions like adding lines to the figure in order to get the solution. Construction is a good technique in many geometrical problems. But construction should be made at appropriate places. Thus proof by construction requires fore-sightedness on the part of the problem solver. 6

131 Example 0: 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. Figure 5. Exercise 5.. What is called a conjecture?. What is called a theorem?. 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 numbers such that x + y, then either x or y. 5. Mathematical Models Mathematical models are created to understand the real world problems. These models are framed by using subjects such as Algebra and Geometry. If mathematical models are formulated based upon geometric ideas, then such models are called geometric models. If algebraic ideas are used in framing mathematical models, then such models are called algebraic models. Some times mathematical models are analysed before conducting experiments. Conducting an experiment may be very expensive, time consuming and risky. Mathematical models give proof for the experimental works. Many of mathematical ideas were developed when attempts were made to study practical problems. Now we proceed to study some algebraic models and geometric models. We first consider an example. If the cost of note book is Rs., then the cost of 0 note books 0 Rs. 0 and the cost of 0 note books 0 Rs. 40. The cost of 5 note books 5 60 and the cost of note books is Rs. 6. The table below provides the variation of cost of note books with respect to the number of note books bought. 7

132 Number of Note books (x) Cost of note books (y) 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 0 through 5 and the number y among them attending high schools in a town for the years 999 to 00. The recordings are tabulated as given below: x y Here x stands for the number of people of age group 0 through 5 years and y stands for the number of them enrolled in high schools in a year. We observe that, for each year, the ratio y x 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 5,000 people of the age group 0 through 5 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 0 through 5 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 0 through 5 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 8

133 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 60 km, then the speed of the train is k.m.p.h. 4 If the train takes hours to cover the distance of 60 k.m., then the speed of the train is 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 60 k.m., the speed v is 60 0 k.m.p.h. 8 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 Speed (v) k.m From the above table, we observe that, t v 80 60, t v , t v 5 60, t v which means 60 tv 60 or v. t 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 60, 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. 9

134 Figure 5.5 In physics we come across the equation PV constant. This is an example of inverse variation. Under constant temperature, as the pressure (P) increases, the volume (V) decreases and if volume increases, the pressure decreases. The inverse variation can also be understood by considering a rectangle of given area. Suppose x and y are the length and breadth of the rectangle whose area is A. Then xy A. y x A or x y A. As the breadth y increases, the length x decreases; and as the length x increases, the breadth y decreases.. What is a mathematical model?. Explain direct variation.. Explain inverse variation. Exercise 5. 0

135 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 0 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. 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.. 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.

136 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 AC 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 BAC or CAB. Here the segments 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 Figure 6. starting from one arm to the other arm (see Figure 6.). 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 Figure 6. angle of measure 60 degrees (written as 60º) is formed. (see Figure 6.) Here the two arms of the angle coincide. Measurement of all other angles are made based on a 60º angle. When a ray makes th part of one complete rotation, an angle of measure 4

137 (60º) 90º is formed. When a ray makes th part of one complete rotation, an angle of 4 6 measure (60º) 60º is formed. When a ray makes th part of one complete rotation, 6 60 an angle of measure (60) º is formed. 60 Figure 6. Figure 6.4 Figure 6.5 If m BAC 90º then BAC is called a right angle (see Figure 6.). If m BAC is greater than 90º, then BAC is called an obtuse angle (see Figure 6.4). If m BAC is less than 90º, then BAC is called an acute angle (see Figure 6.5). If m BAC 80º, then BAC is called a straight angle (see Figure 6.6). If BAC is a straight angle, then BC is a line segment and the points A, B and C are collinear; that is, Figure 6.6 Figure 6.7 they lie on a straight line. When two lines BD and CE intersect at the point A, the two angles BAC and DAE are called vertically opposite angles. We observe that CAD and BAE are also vertically opposite angles (see Figure 6.7). If m BAC is greater than 80º but less than 60º, then the angle is called a reflex angle (see Figure 6.8). Figure 6.8 Figure 6.9 Figure 6.0

138 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 DAC are adjacent angles since they have the common vertex A and lie on the opposite sides of the common arm 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 6.0). In Figure 6.0, m BAD + m DAC 90º and so BAD and DAC are complementary angles. Here m BAD 90º m DAC and we say that one angle is the complement of the other. If the sum of the measures of two adjacent angles is 80º, then the two angles are called supplementary angles and we say that one angle is the supplement of the other (see Figure 6.). In figure BAD and DAC are supplementary angles and we observe that m DAC 80º m BAD. Figure 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. Figure 6. Through A, we can draw several (infinite number of ) lines (see Figure 6.). Of these lines, there is one and only one line, namely AB which passes through the point B. Similarly, through B, we can draw infinite number of lines. Of these lines, there is one and only one line, namely the line AB which passes through A. From the above activity, we postulate the following property: Property : Given any two points on a plane, there is one and only one line containing them. 4

139 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.). Figure 6. From the above activity, we enunciate the following property. Property : Two distinct lines cannot have more than one point in common. Note: If two distinct lines have a common point, then the lines are called intersecting lines. If two distinct lines in a plane have no point in common, then the two lines are called non-intersecting lines. Two non-intersecting lines are called parallel lines. Activity: Draw a line AB and mark a point P not on the line. Draw lines through P. We observe that there is a unique line passing through P and parallel to AB (see Figure 6.4). From this activity, we have the following property. Figure 6.4 Property : Given a line and a point not on it, there is one and only one line that passes through the given point and is parallel to the given line. Note: If three or more lines pass through the same point, then the lines are called concurrent lines. 5

140 Activity: Draw two intersecting lines AB and CD. Make the point of intersection as O. Measure the angles AOC, BOD, AOD and BOC (see Figure 6.5 We observe that m AOD m BOC, m AOC m BOD. The angles AOC and BOD are said to form a pair of vertically opposite angles. Similarly BOC and AOD are vertically opposite angles. From the above activity, we get the following: Property 4: If two lines intersect, then the vertically opposite angles are equal. Figure 6.5 Note: Let m AOC xº and m AOD yº. Then, using the above axiom, m BOD xº and m BOC yº. But m AOC + m BOC + m BOD + m AOD 60º or xº + yº + xº + yº 60º or xº + yº 60º or xº + yº 80º. So xº and yº are supplementary angles. Thus, we observe that m AOC + m BOC 80º and m BOD + m AOD 80º, m AOC + m AOD 80º and m BOC + m BOD 80º. That is, AOC and BOC, BOD and AOD, AOC and AOD and BOC and BOD form pairs of supplementary angles. Activity: Draw two parallel lines AB and CD. Draw a line PQ not parallel to AB (see Figure 6.6). Figure 6.6 We observe that PQ intersects AB and CD. Mark the common point of AB and PQ as L and the common point of CD and PQ as M. Measure the angles PLB, PLA, BLM, ALM, LMD, LMC, CMQ and QMD. We observe that m PLB m LMD, m BLM m DMQ, m PLA m LMC, m ALM m CMQ. From the above activity, we get the following property. 6

141 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 LMC are alternate interior angles. In the above activity, we observe that BLM LMC and ALM LMD. We state this fact as a theorem. Theorem : If a transversal intersects two parallel lines, then any pair of alternate interior angles are equal. (ii) The angles PLA and DMQ are said to form a pair of alternate exterior angles. The angles BLP and CMQ are alternate exterior angles. We observe from the activity that m PLA m DMQ, m BLP m CMQ. We state this fact as a theorem. Theorem : 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 80,m ALM + m LMC 80. We state this fact as a theorem. Theorem : If a transversal intersects two parallel lines, then any pair of interior angles are supplementary. Activity: Draw a line AB. Draw another line PQ which is not parallel to AB. Mark the intersecting point of AB and PQ as L. Take a point M on PQ other than L.. Measure the angle ALM. Draw a Figure 6.7 line CD through M such that m ALM m LMD (see Figure 6.7). We observe that PQ is a transversal of the lines AB and CD, and the two lines AB and CD do not meet at all. From this activity, we have the following property. Property 6: If a transversal cuts two lines such that a pair of alternate angles are equal, then the lines are parallel. As a consequence of the above property, the following theorems can be proved. Theorem 4: If a transversal intersects two lines such that a pair of corresponding angles are equal, then the lines are parallel. Theorem 5: If a transversal cuts two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the lines are parallel. 7

142 Example : In Figure 6.8, the line L is parallel to the line L and the line L is the transversal of the lines L and L. If the measures of the angles and are in the ratio 4:5, find the measures of the angles,,, 4, 5, 6, 7, and 8. Figure 6.8 m 4 4 Solution: m : m 4 : 5 m m. m 5 5 But m + m 80 since they are supplementary angles. 4 4 m + 5 m m + m 80 or or 9 m 80 5 or m 00. m 80 m Now m m (vertically opposite angles) 80, m 4 m (vertically opposite angles) 00, m 5 m (corresponding angles) 80, m 6 m (corresponding angles) 00, m 7 m 5 (vertically opposite angles) 80, m 8 m 6 (vertically opposite angles) 00. Example : In Figure 6.9 prove that AB is parallel to CD and AC is parallel to BD. Solution: Figure 6.9 CD is a transversal to the lines AC and BD and m ACD + m CDB So the two angles ACD and CDB which are on the same side of the transversal are 8

143 supplementary. Hence AC and BD are parallel. Then the corresponding angles XCA and CDB are equal. So m XCA 0. Now, we observe that AC is a transversal to the lines AB and CAB CD and m XCA m CAB 0. That is, the alternate angles XCA and are equal. Hence the lines AB and 6.. Axioms and Theorems on a triangle CD 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 6.0 DE, XY and PQ are three lines. The lines PQ and XY intersect at A, the lines DE and XY intersect at B, and the lines PQ and DE intersect at C. We name are parallel. Figure 6.0 the shape bounded by the line segments AB, BC, CA as the triangle ABC. The line segments AB, BC and CA are called the sides of the triangle ABC. The points A, B and C are called the vertices of the triangle ABC. The angles BAC, ABC and BCA are called the interior angles or simply the angles of the triangle ABC. They are also simply denoted by A, B and C respectively. The angle PAB is called an exterior angle of the triangle ABC. We observe that the angles DBA, YBC, BCQ, ECA and CAX are exterior angles of the triangle ABC. From what we have studied in our earlier classes, we recall the following definitions: (i) If no two sides of a triangle are of equal length, then the triangle is called a scalene triangle. (ii) If two sides of a triangle are of equal length, then the triangle is called an isosceles triangle. (iii) If all the sides of a triangle are of equal length, then the triangle is called an equilateral triangle. (iv) If each of the three angles of a triangle is an acute angle, then the triangle is termed an acute angled triangle or simply acute triangle. (v) If one angle of a triangle is an obtuse angle, then the triangle is called an obtuse angled triangle or simply obtuse triangle. 9

144 (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 80. 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.. Now m ACX is an external angle and A and B are interior opposite angles. We have to show that m ACX A + B. Now ACX and C are supplementary. Figure 6. ACX + C 80. But A + B + C 80. ACX + C A + B + C. Cancelling C on both sides, we get ACX A + B. Activity: Draw a triangle PQR as in Figure 6.. Measure the lengths of the sides PQ, QR and RP.Compute the sums PQ + QR, QR + PR and PR + PQ. We observe that (i) PQ + QR > PR, (ii) QR + PR > PQ, (iii) PR + PQ > QR. From the above activity, we have the following property. Property 7: The sum of any two sides of a triangle is greater than the third side. Note: We observe that in any triangle ABC, (i) AB < BC + CA (ii) BC < CA + AB (iii) CA < AB + BC. These inequalities are called triangle inequalities. Activity: Draw a triangle ABC (see Figure 6.). Measure the angles A, B and C. Measure also the lengths of the sides AB, BC and CA. Compare the measures of the angles and find out which angle has the greater measure. Compare also the lengths of the sides and find out which side has the larger length. We observe that the greater angle has the larger side opposite to it, where we have named the angle of greater measure, the greater angle and the side of greater length, the greater side. The above fact is put as the following property. Figure 6. Figure 6. 40

145 Property 8: In any triangle, the largest side has the greatest angle opposite to it. Note: Consider a line l and a point P not lying on l see Figure 6.4. Draw the perpendicular linesegment PL to the line l. The point L is called the foot of the perpendicular from P to l. Take any Figure 6.4 point M other than L on l. Now the triangle PLM is a right triangle. Since PL is perpendicular to l, m PLM 90. Since m PLM + m LMP + m LPM 80, we get 90 + m LMP + m LPM 80 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 PLM 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 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 6.5). Figure 6.5 4

146 where AB DX, BC PX, CA PD; B X, C P, A D. 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, A P 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 6.6). Figure 6.6 We are able to adjust them and we observe that the triangles fit exactly such that the vertex P coincides with A, Q coincides with B and R coincides with C. The A is called the included angle between the sides AB and AC. Similarly, P is the included angle between the sides PQ and PR. Thus, we have the following property. Property 9: If any two sides and the included angle of one triangle are equal to any two sides and the included angle of another triangle, then the two triangles are congruent. 4

147 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 B Q, BC QR and C R (see Figure 6.7) Figure 6.7 Now cut out the two triangles and place them one over the other and adjust them in such a manner that they fit exactly as one triangle. From this activity, we state the following criterion which can be proved. However, we do not give the proof here and simply state the important criterion as a theorem. Theorem 8: Two triangles are congruent if any two angles and the included side of one triangle are equal to the two angles and the included side of the other triangle. Note: The above criterion is known as Angle-Side-Angle criterion or simply as ASA criterion. We observe that in this criterion, a side and the angles on this side of one triangle should correspond to a side and the angles on it of another triangle for congruency. Activity: Draw two triangles ABC and DEF such that A D, B E and BC EF (see Figure 6.8). Figure 6.8 Since ABC is a triangle, A + B + C 80 () Since DEF is a triangle, D + E + F 80 But D A, E B. A + B + F 80 () From () and (), we get A + B + C A + B + F. C F. 4

148 Now, in triangles ABC and DEF, we observe that the side BC and the angles B and C on it correspond to the side EF and the angles 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 6.9). Figure 6.9 Cut the triangle DEF and place it over ABC and adjust such that they fit exactly as one. In this position D stands on A, E stands on B and F stands on C and ABC DEF. The result of the activity is stated as the theorem given below. Theorem 0: Two triangles are congruent if the three sides of one triangle are equal to the three sides of the other triangle. Note: The above criterion is known as Side-Side-Side or SSS criterion for congruence of two triangles. Now, we shall examine whether we can have SSA criterion or AAA criterion for congruency of two triangles. Activity: Draw a triangle ABC and a line XY parallel to the side BC (see Figure 6.0). Mark the point of intersection of line XY and the side AB as D, and the point of intersection of the line XY and the side AC as E. Since the side AB and the side AC are transversal of the parallel line segments XY and BC, D B, E C (corresponding angles). Figure

149 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.). Figure 6. In triangles ABC and ABD, we have SSA correspondence. But AC AD. So ABC ABD. Hence, we conclude that SSA correspondence cannot be taken as a criterion for congruency of two triangles. Activity: Draw line segments BC and QR such that BC QR a units. Mark the mid points of BC and QR as X and Y respectively. Draw circles with X and Y as centres and radius a units. Now cut arcs of radius b(< a) units with centres at B and Q. Mark the points of intersection of these arcs with the circles as A and P respectively (see Figure 6.). Join AB, AC, PQ and PR. We observe that AC PR. Hence ABC PQR. From the above activity, we conclude the following theorem. Figure 6. Theorem : Two right triangles are congruent if the hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle. Note: The above criterion is known as the Right-Hypotenuse-Side or RHS criterion for congruence of right triangles. 45

150 6..5 Properties of parallelogram We know that a quadrilateral is a closed figure formed by four line segments and a parallelogram is a quadrilateral in which the opposite sides are parallel to each other. Figure 6.4 Figure 6. In Figure 6., ABCD is a quadrilateral. In Figure 6.4. PQRS is a parallelogram in which PQ SR (i.e., PQ is parallel to SR) and PS QR. Property : In a parallelogram, the opposite sides are of equal length. Proof: Let ABCD be a parallelogram. Join BD (see Figure 6.5). Consider the triangles ABD and BDC. Since AB CD and BD is a transversal of AB and CD, m ABD m BDC. Since AD BC and BD is a transversal of AD and BC, m ADB m DBC. The side BD is common to both ABD and BDC. Hence, by AAS property, ABD BDC. The corresponding sides are equal. Hence AB CD and AD BC. Property : In a parallelogram, the opposite angles are Figure 6.5 of equal measure. Proof: Let ABCD be the parallelogram. Join BD (see Figure 6.6). Since AB DC and BD is a transversal to AB and DC, m ABD m BDC. Since AD BC and BD is a transversal to AD and BC, m ADB m CBD. m ABC m ABD + m DBC m BDC + m ADB m ADC. Similarly, m BAD m BCD. Figure 6.6 Property : The diagonals of a parallelogram bisect each other. Proof: ABCD is a parallelogram. AC and BD are diagonals. By ASA criterion, AMB CMD (see Figure 6.7). AM CM, BM DM. The diagonals bisect each other. Figure

151 Property 4: If the opposite sides of a quadrilateral are of equal length, then the quadrilateral 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 6.8). Then m BAC m ACD, m CAD m ACB. AB CD and AD BC. Hence ABCD is a parallelogram. Figure 6.8 Property 5: If the opposite angles in a quadrilateral are of equal measure, then the quadrilateral is a parallelogram. Proof: ABCD is a quadrilateral (see Figure 6.9). m BAD m BCD. and m ABC m ADC. Join BD. Consider the triangles ABD and CDB. Now, m + m + m BCD 80 and m + m 4 + m BAD 80 m + m + m BCD m + m 4 + m BAD Figure 6.9 m BCD m BAD m + m m + m 4 () But m + m m ABC, m + m 4 m ADC m + m m + m 4 i.e., m m m 4 m ) () + () m m 4 m m 4. AD BC. () () m m m m AB CD. Hence ABCD is a parallelogram. Property 6: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 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. m AMB m CMD, m AMD m BMC. By SAS criterion, AMB CMD, AMD CMB AB CD, AD BC. Hence ABCD is a parallelogram. Figure

152 Theorem : 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. To prove: ABCD is a parallelogram. Construction: Draw the diagonal AC (see Figure 6.4). 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. Figure 6.4 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 l, l, l are parallel to each other. PQ and XY are transversals to l, l, l, AC CE. We have to prove that BD DF. Draw AG BD and CH DF (see Figure 6.4). We observe that AGDB and CHFD are parallelograms. AG BD, CH DF () 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 () From () and () BD DF. Figure 6.4 Property 8: In a triangle, the line joining the mid points of two sides is parallel to the third side and is equal to one half of it. Proof: ABC is a triangle. D and E are mid points of AB and AC respectively. To prove DE BC and DE ( BC ). Draw CF BD to meet DE extended at F (see Figure 6.4). Consider the triangles ADE and CFE. 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. 48

153 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 DE BC. Thus DE BC and DE ( BC ). Figure 6.4 Property 9: In a triangle, the line drawn through the mid-point of one side, parallel to another side, bisects the third side. Proof: Figure 6.44 Let ABC be the given triangle and D be the mid point of AB. Draw DE parallel to BC to meet AC at E (see Figure 6.44). We have to prove that E is the mid point of CA. BC 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 () and (), m ABC m EFC, (4) AD CF and AC is a transversal to AD and CF. m BAC m ACF. (5) Now m BCF m BCE + m ECF m AED + ACF by () m AED + m DAE by (5) m BDE (since ext. angle sum of int. opp. angles) i.e., m BCF m BDE (6) 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. 49

154 6..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 Figure 6.45 theorem. Theorem : 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. Figure 6.46 From the activity, we understand the following theorem. Theorem 4: The altitudes of a triangle are concurrent. Note: The point of concurrence of the altitudes of a triangle is called the orthocenter of the triangle and is denoted by the letter H. 50

155 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 5: 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. Figure 6.48 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, Figure 6.49 BG, GE, CG, GF. We observe that AG GD BG GE CG GF. From the above activity, we understand the following theorem. Theorem 6: The medians of a triangle are concurrent and the point of concurrency divides each median in the radio :. 5

156 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. Figure 6.50 Theorem 7: Parallelograms on the same base and between the same parallels are equal in area. 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 6.5). We find the area of the triangle ABC as base height AB AL. Mark another point P on l. We find the area of ABP as base height AB AL. 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. Figure 6.5 Theorem 8: Triangles on the same base and between the same parallels are equal in area. Exercise 6. Which of the following statements are true and which are false:. If two lines are intersected by a transversal, then the alternate angles are equal.. If two parallel lines are intersected by a transversal, then the corresponding angles are equal.. If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal. 5

157 6. 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 9: If a ray stands on a line, then the sum of the adjacent angles so formed is 80. Given: The ray PQ stands on the line XY. To Prove: m QPX + m YPQ 80. Construction: Draw PE perpendicular to XY. Proof: m QPX m QPE + m EPX m QPE + 90 () m YPQ m YPE m QPE 90 m QPE () Figure 6.5 () + () m QPX + m YPQ (m QPE + 90 ) + (90 m QPE) 80. Thus the theorem is proved. Theorem 0: If two lines intersect, then the vertically opposite angles are of equal measure. Given: Two lines AB and CD intersect at the point O (see Figure 6.5). To prove: m AOC m BOD, m BOC m AOD. Proof: The ray OB stands on the line CD. m BOD + m BOC 80 () The ray OC stands on the line AB. m BOC + m AOC 80 () From () and (), 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 80 () Figure 6.5 From () and (), we get m BOC + m AOC m AOC + m AOD m BOC m AOD. Hence the theorem is proved. Theorem : The sum of the three angles of a triangle is 80. Given: ABC is a triangle (see Figure 6.54). To prove: A + B + C 80. Construction: Through the vertex A, draw the line XY parallel to the side BC. Proof: XY BC. Figure

158 Now, AB is a transversal to the lines XY and BC. m XAB m ABC (alternate angles). B. () Next, AC is a transversal to the parallel lines XY and BC. m YAC m ACB (alternate angles) C. () We also have m BAC m A. () () + () + () 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 80 m A + m B + m C. Hence the theorem is proved. Theorem : The angles opposite to equal sides of a triangle are equal. Given: 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. Figure 6.55 Theorem : The side opposite to the larger of two angles in a triangle is longer than the side opposite to the smaller angle. Given: ABC is a triangle, where B is larger than C, that is m B > m C. To prove: The length of the side AC is longer than the length of the side AB. 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 Figure 6.56 Case (i) Suppose that AC < AB. Then the side AB has longer length than the side AC. So the angle C which is opposite to AB is larger measure than that of B which is opposite to the shorter side AC. That is, m C > m B. This contradicts the given fact that m B > m C. Hence the assumption that AC < AB is wrong. AC < AB. Case (ii) Suppose that AC AB. Then the two sides AB and AC are equal. So the angles opposite to these sides are equal. That is B C. This is again a contradiction to the given fact that B > C. Hence AC AB is impossible. Now Case (iii) remains alone to be true. Hence the theorem is proved. 54

159 Theorem 4: 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 : Find the complement of the following angles: (i) 0 (ii) 45 (iii) 55 (iv) 8 Solution: Since the sum of complementary angles is 90 (i) the complement of 0 is (ii) the complement of 45 is (iii) the complement of 55 is (iv) the complement of 8 is Example 4: Find the supplement of the following angles: (i) 70 (ii) 45 (iii) 0 (iv) 55 Solution: Since the sum of supplementary angles is 80, (i) the supplement of 70 is (ii) the supplement of 45 is (iii) the supplement of 0 is (iv) the supplement of 55 is 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 0 more than the other (iii) The angles are adjacent and form an angle of 0. The larger is 0 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 x. Since the two angles are supplementary x + x 80 or x 80 or x 60. smaller angle 60, larger angle 0. Figure

160 (ii) Let x be the smaller angle. Then the larger angle x + 0. Since the angles are complementary, x + (x + 0 ) 90 or x 70 or x 5. smaller angle 5, larger angle (iii) Let x be the smaller angle. Then the larger angle x 0 The angles are adjacent and form an angle of 0. x + (x 0 ) 0. 4x 40 or x 5. smaller angle 5, larger angle (iv) Let the vertically opposite angles be x each. Since they are complementary, x + x 90 or x 90 or x 45. The angles are 45 and 45. Figure 6.59 Figure 6.60 Figure 6.6 Example 6: In Figure 6.6 the line l is a transversal to the parallel lines l and l. Find the angles x and y. Solution: Alternate angles are equal. x 0. Interior angles on the same side of the transversal are supplementary. Figure 6.6 y or y Example 7: Find x and y in Figure 6.6 where the line l 4 is a transversal to the parallel lines l, l and l. Solution: Corresponding angles are equal. x 75. Interior angles on the same side of the transversal are supplementary. y or y Figure

161 Example 8: Find the angles x and y in Figure 6.64 where the lines l and l are parallel and l is a transversal to l and l. Solution: 4y (since the interior angles on the same side of the transversal are supplementary). 4y or y. Now, since the corresponding angles are equal, x + y 9 x x Example 9: If the angles of a triangle are in the ratio : 4 : 5, find them. Solution: Let the angles be x, 4x, 5x. Then x + 4x + 5x 80 or x 80 or x 5. The angles are 5, 4 5, 5 5, or 45, 60, 75. Figure 6.64 Example 0: Find the angles x and y marked in Figure Solution: In ABC, x or x or x In BDC, x + y or 5 + y or y or y Example : Find x and y from the following figures: Figure 6.65 (i) (ii) Figure 6.66 Solution: (i) AD BC, AB CD. ABCD is a parallelogram. x 4, y 60 (alternate angles are equal) x, y 0. Figure 6.67 (ii) In triangles ACD and ACB, AD AB, CD BC and AC is common. ADC ABC. Corresponding angles are equal. x + 0 6, y 5 4 or x 6 0, y or x 6, y

162 Example : Prove that the bisector of the vertex angle of an isosceles triangle is a median to the base. Solution: Let ABC be an isosceles triangle where AB AC. Let AD be the bisector of the vertex angle A. We have to prove that AD is the median of the base BC. That is, we have to prove that D is the mid point of BC. In the triangles ADB and ADC, 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. Figure 6.68 Example : ABC is a triangle and D is the mid point of BC. DA is drawn. If DA DC, prove that BAC is a right angle. Solution: Given DA DC. Since D is the mid point of BC, BD DC. The triangles ABD and ACD are isosceles. DAB DBA () DAC DCA () () + () DAB + DAC DBA + DCA BAC DBA + DCA BAC CBA + BCA () But BAC + CBA + BCA 80 (4) (4) BAC + ( CBA + BCA) 80 Figure 6.69 BAC + BAC 80 (using ()) BAC 80 BAC 90. Example 4: Prove that the sum of the four angles of a quadrilateral is 60. Solution: Let ABCD be the given quadrilateral. We have to prove that A + B + C + D 60. Draw the diagonal AC. From the triangles ACD and ABC, we get DAC + D + ACD 80 () CAB + B + ACB 80 () () + () DAC + D + ACD + CAB + B + ACB 60 Figure 6.70 ( DAC + CAB) + B + ( ACD + ACB)+ D 60 A + B + C + D

163 Example 5: ABC is an isosceles triangle with AB AC. D is a point inside the triangle ABC such that DBC DCB. Prove that AD bisects A. Solution: Since DBC DCB, the triangle DBC is an isosceles triangle. BD DC. Already we have AB AC and AD is common. So by the SSS criterion, ADB ADC. In particular, BAD CAD. AD bisects A. Figure 6.7 Example 6: AD and BE are two altitudes of a triangle ABC such that AE BD. Prove that AD BE. Solution: In triangles ADB and AEB, we have (i) ADB AEB 90 (ii) AB is common. (iii) BD AE. By RHS criterion, ADB AEB. AD BE. Figure 6.7 Example 7: In a rectangle ABCD, E is the mid point of BC. Prove that AE ED (see Figure 6.7) Solution: In triangles ABE and DCE, we have (i) ABE DCE 90 (ii) BE CE (since E is the mid point of BC) (iii) AB CD (ABCD is a rectangle) By SAS criterion, ABE DCE. AE ED. Figure 6.7 Example 8: In a rhombus, prove that the diagonals bisect each other at right angles. Solution: Let ABCD be a rhombus, Draw the diagonals AC and BD. Let them meet at O. We have to prove that O is the mid point of both AC and BD and that AC is perpendicular ( ) to BD. Since a rhombus is a parallelogram, the diagonals AC and BD bisect each other. OA OC, OB OD. In triangles AOB and BOC, we have (i) AB BC (ii) OB is common (iii) OA OC AOB BOC, by SSS criterion. Figure 6.74 AOB BOC. Similarly, we can get BOC COD, COD DOA. AOB BOC COD DOA x (say) But AOB + BOC + COD + DOA

164 x + x + x +x x 60 or x The diagonals bisect each other at right angles. Example 9: 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) () But AD CD (since ABCD is a rhombus) ADC is isosceles. ACD DAC (angles opposite to the equal sides are equal) () From () and (), we get BAC DAC Figure 6.75 i.e., AC bisects the angle A. Similarly we can prove that AC bisects C, BD bisects B and BD bisects D. Example 0: 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 5. Find POQ Solution: Produce PO to meet CD at X. Produce QO to meet AB at Y. Since AB CD and PX is a transversal to AB and CD, OXQ OPY 45 (alternate angles). Figure 6.76 In the triangle OXQ, POQ is the exterior angle and it is equal to the sum of the interior opposite angles OXQ and OQX. So POQ OXQ + OQX Example : 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, C 80 C From ABC, A + B + C 80 A or A Figure

165 Example : ABCD is a trapezium is which AB is parallel to CD. If AD BC, prove that ADC BCD. Solution: Draw BE parallel to AD. Since ABED is a parallelogram, BE AD. But AD BC. BC BE. So the triangle BEC is isosceles. BCE BEC. () But AD BE and DEC is a transversal to AD and BE. ADC BEC (corresponding angles) () From () and (), we get BCE ADC or BCD ADC. Figure 6.78 Exercise 6.. Which of the following statements are true and which are false: (i) If a ray stands on a line, then the sum of the two adjacent angles so formed is 80 (ii) If two lines intersect, then vertically opposite angles are equal. (iii) A triangle can have two obtuse angles. (iv) The sum of the angles of a quadrilateral is 80 (v) If ABC PQR, then A Q (vi) If DEF XYZ, then DE XY (vii) In a parallelogram, the diagonals bisect each other.. Find the complement of the following: (i) 0 (ii) 65 (iii) 70 (iv) 78. Find the supplement of the following? (i) 50 (ii) 0 (iii) 80 (iv) Find the angles in each of the following: (i) The angles are complementary and the smaller is 40 less than the other. (ii) The angles are complementary and the larger is 4 times the smaller. (iii) The angles are supplementary and the larger is 58 more than the smaller. (iv) The angles are supplementary and the larger is 0 less than three times the smaller. (v) The angles are adjacent and form an angle of 40. The smaller is 8 less than the larger. (vi) The angles are vertically opposite and supplementary. 6

166 5. Find x and y from the following figures: (i) (ii) Figure 6.80 Figure 6.79 (iii) (iv) Figure 6.8 Figure In each of the following, find x and y: (i) (ii) Figure 6.8 (iii) Figure 6.84 Figure

167 Answers Exercise 6.. F. T. F Exercise 6.. (i) T (ii) T (iii) F (iv) F (v) F (vi) F (vii) T. (i) 70 (ii) 5 (iii) 0 (iv). (i) 0 (ii) 50 (iii) 00 (iv) 8 4. (i) 5, 65 (ii) 8 (iii) 6, 9 (iv) 50, 0 (v) 56, 84 (vi) 90, (i) x 0, y 50 (ii) x 80, y 70 (iii) x 0, y 0 (iv) x 50, y 0 6. (i) x 9, y 8 (ii) x 48, y (iii) x 6, y. 6

168 7. ALGEBRAIC GEOMETRY In Chapter, 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. 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.. However, they can be drawn in any way as indicated in Figure 7.. These two lines intersect at the point named as O and called the origin. The point O is fixed since the perpendicular lines are fixed. Now, let us scale the lines with the point O Figure 7. Figure 7. representing the number 0 for both the lines. We use the same scaling on the two lines. Now the two perpendicular lines become two perpendicular number lines. The positive numbers for 64

169 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.. 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.. 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.). We write the numbers a and b within the parentheses (, ) separated by a comma as (a, b) and call it the ordered pair of a and b. It is called an ordered pair because the number to the left of the comma is the x coordinate and the number to the right of the comma is the y coordinate of the point P. The ordered pair (a, b) is unique for the point P. That is, there is no other ordered pair of numbers for the same point P. The point P is represented as P(a, b) or simply (a, b). We say that P has coordinates (a, b). Thus, every point in the plane is Figure 7. represented as an ordered pair of real numbers. The plane now is called the Cartesian plane to honour the great work of Rene Descartes. It is also called the rectangular coordinate plane or the xy plane. The system of representation of points in the plane by ordered pairs of numbers is called the Cartesian or rectangular or xy- coordinate system. The two axes are called rectangular or coordinate axes. 65

170 We observe that, (i) The origin O has coordinates (0, 0). (ii) Any point on the x axis has its y coordinate 0. (iii) Any point on the y axis has its x coordinate 0. (iv) Whenever an ordered pair of real numbers is given, we can locate a unique point in the Cartesian plane and plot it by a dot at an appropriate place in the plane. (v) If a point lies within the I quadrant, then both of its coordinates are positive. If the point lies within the II quadrant, then its x coordinate is negative and y coordinate is positive. If the point lies within the III quadrant, then both of its coordinates are negative. If the point lies within the IV quadrant, then its x coordinate is positive and the y coordinate is negative. The algebraic signs of the coordinates of any point are as shown in Figure 7.4. Figure 7.4 (vi) All points on a line parallel to x-axis have the same y-coordinate(see Figure 7.5) Figure 7.5 (vii) All points on a line parallel to y-axis have the same x-coordinate (see Figure 7.6). Figure

171 If A(x,,y ) and B(x, y ) are any two points in the Cartesian plane, then the horizontal distance between A and B is x x if x > x, x x if x > x, 0 if x x and the vertical distance between A and B is y y if y > y, y y if y > y, 0 if y y. Figure 7.7 They are respectively denoted by x x and y y. 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 ( x ) + (x ) x x, AN AL + LN y + MB y +( y ) y y. Similarly, in the Figure 7.8, the horizontal and vertical distances between A(x, y ) and B(x, y ) are respectively BN ML OM OL x ( x ) x x, AN LN AL BM ( y ) ( y ) + y y y. Figure 7.8 Example : Plot the points A (, 0), B (0, ), C (4, 4), D (, ), E (.5, ), F (, ), G (, 0) and H (0, 4). Also specify the quadrant in which each point lies. Solution: The points are plotted in the coordinate plane as in Figure 7.9. The point A lies on the positive side of the x axis, the point B lies on the positive side of the y axis, the point C lies within the IV quadrant, the point D lies in the I quadrant, the point E lies in the II quadrant, the point F lies in the III quadrant, the point G lies on the negative side of the x axis and the point H lies on the negative side of the y axis. Figure

172 Example : Find the horizontal and the vertical distances between the points (, 4) and ( 9, ). Solution: The horizontal distance between the points (, 4) and ( 9, ) is a distance between the point corresponding to x coordinates and 9 on the number line x- axis; i.e., ( ) ( 9) 9 6. and the vertical distances between (, 4) and ( 9, ) is the distance between the points corresponding to y co-ordinates 4 and on the number line y axis; i.e., () ( 4) 5. Exercise 7.. Plot the following points and specify in which quadrant each point lies. (i) (, ) (ii) (7, 6) (iii) (, ) (iv) (6, ) (v) ( 9, 0) (vi) (5, 0) (vii) (0,) (viii) (, ). Answer true or false (i) (9, ) lies in the II quadrant. (ii) (, 0) lies on the y axis.. (iii) (,) lies to the right of y axis. (iv) (, ) lies below the x axis. (v) (0, 0) is the point of intersection of the coordinate axes. (vi) (, ) lies in the II quadrant. (vii) ( π, ) lies in the III quadrant. (viii) (, ) lies in the IV quadrant. (ix) (0, ) 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.. Find the horizontal and vertical distances between (i) (, 4) and (, 5). (ii) (, ) and (4, 6). (iii) (, 5) and (7,). (iv) (, ) and ( 4, ). 7. 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 7.0 or Figure 7.). Figure 7.0 Figure 7. 68

173 As P moves in one particular direction along the line, the x-coordinate of P increases (see Figures 7. and 7.). We call this particular direction, the positive direction of the line. Figure 7. Figure 7. The other direction is called the negative direction of the line. We observe that if P moves in the negative direction, then its x-coordinate decreases. Let a point P move along the line in the positive direction from the point P (x, y ) to the point P (x, y ).Then x > x. We observe that the x-coordinate of P changes from the value x to x and the y-coordinate of P correspondingly changes from the value y to the value y. The change in the x-coordinate value is x x and is called the run of the moving point P. The corresponding change in the y-coordinate values is y y and is called the rise of the moving point P. We observe that the run x x is positive (see Figures 7. and 7.) In Figure 7., the point P is moving up the line, that is it is rising up the line from P to P and the point P is at a higher position than the point P. So the rise y y is positive and y y hence the ratio is positive. In Figure 7., the point P is moving down the line; that x x is, it is falling down the line from P to P and the point P is at a lower position than the y y point P. So the rise y y is negative and hence the ratio is negative. Thus, the ratio x x y y is positive for a rising line and it is negative for a falling line. The ratio x x called the slope of the line. y x y x is Next, we shall examine the slope of a line parallel to the x-axis. In Figure 7.4, the line LL is parallel to the x-axis and we observe that all points on the line have the same y- coordinate. If P (x, y ) and P (x, y ) are two points on the parallel line, then y y and so the rise y y 0. y y Hence the slope 0. x x Figure

174 Next, let us examine the slope of a line parallel to the y-axis. Let P (x, y ) and P (x, y ) be any two points on the parallel line (see Figure 7.5). Then x x. So, the run x x 0. Since P and P are distinct, y y. Hence, the slope rise y m y y y. run x x 0 This is undefined. The slope of a line perpendicular to x-axis is undefined. Figure 7.5 y x y x y x y x y x y x y x y x > 0 for rising line. < 0 for falling line 0 for the line parallel to x-axis. is undefined for the line parallel to the y-axis. We note that y x y x ( y y ) ( x x ) y y x x y y y y points (x, y ) and (x, y ) is. x x x x From this, we observe that the slope is independent of the direction of the line. The y y slope of the line is also independent of x x the particular choice of the points P and P. To understand this fact, consider any other two points P (x, y ) and P 4 (x 4, y 4 ) on the line (see Figure 7.6). Then the slope of the line y4 y from P to P 4 is. x4 x y y The slope of the line from P and P is. x x.thus, the slope of the line joining the two Figure

175 But the triangles P AP and P BP 4 are similar. P A AP AP or BP 4. P B BP P A P B 4 y y y 4 y. x x x4 x That is, the slope is independent of the positions of two points on the line. Note: Through two given points (x, y ) and (x, y ) one and only one straight line can be y y drawn. The slope of the line is. x x Example : Find the slope of the line passing through (5,6) and (5,9) and state whether the line is rising up or falling down. Solution: Take (5,6) as (x, y ) and (5, 9) as (x, y ). Then the slope of the line is y y 9 6 m.. x x The slope is a positive number and so the line is rising up as shown in Figure 7.7. Figure 7.7 Example 4: Find the slope of the line passing through ( 6, 9) and (40, 6) and state whether the line is rising up or falling down. Solution: The slope of the line is m. 40 ( 6) 56 8 Since m is a negative number, the line is falling down as indicated in Figure 7.8. Example 5: Interpret the slopes of the following lines joining (i) (6,4) and ( 7, 4) (ii) (,8) and (, 7). Solution: y y (i) Slope of the line 0, x x 7 6 The line is parallel to the x-axis. Figure 7.8 7

176 (ii) Slope of the line y x y x The line is perpendicular to the x-axis. not defined. Example 6: Find another point on the line with slope (, ). Solution: First we shall write the slope as 5 number). Designate the given point (, ) as P (see Figure 7.9). From P, move 5 units to the right (since the run 5) to reach the point Q ( + 5, ); i.e., Q (,). From Q, move units down (since the rise ) to reach the point R (, + ( )); i.e., R (,0). The point R (,0) is another point on the line. We can verify that the slope of the line joining P and R is 0 or. ( ) The equation of a straight line which passes through the point 5 (i.e., with the denominator as a positive Figure 7.9 Let P (x, y) be a variable point on a given straight line. Then an algebraic equation connecting the variables x and y is called the equation of the straight line. The co-ordinates x and y of any point on the straight line satisfy the equation of the line. By plotting the ordered pairs (x, y) as points in the Cartesian plane, we get the graph of the straight line. A straight line is hereafter called simply a line. The graph crosses the x-axis at a unique point A and it crosses the y-axis at a unique point B. Since A lies on the x-axis its y-coordinate is 0. If a is the x-coordinate of A, then (a, 0) should satisfy the equation of the line. Substituting a for x and 0 for y in the equation of the line, we can solve for the value of a. This value of a is called the x-intercept of the line. That is, the x-intercept of the line is the x-coordinate of the point where the line crosses the x-axis. Similarly, since B lies on the y-axis, its x-coordinate is 0. So, if b is the y-coordinate of B, then (0, b) should satisfy the equation of the line. Replacing x by 0 and y by b in the equation of the line, we can solve for b. This value b is called the y-intercept of the line. Thus, the y-intercept of the line is the y-coordinate of the point where the line crosses the y-axis. 7

177 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 P (0, c) is the point at which the line crosses the y-axis (see Figure 7.0). Let P (x, y) be any point on the line. Then the slope of the line is y c y c or. x 0 x But the slope of the line is given to be m. y c m or y c mx or y mx + c. x The above equation is called the slope intercept formula for the equation of a line. Figure 7.0 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 and y-intercept. Solution: Applying the slope-intercept formula, the equation of the line is y x + ( ) m or y x 6 or x y 6 0. c y mx + c Example 8: Find the slope and the y-intercept of the line whose equation is x + 4y Solution: Rewriting the equation, we get 4y x 5 or y x Comparing this equation with y mx + c, we get slope m and y-intercept c. 4 4 Exercise 7.. Find the slope of the line joining the two given points (i) ( 4,) and ( 5, ). (ii) (4, 8) and (5, ). (iii) ( 5,0) and (0, 8). (iv) (0,0) and (, ). (v) (a, b) and (a, b). (vi) (a, 0) and (0, b).. Find another point on the line (i) through (5, 6) with slope. 7

178 (ii) through (0, 4) with slope 4. (iii) through (, ) with slope. (iv) through (, ) with slope 4. (v) through (, 4) with slope 7.. Find the equation of the line whose slope and y-intercept are (i) and 7. (ii) 5 and 9. (iii) and 5. (iv) 6 and. (v) and. 5 (vi) and Find the slope and y-intercept of the line whose equation is (i) x + y 4. (ii) x y. (iii) x y 0. (iv) 5x 4y The distance between any two points (x, y ) and (x, y ) The distance between two points is a basic concept in geometry. We now give an algebraic expression for the same. Let P (x, y ) and P (x, y ) be two distinct points in the Cartesian plane and denote the distance between P and P by d(p, P ) or by P P. Draw the line segment P P. Three cases arise. Case (i): The segment P P is parallel to the x-axis (see Figure 7.). Then y y. Draw P L and P M, perpendicular to the x-axis. Then d(p,p ) is equal to the distance between L and M. But L is (x, 0) and M is (x, 0). So the length LM x x. Hence d (P, P ) x x. Figure 7. Case (ii): The segment P P is parallel to the y-axis (see Figure 7.). Then x x.draw P L and P M, perpendicular to the y-axis. Then d(p, P ) is equal to the distance between L and M. But L is (0, y ) and M is (0, y ). So the length LM y y. Hence d(p, P ) y y. Figure 7. 74

179 Case (iii): The line segment P P is neither parallel to the x-axis nor parallel to the y-axis (see Figure 7.). Draw a line through P parallel to x-axis and a line through P parallel to y-axis. Let these lines intersect at the point P. Then P (x, y ). The length of the line segment P P is x x and the length of the segment P P is y y the triangle P P P is a right triangle.. We observe that Figure 7. [ d( P P )] [ d( P, P )] + [ d( P P )],, x x + y y (x x ) + (y y ) (x x ) + (y y ). d(p, P ) ( x y. x) + ( y ) This is called the distance formula which gives the distance d between the two given points (x, y ) and (x, y ). We observe that d(p, P ) d(p, P ). 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 P and P lie on the same horizontal line, then y y and so d(p, P ) ( x y x) + ( y ) x + 0 x x x. When P and P lie on the same vertical line, then x x and so d(p, P ) ( x y x ) + ( y ) 0 y y + y. y Note: Since the origin O is (0,0), then, for any point P (x, y), we have OP ( x 0) + ( y 0) x + y. This distance x + y is called the radius vector of the point (x, y) from the origin. (i) (ii) Using the distance formula, we can show whether three given points are collinear or form a right triangle, isosceles triangle or equilateral triangle. four given points form a parallelogram, rectangle, square or rhombus. 75

180 Example 9: Find the distance between the points A( 5, ) and B (7, ). Solution: Let d be the distance between A and B. Then d (A, B) ( x x) + ( y y) ( 7 + 5) + ( + ) (x, y ) ( 5, ) (x, y ) (7, ) Example 0: Show that the points ( 4, 9), (, 0) and (4, ) are collinear. Solution: Let A, B and C be the given points respectively. Then A ( 4, 9) B (, 0) AB ( + 4) + (0 + 9) B (, 0) C (4, ) Rough Sketch Figure 7.4 BC ( 4 ) + ( 0) AC ( 4 + 4) + ( + 9) We observe that AB + BC AC A, B and C are collinear. Example : Show that the points (, ), (, 5) and (8, 7) form an isosceles triangle. Solution: Let the given points be P, Q and R respectively. One way of proving that PQR is an isosceles triangle is to show that two of its sides are of equal length. Here we have Figure

181 d(p,q) ) + (5 ) ( + d (Q, R) (8 ) + ( 7 5) d (R, P) (8 ) + ( 7 + ) 5 + ( 5) d (P, Q) d (R, P) d (Q, R). PQR is an isosceles triangle but not an equilateral triangle. Example : Show that the points (0, ), (0,) and (,) triangle. Solution: Let the points be A, B and C respectively. One way of showing that ABC is an equilateral triangle is to show that all its sides are of equal length. Here we find that d (A, B) (0 0) + ( ) 0 + ( ) 4. are the vertices of an equilateral d (B, C) ( 0) + ( ) + 4. d (C, A) (0 ) + ( ) + 4. d (A, B) d (B, C) d (C, A). ABC is an equilateral triangle. Figure 7.6 Example : Examine whether the points P (7, ), Q ( 4, ) and R (4,5) are the vertices of a right triangle. Solution: The points P,Q, R form a triangle. To show that PQR is a right triangle, we have to show that one vertex angle is 90. This is done by showing that the lengths of the sides of the triangle satisfy Pythagoras theorem. Here PQ ( 4 7) + ( ) QR (4 + 4) + (5 + ) PR (4 7) + (5 ) PQ 5, QR 00 and PR 5. We observe that QR + PR PQ. The Pythagoras formula is satisfied. PQR is a right triangle with right angle at R. Figure 7.7 Example 4: Show that the points (, ), (, ), (5, ) and (4, 6) taken in order form a parallelogram. Is it a rectangle? Justify. Solution: Let the points be P, P,P and P 4 respectively. One way of showing that P P P P 4 is a parallelogram is to show that the opposite sides are of equal length. Here we find P P ( ) + ( ) P P (5 ) + ( + ) Figure 7.8

182 P P 4 (4 5) + (6 ) P 4 P (4 ) + (6 ) P P P P 4 0 and P P P 4 P 5. P P P P 4 is a parallelogram. Since P P (5 ) + ( ) and (P P ) + (P P ) , (P P ) 7, (P P ) + (P P ) (P P ). P P P is not a right triangle. P P P is not a right angle. P P P P 4 is not a rectangle. Example 5: Show that the points (0, ), (, ), (6, 7) and (8, ), 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 ( 0) + ( + ) BC (6 + ) + (7 ) CD (8 6) + ( 7) AD (8 0) + ( + ) AC (6 0) + (7 + ) We observe that AB CD 5, BC AD 4 5 and AB + BC AC ABCD is a rectangle but not a square. Figure 7.9 Example 6: Show that the points (0, ), (, ) (0, ) and (, ) taken in order form the vertices of a square. Solution: Let A, B, C, D be the given points respectively. One way of showing that ABCD is a square is to show that all its sides are of equal length and the diagonals are of equal length. AB ( 0) + ( + ) , BC (0 ) + ( ) , CD ( 0) + ( ) , AD ( 0) + ( + ) , Figure

183 BD ( ) + ( ) , AC (0 0) + ( + ) We observe here that AB BC CD AD and BD AC 4. ABCD is a square. Example 7: Prove that the points A(, ), B(6, 5), C(, ) 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 BC AC BD CD AD ( 6 ) + ( ) ( 6) + ( ( + 5) ) + ( ) Figure ( 6 6) + ( ( 6 + ) + ( ( 6 ) + ( AB BC CD AD, AC BD. ABCD is a rhombus but not a square. 5) ) ) Exercise 7.. Find the distance between the following pair of points: (i) (, ) and (4, ) (vi) (a, b) and ( b, a) (ii) (, 4) and ( 7, ) (vii) ( +,) and (, ) 5 (iii) ( 7, ) and (, ) (viii), 4 and (, ) 79

184 (iv) (4, 5) and ( 4, 5) (ix) (, 0) and (5, 4) (v) (a, b) and (b, a) (x) (, ) and (, 5). Examine whether the following points are collinear: (i) (5, ), (, ) and (8, 8) (ii) (, ), (, ) and (0, ) (iii) (, 4), (, ) and (, 6) (iv) ( 4, 8), (, 4) and (, 6) (v) (8, 4), (5, ) and (9, 6).. Examine whether the following points form an isosceles triangle: (i) (5, 4), (, 0) and (, ). (ii) (6, 4), (, 4) and (, 0). (iii) (, ), ( 4, ) and (, 5). 4. Examine whether the following points form an equilateral triangle: (i) (, ), (, ) and (, 4). (ii) (, ), (0, ) and (0, ). (iii) (0, ) (0, 5) and (, 4). 5. Examine whether the following points are the vertices of a right triangle: (i) (4, 4), (, 5) and (, ). (ii) (, 0), (, ) and (, 5). 6. Find the type of the triangle whose vertices are given below: (i) (, 7), ( 4, 0) and ( 0, 8). (ii) ( 5, ), (0, 6) and (8, ). 7. Examine whether the following points taken in order form a parallelogram: (i) (, 5), ( 5, 4), (7, 0) and (5, 9). (ii) (5, 8), (6, ), (, ) and (, 6). (iii) (6, ), (5, 6), ( 4, ) and (, ). (iv) (0, ), (4, 4), (6, ) and (, ). 8. Examine whether the following points taken in order form a rectangle: (i) (8, ), (0, ), (, ) and (6, 7) (ii) (, 7), (5, 4) (, 0) and ( 8, 7). (iii) (, 0), (, ), (5, 6) and (, 8). (iv) (, ), (0, 0) (, ) and (, 4) 9. Examine whether the following points taken in order form a square: (i) (, ), (, ), (, ) and (, ). (ii) (, 8), (4, 6), (, ) and (, ). (iii) (, ), (0, 4), (7, ) and (8, 0). (iv) (, 9), (0, 6), (5, 4) and (, ). (v) (, ), (, 0), (, 4) and (, ). 80

185 0. Examine whether the following points taken in order form the vertices of a rhombus: (i) (0, 0), (, 4), (0, 8) and (, 4). (ii) (, ), (6, 5), (, ) and ( 6, 7). (iii) (, 4), (5, ), (, ) and (, ) Answers Exercise 7.. (i) I (ii) I (iii) III (iv) IV (v) No quadrant (vi) No quadrant (vii) No quadrant (viii) II. (i) F (ii) F (iii) F (iv) T (v) T (vi) T (vii) T (viii) F (ix) F (x) F (xi) F (xii) T. (i) (, ) (ii) (6, 9) (iii) (0, 7) (iv) (, ) Exercise 7.. (i) (ii) 6 (iii) 8 b (iv) (v) 4 5 a (vi). (i) (6, 7) (ii) (4, 5) (iii) (, ) (iv) (, ) (v) (, ) b a. (i) x + y (ii) 5x y (iii) x + y 5 0 (iv) 6x y 0 (v) x + 5y 5 0 (vi) x + 5y (i), 5 (ii) (, 0) (iii) (, ) (iv), 4 Exercise 7.. (i) 0 (ii) 6 (iii) 0 (iv) 4 (v) (a b) 48 (vi) (a + b) (vii) 6 (viii). (i) Collinear (ii) Non-collinear (iii) Collinear (iv) Non-Collinear (v) Non-Collinear. (i) Isosceles (ii) Isosceles (iii) Isosceles (ix) 5 (x) 65. 8

186 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 0. (i) Rhombus (ii) Rhombus (iii) Rhombus 8

187 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. Figure. 8. 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.) or in the clockwise direction (see Figure 8.). If OAand OB are the initial and terminal sides of an angle, then the angle is Figure 8. 8

188 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.) 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 60 degrees (written as 60 ) is formed. Measurement of all other angles are based on a 60 angle. When a ray makes no rotation, we say that an angle of measure 0 is formed. For example, when a ray makes th of one complete rotation in the anticlockwise direction, an angle of measure 4 4 (60 ) 90 is formed. When a ray makes 4 th of one complete rotation in the clockwise direction, an angle of measure 4 (60 ) 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 80 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 80, 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, AC AB + BC. This is known as Pythagoras Theorem. Figure

189 8. Trigonometric Ratios Let us consider any acute angle AOB and denote it by the Greek letter θ. Let P be a point on the ray OB and PQ be drawn perpendicular to the ray OA. Then the triangle OQP is a right triangle having right angle at the vertex Q. The side OP is the hypotenuse side of OQP. The side PQ is opposite to the angle θ and the side OQ is the adjacent to the angle θ. We shall denote the lengths of these sides by OP, PQ, OQ respectively. Using these lengths, we define the six trigonometric ratios as follows: Figure 8.5 length of opposite side PQ sine θ, length of hypotenuse side OP length of adjacent side OQ cosine θ, length of hypotenuse side OP length of opposite side PQ tangent θ, length of adjacent side OQ length of hypotenuse side OP cosecant θ, length of opposite side PQ length of hypotenuse side OP secant θ, length of adjacent side OQ length of adjacent side OQ cotangent θ. length of opposite side PQ We abbreviate the names of the above ratios as sinθ, cosθ, tanθ, cosecθ, secθ, cotθ respectively. The values of the above ratios do not depend on the size of the right triangle OQP. To know this let P be any other point on the ray OB and P Q be drawn perpendicular to the ray OA (see Figure 8.5). Since the right triangles OQP and OQ P are similar, we get PQ P Q OQ OQ From this, we get or OP OP. PQ P Q OQ OQ PQ P Q,, OP OP OP OP OQ OQ OP OP OP OP OQ OQ,, PQ P Q OQ OQ PQ P Q 85

190 Thus the six ratios have the same value regardless of the position of the point P on the ray OB. From the above six ratios, we find that PQ OP sin θ cosecθ, cos ecθ, sin θ. OP PQ sin θ cos ecθ OQ OP cos θ secθ, sec θ, cosθ. OP OQ cosθ secθ PQ OQ tan θ cotθ. cot θ, tan θ. OQ PQ tan θ cot θ PQ sinθ We also note that, OP PQ OP cosθ OQ OP OQ OP cosθ Taking reciprocals, we get cotθ. sinθ tanθ sinθ cosθ Thus, we have tanθ, cotθ. cosθ sinθ PQ OQ tanθ. 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 : 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) length of hypotenuse side Figure 8.6 length of opposite side 6 sin θ, length of hypotenuse side 0 5 length of adjacent side 8 4 cos θ, length of hypotenuse side 0 5 length of opposite side 6 tan θ, length of adjacent side 8 4 length of hypotenuse side 0 5 cosec θ, length of opposite side 6 length of hypotenuseside 0 5 sec θ, length of adjacent side 8 4 length of adjacent side 8 4 cot θ. length of opposite side 6 86

191 Example : In ABC, m B 90, AB 8cm, AC 7cm. 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, AC AB + BC BC AC AB BC 5 5. Hence we have BC 5 AB 8 BC 5 sin A, cos A, tan A, AC 7 AC 7 AB cot A, sec A, cos ec A, tan A 5 cos A 8 sin A 5 AB 8 BC 5 AB 8 sin C, cos C, tan C, AC 7 AC 7 BC cot C, sec C, cos ec C, tan C 8 cos C 5 sin C 8 Figure 8.7 Note: In the above problem, we observe that sin C cos A, cos C sin A, tan C cot A,. This is so because the angles A and C are complementary. 7 Example : If sin θ, find the other trigonometric ratios. 5 7 Solution: Since sin θ, let us consider a right triangle ABC in which m ABC 90, 5 m ACBθ, AB 7 and AC 5 (see Figure 8.8). By Pythagoras formula, AC AB + BC BC or BC BC BC BC 4 Hence, cosθ, AC 5 AB 7 tan θ, BC 4 5 cosecθ, sin θ 7 5 secθ, cosθ 4 4 cot θ. tan θ 7 Figure 8.8 Example 4: If cosec A, find (i) sin A + cos A (ii) tan A + cot A. Solution: Since cosec A length of hypotenuse side, length of opposite side we consider a right triangle PQR where m QRP A, 87

192 PR and PQ (see Figure 8.9). By Pythagoras theorem, PR PQ + QR ( ) () + QR + QR QR QR. Hence we get PQ sin A, PR PQ tan A, QR QR cos A, PR QR cot A. PQ (i) sin A + cos A +, (ii) tan A + cot A +. Figure 8.9 Note : Whenever we are asked to prove an equation, we adopt any one of the following methods: Method : Simplify the expression in the L.H.S. or R.H.S and obtain the expression on the other side. Method : Simplify the expression on the L.H.S to a form (). Next simplify the R.H.S to a form (). Show () (). tan A+ tan B sin Acos B + cos Asin B Example 5: Prove that. tan A tan B cos Acos B sin Asin B Solution: sin A sin B sin Acos B + cos Asin B + tan A+ tan B L.H.S cos A cos B cos Acos B tan Atan B sin A sin B cos Acos B sin Asin B cos A cos B cos Acos B (sin Acos B + cos Asin B) cos Acos B cos Acos B cos Acos B sin Asin B sin Acos B + cos Asin B R.H.S. cos Acos B sin Asin B Example 6: Prove that tan A+ cot B Solution: L.H.S cot A + tan B tan A+ cot B tan A. cot A + tan B tan B tan A + tan B tan B + tan A tan Atan B + tan B + tan Atan B tan A 88

193 (tan A tan B + ) tan A. tan B ( + tan A tan B) tan A R.H.S. tan B + tan θ sin θ + tan θ Example 7: Show that. + cot θ + cosθ + tan θ + tan θ Solution: L.H.S. + cot θ + tanθ ( + tan θ) ( + tan θ) tan θ tan θ () tan θ + ( tan θ + ) tan θ sin θ sin θ + sin θ + tan θ R.H.S. cosθ + cosθ + cosθ (sin θ cosθ + sin θ) sin θ (cosθ + ) cosθ ( + cosθ) cosθ (+ cosθ) sin θ tanθ` () cos θ From () and (), 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 0, 45 and 60. (i) Trigonometric ratios of 0 and 60 angles We consider an equilateral triangle ABC with sides of length (see Figure 8.0) and draw CD perpendicular to AB. Then D bisects the side AB. Now AD, AC. m DAC 60, m ACD 0 and ADC is a right triangle. By Pythagoras theorem, AC AD + DC + DC DC DC. Figure

194 Here from the right triangle ADC, (see Figure 8.) we get sin 60 DC AC DA cos 60 AC DC tan 60 DA cot 60 DA DC AC sec 60 DA cosec 60 AC DC sin 0 cos 0 tan 0 AD AC DC AC AD DC DC cot 0 AD sec 0 AC DC AC cosec 0 AD Figure 8. (ii) Trigonometric ratios of a 45 angle. We consider the isosceles right angle ABC where m B 90, AB BC (see Figure 8.). Then m CAB 45 and m BCA 45. Now from the right triangle ABC, we get, by Pythagoras theorem, AC AB + BC + and so AC, AB sin 45 AC AB cos 45 AC BC tan 45 AB AB cot 45 BC AC sec 45 AB Figure 8. AC cosec 45 BC (iii) Trigonometric ratios of a 0 angle and a 90 angle To get the trigonometric ratios of these two angles, we consider a circle of radius r with center at the origin in the Cartesian coordinate plane. Let P be any point on the arc of the circle in the positive quadrant of the coordinate plane (see Figure 8.). Let PM be drawn perpendicular to the x axis. Let the coordinates of P be x and y. Figure 8. 90

195 Then OM x and PM y. By applying Pythagoras theorem in the right triangle OMP, we get x + y r ; r x + y. Let MOP θ. Then θ is an acute angle and y x sin[θ, cos θ. r r Choosing P at different positions on the arc, we note that as the ray OP turns from the position OAto 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, r x decreases from to 0 and y increases from 0 to. That is, as θ increases from 0 to 90, cos θ decreases from to 0 r and sin θ increases from 0 to. 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 o 0 o r 0, x r and y 0. So we have sin 0 0, cos 0. When OP is in the position r r o r o 0 OB, θ 90, x 0 and y r. So, we have sin 90, cos r r o o o sin 0 0 o cos0 Now, we have tan 0 0, cot 0 not defined, o o cos 0 sin 0 0 o o sec 0 ; cos ec 0 not defined, o o cos 0 sin 0 0 o o o sin 90 o cos90 0 tan 90 not defined, cot 90 0, o o cos90 0 sin 90 o o sec 90 not defined, cos ec90. o o cos90 0 sin 90 All the trigonometric ratios for angle of measures 0, 0, 45, 60 and 90 are provided in the following table: θ sin θ 0 cos θ tan θ 0 cot θ Not defined sec θ cosec θ Not defined Table - 0 Not defined 0 Not defined 9

196 Notation. We write (sinθ) as sin θ but not as sin θ. Similarly (tan θ) is written as tan θ. Example 8: Evaluate cos 0 tan 60 sec 45 sin 60. Solution: cos 0, tan 60, sec 45, sin 60. cos 0 tan 60 sec 45 sin 60 ( ) ( ) o sin 60 Example 9: Find the acute angle A if tan A. o + cos60 Solution: sin 60, cos 60 tan A. But tan 0. A 0. + Example 0: If sin (A + B) and cos B, find A and B. Solution: Since sin (A + B), we get sin (A + B). But sin 60. So A + B 60. () Since cos B, we get cos B. But cos 45. So B 45. () Solving () and (), A 5. 9

197 Exercise 8. In problems to 4 find the indicated trigonometric ratios in the given right triangle... sin B, cos C, tan B sec X, cot Z, cosec Z Figure 8.4 Figure cos Q, tan R, cot Q tan M, sec N, cosec N Figure 8.6 Figure 8.7 In problems 5 to 0 find the other trigonometric ratios. 5. cos θ 6. sin θ 5 8. cosec θ 0 9. cot θ 7 7. sec θ 0. tan θ 5. If 5 cos A, find 7 sec A+ tan A sec A tan A cos ecθ. If sin θ, find 5 cot θ cosθ. If cosec θ, find the value of cot θ + sin θ. + cosθ 9

198 4. If cot θ cos θ, show that. sin θ 5 sin θ + cosθ 5. If cot θ 4, find the value of. secθ + cos ecθ 6. Evaluate (i) cosec 45 cot 0 + sin 60 sec 0 (ii) cos 0 sin 0 cos 60 (iii) 8 sin 60 cos 60 (iv) o tan 45 o o tan 0 + tan Verify the following: (i) sin 0 + cos 0 (ii) sec 60 tan 60 (iii) + cot 0 cosec 0 8. If sin (A+B) sin (A B), find A and B. 8. 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 8.8). Let PQ be drawn perpendicular to x axis. Then OQ x, PQ y. Let OP r. Applying Pythagoras identity in the right triangle OQP, we get x + y r. Dividing both sides by r, x r + r y x + y r. r r (or) But sin θ r y, cos θ r x. (cos θ) + (sin θ). Figure 8.8 or cos θ + sin θ () Dividing both sides of () by cos θ, we get cos θ + sin cos θ θ cos θ sin θ cos cos + θ θ (or) cos cos θ θ sin + cos (or) +(tan θ) (sec θ) θ θ cos θ 94

199 Dividing both sides of () by sin θ, we get PQ OQ OP P Q O Q O P (or) cos sin θ θ (or) + tan θ sec θ () sin θ + (cosecθ) (or) cot θ + cosec θ sin θ sin θ (or) + cot θ cosec θ () The three identities (), () and () are based on the Pythagoras identity. We deduce some more identities from them. Using the fundamental identity (), we have (i) sin θ (sin θ + cos θ) cos θ cos θ. (ii) cos θ (cos θ + sin θ) sin θ sin θ. Using the fundamental identity (), we get (i) tan θ ( + tan θ) sec θ (ii) sec θ tan θ ( + tan θ) tan θ. Using fundamental identity (), we get (i) cot θ ( + cot θ) cosec θ (ii) cosec θ cot θ ( + cot θ) cot θ. We list the identities in the following table sin θ + cos θ + tan θ sec θ + cot θ cosec θ sin θ cos θ cos θ sin θ tan θ sec θ sec θ tan θ cot θ cosec θ cosec θ cot θ We again mention here that an identity is applied in both ways, left to right or right to left. Example : Prove that sin 4 θ + cos 4 θ sin θ cos θ. Solution: L.H.S. sin 4 θ + cos 4 θ (sin θ) + (cos θ) [sin θ + cos θ] (sin θ)(cos θ) ( a + b (a + b) ab) () sin θ cos θ sin θ cos θ R.H.S. 95

200 cosθ Example : Prove that secθ tan θ. + sin θ Solution: cosθ cosθ sin θ L.H.S. + sin θ + sin θ sin θ cosθ ( sin θ) cosθ ( sin θ) sin θ cos θ sin θ sin θ cos θ cosθ cosθ sec θ tan θ R.H.S. + cos A Example : Prove that (cos ec A+ cot A). cos A Solution : + cos A + cos A L.H.S cos A + cos A ( + cos A) ( + cos A) + cos A + cos A sin A sin A sin A (cosec A + cot A) R.H.S. Alternately, R.H.S. (cosec A + cot A) cos A + cos A + sin sin A A sin A AC (+ cos A) AD cos A (+ cos A) + cos A (+ cos A)( cos A) cos A L.H.S. Example 4: Prove that sin 4 θ cos 4 θ sin θ cos θ. Solution: L.H.S. sin 4 θ cos 4 θ (sin θ) (cos θ) (sin θ + cos θ) (sin θ cos θ ) () (sin θ cos θ) sin θ cos θ R.H.S. Example 5: Prove that sec A tan A. sec A + tan A Solution : sec A tan A R.H.S. sec A + tan A sec A+ tan A sec A tan A sec A tan A sec A tan A sec A tan A L.H.S. sec A tan A cos sin A A 96

201 Example 6: Prove that (sec θ + cosθ) (secθ cosθ) tan θ + sin θ. Solution : L.H.S. (sec θ + cosθ) (secθ cosθ) sec θ cos θ ( + tan θ) cos θ tan θ + ( cos θ) tan θ + sin θ R.H.S. Example 7: Prove that + cosec θ. + cosθ cosθ Solution : ( cosθ ) + ( + cosθ ) L.H.S. + + cosθ cosθ ( + cosθ )( cosθ ) cosθ + + cosθ cos θ sin θ cosec θ R.H.S. Example 8: Prove that sin A sin B + cos A cos B + sin A cos B + cos A sin B. Solution : L.H.S. (sin A sin B + sin A cos B) + (cos A cos B + cos A sin B) sin A (sin B + cos B) + cos A (cos B + sin B) sin A() + cos A () sin A + cos A R.H.S Example 9: If m tan A + sin A, n tan A sin A. Prove that m n 4 mn. Solution : L.H.S. m n (tan A + sin A) (tan A sin A) tan A + sin A + tan A sin A (tan A + sin A tan A sin A) 4 tan A sin A R.H.S 4 mn y r 4 sin tan A sin A 4 A sin A cos A 4 sin A sin Acos A sin A( cos A) 4 cos A cos A 4 sin A tan A 4 sin A tan A L.H.S. Example 0: Prove that cos 6 θ + sin 6 θ cos θ sin θ Solution : L.H.S cos 6 θ + sin 6 θ (cos θ) + (sin θ) (cos θ + sin θ) (cos 4 θ cos θ sin θ + sin 4 θ) () (cos 4 θ + sin 4 θ cos θ sin θ) [(cos θ) + (sin θ) ] cos θ sin θ [(cos θ + sin θ) cos θ sin θ] cos θ sin θ () cos θ sin θ cos θ sin θ R.H.S. 97

202 Example : Prove that + sinθ cosθ sinθ + cosθ Solution: L.H.S sin θ + cos θ + sin θ cosθ sin θ + cos ( sin θ + cosθ) sinθ + cosθ R.H.S θ Exercise 8.. Prove that sec A sin Asec A.. Prove that (sin A + cos A) + (sin A cos A). cot θ cosec θ. Simplify :. sec θ tan θ sec A+ tan A + sin A 4. Prove that. sec A tan A sin A 5. Prove that + sec θ. + sin θ sin θ 6. If x r sin A sin B, y r sin A cos B, z r cos A, find the value of x + y +z. 7. Show that tan A + cot A cosec A sec A. tan A 8. Prove that cos A. + tan A 9. Prove that cosec θ + cot θ. cos ecθ cot θ 0. Prove that (tan A + cot A) sec A+ cosec A. cos A sin A. Prove that + sin A+ cos A. tan A cot A tan A+ sec A + sin A. Prove that. tan A sec A+ cos A. Prove that (tan A tan B) + ( + tan A tan B) sec A sec B. tan θ cot θ 4. Prove that + sec θ cosec θ +. cot θ tan θ + sin θ cosθ cosθ 5. Prove that sin θ cosθ cosθ 6. Prove that (sin θ + cosec θ) + (cos θ + sec θ) 7 + tan θ + cot θ. cos θ + sin θ cos θ sin θ 7. Prove that +. cosθ + sin θ cosθ sin θ 4 4 cos θ + sin θ 8. Prove that tan θ. 4 4 sin θ + cos θ 98

203 8. Trigonometric Ratios For Complementary Angles We have already learnt about complementary angles in a right triangle. In the right triangle OQP (see figure 8.9), 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 PQ OQ PQ sin θ, cosθ, tanθ, OP OP OQ Figure () OP OP OQ cos ecθ, secθ, cotθ PQ OQ PQ Similarly, for the angle 90 θ, we get o PQ o OQ sin (90 - θ) cos(90 θ ), tan(90 θ ), 4 OP PQ.. () o OP o OP o PQ cosec(90 θ ), sec(90 θ ),cot (90 θ ) OQ PQ OQ Comparing () and (), we get PQ o sinθ cos (90 θ) OP OQ cos θ sin (90 θ) OP PQ tan θ cot (90 θ) OQ OP cosec θ sec (90 θ) PQ OP sec θ cosec (90 θ) OQ OQ cot θ tan (90 θ). PQ Hence, we obtain the following table: sin (90 θ) cosθ cos (90 θ) sin θ tan (90 θ) cot θ cot (90 θ) tan θ sec (90 θ) cosec θ cosec (90 θ) sec θ 99

204 tan 65 Example : Evaluate cot 5 Solution : tan 65 tan(90 5 ) cot 5 o o tan 65 cot 5. o o cot 5 cot 5 Example : Evaluate sin 0 tan 60 sec 70 Solution : sec 70 sec(90 0 ) cosec 0 o o. sin 0 sin 0 tan 60 sec 70 sin 0 tan 60 cosec 0 sin 0. o sin 0 Example 4: Find x if cosec x sec 5. Solution: Since cosec x sec(90 x ),we have sec(90 x ) sec x 5. x 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.. Evaluate (i) o o sin6 tan5 (ii) o o cos54 cot55 (iii) sin θ sec (90 θ). Simplify: (i) o o o tan sin 4 sec5 + + o o o cot 57 cos 48 cos ec9. (ii) o o sin sec o o cos67 cosec4.. Find x if (i) sin 60 cos x (ii) cosec x cos54 (iii) sec x cosec 5 (iv) tan x tan 5 o Answers Exercise , 5 5, 4, tan 5. 5,, 5. 8, sin θ θ, cos ecθ, secθ, cot θ. 4, ,, 00

205 cos θ, tan θ, cos ecθ, secθ, cot θ sin θ, cosθ, tan θ, cosecθ, cot θ sin θ, cosθ, tan θ, secθ, cot θ sin θ, cosθ, tan θ 7, cosecθ, secθ sin θ, cosθ, cosecθ, secθ, cot θ (i) 7 (ii) 0 (iii) (iv) 4 8. A 60, B r. (i) (ii) (iii). (i) (ii) 7 Exercise 8. Exercise 8.. (i) 0 (ii) 6 (iii) 65 (iv) 55 0

206 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. 0

207 9. 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 :. (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.. 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 :. 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 : Draw the given triangle ABC. Step : Locate the mid point D of the side BC and draw the median AD. Step : 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 : 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.). Now we proceed to locate the centroid. The steps are given below: 0

208 Step : Draw the line segment BC with BC 5.5 cm. With B as centre, cut an arc with radius 7 cm. Similarly with C as centre, cut an arc of radius 7.5 cm. These two arcs intersect at A. Now draw AB and AC. The triangle ABC is drawn. Step : We locate the mid point D of BC and the mid point E of AC by the perpendicular bisector method. Step : Draw AD and BE. They meet at the point G. G is Figure 9. 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.5cm. We observe AG 4.5 that. That is G divides AD in the ratio :. GD.5 Figure Circumcentre The perpendicular bisectors of the sides of a triangle are concurrent at a point. This point is called the circumcentre of the triangle and is usually denoted by the letter S. It is at an equal distance R from the vertices of the triangle. The circle drawn with S as the centre and the equidistance R as radius passes through the vertices of the triangle. This circle is called the circumcircle of the triangle and R is called its circumradius. To locate the circumcentre and the circumcircle, we adopt the following procedure: Step : Draw the triangle ABC. Step : Draw the perpendicular bisectors of BC and AC. Step : Mark the point of intersection of the perpendicular bisectors of BC and AC as S. This point S is the circumcentre of ABC. 04

209 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 : 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 : Draw the perpendicular bisectors of BC and AB. Step : Mark the meeting point S of the perpendicular bisectors. S is the circumcentre. Step : Measure the lengths SA, SB and SC. We find SA SB SC.6 cm Step 4: Draw a circle with S as the centre and SA as Figure 9. 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 drawn, we say that the triangle is circumscribed. 9.. 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 Figure 9.5 circle and it is denoted by r. The circle drawn with I as centre and r as radius touches all the sides of the triangle internally. The circle is said to be inscribed in the triangle and it is called the incircle of the triangle (see Figure 9.5) 05

210 To locate the incentre, to measure the inradius and to draw the incircle of a triangle, the steps are given below: Step : Draw the triangle ABC. Step : Draw the internal bisectors of the angles B and C. Step : 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 : 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 : Draw the PQR using ASA method. Step : Draw the angle bisector of P. Step : 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. Figure 9.6 Step 6: Measure the length of ID. This length is the inradius of the triangle. We find ID cm. Step 7: With I as centre and ID as radius, draw a circle. This is the incircle of the triangle. Figure Orthocentre We recall that an altitude of a triangle is a perpendicular line segment drawn from a 06

211 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. The steps for locating the orthocentre H are given below: Step : Draw the triangle ABC with the given measurements. Step : Draw the altitudes AL, and BM. Figure 9.8 Step : 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 : Draw the triangle XYZ using SSS Rough Figure procedure. Step : Draw the altitudes XL, YM. Step 4: Mark the point of intersection of XL and YM as H. H is the orthocentre of XYZ. Step 5: Join ZH and produce it to meet XY at N. We observe that ZN is the altitude through Z. Figure 9.9 Figure 9.0 If we locate the centroid, circumcentre, incentre and orthocentre of various types of triangles, we observe that their positions are as given below: 07

212 Table Point of Location of the point of Type of triangle concurrence concurrence Centroid Any type of triangle Inside the triangle Acute angled triangle Inside the triangle Circumcentre Right triangle Mid point of the hypotenuse Obtuse angled triangle Outside the triangle Incentre Any type of triangle Inside the triangle Acute angled triangle Inside the triangle Orthocentre Right triangle Vertex of the right angle Obtuse angled triangle Outside the triangle Exercise 9. In problems to 4, locate the centroid G of the given triangle.. ABC, where BC 6cm, m B 40º, m C 60º.. ABC, where all sides are of 6.5 cm long.. PQR, where m R 90º, PQ 7 cm, PR 6 cm. 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 00º. 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, draw the incircle and measure its inradius. 9. ABC, where AB 9cm, BC 7cm, CA 5cm. 0. XYZ, where XY YZ ZX 8cm.. PQR, where PQ 0cm, m P 90º, m Q 60º.. ABC, where AB 5.4 cm, m A 50º, AC 5cm. In problems to 6, locate the orthocentre of the triangle.. ABC, where BC 5.6cm, m B 55º, m C 65º. 4. PQR, where m P 90º, m Q 0º, PQ 4.5 cm. 5. LMN, where, LM 7cm, m M 0º, MN 6 cm. 6. XYZ, where XY 7cm, YZ 5 cm, ZX 6 cm. 08

213 9. Geometrical Interpretation of Averages We shall now study geometrical construction methods to find the arithmetic mean and geometric mean between two positive numbers Arithmetic mean a + b The Arithmetic mean between two numbers a and b is. To find the number a + b geometrically, we proceed as follows: Step : Take a line segment PQ whose length is a + b. Step : Draw the perpendicular bisector of PQ. Step : The meeting point of PQ and its perpendicular bisector is marked and its distance a + b from P or Q is measured. This distance gives. The reason is that if a line segment XY l is of length l, then the midpoint of XY is at a distance from X or Y. Example 5: Find the arithmetic mean between 6 and 9. Solution: The procedure is given below: Step : Draw a line segment sufficiently long. Cut off a line segment AB on it such that AB is of length 6 cm. Step : Cut off a line segment BC on AX to the right of B such that BC is of length 9 cm. Step : Draw the perpendicular bisector AC. Mark the meeting point of this bisector with AC. Name the meeting point as M. Step 4: Measure the distance of M from A or C. We observe that AM is of length 7.5 cm. This gives the arithmetic mean between 6 and 9. Figure The geometric mean or mean proportional between two numbers a and b Let a and b be two positive numbers. Then the geometric mean of a and b is ab. If x ab, then x a x ab or x x ab or or a : x x : b. So x is also called the mean x b proportional between a and b. (If a number x is such that a : x x : b, then x is called the mean proportional between a and b). To find ab through geometrical constructions, we proceed as follows: 09

214 Step : Draw a line segment sufficiently long. Step : Along it, cut off the line segments AB and BC of lengths a and b respectively. Step : Draw a circle with AC as diameter. Step 4: Draw a chord DE through B perpendicular to AC. Step 5: Measure the length of BD or BE. This length gives the mean proportion between a and b. Let us now understand how the length of BD gives the mean proportional between the lengths of AB and BC. AC is a diameter and D is a point on the circle. Figure 9. So ADC 90º. BD is perpendicular to AC. B and the side BD are common to the right triangle ADB and DCB. So these two triangles are similar. Hence the sides are proportional. AB BD Then or BD AB BC or BD BC x ab or x ab. Example 6: Find the geometric mean between two segments of lengths 9 cm and cm. Solution: Step : Draw a line segment AX sufficiently long. Step : Cut off from it line segments AB and BC of lengths 9 cm and cm respectively. Step : 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. We measure BD and find BD 5. cm. Figure 9. 0

215 Example 7: Find the mean proportional between 4 and 9. Solution: We know that the mean proportional between 4 and 9 is We find this value 6 through geometrical construction. Step : First draw a line segment AX. Step : On AX cut off line segment AB of length 4 cm. Step : Starting with B, cut off BC of length 9 cm. to the right of B. Step 4: Mark the mid point of AC as O. Step 5: Draw a circle with O as centre and OA as radius. Step 6: Draw the chord DE through B, perpendicular to AC. Step 7: BD represents the mean proportional between AB and BC. Step 8: Measure the length of BD. We find BD 6 cm. The number 6 is the mean proportional between the numbers 4 and 9. Figure 9.4 Example 8: Find geometrically the value of. Solution: We observe that 4. So can be considered as the mean proportional between 4 and. Applying the geometrical construction for getting the mean proportional between 4 and, we get Step : Cut off line segments AB and BC on a line AX Such that AB 4 and BC. Step : Draw the circle with AC as one of its diameter. Step : Draw the chord DE through B perpendicular to AC. Step 4: Measure the length of BD. We find that it is.4 cm. Figure 9.5

216 Exercise 9. In problems to 4, find the arithmetic mean between the given numbers.. 6 and 4. 0 and 5. 9 and 4..5 and 6.5 In problems 5 to 8, find the geometric mean between the given numbers. 5.. and and and and 4 In problems 9 to, find the square root of the given number Answers Exercise cm 6..7 cm cm 8.. cm cm 0.. cm..7 cm..5 cm Exercise

217 0. 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 0 students in mathematics given below Let us arrange the given marks in the ascending order. Then, we get 0,,,, 7, 7, 7, 9, 9, 40, 40, 4, 4, 4, 4, 4, 4, 45, 45, 46, 46, 46, 48, 48, 48, 49, 50, 50, 56, 56 In the above list, the mark 0 appears once, thrice, 7 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 f 4

218 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 0. So, The range The largest value The smallest value We shall form intervals called class intervals to include the given marks. The first class interval is 0-4. This interval includes the marks 0,,, and 4. The next class interval is 5-9. Proceeding in this way the last class interval is 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 Frequency Total 0 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 5 th 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 4.5, 9.5. To cover such situations, we can alter the intervals as , ,, with the convention that each interval does not include its upper limit. The modified frequency table is presented below. Class Interval Tally bars Frequency Total 0 4

219 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 , the size is 5 and the class mark is 7, the mid value of the interval. From the above table we observe that the number of students who have obtained marks below 4.5 is 4, the number of students who have obtained marks below 9.5 is , the number of students who have obtained marks below 44.5 is , the number of students who have obtained marks below 49.5 is , the number of students who have obtained marks below 54.5 is and the number of students who have obtained marks below 59.5 is 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.f Total 0 0. 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. Arithmetic mean. Median. Mode. 0.. Arithmetic mean Consider the observations,, 7,, 7. If we subtract 0 from each of the observations, we get 9,,,, 7. Adding all these differences, we get 0. This means that the number 0 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 x, x,, x n 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 x, x,, x n is given by the equation (x x ) +(x x ) (x n x ) 0 or ( x + x xn ) n x 0. 5

220 Hence x x x +... n + + In mathematics, the symbol, called sigma notation is used to represent summation. With this symbol, the sum n + x +... xn is denoted as x i or simply as i x + x n. x. Then, we have i x i x. n If the observations are represented in the form of a frequency table, the mean x is given by fx + f x f n xn x, f + f f where x, x,, x n are the individual values or the mid-values of the class intervals whose frequencies are f, f,,f n. In this case, with sigma notation, we have f i xi x, where N f + f f. N n If the observed values x, x,, x n are numerically large, the mean can be calculated by a short-cut method. Let A be a suitably chosen number. We form the deviations x A, x A,, x n A. If these deviations have a common factor c, then form the ratios x A x A x A,,..., n. c c c Let them be d, d,,d n.. Then x A x A xn A f idi f d + f d +..+ f n d n f + f f n. c c c [( f x fa) + ( f x f A) ( f n xn f n A) ] c [( f x + f x f n xn ) A( f + f f n )] [ f i xi A N ]. c c f xi A N c f i i d i f i xi or f i x A N + c fid or i x i A f id i + c. N N Example : Calculate the mean of the data 9,,, 5, 7, 9. x i Solution : x 4. N 6 Example : Compute the A.M. of the following data: x f n 6

221 Solution: Direct Method: x f f x Total N 0 fx 40 Short-cut Method: Take A 4, c, d x A x f d f d Total N0 fd x f i xi N 0 fd 0 x A+ c 4 + N Example : Calculate the A.M. for the following data: Marks No. of students x A Solution: Take A 90, c 5 and d. c X f d f d N 0 fd A.M. x A+ c N fd Example 4: Calculate A.M for the following data: Class Interval Marks

222 Solution: Direct Method: Class Mid-value x Frequency f f x N 00 fx 800 From the table, we get N the total frequency 00, fx 800. fx 800 x 8. N 00 Short-cut Method: Let A 0, c 5 and d x A. c Class Mid-value x d Frequency f f d N 00 fd fd A.M. x A+ c N Exercise 0... Calculate the mean of 7,, 8, 4, 9, 0.. If in a class of 5 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.. Calculate the Arithmetic mean of the following data: 8

223 4. Obtain the A.M. of the following: X f Marks No. of students Find the A. M. of the following data: Variable Frequency Find the Arithmetic mean from the following table: 0.. Median Class interval Frequency 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 4, 8, 0, 9, 8, 5, 6, 7, 6. Arranging them in the ascending order, we get 4,7,8,0,5,6,8,9,6. We observe that 5 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, 6, 40,, 48, 5. Arranging them in the ascending order, we get 6,, 40, 48, 5, 57, 59, 66, 79, 85. Since there are even number of observations, we get 5 and 57 centrally located in the series. Hence, we take their average, namely 55 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 N + th observations. If N is an odd integer, then there is only one middle term and it is the term of the given set of observations. This is the median. If N is an even integer, there are two 9

224 N middle terms. They are the th term and these two terms. N + th term. Hence the median is the average of 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 N 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 to N or just greater than N. 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, 5, 9, 0, 9. Solution: The given values are already in the ascending order. No. of observations N 5. N + This is an odd number. So the median th 5 + term th term rd term 9. Median 9. Example 6: Find the median of, 4, 0,, 7,, 4, 49, 50, 55, 60, 6, 7, 75, 80. Solution: The given values are already in the ascending order. N No. of observations 5, odd number. N + Median th 5 + term th term 8 th term 49. Example 7: Find the median of 9,, 5, 9, 0, 5, 8. Solution: Arranging the observations in the ascending order, we get, 5, 5, 8, 9, 9, 0. N Number of observations 7, odd integer. N + th 7 + Median term th term 4 th term 8. Example 8: Find the median of 6, 5, 9,, 5, 9, 0, 5, 8, 0. Solution: Arranging the observations in the ascending order, we get, 5, 5, 5, 6, 8, 9, 9, 0, 0. N No. of observations 0, an even integer. N Median average of th N and + th terms average of 5 th and 6 th terms average of 6 and

225 Example 9: Calculate the median of the following table: Solution: Variable ( x) Frequency( f ) x f Cumulative frequency Total frequency N f and so N 6. N The median is the th value 6 th value. But the 6 th value occurs in the class whose cumulative frequency is 9. The corresponding value of the variate is 5. Hence, the median 5. Exercise 0... Find the median of the following set of variables: (i) 66, 6, 55, 60, 46, 0 (ii) 5, 9, 6, 4, 8, 7, 45, 4 (iii) 60, 6, 60, 58, 57, 59, 70 (iv) 4, 45, 6, 7, 4, 45, 4, 6. Find the median for the marks of 40 students Marks No. of students Find the median for the following data x f The wages of 4 employees are given. Find the median. Wage No. of Employees

226 0.. 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 0: Find the mode of 7, 4, 5,, 7,, 4, 6,7. Solution: Arranging the data in the ascending order, we get,, 4, 4, 5, 6, 7, 7, 7. In the above data 7 occurs maximum number of times. Hence mode 7. Example : Find the mode of 9, 0,, 4, 7, 0. 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 : Find the mode for, 5,,, 9, 5, 4, 7, 0,, 9, 5. Solution: Arranging the data in the ascending order, we get,,,, 5, 5, 5, 9, 9, 0, 4, 7. Here occurs times and 5 also occurs times. both and 5 are the modes for this data. We observe that there are two modes for the given data. Example : Find the mode from the following frequency table: Wage No. of Employees Solution: We observe from the table that the maximum frequency is 4. The value of the variate (wage) corresponding to the maximum frequency 4 is 55. This is the mode of the data. Exercise 0... Find the mode of the following data: (i) 84, 9, 7, 68, 87, 84 (ii) 65, 6, 7, 8, 5, (iii) 8,,, 0,, 6, 5, 0 (iv) 5,, 8,,, 9,

227 . Find the mode for the following distribution x f Find the mode for the following table. x f Answers Exercise Exercise 0... (i) 57.5 (ii) 5.5 (iii) 60 (iv) Exercise 0... (i) 84 (ii) No mode (iii) 0, (iv). 0. 6

228 . 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.. 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 : By substituting two different values for x in the equation y mx + c, we get two values for y. Thus we get two points (x, y ) and (x, y ) on the line. Step : 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. If the coordinate values are large, then cm along the axes may be taken to represents large number of units. Step : Plot the two points (x, y ) and (x, y ) 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. 4

229 Example : Draw the graph of the line joining the points (, ) and ( 4, ). Solution: Draw the x-axis and y-axis on a graph paper and take cm unit on both the axes. Let A and B be the points (, ) and ( 4, ). We mark these points on the graph paper. We join the points A and B by a line segment and extend it along the two directions. The required graph is now obtained (see Figure.) Figure. Example : Draw the graph of y x. Solution: Since the equation of the line is y x, the line passes through the origin. Substituting x, 0, in the equation of the line, we get correspondingly y, 0,. We form the table as given below: x 0 y 0 We plot the points (, ), (0, 0) and (, ) in the graph sheet by taking cm unit for both Figure. the axes. We join the points by a line segment and extend it in both directions. We get the required linear graph (see Figure.). Example : Draw the graph y x. Solution: Substituting x, 0, in the equation of the line, we get y 4,, correspondingly. In a graph, plot the points (, 4), (0, ) and (, ). x 0 y 4 Join the points by a line segment and extend it in both directions. Thus we get the required linear graph ( see Figure.). Figure. 5

230 Example 4: Draw the graph of the line whose slope is Solution: The equation of the line is y mx + c or y x + ( ) or y x Substituting x, 0,, we get y 0,, 6 respectively. Plot the points (, 0), (0, ) and (, 6) in the graph paper. x 0 y 0 6 and y-intercept is. Join the points by a line segment and extend it in both directions. We now get the required linear graph (see Figure.4). Figure.4 Example 5: Draw the graph of the line x + y. Solution: The given equation is rewritten as y x + or y x + 4. Substituting x, 0,, we get y 6, 4, respectively. Plot (, 6), (0, 4) and (, ) in the graph sheet. x 0 y 6 4 Join the points by a line segment and extend it in both the directions. This is the required linear graph.(see Figure.5). Figure.5 Example 6: Draw the graph x. Solution: We observe that y is not specified in the equation x. So, any value of y gives x. Choose any two values say and for y and get two points (, ) and (, )on the line x. x y Figure.6 6

231 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.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,. Then we get two points (, 4) and (, 4) on the line y 4. Plot these two points in the graph paper. x y 4 4 Figure.7 Join the two points and extend it in the both directions. The required graph is the resulting graph(see Figure.7) The line is parallel to the x axis. Exercise.. Draw the linear graph through the points (i) (, ) and (4, 6) (ii) (, 0) and (, 5) (iii) (, ) and (5, ) (iv) (, ) and (5, 4). Draw the graph of the following: (i) y x (ii) y x (iii) x 5y (iv) x 4y. Draw the graphs of the following: (i) x (ii) y 5 (iii) x 5 (iv) y 4 (v) x + 0. (vi) + y Draw the graph for y mx + c when (i) m and c 4. (ii) m and c. (iii) m and c 4 (iv) m and c 5 5. Draw the graph of the following equations: (i) x + y. (ii) x 5y 0. (iii) y + x 5 0. (iv) x y

232 . 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 x + y and 4x + y. Solution: We graph the two equations together. Line : y x + Line : y 4x + i.e., y x + x y x y 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.8). Any point on one line is also a point on the other. That is, there are infinitely many points common to both the equation. So there are infinitely many solutions for the given simultaneous equations. Figure.8 8

233 Example 9: Draw the graphs x y 4 and x y 6 and hence solve the simultaneous equations. Solution: We find points for plotting the two lines. Line : x y 4 or y x 4 or y x X 0 Y Line : x y 6 or y x + 6 or y x + x 0 y 4 We plot the points (0, ) and (, ) in the graph paper and draw the line through them. Next, we plot the points (0, ) and (, 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.9). Figure.9 Example 0: Solve graphically the simultaneous equations x + y 5, x y. Solution: We draw the graphs for the two equations in the same graph sheet. Line (): x + y 5 or y x + 5 Line () : x y or y x X Y 7 6 Plot the points (, 7), (, 6) and (, ) on the graph paper. Draw the line passing through them. This is the linear graph for line (). Next, plot the points (, ) (0, ) and (, 0) in the same graph paper. Draw the line passing through these points. This is the linear graph for line (). The two linear graphs intersect at the point P (4, ) (see Figure.0). Since this point lies on both the lines, the solution of the simultaneous equation is x 4, y. x 0 y 0 Figure.0 9

234 Exercise. In problems to 0, solve graphically the following system of equations:. x + y 0, x x + y, 4x + y.. x y 0, y. 7. x + y 4, x + y 6.. x + y, x y. 8. x y 4, x + y. 4. x y 6, x + y x + y, 6x y x + y 5, x y. 0. x + 0, 4x + y Answers Exercise.. (i). (ii). (iii). (iv) (i) x 0 y 0. (ii) x 0 y 0 0

235 . (iii). (iv) x y 0 x y 0. (i) (ii) x y x y (iii) x y 0 (iv) x 0 y 4 4 4

236 (v) x y 0 (vi) x 0 y (i) (ii) x 0 y 4 7 x 0 y 5 (iii) x 0 y 4 7 (iv) x 0 y 5

237 5 (i) (ii) x 0 y 6 4 x y (iii) x 0 y 7 5 (iv) x 5 y Exercise.. x 0 y 0 x y 0 Solution is x 4; y 4.. x 0 y 0 x 0 y Solution is x ; y.

238 . x 0 y 0 x 4 5 y Solution is x ; y x 4 5 y x 4 y 5 Solution is x 5; y. 5. x 4 y x 4 y 6. Solution is x ; y. x - 0 y x 0 y Infinitely many solutions. 7. x 0 4 y 0 x 0 4 y No solution. 4

239 8. x 4 y 0 x y 0 Solution is x ; y. 9. x 0 y 4 x 0 y 7 5 Solution is x ; y. 0. x y 0 x 0 y Solution is x ; y. 5

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