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Resonance Pre-foundation Career Care Programmes (PCCP) Division WORKSHOP TAPASYA SHEET MATHEMATICS COURSE : NTSE (STAGE-I) CLASS-X NTSE Subject : Mathematics S. No. Topics Page No. 1. Number System 1 - 10 2. Polynomial 11 - 18 3. Linear Equation (2 variables) 19 - 24 4. Quadratic equation 25 - 29 5. Progression 30 - 33 6. Lines and Angles, Triangles 34 - 42 7. Quadrilateral 43 - 47 8. Circle 48 - 56 © Copyright reserved. All right reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to only the enrolled student of RESONANCE. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only. .13RPCCP NUMBER SYSTEM (viii) Real numbers : Numbers which can represent INTRODUCTION actual physical quantities in a meaningful way are Number System is a method of writing numerals to represent numbers. known as real numbers. These can be represented on the number line. Number line is geometrical straight Ten symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 are used to represent any number (however large it may be) in our number system. line with arbitrarily defined zero (origin). (ix) Prime numbers : All natural numbers that have Each of the symbols 0,1, 2, 3, 4, 5, 6, 7, 8, 9 is called a digit or a figure. one and itself only as their factors are called prime numbers i.e. prime numbers are exactly divisible by 1 and themselves. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23,...etc. If P is the set of prime number then P = {2, 3, 5, 7,...}. (i) Natural numbers : Counting numbers are known as natural numbers. N = { 1, 2, 3, 4, ... }. (x) Co-prime Numbers : If the H.C.F. of the given numbers (not necessarily prime) is 1 then they are known as co-prime numbers. e.g. 4, 9 are co-prime (ii) Whole numbers : All natural numbers together with 0 form the collection of all whole numbers. W = { 0, 1, 2, 3, 4, ... }. as H.C.F. of (4, 9) = 1. Any two consecutive numbers will always be co-prime. (iii) Integers : All natural numbers, 0 and negative of natural numbers form the collection of all integers. I or Z = { ..., – 3, – 2, – 1, 0, 1, 2, 3, ... }. (xi) Twin primes : Pairs of prime numbers which have (iv) Even Numbers : All integers which are divisible by 2 are called even numbers. Even numbers are denoted by the expression 2n, where n is any integer. So, if E is a set of even numbers, then E = { ..., – 4, –2, 0, 2, 4,...}. 3, 5 ; 5, 7 ; 11,13 ; 17, 19 ; 29, 31 ; 41, 43 ; 59, 61 and (v) Odd Numbers : All integers which are not divisible by 2 are called odd numbers. Odd numbers are denoted by the general expression 2n –1 where n is any integer. If O is a set of odd numbers, then O = {..., –5, –3, –1, 1, 3, 5,...}. are not prime are composite numbers. If C is the set (vi) Rational numbers : These are real numbers which p can be expressed in the form of , where p and q are q integers and q 0 . e.g. 2/3, 37/15, –17/19. All natural numbers, whole numbers and integers are rational numbers. Rational numbers include all Integers (without any decimal part to it), terminating fractions ( fractions in which the decimal parts are terminating e.g. 0.75, – 0.02 etc.) and also non-terminating but recurring decimals e.g. 0.666....., – 2.333...., etc. (vii) Irrational Numbers : All real number which are not rational are irrational numbers. These are nonrecurring as well as non-terminating type of decimal numbers. For Ex. : 2, 3 4 , 2 3 , 2 3 , 47 3 etc. only one composite number between them are called Twin primes. 71, 73 etc. are twin primes. (xii) Composite numbers : All natural numbers, which of composite number then C = {4, 6, 8, 9, 10, 12,...}. 1 is neither prime nor composite number. (xiii) Imaginary Numbers : All the numbers whose square is negative are called imaginary numbers. e.g. 3i, -4i, i, ... ; where i = - 1. IDENTIFICATION OF PRIME NUMBER Step (i) : Find approximate square root of given number. Step (ii) : Divide the given number by prime numbers less than approximately square root of number. If given number is not divisible by any of these prime number then the number is prime otherwise not. Ex.1 Is 673 a prime number ? Sol. Approximate square root = 26 Prime number < 26 are 2, 3, 5, 7, 11, 13, 17, 19, 23. But 673 is not divisible by any of these prime number. So, 673 is a prime number. PAGE # 11 FRACTIONS (a) Common fraction : Fractions whose denominator is not 10. (b) Decimal fraction : Fractions whose denominator is 10 or any power of 10. (c) Proper fraction : Numerator < Denominator i.e. (d) Improper fraction : Numerator > Denominator i.e. 3 5 . 5 . 3 (e) Mixed fraction : Consists of integral as well as fractional part i.e. 3 Short cut method for mixed recurring decimal : Form a fraction in which numerator is the difference between the number formed by all the digits after the decimal point taking the repeated digits only once and that formed by the digits which are not repeated and the denominator is the number formed by as many nines as there are repeated digits followed by as many zeros as the number of non-repeated digits. Ex.2 Change 2.76 45 in the form of p/q. Sol. 2.7645 = 3041 27645 276 27369 = = . 9900 9900 1100 2 . 7 (f) Compound fraction : Fraction whose numerator and The absolute value of a real number I x I is defined as 2/3 denominator themselves are fractions. i.e. . 5/7 Improper fraction can be written in the form of mixed fraction. x, if x 0 |x|= x, if x 0 For example : | 2 |= 2 ; 2 > 0 and | – 2 | = – (– 2 ) = 2; – 2 < 0. COMPARISON OF FRACTIONS Suppose some fractions are to be arranged in ascending or descending order of magnitude. Then, convert each one of the given fractions in the decimal form, and arrange them accordingly. Now, 3 6 7 = 0.6, = 0.857, = 0.777..... 5 7 9 Since 0.857 > 0.777....> 0.6, so 6 7 3 > > . 7 9 5 PURE RECURRING DECIMAL If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called recurring decimal. Thus, 1 = 0.333...... = 0.3 , 3 22 = 3.142857142857..... 7 = 3. 142857 . Conversion of decimal numbers into rational numbers of the form p/q. Short cut method for pure recurring decimal: Write the repeated digit or digits only once in the numerator and take as many nines in the denominator as there are repeating digits in the given number. BODMAS RULE This rule depicts the correct sequence in which the operations are to be executed so as to find out the value of a given expression. Here,‘B’ stands for ‘Bracket’, ‘O’ for ‘Of’, ‘D’ for ‘Division’, ‘M’ for ‘Multiplication’, ‘A’ for ‘Addition’ and ‘S’ for ‘Subtraction’. Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order ( ), { } and [ ]. After removing the brackets, we must use the following operations strictly in the order : (i) Of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction. Vinculum (or Bar) : When an expression contains Vinculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Vinculum. MIXED RECURRING DECIMAL A decimal is said to be a mixed recurring decimal if there is at least one digit after the decimal point, which is not repeated. 3 5 1 7 6 Sol. 5 – 4 2 2 42 e.g., (i) 0. 3 = 3/9 or 1/3 (ii) 0. 387 = 387/999 3 1 1 1 Ex.3 Simplify : 5 – 4 2 2 0.5 6 7 : 3 5 1 3 5 22 1 = 5 – 4 2 2 42 3 83 = 5 – 4 2 42 = 5 – 4 42 = 5– 420 229 229 191 23 = = =2 . 84 84 84 84 PAGE # 22 Properties of Square Roots : Squares : When a number is multiplied by itself then the product is called the square of that number. Perfect Square : A natural number is called a perfect square if it is the square of any other natural number e.g. 1, 4, 9,... are the squares of 1, 2, 3,... respectively. Ex.4 Find the smallest number by which 180 must be multiplied so that the product is a perfect square. Sol. Given number is 180, first we resolve it into prime factors. (i) If the unit digit of a number is 2, 3, 7 or 8, then it does not have a square root in N. (ii) If a number ends in an odd number of zeros, then it does not have a square root in N. (iii) The square root of an even number is even and square root of an odd number is odd.e.g. 81 = 9, 256 = 16, 324 = 18 ...etc. (iv) Negative numbers have no square root in set of real numbers. Ex.6 Find the square root of 5 + Sol. Let 2 180 2 90 3 45 3 15 5 5 1 5+ 5 3 = p + 3 =p+q+2 180 = 2 × 2 × 3 × 3 × 5 Clearly 5 has no pair. Thus if we multiply it by 5 then product will be a perfect square. 2 pq = 3 ...(ii) 4pq = 3 ...(iii) (p – q)2 = (p + q)2 – 4 pq (p – q)2 = 25 – 3 (p – q)2 = 22 p= 1 5 22 2 q= 1 5 22 2 Required smallest number is 5. Ex.5 Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square. ...(i) p – q = 22 p+q=5 Sol. Given number is 28812, first we write it as the product q pq p+q=5 5 [By squaring both sides] [By equating the parts] [By squaring both sides ] ...(iv) [By eqn (i)] 3 = of prime factors. 3. [On adding (i) & (iv)] [On subtracting (i) & (iv)] 1 5 22 5 22 2 2 28812 2 14406 Cube : If any number is multiplied by itself three times then the result is called the cube of that number. 3 7203 7 2401 7 343 7 49 7 7 Ex.7 What is the smallest number by which 392 must be multiplied so that the product is a perfect cube. 1 Sol. Resolving 392 into prime factor, we get Perfect cube : A natural number is said to be a perfect cube if it is the cube of any other natural number. 28812 = 2 × 2 × 3 × 7 × 7 × 7 × 7 Clearly, 3 has no pair, so if we divide it by 3 then quotient become a perfect square. Square roots : The square root of a number x is that number which when multiplied by itself gives x as the product. As we say square of 3 is 9, then we can also say that square root of 9 is 3. The symbol use to indicate the square root of a number is ‘ ’ , i.e. 81 = 9, 225 = 15 ...etc. We can calculate the square root of positive numbers only. However the square root of a positive number may be a positive or a negative number. e.g. 25 = + 5 or – 5. 392 = {2 × 2 × 2} × 7 × 7 Grouping the factor in triplets of equal factors, we get 392 = {2 × 2 × 2} × 7 × 7 We find that 2 occurs as a prime factor of 392 thrice but 7 occurs as a prime factor only twice. Thus, if we multiply 392 by 7,7 will also occur as a prime factor thrice and the product will be 2 × 2 × 2 × 7 × 7 × 7, which is a perfect cube. Hence, we must multiply 392 by 7 so that the product becomes a perfect cube. PAGE # 33 Ex.8 What is the smallest number by which 3087 must be divided so that the quotient is a perfect cube ? Sol. Resolving 3087 into prime factors, we get 3087 = 3 × 3 × 7 × 7 × 7 Grouping the factors in triplets of equal factors, we get 3087 = 3 × 3 × {7 × 7 × 7} Clearly, if we divide 3087 by 3 × 3 = 9, the quotient would be 7 × 7 × 7 which is a perfect cube. Therefore, we must divide 3087 by 9 so that the quotient is a perfect cube. Factors : ‘a’ is a factor of ‘b’ if there exists a relation such that a × n = b, where ‘n’ is any natural number. 1 is a factor of all numbers as 1 × b = b. Factor of a number cannot be greater than the number (infact the largest factor will be the number itself). Thus factors of any number will lie between 1 and the number itself (both inclusive) and they are limited. Multiples : ‘a’ is a multiple of ‘b’ if there exists a relation of the type b × n = a. Thus the multiples of 6 are 6 × 1 =6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, and so on. The smallest multiple will be the number itself and the number of multiples would be infinite. NOTE : To understand what multiples are, let’s just take an example of multiples of 3. The multiples are 3, 6, 9, 12,.... so on. We find that every successive multiples appears as the third number after the previous. So if one wishes to find the number of multiples of 6 less than 255, we could arrive at the number through 255 = 42 (and the remainder 3). The remainder is of 6 no consequence to us. So in all there are 42 multiples. 255 =7 36 (and the remainder is 3). Hence, there are 7 multiples of 36. Factorisation : It is the process of splitting any number into form where it is expressed only in terms of the most basic prime factors. For example, 36 = 22 × 3 2. 36 is expressed in the factorised form in terms of its basic prime factors. Number of factors : For any composite number C, which can be expressed as C = ap × bq × cr ×....., where a, b, c ..... are all prime factors and p, q, r are positive integers, then the number of factors is equal to (p + 1) × (q + 1) × (r + 1).... e.g. 36 = 22 × 32. So total no. of factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9. If one wishes to find the multiples of 36, find Ex.9 Directions : (i to v) Read the following information carefully and answer the questions given below. In a big hostel, there are 1,000 rooms. In that hostel only even numbers are used for room numbers, i.e. the room numbers are 2, 4, 6, ...., 1998, 2000. All the rooms have one resident each. One fine morning, the warden calls all the residents and tells them to go back to their rooms as well as multiples of their room numbers. When a guy visits a room and finds the door open, he closes it, and if the door is closed, he opens it, All 1,000 guys do this operation. All the doors were open initially. (i) The last room that is closed is room number ? (ii) The 38th room that is open is room number ? (iii) If only 500 guys, i.e. residents of room number 2 to 1000 do the task, then the last room that is closed is room number ? (iv) In the case of the previous question, how many rooms will be closed in all ? (v) If you are a lazy person, you would like to stay in a room whose number is ? Sol (i) If a room has an odd number of visitors, it will be closed. Any room number that is twice a perfect square will have an odd number of visitors. The room with the largest such number (twice a perfect square) will be the last room to have an odd number of visitors. Hence the last number is 2 × 961 = 1922. (ii) Note that the 38th room with an open door will be the 38th room whose number is not twice a perfect square, which happens to be 88. (iii) Note that if only occupants of 2 to 1000 do the task, then only 1 person, i.e. the occupant of number 1000 will go to room number 2000, (since 2000 is a multiple of 1000), and this room will therefore be closed. (Since it is initially open). (iv) Number of room closed upto 1000 = 22, since there are 22 twice a perfect square numbers. Number of remaining room = 500. Number of twice a perfect number between 1000 and 2000 = 9. All rooms with twice a perfect square between 1000 and 2000 will be open. Number of room closed = 22 + 500 – 9 = 513. (v) As regards questions 34, anyone who lives in a room with a number greater than 1000 will obviously have to visit only that particular room, as it will not have any multiple which is not greater than 2000. Hence, the answer is greater than 1000. LCM (least Common Multiple) : The LCM of given numbers, as the name suggests is the smallest positive number which is a multiple of each of the given numbers. PAGE # 44 HCF (Highest Common factor) : The HCF of given numbers, as the name suggests is the largest factor of the given set of numbers. Consider the numbers 12, 20 and 30. The factors and the multiples are : Factors 1, 2, 3, 4, 6, 12 1, 2, 4, 5, 10, 20 1, 2, 3, 5, 6, 10, 15, 30 Given numbers 12 20 30 Multiples 12, 24, 36, 48, 60, 72, 84, 96, 108, 120.... 20, 40, 60, 80, 100, 120..... 30, 60, 90, 120.... The common factors are 1 and 2 and the common multiples are 60, 120... Thus the highest common factor is 2 and the meaning of HCF is that, HCF is the largest number that divides all the given numbers. Also since a number divides its multiple, the meaning of LCM is that it is smallest number which can be divided by the given numbers. HCF will be lesser than or equal to the least of the numbers and LCM will be greater than or equal to the greatest of the numbers. There are two methods of finding the H.C.F. of a given set of numbers : (i) Factorization Method : Express each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F. and product of highest powers of common prime factors gives L.C.M. Some important points : Ex.12 Find the greatest number that will exactly divide 65, 52 and 78. Sol. Required number = HCF of 65, 52 and 78 = 13. (ii) Division Method : Suppose we have to find the H.C.F. of two given numbers. Divide the larger number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding divisor by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F. Ex.11 Find the HCF of (72, 108 ). Sol. So, HCF of (72, 108 ) = 36. Finding the H.C.F. of more than two numbers : Suppose we have to find the H.C.F. of three numbers. Then, H.C.F. of [(H.C.F. of any two) and (the third number) gives the H.C.F. of three given numbers. Similarly, the H.C.F. of more than three numbers may be obtained. If we have to find the greatest number that will divide p, q and r leaving remainders a, band c respectively, then the required number = HCF of (p – a), (q – b) and (r – c). Ex.13 Find the greatest number that will divide 65, 52 and 78 leaving remainders 5, 2 and 8 respectively. Sol. Required number = HCF of (65 - 5), (52 - 2) and (78 - 8) = HCF of 60, 50 and 70 = 10. If we have to find the greatest number that will divide p, q and r leaving the same remainder in each case, then required number = HCF of the absolute values of (p – q), (q – r) and (r – p). Ex.14 Find the greatest number that will divide 65, 81 and 145 leaving the same remainder in each case. Sol. Required number = HCF of (81 – 65) (145 – 81) and (145 – 65) = HCF of 16, 64 and 80 = 16. / Ex.10 Find the HCF of (72, 108 ). Sol. HCF of (72, 108 ) = HCF of (23 × 32, 22 × 33 ) = 22 × 32 = 4 × 9 = 36. If we have to find the greatest number that will exactly divide p, q and r, then required number = HCF of p, q and r. Least Common Multiple (L.C.M.) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M. (i) Factorization Method of Finding L.C.M. : Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors. Ex.15 Find the LCM of (72, 108 ) Sol. The LCM of (72, 108 ) The LCM of (23 × 32, 22 × 33) = 23 × 33 = 8 × 27 = 216. (ii) Common Division Method (Short-cut Method) of finding L.C.M. : Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers. Ex.16 Find the LCM of (72, 108). Sol. 2 72, 108 2 36, 2 18, 54 27 3 9, 27 3 3, 9 3 1, 3 1, 1 So, LCM of (72, 108) = 2 × 2 × 2 × 3 × 3 × 3 = 216. PAGE # 55 Some important points : If we have to find the least number which is exactly divisible by p, q and r, then the required number = LCM of p, q and r . Ex.17 Find the least number that is exactly divisible by 6, 5 and 7. Sol. Required number = LCM of 6, 5 and 7 = 210. Ex.21 In a school 437 boys and 342 girls have been divided into classes, so that each class has the same number of students and no class has boys and girls mixed. What is the least number of classes needed? Sol. We should have the maximum number of students in a class. So we have to find HCF (437, 342) = 19. HCF is also the factor of difference of the number. If we have to find the least number which when divided by p, q and r leaves the remainders a, b and c respectively then if it is observed that (p – a) = (q – b) = (r – c) = K (say), then the required number = (LCM of p, q and r) – (K). Ex.18 Find the least number which when divided by 6, 7 and 9 leaves the remainders 1, 2 and 4 respectively. Sol. Here, (6 – 1) = (7 – 2) = (9 – 4) = 5. Required number = (LCM of 6, 7 and 9) – 5 = 126 – 5 = 121. If we have to find the least number which when divided by p, q and r leaves the same remainder 'a' in each case, then required number = (LCM of p, q and r) + a. L.C.M. of Numerators L.C.M. = H.C.F. of Deno min ators . HCF of ( HCF of (a, c, e ) a c e , , )= LCM of (b, d, f ) b d f LCM of ( LCM of (a, c, e ) a c e , , )= HCF of (b, d, f ) b d f H.C.F. and L.C.M. of Decimal Fractions : In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers. 30 1 Therefore in 30 min they will toll together 2 = 16 times 1 is added as all the bells are tolling together at the start also, i.e. 0th second. Ex.23 LCM of two distinct natural numbers is 211. What is their HCF ? Sol. 211 is a prime number. So there is only one pair of distinct numbers possible whose LCM is 211, i.e. 1 and 211. HCF of 1 and 211 is 1. Ex.24 An orchard has 48 apple trees, 60 mango trees and 96 banana trees. These have to be arranged in rows such that each row has the same number of trees and all are of the same type. Find the minimum number of such rows that can be formed. Sol. Total number of trees are 204 and each of the trees are exactly divisible by 12. HCF of (48, 60, 96). = 500 2060 1025 , , ) 1000 1000 1000 5 = 0.005 1000 204 = 17 such rows are possible. 12 Division Algorithm : General representation of result is, Dividend Re mainder Quotient Divisor Divisor Ex.20 The HCF of (0.5, 2.06, 1.025 ) Sol. HCF of (0.500, 2.060, 1.025 ) = HCF of ( For any two numbers x and y : x × y = HCF (x, y) × LCM (x, y). Make sure the fractions are in the most reducible form. Ex.22 Six bells start tolling together and they toll at intervals of 2, 4, 6, 8, 10, 12 sec. respectively, find. (i) After how much time will all six of them toll together ? (ii) How many times will they toll together in 30 min ? Sol. The time after which all six bells will toll together must be multiple of 2, 4, 6, 8, 10, 12. Therefore, required time = LCM of time intervals. = LCM (2, 4, 6, 8, 10, 12) = 120 sec. Therefore after 120 s all six bells will toll together. After each 120 s, i.e. 2 min, all bells are tolling together. Ex.19 Find the least number which when divided by 15, 20 and 30 leaves the remainder 5 in each case. Sol. Required number = (LCM of 15, 20 and 30) + 5 = 60 + 5 = 65 H.C.F and L.C.M. of fractions : H.C.F. of Numerators H.C.F = L.C.M. of Deno min ators 437 342 + = 23 + 18 19 19 = 41 classes. Number of classes = Dividend = (Divisor × Quotient ) + Remainder NOTE : (i) (xn – an) is divisible by (x – a) for all the values of n. (ii) (xn – an) is divisible by (x + a) and (x – a) for all the even values of n. (iii) (xn + an) is divisible by (x + a) for all the odd values of n. PAGE # 66 Test of Divisibility : No. Divisiblity Test 2 Unit digit should be 0 or even 3 The sum of digits of no. should be divisible by 3 4 The no formed by last 2 digits of given no. should be divisible by 4. 5 Unit digit should be 0 or 5. 6 No should be divisible by 2 & 3 both 8 The number formed by last 3 digits of given no. should be divisible by 8. 9 Sum of digits of given no. should be divisible by 9 The difference between sums of the digits at even & at odd places 11 should be zero or multiple of 11. 25 Last 2 digits of the number should be 00, 25, 50 or 75. Rule for 7 : Double the last digit of given number and subtract from remaining number the result should be zero or divisible by 7. Ex.25 Check whether 413 is divisible by 7 or not. Sol. Last digit = 3, remaining number = 41, 41 – (3 x 2) = 35 (divisible by 7). i.e. 413 is divisible by 7. This rule can also be used for number having more than 3 digits. Ex.26 Check whether 6545 is divisible by 7 or not. Sol. Last digit = 5, remaining number 654, 654 – (5 x 2) = 644; 64 – (4 x 2) = 56 divisible by 7. i.e. 6545 is Ex.31 Find the largest four digit number which when reduced by 54, is perfectly divisible by all even natural numbers less than 20. Sol. Even natural numbers less than 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18. Their LCM = 2 × LCM of first 9 natural numbers = 2 × 2520 = 5040 This happens to be the largest four-digit number divisible by all even natural numbers less than 20. 54 was subtracted from our required number to get this number. Hence, (required number – 54) = 5040 Required number = 5094. Ex.32 Ajay multiplied 484 by a certain number to get the result 3823a. Find the value of ‘a’. Sol. 3823a is divisible by 484, and 484 is a factor of 3823a. 4 is a factor of 484 and 11 is also a factor of 484. Hence, 3823a is divisible by both 4 and 11. To be divisible by 4, the last two digits have to be divisible by 4. ‘a’ can take two values 2 and 6. 38232 is not divisible by 11, but 38236 is divisible by 11. Hence, 6 is the correct choice. divisible by 7. Rule for 13 : Four times the last digit and add to remaining number the result should be divisible by 13. Ex.27 Check whether 234 is divisible by 13 or not . Sol. 234, (4 x 4) + 23 = 39 (divisible by 13), i.e. 234 is divisible by 13. Ex.33 Which digits should come in place of and $ if the number 62684$ is divisible by both 8 and 5 ? Sol. Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 in never divisible by 8. So, 0 will replace $. Now, the number formed by the last three digits is 40, which becomes divisible by 8, if is replaced by 4 or 8. Hence, digits in place of and $ are 4 or 8 and 0 respectively. Rule for 17 : Five times the last digit of the number and subtract from previous number the result obtained should be either 0 or divisible by 17. Ex.28 Check whether 357 is divisible by 17 or not. Sol. 357, (7 x 5) – 35 = 0, i.e. 357 is divisible by 17. Rule for 19 : Double the last digit of given number and add to remaining number The result obtained should be divisible by 19. Ex.29 Check whether 589 is divisible by 19 or not. Ex.34 On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor. Dividend Re mainder 15968 37 Sol. Divisor = = 179. Quotient 89 Ex.35 How many numbers between 200 and 600 are divisible by 4, 5, and 6 ? Sol. Every such number must be divisible by L.C.M. of 4, 5, 6, i.e.60. Such numbers are 240, 300, 360, 420, 480, 540 Clearly, there are 6 such numbers. Sol. 589, (9 x 2) + 58 = 76 (divisible by 19), i.e. the number is divisible by 19. The method of finding the remainder without actually performing the process of division is termed as Remainder Theorem. Ex.30 Find the smallest number of six digits which is exactly divisible by 111. Sol. Smallest number of 6 digits is which is 100000. On dividing 100000 by 111, we get 100 as remainder. Number to be added = (111 – 100) = 11. Hence, required number = 100011. Remainder should always be positive. For example if we divide –22 by 7, generally we get –3 as quotient and –1 as remainder. But this is wrong because remainder is never be negative hence the quotient should be –4 and remainder is +6. We can also get remainder 6 by adding –1 to divisor 7 ( 7–1 = 6). PAGE # 77 Ex.36 Two numbers, x and y, are such that when divided by 6, they leave remainders 4 and 5 respectively. Find the remainder when (x2 + y2) is divided by 6. Sol. Suppose x = 6k1 + 4 and y = 6k2 + 5 x2 + y2 = (6k1 + 4)2 + (6k2 + 5)2 = 36k12 + 48k1 + 16 + 36k22 + 60k2 + 25 = 36k12 + 48k1 + 36k22 + 60k2 + 41 Obviously when this is divided by 6, the remainder will be 5. Ex.37 A number when divided by 259 leaves a remainder 139. What will be the remainder when the same number is divided by 37 ? Sol. Let the number be P. So, P – 139 is divisible by 259. P 139 Let Q be the quotient then, =Q 259 P = 259Q + 139 P 259 Q 139 = 37 37 259 is divisible by 37, When 139 divided by 37, leaves a remainder of 28. Ex.38 A number being successively divided by 3, 5 and 8 leaves remainders 1, 4 and 7 respectively. Find the respective remainders if the order of divisors be reversed. 3 x 5 y 1 Sol. 8 z 4 1 7 z = (8 × 1 + 7) = 15 ; y = (5z + 4) = (5 × 15 + 4) = 79 ; x = (3y + 1) = (3 × 79 + 1) = 238. 8 238 5 29 6 Now, 3 5 4 1 2 Respective remainders are 6, 4, 2. Ex.39 A number was divided successively in order by 4, 5 and 6. The remainders were respectively 2, 3 and 4. Then find out the number. 4 x Sol. 5 y 2 6 z 3 1 4 z = (6 × 1 + 4) = 10 y = (5 × z + 3) = (5 × 10 + 53) = 53 x = (4 × y + 2) = (4 × 53 + 2) = 214 Hence, the required number is 214. Ex.40 In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors of 585) and got the remainders 4, 8 and 12. If he had divided number by 585, then find out the remainder. Sol. 5 x 9 y 4 13 z 8 1 12 Now, 1169 when divided by 585 gives remainder = 584. We are having 10 digits in our number systems and some of them shows special characterstics like they, repeat their unit digit after a cycle, for example 1 repeat its unit digit after every consecutive power. So, its cyclicity is 1 on the other hand digit 2 repeat its unit digit after every four power, hence the cyclicity of 2 is four. The cyclicity of digits are as follows : Digit Cyclicity 0, 1, 5 and 6 1 4 and 9 2 2, 3, 7 and 8 4 So, if we want to find the last digit of 245, divide 45 by 4. The remainder is 1 so the last digit of 245 would be same as the last digit of 21 which is 2. Ex.41 Find the last digit of (i) 357 (ii) 1359 57 gives the remainder 4 57 1. So, the last digit of 3 is same as the last digit of 31, i.e. 3. (ii) The number of digits in the base will not make a difference to the last digit. It is last digit of the base which decides the last digit of the number itself. For Sol. (i) The cyclicity of 3 is 4. Hence, 59 which gives a remainder 3. So the 4 last digit of 1359 is same as the last digit of 33, i.e. 7. 1359, we find Ex.42 Find the last digit of the product 723 x 813. Sol. Both 7 and 8 exhibit a cyclicity of 4. The last digits are 71 = 7 81 = 8 72 = 9 82 = 4 3 7 =3 83 = 2 4 7 =1 84 = 6 5 7 =7 85 = 8 The cycle would repeat itself for higher powers. 723 ends with the same last digit as 73, i.e. 3. 813 ends with the same last digit as 81, i.e. 8. Hence, the product of the two numbers would end with the same last digit as that of 3 × 8, i.e. 4. Ex.43 Find unit’s digit in y = 717 + 734. Sol. Unit digit of (7 17 + 7 34 ) is same as unit digit of (71 + 72 = 56), Hence the unit digit is 6. Ex.44 What will be the last digit of (73 )75 Sol. Let (73 )75 6476 = (73)x where x = 75 64 6476 76 = (75)even power Cyclicity of 3 is 4 To find the last digit we have to find the remainder when x is divided by 4. x = (75)even power = (76 – 1)even power , where n is divided by 4 so remainder will be 1. Therefore, the last digit of (73 )75 6476 will be 31 = 3. PAGE # 88 Ex.45 What will be the unit digit of (87 )75 55 75 63 6355 . 55 Sol. Let (87 ) = (87)x where x = 75 63 = (75)odd Cyclicity of 7 is 4. To find the last digit we have to find the remainder when x is divided by 4. x = (75)odd power = (76 – 1)odd power where x is divided by 4 so remainder will be –1 or 3, but remainder should be positive always therefore, the last digit of (87 )75 6355 Hence, the last digit is of (87 ) digits, then A + B + C = ? Sol. As the unit digit of the product is 5, therefore, the unit digit of one of the numbers is 5 and the unit digit of the other number is odd. Therefore, AB5 x 5BA = 65125, where A = 1, 3, 5, 7 or 9. As the product of two three-digit numbers is a five-digit number, and not a six-digit number, A can only be equal will be 73 = 343. to 1. 1B5 x 5B1 = 65125. is 3. The digit sums of both numbers, 1B5 and 5B1 will be 55 75 63 Ex.49 If ABC x CBA = 65125, where A, B and C are single same. Therefore, the product would give digit sum of a ALPHA NUMERIC PUZZLES perfect square. The digit sum on the R.H.S. is 1. Therefore, the digit sum of each number can be 1or 8. Ex.46 If a – b = 2, Correspondingly B will be 4 or 2 (as digit sum cannot a a and be equal to 1). b b _____ then find the value of a, b and c. cc 0 Sol. These problems involve basic number (i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers. Hence, their sum cannot exceed 198. So, c must be 1. (iii) Hence, cc0 = 110. This implies a + b = 10 or a = 6 and b = 4. Such problems are part of a category of problems called alphanumerics. a 3b Ex.47 If a c _____ a a 9 Rules on Counting Numbers : (i) Sum of first n natural numbers = n(n 1) 2 (ii) Sum of first n odd numbers = n2 (iii) Sum of first n even number = n(n + 1) (iv) Sum of the squares of first n natural numbers = n(n 1)(2n 1) 6 (v) Sum of the cubes of first n natural number then find a, b and c if each of them is distinctly different digit. Sol. (i) since the first digit of (a 3 b) is written as it is after subtracting ac carry over from a to 3. (ii) there must be a carry over from 3 to b, because if no carry over is there, it means 3 – a = a. 2a = 3 3 a= 2 which is not possible because a is a digit. For a carry over 1, 2 – a = a a=1 (iii) it means b and c are consecutive digit (2, 3), (3, 4),.... (8, 9). Ex.48 If Keeping B = 2, we can see that 125 x 521 = 65125. 1a4 x3b 8c8 s 72 t 5d8 then, find the value of a, b, c, d, s and t, where all of them are different digits. Sol. Let us consider 1 a 4 × 3 = s72. a × 3 results in a number ending in 6. As 16 and 26 is ruled out, a is 2. Thus, s = 3, t, 4 Now 1 a 4 × b = 8c8 ; b = 2 or 7 Again 2 is ruled out because in that case, product would be much less than 800. b = 7. Hence, a = 2, b = 7, c = 6, d = 8, s = 3 and t = 4. 2 n(n 1) = . 2 LOGARITHM If ‘a’ is a positive real number, other than 1 and x is a rational number such that ax = N, then x is the logarithm of N to the base a. If ax = N then loga N = x. [ Remember N will be +ve] Systems of Logarithm : There are two systems of logarithm which are generally used. (i) Common logarithm : In this system base is always taken as 10. (ii) Natural logarithm : In this system the base of the logarithm is taken as ‘e’. Where ‘e’ is an irrational number lying between 2 and 3. (The approximate value of e upto two decimal places is e = 2.73) Some Useful Results : (i) If a > 1 then (a) loga x < 0 [for all x satisfying 0 < x < 1] (b) loga x = 0 for x = 1 (c) loga x > 0 for x > 1 (d) x > y loga x > loga y i.e. logax is an increasing function. Graph of y = loga x, a > 1 PAGE # 99 Ex.50 If log3a = 4, find value of a. y Sol. y = logax, a > 1 0 x' (1,0) log3a = 4 a = 34 a = 81. x Ex.51 Find the value of log Sol. Given : y' (ii) If 0 < a < 1, then (a) loga x < 0 for all x > 1 (b) loga x = 0 for x = 1 (c) logax > 0 for all x satisfying 0 < x < 1 (d) x > y 9 27 3 – log log 8 32 4 logax < loga y i.e. loga x is a decreasing log 3 9 27 3 9 27 – log log log log 4 8 32 4 8 32 9 32 3 log 8 27 4 function. Graph of y = loga x, 0 < a < 1. y = logax, 0 < a < 1. 0 Sol. Given 2log4x = 1 + log4(x – 1) log4x2 – log4(x – 1) = 1 (1,0) x y' Fundamental Laws of Logarithm : Logarithm to any base a (where a > 0 and a 1 ). (i) loga a = 1 x2 =1 x –1 41 = x2 = 4x – 4 x2 – 4x + 4 = 0 (x – 2)2 = 0 x = 2. Ex.53 Evaluate : 3 2 – log3 5 . Sol. Given 3 2 – log3 5 = 3 2.3 – log3 5 = 9. 3 m (v) loga = logam – logan n (vi) loga(m)n = n logam log m b = log3 5 –1 9 . 5 Ex.54 If A = log27625 + 7 log11 13 Sol. A = log27625 + 7 or, A = log11 13 (ix) If ‘a’ is a positive real number and ‘n’ is a positive rational number, then or, B = log 3 2 5 3 + 7 log11 13 3 log 13 log35 + 7 11 2 By (i) and (ii) we have, or, B = A– q (xi) ploga q qloga p ...(ii) 3 4 log35 = B – log35 3 2 4 3 log35 < log35 3 2 A < B. (xii) logax = logay x = y ....(i) log11 7 and,B = log9125 + 13 loga q np p log n a log11 7 4 log 13 = log 3 3 5 + 7 11 4 log 13 log35 + 7 11 3 (viii)logam . logma = 1 (x) If ‘a’ is a positive real number and ‘n’ is a positive rational number, then and B = log9125 + 13 then find the relation between A and B. log a b a loga n n [ am + n = am.an] = 9 × 5–1 (iii) loga (–ve no.) = not defined. [As in loga N, N will always be (+ ve)] (iv) loga (mn) = loga m + logan [Where m and n are +ve numbers] x2 x –1 log4 (ii) loga 0 = not defined [As an = 0 is not possible, where n is any number] (vii) logam [ loga1 = 0] Ex.52 If 2log4x = 1 + log4(x – 1), find the value of x. y x' = log1 = 0. PAGE # 1010 , POLYNOMIALS An algebraic expression f(x) 2 of the (B) Linear Polynomial : A polynomial of degree one is called a linear polynomial. The general form of linear polynomial is ax + b, where a and b are any real constant and a 0. form n f(x) = a0 + a1x + a2x +...........+ anx , where a0, a1, a2,............,an are real numbers and all the index of ‘x’ (C) Quadratic Polynomial : A polynomial of degree two is called a quadratic polynomial. The general form of a quadratic polynomial is ax2 + bx + c, where a 0. are non-negative integers is called a polynomials in x. Identification of Polynomial : (D) Cubic Polynomial : A polynomial of degree three is called a cubic polynomial. The general form of a cubic polynomial is ax3 + bx2 + cx + d, where a 0. For this, we have following examples : (i) 3 x2 + x – 5 is a polynomial in variable x as all the exponents of x are non negative integers. (ii) 3 x2 + (E) Biquadratic (or quartic) Polynomials : A polynomial of degree four is called a biquadratic (quartic) polynomial. The general form of a biquadratic polynomial is ax4 + bx3 + cx2 + dx + e, where a 0. x – 5x is not a polynomial as the exponent of second term ( x = x1/2) is not a non–negative integer. (iii) 5x3 + 2x2 + 3x – 5 + 6 is not a polynomial as the x NOTE : A polynomial of degree five or more than five does not have any particular name. Such a polynomial usually called a polynomial of degree five or six or ....... etc. (ii) Based on number of terms There are three types of polynomials based on number of terms. These are as follows : 5 1 exponent of fourth term 5 x is not x non–negative integer. (A) Monomial : A polynomial is said to be a monomial if it has only one term. e.g. x, 9x2, 5x3 all are monomials. (a) Degree of the Polynomial : (B) Binomial : A polynomial is said to be a binomial if it contains two terms. e.g. 2x2 + 3x, all are binomials. Highest index of x in algebraic expression is called the degree of the polynomial, here a0, a1x, a2x2,............., anxn, it contains three terms. e.g. 3x3 – 8x + polynomial f(x). For example: (ii) q(x) = 5x4 + 2x5 – 6x6 – 5 is a polynomial of degree 6 x + 5x3, – 8x3 + 3, (C) Trinomials : A polynomial is said to be a trinomial if are called the terms of the polynomial and a 0, a 1, a2,.........., a n are called various coefficients of the (i) p(x) = 3x4 – 5x2 + 2 is a polynomial of degree 4 1 2 5 , 5 – 7x + 8x9, 2 7 x10 + 8x4 – 3x2 are all trinomials. NOTE : A polynomial having four or more than four terms does not have particular name. These are simply called polynomials. (iii) f(x) = 2x3 + 7x – 5 is a polynomial of degree 3. (b) Different Types of Polynomials : Some important identities are : Generally, we divide the polynomials in the following categories. (i) (a + b)2 = a2 + 2ab + b2 (i) Based on degrees : There are four types of polynomials based on degrees. (iii) a2 – b2 = (a + b) (a – b) These are listed below : (A) Zero degree polynomial : Any non-zero number (constant) is regarded as a polynomial of degree zero or zero degree polynomial. i.e. f(x) = a, where a 0 is a zero degree polynomial, since we can write f(x) = a as f(x) = axo. (ii) (a – b)2 = a2 – 2ab + b2 (iv) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2 bc + 2ca (v) a3 + b3 = (a + b) (a2 – ab + b2) (vi) a3 – b3 = (a – b) (a2 + ab + b2) (vii) (a + b)3 = a3 + b3 + 3ab (a + b) (viii) (a – b)3 = a3 – b3 – 3ab (a – b) (ix) a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac) Special case : if a + b + c = 0 then a3 + b3 + c3 = 3abc. PAGE # 1111 Ex.3 If a + b + c = 15, a2 + b2 + c2 = 83, then find the value of Value Form : (i) a2 + b2 = (a + b)2 – 2ab, if a + b and ab are given. 2 2 2 (ii) a + b = (a – b) + 2ab, if a – b and ab are given. (iii) a + b = a b 2 4ab , if a – b and ab are given. (iv) a – b = a b 2 4ab , if a + b and ab are given. 2 1 (v) a2 + 1 1 = a – 2, if a + is given. a a a2 (vi) a2 + 2 1 1 = a + 2, if a – is given. a a 1 a 2 a3 + b3 + c3 – 3abc. Sol. a + b + c = 15 a2 + b2 + c2 = 83 To find the value of : a3 + b3 + c3 – 3abc a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca) (15)2 = 83 + 2 (ab + bc + ca) 2 (ab + bc + ca) = 225 – 83 2 (ab + bc + ca) = 142 (ab + bc + ca) = 71 a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = (15) (83 – 71) (vii) a3 + b3 = (a + b)3 – 3ab (a + b), if (a + b) and ab are = (15) (12) = 180. given. (viii) a3 – b3 = (a – b)3 + 3ab (a – b), if (a – b) and ab are Ex.4 Prove that a2 + b2 + c2 – ab – bc – ca is always non - negative for all values of a, b & c. given. (ix) a3 (x) a 3 3 1 1 1 = a – 3 a , if a + is given. a a a 1 a3 3 1 1 1 = a + 3 a , if a – is given. 3 a a a a Sol. a2 + b2 + c2 – ab – bc – ca = 1 [ 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca ] 2 = 1 [(a – b)2 + (b – c)2 + (c – a)2] 2 1 (xi) a4 – b4 = (a2 + b2) (a2 – b2) = [(a + b)2 – 2ab](a + b) (a – b). We know square of any negative or non negative Ex.1 Evaluate : 1 (i) (5x + 4y) (ii) (4x – 5y) (iii) 2x x 2 2 2 Sol. (i) (5x + 4y) = (5x) + (4y) + 2(5x)(4y) 2 2 2 value is positive and sum of then will also positive thus we can conclude given eqn will be non–negative for all values of a, b, c. = 25x2 + 16y2 + 40xy 2 (ii) (4x – 5y) = (4x)2 + (5y)2 – 2(4x)(5y) To express a given polynomial as the product of = 16x2 + 25y2 – 40xy 1 1 1 (iii) 2x = (2x)2 + – 2 (2x) x x x 1 = 4x2 + 2 – 4. x Ex.2 If a + b = 10 and a2 + b2 = 58, find the value of a3 + b3. polynomials, each of degree less than that of the given Sol. Given a + b = 10 ... (i) divide each term by this factor and take out as a multiple. and a2 + b2 = 58 ... (ii) 2 2 we know a3 + b3 = (a + b)(a2 + b2 – ab) We know (a + b)2 = a2 + b2 + 2ab ... (iii) polynomial such that no such a factor has a factor of lower degree, is called factorization. (a) Factorization by taking out the common factor : When each term of an expression has a common factor, Ex.5 Factorize : 6x3 + 8x2 – 10x Sol. 6x3 + 8x2–10x = 2x (3x2 + 4x – 5) (b) Factorization by grouping : Thus 2ab = (a + b)2 – (a2 + b2) Put values from eqn (i) & eqn (ii) Ex.6 Factorize : ax + by + ay + bx 2ab = (10)2 – (58) Sol. ax + by + ay + bx = ax + ay + bx + by = a (x + y) + b(x+y) = 100 – 58 = 42 ab = 21 a3 + b3 = (a + b)(a2 + b2 – ab) = 10 × (58 – 21) = 10 × 37 = 370. = (x + y) (a + b) (c) Factorization by making a perfect square : Ex.7 Factorize : 4x2 + 12x + 9 Sol. 4x2 + 12x + 9 = (2x)2 + 2 (2x) (3) + 32 = (2x + 3)2 PAGE # 1212 (d) Factorization the difference of two squares : 2 Ex.8 Factorize : 4x – 25. Sol. 4x2 – 25 = (2x)2 – (5)2 = (2x – 5) (2x + 5) (e) Factorization of a quadratic polynomial by splitting the middle term : Ex.9 Factorize : x2 + 4 2 x + 6 Sol. x2 + 4 2 x + 6 = x2 + 3 2 x + 2 x+6 = x(x + 3 2 ) + 2 (x + 3 2 ) = (x + 3 2 )(x + 2 ) (f) Factorization of a algebraic expression as the sum or difference of two cubes : Ex.10 Factorize : 16a3b–250b4. 16a3b–250b 4=2b(8a 3–125b 3) Sol. 3 3 = 2b{(2a) –(5b) } = 2b{(2a–5b)(4a2+25b2+10ab)} (g) Factorization of a algebraic expression of the form a3 + b3 + c3 – 3abc : L.C.M. (Least Common Multiple ) : A polynomial P(x) is called the LCM of two or more given polynomials, if it is a polynomial of smallest degree which is divided by each one of the given polynomials. For any two polynomials P(x) and Q(x). We have : P(x) × Q(x) = [HCF of P(x) and Q(x)] × [LCM of P(x) and Q (x)] Ex.12 If p(x) = (x + 2)(x2 – 4x–21), Q(x) = (x– 7) (2x2 + x – 6) find the HCF and LCM of P(x) and Q(x). Sol. P(x) = (x + 2) (x2 – 4x – 21) = (x + 2) (x2 – 7x + 3x – 21) = (x + 2) (x – 7) (x + 3) Q(x) = (x – 7) (2x2 + x – 6) = (x – 7)(2x2 + 4x – 3x – 6) = (x – 7) [2x (x + 2) – 3 (x + 2)] = (x – 7) (2x – 3) (x + 2) HCF = (x + 2)(x– 7) LCM = (x + 3)(x – 7) (2x – 3) (x + 2). Ex.13 If HCF & LCM of P(x) and Q(x) are (x + 2) and (x + 3) (x2 + 9x + 14) respectively if P(x) = x2 + 5x + 6, find Q(x). Sol. P(x) = (x2 + 5x + 6) = (x + 2) (x + 3) LCM = (x + 3) (x2 + 9x + 14) = (x + 3)(x + 7)(x + 2) We know that HCF LCM = P(x) Q(x) Q(x) = Ex.11 Factorize : ( x 2)( x 3)( x 7)( x 2) ( x 2)( x 3) = (x + 7) (x+2) = x2 + 9x + 14. (i) 2 2a 3 8b 3 27c 3 18 2abc (ii) (x – y)3 + (y – z)3 + (z –x)3 Sol. (i) 2 2a 3 8b 3 27c 3 18 2abc = 3 3 2a (2b) (3c ) = 2a 2b 3c 2a 2 2b 2 3c 2 = 2a 2b 3c 2a 2 3 (a) Value of a Polynomial : The value of a polynomial f(x) at x = is obtained by 3( 2a)(2b )(3c ) substituting x = in the given polynomial and is denoted by f(). Consider the polynomial f(x) = x3 – 6x2 + 11x – 6, 2a2b 2b 3c 3c 2a 4b 2 9c 2 2 2ab 6bc 3 2ac If we replace x by – 2 everywhere in f(x), we get (ii) (x – y)3 + (y – z)3 + (z –x)3 Let a =x – y, b = y–z and c = z–x So, (x – y)3 + (y – z)3 + (z –x)3 = a3 + b3 + c3 Now, a + b + c = x – y + y – z + z – x = 0 So, a3 + b3 + c3 = 3abc 3 3 3 (x – y) + (y – z) + (z –x) = 3(x – y) (y – z)(z –x) H.C.F. AND L.C.M. OF POLYNOMIALS A polynomial D(x) is a divisor of the polynomial P(x) if it is a factor of P(x). Where Q(x) is another polynomial such that P(x) = D(x) × Q(x) HCF/GCD (Greatest Common Divisor) : A polynomial h(x) is called the HCF or GCD of two or more given polynomials, if h(x) is a polynomial of highest degree dividing each of one of the given polynomials. f(– 2) = (– 2)3 – 6(– 2)2 + 11(– 2) – 6 f(– 2) = – 8 – 24 – 22 – 6 f(– 2) = – 60 0. So, we can say that value of f(x) at x = – 2 is – 60. (b) Zero or root of a Polynomial : The real number is a root or zero of a polynomial f(x), if f( = 0. Consider the polynomial f(x) = 2x3 + x2 – 7x – 6, If we replace x by 2 everywhere in f(x), we get f(2) = 2(2)3 + (2)2 – 7(2) – 6 = 16 + 4 – 14 – 6 = 0 Hence, x = 2 is a root of f(x). Let ‘p(x)’ be any polynomial of degree greater than or equal to one and a be any real number and If p(x) is divided by (x – a), then the remainder is equal to p(a) PAGE # 1313 Ex.14 If x51 + 51 is divided by (x + 1) then find the remainder. Sol. Let P(x)=x51 + 51 In algebraic or in set theoretic language the graph When P(x) is divided by (x+1) of a polynomial f(x) is the collection (or set) of all Then remainder = P (–1) points (x, y), where y = f(x). In geometrical or in = (–1 )51 + 51 graphical language the graph of a polynomial f(x) is = – 1 + 51 a smooth free hand curve passing through points (x 1, y1), (x 2, y2), (x 3, y3), ... etc, where y1, y2, y3, ... are = 50 3 2 3 2 Ex.15 If the polynomial 2x + ax + 3x – 5 and x + x – 4x + a the values of the polynomial f(x) at x 1, x 2, x 3, ... leave the same remainder when divided by x – 2, find respectively. the value of a. In order to draw the graph of a polynomial f(x), follow Sol. Let p(x) = 2x3 + ax2 + 3x – 5 and q(x) = x3 + x2 – 4x + a When divided by x – 2 leave same remainder therefore p(a) = q(a) the following algorithm. ALGORITHM : Step (i) Find the values y1, y2, ...., yn, .... of polynomial or p(2) = q(2) 2(2)3 + a(2)2 + 3(2) – 5 = (2)3 + (2)2 – 4 × 2 + a 16 + 4a + 6 – 5 = 8 + 4 – 8 + a f(x) on different points x 1, x 2, ...., x n, .... and prepare a table that gives values of y or f(x) for various values of x. 3a + 13 = 0 x: x1 x2 y = f(x) : y1 = f(x1) y2 = f(x2) 13 a= . 3 ... ... xn yn = f(xn) xn + 1 ... yn + 1 = f(xn + 1) ... Step (ii) Plot that points (x 1, y1), (x 2, y2), (x 3, y3), ... (x n , yn), ... on rectangular co-ordinate system. FACTOR THEOREM In plotting these points use different scales on the Let p(x) be a polynomial of degree greater than or equal X and Y axes. to 1 and ‘a’ be a real number such that p(a) = 0, then (x – a) is a factor of p(x). Conversely, if (x – a) is a factor Step (iii) Draw a free hand smooth curve passing of p(x), then p(a) = 0. through points plotted in step 2 to get the graph of Ex.16 If x2 – 4 is a factor of 2x3 + ax2 + bx + 12, where a and b are constant. Then the values of a and b are : the polynomial f(x). (a) Graph of a Linear Polynomial : Sol. x2 – 4 is a factor of 2x3 + ax2 + bx + 12 x2 – 4 is factor of given expression Consider a linear polynomial f(x)= ax + b, a 0. Let x – 4 = 0 Graph of y = ax + b is a straight line. That is why x=±2 f(x) = ax + b is called a linear polynomial. Since two f (x) = 2x3 + ax2 + bx + 12 points determine a straight line, so only two points f (2) = 0 need to plotted to draw the line y = ax + b. The line 2 16 + 4a + 2b + 12 = 0 2a + b = – 14 represented by y = ax + b crosses the ... (i) f (–2) = 0 – 16 + 4a – 2b + 12 = 0 2a – b = 2 ... (ii) Solve eqn (i) & eqn (ii) (add them) X-axis at b exactly one point, namely , 0 . a Ex.17 Draw the graph of y = x – 4. Sol. y = x – 4 y=x–4 2a + b = – 14 2a – b = 2 x 0 4 5 4a = – 12 y –4 0 1 a = – 3 and b = – 14 – 2a = – 14 + 6 = – 8. (a, b) (– 3, – 8) PAGE # 1414 Conversely, if the parabola y = ax2 + bx + c intersects the X-axis at a point (, 0), then (, 0) satisfies the equation y = ax2 + bx + c a2 + b + c = 0 [ is a real root of ax2 + bx + c = 0] Thus, the intersection of the parabola y = ax2 + bx + c with X-axis gives all the real roots of ax2 + bx + c = 0. Following conclusions may be drawn : (i) If D > 0, the parabola will intersect the x-axis in two distinct points and vice-versa. The parabola meets x-axis at (b) Graph of a Quadratic Polynomial : –b– D and 2a –b D . 2a Let a, b, c be real numbers and a 0. Then f(x) = ax2 + bx + c is known as a quadratic polynomial in x. Graph of the quadratic polynomial i.e. the curve whose equation is y = ax2 + bx + c, a 0. Graph of a quadratic polynomial is always a parabola. Let y = ax2 + bx + c, where a 0. (ii) If D = 0, the parabola will just touch the x-axis at one point and vice-versa. 2 2 4ay = 4a x + 4abx + 4ac 4ay = 4a2x2 + 4abx + b2 – b2 + 4ac 4ay = (2ax + b)2 – (b2 – 4ac) 4ay + (b2 – 4ac) = (2ax + b)2 4ay + (b2 – 4ac) = 4a2(x + b/2a)2 4ay y b 2 – 4ac b 2 4a x 4a 2a D b a x 4a 2a 2 2 . ....(i) (iii) If D < 0, the parabola will not intersect x-axis at all and vice-versa. where, D b2 – 4ac is the discriminant of the quadratic equation. REMARKS : D b ,– Shifting the origin at – , we have 4a 2a (–D) b and Y = y – . X = x – – 2a 4a Substituting these values in (i), we obtain Y = aX2 ...(ii) which is the standard equation of parabola. Clearly, this is the equation of a parabola having its D b ,– vertex at – . 2 a 4 a The parabola opens upwards or downwards according as a > 0 or a < 0. Let be a real root of ax 2 + bx + c = 0. Then, a2 + b + c = 0. Point (, 0) lies on y = ax2 + bx + c. Thus, every real root of ax2 + bx + c = 0 represents a point of intersection of the parabola with the X-axis. REMARKS : x R, y > 0 only if a > 0 & D b² 4ac < 0 x R, y < 0 only if a < 0 & D b² 4ac < 0 Ex.18 Draw the graph of the polynomial f(x) = x2 – 2x – 8. Sol. Let y = x2 – 2x – 8. The following table gives the values of y or f(x) for various values of x. x –4 –3 –2 –1 0 1 2 3 4 5 6 y = x – 2x – 8 16 7 0 –5 –8 –9 –8 –5 0 7 16 2 PAGE # 1515 Let us plot the points (–4,16), (–3, 7), (–2, 0), (–1, –5), (0, –8), (1, –9), (2, –8), (3, –5), (4, 0), (5, 7) and (6, 16) on a graphs paper and draw a smooth free hand curve passing through these points. The Ex.19 Draw the graphs of the polynomial f(x) = x3 – 4x. Sol. Let y = f(x) or, y = x3 – 4x. The values of y for variable value of x are listed in the following table : x –3 –2 –1 0 1 2 3 y = x3 – 4x – 15 0 3 0 –3 0 15 curve thus obtained represents the graphs of the polynomial f(x) = x2 – 2x – 8. This is called a parabola. Thus, the curve y = x3 – 4x passes through the points (–3, –15), (–2, 0), (–1, 3), (0, 0), (1, –3), (2, 0), (3, 15), (4, 48) etc. Plotting these points on a graph paper and drawing a free hand smooth curve through these points, we obtain the graph of the given polynomial as shown figure. The lowest point P, called a minimum point, is the vertex of the parabola. Vertical line passing through P is called the axis of the parabola. Parabola is symmetric about the axis. So, it is also called the line of symmetry. Observations : From the graphs of the polynomial f(x) = x2 – 2x – 8, following observations can be drawn : (i) The coefficient of x2 in f(x) = x2 – 2x – 8 is 1 (a positive real number) and so the parabola opens upwards. (ii) D = b 2 – 4ac = 4 + 32 = 36 > 0. So, the parabola cuts X-axis at two distinct points. (iii) On comparing the polynomial x2 – 2x – 8 with ax2 + bx + c, we get a = 1, b = – 2 and c = – 8. The vertex of the parabola has coordinates (1, – 9) b D , , where D b 2 – 4ac. i.e. 2a 4a (iv) The polynomial f(x) = x2 – 2x – 8 = (x – 4) (x + 2) is factorizable into two distinct linear factors (x – 4) and (x + 2). So, the parabola cuts X-axis at two d is t inc t poi nts (4 , 0 ) a nd (– 2, 0) . The x-coordinates of these points are zeros of f(x). Graphs of a cubic polynomial does not have a fixed standard shape. Cubic polynomial graphs will always cross X-axis at least once and at most thrice. Observations : From the graphs of the polynomial f(x) = x3 – 4x, following observations are as follows : (i) The polynomial f(x) = x 3 – 4x = x (x 2 – 4) = x (x – 2) (x + 2) is factorisable into three distinct linear factors. The curve y = f(x) also cuts X-axis at three distinct points. (ii) We have, f(x) = x (x – 2) (x + 2) Therefore, 0, 2 and –2 are three zeros of f(x). The c u rve y = f (x) c u ts X- axi s a t t hre e p oin ts O (0, 0), P (2, 0) and Q (–2, 0). Let and be the zeros of a quadratic polynomial f(x) = ax2 + bx + c. By factor theorem (x – ) and (x – ) are the factors of f(x). f(x) = k (x – ) (x – ) are the factors of f(x) ax2 + bx + c = k{x2 – ( + ) x + } ax2 + bx + c = kx2 – k ( + ) x + k PAGE # 1616 Comparing the coefficients of x2, x and constant terms on both sides, we get a = k, b = – k ( + ) and c = k c b +=– and = a a + = – and = Let , , be the zeros of a cubic polynomial Coefficient of x f(x) = ax3 + bx2 + cx + d, a 0. Then, by factor theorem, 2 x – , x – and x – are factors of f(x) . Also, f(x) being a Coefficient of x cubic polynomial, cannot have more than three linear Constant term Coefficien t of x 2 factors. Hence, Coefficient of x b Sum of the zeros = – =– Coefficient of x 2 a Constant term c Product of the zeros = = . a Coefficient of x 2 REMARKS : ax3 + bx2 + cx + d = k(x – ) (x – ) (x – ) ax3 + bx2 + cx + d = k{x3 – ( + + ) x2 + ( + + ) x – } ax3 +bx2 +cx+d =k x3 –k (+ + )x2 +k ( ++)x– k a = k, b = – k ( + + ), c = k ( + + ) and d = – k () the zeros and their coefficients. Sol. p (x) = 4x2 + 24x + 36, Here a = 4, b = 24, c = 36 p(x) = 4x2 + 12x + 12x + 36 = 4x (x + 3) + 12 (x + 3) = (x + 3) (4x + 12) = 4 (x + 3)2 So, the zeros of quadratic polynomial is – 3 and – 3. Let = – 3 and = – 3 Sum of zeros = + =–3–3 =–6 ++ =– + + = b –24 = =–6 a 4 Coefficient of x c = a Coefficient of x 3 Product of the zeros = – d Constant term =– a Coefficient of x 3 given by f(x) = k (x – ) (x – ) (x – ) f(x) = k{x3 – ( + + ) x2 + ( + + ) x – }, where k is any non-zero real number. cons tan t term So, product of zeros = = Coefficien t of x 2 b =– a Coefficien t of x 3 Sum of the products of the zeros taken two at a Cubic polynomial having , and as its zeros is coefficien t of x coefficien t of x c 36 = = =9 a 4 d a REMARKS : coefficien t of x 2 product of zeros = = (– 3) (– 3) = 9 Also, product of zeros = c a Sum of the zeros = – time = coefficien t of x 2 b a And,= – coefficien t of x So, sum of zeros = + = – terms on both sides, we get Ex.20 Find the zeros of quadratic polynomial p(x) = 4x2 + 24x + 36 and verify the relationship between = – f(x) = k(x – ) (x – ) (x – ) Comparing the coefficients of x3, x2, x and constant If and are the zeros of a quadratic polynomial f(x). Then, the polynomial f(x) is given by f(x) = k{x2 – ( + ) x + } or, f(x) = k{x2 – (Sum of the zeros) x + Product of the zeros} Also, sum of zeros = – 1 3 Ex.22 Verify that 5, , are zeros of cubic polynomial 2 4 8x3 + 30x2 – 47x + 15. Also verify the relationship 2 cons tan t term 2 . coefficien t of x Ex.21 Find a quadratic polynomial whose zeros are 3 5 and 3 5 . Sol. Given : = 3 + 5 and = 3 – 5 Sum of zeros = + =3+ 5 +3– 5=6 Products of zeros = = (3 + 5 ) (3 – 5 ) =9–5=4 So, quadratic polynomial isk {x2 – 6x + 4}, where k is any constant. between the zeros and the coefficients. Sol. Given : f (x) = 8x3 + 30x2 – 47x + 15 Here, a = 8, b = 30, c = – 47, d = 15 f (– 5) = 8 (– 5)3 + 30 (– 5)2 – 47 (– 5) + 15 = – 1000 + 750 + 235 + 15 = 0 f( 1 )= 2 = 8 = 3 2 1 1 1 8 30 – 47 15 2 2 2 1 1 47 30 – 15 2 8 4 8 60 – 188 120 8 =0 PAGE # 1717 3 and f ( = 8 Also, sum of product of zeros taken two at a time 2 3 3 3 3 ) = 8 30 – 47 15 4 4 4 4 coefficient of x –47 c = = coefficient of x 3 a 8 So, sum of product of zeros taken two at a time = 27 9 141 30 – 15 4 64 16 coefficien t of x 27 135 141 – 15 = 4 8 8 = = + + = Product of zeros = 27 135 – 282 120 =0 8 So, – 5, = (– 5) ( 1 3 and are the zeros of the polynomial f (x). 2 4 = – = – coefficient of x 2 coefficient of x 3 –30 –15 b = = 8 4 a So, sum of zeros = + + = – coefficient of x 2 coefficient of x 3 Sum of product of zeros taken two at a time = + + = (– 5) ( 1 1 3 3 ) + ( ) ( ) + ( ) (– 5) 2 2 4 4 = –5 3 15 – 2 8 4 = –20 3 – 30 –47 = 8 8 cons tan t term coefficient of x 3 15 d = – . 8 a So, product of zeros = 1 3 –15 = –5 = 2 4 4 Also, sum of zeros = – 1 3 15 )( )= – 2 4 8 Also, product of zeros = – 1 3 Let, = – 5, = and = 2 4 Sum of zeros = + + coefficien t of x 3 = – cons tan t term coefficient of x 3 Ex.23 , , are zeros of cubic polynomial x3 – 12x2 + 44x + c. If , , are in A.P., find the value of c. Sol. f (x) = x3 – 12x2 + 44x + c. , , are in A.P. Let = a – d, = a, = a + d Sum of zeros = + + =a–d+a+a+d = 3a So, 3a = – (– 12) a = 4. Sum of product of zeros taken two at a time + + = 44 (a – d) a + (a) (a + d) + (a + d) (a – d) = 44 a2 – ad + a2 + ad + a2 – d2 = 44 3a2 – d2 = 44 3 (4)2 – d2 = 44 d2 = 4 d=±2 product of zeros = = – c (a – d) a (a + d) = – c (4 – 2) (4) (4 + 2) = – c (2) (4) (6) = – c c = – 48. PAGE # 1818 LINEAR EQUATIONS IN TWO VARIABLES (b) Elimination by Equating the Coefficients : An equation of the form Ax + By + C = 0 is called a linear equation. Where A is called coefficient of x, B is called coefficient of y and C is the constant term (free from x & y) A, B, C, R [ belongs to, R Real No.] But A and B can not be simultaneously zero. f A 0, B = 0 equation will be of the form Ax + C = 0. [Line || to Y-axis] f A = 0, B 0 , equation will be of the form By + C = 0. [Line || to X-axis] f A 0 , B 0 , C = 0 equation will be of the form A x + By = 0. [Line passing through origin] f A 0 , B 0 , C 0 equation will be of the form A x + By + C = 0. t is called a linear equation in two variable because the two unknowns (x & y) occurs only in the first power, and the product of two unknown quantities does not occur. Since it involves two variables therefore a single equation will have infinite set of solution i.e. indeterminate solution. So we require a pair of equation i.e. simultaneous equations. Standard form of linear equation : (Standard form refers to all positive coefficients) a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 ....(i) ....(ii) For solving such equations we have three methods. (a) Elimination by substitution (b) Elimination by equating the coefficients (c) Elimination by cross multiplication. (a) Elimination By Substitution : Ex.1 Solve for x and y : 2x + y = 7, 4x – 3y + 1 = 0 Sol. The given system of equations is 2x + y = 7 4x – 3y = – 1 ....(i) ....(ii) From (i), we get : y = (7 – 2x) substituting y = (7 –2x) in (ii) , we get : 4x – 3 (7 – 2x) = – 1 4x – 21 + 6x = – 1 10x = 20 x=2 Substituting x = 2 in (i), we get 2 2 + y = 7 y = (7 – 4) = 3 Hence, the solution is x = 2, y = 3. Ex.2 Solve for x and y : 10x + 3y = 75, 6x – 5y = 11. Sol. The given equations are 10x + 3y = 75 ..(i) 6x – 5y = 11 ..(ii) Multiplying (i) by 5 and (ii) by 3, we get : 50x + 15y = 375 ..(iii) 18x – 15y = 33 ..(iv) Adding (iii) and (iv), we get 68x = 408 x = 408 x=6 68 Putting x = 6 in (i), we get : (10 6) + 3y = 75 60 + 3y = 75 3y = (75 – 60) 3y = 15 y = 5 x = 6, y = 5. Ex.3 Solve for x and y : ax + by – a + b = 0, bx – ay – a – b = 0 Sol. The given equation may be written as ax + by = a – b ...(i) bx – ay = a + b ... (ii) Multiplying (i) by a and (ii) by b, we get a2x + aby = a2 – ab ... (iii) b2x – aby = ab + b2 ...(iv) Adding (iii) and (iv) , we get (a2 + b2) x = (a2 + b2) x = (a 2 b 2 ) (a 2 b 2 ) x=1 Putting x = 1 in (i) we get (a 1) + by = a – b a + by = a – b by = a – b – a by = – b y= b =–1 b y=–1 x = 1 and y = – 1 is the required solution Ex.4 Solve for x and y : 15 22 40 55 + = 5, + = 13, x y and x –y xy xy xy xy Sol. Putting 1 1 = u and = v, the given equation xy xy become 15u + 22v = 5 40u + 55v = 13 ` Multiplying (i) by 5 and (ii) by 2, we get 75u + 110v = 25 80u + 110v = 26 ..(i) ..(ii) ..(iii) ..(iv) PAGE # 1919 On subtracting (iii) from (iv), we get 1 5u = 1 u = 5 1 Putting u = in (iii), we get 5 x = 15 + 110v = 25 110v = 10 v= Hence, x = 4 and y = 3 is the required solution. (a) Unique Solution : Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, if the denominator a1b2 – a2 b1 0 then the given system of equations have unique solution (i.e. only one solution) and solutions are said to be consistent. 1 1 u 5 and v 11 x – y = 5 and x + y = 11 2x = 16 and 2y = 6 x = 8 and y = 3 x = 8 and y = 3 is the required solution. (c) Elimination by Cross Multiplication : a1x + b1y + c1 = 0 a1 b1 a2 b2 a2x + b2y + c2= 0 a1b2 – a2 b1 0 a1 b 1 a2 b2 (b) No Solution : Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, if the denominator a1b2 – a2 b1 = 0 then the given system of equations have no solution and solutions are said to be inconsistent. i.e. a1 b1 c1 = a2 b2 c2 b1 c1 a1 b1 (c) Many Solution (Infinite Solutions) : Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, b2 c2 a2 b2 if [Write the coefficient in this manner] x y 1 = = b1c 2 – b 2c 1 a 2c 1 – a1c 2 a1b 2 – a 2b1 x 1 = b1c 2 – b 2c 1 a1b 2 – a 2b1 b1c 2 – b 2c 1 x= a b –a b 1 2 2 1 a1 b1 c1 = = then system of equations has many a2 b2 c2 solution and solutions are said to be consistent. Ex.6 Find the value of 'p' does the pair of equations given below has unique solution? 4x + py + 8 = 0 & 2x + 2y + 2 = 0 Sol. 4x + py + 8 = 0 2x + 2y + 2 = 0 Here a1 = 4, b1 = p, c1 = 8 a2 = 2, b2 = 2, c2 = 2 For unique solution a1 b1 a2 b2 y 1 Also a c – a c = a b – a b 2 1 1 2 1 2 2 1 a 2c 1 – a1c 2 y= a b –a b 1 2 2 1 Ex.5 Solve the system of equations 2x + 3y = 17, 3x – 2y = 6 by the method of cross multiplication. Sol. The given equations may be written as 2x + 3y – 17 = 0 ..(i) 3x – 2y – 6 = 0 ..(ii) By cross multiplication, we have y x 1 17 3 3 2 2 52 39 =4,y= =3 13 13 1 11 1 1 1 1 = and = xy xy 5 11 x y 1 = = 52 39 13 6 3 2 x y = {3 ( 6) ( 2) ( 17)} {( 17 ) 3 ( 6 ) 2} 1 = {2 ( 2) 3 3} x y 1 = = ( 18 34) ( 51 12) ( 4 9) 4 p 2 2 p 4. Ex 7.If the system of the given equation has no solution then find the value of k. 2x + ky = 7, 2kx + 3ky = 20 Sol. 2x + ky = 7 2kx + 3ky = 20 For no solution a1 b1 c 1 = a2 b2 c 2 2 k 7 2k 3k 20 2 k 2 7 = and 2k 3k 2k 20 k =3 and k 20 7 So, the value of k is 3. PAGE # 2020 Ex.8 Find the values of p and q for which the following system of equations has infinite number of solution : 2x + 3y = 7 (p + q)x + (2p – q)y = 3(p + q + 1) Sol. 2x + 3y = 7 (p + q)x + (2p – q)y = 3(p + q + 1) For infinitely many solution (iii) x = 0 a1 b1 c1 = = a2 b2 c 2 Graphs of the type (ii) ay = b. 2 3 7 = = pq 2p – q 3(p q 1) 2 3 = pq 2p – q Ex.10 Draw the graph of following : (i) y = 0 (ii) y – 2 = 0 (iii) 2y + 4 = 0 Sol. (i) y = 0 4p – 2q = 3p + 3q p = 5q ... (i) and 3 7 = 2p – q 3(p q 1) 4p – 3p = 5q Put p = 5q 3 7 = 10q – q 3(5q q 1) (ii) y – 2 = 0 3 7 = 9q 18q 3 21q = 18q + 3 q = 1. p = 5 × 1 = 5. 3q = 3 (iii) 2y + 4 = 0 y = – 2 Graphs of the type (i) ax = b Ex.9 Draw the graph of following : (i) x = 2 (ii) x + 4 = 0 (iii) x = 0 Sol. (i) x = 2 Graphs of the type (iii) ax + by = 0 (Passing through origin) (ii) x + 4 = 0 x = – 4 Ex.11 Draw the graph of following : (i) x = y (ii) x = – y Sol. (i) x = y x 1 4 –3 0 y 1 4 –3 0 PAGE # 2121 (ii) x = –y x 1 –2 0 y –1 2 0 Let equations of two lines are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. (i) Lines are consistent (unique solution) i.e. they meet at one point and condition is a1 b 1 a2 b2 (ii) Lines are inconsistent (no solution) i.e. they do not meet at one point and condition is a1 b c = 1 1 a2 b2 c2 Graphs of the Type (iv) ax + by + c = 0. (Making Interception x-axis , y-axis) Ex.12 Solve the following system of linear equation graphically : 4x – 3y + 4 = 0, 4x + 3y – 20 = 0. Shade the Area bounded by the above two equation and the x – axis, also find its area. Sol. 4x – 3y + 4 = 0 ...... (i) 4x + 3y – 20 = 0 ...... (ii) In (i) 4x + 4 = 3y y= 4x 4 = 3 x y 2 4 1 0 4 4 5 0 8 4 In (ii) 3y = 20 × 4x 20 4 x y= = 3 x y 2 4 Drawing the equations, we get : Since two lines are meeting at (2,4) their solution is x = 2, y = 4 Area = 1 1 BC × h = 6 4 2 2 = 12 sq.unit So the solution of equation is x = 2 and y = 4 and the area between the two lines and x-axis is 12 sq. unit (iii) Lines are coincident (infinite solution) i.e. overlapping lines (or they are on one another) and condition is a1 b1 c 1 = = . a2 b2 c 2 Ex.13 Show graphically that the system of equations 2x + 4y = 10 and 3x + 6y = 12 has no solution. Sol. Graph of 2x + 4y = 10 We have, 2x + 4y = 10 4y = 10 – 2x y = 5x 2 When x = 1, we have : y = 5 1 =2 2 PAGE # 2222 Type of Problems : 53 =1 2 Thus, we have the following table : When x = 3, we have y = (i) Determining two numbers when the relation between them is given. x 1 3 (ii) Problems regarding fractions, digits of a number, ages of persons. y 2 1 (iii) Problems regarding current of a river, regarding time & distance. Plot the points A (1,2) and B (3,1) on a graph paper. Join A and B and extend it on both sides to obtain the graph of 2x + 4y = 10 as shown in figure. Graph of 3x + 6y = 12 We have, 3x + 6y = 12 6y = 12 – 3x y = When x = 2, we have : y = 4x 2 42 =1 2 40 =2 2 Thus, we have the following table : When x = 0, we have : y = x 2 0 y 1 2 Plot the point C (2,1) and D (0,2) on the same graph paper. Join C and D and extend it on both sides to obtain the graph of 3x + 6y = 12 as shown in figure. (iv) Problems regarding mensuration and geometry. (v) Problems regarding time & work. (vi) Problems regarding mixtures, cost of articles, profit & loss, discount etc. Ex.14 3 bags and 4 pens together cost Rs.257, where as 4 bags and 3 pens cost Rs. 324, find the cost of a bag. Sol.Let cost of a bag & a pen be x & y respectively 3x + 4y = 257 ...(i) 4x + 3y = 324 ...(ii) Add (i) &(ii) 7x + 7y = 581 or x + y = 83 ...(iii) Subtract (i) from (ii) x – y = 67 ...(iv) Add (iii) & (iv) 2x = 150 x = Rs.75 y = Rs.8 Hence the cost of 1 bag is Rs.75. Ex15. The sum of numerator and denominator of a fraction is 8. If 3 is added to both the numerator and denominator the fraction becomes 3 , find the fraction. 4 Sol. Let numerator be x and denominator be y. So, the fraction is x y. x+y=8 .... (i) x3 3 and y3 4 4x + 12 = 3y + 9 4x – 3y = – 3 Solving equation (i) and (ii) x = 3 and y = 5 so, the fraction is We find the lines represented by given equations are parallel. So, the two lines have no common point. Hence, the given system of equations has no solution. For solving daily – life problems with the help of simultaneous linear equation in two variables or equations reducible to them proceed as :(i) Represent the unknown quantities by some variable x and y, which are to be determined. (ii) Find the conditions given in the problem and translate the verbal conditions into a pair of simultaneous linear equation. (iii) Solve these equations & obtain the required quantities with appropriate units. .... (ii) 3 . 5 Ex16. Seven times a positive two digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number. Sol. Let the unit digit and ten’s of two digit number be y, x respectively the two digit positive number is 10x + y. Reversed two digit number = 10 y + x 7(10x + y) = 4(10y + x) 66x – 33y = 0 y = 2x ...(i) And y – x = 3 2x – x = 3 from (i) x=3 And y = 6 the two digit positive number = 10x + y = 10 × 3 + 6 = 30 + 6 = 36. PAGE # 2323 Ex.17A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by man and his original rate of walking. Sol. Let the speed man = x km/hr and Time taken = y hr 1 x (y – 1) = xy 2 xy – x + –x+ 1 1 y– = xy 2 2 1 1 y= 2 2 or – 2x + y = 1 ....(i) And (x – 1) (y + 3) = xy xy + 3x – y – 3 = xy 3x – y = 3 ....(ii) Solving equation (i) & (ii) x = 4 and y = 9 So, total distance covered = xy km = 4 × 9 = 36 km and original rate of walking = 4 km/hr Ex.18 If 4x + 3y = 120, find how many positive integer solutions are possible ? Sol. We can write the equation in another form. 4(x – 3) +3 (y + 4) = 120 Now, we make a table x 30 27 24 21 18 15 12 9 6 3 0 y 0 4 8 12 16 20 24 28 32 36 40 W e observe a pattern that x reduces by 3 & y increases by 4. But, both x & y cannot be negative or 0. So, the value of x = 0 or y = 0 is ruled out. Hence, nine such positive integer solutions are possible. Alternately, we can write the given equation as 4(x + 3) + 3(y – 4) = 120. We get the same values of x & y. PAGE # 2424 QUADRATIC EQUATIONS If P(x) is quadratic expression in variable x, then P(x) = 0 is known as a quadratic equation. (a) General form of a Quadratic Equation : The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, c are real numbers and a 0. Since a 0, quadratic equations, in general, are of the following types:(i) b = 0, c 0, i.e., of the type ax2 + c = 0. Consider the quadratic equation, a x2 + b x + c = 0 having and as its roots and b2 4ac is called discriminant of roots of quadratic equation. It is denoted by D or . Roots of the given quadratic equation may be (i) Real and unequal (ii) Real and equal (iii) Imaginary and unequal. Let the roots of the quadratic equation ax2 + bx + c = 0 (where a 0, b, c R) be and then (ii) b 0, c = 0, i.e., of the type ax2 + bx = 0. = (iii) b = 0, c = 0, i.e., of the type ax2 = 0. (iv) b 0, c 0, i.e., of the type ax2 + bx + c = 0. The value of x which satisfies the given quadratic equation is known as its root. The roots of the given equation are known as its solution. General form of a quadratic equation is : ax2 + bx + c = 0 or or or or 4a2x2 + 4abx + 4ac = 0 [Multiplying by 4a] 4a2x2 + 4abx = – 4ac [By adding b2 both sides] 2 2 2 2 4a x + 4abx + b = b – 4ac (2ax + b)2 = b2 – 4ac Taking square root of both the sides 2ax + b = ± or x= b 2 4ac b b 2 4ac 2a b b 2 4ac b b 2 4ac and . 2a 2a REMARKS : b b 2 4ac ... (ii) 2a The nature of roots depends upon the value of expression ‘b2 – 4ac’ with in the square root sign. This is known as discriminant of the given quadratic equation. and = Consider the Following Cases : Case-1 When b2 – 4ac > 0, (D > 0) In this case roots of the given equation are real and distinct and are as follows = b b 2 4ac b b 2 4ac and = 2a 2a (ii) When a( 0), b, c Q and b2 – 4ac is not a perfect square In this case both the roots are irrational and distinct. [See remarks also] Case-2 When b2 – 4ac = 0, (D = 0) In this case both the roots are real and equal to – A quadratic equation is satisfied by exactly two values of ' x ' which may be real or imaginary. The equation, a x2 + b x + c = 0 is: A quadratic equation if a 0. [Two roots] A linear equation if a = 0, b 0. [One root] A contradiction if a = b = 0, c 0. [No root] An identity if a = b = c = 0. A quadratic equation cannot have more than two roots. ... (i) (i) When a( 0), b, c Q and b2 – 4ac is a perfect square In this case both the roots are rational and distinct. Hence, roots of the quadratic equation ax2 + bx + c = 0 are b b 2 4ac 2a [Infinite roots] It follows from the above statement that if a quadratic equation is satisfied by more than two values of x, then it is satisfied by every value of x and so it is an identity. b . 2a Case-3 When b2 – 4ac < 0, (D < 0) In this case b2 – 4ac < 0, then 4ac – b2 > 0 = and = or = b ( 4ac b 2 ) 2a b ( 4ac b 2 ) 2a b i 4ac b 2 2a b i 4ac b 2 [ 1 = i ] 2a i.e. in this case both the roots are imaginary and distinct. and = 25 25 PAGE # 25 25 REMARKS : If a, b, c Q and b2 – 4ac is positive (D > 0) but not a perfect square, then the roots are irrational and they (ii) If only one root is common, then the common root ' ' will be : = always occur in conjugate pairs like 2 + 3 and 2 – 3 . However, if a, b, c are irrational numbers and b2 – 4ac is positive but not a perfect square, then the roots may not occur in conjugate pairs. Hence the condition for one common root is: c 1 a 2 c 2 a1 2 = a1 b 2 a 2 b1 b1 c 2 b 2 c 1 If b2 – 4ac is negative (D < 0), then the roots are complex conjugate of each other. In fact, complex roots of an equation with real coefficients always occur in conjugate pairs like 2 + 3i and 2 – 3i. However, this may not be true in case of equations with complex coefficients. For example, x2 – 2ix – 1 = 0 has both roots equal to i. If a and c are of the same sign and b has a sign opposite to that of a as well as c, then both the roots are positive, the sum as well as the product of roots is positive (D 0). If a, b, c are of the same sign then both the roots are negative, the sum of the roots is negative but the product of roots is positive (D 0). Relation Between Roots & Coefficients (i) The solutions of quadratic equation a x2 + b x + c = 0 are given by x= b b2 4 a c 2a 2 (ii) The expression b 4 a c D is called discriminant of the quadratic equation a x2 + b x + c = 0. If , are the roots of the quadratic equation a x2 + b x + c = 0, then coefficien t of x (a) Sum of the roots = – + = coefficien t of x 2 b a (b) Product of the roots = NOTE : If f(x) = 0 & g(x) = 0 are two polynomial equation having some common root(s) then those common root(s) is/are also the root(s) of h(x) = a f(x) + bg (x) = 0. Ex.1 If x2 – ax + b = 0 and x2 – px + q = 0 have a root in common and the second equation has equal roots, ap . 2 Sol. Given equations are : x2 – ax + b= 0 ..... (i) and x2 – px + q = 0 .... (ii) Let be the common root. Then roots of equation (ii) will be and . Let be the other root of equation (i). Thus roots of equation (i) are , and those of equation (ii) are , . Now +=a ...(iii) = b ...(iv) 2 = p ...(v) 2 = q ...(vi) L.H.S. = b + q = + 2 = ( + ) ...(vii) show that b + q = ap ( ) 2 = = ( + ) 2 2 From (vii) and (viii), L.H.S. = R.H.S. And R.H.S. = c a b = = 9 1 2 a:b:c=1:2:9 (a) By Factorisation : ALGORITHM : Condition for common Root 2 Consider two quadratic equations, a1 x + b1 x + c1 = 0 & a 2 x2 + b 2 x + c 2 = 0. (i) If two quadratic equations have both roots common, then the equations are identical and their coefficients are in proportion. i.e. a1 b c = 1 = 1 . a2 b2 c2 ... (viii) Ex.2 If a, b, c R and equations ax2 + bx + c = 0 and x2 + 2x + 9 = 0 have a common root, show that a : b : c = 1 : 2 : 9. Sol. Given equations are : x2 + 2x + 9 = 0 ...(i) and ax2 + bx + c = 0 ...(ii) Clearly roots of equation (i) are imaginary since equation (i) and (ii) have a common root, therefore common root must be imaginary and hence both roots will be common. Therefore equations (i) and (ii) are identical constant term coefficien t of x 2 c = a (iii) A quadratic equation whose roots are and is (x ) (x ) = 0 i.e. x2 (sum of roots) x + (product of roots) = 0. c 1 a 2 c 2 a1 b1 c 2 b 2 c 1 = a1 b 2 a 2 b1 c 1 a 2 c 2 a1 Step (i) Factorise the constant term of the given quadratic equation. Step (ii) Express the coefficient of middle term as the sum or difference of the factors obtained in step 1. Clearly, the product of these two factors will be equal to the product of the coefficient of x2 and constant term. Step (iii) Split the middle term in two parts obtained in step 2. Step (iv) Factorise the quadratic equation obtained in step 3. 26 26 PAGE # 26 26 Ex.3 Solve the following quadratic equation by factorisation method : x2 – 2ax + a2 – b2 = 0. Sol. Here, factors of constant term (a2 – b2) are (a – b) and (a + b). Sol. We have, x 3 x – 3 2x – 3 x –1 x2 x–2 Also,Coefficient of the middle term = – 2a = – [(a – b) + (a + b)] x2 – 2ax + a2 – b2 = 0 2 x – {(a – b) + (a + b)} x + (a – b) (a + b) = 0 x2 – (a – b) x – (a + b) x + (a – b) (a + b) = 0 x{x – (a – b)} – (a + b) {x – (a – b)} = 0 {x – (a – b)} {x – (a + b)} = 0 x – (a – b) = 0 or, x – (a + b) = 0 x = a – b or x = a + b ( x 2) 1 ( x – 2) – 1 2( x – 1) – 1 x–2 x –1 x2 1 1 1 1 1– 2– x–2 x –1 x2 1 1 1 – – x2 x–2 x –1 x–2–x–2 1 – 2 x –1 x –4 –4 –1 x2 – 4 x – 1 Ex.4 Solve 9x2 – 49 = 0 Sol. We have 9x2 – 49 = 0 4(x – 1) = x2 – 4 or (3x)2 – (7)2 = 0 or (3x + 7)(3x – 7) = 0 x2 – 4x = 0 i.e., 3x + 7 = 0 or 3x – 7 = 0. x(x – 4) = 0 This gives x = – Thus, x = – x = 0, 4. 7 7 or . 3 3 7 3 or x 3 x – 3 2x – 3 . x –1 x2 x–2 Ex.8 Solve the equation : 7 3 Hence, the roots of the given equation are 0, 4. are solutions of the given (b) By the Method of Completion of Square : equation. ALGORITHM : 2 Ex.5 Solve the quadratic equation : 3x – 15x = 0. Sol. The given equation may be written as 3x(x – 5) = 0 Step-(i) Obtain the quadratic equation. Let the quadratic equation be ax2 + bx + c = 0, a 0. This gives x = 0 or x = 5. Step-(ii) Make the coefficient of x2 unity, if it is not unity. x = 0, 5 are the required solutions. i.e., obtain x2 + Ex. 6 Solve the quadratic equation : 48y2 – 13y – 1 = 0. Step-(iii) Shift the constant term Sol. 48y2 – 13y – 1 = 0 48 y2 – 16 y + 3y – 1 = 0 16 y (3y – 1) + 1(3y – 1) = 0 (16y + 1) (3y – 1) = 0 y= b c x+ = 0. a a c on R.H.S. to get a b c x=– a a x2 + Step-(iv) Add square of half of the coefficient of x.i.e. 2 b on both sides to obtain 2a –1 1 ,y= 16 3 2 2 b b b c x +2 x+ = – 2a 2a 2a a 2 2 Ex.7 Solve the quadratic equation : 6 2 x 5 x – 3 2 0 . Sol. 6 2 x2 + 5x – 3 2 =0 Step-(v) Write L.H.S. as the perfect square of a binomial expression and simplify R.H.S. to get 2 6 2 x2 + 9x – 4x – 3 2 = 0 3x (2 2 x + 3) – 2 (2 2 x + 3)= 0 (3x – 2 ) (2 2 x + 3) = 0 x= 2 –3 or x = 3 2 2 b b 2 4ac x = . 2a 4a 2 Step-(vi) Take square root of both sides to get x+ b =± 2a b 2 4ac 4a 2 Step (vii) Obtain the values of x by shifting the constant term b on RHS. 2a 27 27 PAGE # 27 27 Ex.9 Solve : x2 + 5x + 6 = 0. Sol. We have x2 + 5x + 6 = 0 Add and subtract ( 5 2 2 1 coefficient of x)2 in L.H.S. and get 2 5 2 2 by x = x2 + 5x + 6 + – = 0 2 2 5 5 5 x2 + 2 x + – + 6 = 0 2 2 2 5 1 x – 0 2 4 5 1 x 2 4 x 2 2 5 1 2 2 This gives x = –3 or x = – 2 Therefore, x =–3,–2 are the solutions of the given equation. Ex.10 By using the method of completing the square, show that the equation 4x2 + 3x + 5 = 0 has no real roots. Sol. We have, 4x2 + 3x + 5 = 0 3 5 x+ =0 4 4 x2 + 3 5 x2 + 2 x = – 8 4 3 x2 + 2 x + 8 3 71 x = – 8 64 Ex11 Solve the quadratic equation x2 – 7x – 5 = 0. Sol. Comparing the given equation with ax2 + bx + c = 0, a = 1, b = –7 and c = –5. Therefore, D = (–7)2 – 4 × 1 × (–5) = 49 + 20 = 69 > 0 Since D is positive, the equation has two roots given 2 2 3 3 5 = – 8 8 4 Clearly, RHS is negative. 2 3 But, x cannot be negative for any real value of x. 8 Hence, the given equation has no real roots. 7 69 7 – 69 , are the required solutions. 2 2 Ex.12 Find the value of P, If the quadratic equation x2 – 2x (1 + 3P) + 7(3 + 2P) = 0 has real & equal roots. Sol. x2 – 2x (1 + 3P) + 7 (3 + 2P) = 0 If it has real and equal roots, then D = 0 D = b2 – 4ac a = 1, b = – 2 (1 + 3P), c = 7 (3 + 2P) D = 4 (1 + 3P)2 – 4 × 1 × 7 (3 + 2P) = 4[(1 + 3p)2 – 7(3 + 2P)] = 4 [1 + 6P + 9P2 – 21 – 14 P] D = 4 [9P2 – 8P – 20] Now, D = 0 [according to condition] 4[9P2 – 8P – 20] = 0 9P2 – 8P – 20 = 0 9p2 – 18p + 10p – 20 = 0 9P (P – 2) + 10 (P – 2) = 0 (9P + 10) (P – 2) = 0 So, 9P + 10 = 0 or p – 2 = 0 P = – 10/9 or p = 2. 2 1 1 . b a c Sol. Since the roots of the given equations are equal, so discriminant will be equal to zero. b2(c – a)2 – 4a(b – c)c(a – b) = 0 b2(c2 + a2 – 2ac) – 4ac(ba – ca – b2 + bc) = 0, a2b2 + b2c2 + 4a2c2 + 2b2ac – 4a2bc – 4abc2 = 0 (ab + bc – 2ac)2 = 0 ab + bc – 2ac = 0 ab + bc = 2ac 1 1 2 + = c a b 2 1 1 = + . b a c (c) By Using Quadratic Formula : Solve the quadratic equation in general form viz. ax2 + bx + c = 0. Step (i) By comparison with general quadratic equation, find the value of a, b and c. Step (ii) Find the discriminant of the quadratic equation. D = b2 – 4ac Step (iii) Now find the roots of the equation by given equation x= Ex.13 If the roots of the equation a(b – c) x2 + b(c – a)x + c (a – b) = 0 are equal, show that 2 x= 7 69 7 – 69 , 2 2 b D b D , 2a 2a REMARK : If b2 – 4ac < 0, i.e., negative, then b 2 – 4ac is not real and therefore, the equation does not have any real roots. Hence Proved. Ex.14 For what value of K does the equation (k + 3)x2 – (2k + 5)x + k + 1 = 0 have no real roots. Sol. (k + 3)x2 – (2k + 5) x + k + 1 = 0 If it has no real roots, then D < 0 D = b2 – 4ac {– (2k + 5)}2 – 4 × (k + 3) × (k + 1) = 4k2 + 20 k + 25 – 4 [k2 + k + 3k + 3] = 4k + 13 SInce D < 0 4k + 13 < 0 Hence , k < 13 . 4 28 28 PAGE # 28 28 4500 = x2 – 15x x2 – 15x – 4500 = 0 The method of problem solving consists of the following three steps : x= 15 225 18000 2 Step (i) Translating the word problem into symbolic language (mathematical statement) which means identifying relationships existing in the problem and then forming the quadratic equation. Step (ii) Solving the quadratic equation thus formed. Step (iii) Interpreting the solution of the equation, which means translating the result of mathematical statement into verbal language. x= 15 135 = 75, – 60 2 ALGORITHM : Ex.15 Sum of a number and its reciprocal is 2 1 . Find the 30 number. Sol. Let the number be x. Then according to given condition x+ expenses = 360 x4 Now, according given condition 30x2 + 30 = 61x 30x2 – 61x + 30 = 0 By using quadratic formula 360 360 – =3 x x4 61 3721 – 3600 60 61 11 x= 60 61 11 72 6 x= = = 60 60 5 x= 1440 = 3x2 + 12x x2 + 4x – 480 = 0 (x + 24) (x – 20) = 0 x = 20 So, the original duration of the tour = 20 days. 61 11 50 5 x= = = 60 60 6 Hence the required number is 5 6 or . 6 5 Ex.16 A fast train takes 3 hours less than a slow train for a journey of 900 km. If the speed of slow train is 15 km/ hr. less than that the fast train. Find the speeds of the two trains. Sol. Let the speed of fast train = x km/h and the speed slow train = (x – 15) km/h 900 hr x 900 the time taken by slow train = hr x – 15 then according the given the condition 300 300 3 – =3 x – 15 x Total money 360 = Number of day x when he extend his tour by 4 days, then his daily x 2 1 61 = 30 x 900 900 – =3 x – 15 x Ex.17 A person on tour has Rs. 360 for his expenses. If he extends his tour 4 days, he has to cut down his daily expenses by Rs. 3, Find the original duration of the tour. Sol. let x be the number of days So his daily expenses = 61 1 = 30 x the time taken by fast train = x – 15 = 60 x = 75 So speed of fast train is 75 km/hr and the speed of slow train (75 – 15) km/hr = 60 km/hr. Ex.18 A two digit number is such that the product of its digits is 10 when 27 is subtracted from the number, the digit interchange their places. Find the number. Sol. Let the tens and unit digit of required number be x and y respectively. Then according given condition xy = 10 y = 10 x and 10x + y – 27 = 10y + x 10x + 10 10 +x – 27 = 10 x x 10 x 2 10 27 x 100 x 2 = x x 9x2 – 27x – 90 = 0 9(x2 – 3x – 10) = 0 x2 – 3x – 10 = 0 x2– 5x + 2x – 10 = 0 x(x – 5) + 2(x – 5) = 0 (x – 5)(x + 2) = 0 x = – 2 or 5 So, x = 5, y = 10 10 = =2 x 5 So the required number is 10x + y = 10 × 5 + 2 = 52. 29 29 PAGE # 29 29 PROGRESSIONS SEQUENCE A sequence is an arrangement of numbers in a definite order according to some rule. e.g. (i) 2, 5, 8, 11, ... (ii) 4, 1, – 2, – 5, ... (iii) 3, –9, 27, – 81, ... Types of Sequence On the basis of the number of terms there are two types of sequence : (i) Finite sequences : A sequence is said to be finite if it has finite number of terms. (ii) Infinite sequences : A sequence is said to be infinite if it has infinite number of terms. Ex.1 Write down the sequence whose n th term is n 2+1 : Sol. Let tn = n 2+1 Put n = 1,2,3,4,.............. we get t1 =2, t2 = 5, t3 = 10,.... So the sequence is 2,5,10,17,..... PROGRESSIONS Those sequence whose terms follow certain patterns are called progressions. Generally there are three types of progressions. (i) Arithmetic Progression (A.P.) (ii) Geometric Progression (G.P.) (iii) Harmonic Progression (H.P.) ARITHMETIC PROGRESSION A sequence is called an A.P., if the difference of a term and the previous term is always same. i.e. d = tn + 1 – tn = Constant for all n N. The constant difference, generally denoted by ‘d’ is called the common difference. Ex.2 Find the common difference of the following A.P. : 5,9,13,17,21,..... Sol. 9 – 5 = 13 – 9 = 17 – 13 = 21 – 17 = 4 (constant). Common difference (d) = 4. GENERAL FORM OF AN A.P. If we denote the starting number i.e. the 1st number by ‘a’ and a fixed number to be added is ‘d’ then a, a + d, a + 2d, a + 3d, a + 4d,........... forms an A.P. Ex.3 Find the A.P. whose 1st term is 10 & common difference is 5. Sol. Given :First term (a) = 10 & Common difference (d) = 5. A.P. is 10, 15, 20, 25, 30,....... th n TERM OF AN A.P. Let A.P. be a, a + d, a + 2d, a + 3d,........... Then, First term (a1) = a + 0.d Second term (a2) = a + 1.d Third term (a3) = a + 2.d . . . . . . nth term (an) = a + (n – 1) d an = a + (n – 1) d is called the nth term. Ex.4. Which term of the A.P. 21, 42, 63, 84, .............. is 420? Sol. 21, 42, 63, ........ 420 first term (a) = 21, common difference (d) = 21 Let 420 will come at an term. an=a+(n–1)d 420 = 21 + (n – 1) 21 420 = 21 + 21n – 21 21n = 420 n = 20 Ex.5. The 14th term of an A.P. is twice its 7th term. If its 5th term is 10. Find its first term and the common difference. Also find its 53 rd term. Sol. a14 = 2 a7 a + 13 d = 2 ( a + 6 d) a + 13 d = 2a + 12 d a=d and a5 = 10 a + 4d = 10 a + 4a = 10 [a = d ] 5 a = 10 a = 2. So, d = 2. 1st term = 2, common difference = 2 a53 = a + 52 d a53 = 2 + 52 × 2 a53 = 2 + 104 a53 = 106. Ex.6 Is 51 a term of the A.P. 5,8,11,14,.............? Sol. 5, 8, 11, 14 ............. first term (a)= 5 common difference (d) = 3 Let 51 be the nth term a term of AP and we know an=a+(n–1)d 51 = 5 + (n – 1)3 51 = 5 + 3n – 3 51 = 2 + 3n 49 = 3n n= 49 3 But n can never be a fraction in an AP 51 is not a term of AP. PAGE # 30 Ex.9 Find the sum of 16 terms of the A.P. 3,8,13,18,....... Sol. a = 3, d = 5 th m TERM OF AN A.P. FROM THE END Let ‘a’ be the 1st term and ‘d’ be the common difference of an A.P. having n terms. Then mth term from the end is (n – m + 1)th term from beginning or {n – (m – 1)}th term from beginning. Ex.7 Find 12th term from the end of an A.P. 5,12,19,... 495. Sol. 495 = 5 + (n – 1) 7 n = 71 12h term from end m = 12 a71 – (12 – 1) = a71 – 11 = a60 from the beginning. a60 = 5 + (60 – 1) 7 = 418. SELECTION OF TERMS IN AN A.P. Sometimes we require certain number of terms in A.P. The following ways of selecting terms are generally very convenient. No. of Terms Terms Common Difference For 3 terms a – d, a, a + d d For 4 terms a – 3d, a – d, a + d, a + 3d 2d For 5 terms a – 2d, a – d, a, a + d, a + 2d d For 6 terms a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d 2d Ex.8 The sum of three numbers in A.P. is –3 and their product is 8. Find the numbers. Sol. Let three no.’s in A.P. be a – d, a, a + d a – d + a + a + d = –3 3a = –3 a = –1 & (a – d) a (a + d) = 8 a (a2 – d2) = 8 (–1) (1 – d2) = 8 1 – d2 = – 8 d2 = 9 d= 3 If a = – 1 & d = 3 numbers are – 4, –1, 2. If a = – 1 & d = –3 numbers are 2, –1, – 4. SUM OF n TERMS OF AN A.P. S16 = n [2a + (n – 1)d] 2 16 [2(3) + (16 – 1)5] 2 = 648. Ex.10. Find the value of x when 1 + 6 + 11 + 16 + ........ + x = 148. Sol. Clearly, terms of the given series form an A.P. with first tern a = 1 and common difference d = 5. Let there be n terms in this series. Then, 1 + 6 + 11 + 16 + ...+ x = 148. Sum of n terms = 148 n [ 2+ (n – 1) × 5] = 148 2 5n2 – 3n – 296 = 0 (n – 8) (5n + 37) = 0 n=8 Now, x = nth term x = a + (n – 1)d x = 1 + (8 – 1) × 5 = 36 [ n is not negative] [ a =1, d = 5, n = 8] PROPERTIES OF A.P. (i) For any real numbers a and b, the sequence whose nth term is an = an + b is always an A.P. with common difference ‘a’ (i.e. coefficient of term containing n). (ii) If any nth term of a sequence is a linear expression in n then the given sequence is an A.P. (iii) If a constant term is added to or subtracted from each term of an A.P. then the resulting sequence is also an A.P. with the same common difference. (iv) If each term of a given A.P. is multiplied or divided by a non-zero constant K, then the resulting sequence is also an A.P. with common difference d respectively. Where d is the common K difference of the given A.P. Let A.P. be a, a + d, a + 2d, a + 3d,............., a + (n – 1)d Then, Sn = a + (a + d) +...+ {a + (n – 2) d} + {a + (n – 1) d} ..(i) also Sn= {a + (n – 1) d} + {a + (n – 2) d} +....+ (a + d) + a ..(ii) Add (i) & (ii) Kd or 2Sn = 2a + (n – 1)d + 2a + (n – 1)d +....+ 2a + (n – 1)d (vi) If three numbers a, b, c are in A.P. , then 2b = a + c. 2Sn = n [2a + (n – 1) d] Sn n 2a ( n 1) d 2 Sn = n [a + a + (n – 1)d] 2 n [a + ] 2 n S n a where, is the last term. 2 rth term of an A.P. when sum of first r terms is Sr is given by, tr = Sr – Sr – 1. = Sn = (v) In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of 1st and last term. Ex.11 Check whether an = 2n2 + 1 is an A.P. or not. Sol. an = 2n2 + 1 Then, an + 1 = 2(n + 1)2 + 1 an + 1 – an = 2(n2 + 2n + 1) + 1 – 2n2 – 1 = 2n2 + 4n + 2 + 1 – 2n2 – 1 = 4n + 2, which is not constant The above sequence is not an A.P.. Arithmetic Mean (Mean or Average) (A.M.) If three terms are in A.P. then the middle term is called the A.M. between the other two, so if a, b, c are in A.P., b is A.M. of a & c. PAGE # 31 (ii) Sum of the first n terms. A.M. for any n number a 1, a 2,..., a n is; A= a1 a 2 a 3 ..... a n . n n - Arithmetic Numbers : Means Between Two If a, b are any two given numbers & a, A1, A2,...., An, b are in A.P. then A1, A2,... An are the n-A.M.’s between a & b. Total terms are n + 2. ba . n1 2(b a ) = a + ,. .. ... ... .. ., n 1 Last term b = a + (n+2–1)d.Now, d = ba , A2 n 1 n (b a) An = a + . n 1 A1 = a + NOTE : Sum of all n-A.M.’s inserted between a & b is equal to n times the single A.M. between a & b. n i.e. r 1 Ar = nA where A is the single A.M. between a & b. 13 , an even 6 number of A.M.s is inserted, the sum of these means exceeds their number by unity. Find the number of means. Sol. Let a and b be two numbers and 2n A.M.s are inserted between a and b then 2n (a + b) = 2n + 1. 2 Ex.12 Between two numbers whose sum is a rn 1 r 1 Sn = a 1 rn 1 r 13 13 n = 2n + 1. Given a b 6 6 n = 6. Number of means = 12. Ex.13 Insert 20 A.M. between 2 and 86. Sol. Here 2 is the first term and 86 is the 22 nd term of A.P. So, 86 = 2 + (21)d d=4 So, the series is 2, 6, 10, 14,......., 82, 86 required means are 6, 10, 14,...82. , r 1 , r 1 (iii) Sum of an infinite G.P. when r < 1. When n rn 0 if r < 1 therefore, a (| r | 1) . S = 1– r Ex.14 Find 10th term of the G.P. 7, 14, 28, ..... Sol. In this G.P. we have a = 7 and r = 2 10th term = ar(10–1) = ar9 = 7 29 = (7 512) = 3584. Ex.15 How many terms are there in the G.P. 16,8,4....., Sol. In this G.P. we have a = 16, r = 8 1 = 16 2 Let the number of terms be n. Then, 1 16 2 1 2 n 1 = n 1 = 1 16 1 1 = 16 16 2 8 n – 1 = 8 or n = 9 Ex.16 A boy saves Re 1 on first day of the month, Rs. 2 on 2nd day of the month, Rs. 4 on third day, Rs. 8 on 4th day, Rs. 16 on 5th day and so on. How much does he save in 12 days ? Sol. His savings are 1,2,4,8,16, .... upto terms. This is a G.P. with a = 1 and r = 2 =2 1 Savings = Sum of above G.P. upto 12 terms. G.P. is a sequence of numbers whose first term is non zero & each of the succeeding terms is equal to the preceding terms multiplied by a constant. Thus in a G.P. the ratio of successive terms is c o ns t ant . Thi s c ons tan t f ac t or is c a lle d t he common ratio of the series & is obtained by dividing any term by that which immediately precedes it. 1 1 1 1 Example : 2, 4, 8, 16 ... & , , , ...are in G.P. .P. 3 9 27 81 (i) Therefore a, ar, ar2, ar3, ar4,...... is a G.P. with ‘a’ as the first term and ‘r’ as common ratio. nth term = a rn1 1 ? 16 = a(r n 1) 1 ( 212 1) = = 4095. (r 1) (2 1) Ex.17 Find the sum of 10 terms of the sequence 5, 5.5, 5.55, 5.555, ...... Sol. 5, 5.5., 5.55, 5.555, ..... 10 terms = 5(1 + 1.1 + 1.11 + ..... 10 terms) = 5 (9 + 9.9 + 9.99 + .... 10 terms) 9 = 5 (10 – 1 + 10 – 0.1 + 10 – 0.01 + ..... 10 terms) 9 = 5 (10 + 10 + 10 +.....10 terms) – (1 + 0.1 + 0.01 9 +.....10 terms) PAGE # 32 = 5 (100 – sum of G.P. of 10 terms with a = 1 & r = 0.1) 9 Sol. Let G1, G2, G3 be three geometric means between 4 10 5 = 9 100 = 5 Ex.18 Find three geometric means between 4 and 64. 1 (1 (0.1) ) 5 10 1 100 1 10 1 0.1 9 9 10 and 64. Then 4, G1, G2, G3, G4, 64 are in G.P. 5th term = 64 Let common ratio be r. 5 (890 10 9 ) . 81 Then, 4 r4 = 64 r4 = 16 = 24 r=2 G1 = (4 2) = 8, G2 = (8 2) = 16 & G3 = (16 2) = 32 (i) If a, b, c are in G.P. then b 2 = ac, in general if a 1, a 2, a 3, a 4,......... a n – 1 , a n are in G.P., then a 1a n = a 2a n – 1 = a 3 a n – 2 = .............. So, the required geometric means are 8,16, 32. Important results : n(n 1) 2 (ii) Any three consecutive terms of a G.P. can be i. Sum of the first n natural numbers = a , a , ar.. r (iii) Any four consecutive terms of a G.P. can be taken a a as 3 , , ar, ar3. r r ii. Sum of the squares of first n numbers taken as (iv) If each term of a G.P. be multiplied or divided or raised to power by the some nonzero quantity, the resulting sequence is also a G.P. = n(n 1)(2n 1) 6 n(n 1) iii. Sum of the cubes of first n numbers = 2 iv. Sum of the first n even numbers = n(n + 1) v. Sum of the first n odd numbers = n2 If a, b, c are in G.P., b is the G.M. between a & c. b² = ac, therefore b = a c ; a > 0, c > 0. n-Geometric Means Between a, b : If a, b are two given numbers & a, G1, G2,....., Gn, b are in G.P.. Then, G1, G2, G3,...., Gn are n-G.M.s between a & b. G1 = a(b/a)1/n+1, G2 = a(b/a)2/n+1,......, Gn = a(b/a)n/n+1 NOTE : The product of n G.M.s between a & b is equal to the nth power of the single G.M. between a & b n i.e. n G = ab = G r n where G is the single G.M. r 1 between a & b. PAGE # 33 2 LINES AND ANGLES, TRIANGLES (iii) Obtuse angle : An angle whose measure is more than 90º but less than 180º is called an obtuse angle. A line has length but no width and no thickness. An angle is the union of two non-collinear rays with a common initial point. The common initial point is called the ‘vertex’ of the angle and two rays are called the ‘arms’ of the angles. 90º < AOB < 180º. (iv) Straight angle : An angle whose measure is 180º is called a straight angle. (v) Reflex angle : An angle whose measure is more than 180º is called a reflex angle. REMARK : Every angle has a measure and unit of measurement is degree. One right angle = 90º 1º = 60’ (minutes) 1’ = 60” (Seconds) Angle addition axiom : If X is a point in the interior of BAC, then m BAC = m BAX + m XAC. (a) Types of Angles : (i) Right angle : An angle whose measure is 90º is 180º < AOB < 360º. (vi) Complementary angles : Two angles, the sum of whose measures is 90º are called complementary angles. AOC & BOC are complementary as their sum is 90 º. called a right angle. (vii) Supplementary angles : Two angles, the sum of whose measures is 180 º , are called the supplementary angles. (ii) Acute angle : An angle whose measure is less AOC & BOC are supplementary as their sum is 180 º. than 90 is called an acute angle. º (viii) Angle Bisectors : A ray OX is said to be the bisector of AOB , if X is a point in the interior of AOB, and AOX = BOX. B O 0 0 < BOA < 90 A 0 PAGE # 3434 (ix) Adjacent angles : Two angles are called adjacent angles, if (A) they have the same vertex, (B) they have a common arm, (C) non common arms are on either side of the common arm. AOX and BOX are adjacent angles, OX is common arm, OA and OB are non common arms and lies on either side of OX. (x) Linear pair of angles : Two adjacent angles are said to form a linear pair of angles, if their non common arms are two opposite rays. AOC + BOC = 180º. (xi) Vertically opposite angles : Two angles are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays. AOC & BOD form a pair of vertically opposite (i) Corresponding angles : Two angles on the same side of a transversal are known as the corresponding angles if both lie either above the two lines or below the two lines, in figure 1 & 5, 4 & 8, 2 & 6, 3 & 7 are the pairs of corresponding angles. If a transversal intersects two parallel lines then the corresponding angles are equal i.e. 1 = 5, 4 = 8, 2 = 6 and 3 = 7. (ii) Alternate interior angles : 3 & 5, 2 & 8, are the pairs of alternate interior angles. If a transversal intersects two parallel lines then the each pair of alternate interior angles are equal i.e. 3 = 5 and 2 = 8. (iii) Co- interior angles : The pair of interior angles on the same side of the transversal are called pairs of consecutive or co - interior angles. In figure 2 &5, 3 & 8, are the pairs of co-interior angles. If a transversal intersects two parallel lines then each pair of consecutive interior angles are supplementary i.e. 2 + 5 = 180º and 3 + 8 = 180º. Ex.1. Find the measure of an angle, if six times its complement is 12º less than twice its supplement. Sol. Let the measure of the required angle be xº. Then, measure of its complement = (90 – x)º. Measure of its supplement = (180 – x)º. 6 (90 – x) = 2 (180 – x) – 12 540 – 6x = 360 – 2x – 12 4x = 192 x = 48º. Hence, the required angle is 48º. Ex.2. In the adjoining figure, ABC = 100º, EDC = 120º and AB || DE. Then, find BCD . angles. Also AOD & BOC form a pair of vertically opposite angles. D If two lines intersect, then the vertically opposite angles are equal i.e. AOC = BOD and BOC = AOD. A xº (b) Angles Made by a Transversal with two Parallel Lines : C Transversal : A line which intersects two or more given parallel lines at distinct points is called a transversal of the given lines. E B D Sol. A B E F xº C Produce AB to meet CD at F. BFD = EDF = 120° [alternate interior s] BFC = (180° – 120°) = 60° CBF = (180° – 100°) = 80° BCF = 180°– (60° + 80°) = 40°. PAGE # 3535 POLYGON A closed plane figure bounded by line segments is called a polygon. A polygon is named according to the number of sides it has : No. of sides Figure 3 4 5 6 7 8 Ex.3 If the sum of interior angles of a polygon is 1620º, find its number of sides. Sol. We know that sum of all interior angles of a n sided polygon = (2n – 4) right angles = [(2n – 4) 900] but given sum of interior angles = 1620º (2n – 4) 900 = 1620º 10 2n – 4 = Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Decagon 2n – 4 = 18 2n = 22 n = 11. In general, a polygon having n sides is called 'n' sided polygon. Diagonal of Polygon : Line segment joining any two non-consecutive vertices of a polygon is called its diagonal. TRIANGLE A plane figure bounded by three lines in a plane is called a triangle. Every triangle have three sides and three angles. If ABC is any triangle then AB, BC & CA are three sides and A,B and C are three angles. Convex Polygon : If all the interior angles of a polygon are less than 1800, it is called a convex polygon. 1620 º 90 º Concave Polygon : If one or more of the interior angles of a polygon is greater than 1800 i.e. reflex, it is called a concave polygon. Types of triangles : A. On the basis of sides we have three types of triangle. 1. Scalene triangle – A triangle in which no two sides are equal is called a scalene triangle. 2. Isosceles triangle – A triangle having two sides equal is called an isosceles triangle. 3. Equilateral triangle – A triangle in which all sides are equal is called an equilateral triangle. Regular Polygon : A polygon is called a regular polygon if all its sides have equal length and all the angles have equal measure. B. On the basis of angles we have three types : 1. Right triangle – A triangle in which any one angle is right angle is called right triangle. 2. Acute triangle – A triangle in which all angles are acute is called an acute triangle. 3. Obtuse triangle – A triangle in which any one angle is obtuse is called an obtuse triangle. REMARKS : (1) The sum of the interior angles of a convex polygon of n sides is (2n – 4) right angles or (2n – 4) 90º. (2) The sum of the exterior angles of a convex polygon is 4 right angles or 360º. (3) Each interior angle of a n-sided regular polygon is 2n 4 90º n (4) Each exterior angle of a regular polygon of n sides 3600 = n (5) If a polygon has n sides, then the number of diagonals nn 3 of the polygon = . 2 SOME IMPORTANT THEOREMS : Theorem : The sum of interior angles of a triangle is 180º. Theorem : If the bisectors of angles ABC and ACB of a triangle ABC meet at a point O, then 1 BOC = 90º + A. 2 Exterior Angle of a Triangle : If the side of the triangle is produced, the exterior angle so formed is equal to the sum of two interior opposite angles. Corollary : An exterior angle of a triangle is greater than either of the interior opposite angles. Theorem : The sides AB and AC of a ABC are produced to P and Q respectively. If the bisectors of PBC and 1 QCB intersect at O, then BOC = 90º – A. 2 PAGE # 3636 Ex.4 In figure, TQ and TR are the bisectors of Q and R respectively. If QPR = 80º and PRT = 30º, determine TQR and QTR. Sol. Since the bisectors of Q and R meet at T. (b) Sufficient Conditions for Congruence of two Triangles : (i) SAS Congruence Criterion : A P P 80º T B Q C R Two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle. 30 º Q R 1 QTR = 90º + QPR 2 1 QTR = 90º + (80º) 2 QTR = 90º + 40º = 130º In QTR, we have TQR + QTR + TRQ = 180º TQR + 130º + 30º = 180º [ TRQ = PRT = 30º ] TQR = 20º Thus, TQR = 20º and QTR = 130º. (ii) ASA Congruence Criterion : A P B Q C R Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle. The figures are called congruent if they have same shape and same size. In other words, two figures are called congruent if they are having equal length, width and height. Fig.(i) (iii) AAS Congruence Criterion : A P B Q C R If any two angles and a non included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent. Fig.(ii) (iv) SSS Congruence Criterion : In the above figures {fig.(i) and fig.(ii)} both are equal in length, width and height, so these are congruent figures. A P (a) Congruent Triangles : Two triangles are congruent if and only if one of them can be made to superimposed on the other, so as to cover it exactly. B C Q R Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle. ( v ) RHS Congruence Criterion : A If two triangles ABC and DEF are congruent then A = D, B = E, C = F and AB = DE, BC = EF, AC = DF. If two ABC & DEF are congruent then we write B Q C R Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle. ABC DEF, we can not write as ABC DFE or ABC EDF. Hence, we can say that “two triangles are congruent if and only if there exists a one-one correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal. P NOTE : If two triangles are congruent then their corresponding sides and angles are also congruent by CPCT (corresponding parts of congruent triangles are also congruent). Theorem : If the bisector of the vertical angle bisects the base of the triangle, then the triangle is isosceles. PAGE # 3737 Ex.5 In figure, line m is the bisector of an angle A and O is any point on m. OQ and OP are perpendiculars from O to the arms of A. Prove that : In triangle ABC, AB = c, BC = a & CA = b then, A P m O c b B A Q (i) AOQ AOP (ii) OQ = OP. Sol. (i) In AOQ and AOP, OAQ = OAP [ line m is the bisector of A] AQO = APO = 90º [Given] and AO = AO [Common] By AAS congruency AOQ AOP (ii) OQ = OP [C.P.C.T.] Ex.6 If D is the mid-point of the hypotenuse AC of a right 1 triangle ABC, prove that BD = AC. 2 Sol. Given : ABC is a right triangle such that B = 900 and D is mid point of AC. 1 To prove : BD = AC. 2 Construction : Produce BD to E such that BD = DE and join EC. Proof : In ADB and CDE AD = DC [Given] BD = DE [By construction] And, ADB = CDE [Vertically opposite angles] By SAS criterion of congruence we have ADB CDE EC = AB and CED = ABD ... (i) [By CPCT]] But CED & ABD are alternate interior angles CE || AB ABC + ECB = 1800 [Consecutive interior angles] 90 + ECB = 1800 ECB = 900. Now, In ABC & ECB we have AB = EC [By (i)] BC = BC [Common] And, ABC = ECB = 900 By SAS criterion of congruence ABC ECB [By CPCT] AC = EB 1 1 AC = EB 2 2 1 BD = AC. 2 Hence Proved. a C (i) The sum of any two sides of a triangle is greater than its third side. i.e. in ABC, (A) a + b > c (B) b + c > a (C) a + c > b (ii) If two sides of a triangle are unequal, then the longer side has greater angle opposite to it i.e. in ABC, if AB > AC then C > B. (iii) Of all the line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest. i.e. in PMN, P M N (A) PM < PN. (iv) The difference between any two sides of a triangle is less than its third side. i.e. in ABC, (A) a – b < c (B) b – c < a (C) a – c < b (v) In a right angle triangle the sum of squares of two smaller sides is equal to the square of its third side. i.e. in ABC, a2 + b2 = c2. (vi) If sum of squares of two smaller sides is greater than the square of its third side then that triangle is acute angled triangle. i.e in ABC, a2 + b2 > c2. (vii) If sum of squares of two smaller sides is lesser than the square of its third side then that triangle is obtuse angled triangle. i.e in ABC, a2 + b2 < c2. Ex.7 From which given triplet we can make the sides a triangle. (i) {15, 7, 8} (ii) {3.5, 4.5, 5.5} Sol. As we know sum of two sides is always greater then third side. (i) we can not make the triangle because here the sum of two side is equal to third side i.e 7+8=15 (ii) We can make the triangle because sum of two sides is always greater then third side 3.5 +4.5 >5.5 3.5 +5.5 >4.5 4.5 +5.5 >3.5 PAGE # 3838 Ex.8 Prove that any two sides of the triangle are together greater than twice the median drawn to the third side. Sol. Given :ABC and AD is the median. To prove : AB + AC > 2AD A (ii) (SSS Similarity) If the corresponding sides of two triangles are proportional, then they are similar. (iii) (SAS Similarity) If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar. D B (b) Results Based Upon Characteristic Properties C of Similar Triangles : (i) If two triangles are equiangular, then the ratio of the corresponding sides is the same as the ratio of the corresponding medians. E Construction : Produce AD to E such that AD = DE. Join EC. (ii) If two triangles are equiangular, then the ratio of the Proof : In ADB and CDE corresponding sides is same as the ratio of the AD = DE [By construction] BD = DC [AD is the median] ADB = CDE ADB CDE [Vertically opposite angles] [By SAS congruency] corresponding angle bisector segments. (iii) If two triangles are equiangular then the ratio of the corresponding sides is same as the ratio of the So, by CPCT corresponding altitudes. AB = EC (iv) If one angle of a triangle is equal to one angle of In AEC another triangle and the bisectors of these equal AC + EC > 2AD. angles divide the opposite side in the same ratio, then [Sum of two sides of a triangle is always greater than the third side] the triangles are similar. So, (v) If two sides and a median bisecting the third side of AC + AB > 2AD. [As EC = AB]. a triangle are respectively proportional to the corresponding sides and the median of another Two triangles ABC and DEF are said to be similar if their (i) Corresponding angles are equal. triangle, then two triangles are similar. (vi) If two sides and a median bisecting one of these sides of a triangle are respectively proportional to the i.e. A = D, B = E, C = F And, two sides and the corresponding median of another triangle, then the triangles are similar. Statement : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. i.e., In ABC in which a line parallel to side BC intersects (ii) Corresponding sides are proportional. other two sides AB and AC at D and E respectively. AB BC AC i.e. = = . DE EF DF Then, AE AD = . EC DB A (a) Characteristic Properties of Similar Triangles : (i) (AAA Similarity) If two triangles are equiangular, then they are similar. D B E C PAGE # 3939 Ex.9 In a trapezium ABCD, AB || DC and DC = 2AB. EF drawn parallel to AB cuts AD in F and BC in E such that Corollary : If in a ABC, a line DE || BC, intersects AB in D and AC in E, then (i) DB EC AD AE BE 3 . Diagonal DB intersects EF at G. Prove that EC 4 7FE = 10AB. Sol. In DFG and DAB, 1 = 2 AB AC (ii) AD AE A B 2 F E D AB AC DB EC (v) G 1 AD AE (iii) AB AC (iv) [Corresponding s AB ||FG] DB EC AB AC Converse of Basic Proportionality Theorem : Statement : If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. Some Important Results and Theorems : (i) The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. (ii) In a triangle ABC, if D is a point on BC such that C FDG = ADB [Common] DFG ~ DAB [By AA rule of similarity] DF FG DA AB Again in trapezium ABCD EF ||AB ||DC AF BE DF EC … (i) BE 3 EC 4 ( given ) AF 3 DF 4 AF 3 1 1 DF 4 AF DF 7 DF 4 AD 7 DF 4 DF 4 AD 7 D divides BC in the ratio AB : AC, then AD is the bisector of A. From (i) and (ii), we get FG 4 4 i.e., FG = AB … (iii) AB 7 7 (iii) The external bisector of an angle of a triangle divides the opposite sides externally in the ratio of the sides containing the angle. (iv) The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side. (v) The line joining the mid-points of two sides of a In BEG and BCD, we have BEG = BCD [Corresponding angle EG || CD] GBE = DBC [Common] BEG ~ BCD [By AA rule of similarity] BE EG BC CD triangle is parallel to the third side. (vi) The diagonals of a trapezium divide each other proportionally. (vii) If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium. (viii) Any line parallel to the parallel sides of a trapezium divides the non-parallel s ides proportionally. 3 EG 7 CD EC 4 EC BE 4 3 BC 7 BE 3 EG 7 i.e., BE 3 BE 3 BE 3 3 3 EG = CD (2 AB) CD 2AB (given) 7 7 6 EG = AB ... (iv) 7 Adding (iii) and (iv), we get FG + EG = (ix) If three or more parallel lines are intersected by two transversals, then the intercepts made by them on the transversals are proportional. … (ii) EF = 4 6 10 AB AB AB 7 7 7 10 AB i.e., 7EF = 10AB. 7 Hence proved. PAGE # 4040 Ex.10 In the given figure, PA, QB and RC are each perpendicular to AC. Prove that 1 1 1 . x z y Properties of Areas of Similar Triangles : (i) The areas of two similar triangles are in the ratio of the squares of corresponding altitudes. Sol. In PAC, we have BQ || AP (ii) The areas of two similar triangles are in the ratio of BQ CB AP CA [ CBQ ~ CAP] y CB x CA … (i) the squares of the corresponding medians. (iii) The area of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments. In ACR, we have BQ || CR BQ AB CR AC Ex.11 Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal. Sol. Given : A square ABCD. Equilateral triangles BCE and ACF have been described on side BC and diagonal AC respectively. ABQ ~ ACR y AB … (ii) z AC Adding (i) and (ii), we get 1 . Area (ACF) 2 Proof : Since BCE and ACF are equilateral. Therefore, they are equiangular ( each angle being equal to 60º) and hence BCE ~ ACF. P To Prove : Area (BCE) = R Q x z y A C B y y CB AB x z AC AC y y AB BC x z AC y y AC x z AC y y 1 x z 1 1 1 . x z y Area( BCE) BC 2 = Area( ACF) AC 2 Area( BCE) = Area( ACF) Area( BCE) 1 = . Area( ACF) 2 Hence Proved. Statement :The ratio of the areas of two similar triangles is equal to the square of the ratio of their BC 2 2 2BC = 1 2 Hence Proved. corresponding sides. i.e., In 2 triangles ABC and PQR PYTHAGORA S THEOREM such that ABC ~ PQR [Shown in the figure] Statement : In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the P A other two sides. i.e., In a right triangle ABC, right angled at B. B M C Q 2 Then, R N 2 2 ar( ABC) AB BC CA = = . = ar(PQR) PQ QR RP Then, AC2 = AB2 + BC2 PAGE # 4141 Converse of Pythagoras Theorem : , Statement : In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, Ex.12 In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD2 = 7AB2. Sol. ABC be an equilateral triangle and D be point on BC then the angle opposite to the first side is a right angle. such that BD = Some Results Deduced From Pythagoras Theorem : (i) In the given figure ABC is an obtuse triangle, obtuse angled at B. If AD CB, then AC2 = AB2 + BC2 + 2BC. BD 1 BC (Given) 3 Draw AE BC, Join AD. BE = EC ( Altitude drown from any vertex of an equilateral triangle bisects the opposite side) A D A B C (ii) In the given figure, if B of ABC is an acute angle and AD BC, then AC2 = AB2 + BC2 – 2BC . BD B So, BE = EC = D E BC 2 In ABC AB2 = AE2 + EB2 AD2 = AE2 + ED2 From (i) and (ii) AB2 = AD2 – ED2 + EB2 (iii) In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. (iv) Three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle. C ......(i) .....(ii) BC2 BC2 + 36 4 BC BC BC [ BD + DE = + DE = DE = 3 2 2 BC 2 BC2 AB2 + – = AD2 [ EB = 36 4 AB2 = AD2 – AB2 + AB 2 AB2 – = AD2 36 4 BC ] 6 BC ] 2 [ AB = BC] 36 AB 2 AB 2 9 AB 2 = AD2 36 28 AB 2 = AD2 7AB2 = 9AD2. 36 PAGE # 4242 QUADRILATERALS QUADRILATERAL A quadrilateral is a four sided closed figure. D (ii) Rectangle : A rectangle is a parallelogram, in which each of its angle is a right angle. If ABCD is a rectangle then A = B = C = D = 90°, AB = CD, BC = AD and diagonals AC = BD. C D A C 900 A B (iii) Rhombus : A rhombus is a parallelogram in which all its sides are equal in length. If ABCD is a rhombus then, AB = BC = CD = DA. B Let A, B, C and D be four points in a plane such that : (i) No three of them are collinear. (ii) The line segments AB, BC, CD and DA do not intersect except at their end points, then figure obtained by joining A, B, C & D is called a quadrilateral. Convex and Concave Quadrilaterals : (i) A quadrilateral in which the measure of each interior angle is less than 180° is called a convex quadrilateral. In figure, PQRS is convex quadrilateral. R The diagonals of a rhombus are perpendicular to each other. (iv) Square : A square is a parallelogram having all sides equal and each angle equal to right angle. If ABCD is a square then AB = BC = CD = DA, diagonal AC = BD and A = B = C = D = 90°. S Q P (ii) A quadrilateral in which the measure of one of the interior angles is more than 180° is called a concave quadrilateral. In figure, ABCD is concave quadrilateral. (v) Trapezium : A trapezium is a quadrilateral with only one pair of opposite sides parallel. In figure, ABCD is a trapezium with AB || DC. B D The diagonals of a square are perpendicular to each other. C D C A Special Quadrilaterals : (i) Parallelogram : A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. In figure, AB || DC, AD || BC therefore, ABCD is a parallelogram. D A B (vi) Kite : A kite is a quadrilateral in which two pairs of adjacent sides are equal. If ABCD is a kite then AB = AD C and BC = CD. C B A B Properties : (a) A diagonal of a parallelogram divides it into two congruent triangles. D A (vii) Isosceles trapezium : A trapezium is said to be an isosceles trapezium, if its non-parallel sides are equal. Thus a quadrilateral ABCD is an isosceles trapezium, if AB || DC and AD = BC. (b) In a parallelogram, opposite sides are equal. (c) The opposite angles of a parallelogram are equal. (d) The diagonals of a parallelogram bisect each other. In isosceles trapezium A = B and C =D. PAGE # 4343 Ex.2. In the figure below, P and Q are the mid-points of the PROPERTIES sides AB and BC of the rectangle ABCD Theorem 1 : The sum of the four angles of a quadrilateral is 360°. D C Theorem 2 : A diagonal of a parallelogram divides it into two congruent triangles. Q Theorem 3 : In a parallelogram, opposite sides are equal. A Theorem 6 : Each of the four angles of a rectangle is a right angle. Theorem 7 : Each of the four sides of a rhombus is of the same length. Theorem 8 : Each of the angles of a square is a right angle and each of the four sides is of the same length. Theorem 9 : The diagonals of a rectangle are of equal length. what is the area of the whole rectangle ? Sol. D C Q A P B In QAB, QP is median So, ar QAB = 2 area APQ = 2 × 1 = 2 cm2 In ABC, AQ is median Area ABC = 2 area ABQ Theorem 10 : The diagonals of a rhombus are perpendicular to each other. Theorem 11 : The diagonals of a square are equal and perpendicular to each other. Theorem 12 : Parallelogram and Triangles on the same base (or equal bases) and between the same parallels, then area of parallelogram is twice the area of triangle. = 2 × 2 = 4 cm2 Area of rectangle ABCD = 2 ar. ABC = 2 × 4 = 8 cm2. Ex.3 In the adjoining figure, ABCD is a parallelogram and the bisector of A bisect BC at X. Prove that AD = 2AB. A quadrilateral become a parallelogram when : (i) Opposite angles are equal. (ii) Both the pair of opposite sides are equal (iii) A pair of opposite side is equal as well as parallel (iv) Diagonals of quadrilateral bisect each other. REMARK : (i) Square, rectangle and rhombus are all parallelograms. (ii) Kite and trapezium are not parallelograms. (iii) A square is a rectangle. (iv) A square is a rhombus. (v) A parallelogram is a trapezium. Ex.1 In the parallelogram, Find the value of “x”. 80º xº D C X 2 A 1 B Sol. ABCD is a parallelogram. AD || BC and AX cuts them. BXA = DAX = 1 A [Alternate interior angles] 2 1 A. 2 1 Also, 1 = A 2 2 = 2 = 1 150º Sol. A = C = 80º (Opposite angle of a parallelogram) FEB = x + A (Exterior angle of a triangle is equal to interior opposite angle) D C 150° = x + 80 80 x = 70° F x 150 A B E B If the area of the triangle APQ is 1 square centimeter, Theorem 4 : The opposite angles of a parallelogram are equal. Theorem 5 : The diagonals of a parallelogram bisect each other. P AB = BX AB = 1 BC 2 1 AB = AD 2 AD = 2 AB. PAGE # 4444 In ADC, SR || AC and SR = In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and is half of it. i.e., In a triangle ABC in which P is the mid-point of side AB and Q is the mid-point of side AC. A – P 1 AC ... (ii) 2 [By mid-point theorem] PQ = SR and PQ || SR [From (i) and (ii)] PQRS is a parallelogram. Now, PQRS will be a rectangle if any angle of the parallelogram PQRS is 90º. Q – C B then, PQ is parallel to BC and is half of it i.e., PQ || BC 1 and PQ = BC. 2 PQ || AC [By mid-point theorem] QR || BD [By mid-point theorem] But, AC BD [Diagonals of a rhombus are perpendicular to each other] PQ QR Converse of the Mid-Point Theorem : The line drawn through the mid-point of one side of a triangle parallel to the another side; bisects the third side. i.e., A triangle ABC in which P is the mid-point of side AB and PQ is parallel to BC. A P [Angle between two lines = angle between their parallels] PQRS is a rectangle. Hence Proved. Ex.5 In the given figure, E and F are respectively, the mid-points of non-parallel sides of a trapezium ABCD. Prove that : (i) EF || AB Q B (ii) EF = 1 (AB + DC). 2 C then, PQ bisects the third side AC i.e., AQ = QC. REMARK : In quadrilateral ABCD, if side AD is parallel to side BC; ABCD is a trapezium. Sol. Join BE and produce it to intersect CD produced at point P. In AEB and DEP, AB || PC and BP is transversal. Now, P and Q are the mid-points of the non-parallel 1 sides of the trapezium; then PQ = (AD + BC). 2 i.e. The length of the line segment joining the mid-points of the two non-parallel sides of a trapezium is always equal to half of the sum of the lengths of its two parallel sides. Ex.4 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Prove that the quadrilateral PQRS is a rectangle. Sol. ABE = DPE [Alternate interior angles] AEB = DEP [Vertically opposite angles] And AE = DE [E is mid-point of AD] So, AEB DEP [By ASA congruency] BE = PE [By CPCT] And AB = DP [By CPCT] Since, the line joining the mid-points of any two sides of a triangle is parallel and half of the third side, Therefore, in BPC, E is mid-point of BP [As, BE = PE] and F is mid-point of BC [Given] 1 PC 2 EF || DC and EF = 1 (PD + DC) 2 1 EF || AB and EF = (AB + DC) 2 [As, DC || AB and PD = AB] EF || PC and EF = In ABC, PQ || AC and PQ = 1 AC ... (i) 2 [By mid-point theorem] Hence Proved. PAGE # 4545 Ex.6 AD and BE are medians of ABC and BE || DF. Prove that CF = A 1 AC. 4 then, ar(||gm ABCD) = ar(||gm ABEF) Theorem : The area of parallelogram is the product of its base and the corresponding altitude. i.e., In a ||gm ABCD in which AB is the base and AL is the corresponding height. E F Sol. B D C In BEC, DF is a line through the mid - point D of BC and parallel to BE intersecting CE at F. Therefore, F is the midpoint of CE. Because the line drawn through the mid point of one side of a triangle and parallel to another sides bisects the third side. Now, F is the mid point of CE 1 CE 2 1 1 CF = ( AC) 2 2 CF = 1 AC. 4 CF = Ex.7 ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ADX) = ar(ACY). Sol. D C Y A (a) Base and Altitude of a Parallelogram : (i) Base : Any side of a parallelogram can be called its base. (ii) Altitude : The length of the line segment which is perpendicular to the base from the opposite side is called the altitude or height of the parallelogram corresponding to the given base. D B X Join CX, DX and AY. Clearly, triangles ADX and ACX are on the same base AX and between the parallels AB and DC. ar (ADX) = ar (ACX) ... (i) Also, ACX and ACY are on the same base AC and between the parallels AC and XY. ar (ACX) = ar (ACY) ...(ii) From (i) and (ii), we get ar (ADX) = ar (ACY). C M A then, area (||gm ABCD) = AB × AL. B L gm (i) DL is the altitude of || ABCD, corresponding to the base AB. (ii) DM is the altitude of ||gm ABCD, corresponding to the Theorem : Two triangles on the same base (or equal bases) and between the same parallels are equal in area. i.e., If two triangles ABC and PBC are on the same base BC and between the same parallel lines BC and AP. then, ar( ABC) = ar( PBC). base BC. Theorem : A diagonal of a parallelogram divides it into two triangles of equal area. between the same parallels are equal in area. i.e., If two ||gms ABCD and ABEF are on the same base Theorem : The area of a trapezium is half the product of its height and the sum of the parallel sides. i.e., In a trapezium ABCD in which AB || DC, AL DC, AB and between the same parallels AB and FC. CN AB and AL = CN = h (say), AB = a, DC = b. Theorem : Parallelograms on the same base and b L D C h A h a N B PAGE # 4646 then, ar(trapezium ABCD) = 1 h × (a + b). 2 Theorem : Median of a triangle divides it into two triangles of equal area. i.e., In a ABC in which AD is the median. A B D then, ar (ABD) = ar(ADC) = C 1 ar(ABC) 2 Ex.8 ABC is a triangle in which D is the mid-point of BC and E is the mid-point of AD. Prove that the area of BED 1 area of ABC. 4 Sol. Given : A ABC in which D is the mid-point of BC and = E is the mid-point of AD. To prove : ar( BED) = 1 ar( ABC). 4 Proof : AD is a median of ABC. 1 ar ( ABC) ... (i) 2 [ Median of a triangle divides it into two triangles of equal area] Again, ar ( ABD) = ar( ADC) = BE is a median of ABD. 1 ar ( ABD) 2 [ Median of a triangle divides it into two triangles of equal area] 1 And ar( BED) = ar( ABD) 2 ar ( BEA) = ar( BED) = = ar ( BED) = 1 1 × ar ( ABC) 2 2 1 ar( ABC). 4 [From (i)] Hence Proved. PAGE # 4747 CIRCLES Circumference : The length of the complete circle is called its circumference. Circle : The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle. Segment : The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle. There are two types of segments which are the major segment and the minor segment (as in figure). O Major segment P P In figure, O is the centre and the length OP is the radius of the circle. So the line segment joining the centre and any point on the circle is called a radius of the circle. Chord : If we take two points P and Q on a circle, then the line segment PQ is called a chord of the circle. Sector : The region between an arc and the two radii, joining the centre to the end points of an arc is called a sector. Minor arc corresponds to the minor sector and the major arc corresponds to the major sector. When two arcs are equal, then both segments and both Q P Major sector O Diameter : The chord which passes through the centre of the circle, is called the diameter of the circle. P Q Q Theorem : Equal chords of a circle subtend equal angles at the centre. O A Given : AB and CD are the two equal chords of a circle with centre O. A diameter is the longest chord and all diameters of same circle have the same length, which is equal to two times the radius. In figure, AOB is a diameter of circle. To Prove : AOB = COD. A arc PQ is also denoted by PQ and the major arc PQ by QP . When P and Q are ends of a diameter, then both arcs are equal and each is called a semi circle. O D ` Arc : A piece of a circle between two points is called an arc. The longer one is called the major arc PQ and the shorter one is called the minor arc PQ. The minor B C Proof : In AOB and COD, OA = OC OB = OD [Radii of a circle] [Radii of a circle] AB = CD [Given] AOB COD [By SSS congruency] Hence Proved. AOB = COD. [By CPCT] R Major arc PQ P Semicircular region O Semicircular region Minor sector P B Q sectors become the same and each is known as a semicircular region. O Minor segment Converse : Q P Minor arc PQ Q If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal. PAGE # 4848 Theorem : The perpendicular from the centre of a circle to a chord bisects the chord. O A B M Ex.1The radius of a circle is 13 cm and the length of one of its chords is 10 cm. Find the distance of the chord from the centre. Sol. Let O be the center of the circle of radius 13 cm and AB is the chord of length 10 cm OC AB AB 10 = =5 2 2 Given : A circle with centre O. AB is a chord of this circle. AC = OM AB. To Prove : MA = MB. Construction : Join OA and OB. Proof : In right triangles OMA and OMB, OA = OB [Radii of a circle] OM = OM [Common] OMA = OMB [90º each] [Line perpendicular from centre to chord bisect the chord] In AOC (OC)2 + (AC)2 = (AO)2 (OC)2 = (13)2 – (5)2 OC = 12 cm. OMA OMB MA = MB [By RHS] [By cpctc] Hence Proved. Converse : The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Theorem : There is one and only one circle passing through three given non-collinear points. Proof : Take three points A, B and C, which are not in the same line, or in other words, they are not collinear [as in figure ]. Draw perpendicular bisectors of AB and BC say, PQ and RS respectively. Let these perpendicular bisectors intersect at one point O.(Note that PQ and RS will intersect because they are not parallel) [as in figure]. C P R O S A Q B O lies on the perpendicular bisector PQ of AB. OA = OB [ Every point on the perpendicular bisector of a line segment is equidistant from its end points] Similarly, O lies on the perpendicular bisector RS of BC. OB = OC [ Every point on the perpendicular bisector of a line segment is equidistant from its end points] So, OA = OB = OC i.e., the points A, B and C are at equal distances from the point O. So, if we draw a circle with centre O and radius OA it will also pass through B and C. This shows that there is a circle passing through the three points A, B and C. We know that two lines (perpendicular bisectors) can intersect at only one point, so we can draw only one circle with radius OA. In other words, there is a unique Hence Proved. circle passing through A, B and C. Ex.2 In figure, and O is the centre of the circle. Prove that OA is the perpendicular bisector of BC. Sol. Given : In figure, and O is the centre of the circle. To Prove : OA is the perpendicular bisector of BC. Construction : Join OB and OC. Proof : [Given] chord AB = chord AC. [ If two arcs of a circle are congruent, then their corresponding chords are equal] AOB = AOC ...(i) [ Equal chords of a circle subtend equal angles at the centre] In OBD and OCD, DOB = DOC [From (i)] OB = OC [Radii of the same circle] OD = OD [Common] OBD OCD [By SAS congruency] [By CPCT] ODB = ODC ...(ii) And, BD = CD ...(iii) [By CPCT] But BDC = 180º ODB + ODC = 180º ODB + ODB = 180º [From equation(ii)] 2ODB = 180º ODB = 90º ODB = ODC = 90º ...(iv) [From (ii)] So, by (iii) and (iv), OA is the perpendicular bisector of BC. Hence Proved. PAGE # 4949 Theorem : Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres). Angle Subtended by an Arc of a Circle : In figure, the angle subtended by the minor arc PQ at O C A O is POQ and the angle subtended by the major arc PQ N at O is reflex angle POQ. M D B O Given : A circle have two equal chords AB & CD. i.e. AB = CD and OM AB, ON CD. To Prove : OM = ON Construction : Join OB & OD Proof : AB = CD (Given) [ The perpendicular drawn from the centre of a circle to a chord bisects the chord] P Q Theorem : The angle subtended by an arc at the centre is double the angle subtended by it at any 1 1 AB = CD 2 2 BM = DN point on the remaining part of the circle. In OMB & OND OMB = OND = 90º OB = OD BM = DN of the circle. Given : An arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part OMB OND OM = ON REMARK : To Prove : POQ = 2PAQ. [Given] [Radii of same circle] [Proved above] Construction : Join AO and extend it to a point B. [By R.H.S. congruency] [By CPCT] Hence Proved. REMARK : Chords equidistant from the centre of a circle are equal in length. Ex.3 AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. Prove that EB = ED. Sol. Given : AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. To Prove : EB = ED. Construction : From O draw OP AB and OQ CD. Join OE. Proof : AB = CD [ Given] OP = OQ [ Equal chords of a circle are equidistant from the centre] A (A) (A) arc PQ is minor (B) arc PQ is a semi-circle (C) arc PQ is major. In all the cases, BOQ = OAQ + AQO the two interior opposite angles] In OAQ, OA = OQ E ...(ii) From (i) and (ii) BOQ = 2OAQ [By RHS congruency] [By CPCT] EB = ED. Hence Proved. [ AB = CD (Given)] ...(iii) Similarly, [Each 90º] [ Common] [ Proved above] OPE OQE PE QE 1 1 PE – AB = QE – CD 2 2 PE – PB = QE – QD [Radii of a circle] OQA = OAQ [Angles opposite equal sides of a triangle are equal] D Q Now in OPE and OQE, OPE = OQE OE = OE OP = OQ ...(i) [ An exterior angle of a triangle is equal to the sum of B C (C) Proof : There arises three cases:- P O (B) BOP = 2OAP ...(iv) Adding (iii) and (iv), we get BOP + BOQ = 2(OAP + OAQ) POQ = 2PAQ. ...(v) NOTE : For the case (C), where PQ is the major arc, (v) is replaced by reflex angles. Thus, reflex POQ = 2PAQ. PAGE # 5050 Theorem : Angles in the same segment of a circle are equal. Proof : Let P and Q be any two points on a circle to form a chord PQ, A and C any other points on the remaining part of the circle and O be the centre of the circle. Then, Ex.5. In the given figure, the chord ED is parallel to the diameter AC. Find CED. B 50º A O P 1 C 2 3 E POQ = 2PAQ ...(i) And POQ = 2PCQ ...(ii) [Angle subtended at the centre is double than the angle subtended by it on the remaining part of the circle] From equation (i) & (ii) 2PAQ = 2PCQ PAQ = PCQ. D Sol. CBE = 1 [Angles in the same segment] 1 = 50º …(i) [ CBE = 50º] AEC = 90º ..(ii) [Angle in a semicircle is a right angle] Hence Proved. Theorem : Angle in the semicircle is a right angle. Now, in AEC, 1 + AEC + 2 = 180º 50º + 90º + 2 = 180º Proof : PAQ is an angle in the segment, which is a semicircle. 1 1 PAQ = POQ = × 180º = 90º 2 2 [ POQ is straight line angle or POQ = 180º] If we take any other point C on the semicircle, then again we get Also, ED || AC 2=3 1 1 POQ = × 180º = 90º. 2 2 Hence Proved. Ex.4. A chord of a circle is equal to the radius of the circle, find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. Sol. 2 = 180º – 140º = 40º Thus 2 = 40º ...(iii) [Given] [Alternate angles] 3 = 40º Hence, CED = 40º. PCQ = A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle. C O B A D Let AB be a chord of the circle with centre at O such that OA = OB = AB. OAB is equilateral. AOB = 60º 1 ACB = AOB = 30º 2 Consider arc ACB. Clearly, it makes 360º – 60º = 300º at the centre O. Theorem : The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. Given : A cyclic quadrilateral ABCD. D C A B PAGE # 5151 To Prove : A + C = B + D = 180º. Construction : Join AC and BD. Proof : ACB = ADB ...(i) And BAC = BDC ...(ii) [Angles of same segment of a circle are equal] Adding equation (i) & (ii) ACB + BAC = ADB + BDC ACB + BAC = ADC. Adding ABC to both sides, we get ACB + BAC + ABC = ADC + ABC. ADC + ABC = 180º i.e., D + B = 180º A + C = 360º – (B + D) = 180º [ A B C D 360 º ] Hence Proved. Ex.6 If a side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. Sol. Let ABCD be a cyclic quadrilateral inscribed in a circle with centre O. The side AB of quadrilateral ABCD is produced to E. Then, we have to prove that CBE = ADC. Ex.8 If the nonparallel side of a trapezium are equal, prove that it is cyclic. Sol. Given : ABCD is a trapezium whose two non-parallel sides AD and BC are equal. To Prove : Trapezium ABCD is a cyclic. Construction : Draw BE || AD. Proof : AB || DE [Given] and AD || BE [By construction] Quadrilateral ABED is a parallelogram. [Opp. angles of a ||gm] BAD = BED ...(i) And , AD = BE ...(ii) [Opp. sides of a ||gm] But AD = BC ...(iii) [Given] From (ii) and (iii), BE = BC BCE = BEC ...(iv) [Angles opposite to equal sides] BEC + BED = 180º [Linear Pair Axiom] BCE + BAD = 180º [From (iv) and (i)] Trapezium ABCD is cyclic. [ If a pair of opposite angles of a quadrilateral is 180º, then the quadrilateral is cyclic] Hence Proved. Ex.9 In figure, O is the centre of the circle. Prove that x + y = z. Since the sum of opposite pairs of angles of a cyclic quadrilateral is 180° ABC + ADC = 180° But ABC + CBE = 180° [ ABC and CBE form a linear pair] ABC + ADC = ABC + CBE ADC = CBE or CBE = ADC. Hence proved 1 1 EOF = z 2 2 [ Angle subtended by an arc of a circle at the centre is Sol. EBF = twice the angle subtended by it at any point of the remaining part of the circle] Ex.7 Find the value of a & b. C D 130º A b 1 z ...(i) [Linear Pair Axiom] 2 1 1 EDF = EOF = z 2 2 [ Angle subtended by any arc of a circle at the centre ABF = 180º – a O B Sol. In ADC AD = DC DAC = DCA = x (say) 130º + x + x = 180º 2x = 50º x = 25º ADC + ABC = 180º [Opposite angles in a cyclic quadrilateral are supplementary] 130º + ABC = 180º ABC = 50º b = 90º [Angle in a semicircle] a = 180º – ABC – b = 180º – 50º – 90º a = 180º– 140º = 40º. So, a = 40º and b = 90º. is twice the angle subtended by it at any point of the remaining part of the circle] ADE = 180º – 1 z 2 ...(ii) [Linear Pair Axiom] BCD = ECF = y [Vertically Opp. Angles] BAD = x In quadrilateral ABCD ABC + BCD + CDA + BAD = 2 × 180º 180º – 1 1 z + y + 180º – z + x = 2 × 180º 2 2 x + y = z. Hence Proved. PAGE # 5252 Ex.10 AB is a diameter of the circle with centre O and chord CD is equal to radius OC. AC and BD produced meet at P. Prove that CPD = 60º. Sol. Given : AB is a diameter of the circle with centre O and chord CD is equal to radius OC. AC and BD produced meet at P. To Prove : CPD = 60º. O A B Construction : Join AD. Proof : In OCD, C OC = OD ...(i) [Radii of the same circle] OC = CD ...(ii) [Given] From (i) and (ii), OC = OD = CD OCD is an equilateral triangle. COD = 60º D 1 1 COD = (60º) = 30º 2 2 [ Angle subtended by any arc of a circle at the centre is twice the angle subtended by it at any point of the remaining part of the circle] PAD = 30º ...(iii) ...(iv) [Angle in a semi-circle] ADB + ADP = 180º [Linear Pair Axiom] 90º + ADP = 180º [From (iv)] ADP = 90º ...(v) In ADP, APD + PAD + ADP = 180º Lengths of two tangents drawn from an external point to a circle are equal. Given : AP and AQ are two tangents drawn from a point A to a circle C (O, r). To prove : AP = AQ Construction : Join OP, OQ and OA. Proof : In AOQ and APO OQA = OPA [Tangent at any point of a circle is perp. to radius through the point of contact] AO = AO [Common] OQ = OP [Radius] So, by R.H.S. criterion of congruency AOQ AOP AQ = AP [By CPCT] Hence Proved. RESULTS : APD + 30º + 90º = 180º [From (iii) and (v)] APD + 120º = 180º APD = 180º – 120º = 60º CPD = 60º. THEOREM P CAD = And,ADB = 90º Clearly OP = OR (Radius) Now, OQ = OR + RQ OQ > OR OQ > OP ( OP = OR) Thus, OP is shorter than any other segment joining O to any point of AB. Hence, OP AB. Hence Proved. THEOREM A tangent to a circle is perpendicular to the radius (i) If two tangents are drawn to a circle from an external point, then they subtend equal angles at the centre. QOA = POA [By CPCT] (ii) If two tangents are drawn to a circle from an external point, they are equally inclined to the segment, joining the centre to that point OAQ = OAP [By CPCT] Ex.11 In figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD. through the point of contact. Given : A circle C (O, r) and a tangent AB at a point P. To prove : OP AB Construction : Take any point Q, other than P on the tangent AB. Join OQ. Suppose OQ meets the circle at R. Proof : Among all line segments joining the point O to a point on AB, the shortest one is perpendicular to AB. So, to prove that OP AB, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB. Sol. PB = BQ, CR = CQ, DR = DS, AP= AS [ Length of tangents drawn from external point are equal] Add PB + CR + DR + AP = BQ + CQ + DS + AS (PB + AP) + (CR + DR) = (BQ + CQ) + (DS + AS) AB + CD = BC + AD AD = AB + CD – BC AD = 6 + 4 – 7 AD = 3 cm. PAGE # 5353 Ex.12 Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that PTQ = 2OPQ. Sol. We know that lengths of tangents drawn from an external point to a circle are equal. Proof : Case 1. When AB and CD intersect at P inside the circle. In ’s APC and BPD, APC = BPD [Vertically opposite angles] TPQ is an isosceles triangle. PAC = PDB [Angles in the same segment] APC ~ BPD TPQ = TQP Hence, TP = TQ PA PC = ` PD PB PA × PB = PC × PD. Case 2. When AB and CD when produced intersect at P outside the circle. In PBD and PAC, PBD = ACP [ABCD is a cyclic quadrilateral and exterior angle is In TPQ, we have TPQ + TQP + PTQ = 180º 2TPQ = 180º – PTQ 1 TPQ = 90º – PTQ 2 1 PTQ = 90º – TPQ 2 Since, equal to the interior opposite angle] PDB = CAP [ABCD is a cyclic quadrilateral and exterior angle is equal to the interior opposite angle] PBD ~ PCA PB PD Hence, = ` PC PA PA × PB = PC × PD. ....(i) OP TP OPT = 90º THEOREM OPQ + TPQ = 90º OPQ = 90º – TPQ ...(ii) From (i) and (ii), we get 1 PTQ = OPQ 2 PTQ = 2OPQ. SEGMENTS OF A CHORD Let AB be a chord of a circle, and let P be a point on AB If PAB is a secant to a circle intersecting the circle at A and B and PT is a tangent segment, then PA × PB = PT2. Given : A circle with centre O; PAB is a secant intersecting the circle at A and B and PT is a tangent segment to the circle. To prove : PA × PB = PT2. Construction : Join OA, OP and OT. Draw OD AB. T inside the circle. Then, P is said to divide AB internally into two segments PA and PB. O THEOREM P If two chords of a circle intersect inside or outside the circle when produced, the rectangle formed by two segments of one chord is equal in area to the rectangle formed by the two segments of another chord. D A P B C Fig. (i) Given : Two chords AB and CD of a circle C(O, r) intersecting at P, inside in fig. (i) and outside in fig.(ii). To prove : PA . PB = PC. PD Construction : Join AC and BD. A D B Proof : Since OD chord AB AD = DB. Now PA × PB = (PD – AD) (PD + DB) = (PD – AD)(PD + AD) [ AD = DB] = PD2 – AD2 = (OP2 – OD2) – AD2 [From right-angled ODP] = OP2 – (OD2 + AD2) = OP2 – OA2 [From right-angled ODA] = OP2 – OT2 = PT2 OA = OT [Radii] [From right-angled OTP] Hence, PA × PB = PT2. PAGE # 5454 BEA = BAT But BEA = ACB Ex.13 Two chords AB and CD of a circle intersect each other at O internally. If AB = 11 cm, CO =6 cm and DO = [Angles in the same segment] ACB = BAT i.e. BAT = ACB. This proves the first part. Again, since ADBC is a cyclic quadrilateral. ACB + ADB. = 180º Also BAT + BAP = 180º [Linear pair] ACB + ADB =BAT + BAP BAT + ADB = BAT + BAP [ ACB = BAT] ADB = BAP BAP = ADB Hence, BAT = ACB and BAP = ADB. 4 cm, find OA and OB. Sol. Let OA = x cm, then OB = (11– x) cm We have, AO × OB = CO × OD (x) × (11–x) = 6 × 4 = 24 11x – x2 = 24 x2–11x+24 = 0 x–3x–8 x= 3 or 8 so, if x = 3 then OA = 3cm, OB = 8cm but if x = 8 then OA = 8cm, OB = 3cm Ex.14 ABCD is a cyclic quadrilateral and PQ is a tangent to the circle at C. If BD is a diameter and DCQ = 40º and ABD = 60º ; find the measure of the following angles : (i) DBC (ii) BCP (iii) ADB. A ANGLES IN THE ALTERNATE SEGMENTS Let PAQ be a tangent to a circle at point A and AB be a chord. Then, the segment opposite to the angle formed by the chord of a circle with the tangent at a point is called the alternate segment for that angle. THEOREM A line touches a circle and from the point of contact a chord is drawn. Prove that the angles which the chord makes with the given line are equal respectively to angles formed in the corresponding alternate segments. Given : A circle with centre O and PAT is a tangent to the circle at A. A chord AB is drawn. Let C and D be the points on the circumference on opposite sides of AB. To prove : BAT = ACB and BAP = ADB. Construction : Draw the diameter AOE and Join EB. Proof : Since AE is the diameter, ABE = 90º [Angle in semi circle] BEA + EAB = 90º ...(i) [Remaining s of BEA] But since EA AT [ Radius is to the tangent] EAB + BAT = 90º ...(ii) Hence from (i) and (ii) BEA + EAB = EAB + BAT ...(iii) 60º B D Oº Sol. 40º P C Q Since BD is a diameter of the circle. BAD = 90º and also BCD = 90º (i)DBC = DCQ = 40º [s is the alternate segments] (ii) BCP + BCD + DCQ = 180º BCP + 90º + 40º = 180º BCP = 50º (iii) Similarly from BAD, 60º + BAD + ADB = 180º 60º + 90º + ADB = 180º ADB = 30º. Ex.15 In a right triangle ABC, the perpendicular BD on the hypotenuse AC is drawn. Prove that (i) AC × AD = AB2 (ii) AC × CD = BC2 Sol. We draw a circle with BC as diameter. Since BDC = 90º. The circle on BC as diameter will pass through D. Again BC is a diameter and AB BC. AB is a tangent to the circle at B. Since AB is a tangent and ADC is a secant to the circle. PAGE # 5555 AC × AD = AB2 This proves (i) Again AC × CD = AC × (AC – AD) = AC2 – AC × AD = AC2 – AB2 [Using (i)] 2 = BC [ABC is a right triangle] 2 Hence, AC × CD = BC . This proves (ii). Ex.16 Two circles touch externally at P and a common tangent touches them at A and B. Prove that (i) the common tangent at P bisects AB. (ii) AB subtends a right angle at P. Sol. Let PT be the common tangent at any point P. Since the tangent to a circle from an external point are equal, TA = TP, TB = TP TA = TB i.e. PT bisects AB at T TA = TP gives TAP = TPA (from PAT) (a) In fig. (i) d > r1 + r2 i.e. two circles do not intersect. In this case, four common tangents are possible. The tangent lines l and m are called direct common tangents and the tangent lines p and q are called indirect (transverse) common tangents. TB = TP gives TBP = TPB [from PBT] (b) In fig. (ii), d = r1 + r2. In this case, two circles touch TAP + TBP = TPA + TPB = APB externally and there are three common tangents. TAP + TBP + APB = 2 APB (c) In fig.(iii) d < r1 + r2. In this case two circles intersect 2 APB = 180º APB = 90º [sum of s of a = 180º] Hence proved in two distinct points and there are only two common tangents. (d) In fig. (iv), d = r1 – r2 (r1 > r2), in this case, two circles Definition : A line which touches the two given circles is touch internally and there is only one common tangent. called common tangent to the two circles. Let (e) In fig. (v), the circle C(O2, r2) lies wholly in the circle C(O1, r1), C(O2, r2) be two given circles. Let the distance C(O1, r1) and there is no common tangent. between centres O1 and O2 be d i.e., O1O2 = d. Ex.17 Two circles of radii R and r touch each other externally and PQ is the direct common tangent. Then show that PQ2 =4rR Sol. Draw O'S|| PQ, O'SPQ is rectangle O'S=PQ, PS=QO'=r OO'=R+r OS=OP–PS=R–r In O'OS (O'S)2 = (OO')2–(OS)2 PQ2=(R+r)2–(R–r)2 PQ2=R2+r2+2Rr–(R2+r2–2Rr) PQ2=R2+r2+2Rr–R2–r2+2Rr PQ2=4rR PAGE # 5656