ALWAYS WERE ANXIOUS ABOUT SOLVING CBSE CLASS X MATHS QUESTIONS - PLEASE GO THROUGH THESE SOLUTIONS AND ALSO THE RELEVANT NOTES FOR INDIVIDUAL CHAPTERS...

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CBSE CLASS X MATHEMATICS CHAPTER 1 REAL NUMBERS RECAP FROM PREVIOUS CLASS/ES 1. NUMBER LINE: Representation of fractions on a number line 2. Whole Numbers (W): 0, 1, 2, 3, ……….. 3. Natural Numbers (N): 1, 2, 3, ……….. 4. Integers (Z): …….. –3, -2, -1, 0, 1, 2, 3, ……. (Comes from German word, ‘Zahlen’, which means ‘To Count’. 5. Rational Numbers (r): Can be written in the form of p/q, where, q ≠ 0, As, anything divided by 0 is not defined. ( These Include N, W & Z). p & q have no common factors other than 1, ie. They are co-prime. Equivalent Rational Numbers or Fractions: Their Standard form is same. Eg. 1/2 = 2/4 = 10/20. Finding a Rational Number between Two Given Rational Numbers, a & b: (a + b)/2, is the required number; Proceeding in the same way, we may find more Rational Numbers between the given two Rational Numbers. There are infinitely many Rational Numbers between the given two Rational Numbers. The other way is to convert the given rational numbers into like fractions ( ie. With the same denominators). 6. Irrational Numbers (s) are the ones which can’t be written in the form of p/q, where p & q are integers, and, q≠0. Pythagoreans were the first ones to discover irrational numbers. 7. Real Numbers (R): The Rational and Irrational Numbers together form the Real Numbers. A Real Number is either a Rational or Irrational Number. Every Real Number is represented by a unique point on a number line. And, also, every point on a number line, represents a unique real number. (Shown by German Scientists Cantor & Dedekind.) Thus, the number line is also called ‘Real Number Line’. The decimal expansion of a rational number is either ‘terminating’ or ‘non - terminating recurring’ (which can be pure recurring or mixed recurring - Explained Later) ; & vice - versa. The decimal expansion of a irrational number is ‘non - terminating non recurring’; & vice - versa. Let a > 0, be a real number, and, n be a positive integer, then,

√n a = b, if bn= a. The Symbol √ is called the

‘Radical Sign’. 8. Laws of Indices: Let a and b be positive real numbers, then, (a, n, m are natural numbers; ‘a’ is called the base; ‘m’ & ‘n’ are the exponents) Now, a = 1; so, 1/a = a / a = a = a . 1/√2 = √2 / 2; which is half of √2; and Thence 1/√2 can be represented on a Number Line. When the denominator of an expression contains a term with a square root (or a number with a radical sign), the process of converting it to an equivalent expression whose denominator is a rational number is called ‘Rationalising the Denominator’. 0

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9. FEW RELEVANT POINTS RELATED TO THE CHAPTER: Note the Following points about the Real Numbers: 1. Sum or Difference of a rational and an irrational number is irrational [eg. (3 + √7) or (√7 - 3) is irrational.] 2. The Product or Quotient of a non - zero rational number with an irrational number is irrational [eg. The Expressions 2√2 OR 3÷√2 are irrational.] 3. If we add, subtract, multiply or divide two irrationals, the result may be rational or irrational [eg. √2 - √2 is 0, which is rational; but, √7 - √2 is irrational]. Square

root of all positive integers which are not perfect squares; cube roots of all integers which are not perfect cubes, and so on, are all irrational numbers. Pure Recurring Decimals are the ones, in which, all the digits after the decimal point are repeated; eg. 0.32323232… = 0.32. Mixed Recurring Decimals are the ones, in which, at least one digit after the decimal point is not repeated; eg. 18.33249494949… = 18.33249 Prime Numbers are the numbers, other than 1, whose only factors are1 and the number itself; Eg. 3, 13, 17… Composite Numbers are the numbers which have more than 2 common factors; Eg. 12, 15 etc. Co - Prime Numbers are those two Natural Numbers (Not necessary prime numbers), ehich have their Highest Common Factor as ‘One’. Eg. (3, 10); (15, 33) etc. 1 (One) is a Unique Number; ie. It is neither a Prime or a Composite Number. ———————————————————————————————————————CBSE CLASS X CHAPTER ONE: REAL NUMBERS 1. A non-zero integer ‘a’ is said to divide an integer ‘b’, if there exists an integer ‘c’ such that: b = ac Euclid’s Division Lemma: Let there be two positive integers a and b. Then, there exist unique integers q and r such that a = bq + r, 0 ≤ r < b. ‘Euclid’s Division Lemma’ is a restatement of the long division process, and the integers ‘q’ & ‘r’ are called the ‘Quotient’ and ‘Remeinder’. ((Find integers q and r for follg. pairs of positive integers a and b: (i) 10, 3 (ii) 4, 19 (iii) 81, 3)) The Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes (or the powers of primes) in a unique way (ie. The factorization is unique); apart from the order in which the prime factors occur. Composite Number = Product of Primes. This Theorum is used: (i) to prove the irrationality of many of the numbers, and, (ii) to find out when exactly the decimal expansion of a rational number is terminating, and when it is nonterminating, repeating. An algorithm is a series of well defined steps which gives a procedure for solving a type of problem. A lemma is a proven statement used for proving another statement. 2. Finding HCF & LCM by prime Factorisation Method: We can find HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic. This method is also called the ‘prime factorisation method’. HCF = Product of the smallest power of each common prime factor in the numbers. LCM = Product of the greatest power of each prime factor, involved in the numbers. For any two positive integers (a & b), HCF (a, b) X LCM (a, b) = a X b

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Let x

= p/q be a rational Number (where ‘p’ ‘q’ are co-prime): If ‘q’ = 2 5 , then p/q is a Terminating Decimal Expansion. If ‘q’ ≠ 2 5 , then p/q is a Non-Terminating Repeating Decimal Expansion. (‘n’& ‘m’ are non-negative integers) n

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3. Finding HCF of two Positive Integers using Euclid’s Division Lemma: Let ‘a’ & ‘b’ be two positive integers. Obtain two whole numbers, q & r , such that a = bq + r ; 0 ≤ r < b. If r = 0, b is the HCF of ‘a’ & ‘b’. If r ≠ 0, Apply Euclid’s Division Lemma to ‘b’ & ‘r ’, and Obtain two whole numbers, q & r , such that b = rq + r; Proceed as such, till the remainder becomes ’zero’. The ’divisor’ at this stage is the HCF of ‘a’ & ‘b’. 1

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4. Theorum: Let ‘p’ be a prime number. If ‘p’ divides a , then ‘p’ divides ‘a’, where ‘a’ is a positive integer. Proof: Let P , P , P …...P be factors of ‘a’. Thus, a = P .P .P …...P ; where, P , P , P …...P are primes, not necessarily all distinct. Thus, a = (P .P .P …...P ). (P .P .P …...P ); Thus, a = P .P .P …...P ‘P’ divides a (i) => ’P’ is a Prime factor of a , from The Fundamental Theorem of Arithmetic. Prime factors of a are only (P .P .P …...P ) (ii) (Uniqueness of factors from Fund. Th. Of Arith.) From (i) & (ii), we have ‘P’ is a prime factor of (P .P .P …...P ) => P divides a. 2

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HANDLING INDICES 1. Introduction: Why

Exponents were Invented: 10,000,000,000,000 = Ten trillion, or simply 10 ; 7x7x7x7x7x7x7x7 = 7 ; = 6 x 1,000,000 = 6 x 10 13

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presented as such: √2 = 2 [A ‘second root of 2’, or ‘Square root of 2’, or simply, under-root of 2]; 3√3 = 3 [A ‘Third root of 3’, or ‘Cube root of 3’]; n√x = x [ n root of x] 1/2

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2. Laws of Indices: a .a = a [3 . 3 = 3x3x3x3x3 = 3 = 3 ]; a /a = a [5 / 5 = (5x5x5x5x5x5) / (5x5x5x5) = 5x5 = 5 = 5 ]; (a ) = a [(7 ) = (7x7x7) = 7x7x7 x 7x7x7 = 7 = 7 ]; (ab) = a x b [(2x5) = 2x5 x 2x5 x 2x5 = 2x2x2x5x5x5 = 2 x 5 ]; (a/b) = a / b [ (3/5) = 3/5 x 3/5 = (3x3) / (5x5) = 3 / 5 ] [ 1/a = a / a = a = a ] m

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3. Applying Indices to Roots: √ab = √a √b [(ab) = a . b = √a . √b]; √a/b = √a / √b [(a/b) = a / b = √a / √b]; (√a - √b) x (√a + √b) = a - b [ (√a - √b) x (√a + √b) = (√a) - (√b) = (a ) - (b ) = a - b = a - b = a - b] 1/2

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CBSE CLASS X MATHEMATICS CHAPTER 1 REAL NUMBERS - NCERT EXERCISES SOLUTIONS EXERCISE 1.1 (P. 7) Q1. Use Euclid’s division algorithm to find the HCF of :(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255. A. (i) 225 = 135 X 1 + 90; 135 = 90 X 1 + 45; 90 = 45 X 2 + 0; HCF = 45 (ii) HCF = 196 (iii) 867 = 255 X 3 + 102; 255 = 102 X 2 + 51; 102 = 51 X 2 + 0; HCF (867, 255) = HCF (255, 102) = HCF (102, 51) = 51. Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. A. By Euclid’s division algorithm, a = bq + r (i), 0 ≤ r < b (The value of ‘r’ will always be less than the value of ‘b’; ie. Less than the value of divisor) (Also. The degree of ‘r’ is always less than the degree of the ‘divisor’) Putting ‘b’ = 6 in (i), we get a = 6q + r [ 0 ≤ r < 6; ie. r = 0, 1, 2, 3, 4, 5] If ‘r’ = 0, a = 6q, 6q is divisible by 6 => 6q is even. (Divisibility Criteria: If a number is divided by 2 & 3, it is divisible by 6!) If ‘r’ = 1, a = 6q + 1, 6q +1 is not divisible by 2. If ‘r’ = 2, a = 6q + 2, 6q + 2 is even and is divisible by 2. If ‘r’ = 3, a = 6q + 3, 6q + 3 is not divisible by 2. If ‘r’ = 4, a = 6q + 4, 6q + 4 is even and is divisible by 2. If ‘r’ = 5, a = 6q + 5, 6q + 5 is not divisible by 2. As 6q, 6q + 2, 6q + 4 are even; therefore, 6q + 1, 6q + 3, 6q + 5 are odd. Q3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? A. To find the maximum number of columns, we have to find the HCF of 616 and 32. By Euclid’s division Lemma, HCF of 616 & 32 is 8. Hence, maximum number of columns is 8. Q4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. A. By Euclid’s division algorithm, a = bq + r, 0 ≤ r < b (i); On putting, b = 3 in (i), we get a = 3q + r, 0 ≤ r < 3, ie. R = 0, 1, 2. If r = 0, a = 3q => a = 9q ; (ii) If r = 1, a = 3q + 1 => a = 9q + 6q + 1; (iii) If r = 2, a = 3q + 2 => a = 9q + 12q + 4; (iv) 2

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From (ii), 9q is a square of the form 3m, where m = 3q ; From (iii), 9q + 6q + 1, ie. 3(3q + 2q) + 1 is a square of the form 3m + 1, where m = 3q + 2q; From (iv), 9q + 12q + 4, ie. 3(3q + 4q + 1) + 1 is a square of the form 3m + 1, where m = 3q + 4q + 1. Thus, square of any positive integer is either of the form ‘3m’ or ‘3m + 1’. Q5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. A. Let ‘x’ be any positive integer. Then, it is of the form 3m, 3m + 1, 3m + 2. now, we have to prove that the cube of each of these can be rewritten in the form 9q, 9q + 1, 9q + 8. Now, (3m) = 27m = 9(3m ) = 9q, where q = 3m ; (3m + 1) = 27m + 27m + 9m + 1 = 9(3m + 3m + m) + 1 = 9q + 1, where ‘q’ = 3m + 3m + m; (3m + 2) = 27m + 54m + 36m + 8 = 9(3m + 3m + m) + 8 = 9q + 8, where 2

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EXERCISE 1.2 (P. 11) Q1. Express each number as a product of its prime factors:(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429. (Use Division Method) A. (i) 140 = 2 X 5 X 7 (ii) 156 = 2 X 3 X 13 (iii) 3825 = 3 x 5 x 17 (iv) 5005 = 5 X 7 X 11 X 13 (v) 7429 = 17 X 19 X 23 2

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Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 A. Do this solution yourself. Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25 A. (i) 12 = 2X2X3; 15 = 3X5; 21 = 3X7; HCF = 3; LCM = 2x2x3x5x7 = 420 (ii) The given three numbers don't have any common factor; They are primes. HCF = 1; LCM = 17X23X29 = 11339. (iii) Do this yourself! Q4. Given that HCF (306, 657) = 9, find LCM (306, 657). A. For any two positive integers a and b, HCF (a, b) × LCM (a, b) = The product of the numbers (a × b) Q5. Check whether 6 can end with the digit 0 for any natural number n. A. If the number 6 ends with the digit zero; then, it is divisible by 5. Therefore, the prime factorisation of 6 contains the prime number 5. This is not possible, as the only prime in the factorisation of 6 is 2 & 3; and, the uniqueness of the fundamental theorum of arithmetic guarantees that there are no other prime in the factorisation of 6 . So, there is no value of n in natural numbers for which 6 ends with the digit zero. n

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Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. A. (i) 7 × 11 × 13 + 13 = 1001 + 13 = 1014; and, 1014 = 2X3X13X13. Thus, 1014 is the product of prime factors. Hence, it is a composite number. (ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5040 + 5 = 5045 = 5 X 1009. Thus, it is a product of prime factors 5 and 1009. Hence, it is a composite number. Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? A. They will be again at the starting point after common multiples of 18 & 12. 18 = 2x3x3; 12 = 2x2x3; Thus, LCM of 18 & 12 = 36. In 36 mts. Ravi arrives at the starting point after making 3 rounds, as he takes 12 mts. to drive one round & Sonia also arrives at the starting point after making 2 rounds, as he takes 18 mts. to drive one round. Thus, they will meet again at the starting point after 36 minutes. EXERCISE 1.3 (P. 14) Q1. Prove that √5 is irrational. A. Let √5 is a rational number. Then, √5 can be written in the form p/q, where p, q are integers and have no common factor (other than 1), q ≠ 0. ie. √5 = p/q; Squaring both sides: 5 = p / q => p = 5q (i)=> 5 is a factor of p , & thus divides p . This implies that 5 divides p (ii). Now, p = 5m => p = 25m . Putting this in equation (i), we get 25m = 5q => 5m = q => 5 divides q => 5 divides q (iii). 2

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From (ii), 5 divides p, and from (iii) 5 divides q. This means 5 is a common factor of p and q. This contradicts the supposition that there is no common factor of p and q. Hence, √5 is an irrational number. Q2. Prove that 3 + 2√5 is irrational. A. Let 3 + 2√5 be a rational number. Now, let 3 + 2√5 = a/b, where a and b are co-prime & b ≠ 0. So, 2√5 = a/b - 3; Or, √5 = a/(2b) - 3/2. Since, a and b are integers, therefore a/(2b) - 3/2 is a rational number. And, thus √5 is a rational number. But √5 is an irrational number. Thus, our supposition is wrong. Hence, 3 + 2√5 is an irrational number. Q3. Prove that the following are irrationals : (i) 1/√2, (ii) 7√5, (iii) 6 + √2 (From Previous Class: Square root of all the numbers which are not perfect squares; and, Cube root of all the numbers which are not perfect cubes are Irrational Numbers.) A. (i) Let 1/√2 be a rational number. Now, let 1/√2 = a/b, where a and b are co-prime & b ≠ 0. Or, (1 X √2 )/(√2 X√2 ) = a / b; Or, √2 / 2 = a / b; Or, √2 = 2a/b. ((Always try to isolate the Irrational Component)) Since a and b are integers, 2a/b is rational and so √2 is rational. But, √2 is irrational. Thus, 1/√2 is Irrational. (ii) Let 7√5 be a rational number. Now, let 7√5= a/b, where a and b are co-prime & b ≠ 0. Or, √5= a / 7b; ((Always try to isolate the Irrational Component)) Since a and b are integers, a/7b is rational and so √5 is rational. But, √5 is irrational. Thus, 7√5is Irrational. (iii) Let 6 + √2 be a rational number. Now, let 6 + √2= a/b, where a and b are co-prime & b ≠ 0. Or, a/b - 6 = √2; ((Always try to isolate the Irrational Component)) Since a and b are integers, a/b - 6 is rational and so √2 is rational. But, √2 is irrational. Thus, 6 + √2 is Irrational. ====================================================================== EXERCISE 1.4 (P. 17) Q1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: i. 13 / 3125. 3125 = 2 x 5 . Thus, Denominator is of the form 2 X 5 . Thus, 13 / 3125 is a Terminating Decimal. ii. 17 / 8. 8 = 2 x 5 . Thus, Denominator is of the form 2 X 5 . Thus, 8 is a Terminating Decimal. iii. 64 / 455. 455 = 5 X 7 X 13. Thus, Denominator is not of the form 2 X 5 . Thus, 64 / 455 is a Non Terminating Repeating Decimal. iv. 15 / 1600. 1600 = 2 x 5 . Thus, Denominator is of the form 2 X 5 . Thus, 15 / 1600 is a Terminating Decimal. v. 29 / 343. Denominator is not of the form 2 X 5 . Thus, 64 / 455 is a Non - Terminating Repeating Decimal. (As 343 is a odd number, it is not divisible by 2, and hence its denominator’s factors can’t have powers of 2) vi. 23 / 2 5 . Denominator is of the form 2 X 5 . Thus, 23 / 2 5 is a Terminating Decimal. vii. 129 / 2 5 7 . Denominator is not of the form 2 X 5 . Thus, 129 / 2 5 7 is a Non - Terminating Repeating Decimal. viii. 6/15 = 2/5. 5 = 2 x 5 . Thus, Denominator is of the form 2 X 5 . Thus, 6 / 15 is a Terminating Decimal. ix. 35 / 50 = 7 / 10 = 7 / (2 x 5 ). Thus, Denominator is of the form 2 X 5 . Thus, 35 / 50 is a Terminating Decimal. x. 77 / 210. 210 = 7x3x2x5. Thus, Denominator is not of the form 2 X 5 . Thus, 77 / 210 is a Non Terminating Repeating Decimal. 0

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Q2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. A. (i) 13 / 3125 = 13 / (5x5x5x5x5) = (13x2x2x2x2x2) / (5x2x5x2x5x2x5x2x5x2) = 416/100000 = 0.00416. (ii) 17 / 8 = 17 / (2 x 5 ) = (17x5 ) / (2 x 5 ) = (17x125)/10 = 2125/1000 = 2.125. (iii) Non - Terminating Repeating. (iv) 15 / 1600 = 15 / (2 x 5 ) = 15 / (2 x2 x5 ) = (15x5 )/(2 x5 x2 x5 ) = (15x625)/10 = 9375/1000000 = 0.009375. (v) Non - Terminating Repeating. (vi) 23 / 2 5 = (23x5)/2x5x2 x5 ) = 115/10 = 0.115 (vii) Non - Terminating Repeating. (viii) 6/15 = 2/5 = (2x2)/(5x2) = 4/10 = 0.4. (ix) 35/50 = 35/(5x10) = (35x2)/(2x5x10) = 70/100 = 0.7 (x) Non - Terminating Repeating. 3

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CBSE CLASS X MATHEMATICS CHAPTER 2 POLYNOMIALS RECAP FROM PREVIOUS CLASS/ES 1. A Function p(x) of the form p(x) = a + a x + a x + a x +…….+ a x is called a polynomial, where (i) a a a a are real numbers, and, (ii) n is a non - negative (ie. 0 or positive) integer. ((Thus, (2/x + 3), (x + 2x + 4/x)… are not Polynomials. a a a a are called co - efficient of the polynomial. The expression of this form is called ‘Polynomial in one variable’. We can also have polynomials in more than one variable, eg. x + y + xyz is a polynomial in three variables. 2. Terms: The polynomial 3x + 4x + 5 has three terms. 3. Co - efficient: Each term of the polynomial has a co-efficient. In 3x + 4x + 5, the co-efficients of x , x and x are 3, 4, and 5. 4. Constant Polynomial: -3, 0, 2, 5….. are examples of Constant Polynomials. The Constant Polynomial 0 is called a ‘Zero Polynomial’. A Polynomial written either in the descending or ascending powers of ‘x’ is called the ‘Standard Form’ of a Polynomial. A Polynomial can be ‘Monomial’, Binomial’ and so on. 5. Degree of a Polynomial: The highest exponent in various terms of one variable is called its degree. Eg. A Polynomial can be ‘Linear’, ’Quadratic’, ’Cubic’, ’Biquadratic’, etc. If a = a = a = =a = 0, ie. If all the constants are zero, we get a ‘Zero Polynomial’, which is denoted by 0. ‘The Degree of a Zero Polynomial is not defined’. 6. Zeroes of a Polynomial: The value of a polynomial p(x) at x = a, is p(a), obtained on replacing ‘x’ by ‘a’. In general, a Zero of a Polynomial p(x) is ‘a’, such that p(a) = 0. Consider a polynomial p(x) = x - 1; Now, p(1) = 0, ie. 1 is a Zero (or root) of the Polynomial p(x), which gives x=1 Now, consider a constant polynomial 5. It has no zero because replacing x by any number in 5x , still gives 5. Thus, a non - zero constant polynomial has no zero. We will be dealing with Polynomials in one variable only. Zeroes of a Zero (0)Polynomial: By Convention, every Real Number is a zero of the Zero Polynomial. Number of Zeroes of a polynomial of degree ‘n’ is ‘n’. 7. Division of a Polynomial: Say, on dividing polynomial p(x) by another polynomial g(x), we get quotient q(x) & remainder r(x). Dividend = ( Divisor X Quotient) + Remainder; ie. P(x) = g(x) . Q(x) + r(x). The degree of the remainder r(x) is always less than the degree of the divisor g(x), ie. Degree of r(x) < Degree of g(x). If the divisor is linear (ie. One degree), then the degree of remainder will be 0, ie. The remainder will be a constant. 2x + x + x = x(2x + x + 1), We say that x and (2x + x + 1) are ‘Factors’ of (2x + x + x ), and (2x + x + x ) is a multiple of x as well as (2x + x + 1) 8. Remainder Theorum: Let p(x) be any polynomial of degree greater than or equal to one, and, let ‘a’ be any real number. If p(x) is divided by a linear polynomial (x-a), then the remainder is p(a). 0

0,

1,

2, …..

1

2

2

3

3

n

n

n

2

0,

1,

2, …..

n

2

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0

0

1

2

…..

n

0

3

2

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2

====================================================================== Remember The Following Points: A zero of a polynomial need not be zero. Every Linear Polynomial has one and only one zero. A polynomial can have more than one zero. Number of Zeroes is the degree of the polynomial. [Division of a polynomial by other polynomial using long division method has to be thoroughly practiced by the student.] Proof of Remainder Theorum: Let p(x) be any polynomial with degree ≥ 1. Let’s say, when p(x) is divided by (x-a), the quotient is q(x), and, the remainder is r(x); ie. P(x) = (x-a) . q(x) + r(x). Since, the degree of (x-a) is one, and the degree of r(x) is less than the degree of (x-a), the degree of Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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r(x) = 0. This means that r(x) is a constant, say ‘r’. Thus, for every value of x, r(x) = r. Therefore, p(x) = (x-a) . q(x) + r. In particular, if x=a, this equation gives us p(a) = (a-a) . q(a) + r = r; which proves the theorem. 9. Factor Theorum: If p(x) is a polynomial of degree n ≥ 1, and ‘a’ is any real number, then: (i) (x-a) is a factor of p(x), if p(a) = 0, and, (ii) p(a) = 0, if (x-a) is a factor of p(x). This actually follows from the Remainder Theorum. 10. Factorisation of Polynomials: A) Splitting the Middle Term: (ie. Quadratic Equations): To factorise ax + bx + c, we have to write ‘b’ as the sum of two numbers whose product is ‘ac’. Let p & q be those two numbers, ie. p+q = b, and, p.q = c. [If on multiplying a & c, we get + sign, we’ll get the middle term by adding; Sign will be that of the middle term.] 2

B) By Factor Theorum: Let f(x) be a given polynomial. 1. Proceed as follows: (Factorising equations of Degree 3) Write separately, the constant term of the polynomial. Write down its factors. If the number is big, and has many factors, write down the smallest, first few factors. Let ‘a’ be the zero of polynomial; ie. (x-a) is factor of polynomial f(x). Divide the given polynomial f(x) by (x-a), and get the Quotient. Factorise the Quotient by splitting the middle term. You would thus obtain the factors of the given polynomial. READ THE FOLLOWING EXAMPLE: Let f(x) = x - 3x - 9x - 5. The constant term is 5, so, its factors are ±5, ±1 Now, p(-1) = 0; Thus (x+1) is a factor of p(x). Now, do the long division. 3

2

= (x - 4x - 5) (Remainder is zero); 2

Now, by splitting the middle term, (x - 4x - 5) = (x+1) (x - 5). Thus, x - 3x - 9x - 5 = (x+1)(x+1) (x - 5) = (x+1) (x-5) In Regards to this method, remember the following important points: If the sum of coefficients of odd powers of any polynomial p(x) is the same as that the sum of its coefficients of even powers, then, (x+1) is always the factor of p(x). 2

3

2

2

If the sum of the coefficients of given expression p(x) is zero, then, (x-1) is always the factor of p(x).

C) Factorisation by Algebric Identities: Read the following identies. Understand them, prove them if you have to, and Remember them. (a + b) = a + 2ab + b (a - b) = a - 2ab + b a − b = (a + b) · (a − b) (x + a)(x + b) = x + (a + b)x + ab (a + b + c) = a + b + c + 2ab + 2ac + 2bc (a + b) = a + b + 3ab(a+b) (a - b) = a - b - 3ab(a-b) a + b = (a + b) · (a − ab + b ) a − b = (a − b) · (a + ab + b ) 2

2

2

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a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca); Thus, a3 + b3 + c3 = 3abc, if (a + b + c) = 0.

11. Least Common Multiple (LCM) of Polynomials = (Highest Power of Common Term) X Remaining Terms. ———————————————————————————————————————--

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CBSE CLASS X CHAPTER TWO: POLYNOMIALS 1. A Quadratic Polynomial can have either two distinct zeroes, two equal zeroes or no zeroes; ie. A polynomial of degree 2 has at the most two zeroes. Similarly, a Cubic Polynomial can have at the most 3 zeroes. Consider the following: If ‘k’ is a zero of p(x) = ax + b; then p(k) = ak + b = 0, ie. k = (-b) / a. Thus, the zero of the linear polynomial [ak + b] is [(-b) / a] = (-)Constant Term / Coefficient of x. Type of Polynomial

General Form

Linear

ax + b, a≠0

Quadratic

No. of Zeroes 1

ax + bx + c; a≠0 2

2

(Let α, β be two zeroes) Cubic

Polynomial

Relationship between Zeroes −b k= a = Constant term

Co-efficient of ‘x’ −b Sum of Zeroes (α + β)= a Product of Zeroes (α.β) =

ax + bx + cx + d 3

2

(Let α, β, γ be three zeroes)

3

c a

Sum of Zeroes (α + β+γ) −b = = -Co-efficient of x a

.

2

Co-efficient of x −d Product of Zeroes (α.β.γ) = a

=

3

−Constant term 3 Coefficient of x

Sum of the product of zeroes taken two at a time (For Cubic Polynomial) Coefficient of x c = (α.β + β.γ + γα) = Coefficient of x 3 = a 2. To form a Quadratic Polynomial with its given zeroes: Let α, β be the zeroes of a Quadratic Polynomial. ∴ x = α, => x - α = 0; x = β, => x - β = 0. Thus, (x - α).(x - β) is the Quadratic Polynomial, ie. x - (α + β)x + αβ; ie. x - (Sum of Zeroes) x + Product of Zeroes. 2

2

3. Division Algorithm for Polynomial: If p(x) & g(x) are any two polynomials with g(x) ≠0, then we can find polynomials q(x) & r(x) such that p(x) = g(x) X q(x) + r(x); where r(x) = 0, or degree of r(x) is less than the degree of g(x). Dividend = Divisor X Quotient + Remainder. 4. Geometric Meaning of the zeroes of the Polynomial. a. Graph of y = ax + b is a straight line which intersects the x-axis at (

−b a

, 0). This

x-Coordinate (of the point of intersection of the graph with x-axis) is the zero of the polynomial y = ax + b . b. Graph of the Quadratic Equation y = ax + bx + c, a ≠ 0, can have two distinct zeroes (graph cutting the x2

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axis at two distinct points), one zero (or two equal zeroes), or no zeroes (graph not cutting the x-axis at all (in which case, it would not be possible to factorise the quadratic polynomial). In fact, for any quadratic polynomial ax2 + bx + c, a ≠ 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards or open downwards depending on whether a > 0 or a < 0. (These curves are called parabolas.) c. A cubic polynomial of the form y = a can have at the most 3 zeroes. In general, a polynomial p(x) of degree n has at the most n zeroes. N.B.: Formation of a cubic polynomial: Let α, β, γ be three zeroes of the Polynomial. Then, the required cubic 3

polynomial is (x - α) (x - β)(x - γ). CBSE CLASS X MATHEMATICS CHAPTER 2 POLYNOMIALS - NCERT EXERCISES SOLUTIONS EXERCISE 2.1 (P. 28) Q1. The graphs of y = p(x) are given in the Fig. below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. A. Self Explanatory. ———————————————————————————————————————————EXERCISE 2.2 (P. 33) Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. Compare the quadratic equations given with the general quadratic equation, ie. ax + bx + c. A. (i) x – 2x – 8 = (x-4)(x+2). The zeroes are x = 4, -2 Sum of Zeroes = 4 - 2 = 2 = -(-2)/1 = - Coefficient of x = (-b)/a Coefficient of x Product of zeroes = -8 = Constant Term Coefficient of x (ii) 4s – 4s + 1 (iii) 6x – 3 – 7x (iv) 4u + 8u (v) t – 15 (vi) 3x – x – 4 2

2

2

2

2

2

2

2

2

Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. A. Let the polynomial be ax + bx + c, and its zeroes be α & β. (i) 1/4, -1; α + β = 1/4; α . Β = -1 The polynomial formed is x - (Sum of Zeroes) x + Product of Zeroes = x - 1/4 x + (-1) = x - x/4 - 1. The other possible polynomials would be k(x - x/4 - 1). If ‘k’ = 4, then the polynomial is 4x - x - 4. 2

2

2

2

2

2

(ii) √2, 1/3; α + β = √2; α . Β = 1/3 The polynomial formed is x - (Sum of Zeroes) x + Product of Zeroes = x - √2x + 1/3. The other possible polynomials would be k(x - √2x + 1/3). If ‘k’ = 3, then the polynomial is 3x - 3√2x + 1. 2

2

2

2

(iii) 0, √5; α + β = 0; α . Β = √5 The polynomial formed is x - (Sum of Zeroes) x + Product of Zeroes = x - 0.x + √5 = x + √5. 2

2

2

(iv) 1, 1; α + β = 1; α . Β = 1 The polynomial formed is x - (Sum of Zeroes) x + Product of Zeroes = x - 1.x + 1 = x - x + 1. 2

2

2

(v) -1/4, 1/4; α + β = -1/4; α . Β = 1/4 The polynomial formed is x - (Sum of Zeroes) x + Product of Zeroes = x - (-1/4)x + 1/4 = x + x/4 + 1/4. The other possible polynomials would be k(x + x/4 + 1/4). 2

2

2

2

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If ‘k’ = 4, then the polynomial is 4x + x + 1. 2

(vi) 4, 1; α + β = 4; α . Β = 1 The polynomial formed is x - (Sum of Zeroes) x + Product of Zeroes = x - 4x + 1. ———————————————————————————————————————-2

2

EXERCISE 2.3 (P. 36) Q1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following : (i) p(x) = x – 3x + 5x – 3, g(x) = x – 2 (x – 3x + 5x – 3) = (x – 2).(x-3) + (7x - 9). 3

3

2

2

2

2

(ii) p(x) = x - 3x + 4x + 5, g(x) = x – x + 1; (x + 0x - 3x + 4x + 5) = (x – x + 1).(x + x - 3) + (8). 4

2

2

4

3

2

2

2

(iii) p(x) = x + 0.x - 5x + 6, g(x) = 2 - x -x + 2 (x + 0.x - 5x + 6) = (-x + 2).(-x - 2) + (-5x + 10). 4

4

2

2

2

2

2

=

2

Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: Remainder is ‘0’, So,(t - 3) is a factor of (2t + 3t - 2t - 9t - 12). 2

4

3

2

(ii) x + 3x + 1; 3x + 5x - 7x + 2x + 2 Remainder is ‘0’, So,(x + 3x + 1) is a factor of (3x + 5x - 7x + 2x + 2). 2

4

3

2

2

4

3

2

(iii) x - 3x + 1; x - 4x + + x + 3x + 1 As Remainder is not ‘0’, So,(x - 3x + 1) is not a factor of (x - 4x + + x + 3x + 1). 3

5

4

2

3

5

4

2

Q3. Obtain all other zeroes of 3x + 6x – 2x – 10x – 5, if two of its zeroes are √(5/3), and - √(5/3) 4

A. Since the two zeroes are ,

3

2

therefore,

is a factor of p(x). Now, applying division algorithm to the given polynomial and 3x - 5. 2

Now, x + 2x + 1 = (x + 1) . Zeroes of (x + 1) are -1, -1. Hence, all its zeroes are √(5/3), (-)√(5/3), -1,-1. 2

2

2

Q4. On dividing x – 3x + x + 2 by a polynomial g(x), the quotient and remainder were (x – 2) and (–2x + 4), respectively. Find g(x). A. By Division Algorithm, p(x) = q(x).g(x) + r(x) 3

2

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Thus, (x – 3x + x + 2) = (x – 2).g(x) + (–2x + 4); Or, (x – 2).g(x) = (x – 3x + x + 2) - (–2x + 4) = (x – 3x + 3x - 2) Thus, g(x) = (x – 3x + 3x - 2) ÷ (x – 2). This gives g(x) as x - x + 1 3

2

3

3

2

3

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Q5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0 A. According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x) Degree of a polynomial is the highest power of the variable in the polynomial (i) deg p(x) = deg q(x) Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant) Let p(x) = 12x + 8x + 24; q(x) = 3x + 2x + 6; g(x) = 4; & r(x) =0. Degree of p(x) and q(x) is the same i.e., 2. Checking for division algorithm, p(x) = g(x) × q(x) + r(x); (12x + 8x + 24) = 4(3x + 2x + 6) + 0; Thus, the division algorithm is satisfied. 2

2

2

2

(ii) deg q(x) = deg r(x) (Here, we have to remember that the degree of r(x) is always less than the degree of g(x) ) Let us assume the division of x + x by x , Here, p(x) = x + x g(x) = x q(x) = x and r(x) = x Clearly, the degree of q(x) and r(x) is the same i.e., 1. Checking for division algorithm, p(x) = g(x) × q(x) + r(x) x +x=x ×x+x x +x=x +x Thus, the division algorithm is satisfied. 3

2

3

2

3

2

3

3

(iii) deg r(x) = 0 Degree of remainder will be 0 when remainder comes to a constant. Let us assume the division of x + 1by x . Here, p(x) = x + 1 g(x) = x q(x) = x and r(x) = 1 Clearly, the degree of r(x) is 0. Checking for division algorithm, p(x) = g(x) × q(x) + r(x) x + 1 = (x ) × x + 1 x +1=x +1 Thus, the division algorithm is satisfied. 3

2

3

2

3 3

2

3

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CBSE CLASS X MATHEMATICS CHAPTER 3 LINEAR EQUATIONS IN TWO VARIABLES RECAP FROM PREVIOUS CLASS/ES 1. An ‘Equation’ is a statement of equality of two algebraic expressions involving one or more unknown quantities called the variables. The ‘Equations’ involving only one variable are called equations in one variable, eg. 3x - 6=0; y - 8 =0. An ‘Equation’ involving only linear polynomial is called a linear equation; eg. 3x+7=0. The value of the variable, which when substituted for the variable (in the equation), makes both sides of the given equation equal, is called a ‘solution’, or ‘root’ of the equation; eg. x=3 is root of the equation 3x+10 = 19. 2. Properties of Equation: We can add, subtract, to both sides of the equation by the same number, We can perform multiplication or division to both sides of the equation by the same non-zero number, without changing the equality. 3. Linear equation in two variables: An Equation of the form ax + by + c = 0, where a, b, c are real numbers, and a ≠ 0, b ≠ 0, is called a linear equation in two variables, eg. x - y = 0. A linear equation in two variables has infinitely many solutions . Solutions to a ‘Linear Equation’ lie on a straight line. Often, the condition, ‘a’ & ‘b’ are not both zero is denoted by “ a + b ≠ 0” The reason that a degree one polynomial equation ax + by + c = 0 is called a linear equation is that its’ geometrical representation is a straight line. 4. Drawing graph of the equation ax + by + c = 0: Express the equation in the form y= - ax + c ; b Give integral values to x, and find corresponding values of ‘y’; Plot the points (a1,b1), (a2,b2), (a3,b3) so obtained on the graph paper; Join the points to get a line which represents the equation ‘ax + by + c = 0’. Each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa. 3

2

2

CBSE CLASS X CHAPTER THREE: LINEAR EQUATIONS IN TWO VARIABLES 1. The general form for a pair of linear equations in two variables x and y is a x + b y + c = 0and a x + b y + c = 0, where, a b , c , a , b , c are all real numbers, and “ a + b ≠ 0”; “ a + b ≠ 0” For any two given lines in a plane, only one of the following three possibilities can happen: The two lines will intersect at one point, The two lines will not intersect, i.e., they are parallel, & The two lines will be coincident. (( When solving the linear equation in two variable, when one of the variable is put to zero, the equation reduces to a linear equation in one variable, which is solved easily.)) A pair of linear equations in two variables is said to form a system of “Simultaneous Linear 1

1 ,

1

1

1

2

2

1

2

2

2

2

2

1

2

1

2

2

2

2

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Equations”. A pair of values of ‘x’ & ‘y’ satisfying each one of the equation in ‘x’ & ‘y’ is called a solution of the system. 2. Graphical Method of Solution of a Pair of Linear Equations: A pair of linear equations which has no solution, is called an inconsistent pair of linear equations. A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent. Lines representing a pair of linear equations in two variables: The lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations with unique solution), The lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations with infinite solutions], The lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations with no solution).

Summarizing the above facts: Again, consider the general form for a pair of linear equations in two variables x and y a x + b y + c = 0and a x + b y + c = 0, where, a b , c , a , b , c are all real numbers, and “ a + b ≠ 0”; “ a + b ≠ 0” (‘a’ & ‘b’ are not both zero) Intersecting lines (Consistent pair of equations with unique solution): a /a ≠ b /b Coincident lines (Consistent pair of equations with infinite solutions): a /a = b /b = c /c Parallel Lines (Inconsistent pair of equations with no solution): a /a = b /b ≠ c /c 1

1 ,

1

1

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1

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3. Algebraic solution of a system of linear equations: The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like ( √3, 2√7 ), (–1.05, 3.8) etc. Three algebraic methods available to solve a pair of linear Equations are: 3.1 Substitution Method: Consider the following steps to understand this method of solving a given pair of equations: Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient. Step 2 : Substitute this value of y in the other equation. This reduces it to an equation in one variable, i.e., in terms of x, which can now be solved. Sometimes, we may get statements with no variable. If the statement obtained is true (eg. 23 = 23), we can conclude that the pair of linear equations has infinitely many solutions. If the statement obtained is false, then the pair of linear equations is inconsistent (& has no solutions). Step 3 : Substitute the value of x obtained in Step 2 in any of the original equation or the one obtained in Step 1 to obtain the value of the other variable. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Remark : We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method. 3.2 Elimination Method: This is the method of eliminating (i.e., removing) one variable. The steps involved are: Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal. Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. If we get an equation in one variable, go to Step 3. If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions. If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent. Step 3: Solve the equation in one variable (x or y) so obtained to get its value. Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable. 3.3 Cross - Multiplication Method: Consider the two linear equations in their general form: a x + b y + c = 0 (i), and, a x + b y + c = 0 (ii). Follow the following steps: Step 1: Multiply equation (i) by b , & Equation (ii) by b . We get: b a x + b b y + b c = 0 (iii), and, b a x + b b y + b c = 0 (iv). Step 2: Subtracting (iv) from (iii), we get: x=(b c - b c )/(a b - a b ); where, a b - a b ≠ 0 (v) Step 3: Substituting this value of x in (i) or (ii), we get, y=(c a - c a )/(a b - a b ) (vi); Now, Here two cases may arise: Case 1. a b - a b ≠ 0. Thus, a /a ≠ b /b => The pair of linear equations has a unique solution. Case 2. a b - a b = 0. If we write a /a = b /b = k, then, a = ka ; b = kb . Putting the values of a & a in (i), k(a x + b y) + c = 0 (vii) Now, Equations (vii) & (ii) can both be satisfied only if c = kc => c /c = k. If c = kc , , any solution of Equation (ii) will satisfy the Equation (i), and vice versa. So, if, a /a = b /b = c /c = k, then there are infinitely many solutions to the pair of linear equations given by (i) & (ii). If c ≠ kc , then any solution of Equation (i) will not satisfy Equation (ii) and vice versa. Therefore the pair has no solution. Thus, we have: When a /a ≠ b /b ; we get a unique solution; When a /a = b /b = c /c ; we get infinitely many solutions; When a /a = b /b ≠ c /c ; we don't get any solutions (Inconsistent pair of equations). ==> 1

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The solutions given in equations (v) & (vi), may be written as: x = y = 1 (viii) bc - bc ca - ca ab -ab Use the following diagram to memorise the above result.: x y 1 b c a b 1

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b c a b Thus, we follow the following steps, in solving a pair of linear equations by the cross-multiplication method: Step 1 : Write the given equations in the form (i) and (ii), Step 2 : Taking the help of the diagram above, write Equations as given in (viii), Step 3 : Find x and y, provided a b - a b ≠ 0 2

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CBSE CLASS X MATHEMATICS CHAPTER 3 LINEAR EQUATIONS IN TWO VARIABLES - NCERT EXERCISES SOLUTIONS

EXERCISE 3.1 (P. 44) Q1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. A. Let the present age of Aftab be x, and, present age of his daughter = y Seven years ago: Age of Aftab = x − 7, and, Age of his daughter = y − 7 Three years hence, Age of Aftab = x + 3 Age of his daughter = y + 3 Q2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. A. Let the cost of a bat be Rs x, and of ball be Rs y According to the question, Preparing solution table for the above two equations: y = (3900 - 3x)/6; and, y = (1300 - x)/2 Q3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. A. Let the cost of 1 kg of apples be Rs x, & cost of 1 kg of grapes = Rs y. Acc. to the question, Preparing solution table for the above two equations: y = 160 - 2x; and, y = (300 - 4x)/2, we get the above graph.

EXERCISE 3.2 (P. 49) Q1. Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. A. Let the number of girls be x and the number of boys be y. According to the question, x + y = 10 (ie. y = 10 - x); and x − y = 4 (ie. y = x - 4). Drawing the solution table for the two equations, we get the required graph: (ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen. A. Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y. According to the question, 5x + 7y = 50 (ie. y = (50 - 5x)/7); and, 7x + 5y = 46 (ie. y = (46 - 7x)/5) Drawing the solution table for the two equations, we get the required graph: Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Q2. On comparing the ratios a /a , b /b & c /c , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0 Comparing the equations with general form, we have Since, a /a ≠ b /b ; Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point. (ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0 Comparing the equations with general form, we have Since, a /a = b /b = c /c ; Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations. (iii) 6x – 3y + 10 = 0 2x – y + 9 = 0 Comparing the equations with general form, we have Since, a /a = b /b ≠ c /c ; Hence, the given pair of Equations are inconsistent (Parallel Lines) with no solution. 1

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Q3. On comparing the ratios a /a , b /b & c /c , find out whether the following pair of linear equations are consistent, or inconsistent. (i) 3x + 2y = 5; 2x − 3y = 7 Comparing the equations with general form, we have Since, a /a ≠ b /b ; Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point. 1

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(ii) 2x – 3y = 8; 4x – 6y = 9 Comparing the equations with general form, we have Since, a /a = b /b ≠ c /c ; Hence, the given pair of Equations are inconsistent (Parallel Lines) with no solution. 1

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(iii) 3/2 x + 5/3 y = 7; ; 9x – 10y = 14 Comparing the equations with general form, we have Since, a /a ≠ b /b ; Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point. Hence, the pair of linear equations is consistent. 1

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(iv) 5x – 3y = 11; – 10x + 6y = –22 Comparing the equations with general form, we have Since, a /a = b /b = c /c ; Hence, the lines representing the given pair of equations are Coincident lines (ie. Consistent pair of equations with infinite solutions). 1

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(v) 4/3 x + 2y = 8; 2x + 3y = 12 Comparing the equations with general form, we have Since, a /a = b /b = c /c ; Hence, the lines representing the given pair of equations are Coincident lines (ie. Consistent pair of equations with infinite solutions). 1

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Q4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: A. (i) x + y = 5; 2x + 2y = 10 Comparing the equations with general form, Since, a /a = b /b = c /c ; Hence, the lines representing the given pair of equations are Coincident lines (ie. Consistent pair of equations with infinite solutions). x + y = 5 => y = 5 - x; 2x + 2y = 10 => y = (10 - 2x)/2 Drawing the solution table for the two equations, we get the required graph: It can be seen that the two lines are overlapping each other. (ii) x – y = 8, 3x – 3y = 16 Comparing the equations with general form, we have 1

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Since, a /a = b /b ≠ c /c ; Hence, the given pair of Equations are inconsistent (Parallel Lines) with no solution. 1

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(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 Comparing the equations with general form, we have Since, a /a ≠ b /b ; Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point (ie. They are consistent). 2x + y – 6 = 0, ie. y = - 2x + 6; 4x – 2y – 4 = 0, ie. y = (4x - 4)/2 Drawing the solution table for the two equations, we get the required graph: 1

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(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0 Since, a /a = b /b ≠ c /c ; Hence, the given pair of Equations are inconsistent (Parallel Lines) with no solution. (To demonstrate by drawing a graph, that the two lines are indeed parallel.) 1

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Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. A. Let the length of the garden be x, & the width be y. According to the question, x=y+4 (1) y + x = 36 (2) Drawing the solution table for the two equations, we will get the required graph. The point where the two lines will intersect, will give us the required dimensions of the garden. Length = x = 20m; Width = y = 16m. Q6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines (ii) parallel lines (iii) coincident lines A. (i) For intersecting lines: 6x + 12y - 8 = 0. (ii) For parallel lines: 4x + 6y − 8 = 0. (iii) For coincident lines: 6x + 9y − 24 = 0. Q7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. A. x – y + 1 = 0, ie. y = x + 1; and, 3x + 2y – 12 = 0, ie. y = -3x + 12. Drawing the solution table for the two equations, we will get the required graph: From the figure, the three lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0). ———————————————————————————————————————-EXERCISE 3.3 (P. 53) Q1. Solve the following pair of linear equations by the substitution method. (i) x + y = 14 (1); x - y = 4 (2). From (1), we obtain y = 14 - x (3); Putting this value in eq. (2), x - (14 - x) = 4; ie. x = 9. Putting the value of ‘x’ in eq. (3), we get y = 5. (ii) s – t = 3 (1); s/3 + t/2 = 6 (2). From (1), we obtain s = t + 3 (3); Putting this value in eq. (2), (t + 3)/3 + t/2 = 6; ie. t = 6. Putting the value of ‘t’ in eq. (3), we get s = 9. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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(iii) 3x – y = 3 (1); 9x – 3y = 9 (2). From (1), we obtain y = 3x − 3 (3 ); Putting this value in eq. (2), 9x - 3(3x − 3) = 9; ie. 9 = 9. This is always true. Hence, the given pair of equations has infinite possible solutions One of its possible solutions is x = 0, y = -3. (iv) 0.2x + 0.3y = 1.3 (1); 0.4x + 0.5y = 2.3 (2). From (1), we obtain y = (1.3 - 0.2x)/0.3 (3); Putting this value in eq. (2), 0.4x + 0.5(1.3 - 0.2x)/0.3 = 2.3; ie. x = 2. Putting the value of ‘x’ in eq. (3), we get y = 3. (v) √2 x + √3 y = 0 (1); √3 x - √8 y = 0 (2). From (1), we obtain y = (-√2 x)/√3 (3); Putting this value in eq. (2), √3 x - √8 ((-√2 x)/√3) = 0; ie. x = 0. Putting the value of ‘x’ in eq. (3), we get y = 0. (vi) 3x/2 - 5y/3 = -2 (1); x/3 + y/2 = 13/6 (2). DO IT ON YOUR OWN!! x = 2, y = 3. Q2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3. A. 2x + 3y = 11 (1); 2x - 4y = -24 (2). Solving for ‘x’ & ‘y’, using substitution method, we get x = -2; y = 5. y = mx + 3; ie. 5 = m (-2) + 3; or, 2 = -2m; or, m = -1. Q3. Form the pair of linear equations for the following problems and find their solution by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. Find them. A. Let the first number be x and the other number be y such that y > x. According to given info., y = 3x (1); y - x = 26 (2) Substituting the value of y from equation (1) into equation (2), 3x - x = 26 => x = 13 (3) Putting this value in eq. (1), y = 3 x 13 = 39. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. A. Let the larger angle be x and smaller angle be y. As per question. x - y = 18 (1), and as the two are supplementary angles, x + y = 180 (2) From (1) y = x - 18 (3). Put this value in eq. (2), x + x - 18 = 180 ; Thus, x = 99 . Putting the value of ‘x’ in eq. (3), we get y = 99 - 18 = 81 . 0

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(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball. A. Let the cost of a bat and a ball be x and y. According to the given info. 7x + 6y = 3800 (1); and, 3x + 5y = 1750 (2) From (1), y = (3800 - 7x)/6 (3); Putting this value in eq. (2) => 3x + 5 [(3800 - 7x)/6] = 1750, which gives x = 500 (4). Putting this value in eq. (3), we get y = 50. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? A. Let the fixed charge be Rs x and per km charge be Rs y. As per the question, x + 10y = 105 - (i); and, x + 15y = 155 - (ii) From (i), we get, x = 105 - 10y - (iii). Putting this in eq. (ii), we get, 105 - 10y + 15y = 155, Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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which gives, y = 10 - (iv) Putting this in eq. (iii), we get, x = 5. Hence, fixed charge = Rs 5, and, Per km charge = Rs 10. Charge for 25 km = x + 25y = Rs 255 (v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. A. Let the fraction be x/y. As per the given information, (x+2)/(y+2) = 9/11 => 11x - 9y = (-)4 - (i); (x+3)/(y+3) = 5/6 => 6x - 5y = (-)3 - (ii) From (i), x = (-4 + 9y)/11 - (iii). Substituting this in equation (ii), we get, y = 9 - (iv) Substituting y = 9 in equation (iii), we get, x = 7. Thus the fraction is 7/9 (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages? A. Let the age of Jacob be x and the age of his son be y. According to the given information, (x + 5) = 3(y + 5) => x - 3y = 10 - (i) (x - 5) = 7(y - 5) => x - 7y = -30 - (ii) From, (i), x = 3y + 10 - (iii). Substituting this in equation (ii), we get, y = 10 - (iv) Substituting y = 10 in equation (iii), we get, x = 40. Hence, the present age of Jacob is 40 years, and the present age of his son is 10 years. ———————————————————————————————————————-EXERCISE 3.4 (P. 56) Q1. Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2x – 3y = 4 By Elimination Method: x + y = 5 (i); 2x – 3y = 4 (ii) Multiplying (i) by 2, we get, 2x + 2y = 10 (iii) Subtracting (ii) from (iii), we get, y = 6/5 (iv). Putting this in eq. (i), we get, x = 19/5 By Substitution Method: x + y = 5 (i); 2x – 3y = 4 (ii) x + y = 5 (i), gives, x = 5 - y (iii). Putting this in (ii), we get, y = 6/5. Putting y = 6/5 in (iii), we get, x = 19/5 (ii) 3x + 4y = 10 and 2x – 2y = 2; (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7; and, (iv) x/2 + 2y/3 = -1 and x - y/3 = 3 These three pair of equations can be solved following the method described above. Q2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction? A. Let the fraction be x/y. According to given information, (x + 1) / (y - 1) = 1=> x - y = -2 (i) x / (y + 1) = 1/2 => 2x - y = 1 (ii) Subtracting (i) from (ii), we get x = 3 (iii). Putting this in (i), we get y = 5. Thus, the fraction is 3/5 (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? A. Do it yourself! Age of Nuri = 50 years; and, Age of Sonu = 20 years. (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. Let the unit digit and tens digits of the number be x and y respectively. Then, number (written as ‘yx’) = 10y + x Number after reversing the digits = 10x + y According to the given information, x + y = 9 (i); 9(10y + x) = 2(10x + y) => 88y − 11x = 0 => − x + 8y =0 (ii) Adding equation (i) and (ii), we get 9y = 9, or, y = 1 (iii) Substituting the value in equation (1), we get, x = 8 Hence, the number is 10y + x = 10 × 1 + 8 = 18. (iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received. A. Let the number of Rs 50 and Rs 100 notes be x and y. According to the given information, x + y = 25 (i); 50x + 50y = 2000 (ii) Multiplying equation (i) by 50, we get, 50x + 50y = 1250 (iii) Subtracting equation (iii) from equation (ii), we get, 50y = 750 => y = 15. Substituting y = 15, in equation (i), we get, x = 10 Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. A. Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y . According to the given information, x + 4y = 27 (i); x + 2y = 21 (ii) Subtracting equation (ii) from equation (i), we get, 2y = 6 => y = 3 (iii) Substituting in equation (i), we get, x + 12 = 27 => x = 15 Hence, fixed charge = Rs 15, and, Charge per day = Rs 3 ———————————————————————————————————————-EXERCISE 3.5 (P. 62) Q1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method. (i) x – 3y – 3 = 0; 3x – 9y – 2 = 0 a / a = 1/3; b / b = 1/3; c / c = 3/2. Therefore, a / a = b / b ≠ c / c Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations. 1

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(ii) 2x + y = 5; 3x + 2y = 8 a / a = 2/3; b / b = 1/2; c / c = 5/8. Therefore, a / a ≠ b / b . Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. By cross-multiplication method, x = y = 1____ bc - bc ca - ca ab - ab This gives, x/2 = y/1 = 1. Therefore, x = 2, and, y = 1 1

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(iii) 3x – 5y = 20; 6x – 10y = 40 a / a = 1/2; b / b = 1/2; c / c = 1/2. Therefore, a / a = b / b = c / c Therefore, the given sets of lines are coincident to each other and thus, there are infinite solutions possible for these equations. 1

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(iv) x – 3y – 7 = 0; 3x – 3y – 15 = 0 a / a = 1/3; b / b = 1; c / c = 7/15. Therefore, a / a ≠ b / b Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. By cross-multiplication, x = y = 1____ 1

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bc - bc ca - ca ab - ab Thus, x = y = 1____ 45 - 21 -21 - (-15) -3 - (-9) Or, x / 24 = y / (-6) = 1/6; This gives, x = 4 & y = -1 1

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Q2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2 A. a / a = 2/(a - b); b / b = 3/(a + b); c / c = 7/(3a +b - 2). For infinitely many solutions, a / a = b / b = c / c 2/(a - b) = 7/(3a +b - 2), which gives, a - 9b = -4 (i) 2/(a - b) = 3/(a + b), which gives, a - 5b = 0 (ii) Solving equations (i), and (ii), we get, a = 5, and, b = 1; for which the given equations will have infinitely many solutions. 1

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Q2. (ii) For which values of k will the following pair of linear equations have no solutions? 3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1 A. a / a = 3/(2k - 1); b / b = 1/(k - 1); c / c = 1/(2k + 1). For no solution, a / a = b / b ≠ c / c ; which gives k = 2. Hence, for k = 2, the given equation will have no solution. 1

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Q3. Solve the following pair of linear equations by the substitution and cross-multiplication methods : 8x + 5y = 9; 3x + 2y = 4 A. Do It Yourself! Q4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day. A. Let x be the fixed charge of the food and y be the charge for food per day. According to the given information, x + 20y = 1000 (i); x + 26y = 1180 (ii) Subtracting equation (i) from equation (ii), we get, 6y = 180 => y = 30 Substituting this value in equation (i), we get, x = 400 Hence, fixed charge = Rs 400; and charge per day = Rs 30 Q4.(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction. A. Let the fraction be x/y. According to the given info., (x - 1) / y = 1/3 => 3x - y = 3 (i); x / (y + 8) = 1/4 => 4x - y = 8 (ii) Subtracting equation (i) from equation (ii), we get, x = 5. Putting this (i), we get, y = 12 Hence, the fraction is 5/12. Q4.(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test? A. Let the number of right answers and wrong answers be x and y. According to the given information, 3x - y = 40 (i); 4x - 2y = 50 => 2x - y = 25 (ii) Subtracting equation (ii) from equation (i), we get, x = 15 (iii). Putting this in (ii), y = 5 Therefore, number of right answers = 15; and, number of wrong answers = 5. Total number of questions = 20 Q4.(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? A. Let the speed of 1st car and 2nd car be u km/h and v km/h. Relative speed of both cars while they are travelling in same direction = (u - v) km/h. (u > v) Relative speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (u + v) km/h According to the given information, 5(u - v) = 100 => u - v = 20 (i); and, 1(u + v) = 100 (ii) Adding both the equations, we get, u = 60 km/h (iii) Substituting this value in equation (2), we get, v = 40 km/h Hence, speed of one car = 60 km/h and speed of other car = 40 km/h. Q4.(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle. A. (v) Let length and breadth of rectangle be x unit and y unit. Area = xy. Acc. to the question, (x - 5)(y + 3) = xy - 9 => 3x - 5y - 6 = 0 (i); (x + 3)(y + 2) = xy + 67 => 2x + 3y - 61 = 0 (ii) By cross-multiplication method, we get, x = y = 1____ ; bc - bc ca - ca ab - ab Or, x = y = 1___; 305 - (-18) -12 - (-183) 9 - (-10) Or, x / 323 = y / 171 = 1 / 19 => x = 17, and, y = 9 Hence, the length and breadth of the rectangle are 17 units and 9 units respectively. 1

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EXERCISE 3.6 (P. 67) Q1. Solve the following pairs of equations by reducing them to a pair of linear equations: Q1. (i) 1 + 1 = 2; 1 + 1 = 13/6; 2x 3y 3x 2y Let 1/x = p, & 1/y = q; The equations thus become, p/2 + q/3 = 2 => 3p + 2q - 12 = 0 (i) p/3 + q/2 = 13/6 => 2p + 3q - 13 = 0 (ii) Using cross multiplication method, we get, p = 2, and, q = 3; which gives, x = 1/2, and, y = 1/3. Q1. (ii) 2/(√x) + 3/(√y) = 2; /(√x) - 9/(√y) = -1 Putting 1/(√x) = p, and, 1/(√y) = q. The equations thus become, 2p + 3q = 2 (i); 4p - 9q = -1 (ii) Solving, we get, p (=1/(√x)) = 1/2; q (=1/(√y)) = 1/3 Solving for ‘x’, and ‘y’, we get, x = 4; y = 9 Q1. (iii) 4/x + 3y = 14; 3/x - 4y = 23 Putting 1/x = p, the equations become, 4p + 3y = 14 (i); 3p - 4y = 23 (ii) Solving, we get, p (=1/x) = 5, => x = 1/5; y = -2 Q1. (iv) 5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 2) = 1 Putting 1/(x - 1) = p, and, 1/(y - 2) = q. The equations thus become, 5p + q = 2 (i); 6p - 3q = 1 (ii) Solving, we get, p (=1/(x - 1) ) = 1/3; which gives, x = 4 Also, we get, q (=1/(y - 2)) = 1/3; which gives, y = 5 Therefore, x = 4, and, y = 5 Q1. (v) (7x - 2y)/xy = 5 => 7/y - 2/x = 5 (i); (8x + 7y)/xy = 15 => 8/y + 7/x = 15 (ii) Putting 1/x = p, and, 1/y = q. The equations thus become, Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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-2p + 7q = 5 (iii); 7p + 8q = 15 (iv) Solving, we get, p (=1/x) = 1; which gives, x = 1; Also, we get, q (=1/y) = 1; which gives, y = 1 Therefore, x = 1, and, y = 1 Q1. (vi) 6x + 3y = 6xy => 6/y + 3/x = 6 (i); 2x + 4y = 5xy => 2/y + 4/x = 5 (ii) Putting 1/x = p, and, 1/y = q, the equations thus become, 3p + 6q - 6 = 0 (iii); 4p + 2q - 5 = 0 (iv) Solving, using cross multiplication method, we get, p (=1/x) = 0 => x = 1; q (=1/y) = 1/2 => y = 2 Q1. (vii) 10/(x + y) + 2/(x - y) = 4 (i); 15/(x + y) - 5/(x - y) = -2 (ii) Putting 1/(x+y) = p, and, 1/(x-y) = q, the equations become, 10p + 2q = 4(iii); 15p - 5q = -2 (iv) Solving, using cross multiplication method, we get, p (=1/(x+y)) = 1/5 => x + y = 5 (v); and, q (= 1/(x - y) = 1 => x - y = 1 (vi) Solving equations (v) and (vi), we get, x = 3, and y = 2 Q1. (viii) 1/(3x + y) + 1/(3x - y) = 3/4 (i); 1/[2(3x + y)] - 1/[2(3x - y) = -1/8 (ii) Putting 1/(3x+y) = p, and, 1/(3x-y) = q, the equations become, p + q = 3/4(iii); p/2 - q/2 = -1/8 => p - q = -1/4 (iv) Solving, we get, p [= 1/(3x+y) ] = 1/4 => 3x + y = 4 (v), and, q (= 1/(3x - y) = 1/2 => 3x - y = 2 (vi) Solving equations (v) and (vi), we get, x = 1, and y = 1 Q2. Formulate the following problems as a pair of equations, and hence find their solutions: (i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. A. Let the speed of Ritu in still water be ‘x’ km/h, and the speed of stream be ‘y’ km/h. Speed of Ritu while rowing: Upstream = (x - y)km/h; Downstream = (x + y)km/h According to question: 2(x + y) = 20 => x + y = 10 (i); 2(x - y) = 4 => x - y = 2 (ii) solving the equations, we get, x = 6, and, y = 4. Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h. (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. A. Let the number of days taken by a woman and a man, to complete the work be x and y. Therefore, work done by a woman in 1 day = 1/x, and, Work done by a man in 1 day = 1/y According to the question: 2/x + 5/y = 1/4 (i); and, 3/x + 6/y = 1/3 (ii) Putting 1/x = p, and 1/y = q in these equations, we get, 2p + 5q = 1/4 => 8p + 20q = 1 (iii), and, 3p + 6q = 1/3 => 9p + 18q = 1 (iv) Solving equations (iii) & (iv), by cross-multiplication, we get, p (= 1/x) = 1/18 => x = 18; and, q (= 1/y) = 1/36 => y = 36 Hence, number of days taken by a woman to finish the work = 18, and, Number of days taken by a man to finish the work= 36 (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately. A. Let the speed of train and bus be u km/h and v km/h. According to the given information, 60/u + 240/v = 4 (i); [4Hrs10Mts = 25/6Hrs]; 100/u + 200/v = 25/6 (ii). Putting 1/u = p, and 1/v = q, in these equatns, we get,60p + 240q = 4 (iii), 600p + 1200q = 25 (iv) Solving equations (iii) and (iv), we get, p = 1/60; q = 1/80 Hence, Speed of train = u = 60km/h, and, Speed of bus = 80km/h.

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CBSE CLASS X MATHEMATICS CHAPTER 4 QUADRATIC EQUATIONS 1. Introduction: Quadratic polynomial is of the form ax2 + bx + c, a ≠ 0. When we equate this polynomial to zero, we get a quadratic equation. 2. Quadratic Equations: A quadratic equation in the variable x is an equation of the form ax + bx + c = 0, where a, b, c are real numbers, a ≠ 0. For example, 2x + x – 300 = 0 is a quadratic equation. In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation. 2

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Q. Check whether the following are quadratic equations: (i) (x – 2) + 1 = 2x – 3; (ii) x(x + 1) + 8 = (x + 2) (x – 2); (iii) x (2x + 3) = x + 1; (iv) (x + 2) = x – 4 2

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3. Solution of a Quadratic Equation by Factorisation: A real number ‘α’ is called a root of the quadratic equation ax + bx + c = 0, a ≠ 0 if aα + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that ‘α’ satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax + bx + c and the roots of the quadratic equation ax + bx + c = 0 are the same. Any quadratic equation can have at-most two roots. We can find the roots of a quadratic equation by factorising it into two linear factors (by splitting the middle term) and equating each factor to zero. Ex. Find the roots of the quadratic equation 6x – x – 2 = 0. A. 6x – x – 2 = (3x – 2)(2x + 1) The roots of 6x – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0. Therefore, 3x – 2 = 0 or 2x + 1 = 0, ie. x = 2/3; or, x = -1/2 We verify the roots, by checking that 2/3 & -1/2 satisfy 6x – x – 2 = 0 2

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4. Solution of a Quadratic Equation by Completing the Square Herein, the term containing x is brought completely inside a square, and thence, the roots are found easily by taking the square roots. We can convert any quadratic equation to the form (x + a) – b = 0 and then we can easily find its roots. The process is: x + 4x = (x + 4/2) - 4 ie. Make the term containing x , a perfect square (by dividing or multiplying appropriate number); Half the co-efficient of x, and bring it in the term containing perfect square. Few Examples: (i) Solve the equation 3x - 5x + 2 = 0. Dividing both sides by 3; x - 5/3x + 2/3 = 0 => (x - 5/6) + 2/3 - 25/36 = 0 Solving we get, (x - 5/6) = (1/6) => x = 1, or x = 2/3; which are the roots of the given equation. (ii) Solve the equation 2x - 5x + 3 = 0 => x - 5/2 x + 3/2 = 0 => (x - 5/4) + 3/2 - 25/16 = 0; 2

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Or, (x - 5/4) = 1/16 => x - 5/4 = ± 1/4 => x = 3/2 or x = 1 2

(iii) Find the roots of 4x + 3x + 5 = 0 by the method of completing the square. A. 4x + 3x + 5 = 0 => x + 3/4x + 5/4 = 0 => (x + 3/8) - 9/64 + 5/4 = 0 => (x + 3/8) = (-71)/64 < 0. But, (x + 3/8) can’t be negative for any real value of x (as –ve x –ve = +ve). So, there is no real value of ‘x’ satisfying the given equation. Therefore, the given equation has no real roots. 2

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Generalising the above method of ‘completing the squares’: Consider the quadratic equation ax + bx + c = 0 (a ≠ 0). Dividing throughout by ‘a’, we get x + b/a x + c/a = 0=> (x + b/2a) - (b/2a) + c/a =0=> (x + b ) - (b - 4ac) => 2a 4a => (x + b ) = b - 4ac (i) 2a 4a If b – 4ac ≥ 0, then by taking the square roots in (i), we get (x + b/2a) = ± (√(b - 4ac) / 2a Thus, x = -b ± (√(b - 4ac) [which are the two roots of quadratic equation ax + bx + c = 0 (a ≠ 0) 2a if √(b - 4ac) ≥ 0. This is also known as the Quadratic Formula.] Solve the following equations by quadratic formula (find the roots if they exist): (i) x + 2x – 143 = 0 [ x = 11, or x = -13 ] (ii) x - 3x - 4 = 0 [ x = 4, or x = -1 ] (iii) x + 4x + 5 = 0 [ b – 4ac < 0 ] (iv) 2x – 2√2 x + 1 = 0 [ x = 1/√2, or, x = 1/√2 ] 2

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5. Nature of Roots: The roots of the equation ax + bx + c = 0 are given by x = - b ± √(b - 4ac). 2a If ‘b - 4ac’ > 0, we get two distinct real roots: x = - b + √(b - 4ac), and, x = - b - √(b - 4ac). 2a 2a If ‘b - 4ac’ = 0, then x = -b/2a ± 0, ie. x = -b/2a or –b/2a. So, the roots of the equation ax + bx + c = 0 are both –b/2a. Thus, the quadratic equation ax + bx + c = 0 has two equal real roots in this case. If b – 4ac < 0, then there is no real number whose square is b2 – 4ac. Therefore, there are no real roots for the given quadratic equation in this case. Since b – 4ac determines whether the quadratic equation ax + bx + c = 0 has real roots or not, b – 4ac is called the discriminant of this quadratic equation. 2

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So, a quadratic equation ax + bx + c = 0 has (i) two distinct real roots, if b – 4ac > 0, (ii) two equal real roots, if b – 4ac = 0, (iii) no real roots, if b – 4ac < 0. Eg. Find the ‘Discriminant’ of the equation 3x - 2x + 1/3 = 0, and hence find the nature of its roots. Find 2

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them if the are real. Ans. a = 3, b = -2, c = 1/3. Therefore, ‘Discriminant’ b – 4ac = 0 Hence, the given quadratic equation has two equal real roots. The roots are –b/2a, -b/2a; ie. 1/3, 1/3 2

CBSE CLASS X MATHEMATICS CHAPTER 4 QUADRATIC EQUATIONS- NCERT EXERCISES SOLUTIONS

EXERCISE 4.1 (P. 73): Q1. Check whether the following are quadratic equations :( i) (x + 1) = 2(x – 3) => x + 7 = 0. It is of the form ax + bx + c = 0; and hence is a quadratic equation. (ii) x – 2x = (–2) (3 – x) => x – 4x + 6 = 0. It is of the form ax + bx + c = 0; and hence is a quadratic equation. (iii) (x – 2)(x + 1) = (x – 1)(x + 3) => 3x - 1 = 0 It is not of the form ax + bx + c = 0; and hence is not a quadratic equation. (iv) (x – 3)(2x +1) = x(x + 5) => x – 10x - 3 = 0. It is of the form ax + bx + c = 0; and hence is a quadratic equation. (v) (2x – 1)(x – 3) = (x + 5)(x – 1) => x – 11x + 8 = 0. It is of the form ax + bx + c = 0; and hence is a quadratic equation. (vi) x + 3x + 1 = (x – 2) => 7x - 3 = 0. It is not of the form ax + bx + c = 0; and hence is not a quadratic equation. (vii) (x + 2) = 2x (x – 1) => x - 6x - 14x - 8 = 0. It is not of the form ax + bx + c = 0; and hence is not a quadratic equation. (viii) x – 4x – x + 1 = (x – 2) => 2x - 13x + 9 = 0. It is of the form ax + bx + c = 0; and hence is a quadratic equation. 2

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Q2. Represent the following situations in the form of quadratic equations : (Assume what is to be found out in the given question) (i) The area of a rectangular plot is 528 m . The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot. A. Let the breadth of the plot be x m => length of the plot is (2x + 1) m. Area of a rectangle = Length × Breadth; ∴ 528 = x (2x + 1) => 2x + x - 528 = 0 2

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(ii) The product of two consecutive positive integers is 306. We need to find the integers. A. Let the consecutive integers be x and x + 1. It is given that their product is 306. ∴ x(x + 1) = 306 => x + x - 306 = 0. 2

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age. A. Let Rohan’s age be x. Hence, his mother’s age = x + 26. Three years hence: Rohan’s age = x + 3; Mother’s age = x + 26 + 3 = x + 29 It is given that the product of their ages after 3 years is 360. ∴ (x + 3)(x + 29) = 360 => x + 32x - 273 = 0. 2

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train. A. Let the speed of train be x km/h. Time taken to travel 480 km = 480/x hrs. If speed = (x - 8) km/h; time taken to cover 480 kms is (480/x + 3) hrs. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Now, distance = speed X time => 480 = (x - 8) X (480/x + 3) => x - 8x + 1280 = 0. 2

EXERCISE 4.2 (P. 76): Q1. Find the roots of the following quadratic equations by factorisation: (ie. By factorizing the middle term) (i) x – 3x – 10 = 0; or, (x - 5)(x + 2) = 0. Roots of this equation are the values for which (x - 5)(x + 2) = 0 => x - 5 = 0; or x + 2 = 0. Thus, i.e., x = 5 or x = −2 2

(ii) 2x + x – 6 = 0; or, (x + 2)(2x - 3) = 0. Roots of this equation are the values for which (x + 2)(2x - 3) = 0; => (x + 2) = 0; or, (2x - 3) = 0 i.e., x = −2 or x = 3/2 2

(iii) √2 x + 7x + 5√2 = 0; or, (√2x + 5)(x + √2) = 0. Roots of this equation are the values for which (√2x + 5)(x + √2) = 0; => (√2x + 5)= 0; or, (x + √2) = 0; i.e., x = −5/√2 or x = -√2 2

(iv) 2x - x + 1/8 = 0; or, 1/8(16x - 8x + 1) = 1/8 (4x - 1) = 0 Roots of this equation are the values for which (4x - 1) = 0; => (4x - 1)= 0; or, (4x - 1) = 0; i.e., x = 1/4, or x = 1/4 2

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(v) 100x – 20x + 1 = 0; or, (10x - 1) = 0. Roots of this equation are the values for which (10x - 1) = 0; => (10x - 1)= 0; or, (10x - 1) = 0; i.e., x = 1/10, or x = 1/10 2

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Q2. Solve the problems given in Example 1. Example 1: (i) Required representation of the problem: x - 45x + 324 = 0. (Solve by splitting the middle term) (ii) Required representation of the problem: x - 55x + 750 = 0. (Solve by splitting the middle term) 2

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Q3. Find two numbers whose sum is 27 and product is 182. A. Let the first number be x; then, the second number is 27 − x. As per question, x(27 − x) = 182 => x - 27x + 182 = 0 => (x - 13)(x - 14) = 0. Thus, either (x - 13) = 0; or, (x - 14) = 0; ie. x = 13, or x = 14. If x = 13, the other number is 14, and vice versa. Thus, the numbers are 13 and 14. 2

Q4. Find two consecutive positive integers, sum of whose squares is 365. A. Let the consecutive positive integers be x and x + 1. Thus, x + (x + 1) = 365 => (x + 14)(x - 13) = 0; Thus, x = -14; or, x = 13. Since the integers are positive, x = 13. Thus, the two consecutive positive integers are 13 and 14. 2

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Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. A. Let the base of the right triangle be x cm. Then, its’ altitude = (x − 7) cm. By Pythagoras Theorum, x + (x - 7) = 13 => (x - 12)(x + 5) = 0 Thus, x = 12, or, x = -5. As sides can’t be negative, x = 12. Thus, the base of the triangle is 12cm, and the altitude is 5cm. 2

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Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. Let the number of articles produced be x. Therefore, cost of production of each article = Rs (2x + 3) It is given that the total production is Rs 90. Thus, x(2x + 3) = 90 => (2x + 15)(x - 6) = 0 Thus, x = -15/2, or, x = 6. As the number of articles produced can only be a positive integer, therefore, x can only be 6. Hence, number of articles produced = 6, and, Cost of each article = 2 × 6 + 3 = Rs 15

EXERCISE 4.3 (P. 87): Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2x – 7x + 3 = 0 => x - 7/2x + 3/2 = 0 => (x - 7/4) - 49/16 + 3/2 = 0 => (x - 7/4) = 49/16 - 3/2 => (x - 7/4) = ± 5/4 => x = 3, or x = 1/2 (ii) 2x + x – 4 = 0 => x + x/2 - 2 = 0 => (x + 1/4) - 1/16 - 2 = 0 => (x + 1/4) = 1/16 + 2 => (x + 1/4) = 33/16. Solving, x = (√33 - 1)/4 or ( - √33 - 1)/4 (iii) 4x + 4√3 x + 3 = 0; (x = -√3/2; x = -√3/2) (iv) 2x + x + 4 = 0; (No real roots, as the square of a number cannot be negative) 2

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Q2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula. A. Compare the equation with ax + bx + c = 0. Get the value of ‘a’, ‘b’, and ‘c’. Find the value of √(b - 4ac), and proceed. Part (iv) is being solved: (iv) 2x + x + 4 = 0. Using quadratic formula, x - b ± √(b - 4ac), we get, 2a x -1 ± √(-31). But, the square of a number can’t be negative. Thus, there are no real roots for the 4 given equation. 2

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Q3. Find the roots of the following equations: (i) x - 1/x = 3, x ≠ 0. Use quadratic formula, x = (3 ± √13) / 2 Form an equation, factorise it by splitting the middle term. We get x = 1 or x = 2 Q4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age. A. Let ‘x’ be Rehman’s present age. As per question, 1/(x - 3) + 1/(x + 5) = 1/3. Forming a quadratic equation, and factorizing it by splitting the middle term, we get, x = 7, or x = -3. But, as age can’t be negative, x = 7. Q5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects. A. Let the marks in Maths be x. Then, the marks in English will be 30 − x. Acc. to the question, (x + 2)(30 - x - 3) = 210. Factorizing we get, (x - 12)(x - 13) = 0 => x = 12, or x = 13. If the marks in Maths are 12, then marks in English will be 30 − 12 = 18 If the marks in Maths are 13, then marks in English will be 30 − 13 = 17 ==> Q6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field. A. Let the shorter side of the rectangle be x m. Then, larger side of the rectangle = (x + 30) m. Diagonal is given to be 60 more than the shorter side.

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However, side cannot be negative. Therefore, the length of the shorter side will be 90 m. Hence, length of the larger side will be (90 + 30) m = 120 m.

Q7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers. A. Let the larger and smaller number be x and y respectively.

According to the given question, However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will also be negative which is not possible. Therefore, the larger number is 18. Thus, y = 144 => y = ± 12. Thus, the numbers are 18 & 12; or, 18 & -12. 2

Q8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. A. Let the speed of the train be x km/hr. Time taken to cover 360 km = 360/x hrs According to the given question,

However, speed cannot be negative. Therefore, the speed of train is 40 km/h. Q9. Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. A. Let the time taken by the smaller pipe to fill the tank be x hr. Time taken by the larger pipe = (x − 10) hr Part of tank filled by smaller pipe in 1 hour = 1/x Part of tank filled by larger pipe in 1 hour = 1 / (x - 10) It is given that the tank can be filled in 9 3/8 = 75/8hours by both the pipes together. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Therefore, Part of tank filled by both the pipes together in 1 hour = 8/75 Thus, 1/x + 1/(x - 10) = 8/75 => (x - 25)(8x - 30) = 0 => x = 25, or, x = 30/8 Time taken by the smaller pipe cannot be 30/8 = 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible. Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 − 10 =15 hours respectively. Q10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains. A. Let the average speed of passenger train be x km/h. Average speed of express train = (x + 11) km/h It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

Now, Speed cannot be negative. Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h. Q11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. A. Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively and their areas will be x and y respectively. It is given that, 4x − 4y = 24 => x − y = 6 => x = y + 6 2

2

Thus, y = -18; y = 12

However, side of a square cannot be negative. Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m ---------------------------------------------------------------------------------------------------------------------------------EXERCISE 4.4 (P. 91): Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: A. For a quadratic equation ax + bx + c = 0, discriminant is b − 4ac. (A) If b − 4ac > 0 → two distinct real roots (B) If b − 4ac = 0 → two equal real roots (C) If b − 4ac < 0 → no real roots (i) 2x −3x + 5 = 0 Comparing this equation with ax + bx + c = 0, we obtain, a = 2, b = −3, c = 5 2

2

2

2 2

2

2

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Discriminant = b − 4ac = (− 3) − 4 (2) (5) = 9 − 40 = −31 As b2 − 4ac < 0. Therefore, no real root is possible for the given equation. 2

2

(ii) 3x - 4√3 x + 4 = 0 Comparing this equation with ax + bx + c = 0, we obtain, a = 3, b = - 4√3, c = 4 Discriminant = b − 4ac = (- 4√3) − 4 (3) (4) = 48 − 48 = 0 As b2 − 4ac = 0. Therefore, real roots exist for the given equation and they are equal to each other. The roots are: -b/2a, and –b/2a. [ x = -b/2a = -(- 4√3)/(2x3) = 2/√3]; that is 2/√3 & 2/√3 2

2

2

2

(iii) 2x - 6x + 3 = 0 Comparing this equation with ax + bx + c = 0, we obtain, a = 2, b = −6, c = 3 Discriminant = b − 4ac = (− 6) − 4 (2) (3) = 36 − 24 = 12 As b − 4ac > 0, therefore, distinct real roots exist for this equation as follows: Therefore, the roots are, (3 + √3)/2, or (3 - √3)/2 2

2

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Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots. A. If an equation ax + bx + c = 0 has two equal roots, its discriminant (b − 4ac) will be 0. (i) 2x + kx + 3 = 0 Comparing this equation with ax + bx + c = 0, we obtain, a = 2, b = k, c = 3 Discriminant = b − 4ac =(k) − 4(2) (3) = k − 24. For equal roots, Discriminant = 0 => k − 24 = 0 => k = 24 => k = ± √24 = ± 2√6 2

2

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(ii) kx (x – 2) + 6 = 0 => kx - 2kx + 6 + 0 Comparing this equation with ax + bx + c = 0, we obtain, a = k, b = −2k, c = 6 Discriminant = b − 4ac = (− 2k) − 4 (k) (6) = = 4k − 24k For equal roots, b − 4ac = 0 => 4k − 24k = 0 => 4k(k − 6) = 0 => Either 4k = 0 or k = 6 = 0 => k = 0 or k = 6 However, if k = 0, then the equation will not have the terms ‘x ’ and ‘x’. Therefore, if this equation has two equal roots, k should be 6 only. 2

2

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Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800m2? If so, find its length and breadth. A. Let the breadth of mango grove be x. Thus, length of mango grove = 2x. Area of mango grove = (2x)(x) = 2x . Given, 2x = 800 => x - 400 = 0. Comparing this with ax + bx + c = 0, we obtain, a = 1, b = 0, c = 400 Discriminant = b − 4ac = (0) − 4 × (1) × (− 400) = 1600 Here, b − 4ac > 0 => The equation will have real roots. And hence, the desired rectangular mango grove can be designed. x = ± 20. However, length cannot be negative. Therefore, breadth of mango grove = 20 m Length of mango grove = 2 × 20 = 40 m 2

2

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Q4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. A. Let the age of one friend be x years => Age of the other friend = (20 − x) years. Four years ago, age of 1st friend = (x − 4) yrs; &, age of 2nd friend = (20 − x − 4) = = (16 − x) yrs Given that, (x − 4) (16 − x) = 48 => 16x - x - 64 + 4x = 48 => − x + 20x − 112 = 0 => x − 20x + 112 = 0. Comparing this equation with ax + bx + c = 0, we obtain, a = 1, b = −20, c = 112 Discriminant = b − 4ac = (− 20) − 4 (1) (112) = 400 − 448 = −48 As b − 4ac < 0, no real root is possible for this equation and hence, this situation is not possible. Q5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth. 2

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A. Let the length and breadth of the park be x and y. Perimeter = 2(x + y) = 80 => x + y= 40 => y = 40 - x Area = x × y = x(40 − x) => Area = 40x − x As per the question, 40x − x = 400 => x - 40x + 400 = 0 Comparing this equation with ax + bx + c = 0, we obtain, a = 1, b = −40, c = 400 Discriminate = b − 4ac = (− 40)2 −4 (1) (400) = 1600 − 1600 = 0 As b − 4ac = 0, therefore, this equation has equal real roots. And hence, this situation is possible. Root of this equation are: x = -b/2a = - (-40)/2(1) = 20. Therefore, length of park, x = 20 m, and, breadth of park, y = 40 − x = 40 − 20 = 20 2

2

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2

CBSE CLASS X MATHEMATICS CHAPTER 5 ARITHMETIC PROGRESSIONS 1. Introduction: Some of the patterns observed in everyday life are such that, the succeeding terms are obtained by adding a fixed number, or by multiplying with a fixed number, or we find that they are squares of consecutive numbers, and so on. Here-in, we shall learn about a pattern in which succeeding terms are obtained by adding a fixed number to the preceding terms. And, also, to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems. 2. Arithmetic Progressions: Consider the following lists of numbers : 1, 2, 3, 4, . . . 100, 70, 40, 10, . . . Each of the numbers in the list is called a term. In each of the above two lists, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an Arithmetic Progression (AP). So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP, which can be positive, negative or zero. Let us denote the first term of an AP by a , second term by a , . . ., nth term by a and the common difference by ‘d’. Then the AP becomes a , a , a , . . ., a . So, a – a = a – a = . . . = a – a = d. a, a + d, a + 2d, a + 3d, . . . represents an arithmetic progression where ‘a’ is the first term and ‘d’ the common difference. This is called the general form of an AP. A finite AP has finite number of terms (The last term is known). An infinite Arithmetic Progression does not have a last term. 1

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3. n term of an A.P.: Let a , a , a , . . . be an AP whose first term a is ‘a’ and the common difference is ‘d’. Then, the second term a = a + d = a + (2 – 1) d the third term a = a + d = (a + d) + d = a + 2d = a + (3 – 1) d the fourth term a = a + d = (a + 2d) + d = a + 3d = a + (4 – 1) d ……. ……. Looking at the pattern, we arrive at the nth term, a = a + (n – 1) d. So, the nth term an of the AP with first term ‘a’ and common difference ‘d’ is given by a = a + (n – 1) d. a is also called the general term of the AP. If there are m terms in the AP, then a represents the last term which is sometimes also denoted by l. An Example: Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62. A. Here, a = 10, d = 7 – 10 = – 3, l = – 62, where l = a + (n – 1) d. Putting the given values in the given in the above formula, we get n = 25. If we write the given AP in the reverse order, then a = – 62 and d = 3. th

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So, the question now becomes finding the 11th term with these a and d. So, a = – 62 + (11 – 1) × 3 = – 62 + 30 = – 32, which is the required term. 11

4. Sum of First n Terms of an AP: Let the given A.P. be a, a+ d, a + 2d, …... The nth term of this AP is a + (n – 1) d. Let S denote the sum of the first n terms of the AP. We have: S = a + (a + d ) + (a + 2d ) + . . . + [a + (n – 1) d ] (1) Rewriting the terms in reverse order, we have S = [a + (n – 1) d ] + [a + (n – 2) d ] + . . . + (a + d ) + a (2) On adding (1) and (2), term-wise. we get 2S = [2a + (n - 1)d] + [2a + (n - 1)d] + ……. + [2a + (n - 1)d] ( The term [2a + (n - 1)d] comes ‘n’ times.) Thus, 2S = n[2a + (n - 1)d] Or, S = The sum of first n terms of an AP = n [2a + (n - 1)d] = n[a + a + (n - 1)d] = n[a + a ] (3) 2 2 2 n

Now, if there are only n terms in an AP, then a = ‘l’, the last term. From (3), we get: S = n/2 (a + l) n

This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given. We also use S in place of S to denote the sum of first n terms of the AP. We write S to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth. Remark: The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., a = S – S . n

n

n

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n–1

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CBSE CLASS X MATHEMATICS CHAPTER 5 ARITHMATIC PROGRESSIONS- NCERT EXERCISES SOLUTIONS

EXERCISE 5.1 (P. 99): Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km. A. Taxi fare for 1st km = 15 Taxi fare for first 2 km = 15 + 8 = 23 Taxi fare for first 3 km = 23 + 8 = 31 Taxi fare for first 4 km = 31 + 8 = 39 Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. A. Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 - 1/4 = 3/4 th part of air will remain. Therefore, volumes will be V, (3V/4), (3V/4) , (3V/4) …. 2

3

Thus, adjacent terms of this series don’t have same difference between them. Thus, this isn’t an A.P.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre. A. Cost of digging for first metre = 150 Cost of digging for first 2 metres = 150 + 50 = 200 Cost of digging for first 3 metres = 200 + 50 = 250 Cost of digging for first 4 metres = 250 + 50 = 300 Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term. (iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum. r n P 1+ A. Rs P deposited at r% compound interest per annum for n years, becomes after n years. 100

(

)

Therefore, after every year, our money will be

Thus, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P. Q2. Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) a = 10, d = 10 A. Let the series be a , a , a , a , a … Therefore, a = a = 10; a = a + d = 10 + 10 = 20;a = a + d = 20 + 10 = 30; 1

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a = a + d = 30 + 10 = 40; a = a + d = 40 + 10 = 50 Therefore, the series will be 10, 20, 30, 40, 50 … First four terms of this A.P. will be 10, 20, 30, and 40. (ii) a = -2, d = 0 A. Let the series be a , a , a , a … a = a = −2 a = a + d = − 2 + 0 = −2 a = a + d = − 2 + 0 = −2 a = a + d = − 2 + 0 = −2 Therefore, the series will be −2, −2, −2, −2 …; which is an A.P. , as the difference between the two consecutive terms is constant, ie. 0 First four terms of this A.P. will be −2, −2, −2 and −2. 4

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(iii) a = 4, d = -3 A. Let the series be a , a , a , a … a =a=4 a =a +d=4-3=1 a = a + d = 1 - 3 = −2 a = a + d = −2 - 3 = −5 Therefore, the series will be 4, 1, -2, -5 …; which is an A.P. , as the difference between the two consecutive terms is constant, ie. 0 First four terms of this A.P. will be 4, 1, -2, and -5 . 1

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(iv) a = -1, d = 1/2 A. Let the series be a , a , a , a … a = a = -1 a = a + d = -1 + 1/2 = -1/2 a = a + d = -1/2 + 1/2 = 0 a = a + d = 0 + 1/2 = 1/2 Therefore, the series will be -1, -1/2, 0, 1/2...; which is an A.P. , as the difference between the two consecutive terms is constant, ie. 1/2 First four terms of this A.P. will be -1, -1/2, 0, and 1/2. 1

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(v) a = -1.25, d = -0.25 A. Let the series be a , a , a , a … a = a = -1.25 a = a + d = -1.25 - 0.25 = -1.5 a = a + d = -1.5 - 0.25 = -1.75 a = a + d = -1.75 - 0.25 = -2.0 Therefore, the series will be -1.25, −1.50, −1.75, −2.00 …… ; which is an A.P. , as the difference between the two consecutive terms is constant, ie. –0.25 First four terms of this A.P. will be -1.25, −1.50, −1.75, and −2.00 1

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Q3. Please Do It Yourself! Q4. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. Read the following AND Please Do The Question Yourself! A. An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP, which can be positive, negative or zero. (xii) √2, √8, √18, √32…. A. Here a - a = √2 = a - a = a - a … i.e., a − a is same every time. Therefore, the given numbers are in A.P. And, d = √2. Three more terms are:…… 2

k+1

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_______________________________________________________________________________________ EXERCISE 5.2 (P. 105) Q1. Do It Yourself!! Q2. Do It Yourself!! Q3. In the following APs, find the missing terms in the boxes : A. (i) For this A.P., a = 2, a = 26. Now, a = a + (n − 1) d 3

n

a = 2 + (3 − 1) d => 26 = 2 + 2d => d = 12 => a = 2 + (2 − 1) 12 = 14 Therefore, 14 is the missing term. 3

2

(ii) For this A.P., a = 13 and a = 3. Now, a = a + (n − 1) d 2

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n

a = a + (2 − 1) d => 13 = a + d (i) a = a + (4 − 1) d => 3 = a + 3d (ii) Solving, we get. d = -5; and, a = 18 a = 18 + (3 − 1) (−5) = 18 + 2 (−5) = 18 − 10 = 8 Therefore, the missing terms are 18 and 8 respectively 2 4

3

(iii) Please do the rest of the parts yourself! Q4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78? A. a, d, and a is given. Find the value of ‘n’. n

Q5. Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205 (ii) 18, 15 1/2, 13, …, -47 A. In both the above parts ‘a’, ‘d’, and, ‘a ’ are given; Using the appropriate formula, find the value of ‘n’. n

Q6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . . Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. A. Do both the questions yourself! Q8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term. A. Given that, a = 12, a = 106. Now, a = a + (n − 1) d a = a + (3 − 1) d => 12 = a + 2d (i) Similarly, a = a + (50 − 1) d => 106 = a + 49d (ii) Solving, d = 2, and, a = 8 a = a + (29 − 1) d => a = 8 + (28)2 = 64. Therefore, 29th term is 64. 3

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Q9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero? A. Given that, a = 4, a = −8. Now, a = a + (n − 1) d; Thus, a = a + (3 − 1) d => 4 = a + 2d (i), and, a = a + (9 − 1) d => −8 = a + 8d (ii). Solving, d = −2, and, a = 8. Let nth term of this A.P. be zero. Thus, a = a + (n − 1) d => 0 = 8 + (n − 1) (−2) => n = 5. Hence, 5th term of this A.P. is 0. 3

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Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. A. For an A.P., a = a + (n − 1) d Thus, a = a + (17 − 1) d => a = a + 16d, and, n

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a = a + 9d. It is given that a − a = 7 => (a + 16d) − (a + 9d) = 7 => d = 1. Therefore, the common difference is 1. 10

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Q11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term? A. Given A.P. is 3, 15, 27, 39, … Thus, a = 3, d = a − a = 15 − 3 = 12 a = a + (54 − 1) d = 3 + (53) (12) = 3 + 636 = 639 Now, 132 + 639 = 771 We have to find the term of this A.P. which is 771. Let nth term be 771 => a = a + (n − 1) d => 771 = 3 + (n − 1) 12 => 768 = (n − 1) 12 => n = 65 Therefore, 65th term was 132 more than 54th term. Alternatively: Let nth term be 132 more than 54th term. Thus, n = 54 + 132/12 = 65 term. 2

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Q12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms? A. Let the first term of these A.P.s be a and a and the common difference of these A.P.’s be d. For first A.P., a = a + (100 − 1) d = a + 99d a = a + (1000 − 1) d = a + 999d For second A.P., a = a + (100 − 1) d = a + 99d a = a + (1000 − 1) d = a + 999d Given that, difference between 100th term of these A.P.s = 100 Therefore, (a + 99d) − (a + 99d) = 100 => a − a = 100 (i) Difference between 1000th terms of these A.P.s = (a + 999d) − (a + 999d) = a − a From equation (i); This difference, a − a = 100 Hence, the difference between 1000 terms of these A.P. will be 100. 1

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Q13. How many three-digit numbers are divisible by 7? A. First three-digit number that is divisible by 7 = 105 Next number = 105 + 7 = 112 Therefore, 105, 112, 119, …; all are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7. The maximum possible three-digit number is 999. Dividing it by 7, the remainder is 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7. Thus the series is as follows: 105, 112, 119, …, 994 Let 994 be the nth term of this A.P. Thus, a = 105, d = 7, a = 994, n = ? a = a + (n − 1) d => 994 = 105 + (n − 1) 7 => 889 = (n − 1) 7 => n = 128 Therefore, 128 three-digit numbers are divisible by 7. n

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Q14. How many multiples of 4 lie between 10 and 250? A. First multiple of 4 that is greater than 10 is 12. Next will be 16. Therefore, 12, 16, 20, 24, …, all are divisible by 4 and thus, are terms of an A.P. with first term as 12 and common difference as 4. When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4, and is the maximum number before 250, which is a multiple of 4. The series is as follows. 12, 16, 20, 24, …, 248 Let 248 be the nth term of this A.P. Thus, a = 12, d = 4, a = 248. a = a + (n - 1)d => 248 = 12 + (n - 1)4 => n = 60. Therefore, there are 60 multiples of 4 between 10 and 250. n

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Q15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal? Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. 63, 65, 67, … => a = 63, and, d = a − a = 65 − 63 = 2. nth term of this A.P. = a = a + (n − 1) d = 63 + (n − 1) 2 = 63 + 2n − 2 => a = 61 + 2n (i) 3, 10, 17, … => a = 3, and, d = a − a = 10 − 3 = 7 nth term of this A.P. = a = 3 + (n − 1) 7 => a = 7n − 4 (ii) It is given that, nth term of these A.P.s are equal to each other. Equating both these equations, we get, 61 + 2n = 7n − 4 => n = 13. Therefore, 13th terms of both these A.P.s are equal to each other. 2

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Q16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. A. a = 16 => a + (3 − 1) d = 16 => a + 2d = 16 (i) As, a − a = 12 => [a+ (7 − 1) d] − [a + (5 − 1) d]= 12 => (a + 6d) − (a + 4d) = 12 => d = 6 Putting this in equation (i), we get, a = 4 Therefore, the required A.P. is 4, 10, 16, 22, … 3

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Q17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253. A. Given A.P. is 3, 8, 13, …, 253 ‘d’ for this A.P. is 5; thus, this A.P. can be written in reverse order as 253, 248, 243, …, 13, 8, 5 For this A.P., a = 253, d = 248 − 253 = −5, n = 20 a = a + (20 − 1) d => a = 253 + (19) (−5) = 253 − 95 = 158 Therefore, 20th term from the last term is 158. 20

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Q18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP. A. We know that, a = a + (n − 1) d => a = a + (4 − 1) d => a = a + 3d Similarly, a = a + 7d; a = a + 5d; a = a + 9d Given that, a + a = 24 => a + 3d + a + 7d = 24 => 2a + 10d = 24 => a + 5d = 12 (i) a + a = 44 => a + 5d + a + 9d = 44 => 2a + 14d = 44 => a + 7d = 22 (ii) Solving, we get, a = -13, and d = 5 Thus, a = a + d = − 13 + 5 = −8, and, a = a + d = − 8 + 5 = −3 Therefore, the first three terms of this A.P. are −13, −8, and −3. n

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Q19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000? A. It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200. Therefore, the salaries of each year starting 1995 are 5000, 5200, 5400, … Here, a = 5000, and, d = 200 Let after nth year, his salary be Rs 7000. Therefore, a = a + (n − 1) d => 7000 = 5000 + (n − 1) 200 => n = 11. Therefore, in 11th year, his salary will be Rs 7000. n

Q20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n. A. Given that, a = 5, d = 1.75, a = 20.75, n = ? a = a + (n − 1) d => 20.75 = 5 + (n - 1)1.75 => n = 10. Hence, n = 10. n

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_______________________________________________________________________________________ EXERCISE 5.3 (P. 112): Q1. Find the sum of the following APs: (i) 2, 7, 12, . . ., to 10 terms. A. For this A.P., a = 2, d = a − a = 7 − 2 = 5, n = 10. We know that, S = n/2(a + a ) = n/2(a + a + (n - 1)d) = n/2(2a + (n - 1)d) = 10/2(2x2 + (10 - 1)x5) Or, S = 5(4 + 9x5) = 5 x 49 = 245. 2

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Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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(ii) –37, –33, –29, . . ., to 12 terms. (iii) 0.6, 1.7, 2.8, . . ., to 100 terms. (iv) 1/15, 1/12, 1/10…, to 11 terms. A. Do parts (ii), (iii), and (iv) yourself. Q2. Find the sums given below : (i) 7 + 10 1/2 + 14 + ……. + 84 A. For this A.P., a = 7, l = 84, d = a - a = 10 1/2 - 7 = 7/2 Let 84 be the nth term of this A.P.; Thus, l = a = a + (n − 1)d => 84 = 7 + (n - 1)x7/2 => n = 23. We know that, S = n/2(a + l) = 23/2(7 + 84) = 1046 1/2 2

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Q2 (ii) 34 + 32 + 30 + . . . + 10 A. For this A.P., a = 34, d = a − a = 32 − 34 = −2, l = 10 Let 10 be the nth term of this A.P. Now, l = a + (n − 1) d, Or, 10 = 34 + (n − 1) (−2) => n = 13. S = n/2 (a + l) = 13/2 (34 + 10) = 286. 2

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(iii) –5 + (–8) + (–11) + . . . + (–230) A. For this A.P., a = −5, l = −230, d = a − a = (−8) − (−5) = -3 Let −230 be the nth term of this A.P., ie. l = a + (n − 1)d => −230 = − 5 + (n − 1) (−3) => n = 76 Now, S = n/2(a + l) = 76/2 [(-5) + (-230)] = -8930 2

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Q3. In an AP: (i) given a = 5, d = 3, a = 50, find n and S . (ii) given a = 7, a = 35, find d and S . (iii) given a = 37, d = 3, find a and S . (iv) given a = 15, S = 125, find d and a . (v) given d = 5, S = 75, find a and a . (vi) given a = 2, d = 8, S = 90, find n and a . (vii) given a = 8, a = 62, S = 210, find n and d. (viii) given a = 4, d = 2, S = –14, find n and a. (ix) given a = 3, n = 8, S = 192, find d. (x) given l = 28, S = 144, and there are total 9 terms. Find a. A. Follow the trick provided in the class, and do the questions yourself! n

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Q4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636? A. Do it Yourself! Q5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. A. Do it Yourself! Q6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? A. Do it Yourself! Q7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. A. Do it Yourself! Q8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. A. Do it Yourself! Q9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. Do it Yourself! Q10. Show that a , a , . . ., a , . . . form an AP where a is defined as below : (i) a = 3 + 4n (ii) a = 9 – 5n Also find the sum of the first 15 terms in each case. A. (i) a = 3 + 4(1) = 7; a = 3 + 4(2) = 11; a = 3 + 4(3) = 15; a = 3 + 4(4) = 19 It can be observed that a −a =a −a =a −a =4 i.e., a − a is same every time. Therefore, this is an AP with ‘d’ = 4 and ‘a’ = 7. Now, S = n/2[2a + (n - 1)d] => S = 15/2[2(7) + 14x4] = 525 (ii) a = 9 – 5n A. a = 9 − 5 × 1 = 4; a = 9 − 5 × 2 = −1; a = 9 − 5 × 3 = −6; a = 9 − 5 × 4 = −11 It can be observed that a − a = a − a = a − a = -5. i.e., a − a is same every time. Therefore, this is an AP with ‘d’ = -5 and ‘a’ = 4. Now, S = n/2[2a + (n - 1)d] => S = 15/2[2(4) +(15 - 1)(-5)] = -465 Q11. If the sum of the first n terms of an AP is 4n – n , what is the first term (that is S )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms. A. Given that, Sn = 4n − n First term, a = S = 4(1) − (1) = 3 Sum of first two terms = S = 4(2) − (2)2 = 4 Second term, a = S − S = 1 d = a − a = 1 − 3 = −2 Now, a = a + (n − 1)d = 3 + (n − 1) (−2) = 3 − 2n + 2 = 5 − 2n Therefore, a = 5 − 2(3) = 5 − 6 = −1; and, a = 5 − 2(10) = 5 − 20 = −15 Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively. 1

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Q12. Find the sum of the first 40 positive integers divisible by 6. A. The positive integers that are divisible by 6 are 6, 12, 18, 24 … It can be observed that these are making an A.P. whose first term is 6 and common difference is 6. ie. a = 6; d = 6; S =? Now, S = n/2[2a + (n - 1)d] => S = 20[2(6) + (40 -1)6] = 4920 40

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Q13. Find the sum of the first 15 multiples of 8. => Do It Yourself! Q14. Find the sum of the odd numbers between 0 and 50. => Do It Yourself! Q15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? A. It can be observed that these penalties are in A.P. with ‘a’ as 200 and ‘d’ as 50. Penalty that has to be paid if he has delayed the work by 30 days. S = 30/2[2(200) + (30 - 1)50] = 27750. Therefore, the contractor has to pay Rs 27750 as penalty. 30

Q.16 A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes. A. Let the cost of 1st prize be P. Then, cost of 2nd prize = P − 20, and, cost of 3rd prize = P − 40 Thus, the cost of these prizes are in an A.P. having ‘a’ = P, and, ‘d’ = −20. Thus, a = P; d = −20; Given that, S = 700 => 7/2[2a + (7 - 1)(-20)] => a = 160. Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40. 7

Q17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students? A. Thus, total trees planted (by all the sections) = 234. Q18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in the adjoining figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7) A. Length of successive semicircles are = 0.5π, π, 1.5π, 2π,....13 terms. This forms an A.P. with ‘a’ = 0.5π, and, ‘d’ = ‘0.5π’, and n = 13 Thus, total length of the spiral, made up of 13 consecutive semicircles is S = 13/2[2x0.5π + (13 - 1)0.5π] = 13/2[π + 12/2π] = 13/2 x 7x 22/7 = 13x11 = 143cm. 13

Q19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how may rows are the 200 logs placed and how many logs are in the top row?

A. The number of rows in each row, ie. 20, 19, 18, ….forms an A.P., with ‘a’ = 20; and, ‘d’ = 1 Let there be total ‘n’ rows in total, thus, total of 200 logs are to placed in ‘n’ rows. Now, S = n/2[2a + (n - 1)d] => 200 = n/2[2x20 + (n - 1)(-1)] => 400 = n (40 − n + 1) => 400 = n (41 − n) => 400 = 41n − n => n − 41n + 400 = 0 => (n − 16) (n − 25) = 0 Thus, (n − 16) = 0 or n − 25 = 0 => n = 16 or n = 25 Now, a = a + (n − 1)d. For, n = 16; a = 20 + (16 - 1)(-1) = 20 - 15 = 5 Similarly, for n = 25, a = 20 + (25 - 1)(-1) = 20 - 25 + 1 = -4. As the number of rows can’t be negative, => Number of Rows = n = 16; and, the number of rows in the 16 row = a = 5. Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5. n

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Q.20 In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? A. The distances of potatoes are as follows: 5, 8, 11, 14…; Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are 10, 16, 22, 28, 34,………., which form an A.P., such that, a = 10, d = 16 − 10 = 6, and, n = 10. S =?; S = 10/2[2x10 + (10 - 1)x6] = 5[20 + 54] = 370. Therefore, the competitor will run a total distance of 370 m. 10

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_______________________________________________________________________________________ EXERCISE 5.4 (P. 115) Q1. Which term of the AP : 121, 117, 113, . . ., is its first negative term? A. [Hint : Find n for a < 0; n> 125/4 or n > 31 1/4 ; => n = 32nd term, which is the firstnegative term.] Q 2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. A. let the AP be a - 4d, a - 3d, a - 2d………; a = 3; d = ±1/2; S = 20, 76. Q3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2 1/2 m apart, what is the length of the wood required for the rungs? A. n = number of rungs = ( 2 1/2m/25cm = 250cm/25cm = 10) Length of wood required for rungs = Sum of 10 rungs = 10/2 (25 + 45) = 350 cm. n

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Q4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. A. a = 1; d = 1; S = 49x25; S = S - S ; x = 35. Q5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2m. Calculate the total volume of concrete required to build the terrace. A. Volume of concrete req’d to build the first, second, third step…. (in m ) is 1/4 x 1/2 x 50, (2 x 1/4) x 1/2 x 50, (3 x 1/4) x 1/2 x 50….. ie. 50/8, 2 x 50/8, 3 x 50/8….. Volume req’d = 50/8 + 2 x 50/8 + 3 x 50/8….. = 50/8 [1 + 2 + 3 ...15 terms] (a = 1, d = 1) = 750m . 49

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CBSE CLASS X MATHEMATICS CHAPTER 6 TRIANGLES RECAP FROM PREVIOUS CLASS/ES LINES AND ANGLES:

> An Angle is formed by two rays, with common initial point. Vertex is the common initial point of the two rays. Arms or Sides are the rays forming the angle. > Congruent or Equal Angles have same measure. Types of Angles: > Right angles > Straight Angle > Reflex Angle: > 180 and < 360 . > Acute angle > Obtuse angle Two angles are Complementary / Supplementary if their sum is 90 / 180 . Two angles are Adjacent Angles if : > They have a common vertex, > They have a common arm, > Their non-common arms lie on opposite sides of the common arm. Linear Pair: If OA & OB are opposite rays, and OC is any other ray, then ∠BOC & ∠AOC form a linear pair. Axiom 1: If a ray stands on a line, then the sum of two adjacent angles is 180 . the non-common arm of the angles are the two opposite rays. Axiom 2: If the sum of two adjacent angles is 180 , then the non-common arms of the angles form a line. These two axioms together are called Linear Pair Axiom. A Transversal is a line which intersects two or more given lines in distinct points, and, forms the following angles: Corresponding Angles, Alternate Interior (or just ‘Alternate’) angles, Alternate Exterior Angles, and CoInterior (or Consecutive Interior) Angles [ These are interior angles on the same side of transversal]. Parallel Lines and Transversal: When a transversal intersects two parallel lines, then, > Each pair of corresponding angles are equal, ie. ∠1=∠5; ∠2=∠6; ∠3=∠7; ∠4=∠8, (Axiom) > Each pair of alternate (interior) angles are equal, ie. ∠3=∠6; ∠4=∠5, (Theorum) > Each pair of alternate (exterior) angles are equal, ie. ∠1=∠8; ∠2=∠7, > Each pair of co-interior angles are supplementary, ie. ∠3 + ∠5 = 180 , ∠4 + ∠6 = 180 . The converse of each of the above four statements is also true. 0

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Axiom 3: If a transversal intersects two parallel lines, then each pair of corresponding angles are equal. (This Axiom is also called Corresponding Angles Axiom); Axiom 4: If a transversal intersects two lines, such that a pair of corresponding angles are equal, then the two lines are parallel to each other. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

46 THEORUMS:

> Theorem 1: The sun of all the angles around a point is 360 (Use Linear Pair Axiom); > Theorem 2: If two lines intersect each other, then the vertically opposite angles are equal. (Use Linear Pair Axiom); > Theorem 3: Lines which are parallel to the same line, are parallel to each other (Use Corresponding Angles Axiom); > Theorem 4: The sum of angles of a triangle is 180 (From a Vertex, draw a line parallel to the opposite side of the triangle); > Theorem 5: If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles; Thus, The exterior angle of a triangle is greater than either of its interior opposite angles. (Exterior Angle = Sum of Interior Opp. Angles = 180 - (The Third Angle). TRIANGLE (RECAP FROM PREVIOUS CLASSES) Congruence: Two geometric figures are said to be congruent, if they are exactly of the same size and shape; Congruent Triangles: Two triangles are congruent if and only if three sides and three angles of one are congruent to the corresponding sides and angles of the other (In total, there are 6 elements in a Triangle); CPCT: Corresponding Parts of Congruent Triangles; Different conditions for Congruence of two triangles: > SAS (This is an axiom and can be used without proof), > SSS, > AAS, > ASA, > RHS THEORUMS: > Theorum: Let AB be a line segment, and line CD is its perpendicular bisector. A point ‘D’ on the perpendicular bisector (ie. CD) is equidistant from A & B (SAS); > Theorum: Angles opposite to equal sides of an Isosceles Triangle are equal, (Given: AC = BC; Draw CD ⊥’r to AB; Use RHS); > Theorum: (Converse of above) If two angles of a triangle are equal, then the sides opposite to them are also equal; (Draw RS ⊥’r to PQ; By AAS, ∆PSR ≅ ∆OSR; Use CPCT); > Theorum: If two sides of a triangle are unequal, the angle opposite to the larger side is larger (Experimental Verification); > Theorum: In any ∆, the side opp. to the larger angle is longer (Experimental Verification); > Theorum: The sum of any two sides of a ∆ is greater than the third side. To Prove: AB + AC > BC Cons.: Produce BA to D such that AD = AC, & join CD. Proof: In ∆ ACD, AC = AD (by cons.); Thus, ∠1 = ∠2 (Th.: Angles opp. to eq. sides……) 0

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But, ∠3 > ∠1 (As ∠1 is part of ∠3). Thus, ∠3 > ∠2. In ∆ BCD, BD > BC ( Th.: In any ∆, the side opp. to the larger angle is longer. ); ie. AB + AD > BC; Or, AB + AC > BC (As AD = AC (by cons.)) > Theorum: (Corollary of above) In a Right Angled Triangle, the Hypotenuse is the Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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longest side, N.B.: Two non-parallel lines always intersect. If two or more lines pass through the same point, then they are said to be concurrent, and, the common point is called the point of concurrency. > Theorum: A point on an angular bisector of an angle, is equidistant from the arms of the angle.

> Theorum: The angular bisectors of a triangle pass through the same point, ie. They are concurrent. Given: ∆NMO, bisectors of angles M & O intersect at K; KN is joined, To Prove: KN bisects ∠N, Const,: Draw KV⊥ MO, KU ⊥ NO, KT ⊥ MN,

Proof: As K lies on the bisector of ∠M, KV = KT -(i) As K lies on the bisector of ∠O, KV = KU -(ii) From (i), & (ii), KU = KT => K is equidistant from MN & NO; ∴ KN bisects ∠N. ∴ MK, OK, NK are concurrent, & the point of concurrency K is the incentre of the triangle NMO. Incentre: The point of intersection K of the angle bisectors is called the incentre of the Triangle MNO. KV = KU = KT = Inradius. The point K is equidistant from the three sides of the triangle MNO.

Incircle: If we draw a circle with centre K & Radius KV, it will pass through T & U. This circle is called Incircle of ∆NMO. > Theorum: The perpendicular bisectors of the sides of a ∆ pass through the same point. Given: ∆ABC, in which ⊥’r bisectors of sides AB & AC intersect at O. Also OX ⊥ BC is drawn. To Prove: OX is perpendicular bisectors of BC. Const.: Join OA, OB & OC.

Proof: Since O lies on ⊥’r bisector of AB, OA = OB - (i) Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Since O lies on ⊥’r bisector of AC, OA = OC - (ii) From (i) and (ii), we get, OB = OC (iii) => O lies on ⊥’r bisector of BC; Also, OX ⊥ BC (by const.); ∴ OX is perpendicular bisectors of BC. Circumcentre:

The Perpendicular bisectors of the sides of a Triangle pass through the same point, and this common point is called the circumcentre of the triangle. OA = OB = OC = Circumradius => The point O is equidistant from the three vertices A, B, & C of ∆ABC.

Median: A line segment

joining the vertex of a triangle to the mid-point of the opposite side is called a median of the triangle. Note the following two (provable) points about medians: > The medians of an equilateral triangle are equal, > The medians of an equilateral triangle coincide with the altitudes. Theorem: Medians of a triangle pass through the same point which divides each of the medians in the ratio 2:1. Centroid: The point of intersection of the three medians of a triangle is called the ‘Centroid’ of the triangle. ‘Centroid’ divides each of the median in the ratio 2:1. Altitude:

The line segment from a vertex of a triangle, perpendicular to the line containing the opposite side, is called an altitude of the triangle. Theorem: In a triangle, the three altitudes pass through the same point. This common point is called the ortho-centre of the triangle. The ortho-centre of a right-angled triangle is the vertex of the right angle.

The Locus

is a set of all the points that satisfy a particular condition; or, share a property. Eg. A Circle is "the locus of points on a plane that are a certain distance from a central point".

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ENOUGH OF REVISION – CLASS X TRIANGLES NOW! 1. Introduction: Two figures are congruent, if they have the same shape and the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. Techniques of Indirect measurements are based on the principle of similarity of figures. All congruent figures are similar but the similar figures need not be congruent. Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). ‘Regular Polygons’ are the ones in which all the sides and all the angles are equal to each other. The same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons. Circles of equal radii are congruent. All the circles are similar. 2. Similar Triangles and Their Properties: Two triangles are said to be similar to each other, if Their corresponding angles are equal, and Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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corresponding sides are proportional. For triangles, these two conditions are dependent on each other. ie. If one condition holds good, the other holds good automatically. This is not the case with Quadrilaterals, for which the two conditions are independent. That is, if the one condition holds good, it is not necessary, that the other condition will also hold good. ‘Thales’ stated that the ratio of any two corresponding sides in equiangular triangles is always the same irrespective of their sizes. That is, if in two triangles, the corresponding angles are equal, the corresponding sides are automatically proportional to each other. Activity to prove ‘Basic Proportionality Theorum’: Draw any angle XAY and on AX, mark points P, Q, D, R and B such that AP = PQ = QD = DR = RB. Thus, AD / DB = 3 / 2 (by construction). Now, through B, draw any line intersecting arm AY at C. Through D, draw a line parallel to BC to intersect AC at E. On measuring AE & EC, we find AE/EC = 3/2. Thus, AD/DB = AE/EC. Thus, in ∆ABC, DE is parallel to BC, and AD/DB = AE/EC. 3. THEOREMS: Theorem 1: (‘Basic Proportionality’ or ‘Thales Theorum’) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Given: ∆ABC in which DE ∥ BC. To Prove: AD/DB = AE/EC Cons.: Draw DM ⊥ AC and EN ⊥ AB. Proof: Ar (∆ADE) = 1/2. AD. EN; Ar (∆BDE) = 1/2. BD. EN; Thus, Ar (∆ADE) / Ar (∆BDE) = AD/BD — (i) Also, Ar (∆ADE) = 1/2. AE. DM; Ar (∆CED) = 1/2. CE. DM; Thus, Ar (∆ADE) / Ar (∆CED) = AE/CE — (ii) Now, Ar (∆BDE) = Ar (∆CED) (∆’s on the same base & between the same parallels) Thus, from (i) & (ii), we have, AD/BD = AE/CE Corollary (Conclusion / Inference) from above Theorum: If in a Triangle ∆ABC, DE ∥ BC, then (i) AB/AD = AC/AE; (ii) AB/DB = AC/EC (i) From the ‘Basic Proportionality Theorum’, AD/DB = AE/EC; Or, DB/AD = EC/AE; DB EC AB AC +1= +1 ;∨, = Or, AD AE AD AE (ii) From the ‘Basic Proportionality Theorum’, AD/DB = AE/EC; Or, AD/DB + 1 = AE/EC + 1; AB AC Or, DB = EC Theorem 2: (Converse of ‘Basic Proportionality’ or ‘Thales Theorem’) If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. Given: ∆ABC, and a line ‘l’ intersecting AB in D, & AC in E, such that AD AE = DB EC . To Prove: DE ∥ BC. Proof: Let DE not be parallel to BC; and then let DE’ be another line parallel to BC. Now, since DE’ ∥ BC, By ‘Basic Proportionality Theorum’, AD A E ' = DB E' C - (i) Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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But it is given that

AD AE = DB EC

- (ii)

From (i) & (ii), AE’/E’C = AE/EC; Or,

A E' AE +1= + 1 ; Or, AC/E’C = AC/EC ' EC EC

Thus, E’C = EC; This is only possible if E & E’ coincide. Thus, DE ∥ BC. 4. Criteria for Similarity of Triangles: Two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). That is, in Δ ABC and Δ DEF, if (i) ∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F and (ii) AB BC CA , then the two triangles are similar. DE EF FD As in the case of congruency of two triangles, the similarity of two triangles should also be expressed symbolically, using correct correspondence of their vertices. Eg., for the triangles ABC and DEF (Abv. Fig.), we cannot write Δ ABC ~ Δ EDF or Δ ABC ~ Δ FED. However, we can write Δ BAC ~ Δ EDF. =

=

4.1 Theorems: Theorem 1: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio ( or proportion) and hence the two triangles are similar (AAA Criterion). Given: Two triangles ABC and DEF such that ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F. Cons.: Cut DP = AB and DQ = AC and join PQ. Proof: From the above construction, Δ ABC ≅Δ DPQ (SAS). This gives ∠B = ∠P= ∠E and PQ || EF (Δ ABC ≅Δ DPQ; ∠B = ∠ P (by cpct); ∠ B = ∠ E (given); ie. ∠P = ∠E (corresponding angles); and, hence PQ || EF ) Thus, DP/DE = DQ/DF (Corollary of Basic Proportionality Theorem); Or AB/DE = AC/DF (As, DP=AB, DQ=AC) Similarly, (ie. By similar construction) AB/DE = BC/EF. Thus, AB/DE=BC/EF=AC/DF. N.B.:If two angles of a Δ are respectively equal to two angles of another Δ, then by the angle sum property of a Δ their third angles will also be equal. Therefore, AAA similarity criterion can also be stated as follows: If two angles of one Δ are respectively equal to two angles of another Δ, then the two Δ’s are similar (AA similarity criterion). Theorem

2: If in two triangles, sides of one triangle are proportional to (i.e., in the same

ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar (SSS Criterion).

Given: Consider two triangles ABC and DEF such that AB/DE = BC/EF = AC/DF (<1). To Prove: Corresponding angles are equal, ie. ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Cons.: Cut DP = AB and DQ = AC and join PQ. Proof: DP/PE = DQ/QF & PQ || EF (Basic Proportionality Theorem) (As AB=DP; AC=DQ) Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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So, ∠P = ∠E and ∠Q = ∠F. (Corresponding Angles) Thus, ΔDPQ ~ ΔDEF (AA Criterion) Thus, DP/DE = DQ/DF = PQ/EF (Corollary of Basic Proportionality Theorem); Now, Given AB/DE (=DP/DE, As, DP=AB by Cons.)= BC/EF = AC/DF=PQ/EF Thus, PQ/EF = BC/EF; Thus, PQ=BC Thus, by SSS ΔABC ≅ ΔDPQ. Thus, ∠A = ∠D, ∠B = ∠P and ∠C = ∠Q; Or, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F (As, ∠P = ∠E and ∠Q = ∠F (Corresponding Angles)). Thus, ΔABC ~ ΔDEF

Theorem

3: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. (SAS Criterion) Given: Consider two triangles ABC and DEF such that AB/DE = AC/DF (<1); & ∠A = ∠D To Prove: Corresponding angles are equal, ie. ∠A = ∠D, ∠B = ∠E and ∠C = ∠F; ie. ABC ~ DEF Cons.: Cut DP = AB and DQ = AC and join PQ. Proof: In ΔABC & ΔDPQ; AB = DP (By Cons.); ∠A = ∠D (Given); AC = DQ (By Cons.) Thus, ΔABC ≅ ΔDPQ (By SAS) - (i) Now, AB/DE = AC/DF (Given) => DP/DE = DQ/DF (By Cons. AB=DP; AC=DQ) =>PQ || EF (Converse of Basic Proportionality Th.). Thus, ∠P=∠E and ∠Q=∠F (Corres. Angles). In ΔDPQ & ΔDEF, ∠P=∠E, ∠Q=∠F & ∠D=∠D. Therefore, ΔDPQ ~ ΔDEF—(ii). Thus, ΔABC ~ ΔDEF

5. Areas of Similar Triangles: (Area is measured in square units. So, we may expect that this ratio is the square of the ratio of their corresponding sides!)

Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Given: Two similar triangles ABC & PQR (ie. ∠A=∠P, ∠B = ∠Q, and, ∠C = ∠R) To Prove: Ar(ΔABC) AB / PQ = BC /QR = CA /RP Ar(ΔPQR) Cons.: Draw Altitudes AM (⊥BC), and PN (⊥QR) of the triangles. Proof: Ar(ΔABC) = 1/2 x BC x AM; and, Ar(ΔPQR) = 1/2 x QR x PN BC x AM Thus, Ar(ΔABC) / Ar(ΔPQR) = QR x PN - (i) =

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Now, In Δ’s ABM & PQN: ∠B = ∠Q; ∠M = ∠N (=90 ). Thus, by AA criterion, ΔABM~ΔPQN. Therefore, AM/PN = AB/PQ - (ii) Also, ABC ~ PQR (Given); Thus, AB/PQ = BC/QR = AC/PR (iii) 2 2 Ar ( ∆ ABC ) BCxAM ABxAM AB BC AC = = = = Thus, Ar ( ∆ PQR ) QRxPN PQxPN (From (i) & (iii)); which equals, PQ QR PR 0

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( ) ( ) ( )

(From (ii) & (iii) Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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.1 Some Important Results based on Areas: Result 1. The Areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. Given: ΔABC ~ ΔDEF, and AL⊥BC, DM⊥EF. To Prove: Ar(ΔABC) AL / DM Ar(ΔDEF) Proof: Since, ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Therefore, Ar(ΔABC)/Ar(ΔDEF) = AB /DE - (i) In ΔALB & ΔDME, ∠ALB = ∠DME = 90 , and ∠B = ∠E (ΔABC ~ ΔDEF) Thus, ΔALB ~ ΔDME (AA Criterion). Thus, AB/DE = AL/DM, Or, AB /DE = AL /DM - (ii) Putting this in Eq. (i), we get: Ar(ΔABC) / Ar(ΔDEF) AL / DM 2

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Result 2. The Areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments. Given: ΔABC ~ ΔDEF, and AX, DY are the bisectors of ∠A & ∠D. To Prove: Ar(ΔABC) AX / DY Ar(ΔDEF) Proof: Since, ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Therefore, Ar(ΔABC)/Ar(ΔDEF) = AB /DE - (i) Now, ΔABC ~ ΔDEF => ∠A = ∠D => ∠A = ∠D => ∠BAX = ∠EDY - (ii) 2 2 In ΔABX & ΔDEF, from (ii) ∠BAX = ∠EDY, and ∠B = ∠E ( ∵ ΔABC ~ ΔDEF). ∴ ΔABX ~ ΔDEY (AA Corollary). ∴ AB/DE = AX/DY => AB /DE = AX /DY - (iii). Putting this in (i), we get Δ≝¿ ¿ Ar ¿ Ar ( Δ ABC ) ¿ =

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6. Pythagoras Theorem: This Theorem was given by Pythagoras, but, earlier to him, it was given by an ancient Indian mathematician Baudhayan (about 800 B.C.) in the following form: The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth). For this reason, this theorem is sometimes also referred to as the Baudhayan Theorem. Theorem: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other. Given: A right angled triangle ABC, right angled at B. Let BD be the perpendicular to the hypotenuse AC. To Prove: Δ ADB ~ Δ ABC; Δ BDC ~ Δ ABC; ΔADB ~ ΔBDC Proof: In Δ ADB and Δ ABC; ∠ A = ∠ A (Common), ∠ ADB = ∠ ABC (each 90 ). Thus, ΔADB ~ ΔABC (AA Criterion) - (i) In ΔBDC and Δ ABC; ∠C = ∠C, ∠BDC = ∠ABC. Thus, ΔBDC ~ ΔABC (AA Criterion) - (ii) From (i) & (ii), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC. Also, from (i) & (ii), ΔADB ~ ΔBDC - (iii) Also prove, that the square of the perpendicular (ie. a perpendicular drawn from the vertex of the right angle of a right triangle to the hypotenuse) is equal to the product of the lengths of the two parts of the hypotenuse. ie. To Prove: BD = AD.DC Proof: From (iii) above, we have ΔADB ~ ΔBDC ; Or, AD/BD = DB/DC => BD = AD.DC 0

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Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given: Triangle ABC right angled at B. To Prove: AC = AB + BC Cons. : Draw BD ⊥ AC Proof: Δ ADB ~ Δ ABC (Above Theorem); Thus, AD/AB = AB/AC; Or, AD.AC = AB (i) Also, Δ BDC ~ Δ ABC (Above Theorem); Thus, CD/BC = BC/AC; Or, CD.AC = BC (ii) Adding (i) & (ii), AB + BC = AC(AD+CD) = AC 2

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Theorem: (Converse of Pythagoras Theorem) In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Given: Triangle ABC in which AC = AB + BC To Prove: ∠B = 90°. Cons. : Construct a ΔPQR right angled at Q, such that PQ = AB, and, QR = BC. Proof: From Δ PQR, PR = PQ + QR (Pythagoras Theorem, as ∠ Q = 90°) Or, PR = AB + BC (By construction) (i) But, AC = AB + BC (Given) (ii); Thus, AC = PR (iii) (From (i) & (ii). Now, in ΔABC and ΔPQR, AB = PQ (By construction), BC = QR (By construction), AC = PR [Proved in (3) above]. So, Δ ABC ≅ Δ PQR (SSS congruence). Therefore, ∠B = ∠Q (CPCT); But ∠ Q = 90° (By construction) So, ∠ B = 90°. H.P. 2

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6.1 Some Important Results based on Pythagoras Theorum: Result 1: In an obtuse triangle, the square of the side opposite to obtuse angle is equal to the sum of the squares of the other two sides plus twice the product of one side and the projection of other on first; OR, To Prove: AC = AB + BC + 2BC.BD Given: An obtuse Triangle ABC, obtuse angles at B. AD⊥CB. Proof: Δ ABC is a right triangle, right angled at D. Thus, AC = AD + DC = AD + (DB+BC) = AD + DB +BC + 2DB.BC = AB + BC + 2DB.BC = AB + BC + 2BC.BD (AD + DB = AB ; ΔADB is a right triangle, right angled at D). 2

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Result 2: Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half the third side together with twice the square of the medium which bisects the third side; Or, In ΔABC, if BD is the medium, show that BC + AB = 2(BD + CD ) Given: ABC is a triangle, & BD is a median To Prove: (i) BC + AB = 2(BD + CD ); And (ii) BC + BA = 2BD + AC /2 Proof: (i) BC = CE + BE (i); AB = AE + BE (ii) [Rt. ∠’d Δ’s BCE & BAE.] Adding (i) & (ii), BC + AB = 2BE + CE + AE = 2BE + (CD - ED) + (DA + ED) = 2BE + CD + ED - 2.CD.ED + DA + ED + 2.DA.ED [CD = AD; and, In ΔBED, BE2 + ED2 = BD2] = 2BE + 2ED + 2CD = 2BD + 2CD = 2(BD + CD ) Thus, BC + AB = 2(BD + CD ) [Part (i) is proved.] (ii) From (i), BC + AB =2(BD + CD ) = 2BD + 2CD = 2BD + 2 x AC /4 = 2BD + AC /2. H.P. 2

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Result 3: Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle. Given: ABC is a triangle, & AD, BE & CF are three medians. To Prove: 3(AB + BC + CA ) = 4(AD + BE + CF ) Proof: From Result 2, AB + AC = 2(AD + BD ); Or, AB + AC = 2AD + 2xBC /4; Or, AB + AC = 2AD + BC /2 - (i) Similarly, for medians BE & CF, we have, BC + AB = 2BE + AC /2 - (ii) AC + BC = 2CF + AB /2 - (iii) Adding Equations (i), (ii) and (iii), we have, 2 (AB + BC + CA ) = 2AD + 2BE + 2CF + 1/2 (BC + AC + AB ); Or, 2 x 2 (AB + BC + CA ) = 4(AD + BE + CF ) + (BC + AC + AB ); Or, 3(AB + BC + CA ) = 4(AD + BE + CF ) 2

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CBSE CLASS X MATHEMATICS CHAPTER 6 TRIANGLES - NCERT EXERCISES SOLUTIONS

EXERCISE 6.1 (P. 122): The exercise is self explanatory and checks if the student has understood the meaning of Congruent and Similar Figures’. EXERCISE 6.2 (P. 128) Q1. In Adj. Figs. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). A. Do it yourself! Q2. E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. Do it yourself! Q3. If LM || CB and LN || CD, prove that AM/AB = AN/AD. A. Do it yourself! Q4. In the adjoining figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC. A. Do it yourself! Q5. 5. In the adjoining figure, DE || OQ and DF || OR. Show that EF || QR. A. Do it yourself! (Basic Proportionality Theorem) Q6.In the adjoining figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. A. Do it yourself! (Basic Proportionality Theorem) Q7. Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. A. Consider the adjoining figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ || BC. By using Basic Proportionality Theorem, (Because PQ || BC), We have, AP/PB = AQ/QC. Now, ‘P’ is the mid-point of AB, thus, AP = PB. Therefore, AP/PB = 1 = AQ/QC. Thus, AQ = QC. Thus, ‘Q’ is the mid-point AC; and, thus, ‘PQ’ bisects ‘AC’. Q8. Using Converse of Basic Proportional Theorem , prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. ( Converse of Basic Proportional Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side). A. Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively; i.e., AP = PB and AQ = QC. It can be observed that AP/PB = 1, and, AQ/QC = 1 Thus, AP/PB = AQ/QC. Thus, by Converse of Basic Proportional Theorem PQ || BC.

Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO. A. Given: Adj. figure (A Trapezium) ABCD, with AB || DC To Prove: AO/BO = CO/DO Proof: Draw a line EF through point O, such that EF || CD [ Hence, AB || EF || DC] In ΔADC, EO || DC, Hence, By Basic Proportional Theorem, we get AE/ED = AO/OC - (i) In ΔABD, OE || BA, Hence, By Basic Proportional Theorem, we get ED/AE = OD/BO; Or, AE/ED = BO/DO - (ii) From (i) & (ii), we get, AO/OC = BO/DO; Or, AO/BO = CO/DO. Q10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium. A. Consider a Trapezium ABCD. Draw OE || AB. In ΔABD, OE || AB (By Cons.); By basic proportionality theorem, we obtain AE/ED = BO/OD (i). Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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But, it is given that AO/BO = CO/DO; Or, it is given that AO/CO = BO/OD - (ii) From (i) & (ii), we get, AE/ED = AO/CO. This implies that EO || DC (By Basic Proportional Theorem). This implies that AB || OE || DC. Thus, AB || DC. Thus, ABCD is a Trapezium. EXERCISE 6.3 (P. 138) Q1. State which pairs of triangles in the given figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : A. Do it yourself!

EXERCISE 6.3 (P. 138) (cont’d) Q2. In the adjacent figure, ΔODC ~ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB. A. Do it yourself! Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD. A. Do the question yourself. Q4. In the adjacent figure, QR/QS = QT/PR and ∠ 1 = ∠ 2. Show that Δ PQS ~ Δ TQR. A. QR/QS = QT/PR (Given); Or, QT/PR = QR/QS; Or, QT/QR = PR/QS (i) Also, ∠ 1 = ∠ 2 => PR = PQ (ii) ( Sides opp. to equal angles are equal) From (i) & (ii), we have, QT/QR = PQ/QS; Or, QT/QP = QR/QS In Δ PQS & Δ TQR, QT/QP = QR/QS, & ∠ PQS = ∠TQR. Thus, by SAS, the two triangles are similar. Q5. S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS. Show that Δ RPQ ~ Δ RTS. A. Draw the figure, and use AA Criterion.

Q6. In the adjoining figure, if Δ ABE ≅ Δ ACD, show that Δ ADE ~ Δ ABC. A. Δ ABE ≅ Δ ACD (Given); Thus, AB=AC (i); and, AE=AD (ii)(By cpct) Eq. (ii) can be written as AD = AE (iii) Dividing eq. (iii) by eq. (i) => AB/AD = AC/AE - (iv) Now, In Δ ADE & Δ ABC, AB/AD = AC/AE from eq. (iv); and, ∠DAE = ∠BAC; Thus, by SAS, Δ ADE ~ Δ ABC Q7. In the adjoining figure, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that: (i) Δ AEP ~ Δ CDP (ii) Δ ABD ~ Δ CBE (iii) ΔAEP ~ ΔADB, and (iv) Δ PDC ~ Δ BEC A. (i) To Prove: ΔAEP ~ ΔCDP Proof: In ΔAEP & ΔCDP; ∠AEP = ∠CDP = 90 (AD⊥BC; CE⊥AB); ∠APE = ∠CPD (VOP). Thus, By AA, ΔAEP ~ ΔCDP. (ii) To Prove: ΔABD ~ ΔCBE Proof: In ΔABD ~ ΔCBE; ∠ADB = ∠CEB = 90 (AD⊥BC; CE⊥AB); ∠ABD = ∠CBE (Common). Thus, By AA, ΔABD ~ ΔCBE. (iii) To Prove: ΔAEP ~ ΔADB 0

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Proof: In ΔAEP ~ ΔADB; ∠AEP = ∠ADB = 90 (AD⊥BC; CE⊥AB); ∠EAP = ∠DAB (Common). Thus, By AA, ΔAEP ~ ΔADB. (iv) To Prove: (iv) Δ PDC ~ Δ BEC Proof: In ΔPDC & ΔBEC; ∠PDC = ∠BEC = 90 (AD⊥BC; CE⊥AB); and, ∠PCD = ∠BCE (Common). Thus, By AA, Δ PDC ~ Δ BEC. 0

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Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB. A. Do it Yourself! Q9. In the adjacent figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) Δ ABC ~ Δ AMP; (ii) CA/PA = BC/MP (i) In ΔABC and ΔAMP, ∠ABC = ∠AMP (Each 90°) ∠A = ∠A (Common) ∴ ΔABC ∼ ΔAMP (By AA similarity criterion). (ii) As, ΔABC ∼ ΔAMP ∴, CA/PA = BC/MP Q10. CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that: (i) CD/GH = AC/FG, (ii) ΔDCB ~ ΔHGE, (iii) ΔDCA ~ ΔHGF. A. Given: ΔABC ∼ ΔFEG; and, CD & GH are bisectors of ∠C & ∠G. To Prove: (i) CD/GH = AC/FG It is given that ΔABC ∼ ΔFEG ∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE ∴ ∠ACD = ∠FGH (Angle bisector), and, ∠DCB = ∠HGE (Angle bisector) In ΔACD and ΔFGH, ∠A = ∠F (Proved above); ∠ACD = ∠FGH (Proved above). ∴ ΔACD ∼ ΔFGH (By AA similarity criterion). Thus, CD/GH = AC/FG (ii) ΔDCB ~ ΔHGE In ΔDCB and ΔHGE, ∠DCB = ∠HGE (Proved above), and, ∠B = ∠E (Proved above) ∴ ΔDCB ∼ ΔHGE (By AA similarity criterion) (iii) ΔDCA ~ ΔHGF In ΔDCA and ΔHGF, ∠ACD = ∠FGH (Bisectors of equal angles ∠ACB = ∠FGE); ∠A = ∠F (Proved above); ∴ ΔDCA ∼ ΔHGF (By AA similarity criterion).

Q11. In the adjoining figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF. A. In ΔABC, AB = AC => ∠ABD = ∠C = ∠ECF (Angles opp. to equal sides are equal). Now, in Δ ABD & Δ ECF >> ∠ADB = ∠EFC = 90 ; ∠ABD = ∠ECF (Proved above). Thus, ΔABD ~ ΔECF (By AA Criterion). 0

Q12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR (see Fig. 6.41). Show that Δ ABC ~ Δ PQR. A. Given: AB/PQ = BC/QR = AD/PM Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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To Prove: Δ ABC ~ Δ PQR Proof: A median divides the opp. side into two equal halves; thus, BD = BC/2, & QM = QR/2. Now, AB/PQ = BC/QR = AD/PM => AB/PQ = (BC/2)/(QR/2) = AD/PM; Thus, AB/PQ = BD/QM = AD/PM. Thus, by SSS Criterion, Δ ABD ~ Δ PQM. Thus, ∠B = ∠Q. In Δ ABC ~ Δ PQR; AB/PQ = BC/QR (Given); and, ∠B = ∠Q. Thus, by SAS, Δ ABC ~ Δ PQR. Q13. D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA = CB.CD. A. In ΔADC and ΔBAC, ∠ADC=∠BAC (Given), ∠ACD=∠BCA (Common) ∴ ΔADC ∼ ΔBAC (By AA similarity criterion). Thus, DC/AC = AC/BC; Or, DC.BC = AC ; Or, CA = CB.CD 2

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Q14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR. A. Given: AB/PQ = AC/PR = AD/PM To Prove: ΔABC ~ ΔPQR Cons: Extend AD and PM up to E and L, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L. Proof: A median divides the opp. side into two equal halves. Therefore, BD = DC and QM = MR Also, AD = DE, and, PM = ML (By construction). In quadrilateral ABEC, diagonals AE and BC bisect each other at D. Therefore, quadrilateral ABEC is a parallelogram. ∴ AC = BE and AB = EC (Opposite sides of a parallelogram). Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR It is given that AB/PQ = AC/PR = AD/PM => CE/RL = AC/PR = 2AD/2PM => AC/PR = CE/RL = AE/PL => ΔACE ~ ΔPRL (By SSS). Thus, ∠CAE = ∠RPL (i) Similarly, we can prove, ΔABE ~ ΔPQL (By SSS) => ∠BAE = ∠QPL (ii) Adding eq. (i) & (ii); we get, ∠BAC = ∠QPR (iii) Now, in ΔABC & ΔPQR; AB/PQ = AC/PR (Given); and, ∠BAC = ∠QPR (From (iii)). Thus, ΔABC ~ ΔPQR (By SAS Similarity Criterion).

Q15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. A. Let AB and CD be a tower and a pole. Let BE and DF be the shadow of AB (tower) and CD (pole). Light rays from the sun are falling on the tower and the pole at the same angle. Therefore, ∠DCF = ∠BAE, and, ∠DFC = ∠BEA also, ∠CDF = ∠ABE (Tower and pole are vertical to the ground) ∴ ΔABE ∼ ΔCDF (AAA similarity criterion) => AB/CD = BE/DF => AB/6 = 28/4 => AB = 42m. Therefore, the height of the tower is 42 metres. Q16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that AB/PQ=AD/PM. A. ΔABC ~ ΔPQR => AB/PQ=AC/PR=BC/QR - (i) Also, ∠A = ∠P, ∠B=∠Q, ∠C=∠R - (ii) As AD & PM are medians, BD=BC/2, QM=QR/2 - (iii)

From equations (i) & (iii), AB/PQ = BD/QM - (iv) Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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In ΔABD and ΔPQM, ∠B = ∠Q [Using (ii)], and, AB/PQ = BD/QM [Using (iv)]. ∴ ΔABD ∼ ΔPQM (By SAS similarity criterion) => AB/PQ = AD/PM. [If the two triangles are similar, then their corresponding angles are equal, and, their corresponding sides are in proportion to each other.] ————————————————————————————————————————————EXERCISE 6.4 (P. 143) Q1. Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm and 121 cm . If EF = 15.4 cm, find BC. A. Δ ABC ~ Δ DEF => ar (ΔABC) / ar (ΔDEF) = BC / EF => 64 / 121 = BC / 15.4 This gives, BC = 11.2 cm. Q2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD. A. Since AB || CD, ∴ ∠OAB=∠OCD and ∠OBA=∠ODC (Alternate angles) In ΔAOB and ΔCOD, 2

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∠AOB = ∠COD (VOA), ∠OAB = ∠OCD (Alternate angles), and, ∠OBA = ∠ODC (Alternate angles)

∴ ΔAOB ∼ ΔCOD (By AAA similarity criterion. ∴ Ar (ΔAOB) / Ar (ΔCOD) = AB / CD => Ar (ΔAOB) / Ar (ΔCOD) = 4 : 1 2

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Q3. In the adjoining figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar (Δ ABC) / ar (Δ DBC) = AO/ DO. A. Given: The adjoining figure. To Prove: ar (Δ ABC) / ar (Δ DBC) = AO/ DO Cons. : Draw AP ⊥ BC, & DM ⊥ BC. Proof: ar(Δ ABC) / ar(Δ DBC) = [1/2 x BC x AP] / [1/2 x BC x DM] = AP/DM. In ΔAPO and ΔDMO, ∠APO = ∠DMO (Each = 90°), ∠AOP = ∠DOM (VOA) ∴ ΔAPO ∼ ΔDMO (By AA similarity criterion) ∴ AP/DM = AO/DO => ar(Δ ABC) / ar(Δ DBC)=AP/DM = AO/DO Q4. If the areas of two similar triangles are equal, prove that they are congruent. A. Consider two triangles Δ ABC & Δ PQR, which are similar to each other. Thus, ar(Δ ABC) / ar(Δ PQR) = (AB/PQ) = (BC/QR) = (AC/PR) - (i) It is given that ar(Δ ABC) = ar(Δ PQR) => ar(Δ ABC) / ar(Δ PQR) = 1. Putting this in Eq. (i), 1 = (AB/PQ) = (BC/QR) = (AC/PR) => AB = PQ, BC = QR, AC = PR. Thus, By SSS, Δ ABC ≅ Δ PQR Q5. D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find the ratio of the areas of Δ DEF and Δ ABC. A. D & E are midpoints of sides AB & BC; Therefore, DE || AC, and DE = AC/2. Now, in ΔBED & ΔBCA ∠BED = ∠BCA (Corresponding Angles), ∠BDE = ∠BAC (Corresponding Angles), and, ∠EBD = ∠CBA (Common). ∴ ΔBED ∼ ΔBCA (By AAA similarity criterion) ar(ΔBED)/ar(ΔBCA) = (DE/AC) = 1/4 => ar(ΔBED) = 1/4 ar(ΔBCA) Similarly, ar(ΔCFE) = ar(ΔADF) = 1/4 ar(ΔABC). Also, ar(ΔDEF) = ar(ΔABC) - [ar(ΔBED) + ar(ΔCFE) + ar(ΔADF)] = ar(ΔABC) - 3/4 ar(ΔABC) = 1/4 ar(ΔABC) ==> ar(ΔDEF) / ar(ΔABC) = 1/4 Q6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. A. Consider 2 similar triangles ΔABC ∼ ΔPQR. Let AD & PS be the medians of these 2

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triangles. Because, ΔABC ∼ ΔPQR; ∴ AB/PQ = BC/QR = AC/RP - (i) ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R - (ii) As AD & PS are medians, ∴ BD = DC = BC/2; and, QS = SR = QR/2. Putting this in equation (i) AB/PQ = BD/QS = AC/PR - (iii) In ΔABD and ΔPQS, ∠B = ∠Q [From (i)]; and, AB/PQ = BD/QS[From (iii)] Thus, ΔABD ~ ΔPQS (By SAS Criterion). Therefore, AB/PQ = BD/QS = AD/PS - (iv) Now, ar(Δ ABC) / ar(Δ PQR) = (AB/PQ) = (BC/QR) = (AC/PR) . Using eq. (i) & (iv) ar(Δ ABC) / ar(Δ PQR) = [AD/PS] 2

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Q7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. A. Let ABCD be a square of side a. Therefore, its diagonal = √2 a Two desired equilateral triangles are formed as ΔABE and ΔDBF. Side of an equilateral triangle, ΔABE, described on one of its sides = a Side of an equilateral triangle, ΔDBF, described on one of its diagonal = √2 a We know that equilateral triangles have all its angles as 60º. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. Thus,

Q8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 A. We know that equilateral triangles have all its angles as 60º . Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. Let side of ΔABC = x. Therefore, side of triangle BDE = x/2 ∴ area (ΔABC) / area (ΔBDE) = [x ÷ x/2] = 4/1 = 4:1 Hence, the correct answer is (C). 2

Q9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81 A. If two triangles are similar to each other, then the ratio of the areas of these triangles is equal to the square of the ratio of the corresponding sides of these triangles. It is given that the sides are in the ratio 4:9. Therefore, ratio between areas of these triangles = (4/9) = 16/81 Hence, the correct answer is (D). ———————————————————————————————————————-2

EXERCISE 6.5 (P. 150) Q1. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm; (ii) 3 cm, 8 cm, 6 cm; (iii) 50 cm, 80 cm, 100 cm; (iv) 13 cm, 12 cm, 5 cm A. (i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides, we obtain 49, 576, and 625. 49 + 576 = 625; Or, 7 + 24 = 25 The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a right triangle. Now, the hypotenuse is the longest side of a right triangle. Therefore, the length of the hypotenuse of this triangle is 25 cm. 2

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(ii) 3 cm, 8 cm, 6 cm A. It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Squaring the lengths of three sides, we get 9, 64, & 36. However, 9 + 36 ≠ 64, Or, 3 + 6 ≠ 8 Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side. Therefore, the given triangle is not satisfying Pythagoras theorem. Hence, it is not a right triangle. Do (iii) & (iv) on your own; Remember, that the hypotenuse is the longest side of the right angled triangle. 2

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Q2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM = QM . MR. A. Let ∠MPR = ‘x’; Thus, in ΔMPR, ∠MRP = 90 - x. Similarly, ∠MPQ = 90 - x; ∠MQP = 180 - (90 + 90 - x) = x In ΔQMP & ΔPMR: ∠PQM = ∠RPM; ∠QMP = ∠PMR = 90 ; Thus, by AA, ΔQMP ~ ΔPMR; => MP/MR = QM/PM => PM/MR = QM/PM => PM = QM . MR 2

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Q3. ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB = BC × BD; (ii) AC = BC × DC; (iii) AD = BD × CD A. (i) In ΔBAD & ΔBCA; ∠BAD = ∠BCA = 90 , ∠ABD =∠CBA (Common) Thus, by AA Criterion, ΔBAD ~ ΔBCA => BA/CB = BD/BA =>AB = BC x BD. (ii) Let ∠CAB = x; => ∠CBA = 90 - x, ∠CAD = 90 - x, and, ∠CDA = x In ΔCAB & ΔCDA; ∠ACB = ∠DCA (=90 ), ∠CAB = ∠CDA (=x); Thus, by AA, ΔCAB ~ ΔCDA => AC/DC = BC/AC => AC = BC × DC (iii) Consider ΔABD & ΔCAD 2

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Q4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2. A. Do it yourself! Q5. ABC is an isosceles triangle with AC = BC. If AB = 2AC , prove that ABC is a right triangle. A. AB = 2AC => AB = AC + AC => AC + BC Thus, the triangle is satisfying Pythagoras Theorem. Thus, the given triangle is a right angled triangle. 2

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Q6. ABC is an equilateral triangle of side 2a. Find each of its altitudes. A. Let AD be the altitude of an equilateral triangle, ΔABC. Now, altitude of an equilateral triangle bisects the opposite side, and, all the three altitudes of an equilateral triangle are of equal lengths. Thus, BD = DC = a. In ΔADB, AD + DB = AB , => AD = a√3 = Length of each of the altitudes of an equilateral triangle. 2

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Q7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. A. Applying Pythagoras theorem in ΔAOB, ΔBOC, ΔCOD, ΔAOD, we get, AB = AO + BO - (i); BC = BO + OC - (ii); CD = CO + OD - (iii); AD = OD + AO - (iv) Adding the above four equations, we get, AB + BC + CD + AD = 2(OA + OB + OC + OD ) = 2[(AC/2) + (BD/2) + (AC/2) + (BD/2) (Diagonals of Parallelogram bisect each other) = 2 x 2 [(AC/2) + (BD/2) ] = AC + BD . 2

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Q8. In the adjoining figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that (i) OA + OB + OC – OD – OE – OF = AF + BD + CE , 2

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(ii) AF + BD + CE = AE + CD + BF . A. Cons.: Join OA, OB, and OC (i) Applying Pythagoras Theorem in ΔAOF, ΔBOD, & ΔCOE, we get OA = OF + AF ; OB = OD + BD ; and, OC = OE + EC ; Adding the above three equations, we get, OA + OB + OC = OF + AF + OD + BD + OE + EC ; Or, OA + OB + OC – OD – OE – OF = AF + BD + CE 2

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(ii) From the above result, AF + BD + EC = (OA - OE )+ (OC - OD ) + (OB - OF ); Or, AF + BD + CE = AE + CD + BF . 2

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Q9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall. A. Use the adjoining figure, and use Pythagoras Theorem, OB = 6m. Q10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut? A. Let OB be the pole and AB be the wire. Use Pythagoras theorem, OA = Distance of stack from the base of the pole = 6√7 m. Q11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after hours 1 1/2 hours? A. Distances travelled by the plane flying towards north, and the plane flying towards west in 1 1/2 hours are 1500kms (=OA) & 1800kms (=OB). Use Pythagoras theorem, AB = Distance between the planes after 1.5 hrs = 300√61 km. Q12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops. A. Let AB & CD be the poles of height 6m and 11m. ∴, CP = 11−6=5 m From the figure, AP = 12m Apply Pythagoras theorem in ΔAPC, we get, AC = Distance between the tops of the two poles = 13m.

Q13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE + BD = AB + DE . A. Applying Pythagoras theorem in ΔACE, we get, AC + CE = AE - (i) Applying Pythagoras theorem in ΔBCD, we get, BC + CD = BD - (ii) Adding equations (i) & (ii), we get, AC + CE + BC + CD = AE + BD - (iii) From ΔCDE, DE = CD + CE ; and, From ΔABC, AB = AC + CB => Putting these values in eq. (iii), we get, DE + AB = AE + BD 2

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Q14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD. Prove that 2AB = 2AC + BC . A. Applying Pythagoras theorem for ΔACD, we get, AC = AD + CD => AD = AC - CD - (i) Applying Pythagoras theorem for ΔABD, we get, 2

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AB = AD + BD => AD = AB - BD - (ii) From equations (i) & (ii), we have, AC - CD = AB - BD - (iii) Now, it is given, DB = 3CD => We convert DB & DC in the form of BC; ie. DB = 3BC/4; and, CD = BC/4. Putting these values in equation (iii), we get AC - (BC/4) = AB - (3BC/4) . Thus, we get, 2AB = 2AC + BC . 2

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Q15. In an equilateral triangle ABC, D is a point on side BC such that BD = BC/3. Prove that 7AB . A. Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC. (An Altitude and a Median from a vertex to the opposite side is the same in case of equilateral triangles). Thus, BE = EC = BC/2 = a/2; and, AE = (a√3)/2. Also, given that BD = BC/3 = a/3. Thus, DE = BE - BD = a/2 - a/3 = a/6. Applying Pythagoras theorem in ΔADE, we get, AD = AE + DE = [(a√3)/2] + [a/6] = 28a /36 = 7/9 AB => 9AD = 7AB

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Q16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. A. Let the side of an equilateral triangle be a, & AE be the altitude of ΔABC. ∴ BE = EC = a/2. Applying Pythagoras theorem in ΔABE, we get, AB = AE + BE ; Thus, a = AE + (a/2) ; which gives, AE = 3a /4 => 4AE = 3a Thus, 4 × (Square of altitude) = 3 × (Square of one side) 2

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Q17. Tick the correct answer and justify: In ΔABC, AB = 6√3cm, AC = 12 cm and BC = 6 cm. The angle B is: (A) 120° (B) 60° (C) 90° (D) 45° A. AB = 6√3cm => AB = 108 cm ; AC = 12 cm => AC = 144 cm ; and, BC = 6 cm => BC = 36 cm . Thus, AB + BC = AC . Thus, the given triangle, ΔABC, is satisfying Pythagoras Theorem. Thus, ΔABC is a right triangle, right angled at B. Therefore, ∠B = 90 . Thus, the right answer is ( C ). ———————————————————————————————————————EXERCISE 6.6 (P. 150): Optional – 10 Questions. 2

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CBSE CLASS X MATHEMATICS CHAPTER 7 COORDINATE GEOMETRY RECAP FROM PREVIOUS CLASSES The co-ordinate geometry is the synthesis of Algebra and Geometry after the work of Rene Descartes (French). The position of any object lying in a plane can be represented with the help of two perpendicular lines. The system used for describing the position of a point in a plane is also known as ‘Cartesian System’. On a ‘Number Line’, distances from a fixed point (origin) are marked in equal units positively in one direction and negatively in the other. ‘Descartes’ placed two such lines perpendicular to each other on a plane & located points on the plane. The planes on which these lines are graphed is known as ‘Cartesian Plane’. ‘Co-ordinate Geometry’ is a branch of Geometry which sets up a definite correspondence between the position Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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of a point in a plane and a pair of algebraic numbers, called co-ordinates. (P(x,y); x=10, y=6) Values of points written as (x1,y1), (x2,y2), (x3,y3)… are called ordered pairs. The two perpendicular lines (or perpendicular number lines) through the origin are called Cartesian or Rectangular Co-ordinate axis. X Co-ordinate => Abscissa of Point P, Y Co-ordinate => Ordinate of Point P. OX & OY are called x-axis & y-axis, and both taken together are called axis of co-ordinates; ‘O’ (origin) is the intersection of the axis of co-ordinates. The equation of x-axis is y=0; and, The equation of y-axis is x=0. The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y). N.B.: The word ‘Regular’ in ‘Regular Pentagon’ or ‘Regular Hexagon’ implies that all the sides and angles of the given shape are equal, While drawing a graph (or line) for a given linear equation, we must always take at least 3 points. Though, through two points, one and only one line can be drawn, but we take at least the third point for verification purposes, that weather the line we have drawn is correct or not. ———————————————————————————————————————--1. Introduction: Now, a linear equation in two variables of the form ax + by + c = 0, (a, b are not simultaneously zero), when represented graphically, gives a straight line. Also, the graph of y = ax + bx + c (a ≠ 0), is a parabola. Coordinate geometry has been developed as an algebraic tool for studying geometry of figures. It helps us to study geometry using algebra, and understand algebra with the help of geometry. 2. Distance Formula: Finding the distance between any two points P(x , y ) and Q(x , y ): Draw PR and QS perpendicular to the x-axis. A perpendicular from the point P on QS is drawn to meet it at the point T. Then, OR = x , OS = x . So, RS = x – x = PT. Also, SQ = y , ST = PR = y . So, QT = y – y . Now, applying the Pythagoras theorem in Δ PTQ, we get, PQ = PT + QT = (x – x ) + (y – y ) . Therefore, PQ = [(x – x ) + (y – y ) ] which is called the distance formula; which can also be written as: PQ = [(x – x ) + (y – y ) ] Points to Remember: In particular, the distance of a point P(x, y) from the origin O(0, 0) is, OP = [x + y ] . A quadrilateral with all four sides equal and one angle 90° is a square. 2

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Consider three points A, B & C. If AB + BC = BC, we can say that the three points are

collinear. A point

which is equidistant from the two end points ‘A’ & ‘B’, of a line AB, lies on the perpendicular bisector of AB.

segment

3. Section Formula: It is used when we need to identify the position of a point between the given point (ie. Coordinates of the two points are known). Consider any two points A(x1, y1) andB(x2, y2) and assume that P (x, y) divides AB internally in the ratio m1 : m2, i.e., PA/PB = m /m Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC parallel to the x-axis. Then, by the AA similarity criterion, Δ PAQ ~ Δ BPC Thus, PA/BP = AQ/PC = PQ/BC (i) Now, AQ = RS = OS – OR = x – x PC = ST = OT – OS = x – x PQ = PS – QS = PS – AR = y – y BC = BT– CT = BT – PS = y – y. Substituting these values in (i), we get m / m = (x – x )/(x – x) = (y – y )/(y – y) Taking, m / m = (x – x )/(x – x), we get, x = (m x + m x ) / (m + m ) Similarly taking, m / m = (y – y )/(y – y), we get, 1

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y = (m y + m y ) / (m + m ) So, the coordinates of the point P(x, y) which divides the line segment joining the points A(x , y ) and B(x , y ), internally, in the ratio m : m are (m x + m x )/(m + m ), and, (m y + m y )/(m + m ) This is known as the section formula. Special Case: The mid-point of a line segment divides the line segment in the ratio 1 : 1. Therefore, the coordinates of the mid-point P of the line segment joining the two points A(x , y ) and B(x , y ) is: [(1.x + 1.x )/(1 + 1), (1.y + 1.y )/(1 + 1)] = [(x + x )/2, (y + y )/2] 1

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4. Area of a Triangle: From previous classes: Area of Triangle = 1/2 x base x height. Heron’s Formula [s(s - a)(s - b)(s - c)] Area of a Trapezium = 1/2 x (Sum of Parallel Sides) x (Perpendicular Dis tance (or Height) between them). 1/2

Using coordinate geometry to find the area of a triangle: Let ABC be any triangle whose vertices are A(x , y ), B(x , y ) and C(x , y ). Draw AP, BQ and CR perpendiculars from A, B and C, respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapezia. Now, from the figure, it is clear that the area of 1

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Δ ABC = area of trapezium ABQP + area of trapezium APRC – area of trapezium BQRC. Also, we know, Area of a Trapezium = 1/2 x (Sum of Parallel Sides) x (Perpendicular Distance between them). Thus, area of ΔABC = 1/2 (BQ + AP) QP + 1/2 (AP + CR) PR - 1/2 (BQ + CR) QR = 1/2 (y + y ) (x - x ) + 1/2 (y + y ) (x - x ) - 1/2 (y + y ) (x - x ) = 1/2 [x (y - y ) + x (y - y ) + x (y - y ) Thus, the area of Δ ABC is the numerical value of the expression 1/2 [x (y - y ) + x (y - y ) + x (y - y ) N.B.: If the area of a triangle is 0 square units, then its vertices will be collinear. N.B.: To find the area of a polygon, we divide it into triangular regions, which have no common area, and add the areas of these regions. 2

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CBSE CLASS X MATHEMATICS CHAPTER 7 COORDINATE GEOMETRY - NCERT EXERCISES SOLUTIONS

EXERCISE 7.1 (P. 161) Q1. Find the distance between the following pairs of points :(i) (2, 3), (4, 1) (ii) (– 5, 7), (– 1, 3) (iii) (a, b), (– a, – b). A. Do It Yourself! Q2. Do It Yourself! Q3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear. A. Do It Yourself! (AB + BC = BC => Points are collinear.) Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. A. Find the three sides of the given triangle; If any of the two sides of the given triangle are equal, the triangle is isosceles. Q5. Verify if ABCD is a square or not. A. Can be done in two ways: Find the four sides, and the two diagonals; or, Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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the four sides, apply Pythagoras Theorem, to verify if an angle is 90 or not. If it is so, then we have a figure, with four sides equal, and an angle of 90 => The given figure is a square. 0

0

Q6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0); (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4); (iii) (4, 5), (7, 6), (4, 3), (1, 2) A. Find the four sides and the two diagonals; and, thereby state, if the given figure is of any particular type of a quadrilateral or not. Q7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). A. Do It Yourself! Q8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units. A. Do It Yourself!

Q9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR. A. Do It Yourself! [You will get the value of ‘x’ as ±4; and hence, the point ‘R’ can be (4, 6), or, (-4, 6). Find the corresponding values of PR & QR, for the two values of ‘R’.]

Q10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4). A. Do It Yourself! [ The equation obtained will be 3x + y = 5] EXERCISE 7.2 (P. 167): Q1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. A. Let P(x, y) be the required point. Using the section formula, we get, x = (2x4 + 3x(-1)/(2 + 3) = 1; y = (2x(-3) + 3x7)/(2 + 3) = 3 Therefore, the point is (1, 3).

Q2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). A. Let P (x , y ) and Q (x , y ) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB Therefore, Point P divides AB internally in the ratio 1:2 Thus, x = (1x(-2) + 2x4)/(1 + 2) = 2; y = (1x(-3) + 2x(-1))/(1 + 2) = (-3-2)/3 = -5/3 Thus, P((x , y ) = (2, -5/3) Point Q divides AB internally in the ratio 2:1 Thus, x = (2x(-2) + 1x4)/(2 + 1), and, y = (2x(-3) + 1x(-1))/(2 + 1); Or, Point Q is (-4 +4)/3; and, ((-6-1)/3) => Q is (0, -7/3) 1

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Q3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? Benefitted: My Paypal a/c - m[email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. Niharika posted the green flag at 1/4 of the distance AD i.e., 1/4 X 100 = 25m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25). (Assuming ‘A’ as the origin (0,0)). Similarly, Preet posted red flag at 1/5 of the distance AD i.e., 1/5 X 100 = 20m from the starting point of 8th line. Therefore, the coordinates of this point R is (8, 20). Distance between these flags (using distance formula) = GR = The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y); such that, x = (2 + 8)/2; and, y = (25 + 20)/2. Hence, A (x, y) = (5, 22.5). Therefore, Rashmi should post her blue flag at 22.5m on 5th line. th

th

Q4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

A. Let the ratio be m :m ; Thus, -1 = (6m - 3m )/(m + m ) => -m - m = 6m - 3m => 2m = 7m => m : m = 2 : 7 The easier method would be to assume that the line segment joining (−3, 10) and (6, −8) is divided by the point (−1, 6) in the ratio of k : 1. Q5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division. A. Let the ratio in which the line segment joining A (1, −5) & B (−4, 5) is divided by x-axis be k:1 Thus, coordinate of the point of division is (-4k + 1)/(k + 1), and, (5k + (-5))/(k + 1) = (-4k + 1)/(k + 1), and, (5k - 5)/(k + 1) Now, the ‘y’ coordinate of any point on ’x’ axis is zero => (5k - 5)/(k + 1) = 0 => 5k - 5 = 0 => k = 1. Thus, the ‘x’ axis divides the given line segment in the ratio of 1 : 1 The coordinates of the given point would be: (-4 + 1)/2, and, 0; Or, (-3/2, 0) 1

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Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. A. The diagonals of a parallelogram bisect each other. Hence, we would find the midpoint of the given two diagonals, and equate them to each other. As the vertices of the parallelogram are given in order, we name them as: A(1, 2), B(4, y), C(x, 6) and D(3, 5). Coordinates of mid-point of AC => [(1 + x)/2, (2 + 6)/2] => [(1 + x)/2, 4] Coordinates of mid-point of BD => [(4 + 3)2, (y + 5)/2] => [7/2, (y + 5)/2]] Now, equating the coordinates, we get, (1 + x)/2 = 7/2 => x = 6; and, 4 = (y + 5)/2 => y = 3.

Q7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4). A. Let the coordinates of point A be (x, y). Now, the centre of the circle (2, -3) is the mid-point of its’ diameter (AB). Thus, (2, -3) = [(x + 1)/2, ((y + 4)/2 => (x + 1)/2 = 2, and, (y + 4)/2 = -3 => x = 3, and, y = -10 Therefore, the coordinates of point ‘A’ is (3, -10) Q8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB. A. Given: AP : AB = 3 : 7 => AP:PB = 3:4; Thus, point ‘P’ divides AB in the ratio of 3 : 4 Coordinates of ‘P’: [(3x2 + 4x(-2))/7; (3x(-4) + 4x(-2))/7] = [(6 - 8)/7; (-12 - 8)/7] = (-2/7, -20/7) Q9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts. A. From the figure, it can be seen that points X, Y, Z are dividing the line segment ‘AB’ in the ratio Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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1:3, 1:1, 3:1 respectively. Thus, coordinates of ‘X’ are: [(1x2 + 3x(-2))/4; (1x8 + 3x2)/4] = [(2 - 6)/4; (8 + 6)/4] = (-1, 7/2) Coordinates of ‘Y’ are: [(2 + (-2))/2; (8 + 2)/2] = [0, 5] Coordinates of ‘Z’ are: [(3x2 + 1x(-2))/4; (3x8 + 1x2)/4] = [1, 13/2] Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

A. Area of Rhombus = 1/2 x (Product of Its’ diagonals) Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.

EXERCISE 7.3 (P. 170): Q1. Find the area of the triangle whose vertices are :(i) (2, 3), (–1, 0), (2, – 4) (ii) (–5, –1), (3, –5), (5, 2) A. Do it Yourself! Q2. In each of the following find the value of ‘k’, for which the points are collinear. (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, –5) A. Hint: For collinear points, area of triangle formed by them is zero. Solving, for (i), k = 4; and, for (ii), k = 3 Q3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle. A.

(i) (7, –2),

Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3). Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by…. Refer Above. Q4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3). A. Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

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Q5. You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for Δ ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2). A. Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2). Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

However, area cannot be negative. Therefore, area of ΔABD is 3 square units. (Negative sign here signifies that the triangle is below the ‘x’ axis)

However, area cannot be negative. Therefore, area of ΔADC is 3 square units. (Negative sign here signifies that the triangle is below the ‘x’ axis) Clearly, median AD has divided ΔABC in two triangles of equal areas.

EXERCISE 7.4 (Optional)* Q1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).. A. 2:9 internally Q2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear. A. The area formed by them is 0.; x + 3y - 7 = 0 Q3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3). A. OA = OB & OB = OC will give two equations. The centre is (3, -2) Q4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices. A. (1, 0), (1, 4). Q 5. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown. The students are to sow Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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seeds of flowering plants on the remaining area of the plot. (i) Taking A as origin, find the coordinates of the vertices of the triangle. (ii) What will be the coordinates of the vertices of Δ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe? A. Taking A as origin, coordinates of P, Q and R are (4, 6), (3, 2) & (6, 5). The area of the triangle is 9/2 square units. Taking C as origin, coordinates of P, Q and R are (12, 2), (13, 6) & (10, 3). The area of the triangle is 9/2 square units. Thus, the area is same in both the cases. Q6. The vertices of a Δ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of the Δ ADE and compare it with the area of Δ ABC. A. D divides AB in the ratio 1:3, thus, coordinates of D are (13/4, 23/4) Similarly, coordinate of E are (19/4, 5) Area of triangle ABC = 15/2 sq units; and area of triangle ABC = 15/32 sq units. Thus, area ∆ADE : area ∆ABC = 1:16 ANOTHER WAY: Ratio of areas of two similar triangles equals ratio of the squares of any two corresponding sides => area ∆ADE : area ∆ABC = AD : AB = 1:16 2

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Q7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of Δ ABC. (i) The median from A meets BC at D. Find the coordinates of the point D. (ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1 A. (i) Coordinates of D (mid-point of BC) = (7/2, 9/2) (ii) P (11/3, 11/3) (iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1. (iv) What do yo observe? (v) If A(x , y ), B(x , y ) and C(x , y ) are the vertices of Δ ABC, find the coordinates of the centroid of the triangle. A. (iii) E (5/3, 3); Q (11/3, 11/3); F (5, 7/2); R (11/3, 11/3) (iv) We observe that points P, Q & R coincide, ie. the medians AD, BE & CF are concurrent at point (11/3, 11/3). This point is called the centroid of the triangle. (v) Let A, B, C be vertices of triangle ABC, whose medians are AD, BE & CF. Coordinate of point dividing AD in the ratio 2:1 is (x1 + x2 + x3)/3, (y1 + y2 + y3)/3. Similarly, the coordinates of point dividing BE & CF in the ratio 2:1 is same. Thus, medians of a triangle is concurrent and the coordinates of the centroid is (x1 + x2 + x3)/3, (y1 + y2 + y3)/3. 1

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Q8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer. A. All the sides are equal, but the diagonals are not equal to each other => PQRS is a rhombus.

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CBSE CLASS X MATHEMATICS CHAPTER 8 TRIGNOMETRY 1. Introduction: The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle. We will confine ourselves to the study of some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. However, these ratios can be extended to other angles also. 2. Trigonometric Ratios: Here, ∠CAB (or angle A) is an acute angle. Side BC is the side opposite to angle A (called as a ‘perpendicular’). AC is the hypotenuse of the right triangle and the side AB is the side adjacent to angle A (called as a ‘base’). Certain ratios, called trignometric ratios, involving the sides of a right triangle, are: The trigonometric ratios of the angle A in right triangle ABC : Sine of ∠A = Perpendicular = BC Hypotenuse AC Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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of ∠A =

Base = AB Hypotenuse AC Tangent of ∠A = Perpendicular = BC (=(BC/AC)/(AB/AC) = Sin A/Cos A) Base AB Cosecant of ∠A = 1 = Hypotenuse = AC Sine of ∠A Perpendicular BC Secant of ∠A = 1 = Hypotenuse = AC Cosine of ∠A Base AB Cotangent of ∠A = 1 = Base = AC (=(AC/AB)/(BC/AB)=Cos A/Sin A) Tangent of ∠A Perpendicular BC So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides. N.B.: ‘Sin A’ is ‘the sine of the angle A’; and, is not the product of ‘sin’ & ‘A’. The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same. This is true, if the triangles considered are similar, and the length of the sides vary in proportion. N.B.: For the sake of convenience, we may write sin A, cos A, etc., in place of (sin A) , (cos A) , etc., respectively. But cosec A = (sin A) ≠ sin A (which is called sine inverse A). Also, sometimes, the Greek letter θ (theta) is also used to denote an angle. 2

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If we know any one of the ratios, we can obtain the other ratios. Say, for example, it is given, (i) Sin A = 1/3, find all the other trigonometric ratios of the angle A; (ii) Tan A = 4/3, find all the other trigonometric ratios of the angle A. Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1). 3. Trigonometric Ratios of Some Specific Angles: Draw a right - angled triangle, and attempt to find trignometric ratios of 30 , 45 , 60 . Trigonometric Ratios of 0° and 90°: 0

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As ∠ A gets smaller and smaller, the length of the side BC decreases. The point C gets closer to point B, and finally when ∠ A becomes very close to 0°, AC becomes almost the same as AB. When ∠A is very close to 0°, BC gets very close to 0 and so the value of sin A = BC/AC is very close to 0. Also, when ∠ A is very close to 0°, AC is nearly the same as AB and so the value of cos A = AB/AC, is very close to 1. This gives, sin0 = 0, and cos 0 = 1. The other trignometric ratios are: tan 0 = 0, sec 0 = 0, cosec 0 & cot 0 are not defined. Consider the trigonometric ratios of ∠A, as ∠A is made larger and larger in Δ ABC, till it becomes 90°. As ∠A gets larger and larger, ∠C gets smaller and smaller. Therefore, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally, when ∠A is very close to 90°, ∠C becomes very close to 0° and the side AC almost coincides with side BC. When ∠ C is very close to 0°, ∠ A is very close to 90°, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when ∠ A is very close to 90°, ∠ C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0. 0

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So, we define : sin 90° = 1 and cos 90° = 0; from which we may find the other trignometric ratios. Remember the following table: From the table, we observe that as ∠ A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0.

4. Trigonometric Ratios of Complementary Angles: Since ∠ A + ∠ C = 90°, they form a pair of complementary angles. We have:

Trigonometric ratios for ∠ C = 90° – ∠ A.

Comparing the ratios in (1) and (2), we get, So, sin (90° – A) = cos A, cos (90° – A) = sin A, tan (90° – A) = cot A, cot (90° – A) = tan A, sec (90° – A) = cosec A, cosec (90° – A) = sec A, for all values of angle A lying between 0° and 90°. Note : tan 0° = 0 = cot 90°, sec 0° = 1 = cosec 90° and sec 90°, cosec 0°, tan 90° and cot 0° are not defined. 5. Trigonometric Identities: An equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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In Δ ABC, right-angled at B, we have:AB + BC = AC (i) Dividing each term of (i) by AC , we get, cos A + sin A = 1 (ii) This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity. Now dividing (i) by AB , we get, 1 + tan A = sec A (iii) This equation is true for A = 0°. And, tan A and sec A are not defined for A = 90°. So, (iii) is true for all A such that 0° ≤ A < 90°. Dividing (i) by BC ,we get, 1 + cot A = cosec A (iv) Note that cosec A and cot A are not defined for A = 0°. Therefore (iv) is true for all A such that 0° < A ≤ 90°. Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios. 2

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CBSE CLASS X MATHEMATICS CHAPTER 8 TRIGNOMETRY - NCERT EXERCISES SOLUTIONS

EXERCISE 8.1 (P. 181) Q1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A A. Use Pythagoras Theorem to find AC, and, proceed. (ii) sin C, cos C A. Same as above. Q2. In the adjoining figure, find tan P – cot R. A. Use Pythagoras Theorem to find QR (=5cm), and, proceed. tan P – cot R = 0 Q3. If sin A = 3/4, calculate cos A and tan A. A. Do it yourself! Q4. Given 15 cot A = 8, find sin A and sec A. A. Do it yourself! Q5. Given sec θ = 13/12, calculate all other trigonometric ratios. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. Consider a right-angle triangle ΔABC, right-angled at point B. Sec θ = 13/12 = Hypotenuse/Adjacent side = AC/AB Let constant of proportionality be ‘k’, and, using Pythagoras Theorem, we get BC = 5k. Thus, proceed to find all other trignometric ratios. Q6. If ∠A and ∠B are acute angles such that cos A=cos B, then show that ∠A=∠B. A. Given: Triangle ABC in which CD⊥AB; cos A = cos B => AD/AC = BD/BC (i) To Prove: ∠A = ∠B. Cons.: Draw AD ⊥AB, and, extend AC to P such that BC = CP. Proof: From Eq. (i), we get, AD/BD = AC/BC => AD/BD = AC/CP (By Cons.) (ii) Thus, CD||BP (Converse of Basic Proportionality Theorem), ⇒∠ACD = ∠CPB (Corres. angles) (iii); and, ∠BCD = ∠CBP (Alt. angles) (iv). By construction, BC = CP; ∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (v) From equations (iii), (iv), and (v), we get, ∠ACD = ∠BCD … (vi) In ΔCAD and ΔCBD, ∠ACD = ∠BCD [Using equation (vi)] ∠CDA = ∠CDB [Both 90°] Therefore, the third angle of the triangle is also equal (angle sum property). ∴∠CAD = ∠CBD ⇒ ∠A = ∠B [ ( 1+ sin ) (1−sin ) ] Q 7: If cot θ = 7/8, evaluate: (i) [ ( 1+cos ) (1−cos ) ] ; (ii) cot θ 2

A. Consider a right triangle ABC, right-angled at point B. Assume ’k’ as a constant of proportionality; use Pythagoras Theorem to find the third side, and proceed.

Q8. If 3 cot A = 4, check whether

or not.

A. It is given that 3cot A = 4; Or, cot A = 4/3. Consider a right triangle ABC, right-angled at B. cot A = base/perpendicular = AB/BC = 4/3. Let ‘k’ be constant of proportionality; then, AB = 4k & BC = 3k. In ΔABC, (AC) = (AB) + (BC) = (4k) + (3k) = 16k + 9k = 25k => AC = 5k. Thus, cos A = AB/AC = 4/5; Sin A = BC/AC = 3/5; Tan A= BC/AB = 3/4. Now, L.H.S. = (1 - tan A) / (1 + tan A) = 7/25; R.H.S. = cos A − sin A = (4/5) - (3/5) = 7/25 ∴ L.H.S. = R.H.S. 2

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Q9. In triangle ABC, right-angled at B, if tan A = 1/√3 , find the value of: (i) sin A cos C + cos A sin C; (ii) cos A cos C – sin A sin C A. Consider a right triangle ABC, right-angled at B. tan A = Perp./Base = BC/AB = 1/√3 Proceeding as above, find the value of third side (in terms of ‘k’); thence, find the value of sin A, cos A, sin C, cos C . Thus, proceed to find the value of (i) & (ii). Q10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. A. Given, PR + QR = 25, and, PQ = 5. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Let PR be x. ∴, QR = 25 − x Applying Pythagoras theorem in ΔPQR, we obtain PR = PQ + QR ; ∴ x = (5) + (25 − x) ; Or, 50x = 650; Or, x = 13 Therefore, PR = 13 cm; QR = (25 − 13) cm = 12 cm; Thus, sin P = QR/PR = 12/13; cos P = PQ/PR = 5/13; tan P = QR/PQ = 12/5. Q11. State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1. A. Consider a right triangle ABC, right-angled at B, for all the following parts. (i) Sin A increases from 0 to 1, as ‘A’ increases from 0 to 90 ; and, Cos A decreases from 1 to 0, as ‘A’ increases from 0 to 90 ; For angles 45 - 90 , sin A is > 1/√2; and cos A is < 1/√2; Hence, tan A for angles 45 - 90 = sin A/cos A, is greater than 1; Hence the given statement is false. (ii) sec A = 12/5 for some value of angle A. A. sec A = Hyp. / Base = AC / AB = 12/5. Let ‘k’ be constant of proportionality. Let AC = 12k, AB = 5k. Applying Pythagoras Theorem, we get, BC = 10.9 k. Now, for any triangle, any side is ‘more’ than the difference of the other two sides, and ‘less’ than the sum of the other two sides. Thus, BC should be such that, AC − AB < BC < AC + AB; Or, 12k − 5k < BC < 12k + 5k; Or, 7k < BC < 17 k However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible. Hence, the given statement is true. 2

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Q11(iii) cos A is the abbreviation used for the cosecant of angle A. A. Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false. Q11(iv) cot A is the product of cot and A. A. cot A is not the product of cot and A. It is the cotangent of ∠A. Hence, the given statement is false. Q11(v) sin θ = 4/3 for some angle θ. A. In a right-angled triangle, sin θ = (Side opposite to θ) / Hypotenuse. In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible. Hence, the given statement is false. ———————————————————————————————————————-EXERCISE 8.2 (P. 187) Q1. Evaluate the following : (i) sin 60° cos 30° + sin 30° cos 60° A. Answer is ‘1’. (ii) 2 tan 45° + cos 30° – sin 60° A. Answer is ‘2’. (iii) cos 45 / (sec 30 + cosec 30 ) A. Answer is ‘(3√2 - √6) / 8’. 2

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(iv) A. Answer is ‘(43 - 24√3) / 11’. (v) A. Answer is ‘67 / 12’. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Q2. Choose the correct option and justify your choice : (i) (A). sin60°, (B). cos60°, (C). tan60°, (D). sin30° A. =√3 / 2 = sin 60 . Hence, (A) is correct. 0

Q2 (ii) (A) tan 90°; (B) 1; (C) sin 45°; (D) 0 A. Correct answer is (D). (iii) sin 2A = 2 sin A is true when A = (A) 0°; (B) 30°; (C) 45°; (D) 60° A. Out of the given alternatives, only A = 0° is correct. As sin 2A = sin 0° = 0 2 sinA = 2sin 0° = 2(0) = 0 Hence, (A) is correct. (iv) (A) cos 60°; (B) sin 60°; (C) tan 60°; (D) sin 30° A. The correct answer is ( C ). Given expression = √3 = tan 60 0

Q3. If tan (A + B) = √3 and tan (A - B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B. A. tan (A + B) = √3 = tan 60 => A + B = 60 (i) tan (A - B) = 1/√3 = tan 30 => A - B = 30 (ii) Solving (i) & (ii), we get, A = ∠A = 45 & ∠B = 15 . Q4. State whether the following are true or false. Justify your answer. (i) sin (A + B) = sin A + sin B. A. Let A = 30° and B = 60°; then, sin (A + B) = sin (30° + 60°) = = sin 90° = 1. sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = (1 + √3)/2. Clearly, sin (A + B) ≠ sin A + sin B; Hence, the given statement is false. 0

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(ii) The value of sin θ increases as θ increases. A. The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as: Sin 0 = 0; sin 30 = 0.5; sin 60 = √3/2 = 0.866; sin 90 = 1. Hence, the given statement is true. 0

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(iii) The value of cos θ increases as θ increases. A. The value of cos θ decreases as θ increases in the interval of 0° < θ < 90° as: cos 0 = 1; cos 30 = √3/2 = 0.866; cos 45 = 1/√2 = 0.707; cos 60 = 1/2 = 0.5; sin 90 = 1. Hence, the given statement is false. 0

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(iv) sin θ = cos θ for all values of θ. A. The given statement is false, as sin θ = cos θ; only when θ = 45 . 0

(v) cot A is not defined for A = 0°. A. The given statement is true; as, cot A = cosA/sinA = cos0 / sin0 = 1/0 = Not-defined. 0

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EXERCISE 8.3 (P. 189) Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Q1. Evaluate : (i) sin18 / cos72 = sin (90 - 72 )/cos72 = 1 0

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(ii) tan 26 / cot 64 = tan (90 - 64 )/cot 64 = cot 64 /cot 64 = 1 0

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(iii) cos 48° – sin 42° = cos (90 - 42 ) - sin 42 = sin 42 - sin 42 = 0 0

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(iv) cosec 31° – sec 59° = cosec (90 - 59 ) - sec 59 = sec 59 - sec 59 = 0 0

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Q2. Show that : (i) tan 48° tan 23° tan 42° tan 67° = 1 A. tan 48° tan 23° tan 42° tan 67° = tan (90° − 42°) tan (90° − 67°) tan 42° tan 67° = cot 42° cot 67° tan 42° tan 67° = (cot 42° tan 42°) (cot 67° tan 67°) = (1) (1) = 1 (ii) cos 38° cos 52° – sin 38° sin 52° = 0 A. cos 38° cos 52° − sin 38° sin 52° = cos (90° − 52°) cos (90°−38°) − sin 38° sin 52° = sin 52° sin 38° − sin 38° sin 52° = 0 Q3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. A. Given that, tan 2A = cot (A− 18°) => cot (90° − 2A) = cot (A −18°) => 90° − 2A = A− 18° => 108° = 3A => A = 36° Q4. If tan A = cot B, prove that A + B = 90°. A. Given that, tan A = cot B => tan A = tan (90° − B) => A = 90° − B => A + B = 90° Q5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A. A. Given that, sec 4A = cosec (A − 20°) => cosec (90° − 4A) = cosec (A − 20°) => 90° − 4A= A− 20° => 110° = 5A => A = 22° Q6. If A, B and C are interior angles of a triangle ABC, then show that sin((B + C)/2) = Cos (A/2) A. For a triangle ABC, ∠A + ∠B + ∠C = 180° => ∠B + ∠C = 180 - ∠A => ∠B + ∠C = 90 - ∠A Thus, sin((B + C)/2) = sin (90 - A/2) = cos A/2. 2 2 Q7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. A. sin 67° + cos 75° = sin (90° − 23°) + cos (90° − 15°) = = cos 23° + sin 15° 0

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EXERCISE 8.4 (P. 193) Q1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. A. (i) cosec = 1 + cot A. Thus, 1/cosec = 1/(1 + cot A) => sin A = 1/(1 + cot A) => sin A = ± 1/√(1 + cot A) Now, √(1 + cot A) will always be positive as we are adding two positive quantities. Thus, sin A = 1/√(1 + cot A) (ii) tan A = 1/cot A. Also, sec A = 1 + tan A = 1 + 1/cot A = cot A + 1 cot A Thus, sec A = [√(1 + cot A)]/cotA (iii) tan A = 1/cot A. 2

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Q2. Write all the other trigonometric ratios of ∠A in terms of sec A. A. (i) cos A = 1/sec A (ii) sin A + cos A = 1 2

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Thus, sin A = (iii) 1 + tan A = sec A = (iv) cot A = 1/tan A = (v) cosec A = 1/sin A = 2

2

DO IT YOURSELF! Q3. Evaluate: (i)

=

[sin(90 - 27)] + sin 27 = 1/1 =1 [cos(90 - 73)] + cos 73 2

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(ii) sin 25° cos 65° + cos 25° sin 65° = 1 (Ans.) Q4. Choose the correct option. Justify your choice. (i) 9 sec A – 9 tan A = (A) 1 ;(B) 9 ;(C) 8 ;(D) 0 A. (i) 9 sec A − 9 tan A = 9 (sec A − tan A) = 9 (1) [As sec2 A − tan2 A = 1] = 9. Hence, alternative (B) is correct. 2

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(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ) = (A) 0; (B) 1; (C) 2; (D) −1 A. Convert all the given trignometric ratios in the form of sin A & cos A. The correct answer is ‘2’. Q4. (iii) (sec A + tan A) (1 – sin A) = (A) sec A; (B) sin A; (C) cosec A; (D) cos A A. Convert all the given trignometric ratios in the form of sinA & cosA. Correct answer is ‘cos A’. (iv) (1 + tan A) / (1 + cot A) = (A) sec A; (B) –1; (C) cot A; (D) tan A A. Convert all the given trignometric ratios in the form of sinA & cosA. Correct answer is ‘tan A’. 2

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Q5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) (ii) A. Take L.C.M., use the identity: sin A + cos A = 1 2

2

(iii) A. Do it yourself. (iv) A. Can be done in two ways: Start from L.H.S.; use sin A + cos A = 1 => sin A = 1 - cos A; & you would reach R.H.S.. Simplify LHS and RHS separately] 2

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(v)

using the identity 1 + cot A = cosec A. 2

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A. Do it yourself! (vi) A. Do it yourself! (vii) (viii) (ix)

(x)

CBSE CLASS X MATHEMATICS CHAPTER 9 SOME APPLICATIONS OF TRIGNOMETRY 1. Heights and Distances: The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer. The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e., the case when we raise our head to look at the object.

The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e., the case when we lower our head to look at the point being viewed.

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CBSE CLASS X MATHEMATICS CHAPTER 9 SOME APPLICATIONS OF TRIGNOMETRY - NCERT EXERCISES SOLUTIONS

EXERCISE 9.1 (P. 203): Q1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.

A. Do it Yourself! Q2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. A. Firstly draw the diagram, and solve the question yourself. Q3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? A. The two diagrams formed would be as the ones adjacently drawn. Proceed and do the question yourself! Q4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. tan θ = Perpendicular/Base. Solve the Question Yourself! Q5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. A. Draw the diagram (Adjacent figure), and, Solve the question yourself! Q6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. A. Draw the diagram (Adjacent figure), and, Solve the question yourself! Q7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. A. Draw the diagram (Adjacent figure), and, Solve the question yourself! Q8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. A. Draw the diagram (Adjacent figure), and, Solve the question yourself! Q9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. A. Let AB be the building and CD be the tower. Draw the diagram (Adjacent figure), and, Solve the question yourself! Q10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. A. Draw the diagram (Adjacent figure), and, Solve the question yourself! Q11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see adjacent figure). Find the height of the tower and the width of the canal. A. Draw the diagram; B & C are the two banks of the canal; AB is the TV Tower; and, ‘D’ is a point 20m away from the bank ‘C’ of the canal. Q12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. A. Draw the diagram (Adjacent figure ~ ‘AB’ is the building; and, ‘CD’ is the cable tower), and, Solve the question yourself! Hence, the given statement is true.

Q13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. A. Draw the diagram (Adjacent figure ~ ‘AB’ is the light house; ’C’ & ’D’ are the two ships ), and, Solve the question yourself! Q14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval. A. Let the initial position A of balloon change to B after some time and CD be the girl (1.2m high) Draw the diagram, and, Solve the question yourself! Q15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. A. 6secs & 3 secs. Q16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

CBSE CLASS X MATHEMATICS CHAPTER 10 CIRCLES (RECAP FROM PREVIOUS CLASSES) 1. Introduction: 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. Arc is a continuous piece of a circle. Degree measure of a

minor arc is the measure of the central angle subtended by the arc.

Degree measure of a major arc is “3600 - The Degree measure of the corresponding minor arc”. The Degree measure of the circumference of the circle is always 3600. A Sector is that region of a circular disc which lies between an arc, and the two radii joining the extremities

of the arc and the centre. A Quadrant is ‘One-Fourth’ of the Circular Disc. Concentric Circles are the ones which have the same centre.

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2. Theorums: Following theorems have to be remembered along with their proofs: Theorum 1: Equal chords of a circle subtend equal angles at the centre. (Use SSS) Theorum 2: If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal. (Use SAS) Theorum 3: The Perpendicular from the centre of a circle to a chord bisects the chord. (Use RHS) Theorum 4: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Theorum 5: There is one and only one circle passing through three given non-collinear points. Proof: Let perpendicular bisectors of AB & BC meet at ‘O’. Thus, OA = OB = OC = ‘r’ (say). So, with ‘O’ as centre, and ‘r’ as radius, we can draw a circle passing through the three points ‘A’, ‘B’ & ‘C’. If possible, let there be another circle C(O’, r), passing through the three points ‘A’, ‘B’ & ‘C’, then, O’ will be on the perpendicular bisectors of AB & BC. Since, two lines can’t intersect at more than one point, O’ must coincide with O. Thus, There is one and only one circle passing through three given non-collinear points. N.B.: If ABC is a triangle, then, by the above Theorum, there is a unique circle passing through the three vertices A, B & C of a Triangle. This circle is called the ‘Circumcircle’ of the ∆ABC. Its centre and radius are called the ‘Circumcentre’ and the ‘Circumradius’ of the triangle.

Theorum

6: Equal chords of a circle (or, of Congruent Circles) are equidistant from the centre (or centres). (Use RHS) Theorum 7: Chords equidistant from the centre of the circle are equal in length. N.B.: If two chords of a circle are equal, then their corresponding arcs are congruent, and conversely, if two arcs are congruent, then their corresponding chords are equal. The angle subtended by the minor arc PQ at O is ∠POQ, and the angle subtended by the major arc PQ at O is ‘Reflex Angle POQ’. Congruent arcs of a circle subtend equal angles at the centre. Theorum 8: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. ((While dealing with an angle subtended by an arc at the centre of the circle, we see the angle subtended by the arc, in front of the arc.)) Proof: (Consider fig. (i)) ∠APO = ∠PAO; ∠AQO = ∠QAO (Angles opposite to the equal sides) ∠POB = 2∠PAO - (i) ∠QOB = 2∠QAO - (ii) (The above two equations are due to Exterior Angle Property) Adding Eqs. (i0 & (ii), we get the desired result.

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Theorum

9: Angles in the same segment of a circle are equal. Proof: (Again, consider fig. (i) above) An Important Result: The Angle in a Semi-Circle is a Right Angle. (Using the above Theorum and figure (ii) above) Theorem 10: If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a same circle (i.e. they are concyclic). (Very Important) Proof: AB is a line segment, which subtends equal angles at two points C and D. That is, ∠ ACB = ∠ ADB.

To show that the points A, B, C and D lie on a circle let us draw a circle through the points A, C and B. Suppose it does not pass through the point D. Then it will intersect AD (or extended AD) at a point, say E (or E′). If points A, C, E and B lie on a circle, ∠ ACB = ∠ AEB, But it is given that ∠ ACB = ∠ ADB. Therefore, ∠ AEB = ∠ ADB. This is not possible unless E coincides with D. Similarly, E′ should also coincide with D.

An Important Result: Any Angle subtended by a minor arc in the Alternate Segment (ie. Major Segment) is acute, while any angle subtended by a major arc in the Alternate Segment (ie. Minor Segment) is obtuse: Proof: (i) Here ∠ACB is an angle subtended by minor arc; Thus, ∠AOB = 2∠ACB. As, ∠AOB is angle subtended at centre by minor arc; ∠AOB < 180 ; Or, 1/2 ∠AOB is < 90 . Thus, ∠ACB is < 90 ; Or, ∠ACB is Acute. (ii) Here ∠ADB is an angle subtended by major arc; Thus, Reflex ∠AOB = 2∠ADB. Also, Reflex ∠AOB > 2 Right Angles or > 180 , Thus, 2∠ADB > 180 , or, ∠ADB > 90 Thus, ∠ADB is Obtuse. 0

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Cyclic Quadrilaterals: A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle. Theorem 10: The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. Proof: x = 2 ∠A (i) y = 2 ∠C (ii) Adding eq. (i) and (ii), x + y = 2 (∠A + ∠C ); Or, 360 = 2 (∠A + ∠C ); ie. (∠A + ∠C ) = 180 . Similarly, we can prove the result for the other pair of angles. 0

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11: If the sum of a pair of opposite angles of a quadrilateral is 180º, then the quadrilateral is Cyclic. (Proof beyond our scope) RECAP OVER! NOW MATHEMATICS FOR CLASS X! CBSE CLASS X MATHEMATICS CHAPTER 10 CIRCLES

1. Introduction: Consider a circle and a line PQ. There can be three possibilities as given: (i) PQ is a non-intersecting line; (ii) PQ is a Secant of the circle; (iii) PQ is a Tangent to the circle - There is only one point ‘A’, common to the line PQ & the circle.

2. A Tangent to a Circle is a line that intersects the circle at only one point. Activity: There is only one tangent that can be drawn at a point of the circle. The tangent to a circle is a special case of the secant, when the two end points of its corresponding chord coincide. The common point of the tangent and the circle is called the point of contact and the tangent is said to touch the circle at the common point.

Theorem 1.: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Proof: We are given a circle with centre O and a tangent XY to the circle at a point P. We need to prove that OP is perpendicular to XY.

The point Q must lie outside the circle. (Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius OP of the circle. That is, OQ > OP. Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY. Remarks : 1. By theorem above, we can also conclude that at any point on a circle there can be one and only one tangent. 2. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point.

3. Number of Tangents from a Point on a Circle: Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Activity: (i) All the lines through ’P’ (a point inside the circle) intersect the circle in two points. So, it is not possible

to draw any tangent to a circle through a point inside it. (ii) There is only one tangent that can be drawn to the circle at a point (say ‘P’).

(iii) Now taking a point P outside the circle and drawing tangents to the circle from this point. We find that we can draw exactly two tangents to the circle through this point (Say ‘P’). T1 and T are the points of contact of the tan gents PT & PT respectively. The length of the segment of the tangent from the external point P and the point of contact with the circle is called the length of the tangent from the point P to the circle. 2

1

2

Theorem 2.: The lengths of tangents drawn from an external point to a circle are equal. Proof: We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P. We are required to prove that PQ = PR. For this, we join OP, OQ and OR. Then ∠OQP and ∠ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 1. they are right angles. Now in right triangles OQP and ORP, OQ = OR (Radii of the same circle); OP = OP (Common); Thus, ΔOQP ≅ ΔORP (by RHS). Thus, by cpct, PQ = PR. Remarks: The theorem can also be proved by using the Pythagoras Theorem as follows: PQ = OP – OQ = OP – OR = PR (As OQ = OR), which gives PQ = PR; Also, ∠OPQ = ∠OPR. Therefore, OP is the angle bisector of ∠ QPR, ie. i.e., the centre lies on the bisector of the angle between the two tangents. 2

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CBSE CLASS X MATHEMATICS CHAPTER 10 CIRCLES - NCERT EXERCISES SOLUTIONS EXERCISE 10.1: Q1. How many tangents can a circle have?

A. A circle can have infinite tangents. Q2. Fill in the blanks : (i) A tangent to a circle intersects it in one point (s). (ii) A line intersecting a circle in two points is called a Secant . (iii) A circle can have two parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called Point of Contact . Q3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is: (A) 12 cm (B) 13 cm (C) 8.5 cm (D)√119 cm A. A line drawn from the centre of the circle to the tangent is perpen dicular to the tangent. Thus, OP PQ. Applying Pythagoras theorem in ΔOPQ, PQ = √119 cm Q4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. From the figure, AB & CD are two parallel lines. Line AB is intersecting the circle at exactly two points, P and Q. Therefore, line AB is the secant of this circle. Since line CD is intersecting the circle at exactly one point, R, line CD is the tangent to the circle.

EXERCISE 10.2:

Q1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm A. By Pythagoras Theorum, OP = 7cm. Q2. If TP and TQ are the two tangents to a circle with centre O so that ∠ POQ = 110° then ∠ PTQ is equal to (A) 60° (B) 70° (C) 80° (D) 90° A. TP and TQ are tangents => Radius drawn to these tangents are perpendicular to the tangents. Thus, OP⊥TP & OQ⊥TQ => ∠OPT = 90º & ∠OQT = 90º In quadrilateral POQT, Sum of all interior angles = 360 => ∠OPT + ∠POQ +∠OQT + ∠PTQ = 360 ⇒ 90 + 110º + 90 +∠PTQ = 360 ⇒ PTQ = 70 . 0

0

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Q3.If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to (A) 50° (B) 60° (C) 70° (D) 80° A. 90 + 80 + 90 + ∠BOA = 360 => ∠BOA = 100 . By RHS, ΔOPB ≅ ΔOPA => ∠BOP = ∠AOP => ∠AOP = ∠ POA = 50 . 0

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Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. A. If the co-Interior angles between the two lines are supplementary, then the lines are parallel. Q5. Prove that the perpendicular at the point of contact to the tangent of a circle passes through the centre. A. To Prove: The line perpendicular to AB at P passes through centre O. Proof: Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P. As perpendicular to AB at P passes through O’, therefore, ∠O’PB = 90 (i) O is the centre of the circle and P is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other. ∴ ∠OPB = 90° (ii) Comparing equations (1) and (2), we obtain ∠O’PB = ∠OPB (iii) This is only possible, when the line O’P coincides with OP. Therefore, the perpendicular to AB through P passes through centre O. Q6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. A. Let us consider a circle centered at point O. AB is a tangent drawn on this circle from point A. Given that, OA = 5cm and AB = 4 cm. In ΔABO, OB ⊥ AB (Radius ⊥ tangent at the point of contact). Applying Pythagoras theorem in ΔABO, we obtain AB + BO = OA => 42 + BO = 52 => BO = 3. Hence, the radius of the circle is 3 cm. 2

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Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. A. Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle => OA ⊥ PQ (As OA is the radius of the circle) Applying Pythagoras theorem in ΔOAP, we obtain OA + AP = OP => 32 + AP = 52 => AP = 4 In ΔOPQ, OA ⊥ PQ, AP = AQ (Perpendicular from the center of the circle bisects n the chord). Thus, PQ = 2AP = 2 × 4 = 8 => The length of the chord of the larger circle is 8 cm. 2

2

2

2

Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. A. It can be observed that DR = DS (Tangents on the circle from point D) … (i) CR = CQ (Tangents on the circle from point C) … (ii) BP = BQ (Tangents on the circle from point B) … (iii) AP = AS (Tangents on the circle from point A) … (iv) Adding all the equations, we get, DR + CR + BP + AP = DS + CQ + BQ + AS Thus, (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) => CD + AB = AD + BC Q9. XY and X' Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠ AOB = 90°. A. Join O to C. In ΔOPA and ΔOCA, OP = OC (Radii of the same circle) AP = AC (Tangents from point A) AO = AO (Common side) Thus, ΔOPA ΔOCA (SSS congruence criterion) Therefore, ∠POA = ∠COA … (i) Similarly, ΔOQB ΔOCB (by SSS) Thus, ∠QOB = ∠COB … (ii) Since POQ is a diameter of the circle, it is a straight line. Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º From equations (i) & (ii), we get, 2∠COA + 2∠COB = 180º =>∠COA + ∠COB = 90º =>∠AOB = 90° Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. A. Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle. It can be observed that OA (radius) ⊥ PA (tangent) => ∠OAP = 90° Similarly, OB (radius) ⊥ PB (tangent) => ∠OBP = 90° In quadrilateral OAPB, Sum of all interior angles = 360º Thus, ∠OAP +∠APB+∠PBO +∠BOA = 360º => 90º + ∠APB + 90º + ∠BOA = 360º Thus, ∠APB + ∠BOA = 180º Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Q11. Prove that the parallelogram circumscribing a circle is a rhombus. A. Let ABCD be the parallelogram, => AB = CD (i), & BC = AD (ii) It can be observed that, DR = DS (Tangents on the circle from point D), CR = CQ (Tangents on the circle from point C), Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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BP = BQ (Tangents on the circle from point B), AP = AS (Tangents on the circle from point A), Adding all these equations, we get, DR + CR + BP + AP = DS + CQ + BQ + AS => (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) => CD + AB = AD + BC Putting values of equations (i) and (ii) in this equation, we get, 2AB = 2BC => AB = BC (iii) Comparing equations (1), (2), and (3), we obtain AB = BC = CD = DA => Hence, ABCD is a rhombus. Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively . Find the sides AB and AC. A. Let the given circle touch the sides AB and AC of the triangle at point E and F, and, the length of the line segment AF be x. In Δ ABC, CF = CD = 6cm (Tangents on the circle from point C) BE = BD = 8cm (Tangents on the circle from point B) AE = AF = x (Tangents on the circle from point A) AB = AE + EB = x + 8; BC = BD + DC = 8 + 6 = 14; CA = CF + FA = 6 + x; 2s = AB + BC + CA (where ‘s’ = Semi-Perimeter) = x + 8 + 14 + 6 + x => s = 14 + x

Area of ΔOBC = 1/2 x 4 x 14 = 28 cm ; Area of ΔOCA = 1/2 x 4x(6 + x)=(12 + 2x)cm ;&, Area of ΔOAB=1/2 x 4x(8 + x)=16 + 2x cm ; Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB 2

2

Thus,

2

= 28 + 12 + 2x + 16 + 2x => (x + 14)(x - 7) = 0 => x = -14; or x = 7.

However, x = −14 is not possible as the length of the sides can’t be negative. Therefore, x = 7 => AB = x + 8 = 7 + 8 = 15 cm; and, CA = 6 + x = 6 + 7 = 13 cm. Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. A. To Prove: (∠1 + ∠2) + (∠5 + ∠6) = 180º & (∠3 + ∠4) + (∠7 + ∠8) = 180º Proof: Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at points P, Q, R, & S. Join the vertices of the quadrilateral ABCD to the center of the circle. Consider ΔOAP and ΔOAS, AP = AS (Tangents from the same point) OP = OS (Radii of the same circle) OA = OA (Common side) Thus, ΔOAP ≅ ΔOAS (SSS congruence criterion) => ∠POA = ∠AOS (cpct) => ∠1 = ∠8 Similarly, ∠2 = ∠3; ∠4 = ∠5; & ∠6 = ∠7. Now, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º => 2∠1 + 2∠2 + 2∠5 + 2∠6 = 360º => 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º => (∠1 + ∠2) + (∠5 + ∠6) = 180º => ∠AOB + ∠COD = 180º Similarly, we can prove that ∠BOC + ∠DOA = 180º Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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of the circle.

CBSE CLASS X MATHEMATICS CHAPTER 11 CONSTRUCTIONS 1. Division of a Line Segment: 1. To divide a line segment in a given ratio: Given a line segment AB, we want to divide it in the ratio m : n, where both m and n are positive integers. Let, m = 3 and n = 2. Method 1: Steps: Draw any ray AX, making an acute angle with AB, Locate 5 (= m + n) points on AX, such that, AA1 = A1A2 = A2A3 …., Join BA5, Through A3 (m = 3), draw a line parallel to BA5 (by making ∠AA3C Then, AC:CB = 3:2

= ∠AA B), intersecting AB at C, 5

Proof: Since A C is parallel to A B, thus, AA / A A = AC / CB (by BPT), By construction, AA / A A = 3/5; Thus, AC / CB = 3/5. Thus, ‘C’ divides AB in the ratio 3 : 2. Method 2: Steps: Draw any ray AX, making an acute angle with AB, 3

5

3

3

3

3

5

5

Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

92 Draw a ray BY parallel to AX by making ∠ABY equal to ∠ BAX, Locate the points A1, A2, A3 (m = 3) on AX and B1, B2 (n = 2) on BY such

that

AA = A A = A A = BB = B B , Let it intersect AB at a point C. Then AC : CB = 3 : 2. Proof: Δ AA C is similar to Δ BB C (by AAA), Thus, AA /BB =AC/BC, Now, by construction, AA /BB =3/2 => AC/BC=3/2 This method may be used for dividing the line segment in any ratio. 2. To construct a triangle similar to a given triangle as per given scale factor: The two different cases involved are that the triangle to be constructed is smaller than the given triangle, or, the triangle to be constructed larger than the given triangle. The scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle. Consider the following example: Ex. Construct a triangle similar to a given triangle ABC with its sides equal to 3/4 of the corresponding sides of the triangle ABC (i.e., of scale factor 3/4). A. Steps: 1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A, 2. Locate 4 points B , B , B and B on BX so that BB = B B = B B = B B , 3. Join B C and draw a line through B parallel to B C to intersect BC at C′, 4. Draw a line through C′ parallel to the line CA to intersect BA at A′ Then, Δ A′BC′ is the required triangle. 1

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Join A3B2. 3

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Proof: By construction, BC’/ C’C = 3/1, Thus, BC/BC’= (BC’ + C’C)/BC’ =

1 + C’C/BC’ = 1 + 1/3 = 4/3; ie. BC’/BC = 3/4 Also C′A′ is parallel to CA. Therefore, Δ A′BC′ ~ Δ ABC (by AA Criteria). Thus, A’B/AB = BC’/BC = A’C’/AC = 3/4.

Ex. Construct a triangle similar to a given triangle ABC with its sides equal to 5/3 of the corresponding sides of the triangle ABC (ie. Of scale factor 5/3).

A. Steps: 1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A,

2. Locate 5 points (the greater of 5 & 3 in 5/3) B , B , B , B & B on BX so that BB = B B = B B = B B = B B 3. Join B (the 3 point, 3 being smaller of 3 & 5 in 5/3) to C draw a line through B parallel to B C, intersecting the extended line segment BC at C’, 4. Draw a line through C’, parallel to CA intersecting the extended line segment 1

1

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rd

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5,

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BA at A’, Then, A’BC’ is the required triangle.

Proof: ΔABC ~ ΔA’BC’ (AA Criteria), Thus, AB/A’B = BC/BC’ = AC/A’C’, But, BC/BC’ = BB /BB = 3/5 => BC’/BC = 5/3; Thus, A’B/AB = BC’/BC = A’C’/AC = 5/3. In both the above examples, we could take a ray making an acute angle with AB or AC & proceed similarly. 3

5

2. Construction of Tangents to a Circle: Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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If a point lies on the circle, then there is only one tangent to the circle at this point and it is perpendicular to the radius through this point. Also, if the point lies outside the circle, there will be two tangents to the circle from this point. 1. To construct the tangents to a circle from a point outside it.

We are given a circle with centre O, and a point ‘P’ outside it. We have to construct the two tangents from ‘P’ to the circle. Steps: Join PO and bisect it. Let M be the midpoint of PO, Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at the points Q and R, Join PQ & PR, PQ & PR are the required two tangents. Proof: Join OQ. Then, ∠PQO is an angle in the semicircle and, therefore, ∠PQO = 90°, Thus, PQ ⊥ OQ, Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle. Note: If centre of the circle is not given, we may locate its centre first by taking any two non-parallel chords and then finding the point of intersection of their perpendicular bisectors. Then we could proceed as above.

CBSE CLASS X MATHEMATICS CHAPTER 11 CONSTRUCTIONS - NCERT EXERCISES SOLUTIONS EXERCISE 11.1: (P.219)

Q1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts. A. A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows: Steps: Draw line segment AB of 7.6 cm & draw a ray AX making an acute angle with line segment AB, Locate 13 (= 5 + 8) points, A , A , A , A …….. A , on AX such that AA = A A = A A …, Join BA , Through the point A , draw a line parallel to BA (by making an angle equal to ∠AA B) at A intersecting AB at point C. C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8. The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively. 1

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Proof: By construction, A C || A B. By applying Basic proportionality theorem for the triangle AA B, we get, 5

13

13

… (i) figure, it can be observed that AA and A A contain 5 and 8 equal divisions of line segments respectively.

From the

5

5

13

… (ii) On comparing equations (i) and (ii), we obtain This justifies the construction. Q2 & Q3: Please Do It Yourself! Q4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle. A. Let ΔABC be an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm. A ΔAB'C' whose sides are 3/2 times of ΔABC is drawn as: Steps: Draw AB of 8 cm. Draw perpendicular bisector of AB. Let OO' intersect AB at D; Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles

ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm; Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C; Locate 3 points (as 3 is greater between 3 and 2) A1, A2, and A3 on AX such that AA1= A1A2 = A2A3; Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment AB at point B'; Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. ΔAB'C'

is the

required triangle.

Proof: To justify the construction, we need to prove that, AB’ = 3/2AB, B’C’ = 3/2BC, AC’ = 3/2AC, In

ΔABC and ΔAB'C', ∠ABC = ∠AB'C' (Corres. angles), ∠BAC = ∠B'AC' (Com’n), ∴ ΔABC ∼ ΔAB'C' (AA crit.); => AB/AB’ = BC/B’C’ = AC/AC’ (i) In ΔAA B and ΔAA B', ∠A AB = ∠A AB' (Common), ∠AA B = ∠AA B' (Corresponding angles), ∴ ΔAA B ∼ ΔAA B' (AA similarity criterion) => AB/AB’ = AA /AA => AB/AB’ = 2/3 (ii) Comparing (i) & (ii), we get, 2

3

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3

3

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3

⇒ Q5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of triangle ABC. A. A ΔA'BC' whose sides are 3/4 of the corresponding sides of ΔABC can be drawn as follows: Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Steps: Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°, Draw BX making an acute angle with BC on the opposite side of vertex A, Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on BX’ Join B4C and draw a line through B3, parallel to B4C intersecting BC at C'‘ Draw a line through C' parallel to AC intersecting AB at A'. Then, ΔA'BC' is the required triangle. Proof: The construction can be justified by proving In ΔA'BC' and ΔABC, ∠A'C'B = ∠ACB (Corresponding angles), ∠A'BC' = ∠ABC (Common) ∴ ΔA'BC' ∼ ΔABC (AA crit.) => A’B/AB = BC’/BC = A’C’/AC (i) In ΔBB3C' and ΔBB4C, ∠B3BC' = ∠B4BC (Common), ∠BB3C' = ∠BB4C (Corres. angles) ∴ ΔBB3C' ∼ ΔBB4C (AA similarity criterion) => BC’/BC = BB3/BB4 (ii)(which equals 3/4, by construction) From equations (i) and (ii), we obtain ⇒

Q6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC. A. ∠B = 45°, ∠A = 105° => ∠C = 30° The required triangle can be drawn as follows: Steps: Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°, Draw a ray BX making an acute angle with BC on the opposite side of vertex A, Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX, Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C', Through C', draw a line parallel to AC intersecting extended line segment at A'. ΔA'BC' is the required triangle.

Proof: The construction can be justified by proving that In ΔABC and ΔA'BC', ∠ABC = ∠A'BC' (Common) ∠ACB = ∠A'C'B (Corresponding angles) ∴ ΔABC ∼ ΔA'BC' (AA similarity Crit.) => AB/A’B = BC/BC’ = AC/A’C’

(i)

In ΔBB3C and ΔBB4C', ∠B3BC = ∠B4BC' (Common), ∠BB3C = ∠BB4C' (Corresponding angles), ∴ ΔBB3C ∼ ΔBB4C' (AA similarity crit.) … (ii)

On comparing equations (i) and (ii), we obtain =>

Q7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. It is given that sides other than hypotenuse are 4 cm and 3 cm. Clearly, these will be perpendicular to each other (As Hypotenuse is the longest side). The required triangle can be drawn as: Steps: Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it, Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle, Draw a ray AX making an acute angle with AB, opposite to vertex C, Locate 5 points (as 5 is greater in 5 and 3), A1, A2, A3, A4, A5, on line segment AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5, Join A3B. Draw a line through A5 parallel to A3B intersecting extended line segment AB at B', Through B', draw a line parallel to BC intersecting extended line segment AC at C'. ΔAB'C' is the required triangle.

Proof: The construction can be justified by proving that In ΔABC and ΔAB'C', ∠ABC = ∠AB'C' (Corresponding angles), ∠BAC = ∠B'AC' (Common), ∴ ΔABC ∼ ΔAB'C' (AA similarity crit.) => AB/AB’ = BC/B’C’ = AC/AC’ (i) In ΔAA3B and ΔAA5B', ∠A3AB = ∠A5AB' (Common), ∠AA3B = ∠AA5B' (Corresponding angles) ∴ ΔAA3B ∼ ΔAA5B' (AA similarity criterion) => AB/AB’ = AA3/AA5 = 3/5 (by cons.)=> AB/AB’ = 3/5 (ii). On comparing equations (i) and (ii), we get, =>

EXERCISE 11.2: (P.221) Q1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. A. A pair of tangents to the given circle can be constructed as follows. Steps: Taking any point O of the given plane as centre, draw a circle of 6 cm radius. Locate a point P, 10 cm away from O. Join OP, Bisect OP. Let M be the mid-point of PO, Taking M as centre and MO as radius, draw a circle, Let this circle intersect the previous circle at point Q and R, Join PQ and PR. PQ and PR are the required tangents. The lengths of tangents PQ and PR are 8 cm each. Proof: The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 6 cm). For this, join OQ and OR, ∠PQO is an angle in the semi-circle, ∴ ∠PQO = 90° ⇒ OQ ⊥ PQ Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle. Q2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation. A. Tangents on the given circle can be drawn as follows:

Steps: Draw a circle of 4 cm radius with centre as O on the given plane, Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle and join OP,

Bisect OP. Let M be the mid-point of PO, Taking M as its centre and MO as its radius, draw a circle. Let it intersect the given circle at the points Q and R, Join PQ and PR. PQ and PR are the required tangents. It can be observed that PQ and PR are of length 4.47 cm each. In ΔPQO, Since PQ is a tangent, ∠PQO = 90°, PO = 6 cm, QO = 4 cm Applying Pythagoras theorem in ΔPQO, we get, PQ2 + QO2 = PQ2 =>PQ2 = 20 => PQ = 2√5 = 4.47 cm Justification: The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ and OR. ∠PQO is an angle in the semi-circle; Thus, ∠PQO = 90° ⇒ OQ ⊥ PQ Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle. Q3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. A. The tangent can be constructed on the given circle as: Steps: Taking any point O on the given plane as centre, draw a circle of 3 cm radius, Take one of its diameters, PQ, and extend it on both sides. Locate two points on this diameter such that OR = OS = 7 cm, Bisect OR and OS. Let T and U be the mid-points of OR and OS respectively, Taking T and U as its centre and with TO and UO as radius, draw two circles. These two circles will intersect the circle at point V, W, X, Y respectively. Join RV, RW, SX, and SY. These are the required tangents. Justification: The construction can be justified by proving that RV, RW, SY, and SX are the tangents to the circle (whose centre is

O and radius is 3 cm). For this, join OV, OW, OX, and OY, ∠RVO is an angle in the semi-circle, ∴ ∠RVO = 90° ⇒ OV ⊥ RV Since OV is the radius of the circle, RV has to be a tangent of the circle. Similarly, OW, OX, and OY are the tangents of the circle. Q4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. A. The tangents can be constructed as: Steps: Draw a circle of radius 5 cm and with centre as O, Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A, Draw a radius OB, making an angle of 120° (180° − 60°) with OA, Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°. Justification: The construction can be justified by proving that ∠APB = 60° By our construction ∠OAP = 90°; ∠OBP = 90°; And ∠AOB = 120° Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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We know that the sum of all interior angles of a quadrilateral = 360° Thus, ∠OAP + ∠AOB + ∠OBP + ∠APB = 360° 90° + 120° + 90° + ∠APB = 360° => ∠APB = 60° This justifies the construction.

Q5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. A. The tangents can be constructed on the given circles as follows:

Steps: Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius, Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AR. These are the required tangents. Justification: The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR. ∠ASB is an angle in the semi-circle. ∴ ∠ASB = 90°; ⇒ BS ⊥ AS Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents. Q6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. A. Consider the following situation. If a circle is drawn through B, D, and C, BC will be its diameter as ∠BDC is of measure 90°. The centre E of this circle will be the mid-point of BC. The required tangents can be constructed on the given circle as follows: Steps: Join AE and bisect it. Let F be the mid-point of AE, Taking F as centre and FE as its radius, draw a circle which will intersect the circle at point B and G. Join AG. AB and AG are the required tangents. Justification: The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG. ∠AGE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. ∴ ∠AGE = 90° ⇒ EG ⊥ AG Since EG is the radius of the circle, AG has to be a tangent of the circle. Already, ∠B = 90° ⇒ AB ⊥ BE Since BE is the radius of the circle, AB has to be a tangent of the circle.

Q7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle. A. The required tangents can be constructed on the given circle as follows: Draw a circle with the help of a bangle, Take a point P outside this circle and take two chords QR and ST, Draw perpendicular bisectors of these chords. Let them intersect each other

at point O, Join PO and bisect it. Let U be the mid-point of PO. Taking U as centre,

draw a circle of radius OU, which will intersect the circle at V and W. Join PV and PW. PV and PW are the required tangents. Justification: Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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The construction can be justified by proving that PV and PW are the tangents to the circle. For this, first of all, it has to be proved that O is the centre of the circle. Let us join OV and OW. We know that perpendicular bisector of a chord passes through the centre. Therefore, the perpendicular bisector of chords QR and ST pass through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. ∠PVO is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. ∴ ∠PVO = 90° ⇒ OV ⊥ PV Since OV is the radius of the circle, PV has to be a tangent of the circle. Similarly, PW is a tangent of the circle.

CBSE CLASS X MATHEMATICS CHAPTER 12 AREAS RELATED TO CIRCLES 1. Perimeter and Area of a Circle: The distance covered by travelling once around a circle is its perimeter, usually called its circumference. Circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter p (read as ‘pi’). Thus, circumference / diameter = , or, circumference = × diameter = × 2r (where r is the radius of the circle) =2r is an irrational number and its decimal expansion is non-terminating and non-recurring (nonrepeating). However, for practical purposes, we generally take the value of as 22/7 or 3.14 approximately. Area of a circle is r , where r is the radius of the circle. As shown, we cut a circle into a number of sectors and rearranging them. 2

We get a rectangle of length 1/2 x 2r and breadth ‘r’, which gives the area of the circle as r

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2. Areas of Sector and Segment of a Circle: The portion (or part) of the circular region enclosed by two Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle. Thus, the shaded region OAPB is a sector of the circle with centre O. ∠AOB is called the angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of the circle. OAPB is called the minor sector and OAQB is called the major sector. The angle of the major sector is 360° – ∠AOB. AB is a chord of the circle with centre O. So, shaded region APB is a segment of the circle (called the minor segment). You can also note that the un-shaded region AQB is another segment of the circle (called the major segment) formed by the chord AB. ((Unless otherwise specified, ‘Segment’ & ‘Sector’ would mean ’Minor Segment’ & ’Minor Sector’.)) Let OAPB be a sector of a circle with centre O and radius r. Let the degree measure of ∠AOB be θ. Now, area of a circle = r . In a way, we can consider this circular region to be a sector forming an angle of 360° (i.e., of degree measure 360) at the centre O. Now, consider this: When degree measure of the angle at the centre is 360, area of the sector = r So, when the degree measure of the angle at the centre is 1, area of the sector = r / 360 Thus, if the degree measure of the angle at the centre is θ, area of the sector = r /360 x θ, Or, θ/360 x r . Again, by applying the Unitary Method and taking the whole length of the circle (of angle 360°) as 2r, we get the required length of the arc APB as θ/360 x 2r. Area of the segment APB = Area of sector OAPB – Area of ΔOAB = θ/360 x r - Area of ΔOAB 2

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Area of combination of figures - An Example: Ex. Find the area of the shaded design, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. A. Let us mark the four un-shaded regions as I, II, III and IV. Area of I + Area of III = Area of ABCD – Areas of two semicircles of each of radius 5 cm = 10 x 10 - 2 x 1/2 x x 5 = 21.5 cm . Similarly, Area of II + Area of IV = 21.5 cm. Thus, area of the shaded design = Area of ABCD – Area of (I + II + III + IV) = (100 – 2 × 21.5) cm = 57 cm . 2

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CBSE CLASS X MATHEMATICS CHAPTER 12 AREAS RELATED TO CIRCLES - NCERT EXERCISES SOLUTIONS EXERCISE 12.1:

Q1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. A. 2r = 2x 19 +2 x 9 => ‘r’ = 28cm. Q2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles. A. Do It Yourself! Q3. An archery target is marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions. A. Do It Yourself! Q4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? A. Do It Yourself! Q5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is (A) 2 units (B) units (C) 4 units (D) 7 units A. Do It Yourself! Correct answer is 2. EXERCISE 12.2(Page 230): Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

A. Do It Yourself! (Ans. 132/7 cm ) 2

Q2. Find the area of a quadrant of a circle whose circumference is 22 cm. A. Do It Yourself! (Ans. 77/8 cm ) 2

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. A. Do It Yourself! (Ans. 154/3 cm ) 2

Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use = 3.14)

A. Area of major sector OADB [(360 - 90 )/360 ]x r = 235.5cm Area of minor sector OADB [90 /360 ]x r = 235.5cm = 78.5cm Area of ΔOAB = 1/2 x 10 x 10 = 50cm Area of minor segment ACB = 78.5 - 50 = 28.5cm 0

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Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) The length of the arc (ii) Area of the sector formed by the arc (iii) Area of the segment forced by the corresponding chord A. Length of arc ACB = 2r 360 x θ = 22cm Area of sector OACB = r /360 x θ = 231 cm In ΔOAB, ∠OAB = ∠OBA (As OA = OB); ∠OAB + ∠AOB + ∠OBA = 180°=> 2∠OAB + 60° = 180° => ∠OAB = 60° Therefore, ΔOAB is an equilateral triangle. Area of Equilateral ΔOAB = √3/4 x (side) = 441√3/4 cm Area of segment ACB=Area of sector OACB−Area of ΔOAB=[231 - 441√3/4] 0

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Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. A. Same as Above. Q7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. A. Draw Perpendicular OV on ST; and, use Cos60 & Sin60 to find OV & SV Area of segment SUT = Area of sector OSUT − Area of ΔOST = 150.72 − 62.28 = 88.44 cm 0

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Q8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find (i) the area of that part of the field in which the horse can graze, (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. A. Area that can be grazed by horse = Area of sector OACB = 19.625 m Area that can be grazed by the horse when length of rope is 10 m long = 90 /360 x r = 78.5m Increase in grazing area = (78.5 − 19.625) m = 58.875 m 2

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Q9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find : (i) the total length of the silver wire required, (ii) the area of each sector of the brooch. A. Length of wire required =Circumference + 5xdia. = 110 x 5 x 35 = 285mm Each of 10 sectors of the circle is subtending 36° at the centre of the circle, thus, area of Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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each sector = r /360 x 36 = 385/4 mm . 2

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Q10. An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. A. There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending 360 /8 = 45 at the centre of the assumed flat circle. Area between two consecutive ribs of circle = r /360 x θ = 22275/28 cm . 0

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Q11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. A. Area of each Sector = 158125/252 cm ; Area swept by two blades = 2 x 158125/252 = 158125/126 cm . 2

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Q12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. A. 189.97 km 2

Q13. A round table cover has six equal designs. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm A. From the figure, it can be seen, that, these designs are segments of a circle. Each chord is a side of a hexagon, and will subtend 60 at the centre of the circle. First prove that ΔOAB is an equilateral triangle. Area of ΔOAB = √3/4 x (side) = 196√3 cm . Area of Sector OAPB = 1232/3 cm . Now, first find ‘Area of segment APB’, and thence, ‘The Area of Designs = 6 x Area of segment’ = 464.8 cm . Cost of making the designs = 464.8 x 0.35 = Rs. 162.68 2 .

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Q14. Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is => A. r /360 x θ = 2r /720 x p = p/720 x 2r 2

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EXERCISE 12.3(Page 234): Q1. Find the area of the shaded region, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. A. Area of shaded region = Area of Semi-Circle - Area of Triangle QRP Now, in ΔQRP, using Pythagoras Theorum, RQ = RP + PQ = 625 cm. Thus, Area of shaded region = 1/2 x 2

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22/7 x (25/2) - 1/2 x 7 x 24 = 4523/28 cm . 2

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Q2. Find the area of the shaded region, if radii of the two concentric circles with centre O are 7 cm and 14 cm, and ∠ AOC = 40°. A. Area of shaded region = Area of sector OAFC − Area of sector OBED = 154/3 cm . 2

Q3. Find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles. A. Area of the shaded region = Area of square ABCD − Area of semi-circle APD − Area of semi-circle BPC

= 196 − 77 − 77 = 196 − 154 = 42 cm

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Q4. Find the area of the shaded region, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre. A. Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Q5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut. Find the area of the remaining portion of the square. A. Area of the shaded region = Area of square − Area of circle − 4 × Area of quadrant = 16 - 22/7 - 4 x 22/8 = 68/7 cm . 2

Q6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the design (shaded region). A. Area of design = Area of circle − Area of ΔABC = 22528/7 - 768√3 Q7. ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region. A. Area of shaded portion = Area of square ABCD − 4 × Area of each sector = 196 - 4 x 77/2 = 196 - 154 = 42 cm = Area of the Shaded Portion. 2

Q8. The figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find : (i) the distance around the track along its inner edge, (ii) the area of the track. A. (i) Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA = 106 + 1/2 x 2 r + 106 + 1/2 x 2 r = 2804/7 m. (ii) Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semi-circle HKI − Area of semi-circle BEC) + (Area of semi-circle GLJ − Area of semi-circle AFD) = 4320 m . 2

Q9. AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region. A. Area of the shaded region = Area of smaller circle + Area of semi-circle AECFB − Area of ΔABC

= 77/2 + 77 - 49 = 66.5 cm . 2

Q10. The area of an equilateral triangle ABC is 17320.5 cm . With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. A. Area of equilateral triangle = √3/4(a) = 17320.5 cm => a =200cm. Area of sector ADEF = 60 /360 x r = 15700/3 cm . 2

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Area of shaded region = Area of equilateral triangle − 3 × Area of each sector = 17320.5 - 15700 = 1620.5 cm2.

Q11. On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief. A. From the figure, side of the square = 42 cm. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Area of square = (Side) = 1764 cm Area of each circle = πr = 22/7 x (7) = 154 cm . Area of 9 circles = 9 × 154 = 1386 cm Thus, area of the remaining portion of the handkerchief = 1764 − 1386 = 378 cm . 2

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Q12. OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region. A. (i) Since OACB is a quadrant, it will subtend 90° angle at O. Thus, Area of quadrant OACB=90 /360 x r =1/4 x 22/7x(3.5) =77/8 cm (ii) Area of ΔOBD = 1/2 x OB x OD = 1/2 x 3.5 x 2 = 7/2 cm . Area of the shaded region = Area of quadrant OACB − Area of ΔOBD = 77/8 - 7/2 = 49/8 cm . Q13. A square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. A. In ΔOAB, OB = OA + AB = (20) + (20) => OB = 20√2 cm = Radius ( r ) of the circle. Area of quadrant OPBQ = 90 /360 x x (20√2) = 628 cm . Area of Square OABC = (Side) = (20) = 400 cm . Area of shaded region = Area of quadrant OPBQ − Area of OABC = (628 − 400) cm = 228 cm . 0

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Q14. AB and CD are arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ∠ AOB = 30°, find the area of the shaded region. A. Area of the shaded region = Area of sector OAEB − Area of sector OCFD = 30 /360 x 22/7 x (21 - 7 ) = 308/3 cm . 0

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Q15. ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. A. As ABC is a quadrant of the circle, ∠BAC is 90º. In ΔABC, BC = AC + AB = (14) + (14) => BC = 14√2 cm Radius (r ) of semi-circle drawn on BC = 14√2 /2 = 7√2 cm. Area of ΔABC = 1/2 x AC x AB = 1/2 x 14 x 14 = 98 cm . 2

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Q16. Calculate the area of the designed region common between the two quadrants of circles of radius 8 cm each. A. Area of Sector BAEC = 90 /360 x 22/7 x 8 = 352/7 cm . Area of ΔBAC = 1/2 x 8 x 8 = 32 cm . Area of the designed portion = 2 × (Area of segment AEC) = 2 × (Area of sector BAEC − Area of ΔBAC) = 2 x [352/7 - 32] = 256/7 cm . 0

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CBSE CLASS X MATHEMATICS CHAPTER 13 SURFACE AREA AND VOLUME Surface Area

& Volume of a Cuboid & a Cube: TSA (Cuboid) = 2(lb + bh + lh) Lateral Surface Area of Cuboid (LSA) = 2(l + b)h Volume (Cuboid) = lbh

TSA (Cube) 6a2 Lateral Surface Area of a cube = 4side . 2

Surface Area

& Volume of a Right Circular Cylinder: CSA = 2rh Total Surface Area of the Cylinder = 2πrh + 2πr = 2πr(r + h) Volume of a Cylinder = base area (ie. Area of circular base) x height = πr h Volume of the material of the hollow Cylinder = External Volume - Internal Volume = πR h - πr h = πh(R - r ) Surface Area & Volume of a Right Circular Cone: Slant Height (l) of a cone = (h + r ) CSA = rl TSA = r(l + r) Volume = 1/3 r2h 2

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Frustum of a Cone is the solid obtained after removing the upper portion of the cone by a plane parallel to the base. The lower portion is the frustum of the cone. Let h be the height, l the slant height and r and r the radii of the ends (r > r ) of the frustum of a cone. Remember the following formulae: Volume of the frustum of the cone = 1/3 h (r + r + r r ); CSA of the frustum of the cone = r + r )l, where, 1

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frustum of the cone = r + r )l + r +r , where, l is same as above

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& Volume of a Sphere:

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Volume of a Sphere = 4/3 πr ; Volume of the Hemisphere = 2/3 πr ‘Volume of the Material’, forming a hollow sphere = 4/3 π(R - r ) 3

3

3

3

Exercise 13.1: Do it yourself Exercise 13.2: Do it yourself Exercise 13.3: Q8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed? A. Area of cross-section of the canal = 6 × 1.5 = 9 m Speed of water = 10 km/h = 10000/60 metre/min. Volume of water that flows in 1 minute from canal = 9 x 10000/60 =1500 m . Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m . Let the irrigated area be A. Volume of water irrigating the required area will be equal to the volume of water that flowed in 30 minutes from the canal. Vol. of water flowing in 30 minutes from canal = Vol. of water irrigating the reqd. area => A = 562500 m , Therefore, area irrigated in 30 minutes is 562500 m . 2

3

3

2

2

Exercise 13.4: Do it yourself

EXERCISE 13.5 (Optional) Q 1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm . A. SKIP IT Length of wire = ‘l’; area of wire = l x 3/10 cm - NOT PRACTICABLE IF THE SHAPE OF THE WIRE IS CYLINDRICAL! Q2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. A. Use similar triangle criterion; Volume of double cone = 10.86 + 19.31 = 30.17 cm . Surface area of double cone = 22.63 + 30.17 = 52.80 cm . Q 3. A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm? A. 129600 + n x (22.5 cm × 7.5 cm × 6.5 cm)(1 - 1/17) = 150 x 120 x 110 => n = 1792 Q 4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km , show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep. A. DO IT YOURSELF! Q 5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel. A. DO IT YOURSELF! Q 6. Derive the formula for the curved surface area and total surface area of the frustum of a cone. A. SKIP IT PLEASE! Q 7. Derive the formula for the volume of the frustum of a cone A. SKIP IT PLEASE! 3

2

3

2

3

2

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CBSE CLASS X MATHEMATICS CHAPTER 14 STATISTICS 1. Introduction: ‘Statistics’ is defined as the science of ‘Collection’, ‘Organisation’, Presentation’, ‘Analysis’ and, ‘Interpretation’ of numerical data. 2. Series: There are two types of Series: Inclusive or Discrete Series: When the Class-Intervals are so fixed that the upper limit of the class is included in the same class interval, such series are called ’Inclusive Series’. Exclusive or Continuous Series: The Upper Limit of one class is the Lower Limit of the next Class. The common point of the two classes is included in the Higher Class. Converting a Discontinuous Series into a Continuous Series: Let, ‘h’ = Lower Limit of 2 Class Interval - Upper Limit of 1 Class Interval. nd

st

Then, Subtract h/2 from the Lower Limits of each Class, and Add h/2 to the Upper Limits of each Class.

3. Arithmetic Mean of Grouped or Un-grouped Data: The sum of the values of all the observations = f x + f x +….+ f x , and, the number of observations = f + f +….+ f So, the mean x-bar of the data is given by, x-bar = (f x + f x +….+ f x )/(f + f +….+ f ); Or, ……. 1

1

2

1

2

2

n

n

n

1

1

2

2

n

n

1

2

n

We may convert the Un-grouped Data into Grouped Data by forming Class Intervals of suitable width (Always remembering, that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class). Arithmetic Mean by Direct Method: Used when data is grouped and the classes are provided. It is assumed that the frequency of each class interval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. Class mark (or mid-point of a class) = (Upper class limit + Lower class limit)/2

Arithmetic Mean by Assumed Mean (or, Deviation) Method: Used when the numerical values of x & f are large. May be used, weather the data is grouped or ungrouped. For grouped data, choose class--mark as x . Follow the following steps: 1

1

i

Choose one among the x’s as the assumed mean, and denote it by ‘a’, We may take ‘a’ to be that xi which lies in the centre of x1, x2, …… xn , Find the difference di (= xi - a) between a and each of the xi’s, that is, the deviation Find the product of di with the corresponding fi, and take the sum of all the fidi’s, Find

of ‘a’ from each of the x,

the mean of the deviation, as, In

obtaining d , we subtracted ‘a’ from each x , so, in order to get the mean x-bar , we need to add ‘a’ to d . Mathematical Explanation follows: ==> It can be verified, that the value of the mean obtained does not i

i

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Arithmetic Mean by Step Deviation Method: While solving a question, it will be observed that the column (d = x - a) has a common factor ‘h (the class size)’; and, hence dividing the values in the column (d = x - a), by ‘h’, we will get smaller numbers to multiply with f . Let u = (x - a)/h, where, where a is the assumed mean and h is the class size. i

i

i

i

i

i

i

Mathematical Explanation: We have, u = (xi - a)/h.

Let ,

i

Consider the following: The step-deviation method will be convenient to apply if all the d ’s have a common factor, The mean obtained by all the three methods is the same, The assumed mean method and step-deviation method are just simplified forms of the direct method,

Weighted Mean: The term ‘weight’ stands for relative importance of different items. ‘Rate’ assigned to an ‘item’ is proportional to the importance of the ‘item’. If x , x , …., x denote ‘n’ values of a variable ‘x’, and, w , w , …. ,w denote their weights, then …. ……… 1

2

n

1

2

n

Combined Mean: If x -bar & x -bar are means of two groups computed from n & n , then, combined mean x-bar of variate values of two groups taken together is computed as: 1

2

1

2

x-bar = (n x -bar + n x -bar)/(n +n ) 1

1

2

2

1

2

4. Measures of Central Tendency: 1. Mean: Done Above,

2. Mode (of grouped data) is the size of variable which occurs most frequently. It is possible that more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal. In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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mode is a value inside the modal class, and is given by the formula: ; where, l = lower limit of the modal class, h = size of the class interval (assuming all class sizes to be equal), f = frequency of the modal class, f = frequency of the class preceding the modal class, f = frequency of the class succeeding the modal class. 1

0

2

Note: It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the average of the marks obtained by most of the students. In the first situation, the mean is required and in the second situation, the mode is required.

3. Median is a measure of central tendency which gives the value of the middle-most observation in the data. For finding the median of ungrouped data, we first arrange the data values of the observations in th n+1 ascending or in descending order. Then, if n is odd, the median is the observation. And, if ‘n’ 2

( )

is even, then the median will be the average of the

n th and the 2

th

[( ) ] n +1 2

observations.

Median of Grouped data: Consider a grouped frequency distribution of marks obtained, out of 100, by 53 students, in a certain examination: Computing the cumulative frequency of the classes (Less Than Type), we get:

Similarly, we can compute cumulative frequency distribution of the more than type. To find the median of a grouped data, we can make use of any of these cumulative frequency distributions. Now in a grouped data, we may not be able to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves. n . To find this class, we find the cumulative frequencies of all the classes and 2 We now locate the class whose cumulative frequency is nearest to

n . 2

This is called the ‘median class’.

In the distribution above, n = 53. So, n/2 = 26.5. Now 60 – 70 is the class whose cumulative frequency 29 is nearest to n/2, i.e., 26.5. Therefore, 60 – 70 is the median class. Thence, we use the following formula to find the median:

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Where, l = lower limit of median class, n = number of observations, cf = cumulative frequency of class preceding the median class, f = frequency of median class, h = class size (assuming class size to be equal). Computing , we get median as 66.4. Hence, about half the students have scored marks less than 66.4, and the other half have scored marks more 66.4. 4A. Which measure would be best suited for a particular requirement: The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes, i.e., the largest and the smallest observations of the entire data. It also enables us to compare two or more distributions. For example, by comparing the average (mean) results of students of different schools of a particular examination, we can conclude which school has a better performance. However, extreme values in the data affect the mean. For example, the mean of classes having frequencies more or less the same is a good representative of the data. But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20, 21, 18, then the mean will certainly not reflect the way the data behaves. So, in such cases, the mean is not a good representative of the data. In problems where individual observations are not important, and we wish to find out a ‘typical’ observation, the median is more appropriate, e.g., finding the typical productivity rate of workers, average wage in a country, etc. These are situations where extreme values may be there. So, rather than the mean, we take the median as a better measure of central tendency. In situations which require establishing the most frequent value or most popular item, the mode is the best choice, e.g., to find the most popular T.V. programme being watched, the consumer item in greatest demand, the colour of the vehicle used by most of the people, etc. Note 1: 3 Median = Mode + 2 Mean Note 2: The median of grouped data with unequal class sizes can also be calculated. However, that is beyond the scope of this class. 5. Graphical Representation of Cumulative Frequency Distribution: Consider the following (cumulative frequency distribution) table:

The values 10, 20, 30, . . ., 100 are the upper limits of the respective class intervals. To represent the data in the table graphically, we mark the upper limits of the class intervals on the ‘x-axis’ and their corresponding cumulative frequencies on the ‘y-axis’, choosing a convenient scale. The scale may not be the same on both the axis. Let us now plot the points corresponding to the ordered pairs given by (upper limit, corresponding cumulative frequency), i.e., (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45), (100, 53) on a graph paper and join them by a free hand smooth curve. The curve we get is called a cumulative frequency curve, or an ogive (of the less than type).

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Now, Consider the following (cumulative frequency distribution) table: 0, 10, 20, . . ., 90 are the lower limits of the respective class intervals 0 - 10, 10 - 20, . . ., 90 - 100.

To represent ‘the more than type’ graphically, we plot the lower limits on the x-axis and the corresponding cumulative frequencies on the y-axis. Then we plot the points (lower limit, corresponding cumulative frequency), i.e., (0, 53), (10, 48), (20, 45), (30, 41), (40, 38), (50, 35), (60, 31), (70, 24), (80, 15), (90, 8), on a graph paper, and join them by a free hand smooth curve. The curve we get is a cumulative frequency curve, or an ogive (of the more than type).

Summary on how to go about drawing a Cumulative Frequency Curve: Cumulative Frequency Curve (Ogive) is the graphical representation of a cumulative frequency distribution. It can be drawn in the two ways: ‘Less than Method’, and, ‘More than Method’. Less than Method: If the frequency is in inclusive form, convert it into exclusive form, Construct a cumulative frequency table, Mark upper class limits along the ‘x’ axis, Mark the corresponding cumulative frequency along the ‘y’ axis, Plot the points and join them by a free hand curve, The lower limit of the first class interval becomes the upper limit of the imagined class with frequency ‘0’. Join the imagined point (lower limit of first, 0) with the first point of the curve and so on. We get the required curve called ‘Ogive’. More than Method: Convert the frequency distribution into more than type cumulative frequency distribution by subtracting the frequency of each class from the total frequency, Mark the lower class limits on x-axis, Mark the corresponding cumulative frequency on ‘y’ axis, Plot the points and join them by a free hand curve. Obtaining Median from the Cumulative Frequency Curves: First Method: Locate n/2 = 53/2 = 26.5 on ‘y’ axis. From this point, draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis determines the median of the data. Second Method: Draw both ogives (i.e., of the less than type and of the more than type) on the same axis. The two ogives will intersect each other at a point. From this point, if we draw a perpendicular on the x-axis, the point at which it cuts the x-axis gives us the median. N.B.: For drawing ogives, it should be ensured that the class intervals are Continuous.

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CBSE CLASS X MATHEMATICS CHAPTER 14 STATISTICS - NCERT EXERCISES SOLUTIONS EXERCISE 14.1 (P. 270): Q1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. (Table not drawn) Which method did you use for finding the mean, and why? A. Class Mark (x )= (Upper Class Limit + Lower Class Limit)/2.We thus find x & f i

i

i:

From the table: ∑f = 20; ∑fx = 162; x-bar = 162/20 = 8.1 Therefore, mean number of plants per house is 8.1. Here, direct method has been used as the values of class marks (x ) and f are small. i

i

Q2. Consider the following distribution of daily wages of 50 workers of a factory. (Table not drawn). Find the mean daily wages of the workers of the factory by using an appropriate method. A. Class Mark x = (Upper Class Limit + Lower Class Limit)/2. Class size (h) of this data = 20. Here, ‘a’ (assumed mean) = 150. d = (x - a); ∑f = 50; ∑f u = -12 i

i

i

i

i

Therefore, the mean daily wage of the workers of the factory is Rs 145.20.

Q3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. (Table not drawn). Find the missing frequency f. A. From the table, 18 = (752 + 20f) / (44 + f) => 792 + 18f = 752 + 20f => 40 = 2f = f = 20. NOW, PLEASE REDO THE QUESTION BY DEVIATION METHOD ALSO. Q4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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recorded and summarised as follows. (Table not drawn). Find the mean heart beats per minute for these women, choosing a suitable method. A.

Number of heart beats per minute

Number of women fi

xi (Class Mark)

di = xi − 75.5

65 − 68

2

66.5

−9

−3

−6

68 − 71

4

69.5

−6

−2

−8

71 − 74

3

72.5

−3

−1

−3

74 − 77

8

75.5

0

0

0

77 − 80

7

78.5

3

1

7

80 − 83

4

81.5

6

2

8

83 − 86

2

84.5

9

3

6

Total

30

fu i

i

4

Class Size (h) = 3; Let assumed mean = ‘a’;

Therefore, mean hear beats per minute for these women are 75.9 beats per minute. Q5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? A. The Class-Intervals are not continuous. Here, ‘g (gap between the two intervals)’ = 1; ‘g/2’ = 1/2, which we subtract from the lower class, and add to the upper class. Class-Size (h) = 3. Let, Assumed Mean (a) = 57. Class interval

f

i

x (Class - Mark) i

d=x -a = x − 57 i

i

fu i

i

i

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115 49.5 − 52.5

15

51

−6

−2

− 30

52.5 − 55.5

110

54

−3

−1

− 110

55.5 − 58.5

135

57

0

0

0

58.5 − 61.5

115

60

3

1

115

61.5 − 64.5

25

63

6

2

50

Total

400

25

Mean number of mangoes kept in a packing box is 57.19. Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all di. Q6. The table below shows the daily expenditure on food of 25 households in a locality. (Table not drawn). Find the mean daily expenditure on food by a suitable method. A. Class Size = 50. Assumed Mean(a) = 225. Mean daily expenditure on food is Rs 211. Q7. To find out the concentration of SO in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: (table not drawn) Find the mean concentration of SO in the air. 2

2

A. Mean concentration of SO in air = x - bar = ∑f x / ∑f = 0.09867 = 0.099 ppm. 2

i

i

Do Q.8 & Q.9 yourself. EXERCISE 14.2 (P. 275): Q1. The following table shows the ages of the patients admitted in a hospital during a year: (Table not Drawn). Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. A. Let Assumed Mean (a) = 30; Mean = 35.38; Mode = age of maximum number of patients admitted in hospital is 36.8 years. Q2, Q3, Q4….. Do It Yourself! Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data A. Do it Yourself! Mode of the data = 4608.695 ~ 4608.7 runs Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data : (Table not Drawn) A. Do it Yourself! Mode of the data = 44.7 cars. EXERCISE 14.3 (P. 275): Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. A. Finding Mean: Let, Assumed Mean (a) = 135. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Using Step-Deviation Method = x-bar = 137.058 Finding Mode: Modal Class = 125 - 145; l = 125; h = 20; f = 13; f = 20; f = 14. Solving, we get, mode = 135.76 Finding Median: For this, we first find the cumulative frequency – Draw the table. Thus, n = 68; Cumulative frequency (cf) just greater than n/2 (ie., 68/2 = 34) is 42, belonging to class 125-145; (Lower Limit of the median class)l = 125; Class size (h) = 20; Frequency (f) of median class = 20; ‘cf’ of class preceding median class =22 ……….. Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively. The three measures are approximately the same in this case. 0

1

2

Q2. If the median of the distribution given below is 28.5, find the values of x and y. A. From the table, n = 60 Also, 45 + x + y = 60, => x + y = 15 (i), Median of the data (28.5) lies in the interval 20 - 30; which, thus, is the median class.l = 20; cf of class preceding the median class = 5 + x; Frequency (f) of the median class = 20; Class size (h) = 10; Median = ……. Solving, x = 8; Putting this in (i), we get, 8 + y = 15 => y = 7. Q3 to Q7: DO IT YOURSELF!

EXERCISE 14.4 (P. 293): Q1 – Q3: DO IT YOURSELF!

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CBSE CLASS X MATHEMATICS CHAPTER 15 PROBABILITY The uncertainty of ‘probably’ etc can be measured numerically by means of ‘probability’ in many cases.

‘Outcome’ signifies the possible outcomes in an activity. A ‘trial’ is an action which results in one or several outcomes.

An ‘Event’ for an experiment is the collection of some outcomes of the experiment. The chance of happening of an event when expressed quantitatively is called ‘Probability’. We have two definitions of probability: Statistical or Empirical; Mathematical. ‘Statistical or Empirical Probability’ is found on the basis of what we directly observe as the outcomes of our trial. It is based on the results of actual experiments and adequate recordings of the happening of the events. Moreover, these probabilities are only ‘estimates’. If we perform the same experiment for another 1000 times, we may get different data giving different probability estimates. ‘Mathematical Probability’ is the numerical measure of the degree of Uncertainty/Certainty of the occurrence of events. Let n be the total number of trials. The empirical probability P(E) (or just ‘Probability’) of an event E happening, is P(E) = Number of trials in which the event happened ; The total number of trials

Number of outcomes favourable ¿ Or, P ( E )=¿ E Number of all possible outcomes of the experiment

N.B. => The probability of each event lies between 0 and 1; ie. 0 ≤ P(E) ≤ 1 => The sum of all the probabilities is 1. Understand & Remember the meaning of the following terms: Equally Likely Terms (Each outcome is as likely to occur as the other), Impossible Event: Probability of an Impossible event is always zero, Sure Event: Probability of an Impossible event is always ‘One’.

Note the following: In general, it is true that for an event E, P( E ) = 1 – P(E); The event E , representing ‘not E’, is called the complement of the event E. We also say that E and E are complementary events.

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CBSE CLASS X MATHEMATICS CHAPTER 15 PROBABILITY - NCERT EXERCISES SOLUTIONS EXERCISE 15.1 (P. 308): Q1. Complete the following statements: DO IT YOURSELF! Q2. Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. A. It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same. (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. A. It is not an equally likely event, as it depends on the player’s ability and there is no information given about that. (iii) A trial is made to answer a true-false question. The answer is right or wrong A. It is an equally likely event. (iv) A baby is born. It is a boy or a girl. A. It is an equally likely event. Q3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? A. When we toss a coin, the possible outcomes are only two, head or tail, which are equally likely outcomes. Therefore, the result of an individual toss is completely unpredictable. Q4. Which of the following cannot be the probability of an event? (i) 2/3; (ii) - 1.5; (iii) 15%; (iv) 0.7 A. Probability of an event (E) is always greater than or equal to 0. Also, it is always less than or equal to one. This implies that the probability of an event cannot be negative or greater than 1. Therefore, out of these alternatives, −1.5 cannot be a probability of an event. Hence, (ii). Q5. If P(E) = 0.05, what is the probability of ‘not E’? A. The probability of ‘not E’ is 0.95. Q6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy? A. (i) The bag contains lemon flavoured candies only. It does not contain any orange flavoured candies. This implies that every time, she will take out only lemon flavoured candies. Therefore, event that Malini will take out an orange flavoured candy is an impossible event. Hence, P (an orange flavoured candy) = 0 (ii)As the bag has lemon flavoured candies, Malini will take out only lemon flavoured candies. Therefore, event that Malini will take out a lemon flavoured candy is a sure event. P (a lemon flavoured candy) = 1. Q7 – Q13: DO IT YOURSELF! Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour A. P (getting a king of red colour) = 2/52 = 1/26 (ii) a face card A. Total number of face cards = 12 P (getting a face card)=(No. of favourable outcomes)/(No. of ttl possible outcomes)=12/52 = 3/13 (iii) a red face card Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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A. P (getting a red face card) = 6/52 = 3/26 (iv) the jack of hearts A. P (getting a jack of hearts) = 1/52 (v) a spade A. P (getting a spade) = 13/52 = 1/4 (vi) the queen of diamonds A. P (getting a queen of diamonds) = 1/52. Q15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? A. Total number of cards = 5; Total number of queens = 1; P (getting a queen) = 1/5 (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen? A. (a) P (getting an ace) = 1/4; (b) P (getting an queen) = 0/4 = 0 Q16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one. A. Total number of pens = 12 + 132 = 144; Total number of good pens = 132 P (getting a good pen) = 132/144 = 11/12 Q17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? A. P (getting a defective bulb) = 4/20 = 1/5; (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ? A. Remaining total number of bulbs = 19; Remaining total number of non-defective bulbs = 16 − 1 = 15; thus, P (getting a not defective bulb) = 15/19 Q18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number A. 81/90; (ii) a perfect square number A. Favourable outcomes = 1,4,9,16,25,36,49,64,81 => P (getting a perfect square number) = 9/90 = 1/10 (iii) a number divisible by 5. A. Favourable outcomes = 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90 => P (getting a number divisible by 5) = 18/90 = 1/5 Q19. A child has a die whose six faces show the letters as given below: The die is thrown once. What is the probability of getting (i) A? (ii) D? A. (i) P (getting A) = 2/6 = 1/3 (ii) P (getting D) = 1/6 Q20. Suppose you drop a die at random on the rectangular region as shown. What is the probability that it will land inside the circle with diameter 1m? A. Area of rectangle = l × b = 3 × 2 = 6 m Area of circle (of diameter 1 m) = / 4 m . P (die will land inside the circle) =(/4)/6 = /24. 2

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Q21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it ? A. Total no. of pens = 144; Total no. of defective pens = 20 => Total number of good pens = 124 (i) Probability of getting a good pen = 124/144 = 31/36. Thus, P (Nuri buys a pen) = 31/36. (ii) She will not buy it ? A. P (Nuri will not buy a pen) = 1 - 31/36 = 5/36 Q22. Two dice, one blue and one grey, are thrown at the same time. (i) Write down all the possible outcomes and complete the following table: Event: Sum of two dice

2

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11

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(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify. A. Probability of each of these sums will not be 1/11 as these sums are not equally likely. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die:

Event: Sum of two dice

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Q23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. A. The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT} Number of total possible outcomes = 8 Number of favourable outcomes = 2 {i.e., TTT and HHH} P (Hanif will win the game) = 2/8 = 1/4 P (Hanif will lose the game) = 1 - 1/4 = 3/4. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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Q24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? A. Total number of outcomes = 6 × 6 = 36 Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5). Hence, total number of favourable cases = 11 P (5 will come up either time) = 11/36. P (5 will not come up either time) = 1 - 11/36 = 25/36. (ii) 5 will come up at least once? A. Total no. of cases, when 5 can come at least once=11=> P(5 will come at least once)=11/36. Q25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes —two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3. A. Incorrect; When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways − (H, T), (T, H). Therefore, the probability of getting two heads is 1/4, the probability of getting two tails is 1/4, and the probability of getting one of each is 1/2.

Thus, for each outcome, the probability is not 1/3. (ii) If a die is thrown, there are two possible outcomes−−an odd number or an even number. Therefore, the probability of getting an odd number is 1/2. A. Correct; When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these, 1, 3, 5 are odd and 2, 4, 6 are even numbers. Therefore, the probability of getting an odd number is 1/2.

EXERCISE 15.2 (Optional) Q1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) different days ? (iii) consecutive days? Ans. Total possible ways of visiting shop by them = 5 x 5 = 25 (Prepare a matrix) (i) They can visit the shop on all week days Tuesday to Saturday. Favourable outcomes of visiting shop by them on the same day = 5 P (visiting shop same day) = 5/25 = 1/5 (ii) Favourable outcomes of visiting shop on different days = 25 - 5 = 20 => P = 20/25 = 4/5 (iii) Favourable outcomes of visiting shop by them on consecutive days: SHYAM Tue Wed Thu Fri EKTA Tue Wed Thu Fri EKTA Wed Thu Fri Sat SHYAM Wed Thu Fri Sat Thus, favourable outcomes = 8 Thus, P(visiting shop on consecutive days) = 8/25 Q 2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws: (COMPLETE THE TABLE!)

What is the probability that the total score is (i) even? (ii) 6? (iii) at least 6? A. (i) P (even) = 18/36 = 1/2 (ii) P (getting total score 6) = 4/36 = 1/9 (iii) P (getting at least 6 => getting 6 or more) = 15/36 = 5/12 Q 3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP

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a red ball, determine the number of blue balls in the bag. A. Let no. of blue balls = x => Total no. of balls = 5 + x P (getting blue ball) = x/(5 + x); P (getting red ball ) = 5/(5 + x) As per question, x/(5 + x) = 2 x 5/(5 + x) => x = 10 = no. of blue balls. Q 4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x. A. (i) x/12; (ii) (x + 6)/18 = x/6 => x = 3 Q 5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue balls in the jar. A. Number of green marbles = 16; and, Number of blue marbles = 8.

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