CBSE - Class x Maths - Notes & Solution

1CBSE 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 n 0 n 0-n -n  Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP 2 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 Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP 3  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 m n m 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 1 1 1 1 1 1 1 1 2 2 2 2 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 1 2 1 2 2 3 3 n n 1 1 2 3 2 n 2 2 2 1 3 2 2 n 2 3 1 2 3 n 2 n 2 2 1 2 3 n 1 2 3 n Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP 000 6  In a or a . x = 1/x . 7 = 7 …… 1 1 1  A ‘Negative Index’ implies the Reciprocal of a number.000. ‘The Exponent’ is taken to be as 1. √b]. a = 1/a .  (√a .  (ab) = a x b [(2x5) = 2x5 x 2x5 x 2x5 = 2x2x2x5x5x5 = 2 x 5 ]. ie.a = a [3 .4 HANDLING INDICES 1. ie. x = x . ie. 3√3 = 3 [A ‘Third root of 3’.  a0 = 1. Thus. m n 0 0 0  When no exponent is mentioned on to a number. or ‘Square root of 2’. x =1. or simply. Anything raised to the power 0 is One (1). a = 1/a .(√b) = (a ) . -2 2 -4 4 -3 3 n -n -n n  A ‘root’ is presented as such: √2 = 2 [A ‘second root of 2’. Residence: Ghaziabad-UP . 3 = 1/3 = 1/9. a /a = a [5 / 5 = (5x5x5x5x5x5) / (5x5x5x5) = 5x5 = 5 = 5 ]. or ‘Cube root of 3’].  (a ) = a [(7 ) = (7x7x7) = 7x7x7 x 7x7x7 = 7 = 7 ].  √a/b = √a / √b [(a/b) = a / b = √a / √b]. 3 = 3 .b = a .b = a .√b) x (√a + √b) = (√a) .000 = Ten trillion. Mobile (where u may transfer cash!) . 7x7x7x7x7x7x7x7 = 7 .000.09871823473. 3 =1. ‘a’ is the ‘base’.com. n√x = x [ n root of x] 1/2 1/3 1/n th 2. Laws of Indices: a . Applying Indices to Roots:  √ab = √a √b [(ab) = a .  (a/b) = a / b [ (3/5) = 3/5 x 3/5 = (3x3) / (5x5) = 3 / 5 ] [ 1/a = a / a = a = a ] m n m+n m n m n 2 m-n 6 mn m m 3 5 4 2 3 2 m m 2+3 2 m 3 m 2 6 3x2 3 2 n 0 6-4 n 3 2 0-n -n 3. Introduction:  Why Exponents were Invented: 10. 10 =1….manojarora23@gmail. under-root of 2]. b = √a .000.b [ (√a . = 6 x 1.b] 1/2 1/2 1/2 1/2 1/2 1/2 2 1/2 x 2 1/2 x 2 1 2 1/2 2 1/2 2 1 Benefitted: My Paypal a/c . 2 = 1/2 = 1/8.√b) x (√a + √b) = a .000. 3 = 3x3x3x3x3 = 3 = 3 ]. or simply 10 .000 = 6 x 10 13 8 6. and ‘m & n’ are called the exponents.000.(b ) = a . where m = 3q + 2q. maximum number of columns is 8. (Divisibility Criteria: If a number is divided by 2 & 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. 3m + 1. R = 0. 6q + 5 is not divisible by 2. where m = 3q + 4q + 1. 1. Q2. 6q +1 is not divisible by 2. ie. r = 0. (3m) = 27m = 9(3m ) = 9q. The degree of ‘r’ is always less than the degree of the ‘divisor’) Putting ‘b’ = 6 in (i). therefore. 6q + 2 is even and is divisible by 2.5 CBSE CLASS X MATHEMATICS CHAPTER 1 REAL NUMBERS . Mobile (where u may transfer cash!) . By Euclid’s division Lemma. Use Euclid’s division algorithm to find the HCF of :(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255. 4. From (iv). A. 90 = 45 X 2 + 0. we get a = 3q + r. 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. or 6q + 5. Show that any positive odd integer is of the form 6q + 1. 6q is divisible by 6 => 6q is even. 6q + 3 is not divisible by 2. 0 ≤ r < b (i). Let ‘x’ be any positive integer. 3(3q + 4q + 1) + 1 is a square of the form 3m + 1. (3m + 1) = 27m + 27m + 9m + 1 = 9(3m + 3m + m) + 1 = 9q + 1. 102 = 51 X 2 + 0. By Euclid’s division algorithm. Q4. 9q + 12q + 4. (iii) If r = 2. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m. a = 6q. or 6q + 3. Q3. we have to prove that the cube of each of these can be rewritten in the form 9q. we have to find the HCF of 616 and 32. From (iii).1 (P. ie. HCF of 616 & 32 is 8. a = 3q => a = 9q . (ii) If r = 1. 2. where ‘q’ = 3m + 3m + m. 255 = 102 X 2 + 51. 9m + 1 or 9m + 8. (3m + 2) = 27m + 54m + 36m + 8 = 9(3m + 3m + m) + 8 = 9q + 8. 102) = HCF (102. 1. 7) Q1.com. 9q + 6q + 1. HCF = 45 (ii) HCF = 196 (iii) 867 = 255 X 3 + 102. it is of the form 3m. Now. 0 ≤ r < 3. (i) 225 = 135 X 1 + 90. a = 6q + 4. 51) = 51. 6q + 5 are odd. a = 6q + 5. a = 6q + 1. where 2 2 2 2 2 2 2 3 3 3 3 3 3 3 2 3 2 3 2 2 3 2 3 2 Benefitted: My Paypal a/c . a = bq + r. 9q + 1. If ‘r’ = 4. A. The two groups are to march in the same number of columns. Q5. By Euclid’s division algorithm. 6q + 1. What is the maximum number of columns in which they can march? A. Residence: Ghaziabad-UP . now. ie. a = 6q + 3. (iv) 2 2 2 2 2 2 From (ii). If ‘r’ = 3. a = 3q + 1 => a = 9q + 6q + 1.NCERT EXERCISES SOLUTIONS EXERCISE 1. Less than the value of divisor) (Also. If r = 0. Thus. 135 = 90 X 1 + 45. 6q + 4 is even and is divisible by 2. A. a = 3q + 2 => a = 9q + 12q + 4. HCF (867. ie. 6q + 4 are even. 3. As 6q. Then. Hence. 3(3q + 2q) + 1 is a square of the form 3m + 1. On putting. where m = 3q . To find the maximum number of columns. If ‘r’ = 2. 2. 255) = HCF (255. If ‘r’ = 5. 0 ≤ r < b (The value of ‘r’ will always be less than the value of ‘b’. a = 6q + 2. ie. where q = 3m . 6q + [email protected]. we get a = 6q + r [ 0 ≤ r < 6. square of any positive integer is either of the form ‘3m’ or ‘3m + 1’. 6q + 2. 3m + 2. 9q is a square of the form 3m. where q is some integer. 5] If ‘r’ = 0. 9q + 8. b = 3 in (i). it is divisible by 6!) If ‘r’ = 1. a = bq + r (i). A. If the number 6 ends with the digit zero. q ≠ 0.09871823473. while Ravi takes 12 minutes for the same. Ravi arrives at the starting point after making 3 rounds. there is no value of n in natural numbers for which 6 ends with the digit zero. HCF = 1. Q7. and go in the same direction. 14) Q1. Let √5 is a rational number. where p. Suppose they both start at the same point and at the same time.6 ‘q’ = 3m + 6m + 4m. Check whether 6 can end with the digit 0 for any natural number n. & thus divides p . This implies that 5 divides p (ii). 2 2 2 2 2 2 2 2 2 2 2 2 2 Benefitted: My Paypal a/c . For any two positive integers a and b. to drive one round. as he takes 12 mts. 12 = 2x2x3. Do this solution yourself. and. 657) = 9. √5 can be written in the form p/q. the uniqueness of the fundamental theorum of arithmetic guarantees that there are no other prime in the factorisation of 6 . it is divisible by 5. Thus. EXERCISE 1. b) × LCM (a. b) = The product of the numbers (a × b) Q5. it is a composite number. 3 2 EXERCISE 1. Thus. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 A. it is a composite number. Squaring both sides: 5 = p / q => p = 5q (i)=> 5 is a factor of p . p = 5m => p = 25m . 11) Q1. LCM of 18 & 12 = 36. Given that HCF (306. Express each number as a product of its prime factors:(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429. we get 25m = 5q => 5m = q => 5 divides q => 5 divides q (iii). Find the LCM and HCF of the following integers by applying the prime factorisation method. HCF = 3. they will meet again at the starting point after 36 minutes. find LCM (306. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. as the only prime in the factorisation of 6 is 2 & 3. and. LCM = 2x2x3x5x7 = 420 (ii) The given three numbers don't have any common factor. Mobile (where u may transfer cash!) . Sonia takes 18 minutes to drive one round of the field.3 (P. (i) 7 × 11 × 13 + 13 = 1001 + 13 = 1014. Hence. 1014 is the product of prime factors. Then. So. A. HCF (a. Residence: Ghaziabad-UP . A. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. to drive one round & Sonia also arrives at the starting point after making 2 rounds. 23 and 29 (iii) 8. This is not possible. Q3. (ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5040 + 5 = 5045 = 5 X 1009. Putting this in equation (i). Therefore. Prove that √5 is irrational. n n n n n n Q6. q are integers and have no common factor (other than 1). it is a product of prime factors 5 and 1009. (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 2 2 2 Q2. 1014 = 2X3X13X13. A. (iii) Do this yourself! Q4. 21 = 3X7. They will be again at the starting point after common multiples of 18 & 12. LCM = 17X23X29 = 11339. (Use Division Method) A. Hence. There is a circular path around a sports field. (i) 12. (i) 12 = 2X2X3. Thus.manojarora23@gmail. They are primes. 657). 18 = 2x3x3. After how many minutes will they meet again at the starting point? A. In 36 mts. Thus. ie. Now. 9 and 25 A. as he takes 18 mts. 15 = 3X5.com. the prime factorisation of 6 contains the prime number 5. 15 and 21 (ii) 17. √5 = p/q. then.2 (P. thus √5 is a rational number. Thus. a/b .6 = √2. Since. ====================================================================== EXERCISE 1. iv. Thus. But. 13 / 3125. 23 / 2 5 is a Terminating Decimal. (ii) 7√5. let 7√5= a/b. Thus. vii. √2 is irrational. 7√5is Irrational. Now. Denominator is of the form 2 X 5 . 17 / 8. ((Always try to isolate the Irrational Component)) Since a and b are integers. (iii) Let 6 + √2 be a rational number. 13 / 3125 is a Terminating Decimal. Thus. let 3 + 2√5 = a/b. But. So. Thus. 35 / 50 is a Terminating Decimal. ((Always try to isolate the Irrational Component)) Since a and b are integers. (i) Let 1/√2 be a rational number. ii. Thus. therefore a/(2b) .manojarora23@gmail. (ii) Let 7√5 be a rational number. where a and b are co-prime & b ≠ 0. 3125 = 2 x 5 . Denominator is of the form 2 X 5 . Thus.3/2. 77 / 210. 29 / 343. 1/√2 is Irrational. 129 / 2 5 7 is a Non . v. Hence. (As 343 is a odd number. A. ((Always try to isolate the Irrational Component)) Since a and b are integers. where a and b are co-prime & b ≠ 0. Thus. 6/15 = 2/5. 129 / 2 5 7 . Q3. 64 / 455 is a Non Terminating Repeating Decimal. 77 / 210 is a Non Terminating Repeating Decimal. √5 is an irrational number. state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: i. and from (iii) 5 divides q. x. 17) Q1.09871823473. Denominator is not of the form 2 X 5 . Mobile (where u may transfer cash!) . Or. ix. Now. 15 / 1600. Prove that the following are irrationals : (i) 1/√2. Thus. 35 / 50 = 7 / 10 = 7 / (2 x 5 ). Denominator is not of the form 2 X 5 . Denominator is not of the form 2 X 5 . 6 / 15 is a Terminating Decimal. Denominator is of the form 2 X 5 . √2 = 2a/b. Thus. 5 = 2 x 5 . Thus. Let 3 + 2√5 be a rational number. √5= a / 7b. √5 is irrational. Thus. Thus. Thus. Or. Thus. 64 / 455. a and b are integers. 5 divides p. √5 = a/(2b) . Hence. Now. let 6 + √2= a/b. a/b . Thus. This means 5 is a common factor of p and q. it is not divisible by 2. And. 0 3 5 m 0 m n n m 6 2 m m 3 2 2 m 7 5 1 1 n n n m 0 n 3 2 n 2 m 7 5 n 1 m m n n Benefitted: My Paypal a/c . Without actually performing the long division. Prove that 3 + 2√5 is irrational. iii. Thus. 3 + 2√5 is an irrational number. Or. 210 = 7x3x2x5.) A. viii. Thus. Now. Residence: Ghaziabad-UP . Or. Cube root of all the numbers which are not perfect cubes are Irrational Numbers. 64 / 455 is a Non . But.4 (P. 1600 = 2 x 5 .Terminating Repeating Decimal. Denominator is of the form 2 X 5 . √2 is irrational. 6 + √2 is Irrational. √2 / 2 = a / b. 8 is a Terminating Decimal. Thus. (iii) 6 + √2 (From Previous Class: Square root of all the numbers which are not perfect squares.3. Denominator is of the form 2 X 5 . where a and b are co-prime & b ≠ 0. Thus. (1 X √2 )/(√2 X√2 ) = a / b. This contradicts the supposition that there is no common factor of p and q. where a and b are co-prime & b ≠ 0. and hence its denominator’s factors can’t have powers of 2) vi. a/7b is rational and so √5 is rational. 455 = 5 X 7 X 13. But √5 is an irrational number. Q2. 2√5 = a/b . Or. 15 / 1600 is a Terminating Decimal. Denominator is of the form 2 X 5 . 23 / 2 5 .6 is rational and so √2 is rational. Or. 8 = 2 x 5 . Denominator is not of the form 2 X 5 . 2a/b is rational and so √2 is rational.Terminating Repeating Decimal.7 From (ii).com. Thus. let 1/√2 = a/b.3/2 is a rational number. and. our supposition is wrong. In each case. (i) 13 / 3125 = 13 / (5x5x5x5x5) = (13x2x2x2x2x2) / (5x2x5x2x5x2x5x2x5x2) = 416/100000 = 0. .7 (x) Non . (iv) 15 / 1600 = 15 / (2 x 5 ) = 15 / (2 x2 x5 ) = (15x5 )/(2 x5 x2 x5 ) = (15x625)/10 = 9375/1000000 = 0.Terminating Repeating. 1000000000 = 9 9 9 (ii) 0.4. 43. . If they are rational. 43.009375. Hence.123456789 (ii) 0. and hence is Rational. Residence: Ghaziabad-UP . and hence is a Rational Number.Terminating Repeating.123456789 is terminating.123456789 = 43123456789 p/[email protected] A. q = 1000000000 = 10 = 2 x5 . (vi) 23 / 2 5 = (23x5)/2x5x2 x5 ) = 115/10 = 0. . .Repeating. Mobile (where u may transfer cash!) . (iii) Non . 999999999 = Benefitted: My Paypal a/c . The following real numbers have decimal expansions as given below.115 (vii) Non .Terminating but Repeating.123456789 is Non . is Non . 3 0 6 3 2 3 2 3 4 2 2 3 2 3 2 4 4 4 2 2 6 3 Q3.Terminating Repeating. what can you say about the prime factors of q? (i) 43.125. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.com. and hence is Irrational. (iii) 43. (iii) 43.8 Q2.123456789 = 4312345646 p/q.Terminating Non . A.Terminating Repeating.120120012000120000. decide whether they are rational or not. q = 999999999. (ix) 35/50 = 35/(5x10) = (35x2)/(2x5x10) = 70/100 = 0. and of the form p/q.120120012000120000. (i) 43.00416. (ii) 17 / 8 = 17 / (2 x 5 ) = (17x5 ) / (2 x 5 ) = (17x125)/10 = 2125/1000 = 2. (v) Non . (viii) 6/15 = 2/5 = (2x2)/(5x2) = 4/10 = 0.09871823473. + a x is called a polynomial. If a = a = a = =a = 0. A Polynomial can be ‘Monomial’. ie. 5…. We say that x and (2x + x + 1) are ‘Factors’ of (2x + x + x ). x and x are 3. The expression of this form is called ‘Polynomial in one variable’. 0. ie. Dividend = ( Divisor X Quotient) + Remainder. P(x) = (x-a) . 7. q(x) + r(x). 2. ((Thus. A Polynomial written either in the descending or ascending powers of ‘x’ is called the ‘Standard Form’ of a Polynomial. when p(x) is divided by (x-a). the remainder is r(x). and (2x + x + x ) is a multiple of x as well as (2x + x + 1) 8. Since. 2x + x + x = x(2x + x + 1). Let’s say. Constant Polynomial: -3. we get quotient q(x) & remainder r(x). Consider a polynomial p(x) = x . In general. Degree of a Polynomial: The highest exponent in various terms of one variable is called its degree.com. and. on dividing polynomial p(x) by another polynomial g(x). 2. still gives 5. ie.manojarora23@gmail. eg.efficient of the polynomial. 5. Zeroes of a Polynomial: The value of a polynomial p(x) at x = a. 2. 3. such that p(a) = 0. Binomial’ and so on. 1 is a Zero (or root) of the Polynomial p(x). Zeroes of a Zero (0)Polynomial: By Convention. a non . Remainder Theorum: Let p(x) be any polynomial of degree greater than or equal to one. The degree of the remainder r(x) is always less than the degree of the divisor g(x). [Division of a polynomial by other polynomial using long division method has to be thoroughly practiced by the student. Number of Zeroes of a polynomial of degree ‘n’ is ‘n’.] Proof of Remainder Theorum: Let p(x) be any polynomial with degree ≥ 1. is p(a). 1.1.  Every Linear Polynomial has one and only one zero. and. 4.  A polynomial can have more than one zero. Mobile (where u may transfer cash!) . If all the constants are zero. If p(x) is divided by a linear polynomial (x-a).. we get a ‘Zero Polynomial’. 0 0. A Function p(x) of the form p(x) = a + a x + a x + a x +……. which gives x=1 Now. The Constant Polynomial 0 is called a ‘Zero Polynomial’. ‘The Degree of a Zero Polynomial is not defined’. A Polynomial can be ‘Linear’. 4. 0 or positive) integer. ie. It has no zero because replacing x by any number in 5x . (x + 2x + 4/x)… are not Polynomials. ’Quadratic’. One degree). If the divisor is linear (ie. n 0 3 2 2 2 2 3 2 3 2 ====================================================================== Remember The Following Points:  A zero of a polynomial need not be zero. obtained on replacing ‘x’ by ‘a’. Degree of r(x) < Degree of g(x). every Real Number is a zero of the Zero Polynomial. In 3x + 4x + 5. Thus. which is denoted by 0. Division of a Polynomial: Say. Now. then the degree of remainder will be 0. ie. ie.  Number of Zeroes is the degree of the polynomial. where (i) a a a a are real numbers. and the degree of r(x) is less than the degree of (x-a). 2.. let ‘a’ be any real number. We will be dealing with Polynomials in one variable only. n 2 2 2 2 2 0 0 1 2 …. x + y + xyz is a polynomial in three variables. 1. consider a constant polynomial 5.efficient: Each term of the polynomial has a co-efficient. the co-efficients of x . ’Cubic’. Co . the degree of Benefitted: My Paypal a/c . …. We can also have polynomials in more than one variable. …. and 5. Terms: The polynomial 3x + 4x + 5 has three terms. are examples of Constant Polynomials. a a a a are called co . The remainder will be a constant. Q(x) + r(x). Eg. the degree of (x-a) is one. P(x) = g(x) .09871823473.. 6.9 CBSE CLASS X MATHEMATICS CHAPTER 2 POLYNOMIALS RECAP FROM PREVIOUS CLASS/ES 1. a Zero of a Polynomial p(x) is ‘a’. etc. and.negative (ie. then the remainder is p(a). p(1) = 0. ’Biquadratic’. (2/x + 3).zero constant polynomial has no zero. 1 2 2 3 3 n n n 2 0. the quotient is q(x). Residence: Ghaziabad-UP .. (ii) n is a non . 4x .5. which proves the theorem. C) Factorisation by Algebric Identities: Read the following identies. Factor Theorum: If p(x) is a polynomial of degree n ≥ 1. and ‘a’ is any real number. 3 2 = (x . and. The constant term is 5. (x+1) is always the factor of p(x). Write down its factors.4x .com.5) = (x+1) (x . then. and has many factors. p. Therefore. Residence: Ghaziabad-UP .ab . 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.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 . Factorisation of Polynomials: A) Splitting the Middle Term: (ie. 2 Now. Thus. this equation gives us p(a) = (a-a) . (x . the constant term of the polynomial.] 2 B) By Factor Theorum: Let f(x) be a given polynomial. prove them if you have to. we have to write ‘b’ as the sum of two numbers whose product is ‘ac’. Least Common Multiple (LCM) of Polynomials = (Highest Power of Common Term) X Remaining Terms.09871823473.3x . then: (i) (x-a) is a factor of p(x).  (a + b) = a + 2ab + b  (a . then. and get the Quotient. Now. say ‘r’.b) = a . p(-1) = 0. if p(a) = 0.9x . Quadratic Equations): To factorise ax + bx + c. 9. by splitting the middle term. q(a) + r = r. Thus.bc . ±1 Now. This actually follows from the Remainder Theorum.ca).3x . This means that r(x) is a constant. (ii) p(a) = 0.b . and. 2 3 2 2  If the sum of the coefficients of given expression p(x) is zero. 1. p(x) = (x-a) .10 r(x) = 0.manojarora23@gmail. Sign will be that of the middle term. (x-a) is factor of polynomial f(x). for every value of x. READ THE FOLLOWING EXAMPLE: Let f(x) = x . if x=a. if (a + b + c) = 0. [If on multiplying a & c. we’ll get the middle term by adding.3ab(a-b)  a + b = (a + b) · (a − ab + b )  a − b = (a − b) · (a + ab + b ) 2 2 2 2 2 2 2 2 2 2 3 3 2 3 3 2 2 3 3 3 3 2 2 3 3 2 2  a3 + b3 + c3 . 10.  Divide the given polynomial f(x) by (x-a). Understand them. x . a3 + b3 + c3 = 3abc.q = c.5) (Remainder is zero).5) = (x+1) (x-5) In Regards to this method. we get + sign.5 = (x+1)(x+1) (x .  Let ‘a’ be the zero of polynomial. r(x) = r.  Factorise the Quotient by splitting the middle term.9x . ie. do the long division.5). write down the smallest. Proceed as follows: (Factorising equations of Degree 3) Write separately.  You would thus obtain the factors of the given polynomial. If the number is big. its factors are ±5.3abc = (a + b + c)(a2 + b2 + c2 . 11.b) = a . Thus. (x-1) is always the factor of p(x). Let p & q be those two numbers. so. Mobile (where u may transfer cash!) . p+q = b. first few factors. Thus (x+1) is a factor of p(x). In particular. ie. q(x) + r. ———————————————————————————————————————-- Benefitted: My Paypal a/c . if (x-a) is a factor of p(x). and Remember them. can have two distinct zeroes (graph cutting the x2 Benefitted: My Paypal a/c . Graph of y = ax + b is a straight line which intersects the x-axis at ( −b a . a≠0 2 2 (Let α.(Sum of Zeroes) x + Product of Zeroes. Consider the following: If ‘k’ is a zero of p(x) = ax + b.α = 0. 0). the zero of the linear polynomial [ak + b] is [(-b) / a] = (-)Constant Term / Coefficient of x. Geometric Meaning of the zeroes of the Polynomial. Division Algorithm for Polynomial: If p(x) & g(x) are any two polynomials with g(x) ≠0.(α + β)x + αβ.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. Type of Polynomial General Form  Linear ax + b.manojarora23@gmail.β) is the Quadratic Polynomial.11 CBSE CLASS X CHAPTER TWO: POLYNOMIALS 1. A polynomial of degree 2 has at the most two zeroes.(x . a Cubic Polynomial can have at the most 3 zeroes. (x .β) = ax + bx + cx + d 3 2 (Let α. then p(k) = ak + b = 0. 2 Co-efficient of x −d Product of Zeroes (α. ie. To form a Quadratic Polynomial with its given zeroes: Let α.γ + γα) = Coefficient of x 3 = a 2. a ≠ 0. x = β. β be the zeroes of a Quadratic Polynomial. Residence: Ghaziabad-UP . then we can find polynomials q(x) & r(x) such that p(x) = g(x) X q(x) + r(x).09871823473. of Zeroes 1 ax + bx + c. ie. ie. b.α). or degree of r(x) is less than the degree of g(x). k = (-b) / a. Thus.β. A Quadratic Polynomial can have either two distinct zeroes. a. γ be three zeroes) 3 c a Sum of Zeroes (α + β+γ) −b = = -Co-efficient of x a . => x . ∴ x = 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. 4.β + β. where r(x) = 0. Mobile (where u may transfer cash!) . Graph of the Quadratic Equation y = ax + bx + c. Dividend = Divisor X Quotient + Remainder. two equal zeroes or no zeroes. ie. This x-Coordinate (of the point of intersection of the graph with x-axis) is the zero of the polynomial y = ax + b . x . x . Thus. Similarly. 2 2 3. a≠0  Quadratic No.β = 0.com. => x . α + β = 1.1). A cubic polynomial of the form y = a can have at the most 3 zeroes.x + 1. (i) x – 2x – 8 = (x-4)(x+2). If ‘k’ = 4. Mobile (where u may transfer cash!) . α . 2 2 2 2 Benefitted: My Paypal a/c . CBSE CLASS X MATHEMATICS CHAPTER 2 POLYNOMIALS .√2x + 1/3). -2 Sum of Zeroes = 4 . Β = 1/4 The polynomial formed is x .√2x + 1/3. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. Residence: Ghaziabad-UP . Let the polynomial be ax + bx + c. α + β = -1/4. then the polynomial is 3x . or no zeroes (graph not cutting the x-axis at all (in which case. in each case.γ). 1/3.4.x . ie. one zero (or two equal zeroes). Β = 1/3 The polynomial formed is x . 2 2 2 2 2 2 (ii) √2.x + √5 = x + √5. In fact. α . then the polynomial is 4x . a ≠ 0. 28) Q1. α + β = 1/4. the required cubic 3 polynomial is (x .1/4 x + (-1) = x . α .β)(x . 33) Q1. (These curves are called parabolas. -1. α + β = √2. ax + bx + c.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. β. A.3√2x + 1.x/4 . for some polynomials p(x). Compare the quadratic equations given with the general quadratic equation.0. a polynomial p(x) of degree n has at the most n zeroes.(Sum of Zeroes) x + Product of Zeroes = x . Β = -1 The polynomial formed is x . 1.x + 1 = x . A. Find the number of zeroes of p(x). N. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 2 2 2 (v) -1/4. α . for any quadratic polynomial ax2 + bx + c. In general.1.2 (P.manojarora23@gmail. Self Explanatory.: Formation of a cubic polynomial: Let α. 1/4. Β = √5 The polynomial formed is x .(Sum of Zeroes) x + Product of Zeroes = x . 2 2 2 (iv) 1. A.(Sum of Zeroes) x + Product of Zeroes = x .NCERT EXERCISES SOLUTIONS EXERCISE 2. below. If ‘k’ = 3. γ be three zeroes of the Polynomial.α) (x . The other possible polynomials would be k(x + x/4 + 1/4).x/4 . 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.B.(Sum of Zeroes) x + Product of Zeroes = x . The other possible polynomials would be k(x .1 (P.12 axis at two distinct points). α + β = 0. α . and its zeroes be α & β.(-1/4)x + 1/4 = x + x/4 + 1/4. 2 2 2 2 (iii) 0. (i) 1/4. it would not be possible to factorise the quadratic polynomial).com. Then. The graphs of y = p(x) are given in the Fig. The zeroes are x = 4.(Sum of Zeroes) x + Product of Zeroes = x . ———————————————————————————————————————————EXERCISE 2.1. Β = 1 The polynomial formed is x .2 = 2 = -(-2)/1 = .09871823473. The other possible polynomials would be k(x .) c. √5. (x-3) + (7x .4x + + x + 3x + 1).3x + 4x + 5) = (x – x + 1). 2 2 2 Q4. So.3x + 4x + 5. 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. -1.x -x + 2 (x + 0.3x + 1. is a factor of p(x). By Division Algorithm.7x + 2x + 2 Remainder is ‘0’.(x + x . 2 Now. 3 3 2 2 2 2 (ii) p(x) = x . Mobile (where u may transfer cash!) . if two of its zeroes are √(5/3). (x + 0x .3) + (8).(t . Residence: Ghaziabad-UP . 2 (vi) 4.√(5/3) 4 A. α + β = 4. Hence.(x . Obtain all other zeroes of 3x + 6x – 2x – 10x – 5. So. ———————————————————————————————————————-2 2 EXERCISE 2. 3 5 4 2 3 5 4 2 Q3.5.09871823473. A. 1. Β = 1 The polynomial formed is x .x . the quotient and remainder were (x – 2) and (–2x + 4). On dividing x – 3x + x + 2 by a polynomial g(x).13 If ‘k’ = 4.9).4x + 1. and .(-x . Zeroes of (x + 1) are -1.3 (P. 3x + 5x .4x + + x + 3x + 1 As Remainder is not ‘0’. -1. 36) Q1.5x + 6) = (-x + 2).manojarora23@gmail. x .(Sum of Zeroes) x + Product of Zeroes = x . 2 4 3 2 2 4 3 2 (iii) x . 3 2 therefore.-1.7x + 2x + 2).12).3) is a factor of (2t + 3t . g(x) = x – 2 (x – 3x + 5x – 3) = (x – 2). all its zeroes are √(5/3). (-)√(5/3). p(x) = q(x).x . g(x) = x – x + 1. then the polynomial is 4x + x + 1. Since the two zeroes are . Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: Remainder is ‘0’.g(x) + r(x) 3 2 Benefitted: My Paypal a/c . Now. x + 2x + 1 = (x + 1) .5x + 6. So. α . g(x) = 2 .3x + 1) is not a factor of (x . respectively.9t . 4 2 2 4 3 2 2 2 (iii) p(x) = x + 0. Find g(x).com. 2 4 3 2 (ii) x + 3x + 1.2t . 4 4 2 2 2 2 2 = 2 Q2.(x + 3x + 1) is a factor of (3x + 5x . applying division algorithm to the given polynomial and 3x .2) + (-5x + 10). 14 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 2 2 2 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 Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP 15 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 Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP 16 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 1 1 2 2 2 2 2 2 2 1 2 1 2 2 2 2 1 1 1 2 2 2 1 2 1 2 1 2 1 1 2 2 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 a x + b y + c = 0 (ii). y=(c a .: x y 1 b c a b 1 2 2 1 1 1 2 2 1 1 2 2 1 1 1 1 b c a b Thus. If we get an equation in one variable. a b . If c = kc .ca ab -ab Use the following diagram to memorise the above result. where. We get: b a x + b b y + b c = 0 (iii). Putting the values of a & a in (i).  Step 2 : Taking the help of the diagram above. may be written as: x = y = 1 (viii) bc .  When a /a = b /b ≠ c /c [email protected]. we follow the following steps. Follow the following steps:  Step 1: Multiply equation (i) by b . we don't get any solutions (Inconsistent pair of equations).. 3. a b .e. Residence: Ghaziabad-UP . If c ≠ kc .2 Elimination Method: This is the method of eliminating (i. Mobile (where u may transfer cash!) . we obtain a false statement involving no variable. Here two cases may arise: Case 1. b a x + b b y + b c = 0 (iv). a = ka .  Step 3 : Find x and y. 3. provided a b .a b ≠ 0 (v)  Step 3: Substituting this value of x in (i) or (ii).a b ) (vi). it is inconsistent. a /a ≠ b /b => The pair of linear equations has a unique solution. If in Step 2.  Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. k(a x + b y) + c = 0 (vii) Now. then the original pair of equations has infinitely many solutions. That is why the method is known as the substitution method. we get infinitely many solutions. go to Step 3. removing) one variable. If we write a /a = b /b = k. write Equations as given in (viii)..  When a /a = b /b = c /c . i. 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. we get a unique solution. and. . then there are infinitely many solutions to the pair of linear equations given by (i) & (ii). and.09871823473.b c )/(a b . we get: x=(b c .e. a b . & Equation (ii) by b .17 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. we have:  When a /a ≠ b /b .Multiplication Method: Consider the two linear equations in their general form: a x + b y + c = 0 (i). we obtain a true statement involving no variable. then any solution of Equation (i) will not satisfy Equation (ii) and vice versa. then the original pair of equations has no solution. Therefore the pair has no solution. 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 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable. Case 2.a b ≠ 0 2 2 2 1 2 2 2 1 Benefitted: My Paypal a/c . So. Now. ==> 1 1 1 2 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 1 2 2 1 1 2 2 1 1 1 2 2 2 2 1 1 2 2 1 1 2 1 1 2 1 1 2 1 1 2 2 1 2 2 1 2 2 2 1 2 1 1 2 2 1 1 1 1 2 2 1 2 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 The solutions given in equations (v) & (vi). if. a /a = b /b = c /c = k. Thus.bc ca . any solution of Equation (ii) will satisfy the Equation (i).a b = 0.  Step 3: Solve the equation in one variable (x or y) so obtained to get its value. then. If in Step 2.3 Cross . and vice versa. Equations (vii) & (ii) can both be satisfied only if c = kc => c /c = k.a b ≠ 0.  Step 2: Subtracting (iv) from (iii).c a )/(a b . Thus. we get.a b ). b = kb . I shall be three times as old as you will be. Acc. Mobile (where u may transfer cash!) . and. and of ball be Rs y According to the question. Form the pair of linear equations in the following problems. three years from now. whereas 7 pencils and 5 pens together cost Rs 46.5x)/7). & cost of 1 kg of grapes = Rs y. y = (300 . Also. and x − y = 4 (ie.” (Isn’t this interesting?) Represent this situation algebraically and graphically. she buys another bat and 2 more balls of the same kind for Rs 1300. y = (46 . Age of Aftab = x + 3 Age of his daughter = y + 3 Q2. and. EXERCISE 3. Preparing solution table for the above two equations: y = 160 . Represent this situation algebraically and geometrically.4x)/2. 7x + 5y = 46 (ie. Age of his daughter = y − 7 Three years hence.x)/2 Q3.com.x).manojarora23@gmail. After a month. Aftab tells his daughter. Let the cost of 1 kg of apples be Rs x. the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Later. 49) Q1. Let the number of girls be x and the number of boys be y. Residence: Ghaziabad-UP . Find the cost of one pencil and that of one pen. find the number of boys and girls who took part in the quiz. y = (50 . A. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. 44) Q1. 5x + 7y = 50 (ie. I was seven times as old as you were then. According to the question.09871823473. A. Represent the situation algebraically and geometrically.4). Preparing solution table for the above two equations: y = (3900 . A. y = 10 . and. If the number of girls is 4 more than the number of boys. (i) 10 students of Class X took part in a Mathematics quiz. y = x . we get the above graph.1 (P. and.NCERT EXERCISES SOLUTIONS EXERCISE 3. “Seven years ago. Let the present age of Aftab be x.3x)/6. A. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. x + y = 10 (ie. we get the required graph: (ii) 5 pencils and 7 pens together cost Rs 50. According to the question.2 (P.18 CBSE CLASS X MATHEMATICS CHAPTER 3 LINEAR EQUATIONS IN TWO VARIABLES . A. Let the cost of a bat be Rs x. present age of his daughter = y Seven years ago: Age of Aftab = x − 7.2x. Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y. we get the required graph: Benefitted: My Paypal a/c . and find their solutions graphically. and. Drawing the solution table for the two equations.7x)/5) Drawing the solution table for the two equations. to the question. y = (1300 . 1 2 1 2 1 2 (v) 4/3 x + 2y = 8. (ii) 9x + 3y + 12 = 0. 3x – 3y = 16 Comparing the equations with general form. . a /a = b /b = c /c . the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point. Hence. 1 1 2 1 1 2 1 2 1 2 1 2 2 1 2 1 2 2 1 2 1 2 Q3. 18x + 6y + 24 = 0 Comparing the equations with general form. 4x – 6y = 9 Comparing the equations with general form. 7x + 6y – 9 = 0 Comparing the equations with general form. the lines representing the given pair of equations are Coincident lines (ie. the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point. we have Since. the lines representing the given pair of equations are Coincident lines (ie. are parallel or coincident: (i) 5x – 4y + 8 = 0. Hence. (iii) 6x – 3y + 10 = 0 2x – y + 9 = 0 Comparing the equations with general form. On comparing the ratios a /a . Mobile (where u may transfer cash!) . find out whether the following pair of linear equations are consistent. or inconsistent. Hence. On comparing the ratios a /a . the lines representing the given pair of equations are Coincident lines (ie. we have Since. we have Since. Hence. we have 1 2 1 2 1 2 Benefitted: My Paypal a/c . 1 2 1 2 1 2 Q4. the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations. a /a = b /b = c /c . Consistent pair of equations with infinite solutions). b /b & c /c . a /a = b /b = c /c . the given pair of Equations are inconsistent (Parallel Lines) with no solution.com. find out whether the lines representing the following pairs of linear equations intersect at a point. a /a ≠ b /b .09871823473. Since. obtain the solution graphically: A. the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point. we have Since. Consistent pair of equations with infinite solutions). – 10x + 6y = –22 Comparing the equations with general form.2x)/2 Drawing the solution table for the two equations. we have Since. the pair of linear equations is consistent.19 Q2. x + y = 5 => y = 5 . we have Since. 1 1 2 1 2 1 2 1 2 2 (ii) 2x – 3y = 8. a /a ≠ b /b . (i) x + y = 5. Residence: Ghaziabad-UP . the given pair of Equations are inconsistent (Parallel Lines) with no solution. we have Since. a /a = b /b = c /c . Hence. 2x − 3y = 7 Comparing the equations with general form.x. a /a ≠ b /b . Which of the following pairs of linear equations are consistent/inconsistent? If consistent. b /b & c /c . Hence. 9x – 10y = 14 Comparing the equations with general form. Hence. Hence. (i) 3x + 2y = 5. 1 2 1 2 1 2 (iii) 3/2 x + 5/3 y = 7. a /a = b /b ≠ c /c . 2x + 3y = 12 Comparing the equations with general form. 2x + 2y = 10 => y = (10 . 2x + 2y = 10 Comparing the equations with general form. we get the required graph: It can be seen that the two lines are overlapping each other.manojarora23@gmail. Hence. we have Since. (ii) x – y = 8. 1 2 1 2 (iv) 5x – 3y = 11. Hence. a /a = b /b ≠ c /c . Consistent pair of equations with infinite solutions). ie. the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point (ie. the vertices of the triangle are (2.) 1 2 1 2 1 2 Q5. Drawing the solution table for the two equations.2x + 6. a /a = b /b ≠ c /c . (−1. ie. 53) Q1. 4x – 4y – 5 = 0 Since. Therefore.manojarora23@gmail. a /a = b /b ≠ c /c . ie. we get the required graph: 1 2 1 2 (iv) 2x – 2y – 2 = 0. y = . According to the question. y = (4x . 0). Half the perimeter of a rectangular garden. (t + 3)/3 + t/2 = 6. a /a ≠ b /b . Hence. (2). (i) For intersecting lines: 6x + 12y . 2x + y – 6 = 0. we get y = 5. Let the length of the garden be x. the three lines are intersecting each other at point (2. x . 0). Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis. we will get the required graph: From the figure. A. is 36 m. we will get the required graph. and shade the triangular region. They are consistent). whose length is 4 m more than its width.(14 . (ii) s – t = 3 (1). x – y + 1 = 0. ie. (3). 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. Find the dimensions of the garden. y = -3x + 12. A. x .com. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Hence. we obtain y = 14 . 3) and x-axis at (−1. (3).x) = 4. that the two lines are indeed parallel. From (1).09871823473. we get s = 9.8 = 0. will give us the required dimensions of the garden. Putting this value in eq. Benefitted: My Paypal a/c . (To demonstrate by drawing a graph. and (4. ———————————————————————————————————————-EXERCISE 3. 4x – 2y – 4 = 0.y = 4 (2). Hence. Residence: Ghaziabad-UP . (ii) For parallel lines: 4x + 6y − 8 = 0.4)/2 Drawing the solution table for the two equations. t = 6. Putting this value in eq. Mobile (where u may transfer cash!) . (2). (i) x + y = 14 (1).20 Since. 0) and (4. Putting the value of ‘t’ in eq. x=y+4 (1) y + x = 36 (2) Drawing the solution table for the two equations. From (1). we have Since. Given the linear equation 2x + 3y – 8 = 0. the given pair of Equations are inconsistent (Parallel Lines) with no solution. we obtain s = t + 3 (3). 1 2 1 2 1 2 (iii) 2x + y – 6 = 0. Q6. s/3 + t/2 = 6 (2). Q7. 3). Putting the value of ‘x’ in eq. Width = y = 16m. x = 9. & the width be y. (iii) For coincident lines: 6x + 9y − 24 = 0. 4x – 2y – 4 = 0 Comparing the equations with general form. y = x + 1. Length = x = 20m. 3x + 2y – 12 = 0. The point where the two lines will intersect. ie.3 (P. Solve the following pair of linear equations by the substitution method.x (3). 0). the given pair of Equations are inconsistent (Parallel Lines) with no solution. ie. and. Thus. A. Hence. (3).4x + 0. y = mx + 3. (3).2x)/0. Putting this value in eq.manojarora23@gmail. Find the cost of each bat and each ball. Form the pair of linear equations for the following problems and find their solution by substitution method. As per question. Benefitted: My Paypal a/c .. x = 99 . 5 = m (-2) + 3. This is always true.3 (2). y = (3800 . For a distance of 10 km.0. 3x . using substitution method.4y = -24 (2). Find them. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. DO IT ON YOUR OWN!! x = 2. or. Let the fixed charge be Rs x and per km charge be Rs y. the charge paid is Rs 155.10y + 15y = 155.3 (1). ie. we get y = 50. and as the two are supplementary angles. x + y = 180 (2) From (1) y = x . ie.x = 26 => x = 13 (3) Putting this value in eq. (1). 3x + 5y = 1750 (2) From (1). x = 2.(iii). 7x + 6y = 3800 (1). Putting the value of ‘x’ in eq. A. x + 10y = 105 . y = 3x (1). Let the cost of a bat and a ball be x and y. we get x = -2.18 (3). or. x + 15y = 155 . 9x .3. (ii). ie. x + x . which gives x = 500 (4).x = 26 (2) Substituting the value of y from equation (1) into equation (2).com. Mobile (where u may transfer cash!) .18 = 180 . m = -1. x . (vi) 3x/2 .3 . Let the first number be x and the other number be y such that y > x.√8 y = 0 (2). Putting this value in eq. Putting the value of ‘x’ in eq. we get y = 0. (2). (2) => 3x + 5 [(3800 . √3 x .y = 18 (1). we obtain y = 3x − 3 (3 ).√8 ((-√2 x)/√3) = 0.5y = 2. A. and. Putting the value of ‘x’ in eq. and.3y = 1. Putting this value in eq. 0 0 0 0 0 0 0 0 0 (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. A. Putting this value in eq. the given pair of equations has infinite possible solutions One of its possible solutions is x = 0.21 (iii) 3x – y = 3 (1). ie. From (1).2x)/0. we obtain y = (-√2 x)/√3 (3). As per the question. (2). Later. the charge paid is Rs 105 and for a journey of 15 km. 2 = -2m. she buys 3 bats and 5 balls for Rs 1750. Find them.5(1.4x + 0. 0. we obtain y = (1. Put this value in eq. we get y = 99 . x = 0. From (1). y . Putting this in eq. According to given info.09871823473. y = 5. y = 3 x 13 = 39. 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. (2).3 . y = 3. x = 105 . 105 . Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3. we get. 0. Putting this value in eq.3 (3).(ii) From (i). 9 = 9. According to the given info.3(3x − 3) = 9. we get. Q3.10y . (3). 2x . 9x – 3y = 9 (2). Solving for ‘x’ & ‘y’.2x + 0. (v) √2 x + √3 y = 0 (1). (i) The difference between two numbers is 26 and one number is three times the other. (iv) 0. Q2.0. Let the larger angle be x and smaller angle be y. (3). √3 x .5y/3 = -2 (1). From (1).7x)/6 (3). we get y = 3. x/3 + y/2 = 13/6 (2). Residence: Ghaziabad-UP .18 = 81 .7x)/6] = 1750. (2). 2x + 3y = 11 (1). y = -3. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees.3 = 2.(i). Hence. x = 19/5 By Substitution Method: x + y = 5 (i). (i). (x + 1) / (y .y = 1 (ii) Subtracting (i) from (ii). Q2.y/3 = 3 These three pair of equations can be solved following the method described above. Let the fraction be x/y. we get. Let the fraction be x/y. Putting this in (i). and. Putting this in eq.y = -2 (i) x / (y + 1) = 1/2 => 2x . x = (-4 + 9y)/11 .(ii) From (i).5y = (-)3 . A. x = 19/5 (ii) 3x + 4y = 10 and 2x – 2y = 2. we get. Putting y = 6/5 in (iii). we get. we get. x = 5. Also. we get. Do it yourself! Age of Nuri = 50 years. and. the age of Jacob will be three times that of his son. x = 3y + 10 . What is the fraction? A.9y = (-)4 . 56) Q1. Find the number. Residence: Ghaziabad-UP .(iii).com.7y = -30 . Benefitted: My Paypal a/c . Charge for 25 km = x + 25y = Rs 255 (v) A fraction becomes 9/11. the present age of Jacob is 40 years. the fraction is 3/5 (ii) Five years ago.(ii) From.5) = 7(y . and. Thus.1) = 1=> x . 2x – 3y = 4 (ii) x + y = 5 (i). x = 5 . (x+3)/(y+3) = 5/6 => 6x . gives. y = 6/5.4 (P. y = 10 .5) => x . Nuri was thrice as old as Sonu. Let the age of Jacob be x and the age of his son be y. we get. As per the given information. 2x + 2y = 10 (iii) Subtracting (ii) from (iii). x = 40.22 which gives. Per km charge = Rs 10. What are their present ages? A. 3 is added to both the numerator and the denominator it becomes 5/6. ———————————————————————————————————————-EXERCISE 3. Form the pair of linear equations in the following problems. (iii). if 2 is added to both the numerator and the denominator. y = 9 . we get. Find the fraction. we get. we get. a fraction reduces to 1.(iv) Substituting y = 10 in equation (iii). we get y = 5. (i).manojarora23@gmail. and the present age of his son is 10 years.(i). 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. (iv) x/2 + 2y/3 = -1 and x . Ten years later. Substituting this in equation (ii). How old are Nuri and Sonu? A. (x + 5) = 3(y + 5) => x . 2x – 3y = 4 (ii) Multiplying (i) by 2. 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).09871823473.(i) (x . fixed charge = Rs 5. we get. Jacob’s age was seven times that of his son. Thus the fraction is 7/9 (vi) Five years hence.(iv) Substituting y = 9 in equation (iii). Substituting this in equation (ii). Age of Sonu = 20 years. y = 6/5 (iv).(iv) Putting this in eq. x = 7. we get x = 3 (iii).(iii). (x+2)/(y+2) = 9/11 => 11x . It becomes 1/2 if we only add 1 to the denominator. Hence.3y = 10 . (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7. y = 10 . Mobile (where u may transfer cash!) . According to the given information. Five years ago. (iii) The sum of the digits of a two-digit number is 9. nine times this number is twice the number obtained by reversing the order of the digits.y (iii). According to given information. If. Nuri will be twice as old as Sonu. Putting this in (ii). they will intersect each other at a unique point and thus. no solution. (i) x – 3y – 3 = 0. x + 4y = 27 (i). (iv) Meena went to a bank to withdraw Rs 2000. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. b / b = 1/2.manojarora23@gmail. c / c = 5/8. b / b = 1/3. the given sets of lines are coincident to each other and thus. and. fixed charge = Rs 15. Meena has 10 notes of Rs 50 and 15 notes of Rs 100. Let the unit digit and tens digits of the number be x and y respectively. 3x + 2y = 8 a / a = 2/3. there are infinite solutions possible for these equations. the given sets of lines are parallel to each other. Meena got 25 notes in all. y = 1 1 2 1 2 1 2 1 1 2 2 1 2 1 2 1 2 1 2 2 1 2 1 (iii) 3x – 5y = 20. A. Substituting y = 15. b / b = 1/2. 1 2 1 2 1 2 1 2 1 2 1 2 (ii) 2x + y = 5. Charge per day = Rs 3 ———————————————————————————————————————-EXERCISE 3.ab This gives. 6x – 10y = 40 a / a = 1/2. there will be a unique solution for these equations. Mobile (where u may transfer cash!) . or. a / a = b / b ≠ c / c Therefore. we get. number (written as ‘yx’) = 10y + x Number after reversing the digits = 10x + y According to the given information. we get.23 A. c / c = 1/2. Saritha paid Rs 27 for a book kept for seven days. there will not be any solution for these equations. we get. Therefore. x + 12 = 27 => x = 15 Hence. the number is 10y + x = 10 × 1 + 8 = 18. 3x – 3y – 15 = 0 a / a = 1/3.bc ca . find it by using cross multiplication method. x/2 = y/1 = 1. or infinitely many solutions. A. a / a ≠ b / b .ca ab . x + y = 9 (i). Therefore. x + y = 25 (i). they will intersect each other at a unique point and thus. there will be a unique solution for these equations. Therefore. x = y = 1____ 1 2 1 2 1 2 1 2 1 2 Benefitted: My Paypal a/c . b / b = 1. By cross-multiplication. Which of the following pairs of linear equations has unique solution. in equation (i). x = y = 1____ bc . x + 2y = 21 (ii) Subtracting equation (ii) from equation (i). Then. Find the fixed charge and the charge for each extra day. we get 9y = 9. we get. Find how many notes of Rs 50 and Rs 100 she received. Therefore. and. According to the given information. According to the given information. 50y = 750 => y = 15.5 (P. Let the number of Rs 50 and Rs 100 notes be x and y.09871823473.com. we get. Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y . x = 10 Hence. 50x + 50y = 2000 (ii) Multiplying equation (i) by 50. 9(10y + x) = 2(10x + y) => 88y − 11x = 0 => − x + 8y =0 (ii) Adding equation (i) and (ii). 2y = 6 => y = 3 (iii) Substituting in equation (i). x = 8 Hence. 50x + 50y = 1250 (iii) Subtracting equation (iii) from equation (ii). 3x – 9y – 2 = 0 a / a = 1/3. 62) Q1. a / a ≠ b / b Therefore. c / c = 7/15. She asked the cashier to give her Rs 50 and Rs 100 notes only. while Susy paid Rs 21 for the book she kept for five days. Therefore. Residence: Ghaziabad-UP . c / c = 3/2. 1 2 1 2 1 2 1 2 1 2 1 2 (iv) x – 3y – 7 = 0. we get. a / a = b / b = c / c Therefore. y = 1 (iii) Substituting the value in equation (1). By cross-multiplication method. Therefore. x = 2. Therefore. In case there is a unique solution. they will not intersect each other and thus. 5b = 0 (ii) Solving equations (i). Hence. Do It Yourself! Q4. fixed charge = Rs 400.2). x = 15 (iii). c / c = 1/(2k + 1). Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer.(iii) Yash scored 40 marks in a test. A. we get. Let the fraction be x/y. How many questions were there in the test? A.1). we get. Mobile (where u may transfer cash!) . a . Let x be the fixed charge of the food and y be the charge for food per day.y = 25 (ii) Subtracting equation (ii) from equation (i).1).2). a / a = 3/(2k . y = 12 Hence.(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. 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.ca ab . (ii) For which values of k will the following pair of linear equations have no solutions? 3x + y = 1.21 -21 . and. y = 5 Therefore.com. the given equation will have no solution.1) / y = 1/3 => 3x . for which the given equations will have infinitely many solutions. pays Rs 1180 as hostel charges. x = 400 Hence. According to the given information. (x . Putting this (i). which gives. x + 26y = 1180 (ii) Subtracting equation (i) from equation (ii). getting 3 marks for each right answer and losing 1 mark for each wrong answer. number of wrong answers = 5. for k = 2. x = 4 & y = -1 1 2 2 1 1 2 2 1 1 2 2 1 Q2. x = 5. For no solution. we get. 6y = 180 => y = 30 Substituting this value in equation (i). Let the number of right answers and wrong answers be x and y. Q4. According to the given info. x / (y + 8) = 1/4 => 4x . b / b = 3/(a + b). number of right answers = 15.y = 40 (i). x = y = 1____ 45 . Residence: Ghaziabad-UP .24 bc . (2k – 1) x + (k – 1) y = 2k + 1 A. A. then Yash would have scored 50 marks.y = 8 (ii) Subtracting equation (i) from equation (ii). c / c = 7/(3a +b .09871823473. x + 20y = 1000 (i). and charge per day = Rs 30 Q4. b / b = 1/(k . a .b) = 7/(3a +b . One car starts from A and another from B at the Benefitted: My Paypal a/c . 1 2 1 2 1 1 2 2 1 2 1 2 Q2.y = 3 (i). Find the fixed charges and the cost of food per day. According to the given information. 1 2 1 1 2 2 1 1 2 1 2 2 Q3. which gives k = 2. and (ii).b). 4x . the fraction is 5/12. For infinitely many solutions. and. a / a = b / b ≠ c / c . a = 5. we get. 3x . we get.ab Thus. we get.b) = 3/(a + b). a / a = b / b = c / c 2/(a . Total number of questions = 20 Q4. Solve the following pair of linear equations by the substitution and cross-multiplication methods : 8x + 5y = 9. This gives.(-9) Or.(iv) Places A and B are 100 km apart on a highway. x / 24 = y / (-6) = 1/6.. b = 1. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7.(-15) -3 .9b = -4 (i) 2/(a . which gives. 3x + 2y = 4 A. who takes food for 26 days. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B. a / a = 2/(a . Putting this in (ii). (a – b) x + (a + b) y = 3a + b – 2 A.bc ca .2y = 50 => 2x .manojarora23@gmail. Find the fraction. x = y = 1___. y = 5 Q1. p (=1/(√x)) = 1/2. Acc. (8x + 7y)/xy = 15 => 8/y + 7/x = 15 (ii) Putting 1/x = p. which gives. x = 4. Residence: Ghaziabad-UP . y = -2 Q1. Q1. q (=1/(y . 1 2 2 1 1 2 2 1 1 2 2 1 EXERCISE 3.3/(y . 1/y = q. x / 323 = y / 171 = 1 / 19 => x = 17. Let the speed of 1st car and 2nd car be u km/h and v km/h. The equations thus become. (v) (7x . and.(-183) 9 .bc ca . the equations become. and ‘y’. The equations thus become. Mobile (where u may transfer cash!) . we get. x = y = 1____ .3q = 1 (ii) Solving. (v) Let length and breadth of rectangle be x unit and y unit. which gives. 67) Q1.4y = 23 (ii) Solving. travelling towards each other = (u + v) km/h According to the given information.v) km/h. 3p . the area increases by 67 square units.12 = 0 (i) p/3 + q/2 = 13/6 => 2p + 3q . we get.2) = 2. (iv) 5/(x . and. we get. we get. speed of one car = 60 km/h and speed of other car = 40 km/h. p (=1/(x .2) = q. and. y = 9 Q1. we get. we get.5y . Find the dimensions of the rectangle. the length and breadth of the rectangle are 17 units and 9 units respectively.25 same time. they meet in 5 hours. (x . What are the speeds of the two cars? A. 4p + 3y = 14 (i).1) + 1/(y .13 = 0 (ii) Using cross multiplication method.v = 20 (i).ca ab .5)(y + 3) = xy . they meet in 1 hour. (iii) 4/x + 3y = 14. if its length is reduced by 5 units and breadth is increased by 3 units.2)) = 1/3. Benefitted: My Paypal a/c .1) . we get.(v) The area of a rectangle gets reduced by 9 square units. & 1/y = q. and. v = 40 km/h Hence.. The equations thus become.2) = 1 Putting 1/(x . x = 4 Also. we get. 2p + 3q = 2 (i). The equations thus become. p = 2. 5(u . 6/(x . A.v) = 100 => u . (x + 3)(y + 2) = xy + 67 => 2x + 3y . /(√x) .(-18) -12 . Relative speed of both cars while they are travelling in same direction = (u . 1/(y .com.9q = -1 (ii) Solving. If we increase the length by 3 units and the breadth by 2 units. Area = xy.1) = p. which gives. and. 3/x .6 (P.e. => x = 1/5. p (=1/x) = 5. and.4y = 23 Putting 1/x = p. If they travel towards each other.2y)/xy = 5 => 7/y . and.1) ) = 1/3. Q4.ab Or. Solve the following pairs of equations by reducing them to a pair of linear equations: Q1. we get. q (=1/(√y)) = 1/3 Solving for ‘x’. u = 60 km/h (iii) Substituting this value in equation (2).2/x = 5 (i).(-10) Or. p/2 + q/3 = 2 => 3p + 2q . y = 9 Hence. If the cars travel in the same direction at different speeds. x = 1/2. 2x 3y 3x 2y Let 1/x = p.9 => 3x . bc . 4p . y = 5 Therefore.manojarora23@gmail. 305 . 6p . 5p + q = 2 (i). to the question. (ii) 2/(√x) + 3/(√y) = 2. y = 1/3. 1(u + v) = 100 (ii) Adding both the equations. 1/(√y) = q.9/(√y) = -1 Putting 1/(√x) = p.61 = 0 (ii) By cross-multiplication method. (u > v) Relative speed of both cars while they are travelling in opposite directions i. (i) 1 + 1 = 2. q = 3. x = 4.6 = 0 (i). and.09871823473. 1 + 1 = 13/6. and y = 2 Q1.y) = 3/4 (i). y = 1 Q1. 2(x . Benefitted: My Paypal a/c . Speed of bus = 80km/h. q = 1/80 Hence. 10p + 2q = 4(iii). q (=1/y) = 1/2 => y = 2 Q1. and. number of days taken by a woman to finish the work = 18. p (=1/x) = 0 => x = 1.y) = 4 => x . we get. which gives. (viii) 1/(3x + y) + 1/(3x . According to the given information. we get. y = 4. [4Hrs10Mts = 25/6Hrs]. and. Formulate the following problems as a pair of equations. Downstream = (x + y)km/h According to question: 2(x + y) = 20 => x + y = 10 (i). q (=1/y) = 1. 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.y = 1 (vi) Solving equations (v) and (vi). and. p + q = 3/4(iii).q = -1/4 (iv) Solving. 2x + 4y = 5xy => 2/y + 4/x = 5 (ii) Putting 1/x = p.09871823473.y) = 4 (i). A.60p + 240q = 4 (iii). q (= 1/(x . 2p + 5q = 1/4 => 8p + 20q = 1 (iii).y) = 1 => x . using cross multiplication method. and. 100/u + 200/v = 25/6 (ii). Putting 1/u = p. 4p + 2q . Also. and. and 1/y = q in these equations. the equations thus become. and y = 1 Q2.6 = 0 (iii).y) = 1/2 => 3x . p (=1/x) = 1. 1/(x-y) = q. Work done by a man in 1 day = 1/y According to the question: 2/x + 5/y = 1/4 (i). 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. 600p + 1200q = 25 (iv) Solving equations (iii) and (iv).5/(x . Let the speed of train and bus be u km/h and v km/h.y) = -1/8 (ii) Putting 1/(3x+y) = p. Find the time taken by 1 woman alone to finish the work. we get. the equations become. by cross-multiplication. we get. Speed of train = u = 60km/h. while 3 women and 6 men can finish it in 3 days. 3/x + 6/y = 1/3 (ii) Putting 1/x = p. we get.q/2 = -1/8 => p . Hence. we get. she takes 10 minutes longer. and hence find their solutions: (i) Ritu can row downstream 20 km in 2 hours. and. (ii) 2 women and 5 men can together finish an embroidery work in 4 days. using cross multiplication method. we get. and. and. Therefore. we get. work done by a woman in 1 day = 1/x.1/[2(3x . x = 1. p [= 1/(3x+y) ] = 1/4 => 3x + y = 4 (v).5q = -2 (iv) Solving. p = 1/60.5 = 0 (iv) Solving. 3p + 6q . Mobile (where u may transfer cash!) . the equations become. A. we get. Let the number of days taken by a woman and a man. and. y = 1 Therefore. 7p + 8q = 15 (iv) Solving. and. 1/(3x-y) = q. q (= 1/(3x . 1/[2(3x + y)] . and.y) = -2 (ii) Putting 1/(x+y) = p. Find the speed of the train and the bus separately. in these equatns. x = 6. x = 1. p (= 1/x) = 1/18 => x = 18.y = 2 (vi) Solving equations (v) and (vi). q (= 1/y) = 1/36 => y = 36 Hence. Let the speed of Ritu in still water be ‘x’ km/h. and upstream 4 km in 2 hours. 15p . Residence: Ghaziabad-UP .y)km/h. 3p + 6q = 1/3 => 9p + 18q = 1 (iv) Solving equations (iii) & (iv). and.com. we get. 1/y = q. to complete the work be x and y. Find her speed of rowing in still water and the speed of the current. Speed of Ritu while rowing: Upstream = (x .26 -2p + 7q = 5 (iii). (vi) 6x + 3y = 6xy => 6/y + 3/x = 6 (i). we get. 60/u + 240/v = 4 (i).manojarora23@gmail. and 1/v = q. we get. x = 3. p/2 . and also that taken by 1 man alone. 15/(x + y) .y = 2 (ii) solving the equations. p (=1/(x+y)) = 1/5 => x + y = 5 (v). A. and. and the speed of stream be ‘y’ km/h. Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h. which gives. (vii) 10/(x + y) + 2/(x . x = 1. 2 2 2 Q. ax + bx + c = 0. ie. (ii) x(x + 1) + 8 = (x + 2) (x – 2). a ≠ 0 is called the standard form of a quadratic equation.09871823473. We also say that x = α is a solution of the quadratic equation. the roots are found easily by taking the square roots.4 ie. (iii) x (2x + 3) = x + 1. Check whether the following are quadratic equations: (i) (x – 2) + 1 = 2x – 3. For example. a perfect square (by dividing or multiplying appropriate number). or. we get a quadratic equation. x = 2/3. 2.5/2 x + 3/2 = 0 => (x . c are real numbers. But when we write the terms of p(x) in descending order of their degrees. b. (iv) (x + 2) = x – 4 2 2 3 3 3.com.5x + 2 = 0. then we get the standard form of the equation. Make the term containing x . and thence. A. x = -1/2 We verify the roots. Find the roots of the quadratic equation 6x – x – 2 = 0. by checking that 2/3 & -1/2 satisfy 6x – x – 2 = 0 2 2 2 2 2 2 2 2 4. (ii) Solve the equation 2x .5/4) + 3/2 .25/36 = 0 Solving we get.5/6) = (1/6) => x = 1. Solution of a Quadratic Equation by Completing the Square Herein. is a quadratic equation. Ex. Residence: Ghaziabad-UP . a ≠ 0 if aα + bα + c = 0. x . which are the roots of the given equation.5/3x + 2/3 = 0 => (x . Dividing both sides by 3.5/6) + 2/3 . Mobile (where u may transfer cash!) . a ≠ 0. That is. Quadratic Equations: A quadratic equation in the variable x is an equation of the form ax + bx + c = 0.5x + 3 = 0 => x . When we equate this polynomial to zero. and bring it in the term containing perfect square. Any quadratic equation can have at-most two roots. The process is: x + 4x = (x + 4/2) . 3x – 2 = 0 or 2x + 1 = 0. 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.manojarora23@gmail. (x . Therefore. the term containing x is brought completely inside a square. any equation of the form p(x) = 0. 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.27 CBSE CLASS X MATHEMATICS CHAPTER 4 QUADRATIC EQUATIONS 1. where p(x) is a polynomial of degree 2. Introduction: Quadratic polynomial is of the form ax2 + bx + c. or x = 2/3. Few Examples: (i) Solve the equation 3x . 2 2 2 2 2 2 2 2 2 2 2 2 2 Benefitted: My Paypal a/c . Solution of a Quadratic Equation by Factorisation: A real number ‘α’ is called a root of the quadratic equation ax + bx + c = 0. where a. We can convert any quadratic equation to the form (x + a) – b = 0 and then we can easily find its 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. a ≠ 0.25/16 = 0. Half the co-efficient of x. or that ‘α’ satisfies the quadratic equation. In fact. 2x + x – 300 = 0 is a quadratic equation. then by taking the square roots in (i). then there is no real number whose square is b2 – 4ac. then x = -b/2a ± 0.b .4ac’ > 0. Therefore.  If b – 4ac < 0. if b – 4ac > 0. a quadratic equation ax + bx + c = 0 has  (i) two distinct real roots. Since b – 4ac determines whether the quadratic equation ax + bx + c = 0 has real roots or not.4ac) => 2a 4a => (x + b ) = b . Find 2 2 2 2 2 Benefitted: My Paypal a/c . there is no real value of ‘x’ satisfying the given equation. Mobile (where u may transfer cash!) .(b/2a) + c/a =0=> (x + b ) . Thus.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.09871823473.  (ii) two equal real roots. 4x + 3x + 5 = 0 => x + 3/4x + 5/4 = 0 => (x + 3/8) . x = 1/√2 ] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 5. 2 2 2 2 2 2 Generalising the above method of ‘completing the squares’: Consider the quadratic equation ax + bx + c = 0 (a ≠ 0). and. there are no real roots for the given quadratic equation in this case. Dividing throughout by ‘a’. and hence find the nature of its roots.b + √(b . the given equation has no real roots.4ac) ≥ 0.  (iii) no real roots.4ac (i) 2a 4a If b – 4ac ≥ 0.3x .4ac).(b .4 = 0 [ x = 4. x = . we get two distinct real roots: x = .4ac) [which are the two roots of quadratic equation ax + bx + c = 0 (a ≠ 0) 2a if √(b . we get (x + b/2a) = ± (√(b . Therefore. if b – 4ac = 0.manojarora23@gmail. Residence: Ghaziabad-UP . (x .4ac).4ac). 2a  If ‘b . This is also known as the Quadratic Formula. the quadratic equation ax + bx + c = 0 has two equal real roots in this case. b – 4ac is called the discriminant of this quadratic equation. or x = -1 ] (iii) x + 4x + 5 = 0 [ b – 4ac < 0 ] (iv) 2x – 2√2 x + 1 = 0 [ x = 1/√2. So. ie. (x + 3/8) can’t be negative for any real value of x (as –ve x –ve = +ve).4ac’ = 0.28 Or. 2a 2a  If ‘b . or. Nature of Roots: The roots of the equation ax + bx + c = 0 are given by x = .2x + 1/3 = 0.com. 2 2 2 2 2 2 2 2 2 2 2 2 So. we get x + b/a x + c/a = 0=> (x + b/2a) .] Solve the following equations by quadratic formula (find the roots if they exist): (i) x + 2x – 143 = 0 [ x = 11. Find the ‘Discriminant’ of the equation 3x . A.5/4) = 1/16 => x .√(b . Eg. But. if b – 4ac < 0.9/64 + 5/4 = 0 => (x + 3/8) = (-71)/64 < 0. So. x = -b/2a or –b/2a. x = -b ± (√(b . the roots of the equation ax + bx + c = 0 are both –b/2a.b ± √(b .4ac) / 2a Thus. or x = -13 ] (ii) x . 1/3. (vii) (x + 2) = 2x (x – 1) => x . a = 3.29 them if the are real. (viii) x – 4x – x + 1 = (x – 2) => 2x . Mobile (where u may transfer cash!) . If the speed had been 8 km/h less. It is given that their product is 306. 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 not a quadratic equation. Ans. Mother’s age = x + 26 + 3 = x + 29 It is given that the product of their ages after 3 years is 360. then it would have taken 3 hours more to cover the same distance. ∴ 528 = x (2x + 1) => 2x + x . If speed = (x . ie. and hence is a quadratic equation.09871823473. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 3 2 2 3 2 3 2 2 Q2. Area of a rectangle = Length × Breadth. We would like to find Rohan’s present age. his mother’s age = x + 26. time taken to cover 480 kms is (480/x + 3) hrs. We need to find the speed of the train. Hence. 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.3 = 0. the given quadratic equation has two equal real roots.306 = 0. It is not of the form ax + bx + c = 0. Benefitted: My Paypal a/c . 1/3 2 CBSE CLASS X MATHEMATICS CHAPTER 4 QUADRATIC EQUATIONS. and hence is a quadratic equation.8) km/h. Therefore. (ii) x – 2x = (–2) (3 – x) => x – 4x + 6 = 0. ∴ x(x + 1) = 306 => x + x .14x .manojarora23@gmail. and hence is a quadratic equation. The length of the plot (in metres) is one more than twice its breadth. ‘Discriminant’ b – 4ac = 0 Hence. The product of their ages (in years) 3 years from now will be 360.13x + 9 = 0. 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 . and hence is not a quadratic equation.1 (P. The roots are –b/2a. 2 (iii) Rohan’s mother is 26 years older than him. (vi) x + 3x + 1 = (x – 2) => 7x . (iii) (x – 2)(x + 1) = (x – 1)(x + 3) => 3x . It is of the form ax + bx + c = 0. Three years hence: Rohan’s age = x + 3. ∴ (x + 3)(x + 29) = 360 => x + 32x . We need to find the integers. and hence is not a quadratic equation. Let the speed of train be x km/h.6x . It is of the form ax + bx + c = 0. 73): Q1. A. Residence: Ghaziabad-UP .8 = 0.3 = 0.273 = 0.1 = 0 It is not of the form ax + bx + c = 0. (v) (2x – 1)(x – 3) = (x + 5)(x – 1) => x – 11x + 8 = 0. b = -2. A. and hence is a quadratic equation. 2 (iv) A train travels a distance of 480 km at a uniform speed. It is not of the form ax + bx + c = 0. Let Rohan’s age be x. It is of the form ax + bx + c = 0. Let the consecutive integers be x and x + 1.528 = 0 2 2 (ii) The product of two consecutive positive integers is 306.NCERT EXERCISES SOLUTIONS EXERCISE 4. and hence is a quadratic equation. It is of the form ax + bx + c = 0.com. c = 1/3. Time taken to travel 480 km = 480/x hrs. A. -b/2a. (iv) (x – 3)(2x +1) = x(x + 5) => x – 10x . 30 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 2 2 2 (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 2 2 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 2 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 2 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 2 2 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 31 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 2 2 2 2 2 2 2 2 2 2 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 2 2 = 2 = 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. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP 32 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 Find the nature of the roots of the following quadratic equations. Part of tank filled by both the pipes together in 1 hour = 8/75 Thus. 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. the sides of the squares are 12 m and (12 + 6) m = 18 m ---------------------------------------------------------------------------------------------------------------------------------EXERCISE 4. If the difference of their perimeters is 24 m. the speed of the passenger train will be 33 km/h and thus. a = 2. (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. Therefore. Residence: Ghaziabad-UP . A. Therefore.com.30) = 0 => x = 25.25)(8x . If the real roots exist. A. side of a square cannot be negative. Hence. Q11.33 Therefore. It is given that. For a quadratic equation ax + bx + c = 0.4 (P.09871823473. b = −3. which is logically not possible. find the sides of the two squares. we obtain. 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). 4x − 4y = 24 => x − y = 6 => x = y + 6 2 2 Thus. find the average speed of the two trains. the speed of the express train will be 33 + 11 = 44 km/h. the time taken by the larger pipe will be negative. discriminant is b − 4ac. time taken individually by the smaller pipe and the larger pipe will be 25 and 25 − 10 =15 hours respectively. Speed cannot be negative. Q10.manojarora23@gmail. 1/x + 1/(x . Now. c = 5 2 2 2 2 2 2 2 Benefitted: My Paypal a/c . or. Sum of the areas of two squares is 468 m2. If the average speed of the express train is 11km/h more than that of the passenger train. y = -18. 91): Q1. x = 30/8 Time taken by the smaller pipe cannot be 30/8 = 3. Let the average speed of passenger train be x km/h. Therefore. As in this case.10) = 8/75 => (x . Mobile (where u may transfer cash!) .75 hours. their perimeter will be 4x and 4y respectively and their areas will be x and y respectively. Let the sides of the two squares be x m and y m. y = 12 However. find them: A. if this equation has two equal roots. and the area is 800m2? If so. k should be 6 only. then the equation will not have the terms ‘x ’ and ‘x’. c = 6 Discriminant = b − 4ac = (− 2k) − 4 (k) (6) = = 4k − 24k For equal roots.6x + 3 = 0 Comparing this equation with ax + bx + c = 0. a = k. Given.400 = 0. age of 2nd friend = (20 − x − 4) = = (16 − x) yrs Given that. Four years ago. a = 2.4√3 x + 4 = 0 Comparing this equation with ax + bx + c = 0. Area of mango grove = (2x)(x) = 2x . Therefore. x = ± 20. if k = 0. determine their present ages. we obtain.2kx + 6 + 0 Comparing this equation with ax + bx + c = 0.x . A. b = −20. a = 3. age of 1st friend = (x − 4) yrs. distinct real roots exist for this equation as follows: Therefore. If an equation ax + bx + c = 0 has two equal roots. 2 2 2 2 2 2 2 Benefitted: My Paypal a/c . The roots are: -b/2a. a = 1. Therefore. the desired rectangular mango grove can be designed. a = 2. A.4√3) − 4 (3) (4) = 48 − 48 = 0 As b2 − 4ac = 0. 2 2 2 2 2 2 2 2 Q3. c = 3 Discriminant = b − 4ac =(k) − 4(2) (3) = k − 24. therefore. c = 400 Discriminant = b − 4ac = (0) − 4 × (1) × (− 400) = 1600 Here. length cannot be negative. Comparing this with ax + bx + c = 0. &. that is 2/√3 & 2/√3 2 2 2 2 (iii) 2x . we obtain. its discriminant (b − 4ac) will be 0. find its length and breadth.4√3. no real root is possible for the given equation.64 + 4x = 48 => − x + 20x − 112 = 0 => x − 20x + 112 = 0.34 Discriminant = b − 4ac = (− 3) − 4 (2) (5) = 9 − 40 = −31 As b2 − 4ac < 0. Four years ago. Q5. Is the following situation possible? If so. we obtain. this situation is not possible. Discriminant = 0 => k − 24 = 0 => k = 24 => k = ± √24 = ± 2√6 2 2 2 2 2 2 2 2 2 (ii) kx (x – 2) + 6 = 0 => kx . and –b/2a. Find the values of k for each of the following quadratic equations. 2 2 (ii) 3x . (i) 2x + kx + 3 = 0 Comparing this equation with ax + bx + c = 0. Therefore. (3 + √3)/2. And hence. Let the breadth of mango grove be x. we obtain. we obtain. b = −2k. c = 3 Discriminant = b − 4ac = (− 6) − 4 (2) (3) = 36 − 24 = 12 As b − 4ac > 0. Therefore. (x − 4) (16 − x) = 48 => 16x . b − 4ac = 0 => 4k − 24k = 0 => 4k(k − 6) = 0 => Either 4k = 0 or k = 6 = 0 => k = 0 or k = 6 However. or (3 . A. Let the age of one friend be x years => Age of the other friend = (20 − x) years. find its length and breadth. b = 0. the product of their ages in years was 48. 2x = 800 => x . length of mango grove = 2x. However. no real root is possible for this equation and hence. we obtain.manojarora23@gmail. Residence: Ghaziabad-UP . Thus. breadth of mango grove = 20 m Length of mango grove = 2 × 20 = 40 m 2 2 2 2 2 2 2 Q4. Comparing this equation with ax + bx + c = 0. real roots exist for the given equation and they are equal to each other.com. b = −6. c = 112 Discriminant = b − 4ac = (− 20) − 4 (1) (112) = 400 − 448 = −48 As b − 4ac < 0. the roots are. The sum of the ages of two friends is 20 years. b = k. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so. c = 4 Discriminant = b − 4ac = (.√3)/2 2 2 2 2 2 Q2. so that they have two equal roots. [ x = -b/2a = -(. For equal roots.4√3)/(2x3) = 2/√3]. Is it possible to design a rectangular mango grove whose length is twice its breadth. b − 4ac > 0 => The equation will have real roots.09871823473. a = 1. Mobile (where u may transfer cash!) . b = . the succeeding terms are obtained by adding a fixed number. Introduction: Some of the patterns observed in everyday life are such that. Therefore. . this equation has equal real roots. which can be positive. . Looking at the pattern.09871823473. c = 400 Discriminate = b − 4ac = (− 40)2 −4 (1) (400) = 1600 − 1600 = 0 As b − 4ac = 0. 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 ……. So. Then the AP becomes a .(-40)/2(1) = 20. b = −40. 3.40x + 400 = 0 Comparing this equation with ax + bx + c = 0. . A finite AP has finite number of terms (The last term is known). we see that successive terms are obtained by adding a fixed number to the preceding terms. . a + 2d.x Area = x × y = x(40 − x) => Area = 40x − x As per the question. An Example: Find the 11th term from the last term (towards the first term) of the AP : 10. represents an arithmetic progression where ‘a’ is the first term and ‘d’ the common difference. a = 1. a + 3d. or we find that they are squares of consecutive numbers. a = a + (n – 1) d. Here-in. And hence. Perimeter = 2(x + y) = 80 => x + y= 40 => y = 40 .  100. = a – a = d. If there are m terms in the AP. . . length of park. 7. . breadth of park. 1 1 2 1 3 2 n 2 2 3 n n n-1 3. 10. and use this knowledge in solving some daily life problems. or by multiplying with a fixed number. x = 20 m. then a = – 62 and d = 3. . also. A. Then. a . this situation is possible. So. a is also called the general term of the AP. Putting the given values in the given in the above formula. . 2. If we write the given AP in the reverse order. 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. negative or zero. – 62. d = 7 – 10 = – 3.: Let a . . to find their nth terms and the sum of n consecutive terms. 70. . a. we shall learn about a pattern in which succeeding terms are obtained by adding a fixed number to the preceding terms. we get n = 25. . .35 A. And. . . Root of this equation are: x = -b/2a = . . a . a = 10. Each of the numbers in the list is called a term. Here. This fixed number is called the common difference of the AP. a – a = a – a = . This is called the general form of an AP. l = – 62. a + d. So. y = 40 − x = 40 − 20 = 20 2 2 2 2 2 2 CBSE CLASS X MATHEMATICS CHAPTER 5 ARITHMETIC PROGRESSIONS 1. therefore.P. Let the length and breadth of the park be x and y. a . . th 1 2 3 1 2 3 2 4 3 n n n m Benefitted: My Paypal a/c . . we arrive at the nth term. second term by a . Arithmetic Progressions: Consider the following lists of numbers :  1. where l = a + (n – 1) d. . 2. then a represents the last term which is sometimes also denoted by l. 4. the nth term an of the AP with first term ‘a’ and common difference ‘d’ is given by a = a + (n – 1) d. nth term by a and the common difference by ‘d’. 40x − x = 400 => x . Residence: Ghaziabad-UP . 4. . An infinite Arithmetic Progression does not have a last term. and. a . Mobile (where u may transfer cash!) .com. Let us denote the first term of an AP by a . a . . . In each of the above two lists... 40. be an AP whose first term a is ‘a’ and the common difference is ‘d’. ……. we obtain.manojarora23@gmail. and so on. Such list of numbers is said to form an Arithmetic Progression (AP).. . n term of an A. 1)d] comes ‘n’ times.P. Residence: Ghaziabad-UP .) Thus.1)d] = n[a + a ] (3) 2 2 2 n Now. if there are only n terms in an AP. If we know any three of them. a = S – S .1)d] ( The term [2a + (n .manojarora23@gmail. + (a + d ) + a (2) On adding (1) and (2). S = The sum of first n terms of an AP = n [2a + (n . 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. The formula for the sum of the first n terms involves four quantities S. the last term. d and n. Sum of First n Terms of an AP: Let the given A. …..1)d] + [2a + (n . a + 2d. a.e. be a.09871823473. then a = ‘l’. which is the required term.1)d] = n[a + a + (n .1)d] Or. So.. term-wise. + [2a + (n ..1)d] + ……. 2S = n[2a + (n . we can find the fourth.36 So.com. We also use S in place of S to denote the sum of first n terms of the AP. Let S denote the sum of the first n terms of the AP. n n n 20 n–1 Benefitted: My Paypal a/c . + [a + (n – 1) d ] (1) Rewriting the terms in reverse order. The nth term of this AP is a + (n – 1) d. Mobile (where u may transfer cash!) . i. . 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. We write S to denote the sum of the first 20 terms of an AP. . . we get 2S = [2a + (n . a = – 62 + (11 – 1) × 3 = – 62 + 30 = – 32. . a+ d. From (3). we have S = [a + (n – 1) d ] + [a + (n – 2) d ] + . We have: S = a + (a + d ) + (a + 2d ) + . 11 4. the question now becomes finding the 11th term with these a and d. In which of the following situations. 39 … forms an A. only 1 . a . 2 3 Thus.P.1/4 = 3/4 th part of air will remain. after every year. a = a + d = 10 + 10 = 20. adjacent terms of this series do not have the same difference between them. Let the series be a .NCERT EXERCISES SOLUTIONS EXERCISE 5. r n P 1+ A. 31. A. a = a = 10. a … Therefore. 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. Rs P deposited at r% compound interest per annum for n years. 23. (iii) The cost of digging a well after every metre of digging. this is not an A. this isn’t an A. a . 300 … forms an A. volumes will be V. 150. Mobile (where u may transfer cash!) .1 (P. 99): Q1. 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. (3V/4). Let the initial volume of air in a cylinder be V lit. when the first term a and the common difference d are given as follows: (i) a = 10. does the list of numbers involved make an arithmetic progression. In other words. becomes after n years. the vacuum pump removes 1/4 of air remaining in the cylinder at a time. Write first four terms of the AP. 200.P. Residence: Ghaziabad-UP .a = a + d = 20 + 10 = 30. when Rs 10000 is deposited at compound interest at 8 % per annum.com. when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.37 CBSE CLASS X MATHEMATICS CHAPTER 5 ARITHMATIC PROGRESSIONS. (iv) The amount of money in the account every year. a . d = 10 A. In each stroke.P. adjacent terms of this series don’t have same difference between them.manojarora23@gmail. (3V/4) . (3V/4) …. A. because every term is 8 more than the preceding term. because every term is 50 more than the preceding term. 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. 250. 1 1 2 2 3 4 1 5 3 2 Benefitted: My Paypal a/c . A. our money will be Thus. 100 ( ) Therefore. Thus. Q2. after every stroke.P. Therefore.09871823473. Therefore. (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. 25.. Let the series be a . Mobile (where u may transfer cash!) . 0.P. -5 …. d = 0 A. . a = a + d = 40 + 10 = 50 Therefore. 1. Residence: Ghaziabad-UP . a … a =a=4 a =a +d=4-3=1 a = a + d = 1 . will be -1. −2. 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.0 Therefore. −1.38 a = a + d = 30 + 10 = 40. Please Do It Yourself! Q4. d = -3 A.P. 20. ie. −2. 50 … First four terms of this A. 0 First four terms of this A. the series will be 10. √8. a .25. -1/2.manojarora23@gmail. 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. Let the series be a . –0. ie. as the difference between the two consecutive terms is constant. and 1/2. -1/2.. will be -1.P. the series will be 4. which is an A. and −2. 20.com. −2 and −2. as the difference between the two consecutive terms is constant. Therefore.a = √2 = a .P. 40. 0 First four terms of this A. −1.3 = −2 a = a + d = −2 .00 1 2 3 4 1 2 1 3 2 4 3 Q3.P.P. as the difference between the two consecutive terms is constant.P. 0.5 . −2 …. d = -0. which can be positive. a . which is an A. Let the series be a . ie. 4 3 5 1 2 3 4 4 1 2 1 3 2 4 3 (iii) a = 4.50. Read the following AND Please Do The Question Yourself! A.25 a = a + d = -1. Let the series be a .0. √32…. will be 4. .75 . 1/2 First four terms of this A.25 . -2. (ii) a = -2.50. . And. a … a = a = −2 a = a + d = − 2 + 0 = −2 a = a + d = − 2 + 0 = −2 a = a + d = − 2 + 0 = −2 Therefore. a . 1 2 3 4 1 2 1 3 2 4 3 (v) a = -1. the series will be -1.0. d = 1/2 A. which is an A. −2.25 = -2.. find the common difference d and write three more terms. as the difference between the two consecutive terms is constant.25 First four terms of this A. Here a . 30. will be 10.a … i. 30. a .75 a = a + d = -1. the given numbers are in A. a .25 A. −1. will be −2. the series will be −2.09871823473.P.25 = -1. 1/2. a .25 = -1. . the series will be -1.75. A.3 = −5 Therefore. a − a is same every time. a … a = a = -1. a .75. √18. -2. which is an A. −1. a .25.00 …… . 1. negative or zero. and 40.P.0. d = √2. Which of the following are APs ? If they form an AP. −2. 1 2 3 4 1 2 1 3 2 4 3 (iv) a = -1.. Three more terms are:…… 2 k+1 1 3 2 4 3 k Benefitted: My Paypal a/c . This fixed number is called the common difference of the AP.5 a = a + d = -1.P.a = a . ie.e. (xii) √2. and -5 . n 17 17 Benefitted: My Paypal a/c . a = a + (n − 1) d 2 4 n a = a + (2 − 1) d => 13 = a + d (i) a = a + (4 − 1) d => 3 = a + 3d (ii) Solving. Therefore. Given that. Solving. Now. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the common difference. Residence: Ghaziabad-UP . Thus. 15 1/2. Mobile (where u may transfer cash!) . a = a + (17 − 1) d => a = a + 16d. a = a + (9 − 1) d => −8 = a + 8d (ii). 3 9 n 3 9 n Q10. a = a + (3 − 1) d => 4 = a + 2d (i). is 0. find the value of ‘n’. Do It Yourself!! Q2. 8. a = 4. a. Do It Yourself!! Q3. 5th term of this A. d. . ‘a ’ are given.com. and. and.P. ‘d’. a = a + (n − 1) d Thus. In both the above parts ‘a’. Do both the questions yourself! Q8. 2 . and. . . a = a + (n − 1) d a = a + (3 − 1) d => 12 = a + 2d (i) Similarly. 13. Now. a = 18 a = 18 + (3 − 1) (−5) = 18 + 2 (−5) = 18 − 10 = 8 Therefore. a = 13 and a = 3.. a = 2. Given that. Using the appropriate formula. . 13. a = a + (n − 1) d. which term of this AP is zero? A. 5.P. a = 12. A. A. we get. a = 106. Now. and. 29th term is 64. . and. d = −2. The 17th term of an AP exceeds its 10th term by 7. Find the number of terms in each of the following APs : (i) 7.09871823473. a = a + (n − 1) d 3 n a = 2 + (3 − 1) d => 26 = 2 + 2d => d = 12 => a = 2 + (2 − 1) 12 = 14 Therefore. 19. . 105) Q1. Thus. Hence.2 (P.. Find the value of ‘n’. Q7..manojarora23@gmail. . find the missing terms in the boxes : A. 8. Check whether – 150 is a term of the AP : 11. n Q6. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. Let nth term of this A. a = −8. and a is given.P. be zero. 205 (ii) 18. a = 8 a = a + (29 − 1) d => a = 8 + (28)2 = 64. Find the 29th term.P. d = -5. n Q5.P. 14 is the missing term. 3 2 (ii) For this A. a = 26. -47 A. A. 13. . 3 50 n 3 50 29 29 Q9. a = 8. .is 78? A. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively. d = 2. …. and. 18. For an A. .39 _______________________________________________________________________________________ EXERCISE 5. the missing terms are 18 and 8 respectively 2 4 3 (iii) Please do the rest of the parts yourself! Q4. (i) For this A. Which term of the AP : 3. Now. a = a + (n − 1) d => 0 = 8 + (n − 1) (−2) => n = 5. a = a + (50 − 1) d => 106 = a + 49d (ii) Solving. In the following APs. a = a + (n . .P. with first term as 12 and common difference as 4. 17.s = 100 Therefore. n = ? a = a + (n − 1) d => 994 = 105 + (n − 1) 7 => 889 = (n − 1) 7 => n = 128 Therefore. a = a + (100 − 1) d = a + 99d a = a + (1000 − 1) d = a + 999d For second A. a = 3. Two APs have the same common difference. Let nth term be 771 => a = a + (n − 1) d => 771 = 3 + (n − 1) 12 => 768 = (n − 1) 12 => n = 65 Therefore. 105. the common difference is 1. … Thus. 2 1 54 n th Q12.com. 27. 248 Let 248 be the nth term of this A. difference between 100th term of these A. Residence: Ghaziabad-UP . 10 17 10 Q11. Alternatively: Let nth term be 132 more than 54th term. …. When we divide 250 by 4. Given A. 128 three-digit numbers are divisible by 7. 67. n = 54 + 132/12 = 65 term. 15. 20.P. the remainder will be 2. which is a multiple of 4.1)4 => n = 60.’s be d. Therefore.09871823473. Dividing it by 7. 112. n n Q15. For what value of n.s be a and a and the common difference of these A. Thus. n n Q14. First multiple of 4 that is greater than 10 is 12. How many multiples of 4 lie between 10 and 250? A. Mobile (where u may transfer cash!) . Clearly.P. 132 + 639 = 771 We have to find the term of this A.P. Thus the series is as follows: 105. a = 994. and 3. all are divisible by 4 and thus. are terms of an A. 16. will be 132 more than its 54th term? A. . Therefore. 65th term was 132 more than 54th term. a − a = 100 Hence. the difference between 1000 terms of these A. there are 60 multiples of 4 between 10 and 250.P. . 20. 16. a = 105.manojarora23@gmail. which is 771. Therefore. having first term as 105 and common difference as 7. …. 119.P.. 65. The series is as follows.s = (a + 999d) − (a + 999d) = a − a From equation (i). . . Next will be 16. Which term of the AP : 3.. (a + 99d) − (a + 99d) = 100 => a − a = 100 (i) Difference between 1000th terms of these A. This difference. . will be 100.1)d => 248 = 12 + (n . The difference between their 100th terms is 100. 112. what is the difference between their 1000th terms? A. The maximum possible three-digit number is 999.40 a = a + 9d. 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7. the remainder is 5. 24. d = 7. 39. Let the first term of these A. 119. . It is given that a − a = 7 => (a + 16d) − (a + 9d) = 7 => d = 1. First three-digit number that is divisible by 7 = 105 Next number = 105 + 7 = 112 Therefore.P.P. Thus. ….P. 12. Therefore. all these are terms of an A. a = 248. d = 4. …. Thus. are the nth terms of two APs: 63.P.P. 994 Let 994 be the nth term of this A. equal? Benefitted: My Paypal a/c . How many three-digit numbers are divisible by 7? A. 15. . d = a − a = 15 − 3 = 12 a = a + (54 − 1) d = 3 + (53) (12) = 3 + 636 = 639 Now. 250 − 2 = 248 is divisible by 4. 12. a = a + (100 − 1) d = a + 99d a = a + (1000 − 1) d = a + 999d Given that. and is the maximum number before 250. 10. 39. For first A. . 27. 24. all are three digit numbers which are divisible by 7 and thus.P. 1 100 1000 1 1000 2 1 2 1 1 100 2 2 2 1 2 1 2 1 1 2 1 2 2 th Q13. is 3. a = 12.P. a = a + 9d Given that. 8. a = a + d = − 13 + 5 = −8. and. we get. 12. Mobile (where u may transfer cash!) . . 20th term from the last term is 158. = a = a + (n − 1) d = 63 + (n − 1) 2 = 63 + 2n − 2 => a = 61 + 2n (i) 3. thus. her weekly savings become Rs 20. nth term of this A. 16. For this A. … Here. n n _______________________________________________________________________________________ EXERCISE 5. are −13. If in the nth week. … => a = 3. Therefore. we get. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44.75 = 5 + (n . d = 200 Let after nth year. 5400. is 5.1)1. a = 16 => a + (3 − 1) d = 16 => a + 2d = 16 (i) As.P. ….P. in 11th year. 63. . 253 ‘d’ for this A. Therefore. 13. d = 1. 5200. d = a − a = 10 − 3 = 7 nth term of this A. …. 8. A.P. is 3. d = a − a = 65 − 63 = 2. . n = 10. d = a − a = 7 − 2 = 5.P. a = 5. 17. and. a = 20. is 4.75. this A. a = a + 5d.75. a = 2.s are equal to each other. a = 5000.P. Therefore..1)d) = 10/2(2x2 + (10 . 67. Given A. 5 For this A. S = n/2(a + a ) = n/2(a + a + (n . S = 5(4 + 9x5) = 5 x 49 = 245. We know that.P. to 10 terms. A. a = a + (n − 1) d => 7000 = 5000 + (n − 1) 200 => n = 11. A. a = a + (n − 1) d => a = a + (4 − 1) d => a = a + 3d Similarly. can be written in reverse order as 253.09871823473.3 (P. and −3. 8.. . n 4 8 6 4 8 6 10 4 10 2 3 2 Q19. In which year did his income reach Rs 7000? A. 2 1 n n 2 1 n n Q16. a = a + d = − 8 + 5 = −3 Therefore. 248.41 A. 2 n 1 n n Benefitted: My Paypal a/c . a = 4 Therefore.75 => n = 10. 13. 112): Q1. It can be observed that the incomes that Subba Rao obtained in various years are in A.com.P. 253. his salary will be Rs 7000. 243. … 3 7 5 Q17.P.1)x5) Or.75. Find the 20th term from the last term of the AP : 3. .75.P. and d = 5 Thus. the first three terms of this A. 61 + 2n = 7n − 4 => n = 13. 7. Find the first three terms of the AP. the required A. n = 10. We know that. his salary is increased by Rs 200.. −8. .. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. Hence. the salaries of each year starting 1995 are 5000.manojarora23@gmail. and. 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. 65. … => a = 63.s are equal to each other. d = 248 − 253 = −5. find n. Equating both these equations. a − a = 12 => [a+ (7 − 1) d] − [a + (5 − 1) d]= 12 => (a + 6d) − (a + 4d) = 12 => d = 6 Putting this in equation (i). 10.1)d) = n/2(2a + (n . 13th terms of both these A. n = ? a = a + (n − 1) d => 20. n Q20.P. and. 22. Residence: Ghaziabad-UP . Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1. A. a = -13. 10. = a = 3 + (n − 1) 7 => a = 7n − 4 (ii) It is given that. as every year. a = a + 7d. A. nth term of these A. his salary be Rs 7000. we get. n = 20 a = a + (20 − 1) d => a = 253 + (19) (−5) = 253 − 95 = 158 Therefore. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. Therefore.P. Given that.P. Find the sum of the following APs: (i) 2. 13. a = 253. 20 20 Q18. . S = 125. For this A. S = 75. (vi) given a = 2.. (ix) given a = 3. Now. to 11 terms.P. Find the number of terms and the common difference. Do parts (ii). . S = n/2 (a + l) = 13/2 (34 + 10) = 286. If the common difference is 9.com. .09871823473.. the last term is 45 and the sum is 400... + 10 A. . Do it Yourself! Q8. d = 8. A. (vii) given a = 8. In an AP: (i) given a = 5. n = 8. a = −5. For this A. . 1. Find the sums given below : (i) 7 + 10 1/2 + 14 + …….7. .a = 10 1/2 . S = 90. how many terms are there and what is their sum? A. We know that. 25. d = a . Find a. Do it Yourself! Q5.. (iii). . l = a = a + (n − 1)d => 84 = 7 + (n . Residence: Ghaziabad-UP . S = –14. d = 2. A. l = 10 Let 10 be the nth term of this A. to 100 terms. A. find n and a . S = 144. Q2. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. l = 84. A. (ii) given a = 7.1)x7/2 => n = 23. (iv) given a = 15. A. How many terms of the AP : 9.P. find n and d. l = a + (n − 1) d. (x) given l = 28. find d and a .42 (ii) –37. (iv) 1/15. 10 = 34 + (n − 1) (−2) => n = 13. find d. a = 7.P. a = 35. (iii) given a = 37.manojarora23@gmail. + 84 A. find n and S . 2. find a and a . find n and a. . find d and S . S = n/2(a + l) = 23/2(7 + 84) = 1046 1/2 2 1 n n Q2 (ii) 34 + 32 + 30 + . d = a − a = (−8) − (−5) = -3 Let −230 be the nth term of this A. Follow the trick provided in the class. Thus. (viii) given a = 4. The first and the last terms of an AP are 17 and 350 respectively. For this A. . Or.7 = 7/2 Let 84 be the nth term of this A. + (–230) A. Do it Yourself! Q7. . 1/12. –33. l = a + (n − 1)d => −230 = − 5 + (n − 1) (−3) => n = 76 Now. d = 3.6. find the sum of first n terms. S = n/2(a + l) = 76/2 [(-5) + (-230)] = -8930 2 1 n Q3. and (iv) yourself. Mobile (where u may transfer cash!) . . find a and S . 17.8. . (iii) 0. S = 210. d = 3. and there are total 9 terms. a = 62. to 12 terms.. 2 1 n (iii) –5 + (–8) + (–11) + . Do it Yourself! Q6. a = 34. S = 192. ie.P. –29. l = −230. 1/10…. (v) given d = 5. The first term of an AP is 5.. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. and do the questions yourself! n n 13 13 12 12 3 10 10 9 9 n n n n n n Q4.P. Do it Yourself! Q9. must be taken to give a sum of 636? A. a = 50. . d = a − a = 32 − 34 = −2.P. Benefitted: My Paypal a/c . ‘d’ = −20. If each prize is Rs 20 less than its preceding prize. whose first term is 6 and common difference is 6.. a = 9 − 5 × 2 = −1. S = n/2[2a + (n . Residence: Ghaziabad-UP .1)d] => S = 20[2(6) + (40 -1)6] = 4920 40 n 40 Q13. Find the sum of the first 40 positive integers divisible by 6. a = 5 − 2(3) = 5 − 6 = −1. and Rs 40. and. Therefore. Mobile (where u may transfer cash!) . d = −20. Rs 80. a = S = 4(1) − (1) = 3 Sum of first two terms = S = 4(2) − (2)2 = 4 Second term. Therefore. S = 30/2[2(200) + (30 . . a = 9 − 5 × 4 = −11 It can be observed that a − a = a − a = a − a = -5. cost of 3rd prize = P − 40 Thus. and 5 − 2n respectively. Rs 300 for the third day. Let the cost of 1st prize be P. Rs 140. a = 6. How much money the contractor has to pay as penalty. the cost of these prizes are in an A.P. Thus.1)d] => S = 15/2[2(4) +(15 . Penalty that has to be paid if he has delayed the work by 30 days. . Find the sum of the first 15 multiples of 8. and.1)(-5)] = -465 Q11. Now. etc. The positive integers that are divisible by 6 are 6. Rs 60. a = 3 + 4(4) = 19 It can be observed that a −a =a −a =a −a =4 i.e. Do it Yourself! Q10. a − a is same every time. Rs 100. Therefore. a − a is same every time. a = 9 − 5 × 1 = 4.43 A. with ‘a’ as 200 and ‘d’ as 50.. 12. => Do It Yourself! Q14. the value of each of the prizes was Rs 160. In a school. 18. if he has delayed the work by 30 days? A. Find the sum of the odd numbers between 0 and 50. what is the first term (that is S )? What is the sum of first two terms? What is the second term? Similarly. . this is an AP with ‘d’ = -5 and ‘a’ = 4. . a = 3 + 4(2) = 11. ie. S = 700 => 7/2[2a + (7 . 10th. 24 … It can be observed that these are making an A. 7 Q17. i.P. −15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day. A. a = a + (n − 1)d = 3 + (n − 1) (−2) = 3 − 2n + 2 = 5 − 2n Therefore. A. Given that. find the value of each of the prizes. Rs 120.. the contractor has to pay Rs 27750 as penalty.e. d = 6. the 10th and the nth terms. It can be observed that these penalties are in A.. the penalty for each succeeding day being Rs 50 more than for the preceding day. a = 9 − 5 × 3 = −6. Sn = 4n − n First term. this is an AP with ‘d’ = 4 and ‘a’ = 7. a . Given that. S = n/2[2a + (n . . Now. the sum of first two terms is 4. Therefore.1)50] = 27750. A. If the sum of the first n terms of an AP is 4n – n . students thought of planting trees in and around the school to reduce air pollution.com. cost of 2nd prize = P − 20. The second term is 1. (i) a = 3 + 4(1) = 7. S =? Now. 1 2 n n 1 2 n n 2 1 3 k+1 2 4 3 4 3 k n 15 n 1 2 3 2 k+1 1 3 2 4 4 3 k n 15 2 1 2 2 1 2 2 2 2 1 1 n 3 10 Q12.1)(-20)] => a = 160. Rs 250 for the second day. Then. a = 3 + 4(3) = 15. having ‘a’ = P. . find the 3rd. 30 Q. a .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. and nth terms are −1. S = n/2[2a + (n .09871823473.P. 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 = S − S = 1 d = a − a = 1 − 3 = −2 Now. a = P. => Do It Yourself! Q15.manojarora23@gmail. It was Benefitted: My Paypal a/c . 3rd. and. a = 5 − 2(10) = 5 − 20 = −15 Hence. Show that a . A.1)d] => S = 15/2[2(7) + 14x4] = 525 (ii) a = 9 – 5n A. 1)x6] = 5[20 + 54] = 370. total of 200 logs are to placed in ‘n’ rows. 10 10 _______________________________________________________________________________________ EXERCISE 5.20 In a potato race. total trees planted (by all the sections) = 234.forms an A. 34.. . The number of rows in each row.. a bucket is placed at the starting point. . 14…. and.1)(-1) = 20 . Residence: Ghaziabad-UP . 1.4 (P. runs back to pick up the next potato.0 cm.1)d] => 200 = n/2[2x20 + (n .. A ladder has rungs 25 cm apart.P. will be the same as the class.P. which form an A. => Number of Rows = n = 16. 113. of radii 0.. which is the firstnegative term. The houses of a row are numbered consecutively from 1 to 49. A spiral is made up of successive semicircles.] Q 2. For. . a . the number of rows in the 16 row = a = 5.4d. The distances of potatoes are as follows: 5. Q3..13 terms.1)0. 2π. 115) Q1.1)(-1) = 20 .. and. a = 20 + (16 .. and she continues in the same way until all the potatoes are in the bucket. picks up the nearest potato. …. a = a + (n − 1)d. What is the total distance the competitor has to run? A. 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5. n> 125/4 or n > 31 1/4 . 8. e. Mobile (where u may transfer cash!) . as shown in the adjoining figure. 18 in the row next to it and so on.g.5π. S = 20.P. with ‘a’ = 20. drops it in the bucket.0 cm.………. A competitor starts from the bucket. made up of 13 consecutive semicircles is S = 13/2[2x0. d = 16 − 10 = 6. such that. is its first negative term? A. . ie. 76. a section of Class II will plant 2 trees and so on till Class XII. => n = 32nd term. a = 20 + (25 . 1. Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. ‘d’ = ‘0. n 2 n 2 16 25 th 16 Q. 19. 200 logs are stacked in the following manner: 20 logs in the bottom row. S = 10/2[2x10 + (10 .. 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. and the other potatoes are placed 3 m apart in a straight line. This forms an A. Now. n = 10. Thus. . Show that there is a value of x such Benefitted: My Paypal a/c . 20.2d………. 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. a = 3. a .5π + (13 . The sum of the third and the seventh terms of an AP is 6 and their product is 8. S =?. [Hint : Find n for a < 0. and. As the number of rows can’t be negative. There are ten potatoes in the line. How many trees will be planted by the students? A. a = 10. π. n 16 Q4.5π] = 13/2[π + 12/2π] = 13/2 x 7x 22/7 = 13x11 = 143cm. that each section of each class will plant. let the AP be a . a section of Class I will plant 1 tree.1)(-1)] => 400 = n (40 − n + 1) => 400 = n (41 − n) => 400 = 41n − n => n − 41n + 400 = 0 => (n − 16) (n − 25) = 0 Thus.3d. There are three sections of each class. Find the sum of first sixteen terms of the AP. 13 Q19. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. and.44 decided that the number of trees.manojarora23@gmail. 117.15 = 5 Similarly.25 + 1 = -4. 11. n = 16. what is the length of the wood required for the rungs? A. 16. in which they are studying.5 cm. ‘d’ = 1 Let there be total ‘n’ rows in total. Therefore. If the top and the bottom rungs are 2 1/2 m apart. A. 1. which is 5 m from the first potato. Q18. In how may rows are the 200 logs placed and how many logs are in the top row? A. Therefore..5π. starting with centre at A. 22. Therefore. for n = 25. S = n/2[2a + (n .5 cm. d = ±1/2. total length of the spiral. runs to the bucket to drop it in. runs back with it. distances to be run are 10. 2. Which term of the AP : 121.5π’. the competitor will run a total distance of 370 m.com.5π. with ‘a’ = 0. and n = 13 Thus. 19 in the next row. (n − 16) = 0 or n − 25 = 0 => n = 16 or n = 25 Now. thus. 28. with centres alternately at A and B. 18. .09871823473. ∠1=∠5. the non-common arm of the angles are the two opposite rays. S = 49x25. Benefitted: My Paypal a/c . third step…. ∠3=∠7.15 terms] (a = 1. A. (2 x 1/4) x 1/2 x 50. Find this value of x.S . Axiom 1: If a ray stands on a line. Each step has a rise of 1/4 m and a tread of 1/2m. 49 x—1 49 x 3 3 CBSE CLASS X MATHEMATICS CHAPTER 6 TRIANGLES RECAP FROM PREVIOUS CLASS/ES  LINES AND ANGLES: > An Angle is formed by two rays. such that a pair of corresponding angles are equal. = 50/8 [1 + 2 + 3 . A.. > Acute angle > Obtuse angle  Two angles are Complementary / Supplementary if their sum is 90 / 180 .09871823473. ie. ie. ∠3 + ∠5 = 180 . ie. > They have a common arm. then. ∠3=∠6. then the two lines are parallel to each other. Axiom 2: If the sum of two adjacent angles is 180 . ∠2=∠6. 3 x 50/8…. ∠4=∠8. (This Axiom is also called Corresponding Angles Axiom). then the sum of two adjacent angles is 180 . Alternate Exterior Angles.  Types of Angles: > Right angles > Straight Angle > Reflex Angle: > 180 and < 360 . Residence: Ghaziabad-UP . 2 x 50/8. These two axioms together are called Linear Pair Axiom. d = 1) = 750m . (Axiom) > Each pair of alternate (interior) angles are equal.. ie.  Parallel Lines and Transversal: When a transversal intersects two parallel lines. Volume req’d = 50/8 + 2 x 50/8 + 3 x 50/8…. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Axiom 4: If a transversal intersects two lines. Linear Pair: If OA & OB are opposite rays. ∠4=∠5.com. Mobile (where u may transfer cash!) . Calculate the total volume of concrete required to build the terrace.  Two angles are Adjacent Angles if : > They have a common vertex.manojarora23@gmail.. ∠4 + ∠6 = 180 . d = 1. > Their non-common arms lie on opposite sides of the common arm. > Each pair of co-interior angles are supplementary. 50/8. ∠1=∠8. then ∠BOC & ∠AOC form a linear pair. > Each pair of corresponding angles are equal. 0 0 0 0 0 0 0 0 Axiom 3: If a transversal intersects two parallel lines. then the non-common arms of the angles form a line. > Congruent or Equal Angles have same measure. a = 1. and OC is any other ray. x = 35.. S = S . ie. forms the following angles: Corresponding Angles. The converse of each of the above four statements is also true. and. Q5. (in m ) is 1/4 x 1/2 x 50. then each pair of corresponding angles are equal. Arms or Sides are the rays forming the angle. (Theorum) > Each pair of alternate (exterior) angles are equal..  A Transversal is a line which intersects two or more given lines in distinct points. and CoInterior (or Consecutive Interior) Angles [ These are interior angles on the same side of transversal]. ∠2=∠7. (3 x 1/4) x 1/2 x 50…. Alternate Interior (or just ‘Alternate’) angles. second. Volume of concrete req’d to build the first. Vertex is the common initial point of the two rays.45 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. with common initial point. Thus.: In any ∆. Angles = 180 . AC = AD (by cons. Thus. Use RHS). to the larger angle is longer. (Exterior Angle = Sum of Interior Opp. ∠1 = ∠2 (Th. > ASA. TRIANGLE (RECAP FROM PREVIOUS CLASSES)  Congruence: Two geometric figures are said to be congruent. ∆PSR ≅ ∆OSR. ∠3 > ∠2. if they are exactly of the same size and shape. Use CPCT). (Use Linear Pair Axiom). then the vertically opposite angles are equal. CD) is equidistant from A & B (SAS).09871823473. Thus. the side opp. The exterior angle of a triangle is greater than either of its interior opposite angles. and line CD is its perpendicular bisector. AB + AD > BC. Residence: Ghaziabad-UP . Or. In ∆ BCD. > Theorem 5: If a side of a triangle is produced. (Given: AC = BC. > Theorum: In any ∆. draw a line parallel to the opposite side of the triangle).: Angles opp.  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. A point ‘D’ on the perpendicular bisector (ie. > Theorem 2: If two lines intersect each other. > Theorum: If two sides of a triangle are unequal. BD > BC ( Th.(The Third Angle). ie. there are 6 elements in a Triangle). then the exterior angle so formed is equal to the sum of the two interior opposite angles. To Prove: AB + AC > BC Cons.: Produce BA to D such that AD = AC. > RHS  THEORUMS: > Theorum: Let AB be a line segment.). the angle opposite to the larger side is larger (Experimental Verification). > SSS. > Theorum: The sum of any two sides of a ∆ is greater than the third side. (Draw RS ⊥’r to PQ. to eq. then the sides opposite to them are also equal.  CPCT: Corresponding Parts of Congruent Triangles. Draw CD ⊥’r to AB. AB + AC > BC (As AD = AC (by cons. the Hypotenuse is the Benefitted: My Paypal a/c . ∠3 > ∠1 (As ∠1 is part of ∠3).46  THEORUMS: > Theorem 1: The sun of all the angles around a point is 360 (Use Linear Pair Axiom). are parallel to each other (Use Corresponding Angles Axiom). > Theorum: (Converse of above) If two angles of a triangle are equal. to the larger angle is longer (Experimental Verification). > Theorem 4: The sum of angles of a triangle is 180 (From a Vertex. > Theorum: Angles opposite to equal sides of an Isosceles Triangle are equal. sides……) 0 0 0 But. ).  Different conditions for Congruence of two triangles: > SAS (This is an axiom and can be used without proof). Proof: In ∆ ACD. & join CD. > AAS.com. Mobile (where u may transfer cash!) . By AAS. > Theorem 3: Lines which are parallel to the same line.)) > Theorum: (Corollary of above) In a Right Angled Triangle.manojarora23@gmail. the side opp. com. Proof: As K lies on the bisector of ∠M.09871823473. OB & OC. & (ii). Proof: Since O lies on ⊥’r bisector of AB. & the point of concurrency K is the incentre of the triangle NMO. Incircle: If we draw a circle with centre K & Radius KV. KV = KU -(ii) From (i).(i) Benefitted: My Paypal a/c . and.manojarora23@gmail. ∴ MK. NK are concurrent. > Theorum: The perpendicular bisectors of the sides of a ∆ pass through the same point. They are concurrent. bisectors of angles M & O intersect at K. KV = KU = KT = Inradius.47 longest side. KN is joined. To Prove: KN bisects ∠N. then they are said to be concurrent. N. Also OX ⊥ BC is drawn. OA = OB . KV = KT -(i) As K lies on the bisector of ∠O.B. it will pass through T & U. Mobile (where u may transfer cash!) . Residence: Ghaziabad-UP . KT ⊥ MN. OK. > Theorum: A point on an angular bisector of an angle.: Two non-parallel lines always intersect. This circle is called Incircle of ∆NMO. the common point is called the point of concurrency. in which ⊥’r bisectors of sides AB & AC intersect at O. The point K is equidistant from the three sides of the triangle MNO. > Theorum: The angular bisectors of a triangle pass through the same point. Incentre: The point of intersection K of the angle bisectors is called the incentre of the Triangle MNO. To Prove: OX is perpendicular bisectors of BC. Const. Const. ∴ KN bisects ∠N. KU ⊥ NO. KU = KT => K is equidistant from MN & NO. If two or more lines pass through the same point.: Join OA. Given: ∆ABC.: Draw KV⊥ MO. ie. is equidistant from the arms of the angle. Given: ∆NMO. ‘Regular Polygons’ are the ones in which all the sides and all the angles are equal to each other. OB = OC (iii) => O lies on ⊥’r bisector of BC. Theorem: In a triangle. is called an altitude of the triangle. Residence: Ghaziabad-UP . if  Their corresponding angles are equal. & C of ∆ABC. 2. or. Also. perpendicular to the line containing the opposite side. OA = OB = OC = Circumradius => The point O is equidistant from the three vertices A. OA = OC . All the circles are similar. This common point is called the ortho-centre of the triangle. ‘Centroid’ divides each of the median in the ratio 2:1. Centroid: The point of intersection of the three medians of a triangle is called the ‘Centroid’ of the triangle. and Benefitted: My Paypal a/c . ∴ OX is perpendicular bisectors of BC.  The Locus is a set of all the points that satisfy a particular condition. ————————————–——————————————--————————————————————————— ENOUGH OF REVISION – CLASS X TRIANGLES NOW! 1. the three altitudes pass through the same point. Eg. Theorem: Medians of a triangle pass through the same point which divides each of the medians in the ratio 2:1. Introduction: Two figures are congruent. if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).). Two polygons of the same number of sides are similar.(ii) From (i) and (ii).48 Since O lies on ⊥’r bisector of AC.09871823473.manojarora23@gmail. and this common point is called the circumcentre of the triangle. The ortho-centre of a right-angled triangle is the vertex of the right angle. if they have the same shape and the same size.  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. we get. A Circle is "the locus of points on a plane that are a certain distance from a central point". Two figures having the same shape (and not necessarily the same size) are called similar figures. Circles of equal radii are congruent. All congruent figures are similar but the similar figures need not be congruent. Mobile (where u may transfer cash!) .  Circumcentre: The Perpendicular bisectors of the sides of a Triangle pass through the same point.  Altitude: The line segment from a vertex of a triangle. Techniques of Indirect measurements are based on the principle of similarity of figures. > The medians of an equilateral triangle coincide with the altitudes. Similar Triangles and Their Properties: Two triangles are said to be similar to each other. The same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons. share a property.com. Note the following two (provable) points about medians: > The medians of an equilateral triangle are equal. OX ⊥ BC (by const. B. To Prove: DE ∥ BC. if in two triangles. BD. On measuring AE & EC. we find AE/EC = 3/2. AB AC Or. DE ∥ BC. DM. 3. AD AE AD AE (ii) From the ‘Basic Proportionality Theorum’. it is not necessary. Residence: Ghaziabad-UP .∨. DE is parallel to BC. then (i) AB/AD = AC/AE. Now. Proof: Let DE not be parallel to BC. if the one condition holds good. draw any line intersecting arm AY at C. and a line ‘l’ intersecting AB in D.09871823473.manojarora23@gmail. Q. Thus. (ii) AB/DB = AC/EC (i) From the ‘Basic Proportionality Theorum’. we have. AD/DB = AE/EC. EN. = Or. DB/AD = EC/AE. AD/DB = AE/EC. For triangles. To Prove: AD/DB = AE/EC Cons. This is not the case with Quadrilaterals. D. the corresponding sides are automatically proportional to each other. Ar (∆BDE) = 1/2. that the other condition will also hold good. AD A E ' = DB E' C . mark points P. AE. AD/DB = AE/EC. the corresponding angles are equal. 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. from (i) & (ii). such that AD AE = DB EC . then the line is parallel to the third side. Now. the other two sides are divided in the same ratio. EN. DM. through B. Ar (∆ADE) = 1/2. Activity to prove ‘Basic Proportionality Theorum’: Draw any angle XAY and on AX. draw a line parallel to BC to intersect AC at E.: Draw DM ⊥ AC and EN ⊥ AB. AD/BD = AE/CE  Corollary (Conclusion / Inference) from above Theorum: If in a Triangle ∆ABC. Thus. AD / DB = 3 / 2 (by construction). and AD/DB = AE/EC. Ar (∆ADE) / Ar (∆BDE) = AD/BD — (i) Also.49  Their corresponding sides are proportional. CE. Through D. Thus. 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. Proof: Ar (∆ADE) = 1/2. Given: ∆ABC. Ar (∆CED) = 1/2.com. ‘Thales’ stated that the ratio of any two corresponding sides in equiangular triangles is always the same irrespective of their sizes. AD/DB + 1 = AE/EC + 1. If one condition holds good. Thus. Ar (∆BDE) = Ar (∆CED) (∆’s on the same base & between the same parallels) Thus. Or. R and B such that AP = PQ = QD = DR = RB. these two conditions are dependent on each other. By ‘Basic Proportionality Theorum’. Thus. for which the two conditions are independent. DB EC AB AC +1= +1 . the other holds good automatically. Mobile (where u may transfer cash!) . in ∆ABC. That is. and then let DE’ be another line parallel to BC. That is. AD. Or. Given: ∆ABC in which DE ∥ BC. ie.(i) Benefitted: My Paypal a/c . & AC in E. Ar (∆ADE) / Ar (∆CED) = AE/CE — (ii) Now. since DE’ ∥ BC. ∠A = ∠D. Δ ABC ≅Δ DPQ (SAS).09871823473. we can write Δ BAC ~ Δ EDF.com. This is only possible if E & E’ coincide. then their corresponding sides are in the same ratio ( or proportion) and hence the two triangles are similar (AAA Criterion). ∠ C = ∠ F and (ii) AB BC CA . Mobile (where u may transfer cash!) . Eg. Or AB/DE = AC/DF (As. Proof: DP/PE = DQ/QF & PQ || EF (Basic Proportionality Theorem) (As AB=DP. Given: Two triangles ABC and DEF such that ∠ A = ∠ D. E’C = EC. To Prove: Corresponding angles are equal.:If two angles of a Δ are respectively equal to two angles of another Δ. This gives ∠B = ∠P= ∠E and PQ || EF (Δ ABC ≅Δ DPQ. 4.manojarora23@gmail. Criteria for Similarity of Triangles: Two triangles are similar. ∠ B = ∠ E and ∠ C = ∠ F. Therefore.B. DQ=AC) Similarly.). Proof: From the above construction. if (i) ∠ A = ∠ D. ie. Cons. DE ∥ BC. (ie. then by the angle sum property of a Δ their third angles will also be equal. N. Fig.50 But it is given that AD AE = DB EC . Or. ∠ B = ∠ E (given).. for the triangles ABC and DEF (Abv..  Theorem 2: If in two triangles. the similarity of two triangles should also be expressed symbolically.: Cut DP = AB and DQ = AC and join PQ. AAA similarity criterion can also be stated as follows: If two angles of one Δ are respectively equal to two angles of another Δ.(ii) From (i) & (ii). DP=AB. ∠B = ∠E and ∠C = ∠F. sides of one triangle are proportional to (i.: Cut DP = AB and DQ = AC and join PQ. Cons. However. in the same ratio of ) the sides of the other triangle. AC/E’C = AC/EC ' EC EC Thus. then the two Δ’s are similar (AA similarity criterion). ∠ B = ∠ E. ie. Thus. By similar construction) AB/DE = BC/EF. corresponding angles are equal. we cannot write Δ ABC ~ Δ EDF or Δ ABC ~ Δ FED. using correct correspondence of their vertices. if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). Thus. ∠B = ∠ P (by cpct). = = 4.e. then their corresponding angles are equal and hence the two triangles are similar (SSS Criterion). DE EF FD As in the case of congruency of two triangles. Or. in Δ ABC and Δ DEF. hence PQ || EF ) Thus. AE’/E’C = AE/EC. DP/DE = DQ/DF (Corollary of Basic Proportionality Theorem). A E' AE +1= + 1 . That is. Given: Consider two triangles ABC and DEF such that AB/DE = BC/EF = AC/DF (<1). AC=DQ) Benefitted: My Paypal a/c . and. AB/DE=BC/EF=AC/DF. then the two triangles are similar. Residence: Ghaziabad-UP . ∠P = ∠E (corresponding angles).1 Theorems:  Theorem 1: If in two triangles. So. ∠B = ∠Q. ∠P=∠E. Thus. which equals. In ΔDPQ & ΔDEF. ie. Therefore. (SAS Criterion) Given: Consider two triangles ABC and DEF such that AB/DE = AC/DF (<1). PQ=BC Thus. Mobile (where u may transfer cash!) . ∠P = ∠E and ∠Q = ∠F. AB/DE = AC/DF (Given) => DP/DE = DQ/DF (By Cons. ABC ~ DEF Cons. AB=DP. ie. Angles). 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.). Thus.: Cut DP = AB and DQ = AC and join PQ. Thus.(ii) Also.). and PN (⊥QR) of the triangles. Ar(ΔPQR) = 1/2 x QR x PN BC x AM Thus. Therefore. Ar ( ∆ PQR ) QRxPN PQxPN (From (i) & (iii)). AM/PN = AB/PQ .(i) = 2 2 2 2 2 2 Now. PQ QR PR 0 2 ( ) ( ) ( ) (From (ii) & (iii) Benefitted: My Paypal a/c . then the two triangles are similar. Ar(ΔABC) / Ar(ΔPQR) = QR x PN . Proof: Ar(ΔABC) = 1/2 x BC x AM. AC=DQ) =>PQ || EF (Converse of Basic Proportionality Th.: Draw Altitudes AM (⊥BC). As.) Thus. PQ/EF = BC/EF. Thus. Thus. Residence: Ghaziabad-UP . ∠C = ∠R) To Prove: Ar(ΔABC) AB / PQ = BC /QR = CA /RP Ar(ΔPQR) Cons. Or. DP=AB by Cons.manojarora23@gmail. ΔDPQ ~ ΔDEF (AA Criterion) Thus. Thus. ∠A = ∠D (Given). AB = DP (By Cons. Now. ΔABM~ΔPQN. by AA criterion. AC = DQ (By Cons. ΔDPQ ~ ΔDEF—(ii). & ∠A = ∠D To Prove: Corresponding angles are equal.com. ∠B = ∠E and ∠C = ∠F (As.09871823473.(i) Now. ∠A=∠P. Given: Two similar triangles ABC & PQR (ie. (Corresponding Angles) Thus. Areas of Similar Triangles: (Area is measured in square units. and. ∠B = ∠E and ∠C = ∠F.51 So. ∠B = ∠P and ∠C = ∠Q. ΔABC ≅ ΔDPQ (By SAS) . ∠Q=∠F & ∠D=∠D. by SSS ΔABC ≅ ΔDPQ. In Δ’s ABM & PQN: ∠B = ∠Q.)= BC/EF = AC/DF=PQ/EF Thus. ΔABC ~ ΔDEF 5. ∠P = ∠E and ∠Q = ∠F (Corresponding Angles)). ∠M = ∠N (=90 ). ∠A = ∠D. AB/PQ = BC/QR = AC/PR (iii) 2 2 Ar ( ∆ ABC ) BCxAM ABxAM AB BC AC = = = = Thus. Given AB/DE (=DP/DE. ABC ~ PQR (Given). Thus. and. ∠P=∠E and ∠Q=∠F (Corres. ∠A = ∠D. 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. ∠A = ∠D. Proof: In ΔABC & ΔDPQ. DP/DE = DQ/DF = PQ/EF (Corollary of Basic Proportionality Theorem). Mobile (where u may transfer cash!) . Δ BDC ~ Δ ABC. and ∠B = ∠E ( ∵ ΔABC ~ ΔDEF). Thus. The Areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments. Given: ΔABC ~ ΔDEF. ∴ ΔABX ~ ΔDEY (AA Corollary).) in the following form: The diagonal of a rectangle produces by itself the same area as produced by its both sides (i. Or. To Prove: Δ ADB ~ Δ ABC. and AX.(i) In ΔALB & ΔDME. Let BD be the perpendicular to the hypotenuse AC. Given: ΔABC ~ ΔDEF. we have ΔADB ~ ΔBDC .. and ∠B = ∠E (ΔABC ~ ΔDEF) Thus. ΔBDC ~ ΔABC (AA Criterion) . DM⊥EF. Therefore.1 Some Important Results based on Areas: Result 1. AD/BD = DB/DC => BD = AD. Putting this in (i). AB/DE = AL/DM. To Prove: Ar(ΔABC) AX / DY Ar(ΔDEF) Proof: Since. ∴ AB/DE = AX/DY => AB /DE = AX /DY . ∠C = ∠C.manojarora23@gmail. ∠BDC = ∠ABC. 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.DC Proof: From (iii) above. Ar(ΔABC)/Ar(ΔDEF) = AB /DE . The Areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. ΔALB ~ ΔDME (AA Criterion).com. Residence: Ghaziabad-UP . we get: Ar(ΔABC) / Ar(ΔDEF) AL / DM 2 = 2 2 2 0 2 = 2 2 2 2 2 Result 2. (i).DC 0 2 2 Benefitted: My Paypal a/c . ΔABC ~ ΔDEF => ∠A = ∠D => ∠A = ∠D => ∠BAX = ∠EDY .(ii) 2 2 In ΔABX & ΔDEF.(i) Now. ΔADB ~ ΔBDC Proof: In Δ ADB and Δ ABC. triangles on both sides of the perpendicular BD are similar to the whole triangle ABC. Given: A right angled triangle ABC.e. ∠ ADB = ∠ ABC (each 90 ). from (i) & (ii).(iii). To Prove: BD = AD.09871823473.52 . earlier to him. ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.(i) In ΔBDC and Δ ABC. ie. right angled at B. Therefore. Thus.(iii) Also prove. ΔADB ~ ΔABC (AA Criterion) . 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. this theorem is sometimes also referred to as the Baudhayan Theorem. from (ii) ∠BAX = ∠EDY. For this reason. we get Δ≝¿ ¿ Ar ¿ Ar ( Δ ABC ) ¿ = 2 2 2 2 2 2 2 2 6. Or. length and breadth). it was given by an ancient Indian mathematician Baudhayan (about 800 B. but. DY are the bisectors of ∠A & ∠D. Thus.C. Also. that the square of the perpendicular (ie. Pythagoras Theorem: This Theorem was given by Pythagoras. ∠ALB = ∠DME = 90 . AB /DE = AL /DM . ΔADB ~ ΔBDC .(ii) Putting this in Eq. and AL⊥BC.(ii) From (i) & (ii). ∠ A = ∠ A (Common). ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Ar(ΔABC)/Ar(ΔDEF) = AB /DE . To Prove: Ar(ΔABC) AL / DM Ar(ΔDEF) Proof: Since. as ∠ Q = 90°) Or. Residence: Ghaziabad-UP .BC = AB + BC + 2BC. Thus. AC = PR [Proved in (3) above]. But ∠ Q = 90° (By construction) So. : Construct a ΔPQR right angled at Q. Given: Triangle ABC right angled at B. Proof: Δ ABC is a right triangle. Cons.P. 2 2 2 2 2 2 2 2 2 2 2 2 6. and. AD/AB = AB/AC. Given: Triangle ABC in which AC = AB + BC To Prove: ∠B = 90°. ∠ B = 90°. BE2 + ED2 = BD2] = 2BE + 2ED + 2CD = 2BD + 2CD = 2(BD + CD ) Thus.1 Some Important Results based on Pythagoras Theorum: Result 1: In an obtuse triangle. the square of the hypotenuse is equal to the sum of the squares of the other two sides. In ΔBED. To Prove: AC = AB + BC Cons. if BD is the medium.AC = AB (i) Also. To Prove: AC = AB + BC + 2BC. PR = AB + BC (By construction) (i) But.ED + DA + ED + 2. And (ii) BC + BA = 2BD + AC /2 Proof: (i) BC = CE + BE (i). Mobile (where u may transfer cash!) . [email protected] Given: An obtuse Triangle ABC. AD.ED [CD = AD. 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. BC + AB =2(BD + CD ) = 2BD + 2CD = 2BD + 2 x AC /4 = 2BD + AC /2. AC = AD + DC = AD + (DB+BC) = AD + DB +BC + 2DB. 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. Proof: From Δ PQR. CD/BC = BC/AC. then the angle opposite the first side is a right angle.CD. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Benefitted: My Paypal a/c . AB = PQ (By construction). BC = QR (By construction). QR = BC.] Adding (i) & (ii).09871823473.P. right angled at D). AC = PR (iii) (From (i) & (ii). ∠’d Δ’s BCE & BAE. So. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Result 2: Prove that in any triangle. AB = AE + BE (ii) [Rt. BC + AB = 2(BD + CD ) [Part (i) is proved. OR. show that BC + AB = 2(BD + CD ) Given: ABC is a triangle. AD⊥CB. obtuse angles at B.BC = AB + BC + 2DB. ∠B = ∠Q (CPCT). : Draw BD ⊥ AC Proof: Δ ADB ~ Δ ABC (Above Theorem). Thus.DA. & BD is a median To Prove: (i) BC + AB = 2(BD + CD ). Δ BDC ~ Δ ABC (Above Theorem). Thus.ED) + (DA + ED) = 2BE + CD + ED . AB + BC = AC(AD+CD) = AC 2 2 2 2 2 2 2 2 Theorem: (Converse of Pythagoras Theorem) In a triangle. such that PQ = AB. Or. H. right angled at D.2. if square of one side is equal to the sum of the squares of the other two sides. AC = AB + BC (Given) (ii).] (ii) From (i).53 Theorem: In a right triangle. In ΔABC. and. in ΔABC and ΔPQR. Δ ABC ≅ Δ PQR (SSS congruence). CD. H.BD (AD + DB = AB . Now. Or. Therefore. ΔADB is a right triangle. BC + AB = 2BE + CE + AE = 2BE + (CD .AC = BC (ii) Adding (i) & (ii). Thus. PR = PQ + QR (Pythagoras Theorem.com. com.(ii) AC + BC = 2CF + AB /2 .4 cm (ii) PE = 4 cm.28 cm. 122): The exercise is self explanatory and checks if the student has understood the meaning of Congruent and Similar Figures’.56 cm. Or.NCERT EXERCISES SOLUTIONS EXERCISE 6. we have. PE = 0. PR = 2.6 cm and FR = 2. (ii) and (iii). Or. QE = 4. state whether EF || QR : (i) PE = 3. Given: ABC is a triangle. EQ = 3 cm. AB + AC = 2AD + BC /2 . To Prove: 3(AB + BC + CA ) = 4(AD + BE + CF ) Proof: From Result 2. BE & CF are three medians. Find EC in (i) and AD in (ii).09871823473. (i) and (ii).9 cm.36 cm Benefitted: My Paypal a/c . A.2 (P. for medians BE & CF.5 cm.manojarora23@gmail. Or. Do it yourself! Q2.(iii) Adding Equations (i). EXERCISE 6. Or. & AD.54 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. Figs.(i) Similarly. Mobile (where u may transfer cash!) . For each of the following cases. BC + AB = 2BE + AC /2 . PF = 8 cm and RF = 9 cm (iii) PQ = 1. we have. AB + AC = 2AD + 2xBC /4. 2 (AB + BC + CA ) = 2AD + 2BE + 2CF + 1/2 (BC + AC + AB ).18 cm and PF = 0. Residence: Ghaziabad-UP .1 (P. PF = 3. 3(AB + BC + CA ) = 4(AD + BE + CF ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 CBSE CLASS X MATHEMATICS CHAPTER 6 TRIANGLES . 128) Q1. DE || BC. E and F are points on the sides PQ and PR respectively of a Δ PQR. In Adj. AB + AC = 2(AD + BD ). 2 x 2 (AB + BC + CA ) = 4(AD + BE + CF ) + (BC + AC + AB ). B and C are points on OP. we get ED/AE = OD/BO. By Basic Proportional Theorem. AB || EF || DC] In ΔADC. Or. In the adjoining figure. then the line is parallel to the third side). The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. AP = PB and AQ = QC. AP/PB = AQ/QC. prove that AM/AB = AN/AD. Therefore. Q9. Thus. Hence. Show that BC || QR. ‘P’ is the mid-point of AB. Show that EF || QR. Do it yourself! Q4. by Converse of Basic Proportional Theorem PQ || BC. OE || BA. Consider a Trapezium ABCD. thus. and. AE/ED = BO/DO . A.55 A. 5.). Mobile (where u may transfer cash!) . OE || AB (By Cons. i. Benefitted: My Paypal a/c . Using Converse of Basic Proportional Theorem . we obtain AE/ED = BO/OD (i). figure (A Trapezium) ABCD. We have. Show that AO/BO = CO/DO. A. Consider the adjoining figure in which PQ is a line segment drawn through the mid-point P of line AB. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. thus. In the adjoining figure. Do it yourself! Q3.(ii) From (i) & (ii). AQ/QC = 1 Thus.manojarora23@gmail. DE || AC and DF || AE.com. such that PQ || BC. A. such that EF || CD [ Hence. AP/PB = 1 = AQ/QC. ‘Q’ is the mid-point AC. Given: Adj. Show that ABCD is a trapezium.e. Thus.(i) In ΔABD. Prove that BF/FE = BE/EC. Now. AO/BO = CO/DO. AP = PB. we get. AQ = QC.09871823473. with AB || DC To Prove: AO/BO = CO/DO Proof: Draw a line EF through point O. we get AE/ED = AO/OC . By basic proportionality theorem. Residence: Ghaziabad-UP . A. By using Basic Proportionality Theorem. DE || OQ and DF || OR. It can be observed that AP/PB = 1.. Thus. Or. Do it yourself! (Basic Proportionality Theorem) Q7. ‘PQ’ bisects ‘AC’. and. Using Basic Proportionality Theorem. A. (Because PQ || BC). By Basic Proportional Theorem. A. Q8. ( Converse of Basic Proportional Theorem: If a line divides any two sides of a triangle in the same ratio. In ΔABD. Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively. If LM || CB and LN || CD. Hence. AO/OC = BO/DO. prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. Do it yourself! (Basic Proportionality Theorem) Q6. EO || DC. Do it yourself! Q5. Q10.In the adjoining figure. A. A. Draw OE || AB. prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. AP/PB = AQ/QC. A. OQ and OR respectively such that AB || PQ and AC || PR. QT/PR = QR/QS. AB || DC. Q6. In the adjacent figure. State which pairs of triangles in the given figure are similar. Or. ∠ DCO and ∠ OAB. AB=AC (i).(iv) Now. In the adjoining figure. it is given that AO/CO = BO/OD . Thus. 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! Q3. By AA. S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS. altitudes AD and CE of Δ ABC intersect each other at the point P. we get. QT/QR = PR/QS (i) Also.3 (P. 138) (cont’d) Q2. Δ ADE ~ Δ ABC Q7.(ii) From (i) & (ii). (iii) by eq. This implies that AB || OE || DC. Q4. Show that Δ RPQ ~ Δ RTS. the two triangles are similar. Using a similarity criterion for two triangles.56 But. Δ ABE ≅ Δ ACD (Given). show that Δ ADE ~ Δ ABC. AB/AD = AC/AE from eq. by SAS. In the adjacent figure. A. ∠APE = ∠CPD (VOP). In Δ ADE & Δ ABC. Find ∠ DOC. & ∠ PQS = ∠TQR. CE⊥AB). Thus. Or. In the adjoining figure. QR/QS = QT/PR (Given). show that OA/OC = OB/OD. QT/QP = QR/QS. ΔODC ~ ΔOBA. Thus. A. A.manojarora23@gmail. Q5. Or. ΔAEP ~ ΔCDP. ΔABD ~ ΔCBE.com. AE/ED = AO/CO. if Δ ABE ≅ Δ ACD. ∠ BOC = 125° and ∠ CDO = 70°. Or. QT/QR = PQ/QS. (i) => AB/AD = AC/AE . ABCD is a Trapezium. Thus. Show that: (i) Δ AEP ~ Δ CDP (ii) Δ ABD ~ Δ CBE (iii) ΔAEP ~ ΔADB. it is given that AO/BO = CO/DO. (i) To Prove: ΔAEP ~ ΔCDP Proof: In ΔAEP & ΔCDP. A. Mobile (where u may transfer cash!) . ∠DAE = ∠BAC.09871823473. ∠ 1 = ∠ 2 => PR = PQ (ii) ( Sides opp. QR/QS = QT/PR and ∠ 1 = ∠ 2. and use AA Criterion. Residence: Ghaziabad-UP . (ii) To Prove: ΔABD ~ ΔCBE Proof: In ΔABD ~ ΔCBE. and. ∠AEP = ∠CDP = 90 (AD⊥BC. ∠ABD = ∠CBE (Common). QT/QP = QR/QS In Δ PQS & Δ TQR. and. This implies that EO || DC (By Basic Proportional Theorem).3 (P. (iii) To Prove: ΔAEP ~ ΔADB 0 0 Benefitted: My Paypal a/c . CE⊥AB). By AA. Do the question yourself. to equal angles are equal) From (i) & (ii). and (iv) Δ PDC ~ Δ BEC A. Thus. (ii) can be written as AD = AE (iii) Dividing eq. AE=AD (ii)(By cpct) Eq. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Thus. by SAS. Draw the figure. EXERCISE 6. Show that Δ PQS ~ Δ TQR. ∠ADB = ∠CEB = 90 (AD⊥BC. Thus. 138) Q1. we have. A. Do it yourself! EXERCISE 6. (iv). Prove that: (i) Δ ABC ~ Δ AMP. 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. and. 0 0 Q8. Do it Yourself! Q9. 0 Q12. Now. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. ΔAEP ~ ΔADB. Thus. CE⊥AB). prove that Δ ABD ~ Δ ECF. AB = AC => ∠ABD = ∠C = ∠ECF (Angles opp.manojarora23@gmail. In the adjoining figure.09871823473. Δ PDC ~ Δ BEC. (ii) As. ∠EAP = ∠DAB (Common). in Δ ABD & Δ ECF >> ∠ADB = ∠EFC = 90 . ΔABD ~ ΔECF (By AA Criterion).41). ∠B = ∠E (Proved above) ∴ ΔDCB ∼ ΔHGE (By AA similarity criterion) (iii) ΔDCA ~ ΔHGF In ΔDCA and ΔHGF. Given: AB/PQ = BC/QR = AD/PM Benefitted: My Paypal a/c . Show that Δ ABE ~ Δ CFB. (iii) ΔDCA ~ ΔHGF. (iv) To Prove: (iv) Δ PDC ~ Δ BEC Proof: In ΔPDC & ΔBEC. By AA. ∴ ΔACD ∼ ΔFGH (By AA similarity criterion). ∠ACD = ∠FGH (Proved above). A. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. ABC and AMP are two right triangles. (ii) ΔDCB ~ ΔHGE. CE⊥AB). Thus. A. right angled at B and M respectively.com. ∠PCD = ∠BCE (Common). ∠B = ∠E. Residence: Ghaziabad-UP . ∠A = ∠F (Proved above). 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. By AA.57 Proof: In ΔAEP ~ ΔADB. To Prove: (i) CD/GH = AC/FG It is given that ΔABC ∼ ΔFEG ∴ ∠A = ∠F. Q11. 6. In ΔABC. Show that Δ ABC ~ Δ PQR. A. and. If AD ⊥ BC and EF ⊥ AC. ∠ACD = ∠FGH (Bisectors of equal angles ∠ACB = ∠FGE). to equal sides are equal). Given: ΔABC ∼ ΔFEG. CD/GH = AC/FG (ii) ΔDCB ~ ΔHGE In ΔDCB and ΔHGE. ∠AEP = ∠ADB = 90 (AD⊥BC. and. (ii) CA/PA = BC/MP (i) In ΔABC and ΔAMP. If Δ ABC ~ Δ FEG. CD & GH are bisectors of ∠C & ∠G. ΔABC ∼ ΔAMP ∴. ∠ABC = ∠AMP (Each 90°) ∠A = ∠A (Common) ∴ ΔABC ∼ ΔAMP (By AA similarity criterion). ∠PDC = ∠BEC = 90 (AD⊥BC. Thus. and ∠ACB = ∠FGE ∴ ∠ACD = ∠FGH (Angle bisector). ∠DCB = ∠HGE (Proved above). show that: (i) CD/GH = AC/FG. In the adjacent figure. Mobile (where u may transfer cash!) . ∠ABD = ∠ECF (Proved above). CA/PA = BC/MP Q10. ∠DCB = ∠HGE (Angle bisector) In ΔACD and ΔFGH. ∴ ΔDCA ∼ ΔHGF (By AA similarity criterion). and. ∠A = ∠F (Proved above). A. Thus. 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.(iv) Benefitted: My Paypal a/c . diagonals AE and BC bisect each other at D. and. Now.(ii) As AD & PM are medians. AB/PQ = BD/QM . 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. AB/PQ = BD/QM = AD/PM. D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. DC/AC = AC/BC. A. Therefore. Therefore. side into two equal halves. Thus. (i) & (ii).BC = AC . side into two equal halves.manojarora23@gmail. In quadrilateral ABEC. by SSS Criterion. ∠B = ∠Q. ∴ AC = BE and AB = EC (Opposite sides of a parallelogram). Thus. ΔABC ~ ΔPQR => AB/PQ=AC/PR=BC/QR . ∠A = ∠P. ∠C=∠R . ∠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. Thus. Show that ΔABC ~ ΔPQR. and. ∠CAE = ∠RPL (i) Similarly.(iii) From equations (i) & (iii). Therefore.CD 2 2 2 Q14. Show that CA = CB. Δ ABD ~ Δ PQM.58 To Prove: Δ ABC ~ Δ PQR Proof: A median divides the opp. join B to E. A. ∠BAC = ∠QPR (iii) Now. prove that AB/PQ=AD/PM. BD=BC/2.(i) Also. ∠B=∠Q. by SAS. Find the height of the tower. we get. A. BD = DC and QM = MR Also. Q13. AB/PQ = AC/PR (Given). Q15. thus. Proof: A median divides the opp. Light rays from the sun are falling on the tower and the pole at the same angle. ∠BAC = ∠QPR (From (iii)). QM=QR/2 . we can prove that quadrilateral PQLR is a parallelogram and PR = QL. respectively where ΔABC ~ ΔPQR. In Δ ABC ~ Δ PQR. such that AD = DE and PM = ML. Δ ABC ~ Δ PQR. Q to L. and. & QM = QR/2. Thus. In ΔADC and ΔBAC. ∠ADC=∠BAC (Given). ΔABC ~ ΔPQR (By SAS Similarity Criterion). Thus. the height of the tower is 42 metres. ∠DFC = ∠BEA also. ΔABE ~ ΔPQL (By SSS) => ∠BAE = ∠QPL (ii) Adding eq. ∠ACD=∠BCA (Common) ∴ ΔADC ∼ ΔBAC (By AA similarity criterion). AB/PQ = BC/QR (Given). DC. in ΔABC & ΔPQR. Or. A. 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). and R to L. quadrilateral ABEC is a parallelogram. AD = DE. CA = CB. ∠B = ∠Q. Then. Let AB and CD be a tower and a pole. Similarly. Let BE and DF be the shadow of AB (tower) and CD (pole).09871823473. Or. Q16. Thus. we can prove. ∠DCF = ∠BAE. Mobile (where u may transfer cash!) .com. Therefore. AB/PQ = BC/QR = AD/PM => AB/PQ = (BC/2)/(QR/2) = AD/PM. Given: AB/PQ = AC/PR = AD/PM To Prove: ΔABC ~ ΔPQR Cons: Extend AD and PM up to E and L. Thus. and.CD. PM = ML (By construction). Residence: Ghaziabad-UP . C to E. BD = BC/2. If AD and PM are medians of triangles ABC and PQR. Now. and. If AB = 2 CD. Consider 2 similar triangles ΔABC ∼ ΔPQR. and. If AD intersects BC at O.59 In ΔABD and ΔPQM. Q2. A. Δ ABC ≅ Δ PQR Q5. : Draw AP ⊥ BC.4 (P. AC = PR. DE || AC. If the areas of two similar triangles are equal. Let AD & PS be the medians of these 2 2 2 2 2 2 2 Benefitted: My Paypal a/c .com. ∴ ΔBED ∼ ΔBCA (By AAA similarity criterion) ar(ΔBED)/ar(ΔBCA) = (DE/AC) = 1/4 => ar(ΔBED) = 1/4 ar(ΔBCA) Similarly.manojarora23@gmail. Let Δ ABC ~ Δ DEF and their areas be.] ————————————————————————————————————————————EXERCISE 6. In ΔAPO and ΔDMO. ar(ΔDEF) = ar(ΔABC) . in ΔBED & ΔBCA ∠BED = ∠BCA (Corresponding Angles). ∠AOP = ∠DOM (VOA) ∴ ΔAPO ∼ ΔDMO (By AA similarity criterion) ∴ AP/DM = AO/DO => ar(Δ ABC) / ar(Δ DBC)=AP/DM = AO/DO Q4. In the adjoining figure. If EF = 15. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.[ar(ΔBED) + ar(ΔCFE) + ar(ΔADF)] = ar(ΔABC) . ∠BDE = ∠BAC (Corresponding Angles). their corresponding sides are in proportion to each other.4 cm. ∴ ΔABD ∼ ΔPQM (By SAS similarity criterion) => AB/PQ = AD/PM. respectively. ∠APO = ∠DMO (Each = 90°). Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. ∠OAB = ∠OCD (Alternate angles). BC = QR. 64 cm and 121 cm . ∴ Ar (ΔAOB) / Ar (ΔCOD) = AB / CD => Ar (ΔAOB) / Ar (ΔCOD) = 4 : 1 2 2 Q3. and. find the ratio of the areas of triangles AOB and COD. By SSS. D.2 cm. Therefore. ∴ ∠OAB=∠OCD and ∠OBA=∠ODC (Alternate angles) In ΔAOB and ΔCOD. Consider two triangles Δ ABC & Δ PQR. Given: The adjoining figure. Residence: Ghaziabad-UP . ar(ΔCFE) = ar(ΔADF) = 1/4 ar(ΔABC). ∠EBD = ∠CBA (Common). and. Find the ratio of the areas of Δ DEF and Δ ABC. 1 = (AB/PQ) = (BC/QR) = (AC/PR) => AB = PQ. Δ ABC ~ Δ DEF => ar (ΔABC) / ar (ΔDEF) = BC / EF => 64 / 121 = BC / 15. 143) Q1.3/4 ar(ΔABC) = 1/4 ar(ΔABC) ==> ar(ΔDEF) / ar(ΔABC) = 1/4 Q6. 2 2 2 2 2 2 ∠AOB = ∠COD (VOA). show that ar (Δ ABC) / ar (Δ DBC) = AO/ DO. find BC. Also. BC = 11. ar(Δ ABC) / ar(Δ PQR) = (AB/PQ) = (BC/QR) = (AC/PR) . ABC and DBC are two triangles on the same base BC. then their corresponding angles are equal. Thus.4 This gives. BC and CA of Δ ABC. D & E are midpoints of sides AB & BC. Thus. Proof: ar(Δ ABC) / ar(Δ DBC) = [1/2 x BC x AP] / [1/2 x BC x DM] = AP/DM. which are similar to each other. A. To Prove: ar (Δ ABC) / ar (Δ DBC) = AO/ DO Cons. (i). Mobile (where u may transfer cash!) . Since AB || CD. A. A. and DE = AC/2. AB/PQ = BD/QM [Using (iv)].09871823473. [If the two triangles are similar. Putting this in Eq. A. ∠OBA = ∠ODC (Alternate angles) ∴ ΔAOB ∼ ΔCOD (By AAA similarity criterion. prove that they are congruent.(i) It is given that ar(Δ ABC) = ar(Δ PQR) => ar(Δ ABC) / ar(Δ PQR) = 1. & DM ⊥ BC. ∠B = ∠Q [Using (ii)]. A. E and F are respectively the mid-points of sides AB. AB/PQ = BD/QS[From (iii)] Thus. Mobile (where u may transfer cash!) . Or. the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. 8 cm. 5 cm A.(iii) In ΔABD and ΔPQS. ΔDBF. 12 cm. described on one of its diagonal = √2 a We know that equilateral triangles have all its angles as 60º. 8 cm. and. it is a right triangle. its diagonal = √2 a Two desired equilateral triangles are formed as ΔABE and ΔDBF. ∴ BD = DC = BC/2. A. We know that equilateral triangles have all its angles as 60º .(ii) As AD & PS are medians. Thus. Let side of ΔABC = x. Therefore. AB/PQ = BD/QS = AD/PS . Q8. 80 cm. 150) Q1. 7 + 24 = 25 The sides of the given triangle are satisfying Pythagoras theorem. ∠B = ∠Q. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81 A. ar(Δ ABC) / ar(Δ PQR) = (AB/PQ) = (BC/QR) = (AC/PR) . (i) & (iv) ar(Δ ABC) / ar(Δ PQR) = [AD/PS] 2 2 2 2 Q7. the hypotenuse is the longest side of a right triangle. ΔABC ∼ ΔPQR. Side of an equilateral triangle. write the length of its hypotenuse. Let ABCD be a square of side a. Ratio of the areas of triangles ABC and BDE is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 A. and 25 cm. It is given that the sides of the triangle are 3 cm. 24 cm. (i) It is given that the sides of the triangle are 7 cm. 24 cm. Therefore. Using eq. ΔABD ~ ΔPQS (By SAS Criterion). ∠B = ∠Q [From (i)]. Therefore. and 6 cm. If two triangles are similar to each other. It is given that the sides are in the ratio 4:9. ratio between areas of these triangles = (4/9) = 16/81 Hence. 6 cm. QS = SR = QR/2. Putting this in equation (i) AB/PQ = BD/QS = AC/PR . the correct answer is (D). ∴ AB/PQ = BC/QR = AC/RP . 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. and 625. ∠C = ∠R .(iv) Now. and.(i) ∠A = ∠P. Sides of two similar triangles are in the ratio 4 : 9. (iv) 13 cm. all equilateral triangles are similar to each other. 25 cm. 49 + 576 = 625. the length of the hypotenuse of this triangle is 25 cm. Therefore. 576. the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. 100 cm. all equilateral triangles are similar to each other. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Therefore. Therefore. Now. Hence.09871823473. Because. 2 Q9. described on one of its sides = a Side of an equilateral triangle. we obtain 49. Determine which of them are right triangles. 2 2 2 (ii) 3 cm. Sides of triangles are given below. Squaring the lengths of these sides. Therefore. 6 cm A.60 triangles. Residence: Ghaziabad-UP . the correct answer is (C). (i) 7 cm.5 (P.com. then the ratio of the areas of these triangles is equal to the square of the ratio of the corresponding sides of these triangles. (ii) 3 cm. ΔABE. Hence. Benefitted: My Paypal a/c . Therefore. side of triangle BDE = x/2 ∴ area (ΔABC) / area (ΔBDE) = [x ÷ x/2] = 4/1 = 4:1 Hence. In case of a right triangle. 8 cm.manojarora23@gmail. ———————————————————————————————————————-2 EXERCISE 6. (iii) 50 cm. 2 2 2 Q2. Let AD be the altitude of an equilateral triangle.x) = x In ΔQMP & ΔPMR: ∠PQM = ∠RPM. ∠CAD = 90 . (iii) AD = BD × CD A. Hence. O is a point in the interior of a triangle ABC. ABC is an isosceles triangle right angled at C. => MP/MR = QM/PM => PM/MR = QM/PM => PM = QM . MR 2 0 0 0 0 0 0 2 Q3. Residence: Ghaziabad-UP . Show that (i) AB = BC × BD. 64. the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side. all the three altitudes of an equilateral triangle are of equal lengths. by AA. Thus. 2 2 2 2 2 2 2 2 2 Benefitted: My Paypal a/c . ABC is an isosceles triangle with AC = BC. it is not a right triangle. Show that (i) OA + OB + OC – OD – OE – OF = AF + BD + CE . ΔABC. the given triangle is not satisfying Pythagoras theorem. Similarly. prove that ABC is a right triangle. ∠MPQ = 90 . 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Q8. (ii) AC = BC × DC. Let ∠MPR = ‘x’. If AB = 2AC . In ΔADB. ∠ABD =∠CBA (Common) Thus.x. the triangle is satisfying Pythagoras Theorem. Find each of its altitudes.com.x. ABC is an equilateral triangle of side 2a. ∠MRP = 90 . the given triangle is a right angled triangle. A.x. ∠CAB = ∠CDA (=x). MR. In the adjoining figure. Thus.manojarora23@gmail. 2 2 2 Q7. ΔBOC. ΔAOD. CD = CO + OD . we get 9. OE ⊥ AC and OF ⊥ AB. ∠MQP = 180 . AB = AO + BO . Show that PM = QM . ∠QMP = ∠PMR = 90 . Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. we get. Applying Pythagoras theorem in ΔAOB. ΔQMP ~ ΔPMR. Prove that AB2 = 2AC2. => ∠CBA = 90 .61 Squaring the lengths of three sides. AD = OD + AO . Do (iii) & (iv) on your own. (ii) Let ∠CAB = x.(iii).09871823473. altitude of an equilateral triangle bisects the opposite side.x. Thus. A. AB = 2AC => AB = AC + AC => AC + BC Thus. BD = DC = a. Remember. (i) In ΔBAD & ΔBCA. 9 + 36 ≠ 64. ΔBAD ~ ΔBCA => BA/CB = BD/BA =>AB = BC x BD. Thus. Thus. Now. in ΔMPR. ΔCOD. OD ⊥ BC. Therefore.(i).(90 + 90 . we get. A. ∠BAD = ∠BCA = 90 . Or. by AA. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. AD + DB = AB . by AA Criterion.(ii). ∠CDA = x In ΔCAB & ΔCDA. 3 + 6 ≠ 8 Clearly. ∠ACB = ∠DCA (=90 ). and. Mobile (where u may transfer cash!) . ΔCAB ~ ΔCDA => AC/DC = BC/AC => AC = BC × DC (iii) Consider ΔABD & ΔCAD 2 2 2 0 2 0 0 0 2 Q4. ABD is a triangle right angled at A and AC ⊥ BD. 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 . that the hypotenuse is the longest side of the right angled triangle. BC = BO + OC . and. & 36. A. 2 2 2 2 2 2 2 2 2 Q6.(iv) Adding the above four equations. A. => AD = a√3 = Length of each of the altitudes of an equilateral triangle. Do it yourself! Q5. However. OC = OE + EC . BC + CD = BD .OF ). Applying Pythagoras theorem for ΔACD. AB = AC + CB => Putting these values in eq.com. Q12. OA + OB + OC = OF + AF + OD + BD + OE + EC . ∴. find the distance between their tops. Use Pythagoras theorem. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD. OB = 6m. Cons. Use Pythagoras theorem. OB = OD + BD . and OC (i) Applying Pythagoras Theorem in ΔAOF. How far from the base of the pole should the stake be driven so that the wire will be taut? A. and the plane flying towards west in 1 1/2 hours are 1500kms (=OA) & 1800kms (=OB).OE )+ (OC . 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Q9. Q13. AB = Distance between the planes after 1. OB. Q10. we get.62 (ii) AF + BD + CE = AE + CD + BF .5 hrs = 300√61 km. AC + CE + BC + CD = AE + BD . Adding the above three equations.200 km per hour. AF + BD + EC = (OA . How far apart will be the two planes after hours 1 1/2 hours? A. Residence: Ghaziabad-UP . Applying Pythagoras theorem in ΔACE. we get. & ΔCOE. AP = 12m Apply Pythagoras theorem in ΔAPC. Prove that 2AB = 2AC + BC . Or. From ΔABC. we get. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. and. OA = Distance of stack from the base of the pole = 6√7 m. Distances travelled by the plane flying towards north. Let AB & CD be the poles of height 6m and 11m. A. AC = Distance between the tops of the two poles = 13m. we get. we get OA = OF + AF .09871823473. Or. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. If the distance between the feet of the poles is 12 m. Find the distance of the foot of the ladder from base of the wall. A. AC = AD + CD => AD = AC .(i) Applying Pythagoras theorem in ΔBCD. A. DE + AB = AE + BD 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Q14. we get. DE = CD + CE . A. Mobile (where u may transfer cash!) . A ladder 10 m long reaches a window 8 m above the ground.(iii) From ΔCDE. ΔBOD. 2 2 2 2 2 2 2 2 2 Benefitted: My Paypal a/c . At the same time. we get. AC + CE = AE . Use the adjoining figure.(ii) Adding equations (i) & (ii). Let OB be the pole and AB be the wire. 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. and use Pythagoras Theorem. Two poles of heights 6 m and 11 m stand on a plane ground. AF + BD + CE = AE + CD + BF . we get. another aeroplane leaves the same airport and flies due west at a speed of 1. A. we get.CD . and. Q11.manojarora23@gmail. OA + OB + OC – OD – OE – OF = AF + BD + CE 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 (ii) From the above result.OD ) + (OB .(i) Applying Pythagoras theorem for ΔABD.: Join OA. Prove that AE + BD = AB + DE . (iii). CP = 11−6=5 m From the figure. given that BD = BC/3 = a/3. The planes on which these lines are graphed is known as ‘Cartesian Plane’. we get. BC = 6 cm => BC = 36 cm . we get. a = AE + (a/2) . Residence: Ghaziabad-UP .6 (P. Tick the correct answer and justify: In ΔABC. we get. AE = (a√3)/2. ΔABC is a right triangle.BD .63 AB = AD + BD => AD = AB . AC = 12 cm => AC = 144 cm . The angle B is: (A) 120° (B) 60° (C) 90° (D) 45° A. 150): Optional – 10 Questions. ———————————————————————————————————————EXERCISE 6.(ii) From equations (i) & (ii). and AE be the altitude of ΔABC.09871823473. & AE be the altitude of ΔABC. AC = 12 cm and BC = 6 cm. DB = 3CD => We convert DB & DC in the form of BC. Also. Thus. and. ‘Co-ordinate Geometry’ is a branch of Geometry which sets up a definite correspondence between the position Benefitted: My Paypal a/c . ∴ BE = EC = a/2. D is a point on side BC such that BD = BC/3. On a ‘Number Line’. Applying Pythagoras theorem in ΔABE. 4 × (Square of altitude) = 3 × (Square of one side) 2 2 2 2 2 2 2 2 2 2 Q17. it is given.com.(3BC/4) . ∠B = 90 .BD . and. CD = BC/4. ΔABC. we get AC . Thus. ‘Descartes’ placed two such lines perpendicular to each other on a plane & located points on the plane. right angled at B. DE = BE . 2AB = 2AC + BC . Thus. A. The system used for describing the position of a point in a plane is also known as ‘Cartesian System’. the right answer is ( C ). The position of any object lying in a plane can be represented with the help of two perpendicular lines. DB = 3BC/4.CD = AB . is satisfying Pythagoras Theorem. Therefore.(iii) Now. Applying Pythagoras theorem in ΔADE. Putting these values in equation (iii). Thus. AB = 6√3cm => AB = 108 cm . distances from a fixed point (origin) are marked in equal units positively in one direction and negatively in the other. the given triangle. Mobile (where u may transfer cash!) . Thus.a/3 = a/6. Prove that 7AB . A. AE = 3a /4 => 4AE = 3a Thus. (An Altitude and a Median from a vertex to the opposite side is the same in case of equilateral triangles). In an equilateral triangle. AB = AE + BE . BE = EC = BC/2 = a/2.manojarora23@gmail. AB = 6√3cm. Thus. AC . ie. which gives. and. Thus.(BC/4) = AB . AB + BC = AC . 2 2 2 2 2 2 2 2 2 0 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). Thus.BD = a/2 . we have. Let the side of the equilateral triangle be a. AD = AE + DE = [(a√3)/2] + [a/6] = 28a /36 = 7/9 AB => 9AD = 7AB 9AD = 2 2 2 2 2 2 2 2 2 2 2 Q16. prove that three times the square of one side is equal to four times the square of one of its altitudes. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Q15. In an equilateral triangle ABC. Let the side of an equilateral triangle be a. PQ = [(x – x ) + (y – y ) ] which is called the distance formula. x = (m x + m x ) / (m + m ) Similarly taking.com. 2.y2).y). Substituting these values in (i). RS = x – x = PT. one and only one line can be drawn. The two perpendicular lines (or perpendicular number lines) through the origin are called Cartesian or Rectangular Co-ordinate axis. Consider any two points A(x1. we get m / m = (x – x )/(x – x) = (y – y )/(y – y) Taking. the distance of a point P(x. 2 1 1 2 2 1 2 2 2 1 2 2 2 1 1 2 2 1 2 1 2 1 2 2 1 2 2 2 1 2 1 2 2 1/2 2 1/2 2 2 1/2  Consider three points A. If AB + BC = BC. PA/BP = AQ/PC = PQ/BC (i) Now.  A point which is equidistant from the two end points ‘A’ & ‘B’. Therefore. ———————————————————————————————————————--1. PS and BT perpendicular to the x-axis. by the AA similarity criterion. Now. we get. y ) and Q(x . segment 3. Coordinate geometry has been developed as an algebraic tool for studying geometry of figures. and. 0) is. m / m = (x – x )/(x – x). Also. SQ = y . called co-ordinates. The coordinates of a point on the x-axis are of the form (x. (a. X Co-ordinate => Abscissa of Point P. is a parabola. Draw AQ and PC parallel to the x-axis. Y Co-ordinate => Ordinate of Point P..B. y) divides AB internally in the ratio m1 : m2. when represented graphically. of a line AB. y) from the origin O(0. and of a point on the y-axis are of the form (0. lies on the perpendicular bisector of AB. OR = x . y ): Draw PR and QS perpendicular to the x-axis. Δ PAQ ~ Δ BPC Thus.:  The word ‘Regular’ in ‘Regular Pentagon’ or ‘Regular Hexagon’ implies that all the sides and angles of the given shape are equal. and both taken together are called axis of co-ordinates. y=6) Values of points written as (x1. that weather the line we have drawn is correct or not. The equation of y-axis is x=0. which can also be written as: PQ = [(x – x ) + (y – y ) ] Points to Remember:  In particular.e. Then. gives a straight line. Also. Section Formula: It is used when we need to identify the position of a point between the given point (ie. 0). we can say that the three points are collinear.y1). x=10. Coordinates of the two points are known). but we take at least the third point for verification purposes. Introduction: Now.09871823473. Distance Formula: Finding the distance between any two points P(x . y). m / m = (y – y )/(y – y).  While drawing a graph (or line) for a given linear equation. OS = x . 1 2 1 2 1 2 1 2 1 1 2 2 1 1 2 1 2 1 2 2 1 2 2 1 1 2 Benefitted: My Paypal a/c . (x3. OX & OY are called x-axis & y-axis. Mobile (where u may transfer cash!) . A perpendicular from the point P on QS is drawn to meet it at the point T. The equation of x-axis is y=0. through two points.  A quadrilateral with all four sides equal and one angle 90° is a square. So. i. y1) andB(x2. PA/PB = m /m Draw AR. a linear equation in two variables of the form ax + by + c = 0. we must always take at least 3 points. So. applying the Pythagoras theorem in Δ PTQ. ‘O’ (origin) is the intersection of the axis of co-ordinates. Though. Residence: Ghaziabad-UP .manojarora23@gmail. (x2. we get. QT = y – y . b are not simultaneously zero). PQ = PT + QT = (x – x ) + (y – y ) . we get. the graph of y = ax + bx + c (a ≠ 0). 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. OP = [x + y ] . B & C. and understand algebra with the help of geometry. y2) and assume that P (x. Then. (P(x.y3)… are called ordered pairs. ST = PR = y . N.64 of a point in a plane and a pair of algebraic numbers. It helps us to study geometry using algebra. 3) and (– 2.y ) N. the area of Δ ABC is the numerical value of the expression 1/2 [x (y . Draw AP.NCERT EXERCISES SOLUTIONS EXERCISE 7.65 y = (m y + m y ) / (m + m ) So. Area of a Trapezium = 1/2 x (Sum of Parallel Sides) x (Perpendicular Distance between them).y ) + x (y . Also.y ) + x (y . If any of the two sides of the given triangle are equal. and add the areas of these regions. which have no common area. internally. Heron’s Formula [s(s .1/2 (BQ + CR) QR = 1/2 (y + y ) (x . (1. 161) Q1.B. 5). Clearly ABQP. A. N. Special Case: The mid-point of a line segment divides the line segment in the ratio 1 : 1. y ). Therefore. Mobile (where u may transfer cash!) . Thus. Now. y) which divides the line segment joining the points A(x .y ) + x (y . y ). the coordinates of the mid-point P of the line segment joining the two points A(x . – 11) are collinear.: If the area of a triangle is 0 square units.b)(s . area of ΔABC = 1/2 (BQ + AP) QP + 1/2 (AP + CR) PR . Do It Yourself! (AB + BC = BC => Points are collinear. B(x . and. – 2) are the vertices of an isosceles triangle.c)] Area of a Trapezium = 1/2 x (Sum of Parallel Sides) x (Perpendicular Dis tance (or Height) between them). Find the distance between the following pairs of points :(i) (2.1/2 (y + y ) (x .x ) + 1/2 (y + y ) (x . (4.1 (P.com. Do It Yourself! Q2. y ). Check whether (5. y ) is: [(1.y ) + x (y .x ) = 1/2 [x (y .x )/(1 + 1). Residence: Ghaziabad-UP . (– a.: To find the area of a polygon. Area of a Triangle: From previous classes: Area of Triangle = 1/2 x base x height. (m y + m y )/(m + m ) This is known as the section formula. Verify if ABCD is a square or not. Determine if the points (1. APRC and BQRC are all trapezia. Do It Yourself! Q3. respectively. A. B and C. 2 1 1 1 2 3 2 3 1 2 1 3 3 2 2 1 1 3 1 3 3 1 3 1 2 3 3 2 2 2 CBSE CLASS X MATHEMATICS CHAPTER 7 COORDINATE GEOMETRY . Find the three sides of the given triangle.a)(s . the coordinates of the point P(x.y )/(1 + 1)] = [(x + x )/2.09871823473.y + 1. we divide it into triangular regions. and the two diagonals. to the x-axis.B.) Q4. Q5. b). (– 1.x + 1. from the figure. 7). it is clear that the area of 1 2 2 3 1 3 Δ ABC = area of trapezium ABQP + area of trapezium APRC – area of trapezium BQRC. A. in the ratio m : m are (m x + m x )/(m + m ). Can be done in two ways:  Find the four sides.y ) Thus. y ) and C(x .manojarora23@gmail. (6. 3) (iii) (a. A. we know. 1) (ii) (– 5. 3). 4) and (7. then its vertices will be collinear. 1/2 Using coordinate geometry to find the area of a triangle: Let ABC be any triangle whose vertices are A(x . – 2). Benefitted: My Paypal a/c .x ) . the triangle is isosceles. y ) and B(x . (2. BQ and CR perpendiculars from A. or. – b). (y + y )/2] 1 2 2 1 1 2 1 1 2 1 2 2 1 1 1 2 2 1 2 2 2 1 1 2 1 2 1 2 1 1 2 1 1 2 2 2 4. y ) and B(x . the point ‘R’ can be (4. Find the point on the x-axis which is equidistant from (2. where should she post her flag? Benefitted: My Paypal a/c . (ii) (–3. if any. the point is (1. 1) is equidistant from P(5. 0 0 Q6. –5) and (–2. apply Pythagoras Theorem. AP = PQ = QB Therefore. and give reasons for your answer: (i) (– 1. 4). 3). 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. (iii) (4. Find a relation between x and y such that the point (x. 6). (3. 7) and (4. Let P(x. (-4. Using the section formula. 6) and (– 3. Point Q is (-4 +4)/3. –3). y = (2x(-3) + 1x(-1))/(2 + 1). then we have a figure.com. or. Let P (x . with four sides equal. 2). we get. Do It Yourself! [ The equation obtained will be 3x + y = 5] EXERCISE 7. y) be the required point. 100 flower pots have been placed at a distance of 1m from each other along AD. Residence: Ghaziabad-UP . Find the corresponding values of PR & QR. Find the coordinates of the point which divides the join of (–1. and. and. P((x . find the values of x. and hence. 5). To conduct Sports Day activities. if the given figure is of any particular type of a quadrilateral or not. (0. x = (2x(-2) + 1x4)/(2 + 1). – 4). 6). Preet runs 1/5 th the distance AD on the eighth line and posts a red flag.2 (P.manojarora23@gmail. 167): Q1. Name the type of quadrilateral formed. – 2). and. A. (7. lines have been drawn with chalk powder at a distance of 1m each. 1). as shown. A.09871823473. Q2. y ) and Q (x . -5/3) Point Q divides AB internally in the ratio 2:1 Thus. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. and an angle of 90 => The given figure is a square. ((-6-1)/3) => Q is (0. (– 3. 3).e. A. –3) and R(x. thereby state. Point P divides AB internally in the ratio 1:2 Thus. 0). y ) = (2. – 3) and Q(10. x = (1x(-2) + 2x4)/(1 + 2) = 2. Do It Yourself! Q9. If it is so. Find the values of y for which the distance between the points P(2. 2) A. If Q(0. A. in your rectangular shaped school ground ABCD. (–1. Q7. (– 1.66  Find the four sides. y = (1x(-3) + 2x(-1))/(1 + 2) = (-3-2)/3 = -5/3 Thus. x = (2x4 + 3x(-1)/(2 + 3) = 1. by the following points.. y ) are the points of trisection of the line segment joining the given points i. Do It Yourself! [You will get the value of ‘x’ as ±4. 9).] Q10. y = (2x(-3) + 3x7)/(2 + 3) = 3 Therefore. A. A. for the two values of ‘R’. 5). Also find the distances QR and PR. (1. to verify if an angle is 90 or not. Mobile (where u may transfer cash!) . (4. 0). y) is 10 units. y) is equidistant from the point (3. Find the coordinates of the points of trisection of the line segment joining (4. (1. 6). 6). Do It Yourself! Q8. Find the four sides and the two diagonals. 3). -7/3) 1 1 1 1 1 2 2 2 1 2 Q3. –1) and (–2. Or. –3) in the ratio 2 : 3. (2. Q7. – 5) and B(– 4. y). From the figure. equating the coordinates. and. Find the ratio in which the line segment joining A(1. 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. If A and B are (– 2. A (x. Let this point be A (x. y). 6) in the ratio of k : 1. x = (2 + 8)/2. Thus. (4. Thus. 4 = (y + 5)/2 => y = 3. – 8) is divided by (– 1. the ‘y’ coordinate of any point on ’x’ axis is zero => (5k . Let the coordinates of point A be (x.3m => 2m = 7m => m : m = 2 : 7 The easier method would be to assume that the line segment joining (−3. Therefore. Coordinates of mid-point of AC => [(1 + x)/2. and. -3) = [(x + 1)/2. A. Hence. (Assuming ‘A’ as the origin (0. and.8)/7. Rashmi should post her blue flag at 22. B(4. and. A. Niharika posted the green flag at 1/4 of the distance AD i. Residence: Ghaziabad-UP . 2).3m )/(m + m ) => -m . Given: AP : AB = 3 : 7 => AP:PB = 3:4. A. it can be seen that points X. (5k . 5) is divided by the x-axis.8)/7] = (-2/7. 6) and (3. y = -10 Therefore. we get. -1 = (6m . 20).. 2). – 2) and (2. the coordinates of this point R is (8. – 4). 6) and D(3. (y + 5)/2]] Now. Z are dividing the line segment ‘AB’ in the ratio Benefitted: My Paypal a/c . we would find the midpoint of the given two diagonals.com. Therefore. Hence.0)). Or. find x and y. Find the coordinates of a point A. A. and equate them to each other. y).5 = 0 => k = 1. Y. Now. (1 + x)/2 = 7/2 => x = 6.. The diagonals of a parallelogram bisect each other.67 A. Thus. Mobile (where u may transfer cash!) . 1/4 X 100 = 25m from the starting point of 2nd line. the coordinates of point ‘A’ is (3. Let the ratio be m :m . (y + 4)/2 = -3 => x = 3. the centre of the circle (2. (x.5)/(k + 1) Now. 0. Find the coordinates of the points which divide the line segment joining A(– 2. 1/5 X 100 = 20m from the starting point of 8th line. 22. -3) is the mid-point of its’ diameter (AB). As the vertices of the parallelogram are given in order. coordinate of the point of division is (-4k + 1)/(k + 1). 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. 4] Coordinates of mid-point of BD => [(4 + 3)2. and. we name them as: A(1. 5) are the vertices of a parallelogram taken in order. 2) and B(2.manojarora23@gmail. 10) and (6. If (1. y). Find the ratio in which the line segment joining the points (– 3. −8) is divided by the point (−1. (5k + (-5))/(k + 1) = (-4k + 1)/(k + 1). th th Q4. (-3/2. Q5. (3x(-4) + 4x(-2))/7] = [(6 . −5) & B (−4.m = 6m . and. C(x. ((y + 4)/2 => (x + 1)/2 = 2. point ‘P’ divides AB in the ratio of 3 : 4 Coordinates of ‘P’: [(3x2 + 4x(-2))/7. respectively. Therefore. y) = (5. -20/7) Q9. (y + 5)/2] => [7/2. (2 + 6)/2] => [(1 + x)/2. Let the ratio in which the line segment joining A (1. 5). 4). 8) into four equal parts. y = (25 + 20)/2.5m on 5th line.09871823473.5)/(k + 1) = 0 => 5k . – 3) and B is (1. Similarly. -10) Q8. find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.e. and. 25). Thus. the coordinates of this point G is (2. Preet posted red flag at 1/5 of the distance AD i.5). such that. 5) is divided by x-axis be k:1 Thus. 6). where AB is the diameter of a circle whose centre is (2. (-12 . Also find the coordinates of the point of division. 10) and (6. A. 0) 1 2 1 1 2 1 2 1 2 1 2 1 2 2 Q6.e. A. 5] Coordinates of ‘Z’ are: [(3x2 + 1x(-2))/4. 2) A. (– 1. 1). –5). −1) are the vertices A. – 4). (3. k = 4. and F are given by…. –2). coordinates of ‘X’ are: [(1x2 + 3x(-2))/4. Find the area of the triangle whose vertices are :(i) (2. – 2) and (2. (2. Find the ratio of this area to the area of the given triangle. (2. Let the vertices of the triangle be A (0. 3). (– 3. (i) (7. (2. – 2). for (ii). (4. k) (ii) (8. 0). D of a rhombus ABCD. and D (2. 3). 3). –1). −5). –5) A. Hint: For collinear points. for (i). −2).3 (P.09871823473. Benefitted: My Paypal a/c . (8 + 2)/2] = [0. C (0. EXERCISE 7.manojarora23@gmail. (4. area of triangle formed by them is zero. 3:1 respectively. B (−3. Residence: Ghaziabad-UP . Area of Rhombus = 1/2 x (Product of Its’ diagonals) Let (3. −2). Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0. E. E. (−1. –1). B (2. −1). 1). 5). (–1. Find the area of a rhombus if its vertices are (3. – 5). 1) and (0. (8 + 6)/4] = (-1. C. B. are (– 4. Join AC to form two triangles ΔABC and ΔACD. In each of the following find the value of ‘k’. A. 3). taken in order. Find the area of the quadrilateral whose vertices. 0). Coordinates of D. Mobile (where u may transfer cash!) . for which the points are collinear. (k. 3). 5). k = 3 Q3. Solving. A. 1:1. (3. (1x8 + 3x2)/4] = [(2 . (3. Let the vertices of the quadrilateral be A (−4.com. F be the mid-points of the sides of this triangle. 13/2] Q10. 1). (3x8 + 1x2)/4] = [1. 0). 4) and (– 2. – 1) taken in order. 170): Q1. Let D.68 1:3. A. and. Thus. (5. Do it Yourself! Q2.6)/4. – 4) (ii) (–5. Q4. (5. 7/2) Coordinates of ‘Y’ are: [(2 + (-2))/2. C (3. Refer Above. 4) and (−2. Let the vertices of the triangle be A (4. x + 3y . area of ΔADC is 3 square units.. (3. AD is the median in ΔABC. 2). The centre is (3. and C (5. The two opposite vertices of a square are (–1. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. 0). Mobile (where u may transfer cash!) . (Chapter 9. 3). A. – 6). OA = OB & OB = OC will give two equations. – 6). Therefore. – 2) and B(3. area of ΔABD is 3 square units. Find a relation between x and y if the points (x. Find the centre of a circle passing through the points (6. that a median of a triangle divides it into two triangles of equal areas. EXERCISE 7.manojarora23@gmail. median AD has divided ΔABC in two triangles of equal areas. A. −6). There is a triangular grassy lawn in the plot as shown. –2) and C(5.69 Q5. 0) are collinear. 7). area cannot be negative. Find the coordinates of the other two vertices. Residence: Ghaziabad-UP . Therefore. 4). Therefore. −2). Let D be the mid-point of side BC of ΔABC. 2:9 internally Q2.4 (Optional)* Q1. (1. B(3. A. The students are to sow Benefitted: My Paypal a/c .09871823473.. B (3. (Negative sign here signifies that the triangle is below the ‘x’ axis) However. 2) and (7.7 = 0 Q3.com. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2. A. Verify this result for Δ ABC whose vertices are A(4. Example 3). The area formed by them is 0. area cannot be negative. (Negative sign here signifies that the triangle is below the ‘x’ axis) Clearly. 2) and (3. – 7) and (3. y). However. You have studied in Class IX. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. A. -2) Q4. 2). 2). (1. (1. Q 5. All the sides are equal. Q. (3. Coordinate of point dividing AD in the ratio 2:1 is (x1 + x2 + x3)/3. and area of triangle ABC = 15/32 sq units. – 1). The area of the triangle is 9/2 square units. 2). such that AD/AB = AE/AC = 1/4. but the diagonals are not equal to each other => PQRS is a rhombus. y ) and C(x . 5) and C(1. Calculate the area of the Δ ADE and compare it with the area of Δ ABC. The vertices of a Δ ABC are A(4. 11/3). CD and DA respectively. A. 5). whose medians are [email protected] seeds of flowering plants on the remaining area of the plot. Thus. Thus. Q and R are (12. (i) The median from A meets BC at D. C be vertices of triangle ABC. 7/2). (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. A line is drawn to intersect sides AB and AC at D and E respectively. This point is called the centroid of the triangle. BC. F (5. 1 1 2 2 3 3 Q8. (13. A.com. (i) Taking A as origin.09871823473. (y1 + y2 + y3)/3. R and S are the mid-points of AB. coordinates of D are (13/4. (i) Coordinates of D (mid-point of BC) = (7/2. Taking C as origin. 4) and D(5. C(5. Benefitted: My Paypal a/c . –1). Residence: Ghaziabad-UP . ABCD is a rectangle formed by the points A(–1. P. find the coordinates of the centroid of the triangle. R (11/3. the coordinates of point dividing BE & CF in the ratio 2:1 is same. B(1. BE & CF are concurrent at point (11/3. 5) and C(7. 4) be the vertices of Δ ABC. The area of the triangle is 9/2 square units. Q (11/3. (iii) E (5/3. coordinates of P. A. 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 2 Q7. (ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1 A. Thus. Q & R coincide. y ). What do you observe? A. thus. Let A (4. 5) Area of triangle ABC = 15/2 sq units. coordinates of P. (iv) What do yo observe? (v) If A(x . 4). Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer. y ) are the vertices of Δ ABC. B(– 1. 3). 2) & (6. 2). Find the coordinates of the point D. 2). ie. medians of a triangle is concurrent and the coordinates of the centroid is (x1 + x2 + x3)/3. Q and R are (4. (y1 + y2 + y3)/3. the medians AD. BE & CF. B(x . find the coordinates of the vertices of the triangle. Mobile (where u may transfer cash!) . 3). Q6. Similarly. 6). (v) Let A. 6). 9/2) (ii) P (11/3. coordinate of E are (19/4. B(6. 11/3) (iv) We observe that points P. Taking A as origin. D divides AB in the ratio 1:3. the area is same in both the cases. B. 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. 23/4) Similarly. 6) & (10. involving the sides of a right triangle. ∠CAB (or angle A) is an acute angle.manojarora23@gmail. AC is the hypotenuse of the right triangle and the side AB is the side adjacent to angle A (called as a ‘base’). Residence: Ghaziabad-UP . Side BC is the side opposite to angle A (called as a ‘perpendicular’). 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.71 CBSE CLASS X MATHEMATICS CHAPTER 8 TRIGNOMETRY 1.09871823473. called trignometric ratios. are: The trigonometric ratios of the angle A in right triangle ABC :  Sine of ∠A = Perpendicular = BC Hypotenuse AC Benefitted: My Paypal a/c . Mobile (where u may transfer cash!) . 2.com. Certain ratios. called trigonometric ratios of the angle. Trigonometric Ratios: Here. these ratios can be extended to other angles also. We will confine ourselves to the study of some ratios of the sides of a right triangle with respect to its acute angles. However. for example. we can obtain the other ratios. This is true. and attempt to find trignometric ratios of 30 . N. in place of (sin A) . 2 -1 2 2 2 -1 If we know any one of the ratios. find all the other trigonometric ratios of the angle A. sin0 = 0. the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides. when ∠ A is very close to 0°. cosec 0 & cot 0 are not defined. Consider the trigonometric ratios of ∠A.09871823473. (cos A) . When ∠ C is very close to 0°. (ii) Tan A = 4/3. N. it is given. When ∠A is very close to 0°. and so sin A is very close to 1.. Therefore. and finally when ∠ A becomes very close to 0°. ∠C gets smaller and smaller. sec 0 = 0. The point C gets closer to point B. Say. Residence: Ghaziabad-UP . AC is nearly the same as AB and so the value of cos A = AB/AC. The other trignometric ratios are: tan 0 = 0.angled triangle. Trigonometric Ratios of 0° and 90°: 0 0 0 As ∠ A gets smaller and smaller. Finally. and cos 0 = 1. the length of the side BC decreases. if the triangles considered are similar. 0 0 0 0 0 0 Benefitted: My Paypal a/c . as ∠A is made larger and larger in Δ ABC. etc. 60 .com. AC becomes almost the same as AB.72  Cosine 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.B. Also.B. the Greek letter θ (theta) is also used to denote an angle. ∠ A is very close to 90°. the value of sin A or cos A is always less than 1 (or. cos A. The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle. respectively.. Mobile (where u may transfer cash!) . if the angle remains the same. Trigonometric Ratios of Some Specific Angles: Draw a right . is very close to 1. 45 .manojarora23@gmail. BC gets very close to 0 and so the value of sin A = BC/AC is very close to 0. when ∠A is very close to 90°. we may write sin A. so cos A is very close to 0. But cosec A = (sin A) ≠ sin A (which is called sine inverse A). equal to 1). Also when ∠ A is very close to 90°. 3. till it becomes 90°. and. As ∠A gets larger and larger. etc. the length of the side AB goes on decreasing. sometimes. side AC is nearly the same as side BC. This gives. Remark : Since the hypotenuse is the longest side in a right triangle. ∠ C is very close to 0°. and the length of the sides vary in proportion.: For the sake of convenience.: ‘Sin A’ is ‘the sine of the angle A’. find all the other trigonometric ratios of the angle A. is not the product of ‘sin’ & ‘A’. in particular. and the side AB is nearly zero. ∠C becomes very close to 0° and the side AC almost coincides with side BC. Also. The point A gets closer to point B. (i) Sin A = 1/3. Benefitted: My Paypal a/c . Similarly. for all values of angle A lying between 0° and 90°. sec 0° = 1 = cosec 90° and sec 90°. We have: Trigonometric ratios for ∠ C = 90° – ∠ [email protected] So. sec (90° – A) = cosec A. sin (90° – A) = cos A. they form a pair of complementary angles. 5. we get. Remember the following table: From the table. tan (90° – A) = cot A. Trigonometric Identities: An equation is called an identity when it is true for all values of the variables involved. 4. sin A increases from 0 to 1 and cos A decreases from 1 to 0. Mobile (where u may transfer cash!) . Residence: Ghaziabad-UP . Note : tan 0° = 0 = cot 90°. So. cos (90° – A) = sin A. tan 90° and cot 0° are not defined. from which we may find the other trignometric ratios. an equation involving trigonometric ratios of an angle is called a trigonometric identity. we observe that as ∠ A increases from 0° to 90°.com.09871823473. if it is true for all values of the angle(s) involved. cot (90° – A) = tan A. cosec 0°. we define : sin 90° = 1 and cos 90° = 0. cosec (90° – A) = sec A. Trigonometric Ratios of Complementary Angles: Since ∠ A + ∠ C = 90°. Comparing the ratios in (1) and (2). we get. Use Pythagoras Theorem to find AC.manojarora23@gmail. A. we can also determine the values of other trigonometric ratios. we get.. If sin A = 3/4. Use Pythagoras Theorem to find QR (=5cm). Same as above. So. (iii) is true for all A such that 0° ≤ A < 90°. calculate all other trigonometric ratios. And.1 (P. cos C A. Given 15 cot A = 8. Given sec θ = 13/12. cos A A. 1 + tan A = sec A (iii) This equation is true for A = 0°. if any one of the ratios is known.e. find tan P – cot R. we can express each trigonometric ratio in terms of other trigonometric ratios. Q2. i. In the adjoining figure.we get. BC = 7 cm. AB = 24 cm. tan P – cot R = 0 Q3. right-angled at B. Benefitted: My Paypal a/c . proceed. Mobile (where u may transfer cash!) . this is a trigonometric identity. Dividing (i) by BC . we have:AB + BC = AC (i) Dividing each term of (i) by AC . right-angled at B. A. Using these identities. find sin A and sec A. tan A and sec A are not defined for A = 90°. Do it yourself! Q5. Determine : (i) sin A. 1 + cot A = cosec A (iv) Note that cosec A and cot A are not defined for A = 0°.74 In Δ ABC. (ii) sin C. cos A + sin A = 1 (ii) This is true for all A such that 0° ≤ A ≤ 90°. 181) Q1. A. Therefore (iv) is true for all A such that 0° < A ≤ 90°. and. In Δ ABC.com. Now dividing (i) by AB .09871823473. and. calculate cos A and tan A. So. Do it yourself! Q4. proceed. 2 2 2 2 2 2 2 2 2 2 2 2 CBSE CLASS X MATHEMATICS CHAPTER 8 TRIGNOMETRY .NCERT EXERCISES SOLUTIONS EXERCISE 8. Residence: Ghaziabad-UP . ∠ACD = ∠BCD … (vi) In ΔCAD and ΔCBD. = cos A − sin A = (4/5) . right-angled at point B. thence.(3/5) = 7/25 ∴ L. Assume ’k’ as a constant of proportionality.75 A. check whether or not. Q10. Consider a right triangle ABC. ⇒∠ACD = ∠CPB (Corres. if tan A = 1/√3 . and (v). we get. = (1 . find the value of: (i) sin A cos C + cos A sin C. L. proceed to find all other trignometric ratios. we get BC = 5k. PR + QR = 25 cm and PQ = 5 cm. angles) (iii). Consider a right-angle triangle ΔABC. we get. cos A = cos B => AD/AC = BD/BC (i) To Prove: ∠A = ∠B. If 3 cot A = 4. Q8.com. right-angled at point B. using Pythagoras Theorem. use Pythagoras Theorem to find the third side. Thus. Tan A= BC/AB = 3/4. and proceed. BC = CP. find the value of sin A. PQ = 5. right-angled at Q. and. the third angle of the triangle is also equal (angle sum property). [email protected]. (iv).H. Consider a right triangle ABC. cos C . CD||BP (Converse of Basic Proportionality Theorem). Cons.09871823473. In Δ PQR. If ∠A and ∠B are acute angles such that cos A=cos B. (i). and. Consider a right triangle ABC. Determine the values of sin P. Now. Given. A.tan A) / (1 + tan A) = 7/25. extend AC to P such that BC = CP. (ii) cos A cos C – sin A sin C A. Let ‘k’ be constant of proportionality.H. cos A = AB/AC = 4/5.: Draw AD ⊥AB. Sec θ = 13/12 = Hypotenuse/Adjacent side = AC/AB Let constant of proportionality be ‘k’. proceed to find the value of (i) & (ii). Benefitted: My Paypal a/c . (AC) = (AB) + (BC) = (4k) + (3k) = 16k + 9k = 25k => AC = 5k. It is given that 3cot A = 4. ∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (v) From equations (iii). Given: Triangle ABC in which CD⊥AB. cos A. Q6. = R. and. right-angled at B. and. sin C. find the value of third side (in terms of ‘k’). evaluate: (i) [ ( 1+cos ) (1−cos ) ] . cot A = 4/3. PR + QR = 25.S./Base = BC/AB = 1/√3 Proceeding as above. ∠ACD = ∠BCD [Using equation (vi)] ∠CDA = ∠CDB [Both 90°] Therefore. A. R. tan A = Perp. angles) (iv). 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Q9. ∠BCD = ∠CBP (Alt.S. cos P and tan P. ∴∠CAD = ∠CBD ⇒ ∠A = ∠B [ ( 1+ sin ) (1−sin ) ] Q 7: If cot θ = 7/8.) (ii) Thus. cot A = base/perpendicular = AB/BC = 4/3. Mobile (where u may transfer cash!) . By construction. A. right-angled at B.S. then show that ∠A=∠B. In ΔABC. Thus. AB = 4k & BC = 3k. Residence: Ghaziabad-UP .H. Thus. In triangle ABC. Sin A = BC/AC = 3/5. Proof: From Eq. AD/BD = AC/BC => AD/BD = AC/CP (By Cons. Or.H. (ii) cot θ 2 A. right-angled at B. the given statement is false.manojarora23@gmail. (iii) cos 45 / (sec 30 + cosec 30 ) A. Cos A decreases from 1 to 0. Hence.09871823473. It is the cotangent of ∠A. Clearly. / Base = AC / AB = 12/5. (ii) 2 tan 45° + cos 30° – sin 60° A. Benefitted: My Paypal a/c . A. such value of sec A is possible. 187) Q1. Hence. Answer is ‘(43 . A. Therefore. as ‘A’ increases from 0 to 90 . x = 13 Therefore. the given statement is true. AC − AB < BC < AC + AB. (i) Sin A increases from 0 to 1. ———————————————————————————————————————-EXERCISE 8.76 Let PR be x. such a triangle is possible and hence. Answer is ‘2’. A.9k. Let AC = 12k.90 . Or. Or. Hence the given statement is false. the given statement is false. 7k < BC < 17 k However. and cos A is < 1/√2. Abbreviation used for cosecant of angle A is cosec A. BC = 10. A. Q11(v) sin θ = 4/3 for some angle θ. BC = 10.2 (P. For angles 45 . Mobile (where u may transfer cash!) . (ii) sec A = 12/5 for some value of angle A. for all the following parts. Evaluate the following : (i) sin 60° cos 30° + sin 30° cos 60° A. Applying Pythagoras Theorem. cos P = PQ/PR = 5/13. Hence. 50x = 650. and. ∴ x = (5) + (25 − x) .9 k. we obtain PR = PQ + QR . Or. Q11. we get. Q11(iv) cot A is the product of cot and A. Answer is ‘67 / 12’. sec A = Hyp. Consider a right triangle ABC. BC should be such that. 2 2 0 2 0 0 (iv) A. Hence.24√3) / 11’. QR = 25 − x Applying Pythagoras theorem in ΔPQR. Thus. for any triangle. and ‘less’ than the sum of the other two sides. as ‘A’ increases from 0 to 90 . is greater than 1. sin θ = (Side opposite to θ) / Hypotenuse. In a right-angled triangle. sin P = QR/PR = 12/13. the given statement is false. 2 2 2 2 2 2 0 0 0 0 0 0 0 0 Q11(iii) cos A is the abbreviation used for the cosecant of angle A. State whether the following are true or false. any side is ‘more’ than the difference of the other two sides. Residence: Ghaziabad-UP . cot A is not the product of cot and A.√6) / 8’. (i) The value of tan A is always less than 1. PR = 13 cm.com. hypotenuse is always greater than the remaining two sides. And cos A is the abbreviation used for cosine of angle A. 12k − 5k < BC < 12k + 5k. In a right-angled triangle. Now. tan P = QR/PQ = 12/5. A. Thus. Let ‘k’ be constant of proportionality. tan A for angles 45 . right-angled at B. AB = 5k. Hence. ∴. Answer is ‘1’.90 = sin A/cos A. QR = (25 − 13) cm = 12 cm. Answer is ‘(3√2 . (v) A. Justify your answer. Or. such value of sin θ is not possible. sin A is > 1/√2. cot A = cosA/sinA = cos0 / sin0 = 1/0 = Not-defined. (C). find A and B. cos 30 = √3/2 = 0. A. (C) tan 60°. cos60°. the given statement is false. sin 30 = 0. only when θ = 45 . Residence: Ghaziabad-UP . Hence. sin30° A.866. The correct answer is ( C ).707. Clearly. (i) sin (A + B) = sin A + sin B.manojarora23@gmail. tan (A + B) = √3 = tan 60 => A + B = 60 (i) tan (A .866.com. cos 60 = 1/2 = 0.5. sin 90 = 1. 0 0 0 0 (iii) The value of cos θ increases as θ increases. sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = (1 + √3)/2. the given statement is false. Given expression = √3 = tan 60 0 Q3.3 (P. sin (A + B) = sin (30° + 60°) = = sin 90° = 1. Hence. we get. 0 Q2 (ii) (A) tan 90°. then. (D) 60° A. (D) 0 A.09871823473. sin 60 = √3/2 = 0. 0° < A + B ≤ 90°. The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as: Sin 0 = 0. the given statement is true. As sin 2A = sin 0° = 0 2 sinA = 2sin 0° = 2(0) = 0 Hence. A = ∠A = 45 & ∠B = 15 .5. Correct answer is (D).77 Q2.B = 30 (ii) Solving (i) & (ii). If tan (A + B) = √3 and tan (A . Out of the given alternatives. 189) Benefitted: My Paypal a/c . 0 0 0 0 (ii) The value of sin θ increases as θ increases. A. =√3 / 2 = sin 60 . sin (A + B) ≠ sin A + sin B. cos 45 = 1/√2 = 0. 0 (v) cot A is not defined for A = 0°. (B) 1. (B). (iv) (A) cos 60°. (A) is correct. (D). Choose the correct option and justify your choice : (i) (A). Justify your answer. (B) 30°. Hence. The given statement is false. A. Mobile (where u may transfer cash!) .B) = 1/√3. A. The given statement is true. The value of cos θ decreases as θ increases in the interval of 0° < θ < 90° as: cos 0 = 1. 0 0 EXERCISE 8. State whether the following are true or false. sin 90 = 1. (C) sin 45°. as. A. (iii) sin 2A = 2 sin A is true when A = (A) 0°. (C) 45°. Hence. (A) is correct. A. sin60°. (D) sin 30° A. as sin θ = cos θ. tan60°. A > B. only A = 0° is correct. 0 0 0 0 0 (iv) sin θ = cos θ for all values of θ. Q4. Let A = 30° and B = 60°.B) = 1/√3 = tan 30 => A . (B) sin 60°. If tan A = cot B. 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. √(1 + cot A) will always be positive as we are adding two positive quantities. A. sec A and tan A in terms of cot A. Thus.sec 59 = 0 0 0 0 0 0 Q2. sin A = 1/√(1 + cot A) (ii) tan A = 1/cot A. Thus. sec A = 1 + tan A = 1 + 1/cot A = cot A + 1 cot A Thus. 2 2 Q7. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Q2. If sec 4A = cosec (A – 20°). A.∠A Thus.com.78 Q1.∠A => ∠B + ∠C = 90 . find the value of A. prove that A + B = 90°. where 4A is an acute angle. ∠A + ∠B + ∠C = 180° => ∠B + ∠C = 180 .A/2) = cos A/2. tan A = cot B => tan A = tan (90° − B) => A = 90° − B => A + B = 90° Q5. Given that. Given that. A. If tan 2A = cot (A – 18°). A. sin((B + C)/2) = sin (90 . sec A = [√(1 + cot A)]/cotA (iii) tan A = 1/cot A. B and C are interior angles of a triangle ABC.59 ) .42 ) . Mobile (where u may transfer cash!) . A.manojarora23@gmail. where 2A is an acute angle. tan 2A = cot (A− 18°) => cot (90° − 2A) = cot (A −18°) => 90° − 2A = A− 18° => 108° = 3A => A = 36° Q4. For a triangle ABC. Evaluate : (i) sin18 / cos72 = sin (90 . Residence: Ghaziabad-UP . Write all the other trigonometric ratios of ∠A in terms of sec A. Given that. sec 4A = cosec (A − 20°) => cosec (90° − 4A) = cosec (A − 20°) => 90° − 4A= A− 20° => 110° = 5A => A = 22° Q6. Also. If A. sin 67° + cos 75° = sin (90° − 23°) + cos (90° − 15°) = = cos 23° + sin 15° 0 0 0 EXERCISE 8. 1/cosec = 1/(1 + cot A) => sin A = 1/(1 + cot A) => sin A = ± 1/√(1 + cot A) Now.sin 42 = 0 0 0 0 0 0 (iv) cosec 31° – sec 59° = cosec (90 .4 (P.64 )/cot 64 = cot 64 /cot 64 = 1 0 0 0 0 0 0 0 (iii) cos 48° – sin 42° = cos (90 .sec 59 = sec 59 .sin 42 = sin 42 . Show that : (i) tan 48° tan 23° tan 42° tan 67° = 1 A. Express the trigonometric ratios sin A. (i) cosec = 1 + cot A. then show that sin((B + C)/2) = Cos (A/2) A. 193) Q1. (i) cos A = 1/sec A (ii) sin A + cos A = 1 2 2 Benefitted: My Paypal a/c . 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 Q3.09871823473. find the value of A.72 )/cos72 = 1 0 0 0 0 0 (ii) tan 26 / cot 64 = tan (90 . Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. (C) cot A. Q4. 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. (i) 9 sec A – 9 tan A = (A) 1 . (iv) (1 + tan A) / (1 + cot A) = (A) sec A.manojarora23@gmail. Choose the correct option.(C) 8 . where the angles involved are acute angles for which the expressions are defined.H..09871823473. (B) 1.H. Residence: Ghaziabad-UP .27)] + sin 27 = 1/1 =1 [cos(90 . (iii) (sec A + tan A) (1 – sin A) = (A) sec A. (C) 2. Correct answer is ‘tan A’. alternative (B) is correct. use the identity: sin A + cos A = 1 2 2 (iii) A. (D) −1 A. (iv) A. Convert all the given trignometric ratios in the form of sinA & cosA. & you would reach R.S. Convert all the given trignometric ratios in the form of sinA & cosA.S.com. 2 2 2 2 2 2 (ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ) = (A) 0. (D) cos A A. Mobile (where u may transfer cash!) . Prove the following identities. Correct answer is ‘cos A’. Do it yourself. (C) cosec A. Justify your choice. The correct answer is ‘2’.C.. (D) tan A A.) Q4. (B) sin A. Evaluate: (i) = [sin(90 . 2 2 2 2 2 2 Q5. Can be done in two ways:  Start from L. Hence. (B) –1.cos A. (i) 9 sec A − 9 tan A = 9 (sec A − tan A) = 9 (1) [As sec2 A − tan2 A = 1] = 9. use sin A + cos A = 1 => sin A = 1 .M. Take L.(D) 0 A.. (i) (ii) A.79 Thus. Convert all the given trignometric ratios in the form of sin A & cos A.73)] + cos 73 2 2 2 2 (ii) sin 25° cos 65° + cos 25° sin 65° = 1 (Ans.  Simplify LHS and RHS separately] 2 2 2 2 Benefitted: My Paypal a/c .(B) 9 . 2 2 A. 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.e. i.80 (v) using the identity 1 + cot A = cosec A. Benefitted: My Paypal a/c .manojarora23@gmail. Do it yourself! (vi) A.com. Residence: Ghaziabad-UP . Mobile (where u may transfer cash!) ... i. Do it yourself! (vii) (viii) (ix) (x) CBSE CLASS X MATHEMATICS CHAPTER 9 SOME APPLICATIONS OF TRIGNOMETRY 1. 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. 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 case when we lower our head to look at the point being viewed. the case when we raise our head to look at the object.09871823473.e. A.1 (P. A contractor plans to install two slides for the children to play in a park. What should be the length of the slide in each case? A.5 m. and inclined at an angle of 60° to the ground. 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. Q3.manojarora23@gmail. is 30°. A. and solve the question yourself. Find the height of the tree. The angle of elevation of the top of a tower from a point on the ground.NCERT EXERCISES SOLUTIONS EXERCISE 9. Benefitted: My Paypal a/c . whereas for elder children. A circus artist is climbing a 20 m long rope.09871823473. if the angle made by the rope with the ground level is 30°. 203): Q1. which is tightly stretched and tied from the top of a vertical pole to the ground. For the children below the age of 5 years. Firstly draw the diagram. she wants to have a steep slide at a height of 3m. The two diagrams formed would be as the ones adjacently drawn. and is inclined at an angle of 30° to the ground. Residence: Ghaziabad-UP . Find the height of the tower. Proceed and do the question yourself! Q4. Find the height of the pole. Do it Yourself! Q2.81 CBSE CLASS X MATHEMATICS CHAPTER 9 SOME APPLICATIONS OF TRIGNOMETRY . which is 30 m away from the foot of the tower. Mobile (where u may transfer cash!) . The distance between the foot of the tree to the point where the top touches the ground is 8 m. she prefers to have a slide whose top is at a height of 1.com. 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 building.82 A. Let AB be the building and CD be the tower. Draw the diagram (Adjacent figure). From a point between them on the road. and. stands on the top of a pedestal. A TV tower stands vertically on a bank of a canal. Draw the diagram (Adjacent figure). A 1. The string attached to the kite is temporarily tied to a point on the ground.com. A statue. Find the height of the pedestal. A. ‘D’ is a point 20m away from the bank ‘C’ of the canal. A. If the tower is 50 m high. A. Draw the diagram (Adjacent figure ~ ‘AB’ is the building. Solve the question yourself! Q6. B & C are the two banks of the canal. From a point on the other bank directly opposite the tower. Solve the question yourself! Q11. and. A. the given statement is true. Two poles of equal heights are standing opposite each other on either side of the road. From a point on the ground. and. A kite is flying at a height of 60 m above the ground. the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Draw the diagram (Adjacent figure). and. and. Determine the height of the tower. and. As observed from the top of a 75 m high lighthouse from the sea-level.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88. the angles of elevation of the top of the poles are 60° and 30°. 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. ‘CD’ is the cable tower). Residence: Ghaziabad-UP . Find the height of the tower. From the top of a 7 m high building. find the distance between the two ships. Find the distance he walked towards the building. 1. A. which is 80 m wide. and. ’C’ & ’D’ are the two ships ). Draw the diagram (Adjacent figure). Solve the question yourself! Q10. the angle of elevation of the top of the tower is 30° (see adjacent figure). From a point on the ground. 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°.manojarora23@gmail. A. Solve the Question Yourself! Q5. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. 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°. A 1. Solve the question yourself! Q7. Find the length of the string. and. AB is the TV Tower. Find the height of the tower and the width of the canal. Find the height of the poles and the distances of the point from the poles.2 m from Benefitted: My Paypal a/c . tan θ = Perpendicular/Base. Draw the diagram (Adjacent figure). Draw the diagram. and. The inclination of the string with the ground is 60°. Solve the question yourself! Hence. and. A. Draw the diagram (Adjacent figure). Solve the question yourself! Q14. Solve the question yourself! Q9. A. respectively. A. Q12. If one ship is exactly behind the other on the same side of the lighthouse. Q13. the angles of depression of two ships are 30° and 45°.6 m tall. Mobile (where u may transfer cash!) .5 m tall boy is standing at some distance from a 30 m tall building. Solve the question yourself! Q8.09871823473. assuming that there is no slack in the string. Draw the diagram (Adjacent figure ~ ‘AB’ is the light house. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°.  A Quadrant is ‘One-Fourth’ of the Circular Disc.manojarora23@gmail. Find the distance travelled by the balloon during the interval. Solve the question yourself! Q15. which are at a fixed distance from a fixed point in the plane. the angle of elevation reduces to 30° (see Fig. Find the time taken by the car to reach the foot of the tower from this point.2m high) Draw the diagram. 6secs & 3 secs. 9. After some time. Prove that the height of the tower is 6 m. CBSE CLASS X MATHEMATICS CHAPTER 10 CIRCLES (RECAP FROM PREVIOUS CLASSES) 1.The Degree measure of the corresponding minor arc”. A straight highway leads to the foot of a tower. Introduction: The collection of all the points in a plane.  Degree measure of a major arc is “3600 . and. A. 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.09871823473. is called a circle.  Degree measure of a minor arc is the measure of the central angle subtended by the arc. Benefitted: My Paypal a/c . A man standing at the top of the tower observes a car at an angle of depression of 30°.  A Sector is that region of a circular disc which lies between an arc. A.  The Degree measure of the circumference of the circle is always 3600.com. which is approaching the foot of the tower with a uniform speed.  Arc is a continuous piece of a circle. Six seconds later. and the two radii joining the extremities of the arc and the centre. Residence: Ghaziabad-UP .83 the ground. Q16. Let the initial position A of balloon change to B after some time and CD be the girl (1.13).  Concentric Circles are the ones which have the same centre. Mobile (where u may transfer cash!) . the angle of depression of the car is found to be 60°. if two arcs are congruent. Proof: Let perpendicular bisectors of AB & BC meet at ‘O’. then. then the chords are equal. Its centre and radius are called the ‘Circumcentre’ and the ‘Circumradius’ of the triangle. ‘B’ & ‘C’. OA = OB = OC = ‘r’ (say).  Theorum 6: Equal chords of a circle (or.com.(i) ∠QOB = 2∠QAO . (Use SAS)    Theorum 3: The Perpendicular from the centre of a circle to a chord bisects the chord. This circle is called the ‘Circumcircle’ of the ∆ABC. and conversely. then their corresponding arcs are congruent. and the angle subtended by the major arc PQ at O is ‘Reflex Angle POQ’. ∠AQO = ∠QAO (Angles opposite to the equal sides) ∠POB = 2∠PAO . then. we get the desired result. of Congruent Circles) are equidistant from the centre (or centres). Mobile (where u may transfer cash!) . Since.: If two chords of a circle are equal. The angle subtended by the minor arc PQ at O is ∠POQ. Congruent arcs of a circle subtend equal angles at the centre.(ii) (The above two equations are due to Exterior Angle Property) Adding Eqs. two lines can’t intersect at more than one point. (Use RHS)  Theorum 4: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Benefitted: My Paypal a/c . O’ will be on the perpendicular bisectors of AB & BC. B & C of a Triangle. Thus. Residence: Ghaziabad-UP . r). If possible. then their corresponding chords are equal. (Use SSS)    Theorum 2: If the angles subtended by the chords of a circle at the centre are equal.84 2. we can draw a circle passing through the three points ‘A’.09871823473.: If ABC is a triangle. ((While dealing with an angle subtended by an arc at the centre of the circle.B. passing through the three points ‘A’.B. So. in front of the arc.  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. (i)) ∠APO = ∠PAO. Theorums: Following theorems have to be remembered along with their proofs:  Theorum 1: Equal chords of a circle subtend equal angles at the centre. There is one and only one circle passing through three given non-collinear points. (Use RHS)  Theorum 7: Chords equidistant from the centre of the circle are equal in length.)) Proof: (Consider fig. we see the angle subtended by the arc.  Theorum 5: There is one and only one circle passing through three given non-collinear points.manojarora23@gmail. (i0 & (ii). Thus. ‘B’ & ‘C’. N. O’ must coincide with O. let there be another circle C(O’. there is a unique circle passing through the three vertices A. and ‘r’ as radius. N. by the above Theorum. with ‘O’ as centre. B. Major Segment) is acute. That is. 360 = 2 (∠A + ∠C ). Or. x + y = 2 (∠A + ∠C ). ie. Or. say E (or E′).09871823473. As. Proof: x = 2 ∠A (i) y = 2 ∠C (ii) Adding eq. E and B lie on a circle. or. (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. ∠AOB is angle subtended at centre by minor arc. Reflex ∠AOB > 2 Right Angles or > 180 .manojarora23@gmail. C and B. If points A. (Very Important) Proof: AB is a line segment.  Theorem 10: The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. ∠ AEB = ∠ ADB. E′ should also coincide with D. (i) above) An Important Result: The Angle in a Semi-Circle is a Right Angle. (i) and (ii). ∠ ACB = ∠ AEB. ∠AOB < 180 . But it is given that ∠ ACB = ∠ ADB. Thus. ∠ ACB = ∠ ADB. ∠ADB is Obtuse. An Important Result: Any Angle subtended by a minor arc in the Alternate Segment (ie. ∠ACB is Acute. Similarly. they are concyclic).85  Theorum 9: Angles in the same segment of a circle are equal. consider fig. the four points lie on a same circle (i. 0 0 0 0 0 0 Cyclic Quadrilaterals: A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle. Residence: Ghaziabad-UP .com. Minor Segment) is obtuse: Proof: (i) Here ∠ACB is an angle subtended by minor arc. we can prove the result for the other pair of angles. Similarly. Thus. Or. (ii) Here ∠ADB is an angle subtended by major arc. Mobile (where u may transfer cash!) . while any angle subtended by a major arc in the Alternate Segment (ie. Then it will intersect AD (or extended AD) at a point. Suppose it does not pass through the point D. 0 0 Benefitted: My Paypal a/c . which subtends equal angles at two points C and D. ∠ACB is < 90 .e. 2∠ADB > 180 . ∠ADB > 90 Thus. Proof: (Again. ∠AOB = 2∠ACB. Therefore. Also. Reflex ∠AOB = 2∠ADB. Thus. Thus. This is not possible unless E coincides with D. C and D lie on a circle let us draw a circle through the points A. C. 1/2 ∠AOB is < 90 . To show that the points A. (∠A + ∠C ) = 180 . Mobile (where u may transfer cash!) .com.There is only one point ‘A’. Proof: We are given a circle with centre O and a tangent XY to the circle at a point P. (Note that if Q lies inside the circle. 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. By theorem above. OQ > OP. The tangent to a circle is a special case of the secant. (Proof beyond our scope) RECAP OVER! NOW MATHEMATICS FOR CLASS X! CBSE CLASS X MATHEMATICS CHAPTER 10 CIRCLES 1. (iii) PQ is a Tangent to the circle . Introduction: Consider a circle and a line PQ. We need to prove that OP is perpendicular to XY. That is. The point Q must lie outside the circle.manojarora23@gmail. OQ is longer than the radius OP of the circle. when the two end points of its corresponding chord coincide. XY will become a secant and not a tangent to the circle).09871823473.86  Theorem 11: If the sum of a pair of opposite angles of a quadrilateral is 180º. OP is the shortest of all the distances of the point O to the points of XY. Since this happens for every point on the line XY except the point P. Theorem 1. Residence: Ghaziabad-UP . Number of Tangents from a Point on a Circle: Benefitted: My Paypal a/c .: The tangent at any point of a circle is perpendicular to the radius through the point of contact. then the quadrilateral is Cyclic. Remarks : 1. 2. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point. 2. There can be three possibilities as given: (i) PQ is a non-intersecting line. we can also conclude that at any point on a circle there can be one and only one tangent. So OP is perpendicular to XY. Activity: There is only one tangent that can be drawn at a point of the circle. 3. (ii) PQ is a Secant of the circle. common to the line PQ & the circle. Therefore. A Tangent to a Circle is a line that intersects the circle at only one point. How many tangents can a circle have? A. because these are angles between the radii and tangents. Draw a circle and two lines parallel to a given line such that one is a tangent and the other.NCERT EXERCISES SOLUTIONS EXERCISE 10. So. Now in right triangles OQP and ORP. by cpct.5 cm (D)√119 cm A. T1 and T are the points of contact of the tan gents PT & PT respectively. Q2. Q3.com. i.1: Q1. A line drawn from the centre of the circle to the tangent is perpen dicular to the tangent.manojarora23@gmail. Remarks:  The theorem can also be proved by using the Pythagoras Theorem as follows: PQ = OP – OQ = OP – OR = PR (As OQ = OR). Therefore.  Also. Then ∠OQP and ∠ORP are right angles. a point P lying outside the circle and two tangents PQ. OP PQ. Mobile (where u may transfer cash!) . Applying Pythagoras theorem in ΔOPQ. OQ and OR. We find that we can draw exactly two tangents to the circle through this point (Say ‘P’). For this. 2 2 2 2 2 2 CBSE CLASS X MATHEMATICS CHAPTER 10 CIRCLES . Length PQ is: (A) 12 cm (B) 13 cm (C) 8.e. A circle can have infinite tangents. Thus. OQ = OR (Radii of the same circle). (iv) The common point of a tangent to a circle and the circle is called Point of Contact . PQ = PR. (ii) There is only one tangent that can be drawn to the circle at a point (say ‘P’). (iii) A circle can have two parallel tangents at the most. PQ = √119 cm Q4. a secant to the circle.. Thus. OP = OP (Common). 2 1 2 Theorem 2.87 Activity: (i) All the lines through ’P’ (a point inside the circle) intersect the circle in two points. ie. OP is the angle bisector of ∠ QPR. the centre lies on the bisector of the angle between the two tangents. we join OP. (ii) A line intersecting a circle in two points is called a Secant . ∠OPQ = ∠OPR. Benefitted: My Paypal a/c . 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. Proof: We are given a circle with centre O. they are right angles. Residence: Ghaziabad-UP . it is not possible to draw any tangent to a circle through a point inside it.: The lengths of tangents drawn from an external point to a circle are equal. 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. which gives PQ = PR.09871823473. We are required to prove that PQ = PR. Fill in the blanks : (i) A tangent to a circle intersects it in one point (s). (iii) Now taking a point P outside the circle and drawing tangents to the circle from this point. Thus. ΔOQP ≅ ΔORP (by RHS). and according to Theorem 1. PR on the circle from P. OA = 5cm and AB = 4 cm. Find the radius of the circle.2: Q1. 0 0 0 0 0 0 Q3. A. A. ∴ ∠OPB = 90° (ii) Comparing equations (1) and (2). 2 2 2 2 Benefitted: My Paypal a/c . Q5. Let us consider a circle centered at point O. Line AB is intersecting the circle at exactly two points. As perpendicular to AB at P passes through O’. therefore. Proof: Let us assume that the perpendicular to AB at P does not pass through centre O. then ∠ POA is equal to (A) 50° (B) 60° (C) 70° (D) 80° A. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. AB & CD are two parallel lines. AB is a tangent drawn on this circle from point A. Therefore. Join OP and O’P.com. OP = 7cm. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. TP and TQ are tangents => Radius drawn to these tangents are perpendicular to the tangents. OP⊥TP & OQ⊥TQ => ∠OPT = 90º & ∠OQT = 90º In quadrilateral POQT. OB ⊥ AB (Radius ⊥ tangent at the point of contact). From a point Q. A. Q6. when the line O’P coincides with OP. Residence: Ghaziabad-UP . then the lines are parallel. 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. Let it pass through another point O’. Hence. Thus. ∠O’PB = 90 (i) O is the centre of the circle and P is the point of contact. Mobile (where u may transfer cash!) .If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°. 90 + 80 + 90 + ∠BOA = 360 => ∠BOA = 100 . By RHS. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24. From the figure. 0 0 0 0 0 0 Q4. the perpendicular to AB through P passes through centre O. Since line CD is intersecting the circle at exactly one point.manojarora23@gmail. the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. P and Q. we obtain AB + BO = OA => 42 + BO = 52 => BO = 3. If the co-Interior angles between the two lines are supplementary.88 A. Q2. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other. we obtain ∠O’PB = ∠OPB (iii) This is only possible. R. Therefore. line CD is the tangent to the circle. Given that. By Pythagoras Theorum. Prove that the perpendicular at the point of contact to the tangent of a circle passes through the centre. the radius of the circle is 3 cm. ΔOPB ≅ ΔOPA => ∠BOP = ∠AOP => ∠AOP = ∠ POA = 50 .5 cm A. EXERCISE 10. Applying Pythagoras theorem in ΔABO.09871823473. To Prove: The line perpendicular to AB at P passes through centre O. line AB is the secant of this circle. In ΔABO. Sum of all interior angles = 360 => ∠OPT + ∠POQ +∠OQT + ∠PTQ = 360 ⇒ 90 + 110º + 90 +∠PTQ = 360 ⇒ PTQ = 70 . 89 Q7. Prove that AB + CD = AD + BC. A. Benefitted: My Paypal a/c .manojarora23@gmail. PQ = 2AP = 2 × 4 = 8 => The length of the chord of the larger circle is 8 cm. ΔOQB ΔOCB (by SSS) Thus. Prove that ∠ AOB = 90°. ΔOPA ΔOCA (SSS congruence criterion) Therefore. Join O to C. Sum of all interior angles = 360º Thus. & BC = AD (ii) It can be observed that. DR + CR + BP + AP = DS + CQ + BQ + AS Thus. we obtain OA + AP = OP => 32 + AP = 52 => AP = 4 In ΔOPQ. ∠APB + ∠BOA = 180º Hence. Therefore. AP = AQ (Perpendicular from the center of the circle bisects n the chord). Therefore. Let the two concentric circles be centered at point O. PQ is tangent to the smaller circle => OA ⊥ PQ (As OA is the radius of the circle) Applying Pythagoras theorem in ΔOAP. ∠POA = ∠COA … (i) Similarly. A. A. 2∠COA + 2∠COB = 180º =>∠COA + ∠COB = 90º =>∠AOB = 90° Q10. 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. A. => AB = CD (i). DR = DS (Tangents on the circle from point D).com. CR = CQ (Tangents on the circle from point C). we get. 2 2 2 2 Q8. A. A quadrilateral ABCD is drawn to circumscribe a circle. Two concentric circles are of radii 5 cm and 3 cm. Let us consider a circle centered at point O. In ΔOPA and ΔOCA. Q11. OP = OC (Radii of the same circle) AP = AC (Tangents from point A) AO = AO (Common side) Thus. It can be observed that OA (radius) ⊥ PA (tangent) => ∠OAP = 90° Similarly. ∠OAP +∠APB+∠PBO +∠BOA = 360º => 90º + ∠APB + 90º + ∠BOA = 360º Thus. Thus. joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle. we get. ∠QOB = ∠COB … (ii) Since POQ is a diameter of the circle. Residence: Ghaziabad-UP . Find the length of the chord of the larger circle which touches the smaller circle. (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) => CD + AB = AD + BC Q9. 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. And let PQ be the chord of the larger circle which touches the smaller circle at point A. ∠POA + ∠COA + ∠COB + ∠QOB = 180º From equations (i) & (ii).09871823473. Mobile (where u may transfer cash!) . 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. 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. Let ABCD be the parallelogram. 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. OB (radius) ⊥ PB (tangent) => ∠OBP = 90° In quadrilateral OAPB. it is a straight line. OA ⊥ PQ. Prove that the parallelogram circumscribing a circle is a rhombus. R. Adding all these equations.90 BP = BQ (Tangents on the circle from point B). & S. Therefore. ABCD is a rhombus. ∠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. AP = AS (Tangents on the circle from point A). In Δ ABC.manojarora23@gmail. ∠4 = ∠5. Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB 2 2 Thus. However. Area of ΔOAB=1/2 x 4x(8 + x)=16 + 2x cm . BC = BD + DC = 8 + 6 = 14. Area of ΔOCA = 1/2 x 4x(6 + x)=(12 + 2x)cm . x = −14 is not possible as the length of the sides can’t be negative. and. A. Join the vertices of the quadrilateral ABCD to the center of the circle. 2AB = 2BC => AB = BC (iii) Comparing equations (1).&. ΔOAP ≅ ΔOAS (SSS congruence criterion) => ∠POA = ∠AOS (cpct) => ∠1 = ∠8 Similarly. Now. x = 7 => AB = x + 8 = 7 + 8 = 15 cm. Q. CA = 6 + x = 6 + 7 = 13 cm. (2). Mobile (where u may transfer cash!) . the length of the line segment AF be x. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. 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 . 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. & ∠6 = ∠7. Let the given circle touch the sides AB and AC of the triangle at point E and F. we can prove that ∠BOC + ∠DOA = 180º Hence. CA = CF + FA = 6 + x. A. and (3). 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. we get. we get. Consider ΔOAP and ΔOAS.09871823473. 2 = 28 + 12 + 2x + 16 + 2x => (x + 14)(x .com. or x = 7. ∠2 = ∠3.7) = 0 => x = -14. 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. Residence: Ghaziabad-UP . Q13. Q12. opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre Benefitted: My Paypal a/c . Find the sides AB and AC. AP = AS (Tangents from the same point) OP = OS (Radii of the same circle) OA = OA (Common side) Thus. we obtain AB = BC = CD = DA => Hence. 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 . and. AC / CB = 3/5. draw a line parallel to BA5 (by making ∠AA3C  Then. 5 Proof:  Since A C is parallel to A B.  By construction. ‘C’ divides AB in the ratio 3 : 2. Method 1: Steps:  Draw any ray AX. Thus. AC:CB = 3:2 = ∠AA B).91 of the circle. To divide a line segment in a given ratio: Given a line segment AB. m = 3 and n = 2. thus. Method 2: Steps:  Draw any ray AX. AA / A A = AC / CB (by BPT). Division of a Line Segment: 1. AA / A A = 3/5. Let. where both m and n are positive integers. making an acute angle with AB. we want to divide it in the ratio m : n. making an acute angle with AB. intersecting AB at C. 3 5 3 3 3 3 5 5 Benefitted: My Paypal a/c .com. Residence: Ghaziabad-UP .09871823473.. Thus.  Join BA5. such that. CBSE CLASS X MATHEMATICS CHAPTER 11 CONSTRUCTIONS 1.  Through A3 (m = 3).manojarora23@gmail. Mobile (where u may transfer cash!) .  Locate 5 (= m + n) points on AX. AA1 = A1A2 = A2A3 ….  Thus. A2. Thus.09871823473. the triangle to be constructed larger than the given triangle. 2. B . we could take a ray making an acute angle with AB or AC & proceed similarly. Of scale factor 5/3). or. AA /BB =AC/BC. Proof:  Δ AA C is similar to Δ BB C (by AAA). Steps: 1. Residence: Ghaziabad-UP . BC/BC’= (BC’ + C’C)/BC’ = 1 + C’C/BC’ = 1 + 1/3 = 4/3. 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. B & B on BX so that BB = B B = B B = B B = B B 3. A. Join B (the 3 point. In both the above examples. B . A’BC’ is the required triangle. Therefore. Draw a line through C′ parallel to the line CA to intersect BA at A′ Then. A’B/AB = BC’/BC = A’C’/AC = 3/4. 2. The scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle. ie. Draw a line through C’. 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. BC/BC’ = BB /BB = 3/5 => BC’/BC = 5/3. B . Mobile (where u may transfer cash!) . BC’/ C’C = 3/1.manojarora23@gmail. 3 2 3 2 3 1 2 2 3 4 1 4 3 1 2 2 3 3 4 4 Proof:  By construction. of scale factor 3/4). A’B/AB = BC’/BC = A’C’/AC = 5/3. Δ A′BC′ ~ Δ ABC (by AA Criteria). Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.  Thus. Then AC : CB = 3 : 2. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. 4. intersecting the extended line segment BC at C’. by construction.92  Draw a ray BY parallel to AX by making ∠ABY equal to ∠ BAX. B2 (n = 2) on BY such that AA = A A = A A = BB = B B .  Thus. 3 5 2. 3. Ex. Locate 4 points B . Let it intersect AB at a point C. Proof:  ΔABC ~ ΔA’BC’ (AA Criteria). 5 3 BA at A’. AB/A’B = BC/BC’ = AC/A’C’. A.  Thus..com. AA /BB =3/2 => AC/BC=3/2 This method may be used for dividing the line segment in any ratio.  Locate the points A1. BC’/BC = 3/4 Also C′A′ is parallel to CA. Construction of Tangents to a Circle: Benefitted: My Paypal a/c . A3 (m = 3) on AX and B1. Locate 5 points (the greater of 5 & 3 in 5/3) B . 2.e. 3 being smaller of 3 & 5 in 5/3) to C draw a line through B parallel to B C.  But. 1 1 2 2 3 1 1 2  Join A3B2. Δ A′BC′ is the required triangle. Consider the following example: Ex. B and B on BX so that BB = B B = B B = B B . parallel to CA intersecting the extended line segment 1 1 1 2 2 3 3 4 4 rd 3 2 3 4 5 5. Then. Steps: 1. Join B C and draw a line through B parallel to B C to intersect BC at C′. 4.  Now. 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. com. 1 2 3 4 13 1 1 2 2 3 13 5 13 13 5 Benefitted: My Paypal a/c .NCERT EXERCISES SOLUTIONS EXERCISE 11.6 cm & draw a ray AX making an acute angle with line segment AB. C is the point dividing line segment AB of 7. there will be two tangents to the circle from this point.7 cm respectively. A line segment of length 7. Then.219) Q1. OQ is a radius of the given circle. We are given a circle with centre O.93 If a point lies on the circle. Let it intersect the given circle at the points Q and R. Let M be the midpoint of PO. Similarly. The lengths of AC and CB can be measured. and a point ‘P’ outside it. if the point lies outside the circle.6 cm in the required ratio of 5:8. ∠PQO = 90°. PQ ⊥ OQ. Draw a line segment of length 7. draw a circle. Note: If centre of the circle is not given. PR is also a tangent to the circle. A . A .  Join BA .  Taking M as centre and MO as radius.  Thus. Also. Since. then there is only one tangent to the circle at this point and it is perpendicular to the radius through this point. Measure the two parts. A. To construct the tangents to a circle from a point outside it. ∠PQO is an angle in the semicircle and. we may locate its centre first by taking any two non-parallel chords and then finding the point of intersection of their perpendicular bisectors. A .  Join PQ & PR.. A ……. 1. PQ & PR are the required two tangents.1: (P. We have to construct the two tangents from ‘P’ to the circle. It comes out to 2.9 cm and 4. draw a line parallel to BA (by making an angle equal to ∠AA B) at A intersecting AB at point C. Steps:  Join PO and bisect it. Mobile (where u may transfer cash!) .6 cm can be divided in the ratio of 5:8 as follows: Steps:  Draw line segment AB of 7.09871823473. Proof:  Join OQ.  Locate 13 (= 5 + 8) points. A .6 cm and divide it in the ratio 5 : 8.  Through the point A . CBSE CLASS X MATHEMATICS CHAPTER 11 CONSTRUCTIONS .manojarora23@gmail. therefore. PQ has to be a tangent to the circle. Then we could proceed as above. on AX such that AA = A A = A A …. Residence: Ghaziabad-UP . A ΔAB'C' whose sides are 3/2 times of ΔABC is drawn as: Steps:  Draw AB of 8 cm. B’C’ = 3/2BC. 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. AC’ = 3/2AC. Mobile (where u may transfer cash!) . ∴ ΔABC ∼ ΔAB'C' (AA crit. ∠ABC = ∠AB'C' (Corres. Then construct a triangle whose sides are 3/4 of the corresponding sides of triangle ABC. A.  Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C. ΔAB'C' is the required triangle. Let OO' intersect AB at D. An isosceles ΔABC is formed. we obtain This justifies the construction. A ΔA'BC' whose sides are 3/4 of the corresponding sides of ΔABC can be drawn as follows: Benefitted: My Paypal a/c .  Taking D as centre. ∠AA B = ∠AA B' (Corresponding angles). draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. and A3 on AX such that AA1= A1A2 = A2A3. Residence: Ghaziabad-UP . AB = 5 cm and ∠ ABC = 60°. AB’ = 3/2AB. => AB/AB’ = BC/B’C’ = AC/AC’ (i)  In ΔAA B and ΔAA B'.  Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment AB at point B'. Proof:  To justify the construction.09871823473.94 Proof:  By construction. 5 13 13 … (i) figure. we get. A C || A B. and AD is the altitude of 4 cm. angles). we get. it can be observed that AA and A A contain 5 and 8 equal divisions of line segments respectively. By applying Basic proportionality theorem for the triangle AA B. Let ΔABC be an isosceles triangle having CA and CB of equal lengths. ∠A AB = ∠A AB' (Common). 2 3 2 2 3 3 2 3 2 3 ⇒ Q5. having CD (altitude) as 4 cm and AB (base) as 8 cm. Q2 & Q3: Please Do It Yourself! Q4. A2. Draw a triangle ABC with side BC = 6 cm.  Draw a line through B' parallel to BC intersecting the extended line segment AC at C'.  In ΔABC and ΔAB'C'. base AB of 8 cm.  Locate 3 points (as 3 is greater between 3 and 2) A1. ∠BAC = ∠B'AC' (Com’n).  From the 5 5 13 … (ii) On comparing equations (i) and (ii). ∴ ΔAA B ∼ ΔAA B' (AA similarity criterion) => AB/AB’ = AA /AA => AB/AB’ = 2/3 (ii) Comparing (i) & (ii). A.). Draw perpendicular bisector of AB. we need to prove [email protected]. B1. ΔA'BC' is the required triangle. Benefitted: My Paypal a/c .  Locate 4 points (as 4 is greater in 4 and 3). A.95 Steps:  Draw a ΔABC with side BC = 6 cm. ∠ABC = ∠A'BC' (Common) ∠ACB = ∠A'C'B (Corresponding angles) ∴ ΔABC ∼ ΔA'BC' (AA similarity Crit.com. Then. Draw a triangle ABC with side BC = 7 cm. ∠B3BC = ∠B4BC' (Common). B2. B4. ∠A'BC' = ∠ABC (Common) ∴ ΔA'BC' ∼ ΔABC (AA crit. B3. Residence: Ghaziabad-UP . we obtain ⇒ Q6. on BX. ∴ ΔBB3C ∼ ΔBB4C' (AA similarity crit. ∠ A = 105°. ∠B = 45°. ∠A'C'B = ∠ACB (Corresponding angles). Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.09871823473.  Locate 4 points (as 4 is greater in 3 and 4).) => AB/A’B = BC/BC’ = AC/A’C’ (i) In ΔBB3C and ΔBB4C'. ∠B = 45°. B1. Proof: The construction can be justified by proving that In ΔABC and ΔA'BC'. ∠B3BC' = ∠B4BC (Common). ΔA'BC' is the required triangle. ∠BB3C' = ∠BB4C (Corres.) => A’B/AB = BC’/BC = A’C’/AC (i)  In ΔBB3C' and ΔBB4C. on BX’  Join B4C and draw a line through B3.  Through C'. B4. by construction)  From equations (i) and (ii). draw a line parallel to AC intersecting extended line segment at A'.  Draw a ray BX making an acute angle with BC on the opposite side of vertex A. we obtain => Q7. B3. construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.) … (ii) On comparing equations (i) and (ii).  Join B3C. ∠ B = 45°. Proof: The construction can be justified by proving  In ΔA'BC' and ΔABC. Draw a line through B4 parallel to B3C intersecting extended BC at C'.  Draw BX making an acute angle with BC on the opposite side of vertex A. Mobile (where u may transfer cash!) . ∠C = 30°. parallel to B4C intersecting BC at C'‘  Draw a line through C' parallel to AC intersecting AB at A'. ∠BB3C = ∠BB4C' (Corresponding angles).manojarora23@gmail. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. B2. angles) ∴ ΔBB3C' ∼ ΔBB4C (AA similarity criterion) => BC’/BC = BB3/BB4 (ii)(which equals 3/4. AB = 5 cm and ∠ABC = 60°. ∠A = 105° => ∠C = 30° The required triangle can be drawn as follows: Steps:  Draw a ΔABC with side BC = 7 cm. Then. opposite to vertex C. ∠AA3B = ∠AA5B' (Corresponding angles) ∴ ΔAA3B ∼ ΔAA5B' (AA similarity criterion) => AB/AB’ = AA3/AA5 = 3/5 (by cons. PQ and PR are the required tangents. A2.  ∠PQO is an angle in the semi-circle.  For this.com. The required triangle can be drawn as: Steps:  Draw a line segment AB = 4 cm. It is given that sides other than hypotenuse are 4 cm and 3 cm. draw a circle of 6 cm radius.  Through B'. ∴ ∠PQO = 90° ⇒ OQ ⊥ PQ Benefitted: My Paypal a/c . Draw a ray SA making 90° with it.  Join A3B.221) Q1. Locate a point P. => EXERCISE 11.) => AB/AB’ = BC/B’C’ = AC/AC’ (i) In ΔAA3B and ΔAA5B'. ∴ ΔABC ∼ ΔAB'C' (AA similarity crit. ∠BAC = ∠B'AC' (Common).  Locate 5 points (as 5 is greater in 5 and 3). we get. From a point 10 cm away from its centre. A5. A1. The lengths of tangents PQ and PR are 8 cm each. ∠A3AB = ∠A5AB' (Common). Proof: The construction can be justified by proving that  In ΔABC and ΔAB'C'. 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). A. ∠ABC = ∠AB'C' (Corresponding angles).2: (P. Draw a line through A5 parallel to A3B intersecting extended line segment AB at B'. A4.  Draw a ray AX making an acute angle with AB. Join BC. Draw a circle of radius 6 cm.  Taking M as centre and MO as radius.  Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C.09871823473. 10 cm away from O. On comparing equations (i) and (ii). A3. Clearly. A pair of tangents to the given circle can be constructed as follows.96 A. on line segment AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.  Join PQ and PR. Residence: Ghaziabad-UP . Join OP. ΔAB'C' is the required triangle. Steps:  Taking any point O of the given plane as centre.)=> AB/AB’ = 3/5 (ii). draw a line parallel to BC intersecting extended line segment AC at C'.  Bisect OP.manojarora23@gmail. Let M be the mid-point of PO. these will be perpendicular to each other (As Hypotenuse is the longest side).  Let this circle intersect the previous circle at point Q and R. draw a circle. construct the pair of tangents to the circle and measure their lengths. ΔABC is the required triangle. join OQ and OR. Mobile (where u may transfer cash!) . Draw a circle of radius 3 cm. we get.  Draw a perpendicular to OB at point B. Justification:  The construction can be justified by proving that RV. These are the required tangents. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. ∠PQO = 90°. Let both the perpendiculars intersect at point P. Similarly. 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. ∠PQO is an angle in the semi-circle.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). In ΔPQO. Q4. PQ and PR are the required tangents. let us join OQ and OR. Draw tangents to the circle from these two points P and Q. PQ has to be a tangent of the circle. draw a circle. QO = 4 cm Applying Pythagoras theorem in ΔPQO. PQ. ∠RVO is an angle in the semi-circle. OX.  Take one of its diameters. OW. Let M be the mid-point of PO. The tangent can be constructed on the given circle as: Steps:  Taking any point O on the given plane as centre. SX. PR is a tangent of the circle. PR is a tangent of the circle. PQ has to be a tangent of the circle.47 cm each. For this. PA and PB are the required tangents at an angle of 60°. join OV. A. Since PQ is a tangent. Let it intersect the given circle at the points Q and R. Q2. RW. PQ2 + QO2 = PQ2 =>PQ2 = 20 => PQ = 2√5 = 4.  Draw a radius OB. Justification: The construction can be justified by proving that ∠APB = 60° By our construction ∠OAP = 90°. It can be observed that PQ and PR are of length 4. and OY. Q3. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre.  Take a point A on the circumference of the circle and join OA. RW. SY. draw a circle of 3 cm radius.97 Since OQ is the radius of the circle. draw two circles. Residence: Ghaziabad-UP . and SX are the tangents to the circle (whose centre is O and radius is 3 cm). Join RV. OX. Also verify the measurement by actual calculation. Similarly. W. Locate a point P on this circle and join OP. OW. 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.  Bisect OP. Join PQ and PR.09871823473. Y respectively. Mobile (where u may transfer cash!) . Similarly. PO = 6 cm.manojarora23@gmail. and SY. Draw a perpendicular to OA at point A. ∠PQO = 90° ⇒ OQ ⊥ PQ Since OQ is the radius of the circle.  Bisect OR and OS. ∴ ∠RVO = 90° ⇒ OV ⊥ RV Since OV is the radius of the circle. Thus. For this. X. 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. RV has to be a tangent of the circle. A. making an angle of 120° (180° − 60°) with OA. A.  Draw a circle of 6 cm radius taking O as its centre.  Taking M as its centre and MO as its radius. These two circles will intersect the circle at point V. and extend it on both sides. The tangents can be constructed as: Steps:  Draw a circle of radius 5 cm and with centre as O. Locate two points on this diameter such that OR = OS = 7 cm. And ∠AOB = 120° Benefitted: My Paypal a/c .com. and OY are the tangents of the circle. ∠OBP = 90°. 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). Let them intersect each other at point O. AS has to be a tangent of the circle. draw a circle which will intersect the circle at point B and G. ∠AGE is an angle in the semi-circle. join AP. C. These are the required tangents. ∴ ∠AGE = 90° ⇒ EG ⊥ AG Since EG is the radius of the circle. Join PV and PW. Let F be the mid-point of AE. Join AG. draw a circle of radius OU. Residence: Ghaziabad-UP . For this. PV and PW are the required tangents. Let the mid-point of AB be C. R. The centre E of this circle will be the mid-point of BC. Already. AR. and S. Justification: Benefitted: My Paypal a/c . BS. D. Mobile (where u may transfer cash!) . If a circle is drawn through B. AQ. Justification: The construction can be justified by proving that AG and AB are the tangents to the circle. The tangents can be constructed on the given circles as follows: Steps:  Draw a line segment AB of 8 cm. BQ.98 We know that the sum of all interior angles of a quadrilateral = 360° Thus. Taking A and B as centre. ∴ ∠ASB = 90°. ∠B = 90° ⇒ AB ⊥ BE Since BE is the radius of the circle. The required tangents can be constructed on the given circle as follows: Steps:  Join AE and bisect it. which will intersect the circle at V and W.  Draw perpendicular bisectors of these chords. Draw a circle with the help of a bangle.com. and BR. Construct the pair of tangents from this point to the circle. join EG. Draw a line segment AB of length 8 cm. Q5. BC will be its diameter as ∠BDC is of measure 90°. and BQ are the tangents. The required tangents can be constructed on the given circle as follows:  Draw a circle with the help of a bangle. Construct tangents to each circle from the centre of the other circle. For this. AS. AB and AG are the required tangents. Join BP. The circle through B. BD is the perpendicular from B on AC. Let ABC be a right triangle in which AB = 6 cm. Q6. Q7. Taking A as centre. AB has to be a tangent of the circle.  Take a point P outside this circle and take two chords QR and ST. ⇒ BS ⊥ AS Since BS is the radius of the circle. draw another circle of radius 3 cm. Take a point outside the circle. ∠OAP + ∠AOB + ∠OBP + ∠APB = 360° 90° + 120° + 90° + ∠APB = 360° => ∠APB = 60° This justifies the construction.manojarora23@gmail. Let U be the mid-point of PO. draw two circles of 4 cm and 3 cm radius. ∠ASB is an angle in the semi-circle. AG has to be a tangent of the circle. draw a circle of radius 4 cm and taking B as centre.  Taking F as centre and FE as its radius. A. Similarly. Taking C as centre. Construct the tangents from A to this circle. D is drawn.  Join PO and bisect it.  Bisect the line AB. and AR. Consider the following situation. A. BP. and C. Q. We know that an angle in a semi-circle is a right angle. draw a circle of AC radius which will intersect the circles at points P. Taking U as centre. A. BC = 8 cm and ∠ B = 90°.09871823473. for practical purposes.09871823473. ∠PVO is an angle in the semi-circle.14 approximately. which gives the area of the circle as r 2 2. Residence: Ghaziabad-UP . We know that an angle in a semi-circle is a right angle. Areas of Sector and Segment of a Circle: The portion (or part) of the circular region enclosed by two Benefitted: My Paypal a/c . Mobile (where u may transfer cash!) . ∴ ∠PVO = 90° ⇒ OV ⊥ PV Since OV is the radius of the circle. PV has to be a tangent of the circle. usually called its circumference. As shown. We know that perpendicular bisector of a chord passes through the centre. Similarly. CBSE CLASS X MATHEMATICS CHAPTER 12 AREAS RELATED TO CIRCLES 1.99 The construction can be justified by proving that PV and PW are the tangents to the circle. Therefore.com. circumference / diameter = . the perpendicular bisector of chords QR and ST pass through the centre. Area of a circle is r .manojarora23@gmail. first of all. 2 We get a rectangle of length 1/2 x 2r and breadth ‘r’. it has to be proved that O is the centre of the circle. This constant ratio is denoted by the Greek letter p (read as ‘pi’). where r is the radius of the circle. we cut a circle into a number of sectors and rearranging them. However. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. Circumference of a circle bears a constant ratio with its diameter. circumference =  × diameter =  × 2r (where r is the radius of the circle) =2r  is an irrational number and its decimal expansion is non-terminating and non-recurring (nonrepeating). For this. PW is a tangent of the circle. or. we generally take the value of  as 22/7 or 3. Thus. Let us join OV and OW. Perimeter and Area of a Circle: The distance covered by travelling once around a circle is its perimeter. area of the sector = r So. Area of the segment APB = Area of sector OAPB – Area of ΔOAB = θ/360 x r . Now. ∠AOB is called the angle of the sector.e. ((Unless otherwise specified. Let the degree measure of ∠AOB be θ. II.)) Let OAPB be a sector of a circle with centre O and radius r.. ‘Segment’ & ‘Sector’ would mean ’Minor Segment’ & ’Minor Sector’. area of the sector = r /360 x θ. 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. Similarly. Residence: Ghaziabad-UP . unshaded region OAQB is also a sector of the circle. The angle of the major sector is 360° – ∠AOB. we get the required length of the arc APB as θ/360 x 2r.09871823473. Find the area of the shaded design. Again.com. area of the shaded design = Area of ABCD – Area of (I + II + III + IV) = (100 – 2 × 21. III and IV.manojarora23@gmail. shaded region APB is a segment of the circle (called the minor segment). of degree measure 360) at the centre O. Note that in this figure. area of the sector = r / 360 Thus. the shaded region OAPB is a sector of the circle with centre O. Let us mark the four un-shaded regions as I. Area of II + Area of IV = 21.100 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.5) cm = 57 cm . 2 2 2 2 Benefitted: My Paypal a/c . Or. area of a circle = r . if the degree measure of the angle at the centre is θ. Mobile (where u may transfer cash!) .Area of ΔOAB 2 2 2 2 0 0 0 2 0 0 2 Area of combination of figures . Thus. In a way.5 cm . by applying the Unitary Method and taking the whole length of the circle (of angle 360°) as 2r. when the degree measure of the angle at the centre is 1. AB is a chord of the circle with centre O. where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter.5 cm. A. θ/360 x r .An Example: Ex.2 x 1/2 x  x 5 = 21. OAPB is called the minor sector and OAQB is called the major sector. Area of I + Area of III = Area of ABCD – Areas of two semicircles of each of radius 5 cm = 10 x 10 . So. we can consider this circular region to be a sector forming an angle of 360° (i. Now. consider this: When degree measure of the angle at the centre is 360. Thus. Residence: Ghaziabad-UP . Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. Mobile (where u may transfer cash!) . A. Do It Yourself! Q3. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.09871823473. Do It Yourself! Q4.101 CBSE CLASS X MATHEMATICS CHAPTER 12 AREAS RELATED TO CIRCLES . Red. The wheels of a car are of diameter 80 cm each. An archery target is marked with its five scoring areas from the centre outwards as Gold. then the radius of the circle is (A) 2 units (B)  units (C) 4 units (D) 7 units A. Find the radius of the circle having area equal to the sum of the areas of the two circles. A.5 cm wide.manojarora23@gmail. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal. A. 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. 2r = 2x 19 +2 x 9 => ‘r’ = 28cm. The radii of two circles are 8 cm and 6 cm respectively.1: Q1. The radii of two circles are 19 cm and 9 cm respectively.NCERT EXERCISES SOLUTIONS EXERCISE 12. Do It Yourself! Correct answer is 2. Find the area of each of the five scoring regions. Q2. Black and White.2(Page 230): Benefitted: My Paypal a/c . EXERCISE 12.com. Blue. Do It Yourself! Q5. manojarora23@gmail. Find the area swept by the minute hand in 5 minutes.5m Increase in grazing area = (78. In a circle of radius 21 cm. Area of Equilateral ΔOAB = √3/4 x (side) = 441√3/4 cm Area of segment ACB=Area of sector OACB−Area of ΔOAB=[231 . 154/3 cm ) 2 Q4. Do It Yourself! (Ans. A.com. Same as Above.5cm Area of ΔOAB = 1/2 x 10 x 10 = 50cm Area of minor segment ACB = 78. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. 132/7 cm ) 2 Q2. A.5cm 0 0 0 0 0 2 2 2 2 2 2 2 Q5. Do It Yourself! (Ans. A brooch is made with silver wire in the form of a circle with diameter 35 mm. Area that can be grazed by horse = Area of sector OACB = 19.5 .28 = 88. an arc subtends an angle of 60° at the centre.90 )/360 ]x r = 235.875 m 2 0 0 2 2 2 2 Q9. ΔOAB is an equilateral triangle. A. (ii) the area of each sector of the brooch. 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. Residence: Ghaziabad-UP . Mobile (where u may transfer cash!) . A. = 110 x 5 x 35 = 285mm Each of 10 sectors of the circle is subtending 36° at the centre of the circle. The length of the minute hand of a clock is 14 cm.625 m Area that can be grazed by the horse when length of rope is 10 m long = 90 /360 x r = 78. 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. Do It Yourself! (Ans.72 − 62.09871823473. Area of major sector OADB [(360 .44 cm 0 0 2 Q8.441√3/4] 0 2 0 2 2 2 Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find : (i) the total length of the silver wire required. A.102 Q1. Find the area of the corresponding segment of the circle. A chord of a circle of radius 10 cm subtends a right angle at the centre. (Use  = 3. Find the area of a quadrant of a circle whose circumference is 22 cm. use Cos60 & Sin60 to find OV & SV Area of segment SUT = Area of sector OSUT − Area of ΔOST = 150. Find the area of the corresponding : (i) minor segment (ii) major sector.14) A. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. 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. area of Benefitted: My Paypal a/c . thus. Find the areas of the corresponding minor and major segments of the circle. and. 77/8 cm ) 2 Q3.5 − 19. ∠OAB = ∠OBA (As OA = OB). Q7. Length of wire required =Circumference + 5xdia. Length of arc ACB = 2r 360 x θ = 22cm Area of sector OACB = r /360 x θ = 231 cm In ΔOAB. Draw Perpendicular OV on ST.5cm Area of minor sector OADB [90 /360 ]x r = 235.50 = 28. ∠OAB + ∠AOB + ∠OBA = 180°=> 2∠OAB + 60° = 180° => ∠OAB = 60° Therefore.625) m = 58.5cm = 78. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. A. There are 8 ribs in an umbrella. Area of each Sector = 158125/252 cm .103 each sector = r /360 x 36 = 385/4 mm . and will subtend 60 at the centre of the circle. Area swept by two blades = 2 x 158125/252 = 158125/126 cm . it can be seen. PR = 7 cm and O is the centre of the circle. Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE Benefitted: My Paypal a/c . A. Now. first find ‘Area of segment APB’. A. and thence. Find the area of the shaded region. r /360 x θ = 2r /720 x p = p/720 x 2r 2 0 2 0 0 2 EXERCISE 12. 162. Thus. Assuming umbrella to be a flat circle of radius 45 cm. A. Area of shaded region = 1/2 x 2 2 2 22/7 x (25/2) . Area of ΔOAB = √3/4 x (side) = 196√3 cm . these designs are segments of a circle. using Pythagoras Theorum. Find the area of the shaded region. A. Find the area of the shaded region.Area of Triangle QRP Now. 2 Q3. RQ = RP + PQ = 625 cm. Find the area of the sea over which the ships are warned.8 cm . 189. A round table cover has six equal designs. find the area between the two consecutive ribs of the umbrella.3(Page 234): Q1. Find the total area cleaned at each sweep of the blades. 0 0 2 0 2 Q11. if radii of the two concentric circles with centre O are 7 cm and 14 cm. Find the area of the shaded region. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. First prove that ΔOAB is an equilateral triangle. ‘The Area of Designs = 6 x Area of segment’ = 464. The area between two consecutive ribs is subtending 360 /8 = 45 at the centre of the assumed flat circle. A. a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.35 per cm A.1/2 x 7 x 24 = 4523/28 cm .09871823473. 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 2 Q4. From the [email protected] km. To warn ships for underwater rocks. If the radius of the cover is 28 cm. An umbrella has 8 ribs which are equally spaced.com. Area of shaded region = Area of sector OAFC − Area of sector OBED = 154/3 cm . that.68 2 . 2 2 Q12. Area of Sector OAPB = 1232/3 cm . 2 0 2 Q10. A car has two wipers which do not overlap. A. A. Area between two consecutive ribs of circle = r /360 x θ = 22275/28 cm .97 km 2 Q13. if PQ = 24 cm. if ABCD is a square of side 14 cm and APD and BPC are semicircles. and ∠ AOC = 40°. Residence: Ghaziabad-UP . Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is => A. Area of shaded region = Area of Semi-Circle . 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. 2 2 Q2. in ΔQRP.35 = Rs. Cost of making the designs = 464. Mobile (where u may transfer cash!) .8 x 0. 0 2 2 2 2 Q14. Each chord is a side of a hexagon. find the cost of making the designs at the rate of Rs 0. Find the area of the shaded region. If OA = 7 cm.15700 = 1620. Benefitted: My Paypal a/c . 2 Q6. A. A. From the figure. Area of equilateral triangle = √3/4(a) = 17320. 2 2 0 0 2 2 2 Area of shaded region = Area of equilateral triangle − 3 × Area of each sector = 17320. Area of shaded portion = Area of square ABCD − 4 × Area of each sector = 196 . The figure depicts a racing track whose left and right ends are semicircular.5 cm2. find the area of the shaded region. 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. 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. A.5 cm => a =200cm. 2 Q10.5 cm . (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 . Area of sector ADEF = 60 /360 x r = 15700/3 cm . nine circular designs each of radius 7 cm are made. Mobile (where u may transfer cash!) . A. Area of the shaded region = Area of smaller circle + Area of semi-circle AECFB − Area of ΔABC = 77/2 + 77 . A.49 = 66. a circle is drawn with radius equal to half the length of the side of the triangle. 2 Q9. Q11. A.4 x 22/8 = 68/7 cm . Area of the shaded region = Area of square − Area of circle − 4 × Area of quadrant = 16 .154 = 42 cm = Area of the Shaded Portion. Residence: Ghaziabad-UP . Find the area of the remaining portion of the square. Find the area of the remaining portion of the handkerchief. B. A. If the track is 10 m wide. Area of design = Area of circle − Area of ΔABC = 22528/7 . On a square handkerchief.104 Q5.768√3 Q7. With each vertex of the triangle as centre. Find the area of the shaded region. With centres A. The area of an equilateral triangle ABC is 17320. ABCD is a square of side 14 cm.manojarora23@gmail. C and D. a design is formed leaving an equilateral triangle ABC in the middle. The distance between the two inner parallel line segments is 60 m and they are each 106 m long.09871823473.5 cm . In a circular table cover of radius 32 cm. four circles are drawn such that each circle touch externally two of the remaining three circles.5 . Find the area of the design (shaded region).22/7 . (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. find : (i) the distance around the track along its inner edge. 2 Q8. side of the square = 42 cm. (ii) the area of the track.4 x 77/2 = 196 .com. A. 2 2 2 2 2 1 2 Area of sector ABDC= 90 /360 x r =1/4 x 22/7 x 14x14 = 154 cm . In ΔABC. Thus. Area of ΔABC = 1/2 x AC x AB = 1/2 x 14 x 14 = 98 cm . OACB is a quadrant of a circle with centre O and radius 3.105 Area of square = (Side) = 1764 cm Area of each circle = πr = 22/7 x (7) = 154 cm .7 ) = 308/3 cm . find the area of the shaded region. (i) Since OACB is a quadrant. A. Area of Square OABC = (Side) = (20) = 400 cm . Area of the shaded region = Area of sector OAEB − Area of sector OCFD = 30 /360 x 22/7 x (21 . it will subtend 90° angle at O.Area of Sector ABDC + Area of ΔABC = 154 − 154 + 98 = 98 cm . area of the remaining portion of the handkerchief = 1764 − 1386 = 378 cm . If OA = 20 cm. 0 0 2 2 2 2 2 2 2 2 2 0 2 0 2 2 2 2 2 2 2 Q14. Area of Sector BAEC = 90 /360 x 22/7 x 8 = 352/7 cm . Area of quadrant OPBQ = 90 /360 x     x (20√2) = 628 cm . AB and CD are arcs of two concentric circles of radii 21 cm and 7 cm and centre O. A. 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 Semi-Circle drawn on BC = 1/2 r = 1/2 x 22/7 x (7√2)2 = 154 cm2.09871823473. find the area of the shaded region. Area of quadrant OACB=90 /360 x r =1/4 x 22/7x(3.32] = 256/7 cm . A square OABC is inscribed in a quadrant OPBQ. A. Area of 9 circles = 9 × 154 = 1386 cm Thus.5 cm. 2 2 2 2 2 2 2 Q12. Find the area of the shaded region. ∠BAC is 90º. (ii) shaded region.com. Area of the shaded region = Area of quadrant OACB − Area of ΔOBD = 77/8 . In ΔOAB.5 x 2 = 7/2 cm . If ∠ AOB = 30°. 0 0 2 2 1 2 Area of Shaded Region = Area of Semi-Circle . Area of ΔBAC = 1/2 x 8 x 8 = 32 cm .5) =77/8 cm (ii) Area of ΔOBD = 1/2 x OB x OD = 1/2 x 3. 0 0 2 2 2 2 Benefitted: My Paypal a/c .7/2 = 49/8 cm .manojarora23@gmail. If OD = 2 cm. Residence: Ghaziabad-UP . Calculate the area of the designed region common between the two quadrants of circles of radius 8 cm each. Area of shaded region = Area of quadrant OPBQ − Area of OABC = (628 − 400) cm = 228 cm . As ABC is a quadrant of the circle. Mobile (where u may transfer cash!) . find the area of the (i) quadrant OACB. Area of the designed portion = 2 × (Area of segment AEC) = 2 × (Area of sector BAEC − Area of ΔBAC) = 2 x [352/7 . OB = OA + AB = (20) + (20) => OB = 20√2 cm = Radius ( r ) of the circle. Q13. 2 Q16. ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. A. 0 0 2 2 2 Q15. 1 2 2 2 1 1  TSA of 2 1 1 2 2 2 frustum of the cone = r + r )l + r +r . where.  CSA of the frustum of the cone = r + r )l. Let h be the height. l is same as above  Surface Area 1 2 2 1 2 2 & Volume of a Sphere: TSA = 4r2 Benefitted: My Paypal a/c . Area of circular base) x height = πr h Volume of the material of the hollow Cylinder = External Volume . Mobile (where u may transfer cash!) . Remember the following formulae:  Volume of the frustum of the cone = 1/3 h (r + r + r r ).manojarora23@gmail.πr h = πh(R .Internal Volume = πR h . l the slant height and r and r the radii of the ends (r > r ) of the frustum of a cone. Residence: Ghaziabad-UP . where. 2  Surface Area & Volume of a Right Circular Cylinder: CSA = 2rh Total Surface Area of the Cylinder = 2πrh + 2πr = 2πr(r + h) Volume of a Cylinder = base area (ie. The lower portion is the frustum of the cone.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 2 2 2 2 2 2 2 1/2 Frustum of a Cone is the solid obtained after removing the upper portion of the cone by a plane parallel to the base.106 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 .com.09871823473. show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long. Find the volume and surface area of the double cone so formed. A cistern. Find the length and mass of the wire.5 cm × 7. each brick being 22. How much area will it irrigate in 30 minutes. 3 mm in diameter. How many bricks can be put in without overflowing the water. DO IT YOURSELF! Q 5.09871823473.3: Q8. Volume of double cone = 10. diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm. so as to cover the curved surface of the cylinder. of water flowing in 30 minutes from canal = Vol.5 cm × 6.5 cm? A.4: Do it yourself EXERCISE 13. assuming the density of copper to be 8. 129600 + n x (22. find the area of the tin sheet required to make the funnel. Volume of water that flows in 1 minute from canal = 9 x 10000/60 =1500 m . Therefore. Derive the formula for the volume of the frustum of a cone A. A right triangle.63 + 30.r ) 3 3 3 3 Exercise 13. Area of cross-section of the canal = 6 × 1.NOT PRACTICABLE IF THE SHAPE OF THE WIRE IS CYLINDRICAL! Q2.5 (Optional) Q 1. area => A = 562500 m .80 cm .5 cm)(1 . Use similar triangle criterion. SKIP IT PLEASE! Q 7. Porous bricks are placed in the water until the cistern is full to the brim. Q 3. Water in a canal.31 = 30. DO IT YOURSELF! Q 6. A.88 g per cm .2: Do it yourself Exercise 13.manojarora23@gmail. internally measuring 150 cm × 120 cm × 110 cm. if 8 cm of standing water is needed? A. is wound about a cylinder whose length is 12 cm. SKIP IT Length of wire = ‘l’. A. SKIP IT PLEASE! 3 2 3 2 3 2 Benefitted: My Paypal a/c . A copper wire.5 m deep.17 = 52. If the area of the valley is 97280 km . whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse.107 Volume of a Sphere = 4/3 πr . A. Volume of the Hemisphere = 2/3 πr ‘Volume of the Material’. A. area irrigated in 30 minutes is 562500 m . Residence: Ghaziabad-UP .5 cm × 7. Let the irrigated area be A. A.86 + 19. Derive the formula for the curved surface area and total surface area of the frustum of a cone. 6 m wide and 1. Mobile (where u may transfer cash!) . forming a hollow sphere = 4/3 π(R .1/17) = 150 x 120 x 110 => n = 1792 Q 4. area of wire = l x 3/10 cm .1: Do it yourself Exercise 13.17 cm .5 cm × 6. Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m . Surface area of double cone = 22.5 = 9 m Speed of water = 10 km/h = 10000/60 metre/min. 2 3 3 2 2 Exercise 13. In one fortnight of a given month.com. Each brick absorbs one-seventeenth of its own volume of water. has 129600 cm of water in it. there was a rainfall of 10 cm in a river valley. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. Vol. If the total height is 22 cm. is flowing with a speed of 10 km/h. Volume of water irrigating the required area will be equal to the volume of water that flowed in 30 minutes from the canal. of water irrigating the reqd. and diameter 10 cm. 75 m wide and 3 m deep. For grouped data. Subtract h/2 from the Lower Limits of each Class. students falling in any upper class-limit would be considered in the next class). and Add h/2 to the Upper Limits of each Class. Or.Upper Limit of 1 Class Interval. x2. ‘Organisation’.  Find of ‘a’ from each of the x. 2. weather the data is grouped or ungrouped. the number of observations = f + f +…. Mobile (where u may transfer cash!) . that the value of the mean obtained does not i i depend on the choice of ‘a’.108 CBSE CLASS X MATHEMATICS CHAPTER 14 STATISTICS 1. we subtracted ‘a’ from each x . Arithmetic Mean of Grouped or Un-grouped Data: The sum of the values of all the observations = f x + f x +….com. Class mark (or mid-point of a class) = (Upper class limit + Lower class limit)/2 Arithmetic Mean by Assumed Mean (or.  We may take ‘a’ to be that xi which lies in the centre of x1.+ f x )/(f + f +…. ‘Analysis’ and.+ f So.manojarora23@gmail. and denote it by ‘a’. ‘h’ = Lower Limit of 2 Class Interval . that is. the mean of the deviation. Introduction: ‘Statistics’ is defined as the science of ‘Collection’. The common point of the two classes is included in the Higher Class. 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. while allocating frequencies to each class-interval. Benefitted: My Paypal a/c . Residence: Ghaziabad-UP . 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. …… xn .a) between a and each of the xi’s. 3.+ f ). the deviation  Find the product of di with the corresponding fi. ‘Interpretation’ of numerical data. Arithmetic Mean by Direct Method: Used when data is grouped and the classes are provided. the mean x-bar of the data is given by. Deviation) Method: Used when the numerical values of x & f are large. nd st Then. we need to add ‘a’ to d . Converting a Discontinuous Series into a Continuous Series: Let. It is assumed that the frequency of each class interval is centred around its mid-point. that.09871823473. such series are called ’Inclusive Series’. x-bar = (f x + f x +…. Presentation’.  Exclusive or Continuous Series: The Upper Limit of one class is the Lower Limit of the next Class. in order to get the mean x-bar .  Find the difference di (= xi . and take the sum of all the fidi’s.+ f x . In obtaining d . choose class--mark as x . Follow the following steps: 1 1 i  Choose one among the x’s as the assumed mean. as. ……. and. so. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. May be used. Mathematical Explanation follows: ==> It can be verified.  The assumed mean method and step-deviation method are just simplified forms of the direct method. u = (xi .a)/h. we can only locate a class with the maximum frequency. by ‘h’.109 Arithmetic Mean by Step Deviation Method: While solving a question. x .a). 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. ……… 1 2 n 1 2 n Combined Mean: If x -bar & x -bar are means of two groups computed from n & n . In such situations. ….a)/h. Weighted Mean: The term ‘weight’ stands for relative importance of different items. it will be observed that the column (d = x . hence dividing the values in the column (d = x .. The Benefitted: My Paypal a/c . then. Residence: Ghaziabad-UP . and. Measures of Central Tendency: 1. and. Mode (of grouped data) is the size of variable which occurs most frequently. In a grouped frequency distribution. we will get smaller numbers to multiply with f . w .09871823473.com. i i i i i i i Mathematical Explanation: We have. it is not possible to determine the mode by looking at the frequencies. Mobile (where u may transfer cash!) . Here.w denote their weights. x denote ‘n’ values of a variable ‘x’. the data is said to be multimodal.manojarora23@gmail. where. ‘Rate’ assigned to an ‘item’ is proportional to the importance of the ‘item’. . …. 2.  The mean obtained by all the three methods is the same. called the modal class. i Consider the following:  The step-deviation method will be convenient to apply if all the d ’s have a common factor. If x . Mean: Done Above. Let u = (x .a) has a common factor ‘h (the class size)’. It is possible that more than one value may have the same maximum frequency. w . where a is the assumed mean and h is the class size. Let . Thence.. Now in a grouped data. n/2 = 26. Now 60 – 70 is the class whose cumulative frequency 29 is nearest to n/2. and is given by the formula: . Therefore. In the distribution above. Median is a measure of central tendency which gives the value of the middle-most observation in the data.com. Then. then the median will be the average of the n th and the 2 th [( ) ] n +1 2 observations. Mobile (where u may transfer cash!) . And. by 53 students. 26. the median is the observation. f = frequency of the modal class. where. i. therefore.manojarora23@gmail. we get: Similarly. f = frequency of the class succeeding the modal class. in a certain examination: Computing the cumulative frequency of the classes (Less Than Type).e. 3. To find the median of a grouped data. if ‘n’ 2 ( ) is even.5. 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. 60 – 70 is the median class. Median of Grouped data: Consider a grouped frequency distribution of marks obtained. we can make use of any of these cumulative frequency distributions. n .5. the mean is required and in the second situation. if n is odd. we can compute cumulative frequency distribution of the more than type. For finding the median of ungrouped data. So. 2 This is called the ‘median class’. h = size of the class interval (assuming all class sizes to be equal). f = frequency of the class preceding the modal class. To find this class. n = 53. the mode is required. we first arrange the data values of the observations in th n+1 ascending or in descending order. In the first situation. we use the following formula to find the median: Benefitted: My Paypal a/c . we find the cumulative frequencies of all the classes and 2 We now locate the class whose cumulative frequency is nearest to n . 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. l = lower limit of the modal class.09871823473. out of 100. Residence: Ghaziabad-UP .110 mode is a value inside the modal class. necessary to find the value inside a class that divides the whole distribution into two halves. It is.  In problems where individual observations are not important. the mode is the best choice. we take the median as a better measure of central tendency. 21. 20. the median is more appropriate. Residence: Ghaziabad-UP . l = lower limit of median class. (20. (90. 4A. 20. 18). Note 1: 3 Median = Mode + 2 Mean Note 2: The median of grouped data with unequal class sizes can also be calculated. etc. corresponding cumulative frequency). 15). or an ogive (of the less than type). e. (70.g. e.09871823473. to find the most popular T.e. say 2.V.e. (80. we get median as 66. cf = cumulative frequency of class preceding the median class.com. So. The scale may not be the same on both the axis. (100. 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. that is beyond the scope of this class. 8).. about half the students have scored marks less than 66. 30. These are situations where extreme values may be there. f = frequency of median class. and the other half have scored marks more 66. the mean of classes having frequencies more or less the same is a good representative of the data. we can conclude which school has a better performance. 29). (30. the largest and the smallest observations of the entire data.4.. However. 25. . extreme values in the data affect the mean. the colour of the vehicle used by most of the people. 12). Let us now plot the points corresponding to the ordered pairs given by (upper limit.. The curve we get is called a cumulative frequency curve. Benefitted: My Paypal a/c . in such cases. . etc. i. finding the typical productivity rate of workers. 100 are the upper limits of the respective class intervals. It also enables us to compare two or more distributions. For example.  In situations which require establishing the most frequent value or most popular item. programme being watched. average wage in a country. 45). 22). i.g.manojarora23@gmail. Mobile (where u may transfer cash!) . and the five others have frequency 20. 5). (60. 53) on a graph paper and join them by a free hand smooth curve. h = class size (assuming class size to be equal). the mean is not a good representative of the data. if one class has frequency. But. 5. (40. rather than the mean. 18.. we mark the upper limits of the class intervals on the ‘x-axis’ and their corresponding cumulative frequencies on the ‘y-axis’.4. For example. Hence. n = number of observations.111 Where. . then the mean will certainly not reflect the way the data behaves. by comparing the average (mean) results of students of different schools of a particular examination.4. the consumer item in greatest demand. Computing . However. To represent the data in the table graphically. (10. So. (50. 38). choosing a convenient scale.. Graphical Representation of Cumulative Frequency Distribution: Consider the following (cumulative frequency distribution) table: The values 10. and we wish to find out a ‘typical’ observation. and lies between the extremes. the point at which it cuts the x-axis gives us the median.: For drawing ogives.  Plot the points and join them by a free hand curve.5 on ‘y’ axis. 10 . The two ogives will intersect each other at a point. and. 31). (80. (50. 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. Consider the following (cumulative frequency distribution) table: 0. . 41). we plot the lower limits on the x-axis and the corresponding cumulative frequencies on the y-axis.112 Now.100. it should be ensured that the class intervals are Continuous. of the less than type and of the more than type) on the same axis.  Construct a cumulative frequency table. 90 are the lower limits of the respective class intervals 0 . Obtaining Median from the Cumulative Frequency Curves: First Method: Locate n/2 = 53/2 = 26.10.manojarora23@gmail.  Mark upper class limits along the ‘x’ axis. (10. if we draw a perpendicular on the x-axis. 8). Mobile (where u may transfer cash!) .. convert it into exclusive form. .  Mark the corresponding cumulative frequency on ‘y’ axis. Join the imagined point (lower limit of first. draw a line parallel to the x-axis cutting the curve at a point. 15). (90..B. . It can be drawn in the two ways: ‘Less than Method’. We get the required curve called ‘Ogive’. N.  Plot the points and join them by a free hand curve. From this point. (30. Second Method: Draw both ogives (i. 35).. (0. i. Then we plot the points (lower limit. and join them by a free hand smooth curve. draw a perpendicular to the x-axis. Less than Method:  If the frequency is in inclusive form. corresponding cumulative frequency). 53). The point of intersection of this perpendicular with the x-axis determines the median of the data. From this point. (60.  Mark the corresponding cumulative frequency along the ‘y’ axis. . From this point. 24). 45). 0) with the first point of the curve and so on. 48).. or an ogive (of the more than type). Residence: Ghaziabad-UP .09871823473.  The lower limit of the first class interval becomes the upper limit of the imagined class with frequency ‘0’. To represent ‘the more than type’ graphically. Benefitted: My Paypal a/c .com. The curve we get is a cumulative frequency curve.20. 38). 10.e. Summary on how to go about drawing a Cumulative Frequency Curve: Cumulative Frequency Curve (Ogive) is the graphical representation of a cumulative frequency distribution.  Mark the lower class limits on x-axis. . (20. (70. . on a graph paper.e. (40. ‘More than Method’. 20. 90 . ∑f u = -12 i i i i i Therefore. and why? A.com.1 (P.113 CBSE CLASS X MATHEMATICS CHAPTER 14 STATISTICS . the mean daily wage of the workers of the factory is Rs 145. A. i i Q2. d = (x . Find the mean number of plants per house. Class Mark x = (Upper Class Limit + Lower Class Limit)/2. Class size (h) of this data = 20.a). Q4.20. Mobile (where u may transfer cash!) . Here.09871823473. Find the mean daily wages of the workers of the factory by using an appropriate method. (Table not drawn). in which they collected the following data regarding the number of plants in 20 houses in a locality. Q3. x-bar = 162/20 = 8.1 Therefore. ‘a’ (assumed mean) = 150. A. The mean pocket allowance is Rs 18.We thus find x & f i i i: From the table: ∑f = 20. NOW. PLEASE REDO THE QUESTION BY DEVIATION METHOD ALSO. mean number of plants per house is 8. Find the missing frequency f. Class Mark (x )= (Upper Class Limit + Lower Class Limit)/2. ∑f = 50. Residence: Ghaziabad-UP . ∑fx = 162. 270): Q1. From the table. Here. (Table not drawn). direct method has been used as the values of class marks (x ) and f are small. The following distribution shows the daily pocket allowance of children of a locality. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were Benefitted: My Paypal a/c .1.manojarora23@gmail. Consider the following distribution of daily wages of 50 workers of a factory. A survey was conducted by a group of students as a part of their environment awareness programme. (Table not drawn) Which method did you use for finding the mean. 18 = (752 + 20f) / (44 + f) => 792 + 18f = 752 + 20f => 40 = 2f = f = 20.NCERT EXERCISES SOLUTIONS EXERCISE 14. (Table not drawn). Therefore.5 3 1 7 80 − 83 4 81. Which method of finding the mean did you choose? A.5 −9 −3 −6 68 − 71 4 69.com. Find the mean heart beats per minute for these women. Number of heart beats per minute Number of women fi xi (Class Mark) di = xi − 75.9 beats per [email protected]) i d=x -a = x − 57 i i fu i i i Benefitted: My Paypal a/c . ‘g (gap between the two intervals)’ = 1. Let. Find the mean number of mangoes kept in a packing box.5 65 − 68 2 66.09871823473. and add to the upper class.5 0 0 0 77 − 80 7 78. Residence: Ghaziabad-UP . Q5.114 recorded and summarised as follows.5 6 2 8 83 − 86 2 84.5 −6 −2 −8 71 − 74 3 72. The Class-Intervals are not continuous. Let assumed mean = ‘a’. ‘g/2’ = 1/2. Assumed Mean (a) = 57. Here. Class-Size (h) = 3. which we subtract from the lower class. The following was the distribution of mangoes according to the number of boxes. mean hear beats per minute for these women are 75. These boxes contained varying number of mangoes.5 9 3 6 Total 30 fu i i 4 Class Size (h) = 3. fruit vendors were selling mangoes kept in packing boxes. In a retail market.5 −3 −1 −3 74 − 77 8 75. choosing a suitable method. Class interval f i x (Class . Mobile (where u may transfer cash!) . A. 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 116 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! Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP 117 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. Benefitted: My Paypal a/c - [email protected]; Mobile (where u may transfer cash!) - 09871823473; Residence: Ghaziabad-UP (iii) 15%.7 A. of ttl possible outcomes)=12/52 = 3/13 (iii) a red face card Benefitted: My Paypal a/c .manojarora23@gmail. It does not contain any orange flavoured candies.5. (i) A driver attempts to start a car. Hence. Complete the following statements: DO IT YOURSELF! Q2. (ii) . Hence. event that Malini will take out a lemon flavoured candy is a sure event. 308): Q1. (iv) 0. which are equally likely outcomes. It is not an equally likely event. A. 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.5 cannot be a probability of an event. what is the probability of ‘not E’? A. (ii) A player attempts to shoot a basketball. This implies that the probability of an event cannot be negative or greater than 1. Therefore. (ii).118 CBSE CLASS X MATHEMATICS CHAPTER 15 PROBABILITY . Total number of face cards = 12 P (getting a face card)=(No. P (a lemon flavoured candy) = 1. Malini takes out one candy without looking into the bag. the result of an individual toss is completely unpredictable. And factors for both the conditions are not the same.09871823473. the possible outcomes are only two. It is an equally likely event. The probability of ‘not E’ is 0. (i) The bag contains lemon flavoured candies only. A. Which of the following experiments have equally likely outcomes? Explain. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy? A.05. Also. Therefore. If P(E) = 0. This implies that every time. P (getting a king of red colour) = 2/52 = 1/26 (ii) a face card A. A bag contains lemon flavoured candies only. head or tail. Q4. Residence: Ghaziabad-UP . event that Malini will take out an orange flavoured candy is an impossible event. A.1 (P. Probability of an event (E) is always greater than or equal to 0. it is always less than or equal to one. It is not an equally likely event. Which of the following cannot be the probability of an event? (i) 2/3.95. P (an orange flavoured candy) = 0 (ii)As the bag has lemon flavoured candies. It is an equally likely event. as it depends on various factors such as whether the car will start or not. Q3. The answer is right or wrong A. Q7 – Q13: DO IT YOURSELF! Q14. Q5. (iv) A baby is born. The car starts or does not start. Therefore. she will take out only lemon flavoured candies. Malini will take out only lemon flavoured candies. Therefore. as it depends on the player’s ability and there is no information given about that. −1. When we toss a coin. Mobile (where u may transfer cash!) . Find the probability of getting (i) a king of red colour A. Q6. She/he shoots or misses the shot. One card is drawn from a well-shuffled deck of 52 cards. out of these alternatives.1.NCERT EXERCISES SOLUTIONS EXERCISE 15.com. (iii) A trial is made to answer a true-false question. It is a boy or a girl. of favourable outcomes)/(No. P (getting a red face card) = 6/52 = 3/26 (iv) the jack of hearts A. Favourable outcomes = 5.75. P (getting a jack of hearts) = 1/52 (v) a spade A. Q15. (ii) a perfect square number A. Total number of cards = 5. What is the probability that this bulb is defective? A. Total number of queens = 1. A child has a die whose six faces show the letters as given below: The die is thrown once. P (getting a defective bulb) = 4/20 = 1/5.9. P (getting a spade) = 13/52 = 1/4 (vi) the queen of diamonds A. Determine the probability that the pen taken out is a good one. One card is then picked up at random.36. Favourable outcomes = 1. Residence: Ghaziabad-UP . (b) P (getting an queen) = 0/4 = 0 Q16.10. One pen is taken out at random from this lot. Total number of good pens = 132 P (getting a good pen) = 132/144 = 11/12 Q17.49. (i) What is the probability that the card is the queen? A. find the probability that it bears (i) a two-digit number A.85. Total number of pens = 12 + 132 = 144. Now one bulb is drawn at random from the rest.90 => P (getting a number divisible by 5) = 18/90 = 1/5 Q19. P (getting a queen of diamonds) = 1/52. (i) A lot of 20 bulbs contain 4 defective ones. jack.20. P (getting a not defective bulb) = 15/19 Q18. One bulb is drawn at random from the lot.64.com. 81/90. what is the probability that the second card picked up is (a) an ace? (b) a queen? A. A.119 A. What is the probability of getting (i) A? (ii) D? A.60. Remaining total number of non-defective bulbs = 16 − 1 = 15. Suppose you drop a die at random on the rectangular region as shown. 2 2 Benefitted: My Paypal a/c . Remaining total number of bulbs = 19.15.25.25. (i) P (getting A) = 2/6 = 1/3 (ii) P (getting D) = 1/6 Q20. Area of rectangle = l × b = 3 × 2 = 6 m Area of circle (of diameter 1 m) =  / 4 m .40.65. thus. 12 defective pens are accidentally mixed with 132 good ones.35. A. P (die will land inside the circle) =(/4)/6 = /24.50.70. What is the probability that it will land inside the circle with diameter 1m? A. queen.4. Five cards—the ten. What is the probability that this bulb is not defective ? A. (a) P (getting an ace) = 1/4. king and ace of diamonds. (ii) Suppose the bulb drawn in (i) is not defective and is not replaced.16.80. A box contains 90 discs which are numbered from 1 to [email protected]. are well-shuffled with their face downwards.55.45. If one disc is drawn at random from the box. Mobile (where u may transfer cash!) . It is not possible to just look at a pen and tell whether or not it is defective. P (getting a queen) = 1/5 (ii) If the queen is drawn and put aside.81 => P (getting a perfect square number) = 9/90 = 1/10 (iii) a number divisible by 5. but will not buy if it is defective. Benefitted: My Paypal a/c . Residence: Ghaziabad-UP . The possible outcomes are {HHH. Do you agree with this argument? Justify. A game consists of tossing a one rupee coin 3 times and noting its outcome each time.com. HTT} Number of total possible outcomes = 8 Number of favourable outcomes = 2 {i.. TTH. Calculate the probability that Hanif will lose the game. A.e.manojarora23@gmail. 4. THH. 5. 3. 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 2 3 4 5 6 7 8 9 10 11 12 Probability Q23. Nuri will buy a pen if it is good. HHT. What is the probability that (i) She will buy it ? A. Therefore. of defective pens = 20 => Total number of good pens = 124 (i) Probability of getting a good pen = 124/144 = 31/36. HTH. 6. 11 and 12.1/4 = 3/4. and loses otherwise. are thrown at the same time. 10. Thus. (i) Write down all the possible outcomes and complete the following table: Event: Sum of two dice 2 3 4 5 6 7 8 9 10 11 12 Probability (ii) A student argues that ‘there are 11 possible outcomes 2. TTT and HHH} P (Hanif will win the game) = 2/8 = 1/4 P (Hanif will lose the game) = 1 . 9. A lot consists of 144 ball pens of which 20 are defective and the others are good.31/36 = 5/36 Q22. P (Nuri will not buy a pen) = 1 . Hanif wins if all the tosses give the same result i.. three heads or three tails. of pens = 144. Total no. P (Nuri buys a pen) = 31/36. The shopkeeper draws one pen at random and gives it to her.e. Total no. A. one blue and one grey.09871823473. The possible outcomes of the experiment are listed in the table below. Two dice. each of them has a probability 1/11. Mobile (where u may transfer cash!) .120 Q21. Probability of each of these sums will not be 1/11 as these sums are not equally likely. 7. 8. TTT. (ii) She will not buy it ? A. THT. (5. Hence. the possible outcomes are (H. Therefore. 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. (5. 5). the probability is not 1/3. 5). the probability of getting an odd number is 1/2.com.121 Q24.2 (Optional) Q1. 2. Q25. 5 are odd and 2. Total number of outcomes = 6 × 6 = 36 Total number of outcomes when 5 comes up on either time are (5. Which of the following arguments are correct and which are not correct? Give reasons for your answer. 1. (H. 3. 4. 2. (5. Each is equally likely to visit the shop on any day as on another day. A die is thrown twice. What is the probability that (i) 5 will not come up either time? A. 6). (5. If the probability of drawing a blue ball is double that of Benefitted: My Paypal a/c . Out of these. T). Thus. 2. T). (ii) 5 will come up at least once? A. total number of favourable cases = 11 P (5 will come up either time) = 11/36. of cases. 3. 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 . 1). two tails or one of each. 5). the probability is 1/3. A die is numbered in such a way that its faces show the numbers 1. 5. 5). 6.09871823473. the probability of getting two heads is 1/4. When a dice is thrown. A. 5). (5. (4. for each outcome. 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. Therefore. (6. (1. (T. 4. It is thrown two times and the total score in two throws is noted. when 5 can come at least once=11=> P(5 will come at least once)=11/36. Therefore. EXERCISE 15.manojarora23@gmail. H). Correct. for each of these outcomes. P(visiting shop on consecutive days) = 8/25 Q 2. A. the possible outcomes are 1.11/36 = 25/36. 3. 4). P (5 will not come up either time) = 1 . and the probability of getting one of each is 1/2. (T. 3). A bag contains 5 red balls and some blue balls. (2. H). and (T. the probability of getting two tails is 1/4. Mobile (where u may transfer cash!) . 2). Therefore. there are two possible outcomes−−an odd number or an even number. (ii) If a die is thrown. H).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. 5). (3. Total no. (i) If two coins are tossed simultaneously there are three possible outcomes —two heads. T). the probability of getting an odd number is 1/2. It can be observed that there can be one of each in two possible ways − (H. Residence: Ghaziabad-UP . When two coins are tossed. 3. 6 are even numbers. (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. Incorrect. What is the probability that both will visit the shop on (i) the same day? (ii) different days ? (iii) consecutive days? Ans. and 6. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). of blue balls.com. Let no. the probability that it is green is 2/3. (i) x/12. Residence: Ghaziabad-UP . determine the number of blue balls in the bag. some are green and others are blue. A. Number of blue marbles = 8. Find x. P (getting red ball ) = 5/(5 + x) As per question. A box contains 12 balls out of which x are black. Q 4.09871823473.manojarora23@gmail. of balls = 5 + x P (getting blue ball) = x/(5 + x). Number of green marbles = 16. (ii) (x + 6)/18 = x/6 => x = 3 Q 5. A. If one ball is drawn at random from the box. of blue balls = x => Total no. Mobile (where u may transfer cash!) . If a marble is drawn at random from the jar. the probability of drawing a black ball is now double of what it was before. what is the probability that it will be a black ball? If 6 more black balls are put in the box. and. A. A jar contains 24 marbles. Benefitted: My Paypal a/c .122 a red ball. Find the number of blue balls in the jar. x/(5 + x) = 2 x 5/(5 + x) => x = 10 = no.