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Exaguru

Published by Shalu Kumari, 2021-11-25 15:08:54

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2. If 2 + 2 y 2 = 15 and − 4 y = 3, then the values of x and y, respectively are xx 2 1 11 (a) , 2 (b) 3, (c) 4, (d) , 4 11 3 44 3. If 1 = u and 1 = v, then 23 = 13 becomes + xy xy (a) 3u + 2v = 13 (b) 2u + 3v = 13 (c) 2 + 3 = 13 23 uv (d) v + u = 13 51 4. Equation + = 2 is reduced in linear equation as 5p + q = 2. Then the values of p and q x −1 y−2 respectively are 11 11 , , (a) x − 1 y − 2 (b) (c) x – 1, y – 2 (d) x + 1, y + 2 y −1 x −2 23 54 5. If + = 13 and − = –2, then the values of x and y respectively are xy xy 11 (b) 11 (c) 11 (d) 2, 3 (a) , , , 23 34 25 11 = 3 and 11 = −1 , where 3x + y ≠ 0, 3x – y ≠ 0, then 6. 3x + y + 3x − y 4 2(3x + y) − 2(3x − y) 8 (a) 1, 2 (b) 1, 1 (c) 1, 3 (d) 1, 4 7. If 3xa − 2b + 5 = 0 and a + 3b – 2 = 0, then y x y (a) x = 1 y= 1 (b) x = 1 y= 1 (c) x – a, y = b (d) x = a, y = –b , , ba ab 11 11 8. If − = –1 and + = 8, (x ≠ 0, y ≠ 0), then 2x y x 2y 11 11 11 11 (a) x = , y = (b) x = , y = (c) x = , y = (d) x = , y = 68 64 42 35 Answers and Hints A. Multiple Choice Questions (MCQs) Then the given equation become 5 2 2p + 3q = 13 ...(i) 1. (c) 2. (a) , 2 and 5p – 4q = –2 ...(ii) 6 11 11 Multiplying equation (i) by 5 and equation 3. (b) 2u + 3v = 13 4. (a) , y−2 (ii) by 2, we get 11 x −1 10p + 15q = 65 ...(iii) 5. (a) , 10p – 8q = –4 23 – + – Let 1 = p and 1 = q. 23q = 69 q = 3 xy 50 Mathematics–10

F rom (i), 2p + 9 = 13 p = 2 Subtracting (vi) from (v) But p = 1 1 = 2 x = 1 3x + y = 4 x x2 3x – y = 2 and q = 1 1 1 (–) (+) (–) y= = 3 2 y y3 2y = 2 ﬁ y= 2 =1 6. (b) 1, 1  x = 1  y = 1 1 13 Thus, the required solution is:  . + = ...(i) (3x + y) (3x − y) 4 11 = −1 ...(ii) 7. (c) x = –a, y = b − 2(3x + y) 2(3x − y) 8 11 8. (d) x = , y = 64 11 Given equations are Let = p and =q (3x + y) (3x − y) 11 ...(i) \\ (i) and (ii) can be expressed as 2x – y = –1 p + q = 3 ...(iii) 11 ...(ii) 4 x + 2y = 8 p − q = − 1 ...(iv) 1 2 2 8 Multiplying eqn. (i) by and then adding the result to (ii), we get 2 1 Multiplying equation (iii) by and adding 2 it to (iv), we get 11 1 4x – 2y = – 2 p + q = 3 22 8 1 1 x + 2y = 8 p − q = − 1 1 1 1 22 8 4x x = 8 − 2 +  p + p =  3 − 1 ﬁ 21  1 + 4 16 – 1  2 2   8 8 p= = 4x = 2 84 From (iii),  5 15 4x = 2 13 + q = 1 x = 6 44 31 2 1  q = − = = ﬁ 11 44 4 2 Now, x + 2y = 8 [From (ii)] 1 11 But p = 3x + y ﬁ 3x + y = 4  11 = 8 ;a x = 1 E + 6 1 2y ﬁ 3x + y = 4 ...(v) c 6 m ...(vi) 1 11  1 and q = 3x − y ﬁ 3x − y = 2 6 + 2y = 8 ﬁ 3x – y = 2 1 2y = 2 Adding (v) and (vi)  3x + y = 4 1 y = 4 3x – y = 2  6x = 6 ﬁ x = 6 = 1 11 6 So, x = 6 and y = 4 Pair of Linear Equations in Two Variables 51

4. Solving Word Problems on Pair of Linear Equations in Two Variables To solve the word problems, we should consider the following points: (i) Reduce the given situation in terms of x and y as mathematical statements. (ii) Solve the linear equations thus formed using any algebraic method. Example 1. Read the following problem. Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. The pair of linear equations representing the above situation is (a) x + y = 10, x + y = 2 (b) x + y = 10, x – y = 2 (c) x – y = 10, x + y = 2 (d) x – y = 10, x – y = 2 Solution. Let her speed of rowing in still water be x km/h. and Speed of the current = y km/h. \\ Upstream speed = (x – y) km/h and Downstream speed = (x + y) km/h Since Distance = Time Speed \\ 20 = 2 ⇒ x + y = 10 ...(i) x+ y and 4 = 2 ⇒ x – y = 2 ...(ii) x− y Hence, option (b) is the correct answer. Example 2. A part of monthly hostel charges in a college is fixed and the remaining depends on the number of days one has taken food in the mess. When a student ‘A’ takes food for 22 days, he has to pay ` 1380 as hostel charges; whereas a student ‘B’, who takes food for 28 days, pays ` 1680 as hostel charges. The fixed charges and the cost of food per day respectively are (a) ` 280, ` 50 (b) ` 240, ` 60 (c) ` 290, ` 100 (d) ` 170, ` 210 Solution. Let the fixed hostel charges be ` x and the cost of food per day be ` y. According to the question, we get x + 22y = 1380 ...(i) and x + 28y = 1680 ...(ii) Subtracting (i) from (ii), we get 6y = 300 ⇒ y = 300 ÷ 6 = 50 Putting y = 50 in (i), we get x + 22(50) = 1380 ⇒ x + 1100 = 1380 ⇒ x = 280 \\ Fixed hostel charges = ` 280 and cost of the food per day = ` 50. Hence, option (a) is the correct answer. 52 Mathematics–10

Example 3. Atul sold a television set and a mobile phone for ` 10500, thereby making a profit of 10% on the television set and 25% on the mobile phone. If he had taken a profit of 25% on the television set and 10% on the mobile phone, he would have got ` 10650. The cost of each item is (a) TV : ` 7000, Mobile : ` 5000 (b) TV : ` 6000, Mobile : ` 4000 (c) TV : ` 4000, Mobile : ` 5000 (d) TV : ` 5000, Mobile : ` 4000 Solution. Let the cost price of a television set be ` x and the cost price of the mobile phone be ` y. The selling price of the television set when it is sold at a profit of 10% = `  x + 10 x = ` 110 x 100 100 The S.P. of the mobile phone when it is sold at a profit of 25%. = `  y + 25 y = ` 125 y  100 100 So, 110x + 125y = 10500 ...(i) 100 100 Similarly, 125x + 110 y = 10650 ...(ii) 100 100 From equations (i) and (ii), we get 110x + 125y = 1050000 ...(iii) and 125x + 110y = 1065000 ...(iv) On adding equations (iii) and (iv), we get 235x + 235y = 2115000 ﬁ x + y = 9000 ﬁ x = 9000 – y Putting this value of x in (i), we get 110 (9000 – y) 125 y 100 + 100 = 10500 ﬁ 990000 – 110 y + 125 y = 1050000 ﬁ 15 y = 60000 \\ y = 4000 and x = 9000 – 4000 = 5000 So, the cost price of the television set is ` 5000 and cost price of the mobile phone is ` 4000. Hence, option (d) is the correct answer. Example 4. Base of an isosceles triangle is 2 times its congruent sides. Perimeter of the triangle is 3 32 cm. The length of each side of that triangle is (a) 8 cm, 8 cm, 6 cm (b) 12 cm, 12 cm, 10 cm (c) 10 cm, 10 cm, 12 cm (d) None of these Pair of Linear Equations in Two Variables 53

Solution. Let each of the congruent sides of an isosceles triangle be x cm long and base be y cm long. According to the first condition, ...(i) 2 y = 3 x ⇒ 2x – 3y = 0 According to the second condition, x + x + y = 32 ⇒ 2x + y = 32 ...(ii) Subtracting (i) from (ii), we get 4y = 32 ⇒ y = 32 ÷ 4 = 8 Putting y = 8 in equation (i), we get 2x – 3 (8) = 0 ⇒ 2x = 24 ⇒ x = 24 ÷ 2 = 12 \\ Each of the congruent sides is 12 cm long and base 8 cm long. Hence, option (b) is the correct answer. Example 5. Places A and B are 80 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in same direction they meet in 8 hrs and if they move in opposite directions they meet in 1 hr 20 minutes. The speeds of cars started from places A and B respectively are (a) 40 km/hr, 55 km/hr (b) 20 km/hr, 30 km/hr (c) 35 km/hr, 25 km/hr (d) 30 km/hr, 28 km/hr Solution. Let the speed of car starts from A or car A be x km/hr and the speed of car starts from B or car B be y km/hr Case I: A 80 km B 8x car A car B 8y After 8 hours, distance covered by car A = 8x as speed = distance  time  and distance covered by car B = 8y So, 8x – 8y = 80 ﬁ x – y = 10 ...(i) Case II: A 80 km B 4 x 4 y 3 3 car A car B 44 After 1 hr 20 minutes, i.e. hrs, distance covered by car A = x 33 4 and distance covered by car B = y 3 44 ...(ii) So, 3 x + 3 y = 80 ﬁ x + y = 60 On solving equations (i) and (ii), we get x = 35 and y = 25 Thus, speed of car A = 35 km/hr and speed of car B = 25 km/hr. Hence, option (c) is the correct answer. 54 Mathematics–10

Example 6. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. The speed of the stream and that of the boat in still water, respectively are (a) 10 km/hr, 12 km/hr (b) 8 km/hr, 3 km/hr (c) 5 km/hr, 4 km/hr (d) 4 km/hr, 6 km/hr Solution. Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h. \\ Speed of the boat going upstream = (x – y) km/h and speed of the boat going downstream = (x + y) km/h Distance We know that Time = Speed \\ Time taken in going 30 km upstream = 30 h x− y 44 h Time taken in going 44 km downstream = x+ y 30 44 ...(i) (From given condition) \\ x − y + x + y = 10 Again, time taken in going 40 km upstream = 40 h x− y 55 and time taken in going 55 km downstream = x + y h . 40 55 ...(ii) (From given condition) \\ x − y + x + y = 13 Now, putting x 1 y = a and 1 = b , we get − x+ y 30a + 44b = 10 ...(iii) ...(iv) 40a + 55b = 13 ...(v) Multiplying equation (iii) by 4 and equation (iv) by 3 and subtracting, we get ...(vi) 120a + 176b = 40 120a + 165b = 39 – – – 11b = 1 ⇒ b = 1 11 \\ From (iii), 1 30a + 44 × = 10 ⇒ 30a + 4 = 10 ⇒ 30a = 10 – 4 = 6 11 61 ⇒ a = 30 = 5  1 11 a = 5 ⇒ x – y = 5 ⇒ x – y = 5 and b = 1 ⇒ 11 11 x + y = 11 ⇒ x + y = 11 On solving (v) and (vi), we get x = 8, y = 3 Thus, speed of the boat in still water = 8 km/h and speed of the stream = 3 km/h Hence, option (b) is the correct answer. Pair of Linear Equations in Two Variables 55

Exercise 3.4 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. Aruna has only ` 1 and ` 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ` 75, then the number of ` 1 and ` 2 coins are, respectively (a) 35 and 15 (b) 35 and 20 (c) 15 and 35 (d) 25 and 25 2. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, (in years) of the son and the father are, respectively (a) 4 and 24 (b) 5 and 30 (c) 6 and 36 (d) 3 and 24 3. A purse contains 25 paise and 10 paise coins. The total amount in the purse is ` 8.25. If the number of 25 paise coins is one-third of the number of 10 paise coins in the purse, then the total number of coins in the purse is (a) 60 (b) 40 (c) 80 (d) 72 4. A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, the number of hens will be (a) 18 (b) 26 (c) 32 (d) 40 5. If 3 chairs and 1 table costs ` 1500 and 6 chairs and 1 table costs ` 2400, the pair of linear equations to represent this situation is (a) 6x + y = 1500, 3x + y = 2400 (b) x + y = 1500, x + y = 2400 (c) 3x + y = 1500, 6x + y = 2400 36 (d) None of these 11 6. A fraction becomes 3 when 2 is subtracted from the numerator and it becomes 2 when 1 is subtracted from the denominator. The fraction is (a) 2 (b) 5 (c) 4 (d) 7 5 18 13 15 7. The difference between two numbers is 26 and the larger number exceeds thrice of the smaller number by 4. The numbers are (a) 39, 13 (b) 12, 38 (c) 37, 11 (d) None of these 8. Meena went to a bank to withdraw ` 2,000. She asked the cashier to give her ` 50 and ` 100 notes only. Meena got 25 notes in all. How many notes of ` 50 and ` 100 she received? (a) ` 50 : 10, ` 100 : 15 (b) ` 50 : 12, ` 100 : 10 (c) ` 50 : 15, ` 100 : 10 (d) None of these 9. A motor boat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. The speed of the stream is (a) 60 km/hr (b) 6 km/min (c) 6 km/s (d) 6 km/hr 10. The two consecutive odd positive integers, sum of whose squares is 290 are (a) 5, 13 (b) 11, 13 (c) 13, 17 (d) None of these 11. A two-digit number is obtained by either multiplying the sum of digits by 8 and then subtracting 5 or by multiplying the difference of digits by 16 and adding 3. The number is (a) 23 (b) 34 (c) 83 (d) 119 56 Mathematics–10

12. The sum of a two-digit number and the number obtained by interchanging the digits is 132. If the two digits differ by 2, the number is (a) 45 (b) 75 (c) 85 (d) 115 13. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. The present age of father and son, respectively are (a) 40 years, 12 years (b) 30 years, 6 years (c) 32 years, 8 years (d) 42 years, 10 years Answers and Hints A. Multiple Choice Questions (MCQs) From (i), x = 26 + 11 = 37 1. (d) 25 and 25 2. (c) 6 and 36 S o, the larger number is 37 and the smaller 3. (a) 60 4. (b) 26 number is 11. 8. (a) ` 50 : 10, ` 100 : 15 5. (c) 3x + y = 1500, 6x + y = 2400 Let the cost of 1 chair = ` x Let Meena has received x no. of ` 50 notes And the cost of 1 table = ` y and y no. of ` 100 notes. 3x + y = 1500 So, 50x + 100y = 2000 ... (i) 6x + y = 2400 x + y = 25 ... (ii) 6. (d) 7 Multiply (ii) by 50, we get 15 50x + 100y = 2000 Let the numerator be x and denominator be y 50x + 50y = 1250 x Then, fraction = y –– – x − 2 1 —————— —— y 3 ATQ, = 50y = 750 ⇒ 3x – 6 = y ⇒ 3x – y = 6 ...(i) ⇒ y = 15 x 1 Putting y = 15 in equation (ii), we get y −1 and = ⇒ 2x = y – 1 x + 15 = 25 2 ⇒ 2x – y = –1 ...(ii) ⇒ x = 10 M eena has received 10 pieces of ` 50 notes Subtracting (ii) from (i), we have and 15 pieces of ` 100 notes. 3x – y = 6 2x – y = – 1 9. (d) 6 km/hr – + + Given: Speed of boat = 18 km/hr, x = + 7 Distance = 24 km From equation (i) Let x be the speed of stream. 3 × 7 – y = 6 ⇒ y = 21 – 6 = 15 Let t1 and t2 be the time for upstream and downstream. 7 Thus required fraction = . 15 As we know that, 7. (c) 37, 11 distance L et the larger number be x and the smaller ⇒ time = speed number be y. Then, x – y = 26 ...(i) For upstream, Speed = (18 – x) km/hr, Time = t1 and x – 3y = 4 ...(ii) Therefore, 24 – + – t1 = 18 − x 2y = 22 ⇒ y = 11 Pair of Linear Equations in Two Variables 57

For downstream,S peed = (18 + x) km/hr, 10. (b) 11, 13 Time = t2 L et one of the odd positive integer be x then Therefore, 24 the other odd positive integer is x + 2 t2 = 18 + x ATQ, x2 + (x + 2)2 = 290 ⇒ x2 + x2 + 4x + 4 = 290 Now according to the question: ⇒ 2x2 + 4x – 286 = 0 t1 = t2 + 1 ⇒ 2(x2 + 2x – 143) = 0 24 = 24 + 1 18 − x 18 + x ⇒ x2 + 2x – 143 = 0 ⇒ x2 + 13x – 11x – 143 = 0 ⇒ 48x = (18 – x)(18 + x) ⇒ x(x + 13) – 11(x + 13) = 0 ⇒ 48x = 324 + 18x – 18x – x2 ⇒ x2 + 48x – 324 = 0 ⇒ (x – 11)(x + 13) = 0 ⇒ x2 + 54x – 6x – 324 = 0 ⇒ (x – 11) = 0, (x + 13) = 0 ⇒ x(x + 54) – 6(x + 54) = 0 ⇒ (x + 54)(x – 6) = 0 Therefore, x = 11 or –13 ⇒ x = –54 or x = 6 Since speed cannot be negative. According to question, x is a positive odd ⇒ x = –54willberejected integer. \\ x = 6 Thus, the speed of stream is 6 km/hr. H ence, we take positive value of x So, x = 11 and (x + 2) = 11 + 2 = 13 Therefore, the odd positive integers are 11 and 13. 11. (c) 83 12. (b) 75 13. (d) 42 years, 10 years Case Study Based Questions I. Rupesh purchased 3 chairs and one table for `1600 and his friend purchased 5 chairs and 2 tables for `2900. If in both the cases the price of chairs and tables are the same, Rupesh wants to know some answers of his question what are in his mind. Help him to solve his problems. 1. Using the cost price of one chair as ‘x’ and the cost price of one table as ‘y’, the linear pair of equations of the above statement are as follows (a) 3x + y = 1600, 5x + 2y = 2900 (b) 3x – y = 1600, 5x – 2y = 2900 (c) 3x + y = 1600, 5x – 2y = 2900 (d) 3x – y = 1600, 5x + 2y = 2900 2. The linear pair of equations formed above by the given information are (a) Consistent (b) Inconsistent (c) Dependent (d) Both (a) and (c) 3. If lines are drawn for the linear pair of equations formed by the above information, they will (a) intersect at one point (b) be parallel to each other (c) be coincident (d) not exist 4. The solutions of the linear pair of equations formed by the above information are (a) `300, `700 (b) `500, `600 (c) `450, `700 (d) `200, `1100 58 Mathematics–10

5. If Rupesh had purchased 2 chairs and 2 tables the sum of total cost will be (a) `2200 (b) `1800 (c) `2000 (d) `2400 Ans. 1. (a) 3x + y = 1600, 5x + 2y = 2900 2. (d) Both (a) and (c) 3. (a) intersect at one point 4. (a) `300, `700 5. (a) `2000 II. Travelling by a car on a highway gives lots of fun and thrills. People drive and enjoy the moment of thrills. x km/h 150 km/h y km/h Destination A B On morning two friends who are living at a distance of 150 km apart decided to meet at another place. So, they drive their cars from point A and point B at the same time and meet after 15 hours on a highway. In an another incident they decide to meet in a hurry. So, they drive their cars in opposite directions and meet in one hour. 1. Using the speed of car at A, x km/h and the speed of car at B, y km/h, the pair of linear equations representing the situation is (a) 15x + 15y = 150, x + y = 150 (b) 15x – 15y = 150, x + y = 150 (c) 15x – 15y = 150, x – y = 150 (d) 15x + 15y = 150, x – y = 150 2. On comparing the coefficients of the pair of linear equations formed by the above situations, following conditions can arise: (a) a1 ≠ b1 (b) a1 = b1 ≠ c1 (c) a1 = b1 = c1 (d) None of the above a2 b2 a2 b2 c2 a2 b2 c2 3. The lines drawn on the graph paper for the pair of linear equations formed due to above situations can have one of following possibilities. (a) The two lines intersect at one point (b) The two lines do not intersect (c) The two lines coincide (d) Can not be said anything 4. The solutions of the pair of linear equations formed by above situation is (a) a unique solution (b) no solution (c) infinitely many solution (d) not to be determined 5. The speeds of car at A and car at B are (a) 70 km/h, 80 km/h (b) 90 km/h, 60 km/h (c) 80 km/h, 70 km/h (d) 60 km/h, 90 km/h Ans. 1. (b) 15x – 15y = 150, x + y = 150 2. (a) a1 ≠ b1 a2 b2 3. (a) The two lines intersect at one point 5. (c) 80 km/h, 70 km/h 4. (a) a unique solution Pair of Linear Equations in Two Variables 59

Experts’ Opinion Questions based on following types are very important for exams. So, students are advised to revise them thoroughly. 1. To find the value of unknown for which the given pair of linear equations has unique solution, infinitely many solutions or no solutions. 2. To find out whether the lines representing the given pairs of linear equations intersect at a point, are a1 b1 c1 parallel or coincident by comparing the ratios a2 , b2 and c2 . 3. To check whether the given pairs of linear equations are consistent / inconsistent and obtain the solution graphically if consistent. 4. To draw the graphs of given equations and determine the coordinates of the vertices of the triangle formed by the lines and the x-axis. 5. To solve the given pair of linear equations by the substitution method. 6. To solve the given pair of linear equations by the elimination method and substitution method. 7. To solve the given pair of equations by reducing them to a pair of linear equations. 8. To formulate the given problem as a pair of linear equations and hence find its solution. IMPORTANT FORMULAE For the system of two linear equations a1x + b1y + c1 = 0; a2x + b2y + c2 = 0 Compare the ratios Graphical representations Algebraic interpretation a1 ≠ b1 Intersecting lines Exactly one solution (Unique) a2 b2 a1 = b1 = c1 Coincident lines Infinitely many solutions a2 b2 c2 a1 = b1 ≠ c1 Parallel lines No solution a2 b2 c2 60 Mathematics–10

QUICK REVISION NOTES •• Two linear equations in the same two variables are called a pair of linear equations in two variables. The general form of a pair of linear equations is a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 where a1, a2, b1, b2, c1, c2 are real numbers, such that a12 + b12 ≠ 0, a22 + b22 ≠ 0. •• A pair of linear equations in two variables can be represented and solved by the (i) Graphical Method (ii) Algebraic Method Graphical Method: The graph of a pair of linear equations in two variables is represented by two lines. (i) If the lines intersect at a point, then the point gives the unique solution of the two equations. The pair of equations is consistent. (ii) If the lines are parallel, the pair of lines of equations has no solution. The pair of equations are inconsistent. (iii) If the lines coincide, then there are infinitely many solutions. The pair of equations is consistent. Algebraic Method: The following are the algebraic methods: (i) Substitution Method (ii) Elimination Method •• If a pair of linear equations is given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the following situations can arise: (i) a1 ≠ b1 ﬁ Equations are consistent. (ii) a1 = b1 ≠ c1 ﬁ Equations are inconsistent. a2 b2 a2 b2 c2 (iii) a1 = b1 = c1 ﬁ Equations are dependent and are consistent. a2 b2 c2 •• There are many situations which can be mathematically represented by two equations that are not linear to start with. But, when we alter them, they are reduced to a pair of linear equations. COMMON ERRORS Errors Corrections (i) Drawing axes as line segments. (i) Do not forget to mark arrow heads for axes (lines). (ii) Incorrectly transposing the terms. (ii) Be careful about sign of terms while transposing. (iii) Writing incorrectly that if a1 = b1 ≠ c1 , there ( iii) If a1 = b1 ≠ c1 , then the pair of linear equations a2 b2 c2 a2 b2 c2 are infinitely many solutions. is inconsistent and so has no solution. (iv) Making mistake in the problem of upstream and (iv) The speed of the boat downstream downstream. = (x + y) km / h For example, let the speed of the boat in still and the speed of the boat upstream water be x km / h and speed of the stream be = (x – y) km / h y km / h. Then, the speed of the boat downstream = (x – y) km / h and the speed of the boat upstream = (x + y) km / h Pair of Linear Equations in Two Variables 61

4 Coordinate Geometry Topics Covered 2. Section Formula 1. Distance Formula 1. Distance Formula • Distance between points A(x1, y1) and B(x2, y2) is given by AB = (x2 – x1)2 + (y2 – y1)2 This formula is called the distance formula. • Distance of any point Q (a, b) from origin ‘O’ is given by OQ = a2 + b2 • Application of Distance Formula: (a) To show collinearity of three points say, A, B, C:. Find AB, BC and CA and then show that the sum of the shorter lengths is equal to the largest length. i.e., AB + BC = AC or BC + CA = AB or BA + AC = BC If sum of shorter lengths is not equal to the largest length, then points are not collinear. (b) To show two points A and B are equidistant from a point P: Show PA = PB. (c) To prove given three points A, B, C form: (i) a right angled triangle: Show that sum of squares of any two sides is equal to square of third longest side i.e., AB2 = AC2 + CB2 or AC2 = AB2 + BC2 or BC2 = BA2 + AC2. (ii) a triangle: Show that sum of lengths of any two sides is greater than third side, i.e., AB + BC > AC or BC + CA > BA or BA + AC > BC. (iii) an equilateral triangle: Show that all the three sides are equal, i.e., AB = BC = AC. (iv) an isosceles triangle: Show that any two sides are equal, i.e., AB = AC or BC = BA or CA = CB. (d) To prove given four points A, B, C, D form: (i) a parallelogram: Show both pair of opposite sides are equal, i.e., AD = BC and AB = DC. (ii) a parallelogram but not rectangle: Show that both pairs of opposite sides are equal but diagonals are not equal, i.e., AD = BC and AB = DC but AC π DB. (iii) a rhombus: Show that four sides are equal, i.e., AB = BC = CD = AD. (iv) a rhombus but not a square: Show that all sides are equal but diagonals are not equal, i.e., AB = BC = CD = DA but AC π DB. (v) a square: Show that four sides are equal and the diagonals are also equal, i.e., AB = BC = CD = DA and AC = DB. (vi) a rectangle: Show that both pairs of opposite sides are equal and the diagonals are also equal, i.e., AB = CD, BC = DA and AC = DB. 62

Example 1. The distance between the points (2, 3) and (4, 1) is (a) 2 (b) 2 2 (c) 3 (d) 3 3 Solution. Here, x1 = 2, y1 = 3 and x2 = 4, y2 = 1 Distance between the points (2, 3) and (4, 1) = (x2 – x1)2 + (y2 – y1)2 = (4 – 2)2 + (1 – 3)2 = (2)2 + (–2)2 = 4 + 4 = 2 2 Hence, option (b) is the correct answer. Example 2. The distance between the points (–5, 7) and (–1, 3) is (a) 4 3 (b) 4 5 (c) 4 2 (d) 4 7 Solution. Here, x1 = –5, y1 = 7 and x2 = –1, y2 = 3 D istance between points (–5, 7) and (–1, 3) = (x2 – x1)2 + (y2 – y1)2 = (–1 + 5)2 + (3 – 7)2 = 16 + 16 = 4 2 H ence, option (c) is the correct answer. Example 3. The value of a, so that the point (4, a) lies on the line 3x – 2y = 5 is (a) 2 (b) 3 75 (c) 2 (d) 2 Solution. Since (4, a) lies on the line 3x – 2y = 5 Then 3(4) – 2(a) = 5 ﬁ 12 – 2a = 5 7 ﬁ 2 = a H ence, option (c) is the correct answer. Example 4. (5, –2), (6, 4) and (7, –2) are the vertices of a/an (a) Scalene triangle (b) Equilateral triangle (c) Isosceles triangle (d) None of these Solution. Let A (5, –2), B (6, 4) and C (7, –2) be the vertices of a triangle. Now AB = (6 – 5)2 + (4 + 2)2 = 37 A (5, – 2) BC = (7 – 6)2 + (–2 – 4)2 = 37 AC = (7 – 5)2 + (–2 + 2)2 = 4 = 2 Here AB = BC \\ DABC is an isosceles triangle. C (7, –2) B (6, 4) H ence, option (c) is the correct answer. Example 5. The point on the y-axis which is equidistant from (2, –5) and (–2, 9) is (a) (0, 3) (b) (0, 2) (c) (0, 5) (d) None of these Solution. Let P (0, y) be any point on the y-axis. Since P (0, y) is equidistant from A (2, –5) and B (–2, 9) So PA = PB ﬁ PA2 = PB2 ﬁ (2)2 + (y + 5)2 = (2)2 + (y – 9)2 ﬁ y2 + 10 y + 25 = y2 – 18y + 81 ﬁ 28y = 81 – 25 ﬁ 28y = 56 56 y = 28 = 2 So, the point is (0, 2) H ence, option (b) is the correct answer. Coordinate Geometry 63

Exercise 4.1 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. The distance of the point P (–3, –4) from the x-axis (in units) is (a) 3 (b) –3 (c) 4 (d) 5 (d) 3 units 2. The distance of the point P (3, –4) from the origin is (a) 7 units (b) 5 units (c) 4 units 3. The distance between the points (3, –2) and (–3, 2) is (a) 52 units (b) 4 10 units (c) 2 10 units (d) 40 units 4. The distance of a point from the origin is (a) x2 + y2 (b) x2 – y2 (c) x2 + y2 (d) None of these 5. The distance between the points c– 8 , 2m and c 2 2m is 5 , 5 (a) 0 units (b) 1 unit (c) 2 units (d) 5 units (d) 8 6. The perpendicular distance of A(5, 12) from the y-axis is (a) 4 (b) 5 (c) 7 7. The distance between ( 2 + 1, 2) and (1, 2 − 2) is (a) 2 (b) 3 (c) 11 (d) 17 8. The value of k for which the point (0, 2) is equidistant from two points (3, k) and (k, 5) is (a) 1 (b) 2 (c) 5 (d) 9 9. If points (a, 0), (0, b) and (1, 1) are collinear, then the value of 11 is + a b (a) 1 (b) 2 (c) 5 (d) 9 10. If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k? (a) k = 4 (b) k = –4 (c) k = ± 4 (d) None of these 11. The value(s) of x, if the distance between the points A(0, 0) and B(x, – 4) is 5 units, is (a) ± 2 (b) ± 3 (c) ± 4 (d) ± 5 12. If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), the value of p is (a) 5 units (b) 8 units (c) 10 units (d) None of these 13. Points A(2, –1), B(3, 4), C(–2, 3), D(–3, –2) are the vertices of a: (a) Rectangle (b) Square (c) Rhombus (d) None of these 14. The value(s) of y, if the distance between the points (2, y) and (– 4, 3) is 10, is (a) 11 (b) –5 (c) Both (a) and (b) (d) None of these 15. The points (a, a), (–a, –a) and (− 3a, 3a) are the vertices of a/an (a) Equilateral triangle (b) Isosceles triangle (c) Scalene triangle (d) None of these 16. If the distance of P(x, y) from A(5, 1) and B(–1, 5) are equal, then 3x equals (a) y (b) y (c) 2y (d) 2 2y 17. The point on x-axis which is equidistant from the points (2, –2) and (–4, 2) is (a) (1, 0) (b) (2, 0) (c) (0, 2) (d) (–1, 0) 64 Mathematics–10

B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): The point (0, 4) lies on y-axis. Reason (R): The x-coordinate on the point on y-axis is zero. 2. Assertion (A): The value of y is 6, for which the distance between the points P(2, –3) and Q(10, y) is 10. Reason (R): Distance between two given points A (x1, y1) and B (x2, y2) is given by AB = (x2 − x1)2 + ( y2 − y1)2 Answers and Hints A. Multiple Choice Questions (MCQs) 15. (a) Equilateral triangle 1. (c) 4 2. (b) 5 units Let P(a, a), Q(– a, – a), R(− 3a, 3a) are the vertices of an equilateral triangle. 3. (a) 52 units 4. (c) x2 + y2 5. (c) 2 units 6. (b) 5 PQ = (a + a)2 + (a + a)2 = 2 2a 7. (a) 2 8. (a) 1 QR = (−a + 3a)2 + (−a − 3a)2 9. (a) 1 10. (c) k = ±4 = 2 2a RP = (a + 3a)2 + (a − 3a)2 (4 −1)2 + (k − 0)2 = 5 ﬁ k = ±4 11. (b) ±3 = 2 2a Here, PQ = QR = RP. AB = ]x – 0g2 + ]– 4 – 0g2 \\ P, Q, R are vertices of an equilateral triangle. ⇒ 5 = x2 + 16 16. (c) 2y ⇒ (5)2 = ^ x2 + 16h2 PA2 = PB2 ⇒ 25 = x2 + 16 ⇒ (x – 5)2 + (y – 1)2 = (x + 1)2 + (y – 5)2 ⇒ x2 + 16 – 25 = 0 ⇒ x2 – 9 = 0 ⇒ 12x = 8y ⇒ 3x = 2y 17. (d) (–1, 0) ⇒ x2 = 9 Let P(x, 0) be a point on x-axis. ⇒ x = ±3 PA = PB Hence, x = 3 or –3. PA2 = PB2 12. (c) 10 units (x – 2)2 + (0 + 2)2 = (x + 4)2 + (0 – 2)2 Here, AB = AC [Given] x2 + 4 – 4x + 4 = x2 + 16 + 8x + 4 ﬁ AB2 = AC2 –4x + 4 = 8x + 16 ﬁ (3 – 0)2 + (p – 2)2 = (p – 0)2 + (5 – 2)2 x = –1 ﬁ p = 1 P(–1, 0) So, point B is (3, 1); point C is (1, 5) B. Assertion-Reason Type Questions 1. (a) Both assertion (A) and reason (R) are true Now, length of AB = (3 − 0)2 + (1 − 2)2 and reason (R) is the correct explanation of = 9 +1 = 10 units assertion (A). 13. (c) Rhombus 14. (c) Both (a) and (b) 2. (d) Assertion (A) is false but reason (R) is true. Coordinate Geometry 65

2. Section Formula • Internal Division: If a point P(x, y) divides the line segment XY in the ratio m : n internally, then the coordinate of point P is given by  mx2 + nx1 , my2 + ny1  , where coordinates of points X and Y are  m +n m + n  X (x1, y1) and Y (x2, y2). This is known as section formula. • Mid-point formula: If P (x, y) is the mid-point of AB, A (x1, y1), B (x2, y2), then coordinates of P (x, y) = P  x1 + x2 , y1 + y2  .  2 2  • External Division: If a point P (x, y) divides the line segment AB, A (x1, y1), B (x2, y2) externally in the ratio m : n, then the coordinates of P (x, y) are P  mx2 − nx1 , my2 − ny1  .  m − n m − n  • Coordinate of centroid of a triangle ABC, A (x1, y1), B (x2, y2), C (x3, y3) is  x1 + x2 + x3 , y1 + y2 + y3  .  3 3  • To find the ratio in which the join of two points is divided according to some given condition, then we take ratio as k : 1 as  kx2 + x1 , ky2 + y1  . If k is positive, then it is internal division.  k +1 k + 1  • Application of Section Formula: To show the collinearity of three points say A, B, C by using section formula, we first assume that any one point say B divides AC in the ratio k : 1. Then, we find coordinates of B using Section Formula and then equate them with x and y-coordinates of B. If value of k from both equations are same, then given points i.e., A, B and C are collinear otherwise not. Example 1. The coordinates of the points P and Q are respectively (4, –3) and (–1, 7). The x-coordinate PR 3 (abscissa) of a point R on the line segment PQ such that PQ = 5 , is (a) 0 (b) 1 (c) 2 (d) 3 Solution. Since PQ 5 32 PR = 3 PQ – PR 5 – 3 RQ 2 P R Q ﬁ PR = 3 ﬁ PR = 3 (4, –3) (–1, 7) ﬁ R divides PQ in the ratio 3 : 2. \\ x-coordinate (abscissa) of R = 3 (–1) + 2 (4) –3 + 8 = =1 H ence, option (b) is the correct answer. 3+2 5 Example 2. The ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A (2, –2) and B (3, 7) is [Imp.] (a) 3 : 5 (b) 2 : 9 (c) 5 : 7 (d) 4 : 5 Solution. Let P(x, y) be the point on the line 2x + y – 4 = 0 dividing the line segment joining the points A(2, –2) and B(3, 7) in the ratio k : 1. KJKKL OONOP \\ The coordinate of P are 3k + 2 , 7k – 2 k+1 k+1 2KLKKJ 3k + 2 POOON KKKJL 7k – 2 NOOPO Since, point (x, y) lies on the line 2x + y = 4. ⇒ k+1 + k+1 =4 2 2 ⇒ 6k + 4 + 7k – 2 = 4 ⇒ 13k + 2 = 4k + 4 ⇒ k = 9 So, ratio is 9 : 1 or 2 : 9. k+1 H ence, option (b) is the correct answer. 66 Mathematics–10

Example 3. Let P and Q be the points of trisection of the line segment joining the points A (2, – 2) and B (–7, 4) such that P is nearer to A. The coordinates of P and Q respectively are (a) (3, 2), (1, 9) (b) (–4, 3), (5, 0) (c) (–3, 7), (2, 7) (d) (–1, 0), (–4, 2) Solution. As per the question, the diagram is as follows: A P QB (2, –2) (–7, 4) \\ P divides AB in the ratio 1 : 2. KKKJL NOOO KLKKJ 4 NOPOO P –7 (1) + 2 (2) 4 (1) + (–2) 2 –7 + 4 4 – 3 , 3 3 3 \\ Coordinates of P = = , = (–1, 0) Also mid-point of PB. Q is the iCs othoerdcionrarteecstoafnQsw =e r.KKLJK OOONP –1 + (–7) 0 + 4 KJLKK –8 2OOOPN 2 2 2 \\ , = , = (–4, 2) H ence, option (d) Example 4. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, x and y respectively are (a) x = 6, y = 3 (b) x = 3, y = 6 (c) x = –6, y = –3 (d) x = –3, y = –6 Solution. Let A (1, 2), B (4, y), C (x, 6) and D (3, 5) be the vertices of a parallelogram ABCD. Since, the diagonals of a parallelogram bisect each other \\ JLKKK x + 1 6 + 2 OPNOO = JKKK 3 + 45 + y NOOO D (3, 5) C (x, 6) ﬁ 2 , 2 = L 2 , 2 P P x x+1 7 ﬁ =6 2 2 and 5+y ﬁ y = 3 A (1, 2) B (4, y) Thus, x = 6 and y = 3. 2 = 4 H ence, option (a) is the correct answer. Exercise 4.2 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. The point which lies on the perpendicular bisector of the line segment joining the points A(– 2, – 5) and B(2, 5) is (a) (0, 0) (b) (0, 2) (c) (2, 0) (d) (– 2, 0) 2. The mid-point of the line segment joining the points (–5, 7) and (–1, 3) is (a) (–3, 7) (b) (–3, 5) (c) (–1, 5) (d) (5, –3) 3. If (3, –6) is the mid-point of the line segment joining (0, 0) and (x, y), then the point (x, y) is (a) (–3, 6) C(–1, 2) (b) (6, –6) the line (c) (6, –12) A(2, 5) (d) B32i,n−r3atio 3 : 4, the 4. If the point divides internally segment joining and coordinates of B are (a) B(5, 2) (b) B(–5, –2) (c) B(–7, 2) (d) B(7, –2) 5. The point which divides the line segment joining the points A(0, 5) and B(5, 0) internally in the ratio 2 : 3 is (a) (2, 3) (b) (3, 4) (c) (–2, 3) (d) (3, –5) Coordinate Geometry 67

6. The coordinates of a point A, where AB is a diameter of the circle with centre (–2, 2) and B is the point with coordinates (3, 4) are (a) A(7, 0) (b) A(5, 0) (c) A(–5, 0) (d) A(–7, 0) 7. The ratio in which the line segment joining the points (6, 4) and (1, –7) is divided by x-axis is (a) 4 : 7 (b) 4 : 5 a (c) 4 : 9 (d) 4 : 11 3 8. The value of a, for which point P ( , 2) is the mid-point of the line segment joining the points Q(–5, 4) and R(–1, 0) is (a) –5 (b) –7 (c) –9 (d) –11 9. The ratio in which the y-axis divides the line segment joining the points (5, –6) and (–1, –4) is (a) 1 : 5 (b) 5 : 1 (c) 1 : 7 (d) 7 : 1 10. If A(1, 2), B(4, 3) and C(6, 6) are three vertices of parallelogram ABCD, coordinates of D are (a) (3, 5) (b) (2, 7) (c) (4, 9) (d) (3, 8) 11. The coordinates of the point P which divides the join of A(–2, 5) and B(3, –5) in the ratio 2 : 3 are (a) (1, 0) (b) (2, 0) (c) (3, 0) (d) (0, 1) 12. The centre of a circle is (2a, a – 7). The value of a if the circle passes through the point (11, –9) and has diameter 10 2 units is (a) a = 3 or 6 (b) a = 5 or 3 (c) a = 7 or 4 (d) None of these 13. The value(s) of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units, is (a) 15 or 2 (b) 10 or 9 (c) 17 or 1 (d) None of these 14. P(–2, 5) and Q(3, 2) are two points. The coordinates of the point R on PQ such that PR = 2QR are (a) R  4 3 (b) R  2 5 (c) R  1 7 (d) None of these  ,  ,  , 3 3 3 15. The coordinates of the point which divides the line segment joining the points (4, –3) and (8, 5) in the ratio 3 : 1 internally are (a) P(4, 3) (b) P(7, 3) (c) P(3, 5) (d) P(7, 3) 16. If the point P(m, 3) lies on the line segment joining the points A  − 2 , 6 and B(2, 8), the value of m is  5 (a) 3 (b) 2 (c) –3 (d) –4 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): The point (–1, 6) divides the line segment joining the points (–3, 10) and (6, –8) in the ratio 2 : 7 internally. Reason (R): Given three points, i.e. A, B, C form an equilateral triangle, then AB = BC = AC. 2. Assertion (A): Mid-point of a line segment divides line in the ratio 1 : 1. Reason (R): The ratio in which the point (–3, k) divides the line segment joining the points (–5, 4) and (–2, 3) is 1 : 2. 68 Mathematics–10

Answers and Hints A. Multiple Choice Questions (MCQs) 12. (b) a = 5 or 3 1. (a) (0, 0) 2. (b) (–3, 5) 13. (c) 17 or 1 3. (c) (6, –12) 4. (b) B (–5, –2) Given that, PQ = 10 units 5. (a) (2, 3) ﬁ PQ2 = 100 6. (d) A = (–7, 0) (x – 9)2 + (4 – 10)2 = 100 Let AB be a diameter of circle with centre O. Solve to get x = 17 or x = 1. Since centre is the mid-point of diameter AB. 14. (a) R  4 , 3  3 So, coordinates of O =c x+3, y+4 m 2 2 PR : QR = 2 : 1 R 1(−2) + 2(3) 1(5) + 2(2),   2+1 2+1  R  4 , 3  3 ﬁ (–2, 2) = c x + 3 , y + 4 m 15. (d) P(7, 3) 2 2 Let P(x, y) be the required point. Using Now, x + 3 2 = –2 ﬁ x=–4–3 section-formula y+4  m1x2 + m2 x1 , m1 y2 + m2 y1  = (x, y) 2  m1 + m2 m1 + m2  ﬁ x = – 7 and =2   ﬁ y = 4 – 4 ﬁ y = 0 x = 3(8) + 1(4) , 3+1 Thus, coordinates of point A = (–7, 0) 7. (a) 4 : 7 y = 3(5) + 1(−3) 3+1 8. (c) –9 x = 7 c –5 + (–1) , 4 + 0 m = c a , 2m y = 3 2 2 3 Thus, (7, 3) is the required point. a = –6 & a = –9 32 16. (d) – 4 9. (b) 5 : 1 Let point P divides AB in ratio k : 1. 10. (a) (3, 5) Let coordinates of D be (x, y) and P is mid- point of AC and BD. [Diagonals of a parallelogram bisect each \\ 3 = 8 × k + 6 × 1 ﬁ k= −3 k + 1 5 other] \\  x+4, y + 3 = 1+ 6 , 2 + 6  2   3 2  2 2   2 2   5   5 5 2k + − 2 − − ﬁ x+4 7 y+3 = 8 and m = k +1 = −3 +1 = , 2 22 2 5 ﬁ x + 4 = 7; y + 3 = 8 ﬁ m = –4 ﬁ x = 3; y = 5 B. Assertion-Reason Type Questions 1. (b) Both assertion (A) and reason (R) are true \\ Coordinates of D are (3, 5). but reason (R) is not the correct explanation 11. (d) (0, 1) of assertion (A). 2. (c) Assertion (A) is true but reason (R) is false. 6–6 –10 + 15 x = 5 = 0, y = 5 = 1 Coordinate Geometry 69

Case Study Based Questions I. Class X students of a secondary school in Krishnagar have been allotted a rectangular plot of a land for gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot. Considering A as origin, answer question (1) to (5). 1. Considering A as the origin, what are the coordinates of A? (a) (0, 1) (b) (1, 0) (c) (0, 0) (d) (–1, –1) 2. What are the coordinates of P? (a) (4, 6) (b) (6, 4) (c) (4, 5) (d) (5, 4) 3. What are the coordinates of R? (a) (6, 5) (b) (5, 6) (c) (6, 0) (d) (7, 4) 4. What are the coordinates of D? (a) (16, 0) (b) (0, 0) (c) (0, 16) (d) (16, 1) 5. What are the coordinates of P if D is taken as the origin? (a) (12, 2) (b) (–12, 6) (c) (12, 3) (d) (6, 10) Ans. 1. (c) (0, 0) 2. (a) (4, 6) 3. (a) (6, 5) 4. (a) (16, 0) 5. (b) (–12, 6) II. 10 9 8 B 7 6 5 C A 4 3 2 D 1 (0, 0) 1 2 3 4 5 6 7 8 9 10 In a classroom, 4 friends have decided to seat at the points A, B, C and D as shown in fig. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli some questions which are as follows: 1. What is the coordinates of A and B? (a) A(3, 4), B(6, 7) (b) A(4, 3), B(7, 6) (c) A(–3, –4), B(–6, –7) (d) A(–3, –4), B(–6, –7) 70 Mathematics–10

2. What is the coordinate of C and D? (b) C(9, 4), D(6, 1) (a) C(4, 9), D(1, 6) (d) C(9, –4), D(6, –1) (c) C(–9, –4), D(–6, –1) 3. What is the distance between AB? (a) 2 (b) 3 (c) 2 3 (d) 3 2 4. What is the distance between AC? (a) 6 (b) 4 (c) 8 (d) 10 5. What is the distance between BD? (a) 4 (b) 6 (c) 8 (d) 10 2. (b) C(9, 4), D(6, 1) 3. (d) 3 2 Ans. 1. (a) A(3, 4), B(6, 7) 5. (b) 6 4. (a) 6 III. To conduct sports day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance 1 of 1 m from each other along AD, as shown in figure given below. Niharika runs th the distance 14 AD on the 2nd line and posts a green flag. Preet runs th the distance AD on eighth line and posts 5 a red flag. DC 10 9 8 7 6 5 Niharika Rashmi N Preet 4 3R 2 Green Blue P flag flag Red 1 flag A 1 2 3 4 5 6 7 8 9 10 B Answer questions (1) to (5) 1 Write the coordinates of point N. (a) (1, 25) (b) (2, 25) (c) (4, 25) (d) (3, 25) (d) (2, 8) 2. Write the coordinates of point P. (a) (2, 80) (b) (8, 2) (c) (8, 20) 3. What is the distance between green and red flags? (a) 63 m (b) 26 m (c) 62 m (d) 61 m 4. What is the coordinates of R? (a) (5, 22) (b) (5, 22.5) (c) (22.5, 5) (d) (22, 5) Coordinate Geometry 71

5. Distance between Niharika and Rashmi is (a) 3.90 m (b) 4.01 m (c) 3.06 m (d) 5.10 m Ans. 1. (b) (2, 25) 2. (c) (8, 20) 3. (d) 61 m 4. (b) (5, 22.5) 5. (a) 3.90 m IV. In a classroom, 4 friends are seated at the points A, B, C and D as shown is given fig. Rupa and Rupali walk into the class and after observing for a few minutes Rupa asks Rupali, Some questions which are as follows: 10 9 8 B 7 6 C 5 A 4 3 2 D 1 (0, 0) 1 2 3 4 5 6 7 8 9 10 1. Coordinate of A is (b) (3, 0) (c) (1, 4) (d) (3, 4) (a) (4, 3) 2. Coordinate of D is (b) (1, 6) (c) (2, 6) (d) (6, 2) (a) (6, 1) 3. Distance AB is (a) 3 + 2 (b) 2 (c) 2 2 (d) 3 2 4. Diagonal AC is (a) 3 (b) 4 (c) 6 (d) 6 5. ABCD is a (a) rectangle (b) square (c) rhombus (d) trapezium Ans. 1. (d) (3, 4) 2. (a) (6, 1) 3. (d) 3 2 4. (c) 6 5. (b) square V. In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, few questions, which you are required to answer. They are as follows: 10 9 8 B 7 Rows 6 A C 5 4 3 2 1D 0 1 2 3 4 5 6 7 8 9 10 Columns 72 Mathematics–10

1. What are the coordinates of A and B respectively? (a) (3, 4) and (6, 7) (b) (4, 3) and (7, 6) (c) (–3, 4) and (–6, 7) (d) (4, –3) and (–7, 6) 2. What are the coordinates of C and D respectively? (a) (9, 4) and (6, 1) (b) (4, 9) and (1, 9) (c) (–9, –4) and (–6, –1) (d) (–9, 4) and (–6, 1) 3. What is the distance between AB and CD? (a) 4 2 units (b) 3 2 units (c) 5 2 units (d) 2 2 units 4. Distance between BC and AD is (d) 5 2 units (a) 2 2 units (b) 4 2 units (c) 3 2 units (d) 6 units (c) 8 units 3. (b) 3 2 units 5. What is the distance between A and C? (a) 4 units (b) 5 units Ans. 1. (a) (3, 4) and (6, 7) 2. (a) (9, 4) and (6, 1) 4. (c) 3 2 units 5. (d) 6 units VI. Student of a school are standing in rows and columns in a playground for a drill practice. A, B, C, D are the positions of four students as shown in the figure. 1. The coordinates of A and B are respectively (a) (3, 5) and (7, 9) (b) (5, 3) and (9, 7) (c) (–3, –5) and (7, 9) (d) (3, 5) and (–7, –9) 2. The coordinates of the points C and D are respectively (a) (5, 11) and (1, 7) (b) (11, 5) and (7, 1) (c) (–5, –11) and (–1, –7) (d) (5, 11) and (–1, –7) 3. The distance between A and B is (a) 4 2 units (b) 3 2 units (c) 5 2 units (d) 2 2 units 4. The distance between A and C is (a) 8 units (b) 8 2 units (c) 5 2 units (d) 3 2 units 5. The coordinates of the point which is equidistant from each of the four students A, B, C and D are (a) (5, 7) (b) (–5, –7) (c) (–7, –5) (d) (7, 5) Ans. 1. (a) (3, 5) and (7, 9) 2. (b) (11, 5) and (7, 1) 3. (a) 4 2 units 4. (a) 8 units 5. (d) (7, 5) Coordinate Geometry 73

VII. Three friends are playing in a ground to form a triangle. They are standing at the vertex of the triangle having coordinates A(2a, 4a), B(2a, 6a) and C(2a + a 3 , 5a). Teacher wants to check the some measurements of the triangle and ask few questions. Answer them. y x 1. Length of AB is (a) 2a (b) 4a (c) 8a (d) 6a 2. Length between the points B and C is (a) a (b) 2a (c) 4a (d) 6a 3. Length between the point A and C is (a) a (b) 2a (c) 3a (d) 4a 4. Mid-point of AB is (a) (2a, 5a) (b) (4a, 5a) (c) (2a, 3a) (d) (5a, 4a) 5. Height of triangle = (a) 2a (b) 3a (c) 3a (d) 5a Ans. 1. (a) 2a 2. (b) 2a 3. (b) 2a 4. (a) (2a, 5a) 5. (c) 3a VIII. Kartikeya starts walking from his house to office. Instead of going to the office directly, he goes to the bank first, from there to his daughter’s school and then reaches the office. Assuming that all distances covered are in straight lines and if the coordinates of house at (2, 4), bank at (5, 8), school at (13, 14) and office (13, 26) are represented in km. Then find answer from question (1) to question (5). House (2, 4) Bank (5, 8) A B D C Office (13, 26) School (13, 14) 74 Mathematics–10

1. Total distance covered by Kartikeya is (a) 27 km (b) 24 km (c) 30 km (d) 18 km (d) 4 km 2. Extra distance travelled by Kartikeya in reaching his office is (d) 10 km (d) 24.59 km (a) 1.5 km (b) 3.7 km (c) 2.41 km (d) 18 km 3. Distance travelled by Kartikeya in reaching his daughter’s school is 3. (a) 14 km (a) 14 km (b) 20 km (c) 15 km 4. The shortest distance between his house and the office is (a) 12.53 km (b) 20.53 km (c) 14.53 km 5. The distance between his daughter’s school and his office is (a) 10 km (b) 12 km (c) 16 km Ans. 1. (a) 27 km 2. (c) 2.41 km 4. (d) 24.59 km 5. (b) 12 km Experts’ Opinion Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly. 1. To use distance formula for solving different problems. 2. To use section formula for solving different problems. 3. To use mid-point formula for solving related problems. IMPORTANT FORMULAE •• The distance between P(x1, y1) and Q(x2, y2) = (x2 − x1)2 + ( y2 − y1)2 . •• The distance of a point P(x, y) from the origin = x2 + y2 . •• The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 are  m1x2 + m2 x1 , m1 y2 + m2 y1  .  m1 + m2 m1 + m2  •• The mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) is  x1 + x2 , y1 + y2  .  2 2  COMMON ERRORS Errors Corrections (i) Incorrectly interpreting the internal division (i) In case of internal division, point dividing the line and external division of a line segment and segment lie on it. But in external division, point so using wrong formula. dividing the line segment does not lie on it. So, use correct formula depending upon situations. (ii) Making mistakes in use of distance formula (ii) To show collinearity of three points, say A, B and C and condition of collinearity. • using distance formula: find AB, BC and CA and then show the sum of two shorter lengths is equal to the longest length. ( iii) Exchanging the ordered pairs and ignoring ( iii) Be careful. Do not exchange the order pair and must the negative sign while solving problems. take negative sign if any. Coordinate Geometry 75

5 Triangles Topics Covered 1. Similarity of Triangles 2. Criteria for Similarity of Triangles and Areas of Similar Triangles 3. Pythagoras Theorem 1. Similarity of Triangles • Two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportional). In the triangles ABC and A′B′C′ shown below, ∠A = ∠A′, ∠B = ∠B′, ∠C = ∠C′ and AB = AC BC A′B′ A′C′ = . Therefore, DABC and DA′B′C′ satisfy both the conditions necessary for them to be similar. B′C′ Hence, DABC ~ DA′B′C′. Note: All congruent triangles are similar but the similar triangles need not be congruent. • Basic Proportionality Theorem or Thales Theorem: In a triangle, a line drawn parallel to one side, to intersect the other two sides in distinct points, divides the two sides in the same ratio. • Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. Example 1. In the given figure, AP = 3 cm, AR = 4.5 cm, AQ = 6 cm, AB = 5 cm, AC = 10 cm. The length of AD is A PR Q BD C (a) 4.5 cm (b) 6.5 cm (c) 7 cm (d) 7.5 cm Solution. Given: AP = 3 cm, AR = 4.5 cm, AQ = 6 cm, AB = 5 cm, AC = 10 cm. To find: The length of AD 76

Proof: In DABC, AP 3 AB = 5 ...(i) [BPT] AQ 6 3 AC = 10 = 5 ...(ii) [BPT] From (i) and (ii), we get AP AQ ﬁ PQ  BC [Converse of BPT] AB = AC In DABD, PR  BD ﬁ AP AR 3 4.5 AB = AD ﬁ = 5 AD 4.5 × 5 ﬁ AD = 3 = 7.5 cm Hence, option (d) is the correct answer. Example 2. In the given figure, MN || QR. If PM = x cm, MQ = 10 cm, PN = (x – 2) cm, NR = 6 cm, then the value of x is P x x–2 10 MN cm 6 cm QR (a) 5 cm (b) 7 cm (c) 8 cm (d) 12 cm [Basic proportionality theorem] Solution. Since in DPQR, MN || QR PM PN ⇒ MQ = NR x x − 2 10 6 ⇒ = ⇒ 6x = 10x – 20 ⇒ 4x = 20 ⇒ x = 5 cm Hence, option (a) is the correct answer. Example 3. If in the given figure, AB  DE and BD  EF, then DC2 = C F D E AB (a) CF + AC (b) CF × AC (c) CF + 2AC (d) None of these Solution. Given: DABC in which DE  AB and BD  EF. In DABC, DE  AB ﬁ CADC = CE ...(i) [Basic proportionality theorem] BC Again in DCDB, EF  BD ﬁ CF CE CD = CB ...(ii) [BPT] Triangles 77

From (i) and (ii), we get CD = CF ﬁ CD2 = CF × AC AC CD Hence, option (b) is the correct answer. Exercise 5.1 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. If in two triangles DEF and PQR, –D = –Q and –R = –E, then which of the following is not true? EF DF DE EF DE DF EF DE (a) PR = PQ (b) PQ = RP (c) QR = PQ (d) RP = QR 2. In DABC and DDEF, –B =–E, –F = –C and AB = 3DE. Then, the two triangles are (a) congruent but not similar (b) similar but not congruent (c) neither congruent nor similar (d) congruent as well as similar 3. In the figure shown along side, DE || BC, ∠ADE = 70° A and ∠BAC = 50°, then ∠BCA = DE (a) 30° (b) 60° (c) 40° (d) 45° B C 4. In the figure, if ∠ACB = ∠CDA, AC = 6 cm and AD = 3 cm, then the length of AB is (a) 8 cm (b) 10 cm (c) 12 cm (d) 16 cm 5. In the given figure, if DE || BC, the EC equals (a) 1 cm (b) 2 cm (c) 4 cm (d) 6 cm 6. In the given figure, D and E are points on AB and AC A DE 1 BC respectively such that DE || BC. If AD = 3 BD and AE = 4.5 cm, then AC is equal to (a) 12 cm (b) 14 cm (c) 16 cm (d) 18 cm B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). 78 Mathematics–10

(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): If ∆ABC and ∆PQR are congruent triangles, S then they are also similar triangles. QR P Reason (R): All congruent triangles are similar but the similar triangles need not be congruent. 2. Assertion (A): In the given figure, PA || QB || RC || SD. C Dl Reason (R): If three or more line segments are perpendiculars A B to one line, then they are parallel to each other. Answers and Hints A. Multiple Choice Questions (MCQs) AD 1 \\ BD = 3 ...(i) 1. (b) DE EF = PQ RP 2. (b) similar but not congruent DE || BC, 3. (b) 60° AD AE [By BPT] 4. (c) 12 cm \\ = DACB ~ DADC (AA criterion) BD EC AC AB So, 1 4.5 [From (i)] ⇒ = ∴ AB = 12 cm 3 = EC ⇒ AD AC 5. (b) 2 cm EC = 13.5 cm ⇒ AC = AE + EC AD AE = 4.5 + 13.5 = 18 cm = [By BPT] BD EC B . Assertion-Reason Type Questions ⇒ 1.5 1 A ∴ = 1. (a) B oth assertion (A) and reason (R) are true and reason (R) is the correct explanation 3 EC of assertion (A) EC = 2 cm 2. (a) Both assertion (A) and reason (R) are true 6. (d) 18 cm DE and reason (R) is the correct explanation of assertion (A) 1 B C  AD = 3 BD 2. Criteria for Similarity of Triangles and Areas of Similar Triangles There are some characteristic properties of similar triangles which will help us establishing similarity of two triangles. • AAA Similarity: If in two triangles, corresponding angles are equal or the two corresponding angles are equal, then the triangles are similar. • SSS Similarity: If the corresponding sides of two triangles are proportional, then they are similar. • SAS Similarity: If in two triangles one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar. • The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Example 1. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, the length of her shadow after 4 seconds is (a) 1.1 cm (b) 1.6 cm (c) 2.3 cm (d) 3.5 cm Triangles 79

Solution. Let AB be the lamp-post and CD be the girl after walking for 4 seconds away from the lamp- post. DE is the shadow of the girl. Let DE = x cm. BD = Distance covered by girl in 4 seconds = 1.2 × 4 = 4.8 m A Consider DABE and DCDE, (Each 90°) 3.6 m C ∠B = ∠D ∠E = ∠E(Common) 0.9 m \\ DABE ~ DCDE (AA Similarity) B DxE ⇒ BE AB BD + DE = 3.6 DE = CD ⇒ DE 0.9 ⇒ 4.8 + x = 3.6 ⇒ 4.8 + x = 4x ⇒ 3x = 4.8 ⇒ x = 1.6 x 0.9 ⇒ Shadow of girl after walking for 4 seconds is 1.6 m long. Hence, option (b) is the correct answer. Example 2. Two poles of height a and b (b > a) are c metres apart. The height h (in metres) of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is (a) ab (b) ab + 2ab (c) ab (d) None of these a+b Solution. Given: Two poles, AB = a, CD = b (b > a), BD = c and AD and BC intersect at E (say). Consider DABD and DEFD C Here, ∠ABD = ∠EFD = 90° ∠ADB = ∠EDF(Common) \\ DABD ~ DEFD (AA similarity) E b h A EF FD \\ = AB BD h c − x a a c ⇒ = ...(i) Similarly, DCDB ~ DEFB BxF c–x D (AA similarity) c \\ EF = BF ⇒ h = x ⇒ x= hc ...(ii) CD BD b c b From (i) and (ii), − hc b h c b − h ab a c b a+b = = ⇒ bh = ab – ah ⇒ (a + b)h = ab ⇒ h= Hence, option (c) is the correct answer. Example 3. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Then area (DADE) : area (DABC) = A (a) 1 : 2 (b) 2 : 5 (c) 3 : 4 (d) 3 : 7 E Solution. Given: AD ^ BC in equilateral triangle ABC and DADE is another equilateral triangle. Let BC = x units B C D 80 Mathematics–10

3 \\ AD = 2 × x units  DABC and DADE are equilateral triangle. \\ DABC ~ DADE area DADE AD2 KKJLKKKK 3 x OOPOOOON2 3 area DABC BC2 4 ﬁ = = 2 = x \\ area (DADE) : area (DABC) = 3 : 4 Hence, option (c) is the correct answer. Example 4. In the given figure, DABC ∼ DDEF, BC = 3 cm, EF = 4 cm and area of DABC = 54 cm2. Then the area of DDEF is (a) 54 cm2 (b) 88 cm2 (c) 96 cm2 (d) 108 cm2 Solution. Given: DABC ∼ DDEF D area of DABC BC2 A \\ area of DDEF = EF2 54 32 9 ﬁ area (∆DEF) = 42 = 16 ﬁ area (DDEF) = 54 ×16 = 96 cm2 B CE 4 cm F 9 3 cm Hence, option (c) is the correct answer. Exercise 5.2 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: AB BC CA 1. If in two triangles ABC and PQR, = = , then QR PR PQ (a) DPQR ~ DCAB (b) DPQR ~ DABC (c) DCBA ~ DPQR (d) DBCA ~ DPQR 2. If DPQR ~ DXYZ, ∠Q = 50° and ∠R = 70°, then ∠X + ∠Y is equal to (a) 70° (b) 110° (c) 120° (d) 50° 3. Corresponding sides of two similar triangles are in the ratio 9 : 5. Areas of these triangles are in the ratio (a) 21 : 85 (b) 81 : 25 (c) 9 : 5 (d) 5 : 9 4. The perimeters of two similar triangles DABC and DPQR are 35 cm and 45 cm respectively, then the ratio of the areas of the two triangles is (a) 7 : 9 (b) 28 : 45 (c) 14 : 27 (d) 49 : 81 5. The sides of two similar triangles are in the ratio 2 : 3, then the areas of these triangles are in the ratio (a) 2 : 5 (b) 4 : 9 (c) 7 : 16 (d) None of these 6. The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm, then the corresponding side of second triangle is (a) 5.4 cm (b) 8 cm (c) 9.5 cm (d) 10 cm 7. Given DABC ~ DPQR, if AB = 1 ar(∆ABC) = , then PQ 3 ar(∆PQR) (a) 1 : 3 (b) 1 : 6 (c) 1 : 9 (d) 1 : 5 Triangles 81

8. In the given figures, DPQR ∼ DXYZ. If PQ = 4 cm, P Z QR = 5 cm and XY = 6 cm, then YZ equals. R (a) 4.5 cm 4 cm 5 cm (b) 7.5 cm (c) 6.4 cm XY 6 cm (d) None of these Q 9. Let DABC ∼ DDEF, ar (DABC) = 169 cm2 and ar (DDEF) = 121 cm2. If AB = 26 cm, then DE equals (a) 11 cm (b) 22 cm (c) 33 cm (d) 44 cm 10. In the given figure, DE || BC, AD = 1 cm and BD = 2 cm. What is the ratio of the ar (∆ABC) to the ar (∆ADE)? (a) 1 : 9 (b) 1 : 7 (c) 9 : 1 (d) 7 : 1 11. DABC is isosceles with AC = BC. If AB2 = 2AC2, then the measure of ∠C is (a) 30° (b) 60° (c) 45° (d) 90° 12. In the given figure, DEFG is a square and ∠BAC = 90°. Then FG2 = (a) BG × FC (b) BG + FC BG (d) None of these (c) FC 13. In the given figure, M is mid-point of side CD of a parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced at E. Then EL = AD E (a) BL 1 LM (b) 2 BL (c) 2BL (d) None of these C B 14. In a trapezium ABCD, AC and BD intersecting at O, AB || DC and AB = 2CD, if area of DAOB = 84 cm2, the area of DCOD is (a) 12 cm2 (b) 18 cm2 (c) 21 cm2 (d) 27 cm2 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): In the given figures, ∆ABC ~ ∆GHI. Reason (R): If the corresponding sides of two triangles are proportional, then they are similar. 2. Assertion (A): The sides of two similar triangles are in the ratio 2 : 5, then the areas of these triangles are in the ratio 4 : 25. Reason (R): The ratio of the areas of two similar triangles is equal to the square of the ratio of their sides. 82 Mathematics–10

Answers and Hints A. Multiple Choice Questions (MCQs) ⇒ GD = FE = FG = DE So, equation (v) becomes FG × FG = FC × GB 1. (a) DPQR ~ DCAB 2. (b) 110° ⇒ FG2 = FC × GB 13. (c) 2BL 3. (b) 81 : 25 4. (d) 49 : 81 14. (c) 21cm2 In DAOB and DCOD, 5. (b) 4 : 9 6. (a) 5.4 cm ∠1 = ∠4 ∠2 = ∠3 [Alternate angles] 7. (c) 1 : 9 8. (b) 7.5 cm \\ DAOB ~ DCOD 9. (b) 22 cm 10. (c) 9 : 1 11. (d) 90° 12. (a) BG × FC Since, DEFG is a square ⇒ DE || BC ar (∆AOB) = (AB)2 ⇒ ar (∆COD) (CD)2 Also, ∠ADE = ∠GBD...(i) (Corresponding angles) 84 (2CD)2 and ∠AED = ∠FCE...(ii) ⇒ = ar (Corresponding angles) (∆COD) (CD)2 In DADE and DGBD, 84 4(CD)2 ∠ADE = ∠GBD [From (i)] ⇒ = ∠DAE = ∠DGB [Both 90°] ar (∆COD) (CD)2 DADE ~ DGBD (AA Similarity) ...(iii) 84 4 In DADE and DFEC, ⇒ (∆COD) = 1 ar ∠AED = ∠FCE [From (ii)] \\ ar DCOD = 21 cm2 ∠DAE = ∠EFC [Both 90°] DADE ~ DFEC (AA Similarity) ...(iv) B. Assertion-Reason Type Questions 1. (a) Both assertion (A) and reason (R) are true From (iii) and (iv), and reason (R) is the correct explanation of DGBD ~ DFEC assertion (A) 2. (a) Both assertion (A) and reason (R) are true GD GB and reason (R) is the correct explanation of ⇒ = assertion (A) FC FE ⇒ GD × FE = GB × FC ...(v) Since DEFG is a square 3. Pythagoras Theorem • Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. • Converse of Pythagoras Theorem: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. Example 1. If two sides of a right triangle are 9 cm and 12 cm, then its third side will be (a) 21 cm (b) 15 cm (c) 3 cm (d) None of these Solution: The given triangle is a right triangle. (9 cm)2 + (12 cm)2 = (81 + 144)2 = 225 cm2 Triangles 83

and (15 cm)2 = 225 cm2 Hence, option (b) is the correct answer. Example 2. In an equilateral triangle ABC, D is a point on the side BC such that BD = 1 BC. Then 9AD2 = 3 (a) 7 AB2 (b) 5 AB2 (c) 8 AB2 (d) 11 AB2 Solution:  AL ^ BC ⇒ 1 BL = LC = BC A 2 (In an equilateral triangle, the perpendicular from the vertex bisects the base.) Also, 12 BD = 3 BC and DC = 3 BC Consider right triangle ALD. AD2 = AL2 + DL2 ...(i) and in right triangle ALB, AB2 = AL2 + BL2 ...(ii) B C DL Subtracting (i) from (ii), we get AB2 – AD2 = BL2 – DL2 = (BL + DL) (BL – DL) ⇒ AB2 – AD2 =  1 BC + BL − BD BD  2 ⇒ AB2 – AD2 =  1 BC + BC − BC  BC = 2 BC = 2 BC2  2 2 3  3 BC × 3 9 3 ⇒ 9AB2 – 9AD2 = 2AB2 (BC = AB) ⇒ 7AB2 = 9AD2 Hence, option (a) is the correct answer. Important result: In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. In DABC, AD is a median so that AB2 + AC2 = 2 (BD2 + AD2) Exercise 5.3 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. If in ΔABC, AB = 9 cm, BC = 40 cm and AC = 41 cm, then the ΔABC is a/an (a) Acute angled triangle (b) Right triangle (c) Obtuse angled triangle (d) Isosceles triangle 2. In ΔABC, ∠B = 90° and BD ⊥ AC. If AC = 9 cm and AD = 3 cm, then BD is equal to (a) 2 2 cm (b) 3 2 cm (c) 2 3 cm (d) 3 3 cm 3. In DABC, AB = 6 3 cm , AC = 12 cm and BC = 6 cm, then ∠B = (a) 30° (b) 60° (c) 90° (d) 45° 4. If in an equilateral triangle, the length of the median is 3 cm, then the length of the side of equilateral triangle is (a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm 5. A man goes 12 m due west and then 9 m due north. How far is he from the starting point? (a) 12 m (b) 15 m (c) 18 m (d) 24 m 84 Mathematics–10

6. In an equilateal triangle of side 3 3 cm, the length of the altitude is (a) 3.5 cm (b) 4 cm (c) 4.5 cm (d) 6 cm 7. A ladder is placed against a wall such that its foot is at distance of 5 m from the wall and its top reaches a window 5 3 m above the ground. The length of the ladder is (a) 10 m (b) 15 m (c) 18 m (d) 24 m 8. ABC is an isosceles triangle right-angled at C. The AB2 is equal to (a) AC2 (b) 2AC (c) AC2 (d) AC 2 9. In DABC, ∠ABC = 90°. AD and CE are two medians drawn from A and C, respectively. If AC = 5 cm 35 and AD = cm , the length of CE is 2 (a) 2 5 cm (b) 3 5 cm (c) 4 5 cm (d) 5 cm 10. In an isosceles triangle PQR, PQ = QR and PR2 = 2PQ2. Then ∠Q is (d) None of these (a) 30° (b) 60° (c) 90° B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): If two sides of a right angle are 7 cm and 8 cm, then its third side will be 9 cm. Reason (R): In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 2. Assertion (A): In the ∆ABC, AB = 24 cm, BC = 10 cm and AC = 26 cm, then ∆ABC is a right angle triangle. Reason (R): If in two triangles, their corresponding angles are equal, then the triangles are similar. Answers and Hints A. Multiple Choice Questions (MCQs) ⇒ 3a2 = 12 ⇒ a = 2 cm 1. (b) Right triangle 5. (b) 15 m 2. (b) 3 2 cm Let A be the initial position of the man. He 3. (c) 90° goes from A to B and then from B to C such 4. (b) 2 cm that AB = 12 m, BC = 9 m and ∠ABC = 90°. Let a be the side of equilateral triangle. Median is also the altitude in an equilateral triangle. \\ (Altitude)2 +  a2 = (a)2  2  ⇒ ( 3)2 + a2 = a2 3 cm ] In right triangle ABC, we have ⇒ 4 AC2 = AB2 + BC2 [ Altitude = Median = [By Pythagoras Theorem] 12 + a2 = a2 ⇒ 12 + a2 = 4a2 4 Triangles 85

⇒ AC2 = (12 m)2 + (9 m)2 So, by Pythagoras theorem, AC2 = AB2 + BC2 = (144 + 81) m2 = 225 m2 ⇒ AC = 15 m = 52 + (5 3)2 = 25 + 75 Hence, the man is 15 m away from the starting AC2 = 100 ⇒ AC = 100 point. ⇒ AC = 10 m 6. (c) 4.5 cm Hence, length of the ladder = 10 m. 8. (c) AC2 In DABC, AD is altitude 9. (a) 2 5 cm In right-triangle ABD, ∠B = 90° AD2 = AB2 + BD2 \\ [By Pythagoras Theorem]  BC  2  2  ⇒ AD2 = AB2 + [ BD = DC] AD is median also [ DABC is equilateral] 33 So, BD = 2 cm In DABD, ⇒ (AD)2 + (BD)2 = (AB)2  3 3 2 ⇒  2  ⇒ (AD)2 + = (3 3)2 AD2 = AB2 + BC 2 ...(i) 4 ⇒ AD2 + 27 = 27 Now, in right triangle EBC, ∠B = 90°, 4 CE2 = BC2 + BE2 ⇒ AD2 = 27 – 27 [By Pythagoras Theorem] ⇒ 4 CE2 = BC2 +  1 AB 2 108 − 27 ⇒  2  [ BE=AE] 4 AD2 = CE2 = BC2 + AB2 ...(ii) 4 ⇒ AD2 = 81 ⇒ 4 Adding (i) and (ii), we get ⇒ 9 AD2 + CE2 = AB2 + BC2 +BC2 + AB2 AD = = 4.5 cm 44 2 Altitude = 4.5 cm 5 4 7. (a) 10 m AD2 + CE2 = (AB2 +BC2 )  ABC is a right triangle, right angled at B. AD2 + CE2 = 5 AC2 4 [ By Pythagoras Theorem in DABC, AC2 = AB2 + BC2] 3 5  2  2  + CE2 = 5 × 25 4 86 Mathematics–10

CE2 = 125 − 45 4 4 CE2 = 20 \\ CE = 20 = 2 5 cm 10. (c) 90° B. Assertion-Reason Type Questions PR2= 2PQ2[Given] 1. (d) Assertion (A) is false but reason (R) is true. = PQ2 + PQ2 2. (b) Both assertion (A) and reason (R) are true ⇒ PR2 = PQ2 + QR2[ PQ = QR] but reason (R) is not the correct explanation ⇒ DPQR is right angled at Q of assertion (A). [Using converse of Pythagoras Theorem] \\ ∠Q is a right angle. Case Study Based Questions I. Rahul is studying in X Standard. He is making a kite to fly it on a Sunday. Few questions came to his mind while making the kite. Give answers to his questions by looking at the figure. 1. Rahul tied the sticks at what angles to each other? (a) 30° (b) 45° (c) 90° (d) 60° 2. Which is the correct similarity criteria applicable for smaller triangles at the upper part of this kite? (a) RHS (b) SAS (c) SSA (d) AAS 3. Sides of two similar triangles are in the ratio 4:9. Corresponding medians of these triangles are in the ratio (b) 4:9 (c) 81:16 (d) 16:81 (a) 2:3 4. In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. This theorem is called as, (a) Pythagoras theorem (b) Thales theorem (c) Converse of Thales theorem (d) Converse of Pythagoras theorem 5. What is the area of the kite, formed by two perpendicular sticks of length 6 cm and 8 cm? (a) 48 cm2 (b) 14 cm2 (c) 24 cm2 (d) 96 cm2 A ns. 1. (c) 90° 2. (b) SAS 3. (b) 4 : 9 4. (d) Converse of Pythagoras theorem 5. (a) 48 cm2 Triangles 87

II. For going to city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which is directly connect the two cities A and B. In solving such problem authority of the cities have some questions in his mind. Give answer to his following questions: City C 2x 2(x + 7) km City A 26 km City B 1. Which theorem will you use to solve this problem? (a) Pythagoras theorem (b) Basic proportionality theorem (c) Factor theorem (d) Fundamental Theorem of Arithmetic 2. What is the distance of AC? (a) 15 km (b) 18 km (c) 10 km (d) 20 km 3. What is the distance of BC? (a) 12 km (b) 10 km (c) 20 km (d) 24 km 4. Find how much distance will be saved in reaching city B from city A after the construction of the highway. (a) 6 km (b) 8 km (c) 10 km (d) 12 km 5. If ∠A is supposed to be 30°, then ∠B = ? (a) 30° (b) 45° (c) 60° (d) 90° Ans. 1. (a) Pythagoras theorem 2. (c) 10 km 3. (d) 24 km 4. (b) 8 km 5. (c) 60° III. Ravi is a student of Xth standard. He is drawing a square on a sheet of paper considering each side of the square as 8 cm. Now, he is drawing two equilateral triangles, one by taking side of the square as one of its side and another by taking diagonal of square as sides of another equilateral triangle. Later, few questions came to his mind and he wants to solve them. Give answers to his questions by drawing your own figure. E A 8 cm B 1. The diagonal of the square is F (a) 8 2 cm (b) 8 3 cm (c) 8 5 cm (d) 8 6 cm 88 Mathematics–10

2. The area of the square is (a) 64 sq. cm (b) 81 sq. cm (c) 121 sq. cm (d) 49 sq. cm 3. The area of equilateral triangle by taking side of square as side of the triangle is (a) 8 3 sq. cm (b) 16 3 sq. cm (c) 32 3 sq. cm (d) 24 3 sq. cm 4. The area of equilateral triangle by taking diagonal of square as the side of the equilateral triangle is (a) 8 3 sq. cm (b) 16 3 sq. cm (c) 32 3 sq. cm (d) 64 3 sq. cm 5. Which is true order in area? (a) Equilateral triangle on side < Square < Equilateral triangle with diagonal (b) Equilateral triangle on side > Square > Equilateral triangle on diagonal (c) Equilateral triangle on diagonal > Equilateral triangle on side > Square (d) Square > Equilateral triangle on diagonal > Equilateral triangle on side A ns. 1. (a) 8 2 cm 2. (a) 64 sq. cm 3. (b) 16 3 sq. cm 4. (c) 32 3 sq. cm 5. (d) Square > Equilateral triangle on diagonal > Equilateral triangle on side Experts’ Opinion Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly. 1. To apply Basic Proportionality Theorem and its converse in triangles and quadrilaterals. 2. To apply criteria for similar triangles for solving problems. 3. Use the theorem “The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.” 4. To use Pythagoras theorem and its converse. QUICK REVISION NOTES •• Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal, (ii) their corresponding sides are in the same ratio. •• Similar Triangles: Two triangles are said to be similar if, (i) their corresponding angles are equal, (ii) their corresponding sides are proportional. •• Basic Proportionality Theorem: In a triangle, a line drawn parallel to one side to intersect the other two sides in distinct points, divides the two sides in the same ratio. •• Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side. •• AAA Similarity: If in two triangles, corresponding angles are equal or the two corresponding angles are equal, then the triangles are similar. •• SSS Similarity: If the corresponding sides of two triangles are proportional, then they are similar. •• SAS Similarity: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar. •• The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. •• Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. •• Converse of Pythagoras Theorem: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. Triangles 89

COMMON ERRORS Errors Corrections (i) Writing incorrectly the similarity of two triangles (i) Similarity of two triangles should be expressed in symbolic form. symbolically, using correct correspondence of their vertices. For example, in triangle ABC and DEF if A corresponds to D, B corresponds to E and C corresponds to F, we write similarity of these two triangles as DABC ∼ DDEF. (ii) Applying incorrect similarity criterion for (ii) Read and understand questions properly and then deciding the similarity of two triangles. use correct similarity criterion for deciding the similarity of two triangles. (iii) Incorrectly interpreting that similar figures are (iii) Remember that all the congruent figures are also congruent. similar but the converse is not true. (iv) Incorrectly taking that similar figures are (iv) Congruent figures have equal areas but not for congruent in areas. similar figures. (v) Taking incorrect relation A (v) In a right-angled triangle, the square of hypotenuse such as a2 + b2 = c2 while proving a triangle of given c is equal to the sum of the squares of the other two sides to be a right triangle. b sides. b2 = a2 + c2 B a Thus, C 90 Mathematics–10

6 Introduction to Trigonometry Topics Covered 2. Trigonometric Identities 1. Trigonometric Ratios 1. Trigonometric Ratios Understanding Trigonometric Ratios • Let ABC be a right triangle in which ∠A = 90°, side adjacent to ∠B = AB, side opposite to ∠B = AC and hypotenuse = BC. With reference to angle B, we define the following ratios known as trigonometric ratios. C (i) sine of ∠B = sin B = Side opposite to ∠B = AC Side opposite to ∠B Hypotenuse BC (ii) cosine of ∠B = cos B = Side adjacent to ∠B = AB Hypotenuse Hypotenuse BC (iii) tangent of ∠B = tan B = Side opposite to ∠B = AC Side adjacent to ∠B AB (iv) cosecant of ∠B = cosec B = 1 = Hypotenuse ∠B = BC AB sin B Side opposite to AC Side adjacent to ∠B (v) secant of ∠B = sec B = 1 B = Hypotenuse ∠B = BC cos Side adjacent to AB (vi) cotangent of ∠B = cot B = 1 B = Side adjacent to ∠B = AB tan Side opposite to ∠B AC (vii) tan B = sin B (viii) cot B = cos B cos B sin B The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same. • If one trigonometric ratio of an angle is given, the other trigonometric ratios of the angle can be determined. 4 Example 1. If cos A = 5 , then sin A and sec A respectively are 34 53 B (a) , (b) , 43 34 35 5k 3k (c) , (d) None of these 54 4 Solution. Given: cos A = 5 Since, 4 CA A 4k C cos A = 5 = AB \\ CA = 4 k, AB = 5 k, where k is a positive number. 91

In right-angled triangle ACB, BC = 25k 2 − 16k 2 = 9k 2 = 3k So BC = 3k = 3 1 5 sin A = AB 5k 5 , sec A = = 4 cos A Hence, option (c) is the correct answer. Example 2. If 4 tan q = 3, then the value of 4sin θ − cos θ is 4sin θ + cos θ 11 1 1 (a) 2 (b) 3 (c) 4 (d) 5 Solution.  3 AB Let 4 tan q = 3 ﬁ tan q = 4 ﬁ tan q = BC AB = 3k, BC = 4k A Then, AC = 9k 2 + 16k 2 = 5k (using Pythagoras theorem) 5k AB 3k = 3 and cos q = BC = 4k 4 3k θ \\ sin q = AC = 5k 5 AC 5k =5 B 4k C 34 12 − 4 4× − 4sinθ − cos θ 5 5 5 8 5 8 1 \\ 4sin θ + cos θ = 3 4 = 12 + 4 = 5 × = = 2 4 × + 16 16 55 5 Hence, option (a) is the correct answer. Trigonometric Ratios of Some Specific Angles C • Trigonometric Ratios of 45°: 11 (i) sin 45° = (ii) cos 45° = 22 (iii) tan 45° = 1 (iv) cosec 45° = 2 (v) sec 45° = 2 (vi) cot 45° = 1 • Trigonometric Ratios of 30° and 60°: 45° 1 3A B (i) sin 30° = 2 (ii) cos 30° = 2 A 30° 1 (iii) tan 30° = (iv) cosec 30° = 2 3 2 (v) sec 30° = (vi) cot 30° = 3 3 31 (vii) sin 60° = 2 (viii) cos 60° = 2 2 60° C (ix) tan 60° = 3 (x) cosec 60° = B 3 (xi) sec 60° = 2 1 (xii) cot 60° = 3 92 Mathematics–10

• The value of sin q or cos q never exceeds 1, whereas the value of sec q or cosec q is always greater than or equal to 1. Example 3. The value of (sin 30° + cos 60°) (a) 1 (b) 2 (c) 0 (d) None of these Solution. sin 30° + cos 60° = 1 + 1 = 1+1 = 2 =1 2 2 2 2 Hence, option (a) is the correct answer. Example 4. The value of 5cos2 60° + 4sec2 30° − tan2 45° is sin2 30° + cos2 30° 32 14 67 19 (a) 35 (b) 55 (c) 12 (d) 33 Solution. 5cos2 60° + 4sec2 30° − tan2 45° sin2 30° + cos2 30° 5  1 2 + 4 ×  2 2 − (1)2 5 16 15 + 64 − 12 2  3 4 + −1 12   3 4 67 67 =  = 13 = = 12 = 12  1 2 +  3 2 + 2  2  44 41  Hence, option (c) is the correct answer. Example 5. The value of 3sin 30° + 4 cos2 45° − cot2 30° is cos2 30° + sin2 30° (a) 1 (b) 1 (c) 2 (d) 3 23 5 8 1  1 2 2  2  3sin 30° + 4 cos2 45° − cot2 30° 3 × + 4 × − ( 3)2 34 cos2 30° + sin2 30°  + −3  2 2 Solution. = 32 +  12 2 = 3 1 2  + 44 3+4–6 2 1 Hence, option (a) is the correct answer. = 4 = 2 4 33 Example 6. Given sin (A – B) = 2 and cos (A + B) = 2 . Then A and B respectively are (a) 30°, 45° (b) 45°, –15° (c) 60°, 45° (d) None of these 33 Solution. Since, sin (A – B) = 2 and cos (A + B) = 2 ⇒ sin (A – B) = sin 60° and cos (A + B) = cos 30° Introduction to Trigonometry 93

⇒ A – B = 60° ...(i) and A + B = 30° ...(ii) Solving (i) and (ii), we get 2A = 90° ⇒ A = 45°, B = – 15° Hence, option (b) is the correct answer. Exercise 6.1 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. The value of sin 60° ⋅ cos 30° + sin 30° ⋅ cos 60° is (a) 0 (b) 1 (c) 2 (d) 8 (d) 9 2. Value of cos 0° ⋅ cos 30° ⋅ cos 45° ⋅ cos 60° ⋅ cos 90° is (a) 0 (b) 1 (c) 2 3. The value of  sin 2 1  =  θ = 1 + tan2 θ  (a) 0 (b) 1 (c) 2 (d) 5 (d) None of these 4. The value of (1 + tan2 q)(1 – sin q)(1 + sin q) = (a) 0 (b) 1 (c) 2 1 5. If tan(A + B) = 3 and tan(A – B) = , A > B, then the value of A is 3 (a) A = 30° (b) A = 60° (c) A = 90° (d) A = 45° 6. The value of sin 60° cos 30° + sin 30° cos 60° is (a) 1 (b) 2 (c) 11 (d) 0 7. 2 tan2 45° + cos2 30° – sin2 60° equals (a) 1 (b) 2 (c) 5 (d) 6 8. If sin q = x and sec q = y, then the value of cot q is (a) xy (b) 2xy (c) 1 (d) x + y xy 3 9. If (1 + cos A)(1 – cos A) = , the value of sec A is 4 (a) 2 (b) –2 (c) ±2 (d) 0 10. If 15 cot A = 8, then the value of cosec A is (a) 15 (b) 13 (c) 4 (d) 17 12 15 15 15 11. Evaluate: 4 sin2 60° + 3 tan2 30° – 8 sin 45° cos 45° (a) 0 (b) 1 (c) 2 (d) 5 12. Evaluate: sin 30° + tan 45° − cosec 60° . sec 30° + cos 60° + cot 45° (a) 3 3 + 2 (b) 3 3 − 4 (c) 3 3 + 8 (d) None of these 3 3−2 3 3+4 3 3−9 94 Mathematics–10

13. The value of cos 30° + sin 60° is 1 + cos 60° + sin 30° (a) 3 (b) 2 1 (d) 0 (c) 2 32 14. ABC is a triangle right angled at C and AC = 3 BC. Then ∠ABC = (a) 30° (b) 60° (c) 90° (d) 0° B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 3 1. Assertion (A): In a right-angled triangle, if tan q = , the greatest side of the triangle is 5 units. 4 Reason (R): (Greatest side)2 = (Hypotenuse)2 = (Perpendicular)2 + (Base)2. 13 2. Assertion (A): In a right-angled triangle, if cos q = 2 and sin q = 2 , then tan q = 3 . Reason (R): sin θ cos θ Answers and Hints A. Multiple Choice Questions (MCQs) ⇒ sec A = ±2 10. (d) 1. (b) 1 2. (a) 0 17 3. (b) 1 4. (b) 1 11. (a) 0 cosec A = 5. (d) A = 45° 6. (a) 1 7. (b) 2 15 1 3 3–4 8. (c) xy 12. (b) 3 3 +4 Given sin q = x and sec q = y 3 13. (a) 2 1 14. (b) 60° ⇒ cos q = y Here, in DABC, ∠C = 90° and AC = 3 BC cos θ 1 [given] Now, cot q = sin θ = xy ⇒ AC 1 = 3 ...(i) Hence, cot q = xy 9. (c) ±2 BC 3 AC (1 + cos A)(1 – cos A) = 4 Also, tan B = BC ⇒ 1 – cos2 A = 3 ⇒ 1 – 3 = cos2 A ⇒ tan B = 3 [using (i)] ⇒ B = 60° 44 1 = cos2 A ⇒ sec2 A = 4 ⇒ 4 Introduction to Trigonometry 95

B. Assertion-Reason Type Questions 2. (a) Both assertion (A) and reason (R) are true 1. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of and reason (R) is the correct explanation of assertion (A). assertion (A). 2. Trigonometric Identities An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved. Some trigonometric ratios are listed below: (i) sin2 A + cos2 A= 1, 0° £ A £ 90° (ii) 1 + tan2 A = sec2 A, 0° £ A < 90° (iii) cot2 A + 1 = cosec2 A, 0° < A £ 90° Example 1. The value of sin6 q + cos6 q + 3sin2 q cos2 q is (a) 0 (b) 1 (c) 2 (d) 1 4 Solution. sin6 q + cos6 q + 3sin2 q cos2 q = (sin2 q + cos2 q)3 – 3 sin4 q cos2 q – 3 sin2 q cos4 q + 3 sin2 q cos2 q = 1 – 3 sin2 q cos2 q (sin2 q + cos2 q) + 3 sin2 q cos2 q = 1 – 3 sin2 q cos2 q + 3 sin2 q cos2 q = 1 Hence, option (b) is the correct answer. Example 2. The value of (sin4q – cos4q + 1) cosec2q is (a) 0 (b) 1 (c) 2 (d) 5 Solution. (sin4 q – cos4 q + 1) cosec2 q = {(sin2 q + cos2 q) (sin2 q – cos2 q) + 1} cosec2 q = (sin2 q – cos2 q + 1) cosec2 q = (sin2 q + sin2 q) cosec2 q = (2 sin2 q) cosec2 q = 2 sin 2 θ × 1 =2 sin 2 θ Hence, option (c) is the correct answer. Example 3. If tan A = n tan B and sin A = m sin B, then cos2 A = (a) m2 − 1 (b) m2 + 1 (c) 1− m2 (d) 1+ m2 n2 −1 n2 +1 1+ m2 1− m2 Solution. tan A = n tan B ⇒ tan B = 1 tan A ⇒ cot B = n n tan A and sin A = m sin B ⇒ sin B = 1 sin A ⇒ cosec B = m m sin A Now, cosec2 B – cot2 B = 1 ⇒ m2 − n2 = 1 ⇒ m2 − n2 cos2 A = 1 ⇒ m2 − n2 cos2 A =1 sin2 A tan 2 sin2 A sin2 A sin 2 A A ⇒ m2 – n2 cos2 A = sin2 A ⇒ m2 – 1 = n2 cos2 A – cos2 A ⇒ m2 – 1 = (n2 – 1) cos2 A ⇒ cos2 A = m2 − 1 n2 − 1 Hence, option (a) is the correct answer. 96 Mathematics–10

Exercise 6.2 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. If 4 tan q = 3, then  4 sin θ − cos θ  is equal to  4 sin θ + cos θ  21 1 3 (a) 3 (b) 3 (c) 2 (d) 4 2. If sin q – cos q = 0, then the value of (sin4 q + cos4 q) is 311 (a) 1 (b) (c) (d) 4 2 4 3. If 2 sin θ = 3 , then θ = (a) 30° (b) 60° (c) 45° (d) 90° 4. The value of (1 + tan2 q)(1 – sin q)(1 + sin q) is (a) 0 (b) 1 (c) 8 (d) 17 5. The value of cot2 q – 1 θ is sin 2 (a) 0 (b) –1 (c) 2 (d) –8 6. If cosec q + cot q = x, the value of cosec q – cot q is (a) x (b) 2x (c) x (d) 1 2x 7. The magnitude of q in the equation cos2 θ = 3 is cot2 θ − cos2 θ (a) 0° (b) 30° (c) 60° (d) 90° (d) 4 8. The value of (1 + cot q – cosec q) (1 + tan q + sec q) is equal to (a) 1 (b) 2 (c) 3 9. If 7 sin2A + 3 cos2A = 4, then tan A = (a) 1 (b) 1 (c) 1 (d) 1 23 23 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): sin2 67° + cos2 67° = 1. Reason (R): For any value of q, sin2 q + cos2 q = 1. 2. Assertion (A): The value of sec2 10° – cot2 80° is 1. 1 Reason (R): The value of sin 30° = 2 . Introduction to Trigonometry 97

Answers and Hints A. Multiple Choice Questions (MCQs) 1 ⇒ cot2θ = 3 1 1 1. (c) 2. (c) ⇒ tan2q = 3 2 2 ⇒ tan q = 3 ⇒ q = 60° 3. (b) 60° 4. (b) 1 5. (b) – 1 8. (b) 2 cot2 θ – 1 = cot2 θ – cosec2 θ 1 sin2 θ 9. (d) 3 = – 1 Given, 7sin2A + 3cos2A = 4 Dividing both sides by cos2A, we get 6. (d) 1 cosec q + cot q = x 7 tan2A + 3 = 4 sec2A [Q sec2q = 1 + tan2q] x cosec2 q – cot2 q = 1 ⇒ 7 tan2A + 3 = 4(1 + tan2A) As we know that ⇒ 7 tan2A + 3 = 4 + 4 tan2A ⇒ 3tan2A = 1 ⇒ (cosec q – cot q)(cosec q + cot q) = 1 ⇒ tan2A = 1 ⇒ (cosec q – cot q)x = 1 3 ⇒ cosec q – cot q = 1 1 7. (c) 60° x ⇒ tan A = 3 cos2 θ = 3 cot2 θ − cos2 θ B. Assertion-Reason Type Questions ⇒ cos2 θ = 3 1. (a) Both assertion (A) and reason (R) are true  1 1 cos2 θ  sin 2 θ − and reason (R) is the correct explanation of assertion (A). ⇒ 1 2. (b) Both assertion (A) and reason (R) are true cosec2θ −1 = 3 but reason (R) is not the correct explanation of assertion (A). Case Study Based Questions I. Children were playing a game by making some right angled triangles on the C 13 cm A plane sheet of paper. They assumed angles and their corresponding sides. 12 cm B They wanted to find all trigonometric ratio so they took a right angled triangle with two of its sides AC = 13 cm, BC = 12 cm and –ABC = 90º. They are unable to find trigonometric ratio. So help them to do so. 1. Using the above data, the value of sin A is 12 13 5 13 (a) 13 (b) 12 (c) 13 (d) 5 2. Using the above data, the value of sin C is 12 13 5 13 (a) 13 (b) 12 (c) (d) 13 5 98 Mathematics–10

3. Using the above data, the value of tan C is 12 12 13 5 (a) (b) (c) (d) 5 13 12 12 4. Using the above data, the value of cos A is 12 13 5 12 (a) 5 (b) 5 (c) 13 (d) 13 5. Using the above data, the value of tan A + sin A is tan A − sin A 9 4 12 9 (a) (b) (c) (d) 4 9 9 13 12 5 5 5 9 Ans. 1. (a) 13 2. (c) 13 3. (d) 12 4. (c) 5. (a) 13 4 II. Some students of class-X of a school were asked to A show trigonometric ratio of 60° on the top of the table using some sticks. The students were excited to do this 2a 2a case. So, they took 4 sticks. Three of them were equal and the 4th stick was smaller than the three. They put the sticks as shown in the given figure ∆ABC and height AD. Q AB = BC = CA = 2a (say) 60º C AB2 = AD2 + BD2 BaDb (2a)2 = AD2 1 2 2a  2 + BC 4a2 = AD2 + 1 2  2 2a 4a2 = AD2 + a2 ∴ AD2 = 3a2 ∴ AD = 3 a On the basis of above information, students answered their teacher’s questions, which are as follows: 1. What is the value of sin 60°? (a) 3 (b) 2 (c) 1 (d) 1 2 32 3 2. The value of cos 60° is (a) 1 (b) 3 (c) 1 (d) 1 3 22 Introduction to Trigonometry 99

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