NCERT Solutions For Class 7th Maths http://freehomedelivery.net/CH  1 Integers CBSE Answers PDF Download 2018 2019 NCERT Solutions For Class 7th Maths http://freehomedelivery.net/CH  1 Integers CBSE Answers PDF Download 2018 2019 Class 7th Maths CH  1 Integers CBSE Answers NCERT Solutions For PDF Download 2018 2019 Exercise 1.1 Question 1: Following number line shows the temperature in degree celsius (°C) at different places on a particular day.
(a) Observe this number line and write the temperature of the places marked on it. (b) What is the temperature difference between the hottest and the coldest places among the above? (c) What is the temperature difference between Lahulspiti and Srinagar? (d) Can we say temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar? Answer: (a) By observing the given data, the temperatures of these cities are as follows. Lahulspiti : −8°C Srinagar : −2°C Shimla : 5°C Ooty : 14°C Bangalore : 22°C (b) Temperature at the hottest place, i.e., Bangalore = 22°C Temperature at the coldest place, i.e., Lahulspiti = −8°C Temperature difference = 22°C − (−8°C) = 30°C Therefore, the temperature difference between the hottest and the coldest places is 30ºC. (c) Temperature at Lahulspiti = −8°C Temperature at Srinagar = −2°C Temperature difference = −2°C − (−8°C) = 6°C Therefore, the temperature difference between Lahulspiti and Srinagar is 6ºC. (d) Temperature at Srinagar = −2°C Temperature at Shimla = 5°C Temperature of Srinagar and Shimla taken together = −2°C + 5°C = 3°C 3°C < 5°C 3°C < Temperature of Shimla Yes, the temperature of Srinagar and Shimla taken together is less than the temperature of Shimla. However, 3°C > −2°C Hence, the temperature of Srinagar and Shimla taken together is not less than the temperature of Srinagar. Question 2: In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If jack’s scores in five successive rounds were 25, − 5, − 10, 15 and 10, what was his total at the end? Answer: Jack’s scores in five successive rounds are 25, −5, −10, 15, and 10. Total score of Jack at the end will be the sum of these scores. Therefore, Jack’s total score at the end = 25 − 5 − 10 + 15 + 10 = 35 Question 3: At Srinagar temperature was − 5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day? Answer:
Temperature on Monday = −5°C Temperature on Tuesday = Temperature on Monday −2°C = −5°C − 2°C = −7°C Temperature on Wednesday = Temperature on Tuesday + 4°C = −7°C + 4°C = −3°C Therefore, the temperature on Tuesday and Wednesday was −7ºC and −3ºC respectively. Question 4: A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?
Answer: Height of plane = 5000 m Depth of submarine = −1200 m Distance between plane and submarine = 5000 m − (−1200) m = 5000 m + 1200 m = 6200 m Question 5: Mohan deposits Rs 2,000 in his bank account and withdraws Rs 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal. Answer: Since the amount withdrawn is represented by a negative integer, the amount deposited will be represented by a positive integer. Amount deposited = Rs 2000 Amount withdrawn = −Rs 1642 Balance in Mohan’s account = Money deposited + Money withdrawn = 2000 + (−1642) = 2000 − 1642 = 358 Therefore, balance in Mohan’s account after withdrawal is Rs 358. Question 6: Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A?
Answer: Since the distance towards east is represented by a positive integer, the distance travelled towards west will be represented by a negative integer. Distance travelled in east direction = 20 km Distance travelled in west direction = −30 km Distance travelled from A = 20 + (−30) = −10 km Therefore, we will represent the distance travelled by Rita from point A by a negative integer, i.e., −10 km (i.e., Rita is now in west direction). Question 7: In a magic square each row, column and diagonal have the same sum. Check which of the following is a magic square.
Answer: It can be observed that in square (i), every row and column add up to give 0. However, the sum of one of its diagonals is not 0. As − 4 − 2 = −6 ≠ 0, Therefore, (i) is not a magic square. Similarly, in square (ii), each row, column, and diagonal add up to give −9. Therefore, (ii) is a magic square. Question 8: Verify a − (−b) = a + b for the following values of a and b. (i) a = 21, b = 18 (ii) a = 118, b = 125 (iii) a = 75, b = 84 (iv) a = 28, b = 11 Answer: (i) a = 21, b = 18 a − (−b) = 21 − (−18) = 21 + 18 = 39 a + b = 21 + 18 = 39 ∴a − (−b) = a + b = 39 (ii) a = 118, b = 125 a − (−b) = 118 − (−125) = 118 + 125 = 243 a + b = 118 + 125 = 243 ∴a − (−b) = a + b = 243 (iii) a = 75, b = 84 a − (−b) = 75 − (−84) = 75 + 84 = 159 a + b = 75 + 84 = 159 ∴ a − (−b) = a + b = 159 (iv) a = 28, b = 11 a − (−b) = 28 − (−11) = 28 + 11 = 39 a + b = 28 + 11 = 39 ∴ a − (−b) = a + b = 39 Question 9: Use the sign of >, < or = in the box to make the statements true.
Question 10: A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.
(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?
(ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step? (iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following; (a) − 3 + 2 − … = − 8 (b) 4 − 2 + … = 8. In (a) the sum (− 8) represents going down by eight steps. So, what will the sum 8 in (b) represent? Answer: Let the steps moved down be represented by positive integers and the steps moved up be represented by negative integers. (i) Initially, the monkey was at step = 1 After 1st jump, the monkey will be at step = 1 + 3 = 4 After 2nd jump, the monkey will be at step = 4 + (−2) = 2 After 3rd jump, the monkey will be at step = 2 + 3 = 5 After 4th jump, the monkey will be at step = 5 + (−2) = 3 After 5th jump, the monkey will be at step = 3 + 3 = 6 After 6th jump, the monkey will be at step = 6 + (−2) = 4 After 7th jump, the monkey will be at step = 4 + 3 = 7 After 8th jump, the monkey will be at step = 7 + (−2) = 5 After 9th jump, the monkey will be at step = 5 + 3 = 8 After 10th jump, the monkey will be at step = 8 + (−2) = 6 After 11th jump, the monkey will be at step = 6 + 3 = 9 Clearly, the monkey will be at water level (i.e., 9th step) after 11 jumps. (ii) Initially, the monkey was at step = 9 After 1st jump, the monkey will be at step = 9 + (−4) = 5 After 2nd jump, the monkey will be at step = 5 + 2 = 7 After 3rd jump, the monkey will be at step = 7 + (− 4) = 3 After 4th jump, the monkey will be at step = 3 + 2 = 5 After 5th jump, the monkey will be at step = 5 + (− 4) = 1 Clearly, the monkey will reach back at the top step after 5 jumps. (iii) If steps moved down are represented by negative integers and steps moved up are represented by positive integers, then his moves will be as follows. Moves in part (i) − 3 + 2 − 3 + 2 − 3 + 2 − 3 + 2 − 3 + 2 − 3 = −8 Moves in part (ii) 4−2+4−2+4=8 Moves in part (ii) represent going up 8 steps. Exercise 1.2 Question 1: Write down a pair of integers whose: (a) sum is − 7 (b) difference is − 10 (c) sum is 0 Answer: (a) − 8 + (+1) = −7 (b) − 12 − (−2) = −10 (c) 5 + (−5) = 0 Question 2: (a) Write a pair of negative integers whose difference gives 8. (b) Write a negative integer and a positive integer whose sum is − 5. (c) Write a negative integer and a positive integer whose difference is − 3. Answer: (a) −2 − (−10) = 8 (b) −8 + 3 = −5 (c) −2 − (+1) = −3
Question 3: In a quiz, team A scored − 40, 10, 0 and team B scored 10, 0 − 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order? Answer: Team A scored − 40, 10, 0. Total score = − 40 + 10 + 0 = −30 Team B scored 10, 0, −40. Total score = 10 + 0 + (−40) = −30 ∴ The scores of both teams are equal. Yes, we can add integers in any order. We had observed that the scores obtained by both teams in successive rounds were numerically equal but different in order. Yet, the total score of both teams were equal. Question 4: Fill in the blanks to make the following statements true: (i) (− 5) + (− 8) = (− 8) + (…) (ii) − 53 + … = − 53 (iii) 17 + … = 0 (iv) [13 + (− 12)] + (…) = 13 + [(− 12) + (− 7)] (v) (− 4) + [15 + (− 3)] = [(− 4) + 15] + Answer:
Exercise 1.3 Question 1: Find each of the following products: (a) 3 × (− 1) (b) (− 1) × 225 (c) (− 21) × (− 30) (d) (− 316) × (− 1) (e) (− 15) × 0 × (−18) (f) (− 12) × (− 11) × (10) (g) 9 × (− 3) × (− 6) (h) (− 18) × (− 5) × (− 4) (i) (− 1) × (−2) × (− 3) × 4 (j) (− 3) × (− 6) × (− 2) × (− 1) Answer: (a) 3 × (−1) = −3 (b) (−1) × 225 = −225 (c) (−21) × (−30) = 630 (d) (−316) × (−1) = 316 (e) (−15) × 0 × (−18) = 0 (f) (−12) × (−11) × 10 = 1320 (g) 9 × (−3) × (−6) = 162 (h) (−18) × (−5) × (−4) = −360 (i) (−1) × (−2) × (−3) × 4 = −24 (j) (−3) × (−6) × (−2) × (−1) = 36 Question 2: Verify the following:
Answer: (a) L.H.S. = 18 × [7 + (− 3)] = 18 × [7 − 3] = 18 × 4 = 72 R.H.S. = [18 × 7] + [18 × (− 3)] = 126 + (− 54) = 72
Hence, (b) L.H.S. = (−21) × [(−4) + (−6)] = (−21) × [− 4 − 6] = (−21) × [−10] = 210 R.H.S. = [(−21) × (−4)] + [(−21) × (−6)] = 84 + 126 = 210
Hence, Question 3: (i) For any integer a, what is (− 1) × a equal to? (ii) Determine the integer whose product with (− 1) is (a) − 22 (b) 37 (c) 0 Answer: (i) (−1) × a = −a
Question 4: Starting from (− 1) × 5, write various products showing some pattern to show (− 1) × (− 1) =1. Answer: −1 × 5 = −5 −1 × 4 = −4 = − 5 + 1 −1 × 3 = −3 = − 4 + 1 −1 × 2 = −2 = − 3 + 1 −1 × 1 = −1 = − 2 + 1 −1 × 0 = 0 = − 1 + 1 Therefore, −1 × (−1) = 0 + 1 = 1 Question 5: Find the product, using suitable properties: (a) 26 × (− 48) + (− 48) × (− 36) (b) 8 × 53 × (−125) (c) 15 × (− 25) × (− 4) × (− 10) (d) (− 41) × 102 (e) 625 × (− 35) + (− 625) × 65 (f) 7 × (50 − 2) (g) (− 17) × (− 29) (h) (− 57) × (− 19) + 57 Answer: (a) 26 × (−48) + (−48) × (−36) = (−48) × 26 + (−48) × (−36) (b × a = a × b) = (−48) [26 − 36] (a × b + a × c) = a (b + c) = (−48) × (−10) = 480 (b) 8 ×53 × (−125) = 8 × [53 × (−125)] = 8 × [(−125) × 53] (b × a = a × b) = [8 × (−125)] ×53 a × (b × c) = (a × b) × c = [−1000] × 53 = −53000 (c) 15 × (−25) × (−4) × (−10) = 15 × [(−25) × (−4)] × (−10)
= 15 × [100] × (−10) = 15 × (−1000) = −15000 (d) (−41) × 102 = (−41) × (100 + 2) = (−41) × 100 + (−41) × 2 a × (b + c) = (a × b) + (a × c) = − 4100 − 82 = −4182 (e) 625 × (−35) + (−625) × 65 = 625 × [(−35) + (−65)] (a × b) + (a × c) = a × (b + c) = 625 × [−100] = −62500 (f) 7 × (50 − 2) = (7 × 50) − (7 × 2) a × (b − c) = (a × b) − (a × c) = 350 − 14 = 336 (g) (−17) × (−29) = (−17) × [−30 + 1] = [(−17) × (−30)] + [(−17) × 1] a × (b + c) = (a × b) + (a × c) = [510] + [−17] = 493 (h) (−57) × (−19) + 57 = 57 × 19 + 57 × 1 = 57 [19 + 1] (a × b) + (a × c) = a × (b + c) = 57 × 20 = 1140 Question 6: A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins? Answer: Initial temperature = 40°C Change in temperature per hour = −5°C Change in temperature after 10 hours = (−5) × 10 = −50°C Final temperature = 40ºC + (−50ºC) = −10°C Question 7: In a class test containing 10 questions, 5 marks are awarded for every correct answer and (− 2) marks are awarded for every incorrect answer and 0 for questions not attempted. (i) Mohan gets four correct and six incorrect answers. What is his score? (ii) Reshma gets five correct answers and five incorrect answers, what is her score? (iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score? Answer: (i) Marks given for 1 correct answer = 5 Marks given for 4 correct answers = 5 × 4 = 20 Marks given for 1 wrong answer = −2 Marks given for 6 wrong answers = −2 × 6 = −12 Score obtained by Mohan = 20 − 12 = 8 (ii) Marks given for 1 correct answer = 5 Marks given for 5 correct answers = 5 × 5 = 25 Marks given for 1 wrong answer = −2 Marks given for 5 wrong answers = −2 × 5 = −10 Score obtained by Reshma = 25 − 10 = 15 (iii) Similarly, Marks given for 2 correct answers = 5 × 2 = 10 Marks given for 5 wrong answers = −2 × 5 = −10 Score obtained by Heena = 10 − 10 = 0 Question 8: A cement company earns a profit of Rs 8 per bag of white cement sold and a loss of Rs 5 per bag of grey cement sold. (a) The company sells 3, 000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or
loss? (b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags. Answer: Profit is denoted by a positive integer and loss is denoted by a negative integer. (a) Profit earned while selling 1 bag of white cement = Rs 8 Profit earned while selling 3000 bags of white cement = 8 × 3000 = 24000 Loss incurred while selling 1 bag of grey cement = −Rs 5 Loss incurred while selling 5000 bags of grey cement = −5 × 5000 = −25000 Total profit/loss earned = Profit + Loss = 24000 + (−25000) = −1000 Therefore, a loss of Rs 1000 will be incurred by the company. (b) Loss incurred while selling 1 bag of grey cement = −Rs 5 Loss incurred while selling 6400 bags of grey cement = (−5) × 6400 = −32000 Let the number of bags of white cement to be sold be x. Profit earned while selling 1 bag of white cement = Rs 8 Profit earned while selling x bags of white cement = x × 8 = 8x In condition of no profit no loss, Profit earned + Loss incurred = 0 8x + (−32000) = 0 8x = 32000 x = 4000 Therefore, 4000 bags of white cement must be sold. Question 9: Replace the blank with an integer to make it a true statement.
Exercise 1.4 Question 1: Evaluate each of the following:
Answer: (a) (−30) ÷ 10 = −3 (b) 50 ÷ (−5) = −10 (c) (−36) ÷ (−9) = 4 (d) (−49) ÷ 49 = −1 (e) 13 ÷ [−2 + 1] = 13 ÷ [−1] = −13 (f) 0 ÷ (−12) = 0
(g) (−31) ÷ [(−30) + (−1)] = (−31) ÷ (−31) = 1 (h) [(−36) ÷ 12] ÷ 3 = [−3] ÷ 3 = −1 (i) [− 6 + 5] ÷ [− 2 + 1] = (−1) ÷ (−1) = 1 Question 2: Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c. (a) a = 12, b = −4, c = 2 (b) a = (− 10), b = 1, c = 1 Answer: (a) a = 12, b = −4, c = 2 a ÷ (b + c) = 12 ÷ (− 4 + 2) = 12 ÷ (−2) = −6 (a ÷ b) + (a ÷ c) = [12 ÷ (−4)] + [12 ÷ 2] = −3 + 6 = 3 Hence, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) (b) a = −10, b = 1, c = 1 a ÷ (b + c) = (−10) ÷ (1 + 1) = (−10) ÷ 2 = −5 (a ÷ b) + (a ÷ c) = [(−10) ÷ 1] + [(−10) ÷ 1] = − 10 − 10 = −20 Hence, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) Question 3: Fill in the blanks:
Question 4: Write five pairs of integers (a, b) such a ÷ b = − 3. One such pair is (6, − 2) because 6 ÷ (− 2) = (− 3). Answer: (i) (3, −1) Because 3 ÷ (−1) = −3 (ii) (−3, 1) Because (−3) ÷ 1 = −3 (iii) (9, −3) Because 9 ÷ (−3) = −3 (iv) (−9, 3) Because (−9) ÷ 3 = −3 (v) (12, −4) Because 12 ÷ (−4) = −3 Question 5: The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at midnight? Answer: Initial temperature i.e., at 12 noon = 10°C Change in temperature per hour = −2°C Temperature at 1:00 PM = 10ºC + (−2ºC) = 8ºC Temperature at 2:00 PM = 8ºC + (−2ºC) = 6ºC Temperature at 3:00 PM = 6ºC + (−2ºC) = 4ºC Temperature at 4:00 PM = 4ºC + (−2ºC) = 2ºC Temperature at 5:00 PM = 2ºC + (−2ºC) = 0ºC Temperature at 6:00 PM = 0ºC + (−2ºC) = −2ºC Temperature at 7:00 PM = −2ºC + (−2ºC) = −4ºC Temperature at 8:00 PM = −4ºC + (−2ºC) = −6ºC Temperature at 9:00 PM = −6ºC + (−2ºC) = −8ºC Therefore, the temperature will be 8°C below zero at 9:00 PM. It will take 12 hours to be midnight (i.e., 12:00 AM) after 12:00 noon. Change in temperature in 12 hours = −2°C × 12 = −24ºC
At midnight, the temperature will be = 10 + (−24) = −14°C Therefore, the temperature at midnight will be 14ºC below 0. Question 6: In a class test (+ 3) marks are given for every correct answer and (−2) marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores − 5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly? (iii) Rakesh scores 18 marks by attempting 16 questions. How many questions has he attempted correctly and how many has he attempted incorrectly? Answer: Marks obtained for 1 right answer = +3 Marks obtained for 1 wrong answer = −2 (i) Marks scored by Radhika = 20 Marks obtained for 12 correct answers = 12 × 3 = 36 Marks obtained for incorrect answers = Total score − Marks obtained for 12 correct answers = 20 − 36 = −16 Marks obtained for 1 wrong answer = −2 Thus, number of incorrect answers = (−16) ÷(−2) = 8 Therefore, she attempted 8 questions wrongly. (ii) Marks scored by Mohini = −5 Marks obtained for 7 correct answers = 7 × 3 = 21 Marks obtained for incorrect answers = Total score − Marks obtained for 12 correct answers = − 5 − 21 = −26 Marks obtained for 1 wrong answer = −2 Thus, number of incorrect answers = (−26) ÷(−2) = 13 Therefore, she attempted 13 questions wrongly. (iii) Total marks scored by Rakesh = 18 Number of questions attempted = 16 (Number of correct answers)(3) + (Number of incorrect answers)(−2) = 18 ⇒(Number of correct answers)(3) + (16 − Number of correct answers)(−2) = 18 ⇒(Number of correct answers)(3) + −32 + 2(Number of correct answers) = 18 ⇒(Number of correct answers)(5) + −32 = 18 ⇒(Number of correct answers)(5) = 18 + 32 = 50 ⇒ Number of correct answers = 10 ∴Number of incorrect answers = 16 − 10 = 6 ∴Total number of correct and incorrect answers scored by Rakesh is 10 and 6 respectively. Question 7: An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach − 350 m. Answer: Distance descended is denoted by a negative integer. Initial height = +10 m Final depth = −350 m Total distance to be descended by the elevator = (−350) − (+10) = −360 m Time taken by the elevator to descend −6 m = 1 min Thus, time taken by the elevator to descend −360 m = (−360) ÷ (−6) = 60 minutes = 1 hour
NCERT Solutions For Class 7th Maths http://freehomedelivery.net/CH  3 Data Handling CBSE Answers PDF Download 2018 2019 NCERT Solutions For Class 7th Maths http://freehomedelivery.net/CH  3 Data Handling CBSE Answers PDF Download 2018 2019 Class 7th Maths CH  3 Data Handling CBSE Answers NCERT Solutions For PDF Download 2018 2019
Exercise 3.1
Question 1: Find the range of heights of any ten students of your class. Answer: Let the heights (in cm) of 10 students of our class be 125, 127, 132, 133, 134, 136,138, 141, 144, 146 Highest value among these observations = 146 Lowest value among these observations = 125 Range = Highest value − Lowest value = (146 − 125) cm = 21 cm
Question 2: Organise the following marks in a class assessment, in a tabular form. 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7 (i) Which number is the highest? (ii) Which number is the lowest? (iii) What is the range of the data? (iv) Find the arithmetic mean. Answer:
Marks
Tally marks
1

2

3

4

7

8

9

(i) Highest number = 9 (ii) Lowest number = 1 (iii) Range = (9 − 1) = 8 (iv) Sum of all the observations = 4 + 6 + 7 + 5 + 3 + 5 + 4 + 5 + 2 + 6 + 2 +5+1+9+6+5+8+4+6+7 = 100 Total number of observations = 20
Question 3: Find the mean of the first five whole numbers. Answer: First five whole numbers are 0, 1, 2, 3, and 4.
Therefore, the mean of first five whole numbers is 2. Question 4: A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100 Find the mean score. Answer: Runs scored by the cricketer are 58, 76, 40, 35, 46, 45, 0, and 100.
Therefore, mean score is 50.
Question 5: Following table shows the points of each player scored in four games:
Player
Game 1
Game 2
Ga
A
14
16
10
B
0
8
6
C
8
11
Now answer the following questions: (i) Find the mean to determine A’s average number of points scored per game. (ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why? (iii) B played in all the four games. How would you find the mean? (iv) Who is the best performer? Answer:
(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played 3 games.
(iv) The best performer will have the greatest average among all. Now we can observe that the average of A is 12.5 which is more than that of B and C. Therefore, A is the best performer among these three.
Question 6: The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Find the: (i) Highest and the lowest marks obtained by the students. (ii) Range of the marks obtained. (iii) Mean marks obtained by the group. Answer: The marks obtained by the group of students in a science test can be arranged in an ascending order as follows. 39, 48, 56, 75, 76, 81, 85, 85, 90, 95 (i) Highest marks = 95 Lowest marks = 39 (ii) Range = 95 − 39 = 56
Did
Question 7: The enrolment in a school during six consecutive years was as follow: 1555, 1670, 1750, 2013, 2540, 2820 Find the mean enrolment of the school for this period.
Question 8: The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
Days Rain
fall (in mm
Monday
0.0
Tuesday
12.2
Wednesday
2.1
Thursday
0.0
Friday
20.5
Saturday
5.5
Sunday
1.0
(i) Find the range of the rainfall in the above data. (ii) Find the mean rainfall for the week. (iii) On how many days was the rainfall less than the mean rainfall. Answer: (i) Range = (20.5 − 0.0) mm
= 20.5 mm
(iii) For 5 days (i.e., Monday, Wednesday, Thursday, Saturday, Sunday), the rainfall was less than the average rainfall.
Question 9: The heights of 10 girls were measured in cm and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141 (i) What is the height of the tallest girl? (ii) What is the height of the shortest girl? (iii) What is the range of the data? (iv) What is the mean height of the girls? (v) How many girls have heights more than the mean height. Answer: Arranging the heights of 10 girls in an ascending order, 128, 132, 135, 139, 141, 143, 146, 149, 150, 151 (i) Height of the tallest girl = 151 cm (ii) Height of the shortest girl = 128 cm (iii) Range = (151 − 128) cm = 23 cm
(v) The heights of 5 girls are greater than the mean height (i.e., 141.4 cm) and these heights are 143, 146, 149, 150, and 151 cm.
Exercise 3.2
Question 1: The scores in mathematics test (out of 25) of 15 students is as follows: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 Find the mode and median of this data. Are they same? Answer: Scores of 15 students in mathematics test are 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 Arranging these scores in an ascending order, 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25 Mode of a given data is that value of observation which occurs for the most number of times. Median of a given data is the middle observation when the data is arranged in an ascending or descending order.
As there are 15 terms in the given data, therefore, the median of this data will be the 8th observation. Hence, median = 20 Also, it can be observed that 20 occurs 4 times (i.e., maximum number of times). Therefore, mode of this data = 20 Yes, both are same.
Question 2: The run scored in a cricket match by 11 players is as follows: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15 Find the mean, mode and median of this data. Are the three same? Answer: The runs scored by 11 players are 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15 Arranging these scores in an ascending order, 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
Mode of a given data is that value of observation which occurs for the most number of times and the median of the given data is the middle observation when the data is arranged in an ascending or descending order. As there are 11 terms in the given data, therefore, the median of this data will be the 6th observation. Median = 15 Also, it can be observed that 15 occurs 3 times (i.e., maximum number of times). Therefore, mode of this data = 15 No, these three are not same.
Question 3: The weights (in kg.) of 15 students of a class are: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47 (i) Find the mode and median of this data. (ii) Is there more than one mode? Answer: The weights of 15 students are 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47 Arranging these weights in ascending order, 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50 (i) Mode of a given data is that value of observation which occurs for the most number of times and the median of the given data is the middle observation when the data is arranged in an ascending or descending order. As there are 15 terms in the given data, therefore, the median of this data will be the 8th observation. Hence, median = 40 Also, it can be observed that 38 and 43 both occur 3 times (i.e., maximum number of times). Therefore, mode of this data = 38 and 43 (ii) Yes, there are 2 modes for the given data.
Question 4: Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14 Answer: The given data is 13, 16, 12, 14, 19, 12, 14, 13, 14 Arranging the given data in an ascending order, 12, 12, 13, 13, 14, 14, 14, 16, 19 Mode of a given data is that value of observation which occurs for the most number of times and the median of the given data is the middle observation when the data is arranged in an ascending or descending order. As there are 9 terms in the given data, therefore, the median of this data will be the 5th observation. Hence, median = 14 Also, it can be observed that 14 occurs 3 times (i.e., maximum number of times). Therefore, mode of this data = 14
Question 5: Tell whether the statement is true or false: (i) The mode is always one of the numbers in a data. (ii) The mean is one of the numbers in a data. (iii) The median is always one of the numbers in a data. (iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9. Answer: (i) True Mode of a given data is that value of observation which occurs for the most number of times. Therefore, it is one of the observations given in the data. (ii) False Mean may or may not be one of the numbers in the data. (iii) True The median of the given data is the middle observation when the data is arranged in an ascending or descending order. (iv) False The given data is 6, 4, 3, 8, 9, 12, 13, 9
Exercise 3.3
Question 1: Use the bar graph (see the given figure) to answer the following questions. (a) Which is the most popular pet? (b) How many children have dog as a pet?
Answer: (a) Since the bar representing cats is the tallest, cat is the most popular pet. (b) The number of children having dog as a pet are 8.
Question 2: Read the bar graph (see the given figure) which shows the number of books sold by a bookstore during five consecutive years and answer the questions that follow:
(i) About how many books were sold in 1989? 1990? 1992? (ii) In which year were about 475 books sold? About 225 books sold? (iii) In which years were fewer than 250 books sold? (iv) Can you explain how you would estimate the number of books sold in 1989? Answer: (i) In 1989, 175 books were sold. In 1990, 475 books were sold. In 1992, 225 books were sold. (ii) From the graph, it can be concluded that 475 books were sold in the year 1990 and 225 books were sold in the year 1992. (iii) From the graph, it can be concluded that in the years 1989 and 1992, the number of books sold were less than 250. (iv) From the graph, it can be concluded that the number of books sold in the year 1989 is about 1
and We know that the scale is taken as 1 cm = 100 books.
part of 1 cm.
Therefore, about 175 books were sold in the year 1989.
Question 3: Number of children in six different classes are given below. Represent the data on a bar graph.
Class
Fifth
Sixth
Seve
Number of children
135
120
95
(a) How would you choose a scale? (b) Answer the following questions: (i) Which class has the maximum number of children? And the minimum? (ii) Find the ratio of students of class sixth to the students of class eight. Answer:
(a) We will choose a scale as 1 unit = 10 children because we can represent a more clear difference between the number of students of class 7th and that of class 9th by this scale. (b) (i) Since the bar representing the number of children for class fifth is the tallest, there are maximum number of children in class fifth. Similarly, since the bar representing the number of children for class tenth is the smallest, there are minimum number of children in class tenth. (ii) The number of students in class sixth is 120 and the number of students in class eighth is 100. Therefore, the ratio between the number of students of class sixth and the number of
Question 4: The performance of students in 1st Term and 2nd Term is given. Draw a double bar graph choosing appropriate scale and answer the following:
Subject
English
Hindi
1st Term (M.M. 100)
67
72
2nd Term (M.M. 100)
70
65
(i) In which subject, has the child improved his performance the most? (ii) In which subject is the improvement the least? (iii) Has the performance gone down in any subject? Answer: A double bar graph for the given data is as follows.
(i) There was a maximum increase in the marks obtained in Maths. Therefore, the child has improved his performance the most in Maths. (ii) From the graph, it can be concluded that the improvement was the least in S. Science. (iii) From the graph, it can be observed that the performance in Hindi has gone down.
Question 5: Consider this data collected from a survey of a colony.
Favourite sport
Cricket
Basket Ball
Watching
1240
470
Participating
620
320
(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph? (ii) Which sport is most popular? (iii) Which is more preferred, watching or participating in sports? Answer: (i)A double bar graph for the given data is as follows.
The double bar graph represents the number of people who like watching and participating in different sports. It can be observed that most of the people like watching and participating in cricket while the least number of people like watching and participating in athletics. (ii) From the bar graph, it can be observed that the bar representing the number of people who like watching and participating in cricket is the tallest among all the bars. Hence, cricket is the most popular sport. (iii) The bars representing watching sport are longer than the bars representing participating in sport. Hence, watching different types of sports is more preferred than participating in the sports.
Question 6: Take the data giving the minimum and the maximum temperature of various cities given in the following table: Temperatures of the cities as on 20.6.2006
City
Max.
Ahmedabad
38 ºC
Amritsar
37 ºC
Banglore
28 ºC
Chennai
36 ºC
Delhi
38 ºC
Jaipur
39 ºC
Jammu
41 ºC
Mumbai
32 ºC
Plot a double bar graph using the data and answer the following: (i) Which city has the largest difference in the minimum and maximum temperature on the given date? (ii) Which is the hottest city and which is the coldest city? (iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other. (iv) Name the city which has the least difference between its minimum and the maximum temperature. Answer: A double bar graph for the given data is constructed as follows.
(i) From the graph, it can be concluded that Jammu has the largest difference in its minimum and maximum temperatures on 20.6.2006. (ii) From the graph, it can be concluded that Jammu is the hottest city and Bangalore is the coldest city.
(iii) Bangalore and Jaipur, Bangalore and Ahmedabad For Bangalore, the maximum temperature was 28ºC, while minimum temperature of both cities, Ahmedabad and Jaipur, was 29ºC. (iv) From the graph, it can be concluded that the city which has the least difference between its minimum and maximum temperatures is Mumbai.
Exercise 3.4
Question 1: Tell whether the following is certain to happen, impossible, can happen but not certain. (i) You are older today than yesterday. (ii) A tossed coin will land heads up. (iii) A die when tossed shall land up with 8 on top. (iv) The next traffic light seen will be green. (v) Tomorrow will be a cloudy day. Answer: (i) Certain (ii) Can happen but not certain iii. Impossible as there are only six faces on a dice marked as 1, 2, 3, 4, 5, 6 on it. (iv) Can happen but not certain (v) Can happen but not certain
Question 2: There are 6 marbles in a box with numbers from 1 to 6 marked on each of them. (i) What is the probability of drawing a marble with number 2? (ii) What is the probability of drawing a marble with number 5? Answer:
Question 3: A coin is flipped to decide which team starts the game. What is the probability that your team will start? Answer: A coin has two faces − Head and Tail. One team can opt either Head or Tail.
Question 4: A box contains pairs of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that it will make a pair? Answer: It can be observed that while closing the eyes, one can draw either a black sock or a white sock. Therefore, there are two possible cases.
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Exercise 5.1
Question 1: Find the complement of each of the following angles:
Answer: The sum of the measures of complementary angles is 90º. (i) 20° Complement = 90° − 20° = 70° (ii) 63° Complement = 90° − 63° = 27° (iii) 57° Complement = 90° − 57° = 33°
Question 2: Find the supplement of each of the following angles:
Answer: The sum of the measures of supplementary anglesis 180º. (i) 105° Supplement = 180° − 105° = 75° (ii) 87° Supplement = 180° − 87° = 93° (iii) 154° Supplement = 180° − 154° = 26°
Question 3: Identify which of the following pairs of angles are complementary and which are supplementary. (i) 65°, 115° (ii) 63°, 27° (iii) 112°, 68° (iv) 130°, 50° (v) 45°, 45° (vi) 80°, 10° Answer: The sum of the measures of complementary angles is 90º and that of supplementary anglesis 180º. (i) 65°, 115° Sum of the measures of these angles = 65º + 115º = 180° ∴ These angles are supplementary angles. (ii) 63°, 27° Sum of the measures of these angles = 63º + 27º = 90° ∴ These angles are complementary angles. (iii) 112°, 68° Sum of the measures of these angles = 112º + 68º = 180° ∴ These angles are supplementary angles. (iv) 130°, 50° Sum of the measures of these angles = 130º + 50º = 180° ∴ These angles are supplementary angles. (v) 45°, 45° Sum of the measures of these angles = 45º + 45º = 90° ∴ These angles are complementary angles. (vi) 80°, 10° Sum of the measures of these angles = 80º + 10º = 90° ∴ These angles are complementary angles.
Question 4: Find the angle which is equal to its complement. Answer: Let the angle be x. Complement of this angle is also x. The sum of the measures of a complementary angle pair is 90º. ∴ x + x = 90° 2x = 90°
Question 5: Find the angle which is equal to its supplement. Answer: Let the angle be x. Supplement of this angle is also x. The sum of the measures of a supplementary angle pair is 180º. ∴ x + x = 180° 2x = 180° x = 90°
Question 6: In the given figure, ∠1 and ∠2 are supplementary angles. If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary.
Answer: ∠1 and ∠2 are supplementary angles. If ∠1 is reduced, then ∠2 should be increased by the same measure so that this angle pair remains supplementary.
Question 7: Can two angles be supplementary if both of them are: (i) Acute? (ii) Obtuse? (iii) Right? Answer:
(i) No. Acute angle is always lesser than 90º. It can be observed that two angles, even of 89º, cannot add up to 180º. Therefore, two acute angles cannot be in a supplementary angle pair. (ii) No. Obtuse angle is always greater than 90º. It can be observed that two angles, even of 91º, will always add up to more than 180º. Therefore, two obtuse angles cannot be in a supplementary angle pair. (iii) Yes. Right angles are of 90º and 90º + 90º = 180° Therefore, two right angles form a supplementary angle pair together.
Question 8: An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°? Answer: Let A and B are two angles making a complementary angle pair and A is greater than 45º. A + B = 90º B = 90º − A Therefore, B will be lesser than 45º.
Question 9: In the adjoining figure: (i) Is ∠1 adjacent to ∠2? (ii) Is ∠AOC adjacent to ∠AOE? (iii) Do ∠COE and ∠EOD form a linear pair? (iv) Are ∠BOD and ∠DOA supplementary? (v) Is ∠1 vertically opposite to ∠4? (vi) What is the vertically opposite angle of ∠5?
Answer: (i) Yes. Since they have a common vertex O and also a common arm OC. Also, their noncommon arms, OA and OE, are on either side of the common arm. (ii) No. They have a common vertex O and also a common arm OA. However, their noncommon arms, OC and OE, are on the same side of the common arm. Therefore, these are not adjacent to each other. (iii) Yes. Since they have a common vertex O and a common arm OE. Also, their noncommon arms, OC and OD, are opposite rays. (iv) Yes. Since ∠BOD and ∠DOA have a common vertex O and their noncommon arms are opposite to each other. (v) Yes. Since these are formed due to the intersection of two straight lines (AB and CD). (vi) ∠COB is the vertically opposite angle of ∠5 as these are formed due to the intersection of two straight lines, AB and CD.
Question 10: Indicate which pairs of angles are: (i) Vertically opposite angles. (ii) Linear pairs.
Answer: (i) ∠1 and ∠4, ∠5 and ∠2 +∠3 are vertically opposite angles as these are formed due to the intersection of two straight lines. (ii) ∠1 and ∠5, ∠5 and ∠4 as these have a common vertex and also have noncommon arms opposite to each other.
Question 11: In the following figure, is ∠1 adjacent to ∠2? Give reasons.
Answer: ∠1 and ∠2 are not adjacent angles because their vertex is not common.
Question 12: Find the value of the angles x, y, and z in each of the following:
Answer: (i) Since ∠x and ∠55° are vertically opposite angles, ∠x = 55° ∠x + ∠y = 180° (Linear pair) 55° + ∠y = 180° ∠y = 180º − 55º = 125° ∠y = ∠z (Vertically opposite angles) ∠z = 125° (ii) ∠z = 40° (Vertically opposite angles) ∠y + ∠z = 180° (Linear pair) ∠y = 180° − 40° = 140° 40° + ∠x + 25° = 180° (Angles on a straight line) 65° + ∠x = 180° ∠x = 180° − 65° = 115°
Question 13: Fill in the blanks: (i) If two angles are complementary, then the sum of their measures is _______. (ii) If two angles are supplementary, then the sum of their measures is _______. (iii) Two angles forming a linear pair are _______. (iv) If two adjacent angles are supplementary, they form a _______. (v) If two lines intersect at a point, then the vertically opposite angles are always _______. (vi) If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are _______. Answer: (i) 90° (ii) 180° (iii) Supplementary (iv) Linear pair (v) Equal (vi) Obtuse angles
Question 14: In the adjoining figure, name the following pairs of angles.
(i) Obtuse vertically opposite angles (ii) Adjacent complementary angles (iii) Equal supplementary angles (iv) Unequal supplementary angles (v) Adjacent angles that do not form a linear pair Answer: (i) ∠AOD, ∠BOC (ii) ∠EOA, ∠AOB (iii) ∠EOB, ∠EOD (iv) ∠EOA, ∠EOC (v) ∠AOB and ∠AOE, ∠AOE and ∠EOD, ∠EOD and ∠COD
Exercise 5.2
Question 1: State the property that is used in each of the following statements? (i) If ab, then ∠1 = ∠5 (ii) If ∠4 = ∠6, then ab (iii) If ∠4 + ∠5 = 180°, then ab
Answer: (i) Corresponding angles property (ii) Alternate interior angles property (iii) Interior angles on the same side of transversal are supplementary.
Question 2: In the adjoining figure, identify (i) The pairs of corresponding angles (ii) The pairs of alternate interior angles (iii) The pairs of interior angles on the same side of the transversal (iv) The vertically opposite angles
Answer: (i) ∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7, ∠4 and ∠8 (ii) ∠2 and ∠8, ∠3 and ∠5 (iii) ∠2 and ∠5, ∠3 and ∠8 (iv) ∠1 and ∠3, ∠2 and ∠4, ∠5 and ∠7, ∠6 and ∠8
Question 3: In the adjoining figure, p  q. Find the unknown angles.
Answer: ∠d = 125° (Corresponding angles) ∠e = 180° − 125° = 55° (Linear pair) ∠f = ∠e = 55° (Vertically opposite angles) ∠c = ∠f = 55° (Corresponding angles) ∠a = ∠e = 55° (Corresponding angles) ∠b = ∠d = 125° (Vertically opposite angles)
Question 4: Find the value of x in each of the following figures if l  m.
Answer: (i)
∠y = 110° (Corresponding angles) ∠x + ∠y = 180° (Linear pair) ∠y = 180° − 110° = 70° (ii)
∠x = 100° (Corresponding angles)
Question 5: In the given figure, the arms of two angles are parallel. If ∠ABC = 70°, then find (i) ∠DGC (ii) ∠DEF
Answer: (i) Consider that AB DG and a transversal line BC is intersecting them. ∠DGC = ∠ABC (Corresponding angles) ∠DGC = 70° (ii) Consider that BC EF and a transversal line DE is intersecting them.
∠DEF = ∠DGC (Corresponding angles) ∠DEF = 70°
Question 6: In the given figures below, decide whether l is parallel to m.
Answer: (i)
Consider two lines, l and m, and a transversal line n which is intersecting them. Sum of the interior angles on the same side of transversal = 126º + 44º = 170° As the sum of interior angles on the same side of transversal is not 180º, therefore, l is not parallel to m. (ii)
x + 75° = 180° (Linear pair on line l) x = 180° − 75° = 105° For l and m to be parallel to each other, corresponding angles (∠ABC and ∠x)should be equal. However, here their measures are 75º and 105º respectively. Hence, these lines are not parallel to each other. (iii)
∠x + 123° = 180° (Linear pair) ∠x = 180° − 123º = 57° For l and m to be parallel to each other, corresponding angles (∠ABC and ∠x)should be equal. Here, their measures are 57º and 57º respectively. Hence, these lines are parallel to each other. iv.
98 + ∠x = 180° (Linear pair) ∠x = 82° For l and m to be parallel to each other, corresponding angles (∠ABC and ∠x)should be equal. However, here their mea
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Exercise 7.1
Question 1: Complete the following statements: (a) Two line segments are congruent if __________. (b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is __________. (c) When we write ∠A = ∠ B, we actually mean __________. Answer: (a) They have the same length (b) 70° (c) m ∠A = m ∠B
Question 2: Give any two reallife examples for congruent shapes. Answer: (i) Sheets of same letter pad (ii) Biscuits in the same packet
Question 3: If /ABC ≅ /FED under the correspondence ABC ↔ FED, write all the Corresponding congruent parts of the triangles. Answer: If these triangles are congruent, then the corresponding angles and sides will be equal to each other. ∠A ↔ ∠F ∠B ↔ ∠E
∠C ↔ ∠D
Question 4: If /DEF ≅ /BCA, write the part(s) of /BCA that correspond to
Answer: (i) ∠C
Exercise 7.2
Question 1: Which congruence criterion do you use in the following? (a) Given: AC = DF AB = DE BC = EF So, /ABC ≅ /DEF
(b) Given: ZX = RP RQ = ZY ∠PRQ = ∠XZY So, /PQR ≅ /XYZ
(c) Given: ∠MLN = ∠FGH ∠NML = ∠GFH ML = FG So, /LMN ≅ /GFH
(d) Given: EB = DB AE = BC ∠A = ∠C = 90° So, /ABE ≅ /CDB
Answer: (a) SSS, as the sides of /ABC are equal to the sides of /DEF. (b) SAS, as two sides and the angle included between these sides of /PQR are equal to two sides and the angle included between these sides of /XYZ. (c) ASA, as two angles and the side included between these angles of /LMN are equal to two angles and the side included between these angles of /GFH. (d) RHS, as in the given two rightangled triangles, one side and the hypotenuse are respectively equal.
Question 2: You want to show that /ART ≅ /PEN, (a) If you have to use SSS criterion, then you need to show (i) AR = (ii) RT = (iii) AT = (b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have (i) RT = and (ii) PN = (c) If it is given that AT = PN and you are to use ASA criterion, you need to have (i) ? (ii) ?
Answer: (a) (i) AR = PE (ii) RT = EN (iii) AT = PN (b) (i) RT = EN (ii) PN = AT (c) (i) ∠ATR = ∠PNE (ii) ∠RAT = ∠EPN
Question 3: You have to show that /AMP ≅ AMQ. In the following proof, supply the missing reasons.
Answer: (i) Given (ii) Given (iii) Common
(iv) SAS, as the two sides and the angle included between these sides of /AMP are equal to two sides and the angle included between these sides of /AMQ.
Question 4: In /ABC, ∠A = 30°, ∠B = 40° and ∠C = 110° In /PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110° A student says that /ABC ≅ /PQR by AAA congruence criterion. Is he justified? Why or why not? Answer: No. This property represents that these triangles have their respective angles of equal measure. However, this gives no information about their sides. The sides of these triangles have a ratio somewhat different than 1:1. Therefore, AAA property does not prove the two triangles congruent.
Question 5: In the figure, the two triangles are congruent. The corresponding parts are marked. We can write /RAT ≅ ?
Answer: It can be observed that, ∠RAT = ∠WON ∠ART = ∠OWN AR = OW Therefore, /RAT /WON, by ASA criterion.
Question 6: Complete the congruence statement: /BCA ≅? /QRS ≅?
Answer: Given that, BC = BT TA = CA BA is common. Therefore, /BCA /BTA Similarly, PQ = RS TQ = QS PT = RQ Therefore, /QRS /TPQ
Question 7: In a squared sheet, draw two triangles of equal areas such that (i) The triangles are congruent. (ii) The triangles are not congruent. What can you say about their perimeters? Answer: (i)
Here, /ABC and /PQR have the same area and are congruent to each other also. Also, the perimeter of both the triangles will be the same. (ii)
Here, the two triangles have the same height and base. Thus, their areas are equal. However, these triangles are not congruent to each other. Also, the perimeter of both the triangles will not be the same.
Question 9: If /ABC and /PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
Answer: BC = QR /ABC /PQR (ASA criterion)
Question 10: Explain, why /ABC ≅ /FED
Answer: Given that, ∠ABC = ∠FED (1) ∠BAC = ∠EFD (2) The two angles of /ABC are equal to the two respective angles of /FED. Also, the sum of all interior angles of a triangle is 180º. Therefore, third angle of both triangles will also be equal in measure. ∠BCA = ∠EDF (3)
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Exercise 10.1
Question 1: Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only. Answer: The steps of construction are as follows. (i)Draw a line AB. Take a point P on it. Take a point C outside this line. Join C to P.
(ii)Taking P as centre and with a convenient radius, draw an arc intersecting line AB at point D and PC at point E.
(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at H.
(iv) Adjust the compasses up to the length of DE. Without changing the opening of compasses and taking H as the centre, draw an arc to intersect the previously drawn arc FG at point I.
(v) Join the points C and I to draw a line ‘l’.
This is the required line which is parallel to line AB.
Question 2: Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l. Answer: The steps of construction are as follows. (i) Draw a line l and take a point P on line l. Then, draw a perpendicular at point P.
(ii) Adjusting the compasses up to the length of 4 cm, draw an arc to intersect this perpendicular at point X. Choose any point Y on line l. Join X to Y.
(iii) Taking Y as centre and with a convenient radius, draw an arc intersecting l at A and XY at B.
(iv) Taking X as centre and with the same radius as before, draw an arc CD cutting XY at E.
(v)Adjust the compasses up to the length of AB. Without changing the opening of compasses and taking E as the centre, draw an arc to intersect the previously drawn arc CD at point F.
(vi) Join the points X and F to draw a line m.
Line m is the required line which is parallel to line l.
Question 3: Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose? Answer: The steps of construction are as follows. (i)Draw a line l and take a point A on it. Take a point P not on l and join A to P.
(ii) Taking A as centre and with a convenient radius, draw an arc cutting l at B and AP at C.
(iii)Taking P as centre and with the same radius as before, draw an arc DE to intersect AP at F.
(iv) Adjust the compasses up to the length of BC. Without changing the opening of compasses and taking F as the centre, draw an arc to intersect the previously drawn arc DE at point G.
(v)Join P to G to draw a line m. Line m will be parallel to line l.
(vi)Join P to any point Q on line l. Choose another point R on line m. Similarly, a line can be drawn through point R and parallel to PQ.
Let it meet line l at point S. In quadrilateral PQSR, opposite lines are parallel to each other. PQ  RS and PR  QS Thus, &mnSq1 PQSR is a parallelogram.
Exercise 10.2
Question Construct XYZ in which XY = Answer: The rough figure of this triangle is as follows.
The required triangle (i) Draw a line segment YZ of length 5 cm.
4.5
cm,
is
YZ
=
5
cm
constructed
and
ZX
as
=
6
1: cm.
follows.
(ii) Point X is at a distance of 4.5 cm from point Y. Therefore, taking point Y as centre, draw an arc of 4.5 cm radius.
(iii) Point X is at a distance of 6 cm from point Z. Therefore, taking point Z as centre, draw an arc of 6 cm radius. Mark the point of intersection of the arcs as X. Join XY and XZ.
XYZ is the required triangle.
Question 2: Construct an equilateral triangle of side 5.5 cm. Answer: An equilateral triangle of side 5.5 cm has to be constructed. We know that all sides of an equilateral triangle are of equal length. Therefore, a triangle ABC has to be constructed with AB = BC = CA = 5.5 cm. The steps of construction are as follows. (i) Draw a line segment BC of length 5.5 cm.
(ii)
Taking
point
B
as
centre,
draw
an
arc
of
5.5
(iii) Taking point C as centre, draw an arc of 5.5 cm radius to meet the previous arc at point A.
cm
radius.
(iv)
Join
A
to
B
and
C.
ABC is the required equilateral triangle.
Question 3: Draw PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this? Answer: The steps of construction are as follows. (i) Draw a line segment QR of length 3.5 cm.
(ii)
Taking
point
Q
as
centre,
draw
an
arc
of
4
cm
radius.
(iii) Taking point R as centre, draw an arc of 4 cm radius to intersect the previous arc at point P.
(iv) Join P to Q and R.
PQR is the required triangle. As the two sides of this triangle are of the same length (PQ = PR), therefore, ?PQR is an isosceles triangle.
Question Construct ABC such that AB = 2.5 Answer: The steps of (i) Draw a line segment BC of length 6 cm.
(ii)
Taking
point
C
as
cm,
BC
=
6
cm
construction
centre,
draw
and
AC
=
are
an
arc
6.5
cm.
Measure
as
of
6.5
4: ∠B.
follows.
cm
radius.
(iii) Taking point B as centre, draw an arc of radius 2.5 cm to meet the previous arc at point A.
(iv)
Join
A
to
B
ABC is the required triangle. ∠B can be measured with the help of protractor. It comes to 90º.
Exercise 10.3
and
C.
Question Construct DEF such that DE = 5 Answer: The rough sketch of the required ?DEF is as follows.
The (i)Draw
(ii)
At
steps a
point
of
draw
DF
construction segment DE
line
D,
cm,
a
ray
DX
making
=
3
cm
and
are of
an
m∠EDF
as length
angle
=
of
follows. cm.
5
90°
1: 90°.
with
DE.
(iii)
(iv)
Taking
D
Join
as
centre,
draw
F
to
an
arc
E.
of
3
cm
?DEF
radius.
is
It
will
the
intersect
DX
required
at
point
F.
triangle.
Question 2: Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110°. Answer: An isosceles triangle PQR has to be constructed with PQ = QR = 6.5 cm. A rough sketch of the required triangle can be drawn as follows.
The steps of (i) Draw the line segment QR of length 6.5 cm.
construction
are
as
follows.
(ii)
At
(iii)
Taking
point
Q
Q,
as
draw
centre,
draw
a
an
ray
arc
(iv) Join P to R to obtain the required triangle PQR.
QX
of
6.5
making
cm
an
radius.
angle
It
intersects
110°
QX
with
at
point
QR.
P.
Question Construct ?ABC with BC = 7.5 Answer: A rough sketch of the required triangle is as follows.
The steps of (i) Draw a line segment BC of length 7.5 cm.
cm,
AC
construction
=
5
are
cm
and
m∠C
as
=
3: 60°.
follows.
(ii)
At
point
C,
draw
a
ray
CX
making
60º
with
BC.
(iii) Taking C as centre, draw an arc of 5 cm radius. It intersects CX at point A.
(iv)
Join
A
to
B
to
obtain
triangle
ABC.
Exercise 10.4
Question 1: Construct ?ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm. Answer: A rough sketch of the required ?ABC is as follows.
The steps of (i)Draw a line segment AB of length 5.8 cm.
construction
are
as
follows.
(ii)At point A, draw a ray AX making 60º angle with AB.
(iii)
At
point
B,
draw
a
ray
BY,
making
30º
angle
with
AB.
(iv) Point C has to lie on both the rays, AX and BY. Therefore, C is the point of intersection of these two rays.
This
is
the
Question Construct ?PQR if PQ = (Hint: Recall angle Answer: A rough sketch of
In order to construct ?PQR, According to the ∠PQR + ∠PRQ 105º + 40º + ∠RPQ = 180º
145º ∠RPQ
+ =
180°
required
5
cm, sum the
the angle
m∠PQR = property required
measure of sum +
∠RPQ −
triangle
105°
and of
?PQR
ABC.
m∠QRP a
is
∠RPQ has property ∠RPQ
as
follows.
be of =
calculated. triangles, 180º
=
180º 35°
to
= 145°
2: = 40°. triangle).
The steps of (i) Draw a line segment PQ of length 5 cm.
construction
(ii) At P, draw a ray PX making an angle of 35º with PQ.
are
as
follows.
(iii) At point Q, draw a ray QY making an angle of 105º with PQ.
(iv)Point R has to lie on both the rays, PX and QY. Therefore, R is the point of intersection of these two rays.
This
is
the
required
triangle
PQR.
Question 3: Examine whether you can construct DEF such that EF = 7.2 cm, m∠E = 110° and m∠F = 80°. Justify your answer. Answer: Given that, m∠E = 110° and m∠F = 80° Therefore, m∠E + m∠F = 110° + 80° = 190°
However, according to the angle sum property of triangles, we should obtain m∠E + m∠F + m∠D = 180° Therefore, the angle sum property is not followed by the given triangle. And thus, we cannot construct ?DEF with the given measurements.
Also, it can be observed that point D should lie on both rays, EX and FY, for constructing the required triangle. However, both rays are not intersecting each other. Therefore, the required triangle cannot be formed
Exercise 10.5
Question Construct the right angled ?PQR, Answer: A rough sketch of ?PQR is as follows.
where
m∠Q
=
90°,
QR
=
8
cm
and
PR
=
10
1: cm.
The steps of (i) Draw a line segment QR of length 8 cm.
(ii) At point Q, draw a ray QX making 90º with QR.
construction
are
as
follows.
(iii) Taking R as centre, draw an arc of 10 cm radius to intersect ray QX at point P.
(iv) Join P to R. ?PQR is the required rightangled triangle.
Question 2: Construct a rightangled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long. Answer: A rightangled triangle ABC with hypotenuse 6 cm and one of the legs as 4 cm has to be constructed. A rough sketch of ?ABC is as follows.
The steps of (i) Draw a line segment BC of length 4 cm.
construction
are
as
follows.
(ii) At point B, draw a ray BX making an angle of 90º with BC.
(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray BX at point A.
(iv) Join A to C to obtain the required ?ABC.
Question Construct an isosceles rightangled triangle ABC, where, m∠ACB = 90° and Answer: In an isosceles triangle, the lengths of any two sides Let in ?ABC, AC = BC = 6 cm. A rough sketch of this ?ABC is as follows.
The steps of (i) Draw a line segment AC of length 6 cm.
construction
are
as
AC
=
are
6
3: cm. equal.
follows.
(ii) At point C, draw a ray CX making an angle of 90º with AC.
(iii) Taking point C as centre, draw an arc of 6 cm radius to intersect CX at point B.
(iv) Join A to B to obtain the required ?ABC.