.NCERT SOLUTIONS

MATHS CLASS 10th Ncert Solutions

Please Update App (1MB ONLY) To Latest Version,As All Errors In The App Has Been Fixed Now And Interface Has Also Chnaged

• Chapter 1-Real Numbers
• Chapter 2-Polynomials
• Chapter 3-Pair of Linear Equations in Two Variables
• Chapter 4-Quadratic Equations
• Chapter 5-Arithmetic Progressions
• Chapter 6-Triangles
• Chapter 7-Coordinate Geometry
• Chapter 8-Introduction to Trigonometry
• Chapter 9-Some Applications of Trigonometry
• Chapter 10-Circles
• Chapter 11-Constructions
• Chapter 12-Areas Related to Circles
• Chapter 13-Surface Areas and Volumes
• Chapter 14-Statistics
• Chapter 15- Probability

Chapter 15;Probability

NCERT Solutions for Class 10th: Ch 15 Probability Maths
Page No: 308

Exercise 15.1

1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ___________ .
(ii) The probability of an event that cannot happen is __________. Such an event is called ________ .
(iii) The probability of an event that is certain to happen is _________ . Such an event is called _________ .
(iv) The sum of the probabilities of all the elementary events of an experiment is __________ .
(v) The probability of an event is greater than or equal to  and less than or equal to __________ .

Answer

(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called  impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called sure or certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.


2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.

Answer

(i) It does not have equally likely outcomes as it depends on various reasons like mechanical problems, fuels etc.
(ii) It does not have equally likely outcomes as it depends on the player how he/she shoots.
(iii) It has equally likely outcomes.
(iv)It has equally likely outcomes.

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Answer

Yes, tossing of a coin is a fair way of deciding which team should get the ball at the beginning of a football game because it has only two outcomes either head or tail. A coin is always unbiased.

4. Which of the following cannot be the probability of an event?
(A) 2/3     (B) -1.5     (C) 15%       (D) 0.7

Answer

The probability of an event is always greater than or equal to 0 and less than or equal to 1.
Thus, (B) -1.5 cannot be the probability of an event.

5. If P(E) = 0.05, what is the probability of ‘not E’?

Answer

P(E) = 0.05
also, P(E) + P(not E) = 1
⇒ P(not E) = 1 - P(E)
⇒ P(not E) = 1 - 0.05
⇒ P(not E) = 0.95

6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Answer

(i) Since the bag contains only lemon flavoured.
Therefor, No. of orange flavoured candies = 0
Probability of taking out orange flavoured candies = 0/1 = 0

(ii) The bag only have lemon flavoured candies.
Probability of taking out lemon flavoured candies = 1/1 = 1

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer

Let E be the event of having the same birthday.
P(E) = 0.992
⇒ P(E) + P(not E) = 1
⇒ P(not E) = 1 – P(E)
⇒ 1 - 0.992 = 0.008
The probability that the 2 students have the same birthday is 0.008

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Answer

No. of red balls = 3
No. of black balls = 5
Total no. of balls = 5+3 = 8
(i) Probability of drawing red balls = No. of red balls/Total no. of balls = 3/8

(ii) Probability of drawing black balls = No. of black balls/Total no. of balls = 5/8

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white ? (iii) not green?

Answer

No. of red marbles = 5
No. of white marbles =8
No. of green marbles = 4
Total no. of balls = 5+8+4 = 17

(i) Favourable no. of elementary events = 5
Probability of taking out red marble = 5/17

(ii) Favourable no. of elementary events = 8
Probability of taking out red marble = 8/17

(iii) Favourable no. of elementary events = 4
Probability of taking out red marble = 4/17

Page No: 309

10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a ₹5 coin?

Answer

No. of 50p coins = 100
No. of ₹1 coins = 50
No. of ₹2 coins = 20
No. of ₹5 coins = 10
Total no. of coins = 100 + 50 + 20 + 10 = 180

(i) Favourable no. of elementary events = 100
Probability that it will be 50p coins = 100/180 = 5/9

(ii) Favourable no. of elementary events = 100+50+20 = 170
Probability that it will be 50p coins = 170/180 = 17/18

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?

Answer

No. of male fish in the tank = 5
no. of female fish in the tank = 8
Total number of fish in the tank = 5 + 8 = 13
Favourable number events = 5
Probability of taking out a male fish = 5/13

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Answer

Possible no. of events = 8
(i) Favourable number of events = 1
Probability that it will point at 8 = 1/8

(ii) Odd numbers = 1, 3, 5 and 7
Favourable number of events = 4
Probability that it will be an odd number = 4/8 = 1/2

(iii) Numbers greater than 2 = 3, 4, 5, 6, 7 and 8
Favourable number of events = 6
Probability that a number greater than 4 = 6/8 = 3/4

(iv) Numbers less than 9 = 1,2,3,4,5,6,7,8
Favourable number of events = 8
Probability that a number less than 9 = 8/8 = 1

13. A die is thrown once. Find the probability of getting
(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Answer

Possible numbers of events on throwing a dice = 6
Numbers on dice = 1,2,3,4,5 and 6

(i) Prime numbers = 2, 3 and 5
Favourable number of events = 3
Probability that it will be a prime number = 3/6 = 1/2

(ii) Numbers lying between 2 and 6 = 3, 4 and 5
Favourable number of events = 3
Probability that a number between 2 and 6 = 3/6 = 1/2

(iii) Odd numbers = 1, 3 and 5
Favourable number of events = 3
Probability that it will be an odd number = 3/6 = 1/2

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds

Answer

Possible numbers of events = 52

(i) Numbers of king of red colour = 2
Probability of getting a king of red colour = 2/52 = 1/26

(ii) Numbers of face cards = 12
Probability of getting a face card = 12/52 = 3/13

(iii) Numbers of red face cards = 6
Probability of getting a king of red colour = 6/52 = 3/26

(iv) Numbers of jack of hearts =1
Probability of getting a king of red colour = 1/52

(v) Numbers of king of spade = 13
Probability of getting a king of red colour = 13/52 = 1/4

(vi) Numbers of queen of diamonds = 1
Probability of getting a king of red colour = 1/52

15. Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Answer

Total numbers of cards = 5

(i) Numbers of queen = 1
Probability of picking a queen = 1/5

(ii) When queen is drawn and put aside then total numbers of cards left is 4
(a) Numbers of ace = 1
Probability of picking an ace = 1/4
(a) Numbers of queen = 0
Probability of picking a queen = 0/4 = 0

16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer

Numbers of defective pens = 12
Numbers of good pens = 132
Total numbers of pen = 132 + 12 = 144 pens
Favourable number of events = 132
Probability of getting a good pen = 132/144 = 11/12

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer

(i) Total numbers of bulbs = 20
Numbers of defective bulbs = 4
Probability of getting a defective bulb = 4/20 = 1/5

(ii) One non defective bulb is drawn in (i) then the total numbers of bulb left is 19
Total numbers of events = 19
Favourable numbers of events =  19 - 4 = 15
Probability that the bulb is not defective = 15/19

18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

Answer

Total numbers of discs = 50

(i) Total numbers of favourable events = 81
Probability that it bears a two-digit number = 81/90 = 9/10

(ii) Perfect square numbers = 1, 4, 9, 16, 25, 36, 49, 64 and 81
Favourable numbers of events = 9
Probability of getting a perfect square number = 9/90 = 1/10

(iii) Numbers which are divisible by 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90
Favourable numbers of events = 18
Probability of getting a number divisible by 5 = 18/90 = 1/5

Page No: 310

19. A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?

Answer

Total numbers of events = 6

(i) Total numbers of faces having A on it = 2
Probability of getting A = 2/6 = 1/3

(ii) Total numbers of faces having D on it = 1
Probability of getting A = 1/6

20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?

Answer

Area of the rectangle = (3 × 2) m2= 6m2
Area of the circle = πr2= π(1/2)2m2= π/4 m2
Probability that die will land inside the circle = (π/4) × 1/6 = π/24

21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?

Answer

Total numbers of pens = 144
Numbers of defective pens = 20
Numbers of non defective pens = 144 - 20 = 124

(i) Numbers of favourable events = 124
Probability that she will buy it = 124/144 = 31/36

(ii) Numbers of favourable events = 20
Probability that she will not buy it = 20/144 = 5/36

22. Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.

Answer

Events that can happen on throwing two dices are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total numbers of events : 6 × 6 = 36

(i) To get sum as 2, possible outcomes = (1,1)
To get sum as 3, possible outcomes = (1,2) and (2,1)
To get sum as 4, possible outcomes = (1,3); (3,1);  and (2,2)
To get sum as 5, possible outcomes = (1,4); (4,1); (2,3);  and (3,2)
To get sum as 6, possible outcomes = (1,5); (5,1); (2,4); (4,2);  and (3,3)
To get sum as 7, possible outcomes = (1,6); (6,1); (5,2); (2,5); (4,3);  and (3,4)
To get sum as 8, possible outcomes = (2,6); (6,2); (3,5); (5,3);  and (4,4)
To get sum as 9, possible outcomes = (3,6); (6,3); (4,5);  and (5,4)
To get sum as 10, possible outcomes = (4,6); (6,4) and (5,5)
To get sum as 11, possible outcomes = (5,6) and (6,5)
To get sum as 12, possible outcomes = (6,6)

Event:
Sum on 2 dice
2
3
4
5
6
7
8
9
10
11
12
Probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36

(ii) No, i don't agree with the argument. It is already justified in (i).

23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer

Events that can happen in tossing 3 coins = HHH, HHT, HTH, THH, TTH, HTT, THT, TTT
Total number of events = 8
Hinif will lose the game if he gets HHT, HTH, THH, TTH, HTT, THT
Favourable number of elementary events = 6
Probability of losing the game = 6/8 = 3/4

24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Answer

(i) Consider the following events.
A = first throw shows 5,
B = second throw shows 5
P(A) = 6/36, P(B) = 6/36 and P(notB) = 5/6
⇒ P(notA) = 1– 6/36 = 30/36 = 5/6
Required probability = 5/6 × 5/6 = 25/36

(ii) Number of events when 5 comes at least once = 11
Probability = 11/36

Page No: 311

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

Answer

(i) The statement is incorrect
Possible events = (H,H); (H,T); (T,H) and (T,T)
Probability of getting two heads = 1/4
Probability of getting one of the each = 2/4 = 1/2

(ii) Correct. The two outcomes considered are equally likely.

Chapter 14;Statistics

NCERT Solutions for Class 10th: Ch 14 Statistics Maths
Page No: 270

Exercise 14.1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of Plants
0-2
2-4
4-6
6-8
8-10
10-12
12-14
Number of Houses
1
2
1
5
6
2
3
Which method did you use for finding the mean, and why?

Answer

No. of plants
(Class interval)
No. of houses (fi)
Mid-point (xi)
fixi
0-2
1
1
1
2-4
2
3
6
4-6
1
5
5
6-8
5
7
35
8-10
6
9
54
10-12
2
11
22
12-14
3
13
39
Sum fi= 20

Sum fixi= 162

Mean = x̄ = ∑fixi/∑fi= 162/20 = 8.1
We would use  direct method because the numerical value of fiand xiare small.

2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.)
100-120
120-140
140-160
160-180
180-200
Number of workers
12
14
8
6
10
Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer

Here, the value of mid-point (xi) is very large, so assumed mean A = 150 and class interval is h = 20.
So, ui= (xi- A)/h = ui= (xi- 150)/20

Daily wages
(Class interval)
Number of workers
frequency (fi)
Mid-point (xi)
ui= (xi- 150)/20
fiui
100-120
12
110
-2
-24
120-140
14
130
-1
-14
140-160
8
150
0
0
160-180
6
170
1
6
180-200
10
190
2
20
Total
Sum fi= 50


Sum fiui= -12
Mean = x̄ = A + h∑fiui/∑fi=150 + (20 × -12/50) = 150 - 4.8 = 145.20
Thus, mean daily wage = Rs. 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.


Answer

Here, the value of mid-point (xi)  meanx̄= 18

Class interval
Number of children (fi)
Mid-point (xi)
fixi
11-13
7
12
84
13-15
6
14
84
15-17
9
16
144
17-19
13
18 = A
234
19-21
f
20
20f
21-23
5
22
110
23-25
4
24
96
Total
fi= 44+f

Sum fixi= 752+20f

Mean = x̄ = ∑fixi/∑fi= (752+20f)/(44+f)
⇒ 18 = (752+20f)/(44+f)
⇒ 18(44+f) = (752+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 - 752 = 20f - 18f
⇒ 40 = 2f
⇒ f = 20

Page No: 271

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.


Answer

xi= (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5

Class Interval
Number of women (fi)
Mid-point (xi)
ui= (xi- 75.5)/h
fiui
65-68
2
66.5
-3
-6
68-71
4
69.5
-2
-8
71-74
3
72.5
-1
-3
74-77
8
75.5
0
0
77-80
7
78.5
1
7
80-83
4
81.5
3
8
83-86
2
84.5
3
6
Sum fi= 30


Sum fiui= 4

Mean = x̄ = A + h∑fiui/∑fi= 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3

Class Interval
Number of boxes (fi)
Mid-point (xi)
di= xi- A
fidi
49.5-52.5
15
51
-6
90
52.5-55.5
110
54
-3
-330
55.5-58.5
135
57 = A
0
0
58.5-61.5
115
60
3
345
61.5-64.5
25
63
6
150
Sum fi= 400


Sum fidi= 75

Mean = x̄ = A + ∑fidi/∑fi= 57 + (75/400) = 57 + 0.1875 = 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Answer

Here, assumed mean (A) = 225

Class Interval
Number of households (fi)
Mid-point (xi)
di= xi- A
fidi
100-150
4
125
-100
-400
150-200
5
175
-50
-250
200-250
12
225
0
0
250-300
2
275
50
100
300-350
2
325
100
200
Sum fi= 25


Sum fidi= -350

Mean = x̄ = A + ∑fidi/∑fi= 225 + (-350/25) = 225 - 14 = 211
The mean daily expenditure on food is 211

7. To find out the concentration of SO2in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO2in the air.

Answer

Concentration of SO2(in ppm)
Frequency (fi)
Mid-point (xi)
fixi
0.00-0.04
4
0.02
0.08
0.04-0.08
9
0.06
0.54
0.08-0.12
9
0.10
0.90
0.12-0.16
2
0.14
0.28
0.16-0.20
4
0.18
0.72
0.20-0.24
2
0.20
0.40
Total
Sum fi= 30

Sum (fixi) = 2.96

Mean = x̄ = ∑fixi/∑fi
= 2.96/30 = 0.099 ppm

Page No. 272

8. A class teacher has the following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent.

Number of days
0-6
6-10
10-14
14-20
20-28
28-38
38-40
Number of students
11
10
7
4
4
3
1

Answer

Class interval
Frequency (fi)
Mid-point (xi)
fixi
0-6
11
3
33
6-10
10
8
80
10-14
7
12
84
14-20
4
17
68
20-28
4
24
96
28-38
3
33
99
38-40
1
39
39
Sumfi= 40

Sumfixi= 499

Mean = x̄ = ∑fixi/∑fi
= 499/40 = 12.48 days

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.

Literacy rate (in %)
45-55
55-65
65-75
75-85
85-98
Number of cities
3
10
11
8
3

Answer

Class Interval
Frequency (fi)
(xi)
di=xi- a
ui= di/h
fiui
45-55
3
50
-20
-2
-6
55-65
10
60
-10
-1
-10
65-75
11
70
0
0
0
75-85
8
80
10
1
8
85-95
3
90
20
2
6
Sumfi= 35



Sumfiui= -2

Mean = x̄ = a + (∑fiui/∑fi) х h
= 70 + (-2/35) х 10 = 69.42

Page No. 275
Exercise 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)
5-15
15-25
25-35
35-45
45-55
55-65
Number of patients
6
11
21
23
14
5
Find the mode and the mean of the data given above. Compare and interpret the two
measures of central tendency.

Answer

Modal class = 35 – 45, l = 35, class width (h) = 10, fm= 23, f1= 21 and f2= 14


Calculation of Mean:

Class Interval
Frequency (fi)
Mid-point (xi)
fixi
5-15
6
10
60
15-25
11
20
220
25-35
21
30
630
35-45
23
40
920
45-55
14
50
700
55-65
5
60
300
Sumfi= 80

Sumfixi= 2830

Mean = x̄ = ∑fixi/∑fi
= 2830/80 = 35.37 yr

2. The following data gives the information on the observed lifetimes (in hours) of 225
electrical components :

Lifetime (in hours)
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
10
35
52
61
38
29
Determine the modal lifetimes of the components.

Answer

Modal class of the given data is 60–80.
Modal class = 60-80, l = 60, fm= 61, f1= 52, f2= 38 and h = 20


3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure :

Expenditure
Number of families
1000-1500
24
1500-2000
40
2000-2500
33
2500-3000
28
3000-3500
30
3500-4000
22
4000-4500
16
4500-5000
7

Answer

Modal class = 1500-2000, l = 1500, fm= 40, f1= 24, f2= 33 and h = 500


Calculation for mean:

Class Interval
fi
xi
di = xi - a
ui = di/h
fiui
1000-1500
24
1250
-1500
-3
-72
1500-2000
40
1750
-1000
-2
-80
2000-2500
33
2250
-500
-1
-33
2500-3000
28
2750
0
0
0
3000-3500
30
3250
500
1
30
3500-4000
22
3750
1000
2
44
4000-4500
16
4250
1500
3
48
4500-5000
7
4750
2000
4
28
fi = 200



fiui = -35

Mean = x̄ = a + (∑fiui/∑fi) х h
= 2750 + (35/200) х 500
= 2750 - 87.50 = 2662.50

Chapter 12;Area Related To Circles

NCERT Solutions of Class 10th: Ch 12 Areas Related to Circles

Page No: 230

Exercise: 12.1

Unless stated otherwise, use π =22/7.

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Answer
Let the radius of the third circle be R.
Circumference of the circle with radius R = 2πR
Circumference of the circle with radius 19 cm = 2π × 19 = 38π cm
Circumference of the circle with radius 9 cm = 2π × 9 = 18π cm
Sum of the circumference of two circles = 38π + 18π = 56π cm
Circumference of the third circle = 2πR = 56π
⇒ 2πR = 56π cm
⇒ R = 28 cm
The radius of the circle which has circumference equal to the sum of the circumferences of the two circles is 28 cm.

2.  The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Answer

Let the radius of the third circle be R.
Area of the circle with radius R = πR2
Area of the circle with radius 8 cm = π × 82= 64π cm2
Area of the circle with radius 6 cm = π × 62= 36π cm2
Sum of the area of two circles = 64π cm2+36π cm2= 100πcm2
Area of the third circle = πR2= 100πcm2
⇒ πR2= 100πcm2
⇒ R2= 100cm2
⇒ R = 10cm
Thus, the radius of the new circle is 10 cm.

3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.


Answer

Diameter of Gold circle (first circle) = 21 cm
Radius of first circle, r1= 21/2 cm = 10.5 cm
Each of the other bands is 10.5 cm wide,
∴Radius of second circle, r2= 10.5 cm + 10.5 cm = 21 cm
∴Radius of third circle, r3= 21 cm + 10.5 cm = 31.5 cm
∴Radius of fourth circle, r4= 31.5 cm + 10.5 cm = 42 cm
∴Radius of fifth circle, r5= 42 cm + 10.5 cm = 52.5 cm
Area of gold region = π r12= π (10.5)2= 346.5 cm2
Area of red region = Area of second circle - Area of first circle = π r22- 346.5 cm2
=π(21)2- 346.5 cm2= 1386 - 346.5 cm2= 1039.5 cm2
Area of blue region = Area of third circle - Area of second circle = π r32- 1386 cm2
=π(31.5)2- 1386 cm2= 3118.5 - 1386 cm2= 1732.5 cm2
Area of black region = Area of fourth circle - Area of third circle = π r32- 3118.5 cm2
= π(42)2- 1386 cm2= 5544 - 3118.5 cm2= 2425.5 cm2
Area of white region = Area of fifth circle - Area of fourth circle = π r42- 5544 cm2
= π(52.5)2- 5544 cm2= 8662.5 - 5544 cm2= 3118.5 cm2

4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Answer

Diameter of the wheels of a car = 80 cm
Circumference of wheels = 2πr = 2r × π = 80 π cm
Distance travelled by car in 10 minutes = (66 × 1000 × 100 × 10)/60 = 1100000 cm/s
No. of revolutions = Distance travelled by car/Circumference of wheels
= 1100000/80 π = (1100000 × 7)/(80×22) = 4375
4375 complete revolutions does each wheel make in 10 minutes.

5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units                     (B) π units                  (C) 4 units              (D) 7 units

Answer

Let the radius of the circle be r.
∴ Perimeter of the circle = Circumference of the circle = 2πr
∴ Area of the circle = π r2
A/q,
2πr = π r2
⇒ 2 = r
Thus, the radius of the circle is 2 units. (A) is correct.

Page No: 230

Exercise: 12.2

Unless stated otherwise, use π =22/7.
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Answer

Area of the sector making angle θ = (θ/360°)×π r2
Area of the sector making angle 60° = (60°/360°)×π r2cm2
= (1/6)×62π  = 36/6 π cm2= 6 × 22/7 cm2= 132/7 cm2

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer

Quadrant of a circle means sector is making angle 90°.
Circumference of the circle = 2πr = 22 cm
Radius of the circle = r = 22/2π cm = 7/2 cm
Area of the sector making angle 90° = (90°/360°)×π r2cm2
= (1/4)×(7/2)2π  = (49/16) π cm2= (49/16) × (22/7) cm2= 77/8 cm2
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer

Here, Minute hand of clock acts as radius of the circle.
∴ Radius of the circle (r) = 14 cm
Angle rotated by minute hand in 1 hour = 360°
∴ Angle rotated by minute hand in 5 minutes = 360° × 5/60 = 30°
Area of the sector making angle 30° = (30°/360°)×π r2cm2
= (1/12) × 142π  = 196/12 π cm2= (49/3) × (22/7) cm2= 154/3 cm2
Area swept by the minute hand in 5 minutes = 154/3 cm2
4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14)

Answer

Radius of the circle = 10 cm
Major segment is making 360° - 90° = 270°
Area of the sector making angle 270°
= (270°/360°) × π r2cm2
= (3/4) × 102π  = 75 π cm2
= 75 × 3.14 cm2= 235.5 cm2
∴ Area of the major segment = 235.5 cm2
Height of ΔAOB = OA = 10 cm
Base of ΔAOB = OB = 10 cm
Area of ΔAOB = 1/2 × OA × OB
= 1/2 ×10 × 10 = 50 cm2
Major segment is making  90°
Area of the sector making angle 90°                                                                = (90°/360°) × π r2cm2
= (1/4) × 102π  = 25 π cm2
= 25 × 3.14 cm2= 78.5 cm2
Area of the minor segment = Area of the sector making angle 90° - Area of ΔAOB
= 78.5 cm2-50 cm2= 28.5 cm2

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord

Answer


Radius of the circle = 21 cm

(i) Length of the arc AB = θ/360° × 2πr
= 60°/360° × 2 × 22/7 × 21
= 1/6 × 2 × 22/7 × 21 = 22
The length of the arc is  22 cm.

(ii) Angle subtend by the arc = 60°
Area of the sector making angle 60° = (60°/360°) × π r2cm2
= (1/6) × 212π  = 441/6 π cm2
= 441/6 × 22/7 cm2= 231 cm2
∴ Area of the sector formed by the arc is 231 cm2
(iii) Area of equilateral ΔAOB = √3/4 × (OA)2= √3/4 × 212= (441√3)/4 cm2
Area of the segment formed by the corresponding chord
= Area of the sector formed by the arc - Area of equilateral ΔAOB
= 231 cm2-(441√3)/4 cm2

6.  A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)

Answer

Radius of the circle = 15 cm
ΔAOB is isosceles as two sides are equal.
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠C = 180°
⇒ 2 ∠A = 180° - 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
∴ OA = OB = AB = 15 cm
Area of equilateral ΔAOB = √3/4 × (OA)2= √3/4 × 152
= (225√3)/4 cm2= 97.3 cm2
Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60° = (60°/360°) × π r2cm2
= (1/6) × 152πcm2=225/6 πcm2
=(225/6) × 3.14 cm2= 117.75cm2

Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB
= 117.75cm2- 97.3 cm2= 20.4 cm2
Angle made by Major sector = 360° - 60° = 300°
Area of the sector making angle 300° = (300°/360°) × π r2cm2
= (5/6) × 152πcm2=1125/6 πcm2
=(1125/6) × 3.14 cm2= 588.75cm2

Area of major segment = Area of Minor sector + Area of equilateral ΔAOB
= 588.75cm2+ 97.3 cm2= 686.05 cm2

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

Answer
Radius of the circle, r = 12 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° - (90° + 60°) = 30°
cos 30° = AD/OA
⇒ √3/2 = AD/12
⇒ AD = 6√3 cm
⇒ AB = 2 × AD = 12√3 cm
sin 30° = OD/OA
⇒ 1/2 = OD/12
⇒ OD = 6 cm
Area of ΔAOB = 1/2 × base × height
= 1/2 × 12√3 × 6 = 36√3 cm
= 36 × 1.73 = 62.28 cm2

Angle made by Minor sector = 120°
Area of the sector making angle 120° = (120°/360°) × π r2cm2
= (1/3) × 122πcm2=144/3 πcm2
=  48 × 3.14 cm2= 150.72cm2
∴ Area of the corresponding Minor segment = Area of the Minor sector - Area of ΔAOB
= 150.72cm2-62.28 cm2
= 88.44 cm2
8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m
long rope (see Fig. 12.11). Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Answer

Side of square field = 15 m
Length of rope is the radius of the circle, r = 5 m
Since, the horse is tied at one end of square field, it will graze only quarter of the field with radius 5 m.

(i) Area of circle = π r2= 3.14 × 52= 78.5 m2
Area of that part of the field in which the horse can graze = 1/4 of area of the circle = 78.5/4 = 19.625 m2

(ii) Area of circle if the length of rope is increased to 10 m = π r2=3.14 × 102= 314 m2
Area of that part of the field in which the horse can graze = 1/4 of area of the circle
= 314/4 = 78.5 m2
Increase in grazing area = 78.5 m2- 19.625 m2= 58.875 m2


9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Answer

Number of diameters = 5
Length of diameter = 35 mm
∴ Radius = 35/2 mm

(i) Total length of silver wire required = Circumference of the circle +  Length of 5 diameter
= 2π r + (5×35) mm = (2 × 22/7 × 35/2) + 175 mm
= 110 + 175 mm = 185 mm

(ii) Number of sectors = 10
Area of each sector = Total area/Number of sectors
Total Area = π r2= 22/7 × (35/2)2=1925/2 mm2
∴ Area of each sector = (1925/2) × 1/10 = 385/4 mm2
10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer

Number of ribs in umbrella = 8
Radius of umbrella while flat = 45 cm
Area between the two consecutive ribs of the umbrella =
Total area/Number of ribs
Total Area = π r2= 22/7 × (45)2=6364.29 cm2
∴ Area between the two consecutive ribs = 6364.29/8 cm2
= 795.5 cm2
11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Answer

Angle of the sector of circle made by wiper = 115°
Radius of wiper = 25 cm
Area of the sector made by wiper = (115°/360°) × π r2cm2
=23/72 × 22/7 × 252=23/72 × 22/7 × 625 cm2
= 158125/252 cm2

Total area cleaned at each sweep of the blades = 2 ×158125/252 cm2=158125/126 = 1254.96 cm2

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.
(Use π = 3.14)

Answer

Let O bet the position of Lighthouse.
Distance over which light spread i.e. radius, r = 16.5 km
Angle made by the sector = 80°
Area of the sea over which the ships are warned = Area made by the sector.
Area of sector = (80°/360°) × π r2km2
= 2/9 × 3.14 × (16.5)2km2
=  189.97 km2




13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm2. (Use √3 = 1.7)

Answer


Number of equal designs = 6
Radius of round table cover = 28 cm
Cost of making design = ₹0.35 per cm2
∠O= 360°/6 = 60°
ΔAOB is isosceles as two sides are equal. (Radius of the circle)
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠O = 180°
⇒ 2 ∠A = 180° - 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
Area of equilateral ΔAOB = √3/4 × (OA)2= √3/4 × 282= 333.2 cm2
Area of sector ACB = (60°/360°) × π r2cm2
= 1/6 × 22/7 × 28 × 28 = 410.66 cm2
Area of design = Area of sector ACB - Area of equilateral ΔAOB
= 410.66 cm2- 333.2 cm2= 77.46 cm2

Area of 6 design = 6 × 77.46 cm2= 464.76 cm2
Total cost of making design = 464.76 cm2× ₹0.35 per cm2= ₹ 162.66

14. Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is
(A) p/180 × 2πR             (B) p/180 × π R2(C) p/360 × 2πR           (D) p/720 × 2πR2

Answer

Area of a sector of angle p = p/360 × π R2
=p/360 × 2/2 × π R2
=  2p/720 × 2πR2
Hence, Option (D) is correct.

Page No: 234

Exercise: 12.3

Unless stated otherwise, use π =22/7
1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Answer

PQ = 24 cm and PR = 7 cm
∠P = 90° (Angle in the semi-circle)
∴ QR is hypotenuse of the circle = Diameter of the circle.
By Pythagoras theorem,
QR2= PR2+ PQ2
⇒ QR2= 72+ 242
⇒ QR2= 49+ 576
⇒ QR2= 625
⇒ QR= 25 cm
∴ Radius of the circle = 25/2 cm
Area of the semicircle = (π R2)/2
= (22/7 × 25/2 × 25/2)/2 cm2
= 13750/56 cm2= 245.54 cm2
Area of the ΔPQR = 1/2 × PR × PQ
= 1/2 × 7 × 24 cm2
= 84 cm2
Area of the shaded region = 245.54 cm2- 84 cm2= 161.54 cm2

2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Answer


Radius inner circle = 7 cm
Radius of outer circle = 14 cm
Angle made by sector = 40°
Area of the sector OAC = (40°/360°) × π r2cm2
= 1/9 × 22/7 × 142= 68.44 cm2
Area of the sector OBD = (40°/360°) × π r2cm2
= 1/9 × 22/7 × 72= 17.11 cm2

Area of the shaded region ABDC = Area of the sector OAC - Area of the sector circle OBD
= 68.44 cm2- 17.11 cm2= 51.33 cm2

3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.


Answer

There are two semicircles in the figure.
Side of the square = 14 cm
Diameter of the semicircle = 14 cm
∴ Radius of the semicircle = 7 cm
Area of the square = 14 × 14 = 196 cm2
Area of the  semicircle = (π R2)/2
= (22/7 × 7 × 7)/2 cm2= 77 cm2
Area of two semicircles = 2 × 77 cm2= 154 cm2

Area of the shaded region = 196 cm2-154 cm2= 42 cm2

4. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Answer

OAB is an equilateral triangle with each angle equal to 60°.
Area of the sector is common in both.
Radius of the circle = 6 cm.
Side of the triangle = 12 cm.
Area of the equilateral triangle = √3/4 × (OA)2= √3/4 × 122= 36√3 cm2
Area of the circle = π R2= 22/7 × 62= 792/7 cm2
Area of the sector making angle 60° = (60°/360°) × π r2cm2
= 1/6 × 22/7 × 62cm2= 132/7 cm2
Area of the shaded region = Area of the equilateral triangle + Area of the circle - Area of the sector
= 36√3 cm2+ 792/7 cm2- 132/7 cm2
= (36√3 + 660/7) cm2

5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.

Answers

Side of the square = 4 cm
Radius of the circle = 1 cm
Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.
Area of square = (side)2= 42= 16 cm2
Area of the quadrant = (π R2)/4 cm2= (22/7 × 12)/4 = 11/14 cm2
∴ Total area of the 4 quadrants = 4 × (11/14) cm2= 22/7 cm2
Area of the circle = π R2cm2= (22/7 × 12)= 22/7 cm2
Area of the shaded region = Area of square - (Area of the 4 quadrants + Area of the circle)
= 16 cm2- (22/7 + 22/7) cm2
= 68/7cm2

6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.


Answer

Radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2
Since, AD is the median of the triangle
∴ AO = Radius of the circle = 2/3 AD
⇒ 2/3 AD = 32 cm
⇒ AD = 48 cm
In ΔADB,

By Pythagoras theorem,
AB2= AD2+ BD2
⇒ AB2= 482+ (AB/2)2
⇒ AB2= 2304+ AB2/4
⇒ 3/4 (AB2)= 2304
⇒ AB2= 3072
⇒ AB= 32√3 cm
Area of ΔADB = √3/4 × (32√3)2cm2= 768√3 cm2
Area of circle = π R2= 22/7 × 32 × 32 = 22528/7 cm2
Area of the design = Area of circle - Area of ΔADB
= (22528/7 - 768√3) cm2

Page No: 236

7. In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
Answer

Side of square = 14 cm
Four quadrants are included in the four sides of the square.
∴ Radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 142= 196 cm2
Area of the quadrant = (π R2)/4 cm2= (22/7 × 72)/4 cm2
= 77/2 cm2
Total area of the quadrant = 4 × 77/2 cm2= 154cm2

Area of the shaded region = Area of the square ABCD - Area of the quadrant
= 196 cm2- 154 cm2
= 42 cm2

8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If
the track is 10 m wide, find :
(i) the distance around the track along its inner edge
(ii) the area of the track.

Answer

Width of track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m

DE = CF = 60 m
Radius of inner semicircle, r = OD = O'C
= 60/2 m = 30 m
Radius of outer semicircle, R = OA = O'B
= 30 + 10 m = 40 m
Also, AB = CD = EF = GH = 106 m


Distance around the track along its inner edge = CD + EF + 2 × (Circumference of inner semicircle)
= 106 + 106 + (2 × πr) m = 212 + (2 × 22/7 × 30) m
= 212 + 1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of  outer semicircle) - 2 × (area of  inner                                                                                                                                               semicircle)
= (AB × CD) + (EF × GH) + 2 × (πr2/2) - 2 × (πR2/2) m2
= (106 × 10) + (106 × 10) + 2 × π/2 (r2-R2) m2
= 2120 + 22/7 × 70 × 10 m2
= 4320 m2

9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other
and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Answer


Radius of larger circle, R = 7 cm
Radius of smaller circle, r = 7/2 cm
Height of ΔBCA = OC = 7 cm
Base of ΔBCA = AB = 14 cm
Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm2
Area of larger circle = πR2= 22/7 × 72= 154 cm2
Area of larger semicircle = 154/2 cm2= 77 cm2
Area of smaller circle = πr2= 22/7 × 7/2 × 7/2 = 77/2 cm2

Area of the shaded region = Area of larger circle - Area of triangle - Area of larger semicircle + Area                                               of smaller circle
Area of the shaded region = (154 - 49 - 77 + 77/2) cm2
= 133/2 cm2= 66.5 cm2

10. The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Answer

ABC is an equilateral triangle.
∴ ∠A = ∠B = ∠C = 60°
There are three sectors each making 60°.
Area of ΔABC = 17320.5 cm2
⇒ √3/4 × (side)2=17320.5
⇒ (side)2=17320.5 × 4/1.73205
⇒ (side)2=4 × 104
⇒ side=200 cm
Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°) × π r2cm2
= 1/6 × 3.14 × (100)2cm2
=15700/3cm2
Area of 3 sectors = 3 × 15700/3 = 15700 cm2=
Area of the shaded region = Area of equilateral triangle ABC - Area of 3 sectors
= 17320.5 - 15700 cm2=1620.5cm2

11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.

Answer

Number of circular design = 9
Radius of the circular design = 7 cm
There are three circles in one side of square handkerchief.
∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm
Area of the square = 42 × 42 cm2= 1764 cm2

Area of the circle = π r2= 22/7 × 7 × 7 = 154 cm2
Total area of the design = 9× 154 = 1386 cm2

Area of the remaining portion of the handkerchief = Area of the square - Total area of the design
= 1764 - 1386 = 378 cm2

12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB,                       (ii) shaded region.

Answer

Radius of the quadrant = 3.5 cm = 7/2 cm

(i) Area of quadrant OACB = (πR2)/4 cm2
= (22/7 × 7/2 × 7/2)/4 cm2
= 77/8 cm2
(ii) Area of triangle BOD = 1/2 × 7/2 × 2 cm2
=  7/2cm2
Area of shaded region = Area of quadrant - Area of triangle BOD
= (77/8 - 7/2) cm2= 49/8 cm2
= 6.125 cm2

13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Answer

Side of square = OA = AB = 20 cm
Radius of the quadrant = OB
OAB is right angled triangle
By Pythagoras theorem in ΔOAB ,
OB2= AB2+ OA2
⇒ OB2= 202+ 202
⇒ OB2= 400+ 400
⇒ OB2= 800
⇒ OB= 20√2 cm
Area of the quadrant = (πR2)/4 cm2= 3.14/4 × (20√2)2cm2= 628cm2
Area of the square = 20 × 20 = 400cm2

Area of the shaded region = Area of the quadrant - Area of the square
= 628 - 400 cm2= 228cm2

14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.

Answer

Radius of the larger circle, R = 21 cm
Radius of the smaller circle, r = 7 cm
Angle made by sectors of both concentric circles =  30°
Area of the larger sector = (30°/360°) × π R2cm2
= 1/12 × 22/7 × 212cm2
=231/2cm2
Area of the smaller circle = (30°/360°) × π r2cm2
= 1/12 × 22/7 × 72cm2
=77/6cm2

Area of the shaded region = 231/2 - 77/6 cm2
= 616/6 cm2= 308/3cm2

15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC
as diameter. Find the area of the shaded region.

Answer

Radius of the the quadrant ABC of circle = 14 cm
AB = AC = 14 cm
BC is diameter of semicircle.
ABC is right angled triangle.
By Pythagoras theorem in ΔABC,
BC2= AB2+ AC2
⇒ BC2= 142+ 142
⇒ BC= 14√2 cm
Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of ΔABC = 1/2 × 14 × 14 = 98 cm2
Area of quadrant = 1/4 × 22/7 × 14 × 14 = 154 cm2
Area of the semicircle = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm2

Area of the shaded region =Area of the semicircle + Area of ΔABC - Area of quadrant
= 154 + 98 - 154 cm2= 98cm2

Page No: 238

16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.

Answer

AB = BC = CD = AD = 8 cm
Area of ΔABC = Area of ΔADC = 1/2 × 8 × 8 = 32 cm2
Area of quadrant AECB = Area of quadrant AFCD = 1/4 × 22/7 × 82
=  352/7 cm2

Area of shaded region = (Area of quadrant AECB - Area of ΔABC) +                                           (Area of quadrant AFCD - Area of ΔADC)
= (352/7 - 32) + (352/7 -32) cm2
= 2 × (352/7 -32) cm2
=256/7 cm2

Ch 11 Constructions

NCERT Solutions for Class 10th: Ch 11 Constructions Maths
Page No: 219

Exercise 11.1

In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Answer


Steps of Construction:
Step I:AB = 7.6 cm is drawn.
Step II:A ray AX making an acute angle with                  AB is drawn.
Step III:After that, a ray BY is drawn parallel to               AX making equal acute angle as in the                 previous step.
Step IV:Point A1, A2, A3, A4 and A5 is marked               on AX and point B1, B2.... to B8 is                      marked on BY such that AA1 = A1A2 =              A2A3 =....BB1= B1B2 = .... B7B8
Step V:A5 and B8 is joined and it intersected AB at point C diving it in the ratio 5:8.
AC : CB = 5 : 8

Justification:
ΔAA5C ~ ΔBB8C
∴ AA5/BB8 = AC/BC = 5/8

2.  Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Answer
Steps of Construction:

Step I:AB = 6 cm is drawn.
Step II:With A as a centre and radius equal to 4                cm, an arc is draw.
Step III:Again, with B as a centre and radius                     equal to 5 cm an arc is drawn on same                  side of AB intersecting previous arc at C.
Step IV:AC and BC are joined to form ΔABC.
Step V:A ray AX is drawn making an acute                      angle with AB below it.
Step VI:5 equal points (sum of the ratio = 2 + 3                 =5) is marked on AX as A1 A2....A5
Step VII:A5B is joined. A2B' is drawn parallel                   to A5B and B'C' is drawn parallel to                    BC.
ΔAB'C' is the required triangle

Justification:
∠A(Common)
∠C = ∠C' and ∠B = ∠ B' (corresponding angles)
Thus ΔAB'C' ~  ΔABC by AAA similarity condition
From the figure,
AB'/AB = AA2/AA5 = 2/3
AB' =2/3 AB
AC' = 2/3 AC
3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
7/5 of the corresponding sides of the first triangle.

Answer

Steps of Construction:
Step I:A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.

Step II:A ray BX making an acute angle with BC is drawn                  opposite to vertex A.
Step III:7 points as B1 B2 B3 B4 B5 B6 and B7 are marked                  on BX.
Step IV;Point B5 is joined with C to draw B5C.
Step V:B7C' is drawn parallel to B5C and C'A' is parallel to              CA.
Thus A'BC' is the required triangle.

Justification
ΔAB'C' ~  ΔABC by AAA similarity condition
∴ AB/A'B = AC/A'C' = BC/BC'
and BC/BC' = BB5/BB7 = 5/7
∴A'B/AB = A'C'/AC = = BC'/BC = 7/5

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Answer

Steps of Construction:

Step I:BC = 5 cm is drawn.
Step II:Perpendicular bisector of BC is drawn and it intersect BC at O.
Step III:At a distance of 4 cm, a point A is marked on perpendicular                   bisector of BC.
Step IV:AB and AC is joined to form ΔABC.
Step V:A ray BX is drawn making an acute angle with BC opposite to              vertex A.
Step VI:3 points B1 B2 and B3 is marked BX.
Step VII:B2 is joined with C to form B2C.

Step VIII:B3C' is drawn parallel to B2C and C'A' is drawn parallel to                   CA.
Thus, A'BC' is the required triangle formed.

Justification:

ΔAB'C' ~  ΔABC by AA similarity condition.
∴ AB/AB' = BC/B'C' = AC/AC'
also,
AB/AB' = AA2/AA3 = 2/3
⇒ AB' = 3/2 AB, B'C' = 3/2 BC and AC' = 3/2 AC


5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Answer
Steps of Construction:

Step I:BC = 6 cm is drawn.
Step II:At point B, AB = 5 cm is drawn making an
∠ABC = 60° with BC.
Step III:AC is joined to form ΔABC.
Step IV:A ray BX is drawn making an acute angle with BC                     opposite to vertex A.
Step V:4 points B1 B2 B3 and B4 at equal distance is                              marked on BX.
Step VII:B3 is joined with C' to form B3C'.
Step VIII:C'A' is drawn parallel CA.
Thus, A'BC' is the required triangle.

Justification:

∠A = 60° (Common)
∠C = ∠C'
ΔAB'C' ~  ΔABC by AA similarity condition.
∴ AB/AB' = BC/B'C' = AC/AC'
also,
AB/AB' = AA3/AA4 = 4/3
⇒ AB' = 3/4 AB, B'C' = 3/4 BC and AC' = 3/4 AC

6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.

Answer

Sum of all side of triangle = 180°
∴ ∠A  + ∠B  + ∠C  = 180°
∠C = 180° - 150° = 30°
Steps of Construction:
Step I:BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of  45° with BC.

Step III:At C, a ray making an angle of 30° with                BC is drawn intersecting the previous                  ray at A. Thus, ∠A = 105°.
Step IV:A ray BX is drawn making an acute                     angle with BC opposite to vertex A.
Step V:4 points B1 B2 B3 and B4 at equal                       distance is marked on BX.
Step VI:B3C is joined and B4C' is made parallel              to B3C.
Step VII:C'A' is made parallel CA.
Thus, A'BC' is the required triangle.

Justification:

∠B = 45° (Common)
∠C = ∠C'
ΔAB'C' ~  ΔABC by AA similarity condition.
∴ BC/BC' = AB/A'B' = AC/A'C'
also,
BC/BC' = BB3/BB4 = 34
⇒ AB = 4/3 AB', BC = 4/3 BC' and AC = 4/3 A'C'

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Answer


Steps of Construction:
Step I:BC = 3 cm is drawn.
Step II:At B, A ray making an angle of 90° with BC is drawn.
Step III:With B as centre and radius equal to 4 cm, an arc is                 made on previous ray intersecting it at point A.
Step IV:AC is joined to form ΔABC.
Step V:A ray BX is drawn making an acute angle with BC                  opposite to vertex A.
Step VI:5 points B1 B2 B3 B4 and B5 at equal distance is                           marked on BX.
Step VII:B3C is joined B5C' is made parallel to B3C.
Step VIII:A'C' is joined together.
Thus, ΔA'BC' is the required triangle.

Justification:
As in the previous question 6.

Page No. 221

Exercise 11.2

In each of the following, give also the justification of the construction:
1.  Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Answer


Steps of Construction:
Step I:With O as a centre and radius equal to 6 cm, a circle is                drawn.
Step II:A point P at a distance of 10 cm from the centre O is                   taken. OP is joined.
Step III:Perpendicular bisector OP is drawn and let it                                intersected at M.
Step IV:With M as a centre and OM as a radius, a circle is                        drawn intersecting previous circle at Q and R.
Step V:PQ and PR are joined.
Thus, PQ and PR are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
PQ = PR = 8cm.

Justification:
OQ is joined.
∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.

2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Answer


Steps of Construction:
Step I:With O as a centre and radius equal to 4 cm, a circle is                 drawn.
Step II:With O as a centre and radius equal to 6 cm, a                             concentric circle is drawn.
Step III:P be any point on the circle of radius 6 cm and OP is                   joined.
Step IV:Perpendicular bisector of OP is drawn which cuts it at                 M
Step V:With M as a centre and OM as a radius, a circle is                        drawn which intersect the the circle of radius 4 cm at Q              and R
Step VI: PQ and PR are joined.
Thus, PQ and PR are the two tangents.
Measurement:
OQ = 4 cm (Radius of the circle)
PQ = 6 cm ( Radius of the circle)
∠PQO = 90° (Angle in the semi circle)
Applying Pythagoras theorem in ΔPQO,
PQ2+ QO2= PO2
⇒ PQ2+ 42= 62
⇒ PQ2+ 16 = 36
⇒ PQ2= 36 - 16
⇒ PQ2= 20
⇒ PQ = 2√5

Justification:

∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.

3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Answer


Steps of Construction:
Step I:With O as a centre and radius equal to 3 cm, a                   circle is drawn.
Step II:The diameter of the circle is extended both sides                and an arc is made to cut it at 7 cm.
Step III:Perpendicular bisector of OP and OQ is drawn                 and x and y be its mid-point.
Step IV:With O as a centre and Ox be its radius, a circle                is drawn which intersected the previous circle at                M and N.
Step V:Step IV is repeated with O as centre and Oy as radius and it intersected the circle at R and T.
Step VI:PM and PN are joined also QR and QT are joined.
Thus,  PM and PN are tangents to the circle from P and QR and QT are tangents to the circle from point Q.

Justification:

∠PMO = 90° (Angle in the semi circle)
∴ OM ⊥ PM
Therefor, OM is the radius of the circle then PM has to be a tangent of the circle.
Similarly, PN, QR and QT are tangents of the circle.

4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Answer


We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O)  = 360° - (90° + 90° + 60°) = 120°
Steps of Construction:
Step I:A point Q is taken on the circumference of the circle and OQ              is joined. OQ is radius of the circle.
Step II:Draw another radius OR making an angle equal to 120° with               the previous one.
Step III:A point P is taken outside the circle. QP and PR are joined                  which is perpendicular OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.

Justification:

Sum of all angles in the quadrilateral PQOR = 360°
∠QOR + ∠ORP + ∠OQR + ∠RPQ = 360°
⇒ 120° + 90° + 90° + ∠RPQ = 360°
⇒∠RPQ = 360° - 300°
⇒∠RPQ = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°.

5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Answer


Steps of Construction:
Step I:A line segment AB of 8 cm is drawn.
Step II:With A as centre and radius equal to 4                cm, a circle is drawn which cut the                    line at point O.
Step III:With B as a centre and radius equal to                3 cm, a circle is drawn.
Step IV:With O as a centre and OA as a                         radius, a circle is drawn which                           intersect the previous two circles at P,               Q and R, S.
Step V:AP, AQ, BR and BS are joined.
Thus, AP, AQ, BR and BS are the required tangents.

Justification:
∠BPA = 90° (Angle in the semi circle)
∴ AP ⊥ PB
Therefor, BP is the radius of the circle then AP has to be a tangent of the circle.
Similarly,  AQ, BR and BS are tangents of the circle.
6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Answer


Steps of Construction:
Step I:A ΔABC is drawn.
Step II:Perpendicular to AC is drawn to point B                which intersected it at D.
Step III:With O as a centre and OC as a radius, a               circle is drawn. The circle through B, C,                D is drawn.
Step IV:OA is joined and a circle is drawn with                 diameter OA which intersected the                         previous circle at B and E.
Step V:AE is joined.
Thus, AB and AE are the required tangents to the circle from A.

Justification:

∠OEA = 90° (Angle in the semi circle)
∴ OE ⊥ AE
Therefor, OE is the radius of the circle then AE has to be a tangent of the circle.
Similarly,  AB is tangent of the circle.

Chapter 10:-Circles

NCERT Solutions for Class 10th: Ch 10 Circles
Page No: 209

Exercise: 10.1

1. How many tangents can a circle have?

Answer

A circle can have infinite tangents.

2.  Fill in the blanks :
(i) A tangent to a circle intersects it in ............... point(s).
(ii) A line intersecting a circle in two points is called a .............
(iii) A circle can have ............... parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ............

Answer

(i) one
(ii) secant
(iii) two
(iv) point of contact

3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) √119cm

Answer


The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ
By Pythagoras theorem in ΔOPQ,
OQ2= OP2+PQ2
⇒ (12)2= 52+ PQ2
⇒PQ2= 144 - 25
⇒PQ2= 119
⇒PQ = √119cm
(D) is the correct option.

4. Draw a circle and two lines parallel to a given line such that one is a tangent and the
other, a secant to the circle.

Answer


AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.

Page NO: 213

Exercise: 10.2

In Q.1 to 3, choose the correct option and give justification.

1.  From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A)  7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm

Answer
The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)
By Pythagoras theorem in ΔOPQ,
OQ2= OP2+PQ2
⇒ (25)2= OP2+ (24)2
⇒ OP2= 625 - 576
⇒ OP2= 49
⇒ OP= 7 cm
The radius of the circle is option (A) 7 cm.


2.  In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Answer

OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT  = 360°
⇒ ∠PTQ + 90° + 110° + 90°  = 360°
⇒ ∠PTQ = 70°
∠PTQ is equal to option (B) 70°.

3.  If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Answer


OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°
In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB  = 360°
⇒ ∠AOB + 90° + 90° + 80°  = 360°
⇒ ∠AOB = 100°
Now,
In ΔOPB and ΔOPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)
Thus ∠POB = ∠POA
∠AOB = ∠POB + ∠POA
⇒ 2 ∠POA = ∠AOB
⇒ ∠POA = 100°/2 = 50°
∠POA is equal to option  (A) 50°

Page No: 214

4.  Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and,
∴ OA ⊥ PQ
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º
From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.

5.  Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Answer


Let AB be the tangent to the circle at point P with centre O.
We have to prove that PQ passes through the point O.
Suppose that PQ doesn't passes through point O. Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
PQ intersect CD at R and also intersect AB at P.
AS, CD // AB PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90° (PQ ⊥ AB)
⇒ ∠ORP  = 90°
∠ROP + ∠OPA = 180° (Co-interior angles)
⇒∠ROP + 90° = 180°
⇒∠ROP = 90°
Thus, the ΔORP has 2 right angles i.e. ∠ORP  and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.

6.  The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer

AB is a tangent drawn on this circle from point A.

∴ OB ⊥ AB
OA = 5cm and AB = 4 cm (Given)
In ΔABO,
By Pythagoras theorem in ΔABO,
OA2=AB2+ BO2
⇒ 52= 42+ BO2
⇒ BO2= 25 - 16
⇒ BO2= 9
⇒ BO = 3
∴ The radius of the circle is 3 cm.

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.

Answer


Let the two concentric circles with centre O.
AB be the chord of the larger circle which touches the smaller circle at point P.
∴ AB is tangent to the smaller circle to the point P.
⇒ OP ⊥ AB
By Pythagoras theorem in ΔOPA,
OA2=  AP2+ OP2
⇒ 52= AP2+ 32
⇒ AP2= 25 - 9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
∴ The length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Answer

From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)
BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR)  = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC

9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Answer

We joined O and C

A/q,
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)
Similarly,
ΔOQB  ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°

10.  Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer

Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º
∴ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

11. Prove that the parallelogram circumscribing a circle is a rhombus.

Answer


ABCD is a parallelogram,
∴ AB = CD ... (i)
∴ BC = AD ... (ii)
From the figure, we observe that,
DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC ... (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
⇒ AB = BC ... (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD andDC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Answer


In ΔABC,
Length of two tangents drawn from the same point to the circle are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF =x
We observed that,
AB = AE + EB =x+ 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 +x
Now semi perimeter of circle s,
⇒ 2s = AB + BC + CA
=x+ 8 + 14 + 6 +x
= 28 + 2x
⇒s = 14 +x
Area of ΔABC = √s (s - a)(s - b)(s - c)
= √(14 +x) (14 +x-14)(14 +x-x- 6)(14 +x-x -8)
= √(14 +x) (x)(8)(6)
= √(14 +x) 48x... (i)
also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)
= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]
= 2×1/2 (4x+ 24 + 32) = 56 + 4x... (ii)
Equating equation (i) and (ii) we get,
√(14 +x) 48x= 56 + 4x
Squaring both sides,
48x(14 +x) = (56 + 4x)2
⇒ 48x =[4(14 + x)]2/(14 +x)
⇒ 48x =16 (14 +x)
⇒ 48x =224 + 16x
⇒ 32x =224
⇒x =7 cm
Hence, AB =x+ 8 = 7 + 8 = 15 cm
CA = 6 +x= 6 + 7 = 13 cm

13.  Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒∠1 = ∠8
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Chapter 9;Some Applications Of Trigonometry

NCERT Solutions for Class 10th: Ch 9 Some Applications of Trigonometry Maths
Excercise: 9.1

Page No: 203

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).

Answer
Let AB be the vertical pole Ac be 20 m  long rope tied to point C.
In  right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = AB/20
⇒ AB = 20/2
⇒ AB = 10
The height of the pole is 10 m.

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer

Let AC be the broken part of the tree.
∴ Total height of the tree = AB+AC
In  right ΔABC,
cos 30° = BC/AC
⇒ √3/2 = 8/AC
⇒ AC = 16/√3
Also,
tan 30° = AB/BC
⇒ 1/√3 = AB/8
⇒ AB = 8/√3
Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer

There are two slides of height 1.5 m and 3 m. (Given)
Let AB is 1.5 m and PQ be 3 m slides.

ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at
60° with length PR.
A/q,
In  right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
also,
In  right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.

Page No: 204

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.
A/q,
In  right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer


Let BC be the height of the kite from the ground,
AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.
A/q,
In  right ΔABC,
sin 60° = BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m.

6.  A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.


Answer

Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m
AB = AZ - BZ = (30 - 1.5) = 28.5 m
A/q,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD
⇒ BD = 28.5√3 m
also,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = 28.5/BC
⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3× 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is57/√3 m.

7.  From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer


Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC - BC
A/q,
In  right ΔBCD,
tan 45° = BC/CD
⇒ 1 = 20/CD
⇒ CD = 20 m
also,
In  right ΔACD,
tan 60° = AC/CD
⇒ √3 = AC/20
⇒ AC = 20√3 m
Height of transmission tower = AB = AC - BC = (20√3 - 20) m = 20(√3 - 1) m.

8.  A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer


Let AB be the height of statue.
D is the point on the ground from where the elevation is taken.
Height of pedestal = BC = AC - AB
A/q,
In  right ΔBCD,
tan 45° = BC/CD
⇒ 1 =  BC/CD
⇒ BC = CD.
also,
In  right ΔACD,
tan 60° = AC/CD
⇒ √3 = AB+BC/CD
⇒ √3CD = 1.6 m + BC
⇒ √3BC = 1.6 m + BC
⇒ √3BC - BC = 1.6 m
⇒ BC(√3-1) = 1.6 m
⇒ BC = 1.6/(√3-1) m
⇒ BC = 0.8(√3+1)m
Thus, the height of the pedestal is 0.8(√3+1)m.

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer


Let CD be the height of the tower equal to 50 m (Given)
Let AB be the height of the building.
BC be the distance between the foots of the building and the tower.
Elevation is 30° and 60° from the tower and the building respectively.
A/q,
In  right ΔBCD,
tan 60° = CD/BC
⇒ √3 = 50/BC
⇒ BC = 50/√3
also,
In  right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/BC
⇒ AB = 50/3
Thus, the height of the building is 50/3.

10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer


Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken.
BD is the distance between the poles.
A/q,
AB = CD,
OB + OD = 80 m
Now,
In  right ΔCDO,
tan 30° = CD/OD
⇒ 1/√3 = CD/OD
⇒ CD = OD/√3 ... (i)
also,
In  right ΔABO,
tan 60° = AB/OB
⇒ √3 = AB/(80-OD)
⇒ AB = √3(80-OD)
AB = CD (Given)
⇒ √3(80-OD) = OD/√3
⇒ 3(80-OD) = OD
⇒ 240 - 3 OD = OD
⇒ 4 OD = 240
⇒ OD = 60
Putting the value of OD in equation (i)
CD = OD/√3 ⇒ CD = 60/√3 ⇒ CD = 20√3 m
also,
OB + OD = 80 m ⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.



11.  A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Answer

Here, AB is the height of the tower.
CD = 20 m (given)
A/q,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(20+BC)
⇒ AB = (20+BC)/√3 ... (i)
also,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ AB = √3 BC ... (ii)
From eqn (i) and (ii)
AB = √3 BC = (20+BC)/√3
⇒ 3 BC = 20 + BC
⇒ 2 BC = 20 ⇒ BC = 10 m
Putting the value of BC in eqn (ii)
AB = 10√3 m
Thus, the height of the tower 10√3 m and the width of the canal is 10 m.

12.  From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer


Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°
EC = DE + CD
also, CD = AB = 7 m.
and BC = AD
A/q,
In  right ΔABC,
tan 45° = AB/BC
⇒ 1= 7/BC
⇒ BC = 7 m = AD
also,
In  right ΔADE,
tan 60° = DE/AD
⇒ √3 = DE/7
⇒ DE = 7√3 m
Height of the tower = EC =  DE + CD
= (7√3 + 7) m = 7(√3+1) m.

13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer


Let AB be the lighthouse of height 75 m.
Let C and D be the positions of the ships.
30° and 45° are the angles of depression from the lighthouse.
A/q,
In  right ΔABC,
tan 45° = AB/BC
⇒ 1= 75/BC
⇒ BC = 75 m
also,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 75/BD
⇒ BD = 75√3  m

The distance between the two ships = CD = BD - BC = (75√3 - 75) m = 75(√3-1) m.

Page No: 205


14.  A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m
from the ground. The height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Answer


Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height = 88.2 m - 1.2 m = 87 m
Distance travelled by the balloon =
DE = CE - CD
A/q,
In  right ΔBEC,
tan 30° = BE/CE
⇒ 1/√3= 87/CE
⇒ CE = 87√3 m
also,
In  right ΔADC,
tan 60° = AD/CD
⇒ √3= 87/CD
⇒ CD = 87/√3 m = 29√3 m
Distance travelled by the balloon =  DE = CE - CD = (87√3 - 29√3) m = 29√3(3 - 1) m = 58√3 m.

15.  A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer


Let AB be the tower.
D is the initial and C is the final position of the car respectively.
Angles of depression are measured from A.
BC is the distance from the foot of the tower to the car.
A/q,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ BC = AB/√3 m
also,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(BC + CD)
⇒ AB√3 = BC + CD
⇒ AB√3 = AB/√3 + CD
⇒ CD = AB√3 - AB/√3
⇒ CD = AB(√3 - 1/√3)
⇒ CD = 2AB/√3
Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec.
Time taken by car to travel BC = 6/2 = 3 sec.

16.  The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Answer


Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
A/q,
In  right ΔABC,
tanx= AB/BC
⇒ tanx= AB/4
⇒ AB = 4 tanx... (i)
also,
In  right ΔABD,
tan (90°-x) = AB/BD
⇒ cotx= AB/9
⇒ AB = 9 cotx... (ii)
Multiplying  eqn (i) and (ii)
AB2= 9 cotx× 4 tanx
⇒ AB2= 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.

Go To Chapters


Liked NCERT Solutions and Notes, Share this with your friends::

class10maths