Text Book Solutions All Class

Exercise 14.1

Q1. 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  ________.
Ans:
(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 an impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called a 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.

Q2. Which of the following experiments have equally likely outcomes? Explain. 
(i) A driver attempts to start a car. The car starts or does not start.
(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.
Ans:
(i) Since the car may or may not start, thus the outcomes are not equally likely.
(ii) The player may shoot or miss the shot.
∴ The outcomes are not equally likely.
(iii) In advance, it is known that the answer is to be either right or wrong.
∴ The outcomes right or wrong are equally likely to occur.
(iv) In advance, it is known the newly born baby has to be either a boy or a girl.
∴ The outcomes, either a boy or a girl are equally likely to occur.

Q3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Ans: Since on tossing a coin, the outcomes ‘head’ and ‘tail’ are equally likely, the result of tossing a coin is completely unpredictable and so it is a fairway.

Q4. Which of the following cannot be the probability of an event?
(a) 2/3 
(b) − 1.5 
(c) 15% 
(d) 0.7

Ans: (b)
Solution: Since the probability of an event cannot be negative.
∴ Option (b) −1.5 cannot be the probability of an event.

Q5. If P(E) = 0.05, what is the probability of ‘not E’?
Ans: We know that,

P(E) + P(not E) = 1
∴ 0.05 + P(not E) = 1 ⇒ P(not E) = 1 − 0.05
= 0.95
Thus, the probability of ‘not E’ = 0.95.

Q6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Ans: 
(i) Since, there are lemon flavoured candies only in the bag,
∴ Taking out any orange flavoured candy is not possible.
⇒ Probability of taking out an orange flavoured candy = 0.
(ii) Also, the probability of taking out a lemon flavoured candy = 1.

Q7. 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?
Ans: ∴ Let the probability of 2 students having same birthday = P(SB) And the probability of 2 students not having the same birthday = P(nSB)
∴ P(SB) + P(nSB)=1
⇒ P(SB) + 0.992 = 1
⇒ P(SB)=1 − 0.992 = 0.008
So, the required probability of 2 boys having the same birthday = 0.008.

Q8. 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?
Ans:
Total number of balls = 3 + 5 = 8
∴ Number of all possible outcomes = 8
(i) For red balls
There are 3 red balls.
∴ Number of favourable outcomes = 3

(ii) For not red balls

Q9. 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? and (ii) white? (iii) not green?
Ans:
Total number of marbles = 5 + 8 + 4 = 17
(i) For red marbles
∵ Number of red marbles = 5
∴ Number of favourable outcomes = 5
∴ Probability of red marbles, P(red) = 5/17
(ii) For white balls
∵ Number of white balls = 8
∴ Probability of white balls,
∴ P(white)= 8/17
(iii) For not green balls 
∵ Number of white balls = 4
∴ Number of ‘not green’ balls = 17 − 4 = 13
i.e., Favourable outcomes = 13
P(not green) = 13/17
OR
Number of green marbles = 4
∴ Number of ‘not green balls’ = 17 − 4 = 13
⇒ Favourable outcomes = 13
∴ P(not green) = 13/17

Q10. A piggy bank contains hundred 50p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 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 50p coin? and (ii) will not be Rs 5 coin?
Ans: 
Number of:
50 p coins = 100
Re 1 coins = 50
Rs 2 coins = 20
Rs 5 coins = 10
Total number of coins = 100 + 50 + 20 + 10 = 180
(i) For a 50 p coin:
Favourable events = 100
∴ P(50 p) = 100/180 = 5/9
(ii) For not a Rs 5 coin:
a Number of Rs 5 coins = 10
∴ Number of ‘not Rs 5’ coins = 180 − 10 = 170
⇒ Favourable outcomes = 170
∴ P(not 5 rupee coin) = 170/180 = 17/18.

Q11. 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.). What is the probability that the fish taken out is a male fish?

Ans: 
Number of:
Male fishes = 5
Female fishes = 8
∴ Total number of fishes = 5 + 8 = 13
⇒ Total number of outcomes = 13
For a male fish:
Number of favourable outcomes = 5
∴ P(male fish) = 5/13.

Q12. 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 figure), 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?

Ans: Total numbers marked = 8
(i) When pointer points at 8
Total number of outcomes = 8
Number of favourable outcomes = 1
∴ P(8)

= 1/8

(ii) When pointer points at an odd number
Number of odd numbers from 1 to 8 = 4
[∵ Odd numbers are 1, 3, 5 and 7]
⇒ Number of favourable outcomes = 4
∴ P(odd) = 

(iii) When pointer points at a number greater than 2 
Number of numbers greater than 2 = 6
[∴ The numbers 2, 3, 4, 5, 6, 7 and 8 are greater than 2]
⇒ Number of favourable outcomes = 6

∴ P(greater than 2) = 

(iv) When pointer points a number less than 9: 
Number of numbers less than 9 = 8
[a The numbers 1, 2, 3, 4, 5, 6, 7 and 8 are less than 9]
∴ Number of favourable outcomes = 8

P(greater than 9)

Q13. 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.
Ans: Since, numbers on a die are 1, 2, 3, 4, 5, and 6.
∴ Number of total outcomes = 6
(i) For prime numbers
Since 2, 3, and 5 are prime numbers,
∴ Favourable outcomes = 3
P(prime) =

(ii) For a number lying between 2 and 6 
Since the numbers between 2 and 6 are 3, 4 and 5
∴ Favourable outcomes = 3

(iii) For an odd number
Since 1, 3 and 5 are odd numbers.
⇒ Favourable outcomes = 3

Q14. 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 
Ans: Number of cards in deck = 52
∴ Total number of possible outcomes = 52
(i) For a king of red colour
∵ Number of red colour kings = 2
[∵ Kings of diamond and heart are red]
∴ Number of favourable outcomes = 2
E(red king)

(ii) For a face card 
∵ 4 kings, 4 queens and 4 jacks are face cards
∴ Number of face cards = 12
⇒ Number of favourable outcomes = 12
∴ P(face)

(iii) For a red face card
Since cards of diamond and heart are red
∴ There are [2 kings, 2 queens, 2 jacks] 6 cards are red
⇒ Favourable outcomes = 6
∴ P(red face)

 
(iv) For a jack of hearts 
Since there is only 1 jack of hearts.
∴ Number of favourable outcomes = 1
P(jack of hearts)

(v) For a spade
∵There are 13 spades in a pack of 52 cards:
∴ Favourable outcomes are 13.
P(spade)

(vi) For the queen of diamonds
∵ There is only one queen of a diamond.
∴ Number of favourable outcomes = 1
P(queen of diamonds)

Q 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? and (b) a queen?
Sol: We have five cards.
∴ All possible outcomes = 5
(i) For a queen:
∵ Number of queens = 1

(ii) The queen is drawn and put aside,
∴ Only 5 − 1 = 4 cards are left,
⇒ All possible outcomes = 4
(a) For an ace:
∵ There is only one ace
∴ Number favourable outcomes = 1

(b) For a queen:
Since, the only queen has already been put aside.
∴ Number of possible outcomes = 0

Q 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.
Sol: We have
Number of good pens = 132
Number of defective pens = 1 2
∴ Total number of pens = 132 + 12 = 144
For good pens:
∵ There are 132 good pens
∴ Number of favourable outcomes = 132

Q 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?
Sol: Since, there are 20 bulbs in the lot.
∴ Total number of possible outcomes = 20
(i) ∵ Number defective bulbs = 4
i.e., Favourable outcomes = 4

(ii) ∵ The bulb drawn above is not included in the lot.
∴ Remaining number of bulbs = 20 − 1 = 19.
⇒ Total number of possible outcomes = 19.
∵ Number of bulbs which are not defective = 19 − 4 = 15
⇒ Favourable number of outcomes = 15

Q 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 and (iii) a number divisible by 5.

Sol: We have:
Total number of discs = 90
∴ Total number of possible outcomes = 90
(i) For a two-digit number:
Since the two-digit numbers are 10, 11, 12, ….., 90.
∴ Number of two-digit numbers = 90 − 9 = 81
[∵ 1, 2, 3, 4, 5, 6, 7, 8, and 9 are 1-digit numbers]
⇒ Number of favourable outcomes = 81

(ii) For a perfect square:
Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81
∴ Number of perfect numbers = 9
⇒ Number of favourable outcomes = 9

(iii) For a number divisible by 5:
Numbers divisible by 5 [from 1 to 90] are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
i.e. There are 18 number (1 to 90) which are divisible by 5.
∴ Number of favourable outcomes = 18

Q 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? and (ii) D?
Sol: Since there are six faces of the given die and these faces are marked with letters

∴ Total number of letters = 6
⇒ Number of possible outcomes = 6
(i) For the letter A 
∵ Two faces are having the letter A.
∴ Number of favourable outcomes = 2

(ii) For the letter D:
∵ Number of D’s = 1
∴ Number of possible outcomes = 1

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

Sol: Here, Area of the rectangle = 3 m × 2 m = 6 m²
And, the area of the circle = πr²

∴ Probability for the die to fall inside the circle

Q 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?
Sol: Total number of ball pens = 144
⇒ All possible outcomes = 144
(i) Since there are 20 defective pens
∴ Number of good pens 144 − 20 = 124
⇒ Number of favourable outcomes = 124
∴ Probability that she will buy it

(ii) Probability that she will not buy it
=1 − [Probability that she will buy it]

Q 22: Two dice one blue and one grey, are thrown at the same time. Now 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.
Sol: 

∵ The two dice are thrown together.
∴ Following are the possible outcomes:
(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).
⇒ Number of all possible outcomes is 6 × 6 = 36.
(i) Let the required probability be P(E).
(a) ∵ The sum on two dice is 3 for: (1, 2) and (2, 1)
∴ Favourable outcomes = 2

(b) a The sum on two dice is 4 for: (1, 3), (2, 2) and (3, 1).
∴ Number of favourable outcomes = 3

(c) ∵ The sum on two dice is 5 for:
(1, 4), (2, 3), (3, 2) and (4, 1)
∴ Number of favourable outcomes = 4

(d) The sum on two dice is 6 for:
(1, 5), (2, 4), (3, 3), (4, 2) and (5, 1)
∴ Number favourable outcomes = 5

(e) The sum on two dice is 7 for:
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1)
∴ Number of favourable outcomes = 6

(f) The sum on two dice is 9 for:
(3, 6), (4, 5), (5, 4) and (6, 3)
∴ Number of favourable outcome = 4

(g) The sum on two dice is 10 for:
(4, 6), (5, 5), (6, 4)
∴ Number of favourable outcomes = 3

(h) The sum on two dice is 11 for:
(5, 6) and (6, 5)
∴ Number of favourable outcomes = 2

Thus, the complete table is as under:

(ii) No. The number of all possible outcomes is 36 and not 11.
∴ The argument is not correct.

Q 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.
Sol: Let T denotes the tail and H denotes the head.
∴ All the possible outcomes are:
HHH, HHT, HTT, TTT, TTH, THT, TTH, HTH
∴ Number of all possible outcomes = 8
Let the event that Hanif will lose the game be denoted by E.
∴ Favourable events are:
HHT, HTH, THH, THT , TTH, HTT
⇒ Number of favourable outcomes = 6

Q 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.]
Sol: Since, throwing a die twice or throwing two dice simultaneously is the same.
∴ All possible outcomes 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)
∴ All possible outcomes = 36
(i) Let E be the event that 5 does not come up either time, then
The favourable outcomes are [36 − (5 + 6)] = 25

(ii) Let N be the event that 5 will come up at least once, then Number of favourable outcomes = 5 + 6 = 11

Q 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.
Sol: (i) Argument is incorrect.
The possible outcomes are- (HH), (HT), (TH), (TT)
∴ 

(ii) Argument is correct.
Possible outcomes = 1, 2, 3, 4, 5, 6
Odd numbers are = 1, 3, 5,
Hence P (an odd number) = 3/6 = 1/2
Even numbers are = 2, 4, 6,
Hence P (an even number) = 3/6 = 1/2

Exercise 13.1

Q1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Sol. We can calculate the mean as:

⇒ 
Thus, the mean number of plants per house is 8.1.
Since values of x_i and f_i are small
∴ We have used the direct method.

Q2. Consider the following distribution of daily wages of 50 workers of a factory. 

Find the mean daily wages of the workers of the factory by using an appropriate method.
Sol. 
To find the class mark for each interval, the following relation is used.

Class size (h) of this data = 20
taking 550 as assured mean (a), d_i, u_j and f_iu_j can be calculated as follows.

From the table, it can be observed that

Therefore, the mean daily wage of the workers of the factory is Rs 545.20

Q3. 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.

Sol. Let the assumed mean, a = 16
∵ Class interval h = 2

Now, we have the following table:

Since 
∴ 

Thus, the missing frequency is  84.

Q4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Sol. Let the assumed mean a = 75.5

∴ Class interval h = 3

Now, we have the following table:

Thus, the mean heartbeat per minute is 75.9.

Q5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Sol. Let the assumed mean a = 60
Here, Class interval h = 3

Now, we have the following table:

Thus, the average number of mangoes per box = 57.19.

Q6. 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.
Sol. Let the assumed mean, a = 225
And class interval h = 50

∴ We have the following table:

Thus, the mean daily expenditure of food is Rs 211.

Q7. To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Sol. Let the assumed mean a = 0.14
Here, class interval h = 0.04

∴ We have the following table:

Thus, mean concentration of SO₂ in air is 0.099 ppm.

Q8. 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.

Sol. Using the direct method, we have the following table:

Thus, mean number of days a student remained absent = 12.48 (Approx.)

Q9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Sol. Let  assumed mean a =70
∴ Class interval h =10

Now, we have the following table:

Thus, the mean literacy rate is 69.43%.

Exercise 13.2

Q1: The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Sol: Mode:
Here, the highest frequency is 23.
The frequency 23 corresponds to the class interval 35 − 45.
∴ The modal class is 35 − 45
Now, Class size (h) = 10
Lower limit (l) = 35 Frequency of the modal class (f₁) = 23
Frequency of the class preceding the modal class f₀ = 21
Frequency of the class succeeding the modal class f₂ = 14

Mean
Let assumed mean a = 40
∵  h = 10

∴ Required mean = 35.37 years.

Q2: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.
Sol: Here, the highest frequency = 61.
∵ The frequency 61 corresponds to class 60 − 80
∴ The modal class = 60 − 80
∴ We have: l = 60
h = 20
f₁ = 61
f₀ = 52
f₂ = 38

Thus, the required modal life times of the components are 65.625 hours.

Q3: 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:

Sol: Mode: 
∵ The maximum number of families 40 have their total monthly expenditure is in interval 1500−2000.
∴ Modal class is 1500−2000.
l = 1500, h = 500
f₁ = 40,   f₀ = 24
f₂ = 33

Thus, the required modal monthly expenditure of the families is Rs 1847.83.
Mean: Let assumed mean (a) = 3250
∵ h = 500
∴ We have the following table:

Thus, the mean monthly expenditure = Rs 2662.50.

Q4: The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Sol: Mode: 
Since the class 30 − 35 has the greatest frequency and h = 5
l = 30
f₁ = 10
f₀ = 9
f₂ = 3

Mean:
Let the assumed mean (a) = 37.5
Since, h = 5
∴ We have the following table:

Thus ,the required mean is 29.2

Q5: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.
Sol: The class 4000−5000 has the highest frequency i.e., 18
∴ h = 1000
l = 4000
f₁ = 18
f₀ = 4
f₂ = 9
Now,

Thus, the required mode is 4608.7.

Q6: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Sol: ∵ The class 40 − 50 has the maximum frequency i.e., 20
∴ f₁ = 20,
f₀ = 12,
f₂ = 11 and h = 10
Also
l = 40

= 40 + 4.7 = 44.7
Thus, the required mode is 44.7

Exercise 13.3

Q1: The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Sol: 
Median 
Let us prepare a cumulative frequency table:

Now, we have  
∵ This observation lies in the class 125−145.
∴ 125−145 is the median class.
∴ l = 125, cf = 22
f = 20 and h = 20
Using the formula,

Mean Assumed mean (a) = 135
∵ Class interval (h) = 20

Now, we have the following table:

Mode 
∵ Class 125−145 has the highest frequency.
∴ This is the modal class.
We have:
h = 20
l = 125
f₁ = 20
f₀ = 13
f₂ = 14

We observe that the three measures are approximately equal in this case.

Q2: If the median of the distribution given below is 28.5, find the values of x and y.

Sol: Here, we have n = 60 [∵ ∑f_i = 60]
Now, cumulative frequency table is:

Since, median = 28.5
∴ Median class is 20 − 30
We have: l = 20
h = 10
f = 20
cf = 5 + x

⇒ 57 = 40 + 25 − x
⇒ x = 40 + 25 − 57 = 8
Also 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 − 45 − 8 = 7
Thus, x = 8
y = 7

Q3: A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years on wards but less than 60 years.

Sol: The given table is cumulative frequency distribution. We write the frequency distribution as given below:

∵ The cumulative frequency just greater than n/2 i.e., just greater than 50 is 78.
∴ The median class is 78.
Now n/2 = 50, l = 35, cf = 45, f = 33 and h = 5

Thus, the median age = 35.76 years.

Q4: The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves.
[Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5−126.5, 126.5−135.5, …, 171.5−180.5]
Sol: After changing the given table as continuous classes we prepare the cumulative frequency table:

∑ f_i =40 ⇒ n = 40
Now,

The cumulative frequency just above n/2 i.e., 20 is 29 and it corresponds to the class 144.5−153.5.
So, 144.5−153.5 is the median class.
We have:
n/2 = 20, l = 144.5, f = 12, cf = 17 and h = 9


Q5: The following table gives the distribution of the life time of 400 neon lamps:

Find the median life time of a lamp
Sol: To compute the median, let us write the cumulative frequency distribution as given below:

∑ f_i = 400 ⇒ n = 400

Since, the cumulative frequency just greater than n/2 i.e., greater than 200 is 216.
∴ The median class is 3000−3500
∴ l = 3000, cf = 130, f = 86, h = 500 and n/2 = 200

Thus, median life = 3406.98 hours.

Q6: 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Sol: Median

Since, the cumulative frequency just greater than n/2 i.e., greater than 50 is 76.
∴ The class 7 − 10 is the median class,
We have 
l = 7
cf = 36
f = 40 and  h = 3

Mode
Since the class 7−10 has the maximum frequency.
∴ The modal class is 7−10
So, we have
l = 7, h = 3
f₁ = 40,
f₀ = 30
f₂ = 16

Thus, the required Median = 8.05,
Mean = 8.32 and Mode = 7.88.

Q7: The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Sol: We have

Weight (in kg)

Frequency

(f)

Cumulative

frequency

40-45

2

2 + 0= 2

45-50

3

2 + 3= 5

50-55

8

5 + 8 = 13

55-60

6

13 + 6 = 19

60-65

6

19 + 6 = 25

65-70

3

25 + 3 = 28

70-75

2

28 + 2 = 30

Total

If  = 30

n = 30

The cumulative frequency just more than n/2 i.e., more than 15 is 19, which corresponds to the class 55−60.
n/2 = 15.
l = 55
5f = 6
cf = 13 and h = 5

Thus, the required median weight = 56.67 kg.

Exercise 12.1

Q1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Sol.  Volume of each cube = 64 cm³
∴ Total volume of the two cubes = 2 × 64 cm³ = 128 cm³
Let the edge of each cube = x
∴ x³ = 64 = 4³
⇒ x =4 cm
⇒ Now, the Length of the resulting cuboid ‘l’ = 2x cm = 8cm
⇒ Breadth of the resulting cuboid ‘b’ = x cm = 4 cm
⇒ Height of the resulting cuboid ‘h’ = x cm = 4 cm
∴ Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(8 × 4) + (4 × 4) + (4 × 8)]
= 2 [32 + 16 + 32] cm² = 2 [80] cm² = 160 cm².

Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Sol.  For the cylindrical part
 Radius (r) = 7 cm
 Height (h) = 6 cm
∴ Curved surface area
= 2πrh
= 2 × 227 × 7 × 6 cm² = 264 cm²

For hemispherical part
  Radius (r) = 7 cm
∴ Surface area = 2πr²
= 2 × 227 × 7 × 7 cm² = 308 cm²
∴ Total surface area = CSA of cylinder + CSA of hemisphere
= (264 + 308) cm² = 572 cm².

Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Sol. Here, r = 3.5 cm
∴ h = (15.5 − 3.5) cm = 12.0 cm
⇒ Surface area of the conical part = πrl
⇒ Surface area of the hemispherical part = 2πr²

∴ l² = (12)² + (3.5)²
l² = 144 + 12.25 = 156.25
⇒ l = 12.5 cm

∴ Total surface area of the toy
= πr² + 2πr² = πr (l + 2r) cm²
= 227 × 3510 (12.5 + 2 × 3.5) cm²
= 11 × (12.5 + 7) cm²
= 11 × 19.5 cm² = 214.5 cm²

Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Sol. Side of the block = 7 cm
⇒ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid:
= [Total S.A. of the cubical block] + [S.A. of the hemisphere] − [Base area of the hemisphere]
= (6 × l²) + 2πr² − πr²

= [6(7)]² + 2π(3.5)² − π(3.5)² cm²
= 6(7)² + π(3.5)²
= 6 × 497 + 22 × 7 × 72 × 2 cm²
= 294 + 777 cm² 
= 332.5 cm²

 
Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Sol. Let ‘l’ be the side of the cube.

Now, the diameter of the hemisphere = Edge of the cube = l

So, the radius of the hemisphere = l/2

∴ The total surface area of solid = surface area of cube + CSA of the hemisphere – Area of the base of the hemisphere

The surface area of the remaining solid = 6 (edge)²+2πr²-πr²

= 6l² + πr²

= 6l²+π(l/2)²

= 6l²+πl²/4

= l²/4 (24+π) sq. units

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig.). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Sol. Radius of the hemispherical part

= 5/2 mm = 2.5 mm

∴ Surface area of one hemispherical part = 2πr²

⇒ Surface area of both hemispherical parts
= 2 (2πr²) = 4 × 227 × 2510 ² mm²
= 4 × 227 × 2510 × 2510 mm²
Area of cylindrical part = 2πrh = 2 × 227 × 2.5 × 9 mm² = 2 × 227 × 2510 × 9 mm²
.∴ Total surface area
= 2 × 22 × 25 × 10 × 97 + 4 × 22 × 25 × 25 × 107 mm²
= 2 × 22 × 25 × 107 × 9 + 5070 mm² = 44 × 2510 mm² = 44 × 52 cm² 
= 220 mm²

Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m². (Note that the base of the tent will not be covered with canvas.)
Sol. 

For cylindrical part:
 Radius (r) = 4/2 m = 2 m
 Height (h) = 2.1 m
∴ Curved surface area = 2πrh = 2 × 227 × 2 × 2110 m²
For conical part:
Slant height (l) = 2.8 m
Base radius (r) = 2 m
.∴ Curved surface area = πrl = 227 × 2 × 2810 m²
.∴ Total surface area
= [Surface area of the cylindrical part] + [Surface area of conical part]
= 2 × 22 × 2 × 217 × 10 + 22 × 2 × 287 × 10 m²
= 2 × 22 × 2 × 217 × 10 × 42 + 2810 m²
= 2 × 22 × 257 × 10 m² = 22 × 7010 m² = 44 m²

Cost of the canvas used 
Cost of 1 m² of canvas = Rs 500
∴ Cost of 44 m² of canvas = Rs 500 × 44 = Rs. 22000

Q8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Sol. 

For cylindrical part:
 Height = 2.4 cm
 Diameter = 1.4 cm
⇒ Radius (r) = 0.7 cm
For conical part:
 Base area (r) = 0.7 cm
 Height (h) = 2.4 cm
∴ Slant height (l) = √r² + h²√(0.7)² + (2.4)² = √(0.49 + 5.76) = √6.25 = 2.5 cm
∴ Curved surface area of the conical part = πrl = 227 × 0.7 × 2.5 cm² = 22 × 7 × 25100 cm² = 550100 cm²
Base area of the conical part = πr² = 227 × 710² cm² = 22 × 7100 cm² = 154100 cm²
∴ Total surface area of the remaining solid:
= [Total SA of cylindrical part] + [Curved surface area of conical part] − [Base area of conical part]
= 1364 + 550100 + 154100 cm²
= 1914100 − 154100 = 1760100 cm² = 17.6 cm²

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Sol. Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
∴ Total surface area = 2πrh + 2πr² = 2πr (h + r)

= 2 × 227 × 3510 (10 + 3510) cm²
= 22 × 13510 cm² = 297 cm²

Curved surface area of a hemisphere = 2πr²
.∴ Curved surface area of both hemispheres
= 2 × 2πr² = 4 × 227 × 3510 × 3510 cm² = 154 cm²

Base area of a hemisphere = πr²
.∴ Base area of both hemispheres = 2πr²
= 2 × 227 × (3.5)² = 2 × 22 × 35 × 357 × 100 cm² = 77 cm²

∴ Total surface area of the remaining solid
= 297 cm² + 154 cm² − 77 cm²
= (451 − 77) cm² = 374 cm².

Exercise 12.2

[Unless stated otherwise, take π = 22/7]
Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Sol. Here, r = 1 cm and h = 1 cm.
∵ Volume of the conical part = 1/3 πr²h
Volume of the hemispherical part =2/3 πr³
∴ Volume of the solid shape
= 13 πr²h + 23 πr³ = 13 πr² [h + 2r]
= 13 π (1)² [1 + 2 (1)] cm³ = 13 π × 1 × [3] cm³
= 3π3 cm³ = π cm³

Q2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Sol. Here, diameter = 3 cm
⇒ Radius (r)= 3/2 cm
Total height = 12 cm
Height of a cone (h₁) = 2 cm
∴ Height of both cones = 2 × 2 = 4 cm
⇒ Height of the cylinder (h₂) = (12 − 4) cm = 8 cm.
Now, volume of the cylindrical part = πr²h₂
Volume of both conical parts = 2 [ 13 πr²h₁]
.∴ Volume of the whole model
= πr²h₂ + 23 πr²h₁ = π² [h₂ + 23 h₁]
= 227 × (3/2)³ × 8 + 23 cm³ = 227 × 94 × 24 + 43 cm³
= 227 × 34 × 28 cm³ = 223 cm³ = 66 cm³

Q3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

Sol. Since a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.
Diameter = 2.8 cm
⇒ Radius = 1.4 cm
∴ Length (height) of the cylindrical part = 5 cm − (1.4 + 1.4) cm
= 5 cm − 2.8 cm = 2.2 cm
Now, volume of the cylindrical part
= πr²h
Volume of a hemispherical end = 2/3 πr³
Volume of both the hemispherical ends 

∴ Volume of a gulab jamun

= πr²h + 43 πr³
= πr² [h + 43]
= 227 × (1.4)² [2.2 + 43 (1.4)] cm³
= 227 × 1.96 [2.2 + 43 (1.4)] cm³
= 22 × 2 × 147 × 10 [66 + 56] cm³
= 44 × 14100 cm³ = 122 cm³

⇒ Volume of 45 gulab jamuns = 45 × 44 × 14 × 14 × 122100 × 30 cm³ = 15 × 44 × 14 × 14 × 1221000 cm³
Since the quantity of syrup in gulab jamuns
= 30% of [volume] = 30% of 15 × 44 × 14 × 14 × 1221000 cm³ = 338.184 cm³
= 338 cm³ (approx.)

Q4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig.).

Sol. Dimensions of the cuboid are 15 cm, 10 cm and 3.5 cm.
∴ Volume of the cuboid = 15 × 10 × 3510 cm³ = 525 cm³
Since each depression is conical with base radius (r) = 0.5 cm and depth (h) = 1.4 cm,
.∴ Volume of each depression (cone)
= 13 πr²h = 13 × 227 × 510² × 1410 cm³
Since there are 4 depressions,
.∴ Total volume of 4 depressions
= 4 × 1 × 22 × 5 × 5 × 143 × 7 × 10 × 10 × 10 cm³
= 4 × 1130 cm³ = 4430 cm³
Now, volume of the wood in the entire stand
= [Volume of the wooden cuboid] − [Volume of 4 depressions]
= 525 − 4430 cm³
= 15750 − 4430 cm³ = 1570630 cm³
= 523.53 cm³

Q5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Sol. Height of the conical vessel (h) = 8 cm
Base radius (r) = 5 cm
∴ Volume of the cone = 1/3 πr²h
= 13 × 227 × (5)² × 8 cm³
= 440021 cm³
Since Volume of the cone = [Volume of water in the cone]
.∴ [Volume of water in the cone] = 440021 cm³
Now, Total volume of lead shots = 14 of [Volume of water in the cone]
= 14 × 440021 cm³ = 110021 cm³
Since, radius of a lead shot (sphere) (r) = 0.5 cm,
.∴ Volume of 1 lead shot = 43 πr³ = 43 × 227 × 510 × 510 × 510 cm³
.∴ Number of lead shots = Total volume of lead shotsVolume of 1 lead shot
= 110021 ÷ 4 × 22 × 5 × 5 × 53 × 7 × 1000
= 100
Thus, the required number of lead shots = 100.

Q6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14) 

Sol. Height of the big cylinder (h) = 220 cm
Base radius (r) = 24/2 cm = = 12 cm
∴ Volume of the big cylinder = πr²h = π (12)2 × 220 cm³
Also, height of smaller cylinder (h₁) = 60 cm
Base radius (r₁) = 8 cm
∴ Volume of the smaller cylinder πr₁²h₁ = π (8)2 × 60 cm³
∴ Volume of iron
= [Volume of big cylinder] + [Volume of the smaller cylinder]
= π × 220 × 122 + π × 60 × 8² cm³
= 3.14 [220 × 12 × 12 + 60 × 8 × 8] cm³ = 314/100 [20 × 144 + 60 × 64] cm³

= 314100 × [31680 + 3840] cm³ = 314100 × 35520 cm³
Mass of iron = 8 × 314 × 35520100 g = 89226240100 g = 892262410000 g
= 892.2624 kg 
= 892.26 kg.

Q7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Sol. Height of the conical part = 120 cm.
Base radius of the conical part = 60 cm.
∴ Volume of the conical part = 13 × 227 × 60² × 120 cm³
Radius of the hemispherical part = 60 cm.
.∴ Volume of the hemispherical part = 23 × 227 × 60³ cm³
.∴ Volume of the solid
= [Volume of conical part] + [Volume of hemispherical part]
= 13 × 227 × 60² × 120 + 23 × 227 × 60³ cm³
= 2 × 22 × 60 × 60 × 407 cm³ = 63360007 cm³
Volume of the cylinder = πr²h
= 227 × 60² × 180 cm³ = 22 × 60 × 60 × 1807 cm³
= 142560007 cm³

⇒ Volume of water in the cylinder = 142560007 cm³
.∴ Volume of the water left in the cylinder
= 142560007 − 63360007 cm³
= 79200007 cm³
= 1131428.57142 cm³
= 1131428.571421000000 m³ [∵ 1000000 cm³ = 1 m³]
= 1.13142857142 m³ = 1.131 m³ (approx.)

Q8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Sol. Volume of the cylindrical part
= πr²h = 3.14 × 1² × 8 cm³ [∵ Radius = 2/3 = 1 cm, height (h) = 8 cm]
= 314100 × 8 cm³
Volume of the spherical part
= 43 πr₁³   | Here r₁ = 8.52 cm
= 43 × 314100 × 8520 × 8520 × 8520 cm³
Total volume of the glass-vessel
= 314100 × 8 + 314100 × 43 × 85 × 85 × 858000
= 314100 [8 + 4 × 85 × 85 × 8524000] cm³
= 314100 [8 + 6141256000] cm³
= 314100 × 48000 + 6141256000 cm³
= 314100 × 6621256000 cm³
= 314100 × 529724 cm³
= 157100 × 529724 cm³ = 8316292400 cm³
≈ 346.51 cm³ (approx.)
⇒ Volume of water in the vessel = 346.51 cm³

Since the child finds the volume as 345 cm³
∴ The child’s answer is not correct
⇒ The correct answer is 346.51 cm³.

Page No 158

Use π = 22/7 (unless stated otherwise)
Q1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Sol: Here,
r = 6 cm
θ = 60°

∴ Using, the Area of a sector =  
We have,
Area of the sector with r  = 6 cm and θ = 60°


Q2: Find the area of a quadrant of a circle whose circumference is 22 cm.
Sol: Let the radius of the circle = r
∴ 2πr = 22

⇒  
Here θ = 90°

∴ Area of the quadrant  of the circle,


Q3: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Sol: [Length of minute hand] = [radius of the circle]
⇒ r = 14 cm
∵ Angle swept by the minute hand in 60 minutes = 360°
∴ Angle swept by the minute hand in 5 minutes =  
Now, area of the sector with r = 14 cm and θ = 30°

Thus, the required area swept by the minute hand by 5 minutes = 154/3 cm².

Q4: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major sector. (Use π = 3.14)
Sol: Given the radius of the circle = 10 cm
Angle subtend by chord at centre = 90° …(i)
(i) Area of the minor segment = (Area of the sector OAB) – (Area of ΔAOB formed with radius and chord)


= 3.14 x 25 – 50 = 78.5 – 50 = 28.5 cm² 
(ii) Area of major sector = Area of the circle – Area of the minor sector

=(1- 1/4) πr²

=3/4 πr²

=3/4 x π(10)²

= (3.14×10²)-78.5

= 235.5 cm²

Q5: In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
 (i) the length of the arc
 (ii) area of the sector formed by the arc
 (iii) area of the segment formed by the corresponding 
Sol:
 
Here, radius = 21 cm and θ = 60°

(i) Circumference of the circle = 2πr

(ii) Area of the sector with sector angle 60°

(iii)  Area of the segment formed by the corresponding chord – area of the sector – area of the Δ formed between chord and radius of the circle


Q6:  A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.  (Use π = 3.14 and √3 = 1.73)
Sol:  

 Radius of the circle = 15 cm
Central angle subtends by chord = 60⁰
Area of sector = 
= 117.75 cm²

Area of the triangle formed by radii and chord

Area of the minor segment = Area of the sector -Area of the triangle formed by radii and chord
= 117.75-97.31 =20.44 cm² 
Area of the circle = πr² 
= 3.14 x 15 x 15 = 706.5 cm² 
Area of the major segment = Area of the circle – Area of the minor segment
= 706.5 – 20.44 = 686.06 cm²

Q7: A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)
Sol:  Here, θ = 120° and r = 12 cm

In Δ OAB, ∠O = 120°
⇒∠A + ∠B = 180° − 120 = 60°
∵ OB = OA = 12 cm ⇒∠A = ∠B = 30°
So,  
⇒  

In right Δ AMO, 12² − 6² = AM²
⇒ 144 − 36 = AM²
⇒ 108 = AM²
⇒ 
⇒ 
⇒ 

Now, from (2),

= 36 × 1.73 cm² = 62.28 cm²     …(3)
From (1) and (3)
Area of the minor segment = [Area of minor segment] − [Area of Δ AOB]
= [150.72 cm²] − [62.28 cm²] = 88.44 cm².

Q8:  A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). 
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)

Sol: Here, Length of the rope = 5 m
∴ Radius of the circular region grazed by the horse = 5 m

(i) Area of the circular portion grazed
     [∵ θ = 90° for a square field.]

(ii) When length of the rope is increased to 10 m,
∴ r = 10 m
⇒ Area of the circular region where θ = 90°.

∴ Increase in the grazing area = 78.5 − 19.625 m² = 58.875 m².

Page  No 159
Q9:  A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 
Find: 
(i) the total length of the silver wire required. 
(ii) the area of each sector of the brooch.

Sol: Diameter of the circle = 35 mm

∴  
(i) Diameter (D) = 35 mm

Total number of diameters to be considered= 5

Now, the total length of 5 diameters that would be required = 35×5 = 175

Circumference of the circle = 2πr

Or, C = πD 

= 22/7×35 = 110
∴ Total length of the silver wire = 110 + 175 mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,
∴ Sector angle 
⇒ Area of each sector  


Q10: An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Sol: Here, radius (r) = 45 cm
Since circle is divided in 8 equal parts,
∴ Sector angle corresponding to each part

⇒ Area of a sector (part)

∴ The required area between the two ribs  

Q 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.
Sol:  Here, radius (r) = 25 cm
Sector angle (θ) = 115°
∴ Area cleaned by each sweep of the blades
   [∵ Each sweep will have to and fro movement]

Q12: To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. 
Find the area of the sea over which the ships are warned. (Use π = 3.14)
Sol: Here, Radius (r) = 16.5 km
Sector angle (θ) = 80°
∴ Area of the sea surface over which the ships are warned


Q13: A round table cover has six equal designs as shown in Fig. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm². (Use √3 = 1.7)

Sol:  Here,    r = 28 cm
Since, the circle is divided into six equal sectors.
∴ Sector angle  
∴ Area of the sector with θ = 60° and r = 28 cm

Now, area of 1 design
= Area of segment APB
= Area of sector − Area of ΔAOB  …(2)
In ΔAOB, ∠AOB = 60°, OA = OB = 28 cm
∴   ∠OAB = 60° and ∠OBA = 60°
⇒ ΔAOB is an equilateral triangle.
⇒ AB = AO = BO
⇒ AB = 28 cm

Draw OM ⊥ AB
∴ In right ΔAOM, we have

⇒ 
⇒ 

       …(3)

Now, from (1), (2) and (3), we have:
Area of segment APQ = 410.67 cm² − 333.2 cm² = 77.47 cm²
⇒ Area of 1 design = 77.47 cm²
∴ Area of the 6 equal designs = 6 × (77.47) cm²
= 464.82 cm²
Cost of making the design at the rate of Rs 0.35 per cm²,
= Rs 0.35 × 464.82
= Rs 162.68.

Q14: Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is




Sol: Here, radius (r)= R
Angle of sector (θ)= p°
∴ Area of the sector  
Thus, the option is correct.

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

Q1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is 
(A) 7 cm 
(B) 12 cm 
(C) 15 cm 
(D) 24.5 cm
Sol. From figure
 
OQ² = OP² + PQ²
(25)² = OP² + (24)²
⇒ 625 – 576 = OP²
⇒ 49 = OP²
⇒ 
OP = 7cm
Radius of the circle = 7cm.
Hence, correct option is (a).

Q2. In figure, 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°

Sol. ∠OPT = 90°; ∠OQT = 90°; ∠POQ = 110⁰
TPOQ is a quadrilateral
⇒ ∠PTQ + ∠POQ = 180°
⇒ ∠PTQ + 110⁰ = 180⁰
⇒ ∠PTQ  = 180⁰ – 100⁰ = 70⁰

^{Hence, correct option is (b).}

Q3. Choose the correct option: If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to
(A) 50° 
(B) 60° 
(C) 70° 
(D) 80°
Sol. In ΔOAP and ΔOBP
OA = OB    [Radii]
PA = PB
[Length of tangents from an external point are equal]
OP = OP     [common]



Hence, Option (a) is correct

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Sol. AB is the diameter of the circle, p and q are two tangents
OA ⊥ p and OB ⊥ q and ∠1 = ∠2 = 90⁰
⇒ p ║q    [∠1 and ∠2 are alternate angles]

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Sol. XY tangent to the circle C(O, r) at B and AB ⊥ XY

Join OB
∠ABY = 90º   
∠OBY = 90º
[Radius through the point of contact is ⊥ to the tangent]
∴ ∠ABY + ∠OBY = 180⁰
⇒ ABO is collinear.
∴ AB passes through the centre of the circle.

Q6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Sol. OP = Radius of the circle
OA = 5 cm; AP = 4 cm

∠OPA = 90°   [Radius and tangent are ⊥ar]
OA² = AP² + OP²    [By Pythagoras theorem]
5² = 4² + OP² 
⇒ 25 = 16 + OP²
⇒ 25 – 16 = OP²
⇒ 9 = OP²
⇒ OP = √ 9 = 3
Radius = 3 cm

Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Sol. The radius of larger circle = 5 cm and the radius of a smaller circle = 3 cm
OP ⊥ AB    [Radius of the circle is perpendicular to the tangent]
AB is a chord of the larger circle
∴ OP bisect  AB   AP = BP
In ΔOAP     OA² =    AP² + OP²
⇒    (5)2 =   AP² + (3)²
⇒    AP² =   25 – 9 = 16
⇒  
AB = 2AP =   2 x 4 = 8 cm

Q8. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that: AB + CD = AD + BC

Sol. 
AP = AS    …(i)    [Lengths of tangents from an external point are equal]
BP = BQ    …(ii)
CR = CQ …(iii)
DR = DS   …(iv)
Adding equations (i), (ii), (iii) and (iv), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) +(BQ + CQ
⇒ AB + CD = AD + BC

Q9. In the figure, 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°.

Sol. Given: Two parallel tangents to a circle with centre O. Tangent AB with point of contact C intersects XY at A and X’Y’ at B
To Prove: ∠AOB = 90°
Construction: Join OA, OB and OC
Proof: In ΔAOP and ΔAOC
AP = AC [Lengths of tangents]
OP = OC [Radii]
OA = OA
⇒ ΔAOP ≅ ΔAOC   [SSS congruence rule]
⇒ ∠PAO = ∠CAO    [C.P.C.T]
∠PAC = 2 ∠OAC    …(i)
Similarly ∠QBC = 2 ∠OBC    …(ii)
Adding (i) and (ii), we get
∠PAC + ∠QBC = 2      [∠OAC + ∠OBC]
∠PAC + ∠QBC = 180°
[interior consecutive angle on same side of transversal]
2 = 180° [∠OAC + ∠OBC]
⇒ ∠OAC + ∠OBC = 90°
In ΔAOB, ∠AOB + [∠OAC + ∠OBC] = 180°
⇒ ∠AOB + 90° = 180°
⇒ ∠AOB = 90°

Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Sol. PA and PB are two tangents, and A and B are the points of contact of the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
[Radius and tangent are perpendicular to each other]
In quadrilateral OAPB
(∠OAP + ∠OBP) + ∠APB + ∠AOB = 360⁰

⇒ 180° + ∠APB + ∠AOB = 360°
∠APB + ∠AOB = 360° – 180° = 180°

Q11. Prove that the parallelogram circumscribing a circle is a rhombus. 
Sol. Parallelogram ABCD circumscribing a circle with centre O.

OP⊥ AB and OS ⊥ AD
In ΔOPB and ΔOSD, ∠OPB = ∠OSD [Each 90°]
OB = OD
[Diagonals of ║gm bisect each other]
OP = OS    [Radii]
⇒  ΔOPB ≅ ΔOSD [RHS congruence rule]
PB = SD     [C.P.C.T] ...(i)
AP = AS [Lengths of tangents]... (ii)
Adding (i) and (ii)
AP + PB = AS + DS
⇒   AB AD
Similarly  AB = BC = CD = DA
∴ ║gm ABCD is a rhombus

Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Sol. BD = 8 cm and DC = 6 cm
BE = BD = 8 cm; CD= CF = 6 cm;

Let AE=AF = x cm
In ΔABC   a = 6 + 8 = 14 cm;
b = (x + 6) cm;
c = (x +8)


ar ΔABC = ar ΔOBC + ar ΔOCA + ar ΔOAB

From (i) and (ii)

⇒ 3x(x+14) = (x + 14)² 
⇒ 3x² + 42x = x² + 196 + 28x
⇒ 2x² + 14x – 196 = 0
⇒ x² + 7x – 98 = 0
⇒ x² + 14x – 7x – 98 = 0
⇒ x(x + 14) – 7(x + 14) = 0
⇒ (x – 7)(x + 14) = 0 ⇒ x = 7
AB = x + 8 = 7 + 8 = 15 cm
AC = x + 6 = 7 + 6 = 13 cm

Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Sol. 

AB touches P and BC, and CD and DA touch the circle at Q, R and S.
Construction: Join OA, OB, OC, OD and OP, OO, OR, OS
∴ ∠1 = ∠2    [OA bisects ∠POS]
Similarly


Similarly ∠AOB + ∠COD = 180°
Hence, opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of a circle.

Exercise 10.1

Q1. How many tangents can a circle have?
Sol. A circle can have an infinite number of tangents.

Q2. 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 ………..
Sol. 
(i) Exactly one
(ii) secant
(iii) two
(iv) point of contact.

Q3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length of PQ is:
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) √119 cm

Ans. (d)
Sol. 

Radius of the circle = 5 cm
OQ = 12 cm
∠OPQ = 90⁰
[The tangent to a circle is perpendicular to the radius through the point of contact]
PQ² = OQ² – OP² [By Pythagoras theorem]
PQ² = 12² – 5² = 144 – 25 = 119
PQ = √119 cm
Hence correct option is (d)

Q4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other a secant to the circle.
Sol. A line ‘m’ is parallel to the given line ‘n’ and a line ‘l’ which is secant is parallel to the given line ‘n’.

Exercise 10.2

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

Q1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is 
(A) 7 cm 
(B) 12 cm 
(C) 15 cm 
(D) 24.5 cm
Sol. From figure
 
OQ² = OP² + PQ²
(25)² = OP² + (24)²
⇒ 625 – 576 = OP²
⇒ 49 = OP²
⇒ 
OP = 7cm
Radius of the circle = 7cm.
Hence, correct option is (a).

Q2. In figure, 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°

Sol. ∠OPT = 90°; ∠OQT = 90°; ∠POQ = 110⁰
TPOQ is a quadrilateral
⇒ ∠PTQ + ∠POQ = 180°
⇒ ∠PTQ + 110⁰ = 180⁰
⇒ ∠PTQ  = 180⁰ – 100⁰ = 70⁰

^{Hence, correct option is (b).}

Q3. Choose the correct option: If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to
(A) 50° 
(B) 60° 
(C) 70° 
(D) 80°
Sol. In ΔOAP and ΔOBP
OA = OB    [Radii]
PA = PB
[Length of tangents from an external point are equal]
OP = OP     [common]



Hence, Option (a) is correct

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Sol. AB is the diameter of the circle, p and q are two tangents
OA ⊥ p and OB ⊥ q and ∠1 = ∠2 = 90⁰
⇒ p ║q    [∠1 and ∠2 are alternate angles]

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Sol. XY tangent to the circle C(O, r) at B and AB ⊥ XY

Join OB
∠ABY = 90º   
∠OBY = 90º
[Radius through the point of contact is ⊥ to the tangent]
∴ ∠ABY + ∠OBY = 180⁰
⇒ ABO is collinear.
∴ AB passes through the centre of the circle.

Q6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Sol. OP = Radius of the circle
OA = 5 cm; AP = 4 cm

∠OPA = 90°   [Radius and tangent are ⊥ar]
OA² = AP² + OP²    [By Pythagoras theorem]
5² = 4² + OP² 
⇒ 25 = 16 + OP²
⇒ 25 – 16 = OP²
⇒ 9 = OP²
⇒ OP = √ 9 = 3
Radius = 3 cm

Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Sol. The radius of larger circle = 5 cm and the radius of a smaller circle = 3 cm
OP ⊥ AB    [Radius of the circle is perpendicular to the tangent]
AB is a chord of the larger circle
∴ OP bisect  AB   AP = BP
In ΔOAP     OA² =    AP² + OP²
⇒    (5)2 =   AP² + (3)²
⇒    AP² =   25 – 9 = 16
⇒  
AB = 2AP =   2 x 4 = 8 cm

Q8. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that: AB + CD = AD + BC

Sol. 
AP = AS    …(i)    [Lengths of tangents from an external point are equal]
BP = BQ    …(ii)
CR = CQ …(iii)
DR = DS   …(iv)
Adding equations (i), (ii), (iii) and (iv), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) +(BQ + CQ
⇒ AB + CD = AD + BC

Q9. In the figure, 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°.

Sol. Given: Two parallel tangents to a circle with centre O. Tangent AB with point of contact C intersects XY at A and X’Y’ at B
To Prove: ∠AOB = 90°
Construction: Join OA, OB and OC
Proof: In ΔAOP and ΔAOC
AP = AC [Lengths of tangents]
OP = OC [Radii]
OA = OA
⇒ ΔAOP ≅ ΔAOC   [SSS congruence rule]
⇒ ∠PAO = ∠CAO    [C.P.C.T]
∠PAC = 2 ∠OAC    …(i)
Similarly ∠QBC = 2 ∠OBC    …(ii)
Adding (i) and (ii), we get
∠PAC + ∠QBC = 2      [∠OAC + ∠OBC]
∠PAC + ∠QBC = 180°
[interior consecutive angle on same side of transversal]
2 = 180° [∠OAC + ∠OBC]
⇒ ∠OAC + ∠OBC = 90°
In ΔAOB, ∠AOB + [∠OAC + ∠OBC] = 180°
⇒ ∠AOB + 90° = 180°
⇒ ∠AOB = 90°

Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Sol. PA and PB are two tangents, and A and B are the points of contact of the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
[Radius and tangent are perpendicular to each other]
In quadrilateral OAPB
(∠OAP + ∠OBP) + ∠APB + ∠AOB = 360⁰

⇒ 180° + ∠APB + ∠AOB = 360°
∠APB + ∠AOB = 360° – 180° = 180°

Q11. Prove that the parallelogram circumscribing a circle is a rhombus. 
Sol. Parallelogram ABCD circumscribing a circle with centre O.

OP⊥ AB and OS ⊥ AD
In ΔOPB and ΔOSD, ∠OPB = ∠OSD [Each 90°]
OB = OD
[Diagonals of ║gm bisect each other]
OP = OS    [Radii]
⇒  ΔOPB ≅ ΔOSD [RHS congruence rule]
PB = SD     [C.P.C.T] ...(i)
AP = AS [Lengths of tangents]... (ii)
Adding (i) and (ii)
AP + PB = AS + DS
⇒   AB AD
Similarly  AB = BC = CD = DA
∴ ║gm ABCD is a rhombus

Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Sol. BD = 8 cm and DC = 6 cm
BE = BD = 8 cm; CD= CF = 6 cm;

Let AE=AF = x cm
In ΔABC   a = 6 + 8 = 14 cm;
b = (x + 6) cm;
c = (x +8)


ar ΔABC = ar ΔOBC + ar ΔOCA + ar ΔOAB

From (i) and (ii)

⇒ 3x(x+14) = (x + 14)² 
⇒ 3x² + 42x = x² + 196 + 28x
⇒ 2x² + 14x – 196 = 0
⇒ x² + 7x – 98 = 0
⇒ x² + 14x – 7x – 98 = 0
⇒ x(x + 14) – 7(x + 14) = 0
⇒ (x – 7)(x + 14) = 0 ⇒ x = 7
AB = x + 8 = 7 + 8 = 15 cm
AC = x + 6 = 7 + 6 = 13 cm

Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Sol. 

AB touches P and BC, and CD and DA touch the circle at Q, R and S.
Construction: Join OA, OB, OC, OD and OP, OO, OR, OS
∴ ∠1 = ∠2    [OA bisects ∠POS]
Similarly


Similarly ∠AOB + ∠COD = 180°
Hence, opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of a circle.