ncert solutions for class 9 maths

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NCERT Solutions for Class 9 Maths Chapter 15 Probability

NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1.

NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

Ex 15.1 Class 9 Maths Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Here, the total number of trials = 30
∵ Number of times, the ball touched the boundary =6
∴ Number of times, the ball missed the boundary = 30 – 6 = 24
Let the event of not hitting the boundary be represented by E, then
P(E)=[No.oftimesthebatswomandidnothittheboundary][Totalnumberofballssheplayed]=2430=45
Thus, the required probability = 45 Ex 15.1 Class 9 Maths Question 2.
1500 families with 2 children were selected randomly, and the following data were recorded

Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) no girl
Also, check whether the sum of these probabilities is 1.
Solution:
Here, total number of families = 1500

(i) ∵ Number of families having 2 girls = 475
∴ Probability of selecting a family having 2 girls = 4751500=1960

(ii) ∵ Number of families having 1 girl = 814
∴ Probability of selecting a family having 1 girl = 8141500=407750

(iii) Number of families having no girl = 211
Probability of selecting a family having no girl = 2111500
Now, the sum of the obtained probabilities
=1960+407750+2111500=475+814+2111500=15001500=1
i.e., Sum of the above probabilities is 1. Ex 15.1 Class 9 Maths Question 3.
In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data so obtained.

Find the probability that a student of the class was born in August.
Solution:
From the graph, we have
Total number of students bom in various months = 40
Number of students bom in August = 6
∴ Probability of a student of the Class-IX who was bom in August =640+320 Ex 15.1 Class 9 Maths Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes.

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
Total number of times the three coins are tossed = 200
Number of outcomes in which 2 heads coming up = 72
∴ Probability of 2 heads coming up = 72200=925
∴ Thus, the required probability = 925 Ex 15.1 Class 9 Maths Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below

Suppose a family is chosen. Find the probability that the family chosen is
(i) earning ₹ 10000-13000 per month and owning exactly 2 vehicles.
(ii) earning ₹16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not own any vehicle.
(iv) earning ₹13000-16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Here, total number of families = 2400
(i) ∵ Number of families earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles = 29
∴ Probability of a family earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles = 292400

(ii) ∵ Number of families earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579
∴ Probability of a family earning Rs. 16000 or more per month and owning exactly 1 vehicle = 5792400

(iii) ∵ Number of families earning less than Rs. 7000 per month and do not own any vehicle = 10
∴ Probability of a family earning less than Rs. 7000 per month and does not own any vehicle = 102400 = 1240

(iv) ∵ Number of families earning Rs. 13000 – Rs. 16000 per month and owning more than 2 vehicles = 25
∴ Probability of a family earning Rs. 13000 – Rs. 16000 per month and owning more than 2 vehicles = 252400 = 196

(v) ∵ Number of families owning not more than 1 vehicle
= [Number of families having no vehicle] + [Number of families having only 1 vehicle]
= [10 + 1 + 2 + 1] + [160 + 305 + 535 + 469 + 579] = 14 + 2048 = 2062
∴ Probability of a family owning not more than 1 vehicle = 20622400 = 10311200 Ex 15.1 Class 9 Maths Question 6.
A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows
0 – 20, 20 – 30, …, 60 – 70, 70 – 100. Then she formed the following table

Find the probability that a student chosen at random
(i) likes statistics,
(ii) does not like it.
Solution:
Total number of students whose opinion is obtained = 200
(i) ∵ Number of students who like statistics = 135
∴ Probability of selecting a student who likes statistics = 135200 = 2740

(ii) ∵ Number of students who do not like statistics = 65
∴ Probability of selecting a student who does not like statistics = 60200 = 1340 Ex 15.1 Class 9 Maths Question 8.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows

What is the empirical probability that an engineer lives
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within 12 km from her place of work?
Solution:
Here, total number of engineers = 40

(i) ∵ Number of engineers who are living less than 7 km from their work place = 9
∴ Probability of an engineer who is living less than 7 km from her place of work = 940

(ii) ∵ Number of engineers living at a distance more than or equal to 7 km from their work place = 31
∴ Probability of an engineer who is living at distance more than or equal to 7 km from her place of work = 3140

(iii) ∵ The number of engineers living within 12km from their work place = 0
∴ Probability of an engineer who is living within 12km from her place of work = 040 = 0

Ex 15.1 Class 9 Maths Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler?
Solution:
It is an activity. Students can do it themselves.

Ex 15.1 Class 9 Maths Question 10.
Activity : Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digit is divisible by 3.
Solution:
A class room activity for students.

Ex 15.1 Class 9 Maths Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg)
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags,chosen at random contains more than 5 kg of flour.
Solution:
Here, total number of bags = 11
∵ Number of bags having more than 5 kg of flour = 7
∴ Probability of a bag having more than 5 kg of flour = 711

Ex 15.1 Class 9 Maths Question 12.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows You were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.
Solution:
Here, total number of days = 30
∵ The number of days on which the sulphur dioxide concentration is in the interval 0.12 – 0.16 = 2
∴ Probability of a day on which sulphur dioxide is in the interval 0.12 – 0.16 = 230 = 115

Ex 15.1 Class 9 Maths Question 13.
The blood groups of 30 students of class VIII are recorded as follows
A, B, 0, 0, AB, 0, A, 0, B, A, 0, B, A, 0, 0, A, AB, 0, A, A, 0, 0, AB, B, A, B, 0
You were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Solution:
Here, total number of students = 30
∵ Number of students having blood group AB = 3
∴ Probability of a student whose blood group is AB = 330 = 110.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1

Ex 14.1 Class 9 Maths Question 1.
Give five examples of data that you collect from your day-to-day life.
Solution:
Following are the five examples which are related to day-to-day life :

• Number of girl students in our class.
• Number of computer sets in our computer lab.
• Telephone bills of our house for last two years.
• Number of students appeared in an examination obtained from newspapers.
• Number of female teachers in all the schools in a state obtained from the education department.

Ex 14.1 Class 9 Maths Question 2.
Classify the data in Q.1 above as primary or secondary data.
Solution:
We have,
Primary data: (i), (ii) and (iii)
Secondary data: (iv) and (v)

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2

Ex 14.2 Class 9 Maths Question 1.
The blood groups of 30 students of class VIII are recorded as follows
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
Represent this data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students?
Solution:
The required frequency distribution table is

From the above table, we have The most common blood group is O. The rarest blood group is AB. Ex 14.2 Class 9 Maths Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Solution:
Here, the observation with minimum and maximum values are 2 and 32 respectively.
∴ The class intervals are as follows:
0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25, 25 – 30, 30 – 35
The required frequency distribution table is

From the above table we observe that:
(i) Frequencies of class intervals 5-10 and 10 – 15 are equal, i.e., 11 each. It shows that maximum number of engineers have their residences at 5 to 15 km away from their work place.
(ii) Frequencies of class intervals 20 – 25 and 25 – 30 are also equal, i.e., 1 each. It shows that minimum number of engineers have their residences at 20 to 30 km away from their work place. Ex 14.2 Class 9 Maths Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows

(i) Construct a grouped frequency distribution table with classes 84-86, 86-88 etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
Here, the lowest value of observation = 84.9
The highest value of observation = 99.2
So, class intervals are 84 – 86, 86 – 88, 88 – 90, ……. , 98 – 100

(i) Thus, the required frequency distribution table is

(ii) Since, the relative humidity is high during the rainy season, so, the data appears to be taken in the rainy season.
(iii) Range = (Highest observation) – (Lowest observation) = 99.2 – 84.9 = 14.3 Ex 14.2 Class 9 Maths Question 4.
The heights of 50 students, measured to the nearest centimeters have been found to be as follows

(ii) From the above table, we can conclude that more than 50% of the students are shorter than 165 cm. Ex 14.2 Class 9 Maths Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08 and so on.
(ii) For how many day’s was the the concentration of sulphur dioxide more than 0.11 parts per million ?
Solution:
(i) Here, the lowest value of the observation = 0.01
The highest value of the observation = 0.22
∴ Class intervals are 0.00 – 0.04, 0.04 – 0.08,……., 0.20 – 0.24
The required frequency distribution table is

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 8 days. Ex 14.2 Class 9 Maths Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows

Prepare a frequency distribution table for the data given above.
Solution:
The required frequency distribution table is

Ex 14.2 Class 9 Maths Question 7.
The value of π upto 50 decimal places is given below
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Solution:
(i) The required frequency distribution table

(ii) The most frequently occurring digits are 3 and 9 and the least frequently occurring digit is 0. Ex 14.2 Class 9 Maths Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week.
The results were found as follows

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10.
(ii) How many children watched television for 15 or more hours a week?
Solution:
(i) Here, the lowest value of the observation = 1 and the highest value of the observation = 17
∴ Class intervats are 0 – 5, 5 – 10 ., 15 – 20
The required frequency distribution table is

(ii) Number of children who watched television for 15 or more hours in a week = 2. Ex 14.2 Class 9 Maths Question 9.
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Solution:
Here, the lowest value of the observation = 2.2
and the highest value of the observation = 4.6
∴ Class intervals are 2.0 – 2.5, 2.5 – 3.0, …., 4.5 – 5.0
The required frequency distribution table is

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

Ex 14.3 Class 9 Maths Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %)

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution:
(i) The required graphical representation is shown as follows:

(ii) The major cause of women’s ill health and death worldwide is ‘reproductive health conditions’.
(iii) Two factors may be un education and poor background. Ex 14.3 Class 9 Maths Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss, what conclusions can be arrived at from the graph.
Solution:
(i) The required bar graph is shown below:

(ii) We conclude that number of girls per thousand boys are maximum in scheduled tribe section whereas minimum in urban section. Ex 14.3 Class 9 Maths Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution:
(i) The required bar graph is shown below:

(ii) The political party A won the maximum number of seats. Ex 14.3 Class 9 Maths Question 4.
The length of 40 leaves of a plant measured correct to one millimetre and the obtained data is represented in the following table

(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves 153 mm long and Why?
Solution:
(i) The given frequency distribution table is not continuous. Therefore, first we have to modify it to be continuous distribution. Thus, the modified frequency distribution table is:

Now, the required histogram of the frequency distribution is shown below :

(ii) Yes, other suitable graphical representation is a ‘frequency polygon’.
(iii) No, it is not a correct statement. The maximum number of leaves lie in the class interval 145 – 153. Ex 14.3 Class 9 Maths Question 5.
The following table gives the lifetimes of 400 neon lamps

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a lifetime of more than 700 h?
Solution:
(i) The required histogram is shown below:

(ii) Number of lamps having life time of more than 700 hours = 74 + 62 + 48 = 184. Ex 14.3 Class 9 Maths Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
To draw a frequency polygon, we mark the class marks along x-axis. Therefore, the modified table is:

So, the two frequency polygons are as shown below:

From the above frequency polygon, we can see that more students of section A have secured good Ex 14.3 Class 9 Maths Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below

Represent the data of both the teams on the same graph by frequency polygons.
Solution:
The given class intervals are not continuous. Therefore, we first modify the distribution as continuous.

Ex 14.3 Class 9 Maths Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.
Solution:
Here, the class sizes are different. So, we calculate the adjusted frequencies corresponding to each rectangle i.e., length of the rectangle.
Adjusted frequency or length of the rectangle

Here, the minimum class size = 2 – 1 = 1
∴ We have the following table for adjusted frequencies or length of rectangles

Now, the required histogram is shown below:

Ex 14.3 Class 9 Maths Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Solution:
(i) Since, class intervals of the given frequency distribution are unequal, and the minimum class size = 6 – 4 = 2.
Therefore, we have the following table for length of rectangles.

The required histogram is shown below:

(ii) The maximum frequency is 44, which is corresponding to the class interval 6 – 8.
∴ Maximum number of surnames lie in the class interval 6 – 8.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

Ex 14.4 Class 9 Maths Question 1.
The following number of goals were scored by a team in a series of 10 matches
2, 3, 4, 5, 0, 1, 3, 3, 4, 3.
Find the mean, median and mode of these scores.
Solution:
To find the

Here, n = 10

Thus, mean = 2.8 To find median:
Now arranging the given data in ascending order,
we have 0,1, 2, 3, 3, 3, 3, 4, 4, 5
∵ n = 10, an even number

Thus, median = 3

To find mode:
In the given data, the observation 3 occurs 4 times,
i.e., maximum number of times.
Thus, mode = 3 Ex 14.4 Class 9 Maths Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Solution:
To find the mean:
Here, n = 15

Thus, mean = 54.8 To find median:
Arranging the given data in ascending order,
we have
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96,98
∵ n = 15, an odd number

Thus, median = 52

To find mode:
In the given data, the observation 52 occurs 3 times,
i.e., the maximum number of times.
Thus, mode = 52 Ex 14.4 Class 9 Maths Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Here, the given observations are in ascending order.
Since, n = 10 (an even number)

Since, median = 63 [Given]
∵ x + 1 = 63 ⇒ x = 63 – 1 = 62
Thus, the required value of x is 62.

Ex 14.4 Class 9 Maths Question 4.
Find the mode of 14, 25,14, 28,18,17,18,14, 23, 22,14 and 18.
Solution:
Arranging the given data in ascending order, we have 14, 14, 14, 14, 17, 18, 18, 18, 22, 23 25, 28.
Since the observation 14 is occuring the maximum number of times (i.e. 4 times)
∴ Mode of the given data = 14 Ex 14.4 Class 9 Maths Question 5.
Find the mean salary of 60 workers of a factory from the following table

Solution:

Thus, the required mean salary = Rs. 5083.33

Ex 14.4 Class 9 Maths Question 6.
Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Solution:
(i) Mean height of the students of a class.
(ii) Median weight of a pen, a book, a rubber band, a match box and a chair.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Ex 13.1 Class 9 Maths Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine
(i) The area of the sheet required for making the box.
(ii) The cost of the sheet for it, if a sheet measuring 1m2 costs ₹20.
Solution:
(i) Here, length (l) = 1.5 m, bread th(b) = 1 .25 m
and height (h) = 65 cm = 65100 m = 0.65 m

∵ It is open from the top.
∴ Its surface area
= [Lateral surface area] + [Base area]
= [2(1 + b)h] + [lb]
= [2(1.50 + 1.25)0.65] m2 + [1.50 x 1.25] m2
= [2 x 2.75 x 0.65] m2 + [1.875] m2
= 3.575 m2+ 1.875 m2 = 5.45 m2
∴ Area of the sheet required for making the box = 5.45 m2

(ii) Cost of 1 m2 sheet = Rs. 20
Cost of 5.45 m2 sheet = Rs. (20 x 5.45)
= Rs. 109
Hence, cost of the required sheet = Rs. 109

Ex 13.1 Class 9 Maths Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹17.50 per m2.
Solution:
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
The room is like a cuboid whose four walls (lateral surface) and ceiling are to be white washed.
∴ Area for white washing
= [Lateral surface area] + [Area of the ceiling]
= [2(l + b)h] + [l x b]
= [2(5 + 4) x 3] m2 + [5 x 4] m2 = 54 m2 + 20 m2 = 74 m2
Cost of white washing for 1 m2 area = Rs. 7.50
∴ Cost of white washing for 74 m2 area = Rs. (7.50 x 74) = Rs. 555
Thus, the required cost of white washing = Rs. 555

Ex 13.1 Class 9 Maths Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹10 per m2 is ₹15000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area]
Solution:
A rectangular hall means a cuboid.
Let the length and breadth of the hall be l and b respectively.
∴ Perimeter of the floor = 2(l + b)
⇒ 2(l + b) = 250 m
∵ Area of four walls = Lateral surface area = 2(1 + b) x h, where h is the height of the hall = 250 h m2
Cost of painting the four walls
= Rs. (10 x 250 h) = Rs. 2500h
⇒ 2500 h = 15000 ⇒ h = 150002500 = 6
Thus, the required height of the hall = 6 m Ex 13.1 Class 9 Maths Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container.
Solution:
Total area that can be painted = 9.375 m2
Here, Length of a brick (l) = 22.5 cm
Breadth of a brick (b) = 10 cm
Height of a brick (h) = 7.5 cm
Since a brick is like a cuboid, then
Total surface area of a brick = 2[lb + bh + hl]
= 2[(225 x 1(0) + (10 x 7.5) + (7.5 x 22.5)] cm2
= 2[(225) + (75) + (168.75)] cm2
= 2[468.75] cm2 = 937.5 cm2 = 937.510000 m2
Let the required number of bricks be n
∴ Total surface area of n bricks = n x 937.510000 m2

Thus, the required number of bricks = 100

Ex 13.1 Class 9 Maths Question 5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution:
For the cubical box with edge (a) = 10 cm
Lateral surface area = 4a2 = 4 x 102 cm2
= 400 cm2
Total surface area = 6a2 = 6 x 102 cm2
= 600 cm2
For the cuboidal box with dimensions,
Length (l) = 12.5 cm,
Height (h) = 8 cm
∴ Lateral surface area = 2[l + b] x h = 2[12.5 + 10] x 8 cm2 = 360 cm2
Total surface area = 2[lb + bh + hl]
= 2[(12.5 x 10) + (10 x 8) + (8 x 12.5)] cm2
= 2[125 + 80 + 100] cm2
= 2 cm2
= 610 cm2
(i) A cubical box has the greater lateral surface area by (400 – 360) cm2 = 40 cm2.
(ii) Total surface area of a cubical box is smaller than the cuboidal box by (610 – 600) cm2 = 10 cm2.

Ex 13.1 Class 9 Maths Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Solution:
The herbarium is like a cuboid.
Here, length (l) = 30 cm,
height (h) = 25 cm
(i) Surface area of the herbarium (glass)
= 2[lb + bh + hl]
= 2[(30 x 25) + (25 x 25) + (25 x 30)] cm2 – 2[750 + 625 + 750] cm2
= 2 cm2
= 4250 cm2
Thus, the required area of the glass = 4250 cm2

(ii) Total length of 12 edges = 4l + 4b + 4h
= 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4 x 80 cm = 320 cm
Thus, the required length of tape = 320 cm

Ex 13.1 Class 9 Maths Question 7.
Shanti Sweets Stalll was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
For bigger box:
Length (l) = 25 cm,
Height (h) = 5 cm
Total surface area of a box = 2(lb + bh + hl)
= 2[(25 x 20) + (20 x 5) + (5 x t25)] cm2
= 2 [500 + 100 + 125] cm2
= 2 cm2
= 1450 cm2
Total surface area of 250 boxes = (250 x 1450) cm2 = 362500 cm2

For smaller box:
l = 15 cm, b = 12 cm, h = 5 cm
Total surface area of a box = 2 [lb + bh + hl]
= 2[(15 x 12) + (12 x 5) + (5 x 15)] cm2
= 2[180 + 60 + 75] cm2 = 2 cm2 = 630 cm2
∴ Total surface area of 250 boxes = (250 x 630) cm2 = 157500 cm2
Now, total surface area of both type of boxes = 362500 cm2 +157500 cm2 = 520000 cm2 Area for overlaps = 5% of [total surface area]
= 5100 x 520000 cm2 = 26000 cm2
∴ Total surface area of the cardboard required = [Total surface area of 250 boxes of each type] + [Area for overlaps]
= 520000 cm2 + 26000 cm2 = 546000 cm2
∵ Cost of 1000 cm2 cardboard = Rs. 4
∴ Cost of 546000 cm2 cardboard
= Rs.4×5460001000 = Rs. 2184

Ex 13.1 Class 9 Maths Question 8.
Parveen wanted to make a temporary shelter, for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?
Solution:
Here, length (l) = 4 m,
and height (h) = 2.5 m
The structure is like a cuboid.
∴ The surface area of the cuboid, excluding the base
=[Lateral surface area] + [Area of ceiling]
= [2(l + b)h] + [lb]
= [2(4 + 3) x 2.5] m2 + [4 x 3] m2
= 35 m2 + 12 m2 = 47 m2
Thus, 47 m2 tarpaulin would be required.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Ex 13.2 Class 9 Maths Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
Solution:
Let r be the radius of the cylinder.
Here, height (h) = 14 cm and curved surface area = 88 cm2
Curved surface area of a cylinder = 2πrh
⇒ 2πrh = 88
⇒ 2 x 227 x r x 14 = 88
⇒ r = 88×72×22×14 = 1 cm
∴ Diameter = 2 x r = (2 x 1) cm = 2 cm

Ex 13.2 Class 9 Maths Question 2.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?
Solution:
Here, height (h) = 1 m
Diameter of the base = 140 cm = 1.40 m
Radius (r) = 1.402m = 0.70 m
Total surface area of the cylinder = 2πr (h + r)
= 2 x 227 x 0.70(1 + 0.70)m2
= 2 x 22 x 0.10 x 1.70 m2
= 2 x 22 x 10100 x 170100m2
= 748100m2 = 7.48 m2
Hence, the required sheet = 7.48 m2 Ex 13.2 Class 9 Maths Question 3.
A metal pipe is 77 cm long. The inner ft diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its
(i) inner curved surface area.
(ii) outer curved surface area.
(iii) total surface area.

Solution:
Length of the metal pipe = 77 cm
It is in the form of a cylinder.
∴ Height of the cylinder (h) = 77 cm
Inner diameter = 4 cm
Inner radius (r) = 42 cm = 2 cm
Outer diameter = 4.4 cm
⇒ Outer radius (R) = 4.42 cm = 2.2 cm

(i) Inner curved surface area = 2πrh
= 2 x 227 x 2 x 77 cm2
= 2 x 22 x 2 x 11 cm2 = 968 cm2

(ii) Outer curved surface area = 2πRh (iii)Total surface area = [Inner curved surface area] + [Outer curved surface area] + [Area of two circular ends]
= [2πrh] + [2πRh] + 2[π(R2 – r2)]
= [968 cm2] + [1064.8 cm2] Ex 13.2 Class 9 Maths Question 4.
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
Solution:
The roller is in the form of a cylinder of diameter = 84 cm
⇒ Radius of the roller(r) = 842 cm = 42 cm
Length of the roller (h) = 120 cm
Curved surface area of the roller = 2πrh
= 2 x 227 x 42 x 120 cm2
= 2 x 22 x 6 x 120 cm2 = 31680 cm2
Now, area of the playground levelled in one revolution of the roller = 31680 cm2
= 3168010000m2
∴ Area of the playground levelled in 500
revolutions = 500 x 3168010000m2 = 1584m2

Ex 13.2 Class 9 Maths Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 .
Solution:
Diameter of the pillar = 50 cm
∴ Radius (r) = 502m = 25 m = 14m
and height (h) = 3.5m
Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2
∴ Cost of painting of 1 m2 pillar = Rs. 12.50
∴ Cost of painting of 112 m2 pillar
= Rs. ( 112 x 12.50 )
= Rs. 68.75.

Ex 13.2 Class 9 Maths Question 6.
Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
Let height of the cylinder be h m
Curved surface area of a cylinder = 2πrh
= 2 x 227 x 710 x hm2
But the curved surface area is 4.4 m2. [Given] Ex 13.2 Class 9 Maths Question 7.
he inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of ₹40 per m2.
Solution:
Hans Inner diameter of the well = 3.5 m
Radius of the well (r) = 3.52
Height of the well (h) = 10 m
(i) Inner curved surface area = 2πrh (ii) Cost of plastering per m2 = Rs. 40
∴ Total cost of plastering the area 110 m2
= Rs. (110 x 40) = Rs. 4400

Ex 13.2 Class 9 Maths Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
USD Length of the cylindrical pipe (h) = 28 m
Diameter of the pipe = 5 cm
∴ Radius (r) = 52 cm = 5200 m
Curved surface area of a cylinder = 2πrh Thus, the total radiating surface is 4.4 m2 .

Ex 13.2 Class 9 Maths Question 9.
Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if 112 of the steel actually used was wasted in making the tank.
Solution:
The storage tank is in the form of a cylinder.
∴ Diameter of the tank = 4.2 m
⇒ Radius (r) = 4.22 = 2.1 m
Height (h) = 4.5 m
Now,
(i) Lateral (or curved) surface area of the tank = 2πrh
= 2 x 227 x 2.1 x 4.5 m2
= 2 x 22 x 0.3 x 4.5 m2 59.4 m2 (ii) Total surface area of the tank = 2πr(r + h)
= 2 x 227 x 2.1(2.1 + 4.5)m2
= 44 x 0.3 x 6.6 m2 = 87.13 m2
Let actual area of the steel used be x m2
∴ Area of steel that was wasted = 112 x x m
= x12m2

Area of steel used = x – x12 m2

Thus, the required area of the steel that was actually used is 95.04 m2. Ex 13.2 Class 9 Maths Question 10.
In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution:
The lampshade is in the form of a cylinder,
where radius (r) = 202 cm = 10 cm
and height = 30 cm.
A margin of 2.5 cm is to be added to the top and bottom of the frame.
∴ Total height of the cylinder, (h)
= 30 cm + 2.5 cm + 2.5 cm = 35 cm
Now, curved surface area = 2πrh
= 2 x 227 x 10 x 35 cm2
= 2200 cm2
Thus, the required area of the cloth = 2200 cm2 Ex 13.2 Class 9 Maths Question 11.
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Solution:
Here, the penholders are in the form of cylinder.
Radius of a penholder (r) = 3 cm
Height of a penholder (h) = 10.5 cm
Since, a penholder must be open from the top.
Now, surface area of a penholder = [Lateral surface area] + [Base area]

= [2πrh] + [πr2]

∴ Surface area of 35 penholders
= 35 x 15847 cm2 = 7920 cm2
Thus, 7920 cm2 of cardboard was required to be bought.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Ex 13.3 Class 9 Maths Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
Here, diameter of the base = 10.5 cm
⇒ Radius (r) = 10.52 cm
and slant height (l) =10 cm
Curved surface area of the cone = πrl
= 227 x 10.52 x 10cm2
= 11 x 15 x 1 cm2 = 165cm2 Ex 13.3 Class 9 Maths Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Here, diameter = 24 m 24
∴ Radius (r) = 242 m = 12 m
and slant height (l) = 21 m
∴ Total surface area of a cone = πr(r +1)

Ex 13.3 Class 9 Maths Question 3.
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Solution:
Here, curved surface area = πrl = 308 cm2
Slant height (l) = 14 cm

(i) Let the radius of the base be ‘r’ cm
∴ πrl = 308 ⇒ 227 x r x 14 = 308
r = 308×722×14 = 7cm
Thus, radius of the cone is 7 cm

(ii) Base area = πr2 = 227 x 72 cm2
= 22 x 7 cm2 = 154 cm2
and curved surface area = 308 cm2 [Given]
∴ Total surface area of the cone
= [Curved surface area] + [Base area] = 308 cm2 + 154 cm2
= 462 cm2

Ex 13.3 Class 9 Maths Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹70.
Solution:
Here, height of the tent (h) = 10 m
Radius of the base (r) = 24 m

i) The slant height, l = r2−h2−−−−−−√
= 242+102−−−−−−−−√ m = 576+100−−−−−−−−√ m = 676−−−√ m = 26m
Thus, the required slant height of the tent is 26 m. (ii) Curved surface area of the cone = πrl
∴ Area of the canvas required

Ex 13.3 Class 9 Maths Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14)
Solution:
Here, base radius (r) = 6 m
Height(h) = 8m
∴ Slant height (l) = r2−h2−−−−−−√ = 62−82−−−−−−√ m
= 36+64−−−−−−√ m
= 100−−−√m = 10 m
Now, curved surface area = πrl
= 3.14 x 6 x 10m2
= 314100 x 6 x 10 m2 = 1884 m2
Thus, area of the tarpaulin required to make the tent = 188.4 m2
Let the length of the tarpaulin be L m
⇒ L x 3 = 188.4 ⇒ L = 188.43 = 62.8
Extra length of tarpaulin required for margins = 20cm = 20100m = 0.2m
Thus, total length of tarpaulin required = 62.8 m + 0.2 m = 63 m

Ex 13.3 Class 9 Maths Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m2 .
Solution:
Here, base radius (r) = 142 m = 7 m
Slant height (l) = 25 m
∴ Curved surface area = πrl
= 227 x 7 x 25 m2 = 550 m2
Cost of white-washing for 100 m2 area = Rs. 210
∴ Cost of white-washing for 550 m2 area
= Rs. 210100 x 550 = Rs. 1155

Ex 13.3 Class 9 Maths Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
Radius of the base (r) = 7 cm and height (h) = 24 cm
Slant height (l) = h2−r2−−−−−−√ = 242−72−−−−−−−√cm
= 576+49−−−−−−−√cm = 625−−−√ cm = 25 cm
∴Lateral surface area = πrl = 227 x 7 x 25 cm2 = 550 cm2
∴ Lateral surface area of 10 caps = 10 x 550 cm2
= 5500 cm2
Thus, the required area of the sheet = 5500 cm2 Ex 13.3 Class 9 Maths Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take 104−−−√ = 1.02)
Solution:
Diameter of the base = 40cm

= Rs. 384.336 = Rs. 384.34 (approx.)
Thus, the required cost of painting is Rs. 384.34 (approx.).

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Ex 13.4 Class 9 Maths Question 1.
Find the surface area of a sphere of radius
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
(i) Here, r = 10.5 cm
Surface area of a sphere = 4πr2 (ii) Here, r = 5.6 cm
Surface area of a sphere = 4πr2

(iii) Here, r = 14 cm
Surface area of a sphere = 4πr2

Ex 13.4 Class 9 Maths Question 2.
Find the surface area of a sphere of diameter
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Solution:
(i) Here, diameter = 14 cm

(ii) Here, diameter = 21 cm

(iii) Here, diameter = 3.5 m

Ex 13.4 Class 9 Maths Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
Here, radius (r) = 10 cm
Total surface area of hemisphere = 3πr2
= 3 x 3.14 x 10 x 10 cm2 = 942 cm2

Ex 13.4 Class 9 Maths Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Find the ratio of surface areas of the balloon in the two cases.
BSD Case I: When radius (r1) = 7 cm
Surface area = 4πr12 = 4 x 227 x (7) cm2
= 4 x 22 x 7 cm2 = 616 cm2

Case II: When radius (r2) = 14 cm2
Surface area = 4πr22=4 x 227 x (14)2 cm2
= 4 x 22 x 14 x 2 cm2 = 2464 cm2
∴ The required ratio = 6162464 = 14 or 1 : 4 Ex 13.4 Class 9 Maths Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².
Solution:
Inner diameter of the hemispherical bowl = 10.5 cm

Ex 13.4 Class 9 Maths Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Let the radius of the sphere be r cm.
Surface area of a sphere = 4πr2
∴ 4πr2 = 154

Thus, the required ratio = 1 : 16. Ex 13.4 Class 9 Maths Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Inner radius (r) = 5 cm
Thickness = 0.25 cm

∴ Outer radius (R) [5.00 + 0251 cm = 5.25 cm
∴ Outer curved surface area of the bowl = 2πR2

Ex 13.4 Class 9 Maths Question 9.
A right circular cylinder just encloses a sphere of radius r (see figure). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Solution:
(i) For the sphere, radius = r
∴ Surface area of the sphere = 4πR2 (ii) For the right circular cylinder,
∴Radius of the cylinder = r
Height of the cylinder = Diameter of the sphere
∴ Height of the cylinder (h) 2r
Since, curved surface area of a cylinder = 2πrh
= 2πr(2r) = 4πr2

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Ex 13.5 Class 9 Maths Question 1.
A matchbox measures 4 cm x 2.5 cm x 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Solution:
Since, a matchbox is in the form of cuboid.
Here, length (l) = 4 cm, breadth (b) = 2.5 cm
and height (h) = 1.5 cm
∴ Volume of a matchbox = l x b x h
= 4 x 25 x 1.5 cm3
= 4 x 2510 x 1510cm3
= 15 cm3
⇒ Volume of 12 such boxes = 12 x 15 cm3
= 180 cm3

Ex 13.5 Class 9 Maths Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? ( 1 m3 = 1000L)
Solution:
Length (l) =6 m, breadth (h) =5 m and
depth (h) = 4.5 m
∴ Capacity =l x b x h = 6 x 5 x 4.5m3
= 6 x 5 x 4510m = 3 x 45m3 = 135m3
∵ 1 m3 = 1000 litres
⇒ 135 m3 = 135000 litres
∴ The required amount of water in the tank = 135000 litres.

Ex 13.5 Class 9 Maths Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Solution:
Length (l) = 10 m, breadth (b) = S m
Volume(V) = 380m3
Let height of the cuboidal vessel be ‘h’.
Ex 13.5 Class 9 Maths Volume of the cuboidal vessel = l x b x h
⇒ 10 x 8 x h m3 = 80h m3
⇒ 80h = 380
⇒ h = 38080 = 4.75m
Thus, the required height of the vessel = 4.75 m

Ex 13.5 Class 9 Maths Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m³.
Solution:
Length (i) = 8m
Depth (h) = 3 m
∴ Volume of the cuboidal pit = l x b x h
= 8 x 6 x 3 m3 = 144 m3
Hence, the cost of digging a pit = Rs. (144 x 30)
= Rs. 4320

Ex 13.5 Class 9 Maths Question 5.
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are 2.5 m and 10 m, respectively.
Solution:
ira Length of the tank (l) = 2.5 m
Depth of the tank (h) = 10 m
Let breadth of the tank be b m.
∴ Volume (capacity) of the tank = l x b x h
= 2.5 x b x 10 m3
= 2510 x 10 x bm3
= 25bm3
But the capacity of the tank = 50000 litres
= 50 m3 [ ∵ 1000 litres = 1 m3 ]
∴ 25b = 50 ⇒ b = 5025 = 2
Thus, the breadth of the tank = 2 m

Ex 13.5 Class 9 Maths Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m x 15 m x 6 m. For how many days will the water of this tank last?
Solution:
fcWra Length of the tank (l) = 20 m
Breadth of the tank (b) = 15 m
Height of the tank (h) = 6m
∴ Volume of the tank = l x b x h
= 20 x 15 x 6 m3 = 1800 m3
Since, 1 m3 = 1000 litres
∴ Capacity of the tank = 1800 x 1000 litres = 1800000 litres
Since, 150 litres of water is required per head per day.
∴ Amount of water required by 4000 people per day = 150 x 4000 litres
Let the required number of days be x
∴ 4000 x 150 x x = 1800000
⇒ x = 18000004000×150 = 3
Thus, the required number of days is 3. Ex 13.5 Class 9 Maths Question 7.
A go down measures 40 m x 25 m x 10 m. Find the maximum number of wooden crates each measuring 15 m x 125 m x 0.5 m that can be stored in the go down.
Solution:
Volume of the go down = 40 x 25 x 10 m3
Volume of a wooden crate = 1.5 x 1.25 x 0.5 m3

∴ Maximum number of wooden crates = 10667. Ex 13.5 Class 9 Maths Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Side of the given cube = 12 cm
Volume of the given cube = (side)3 = (12)3 cm3
Let the side of the new cube be n
∴ Volume of new cube = n3
⇒ Volume of 8 new cubes = 8n3
According to question,
8n3 = (12)3 = 12 x 12 x 12
⇒ n3 = 12×12×128 = 6 x 6 x 6
⇒ n3 = 63
⇒ n = 6
Thus, the required side of the new cube is 6 cm.
Now, surface area of the given cube = 6 x (side)2 = 6 x 122 cm2 = 6 x 12 x 12 cm2
Surface area of new cube = 6 x 62 cm2
= 6 x 6 x 6 cm2

Thus, the required ratio = 4 : 1 Ex 13.5 Class 9 Maths Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution:
The water flowing in a river can be considered in the form of a cuboid.

Length (l) = 2 km = 2000 m
depth (h) = 3 m
∴ Volume of water (volume of the cuboid so formed) = l x b x h
= 2000 x 40 x 3 m3
Now, volume of water flowing in 1 hr (60 minutes) = 2000 x 40 x 3 m3
Volume of water that will fall in 1 minute
= [2000 x 40 x 3] + 60 m3
= 4000m3

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

Ex 13.6 Class 9 Maths Question 1.
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000cm³ =1 L.)
Solution:
Let the base radius of the cylindrical vessel be r cm.
∴ Circumference of the base = 2πr
⇒ 2πr = 132 [Circumference = 132 cm, (given)]
= 2 x 227 x r = 132
⇒ r = 132x72x22 cm = 21cm
Since, height of the vessel (h) = 25 cm
Volume of a vessel (h) = πr2h
= 227 x (21)2 x 25cm3
= 227 x 21 x 21 x 25cm3
= 22 x 3 x 21 x 25 cm3
= 34650 cm3
∵ Capaoty of the vessel = Volume of the vsel
∴ Capacity of cylindrical vessel = 34650 cm3
Since, 1000 cm3= 1 litre
⇒ 34650 cm3 = 346501000 litres = 34.65 litres
Thus, the vessel can hold 34.65 litres of water.

Ex 13.6 Class 9 Maths Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Solution:
Inner diameter of the cylindrical pipe = 24cm
⇒ Inrr radius of the pipe (r) = 242cm = 12cm
Outer diameter of the pipe = 28 cm
Outer radius of the pipe(R) = 14cm
Length of the pipe (h) = 35 cm
∴ Amount of wood in the pipe = Outer volume – Inner volume
= πR2h – πr2h
= πh (R + r) (R – r)
[∵ a2 – b2 = (a + b)(a – b)]
= 227 x 35 x (14+12) x (14 – 12)cm3
=22 x 5 x 26 x 2 cm3
Mass of wood in the pipe = (Mass of wood in 1 cm3 of wood) x (Volume of wood in the pipe)
= (0.6g) x (22 x 5 x 26 x 2)cm3
= 610 x 22 x 10 x 26 g = 3432 g
= 34321000 kg = 3.432kg [∵ 1000 g = 1 kg]
Thus, the required mass of the pipe is 3.432 kg.

Ex 13.6 Class 9 Maths Question 3.
A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm.
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Solution:
(i) For rectangular pack,
Length (l) = 5 cm,
Height (h) = 15 cm
Volume = l x b x h = 5 x 4 x 15 cm3 = 300 cm3
∴ Capacity of the rectangular pack = 300 cm3

(ii) For cylindrical pack,
Base diameter = 7 cm
∴ Radius of the base (r) = 72 cm
Height (h) = 10 cm
∴ Volume = πr2h = 227 x (72)2 x 10cm
= 227 x 72 x 72 x 10cm
= 11 x 7 x 5cm3 = 385cm3
∴ Capacity of the cylindrical pack = 385 cm3
So, the cylindrical pack has greater capacity
by (385 – 300) cm3 = 85 cm3

Ex 13.6 Class 9 Maths Question 4.
If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find
(ii) its volume. (Use π = 3.14)
Solution:
Height of the cylinder (h) = 5 cm
Let the base radius of the cylinder he ‘r’. (i) Since lateral surface area o the cylinder = 2πrh,
But lateral surface of the cylinder 94.2 cm2 [given]

Thus, the radius of the cylinder = 3 cm

(ii) Volume of a cylinder = πr2h
⇒ Volume of tlw given cylinder
=3.14 (3)2 x 5cm3
= 314100 x 3 x 3 x 5 cm3
= 141310 cm = 141.3cm3
Thus, the required volume = 141.3 cm3 Ex 13.6 Class 9 Maths Question 5.
It costs ₹2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹20 per m², find
(i) inner curved surface area of the vessel,
(iii) capacity of the vessel.
Solution:

(i) Total cost of painting = Rs. 2200
Cost of painting of area l m2 = Rs. 20
Total cost

= 110 m2

(ii) Let r be the base radius of the cylindrical vessel.
Curved surface area of a cylinder = 2πrh Ex 13.6 Class 9 Maths Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Solution:
Capacity of the cylindrical vessel
= 15.4 litres = 15.4 x 1000 cm3 [1 litre = 10(x) cm3]

Let r m be the radius of the base of the vessel.

Now, total surface area of the cylindrical vessel

Thus, the required metal sheet = 0.4708 m2. Ex 13.6 Class 9 Maths Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of. the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:
Since, 10mm = 1 cm:
∴ 1mm = 110 cm
For graphite cylinder,

∴ Radius of the pencil (R) = 720 cm
Height of the pencil (h) = 14 cm
Volume of the pencil = πR2h

∴ Volume of the wood = [Volume of the pendil] – [Volume of the graphite]
= 5.39 cm3 – 0.11 cm3
= 5.28 cm3
Thus, the required volume of the wood is = 5.28 cm3. Ex 13.6 Class 9 Maths Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
∵ The bowl is cylindrical, where diameter of the base = 7 cm

∴ Volume of the wood = [Volume of the pendil] – [Volume of the graphite]
= 5.39 cm3 – 0.11 cm3
= 5.28 cm3
Thus, the required volume of the wood is = 5.28 cm3. Ex 13.6 Class 9 Maths Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
∵ The bowl is cylindrical, where diameter of the base = 7 cm

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7

Ex 13.7 Class 9 Maths Question 1.
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
Solution:
(i) Here, radius of the cone (r) =6 cm
Height (h) = 7 cm

Ex 13.7 Class 9 Maths Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
Solution:
(i) Here, radius (r) = 7 cm and slant height (l) =25 cm

Thus, the required capacity of the conical vessel is 1.232 litres. (ii) Here, height (h) = 12 cm and slant height (l) = 13 cm

Thus, the required capacity of the conical vessel is 1135 litres. Ex 13.7 Class 9 Maths Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)
Solution:
Here, height of the cone (h) = 15 cm
Volume of the cone = 1570 cm3
Let the radius of the base be ‘r’ cm.

Thus, the required radius of the base is 10 cm. Ex 13.7 Class 9 Maths Question 4.
If the volume of a right circular cone of height 9 cm is 48 cm³, find the diameter of its base.
Solution:
Volume of the cone = 48 it cm3
Height of the cone (h) = 9 cm
Let r cm be its base radius.

Diameter = 2 x radius .
∴ Diameter of the base of the cone = (2 x 4)cm = 8 cm Ex 13.7 Class 9 Maths Question 5.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
Here, diameter of the conical pit = 3.5 m
Radius (r) = 3.52 m = 3520m,
Depth (h) = 12m
⇒ Volume (capacity) = 13 πr2h

Thus, the capacity of the conical pit is 38.5 kl.

Ex 13.7 Class 9 Maths Question 6.
The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
Solution:
Volume of the cone = 9856 cm3
Diameter of the base 28 cm
Radius of the base (r) = 282 = 14 cm

(i) Let the height of the cone be h cm.

Thus, the required height is 48 cm.

(ii) Let the slant height be l cm.
⇒ l2 = r2 +h2
⇒ l2 = 142 + 482 = 196 + 2304 = 2500
∴ l = 50
Thus, the required slant height = 50 cm.

(iii) The curved surface area of a cone = πrl
∴ Curved surface area = 227 x 14 x 50 cm2
= 2200 cm2
Thus, the curved surface area of the cone is 2200 cm2. Ex 13.7 Class 9 Maths Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
Sides of the right triangle ABC are 5 cm, 12 cm and 13cm.
The right angled triangle is revolved about the side 12 cm.

Thus, we have radius of the base of the cone so formed (r) = 5 cm
Height(h) = 12cm
∴ Volume of the cone so obtained = 13πr2h
= 13 x π x (5)2 x 12cm3
= 100 π cm3
Thus, the required volume of the cone is 100πcm3. Ex 13.7 Class 9 Maths Question 8.
If the triangle ABC in Question 7 above is revolved around the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
Since the right triangle is revolved around the side 5 cm.
∴ Height of the cone so obtained (h) = 5 cm
Radius of the cone (r) = 12 cm

Thus, the required ratio is 5 : 12. Ex 13.7 Class 9 Maths Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution:
Here, the heap of wheat is in the form of a cone with base diameter = 10.5 m

Thus, the required volume = 86.625m3
Now, the area of the canvas to cover the heap must be equal to the curved surface area of the conical heap.

Thus the required area of the canvas is 99.825 m2.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Ex 13.8 Class 9 Maths Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 cm
Solution:
(i) Here, radius (r) = 7cm

Thus, the required volume = 143713cm3 (ii) Here, radius (r) = 0.63 m

Thus, the required volume is 1.05 m3 (approx.) Ex 13.8 Class 9 Maths Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Solution:
(i) Diameter of the ball =28 cm
Radius of the ball (r) cm 282cm = 14cm
Volume of the spherical ball = 43πr3

ii) Diameter of the ball = 0.21 m
⇒ Radius(r) = 0.212m = 21200m

Thus, the amount of water displayed = 0.004851 m3. Ex 13.8 Class 9 Maths Question 3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?
Solution:
Diameter of the metallic bait = 4.2 cm
⇒ Radius (r) = 4.22cm = 2.1cm

Thus, the mass of ball is 345.39 g (approx.) Ex 13.8 Class 9 Maths Question 4.
The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What fraction of the volume of the Earth is the volume of the Moon?
Solution:
Let diameter of the earth be 2r.
⇒ Radius of the earth = 2r2 = r
Since, diameter of the moon = 14 (Diameter of the earth)
Radius of the moon = 14 (r) = r4
∴ Volume of the earth = 43πr3 and

Ex 13.8 Class 9 Maths Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
Diameter of the hemispherical bowl = 10.5 cm
⇒ Radius of the hemispherical bowl (r) = 10.52cm = 10520cm

Thus, the capacity of the bowl = 0.303 litres (approx.) Ex 13.8 Class 9 Maths Question 6.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Solution:
Inner radius (r) = 1 m
∵ Thickness = 1 cm = 1100m = 0 .01m

Thus, the required volume of the iron used
= 0.06348 m3 (approx.) Ex 13.8 Class 9 Maths Question 7.
Find the volume of a sphere whose surface area is 154 cm2.
Solution:
Let ‘r’ be the radius of the sphere.
∴Its surface area = 4πr2

Ex 13.8 Class 9 Maths Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹498.96. If the cost of white washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Solution:
(i) Total cct of white-washing = Rs. 498.96
Cost of 1m² of white-washing = Rs. 2
Total cost 498.96 2
∴ Area = TotalCostCostof1m2area=498.962=249.48m2
Thus, the required surface area of the dome is 249.48 m2. (ii) Let ‘r’ be the radius of the hemispherical dome.
∴ Inside surface area of the dome = 2πr2

Now, volume of air inside the dome = Volume of a hemisphere

Thus, the required volume of air inside the dome is 523.9 m3 (approx). Ex 13.8 Class 9 Maths Question 9.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
(i) radius r’ of the new sphere,
(ii) ratio of S and S’.
Solution:
(i) Let the radius of a small sphere be r
∴ Its volume = 43πr3
Volume of 27 small spheres 27 x [ 43πr3]
Let the radius of the new sphere be r’
∴ Volume of the new sphere = 43π(r’)3

Hence, the radius of a new sphere is 3r. (ii) Surface area of a sphere = 4πr2
= S = 4πr2 and S’ = 4π (3r)2 [∵ r’ = 3r]

Thus, S : S’ = 1 : 9 Ex 13.8 Class 9 Maths Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?
Solution:
Diameter of the spherical capsule = 3.5mm
⇒ Radius of the spherical capsule (r) = 3.52 mm = 3520mm

= 22.45833 mm3
= 22.46 mm3 (approx.)
Thus, the required quantity of medicine = 22.46 mm3(approx.)

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Ex 13.9 Class 9 Maths Question 1.
A wooden bookshelf has external dimensions as follows :
Height = 110cm, Depth = 25cm, Breadth = 85cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing-is 20 paise per cm2 and the rate of pointing is 10 paise per cm2, find the total expenses required for palishing and painting the surface of the bookshelf.

Solution:
Here, length (l) = 85 cm,
breadth (b) = 25 cm and height (h) = 110 cm
External surface area = Area of four faces + Area of back + Area of front beading
= [2 (110 + 85) x 25 + 110 x 85 + (110 x 5 x 2) + (75 x 5) x 4] cm2 = 21700 cm2
∴ Cost of polishing external faces = Rs. (21700 x 20100 ) = Rs. 4340
Internal surface area = Area of five faces of 3 cuboids each of dimensions 75 cm x 30 cm x 20 cm
= Total surface area of 3 cuboids of dimensions 75 cm x 30 cm x 20 cm – Area of bases of 3 cuboids of dimensions 75 cm x 30 cm x 20 cm 3(2(75 x 30 + 30 x 20 + 75 x 20)) cm2 – 3 x (75 x 30) cm2

= 6(2250 + 600 + 1500) cm2 – 6750 cm2 = 19350 cm2
∴ Cost of painting inner faces = Rs. 19350 x 10100 = Rs. 1935
Hene, total expenses = Rs. (4340 + 1935)
= Rs. 6275 Ex 13.9 Class 9 Maths Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are-used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.

Solution:
Here, diameter of a sphere = 21 cm
Radius of a sphere (r) = 212 cm
Surface area of a sphere = 4πr2
∴ Surface area of 8 spheres
= 8 x 4 x 227 x (212)2cm2

Hence, the cost of paint required = Rs. 2784.25 Ex 13.9 Class 9 Maths Question 3.
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Solution:
Let the diameter of a sphere be d.
After decreasing, diameter of the sphere
= d – 25100 x d
= d – 14d = 34d
Since, surface area of a sphere = 4πr2 or π(2r)2 or πd2
Surface area of a sphere, when diameter of the sphere is

Now, decrease percentage in curved surface area

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1

Ex 11.1 Class 9 Maths Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Steprf of Construction:
Step I : Draw AB¯¯¯¯¯¯¯¯.
Step II : Taking O as centre and having a suitable radius, draw a semicircle, which cuts OA¯¯¯¯¯¯¯¯ at B.
Step III : Keeping the radius same, divide the semicircle into three equal parts such that BC˘=CD˘=DE˘ .
Step IV : Draw OC¯¯¯¯¯¯¯¯ and OD¯¯¯¯¯¯¯¯.
Step V : Draw OF¯¯¯¯¯¯¯¯, the bisector of ∠COD.

Thus, ∠AOF = 90°
Justification:
∵ O is the centre of the semicircle and it is divided into 3 equal parts.
BC˘=CD˘=DE˘
⇒ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]
And, ∠BOC + ∠COD + ∠DOE = 180°
⇒ ∠BOC + ∠BOC + ∠BOC = 180°
⇒ 3∠BOC = 180°
⇒ ∠BOC = 60°
Similarly, ∠COD = 60° and ∠DOE = 60°
OF¯¯¯¯¯¯¯¯ is the bisector of ∠COD
∴ ∠COF = 12 ∠COD = 12 (60°) = 30°
Now, ∠BOC + ∠COF = 60° + 30°
⇒ ∠BOF = 90° or ∠AOF = 90°

Ex 11.1 Class 9 Maths Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:
Steps of Construction:
Stept I : Draw OA¯¯¯¯¯¯¯¯.
Step II : Taking O as centre and with a suitable radius, draw a semicircle such that it intersects OA¯¯¯¯¯¯¯¯. at B.
Step III : Taking B as centre and keeping the same radius, cut the semicircle at C. Now, taking C as centre and keeping the same radius, cut the semicircle at D and similarly, cut at E, such that BC˘=CD˘=DE˘

Step IV : Draw OC¯¯¯¯¯¯¯¯ and OD¯¯¯¯¯¯¯¯.
Step V : Draw OF¯¯¯¯¯¯¯¯, the angle bisector of ∠BOC.
Step VI : Draw OG¯¯¯¯¯¯¯¯, the ajngle bisector of ∠FOC.

Thus, ∠BOG = 45° or ∠AOG = 45°
Justification:
BC˘=CD˘=DE˘
∴ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]
Since, ∠BOC + ∠COD + ∠DOE = 180°
⇒ ∠BOC = 60°
OF¯¯¯¯¯¯¯¯ is the bisector of ∠BOC.
∴ ∠COF = 12 ∠BOC = 12(60°) = 30° …(1)
Also, OG¯¯¯¯¯¯¯¯ is the bisector of ∠COF.
∠FOG = 12∠COF = 12(30°) = 15° …(2)
Adding (1) and (2), we get
∠COF + ∠FOG = 30° + 15° = 45°
⇒ ∠BOF + ∠FOG = 45° [∵ ∠COF = ∠BOF]
⇒ ∠BOG = 45° Ex 11.1 Class 9 Maths Question 3.
Construct the angles of the following measurements
(i) 30°
(ii) 22 12∘
(iii) 15°
Solution:
(i) Angle of 30°
Steps of Construction:
Step I : Draw OA¯¯¯¯¯¯¯¯.
Step II : With O as centre and having a suitable radius, draw an arc cutting OA¯¯¯¯¯¯¯¯ at B.
Step III : With centre at B and keeping the same radius as above, draw an arc to cut the previous arc at C.
Step IV : Join OC¯¯¯¯¯¯¯¯ which gives ∠BOC = 60°.
Step V : Draw OD¯¯¯¯¯¯¯¯, bisector of ∠BOC, such that ∠BOD = 12∠BOC = 12(60°) = 30°

Thus, ∠BOD = 30° or ∠AOD = 30° (ii) Angle of 22 12∘
Steps of Construction:
Step I : Draw OA¯¯¯¯¯¯¯¯.
Step II : Construct ∠AOB = 90°
Step III : Draw OC¯¯¯¯¯¯¯¯, the bisector of ∠AOB, such that
∠AOC = 12∠AOB = 12(90°) = 45°
Step IV : Now, draw OD, the bisector of ∠AOC, such that
∠AOD = 12∠AOC = 12(45°) = 22 12∘

Thus, ∠AOD = 22 12∘ (iii) Angle of 15°
Steps of Construction:
Step I : Draw OA¯¯¯¯¯¯¯¯.
Step II : Construct ∠AOB = 60°.
Step III : Draw OC, the bisector of ∠AOB, such that
∠AOC = 12∠AOB = 12(60°) = 30°
i.e., ∠AOC = 30°
Step IV : Draw OD, the bisector of ∠AOC such that
∠AOD = 12∠AOC = 12(30°) = 15°

Thus, ∠AOD = 15° Ex 11.1 Class 9 Maths Question 4.
Construct the following angles and verify by measuring them by a protractor
(i) 75°
(ii) 105°
(iii) 135°
Solution:
Step I : Draw OA¯¯¯¯¯¯¯¯.
Step II : With O as centre and having a suitable radius, draw an arc which cuts OA¯¯¯¯¯¯¯¯ at B.
Step III : With centre B and keeping the same radius, mark a point C on the previous arc.
Step IV : With centre C and having the same radius, mark another point D on the arc of step II.
Step V : Join OC¯¯¯¯¯¯¯¯ and OD¯¯¯¯¯¯¯¯, which gives ∠COD = 60° = ∠BOC.
Step VI : Draw OP¯¯¯¯¯¯¯¯, the bisector of ∠COD, such that
∠COP = 12∠COD = 12(60°) = 30°.
Step VII: Draw OQ¯¯¯¯¯¯¯¯, the bisector of ∠COP, such that
∠COQ = 12∠COP = 12(30°) = 15°.

Thus, ∠BOQ = 60° + 15° = 75°∠AOQ = 75° (ii) Steps of Construction:
Step I : Draw OA¯¯¯¯¯¯¯¯.
Step II : With centre O and having a suitable radius, draw an arc which cuts OA¯¯¯¯¯¯¯¯ at B.
Step III : With centre B and keeping the same radius, mark a point C on the previous arc.
Step IV : With centre C and having the same radius, mark another point D on the arc drawn in step II.
Step V : Draw OP, the bisector of CD which cuts CD at E such that ∠BOP = 90°.
Step VI : Draw OQ¯¯¯¯¯¯¯¯, the bisector of BC˘ such that ∠POQ = 15°

Thus, ∠AOQ = 90° + 15° = 105° (iii) Steps of Construction:
Step I : Draw OP¯¯¯¯¯¯¯¯.
Step II : With centre O and having a suitable radius, draw an arc which cuts OP¯¯¯¯¯¯¯¯ at A
Step III : Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II such that AQ˘=QR˘=RS˘ .
StepIV :Draw OL¯¯¯¯¯¯¯, thebisector of RS˘ which cuts the arc RS˘ at T.
Step V : Draw OM¯¯¯¯¯¯¯¯¯, the bisector of RT˘.

Thus, ∠POQ = 135° Ex 11.1 Class 9 Maths Question 5.
Construct an equilateral triangle, given its side and justify the construction.
Solution:
pt us construct an equilateral triangle, each of whose side = 3 cm(say).
Steps of Construction:
Step I : Draw OA¯¯¯¯¯¯¯¯.
Step II : Taking O as centre and radius equal to 3 cm, draw an arc to cut OA¯¯¯¯¯¯¯¯ at B such that OB = 3 cm
Step III : Taking B as centre and radius equal to OB, draw an arc to intersect the previous arc at C.
Step IV : Join OC and BC.

Thus, ∆OBC is the required equilateral triangle.

Justification:
∵ The arcs OC˘ and BC˘ are drawn with the same radius.
OC˘ = BC˘
⇒ OC = BC [Chords corresponding to equal arcs are equal]
∵ OC = OB = BC
∴ OBC is an equilateral triangle.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

Ex 11.2 Class 9 Maths Question 1.
Construct a ∆ ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Solution:
Steps of Construction:
Step I : Draw BX¯¯¯¯¯¯¯¯.
Step II : Along BX¯¯¯¯¯¯¯¯, cut off a line segment BC = 7 cm.
Step III : At B, construct ∠CBY = 75°
Step IV : From BY¯¯¯¯¯¯¯¯, cut off BD = 13 cm (= AB + AC)
Step V : Join DC.
Step VI : Draw a perpendicular bisector of CD which meets BD at A.
Step VII: Join AC.

Thus, ∆ABC is the required triangle. Ex 11.2 Class 9 Maths Question 2.
Construct a ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm.
Solution:
Steps of Construction:
Step I : Draw BX¯¯¯¯¯¯¯¯.
Step II : Along BX¯¯¯¯¯¯¯¯, cut off a line segment BC = 8 cm.
Step III : At B, construct ∠CBY = 45°
Step IV : From BX¯¯¯¯¯¯¯¯, cut off BD = 3.5 cm (= AB – AC)
Step V : Join DC.
Step VI : Draw PQ, perpendicular bisector of DC, which intersects BY¯¯¯¯¯¯¯¯ at A.
Step VII: Join AC.

Thus, ∆ABC is the required triangle. Ex 11.2 Class 9 Maths Question 3.
Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
Solution:
Steps of Construction:
Step I : Draw QX¯¯¯¯¯¯¯¯.
Step II : Along QX¯¯¯¯¯¯¯¯, cut off a line segment QR = 6 cm.
Step III : Construct a line YQY’ such that ∠RQY = 60°.
Step IV : Cut off QS = 2 cm (= PR – PQ) on QY’.
Step V : Join SR.
Step VI : Draw MN, perpendicular bisector of SR, which intersects QY at P.
Step VII: Join PR.

Thus, ∆PQR is the required triangle. Ex 11.2 Class 9 Maths Question 4.
Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm.
Solution:
Steps of Construction:
Step I : Draw a line segment AB = 11 cm = (XY+YZ + ZX)
Step II : Construct ∠BAP = 30°
Step III : Construct ∠ABQ = 90°
Step IV : Draw AR, the bisector of ∠BAP.
Step V : Draw BS, the bisector of ∠ABQ. Let AR and BS intersect at X.
Step VI : Draw perpendicular bisector of AX¯¯¯¯¯¯¯¯, which intersects AB at Y.
Step VII: Draw perpendicular bisector of XB¯¯¯¯¯¯¯¯, which intersects AB at Z.
Step VIII: Join XY and XZ.

Thus, ∆XYZ is the required triangle. Ex 11.2 Class 9 Maths Question 5.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Solution:
Steps of Construction:
Step I : Draw BC = 12 cm.
Step II : At B, construct ∠CBY = 90°.
Step III : Along BY¯¯¯¯¯¯¯¯, cut off a line segment BX = 18 cm.
Step IV : Join CX.
Step V : Draw PQ, perpendicular bisector of CX, which meets BX at A.
Step VI : Join AC.

Thus, ∆ABC is the required triangle.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.1.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.1

Ex 10.1 Class 9 Maths Question 1.
Fill in the blanks.
(i) The centre of a circle lies in ___ of the circle. (exterior/interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle, (exterior/interior)
(iii) The longest chord of a circle is a ____ of the circle.
(iv) An arc is a ____ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and ____ of the circle.
(vi) A circle divides the plane, on which it lies, in ____ parts.
Solution:
(i) interior
(ii) exterior
(iii) diameter
(iv) semicircle
(v) the chord
(vi) three

Ex 10.1 Class 9 Maths Question 2.
(i) Line segment joining the centre to any point on the circle is a , radius of the circle.
(ii) A circle has only finite number of equal chords.

(iii) If a circle is divided into three equal arcs, each is a major arc.

(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.
Solution:
(i) True [∵ All points on the circle are equidistant from the centre]
(ii) False [ ∵ A circle can have an infinite number of equal chords]

(iii) False [∵ Each part will be less than a semicircle]
(iv) True [ ∵ Diameter = 2 x Radius]
(v) False [ ∵ The region between the chord and its corresponding arc is a segment]
(vi) True [ ∵ A circle is drawn on a plane]

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.2

Ex 10.2 Class 9 Maths Question 1.
Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres
Solution:
Given: Two congruent circles with centres O and O’ and radii r, which have chords AB and CD respectively such that AB = CD.
To Prove: ∠AOB = ∠CO’D
Proof: In ∆AOB and ∆CO’D, we have
AB = CD [Given]
OA = O’C [Each equal to r]
OB = O’D [Each equal to r]
∴ ∆AOB ≅ ∆CO’D [By SSS congruence criteria]
⇒ ∠AOB = ∠CO’D [C.P.C.T.]

Ex 10.2 Class 9 Maths Question 2.
Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:
Given: Two congruent circles with centres O & O’ and radii r which have chords AB and CD respectively such that ∠AOB = ∠CO’D.

To Prove: AB = CD
Proof: In ∆AOB and ∆CO’D, we have
OA = O’C [Each equal to r]
OB = O’D [Each equal to r]
∠AOB = ∠CO’D [Given]
∴ ∆AOB ≅ ∆CO’D [By SAS congruence criteria]
Hence, AB = CD [C.P.C.T.]

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3

Ex 10.3 Class 9 Maths Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
Let us draw different pairs of circles as shown below:

We have

Thus, two circles can have at the most two points in common. Ex 10.3 Class 9 Maths Question 2.
Suppose you are given a circle. Give a construction to find its centre.
Solution:
Steps of construction :
Step I : Take any three points on the given circle. Let these points be A, B and C.
Step II : Join AB and BC.
Step III : Draw the perpendicular bisector, PQ of AB.
Step IV: Draw the perpendicular bisector, RS of BC such that it intersects PQ at O.

Thus, ‘O’ is the required centre of the given drcle. Ex 10.3 Class 9 Maths Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution:
We have two circles with centres O and O’, intersecting at A and B.
∴ AB is the common chord of two circles and OO’ is the line segment joining their centres.
Let OO’ and AB intersect each other at M.

∴ To prove that OO’ is the perpendicular bisector of AB,
we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’,
we have
OA = OB [Radii of the same circle]
O’A = O’B [Radii of the same circle]
OO’ = OO’ [Common]
∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence criteria]
⇒ ∠1 = ∠2 , [C.P.C.T.]
Now, in ∆AOM and ∆BOM, we have
OA = OB [Radii of the same circle]
OM = OM [Common]
∠1 = ∠2 [Proved above]
∴ ∆AOM = ∆BOM [By SAS congruence criteria]
⇒ ∠3 = ∠4 [C.P.C.T.]
But ∠3 + ∠4 = 180° [Linear pair]
∴∠3=∠4 = 90°
⇒ AM ⊥ OO’
Also, AM = BM [C.P.C.T.]
⇒ M is the mid-point of AB.
Thus, OO’ is the perpendicular bisector of AB.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4

Ex 10.4 Class 9 Maths Question 1.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:
We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord.
∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord.

∴∠OLP = ∠OLQ = 90° and PL = LQ
Now, in right ∆OLP, we have
PL2 + OL2 = 2
⇒ PL2 + (4 – x)2 = 52
⇒ PL2 = 52 – (4 – x)2
⇒ PL2 = 25 -16 – x2 + 8x
⇒ PL2 = 9 – x2 + 8x …(i)
Again, in right ∆O’LP,
PL2 = PO‘2 – LO‘2
= 32 – x2 = 9 – x2 …(ii)
From (i) and (ii), we have
9 – x2 + 8x = 9 – x2
⇒ 8x = 0
⇒ x = 0
⇒ L and O’ coincide.
∴ PQ is a diameter of the smaller circle.
⇒ PL = 3 cm
But PL = LQ
∴ LQ = 3 cm
∴ PQ = PL + LQ = 3cm + 3cm = 6cm
Thus, the required length of the common chord = 6 cm. Ex 10.4 Class 9 Maths Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Solution:
Given: A circle with centre O and equal chords AB and CD intersect at E.
To Prove: AE = DE and CE = BE
Construction : Draw OM ⊥ AB and ON ⊥ CD.
Join OE.
Proof: Since AB = CD [Given]
∴ OM = ON [Equal chords are equidistant from the centre]
Now, in ∆OME and ∆ONE, we have
∠OME = ∠ONE [Each equal to 90°]
OM = ON [Proved above]
OE = OE [Common hypotenuse]
∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]
⇒ ME = NE [C.P.C.T.]

Adding AM on both sides, we get
⇒ AM + ME = AM + NE
⇒ AE = DN + NE = DE
∵ AB = CD ⇒ 12AB = 12DC
⇒ AM = DN
⇒ AE = DE …(i)
Now, AB – AE = CD – DE
⇒ BE = CE …….(ii)
From (i) and (ii), we have
AE = DE and CE = BE Ex 10.4 Class 9 Maths Question 3.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
Given: A circle with centre O and equal chords AB and CD are intersecting at E.
To Prove : ∠OEA = ∠OED
Construction: Draw OM ⊥ AB and ON ⊥ CD.
Join OE.
Proof: In ∆OME and ∆ONE,
OM = ON
[Equal chords are equidistant from the centre]
OE = OE [Common hypotenuse]

∠OME = ∠ONE [Each equal to 90°]
∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]
⇒ ∠OEM = ∠OEN [C.P.C.T.]
⇒ ∠OEA = ∠OED Ex 10.4 Class 9 Maths Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure).

Solution:
Given : Two circles with the common centre O.
A line D intersects the outer circle at A and D and the inner circle at B and C.
To Prove : AB = CD.
Construction:
Draw OM ⊥ l.
Proof: For the outer circle,
OM ⊥ l [By construction]
∴ AM = MD …(i)
[Perpendicular from the centre to the chord bisects the chord]

For the inner circle,
OM ⊥ l [By construction]
∴ BM = MC …(ii)
[Perpendicular from the centre to the chord bisects the chord]
Subtracting (ii) from (i), we have
AM – BM = MD – MC
⇒ AB = CD Ex 10.4 Class 9 Maths Question 5.
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Solution:
Let the three girls Reshma, Salma and Mandip be positioned at R, S and M respectively on the circle with centre O and radius 5 m such that
RS = SM = 6 m [Given]

Equal chords of a circle subtend equal angles at the centre.
∴ ∠1 = ∠2
In ∆POR and ∆POM, we have
OP = OP [Common]
OR = OM [Radii of the same circle]
∠1 = ∠2 [Proved above]
∴ ∆POR ≅ ∆POM [By SAS congruence criteria]
∴ PR = PM and
∠OPR = ∠OPM [C.P.C.T.]
∵∠OPR + ∠OPM = 180° [Linear pair]
∴∠OPR = ∠OPM = 90°
⇒ OP ⊥ RM
Now, in ∆RSP and ∆MSP, we have
RS = MS [Each 6 cm]
SP = SP [Common]
PR = PM [Proved above]
∴ ∆RSP ≅ ∆MSP [By SSS congruence criteria]
⇒ ∠RPS = ∠MPS [C.P.C.T.]
But ∠RPS + ∠MPS = 180° [Linear pair]
⇒ ∠RPS = ∠MPS = 90°
SP passes through O.
Let OP = x m
∴ SP = (5 – x)m
Now, in right ∆OPR, we have
x2 + RP2 = 52
RP2 = 52 – x2
In right ∆SPR, we have
(5 – x)2 + RP2 = 62
⇒ RP2 = 62 – (5 – x)2 ……..(ii)
From (i) and (ii), we get
⇒ 52 – x2 = 62 – (5 – x)2
⇒ 25 – x2 = 36 – [25 – 10x + x2]
⇒ – 10x + 14 = 0
⇒ 10x = 14 ⇒ x = 1410 = 1.4
Now, RP2 = 52 – x2
⇒ RP2 = 25 – (1.4)2
⇒ RP2 = 25 – 1.96 = 23.04
∴ RP = 23.04−−−−√= 4.8
∴ RM = 2RP = 2 x 4.8 = 9.6
Thus, distance between Reshma and Mandip is 9.6 m. Ex 10.4 Class 9 Maths Question 6.
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Solution:
Let Ankur, Syed and David are sitting at A, S and D respectively in the circular park with centre O such that AS = SD = DA
i. e., ∆ASD is an equilateral triangle.
Let the length of each side of the equilateral triangle be 2x.
Draw AM ⊥ SD.
Since ∆ASD is an equilateral triangle.
∴ AM passes through O.
⇒ SM = 12 SD = 12 (2x)
⇒ SM = x

Now, in right ∆ASM, we have
AM2 + SM2 = AS2 [Using Pythagoras theorem]
⇒ AM2= AS2 – SM2= (2x)2 – x2
= 4x2 – x2 = 3x2
⇒ AM = 3x−−√m
Now, OM = AM – OA= (3x−−√ – 20)m
Again, in right ∆OSM, we have
OS2 = SM2 + OM2 [using Pythagoras theorem]
202 = x2 + (3x−−√ – 20)2
⇒ 400 = x2 + 3x2 – 403x−−√ + 400
⇒ 4x2 = 40 3x−−√
⇒ x = 10√3 m
Now, SD = 2x = 2 x 10√3 m = 20√3 m
Thus, the length of the string of each phone = 20√3 m

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

Ex 10.5 Class 9 Maths Question 1.
In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.

Solution:
We have a circle with centre O, such that
∠AOB = 60° and ∠BOC = 30°
∵∠AOB + ∠BOC = ∠AOC
∴ ∠AOC = 60° + 30° = 90°
The angle subtended by an arc at the circle is half the angle subtended by it at the centre.
∴ ∠ ADC = 12 (∠AOC) = 12(90°) = 45° Ex 10.5 Class 9 Maths Question 2.
A chord of a circle is equal to the radius of the circle, find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
We have a circle having a chord AB equal to radius of the circle.
∴ AO = BO = AB
⇒ ∆AOB is an equilateral triangle.
Since, each angle of an equilateral triangle is 60°.
⇒ ∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle.

Hence, the angle subtended by the chord on the minor arc = 150°.
Similarly, ∠ADB = 12 [∠AOB] = 12 x 60° = 30°
Hence, the angle subtended by the chord on the major arc = 30° Ex 10.5 Class 9 Maths Question 3.
In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:
The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point pn the circumference.
∴ reflex ∠POR = 2∠PQR
But ∠PQR = 100°
∴ reflex ∠POR = 2 x 100° = 200°
Since, ∠POR + reflex ∠POR = 360°
⇒ ∠POR = 360° – 200°
⇒ ∠POR = 160°
Since, OP = OR [Radii of the same circle]
∴ In ∆POR, ∠OPR = ∠ORP
[Angles opposite to equal sides of a triangle are equal]
Also, ∠OPR + ∠ORP + ∠POR = 180°
[Sum of the angles of a triangle is 180°]
⇒ ∠OPR + ∠ORP + 160° = 180°
⇒ 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP]
⇒ ∠OPR = 20∘2 = 10° Ex 10.5 Class 9 Maths Question 4.
In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.

Solution:
In ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 69° + 31° + ∠BAC = 180°
⇒ ∠BAC = 180° – 100° = 80°
Since, angles in the same segment are equal.
∴∠BDC = ∠BAC ⇒ ∠BDC = 80° Ex 10.5 Class 9 Maths Question 5.
In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC.

Solution:
∠BEC = ∠EDC + ∠ECD
[Sum of interior opposite angles is equal to exterior angle]
⇒ 130° = ∠EDC + 20°
⇒ ∠EDC = 130° – 20° = 110°
⇒ ∠BDC = 110°
Since, angles in the same segment are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BAC = 110° Ex 10.5 Class 9 Maths Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution:
Since angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BDC = 30°

lso, ∠DBC = 70° [Given]
In ∆BCD, we have
∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°]
⇒ ∠BCD + 70° + 30° = 180°
⇒ ∠BCD = 180° -100° = 80°
Now, in ∆ABC,
AB = BC [Given]
∴ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal]
⇒ ∠BCA = 30° [∵ ∠B AC = 30°]
Now, ∠BCA + ∠BCD = ∠BCD
⇒ 30° + ∠ECD = 80°
⇒ ∠BCD = 80° – 30° = 50° Ex 10.5 Class 9 Maths Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Since AC and BD are diameters.
⇒ AC = BD …(i) [All diameters of a circle are equal]
Also, ∠BAD = 90° [Angle formed in a semicircle is 90°]
Similarly, ∠ABC = 90°, ∠BCD = 90°
and ∠CDA = 90°

Now, in ∆ABC and ∆BAD, we have
AC = BD [From (i)]
AB = BA [Common hypotenuse]
∠ABC = ∠BAD [Each equal to 90°]
∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria]
Similarly, AB = DC
Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle.
∴ ABCD is a rectangle. Ex 10.5 Class 9 Maths Question 8.
If the non – parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
We have a trapezium ABCD such that AB ॥ CD and AD = BC.
Let us draw BE ॥ AD such that ABED is a parallelogram.
∵ The opposite angles and opposite sides of a parallelogram are equal.
But AD = BC [Given] …(iii)

∴ From (ii) and (iii), we have BE = BC
⇒ ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal]
Now, ∠BED + ∠BEC = 180° [Linear pair]
⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)]
i.e., A pair of opposite angles of a quadrilateral ABCD is 180°.
∴ ABCD is cyclic.
⇒ The trapezium ABCD is cyclic. Ex 10.5 Class 9 Maths Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A,D and P, Q respectively (see figure). Prove that ∠ ACP = ∠QCD.

Solution:
Since, angles in the same segment of a circle are equal.
∴ ∠ACP = ∠ABP …(i)
Similarly, ∠QCD = ∠QBD …(ii)
Since, ∠ABP = ∠QBD …(iii) [Vertically opposite angles]
∴ From (i), (ii) and (iii), we have
∠ACP = ∠QCD Ex 10.5 Class 9 Maths Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
We have ∆ABC, and two circles described with diameter as AB and AC respectively. They intersect at a point D, other than A.
Let us join A and D.

∵ AB is a diameter.
∴∠ADB is an angle formed in a semicircle.
Adding (i) and (ii), we have
i. e., B, D and C are collinear points.
⇒ BC is a straight line. Thus, D lies on BC.

Ex 10.5 Class 9 Maths Question 11.
ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution:
We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC
∴ Both the triangles are in semi-circle. Case – I: If both the triangles are in the same semi-circle.

⇒ A, B, C and D are concyclic.
Join BD.
DC is a chord.
∴ ∠CAD and ∠CBD are formed in the same segment.
⇒ ∠CAD = ∠CBD Case – II : If both the triangles are not in the same semi-circle.

⇒ A,B,C and D are concyclic. Join BD. DC is a chord.
∴ ∠CAD and ∠CBD are formed in the same segment.
⇒ ∠CAD = ∠CBD Ex 10.5 Class 9 Maths Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral.
∴ Sum of its opposite angles is 180°.
⇒ ∠A + ∠C = 180° …(i)
But ∠A = ∠C …(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
∠A = ∠C = 90°

Similarly,
∠B = ∠D = 90°
⇒ Each angle of the parallelogram ABCD is 90°.
Thus, ABCD is a rectangle.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

Ex 10.6 Class 9 Maths Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given : Two circles with centres O and O’ respectively such that they intersect each other at P and Q.
To Prove: ∠OPO’ = ∠OQO’.
Construction : Join OP, O’P, OQ, O’Q and OO’.
Proof: In ∆OPO’ and ∆OQO’, we have

OP = OQ [Radii of the same circle]
O’P = O’Q [Radii of the same circle]
OO’ = OO’ [Common]
∴ AOPO’ = AOQO’ [By SSS congruence criteria]
⇒ ∠OPO’ = ∠OQO’ [C.P.C.T.] Ex 10.6 Class 9 Maths Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Solution:
We have a circle with centre O.
AB || CD and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm.

Let r cm be the radius of the circle.
Let us draw OP ⊥ AB and OQ ⊥ CD such that
PQ = 6 cm
Join OA and OC.
Let OQ = x cm
∴ OP = (6 – x) cm
∵ The perpendicular drawn from the centre of a circle to chord bisects the chord.

Ex 10.6 Class 9 Maths Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ?
Solution:
We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre

Let r cm be the radius of the circle and draw OP ⊥ AB and join OA and OC.
∵ OP ⊥ AB
∴ P is the mid-point of AB.
⇒ AP = 12AB = 12(6cm) = 3 cm
Similarly, CQ = 12CD = 12(8cm)= 4 cm
Now in ∆OPA, we have OA2 = OP2 + AP2
⇒ r2 = 42 + 32
⇒ r2 = 16 + 9 = 25
⇒ r = 25−−√ =5
Again, in ∆CQO, we have OC2 = OQ2 + CQ2
⇒ r2 = OQ2 + 42
⇒ OQ2 = r2 – 42
⇒ OQ2 = 52 – 42 = 25 – 16 = 9 [∵ r = 5]
⇒ OQ
⇒ √9 = 3
The distance of the other chord (CD) from the centre is 3 cm.
Note: In case if we take the two parallel chords on either side of the centre, then

In ∆POA, OA2 = OP2 + PA2
⇒ r2 = 42 + 32 = 52
⇒ r = 5
In ∆QOC, OC2 = CQ2 + OQ2
⇒ OQ2 = 42 + OQ2
⇒ OQ2 = 52 – 42 = 9
⇒ OQ = 3 Ex 10.6 Class 9 Maths Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
Given : ∠ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE.
To prove: ∠ABC = 12 [∠DOE – ∠AOC]
Construction: Join AE.
Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles.
∴ In ∆BAE, we have
∠DAE = ∠ABC + ∠AEC ……(i)
The chord DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.

⇒ ∠ABC = 12 [(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)]
⇒ ∠ABC = 12 [Difference of the angles subtended by the chords DE and AC at the centre] Ex 10.6 Class 9 Maths Question 5.
Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Solution:
We have a rhombus ABCD such that its diagonals AC and BD intersect at O.
Taking AB as diameter, a circle is drawn. Let us draw PQ || DA and RS || AB, both are passing through O.
P, Q, R and S are the mid-points of DC, AB, AD and BC respectively,
∵ Q is the mid-point of AB.
⇒ AQ = QB …(i)
Since AD = BC [ ∵ ABCD is a rhombus]
∴ 12 AD = 12 BC
⇒ RA = SB
⇒ RA = OQ …(ii)

[ ∵ PQ is drawn parallel to AD and AD = BC]
We have, AB = AD [Sides of rhombus are equal]
⇒ 12 AB = 12 AD
⇒ AQ = AR …(iii)
From (i), (ii) and (iii), we have AQ = QB = OQ
i.e. A circle drawn with Q as centre, will pass through A, B and O.
Thus, the circle passes through the of intersection ‘O’ of the diagonals rhombus ABCD. Ex 10.6 Class 9 Maths Question 6.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
We have a circle passing through A, B and C is drawn such that it intersects CD at E.
∴∠AEC + ∠B = 180° …(i)
[Opposite angles of a cyclic quadrilateral are supplementary] But ABCD is a parallelogram. [Given]
∴∠D = ∠B …(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
∠AEC + ∠D = 180° …(iii)
But ∠AEC + ∠AED = 180° [Linear pair] …(iv)

From (iii) and (iv), we have ∠D = ∠AED
i.e., The base angles of AADE are equal.
∴ Opposite sides must be equal.
⇒ AE = AD Ex 10.6 Class 9 Maths Question 7.
AC and BD are chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.
Solution:
Given: A circle in which two chords AC and BD are such that they bisect each other. Let their point’of intersection be O.
To Prove: (i) AC and BD are diameters.
(ii) ABCD is a rectangle.
Construction: Join AB, BC, CD and DA.
Proof: (i) In ∆AOB and ∆COD, we have
AO = CO [O is the mid-point of AC]
BO = DO [O is the mid-point of BD]
∠AOB = ∠COD [Vertically opposite angles]
∴ Using the SAS criterion of congruence,
∆AOB ≅ ∆COD

⇒ AB = CD [C.P.C.T.]
⇒ arc AB = arc CD …(1)
Similarly, arc AD = arc BC …(2)
Adding (1) and (2), we get
arc AB + arc AD = arc CD + arc BC ⇒ BD divides the circle into two equal parts.
∴ BD is a diameter.
Similarly, AC is a diameter.

(ii) We know that ∆AOB ≅ ∆COD
⇒ ∠OAB = ∠OCD [C.P.C.T]
⇒ ∠CAB = ∠ACD
AB || DC
∴ ABCD is a parallelogram.
Since, opposite angles of a parallelogram are equal.
∴ ∠DAB = ∠DCB
But ∠DAB + ∠DCB = 180°
[Sum of the opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠DAB = 90° = ∠DCB Thus, ABCD is a rectangle. Ex 10.6 Class 9 Maths Question 8.
Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – 12 A, 90° – 12 B and 90° – 12 C.
Solution:
Given : A triangle ABC inscribed in a drcle, such that bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, £ and F respectively.
Construction: Join DE, EF and FD.
Proof:

∵ Angles in the same segment are equal.
∴ ∠ED A = ∠FCA …(i)
∠EDA = ∠EBA …(ii)
Adding (i) and (ii), we have
∠FDA + ∠EDA = ∠FCA + ∠EBA
⇒ ∠FDE = ∠FCA + ∠EBA

Ex 10.6 Class 9 Maths Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Solution:
We have two congruent circles such that they intersect each other at A and B. A line segment passing through A, meets the circles at P and Q.
Let us draw the common chord AB.

Since angles subtended by equal chords in the congruent circles are equal.
⇒ ∠APB = ∠AQB
Now, in ∆PBQ, we have ∠AQB = ∠APB
So, their opposite sides must be equal.
⇒ BP = BQ. Ex 10.6 Class 9 Maths Question 10.
In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.
Solution:
∆ABC with O as centre of its circumcirde. The perpendicular bisector of BC passes through O. Join OB and OC. Suppose it cuts circumcirde at P.
In order to prove that the perpendicular bisector of BC and bisector of angle A of ∆ABC intersect at P, it is sufficient to show that AP is bisector of ∠A of ∆ABC

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.1.

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.1

Ex 9.1 Class 9 Maths Question 1
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Solution:
The figures (i), (iii) and (v) lie on the same base and between the same parallels.

NCERT Solutions for Class 9 Maths Chapter 9 Area of ​​Parallelograms and Triangles (समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल) (Hindi Medium) Ex 9.1

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

Ex 9.2 Class 9 Maths Question 1.
In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD

Solution:
BSOWe have, AE ⊥ DC and AB = 16 cm
∵ AB = CD [Opposite sides of parallelogram]
∴ CD = 16 cm
Now, area of parallelogram ABCD = CD x AE
= (16 x 8) cm2 = 128 cm2 [∵ AE = 8 cm]
∴ Area of parallelogram ABCD = AD x CF
⇒ AD x CF = 128 cm
⇒ AD x 10 cm = 128 cm2 [∵ CF= 10 cm]
⇒ AD = 12810 cm = 12.8 cm 10
Thus, the required length of AD is 12.8 cm Ex 9.2 Class 9 Maths Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 12 ar (ABCD).
Solution:
Join GE and HE, where GE || BC || DA and HF || AB || DC
(∵ E, F, G and H are the mid¬points of the sides of a ||gm ABCD).
If a triangle and a parallelogram are on the same base and between the same parallels, then A E U the area of the triangle is equal to half the area of the parallelogram.

Now, ∆EFG and parallelogram EBCG are on the same base EG and between the same parallels EG and BC.
∴ ar(∆EFG) = 12ar(∥gmEBCG) … (1)
Similarly, ar(∆EHG) = 12ar(∥gmAEGD) …(2)
Adding (1) and (2), we get
ar(∆EFG) + ar(∆EHG) = 12ar(∥gmEBCG)+12ar(∥gmAEGD)
= 12ar(∥gmABCD)
Thus, ar(EFGH) = 12ar(ABCD) Ex 9.2 Class 9 Maths Question 3.
P and Q are any two points lying on the sides DC and AD, respectively of a parallelogram ABCD. Show that ar (APB) = ar(BQC).
Solution:
∵ ABCD is a parallelogram.
∴ AB || CD and BC || AD.
Now, ∆APB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
∴ ar(∆APB) = 12ar(∥gmABCD) …….(1)

Also, ∆BQC and parallelogram ABCD are on the same base BC and between the same parallels BGand AD.
∴ ar(∆BQC) = 12ar(∥gmABCD) …(2)
From (1) and (2), we have ar(∆APB) = ar(∆BQC). Ex 9.2 Class 9 Maths Question 4.
In figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = 12ar(ABCD)
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)
Solution:
We have a parallelogram ABCD, i.e., AB || CD and BC || AD. Let us draw EF || AB and HG || AD through P.

(i) ∆APB and ||gm AEFB are on the same base AB and between the same parallels AB and EF.
∴ ar(∆APB) = 12ar(∥gmAEFB) …(1)
Also, ∆PCD and parallelogram CDEF are on the same base CD and between the same parallels CD and EF.
∴ ar(APCD) = 12ar(∥gmCDEF) …(2)
Adding (1) and (2), we have
ar(∆APB) + ar(∆PCD) = 12ar(∥gmAEFB)+12ar(∥gmCDEF)
⇒ ar(∆APB) + ar(∆PCD) = 12ar(∥gmABCD) …(3)

(ii) ∆APD and ||gm ∆DGH are on the same base AD and between the same parallels AD and GH.
Similarly,
ar(∆PBC) = 12ar(∥gmBCGH) …(5)
Adding (4) and (5), we have
ar(∆APD) + ar(∆PBC) = = 12ar(∥gmADGH)+12ar(∥gmBCGH)
⇒ ar(∆APD) + ar(∆PBC) =12ar(∥gmABCD) …….(6)
From (3) and (6), we have
ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD) Ex 9.2 Class 9 Maths Question 5.
In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 12ar(PQRS)

Solution:
(i) Parallelogram PQRS and parallelogram ABRS are on the same base RS and between the same parallels RS and PB.
∴ ar(PQRS) = ar(ABRS)
(ii) AAXS and ||gm ABRS are on the same base AS and between the same parallels AS and BR. *
∴ ar(AXS) = 12ar(ABRS) …(1)
But ar(PQRS) = ar(ABRS) …(2) [Proved in (i) part]
From (1) and (2), we have
ar(AXS) = 12ar(PQRS) Ex 9.2 Class 9 Maths Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it.
Solution:
The farmer is having the field in the form of parallelogram PQRS and a point A is situated on RS. Join AP and AQ.
Clearly, the field is divided into three parts i.e., in ∆APS, ∆PAQ and ∆QAR.

Since, ∆PAQ and pt.
parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.
ar(∆PAQ) = 12ar(∥gmPQRS) …(1)

⇒ ar(||gm PQRS) – ar(∆PAQ) = ar(||gm PQRS) – 12ar(∥gmPQRS)

[ar(APS) + ar(QAR)]

= 12ar(∥gmPQRS) …(2)
From (1) and (2), we have
ar(∆PAQ) = ar[(∆APS) + (∆QAR)]
Thus, the farmer can sow wheat in (∆PAQ) and pulses in [(∆APS) + (∆QAR)] or wheat in [(∆APS) + (∆QAR)] and pulses in (∆PAQ).

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

Ex 9.3 Class 9 Maths Question 1.
In figure, E is any point on median AD of a ∆ABC. Show that ar (ABE) = ar (ACE).

Solution:
We have a ∆ABC such that AD is a median.
∴ ar(∆ABD) = ar(∆ACD) …(1)
[∵ A median divides the triangle into two triangles of equal areas]
Similarly, in ∆BEC, we have
ar(∆BED) = ar(∆DEC) …(2)
Subtracting (2) from (1), we have
ar(∆ABD) – ar(∆BED) = ar(∆ACD) – ar(∆DEC)
⇒ ar(∆ABE) = ar(∆ACE). Ex 9.3 Class 9 Maths Question 2.
In a triangle ABC, E is the mid-point of median AD. Show that ax (BED) = 12ar(ABC).
Solution:
We have a ∆ABC and its median AD.
Let us join B and E.

Since, a median divides the triangle into two triangles of equal area.
ar (∆ABD) = 12arABC) …….(1)
Now, in ∆ABD, BE is a median.
[ ∵ E is the mid-point of AD]
∴ ar(∆BED) = 12arABC) …(2)
From (1) and (2), we have
ar(∆BED) = 12 [12arABC) ]
⇒ ar(∆BED) = 14arABC) Ex 9.3 Class 9 Maths Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:
We have a parallelogram ABCD (say)
such that its diagonals intersect at O.
∵Diagonals of a parallelogram bisect each other.
∴ AO = OC and BO = OD
Let us draw CE ⊥ BD.
Now, ar(∆BOC) = 12BO x CE and
ar(∆DOC) = 12OD x CE

Since, BO = OD
∴ ar(∆BOC) = ar(∆DOC) …(1)
Similarly, ar(∆AOD) = ar(∆DOC) …(2)
and ar(∆AOB) = ar(∆BOC) …(3)
From (1), (2) and (3), we have
ar(∆AOB) = ar(∆BOC) = ar(∆COD) = ar(∆DOA)
Thus, the diagonals of a parallelogram divide it into four triangles of equal area. Ex 9.3 Class 9 Maths Question 4.
In figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar(ABD)

Solution:
we have ∆ABC and ∆ABD are on the same base AB.
∵ CD is bisected at O. [Given]
∴ CO = OD
Now, in ∆ACD, AO is a median
Again, in ∆BCD, BO is a median
∴ ar(∆OBC) = ar(∆ODB) …(2)
Adding (1) and (2), we have
ar(∆OAQ + ar(∆OBQ) = ar(∆OAD) + ar(∆ODB)
⇒ ar(∆ABC) = ar(∆ABD) Ex 9.3 Class 9 Maths Question 5.
D,E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC. Show that
(i) BDEF is a parallelogram.
(ii) ar(DEF) = 14ar(ABC)
(iii) ar(BDEF) = 14ar(ABC)

Solution:
We have ∆ABC such
that D,E and Fare the mid-points of BC, CA and AB respectively.

(i) In ∆ABC, E and F are the mid-points of AC and B D C AB respectively.
∴ EF || BC [Mid-point theorem]
⇒ EF || BD
Also, EF = 12(BC)
⇒ EF = BD [D is the mid – point of BC]
Since BDEF is a quadrilateral whose one pair of opposite sides is parallel and of equal lengths.
∴ BDEF is a parallelogram.

(ii) We have proved that BDEF is a parallelogram.
Similarly, DCEF is a parallelogram and DEAF is also a parallelogram.
Now, parallelogram BDEF and parallelogram DCEF are on the same base EF and between the same parallels BC and EF.
∴ ar(||gm BDEF) = ar(||gm DCEF)
⇒ 12ar(||gm BDEF) = 12ar(||gm DCEF)
⇒ ar(∆BDF) = ar(∆CDE) …(1)
[Diagonal of a parallelogram divides it into two triangles of equal area]
Similarly, ar(∆CDE) = ar(∆DEF) …(2)
and ar(∆AEF) = ar(∆DEF) …(3)
From (1), (2) and (3), we have
ar(∆AEF) = ar(∆FBD) = ar(∆DEF) = ar(∆CDE)
Thus, ar(∆ABC) = ar(∆AEF) + ar(∆FBD) + ar(∆DEF) + ar(∆CDE) = 4 ar(∆DEF)
⇒ ar(∆DEF) = 14ar(∆ABC)

(iii) We have, ar (||gm BDEF) = ar(∆BDF) + ar(∆DEF)
= ar(∆DEF) + ar(∆DEF) [∵ ar(∆DEF) = ar(∆BDF)]
2ar(∆DEF) = 2[14ar(∆ABC)]
= 12ar(∆ABC)
Thus, ar (||gm BDEF) = 12ar(∆ABC) Ex 9.3 Class 9 Maths Question 6.
In figure, diagonals AC and BD of quadrilateral ABCD intersect at 0 such that OB = OD. If AB = CD, then show that
(i) ar(DOC) = ar(AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram

Solution:
We have a quadrilateral ABCD whose diagonals AC and BD intersect at O.
We also have that OB = OD, AB = CD Let us draw DE ⊥ AC and BF ⊥ AC

(i) In ∆DEO and ∆BFO, we have
DO = BO [Given]
∠DOE = ∠BOF [Vertically opposite angles]
∠DEO = ∠BFO [Each 90°]
∴ ∆DEO ≅ ∆BFO [By A AS congruency]
⇒ DE = BF [By C.P.C.T.]
and ar(∆DEO) = ar(∆BFO) …(1)
Now, in ∆DEC and ∆BFA, we have
∠DEC = ∠BFA [Each 90°]
DE = BF [Proved above]
DC = BA [Given]
∴ ∆DEC ≅ ∆BFA [By RHS congruency]
⇒ ar(∆DEC) = ar(∆BFA) …(2)
and ∠1 = ∠2 …(3) [By C.P.C.T.]
Adding (1) and (2), we have
ar(∆DEO) + ar(∆DEC) = ar(∆BFO) + ar(∆BFA)
⇒ ar(∆DOC) = ar(∆AOB)

(ii) Since, ar(∆DOC) = ar(∆AOB) [Proved above]
Adding ar(∆BOC) on both sides, we have
ar(∆DOC) + ar(∆BOC) = ar(∆AOB) + ar(∆BOC)
⇒ ar(∆DCB) = ar(∆ACB)

(iii) Since, ∆DCS and ∆ACB are both on the same base CB and having equal areas.
∴ They lie between the same parallels CB and DA.
⇒ CB || DA
Also ∠1 = ∠2, [By (3)]
which are alternate interior angles.
So, AB || CD
Hence, ABCD is a parallelogram. Ex 9.3 Class 9 Maths Question 7.
D and E are points on sides AB and AC respectively of ∆ ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.
Solution:
We have ∆ABC and points D and E are such that ar(DBC) = ar{EBC)
Since ∆DBC and ∆EBC are on the same base BC and having same area.

∴ They must lie between the same parallels DE and BC.
Hence, DE || BC Ex 9.3 Class 9 Maths Question 8.
XY is a line parallel to side BC of a ∆ ABC. If BE ||AC and CF || AB meet XY at E and F respectively, show that ar (ABE) =ar (ACF)
Solution:
We have a ∆ABC such that XY || BC,
BE || AC and CF || AB.
Since, XY ||BC and BE || CY
∴ BCYE is a paralleloam.

Now, the parallelogram BCYE and ∆ABE are on the same base 8E and between the same parallels BE and AC.
∴ ar(∆ABE) = 12ar(∥gmBCYE) …..(1)
Again, CF || AB [Given]
XY || BC [Given]
CF || BX and XF || BC
∴ BCFX is a parallelogram.
Now, ∆ACF and parallelogram BCFX are on the same base CF and between the same parallels AB and CF.
∴ar(∆ACF) = 12ar(∥gmBCFX) …(2)
Also, parallelogram BCFX and parallelogram BCYE are on the same base BC and between the same parallels BC and EF.
∴ ar(||gm BCFX) = ar(||gm BCYE) ………(3)
From (1), (2) and (3), we get
ar(∆BE) = ar(∆ACF) Ex 9.3 Class 9 Maths Question 9.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then A parallelogram PBQR is completed (see figure).
Show that ax (ABCD) = ar(PBQR).
[Hint Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Solution:
Let us join AC and PQ.
ABCD is a parallelogram [Given]
and AC is its diagonal, we know that diagonal of a parallelogram divides it into two triangles of equal areas.
∴ ar(∆ABC) = 12ar(∥gmABCD) …(1)
Also, PBQR is a parallelogram [Given]
and QP is its diagonal.
∴ ar(∆BPQ) = 12ar(∥gmPBQR) …(2)
Since, ∆ACQ and AAPQ are on the same base AQ and between A the same parallels AQ and CP.

∴ ar(∆ACQ) = ar(∆APQ)
⇒ ar(∆ACQ) – ar(∆ABQ)
= ar(∆APQ) – ar(∆ABQ)
[Subtracting ar(∆ABQ) from both sides]
⇒ ar(∆ABC) = ar(∆BPQ) …(3)
From (1), (2) and (3), we get
12ar(∥gmABCD) = 12ar(∥gmPBQR)
⇒ ar( ||gm ABCD) = ar(||gm PBQR) Ex 9.3 Class 9 Maths Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC)

Solution:
BBlliWWp have a trapezium ABCD having AB || CD and its diagonals AC and BD intersect each other at O.
Since, triangles on the same base and between the same parallels have equal areas.
∆ABD and ∆ABC are on the same base AB and between the same parallels AB and DC
∴ ar(∆ABD) = ar(∆ABC)
Subtracting ar(∆AOB) from both sides, we get

ar(∆ABD) – ar(∆AOB) = ar(∆ABC) – ar(∆AOB)
⇒ ar(∆AOD) = ar(∆BOC)

Ex 9.3 Class 9 Maths Question 11.
In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)

Solution:
We have a pentagon ABCDE in which BF || AC and DC is produced to F.
(i) Since, the triangles between the same parallels and on the same base are equal in area.
∆ACB and ∆ACF are on the same base AC and between the same parallels AC and BF.
∴ ar(∆ACB) = ar(∆ACF)

(ii) Since, ar(∆ACB) = ar(∆ACF) [Proved above]
∴ ar(ABCDE) = ar(AEDF) Ex 9.3 Class 9 Maths Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Solution:
We have a plot in the form of a quadrilateral ABCD.
Let us draw DF || AC and join AF and CF.

Now, ∆DAF and ∆DCF are on the same base DF and between the same parallels AC and DF.
Subtracting ar(∆DEF) from both sides, we get
ar(∆DAF) – ar(∆DEF) = ar(∆DCF) – ar(∆DEF)
The portion of ∆ADE can be taken over by the Gram Panchayat by adding the land (∆CEF) to his (Itwaari) land so as to form a triangular plot,
i.e. ∆ABF.
Let us prove that ar(∆ABF) = ar(quad. ABCD), we have
⇒ ar(∆ABF) = ar (quad. ABCD) Ex 9.3 Class 9 Maths Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ADX) = ar(ACY). [Hint Join IX]
Solution:
We have a trapezium ABCD such that AB || DC.
XY || AC meets AB at X and BC at Y. Let us join CX.

∆ADX and ∆ACX are on the same base AX and between the same parallels AX and DC.
∵∆ACX and ∆ACY are on the same base AC and between the same parallels AC and XY.
∴ ar(∆ACX) = ar(∆ACY) …(2)
From (1) and (2), we have
ar(∆ADX) = ar(∆ACY) Ex 9.3 Class 9 Maths Question 14.
In figure, AP || BQ || CR. Prove that ar(AQC) = ax(PBR).

Solution:
We have, AP || BQ || CR
∵ ∆BCQ and ∆BQR are on the same base BQ and between the same parallels BQ and CR.
∴ ar(∆BCQ) = ar(∆BQR) …(1)
∵ ∆ABQ and ∆PBQ are on the same base BQ and between the same parallels AP and BQ.
∴ ar(∆ABQ) = ar(∆PBQ) …(2)
Adding (1) and (2), we have
ar(∆BCQ) + ar(∆ABQ) = ar(∆BQR) + ar(∆PBQ)
⇒ ar(∆AQC) = ar(∆PBR) Ex 9.3 Class 9 Maths Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ax(AOD) = ar(BOC). Prove that ABCD is a trapezium.
Solution:
We have a quadrilateral ABCD and its diagonals AC and BD intersect at O such that
ar(∆AOD) = ar(∆BOC) [Given]

Adding ar(∆AOB) to both sides, we have
ar(∆AOD) + ar(∆AOB) = ar(∆BOC) + ar(∆AOB)
⇒ ar(∆ABD) = ar(∆ABC)
Also, they are on the same base AB.
Since, the triangles are on the same base and having equal area.
∴ They must lie between the same parallels.
∴ AB || DC
Now, ABCD is a quadrilateral having a pair of opposite sides parallel.
So, ABCD is a trapezium. Ex 9.3 Class 9 Maths Question 16.
In figure ax(DRC) = ar(DPC) and ai(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:
tfclfiftWe have, ar(∆DRC) = ar(∆DPC) [Given]
And they are on the same base DC.
∴ ∆DRC and ∆DPC must lie between the same parallels.
So, DC || RP i.e.r a pair of opposite sides of quadrilateral DCPR is parallel.
∴ Quadrilateral DCPR is a trapezium.
Again, we have
ar(∆BDP) = ar(∆ARC) [Given] …(1)
Also, ar(∆DPC) = ar(∆DRC) [Given] …(2)
Subtracting (2) from (1), we get
ar(∆BDP) – ar(∆DPC) = ar(∆ARQ – ar(∆DRQ
And they are on the same base DC.
∴ ABDC and AADC must lie between the same parallels.
So, AB || DC i.e. a pair of opposite sides of quadrilateral ABCD is parallel.
∴ Quadrilateral ABCD is a trapezium.

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

Ex 9.4 Class 9 Maths Question 1.
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Solution:
We have a parallelogram ABCD and rectangle ABEF such that
ar(||gm ABCD) = ar( rect. ABEF)

AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
⇒ CD = EF
⇒ AB + CD = AB + EF … (1)
BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side] ⇒ (BC + AD) > (BE + AF) …(2)
From (1) and (2), we have
(AB + CD) + (BC+AD) > (AB + EF) + BE + AF)
⇒ (AB + BC + CD + DA) > (AB + BE + EF + FA)
⇒ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF. Ex 9.4 Class 9 Maths Question 2.
In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).

Solution:
Let us draw AF, perpendicular to BC
such that AF is the height of ∆ABD, ∆ADE and ∆AEC

Ex 9.4 Class 9 Maths Question 3.
In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ax(BCF).

Solution:
Since, ABCD is a parallelogram [Given]
∴ Its opposite sides are parallel and equal.
Now, ∆ADE and ∆BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF.
So, ar(∆ADE) = ar(∆BCF). Ex 9.4 Class 9 Maths Question 4.
In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(BPC) = ax(DPQ).[Hint Join AC.]

Solution:
We have a parallelogram ABCD and AD = CQ. Let us join AC.
We know that triangles on the same base and between the same parallels are equal in area.
Since, ∆QAC and ∆QDC are on the same base QC and between the same parallels AD and BQ.
∴ ar(∆QAC) = ar(∆QDC)
Subtracting ar(∆QPC) from both sides, we have
ar(∆QAQ – ar(∆QPC) = ar(∆QDC) – ar(∆QPC)
⇒ ar(∆PAQ = ar(∆QDP) …(1)
Since, ∆PAC and ∆PBC are on the same base PC and between the same parallels AB and CD.
∴ ar(∆PAC) = ar(∆PBC) …(2)
From (1) and (2), we get
ar(∆PBC) = ar(∆QDP) Ex 9.4 Class 9 Maths Question 5.
In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Show that

[Hint Join EC and AD. Show that BE || AC and DE || AB, etc.]
Solution:
Let us join EC and AD. Draw EP ⊥ BC.
Let AB = BC = CA = a, then
BD = a2 = DE = BE

(ii) Since, ∆ABC and ∆BED are equilateral triangles.
⇒ ∠ACB = ∠DBE = 60°
⇒ BE || AC
∆BAE and ∆BEC are on the same base BE and between the same parallels BE and AC.
ar(∆BAE) = ar(∆BEC)
⇒ ar(∆BAE) = 2 ar(∆BDE) [ DE is median of ∆EBC. ∴ ar(∆BEC) = || ar(∆BDE)]
⇒ ar(ABDE) = 12ar(∆BAE)

(iii) ar(∆ABC) = 4 ar(∆BDE)[Proved in (i) part]
ar(∆BEC) = 2 ar(∆BDE)
[ ∵ DE is median of ∆BEC]
⇒ ar(∆ABC) = 2 ar(∆BEC)

(iv) Since, ∆ABC and ∆BDE are equilateral triangles.
⇒ ∠ABC = ∠BDE = 60°
⇒ AB || DE
∆BED and ∆AED are on the same base ED and between the same parallels AB and DE.
∴ ar(∆BED) = ar(∆AED)
Subtracting ar(AEFD) from both sides, we get
⇒ ar(∆BED) – ar(∆EFD) = ar(∆AED) – ar(∆EFD)
⇒ ar(∆BEE) = ar(∆AFD) (v) In right angled ∆ABD, we get

From (1) and (2), we get
ar(∆AFD) = 2 ar(∆EFD)
ar(∆AFD) = ar(∆BEF) [From (iv) part]
⇒ ar(∆BFE) = 2 ar(∆EFD)

(vi) ar(∆AFC) = ar(∆AFD) + ar(∆ADC)
= ar(∆BFE) + 12 ar(∆ABC) [From (iv) part]
= ar(∆BFE) + 12 x 4 x ar(∆BDE) [From (i) part]
= ar(∆BFE) + 2ar(∆BDE)
= 2ar(∆FED) + 2[ar(∆BFE) + ar(∆FED)]
= 2ar(∆FED) + 2[2ar(∆FED) + ar(∆FED)] [From (v) part]
= 2ar(∆FED) + 2[3ar(∆FED)]
= 2ar(∆FED) + 6ar(∆FED)
= 8ar(∆FED)
∴ ar(∆FED) = 18 ar(∆AFC) Ex 9.4 Class 9 Maths Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(APB) x ar(CPD) = ar(APD) x ar(BPC).
[Hint From A and C, draw perpendiculars to BD.]
Solution:
We have a quadrilateral ABCD such that its diagonals AC and BD intersect at P.

Let us draw AM ⊥ BD and CN ⊥ BD.

Ex 9.4 Class 9 Maths Question 7.
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

Solution:
We have a ∆ABC such that P is the mid-point of AB and Q is the mid-point of BC.
Also, R is the mid-point of AP. Let us join AQ, RQ, PC and PC.

(i) In ∆APQ, R is the mid-point of AP. [Given] B

∴RQ is a median of ∆APQ.
⇒ ar(∆PRQ) = 12ar(∆APQ) …(1)
In ∆ABQ, P is the mid-point of AB.
∴ QP is a median of ∆ABQ.
∴ ar(∆APQ) = 12ar(∆ABQ) …(2)

Ex 9.4 Class 9 Maths Question 8.
In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that

(i) ∆MBC = ∆ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ax(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
Solution:
We have a right ∆ABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line segment AX 1 DE is also drawn such that it meets BC at Y.

(i) ∠CBD = ∠MBA [Each90°]
∴ ∠CBD + ∠ABC = ∠MBA + ∠ABC
(By adding ∠ABC on both sides)
or ∠ABD = ∠MBC
In ∆ABD and ∆MBC, we have
AB = MB [Sides of a square]
BD = BC
∠ABD = ∠MBC [Proved above]
∴ ∆ABD = ∆MBC [By SAS congruency]

(ii) Since parallelogram BYXD and ∆ABD are on the same base BD and between the same parallels BD and AX.
∴ ar(∆ABD) = 12ar(||gm BYXD)
But ∆ABD ≅ ∆MBC [From (i) part]
Since, congruent triangles have equal
areas.
∴ ar(∆MBC) = 12ar(||gm BYXD)
⇒ ar(||gm BYXD) = 2ar(∆MBC)

(iii) Since, ar(||gm BYXD) = 2ar(∆MBC) …(1) [From (ii) part]
and or(square ABMN) = 2or(∆MBC) …(2)
[ABMN and AMBC are on the same base MB and between the same parallels MB and NC]
From (1) and (2), we have
ar(BYXD) = ar(ABMN) .

(iv) ∠FCA = ∠BCE (Each 90°)
or ∠FCA+ ∠ACB = ∠BCE+ ∠ACB
[By adding ∠ACB on both sides]
⇒ ∠FCB = ∠ACE
In ∆FCB and ∆ACE, we have
FC = AC [Sides of a square]
CB = CE [Sides of a square]
∠FCB = ∠ACE [Proved above]
⇒ ∆FCB ≅ ∆ACE [By SAS congruency]

(v) Since, ||gm CYXE and ∆ACE are on the same base CE and between the same parallels CE and AX.
∴ ar(||gm CYXE) = 2ar(∆ACE)
But ∆ACE ≅ ∆FCB [From (iv) part]
Since, congruent triangles are equal in areas.
∴ ar (||<gm CYXE) = 2ar(∆FCB)

(vi) Since, ar(||gm CYXE) = 2ar(∆FCB) …(3)
[From (v) part]
Also (quad. ACFG) and ∆FCB are on the same base FC and between the same parallels FC and BG.
⇒ ar(quad. ACFG) = 2ar(∆FCB) …(4)
From (3) and (4), we get

[From (iii) part]
[From (vi) part]

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1

Ex 8.1 Class 9 MathsQuestion 1.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
∴ 3x + 5x + 9x + 13x = 360°
[Angle sum property of a quadrilateral]
⇒ 30x = 360°
⇒ x = 360∘30 = 12°
∴ 3x = 3 x 12° = 36°
5x = 5 x 12° = 60°
9x = 9 x 12° = 108°
13a = 13 x 12° = 156°
⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°. Ex 8.1 Class 9 Maths Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Let ABCD is a parallelogram such that AC = BD.

In ∆ABC and ∆DCB,
AC = DB [Given]
AB = DC [Opposite sides of a parallelogram]
BC = CB [Common]
∴ ∆ABC ≅ ∆DCB [By SSS congruency]
⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)
Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]
∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]
From (1) and (2), we have
∠ABC = ∠DCB = 90°
i.e., ABCD is a parallelogram having an angle equal to 90°.
∴ ABCD is a rectangle. Ex 8.1 Class 9 Maths Question 3.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.

∴ In ∆AOB and ∆AOD, we have
AO = AO [Common]
OB = OD [O is the mid-point of BD]
∠AOB = ∠AOD [Each 90]
∴ ∆AQB ≅ ∆AOD [By,SAS congruency
∴ AB = AD [By C.P.C.T.] ……..(1)
Similarly, AB = BC .. .(2)
BC = CD …..(3)
CD = DA ……(4)
∴ From (1), (2), (3) and (4), we have
AB = BC = CD = DA
Thus, the quadrilateral ABCD is a rhombus.
Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus. Ex 8.1 Class 9 Maths Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Let ABCD be a square such that its diagonals AC and BD intersect at O.

(i) To prove that the diagonals are equal, we need to prove AC = BD.
In ∆ABC and ∆BAD, we have
AB = BA [Common]
BC = AD [Sides of a square ABCD]
∠ABC = ∠BAD [Each angle is 90°]
∴ ∆ABC ≅ ∆BAD [By SAS congruency]
AC = BD [By C.P.C.T.] …(1)

(ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram]
∴ ∠1 = ∠3
[Alternate interior angles are equal]
Similarly, ∠2 = ∠4
Now, in ∆OAD and ∆OCB, we have
AD = CB [Sides of a square ABCD]
∠1 = ∠3 [Proved]
∠2 = ∠4 [Proved]
∴ ∆OAD ≅ ∆OCB [By ASA congruency]
⇒ OA = OC and OD = OB [By C.P.C.T.]
i.e., the diagonals AC and BD bisect each other at O. …….(2)

(iii) In ∆OBA and ∆ODA, we have
OB = OD [Proved]
BA = DA [Sides of a square ABCD]
OA = OA [Common]
∴ ∆OBA ≅ ∆ODA [By SSS congruency]
⇒ ∠AOB = ∠AOD [By C.P.C.T.] …(3)
∵ ∠AOB and ∠AOD form a linear pair.
∴∠AOB + ∠AOD = 180°
∴∠AOB = ∠AOD = 90° [By(3)]
⇒ AC ⊥ BD …(4)
From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles. Ex 8.1 Class 9 Maths Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:
Let ABCD be a quadrilateral such that diagonals AC and BD are equal and bisect each other at right angles.

Now, in ∆AOD and ∆AOB, We have
∠AOD = ∠AOB [Each 90°]
AO = AO [Common]
OD = OB [ ∵ O is the midpoint of BD]
∴ ∆AOD ≅ ∆AOB [By SAS congruency]
⇒ AD = AB [By C.P.C.T.] …(1)
Similarly, we have
AB = BC … (2)
BC = CD …(3)
CD = DA …(4)
From (1), (2), (3) and (4), we have
AB = BC = CD = DA
∴ Quadrilateral ABCD have all sides equal.
In ∆AOD and ∆COB, we have
AO = CO [Given]
OD = OB [Given]
∠AOD = ∠COB [Vertically opposite angles]
So, ∆AOD ≅ ∆COB [By SAS congruency]
∴∠1 = ∠2 [By C.P.C.T.]
But, they form a pair of alternate interior angles.
Similarly, AB || DC
∴ ABCD is a parallelogram.
∴ Parallelogram having all its sides equal is a rhombus.
∴ ABCD is a rhombus.
Now, in ∆ABC and ∆BAD, we have
AC = BD [Given]
AB = BA [Common]
∴ ∆ABC ≅ ∆BAD [By SSS congruency]
∴ ∠ABC = ∠BAD [By C.P.C.T.] ……(5)
Since, AD || BC and AB is a transversal.
∴∠ABC + ∠BAD = 180° .. .(6) [ Co – interior angles]
⇒ ∠ABC = ∠BAD = 90° [By(5) & (6)]
So, rhombus ABCD is having one angle equal to 90°.
Thus, ABCD is a square. Ex 8.1 Class 9 Maths Question 6.
Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.

Solution:
We have a parallelogram ABCD in which diagonal AC bisects ∠A
⇒ ∠DAC = ∠BAC

(i) Since, ABCD is a parallelogram.
∴ AB || DC and AC is a transversal.
∴ ∠1 = ∠3 …(1)
[ ∵ Alternate interior angles are equal]
Also, BC || AD and AC is a transversal.
∴ ∠2 = ∠4 …(2)
[ v Alternate interior angles are equal]
Also, ∠1 = ∠2 …(3)
[ ∵ AC bisects ∠A]
From (1), (2) and (3), we have
∠3 = ∠4
⇒ AC bisects ∠C.

(ii) In ∆ABC, we have
∠1 = ∠4 [From (2) and (3)]
⇒ BC = AB …(4)
[ ∵ Sides opposite to equal angles of a ∆ are equal]
But, ABCD is a parallelogram. [Given]
∴ AB = DC …(6)
From (4), (5) and (6), we have
AB = BC = CD = DA
Thus, ABCD is a rhombus. Ex 8.1 Class 9 Maths Question 7.
ABCD is a rhombus. Show that diagonal AC bisects ∠Aas well as ∠C and diagonal BD bisects ∠B as well AS ∠D.
Solution:
Since, ABCD is a rhombus.
⇒ AB = BC = CD = DA
Also, AB || CD and AD || BC

Now, CD = AD ⇒ ∠1 = ∠2 …….(1)
[ ∵ Angles opposite to equal sides of a triangle are equal]
Also, AD || BC and AC is the transversal.
[ ∵ Every rhombus is a parallelogram]
⇒ ∠1 = ∠3 …(2)
[ ∵ Alternate interior angles are equal]
From (1) and (2), we have
∠2 = ∠3 …(3)
Since, AB || DC and AC is transversal.
∴ ∠2 = ∠4 …(4)
[ ∵ Alternate interior angles are equal] From (1) and (4),
we have ∠1 = ∠4
∴ AC bisects ∠C as well as ∠A.
Similarly, we can prove that BD bisects ∠B as well as ∠D. Ex 8.1 Class 9 Maths Question 8.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.
Solution:
We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.
i.e., ∠1 = ∠4 and ∠2 = ∠3 ……..(1)

(i) Since, every rectangle is a parallelogram.
∴ ABCD is a parallelogram.
⇒ AB || CD and AC is a transversal.
∴∠2 = ∠4 …(2)
[ ∵ Alternate interior angles are equal]
From (1) and (2), we have
∠3 = ∠4
In ∆ABC, ∠3 = ∠4
⇒ AB = BC
[ ∵ Sides opposite to equal angles of a A are equal]
Similarly, CD = DA
So, ABCD is a rectangle having adjacent sides equal.
⇒ ABCD is a square.

(ii) Since, ABCD is a square and diagonals of a square bisect the opposite angles.
So, BD bisects ∠B as well as ∠D. Ex 8.1 Class 9 Maths Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that

Solution:
We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB

(i) Since, AD || BC and BD is a transversal.
∴ ∠ADB = ∠CBD [ ∵ Alternate interior angles are equal]
Now, in ∆APD and ∆CQB, we have
AD = CB [Opposite sides of a parallelogram ABCD are equal]
PD = QB [Given]
∴ ∆APD ≅ ∆CQB [By SAS congruency]

(ii) Since, ∆APD ≅ ∆CQB [Proved]
⇒ AP = CQ [By C.P.C.T.]

(iii) Since, AB || CD and BD is a transversal.
∴ ∠ABD = ∠CDB
⇒ ∠ABQ = ∠CDP
Now, in ∆AQB and ∆CPD, we have
QB = PD [Given]
∠ABQ = ∠CDP [Proved]
AB = CD [ Y Opposite sides of a parallelogram ABCD are equal]
∴ ∆AQB = ∆CPD [By SAS congruency]

(iv) Since, ∆AQB = ∆CPD [Proved]
⇒ AQ = CP [By C.P.C.T.]

Opposite sides are equal. [Proved]
∴ ∆PCQ is a parallelogram. Ex 8.1 Class 9 Maths Question 10.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that

Solution:
(i) In ∆APB and ∆CQD, we have
∠APB = ∠CQD [Each 90°]
AB = CD [ ∵ Opposite sides of a parallelogram ABCD are equal]
∠ABP = ∠CDQ
[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]
∴ ∆APB = ∆CQD [By AAS congruency]

(ii) Since, ∆APB ≅ ∆CQD [Proved]
⇒ AP = CQ [By C.P.C.T.] Ex 8.1 Class 9 Maths Question 11.
In ∆ABC and ∆DEF, AB = DE, AB || DE, BC – EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram

(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF
Solution:
(i) We have AB = DE [Given]
and AB || DE [Given]
i. e., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.
∴ ABED is a parallelogram.

(ii) BC = EF [Given]
and BC || EF [Given]
i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.
∴ BEFC is a parallelogram.

(iii) ABED is a parallelogram [Proved]
[ ∵ Opposite sides of a parallelogram are equal and parallel] Also, BEFC is a parallelogram. [Proved]
BE || CF and BE = CF …(2)
[ ∵ Opposite sides of a parallelogram are equal and parallel]
From (1) and (2), we have

i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.

(v) Since, ACFD is a parallelogram. [Proved]
So, AC =DF [∵ Opposite sides of a parallelogram are equal]

(vi) In ∆ABC and ∆DFF, we have
AB = DE [Given]
BC = EF [Given]
AC = DE [Proved in (v) part]
∆ABC ≅ ∆DFF [By SSS congruency] Ex 8.1 Class 9 Maths Question 12.
ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that
(i )∠A=∠B
(ii )∠C=∠D
(iv) diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].
Solution:
We have given a trapezium ABCD in which AB || CD and AD = BC. (i) Produce AB to E and draw CF || AD.. .(1)

∵ AB || DC
⇒ AE || DC Also AD || CF
∴ AECD is a parallelogram.
[ ∵ Opposite sides of the parallelogram are equal]
But AD = BC …(2) [Given]
By (1) and (2), BC = CF
Now, in ∆BCF, we have BC = CF
⇒ ∠CEB = ∠CBE …(3)
[∵ Angles opposite to equal sides of a triangle are equal]
Also, ∠ABC + ∠CBE = 180° … (4)
[Linear pair]
and ∠A + ∠CEB = 180° …(5)
[Co-interior angles of a parallelogram ADCE]
From (4) and (5), we get
∠ABC + ∠CBE = ∠A + ∠CEB
⇒ ∠ABC = ∠A [From (3)]
⇒ ∠B = ∠A …(6)

(ii) AB || CD and AD is a transversal.
∴ ∠A + ∠D = 180° …(7) [Co-interior angles]
Similarly, ∠B + ∠C = 180° … (8)
From (7) and (8), we get
∠A + ∠D = ∠B + ∠C
⇒ ∠C = ∠D [From (6)]

(iii) In ∆ABC and ∆BAD, we have
AB = BA [Common]
∴ ∆ABC = ∆BAD [By SAS congruency]

(iv) Since, ∆ABC = ∆BAD [Proved]
⇒ AC = BD [By C.P.C.T.]

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2

Question 1.
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that
(i) SR || AC and SR = 12 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution:
(i) In ∆ACD, We have
∴ S is the mid-point of AD and R is the mid-point of CD.
SR = 12AC and SR || AC …(1)
[By mid-point theorem]

(ii) In ∆ABC, P is the mid-point of AB and Q is the mid-point of BC.
PQ = 12AC and PQ || AC …(2)
[By mid-point theorem]
From (1) and (2), we get
PQ = 12AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
PQ = SR and PQ || SR [Proved]
∴ PQRS is a parallelogram. Question 2.
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
Solution:
We have a rhombus ABCD and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Join AC.

In ∆ABC, P and Q are the mid-points of AB and BC respectively.
∴ PQ = 12AC and PQ || AC …(1)
[By mid-point theorem]
In ∆ADC, R and S are the mid-points of CD and DA respectively.
∴ SR = 12AC and SR || AC …(2)
[By mid-point theorem]
From (1) and (2), we get
PQ = 12AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
∴ PQRS is a parallelogram. …….(3)
Now, in ∆ERC and ∆EQC,
∠1 = ∠2
[ ∵ The diagonals of a rhombus bisect the opposite angles]
CR = CQ [ ∵CD2 = BC2]
CE = CE [Common]
∴ ∆ERC ≅ ∆EQC [By SAS congruency]
⇒ ∠3 = ∠4 …(4) [By C.P.C.T.]
But ∠3 + ∠4 = 180° ……(5) [Linear pair]
From (4) and (5), we get
⇒ ∠3 = ∠4 = 90°
Now, ∠RQP = 180° – ∠b [ Y Co-interior angles for PQ || AC and EQ is transversal]
But ∠5 = ∠3
[ ∵ Vertically opposite angles are equal]
∴ ∠5 = 90°
So, ∠RQP = 180° – ∠5 = 90°
∴ One angle of parallelogram PQRS is 90°.
Thus, PQRS is a rectangle.

Question 3.
ABCD is a rectangle and P, Q, R ans S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.
Solution:
We have,
Now, in ∆ABC, we have
PQ = 12AC and PQ || AC …(1)
[By mid-point theorem]
SR = 12AC and SR || AC …(2)
From (1) and (2), we get
PQ = SR and PQ || SR
∴ PQRS is a parallelogram.
Now, in ∆PAS and ∆PBQ, we have
∠A = ∠B [Each 90°]
AP = BP [ ∵ P is the mid-point of AB]
AS = BQ [∵ 12AD = 12BC]
∴ ∆PAS ≅ ∆PBQ [By SAS congruency]
⇒ PS = PQ [By C.P.C.T.]
Also, PS = QR and PQ = SR [∵opposite sides of a parallelogram are equal]
So, PQ = QR = RS = SP i.e., PQRS is a parallelogram having all of its sides equal.
Hence, PQRS is a rhombus. Question 4.
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.

Solution:
We have,

In ∆DAB, we know that E is the mid-point of
AD and EG || AB [∵ EF || AB]
Using the converse of mid-point theorem, we get, G is the mid-point of BD.
Again in ABDC, we have G is the midpoint of BD and GF || DC.
[∵ AB || DC and EF || AB and GF is a part of EF]
Using the converse of the mid-point theorem, we get, F is the mid-point of BC. Question 5.
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.

Solution:
Since, the opposite sides of a parallelogram are parallel and equal.
∴ AB || DC
⇒ AE || FC …(1)
and AB = DC
⇒ 12AB = 12DC
⇒ AE = FC …(2)
From (1) and (2), we have
AE || PC and AE = PC
∴ ∆ECF is a parallelogram.
Now, in ∆DQC, we have F is the mid-point of DC and FP || CQ
[∵ AF || CE]
⇒ DP = PQ …(3)
[By converse of mid-point theorem] Similarly, in A BAP, E is the mid-point of AB and EQ || AP [∵AF || CE]
⇒ BQ = PQ …(4)
[By converse of mid-point theorem]
∴ From (3) and (4), we have
DP = PQ = BQ
So, the line segments AF and EC trisect the diagonal BD. Question 6.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution:
Let ABCD be a quadrilateral, where P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
Join PQ, QR, RS and SP.
Let us also join PR, SQ and AC.

Now, in ∆ABC, we have P and Q are the mid-points of its sides AB and BC respectively.
∴ PQ || AC and PQ = 12 AC …(1)
[By mid-point theorem]
Similarly, RS || AC and RS = 12AC …(2)
∴ By (1) and (2), we get
PQ || RS, PQ = RS
∴ PQRS is a parallelogram.
And the diagonals of a parallelogram bisect each other, i.e., PR and SQ bisect each other. Thus, the line segments joining the midpoints of opposite sides of a quadrilateral ABCD bisect each other. Question 7.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = 12AB
Solution:

we have

(i) In ∆ACB, We have
M is the mid-point of AB. [Given]
MD || BC , [Given]
∴ Using the converse of mid-point theorem,
D is the mid-point of AC.

(ii) Since, MD || BC and AC is a transversal.
∠MDA = ∠BCA
[ ∵ Corresponding angles are equal] As
∠BCA = 90° [Given]
∠MDA = 90°
⇒ MD ⊥AC.

(iii) In ∆ADM and ∆CDM, we have
∠ADM = ∠CDM [Each equal to 90°]
MD = MD [Common]
AD = CD [∵ D is the mid-point of AC]
∴ ∆ADM ≅ ∆CDM [By SAS congruency]
⇒ MA = MC [By C.P.C.T.] .. .(1)
∵ M is the mid-point of AB [Given]
MA = 12AB …(2)
From (1) and (2), we have
CM = MA = 12AB

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Ex 7.2 Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

Ex 7.1 Class 9 Maths Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

Solution:
In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A.
Now, In ∆ABC and ∆ABD,
∠ CAB = ∠ DAB ( AB bisects ∠ CAB)
and AB = AB (Common)
∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)
∴ BC = BD (By CPCT) Ex 7.1 Class 9 Maths Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that

(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠ BAC
Solution:
In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA

(i) In ∆ ABC and ∆ BAC,
∠DAB = ∠CBA (Given)
AB = AB (Common)
∴ ∆ ABD ≅ ∆BAC (By SAS congruence)

(ii) Since ∆ABD ≅ ∆BAC
⇒ BD = AC [By C.P.C.T.]

(iii) Since ∆ABD ≅ ∆BAC
⇒ ∠ABD = ∠BAC [By C.P.C.T.] Ex 7.1 Class 9 Maths Question 3.
AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Solution:
In ∆BOC and ∆AOD, we have
∠BOC = ∠AOD
∠BOC = ∠AOD [Vertically opposite angles]
∴ ∆OBC ≅ ∆OAD [By AAS congruency]
⇒ OB = OA [By C.P.C.T.]
i.e., O is the mid-point of AB.
Thus, CD bisects AB. Ex 7.1 Class 9 Maths Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.

Solution:
∵ p || q and AC is a transversal,
∴ ∠BAC = ∠DCA …(1) [Alternate interior angles]
Also l || m and AC is a transversal,
∴ ∠BCA = ∠DAC …(2)
[Alternate interior angles]
Now, in ∆ABC and ∆CDA, we have
∠BAC = ∠DCA [From (1)]
CA = AC [Common]
∠BCA = ∠DAC [From (2)]
∴ ∆ABC ≅ ∆CDA [By ASA congruency] Ex 7.1 Class 9 Maths Question 5.
Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms ot ∠A.

Solution:
We have, l is the bisector of ∠QAP.
∴ ∠QAB = ∠PAB
∠Q = ∠P [Each 90°]
∠ABQ = ∠ABP
[By angle sum property of A]
Now, in ∆APB and ∆AQB, we have
∠ABP = ∠ABQ [Proved above]
AB = BA [Common]
∠PAB = ∠QAB [Given]
∴ ∆APB ≅ ∆AQB [By ASA congruency]
Since ∆APB ≅ ∆AQB
⇒ BP = BQ [By C.P.C.T.]
i. e., [Perpendicular distance of B from AP]
= [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of ∠A. Ex 7.1 Class 9 Maths Question 6.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ∆ABC and ∆ADE. we have
∠BAC = ∠DAE [Proved above]
AC = AE [Given]
∴ ∆ABC ≅ ∆ADE [By SAS congruency]
⇒ BC = DE [By C.P.C.T.] Ex 7.1 Class 9 Maths Question 7.
AS is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB. (see figure). Show that
(i) ∆DAP ≅ ∆EBP

Solution:
We have, P is the mid-point of AB.
∴ AP = BP
∠EPA = ∠DPB [Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE

(i) Now, in ∆DAP and ∆EBP, we have
AP = BP [Proved above]
∠DPA = ∠EPB [Proved above]
∴ ∆DAP ≅ ∆EBP [By ASA congruency]

(ii) Since, ∆ DAP ≅ ∆ EBP
⇒ AD = BE [By C.P.C.T.] Ex 7.1 Class 9 Maths Question 8.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB

(iv) CM = 12 AB
Solution:
Since M is the mid – point of AB.
∴ BM = AM

(i) In ∆AMC and ∆BMD, we have
CM = DM [Given]
∠AMC = ∠BMD [Vertically opposite angles]
AM = BM [Proved above]
∴ ∆AMC ≅ ∆BMD [By SAS congruency]

(ii) Since ∆AMC ≅ ∆BMD
⇒ ∠MAC = ∠MBD [By C.P.C.T.]
But they form a pair of alternate interior angles.
∴ AC || DB
Now, BC is a transversal which intersects parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180° [Co-interior angles]
But ∠BCA = 90° [∆ABC is right angled at C]
∴ 90° + ∠DBC = 180°
⇒ ∠DBC = 90°

(iii) Again, ∆AMC ≅ ∆BMD [Proved above]
∴ AC = BD [By C.P.C.T.]
Now, in ∆DBC and ∆ACB, we have
BD = CA [Proved above]
∠DBC = ∠ACB [Each 90°]
BC = CB [Common]
∴ ∆DBC ≅ ∆ACB [By SAS congruency]

(iv) As ∆DBC ≅ ∆ACB
DC = AB [By C.P.C.T.]
But DM = CM [Given]
∴ CM = 12DC = 12AB
⇒ CM = 12AB

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

Ex 7.2 Class 9 Maths Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that
(i) OB = OC
(ii) AO bisects ∠A
Solution:
i) in ∆ABC, we have
AB = AC [Given]
∴ ∠ABC = ∠ACB [Angles opposite to equal sides of a A are equal]

⇒ 12∠ABC = 12∠ACB
or ∠OBC = ∠OCB
⇒ OC = OB [Sides opposite to equal angles of a ∆ are equal]

(ii) In ∆ABO and ∆ACO, we have
AB = AC [Given]
∠OBA = ∠OCA [ ∵12∠B = 12∠C]
OB = OC [Proved above]
∆ABO ≅ ∆ACO [By SAS congruency]
⇒ ∠OAB = ∠OAC [By C.P.C.T.]
⇒ AO bisects ∠A. Ex 7.2 Class 9 Maths Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.

Solution:
Since AD is bisector of BC.
∴ BD = CD
Now, in ∆ABD and ∆ACD, we have
BD = CD [Proved above]
∴ ∆ABD ≅ ∆ACD [By SAS congruency]
⇒ AB = AC [By C.P.C.T.]
Thus, ∆ABC is an isosceles triangle. Ex 7.2 Class 9 Maths Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

Solution:
∆ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal]
⇒ ∠BCE = ∠CBF
Now, in ∆BEC and ∆CFB
∠BCE = ∠CBF [Proved above]
∠BEC = ∠CFB [Each 90°]
BC = CB [Common]
∴ ∆BEC ≅ ∆CFB [By AAS congruency]
So, BE = CF [By C.P.C.T.] Ex 7.2 Class 9 Maths Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure).

Show that
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC i.e., ABC is an isosceles triangle.
Solution:
(i) In ∆ABE and ∆ACE, we have
∠AEB = ∠AFC
[Each 90° as BE ⊥ AC and CF ⊥ AB]
∠A = ∠A [Common]
BE = CF [Given]
∴ ∆ABE ≅ ∆ACF [By AAS congruency]

(ii) Since, ∆ABE ≅ ∆ACF
∴ AB = AC [By C.P.C.T.]
⇒ ABC is an isosceles triangle. Ex 7.2 Class 9 Maths Question 5.
ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD.

Solution:
In ∆ABC, we have
AB = AC [ABC is an isosceles triangle]
∴ ∠ABC = ∠ACB …(1)
[Angles opposite to equal sides of a ∆ are equal]
Again, in ∆BDC, we have
BD = CD [BDC is an isosceles triangle]
∴ ∠CBD = ∠BCD …(2)
[Angles opposite to equal sides of a A are equal]
Adding (1) and (2), we have
∠ABC + ∠CBD = ∠ACB + ∠BCD
⇒ ∠ABD = ∠ACD. Ex 7.2 Class 9 Maths Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

Solution:
AB = AC [Given] …(1)
From (1) and (2), we have
Now, in ∆ABC, we have
∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]
⇒ 2∠ACB + ∠BAC = 180° …(3)
[∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)]
Similarly, in ∆ACD,
⇒ 2∠ACD + ∠CAD = 180° …(4)
[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°
⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]
⇒ 2∠BCD = 360° – 180° = 180°
⇒ ∠BCD = 180∘2 = 90°
Thus, ∠BCD = 90° Ex 7.2 Class 9 Maths Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.
Solution:
In ∆ABC, we have AB = AC [Given]
∴ Their opposite angles are equal.

⇒ ∠ACB = ∠ABC
Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]
⇒ 90° + ∠B + ∠C = 180° [∠A = 90°(Given)]
⇒ ∠B + ∠C= 180°- 90° = 90°
But ∠B = ∠C
∠B = ∠C = 90∘2 = 45°
Thus, ∠B = 45° and ∠C = 45° Ex 7.2 Class 9 Maths Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
In ∆ABC, we have

AB = BC = CA
[ABC is an equilateral triangle]
AB = BC
⇒ ∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]
Similarly, AC = BC
⇒ ∠A = ∠B …(2)
From (1) and (2), we have
∠A = ∠B = ∠C = x (say)
Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]
∴ x + x + x = 180o
⇒ 3x = 180°
⇒ x = 60°
∴ ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

Ex 7.3 Class 9 Maths Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that

(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:
(i) In ∆ABD and ∆ACD, we have
AB = AC [Given]
BD = CD [Given]
∴ ∆ABD ≅ ∆ACD [By SSS congruency]

(ii) In ∆ABP and ∆ACP, we have
AB = AC [Given]
∠BAP = ∠CAP [From (1)]
∴ AP = PA [Common]
∴ ∆ABP ≅ ∆ACP [By SAS congruency]

(iii) Since, ∆ABP ≅ ∆ACP
⇒ ∠BAP = ∠CAP [By C.P.C.T.]
∴ AP is the bisector of ∠A.
Again, in ∆BDP and ∆CDP,
we have BD = CD [Given]
DP = PD [Common]
BP = CP [ ∵ ∆ABP ≅ ∆ACP]
⇒ A BDP = ACDP [By SSS congruency]
∴ ∠BDP = ∠CDP [By C.P.C.T.]
⇒ DP (or AP) is the bisector of ∠BDC
∴ AP is the bisector of ∠A as well as ∠D.

(iv) As, ∆ABP ≅ ∆ACP
⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.]
But ∠APB + ∠APC = 180° [Linear Pair]
∴ ∠APB = ∠APC = 90°
⇒ AP ⊥ BC, also BP = CP
Hence, AP is the perpendicular bisector of BC. Ex 7.3 Class 9 Maths Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
Solution:
(i) In right ∆ABD and ∆ACD, we have
AB =AC [Given]

∴ ∆ABD ≅ ∆ACD [By RHS congruency]
So, BD = CD [By C.P.C.T.]
⇒ D is the mid-point of BC or AD bisects BC.

(ii) Since, ∆ABD ≅ ∆ACD,
So, AD bisects ∠A Ex 7.3 Class 9 Maths Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that
(i) ∆ABC ≅ ∆PQR
(ii) ∆ABM ≅ ∆PQN

Solution:
In ∆ABC, AM is the median.
∴BM = 12 BC ……(1)
In ∆PQR, PN is the median.
∴ QN = 12QR …(2)
And BC = QR [Given]
⇒ 12BC = 12QR
⇒ BM = QN …(3) [From (1) and (2)]

(i) In ∆ABM and ∆PQN, we have
AB = PQ , [Given]
AM = PN [Given]
BM = QN [From (3)]
∴ ∆ABM ≅ ∆PQN [By SSS congruency]

(ii) Since ∆ABM ≅ ∆PQN
⇒ ∠B = ∠Q …(4) [By C.P.C.T.]
Now, in ∆ABC and ∆PQR, we have
∠B = ∠Q [From (4)]
AB = PQ [Given]
BC = QR [Given]
∴ ∆ABC ≅ ∆PQR [By SAS congruency] Ex 7.3 Class 9 Maths Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
Since BE ⊥ AC [Given]

∴ BEC is a right triangle such that ∠BEC = 90°
Similarly, ∠CFB = 90°
Now, in right ∆BEC and ∆CFB, we have
BE = CF [Given]
BC = CB [Common hypotenuse]
∠BEC = ∠CFB [Each 90°]
∴ ∆BEC ≅ ∆CFB [By RHS congruency]
So, ∠BCE = ∠CBF [By C.P.C.T.]
or ∠BCA = ∠CBA
Now, in ∆ABC, ∠BCA = ∠CBA
⇒ AB = AC [Sides opposite to equal angles of a ∆ are equal]
∴ ABC is an isosceles triangle. Ex 7.3 Class 9 Maths Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
We have, AP ⊥ BC [Given]

∠APB = 90° and ∠APC = 90°
In ∆ABP and ∆ACP, we have
∠APB = ∠APC [Each 90°]
AB = AC [Given]
AP = AP [Common]
∴ ∆ABP ≅ ∆ACP [By RHS congruency]
So, ∠B = ∠C [By C.P.C.T.]

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

Ex 7.4 Class 9 Maths Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Let us consider ∆ABC such that ∠B = 90°
∴ ∠A + ∠B + ∠C = 180°
⇒ ∠A + 90°-+ ∠C = 180°
⇒ ∠A + ∠C = 90°
⇒∠A + ∠C = ∠B
∴ ∠B > ∠A and ∠B > ∠C

⇒ Side opposite to ∠B is longer than the side opposite to ∠A
i.e., AC > BC.
Similarly, AC > AB.
Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side. Ex 7.4 Class 9 Maths Question 2.
In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:
∠ABC + ∠PBC = 180° [Linear pair]
and ∠ACB + ∠QCB = 180° [Linear pair]
But ∠PBC < ∠QCB [Given] ⇒ 180° – ∠PBC > 180° – ∠QCB
⇒ ∠ABC > ∠ACB
The side opposite to ∠ABC > the side opposite to ∠ACB
⇒ AC > AB. Ex 7.4 Class 9 Maths Question 3.
In figure, ∠B <∠ A and ∠C <∠ D. Show that AD < BC.

Solution: Since ∠A > ∠B [Given]
∴ OB > OA …(1)
[Side opposite to greater angle is longer]
Similarly, OC > OD …(2)
Adding (1) and (2), we have
OB + OC > OA + OD
⇒ BC > AD Ex 7.4 Class 9 Maths Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠ A > ∠C and ∠B >∠D.

Solution:
Let us join AC.

Now, in ∆ABC, AB < BC [∵ AB is the smallest side of the quadrilateral ABCD] ⇒ BC > AB
⇒ ∠BAC > ∠BCA …(1)
[Angle opposite to longer side of A is greater]
Again, in ∆ACD, CD > AD
[ CD is the longest side of the quadrilateral ABCD]
[Angle opposite to longer side of ∆ is greater]
Adding (1) and (2), we get
∠BAC + ∠CAD > ∠BCA + ∠ACD
⇒ ∠A > ∠C
Similarly, by joining BD, we have ∠B > ∠D. Ex 7.4 Class 9 Maths Question 5.
In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ.

Solution:
In ∆PQR, PS bisects ∠QPR [Given]
∴ ∠QPS = ∠RPS
and PR > PQ [Given]
⇒ ∠PQS > ∠PRS [Angle opposite to longer side of A is greater]
⇒ ∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS]
∵ Exterior ∠PSR = [∠PQS + ∠QPS]
and exterior ∠PSQ = [∠PRS + ∠RPS]
[An exterior angle is equal to the sum of interior opposite angles]
Now, from (1), we have
∠PSR = ∠PSQ. Ex 7.4 Class 9 Maths Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:

Let us consider the ∆PMN such that ∠M = 90°

Since, ∠M + ∠N+ ∠P = 180°
[Sum of angles of a triangle is 180°]
∵ ∠M = 90° [PM ⊥ l]
So, ∠N + ∠P = ∠M
⇒ ∠N < ∠M
⇒ PM < PN …(1)
Similarly, PM < PN1 …(2)
and PM < PN2 …(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular line segment is the shortest line segment drawn on a line from a point not on it.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

Ex 7.5 Class 9 Maths Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Solution:
Let us consider a ∆ABC.
Draw l, the perpendicular bisector of AB.
Draw m, the perpendicular bisector of BC.
Let the two perpendicular bisectors l and m meet at O.
O is the required point which is equidistant from A, B and C.

Note: If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle.

Ex 7.5 Class 9 Maths Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Let us consider a ∆ABC.

Draw m, the bisector of ∠C.
Let the two bisectors l and m meet at O.
Thus, O is the required point which is equidistant from the sides of ∆ABC.
Note: If we draw OM ⊥ BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle. Point O is called incentre of the triangle.

Ex 7.5 Class 9 Maths Question 3.
In a huge park, people are concentrated at three points (see figure)

A: where these are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exist.
Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?
[Hint The parlour should be equidistant from A, B and C.]
Solution:
Let us join A and B, and draw l, the perpendicular bisector of AB.
Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O.
The point O is the required point where the ice cream parlour be set up.
Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O.

Ex 7.5 Class 9 Maths Question 4.
Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Solution:
It is an activity.
We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii).
∴ The Fig. (ii) has more triangles.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.1.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.1

Ex 5.1 Class 9 Maths Question 1.
Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In figure, if AB – PQ and PQ = XY, then AB = XY.

Solution:
(i) False
Reason : If we mark a point O on the surface of a paper. Using pencil and scale, we can draw infinite number of straight lines passing
through O.

(ii) False
Reason : In the following figure, there are many straight lines passing through P. There are many lines, passing through Q. But there is one and only one line which is passing through P as well as Q.

(iii) True
Reason: The postulate 2 says that “A terminated line can be produced indefinitely.”

(iv) True
Reason : Superimposing the region of one circle on the other, we find them coinciding. So, their centres and boundaries coincide.
Thus, their radii will coincide or equal.

(v) True
Reason : According to Euclid’s axiom, things which are equal to the same thing are equal to one another.

Ex 5.1 Class 9 MathsQuestion 2.
Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they and how might you define them?
(i) Parallel lines
(ii) Perpendicular lines
(iii) Line segment
(v) Square
Solution:
Yes, we need to have an idea about the terms like point, line, ray, angle, plane, circle and quadrilateral, etc. before defining the required terms.
Definitions of the required terms are given below: (i) Parallel Lines:
Two lines l and m in a plane are said to be parallel, if they have no common point and we write them as l ॥ m.

(ii) Perpendicular Lines:
Two lines p and q lying in the same plane are said to be perpendicular if they form a right angle and we write them as p ⊥ q

(iii) Line Segment:
A line segment is a part of line and having a definite length. It has two end-points. In the figure, a line segment is shown having end points A and B. It is written as AB¯¯¯¯¯¯¯¯ or BA¯¯¯¯¯¯¯¯.

(iv) Radius of a circle :
The distance from the centre to a point on the circle is called the radius of the circle. In the figure, P is centre and Q is a point on the circle, then PQ is the radius.

(v) Square :
A quadrilateral in which all the four angles are right angles and all the four sides are equal is called a square. Given figure, PQRS is a square.

Ex 5.1 Class 9 MathsQuestion 3.
Consider two ‘postulates’ given below
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist atleast three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Solution:
Yes, these postulates contain undefined terms such as ‘Point and Line’. Also, these postulates are consistent because they deal with two different situations as
(i) says that given two points A and B, there is a point C lying on the line in between them. Whereas
(ii) says that, given points A and B, you can take point C not lying on the line through A and B.
No, these postulates do not follow from Euclid’s postulates, however they follow from the axiom, “Given two distinct points, there is a unique line that passes through them.” Ex 5.1 Class 9 Maths Question 4.
If a point C lies between two points A and B such that AC = BC, then prove that AC = 12 AB, explain by drawing the figure.
Solution:
We have,

AC = BC [Given]
∴ AC + AC = BC + AC
[If equals added to equals then wholes are equal]
or 2AC = AB [∵ AC + BC = AB]
or AC = 12AB Ex 5.1 Class 9 Maths Question 5.
In question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Solution:
Let the given line AB is having two mid points ‘C’ and ‘D’.

AC = 12AB ……(i)
Subtracting (i) from (ii), we have
or AD – AC = 0 or CD = 0
∴ C and D coincide.
Thus, every line segment has one and only one mid-point. Ex 5.1 Class 9 Maths Question 6.
In figure, if AC = BD, then prove that AB = CD.

Solution:
Given: AC = BD
⇒ AB + BC = BC + CD
Subtracting BC from both sides, we get
AB + BC – BC = BC + CD – BC
[When equals are subtracted from equals, remainders are equal]
⇒ AB = CD

Ex 5.1 Class 9 MathsQuestion 7.
Why is axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that, the question is not about the fifth postulate.)
Solution:
As statement is true in all the situations. Hence, it is considered a ‘universal truth.’

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.2

Ex 5.2 Class 9 MathsQuestion 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Solution:
We can write Euclid’s fifth postulate as ‘Two distinct intersecting lines cannot be parallel to the same line.’ Ex 5.2 Class 9 Maths Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines ? Explain.
Solution:
Yes. If a straight line l falls on two lines m and n such that sum of the interior angles on one side of l is two right angles, then by Euclid’s fifth postulate, lines m and n will not meet on this side of l. Also, we know that the sum of the interior angles on the other side of the line l will be two right angles too. Thus, they will not meet on the other side also.

∴ The lines m and n never meet, i.e, They are parallel.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1

Ex 4.1 Class 9 MathsQuestion 1.
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be Rs. x and that of a pen to be Rs.y).
Solution:
Let the cost of a notebook = Rs. x
and the cost of a pen = Rs. y
According to the condition, we have
[Cost of a notebook] =2 x [Cost of a pen]
i. e„ (x) = 2 x (y) or, x = 2y
or, x – 2y = 0
Thus, the required linear equation is x – 2y = 0.

Ex 4.1 Class 9 MathsQuestion 2
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35¯¯¯
(ii) xy5−10=0
(iii) – 2x + 3y = 6
(iv) x = 3y
(v) 2x = -5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(viii) 5 = 2x
Solution:
(i) We have 2x + 3y = 9.35¯¯¯
or (2)x + (3)y + (−9.35¯¯¯ ) = 0
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= –9.35¯¯¯ .

(ii) We have xy5−10=0
or x + (- 15) y + (10) = 0
Comparing it with ax + by + c = 0, we get
a =1, b =- 15 and c= -10

(iii) Wehave -2x + 3y = 6 or (-2)x + (3)y + (-6) = 0
Comparing it with ax – 4 – by + c = 0,we get a = -2, b = 3 and c = -6.

(iv) We have x = 3y or (1)x + (-3)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 1, b = -3 and c = 0.
(v) We have 2x = -5y or (2)x + (5)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 2, b = 5 and c = 0.
(vi) We have 3x + 2 = 0 or (3)x + (0)y + (2) = 0 Comparing it with ax + by + c = 0, we get a = 3, b = 0 and c = 2.
(vii) We have y – 2 = 0 or (0)x + (1)y + (-2) = 0 Comparing it with ax + by + c = 0, we get a = 0, b = 1 and c = -2.
(viii) We have 5 = 2x ⇒ 5 – 2x = 0
or -2x + 0y + 5 = 0
or (-2)x + (0)y + (5) = 0
Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1
Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions
Solution:
Option (iii) is true because for every value of x, we get a corresponding value of y and vice-versa in the given equation.
Hence, given linear equation has an infinitely many solutions.

Question 2
Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i) 2x + y = 7
When x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)
When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)
When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9
When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
When x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
When x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))

(iii) x = 4y
When x = 0, 4y = 1 ⇒ y = 0
∴ Solution is (0, 0)
When x = 1, 4y = 1 ⇒ y = 14
∴ Solution is (1,14 )
When x = 4, 4y = 4 ⇒ y = 1
∴ Solution is (4, 1)
When x = 4, 4y = 4 ⇒ y = -1
∴ Solution is (-4, -1)

Question 3
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0,2)
(ii) (2,0)
(iii) (4, 0)
(iv) (√2, 4√2)
(v) (1, 1)
Solution:
(i) (0,2) means x = 0 and y = 2
Puffing x = 0 and y = 2 in x – 2y = 4, we get
L.H.S. = 0 – 2(2) = -4.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ x =0, y =2 is not a solution.

(ii) (2, 0) means x = 2 and y = 0
Putting x = 2 and y = 0 in x – 2y = 4, we get
L.H:S. 2 – 2(0) = 2 – 0 = 2.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (2,0) is not a solution.

(iii) (4, 0) means x = 4 and y = 0
Putting x = 4 and y = o in x – 2y = 4, we get
L.H.S. = 4 – 2(0) = 4 – 0 = 4 =R.H.S.
∴ L.H.S. = R.H.S.
∴ (4, 0) is a solution.

(iv) (√2, 4√2) means x = √2 and y = 4√2
Putting x = √2 and y = 4√2 in x – 2y = 4, we get
L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (√2 , 4√2) is not a solution.

(v) (1, 1)means x =1 and y = 1
Putting x = 1 and y = 1 in x – 2y = 4, we get
LH.S. = 1 – 2(1) = 1 – 2 = -1. But R.H.S = 4
∴ LH.S. ≠ R.H.S.
∴ (1, 1) is not a solution.

Question 4
Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k.
Solution:
We have 2x + 3y = k
putting x = 2 and y = 1 in 2x+3y = k,we get
2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k
Thus, the required value of k is 7.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 1
Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y
Solution:
(i) x + y = 4 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y =4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1
Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions
Solution:
Option (iii) is true because for every value of x, we get a corresponding value of y and vice-versa in the given equation.
Hence, given linear equation has an infinitely many solutions.

Question 2
Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i) 2x + y = 7
When x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)
When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)
When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9
When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
When x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
When x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))

(iii) x = 4y
When x = 0, 4y = 1 ⇒ y = 0
∴ Solution is (0, 0)
When x = 1, 4y = 1 ⇒ y = 14
∴ Solution is (1,14 )
When x = 4, 4y = 4 ⇒ y = 1
∴ Solution is (4, 1)
When x = 4, 4y = 4 ⇒ y = -1
∴ Solution is (-4, -1)

Question 3
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0,2)
(ii) (2,0)
(iii) (4, 0)
(iv) (√2, 4√2)
(v) (1, 1)
Solution:
(i) (0,2) means x = 0 and y = 2
Puffing x = 0 and y = 2 in x – 2y = 4, we get
L.H.S. = 0 – 2(2) = -4.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ x =0, y =2 is not a solution.

(ii) (2, 0) means x = 2 and y = 0
Putting x = 2 and y = 0 in x – 2y = 4, we get
L.H:S. 2 – 2(0) = 2 – 0 = 2.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (2,0) is not a solution.

(iii) (4, 0) means x = 4 and y = 0
Putting x = 4 and y = o in x – 2y = 4, we get
L.H.S. = 4 – 2(0) = 4 – 0 = 4 =R.H.S.
∴ L.H.S. = R.H.S.
∴ (4, 0) is a solution.

(iv) (√2, 4√2) means x = √2 and y = 4√2
Putting x = √2 and y = 4√2 in x – 2y = 4, we get
L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (√2 , 4√2) is not a solution.

(v) (1, 1)means x =1 and y = 1
Putting x = 1 and y = 1 in x – 2y = 4, we get
LH.S. = 1 – 2(1) = 1 – 2 = -1. But R.H.S = 4
∴ LH.S. ≠ R.H.S.
∴ (1, 1) is not a solution.

Question 4
Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k.
Solution:
We have 2x + 3y = k
putting x = 2 and y = 1 in 2x+3y = k,we get
2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k
Thus, the required value of k is 7.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 1
Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y
Solution:
(i) x + y = 4 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y =4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:

Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.

Thus, the line AB is the required graph of x + y = 4 (ii) x – y = 2 ⇒ y = x – 2
If we have x = 0, then y = 0 – 2 = -2
x = 1, then y = 1 – 2 = -1
x = 2, then y = 2 – 2 = 0
∴ We get the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.

Thus, the ime is the required graph of x – y = 2 (iii) y = 3x
If we have x = 0,
then y = 3(0) ⇒ y = 0
x = 1, then y = 3(1) = 3
x= -1, then y = 3(-1) = -3
∴ We get the following table:

Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM as shown.

Thus, the line LM is the required graph of y = 3x. (iv) 3 = 2x + y ⇒ y = 3 – 2x
If we have x = 0, then y = 3 – 2(0) = 3
x = 1,then y = 3 – 2(1) = 3 – 2 = 1
x = 2,then y = 3 – 2(2) = 3 – 4 = -1
∴ We get the following table:

Plot the ordered pairs (0, 3), (1, 1) and (2, – 1) on the graph paper. Joining these points, we get a straight line CD as shown.

Thus, the line CD is the required graph of 3 = 2x + y.

Question 2
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution:
(2, 14) means x = 2 and y = 14
Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0
There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.

Question 3
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
The equation of the given line is 3y = ax + 7
∵ (3, 4) lies on the given line.
∴ It must satisfy the equation 3y = ax + 7
We have, (3, 4) ⇒ x = 3 and y = 4.
Putting these values in given equation, we get
3 x 4 = a x 3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7 = 5 ⇒ a = 53
Thus, the required value of a is 53 Question 4
The taxi fare In a city Is as follows: For the first kilometre, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs.y, write a linear equation for this Information, and draw Its graph.
Solution:
Here, total distance covered = x km and total taxi fare = Rs. y
Fare for 1km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1)km = Rs.5 x(x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to question,
y = 8 + 5(x – 1) = y = 8 + 5x – 5
⇒ y = 5x + 3,
which is the required linear equation representing the given information.
Graph: We have y = 5x + 3
Wben x = 0, then y = 5(0) + 3 ⇒ y = 3
x = -1, then y = 5(-1) + 3 ⇒ y = -2
x = -2, then y = 5(-2) + 3 ⇒ y = -7
∴ We get the following table: Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation y = 5x + 3.

Question 5
From the choices given below, choose the equation whose graphs are given ¡n Fig. (1) and Fig. (2).
For Fig. (1)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x For Fig. (2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6

Solution:
For Fig. (1), the correct linear equation is x + y = 0
[As (-1, 1) = -1 + 1 = 0 and (1,-1) = 1 + (-1) = 0]
For Fig.(2), the correct linear equation is y = -x + 2
[As(-1,3) 3 = -1(-1) + 2 = 3 = 3 and (0,2)
⇒ 2 = -(0) + 2 ⇒ 2 = 2] Question 6
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units
(ii) 0 unit
Solution:
Constant force is 5 units.
Let the distance travelled = x units and work done = y units.
Work done = Force x Distance
⇒ y = 5 x x ⇒ y = 5x
For drawing the graph, we have y = 5x
When x = 0, then y = 5(0) = 0
x = 1, then y = 5(1) = 5
x = -1, then y = 5(-1) = -5
∴ We get the following table:

Ploffing the ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we get a straight line AB as shown.

From the graph, we get
(i) Distance travelled =2 units i.e., x = 2
∴ If x = 2, then y = 5(2) = 10
⇒ Work done = 10 units.

(ii) Distance travelled = 0 unit i.e., x = 0
∴ If x = 0 ⇒ y = 5(0) – 0
⇒ Work done = 0 unit. Question 7
Yamini and Fatima, two students of Class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs.xand Rs.y.) Draw the graph of the same.
Solution:
Let the contribution of Yamini = Rs. x
and the contribution of Fatima Rs. y
∴ We have x + y = 100 ⇒ y = 100 – x
Now, when x = 0, y = 100 – 0 = 100
x = 50, y = 100 – 50 = 50
x = 100, y = 100 – 100 = 0
∴ We get the following table:

Plotting the ordered pairs (0,100), (50,50) and (100, 0) on a graph paper using proper scale and joining these points, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation x + y = 100. Question 8
In countries like USA and Canada, temperature is measured In Fahrenheit, whereas in countries like India, it is measured in Celsius. Here Is a
linear equation that converts Fahrenheit to Celsius:
F = (95 )C + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature Is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what Is the temperature In Fahrenheit and If the temperature is 0°F, what Is the temperature In Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find It.
Solution:
(i) We have
F = (95 )C + 32
When C = 0 , F = (95 ) x 0 + 32 = 32
When C = 15, F = (95 )(-15) + 32= -27 + 32 = 5
When C = -10, F = 95 (-10)+32 = -18 + 32 = 14
We have the following table:

Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.

(ii) From the graph, we have 86°F corresponds to 30°C.
(iii) From the graph, we have 95°F corresponds 35°C.
(iv) We have, C = 0
From (1), we get
F = (95)0 + 32 = 32
Also, F = 0
From (1), we get
0 = (95)C + 32 ⇒ −32×59 = C ⇒ C = -17.8
(V) When F = C (numerically)
From (1), we get
F = 95F + 32 ⇒ F – 95F = 32
⇒ −45F = 32 ⇒ F = -40
∴ Temperature is – 40° both in F and C.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4

Question 1
Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Solution:
(i) y = 3
∵ y = 3 is an equation in one variable, i.e., y only.
∴ y = 3 is a unique solution on the number line as shown below:

(ii) y = 3
We can write y = 3 in two variables as 0.x + y = 3
Now, when x = 1, y = 3
x = 2, y = 3
x = -1, y = 3
∴ We get the following table:

Plotting the ordered pairs (1, 3), (2, 3) and (-1, 3) on a graph paper and joining them, we get aline AB as solution of 0. x + y = 3,
i.e. y = 3.

Question 2
Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables
Solution:
(i) 2x + 9 = 0
We have, 2x + 9 = 0 ⇒ 2x = – 9 ⇒ x = −92
which is a linear equation in one variable i.e., x only.
Theref ore, x = −92 is a unique solution on the number line as shown below:

(ii) 2x +9=0
We can write 2x + 9 = 0 in two variables as 2x + 0, y + 9 = 0
or x=−9−0.y2
∴ When y = 1, x = x=−9−0(1)2 = −92 Thus, we get the following table:

Now, plotting the ordered pairs (−92,3) ,(−92,3) and (−92,3) on a graph paper and joining them, we get a line PQ as solution of 2x + 9 = 0. 