Category: Class 9

Std 9 Science Chapter 10 Information Communication Technology (ICT) Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) Notes, Textbook Exercise Important Questions and Answers. BALRAMCHIN Pivot Point Calculator

Class 9 Science Chapter 10 Information Communication Technology (ICT) Question Answer Maharashtra Board

Class 9 Science Chapter 10 Information Communication Technology (ICT) Textbook Questions and Answers

1. Fill in the blanks to complete the statements. Justify the statements.

a. While working with a computer we can read the information stored in its memory and perform other actions in ………………………. memory.
b. While presenting pictures and videos about the works of scientists, we can use ……………………… .
c. To draw graphs based on the quantitative information obtained in an experiment, one uses ……………………… .
d. The first generation computers used to shut down because of ……………………… .
e. A computer will not work unless ………………………. is supplied to it.

2. Answer the following questions.

a. Explain the role and importance of information communication in science and technology.
Answer:

• ICT plays a key role in creating, displaying, collecting, processing and communicating information in the field of science and technology.
• Following is the importance of ICT in science and technology:
  (a) Access to wide range of information
  (b) Storing of Data
  (c) Processing of Data
  (d) Securing work files
  (e) Proper representation of data

b. Which application software in the computer system did you find useful while studying science, and how?
Answer:

• Microsoft word: To write down the information collected and making a document for further evaluation.
• Microsoft excel : To draw graph based on the obtained numerical information from the experiment.
• Internet explorer: To search for information in finding out the solution or solving the queries by reading the available information.

c. How does a computer work?
Answer:

d. What precautions should be taken while using various types of software on the computer?
Answer:

• Antivirus must be installed.
• Software should be legal and from a trusted place.
• Application should be scanned before using.
• Pirated Software should not be used
• Provide all necessary data to obtain the best possible results.

e. Which are the various devices used in information communication? How are they used in the context of science?
Answer:

• Various devices used in information communication are: Computers, Laptops, Mobiles, Radios, Television, etc.
• Computers, Laptops and Mobiles: Help in accessing, collecting, processing, communicating, sharing and storing of information. It helps in determining the appropriate conclusions in all fields, including the field of science.
• Television: Help in getting information about the new and innovative technology.

3. Using a spreadsheet, draw graphs between distance and time, using the information about the movements of Amar, Akbar and Anthony given in the table on page 4, in the lesson on Laws of Motion. What precautions will you take while drawing the graph?

Answer:
Precautions to be taken while drawing a graph:

• The data should be kept in tabular form.
• Whenever there is ‘drag and fill’ option used, ‘smart tag’ option should be used after ‘drag data’ to fill data as required.
• Entered data should be formatted in the manner required.
• Various types of graphs can be created by using the same data, so appropriate graph should be selected.
• Chart titles and axes titles should be updated as per the data.

4. Explain the differences between the different generations of computers. How did science contribute to these developments?
Answer:
Generation: 1st
Time Period: 1946 – 1956
Development: Vacuum Tubes Characteristics:

• Huge in size
• Expensive
• Lot of electricity consumption
• Heat generation

Generation: 2nd
Time Period: 1956 – 1963
Development: Transistors
Characteristics:

• Frequent shutdowns
• Superior to 1st Generation
• Small in size and fast
• Cheaper as compared to 1st Generation
• Less consumption of electricity

Generation: 3rd
Time Period: 1963 -1971
Development: IC
Characteristics:

• Keyboards and monitors
• OS
• Smaller and still cheaper

Generation: 4th
Time Period: 1971 – 2010
Development: Microprocessor Characteristics:

• Use of Internet
• GUI
• Introduction of portable devices like mobiles, laptops, etc.

Generation: 5th
Time Period: 2010 – Till Date
Development: Artificial Intelligence (AI) Characteristics:

• Voice recognition
• Sensors
• Nano technology

1st Generation computers occupied the entire room, but due to advancement in science and technology, today’s computer fits into our pockets.

Initially computers needed a specific language to interact but today we use voice recognition for the same.

In these ways, science has contributed in making the computers faster, smaller, cheaper and much more useful.

5. What devices will you use to share with others the knowledge that you have?
Answer:
Devices like radios, televisions, pendrives, computers, laptops, mobiles, landlines, hard drives, CDs, memory cards help us in sharing our knowledge with others.

6. Using information communication technology, prepare powerpoint presentations on at least three topics in your textbook. Make a flowchart of the steps you used while making these presentations.
Answer:
Steps for preparation of PowerPoint presentations:

7. Which technical difficulties did you face while using the computer? What did you do to overcome them?
Answer:

• Lagging: Due to lot of applications running at the same time, the computer starts lagging and becomes slow. Closing a few applications helped solve the problem of lagging.
• Viruses and Bugs: Cybercrimes are rising daily, even from single mail the computer can be attacked by viruses. Installing a valid antivirus helps solve the problem of viruses and bugs.
• Breach of Privacy: Confidential information being accessed by anyone is the breach of privacy. Putting privacy setting in place helps solve the problem.
• Physical Damage: Hardware over a period of time might get physically damaged. Taking precautions while using will help to solve the problem.

Class 9 Science Chapter 10 Information Communication Technology (ICT) Intext Questions and Answers

Question 1.
Make a list of various hardware and software items of a computer,
Answer:
Hardware: Mouse, Keyboard, Pendrive, Monitor and other parts of computer.
Software: Operating Systems, Application Programs, Antivirus, etc.

Answer the following questions:

Question 2.
Which devices do we directly or indirectly use for collecting, sharing, processing and communicating information?
Answer:

• Computers
• Laptops
• Mobiles
• Memory Cards
• Pendrives
• Landlines
• Hard disks etc.

Question 3.
How is information communication technology important for dealing with explosion of information?
Answer:

• Information explosion means a situation whère information is available in abundance, in other words, too much information.
• Devices like computers, laptops help us in easier accessment of information that we need from all the data.

Class 9 Science Chapter 10 Information Communication Technology (ICT) Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
…………………….. includes communication devices and the use of those devices as well as the services provided with their help.
(a) Operating System
(b) Office
(c) Computers
(d) Information Communication Technology
Answer:
(d) Information Communication Technology

Question 2.
Computers have gone through …………………….. generations.
(a) 5
(b) 7
(c) 10
(d) 8
Answer:
(a) 5

Question 3.
First generation of Computers were considered to be present in the period of ……………………. .
(a) 2000 – 2001
(b) 1901 – 2001
(c) 1946 -1959,
(d) 1996 – 2001
Answer:
(c) 1946-1959

Question 4.
Full form of RAM is ……………………. .
(a) Roaming Application Memory
(b) Random Accessible Media
(c) Random Access Memory
(d) None of the above
Answer:
(c) Random Access Memory

Question 5.
Full form of ROM is ……………………. .
(a) Roaming Only Memory
(b) Random Output Media
(c) Read Only Memory
(d) None of the above
Answer:
(c) Read Only Memory

Question 6.
…………………….. is raw information.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(c) Data

Question 7.
…………………….. are used for sharing information.
(a) Telephones
(b) Hard disks
(c) RAM
(d) ROM
Answer:
(a) Telephones

Question 8.
Hard disks are used for …………………….. information.
(a) storing
(b) communicating
(c) sharing
(d) all of the above
Answer:
(a) storing

Question 9.
Computers are used for …………………….. information.
(a) storing
(b) managing
(c) sharing
(d) all of the above
Answer:
(d) All of the above

Question 10.
RAM and ROM are 2 types of …………………….. memory.
(a) external
(b) internal
(c) physical
(d) garbage
Answer:
(b) internal

Question 11.
The information stored in ROM is only …………………….., changes cannot be made.
(a) external memory
(b) readable
(c) accessible
(d) physical
Answer:
(b) readable

Question 12.
…………………….. is a group of commands to be given to the computer.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(a) Program

Question 13.
…………………….. communicates between the computer and the person working on it.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(d) Operating System

Match the columns:

Answer:
(1 – b),
(2 – d),
(3 – e),
(4 – a),
(5 – c)

State whether the following statements are true or false and if false, correct the false statements:

(1) RAM and ROM are the types of external memory.
(2) ICT includes communication devices and the use of those devices as well as services provided with their help.
(3) A computer cannot be used without its operating system.
(4) Microsoft Excel is used to make PowerPoint.
(5) Software refers to the set of commands given to the computer.
Answer:
(1) False. RAM and ROM are the types of internal memory.
(2) True
(3) True
(4) False. Microsoft Excel is used to make spreadsheets.
(5) True

Answer the following in one sentence:

Question 1.
Name the computer which was made between 1946 -1959.
Answer:
The ENIAC computer was made in the period of 1946-1959.

Question 2.
Give one example of Input Unit.
Answer:
Keyboard.

Question 3.
Name the 3 major parts of the processing units.
Answer:

• Memory unit
• Control unit
• ALU unit

Question 4.
What precautions need to the taken care while entering formula into the excel?
Answer:
While using a formula, the ‘=’ sign should be typed first. Similarly, no space should be inserted while typing any formula.

Question 5.
What is Internet Explorer?
Answer:
This is a kind of Search Engine. It helps to find the information we want from all the information available on the internet.

Question 6.
What is a PDF?
Answer:
A PDF or Portable Document Format file can be used to view the file to print it or to handle files.

Question 7.
What is C-DAC?
Answer:
C-D AC, is a well-known Centre for Development of Advanced Computing, situated in Pune.

Write the Full forms of the following abbreviation:

Question 1.
ICT
Answer:
Information Communication Technology

Question 2.
OS
Answer:
Operating System

Question 3.
RAM
Answer:
Random Access Memory

Question 4.
ROM
Answer:
Read Only Memory

Question 5.
CPU
Answer:
Central Processing Unit

Question 6.
DOS
Answer:
Disk Operating System

Question 7.
PDF
Answer:
Portable Document Format

Question 8.
ALU
Answer:
Arithmetic Logical Unit

Question 9.
GUI
Answer:
Graphical User Interface

Question 10.
C-DAC
Answer:
Centre for Development of Advanced Computing

Question 11.
ISCII
Answer:
Indian Script Code for Information Interchange

Define the following:

Question 1.
Memory
Answer:
Memory is the place for storing data obtained from the input and also the generated solution or answer by the computer.

Question 2.
RAM
Answer:
RAM is created from electronic components and can function only as long as it is supplied with electricity.

Question 3.
ROM
Answer:
Information stored in ROM can only be read and changes cannot be made to the information originally stored here.

Question 4.
Operating System
Answer:
It is a program which provides a means of communication between the computer and the person working on it. It is called the DOS (Disk Operating System).

Question 5.
Program
Answer:
A program is a group of commands to be given to a computer.

Question 6.
Data and Information
Answer:
Data is information in its raw (unprocessed) form.

Question 7.
Hardware
Answer:
Hardware consists of all the electronic and mechanical parts used in computers.

Question 8.
Software:
Answer:
Software refers to the commands given to the computer, information supplied to it (input) and the results obtained from the computer after analysis (output).

Give scientific reasons:

Question 1.
Computer cannot function without its operating system.
Answer:

• Operating system is like a link between the computer and the person working on it.
• Operating system manages all the activities performed by the computer.
• Without the operating system, the user won’t be able to input any data or run any program. Thus, a computer cannot run without an operating system.

Question 2.
ROM is a Read Only Memory.
Answer:

• ROM also known as Read Only Memory is a part of internal memory of a computer where the information stored can only be read.
• ROM helps store data permanently for a long period of time and the information stored cannot be deleted.
• Thus, data in a ROM can only be read and cannot be altered or modified and hence, it is called as Read Only Memory.

Complete the table:
Answer:

Answer the following questions:

Question 1.
What precautions will you take when entering data?
Answer:

1. As far as possible, the data should be kept in tabular form. Different types of data should be entered in different cells. Data should be entered neatly and in one ‘flow’. Unnecessary space and special characters should not be used.
2. Many times we ‘drag and fill’ data. At such times, the ‘smart tag’ can be used after ‘drag data’ to fill any data in any manner as required.
3. Once the data has been entered, it can be formatted in different ways. Similarly, we can perform different types of calculations, using different formulae.
4. While using a formula, the ‘=’ sign should be typed first. Similarly, no space should be inserted while typing any formula.

Answer in detail:

Question 1.
Write in short about the opportunities in the field of ICT.
Answer:
(i) Software Field: This is an important field. Having accepted the challenge of creating software, many companies have entered this field. The opportunities in the software field can be classified as follows – application program development, software package development, operating systems and utility development, special purpose scientific applications.

(ii) Hardware Field: Today, there are several companies in our country too, which make computers. They sell computers that they have themselves made. Others sell computers brought from outside as well as repair them and take maintenance contracts to keep computers in big companies working efficiently without a break. Plenty of jobs are available here. There are job opportunities in hardware designing, hardware production, hardware assembly and testing, hardware maintenance, servicing and repairs, etc.

(iii) Marketing: There are many establishments which make and sell computers and related accessories. They need good sales personnel who are experienced in the working of computers as well as skilled in marketing.

(iv) Training: The training of new entrants for various jobs is a vast field. It is very important to have dedicated teachers who are competent in the field of computers.

Question 2.
Write in short about the industries conducting research in the field of computers.
Answer:

• C-DAC, the well-known Centre for Development of Advanced Computing, situated in Pune, is the leading institute in India that conducts research in the field of computers.
• The first Indian supercomputer was made with help from this institute. Valuable guidance for making this computer (the Param computer) was received from the senior scientist Vijay Bhatkar. Param means the supreme.
• This computer can perform one billion calculations per second. It is used in many fields like space research, movements in the interior of the earth, research in oil deposits, medicine, meteorology, engineering, military etc.
• C-DAC has also participated in developing the ISCII code for writing different language scripts. (Indian Script Code for Information Interchange).

Question 3.
Use Microsoft Word to create a document and write equations.
Answer:

• Click on the Microsoft word 2010 icon on the desktop.
• Select the ‘New option in the ‘File’ tab, and then select the ‘Blank document’ option.
• Type your material on the blank page on the screen using the keyboard. Use the language, font size, bold, etc. options in the Home tab to make the typed material attractive.
• To type equations in the text, select the ‘Equation’ option in the ‘Insert’ tab.
  
• Select the proper equation and type it using mathematical symbols.
  

Balbharati Maharashtra State Board 9th Std Science Textbook Solutions 

• Information Communication Technology (ICT) Class 9 Science Textbook Solutions
• Reflection of Light Class 9 Science Textbook Solutions
• Study of Sound Class 9 Science Textbook Solutions
• Carbon: An Important Element Class 9 Science Textbook Solutions
• Substances in Common Use Class 9 Science Textbook Solutions
• Life Processes in Living Organisms Class 9 Science Textbook Solutions
• Heredity and Variation Class 9 Science Textbook Solutions
• Introduction to Biotechnology Class 9 Science Textbook Solutions
• Observing Space: Telescopes Class 9 Science Textbook Solutions

9th Standard Maths 2 Practice Set 9.1 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9.1 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
Length, breadth and height of a cuboid shape box of medicine is 20 cm, 12 cm and 10 cm respectively. Find the surface area of vertical faces and total surface area of this box.
Given: For cuboid shape box of medicine,
length (l) = 20 cm, breadth (b) = 12 cm and height (h) = 10 cm.
To find: Surface area of vertical faces and total surface area of the box
Solution:
i. Surface area of vertical faces of the box
= 2(l + b) x h
= 2(20+ 12) x 10
= 2 x 32 x 10
= 640 sq.cm.

ii. Total surface area of the box
= 2 (lb + bh + lh)
= 2(20 x 12+ 12 x 10 + 20 x 10)
= 2(240 + 120 + 200)
= 2 x 560
= 1120 sq.cm.
∴ The surface area of vertical faces and total surface area of the box are 640 sq.cm, and 1120 sq.cm, respectively.

Question 2.
Total surface area of a box of cuboid shape is 500 sq.unit. Its breadth and height is 6 unit and 5 unit respectively. What is the length of that box?
Given: For cuboid shape box,
breadth (b) = 6 unit, height (h) = 5 unit Total surface area = 500 sq. unit.
To find: Length of the box (l)
Solution:
Total surface area of the box = 2 (lb + bh + lh)
∴ 500 = 2 (6l + 6 x 5 + 5l)
∴ \(\frac { 500 }{ 2 }\) = (11l + 30)
∴ 250= 11l + 30
∴ 250 – 30= 11l
∴ 220 = 11l
∴ 220 = l
∴ \(\frac { 220 }{ 11 }\) = l
∴ l = 20 units
∴ The length of the box is 20 units.

Question 3.
Side of a cube is 4.5 cm. Find the surface area of all vertical faces and total surface area of the cube.
Given: Side of cube (l) = 4.5 cm
To find: Surface area of all vertical faces and the total surface area of the cube
Solution:
i. Area of vertical faces of cube = 4l²
= 4 (4.5)² = 4 x 20.25 = 81 sq.cm.
ii. Total surface area of the cube = 6l²
= 6 (4.5)²
= 6 x 20.25
= 121.5 sq.cm.
∴ The surface area of all vertical faces and the total surface area of the cube are 81 sq.cm, and 121.5 sq.cm, respectively.

Question 4.
Total surface area of a cube is 5400 sq. cm. Find the surface area of all vertical faces of the cube.
Given: Total surface area of cube = 5400 sq.cm.
To find: Surface area of all vertical faces of the cube
Solution:
i. Total surface area of cube = 6l²
∴ 5400 = 6l²
∴ \(\frac { 5400 }{ 6 }\) = l²
∴ l² = 900
ii. Area of vertical faces of cube = 4l²
= 4 x 900 = 3600 sq.cm.
∴ The surface area of all vertical faces of the cube is 3600 sq.cm.

Question 5.
Volume of a cuboid is 34.50 cubic metre. Breadth and height of the cuboid is 1.5 m and 1.15 m respectively. Find its length.
Given: Breadth (b) = 1.5 m, height (h) = 1.15 m
Volume of cuboid = 34.50 cubic metre
To find: Length of the cuboid (l)
Solution:
Volume of cuboid = l x b x h
∴ 34.50 = l x b x h
∴ 34.50 = l x 1.5 x 1.15

= 20
∴ The length of the cuboid is 20 m.

Question 6.
What will be the volume of a cube having length of edge 7.5 cm ?
Given: Length of edge of cube (l) = 7.5 cm
To find: Volume of a cube
Solution:
Volume of a cube = l²
= (7.5)³
= 421.875 ≈ 421.88 cubic cm
∴The volume of the cube is 421.88 cubic cm.

Question 7.
Radius of base of a cylinder is 20 cm and its height is 13 cm, find its curved surface area and total surface area, (π = 3.14)
Given: Radius (r) = 20 cm, height (h) = 13 cm
To find: Curved surface area and
the total surface area of the cylinder
Solution:
i. Curved surface area of cylinder = 2πrh
= 2 x 3.14 x 20 x 13
= 1632.8 sq.cm

ii. Total surface area of cylinder = 2πr(r + h)
= 2 x 3.14 x 20(20 + 13)
= 2 x 3.14 x 20 x 33 = 4144.8 sq.cm
∴ The curved surface area and the total surface area of the cylinder are 1632.8 sq.cm and 4144.8 sq.cm respectively.

Question 8.
Curved surface area of a cylinder is 1980 cm² and radius of its base is 15 cm. Find the height of the cylinder. (π = \(\frac { 22 }{ 7 }\))
Given: Curved surface area of cylinder = 1980 sq.cm., radius (r) = 15 cm
To find: Height of the cylinder (h)
Solution:
Curved surface area of cylinder = 2πrh
∴ 1980 = 2 x \(\frac { 22 }{ 7 }\) x 15 x h
∴ \(h=\frac{1980 \times 7}{2 \times 22 \times 15}\)
∴ h = 21 cm
∴ The height of the cylinder is 21 cm.

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 9.2 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9.2 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.
Given: Height (h) = 12 cm, length (l) = 13 cm
To find: Radius of the base of the cone (r)
Solution:
l² = r² + h²
∴ 13² = r² + 12²
∴ 169 = r² + 144
∴169 – 144 = r²
∴ r² = 25
∴ r = √25 … [Taking square root on both sides]
= 5 cm
∴ The radius of base of the cone is 5 cm.

Question 2.
Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm. ( π = \(\frac { 22 }{ 7 }\))
Given: Radius (r) = 28 cm,
Total surface area of cone = 7128 sq.cm
To find: Volume of the cone
Solution:
i. Total surface area of cone = πr (l + r)
∴ 7128= y x 28 x (l + 28)
∴ 7128 = 22 x 4 x(l +28)
∴ l + 28 = \(\frac { 7128 }{ 22\times 4 }\)
∴ l + 28 = 81
∴ l = 81 – 28
∴ l = 53cm

ii. Now, l² = r² + h²
∴ 53² = 28²+ h²
∴ 2809 = 784 + h²
∴ 2809 – 784 = h²
∴ h² = 2025
∴ h = \(\sqrt { 2025 }\) …… [Taking square root on both sides]
= 45 cm

= 22 x 4 x 28 x 15
= 36960 cubic.cm
∴ The volume of the cone is 36960 cubic.cm.

Question 3.
Curved surface area of a cone is 251.2 cm² and radius of its base is 8 cm. Find its slant height and perpendicular height, (π = 3.14)
Given: Radius (r) = 8 cm, curved surface area
of cone = 251.2 cm²
To find: Slant height (l) and the perpendicular height (h) of the cone
Solution:
i. Curved surface area of cone = πrl
∴ 251.2 = 3.14 x 8 x l

∴ l= 10 cm

ii. Now, l² = r² + h²
∴ 10² = 8² + h²
∴ 100 = 64 + h²
∴ 100 – 64 = h²
∴ h² = 36
∴ h = √36 … [Taking square root on both sides]
= 6 cm
∴ The slant height and the perpendicular height of the cone are 10 cm and 6 cm respectively.

Question 4.
What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is ₹ 10 per sq.m?
Given: Radius (r) = 6 m, length (l) = 8 m
To find: Total cost of making the cone
Solution:
i. To find the total cost of making the cone of tin sheet, first we need to find the total surface area of the cone.
Total surface area of the cone = πr (l + r)
= \(\frac { 22 }{ 7 }\) x 6 x (8 + 6)
= \(\frac { 22 }{ 7 }\) x 6 x 14
= 22 x 6 x 2 = 264 sq.m

ii. Rate of making the cone = ₹ 10 per sq.m
∴ Total cost = Total surface area x Rate of making the cone
= 264 x 10
= ₹ 2640
∴ A The total cost of making the cone of tin sheet is ₹ 2640.

Question 5.
Volume of a cone is 6280 cubic cm and base radius of the cone is 20 cm. Find its perpendicular height, (π = 3.14)
Given: Radius (r) = 20 cm,
Volume of cone = 6280 cubic cm
To find: Perpendicular height (h) of the cone
Solution:

∴ The perpendicular height of the cone is 15 cm.

Question 6.
Surface area of a cone is 188.4 sq.cm and its slant height is 10 cm. Find its perpendicular height (π = 3.14).
Given: Length (l) =10 cm, curved surface area of the cone = 188.4 sq.cm
To find: Perpendicular height (h) of the cone
Solution:
i. Curved surface area of the cone = πrl
∴ 188.4 = 3.14 x r x 10

ii. Now, l² = r² + h²
∴ 10² = 6² + h²
∴ 100 = 36 + h²
∴ 100 – 36 = h²
∴ h² = 64
∴ h = \(\sqrt { 64 }\) … [Taking square root on both sides]
= 8 cm
∴ The perpendicular height of the cone is 8 cm.

Question 7.
Volume of a cone is 1232 cm³ and its height is 24 cm. Find the surface area of the cone. (π = \(\frac { 22 }{ 7 }\))
Given: Height (h) = 24 cm,
Volume of cone = 1232 cm³
To find: Surface area of the cone
Solution:

∴ r² = 49
∴ r = \(\sqrt { 49 }\) … [Taking square root on both sides]
= 7 cm

ii. Now, l² = r² + h²
∴ l² = 7² + 24²
= 49 + 576 = 625
∴ l = \(\sqrt { 625 }\) … [Taking square root on both sides]
= 25

iii. Curved surface area of cone = πrl
= \(\frac { 22 }{ 7 }\) x 7 x 25
= 22 x 25
= 550 sq.cm
∴The surface area of the cone is 550 sq.cm.

Question 8.
The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the total surface area of cone. (π = \(\frac { 22 }{ 7 }\))
Given: Length (l) = 50 cm, curved surface area of cone = 2200 sq.cm
To find: Total surface area of the cone
Solution:
i. Curved surface area of cone = πrl

ii. Total surface area of cone = πr (l + r)
= \(\frac { 22 }{ 7 }\) x 14 x (50 + 14)
= \(\frac { 22 }{ 7 }\) x 14 x 64
= 22 x 2 x 64
= 2816 sq.cm
∴ The total surface area of the cone is 2816 sq.cm.

Question 9.
There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq.m, of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.
Given: For the tent,
height (h) = 18m,
number of people in the tent = 25,
area required for each person = 4 sq.m
To find: Volume of the tent
Solution:
i. Every person needs an area of 4 sq.m, of the ground inside the tent.
Surface area of the base of the tent = number of people in the tent × area required for each person
= 25 × 4
= 100 sq.m

ii. Surface area of the base of the tent = πr²
∴ 100 = πr²
∴ πr² = 100

iii. Volume of the tent= \(\frac { 1 }{ 3 }\) πr²h
= \(\frac { 1 }{ 3 }\) x 100 x 18 …….[∵ πr² = 100]
= 100 x 6
= 600 cubic metre
∴ The volume of the tent is 600 cubic metre.

Question 10.
In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m and diameter of base is 7.2 m. Find the volume of the heap of the fodder. If it is to be covered by polythene in rainy se&son then how much minimum polythene
sheet is needed? (π = \(\frac { 22 }{ 7 }\) and \(\sqrt { 17.37 }\) = 4.17 ]
Given: Height of the heap (h) = 2.1 m.
diameter of the base (d) = 7.2 m
∴Radius of the base (r) = \(\frac { d }{ 2 }\) = \(\frac { 7.2 }{ 2 }\) = 3.6 m
To find: Volume of the heap of the fodder and polythene sheet required
Solution:
i. Volume of the heap of fodder = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x (3.6)² x 2.1
= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 3.6 x 3.6 x 2.1
= 1 x 22 x 1.2 x 3.6 x 0.3
= 28.51 cubic metre

ii. Now, l² = r² + h²
= (3.6)² + (2.1)²
= 12.96 + 4.41
∴ l² =17.37
∴ l² = \(\sqrt { 17.37 }\) .. .[Taking square root on both sides]
= 4.17 m

iii. Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap
= πrl
= \(\frac { 22 }{ 7 }\) x 3.6 x 4.17
= 47.18 sq.m
∴ The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it.

Maharashtra Board Class 9 Maths Solutions

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 9.3 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9.3 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
Find the surface areas and volumes of spheres of the following radii
i. 4 cm
ii. 9 cm
iii. 3.5 cm (π = 3.14)
i. Given: Radius (r) = 4 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x 4²
∴ Surface area of sphere = 200.96 sq.cm
Volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x 4²
∴ Volume of sphere = 267.95 cubic cm

ii. Given: Radius (r) = 9 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x 9²
∴ Surface area of sphere = 1017.36 sq.cm

∴ Volume of sphere = 3052.08 cubic cm

iii. Given: Radius (r) = 3.5 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x (3.5)²
∴ Surface area of sphere = 153.86 sq.cm
Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x (3.5)³
∴ Volume of sphere = 179.50 cubic cm

Question 2.
If the radius of a solid hemisphere is 5 cm, then find its curved surface area and total surface area, (π = 3.14)
Given: Radius (r) = 5 cm
To find: Curved surface area and total surface area of hemisphere
Solution:
i. Curved surface area of hemisphere = 2πr²
= 2 x 3.14 x 5²
= 2 x 3.14 x 25
= 50 x 3.14
= 157 sq.cm.

ii. Total surface area of hemisphere = 3πr²
= 3 x 3.14 x 5²
= 235.5 sq.cm.
∴ The curved surface area and totai surface area of hemisphere are 157 sq.cm, and 235.5 sq.cm, respectively.

Question 3.
If the surface area of a sphere is 2826 cm² then find its volume. (π = 3.14)
Given: Surface area of sphere = 2826 sq.cm.
To find: Volume of sphere
Solution:
i. Surface area of sphere = 4πr²
∴ 2826 = 4 x 3.14 x r²
2826 = 282600 = 900
∴ \( r^{2}=\frac{2826}{4 \times 3.14}=\frac{282600}{4 \times 314}=\frac{900}{4}\)
∴ r² = 225
∴ r = \(\sqrt { 225 }\) … [Taking square root on both sides]
= 15 cm

ii. Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x 15³
= \(\frac { 4 }{ 3 }\) x 3.14 x 15 x 15 x 15
= 4 x 3.14 x 5 x 15 x 15
= 14130 cubic cm.
∴ The volume of the sphere is 14130 cubic cm.

Question 4.
Find the surface area of a sphere, if its volume is 38808 cubic cm. (π = \(\frac { 22 }{ 7 }\))
Given: Volume of sphere = 38808 cubic cm.
To find: Surface area of sphere
Solution:

∴ r³ = 441 x 21 = 21 x 21 x 21
∴ r = 21 cm … [Taking cube root on both sides]
ii. Surface area of sphere = 4πr²
= 4 x \(\frac { 22 }{ 7 }\) x 21
= 4 x \(\frac { 22 }{ 7 }\) x 21 x 21
= 4 x 22 x 3 x 21
= 5544 sq.cm.
∴ The surface area of sphere is 5544 sq.cm.

Question 5.
Volume of a hemisphere is 18000π cubic cm. Find its diameter.
Given: Volume of hemisphere = 1 8000π cubic cm.
To find: Diameter of the hemisphere
Solution:
i. Volume of hemisphere = \(\frac { 2 }{ 3 }\) πr³

= 9000 x 3
∴ r³ = 27000
∴ r = \(\sqrt [ 3 ]{ 27000 }\) … [Taking cube root on both sides]
= 30 cm

ii. Diameter = 2r
= 2 x 30 = 60 cm
∴ The diameter of the hemisphere is 60 cm.

Maharashtra Board Class 9 Maths Solutions

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 9 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations? ( π = \(\frac { 22 }{ 7 }\))
Given: For road roller,
diameter (d) = 0.9 m, length (h) = 1.4 m
To find: Area of a field pressed in 500 rotations
Solution:
i. Since, area of field pressed in 1 rotation of road roller = curved surface area of road roller
∴ Curved surface area of the road roller = 2πrh
= πdh ,..[∵ d = 2r]
= \(\frac { 22 }{ 7 }\) x 0.9 x 1.4 7
= 22 x 0.9 x 0.2
= 3.96 sq.m.

ii. Area of land pressed in 1 rotation = 3.96 sq.m.
∴Area of land pressed in 500 rotations = 500 x 3.96
= 1980 sq.m.
∴ 1980 sq.m, land will be pressed in 500 rotations of the road roller.

Question 2.
To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it ?
Given: Thickness of the glass = 2 mm,
outer length of the tank = 60.4 cm,
outer breadth of the tank = 40.4 cm,
outer height of the tank = 40.2 cm
To find: Volume of water fish tank contains
Solution:

i. Thickness oldie glass = 2 mm.
= \(\frac { 2 }{ 10 }\) cm
= 0.2 cm
Outerlengthofthetank = 60.4 cm
∴ Inner length oldie tank (l) = Outer length – thickness oldie glass on both sides
= 60.4 – 0.2 – 0.2
= 60cm
Outer breadth oldie tank = 40.4 cm
∴ Inner breadth of the tank (b) = 40.4 – 0.2 – 0.2
= 40 cm
Outer height of the tank = 40.2 cm
∴Inner height of the tank (h) = 40.2 – 0.2
= 40 cm

ii. Maximum volume of water that can be contained in the tank = volume of the tank
= l x b x h
= 60 x 40 x 40
= 96000 cubic cm.
∴ The fishtank can contain maximum of 96000 cubic cm. water in it.

Question 3.
If the ratio of radius of base and height of a cone is 5 : 12 and its volume is 314 cubic metre. Find its perpendicular height and slant height (π = 3.14).
Given: Ratio of radius of base and height of a cone = 5 : 12,
Volume = 314 cubic metre
To find: Perpendicular height (h) and slant height (l)
Solution:
i. The ratio of radius and height of cone is 5 : 12
Let the common multiple be x.
∴ Radius of base (r) = 5x
Perpendicular height (h) = 12x

∴ x³ = 1
∴ x = 1 … [Taking cube root on both sides]
∴ r = 5x = 5(1) = 5m
h = 12x = 12(1) = 12 m

ii. Now, l² = r² + h²
= 5² + 12²
= 25 + 144
∴l² = 169
∴ l = \(\sqrt { 169 }\) … [Taking square root on both sides]
= 13 m
The perpendicular height and slant height of the cone are 12 m and 13 m respectively.

Question 4.
Find the radius of a sphere if its volume is 904.32 cubic cm. (π = 3.14)
Given: Volume of sphere = 904.32 cubic cm.
To find: Radius of a sphere
Solution:
Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
∴ 904.32 = \(\frac { 4 }{ 3 }\) x 3.14 x r³

= 216
∴ r = \(\sqrt [ 3 ]{ 216 }\) … [Taking cube root on both sides]
= 6 cm
∴ The radius of the sphere is 6 cm.

Question 5.
Total surface area of a cube is 864 sq.cm. Find its volume.
Given: Total surface area of cube = 864 sq. cm
To find: Volume of cube
Solution:
i. Total surface area of cube = 6l²
∴ 864 = 6l²
∴ l²= \(\sqrt [ 864 ]{ 6 }\)
∴ l² = 144
∴ l = \(\sqrt { 144 }\) … [Taking square root on both sides]
= 12 cm

ii. Volume of cube = l²
= 12³
= 1728 cubic cm.
∴ The volume of cube is 1728 cubic cm.

Question 6.
Find the volume of a sphere, if its surface area is 154 sq.cm.
Given: Surface area of sphere = 154 sq. cm.
To find: Volume of sphere
Solution:
i. Surface area of sphere = 4πr²


∴ The volume of sphere is 179.67 cubic cm.

Question 7.
Total surface area of a cone is 616 sq.cm. If the slant ‘height of the cone Is three times the radius of its base, find its slant height.
Given: Total surface area of a cone = 616 sq.cm., slant height of the cone is three times the radius of its base
To find: Slant height (l)
Solution:
i. Let the radius of base be r cm.
∴ Slant height (l) = 3r cm
Total surface area of cone = πr (l + r)
∴ 616 = πr(l + r)
∴ 616 = \(\sqrt [ 22 ]{ 7 }\) x r x (3r + r)
∴ 616 = \(\sqrt [ 22 ]{ 7 }\) x 4r²

∴ r² = 49
∴ r = \(\sqrt { 49 }\) … [Taking square root on both sides]
= 7

ii. Slant height (l) = 3r = 3 x 7 = 21 cm
∴ The slant height of the cone is 21 cm.

Question 8.
The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate ₹ 52 per sq.m.
Given: Inner diameter (d) = 4.2 m,
To find: depth (h) = 10 m,
rate of plastering = ₹ 52 per sq.m.
Inner surface area and total cost of plastering
Solution:
i. Inner curved surface area of the well = 2πrh
= πdh …[∵ d = 2r]
= \(\sqrt [ 22 ]{ 7 }\) x 4.2 x 10
= \(\sqrt [ 22 ]{ 7 }\) x 42
= 22 x 6
= 132 sq.m.

ii. Rate of plastering = ₹52 per sq.m.
∴ Total cost = Curved surface area x Rate of plastering
= 132 x 52 = ₹6864
∴ The cost of plastering the well from inside is ₹6864.

Question 9.
The length of a road roller is 2.1 m and its diameter is 1.4 m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of ₹ 7 per sq.m.
Given: For road roller,
diameter (d) = 1.4 m,
length (h) = 2.1 m
number of rotations required for levelling the ground = 500,
rate of levelling = ₹ 7 per sq. m.
To find: Area of ground leveled by the road roller and cost of levelling
Solution:
i. Since, area of ground levelled in 1 rotation of road roller = curved surface area of road roller
∴Curved surface area of the road roller = 2πrh
= πdh …[∵ d = 2r]
= \(\frac { 22 }{ 7 }\) x 1.4 x 2.1
= 22 x 0.2 x 2.1
= 9.24 sq.m.

ii. Area of ground levelled in 1 rotation = 9.24 sq.m.
∴Area of ground levelled in 500 rotations = 9.24 x 500
= 4620 sq.m.

iii. Rate of levelling ₹ 7 per sq.m.
∴Total cost = Area of ground levelled x Rate of levelling
= 4620 x 7
= ₹32340
∴ The road roller levels 4620 sq.m. land in 500 rotation, and the cost of levelling is ₹32340.

Maharashtra Board Class 9 Maths Chapter 9 Surface Area and Volume Practice Set 9 Intext Questions and Activities

Question 1.
Curved surface area of cone. (Textbook pg. no. 116)

Circumference of base of the cone = 2πr
As shown in the figure (c), make pieces of the net as small as possible. Join them as shown in the figure (d),. By joining the small pieces of net of the cone, we get a rectangle ABCD approximately.
Total length of AB and CD is 2πr.
∴ length of side AB of rectangle ABCD is πr and length of side CD is also πr.
Length of side BC of rectangle = slant height of cone = l.
Curved surface area of cone is equal to the area of the rectangle.
∴ curved surface area of cone = Area of rectangle = AB x BC = πr x l = πrl

Question 2.
Prepare a cylinder of a card sheet, keeping one of its faces open. Prepare an open cone of card sheet which will have the same base-radius and the same height as that of the cylinder. Pour fine sand in the cone till it just fills up the cone. Empty the cone in the cylinder. Repeat the procedure till the cylinder is just filled up with sand. Note how many coneful of sand is required to fill up the cylinder. (Textbook pg, no 117)

Answer:
To fill the cylinder, three coneful of sand is required.

Question 3.
Finding total surface area of sphere. (Textbook pg, no 120)

i. Take a sweet lime (Mosambe), Cut it into two equal parts.

ii. Take one of the parts. Place its circular face on a paper. Draw its circular border. Copy three more such circles. Again, cut each half of the sweet lime into two equal parts.

iii. Now you get 4 quarters of sweet lime. Separate the peel of a quarter part. Cut it into pieces as small as possible. Try to cover one o’f the circles drawn, by the small pieces. Observe that the circle gets nearly covered.
The activity suggests that,
Curved surface area of a sphere = 4πr²

∴ Curved surface area of a sphere = 4 x Area of a circle

Question 4.
Make a cone and a hemisphere of cardsheet such that radii of cone and hemisphere are equal and height of cone is equal to radius of the hemisphere.
Fill the cone with fine sand. Pour the sand in the hemisphere. How many cones are required to fill the hemisphere completely ? (Textbook pg. no. 121)

Answer:
To fill the hemisphere, two coneful of sand is required.

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Problem Set 8 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 8 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Problem Set 8 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for the following multiple choice questions.

i. Which of the following statements is true?
(A) sin θ = cos (90 – θ)
(B) cos θ = tan (90 – θ)
(C) sin θ = tan (90 – θ)
(D) tan θ = tan (90 – θ)
Answer:
(A) sin θ = cos (90 – θ)

ii. Which of the following is the value of sin 90°?
(A) \( \frac { \sqrt { 3 } }{ 2 }\)
(B) 0
(C) \(\frac { 1 }{ 2 }\)
(D) 1
Answer:
(D) 1

iii. 2 tan 45° + cos 45° – sin 45° = ?
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
2 tan 45° + cos 45° – sin
\( =2(1)+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2\)
(C) 2

iv. \( \frac{\cos 28^{\circ}}{\sin 62^{\circ}}\) =?
(A) 2
(B) -1
(C) 0
(D) 1
Answer:
\( \frac{\cos 28^{\circ}}{\sin 62^{\circ}}\)

(D) 1

Question 2.
In right angled ∆TSU, TS = 5, ∠S = 90°, SU = 12, then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

Solution:
i. TS = 5, SU = 12 …[Given]
In ∆TSU, ∠S = 90° … [Given]
∴ TU² = TS² + SU² …[Pythagoras theorem]
= 5² + 12² = 25 + 144 = 169
∴ TU = \(\sqrt { 169 }\) .. .[Taking square root of both sides]
= 13

Question 3.
In right angled ∆YXZ, ∠X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

Solution:
i. XZ = 8 cm, YZ = 17 cm …[Given]
In ∆YXZ, ∠X = 90° … [Given]
∴ YZ² = XY² + XZ² .. .[Pythagoras theorem]
∴ 17² = XY² + 8²
∴ 289 = XY² + 64
∴ XY² = 289 – 64
= 225
∴ x = \(\sqrt { 225 }\) .. .[Taking square root of both sides]
= 15

Question 4.
In right angled ∆LMN, if ∠N = θ, ∠M = 90°, cos θ = \(\frac { 24 }{ 25 }\), find sin θ and tan θ. Similarly, find (sin²θ) and (cos²θ).

Solution:
i. cos θ = \(\frac { 24 }{ 25 }\)
In ∆LMN, ∠M = 90°, ∠N = θ

Let the common multiple be k.
∴ MN = 24k and LN = 25k
Now, LN²= LM² + MN² … [Pythagoras theorem]
∴ (25k)² = LM² + (24k)²
∴ 625 k² = LM² + 576k²
∴ LM² = 625k² – 576k²
∴ LM² = 49k²
∴ LM = \(\sqrt { 49{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 7k

Question 5.
Fill in the blanks.
i. sin 20° = cos
ii. tan 30° x tan  = 1
iii. cos 40° = sin
Solution:
i. sin 20° = cos (90° – 20°) …..[∵ sin θ = cos (90 – θ)]
= cos 70°

ii. tan θ x tan (90 – θ) = 1
Substituting θ = 30°,
tan 30° x tan (90 – 30)° = 1
∴ tan 30° x tan 60° = 1

iii. cos 40° = sin (90° – 40°) …[∵ COS θ = sin (90 – θ)]
= sin 50°

Maharashtra Board Class 9 Maths Solutions Chapter 8 Trigonometry Problem Set 8

Question 1.
Measuring height of a tree using trigonometric ratios. (Textbook pg. no. 101)

This experiment can be conducted on a clear sunny day. Look at the figure given above. Height of the tree is QR, height of the stick is BC.
Thrust a stick in the ground as shown in the figure. Measure its height and length of its shadow. Also measure the length of the shadow of the tree. Using these values, how will you determine the height of the tree?
Solution:
Rays of sunlight are parallel.
So, ∆PQR and ∆ABC are equiangular i.e., similar triangles.
Sides of similar triangles are proportional.
∴ \(\frac { QR }{BC }\) = \(\frac { PR }{ AC }\)
∴ Height of the tree (QR) = \(\frac { BC }{ AC }\) x PR
Substituting the values of PR, BC and AC in the above equation, we can get length of QR i.e., the height of the tree.

Question 2.
It is convenient to do the above experiment between 11:30 am and 1:30 pm instead of doing it in the morning at 8’O clock. Can you tell why? (Textbook pg. no. 101)
Solution:
At 8’O clock in the morning, the sunlight is not very bright. At the same time, the sun is on the horizon and the shadow would by very long. It would be extremely difficult to measure shadow in that case.
Between 11:30 am and 1:30 pm, the sun is overhead and it would be easier to measure the length of shadow.

Question 3.
Conduct the above discussed activity and find the height of a tall tree in your surrounding. If there is no tree in the premises, then find the height of a pole. (Textbook pg. no. 101)

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 8.2 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Practice Set 8.2 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

Solution:
i. cos θ = \(\frac { 35 }{ 37 }\) …(i) )[Given]
In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.
∴ BC = 35k and AC = 37k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (37k)² = AB²+ (35k)²
1369k² = AB² + 1225k²
AB2 = 1369k² – 1225k²
= 144k²
AB = 144k²
AB = \(\sqrt { 2ghK }\)² … [Taking square root of both sides]
= 12k

ii. sin θ = \(\frac { 11 }{ 61 }\) …..(i) [Given]
In right angled ∆ABC, ∠C = θ.


Let the common multiple be k.
AB = 11k and AC = 61k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (61k)² = (11k)² + BC²
∴ 3721k² = 121k² + BC²
∴ BC² = 3721k² – 121k² = 3600k²
BC = \(\sqrt { 3600{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 60k

iii. tan θ = 1 = \(\frac { 1 }{ 1 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 1k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + K²
= 2K²
∴ AC = \(\sqrt { 2{ k }\)

iv. sin θ = \(\frac { 1 }{ 2 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 2k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ 2K² = K² + BC²
∴ 4K² = K² + BC²
∴ BC² = 4K² – K² = 3K²
∴ BC = \(\sqrt { 3{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= \(\sqrt { 3{ k }\)

v. cos θ = \(\frac { 1 }{ \sqrt { 3 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = √3k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (√3K)² = AB² + K²
∴ 3K² = 3K² – K² = 2K²
∴ AB = \(\sqrt { 2{ k }^{ 2 } }\) .. .[Taking square root of both sides]
AB = √2K

vi. cos θ = \(\frac { 21 }{ \sqrt { 20 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 21k and BC = 20k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (21)K² + (20K)²
= 441K² – 400²
= 841K²
∴ AB = \(\sqrt { 841{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 29K

vii. tan θ = \(\frac { 8 }{ 15 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 8k and BC = 15k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (8)K² + (15K)²
= 64K² – 225²
= 289K²
∴ AC = \(\sqrt { 289{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 17K

viii. sin θ = \(\frac { 3 }{ 5 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.



Let the common multiple be k.
∴ AB = 3k and AC = 5k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (5)K²= (3)K² + BC²
∴ 25K² = 9K² – 225²
∴ BC² = 25K² – 9K²
∴ BC = \(\sqrt { 16{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 4K

ix. tan θ = \(\frac { 1 }{ 2\sqrt { 2 } }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and AC = 2√2 k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + (2√2 k )²
= K² – 225²
= 25K² + 8K²
= 9K²
∴ AC = \(\sqrt { 9{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 3K

Question 2.
Find the values of:
i. 5 sin 30° + 3 tan 45°
ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°
iii. 2 sin 30° + cos 0° + 3 sin 90°
iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)
v. cos² 45° + sin² 30°
vi. cos 60° x cos 30° + sin 60° x sin 30°
Solution:
i. sin 30° = \(\frac { 1 }{ 2 }\) and tan 45° = 1

ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°

iii. 2 sin 30° + cos 0° + 3 sin 90°
2 sin 30° + cos0° + 3 sin 90° = 2 (\(\frac { 1 }{ 2 }\)) + 1 + 3(1)
= 1 + 1 + 3
∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)

v. cos² 45° + sin² 30°

vi. cos 60° x cos 30° + sin 60° x sin 30°

Question 3.
If sin θ = \(\frac { 4 }{ 5 }\) , then find cos θ.
Solution:
sin θ = \(\frac { 4 }{ 5 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 4k and AC = 5k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (5 k)² = (4k)² + BC²
∴ 25k² = 16k² + BC²
∴ BC² = 25k² – 16k² = 9k²
∴ BC = \(\sqrt { 9{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 3k

Question 4.
If cos θ = \(\frac { 15 }{ 17 }\) , then find sin θ.
Solution:
cos θ = \(\frac { 15 }{ 17 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ BC = 15k and AC = 17k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (17 k)² = AB² + (15K)²
∴ 289k² = AB² + 225²
∴ AB² = 289k² – 225k²
= 64k²
∴ AB = \(\sqrt { 64{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 8k

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.2 Intext Questions and Activities

Question 1.
In right angled ∆PQR, ∠Q = 900. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.
i. sin θ = cos (90 – θ)
ii. cos θ = sin (90 – θ)
iii. sin 30° = cos (90° – 30°) = cos 60°
iv. cos 30° = sin (90° – 30°) = sin 60° (Textbook pg. no. 107)
Solution:
In ∆PQR, ∠Q = 90°, ∠P = θ
∴ ∠R = 90 – θ

i. sin θ = cos (90 – θ)

ii. cos θ = sin (90 – θ)

iii. Let ∠P = θ = 30°
∴ ∠R = 90° – 30°

sin 30° = cos (90° – 30°) … [From (i) and (ii)]
sin 30° = cos 60°

iv. cos 30° = sin (90° – 30°) = sin 60°

∴ cos 30° = sin (90° – 30°) .,.[From (i) and (ii)]
∴ cos 30° = sin 60°

Question 2.
In right angled ∆PQR, ∠Q = 90°, ∠R = θ and if sin θ = \(\frac { 5 }{ 13 }\), then find cos θ and tan θ. (Textbook pg. no. 110)
Solution:
i. Take the given trigonometric ratio as 13k equation (i).
sin θ = \(\frac { 5 }{ 13 }\) .. .(i)[Given]
By using the definition write the trigonometric ratio of sin O and take it as equation (ii).
In right angled ∆PQR, ∠R = θ


Let the common multiple be k.
∴ PQ = 5k and PR = 13k
Find QR by using Pythagoras theorem.
PR² = PQ² + QR² … [Pythagoras theorem]
∴ (13k)² = (5k)² + QR²
∴ 169k² = 25k² + QR²
∴ QR² = 169k² – 25k²
= 144k²
∴ QR = \(\sqrt { 144{ k }^{ 2 } }\) . . . [Taking square root of both sides]
= 12k

Question 3.
While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Solution:
\(\frac { PQ }{ PR }\) = \(\frac { 5 }{ 13 }\) … [Given]
Here, the ratio of the lengths of sides PQ and PR is 5 : 13.
The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving.

Question 4.
While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Tcxtbook pg. no. 111)
Solution:
Yes, we can take lengths of PQ and PR as 5 and 13.
In that case, we will have to take k = 1 and solve the problem accordingly.

Question 5.
Verify that the equation ‘sin² θ + cos² θ = 1’ is true when θ = 0° or θ = 90°.
(Textbook pg. no. 112)
Solution:
sin² θ + cos² θ = 1
i. lf θ = 0°,
LH.S. = sin² θ + cos² θ
= sin² 0° + cos² 0°
= 0 + 1 …[∵ sin 0° = 0, cos 0° = 1]
= R.H.S.
∴ sin² θ + cos² θ = 1

ii. If θ = 90°,
L.H.S.= sin² θ +cos² θ
= sin² 90° + cos² 90°
= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0]
= 1
= R.H.S.
∴ sin² θ + cos² θ = 1

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 8.1 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Practice Set 8.1 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
In the given figure, ∠R is the right angle of ∆PQR. Write the following ratios.

i. sin P
ii. cos Q
iii. tan P
iv. tan Q
Solution:

Question 2.
In the right angled ∆XYZ, ∠XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios.

i. sin x
ii. tan z
iii. cos x
iv. tan x.
Solution:

Question 3.
In right angled ∆LMN, ∠LMN = 90°, ∠L = 50° and ∠N = 40°. Write the following ratios.

i. sin 50°
ii. cos 50°
iii. tan 40°
iv. cos 40°
Solution:

Question 4.
In the given figure, ∠PQR = 90°, ∠PQS = 90°, ∠PRQ = α and ∠QPS = θ. Write the following trigonometric ratios.
i. sin α, cos α , tan α
ii. sin θ, cos θ, tan θ

Solution:
i. In ∆PQR,

ii. In ∆PQS,

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.1 Intext Questions and Activities

Question 1.
In the figure gIven below, ∆PQR is a right angled triangle. Write the names of sides opposite and adjacent to ∠P and ∠R. (Textbook pg no. 102)

Solution:
In right angled ∆PQR,
i. side opposite to ∠P = QR
ii. side opposite to ∠R = PQ
iii. side adjacent to ∠P = PQ
iv. side adjacent to ∠R = QR

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions

9th Standard Maths 2 Problem Set 7 Chapter 7 Co-ordinate Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

Class 9 Maths Part 2 Problem Set 7 Chapter 7 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for the following questions.

i. What is the form of co-ordinates of a point on the X-axis?
(A) (b,b)
(B) (0, b)
(C) (a, 0)
(D) (a, a)
Answer:
(C) (a, 0)

ii. Any point on the line y = x is of the form _____.
(A) (a, a)
(B) (0, a)
(C) (a, 0)
(D) (a, -a)
Answer:
(A) (a, a)

iii. What is the equation of the X-axis ?
(A) x = 0
(B) y = 0
(C) x + y = 0
(D) x = y
Answer:
(B) y = 0

iv. In which quadrant does the point (-4, -3) lie ?
(A) First
(B) Second
(C) Third
(D) Fourth
Answer:
(C) Third

v. What is the nature of the line which includes the points (-5, 5), (6, 5), (-3, 5), (0, 5)?
(A) Passes through the origin
(B) Parallel to Y-axis
(C) Parallel to X-axis
(D) None of these
Answer:
The y co-ordinate of all the points is the same.
∴ The line which passes through the given points is parallel to X-axis.
(C) Parallel to X-axis

vi. Which of the points P(-1, 1), Q(3, -4), R( -1, -1), S(-2, -3), T (-4, 4) lie in the fourth quadrant?
(A) P and T
(B) Q and R
(C) only S
(D) P and R
Answer:
(B) Q and R

Question 2.
Some points are shown in the adjoining figure. With the help of it answer the following questions :

i. Write the co-ordinates of the points Q and R.
ii. Write the co-ordinates of the points T and M.
iii. Which point lies in the third quadrant ?
iv. Which are the points whose x and y co-ordinates are equal ?
Solution:
i. Q(-2, 2) and R(4, -1)
ii. T(0, -1) and M(3, 0)
iii. Point S lies in the third quadrant.
iv. The x and y co-ordinates of point O are equal.

Question 3.
Without plotting the points on a graph, state in which quadrant or on which axis do the following points lie.
i. (5, -3)
ii. (-7, -12)
iii. (-23, 4)
iv. (-9, 5)
v. (0, -3)
vi. (-6, 0)
Solution:

Question 4.
Plot the following points on one and the same co-ordinate system.
A(1, 3), B(-3, -1), C(1, -4), D(-2, 3), E(0, -8), F(1, 0)
Solution:

Question 5.
In the graph alongside, line LM is parallel to the Y-axis.

i. What is the distance of line LM from the Y-axis?
ii. Write the co-ordinates of the points P, Q and R.
iii. What is the difference between the x co-ordinates of the points L and M?
Solution:
i. Distance of line LM from the Y-axis is 3 units.
ii. P(3, 2), Q (3, -1), R(3, 0)
iii. x co-ordinate of point L = 3
x co-ordinate of point M = 3
∴ Difference between the x co-ordinates of the points L and M = 3 – 3
= 0

Question 6.
How many lines are there which are parallel to X-axis and having a distance 5 units?
Solution:
The equation of a line parallel to the X-axis is y = b.
There are 2 lines which are parallel to X-axis and at a distance of 5 units.
Their equations are y = 5 and y = -5.

Question 7.
If ‘a’ is a real number, what is the distance between the Y-axis and the line x = a?
Solution:
Equation of Y-axis is x = 0.
Since, ‘a’ is a real number, there are two possibilities.
Case I: a > 0
Case II: a < 0 ∴ Distance between the Y-axis and the line x = a = a-0 = a Since, |a| = a, a > 0
= – a, a < 0
∴ Distance between the Y-axis and the line x = a is |a|.

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Problem Set 7 Intext Questions and Activities

Question 1.
As shown in the adjoining figure, ask girls to sit in lines so as to form the X-axis and Y-axis.
i. Ask some boys to sit at the positions marked by the coloured dots in the four quadrants.
i. Now, call the students turn by turn using the initial letter of each student’s name. As his or her initial is called, the student stands and gives his or her own co-ordinates. For example Rajendra (2, 2) and Kirti (-1, 0)
iii. Even as they have fun during this field activity, the students will leam how to state the position of a point in a plane. (Textbook pg. no. 92)

Maharashtra Board Class 9 Maths Solutions

• Co-ordinate Geometry Practice Set 7.1 Class 9 Maths Solutions
• Co-ordinate Geometry Practice Set 7.2 Class 9 Maths Solutions
• Co-ordinate Geometry Problem Set 7 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 7.2 Chapter 7 Co-ordinate Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

Class 9 Maths Part 2 Practice Set 7.2 Chapter 7 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
On a graph paper plot the points A(3, 0), B(3, 3), C(0, 3). Join A, B and B, C. What is the figure formed?
Soiution:

d(O, A) = 3 cm, d(A, B) = 3 cm, d(B, C) = 3 cm, d(O, C) = 3 cm and each angle of □ OABC is 90°
∴ □ OABC is a square.

Question 2.
Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its left.
Solution:
The equation of a line parallel to the Y-axis is x = a.
Since, the line is at a distance of 7 units to the left of Y-axis,
∴ a = -7
∴ x = -1 is the equation of the required line.

Question 3.
Write the equation of the line parallel to the X-axis at a distance of 5 units from it and below the X-axis.
Solution:
The equation of a line parallel to the X-axis is y = b.
Since, the line is at a distance of 5 units below the X-axis.
∴ b = -5
∴ y = -5 is the equation of the required line.

Question 4.
The point Q( -3, -2) lies on a line parallel to the Y-axis. Write the equation of the line and draw its graph.
Solution:
The equation of a line parallel to the Y-axis is x = a.
Here, a = -3
∴ x = -3 is the equation of the required line.

Question 5.
Y-axis and line x = – 4 are parallel lines. What is the distance between them?
Solution:
Equation of Y-axis is x = 0.
Equation of the line parallel to the Y-axis is x = – 4. … [Given]
∴ Distance between the Y-axis and the line x = – 4 is 0 – (- 4) … [0 > -4]
= 0 + 4 = 4 units
∴ The distance between the Y-axis and the line x = – 4 is 4 units.
[Note: The question is modified as X-axis cannot be parallel to the line x = – 4.]

Question 6.
Which of the equations given below have graphs parallel to the X-axis, and which ones have graphs parallel to the Y-axis? [1 Mark each]
i. x = 3
ii. y – 2 = 0
iii. x + 6 = 0
iv. y = -5
Solution:
i. The equation of a line parallel to the Y-axis is x = a.
∴ The line x = 3 is parallel to the Y-axis.

ii. y – 2 = 0
∴ y = 2
The equation of a line parallel to the X-axis is y = b.
∴ The line y – 2 = 0 is parallel to the X-axis.

iii. x + 6 = 0
∴ x = -6
The equation of a line parallel to the Y-axis is x = a.
∴ The line x + 6 = 0 is parallel to the Y-axis.

iv. The equation of a line parallel to the X-axis is y = b.
∴ The line y = – 5 is parallel to the X-axis.

Question 7.
On a graph paper, plot the points A(2, 3), B(6, -1) and C(0, 5). If these points are collinear, then draw the line which includes them. Write the co-ordinates of the points at which the line intersects the X-axis and the Y-axis.
Solution:

From the graph, the line drawn intersects the X-axis at D(5, 0) and the Y-axis at C(0, 5).

Question 8.
Draw the graphs of the following equations on the same system of co-ordinates. Write the co-ordinates of their points of intersection.
x + 4 = 0,
y – 1 = 0,
2x + 3 = 0,
3y – 15 = 0
Solution:
i. x + 4 = 0
∴ x = – 4

ii. y – 1 = 0
∴ y = 1

iii. 2x + 3 = 0
∴2x = -3
∴ x = \(\frac { -3 }{ 2 }\)
∴ x = -1.5

iv. 3y- 15 = 0
3y = 15
y = \(\frac { 15 }{ 3 }\)
∴ y = 5

The co-ordinates of the point of intersection of x + 4 = 0 and y – 1 = 0 are A(-4, 1).
The co-ordinates of the point of intersection ofy – 1 = 0 and 2x + 3 = 0 are B(-1.5, 1).
The co-ordinates of the point of intersection of 3y – 15 = 0 and 2x + 3 = 0 are C(-1.5, 5).
The co-ordinates of the point of intersection of x + 4 = 0 and 3y – 15 = 0 are D(-4, 5).

Question 9.
Draw the graphs of the equations given below.
i. x + y = 2
ii. 3x – y = 0
iii. 2x + y = 1
Solution:
i. x + y = 2
∴ y = 2 – x
When x = 0,
y = 2 – x
= 2 – 0
= 2
When x = 1,
y = 2 – x
= 2 – 1
= 1
When x = 2,
y = 2 – x
= 0

ii. 3x – y = 0
∴ y = 3x
When x = 0,
y = 3x
= 3(0)
= 0

When x = 1,
y = 3x
= 3(1)
= 3

When x = -1,
y = 3x
= 3(-1)
= -3

iii. 2x + y = 1
∴ y = 1 – 2x
When x = 0,
y = 1 – 2x
= 1 – 2(0)
= 1 – o
When x = 1,
y = 1 – 2x
= 1- 2(1)
= 1 – 2
= -1
When x = -1,
y = 1 – 2x
= 1 – 2(-1)
= 1 + 2
= 3

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Practice Set 7.2 Intext Questions and Activities

Question 1.
i. Can we draw a line parallel to the X-axis at a distance of 6 unIts from It and below the X-axis?
ii. Will all of the points (-3,-6), (10,-6), ( \(\frac { 1 }{ 2 }\), -6) be on that line?
iii. What would be the equation of this line?(Textbook pg. no. 94)
Solution:

i. Yes.
This line will pass through the point (0,-6).

ii. Yes.
Here, y co-ordinate of the points (-3, -6), (10,-6), ( \(\frac { 1 }{ 2 }\), -6) is the same, which is -6.
∴ All the above points lie on the same line.

iii. Since, the line is at a distance of 6 units below the X-axis.
∴ b = -6
∴ Equation of the line is y = -6.

Question 2.
i. Can we draw a line parallel to the Y – axis at a distance of 2 units from ¡t and to its right?
ii. Will all of the points (2, 10), (2, 8), (2, -) be on that line?
iii. What would be the equation of this line? (Textbook pg. no. 95)
Solution:

i. Yes.
(2, 10)
This line will pass through the point (2, 0).
(2,8)
ii. Yes.
Here, x co-ordinate of the points (2, 10), (2, 8), (2,-\(\frac { 1 }{ 2 }\) ) is the same, which is 2.
∴ All the above points lie on the same line.

iii. Since, the line is at a distance of 2 units to the right of Y-axis.
a = 2
∴ Equation of the line is x = 2.

Question 3.
On a graph paper, plot the points (0, 1), (1, 3), (2, 5). Are they collinear? If so, draw the line that passes through them.
i. Through which quadrants does this line pass ?
ii. Write the co-ordinates of the point at which it intersects the Y-axis.
iii. Show any point in the third quadrant which lies on this line. Write the co-ordinates of the point. (Textbook pg. no. 96)
Solution:
i. The line passes through the quadrants I, II and III.
ii. The line intersects the Y-axis at (0, 1).
iii. (-1,-1)

Maharashtra Board Class 9 Maths Solutions

• Co-ordinate Geometry Practice Set 7.1 Class 9 Maths Solutions
• Co-ordinate Geometry Practice Set 7.2 Class 9 Maths Solutions
• Co-ordinate Geometry Problem Set 7 Class 9 Maths Solutions