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NCERT Solutions for Class 9 Mathematics Chapter 11 Surface Areas and Volumes in English and Hindi Medium updated for CBSE 2024-25 Exams. The effectiveness of Tiwari Academy or any educational platform vary depending on individual learning styles and needs. According to Class 9 NCERT Maths textbook for CBSE 2024-25 and new syllabus, 9th math chapter 11 has only four exercises.
Class 9 Maths Chapter 11 Solutions for CBSE Board
Class 9 Maths Exercise 11.1 in English
Class 9 Maths Exercise 11.2 in English
Class 9 Maths Exercise 11.3 in English
Class 9 Maths Exercise 11.4 in English
Class 9 Maths Chapter 11 Solutions for State Boards
Class 9 Maths Chapter 11 Exercise 11.1
Class 9 Maths Chapter 11 Exercise 11.2
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Class 9 Maths Chapter 11 Solutions in Hindi Medium
Class 9 Maths Exercise 11.1 in Hindi
Class 9 Maths Exercise 11.2 in Hindi
Class 9 Maths Exercise 11.3 in Hindi
Class 9 Maths Exercise 11.4 in Hindi
NCERT Solutions for Class 9 Maths Chapter 11
| Class: 9 | Mathematics |
| Chapter 11: | Surface Areas and Volumes |
| Number of Exercises: | 4 (Four) |
| Content: | NCERT Exercises Solutions |
| Content Type: | Online Videos and Text Format |
| Academic Session: | CBSE 2024-25 |
| Medium: | English and Hindi Medium |
An Overview of Chapter 11 in Class 9 Mathematics
Surface Area of a Right Circular Cone: The class 9 Maths chapter 11 begins by discussing the surface area of various shapes, including cubes, cuboids, cylinders, and cones. It particularly focuses on the right circular cone, providing activities and examples to help understand its geometry.
Calculations and Examples: Class 9 Maths chapter 11 includes detailed mathematical calculations for finding the surface area of a right circular cone. It explains concepts like the height, radius, slant height of the cone, and how these dimensions are used in calculations. There are also examples with step-by-step solutions to demonstrate how to calculate both the curved surface area and the total surface area of a cone.
Practical Applications: Chapter 11 of 9th mathematics, applies these concepts to real-world objects, such as a corn cob, to illustrate how these mathematical principles are used in everyday life. It also includes exercises and problems for further practice.
Exercises and Problems: 9th Maths chapter 11 contains various exercises and problems related to the surface area of cones, spheres, and other shapes. These problems are designed to test the understanding of the concepts discussed in the document.
Surface Area of a Sphere: This chapter also touches upon the surface area of a sphere, drawing parallels with simpler shapes like circles to help explain the concept.
UP Board Students of Class 9 (High School) are using NCERT Book in session 2024-25, so they can also using these textbooks solutions as UP Board Solutions for Class 9 Maths Chapter 11. Here, they can download Prashnavali 11.1, Prashnavali 11.2, Prashnavali 11.3 and Prashnavali 11.4 in Hindi Medium to use offline for session 2024-25. NCERT Solutions for class 9 chapter 11 is applicable for CBSE Delhi Board, MP Board, UP Board – High School, UK Board, Gujrat Board and other board using NCERT Books. NCERT Solutions for other are also given in downloadable format.
Class 9 Maths Chapter 11 Practice Questions with Solution
What are the formulae for Cuboid?
Cuboid: A cuboid is a solid bounded by six rectangular plane surfaces for example match box, brick, books, etc. are cuboid.
1. Surface area (or total surface area) of cuboid = 2 (lb + bh + hl) square units
2. Lateral surface area of cuboid = 2(l + b)h square units
3. Diagonal of a cuboid = √[l² + b² + h²] units
4. Total length of a edges of a cuboid = 4 (l + b + h) units
5. Volume of cuboid = lbh cubic units
What are the formulae for Cube?
Cube: A cuboid whose length, breadth and height are equal, is called a cube.
1. Surface area (or total surface area) of cuboid = 6a² square units
2. Lateral surface area of cuboid = 4a² square units
3. Diagonal of a cuboid = √3a units
4. Total length of a edges of a cuboid = 12a units
5. Volume of cuboid = a³ cubic units
What is meant by a Cylinder? Write its formulae?
Cylinder: A solid generated by the revolution of a rectangle about one of its sides which is kept fixed is called right circular cylinder.
Curved Surface Area (CSA) = 2πrh square units
Total Surface Area (TSA) = 2πr(r + h) square units
Volume = πr²h cubic units
What do you understand by a Cone? Write its Formulae?
Cone: A right circular cone is solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with a fixed line.
Slant Height = √[r² + h²]
Curved Surface Area (CSA) = πrl square units
Total Surface Area (TSA) = πr(r + l) square units
Volume = 1/3 πr²h cubic units
What are the basic formulae of Sphare?
Sphere: A sphere is three dimensional figure which is made up of all points in the space, which lie at a constant distance, form a fixed point called the centre of the sphere and the constant distant is called its radius.
Curved Surface Area (CSA) = Total Surface Area (TSA) = 4πr² square units
Volume = 4/3 πr³ cubic units
9th Maths Chapter 11 Solutions in English & Hindi Medium
Download free NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes all exercises which are given below in PDF format to download for new academic session 2024-25.
NCERT Solutions as well as NCERT Solutions Offline Apps are updated for new academic session 2024-25 based on latest CBSE Syllabus 2024-25. NCERT Books in English and Hindi are now implemented in Uttar Pradesh also. So, students can download UP Board solutions for class 9 Maths chapter 11 all exercises from here.
Important Questions on 9th Maths Chapter 11
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.
Inner radius of cylindrical pipe r = 24/2 = 12 cm,
outer radius R = 28/2 = 14 cm and
length h = 35 m
Volume of cylindrical wooden pipe
= π(R²-r²)h
= 22/7 × (14²-12²)×35
= 22 × (196 – 144) × 5
= 22 × 52 × 5
= 5720 cm³
Mass of cylindrical wooden pipe
= 5720 × 0.6 g
= 3432 g
= 3.432 kg [∵1 cm³ of wood has a mass of 0.6 g]
Hence, the volume of cylindrical wooden pipe is 3.432 kg.
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 and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Length of tin can l = 5 cm, breadth b = 4 cm and height h = 15 cm
Volume of tin can = lbh = 5 × 4 × 15 = 300 cm³
Radius of plastic cylinder r = 7/2 = 3.5 cm and height H = 10 cm
Volume of plastic cylinder
= πr²H = 22/7 × 3.5 × 3.5 × 10
= 22 × 0.5 × 3.5 × 10
= 385 cm³
Difference between capacities of two packs = 385 – 300 = 85 cm³
Hence, the capacity of plastic cylindrical pack is greater than tin can by 85 cm³.
If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find radius of its base.
Lateral surface area of cylinder C = 94.2 cm² and height h = 5 cm.
Let, the radius of cylinder = r cm
Lateral surface area of cylinder C = 2πrh
⇒ 94.2 = 2 × 3.14 × r × 5
⇒ r = 94.2/(3.14×10) = 3 cm
Hence, the radius of base is 3 cm.
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 inner curved surface area of the vessel.
Cost of painting the inner curved surface of cylindrical vessel = ₹ 2200 and
height h = 10 m.
Let, the inner radius of cylindrical vessel = r m
The inner curved surface area of cylindrical vessel = 2πrh
The cost of painting is at the rate of ₹ 20 per m² = ₹ 20 × 2πrh
According to question,
₹ 20 × 2πrh = ₹ 2200
⇒ 2πrh = 2200/20 = 110
Hence, the inner curved surface area is 110 m².
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?
Radius of cylindrical bowl r = 7/2 = 3.5 and
height of soup inside the cylindrical bowl h = 4 cm
Volume of cylindrical bowl
= πr²h = 22/7 × (3.5)² × 4
= 22/7× 3.5 × 3.5 × 4
= 22 × 0.5 × 3.5 × 4 = 154 cm³
Therefore, the volume of soup per day for 250 patient
= 250 × 154
= 38500 cm³
Hence, hospital has to prepare 38500 cm³ soup daily to serve 250 patients.
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.
Curved surface area of cylinder 88 cm² and height h = 14 cm
Let, the radius of base of cylinder = r cm
Curved surface area of cylinder = 2πrh
⇒ 88 = 2 × 22/7 × r × 14
⇒ 88 = 88r
⇒ r = 1 cm
Hence, the diameter of base of cylinder
= 2r = 2 × 1 = 2 cm
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.
Radius of roller r = 84/2 = 42 cm = 0.42 m and
length h = 120 cm = 1.2 m
Outer curved surface area of roller
= 2πrh = 2 × 22/7 × 0.42 × 1.2 = 2 × 22 × 0.06 × 1.2 = 3.168 m²
Area of ground levelled in on revolution = 3.168 m²
Therefore, area of ground levelled in 500 revolutions
= 500 × 3.168 = 1584 m²
Hence, the area of playground is 1584 m².
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 cylinder 4.4 m² and radius r = 0.7 m
Let, the height of cylinder = h m
Curved surface area of cylinder = 2πrh
⇒ 4.4 = 2 × 22/7 × 0.7 × h
⇒ 4.4 = 4.4h
⇒ h = 1 m
Hence, the height of the cylinder is 1 m.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Radius of hemisphere r = 10 cm
Surface area of hemisphere = 3πr²
= 3 × 3.14 × 10 × 10
= 942 cm²
Hence, the total surface area of hemisphere is 942 cm².
Surface Area:
Surface area of a solid body is the area of all of its surface together and it is always measured in square units. Surface area is also known as total surface area (TSA).
Volume:
Space occupied by an object (solid body) is called the volume of the object. Volume is always measured in cubic units.
How many exercises are there in chapter 11 of 9th Maths?
According to new syllabus, there are total 4 exercises in chapter 11 surface areas and volumes of 9th mathematics. In all exercises only 8 or 9 questions are given based on separate sections. Only 1 or 2 questions in exercise are tricky.
Is chapter 11 of class 9 Maths easy?
Class 9 Maths chapter 11 Surface Areas and Volumes, is easy to solve and understand but need to learn formulae. If a student learn formulae based on areas and volumes of 3-dimensional figures, the questions in each exercise look simple.
