CBSE Class 9 Maths Revision Notes Chapter 13

Class 9 Mathematics Revision Notes for Surface Areas and Volumes of Chapter 13

Extramarks’ Revision Notes of Class 9 Mathematics Chapter 13 assist students in thoroughly revising each and every important concept related to Surface Areas and Volumes. These notes are prepared in accordance with the most recent CBSE guidelines and are easily accessible from the website.

Class 9 Mathematics Revision Notes for Surface Areas and Volumes of Chapter 13 – Free Download

Access Class 9 Mathematics Chapter 13 – Surface Areas and Volumes Notes

Definitions:

Solids:

A solid is any object that has a fixed space and volume, for instance, a cube, cuboid, sphere, cylinder, cone, and so on.

  1. Surface Area of a Solid: 

The area that a solid object occupies is referred to as its surface area.

The surface area is calculated using square units.

Square metre is an example (m2).

  1. Volume of a Solid:

The volume of a solid is the measurement of occupied space.

The cubic unit is the volume unit.

Cubic metre is an example (m3)

Formulas for Different Solids:

  1. Cuboid

A cuboid is a three-dimensional solid with six rectangular faces. A cuboid has 6 rectangular faces, 12 edges, and 8 vertices with equal-sized opposite faces.

Examples of a cuboid are a book, a matchbox, a shoebox, and so on.

Surface Area of Cuboid:

S = 2(lb+bh+lh) 

l  is length, b is breadth and h is the height of the cuboid.

Volume of Cuboid:

V = l×b×h

l is length, b is breadth and h is the height of the cuboid.

  1. Cube  

A cuboid which has equal length, breadth and height is called a cube. 

For example- ice cubes, dice etc.

Surface Area of Cube:

S = 6l2

Where l is the length of each side of the cube.

Volume of Cube:

V=l3

Where l is the length of each side of the cube.

  1. Cylinders

A cylinder is a solid that is created by piling a lot of circular discs together along their diameter. For example, circular pillars, circular pipes, measuring cylinders, soft drink cans, and so on.

Hollow Cylinder

Solids like iron pipes, rubber tubes, etc., are in the shape of hollow cylinders. 

Surface Area of Cylinder: 

  1. a) Curved surface area (CSA): CSA=2πrh
  2. b) Total Surface area (TSA): TSA=2πr (r+h)

r is the radius of the circular top and bottom, and h is the height of the cylinder.

Volume of Cylinder:

V=πr2h

 r is the radius of the circular top and bottom, and h is the height of the cylinder.

  1. Right Circular Cone

A right circular cone is the solid created when a right-angled triangle is rotated around a right-angled side.

Surface Area of Cone:

  1. a) Curved Surface Area (CSA): CSA=πrl
  2. b) Total Surface Area (TSA): TSA=πr (r+l)

r is the radius of the circular part, h is the perpendicular height and l= √r2+h2 is the slant height of the cone.

Volume of Cone:

V=1/3πr2h

r is the radius of the circular part, and h is the perpendicular height of the cone.

  1. Sphere

A sphere is a three-dimensional solid made up of all the points in space that are spaced at regular intervals, or radius, from a fixed point, or centre.

For example- a bowling ball, cricket ball, etc.

Surface Area of Sphere:

SA=4πr2

Where r is the radius of the sphere.

Volume of Sphere:

V=4/3πr3

Where r is the radius of the sphere.

  1. Spherical Shell

A spherical shell is the solid space between two hollow, concentric spheres of differing radii.

For example, a ping-pong ball, football, etc.

Surface Area of Shell:

SA=4πR2

Where R is the radius of the outer sphere.

Volume of Solid Part of Shell:

V=4/3π(R3−r3)

Where R is the radius of the outer sphere and r is the radius of the inner sphere.

  1. Hemisphere

They form a hemisphere when a plane cuts a solid into two equal pieces while passing through the centre.

For example, a ball is divided into equal pieces, the roof of a building with a dome shape, etc.

Surface Area of Hemisphere:

  1. a) Curved Surface Area (CSA): CSA=2πr2
  2. b) Total Surface Area (TSA): TSA=3πr2

Where r is the radius of the circular region.

Volume of Hemisphere:

V=2/3πr3

Where r is the radius of the hemisphere.

CBSE Class 9 Mathematics Notes Chapter 13 Surface Areas and Volumes 

The exercises in CBSE Class 9 Chapter 13 Surface Areas and Volumes extend from 13.1 to 13.9. The solutions to these exercises can be found below:

  • Exercise 13.1: 8 questions
  • Exercise 13.2: 11 questions
  • Exercise 13.3: 8 questions
  • Exercise 13.4, 13.5 and 13.7: 9 questions each
  • Exercise 13.6: 8 questions
  • Exercise 13.8: 10 questions
  • Exercise 13.9: 3 questions

Importance of CBSE Class 9 Mathematics Revision Notes

The Extramarks Revision Notes on Surface Areas and Volumes for Class 9 provides a summary of all the important and relevant topics as well as highlights the significant references from the Surface Areas and Volumes Class 9 NCERT book.

Let us Revise Some Important Concepts and Formulas of Surface Areas and Volumes.

  1. Surface Area

The surface area is the volume that a three-dimensional object takes up in space. It is equal to the sum of the areas of all its two-dimensional faces since three-dimensional objects are comprised of them.

The surface area is classified as follows:

Curved Surface Area (CSA)

Lateral Surface Area (LSA) 

Total Surface Area (TSA)

  1. Volume

The volume of any 3D object is the space it occupies. Because the volume of a solid shape is the product of three dimensions, it is expressed in cubic units.

Surface Area and Volume Formula in a Tabular Form are Given Below

Sr. No Name Abbreviation used Lateral/Curved Surface Area Total Surface Area Volume
1. Cuboid H= height, l= length b= breadth 2h(1+b) 6l² Lxbxh
2. Cube a= length of the sides 4a² 6a²
3. Right Prism .. Perimeter of Base x Height Lateral

Surface Area+2(Area of One End)

Area of

Base x Height

4. Right Circular Cylinder r= radius h= height 2 (π xrxh) 2πr (r + h) πr²h
5. Right Pyramid .. 1/2 (Perimeter of Base x Slant Height) Lateral Surface Area+Area of the Base ⅓ (Area of the Base) Height
6. Right Circular Cone r= radius l= length πrl πr (l+r) ⅓(πr2h)
7. Sphere r= radius 4πr² 4πr² 43 (πr³)
8. Hemisphere r= radius 2πr² 3πr² ⅔ (πr³)

Surface Area and Volumes Notes 

Students can refer to Extramarks’ Class 9 Mathematics Notes of Surface Area and Volumes to perform well in the CBSE Class 9 examination. These notes explain each concept in detail, helping them to revise each and every important concept related to the surface area and volumes.

Why Choose Extramarks for Surface Area and Volumes Notes?

Experts have created Extramarks Chapter Wise CBSE Surface Areas and Volumes Class 9 Notes according to the updated CBSE guidelines to help students perform well in final exams.

Surface Areas and Volumes Class 9 Notes will help students understand concepts better and improve their learning experience. 

Extramarks’ Surface Areas and Volumes Class 9 Notes can be accessed anytime from the website. 

Extramarks Surface Areas and Volumes Class 9 Notes will assist students in solving surface area and volume problems in a systematic manner.

Expert Tips

Effective revision begins with planning and preparation. Surface Areas and Volumes Revision Notes Class 9 Notes are extremely useful for students looking to improve their exam performance. These notes help in thoroughly revising the chapter while also making learning simple.

The Final Words

Extramarks’ subject matter experts prepare Mathematics Class 9 Surface Areas and Volumes Notes based on the revised CBSE syllabus and NCERT guidelines. These simple notes will assist students to revise chapter concepts and thereby, improve their performances in exams. Extramarks provides refined and effective CBSE Class 9 Mathematics Chapter 13 Revision Notes to help them ace their preparation.

FAQs (Frequently Asked Questions)

1. It is necessary to make a closed cylindrical tank of a height of 1m and a base diameter of 140 cm from a metal sheet. How many square metres of the sheet is required for the same? Assume π = 22/7

If we assume that h is the height and r is the radius of a cylindrical tank,

Then,

Height of the cylindrical tank, h = 1m

As we already know,

Radius = half of diameter = (140/2) cm = 70 cm = 0.7m

Therefore, r = 0.7m

So we can say,

Area of sheet required = Total surface area of tank = 2πr(r+h) unit square

 

= 2×(22/7)×0.7(0.7+1)

= 7.48

 

Hence the total area of the sheet required is 7.48m2.