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Surface Areas and Volumes Class 10th Notes - Free NCERT Class 10 Maths Chapter 13 Notes - Download PDF

Surface Areas and Volumes Class 10th Notes - Free NCERT Class 10 Maths Chapter 13 Notes - Download PDF

Updated on Apr 19, 2025 12:34 PM IST

Surface Areas and Volumes is an important chapter which helps us understand the concept of the measurement of three-dimensional shapes. The surface area can be distinguished as Lateral surface area (LSA), Curved surface area (CSA) and Total surface area (TSA). In this article, we will discuss the formulas for volume and surface areas of different shapes, such as cuboid, cube, cylinder, cone, frustum, etc. The formulas we study in this chapter are important for solving practical problems related to geometry and real-life applications like construction, packaging, engineering, etc.

This Story also Contains
  1. NCERT Notes Class 10 Maths Chapter 12 Surface Areas and Volumes
  2. Surface Area And Volume of the Cuboid:
  3. Surface Area and Volume of a Cube
  4. Surface Area and Volume of a Cylinder
  5. Surface Area and Volume of Right Circular Cone
  6. Surface Area and Volume of Sphere
  7. Surface Area and Volume of a Hemisphere
  8. Surface Area and Volume of Frustum of a Cone
  9. Surface Area and Volume of Combination of Solids
  10. Formulas at a Glance:
  11. Class 10 Chapter Wise Notes
  12. NCERT Exemplar Solutions for Class 10
  13. NCERT Solutions for Class 10
  14. NCERT Books and Syllabus

CBSE Class 10 Maths chapter 12 notes contain systematic explanations of topics using examples and exercises. All these topics can be downloaded from the Class 10 Maths chapter 12 notes PDF. The students must practice the NCERT examples and check the NCERT Exemplar for Class 10 Maths Chapter 12 Surface Area and Volumes. Apart from this, the use of NCERT class 10th maths notes provides you with the ability to examine mathematical concepts rigorously. Also, after studying the NCERT Notes, you will have access to short summaries that help you to remember information fast.

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NCERT Notes Class 10 Maths Chapter 12 Surface Areas and Volumes

Surface area: Surface area can be defined as the measurement of the total area that has been occupied by the object.

Volume: Volume is defined as the measurement of space that is occupied by the object.

Now, let's discuss the basic solid shapes and their respective formula.

Surface Area And Volume of the Cuboid:

A cuboid is a three-dimensional shape, also known as a rectangular prism. It has six rectangular faces, eight vertices, and 12 edges.

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The surface area of a cuboid with the roof.
The total surface area of the cuboid (TSA) = The sum of the areas of all its 6 faces.

The surface area of the cuboid = 2(lb+bh+hl) sq. unit.

Here, l denotes the length, b denotes the breadth, and h denotes the height.

The lateral surface area of the Cuboid:
The lateral surface area (LSA) includes all the sides except the top and the bottom faces.

Area of cuboid covering four walls = Area of face AEHD + Area of face BFGC + Area of face ABFE + Area of face DHGC

Lateral Surface Area of the Cuboid = 2(b×h)+2(l×h) = 2h(l+b)

Length of diagonal of a cuboid =(l2+b2+h2)

Volume of cuboid: The volume of a cuboid is defined as the space occupied between its six rectangular faces. It is given by the product of its dimensions, such as length, breadth and height.

Volume = l×b×h cubic units.

Surface Area and Volume of a Cube

A cube is defined as a three-dimensional solid shape with six square faces, twelve edges and eight vertices.

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Surface Area of Cube
As we know, all sides of a cube are equal.
So, Length = Breadth = Height = a.

The total surface area of the cube (TSA) = The sum of the areas of all its six faces.

TSA of Cube = 6 × area of Square = 6a2

The lateral surface area of cube = 2(a×a+a×a) = 4a2

Diagonal of a cube = 3l

Volume of a Cube: In a cube, all sides are the same or equal. So, the volume of the cube is:

Volume of a cube = Base area × Height

As all dimensions of a cube are the same.

So, Volume = a3

Where a is the length of the edge of the cube.

Surface Area and Volume of a Cylinder

A cylinder is a three-dimensional solid shape having two parallel, congruent bases which are connected by a curved surface at a fixed distance from the center of the bases. Therefore, it consists of three faces, in which two are circular and one is lateral. Now, based on these dimensions, we can find the surface area and volume of a cylinder.

Surface Area of Cylinder
As shown in the figure below, assume a cylinder of base radius r and height h units. The curved surface of the cylinder, if it were opened along the diameter (d = 2r) of the circular base, can be changed into a rectangle of length 2πr and height h units.

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Therefore, the cylinder changes to a rectangle.

CSA of a cylinder of base (where radius = r and height = h) = 2π×r×h

TSA of a cylinder of base (where radius = r and height = h) = 2π×r×h + 2πr2 (Area of two circular bases = 2πr2)

= 2πr(h+r)

Volume of a Cylinder:
The volume of a cylinder is defined as the amount of space it occupies in a three-dimensional space. It can be calculated as the product of the area of its circular base and its height, as shown below in a diagram.

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Volume of a Cylinder = πr2×h = πr2h

(where r is the radius of the circular base and h is the height of the cylinder.)

Surface Area and Volume of Right Circular Cone

A cone is defined as a three-dimensional geometric shape that consists of a circular base and a curved surface that gradually decreases to a single point called the vertex.

Surface Area of Cone
Let's consider a right circular cone with slant length l, radius r and height h as shown in the diagram.

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The cone with base radius r and height h

The curved surface area (CSA) of a right circular cone = πrl

The total surface area of the cone (TSA) = πrl+πr2 (where, πr2 = Area of a base)
= πr(l+r)

Volume of a Right Circular Cone
The volume of a right circular cone is defined as the amount of space it occupies in a three-dimensional space, and the volume of a Right circular cone is equal to one-third that of a cylinder of the same height and base.

The volume of a Right circular cone = 13πr2h

Here, r is the radius of the base and h is the height of the cone.

Surface Area and Volume of Sphere

A solid that is round is known as a sphere. All the points that are located on its surface are equidistant from the center.

Surface Area of Sphere:
As shown in the diagram, let's consider a sphere of radius r.

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Curved Surface Area (CSA) = Total Surface Area (TSA) = 4πr2

Volume of Sphere
The volume of a sphere of radius r = 43 πr3

Surface Area and Volume of a Hemisphere

A hemisphere is defined as one-half of a sphere, which is represented as a 3D shape cut by a plane through its center and has one flat surface, as shown in the figure below.

1744089561707

Surface Area of a Hemisphere
As we studied above, the curved surface area (CSA) of a sphere = 4πr2.
Half of a sphere is a hemisphere.

Therefore, the Curved Surface Area (CSA) of a hemisphere of radius r = 2πr2.

Total Surface Area of hemisphere (TSA) = Curved surface area + Area of the base circle ⇒ 2πr2 + πr2

Total Surface Area = 3πr2

Volume of Hemisphere
Half of the volume (V) of the sphere is equal to the volume of a hemisphere.

Therefore, the volume of the hemisphere of radius r = 23πr3

Surface Area and Volume of Frustum of a Cone

A plane cutting through a solid creates a new solid form. A cone's frustum, which is created when a plane cuts a cone parallel to its base, is one example of this type of solid. Now let's talk about its volume and surface area.

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Frustum with radii r1 and r2 and height h
The curved surface area (CSA) of a frustum is π(r1+r2)l, where l = [h2+(r2r1)2].

The total surface area (TSA) of the frustum is the (CSA) + (the areas of the two circular faces) = π(r1+r2)l+π(r12+r22)

Volume of a Frustum
The volume of a frustum of a cone = 13πh(r12+r22+r1r2)

Surface Area and Volume of Combination of Solids

The combination of solids discusses the forms generated when two different solids are joined together. Thus, the surface area and volume for such shapes will be different from the other fundamental solids.

Surface Area of a Combination of Solids

Complex figures' sections can be dissected and examined as more straightforward, well-known shapes. We can get the necessary area of the unknown figure by calculating the areas of four well-known shapes.

Example: Two cubes with volumes of 125cm3 each are connected end to end. Determine the final cuboid's surface area.

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Solution: Length of each cube = 12513 = 5cm
Due to their adjacent joining, these cubes create a cuboid with a length of l = 10cm. However, the width and height will stay at 5cm.

Now, the combination of 2 equal cubes
Therefore, the new surface area, TSA = 2(lb+bh+lh)

TSA = 2(10×5+5×5+10×5)

= 2(50+25+50)

= 2×125

So, the total surface area (TSA) = 250cm2.

Volume of a Combination of Solids

Visualizing complex objects as a mix of shapes of known solids might simplify their volume.

Example: Daanish operates a business out of a shed that resembles a cuboid with a half-cylinder on top. Determine the amount of air that the shed can contain if its base is 7m by 15m and its cuboidal part is 8m high. Additionally, say that there are 20 workers and that the machinery in the shed takes up 300m3, with each worker taking up an average of 0.08m3 of area. How much air is there in the shed, then?

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Solution: When no people or equipment are present, the volume of air within the shed is determined by adding the volumes of air inside the cuboid and the half-cylinder.

The cuboid's current dimensions are 15m for length, 7m for width, and 8m for height.

Additionally, the half cylinder has a height of 15m and a diameter of 7m.

Therefore, the required volume = volume of the cuboid + 12 volume of the cylinder.

= [15×7×8+12×227×72×72×15]m3

= 1128.75m3

Formulas at a Glance:

The students must read the formulas table given below for a quick revision.

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Class 10 Chapter Wise Notes

Students must download the notes below for each chapter to ace the topics.


NCERT Exemplar Solutions for Class 10

Students must check the NCERT Exemplar solutions for class 10 of Mathematics and Science Subjects.

NCERT Solutions for Class 10

Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.

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NCERT Books and Syllabus

To learn about the NCERT books and syllabus, read the following articles and get a direct link to download them.



Frequently Asked Questions (FAQs)

1. What are the important formulas in Chapter 12 Surface Areas and Volumes?

The key formulas from this chapter:

  • Cube:

The total surface area of the cube (TSA) = 6 × area of Square = 6a2
The lateral surface area of cube = 2(a×a+a×a) = 4a2
Diagonal of a cube = 3l
Volume = a3

  • Cuboid:

The total surface area of the cuboid (TSA) =  2(lb+bh+hl) sq. unit.
The lateral surface area of the Cuboid 2(b×h)+2(l×h) = 2h(l+b)
Length of diagonal of a cuboid(l2+b2+h2)
Volume = l×b×h cubic units.

  • Right Circular Cone:

Curved surface area (CSA) of a right circular cone = πrl
Total surface area of the cone (TSA) = πrl+πr2 = πr(l+r)
Volume of a Right circular cone = 13πr2h

  • Sphere:

Curved Surface Area (CSA) = Total Surface Area (TSA) = 4πr2
Volume of Sphere43 πr3

  • Hemisphere:

CSA of a hemisphere of radius r = 2πr2
Volume of the hemisphere of radius r = 23 πr3

2. What is the formula for the volume of a cone in NCERT Class 10 Maths?

The volume of a Right circular cone = 13πr2h

Where r = radius and h = height of the cone

3. How do you calculate the curved surface area of a cylinder?

CSA of a cylinder of base (where radius = r and height = h) = 2π×r×h

4. What is the difference between total surface area and curved surface area?

Curved Surface Area (CSA): The calculation excludes base areas while determining the curved surface area of a 3D figure.
Total Surface Area (TSA): The calculation combines both the curved surface area with the specified bases.

5. How to solve word problems on surface areas and volumes?
  • Identify the given shape.
  • Record all provided measurement values (radius and height, and more).
  • Apply the correct formula.
  • The solution requires step-by-step operations followed by appropriate unit conversions.
  • Check if the answer makes sense logically.

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0.34\; J

Option 2)

0.16\; J

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1.00\; J

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0.67\; J

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2.45×10−3 kg

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 6.45×10−3 kg

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 9.89×10−3 kg

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12.89×10−3 kg

 

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2,000 \; J - 5,000\; J

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200 \, \, J - 500 \, \, J

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2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

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K/2\,

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\; K\;

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zero\;

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K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

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6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

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33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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0.02

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3.125 × 10-2

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1.25 × 10-2

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2.5 × 10-2

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decrease twice

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increase two fold

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remain unchanged

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be a function of the molecular mass of the substance.

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Molality

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twice that in 60 g carbon

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6.023 × 1022

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less than 3

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more than 3 but less than 6

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more than 6 but less than 9

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more than 9

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