CBSE Class 10 Maths Chapter 28: Surface Areas and Volumes NCERT Solutions
This chapter delves into the calculation of surface areas and volumes of combined solid figures. Students will learn to break down complex shapes into simpler ones like cubes, cylinders, cones, and hemispheres, and then apply formulas to find their total surface area and volume. The NCERT Solutions for Class 10 Maths Chapter 28 provide step-by-step explanations for various problems, including finding the surface area of a cuboid formed by joining cubes and calculating the inner surface area of a vessel composed of a cylinder and a hemisphere. These solutions are designed to clarify concepts and build problem-solving skills, making them an invaluable resource for exam preparation and revision.
Quick info
| Board | CBSE |
|---|---|
| Class | Class 10 |
| Subject | Mathematics |
| Session | 2026 |
| Language | English |
| Type | NCERT Solutions |
| Chapter | Chapter 28 |
Chapter summary
Chapter 28 of the NCERT Class 10 Mathematics textbook focuses on Surface Areas and Volumes of combined solids. The exercises cover calculating the surface area of composite shapes formed by joining basic solids. Key concepts include understanding how joining solids affects their dimensions and applying formulas for surface areas of cuboids, cylinders, and hemispheres. The solutions provide a clear, step-by-step approach to solving these problems, reinforcing the application of geometric formulas.
Learning outcomes
- Understand the concept of surface area and volume for combined solids.
- Calculate the dimensions of a cuboid formed by joining two cubes.
- Determine the surface area of a cuboid.
- Identify the components of a composite vessel (cylinder and hemisphere).
- Calculate the inner surface area of a composite vessel.
- Apply formulas for the surface area of cylinders and hemispheres.
Topics covered
Paper topics
- Surface Area of Combined Solids
- Volume of Combined Solids
- Cuboids
- Cubes
- Hemispheres
- Cylinders
- Composite Shapes
- Calculating Surface Area
- Joining Solids
- Vessel Surface Area
Important topics
- Surface Area of Cuboid formed by joining Cubes
- Inner Surface Area of Composite Vessels
- Formulas for Surface Areas of Basic Solids
- Application of Formulas to Combined Shapes
PDF preview
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Questions and Solutions
Question 1
Given the volume of each cube is 64 cm<sup>3</sup>.
The formula for the volume of a cube is $V = \text{edge}^3$.
So, $\text{edge}^3 = 64 \text{ cm}^3$.
Taking the cube root of both sides, we find the edge length: $\text{edge} = \sqrt[3]{64} = 4 \text{ cm}$.
When two such cubes are joined end to end, they form a cuboid. The dimensions of this cuboid will be:
- Length (l) = 4 cm + 4 cm = 8 cm
- Breadth (b) = 4 cm
- Height (h) = 4 cm
The formula for the surface area of a cuboid is $A = 2(lb + bh + lh)$.
Substituting the dimensions:
Therefore, the surface area of the resulting cuboid is 160 cm<sup>2</sup>.
Question 2
The vessel consists of a hollow hemisphere at the bottom and a hollow cylinder on top.
Given the diameter of the hemisphere is 14 cm, the radius (r) of the hemisphere is half of the diameter:
$r = \frac{14}{2} = 7 \text{ cm}$.
Since the cylinder is mounted on the hemisphere, the radius of the cylinder is the same as the radius of the hemisphere, so $r = 7 \text{ cm}$.
The total height of the vessel is 13 cm.
The height of the hemispherical part is equal to its radius, which is 7 cm.
The height of the cylindrical part (h) is the total height minus the height of the hemispherical part:
$h = 13 \text{ cm} - 7 \text{ cm} = 6 \text{ cm}$.
The inner surface area of the vessel is the sum of the curved surface area (CSA) of the cylindrical part and the CSA of the hemispherical part.
CSA of cylinder = $2\pi rh$
CSA of hemisphere = $2\pi r^2$
Inner surface area = $2\pi rh + 2\pi r^2$
Substituting the values with $\pi = \frac{22}{7}$:
Inner surface area = $2 \times \frac{22}{7} \times 7 \times 6 + 2 \times \frac{22}{7} \times 7^2$
Inner surface area = $(2 \times 22 \times 6) + (2 \times 22 \times 7)$
Inner surface area = $264 + 308$
Inner surface area = $572 \text{ cm}^2$.
Therefore, the inner surface area of the vessel is 572 cm<sup>2</sup>.
Common mistakes
- Incorrectly calculating the dimensions of the combined solid.
- Forgetting to add or subtract relevant areas when dealing with composite shapes.
- Using the wrong formula for surface area or volume.
- Errors in substituting values or performing arithmetic calculations.
Revision tips
- Review the formulas for surface areas of basic solids (cubes, cuboids, cylinders, hemispheres).
- Practice visualizing how solids are combined and how this affects their dimensions.
- Work through each step of the provided solutions to understand the logic.
- Attempt to solve the problems independently after reviewing the solutions.
Practice MCQs
Q1. When two cubes of edge 'a' are joined end to end, what are the dimensions of the resulting cuboid?
Explanation: When two cubes are joined end to end, one dimension doubles while the other two remain the same. If the edge is 'a', the dimensions become a, a, and 2a.
Q2. What is the formula for the total surface area of a cuboid with length 'l', breadth 'b', and height 'h'?
Explanation: The total surface area of a cuboid is the sum of the areas of its six faces, which is given by the formula 2(lb + bh + lh).
Q3. In a vessel formed by a hollow hemisphere and a hollow cylinder, if the radius is 'r' and the height of the cylinder is 'h', what is the formula for the inner surface area?
Explanation: The inner surface area is the sum of the curved surface area of the cylinder (2πrh) and the curved surface area of the hemisphere (2πr²).
Q4. If the volume of a cube is 64 cm³, what is the length of its edge?
Explanation: The volume of a cube is edge³. So, edge³ = 64 cm³. Taking the cube root, edge = 4 cm.
Q5. What is the curved surface area of a hemisphere with radius 'r'?
Explanation: The curved surface area of a hemisphere is half the surface area of a sphere, which is 2πr².
Frequently asked questions
What is Chapter 28 of CBSE Class 10 Maths about?
Chapter 28 of CBSE Class 10 Maths covers Surface Areas and Volumes, focusing on calculating these properties for solid figures formed by combining basic shapes like cubes, cylinders, and hemispheres.
How do these NCERT Solutions help with exam preparation?
These solutions provide clear, step-by-step methods to solve problems related to combined solids, helping students understand the concepts, practice calculations, and build confidence for their exams.
What is the key concept in finding the surface area of a cuboid formed by joining two cubes?
When two cubes are joined end to end, the resulting cuboid has dimensions where one length is doubled, and the other two dimensions remain the same as the original cube's edge. The surface area is then calculated using the cuboid formula.
How is the inner surface area of a vessel made of a cylinder and hemisphere calculated?
The inner surface area is the sum of the curved surface area of the cylindrical part and the curved surface area of the hemispherical part. Formulas for these are applied using the given dimensions.
Are the formulas for basic shapes important for this chapter?
Yes, a strong understanding of the formulas for the surface areas of cubes, cuboids, cylinders, and hemispheres is crucial, as these are the building blocks for solving problems involving combined shapes.
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