NCERT Solutions Class 10 Maths Chapter 12 Surface Areas And Volumes

In Class 10 Maths, Chapter 12 focuses on the essential concepts of Surface Areas and Volumes, building on the foundational knowledge of two-dimensional shapes. While we previously learned about calculating the area and perimeter of figures like rectangles and circles, this chapter expands our understanding of three-dimensional objects. Here, students will explore how to determine the surface areas and volumes of various solids, including cuboids, cylinders, cones, and spheres. 

Understanding these concepts is crucial not only for mathematical proficiency but also for practical applications in real life, as they help us quantify the space occupied by objects around us. The NCERT Solutions provide detailed explanations and step-by-step guidance, enabling students to tackle complex problems with confidence. Therefore, in this blog, we have come up with the NCERT solution for class 10 maths chapter 12 and why they are beneficial for us. 

1.0Download NCERT Solutions Class 10 Maths Chapter 12 Surface Area and Volume : Free PDF

Simplify your learning with our downloadable PDF for NCERT Solutions Class 10 Maths Chapter 12, "Surface Areas and Volumes." This comprehensive resource provides all the answers you need to master essential geometric concepts. Perfect for exam preparation, our PDF ensures you have quick access to crucial information, making your study sessions more efficient and effective. Download it now and take the first step toward acing your math exams.

2.0NCERT Class 10 Maths Chapter 12 Surface Area and Volume

In NCERT Class 10 Maths, Chapter 12, "Surface Areas and Volumes," students will study about the measurement of three-dimensional shapes, building on the principles from Class 9. This chapter is important for understanding practical engineering, architecture, and science applications. This chapter of class 10 covers how to calculate the surface areas and volumes of solids such as cylinders, cones, cuboids, and spheres. NCERT solution helps you deal with these concepts, enhancing your confidence and problem-solving skills by providing step-by-step instructions. Along with increasing your mathematical knowledge, this chapter highlights its importance in everyday life. 

3.0Exercise-wise NCERT Solutions Class 10 Maths Chapter 12

NCERT Solution for Class 10 Maths - Chapter 12 consist of some exercises that contain questions covering all the topics under chapter 12, Surface area and volume. By solving these questions, you can easily understand the questions that come up in the exam and topics of high and low weightage. 

Below, we have provided a brief overview of the number of questions in the NCERT Solutions Class 10 Maths Chapter 12 exercises. 

4.0Important Topics - Class 10 Maths Chapter 12 Surface Area and Volume Solutions 

Chapter 12 Surface Area and Volume in Class 10 Math explores the calculation of surface areas and volumes for various 3D shapes, including cubes, cuboids, cylinders, cones, spheres, and hemispheres. Key concepts include understanding and applying formulas to find the surface area and volume of these objects. This chapter also covers practical applications, such as solving real-life problems involving the measurement of spaces and materials, helping you understand the significance of surface area and volume in everyday life.

1. Surface Area and Volume of a Cube
2. Surface Area and Volume of a Cuboid 
3. Surface Area and Volume of a Cylinder 
4. Surface Area and Volume of a Sphere
5. Surface Area and Volume of a Hemisphere
6. Conversion Between Surface Area and Volume
7. Applications of Surface Area and Volume

5.0General Outline for Class 10 Maths Chapter 12 Surface Areas and Volumes

Introduction to 3D Shapes

• Understanding the concepts of three-dimensional objects.Difference between 2D and 3D figures.

Formulas for Surface Areas and Volumes

1. Cube and Cuboid:

• Surface Area of Cube: $6a^2$
• Volume of Cube: $a^3$
• Surface Area of Cuboid: $2(lb+bh+hl)$
• Volume of Cuboid: $l\times b\times h$

2. Cylinder:

• Surface Area of Cylinder: $2\pi r(h+r)$
• Volume of Cylinder: $\pi r^{2}h$

3. Cone:

• Surface Area of Cone: $\pi r(l+r)$, where l is the slant height
• Volume of Cone: $\frac{1}{3}\pi r^{2}h$

4. Sphere:

• Surface Area of Sphere: $4\pi r^2$
• Volume of Sphere: $\frac{4}{3}\pi r^3$

5. Hemispheres:

• Surface Area of Hemisphere: $3\pi r^2$
• Volume of Hemisphere: $\frac{2}{3}\pi r^3$

6.0Sample NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Let l cm be the length of an edge of the cube having volume = 64 cm³.

Then, l³ = 64 = (4)³ => l = 4 cm

Now, the dimensions of the resulting cuboid made by joining two cubes are:

8 cm × 4 cm × 4 cm (i.e., length = 8 cm, breadth = 4 cm and height = 4 cm)

Surface area of cuboid =2(ℓb+bh+hℓ)

=2(8×4+4×4+4×8)

=2(32+16+32)=2×80=160 cm2

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm . Find the inner surface area of the vessel.

Solution:

For hemispherical part,

Radius (r) = 14/2 = 7 cm

Therefore, Curved surface area = 2πr²

= 2 × 22/7 × 7 × 7 cm² = 308 cm²

Total height of vessel = 13 cm

Therefore, Height of cylinder = (13 - 7) cm = 6 cm and radius (r) = 7 cm

Therefore, Curved surface area of cylinder = 2πrh

= 2 × 22/7 × 7 × 6 cm² = 264 cm²

Therefore, Inner surface area of vessel = Curved surface area of hemispherical part + Curved surface area of cylinder = (308 + 264) cm² = 572 cm²

3. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Sol.

Here, r = 1 cm and h = 1 cm.

Volume of the conical part = (1/3)πr²h and volume of the hemispherical part = (2/3)πr³

Therefore, Volume of the solid shape = (Volume of cone + Volume of hemisphere)

= (1/3)πr²h + (2/3)πr³ = (1/3)πr²(h + 2r)

= (1/3)π(1)²[1 + 2(1)] cm³

= (1/3)π × 1 × 3 cm³

= π cm³

4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of the metal sheet used to make the container, if it costs Rs. 8 per 100 cm^2. (Take π = 3.14).

Sol.

We have: r₁ = 20 cm, r₂ = 8 cm and h = 16 cm.

Therefore, Volume of the frustum = (1/3)πh[r₁² + r₂² + r₁r₂]

= (1/3) × (314/100) × 16 [20² + 8² + 20 × 8] cm³

= (1/3) × (314/100) × 16 [400 + 64 + 160] cm³

= (1/3) × (314/100) × 16 × 624 cm³

= (314/100) × 16 × 208 cm³

= (314 × 16 × 208) / 100000 litres

Therefore, Cost of milk = ₹ 20 × (314 × 16 × 208) / 100000 litres

= ₹ 208.998 ≈ ₹ 209

Now, slant height of the given frustum l = √(h² + (r₁ - r₂)²)

= √(16² + (20 - 8)²)

= √(16² + 12²)

= √(256 + 144)

= √400 = 20 cm.

Therefore, Curved surface area = π(r₁ + r₂)l

= 3.14 × (20 + 8) × 20 cm²

= (314/100) × 28 × 20 cm²

= 1758.4 cm²

Area of the bottom = πr₂²

= (314/100) × 8 × 8 cm²

= 200.96 cm²

Therefore, Total area of metal required = CSA + area of bottom

= 1758.4 cm² + 200.96 cm²

= 1959.36 cm²

Cost of metal required = ₹ (8/100) × 1959.36 cm²

= ₹ 156.75

5. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

On a 7 cm × 7 cm base of a cubical block, we can mount a hemisphere having the greatest diameter equal to 7 cm.

Here, the radius of the hemisphere = 3.5 cm.

Now, the surface area of the solid made in the figure is:

The total surface area of the cube + The curved surface area of the hemisphere - The area of the base of the hemisphere.

= {6 × (7)² + 2π × (3.5)² - π × (3.5)²} cm² (Since the part of the top of the cubical part which is covered by the hemisphere is not visible outside)

= {6 × 49 + (22/7) × (35/10) × (35/10)} cm²

= {294 + 11 × (35/10)} cm²

= 332.5 cm²

7.0Benefits of Introduction to Surface Areas and Volumes

The Surface Areas and Volumes chapter in Class 10 Maths offers several key benefits:

1. Real-World Applications: It helps in calculating the surface area and volume of 3D objects, which is essential in fields like engineering, architecture, and manufacturing.
2. Practical Problem-Solving: Students learn to solve practical problems related to packaging, construction, and design.
3. Visualizing 3D Objects: The chapter improves spatial awareness and the ability to visualize and manipulate 3D shapes.
4. Foundation for Higher Studies: It provides a foundation for more advanced topics in geometry and physics, particularly in calculus and mechanics.