NCERT Solutions Class 10 Maths Chapter 12 Surface Areas And Volumes Exercise 12.1

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.1 will help students understand the basic concepts of three-dimensional figures and their combinations. The exercise, in particular, introduces the basic ideas and formulas for the surface areas of different three-dimensional figures and their composites. These formulas are crucial not only for solving these questions but also for analyzing and addressing various practical problems in different fields of study. So, let’s dive deeper into the insights of this important topic via solutions to exercise 12.1. 

1.0Download NCERT Solutions Class 10 Maths Chapter 12 Surface Areas And Volumes Exercise 12.1 : Free PDF 

2.0Introduction to Surface Areas: 

The surface area of a three-dimensional (3D) shape is the sum of the area taken up by the outer surface of the object. It is the total of the areas of all the faces or curved surfaces which constitute the 3D shapes like cylinders, cones, cuboids, and spheres. Surface area refers to the total space covered by the outer surface of an object, which is useful in practical applications such as painting, wrapping, or construction. The surface areas can be further divided into two types, which include: 

• Total Surface Area of Solids: Total surface area refers to the area of all sides of a solid, including its top and bottom. 
• Curved Surface Area of Solids: It refers to the area of the sides of a solid, excluding the top and bottom of the solid. This concept is useful in calculating the surface area of hollow objects with basic 3D shapes. 

3.0Key Concepts of Exercise 12.1: Overview 

Exercise 12.1 is all about utilising the formulas and key concepts of surface areas of 3D objects to solve various questions. So, let’s explore some basic ideas and formulas used in the exercise:

Formulas for Surface Areas of Various Solids: 

Before solving the questions in this exercise, let's review some important formulas for calculating the surface area of various 3D figures.

Surface Area of a Combination of Solids: 

All the questions in this exercise involve combinations of solids. Calculating their surface area can be challenging. But you can easily do the computation by following these steps for the area:

Step 1: Decompose the composite solids into simple individual solids with the formula of known surface area.

Step 2: Compute the surface area of these individual solids by using the above formulas according to the solid to be computed.

Step 3: After calculating, simply add or subtract the total surface area of all the solids according to the requirement of the question. 

Understanding how to calculate the surface areas of combined solids will help you in exams and provide better insight into geometric structures commonly found in daily life. To excel in this topic, begin practicing now from the NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.1.

Also Read CBSE Notes Class 10 Maths Chapter 12 Surface Areas And Volumes

4.0NCERT Class 10 Maths Chapter 12 Surface Areas And Volumes Exercise 12.1 : Detailed Solutions

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 toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

Let r and h be the radius of the cone, hemisphere and height of the cone.

Therefore, h = (15.5 - 3.5) cm = 12.0 cm

Also l² = h² + r²

= 12² + (3.5)²

= 156.25

Therefore, l = 12.5 cm

Curved surface area of the conical part = πrl

Curved surface area of the hemispherical part = 2πr²

Total surface area of the toy = πrl + 2πr²

= πr(l + 2r)

= (22/7) × (35/10) (12.5 + 2 × 3.5) cm²

= 11 × (12.5 + 7) cm² = 11 × 19.5 cm²

= 214.5 cm²

4. 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

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter ℓ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:

Let ℓ be the side of the cube.

Therefore, the greatest diameter of the hemisphere = l

=> Radius of the hemisphere = l/2

Therefore, surface area of hemisphere = 2πr²

= 2 × π × (l/2) × (l/2) = πl²/2

Base area of the hemisphere = π(l/2)² = πl²/4

Surface area of the cube = 6 × l² = 6l²

Therefore, surface area of the remaining solid = 6l² + πl²/2 - πl²/4

= (24l² + 2πl² - πl²) / 4 = (24l² + πl²) / 4

= (l²/4)(24 + π) sq. units.

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

Surface area of the cylindrical part =2π × r × h

= 2π × (5/2) × 9 mm² = 45π mm²

Sum of the curved surface areas of two hemispherical parts.

= 2 {2π × (5/2)²} mm² = 25π mm²

Total surface area of the capsule

= 45π + 25π mm² = 70π mm²

= 70 × (22/7) mm² = 220 mm²

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also find the cost of the canvas of the tent at the rate of ₹ 500 per m² (Note that the base of the tent will not be covered with canvas).

Solution:

Radius of the cylindrical base = 2 m and height = 2.1 m. The curved surface area of the cylindrical part = 2π × (2) × (2.1) m² (i.e., 2πrh) = 4 × (22/7) × 2.1 m² = 26.4 m².

Now, for the conical part, we have r = 2 m and l (slant height) = 2.8 m. The curved surface area of the conical part = πrl = (22/7) × 2 × 2.8 m² = 17.6 m².

Then the area of the canvas = 26.4 m² + 17.6 m² = 44 m².

Total cost of the canvas at the rate of ₹ 500 per m² = ₹ 500 × 44 = ₹ 22000.

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Solution :

For the cylinder part, height = 2.4 cm and diameter = 1.4 cm. Therefore, radius (r) = 0.7 cm.

Total surface area of the cylindrical part = 2 × (22/7) × (7/10) × [2.4 + 0.7] cm² = (44/10) × 3.1 cm² = (44 × 31)/100 cm² = 1364/100 cm².

For conical part,

Base radius (r) = 0.7 cm

and height (h) = 2.4 cm

Therefore, Slant height (l) = √(r² + h²)

= √((0.7)² + (2.4)²)

= √(0.49 + 5.76) = √6.25 = 2.5 cm

Therefore, Curved surface area of the conical part

= πrl = (22/7) × 0.7 × 2.5 cm² = 550/100 cm²

Base area of the conical part

πr² = (22/7) × (7/10)² cm²

= (22 × 7)/100 cm² = 154/100 cm²

Total surface area of the remaining solid

= [(Total surface area of cylindrical part) + (Curved surface area of conical part) - (Base area of the conical part)]

= (1364/100 + 550/100 - 154/100) cm² = 1760/100 cm²

= 17.6 cm².

Hence, total surface area to the nearest cm² is 18 cm².

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:

Radius of the cylinder (r) = 3.5 cm

Height of the cylinder (h) = 10 cm

Therefore, Curved surface area of cylinder = 2πrh

= 2 × (22/7) × (35/10) × 10 cm²

= 220 cm²

Curved surface area of a hemisphere = 2πr²

Therefore, Curved surface area of both hemispheres = 2 × 2πr² = 4πr²

= 4 × (22/7) × (35/10) × (35/10) cm²

= 154 cm²

Total surface area of the remaining solid

= (Curved surface area of cylinder + curved surface area of 2 hemispheres)

= (220 + 154) cm² = 374 cm².

5.0Benefits of NCERT Solutions Class 10 Maths Chapter 12 Surface Areas And Volumes Exercise 12.1

• Helps students understand the fundamental concepts of surface areas and volumes with detailed explanations.
• Provides structured, well-explained answers to problems, making it easier to follow and learn.
• Strengthens analytical thinking and logical reasoning through different types of problems.
• Covers important questions that are likely to appear in board exams, boosting confidence.
• Helps students learn and apply various formulas for different 3D shapes like cubes, cones, spheres, and cylinders.