NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 - Surface Area and Volumes

NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 - Surface Area and Volumes

Updated on 30 Apr 2025, 05:13 PM IST

Suppose a person has a cuboid and you want to paint this cuboid from the outer, and you want to know how much is required to paint this cuboid. For this, the total area of the outer layer should be calculated, and this area is called the surface area of any object. Therefore, surface area is the total outer space of any three-dimensional object. After painting this cuboid, if a person wants to fill this cuboid with water, then how much water is needed to fill this cuboid? To determine the answer to this question person should know the amount of space within this cuboid, and this amount of space is called the volume of the cuboid. Therefore, the amount of space within any three-dimensional object is called the volume.

This Story also Contains

  1. Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1
  2. Assess NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1
  3. Topics Covered in Chapter 12, Surface Areas and Volumes: Exercise 12.1
  4. NCERT Solutions for Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 - Surface Area and Volumes
exercise 13.1

In the NCERT Books exercise 12.1, Class 10 Maths, we have to find the surface area of the combined solids. Some of these solids are cylinders with both sides hemispheres, and cones with a hemisphere on top. Surface area can be categorised into two types of surface areas: the curved surface area and the total surface area. 10th class Maths exercise 12.1 answers are designed as per the students' demand, covering comprehensive, step-by-step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in-depth understanding of concepts. These NCERT solutions Class 10 Maths ex 12.1 present us with the surface area of a combination of Solids. Some solids, such as the cuboid, cone, cylinder, and sphere, are recognisable to you from Class IX. You've also learned how to calculate their volumes and surface areas. In our daily lives, we come across a variety of solids that are made up of two or more of the fundamental solids.

Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

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Assess NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

Q1. 2 cubes, each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Answer:

We are given that volume of the cube = 64 cm3

Also, the volume of a cube is given by = a3 ( here a is the edge of the cube)

Thus: a3 = 64

a = 4 cm

Now, according to the question, we have combined the two cubes so that the edge lengths of the formed cuboid are 4 cm, 4 cm, and 8 cm.

The surface area of a cuboid is : = 2(lb + bh + hl)

or = 2(8×4 +4×4 + 4×8)

or = 2(80)

or = 160 cm2

Thus, the area of the formed cuboid is 160 cm².

Q2 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.

Answer:

Since the vessel consists of a hemisphere and a cylinder, its area is given by :

Area of vessel = Inner area of the cylinder(curved) + Inner area of the hemisphere

The inner surface area of the hemisphere is :

= 2πr2

or = 2×(227)×72

or = 308 cm2

And the surface area of the cylinder is :

= 2πrh

or = 2×227×7×6

or = 264 cm2

Thus, the inner surface area of the vessel is =308+264=572cm2.

Q3. 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.

Answer:

The required surface area of the toy is given by :

Area of toy = Surface area of hemisphere + Surface area of the cone

Firstly, consider the hemisphere:

The surface area of a hemisphere is = 2πr2

or = 2×227×(3.5)2

or = 77 cm2

Now, for the cone we have:

The surface area of a cone = πrl

Thus, we need to calculate the slant of the cone.

We know that:

l2 = h2 + r2

or = 122 + 3.52

or = 6254

or l = 252 = 12.5 cm

Thus surface area of a cone = πrl

or = 227×3.5×12.5

or = 137.5 cm2

Hence, the total surface area of toy = = 77 + 137.5 = 214.5 cm2

Q4. 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.

Answer:

It is given that the hemisphere is mounted on the cuboid, thus, the hemisphere can take on complete as its diameter (which is maximum).

Thus, the greatest diameter of the hemisphere is 7 cm.

Now, the total surface area of the solid = Surface area of the cube + Surface area of the hemisphere - Area of the base of the hemisphere (as this is counted on one side of the cube)

The surface area of the cube is :

= 6a3

= 6×73 = 294 cm2

Now the area of a hemisphere is

= 2πr2

= 2×227×(72)2 = 77 cm2

And the area of the base of a hemisphere is

= πr2 = 227×(72)2 = 38.5 cm2

Hence, the surface area of the solid is =294+7738.5=332.5cm2.

Q5. 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.

Answer:

It is given that the diameter of the hemisphere is equal to the edge length of the cube.

The total surface area of the solid is given by :

The surface area of a solid = Surface area of the cube + Surface area of the hemisphere - Area of the base of the hemisphere

The surface area of the cube = 6l2

And the surface area of the hemisphere:

= 2πr2 = 2π(l2)2

Area of the base of the hemisphere:

= πr2 = π(l2)2

Thus, the area of the solid is:

= 6l2 + π(l2)2 unit2

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig.). The length of the entire capsule is 14 mm, and the diameter of the capsule is 5 mm. Find its surface area.

1636091994484

Answer:

It is clear from the figure that the capsule has a hemisphere and cylinder structure.

The surface area of the capsule = 2 (Area of the hemisphere) + Area of the cylindrical part

Area of hemisphere = 2πr2

or = 2π×(52)2

or = 252π mm2

And the area of the cylinder = 2πrh

or = 2π×52×9

or = 45π mm2

Thus, the area of the solid is :

= 2(252)π + 45π

= 25π + 45π

= 70π

= 220 mm2

Q7. 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 Rs 500 per m 2. (Note that the base of the tent will not be covered with canvas.)

Answer:

The canvas will cover the cylindrical part as well as the conical part.

So, the area of canvas = Area of cylindrical part (curved) + Area of the conical part

Now, the area of the cylindrical part is = 2πrh

or = 2π×2×2.1

or = 8.4π m2

And the area of the cone is = πrl

or = π×2×2.8

or = 5.6π m2

Thus, the area of the canvas = 8.4π + 5.6π

or = 14π = 44 m2

Further, it is given that the rate of canvas per m 2 is = Rs. 500.

Thus, the required money is = 500×44 = Rs. 22,000

Q8. 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 cm 2.

Answer:

Firstly, we need to calculate the slant height of the cone :

l2 = r2 + h2

or = (0.7)2 + (2.4)2

or l2 = 6.25

or l = 2.5 cm

Now, the total surface area of the solid can be calculated as :

The surface area of a solid = Surface area of cylinder + Surface area of cone + Area of base of the cylinder

The surface area of the cylinder is = 2πrh

or = 2π×0.7×2.4

or = 10.56 cm2

Now, the surface area of a cone = πrl

or = π×0.7×2.5

or = 5.50 cm2

And the area of the base of the cylinder is = πr2

or = π×0.7×0.7

or = 1.54 cm2

Thus required area of the solid = 10. 56 + 5.50 + 1.54 = 17.60 cm2 .

Thus total surface area of the remaining solid to the nearest cm2 is 18 cm2.

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. 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.

2%20(1)

Answer:

The required surface area is given by :

The surface area of the article = Surface area of the cylindrical part + 2 (Surface area of the hemisphere)

Now, the area of the cylinder = 2πrh

or = 2π×3.5×10

or = 70π cm2

And the surface area of the hemisphere: = 2πr2

or = 2π×3.5×3.5

or = 24.5π cm2

Thus, the required area = 70π + 2(24.5π) = 374 cm2



Also Read

Topics Covered in Chapter 12, Surface Areas and Volumes: Exercise 12.1

  • Surface Area: The total outer area covered by any object is called the surface area of an object. The surface area includes: curved surface area, lateral surface area, and total surface area.
  • Curved Surface Area: The total outer curved area covered by an object is called the curved surface area, and it does not include any flat or base surface of an object.
  • Lateral Surface Area: The outer area covered by an object, without including the top and base, is called the lateral surface area of an object
  • Total Surface Area: The outer covered area of an object, including the top or curved area and the bottom of the object, is called the total surface area.
  • Volume: The space occupied within the boundaries of an object is called the volume of an object.
  • Surface Area of Combination Object: When two or more solid objects are combined, then finding the surface area of the combination of these objects. For example, combining a cone and a cylinder.

Also See:

NCERT Exemplar Solutions of Class 10 Subject Wise

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

Frequently Asked Questions (FAQs)

Q: What is the most commonly used value of pi used in Exercise 12.1 Class 10 Maths ?
A:

The most commonly used value of pi used in exercise 12.1 Class 10 Maths is 227.

Q: What kind of questions are asked in 10 Maths chapter 12 exercise 12.1?
A:

There are many basic question in which we have to find the total surface of combination of solid or real life objects such as tents , capsules toys etc but some are direct that only ask about the solid  

Q: What is the total number of example that have been solved before the Exercise 12.1 Class 10 Maths.
A:

The total number of examples that have been solved before the class 10 Maths chapter 12 exercise 12.1 is four 

Q: What is the total number of questions in the NCERT solutions for Class 10 Maths chapter 12 exercise?
A:

The total number of questions in the NCERT solutions for Class 10 Maths chapter 12 exercise is nine.

Q: What is the curved surface area of the hemisphere ?
A:

Curved surface area of hemisphere used is =2πr²

Q: What is the curved surface area of the cylinder u?
A:

Curved surface area of hemisphere used in  2πrl

Q: How can we use surface area in real life? Give some examples ?
A:

We can use surface area in many places in our daily life, for example identifying the amount of paint require to paint object , determining the cost of things like clothes , covers and canvas tents 

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