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

Edited By Ramraj Saini | Updated on Nov 27, 2023 05:35 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.1

NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 Surface Area and Volumes are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 Maths ex 13.1 presents us with the surface area of 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.

In exercise 13.1 Class 10 Maths we have to find the surface area of the combined solids .Some of these solids are cylinders with both sides hemisphere , cone with hemisphere on top. Surface area can be categorized into two types of surface areas, the curved surface area and total surface area. 10th class Maths exercise 13.1 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. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths chapter 13 exercise 13.1

Surface Areas and Volumes Class 10 Maths Chapter 13 Exercise: 13.1

We are given that volume of the cube $=\ 64\ cm^3$

Also, the volume of a cube is given by $=\ a^3$ ( here $a$ is the edge of the cube)

Thus : $a^3\ =\ 64$

$a\ =\ 4\ cm$

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

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

or $=\ 2\left ( 8\times 4\ +4\times 4\ +\ 4\times 8 \right )$

or $=\ 2\left ( 80 \right )$

or $=\ 160\ cm^2$

Thus the area of the formed cuboid is 160 cm 2 .

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

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

The inner surface area of the hemisphere is :

$=\ 2\pi r^2$

or $=\ 2\times \left ( \frac{22}{7} \right ) \times 7^2$

or $=\ 308\ cm^2$

And the surface area of the cylinder is :

$=\ 2\pi rh$

or $=\ 2\times \frac{22}{7}\times 7\times 6$

or $=\ 264\ cm^2$

Thus the inner surface area of the vessel is $= 308 + 264 = 572 \:cm^2.$

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 $=\ 2 \pi r^2$

or $=\ 2 \times \frac{22}{7}\times \left ( 3.5 \right )^2$

or $=\ 77\ cm^2$

Now for cone we have :

The surface area of a cone $=\ \pi rl$

Thus we need to calculate the slant of the cone.

We know that :

$l^2\ =\ h^2\ +\ r^2$

or $=\ 12^2\ +\ 3.5^2$

or $=\ \frac{625}{4}$

or $l\ =\ \frac{25}{2}\ =\ 12.5\ cm$

Thus surface area of a cone $=\ \pi rl$

or $=\ \frac{22}{7}\times 3.5\times 12.5$

or $=\ 137.5\ cm^2$

Hence the total surface area of toy = $=\ 77\ +\ 137.5\ =\ 214.5\ cm^2$

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

The surface area of the cube is :

$=\ 6a^3$

$=\ 6\times 7^3\ =\ 294\ cm^2$

Now the area of a hemisphere is

$=\ 2\pi r^2$

$=\ 2\times \frac{22}{7}\times \left ( \frac{7}{2} \right )^2\ =\ 77\ cm^2$

And the area of the base of a hemisphere is

$=\ \pi r^2\ =\ \frac{22}{7}\times \left ( \frac{7}{2} \right )^2\ =\ 38.5\ cm^2$

Hence the surface area of solid is $= 294 + 77 - 38.5 = 332.5 \:cm^2$ .

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

The total surface area of solid is given by :

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

The surface area of the cube $=\ 6l^2$

And surface area of the hemisphere:

$=\ 2\pi r^2\ =\ 2\pi \left ( \frac{l}{2} \right )^2$

Area of base of the hemisphere:

$=\ \pi r^2\ =\ \pi \left ( \frac{l}{2} \right )^2$

Thus the area of solid is:

$=\ 6l^2\ +\ \pi \left ( \frac{l}{2} \right )^2\ unit^2$

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

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

Area of hemisphere $=\ 2\pi r^2$

or $=\ 2\pi\times \left ( \frac{5}{2} \right )^2$

or $=\ \frac{25}{2} \pi\ mm^2$

And the area of the cylinder $=\ 2\pi rh$

or $=\ 2\pi \times \frac{5}{2}\times 9$

or $=\ 45 \pi\ mm^2$

Thus the area of the solid is :

$=\ 2\left ( \frac{25}{2} \right )\pi\ +\ 45 \pi$

$=\ 25\pi\ +\ 45 \pi$

$=\ 70 \pi$

$=\ 220\ mm^2$

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 $=\ 2\pi rh$

or $=\ 2\pi \times 2\times 2.1$

or $=\ 8.4 \pi\ m^2$

And the area of the cone is $=\ \pi rl$

or $=\ \pi \times 2\times 2.8$

or $=\ 5.6\pi\ m^2$

Thus, the area of the canvas $=\ 8.4\pi\ +\ 5.6\pi$

or $=\ 14\pi\ =\ 44\ m^2$

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

Thus the required money is $=\ 500\times 44\ =\ Rs.\ 22,000$

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

$l^2\ =\ r^2\ +\ h^2$

or $=\ (0.7)^2\ +\ (2.4)^2$

or $l^2\ =\ 6.25$

or $l\ =\ 2.5\ cm$

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

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

The surface area of the cylinder is $=\ 2 \pi rh$

or $=\ 2 \pi \times 0.7\times 2.4$

or $=\ 10.56\ cm^2$

Now, the surface area of a cone $=\ \pi rl$

or $=\ \pi \times 0.7\times 2.5$

or $=\ 5.50\ cm^2$

And the area of the base of the cylinder is $=\ \pi r^2$

or $=\ \pi \times 0.7\times 0.7$

or $=\ 1.54\ cm^2$

Thus required area of solid = 10. 56 + 5.50 + 1.54 = 17.60 cm 2 .

Thus total surface area of remaining solid to nearest cm 2 is 18 cm 2 .

The required surface area is given by :

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

Now, the area of the cylinder $=\ 2\pi rh$

or $=\ 2\pi \times 3.5\times 10$

or $=\ 70\pi \ cm^2$

And the surface area of the hemisphere : $=\ 2\pi r^2$

or $=\ 2\pi \times 3.5\times 3.5$

or $=\ 24.5\pi\ cm^2$

Thus the required area $=\ 70\pi\ +\ 2(24.5\pi)\ =\ 374\ cm^2$

In NCERT syllabus for Class 10 Maths chapter 13 exercise 13.1 we have covered that the surface of a solid object ismeasurement of the overall area that the object's surface takes up and the curved surface of an object is that includes the area of all the curved surfaces and not the flat surfaces exampl-cone without the base.

Also Read| Surface Areas and Volumes Class 10 Notes

## Benefits of NCERT Solutions for Class 10 Maths Exercise 13.1

• Class 10 Maths chapter 13 exercise 13.1 will be helpful in further exercise of the chapter 10 surface area and volume
• Class 10 Maths chapter 13 exercise 13.1 covers the knowledge which we can use in our real life
• Surface area and volume are an important part of mechanical , mechtronic , tool , physics engineering
• We can also use Class 10 Maths chapter 13 exercise 13.1 as base for the cellular changes in the volume of cell

## Key Features For Class 10 Maths Chapter 13 Exercise 13.1

• Step-by-Step Solutions: Clear and detailed explanations for each problem in Ex 13.1 class 10, ensuring students understand the methodology.
• Conceptual Understanding: Explanation of area and volume concepts involved in the exercises, aiding students in grasping the underlying principles are discussed in class 10 maths ex 13.1.
• Variety of Problems: In class 10 ex 13.1, diverse types of problems, offering a broad spectrum of challenges to reinforce understanding.
• Visual Aids, Diagrams, and Examples: Use of visual representations, diagrams, and real-life examples to simplify complex concepts and aid visual learners.
• Application-Based Problems: In 10th class maths exercise 13.1 answers, application-oriented problems that relate statistics to real-world scenarios, fostering practical understanding and problem-solving skills.

Also See:

## NCERT Solutions Subject Wise

1. What is the curved surface area of the hemisphere ?

Curved surface area of hemisphere used is =2πr²

2. What is the curved surface area of the cylinder u?

Curved surface area of hemisphere used in  2πrl

3. What is the most commonly used value of pi used in Exercise 13.1 Class 10 Maths ?

Most commonly used value of pi used in exercise 13.1 Class 10 Maths is 22/7

4. How can we use surface area in real life? Give some examples ?

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

5. What is the total number of example that have been solved before theExercise 13.1 Class 10 Maths.

The total number of example  that have been solved before the class 10 Maths chapter 13 exercise 13.1 is four

6. What is the total number of questions in the NCERT solutions for Class 10 Maths chapter 13 exercise?

The total number of questions in the NCERT solutions for Class 10 Maths chapter 13 exercise is nine.

7. What kind of questions are asked in 10 Maths chapter 13 exercise 13.1?

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

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### Questions related to CBSE Class 10th

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Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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