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# RD Sharma Class 10 Solutions Chapter 16 - Exercise 16.3 Last updated date: 07th Dec 2023
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## RD Sharma Class 10 Solutions Chapter 16 - Surface Areas and Volumes (Ex 16.3) Exercise 16.3 - Free PDF

Free PDF download of RD Sharma Class 10 Solutions Chapter 16 - Surface Areas and Volumes Exercise 16.3 solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 16 - Surface Areas and Volumes Ex 16.3 Questions with Solutions for RD Sharma to help you to revise the complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced) and other engineering entrance exams.

Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Register Online for Class 10 Science tuition on Vedantu.com to score more marks in the CBSE board examination. Vedantu.com is the No.1 Online Tutoring Company in India Provides you Free PDF download of NCERT Solutions for Class 10 Maths solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter wise Questions with Solutions to help you to revise the complete Syllabus and Score More marks in your examinations.

## RD Sharma Class 10 Solutions Chapter 16

Surface Areas and Volumes, chapter 16 of Class 10 provides an introduction of Area (region of space occupied by the surface of an object) and Volume (amount of space that is available in an object). We see different objects of various shapes and sizes around us such as cube, cuboid, cone, sphere, cylinder etc. All these objects are three-dimensional figures and occupy an area and have a definite volume which can be calculated with the help of different mathematical expressions that are covered in this chapter. Further in the chapter students will get to know about different types of area (Total surface area and Curved/Lateral surface area) and Volume which comes in handy in finding the dimensions of different objects that we encounter in our day to day life.

The solution for the exercises of this chapter is given in a very simple, concise and step by step method making it easier for the student to understand the problem better. All the formulas are available in the book for the students to memorise along with their easy and effective solutions. After getting introduced to the chapter the students may take a look at the solved questions and then proceed further to solve the questions given in their respective textbooks. This approach will make the concepts crystal clear in the mind of the students and will help them ace the exam. RD Sharma provides a better glance of the problem and provides an accurate easy to understand solution with detailed illustrations adding to the efficiency of the student in solving the problem in a relatively lesser amount of time.

### RD Sharma Class 10 Chapter 16 Exercises

1. Chapter 16 Exercise 16.1

2. Chapter 16 Exercise 16.2

3. Chapter 16 Exercise 16.3

### Solution for Exercise 16.3

After introducing the student to various formulae and notations that will be used, the chapter advances towards the questions along with their solutions. Area and volume of different objects are to be calculated, details like diameter and radius of the object are given and based on this available information necessary dimensions are calculated to know the answer. The area and volume of a cone are to be calculated in different questions, easy to understand and concise solutions are given below each question for better understanding. All the solutions provide a deeper understanding of the topic to the students with illustrations and relevant data. Students can use these solutions to overcome the fear of mathematics and the solutions have been organised in such a way that it enables them to discover easy methods to solve the problems. These solutions will help the students in attaining a better command of mathematics and also enhance their problem-solving skills. Students can through the RD Sharma book exercises for improving their performance in the examination.

## FAQs on RD Sharma Class 10 Solutions Chapter 16 - Exercise 16.3

1. What is the area of an object according to RD Sharma Class 10 solutions?

According to RD Sharma class 10 solutions, The area of an object can be described as the space consumed by the flat surface of an object. The area of an object is the total number of unit squares that cover the total surface of a closed figure. It is measured in square units, for example - Square centimetres, square inches, square feet etc.

2. About class 10 chapter 16 solutions of RD Sharma, What is the volume of an object?

Referring to RD Sharma class 10 chapter 16 solutions, the Volume of an object can be defined as the 3-dimensional space enclosed within a boundary or occupied by an object. In simple words, it is the space that a 3D occupies or contains. It is measured in cubic units, for example - cubic meter, cubic centimetres, cubic inch, cubic foot etc.

3. Where can I find useful study material for RD Sharma class 10 maths solutions?

You can find useful study resources on the Vedantu website or app. The content is designed by top-notch industry experts and professionals and is very accurate and reliable. Students can find important questions, notes, revision material and a lot freer of cost! All you have to do is sign in and you will be able to download these resources in pdf format.

4. How to get better in Class 10 maths while referring to RD Sharma solutions?

As well all know that Math is a subject that demands more and more practice, the more you practice the more you get better at solving math problems quickly. Books like RD Sharma which provide a simple and easy to understand solution can prove to be very helpful. Students need to solve as many questions as possible from this book to perform well in exams because it covers every topic that is asked in the exam.

5. How to find the curved surface area of the frustum of a cone that is described in RD Sharma class 10 chapter 16 solutions?

While taking the reference with class 10 RD Sharma, the Curved surface area of the frustum of a cone can be calculated using this formula - π(r + r)L where,

r = ratio of the top

r = ratio of the bottom

L = slant height of the frustum of a cone

By putting values in the above formula you will get the curved surface area of the frustum of the cone.