RD Sharma Class 9 Solutions Chapter 21 - Surface Area and Volume of Sphere (Ex 21.1) Exercise 21.1 - Free PDF
Free PDF download of RD Sharma Class 9 Solutions Chapter 21 - Surface Area and Volume of Sphere Exercise 21.1 solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 21 - Surface Area and Volume of Sphere Ex 21.1 Questions with Solutions for RD Sharma Class 9 Maths to help you to revise complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced) and other Engineering entrance exams. You can also register Online for Class 9 Science tuition on Vedantu.com to score more marks in your examination. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students.
FAQs on RD Sharma Class 9 Solutions Chapter 21 - Surface Area and Volume of Sphere (Ex 21.1) Exercise 21.1
1. What are the key features of RD Sharma Solutions for Class 9 Maths Chapter 21?
Students can find solutions to all problems correctly based on their intelligence quotient with RD Sharma Solutions for Class 9 Maths Chapter 21. The solutions offer different approaches to solving tricky questions systematically and answer students’ doubts instantly while solving. In analyzing their readiness level, the students can work on the areas in which they need more practice. The textbook contains various sets of questions that will help the students to practice better. Answers to such questions are given in a detailed format so that students can grasp them easily.
2. What is the meaning of the volume of Sphere according to RD Sharma Solutions for Class 9 Maths Chapter 21?
The volume of the sphere is the total area occupied by a sphere. A sphere can be formed by taking a circular disc, pasting a string along its circumference, and then rotating it along the string. The sphere has a volume of three-dimensional radius. It contains three axes, namely, x-axis, y-axis, and z-axis, which define its shape. For examples football and basketball are spheres that have volume. The size of the sphere has to do with its radius or diameter since if we look at its cross-section, it is a circle.
3. How are RD Sharma Solutions for Class 9 Maths Chapter 21 helpful from an exam perspective?
The chapter surface area and volume of the sphere is somewhat difficult than other chapters, thus this requires better understanding and continuous practice. A diligent approach to preparing these solutions helps students ace the final exams and top their class. RD Sharma Solutions for Class 9 Maths Chapter 21 is the most popular learning material used by students. These solutions are designed according to the most up-to-date RD Sharma Solutions for Class 9 Maths Syllabus, covering all the important topics of the respective subject.
4. What is the Surface Area of Sphere according to chapter 21 of Class 9 Maths? Why is the Surface Area of a Sphere 4 Times the Area of a Circle?
In mathematics, the surface area of a sphere is equal to the total area of all its faces. It is always presented in square units. A sphere's surface area depends on its radius and diameter. It is expressed mathematically as 4πr2 square units.
You can check whether the area of a sphere is four times more than that of a circle by wrapping a string around four circles whose surfaces are completely covered by a string. When we talk about the surface area of a sphere, we can write the surface area of a sphere in the following way = 4πr2 = 4(πr2) = 4 × area of a circle.
5. What are the benefits of Referring to Vedantu Solutions for Exercise 1 of Chapter 21?
The solutions provided by Vedantu are given by subject-matter experts of mathematics, thus the solutions given are accurate and explained step-by-step, to help students understand easily. The answers also include shortcuts and tricks wherever possible, thus saving the students time and effort during the exam. The other benefit of referring to the solutions is that you can also download them for future references and can have access to it from anywhere.