Download Free PDF of RD Sharma Class 10 Solutions Chapter 16 - Surface Areas and Volumes (Ex 16.1) Exercise 16.1
FAQs on RD Sharma Class 10 Solutions Chapter 16 - Exercise 16.1
1. What types of problems are covered in RD Sharma Class 10 Maths Solutions for Chapter 16, Exercise 16.1?
This exercise primarily focuses on two key concepts from the CBSE syllabus: calculating the surface area and volume of combined solids (e.g., a toy that is a cone mounted on a hemisphere) and solving problems that involve the conversion of a solid from one shape to another (e.g., a sphere melted down and recast into a wire). The solutions provide detailed, step-by-step methods for tackling these multi-shape problems accurately.
2. How do the RD Sharma solutions for Surface Areas and Volumes help in preparing for the board exams?
RD Sharma solutions for Chapter 16 offer a more extensive collection of problems compared to the NCERT textbook. While NCERT builds the foundation, RD Sharma includes a wider variety of Higher Order Thinking Skills (HOTS) questions and more complex combinations of solids. Practising these solutions helps students master the application of formulas and develop the confidence needed to solve any type of question in the board exams.
3. What is the fundamental principle for solving problems about the conversion of a solid from one shape to another?
The core principle is the conservation of volume. When a solid object is melted, reshaped, or recast into a new form, its overall dimensions and surface area will change, but the amount of material does not. This means the volume of the original solid is always equal to the volume of the new solid. Every solution for such problems is based on equating the volume formulas for the original and final shapes.
4. When calculating the total surface area of a combined solid, why do we subtract the area of the joined faces?
When two solid shapes are joined to form a new object, the surfaces where they connect are no longer part of the exterior. For example, if a hemisphere is placed on top of a cube, the circular base of the hemisphere covers a part of the cube's top face. To find the Total Surface Area (TSA) of the new composite shape, you must calculate the area of all exposed surfaces only. This involves adding the required surface areas of the individual solids and subtracting the area of the common base that is now hidden.
5. What is a common mistake to avoid when solving problems from Exercise 16.1?
A frequent error is confusing Total Surface Area (TSA) with Curved Surface Area (CSA) for combined solids. Students often mistakenly add the full TSAs of the individual shapes. The correct method is to identify and sum up only the exposed surfaces. For a tent shaped like a cone on a cylinder, you must add the CSA of the cone and the CSA of the cylinder, plus the area of the circular base of the cylinder, not their individual total surface areas.
6. How is understanding Surface Areas and Volumes from Chapter 16 applicable in real-world scenarios?
The concepts from this chapter have direct practical applications in various fields. For example:
- Engineering and Manufacturing: Designing components like capsules, boilers, or storage tanks which are often combinations of cylinders and hemispheres to maximise capacity and minimise material cost.
- Architecture: Calculating the amount of paint or plaster required for buildings with complex structures like pillars, domes, and turrets.
- Packaging: Creating efficient and cost-effective packaging by combining shapes like cuboids and cylinders.
7. What are the essential formulas needed to solve the problems in RD Sharma Chapter 16, Exercise 16.1?
To successfully solve problems in this exercise, a student must be fluent with the surface area and volume formulas for basic 3D solids. The most critical ones are:
- Cylinder: CSA = 2πrh; Volume = πr²h
- Cone: CSA = πrl; Volume = (1/3)πr²h
- Sphere: Surface Area = 4πr²; Volume = (4/3)πr³
- Hemisphere: CSA = 2πr²; TSA = 3πr²; Volume = (2/3)πr³

















