NCERT Solutions Class 10 Maths Chapter 13 helps you calculate surface areas and volumes of combined solids like cones, cylinders, spheres, and hemispheres. You’ll learn how to break complex 3D shapes into simpler parts, apply the right formulas for each component, and solve real-world problems involving frustums and composite figures. These solutions show you exactly which formula to use when, with shortcuts to avoid calculation errors in board exams.
Download Complete Chapter 13 Solutions PDF
All exercises with step-by-step solutions | Updated 2025-26 | Free Download
Download PDF (Free)NCERT Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes – Complete Guide
NCERT Class 10 Chapter 13 – Surface Areas and Volumes takes you beyond basic mensuration into the fascinating world of three-dimensional geometry. You’ll explore how to calculate the surface areas and volumes of various solid figures like cylinders, cones, spheres, and hemispheres, and more importantly, learn to handle combinations of these shapes that appear in real-life scenarios. This chapter carries 5 marks in your CBSE board exam and typically includes one long answer question worth 3-4 marks and additional marks in case study or MCQ sections.
📊 CBSE Class 10 Maths Chapter 13 – Exam Weightage & Marking Scheme
| CBSE Board Marks | 5 Marks |
| Unit Name | Mensuration |
| Difficulty Level | Medium |
| Importance | Medium |
| Exam Types | CBSE Board, State Boards |
| Typical Questions | 1-2 questions |
You will learn essential formulas for curved surface area, total surface area, and volume of each solid shape, along with the skill to apply them in complex situations. The chapter emphasizes practical applications such as calculating the capacity of water tanks, material required for manufacturing objects, and conversion of one shape into another while maintaining volume. You’ll also master problems involving frustums of cones, which have important real-world applications in architecture and engineering.
This chapter builds upon your knowledge from Class 9 mensuration and connects directly to coordinate geometry and trigonometry through three-dimensional visualization. The problems range from straightforward formula-based questions (2-3 marks) to challenging combination problems (4-5 marks) that test your analytical thinking. Understanding the relationship between radius, height, and slant height in different solids is particularly important for solving complex problems efficiently.
Quick Facts – Class 10 Chapter 13
| 📖 Chapter Number | Chapter 13 |
| 📚 Chapter Name | Surface Areas and Volumes |
| ✏️ Total Exercises | 3 Exercises |
| ❓ Total Questions | 22 Questions |
| 📅 Updated For | CBSE Session 2025-26 |
With dedicated practice of NCERT solutions and previous year CBSE questions, you can confidently secure full marks from this chapter. Focus on drawing clear diagrams, writing proper formulas, and showing step-by-step calculations to maximize your score in board examinations.
NCERT Solutions Class 10 Maths Chapter 13 – All Exercises PDF Download
Download exercise-wise NCERT Solutions PDFs for offline study
| Exercise No. | Topics Covered | Download PDF |
|---|---|---|
| Exercise 13.1 | Complete step-by-step solutions for 9 questions | 📥 Download PDF |
| Exercise 13.2 | Complete step-by-step solutions for 6 questions | 📥 Download PDF |
| Exercise 13.3 | Complete step-by-step solutions for 7 questions | 📥 Download PDF |
Surface Areas and Volumes – Key Formulas & Concepts
Quick reference for CBSE exams
| Formula | Description | When to Use |
|---|---|---|
| Curved Surface Area of a Cylinder \(2\pi rh\) | Area of the curved surface (excluding top and bottom) of a cylinder. Note: r = radius, h = height. Remember to use consistent units. | Finding the area of the side of a cylindrical object, like a tank or pipe. |
| Total Surface Area of a Cylinder \(2\pi r(h + r)\) | Area of the entire cylinder, including the curved surface and both circular ends. Note: Includes the area of the top and bottom circles. | Finding the total surface area of a closed cylindrical object. |
| Volume of a Cylinder \(\pi r^2h\) | The amount of space a cylinder occupies. Note: r = radius, h = height. Volume is always in cubic units. | Finding the capacity of a cylindrical container, like a water tank or a can. |
| Curved Surface Area of a Cone \(\pi rl\) | Area of the curved surface (excluding the base) of a cone. Note: r = radius, l = slant height. Make sure you use slant height, not the actual height. | Finding the area of the slanted surface of a cone, like an ice cream cone. |
| Total Surface Area of a Cone \(\pi r(l + r)\) | Area of the entire cone, including the curved surface and the circular base. Note: Includes the area of the circular base. | Finding the total surface area of a closed cone. |
| Volume of a Cone \(\frac{1}{3}\pi r^2h\) | The amount of space a cone occupies. Note: r = radius, h = height. Note the 1/3 factor – it’s easy to forget! | Finding the capacity of a cone-shaped container. |
| Slant Height of a Cone \(l = \sqrt{r^2 + h^2}\) | Relates the slant height, radius, and height of a cone (Pythagorean Theorem). Note: l = slant height, r = radius, h = height. | When you know the radius and height of a cone, but need to find the slant height to calculate surface area. |
| Surface Area of a Sphere \(4\pi r^2\) | Area of the outer surface of a sphere. Note: r = radius. Remember it’s 4 times the area of a circle with the same radius. | Finding the area of a spherical object, like a ball. |
| Volume of a Sphere \(\frac{4}{3}\pi r^3\) | The amount of space a sphere occupies. Note: r = radius. Volume is always in cubic units. | Finding the capacity of a spherical container. |
| Surface Area of a Hemisphere \(3\pi r^2\) | Total surface area (curved surface + circular base) of a solid hemisphere. Note: This includes the flat circular top. Curved surface area is \(2\pi r^2\) | Calculating the total area you could paint on a bowl. |
| Curved Surface Area of a Hemisphere \(2\pi r^2\) | Area of the curved surface (excluding the base) of a hemisphere. Note: r = radius. It’s half the surface area of a full sphere. | Finding the area of just the curved part of a hemispherical bowl. |
| Volume of a Hemisphere \(\frac{2}{3}\pi r^3\) | The amount of space a hemisphere occupies. Note: r = radius. It’s half the volume of a full sphere. | Finding the capacity of a hemispherical bowl. |
| Volume of a Frustum of a Cone \(\frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1r_2)\) | Volume of the frustum (the part left after cutting off the top of a cone parallel to the base). Note: r₁ and r₂ are the radii of the two bases, and h is the height of the frustum. | Finding the capacity of a bucket or a lampshade. |
| Curved Surface Area of a Frustum of a Cone \(\pi l(r_1 + r_2)\) | Area of the curved surface of a frustum of a cone. Note: l is the slant height, r₁ and r₂ are the radii of the two bases. | Finding the area to cover the side of a bucket or lampshade. |
| Slant Height of Frustum of a Cone \(l = \sqrt{h^2 + (r_1 – r_2)^2}\) | Calculates slant height of a frustum. Note: r1 and r2 are radii of the ends, h is vertical height. | When you need slant height to calculate curved surface area of frustum but are given the height and radii. |
Frequently Asked Questions – NCERT Class 10 Maths Chapter 13
📚 Related Study Materials – Class 10 Maths Resources
| Resource | Access |
|---|---|
| NCERT Class 10 Maths All Chapters | View Solutions |
| NCERT Class 10 Science Solutions | View Solutions |
| NCERT Class 10 Social Science | View Solutions |
| NCERT Class 10 English Solutions | View Solutions |