NCERT Solutions Class 10 Maths Chapter 13

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

Class 10 is a  turning point in the life of every student. Whether a student wants to pursue subjects of his choice or attend the college of his dreams, a lot of it depends on their performance in the Class 10 board exam. Chapter 13 is one of those chapters that often confuses students, and they end up losing marks in the examination. Practising questions related to Chapter 13 regularly can help students have a better understanding of the chapter. Referring to NCERT Solutions for Class 10 Mathematics Chapter 13 by Extramarks will enable students to solve practice questions given in their NCERT books with precision, and master the topic. Solutions are there not just for Mathematics, but for all other subjects as well so that students don’t have to look elsewhere for any assistance.

NCERT Solutions Class 10 Mathematics Chapter 13 are curated by subject matter experts. These have answers to every question given at the end of NCERT Class 10 Chapter 13 textbook.

When it comes to solving questions from NCERT Class 10 Mathematics Chapter 13, students are expected to understand theoretical concepts like what surface area & volume is, along with other practical elements like calculating them. Extramarks’ NCERT Solutions for Class 10 Mathematics Chapter 13 are  as per the latest syllabus of Class 10 CBSE pattern. . Thus, making it an ideal solution  for their CBSE Class 10 Mathematics board exam preparations.

NCERT Solutions for Class 10 Maths

Mathematics is one of those subjects in Class 10 that helps to raise the overall percentage of students. To increase the chances of scoring full marks in Class 10 Mathematics,  students need not only practise but do so with precision. A good place to begin for Class 10 Mathematics would be with NCERT Solutions  by Extramarks. Every chapter-wise solution comes with solved answers to textbook questions and in-depth explanations of the same. All the chapters are listed below: :

  •       Chapter 1 – Real Numbers
  •       Chapter 2 – Polynomials
  •       Chapter 3 – Pair of Linear Equations in Two Variables
  •       Chapter 4 – Quadratic Equations
  •       Chapter 5 – Arithmetic Progressions
  •       Chapter 6 – Triangles
  •       Chapter 7 – Coordinate Geometry
  •       Chapter 8 – Introduction to Trigonometry
  •       Chapter 9 – Some Applications of Trigonometry
  •       Chapter 10 – Circles
  •       Chapter 11 – Constructions
  •       Chapter 12 – Areas Related to Circles
  •       Chapter 13 – Surface Areas and Volumes
  •       Chapter 14 – Statistics
  •       Chapter 15 – Probability

Chapter 13 – Surface Area and Volumes

Chapter 13 – Surface Area and Volumes of Class 10 Mathematics is a lesson loaded with formulas. It teaches the students how to calculate areas and volumes of different shapes. The chapter comprises 5 exercises 13.1 – 13.5. NCERT Solutions for Class 10 Mathematics Chapter 13 has solved answers to every question in exercise 13.1 to 13.5.

13.1 Introduction

Just like any other introduction from any other chapter, in Chapter 13 of Class 10 Mathematics, you would be required to recall  what you studied in Class 9 about cubes, cylinders, circles, etc. This is why at Extramarks we never recommend memorising anything. We always suggest understanding a particular concept.  The sole reason is that, by memorising things,you are not able to retain them in your mind for a long time. But if you understand a concept, chances are you will remember it forever. That’s the cardinal rule in Mathematics.

13.2 Surface Area of a Combination of Solids

Some items are a combination of two or more solid shapes. At the end of this chapter, students will also learn how to calculate the surface area and volumes of a combination of solids using formulas.

13.3 Volume of a Combination of Solids

For calculating the volume of shapes that are a combination of two or more solid shapes, students must find out the volume area of individual shapes first.

13.4 Conversion of a Solid From One Form to Another

Shape Formula
Total surface area of sphere = curved surface area of sphere 4 π r2
Total surface area of cone πr(r+l)
Curved surface area of cone πrl
Total surface area of cuboid 2(lb+bh+hl)
Total surface area of cylinder 2 πr(h+r)

  Volumes of 3D objects:

Shape Formula
Volume of sphere 4/3 πr³
Volume of hemisphere 2πr3/3
Volume of cone (⅓)πr2h
Volume of cube s3
Volume of cuboid lbh
Volume of cylinder πr2h

 Even if a solid is converted from one form to another, then the volume remains the same. Let us understand this with an example:

If there is water in the cuboid shape (which has dimensions of 20 * 22) which is transferred into a cylinder that has a height of 3.5 m and 2 m then what is the height of the water level in the cuboid shape if the water (once transferred into the cylinder) fills the cylinder to the brim?

Solution – From the theorem on volumes, we know that the volume of the water in the cylinder and the volume of the water in the cuboid would be the same. If the water fills the cylinder to the brim, then:

Volume of water in cylinder = π * r² * h = 22/7 * 1 * 3.5 = 11.

Volume of water in cuboid = l * b * h = 20 * 22 * h.

Since 20 * 22 * h = 3.5 * π, we get h = 11/(20 * 22) = 2.5 cm.

Related Questions:

Q1. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution: It is known that the coins are cylindrical in shape.

So, height (h1) of the cylinder = 2 mm = 0.2 cm

Radius (r) of circular end of coins = 1.75/2 = 0.875 cm

Now, the number of coins to be melted to form the required cuboids be “n”

So, Volume of n coins = Volume of cuboids

n × π × r2 × h1 = l × b × h

n×π×(0.875)2×0.2 = 5.5×10×3.5

Or, n = 400

Q2. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution: Radius (r1) of the upper base = 4/2 = 2 cm

Radius (r2) of lower the base = 2/2 = 1 cm

Height = 14 cm

Now, Capacity of glass = Volume of frustum of cone

So, Capacity of glass = (⅓)×π×h(r12+r22+r1r2)

= (⅓)×π×(14)(22+12+ (2)(1))

∴ The capacity of the glass = 102×(⅔) cm3

Key Features of NCERT Solutions for Class 9 Maths Chapter 13

  • NCERT Solutions for Class 10 Mathematics Chapter 13 is a study material to help Class 10 students solve the exercise questions given in NCERT Chapter 13 with accuracy.
  • NCERT Solutions Class 10 Mathematics Chapter 13 is prepared by experienced faculty as per the latest CBSE Class 10 syllabus.
  • A student doesn’t require any aid from teachers or parents to understand the answers given NCERT Solutions Class 10 Mathematics Chapter 13. The language used in it is uncomplicated and designed in a way that everything becomes self-explanatory.
  • All numerical problems are solved step-wise with appropriate diagrams and explanations as and when required.

NCERT Solutions for Class 10 Maths Related Chapters

Real Numbers

Polynomials

Pair of Linear Equations in Two Variables

Quadratic Equations

Arithmetic Progressions

Triangles

Coordinate Geometry

Introduction to Trigonometry

Some Applications of Trigonometry

Circles

Constructions

Areas Related to Circles

Statistics

Probability

FAQs (Frequently Asked Questions)

The topics covered in NCERT Solutions for Class 10 Mathematics Chapter 13 by Extramarks include:

 –         Introduction

 –         Surface area of a combination of solids

 –         Volume of a combination of solids

 –        Conversion of solid from one shape to another

 –        Frustum of a cone

 –        Summary

You can access NCERT Solutions Class 10 Mathematics Chapter 13 from the Extramarks website or app.