NCERT Solutions Class 9 Maths Chapter 13 Exercise 13.2 - Embibe
• Written By Vidyashree PN
• Written By Vidyashree PN

# NCERT Solutions for Surface Areas and Volumes Exercise 13.2 Class 9 Maths

Exercise 13.2 Class 9 Maths NCERT Solutions: NCERT Solutions for Class 9 Maths Chapter 13 deals with Surface Area and Volume exercises. Embibe’s subject-matter experts solve these NCERT Solutions by keeping in mind the understanding and grasping capabilities of the students. All the solutions for the Class 9 Maths subject are per the NCERT syllabus and updated CBSE guidelines.

Each question of Ex 13.2 Class 9 is solved in a straightforward manner to help the students understand and prepare well for the final exams. Surface Areas and Volumes is an important topic and will help students score good marks in the final exam. Continue reading the article to access the Class 9 Maths Chapter 13 Exercise 13.2 solutions free PDF and other information related to Chapter 13.

## NCERT Solutions for Surface Areas and Volumes Class 9 Maths Ex 13.2 Class 9 Maths: Overview

We have provided the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 PDF for students to learn and perform well in the exam. These answers will help them clear all doubts about the Surface Area of a Right Circular Cylinder. Moreover, they will help to get the basics of the Surface Areas and Volumes to understand its advanced topics in higher classes.

However, Embibe suggests that students should first try to solve the textbook exercise questions independently. If they get stuck while finding the answer, they must refer to these solutions. Before we provide you with the Class 9 Maths Exercise 13.2 Solutions, let us have an overview of Class 13 Maths chapters:

### NCERT Solutions for Class 9 Maths Chapter 13 Ex 13.2: Download PDFs

Surface Area of a Right Circular Cylinder solutions is available here in PDF format for students so that they can download the same and study offline. Many students may not have access to a reliable internet connection, so they can download the PDF and keep a hard copy if required. We recommend students bookmark this page for future reference.

This chapter covers the various concepts of Surface Areas and Volumes and formulas to solve the problems. Students must understand this chapter thoroughly to excel in their future classes. Students can understand the steps to solve the exercise questions by downloading this PDF solution. Please find the solutions to the Class 9 Maths Exercise 13.2 in PDF here.

### Important Questions for NCERT Solutions Class 9 Maths Chapter 13 Ex 13.2

We have given a few questions from Class 9 Maths Exercise 13.2 in the text format for better reference. Candidates can check the problems below:

Q.1: Assume π=227.
The curved surface area of a right circular cylinder height of 14 cm is 88 cm2. If the diameter of the base of the cylinder is d cm, then find the value of d.

Solution: Given, Height h of cylinder=14 cm
Curved surface area of cylinder=88 cm2
Let us consider the diameter of the cylinder as d ⇒2πrh=88 cm2 (r is the radius of the base of the cylinder)
⇒πdh=88 cm2 d=2r ⇒227×d×14 cm=88 cm2
⇒d=2 cm Hence, the diameter of the base of the cylinder is 2 cm. Hence, d=2.

Q.2: In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and a height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. Use π=227.

Solution: Given the height h of the frame of the lampshade =2.5+30+2.5 cm=35 cm (Including margin)

Radius r of the circular end of the frame of lampshade =Diameter=202=10 cm

We know that the curved surface area of the cylinder =2πrh
Therefore, the cloth required for covering the lampshade is 2×227×10×35 cm2=2200 cm2
Hence, for covering the lampshade, 2200 cm2 of cloth is required.

Q.3: The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base using cardboard. Each penholder was to be of radius 3 cm and a height of 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard in cm2 was required to be bought for the competition? Use π=227

7920

Solution: Given the radius r of the circular end of the cylindrical penholder =3 cm and height h of cylindrical penholder =10.5 cm

Surface area of 1 penholder = Curved surface area of penholder + Area of base of penholder =2πrh+πr2

Therefore, surface area of one penholder =2×227×3×10.5+227×3×3 cm2 =132×1.5+1987 cm2 = 15847 cm2

Area of cardboard sheet used by 1 competitor =15847 cm2 Area of cardboard sheet used by 35 competitors =15847×35 cm2=7920 cm2

Hence, 7920 cm2 of cardboard sheet was required to make 35 penholders.

Students may find the direct PDF links for Exercise 13.2 Class 9 Maths below.

### Benefits of Studying from NCERT Solutions for Class 9 Maths Chapter 13

The following is the list of benefits students can get while solving NCERT solutions at Embibe:

• Embibe’s Mathematics subject-matter experts create NCERT Maths Class 9 Solutions, and they provide a deep understanding of the Surface Areas and Volumes chapter.
• Each question of Ex 13.2 Class 9 is solved in a straightforward manner with a step-by-step guide for students to get a good hold of the concepts and formulas.
• Class 9 Maths Chapter 13 Exercise 13.2 is solved in an understandable format to help students understand the usage of formulas and constants. This will, in turn, help them solve problems with ease in their final exam.
• A full assessment of topics and important questions are provided, along with marks and weightage. These features will assist students in Class 9 exam preparation.
• Students can use these NCERT solutions for last-minute preparation and revisions by helping them clarify doubts and confusion.
• Students can also prepare for their future entrance exams as these concepts form the basics. Exams like JEE Main, BITSAT, TS EAMCET, and other state engineering entrance exams have NCERT syllabus. These NCERT solutions will help them score well in the entrance exams.
• The NCERT Class 9 Maths Solutions are provided in the form of PDFs for students to access for free and use while studying offline.
• Along with deep analysis and understanding of the Class 9 Maths Exercise 13.2 Solutions PDF, students are also provided with the chapter-wise mock tests to improve their preparation level for the final exam.
• Embibe will provide feedback analysis after completing each NCERT Class 9 chapter-wise mock test. Through this, students can understand their mistakes and rectify them before the final exam.

### About NCERT Solutions for Class 12 Maths Ex 5.6

NCERT Class 9 Maths Chapter 13 Exercise 13.2 discusses the Surface Areas and Volumes. In this exercise, students can learn about the Surface Area of a Right Circular Cylinder. Let us see some important points discussed in Exercise 13.2 Class 9 Maths:

• Right Circular Cylinder
• Curved Surface Area of the Cylinder
• Total Surface Area of a Cylinder
• π =22/7 or 3.14

### NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2: Chapter Summary

In this exercise, we have discussed the problems related to the Surface Area of a Right Circular Cylinder. The concepts covered in Class 9 Ex 13.2 include the right circular cylinder, the curved surface area of the cylinder, the total surface area of a cylinder, the related formulas, and their usage. By solving the NCERT solutions, students will also know which value of π is used to solve a particular problem.

The important topics discussed in this exercise are as below:

• In this exercise, the word cylinder means right circular cylinder only.
• The radius of a cylinder means the base radius of the cylinder.
• The value of π is approximately equal to 22/7 or 3.14.

Students will learn the following formulas in this Class 9 Maths Chapter 13:

• Surface area of a cuboid = 2(lb + bh + hl)
• Surface area of a cube = 6a2
• Curved surface area of a cylinder = 2πrh
• Total surface area of a cylinder = 2πr(r + h)
• Curved surface area of a cone = πrl
• Total surface area of a right circular cone = πrl + πr2 , i.e., πr (l + r)
• Surface area of a sphere of radius r = 4 πr2
• Curved surface area of a hemisphere = 2πr2
• Total surface area of a hemisphere = 3πr2
• Volume of a cuboid = l × b × h
• Volume of a cube = a3
• Volume of a cylinder = πr2h
• Volume of a cone = 1/3 πr2h
• Volume of a sphere of radius r = 3 43 πr3
• Volume of a hemisphere = 3 23 πr3
[Here, letters l, b, h, a, r, etc., have been used in their usual meaning, depending on the context.]

### FAQs on NCERT Solutions for Class 9 Maths Chapter 13 Ex 13.2

Here are some of the frequently asked questions on Class 9 Maths Chapter 13 Exercise 13.2 Solutions PDF and their answers: