ncert solutions for class 9 maths chapter 13 exercise 13.2

Textbooks for elementary and secondary school students are made and provided by the National Council of Educational Research and Training (NCERT). When describing the importance of the NCERT books for CBSE board exam preparation, the best thing to say is that they place a heavy focus on fundamentals to help students grasp the key concepts. Since NCERT textbooks are thorough and effective in their own right, CBSE expects students to read the NCERT books. To enhance students’ learning abilities and assess their comprehension, analytical thinking, and problem-solving skills, the Department of Education in Science and Mathematics (DESM) and NCERT developed some excellent problems in Science and Mathematics for classes 9 through 12 called “Exemplar Problems.” Students may benefit from completing the book’s self-evaluation exercises and learning the core tactics for handling complex circumstances. For a deeper understanding of Exercise 13.2 Class 9 Maths, students may consult the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2.

Students will learn that they can promptly respond to all of the questions from the previous year’s papers if they diligently study from these books and follow NCERT. The NCERT textbooks served as the source for each and every question in the yearly Class 9 exams. Only a small number of questions and phrases are changed to gauge students’ understanding. Students can refer to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, for assistance if necessary. Students who intend to take competitive exams can utilise the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2. If a student plans to enroll in the Science stream in Grade 11, they should begin preparing in Class 9. Students can access a variety of study materials along with the solutions to Class 9 Chapter 13 Maths Exercise 13.2.

NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.2) Exercise 13.2

CBSE students of Class 10 can easily download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2. They can download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 from Extramarks’ website or mobile application in PDF format so that they can study NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 even in offline mode.

The students must become familiar with these NCERT solutions in order to do well on the Class 9 examination. These NCERT solutions have been addressed by Mathematics specialists from Extramarks. With the guidance of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, students will be able to grasp and resolve a variety of Surface Areas and Volume problems. With the board examination in mind, students should concentrate on Class 9 Mathematics. The students must prepare for each topic. They might consult NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, for clarification. In order to fully comprehend all the topics and subtopics addressed, students in Class 9 must follow the most recent CBSE curriculum.

Understanding how Mathematics is applied in daily life is crucial. It is always wise for students to choose Mathematics as one of their subjects in Class 11. A student must select the ideal subject for them because their stream will influence their career. Students must be very clear about their post-Class 10 goals. Students can also use Extramarks to help them through their highs and lows. Students must know that Mathematics makes the mind sharp, and they will be able to solve questions faster if they practice it daily.

Students can also use the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 to help them solve problems. They must know that Extramarks helps students in every possible way to make their understanding clear and neat. Therefore, students must solve Mathematics questions every day to get smart at understanding questions.

Access NCERT Solutions for Class 9 Mathematics Chapter 13 – Surface Areas and Volume

For a thorough understanding, the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes comprise a carefully crafted variety of completed exercise questions. These Class 9 Mathematics solutions have been created with the most recent CBSE curriculum in mind. The primary goal of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 was to serve as a helpful resource to assist students in resolving questions on their own and effectively. The NCERT Solutions for Class 9 Mathematics are created by subject-matter experts and instructional staff with extensive teaching backgrounds.

The concept-based approach and the precise examination-taking strategy are both taken into consideration when creating the Class 9 NCERT Solutions. For a higher grade, students must refer to NCERT Solutions for Class 9. These solutions are thorough and organised and will help students absorb the knowledge that is based on concepts. For easy access, NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, can be accessed online and downloaded in PDF format. Students study better from NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, if they have a PDF of these solutions. They can study in offline mode without any interruption. Students must solve each and every question of the chapter exercise-wise. They are advised to stay focused while solving questions and must not get distracted while studying. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 will make studying interesting for students.

There are a total of 9 exercises in Chapter 13- Surface Areas and Volume. Students must solve each exercise with precision and understanding. The first exercise, 13.1 has a total number of 9 questions, and the second exercise, 13.2 has 8 questions whose solutions are available in NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2. The third exercise (13.3) has a total of 9 questions, the fourth exercise (13.4) has a total of 5 questions, and the fifth exercise (13.5) has a total of 5 questions as well. Then moving on to the sixth exercise (13.6) has a total of 8 questions and the seventh exercise (13.7) has a total of 9 questions afterwards, the eighth exercise, 13.8 has 10 questions. Lastly, the last ninth exercise (13.9) has a total of 3 questions. Each question has to be solved by the students again and again until they understand the concepts of the questions. While solving Exercise 13.2, NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 will aid students by providing them with easy and simple methods of solving questions. Students must stay concentrated while learning the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2. Each question and exercise has a number of formulas and theorems, and everything has to be retained by students by practising. Hence, Chapter 13: Surface Areas and Volume has many exercises, and each exercise has to be solved with proper understanding. Students must download NCERT Solutions for every exercise so that they can solve the exercises without any hindrance.

NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.2

Everyday life exposes students to a variety of substances and things. The students must be adept at drawing the figures of these items on plain paper as plane figures based on their prior coursework. The surface areas and volumes of cuboids, cylinders, cones, and spheres will be calculated in this section of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2: Surface Areas and Volumes, because knowing how much space a solid occupies in space is an important concept to learn for day-to-day life. The student’s understanding of this idea would be greatly aided by the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, for Chapter 13 Surface Areas and Volumes. When calculating area, students must keep in mind that the unit of area is the square, whereas the unit of volume is the cube. Most of these formulas and information may be found in the activities that follow, as well as in the PDF of the solutions that are provided below.

The right circular cylinder is described in depth in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 so that the students may grasp what it is. This has been clarified using relevant illustrations and straightforward exercises that involve stacking circular cardboard sheets. Additionally, concrete examples have been used to clarify the concepts in the NCERT solutions for Maths Class 9 Chapter 13 Exercise 13.2.

If a certain sheet of paper is needed to cover the cylinder, the length of the sheet will be equal to the circumference of the Base of the Cylinder (2 π r), and its area will give the Surface Area of the Cylinder. The following link will take students to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, which consists of 8 questions primarily focused on formulas:

NCERT Solutions For Class 9

Through NCERT Solutions for Class 9, students have access to many different study materials. The Extramarks website and mobile application offer access to all teaching materials. As a result, it is encouraged that students look up any pertinent questions they may have about the chapter on the Extramarks website. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 is also available for the students. For Class 9 Maths Chapter 13 Exercise 13.2, they can obtain notes, important questions, and sample papers. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, provides students with access to problems from previous years and assistance with their answers.

All the other chapters’ NCERT solutions for Class 9 are also available on Extramarks’ website and mobile application.

Chapter 1 Number System

Chapter 2 Polynomials

Chapter 3 Coordinate Geometry

Chapter 4 Linear Equations in Two Variables

Chapter 5 Introduction to Euclid’s Geometry

Chapter 6 Lines and Angles

Chapter 7 Triangles

Chapter 8 Quadrilaterals

Chapter 9 Areas of Parallelograms and Triangles

Chapter 10 Circles

Chapter 11 Constructions

Chapter 12 Heron’s Formula

Chapter 13 Surface Areas and Volumes

Chapter 14 Statistics

Chapter 15 Probability

CBSE Study Materials For Class 9

Students get access to a range of study resources through NCERT Solutions for Class 9. All of the course materials are available on the Extramarks website and mobile application. As a result, it is advised that students search the Extramarks website for any relevant information. Students may also access the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2. They are able to find notes, important problems, and sample solutions for the Class 9 Surface Areas and Volume. Students can access problems from previous years and get help with their responses by using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2.

The experienced team that created NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2, examined every issue to ensure that every concept was fully and accurately understood. The knowledge needed to approach the problems systematically will be given to the students through these solutions. It will also assist students in performing well on their Mathematics examinations.

CBSE Study Materials

NCERT is the leading management and research institution in the country. Even though it is possible to pass the CBSE examinations with enough marks solely by studying the NCERT books, it can occasionally be challenging for students who follow the CBSE board curriculum to accomplish this. Students can find it challenging to complete the exercises in the NCERT Mathematics Book. Extramarks has adapted the CBSE format. Each chapter ends with a task that has to be finished. With the help of Extramarks, any answer, even the solutions to Class 9 Maths Chapter 13 Exercise 13.2, is accessible.

To aid students in learning and exam preparation, Extramarks offers CBSE study resources that are innovatively created, interesting, and student-friendly. They are written with clarity and precision by subject specialists, so NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 can be trusted by students. The online CBSE study resources, which include syllabuses, books, sample papers, question papers, NCERT answers, NCERT exemplar solutions, significant questions, and CBSE notes, may be helpful to all students. With the help of the CBSE study resources, students may easily get ready for their examinations. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 has proven to be extremely beneficial for students in every way.

Q.1 The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder

Ans

$\begin{array}{l} Height of the cylinder = 14 cm \\ {Curved surface area of the cylinder = 88 cm}^{2} \\ {2πrh = 88 cm}^{2} \\ r = \frac{88}{2× (\frac{22}{7}) ×14} \\ = \frac{88}{88} = 1 \\ Diameter of the cylinder = 2r \\ = 2 (1) = 2 cm \end{array}$

Q.2 It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Ans

$\begin{array}{l} The height of closed cylinderical tank = 1 m \\ The base diameter of cylinderical tank = 140 c m \\ = 1.4 m \\ The base radius of cylinderical tank = \frac{1.4}{2} \\ = 0.7 m \\ Required sheet for cylinder = 2 \times \frac{22}{7} \times 0.7 \times 1 + 2 \times \frac{22}{7} {(0.7)}^{2} \\ = 4.4 m^{2} + 3.08 \\ = 7.48 m^{2} \\ T hus, 7 {.48 m}^{2} sheet is required for closed cylinderical tank . \end{array}$

Q.3 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (figure shown below) . Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Ans

$\begin{array}{l} Length of metal pipe = 77 cm \\ Inner diameter of metal pipe = 4 cm \\ Outer diameter of metal pipe = 4 .4 cm \\ (i) Inner surface area of metal pipe \\ = π \times inner diameter \times h \\ = \frac{22}{7} \times 4 \times 77 \\ = 968 {cm}^{2} \\ (ii) Outer surface area of metal pipe \\ = π \times Outer diameter \times h \\ = \frac{22}{7} \times 4 .4 \times 77 \\ = 1064 .8 {cm}^{2} \end{array}$

$\begin{array}{l} (iii) Tital surface area = Outer surface area \\ +Inner surface area + 2π ({r_{1}}^{2} - {r_{2}}^{2}) \\ =1064.8+968+2× \frac{22}{7} \{{(2.2)}^{2} {-2}^{2}\} [∵ r_{1} = \frac{4.4}{2} {= 2.2, r}_{2} = \frac{4}{2} = 2] \\ = 2032.8+5.28 \\ {= 2038.08 cm}^{2} \\ {Thus, the total surface area of metal pipe is 2038.08 cm}^{2} . \end{array}$

Q.4 The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Ans

$\begin{array}{l} The diameter of a roller = 84 cm \\ The radius of a roller = \frac{84}{2} \\ = 42 cm \\ The length of roller = 120 cm \\ Area of playground covered in one revolution \\ = curved surface area of roller \\ = 2 \times \frac{22}{7} \times 42 \times 120 \\ = 31680 {cm}^{2} \\ Area of playground covered in 500 revolutions \\ = 500 \times 31680 {cm}^{2} \\ = 15840000 {cm}^{2} \\ = \frac{15840000}{10000} m^{2} [∵ 1 {cm}^{2} = \frac{1}{10000} m^{2}] \\ = 1584 m^{2} \\ Thus, the area of plalyground is 1584 m^{2} . \end{array}$

Q.5 A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m².

Ans

$\begin{array}{l} Diameter of a cylindrical pillar = 50 cm \\ = \frac{50}{100} m \\ = \frac{1}{2} m \\ Radius of a cylindrical pillar = \frac{1}{4} m \\ Height of a cylindrical pillar = 3.5 m \\ Curved surface area of pillar = 2πrh \\ = 2× \frac{22}{7} × \frac{1}{4} ×3.5 \\ = \frac{11}{2} m^{2} \\ {Cost of painting of 1m}^{2} area = Rs.12.50 \\ Cost of painting of \frac{11}{2} m^{2} area = Rs.12.50× \frac{11}{2} \\ = Rs.68.75 \\ Therefore, the cost of painting the Curved surface area of the \\ pillar is Rs 68.75. \end{array}$

Q.6 Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Ans

$\begin{array}{l} {Curved surface area of right circular cylinder = 4.4 m}^{2} \\ Radius of the base of right circular cylinder = 0.7 m \\ {Then, 2πrh = 4.4 m}^{2} \\ 2× \frac{22}{7} {×0.7×h = 4.4 m}^{2} \\ h = \frac{4 .4}{4 .4} m \\ = 1 m \\ Thus, the height of right circular cylinder is 1 m. \end{array}$

Q.7 The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs 40 per m².

Ans

$\begin{array}{l} The inner radius of a circular well = \frac{3.5}{2} m \\ The depth of a circular well = 10 m \\ (i) Inner curved surface area of well = 2πrh \\ = 2× \frac{22}{7} × \frac{3.5}{2} ×10 \\ {= 110 m}^{2} \\ (ii) {Cost of plastering 1 m}^{2} of well = ₹ 40 \\ {Cost of plastering 110 m}^{2} of well = ₹ 40×110 \\ = ₹ 4400 \\ Thus, the cost of plastering the inner curved surface of the \\ {well at the rate of ₹ 40 per m}^{2} is ₹4400. \end{array}$

Q.8 In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Ans

$\begin{array}{l} Length of hot water heating system = 28 m \\ Diameter of hot water heating system = 5 cm \\ Radius of hot water heating system = \frac{5}{100} m \\ Total radiating surface of the system = 2 πrh \\ = 2 \times \frac{22}{7} \times \frac{5}{100} \times 28 \\ = 8 .8 m^{2} \\ {Thus, the total radiating surface area of the system is 8.8 m}^{2} . \end{array}$

Q.9 Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if (1/12) of the steel actually used was wasted in making the tank.

Ans

$\begin{array}{l} The diameter of cylinderical petrol tank = 4.2 m \\ The radius of cylinderical petrol tank = 2.1 m \\ The height of The cylinderical petrol tank = 4.5 m \\ (i) The curved surface area of petrol tank = π×diameter×h \\ = \frac{22}{7} ×4.2×4.5 \\ {= 59.4 m}^{2} \\ Thus, the lateral or curved surface area of a closed cylindrical \\ {petrol storage tank 59.4 m}^{2} . \\ (ii) Total surface area of closed cylinderical tank \\ = 2πr (h+r) \\ = 2× \frac{22}{7} ×2.1 (4.5+2.1) \\ = 44×0.3×6.6 \\ {= 87.12 m}^{2} \\ {Let actually used steel in making tank = x m}^{2} \\ Wasted steel in making cylinderical tank = \frac{1}{12} of x \\ Used steel in making cylinderical tank = x - \frac{x}{12} \\ {87.12 m}^{2} = \frac{11x}{12} \\ x = \frac{12}{11} ×87.12 \\ {= 95.04 m}^{2} \\ {Thus, actually used steel in making cylinderical tank is 95.04 m}^{2} . \end{array}$

Q.10 In Figure, you see the frame of lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and 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.

Ans

$\begin{array}{l} Diameter of the base of the frame = 20 cm \\ Radius of the base of the frame = 10 cm \\ Height of the frame = 30 cm \\ Length of margin = 2.5 cm \\ Height of the frame with margin = 2.5+30+2.5 \\ = 35 cm \\ Cloth required for covering the lampshade \\ = 2πrh \\ = 2× \frac{22}{7} ×10×35 \\ {= 2200 cm}^{2} \\ {Thus, 2200 cm}^{2} cloth is required for covering the lampshade. \end{array}$

Q.11 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 height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Ans

$\begin{array}{l} Radius of penholder = 3 cm \\ Height of penholder = 10.5 cm \\ {Total surface area of a penholder = 2πrh+πr}^{2} \\ = πr (2h+r) \\ = \frac{22}{7} ×3 (2×10.5+3) \\ = \frac{66}{7} {×24 cm}^{2} \\ ∴ Cardboard required for one penholder = \frac{66}{7} {×24 cm}^{2} \\ Cardboard required for 35 penholders = \frac{66}{7} {×24 ×35 cm}^{2} \\ {= 7920 cm}^{2} \\ {Thus, the cardboard required for 35 penholders for competition was 7920 cm}^{2} . \end{array}$

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FAQs (Frequently Asked Questions)

Students can get NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 from the Extramarks website and mobile application, and download the PDF format from there. That is simple to understand and beneficial to students preparing for board examinations and other entrance examinations.

Yes, the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2 are solved by keeping in mind the CBSE Class 9 updated syllabus. As Extramarks always keep an observation on new and added topics in the syllabus so that students would not have to suffer. Hence, students can find the updated syllabus on Extramarks’ website as well.

Students may succeed in the CBSE examinations and master the material by practising the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2. These answers are developed using the most recent CBSE syllabus, including all the important subject material. Thus, answering these questions will boost students’ confidence as they prepare for their board examinations. Top scores are based on the topics listed in these solutions. Additionally, it helps the students become accustomed in responding to inquiries of varying degrees of complexity. The use of these solutions for reference and preparation for the CBSE examinations is strongly advised for the students.

The important topics that have been covered in Chapter 13- Surface And Volumes are given below and also discussed in NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.2:

The Surface Area of a Cuboid and a Cube

The Surface Area of a Right Circular Cylinder

The Surface Area of a Right Circular Cone

The Surface Area of a Sphere

The Volume of a Cuboid

The Volume of a Cylinder

The Volume of a Right Circular Cone

The Volume of a Sphere

NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.2) Exercise 13.2

NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.2) Exercise 13.2

Access NCERT Solutions for Class 9 Mathematics Chapter 13 – Surface Areas and Volume