# NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes (Ex 13.3) Exercise 13.3

NCERT (National Council of Educational Research and Training) textbooks are one of the best resources for students to prepare for examinations. These books are written by some of the best academic experts in India, and they provide a comprehensive overview of all the topics that are likely to be covered in the exam. Due to the books’ frequent updating, students may be certain that they are acquiring the most recent information. Additionally, Hindi and English versions of the books are available for students to choose from, depending on which one best matches their needs. The NCERT books are quite brief and have an absence of unnecessary details. They are the perfect learning resource for students with limited time to study. Therefore, a student should ensure that they read the NCERT books if they want to offer themselves the chance of success in the Class 9 examinations.

For students who are beginning from zero and have no prior understanding of the topics that will be included in the test, reading NCERT solutions like the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 is very crucial. Even those who are somewhat familiar with the subjects might gain an advantage from reading the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 and will be able to strengthen their comprehension of the ideas and discover new details.

Whether an individual is a student or an educator, there are a few key elements of the National Council of Educational Research and Training (NCERT) curriculum that they should be aware of. First and foremost, when it comes to researching India’s history and sociopolitical system, NCERT is among the most trustworthy sources of data. Second, the NCERT syllabus is quite well-designed and includes a variety of subjects. Students who wish to comprehend India’s history and culture should find it to be the perfect course. On the other hand, the textbooks include a tonne of reliable material that has been thoroughly examined. Students who wish to know more about India should use them as an exceptional resource. The probability that a student will perform well on an exam can be increased in a variety of ways. The following tips can be of assistance:

- Planning a studying plan and following it is considered the most significant way to ace the examinations.
- Taking practice tests from the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3. By doing so, students will get a better understanding of the test’s structure and the sort of questions that are most likely to be asked in the exam.

By going through these few recommendations students can get higher grades in their exams. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 are suggested by Extramarks on their website.

**NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes (Ex 13.3) Exercise 13.3 **

The NCERT textbook, as well as NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, is a must-read for all students studying in Class 9 or other classes. The syllabus covered in the textbooks is what is tested in the exams. So, it is important to read and understand these books. The questions in the exams are based on the concepts covered in the NCERT books. So, if students want to score well, they need to read these books thoroughly. Overall, the NCERT is an excellent resource for both students and educators alike, and it is important to be aware of the distinct points that it offers. Similarly, Extramarks recommends NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 which are quite advantageous for students to gather a piece of basic knowledge about Surface Areas and Volumes.

NCERT Solutions are booklets prescribed by subject-matter specialists and academic scholars in India. They are used to teach students about the basics as well as advance knowledge of the subjects in the country. The solutions can be found in both English and Hindi. The benefits of downloading NCERT Solutions are many. The most crucial benefit is that it helps students in their studies. NCERT Solutions can be downloaded in PDF format, and they are very broad. They cover all the topics in the NCERT textbooks, and they are upgraded on a regular basis. The best way to use NCERT Solutions is as a guide or reference. Students can read the chapters in the textbooks and refer to NCERT Solutions to check the areas where they are lacking the most. Students can practice difficult problems using NCERT Solutions. The NCERT Solutions are in-depth and will aid students in better conceptual understanding. All of the questions in the textbook have step-by-step answers included as well. Students will gain confidence in their ability to answer the questions independently with the help of these solutions. Students may use NCERT Solutions as a valuable study aid, and it will help them pass their tests successfully.

Extramarks is a one-stop destination where students can download NCERT solutions. Ultimately, using the NCERT Solutions for Class 9 students can be a great way to strengthen students’ ability to understand the course material and promote their test scores. So if students are looking for ways to improve their study habits, they should be sure to give these solutions a try.

**Access NCERT Solutions Class 9 Maths Chapter 13 – Surface Areas And Volumes**

Students who analyse and complete the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, have a better idea of how to handle the Mathematical Syllabus. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, and other relevant solutions are available on the Extramarks website or mobile application, which students must access. Since they are readily available in that format, students can download them in PDF format for later use. The answers, which are quite comprehensive, provide in-depth justifications for every issue. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, may also be used by students as a crucial tool while preparing for exams. In the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, all questions have thorough answers. As it contains a sizable amount of unanswered questions, practice tests, and past years’ papers. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, are advised by Extramarks, and as a result, they will help students adequately prepare for their exams.

**Exercise (13.3)**

In Exercise 13.3 of Class 9 Mathematics, Chapter 13, students will get to know about the Surface Area of a Right Circular Cone, the Curved Surface Area of a Cone, and the Total Surface Area of a Cone. In the Exercise, there are 8 questions mentioned in Exercise 13.3. The questions in the Exercise can be quickly solved with NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3.

Besides that, to earn higher grades in the Mathematics examination, it is vital to keep going over the chapters and doing as many practice problems as possible. Students can boost their preparation by going over the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.3. Students who do not understand the necessary topics can easily find answers to questions by looking through the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.3. By carefully solving the intext questions, using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 can make the studying process comfortable and quick.

**NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.3**

The study of Mathematics is a broad subject with several uses. It is employed in business to make financial decisions and plans. It is used in the Sciences to examine the natural world. In addition, it is engaged in daily life to resolve issues and make choices. Many students think of Mathematics as a tough and complex topic. But it can also be engaging and enjoyable. There are several varieties of Mathematics, and each one is fascinating in its own way. For instance, Geometry is all about forms and patterns, Algebra is all about resolving equations, and Calculus is all about differential equations. Certain kinds of Mathematical problems and questions related to Volumes and Areas of the Surfaces can be practised and solved with the aid of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3.

It is feasible to apply Mathematics as a strong tool in a wide range of disciplines. It may be applied to businesses to determine earnings and losses, project future trends, and more. Calculating doses and treating patients in the medical sector are both possible with the use of Mathematics. Contrary to that, the speed of light, the size of the universe, and many other quantities may be calculated using it in the scientific community. Mathematics is the study of patterns and relationships that may be used to resolve issues, comprehend the environment, and make predictions. It helps in figuring out how things operate and in solving issues. It is possible to forecast how events will turn out using Mathematical models. For instance, people may use Mathematics to create and forecast the weather and to project how it will evolve. People’s daily lives include Mathematics in significant ways. They employ various technologies to measure objects, compute taxes, and organise their days. People live in a world where Mathematics is universal and necessary.

**NCERT Solutions For Class 9**

When it comes to comprehending the ideas presented in the syllabus, NCERT solutions such as NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 are quite beneficial. The answers are specific and straightforward, and they can assist students in clearing up any problematic questions they have about the subject. Extramarks suggest students study with the help of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, which can also be used as a reference for studying and can help them get ready for assessments and quizzes. There is no question that NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 can aid students in their test preparation. Students may have a comprehensive knowledge of the ideas presented in the curriculum by having access to detailed answers to questions in the NCERT textbooks. Further, practising problems given in NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 can help students strengthen their problem-solving abilities and assess their conceptual awareness.

There is certainly no denying that NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, can help students get better results. All of the topics in the curriculum are covered in detail in the solutions. Practice questions are also included with the solutions so that students may evaluate their apprehension and pinpoint any areas they need to improve in. Students may develop their scores and deepen their understanding of the subject by using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3.

The NCERT Solutions may unquestionably help students in learning more efficiently. These solutions can assist students in understanding things they may be having trouble with by giving brief and simple justifications for complicated ideas. There are a few things to bear in mind when using NCERT Solutions. Using an outdated version may not be accurate because NCERT updates its answers regularly. Students must also understand the kinds of questions that will be asked in the test.

The facilities of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 might be useful for studying if students are trying to comprehend difficult topics or if they want to study for an upcoming test. Students may find it easier to prepare for tests if they receive complete answers to their questions, have access to the appropriate study materials, and have other benefits. Students may have a deeper understanding of the ideas presented in their course with the aid of Extramarks’ NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3. The greatest option for students who wish to score higher is to use the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3.

Students will find a wide range of information in NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, that they may apply in a number of different ways to enhance their studies. These solutions can help students identify their areas of weakness so they can concentrate on revising the topics. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 can also assist students in strengthening their academic performance and getting ready for tests. Using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 is an outstanding opportunity for students who desire to enrich their Mathematics skills. Students can use NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.3 to get feedback on their academic performance and identify areas for improvement.

Students can easily download NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.3 in PDF format from the Extramarks website. These solutions include practice questions and papers from past years. Students may find answers to their problems in NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, available on the smartphone application or website of Extramarks. In the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, which also offers solutions to many open-ended questions, the Surface Areas and Volumes of the Class 9 Mathematics textbook, Chapter 13, are well explained. Students may navigate the concepts presented in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, by accessing the Extramarks website. In order to use them as needed for examinations, students may also obtain the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 by downloading them so that they can be used later.

**CBSE Study Materials For Class 9**

The CBSE study materials for Class 9 can be of great help in students’ preparation. However, it is important to make sure that they make the most of these materials. Here are a few suggestions for students on how to do that:

- Students must make sure that they read the materials thoroughly and do not try to take a glance through them or just read the highlights.
- They should try to practice the exercises and questions that are included in the materials.
- Understanding the concepts rather than memorising the facts is very important for students.
- Students must take advantage of the online resources that are available on the internet, like Extramarks’ learning tools.

Following these tips and Extramarks’ NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 will help students make the most of the CBSE study materials for class 9.

**CBSE Study Materials**

In India, the CBSE test is one of the most significant, so preparing for such examinations with the right guidance is critical. Utilising the CBSE study resources available on the Extramarks website or mobile application is one of the best methods to prepare for the CBSE exam. Students can study the curriculum and prepare for the tests with the aid of these resources. The best possible preparation for a CBSE test may be achieved with the use of CBSE study materials. Students may quickly and simply find the solutions for Class 9 Chapter 13 Maths Exercise 13.3 on the online platform of Extramarks. Students can download these NCERT solutions that help in solving Exercise 13.3 Class 9 Maths and use them as needed when the Mathematics exam is about to be held. Utilising the CBSE study materials as a guide is the best approach for students to use to obtain high grades. The CBSE curriculum, which lists all the subjects that will be covered on the test, is the most encouraging supply of CBSE study resources that can be found online at Extramarks’ website. The CBSE revision notes offer a concise description of the subject matter covered in the syllabus. Students may get a prototype of the kinds of questions that will be asked on the test by using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3. Students can practice answering questions in exam settings by using past years’ papers.

**Q.1 **Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

**Ans**

Curved surface area of cone = (22/7) x (10.5/2) x 10

= 165 cm^{2}

Thus, the curved surface area of cone is 165 cm^{2}.

**Q.2** Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

**Ans**

Diameter of the base of cone = 24 m

Radius of the base of cone = 12 m

Slant height of the cone = 21 m

Total surface area of cone =

$\mathrm{\pi}$

r(*l* + r)

= (22/7) x 12(21 + 12)

= (22/7) x 12 x 33

= 1244.57 cm^{2}

Thus, total surface area of the cone is 1244.57 cm^{2}.

**Q.3 **Curved surface area of a cone is 308 cm^{2} and its slant height is 14 cm. Find

(i) radius of the base

(ii) total surface area of the cone.

**Ans**

(i) Curved surface area of cone = 308 cm^{2}

πrl = 308 cm^{2}

(22/7) x r x 14 = 308 cm^{2}

r = (308/14) x (7/ 22)

= 7 cm

Thus, the radius of the base of cone is 7 cm.

(ii) Total surface area of the cone = πr(*l* + r)

= (22/7) x 7(14 + 7)

= 22 x 21

= 462 cm^{2}

Thus, total surface area of the cone is 462 cm^{2}.

**Q.4 **A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m^{2} canvas is Rs 70.

**Ans**

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Height of conical tent}=10\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Radius of conical tent}=\text{24 m}\\ \text{(i)}\text{Slant height of the conical tent}=\sqrt{{\left(10\right)}^{2}+{\left(24\right)}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{100+576}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{676}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=26\text{\hspace{0.17em}}m\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Curved surface area of tent}=\pi rl\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}\times 24\times 26\text{\hspace{0.17em}}{m}^{2}\\ \because {\text{\hspace{0.17em}The cost of 1m}}^{\text{2}}\text{canvas}=Rs.70\\ \therefore \text{The\hspace{0.17em}}\text{cost of the canvas required to make the tent}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.70\times \frac{22}{7}\times 24\times 26\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.\text{\hspace{0.17em}}137280\\ \text{Thus,the cost of the canvas required to make the tent is}\\ \text{}Rs.\text{\hspace{0.17em}}137280.\end{array}$

**Q.5 **What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.

(Use π = 3.14).

**Ans**

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Height of conical tent}=8\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}\hspace{0.17em}Base radius of conical tent}=6\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}Slant height of conical tent}=\sqrt{{8}^{2}+{6}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{100}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=10\text{\hspace{0.17em}}m\\ \text{surface area of conical tent}=\pi rl\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3.14\times 6\times 10\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}=188.4\text{\hspace{0.17em}}{m}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Width of tarpaulin}=3\text{\hspace{0.17em}}m\\ \text{Required length of tarpaulin}=\frac{188.4}{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=62.8\text{\hspace{0.17em}}m\\ \text{Wastage in cutting of tarpaulin}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{20 cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{20}{100}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.2\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Required tarpaulin for tent}=62.8+0.2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=63\text{\hspace{0.17em}}m\\ \text{Thus},\text{the length of the required tarpaulin will be 63 m}.\end{array}$

**Q.6** The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m^{2}.

**Ans**

$\begin{array}{l}\text{The slant height of conical tomb}=25\text{\hspace{0.17em}}m\\ \text{The base diameter of conical tomb}=14\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}The base radius of conical tomb}=7\text{\hspace{0.17em}}m\\ \text{Curved surface area of conical tomb}=\pi rl\\ \text{\hspace{0.17em}}=\frac{22}{7}\times 7\times 25\\ \text{\hspace{0.17em}}=550\text{\hspace{0.17em}}{\text{m}}^{\text{2}}\\ \text{}\mathrm{Rate}{\text{of white-washing 100 m}}^{\text{2}}\text{area}=Rs.210\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{\hspace{0.17em}Rate of white-washing 1 m}}^{\text{2}}\text{area}=Rs.\frac{210}{100}\\ {\text{Rate of white-washing 550 m}}^{\text{2}}\text{area}=Rs.\frac{210}{100}\times 550\\ \text{\hspace{0.17em}}=Rs.1155\\ \text{Thus,\hspace{0.17em}the cost of white-washing the conical tomb is Rs}\text{. 1155}\text{.}\end{array}$

**Q.7** A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

**Ans**

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Radius of joker}\u2019\text{s cap}=\text{7 cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height of joker}\u2019\text{s cap}=\text{24 cm}\\ \text{\hspace{0.17em}}\text{Slant height of joker\u2019s cap}=\sqrt{{7}^{2}+{24}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{49+576}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{625}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=25\text{\hspace{0.17em}}cm\\ \text{Curved surface area of cap}=\pi rl\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}\times 7\times 25\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=550\text{\hspace{0.17em}}c{m}^{2}\\ \text{Area of the sheet required for 10 caps}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=10\times 550\text{\hspace{0.17em}}c{m}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5500\text{\hspace{0.17em}}c{m}^{2}\\ \mathrm{Thus},{\text{the area of required sheet for 10 caps is 5500 cm}}^{\text{2}}\text{.}\end{array}$

**Q.8 **

\begin{array}{l}\text{A bus stop is barricaded from the remaining part of}\text{\hspace{0.17em}}\text{the road,}\\ \text{by using 50 hollow cones made of recycled cardboard}\text{. Each}\\ \text{cone has a}\text{\hspace{0.17em}}\text{base diameter of 40 cm and height 1 m}\text{. If the}\\ \text{outer side of each of the cones is to be}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{painted and the cost}\\ {\text{of painting is Rs 12 per m}}^{\text{2}}\text{, what will be the cost of painting}\\ \text{all}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{these cones? (Use}\pi \text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{3}\text{.14 and take}\sqrt{1.04}=1.02)\end{array}

**Ans**

$\begin{array}{l}\text{Diameter of hollow cone}=40\text{\hspace{0.17em}}\text{cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Radius of hollow cone}=20\text{\hspace{0.17em}}\text{cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.2\text{\hspace{0.17em}}\text{m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Height of hollow cone}=\text{1}\text{\hspace{0.17em}m}\\ \text{Slant height of hollow cone}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{{1}^{2}+{\left(0.2\right)}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{1+0.04}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{1.04}\text{\hspace{0.17em}}\text{m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1.02\text{\hspace{0.17em}}\text{m}\text{\hspace{0.17em}}\left(\text{approx}\text{.}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}\text{C}\text{.S}\text{.A}\text{. of hollow cone}=\pi rl\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3.14\times 0.2\times 1.02\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.64056\text{\hspace{0.17em}}{\text{m}}^{2}\\ \text{C}\text{.S}\text{.A}\text{. of hollow 50 cones}=50\times 0.64056\text{\hspace{0.17em}}{\text{m}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=32.028\text{\hspace{0.17em}}{\text{m}}^{2}\\ \text{Cost of painting}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{of 1 m}}^{\text{2}}\text{\hspace{0.17em}}\text{area}=Rs.12\\ \text{Cost of painting}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{of 50}\text{\hspace{0.17em}}\text{cones}=Rs.12\times 32.028\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.\text{\hspace{0.17em}}\text{\hspace{0.17em}}384.336\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=Rs.\text{\hspace{0.17em}}\text{\hspace{0.17em}}384.34\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{approx}\right)\\ \text{Thus, the cost of painting 50 cones is Rs}.\text{\hspace{0.17em}}\text{384}.\text{34}.\end{array}$

## FAQs (Frequently Asked Questions)

### 1. What steps can a student take to ensure that they are fully prepared for the Mathematics test and that they will not make any mistakes when trying to tackle the questions in the Maths Class 9 Chapter 13 Exercise 13.3?

It is essential for students to continue reviewing the chapters, completing as many practice problems, and going through the sample tests to achieve higher results on the Mathematics examination. Students may improve their willingness by studying the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 which supports students in putting more effort into preparing for their examinations. By reading the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3, students may quickly get the answers to problems in which they are lacking. Students looking for online materials via the internet can boost their test preparation with the help of Extramarks’ NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3. Students may eventually apprehend these solutions by carefully going over the answers to NCERT questions.

### 2. Can students use NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 to help themselves with the preparation for competitive examinations like JEE or NEET?

Students who are attempting to prepare for competitive examinations will find high-level problems in the solutions for Class 9 Maths Chapter 13 Exercise 13.3 proposed by Extramarks. Students can use the sample questions to help them study for their examinations. There are several benefits to using Extramarks’ NCERT Solutions when taking competitive exams. Students may visit Extramarks’ website to practice for competitive exams like NEET, JEE (IIT), and many more. Study guides with practice problems for examinations are available on Extramarks. Extramarks website underlines the relevance of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 in a variety of ways.

### 3. What other services does Extramarks offer other than NCERT Solutions?

With the help of Extramarks, students can get access to practice questions, past years’ papers, and study materials online and offline in print form. One of the many benefits of using the wonderful facilities provided by Extramarks is the chance to receive instructions directly from lecturers and professionals with experience. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.3 are examples of a variety of study tools.