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.

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  1. Planning a studying plan and following it is considered the most significant way to ace the examinations.
  2. 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.

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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.

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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:

  1. 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.
  2. They should try to practice the exercises and questions that are included in the materials.
  3. Understanding the concepts rather than memorising the facts is very important for students.
  4. 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 cm2
Thus, the curved surface area of cone is 165 cm2.

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 =

π

r(l + r)
= (22/7) x 12(21 + 12)
= (22/7) x 12 x 33

= 1244.57 cm2
Thus, total surface area of the cone is 1244.57 cm2.

Q.3 Curved surface area of a cone is 308 cm2 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 cm2
πrl = 308 cm2
(22/7) x r x 14 = 308 cm2
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 cm2
Thus, total surface area of the cone is 462 cm2.

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 m2 canvas is Rs 70.

Ans

                    Height of conical tent=10 m                   Radius of conical tent=24 m(i)Slant height of the conical tent=(10)2+(24)2    =100+576    =676    =26 m(ii)       Curved surface area of tent=πrl    =227×24×26 m2∵ The cost of 1m2 canvas=Rs.70∴The cost of the canvas required to make the tent    =Rs.70×227×24×26    =Rs. 137280Thus,the cost of the canvas required to make the tent is Rs. 137280.

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

             Height of conical tent=8 m  Base radius of conical tent=6 m Slant height of conical tent=82+62                                  =100                                  =10 msurface area of conical tent=πrl                                  =3.14×6×10                                 =188.4 m2              Width of tarpaulin=3 mRequired length of tarpaulin=188.43                                   =62.8 mWastage in cutting of tarpaulin                                   = 20 cm                                  =20100m                                  =0.2 m  Required tarpaulin for tent=62.8+0.2                                  =63 mThus, the length of the required tarpaulin will be 63 m.

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 m2.

Ans

The slant height of conical tomb=25 mThe base diameter of conical tomb=14 m      The base radius of conical tomb=7 mCurved surface area of conical tomb=πrl =227×7×25 =550 m2 Rate of white-washing 100 m2 area=Rs.210       Rate of white-washing 1 m2 area=Rs.210100Rate of white-washing 550 m2 area=Rs.210100×550 =Rs.1155Thus, the cost of white-washing the conical tomb is Rs. 1155.

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

           Radius of joker’s cap=7 cm             Height of joker’s cap=24 cm Slant height of joker’s cap=72+242          =49+576          =625          =25 cmCurved surface area of cap=πrl          =227×7×25          =550 cm2Area of the sheet required for 10 caps          =10×550 cm2          =5500 cm2Thus, the area of required sheet for 10 caps is 5500 cm2.

Q.8

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be  painted and the cost of painting is Rs 12 per m 2 , what will be the cost of painting all  these cones? (Use π = 3.14 and take 1.04 = 1.02) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabgeacaqGGaGaaeOyaiaabwhacaqGZbGaaeiiaiaabohacaqG0bGaae4BaiaabchacaqGGaGaaeyAaiaabohacaqGGaGaaeOyaiaabggacaqGYbGaaeOCaiaabMgacaqGJbGaaeyyaiaabsgacaqGLbGaaeizaiaabccacaqGMbGaaeOCaiaab+gacaqGTbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyzaiaab2gacaqGHbGaaeyAaiaab6gacaqGPbGaaeOBaiaabEgacaqGGaGaaeiCaiaabggacaqGYbGaaeiDaiaabccacaqGVbGaaeOzaiaaykW7caqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaab+gacaqGHbGaaeizaiaabYcacaqGGaaabaGaaeOyaiaabMhacaqGGaGaaeyDaiaabohacaqGPbGaaeOBaiaabEgacaqGGaGaaeynaiaabcdacaqGGaGaaeiAaiaab+gacaqGSbGaaeiBaiaab+gacaqG3bGaaeiiaiaabogacaqGVbGaaeOBaiaabwgacaqGZbGaaeiiaiaab2gacaqGHbGaaeizaiaabwgacaqGGaGaae4BaiaabAgacaqGGaGaaeOCaiaabwgacaqGJbGaaeyEaiaabogacaqGSbGaaeyzaiaabsgacaqGGaGaae4yaiaabggacaqGYbGaaeizaiaabkgacaqGVbGaaeyyaiaabkhacaqGKbGaaeOlaiaabccacaqGfbGaaeyyaiaabogacaqGObaabaGaae4yaiaab+gacaqGUbGaaeyzaiaabccacaqGObGaaeyyaiaabohacaqGGaGaaeyyaiaaykW7caqGIbGaaeyyaiaabohacaqGLbGaaeiiaiaabsgacaqGPbGaaeyyaiaab2gacaqGLbGaaeiDaiaabwgacaqGYbGaaeiiaiaab+gacaqGMbGaaeiiaiaabsdacaqGWaGaaeiiaiaabogacaqGTbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGObGaaeyzaiaabMgacaqGNbGaaeiAaiaabshacaqGGaGaaeymaiaabccacaqGTbGaaeOlaiaabccacaqGjbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgaaeaacaqGVbGaaeyDaiaabshacaqGLbGaaeOCaiaabccacaqGZbGaaeyAaiaabsgacaqGLbGaaeiiaiaab+gacaqGMbGaaeiiaiaabwgacaqGHbGaae4yaiaabIgacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogacaqGVbGaaeOBaiaabwgacaqGZbGaaeiiaiaabMgacaqGZbGaaeiiaiaabshacaqGVbGaaeiiaiaabkgacaqGLbGaaGPaVlaaykW7caqGWbGaaeyyaiaabMgacaqGUbGaaeiDaiaabwgacaqGKbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4yaiaab+gacaqGZbGaaeiDaaqaaiaab+gacaqGMbGaaeiiaiaabchacaqGHbGaaeyAaiaab6gacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaabMgacaqGZbGaaeiiaiaabkfacaqGZbGaaeiiaiaabgdacaqGYaGaaeiiaiaabchacaqGLbGaaeOCaiaabccacaqGTbWaaWbaaSqabeaacaqGYaaaaOGaaeilaiaabccacaqG3bGaaeiAaiaabggacaqG0bGaaeiiaiaabEhacaqGPbGaaeiBaiaabYgacaqGGaGaaeOyaiaabwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogacaqGVbGaae4CaiaabshacaqGGaGaae4BaiaabAgacaqGGaGaaeiCaiaabggacaqGPbGaaeOBaiaabshacaqGPbGaaeOBaiaabEgacaqGGaaabaGaaeyyaiaabYgacaqGSbGaaGPaVlaaykW7caqG0bGaaeiAaiaabwgacaqGZbGaaeyzaiaabccacaqGJbGaae4Baiaab6gacaqGLbGaae4Caiaab+dacaqGGaGaaeiiaiaabIcacaqGvbGaae4CaiaabwgacaqGGaaccaGae8hWdaNaaGjbVlaab2dacaaMe8Uaae4maiaab6cacaqGXaGaaeinaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaaeiDaiaabggacaqGRbGaaeyzaGqabiaa+bcadaGcaaqaaGqaaiaa9fdacaqFUaGaa0hmaiaa9rdaaSqabaGccaqFGaGaa0xpaiaa9bcacaqFXaGaa0Nlaiaa9bdacaqFYaGaa0xkaaaaaa@6CE0@

Ans

Diameter of hollow cone=40 cm     Radius of hollow cone=20 cm     =0.2 m     Height of hollow cone=1 mSlant height of hollow cone  =12+(0.2)2           =1+0.04          =1.04 m          =1.02 m (approx.)  C.S.A. of hollow cone=πrl      =3.14×0.2×1.02      =0.64056 m2C.S.A. of hollow 50 cones=50×0.64056 m2       =32.028 m2Cost of painting  of 1 m2 area=Rs.12Cost of painting  of 50 cones=Rs.12×32.028                                   =Rs.  384.336                                  =Rs.  384.34  (approx)Thus, the cost of painting 50 cones is Rs. 384.34.

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FAQs (Frequently Asked Questions)

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