# NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4

The National Council of Educational Research and Training (NCERT) creates and provides textbooks for students in elementary and high schools. The easiest way to explain the significance of the NCERT books for CBSE board exam preparation is that they put a strong emphasis on the basics to aid students in understanding the important ideas. Since NCERT textbooks are comprehensive and efficient, the CBSE rarely requires students to read any other materials apart from NCERT books. The Department of Education in Science and Mathematics (DESM) and NCERT have created some excellent problems in Science and Mathematics for classes 9-12 that are referred to as “Exemplar Problems” in order to improve students’ learning abilities and assess their comprehension, analytical thinking, and problem-solving skills. NCERT books aid students in making their ideas clear. Students need to learn concepts from the entire book repeatedly after the principles have been understood. Students just need to review the formulae, concepts, and their applications before examinations. NCERT books are regarded as the best for a full and comprehensive study to create an understanding of clear concepts because of this. Students have the advantage of completing the self-evaluation exercises in the book and learning the fundamental strategies for dealing with challenging situations. Students can refer to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, for a more thorough knowledge of Exercise 13.4 for Class 9 Maths.

Students will learn that they can promptly respond to all of the questions from the past year’s papers if they diligently study from these books and follow the NCERT curriculum. The NCERT textbooks served as the source for each and every question in the yearly CBSE Class 9 exam. 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 13.4, for assistance if necessary. Students who intend to take competitive exams can utilise the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4. If a student plans to choose the Science stream in Grade 11, they should only begin preparing in Class 9.

## NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes (Ex 13.4) Exercise 13.4

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, are readily available for download for CBSE Class 9 students. They can download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, in PDF format. Students can access NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, from the Extramarks website or mobile application so they can study it even when they are not connected to the internet.

Students must become familiar with these NCERT solutions to perform well in the Class 9 examination. Mathematics professionals from Extramarks have answered these NCERT questions. Students will be able to understand and answer a range of Surface Areas and Volume questions with the aid of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4. Students should concentrate on Class 9 Mathematics with the board examination in mind. For each topic, the students must prepare well. For an elaborate understanding, they can refer to NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4. The most recent CBSE curriculum must be followed by Class 9 students in order to completely understand all of the topics and subtopics covered.

It is essential to comprehend how Mathematics is used in everyday life. In Class 11, it is generally advisable for students to select Mathematics as their stream. Goals for after Class 10 must be extremely apparent to the students. Additionally, Extramarks supports students as they handle their highs and lows. Students need to understand that practising Mathematics on a regular basis keeps the mind sharp and speeds up problem-solving.

For help with students’ problem-solving, they can download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4. They must be aware that Exramarks supports students in every manner to ensure that their understanding is organised and clear. Therefore, in order to become proficient at understanding questions, students must complete mathematical problems daily.

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

Students may access the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, on the Extramarks website and mobile application. The mobile application would be very easy to use for students who want to give examinations like the Joint Entrance Exam (JEE). This will help students get into prestigious colleges like the National Institutes of Technology (NITs) and the Indian Institute of Technology (IIT) with a foundational understanding. In addition to choosing engineering as a vocation, students can apply for competitive examinations like the Central Universities Entrance Tests (CUET). They will benefit from NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, since the questions will be multiple choice questions (MCQs).

Since it follows the Extramarks-designed structure, students may easily understand the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4. Students can simply download the Class 9 NCERT Solutions in PDF format from Extramarks, a learning platform, making it handy even when they are not connected to the internet. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 should be accessed by students. Since the subject grows more complex after Class 8, the majority of students find Class 9 exams challenging. However, it might be difficult for a student to decide what to do when they go from a basic to an intermediate level. It may be overwhelming for students who will take the science stream when they are unclear about what to do next. Because the Science stream can be difficult at times, students should become acquainted with Class 9 concepts as early as possible. It is advised for students to regularly study and keep their attention on their areas of weakness.

### NCERT Solution Class 9 Maths Of Chapter 13 All Exercises

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, 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 a total number of 8 questions and the solutions for the same are available on the Extramarks’ website. The third exercise, 13.3, has a total of 9 questions; the fourth exercise, 13.4, has a total of 5 questions, and its solution is also available in NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4. For the fifth exercise, 13.5, has a total of 5 questions as well. Then moving on to the sixth exercise, 13.6, which 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 concept of the questions. While solving exercise 13.4, NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 will aid students by providing them with easy and simple methods of solving questions. Students must stay concentrated while learning from NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4. Every formula and theorem in each question and exercise must be remembered by students through practice. As a result, Chapter 13: Surface Areas and Volume has a number of exercises, each of which must be completed correctly. For ease of use in completing the activities, students must download the NCERT Solutions for each exercise.

The Class 9 NCERT Solutions are prepared using both the concept-based methodology and the precise examination-taking technique. Students must consult NCERT Solutions for Class 9 for better grades. This comprehensive and well-organized response will aid students in understanding the information that is based on concepts. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, are available online and in PDF format for convenience. If they have a PDF version of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, students would learn from it more effectively. They will not be interrupted when studying offline. Every single exercise-based question in the chapter must be answered by students.

### NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.4

The NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes include a range of completed practice problems for a better understanding. The most recent CBSE curriculum was taken into consideration when developing these solutions for Class 9 Mathematics. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 are compiled with the   purpose of acting as a resource and a solution to help students solve problems efficiently and on their own. The instructional team and subject-matter specialists that developed the NCERT Solutions for Class 9 Mathematics have a good amount of experience and come from teaching backgrounds.

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 will considerably benefit the students in learning the concepts of the chapter. These NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 will considerably benefit the learner in learning these concepts. When calculating area, students should remember that the square is the unit of area and the cube is the unit of volume. In the activities that come next and in the PDF of the Class 9 Maths Chapter 13 Exercise 13.4 Solutions that are supplied below, students may find more of these formulas and details.

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 provide detailed explanations of the right circular cylinder to help students understand what the chapter is about. This has been made clear using pertinent examples and a simple activity involving the stacking of circular cardboard sheets. The topic in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, has been clarified using practical examples. The following link will take students to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, which consists of 5 questions primarily focused on formulas. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, can be downloaded in PDF format for offline access.

### NCERT Solutions For Class 9

Class 9 Mathematics is an important turning point in a student’s education since it introduces them to certain fundamental ideas that serve as a basis for later classes. Real Numbers, Polynomials, Two Variable Linear Equations, Lines and Angles, Triangles, Quadrilaterals, Parallelograms, Circles, Coordinate Geometry, Perimeters and Areas, Surface Areas and Volumes, Statistics, Probability, and Euclid’s Geometry are among the subjects covered in class 9 Mathematics. Working through the NCERT Solutions for Class 9 Mathematics is one of the greatest ways to thoroughly understand these subjects. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, are available on the Extramarks’ website and mobile application.

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 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 Euclids Geometry

Chapter 6 Lines and Angles

Chapter 7 Triangles

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

Through NCERT Solutions for Class 9, students get access to many different study materials. The Extramarks website and mobile application offers access to all course materials. Students are encouraged to visit the Extramarks website for any pertinent information. Students can also use the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4. For the Class 9 Surface Areas and Volume, they may locate notes, main topics, and example solutions. Students can use the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 to access problems from past years and receive assistance with their replies.

The knowledgeable group behind NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 have carefully considered all the concerns to make sure that each idea was grasped completely and correctly. These solutions will provide the students with the information required to tackle the challenges methodically. Additionally, it will help students do well on their Mathematics examinations.

Students get access to a wide variety of study materials by going through Extramarks’ website and mobile application. Students can register on the Extramarks website to get access to live doubt-solving questions. Students from all classes and streams can access these solutions. Apart from NCERT Solutions, students can also access past years’ papers, sample question papers, and revision notes. Extramarks’ study materials are complete and can be relied upon during exam preparation.

### CBSE Study Materials

NCERT is the leading management and research institution in the country. Even while 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 13.4, are accessible.

NCERT is the greatest resource out of all the study materials offered because it is written by professionals after extensive research on the subjects. NCERT books are also a major source of inspiration for Class 10 board questions, making them the ideal study resources for students. Students should thus practice the questions from the NCERT textbooks. However, if students prefer a different method of studying NCERT solutions, they can switch to Extramarks’ study materials, like NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, which use visual simulations and other cutting-edge teaching resources.

Q.1 Find the surface area of a sphere of radius:

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Ans

$\begin{array}{l}\text{Surface area of a sphere}=4\pi {r}^{2}\\ \text{(i) Surface area of a sphere}=4\pi {\left(\text{1}0.\text{5}\right)}^{2}\text{\hspace{0.17em}}c{m}^{2}\\ \text{ }=4×\frac{22}{7}×10.5×10.5\text{\hspace{0.17em}}c{m}^{2}\\ \text{ }=4×22×1.5×10.5\text{\hspace{0.17em}}c{m}^{2}\\ \text{ }=1386\text{\hspace{0.17em}}c{m}^{2}\\ \left(ii\right)\text{ }Surface\text{area of a sphere}=4\pi {\left(5.\text{6}\right)}^{2}\text{\hspace{0.17em}}c{m}^{2}\\ \text{ }=4×\frac{22}{7}×5.\text{6}×5.\text{6\hspace{0.17em}}c{m}^{2}\\ \text{ }=4×22×0.8×5.\text{6\hspace{0.17em}}c{m}^{2}\\ \text{ }=394.24\text{\hspace{0.17em}}c{m}^{2}\\ \text{(iii) Surface area of a sphere}=4\pi {\left(14\right)}^{2}\text{\hspace{0.17em}}c{m}^{2}\\ \text{ }=4×\frac{22}{7}×14×14\text{\hspace{0.17em}}c{m}^{2}\\ \text{ }=4×22×2×14\text{\hspace{0.17em}}c{m}^{2}\\ \text{ }=2464\text{\hspace{0.17em}}c{m}^{2}\end{array}$

Q.2 Find the surface area of a sphere of diameter:

(i) 14 cm (ii) 21 cm (iii) 3.5 m

Ans

$\begin{array}{l}\begin{array}{l}\mathrm{Surface}\text{area of a sphere}=4{\mathrm{\pi r}}^{2}\\ \left(\mathrm{i}\right)\text{ }\mathrm{Diameter}\text{of sphere}=14\text{\hspace{0.17em}}\mathrm{cm}\end{array}\\ \begin{array}{l}\text{ }\mathrm{Radius}\text{of sphere}=7\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Surface}\text{area of a sphere}=4\mathrm{\pi }{\left(7\right)}^{2}\end{array}\\ \begin{array}{l}\text{ }=4×\frac{22}{7}×7×7\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} }=\text{ }616\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\end{array}\\ \begin{array}{l}\left(\mathrm{ii}\right)\text{ }\mathrm{Diameter}\text{of sphere}=21\text{\hspace{0.17em}}\mathrm{cm}\\ \text{ }\mathrm{Radius}\text{of sphere}=\frac{21}{2}\text{\hspace{0.17em}}\mathrm{cm}\end{array}\\ \begin{array}{l}\mathrm{Surface}\text{area of a sphere}=4\mathrm{\pi }{\left(\frac{21}{2}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} }=4×\frac{22}{7}×\frac{21}{2}×\frac{21}{2}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} }=1386\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \begin{array}{l}\left(\mathrm{iii}\right)\text{ }\mathrm{Diameter}\text{of sphere}=3.5\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{m}\\ \text{ }\mathrm{Radius}\text{of sphere}=\frac{3.5}{2}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{m}\end{array}\\ \mathrm{Surface}\text{area of a sphere}=4\mathrm{\pi }\left(\frac{3.5}{2}\right)\\ \begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} }=4×\frac{22}{7}×\frac{3.5}{2}×\frac{3.5}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=38.5\text{\hspace{0.17em}}{\mathrm{m}}^{2}\end{array}\end{array}$

Q.3 Find the total surface area of a hemisphere of radius 10 cm.

(Use π= 3.14)

Ans

$\begin{array}{l}{\text{Total\hspace{0.17em}Surface area of a hemisphere = 3πr}}^{\text{2}}\\ \text{ = 3π}{\left(\text{10}\right)}^{\text{2}}{\text{\hspace{0.17em}cm}}^{\text{2}}\\ {\text{ = 3×3.14×10×10\hspace{0.17em}cm}}^{\text{2}}\\ {\text{ = 942\hspace{0.17em}cm}}^{\text{2}}\end{array}$

Q.4 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans

$\begin{array}{l}\text{Surface}\mathit{\text{}}\text{area of a sphere}=4\pi {r}^{2}\\ \text{\hspace{0.17em}When radius of balloon is 7 cm}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Surface area of ballon}=4\pi {\left(7\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4×\frac{22}{7}×7×7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=616\text{\hspace{0.17em}}c{m}^{2}\\ \text{\hspace{0.17em}When radius of balloon is 14 cm}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Surface area of ballon}=4\pi {\left(14\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4×\frac{22}{7}×14×14\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2464\text{\hspace{0.17em}}c{m}^{2}\\ \text{The ratio of surface areas of balloon in the two cases}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{616\text{\hspace{0.17em}}c{m}^{2}}{2464\text{\hspace{0.17em}}c{m}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}\\ \text{Thus},\text{the ratio of surface areas of the balloon in the}\\ \text{two cases is 1:4}\text{.}\end{array}$

Q.5 A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}The inner radius of bowl}=\frac{10.5}{2}\text{\hspace{0.17em}}cm\\ \text{Inner surface area of bowl}=2\pi {r}^{2}\\ =2×\frac{22}{7}×\frac{10.5}{2}\text{\hspace{0.17em}}×\frac{10.5}{2}\text{\hspace{0.17em}}\\ =173.25\text{\hspace{0.17em}}c{m}^{2}\\ {\text{Cost of tin-plating 100 cm}}^{\text{2}}=Rs.\text{\hspace{0.17em}}16\\ {\text{Cost of tin-plating 1 cm}}^{\text{2}}=Rs.\text{\hspace{0.17em}}\frac{16}{100}\\ \text{\hspace{0.17em}Cost of tin-plating bowl}=Rs.\text{\hspace{0.17em}}\frac{16}{100}×173.25\\ =Rs.\text{\hspace{0.17em}}27.72\\ \text{Thus},\text{the cost of tin-platting the inner surface area of bowl is}\\ \text{Rs}\text{. 27}\text{.72}\text{.}\end{array}$

Q.6 Find the radius of a sphere whose surface area is 154 cm2.

Ans

$\begin{array}{l}\text{Let radius of sphere be r cm}\text{.}\\ \text{Surface area of sphere}=154\text{\hspace{0.17em}}c{m}^{2}\\ 4\pi {r}^{2}=154\text{\hspace{0.17em}}c{m}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4×\frac{22}{7}×{r}^{2}=154\text{\hspace{0.17em}}c{m}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}=\frac{154×7}{4×22}\text{\hspace{0.17em}}c{m}^{2}\\ =\frac{7×7}{4}\text{\hspace{0.17em}}c{m}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=\frac{7}{2}\text{\hspace{0.17em}}cm\\ \text{Thus},\text{the radius of sphere is 3}\text{.5 cm}\text{.}\end{array}$

Q.7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Ans

$\begin{array}{l}\text{Surface area of a sphere}=4\pi {r}^{2}\\ \text{Let radius of the earth be R}\text{.}\\ \text{Let radius of the moon be}\frac{\text{R}}{4}\text{.}\left[\begin{array}{l}\text{Since},\text{diameter of the moon is}\\ \frac{1}{4}\text{​}\text{\hspace{0.17em}}\text{of diameter of the earth}\text{.}\end{array}\right]\\ \text{Surface area of the earth}=4\pi {R}^{2}\end{array}$

$\begin{array}{l}\text{Surface area of the moon}=4\pi {\left(\frac{R}{4}\right)}^{2}\\ The\text{ratio of surface areas of the moon and the earth}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4\pi {\left(\frac{R}{4}\right)}^{2}}{4\pi {R}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{16}\\ \text{Thus},\text{the ratio of surface areas of the moon and}\\ \text{the earth is 1:16}\text{.}\end{array}$

Q.8 A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Ans

Inner radius of hemispherical bowl = 5 cm
Thickness of hemispherical bowl = 0.25 cm

Outer radius of hemispherical bowl = 5 + 0.25
= 5.25 cm
So, outer curved surface area of bowl = 2

$\mathrm{\pi }$

r2=2(22/7)( 5.25)2
= 173.25 cm2.
Thus, the outer curved surface area of the bowl is 173.25 cm2.

Q.9 A right circular cylinder just encloses a sphere of radius r (see Figure). Find

(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Ans

$\begin{array}{l}\mathrm{Radius}\mathrm{of}\mathrm{sphere}\mathrm{is}\mathrm{r}\mathrm{.}\\ \left(\mathrm{i}\right)\mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{sphere}=4{\mathrm{\pi r}}^{\mathrm{2}}\\ \left(\mathrm{ii}\right)\mathrm{Height}\mathrm{}\mathrm{of}\mathrm{cylinder}=\mathrm{Diameter}\mathrm{of}\mathrm{sphere}\\ =2\mathrm{r}\\ \mathrm{ }\mathrm{Radius} \left(\mathrm{R}\right)\mathrm{of}\mathrm{cylinder}=\mathrm{r}\\ \mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cylinder}\\ =2\mathrm{\pi Rh}\\ =2\mathrm{\pi r}\left(2\mathrm{r}\right)\\ =4{\mathrm{\pi r}}^{\mathrm{2}}\\ \left(\mathrm{iii}\right)\mathrm{ }\mathrm{Ratio}\mathrm{ }\mathrm{of}\mathrm{the}\mathrm{surface}\mathrm{areas}\mathrm{of}\mathrm{sphere}\mathrm{and}\mathrm{cylinder}\\ =\frac{4{\mathrm{\pi r}}^{\mathrm{2}}}{4{\mathrm{\pi r}}^{\mathrm{2}}}\\ =1\\ \mathrm{Thus},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{ratio}\mathrm{ }\mathrm{of}\mathrm{the}\mathrm{surface}\mathrm{areas}\mathrm{of}\mathrm{sphere}\mathrm{and}\mathrm{cylinder}\mathrm{is}1:1\mathrm{.}\end{array}$

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### 1. How can students access NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 for Exercise 13.4 Class 9 Maths?

Students can get NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 from the Extramarks website and mobile application, and download the PDF format for the same from there. That is simple to understand and beneficial to students preparing for board examinations and other entrance examinations.

### 2. Are of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 for Maths Class 9 Chapter 13 Exercise 13.4 updated according to the recent syllabus of Mathematics Class 9?

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. 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 can also access the Class 9 Chapter 13 Maths Exercise 13.4

### 3. How will the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4 assist with CBSE exam preparation?

By studying with the help of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4, students may excel in the CBSE examinations and become subject champions. These solutions were created utilising the most recent CBSE syllabus, which includes all the pertinent subject matter. As a result, responding to these questions will give students more confidence as they get ready for their board examinations. It also helps the students become used to responding to questions with various levels of difficulty. Students are highly encouraged to utilise these solutions as a resource and to be ready for the CBSE examinations.

### 4. What are the important topics of Chapter 13 - Surface Areas and Volume?

The important topics that have been covered in Chapter 13- Surface Areas and  Volumes are given below and also discussed in NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.4:

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