NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.4
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The Central Board of Secondary Education is one of the premier educational boards that operate in India, and it is also one of the most highly reputed educational boards in India as well. TThere are a multitude of schools that are affiliated with it. CBSE follows a specific pattern for its question papers that is entirely based on an objective educational paradigm. An objective educational paradigm entails discussions and questions which are entrenched deep into the fundamental ideas that are discussed in the syllabus. Students must be very careful when they study because if they do not have a very formidable grasp of the ideas that are being discussed in the syllabus, then they will fail to secure higher marks in the exams. The questions that appear in the exams are primarily short answer type questions, but students must never mistake objectivity for simplicity. The way the entire CBSE curriculum is designed helps students develop their analytical and cognitive abilities.This is done so that students can start to hone their skills for their future. Students must learn how to separate the important information in a chapter from the unnecessary information so that thay do not waste their time learning things that will never come up in the exams. The kind of environment that CBSE imbibes is very different from the way education has been in a traditional classroom situation.
CBSE is widely known among students to have a very vast syllabus. The syllabus that CBSE follows is curated based on contemporary educational trends in other parts of the world. Students often get overwhelmed by this distinct study pattern. Although teachers who have been a part of Extramarks for many years have shared their views on it. The syllabus might seem vast, but the vastness, coupled with the objective nature of the CBSE, gives a study plan which ensures that students actually learn something substantial. CBSE employs a system which is just and makes sure that students who have worked hard for their subjects get more marks. The only thing that students must do to get used to the CBSE pattern is to study regularly and consistently. Teachers have shared that if students incorporate little changes in their daily routines and tweak their studying habits a bit, there will be no stopping them from scoring well in the exams.
The National Council of Educational Research and Training (NCERT) is a centrally organised organisation that was founded by the Indian government in 1961. NCERT was founded with the vision that India would finally have an autonomous body which would legislate on matters related to the academic development of the country. NCERT comprises people who are highly trained and experienced in making difficult decisions to further the educational milieu of the country.
NCERT and CBSE share a very symbiotic relationship. CBSE and NCERT follow each other quite closely. NCERT puts out its own sets of rules and regulations. Students are ought to follow these rules and regulations to make sure that they do not lose marks over trivial matters. NCERT even curated their own syllabus and asked all the CBSE schools and the constituent teachers and students to follow the syllabus as well. NCERT has a separate branch in their executive department which publishes its own books. These books strictly follow the syllabus that NCERT has laid out. Students must treat their NCERT textbook as their primary textbook because regular practise with the NCERT textbooks makes sure that students are getting used to the distinct studying pattern. Teachers have reassured students that there is nothing to be afraid of in the CBSE paradigm.If students diligently listen to their teachers, then there is nothing they must fear. Mathematics is a very important subject for a CBSE Class 8 student.
Therefore, getting a good overall grade on the final exam is contingent on a student’s ability to score well in Mathematics. Teachers on the Extramarks team have years of experience teaching mathematics to students.As a result, these educators collaborated to create the NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.4.The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 provides the solutions to every unsolved Mathematics question in the NCERT textbook. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 follow the exact same chronology that can be followed in the NCERT textbook. This is done so that students find no hindrance while they are using the NCERT solutions.
The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 provide very brisk and comprehensive solutions to every unsolved question. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 are made with great patience and greatly experienced teachers. The way the solutions are provided in the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 is very meticulous, which makes it very easy to follow. The solutions are presented in an easytorefer to format that helps students understand difficult concepts.This is done because Extramarks wanted these solutions to be accessible to a majority of students, irrespective of their educational backgrounds. These solutions focus solely on Class 8 Maths Chapter 11 Exercise 11.4.
CBSE has a provision that allows its students to appear in the exam in two separate languages – English and Hindi. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 are primarily written in English. Although, the teachers in Extramarks have translated the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 into Hindi. These teachers are highly qualified, and they are very fluent in the language. This is to ensure that every student taking the Hindi exam can benefit from the NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.4.Students often get used to the English vocabulary of various different subjects through television or other media, but with time and advancement, fewer students opt for appearing for the Exam in Hindi. Therefore, there is a great lack of substantial and reliable resources in Hindi for students to follow. Therefore, the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 in Hindi is an amazing resource for such students. Teachers always encourage students to use the NCERT solutions when they are studying. It is reliable, free of errors, and easily accessible.
When students have NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 in their possession, they do not have to wait at all. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 provide stepbystep explanations for every problem, making sure that no students who use the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 must have any more doubts. NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.4 has become a popular educational resource for students.
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.4) Exercise 11.4
Extramarks has published the NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.4 on their website and mobile app.The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 are provided in PDF format. The solutions can be accessed online as well as through their mobile application. As a result, many students have benefited from the NCERT Solutions for Class 8 Maths, Chapter 11, Exercise 11.4.To access the Class 8 Maths Exercise 11.4 Solution, click on the link provided below.
Access NCERT Solutions for Class 8 Chapter 11 Mensuration
One of the other ways students seem to have been using the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 is while theyare revising. After proceeding with the syllabus and making significant progress in it, often, as students are told, they come back to the old chapters that they had solved long ago. Exercise 11.4 is very useful while they are revising the NCERT Solutions for Class 8 Maths Chapter 11Although, before students start solving the chapter, they must go through the ideas and the solved examples once again to refresh everything that they have learned. Students are told never to use the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 before trying to solve the problem by themselves. If doubt persists after a few attempts, students can refer to the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4. Click on the link below to access the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4.
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (Ex 11.4) Exercise 11.4
The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 provide solutions for the eleventh chapter of the NCERT syllabus for Class 8 students. The eleventh chapter is called Mensuration. Students learn about various kinds of shapes and their properties. Students learn about formulas which would be used by students to calculate quantities like the Area, Perimeter and Surface Area of the shapes. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 are an amazing resource for students who need help in this chapter. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 includes a diagrammatic representation of the problems and therefore the illustrations are well made and extremely useful. Class 8 Maths Exercise 11.4 is a complicated exercise, and students must deal with it with great precaution.
NCERT Solutions for Class 8
Teachers have observed that students use the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 in varied ways. When students are solving a chapter for the first time, the NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.4 comes in handy.When students are introduced to a new chapter, they are exposed to a lot of new information and ideas in a short span of time. Momentarily, students might seem to understand, but in their own time, they seem to have discrepancies in their understanding. Therefore, before starting to solve the problems by themselves, students should go through the chapter once. Students must go through the necessary theories, formulas, and solved examples. Students are only encouraged to use the NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.4 after that.Generally, when students are solving an exercise, they come across various doubts. In situations like this, they wait for their doubts to be resolved. Then they move on to the next question. Since students are solving the chapter for the first time, the original doubts are related to the questions; therefore, waiting for help often costs students a bunch of precious time. Students can access the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 online on the Extramarks website and the mobile application. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 are free of charge and therefore reach out to a huge percentage of students. Students should use these resources for a thorough understanding of all the important concepts. These solutions have been compiled by subjectmatter experts with years of experience and can be relied upon during exam preparation.
Students can access a variety of other learning resources from the Extramarks’ website, like the NCERT Solution, past years’ papers, sample question papers, and revision notes. Each student is preparing for or studying for competitive exams, such as the NEET, JEE, CUET, NDA, or a board exam, such as one for the state board, ICSE, or CBSE. Every exam has a distinct format, level of difficulty, and requirements for preparation. Students should be aware that whether they are in Class 11 or Class 12, the course material for these exams is largely the same. Students can lessen the stress and strain of lastminute preparation by starting their studies early. It offers students more opportunities to prepare for the test.
Q.1 Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
Ans
(a) We will find the volume.
(b) We will find the surface area.
(c) We will find the volume.
Q.2 Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
Ans
$\begin{array}{l}\text{The heights and diameters of these cylinders A and B are}\\ \text{interchanged}.\\ \text{If measures of \u2018r\u2019 and \u2018h\u2019 are same, then the cylinder with}\\ \text{greater radius will have greater volume}.\\ \text{Radius of cylinder A}=\frac{7}{2}\text{cm}\end{array}$ $\begin{array}{l}\text{Radius of cylinder B}=\frac{14}{2}\text{cm}=\text{7 cm}\\ \\ \text{As the radius of cylinder B is greater},\text{therefore},\text{the volume}\\ \text{of cylinder B will be greater}.\\ \\ \text{Let us verify it by calculating the volume of both the cylinders}.\\ \text{Volume of cylinder A}=\mathrm{\pi}{\text{r}}^{\text{2}}\text{h}\\ =(\frac{22}{7}\times \frac{7}{2}\text{\xd7}\frac{7}{2}\text{\xd714}){\text{cm}}^{\text{3}}\\ =539{\text{\hspace{0.17em}cm}}^{\text{3}}\\ \text{Volume of cylinder B}=\mathrm{\pi}{\text{r}}^{\text{2}}\text{h}\\ =(\frac{22}{7}\times 7\text{\xd7}7\text{\xd77}){\text{cm}}^{\text{3}}\\ =1078{\text{\hspace{0.17em}cm}}^{\text{3}}\\ \\ \mathrm{Hence},\text{volume of cylinder B is greater.}\end{array}$ \begin{array}{l}\text{}\end{array} $\begin{array}{l}\text{Surface area of cylinder A=2}\mathrm{\pi}\text{r(h+r)}\\ =\left[\text{2\xd7}\frac{\text{22}}{7}\text{\xd7}\frac{7}{2}\left(\frac{7}{2}\text{+14}\right)\right]{\mathrm{cm}}^{2}\\ =[22\times \left(\frac{7+28}{2}\right)]{\mathrm{cm}}^{2}\\ =385\text{}{\mathrm{cm}}^{2}\\ \\ \text{Surface area of cylinder B=2}\mathrm{\pi}\text{r(h+r)}\\ =\left[\text{2\xd7}\frac{\text{22}}{7}\text{\xd7}7\left(7\text{+7}\right)\right]{\mathrm{cm}}^{2}\\ =[44\times 14]{\mathrm{cm}}^{2}\\ =616\text{}{\mathrm{cm}}^{2}\\ \\ \text{Thus},\text{the surface area of cylinder B is also greater than}\\ \text{the surface area of cylinder A.}\end{array}$
Q.3 Find the height of a cuboid whose base area is 180 cm^{2} and volume is 900 cm^{3}?
Ans
$\begin{array}{l}\text{Base area of the cuboid}=\text{18}0{\text{cm}}^{\text{2}}\\ \Rightarrow \text{Length}\times \text{Breadth}=\text{18}0{\text{cm}}^{\text{2}}\\ \end{array}$ $\begin{array}{l}\text{Volume of cuboid}=\text{Length}\times \text{Breadth}\times \text{Height}\\ \Rightarrow \text{9}00{\text{cm}}^{\text{3}}=\text{18}0{\text{cm}}^{\text{2}}\times \text{Height}\\ \Rightarrow \text{Height}=\frac{900}{180}=5\text{}\mathrm{cm}\\ \\ \text{Thus},\text{the height of the cuboid is 5 cm}.\end{array}$
Q.4 A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Ans
$\begin{array}{l}\text{Volume of cuboid}=\text{6}0\text{cm}\times \text{54 cm}\times \text{3}0\text{cm}=\text{972}00{\text{cm}}^{\text{3}}\\ \\ \text{Side of the cube}=\text{6 cm}\\ \text{Volume of the cube}={\left(\text{6}\right)}^{\text{3}}{\text{cm}}^{\text{3}}={\text{216 cm}}^{\text{3}}\\ \text{Required number of cubes}=\frac{\text{Volume of the cuboid}}{\text{Volume of the cube}}\\ =\frac{\text{972}00{\text{cm}}^{\text{3}}}{{\text{216 cm}}^{\text{3}}}=450\\ \\ \therefore \text{45}0\text{cubes can be placed in the given cuboid}.\end{array}$
Q.5 Find the height of the cylinder whose volume is 1.54 m^{3} and diameter of the base is 140 cm?
Ans
$\begin{array}{l}\mathrm{Given},\\ \mathrm{Volume}\text{}\mathrm{of}\text{}\mathrm{cylinder}=1.54\text{}{\mathrm{m}}^{3}\\ \mathrm{Diameter}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{base}\text{}=\text{}140\text{}\mathrm{cm}\\ \Rightarrow \mathrm{Radius}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{base}=\text{}\left(\frac{140}{2}\right)\text{}\mathrm{cm}=70\text{}\mathrm{cm}=\frac{70}{100}\mathrm{m}\\ \\ \mathrm{Volume}\text{}\mathrm{of}\text{}\mathrm{cylinder}={\mathrm{\pi r}}^{2}\mathrm{h}\\ 1.54\text{}{\mathrm{m}}^{3}=\frac{22}{7}\times \frac{70}{100}\times \frac{70}{100}\times \mathrm{h}\\ \Rightarrow \mathrm{h}=\left(\frac{1.54\times 100}{22\times 7}\right)\mathrm{m}=1\text{}\mathrm{m}\\ \\ \mathrm{Thus},\text{}\mathrm{the}\text{}\mathrm{height}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{cylinder}\text{}\mathrm{is}\text{}1\text{}\mathrm{m}.\end{array}$
Q.6 A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?
Ans
$\begin{array}{l}\mathrm{Radius}\text{}\mathrm{of}\text{}\mathrm{cylinder}\text{}=\text{}1.5\text{}\mathrm{m}\\ \mathrm{Length}\text{}\mathrm{of}\text{}\mathrm{cylinder}\text{}=\text{}7\text{}\mathrm{m}\\ \mathrm{Volume}\text{}\mathrm{of}\text{}\mathrm{cylinder}={\mathrm{\pi r}}^{2}\mathrm{h}\\ =\frac{22}{7}\times 1.5\times 1.5\times 7\\ =49.5\text{}{\mathrm{m}}^{3}\\ \\ \mathrm{We}\text{}\mathrm{know}\text{}\mathrm{that}\text{},1{\mathrm{m}}^{3}=\text{}1000\text{}\mathrm{L}\\ \\ \therefore \mathrm{The}\text{}\mathrm{quantity}\text{}\mathrm{of}\text{}\mathrm{milk}\text{}\mathrm{in}\text{}\mathrm{liters}=(49.5\text{}\times \text{}1000)\mathrm{L}\text{}\\ \text{}=\text{}49500\text{}\mathrm{L}\\ \\ \mathrm{Therefore},\text{}49500\text{}\mathrm{L}\text{}\mathrm{of}\text{}\mathrm{milk}\text{}\mathrm{can}\text{}\mathrm{be}\text{}\mathrm{stored}\text{}\mathrm{in}\text{}\mathrm{the}\text{}\mathrm{tank}.\end{array}$
Q.7 If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Ans
$\begin{array}{l}\text{Let initially the edge of the cube be \u2018x\u2019}.\\ \text{(i})\text{Initial surface area}={\text{6x}}^{\text{2}}\\ \text{If each edge of the cube is doubled},\text{then it becomes 2x}.\\ \end{array}$ $\begin{array}{l}\therefore \text{New surface area}=\text{6}{\left(\text{2x}\right)}^{\text{2}}\\ ={\text{24x}}^{\text{2}}\\ =\text{4}\times {\text{6x}}^{\text{2}}\\ \\ \text{Hence},\text{the surface area will be increased by 4 times}.\\ \\ \left(\text{ii}\right)\text{Initial volume of the cube}={\mathrm{x}}^{\text{3}}\\ \text{When each edge of the cube is doubled},\text{it becomes 2}\mathrm{x}.\\ \text{New volume}={\left(\text{2x}\right)}^{\text{3}}=\text{8}{\mathrm{x}}^{\text{3}}\text{=8}\times {\mathrm{x}}^{\text{3}}\\ \\ \therefore \text{The volume of the cube will be increased by 8 times}.\end{array}$
Q.8 Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m^{3}, find the number of hours it will take to fill the reservoir.
Ans
$\begin{array}{l}\text{Volume of cuboidal reservoir}=\text{1}0{\text{8 m}}^{\text{3}}\\ =(\text{1}0\text{8}\times \text{1}000)\text{L}\\ =\text{1}0\text{8}000\text{L}\\ \\ \text{It is given that water is being poured at the rate of 6}0\text{L per min}.\\ \mathrm{i}.\mathrm{e}.(\text{6}0\times \text{6}0)\text{L}=\text{36}00\text{L per hour}\\ \\ \text{Required number of hours}=\frac{108000}{3600}\text{= 3}0\text{hours}\\ \\ \therefore \mathrm{I}\text{t will take 3}0\text{hours to fill the reservoir}.\end{array}$
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1. What are the other resources available on the Extramarks website for NCERT besides the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4?
The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 provide solutions to one exercise. Similar resources are available for the following courses.
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The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 help students in dealing with –
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The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.4 is a wellresearched resource made by competent teachers and therefore can be highly trusted by students. Solutions like the NCERT Class 8 Maths Chapter 11 Exercise 11.4 in Hindi and English can be accessed through the link below.