# NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.1

When developing NCERT Solutions, the whole curriculum of the Central Board of Secondary Examination (CBSE) for classes 1 to 12 is taken into account. Students are frequently recommended to use NCERT Solutions while they study for their board examinations and competitive tests. Due to the fact that these solutions are curated by subject-matter experts, their relevance, correctness, and comprehensiveness may be trusted. NCERT solutions offer solutions to several academic disciplines.

Students of junior classes should have learned about the Area and Perimeter of various closed planar figures, such as Triangles, Rectangles, Circles, and so on.  In this section, “Mensuration,” students will learn how to deal with Perimeter-related issues as well as those involving other Plane closed forms like quadrilaterals.Mensuration. Students would next learn about the Volume and Surface Area of several solid forms, such as a Cube, Cuboid, and Cylinder. For students of the CBSE Board studying for the annual exams, the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 are regarded as the best choice. There are many exercises in this chapter. Students can struggle to Complete NCERT Class 8 Maths Chapter 11 Exercise 11.1. They can get the Class 8 Maths Exercise 11.1 Solution as a part of the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 which are available on the Extramarks website. Students can get the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 directly from the Extramarks website or mobile application, or they can download these if necessary.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.1) Exercise 11.1

Extramarks, the most popular e-learning website, provides download links for NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1.The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 are offered by Extramakers and are written in a straightforward, plain, and easy-to-understand manner.

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### Access NCERT Solutions For Class 8 Maths Chapter 11 – Mensuration

The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 are available for convenient download by students on the Extramarks website. The solutions compiled by Extramarks, such as the Class 8 NCERT solutions, have been engineered in collaboration with knowledgeable teachers. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 are laid forth in simple, understandable language. There are several benefits to using the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1. The biggest benefit of these solutions is that they serve as guidelines for examination preparation.

### NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (Ex 11.1) Exercise 11.1

The NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 are regarded as the best option for CBSE students preparing for exams.There are several exercises in this chapter. Extramarks provides the NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 in PDF format.Students can study the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 straight from the Extramarks website or mobile application, or they can download these as needed.

Extramarks’ in-house subject matter experts carefully and in accordance with all CBSE regulations solved the problems and questions from the exercise. Any student in Class 8 who is thoroughly familiar with all the ideas from the Mathematics textbook and is sufficiently knowledgeable about all the activities included in it can easily secure remarkable grades on the final exam. Students may readily grasp the types of questions that could be asked in the exam from this chapter and learn the chapter’s weightage in terms of overall grade by using the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 in order to adequately study for the annual Mathematics examination.

There are several exercises in this chapter that contain numerous questions in addition to Exercise 11.1 encapsulated in NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1. As previously noted, Extramarks’ in-house mentors have already resolved or responded to all of these inquiries. Dueto this, these solutions are all guaranteed to be of the highest quality, and anybody may use them to study for exams. It is crucial to comprehend all the ideas in the textbooks and work through the exercises that are provided to receive excellent results in the annual exams.

Students can use the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 to help them remember how to appropriately apply various formulae to solve challenging problems. They will perform well in the test. Students can review these ideas from earlier chapters if they run intodifficulties.

For better test preparation, students are advised to download the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 from the Extramarks website at their earliest convenience. If students already have the Extramarks mobile application on their phones, they may also download these there. The best feature of these solutions is that they can be used offline and online. Important Formulas for the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1

To understand the solutions provided in Class 8 Maths Chapter 11 Exercise 11.1, students must use formulas for the area and perimeter of forms.TheseThese solutions provide a concise summary of the knowledge of Mensuration students have acquired so far. This section discusses some typical forms, including Squares, Rectangles, Triangles, Parallelograms, and Circles.

### NCERT Solutions Class 8 Maths Chapter 11 Other Exercises

The Class 8 Mathematics CBSE syllabus can be revised using the Class 8 Mathematics NCERT problems that are supplied with the relevant solutions. Numerous exercises of all kinds, including hundreds of questions, are included throughout the textbook. There are a series of questions in the corresponding textbook exercise pertaining to Chapter 11, which is titled Exercise 11.2 of the NCERT Solutions Class 8 Maths. The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 contain the solutions to the aforementioned chapter. Students can use the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 to ensure that they fully comprehend each idea in this chapter.

Exercise 11.3 Solutions: 10 Questions

The NCERT textbooks are indispensable resources within the CBSE curriculum, thereby leading to the possibility of the exercise questions provided in the NCERT textbook appearing in the Class 8 annual exams. Clarity of concepts is ensured by the CBSE Class 8 NCERT solutions to aid students with preparation for examinations. Students could find complete responses compiled by Extramarks’ subject-matter experts as a part of the NCERT solutions for Class 8 Maths Chapter 11, Exercise 11.3 available on the Extramarks learning portal. Along with these solutions, Extramarks provides the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 for preparation.

Exercise 11.4 Solutions: 8 Questions

Class 8, Chapter 11 Mensuration problems are offered on Extramarks with thorough solutions, which help review implicit themes and concepts. Subject matter experts use a step-by-step problem-solving technique to resolve the questions in Chapter 11 Exercise 11.4 of the Class 8 CBSE Mathematics textbook. All the materials required to do better in examinations are available on the Extramarks website. Each step of the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 must be completed by students. Additionally, learners should continue to refer to the solutions for Class 8 Maths Exercise 11.1 while preparing for the Mathematics examinations.

### Chapter-wise NCERT Solutions for Class 8 Maths

Extramarks offers NCERT solutions for all the chapters of the Class 8 Mathematics textbook.

Chapter 1 – Rational Numbers

Chapter 2 – Linear Equations in One Variable

Chapter 4 – Practical Geometry

Chapter 5 – Data Handling

Chapter 6 – Squares and Square Roots

Chapter 7 – Cubes and Cube Roots

Chapter 8 – Comparing Quantities

Chapter 9 – Algebraic Expressions and Identities

Chapter 10 – Visualising Solid Shapes

Chapter 11 – Mensuration

Chapter 12 – Exponents and Powers

Chapter 13 – Direct and Inverse Proportions

Chapter 14 – Factorisation

Chapter 15 – Introduction to Graphs

Chapter 16 – Playing with Numbers

### NCERT Solutions for Class 8

Each chapter should be thoroughly studied by students, who are also required to understand the definitions of all the themes and subtopics. Understanding the principles of each chapter is crucial to successfully preparing the chapters. Solving problems is the best way to understand the foundational concepts of mathematics. Students should deal with the easy questions first before moving on to the challenging ones. All students must adequately prepare for the annual exams. Students must become familiar with the syllabus before beginning their preparation for the Mathematics examination. The syllabus is necessary to design a strategy for learning the chapters.

Students are recommended to use the Class 8 NCERT solutions to practise examples and answer practise problems to have a better understanding. Volume and Capacity are important themes in this chapter; students must learn about Capacity and its formula. Students must give detailed answers to respond to any related inquiries. Students must also understand the distinction between Volume and Capacity. Students should work through the exercise questions and then practise some more using the NCERT solutions.

Additionally, students must pay close attention to the examples and sample questions provided along with the exercises. Before attempting to solve the exercise questions, students should resolve queries based on practise problems and applications of various formulae. Students ought to use the NCERT solutions as quality reference materials as they go through the exercise problems, as all the concepts are very well presented in these solutions and will make learning simpler.

Students can use the Extramarks website or mobile application to acquire the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1. A trustworthy resource to use are the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1.

Q.1 A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area? Ans

$\begin{array}{l}\mathrm{Perimeter}\text{}\mathrm{of}\text{}\mathrm{square}=4×\mathrm{side}\\ =\text{}4×60\text{}\mathrm{m}\text{}\\ =\text{}240\text{}\mathrm{m}\\ \mathrm{Perimeter}\text{}\mathrm{of}\text{}\mathrm{rectangle}=2\left(\mathrm{Length}+\mathrm{Breadth}\right)\\ =\text{}2\left(80\text{}\mathrm{m}+\mathrm{Breadth}\right)\\ =\text{}160\text{}\mathrm{m}+2×\mathrm{Breadth}\end{array}$

$\begin{array}{l}\mathrm{Perimeter}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{square}=\mathrm{Perimeter}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{rectangle}\text{}\\ ⇒160\text{}\mathrm{m}+2×\mathrm{Breadth}\text{}=240\text{}\mathrm{m}\\ ⇒\mathrm{Breadth}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{rectangle}=\left(\frac{80}{2}\right)\mathrm{m}=40\text{}\mathrm{m}\\ \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{square}={\left(\mathrm{Side}\right)}^{2}={\left(60\text{}\mathrm{m}\right)}^{2}=3600\text{}{\mathrm{m}}^{2}\\ \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{rectangle}=\mathrm{Length}×\mathrm{Breadth}\\ =\left(80×40\right){\mathrm{m}}^{2}\\ =3200\text{}{\mathrm{m}}^{2}\\ \\ \mathrm{Therefore},\text{}\mathrm{the}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{square}\text{}\mathrm{field}\text{}\mathrm{is}\text{}\mathrm{larger}\text{}\mathrm{than}\\ \mathrm{the}\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{rectangular}\text{}\mathrm{field}.\end{array}$

Q.2 Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m2. Ans

$\begin{array}{l}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{square}\text{}\mathrm{plot}={\left(25\text{}\right)}^{2}{\mathrm{m}}^{2}=625\text{}{\mathrm{m}}^{2}\\ \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{house}=\left(15\text{}\mathrm{m}\right)×\left(20\text{}\mathrm{m}\right)=300\text{}{\mathrm{m}}^{2}\\ \\ \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{remaining}\text{}\mathrm{portion}=\left(\begin{array}{l}\mathrm{Area}\text{}\mathrm{of}\text{}\\ \mathrm{square}\text{}\mathrm{plot}\end{array}\right)-\left(\begin{array}{l}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\\ \text{}\mathrm{house}\end{array}\right)\\ =625\text{}{\mathrm{m}}^{2}-300\text{}{\mathrm{m}}^{2}\\ =\text{}325\text{}{\mathrm{m}}^{2}\\ \\ \therefore \mathrm{Total}\text{}\mathrm{cost}\text{}\mathrm{of}\text{}\mathrm{developing}\text{}\mathrm{a}\text{}\mathrm{garden}\text{}\mathrm{around}\text{}\mathrm{the}\\ \text{}\mathrm{house}\text{}\mathrm{at}\text{}\mathrm{the}\text{}\mathrm{rate}\text{}\mathrm{of}\text{}\mathrm{Rs}\text{}55\text{}\mathrm{per}\text{}{\mathrm{m}}^{2}=₹\left(55×325\right)\\ =\text{}₹\text{}17,875\end{array}$

Q.3 The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres]. Ans

$\begin{array}{l}\text{Length of the rectangle}=\left[\text{2}0-\left(\text{3}.\text{5}+\text{3}.\text{5}\right)\right]\text{metres}=\text{13 m}\\ \text{Circumference of 1 semi}-\text{circular part}=\mathrm{\pi r}=\frac{22}{7}×3.5\text{}=11\text{}\mathrm{m}.\end{array}$ $\begin{array}{l}\therefore \text{}\mathrm{Perimeter}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{garden}=\mathrm{AB}+\mathrm{Length}\text{}\mathrm{of}\text{}\mathrm{both}\text{}\\ \text{}\mathrm{semi}–\mathrm{circular}\text{}\mathrm{regions}+\mathrm{CD}\\ =\text{}13\text{}\mathrm{m}+22\text{}\mathrm{m}+13\text{}\mathrm{m}\text{}\\ =\text{}48\text{}\mathrm{m}\\ \\ \mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{garden}=\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{rectangle}+2×\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{two}\\ \text{}\mathrm{semi}–\mathrm{circular}\text{}\mathrm{regions}\\ =\left[\left(13×7\right)+2×\frac{1}{2}×\frac{22}{7}×{\left(3.5\right)}^{2}\right]{\mathrm{m}}^{2}\\ =\left(91+38.5\right){\mathrm{m}}^{2}\\ =129.5\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \\ \therefore \mathrm{Area}\text{and perimeter}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{garden}\text{}\mathrm{are}\text{}129.5\text{\hspace{0.17em}}{\mathrm{m}}^{2}\text{}\mathrm{and}\text{\hspace{0.17em}48 m.}\end{array}$

Q.4 A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).

Ans

$\begin{array}{l}\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{parallelogram}=\mathrm{Base}×\mathrm{Height}\\ \mathrm{Hence},\text{}\mathrm{area}\text{}\mathrm{of}\text{}\mathrm{one}\text{}\mathrm{tile}=24\text{}\mathrm{cm}×10\text{}\mathrm{cm}=240\text{}{\mathrm{cm}}^{2}\\ \mathrm{Required}\text{}\mathrm{number}\text{}\mathrm{of}\text{}\mathrm{tiles}=\frac{\mathrm{Area}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{}\mathrm{floor}}{\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{each}\text{\hspace{0.17em}}\mathrm{tile}}=\frac{1080\text{}{\mathrm{m}}^{2}}{240\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\text{}}\\ \mathrm{Since}\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\mathrm{m}=100\text{\hspace{0.17em}}\mathrm{cm}\\ ⇒1\text{\hspace{0.17em}}{\mathrm{m}}^{2}=10000\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \therefore 1080\text{}{\mathrm{m}}^{2}=10000×1080\text{}{\mathrm{cm}}^{2}\\ \text{}=10,800,000\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \mathrm{Hence},\mathrm{the}\text{}\mathrm{required}\text{}\mathrm{number}\text{}\mathrm{of}\text{}\mathrm{tiles}\text{}=\frac{10,800,000\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{cm}}^{2}}{240\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\text{}}\\ \text{}=45000\end{array}$

Q.5 An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle. Ans

$\begin{array}{l}\left(\text{a}\right)\text{Radius of semi-circular part}=\left\{\frac{2.8}{2}\right\}\text{cm =1.4 cm}\\ \text{Perimeter of the given figure}=\text{2}.\text{8 cm}+\text{}\mathrm{\pi r}\\ =\text{2}.\text{8 cm}+\left(\frac{22}{7}×1.4\right)\mathrm{cm}\\ =\text{2}.\text{8 cm}+4.4\text{​}\mathrm{cm}\\ =7.2\text{​}\mathrm{cm}\end{array}$

$\begin{array}{l}\left(\text{b}\right)\text{Radius of semi-circular part}=\left\{\frac{2.8}{2}\right\}\text{cm=1.4 cm}\\ \text{Perimeter of the given figure}=\text{1}.\text{5 cm}+\text{2}.\text{8 cm}+\text{1}.\text{5 cm}+\mathrm{\pi }\left(\text{1}.\text{4 cm}\right)\\ =5.\text{8 cm}+\frac{22}{7}×1.4\text{}\mathrm{cm}\\ =5.\text{8 cm}+4.4\text{​}\mathrm{cm}\\ =10.2\text{​}\mathrm{cm}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\text{(c}\right)\text{Radius of semi-circular part}=\left\{\frac{2.8}{2}\right\}\text{cm=1.4 cm}\\ \text{Perimeter of the given figure}=\text{2 cm}+\mathrm{\pi r}+\text{2 cm}\\ =4\text{cm}+\frac{22}{7}×1.4\text{}\mathrm{cm}\\ =4\text{cm}+4.4\text{​}\mathrm{cm}\\ =8.4\text{​}\mathrm{cm}\end{array}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{Thus},\text{the ant will have to take a longer round for the food}\\ \text{piece}\left(\text{b}\right),\text{because the perimeter of the figure given in}\\ \text{option}\left(\text{b}\right)\text{is the greatest among all}.\end{array}$ $\begin{array}{l}\end{array}$

## 1. Why are the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 crucial for better understanding?

As it relates to calculating the areas of various shapes, this chapter is meant to expand on the students’ understanding of Quadrilaterals and Triangles. Additionally, students will study numerous formulas for calculating the area and volume of Cubes and Cuboidal Constructions. Every question in the exercises for this chapter must be answered, therefore students must make sure that they practice on a regular basis. To strengthen their fundamental understanding, learners need to take reference from the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1.  Problems are solved systematically odder with accuracy so students do not face any difficulty. They can download the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 from the Extramarks learning platform. These solutions are also present on the Learning App of Extramarks.

## 2. Are the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 helpful to the students?

The NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 provide comprehensive justifications in simple terms to help students perform well on exams. In accordance with the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1, to calculate the area and volume of Cubes and Cuboidal forms, certain formulae must be applied. The student’s understanding of the concepts will improve thanks to this solution. Students should use the numerous formulas for calculating the area and volume of Cubes and Cuboidal constructions as a quick and efficient review tool.

## 3. What does Mensuration mean?

Mensuration is a branch of Mathematics that focuses on the analysis of various geometrical forms, including their Perimeters, Areas, and Volumes. Geometric calculations and the application of algebraic equations form its whole foundation. There is a fair amount of accuracy in the results of the Mensuration. There is a fair amount of accuracy in the results of the Mensuration.

## 4. What are the important formulas of the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1?

There are a lot of important formulas in the NCERT Solutions For Class 8 Maths Chapter 11 Exercise 11.1 that will help students to understand the major concepts. Some of the important formulas are the following:

• Square

Area = Side x Side

Perimeter = 4 × Side

• Rectangle

Perimeter = 2 (Length + Breadth)

• Triangle

Area = 1/2 x Base x Height

Perimeter Sum of all three sides

• Circle