NCERT Solutions For Class 6 Maths Chapter 10 Mensuration (Ex 10.1) Exercise 10.1
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NCERT Solutions For Class 6 Maths Chapter 10 Mensuration (Ex 10.1) Exercise 10.1
Almost every class in the CBSEaffiliated schools, studies from the NCERT textbooks. The NCERT book is the accepted text for CBSE board students pursuing Maths. The NCERT books are authored by CBSE specialists with considerable care and expert knowledge. The NCERT books are written in a way that makes it easier for students to understand and grasp concepts. In establishing their fundamentals, NCERT textbooks have consistently proven to be of immense assistance to students. Students may get the most out of the NCERT books by using the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1.
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ToggleThere are certain benefits of the NCERT books –

Organised theory:
The theories presented in the NCERT texts are quite organised and complex. It is of great benefit to students to have the theory organised in a manner that makes it easier to understand. The visual aids in the NCERT textbooks make it easier for students to comprehend the theory presented in written form. The balance between written theory and visual representation in the NCERT book is excellent. Too many drawings could prevent students from understanding the theory, and too many textbased explanations could prevent them from understanding it without the right visual representation. Students can use the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 and more to comprehend the content of the NCERT books.

Simpler Questions First:
The questions in the NCERT books are simple to comprehend and may be answered by using the technique covered in class. Consequently, students may have an easier time applying the theory they have learned. They are able to comprehend how to apply the theoretical knowledge they have gained in order to solve realworld problems. It is likely that students would not have been able to find answers to difficult questions if they had been presented at the beginning of the chapter. Consequently, their understanding of the subject would have been impaired. By answering the straightforward questions at the end of NCERT chapters, students can build confidence in their ability to solve maths problems. As students gain a better understanding of how the theory is applied, they can tackle some of the more difficult problems in the latter half of the chapters. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 and other similar resources on the Extramarks website can be used by the students to answer the NCERT questions.

Helps in Building Confidence:
If students have just learned a particular theory and are expected to respond correctly to a difficult question, they will undoubtedly find it difficult to answer, which will discourage them from responding to further questions.With the use of the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1, students may learn more effectively. By encouraging students to begin with simpler questions, the NCERT solutions provide them with the confidence that they are capable of completing the maths problems presented in the textbook. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 and other resources for Chapter 10 of Class 6 Maths also include solved examples.
After solving the challenging questions in the latter part of the chapter, students gain a better understanding of how questions can be framed and how their knowledge of the subject has grown. Many students may find the challenging questions to be daunting, but by practising with the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1, students can become more confident in this area. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 and other resources available on the Extramarks website include the answers to all the questions covered in exercises and examples.

More Indepth Knowledge:
The structured theory and the visuals help the students comprehend the concepts in the book more deeply, as well as the reasons why they are significant and the specific messages they want to convey. The solved problems that follow each idea and the theory presented in the book serve as examples of how a concept can be used to solve difficulties. The solutions to the issues are really helpful to the students. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 and other related resources on the Extramarks website can be used by the students to find the answers.
The students’ acquisition of a little better understanding of the content is aided by the combination of a complete theory, a visual depiction of their theory wherever needed, and working through different types of questions. Many students—even those studying for competitive exams—return to NCERT reference books repeatedly to build their foundation and understand concepts. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 are also useful for students who may be preparing for competitive exams.
If students carefully study the index of the NCERT book, they will find that the themes are presented in a structured order; the curriculum is wellorganized, and the order in which the topics are covered is highly systematic. The prerequisites must be understood by a student before they can learn any subject. The NCERT book tremendously aids this learning process by first presenting the background information required to comprehend the next topic before moving on to it. With the aid of resources like The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1, students may get an even better grasp of the NCERT books. Maths is generally viewed as a subject that isn’t for everyone, and without such indepth, organized study material, students can become lost or find the topic challenging to study.
Utilizing tools like the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 and others available on the Extramarks website can assist students in strengthening their basic understanding of concepts from an examination perspective.
It can be difficult to understand and use Maths, both in theory and in practice. Using NCERT books and resources, such as the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1, students can learn mathematical topics and concepts more effectively. Students can have a thorough understanding of the study material by answering a large number of questions from the NCERT book. Further assistance can be found by students in the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1.
There is no doubt that the NCERT books make learning a difficult and complex subject incredibly enjoyable, which facilitates a successful learning process. Great resources for students who are just starting their Maths studies are the NCERT Maths textbooks. Students not enrolled with the CBSE board can also refer to the Class 6 Maths NCERT Book and the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 to make their learning process better and easier. Clarifying fundamental concepts will assist students in becoming proficient in Maths. Students can use the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 available on the Extramarks website and mobile application.
Access Other Exercises of Class 6 Maths Chapter 10
Chapter 10 – Mensuration Exercises  
Exercise 10.2 
1 Questions & Solutions

Exercise 10.3 
12 Questions & Solutions

NCERT Solutions For Class 6 Maths Chapter 10 Mensuration (Ex 10.1) Exercise 10.1
The field of Maths is one of the most influential and inspiring. It supports students in enhancing their logical and analytical skills. The topics covered in Maths for CBSE Class 6 include Fractions, Decimals, Algebra, Geometry, and Mensuration.
In order to help students develop their mathematical concepts more effectively, the Extramarks website provides significant questions and answers for Class 6 Maths that adhere to the most recent CBSE syllabus. These important questions help students comprehend the subject matter more thoroughly and prepare for exams in a methodical manner. It is possible to achieve this by using the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 which is designed specifically for Class 6th Exercise 10.1. Students of Class 6 can access similar tools for all the exercises of all chapters.
Access NCERT Solutions For Class 6 Maths Chapter 10 – Mensuration
The National Council of Educational Research and Training was founded in 1969. On the Extramarks website, students may find the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 in addition to other solutions to aid students with their preparation. Students of Class 6 preparing for the Maths examination may acquire the required learning resources from Extramarks by registering themselves on the Extramarks website.
The main objective of NCERT and its departments is to conduct, promote, and coordinate research in the field of education. There are a number of resources that the NCERT publishes, including sample textbooks, supplemental resources, newsletters, magazines, and instructional kits. In particular, students whose schools use the NCERT curriculum are encouraged to make use of the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1.
The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 are provided by Extramarks for Class 6 students to help them with their preparation process. The NCERT books provide practise questions at the end of each chapter. These questions provide a deeper understanding of the subject matter discussed in the preceding chapters. On the Extramarks website, students can get the solutions to the NCERT Class 6 Maths Chapter 10 Exercise 10.1 and other NCERT book problems.
When students run into questions that they might find difficult or lengthy to solve, they may copy other students’ solutions rather than trying to work things out on their own. For students to become proficient in Class 6 Maths Chapter 10 Exercise 10.1 and to learn how to solve problems on their own, it is recommended that theystudy the solutions for Class 6th Maths Chapter 10 Exercise 10.1 available on the Extramarks website. In addition to the Class 6 Maths Chapter 10 Exercise 10.1 Solution, there are a number of other study tools and learning materials available on Extramarks to help students prepare for their exams. Regularly practising NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 can really help students gain confidence in their problemsolving skills.
Students must be made aware of the significance of understanding the theoretical foundations of each chapter right away. If students don’t grasp the core ideas of any subject, they could struggle to complete projects, participate in class quizzes, or do well on final exams. The answers in Class 6 Maths Ex 10.1 solutions that the professionals at Extramarks have created should be carefully reviewed by students in order to truly understand the exercise.
The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 are specifically for the Class 6th Math Chapter 10.1. But along with the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1, there are solutions to all the other exercises that are also available on the Extramarks website. Students may learn how to approach various question types and come up with correct answers for all of them by using the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1.
Students should be made aware early on of how important it is for them to understand the theoretical basis of each chapter. Without a clear understanding of each subject’s core ideas, students may have trouble completing their homework and may do poorly on examinations.
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The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 and other solutions of a similar nature have been developed by experts at Extramarks with the needs of students in mind. The materials are therefore suitable for students from a variety of academic backgrounds. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 are just one of the resources that Extramarks’ Maths specialists have developed for the preparatory needs of students in relation to Math Ex 10.1 Class 6.
Students who need help with Maths Class 6 Ex 10.1 can acquire study material from Extramarks for assistance. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 are one of the resources that may be accessed by students from the Extramarks website or mobile application. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 are developed by experts in the concerned field with the preparatory requirements of students in mind.
Q.1 Find the perimeter of each of the following figures:
Ans.
(a) Perimeter of figure = 4 cm + 2 cm + 1 cm
+ 5 cm
= 12 cm
(b) Perimeter of figure = 23 cm + 35 cm + 35 cm
+ 40 cm
= 133 cm
(c) Perimeter of figure = 15 cm + 15 cm + 15 cm
+ 15 cm
= 60 cm
(d) Perimeter of figure = 4 cm + 4 cm + 4 cm
+ 4 cm + 4 cm
= 20 cm
(e) Perimeter of figure = 2.5 cm + 0.5 cm + 4 cm
+ 1 cm + 4 cm + 0.5 cm
+ 2.5 cm
= 15 cm
(f) Perimeter of figure = 4 cm + 1 cm + 3 cm
+ 2 cm+ 3 cm+ 4 cm+ 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm
= 52 cm
Q.2 The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Ans.
Since, lid of rectangular box is like a rectangle. So,
Length of tape = 2(length + Breadth)
= 2(40 + 10)
= 100 cm
= 1 m
Q.3 A tabletop measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?
Ans.
Length of tabletop = 2 m 25 cm
= 2.25 m
Breadth of tabletop = 1 m 50 cm
= 1.5 m
Perimeter of tabletop = 2(Length + Breadth)
= 2(2.25 + 1.50)
= 2(3.75)
= 7.50 m
Q.4 What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Ans.
Length of photograph = 32 cm
Breadth of photograph = 21 cm
Perimeter of photograph = 2(Length + Breadth)
= 2(32 + 21)
= 106 cm
Q.5 A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Ans.
Length of rectangular piece of land = 0.7 km
Breadth of rectangular piece of land = 0.5 km
Perimeter of rectangular piece of land = 2(0.7 + 0.5)
= 2(1.2) km
= 2.4 km
Length of required wire = 4(2.4) km
= 9.6 km
[Since, Length of required wire = 4(Perimeter of land)]
Q.6 Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Ans.
(a) Perimeter of triangle = 3 cm + 4 cm + 5 cm
= 12 cm
(b) Perimeter of equilateral triangle
= 9 cm + 9 cm + 9 cm
= 27 cm
(c) Perimeter of isosceles triangle
= 8 cm + 8 cm + 6 cm
= 22 cm
Q.7 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Ans.
Perimeter of triangle = 10 cm + 14 cm + 15 cm
= 39 cm
Q.8 Find the perimeter of a regular hexagon with each side measuring 8 m.
Ans.
Perimeter of regular hexagon = 6(Length of side)
= 6(8 cm)
= 48 cm
Q.9 Find the side of the square whose perimeter is 20 m.
Ans.
$\begin{array}{l}\mathrm{Perimeter}\text{of square}=\text{4}\times \text{side}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of square}=\frac{\mathrm{Perimeter}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of square}=\frac{20\text{\hspace{0.17em}}\mathrm{m}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\text{\hspace{0.17em}}\mathrm{m}\end{array}$
Q.10 The perimeter of a regular pentagon is 100 cm. How long is its each side?
Ans.
$\begin{array}{l}\mathrm{Perimeter}\text{of regular pentagon}=\text{5}\times \text{side}\\ \text{\hspace{0.17em}Side of regular pentagon}=\frac{\mathrm{Perimeter}}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of regular pentagon}=\frac{100\text{\hspace{0.17em}}}{4}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25\text{\hspace{0.17em}}\mathrm{cm}\end{array}$
Q.11 A piece of string is 30 cm long. What will be the length of each side if the string is used to form :
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Ans.
$\begin{array}{l}\text{Length of a piece of string}=\text{3}0\text{m}\mathrm{}\\ \left(\text{a}\right)\mathrm{Perimeter}\text{of square}=\text{Length of a piece of string}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 4}\times \mathrm{side}\text{of square}=\text{30 m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}side of square}=\frac{30}{4}\text{\hspace{0.17em}}\mathrm{m}\\ =7.5\text{\hspace{0.17em}}\mathrm{m}\\ \left(\text{b}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of an equilateral triangle}\\ =\text{Length of a piece of string}\\ \text{3}\times \mathrm{side}\text{of equilateral triangle}\\ =\text{30 m}\\ \text{Side of equilateral triangle}\\ =\frac{30}{3}\text{\hspace{0.17em}}\mathrm{m}\\ =10\text{\hspace{0.17em}}\mathrm{m}\\ \left(\text{c}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of a regular hexagon}\\ =\text{Length of a piece of string}\\ \text{6}\times \mathrm{side}\text{of regular hexagon}\\ =\text{30 m}\\ \text{Side of regular hexagon}\\ =\frac{30}{6}\text{\hspace{0.17em}}\mathrm{m}\\ =5\text{\hspace{0.17em}}\mathrm{m}\end{array}$
Q.12 Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Ans.
Let third side of triangle be x cm.
Perimeter of triangle = 12 cm + 14 cm + x cm
36 cm = 26 cm + x
x = 36 cm – 26 cm
= 10 cm
Thus, third side of triangle is 10 cm.
Q.13 Find the cost of fencing a square park of side 250 m at the rate of ₹20 per metre.
Ans.
Side of square park = 250 m
Rate of fencing the park = ₹20 per metre
Perimeter of square park = 4(250 m)
= 1000 m
Cost of fencing the square park
= ₹20 x 1000 m
= ₹20,000
Thus, the cost of fencing the square park is ₹20,000
Q.14 Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹12 per metre.
Ans.
Length of a rectangular park = 175 m
Breadth of a rectangular park = 125 m
Rate of fencing per metre = ₹12
Length of fencing of park = 2(L + B)
= 2(175+ 125) m
= 2(300) m
= 600 m
Cost of fencing the park = ₹12(600 m)
= ₹7200
Q.15 Sweety runs around a square park of side 75m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Ans.
Side of square park = 75 m
Perimeter of square park = 4 (75 m)
= 300 m
Distance covered by Sweety =300 m
Length of park = 60 m
Breadth of park = 45 m
Perimeter of rectangular park = 2 (60 m + 45 m)
= 210 m
Distance covered by Bulbul = 210 m
Thus, Bulbul covers less distance.
Q.16 What is the perimeter of each of the following figures? What do you infer from the answers?
Ans.
(a) Perimeter of square = 4(25 cm)
= 100 cm
(b) Perimeter of rectangle = 2(30 cm + 20 cm)
= 2(50 cm)
= 100 cm
(c) Perimeter of rectangle = 2(40 cm + 10 cm)
= 2(50 cm)
= 100 cm
(d) Perimeter of triangle = 30 cm + 30 cm+ 40 cm
= 100 cm
Q.17
Ans.
$\begin{array}{l}\mathrm{Length}\text{of each side of square paving}=\frac{1}{2}\text{\hspace{0.17em}}\mathrm{m}\\ \mathrm{Number}\text{of square paving}=\text{9}\\ \text{When square paving laid in the form of square,}\\ \text{Number of square paving in each side of square form}\\ \text{3}\\ \text{So, the length of each side of new square}\\ \text{\hspace{0.17em}\hspace{0.17em}}=3\times (\mathrm{side}\mathrm{of}\mathrm{square}\mathrm{paving})\\ \text{\hspace{0.17em}\hspace{0.17em}}=3\times \frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2}\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of new square}=4\times \mathrm{side}\\ \text{\hspace{0.17em}\hspace{0.17em}}=4\times \frac{3}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=6\text{\hspace{0.17em}}\mathrm{m}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of fig}\left(\mathrm{ii}\right)=4\times (\mathrm{side}\mathrm{having}\mathrm{single}\mathrm{paving})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}8\times \left(\text{side having double}\mathrm{paving}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=4\times \frac{1}{2}+8\times \frac{1}{2}\times 2\\ \text{\hspace{0.17em}\hspace{0.17em}}=2+8\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\text{\hspace{0.17em}}\mathrm{m}\\ \left(\mathrm{c}\right)\text{Cross figure i.e., fig}\left(\mathrm{ii}\right)\text{has greater perimeter.}\\ \left(\mathrm{d}\right)\text{Yes, if we put all paving in a straight line. It forms}\\ \text{a rectangle whose dimensions are:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length}=9\times \frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=4.5\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Breadth}=\frac{1}{2}\text{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{0.5\hspace{0.17em}cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of rectangle}=2(\mathrm{L}+\mathrm{B})\\ \text{\hspace{0.17em}\hspace{0.17em}}=2(4.5+0.5)\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\text{\hspace{0.17em}}\mathrm{cm}\end{array}$
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FAQs (Frequently Asked Questions)
1. What is the topic of Chapter 10 of Class 6 Maths?
The topic of Maths Chapter 10 of Class 6 is Mensuration. Mathematical mensuration investigates the calculation of geometric figures and their properties, including area, length, volume, lateral surface area, surface area, etc. It addresses all the crucial equations, the attributes of numerous geometric shapes and figures, and the computation concepts. Use of the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 can be really beneficial for the preparation of this chapter. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 are prepared by experts for students’ preparation for Chapter 10.
2. What is Mensuration?
Geometry includes the topic of measurement. Mensuration deals with the dimensions, areas, and densities of various 2D and 3D shapes. For example, square, triangle, rectangle, cube, sphere, cone etc. Students are advised to make use of the NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1, this may enhance their chances of performing well in the examination of Maths. The NCERT Solutions Class 6 Maths Chapter 10 Exercise 10.1 and other resources related to Chapter 10 may be acquired by students of Class 6 from the website or mobile application of Extramarks.