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NCERT Solutions Class 7 Maths 2024 – Download PDF

NCERT Solutions Class 7 Maths: NCERT solutions are a great way to learn the concepts and how to solve problems correctly. Maths is one of the main subjects in the Class 7 curriculum and forms the fundamentals for concepts that will be needed in the higher classes. Referring to them can help them score good marks and prepare for the exam efficiently.

Embibe provides NCERT solutions for Class 7 Maths for all intext and exercise questions. The solutions have a step-by-step approach to solving problems with clear explanations. Apart from this, Embibe provides over 1000+ practice questions for all 15 chapters along with chapter-wise mock tests and full tests. Continue reading to learn more about the NCERT solutions for Class 7 Maths and ace the exam.

NCERT Solutions Class 7 Maths

Class 7 Maths deals with topics like integers, data handling, triangles and properties, and many more. These concepts are all needed to understand complex topics in the later classes. To ace the examination students should learn the Class 7 Maths concepts on Embibe through 3D videos and Embibe explainers.

Additionally, they can also practice questions for better comprehension. Referring to NCERT solutions can help them grasp the concepts firmly and how to solve the problems systematically so that they can tackle any question that comes up in the exam. Before going to the solutions, students can find the chapters for Class 7 Maths in the following table.

NCERT Solutions for Class 7 Maths: Points to Remember

Besides learning and practicing, students can also take the Class 7 Maths mock tests on Embibe to assess their preparation. And while revising they can refer to some important points to quickly recall the concepts. Below are some of the important points that students must remember about the NCERT Solutions for Class 7 Maths:

• On an integer number line, all numbers to the right of 0 are positive integers and all numbers to the left of 0 are negative integers.
• All the properties applicable to whole numbers are applicable to integers in addition, the subtraction operation has the closure property.
• A fraction whose numerator is less than the denominator is called a proper fraction.
• Fractions having the same denominators are called like fractions. Otherwise, they are called unlike fractions.
• Mode is another form of central tendency or representative value. The mode of a set of observations is the observation that occurs most often.
• In an equation, we can move a term from one side to the other side with the opposite sign. This process is known as transposition.
•  The distance between two parallel lines is the same everywhere and is equal to the perpendicular distance between them.
• The three line segments forming a triangle are called the sides of the triangle.
• Two triangles are congruent if the three sides and the three angles of one triangle are respectively equal to the corresponding parts of the other.
• We can also use the idea of ‘equal corresponding angles’ to do the construction.
• The number π is not a rational number.

NCERT Solutions for Class 7 Maths: All Chapters

Referring to NCERT solutions is really helpful to level up for exam preparation. To help students get easy access to NCERT solutions, we have provided the solutions for all chapters of Class 7 Maths. Students can click on the links below and start practicing right away.

• 1st Chapter: Integers
• 2nd Chapter: Fractions and Decimals
• 3rd Chapter: Data Handling
• 4th Chapter: Simple Equations
• 5th Chapter: Lines and Angles
• 6th Chapter: The Triangle and Its Properties
• 7th Chapter: Congruence of Triangles
• 8th Chapter: Comparing Quantities
• 9th Chapter: Rational Numbers
• 10th Chapter: Practical Geometry
• 11th Chapter: Perimeter and Area
• 12th Chapter: Algebraic Expression
• 13th Chapter: Exponents and Powers
• 14th Chapter: Symmetry
• 15th Chapter: Visualizing Solid Shapes

FAQs on NCERT Solutions for Class 7 Maths 2023-24

Some of the frequently asked questions on the NCERT Solutions for Class 7 Maths are as follows:

Ans: Students can find the NCERT solutions for Class 7 Maths on Embibe.

Ans: Embibe provides numerous practice questions for Class 7 Maths.

Ans: There are 15 chapters in NCERT solutions for Class 7 Maths.

Ans: Students can take the Class 7 Maths mock tests on Embibe.

Ans: Referring to the NCERT solutions for Class 7 Maths helps students grasp the concepts and learn the correct approach to solving problems.

We hope this detailed article on NCERT Solutions for Class 7 Maths helps you. For more such articles, stay tuned to Embibe .

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Home » 7th Class » Class 7 Maths Notes for (PDF) – Study Material

Class 7 Maths Notes for (PDF) – Study Material

Class 7 Maths Notes are now available for download here. Here you can get Short Notes for Class 7 Maths. If you are a Class 7th student, then you can use the Class 7 Maths Notes PDF to study all the chapters. You can download the Class 7 notes for Maths subject for all chapters here at aglasem.com.

• Class 7 Maths Notes

Here are the class 6 Maths notes . If you are studying Maths then you can easily download Class 7 Maths notes pdf here to learn about the topics. Maths notes and summary are compiled by expert teachers so that you can study easily with them. All the notes are chapter wise. That’s why you can read and understand about each chapter or topic given in Class 7th Maths book with this free study material.

• Fractions and Decimals
• Data Handling
• Simple Equations
• Lines & Angles
• Triangles & Its Properties
• Comparing Quantities
• Rational Numbers
• Perimeter and Area
• Algebraic Expressions
• Exponents and Powers
• Visualing Solid Shapes

After notes we have also provided important questions of the Class 7 Maths textbook. You can practice the Maths questions and check your answers from the solutions given after questions. One Class 7 Maths test is also there for your practice.

• Class 7 Notes

Not only can you study all the topics with the Maths notes . These are also the Class 7 Maths summary that help you read the portion quickly. Moreover they help you arrive at the Class 7 Maths solutions to the chapter-wise questions answers. All the notes for Class 7 students, subject-wise, are as follows.

• Class 7 Science Notes

Class 7 Maths Notes for All Boards

Likewise the 7th class Maths notes here are useful for other boards also. As you know, the different boards for Class 7 in India are CBSE, CISCE, SEBA, BSE Odisha, CGBSE, HBSE, HPBOSE, KSEEB, MSBSHSE, PSEB, RBSE, TBSE, UPMSP, UBSE, BSEAP, BSEB, GBSHSE, GSEB, JAC, JKBOSE, KBPE, MBOSE, MBSE, MPBSE, NBSE, DGE TN, BSE Telangana, BOSEM, WBBSE . Therefore if you are a student of any of these boards, then you can use the Class 7 / Class 7th Maths notes PDF here to complete your curriculum.

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 7 Maths

If you are looking for NCERT Solutions for Class 7 Maths, you have come to the right place. Experienced LearnCBSE.in Teachers has created detailed CBSE 7th class maths textbook solutions. We provide that are precise to the point and absolutely error free. NCERT class 7 maths solutions includes all the questions provided as per new revised syllabus in NCERT Class 7 Maths textbook. NCERT maths book class 7 solutions pdf can be downloaded in One click without LOGIN. You can also practice  Extra Questions for Class 7 Maths  on LearnCBSE.in

Chapter wise detailed NCERT Solutions for Class 7 Maths are given below.

• Chapter 1 Integers
• Chapter 2 Fractions and Decimals
• Chapter 3 Data Handling
• Chapter 4 Simple Equations
• Chapter 5 Lines and Angles
• Chapter 6 The Triangles and its Properties
• Chapter 7 Congruence of Triangles
• Chapter 8 Comparing Quantities
• Chapter 9 Rational Numbers
• Chapter 10 Practical Geometry
• Chapter 11 Perimeter and Area
• Chapter 12 Algebraic Expressions
• Chapter 13 Exponents and Powers
• Chapter 14 Symmetry
• Chapter 15 Visualising Solid Shapes

You can also download the free NCERT Solutions for Class 7 Maths All Chapters PDF or save the solution images and take the print out to keep it handy for your exam preparation.

NCERT Solutions for Class 7 Maths Chapter 1 Integers

• Class 7 Maths Integers Exercise 1.1 
• Class 7 Maths Integers Exercise 1.2
• Class 7 Maths Integers Exercise 1.3
• Class 7 Maths Integers Exercise 1.4
• Integers Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

• Class 7 Maths Fractions and Decimals Exercise 2.1 
• Class 7 Maths Fractions and Decimals Exercise 2.2
• Class 7 Maths Fractions and Decimals Exercise 2.3
• Class 7 Maths Fractions and Decimals Exercise 2.4
• Class 7 Maths Fractions and Decimals Exercise 2.5
• Class 7 Maths Fractions and Decimals Exercise 2.6
• Class 7 Maths Fractions and Decimals Exercise 2.7
• Fractions and Decimals Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling

• Class 7 Maths Data Handling Exercise 3.1 
• Class 7 Maths Data Handling Exercise 3.2
• Class 7 Maths Data Handling Exercise 3.3
• Class 7 Maths Data Handling Exercise 3.4
• Data Handling Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

• Class 7 Maths Simple Equations Exercise 4.1 
• Class 7 Maths Simple Equations Exercise 4.2
• Class 7 Maths Simple Equations Exercise 4.3
• Class 7 Maths Simple Equations Exercise 4.4
• Simple Equations Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles

• Class 7 Maths Lines and Angles Exercise 5.1
• Class 7 Maths Lines and Angles Exercise 5.2
• Lines and Angles Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and its Properties

• Class 7 Maths The Triangle and Its Properties Exercise 6.1
• Class 7 Maths The Triangle and Its Properties Exercise 6.2
• Class 7 Maths The Triangle and Its Properties Exercise 6.3
• Class 7 Maths The Triangle and Its Properties Exercise 6.4
• Class 7 Maths The Triangle and Its Properties Exercise 6.5
• The Triangles and its Properties Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles

• Class 7 Maths Congruence of Triangles Exercise 7.1
• Class 7 Maths Congruence of Triangles Exercise 7.2
• Congruence of Triangles Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities

• Class 7 Maths Comparing Quantities Exercise 8.1 
• Class 7 Maths Comparing Quantities Exercise 8.2
• Class 7 Maths Comparing Quantities Exercise 8.3
• Comparing Quantities Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

• Class 7 Maths Rational Numbers Exercise 9.1 
• Class 7 Maths Rational Numbers Exercise 9.2
• Rational Numbers Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry

• Class 7 Maths Practical Geometry Exercise 10.1
• Class 7 Maths Practical Geometry Exercise 10.2
• Class 7 Maths Practical Geometry Exercise 10.3
• Class 7 Maths Practical Geometry Exercise 10.4
• Class 7 Maths Practical Geometry Exercise 10.5
• Practical Geometry Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area

• Class 7 Maths Perimeter and Area Exercise 11.1 
• Class 7 Maths Perimeter and Area Exercise 11.2
• Class 7 Maths Perimeter and Area Exercise 11.3
• Class 7 Maths Perimeter and Area Exercise 11.4
• Perimeter and Area Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions

• Class 7 Maths Algebraic Expressions Exercise 12.1 
• Class 7 Maths Algebraic Expressions Exercise 12.2
• Class 7 Maths Algebraic Expressions Exercise 12.3
• Class 7 Maths Algebraic Expressions Exercise 12.4
• Algebraic Expressions Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

• Class 7 Maths Exponents and Powers Exercise 13.1 
• Class 7 Maths Exponents and Powers Exercise 13.2
• Class 7 Maths Exponents and Powers Exercise 13.3
• Exponents and Powers Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry

• Class 7 Maths Symmetry Exercise 14.1 
• Class 7 Maths Symmetry Exercise 14.2
• Class 7 Maths Symmetry Exercise 14.3
• Symmetry Class 7 Extra Questions

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes

• Class 7 Maths Visualising Solid Shapes Exercise 15.1 
• Class 7 Maths Visualising Solid Shapes Exercise 15.2
• Class 7 Maths Visualising Solid Shapes Exercise 15.3
• Class 7 Maths Visualising Solid Shapes Exercise 15.4
• Visualising Solid Shapes Class 7 Extra Questions

NCERT Solutions for Class 7 Maths (Download PDF) Maths NCERT Solutions Maths Formulas for Class 7

Key Features Class 7 Maths NCERT Solutions:

• Class 7 Maths NCERT Solutions were prepared by subject experts.
• Step by step solutions to understand problems better.
• Detailed Explanation to solve problems and formulas are mentioned in between steps to learn effectively.
• Exercise wise Class 7 Maths solutions also given to access easily.
• You can download NCERT Solutions for Class 7 Maths PDF or save the solution images and take the print out to keep it handy for your exam preparation.

We make it a point to assist students in every possible manner we can, and that includes providing solutions for every subject. NCERT textbooks have been reputed to be the best textbooks for school education, and we make the solutions competent for the books.

Mathematics comes across a rather dreadful subject for most students in school, and we believe that a little practice can sort that problem, Therefore, our solutions focus on building the concepts based on the fundamentals, and also exploring alternative methods to solve a particular problem.

Browse all Class 7 Maths NCERT Solutions from your tablet, desktop or mobile and score more marks in your final exams. You can also go through the RD Sharma Class 7 Solutions and RS Aggarwal Class 7 Solutions which will help you in extra practice and exams.

Here is the list of Main Topics from Class 7 Maths NCERT Text Book:

Class 7 Maths Chapter 1 Integers

After an introduction to whole numbers in class 6, this chapter deals with integers, both positive and negative, to give the students a feel of the real numbers. This chapter gives a new perspective to students, in terms of properties and importance of integers. The exercises are made to ensure that the students grasp the whole concept thoroughly.

• 1.1 Introduction
• 1.3 Properties Of Addition And Subtraction Of Integers
• 1.4 Multiplication Of Integers
• 1.5 Properties Of Multiplication Of Integers
• 1.6 Division Of Integers
• 1.7 Properties Of Division Of Integers

Class 7 Maths Chapter 2 Fractions and Decimals

This is not exactly a new concept, rather, a further exploration of the old concepts. The chapter deals with the properties of fractions and decimals, and the operations on the same. They also deal with the portrayal of fractions and decimals on the number line, and their expansions and subtraction.

• 2.1 Introduction
• 2.2 How Well Have You Learnt About Fractions?
• 2.3 Multiplication Of Fractions
• 2.4 Division Of Fractions
• 2.5 How Well Have You Learnt About Decimal Numbers
• 2.6 Multiplication Of Decimal Numbers
• 2.7 Division Of Decimal Numbers

Class 7 Maths Chapter 3 Data Handling

This chapter can be considered as the first step towards statistics, as it deals with data accumulation, the data interpretation, and the plotting, keeping up with real life examples. The chapters also teach how to make a few deductions from the accumulated data. The exercises are kept very close to real life examples, and thus, practicing the same gives a better feel of the same.

• 3.1 Introduction
• 3.2 Collecting Data
• 3.3 Organisation Of Data
• 3.4 Representative Values
• 3.5 Arithmetic Mean
• 3.8 Use Of Bar Graphs With A Different Purpose
• 3.9 Chance And Probability

Class 7 Maths Chapter 4 Simple Equations

As the name suggests, this chapter deals with the formulation and applications of simple equations. From setting up simple equations to solving them, this chapter explores the theory of equations thoroughly. The problems have been extensively discussed in the solutions.

• 4.1 A Mind-Reading Game!
• 4.2 Setting Up Of An Equation.
• 4.3 Review Of What We Know.
• 4.4 What Equation Is?
• 4.5 More Equations.
• 4.6 From Solution To Equation.
• 4.7 Applications Of Simple Equations To Practical Situations.

Class 7 Maths Chapter 5 Lines and Angles

The first chapter of geometry in Class 7, lines and angles starts with the fundamental definitions of line and angle. The chapter will cruise through the concepts of parallel lines, and the associated angles like the alternate interior angles, corresponding angles, vertically opposite angles. This fairly easy chapter is further made interesting with the help of exercises, and the solutions justify the same effectively.

• 5.1 Introduction.
• 5.2 Related Angles.
• 5.3 Pairs Of Lines.
• 5.4 Checking For Parallel Lines.

Class 7 Maths Chapter 6 The triangle and its properties

The second chapter of geometry deals with triangles and their properties. This chapter talks about the types of triangles, the angle sum property, the medians and altitudes, and the Pythagoras theorem. The students will get a feel of what triangles are in general, and the specific applications of the Pythagoras theorem in this chapter.

• 6.1 Introduction.
• 6.2 Medians Of A Triangle.
• 6.3 Altitudes Of A Triangle.
• 6.4 Exterior Angle Of A Triangle And Its Property.
• 6.5 Angle Sum Property Of A Triangle.
• 6.6 Two Special Triangles: Equilateral And Isosceles.
• 6.7 Sum Of The Lengths Of Two Sides Of A Triangle.
• 6.8 Right-Angled Triangles And Pythagoras Property.

Class 7 Maths Chapter 7 Congruence of Triangles

After the general introduction of triangles in Chapter 6, the seventh chapter deals with the specific property of congruence of triangles. The chapter covers all the congruence criteria, and deal with different kinds of problems. The solutions discuss the congruence criteria extensively, using alternative approach wherever possible.

• 7.1 Introduction.
• 7.2 Congruence Of Plane Figures.
• 7.3 Congruence Among Line Segments.
• 7.4 Congruence Of Angles.
• 7.5 Congruence Of Triangles.
• 7.6 Criteria For Congruence Of Triangles.
• 7.7 Congruence Among Right-Angled Triangles.

Class 7 Maths Chapter 8 Comparing Quantities

This chapter can be safely assumed to be one of the most application oriented chapters in the whole Class 7 Mathematics syllabus. As the name suggests, this chapter gives the tool to measure and compare quantities. The tools are primarily percentage, ratios, profit and loss, and interest. The chapter comes in handy in all folds of life, as the calculations learnt here are the ones used in the real world the most.

• 8.1 Introduction.
• 8.2 Equivalent Ratios.
• 8.3 Percentage – Another Way Of Comparing Quantities.
• 8.4 Use Of Percentages.
• 8.5 Prices Related To An Item Or Buying And Selling.
• 8.6 Charge Given On Borrowed Money Or Simple Interest.

Class 7 Maths Chapter 9 Rational Numbers

After discussing integers extensively in the first chapter, this chapter comes back to the numbers, namely rational numbers. The chapter deals with the definitions and the properties of rational numbers.

• 9.1 Introduction
• 9.2 Need For Rational Numbers.
• 9.3 What Are Rational Numbers?
• 9.4 Positive And Negative Rational Numbers.
• 9.5 Rational Numbers On A Number Line.
• 9.6 Rational Numbers In Standard Form.
• 9.7 Comparison Of Rational Numbers.
• 9.8 Rational Numbers Between Two Rational Number.
• 9.9 Operations On Rational Numbers.

Class 7 Maths Chapter 10 Practical Geometry

This chapter deals with the portrayal of geometry on paper, in terms of construction of lines and angles. This is a fairly simpler chapter, that only requires a set procedure to be followed while going through constructions.

• 10.1 Introduction
• 10.2 Construction Of A Line Parallel To A Given Line, Through A Point Not On The Line.
• 10.3 Construction Of Triangles.
• 10.4 Constructing A Triangle When The Lengths Of Its Three Sides Are Known (SSS Criterion)
• 10.5 Constructing A Triangle When The Lengths Of Two Sides And The Measure Of The Angle Between Them Are Known. (SAS Criterion)
• 10.6 Constructing A Triangle When The Measures Of Two Of Its Angles And The Length Of The Side Included Between Them Is Given. (ASA Criterion)
• 10.7 Constructing A Right-Angled Triangle When The Length Of One Leg And Its Hypotenuse Are Given (RHS Criterion).

Class 7 Maths Chapter 11 Perimeter and Area

This chapter brings in the mensuration part of the syllabus. It deals with the areas and perimeters of all the important shapes in Mathematics. The chapter is very simple, without the introduction of any complex shapes.

• 11.1 Introduction.
• 11.2 Squares And Rectangles.
• 11.3 Area Of A Parallelogram.
• 11.4 Area Of A Triangle.
• 11.5 Circles.
• 11.6 Conversion Of Units.
• 11.7 Applications.

Class 7 Maths Chapter 12 Algebraic Equations

This chapter deals with converting simple mathematical statements into algebraic equations and using them to solve certain problems, using the principles of algebra. The mathematical statements are closely related to some of the real life examples, where the algebra can actually be used. The exercises make it double the fun.

• 12.1 Introduction.
• 12.2 How Are Expressions Formed?
• 12.3 Terms Of An Expression.
• 12.4 Like And Unlike Terms.
• 12.5 Monomials, Binomials, Trinomials And Polynomials.
• 12.6 Addition And Subtraction Of Algebraic Expressions.
• 12.7 Finding The Value Of An Expression.
• 12.8 Using Algebraic Expressions – Formulas And Rules.

Class 7 Maths Chapter 13 Exponents and Powers

This chapter deals with the introduction to exponents, the rules of multiplication and division of exponents, the power of a power, decimal system, and the expression of very large numbers into the Standard Form, or the Scientific Notation.

• 13.1 Introduction.
• 13.2 Exponents.
• 13.3 Laws Of Exponents.
• 13.4 Miscellaneous Examples Using The Laws Of Exponents.
• 13.5 Decimal Number System.
• 13.6 Expressing Large Numbers In The Standard Form.

Class 7 Maths Chapter 14 Symmetry

This chapter gives a perspective of symmetrical shapes to the students. Symmetry is exploited extensively by craftsmen and designers to plan out intricate design patterns. This chapter on symmetry is to give the students the general idea of symmetry in the world.

• 14.1 Introduction: Symmetry
• 14.2 Lines Of Symmetry For Regular Polygons.
• 14.3 Rotational Symmetry.
• 14.4 Line Symmetry And Rotational Symmetry.

Class 7 Maths Chapter 15 Visualising Solid Shapes

This chapter deals with the visuals of geometry, by explaining the various geometrical shapes that are incorporated in designing the everyday objects around us. This chapter deals with both plane figures as well as solid shapes.

• 15.1 Introduction: Plane Figures And Solid Shapes.
• 15.2 Faces, Edges, and Vertices.
• 15.3 Nets For Building 3-D Shapes.
• 15.4 Drawing Solids On A Flat Surface.
• 15.5 Viewing Different Sections Of A Solid.

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Class 7 Maths NCERT Solutions are prepared by experts and they give you step by step solutions to understand the problems better. They can be quite handy during your preparation.

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NCERT Solutions Class 7 Maths Chapter 9 Perimeter and Area

NCERT Solutions for Maths Chapter 9 Class 7 Perimeter and Area - FREE PDF Download

NCERT Solutions for Perimeter and Area Class 7 Maths Chapter 9 by Vedantu introduces the concepts of measuring the boundaries and spaces of various geometric shapes. This chapter lays the groundwork for understanding more complex geometric concepts in higher classes, including the volume and surface area of 3D shapes. Utilising diagrams will enhance your understanding of the formulas and relationships between shapes.

Perimeter and area are fundamental concepts in geometry that measure the boundaries and the surface spaces of various shapes. These concepts are crucial not only for academic learning but also have practical applications in everyday life. Vedantu’s NCERT solutions for class 7 Maths provide step-by-step explanations to help you understand these concepts thoroughly.

Glance on Maths Chapter 9 Class 7 - Perimeter and Area

This chapter deals with the concepts of perimeter and area of different shapes such as squares, rectangles, triangles, parallelograms, and circles.

Perimeter is the total length of the boundary of a closed figure.

Area is the measure of the surface enclosed by the boundary of a figure.

The area of a circle is one complete cycle of the radius of the circle. 

The units of perimeter are linear (e.g., meters, centimeters), while the units of area are square units (e.g., square meters, square centimeters).

A triangle is a plane figure that is enclosed by three line segments called a triangle. 

A Quadrilateral is a plane figure that is enclosed by four line segments. 

A parallelogram is a plane figure enclosed by four line segments. The opposite sides in a parallelogram are parallel and equal in length.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 9 - Perimeter and Area, which you can download as PDFs.

There are two exercises (25 fully solved questions) in class 7th maths chapter 9 Perimeter and Area.

Access Exercise wise NCERT Solutions for Chapter 9 Maths Class 7

Exercises Under NCERT Solutions for Class 7 Maths Chapter 9 Perimeter and Area

Exercise 9.1: This exercise focuses on understanding the Perimeter. Calculation of the perimeter for different shapes such as rectangles, squares, and triangles.

Exercise 9.2: This exercise focuses on understanding the Area of a Rectangle and Square. Calculation of the area of rectangles and squares.

Access NCERT Solutions for Class 7 Maths Chapter 9 – Perimeter and Area

Exercise – 9.1.

1. Find the area of each of the following parallelograms:

As the area of parallelogram ${\text{ =  base x height}}$ 

Given base ${\text{ =  7 cm}}$  and height ${\text{ =  4 cm}}$ 

$\therefore $ Area of parallelogram ${\text{ =  7 x 4  =  28 c}}{{\text{m}}^2}$ 

Given base $ = {\text{ 5 cm}}$and height${\text{ =  3 cm}}$ 

$\therefore $ Area of parallelogram ${\text{ =  5 x 3  =  15 c}}{{\text{m}}^2}$ 

Ans:  

Given base ${\text{ =  2}}{\text{.5 cm}}$ and height ${\text{ =  3}}{\text{.5 cm}}$

$\therefore $ Area of parallelogram ${\text{ =  2}}{\text{.5 x 3}}{\text{.5  =  8}}{\text{.75 c}}{{\text{m}}^2}{\text{ }}$ 

Ans : 

Given base ${\text{ =  5 cm}}$and height ${\text{ =  4}}{\text{.8 cm}}$ 

$\therefore $ Area of parallelogram ${\text{ =  5 x 4}}{\text{.8  =  24 c}}{{\text{m}}^2}$ 

Given base ${\text{ =  2 cm}}$ and height ${\text{ =  4}}{\text{.4 cm}}$ 

$\therefore $ Area of parallelogram ${\text{ =  2 x 4}}{\text{.4  =  8}}{\text{.8 c}}{{\text{m}}^2}$  

2. Find the area of each of the following triangles:

As the area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x base x height}}$ 

a. Given, base ${\text{ =  4 cm}}$ and height ${\text{ =  3 cm}}$ 

$\therefore $Area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x 4 x 3  =  6 c}}{{\text{m}}^2}$ 

b. Given, base ${\text{ =  5 cm}}$ and height $ = {\text{ 3}}{\text{.2 cm}}$

$\therefore $ Area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x 5 x 3}}{\text{.2  =  8 c}}{{\text{m}}^2}$ 

c. Given, base ${\text{ =  3 cm}}$ and height ${\text{ =  4 cm}}$ 

$\therefore $ Area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x 3 x 4  =  6 c}}{{\text{m}}^2}$ 

d. Given, base ${\text{ =  3 cm}}$ and height ${\text{ =  2 cm}}$ 

$\therefore $ Area of triangle $ = {\text{ }}\dfrac{1}{2}{\text{ x  3 x 2  =  3 c}}{{\text{m}}^2}$ 

3. Find the missing values:

As we know that, area of parallelogram ${\text{ =  base x height}}$

a. Here,  base ${\text{ =  20 cm}}$ and area ${\text{ =  246 c}}{{\text{m}}^2}$ 

$\Rightarrow 20{\text{ x height  =  246}} $

$\Rightarrow {\text{ height  =  }}\dfrac{{246}}{{20}} $

$\Rightarrow {\text{ height  =  12}}{\text{.3 cm}} $ 

b. Here, height $ = {\text{ 15 cm}}$ and area ${\text{ =  154}}{\text{.5 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ base x 15  =  154}}{\text{.5}} $

$\Rightarrow {\text{ base  =  }}\dfrac{{154.5}}{{15}} $

$\Rightarrow {\text{ base  =  10}}{\text{.3 cm}} $ 

c. Here, height ${\text{ =  8}}{\text{.4 cm}}$ and area ${\text{ =  48}}{\text{.72 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ base x 8}}{\text{.4  =  48}}{\text{.72}} $

$\Rightarrow {\text{ base  =  }}\dfrac{{48.72}}{{8.4}} $

$\Rightarrow {\text{ base  =  5}}{\text{.8 cm}} $   

d. Here, base ${\text{ =  15}}{\text{.6 cm}}$ and area $ = {\text{ 16}}{\text{.38 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ 15}}{\text{.6 x height  =  16}}{\text{.38}} $

$\Rightarrow {\text{ height  =  }}\dfrac{{16.38}}{{15.6}} $

$\Rightarrow {\text{ height  =  1}}{\text{.05 cm}} $ 

Hence, the missing values are:

4. Find the missing values:

a. Given, base ${\text{ =  15 cm}}$ and area ${\text{ =  87 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ }}\dfrac{1}{2}{\text{ x 15 x height  =  87}} $

$\Rightarrow {\text{ height  =  }}\dfrac{{87{\text{ x 2}}}}{{15}} $

$\Rightarrow {\text{ height  =  11}}{\text{.6 cm}} $ 

b. Given, height ${\text{ =  31}}{\text{.4 mm}}$ and area ${\text{ =  1256 m}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ }}\dfrac{1}{2}{\text{ x base x 31}}{\text{.44  =  1256}} $

$\Rightarrow {\text{ base  =  }}\dfrac{{1256{\text{ x 2}}}}{{31.44}} $

$\Rightarrow {\text{ base  =  80 mm}} $ 

c. Given, base ${\text{ =  22 cm}}$ and area $ = {\text{ 170}}{\text{.5 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ }}\dfrac{1}{2}{\text{ x 22 x height  =  170}}{\text{.5}} $

$\Rightarrow {\text{ height  =  }}\dfrac{{170.5{\text{ x 2}}}}{{22}} $

$\Rightarrow {\text{ height  =  15}}{\text{.5 cm}} $

5. PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR ${\text{ =  12 cm}}$ and QM ${\text{ =  7}}{\text{.6 cm}}$.

Find:  a. the area of the parallelogram PQRS 

In given parallelogram PQRS, 

SR ${\text{ =  12 cm}}$, QM $ = {\text{ 7}}{\text{.6 cm}}$ , PS ${\text{ =  8 cm}}$ 

Area of parallelogram ${\text{ =  base x height  =  12 x 7}}{\text{.6  =  91}}{\text{.2 c}}{{\text{m}}^2}{\text{ }}$

b. QN, if PS ${\text{ =  8 cm}}$

Ans: Area of parallelogram ${\text{ =  base x height}}$ 

$\Rightarrow {\text{ 91}}{\text{.2  =  8 x QN}} $

$\Rightarrow {\text{ QN  =  }}\dfrac{{91.2}}{8} $

$\Rightarrow {\text{ QN  =  11}}{\text{.4 cm}} $

6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is $1470{\text{ c}}{{\text{m}}^2}$, AB $ = {\text{ 35 cm}}$ and AD $ = {\text{ 49 cm}}$ , find the length of BM and DL.

In the given parallelogram ABCD,

Area of parallelogram ${\text{ =  1470 c}}{{\text{m}}^2}$ 

Base (AB) ${\text{ =  35 cm}}$ 

Base (AD) $ = {\text{ 49 cm}}$ 

$\because $ area of parallelogram $ = {\text{ base x height}}$ 

$\Rightarrow {\text{ 1470  =  35 x DL}} $

$\Rightarrow {\text{ DL  =  }}\dfrac{{1470}}{{35}} $

$\Rightarrow {\text{ DL  =  42 cm}} $ 

$\Rightarrow {\text{ 1470  =  49 x BM}} $

$\Rightarrow {\text{ BM  =  }}\dfrac{{1470}}{{49}} $

$\Rightarrow {\text{ BM  =  30 cm}} $ 

Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.

7. ∆ ABC is right angled at A. AD is perpendicular to BC. If AB $ = {\text{ 5 cm}}$ , BC ${\text{ =  13 cm}}$  and AC $ = {\text{ 12 cm}}$, find the area of ∆ ABC. Also, find the length of AD.

In right angled triangle BAC,

AB $ = {\text{ 5 cm}}$ and AC $ = {\text{ 12 cm}}$ 

We know that, area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{x base x height}}$ 

$\Rightarrow $ Area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x AB x AC  =  }}\dfrac{1}{2}{\text{ x 5 x 12  =  30 c}}{{\text{m}}^2}$ 

Now, in triangle ABC,

Area of triangle ABC $ = {\text{ }}\dfrac{1}{2}{\text{ x BC x AD}}$

$\Rightarrow {\text{ 30  =  }}\dfrac{1}{2}{\text{ x 13 x AD}} $

$\Rightarrow {\text{ AD  =  }}\dfrac{{30{\text{ x 2}}}}{{13}} $

$\Rightarrow {\text{ AD  =  }}\dfrac{{60}}{{13}}{\text{ cm}} $ 

8. ∆ ABC is isosceles with ${\text{AB  =  AC  =  7}}{\text{.5 cm}}$and BC $ = {\text{ 9 cm}}$. The height AD from A to BC, is  ${\text{6 cm}}$. Find the area of ∆ ABC. What will be the height from C to AB i.e., CE?

in given triangle ABC,

AD $ = {\text{ 6 cm}}$ and BC ${\text{ =  9 cm}}$ 

We know that, area of triangle $ = {\text{ }}\dfrac{1}{2}{\text{ x base x height}}$ 

$ \Rightarrow $  area of triangle $ = {\text{ }}\dfrac{1}{2}{\text{ x BC x AD  =  }}\dfrac{1}{2}{\text{ x 9 x 6  =  27 c}}{{\text{m}}^2}$ 

area of triangle $ = {\text{ }}\dfrac{1}{2}{\text{ x base x height}}$

$\Rightarrow {\text{ 27  =  }}\dfrac{1}{2}{\text{ x AB x CE}} $

$\Rightarrow {\text{ 27  =  }}\dfrac{1}{2}{\text{ x 7}}{\text{.5 x CE}} $

$\Rightarrow {\text{ CE  =  }}\dfrac{{27{\text{ x 2}}}}{{7.5}} $

$\Rightarrow {\text{ CE  =  7}}{\text{.2 cm}} $ 

Thus, the height from C to AB i.e., CE is  ${\text{7}}{\text{.2 cm}}$.

Exercise – 9.2

1. Find the circumference of the circles with the following radius: (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

We know that, circumference of circle  =  2 $\pi r$ 

$\therefore $ Circumference of the given circle ${\text{ =  2 x }}\dfrac{{22}}{7}{\text{ x 14  =  88 cm}}$ 

We know that, circumference of circle =  2 $\pi r$

$\therefore $ Circumference of the given circle ${\text{ =  2 x }}\dfrac{{22}}{7}{\text{ x 28  =  176 mm}}$ 

We know that, circumference of circle 2 $\pi r$

$\therefore $ Circumference of the given circle ${\text{ =  2 x }}\dfrac{{22}}{7}{\text{ x 21  =  132 cm}}$

2. Find the area of the following circles, given that: (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

a. radius $ = {\text{ 14 mm}}$ 

We know that, area of circle $ =  \pi {{\text{r}}^{\text{2}}}$ 

Given, radius $ = {\text{ 14 mm}}$ 

$\therefore $  area of circle ${\text{ =  }}\dfrac{{22}}{7}{\text{ x 14 x 14  =  616 m}}{{\text{m}}^2}$ 

b. diameter ${\text{ =  49 m}}$

We know that, area of circle $ =  \pi {{\text{r}}^{\text{2}}}$

Given, diameter ${\text{ =  49 m}}$ 

$\because {\text{ radius  =  }}\dfrac{{{\text{diameter}}}}{2}{\text{  =  }}\dfrac{{49}}{2}{\text{ m}}$ 

$\therefore $  area of circle ${\text{ =  }}\dfrac{{22}}{7}{\text{ x }}\dfrac{{49}}{2}{\text{ x }}\dfrac{{49}}{2}{\text{  =  1886}}{\text{.5 }}{{\text{m}}^2}$

c. radius ${\text{ =  5 cm}}$

Given, radius ${\text{ =  5 cm}}$ 

$\therefore $  area of circle ${\text{ =  }}\dfrac{{22}}{7}{\text{ x 5 x 5  =  }}\dfrac{{550}}{7}{\text{ c}}{{\text{m}}^2}$

3. If the circumference of a circular sheet is  $154{\text{ m}}$ , find its radius. Also find the area of the sheet. (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

Given, circumference of the circular sheet ${\text{ =  154 m}}$

$\Rightarrow 2 \pi {\text{r}} =  154 $

$\Rightarrow {\text{ 2 x }}\dfrac{{22}}{7}{\text{ x r  =  154}} $

$\Rightarrow {\text{ r  =  }}\dfrac{{154{\text{ x 7}}}}{{2{\text{ x 22}}}} $

$\Rightarrow {\text{ r  =  24}}{\text{.5 m}} $

Now, area of circular sheet $ =  \pi {{\text{r}}^{\text{2}}}{\text{  =  }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 24}}{\text{.5 x 24}}{\text{.5  =  1886}}{\text{.5 }}{{\text{m}}^{\text{2}}}$

Thus, the radius and area of the circular sheet are $24.5{\text{ m}}$ and $1886.5{\text{ }}{{\text{m}}^2}$ respectively.

4. A gardener wants to fence a circular garden of diameter $21{\text{ m}}$ . Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also, find the costs of the rope, if it costs  ₹$4$ per meter. (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

Given, diameter of the circular garden $ = {\text{ 21 m}}$ 

$\therefore $ radius of circular garden $ = {\text{ }}\dfrac{{21}}{2}{\text{ m}}$ 

Now, circumference of the circular garden =2 $\pi r$

$\therefore $ circumference of the circular garden $ = {\text{ 2 x }}\dfrac{{22}}{7}{\text{ x }}\dfrac{{21}}{2}{\text{  =  66 m}}$ 

The gardener makes 2 rounds of fence,

So, the total length of the rope of fencing = 2 $\pi r  =  2 x 66  =  132 m$

$\because $  the cost of $1{\text{ m}}$ rope $ = $ ₹ $4$ 

$\therefore $  the cost of $132{\text{ m}}$ rope $ = $ $4{\text{ x 132  = }}$ ₹$528$ 

5. From a circular sheet of radius $4{\text{ cm}}$ , a circle of radius ${\text{3 cm}}$ is removed. Find the area of the remaining sheet. (Take $\pi {\text{  =  3}}{\text{.14}}$) 

Given, radius of circular sheet(R) $ = {\text{ 4 cm}}$ 

Radius of removed circle(r) $ = {\text{ 3 cm}}$

$\therefore $ area of remaining sheet $ = $ area of circular sheet $ - $ area of removed circle

$\therefore $ area of remaining sheet $=\pi R^{2}-\pi r^{2}$ 

$\therefore $ area of remaining sheet ${\text{ =  (3}}{\text{.14 x 4 x 4)   -  (3}}{\text{.14 x 3 x 3)}}$ 

$\therefore $ area of remaining sheet ${\text{ =  3}}{\text{.14 x (16  -  9)  =  3}}{\text{.14 x 7  =  21}}{\text{.98 c}}{{\text{m}}^2}$

Thus, the area of the remaining sheet is $21.98{\text{ c}}{{\text{m}}^2}$ .

6. Saima wants to put a lace on the edge of a circular table cover of diameter ${\text{1}}{\text{.5 m}}$. Find the length of the lace required and also find its cost if one meter of the lace costs ` ₹$15$. (Take $\pi {\text{  =  3}}{\text{.14}}$)

Given, diameter of the circular table cover $ = {\text{ 1}}{\text{.5 m}}$ 

$\therefore $ radius of the circular table $ = {\text{ }}\dfrac{{1.5}}{2}{\text{ m}}$

Now, circumference of the circular table ${\text{ =  2}}\pi {\text{r  =  2 x 3}}{\text{.14 x }}\dfrac{{1.5}}{2}{\text{  =  4}}{\text{.71 m}}$ 

$\therefore $ the length of required lace is $4.71{\text{ m}}$ 

As, the cost of $1{\text{ m}}$ lace = ₹$15$

The cost of $4.71{\text{ m}}$lace $ = {\text{ 4 x 4}}{\text{.71  =  }}$ ₹$70.65$ 

Hence, the cost of $4.71{\text{ m}}$is ₹$70.65$.

7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter. 

Given, diameter $ = {\text{ 10 cm}}$ 

$\therefore {\text{ radius  =  }}\dfrac{{{\text{diameter}}}}{2}{\text{  =  }}\dfrac{{10}}{2}{\text{  =  5 cm}}$ 

As per the question,

Perimeter of the figure $ = $ circumference of semicircle $ + $ diameter

$\therefore $ perimeter of the figure  $\pi r$  +  $d  =  \dfrac{{22}}{7}{\text{ x 5  +  10 }}$ 

$\therefore $ perimeter of the figure ${\text{ =  }}\dfrac{{110}}{7}{\text{  +  10  =  }}\dfrac{{110 + 70}}{7}{\text{  =  }}\dfrac{{180}}{7}{\text{  =  25}}{\text{.71 cm}}$ 

Hence, the perimeter of the given figure is $25.71{\text{ cm}}$ 

8. Find the cost of polishing a circular table-top of diameter $1.6{\text{ m}}$ , if the rate of polishing is ₹$15{\text{/}}{{\text{m}}^2}$ . (Take $\pi {\text{  =  3}}{\text{.14}}$)

Given, diameter of the circular table top $ = {\text{ 1}}{\text{.6 m}}$ 

$\therefore $ radius of the circular table top $ = {\text{ }}\dfrac{{1.6}}{2}{\text{  =  0}}{\text{.8 m}}$ 

Now, area of the circular table top $ =  \pi {{\text{r}}^{\text{2}}}{\text{  =  3}}{\text{.14 x 0}}{\text{.8 x 0}}{\text{.8  =  2}}{\text{.0096 }}{{\text{m}}^2}$ 

The cost of $1{\text{ }}{{\text{m}}^2}$ of polishing $ = $ ₹$15$ 

$\therefore $ the cost of $2.0096{\text{ }}{{\text{m}}^2}$ of polishing ${\text{ =  15 x 2}}{\text{.0096  =  30}}{\text{.14(approx)}}$ 

Hence the cost of polishing a circular table of $2.0096{\text{ }}{{\text{m}}^2}$is ₹$30.14$ (approx).

9. Shazli took a wire of length $44{\text{ cm}}$  and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

Given, total length of the wire $ = {\text{ 44 cm}}$ 

$\therefore $ circumference of the circle =2 $\pi r$ 

$\Rightarrow { 2\pi r  =  44} $

$\Rightarrow {\text{ 2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x r  =  44}} $

$\Rightarrow {\text{ r  =  }}\dfrac{{{\text{44 x 7}}}}{{{\text{2 x 22}}}} $

$\Rightarrow {\text{r =  7 cm}} $ 

Now, area of circle $ =  \pi {{\text{r}}^{\text{2}}}{\text{  =  }}\dfrac{{22}}{7}{\text{ x 7 x 7  =  154 c}}{{\text{m}}^2}$ 

Now, the wire is converted into square

$\therefore $ the perimeter of square $ = {\text{ 44 cm}}$ 

$\Rightarrow {\text{ 4 x side  =  44}} $

$\Rightarrow {\text{ side  =  }}\dfrac{{44}}{4}{\text{  =  11 cm}} $ 

So, the area of square $ = $ side ${\text{x}}$ side $ = $ $11{\text{ x 11  =  121 c}}{{\text{m}}^2}$ 

Therefore, in comparison, the area of a circle is greater than that of a square, so the circle encloses more area.

10. From a circular card sheet of radius $14{\text{ cm}}$ , two circles of radius $3.5{\text{ cm}}$  and a rectangle of length $3{\text{ cm}}$and breadth ${\text{1 cm}}$are removed (as shown in the adjoining figure). Find the area of the remaining sheet. (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

Given, radius of circular sheet(R) $ = {\text{ 14 cm}}$ 

Radius of smaller circle(r) $ = {\text{ 3}}{\text{.5 cm}}$

Length of rectangle (l) $ = {\text{ 3 cm}}$

Breadth of the rectangle (b) $ = {\text{ 1 cm}}$

As per given,

Area of remaining sheet $ = $ area of circular sheet $ - $ (area of two smaller circle $ + $ area of rectangle)

$=x R^{2}-\left[2\left(\pi r^{2}\right)+(l \times b)\right]$

$= {\text{ [}}\dfrac{{22}}{7}{\text{ x 14 x 14]  -  [(2 x }}\dfrac{{22}}{7}{\text{ x 3}}{\text{.5 x 3}}{\text{.5)  +  (3 x 1)]}} $

${\text{ =  [22 x 14 x 2]  -  [(44 x 0}}{\text{.5 x 3}}{\text{.5)  +  3]}} $

${\text{ =  616  -  80}} $

${\text{ =  536 c}}{{\text{m}}^2} $

Hence, the area of the remaining sheet is $536{\text{ c}}{{\text{m}}^2}$ 

11. A circle of radius ${\text{2 cm}}$  is cut out from a square piece of an aluminium sheet of side $6{\text{ cm}}$. What is the area of the leftover aluminium sheet? (Take $\pi {\text{  =  3}}{\text{.14}}$)

Given, radius of circle $ = {\text{ 2 cm}}$ 

Side of aluminium square sheet $ = {\text{ 6 cm}}$

Area of aluminium sheet left $ = $ total area of square sheet $ - $ area of circle

$=$ side x side $-\pi r^{2}$

${\text{ =  (6 x 6)  -  (3}}{\text{.14 x 2 x 2)}} $

${\text{ =  36  -  12}}{\text{.56  =  23}}{\text{.44 c}}{{\text{m}}^2} $ 

Hence, the area of aluminium sheet left is $23.44{\text{ c}}{{\text{m}}^2}$

12. The circumference of a circle is${\text{31}}{\text{.4 cm}}$. Find the radius and the area of the circle. (Take $\pi {\text{  =  3}}{\text{.14}}$)

Given, the circumference of the circle $ = {\text{ 31}}{\text{.4 cm}}$ 

$\Rightarrow 2\pi r  =  31{\text{.4}} $

$\Rightarrow {\text{ 2 x 3}}{\text{.14 x r  =  31}}{\text{.4}} $

$\Rightarrow {\text{ r  =  }}\dfrac{{31.4}}{{2{\text{ x 3}}{\text{.14}}}} $

$\Rightarrow {\text{ r  =  5 cm}} $ 

Now, area of circle $=\pi r^{2}=3.14 \times 5 \times 5$  $=78.5 \mathrm{~cm}^{2}$

Therefore, the radius and area of the circle are $5{\text{ cm}}$ and $78.5{\text{ c}}{{\text{m}}^2}$ respectively.

13. A circular flower bed is surrounded by a path $4{\text{ m}}$  wide. The diameter of the flower bed is ${\text{66 m}}$. What is the area of this path? (Take $\pi {\text{  =  3}}{\text{.14}}$)

Given, diameter of the circular flower bed $ = {\text{ 66 m}}$ 

$\therefore $ radius of circular flower bed(r) $ = \dfrac{{66}}{2}{\text{  =  33 m}}$ 

$\therefore $ radius of circular flower bed with $4{\text{ m}}$ wide path(R) $ = {\text{ 33  +  4  =  37 m}}$ 

Area of path $ = $ area of bigger circle $ - $ area of smaller circle

$=\pi \mathrm{R}^{\mathrm{2}}-\pi r^{2}=\pi\left({R}^{2}-r^{2}\right)$ 

\[{\text{ =  [(37}}{{\text{)}}^2}{\text{  -  (33}}{{\text{)}}^2}{\text{]}}\]

$= {\text{ 3}}{\text{.14[(37  +  33)(37  -  33)]}} $

${\text{ =  3}}{\text{.14 x 7 x 4  =  879}}{\text{.20 }}{{\text{m}}^2} $ 

$[\because {\text{ }}{{\text{a}}^2} - {\text{ }}{{\text{b}}^2}{\text{  =  (a  +  b)(a  -  b)]}}$ 

Hence, the area of the path is $879.20{\text{ }}{{\text{m}}^2}$ 

14. A circular flower garden has an area of $314{\text{ }}{{\text{m}}^2}$ . A sprinkler at the centre of the garden can cover an area that has a radius of $12{\text{ m}}$. Will the sprinkler water the entire garden?  (Take $\pi {\text{  =  3}}{\text{.14}}$)

We know that, circumference of the circle =  2 $\pi r$ 

Circular area by the sprinkler $ = {\text{ 3}}{\text{.14 x 12 x 12  =  3}}{\text{.14 x 144  =  452}}{\text{.16 }}{{\text{m}}^2}$ 

Given, area of the circular flower garden $ = {\text{ 314 }}{{\text{m}}^2}$ 

As the area of the circular flower garden is smaller than the area by sprinkler, hence sprinkler will water the entire garden. 

15. Find the circumference of the inner and the outer circles, shown in the adjoining figure.  (Take $\pi {\text{  =  3}}{\text{.14}}$)

given, radius of outer circle(r) $ = {\text{ 19 m}}$ 

$\therefore $ circumference of outer circle =  2 $\pi r$ $ = {\text{ 2 x 3}}{\text{.14 x 19  =  119}}{\text{.32 m}}$ 

Now, radius of the inner circle(r`) $ = {\text{ 19  -  10  =  9 m}}$ 

$\therefore $ circumference of inner circle =  2 $\pi r$ $ = {\text{ 2 x 3}}{\text{.14 x 9  =  56}}{\text{.52 m}}$

Hence, the circumference of inner and outer circles are ${\text{56}}{\text{.52 m and 119}}{\text{.32 m respectively}}$ 

16. How many times a wheel of radius $28{\text{ cm}}$ must rotate to go $352{\text{ m}}$? (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

Let the wheel rotate n times of its circumference.

Given, radius of the wheel $ = {\text{ 28 cm}}$ 

And total distance $ = {\text{ 352 m  =  35200 cm}}$ 

$\therefore $ distance covered by wheel $ = {\text{ n x circumference of wheel}}$

$\Rightarrow$ 35200  =  n x $2\pi r $

$\Rightarrow {\text{ 35200  =  n x 2 x }}\dfrac{{22}}{7}{\text{ x 28}} $

$\Rightarrow {\text{ n  =  }}\dfrac{{35200{\text{ x 7}}}}{{2{\text{ x 22 x 28}}}} $

$\Rightarrow {\text{ n  =  200 revolutions}} $ 

Thus wheel must rotate $200$times to go $352{\text{ m}}$ .

17. The minute hand of a circular clock is ${\text{15 cm}}$ long. How far does the tip of the minute hand move in 1 hour? (Take $\pi {\text{  =  3}}{\text{.14}}$)

In 1 hour, a minute hand completes one round means makes a circle.

Given, radius of the circle(r) $ = {\text{ 15 cm}}$ 

Circumference of the circular clock =  2 $\pi r$  =  $2 x 3{\text{.14 x 15  =  94}}{\text{.2 cm}}$ 

Hence, the tip of the minute hand moves $94.2{\text{ cm}}$ in $1$ hour.

Conversion of Units

Note: Since,   1m = 100 cm ∴ 1 m 2 = 10000 cm 2

1 km = 1000 m ∴ 1 km 2 = 10,00,000 m 2

1 hectare (ha) = 100 m x 100m = 10000 m 2

Overview of Deleted Syllabus for CBSE Class 7 Maths Perimeter and Area

Class 7 Maths Chapter 9: Exercises Breakdown

NCERT Solutions for Maths Perimeter and Area Chapter 9 Class 7 Maths by Vedantu equips you with essential tools to measure and compare shapes. This chapter focuses on calculating the distance around a closed figure (perimeter) and the amount of space enclosed by that figure (area). This chapter is crucial for building a strong foundation in geometry. By understanding the concepts of perimeter and area, students can solve various practical problems and prepare themselves for more advanced topics in mathematics. In previous year exams, around 3-4, questions have been asked from this chapter, highlighting its significance in the overall curriculum. By thoroughly practising the problems and understanding the step-by-step solutions provided by Vedantu, you can confidently tackle algebraic expressions and identities.

Other Study Material for CBSE Class 7 Maths Chapter 9

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions Class 7 Maths Chapter 9 Perimeter and Area

1. Answer the Following Questions.

(a). The length and breadth of a rectangle are 10 cm and 8 cm respectively. What is its perimeter?

(b). The radius of the circle is 1 cm. What is its circumference?

a) 36 cm (Perimeter of a rectangle = 2 ( l + b) = 2 ( 10 + 8) = 2 x 18 = 36)

b) 44/7 (first we need to find diameter)

r = 1      ∴ d = 2. So, C = πd = 22/7 x 2 = 44/7

2. Why Should I Opt for Vedantu’s Study Material?

The study material on Vedantu is prepared by the subject-matter experts in an easy-to-understand manner. Vedantu provides updated and comprehensive explanations on the topics covered in the syllabus of Class 7 Maths, as per the guidelines of the board that will help students to prepare better for their exams.

3. What is the use of practising NCERT Solutions for Class 7 Maths Chapter 11?

Practising NCERT Solutions for Class 7 Maths Chapter 11 will help you get a better understanding of each topic of this chapter. Other than that, practising the questions will help you prepare for your exams better. The solutions provided by Vedantu are free of cost. They are also available on the Vedantu Mobile app too to make it easier for the students. Now, you can study hassle-free and fulfil your doubts easily.

4. Are Vedantu solutions for Class 7 NCERT Maths reliable?

NCERT solutions Class 7 Maths for Chapter 11 is reliable. Vedantu is the best website for students to get accurate, to the point answers for any NCERT subject. Students can also download the solutions in the pdf form too. Vedantu’s content is curated by subject matter experts who have years of experience. And, that’s not all. Vedantu also follows NCERT and CBSE guidelines. So, to answer your question, Vedantu is absolutely reliable.

5. What are the Important Topics Covered in Class 7 Maths NCERT Solutions Chapter 11?

The important topics covered in Class 7 Maths NCERT Solutions Chapter 11 are Basic Geometry, Shapes, Length, Breadth, Area And Perimeter and all the formulas related to them. The solutions provided by Vedantu are free of cost. They are also available on the Vedantu Mobile app.

6. What are the Area and Perimeter?

Perimeter is the total distance that encloses a certain area. The area is a part of a plane that is enclosed by the perimeter. The area refers to the region that is enclosed by a certain shape. The shape can be anything- a square, circle, rectangle, triangle, or it can also be irregular. The perimeter refers to the lines or the distance that covers the area. To understand more about Area and Perimeter, you can visit Vedantu.

7. How to calculate the area and perimeter of a square and a rectangle?

A square has two dimensions, and all the sides are of the same length. So the formula for the area of the square is - “side x side (SxS).” The perimeter of the Square is calculated with the following formula- “4 x Side (4S).” The rectangle has two dimensions, too- the length and the breath with different measurements both. The formula for calculation of area is “Length x Breadth (LxB).” The formula for calculation of the perimeter of the rectangle is “ 2(Length + Breadth).”

8. What topics are covered in Class 7th Perimeter and Area Maths Chapter 9?

Class 7 Maths Chapter 9, titled "Perimeter and Area," covers the concepts of measuring the boundaries and surfaces of various geometric shapes. Key topics include the perimeter and area of rectangles, squares, triangles, and circles, as well as composite figures.

9. What are some common mistakes to avoid while solving perimeter and area problems in class 7 Maths chapter 9?

Common mistakes include:

Using incorrect formulas for different shapes.

Confusing the units of measurement (linear units for perimeter and square units for area).

Miscalculating the dimensions or not considering all parts of a composite figure.

10. What is a real-life application of learning in Maths Class 7 perimeter and area?

Real-life applications include

Construction: Calculating the perimeter for fencing a plot of land or the area for tiling a floor.

Gardening: Planning the layout of a garden or determining the amount of soil needed.

Interior Design: Measuring areas to fit furniture or carpets appropriately.

11. What are the benefits of using NCERT Chapter 9 Class 7 Maths Solutions for studying Perimeter and Area Class 7?

NCERT solutions provide step-by-step explanations, making it easier for students to understand and solve problems. They align with the CBSE curriculum, ensuring that students are well-prepared for their exams.

NCERT Solutions for Class 7 Maths

Ncert solutions for class 7.

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CBSE Class 7 Maths

Mathematics marks the foundation for higher-level reasoning and problem-solving abilities, and the CBSE curriculum of Class 7 is curated in such a way that it helps students master fundamental concepts useful in solving advanced-level questions from topics like algebra, geometry, commercial mathematics, etc.

The CBSE curriculum enables students to dive deep into mathematical concepts and practically apply them, which ensures a better grasp of the subject. Each topic is taught according to its practical application in real life, whether it is calculating the perimeter and areas, solving equations, understanding profit and loss, etc. 

Class 7 Maths NCERT Syllabus has been provided to assist students in understanding all the chapters and preparing their study plans. In this blog, we will cover several aspects of the Class 7 Maths Syllabus, including recommended books, chapter-wise weightage and more.

• 1.0 CBSE Class 7 Maths Syllabus

The Topic-wise syllabus of CBSE Class 7 Maths is given below:

• 2.0 CBSE Class 7 Maths Books

Books are essential in shaping a student's learning journey, providing clarity and depth to each concept. For CBSE Class 7, students primarily rely on NCERT textbooks, known for their clear explanations, solved examples, and alignment with the syllabus. These books offer a systematic understanding of concepts, making them the mainstay of CBSE education.

NCERT Books for Class 7

The main source that explains each topic in detail and adheres to the CBSE curriculum is the NCERT Science Textbook for Class 7.

Additional Reference Books for Practice

• RS Aggarwal

These additional reference books are excellent for practising problems and reinforcing concepts covered in the NCERT textbooks.

• 3.0 Class 7 Mock Test

Mock Tests play a very important role in a student's academic journey. Practising regular mock tests can help students learn more about the exam pattern, question types, and time management strategies. Mock Tests help students identify their areas of improvement and provide them with the platform to enhance their problem-solving skills and score better on the exam. Additionally, mock tests reduce exam anxiety by familiarising students with the exam format.

• 4.0 Class 7 Math Preparation Tips
• Analysis of Syllabus: Students must familiarize themselves with the CBSE Class 7 Maths Syllabus to plan their study routine effectively. 
• Studying the NCERT Curriculum: Your primary source of study should be the NCERT Maths textbook. Study each chapter thoroughly and ensure that you understand all the core concepts.
• Practice Math Every Day : Develop the habit of practising math every day. Frequent practice improves your ability to solve problems and broadens your knowledge of various questions.
• NCERT Solutions: To learn the proper technique to solve questions, refer to the NCERT solutions. To increase your understanding, make sure you practice the exercises and examples given in each chapter. 
• Additional Study Material: For more thorough instruction, utilize extra study resources, such as RD Sharma and RS Aggarwal's Math textbooks for Class 7. These books offer a variety of problems at all skill levels, from beginner to advanced, that will boost your confidence.

Table of Contents

• 2.1 NCERT Books for Class 7
• 2.2 Additional Reference Books for Practice

Examine the concepts in detail and use the NCERT Solutions to help you prepare for the topics. Make sure you create an effective study plan so you can easily pass the Class 7 Exam.

The NCERT Textbooks recommended by the CBSE board are more than sufficient for the fundamentals if you want to do well on the Class 7 Math Exam.

You can visit our page and go through the Chapter List for the NCERT Class 7 Maths. Make sure you plan your preparation based on the topics listed under each chapter.

Related Articles:-

Cbse class 7 science - detailed syllabus and marks weightage.

Class 7 is an important year in a student’s academic journey as it embarks on the transition of the foundational to advanced-level concepts.

CBSE Syllabus for Class 7 - All Subjects Syllabus

The CBSE Class 7 syllabus is crucial for organised learning and exam preparation, providing a clear understanding of the topics covered throughout the academic year.

AssignmentsBag.com

Assignments For Class 7 Mathematics

Assignments for Class 7 Mathematics have been developed for Standard 7 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 7 Mathematics from our website as we have provided all topic wise assignments free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 7 Mathematics. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Mathematics book and get good marks in class 7 exams.

Assignments for Class 7 Mathematics as per CBSE and NCERT pattern

More Assignments for Class 7 Mathematics

Assignments for Class 7 Mathematics as per CBSE NCERT pattern

All students studying in Grade 7 Mathematics should download the assignments provided here and use them for their daily routine practice. This will help them to get better grades in Mathematics exam for standard 7. We have made sure that all topics given in your textbook for Mathematics which is suggested in Class 7 have been covered ad we have made assignments and test papers for all topics which your teacher has been teaching in your class. All chapter wise assignments have been made by our teachers after full research of each important topic in the textbooks so that you have enough questions and their solutions to help them practice so that they are able to get full practice and understanding of all important topics. Our teachers at https://www.assignmentsbag.com have made sure that all test papers have been designed as per CBSE, NCERT and KVS syllabus and examination pattern. These question banks have been recommended in various schools and have supported many students to practice and further enhance their scores in school and have also assisted them to appear in other school level tests and examinations. Its easy to take print of thee assignments as all are available in PDF format.

Some advantages of Free Assignments for Class 7 Mathematics

• Solving Assignments for Mathematics Class 7 helps to further enhance understanding of the topics given in your text book which will help you to get better marks
• By solving one assignments given in your class by Mathematics teacher for class 7 will help you to keep in touch with the topic thus reducing dependence on last minute studies
• You will be able to understand the type of questions which are expected in your Mathematics class test
• You will be able to revise all topics given in the ebook for Class 7 Mathematics as all questions have been provided in the question banks
• NCERT Class 7 Mathematics Workbooks will surely help you to make your concepts stronger and better than anyone else in your class.
• Parents will be able to take print out of the assignments and give to their child easily.

All free Printable practice assignments are in PDF single lick download format and have been prepared by Class 7 Mathematics teachers after full study of all topics which have been given in each chapter so that the students are able to take complete benefit from the worksheets. The Chapter wise question bank and revision assignments can be accessed free and anywhere. Go ahead and click on the links above to download free CBSE Class 7 Mathematics Assignments PDF.

You can download free assignments for class 7 Mathematics from https://www.assignmentsbag.com

You can get free PDF downloadable assignments for Grade 7 Mathematics from our website which has been developed by teachers after doing extensive research in each topic.

On our website we have provided assignments for all subjects in Grade 7, all topic wise test sheets have been provided in a logical manner so that you can scroll through the topics and download the worksheet that you want.

You can easily get question banks, topic wise notes and questions and other useful study material from https://www.assignmentsbag.com without any charge

Yes all test papers for Mathematics Class 7 are available for free, no charge has been put so that the students can benefit from it. And offcourse all is available for download in PDF format and with a single click you can download all assignments.

https://www.assignmentsbag.com is the best portal to download all assignments for all classes without any charges.

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• NCERT Solutions
• NCERT Class 7

NCERT Solutions for Class 7

Ncert solutions for class 7 – cbse 2023-24 edition download free pdfs.

NCERT Solutions for Class 7 are highly significant for the students as they help them grasp the basics of the concepts and lay a strong foundation for their higher studies. Class 7 is more of a continuation of Class 6, and the students get to learn moderately advanced topics compared to their lower classes. NCERT books  are the best study material students can have to study the different subjects of Class 7. The textbook adequately explains all the concepts that a Class 7 student has to learn.

Having a firm grip on the answers to all the questions given in the NCERT textbook gives the students an upper hand in their exam preparation. To help the students refer to the best solutions for the problems, we have provided the NCERT solutions for Class 7 Maths and Science. These NCERT solutions offer detailed explanations for every question present in each chapter of the subjects of Maths and Science in Class 7. These extensive explanations are devised by subject experts after thorough research to produce a useful and relevant source of NCERT Solutions  for the students. These NCERT Solutions, when studied meticulously, can help the students in getting rid of all their doubts. Maths and Science are the subjects considered to be the most important as well as difficult in the academic curriculum of any student.

Due to the same reason, it is necessary to have a stronghold on these subjects right from Class 7 to understand the basic concepts of these subjects from scratch. Emphasis on practising the questions and solutions to understand the topics and theories present in the given chapter is mandatory to secure good marks in the examination.

NCERT Class 7 Solutions includes chapter-wise solutions, equipping the students with the key to unlocking their problem-solving skills. A significant impact can be made on the academic career of students by adopting a suitable learning strategy. These solutions can be used by the students not only to find the right method but also to verify if their answers are correct or not. Helping the students to understand the topics and concepts comprehensively is the ultimate aim of providing the perfect NCERT Solutions .

Other NCERT Resources for Class 7

Ncert books for class 7.

NCERT books are the essence of the school system in India. By developing and publishing textbooks, study materials, educational kits and much more, the NCERT is up to providing quality education to school children. The NCERT Class 7 Maths and Science books can be downloaded from the links given in the table below.

NCERT Syllabus for Class 7

The first step towards preparing for the exams is knowing the syllabus. It helps the students plan their study timetable and academic schedule for the entire year in a better manner, providing enough time for exam preparation and revision. Given in the table below is the syllabus for Class 7.

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BYJU’S App helps students in learning all their concepts in a simple as well as exciting way by providing excellent support in their studies. Students across the globe can avail of the app by downloading BYJU’S – The Learning App either on their computers, laptops, tablets or mobile phones. The app helps the students to explore more about the different topics through interactive videos, games and other quizzes. BYJU’S also provides free study materials and question papers and helps in preparing for both school assignments and final exams.

Frequently Asked Questions on NCERT Solutions for Class 7

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Class 7 Mathematics Lines and Angles Assignments

We have provided below free printable Class 7 Mathematics Lines and Angles Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 7 Mathematics Lines and Angles . These Assignments for Grade 7 Mathematics Lines and Angles cover all important topics which can come in your standard 7 tests and examinations. Free printable Assignments for CBSE Class 7 Mathematics Lines and Angles , school and class assignments, and practice test papers have been designed by our highly experienced class 7 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Lines and Angles Class 7 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Lines and Angles Class 7. Students can click on the links below and download all Pdf Assignments for Mathematics Lines and Angles class 7 for free. All latest Kendriya Vidyalaya Class 7 Mathematics Lines and Angles Assignments with Answers and test papers are given below.

Mathematics Lines and Angles Class 7 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 7 Mathematics Lines and Angles . Students and teachers can download and save all free Mathematics Lines and Angles assignments in Pdf for grade 7th. Our expert faculty have covered Class 7 important questions and answers for Mathematics Lines and Angles as per the latest syllabus for the current academic year. All test papers and question banks for Class 7 Mathematics Lines and Angles and CBSE Assignments for Mathematics Lines and Angles Class 7 will be really helpful for standard 7th students to prepare for the class tests and school examinations. Class 7th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 7 Mathematics Lines and Angles Download in Pdf

Advantages of Class 7 Mathematics Lines and Angles Assignments

• As we have the best and largest collection of Mathematics Lines and Angles assignments for Grade 7, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Mathematics Lines and Angles textbooks for Class 7 .
• All Mathematics Lines and Angles assignments for Class 7 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 7 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Mathematics Lines and Angles chapter wise worksheets and assignments for free in Pdf
• Class 7 Mathematics Lines and Angles question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 7 Mathematics Lines and Angles students

At https://www.cbsencertsolutions.com, we have provided the biggest database of free assignments for Mathematics Lines and Angles Class 7 which you can download in Pdf

We provide here Standard 7 Mathematics Lines and Angles chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Mathematics Lines and Angles in Grade 7, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

We have provided here topic-wise Mathematics Lines and Angles Grade 7 question banks, revision notes and questions for all difficult topics, and other study material.

We have provided the best collection of question bank and practice tests for Class 7 for all subjects. You can download them all and use them offline without the internet.

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