assignment of class 7 maths

NCERT Solutions for Class 7 Maths

Chapter Wise Class 7 Maths NCERT Solutions NCERT Solutions for Class 7 Maths in Hindi and English Medium updated for academic session 2024-25 Exams. Get NCERT Solutions for Class 7 Maths in Videos and PDF by Tiwari Academy, prepared from the textbook from NCERT in India.

NCERT creates the textbooks that shape India’s school curriculum. The newly textbooks for 2024-25 have streamlined the 7th Mathematics curriculum to 13 chapters, all of which are covered in our solutions. 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. Comparing Quantities Chapter 8. Rational Numbers Chapter 9. Perimeter and Area Chapter 10. Algebraic Expressions Chapter 11. Exponents and Powers Chapter 12. Symmetry Chapter 13. Visualising Solid Shapes

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NCERT Solutions App for Class 7

Unlock NCERT Solutions for Class 7 Maths at your fingertips with our mobile apps, ensuring learning is never halted. No matter where you are, as long as you have your smartphone or tablet, access to comprehensive solutions is guaranteed. Crafted to assist students in mastering the NCERT Class 7 Mathematics textbook, our solutions offer detailed step-by-step explanations for each chapter and exercise.

Class 7 Maths solutions are widely used by students and teachers in India to prepare for class 7 math exams and to build a strong foundation in mathematics. They can be found in various formats, including printed books, online PDFs, and digital resources.

Class 7 Solution Apps Tiwari Academy has an experienced and dedicated team of teachers who will help students clear all their doubts and help them score high marks in exams. The NCERT Solutions for 7th Maths are available online and can be accessed and downloaded for free in Hindi and English Medium in PDF Format.

These are easy to solve and provide step-by-step instructions. A complete description of each question is given in the solutions which are based on the latest NCERT Books 2024-25.

NCERT Solutions for Class 7 Maths are available in both English Medium and Hindi Medium to download for offline use. These solutions are updated for current session 2024-25. Students can download Class 7 Solutions App containing all subject’s solutions in English and Hindi Medium.

Learning Tips for Class 7 Maths Exams Preparing for Class 7 Maths finals can be a challenging yet rewarding experience. 1. Start by reviewing the syllabus and identifying key topics. 2. Focus on understanding concepts rather than rote memorization. 3. Utilize resources like textbooks, online tutorials, and practice worksheets. 4. Regular practice is essential in mathematics; solve various problems to strengthen your understanding and application skills. 5. Don’t hesitate to seek help from teachers or peers if you encounter difficult topics.

In addition to structured study, develop problem-solving skills. Math is not just about knowing formulas but applying them in different scenarios. Pay attention to the steps and methods used in solving problems, as these are often as important as the final answer. Regularly review mistakes in practice tests to avoid repeating them. Group study can be beneficial, as explaining concepts to others reinforces your own understanding. Maintain a positive attitude and healthy study habits. A well-rested mind is more efficient in processing and recalling information. Balance study sessions with short breaks to avoid burnout. On the day before the exam, do a light review and relax. Read each question carefully, manage your time wisely, and don’t spend too much time on a single problem.

Class 7 Maths Solutions & Main Points of All Chapters

Chapter 1: integers.

Class 7 Maths Chapter 1 in Hindi and English Medium PDF and video solutions are given here. We have done on a number line when we add a positive integer, and we move to the right. Class 7 Maths Exercise 1.1 Class 7 Maths Exercise 1.2 Class 7 Maths Exercise 1.3 Class 7 Maths Exercise 1.1 in Hindi Class 7 Maths Exercise 1.2 in Hindi Class 7 Maths Exercise 1.3 in Hindi

Chapter 2: Fractions and Decimals

In Class 7 Maths Chapter 2 , we will learn about fractions and decimals along with the operations of addition and subtraction on them. We also learnt how to multiply two decimal numbers. Class 7 Maths Exercise 2.1 Class 7 Maths Exercise 2.2 Class 7 Maths Exercise 2.3 Class 7 Maths Exercise 2.4 Class 7 Maths Exercise 2.5 Class 7 Maths Exercise 2.1 in Hindi Class 7 Maths Exercise 2.2 in Hindi Class 7 Maths Exercise 2.3 in Hindi Class 7 Maths Exercise 2.4 in Hindi Class 7 Maths Exercise 2.5 in Hindi

Chapter 3: Data Handling

In Chapter 3 of Class 7 Maths , we will know about the collection, recording and presentation of data. Before collecting data, we need to know that the data that is collected needs to be organised in a proper table so that it becomes easy to understand and interpret. Class 7 Maths Exercise 3.1 Class 7 Maths Exercise 3.2 Class 7 Maths Exercise 3.3 Class 7 Maths Exercise 3.1 in Hindi Class 7 Maths Exercise 3.2 in Hindi Class 7 Maths Exercise 3.3 in Hindi

Chapter 4: Simple Equations

Chapter 4 of Class 7 Maths describes an equation which is a condition on a variable such that two expressions in the variable should have equal value. It also tells about the value of the variable for which the equation is satisfied is called the solution of the equation. Class 7 Maths Exercise 4.1 Class 7 Maths Exercise 4.2 Class 7 Maths Exercise 4.3 Class 7 Maths Exercise 4.1 in Hindi Class 7 Maths Exercise 4.2 in Hindi Class 7 Maths Exercise 4.3 in Hindi

Chapter 5: Lines and Angles

We already know that a line segment has two endpoints and a ray has only one endpoint. In class 7 Maths chapter 5 , we have studied that a line has no endpoints on either side. Class 7 Maths Exercise 5.1 Class 7 Maths Exercise 5.2 Class 7 Maths Exercise 5.1 in Hindi Class 7 Maths Exercise 5.2 in Hindi

Chapter 6: The Triangles and Its Properties

Chapter 6 of Class 7 Maths describes the six elements of a triangle are its three angles and the three sides. It also briefs about three medians and three altitudes of the triangle. Class 7 Maths Exercise 6.1 Class 7 Maths Exercise 6.2 Class 7 Maths Exercise 6.3 Class 7 Maths Exercise 6.4 Class 7 Maths Exercise 6.5 Class 7 Maths Exercise 6.1 in Hindi Class 7 Maths Exercise 6.2 in Hindi Class 7 Maths Exercise 6.3 in Hindi Class 7 Maths Exercise 6.4 in Hindi Class 7 Maths Exercise 6.5 in Hindi

Chapter 7: Comparing Quantities

Comparing quantities means to find the relative ratio between two or more quantities. We are often required to compare two quantities in our daily life. So, NCERT Books of Class 7 Maths Chapter 7 Comparing Quantities will help to find these comparisons. Class 7 Maths Exercise 7.1 Class 7 Maths Exercise 7.2 Class 7 Maths Exercise 7.1 in Hindi Class 7 Maths Exercise 7.2 in Hindi

Chapter 8: Rational Numbers

Rational numbers are very common to us as it is already used in previous classes. We study in Class 7 Maths Chapter 8 that a number that can be expressed in the form p/q, where p and q are integers and q is not equal to 0, is called a rational number. Class 7 Maths Exercise 8.1 Class 7 Maths Exercise 8.2 Class 7 Maths Exercise 8.1 in Hindi Class 7 Maths Exercise 8.2 in Hindi

Chapter 9: Perimeter and Area

Class 7 Maths Chapter 9 includes the mensuration of 2 – D figures. We know that perimeter is the distance around a closed figure whereas area is the part of the plane occupied by the closed figure. Class 7 Maths Exercise 9.1 Class 7 Maths Exercise 9.2 Class 7 Maths Exercise 9.1 in Hindi Class 7 Maths Exercise 9.2 in Hindi

Chapter 10: Algebraic Expressions

In Class 7 Maths Chapter 10 , we will learn how algebraic expressions are formed from variables and constants. We will also use the operations of addition, subtraction, multiplication and division on the variables and constants to form expressions. Class 7 Maths Exercise 10.1 Class 7 Maths Exercise 10.2 Class 7 Maths Exercise 10.1 in Hindi Class 7 Maths Exercise 10.2 in Hindi

Chapter 11: Exponents and Powers

We know that vast numbers are difficult to read, understand, compare and operate upon. Chapter 11 of Class 7 Maths states the way to make all these easier. We use exponents, converting many of the large numbers in a shorter form. Class 7 Maths Exercise 11.1 Class 7 Maths Exercise 11.2 Class 7 Maths Exercise 11.3 Class 7 Maths Exercise 11.1 in Hindi Class 7 Maths Exercise 11.2 in Hindi Class 7 Maths Exercise 11.3 in Hindi

Chapter 12: Symmetry

Chapter 12 of Grade 7 Maths is based on Symmetry. A figure has line symmetry if there is a line about which the figure may be folded so that the two parts of the figure will coincide. We know that a regular polygons have equal sides and equal angles, so, they have multiple lines of Symmetry. Class 7 Maths Exercise 12.1 Class 7 Maths Exercise 12.2 Class 7 Maths Exercise 12.3 Class 7 Maths Exercise 12.1 in Hindi Class 7 Maths Exercise 12.2 in Hindi Class 7 Maths Exercise 12.3 in Hindi

Chapter 13: Visualising Solid Shapes

We know that the circle, the square, the rectangle, the quadrilateral and the triangle are examples of plane figures. Similarly, the cube, the cuboid, the sphere, the cylinder, the cone and the pyramid are examples of solid shapes given in Class 7 Maths Chapter 13 . Class 7 Maths Exercise 13.1 Class 7 Maths Exercise 13.2 Class 7 Maths Exercise 13.3 Class 7 Maths Exercise 13.4 Class 7 Maths Exercise 13.1 in Hindi Class 7 Maths Exercise 13.2 in Hindi Class 7 Maths Exercise 13.3 in Hindi Class 7 Maths Exercise 13.4 in Hindi

If you are still facing problem to understand the solution any question, please notify us through “Discussion Forum” section. Class 7 Maths NCERT Solutions 2024-25 of all exercises are given as separate PDF files. Important questions from U-like, R S Aggarwal, P K Garg, R D Sharma, Assignments, Notes, Sample Papers, Chapter test and other study material will be uploaded very soon. Download 7 Maths Offline App in English Medium and 7 Ganit Offline App in Hindi Medium. Questions are taken from the Latest CBSE Syllabus for 2024-25. Please give feedback and suggestions to improve the contents and quality if possible.

Hindi Medium NCERT Solutions for class 7 Maths is now available for the current academic year 2024-25 onward. If any advice from students or parents is to improve the website, then do give it. Your contribution will be helpful for other students. In this year Hindi and English medium solutions for almost all subjects given here. Your suggestion is always valuable for us in improving website as well as contents. We are working for UP Board and Bihar Board also. Question answers of 7th NCERT Books for all Subjects are available in PDF as well as ZIP format. Learn here how to get good marks in class 7 Maths exams.

How to study Class 7 Maths with NCERT Solutions?

Class 7 Maths NCERT Solutions are modified according the new books published for academic session 2024-25. These provide clear explanations for each topic, helping students understand the concepts and principles in Class 7 mathematics. Class 7 Maths NCERT solutions serve several important purposes for students. These solutions offer step-by-step explanations for solving problems. They break down complex problems into smaller, more manageable steps, making it easier for students to grasp the solution process.

Step 1: Use Class 7 Maths NCERT Solutions for reference during practice.

Step 2: prepare for exams using ncert solution for class 7 mathematics., step 3: get the solutions from online educational free websites like tiwari academy., step 4: make class 7 maths as one of the easiest subjects using ncert solutions., step 5: clarity of concepts with quality content makes easier to do homework..

How many difficult chapters are there in Class 7 Maths?

According to new syllabus and latest textbooks issued for session 2024-25, there are total of 13 chapters in Class 7 Maths NCERT Book. Most of the chapters are easy to understand and solve. Only few chapters like chapter 6- the triangles and its properties, chapter 6 triangles and chapter 10 algebraic expressions need more practice to be confident. Regular practice of math make it easier than ever.

How to complete syllabus of Class 7 Maths quickly?

For quick response, online help is common, now a days. Be regular in studies by making time table. Try to give at least two hours for Maths in class 7. Practice each of your exercise by taking help from online websites and apps. If a student is doing math regularly for 1 to 2 hours, he will be able to complete the entire syllabus of 7th Maths for 2024-25 with in three or four months. Later on he has to revise only during the rest period of academic session.

How to pass in class 7 Maths within 30 days?

There are many chapters in 7th Mathematics, which are quite easy to solve. Just practice these chapters to score more in less time. For example, chapter 3, 4, 9, 11, 12, and 13 in class 7 math are easier to understand and practice too. So target chapters like these to score well in Maths of 7th standard to clear the exams in 30 days.

Is Class 7 Maths easy to understand?

Class 7 Maths NCERT Books contains the chapters which are following the latest CBSE Syllabus. Most of the chapters are easy to practice and solve. Only few are considered as difficult one. But passing in 7th math is quite easy. Just cover the syllabus doing easy chapters properly a student can score a moderate marks to pass in 7th Maths. If someone want to score excellent marks in standard 7th mathematics, he must work hard.

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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.

FAQs on NCERT Solutions for Class 7 Maths

1.  How can I score good marks in the Maths in Class 7?

Go through the Concepts thoroughly and prepare as per the topics by using the NCERT Solutions. Make sure you develop a proper preparation strategy to clear the Class 7 Exam with ease.

2. Which is the best reference book for Class 7 Maths?

For basics, NCERT Textbooks prescribed by the CBSE Board are more than enough to score better grades in the Class 7 Maths Exam.

3.  Why are NCERT Solutions for Class 7 Maths important?

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.

4. How to download NCERT Solutions for Class 7 free?

Aspirants can download the NCERT Solutions for Class 7 free of cost from our site. Make use of them as a reference and aid your preparation.

5. What are the Chapters contained in the NCERT Solutions for Class 7 Maths?

You can go through the Chapter List for the NCERT Class 7 Maths Solutions by referring to our page. Refer to the topics under each chapter and plan your preparation accordingly.

6. Where can I get NCERT Solutions for Class 7th Maths?

You can get NCERT Solutions for Class 7th Maths by referring to our page. Access the direct links to view or download and use them as a reference during your preparation.

We hope you will have great learning experience while using the solutions. You can download the solutions by clicking on the links above in the description.

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Integers class 7 extra questions with answers and PDF Download

Integers class 7

What are Integers ?

Properties of integers: 

• Negative  integers are on the left side of 0
• Positive integers are on the right side of the zero
• 0 is neither +ve  nor -ve.
• Integers are closed under addition. It means, for any two integers a and b, a + b is an integer.
• Integers are also closed under subtraction. Thus, if a and b are two integers then a – b is also an integer.
• Addition is commutative for integers. In general, for any two integers a and b, we can say a + b = b + a

• Addition is associative for integers. In general for any integers a, b and c, we can say a + (b + c) = (a + b) + c
• For any two positive integers a and b, we can say a × (– b) = (– a) × b = – (a × b)
• Integers are closed under multiplication. In general, a × b is an integer, for all integers a and b
• Multiplication is commutative for integers. For any two integers a and b, a × b = b × a
• Distributivity of multiplication over addition is true for integers. In general, for any integers a, b and c, a × (b + c) = a × b + a × c
• Integers are not closed under division. 
• Division is not commutative for integers. 

Questions on integers for class 7 pdf

Sums on integers for class 7(Solved Examples )

Example 1: Arrange the following integers in ascending order   -102, -39, -51, -5 , 0 , -6, 35  and 7 .

Sol.  Ascending order of the given integers are  : -102 <  -51< -39 < -6< -5 < 0< 7< 35

Example 2: Find the sum of  -72 , 237 , 84 , 72, -184 , -37 .

Sol.  (-72) +237+84+72+(-184)+(-37)=(-72) +393 +(-184)+(-37)

                                                                        = -72+393-184-37

                                                                        =393-293 =100

Example 3: Write down a pair of integers whose (a) sum is –3      (b) difference is –5      (c) difference is 3     (d) sum is 0 Sol.  (a) (–1) + (–2) = –3        or       (–5) + 2 = –3     or   (-7)+(4) =-3   

         (b) (–9) – (– 4) = –5       or        (–2) – 3 = –5    or  (-7) -(-2)= -5

         (c) (–7) – (–10) = 3          or       1 – (–2) = 2     or   +7 -(+4)  =3

        (d) (–10) + 10 = 0           or          4 + (–4) = 0    or    3 +(-3) =0    (You can make more pairs.)

Example 4: Solve the following: (i) (-10) × (-5) + (-6) (ii) (-10) × [(-13) + (-10)] (iii) (-5) × [(-6) + 5] Sol.  (i) (-10) × (-5) + (-6) = 50 – 6 = 44

          (ii) (-20) × [(-13) + (-10)] = (-20) × (-23) = 460

          (iii) (-5) × [(-6) + 5]= (-5)  x  (-1)  =5

Example 5: In a class test containing 15 questions, 4 marks are given for every correct answer and (–2) marks are given for every incorrect answer. Anil attempts all questions but only 10 of her answers are correct.  What is his total score?  Sol. (i) Marks given for one correct answer = 4 So, marks given for 10 correct answers = 4 × 10 = 40 Marks given for one incorrect answer = – 2 So, marks given for 5 (= 15 – 10) incorrect answers = (–2) × 5 = –10 Therefore, Gurpreet’s total score = 40 + ( –10) = 30

Example 6:  Is  (–15) × [(–7) + (–1)] = (–15) × (–7) + (–15) × (–1)?

Sol.  (–15) × [(–7) + (–1)]=   (-15) x (-8) = 120

    (–15) × (–7) + (–15) × (–1)=  105 + 15 = 120 

Hence (–15) × [(–7) + (–1)] = (–15) × (–7) + (–15) × (–1)

Example 7: Find each of the following products:      (a)  (–20) × (–2) × (–5) × 7     (b) (–1) × (–5) × (– 4) × (– 6)

Sol.  (a) (–20) × (–2) × (–5) × 7 =– 20 × (–2 × –5) × 7 = [–20 × 10] × 7 = – 1400

           (b) (–1) × (–5) × (– 4) × (– 6) = [(–1) × (–5)] × [(– 4) × (– 6)] = 5 × 24 = 120 

Example 8:  A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

Example 9: Find: (a) 125 ÷ (–25)       (b) 80 ÷ (–5)        

Sol.  (a) 125 ÷ (-25) = (-125) ÷ 25 = =5 

          (b)  80 ÷ (-5) =(-80) ÷ 5 = -16 

Example  10: Find: (a) (–36) ÷ (– 4)         (b) (–201) ÷ (–3) 

Sol.  (a) (-36) ÷ (-4)= 36 ÷ 4 = 9

          (b) (-201) ÷ (-3) = 201÷ 3 = 67 

Example 11:  Solve the following :

(a) 3+2-1 x 4 ÷ 2 

 Sol. 3+2-1 x 4 ÷ 2 = 3+2-1 x 2 = 3+2-2 = 3

(b) 53 x 2 -1 x 6 

Sol. 53 x 2 -1 x 6  = 106-6 =100 

(c) 7 x 3 +8-2

Sol. 7 x 3 +8-2=  21 +8 – 2 =  29-2 = 27

(d) 12+(-3) +5 – (-2)

Sol. 12+(-3) +5 – (-2) = 12-3+5+2 =9+7 = 16 

Unsolved Examples with answers

(i) Complete the following pattern: (a) 7, 3, – 1, – 5, _____, _____, _____. (b) – 2, – 4, – 6, – 8, _____, _____, _____. (c) 15, 10, 5, 0, _____, _____, _____. (d) – 11, – 8, – 5, – 2, _____, _____, _____.

(iii) Solve the following:  (a) (-15) × 8 + (-15) × 4  (b) [32 + 2 × 17 + (-6)] ÷ 15

(iv) a × (b – c) = a × b – …………

(v) (a) For any integer a, what is (–1) × a equal to?   (b) Determine the integer whose product with (–1) is 24

(vi) Replace the blank with an integer to make it a true statement. (a) (–3) × _____ = 24          (b) 5 × _____ = –40 (c) _____ × (– 9) = –63       (d) _____ × (–12) = 132

(vii) For any integer a,   a ÷ 1 = ……………..

(viii) For any integer a,   a ÷ (- 1) =………….

(ix) Evaluate the following :  (a) (-526)-(-217)      (b) (-31) +31         (c) [(-6)  x (-8) ] x 5     (d) -13 x (7-8)      (e) (-3) x 8 x (-5)

(x) Match the following                        Column I                                                      Column II

 (a) a × 1                                                                        (i) Additive inverse of a

(b) 1                                                                               (ii) Additive identity

(c) ( – a) ÷ ( – b)                                                          (iii) Multiplicative identity

(d) a × ( – 1)                                                                  (iv) a ÷ ( – b)

(e) a × 0                                                                          (v) a ÷ b

(f) ( –a) ÷ b                                                                    (vi) a

(g) 0                                                                                (vii) – a

(h) a ÷ (–a)                                                                   (viii) 0

(i) –a                                                                               (ix) –1

Ans. (i) (a) -9, -13, -17, -21   (b) -10, -12, -14, -16    (c) -5, -10, -15, -20    (d) 1, 4, 7, 10

           (ii) (a) <    (b)  <      (c) >     (d) <      (e) >     (iii) (a) -180     (b) 4        (iv) a x c   

           (v) (a) -a    (b) -24      (vi) (a)  -8    (b)  -8   (c)  7  (d) -11    (vii) a      (viii) -a  

           (ix) (a) -309     (b) 0        (c) 240    (d) 13       (e) 120   

           (x) (a) → (vi), (b) → (iii), c → (v), d → (vii), e → (viii), f → (iv)     g → (ii), h → (ix), i → (i)

DOWNLOAD PDF          Integers class 7 worksheet

MCQ questions  for class 7 maths integers

(a) <                 (b) >               (c) =               (d) none of these

(ii) Solve  40-(-39) + (-5) 

(a) 74              (b) 64              (c) 60            (d) 0

(iii)  When the integers 12, 0, 5, – 5, – 8 are arranged in descending or ascending order, then find out which of the following integers always remains in the middle of the arrangement.

(a)   0                (b) 5                (c) – 8            (d) – 5

(iv) Next three consecutive numbers in the pattern 11, 8, 5, 2, –, –, –are (a) 0, – 3, – 6          (b) – 1, – 5, – 8        (c) – 2, – 5, – 8             (d) – 1, – 4, – 7

(v)The  ……………   is an additive identity for integers.

(a)1               (b) 0           (c)-1           (d) 2

(vi) (–1) × (–1) × (–1) × (–1) × (–1) = ………….  

(a) 1        (b) -1        (c) 0       (d) 2

(vii) …………………… is the multiplicative identity for integers.

(a) 0             (b) 1            (c) -1            (d) 2

(viii) 0 ÷ a = …………… for a ≠ 0

(a) 0          (b) 1          (c) -1           (d) not defined

(ix) any integer divided by zero is ……………………………

(a) 0           (b) meaningless         (c) 1         (d) -1

(x)  (– 85) × 43 + 43 × ( – 15)  is equal to ?

(a) -4300        (b) 4300          (c) 430         (d)  -430

Ans. (i) (a)       (ii) (a)          (iii) 0      (iv) (d)     (v) (b)        (vi)( b)         (vii)  (b)      (viii) (a)    (ix)  (b)      (x) (a) 

 Integers Class 7 Maths quiz 

Maths quiz for class 7 integers

•  Find the absolute value of  -5 .

2.  Add the integers +15,  -4 , + 8 and -6  .

3.  Write a negative integer and a positive integer whose sum is  -5 .

5. In a quiz team A scored -40 , 10 and 0 and team B scored 10 , 0 and -30 in three successive round . Which team scored more ?

6.  Subtract the sum of -1878 and 878   from 2000 .

7.  Subtract -134  from the sum of 37 and -87 .

8. If the sum of two integers is -1700 and one of them is 560 . Find the other number.

9. Determine the integer whose product with -17  is  -153 .

10. Find the value (-23){(-5) +25}

11. Anil is standing in the middle of a bridge which is 30 m above the water level of a river. If a 35 m deep river is flowing under the bridge , then the vertical distance between the foot of Anil and bottom level of  the river is?

12.  [(– 10) × (+ 9)] + ( – 10) is equal to

13.  –16 ÷ [8 ÷ (–2)] is equal to

14. (– 35) × 30 = – 30 × _______.

15. .Write a pair of integers whose product is – 36 and whose difference is 15.

Your score is

The average score is 71%

Restart quiz

Integers class 7 (extra questions  with answers) 

A. State whether the following statements are correct or incorrect. 

(i) When two positive integers are added we get a positive integer.

(ii) When two negative integers are added we get a positive integer.

(iii) When a positive integer and a negative integer are added, we always get a negative integer.

(iv) Additive inverse of an integer 8 is (– 8) and additive inverse of (– 8) is 8.

(v) For subtraction, we add the additive inverse of the integer that is being subtracted, to the other integer.

(vi) (–10) + 3 = 10 – 3

(vii) 8 + (–7) – (– 4) = 8 + 7 – 4

(viii) 25 × (–21) = (–25) × 21

(ix) –1 is a multiplicative identity of integers?

(x)  The distributivity of multiplication over addition is true for integers. 

(xi)  Is division associative for integers? 

(xii) When we change the order of integers, their sum remains the same.

(xiii) When we change the order of integers their difference remains the same.

(xiv) a ÷ b = b ÷ a

(xv)  a – b = b – a

B . Answer the following questions: 

(i) Write a pair of integers whose sum is zero (0) but difference is 8.

(ii) Write a pair of integers whose product is – 15 and whose difference is 8.

(iii) On multiplying or dividing two integers, If the signs of both the integers are  same, the sign of the answer is  ………………..

(iv) On multiplying or dividing two integers, If the signs of both the integers are different, the sign of the answer is  ………………

(v) If a, b and c are integers then a x (b+c)= ………….+…………..

(vi)  The product of three negative integers is a ……………..  integer.

(vii) The product of four negative integers is a ………………….  integer.

(viii) If the number of negative integers in a product is even, then the product is a ………………..  integer; if the number of negative

integers in a product is odd, then the product is a ………………….  integer.

(ix) What will be the sign of the product if we multiply together: (a) 8 negative integers and 3 positive integers?

(b) 5 negative integers and 4 positive integers?

(x)  (-5)  x (-10)  =………… x   (-5)

(xi) The sum of two integers is 116. If one of them is -79, find the other integers.

(xii) The product of three integers does not depend upon the grouping of integers and this is called the ……………………..  for multiplication of integers

Ans. A.  (i) True   (ii) False   (iii) False   (iv) True    (v) True    (vi) False    (viii) False   (viii) True   (ix) False  (x)  True    (xi) False  (xii) True      (xiii) False   (xiv) False     (v) False 

B. (i) 4 and -4      (ii)5,  -3 and  3 ,-5.        (iii) positive        (iv) negative       (v) (a x b)+(b x c)     (vi) negative      (vii) positive    (viii) positive, negative  (ix) (a) positive (b) negative   (x) -10    (xi) 195      (xii) associative property 

Integers class 7 extra questions word problems 

(i) A spacecraft is at 4525km above the earth’s surface.if it descends at the rate of 5km per minute. What will be its position after 9 hours.

(ii)A fruit merchant earns a profit of Rs. 66 per bag of orange sold and a loss of Rs. 44 per bag of grapes sold.

(a)Merchant sells 1800 bags of orange and 2500 bags of grapes. What is the profit or   loss?

(b) What is the number of bags of oranges to be sold to have neither profit nor loss if the number   of grapes bags are sold is 900 bags?​

(iii) Nathan ended round one of a quiz with 200 points. In round two, he scored -300 points and in the third round, he gained 200 points. What was his total score at the end of the third round?

(iv) A submarine submerges from the surface of the sea at the rate of 10 m/min. How long will it take to reach 300 m below the sea level?

(v)Roman civilization began in 509 B.C. and came to an end in 476 A.D. How long did Roman civilization last?

(vi)Metal mercury is a liquid at room temperature. Its melting point is -39°C. The freezing point of alcohol is -114°C. How much warmer is the melting point of mercury than the freezing point of alcohol?

(vii)The temperature in city-X was 10°C in the morning which dropped by 15°C in the evening. What is the temperature in the evening?

(viii)Everest, the highest peak in Asia, is 29,028 feet above sea level. The Dead Sea is 1,312 feet below sea level. What is the level difference between these two places? 7.

(ix)The temperature recorded in a city on Monday is 50°. If the temperature increases by 10° on Tuesday and falls by 11° on Wednesday. Find the temperature recorded on Wednesday

(x)In an examination,3 marks are awarded for every correct answer and1mark is deducted for every incorrect answer. If a student gets 15questions correct and7questions incorrect, find the marks scored by him.

(xi)Andrew has deposited $5500 and has withdrawn $4800 two times consecutively. Find the amount in his account.

(xii) A bird is flying 1200 feet above the sea level. At a particular point, it is directly above a fish floating at 400 feet below the sea level. By how much the bird must descend its flight to be as far as from the sea level, as the fish is?

(xiii)Ryan was playing a game where catching a fish in the basket adds 10 points to his score. If he catches an octopus instead, he loses 5 points. What is his total if he catches 3 fish and 5 octopus?

(xiv)Jason is moving from his hometown to Alaska. He is expecting a big temperature change, and he is getting ready. The hottest it gets in his hometown is 32 degrees celsius. The coldest it gets in Alaska is approximately – 45 degrees celsius. What temperature change will Jason experience?

(xv)The temperature of a hot iron rod drops by 8°𝐶 every hour. If the temperature of the iron rod is 118°𝐶, find the temperature after 4 hours.

(xvi)A submarine submerges at the rate of 5 m/min. If it descends from 20 m above the sea level, how long will it take to reach 250 m below sea level?

(xvii)At Srinagar temperature was − 5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?

(xviii)A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?

(xix)Mohan deposits Rs 2,000 in his bank account and withdraws Rs 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.

(xx)Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A?

(xxi) A cement company earns a profit of Rs 8 per bag of white cement sold and a loss of Rs 5 per bag of grey cement sold.

(a) The company sells 3, 000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?

(b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags.

(xxii) An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach − 350 m.

Ans. (i) 1825 km       (ii)  (a) profit amount is rupees 8800 .   (b) 600 bags of oranges

          (iii) 300    (iv) 30 minutes      (v) 985 degrees     (vi) -75    (vii) -5      (viii) 30340 feet 

          (ix) 49 degrees    (x) 38      (xi) $ -4100     (xii)  800 feet     (xiii)     5 points      (xiv)77            

          degrees    (xv)   86 degrees celcius    (xvi) 54 minutes     (xvii) Monday = −5°C, on     

         tuesday −7°C , on Wednesday −3ºC   (xviii) 6200 m    (xix) rs. 358   (xx)  −10 km (i.e.,  

         Rita is now in west direction)   (xxi) (a) loss of Rs 1000  (b)  4000 bags of white

        cement    (xxii) 1 hour

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• Chapter 6: The Triangles And Its Properties

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

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and Its Properties are available here. For students who feel stressed about searching for the most comprehensive and detailed NCERT Solutions for Class 7 Maths , we at BYJU’S have prepared step-by-step solutions with detailed explanations. We advise students who wish to score good marks in Maths, to go through these solutions and strengthen their knowledge.

Download Exclusively Curated Chapter Notes for Class 7 Maths Chapter – 6 The Triangle and its Properties

Download most important questions for class 7 maths chapter – 6 the triangle and its properties.

Given below are the concepts covered in Chapter 6 – The Triangle and Its Properties of Class 7 Maths NCERT Solutions . The chapter contains 5 exercises covering problems related to these concepts.

• Medians of a Triangle
• Altitudes of a Triangle
• Exterior Angle of a Triangle and Its Property
• Angle Sum Property of a Triangle
• Two Special Triangles: Equilateral and Isosceles
• Sum of the Lengths of Two Sides of a Triangle
• Right-Angled Triangles and Pythagoras Property

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and Its Properties

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Exercise 6.1 Solutions

Exercise 6.2 Solutions

Exercise 6.3 Solutions

Exercise 6.4 Solutions

Exercise 6.5 Solutions

Access answers to NCERT Solutions Class 7 Maths Chapter 6 – The Triangle and Its Properties

Exercise 6.1 Page: 116

An altitude has one endpoint at a vertex of the triangle and another on the line containing the opposite side.

(ii) PD is ___.

A median connects a vertex of a triangle to the mid-point of the opposite side.

(iii) Is QM = MR?

No, QM ≠ MR because D is the mid-point of QR.

2. Draw rough sketches for the following:

(a) In ΔABC, BE is a median.

(b) In ΔPQR, PQ and PR are altitudes of the triangle.

(c) In ΔXYZ, YL is an altitude in the exterior of the triangle.

In the figure, we may observe that for ΔXYZ, YL is an altitude drawn exteriorly to side XZ which is extended up to point L.

3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be the same.

Draw a line segment PS ⊥ BC. It is an altitude for this triangle. Here, we observe that the length of QS and SR is also the same. So PS is also a median of this triangle.

Exercise 6.2 Page: 118

1. Find the value of the unknown exterior angle x in the following diagram:

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50 o + 70 o

= x = 120 o

= x = 65 o + 45 o

= x = 110 o

= x = 30 o + 40 o

= x = 60 o + 60 o

= x = 50 o + 50 o

= x = 100 o

= x = 30 o + 60 o

2. Find the value of the unknown interior angle x in the following figures:

= x + 50 o = 115 o

By transposing 50 o from LHS to RHS, it becomes – 50 o

= x = 115 o – 50 o

= 70 o + x = 100 o

By transposing 70 o from LHS to RHS, it becomes – 70 o

= x = 100 o – 70 o

The given triangle is a right-angled triangle. So, the angle opposite to the x is 90 o .

= x + 90 o = 125 o

By transposing 90 o from LHS to RHS, it becomes – 90 o

= x = 125 o – 90 o

= x + 60 o = 120 o

By transposing 60 o from LHS to RHS, it becomes – 60 o

= x = 120 o – 60 o

= x + 30 o = 80 o

By transposing 30 o from LHS to RHS, it becomes – 30 o

= x = 80 o – 30 o

= x + 35 o = 75 o

By transposing 35 o from LHS to RHS, it becomes – 35 o

= x = 75 o – 35 o

Exercise 6.3 Page: 121

1. Find the value of the unknown x in the following diagrams:

The sum of all the interior angles of a triangle is 180 o .

= ∠BAC + ∠ABC + ∠BCA = 180 o

= x + 50 o + 60 o = 180 o

= x + 110 o = 180 o

By transposing 110 o from LHS to RHS, it becomes – 110 o

= x = 180 o  – 110 o

The given triangle is a right-angled triangle. So, the ∠QPR is 90 o .

= ∠QPR + ∠PQR + ∠PRQ = 180 o

= 90 o + 30 o + x = 180 o

= 120 o + x = 180 o

= x = 180 o  – 120 o

= ∠XYZ + ∠YXZ + ∠XZY = 180 o

= 110 o + 30 o + x = 180 o

= 140 o + x = 180 o

By transposing 140 o from LHS to RHS, it becomes – 140 o

= x = 180 o  – 140 o

= 50 o + x + x = 180 o

= 50 o + 2x = 180 o

= 2x = 180 o  – 50 o

= 2x = 130 o

= x = 130 o /2

= x + x + x = 180 o

= 3x = 180 o

= x = 180 o /3

∴ the given triangle is an equiangular triangle.

= 90 o + 2x + x = 180 o

= 90 o + 3x = 180 o

= 3x = 180 o  – 90 o

= 3x = 90 o

= x = 90 o /3

= 2x = 2 × 30 o = 60 o

2. Find the values of the unknowns x and y in the following diagrams:

= 50 o + x = 120 o

= x = 120 o – 50 o

= x = 70 o 

We also know that,

= 50 o + x + y = 180 o

= 50 o + 70 o  + y = 180 o

= 120 o + y = 180 o

By transposing 120 o from LHS to RHS, it becomes – 120 o

= y = 180 o  – 120 o 

From the rule of vertically opposite angles,

= 50 o + 80 o + x = 180 o

= 130 o  + x = 180 o

By transposing 130 o from LHS to RHS, it becomes – 130 o

= x = 180 o  – 130 o 

= 50 o + 60 o + y = 180 o

= 110 o  + y = 180 o

= y = 180 o  – 110 o 

From the rule of linear pair,

= x + y = 180 o

= x + 70 o = 180 o

= x = 180 o – 70

= 30 o + x + y = 180 o

= 30 o + 60 o + y = 180 o

= 90 o  + y = 180 o

= y = 180 o  – 90 o 

= x + x + y = 180 o

= 2x + 90 o = 180 o

= 2x = 180 o  – 90 o 

= 2x = 90 o

= x = 90 o /2

Exercise 6.4 Page: 126

1. Is it possible to have a triangle with the following sides?

(i) 2 cm, 3 cm, 5 cm

Clearly, we have

(2 + 3) = 5

Thus, the sum of any two of these numbers is not greater than the third.

Hence, it is not possible to draw a triangle whose sides are 2 cm, 3 cm and 5 cm.

(ii) 3 cm, 6 cm, 7 cm

(3 + 6) = 9 > 7

(6 + 7) = 13 > 3

(7 + 3) = 10 > 6

Thus, the sum of any two of these numbers is greater than the third.

Hence, it is possible to draw a triangle whose sides are 3 cm, 6 cm and 7 cm.

(iii) 6 cm, 3 cm, 2 cm

(3 + 2) = 5 < 6

Thus, the sum of any two of these numbers is less than the third.

Hence, it is not possible to draw a triangle whose sides are 6 cm, 3 cm and 2 cm.

2. Take any point O in the interior of a triangle PQR. Is

(i) OP + OQ > PQ?

(ii) OQ + OR > QR?

(iii) OR + OP > RP?

If we take any point O in the interior of a triangle PQR and join OR, OP, OQ.

Then, we get three triangles ΔOPQ, ΔOQR and ΔORP are shown in the figure below.

The sum of the length of any two sides is always greater than the third side.

(i) Yes, ΔOPQ has sides OP, OQ and PQ.

So, OP + OQ > PQ

(ii) Yes, ΔOQR has sides OR, OQ and QR.

So, OQ + OR > QR

(iii) Yes, ΔORP has sides OR, OP and PR.

So, OR + OP > RP

3. AM is a median of a triangle ABC.

Is AB + BC + CA > 2 AM?

(Consider the sides of triangles ΔABM and ΔAMC.)

Now consider the ΔABM,

Then, consider the ΔACM

By adding equations [i] and [ii], we get,

AB + BM + AC + CM > AM + AM

From the figure we have, BC = BM + CM

AB + BC + AC > 2 AM

Hence, the given expression is true.

4. ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD?

Now consider the ΔABC,

Then, consider the ΔBCD

Consider the ΔCDA

Consider the ΔDAB

By adding equations [i], [ii], [iii] and [iv], we get,

AB + BC + BC + CD + CD + DA + DA + AB > CA + DB + AC + DB

2AB + 2BC + 2CD + 2DA > 2CA + 2DB

Take out 2 on both the side,

2(AB + BC + CA + DA) > 2(CA + DB)

AB + BC + CA + DA > CA + DB

5. ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)

Let us consider ABCD as a quadrilateral, and P is the point where the diagonals intersect. As shown in the figure below.

Now consider the ΔPAB,

Then, consider the ΔPBC

Consider the ΔPCD

Consider the ΔPDA

PA + PB + PB + PC + PC + PD + PD + PA < AB + BC + CD + DA

2PA + 2PB + 2PC + 2PD < AB + BC + CD + DA

2PA + 2PC + 2PB + 2PD < AB + BC + CD + DA

2(PA + PC) + 2(PB + PD) < AB + BC + CD + DA

From the figure, we have, AC = PA + PC and BD = PB + PD

2AC + 2BD < AB + BC + CD + DA

2(AC + BD) < AB + BC + CD + DA

6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

From the question, it is given that two sides of the triangle are 12 cm and 15 cm.

So, the third side length should be less than the sum of the other two sides,

12 + 15 = 27 cm

Then, it is given that the third side can not be less than the difference of the two sides, 15 – 12 = 3 cm

So, the length of the third side falls between 3 cm and 27 cm.

Exercise 6.5 Page: 130

1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Let us draw a rough sketch of a right-angled triangle.

By the rule of Pythagoras’ Theorem,

Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of squares on the legs.

In the above figure, RQ is the hypotenuse,

QR 2 = PQ 2 + PR 2

QR 2 = 10 2 + 24 2

QR 2 = 100 + 576

Hence, the length of the hypotenuse QR = 26 cm

2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Let us draw a rough sketch of the right-angled triangle.

AB 2 = AC 2 + BC 2

25 2 = 7 2 + BC 2

625 = 49 + BC 2

By transposing 49 from RHS to LHS, it becomes – 49

BC 2 = 625 – 49

Hence, the length of the BC = 24 cm

3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

15 2 = 12 2 + a 2

225 = 144 + a 2

By transposing 144 from RHS to LHS, it becomes – 144

a 2 = 225 – 144

Hence, the length of a = 9 m

4. Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm

(ii) 2 cm, 2 cm, 5 cm

(iii) 1.5 cm, 2cm, 2.5 cm

In the case of right-angled triangles, identify the right angles.

(i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm

Let us assume the largest value is the hypotenuse side, i.e., b = 6.5 cm.

Then, by Pythagoras’ theorem,

b 2 = a 2 + c 2

6.5 2 = 2.5 2 + 6 2

42.25 = 6.25 + 36

42.25 = 42.25

The sum of squares of two sides of the triangle is equal to the square of the third side,

∴ the given triangle is a right-angled triangle.

The right angle lies on the opposite of the greater side, 6.5 cm.

(ii) Let a = 2 cm, b = 2 cm, c = 5 cm

Let us assume the largest value is the hypotenuse side, i.e. c = 5 cm.

c 2 = a 2 + b 2

5 2 = 2 2 + 2 2

The sum of squares of two sides of the triangle is not equal to the square of the third side,

∴ the given triangle is not a right-angled triangle.

(iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm

Let us assume the largest value is the hypotenuse side, i.e., b = 2.5 cm.

2.5 2 = 1.5 2 + 2 2

6.25 = 2.25 + 4

6.25 = 6.25

The right angle lies on the opposite of the greater side 2.5 cm.

5. A tree is broken at a height of 5 m from the ground, and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Let ABC is the triangle and B be the point where the tree is broken at the height of 5 m from the ground.

Treetop touches the ground at a distance of AC = 12 m from the base of the tree,

By observing the figure, we came to conclude that a right-angle triangle is formed at A.

From the rule of Pythagoras’ theorem,

BC 2 = AB 2 + AC 2

BC 2 = 5 2 + 12 2

BC 2 = 25 + 144

Then, the original height of the tree = AB + BC

6. Angles Q and R of a ΔPQR are 25 o and 65 o .

Write which of the following is true:

(i) PQ 2 + QR 2 = RP 2

(ii) PQ 2 + RP 2 = QR 2

(iii) RP 2 + QR 2 = PQ 2

Given that ∠Q = 25 o , ∠R = 65 o

Then, ∠P =?

We know that sum of the three interior angles of a triangle is equal to 180 o .

∠PQR + ∠QRP + ∠RPQ = 180 o

25 o + 65 o + ∠RPQ = 180 o

90 o + ∠RPQ = 180 o

∠RPQ = 180 – 90

∠RPQ = 90 o

Also, we know that the side opposite to the right angle is the hypotenuse.

∴ QR 2 = PQ 2 + PR 2

Hence, (ii) is true.

7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Let ABCD be the rectangular plot.

Then, AB = 40 cm and AC = 41 cm

According to Pythagoras’ theorem,

From the right angle triangle ABC, we have

= AC 2 = AB 2 + BC 2

= 41 2 = 40 2 + BC 2

= BC 2 = 41 2 – 40 2

= BC 2 = 1681 – 1600

= BC 2 = 81

= BC = 9 cm

Hence, the perimeter of the rectangle plot = 2 (length + breadth)

Where, length = 40 cm, breadth = 9 cm

= 2(40 + 9)

8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Let PQRS be a rhombus, all sides of the rhombus have equal length, and its diagonal PR and SQ are intersecting each other at point O. Diagonals in the rhombus bisect each other at 90 o .

So, PO = (PR/2)

And, SO = (SQ/2)

Then, consider the triangle POS and apply the Pythagoras theorem,

PS 2 = PO 2 + SO 2

PS 2 = 8 2  + 15 2

PS 2 = 64 + 225

Hence, the length of the side of the rhombus is 17 cm

The perimeter of the rhombus = 4 × side of the rhombus

∴ the perimeter of the rhombus is 68 cm.

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