case study questions class 9 maths chapter 7

CBSE Expert

CBSE Class 9 Maths Case Study Questions PDF Download

Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 2023-24. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams.

Case study questions play a pivotal role in enhancing students’ problem-solving skills. By presenting real-life scenarios, these questions encourage students to think beyond textbook formulas and apply mathematical concepts to practical situations. This approach not only strengthens their understanding of mathematical concepts but also develops their analytical thinking abilities.

Table of Contents

CBSE Class 9th MATHS: Chapterwise Case Study Questions

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked. For Class 9 Maths Case Study Questions, there would be 5 case-based sub-part questions, wherein a student has to attempt 4 sub-part questions.

Chapterwise Case Study Questions of Class 9 Maths

• Case Study Questions for Chapter 1 Number System
• Case Study Questions for Chapter 2 Polynomials
• Case Study Questions for Chapter 3 Coordinate Geometry
• Case Study Questions for Chapter 4 Linear Equations in Two Variables
• Case Study Questions for Chapter 5 Introduction to Euclid’s Geometry
• Case Study Questions for Chapter 6 Lines and Angles
• Case Study Questions for Chapter 7 Triangles
• Case Study Questions for Chapter 8 Quadrilaterals
• Case Study Questions for Chapter 9 Areas of Parallelograms and Triangles
• Case Study Questions for Chapter 10 Circles
• Case Study Questions for Chapter 11 Constructions
• Case Study Questions for Chapter 12 Heron’s Formula
• Case Study Questions for Chapter 13 Surface Area and Volumes
• Case Study Questions for Chapter 14 Statistics
• Case Study Questions for Chapter 15 Probability

Checkout: Class 9 Science Case Study Questions

And for mathematical calculations, tap Math Calculators which are freely proposed to make use of by calculator-online.net

The above  Class 9 Maths Case Study Question s will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Study Questions have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

Class 9 Maths Syllabus 2023-24

UNIT I: NUMBER SYSTEMS

1. REAL NUMBERS (18 Periods)

1. Review of representation of natural numbers, integers, and rational numbers on the number line. Rational numbers as recurring/ terminating decimals. Operations on real numbers.

2. Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as √2, √3 and their representation on the number line. Explaining that every real number is represented by a unique point on the number line and conversely, viz. every point on the number line represents a unique real number.

3. Definition of nth root of a real number.

4. Rationalization (with precise meaning) of real numbers of the type

(and their combinations) where x and y are natural number and a and b are integers.

5. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)

UNIT II: ALGEBRA

1. POLYNOMIALS (26 Periods)

Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomial, terms of a polynomial and zero polynomial. Degree of a polynomial. Constant, linear, quadratic and cubic polynomials. Monomials, binomials, trinomials. Factors and multiples. Zeros of a polynomial. Motivate and State the Remainder Theorem with examples. Statement and proof of the Factor Theorem. Factorization of ax2 + bx + c, a ≠ 0 where a, b and c are real numbers, and of cubic polynomials using the Factor Theorem. Recall of algebraic expressions and identities. Verification of identities:

RELATED STORIES

and their use in factorization of polynomials.

2. LINEAR EQUATIONS IN TWO VARIABLES (16 Periods)

Recall of linear equations in one variable. Introduction to the equation in two variables. Focus on linear equations of the type ax + by + c=0.Explain that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers, plotting them and showing that they lie on a line.

UNIT III: COORDINATE GEOMETRY COORDINATE GEOMETRY (7 Periods)

The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations.

UNIT IV: GEOMETRY

1. INTRODUCTION TO EUCLID’S GEOMETRY (7 Periods)

History – Geometry in India and Euclid’s geometry. Euclid’s method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/postulates and theorems. The five postulates of Euclid. Showing the relationship between axiom and theorem, for example: (Axiom)

1. Given two distinct points, there exists one and only one line through them. (Theorem)

2. (Prove) Two distinct lines cannot have more than one point in common.

2. LINES AND ANGLES (15 Periods)

1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180O and the converse.

2. (Prove) If two lines intersect, vertically opposite angles are equal.

3. (Motivate) Lines which are parallel to a given line are parallel.

3. TRIANGLES (22 Periods)

1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).

2. (Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence).

3. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).

4. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. (RHS Congruence)

5. (Prove) The angles opposite to equal sides of a triangle are equal.

6. (Motivate) The sides opposite to equal angles of a triangle are equal.

4. QUADRILATERALS (13 Periods)

1. (Prove) The diagonal divides a parallelogram into two congruent triangles.

2. (Motivate) In a parallelogram opposite sides are equal, and conversely.

3. (Motivate) In a parallelogram opposite angles are equal, and conversely.

4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.

5. (Motivate) In a parallelogram, the diagonals bisect each other and conversely.

6. (Motivate) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it and (motivate) its converse.

5. CIRCLES (17 Periods)

1. (Prove) Equal chords of a circle subtend equal angles at the center and (motivate) its converse.

2. (Motivate) The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.

3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.

4. (Prove) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

5. (Motivate) Angles in the same segment of a circle are equal.

6. (Motivate) If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.

7. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.

UNIT V: MENSURATION 1.

1. AREAS (5 Periods)

Area of a triangle using Heron’s formula (without proof)

2. SURFACE AREAS AND VOLUMES (17 Periods)

Surface areas and volumes of spheres (including hemispheres) and right circular cones.

UNIT VI: STATISTICS & PROBABILITY

STATISTICS (15 Periods)

 Bar graphs, histograms (with varying base lengths), and frequency polygons.

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Benefits of Practicing CBSE Class 9 Maths Case Study Questions

Regular practice of CBSE Class 9 Maths case study questions offers several benefits to students. Some of the key advantages include:

• Deeper Understanding : Case study questions foster a deeper understanding of mathematical concepts by connecting them to real-world scenarios. This improves retention and comprehension.
• Practical Application : Students learn to apply mathematical concepts to practical situations, preparing them for real-life problem-solving beyond the classroom.
• Critical Thinking : Case study questions require students to think critically, analyze data, and devise appropriate solutions. This nurtures their critical thinking abilities, which are valuable in various academic and professional domains.
• Exam Readiness : By practicing case study questions, students become familiar with the question format and gain confidence in their problem-solving abilities. This enhances their readiness for CBSE Class 9 Maths exams.
• Holistic Development: Solving case study questions cultivates not only mathematical skills but also essential life skills like analytical thinking, decision-making, and effective communication.

Tips to Solve CBSE Class 9 Maths Case Study Questions Effectively

Solving case study questions can be challenging, but with the right approach, you can excel. Here are some tips to enhance your problem-solving skills:

• Read the case study thoroughly and understand the problem statement before attempting to solve it.
• Identify the relevant data and extract the necessary information for your solution.
• Break down complex problems into smaller, manageable parts to simplify the solution process.
• Apply the appropriate mathematical concepts and formulas, ensuring a solid understanding of their principles.
• Clearly communicate your solution approach, including the steps followed, calculations made, and reasoning behind your choices.
• Practice regularly to familiarize yourself with different types of case study questions and enhance your problem-solving speed.Class 9 Maths Case Study Questions

Remember, solving case study questions is not just about finding the correct answer but also about demonstrating a logical and systematic approach. Now, let’s explore some resources that can aid your preparation for CBSE Class 9 Maths case study questions.

Q1. Are case study questions included in the Class 9 Maths Case Study Questions syllabus?

Yes, case study questions are an integral part of the CBSE Class 9 Maths syllabus. They are designed to enhance problem-solving skills and encourage the application of mathematical concepts to real-life scenarios.

Q2. How can solving case study questions benefit students ?

Solving case study questions enhances students’ problem-solving skills, analytical thinking, and decision-making abilities. It also bridges the gap between theoretical knowledge and practical application, making mathematics more relevant and engaging.

Q3. How do case study questions help in exam preparation?

Case study questions help in exam preparation by familiarizing students with the question format, improving analytical thinking skills, and developing a systematic approach to problem-solving. Regular practice of case study questions enhances exam readiness and boosts confidence in solving such questions.

Leave a Comment Cancel reply

Save my name, email, and website in this browser for the next time I comment.

Download India's best Exam Preparation App Now.

Key Features

• Revision Notes
• Important Questions
• Previous Years Questions
• Case-Based Questions
• Assertion and Reason Questions

No thanks, I’m not interested!

CBSE Case Study Questions for Class 9 Maths - Pdf PDF Download

CBSE Case Study Questions for Class  9 Maths

CBSE Case Study Questions for Class 9 Maths are a type of assessment where students are given a real-world scenario or situation and they need to apply mathematical concepts to solve the problem. These types of questions help students to develop their problem-solving skills and apply their knowledge of mathematics to real-life situations.

Chapter Wise Case Based Questions for Class 9 Maths

The CBSE Class 9 Case Based Questions can be accessed from Chapetrwise Links provided below:

Chapter-wise case-based questions for Class 9 Maths are a set of questions based on specific chapters or topics covered in the maths textbook. These questions are designed to help students apply their understanding of mathematical concepts to real-world situations and events.

Chapter 1: Number System

• Case Based Questions: Number System

Chapter 2: Polynomial

• Case Based Questions: Polynomial

Chapter 3: Coordinate Geometry

• Case Based Questions: Coordinate Geometry

Chapter 4: Linear Equations

• Case Based Questions: Linear Equations - 1
• Case Based Questions: Linear Equations -2

Chapter 5: Introduction to Euclid’s Geometry

• Case Based Questions: Lines and Angles

Chapter 7: Triangles

• Case Based Questions: Triangles

Chapter 8: Quadrilaterals

• Case Based Questions: Quadrilaterals - 1
• Case Based Questions: Quadrilaterals - 2

Chapter 9: Areas of Parallelograms

• Case Based Questions: Circles

Chapter 11: Constructions

• Case Based Questions: Constructions

Chapter 12: Heron’s Formula

• Case Based Questions: Heron’s Formula

Chapter 13: Surface Areas and Volumes

• Case Based Questions: Surface Areas and Volumes

Chapter 14: Statistics

• Case Based Questions: Statistics

Chapter 15: Probability

• Case Based Questions: Probability

Weightage of Case Based Questions in Class 9 Maths

Why are Case Study Questions important in Maths Class  9?

• Enhance critical thinking:  Case study questions require students to analyze a real-life scenario and think critically to identify the problem and come up with possible solutions. This enhances their critical thinking and problem-solving skills.
• Apply theoretical concepts:  Case study questions allow students to apply theoretical concepts that they have learned in the classroom to real-life situations. This helps them to understand the practical application of the concepts and reinforces their learning.
• Develop decision-making skills:  Case study questions challenge students to make decisions based on the information provided in the scenario. This helps them to develop their decision-making skills and learn how to make informed decisions.
• Improve communication skills:  Case study questions often require students to present their findings and recommendations in written or oral form. This helps them to improve their communication skills and learn how to present their ideas effectively.
• Enhance teamwork skills:  Case study questions can also be done in groups, which helps students to develop teamwork skills and learn how to work collaboratively to solve problems.

In summary, case study questions are important in Class 9 because they enhance critical thinking, apply theoretical concepts, develop decision-making skills, improve communication skills, and enhance teamwork skills. They provide a practical and engaging way for students to learn and apply their knowledge and skills to real-life situations.

Class 9 Maths Curriculum at Glance

The Class 9 Maths curriculum in India covers a wide range of topics and concepts. Here is a brief overview of the Maths curriculum at a glance:

• Number Systems:  Students learn about the real number system, irrational numbers, rational numbers, decimal representation of rational numbers, and their properties.
• Algebra:  The Algebra section includes topics such as polynomials, linear equations in two variables, quadratic equations, and their solutions.
• Coordinate Geometry:  Students learn about the coordinate plane, distance formula, section formula, and slope of a line.
• Geometry:  This section includes topics such as Euclid’s geometry, lines and angles, triangles, and circles.
• Trigonometry: Students learn about trigonometric ratios, trigonometric identities, and their applications.
• Mensuration: This section includes topics such as area, volume, surface area, and their applications.
• Statistics and Probability:  Students learn about measures of central tendency, graphical representation of data, and probability.

The Class 9 Maths curriculum is designed to provide a strong foundation in mathematics and prepare students for higher education in the field. The curriculum is structured to develop critical thinking, problem-solving, and analytical skills, and to promote the application of mathematical concepts in real-life situations. The curriculum is also designed to help students prepare for competitive exams and develop a strong mathematical base for future academic and professional pursuits.

Students can also access Case Based Questions of all subjects of CBSE Class 9

• Case Based Questions for Class 9 Science
• Case Based Questions for Class 9 Social Science
• Case Based Questions for Class 9 English
• Case Based Questions for Class 9 Hindi
• Case Based Questions for Class 9 Sanskrit

Frequently Asked Questions (FAQs) on Case Based Questions for Class 9 Maths

What is case-based questions.

Case-Based Questions (CBQs) are open-ended problem solving tasks that require students to draw upon their knowledge of Maths concepts and processes to solve a novel problem. CBQs are often used as formative or summative assessments, as they can provide insights into how students reason through and apply mathematical principles in real-world problems.

What are case-based questions in Maths?

Case-based questions in Maths are problem-solving tasks that require students to apply their mathematical knowledge and skills to real-world situations or scenarios.

What are some common types of case-based questions in class 9 Maths?

Common types of case-based questions in class 9 Maths include word problems, real-world scenarios, and mathematical modeling tasks.

Top Courses for Class 9

FAQs on CBSE Case Study Questions for Class 9 Maths - Pdf

Objective type Questions

Previous year questions with solutions, past year papers, important questions, extra questions, practice quizzes, cbse case study questions for class 9 maths - pdf, shortcuts and tricks, study material, sample paper, semester notes, mock tests for examination, viva questions, video lectures.

CBSE Case Study Questions for Class 9 Maths - Pdf Free PDF Download

Importance of cbse case study questions for class 9 maths - pdf, cbse case study questions for class 9 maths - pdf notes, cbse case study questions for class 9 maths - pdf class 9, study cbse case study questions for class 9 maths - pdf on the app.

Welcome Back

Create your account for free.

Forgot Password

Unattempted tests, change country, practice & revise.

• Andhra Pradesh
• Chhattisgarh
• West Bengal
• Madhya Pradesh
• Maharashtra
• Jammu & Kashmir
• NCERT Books 2022-23
• NCERT Solutions
• NCERT Notes
• NCERT Exemplar Books
• NCERT Exemplar Solution
• States UT Book
• School Kits & Lab Manual
• NCERT Books 2021-22
• NCERT Books 2020-21
• NCERT Book 2019-2020
• NCERT Book 2015-2016
• RD Sharma Solution
• TS Grewal Solution
• TR Jain Solution
• Selina Solution
• Frank Solution
• Lakhmir Singh and Manjit Kaur Solution
• I.E.Irodov solutions
• ICSE - Goyal Brothers Park
• ICSE - Dorothy M. Noronhe
• Micheal Vaz Solution
• S.S. Krotov Solution
• Evergreen Science
• KC Sinha Solution
• ICSE - ISC Jayanti Sengupta, Oxford
• ICSE Focus on History
• ICSE GeoGraphy Voyage
• ICSE Hindi Solution
• ICSE Treasure Trove Solution
• Thomas & Finney Solution
• SL Loney Solution
• SB Mathur Solution
• P Bahadur Solution
• Narendra Awasthi Solution
• MS Chauhan Solution
• LA Sena Solution
• Integral Calculus Amit Agarwal Solution
• IA Maron Solution
• Hall & Knight Solution
• Errorless Solution
• Pradeep's KL Gogia Solution
• OP Tandon Solutions
• Sample Papers
• Previous Year Question Paper
• Important Question
• Value Based Questions
• CBSE Syllabus
• CBSE MCQs PDF
• Assertion & Reason
• New Revision Notes
• Revision Notes
• Question Bank
• Marks Wise Question
• Toppers Answer Sheets
• Exam Paper Aalysis
• Concept Map
• CBSE Text Book
• Additional Practice Questions
• Vocational Book
• CBSE - Concept
• KVS NCERT CBSE Worksheets
• Formula Class Wise
• Formula Chapter Wise
• JEE Previous Year Paper
• JEE Mock Test
• JEE Crash Course
• JEE Sample Papers
• Important Info
• SRM-JEEE Previous Year Paper
• SRM-JEEE Mock Test
• VITEEE Previous Year Paper
• VITEEE Mock Test
• BITSAT Previous Year Paper
• BITSAT Mock Test
• Manipal Previous Year Paper
• Manipal Engineering Mock Test
• AP EAMCET Previous Year Paper
• AP EAMCET Mock Test
• COMEDK Previous Year Paper
• COMEDK Mock Test
• GUJCET Previous Year Paper
• GUJCET Mock Test
• KCET Previous Year Paper
• KCET Mock Test
• KEAM Previous Year Paper
• KEAM Mock Test
• MHT CET Previous Year Paper
• MHT CET Mock Test
• TS EAMCET Previous Year Paper
• TS EAMCET Mock Test
• WBJEE Previous Year Paper
• WBJEE Mock Test
• AMU Previous Year Paper
• AMU Mock Test
• CUSAT Previous Year Paper
• CUSAT Mock Test
• AEEE Previous Year Paper
• AEEE Mock Test
• UPSEE Previous Year Paper
• UPSEE Mock Test
• CGPET Previous Year Paper
• Crash Course
• Previous Year Paper
• NCERT Based Short Notes
• NCERT Based Tests
• NEET Sample Paper
• Previous Year Papers
• Quantitative Aptitude
• Numerical Aptitude Data Interpretation
• General Knowledge
• Mathematics
• Agriculture
• Accountancy
• Business Studies
• Political science
• Enviromental Studies
• Mass Media Communication
• Teaching Aptitude
• Verbal Ability & Reading Comprehension
• Logical Reasoning & Data Interpretation
• CAT Mock Test
• CAT Important Question
• CAT Vocabulary
• CAT English Grammar
• MBA General Knowledge
• CAT Mind Map
• CAT Study Planner
• CMAT Mock Test
• SRCC GBO Mock Test
• SRCC GBO PYQs
• XAT Mock Test
• SNAP Mock Test
• IIFT Mock Test
• MAT Mock Test
• CUET PG Mock Test
• CUET PG PYQs
• MAH CET Mock Test
• MAH CET PYQs
• NAVODAYA VIDYALAYA
• SAINIK SCHOOL (AISSEE)
• Mechanical Engineering
• Electrical Engineering
• Electronics & Communication Engineering
• Civil Engineering
• Computer Science Engineering
• CBSE Board News
• Scholarship Olympiad
• School Admissions
• Entrance Exams
• All Board Updates
• Miscellaneous
• State Wise Books
• Engineering Exam

CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF

SHARING IS CARING If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

CBSE Class 9 Maths exam 2022-23 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions.

Each question has five sub-questions, each followed by four options and one correct answer. Students can easily download these questions in PDF format and refer to them for exam preparation.

CBSE Class 9 All Students can also Download here Class 9 Other Study Materials in PDF Format.

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 10 Maths
• CBSE Syllabus 2023-24
• Social Media Channels
• Login Customize Your Notification Preferences
• CBSE Class 9th 2023-24 : Science Practical Syllabus; Download PDF 19 April, 2023, 4:52 pm
• CBSE Class 9 Maths Practice Book 2023 (Released By CBSE) 23 March, 2023, 6:16 pm
• CBSE Class 9 Science Practice Book 2023 (Released By CBSE) 23 March, 2023, 5:56 pm
• CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF 10 February, 2023, 6:20 pm
• CBSE Class 9th Maths 2023 : Important Assertion Reason Question with Solution Download Pdf 9 February, 2023, 12:16 pm
• CBSE Class 9th Exam 2023 : Social Science Most Important Short Notes; Download PDF 16 January, 2023, 4:29 pm
• CBSE Class 9th Mathematics 2023 : Most Important Short Notes with Solutions 27 December, 2022, 6:05 pm
• CBSE Class 9th English 2023 : Chapter-wise Competency-Based Test Items with Answer; Download PDF 21 December, 2022, 5:16 pm
• CBSE Class 9th Science 2023 : Chapter-wise Competency-Based Test Items with Answers; Download PDF 20 December, 2022, 5:37 pm

CBSE Class 9th 2023-24 : Science Practical Syllabus; Download PDF

• Second click on the toggle icon

Provide prime members with unlimited access to all study materials in PDF format.

Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding.

Provide prime users with access to exclusive PDF study materials that are not available to regular users.

myCBSEguide

• Mathematics
• CBSE Class 9 Mathematics...

CBSE Class 9 Mathematics Case Study Questions

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you’re looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter. With over 10,000 study notes, solved sample papers and practice questions, it’s got everything you need to ace your exams. Plus, it’s updated regularly to keep you aligned with the latest CBSE syllabus . So why wait? Start your journey to success with myCBSEguide today!

Significance of Mathematics in Class 9

Mathematics is an important subject for students of all ages. It helps students to develop problem-solving and critical-thinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.

CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .

Case studies in Class 9 Mathematics

A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.

Example of Case study questions in Class 9 Mathematics

The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on real-life scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.

The following are some examples of case study questions from Class 9 Mathematics:

Class 9 Mathematics Case study question 1

There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of   XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

• Now he told Raju to draw another line CD as in the figure
• The teacher told Ajay to mark  ∠ AOD  as 2z
• Suraj was told to mark  ∠ AOC as 4y
• Clive Made and angle  ∠ COE = 60°
• Peter marked  ∠ BOE and  ∠ BOD as y and x respectively

Now answer the following questions:

• 2y + z = 90°
• 2y + z = 180°
• 4y + 2z = 120°
• (a) 2y + z = 90°

Class 9 Mathematics Case study question 3

• (a) 31.6 m²
• (c) 513.3 m³
• (b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.

Class 9 Mathematics Curriculum at Glance

At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve real-world problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.

The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.

CBSE Class 9 Mathematics (Code No. 041)

Class 9 Mathematics question paper design

The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problem-solving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.

QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)

myCBSEguide: Blessing in disguise

Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes

Related Posts

• Competency Based Learning in CBSE Schools
• Class 11 Physical Education Case Study Questions
• Class 11 Sociology Case Study Questions
• Class 12 Applied Mathematics Case Study Questions
• Class 11 Applied Mathematics Case Study Questions
• Class 11 Mathematics Case Study Questions
• Class 11 Biology Case Study Questions
• Class 12 Physical Education Case Study Questions

16 thoughts on “CBSE Class 9 Mathematics Case Study Questions”

This method is not easy for me

aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10-?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2-?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3-?2 4+ ?2 2+ ?2 d) 2-?3 the identity applied to solve (?10-?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (a-b)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b

MATHS PAAGAL HAI

All questions was easy but search ? hard questions. These questions was not comparable with cbse. It was totally wastage of time.

Where is search ? bar

maths is love

Can I have more questions without downloading the app.

I love math

Hello l am Devanshu chahal and l am an entorpinior. I am started my card bord business and remanded all the existing things this all possible by math now my business is 120 crore and my business profit is 25 crore in a month. l find the worker team because my business is going well Thanks

I am Riddhi Shrivastava… These questions was very good.. That’s it.. ..

For challenging Mathematics Case Study Questions, seeking a writing elite service can significantly aid your research. These services provide expert guidance, ensuring your case study is well-researched, accurately analyzed, and professionally written. With their assistance, you can tackle complex mathematical problems with confidence, leading to high-quality academic work that meets rigorous standards.

Leave a Comment

Save my name, email, and website in this browser for the next time I comment.

(Education-Your Door To The Future)

CBSE Class 9 Maths Most Important Case Study Based Questions With Solution

Cbse class 9 mathematics case study questions.

In this post I have provided CBSE Class 9 Maths Case Study Based Questions With Solution. These questions are very important for those students who are preparing for their final class 9 maths exam.

All these questions provided in this article are with solution which will help students for solving the problems. Dear students need to practice all these questions carefully with the help of given solutions.

As you know CBSE Class 9 Maths exam will have a set of cased study based questions in the form of MCQs. CBSE Class 9 Maths Question Bank given in this article can be very helpful in understanding the new format of questions for new session.

All Of You Can Also Read

Case studies in class 9 mathematics.

The Central Board of Secondary Education (CBSE) has included case study based questions in the Class 9 Mathematics paper in current session. According to new pattern CBSE Class 9 Mathematics students will have to solve case based questions. This is a departure from the usual theoretical conceptual questions that are asked in Class 9 Maths exam in this year.

Each question provided in this post has five sub-questions, each followed by four options and one correct answer. All CBSE Class 9th Maths Students can easily download these questions in PDF form with the help of given download Links and refer for exam preparation.

There is many more free study materials are available at Maths And Physics With Pandey Sir website. For many more books and free study material all of you can visit at this website.

Given Below Are CBSE Class 9th Maths Case Based Questions With Their Respective Download Links.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT Solutions for Class 9 Maths Chapter 7 Triangles are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 7 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

Class 9 Maths Chapter 7 Triangles NCERT Solutions

Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.3

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5

NCERT Solutions for Class 9 Maths Chapter 7 – Topic Discussion

Below we have listed the topics that have been discussed in this chapter.

• Congruence of triangles.
• Criteria for congruence of triangles (SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule).
• Properties of triangles.
• Inequalities of triangle.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Save my name, email, and website in this browser for the next time I comment.

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

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Ex 7.2 Class 9 Maths  

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 are part of NCERT Solutions for Class 9 Maths . Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

(i) In ∆ ABC and ∆ BAC, AD = BC (Given) ∠DAB = ∠CBA (Given) AB = AB (Common) ∴ ∆ ABD ≅ ∆BAC (By SAS congruence)

(ii) Since ∆ABD ≅ ∆BAC ⇒ BD = AC [By C.P.C.T.]

(iii) Since ∆ABD ≅ ∆BAC ⇒ ∠ABD = ∠BAC [By C.P.C.T.]

(i) Now, in ∆DAP and ∆EBP, we have ∠PAD = ∠PBE [ ∵∠BAD = ∠ABE] AP = BP [Proved above] ∠DPA = ∠EPB [Proved above] ∴ ∆DAP ≅ ∆EBP [By ASA congruency]

(ii) Since, ∆ DAP ≅ ∆ EBP ⇒ AD = BE [By C.P.C.T.]

(i) In ∆AMC and ∆BMD, we have CM = DM [Given] ∠AMC = ∠BMD [Vertically opposite angles] AM = BM [Proved above] ∴ ∆AMC ≅ ∆BMD [By SAS congruency]

(ii) Since ∆AMC ≅ ∆BMD ⇒ ∠MAC = ∠MBD [By C.P.C.T.] But they form a pair of alternate interior angles. ∴ AC || DB Now, BC is a transversal which intersects parallel lines AC and DB, ∴ ∠BCA + ∠DBC = 180° [Co-interior angles] But ∠BCA = 90° [∆ABC is right angled at C] ∴ 90° + ∠DBC = 180° ⇒ ∠DBC = 90°

(iii) Again, ∆AMC ≅ ∆BMD [Proved above] ∴ AC = BD [By C.P.C.T.] Now, in ∆DBC and ∆ACB, we have BD = CA [Proved above] ∠DBC = ∠ACB [Each 90°] BC = CB [Common] ∴ ∆DBC ≅ ∆ACB [By SAS congruency]

(iv) As ∆DBC ≅ ∆ACB DC = AB [By C.P.C.T.] But DM = CM [Given] ∴ CM = \(\frac { 1 }{ 2 }\)DC = \(\frac { 1 }{ 2 }\)AB ⇒ CM = \(\frac { 1 }{ 2 }\)AB

NCERT Solutions for Class 9 Maths Chapter 7 Triangles (त्रिभुज) (Hindi Medium) Ex 7.1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

(ii) In ∆ABO and ∆ACO, we have AB = AC [Given] ∠OBA = ∠OCA [ ∵\(\frac { 1 }{ 2 }\)∠B = \(\frac { 1 }{ 2 }\)∠C] OB = OC [Proved above] ∆ABO ≅ ∆ACO [By SAS congruency] ⇒ ∠OAB = ∠OAC [By C.P.C.T.] ⇒ AO bisects ∠A.

(ii) Since, ∆ABE ≅ ∆ACF ∴ AB = AC [By C.P.C.T.] ⇒ ABC is an isosceles triangle.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

(ii) In ∆ABP and ∆ACP, we have AB = AC [Given] ∠BAP = ∠CAP [From (1)] ∴ AP = PA [Common] ∴ ∆ABP ≅ ∆ACP [By SAS congruency]

(iii) Since, ∆ABP ≅ ∆ACP ⇒ ∠BAP = ∠CAP [By C.P.C.T.] ∴ AP is the bisector of ∠A. Again, in ∆BDP and ∆CDP, we have BD = CD [Given] DP = PD [Common] BP = CP [ ∵ ∆ABP ≅ ∆ACP] ⇒ A BDP = ACDP [By SSS congruency] ∴ ∠BDP = ∠CDP [By C.P.C.T.] ⇒ DP (or AP) is the bisector of ∠BDC ∴ AP is the bisector of ∠A as well as ∠D.

(iv) As, ∆ABP ≅ ∆ACP ⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.] But ∠APB + ∠APC = 180° [Linear Pair] ∴ ∠APB = ∠APC = 90° ⇒ AP ⊥ BC, also BP = CP Hence, AP is the perpendicular bisector of BC.

(ii) Since, ∆ABD ≅ ∆ACD, ⇒ ∠BAD = ∠CAD [By C.P.C.T.] So, AD bisects ∠A

(i) In ∆ABM and ∆PQN, we have AB = PQ , [Given] AM = PN [Given] BM = QN [From (3)] ∴ ∆ABM ≅ ∆PQN [By SSS congruency]

(ii) Since ∆ABM ≅ ∆PQN ⇒ ∠B = ∠Q …(4) [By C.P.C.T.] Now, in ∆ABC and ∆PQR, we have ∠B = ∠Q [From (4)] AB = PQ [Given] BC = QR [Given] ∴ ∆ABC ≅ ∆PQR [By SAS congruency]

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

NCERT Solutions for Class 9 Maths All Chapters

• Chapter 1 Number systems
• Chapter 2 Polynomials
• Chapter 3 Coordinate Geometry
• Chapter 4 Linear Equations in Two Variables
• Chapter 5 Introduction to Euclid Geometry
• Chapter 6 Lines and Angles
• Chapter 7 Triangles
• Chapter 8 Quadrilaterals
• Chapter 9 Areas of Parallelograms and Triangles
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Heron’s Formula
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability
• Class 9 Maths (Download PDF)

We hope the NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1, drop a comment below and we will get back to you at the earliest.

Free Resources

NCERT Solutions

Quick Resources

• Class 6 Maths
• Class 6 Science
• Class 6 Social Science
• Class 6 English
• Class 7 Maths
• Class 7 Science
• Class 7 Social Science
• Class 7 English
• Class 8 Maths
• Class 8 Science
• Class 8 Social Science
• Class 8 English
• Class 9 Maths
• Class 9 Science
• Class 9 Social Science
• Class 9 English
• Class 10 Maths
• Class 10 Science
• Class 10 Social Science
• Class 10 English
• Class 11 Maths
• Class 11 Computer Science (Python)
• Class 11 English
• Class 12 Maths
• Class 12 English
• Class 12 Economics
• Class 12 Accountancy
• Class 12 Physics
• Class 12 Chemistry
• Class 12 Biology
• Class 12 Computer Science (Python)
• Class 12 Physical Education
• GST and Accounting Course
• Excel Course
• Tally Course
• Finance and CMA Data Course
• Payroll Course

Interesting

• Learn English
• Learn Excel
• Learn Tally
• Learn GST (Goods and Services Tax)
• Learn Accounting and Finance
• GST Tax Invoice Format
• Accounts Tax Practical
• Tally Ledger List
• GSTR 2A - JSON to Excel

Are you in school ? Do you love Teachoo?

We would love to talk to you! Please fill this form so that we can contact you

You are learning...

Chapter 7 Class 9 Triangles

Click on any of the links below to start learning from Teachoo ...

Updated for new NCERT Books (for 2023-24 Exams)

NCERT Solutions of Chapter 7 Class 9 Triangles is available free at teachoo. Solutions to all exercise questions, examples and theorems is provided with video of each and every question.

Let's see what we will learn in this chapter. The topics in the chapter are - 

• What is congruency of figures
• Naming of triangles when two triangles are congruent
• What is CPCT - Corresponding Parts of Congruent Triangles
• SAS (Side Angle Side) Congruence Rule (this is without proof)
• ASA (Angle Side Angle) congruence rule with proof (Theorem 7.1)
• Is ASA and AAS congruency the same ?
• Angles opposite to equal sides is equal (Isosceles Triangle Property)
• SSS (Side Side Side) congruence rule with proof (Theorem 7.4)
• RHS (Right angle Hypotenuse Side) congruence rule with proof (Theorem 7.5)
• Angle opposite to longer side is larger, and Side opposite to larger angle is longer
• Triangle Inequality - Sum of two sides of a triangle is always greater than the third side

Click on an exercise link below to study from the NCERT Book.

Or click on a Concept link and study the Teachoo way.

Check it out now!

Serial order wise

Concept wise.

What's in it?

Hi, it looks like you're using AdBlock :(

Please login to view more pages. it's free :), solve all your doubts with teachoo black.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Ncert solutions for class 9 maths chapter 7 triangles| pdf download.

Page No: 118

• Exercise 7.1 Chapter 7 Class 9 Maths NCERT Solutions
• Exercise 7.2 Chapter 7 Class 9 Maths NCERT Solutions
• Exercise 7.3 Chapter 7 Class 9 Maths NCERT Solutions
• Exercise 7.4 Chapter 7 Class 9 Maths NCERT Solutions
• Exercise 7.5 Chapter 7 Class 9 Maths NCERT Solutions

NCERT Solutions for Class 9 Maths Chapters:

How many exercises in Chapter 7 Triangles?

Each of the equal angles of an isosceles triangle is 38°, what is the measure of the third angle, find the measure of each of acute angle in a right angle isosceles triangle., if two angles are (30 ∠ a)º and (125 + 2a)º and they are supplement of each other. find the value of ‘a’., contact form.

Class 9 Maths Case Study Questions of Chapter 2 Polynomials PDF Download

• Post author: studyrate
• Post published:
• Post category: class 9th
• Post comments: 2 Comments

Case study Questions in Class 9 Mathematics Chapter 2  are very important to solve for your exam. Class 9 Maths Chapter 2 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  Class 9 Maths Case Study Questions  Chapter 2 Polynomials

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

These case study questions challenge students to apply their knowledge of polynomials in real-life scenarios, enhancing their problem-solving abilities. This article provides for the Class 9 Maths Case Study Questions of Chapter 2: Polynomials , enabling students to practice and excel in their examinations.

Polynomials Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 2 Polynomials

Case Study/Passage Based Questions

Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p(x) = 4x 2 + 12x + 5, which is the product of their individual shares.

Coefficient of x 2 in the given polynomial is (a) 2 (b) 3 (c) 4 (d) 12

Answer: (c) 4

Total amount invested by both, if x = 1000 is (a) 301506 (b)370561 (c) 4012005 (d)490621

Answer: (c) 4012005

The shares of Ankur and Ranjan invested individually are (a) (2x + 1),(2x + 5)(b) (2x + 3),(x + 1) (c) (x + 1),(x + 3) (d) None of these

Answer: (a) (2x + 1),(2x + 5)

Name the polynomial of amounts invested by each partner. (a) Cubic (b) Quadratic (c) Linear (d) None of these

Answer: (c) Linear

Find the value of x, if the total amount invested is equal to 0. (a) –1/2 (b) –5/2 (c) Both (a) and (b) (d) None of these

Answer: (c) Both (a) and (b)

One day, the principal of a particular school visited the classroom. The class teacher was teaching the concept of a polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked various questions to students. Some of them are given below. Answer them

Which one of the following is not a polynomial? (a) 4x 2 + 2x – 1 (b) y+3/y (c) x 3 – 1 (d) y 2 + 5y + 1

Answer: (b) y+3/y

The polynomial of the type ax 2 + bx + c, a = 0 is called (a) Linear polynomial (b) Quadratic polynomial (c) Cubic polynomial (d) Biquadratic polynomial

Answer: (a) Linear polynomial

The value of k, if (x – 1) is a factor of 4x 3 + 3x 2 – 4x + k, is (a) 1 (b) –2 (c) –3 (d) 3

Answer: (c) –3

If x + 2 is the factor of x 3 – 2ax 2 + 16, then value of a is (a) –7 (b) 1 (c) –1 (d) 7

Answer: (b) 1

The number of zeroes of the polynomial x 2 + 4x + 2 is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (b) 2

Case Study/Passage-Based Questions

Case Study 3. Amit and Rahul are friends who love collecting stamps. They decide to start a stamp collection club and contribute funds to purchase new stamps. They both invest a certain amount of money in the club. Let’s represent Amit’s investment by the polynomial A(x) = 3x^2 + 2x + 1 and Rahul’s investment by the polynomial R(x) = 2x^2 – 5x + 3. The sum of their investments is represented by the polynomial S(x), which is the sum of A(x) and R(x).

Q1. What is the coefficient of x^2 in Amit’s investment polynomial A(x)? (a) 3 (b) 2 (c) 1 (d) 0

Answer: (a) 3

Q2. What is the constant term in Rahul’s investment polynomial R(x)? (a) 2 (b) -5 (c) 3 (d) 0

Answer: (c) 6

Q3. What is the degree of the polynomial S(x), representing the sum of their investments? (a) 4 (b) 3 (c) 2 (d) 1

Answer: (c) 2

Q4. What is the coefficient of x in the polynomial S(x)? (a) 7 (b) -3 (c) 0 (d) 5

Answer: (b) -3

Q5. What is the sum of their investments, represented by the polynomial S(x)? (a) 5x^2 + 7x + 4 (b) 5x^2 – 3x + 4 (c) 5x^2 – 3x + 5 (d) 5x^2 + 7x + 5

Answer: (b) 5x^2 – 3x + 4

Case Study 4. A school is organizing a fundraising event to support a local charity. The students are divided into three groups: Group A, Group B, and Group C. Each group is responsible for collecting donations from different areas of the town.

Group A consists of 30 students and each student is expected to collect ‘x’ amount of money. The polynomial representing the total amount collected by Group A is given as A(x) = 2x^2 + 5x + 10.

Group B consists of 20 students and each student is expected to collect ‘y’ amount of money. The polynomial representing the total amount collected by Group B is given as B(y) = 3y^2 – 4y + 7.

Group C consists of 40 students and each student is expected to collect ‘z’ amount of money. The polynomial representing the total amount collected by Group C is given as C(z) = 4z^2 + 3z – 2.

Q1. What is the coefficient of x in the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 0

Answer: (b) 5

Q2. What is the degree of the polynomial B(y)? (a) 2 (b) 3 (c) 4 (d) 1

Answer: (b) 3

Q3. What is the constant term in the polynomial C(z)? (a) 4 (b) 3 (c) -2 (d) 0

Answer: (c) -2

Q4. What is the sum of the coefficients of the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 17

Answer: (c) 10

Q5. What is the total number of students in all three groups combined? (a) 30 (b) 20 (c) 40 (d) 90

Answer: (c) 40

The Class 9 Maths Case Study Questions of Chapter 2: Polynomials serve as a valuable resource for students seeking to enhance their understanding of polynomial concepts and problem-solving skills. By practicing these case studies, students can strengthen their grasp of polynomials and their applications in real-life scenarios. Embrace the opportunity to engage with practical problems and excel in your mathematical journey.

By Team Study Rate

You Might Also Like

Mcq questions of class 9 social science history chapter 2 socialism in europe and the russian revolution with answers, class 9 mcq questions for chapter 7 diversity in living organisms with answers, mcq questions of class 9 maths chapter 14 statistics with answers, this post has 2 comments.

I didnt understand last question

The number of zeroes of the polynomial x2 + 4x + 2 is The answer is too easy, i.e. 2

Leave a Reply Cancel reply

Save my name, email, and website in this browser for the next time I comment.

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Category: Case Study Questions for Class 9 Maths

Join our Telegram Channel for Free PDF Download

Case Study Questions for Class 9 Maths Chapter 12 Herons Formula

Case study questions for class 9 maths chapter 9 areas of parallelograms and triangles, case study questions for class 9 maths chapter 6 lines and angles, case study questions for class 9 maths chapter 7 triangles, case study questions for class 9 maths chapter 5 introduction to euclid’s geometry, case study and passage based questions for class 9 maths chapter 14 statistics, case study questions for class 9 maths chapter 1 real numbers, case study questions for class 9 maths chapter 4 linear equations in two variables, case study questions for class 9 maths chapter 3 coordinate geometry, case study questions for class 9 maths chapter 15 probability, case study questions for class 9 maths chapter 13 surface area and volume, case study questions for class 9 maths chapter 10 circles, case study questions for class 9 maths chapter 9 quadrilaterals, case study questions for class 9 maths chapter 2 polynomials.

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Exercise 7.2 of Class 9 Maths Chapter 7 consists of properties of Triangles and Theorems related to them. Most students regard this topic as boring as it is more on proving and less on solving. But, if students understand the properties of Triangles by correlating them to the examples, they will start taking an interest in them. Here, we have provided the NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2 for students’ convenience. After going through the solutions, students can easily understand the method of solving the questions.

NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.2

carouselExampleControls112

Previous Next

Access other Exercise Solutions of Class 9 Maths Chapter 7 – Triangles

Chapter 7 consists of a total of 5 exercises. To get the solutions of other exercises, visit the link below.

Exercise 7.1 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Questions)

Exercise 7.3 Solution 5 Questions (3 Short Answer Questions, 2 Long Answer Questions)

Exercise 7.4 Solution 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Exercise 7.5 (Optional) Solution 4 Questions

Access Answers to Chapter 7 – Triangles Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

(i) OB = OC (ii) AO bisects ∠A

AB = AC and

the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

½ ∠B = ½ ∠C

⇒ ∠OBC = ∠OCB (Angle bisectors)

∴ OB = OC (Side opposite to the equal angles are equal)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, ΔAOB ≅ ΔAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects ∠A.

2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.

It is given that AD is the perpendicular bisector of BC

In ΔADB and ΔADC,

AD = AD (It is the Common arm)

∠ADB = ∠ADC

BD = CD (Since AD is the perpendicular bisector)

So, ΔADB ≅ ΔADC by SAS congruency criterion .

AB = AC (by CPCT)

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB, respectively (see Fig. 7.31). Show that these altitudes are equal.

(i) BE and CF are altitudes.

(ii) AC = AB

Triangles ΔAEB and ΔAFC are similar by AAS congruency since

∠A = ∠A (It is the common arm)

∠AEB = ∠AFC (They are right angles)

∴ ΔAEB ≅ ΔAFC and so, BE = CF (by CPCT).

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that

(i) ΔABE ≅ ΔACF

(ii) AB = AC, i.e. ABC is an isosceles triangle.

It is given that BE = CF

(i) In ΔABE and ΔACF,

∠A = ∠A (It is the common angle)

BE = CF (Given in the question)

∴ ΔABE ≅ ΔACF by AAS congruency condition .

(ii) AB = AC by CPCT and so, ABC is an isosceles triangle.

5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

In the question, it is given that ABC and DBC are two isosceles triangles.

We will have to show that ∠ABD = ∠ACD

Triangles ΔABD and ΔACD are similar by SSS congruency since

AD = AD (It is the common arm)

AB = AC (Since ABC is an isosceles triangle)

BD = CD (Since BCD is an isosceles triangle)

So, ΔABD ≅ ΔACD.

∴ ∠ABD = ∠ACD by CPCT.

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

It is given that AB = AC and AD = AB

We will have to now prove ∠BCD is a right angle.

Consider ΔABC,

AB = AC (It is given in the question)

Also, ∠ACB = ∠ABC (They are angles opposite to the equal sides, and so they are equal)

Now, consider ΔACD,

Also, ∠ADC = ∠ACD (They are angles opposite to the equal sides, and so they are equal)

∠CAB + ∠ACB + ∠ABC = 180°

So, ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — (i)

Similarly, in ΔADC,

∠CAD = 180° – 2∠ACD — (ii)

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii), we get,

∠CAB + ∠CAD = 180° – 2∠ACB+180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB-2∠ACD

⇒ 2(∠ACB+∠ACD) = 180°

⇒ ∠BCD = 90°

7. ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

In the question, it is given that

∠A = 90° and AB = AC

⇒ ∠B = ∠C (They are angles opposite to the equal sides, and so they are equal)

∠A+∠B+∠C = 180° (Since the sum of the interior angles of the triangle)

∴ 90° + 2∠B = 180°

⇒ 2∠B = 90°

So, ∠B = ∠C = 45°

8. Show that the angles of an equilateral triangle are 60° each.

Let ABC be an equilateral triangle, as shown below:

Here, BC = AC = AB (Since the length of all sides is the same)

⇒ ∠A = ∠B =∠C (Sides opposite to the equal angles are equal)

Also, we know that

∠A+∠B+∠C = 180°

⇒ 3∠A = 180°

∴ ∠A = ∠B = ∠C = 60°

So, the angles of an equilateral triangle are always 60° each.

• Theorem 7.2: Angles opposite to equal sides of an isosceles triangle are equal.
• Theorem 7.3: The sides opposite to equal angles of a triangle are equal.

These theorems can be proved by the congruence rule that students have studied in the previous exercise, i.e., 7.1. So, go through the proofs in depth to understand them. Also, the questions provided in the exercise are based on these theorems. Students should try to solve the questions by themselves. If they get stuck somewhere, they can then refer to the solutions. The NCERT solutions for Class 9 Maths Chapter 7 are provided to help students in their studies so that they get rid of all their doubts.

We hope this information on “NCERT Solution for Class 9 Maths Chapter 7 Triangles Exercise 7.2” is useful for students. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos, download the BYJU’S App and subscribe to YouTube Channel.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

It’s good platform

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Talk to our experts

1800-120-456-456

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.

• Simple Equations Class 7 Case Study Questions Maths Chapter 4

Last Updated on August 18, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 7 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 7 maths. In this article, you will find case study questions for CBSE Class 7 Maths Chapter 4 Simple Equations. It is a part of Case Study Questions for CBSE Class 7 Maths Series.

Table of Contents

Case Study Questions on Simple Equations

The teacher tells the class that the lowest marks obtained by a student in his class is half the highest marks plus 5. The lowest score is 45. What is the highest score ?

Q. 1. Find the marks which is 20 more than the lowest: (a) 25 (b) 65 (c) 70 (d) None of these

Difficulty Level: Easy

Ans. Option (b) is correct. Explanation: Given lowest marks is 45 and 20 more it is 45 + 20 = 65

Q. 2. Just to pass in the examination border line marks is just 12 less than the lowest marks obtained by the student in the class. Write the required passing marks. (a) 56 (b) 40 (c) 33 (d) 57

Difficulty Level: Medium

Ans. Option (c) is correct. Explanation: Given passing marks = lowest marks – 12 = 45 – 12 = 33

Q. 3. If one student Aavya of another class who scored 9 marks more than the doubled of lowest marks of this class, find the aavya’s marks: (a) 92 (b) 50 (c) 89 (d) 99

Ans. Option (d) is correct. Explanation: Aavya score = 2 × Lowest marks + 9 = 2(45) + 9 = 90 + 9 = 99

Q. 4. Find the highest marks.

Difficulty Level: Hard

Explanation: let the highest marks be $x$ according to the question, lowest marks

$$ \begin{aligned} & =\frac{1}{2} \text { (highest marks) }+5 \\ 45 & =\frac{1}{2}(x)+5 \\ 45-5 & =\frac{1}{2} x \text { [transposing } 5 \text { to LHS] } \\ 40 & =\frac{1}{2} x \\ 40 \times 2 & =x \text { [multiplying both sides by } 2] \\ 80 & =x \\ x & =80 \end{aligned} $$

• Comparing Quantities Class 7 Case Study Questions Maths Chapter 7
• Triangle and its Properties Class 7 Case Study Questions Maths Chapter 6
• Lines and Angles Class 7 Case Study Questions Maths Chapter 5
• Data Handling Class 7 Case Study Questions Maths Chapter 3

Fractions and Decimals Class 7 Case Study Questions Maths Chapter 2

Integers class 7 case study questions maths chapter 1, topics from which case study questions may be asked.

• Linear equation
• Solution of a linear equation
• Transposition

Case study questions from the above given topic may be asked.

• An equation is a statement of equality which involves one or more literal numbers.
• The value of the variable which satisfies an equation is called the solution or the root of an equation.
• A term in an equation can be transposed from one side of an the sign of equality to another side by changing its sign.
• A number that divides a literal or another number on one side of an equation, when transposed multiplies the other side and viceversa.

A linear equation remains the same when the expression in the left and right are interchanged.

Download Customised White Label Study Materials in MS Word Format

We are providing teaching resources to teachers and coaching institute looking for customised study materials in MS word format. Our High-quality editable study material which is prepared by the expert faculties are Highly useful for Teachers, Mentors, Tutors, Faculties, Coaching Institutes, Coaching Experts, Tuition Centers.

Frequently Asked Questions (FAQs) on Simple Equations Case Study

Q1: what are simple equations.

A1: Simple equations are mathematical statements that express equality between two expressions. They usually involve a variable (like x) and can be solved to find the value of the variable that makes the equation true.

Q2: How do you solve a simple equation?

A2: To solve a simple equation, you need to isolate the variable on one side of the equation. This can be done by performing the same mathematical operations on both sides of the equation, such as addition, subtraction, multiplication, or division.

Q3: Can simple equations have more than one solution?

A3: Simple equations usually have a single unique solution. However, certain types of equations, like identities, can have infinite solutions, but those are not typically covered under “simple equations” in Class 7.

Q4: Why is it important to learn simple equations?

A4: Learning simple equations is important because it forms the foundation for more advanced topics in algebra and mathematics. It helps in developing problem-solving skills and understanding how to work with variables and mathematical relationships.

Q5: What are some real-life applications of simple equations?

A5: Simple equations can be used in various real-life situations, such as calculating expenses, determining distances, solving for unknown quantities in recipes, and even in basic physics problems involving speed, distance, and time.

Q6: What common mistakes should students avoid when solving simple equations?

A6: Common mistakes include not performing the same operation on both sides of the equation, incorrectly combining like terms, and forgetting to change the sign when moving terms from one side of the equation to the other.

Q7: What are some tips for mastering simple equations?

A7: To master simple equations, students should focus on understanding the basic principles of balancing equations, practice regularly, and check their solutions by substituting the value back into the original equation to ensure it holds true.

Q8: How can students practice solving simple equations effectively?

A8: Students can practice effectively by solving a variety of problems, starting with simple ones and gradually moving to more complex equations. They can also use online resources, worksheets, and practice tests to reinforce their learning.

Q9: Are there any online resources or tools available for practicing comparing quantities case study questions?

A9: We provide case study questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

Related Posts