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Chapter 6 Class 9 Lines and Angles

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Updated for New NCERT Book - for 2023-24 Edition.

Get NCERT Solutions of all exercise questions and examples of Chapter 6 Class 9 Lines and Angles free at teachoo. Answers to each question has been solved with Video. Theorem videos are also available.

In this chapter, we will learn

• Basic Definitions - Line, Ray, Line Segment, Angles, Types of Angles (Acute, Obtuse, Right, Straight, Reflex), Intersecting Lines, Parallel Lines
• What is Linear Pair of Angles
• Vertically Opposite Angles are equal
• Angles formed by a transversal on parallel lines - Corresponding Angles, Alternate Interior Angles, Alternate Exterior Angles, Interior Angles on the same of transversal. And its properties
• Theorem 6.6 - Lines parallel to the same line are parallel to each other
• Angle Sum Property   of Triangle
• Exterior Angle Property of a Triangle

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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.

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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles in English and Hindi Medium updated for academic session 2024-25. As per new syllabus there are only two exercises in 9th Maths chapter 6 which is given in new NCERT book for CBSE 2024-25. Class 9 Maths Chapter 6 Solutions for CBSE Board Class 9 Maths Exercise 6.1 in English Class 9 Maths Exercise 6.2 in English

Class 9 Maths Chapter 6 Solutions for State Boards Class 9 Maths Chapter 6 Exercise 6.1 Class 9 Maths Chapter 6 Exercise 6.2 Class 9 Maths Chapter 6 Exercise 6.3

Tiwari Academy typically provides detailed and comprehensive solutions to the Class 9 Maths NCERT textbook. These solutions break down complex mathematical concepts into simpler, step-by-step explanations, making it easier for students to understand and follow. Class 9 Maths Chapter 6 Solutions in Hindi Medium Class 9 Maths Exercise 6.1 in Hindi Class 9 Maths Exercise 6.2 in Hindi The solutions offered on Tiwari Academy website and app often focus on building a strong foundation of mathematical concepts. This can help students gain a clear understanding of the subject matter, which is crucial for success in mathematics.

UP Board Solutions for Class 9 Maths Chapter 6 Prashnavali 6.1, and Prashnavali 6.2 of Rekhaen aur Kon in Hindi Medium are given below to free download in PDF file format. All the question answers are updated for academic year 2024-25. 9th Maths Solutions for Chapter 6 are compatible for UP Board and CBSE Board Students. The NCERT Solutions 2024-25 can be used in Video format by all the Boards like CBSE Board, MP Board, Gujrat, Board, UP Board and other board following NCERT (https://ncert.nic.in/) Books in the academic session 2024-25 onward. Chapter 6 of 9th Mathematics includes the concepts of Lines and Angles. It is the first chapter in Geometry in class 9 Maths, which takes time to understand the concepts.

Study Material for Class 9 Maths Chapter 6

• 9th Maths Chapter 6 Study Material in English
• 9th Maths Chapter 6 Study Material in Hindi
• Class 9 Maths NCERT Solutions
• Class 9 all Subjects Solutions

Class 9 Maths Chapter 6 Practice Questions with Solution

What are intersecting lines.

Intersecting lines: Two lines are said to be intersecting, if they intersect each other at one point.

What is meant by Parallel lines?

Parallel lines: Lines in the same plane which do not intersect when produced on either side.

What do you know about Transversal lines?

Transversal lines: It is a straight line which cuts two or more straight lines at distinct point.

What is an Acute angle?

Acute angle: An angle whose measure is less than 90 but more than 0.

What do you understand by an Obtuse angle?

Obtuse angle: An angle whose measure is more than 90 but less than 180.

What is a Right angle?

Right angle: An angle whose measure is 90.

What is the measure of a Straight angle?

Straight angle: The measure of a straight angle is 180.

What is Reflex angl?

Reflex angle: An angle whose measure is more than 180 but less than 360.

Which angle is called Complete angle?

Complete angle: An angle whose measure is 360.

What are Complementary angles?

Complementary angles: Two angles are said to be complementary, if the sum of their measure is 90.

What are Supplementary angles?

Supplementary angles: Two angles are said to be supplementary, if the sum of their measure is 180.

What are Adjacent angles?

Adjacent angles: Two angles are called adjacent angles, if they have a common vertex and common arm.

What do you know about Vertically opposite angles?

Vertically opposite angles: Two angles are said to be a pair of vertically opposite angles, if their arms from two pairs of opposite rays.

NCERT Solutions for class 9 Maths Chapter 6 Exercise 6.1, and 6.2 Lines and angles in English Medium as well as Hindi Medium in PDF form as well as study online options are given for academic session 2024-25. UP Board or CBSE Board or other boards students can ask their doubts and reply to other users through Tiwari Academy Discussion Forum. A platform to share your knowledge.

Important Questions on 9th Maths Chapter 6

What are the different types of angle.

An acute angle measures between 0° and 90°, whereas a right angle is exactly equal to 90°. An angle greater than 90° but less than 180° is called an obtuse angle. A straight angle is equal to 180°. An angle which is greater than 180° but less than 360° is called a reflex angle. Further, two angles whose sum is 90° are called complementary angles, and two angles whose sum is 180° are called supplementary angles.

What is a line-segment?

A part (or portion) of a line with two end points is called a line-segment.

What is a ray?

A part of a line with one end point is called a ray.

What do you mean by collinear points or non-collinear points?

If three or more points lie on the same line, they are called collinear points; otherwise they are called non-collinear points.

What happens If a transversal intersects two parallel lines?

(i) each pair of corresponding angles is equal, (ii) each pair of alternate interior angles is equal, (iii) each pair of interior angles on the same side of the transversal is supplementary.

What is the sum of the three angles of a triangle?

The sum of the three angles of a triangle is 180°.

Important Notes on 9th Maths Chapter 6

If a transversal intersects two parallel lines, then 1. Each of the alternate interior angles is equal. (Converse is also true). 2. Each pair of interior angles on the same side of the transversal is supplementary. (Converse is also true). 3. Each pair of corresponding angles is equal.

What are the uses of Lines and Angles chapter 6 of class 9th Maths in daily life?

Lines and angles are extremely important in many aspects of real life given below:

• Architect/Engineers use angle measurements to construct buildings, bridges, houses, monuments, etc.
• Carpenters use angle measuring devices such as protractors to make furniture like chairs, tables, beds, wooden, etc.
• We can see the angle in the clocks of our homes, made by hands of clocks. For example, if the small hand is in 3 and the big hand is in 12, we get a right angle.
• Athletes use angles to enhance their performance.
• Artists use their knowledge of angles to sketch portraits and paintings.
• Real-world examples of line/line segments are a pencil, a baseball bat, the cord to your cell phone charger, the edge of a table, etc.
• The line is an element of design and an essential part of many forms of art.
• Whenever an architect has to draw a plan for a multistoried building, she has to draw intersecting lines and parallel lines at different angles.
• To find the height of a tower or to find the distance of a ship from the light house, one needs to know the angle formed between the horizontal and the line of sight.

Which topic’s knowledge will students gain after completing chapter 6 of class 9th Maths?

After completing chapter 6 of class 9th Maths, students will gain the knowledge of the following topics:

• Basic terms and definitions (Line-segment, Ray, Collinear points, Non-collinear points, Angles, Arms, Vertex).
• Types of Angles (Acute angle, Obtuse angle, Straight angle, Right angle, Reflex angle).
• Meaning of Complementary, Supplementary Angles, Linear pair of angles, Vertically opposite angles.
• Intersecting Lines and Non-intersecting Lines.
• Parallel Lines and a Transversal.
• Lines which are parallel to the same line are parallel to each other.
• Angle sum property of a triangle.

How many exercises, questions, and examples are there in chapter 6 (Lines and Angles) of class 9th mathematics?

There are three exercises in class 9 mathematics chapter 6 (Lines and Angles). In the first exercise (Ex 6.1), there are six questions. In the second exercise (Ex 6.2), there are six questions. So, there are 12 questions in class 9 mathematics chapter 6 (Lines and Angles). In this chapter, there are six examples also. Examples 1, 2, 3 are based on Ex 6.1, and examples 4, 5, 6 are based on Ex 6.2.

Which theorem can be asked in the school exam for proof from chapter 6 of class 9th Maths?

There are eight theorems in Chapter 6 Lines and Angles of class 9th Maths. All the theorems are important. The theorems whose proof can come in the school exam are:

• 1) Theorem 6.1 (If two lines intersect each other, then the vertically opposite angles are equal)
• 2) Theorem 6.6 (Lines which are parallel to the same line are parallel to each other)
• 3) Theorem 6.7 (Angle Sum Property of a Triangle).

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Chapter 7. triangles ».

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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.

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

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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
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• CBSE Class 9th 2023-24 : Science Practical Syllabus; Download PDF 19 April, 2023, 4:52 pm
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• 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
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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles 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 6 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 6 Lines and Angles 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 6 Exercise 6.1

NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2

NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3

NCERT Solutions for Class 9 Maths Chapter 6 – Topic Discussion

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

• Basic terms and definitions related to a line segment, ray, collinear points, non-collinear points, intersecting and non-intersecting lines,
• Pair of angles (reflex, complementary, supplementary, adjacent, vertical opposite, linear pair)
• Parallel and Transversal Lines and theorems related to them.
• Angle sum property of a triangle.

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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.

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CBSE Class 9 Maths Important Questions for Chapter 6 - Lines and Angles

• Class 9 Important Question
• Chapter 6: Lines And Angles

CBSE Class 9 Maths Important Questions Chapter 6 - Lines and Angles Free PDF Download

Explore the essential queries handpicked by Vedantu's subject experts for CBSE Class 9 Chapter 6 in Mathematics. These important questions, aligned with CBSE standards, encompass all the key topics. Mastering these concepts in Class 9 paves the way for a solid foundation in Class 10. Dive into the realm of "Lines and Angles" and grasp each concept effortlessly using these vital practice questions.

Vedantu goes a step further to enhance your exam readiness. Get a hold of complimentary Class 9 Maths NCERT Solutions , elevating your exam preparation. But that's not all – our platform also offers Class 9 Science NCERT Solutions . Delve into our updated NCERT Book Solutions, a valuable resource spanning diverse subjects. It's all geared to make your academic journey smoother and more effective. Start your journey with Vedantu today!

Download CBSE Class 9 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 9 Maths Important Questions for other chapters:

Important Topics Covered in Class 9 Maths Chapter 6

Introduction

Basic Terms And Definition

Intersecting Lines And Non-Intersecting Lines

Pairs of Angles

Parallel Lines And Transversal Line

Lines Parallel To The Same Line

Angle Sum Property of A Triangle

Study Important Questions for Class 9 Maths Chapter 6 - Lines and Angles

1. Measurement of reflex angle is

(i) ${90^\circ }$

(ii) between ${0^\circ }$ and ${90^\circ }$

(iii) between ${90^\circ }$ and ${180^\circ }$

(iv) between ${180^\circ }$ and ${360^\circ }$

Ans: (iv) between ${180^\circ }$ and ${360^\circ }$

2. The sum of angle of a triangle is

(i) ${0^\circ }$

(ii) ${90^\circ }$

(iii) ${180^\circ }$

(iv) none of these

Ans: (iii) ${180^\circ }$

3. In fig if ${\text{ x}} = {30^\circ }{\text{ }}$ then y = 

(ii) ${180^\circ }$

(iii) ${150^\circ }$

(iv) ${210^\circ }$

Ans: (iii) ${150^\circ }$

4. If two lines intersect each other then

(i) Vertically opposite angles are equal

(ii) Corresponding angle are equal

(iii) Alternate interior angle are equal

(iv) None of these

Ans: (i) Vertically opposite angles are equal

5. The measure of Complementary angle of ${63^\circ }$ is

(a) ${30^\circ }$

(b) ${36^\circ }$

(c) ${27^\circ }$

(d) None of there

Ans: (c) ${27^\circ }$

6. If two angles of a triangle is ${30^\circ }$ and ${45^\circ }$ what is measure of third angle

(a) ${95^\circ }$

(b) ${90^\circ }$

(c) ${60^\circ }$

(d) ${105^\circ }$

Ans: (d) ${105^\circ }$

7. The measurement of Complete angle is

(a) ${0^\circ }$

(c) ${180^\circ }$

(d) ${360^\circ }$

Ans: (d) ${360^\circ }$         

8. The measurement of sum of linear pair is

(a) ${180^\circ }$

(c) ${270^\circ }$

Ans: (a) ${180^\circ }$

9. The difference of two complementary angles is ${40^\circ }$. The angles are

(a) ${65^\circ },{35^\circ }$

(b) ${70^\circ },{30^\circ }$

(c) ${25^\circ },{65^\circ }$

(d) ${70^\circ },{110^\circ }$

Ans: (c) ${25^\circ },{65^\circ }$

10. Given two distinct points ${\text{P}}$ and ${\text{Q}}$ in the interior of $\angle ABC$, then $\overrightarrow {AB} $ will be

(a) In the interior of $\angle ABC$

(b) In the interior of $\angle ABC$

(c) On the $\angle ABC$

(d) On the both sides of $\overrightarrow {BA} $

Ans: (c) On the $\angle ABC$

11. The complement of ${(90 - a)^0}$ is

(a) $ - {a^0}$

(b) ${(90 + 2a)^0}$

(c) ${(90 - a)^0}$

(d) ${a^0}$

Ans: (d) ${a^0}$

12. The number of angles formed by a transversal with a pair of lines is

13. In fig \[{L_1}\parallel {L_2}\] and $\angle 1\, = \,{52^ \circ }$ the measure of $\angle 2$ is.

(A) ${38^\circ }$

(B) ${128^\circ }$

(C) ${52^\circ }$

(D) ${48^\circ }$

Ans: (B) ${128^\circ }$

14. In $fig{\text{ x}} = {30^\circ }$ the value of Y is 

(A) ${10^\circ }$

(B) ${40^\circ }$

(C) ${36^\circ }$

(D) ${45^\circ }$

Ans: (B) ${40^\circ }$

15. Which of the following pairs of angles are complementary angle?

(A) ${25^\circ },{65^\circ }$

(B) ${70^\circ },{110^\circ }$

(C) ${30^\circ },{70^\circ }$

(D) ${32.1^\circ },{47.9^\circ }$

Ans: (A) ${25^\circ },{65^\circ }$

16. In fig the measures of $\angle 1$ is.

(A) ${158^\circ }$

(B) ${138^\circ }$

(C) ${42^\circ }$

Ans: (C) ${42^\circ }$

17. In figure the measure of $\angle a$ is

(b) ${150^0}$

(c) ${15^\circ }$

(d) ${50^\circ }$

Ans: (a) ${30^\circ }$

18. The correct statement is-

F point in common.

Ans: (c) Three points are collinear if all of them lie on a line.

19. One angle is five times its supplement. The angles are-

(a) ${15^\circ },{75^\circ }$

(b) ${30^\circ },{150^\circ }$

(c) ${36^\circ },{144^0}$

(d) ${160^\circ },{40^\circ }$

Ans: (b) ${30^\circ },{150^\circ }$

20. In figure if  and $\angle 1:\angle 2 = 1:2.$ the measure of $\angle 8$ is

(a) ${120^\circ }$

(b) ${60^\circ }$

(c) ${30^\circ }$

(d) ${45^\circ }$

Ans:  (b) ${60^\circ }$

1. In Fig. 6.13, lines ${\text{AB}}$ and ${\text{CD}}$ intersect at $O.$ If $\angle {\text{AOC}} + \angle {\text{BOE}} = {70^\circ }$ and $\angle {\text{BOD}} = {40^\circ }$, find $\angle {\text{BOE}}$ and reflex $\angle {\text{COE}}.$

Ans: According to the question given that, $\angle AOC + \angle BOE = {70^\circ }$ and $\angle BOD = {40^\circ }$ .

We need to find $\angle BOE$ and reflex $\angle COE$ .

According to the given figure, we can conclude that $\angle COB$ and $\angle AOC$ form a linear pair.

As we also know that sum of the angles of a linear pair is ${180^\circ }$ .

So, $\angle COB + \angle AOC = {180^\circ }$

Because, $\angle COB = \angle COE + \angle BOE$ , or

So, $\angle AOC + \angle BOE + \angle COE = {180^\circ }$

$ \Rightarrow {70^\circ } + \angle COE = {180^\circ }$

$ \Rightarrow \angle COE = {180^\circ } - {70^\circ }$

$ = {110^\circ }.$

Reflex $\angle COE = {360^\circ } - \angle COE$

$ = {360^\circ } - {110^\circ }$

$ = {250^\circ }.$

$\angle AOC = \angle BOD$ (Vertically opposite angles), or

$\angle BOD + \angle BOE = {70^\circ }$

But, according to the question given that $\angle BOD = {40^\circ }$ .

${40^\circ } + \angle BOE = {70^\circ }$

$\angle BOE = {70^\circ } - {40^\circ }$

$ = {30^\circ }$ .

Hence, we can conclude that Reflex $\angle COE = {250^\circ }$ and $\angle BOE = {30^\circ }$ .

2. In the given figure, $\angle PQR = \angle PRQ$, then prove that $\angle PQS = \angle PRT$.

(Image will be uploaded soon)

Ans: According to the question we need to prove that $\angle PQS = \angle PRT$ .

According to the question given that $\angle PQR = \angle PRQ$ .

According to the given figure, we can conclude that $\angle PQS$ and $\angle PQR$ , and $\angle PRS$ and $\angle PRT$ form a linear pair.

So, $\angle PQS + \angle PQR = {180^\circ }$ , and(i)

$\angle PRQ + \angle PRT = {180^\circ }$ ..(ii)

According to the equations (i) and (ii), we can conclude that

$\angle PQS + \angle PQR = \angle PRQ + \angle PRT$

But, $\angle PQR = \angle PRQ.$

So, $\angle PQS = \angle PRT$

Hence, the desired result is proved.

3. In the given figure, find the values of ${\text{x}}$ and ${\text{y}}$ and then show that .

Ans: According to the question we need to find the value of $x$ and $y$ in the figure given below and then prove that 

According to the figure, we can conclude that $y = {130^\circ }$ (Vertically opposite angles), and

$x$ and ${50^\circ }$ form a pair of linear pair.

As we also know that the sum of linear pair of angles is ${180^\circ }$ .

$x + {50^\circ } = {180^\circ }$

$x = {130^\circ }$

$x = y = {130^\circ }$

According to the given figure, we can conclude that $x$ and $y$ form a pair of alternate interior angles parallel to the lines AB and CD.

Hence, we can conclude that $x = {130^\circ },y = {130^\circ }$ and.

4. In the given figure, if ${\text{AB}}||{\text{CD}},{\text{CD}}||{\text{EF}}$ and ${\text{y}}:{\text{z}} = 3:7$, find ${\text{x}}$.

Ans: According to the question given that,  and $y:z = 3:7$ .

We need to find the value of $x$ in the figure given below.

As we also know that the lines parallel to the same line are also parallel to each other.

We can determine that .

Assume that, $y = 3a$ and $z = 7a$ .

We know that angles on same side of a transversal are supplementary.

So, $x + y = {180^\circ }.$

$x = z$ (Alternate interior angles)

$z + y = {180^\circ }$ , or $7a + 3a = {180^\circ }$

$ \Rightarrow 10a = {180^\circ }$

$a = {18^\circ }$ .

$z = 7a = {126^\circ }$

$y = 3a = {54^\circ }$

Now, $x + {54^\circ } = {180^\circ }$

$x = {126^\circ }$

Hence, we can determine that $x = {126^\circ }$ .

5. In the given figure, if  and $\angle PRD = {127^\circ }$, find ${\text{x}}$ and ${\text{y}}$.

Ans: According to the question given that,  and $\angle PRD = {127^\circ }$ .

As we need to find the value of $x$ and $y$ in the figure.

$\angle APQ = x = {50^\circ }$ . (Alternate interior angles)

$\angle PRD = \angle APR = {127^\circ }$ . (Alternate interior angles)

$\angle APR = \angle QPR + \angle APQ$

${127^\circ } = y + {50^\circ }$

$ \Rightarrow y = {77^\circ }$

Hence, we can determine that $x = {50^\circ }$ and $y = {77^\circ }$ .

6. In the given figure, sides QP and RQ of $\Delta {\text{PQR}}$ are produced to points ${\text{S}}$ and ${\text{T}}$ respectively. If $\angle {\text{SPR}} = {135^\circ }$ and $\angle {\text{PQT}} = {110^\circ }$, find $\angle {\text{PRQ}}$.

Ans: According to the question given that, $\angle SPR = {135^\circ }$ and $\angle PQT = {110^\circ }$ .

As we need to find the value of $\angle PRQ$ in the figure given below.

According to the given figure, we can determine that $\angle SPR$ and $\angle RPQ$ , and $\angle SPR$ and $\angle RPQ$ form a linear pair.

As we also know that the sum of angles of a linear pair is ${180^\circ }$ .

$\angle SPR + \angle RPQ = {180^\circ }$ , and

$\angle PQT + \angle PQR = {180^\circ }$

${135^\circ } + \angle RPQ = {180^\circ }$ , and

${110^\circ } + \angle PQR = {180^\circ }$ , or

$\angle RPQ = {45^\circ }$ , and

$\angle PQR = {70^\circ }.$

According to the figure, we can determine that

$\angle PQR + \angle RPQ + \angle PRQ = {180^\circ }$ . (Angle sum property)

$ \Rightarrow {70^\circ } + {45^\circ } + \angle PRQ = {180^\circ }$

$ \Rightarrow {115^\circ } + \angle PRQ = {180^\circ }$

$ \Rightarrow \angle PRQ = {65^\circ }$ .

Hence, we can determine that $\angle PRQ = {65^\circ }$ .

7. In the given figure, $\angle {\text{X}} = {62^\circ },\angle {\text{XYZ}} = {54^\circ }.$ If ${\text{YO}}$ and ${\text{ZO}}$ are the bisectors of $\angle {\text{XYZ}}$ and $\angle {\text{XZY}}$ respectively of $\Delta {\mathbf{XYZ}}$, find $\angle {\text{OZY}}$ and $\angle {\text{YOZ}}$.

Ans: According to the question given that, $\angle X = {62^\circ },\angle XYZ = {54^\circ }$ and YO and ZO are bisectors of $\angle XYZ$ and $\angle XZY$ , respectively.

As we need to find the value of $\angle OZY$ and $\angle YOZ$ in the figure.

According to the given figure, we can determine that in $\Delta XYZ$

$\angle X + \angle XYZ + \angle XZY = {180^\circ }$ (Angle sum property)

$ \Rightarrow {62^\circ } + {54^\circ } + \angle XZY = {180^\circ }$

$ \Rightarrow {116^\circ } + \angle XZY = {180^\circ }$

$ \Rightarrow \angle XZY = {64^\circ }$ .

According to the question given that, OY and OZ are the bisectors of $\angle XYZ$ and $\angle XZY$ , respectively.

$\angle OYZ = \angle XYO = \dfrac{{{{54}^\circ }}}{2} = {27^\circ}$ , and

$\angle OZY = \angle XZO = \dfrac{{{{64}^\circ }}}{2} = {32^\circ }$

According to the figure, we can determine that in $\Delta OYZ$

$\angle OYZ + \angle OZY + \angle YOZ = {180^\circ }$ . (Angle sum property)

${27^\circ } + {32^\circ } + \angle YOZ = {180^\circ }$

$ \Rightarrow {59^\circ } + \angle YOZ = {180^\circ }$

$ \Rightarrow \angle YOZ = {121^\circ }$ .

Hence, we can determine that $\angle YOZ = {121^\circ }$ and $\angle OZY = {32^\circ }{\text{. }}$

8. In the given figure, if ${\text{AB}}||{\text{DE}},\angle {\text{BAC}} = {35^\circ }$ and $\angle {\text{CDE}} = {53^\circ }$, find $\angle {\text{DCE}}$.

Ans: According to the question given that,  and $\angle CDE = {53^\circ }$ .

As we need to find the value of $\angle DCE$ in the figure given below.

According to the given figure, we can determine that

$\angle BAC = \angle CED = {35^\circ }$ (Alternate interior angles)

According to the figure, we can determine that in $\Delta DCE$

$\angle DCE + \angle CED + \angle CDE = {180^\circ }$ . (Angle sum property)

$\angle DCE + {35^\circ } + {53^\circ } = {180^\circ }$

$ \Rightarrow \angle DCE + {88^\circ } = {180^\circ }$

$ \Rightarrow \angle DCE = {92^\circ }$ .

Hence, we can determine that $\angle DCE = {92^\circ }$ .

9. In the given figure, if lines PQ and RS intersect at point ${\text{T}}$, such that $\angle {\text{PRT}} = {40^\circ }$, $\angle {\text{RPT}} = {95^\circ }$ and $\angle {\text{TSQ}} = {75^\circ }$, find $\angle {\text{SQT}}$.

Ans: According to the question given that, $\angle PRT = {40^\circ },\angle RPT = {95^\circ }$ and $\angle TSQ = {75^\circ }$ .

As we need to find the value of $\angle SQT$ in the figure.

According to the given figure, we can determine that in $\Delta RTP$

$\angle PRT + \angle RTP + \angle RPT = {180^\circ }$ (Angle sum property)

${40^\circ } + \angle RTP + {95^\circ } = {180^\circ }$

$ \Rightarrow \angle RTP + {135^\circ } = {180^\circ }$

$ \Rightarrow \angle RTP = {45^\circ }$ .

$\angle RTP = \angle STQ = {45^\circ }.$ (Vertically opposite angles)

According to the figure, we can determine that in $\Delta STQ$

$\angle SQT + \angle STQ + \angle TSQ = {180^\circ }$ . (Angle sum property)

$\angle SQT + {45^\circ } + {75^\circ } = {180^\circ }$

$ \Rightarrow \angle SQT + {120^\circ } = {180^\circ }$

$ \Rightarrow \angle SQT = {60^\circ }$ .

Hence, we can determine that $\angle SQT = {60^\circ }$ .

10. In fig lines ${\text{XY}}$ and ${\text{MN}}$ intersect at O If $\angle $ POY $ = {90^\circ }$ and ${\text{ab}} = 2:3$ find ${\text{c}}$

Ans: According to the given figure $\angle {\text{POY}} = {90^\circ }$

Assume that, $a = 2x$ and $b = 3x$

${\text{a}} + {\text{b}} + \angle {\text{POY}} = {180^\circ }(\because {\text{XOY}}$ is a line $)$

$\Rightarrow$ $2{\text{x}} + 3{\text{x}} + {90^\circ } = {180^\circ }$

$\Rightarrow$ $5x = {180^\circ } - {90^\circ }$

$\Rightarrow$ $5{\text{x}} = {90^\circ }$

$\Rightarrow$ $x = \dfrac{{{{90}^\circ }}}{5} = {18^\circ }$

So, $a = {36^\circ },\quad b = {54^\circ }$

MON is a line.

${\text{b}} + {\text{c}} = {180^\circ }$

$\Rightarrow$ ${54^\circ } + {{\text{c}}^\circ } = {180^\circ }$

$\Rightarrow$ $c = {180^\circ } - {54^\circ } = {126^\circ }$

Hence, the value of $c = {126^\circ }$ .

11. In fig find the volume of ${\text{x}}$ and ${\text{y}}$ then Show that $\mathrm{AB} \| \mathrm{CD}$

Ans: Accordign to the given figure, ${50^\circ } + x = {180^\circ }$

(by linear pair)

$x = {180^\circ } - {50^\circ }$

So, $x = {130^\circ }$

$y = {130^\circ }$ (Because vertically opposite angles are equal)

${\text{x}} = {\text{y}}$ as they are corresponding angles. 

So, AB || CD

Hence proved.

12. What value of ${\mathbf{y}}$ would make AOB a line if $\angle {\text{AOC}} = 4{\text{y}}$ and $\angle {\text{BOC}} = 6{\text{y}} + {30^\circ }$

Ans: According to the question given that, $\angle {\text{AOC}} = 4{\text{y}}$ and $\angle {\text{BOC}} = 6{\text{y}} + {30^\circ }$

$\angle {\text{AOC}} + \angle {\text{BOC}} = {180^\circ }$     (By linear pair)

$4y + 6y + {30^\circ } = {180^\circ }$

$\Rightarrow$ $10y = {180^\circ } - {30^\circ }$

$\Rightarrow$  $10y = {150^\circ } $

$\Rightarrow$  $y = {15^\circ }$

13. In fig ${\text{POQ}}$ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $\angle {\text{ROS}} = \dfrac{1}{2}(\angle {\text{QOS}} - \angle {\text{POS}})$

Ans: According to the question,

$R.H.S = \dfrac{1}{2}(\angle QOS - \angle POS)$

$ = \dfrac{1}{2}(\angle {\text{ROS}} + \angle {\text{QOR}} - \angle {\text{POS}})$

$ = \dfrac{1}{2}\left( {\angle {\text{ROS}} + {{90}^\circ } - \angle {\text{POS}}} \right) \ldots  \ldots ..$ (1)

Because, $\angle {\text{POS}} + \angle {\text{ROS}} = {90^\circ }$

So, by equation 1

$ = \dfrac{1}{2}(ROS + \angle POS + \angle ROS - \angle POS)$ (by equation 1)

$ = \dfrac{1}{2} \times 2\angle {\text{ROS}} = \angle {\text{ROS}}$

14. In fig lines l 1 and l 2 intersected at O , if ${\text{x}} = {45^\circ }$ find ${\text{x}},{\text{y}}$ and ${\text{u}}$

Ans: According to the question given that,

$x = {45^\circ }$

So, ${\text{z}} = {45^\circ }$ (Because vertically opposite angles are equal)

${\text{x}} + {\text{y}} = {180^\circ }$

${45^\circ } + y = {180^\circ }$ (By linear pair)

$y = {180^\circ } - 45$

$y = {135^\circ }$

${\text{y}} = {\text{u}}$

Hence, the value of ${\text{u}} = {135^\circ }$ (Vertically opposite angles)

15. The exterior angle of a triangle is ${110^\circ }$ and one of the interior opposite angle is ${35^\circ }$. Find the other two angles of the triangle.

Ans: As we all know that the exterior angle of a triangle is equal to the sum of interior opposite angles.

So, $\angle {\text{ACD}} = \angle {\text{A}} + \angle {\text{B}}$

${110 ^\circ}= \angle {\text{A}} + {35^\circ }$

$\Rightarrow$ $\angle {\text{A}} = {110^\circ } - {35^\circ }$

$\Rightarrow$ $\angle {\text{A}} = {75^\circ }$

$\Rightarrow$ $\angle {\text{C}} = 180 - (\angle {\text{A}} + \angle {\text{B}})$

$\Rightarrow$ $\angle {\text{C}} = 180 - \left( {{{75}^\circ } + {{35}^\circ }} \right)$

$\angle {\text{C}} = {70^\circ }$

16. Of the three angles of a triangle, one is twice the smallest and another is three times the smallest. Find the angles.

Ans: Assume that the smallest angle be ${x^\circ }$

Then the other two angles are $2{x^\circ }$ and $3{x^\circ }$

${x^\circ } + 2{x^\circ } + 3{x^\circ } = {180^\circ }$ As we know that the sum of three angle of a triangle is $\left. {{{180}^\circ }} \right$

$6{x^\circ } = {180^\circ }$

${\text{x}} = \dfrac{{180}}{6}$

$ = {30^\circ }$

Hence, angles are ${30^\circ },{60^\circ }$ and ${90^\circ }$ .

17. Prove that if one angle of a triangle is equal to the sum of other two angles, the triangle is right angled.

Ans: According to the question given that in $\Delta ABC,\,\,\angle {\text{B}} = \angle {\text{A}} + \angle {\text{C}}$

To prove: $\Delta ABC$ is right angled.

Proof: $\angle A + \angle B + \angle C = {180^\circ } \ldots ..$ (1) (As we know that the sum of three angles of a $\Delta {\text{ABC}}$ is $\left. {{{180}^\circ }} \right)$

$\angle A + \angle C = \angle B \ldots ..$ (2)

From equations (1) and (2),

$\angle B + \angle B = {180^\circ }$

$2\angle {\text{B}} = {180^\circ }$

$\angle {\text{B}} = {90^\circ }$

18. In fig. sides QP and RQ of $\Delta PQR$ are produced to points ${\text{S}}$ and ${\text{T}}$ respectively. If $\angle SPR = {135^\circ }$ and $\angle {\text{PQT}} = {110^\circ }$, find $\angle {\text{PRQ}}$.

Ans: According to the given figure,

${110^\circ } + \angle PQR = {180^\circ }$

$\angle PQR = {180^\circ } - {110^\circ }$

$\angle {\text{PQR}} = {70^\circ }$

Also, $\angle {\text{SPR}} = \angle {\text{PQR}} + \angle {\text{PRQ}}$ (According to the Interior angle theorem)

${135^\circ } = {70^\circ } + \angle PRQ$

$\angle {\text{PRQ}} = {135^\circ } - {70^\circ }$

Hence, the value of $\angle {\text{PRQ}} = {65^\circ }$ .

19. In fig the bisector of $\angle ABC$ and $\angle {\text{BCA}}$ intersect each other at point O prove that $\angle BOC = {90^\circ } + \dfrac{1}{2}\angle A$

Ans: According to the question given that in \(\vartriangle ABC\) such that the bisectors of $\angle ABC$ and $\angle {\text{BCA}}$ meet at a point O.

To Prove $\angle BOC = {90^\circ } + \dfrac{1}{2}\angle A$

Proof: In $\vartriangle BOC$

$\angle 1 + \angle 2 + \angle BOC = {180^\circ }$ (1)

In $\vartriangle ABC$

$\angle A + \angle B + \angle C = {180^\circ }$

$\angle A + 2\angle 1 + 2\angle 2 = {180^\circ }$

(BO and CO bisects $\angle B$ and $\angle {\text{C}}$ )

$ \Rightarrow \dfrac{{\angle A}}{2} + \angle 1 + \angle 2 = {90^\circ }$

$\angle 1 + \angle 2 = {90^\circ } - \dfrac{{\angle A}}{2}$

(Divide forth side by 2)

$\angle 1 + \angle 2 = {90^\circ } - \dfrac{{\angle A}}{2}$ in (i)

Substituting, ${90^\circ } - \dfrac{{\angle A}}{2} + \angle BOC = {180^\circ }$

$ \Rightarrow \angle BOC = {90^\circ } + \dfrac{{\angle A}}{2}$

20. In the given figure $\angle POR$ and $\angle QOR$ form a linear pair if ${\mathbf{a}} - {\mathbf{b}} = {80^\circ }$. Find the

value of 'a' and 'b'.

Ans: $a + b = {180^\circ } \to (1)$ (By line as pair)

$a - b = {80^0} \to (2)$

$2{\text{a}} = {260^\circ }$ (Adding equations (1) and (2))

${\text{a}} = {130^\circ }$

Put ${\text{a}} = {130^\circ }$ in equation (1)

${130^\circ } + b = {180^\circ }$

${\text{b}} = {180^\circ } - {130^\circ } = {50^\circ }$

Hence the value of ${\text{a}} = {130^\circ }$ and ${\text{b}} = {50^\circ }$ .

21. If ray ${\text{OC}}$ stands on a line ${\text{AB}}$ such that $\angle AOC = \angle BOC$, then show that $\angle AOC = {90^\circ }$

$\angle AOC = \angle BOC$  

$\angle {\text{AOC}} + \angle {\text{BOC}} = {180^\circ }$   (By lines pair)

$\angle {\text{AOC}} + \angle {\text{AOC}} = {180^\circ }$

$2\angle {\text{AOC}} = {180^\circ }$

$\angle {\text{AOC}} = {90^\circ } = \angle B{\text{OC}}$

22. In the given figure show that $\mathrm{AB} \| \mathrm{EF}$

Ans: $\angle {\text{BCD}} = \angle {\text{BCE}} + \angle {\text{ECD}}$

$ = {36^\circ } + {30^\circ } = {66^\circ } = \angle ABC$

So,  (Alternate interior angles are equal)

Again, $\angle {\text{ECD}} = {30^\circ }$ and $\angle {\text{FEC}} = {150^\circ }$

So, $\angle {\text{ECD}} + \angle {\text{FEC}} = {30^\circ } + {150^\circ } = {180^\circ }$

Therefore,  (We know that the sum of consecutive interior angle is $\left. {{{180}^\circ }} \right)$

$A B \| C D$ and $\mathrm{CD} \| \mathrm{EF}$

Then $\mathrm{AB} \| \mathrm{EF}$

23. In figure if  and $\angle PRD = {127^\circ }$ Find ${\text{x}}$ and ${\text{y}}$.

Ans: $A B \| C D$ and PQ is a transversal

$\angle {\text{APQ}} = \angle {\text{PQD}}$ (Pair of alternate angles)

${50^\circ } = {\text{x}}$

Also  and ${\text{PR}}$ is a transversal

$\angle {\text{APR}} = \angle {\text{PRD}}$

${50^\circ } + Y = {127^\circ }$

${\text{Y}} = {127^\circ } - {50^\circ } = {77^\circ }$

Hence the value of ${\text{x}}\, = \,{50^\circ }$ and ${\text{Y}} = {77^\circ }$ .

24. Prove that if two lines intersect each other then vertically opposite angler are equal.

Ans: According to the given figure: AB and CD are two lines intersect each other at $O$ .

To prove: (i) $\angle 1 = \angle 2$ and (ii) $\angle 3 = \angle 4$

$\angle 1 + \angle 4 = {180^\circ } \to (i)$ (By linear pair)

$\angle 4 + \angle 2 = {180^\circ }\quad  \to (ii)$

$\angle 1 + \angle 4 = \angle 4 + \angle 2$ (By equations (i) and (ii))

$\angle 1 = \angle 2$

$\angle 3 = \angle 4$

25. The measure of an angle is twice the measure of the supplementary angle. Find measure of angles.

Ans: Assume that the measure be ${x^\circ}$ .

Then its supplement is ${180^\circ } - {x^\circ}$ .

According to question

${x^\circ} = 2\left( {{{180}^\circ } - {x^\circ}} \right)$

$\Rightarrow$ ${x^\circ} = {360^\circ } - 2{x^\circ}$

$\Rightarrow$  $3x = {360^\circ }$

$\Rightarrow$  $x = {120^\circ }$

The measure of the angles are ${120^\circ }$ and ${60^\circ }$ .

26. In fig $\angle PQR = \angle PRQ$. Then prove that $\angle PQS = \angle PRT$.

Ans: $\angle PQS + \angle PQR = \angle PRQ + \angle PRT$ (By linear pair)

$\angle PQR = \angle PRQ$ (Accordign to the question)

27. In the given fig $\angle {\text{AOC}} = \angle {\text{ACO}}$ and $\angle {\text{BOD}} = \angle {\text{BDO}}$ prove that AC || DB.

$\angle AOC = \angle ACO$ and $\angle BOD = \angle BDO$

$\angle AOC = \angle BOD$ (Vertically opposite angles)

$\angle AOC = \angle BOD$ and $\angle BOD = \angle BDO$

$ \Rightarrow \angle ACO = \angle BDO$

So,  (By alternate interior angle property)

Hence AC || DB  proved.

28. In figure if lines ${\text{PQ}}$ and ${\text{RS}}$ intersect at point ${\text{T}}$. Such that $\angle PRT = {40^\circ }$, $\angle RPT = {95^\circ }$ and $\angle TSQ = {75^\circ }$, find $\angle SQT$.

Ans: According to the $\Delta $ PRT

$\angle {\text{P}} + \angle {\text{R}} + \angle 1 = {180^\circ }$ (By angle sum property)

${95^\circ } + {40^\circ } + \angle 1 = {180^\circ }$

$\angle 1 = {180^\circ } - {135^\circ }$

$\angle 1 = {45^\circ }$

$\angle 1 = \angle 2$ (Vertically opposite angle)

$\angle 2 = \angle {45^\circ }$

According to the $\Delta {\text{TQS}}\quad \angle 2 + \angle {\text{Q}} + \angle {\text{S}} = {180^\circ }$

${45^\circ } + \angle Q + {75^\circ } = {180^\circ }$

$\angle {\text{Q}} + {120^\circ } = {180^\circ }$

$\angle {\text{Q}} = {180^\circ } - {120^\circ }$

$\angle {\text{Q}} = {60^\circ }$

Hence, the value of $\angle {\text{SQT}} = {60^\circ }$ .

29. In figure, if $QT \bot PR,\angle TQR = {40^\circ }$ and $\angle SPR = {50^\circ }$ find $x$ and $y$.

Ans: According to the $\Delta {\text{TQR}}$

${90^\circ } + {40^\circ } + x = {180^\circ }$ (Angle sum property of triangle)

So, $x = {50^\circ }$

Now, $y = \angle {\text{SPR}} + x$

So, $y = {30^\circ } + {50^\circ } = {80^\circ }$ .

Hence, the value of $x = {50^\circ }$ and $y = {80^\circ }$ .

30. In figure sides QP and RQ of $\Delta PQR$ are produced to points ${\text{S}}$ and ${\text{T}}$ respectively if $\angle SPR = {135^\circ }$ and $\angle PQT = {110^\circ }$, find $\angle PRQ$.

${110^\circ } + \angle 2 = {180^\circ }$ (By linear pair)

$\angle 2 = {180^\circ } - {110^\circ }$

$\angle 2 = {70^\circ }$

$\angle 1 + {135^\circ } = {180^\circ }$

$\angle 1 + \angle 2 + \angle {\text{R}} = {180^\circ }$ (By angle sum property)

${45^\circ } + {70^\circ } + \angle R = {180^\circ }$

$\angle {\text{R}} = {180^\circ } - {115^\circ }$

$\angle {\text{R}} = {65^\circ }$

31. In figure lines ${\text{PQ}}$ and RS intersect each other at point O. If $\angle POR:\angle ROQ = 5:7$. Find all the angles.

Ans: $\angle POR + \angle ROQ = {180^\circ }$ (Linear pair of angle)

But, $\angle {\text{POR}}:\angle {\text{ROQ}} = 5:7$ (According to the question)

So, $\angle {\text{POR}} = \dfrac{5}{{12}} \times {180^\circ } = {75^\circ }$

Similarly, $\angle {\text{ROQ}} = \dfrac{7}{{12}} \times {180^\circ } = {105^\circ }$

Now, $\angle {\text{POS}} = \angle {\text{ROQ}} = {105^\circ }$ (Vertically opposite angle)

And $\angle {\text{SOQ}} = \angle {\text{POR}} = {75^\circ }$ (Vertically app angle)

1. In Fig. 6.16, if $x + y = w + z$, then prove that AOB is a line.

Ans: As we need to prove that AOB is a line.

According to the question, given that $x + y = w + z$ .

As we know that the sum of all the angles around a fixed point is ${360^\circ }$ .

Hence, we can determine that $\angle AOC + \angle BOC + \angle AOD + \angle BOD = {360^\circ }$ , or

$y + x + z + w = {360^\circ }$

But, $x + y = w + z$ (According to the question).

$2(y + x) = {360^\circ }$

$y + x = {180^\circ }$

According to the given figure, we can determine that $y$ and $x$ form a linear pair.

As we also know that if a ray stands on a straight line, then the sum of the angles of linear pair formed by the ray with respect to the line is ${180^\circ }$ .

$y + x = {180^\circ }.$

Hence, we can determine that AOB is a line.

2. It is given that $\angle XYZ = {64^\circ }$ and XY is produced to point ${\mathbf{P}}$. Draw a figure from the given information. If ray YQ bisects $\angle ZYP$, find $\angle XYQ$ and reflex $\angle QYP$.

Ans: According to the question, given that $\angle XYZ = {64^\circ },XY$ is produced to $P$ and YQ bisects $\angle ZYP$ .

As we can determine the given below figure for the given situation:

As we need to find $\angle XYQ$ and reflex $\angle QYP$ .

According to the given figure, we can determine that $\angle XYZ$ and $\angle ZYP$ form a linear pair.

$\angle XYZ + \angle ZYP = {180^\circ }$

But, $\angle XYZ = {64^\circ }$ .

$ \Rightarrow {64^\circ } + \angle ZYP = {180^\circ }$

$ \Rightarrow \angle ZYP = {116^\circ }$ .

Ray YQ bisects $\angle ZYP$ , or

$\angle QYZ = \angle QYP = \dfrac{{{{116}^\circ }}}{2} = {58^\circ }$

$\angle XYQ = \angle QYZ + \angle XYZ$

$ = {58^\circ } + {64^\circ } = {122^\circ }.$

Reflex $\angle QYP = {360^\circ } - \angle QYP$

$ = {360^\circ } - {58^\circ }$

$ = {302^\circ }$

Hence, we can determine that $\angle XYQ = {122^\circ }$ and reflex $\angle QYP = {302^\circ }$ .

3. In the given figure, If ${\text{AB}}\parallel {\text{CD}}$ ,$EF \bot CD$ and $\angle GED = {126^\circ }$, find $\angle AGE,\angle GEF$ and $\angle FGE$.

Ans: According to the question, given that  and $\angle GED = {126^\circ }$ .

As we need to find the value of $\angle AGE,\angle GEF$ and $\angle FGE$ in the figure given below.

$\angle GED = {126^\circ }$

$\angle GED = \angle FED + \angle GEF$

But, $\angle FED = {90^\circ }$ .

${126^\circ } = {90^\circ } + \angle GEF \Rightarrow \angle GEF = {36^\circ }$

Because, $\angle AGE = \angle GED$ (Alternate angles)

$\angle AGE = {126^\circ }.$

According to the given figure, we can determine that $\angle FED$ and $\angle FEC$ form a linear pair.

As we know that sum of the angles of a linear pair is ${180^\circ }$ .

$\angle FED + \angle FEC = 180$

$ \Rightarrow {90^\circ } + \angle FEC = {180^\circ }$

$ \Rightarrow \angle FEC = {90^\circ }$

But $\angle FEC = \angle GEF + \angle GEC$

So, ${90^\circ } = {36^\circ } + \angle GEC$

$ \Rightarrow \angle GEC = {54^\circ }$ .

$\angle GEC = \angle FGE = {54^\circ }$ (Alternate interior angles)

Hence, we can determine that $\angle AGE = {126^\circ }$ , $\angle GEF = {36^\circ }$ and $\angle FGE = {54^\circ }.$

4. In the given figure, ${\text{PQ}}$ and ${\text{RS}}$ are two mirrors placed parallel to each other. An incident ray ${\text{AB}}$ strikes the mirror ${\text{PQ}}$ at ${\text{B}}$, the reflected ray moves along the path ${\text{BC}}$ and strikes the mirror ${\text{RS}}$ at ${\text{C}}$ and again reflects back along ${\text{CD}}.$ Prove that ${\text{AB}}||{\text{CD}}$.

Ans: According to the question, given that PQ and RS are two mirrors that are parallel to each other.

As we need to prove that  in the given figure.

Now we draw lines BX and CY that are parallel to each other, to get

As we also know that according to the laws of reflection

$\angle ABX = \angle CBX$ and $\angle BCY = \angle DCY.$

$\angle BCY = \angle CBX$ (Alternate interior angles)

As we can determine that $\angle ABX = \angle CBX = \angle BCY = \angle DCY$ .

$\angle ABC = \angle ABX + \angle CBX$ , and

$\angle DCB = \angle BCY + \angle DCY.$

Hence, we can determine that $\angle ABC = \angle DCB$ .

According to the figure, we can determine that $\angle ABC$ and $\angle DCB$ form a pair of alternate interior angles corresponding to the lines AB and CD, and transversal BC.

Hence, we can determine that $\angle A B C=\angle D C B$.

5. In the given figure, if  SR, $\angle SQR = {28^\circ }$ and $\angle QRT = {65^\circ }$, then find the values of ${\text{x}}$ and ${\text{y}}$.

Ans: According to the question, given that  and $\angle QRT = {65^\circ }$ .

As we need to find the values of $x$ and $y$ in the figure.

As we know that "If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles."

$\angle SQR + \angle QSR = \angle QRT$ , or

${28^\circ } + \angle QSR = {65^\circ }$

$ \Rightarrow \angle QSR = {37^\circ }$

$x = \angle QSR = {37^\circ }$ (Alternate interior angles)

According to the figure, we can determine that $\Delta PQS$

$\angle PQS + \angle QSP + \angle QPS = {180^\circ }$ . (Angle sum property)

$\angle QPS = {90^\circ }\quad (PQ \bot PS)$

$x + y + {90^\circ } = {180^\circ }$

$ \Rightarrow x + {37^\circ } + {90^\circ } = {180^\circ }$

$ \Rightarrow x + {127^\circ } = {180^\circ }$

$ \Rightarrow x = {53^\circ }$

Hence, we can determine that $x = {53^\circ }$ and $y = {37^\circ }$ .

6. In the given figure, the side $QR$ of $\Delta $ PQR is produced to a point $S$. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point ${\text{T}}$, then prove that $\angle QTR = \dfrac{1}{2}\angle QPR$.

Ans: As we need to prove that $\angle QTR = \dfrac{1}{2}\angle QPR$ in the figure given below.

As we also know that "If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles."

According to the figure, we can determine that in $\Delta QTR,\angle TRS$ is an exterior angle

$\angle QTR + \angle TQR = \angle TRS$ , or 

$\angle QTR = \angle TRS - \angle TQR$     ……….(i)

$\angle QPR + \angle PQR = \angle PRS$

According to the question given that QT and RT are angle bisectors of $\angle PQR$ and $\angle PRS$ .

$\angle QPR + 2\angle TQR = 2\angle TRS$

$\angle QPR = 2(\angle TRS - \angle TQR)$

As we need to substitute the value of equation (i) in the above equation, to get

$\angle QPR = 2\angle QTR$ , or

$\angle QTR = \dfrac{1}{2}\angle QPR$

Hence, we can determine that the desired result is proved.

7. Prove that sum of three angles of a triangle is ${180^\circ }$

Ans: According to the question given that, $\vartriangle {\text{ABC}}$

To prove that, $\angle A + \angle B + \angle C = {180^\circ }$

Now we draw  through point A.

Proof: Because, 

So, $\angle 2 = \angle 4 \to (1)$

Because, Altemate interior angle

And $\angle 3 = \angle 5 \to (2)$

Now we adding the equation (1) and equation (2)

$\angle 2 + \angle 3 = \angle 4 + \angle 5$

Adding both sides $\angle 1$ ,

$\angle 1 + \angle 2 + \angle 3 = \angle 1 + \angle 4 + \angle 5$

$\angle 1 + \angle 2 + \angle 3 = {180^\circ }\,(Because,\,\,\angle 1,\angle 4$ , and $\angle 5$ forms a line)

$\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} = {180^\circ }$

8.  It is given that $\angle XYZ = {64^\circ }$ and ${\text{XY}}$ is produced to point ${\text{P}}$. Draw a figure from the given information. If ray YQ bisects $\angle ZYP$. Find $\angle XYQ$ and reflex $\angle QYP$.

Ans: According to the figure,

YQ bisects $\angle ZYP$

So, $\angle 1 = \angle 2$

$\angle 1 + \angle 2 + \angle {64^\circ } = {180^\circ }({\text{YX}}$ is a line)

$\angle 1 + \angle 1 + {64^\circ } = {180^\circ }$

$2\angle 1 = {180^\circ } - {64^\circ }$

$2\angle 1 = {116^\circ }$

$\angle 1 = {58^\circ }$

So, $\angle {\text{XYQ}} = {64^\circ } + {58^\circ } = {122^\circ }$

$\angle 2 + \angle {\text{XYQ}} = {180^\circ }$

$\angle 1 = \angle 2 = \angle QYP = {58^\circ }$ 

$\angle 2 + {122^\circ } = {180^\circ }$

$\angle 2 = {180^\circ } - {122^\circ }$

$\angle QYP = \angle 2 = {58^\circ }$

Reflex $\angle Q{\text{YP}} = {360^\circ } - \angle QYP$

Hence, the value of $\angle {\text{XYQ}}\,{\text{ = }}\,{\text{12}}{{\text{2}}^ \circ }$ and reflex $\angle {\text{QYP}} = \,{302^ \circ }$ .

9. In fig if  and $\angle {\text{RST}} = {130^\circ }$ find $\angle {\text{QRS}}$.

Ans: Through point R Draw line XY 

Because,$\text{PQ}\|\text{ST}$

$\text{ST}\|\text{KL,}\quad So,\,\text{PQ}\|\text{KL}$

Because, $\text{PQ}\|\text{KL}$

So, $\angle {\text{PQR}} + \angle 1 = {180^\circ }$

(As we know that the sum of interior angle on the same side of transversal is ${180^\circ }$ ) ${110^\circ } + \angle 1 = {180^\circ }$

$\angle 1 = {70^\circ }$

Similarly $\angle 2 + \angle {\text{RST}} = {180^\circ }$

$\angle 2 + {130^\circ } = {180^\circ }$

$\angle 2 = {50^\circ }$

$\angle 1 + \angle 2 + \angle 3 = {180^\circ }$

${70^\circ } + {50^\circ } + \angle 3 = {180^\circ }$

$\angle 3 = {180^\circ } - {120^\circ }$

$\angle 3 = {60^\circ }$

Hence, the value of $\angle {\text{QRS}} = {60^\circ }$ .

10. The side BC of $\vartriangle ABC$ is produced from ray $BD$. $CE$ is drawn parallel to $AB$, show that $\angle ACD = \angle A + \angle B$. Also prove that $\angle A + \angle B + \angle C = {180^\circ }$.

Ans: As we can see,  $\text{AB}\|\text{CE}$ and ${\text{AC}}$ intersect them

$\angle 1 = \angle 4$     ………. (1) (Alternate interior angles)

Also  and BD intersect them

$\angle 2 = \angle 5$     …………. (2) (Corresponding angles)

Now adding equation (1) and equation (2)

$\angle 1 + \angle 2 = \angle 4 + \angle 5$

$\angle A + \angle B = \angle ACD$

Adding $\angle C$ on both sides, we get

$\angle A + \angle B + \angle C = \angle C + \angle ACD$

Hence, proved.

11. Prove that if a transversal intersect two parallel lines, then each pair of alternate interior angles is equal.

Ans: According to the question given that, line  intersected by transversal ${\text{PQ}}$

To Prove: (i) $\angle 2 = \angle 5$ (ii) $\angle 3 = \angle 4$

Proof: 

$\angle 1 = \angle 2$     ………… (i) (Vertically Opposite angle)

$\angle 1 = \angle 5$     ………….. (ii) (Corresponding angles)

By equations (i) and (ii)

$\angle 2 = \angle 5$

Similarly, $\angle 3 = \angle 4$

Hence Proved.

12. In the given figure $\Delta {\text{ABC}}$ is right angled at $A$. $AD$ is drawn perpendicular to $BC$. Prove that $\angle BAD = \angle ACB$

${\text{AD}} \bot BC$

So, $\angle ADB = \angle ADC = {90^\circ }$

From $\vartriangle {\text{ABD}}$

$\angle {\text{ABD}} + \angle {\text{BAD}} + \angle {\text{ADB}} = {180^\circ }$

$\angle {\text{ABD}} + \angle {\text{BAD}} + {90^\circ } = {180^\circ }$

$\angle {\text{ABD}} + \angle {\text{BAD}} = {90^\circ }$

$\angle {\text{BAD}} = {90^\circ } - \angle ABD \to (1)$

But $\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} = {180^\circ }$ in $\vartriangle {\text{ABC}}$

$\angle {\text{B}} + \angle {\text{C}} = {90^\circ },\quad Because,\,\,\angle {\text{A}} = {90^\circ }$

$\angle {\text{C}} = {90^\circ } - \angle B \to \,(2)$

From equations (1) and (2)

$\angle {\text{BAD}} = \angle {\text{C}}$

$\angle {\text{BAD}} = \angle {\text{ACB}}$  

13. In $\Delta {\text{ABC}}\angle B = {45^\circ },\angle C = {55^\circ }$ and bisector $\angle A$ meets ${\text{BC}}$ at a point ${\text{D}}$. Find

$\angle ADB$ and $\angle ADC$

Ans: In $\vartriangle {\text{ABC}}$

$\angle {\text{A}} + \angle {\text{B}} + \angle {\text{C}} = {180^\circ }$ (As we know that the sum of three angle of a $\Delta $ is $\left. {{{180}^\circ }} \right)$

$ \Rightarrow \angle {\text{A}} + {45^\circ } + {55^\circ } = {180^\circ }$

$\angle {\text{A}} = {180^\circ } - {100^\circ } = {80^\circ }$

${\text{AD}}$ bisects $\angle {\text{A}}$

$\angle 1 = \angle 2 = \dfrac{1}{2}\angle A = \dfrac{1}{2} \times {80^\circ } = {40^\circ }$

Now in $\Delta {\text{ADB}}$ , 

We have, $\angle 1 + \angle {\text{B}} + \angle {\text{ADB}} = {180^\circ }$

$ \Rightarrow {40^\circ } + {45^\circ } + \angle ADB = {180^\circ }$

$ \Rightarrow \angle {\text{ADB}} = {180^\circ } - {85^\circ } = {95^\circ }$

$\angle {\text{ADB}} + \angle {\text{ADC}} = {180^\circ }$

Also ${95^\circ } + \angle ADC = {180^\circ }$

$\angle {\text{ADC}} = {180^\circ } - {95^\circ } = {85^\circ }$

Hence, the value of $\angle {\text{ADB}} = {95^\circ }$ and $\angle {\text{ADC}} = {85^\circ }$

14. In figure two straight lines $AB$ and $CD$ intersect at a point 0 . If $\angle BOD = {x^\circ }$ and $\angle AOD = {(4x - 5)^\circ }$. 

Find the value of ${\mathbf{x}}$ hence find

(a) $\angle AOD$

Ans: $\angle AOB = \angle AOD + \angle DOB$ By linear pair

${180^\circ } = 4x - 5 + x$

${180^\circ } + 5 = 5x$

$5{\text{x}} = 185$

${\text{x}} = \dfrac{{185}}{5} = {37^\circ }$

So, $\angle {\text{AOD}} = 4{\text{x}} - 5$

$ = 4 \times 37 - 5 = 148 - 5$

$ = {143^\circ }$

(b) $\angle BOC$

$\angle {\text{BOC}} = {143^\circ }$

Because, $\angle {\text{AOD}}$ and $\angle {\text{BOC}}$

$\angle {\text{BOD}} = {\text{x}} = {37^\circ }$ (Vertically opposite angles)

(c) $\angle BOC$

$\angle BOD = {37^\circ }$

(d) $\angle AOC$

$\angle AOC = {37^\circ }$

15. The side ${\text{BC}}$ of a $\Delta {\text{ABC}}$ is produced to ${\text{D}}$. The bisector of $\angle {\text{A}}$ meets ${\text{BC}}$ at ${\text{L}}$ as shown if fig. prove that $\angle {\text{ABC}} + \angle {\text{ACD}} = 2\angle {\text{ALC}}$

Ans: In $\Delta {\text{ABC}}$ we have

$\angle {\text{ACD}} = \angle {\text{B}} + \angle {\text{A}} \to (1)$ (Exterior angle property)

$ \Rightarrow \angle {\text{ACD}} = \angle {\text{B}} + 2\;{\text{L}}1$

$(So,\angle {\text{A}}$ is the bisector of $\angle {\text{A}} = 2\;{\text{L}}1)$

In $\Delta {\text{AB}}L$

$\angle {\text{ALC}} = \angle {\text{B}} + \angle {\text{BA}}L$ (Exterior angle property)

$\angle {\text{A}}L{\text{C}} = \angle {\text{B}} + \angle 1$

$ \Rightarrow 2\angle {\text{ALC}} = 2\angle {\text{B}} + 2\angle 1 \ldots (2)$

Subtracting equation (1) from equation (2)

$2\angle {\text{ALC}} - \angle {\text{ACD}} = \angle {\text{B}}$

$2\angle {\text{ALC}} = \angle {\text{B}} + \angle {\text{ACD}}$

$\angle {\text{ACD}} + \angle {\text{ABC}} = 2\angle {\text{ALC}}$

16. In fig PT is the bisector of $\angle QPR$ in $\Delta PQR$ and $PS \bot QR$, find the value of $x$

Ans: Sum of $\angle QPR + \angle Q + \angle R = {180^\circ }$ (According to the angle sum property of triangle)

$\angle QPR = {180^\circ } - {50^\circ } - {30^\circ } = {100^\circ }$

$\angle QPT = \dfrac{1}{2}\angle QPR$

$ = \dfrac{1}{2} \times {100^\circ } = {50^\circ }$

$\angle Q + \angle QPS = \angle PST$ (Exterior angle theorem)

$\angle QPS = {90^\circ } - \angle Q$

$ = {90^\circ } - {50^\circ } = {40^\circ }$

$x = \angle QPT - \angle QPS$

$ = {50^\circ } - {40^\circ } = {10^\circ }$

Hence, the value of $x\, = \,{10^ \circ }$ .

17. The sides $BA$ and $DC$ of a quadrilateral $ABCD$ are produced as shown in fig show that $\angle X + \angle Y = \angle a + \angle b$

Ans: In given fugure join $BD$

$\operatorname{In} \Delta ABD$

$\angle b = \angle ABD + \angle BDA$ (Exterior angle theorem) 

$\operatorname{In} \Delta CBD$

$\angle a = \angle CBD + \angle BDC$

$\angle a + \angle b = \angle CBD + \angle BDC + \angle ABD + \angle BDA$

$ = (\angle CBD + \angle ABD) + (\angle BDC + \angle BDA)$

$ = \angle x + \angle y$

$\angle a + \angle b = \angle x + \angle y$

18. In the ${\text{BO}}$ and ${\text{CO}}$ are Bisectors of $\angle {\text{B}}$ and $\angle {\text{C}}$ of $\Delta {\text{ABC}}$, show that $\angle {\text{BOC}} = {90^\circ } + \dfrac{1}{2}$$\angle {\text{A}}$

$\angle 1 = \dfrac{1}{2}\angle ABC$

And $\angle 2 = \dfrac{1}{2}\angle ACB$

So, $\angle 1 + \angle 2 = \dfrac{1}{2}(\angle ABC + \angle ACB)\,\,\,\,\,\,\,\,... \ldots (1)$

$\angle ABC + ACB + \angle A = {180^\circ }$

So, $\angle ABC + ACB = {180^\circ } - \angle A$

$\dfrac{1}{2}(\angle ABC + ACB) = {90^\circ } - \dfrac{1}{2}\angle A\,\,\,\,\,\,\,\,\,... \ldots .(2)$

From equation (1) and equation (2) we get

$\angle 1 + \angle 2 = {90^\circ } - \dfrac{1}{2}\angle A\,\,\,\,\,\,\,\,... \ldots ..(3)$

$\angle BOC + \angle 1 + \angle 2 = {180^\circ }$ (Angle of a)

Put the value of $\angle 1\, + \,\angle 2$ in the above equation,

$ = {180^\circ } - \left( {{{90}^\circ } - \dfrac{1}{2}\angle A} \right)$

$ = {90^\circ } + \dfrac{1}{2}\angle A$

19. In fig two straight lines PQ and RS intersect each other at o, if $\angle {\text{POT}} = {75^\circ }$ Find the values of a, b and c

Ans: PQ intersect RS at ${\text{O}}$

 So, $\angle QOS = \angle POR$ (vertically opposite angles) 

${\text{A}} = 4\;{\text{b         }}... \ldots .(1)$

$a + b + {75^\circ } = {180^\circ }\,(Because,\,\,POQ$ is a straight lines)

So, $a + b = {180^\circ } - {75^\circ }$

$ = {105^\circ }$

Using, equation (1) $4b + b = {105^\circ }$

$5b = {105^\circ }$

$b = \dfrac{{105}}{5} = {21^\circ }$

So, $a = 4b$

$a = 4 \times 21$

Again, $\angle QOR$ and $\angle QOS$ form a linear pair 

So, $a + 2c = {180^\circ }$

Using, equation (2)

${84^\circ } + 2c = {180^\circ }$

$2c = {180^\circ } - {84^\circ }$

$2c = {96^\circ }$

$c = \dfrac{{{{96}^\circ }}}{2} = {48^\circ }$

Hence, $a = {84^\circ },b = {21^\circ }$ and $c = {48^\circ }$

20. In figure ray OS stands on a line POQ, ray OR and ray OT are angle bisector of

$\angle POS$ and $\angle SOQ$ respectively. If $\angle POS = x$, find $\angle ROT$.

Ans: Ray OS stands on the line POQ

So, $\angle {\text{POS}} + \angle {\text{SOQ}} = {180^\circ }$

But $\angle {\text{POS}} = {\text{x}}$

So, ${\text{x}} + \angle {\text{SOQ}} = {180^\circ }$

$\angle {\text{SOQ}} = {180^\circ } - x$

Now ray OR bisects $\angle {\text{POS}}$ , 

Hence, $\angle {\text{ROS}} = \dfrac{1}{2} \times \angle POS = \dfrac{1}{2} \times x = \dfrac{x}{2}$

Similarly, $\angle {\text{SOT}} = \dfrac{1}{2} \times \angle SOQ = \dfrac{1}{2} \times \left( {{{180}^\circ } - X} \right) = {90^\circ } - \dfrac{x}{2}$

$\angle ROT = \angle ROS + \angle SOT = \dfrac{x}{2} + {90^\circ } - \dfrac{x}{2} = {90^\circ }$

Hence, the value of $\angle ROT\, = \,{90^ \circ }$ .

21. If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.

Ans: According to the question and figure given that, AD is transversal intersect two lines PQ and RS

We have to prove  PQ||RS 

Proof: BE bisects $\angle {\text{ABQ}}$  

$\angle \mathrm{EBQ}= \dfrac{1}{2}\angle ABQ \to (1)$

Similarity CG bisects $\angle {\text{BCS}}$  

So, $\angle 2 = \dfrac{1}{2}\angle BCS \to (2)$

But  and ${\text{AD}}$ is the transversal

So, $\quad \angle 1 = \angle 2$

So, $\dfrac{1}{2}\angle ABQ = \dfrac{1}{2}\angle BCS$ (By equations $(1)$ and $(2))$

$ \Rightarrow \angle {\text{ABQ}} = \angle {\text{BCS}}$ (Because corresponding angles are equal)

So, PQ||RS 

22. In figure the sides ${\text{QR}}$ of $\Delta PQR$ is produced to a point ${\text{S}}$. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point ${\text{T}}$. Then prove that $\angle QRT = \dfrac{1}{2}\angle QPR$

Ans:  Solution, In $\Delta {\text{PQR}}$

$\angle {\text{PRS}} = \angle {\text{Q}} + \angle {\text{P}}$   (By exterior angle theorem)

$\angle 4 + \angle 3 = \angle 2 + \angle 1 + \angle {\text{P}}$

$2\angle 3 = 2\angle 1 + \angle {\text{P}} \to (1)$

So, ${\text{QT}}$ and ${\text{RT}}$ are bisectors of $\angle {\text{Q}}$ and $\angle {\text{PRS}}$

In $\vartriangle {\text{QTR}}$ ,

  $\angle 3 = \angle 1 + \angle {\text{T}} \to $ (2) (By exterior angle theorem) 

By equations (1) and (2) we get 

$2(\angle 1 + \angle T) = 2\angle 1 + \angle {\text{P}}$

$2\angle 1 + 2\angle {\text{T}} = 2\angle 1 + \angle {\text{P}}$

$\angle {\text{T}} = \dfrac{1}{2}\angle P$

$\angle {\text{QTR}} = \dfrac{1}{2}\angle QPR$  

23. In figure PQ and RS are two mirror placed parallel to each other. An incident ray AB striker the mirror PQ at $B$, the reflected ray moves along the path BC and strike the mirror ${\text{RS}}$ at ${\text{C}}$ and again reflects back along ${\text{CD}}$. Prove that $\mathrm{AB} \| \mathrm{CD}$.

Ans: Solution, Draw $MB \bot PQ$ and $NC \bot RS$

$\angle 1 = \angle 2 \to (1)$ (Angle of incident) 

And $\angle 3 = \angle 4 \to (2)$ (is equal to angle of reflection)

Because, $\angle {\text{MBQ}} = \angle {\text{NCS}} = {90^\circ }$

So,  (By corresponding angle property)

Because, $\angle 2 = \angle 3 \to (3)$ (Alternate interior angle)

By equations $(1),(2)$ and (3) 

$\angle 1 = \angle 4$

$\angle 1 + \angle 2 = \angle 4 + \angle 3$

$ \Rightarrow \angle {\text{ABC}} = \angle {\text{BCD}}$

So,  (By alternate interior angles)

1. In fig the side AB and AC of $\vartriangle ABC$ are produced to point E and D respectively. If bisector BO And CO of $\angle {\text{CBE}}$ and $\angle {\text{BCD}}$ respectively meet at point ${\text{O}}$, then prove that $\angle {\text{BOC}} = {90^\circ } - \dfrac{1}{2}\angle {\text{BAC}}$

Ans: Ray BO bisects $\angle {\text{CBE}}$

So, $\angle {\text{CBO}} = \dfrac{1}{2}\angle {\text{CBE}}$

$ = \dfrac{1}{2}\left( {{{180}^\circ } - y} \right)\,\,\,\left( {Because,\,\,\angle {\text{CBE}} + {\text{y}} = {{180}^\circ }} \right)$

$ = {90^\circ } - \dfrac{y}{2} \ldots  \ldots ..$ (1)

Similarly, ray CO bisects $\angle BCD$

$\angle {\text{BCO}} = \dfrac{1}{2}\angle BCD$

$ = \dfrac{1}{2}\left( {{{180}^\circ } - Z} \right)$

$ = {90^\circ } - \dfrac{Z}{2} \ldots  \ldots  \ldots ..$ (2)

In $\vartriangle {\text{BOC}}$

$\angle {\text{BOC}} + \angle {\text{BCO}} + \angle {\text{CBO}} = {180^\circ }$

$\angle {\text{BOC}} = \dfrac{1}{2}(y + z)$

But $x + y + z = {180^\circ }$

$y + z = {180^\circ } - x$

$\angle {\text{BOC}} = \dfrac{1}{2}\left( {{{180}^\circ } - x} \right) = {90^\circ } - \dfrac{x}{2}$

$\angle {\text{BOC}} = {90^\circ } - \dfrac{1}{2}\angle {\text{BAC}}$

2. In given fig. AB  CD. Determine $\angle a$.

Ans: Through O draw a line $l$ parallel to both ${\text{AB}}$ and ${\text{CD}}$

$\angle a = \angle 1 + \angle 2$

$\angle 1 = {38^\circ }$

$\angle 2 = {55^\circ }$ (Alternate interior angles)

$\angle a = {55^\circ } + {38^\circ }$

Hence, the value of $\angle a = {93^\circ }$ .

3. In fig ${\text{M}}$ and ${\text{N}}$ are two plane mirrors perpendicular to each other. Prove that the incident ray CA is parallel to reflected ray BD.

Ans: From the figure, it can be seen that ${\text{AP}} \bot {\text{M}}$ and ${\text{BQ}} \bot {\text{N}}$

So, $BQ \bot N$ and $AP \bot M$ and ${\text{M}} \bot {\text{N}}$

So, $\angle {\text{BOA}} = {90^\circ }$

$ \Rightarrow {\text{BQ}} \bot {\text{AP}}$

In $\vartriangle {\text{BOA}}\angle 2 + \angle 3 + \angle {\text{BOA}} = {180^\circ }$ (By angle sum property)

$ \Rightarrow \angle 2 + \angle 3 + {90^\circ } = {180^\circ }$

So, $\angle 2 + \angle 3 = {90^\circ }$

Also $\angle 1 = \angle 2$ and $\angle 4 = \angle 3$

$ \Rightarrow \angle 1 + \angle 4 = \angle 2 + \angle 3 = {90^\circ }$

So, $(\angle 1 + \angle 4) + (\angle 2 + \angle 3) = {90^\circ } + {90^\circ } = {180^\circ }$

$ \Rightarrow (\angle 1 + \angle 2) + (\angle 3 + \angle 4) = {180^\circ }$

or $\angle {\text{CAB}} + \angle {\text{DBA}} = {180^\circ }$

So,  (By sum of interior angles of same side of transversal)

CBSE Class 9 Maths Chapter-6 Important Questions - Free PDF Download

Maths is considered a tough subject but it can be an extremely high-scoring subject just like other subjects. All it needs is a practice of the concepts learned in the chapter. Unless students understand the practical application of the theoretical knowledge, they will not understand the applicability of the mathematical concepts. This can be achieved by practising as many questions based on a particular concept as possible.

By a careful analysis of the past years’ papers, the updated curriculum, and the CBSE guidelines, experts at Vedantu have curated a list of all the important questions for class 9 maths lines and angles. By practising these questions students can gain confidence about their knowledge in the topic.

Topics Covered by Lines and Angles Class 9 Important Questions

The chapter covers basic terms and definitions like a line segment is a part of a line with two endpoints, part of a line with one endpoint is called ray,  three or more points on the same line are called collinear points, when two rays emerge from the same point it is called an angle, a vertex is a point from which the rays emerge and the rays that make the angle are called the arms of the angle. 

When students practice these questions, they understand the format of answering the different types of questions. They get familiar with the exam pattern as several times questions in the exams are based on the same pattern as these questions. The questions include both short answer type and long answer type questions. 

Other Important Topics Covered are:

Types of angles- Acute angle, obtuse angle, right angle, straight angle, reflex angle, complementary angles, supplementary angles, adjacent angles

Intersecting lines and non-interesting lines- If two lines intersect each other then the angle formed vertically opposite to each other are equal

Pair of angles 

Parallel lines and a transversal- 4 theorems

Lines parallel to the same line- If two lines are parallel to the same line then they are also parallel to each other

Angle sum property of an angle

Students must try to solve all the questions from the lines and angles class 9 important questions as they cover all the important topics that can be missed out by them during their revision. The solutions have been prepared in an easy to understand format so that students can understand the concept well. It helps the students in solving questions of all difficulty levels. When the basics of these concepts are clear to the students they can understand the detailed concepts of geometry in class 10 with ease. 

Class 9 Maths Chapter 6 Extra Questions

If line AB and CD intersect at a point X such that ACX = 40°, CAX = 95°, and XDB   = 75°. Find DBX .

Prove that the sum of the angle of a triangle is 180°. 

Prove that two lines are parallel if a transversal line intersects two lines such that the bisectors of a pair of corresponding angles are parallel.

Can all the angles of a triangle be less than 60°? Support your answer with a reason.

Can a triangle have two obtuse angles? Support your answer with a reason.

The sum of the two angles of a triangle is 120° and their difference is 20°. Find the value of all the three angles of a triangle.

If the exterior angle of a triangle is 110 and the interior angle of a triangle is 35°. Find the values of the other two angles of a triangle.

Prove that lines which are parallel to the same line are parallel to each other.

Prove that when two lines intersect each other, then the vertically opposite angles that are formed are equal.

What is the measure of an acute angle, an obtuse angle, a right angle, a reflex angle, and a complementary angle?

Benefits of Lines and Angles Class 9 Important Questions

Math is a subject that requires regular practice. Lines and angles class 9 important questions provide a question bank to the students on class 9 maths chapter 6

The important questions have been compiled along with their step by step detailed answers to help the students know their mistakes if any

These questions cover all the essential topics from the chapter lines and angles and help the students in their revision

Students can easily download important questions for class 9 maths lines and angles in a pdf format for referring to it any time. 

With the help of the questions, students will get an idea of the pattern of the question they might get in the exams.

Revision can be done not only for the exams but also for the class test

Students can also take the help of these important questions for completing their assignments

These important questions provide 100 percent accuracy in the solutions and help the students in deriving their answers with ease.

We hope students have found this information on CBSE Important Questions for Class 9 Maths Chapter 6 important questions useful for their studies. Along with important questions, students can also access CBSE Class 9 Maths Chapter 6 NCERT Solutions,  CBSE Class 9  Maths Chapter 6 Revision notes , and other related CBSE Class 9 Maths study material. Keep learning and stay tuned with Vedantu for further updates on CBSE Class 9 exams.

Conclusion 

CBSE Class 9 Maths Important Questions for Chapter 6 - Lines and Angles serve as a valuable tool for students seeking to excel in their mathematics examinations. These questions are strategically curated to encapsulate the key concepts and theorems presented in the chapter. By working through them, students can reinforce their understanding of lines, angles, and their properties. These questions not only aid in exam preparation but also foster critical thinking and problem-solving skills. As students tackle these questions, they gain confidence in their geometric knowledge, ultimately paving the way for success in their Class 9 mathematics examinations.

FAQs on CBSE Class 9 Maths Important Questions for Chapter 6 - Lines and Angles

Q1. Why is Practising Extra Questions for CBSE Class 9 Maths Chapter 6 Lines and Angles important for Students?

Ans: Many reputed online learning platforms prepare a repository of important questions for exam preparation. At Vedantu. Students can find important questions for Class 9 Maths Chapter 6 Lines and Angles. Practising these questions help students in scoring well in the subject. Solving extra questions for any chapter is a great way to boost students’ confidence. By solving important questions for Class 9 Maths Chapter 6 Lines and Angles, students will be able to practice the chapter properly during exam time. These questions will also help in revision.

Q2. Where can I find Important Questions for Maths Chapter 6 Lines and Angles?

Ans: Vedantu is India’s leading online learning platform which provides a well-prepared set of Important Questions for Class 9 Maths Chapter 6. Students can find extra questions for other chapters also on Vedantu. The site offers exceptional exam materials and relevant questions that can be asked in the exams. It is available as a free PDF of Important Questions for Class 9 Maths Chapter 6 Lines and Angles with solutions. The solutions to these questions are also provided by subject experts. These are beneficial in exam preparation and revision during exams. 

Q3. What is the angle Sum Property?

Ans: As per the Angle Sum Property, the sum of interior angles of a triangle is equal to 180°. It is an extremely important theorem of maths that helps in many calculations related to a triangle. Students must know how to derive this theorem as it might be asked in the exam.

Q4. Does Vedantu Offer Solutions to the important Questions for Class 9 Maths Chapter 6?

Ans: Yes, of course! At Vedantu, students can find complete solutions for all the questions included in the PDF of important questions for Class 9 Maths Chapter 6. These solutions are prepared by subject matter experts who are well versed in the syllabus and exam guidelines. 100 percent accuracy is maintained in the solutions. 

Q5. Where can I find these important questions?

Ans: CBSE Class 9 Maths Important Questions for Chapter 6 can often be found in study guides, online educational platforms, or by performing a quick internet search. They are designed to aid your revision and test your understanding of the chapter.

Q6. Are these questions sufficient for exam preparation, or should I also study the entire chapter thoroughly?

Ans: While important questions are valuable for focused revision, it's essential to have a strong grasp of the entire chapter's concepts. These questions should complement your overall study plan, not replace it.

Q7. Do these important questions include solutions or answers?

Ans: The availability of solutions may vary depending on the source. Some sets of important questions provide solutions or answers, while others may not. It's a good idea to check if solutions are included when using these questions for practice.

CBSE Class 9 Maths Important Questions

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CBSE Class 9 Maths 30 Most Important Case Study Questions with Answers

Cbse class 9 maths 30 most important case study questions with answers download here free in pdf format..

CBSE Class 9 Maths exam 2023 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 webpage 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. Candidates can easily download these questions in PDF format and refer to them for exam preparation 2023.

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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 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 are part of NCERT Solutions for Class 9 Maths . Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (रेखाएँ और कोण) (Hindi Medium) Ex 6.1

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

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 6 Lines and Angles Ex 6.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1, drop a comment below and we will get back to you at the earliest.

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CBSE Class 9 Mathematics Case Study Questions

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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)

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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.

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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

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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.

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Case Study Questions for Class 9 Maths

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Are you preparing for your Class 9 Maths board exams and looking for an effective study resource? Well, you’re in luck! In this article, we will provide you with a collection of Case Study Questions for Class 9 Maths specifically designed to help you excel in your exams. These questions are carefully curated to cover various mathematical concepts and problem-solving techniques. So, let’s dive in and explore these valuable resources that will enhance your preparation and boost your confidence.

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

CBSE Class 9 Maths Board Exam will have a set of questions based on case studies in the form of MCQs. The CBSE Class 9 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out.  CBSE Case Study Questions for Class 9  will provide you with detailed, latest, comprehensive & confidence-inspiring solutions to the maximum number of Case Study Questions covering all the topics from your  NCERT Text Books !

Table of Contents

CBSE Class 9th – MATHS: Chapterwise Case Study Question & Solution

Case study questions are a form of examination where students are presented with real-life scenarios that require the application of mathematical concepts to arrive at a solution. These questions are designed to assess students’ problem-solving abilities, critical thinking skills, and understanding of mathematical concepts in practical contexts.

Chapterwise Case Study Questions for Class 9 Maths

Case study questions play a crucial role in the field of mathematics education. They provide students with an opportunity to apply theoretical knowledge to real-world situations, thereby enhancing their comprehension of mathematical concepts. By engaging with case study questions, students develop the ability to analyze complex problems, make connections between different mathematical concepts, and formulate effective problem-solving strategies.

• 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 Quadilaterals
• 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

The above  Case studies for Class 9 Mathematics will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Studies have been developed by experienced teachers of schools.studyrate.in for benefit of Class 10 students.

• Class 9 Science Case Study Questions
• Class 9 Social Science Case Study Questions

How to Approach Case Study Questions

When tackling case study questions, it is essential to adopt a systematic approach. Here are some steps to help you approach and solve these types of questions effectively:

• Read the case study carefully: Understand the given scenario and identify the key information.
• Identify the mathematical concepts involved: Determine the relevant mathematical concepts and formulas applicable to the problem.
• Formulate a plan: Devise a plan or strategy to solve the problem based on the given information and mathematical concepts.
• Solve the problem step by step: Apply the chosen approach and perform calculations or manipulations to arrive at the solution.
• Verify and interpret the results: Ensure the solution aligns with the initial problem and interpret the findings in the context of the case study.

Tips for Solving Case Study Questions

Here are some valuable tips to help you effectively solve case study questions:

• Read the question thoroughly and underline or highlight important information.
• Break down the problem into smaller, manageable parts.
• Visualize the problem using diagrams or charts if applicable.
• Use appropriate mathematical formulas and concepts to solve the problem.
• Show all the steps of your calculations to ensure clarity.
• Check your final answer and review the solution for accuracy and relevance to the case study.

Benefits of Practicing Case Study Questions

Practicing case study questions offers several benefits that can significantly contribute to your mathematical proficiency:

• Enhances critical thinking skills
• Improves problem-solving abilities
• Deepens understanding of mathematical concepts
• Develops analytical reasoning
• Prepares you for real-life applications of mathematics
• Boosts confidence in approaching complex mathematical problems

Case study questions offer a unique opportunity to apply mathematical knowledge in practical scenarios. By practicing these questions, you can enhance your problem-solving abilities, develop a deeper understanding of mathematical concepts, and boost your confidence for the Class 9 Maths board exams. Remember to approach each question systematically, apply the relevant concepts, and review your solutions for accuracy. Access the PDF resource provided to access a wealth of case study questions and further elevate your preparation.

Q1: Can case study questions help me score better in my Class 9 Maths exams?

Yes, practicing case study questions can significantly improve your problem-solving skills and boost your performance in exams. These questions offer a practical approach to understanding mathematical concepts and their real-life applications.

Q2: Are the case study questions in the PDF resource relevant to the Class 9 Maths syllabus?

Absolutely! The PDF resource contains case study questions that align with the Class 9 Maths syllabus. They cover various topics and concepts included in the curriculum, ensuring comprehensive preparation.

Q3: Are the solutions provided for the case study questions in the PDF resource?

Yes, the PDF resource includes solutions for each case study question. You can refer to these solutions to validate your answers and gain a better understanding of the problem-solving process.

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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

Ncert solutions for class 9 maths chapter 6 lines and angles| pdf download.

• Exercise 6.1 Chapter 6 Class 9 Maths NCERT Solutions
• Exercise 6.2 Chapter 6 Class 9 Maths NCERT Solutions
• Exercise 6.3 Chapter 6 Class 9 Maths NCERT Solutions

NCERT Solutions for Class 9 Maths Chapters:

How many exercises in Chapter 6 Lines and Angles?

If the supplement of an angle is 4 times of its complement, find the angle., what is the measure of an angle whose measure is 32° less than its supplement, angles ∠p and 100° form a linear pair. what is the measure of ∠p, contact form.

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• Triangle and its Properties Class 7 Case Study Questions Maths Chapter 6

Last Updated on August 19, 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 6 Triangle and its Properties. It is a part of Case Study Questions for CBSE Class 7 Maths Series.

Table of Contents

Case Study Questions on Triangle and its Properties

In a triangle first angle is 10° less than twice of second angle and third angle is 30° more than the second angle.

Q. 1. Find the sum of all first, second and third angle of triangle: (a) 180° (b) 70° (c) 40° (d) None of these

Difficulty Level:  Easy

Ans. Option (a) is correct. Explanation: We know that sum of all angles of triangle is 180°.

Q. 2. Which angle is smaller: (a) First Angle (b) Second Angle (c) Third Angle (d) Can’t Say

Ans. Option (b) is correct. Explanation: Clearly given first and third both angles are greater than second angle (point to remember sum of any two sides of triangle is always greater than the third side)

Q. 3. Find the second angle: (a) 70° (b) 40° (c) 90° (d) 50°

Difficulty Level:  Difficult

Ans. Option (b) is correct. Explanation: let the second angle be x, then according to the question first angle = 2x – 10° third angle = second angle + 30° third angle = x + 30° as we know sum of all angles of triangle is 180°, therefore first angle + second angle + third angle = 180° (2x – 10) + (x) + (x + 30) = 180 4x + 20 = 180 4x = 180 – 20 4x = 160 x = 40° hence required second angle = 40°

Q. 4. Find the first angle:

Ans. 70° Explanation: let the second angle be x, then according to the question first angle = 2x – 10° third angle = second angle + 30° third angle = x + 30° as we know sum of all angles of triangle is 180°, therefore first angle + second angle + third angle = 180° (2x – 10) + (x) + (x + 30) = 180 4x + 20 = 180 4x = 180 – 20 4x = 160 x = 160/4 x = 40° first angle = 2x – 10 = 2(40) – 10 = 80 – 10 = 70°

Q. 5. Write the sum of supplementary angle of first angle and complementary angle of second angle:

Difficulty Level:  Medium

Ans. 160° Explanation: Here, supplementary angle = 180° – (first angle) = 180 – 70 = 110° complementary angle = 90° – (second angle) = 90 – 40 = 50° Required Sum = 110 + 50 = 160°

• Comparing Quantities Class 7 Case Study Questions Maths Chapter 7
• Lines and Angles Class 7 Case Study Questions Maths Chapter 5
• Simple Equations Class 7 Case Study Questions Maths Chapter 4
• 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.

• Complementary Angles
• Supplementary Angles
• Intersecting Lines/Transversal
• Angles Made by a Transversal
• Transversal of Parallel Lines
• Medians of a triangle
• Altitudes of a triangle
• Exterior angle of a triangle
• Angle sum property of a triangle
• Sum of lengths of two sides of a triangle
• Pythagoras theorem

A triangle is a 3-sided closed figure formed by three line segments. The three sides and three angles of a triangle are the six elements of the triangle.

On the basis of its sides, a triangle is equilateral, if all its 3 sides are equal; isosceles, if two of its sides are equal; scalene, if all sides are unequal.

On the basis of the measure of its angles, a triangle is acute-angled, if all 3 angles are acute; right angled, if one of its angle is a right angle; obtuse angled, If one of its angles is obtuse.

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

A triangle has 3 altitudes. In right angled triangle, two altitudes are two of its sides. Intersection point of three altitudes is called ORTHOCENTRE.

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Frequently Asked Questions (FAQs) on Triangle and its Properties Case Study

Q1: what is a triangle, and what are its basic properties.

A1: A triangle is a three-sided polygon with three angles. The sum of the angles in any triangle is always 180°. Triangles are classified based on their sides (scalene, isosceles, equilateral) and angles (acute, obtuse, right-angled).

Q2: What are the different types of triangles based on their sides?

A2: Triangles can be classified based on their sides as follows: Scalene Triangle: All three sides are of different lengths. Isosceles Triangle: Two sides are of equal length, and the third side is different. Equilateral Triangle: All three sides are of equal length.

Q3: What are the different types of triangles based on their angles?

A3: Triangles can be classified based on their angles as follows: Acute Triangle: All three angles are less than 90°. Right-Angled Triangle: One of the angles is exactly 90°. Obtuse Triangle: One of the angles is greater than 90°.

Q4: What is the Pythagoras theorem, and how is it related to triangles?

A4: The Pythagoras theorem applies to right-angled triangles and states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Q5: What is the significance of the triangle inequality property?

A5: The triangle inequality property states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This property is essential in determining whether a given set of three lengths can form a triangle.

Q6: What is an exterior angle of a triangle, and what is its property?

A6: An exterior angle of a triangle is an angle formed by one side of the triangle and the extension of an adjacent side. The exterior angle property states that the measure of an exterior angle is equal to the sum of the two opposite interior angles.

Q7: What is the median of a triangle?

A7: A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side. Every triangle has three medians, and they all intersect at a single point called the centroid, which is the triangle’s center of mass.

Q8: What is an altitude of a triangle?

A8: An altitude of a triangle is a perpendicular segment drawn from a vertex to the line containing the opposite side (or its extension). A triangle has three altitudes, and they intersect at a point called the orthocenter.

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

• NCERT Solutions
• NCERT Class 9
• NCERT 9 Maths
• Chapter 6: Lines And Angles
• Exercise 6.2

NCERT Solutions for Class 9 Maths Chapter 6 - Lines And Angles Exercise 6.2

There is no doubt that NCERT textbooks are the most-trusted study material for students. The Class 9 NCERT Maths book is designed in such a way that students can easily understand the concepts and then be able to solve the problems related to them. The NCERT Class 9 Maths Solutions consists of 15 chapters. Chapter 6 deals with Lines and Angles. Here, we have provided the NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles Exercise 6.2.

NCERT Solutions for Class 9 Maths Chapter 6 – Lines And Angles Exercise 6.2

Access answers to ncert solutions for class 9 maths chapter 6 – lines and angles exercise 6.2.

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Access other Exercise Solutions of Class 9 Maths Chapter 6 – Lines and Angles

Exercise 6.1 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Exercise 6.3 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

1. In Fig. 6.28, find the values of x and y and then show that AB || CD.

We know that a linear pair is equal to 180°.

So, x+50° = 180°

We also know that vertically opposite angles are equal.

So, y = 130°

In two parallel lines, the alternate interior angles are equal. In this,

x = y = 130°

This proves that alternate interior angles are equal and so, AB || CD.

2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

It is known that AB || CD and CD||EF

As the angles on the same side of a transversal line sum up to 180°,

x + y = 180° —–(i)

∠O = z (Since they are corresponding angles)

and, y +∠O = 180° (Since they are a linear pair)

So, y+z = 180°

Now, let y = 3w and hence, z = 7w (As y : z = 3 : 7)

∴ 3w+7w = 180°

Or, 10 w = 180°

So, w = 18°

Now, y = 3×18° = 54°

and, z = 7×18° = 126°

Now, angle x can be calculated from equation (i)

Or, x+54° = 180°

3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Since AB || CD, GE is a transversal.

It is given that ∠GED = 126°

So, ∠GED = ∠AGE = 126° (as they are alternate interior angles)

∠GED = ∠GEF +∠FED

As EF⊥ CD, ∠FED = 90°

∴ ∠GED = ∠GEF+90°

Or, ∠GEF = 126° – 90° = 36°

Again, ∠FGE +∠GED = 180° (transversal)

Putting the value of ∠GED = 126° we get,

∠AGE = 126°

∠GEF = 36° and

4. In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

[Hint : Draw a line parallel to ST through point R.]

First, construct a line XY parallel to PQ.

We know that the angles on the same side of a transversal is equal to 180°.

So, ∠PQR+∠QRX = 180°

Or, ∠QRX = 180°-110°

∴ ∠QRX = 70°

∠RST +∠SRY = 180°

Or, ∠SRY = 180°- 130°

∴ ∠SRY = 50°

Now, for the linear pairs on the line XY,

∠QRX+∠QRS+∠SRY = 180°

Putting their respective values, we get,

∠QRS = 180° – 70° – 50°

Hence, ∠QRS = 60°

5. In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

From the diagram,

∠APQ = ∠PQR (alternate interior angles)

Now, putting the value of ∠APQ = 50° and ∠PQR = x we get,

∠APR = ∠PRD (alternate interior angles)

Or, ∠APR = 127° (as it is given that ∠PRD = 127°)

We know that

∠APR = ∠APQ+∠QPR

Now, putting values of ∠QPR = y and ∠APR = 127° we get,

127° = 50°+ y

Or, y = 77°

Thus, the values of x and y are calculated as:

x = 50° and y = 77°

6. In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

First, draw two lines BE and CF such that BE ⊥ PQ and CF ⊥ RS.

Now, since PQ || RS,

So, BE || CF

We know that,

Angle of incidence = Angle of reflection (By the law of reflection)

∠1 = ∠2 and

We also know that alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at B and C

So, ∠2 = ∠3 (As they are alternate interior angles)

Now, ∠1 +∠2 = ∠3 +∠4

Or, ∠ABC = ∠DCB

So, AB || CD (alternate interior angles are equal)

The concepts learned in the exercise will help in solving the problems of Exercise 6.2. So, first and foremost, the students should know the concepts and theorems thoroughly. After that, they should start solving the problems in the exercises. So, in case the students have any doubts, they must refer to the NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles Exercise 6.2.

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