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Case Study Questions for Class 9 Maths Chapter 6 Lines and Angles

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Here we are providing case study questions for Class 9 Maths Chapter 6 Lines and Angles. Students are suggested to solve the questions by themselves first and then check the answers. This will help students to check their grasp on this particular chapter Triangles.

Case Study Questions:

Questions 1:

ΔABC is an isosceles triangle in which ∠B = ∠C and LM | | BC. If ∠A = 50º.

case study questions maths class 9 lines and angles

(i) The quadrilateral LMCB is (A) Trapezium (B) Square (C) rectangle (D) rhombus

(ii) The value of ∠LMC is: (A) 65º (B) 115º (C) 130º (D) 100º

(iii) The value of ∠ALM is: (A) 130º (B) 80º (C) 65º (D) 100º

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myCBSEguide

Mathematics
CBSE Class 9 Mathematics...

CBSE Class 9 Mathematics Case Study Questions

Table of Contents

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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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14 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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CBSE Maths Important Questions
Class 9 Maths
Chapter 6: Lines Angles

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

CBSE Class 9 Maths Chapter 6 (Lines and Angles) Important Questions with solutions are given here, which can be easily accessed. These questions have been prepared by our experts for students of standard 9 to make them prepare for final exam (2022 – 2023) . All the questions are based on CBSE syllabus and taken in reference from NCERT book. Students can do their revision by practising the Important Questions for CBSE Class 9 Maths Chapter 6: Lines and Angles here and score good marks.

To revise the important questions chapter-wise for 9th Maths , reach us at BYJU’S. The chapter lines and angles will consist of topics such as angles formed after the intersection of two lines, linear pair of angles, complementary angles, etc. Let us solve the questions here to understand all the concepts.

Also Check:

Important 2 Marks Questions for CBSE 9th Maths
Important 3 Marks Questions for CBSE 9th Maths
Important 4 Marks Questions for CBSE 9th Maths

Important Questions & Solutions For Class 9 Chapter 6 (Lines and Angles)

Q.1: In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

From the given figure, we can see;

∠AOC, ∠BOE, ∠COE and ∠COE, ∠BOD, ∠BOE form a straight line each.

So, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD + ∠BOE = 180°

Now, by substituting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get:

70° +∠COE = 180°

∠COE = 110°

110° + 40° + ∠BOE = 180°

Q.2: In the Figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

As we know, the sum of the linear pair is always equal to 180°

∠POY + a + b = 180°

Substituting the value of ∠POY = 90° (as given in the question) we get,

a + b = 90°

Now, it is given that a : b = 2 : 3 so,

Let a be 2x and b be 3x.

∴ 2x + 3x = 90°

Solving this we get

So, x = 18°

∴ a = 2 × 18° = 36°

Similarly, b can be calculated and the value will be

b = 3 × 18° = 54°

From the diagram, b + c also forms a straight angle so,

b + c = 180°

=> c + 54° = 180°

Q.3: In the Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).

In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180°

So, ∠POS + ∠ROS + ∠ROQ = 180° (Linear pair of angles)

Now, ∠POS + ∠ROS = 180° – 90° (Since ∠POR = ∠ROQ = 90°)

∴ ∠POS + ∠ROS = 90°

Now, ∠QOS = ∠ROQ + ∠ROS

It is given that ∠ROQ = 90°,

∴ ∠QOS = 90° + ∠ROS

Or, ∠QOS – ∠ROS = 90°

As ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we get

∠POS + ∠ROS = ∠QOS – ∠ROS

=>2 ∠ROS + ∠POS = ∠QOS

Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).

Q.4: It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Here, XP is a straight line

So, ∠XYZ +∠ZYP = 180°

substituting the value of ∠XYZ = 64° we get,

64° +∠ZYP = 180°

∴ ∠ZYP = 116°

From the diagram, we also know that ∠ZYP = ∠ZYQ + ∠QYP

Now, as YQ bisects ∠ZYP,

∠ZYQ = ∠QYP

Or, ∠ZYP = 2∠ZYQ

∴ ∠ZYQ = ∠QYP = 58°

Again, ∠XYQ = ∠XYZ + ∠ZYQ

By substituting the value of ∠XYZ = 64° and ∠ZYQ = 58° we get.

∠XYQ = 64° + 58°

Or, ∠XYQ = 122°

Now, reflex ∠QYP = 180° + ∠XYQ

We computed that the value of ∠XYQ = 122°. So,

∠QYP = 180° + 122°

∴ ∠QYP = 302°

Q.5: In the Figure, 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° (alternate interior angles)

∠GED = ∠GEF + ∠FED

EF ⊥ CD, ∠FED = 90°

∴ ∠GED = ∠GEF + 90°

Or, ∠GEF = 126° – 90° = 36°

Again, ∠FGE + ∠GED = 180° (Transversal)

Substituting the value of ∠GED = 126° we get,

∠AGE = 126°

∠GEF = 36° and

Q.6: In the Figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

First, construct a line XY parallel to PQ.

Class 9 Maths Chapter 6 Important Question 6 Solution

As we know, the angles on the same side of the transversal are 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°

Substituting their respective values we get,

∠QRS = 180° – 70° – 50°

Or, ∠QRS = 60°

Q.7: 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

BE and CF are normals between the incident ray and reflected ray.

As we know,

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)

Q.8: In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.

As we know, the sum of the interior angles of the triangle is 180°.

So, ∠X +∠XYZ + ∠XZY = 180°

substituting the values as given in the question we get,

62° + 54° + ∠XZY = 180°

Or, ∠XZY = 64°

Now, As we know, ZO is the bisector so,

∠OZY = ½ ∠XZY

∴ ∠OZY = 32°

Similarly, YO is a bisector and so,

∠OYZ = ½ ∠XYZ

Or, ∠OYZ = 27° (As ∠XYZ = 54°)

Now, as the sum of the interior angles of the triangle,

∠OZY +∠OYZ + ∠O = 180°

∠O = 180° – 32° – 27°

Or, ∠O = 121°

Q.9: In the figure, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then find ∠QRS.

According to the given figure, we have

AB || CD || EF

If a transversal intersects two parallel lines, then each pair of alternate exterior angles is equal.

Now, since, PQ || RS

⇒ ∠PQC = ∠BRS

We have ∠PQC = 60°

⇒ ∠BRS = 60° … eq.(i)

We also know that,

If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Now again, since, AB || CD

⇒ ∠DQR = ∠QRA

We have ∠DQR = 25°

⇒ ∠QRA = 25° … eq.(ii)

Using linear pair axiom,

∠ARS + ∠BRS = 180°

⇒ ∠ARS = 180° – ∠BRS

⇒ ∠ARS = 180° – 60° (From (i), ∠BRS = 60°)

⇒ ∠ARS = 120° … eq.(iii)

Now, ∠QRS = ∠QRA + ∠ARS

From equations (ii) and (iii), we have,

∠QRA = 25° and ∠ARS = 120°

Hence, the above equation can be written as:

∠QRS = 25° + 120°

⇒ ∠QRS = 145°

Video Lesson on Constructing Angles

Class 9 Maths Chapter 6 Extra Questions

If two lines intersect, prove that the vertically opposite angles are equal.
Bisectors of interior ∠B and exterior ∠ACD of a Δ ABC intersect at the point T.Prove that ∠ BTC = ½ ∠ BAC.
A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.
In the figure, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that the points A, O and B are collinear.

5. The angles of a triangle are in the ratio 5 : 3: 7. The triangle is

An acute-angled triangle
An obtuse-angled triangle
A right triangle
An isosceles triangle

6. Can a triangle have all angles less than 60°? Give a reason for your answer.

7. Can a triangle have two obtuse angles? Give the reason for your answer.

8. How many triangles can be drawn having its angles as 45°, 64° and 72°? Give the reason for your answer.

9. How many triangles can be drawn having its angles as 53°, 64° and 63°? Give the reason for your answer.

10. Two distinct points in the plane determine a _________________ line.

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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.

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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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Chapter 6 Class 9 Lines and Angles

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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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CBSE Important Questions Class 9 Maths Chapter 6

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Important Questions Class 9 Mathematics Chapter 6 – Lines and Angles

Class 9 Mathematics Chapter 6, Lines And Angles, exposes you to fundamental geometry with a particular emphasis on the characteristics of the angles created when two lines cross one another and when a line intersects two or more parallel lines at different points.

Quick Links

You can prepare for the upcoming board exams and improve your grade in class by using the Chapter 6 Class 9 Mathematics important questions. Extramarks has concentrated on getting you ready for the Class 9 exam using the CBSE curriculum. Your mathematical knowledge will be sharpened by solving these important questions Class 9 Mathematics Chapter 6 , which will also help you grasp the topic better.

These important questions Class 9 Mathematics Chapter 6 provide a sample of the kinds of questions that are frequently asked in board exams. Learning about these can also give you more assurance as you take the examinations. Students can practice all types of questions from the chapters with the aid of these important questions Class 9 Mathematics chapter 6 .

Extramarks experts have created the important questions Class 9 Mathematics Chapter 6 in a well-structured format to offer a variety of potential approaches to solving problems and guarantee a thorough comprehension of the concepts. For their exams, it is advised that the students thoroughly practice all of these solutions. Additionally, it will assist students in laying the groundwork for more challenging courses.

Students are given other online learning resources, like revision notes, sample papers, and previous years question papers in addition to the NCERT Solutions, which are accessible in Extramarks. These materials were created with consideration for the NCERT and CBSE curricula. Additionally, it is suggested that students practice the important CBSE questions in Class 9 Mathematics Chapter 6 to get a sense of the final exam’s question format.

Important Questions Class 9 Mathematics Chapter 6 – With Solutions

The students may easily prepare all the concepts included in the CBSE Syllabus in a much better and more effective method with the help of Extramarks important questions Class 9 Mathematics Chapter 6 . These resources include a thorough explanation, key formulas, are also offered to students to assist them in getting a quick review of all the topics.

A few Important Questions Class 9 Mathematics Chapter 6 are provided here, along with their answers:

Question 1: If one angle of the triangle is equal to the sum of the other two angles, then the triangle is

(A) An equilateral triangle

(B) An obtuse triangle

(D) A right triangle

Solution 1: (D) A right triangle

Explanation:

We suppose the angles of △ABC be ∠A, ∠B and ∠C

Given, ∠A= ∠B+∠C …(equation 1)

But, in any △ABC,

Using the angle sum property, we have,

∠A+∠B+∠C=180o …(equation 2)

From equations (eq1) and (eq2), we get

∠A+∠A=180 o

⇒∠A=180 o /2 = 90 o

Thus, we get that the triangle is a right-angled triangle

Question 2. The exterior angle of the triangle is 105°, and its two interior opposite angles are equal. Each of these equal angles is

Solution 2: (B) 52 ½ o

As per the question,

The exterior angle of the triangle will be = 105°

We suppose the two interior opposite angles of the triangle = x

We know that,

The exterior angle of a triangle will be = the sum of interior opposite angles

Thus, we have,

105° = x + x

Question 3: The angles of the triangle are in the ratio 5 : 3: 7. The triangle is

(A) An acute angled triangle

(B) An isosceles triangle

(D) An obtuse-angled of triangle

Solution 3: (A) An acute angled triangle

The angles of the triangle are in the ratio of 5 : 3: 7

Let the ratio 5:3:7 be 5x, 3x and 7x

Using the angle sum property of the triangle,

5x + 3x +7x =180

Putting the value of x, i.e., x = 12, in 5x, 3x and 7x we have,

5x = 5×12 = 60o

3x = 3×12 = 36o

7x = 7×12 = 84o

As all the angles are less than 90o, the triangle will be an acute-angled triangle.

Question 4: In the given figure, if PQ || RS, then find the measure of angle m.

Solution 4:

Here, PQ || RS, PS is a transversal.

⇒ ∠PSR = ∠SPQ = 56°

Also, ∠TRS + m + ∠TSR = 180°

14° + m + 56° = 180°

⇒ m = 180° – 14 – 56 = 110°

Question 5: In Figure, the lines AB and CD intersect at the point O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and the reflex ∠COE.

Solution 5:

From the diagram, we have

(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forming a straight line.

Then, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°

Now, by substituting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we have,

∠COE = 110° and ∠BOE = 30°

So, reflex ∠COE = 360o – 110o = 250°

Question 6: In the given figure, POQ is the line. The ray OR is the perpendicular to the line PQ. OS is another ray lying between the rays OP and OR. Prove that ∠ROS = 1 2 (∠QOS – ∠POS).

Solution 6:

Given that OR is perpendicular to PQ

⇒ ∠POR = ∠ROQ = 90°

∴ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90° – ∠POS

Adding ∠ROS to both sides, we have

∠ROS + ∠ROS = (90° + ∠ROS) – ∠POS

⇒ 2∠ROS = ∠QOS – ∠POS

⇒ ∠ROS = 1 2 (∠QOS – ∠POS).

Question 7: In Figure, the lines XY and MN intersect at the point O. If ∠POY = 90° and a: b = 2 : 3, find c.

Solution 7:

We know, the sum of the linear pair is always equal to 180°

∠POY +a +b = 180°

Putting the value of ∠POY = 90° (given in the question), we have,

Now, given, a: b = 2 : 3, so,

We suppose a be 2x, and b be 3x

∴ 2x+3x = 90°

Solving this equation, we get

So, x = 18°

∴ a = 2×18° = 36°

In the similar manner, b can be calculated, and the value will be

b = 3×18° = 54°

From the given diagram, b+c also forms a straight angle, so,

c+54° = 180°

Therefore, c = 126°

Question 8: In the Figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution 8:

As ST is a straight line so,

∠PQS+∠PQR = 180° (since it is a linear pair) and

∠PRT+∠PRQ = 180° (since it is a linear pair)

Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°

We know, ∠PQR =∠PRQ (as given in the question)

∠PQS = ∠PRT. (Hence proved).

Question 9: In the Figure, if x+y = w+z, then prove that AOB is a line.

Solution 9:

To prove AOB is a straight line, we will first have to prove that x+y is a linear pair

i.e. x+y = 180°

We know, the angles around a point are 360° so,

x+y+w+z = 360°

In the question, it is given that,

So, (x+y)+(x+y) = 360°

2(x+y) = 360°

∴ (x+y) = 180° (Hence proved).

Question 10: In Figure, POQ is a line. The ray OR is perpendicular to the line PQ. OS is another ray lying between the rays OP and OR. Prove that ∠ROS = ½ (∠QOS – ∠POS).

Solution 10:

Given that (OR ⊥ PQ) and ∠POQ = 180°

Thus, ∠POS+∠ROS+∠ROQ = 180°

Now, ∠POS+∠ROS = 180°- 90° (As ∠POR = ∠ROQ = 90°)

Again, ∠QOS = ∠ROQ+∠ROS

Given, ∠ROQ = 90°,

∴ ∠QOS = 90° +∠ROS

Or, ∠QOS – ∠ROS = 90°

As ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we have

∠POS + ∠ROS = ∠QOS – ∠ROS

2 ∠ROS + ∠POS = ∠QOS

Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).

Question 11: In the figure, find the values of x and y and show that AB || CD.

Solution 11:

We know, a linear pair is equal to 180°.

Thus, x+50° = 180°

We also know, vertically opposite angles are equal.

Thus, y = 130°

In the two parallel lines, the alternate interior angles are equal. Here,

x = y = 130°

This proves that the alternate interior angles are equal, and thus, AB || CD.

Question 12: In the given figure, PQ || RS and EF || QS. If ∠PQS = 60°, then find the value of ∠RFE.

Solution 12:

Given PQ || RS

Thus, ∠PQS + ∠QSR = 180°

⇒ 60° + ∠QSR = 180°

⇒ ∠QSR = 120°

Now, EF || QS ⇒ ∠RFE = ∠QSR [corresponding ∠s]

⇒ ∠RFE = 120°

Question 13: In the figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution 13:

We know, AB || CD and CD||EF

Since the angles on the same side of the transversal line sum up to 180°,

x + y = 180° —–equation (i)

∠O = z (Since corresponding angles)

and, y +∠O = 180° (Since linear pair)

So, y+z = 180°

Now, let y = 3w and thus, z = 7w (As y : z = 3 : 7)

Therefore, 3w+7w = 180°

Or, 10 w = 180°

Thus, w = 18°

Now, y = 3×18° = 54°

and, z = 7×18° = 126°

Now, the angle x can be calculated from equation (i)

Or, x+54° = 180°

Question 14: In the figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution 14:

As AB || CD, GE is a transversal.

Given that ∠GED = 126°

So, ∠GED = ∠AGE = 126° (Since 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° (Since transversal)

Substituting the value of ∠GED = 126° we get,

∠AGE = 126°

∠GEF = 36° and

Question 15: In figure, if AB || CD. If ∠ABR = 45° and ∠ROD = 105°, then find ∠ODC.

Solution 15:

Through the point O, we draw a line ‘l’ parallel to AB.

⇒ line I will also be parallel to CD, then

∠1 = 45°[alternate int. angles]

∠1 + ∠2 + 105° = 180° [straight angle]

∠2 = 180° – 105° – 45°

Now, ∠ODC = ∠2 [alternate int. angles]

= ∠ODC = 30°

Question 16: In the figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

[Hint: Draw a line parallel to ST through the point R.]

Solution 16:

First, we construct a line XY parallel to PQ.

We know, the angles on the same side of the transversal are equal to 180°.

Thus, ∠PQR+∠QRX = 180°

Or, ∠QRX = 180°-110°

∴ ∠QRX = 70°

In the similar manner,

∠RST +∠SRY = 180°

Or, ∠SRY = 180°- 130°

Therefore, ∠SRY = 50°

Now, from the linear pairs on the line XY-

∠QRX+∠QRS+∠SRY = 180°

Putting the values, we have,

∠QRS = 180° – 70° – 50°

Hence, ∠QRS = 60°

Question 17: In the figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution 17:

From the above diagram,

∠APQ = ∠PQR (Since Alternate interior angles)

Now, substituting the value of ∠APQ = 50° and ∠PQR = x, we ,

∠APR = ∠PRD (i.e., alternate interior angles)

Or, ∠APR = 127° (Given ∠PRD = 127°)

∠APR = ∠APQ+∠QPR

Now, substituting the values of ∠QPR = y and ∠APR = 127° we get,

127° = 50°+ y

Or, y = 77°

Thus, the measure of x and y are as follows:

x = 50° and y = 77°

Question 18: In the given figure, p ll q, find the value of x.

Solution 18:

We extend the line p to meet RT at S.

Such that MS || QT

Now, in ARMS, we have

∠RMS = 180° – ∠PMR (Since linear pair]

= 180° – 120°

∠RMS + ∠MSR + ∠SRM = 180° [i.e., by angle sum property of a ∆]

⇒ 60° + ∠MSR + 30o = 180°

⇒ MSR = 90°

Now, PS || QT – ∠MSR = ∠RTQ

⇒ ∠RTQ = x = MSR = 90° (Since corresponding ∠s]

Question 19: In the figure, PQ and RS are the 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.

Solution 19:

Firstly, we draw the two lines, BE and CF, such that BE ⊥ PQ and CF ⊥ RS.

Now, since PQ || RS,

So, BE || CF

The angle of incidence = Angle of reflection (By the law of reflection)

∠1 = ∠2 and

We also know, the alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at points B and C.

So, ∠2 = ∠3 (Since they are alternate interior angles)

Here, ∠1 +∠2 = ∠3 +∠4

Or, ∠ABC = ∠DCB

So, AB || CD (since alternate interior angles are equal)

Question 20: In figure, the sides QP and RQ of ΔPQR are produced to the points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution 20:

Given that TQR is a straight line, and thus, the linear pairs (i.e. ∠TQP and ∠PQR) will add up to 180°

So, ∠TQP +∠PQR = 180°

Now, substituting the value of ∠TQP = 110° we have,

We consider the ΔPQR,

The side QP is extended to the point S, and so ∠SPR forms the exterior angle.

Therefore, ∠SPR (∠SPR = 135°) is equal to the sum of the interior opposite angles. (By triangle property)

Or, ∠PQR +∠PRQ = 135°

Now, substituting the value of ∠PQR = 70° we get,

∠PRQ = 135°-70°

Hence, ∠PRQ = 65°

Question 21: In the figure, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively of Δ XYZ, find ∠OZY and ∠YOZ.

Solution 21:

We know, the sum of the interior angles of the triangle is 180

So, ∠X +∠XYZ +∠XZY = 180°

Putting the given values in the question, we have,

62°+54° +∠XZY = 180°

Or, ∠XZY = 64°

Now, we know that ZO is the bisector, so,

∠OZY = ½ ∠XZY

Therefore, ∠OZY = 32°

In the similar manner, YO is a bisector, and so,

∠OYZ = ½ ∠XYZ

Or, ∠OYZ = 27° (As ∠XYZ = 54°)

Now, the sum of the interior angles of the given triangle,

∠OZY +∠OYZ +∠O = 180°

Putting their respective values, we get,

∠O = 180°-32°-27°

Hence, ∠O = 121°

Question 22: In the figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution 22:

We know, AE is the transversal since AB || DE

Here ∠BAC and ∠AED are the alternate interior angles.

Hence, ∠BAC = ∠AED

Given, ∠BAC = 35°

Now considering the triangle CDE. We know that the sum of the interior angles of the triangle is 180°.

∴ ∠DCE+∠CED+∠CDE = 180°

Putting the values, we get

∠DCE+35°+53° = 180°

Hence, ∠DCE = 92°

Question 23: To protect the poor people from cold weather, Ram Lal. has given his land to make a shelter home for them. In the given figure, ‘the sides QP and RQ of ∆PQR are produced to points S and T, respectively. If ∠PQT = 110° and ∠SPR = 135°, find the value of ∠PRQ.

Solution 23:

∠SPR + ∠QPR = 180° [i.e., a linear pair]

135° + ∠QPR = 180° [∵ ∠SPR = 135°]

⇒ ∠QPR = 180° – 135° = 45°

In ∆PQR, by the exterior angle property, we have

∠QPR + ∠PRQ = ∠PQT

45° + ∠PRQ = 110°

∠PRQ = 110° – 45° = 65°

Question 24: In the figure, if the lines PQ and RS intersect at point T, in such a way that ∠PRT = 40°, ∠RPT = 95°, and ∠TSQ = 75°, find ∠SQT.

Solution 24:

We consider the triangle PRT.

∠PRT +∠RPT + ∠PTR = 180°

Thus, ∠PTR = 45°

Now, ∠PTR will be equal to ∠STQ as they are the vertically opposite angles.

So, ∠PTR = ∠STQ = 45°

Again, in triangle STQ,

∠TSQ +∠PTR + ∠SQT = 180°

Solving this equation, we get,

74° + 45° + ∠SQT = 180°

Question 25: In the figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution 25:

x +∠SQR = ∠QRT (Because they are alternate angles and QR is the transversal)

Thus, x+28° = 65°

It is also known that the alternate interior angles are the same, and so,

∠QSR = x = 37°

∠QRS +∠QRT = 180° (Since linear pair)

Or, ∠QRS+65° = 180°

So, ∠QRS = 115°

Using the angle sum property in Δ SPQ,

∠SPQ + x + y = 180°

90°+37° + y = 180°

y = 1800 – 1270 = 530

Hence, y = 53°

Question 26: In the figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at the point T, then prove that ∠QTR = ½ ∠QPR.

Solution 26:

We consider the ΔPQR. ∠PRS is the exterior angle, and ∠QPR and ∠PQR are the interior angles.

So, ∠PRS = ∠QPR+∠PQR (According to the triangle property)

Or, ∠PRS -∠PQR = ∠QPR ———–equation(i)

Now, considering the ΔQRT,

∠TRS = ∠TQR+∠QTR

Or, ∠QTR = ∠TRS-∠TQR

We know, QT and RT bisect ∠PQR and ∠PRS, respectively.

So, ∠PRS = 2 ∠TRS and ∠PQR = 2∠TQR

Here, ∠QTR = ½ ∠PRS – ½∠PQR

Or, ∠QTR = ½ (∠PRS -∠PQR)

From equation (i), we know, ∠PRS -∠PQR = ∠QPR

Therefore, ∠QTR = ½ ∠QPR (hence proved).

Question 27: For what value of x + y in the figure will ABC be a line? Justify the answer.

Solution 27:

The value of x + y should be 180o for ABC to be a line.

Justification:

From the figure, we can state that,

BD is a ray intersecting AB and BC at point B, which implies

and, ∠DBC = x

If a ray stands on the line, then sum of the two adjacent angles formed will be 180°.

⇒ If the sum of the two adjacent angles is 180°, then a ray stands on the line.

So, for ABC to be a line,

Then, the sum of ∠ABD and ∠DBC should be equal to 180°.

⇒ ∠ABD + ∠DBC = 180°

⇒ x + y = 180°

Thus, the value of x + y should be equal to 180° for ABC to be a line.

Question 28: Can a triangle have all angles less than 60°? Give a reason for your answer.

Solution 28:

No. A triangle cannot have all the angles less than 60°

As per the angle sum property,

We know the sum of all the interior angles of a triangle should be = 180°.

We suppose all the angles are 60o,

Then we get, 60o + 60o + 60o = 180o.

Now, considering angles less than 60o,

We suppose 59o to be the highest natural number, less than 60o.

Then we get,

59 o +59 o + 59 o = 177 o ≠ 180 o

Thus, we can say that if all the angles are less than 60o, the measure of the angles won’t be satisfying the angle sum property.

Therefore, a triangle cannot have all the angles less than 60o.

Question 29: Can a triangle have two obtuse angles? Give a reason for your answer.

Solution 29:

No. A triangle cannot have two obtuse angles.

According to the angle sum property,

We know, the sum of all the interior angles of the triangle should be = 180o.

An obtuse angle is one with a value greater than 90° but less than 180°.

We consider the two angles to be equal to the lowest natural number greater than 90o, i.e., 91o.

If the triangle has two obtuse angles, then there are two angles that would be at least 91° each.

By adding the two angles, we get

Sum of the two angles = 91° + 91°

⇒ Sum of the two angles = 182°

The sum of the two angles already exceeds the sum of the three angles of the triangle, even before taking into consideration the third angle.

Thus, a triangle cannot have two obtuse angles.

Question 30: How many triangles can be drawn having angles as 45°, 64° ,and 72°? Give a reason for your answer.

Solution 30:

No such triangle can be drawn having its angles 45°, 64° and 72°.

We know the sum of all the interior angles of a triangle should be = 180o.

But, as per the question,

We have the angles as 45°, 64° and 72°.

Sum of these angles is = 45° + 64° + 72°

= 181 o , which is greater than 180o.

So, the angles do not satisfy the angle sum property of a triangle.

Therefore, no triangle can be drawn having angles 45°, 64° and 72°.

Question 31: How many triangles can be drawn having their angles as 53°, 64° and 63°? Give a reason for your answer.

Solution 31:

Infinitely many triangles can be drawn having angles as 53°, 64° and 63°.

We know the sum of all the interior angles of the triangle should be = 180o.

We have the angles as 53°, 64°, and 63°.

Sum of these angles = 53° + 64° + 63°

Thus, the angles satisfies the angle sum property of the triangle.

Therefore, infinitely many triangles may be drawn, having their angles as 53°, 64° and 63°.

Question 32: In the figure, OD is the bisector of ∠AOC, and OE is the bisector of ∠BOC and OD ⊥ OE. Show that points A, O and B are collinear.

Solution 32:

According to the question,

In the figure,

OD and OE are the bisectors of ∠AOC and ∠BOC.

To prove: The points A, O and B are collinear.

i.e., AOB is a straight line.

As OD and OE bisect angles ∠AOC and ∠BOC, respectively.

∠AOC = 2∠DOC …(equation 1)

And ∠COB = 2∠COE …(equation 2)

Adding (equation 1) and (equation 2), we have,

∠AOC = ∠COB = 2∠DOC + 2∠COE

∠AOC +∠COB = 2(∠DOC +∠COE)

∠AOC + ∠COB = 2∠DOE

Since OD⊥OE

∠AOC +∠COB = 2×90o

∠AOC +∠COB =180o

So, ∠AOC + ∠COB form a linear pair.

Thus, AOB is a straight line.

Therefore, the points A, O and B are collinear.

Question 33: In the figure, OP bisects ∠BOC and OQ bisects ∠AOC. Prove that ∠POQ = 90°

Solution 33:

∵ OP bisects ∠BOC

∴ ∠BOP = ∠POC = x (say)

Also, OQ bisects. ∠AOC

∠AOQ = ∠COQ = y (say) .

∵ Ray OC stands on ∠AOB

∴ ∠AOC + ∠BOC = 180° [linear pair]

⇒ ∠AOQ + ∠QOC + ∠COP + ∠POB = 180°

⇒ y + y + x + x = 180°.

⇒ 2x + 2y = 180°

⇒ x + y = 90°

Now, ∠POQ = ∠POC + ∠COQ

= x + y = 90°

Question 34: In the figure, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel.

Solution 34:

We have from figure ∠1 = 60° and ∠6 = 120°

As, ∠1 = 60° and ∠6 = 120°

Here, ∠1 = ∠3 [i.e.,vertically opposite angles]

∠3 = ∠1 = 60°

Now, ∠3 + ∠6 = 60° + 120°

⇒ ∠3 + ∠6 = 180°

If the sum of the two interior angles on the same side of l is 180°, then the lines are parallel.

Therefore, m || n

Question 35: AP and BQ are the bisectors of the two alternate interior angles which are formed by the intersection of the transversal t with the parallel lines l and m (figure). Show that AP || BQ.

Solution 35:

l || m and t is the transversal

∠MAB = ∠SBA [alternate angles]

⇒ ½ ∠MAB = ½ ∠SBA

⇒ ∠PAB = ∠QBA

But, ∠2 and ∠3 are alternate angles.

Hence, AP||BQ.

Question 36: If in the Figure, the bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m.

Solution 36:

AP is the bisector of ∠MAB

BQ is the bisector of ∠SBA.

Given: AP||BQ.

So ∠2 = ∠3 [Alternate angles]

⇒ ∠2 + ∠2 = ∠3 +∠3

From the figure, we have ∠1= ∠2and ∠3 = ∠4

⇒ ∠1+ ∠2 = ∠3 +∠4

⇒ ∠MAB = ∠SBA

But we also know that these are the alternate angles.

Therefore, the lines l and m are parallel, i.e., l ||m.

Question 37: In the figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF [Hint: Produce DE to intersect BC at P (say)].

Solution 37:

Construction:

We extend DE to intersect BC at point P.

Given that EF||BC and DP are transversal,

∠DEF = ∠DPC …(equation 1) [Since corresponding angles]

Also given, AB||DP and BC is a transversal,

∠DPC = ∠ABC …(equation 2) [Since Corresponding angles]

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

∠ABC = ∠DEF

Hence, Proved.

Question 38: In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.

Solution 38:

external ∠FEB = ∠A + F

= 90° + 40° = 130°

As AB || CD

Therefore, ∠ECD = FEB = 130°

Hence, ∠ECD = 130°.

Question 39: If the two lines intersect each other, prove that the vertically opposite angles are equal.

Solution 39:

AB and CD intersect each other at the point O.

We suppose the two pairs of vertically opposite angles be,

1st pair – ∠AOC and ∠BOD

2nd pair – ∠AOD and ∠BOC

The vertically opposite angles are equal,

i.e., ∠AOC = ∠BOD, and ∠AOD = ∠BOC

The ray AO stands on the line CD.

If a ray lies on the line, then the sum of the adjacent angles is equal to 180°.

⇒ ∠AOC + ∠AOD = 180° (i.e., By linear pair axiom) … equation (i)

In the similar manner, the ray DO lies on the line AOB.

⇒ ∠AOD + ∠BOD = 180° (i.e.,By linear pair axiom) … equation (ii)

From equations (i) and (ii),

∠AOC + ∠AOD = ∠AOD + ∠BOD

⇒ ∠AOC = ∠BOD – – – – equation (iii)

In the similar manner, the ray BO lies on the line COD.

⇒ ∠DOB + ∠COB = 180° (By using linear pair axiom) – – – – equation (iv)

Also, the ray CO is lying on the line AOB.

⇒ ∠COB + ∠AOC = 180° (By using linear pair axiom) – – – – equation (v)

From equations (iv) and (v),

∠DOB + ∠COB = ∠COB + ∠AOC

⇒ ∠DOB = ∠AOC – – – – equation (vi)

Therefore, from equation (iii) and equation (vi),

∠AOC = ∠BOD, and ∠DOB = ∠AOC

Thus, we get vertically opposite angles are equal.

Hence Proved.

Question 40: Bisectors of the interior ∠B and the exterior ∠ACD of a Δ ABC intersect at point T.

Prove that ∠ BTC = ½ ∠ BAC.

Solution 40:

Given: In△ ABC, we produce BC to D and the bisectors of ∠ABC and ∠ACD meet at point T.

∠BTC = ½ ∠BAC

In △ABC, ∠ACD is an exterior angle.

The exterior angle of the triangle is equal to the sum of two opposite angles,

∠ACD = ∠ABC + ∠CAB

Now, dividing the L.H.S and R.H.S by 2,

⇒ ½ ∠ACD = ½ ∠CAB + ½ ∠ABC

⇒ ∠TCD = ½ ∠CAB + ½ ∠ABC …equation (1)

[∵CT is the bisector of ∠ACD⇒ ½ ∠ACD = ∠TCD]

Then in △ BTC,

∠TCD = ∠BTC +∠CBT

⇒ ∠TCD = ∠BTC + ½ ∠ABC …(2)

[∵BT is bisector of △ ABC ⇒∠CBT = ½ ∠ABC ]

From equations (1) and (2),

½ ∠CAB + ½ ∠ABC = ∠BTC + ½ ∠ABC

⇒ ½ ∠CAB = ∠BTC or ½ ∠BAC = ∠BTC

Hence, proved.

Question 41: A transversal intersects the two parallel lines. Prove that the bisectors of any pair of the corresponding angles so formed are parallel.

Solution 41:

We suppose,

EF be the transversal that passes through the two parallel lines at the point P and Q, respectively.

PR and QS are the bisectors of ∠EPB and ∠PQD.

We know, corresponding angles of the parallel lines are equal,

So, ∠EPB = ∠PQD

½ ∠EPB = ½ ∠PQD

∠EPR = ∠PQS

But we also know that they are the corresponding angles of PR and QS

Since the corresponding angles are equal,

Question 42: The lines AB and CD intersect at the point O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and the reflex ∠COE.

Solution 42:

∠AOC, ∠BOE, ∠COE and ∠COE, ∠BOD, ∠BOE form a straight line each.

Thus, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD + ∠BOE = 180°

Here, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get:

70° +∠COE = 180°

∠COE = 110°

110° + 40° + ∠BOE = 180°

Question 43: The lines XY and MN intersect at O. If ∠POY = 90° and a: b = 2 : 3, find c.

Solution 43:

We know, the sum of a linear pair is always equal to 180°

∠POY + a + b = 180°

Substitutg the value of ∠POY = 90° (as given), we get,

a + b = 90°

Now, we know from the question a: b = 2 : 3, so,

We suppose a be 2x, and b be 3x.

∴ 2x + 3x = 90°

By solving the equation, we get

∴ a = 2 × 18° = 36°

b = 3 × 18° = 54°

Here, b + c also forms a straight angle, so,

b + c = 180°

=> c + 54° = 180°

Question 44: Given, POQ is a line. The ray OR is perpendicular to the line PQ. OS is another ray lying between the rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).

Solution 44:

Given in the question that (OR ⊥ PQ) and ∠POQ = 180°

So, ∠POS + ∠ROS + ∠ROQ = 180° (Since linear pair of angles)

Now, ∠POS + ∠ROS = 180° – 90° (As ∠POR = ∠ROQ = 90°)

Now, ∠QOS = ∠ROQ + ∠ROS

Again, given ∠ROQ = 90°,

Therefore, ∠QOS = 90° + ∠ROS

Since ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we get

=>2 ∠ROS + ∠POS = ∠QOS

Question 45: If AB || CD, EF ⊥ CD and ∠GED = 126°, find the value of ∠AGE, ∠GEF and ∠FGE.

Solution 45:

Since AB || CD GE is a transversal.

It is given that ∠GED = 126°

So, ∠GED = ∠AGE = 126° (alternate interior angles)

∠GED = ∠GEF + ∠FED

EF ⊥ CD, ∠FED = 90°

Again, ∠FGE + ∠GED = 180° (Since it is the transversal)

Question 46: If ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively of Δ XYZ, find ∠OZY and ∠YOZ.

Solution 46:

We know, the sum of the interior angles of the triangle is 180°.

Then, ∠X +∠XYZ + ∠XZY = 180°

Substituting the values given in the equation we get,

62° + 54° + ∠XZY = 180°

Now, ZO is the bisector, so,

∴ ∠OZY = 32°

In the similar manner, YO is the bisector, and thus,

Now, as the sum of interior angles of the triangle,

So, ∠OZY +∠OYZ + ∠O = 180°

∠O = 180° – 32° – 27°

Or, ∠O = 121°

Question 47: If AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then find the value of ∠QRS.

Solution 47:

AB || CD || EF

If the transversal intersects the two parallel lines, then each pair of the alternate exterior angles are equal.

Now, as PQ || RS

⇒ ∠PQC = ∠BRS

We have ∠PQC = 60°

⇒ ∠BRS = 60° … equation (i)

If the transversal intersects the two parallel lines, then each pair of the alternate interior angles is equal.

Now, since AB || CD

⇒ ∠DQR = ∠QRA

We have ∠DQR = 25°

⇒ ∠QRA = 25° … equation (ii)

By using the linear pair axiom,

∠ARS + ∠BRS = 180°

⇒ ∠ARS = 180° – ∠BRS

⇒ ∠ARS = 180° – 60° (From equation (i), ∠BRS = 60°)

⇒ ∠ARS = 120° … equation (iii)

Now, ∠QRS = ∠QRA + ∠ARS

From equations (ii) and (iii), we have,

∠QRA = 25° and ∠ARS = 120°

Hence, the above equation can be written as:

∠QRS = 25° + 120°

⇒ ∠QRS = 145°

Question 48: If an angle is half of its complementary angle, then find its degree measure.

Solution 48:

We suppose that the required angle be x

∴ Its complement = 90° – x

Now, according to the given statement, we obtain

x = 1 2 (90° – x)

⇒ 2x = 90° – x

Hence, the required angle is 30°.

Question 49: The two complementary angles are in the ratio of 1: 5. Find the measures of the angles.

Solution 49:

We suppose that the two complementary angles be x and 5x.

∴ x + 5x = 90°

Hence, the two complementary angles are 15° and 5 × 15°, i.e., 15° and 75°.

Question 50: If an angle is 14 o more than its complement, then find its measure.

Solution 50:

Let the required angle be x

x = 90° – x + 14°

⇒ 2x = 104°

Hence, the required angle is 52 o .

Question 51: If AB || EF and EF || CD, then find the value of x.

Solution 51:

Since EF || CD ∴ y + 150° = 180°

⇒ y = 180° – 150° = 30°

Now, ∠BCD = ∠ABC

x + y = 70°

x + 30 = 70

⇒ x = 70° – 30° = 40°

Hence, the value of x is 40°.

Question 52: In the given figure, the lines AB and CD intersect at O. Find the value of x.

Solution 52:

Here, the lines AB and CD intersect at O.

∴ ∠AOD and ∠BOD form a linear pair

⇒ ∠AOD + ∠BOD = 180°

⇒ 7x + 5x = 180°

⇒ 12x = 180°

Question 53: In the given figure, if x°, y° and z° are the exterior angles of ∆ABC, then find the value of x° + y° + z°.

Solution 53:

We know that an exterior angle of a triangle is equal to the sum of two opposite interior angles.

⇒ x° = ∠1 + ∠3

⇒ y° = ∠2 + ∠1

⇒ z° = ∠3 + ∠2

Adding all these, we have

x° + y° + z° = 2(∠1 + ∠2 + ∠3)

Question 54: In figure., AD and CE are the angle bisectors of ∠A and ∠C, respectively. If ∠ABC = 90°, then find ∠AOC.

Solution 54:

∵ AD and CE are the bisectors of ∠A and ∠C

∠OAC = 1 2 ∠A and

∠OCA 1 2 ∠C

∠OAC + ∠OCA = 1 2 (∠A + ∠C)

= 1 2 (180 0 – ∠B) [ Since, ∠A + ∠B +∠C = 180 0 ]

= 1 2 (180 0 – 90 0 ) [Since, ∠ABC = 90 0 ]

= 1 2 x 90 0 = 45 0

∠AOC + ∠OAC + ∠OCA = 180°

⇒ ∠AOC + 45o = 180°

⇒ ∠AOC = 180° – 45° = 135°.

Question 55: In the given figure, prove that m || n.

Solution 55:

ext. ∠BDM = ∠C + ∠B

= 38° + 25° = 63°

Now, ∠LAD = ∠MDB = 63°

But these are corresponding angles. Hence,

Question 56: In the given figure, two straight lines, PQ and RS, intersect each other at O. If ∠POT = 75°, find the values of a, b, and c.

Solution 56:

Here, 4b + 75° + b = 180° [since a straight angle]

5b = 180° – 75° = 105°

b – 105∘ 5 = 21°

Therefore, a = 4b = 4 × 21° = 84° (i.e., vertically opp. ∠s]

Again, 2c + a = 180° [Since, a linear pair]

⇒ 2c + 84° = 180°

⇒ c = 96° 2 = 48°

Therefore, the values of a, b and c are a = 84°, b = 21° and c = 48°.

Question 57: In the figure, ∠X = 72°, ∠XZY = 46°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively ∆XYZ, find ∠OYZ and ∠YOZ.

Solution 57:

In ∆XYZ, we have

∠X + XY + ∠Z = 180°

⇒ ∠Y + ∠Z = 180° – ∠X

⇒ ∠Y + ∠Z = 180° – 72°

⇒ Y + ∠Z = 108°

⇒ 1 2 ∠Y + 1 2 ∠Z = 1 2 × 108°

∠OYZ + ∠OZY = 54°

[∵ YO and ZO are the bisectors of ∠XYZ and ∠XZY]

⇒ ∠OYZ + 1 2 × 46° = 54°

∠OYZ + 23° = 54°

⇒ ∠OYZ = 54 0 – 23° = 31°

Again, in ∆YOZ, we have

∠YOZ = 180° – (∠OYZ + ∠OZY)

= 180° – (31° + 23°) 180° – 54° = 126°

Question 58: Prove that if the two lines intersect each other, then the bisectors of the vertically opposite angles are in the same line.

Solution 58:

Let AB and CD be the two intersecting lines intersecting each other at the point O.

OP and OQ are the bisectors of ∠AOD and ∠BOC.

∴ ∠1 = ∠2 and ∠3 = ∠4 …equation (i)

Now, ∠AOC = ∠BOD [vertically opp. ∠s] ……equation (ii)

⇒ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 [adding equation (i) and (ii)]

Also, ∠1 + ∠AOC + ∠3 + ∠2 + ∠BOD + ∠4 = 360° (Since ∠s at a point are 360°]

⇒ ∠1 + ∠AOC + ∠3 + ∠1 + ∠AOC + ∠3 = 360° [by using equation (i), (ii)]

⇒ ∠1 + ∠AOC + ∠3 = 180°

or ∠2 + ∠BOD + ∠4 = 180°

Therefore, OP and OQ are in the same line.

Question 59: In the given figure, AB || CD and EF || DG, find ∠GDH, ∠AED and ∠DEF.

Solution 59:

As AB || CD and HE is the transversal.

∴ ∠AED = ∠CDH = 40° [i.e., corresponding ∠s]

Now, ∠AED + ∠DEF + ∠FEB = 180° [since a straight ∠]

40° + CDEF + 45° = 180°

∠DEF = 180° – 45 – 40 = 95°

Again, given, EF || DG and HE is the transversal.

∴ ∠GDH = ∠DEF = 95° [i.e., corresponding ∠s]

Therefore, ∠GDH = 95°, ∠AED = 40° and ∠DEF = 95°

Question 60: In the figure, DE || QR and AP and BP are the bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.

Solution 60:

Here, AP and BP are the bisectors of ∠EAB and ∠RBA, respectively.

⇒ ∠1 = ∠2 and ∠3 = ∠4

Since DE || QR and the transversal n intersects DE and QR at A and B, respectively.

⇒ ∠EAB + ∠RBA = 180°

[Since co-interior angles are supplementary]

⇒ (∠1 + ∠2) + (∠3 + ∠4) = 180°

⇒ (∠1 + ∠1) + (∠3 + ∠3) = 180° (using equation (i)

⇒ 2(∠1 + ∠3) = 180°

⇒ ∠1 + ∠3 = 90°

Now, in ∆ABP, by angle sum property, we have

∠ABP + ∠BAP + ∠APB = 180°

⇒ ∠3 + ∠1 + ∠APB = 180°

⇒ 90° + ∠APB = 180°

⇒ ∠APB = 90°

Question 61: If the two parallel lines are intersected by a transversal, then prove that the bisectors of any of the two corresponding angles are parallel.

Solution 61:

Given: AB || CD and the transversal PQ meet these lines at E and F, respectively. EG and FH are

the bisectors of pair of the corresponding angles ∠PEB and ∠EFD.

To Prove: EG || FH

∵ EG and FH are the bisectors of ∠PEB, respectively.

∠PEG = 1 2 ∠PEB ………equation (i)

And, ∠EFH = 1 2 ∠EFD …..equation (ii)

Since, AB || CD and PQ is a transversal

Therefore, ∠PEB = ∠EFD

1 2 ∠PEB = 1 2 ∠EFD

∠PEG = ∠EFH

Which are the corresponding angles of EG and FH ∴ EG || FH.

Question 62: In the given figure, m and n are the two plane mirrors perpendicular to each other. Show that the incident rays CA is parallel to the reflected ray BD.

Solution 62:

Let the normals at A and B meet at point P.

Since the mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.

Thus, BP ⊥ PA, i.e., ∠BPA = 90°

Therefore, ∠3 + ∠2 = 90° [by angle sum property] …equation (i)

Also, ∠1 = ∠2 and ∠4 = ∠3 [Angle of incidence = Angle of reflection]

Thus, ∠1 + ∠4 = 90° [from equations (i)) …(ii]

Adding equation (i) and (ii), we have

∠1 + ∠2 + ∠3 + ∠4 = 180°

i.e., ∠CAB + ∠DBA = 180°

Hence, CA || BD

Question 63: If the two parallel lines are intersected by a transversal, prove that the bisectors of two pairs of the interior angles form a rectangle.

Solution 63:

Given: AB || CD and the transversal EF cut them at P and Q, respectively and the bisectors of

the pair of interior angles form a quadrilateral PRQS.

To Prove: PRQS is a rectangle.

Proof: Since PS, QR, QS and PR are the bisectors of angles

∠BPQ, ∠CQP, ∠DQP and ∠APQ, respectively.

∴∠1 = 1 2 ∠BPQ, ∠2 = 1 2 ∠CQP,

∠3 = 1 2 ∠DQP and ∠4 = 1 2 ∠APQ

Now, AB || CD and EF is the transversal

∴ ∠BPQ = ∠CQP

⇒ ∠1 = ∠2 (∵∠1 = 1 2 ∠BPQ and ∠2 = 1 2 ∠QP)

But these are the pairs of alternate interior angles of PS and QR

In the similar manner, we can prove ∠3 = ∠4 = QS || PR

∴ PRQS is the parallelogram.

Further ∠1 + ∠3 = 1 2 ∠BPQ + 1 2 ∠DQP = 1 2 (∠BPQ + ∠DQP)

= 1 2 × 180° = 90° (Since ∠BPQ + ∠DQP = 180°)

∴ In ∆PSQ, we have ∠PSQ = 180° – (∠1 + ∠3) = 180° – 90° = 90°

Therefore, PRRS is the parallelogram whose one angle ∠PSQ = 90°.

Hence, PRQS is a rectangle.

Question 64: If in ∆ABC, the bisectors of ∠B and ∠C intersect each other at point O. Prove that ∠BOC = 90° + 1 2 ∠ A.

Solution 64:

We suppose ∠B = 2x and ∠C = 2y

∵OB and OC bisect ∠B and ∠C, respectively.

∠OBC = 1 2 ∠B = 1 2 × 2x = x

and ∠OCB = 1 2 ∠C = 1 2 × 2y = y

Now, in ∆BOC, we have

∠BOC + ∠OBC + ∠OCB = 180°

⇒ ∠BOC + x + y = 180°

⇒ ∠BOC = 180° – (x + y)

Again, in ∆ABC, we have

∠A + 2B + C = 180°

⇒ ∠A + 2x + 2y = 180°

⇒ 2(x + y) = 1 2 (180° – ∠A)

⇒ x + y = 90° – 1 2 ∠A …..equation (ii)

From equation (i) and (ii), we have

∠BOC = 180° – (90° – 1 2 ∠A) = 90° + 1 2 ∠A

Question 65: In the figure, if I || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Solution 65:

Here, ∠1 and ∠4 form a linear pair

∠1 + ∠4 = 180°

(2x + y)° + (x + 2y)° = 180°

3(x + y)° = 180°

As I || m and n is a transversal

(x + 2y)° = (3y + 20)°

2x = 80 = x = 40

From (i), we have

40 + y = 60 ⇒ y = 20

Now, ∠1 = (2 x 40 + 20)° = 100°

∠4 = (40 + 2 x 20)° = 80°

∠8 = ∠4 = 80° [corresponding ∠s]

∠1 = ∠3 = 100° [vertically opp. ∠s]

∠7 = ∠3 = 100° [corresponding ∠s]

Hence, ∠7 = 100° and ∠8 = 80°

Question 66: In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28o and ∠QRT = 65°. Find the values of x, y and z.

Solution 66:

Here, PQ || SR.

⇒ ∠PQR = ∠QRT

⇒ x + 28° = 65°

⇒ x = 65° – 28° = 37°

Now, in ∆SPQ, ∠P = 90°

∴ ∠P + x + y = 180° [i.e.,angle sum property]

∴ 90° + 37° + y = 180°

⇒ y = 180° – 90° – 37° = 53°

Now, ∠SRQ + ∠QRT = 180° [linear pair]

z + 65° = 180°

z = 180° – 65° = 115°

Question 67: In the figure, AP and DP are bisectors of two adjacent angles, A and D, of a quadrilateral ABCD. Prove that 2∠APD = ∠B + ∠C.

Solution 67:

In quadrilateral ABCD, we have

∠A + ∠B + ∠C + ∠D = 360°

1 2 ∠A + 1 2 ∠B + 1 2 ∠C + 1 2 ∠D = 1 2 X360 0

1 2 ∠A + 1 2 ∠D = 180 0 – + 1 2 (∠B + ∠C)

As, AP and DP are the bisectors of ∠A and ∠D.

Therefore, ∠PAD = 1 2 ∠A

and, ∠PDA = 1 2 ∠D

Now, ∠PAD + ∠PDA = 180 0 – 1 2 (∠B + ∠C) (i)

In APD, we have,

∠APD + ∠PAD + ∠PDA = 180 0

∠APD + 180 0 – 1 2 (∠B + ∠C) = 180 0 [ Using (i)]

∠APD = 1 2 (∠B + ∠C)

2 ∠APD = (∠B + ∠C)

Question 68: In the figure, PS is the bisector of ∠QPR; PT ⊥ RQ and Q > R. Show that ∠TPS = 1 2 (∠Q – ∠R).

Solution 68:

As PS is the bisector of ∠QPR

∴∠QPS = ∠RPS = x (suppose)

In ∆PRT, we have

∠PRT + ∠PTR + ∠RPT = 180°

⇒ ∠PRT + 90° + ∠RPT = 180°

⇒ ∠PRT + ∠RPS + ∠TPS = 90°

⇒ ∠PRT + x + ∠TPS = 90°

⇒ ∠PRT or ∠R = 90° – ∠TPS – x

Again in ΔΡQT, we have

∠PQT + ∠PTQ + ∠QPT = 180°

⇒ ∠PQT + 90° + ∠QPT = 180°

⇒ ∠PQT + ∠QPS – TPS = 90°

⇒ ∠PQT + x – ∠TPS = 90° [Since ∠QPS = x]

⇒ ∠PQT or ∠Q = 90° + ∠TPS – x …equation (ii)

Subtracting equation (i) from (ii), we get

⇒ ∠Q – ∠R = (90° + ∠TPS – x) – 190° – ∠TPS – x)

⇒ ∠Q – ∠R = 90° + ∠TPS – X – 90° + ∠TPS + x

⇒ 2∠TPS = 2Q- ∠R

⇒ ∠TPS = 1 2 (Q – ∠R)

Question 69: In a triangle, if ∠A = 2∠B = 6∠C, find the measures of ∠A, ∠B and ∠C, also find the value of ∠A+2∠B 3∠C .

Solution 69:

We consider ∠A = 2∠B = 6∠C = x

2∠B = x = ∠B = 1 2

6 ∠ C = x ∠C = x 6

We know that ∠A + ∠B + ∠C = 180 0

[ using angle sum property of a ]

x + x 2 + x 6 = 180 0

6x + 3x + x = 180 0 x 6

10x = 1080 0 x = 108 0

x = ∠A = 108 0

Also, ∠B = x 2 = 108 2 = 54 0

∠C = x 6 = 108 6 = 18 0

Thus, ∠A = 108 0 , ∠B = 54 0 , ∠C = 18 0

Now, ∠A + 2∠B 3∠C = 108 + 108 3 x 18 = 216 54 = 4 0

Question 70: Students in a school are preparing banners for a rally. The parallel lines I and m are cut by the transversal t. If ∠4 = ∠5 and ∠6 = ∠7, what is the measure of angle 8?

Solution 70:

Here, given, ∠4 = ∠5 and ∠6 = ∠7

Now, I || m and t is the transversal

∴ ∠4 + ∠5 + ∠6 + ∠7 = 180° [Since co-interior angles are supplementary]

∠5 + ∠5 + ∠6 + ∠6 = 180° [using equation (i)]

2(∠5 + ∠6) = 180°

∠5 + ∠6 = 90°

We know, the sum of all the interior angles of a triangle is 180°

∴ ∠8 + ∠5 + ∠6 = 180°

⇒ ∠8 + 90° = 180° [using (ii)]

⇒ ∠8 = 180° – 90° = 90°

Benefits of Solving Important Questions Class 9 Mathematics Chapter 6

The chapter on Lines and Angles is divided into two sections; the first section teaches students about lines, while the second section teaches them about various angles. It’s crucial to master the principles of lines and angles in order to comprehend geometry in classes 9th and 10th.

Below are a few benefits for students to practise from our set of Important Questions Class 9 Mathematics Chapter 6:

It is a one-stop solution, meaning students will get questions from many different sources. Consequently, they will save time during preparation by gathering the questions and responses in advance.
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Since every question and answer is created with the exam in mind, students can confidently approach exams and overcome exam anxiety.

Be in touch with us at Extramarks to get the important exam questions for CBSE Class 9 Mathematics, all chapters, with solutions. Use the additional questions offered here by clicking on the links below to practise your revision.

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Q.1 In a PQR, if 2P = 3Q = 2R, find the measures of P, Q and R.

Marks: 4 Ans

We know that the sum of angles of a triangle is 180. P + Q + R = 180 o … i Given, 2 P = 3 Q = 2 R P = 3 2 Q And, R = 3 2 Q By putting the values of P and R in i , we get 3 2 Q + Q + 3 2 Q = 180 8 2 Q = 180 Q = 360 8 = 45 P = 3 2 Q = 3 2 — 45 = 67 . 5 And, R = 3 2 Q = 3 2 — 45 = 67 . 5 Hence, the measures of P , Q and R are 67 . 5 , 45 and 67 . 5 respectively.

Marks: 3 Ans

Marks: 2 Ans

Given, PQ RS As SQ is transversal y = 46 ° [Alternate interior angles] Again, as SP PQ SPQ = 90 ° In SPQ, x + 90 ° + 46 ° = 180 o [Sum of angles of a triangle is 180 o ] x = 44 ° Hence, in the given figure, x = 44 ° and y = 46 °

Q.4 If one angle of a triangle is equal to the sum of other two, show that the triangle is right-angled-triangle.

Marks: 1 Ans

Let x, y and z be three angles of a ABC. We know that the sum of angles of a triangle is 180 ° x + y + z = 180 ° (i) As per the given condition Let x = y + z y + z + y + z = 180 ° 2(y + z) = 180 ° y + z = 90 ° x = 180 ° – 90 ° = 90 ° [From (i)] ABC is a right-angled-triangle

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Solving important questions Class 9 Mathematics Chapter 6 is the greatest technique for mastering the chapters of Class 9 Math. There are several questions in it that are crucial for the exams. These solved questions are meant to help students understand the steps that must be taken while addressing complex questions Additionally, it improves analytical and logical thinking skills, which are important for performing well in exams.

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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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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles PDF

Selfstudys is an educational website that has taken the initiative to help all the students score good marks in their exams. They provide various study materials, conduct tests, and also help the students to check their Solutions which helps them give them a brief idea. One of the critical study materials is NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles, which allows the students to understand the topic deeply. Our subject matter experts at Selfstudys who have been in the field of education for quite some time have developed the Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions after devoting their time and energy. They have included all the topics which not only make the student confident before the exam but also help them score good marks in the exams. These Solutions are also created as per the latest curriculum of the CBSE which also keeps the students updated about the latest syllabus.

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Illustrations and examples: The NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles consists of illustrations and examples to clarify the concepts further. These examples help students enhance various skills and one of them is problem-solving skills.

What are the Benefits of the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles?

The Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions has several benefits to offer to the students. Some of the most important of them are:

Conceptual understanding: The NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles provides a comprehensive explanation of the concepts covered in Chapter 6 Lines and Angles. They help students develop a strong conceptual understanding of various maths concepts.
Accuracy: The Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions are prepared by subject matter experts, ensuring accuracy and correctness. They provide clear and step-by-step explanations, making it easier for students to grasp the concepts and solve problems correctly.
Time management: The Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions include various time management techniques and strategies which can be a complete game changer for all the students. By following the step-by-step Solutions, students can learn how to manage their time effectively during exams and solve problems more efficiently.
Self-assessment: The NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles allows students to self-assess their progress and performance. By comparing their Solutions with the provided Solutions, students can identify their strengths and weaknesses, and work on improving their understanding of the concepts.
Increases Confidence: Regular practicing of the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles helps the students to increase their confidence which can help them to score good marks in their exams.
Helps in clearing doubts: The Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions helps all the students to clear all their doubts while studying them. If a student comes across a hard question, they can refer to these Solutions to understand the correct approach and clear all their doubts.

What are the tips to study from the Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions?

There are various tips that can help the students to study effectively from the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles:

Read the chapter thoroughly: Begin by reading the chapter from your NCERT textbook to get to know about the concepts and topics covered. Look at the explanations, examples, and details provided in the Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions.
Understand the concepts: Before going through the Solutions, ensure you have a clear understanding of the concepts and definitions discussed in the chapter. If you come across any terms which you are not aware of, you can refer to the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles.
Use the Solutions as a guide: The Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions acts as a guide to help you understand the step-by-step process of solving problems. Instead of simply copying the Solutions, aim to understand each step and the reasoning behind it. This will enhance your problem-solving skills.
Compare your Solutions: After attempting the exercise questions, compare your answers with the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles. Analyze any differences and identify any mistakes or errors you made. This will allow you to learn from your mistakes and improve your problem-solving approach.
Do regular practice of the Solutions: To strengthen your understanding and skills, practice regularly with the exercises and additional practice questions provided in the Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions. The more you practice, the better you'll become at applying the concepts and solving problems.

What Are the Methods To Study From NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles?

Studying for the exam requires focus and concentration. Below we will discuss various methods of how students can study for an Exam with the help of NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles:

Read the Solutions: Start by reading the Maths Class 9 Chapter 6 Lines and Angles PDF NCERT Solutions as it can be a complete game changer for them. It can help the students in scoring good marks in the exams. Reading the Solutions on a regular basis will help the students to stick the concepts in their minds.
Understand the NCERT Solutions: After reading the Solutions, all the students should understand the Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions thoroughly as it helps to create a base for all the students. It can help them to strengthen their concepts which helps them to score good marks in exams. Students can understand all the concepts with in-depth knowledge.
Practice the questions regularly: It is advisable for all the students to regularly practice the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles as it helps the students to get familiar with all the concepts and also stick the concepts in their minds.
Identify your strengths and weaknesses: All the students should identify their strengths and weaknesses while solving the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles and work on them as it will help them to know where they are lacking behind. This increases their chances of scoring well in the exams.
Stay Relaxed: This is the most important method of studying the Class 9 Maths Chapter 6 Lines and Angles Miscellaneous Solutions and all the students should follow it. They should stay relaxed and not panic while going through the NCERT Solutions as being panicked affects their preparation.

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CBSE Important Questions for Class 9 Maths pdf

Important questions for class 9 maths chapter-6 lines and angles.

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Jul 21, 2022, 16:45 IST

CBSE Important Questions for class 9 maths chapter-6 Lines and Angles

Find below Important questions of class 9 maths of chapter-Lines and Angles prepared by academic team of Entrancei. All important questions of chapter Lines and Angles class 9 maths are uploaded in pdf form with detail step by step solutions.

Do solve NCERT questions and for reference use NCERT solutions for class 9 prepared by Entrancei team. For additional information related to the subject you can check the Maths Formula section.

Download free pdf for CBSE Important Questions for class 9 maths chapter Lines and Angles

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Test: Lines & Angles- Case Based Type Questions - EmSAT Achieve MCQ

10 questions mcq test - test: lines & angles- case based type questions, direction: bse stands for a disease called bovine spongiform encephalopathy. “bovine” means that the disease affects cows, “spongiform” refers to the way the brain from a sick cow looks spongy under a microscope, and “encephalopathy” indicates that it is a disease of the brain. this disease is commonly called “mad cow disease.” a farmer has a field abcd formed by two pair of parallel roads as shown below in which l ||m and p || q. his four cows suffering from bse. thus, he tied them at four corners of the field abcd. q. if ∠bac = 30°, find ∠acd..

Direction: BSE stands for a disease called Bovine Spongiform Encephalopathy. “Bovine” means that the disease affects cows, “spongiform” refers to the way the brain from a sick cow looks spongy under a microscope, and “encephalopathy” indicates that it is a disease of the brain. This disease is commonly called “mad cow disease.” A farmer has a field ABCD formed by two pair of parallel roads as shown below in which l ||m and p || q. His four cows suffering from BSE. Thus, he tied them at four corners of the field ABCD. Q. ∠ABC + ∠BCD = 180° as :

A. Corresponding angles are supplementary.
B. Alternate interior angles are supplementary.
C. Alternate exterior angles are supplementary.
D. Angles on the same side of a transversal are supplementary.

Direction: BSE stands for a disease called Bovine Spongiform Encephalopathy. “Bovine” means that the disease affects cows, “spongiform” refers to the way the brain from a sick cow looks spongy under a microscope, and “encephalopathy” indicates that it is a disease of the brain. This disease is commonly called “mad cow disease.” A farmer has a field ABCD formed by two pair of parallel roads as shown below in which l ||m and p || q. His four cows suffering from BSE. Thus, he tied them at four corners of the field ABCD. Q. If cow at C and cow at D is 2 km apart, then what is the distance between cow at A and cow at B?

So, AB = CD

Given, distance between cow at C and cow at D = CD = 2 km

⇒ AB = 2 km

Hence distance between cow at A and cow at B is 2 km.

Direction: BSE stands for a disease called Bovine Spongiform Encephalopathy. “Bovine” means that the disease affects cows, “spongiform” refers to the way the brain from a sick cow looks spongy under a microscope, and “encephalopathy” indicates that it is a disease of the brain. This disease is commonly called “mad cow disease.”

A farmer has a field ABCD formed by two pair of parallel roads as shown below in which l ||m and p || q. His four cows suffering from BSE. Thus, he tied them at four corners of the field ABCD.

Q. If ∠B = 45°, then ∠D = _________ .

(opposite angles of a parallelogram are equal)

Q. If we join BD such that BD meet AC at O and ∠BOC = 30°, then what is the measure of ∠AOD?

(vertically opposite angles are equal)

Direction: A plane mirror is a mirror with a flat reflective surface.

An incident ray is a ray of light that strikes a surface. The reflected ray corresponding to a given incident ray, is the ray that represents the light reflected by the surface. In figure, m and n are two plane mirrors perpendicular to each other.

Q. Parallel lines :

A. meet at a point
B. never meet
C. sometimes meet
D. meet at right angle

Q. If BO = 3 cm, AB = 5 cm then, AO =

So, AO is perpendicular to BO.

Thus, triangle AOB is a right-angled triangle.

OA 2 + OB 2 = AB 2

OA 2 + 3 2 = 5 2

OA 2 + 9 = 25

OA 2 = 25 – 9 = 16

Q. Which statement is incorrect ?

A. Corresponding angles formed at corresponding corners.
B. Alternate interior angles are equal
C. Angles on the same side of the transversal are complementary
D. Vertically opposite angles are equal

Q. ∠DBA + ∠BAC

Therefore, they are supplementary.

Hence, ∠DBA + ∠BAC = 180°

Q. Incident ray CA is :

A. parallel to AB
B. perpendicular to BD
C. parallel to BD
D. perpendicular to AO

As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.

So, BP ⊥ PA, i.e., ∠ BPA = 90°

Therefore, ∠ 3 + ∠ 2 = 90° (Angle sum property) (1)

Also, ∠1 = ∠2 and ∠4 = ∠3 (Angle of incidence = Angle of reflection)

Therefore, ∠1 + ∠4 = 90° [From (1)] (2) Adding (1) and (2), we have