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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.

grade 9 lines and angles case study questions

(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

myCBSEguide App

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

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

Significance of Mathematics in Class 9

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

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

Case studies in Class 9 Mathematics

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

Example of Case study questions in Class 9 Mathematics

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

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

Class 9 Mathematics Case study question 1

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

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

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

Now answer the following questions:

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

Class 9 Mathematics Case study question 3

(a) 31.6 m²
(c) 513.3 m³
(b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

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

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

Class 9 Mathematics Curriculum at Glance

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

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

CBSE Class 9 Mathematics (Code No. 041)


I	NUMBER SYSTEMS	10
II	ALGEBRA	20
III	COORDINATE GEOMETRY	04
IV	GEOMETRY	27
V	MENSURATION	13
VI	STATISTICS & PROBABILITY	06

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)


1.	Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers. Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas	43	54
2.	Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.	19	24
3.	Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions	18	22
		80	100

myCBSEguide: Blessing in disguise

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

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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 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:

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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Maths Questions

Lines and Angles Class 9 Questions

Lines and angles Class 9 questions and solutions are available here to cover various problems on lines and angles. These questions are given as per Class 9 Maths Chapter 6 of the NCERT curriculum. Go through the lines and angles questions for Class 9 given below, along with their detailed solutions.

Definition of Lines and Angles

In mathematics and particularly in geometry, lines and angles are two related terms. Lines are straight one-dimensional figures extending endlessly in both directions, and do not have a thickness. The angles formed when lines meet at a common point is called the vertex.

Also, check: Lines and Angles Class 9 Notes

Lines and Angles Class 9 Questions and Answers

1. In the figure, lines PQ and RS intersect at point O. If ∠POR : ∠ROQ = 5 : 7, find all the angles.

From the given figure,

∠POR +∠ROQ = 180° (linear pair of angles since PQ is a straight line)

Also, given that,

∠POR : ∠ROQ = 5 : 7

Therefore, ∠POR = (5/12) × 180° = 75°

Similarly, ∠ROQ = (7/12) × 180° = 105°

Now, ∠POS = ∠ROQ = 105° (vertically opposite angles)

∠SOQ = ∠POR = 75° (vertically opposite angles)

2. In the figure, if x + y = w + z, then prove that AOB is a straight line.

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

i.e. x + y = 180°.

We know that the angles around a point are 360°.

So, x + y + w + z = 360°

x + y = w + z (given)

Thus, (x + y) + (x + y) = 360°

2(x + y) = 360°

∴ (x + y) = 180°

That means AOB is a straight line.

Hence proved.

3. 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.

PQ and RS are the two lines and a transversal AD intersects these two lines at points B and C, respectively.

Ray BE is the bisector of ∠ABQ and ray CG is the bisector of ∠BCS, and BE || CG.

To prove: PQ || RS

Given that ray, BE is the bisector of ∠ABQ.

Therefore, ∠ABE = (1/2) ∠ABQ….(1)

Similarly, ray CG is the bisector of ∠BCS.

Therefore, ∠BCG = (1/2) ∠BCS….(2)

As we know, BE || CG and AD is the transversal.

Therefore, ∠ABE = ∠BCG (corresponding angles axiom)…..(3)

Substituting (1) and (2) in (3), we get;

(1/2) ∠ABQ = (1/2) ∠BCS

⇒ ∠ABQ = ∠BCS

These are the corresponding angles formed by transversal AD with PQ and RS; they are equal.

Thus, by the converse of corresponding angles axiom, we have;

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

In the given figure, we can see that ∠APQ and ∠PQR are alternate interior angles.

That means, ∠APQ = ∠PQR = 50°

i.e., ∠PQR = x = 50°

∠APR = ∠PRD (alternate interior angles)

∠PRD = 127° (given)

⇒ ∠APR = 127°

We know that

∠APR = ∠APQ +∠QPR

⇒ 127° = 50°+ y

⇒ y = 127° – 50° = 77°

Therefore, x = 50° and y = 77°

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

In the given figure, AE is a transversal and AB || DE.

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

⇒ ∠BAC = ∠AED

⇒ ∠AED = 35° (since it is given that ∠BAC = 35°)

In the triangle CDE,

∠DCE + ∠CED + ∠CDE = 180° (angle sum property of the triangle)

Substituting the values, we get;

∠DCE + 35° + 53° = 180°

∠DCE + 88° = 180°

∠DCE = 180° – 88° = 92°

Therefore, ∠DCE = 92°

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

Construction:

Extend DE to intersect BC at point P.

Given that EF||BC and DP is the transversal,

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

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

∠DPC = ∠ABC …(2) [corresponding angles]

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

∠ABC = ∠DEF

7. An exterior angle of a triangle is 105°, and its two interior opposite angles are equal. Find the measure of each of these equal angles.

The exterior angle of triangle = 105°

Let x be the measure of two interior opposite angles of the triangle.

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

105° = x + x

⇒ 2x = 105°

⇒ x = 105°/2

⇒ x = 52½° = 52.5°

8. Bisectors of interior ∠B and exterior ∠ACD of a Δ ABC intersect at point T. Prove that ∠ BTC = ½ ∠ BAC.

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

To prove: ∠BTC = ½ ∠BAC

In △ABC, ∠ACD is an exterior angle.

We know that,

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

∠ACD = ∠ABC + ∠CAB

Dividing both sides of the equation by 2, we get;

⇒ ½ ∠ACD = ½ ∠CAB + ½ ∠ABC

⇒ ∠TCD = ½ ∠CAB + ½ ∠ABC .…(1)

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

As we know, the exterior angle of a triangle is equal to the sum of two opposite angles.

∠TCD = ∠BTC + ∠CBT

⇒ ∠TCD = ∠BTC + ½ ∠ABC ….(2)

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

½ ∠CAB + ½ ∠ABC = ∠BTC + ½ ∠ABC

⇒ ½ ∠CAB = ∠BTC

½ ∠BAC = ∠BTC

9. In the below figure, AB, CD and EF are three concurrent lines intersecting at O. Find the value of y.

∠AOE = ∠BOF = 5y (vertically opposite angles)

CD is a straight line.

Thus, ∠COE + ∠AOE + ∠AOD = 180°

2y + 5y + 2y = 180°

Hence, the value of y is 20°.

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

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

∠PRS = ∠QPR + ∠PQR

⇒ ∠PRS – ∠PQR = ∠QPR ———–(i)

In the ΔQRT,

∠TRS = ∠TQR + ∠QTR

⇒ ∠QTR = ∠TRS – ∠TQR

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

⇒ ∠PRS = 2 ∠TRS

Similarly, ∠PQR = 2∠TQR

∠QTR = ½ ∠PRS – ½∠PQR

⇒ ∠QTR = ½ (∠PRS -∠PQR)

⇒ ∠QTR = ½ ∠QPR [From (i)]

Practice Problems on Lines and Angles Class 9

Prove that two lines that are perpendicular to two intersecting lines intersect each other.
If the ratio of angles of a triangle is 2 : 4 : 3, find the angles.
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, find the greater of the two angles.

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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.

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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.

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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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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.

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Download CBSE Class 9 Maths Important Questions 2024-25 PDF

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

CBSE Class 9 Maths Important Questions
Sl.No	Chapter No	Chapter Name
1	Chapter 1
2	Chapter 2
3	Chapter 3
4	Chapter 4
5	Chapter 5
6	Chapter 6	Lines and Angles
7	Chapter 7
8	Chapter 8
9	Chapter 9
10	Chapter 10
11	Chapter 11
12	Chapter 12
13	Chapter 13
14	Chapter 14
15	Chapter 15

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) ${36^\circ }$

(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) ${90^\circ }$

(d) ${105^\circ }$

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

7. The measurement of Complete angle is

(a) ${0^\circ }$

(d) ${360^\circ }$

Ans: (d) ${360^\circ }$

8. The measurement of sum of linear pair is

(a) ${180^\circ }$

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

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

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

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

(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 $\angle ABC$

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

(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) ${128^\circ }$

(D) ${48^\circ }$

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

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

(A) ${10^\circ }$

(B) ${40^\circ }$

(D) ${45^\circ }$

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

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

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

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

(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) ${138^\circ }$

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

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

(b) ${150^0}$

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

(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) ${60^\circ }$

(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)

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

(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.

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.

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Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Solutions Answers

May 25, 2022 by Prasanna

Here we are providing Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Lines and Angles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 6 Lines and Angles with Solutions Answers

Lines and Angles Class 9 Extra Questions Very Short Answer Type

Question 1. If an angle is half of its complementary angle, then find its degree measure. Solution: Let the required angle be x ∴ Its complement = 90° – x Now, according to given statement, we obtain x = $\frac{1}{2}$(90° – x) ⇒ 2x = 90° – x ⇒ 3x = 90° ⇒ x = 30° Hence, the required angle is 30°.

Question 2. The two complementary angles are in the ratio 1 : 5. Find the measures of the angles. Solution: Let the two complementary angles be x and 5x. ∴ x + 5x = 90° ⇒ 6x = 90° ⇒ x = 15° Hence, the two complementary angles are 15° and 5 × 15° i.e., 15° and 75°.

Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Solutions Answers 1

Question 4. If an angle is 14o more than its complement, then find its measure. Solution: Let the required angle be x ∴ Its complement = 90° – x Now, according to given statement, we obtain x = 90° – x + 14° ⇒ 2x = 104° ⇒ x = 52° Hence, the required angle is 52o.

Solving the variable value in an arithmetic equation is not difficult anymore with our free online find the Value of X Calculator tool.

Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Solutions Answers 2

Lines and Angles Class 9 Extra Questions Short Answer Type 1

Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Solutions Answers 6

Lines and Angles Class 9 Extra Questions Short Answer Type 2

Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Solutions Answers 14

Lines and Angles Class 9 Extra Questions Long Answer Type

Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Solutions Answers 22

Lines and Angles Class 9 Extra Questions HOTS

Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Solutions Answers 29

Lines and Angles Class 9 Extra Questions Value Based (VBQs)

Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Solutions Answers 32

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

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Question 1. If a ray CD stands on a line AB, then prove that ∠ACD + ∠BCD = 180º. Solution: Let us draw CE ⊥ AB. ∴ ∠ACE = 90º and ∠BCE = 90º

Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Now, ∠ACD = ∠ACE + ∠ECD = 90º + ∠ECD …(1) ∠BCD = ∠BCE - ∠ECD = 90º - ∠ECD …(2) Adding (1) and (2), we have: ∠ACD + ∠BCD = [90º + ∠ECD] + [90º - (∠ECD)] = 90º + 90º + ∠ECD - ∠ECD = 180º

Note: Above example may be stated as: “The sum of the angles of linear pair is 180°.” OR “Sum of all the angles formed on the same side of a line at a given point on the line is 180°.”

Question 2. Two lines AB and CD intersect at a point O. Prove that: ∠AOD = ∠ BOC.

Solution: Since, OA stands on the given line CD. ∴ ∠AOC + ∠AOD = 180º ...(1) Again, OD stands on the given line AB. ∴ ∠AOD + ∠BOD = 180º ...(2) From (1) and (2), we have: ∠AOC + ∠AOD = ∠AOD + ∠BOD or ∠AOC = ∠BOD

Question 3. In the following figure, AOB is a straight line. Find ∠AOC and ∠BOD.

Solution: Since AOB is a straight line. ∴ The sum of all the angles on the same side of AOB at a point on it is 180º. ∠AOC + ∠COD + ∠DOB = 180º ∴ x + 60º + (2x - 15)º = 180º or 3x + 60º - 15º = 180º or 3x = 180º - 60º + 15º = 135º or x = (135°/3) = 45º Now 2x - 15 = 2(45) - 15 = 75º or ∠AOC = 45º and ∠BOD = 75º

Question 4. In the following figure, p: q: r = 2: 3: 4. If AOB is a straight line, then find the values of p, q and r.

Solution: Let p = 2x, q = 3x and r = 4x [∵ p : q : r = 2 : 3 : 4] ∵ AOB is a straight line. ∴ ∠AOC + ∠COD + ∠DOB = 180º or 2x + 3x + 4x = 180º or 9x = 180º or x = (180°/9) = 20º So, 2x = 2 x 20º = 40º ∴ 3x = 3 × 20° = 60° 4x = 4 × 20° = 80° Thus, p = 40°, q = 60°, r = 80°

Question 5. In the figure, AB || CD. GE and HF are the bisectors of ∠AEF and ∠EFD respectively. Show that GE || FH

If two parallel lines are intersected by a transversal, then show that the bisectors of a pair of alternate interior angles are parallel. Solution: ∵ AB || CD and EF is a transversal ∴ ∠AEF = ∠EFD [Interior alternate angles] ⇒ (1/2) ∠AEF =(1/2)∠EFD ⇒ ∠GEF = ∠HFE [∵ GF and HF are angle bisectors (given)] But they are the angles of a pair of interior alternate angles. ∴ GE || FH.

Question 6. In the figure, AB || CD. EG and FH are bisectors of ∠PEB and ∠EFD respectively. Show that EG || FH.

If two parallel lines are intersected by a transversal then prove that the bisectors of any pair of corresponding angles are parallel. Solution : ∵ AB || CD and EF is a transversal. ∴ ∠BEP = ∠EFD [corresponding angles] ⇒ (1/2) ∠BEP = (1/2)∠EFD ⇒ ∠PEG = ∠EFH [∵ EG and FH are the angle bisectors of ∠BEP and ∠EFD respectively] But they form pair of corresponding angles. ∴ EG || FH.

Question 7. If the arms of an angle are respectively parallel to the arms of another angle, then show that the two angles are either equal or supplementary. Solution: We have two angles ∠ABC and ∠DEF such that BA || ED and BC || EF in the same sense or in the opposite sense. ∴ We can have the following three cases: Case I: [Both pairs of arms are parallel in the same sense.]

∵ BA || ED and BC is a transversal, ∴ ∠1= ∠2 [Corresponding angles] …(1) Again, BC || EF and DE is a transversal, ∴ ∠3= ∠2 [Corresponding angles] …(2) From (1) and (2), we have ∠1= ∠3 i.e. ∠ABC = ∠DEF.

Case II: [Both pairs of arms are parallel in the opposite sense.]

∵ BA || ED and BC is a transversal, ∴ ∠1= ∠2 [Corresponding angles] …(1) Again BC || EF and ED is a transversal, [Alternate interior angles] ∴ ∠3= ∠2 …(2) From (1) and (2), we have ∠1= ∠3 i.e. ∠ABC = ∠DEF Case III: [One pair of arms are parallel in the same sense and another pair parallel in an opposite sense.]

∵ BA || ED and BC is a transversal. ∴ ∠1= ∠2 [Interior alternate angles] …(1) Again BC || EF and DE is a transversal. ∴ ∠3 + ∠2 = 180º [Sum of interior opposite angles] ⇒ ∠3 + ∠1 = 180º [From (1)] i.e. ∠DEF + ∠ABC = 180º Hence, ∠ABC and ∠DEF are supplementary.

Question 8. Prove that the sum of the angles of a triangle is 180º. Solution: Let us consider ΔABC and through A draw DE || BC.

∵ BC || DE and AB is a transversal. ∴ ∠4= ∠1 [Interior alternate angles] …(1)

Again, BC || DF and AC is a transversal. ∴ ∠5= ∠2 [Interior alternate angles] …(2) Adding (1) and (2), we have ∠4 + ∠5= ∠1 + ∠2 Adding ∠3 on both sides, we have ∠4 + ∠5 + ∠3 = ∠1 + ∠2 + ∠3 Since, ΔAF is a straight line, ∴ ∠4 + ∠3 + ∠5 = 180º ,∠1 + ∠2 + ∠3 = 180° i.e. ∠ABC + ∠BCA + ∠BAC = 180º ∴ The sum of angles of a triangle is 180º.

Question 9. Prove the following statement: “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.” Solution: Let us consider a triangle ABC such that its side BC is produced to D, forming exterior ∠ACD.

∵ Sum of the angles of a triangle = 180º ∴ ∠1 + ∠2 + ∠3 = 180º …(1) Since BCD is a straight line. ∴ ∠2 + ∠4 = 180º …(2) From (1) and (2), we have ∠2 + ∠4= ∠1 + ∠2 + ∠3 ⇒ ∠4= ∠1 + ∠3 i.e. Exterior ∠ACD = [Sum of the interior opposite angles]

Remember: An exterior angle of a triangle is greater than either of the interior opposite angles.

Question 10. The angles of a triangle are in the ratio 2 : 3 : 5. Find the measure of each angle of the triangle. Solution: Let the angles of the given triangle measure (2x)º, (3x)º and (5x)º. ∵ Sum of the angles of a triangle is 180º. ∴ (2x)º + (3x)º + (5x)º = 180º ⇒ 10x = 180º or x = (180°/10)= 18º ∴ 2x = 2 x 18 = 36º 3x = 3 x 18 = 54º 5x = 5 x 18 = 90º ∴ The measures of the angles of the given triangle are 36º, 54º and 90º.

Question 11. In a triangle, the bisectors of ∠B and ∠C intersect each other at a point O. Prove that ∠BOC = 90º + 1 2 ∠A. Solution: In a ΔABC, we have: ∠A + ∠B + ∠C = 180º [By angle sum property]

Again, in ΔOBC, we have ∠1 + ∠2 + ∠BOC = 180º [By angle sum property] ⇒ (∠1 + ∠2) + ∠BOC = 180º

Question 12. In a ΔABC, if ∠A + ∠B = 150º and ∠B + ∠C = 100. Find the measure of each angle of the triangle. Solution: We have ∠ A + ∠ B = 150º …(1) ∠B + ∠C = 100º …(2) Adding (1) and (2), we get ∴ ∠A + 2∠B + ∠C = 150 + 100º = 250º ⇒ (∠A + ∠B + ∠C) + ∠B = 250º ⇒ 180º + ∠B = 250º [∵ ∠ A + ∠B + ∠ C = 180º] ∴ ∠B = 250º - 180º = 70º. Now, ∠A + ∠B = 150º ⇒ ∠A = 150º - ∠B = 150º - 70º = 80º Also ∠B + ∠C = 100º ⇒ ∠C = 100 - ∠B = 100 - 70º = 30º Thus, ∠A = 80º, ∠B = 70º and ∠C = 30º.

Question 13. In a triangle, if ∠A = 2∠B = 6∠C, find the measures of ∠A, ∠B and ∠C. Solution: Let ∠A = 2∠B = 6∠C = x ∴ ∠A= x 2∠B= x ⇒ ∠B = x 2 6∠C = x ⇒ ∠C = x 6 We know that ∠A + ∠B + ∠C = 180º (using angle sum property)

or 6x 0 + 3x o + x o = 6 x 180 o

Question 14: The two complementary angles are in the ratio 1 : 5. Find the measures of the angles. Solution: Let the two complementary angles be x and 5x. ∴ x + 5x = 90° ⇒ 6x = 90° ⇒ x = 15° Hence, the two complementary angles are 15° and 5 × 15° i.e., 15° and 75°.

Solution: Here, PQ || RS, PS is a transversal. ⇒ ∠PSR = ∠SPQ = 56° Also, ∠TRS + m + ∠TSR = 180° 14° + m + 56° = 180° ⇒ m = 180° – 14 – 56 = 110°

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FAQs on Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

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2. How do you identify different types of angles?

3. What is the relationship between vertical angles?

4. How can we determine if two lines are parallel?

5. What is the sum of interior angles in a triangle?

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Unit 9: Lines and angles

Angle pairs.

Complementary & supplementary angles (Opens a modal)
Vertical angles (Opens a modal)
Identifying supplementary, complementary, and vertical angles Get 5 of 7 questions to level up!
Finding angle measures between intersecting lines Get 3 of 4 questions to level up!

Transversal

Angles formed between transversals and parallel lines (Opens a modal)
Missing angles with a transversal (Opens a modal)
Angle relationships with parallel lines Get 5 of 7 questions to level up!
Equation practice with angles Get 3 of 4 questions to level up!

MCQ Questions of Class 9 Maths Chapter 6 Lines and Angles with Answers

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Free PDF Download of CBSE Class 9 Maths Chapter 6 Lines and Angles Multiple Choice Questions with Answers.Based on Latest Exam Pattern. Students can solve NCERT Class 9 Maths Chapter 6 Lines and Angles Multiple Choice Questions with Answers to know their preparation level.

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Lines and Angles Class 9 MCQs Questions with Answers

MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers

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1. Angles of a triangle are in the ratio 2: 4 : 3. The smallest angle of the triangle is

3. Two angles whose sum is equal to 180° are called:

4. Each angle of an equilateral triangle is

5. If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be

7. If one angle of triangle is equal to the sum of the other two, then the triangle is

8. The angle of a triangle are in the ratio 5 : 3 : 7, the triangle is

9. Intersecting lines cut each other at:

11. If two lines intersect each other, then the vertically opposite angles are:

12. A straight angle is equal to:

13. An acute angle is:

14. A reflex angle is:

15. An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is

16. An exterior angle of a triangle is 80° and the interior opposite angles are in the ratio 1 : 3, measure of interior opposite angles are

17. In ΔABC, the bisectors of ∠ABC and ∠BCA intersect each other at O. The measure of ∠BOC is

18. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the greater of the two angles is

19. One of the angles of a triangle is 75°. If the difference of other two is 35°, then the largest angle of other two angles has a measure

20. A line joining two endpoints is called

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