# lines and angles class 9 solutions

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[∵ ∠XYZ = 64° (given)] Since XY and MN interstect at O, Solution: ∴ Reflex ∠COE = 360° – 110° = 250° But ∠PQR = ∠PRQ [Given] While practising the model solutions from this chapter, you will also learn to use the angle sum property of a triangle while solving problems. ∴ ∠XYZ + ∠ZYQ + ∠QYP = 180° ⇒ $$\frac { 1 }{ 2 }$$∠P = ∠T ⇒ ∠QYP = $$\frac { { 116 }^{ \circ } }{ 2 }$$ = 58° ⇒ ∠ABL = ∠MCD …(2) [By (1)] Thus, ∠XYQ = 122° and reflex ∠QYP = 302°. Students can also refer to NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles for better exam preparation and score more marks. ∴ ∠PQS = ∠RSQ = 37° Download free printable worksheets for CBSE Class 9 Lines and Angles with important topic wise questions, students must practice the NCERT Class 9 Lines and Angles worksheets, question banks, workbooks and exercises with solutions which will help them in revision of important concepts Class 9 Lines and Angles. Again ST || EF and RS is a transversal ⇒ ∠YZX = 180° – 54° – 62° = 64° NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 are part of NCERT Solutions for Class 9 Maths. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Solution: ⇒ ∠YOZ + 27° + 32° = 180° Toppers Bulletin Menu. Solution: I n the figure, lines AB and CD intersect at O. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1. 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. NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 Lines and Angles in both Hindi Medium and English Medium. ⇒ ∠QRS = 110° – 50° = 60° ⇒ x + y = 180° [Co-interior angles] ∠PQS + ∠PQR = ∠PRT + ∠PRQ ∴ ∠PTR = ∠QTS ⇒ x = 37° Get NCERT Solutions of all exercise questions and examples of Chapter 6 Class 9 Lines and Angles free at teachoo. NCERT Solutions Class 9 Maths Chapter 6 LINES AND ANGLES. ∴ AB || CD. ∵ PQ || RS ⇒ BL || CM ∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30° z = $$\frac { 7 }{ 3 }$$ y = $$\frac { 7 }{ 3 }$$(180°- z) [By (2)] Also, AB and CD intersect at O. 6.15, PQR = PRQ, then prove that PQS = PRT. In figure, lines AB and CD intersect at 0. ∴ ∠AGE = ∠GED [Alternate interior angles] The RS Aggarwal Solutions for Class 9 Chapter-7 Lines and Angles Solutions Maths have been provided here for the benefit of the CBSE Class 9 students. Viz. So, ∠BAC = ∠AED We also know that vertically opposite angles are equal. [Angle sum property of a triangle] In figure, lines XY and MN intersect at 0. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE. {Angle sum property of a triangle] In Fig. ⇒ ∠SRF = 180° – 130° = 50° In Fig. Ex 6.1 Class 9 Maths Question 6. b = $$\frac { 3 }{ 2 }$$ x 36° = 54° MCQs from CBSE Class 9 Maths Chapter 6: Lines and Angles 1. So, PRS = QPR+PQR (According to triangle property). Pair of angles (reflex, complementary, supplementary, adjacent, vertical opposite, linear pair). [Linear pair] 4. ⇒ 64° + ∠ZYQ + ∠QYP = 180° Now, as the sum of the interior angles of the triangle. ⇒ ∠POS + ∠ROS + 90° = 180° ∴ ∠POS + ∠ROS + ∠ROQ = 180° These solutions help students prepare for their upcoming Board Exams by covering the whole syllabus, in accordance with the NCERT guidelines. ∴ Its complement = 90° – x. Now PTR will be equal to STQ as they are vertically opposite angles. Intersecting lines cut each other at: a) […] ⇒ 2∠QYP = 180° – 64° = 116° In this chapter 6″ lines and angles class 9 ncert solutions pdf” section you studied the following points: 1. OS is another ray lying between rays OP and OR. Lines and Angles Class 9 Extra Questions Very Short Answer Type. Now, AB || CD and GE is a transversal. NCERT Solutions for Class 9 Maths Chapter 6 are created by the BYJU’S expert faculty to help students in the preparation of their examinations. Sum of all the angles at a point = 360° Answer : Q2 : In the given figure, lines XY and MN intersect at O. It will help you to solve the questions in an easy way. Thus, ∠QRS = 60°. Ex 6.2 Class 9 Maths Question 4. OS is another ray lying between rays OP and OR. Now, from (1), we have ∠QRS + 50° = 110° or c = 36° + 90° = 126° Putting the value of POY = 90° (as given in the question) we get, Similarly, b can be calculated and the value will be. NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1. In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT. Also a : b = 2 : 3 ⇒ b = $$\frac { 3a }{ 2 }$$ …(ii) Solution: Prove that ROS = ½ (QOS – POS). Solution: ∠ABL = ∠LBC and ∠MCB = ∠MCD Solution: ∠GEF = 126° -90° = 36° You can download the complete solution pdf of NCERT Chapter 6 Line and Angles of Class 9 by clicking on the link below: List of Exercises in class 9 Maths Chapter 6, Exercise 6.1 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)Exercise 6.2 Solutions 6 Questions (3 Short Answer Questions, 3 Long Answer Question)Exercise 6.3 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question). x = 126°. Home; Maths; Subjects. ∴ 53° + 35° + ∠DCE =180° ∠TRS = ∠TQR + ∠T …(2) Now, putting the value of TQP = 110° we get. 6.40, X = 62°, XYZ = 54°. First, construct a line XY parallel to PQ. In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180° 1. ⇒ ∠ROS + 90° = ∠QOS Thus, ∠SQT = 60°, Ex 6.3 Class 9 Maths Question 5. (AOC +BOE +COE) and (COE +BOD +BOE) forms a straight line. ∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360° Solution: NCERT Solutions for Class 9th: Ch 6 Lines and Angles Maths. Lines and Angles Class 7 NCERT Book: If you are looking for the best books of Class 7 Maths then NCERT Books can be a great choice to begin your preparation. 6.13, lines AB and CD intersect at O. ∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°] In Fig. and ∠OZY = $$\frac { 1 }{ 2 } \angle YZX$$ = $$\frac { 1 }{ 2 }$$(64°) = 32° These Worksheets for Grade 9 Lines and Angles, class assignments and practice … But (x + y) = (⇒ + w) [Given] But ∠GED = 126° [Given] These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook. [YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ] In Fig. All questions and answers from the Rs Aggarwal 2018 Book of Class 9 Math Chapter 7 are provided here for you for free. Parallel and Transversal Lines and theorems related to them. ∴∠QRF = ∠QRS + ∠SRF = 110° …(1) Now, putting the value of APQ = 50° and PQR = x we get, Or, APR = 127° (As it is given that PRD = 127°). Solution: We know that a linear pair is equal to 180°. Ex 6.1 Class 9 Maths Question 2. Solution: All the solutions of Lines and Angles - Mathematics explained in detail by experts to help students prepare for their CBSE exams. rara POQ is a straight line. Ex 6.1 Class 9 Maths Question 1. Telangana SCERT Class 9 Math Chapter 4 Lines and Angles Exercise 4.3 Math Problems and Solution Here in this Post. But ∠XYZ = 54° and ∠ZXY = 62° Question 1. YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively. Now, you must be wondering why we are studying Lines and Angles. LINES AND ANGLES 91 An acute angle measures between 0° and 90°, whereas a right angle is exactly equal to 90°. Ex 6.2 Class 9 Maths Question 2. Here BAC and AED are alternate interior angles. ∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair] Thus, the values of x and y are calculated as: 6. Thus, these are some questions for the different chapters starting from Class 9 Chapter 8 Introduction to Lines and Angles. In figure, lines AB and CD intersect at 0. Ex 6.1 Class 9 Maths Question 1. ∴ AB || EF In this. [Exterior angle property of a triangle] ∴ ∠AED = 35° 2. Ex 6.3 Class 9 Maths Question 1. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles. All the chapter wise questions with solutions to help you to revise the complete CBSE syllabus and score more marks in Your board examinations. 6. [Vertically opposite angles] ∴∠AGE = 126° 5. TQP and PQR) will add up to 180°. Solve all the exercise problems of Lines and Angles. [Given] ∴∠COA = ∠BOD [Vertically opposite angles] NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1, Exercise 6.2 and Exercise 6.3 in English Medium as well as Hindi Medium updated for new academic session 2020-2021 based on latest NCERT Books. 4. Finance. ⇒ 10z = 7 x 180° So, you can easily score marks if you have a thorough understanding of this topic. Thus, ∠DCE = 92°, Ex 6.3 Class 9 Maths Question 4. In Fig. We know that the angles around a point are 360° so. As they are pair of alternate interior angles. [ ∵ ∠P = 95°, ∠R = 40° (given)] Ray OR is perpendicular to line PQ. 2 ∠ROS = (∠QOS – ∠POS) In two parallel lines, the alternate interior angles are equal. But y : z = 3 : 7 Since, PQ || SR and QS is a transversal. NCERT Solutions for Class 9 Maths Chapter 6 are useful for students as it helps them to score well in the class exams. ∴ y = 130° …(1) ⇒ ∠PQR = 180° – 110° = 70° ⇒ 95° + 40° + ∠PTR =180° If and find ∠BOE and reflex ∠COE. Since, angle of incidence = Angle of reflection Since PQ || ST [Given] ∠AEP + ∠AEQ = 180° [Linear pair] The sum of the three angles of a triangle is 180 degree. we have ∠P + ∠PQS + ∠PSQ = 180° Now, according to given statement, we obtain. In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y. We, in our aim to help students, have devised detailed chapter wise solutions for them to understand the concepts easily. In the question, it is given that (OR ⊥ PQ) and POQ = 180°, Now, POS+ROS = 180°- 90° (Since POR = ROQ = 90°), As POS + ROS = 90° and QOS – ROS = 90°, we get. So. NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1. ⇒ 70° + ∠PRQ = 135° [∠PQR = 70°] [Exterior angle property of a triangle] Theorem videos are also available.In this chapter, we will learnBasic Definitions- Line, Ray, Line Segment, Angles, Types of Angles We know that QT and RT bisect PQR and PRS respectively. Solution: Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 9 so that you can refer them as and when required. In figure, find the values of x and y and then show that AB || CD. So, GED = AGE = 126° (As they are alternate interior angles). We know that the sum of the interior angles of a triangle is 180°. We have AB || CD and PQ is a transversal. [∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.] Ex 6.2 Class 9 Maths Question 6. Since, the side QP of ∆PQR is produced to S. But ∠BOD = 40° [Given] Lines and Angles NCERT solution. CBSETuts.com provides you Free PDF download of NCERT Exemplar of Class 9 Maths Chapter 6 Lines And Angles solved by expert teachers as per NCERT (CBSE) Book guidelines. ∴ (x + y) + (x + y) = 360° or, Your email address will not be published. 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE. From (ii), we get If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ. We computed that the value of XYQ = 122°. 6.14, lines XY and MN intersect at O. ∴ b+a+∠POY= 180° 1. 6.30, if AB CD, EF ⊥ CD and GED = 126°, find AGE, GEF and FGE. [Angle sum property of a triangle] If ray YQ bisects ZYP, find XYQ and reflex QYP. Ex 6.1 Class 9 Maths Question 4. Solution: Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair] If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. ∴ ∠FGE + ∠GED = 180° [Co-interior angles] 1. Lines and Angles Class 9 MCQs Questions with Answers. 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