⇒ 64° + 2∠QYP = 180° ⇒ ∠PQR + ∠PRQ = 135° Thus, x = 37° and y = 53°, Ex 6.3 Class 9 Maths Question 6. Mathematics NCERT Grade 9, Chapter 6: Lines and Angles: In this chapter students will study the properties of the angle formed when two lines intersect each other and properties of the angle formed when a line intersects two or more parallel lines at distinct points.The chapter starts from zero level, the first topic of the chapter being Basic Terms and Definitions. If ∠POY = 90° and a : b = 2 : 3, find c. 3. 1. Finance. NCERT Solutions for Class 9th: Ch 6 Lines and Angles Maths. ∴ ∠APR = ∠PRD [Alternate interior angles] 2. Prove that AB CD. 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). Solution: These Worksheets for Grade 9 Lines and Angles, class assignments and practice … This proves that alternate interior angles are equal and so, AB CD. In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT. In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x. In Fig. 2. lines which are parallel to a given lines are parallel to each other. So, PRS = QPR+PQR (According to triangle property). 6.40, X = 62°, XYZ = 54°. Draw a line EF parallel to ST through R. Lines and Angles Class 9 Exercise 6.1 : Solutions of Questions on Page Number : 96 Q1 : In the given figure, lines AB and CD intersect at O. ∴ PQ || EF and QR is a transversal [Angle sum property of a triangle] (Triangle property). There are 3 exercises present in NCERT Solutions for Class 9 Maths Chapter 6. or, (x + y) = \(\frac { { 360 }^{ \circ } }{ 2 }\) = 180° From the diagram, b+c also forms a straight angle so. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Ex 6.1 Class 9 Maths Question 1. Ex 6.1 Class 9 Maths Question 1. Also, recall that a straight angle is equal to 180°. 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. Now, putting the value of TQP = 110° we get. These solutions for Lines And Angles are extremely popular among Class 9 students for Math Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. ⇒ ∠POS + ∠ROS + 90° = 180° From (ii), we get or ∠FGE = 180° – 126° = 54° In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y. or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)] 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS. Ray OR is perpendicular to line PQ. In Fig. In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE. In this. Thus, the values of x and y are calculated as: 6. We know that the sum of the interior angles of the triangle. 6.14, lines XY and MN intersect at O. Thus, SPR (SPR = 135°) is equal to the sum of interior opposite angles. XYP is a straight line. Now PTR will be equal to STQ as they are vertically opposite angles. ∴ ∠ROQ = 90° Sum of all the angles at a point = 360° ∴ ∠AOC + ∠COE + ∠EOB = 180° 6.17, POQ is a line. What are the real-life applications of it? Stay tuned for further updates on CBSE and other competitive exams. 4. ∠PRS = ∠P + ∠PQR x = (90° – x) ⇒ 2x = 90° – x. If POY = 90° and a : b = 2 : 3, find c. We know that the sum of linear pair are always equal to 180°. NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles deals with the questions and answers related to the chapter Lines and Angles. Solution: (i) Angle – When two rays originate from the same end point, then an angle is formed. OS is another ray lying between rays OP and OR. First, construct a line XY parallel to PQ. ⇒ ∠POS + ∠ROS = 90° ⇒ ∠ROS + 90° = ∠QOS AB || CD and GE is a transversal. ⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°] In this chapter 6″ lines and angles class 9 ncert solutions pdf” section you studied the following points: 1. Lines and Angles Class 9 Extra Questions Maths Chapter 6. Thus, these are some questions for the different chapters starting from Class 9 Chapter 8 Introduction to Lines and Angles. But ∠RQS = 28° and ∠QRT = 65° ⇒ ∠ROS = ∠QOS – 90° ……(2) In ∆ QRS, the side SR is produced to T. We know that the angles around a point are 360° so. In Fig. Also a : b = 2 : 3 ⇒ b = \(\frac { 3a }{ 2 }\) …(ii) As they are pair of alternate interior angles. Question 1. ∴ ∠QTS = 45° [ ∵ ∠PTR = 45°] Basic terms and definitions related to a line segment, ray, collinear points, non-collinear points, intersecting and non-intersecting lines. NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. [Vertically opposite angles] ⇒ 50° = x [ ∵ ∠APQ = 50° (given)] 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. 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. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP. 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