ST is a straight line. Ex 6.1 Class 9 Maths Question 4. Now, in ∆PQS, Also, ∠GEF + ∠FED = ∠GED Lines and Angles Class 9 MCQs Questions with Answers. For proving AOB is a straight line, we will have to prove x+y is a linear pair. Question 1. Since, angle of incidence = Angle of reflection ⇒ 64° + 2∠QYP = 180° Question 1. Ex 6.2 Class 9 Maths Question 3. ∴ ∠FGE + ∠GED = 180° [Co-interior angles] This topic introduces you to the basic Geometry primarily focusing on the properties of the angles formed i) when two lines intersect each other and ii) when a line intersects two or more parallel lines at distinct points. It will make your concepts more clear. ∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360° Now from (i) and (ii), we get 5. ∠GEF = 126° -90° = 36° 6.28, find the values of x and y and then show that AB CD. Solution: In Fig. Here, BE ⊥ CF and the transversal line BC cuts them at B and C, So, 2 = 3 (As they are alternate interior angles), So, AB CD alternate interior angles are equal). Solve all the exercise problems of Lines and Angles. ⇒ ∠DCE = 180° – 53° – 35° = 92° Lines and Angles NCERT solution. Required fields are marked *. Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 6. ∴ ∠QTS = 45° [ ∵ ∠PTR = 45°] In Fig. ⇒ y = 127°- 50° = 77° So, GED = AGE = 126° (As they are alternate interior angles). and ∠BAC = 35° [Given] In the figure, we have CD and PQ intersect at F. Again, AB || CD ⇒ ∠PQR + ∠PRQ = 135° Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180° In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS. Now, BL || CM and BC is a transversal. 1. NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1. ⇒ ∠PTR = 180° – 95° – 40° = 45° ⇒ a = \(\frac { { 90 }^{ \circ } }{ 5 } \times 2\quad =\quad { 36 }^{ \circ }\) = 36° ⇒ ∠YOZ = 180° -27° – 32° = 121° It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. Answer : Q2 : In the given figure, lines XY and MN intersect at O. Now, for the linear pairs on the line XY-. ∴ AB || EF RS Aggarwal Solutions for Class 9 Maths Chapter 7 – Lines and Angles Exercise 7(A) PAGE: 198 1. We also know that vertically opposite angles are equal. Now, in ∆QRT, we have An angle greater than 90° but less than 180° is called an obtuse angle. Class 9 Maths Notes Chapter 6 Lines and Angles. Solution: Again ST || EF and RS is a transversal ∴ ∠OYZ = \(\frac { 1 }{ 2 } \angle XYZ\) = \(\frac { 1 }{ 2 }\)(54°) = 27° Solution: Solution: If POY = 90° and a : b = 2 : 3, find c. We know that the sum of linear pair are always equal to 180°. Get NCERT Solutions of all exercise questions and examples of Chapter 6 Class 9 Lines and Angles free at teachoo. These Worksheets for Grade 9 Lines and Angles, class assignments and practice … ∴ AB || CD. (AOC +BOE +COE) and (COE +BOD +BOE) forms a straight line. Lines and Angles Class 9 Solutions are prepared by highly qualified and professional teachers at Vedantu. ∴ Reflex ∠COE = 360° – 110° = 250° We know that AE is a transversal since AB DE. Skip to content. Thus, the required measure of c = 126°. Solution: These solutions help students prepare for their upcoming Board Exams by covering the whole syllabus, in accordance with the NCERT guidelines. 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). We know that QT and RT bisect PQR and PRS respectively. ∴ ∠APR = ∠PRD [Alternate interior angles] An angle which is greater than 180° but less than 360° is called a reflex angle.Further, two angles whose sum is 90° are ∠LBC + ∠ABL = ∠MCB + ∠MCD ⇒ 95° + 40° + ∠PTR =180° Ex 6.1 Class 9 Maths Question 1. ⇒ 110° + ∠PQR = 180° Since XY and MN interstect at O, In this. ∴ ∠ROS = \(\frac { 1 }{ 2 } (\angle QOS-\angle POS)\). ⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP] In figure, PQ and RS are two mirrors placed parallel to each other. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. Ex 6.1 Class 9 Maths Question 1 Now, AB || CD and GE is a transversal. ∠TRS = ∠TQR + ∠T …(2) Q 1. ⇒ ∠ABC = ∠BCD 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE. [Vertically opposite angles] TQP and PQR) will add up to 180°. ∴ ∠AOC + ∠COE + ∠EOB = 180° Thus, ∠XYQ = 122° and reflex ∠QYP = 302°. [Given] The architecture uses lines and angles to design the structure of a building. We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 help you. 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By going through these solutions students will get to learn about the basic concepts of a ray, line segment, intersecting, collinear and non-collinear points, and more. Ex 6.1 Class 9 Maths Question 1. Toppers Bulletin Menu. After solving the Line and Angles chapter of Class 9 Maths, you will get to know the following points: We hope this information on “NCERT Solution for Class 9 Maths Chapter 6 Lines and Angles” is useful for students. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1. In Fig. Videos related to exercise 6.2 in Hindi and English are also given for better understanding. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions. ⇒ ∠SRF = 180° – 130° = 50° Prove that ROS = ½ (QOS – POS). ∴ AB || CD. ∴ ∠XYZ + ∠ZYQ + ∠QYP = 180° 2(x + y) = 360° 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. Now, in ∆OYZ, we have ⇒ 2∠QYP = 180° – 64° = 116° 6.33, PQ and RS are two mirrors placed parallel to each other. Now consider the triangle CDE. 2 ∠ROS = (∠QOS – ∠POS) Angle of incidence = Angle of reflection (By the law of reflection), We also know that alternate interior angles are equal. Since ∠PQR =∠PRQ (as given in the question). ⇒ 64° + ∠ZYQ + ∠QYP = 180° Solution: Here, ∠ AOC and ∠ BOD are vertically opposite angles. ⇒ ∠YOZ + 27° + 32° = 180° This proves that alternate interior angles are equal and so, AB CD. 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. Lines and Angles Class 9 Extra Questions Maths Chapter 6. In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. ∴ (x + y) + (x + y) = 360° or, So, 28° + ∠RSQ = 65° 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. ⇒ z + y = 180° … (2) [By (1)] Students can also refer to NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles for better exam preparation and score more marks. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. We know that the angles on the same side of transversal is equal to 180°. Solution: (i) Angle – When two rays originate from the same end point, then an angle is formed. We hope the given RBSE Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Ex 5.2 will help you. Thus, x = 37° and y = 53°, Ex 6.3 Class 9 Maths Question 6. ∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair] From the diagram, b+c also forms a straight angle so. Also, AB and CD intersect at O. 1. Again, PQ ⊥ PS ⇒ AP = 90° ∴ x = z [Alternate interior angles] …. 4. Also, ∠AOC + ∠BOE = 70° [Angle sum property of a triangle] 6.13, lines AB and CD intersect at O. 6.42, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT. So, you can easily score marks if you have a thorough understanding of this topic. Lines and Angles (Mathematics) Class 9 - NCERT Questions. In Fig. Ex 6.1 Class 9 Maths Question 3. From (1) and (2), Refer to the NCERT Solutions of Class 9 provided by our Experts below. Putting the values as given in the question we get. Your email address will not be published. But (x + y) = (⇒ + w) [Given] In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x. When you stop at a signal and then move on when the signal light is green, then you either take a left angle turn or right-angle turn or move in a straight line. Out of which Geometry constitute a total of 22 marks which includes Introduction to Euclid’s Geometry, Lines and Angles, Triangles, Quadrilaterals, Areas, Circles, Constructions. We know that the sum of the interior angles of the triangle. ⇒ \(\frac { 1 }{ 2 }\)∠QPR = ∠QTR or ∠QTR = \(\frac { 1 }{ 2 }\)∠QPR. While practising the model solutions from this chapter, you will also learn to use the angle sum property of a triangle while solving problems. ∴ c = [a + ∠POY] [Vertically opposite angles] ⇒ ∠QRS = 110° – 50° = 60° ∴ b+a+∠POY= 180° PDF download free. Before starting to solve the exercise problems, you must first read the theory part and get to know the basic terms, definitions and theorems. In Fig. 6.43, if PQ ⊥ PS, PQ SR, SQR = 28° and QRT = 65°, then find the values of x and y. x +SQR = QRT (As they are alternate angles since QR is transversal). [Exterior angle property of a triangle] ⇒ 10z = 7 x 180° It is given the TQR is a straight line and so, the linear pairs (i.e. NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. rara POQ is a straight line. ∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°] ⇒ \(\frac { 1 }{ 2 }\)∠PRS = \(\frac { 1 }{ 2 }\)∠P + \(\frac { 1 }{ 2 }\)∠PQR We have, ∠TQP + ∠PQR = 180° Now, by putting the values of AOC+BOE = 70° and BOD = 40° we get. YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively. XYP is a straight line. ⇒ ∠ROS = 90° – ∠POS … (1) RD Sharma Solution for Class 9 Chapter 8 includes several exercises of Lines and Angles to help the students practice the concepts more effectively. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP. we have ∠P + ∠PQS + ∠PSQ = 180° All the chapter wise questions with solutions to help you to revise the complete CBSE syllabus and score more marks in Your board examinations. Solution: Let the required angle be x. 6.29, if AB CD, CD EF and y : z = 3 : 7, find x. RS Aggarwal Solutions Class 9 Chapter 4 Angles, Lines and Triangles. [Linear pair] or 50° + x = 180° AB || DE and AE is a transversal. 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. ⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°] 6.40, X = 62°, XYZ = 54°. NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1. ⇒ x + y = 180° [Co-interior angles] After that go through the solved examples of Lines and Angles that are given in the Class 9 NCERT Book. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. The sum of the three angles of a triangle is 180 degree. {Angle sum property of a triangle] If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Question 1: (i) Angle: Two rays having a common end point form an angle. 6.44, 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. ⇒ ∠PQR = 180° – 110° = 70° In this chapter 6″ lines and angles class 9 ncert solutions pdf” section you studied the following points: 1. If ∠POY = 90° , and a : b = 2 : 3. find c. NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles are part of NCERT Solutions for Class 9 Maths. Solution: First, draw two lines BE and CF such that BE ⊥ PQ and CF ⊥ RS. Get clarity on concepts like linear pairs, vertically opposite angles, co-interior angles, alternate interior angles etc. ∴ ∠ROQ = 90° Thus, x = 50° and y = 77°. NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles deals with the questions and answers related to the chapter Lines and Angles. 3. The chapter deals with lines and angles, its different types and formulas etc. [Vertically opposite angles] or c = 36° + 90° = 126° ∴ AOB is a straight line. Since, the side QP of ∆PQR is produced to S. Consider the ΔPQR. Pair of angles (reflex, complementary, supplementary, adjacent, vertical opposite, linear pair). Again, PQ is a straight line and EA stands on it. RD Sharma Solutions for Class 9 Mathematics CBSE, 10 Lines and Angles. In ∆ QRS, the side SR is produced to T. In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that Now, as the sum of the interior angles of the triangle. If and find ∠BOE and reflex ∠COE. or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)] PRS is the exterior angle and QPR and PQR are interior angles. ∠ABL = ∠LBC and ∠MCB = ∠MCD ⇒ ∠YZX = 180° – 54° – 62° = 64° From (1) and (2), we have In Fig. In figure, find the values of x and y and then show that AB || CD. In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y. ∠PQS + ∠PQR = ∠PRT + ∠PRQ Thus, ∠OZY = 32° and ∠YOZ = 121°, Ex 6.3 Class 9 Maths Question 3. Session 2020-2021 of an aircraft, then prove that AOB is a transversal are... Exercise problems with the questions and answers related to exercise 6.2 Lines and Angles an. Called an obtuse angle wise Solutions for Class 9 Maths Chapter 5 Plane Geometry and line and so PRS... 6, 7, find ∠QRS other at: a ) PAGE: 198 1 theorems related exercise! 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