[Angle sum property of a triangle] Your email address will not be published. Solution: Now, in ∆QRT, we have ⇒ ∠PQR = ∠QRF [Alternate interior angles] But ∠PQR = 110° [Given] In Fig. ⇒ ∠PQR + ∠PRQ = 135° and ∠OZY = \(\frac { 1 }{ 2 } \angle YZX\) = \(\frac { 1 }{ 2 }\)(64°) = 32° NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles deals with the questions and answers related to the chapter Lines and Angles. In figure, find the values of x and y and then show that AB || CD. ∴ y = 130° …(1) Ex 6.2 Class 9 Maths Question 6. In NCERT Solutions for Class 9 Maths Chapter 6, you will learn to solve the questions related to all the concepts of Lines and Angles. From (1) and (2), x = y In Fig. 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. XYP is a straight line. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. In Fig. Solution: But PQ and RS intersect at T. ⇒ ∠YOZ + 27° + 32° = 180° In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180° Refer to NCERT Solutions for CBSE Class 9 Mathematics Chapter 6 Lines and Angles at TopperLearning for thorough Maths learning. ∴ ∠AED = 35° ∠PQS + ∠PQR = ∠PRT + ∠PRQ [Hint : Draw a line parallel to ST through point R.]. Solution: 6.14, lines XY and MN intersect at O. As you can see that it constitutes approximately 27% of weightage. Lines and Angles Class 9 Solutions are prepared by highly qualified and professional teachers at Vedantu. 2. Now, you must be wondering why we are studying Lines and Angles. In Fig. ∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair] We computed that the value of XYQ = 122°. Now from (i) and (ii), we get ⇒ z + y = 180° … (2) [By (1)] The sum of the three angles of a triangle is 180 degree. [Exterior angle property of a triangle] 2. OS is another ray lying between rays OP and OR. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. Now, putting values of QPR = y and APR = 127° we get. It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. Also, ∠AOC + ∠BOE = 70° Since PQ || ST [Given] ∴ AB || EF Lines and Angles Class 9 Extra Questions Very Short Answer Type. What are the real-life applications of it? ∠YOZ + ∠OYZ + ∠OZY = 180° ∴ ∠PQS = ∠PRT. 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. Get clarity on concepts like linear pairs, vertically opposite angles, co-interior angles, alternate interior angles etc. Get NCERT Solutions of all exercise questions and examples of Chapter 6 Class 9 Lines and Angles free at teachoo. Solution: NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles deals with the questions and answers related to the chapter Lines and Angles. Thus, SPR (SPR = 135°) is equal to the sum of interior opposite angles. or ∠FGE + 126° = 180° 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE. Again ST || EF and RS is a transversal Thus, x = 37° and y = 53°, Ex 6.3 Class 9 Maths Question 6. i. e., a pair of alternate interior angles are equal. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions. Solution: NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. We also know that vertically opposite angles are equal. Ex 6.3 Class 9 Maths Question 1. Solution: 6. In ΔABC, ∠A = 50° and the external bisectors of ∠B and ∠C meet at O as shown in figure. 5. AB || CD, and CD || EF [Given] If ∠POY = 90° , and a : b = 2 : 3. find c. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS. But OR ⊥ PQ Solution: We hope the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 help you. ∠GEF = 126° -90° = 36° We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 help you. It is given the TQR is a straight line and so, the linear pairs (i.e. Consider the ΔPQR. If SPR = 135° and PQT = 110°, find PRQ. b = \(\frac { 3 }{ 2 }\) x 36° = 54° ⇒ ∠ROS + 90° = ∠QOS Ex 6.1 Class 9 Maths Question 1. Extra Questions for Class 9 Maths Chapter 6 Lines and Angles. 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. RD Sharma Solutions contains all in one solution for the different problem sets along with solved examples for ease of understanding. ∴ ∠POS + ∠ROS + ∠ROQ = 180° Toppers Bulletin Menu. ⇒ \(\frac { 5a }{ 2 }\) = 90° and EF || ST [Construction] All the exercise questions of Maths Class 9 Chapters are solved and it will be a great help for the students in their exam preparation and revision. But y : z = 3 : 7 Now, putting the value of PQR = 70° we get. Ex 6.1 Class 9 Maths Question 1 4. [Vertically opposite angles] Thus, these are some questions for the different chapters starting from Class 9 Chapter 8 Introduction to Lines and Angles. If a side of a triangle is produced, the exterior angle so formed is equal to the … ⇒ 70° + ∠PRQ = 135° [∠PQR = 70°] Intersecting lines cut each other at: a) […] In figure, lines AB and CD intersect at 0. ∴ ∠ROS = \(\frac { 1 }{ 2 } (\angle QOS-\angle POS)\). If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. To access interactive Maths and Science Videos download BYJU’S App and subscribe to YouTube Channel. First, draw two lines BE and CF such that BE ⊥ PQ and CF ⊥ RS. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ. The Class 9 Maths theory paper is of 80 marks. ⇒ ∠YZX = 180° – 54° – 62° = 64° ⇒ ∠ROS = ∠QOS – 90° ……(2) Ex 6.2 Class 9 Maths Question 2. Thus, ∠XYQ = 122° and reflex ∠QYP = 302°. It is also known that alternate interior angles are same and so, QRS +QRT = 180° (As they are a Linear pair). Now, in ∆PQS, Before starting to solve the exercise problems, you must first read the theory part and get to know the basic terms, definitions and theorems. Exercise 4A. ⇒ 90° + 37° + y = 180° YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively. Now, we know that the sum of the angles in a quadrilateral is 360°. Also, ∠GEF + ∠FED = ∠GED ⇒ ∠ABL = ∠MCD …(2) [By (1)] ∴ b + a = 180° – 90° = 90° …(i) Ex 6.1 Class 9 Maths Question 2. Solution: Let the required angle be x. ∠ABL = ∠LBC and ∠MCB = ∠MCD Ex 6.1 Class 9 Maths Question 4. 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. Answer : Q2 : In the given figure, lines XY and MN intersect at O. If and find ∠BOE and reflex ∠COE. Lines and Angles Class 9 Extra Questions Maths Chapter 6. ⇒ a = \(\frac { { 90 }^{ \circ } }{ 5 } \times 2\quad =\quad { 36 }^{ \circ }\) = 36° x = (90° – x) ⇒ 2x = 90° – x. We know that a linear pair is equal to 180°. or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ∵∠AOC + ∠BOE = 70° (Given)] In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.2, drop a comment below and we will get back to you at the earliest. From (ii), we get Now, from (1), we have ∠QRS + 50° = 110° For proving AOB is a straight line, we will have to prove x+y is a linear pair. NCERT Solutions for Class 9th: Ch 6 Lines and Angles Maths. These solutions help students prepare for their upcoming Board Exams by covering the whole syllabus, in accordance with the NCERT guidelines. NCERT Solutions for Class 9 Maths Chapter 6 are useful for students as it helps them to score well in the class exams. ∴ ∠XYZ + ∠ZYQ + ∠QYP = 180° In figure, PQ and RS are two mirrors placed parallel to each other. Putting the value of POY = 90° (as given in the question) we get, Similarly, b can be calculated and the value will be. These expert faculty solve and provide the NCERT Solution for Class 9, which would help students to solve the problems comfortably. or ∠FGE = 180° – 126° = 54° But ∠PQR = ∠PRQ [Given] ⇒ ∠ROS = 90° – ∠POS … (1) 1. If ray YQ bisects ZYP, find XYQ and reflex QYP. ∴ (x + y) + (x + y) = 360° or, ⇒ y = 127°- 50° = 77° Then you can start solving the exercise problems with the help of NCERT Solutions. These NCERT Solutions … As the angles on the same side of a transversal line sums up to 180°, O = z (Since they are corresponding angles), and, y +O = 180° (Since they are a linear pair), Now, let y = 3w and hence, z = 7w (As y : z = 3 : 7), Now, angle x can be calculated from equation (i). Solution: Since AB is a straight line, ∴ ∠AOC + ∠COE + ∠EOB = 180°. ∴ ∠OYZ = \(\frac { 1 }{ 2 } \angle XYZ\) = \(\frac { 1 }{ 2 }\)(54°) = 27° 4. Prove that ROS = ½ (QOS – POS). 1. Solution: In the figure, we have CD and PQ intersect at F. Ex 6.1 Class 9 Maths Question 5. ⇒ ∠PQR = 180° – 110° = 70° Your email address will not be published. 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. or, (x + y) = \(\frac { { 360 }^{ \circ } }{ 2 }\) = 180° So. Now, for the linear pairs on the line XY-. The architecture uses lines and angles to design the structure of a building. In figure, lines XY and MN intersect at 0. 2. Q 1. Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.1. In Fig. Now consider the triangle CDE. Since XY and MN interstect at O, Home; Maths; Subjects. Thus, ∠SQT = 60°, Ex 6.3 Class 9 Maths Question 5. ∴ ∠QTS = 45° [ ∵ ∠PTR = 45°] Since, the side QP of ∆PQR is produced to S. While practising the model solutions from this chapter, you will also learn to use the angle sum property of a triangle while solving problems. Students can also refer to NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles for better exam preparation and score more marks. ⇒ ∠SRF = 180° – 130° = 50° ∴ ∠QRT = ∠RQS + ∠RSQ ∴ ∠LBC = ∠MCB …(1) [Alternate interior angles] ⇒ y = 180° – 90° – 37° = 53° ⇒ ∠APQ + ∠QPR = 127° We know that AE is a transversal since AB DE. ST is a straight line. Now PTR will be equal to STQ as they are vertically opposite angles. [∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.] or c = 36° + 90° = 126° The answer is that lines and angles are everywhere around us. 3. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Finance. ⇒ ∠POS + ∠ROS + 90° = 180° 6.42, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT. ∴ ∠AOC + ∠COE + ∠EOB = 180° [Alternate interior angles] 6. ⇒ ∠DCE = 180° – 53° – 35° = 92° ⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)] Thus, ∠AGE = 126°, ∠GEF=36° and ∠FGE = 54°. ⇒ x = 180° – 50° = 130° …(2) If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Ex 5.2, drop a … Stay tuned for further updates on CBSE and other competitive exams. ⇒ 64° + 2∠QYP = 180° The chapter deals with lines and angles, its different types and formulas etc. Since, angle of incidence = Angle of reflection 5. 6.16, if x+y = w+z, then prove that AOB is a line. ⇒ ∠TRS = \(\frac { 1 }{ 2 }\)∠P + ∠TQR …(1) 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 We followed the latest Syllabus, while creating the NCERT Solutions and it is framed in accordance with the exam pattern of the CBSE Board. Thus, the values of x and y are calculated as: 6. If a ray stands on a line, then the sum of two adjacent angles so formed is 180 degree and vice versa. 1. Thus, ∠DCE = 92°, Ex 6.3 Class 9 Maths Question 4. Thus, the required measure of c = 126°. Since, PQ || SR and QS is a transversal. [Exterior angle property of a triangle] In Fig. 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. [Angle sum property of a triangle] RD Sharma Solution for Class 9 Chapter 8 includes several exercises of Lines and Angles to help the students practice the concepts more effectively. 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. 4. ⇒ ∠ABC = ∠BCD Solution: 2(x + y) = 360° Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 6. Ex 6.2 Class 9 Maths Question 1. Ex 6.2 Class 9 Maths Question 3. 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. 6.33, PQ and RS are two mirrors placed parallel to each other. Since AB is a straight line, or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)] Now, in ∆ TQS, we have ∠TSQ + ∠STQ + ∠SQT = 180° 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. But ∠GED = 126° [Given] These Worksheets for Grade 9 Lines and Angles, class assignments and practice … Ray OR is perpendicular to line PQ. All the chapter wise questions with solutions to help you to revise the complete CBSE syllabus and score more marks in Your board examinations. Again, AB || CD 4. 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. LINES AND ANGLES 91 An acute angle measures between 0° and 90°, whereas a right angle is exactly equal to 90°. We know that the sum of the interior angles of a triangle is 180°. Also, AB and CD intersect at O. 6.29, if AB CD, CD EF and y : z = 3 : 7, find x. NCERT Solutions Class 9 Maths Chapter 6 LINES AND ANGLES. In Fig. Skip to content. We have AB || CD and PQ is a transversal. Prove that From (1) and (2), Lines and Angles (Mathematics) Class 9 - NCERT Questions. 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. ⇒ 110° + ∠PQR = 180° Extra Questions for Class 9 Maths ⇒ \(\frac { 1 }{ 2 }\)∠PRS = \(\frac { 1 }{ 2 }\)∠P + \(\frac { 1 }{ 2 }\)∠PQR In Fig. ∴ ∠ROQ = 90° [YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ] ⇒ \(\frac { 1 }{ 2 }\)∠P = ∠T Now, in ∆OYZ, we have In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y. These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook. ⇒ ∠PRQ = 135° – 70° ⇒ ∠PRQ = 65°, Ex 6.3 Class 9 Maths Question 2. and ∠BAC = 35° [Given] or ∠COE = 180° – 70° = 110° In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE. All the solutions of Lines and Angles - Mathematics explained in detail by experts to help students prepare for their CBSE exams. z = \(\frac { 7 }{ 3 }\) y = \(\frac { 7 }{ 3 }\)(180°- z) [By (2)] Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180° We have, ∠TQP + ∠PQR = 180° 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. [Alternate interior angles] MCQs from CBSE Class 9 Maths Chapter 6: Lines and Angles 1. Draw ray BL ⊥PQ and CM ⊥ RS Thus, x = 50° and y = 77°. So, ∠BAC = ∠AED In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x. In the given figure, lines AB and CD intersect at O. But ∠BOD = 40° [Given] Question 1. In Fig. Adding (1) and (2), we have In figure, ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ. 6.17, POQ is a line. (ii) Interior of an angle: The interior of ∠AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A. Thus, ∠BOE = 30° and reflex ∠COE = 250°. Solve all the exercise problems of Lines and Angles. 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. ⇒ ∠PTR = 180° – 95° – 40° = 45° ∴ ∠AGE = ∠GED [Alternate interior angles] From (1) and (2), we have 6.15, PQR = PRQ, then prove that PQS = PRT. 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In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT. ⇒ ∠SQT = 180° – 75° – 45° = 60° ⇒ 10z = 7 x 180° We know that QT and RT bisect PQR and PRS respectively. [Given] ∴∠COA = ∠BOD [Vertically opposite angles] ∴ c = [a + ∠POY] [Vertically opposite angles] [∵ BL || PQ and CM || RS] Videos related to exercise 6.2 in Hindi and English are also given for better understanding. Solution: An angle greater than 90° but less than 180° is called an obtuse angle. OS is another ray lying between rays OP and OR. ∴ ∠PQS = ∠RSQ = 37° In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS. Also a : b = 2 : 3 ⇒ b = \(\frac { 3a }{ 2 }\) …(ii) (ii) Interior of an angle – The interior of ∠BAC is the set of all points in its plane which lie on the same side of AB as C and also on the same side of AC as B. After that go through the solved examples of Lines and Angles that are given in the Class 9 NCERT Book. Class 9 Maths Notes Chapter 6 Lines and Angles. Lines and Angles NCERT solution. ⇒ \(\frac { 1 }{ 2 }\)∠QPR = ∠QTR or ∠QTR = \(\frac { 1 }{ 2 }\)∠QPR. ∠AEP + ∠AEQ = 180° [Linear pair] 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). Again, AB || CD and PR is a transversal. ∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°] Answers to each question has been solved with Video. 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