Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1. Solution: Let the required angle be x. So. Again, PQ ⊥ PS ⇒ AP = 90° ∴ 53° + 35° + ∠DCE =180° The architecture uses lines and angles to design the structure of a building. Telangana SCERT Class 9 Math Chapter 4 Lines and Angles Exercise 4.3 Math Problems and Solution Here in this Post. (1) [YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ] 6.30, if AB CD, EF ⊥ CD and GED = 126°, find AGE, GEF and FGE. In Fig. In Fig. Now, by putting the values of AOC+BOE = 70° and BOD = 40° we get. After that go through the solved examples of Lines and Angles that are given in the Class 9 NCERT Book. In Fig. These solutions help students prepare for their upcoming Board Exams by covering the whole syllabus, in accordance with the NCERT guidelines. ∠TRS = ∠TQR + ∠T …(2) 6.15, PQR = PRQ, then prove that PQS = PRT. Stay tuned for further updates on CBSE and other competitive exams. NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 are part of NCERT Solutions for Class 9 Maths. They give a detailed and stepwise explanation to the problems given in the exercises in the NCERT Solutions for Class 9. As they are pair of alternate interior angles. \(\frac { 3a }{ 2 }\) + A = 90° In Fig. First, construct a line XY parallel to PQ. 6.16, if x+y = w+z, then prove that AOB is a line. ⇒ x = 180° – 50° = 130° …(2) But ∠XYZ = 54° and ∠ZXY = 62° In figure, if x + y = w + ⇒, then prove that AOB is a line. ∴ ∠ROS = \(\frac { 1 }{ 2 } (\angle QOS-\angle POS)\). 6.13, lines AB and CD intersect at O. [∵ ∠XYZ = 64° (given)] 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 2. lines which are parallel to a given lines are parallel to each other. [Given] If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Toppers Bulletin Menu. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE. 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. ⇒ ∠ABC = ∠BCD ∴ ∠PTR = ∠QTS Solution: In this. Get clarity on concepts like linear pairs, vertically opposite angles, co-interior angles, alternate interior angles etc. Lines and Angles NCERT solution. NCERT Solutions for Class 9 Maths Chapter 6 are useful for students as it helps them to score well in the class exams. {Angle sum property of a triangle] To access interactive Maths and Science Videos download BYJU’S App and subscribe to YouTube Channel. ∴ ∠PQS = ∠RSQ = 37° Viz. But ∠GED = 126° [Given] It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. Again, PQ is a straight line and EA stands on it. Thus, the required measure of c = 126°. 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. Now, in ∆PQS, 1. In this chapter 6″ lines and angles class 9 ncert solutions pdf” section you studied the following points: 1. Question 1. ⇒ ∠ROS = ∠QOS – 90° ……(2) Students can also refer to NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles for better exam preparation and score more marks. Ex 6.2 Class 9 Maths Question 1. 2. ∴ ∠APR = ∠PRD [Alternate interior angles] Now, in ∆OYZ, we have We have, ∠TQP + ∠PQR = 180° Also, ∠AOC + ∠BOE = 70° We hope the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 help you. Ex 6.2 Class 9 Maths Question 5. ∴ ∠FGE + ∠GED = 180° [Co-interior angles] Here BAC and AED are alternate interior angles. 6.42, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT. Ex 6.1 Class 9 Maths Question 4. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles. 6.29, if AB CD, CD EF and y : z = 3 : 7, find x. 5. 3. AB || CD and GE is a transversal. Solution: [∵ BL || PQ and CM || RS] So, ∠BAC = ∠AED It will help you to solve the questions in an easy way. i. e., a pair of alternate interior angles are equal. In Fig. It will make your concepts more clear. 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. 6.17, POQ is a line. Thus, x = 50° and y = 77°. RD Sharma Solutions for Class 9 Mathematics CBSE, 10 Lines and Angles. ⇒ a = \(\frac { { 90 }^{ \circ } }{ 5 } \times 2\quad =\quad { 36 }^{ \circ }\) = 36° We also know that vertically opposite angles are equal. 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. 6.14, lines XY and MN intersect at O. [Exterior angle property of a triangle] (Triangle property). From (ii), we get Ex 6.1 Class 9 Maths Question 1. ⇒ \(\frac { 1 }{ 2 }\)∠P = ∠T Since XOY is a straight line. NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles deals with the questions and answers related to the chapter Lines and Angles. Refer to NCERT Solutions for CBSE Class 9 Mathematics Chapter 6 Lines and Angles at TopperLearning for thorough Maths learning. x = 126°. 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. Thus, ∠XYQ = 122° and reflex ∠QYP = 302°. Ex 6.2 Class 9 Maths Question 2. 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. ∴ ∠XYZ + ∠ZYQ + ∠QYP = 180° In ∆PQR, side QR is produced to S, so by exterior angle property, Question 1. In ∆ QRS, the side SR is produced to T. Now, we know that the sum of the angles in a quadrilateral is 360°. 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. Ex 6.3 Class 9 Maths Question 1. or ∠COE = 180° – 70° = 110° ∴ ∠AGE = ∠GED [Alternate interior angles] 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. Parallel and Transversal Lines and theorems related to them. 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. We hope the given RBSE Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Ex 5.2 will help you. Telanagana SCERT Class 9 Math Solution Chapter 4 Lines and Angles Exercise 4.3 ∴ (x + y) + (x + y) = 360° or, If ray YQ bisects ZYP, find XYQ and reflex QYP. ∠GEF = 126° -90° = 36° Angle of incidence = Angle of reflection (By the law of reflection), We also know that alternate interior angles are equal. or, (x + y) = \(\frac { { 360 }^{ \circ } }{ 2 }\) = 180° YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively. NCERT Solutions Class 9 Maths Chapter 6 LINES AND ANGLES. ⇒ \(\frac { 5a }{ 2 }\) = 90° ∠YOZ + ∠OYZ + ∠OZY = 180° Also, AB and CD intersect at O. Question 1. Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair] [Angle sum property of a triangle] Since ∠XYQ = ∠XYZ + ∠ZYQ Again, AB || CD NCERT Solutions for Class 9th: Ch 6 Lines and Angles Maths. [Alternate interior angles] Answers to each question has been solved with Video. We computed that the value of XYQ = 122°. ⇒ ∠PTR = 180° – 95° – 40° = 45° ⇒ ∠ABL = ∠MCD …(2) [By (1)] [Angle sum property of a triangle] and EF || ST [Construction] ⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°] Now PTR will be equal to STQ as they are vertically opposite angles. If ∠POY = 90° and a : b = 2 : 3, find c. 3. 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. Ex 6.1 Class 9 Maths Question 3. ⇒ ∠ROS = 90° – ∠POS … (1) Thus, ∠BOE = 30° and reflex ∠COE = 250°. we have ∠TQR + \(\frac { 1 }{ 2 }\)∠P = ∠TQR + ∠T Q 1. In two parallel lines, the alternate interior angles are equal. In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y. [Vertically opposite angles] ⇒ ∠SRF = 180° – 130° = 50° Ex 6.1 Class 9 Maths Question 1 ∴ ∠AOC + ∠COE + ∠EOB = 180° In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS. MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers MCQs from Class 9 Maths Chapter 6 – Lines and Angles are provided here to help students prepare for their upcoming Maths exam. Since, angle of incidence = Angle of reflection In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE. Solution: ⇒ 64° + 2∠QYP = 180° By putting the value of XYZ = 64° and ZYQ = 58° we get. ⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP] Question 1: (i) Angle: Two rays having a common end point form an angle. ∠PRS = ∠P + ∠PQR Now, BL || CM and BC is a transversal. OS is another ray lying between rays OP and OR. ∴ AB || CD. ⇒ ∠YZX = 180° – 54° – 62° = 64° Adding (1) and (2), we have So, 28° + ∠RSQ = 65° 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. Since PQ || ST [Given] ⇒ 95° + 40° + ∠PTR =180° 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. In Fig. Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180° In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. Solution: ⇒ y = 180° – 90° – 37° = 53° 6. ∴ Its complement = 90° – x. ∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30° In Fig. We know that the sum of the interior angles of a triangle is 180°. Ex 6.2 Class 9 Maths Question 6. We know that the angles on the same side of transversal is equal to 180°. [Exterior angle property of a triangle] UP board high school students also use these solutions as UP Board Solutions updated for academic session 2020-2021. In the figure, we have CD and PQ intersect at F. Solution: Here, the side QP is extended to S and so, SPR forms the exterior angle. ⇒ ∠PRQ = 135° – 70° ⇒ ∠PRQ = 65°, Ex 6.3 Class 9 Maths Question 2. ∴ 54° + ∠YZX + 62° = 180° 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). As you can see that it constitutes approximately 27% of weightage. ∠AEP + ∠AEQ = 180° [Linear pair] But ∠RQS = 28° and ∠QRT = 65° For proving AOB is a straight line, we will have to prove x+y is a linear pair. ⇒ 50° = x [ ∵ ∠APQ = 50° (given)] [Alternate interior angles] 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE. Now, you must be wondering why we are studying Lines and Angles. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Since, the side QP of ∆PQR is produced to S. In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180° ∴ ∠QRT = ∠RQS + ∠RSQ In Fig. Students can now freely access RD Sharma Class 9 Maths solutions for chapter 8 here. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Solution: Now, putting values of QPR = y and APR = 127° we get. Also, recall that a straight angle is equal to 180°. Ex 6.1 Class 9 Maths Question 6. Draw a line EF parallel to ST through R. If and find ∠BOE and reflex ∠COE. 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 … In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT. ∴ AOB is a straight line. Solution: Thus, x = 37° and y = 53°, Ex 6.3 Class 9 Maths Question 6. Exercise 4A. 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. Solution: Solution: If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1, drop a comment below and we will get back to you at the earliest. ⇒ ∠DCE = 180° – 53° – 35° = 92° Thus, ∠AGE = 126°, ∠GEF=36° and ∠FGE = 54°. 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). PRS is the exterior angle and QPR and PQR are interior angles. We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 help you. Videos related to exercise 6.2 in Hindi and English are also given for better understanding. Solution: Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1. In figure, find the values of x and y and then show that AB || CD. 2. ∴ AB || EF Ex 6.1 Class 9 Maths Question 2. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS. It is given the TQR is a straight line and so, the linear pairs (i.e. rara POQ is a straight line. But ∠POY = 90° [Given] But (x + y) = (⇒ + w) [Given] ⇒ ∠PQR + ∠PRQ = 135° ∴ ∠AED = 35° In Fig. The chapter deals with lines and angles, its different types and formulas etc. If ∠POY = and ... Read more . 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