NCERT Solutions for Class 9 Maths Chapter 6 – Lines And Angles Exercise 6.2

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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.2) Exercise 6.2 

The Ex 6.2 Class 9 Maths NCERT Solutions provides students to enhance their analytical and Mathematical skills by practising the unanswered questions given in Exercise 6.2 Class 9 Solutions. Downloading and practising the sample questions from the NCERT Solutions, for example, NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 regulates aid to students as well as teachers.

By preparing from the Class 9 Maths Exercise 6.2 Solution, students can effortlessly solve the questions related to the Parallel lines and Transversal mentioned in the Exercise. The Extramarks website acknowledges students to download the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2, and other solutions in PDF format to prepare for the Mathematics examination. Extramarks’ application or website makes it very easy for students to acquire the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 on their platform. Using the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2, as emphasised by Extramarks, students may fully prepare for their exams. Students who make use of NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 will be better able to understand the subject and apply Mathematics on a regular basis. To get in-depth pedagogy on how to solve Lines and Angles related questions with the use of various Theorems and Axioms, students must download the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 in a PDF file from the Extramarks website.

Access NCERT Solutions for Class 9 Mathematics Chapter 6 – Lines and Angles

Examining and achieving the Class 9 Maths Chapter 6 Exercise 6.2 NCERT Solutions helps students have a better understanding of how to cope with the Mathematical Syllabus. Students must browse through the Extramarks website or smartphone application to find the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 and other necessary solutions. Students can download them in PDF form for later use because they are readily available in that format. There are thorough explanations for every problem in the solutions, which are highly extensive. It is also possible for students to use the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2, as an important way to get ready for tests.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.2) Exercise 6.2

The NCERT Class 9 Mathematics textbook is a comprehensive book that covers all the relevant topics and concepts in Mathematics. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2, is specially designed for students studying in Class 9 and has the syllabus prescribed by the NCERT. The NCERT textbook is heavily relied upon by students and teachers and is considered one of the most significant textbooks in the field of Mathematics. Class 9 Mathematics is important as it lays the foundation for more advanced Mathematics courses in high school and college. The textbook is easy to grasp and includes a wealth of practice problems to help students master the material. Therefore, if a student of Class 9 is searching for appropriate help for the in-text Mathematics queries, they can explore the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 assisted by Extramarks. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2, are considered an important element in a student’s academic life and are also reviewed and encouraged by scholars and subject-matter professionals.

In Exercise 6.2 of Class 9, Chapter Lines and Angles, students will get to know about the following:

1) Parallel Lines and a Transversal – Interior and Exterior Angles, Consecutive interior or Allied Angles or Co-Interior Angles, Axiom of Corresponding Angles, and usage of Theorems.

2) Lines Parallel to the Same Line.

The questions given in the Exercise 6.2 are in total six questions that questions students regarding:

  • Finding the values of the mentioned integers
  • Finding the angles
  • Finding the angles with the help of drawing the parallel line and
  • Proving whether the line is parallel to others or not.

NCERT Solutions for Class 9

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Q.1 In Fig. 6.28, find the values of x and y and then show that AB || CD.


Ans

x+50°=180°          [Linear pair of angles]⇒x=180°−50°    =130°   130°=y              [Vertical opposite angles.]Since,​      x=y∴      AB∥CD                [Alternate interior angles]

Q.2 In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.


Ans

Since, y:z=3:7Let y=3a and z=7aSince, AB∥CD and CD∥EFso, AB∥EFx=z[Alternate interior angles.]∵      AB∥CDx+y=180°[Cointerior angles]z+y=180°[∵x=z]     7a+3a=180°           10a=180°⇒      a=180°10=18°So, x=7a=7(18°)=126°

Q.3 In Fig. 6.30, if AB||CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.


Ans

Given:AB∥CD,  EF⊥CD and ∠GED=126°.To find: ∠AGE, ∠GEF and ∠FGEAB∥CD∠AGE=∠GED    =126°∠GED=∠GEF+∠FED 126°=∠GEF+90°∠GEF=126°−90°    =36°∠AGE+∠FGE=180°   126°+∠FGE=180°       ∠FGE=180°−126°=54°

Q.4 In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.


Ans

Given:PQ∥ST, ∠PQR=110°,  ∠RST=130°To prove: ∠QRSConstruction: Draw line l paralllel to ST through R.Since, ST∥l↔, then∠ TSR+∠ SRW=180°           [Co−interior​ angles]    130°+∠ SRW=180°

            ∠ SRW=180°−130°    =50°Since,  PQ∥ST and ST∥l↔So,     PQ∥l↔∠PQR=∠SRW[Alternate interior angles.]   110°=∠QRS+∠SRW    =∠QRS+50°∠QRS=110°−50°    =60°

Q.5 In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Ans

Given:AB∥CD,  ∠APQ=50° and  ∠PRD = 127°To find: x and y Since, AB∥CD So, ∠APR=∠PRD        50°+y=127°⇒y=127°−50°    =77°And,       ∠APQ=∠PQR     50°=xThus, x=50° and y=77°.

Q.6 In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. 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. Prove that AB || CD.


Ans

Given:PQ∥RS, AB is incident ray to PQ and BC is incident ray       to RS.To Prove: AB∥CDConstruction: Draw MB⊥PQ and NC⊥RS.Proof: ∠ABM=∠MBC...(i)      [incident angle= reflected angle]       ∠BCN=∠NCD...(ii)      [incident angle= reflected angle]

Since, perpendiculars to parallel lines are parallel.So,   MB∥NC∴     ∠MBN=∠BCN 12∠ABC=12∠BCD⇒    ∠ABC=∠BCDSince, alternate interior angles are equal, so  AB∥CD.

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FAQs (Frequently Asked Questions)

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