NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles Exercise 6.2

NCERT Class 9 Maths Chapter 6 - Lines and Angles Exercise 6.2 has been developed to create problems related to the properties of parallel lines and the angles formed with transversals. NCERT Solutions Class 9 Maths provides the basis for having a concrete understanding of angle relationships that are important as we move further into mathematics.

1.0Download Class 9 Maths Chapter 6 Ex 6.2 NCERT Solutions PDF

You can download the NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles Ex 6.2 PDF which contains organised answers, shortcuts and an explanation of the Concepts used for the CBSE based question.

2.0Key Concepts of Exercise 6.2

• Transversal and Parallel Lines
• Alternate Interior Angles
• Corresponding Angles
• Vertically Opposite Angles
• Linear Pair Axiom
• Angles on Same Side of Transversal (Co-Interior Angles)
• This exercise uses these properties to find unknown angles using some logic.

3.0CBSE Class 9 Chapter 6 Linear Equations Exercise 6.2 Comprises

1. Identifying Angle Relationships on Parallel Lines 

Students will be given angle problems based on three properties of angles formed when a transversal crosses two parallel lines: 

• Corresponding Angles 
• Alternate Interior Angles 
• Alternate Exterior Angles 
• Co-Interior (Same Side Interior) Angles 

2. Using Axioms and Theorems within Geometry 

Using the below properties to find unknown angles: 

• Corresponding angles are equal if lines are parallel 
• Equivalent Alternate interior Angles 
• Co-interior angles = 180° (Supplementary) 
• Vertically Opposite Angles are equal 
• Linear Pair of Angles = 180° 

3. Problem Solving using Diagrams 

• To write down or read geometrical figures and say what is the type of angles 
• To mark angles on diagrams appropriately 
• To give the proper reasoning for every step (e.g., “∠A = ∠B because they are corresponding” ) 

4. Improving Logical and Reasoning Skill 

• Utilizing multi-step logic to solve problems 
• Combining multiple angle rules in one question 
• Writing answers in a structured manner (statements) 

5. Developing the basis of Proof Writing 

• It provides students the foundation to take proofs in geometry in higher grades 
• It will assist students in developing a transition from a calculation to justification thinking

4.0NCERT Solutions Class 9 Maths Chapter 6: All Exercises

Here is a quick overview and access to solutions for all exercises from Chapter 6, Lines and Angles.

5.0Detailed CBSE Class 9 Chapter 6 Exercise 6.2 Solutions

1. In figure, find the values of x and y and then show that AB∥CD.

Sol. x+50°=180° (Linear pair of angles)

=> x=130°

y=130° (Vertically opposite angles)

Now, x=y and the two angles form a pair of alternate angles made by a transversal intersecting the lines AB and CD.

Therefore, AB∥CD.

2. In figure, if AB∥CD, CD∥EF and y:z=3:7, find x.

Sol. Given AB∥CD and CD∥EF.

This implies AB∥EF.

So, x=z (Alternate angles)

Now, x+y=180° (Pair of interior angles on the same side of the transversal)

Substituting x=z:

z+y=180°

i.e., y+z=180°

Also, we are given that, y:z=3:7.

Let y=3k, z=7k.

3k+7k=180°

10k=180°

k=18°

Then y = 3 × 18° = 54°

And z = 7 × 18° = 126°

Since x=z, we have x=126°.

Therefore, x=126°.

3. In figure, if AB∥CD, EF⊥CD and ∠GED=126°, find ∠AGE,∠GEF and ∠FGE.

Sol. Given AB∥CD.

∠AGE=∠GED=126° (Alternate angles)

We know that ∠GED = ∠GEF + ∠FED.

Given EF⊥CD, so ∠FED = 90°.

=> ∠GEF+90°=126°

=> ∠GEF=36°

Also, angles on a straight line:

∠GEC+∠GED=180°

∠GEC+126°=180°

∠GEC=180°−126°=54°

∠FGE=∠GEC=54° (Alternate angles)

4. In figure, if PQ∥ST, ∠PQR=110° and ∠RST=130°, find ∠QRS.

Sol. Through R, we draw a line XRY parallel to PQ.

Since PQ∥ST and XRY∥PQ, it implies XRY∥ST.

For transversal QR intersecting parallel lines PQ and XRY:

∠PQR+∠QRX=180° (Interior angles on the same side of transversal)

110°+∠QRX=180°

=> ∠QRX=70°

For transversal SR intersecting parallel lines XRY and ST:

∠YRS+∠RST=180° (Interior angles on the same side of transversal)

∠YRS+130°=180°

=> ∠YRS=50°

Now, ∠QRX+∠QRS+∠YRS=180° (Angles on a straight line XRY)

=> 70°+∠QRS+50°=180°

=> 120°+∠QRS=180°

=> ∠QRS=60°

5. In figure, if AB∥CD, ∠APQ=50° and ∠PRD=127°, find x and y.

Sol. Given AB∥CD.

x = ∠APQ = 50° (Alternate angles)

For parallel lines AB and CD and transversal PR:

∠APR = ∠PRD (Alternate angles)

∠APQ + ∠QPR = ∠PRD

50° + y = 127°

y = 127° − 50° = 77°

6. In figure, 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.

Sol. Draw BE perpendicular to RS, then BE is also perpendicular to PQ (since PQ∥RS).

Draw CF perpendicular to PQ. Here, also CF is perpendicular to RS.

Consider transversal BC intersecting lines BE and CF.

We know that Angle of incidence = Angle of reflection.

So, ∠ABE=∠CBE (from mirror PQ) ... (1)

And ∠BCF=∠FCD (from mirror RS) ... (2)

Since BE⊥PQ and CF⊥PQ (or BE⊥RS and CF⊥RS), and PQ∥RS, then BE∥CF.

Since BE∥CF and BC is a transversal:

∠CBE = ∠BCF (Alternate interior angles) ... (3)

From (1), (2), and (3):

∠ABE = ∠CBE = ∠BCF = ∠FCD

Since ∠CBE = ∠BCF, substituting this into the sum of angles:

∠ABC = ∠ABE + ∠CBE = ∠CBE + ∠CBE = 2∠CBE

∠BCD = ∠BCF + ∠FCD = ∠BCF + ∠BCF = 2∠BCF

Since ∠CBE = ∠BCF (from alternate angles), then 2∠CBE = 2∠BCF.

Therefore, ∠ABC = ∠BCD.

Since these are alternate interior angles and they are equal, AB∥CD.

6.0Key Features and Benefits: Class 9 Maths Chapter 6 Exercise 6.2

• Complete, step-by-step solutions for each question
• Helpful diagrams making it easier to visualize solutions
• Enhances analytical and logical thinking
• Completely based on the latest CBSE syllabus
• Good for quick revision and exam preparation
• Enhances accuracy when answering questions on geometry.