NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1

Throughout Class 9 Maths Chapter 6 - Lines and Angles, we will delve into various properties of lines, angles and the interrelationships between them to begin to understand geometry better. In NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1, we get to the basic definitions and axioms about lines and angles, so students learn to think logically and based on mathematical proof.

1.0Download Class 9 Maths Chapter 6 Ex 6.1 NCERT Solutions PDF

Our comprehensive yet easy to understand Exercise 6.1 of NCERT Solutions for Class 9 Maths Chapter 6 are available for download as a PDF. Click the link below to download and use all of our material anywhere, anytime. 

2.0Key Concepts of Exercise 6.1

• Basic terminology and definitions: point, line, line segment, ray, angle
• Axioms and postulates pertaining to lines and angles
• Properties of angles created by the intersection of lines
• Linear pair axiom
• Supplementary and complementary angles
• Angle sum property on a straight line

3.0CBSE Class 9 Chapter 6 Linear Equations Exercise 6.1 Comprises

1. Linear Pair Axiom: Understanding the concept that:

• If two angles form a linear pair, then they are supplementary (sum is 180°).
• There will be a number of questions asking the students to use this axiom to calculate unknown angle measurements.

2. Vertically Opposite Angles: Understanding that:

• When two lines intersect, then vertically opposite angles are equal.
• Students will work on exercises based on the above principle.

3. Angle Relationships: Understanding different angle relationships like:

• Adjacent angles
• Supplementary and complementary angles.
• Using these relationships to help and solve Unknown values.

4. Logical Reasoning and Proof: Students should be encouraged to:

• Show every step using axioms or properties.
• Begin to formulate a systematic solution pathway—this is an important skill in geometry.

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 6.1 Solutions

1. In figure, lines AB and CD intersect at O. If ∠AOC+∠BOE=70° and ∠BOD=40°, find ∠BOE and reflex ∠COE.

Sol. ∠AOC=∠BOD (Vertically opposite angles)

=> ∠AOC=40° (Since ∠BOD=40° is given)

Now, ∠AOC+∠BOE=70° (Given)

=> 40°+∠BOE=70°

=> ∠BOE=30°

∠AOE+∠BOE=180° (Linear pair of angles)

=> ∠AOE+30°=180°

=> ∠AOE=150°

=> ∠AOC+∠COE=150°

=> 40°+∠COE=150°

=> ∠COE=110°

Reflex ∠COE=360°−110°=250°

2. In figure, lines XY and MN intersect at O. If ∠POY=90° and a:b=2:3, find c.

Sol. Ray OP stands on line XY.

∠POX+∠POY=180°

∠POX+90°=180°

∠POX=90°

∠POM+∠XOM=90°

a+b=90°

Given a:b=2:3. Let a=2k, b=3k.

3k+2k=90°

5k=90°

k=18°

=> a=36°, b=54°

Now, Ray OX stands on line MN.

∠XOM+∠XON=180°

b+c=180°

54°+c=180°

=> c=126°

3. In figure, ∠PQR=∠PRQ, then prove that ∠PQS=∠PRT.

Sol. Let ∠PQR=∠PRQ=x (say) ... (1)

Now, ∠PQS+∠PQR=180° (Linear pair of angles)

And ∠PRT+∠PRQ=180° (Linear pair of angles)

=> ∠PQS+∠PQR=∠PRT+∠PRQ (Since each = 180°)

=> ∠PQS+x=∠PRT+x (By (1))

=> ∠PQS=∠PRT

4. In figure, if x+y=w+z, then prove that AOB is a line.

Sol. Given x+y=w+z.

We know that the sum of angles around a point is 360°.

So, x+y+w+z=360° (Complete angle)

Substitute w+z with x+y from the given condition:

=> (x+y)+(x+y)=360°

=> 2(x+y)=360°

=> x+y=180°

Since x+y forms a linear pair and sums to 180°, AOB is a line.

5. In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS=(1/2)(∠QOS−∠POS).

Sol. ∠POR=∠QOR=90° (Since OR⊥PQ at O)

Now, ∠QOS=∠QOR+∠ROS => ∠QOS=90°+∠ROS ... (2)

∠POS+∠ROS=∠POR => ∠POS=∠POR−∠ROS => ∠POS=90°−∠ROS ... (3)

Subtracting (3) from (2):

∠QOS−∠POS = (90°+∠ROS) − (90°−∠ROS)

∠QOS−∠POS = 90°+∠ROS − 90°+∠ROS

∠QOS−∠POS = 2 × ∠ROS

i.e., ∠ROS=(1/2){∠QOS−∠POS}

6. It is given that ∠XYZ=64° and XY is produced to point P. Draw a figure from the given information. if ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Sol. ∠XYZ+∠ZYP=180° (Linear pair)

=> 64°+∠ZYP=180°

=> ∠ZYP=116°

Ray YQ bisects angle ∠ZYP

=> ∠PYQ=∠ZYQ=116°/2 = 58°

Reflex ∠QYP=360°−58°=302°

∠XYQ=∠XYZ+∠ZYQ=64°+58°=122°

6.0Key Features and Benefits: Class 9 Maths Chapter 6 Exercise 6.1

• Clear and step-by-step NCERT Solutions to improve understanding
• Ideal for daily practice and CBSE exam preparation
• Helps in developing logical reasoning and accuracy
• Adheres strictly to CBSE guidelines
• Supports students in building a strong base in geometry