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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
- 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.
- 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.
- Angle Relationships: Understanding different angle relationships like:
- Adjacent angles
- Supplementary and complementary angles.
- Using these relationships to help and solve Unknown values.
- 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