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

Chapter 6 of Class 9 Maths, "Lines and Angles," introduces fundamental geometric concepts. Exercise 6.1 specifically focuses on the basic relationships formed when lines intersect.The NCERT Solution class 9 maths chapter 6 exercise 6.1 are designed to help students master these foundational concepts.

They provide reliable, step-by-step guidance for every question. Using these solutions ensures clarity on key angle properties, which is crucial for tackling subsequent complex exercises and securing good marks.

What Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1 covers?

Exercise 6.1 acts as the gateway to the "Lines and Angles" chapter. It covers the basic definitions and theorems related to intersecting lines. The core concepts include understanding:

• Linear Pair of Angles: Two adjacent angles whose non-common arms form a line, summing up to 180 degree.

• Vertically Opposite Angles: Angles formed by two intersecting lines, which are always equal.

• Reflex Angle: An angle whose measure is between 180 and 360 degree.
  The problems require students to apply these properties to find unknown angles given certain conditions. 

Class 9 Chapter 6 Exercise 6.1 Questions and NCERT Solutions

Below we have provided NCERT Solutions for Class 9 Maths Chapter 6 -

1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:

From the diagram, we have (∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line. So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180° Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get ∠COE = 110° and ∠BOE = 30° So, reflex ∠COE = 360 o – 110 o = 250 o

2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Solution:

We know that the sum of linear pair is always equal to 180° So, ∠POY +a +b = 180° Putting the value of ∠POY = 90° (as given in the question), we get, a+b = 90° Now, it is given that a:b = 2:3, so Let a be 2x and b be 3x ∴ 2x+3x = 90° Solving this, we get 5x = 90° So, x = 18° ∴ a = 2×18° = 36° Similarly, b can be calculated, and the value will be b = 3×18° = 54° From the diagram, b+c also forms a straight angle, so b+c = 180° c+54° = 180° ∴ c = 126°

3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:

Since ST is a straight line, so ∠ PQS+ ∠ PQR = 180° (linear pair) and ∠ PRT+ ∠ PRQ = 180° (linear pair) Now, ∠ PQS + ∠ PQR = ∠ PRT+ ∠ PRQ = 180° Since ∠ PQR = ∠ PRQ (as given in the question) ∠ PQS = ∠PRT. (Hence proved).

4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.

Solution:

To prove AOB is a straight line, we will have to prove x+y is a linear pair i.e. x+y = 180° We know that the angles around a point are 360°, so x+y+w+z = 360° In the question, it is given that, x+y = w+z So, (x+y)+(x+y) = 360° 2(x+y) = 360° ∴ (x+y) = 180° (Hence proved).

5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = ½ (∠QOS – ∠POS).

Solution:

In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180° We can write it as ∠ROP = ∠ROQ = 90 0 We know that ∠ROP = ∠ROQ It can be written as ∠POS + ∠ROS = ∠ROQ ∠POS + ∠ROS = ∠QOS – ∠ROS ∠SOR + ∠ROS = ∠QOS – ∠POS So we get 2∠ROS = ∠QOS – ∠POS Or, ∠ROS = 1/2 (∠QOS – ∠POS)(Hence proved).

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.

Solution:

Here, XP is a straight line So, ∠XYZ +∠ZYP = 180° Putting the value of ∠XYZ = 64°, we get 64° +∠ZYP = 180° ∴ ∠ZYP = 116° From the diagram, we also know that ∠ZYP = ∠ZYQ + ∠QYP Now, as YQ bisects ∠ZYP, ∠ZYQ = ∠QYP Or, ∠ZYP = 2∠ZYQ ∴ ∠ZYQ = ∠QYP = 58° Again, ∠XYQ = ∠XYZ + ∠ZYQ By putting the value of ∠XYZ = 64° and ∠ZYQ = 58°, we get. ∠XYQ = 64°+58° Or, ∠XYQ = 122° Now, reflex ∠QYP = 180°+XYQ We computed that the value of ∠XYQ = 122°. So, ∠QYP = 180°+122° ∴ ∠QYP = 302°

How to Prepare using NCERT Solutions Class 9 Chapter 6 Ex 6.1?

To maximize learning from the NCERT Solutions Class 9 Chapter 6 Ex 6.1, students should focus on method over just the final answer. First, attempt all the questions yourself using textbook concepts. Then, use the solutions to check the accuracy of your logic and calculation. This approach reinforces conceptual understanding.

• Revise Key Definitions: Ensure you know the difference between a reflex angle, a linear pair, and vertically opposite angles before starting the exercise.

• Draw and Label: Always draw the diagram and correctly label the given and unknown angles for every problem. The visual aid prevents errors.

• Justify Every Step: Pay close attention to how the solutions justify the equality or sum of angles. Practice writing the correct reason for each step.

• Practice Without Looking: Try to solve the problems completely from scratch without referring to the solutions. Only use the solutions for verification.