Lines and Angles Class 9 Maths Exercise 6.1​ NCERT Solutions

Chapter 6, "Lines and Angles," is vital for building a strong foundation in geometry. Exercise 6.2 specifically deals with the intricate relationship between parallel lines and a transversal line intersecting them. The NCERT Solution for class 9 maths chapter 6 exercise 6.2 offers clear and accurate explanations.

They break down the logical steps required to solve every problem in the exercise. This resource is key to understanding geometric proofs and excelling in your Class 9 math assessments.

What Class 9 Maths Chapter 6 Lines and Angles exercise 6.2 covers?

Exercise 6.2 focuses entirely on the angles formed when a transversal line cuts two parallel lines. It covers three main theorems and their converses.

The core concepts include Corresponding Angles Axiom, which states corresponding angles are equal.

It also explores the theorems that prove Alternate Interior Angles are equal and Consecutive Interior Angles (or co-interior angles) are supplementary (sum to 180 degree). Students learn to use these properties to find unknown angles and prove that lines are parallel. 

Class 9 Lines and Angles Exercise 6.1​ NCERT Solutions

The NCERT Solutions provide neat, step-by-step derivations for each problem. They clearly state which axiom or theorem is being used at every stage of the solution. This systematic approach helps students write organized, fully justified answers in their exams.

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

Solution:

It is known that AB || CD and CD||EF As the angles on the same side of a transversal line sum up to 180°, x + y = 180° —–(i) Also, ∠O = z (Since they are corresponding angles) and, y +∠O = 180° (Since they are a linear pair) So, y+z = 180° Now, let y = 3w and hence, z = 7w (As y : z = 3 : 7) ∴ 3w+7w = 180° Or, 10 w = 180° So, w = 18° Now, y = 3×18° = 54° and, z = 7×18° = 126° Now, angle x can be calculated from equation (i) x+y = 180° Or, x+54° = 180° ∴ x = 126°

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

Solution:

Since AB || CD, GE is a transversal. It is given that ∠GED = 126° So, ∠GED = ∠AGE = 126° (As they are alternate interior angles) Also, ∠GED = ∠GEF +∠FED As EF⊥ CD, ∠FED = 90° ∴ ∠GED = ∠GEF+90° Or, ∠GEF = 126° – 90° = 36° Again, ∠FGE +∠GED = 180° (Transversal) Putting the value of ∠GED = 126°, we get ∠FGE = 54° So, ∠AGE = 126° ∠GEF = 36° and ∠FGE = 54°

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

[Hint : Draw a line parallel to ST through point R.]

Solution:

First, construct a line XY parallel to PQ.We know that the angle on the same side of the transversal is equal to 180°. So, ∠PQR+∠QRX = 180° Or, ∠QRX = 180°-110° ∴ ∠QRX = 70° Similarly, ∠RST +∠SRY = 180° Or, ∠SRY = 180°- 130° ∴ ∠SRY = 50° Now, for the linear pairs on the line XY- ∠QRX+∠QRS+∠SRY = 180° Putting their respective values, we get ∠QRS = 180° – 70° – 50° Hence, ∠QRS = 60°

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

Solution:

From the diagram, ∠APQ = ∠PQR (Alternate interior angles) Now, putting the value of ∠APQ = 50° and ∠PQR = x, we get x = 50° Also, ∠APR = ∠PRD (Alternate interior angles) Or, ∠APR = 127° (As it is given that ∠PRD = 127°) We know that ∠APR = ∠APQ+∠QPR Now, putting values of ∠QPR = y and ∠APR = 127°, we get 127° = 50°+ y Or, y = 77° Thus, the values of x and y are calculated as: x = 50° and y = 77°

5. 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.

Solution:

First, draw two lines, BE and CF, such that BE ⊥ PQ and CF ⊥ RS. Now, since PQ || RS, So, BE || CF

We know that, Angle of incidence = Angle of reflection (By the law of reflection) So, ∠1 = ∠2 and ∠3 = ∠4 We also know that alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at B and C So, ∠2 = ∠3 (As they are alternate interior angles) Now, ∠1 +∠2 = ∠3 +∠4 Or, ∠ABC = ∠DCB So, AB || CD (alternate interior angles are equal)

How to Prepare Using Class 9 Maths Lines and Angles exercise 6.1?

Effective preparation involves using the NCERT Solutions Class 9 Chapter 6 Exercise 6.2 for conceptual verification and mistake analysis. Do not simply view them as final answers. First, attempt all exercise questions on your own, ensuring you draw accurate diagrams. Then, use the solutions to meticulously check your application of theorems and algebraic calculations.

• Identify Angle Types: Before solving, clearly identify the relationship between the given angles (e.g., corresponding, alternate interior, linear pair). The solutions confirm your initial assessment.

• Practice Proof Writing: Pay close attention to the format of the solutions when they prove a line is parallel. Learn the precise way to state the converse of a theorem.

• Master Converses: Be sure to understand the difference between the theorem (e.g., if lines are parallel, then alternate angles are equal) and its converse (e.g., if alternate angles are equal, then lines are parallel).

• Work Backwards: For challenging problems, look at the final required result in the solution and trace the steps backward. This helps build problem-solving intuition.