NCERT Solutions for Class 9 Maths Chapter 6 - Lines And Angles Exercise 6.2

There is no doubt that NCERT textbooks are the most-trusted study material for students. The Class 9 NCERT Maths book is designed in such a way that students can easily understand the concepts and then be able to solve the problems related to them. The NCERT Class 9 Maths Solutions consists of 15 chapters. Chapter 6 deals with Lines and Angles. Here, we have provided the NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles Exercise 6.2.

The solutions are provided by experienced teachers at BYJU’S. These solutions will not only help students to score good marks in the board exams but will also help them understand the concepts. The NCERT solutions are explained in detail by covering each and every step. In case a student has missed the class, they can easily understand the solution by taking a glance at it.

NCERT Solutions for Class 9 Maths Chapter 6 – Lines And Angles Exercise 6.2

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Access other Exercise Solutions of Class 9 Maths Chapter 6 – Lines and Angles

Exercise 6.1 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Exercise 6.3 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Access Answers to NCERT Solutions for Class 9 Maths Chapter 6 – Lines And Angles Exercise 6.2

1. In Fig. 6.28, find the values of x and y and then show that AB || CD.

Solution:

We know that a linear pair is equal to 180°.

So, x+50° = 180°

∴ x = 130°

We also know that vertically opposite angles are equal.

So, y = 130°

In two parallel lines, the alternate interior angles are equal. In this,

x = y = 130°

This proves that alternate interior angles are equal and so, AB || CD.

2. 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°

3. 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°

4. 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 angles on the same side of a 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°

5. 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°

6. 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)

In Exercise 6.2 of Chapter 6 Maths, students of Class 9 will recall the concepts of Corresponding angles, alternate interior angles, alternate exterior angles, Interior angles on the same side of the transversal, etc. They will also learn the new theorems related to transversal lines, corresponding angles axiom and so on. Along with this, students will also be introduced to the concepts of “Lines Parallel to the Same Line” and the theorems related to it.

The concepts learned in the exercise will help in solving the problems of Exercise 6.2. So, first and foremost, the students should know the concepts and theorems thoroughly. After that, they should start solving the problems in the exercises. So, in case the students have any doubts, they must refer to the NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles Exercise 6.2.

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