Exercise 6.1

1. As per the figure below, lines, AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, you need to find what is ∠BOE and reflex ∠COE.

Solution:

By looking at the above figure, you know that-

(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) will create a straight line.

Therefore, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°

Thus, now you need to put the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° you will derive

∠COE = 110° and ∠BOE = 30°

This, reflex ∠COE = 360° – 110° = 250°.

2.  As per the figure below, you can see lines XY and MN intersect at O. If ∠POY = 90° and a: b = 2 : 3, what is c?

Solution:

You need to know that the sum of linear pair is always equal to 180°

Therefore,

∠POY + a +b = 180°

Now, put the value of ∠POY = 90° (provided in the question) you will get,

a+b = 90°

As per the question, you have the data that a : b = 2 : 3 hence,

Assume, a as 2x and b as 3x

$\therefore$ 2x+3x = 90°

Solve this equation; you will get-

5x = 90°

Hence, x = 18°

$\therefore$ a = 2$\times$18° = 36°

Likewise, you can calculate b, and its value will be

b = 3$\times$18° = 54°

As per the figure above, b+c also creates a straight angle, thus,

b+c = 180°

c+54° = 180°

Hence, c = 126°

3. As per the figure below, ∠PQR = ∠PRQ; you need to prove that ∠PQS = ∠PRT.

Solution:

As you can see that ST is a straight line, therefore,

∠PQS+∠PQR = 180° (linear pair) and

∠PRT + ∠PRQ = 180° (linear pair)

So, ∠PQS + ∠PQR = ∠PRT + ∠PRQ = 180°

As ∠PQR = ∠PRQ (as per the details in the question)

∠PQS = ∠PRT.

Hence proved.

4. As per the figure, if x+y = w+z, you need to prove that AOB is a line.

Solution:

If you want to prove that AOB is a straight line, you will first have to prove that x+y is a linear pair.

Which means x+y = 180°

You should know that the angles around a point are 360°, thus,

x+y+w+z = 360°

As per the question-

x+y = w+z

Therefore, (x+y)+(x+y) = 360°

2(x+y) = 360°

$\therefore$ (x+y) = 180°.

5. As per the diagram below, POQ is a line. The Ray OR you see is perpendicular to the line PQ. There is another ray, OS lying between rays OP and OR. You need to prove that ∠ROS = ½ (∠QOS – ∠POS).

Solution:

As per the question, the information states that (OR ⊥ PQ) and ∠POQ = 180°

Therefore, ∠POS + ∠ROS + ∠ROQ = 180°

This means, ∠POS+∠ROS = 180°- 90° (As, ∠POR = ∠ROQ = 90°)

Therefore, ∠POS + ∠ROS = 90°

It makes, ∠QOS = ∠ROQ+∠ROS

As per the question, ∠ROQ = 90°,

$\therefore$ ∠QOS = 90° + ∠ROS

Or you can also say that, ∠QOS – ∠ROS = 90°

Now, ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, this will you will get

∠POS + ∠ROS = ∠QOS – ∠ROS

2 ∠ROS + ∠POS = ∠QOS

Or, ∠ROS = ½ (∠QOS – ∠POS).

6. If ∠XYZ = 64° and XY is created to point P. You need to create a figure from this information. You need to assume that ray YQ bisects ∠ZYP; as per this scenario, you need to find ∠XYQ and reflex ∠QYP.

Solution:

As per the above diagram, you can see that XP is a straight line

Thus, ∠XYZ + ∠ZYP = 180°

Put the value of ∠XYZ = 64° you will get,

64° +∠ZYP = 180°

Therefore, ∠ZYP = 116°

As per the figure, you also get the information that ∠ZYP = ∠ZYQ + ∠QYP

So, as YQ bisects ∠ZYP,

∠ZYQ = ∠QYP

Or, ∠ZYP = 2∠ZYQ

Hence, ∠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

You computed earlier that the value of ∠XYQ = 122°.

Hence,

∠QYP = 180°+122°

$\therefore$ ∠QYP = 302°

Exercise 6.2

1. As per the diagram below, you need to find the values of x and y; after that, you are also required to show that AB II CD.

Solution:

After looking at the figure, you can make out that,

50°+ x= 180° (Linear Pair)

$\Rightarrow$ x=130°………. (1)

Now, from the above diagram, it is pretty evident that y=130° (as vertically opposite angles are equivalent)

Also, you can see that x and y are alternate interior angles for lines AB and CD, so measures of these angles are equal.

Hence, line AB || CD.

Q1. How will NCERT solutions for class 9 maths, chapter 6 help students prepare for exams?

A1. Since every chapter is equally important, it is essential for you to clarify its concepts and the basics, especially chapter 6. The complete weightage of the Geometry unit in class 9 Maths is 22 marks, out of which chapter 6 accounts for 6 marks. These solutions are vital because you will be asked similar types of questions in the question paper.

The solutions for this chapter will provide students with comprehensive learning. The best part is that you will be able to understand challenging concepts, and track your weak areas with the help of NCERT solutions.

Q2. How many exercises are there in lines and angles class 9th NCERT solutions?

A2. Overall, you will see that there are 3 exercises, which are as follows-

• In exercise 6.1 Solutions, there are 6 Questions out of which, 5 are short answer questions, and 1 is a long answer type question.
• In exercise 6.2 Solutions, you will find 6 Questions, out of which, 3 are short answer questions, and 3 are long answer type questions.
• In exercise 6.3 Solutions, there are 6 Questions out of which, 5 are short answer questions, and 1 is a long answer type question.

 

Q3. Is every question essential to study in lines and angles for class 9 ncert?

A3. On MVSgo, every question of chapter 6 is designed to make it easier for the students to study and clear their basics. It would help if you studied all the exercises as it will help make the understanding better. You can get any question in the exams. Practise every exercise to score well; it is recommended that you do not skip any question. You will find a solution to all the exercises on MVSgo app easily.

Q4. Explain the types of lines explained in maths class 9 chapter 6?

A4. In the chapter, it is explained that there are two types of lines in geometry: curved & straight.

Straight Lines are separated into two sorts; horizontal and vertical. Some of the examples of straight lines are parallel, intersecting and perpendicular lines.

It is essential to have knowledge of lines and angels as they will be helpful for you in the coming classes.

Maths class 9 chapter 6 explains about the principles and the basic concepts of lines and angels. If you want to learn them in a fun and easy way, you should download the MVSgo app, as they have a streamlined and structured way to make studies easier for students. You can play different maths games on it, plus you can apply for the interschool challenge.