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NCERT Solutions for Exercise 6.1 Class 9 Maths Chapter 6 - Lines and Angles

NCERT Solutions for Exercise 6.1 Class 9 Maths Chapter 6 - Lines and Angles

Edited By Safeer PP | Updated on Jul 18, 2022 02:16 PM IST

A line with two endpoints is referred to as a line segment, and a portion of a line with one endpoint is referred to as a ray.

Collinear points are those that have three or more points on the same line; otherwise, they are non-collinear points. When two rays originate from the same endpoint, an angle is formed.

The rays that form an angle are referred to as the angle's arms, and the endpoint is referred to as the vertex of the angle.

NCERT Solutions Class 9 Maths exercise 6.1 – In NCERT solutions for Class 9 Maths chapter 6 exercise 6.1 we get to know that an acute angle has a range of 0° to 90°, whereas a right angle has a value of 90°. An obtuse angle is one that is greater than 90° but less than 180°.

Also, keep in mind that a straight angle equals 180°. A reflex angle is an angle that is greater than 180° but less than 360°. Furthermore, two angles with a sum of 90° are referred to as complementary angles, while two angles with a sum of 180° are referred to as supplementary angles.

Along with Class 9 Maths, chapter 8 exercise 6.1 the following exercises are also present.

Lines and Angles Class 9 Maths Chapter 6 Exercise: 6.1

Q1 In Fig. 6.13, lines AB and CD intersect at O. If \angle AOC + \angle BOE = 70° and \angle BOD = 40°, find \angle BOE and reflex \angle COE.

1640149435860

Answer:

15956099379961595609934860
Given that,
AB is a straight line. Lines AB and CD intersect at O. \angle AOC + \angle BOE = 70^0 and \angle BOD = 40^0
Since AB is a straight line
\therefore \angle AOC + \angle COE + \angle EOB = 180^0
\Rightarrow \angle COE = 180^0-70^0=110^0 [since \angle AOC + \angle BOE = 70^0 ]

So, reflex \angle COE = 360^0-110^0 = 250^0
It is given that AB and CD intersect at O
Therefore, \angle AOC = \angle BOD [vertically opposite angle]
\Rightarrow \angle COA = 40^0 [ GIven \angle BOD = 40^0 ]
Also, \angle AOC + \angle BOE = 70^0
So, \angle BOE = 30^0

Q2 In Fig. 6.14, lines XY and MN intersect at O. If \angle POY = 90^o and a : b = 2 : 3, find c.

1640149490726

Answer:

15956102104631595610206410
Given that,
Line XY and MN intersect at O and \angle POY = 90^0 also a:b = 2:3 \Rightarrow b = \frac{3a}{2} ..............(i)

Since XY is a straight line
Therefore, \\a+b+\angle POY = 180^0\\ a+b = 180^0-90^0 = 90^0 ...........(ii)
Thus, from eq (i) and eq (ii), we get
\\\Rightarrow \frac{3a}{2}+a = 90^0\\
\\\Rightarrow a = 36^0\\
So, b = 54^0\\
Since \angle MOY = \angle c [vertically opposite angles]
\angle a + \angle POY = c
126^0 =c

Q3 In Fig. 6.15, \angle PQR = \angle PRQ, then prove that \angle PQS = \angle PRT.

1640149622015

Answer:

15956102948431595610292259
Given that,
ABC is a triangle such that \angle PQR = \angle PRQ and ST is a straight line.
Now, \angle PQR + \angle PQS = 180^0 {Linear pair}............(i)
Similarly, \angle PRQ + \angle PRT = 180^0 ..................(ii)

equating the eq (i) and eq (ii), we get

\angle PQR +\angle PQS =\angle PRT + \angle PRQ {but \angle PQR = \angle PRQ }
Therefore, \angle PQS = \angle PRT
Hence proved.

Q4 In Fig. 6.16, if x + y = w + z , then prove that AOB is a line.

1640149658149

Answer:

15956103888821595610382979
Given that,
x+y = z+w ..............(i)
It is known that, the sum of all the angles at a point = 360^0
\therefore x+y+z+w=360^0 ..............(ii)

From eq (i) and eq (ii), we get

\\2(x+y)=360^0\\ x+y = 180^0

Hence proved AOB is a line.

Q5 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 \angle \textup{ROS} = \frac{1}{2}(\angle \textup{QOS} - \angle \textup{POS})

1640149689300

Answer:

15956104575791595610452502
Given that,
POQ is a line, OR \perp PQ and \angle ROQ is a right angle.
Now, \angle POS + \angle ROS + \angle ROQ = 180^0 [since POQ is a straight line]
\\\Rightarrow \angle POS + \angle ROS = 90^0\\ \Rightarrow \angle ROS = 90^0-\angle POS .............(i)
and, \angle ROS + \angle ROQ = \angle QOS
\angle ROS = \angle QOS -90^0 ..............(ii)

Add the eq (i ) and eq (ii), we get

\angle \textup{ROS} = \frac{1}{2}(\angle \textup{QOS} - \angle \textup{POS})

hence proved.

Q6 It is given that \angle XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects \angle ZYP, find \angle XYQ and reflex \angle QYP.

Answer:

15956115058161595611502732
Given that,
\angle XYZ = 64^0 and XY produced to point P and Ray YQ bisects \angle ZYP \Rightarrow \angle QYP = \angle ZYQ
Now, XYP is a straight line
So, \angle XYZ + \angle ZYQ + \angle QYP = 180^0
\Rightarrow 2(\angle QYP )=180^0 - 64^0 = 116^0
\Rightarrow (\angle QYP )= 58^0

Thus reflex of \angle QYP = 360^0- 58^0 = 302^0

Since \angle XYQ = \angle XYZ + \angle ZYQ [ \because \angle QYP = \angle ZYQ


\angle XYQ = 64^0+58^0 = 122^0

More About NCERT Solutions for Class 9 Maths Exercise 6.1

In NCERT solutions for Class 9 Maths exercise 6.1, we get to know that if two angles share a common vertex, a common arm, and their non-common arms are on opposite sides of the common arm, they are adjacent.

A linear pair is formed when two adjacent angles are supplementary, i.e. they form a straight line (add up to 180).

When two lines intersect at a point, they form equal angles that are vertically opposite.

The sum of two adjacent angles formed by a ray on a line is 180°.

When the sum of two adjacent angles equals 180°, the angles' non-common arms form a line.

Also Read| Lines And Angles Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 6.1

  • Exercise 6.1 Class 9 Maths, is based on the concepts of lines and angles

  • From Class 9 Maths chapter 6 exercise 6.1 we learn about parallel lines and pairs of angles

  • Understanding the concepts from Class 9 Maths chapter 6 exercise 6.1 will allow us to understand the concepts related to angle sum properties of LINES AND ANGLES.

This Story also Contains
  1. Lines and Angles Class 9 Maths Chapter 6 Exercise: 6.1
  2. More About NCERT Solutions for Class 9 Maths Exercise 6.1
  3. Benefits of NCERT Solutions for Class 9 Maths Exercise 6.1
  4. NCERT Solutions of Class 9 Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

Also see

NCERT Solutions of Class 9 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 6.1?

This exercise is about lines and angles, types of angles like vertically opposite angles and axioms and concepts like parallel lines and pair of angles.

2. What is an angle?

An angle is a figure which is formed by two rays, called the sides of the angle, both of which shares a common endpoint, called the vertex of the angle.

3. What are collinear points?

If we have three or more points that lie on the same line then they are called collinear points.

4. Name the different types of angles?

There are various types of angles are:

  • Acute angle

  • Right angle

  • Obtuse angle

  • Straight angle

  • Reflex angle

5. What do you understand by reflex angles?

A reflex angle is an angle greater than 180° and less than 360°.

6. What are intersecting lines?

When we have two or more lines meeting or crossing each other in a plane, they are termed as intersecting lines.

7. What is pair of angles?

When we have two lines sharing a common endpoint, called Vertex then an angle that is formed between these two lines is known as the pair of angles.

8. What do you understand by vertical opposite angles?

When we have two lines intersecting each other, the opposite angles are equal. These angles are known as the vertically opposite angles because they are equal and opposite to each other at the vertex.

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