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**NCERT Solutions For Class 9 Maths Chapter 6 Lines And Angles **are provided here. These NCERT solutions are prepared by expert team at Careers360 considering latest syllabus of CBSE 2023 in simple, comprehensive way. therefore practicing these solutions are beneficial for students. In this Class 9 NCERT syllabus chapter, we will learn concepts of vertically opposite angles, supplementary angles, and complementary angles and also study properties of parallel lines and transversal lines, intersecting lines, non-intersecting Lines, and angle sum property of the triangle.

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This Story also Contains

- NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles
- Lines And Angles Class 9 Questions And Answers PDF Free Download
- Lines And Angles Class 9 Solutions - Important Points
- Lines And Angles Class 9 NCERT Solutions (Intext Questions and Exercise)
- NCERT Lines And Angles Class 9 Solutions - Topics
- Summary Of Lines And Angles Class 9 Solutions
- NCERT Solutions For Class 9 Maths - Chapter Wise
- NCERT Solutions For Class 9 - Subject Wise
- NCERT Books and NCERT Syllabus

Lines and angles class 9 questions and answers will also be helpful in the preparation of NTSE, Indian National Olympiad (INO), etc. Here you will get NCERT solutions for class 9 Maths also.

**Also Read**

- Lines And Angles Class 9 NCERT Chapter Notes
- NCERT Exemplar Solutions Class 9 NCERT Maths Chapter Lines And Angles

Types of Angles:

Acute Angle: An acute angle measures between 0° and 90°.

Right Angle: A right angle is exactly equal to 90°.

Obtuse Angle: An angle that is greater than 90° but less than 180°.

Straight Angle: A straight angle is equal to 180°.

Reflex Angle: A reflex angle is greater than 180° but less than 360°.

Complementary Angles (x, y): x + y = 90°

Supplementary Angles (x, y): x + y = 180°

Adjacent Angles:

Adjacent angles are two angles that share a common side and a common vertex (corner point) without overlapping.

Linear Pair:

A linear pair of angles is formed when two lines intersect. Two angles are considered linear if they are adjacent angles formed by the intersection of two lines. The measure of a straight angle is 180°, so a linear pair of angles must add up to 180°.

Vertically Opposite Angles:

Vertically opposite angles are formed when two lines intersect at a point. Vertically opposite angles are always equal.

Transversal:

A transversal is a line that intersects two or more given lines at distinct points. It is used to create various types of angles, including:

Corresponding Angles

Alternate Interior Angles

Alternate Exterior Angles

Interior Angles on the Same Side of the Transversal

Free download **NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles **for CBSE Exam.

**Class 9 lines and angles NCERT solutions - exercise: 6.1**

** Answer: **

Given that,

AB is a straight line. Lines AB and CD intersect at O. and BOD =

Since AB is a straight line

AOC + COE + EOB =

[since ]

So, reflex COE =

It is given that AB and CD intersect at O

Therefore, AOC = BOD [vertically opposite angle]

[ GIven BOD = ]

Also,

So, BOE =

** Q2 ** In Fig. 6.14, lines XY and MN intersect at O. If and a : b = 2 : 3, find c.

** Answer: **

Given that,

Line XY and MN intersect at O and POY = also ..............(i)

Since XY is a straight line

Therefore, ...........(ii)

Thus, from eq (i) and eq (ii), we get

So,

Since MOY = c [vertically opposite angles]

a + POY = c

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

** Answer: **

Given that,

ABC is a triangle such that PQR = PRQ and ST is a straight line.

Now, PQR + PQS = {Linear pair}............(i)

Similarly, PRQ + PRT = ..................(ii)

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

{but PQR = PRQ }

Therefore, PQS = PRT

Hence proved.

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

** Answer: **

Given that,

..............(i)

It is known that, the sum of all the angles at a point =

..............(ii)

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

Hence proved AOB is a line.

** Answer: **

Given that,

POQ is a line, OR PQ and ROQ is a right angle.

Now, POS + ROS + ROQ = [since POQ is a straight line]

.............(i)

and, ROS + ROQ = QOS

..............(ii)

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

hence proved.

** Answer: **

Given that,

XYZ = and XY produced to point P and Ray YQ bisects ZYP

Now, XYP is a straight line

So, XYZ + ZYQ + QYP =

Thus reflex of QYP =

Since XYQ = XYZ + ZYQ [

=

** ** ** Class 9 maths chapter 6 question answer - exercise: 6.2 **

** Q1 ** In Fig. 6.28, find the values of * x * and * y * and then show that AB CD.

** Answer: **

Given that,

In the figure, CD and PQ intersect at F

Therefore, (vertically opposite angles)

PQ is a straight line. So,

Hence AB || CD (since and are alternate interior angles)

** Q2 ** In Fig. 6.29, if AB CD, CD EF and * y : z = 3 : 7 * , find * x * .

** Answer: **

Given AB || CD and CD || EF and

therefore, AB || EF and (alternate interior angles)..............(i)

Again, CD || AB

.............(ii)

Put the value of in equation (ii), we get

Then

By equation (i), we get the value of

** Q3 ** In Fig. 6.30, if AB CD, EF CD and GED = 126°, find AGE, GEF and FGE.

** Answer: **

Given AB || CD, EF CD and GED =

In the above figure,

GE is transversal. So, that AGE = GED = [Alternate interior angles]

Also, GEF = GED - FED

=

Since AB is a straight line

Therefore, AGE + FGE =

So, FGE =

** Answer: **

Draw a line EF parallel to the ST through R.

Since PQ || ST and ST || EF

EF || PQ

PQR = QRF = (Alternate interior angles)

QRF = QRS + SRF .............(i)

Again, RST + SRF = (Interior angles of two parallels ST and RF)

( RST = , given)

Thus, QRS =

** Q5 ** In Fig. 6.32, if AB CD, APQ = 50° and PRD = 127°, find * x * and * y * .

** Answer: **

Given, AB || CD, APQ = and PRD =

PQ is a transversal. So,

APQ = PQR= (alternate interior angles)

Again, PR is a transversal. So,

y + = (Alternate interior angles)

** Answer: **

Draw a ray BL PQ and CM RS

Since PQ || RS (Given)

So, BL || CM and BC is a transversal

LBC = MCB (Alternate interior angles).............(i)

It is known that, angle of incidence = angle of reflection

So, ABL = LBC and MCB = MCD

..................(ii)

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

ABC = DCB

Both the interior angles are equal

Hence AB || CD

** Class 9 maths chapter 6 NCERT solutions - Exercise: 6.3 **

** Answer: **

Given,

PQR is a triangle, SPR = , PQT =

Now, TQP + PQR = (Linear pair)

So, PQR =

Since the side of QP of the triangle, PQR is produced to S

So, PQR + PRQ = (Exterior angle property of triangle)

** Answer: **

We have

X = , XYZ =

YO and ZO bisects the XYZ and XZY

Now, In XYZ, by using angle sum property

XYZ + YZX + ZXY =

So, YZX =

YZX =

and, OYZ = also, OZY =

Now, in OYZ

Y + O + Z = [ Y = and Z = ]

So, YOZ =

** Q3 ** In Fig. 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE.

** Answer: **

We have,

AB || DE, BAC = 35° and CDE = 53°

AE is a transversal so, BAC = AED =

Now, In CDE,

CDE + DEC + ECD = (By angle sum property)

Therefore, ECD =

** Answer:**

We have,

lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°

In PRT, by using angle sum property

PRT + PTR + TPR =

So, PTR =

Since lines, PQ and RS intersect at point T

therefore, PTR = QTS (Vertically opposite angles)

QTS =

Now, in QTS,

By using angle sum property

TSQ + STQ + SQT =

So, SQT =

** Q5 ** In Fig. 6.43, if PQ PS, PQ SR, SQR = 28° and QRT = 65°, then find the values of * x * and * y * .

** Answer: **

We have,

PQ PS, PQ || SR, SQR = 28° and QRT = 65°

Now, In QRS, the side SR produced to T and PQ || RS

therefore, QRT = =

So,

Also, QRT = RSQ + SQR (By exterior angle property of a triangle)

Therefore, RSQ = QRT - SQR

Now, in PQS,

P + PQS + PSQ =

** Answer: **

We have,

PQR is produced to a point S and bisectors of PQR and PRS meet at point T,

By exterior angle sum property,

PRS = P + PQR

Now,

................(i)

Since QT and QR are the bisectors of PQR and PRS respectively.

Now, in QRT,

..............(ii)

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

Hence proved.

- Basic Terms and Definitions
- Intersecting Lines and Non-intersecting Lines
- Pairs of Angles
- Parallel Lines and a Transversal
- Lines Parallel to the Same Line
- Angle Sum Property of a Triangle

- Lines are basic concepts in geometry, and they can be defined as a straight path that extends infinitely in both directions.
- Angles are formed by the intersection of two lines or line segments, and they are measured in degrees.
- Types of angles include acute, right, obtuse, straight, reflex, and complementary angles.
- The sum of the angles in a triangle is always 180 degrees.
- Parallel lines are lines that never intersect, and they have the same slope.
- Perpendicular lines are lines that intersect at a 90-degree angle.
- Transversals are lines that intersect two parallel lines, and they form eight angles.
- Corresponding angles are congruent when two parallel lines are intersected by a transversal.
- Alternate interior angles and alternate exterior angles are congruent when two parallel lines are intersected by a transversal.
- The angle sum property states that the sum of the measures of the angles of a polygon with n sides is (n-2) x 180 degrees.

NCERT solutions for maths chapter 6 class 9 Lines and Angles covers the solutions from every concept. This chapter contains 3 exercises with a total of 18 questions. This chapter carries a detailed explanation of all 18 questions in practice exercises. Lines and angles are everywhere around us. To design the structure of the building, the architecture uses both lines and angles. When we have to find the height of a building or a tower, then we need to know the angles.

Also practice class 9 maths ch 6 question answer using the exercise given below.

- Lines And Angles Class 9 NCERT Exercise 6.1
- Lines And Angles Class 9 NCERT Exercise 6.2
- Lines And Angles Class 9 NCERT Exercise 6.3

Chapter No. | Chapter Name |

Chapter 1 | |

Chapter 2 | |

Chapter 3 | |

Chapter 4 | |

Chapter 5 | |

Chapter 6 | Lines And Angles |

Chapter 7 | |

Chapter 8 | |

Chapter 9 | |

Chapter 10 | |

Chapter 11 | |

Chapter 12 | |

Chapter 13 | |

Chapter 14 | |

Chapter 15 |

** How To Use NCERT Solutions For Class 9 Maths Chapter 6 Lines And Angles **

- Before coming to this chapter, please ensure that you know the concepts of the previous chapter.
- First, go through some basic terminologies and observations given in the NCERT textbook.
- learn the usage of these concepts in the questions.
- Use these concepts while solving the practice exercises and for the help, you can use NCERT solutions for class 9 maths chapter 6 Lines and Angles.

*Keep Working Hard and Happy Learning!*

1. What are the important topics in NCERT solution for maths chapter 6 class 9?

Pair of angles, Parallel and Transversal Lines, Angle sum property of a triangle are the important topics covered in this chapter. Students can practice NCERT solutions for ch 6 maths class 9 to command these concepts.

2. How many chapters are there in the CBSE class 9 maths ?

There are 15 chapters starting from the number system to probability in the CBSE class 9 maths. which are listed below

- Number Systems
- Polynomials
- Coordinate Geometry
- Linear Equations in Two Variables
- Introduction to Euclid's Geometry
- Lines and Angles
- Triangles
- Quadrilaterals
- Areas of Parallelograms and Triangles
- Circles
- Constructions
- Heron's Formula
- Surface Areas and Volumes
- Statistics
- Probability

3. Where can I find the complete NCERT solutions for class 9 maths chapter 6 ?

Here you will get the detailed NCERT solutions for class 9 maths. For ease students can study lines and angles class 9 pdf both online and offline mode. practicing these solutions help them indepth understanding the cocepts which ultimately lead to score well in the exam.

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