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NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

Edited By Ramraj Saini | Updated on Sep 27, 2023 10:31 PM IST

NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

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.

This Story also Contains
  1. NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles
  2. Lines And Angles Class 9 Questions And Answers PDF Free Download
  3. Lines And Angles Class 9 Solutions - Important Points
  4. Lines And Angles Class 9 NCERT Solutions (Intext Questions and Exercise)
  5. NCERT Lines And Angles Class 9 Solutions - Topics
  6. Summary Of Lines And Angles Class 9 Solutions
  7. NCERT Solutions For Class 9 Maths - Chapter Wise
  8. NCERT Solutions For Class 9 - Subject Wise
  9. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles
NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

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.

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Lines And Angles Class 9 Questions And Answers PDF Free Download

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Lines And Angles Class 9 Solutions - Important Points

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.

Lines And Angles Class 9 NCERT Solutions (Intext Questions and Exercise)

Class 9 lines and angles NCERT solutions - 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

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.

1640149711922

Answer:

15956105842171595610559774
Given that,
In the figure, CD and PQ intersect at F
Therefore, y =130^0 (vertically opposite angles)

PQ is a straight line. So,
\\x + 50^0 =180^0\\ x = 130^0

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


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

1640149786744

Answer:

15956106561461595610652079
Given AB || CD and CD || EF and y:z = 3:7 \Rightarrow z = \frac{7}{3}y
therefore, AB || EF and x =z (alternate interior angles)..............(i)

Again, CD || AB
\\\Rightarrow x+y =180^0\\ \Rightarrow z+y =180^0 .............(ii)

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

\frac{10}{3}y =180^0 \Rightarrow y =54^0
Then z=180^0-54^0=126^0

By equation (i), we get the value of x=126^0

Q3 In Fig. 6.30, if AB || CD, EF \bot CD and \angle GED = 126°, find \angle AGE, \angle GEF and \angle FGE.

Answer:

15956107474541595610742595
Given AB || CD, EF \perp CD and \angle GED = 126^0
In the above figure,
GE is transversal. So, that \angle AGE = \angle GED = 126^0 [Alternate interior angles]
Also, \angle GEF = \angle GED - \angle FED
= 126^0-90^0 = 36^0
\angle GEF = 36^0

Since AB is a straight line
Therefore, \angle AGE + \angle FGE = 180^0
So, \angle FGE = 180^0-\angle AGE = 180^0 - 126^0
\Rightarrow \angle FGE =54^0

Q4 In Fig. 6.31, if PQ || ST, \angle PQR = 110° and \angle RST = 130°, find \angle QRS. [ Hint : Draw a line parallel to ST through point R.]

1640149893960

Answer:

Draw a line EF parallel to the ST through R.
Since PQ || ST and ST || EF
\Rightarrow EF || PQ
15956108212371595610815150
\angle PQR = \angle QRF = 110^0 (Alternate interior angles)
\angle QRF = \angle QRS + \angle SRF .............(i)

Again, \angle RST + \angle SRF = 180^0 (Interior angles of two parallels ST and RF)
\Rightarrow \angle SRF =180^0-130^0 = 50^0 ( \angle RST = 130^0 , given)

Thus, \angle QRS = 110^0-50^0 = 60^0

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

1640150269148

Answer:

15956108715001595610869627
Given, AB || CD, \angle APQ = 50^0 and \angle PRD = 127^0
PQ is a transversal. So,
\angle APQ = \angle PQR= 50^0 (alternate interior angles)
\therefore x = 50^0

Again, PR is a transversal. So,
\angle y + 50^0 = 127^0 (Alternate interior angles)
\Rightarrow y = 77^0

Q6 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.

1640150293845

Answer:

15956109561241595610951432
Draw a ray BL \perp PQ and CM \perp RS
Since PQ || RS (Given)
So, BL || CM and BC is a transversal
\therefore \angle LBC = \angle MCB (Alternate interior angles).............(i)

It is known that, angle of incidence = angle of reflection
So, \angle ABL = \angle LBC and \angle MCB = \angle MCD
\Rightarrow \angle ABL =\angle MCD ..................(ii)

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

\angle ABC = \angle DCB
Both the interior angles are equal
Hence AB || CD

Class 9 maths chapter 6 NCERT solutions - Exercise: 6.3

Q1 In Fig. 6.39, sides QP and RQ of \Delta PQR are produced to points S and T respectively. If \angle SPR = 135° and \angle PQT = 110°, find \angle PRQ.

1640150386912

Answer:

15956110080781595611005967
Given,
PQR is a triangle, \angle SPR = 135^0 , \angle PQT = 110^0

Now, \angle TQP + \angle PQR = 180^0 (Linear pair)
So, \angle PQR = 180^0-110^0 = 70^0

Since the side of QP of the triangle, PQR is produced to S
So, \angle PQR + \angle PRQ = 135^0 (Exterior angle property of triangle)
\therefore \angle PRQ = 135^0 - 70^0 = 65^0

Q2 In Fig. 6.40, \angle X = 62°, \angle XYZ = 54°. If YO and ZO are the bisectors of \angle XYZ and \angle XZY respectively of \Delta XYZ, find \angle OZY and \angle YOZ.

15956111030681595611095537
Answer:

We have
\angle X = 62^0 , \angle XYZ = 54^0
YO and ZO bisects the \angle XYZ and \angle XZY
Now, In \Delta XYZ, by using angle sum property
\angle XYZ + \angle YZX + \angle ZXY = 180^0

So, \angle YZX = 180^0-54^0-62^0
\angle YZX = 64^0

and, \angle OYZ = 54^0/2 = 27^0 also, \angle OZY = 32^0

Now, in \Delta OYZ
\angle Y + \angle O + \angle Z = 180^0 [ \angle Y = 32^0 and \angle Z = 64^0 ]
So, \angle YOZ = 121^0

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

15956111843421595611181275
Answer:

We have,
AB || DE, \angle BAC = 35° and \angle CDE = 53°

AE is a transversal so, \angle BAC = \angle AED = 35^0

Now, In \Delta CDE,
\angle CDE + \angle DEC + \angle ECD = 180^0 (By angle sum property)
Therefore, \angle ECD = 180^0-53^0-35^0
\angle ECD = 92^0

Q4 In Fig. 6.42, if lines PQ and RS intersect at point T, such that \angle PRT = 40°, \angle RPT = 95° and \angle TSQ = 75°, find \angle SQT.

15956112332891595611230995
Answer:

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

In \Delta PRT, by using angle sum property
\angle PRT + \angle PTR + \angle TPR = 180^0
So, \angle PTR = 180^0 -95^0-40^0
\Rightarrow \angle PTR = 45^0

Since lines, PQ and RS intersect at point T
therefore, \angle PTR = \angle QTS (Vertically opposite angles)
\angle QTS = 45^0

Now, in \Delta QTS,
By using angle sum property
\angle TSQ + \angle STQ + \angle SQT = 180^0
So, \angle SQT = 180^0-45^0-75^0
\therefore \angle SQT = 60^0

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

15956112915331595611289597
Answer:

We have,
PQ \bot PS, PQ || SR, \angle SQR = 28° and \angle QRT = 65°
Now, In \Delta QRS, the side SR produced to T and PQ || RS
therefore, \angle QRT = 28^0+x = 65^0
So, x = 37^0

Also, \angle QRT = \angle RSQ + \angle SQR (By exterior angle property of a triangle)
Therefore, \angle RSQ = \angle QRT - \angle SQR
\angle RSQ = 65^0-28^0 = 37^0

Now, in \Delta PQS,
\angle P + \angle PQS + \angle PSQ = 180^0
y=180^0-90^0-37^0
y=52^0

Q6 In Fig. 6.44, the side QR of \Delta PQR is produced to a point S. If the bisectors of \angle PQR and \angle PRS meet at point T, then prove that \angle QTR = \frac{1}{2} \angle QPR.

15956113451601595611343477
Answer:

We have,
\Delta PQR is produced to a point S and bisectors of \angle PQR and \angle PRS meet at point T,
By exterior angle sum property,
\angle PRS = \angle P + \angle PQR
Now, \frac{1}{2}\angle PRS =\frac{1}{2}\angle P + \frac{1}{2} \angle PQR
\Rightarrow \angle TRS = \frac{1}{2}\angle P + \angle TQR ................(i)

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

Now, in \Delta QRT,
\angle TRS = \angle T + \angle TQR ..............(ii)

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

\frac{1}{2}\angle P= \angle T
\Rightarrow \frac{1}{2}\angle QPR= \angle QTR
Hence proved.

NCERT Lines And Angles Class 9 Solutions - Topics

  • 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

Summary Of Lines And Angles Class 9 Solutions

  • 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.

NCERT Solutions For Class 9 Maths - Chapter Wise

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

NCERT Solutions For Class 9 - Subject Wise

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.

NCERT Books and NCERT Syllabus

Keep Working Hard and Happy Learning!

Frequently Asked Questions (FAQs)

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

  1. Number Systems
  2. Polynomials
  3. Coordinate Geometry
  4. Linear Equations in Two Variables
  5. Introduction to Euclid's Geometry
  6. Lines and Angles
  7. Triangles
  8. Quadrilaterals
  9. Areas of Parallelograms and Triangles
  10. Circles
  11. Constructions
  12. Heron's Formula
  13. Surface Areas and Volumes
  14. Statistics
  15. 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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