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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 Komal Miglani | Updated on May 02, 2025 08:13 PM IST

Have you ever looked at a long, straight road with no visible ending into the distance and wondered where it might end? Or noticed the hands of a clock at exactly 3 o’clock forming a perfect 90-degree angle? These are everyday examples where we unknowingly observe lines and angles around us. In class 9 Maths NCERT chapter 6, students will study lines, angles and their properties. The purpose of this article is to make learning easier for students and explain the NCERT Solutions thoroughly.

This Story also Contains
  1. Lines And Angles Class 9 Questions And Answers PDF Free Download
  2. Lines And Angles Class 9 Solutions: Important Points
  3. NCERT Solutions for Class 9 Maths Chapter 6: Exercise Questions
  4. Class 9 Maths NCERT Chapter 6: Extra Question
  5. Approach to Solve Questions of Lines and Angles Class 9
  6. NCERT Solutions for Class 9 Mathematics: Chapter-wise
  7. NCERT Solutions For Class 9: Subject Wise
  8. 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

Careers360 experts, experienced in this area, prepared the Class 9 NCERT solutions for lines and angles and adhered to the latest CBSE syllabus. There are various applications of lines and angles in our daily lives. So, having the concepts clear will be beneficial not only for the Class 9 board exams but also for day-to-day life. After going through all the NCERT Solutions for Class 9 Maths, students can practice the NCERT Exemplar Solutions Class 9 NCERT Maths Chapter Lines And Angles for a deeper understanding of the concepts. Along with this, using the NCERT Solutions for Class 9, learners can practise diagram-based questions and prove angle-based properties confidently.

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.

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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

NCERT Solutions for Class 9 Maths Chapter 6: Exercise Questions

Class 9 Maths Chapter 6 Question Answer: Exercise: 6.1
Total Questions: 6
Page number: 76-77

Question 1: In Fig. 6.13, lines AB and CD intersect at O. If AOC + BOE = 70° and BOD = 40°, find BOE and reflex COE.

1640149435860

Answer:

1595609937996
Given that,

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

AOC+BOE=70° and BOD = 40°

Since AB is a straight line,

AOC + COE + EOB = 180°

COE=180°70°=1100

[since AOC+BOE=70° ]

So, reflex COE = 360°1100=250°

It is given that AB and CD intersect at O.

Therefore, AOC = BOD [vertically opposite angle]

COA=40° [ GIven BOD = 400 ]

Also, AOC+BOE=70°

So, BOE = 30°

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

1640149490726

Answer:

1595610210463

Given that,

Line XY and MN intersect at O and POY = 900

Also, a:b=2:3b=3a2 ..............(I)

Since XY is a straight line

Therefore, a+b+POY=1800

a+b=1800900=900 ...........(ii

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

3a2+a=900

a=360

So, b=540

Since MOY = c [vertically opposite angles]

a + POY = c

1260=c

Question 3: In Fig. 6.15, PQR = PRQ, then prove that PQS = PRT.

1640149622015

Answer:

1595610294843

Given that,

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

Now, PQR + PQS = 1800 {Linear pair}............(I)

Similarly, PRQ + PRT = 1800 ..................(ii)

Equating the equation (i) and eq (ii), we get,

PQR+PQS=PRT+PRQ {but PQR = PRQ }

Therefore, PQS = PRT

Hence, it is proved.

Question 4: In Fig. 6.16, ifx + y = w + z, then prove that AOB is a line.

1640149658149

Answer:

1595610388882

Given that,

x+y=z+w ..............(I)

It is known that the sum of all the angles at a point = 3600

x+y+z+w=3600 ..............(ii)

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

2(x+y)=3600

x+y=1800

Hence, proven that AOB is a line.

Question 5: 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 ROS=12(QOSPOS)

1640149689300

Answer:

1595610457579

Given that,

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

Now, POS + ROS + ROQ = 1800 [since POQ is a straight line]

POS+ROS=900

ROS=900POS .............(I)

and, ROS + ROQ = QOS

ROS=QOS900 ..............(ii)

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

ROS=12(QOSPOS)

Hence, proved.

Question 6: It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ZYP, find XYQ and reflex QYP.

Answer:

1595611505816

Given that,

XYZ = 640 and XY produced to point P and Ray YQ bisects ZYP QYP=ZYQ

Now, XYP is a straight line.

So, XYZ + ZYQ + QYP = 1800

2(QYP)=1800640=1160

(QYP)=580

Thus reflex of QYP = 3600580=3020

Since XYQ = XYZ + ZYQ

[ QYP=ZYQ

XYQ = 640+580=1220

Class 9 Maths Chapter 6 Question Answer: Exercise: 6.2
Total Questions: 5
Page number: 80-81

Question 1: In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7 , find x .


1640149786744

Answer:

1595610656146

Given AB || CD and CD || EF and y:z=3:7

z=73y

Therefore, AB || EF and x=z (alternate interior angles)..............(i)

Again, CD || AB

x+y=1800

z+y=1800 .............(ii)

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

103y=1800

y=540

Then z=1800540=1260

By equation (i), we get the value of x=1260

Question 2: In Fig. 6.30, if AB || CD, EF CD and GED = 126°, find AGE, GEF and FGE.

Answer:

1595610747454

Given AB || CD, EF CD and GED = 1260

In the above figure,

GE is transversal.

So, that AGE = GED = 1260

[Alternate interior angles]

Also, GEF = GED - FED = 1260900=360

GEF=360

Since AB is a straight line.

Therefore, AGE + FGE = 1800

So, FGE = 1800AGE=18001260

FGE=540

Question 3: In Fig. 6.31, if PQ || ST, PQR = 110° and RST = 130°, find 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

EF || PQ
1595610821237

PQR = QRF = 1100(Alternate interior angles)

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

Again, RST + SRF = 1800 (Interior angles of two parallels ST and RF)

SRF=18001300=500 ( RST = 1300 , given)

Thus, QRS = 1100500=600

Question 4: In Fig. 6.32, if AB || CD, APQ = 50° and PRD = 127°, find x and y.

1640150269148

Answer:

1595610871500

Given, AB || CD, APQ = 500 and PRD = 1270

PQ is a transversal.

So,

APQ = PQR= 500 (alternate interior angles)

x=500

Again, PR is a transversal.

So,

y + 500 = 1270 (Alternate interior angles)

y=770

Question 5: In Fig. 6.27, 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 along CD. Prove that AB || CD.

Answer:

1595610956124

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

ABL=MCD ..................(ii)

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

ABC = DCB

Both the interior angles are equal

Hence AB || CD


Class 9 Maths NCERT Chapter 6: Extra Question

Question: In the figure, POR and QOR form a linear pair. If a° - b° = 90° find the values of a and b.


Answer:

Given: POR + QOR = 180°

It means a° + b° = 180°..................(i)

And, a° - b° = 90° ..................(ii)

Now, add (i) and (ii), We get;

2 a° = 270°

Therefore, a° = 135°

And, thus b° = 180° - 135° = 45°

Approach to Solve Questions of Lines and Angles Class 9

1. Understand different types of angles: Acute, obtuse, straight, reflex, and right angles need revision as a starting point to understand angle relationships in lines.

2. Learn angle pair relationships: Students need to understand complementary angles along with supplementary angles, whereas adjacent angles and linear pairs and vertically opposite angles are necessary for solving angle problems.

3. Know the angle properties of intersecting lines: When two lines intersect, the resulting vertically opposite angles will always be equal in measure to each other.

4. Apply properties of parallel lines with a transversal: You will discover the required behaviours of alternate interior angles, corresponding angles, and consecutive interior angles in cases where parallel lines intersect with a transversal.

5. Use angle theorems for reasoning and proof: When proving lines parallel, it is essential to use angle theorems, especially the “If two lines are parallel, alternate interior angles are equal” theorem, together with its converse proof.

6. Solve problems step-by-step using angle relationships: The solution of angle-based questions progresses easily through known angle pairs combined with postulates to create exact answers.

NCERT Solutions for Class 9 Mathematics: Chapter-wise

NCERT Solutions For Class 9: Subject Wise

These are links to the solutions of other subjects, which students can check to revise and strengthen those concepts.

NCERT Books and NCERT Syllabus

Before planning a study schedule, always analyse the latest syllabus. Here are the links to the latest NCERT syllabus and some of the important books that will help students in this cause.

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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Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

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0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

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12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

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K/2\,

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\; K\;

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zero\;

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K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

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33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

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0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

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2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

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increase two fold

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remain unchanged

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be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

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Weight fraction of solute

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Fraction of solute present in water

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twice that in 60 g carbon

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6.023 × 1022

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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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less than 3

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more than 3 but less than 6

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more than 6 but less than 9

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more than 9

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