NCERT Solutions for Exercise 6.1 Class 9 Maths Chapter 6

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:

1595609937996
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:

1595610210463
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:

1595610294843
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:

1595610388882
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:

1595610457579
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:

1595611505816
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 [ $\because$ $\angle QYP = \angle ZYQ$

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

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.

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

Don't Miss: JEE Main 2027: Narayana Scholarship Test Preparation Kit for Class 9

This Story also Contains

Lines and Angles Class 9 Maths Chapter 6 Exercise: 6.1
More About NCERT Solutions for Class 9 Maths Exercise 6.1
Benefits of NCERT Solutions for Class 9 Maths Exercise 6.1
NCERT Solutions of Class 9 Subject Wise
Subject Wise NCERT Exemplar Solutions

Also see

NCERT Solutions of Class 9 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (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.

Also Read

NCERT Solutions for Class 9 - Subject-wise PDFs Updated for 2024-25

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Maths

NCERT Syllabus for Class 9 Maths 2023 - Download Syllabus Pdf

NCERT Solutions For Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

NCERT Exemplar Class 9 Maths Solutions - Chapter Wise Problems with Solutions

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

NCERT Exemplar Class 9 Science Solutions Chapter 14 Natural Resources

NCERT Exemplar Class 9 Science Solutions Chapter 13 Why do we fall ill?

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

Articles

MPSOS 10th Result 2024 - Check MPSOS Class 10 December 2024 Exam Result @mpsos.nic.in

Oct 29, 2024

Education Loan for School Students in India – Eligibility, Benefits & Application Process

Oct 29, 2024

How to Track NSP Payment Status – Step-by-Step Guide for Scholarship Updates

Oct 29, 2024

TOSS SSC Time Table 2025, Telangana Open School Class 10 Exam Dates @telanganaopenschool.org

Oct 29, 2024

IB Primary Years Programme (PYP) – Curriculum Overview and Key Benefits

Oct 28, 2024

Importance of Internships in Student Life - 10 Points, 100 Words, 200 Words, 500 Words

Oct 28, 2024

Cambridge Advanced Curriculum: AS and A levels (Grade 11 & 12)

Oct 28, 2024

Cambridge School Fees 2024-25 | Admission, Tuition Fee for Cambridge IGCSE Schools

Oct 25, 2024

Online Courses After 10th

Sep 17, 2024

Pathology Lab Technician Courses - Duration, Fees, Eligibility

Sep 16, 2024

Upcoming School Exams

Himachal Pradesh Board of Secondary Education 10th Examination

Application Date:09 September,2024 - 14 November,2024

Exam Pattern

Admit Card

Result

Himachal Pradesh Board of Secondary Education 12th Examination

Application Date:09 September,2024 - 14 November,2024

Result

Exam Pattern

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Mock Test

National Means Cum-Merit Scholarship

Application Date:05 October,2024 - 04 November,2024

Eligibility Criteria

Application Process

Exam Pattern

View All School Exams

Get answers from students and experts

Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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 3.8×10⁷J 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

Option 4)

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

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

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

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ 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?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 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	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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