# NCERT Solutions For Class 9 Maths Chapter 6 Lines And Angles (Ex 6.1) Exercise 6.1

Class 9 students do their best to ace mathematics examinations. There are a lot of tips that can help make the process a bit easier. One of the most important things is to be sure to understand how to use the theorems and formulas in the particular question that are covered in the textbook. This can be quickly done by reviewing the materials recommended by Extramarks as well as doing any practice problems that are available in the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1. In addition, following the procedures and methods that are used in the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, can be very beneficial for students who want to score high in their exams.

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## NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles (Ex 6.1) Exercise 6.1

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### NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.1) Exercise 6.1

NCERT Solutions are a collection of solutions created by Extramarks’ experts to aid students in comprehending the ideas presented in the council-recommended textbooks. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 are created in line with the most recent CBSE syllabus and are accessible for students studying in classes 6 through 12. Both Hindi and English versions are available for them.

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### NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.1) Exercise 6.1

Students will study Basic terms and definitions of Lines and Angles as well as other vital topics in Chapter 6, Exercise 6.1 of Class 9 in Mathematics subject. The following themes should be well prepared by students while solving Exercise 6.1:

1. Terms and Definitions – Line Segment, Ray, Collinear and Non-collinear points
2. Angles – Arms and Vertexes
3. Types of Angles (7) – Acute Angle, Right Angle, Obtuse Angle, Straight Angle, Reflex Angle, Complementary Angle, and Supplementary Angle
4. Linear Pair of Angles – Vertically opposite angles
5. Intersecting Lines and Non-Intersecting Lines
6. Pair of Angles – Linear (Theorems and Examples)

Exercise 6.1 has 6 questions in total that can be solved with the guidance of NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 emphasised by Extramarks.

### Chapter Wise NCERT Solutions For Class 9 Maths

The chapter-wise NCERT Solutions for Class 9 can be found by students on the Extramarks mobile application or website. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, are a great resource for students to understand the concepts of Mathematics and to prepare for exams. These NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, are easy to follow and will help students score better grades. For students’ ease, Extramarks has provided the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 in PDF format, so that, students can download them and refer to them whenever required. Exercise 6.1 Class 9 Solutions, assists students in understanding the questions mentioned in Class 9 Maths Chapter 6 Exercise 6.1. The chapters that are covered in the Class 9 Mathematics subject course are as follows:

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

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### Class 9 Maths Chapter 6 Includes:

In Class 9, Chapter 6 of Mathematics is all about Lines and Angles as the name of the chapter suggests. In this chapter, students will learn how to solve questions regarding Line segments, Types and Pairs of Angles, Intersecting and Non-Intersecting Lines, Parallel Lines and Transversal, Lines Parallel to the same lines, and other topics. These topics have several exercises incorporated into them that students can practice and get a deeper understanding of it. Similarly, if students are not able to solve any particular question, they can take help from the NCERT Solutions, like the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 advised by Extramarks.

### NCERT Solutions for Class 9

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

Ans

$\begin{array}{l}\mathrm{Given}:\text{Lines AB and CE itersect at O and}\\ \angle \mathrm{AOC}+\angle \mathrm{BOE}=70\mathrm{°}\text{and}\angle \mathrm{BOD}=40\mathrm{°}\\ \mathrm{To}\text{find:}\angle \mathrm{BOE}\text{and reflex\hspace{0.17em}}\angle \mathrm{COE}.\\ \angle \mathrm{AOB}\text{is a straight angle, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=180\mathrm{°}\\ \angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}70\mathrm{°}+\angle \mathrm{COE}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{COE}=180\mathrm{°}-70\mathrm{°}=110\mathrm{°}\\ \angle \mathrm{COD}\text{is a straight angle, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{COD}=180\mathrm{°}\\ \angle \mathrm{COE}+\angle \mathrm{EOB}+\angle \mathrm{BOD}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}110\mathrm{°}+\angle \mathrm{BOE}+40\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOE}=180\mathrm{°}-150\mathrm{°}\\ =30\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Reflex}\text{}\angle \mathrm{COE}=360\mathrm{°}-\angle \mathrm{COE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=360\mathrm{°}-110\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=250\mathrm{°}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{Given}:\text{​ Lines XY and MN intersect at O.}\angle \text{POY}=90\mathrm{°}\text{and a:b}=2:3\\ \mathrm{To}\text{find: c}\\ \text{Since, a:b = 2:3}\\ \text{so, let a}=\text{2x, b}=\text{3x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{POX}+\angle \mathrm{POY}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{b}+90\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+3\mathrm{x}+90\mathrm{°}=180\mathrm{°}\\ 5\mathrm{x}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{180\mathrm{°}}{5}=30\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=2\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2×30\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{°}\\ \mathrm{MN}\text{is a straight line, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{MOX}+\angle \mathrm{XON}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}60\mathrm{°}+\mathrm{c}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=180\mathrm{°}-60\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=120\mathrm{°}\end{array}$

Q.3 In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠PQS = ∠PRT.

Ans

$\begin{array}{l}\mathrm{Given}:\text{In}\mathrm{\Delta }\text{PQR,\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR=}\angle \text{PRQ}\\ \text{To Prove:}\angle \text{PQS=}\angle \text{PRT}\\ \text{Proof}:\text{​}\angle \text{PQR}+\angle \text{PQS}=180\mathrm{°}\left[\mathrm{Linear}\text{pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}=180\mathrm{°}-\angle \text{PQS\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PRQ}+\angle \text{PRT}=180\mathrm{°}\left[\mathrm{Linear}\text{pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PRQ}=180\mathrm{°}-\angle \text{PRT\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{But}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}=\angle \text{PRQ}\left[\mathrm{Given}\right]\\ ⇒180\mathrm{°}-\angle \text{PQS}=180\mathrm{°}-\angle \text{PRT}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQS}=\angle \text{PRT}\\ \text{Hence Proved.}\end{array}$

Q.4 In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Ans

$\begin{array}{l}\mathrm{Given}:\text{x}+\mathrm{y}=\mathrm{w}+\text{z}\\ \mathrm{To}\text{Prove: AOB is a line.}\\ \text{Proof:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\because \mathrm{x}+\mathrm{y}+\mathrm{w}+\mathrm{z}=360\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Complete}\text{​​ angle}\right]\\ \mathrm{x}+\mathrm{y}+\mathrm{x}+\mathrm{y}=360\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because \text{x}+\mathrm{y}=\mathrm{w}+\text{z}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\left(\mathrm{x}+\mathrm{y}\right)=360\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\mathrm{y}=\frac{360\mathrm{°}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AOB\hspace{0.17em}}=180\mathrm{°}\\ ⇒\mathrm{Since},\angle \text{AOB is a straight angle, so AOB is a straight line.}\end{array}$

Q.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 = (1/2)(∠QOS – ∠POS).

Ans

$\begin{array}{l}\mathrm{Given}:\text{​ POQ is a line.}\stackrel{\to }{\mathrm{OR}}\perp \mathrm{PQ}\text{and}\stackrel{\to }{\mathrm{OS}}\text{is another ray between}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OR}}\text{​\hspace{0.17em}and}\stackrel{\to }{\mathrm{OP}}.\\ \mathrm{To}\text{Prove:}\angle \mathrm{ROS}\text{}=\text{}\frac{1}{2}\left(\angle \mathrm{QOS}-\angle \mathrm{POS}\right)\\ \text{Proof:R.H.S.}=\text{}\frac{1}{2}\left(\angle \mathrm{QOS}-\angle \mathrm{POS}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left\{\left(\angle \mathrm{QOR}+\angle \mathrm{ROS}\right)-\left(\angle \mathrm{ROP}-\angle \mathrm{ROS}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left\{\left(90\mathrm{°}+\angle \mathrm{ROS}\right)-\left(90\mathrm{°}-\angle \mathrm{ROS}\right)\right\}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\angle \mathrm{QOR}=\angle \mathrm{ROP}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{°}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(\angle \mathrm{ROS}+\angle \mathrm{ROS}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(2\angle \mathrm{ROS}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\angle \mathrm{ROS}\end{array}$

Q.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.

Ans

$\begin{array}{l}\mathrm{Given}:\text{}\angle \text{XYZ}=\text{64°, XY is}\mathrm{}\text{produced to point P.Ray YQ bisects}\\ \angle \text{ZYP.}\\ \text{To find:}\angle \text{XYQ and reflex}\angle \text{QYP}.\\ \text{​}\mathrm{Since},\text{}\stackrel{\to }{\text{YQ}}\text{bisects}\angle \text{ZYP, so}\angle \text{ZYQ}=\angle \text{PYQ}=\mathrm{a}\left(\mathrm{let}\right)\\ \text{\hspace{0.17em}}\angle \text{XYZ}+\angle \text{ZYP}=\text{180°\hspace{0.17em}}\left[\mathrm{Linear}\text{pair of angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}64\mathrm{°}+2\mathrm{a}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}=180\mathrm{°}-64\mathrm{°}\\ =116\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{116\mathrm{°}}{2}\\ =58\mathrm{°}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{XYQ}=64\mathrm{°}+\mathrm{a}\\ =64\mathrm{°}+58\mathrm{°}\\ =122\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}reflex}\angle \text{QYP}=360\mathrm{°}-\angle \text{QYP}\\ =360\mathrm{°}-58\mathrm{°}\\ =302\mathrm{°}\end{array}$