# NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

**NCERT solutions for class 9 maths chapter 6 Lines And Angles- **When two points are joined together, the resultant is called line. An angle is formed when two unparalleled lines are connecting at a point. These two lines are called the sides of the angle. In the previous chapter, you have already studied some basic formations. In this 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. NCERT solutions for class 9 maths chapter 6 Lines and Angles covers the solutions from every concept. This chapter contains 3 exercises with a total of 18 questions. NCERT solutions for class 9 maths chapter 6 Lines And Angles carry a detailed explanation to all the 18 questions in practice exercises. Lines and angles are everywhere around us. To design the structure of the building, an architecture uses both lines and angles. When we have to find the height of a building or a tower, then we need to know angles. NCERT solutions for class 9 maths chapter 6 Lines and Angles will also be helpful in the preparation NTSE, Indian National Olympiad (INO), etc. Finally, it can be concluded that the NCERT solutions can play a vital role in increasing your scores. Here you will get NCERT solutions for class 9 also.

**Some basic terms and their figures that we are going to discuss in NCERT Class 9 Maths Chapter Lines And Angles are **

Terms | Figures |

Acute angle | |

Right angle | |

Obtuse angle | |

Straight angle | |

Reflex angle |

## ** NCERT solutions for class 9 maths chapter 6 Lines and Angles Excercise: 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 [

=

## ** NCERT solutions for class 9 maths chapter 6 Lines and Angles Excercise: 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

## ** NCERT solutions for class 9 maths chapter 6 Lines and Angles Excercise: 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.

** NCERT solutions for class 9 maths chapter wise **

** 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.

* Keep Working Hard and Happy Learning! *

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

**Question: **What are the important topics in chapter Lines And Angles ?

**Answer: **

Pair of angles, Parallel and Transversal Lines, Angle sum property of a triangle are the important topics covered in this chapter.

**Question: **Does CBSE class 9 maths is tough ?

**Answer: **

No, CBSE class 9 maths is easy but some students find it tough because we study maths in previous classes is very simple and some concepts in class 9 maths are new.

**Question: **Where can I find the complete solutions of NCERT for class 9 maths ?

**Answer: **

Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link.

**Question: **Which is the official website of NCERT ?

**Answer: **

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

**Question: **Where can I find the complete solutions of NCERT for class 9 ?

**Answer: **

Here you will get the detailed NCERT solutions for class 9 by clicking on the link.

**Question: **How many chapters are there in the CBSE class 9 maths ?

**Answer: **

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

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