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, nonintersecting Lines and angle sum property of the triangle. Solutions of NCERT 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. CBSE 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.
Terms 
Figures 
Acute angle 

Right angle 

Obtuse angle 

Straight angle 

Reflex angle 

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.
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.
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.
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.
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.
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 [
=
Q1 In Fig. 6.28, find the values of x and y and then show that AB CD.
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.
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.
Given AB  CD, EFCD 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 =
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.
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)
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
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)
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.
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 =
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.
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 =
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.
Chapter No. 
Chapter Name 
Chapter 1 

Chapter 2 
CBSE NCERT solutions for class 9 maths chapter 2 Polynomials 
Chapter 3 
Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry 
Chapter 4 
NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables 
Chapter 5 
CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry 
Chapter 6 
NCERT solutions for class 9 maths chapter 6 Lines And Angles 
Chapter 7 

Chapter 8 
CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals 
Chapter 9 
Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles 
Chapter 10 

Chapter 11 
CBSE NCERT solutions for class 9 maths chapter 11 Constructions 
Chapter 12 

Chapter 13 
NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes 
Chapter 14 
CBSE NCERT solutions for class 9 maths chapter 14 Statistics 
Chapter 15 
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!
1. In Fig. 6.39, sides QP and RQ of PQR are produced to points S and T respectively. If SPR = 135° and PQT = 110°, find PRQ.
6. 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.
5. In Fig. 6.43, if PQ PS, PQ SR, SQR = 28° and QRT = 65°, then find the values of x and y.