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. 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.

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 0^o < x < 90^o

Acute angle

Right angle y = 90^o

Right angle

Obtuse angle 90^o < z < 180^o

Obtuse angle

Straight angle s = 180^o

Straight angle

 

Reflex angle   180^o< t < 360^o

Reflex angle

 

NCERT solutions for class 9 maths chapter 6 Lines and Angles Excercise: 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 \angleBOE and reflex \angleCOE.

                        

Answer:


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 \angleAOC + \angleCOE + \angleEOB = 180^0
\Rightarrow \angle COE = 180^0-70^0=110^0 [since \angle AOC + \angle BOE = 70^0]

So, reflex \angleCOE = 360^0-110^0 = 250^0
It is given that AB and CD intersect at O
Therefore, \angleAOC  = \angleBOD  [vertically opposite angle]
\Rightarrow \angle COA = 40^0 [ GIven \angle BOD = 40^0]
Also, \angle AOC + \angle BOE = 70^0
So,  \angleBOE = 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.

                

Answer:


Given that,
Line XY and MN intersect at O and \anglePOY = 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 \angleMOY = \anglec [vertically opposite angles]
          \anglea + \anglePOY = c
           126^0 =c

Q3 In Fig. 6.15, \angle PQR = \angle PRQ, then prove that \angle PQS = \angle PRT.

                

Answer:


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.

            

Answer:


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})

                

Answer:


Given that,
POQ is a line, OR \perp PQ and \angle ROQ is a right angle.
Now, \angle POS + \angleROS +  \angleROQ = 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:


Given that,
\angleXYZ = 64^0 and XY produced to point P and Ray YQ bisects \angleZYP  \Rightarrow \angle QYP = \angle ZYQ
Now, XYP is a straight line
So, \angleXYZ + \angleZYQ + \angleQYP = 180^0
\Rightarrow 2(\angle QYP )=180^0 - 64^0 = 116^0
\Rightarrow (\angle QYP )= 58^0

Thus reflex of \angleQYP = 360^0- 58^0 = 302^0

Since \angleXYQ = \angleXYZ + \angleZYQ  [\because \angle QYP = \angle ZYQ


    \angle XYQ64^0+58^0 = 122^0

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, y =130^0  (vertically opposite angles)

PQ is a straight line. So, 
\\x + 50^0 =180^0\\ x = 130^0

Hence AB || CD (since x and are y 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 y:z = 3:7 \Rightarrow z = \frac{7}{3}y
therefore, AB || EF and x =z (alternate interior angles)..............(i)

Again, CD || AB 
\\\Rightarrow x+y =180^0\\ \Rightarrow z+y =180^0.............(ii)

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

\frac{10}{3}y =180^0 \Rightarrow y =54^0
Then z=180^0-54^0=126^0

By equation (i), we get the value of x=126^0

Q3 In Fig. 6.30, if AB || CD, EF \bot CD and \angle GED = 126°, find \angle AGE, \angle GEF and \angle FGE.

                

Answer:


Given AB || CD, EF\perpCD and \angleGED = 126^0
In the above figure, 
GE is transversal. So, that \angleAGE = \angleGED =  126^0 [Alternate interior angles]
Also, \angleGEF = \angleGED - \angleFED 
                     = 126^0-90^0 = 36^0
        \angle GEF = 36^0 

Since AB is a straight line 
Therefore, \angleAGE  + \angleFGE =  180^0
So, \angleFGE = 180^0-\angle AGE = 180^0 - 126^0
\Rightarrow \angle FGE =54^0

Q4 In Fig. 6.31, if PQ || ST, \angle PQR = 110° and \angle RST = 130°, find \angle QRS. [Hint : Draw a line parallel to ST through point R.]

                

Answer:

Draw a line EF parallel to the ST through R.
Since PQ || ST and ST || EF 
\Rightarrow EF || PQ

\anglePQR = \angleQRF = 110^0  (Alternate interior angles)
\angleQRF = \angleQRS + \angleSRF .............(i)

Again, \angleRST + \angleSRF = 180^0 (Interior angles of two parallels ST and RF)
\Rightarrow \angle SRF =180^0-130^0 = 50^0  (\angleRST = 130^0, given)

Thus, \angleQRS = 110^0-50^0 = 60^0

Q5 In Fig. 6.32, if AB || CD, \angle APQ = 50° and \angle PRD = 127°, find x and y.

                

Answer:


Given, AB || CD, \angleAPQ = 50^0 and \anglePRD = 127^0
PQ is a transversal. So, 
\angleAPQ = \anglePQR= 50^0 (alternate interior angles)
\therefore x = 50^0

Again, PR is a transversal. So, 
\angley +50^0 = 127^0 (Alternate interior angles)
\Rightarrow y = 77^0

Q6 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.

 

            

Answer:


Draw  a ray BL \perpPQ and CM \perp RS
Since PQ || RS (Given)
So, BL || CM and BC is a transversal
\therefore \angleLBC =  \angleMCB (Alternate interior angles).............(i)

It is known that, angle of incidence  = angle of reflection
So, \angleABL = \angleLBC and \angleMCB =  \angleMCD
\Rightarrow \angle ABL =\angle MCD..................(ii)

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

\angleABC = \angleDCB
Both the interior angles are equal
Hence AB || CD

NCERT solutions for class 9 maths chapter 6 Lines and Angles Excercise: 6.3

Q1 In Fig. 6.39, sides QP and RQ of \DeltaPQR are produced to points S and T respectively. If \angle SPR = 135° and \angle PQT = 110°, find \angle PRQ.

                

Answer:


Given,
PQR is a triangle, \angleSPR =135^0,  \anglePQT = 110^0

Now, \angleTQP + \anglePQR = 180^0 (Linear pair)
So, \anglePQR = 180^0-110^0 = 70^0

Since the side of QP of the triangle, PQR is produced to S
So, \anglePQR + \anglePRQ = 135^0 (Exterior angle property of triangle)
\therefore \angle PRQ = 135^0 - 70^0 = 65^0

Q2 In Fig. 6.40, \angle X = 62°, \angle XYZ = 54°. If YO and ZO are the bisectors of \angle XYZ and \angle XZY respectively of \Delta XYZ, find \angle OZY and \angle YOZ.

                

Answer:


We have 
\angleX = 62^0\angleXYZ = 54^0
YO and ZO bisects the \angleXYZ and \angleXZY
Now, In \DeltaXYZ, by using angle sum property
\angleXYZ + \angleYZX + \angleZXY = 180^0

So, \angleYZX = 180^0-54^0-62^0
      \angleYZX = 64^0

and, \angleOYZ = 54^0/2 = 27^0 also, \angleOZY = 32^0

Now, in \DeltaOYZ 
\angleY + \angleO + \angleZ = 180^0  [\angleY = 32^0 and \angleZ = 64^0]
So, \angleYOZ = 121^0

Q3 In Fig. 6.41, if AB || DE, \angle BAC = 35° and \angle CDE = 53°, find \angle DCE.

                    

Answer:


We have, 
AB || DE, \angle BAC = 35° and \angle CDE = 53°

AE is a transversal so, \angle BAC = \angle AED = 35^0

Now, In \Delta CDE,
\angleCDE + \angleDEC + \angleECD = 180^0 (By angle sum property)
Therefore, \angleECD = 180^0-53^0-35^0
                 \angle ECD = 92^0

Q4 In Fig. 6.42, if lines PQ and RS intersect at point T, such that \angle PRT = 40°, \angle RPT = 95° and \angle TSQ = 75°, find \angle SQT.

                    

Answer:


We have,
lines PQ and RS intersect at point T, such that \angle PRT = 40°, \angle RPT = 95° and \angle TSQ = 75°

In \DeltaPRT, by using angle sum property
\anglePRT + \anglePTR + \angleTPR = 180^0
So, \anglePTR  = 180^0 -95^0-40^0
  \Rightarrow \angle PTR = 45^0

Since lines, PQ and RS intersect at point T
therefore, \anglePTR = \angleQTS (Vertically opposite angles)
                \angleQTS = 45^0

Now, in \DeltaQTS,
By using angle sum property
\angleTSQ + \angleSTQ + \angleSQT = 180^0
So, \angleSQT = 180^0-45^0-75^0
\therefore \angle SQT = 60^0

Q5 In Fig. 6.43, if PQ \bot PS, PQ ||SR, \angle SQR = 28° and \angle QRT = 65°, then find the values of x and y.

                        

Answer:


We have, 
PQ \bot PS, PQ || SR, \angle SQR = 28° and \angle QRT = 65°
Now, In \Delta QRS, the side SR produced to T and PQ || RS
therefore, \angleQRT = 28^0+x = 65^0
 So, x = 37^0

Also, \angleQRT = \angleRSQ + \angleSQR (By exterior angle property of a triangle)
Therefore, \angleRSQ = \angleQRT - \angleSQR 
                  \angle RSQ = 65^0-28^0 = 37^0

Now, in \Delta PQS,
\angleP + \anglePQS + \anglePSQ = 180^0
 y=180^0-90^0-37^0
y=52^0

Q6 In Fig. 6.44, the side QR of \Delta PQR is produced to a point S. If the bisectors of \angle PQR and \angle PRS meet at point T, then prove that  \angle QTR = \frac{1}{2}   \angleQPR.

           

Answer:


We have, 
\Delta PQR is produced to a point S and  bisectors of \angle PQR and \angle PRS meet at point T,
By exterior angle sum property,
\angle PRS = \angleP + \anglePQR
Now, \frac{1}{2}\angle PRS =\frac{1}{2}\angle P + \frac{1}{2} \angle PQR
\Rightarrow \angle TRS = \frac{1}{2}\angle P + \angle TQR................(i)

Since QT and QR are the bisectors of  \angle PQR and \angle PRS respectively.

Now, in \DeltaQRT,
\angle TRS = \angle T + \angle TQR..............(ii)

From eq (i) and eq (ii),  we get

\frac{1}{2}\angle P= \angle T
\Rightarrow \frac{1}{2}\angle QPR= \angle QTR
Hence proved.

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

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

NCERT solutions for class 9 maths chapter 7 Triangles

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

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

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

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

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!

 

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