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NCERT Solution for Class 9 Mathematics Chapter 6 - Lines And Angles

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NCERT Textbook Solutions are considered extremely helpful when preparing for your CBSE Class 9 Mathematics exams. TopperLearning study resources infuse profound knowledge, and our Textbook Solutions compiled by our subject experts are no different. Here you will find all the answers to the NCERT textbook questions of Chapter 6 - Lines And Angles.

All our solutions for Chapter 6 - Lines And Angles are prepared considering the latest CBSE syllabus, and they are amended from time to time. Our free NCERT Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

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NCERT Solution for Class 9 Mathematics Chapter 6 - Lines And Angles Page/Excercise 6.1

Solution 1

Solution 2

Let common ratio between a and b is x,  a = 2x and b = 3x.

XY is a straight line, OM and OP rays stands on it.

XOM + MOP + POY = 180    b + a + POY = 180

3x + 2x + 90 = 180

               5x  = 90

                 x = 18 

a = 2x

   = 2 * 18

   = 36

b = 3x

   = 3 * 18

   = 54

 

Now, MN is a straight line. OX ray stands on it. 

 b + c = 180

54 + c = 180

c = 180 54   = 126 

          c = 126

Solution 3

In the given figure, ST is a straight line and QP ray stand on it.
     PQS + PQR = 180            (Linear Pair)     PQR = 180 - PQS             (1)
    PRT + PRQ = 180            (Linear Pair)
    PRQ = 180 - PRT            (2)
    Given that PQR = PRQ. Now, equating equations (1) and (2), we have
    180 - PQS = 180  - PRT
                      PQS = PRT

Solution 4

We may observe that
    x + y + z + w = 360                (Complete angle)
    It is given that
x + y = z + w
     x + y + x + y = 360
    2(x + y) = 360
    x + y = 180
    Since x and y form a linear pair, thus AOB is a line.

Solution 5

Given that OR PQ     POR = 90           POS  + SOR = 90   ROS = 90 - POS                ... (1)       QOR = 90                     (As OR PQ)       QOS - ROS = 90       ROS = QOS - 90             ... (2)       On adding equations (1) and (2), we have       2 ROS = QOS - POS

Solution 6

                                           Given that line YQ bisects PYZ.
 Hence, QYP = ZYQ
 Now we may observe that PX is a line. YQ and YZ rays stand on it. XYZ + ZYQ + QYP = 180            64 + 2QYP = 180        2QYP = 180 - 64 = 116        QYP = 58       Also, ZYQ = QYP = 58       Reflex QYP = 360o - 58o = 302o       XYQ = XYZ + ZYQ            = 64o + 58o = 122o

NCERT Solution for Class 9 Mathematics Chapter 6 - Lines And Angles Page/Excercise 6.2

Solution 1

We may observe that
50 + x = 180                   (Linear pair)
x = 130             ... (1)
Also, y = 130                    (vertically opposite angles)
As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, so line AB || CD

Solution 2

  Given that AB || CD and CD || EF      AB || CD || EF    (Lines parallel to a same line are parallel to each other)      Now we may observe that
   x = z             (alternate interior angles)    ... (1)
   Given that y: z = 3: 7
   Let common ratio between y and z be a
    y = 3a and z = 7a
    Also x + y = 180     (co-interior angles on the same side of the transversal)     z + y = 180             [Using equation (1)]     7a + 3a = 180     10a = 180         a = 18
    x = 7 a = 7  18 = 126

Solution 3

  It is given that
AB || CD                         EF    CD GED = 126
GEF + FED = 126 GEF + 90 = 126    
GEF = 36
Now, AGE and GED are alternate interior angles AGE = GED = 126     But AGE +FGE = 180      (linear pair) 126 + FGE = 180 FGE = 180 - 126 = 54 AGE = 126, GEF = 36, FGE = 54

Solution 4

                  Let us draw a line XY parallel to ST and passing through point R.
PQR + QRX = 180     (co-interior angles on the same side of transversal QR)
110 + QRX = 180
QRX = 70
Now,
RST +SRY = 180    (co-interior angles on the same side of transversal SR)
130 + SRY = 180
SRY = 50
XY is a straight line. RQ and RS stand on it.
QRX + QRS + SRY = 180     
70 + QRS + 50 = 180
QRS = 180 - 120 = 60

Solution 5

APR = PRD                 (alternate interior angles)
50 + y = 127
           y = 127 - 50
           y = 77
Also APQ = PQR         (alternate interior angles)
             50 = x x = 50 and y = 77

Solution 6

                                    Let us draw BM   PQ and CN  RS.
 As PQ || RS
So, BM || CN

 Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.   2 = 3                               (alternate interior angles)   But 1 = 2 and 3 = 4      (By laws of reflection) 1 = 2 = 3 = 4 Now, 1 + 2 = 3 + 4 ABC = DCB   But, these are alternate interior angles   AB || CD

NCERT Solution for Class 9 Mathematics Chapter 6 - Lines And Angles Page/Excercise 6.3

Solution 1

Given that
    SPR = 135 and PQT = 110
    Now, SPR + QPR = 180             (linear pair angles)
     135 + QPR = 180
     QPR = 45                         
    Also, PQT + PQR = 180             (linear pair angles)
     110 + PQR = 180
    PQR = 70     
    As we know that sum of all interior angles of a triangle is 180, so, for PQR  
QPR + PQR + PRQ = 180
45 + 70 + PRQ = 180
PRQ = 180 - 115
PRQ = 65

Solution 2

As we know that sum of all interior angles of a triangle is 180, so for XYZ X + XYZ + XZY = 180    
62 + 54 + XZY = 180
XZY = 180 - 116
XZY = 64 OZY =   = 32         (OZ is angle bisector of XZY)
Similarly, OYZ =  = 27
Using angle sum property for OYZ, we have
OYZ + YOZ + OZY = 180º
27 + YOZ + 32 = 180
YOZ = 180 - 59
YOZ = 121

Solution 3

AB || DE and AE is a transversal                    
BAC =CED                 (alternate interior angle)
CED = 35
In CDE,
CDE + CED + DCE = 180         (angle sum properly of a triangle)
53 + 35 + DCE = 180
DCE = 180 - 88
DCE = 92

Solution 4

Using angle sum property for PRT, we have
PRT + RPT + PTR = 180
40 + 95 + PTR = 180
PTR = 180 - 135
PTR = 45
STQ = PTR = 45             (vertically opposite angles)
STQ = 45
By using angle sum property for STQ, we have
STQ + SQT + QST = 180
45 + SQT + 75 = 180
SQT = 180 - 120
SQT = 60  

Solution 5

Given that PQ || SR and QR is a transversal line
PQR = QRT             (alternate interior angles)
x + 28 = 65
x = 65 - 28
x = 37
By using angle sum property for SPQ, we have
SPQ + x + y = 180
90 + 37 + y = 180
y = 180 - 127
y = 53
 x = 37 and y = 53.

Solution 6

In QTR, TRS is an exterior angle.
 QTR + TQR = TRS
QTR = TRS - TQR        (1)
For PQR, PRS is external angle QPR + PQR = PRS QPR + 2TQR = 2TRS    (As QT and RT are angle bisectors)
QPR = 2(TRS - TQR)
QPR = 2QTR            [By using equation (1)]
QTR =  QPR

TopperLearning provides step-by-step solutions for each question in each chapter in the NCERT textbook. Access Chapter 6 - Lines And Angles here for free.

Our NCERT Solutions for Class 9 Mathematics are by our subject matter experts. These NCERT Textbook Solutions will help you to revise the whole chapter, and you can increase your knowledge of Mathematics. If you would like to know more, please get in touch with our counsellor today!

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