Chapter 6 : Lines And Angles - Ncert Solutions for Class 9 Maths CBSE

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Chapter 6 - Lines And Angles Excercise Ex. 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

Chapter 6 - Lines And Angles Excercise Ex. 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

Chapter 6 - Lines And Angles Excercise Ex. 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

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