NCERT Class 9 Maths Solutions Chapter - 6 Lines And Angles, Ex - 6.1

Ex - 6.1

Question 1.  In the given figure, lines AB and CD intersect at O. If AOC+BOE=70�  and BOD=40�   find BOE and reflex COE.


Solution



Question 2.  In the given figure, lines XY and MN intersect at O. If POY = 90^o  and  a:b = 2 : 3, find c.



Solution

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�

Question 3.  In the given figure, PQR = PRQ, then prove that PQS = PRT.   

Solution

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

Question 4.  In the given figure, if  x + y = w + z then prove that AOB is a line.  

Solution

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.

Question 5.  In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

                    

         

Solution

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

Question 6.  It is given that XYZ = 64� and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ZYP, find XYQ and reflex QYP.

Solution

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 = 360^o - 58^o = 302^o

    XYQ = XYZ + ZYQ

         = 64^o + 58^o = 122^o

^Ex-6.2