NCERT Class 9 Maths Solutions Chapter - 6 Lines And Angles, Ex - 6.2

Ex - 6.2

Question 1.  In the given figure, find the values of x and y and then show that



Solution

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

Question 2.  In the given figure, if AB || CD, CD || EF and y: z = 3: 7, find x.


Solution

  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�

Question 3.  In the given figure, If AB || CD, EF  CD and GED = 126�, find AGE, GEF and FGE.

Solution

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�

Question 4.  In the given figure, if PQ || ST, PQR = 110� and RST = 130�, find QRS.

Solution

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�

Question 5.  In the given figure, if AB || CD, APQ = 50� and PRD = 127�, find x and y.

Solution

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�

Question 6.  In the given figure, 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.

Solution

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

Ex-6.3