NCERT Solution for Class 9 Mathematics Chapter 6 - Lines And Angles Page RR ACADEMY MEERUT

NCERT Solution for Class 9 Mathematics Chapter 6 - Lines And Angles Page/Excercise 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 1

Question 2

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

                        

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�

Question 3

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

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

Question 4

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

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.

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 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 

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 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

Question 1

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

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 

Question 2

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

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�

Question 3

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

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�

Question 4

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

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�

Question 5

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

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�

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 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

Question 1

In the given figure, sides QP and RQ of PQR are produced to points S and T respectively. If SPR = 135� and PQT = 110�, find PRQ.                                           

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�

Question 2

In the given figure, X = 62�, XYZ = 54�. If YO and ZO are the bisectors of XYZ and XZY respectively of XYZ, find OZY and YOZ                                     

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�

Question 3

In the given figure, if AB || DE, BAC = 35� and CDE = 53�, find DCE.                                            

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�

Question 4

In the given figure, if lines PQ and RS intersect at point T, such that PRT = 40�, RPT = 95� and TSQ = 75�, find SQT.                                           

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�  

Question 5

In the given figure, if PQ  PS, PQ || SR, SQR = 28� and QRT = 65�, then find the values of x and y.                                       

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�.

Question 6

In the given figure, the side QR of PQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR= QPR. 

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