Lesson: Lines and Angles

Exercise 6.1

Question: 1

In Fig. 6.13, lines AB and CD intersect at O.

AOC BOE 70   



and

BOD 40 , 



find

BOE

and reflex

COE

Solution

Given,

AOC BOE 70   



and

BOD 40 



AOC, BOE 

and

COE

form a straight line.

Hence,

AOC BOE COE 180     



70 COE 180   

 

COE 110  



Also,

COE BOD BOE 180   



(They form a straight line.)

110 40  BOE 180   

  

150 BOE 180   

 

BOE 30  



Reflex of

COE 360 COE 360 110 250      

   

Question: 2

In Fig. 6.14, lines XY and MN intersect at O.

POY 90 



and a : b = 2 : 3, find c.

Solution

Given,

POY 90 



and a : b = 2 : 3

POY

a, and b forms a straight line.

Hence,

POY a b 180   



90 180   

 

a b

90  



a b

As a : b = 2 : 3 so a = 2x and b = 3x

2 3 90 



x x

5 90 



18 



2 18 36   

 

and

3 18 54  

 

also,

180 



b c

(linear pair)

54 180  

 

126 



Question: 3

In Fig. 6.15,

PQR PRQ  

and ST is a straight line.

Prove that

PQS PRT  

Solution

Given,

PQR PRQ  

To prove,

PQS PRT  

PQR

and

PQS

form a straight line.

Hence,

PQR PQS 180  



 PQS 180 PQR   



… (i)

also,

PRQ PRT 180   



(Linear Pair)

PRT 180 PRQ    



PRT 180 PQR    



… (ii)

PQR RQ( )P  

From (i) and (ii)

PQS PRT  

Therefore,

PQS PRT  

Question: 4

In Fig. 6.16, if

,x y w z  

then prove that AOB is

a line.

Solution

Given,

x y w z  

To prove,

AOB is a line or

180 



x y

(linear pair)

x, y, w and z are angles around a point.

Hence,

 360   



x y w z

( ) ( ) 360    



x y w z

( ) ( ) 360    



x y x y

( )  Given x y w z

( )2 360  



x y

( ) 180  



x y

Hence, x and y make a linear pair.

Therefore, AOB is a straight line.

Question: 5

In Fig. 6.17, POQ is a line. Ray OR is perpendicular to

line PQ. OS is another ray lying between rays OP and

OR. Prove that

(

ROS  QOS POS

)    

Solution

Given,

OR is perpendicular to line PQ.

To prove,

(

ROS  QOS POS

)    

Proof:

POR ROQ 90   



(Perpendicular)

QOS ROQ  ROS 90 ROS      



… (i)

POS POR ROS 90 ROS       



… (ii)

Subtracting (ii) from (i),

QOS POS 90 ROS 90 OS( )R        

 

QOS POS 90 ROS 90 ROS         

 

QOS POS 2 ROS     

ROS  QOS POS

( )     

Hence, it is proved.

Question: 6

It is given that

XYZ = 64



and XY is produced on

point P. Draw a figure from the given information. If

ray YQ bisects

ZYP

, find

XYQ

and reflex

QYP

Solution

Given,

XYZ = 64



YQ bisects

ZYP

XYZ ZYP 180   



(linear pair)

64 ZYP 180   

 

 116   ZYP

also,

ZYP ZYQ QYP   

ZYQ QYP  

(YQ bisects

ZYP

)

ZYP 2 ZYQ   

2 ZYQ 116  



ZYQ 58 QYP    



Now,

XYQ XYZ ZYQ   

XYQ 64 58   

 

XYQ 122  



also,

reflex

QYP 180 XYQ   



QYP 180  122  

 

QYP 302  



Exercise 6.2

Question: 1

In Fig. 6.28, find the values of x and y and then show

that

AB CD

Solution

50 180 

 

(Linear pair)

130 



also,

130



(Vertically opposite)

Now,

130 



x y

(Alternate interior angles)

Alternate interior angles are equal.

Therefore,

AB CD

Question: 2

In Fig. 6.29, if

AB CD,CD  EF

and y : z = 3 : 7, find x.

Solution

Given,

AB CD

and

CD EF

y : z = 3 : 7

Now,

 180 



x y

(The angles on the same side of

transversal are supplementary as

AB CD

also,

GOD  z

(Corresponding angles are equal as

CD EF

)

and,

 GOD 180 



(Linear pair)

180  



y z

As y : z = 3 : 7, we get

y = 3w and z = 7w

3 7 180 



w w

10 180 



18 



3 18 54   

 

and,

7 18 126  

 

Now,

 180 



x y

54 180  

 

126 



Question: 3

In Fig. 6.30, if

AB CD, EF CD 

and

GED 126 



find

AGE, GEF 

and

FGE

Solution

Given,

AB CD

EF CD

GED 126 



FED 90 





 

EF CD

Now,

AGE GED  

(Since,

AB CD

and GE is

transversal, the alternate interior angles are equal.)

AGE 126  





Also,

GEF GED FED    

GEF 126 90   

 

GEF 36  



Now,

FGE AGE 180  



(linear pair)

FGE 180 126   

 

FGE 54  



Question: 4

In Fig. 6.31, if

PQ ST, PQR 110 





and

RST 130 , 



find

QRS

[Hint: Draw a line parallel to ST through point R.]

Solution

Given,

PQ ST, PQR 110 



 

and

RST 130 



Construction,

A line XY parallel to PQ and ST is drawn.

PQR QRX 180  



(The angles on the same side of

the transversal are supplementary.)

110 QRX 180   

 

QRX 70  



Also,

RST SRY 180   



(The angles on the same side of

the transversal are supplementary.)

130 SRY 180   

 

SRY 50  



Now,

QRX SRY  QRS 180   



70  50  QRS 180   

  

QRS 60  



Question: 5

In Fig. 6.32, if

AB CD, APQ 50 





and

PRD 127 



, find x and y.

Solution

Given,

AB CD, APQ 50 





and

PRD 127 



50



(The alternate interior angles are equal when

AB CD

PRD RPB 180  



(The angles on the same side of

the transversal are supplementary.)

127 RPB 180   

 

RPB 53  



Now,

 50 RPB 180   

 

(AB is a straight line.)

50 53 180   

  

103 180  

 

77 



Exercise: 6.3

Question: 1

In Fig. 6.39, sides QP and RQ of

PQR

are produced

on points S and T respectively. If

SPR=135



and

PQT =110



, find

PQR

Solution

Given,

SPR 135 



and

PQT 110 



SPR QPR 180  



(SQ is a straight line.)

135 QPR 180   

 

QPR 45  



also,

PQT PQR 180  



(TR is a straight line.)

110 PQR 180   

 

PQR 70  



Now,

PQR QPR PRQ 180   



(The sum of the interior angles of a triangle is 180

70 45 PRQ 180    

  

115 PRQ 180   

 

 PRQ 65  



Question: 2

In Fig. 6.40,

X 62 , XYZ 54   

 

. If YO and ZO are

the bisectors of

 XYZ

and

 XZY

respectively of

 XYZ

,find

 OZY

and

 YOZ

Solution

Given,

X 62 , XYZ 54   

 

YO and ZO are the bisectors of

XYZ

and

XZY

respectively.

X XYZ  XZY 180    



(The sum of the interior

angles of a triangle is

180



62  54 XZY 180    

  

116 XZY 180   

 

XZY 64  



Now,

OZY XZY

  

(ZO is the bisector.)

OZY 32  



also,

OYZ XYZ

  

(YO is the bisector.)

OYZ 27  



Now,

OZY OYZ  O 180   



(The sum of the interior

angles of the triangle)

32  27 180    

  

59 180   

 

121  



Question: 3

In Fig. 6.41, if

AB DE, BAC 35 





and

CDE 53 , 



find

DCE.

Solution

Given,

AB DE, BAC 35 





and

CDE 53 



BAC CED  

(The alternate interior angles are

equal when two lines are parallel.)

 CED 35  



Now,

DCE CED CDE 180   



(The sum of the interior angles of a triangle is

180



DCE 35 53 180    

  

DCE 88 180   

 

DCE 92  



Question: 4

In Fig. 6.42, if lines PQ and RS intersect at point T,

such that

PRT 40 , RPT 95   

 

and

TSQ 75 , 



find

SQT

Solution

Given,

PRT 40 , RPT 95   

 

and

TSQ 75 , 



in triangle PRT

PRT RPT PTR 180   



(The sum of the interior

angles of the triangle is

180



40 95  PTR 180   

  

40  95 PTR 180    

  

135  PTR 180  

 

PTR 45  



PTR STQ 45   



(Vertically opposite angles are

equal.)

Now, in triangle SQT,

TSQ STQ  SQT 180    



(The sum of the interior

angles of the triangle is

180



75  45 SQT 180    

  

120 SQT 180   

 

SQT 60  



Question: 5

In Fig. 6.43, if

PQ PS, PQ SR, SQR 28 



  

and

QRT 65 



Find the values of x and y.

Solution

Given,

PQ PS, PQ RS, SQR 28  



 

and

QRT 65 , 



PQR QRT  

(The alternate angles are equal as

QR is a transversal intersecting the parallel lines PQ

and SR.)

   x SQR QRT

28 65  

 

37 



In triangle PQS,

PQS PSQ QPS 180     



(The sum of the

angles of a triangle is

180



 

37 90 180 PQ PS    

  

127 180  

 

53 



Question: 6

In Fig. 6.44, the side QR of

PQR

is produced on a

point S. If the bisectors of

PQR

and

PRS

meet at

point T, then prove that

QTR QPR

  

Solution

Given,

bisectors of

PQR

and

PRS

meet at point T.

To prove:

QTR QPR

  

Proof:

TRS TQR QTR   

(The exterior angle of a

triangle is equal to the sum of the two interior opposite

angles.)

QTR TRS TQR     

… (i)

also,

 SRP QPR PQR   

(The exterior angle of a

triangle is equal to the sum of the two interior opposite

angles.)

2 TRS QPR 2 TQR     

(TR and TQ are

bisectors of

SRP

and

PQR

respectively)

QPR 2 TRS 2 TQR     

QPR  TRS TQR

     

… (ii)

Equating (i) and (ii),

QTR QPR

  

Hence, it is proved.