Lesson: Lines and Angles
Exercise 6.1
Question: 1
In Fig. 6.13, lines AB and CD intersect at O.
If
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
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.
If
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
x
18
x
2 18 36
a
and
3 18 54
b
also,
180
b c
(linear pair)
126
c
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
(
1
ROS QOS POS
2
)
.
Solution
Given,
OR is perpendicular to line PQ.
To prove,
(
1
ROS QOS POS
2
)
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
1
ROS QOS POS
2
( )
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
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
x
(Linear pair)
130
x
also,
130
y
(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
y
(Linear pair)
180
y z
As y : z = 3 : 7, we get
y = 3w and z = 7w
3 7 180
w w
10 180
w
18
w
3 18 54
y
and,
7 18 126
z
Now,
180
x y
54 180
x
126
x
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
x
(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
y
(AB is a straight line.)
50 53 180
y
103 180
y
77
y
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
o
.)
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,
1
OZY XZY
2
(ZO is the bisector.)
OZY 32
also,
1
OYZ XYZ
2
(YO is the bisector.)
OYZ 27
Now,
OZY OYZ O 180 
(The sum of the interior
angles of the triangle)
32 27 180
O
59 180
O
121
O
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
x
37
x
In triangle PQS,
PQS PSQ QPS 180
(The sum of the
angles of a triangle is
180
.)
37 90 180 PQ PS
y
127 180
y
53
y
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
1
QTR QPR
2
.
Solution
Given,
bisectors of
PQR
and
PRS
meet at point T.
To prove:
1
QTR QPR
2
.
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
1
QPR TRS TQR
2
(ii)
Equating (i) and (ii),
1
QTR QPR
2
.
Hence, it is proved.