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

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.

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)

54 180

c

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

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

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.