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NCERT Exemplar Class 9 Maths

Chapter 6- Lines and Angles

Chapter 6

Lines and Angles

Exercise 6.1 (8 Multiple Choice Questions and Answers)

Question: 1

In Fig. 6.1, if AB CD EF, PQ RS, RQD 25 and

CQP 60 , then QRS is equal to:

(a) 85

(b) 135

(c) 145

(d) 110

Chapter 6

Lines and Angles

Solution:

c

Given that,

PQ RS

PQC BRS 60

[Alternate exterior angles and]

and DQR QRA 25 [Alternate interior angles and

DQR 25 ]

QRS QRA ARS QRA 180 B P( RS Q) C 60

[Linear pair]

25 180 60

205 60

145

Hence, option (c) is correct.

Question: 2

If one angle of a triangle is equal to the sum of the other

two angles, then the triangle is:

(a) an isosceles triangle

Chapter 6

Lines and Angles

(b) an obtuse triangle

(c) an equilateral triangle

(d) a right triangle

Solution:

d

Question: 3

An exterior angle of a triangle is and its two interior

opposite angles are equal. Each of these equal angles is:

(a) 1

372

105

(b) 1

522

(c) 1

722

(d) 75

Solution:

b

Given,

Chapter 6

Lines and Angles

Exterior angle = 105

Let one of interior angle be x .

Sum of two opposite interior angles Exterior angle

x x 105

2x 105

x2

105

1

x 522

Hence, each equal angle of triangle is 1

522

Question 4

The angles of a triangle are in the ratio 5 : 3 : 7. The

triangle is:

(a) an acute angled triangle

(b) an obtuse angled triangle

(c) a right triangle

(d) an isosceles triangle

Chapter 6

Lines and Angles

Solution:

a

Given that,

The ratio of angles of a triangle is 5 : 3 : 7.

Let the angles of a triangle be A, B and C .

Let, A 5x B 3x , and C 7x

In ABC, A B C 180

( The sum of the angles of a triangle is180.)

5x 3x 7x 180

1 5x 180

x 12

A 5x 5 12 60

B 3x 3 12 36

and C 7x 7 12 84

Since all the angles are less than 90° ,the triangle is an

acute angled triangle.

Chapter 6

Lines and Angles

Question: 5

If one of the angles of a triangle is 130°, then the angle

formed when the bisectors of the other two angles meet

can be:

(a) 50

(b) 65

(c) 145

(d) 155

Solution:

d

Let the angles of ABC be A B , and C .

In ABC ,

A B C 180

Chapter 6

Lines and Angles

(The sum of the angles of a triangle is180.)

1 1 1 180 A B C 2 2 2 2

(Dividing both the sides by 2)

1 1 180 1 B C A2 2 2 2

1 1 1 1 1 1 B C 90 A B C 90 A2 2 2 2 2 2

Now, OB and OC are the angle bisectors of

ABC and ACB

Hence, 1

OBC B.2

And1

BCO C2

.

In OBC, OBC BCO COB 180

1 1 B + C 180 COB2 2

1 90 A 180 COB

2 (From equation i)

Chapter 6

Lines and Angles

1COB 90 A

2

130COB 90

2

COB 90 65

COB 155

Question: 6

In Fig. 6.2, POQ is a line. The value of x is:

(a) 20

(b) 25

(c) 30

(d) 35

Chapter 6

Lines and Angles

Solution:

a

Given,

POQ is a line segment.

Hence, POQ 180

POA AOB BOQ 180

Putting POA 40 , AOB 4 , x and BOQ 3 x

40 4 3 180 x x

x 7 140

x 7 140

x 20

Question 7

In Fig. 6.3, if OP RS O, PQ 110 and QRS 130 ,

then PQR is equal to:

Chapter 6

Lines and Angles

(a) 40

(b) 50

(c) 60

(d) 70

Solution:

c

producing OP such that it intersects RQ at point x.

Now, OP RS and RX is transversal.

Hence, RXP QRS

[The alternate interior angles are equal.]

RXP 130 … (i) [ QRS 130 ]

Now, RQ is a line segment.

Chapter 6

Lines and Angles

So, PXQ RXP 180

PXQ 180 RXP

PXQ 180 130 [From eqn (i)]

PXQ 50

In PQX, OPQ is an exterior angle.

Hence, OPQ PXQ PQX

[∴ Exterior angle = sum of two opposite interior angles.]

110 50 PQX

PQX 110 50

PQX 60

Question 8

Angles of a triangle are in the ratio 2 : 4 : 3.

The smallest angle of the triangle is

(a) 60

(b) 40

(c) 80

Chapter 6

Lines and Angles

(d) 20

Solution:

b

Given that,

The ratio of angles of a triangle is 2 : 4 : 3.

Let angles of a triangle be A, B and C .

A 2x B 4x , and C 3x

In ABC, A B C 180

[The sum of the angles of a triangle is180]

2x 4x 3x 180

x 9 180

x 20

x A 2 2 20 40

B 4x 4 20 80

and C 3x 3 20 60

Hence, option (b) is correct.

Chapter 6

Lines and Angles

Exercise 6.2

Question 1

For what value of x y in Fig. 6.4 will ABC be a line?

Justify your answer.

Solution:

For ABC to be a line, the sum of the two adjacent angles

must be 180 or ABD and DBC must form a linear pair.

i.e., x y 180 .

Question 2

Can a triangle have all the angles less than60? Give

reason for your answer.

Chapter 6

Lines and Angles

Solution:

No, a triangle cannot have all angles less than60 ,

because if all angles are less than60 , then their sum will

be less than 180 (not equal to180).

Hence, it will not be a triangle.

Question: 3

Can a triangle have two obtuse angles? Give reason for

your answer.

Solution:

No, because if a triangle has two obtuse angles i.e., more

than 90 angle, then the sum of all three angles of a

triangle will be greater than 180

(not equal to180).

Hence, it will not be a triangle.

Chapter 6

Lines and Angles

Question: 4

How many triangles can be drawn with angles

measuring as45, 64and 72? Give reason for your

answer.

Solution:

None.

The sum of given angles 45 64 72 181 180 .

Hence, we see that sum of all three angles is not equal to

180. So, we cannot draw a triangle with the given angles.

Question: 5

How many triangles can be drawn with angles as 53,

64and63? Give reason for your answer.

Solution:

The sum of the given angles 53 64 63 180 .

Since, the sum of all interior angles of the triangle is 180,

infinitely many triangles can be drawn.

Chapter 6

Lines and Angles

Question: 6

In Fig. 6.5, find the value of x for which the lines l and m

are parallel.

Solution:

In the given figure,l m .

Using the properties of parallel line (if a transversal

intersects two parallel lines, then the sum of interior

angles on the same side of a transversal is

supplementary), we have:

x 44 180

x 180 44

x 136

Chapter 6

Lines and Angles

Question: 7

Two adjacent angles are equal. Is it necessary that each

of these angles will be a right angle? Justify your

answer.

Solution:

No, because each of these will be a right angle only when

they form a linear pair.

Question: 8

If one of the angles formed by two intersecting lines is a

right angle, what can you say about the other three

angles? Give reason for your answer.

Solution:

Let two intersecting lines be l and m. As one of the angles

is a right angle, then it means that lines l and m are

perpendicular to each other. By using linear pair axiom,

other three angles will be right angles.

Chapter 6

Lines and Angles

Question: 9

In Fig. 6.6, which of the two lines are parallel and why?

Solution:

In case (i),

The sum of two interior angles132 48 180 .

Here, the sum of two interior angles on the same side of

line n is180. Hence, l and m are the parallel lines.

In case (ii)

The sum of two interior angles73 106 179 180 .

Chapter 6

Lines and Angles

Here, the sum of two interior angles on the same side of

the transversal is not equal to180. Hence, lines p and q

are not the parallel lines.

Question: 10

Two lines l and m are perpendicular to the same line n.

Are l and m perpendicular to each other? Give reason for

your answer.

Solution:

No.

Let lines l and m be two lines which are perpendicular to

the line n.

1 2 90 90 180

[ l n andm n ]

Chapter 6

Lines and Angles

Exercise: 6.3

Question: 1

In Fig. 6.9, OD is the bisector of AOC , OE is the

bisector of BOC andOD OE . Show that the points A,

O and B are collinear.

Solution:

Given:

In figure, OD OE and OD is the bisector of AOC and OE

is the bisector of BOC .

To prove:

Points A, O and B are collinear i.e., AOB is a straight line.

Proof:

OD and OE bisect angles AOC and BOC respectively.

Chapter 6

Lines and Angles

AOC = 2 DOC …….. (i) and

BOC = 2 COE …….. (ii)

On adding equations (i) and (ii), we get

AOC + COB = 2 DOC + 2 COE

AOC + COB = 2( DOC + COE)

AOC + COB = 2 DOE

AOC + COB = 2 90 OD OE][

AOC + COB =180

AOB =180

So, AOC and COB are linear pairs or AOB is a straight

line.

Hence, points A, O and B are collinear.

Question: 2

In Fig. 6.10, 1 = 60 and 6 = 120 .Show that the lines m

and n are parallel.

Chapter 6

Lines and Angles

Solution:

Given:

In figure, 1 = 60 and 6 = 120

To prove:

m n .

Proof:

Since, 1 = 60 and 6 = 120

Here, 1 = 3 [Vertically opposite angles]

3 = 1 = 60

Now, 3 + 6 = 60 120

3 + 6 = 180

Chapter 6

Lines and Angles

The sum of two interior angles on the same side of the

transversal is180.

Hence, the lines are parallel orm n .

Question: 3

AP and BQ are the bisectors of the two alternate interior

angles formed by the intersection of a transversal t with

parallel lines l and m (Fig. 6.11). Show thatAP BQ .

Chapter 6

Lines and Angles

Solution:

Given:

In figure, l m , AP and BQ are the bisectors of

EAB and ABH respectively.

To prove:

AP BQ

Proof:

Since, l mand t is transversal,

Chapter 6

Lines and Angles

therefore, EAB = ABH [Alternate interior angles]

EAB = ABH1 1

2 2 [Dividing both sides by 2]

PAB = ABQ

[AP and BQ are the bisectors of EAB and ABH ]

Since, PAB and ABQ are alternate interior angles

formed when transversal AB intersects lines AP and BQ,

hence,AP BQ .

Question: 4

If in Fig. 6.11, bisectors AP and BQ of the alternate

interior angles are parallel, then show that l m .

Solution:

Given:

Chapter 6

Lines and Angles

In figure, AP BQ , AP and BQ are the bisectors of

alternate interior angles EAB and ABH .

To prove:

l m

Proof:

Since, AP BQand t is the transversal, therefore

PAB = ABQ [Alternate interior angles].

2 PAB = 2 ABQ [Multiplying both sides by 2]

EAB = ABH

As the alternate interior angles EAB and ABH are

equal, the lines l and m are parallel. [If two alternate

interior angles are equal, then the lines are parallel.]

Chapter 6

Lines and Angles

Question: 5

In Fig. 6.12, BA ED andBC EF . Show that

ABC = DEF

[Hint: Produce DE to intersect BC at P (say)].

Solution:

Given:

BA ED , andBC EF .

To prove: ABC = DEF

Construction:

Produce DE which intersects BC at P.

Chapter 6

Lines and Angles

Proof:

In figure,

BA ED BA DP

ABP = EPC [Corresponding angles]

ABC = EPC … (i)

Again, BC EF PC EF

EPC = DEF … (ii)

[Corresponding angles]

From equations (i) and (ii), we get

ABC= DEF

Hence, it is proved.

Chapter 6

Lines and Angles

Question: 6

In Fig. 6.13, BA ED and BC EF .

Show that ABC+ DEF = 180 .

Solution:

Given:

In the figure,BA ED andBC EF .

To prove:

ABC+ DEF = 180 .

Construction:

Draw a ray PE opposite to ray EF.

Chapter 6

Lines and Angles

Proof:

In figure, BC EFor BC PF

EPB+ PBC = 180 ... (i)

[The sum of the interior angles on the same side of the

transversal is 180°.]

Now, AB ED and PE is a transversal line.

EPB= DEF ... (ii) [Corresponding angles]


DEF+ PBC = 180

ABC+ DEF = 180 [ PBC= ABC ]


Chapter 6

Lines and Angles

Question: 7

In Fig. 6.14, DE QR and AP and BP are the bisectors of

EAB and RBA , respectively. Find APB .

Solution:

Given:

DE QR , and AP and PB are the bisectors of EAB and

RBA , respectively.

To find:

APB

Proof:

As DE QR and AB is a transversal, the sum of the interior

angles on the same side of the transversal is

supplementary.

Chapter 6

Lines and Angles

EAB+ RBA = 180

EAB + RBA =1 1 180

2 2 2

[Dividing both sides by 2]

EAB + RBA = 901 1

2 2

… (i)

Since AP and BP are the bisectors of EAB and RBA

respectively,

BAP = EAB1

2 … (ii)

and ABP = RBA1

2 … (iii)

On adding equations (ii) and (iii), we get

BAP + ABP= EAB + RBA1 1

2 2

Using equation (i), we get

BAP + ABP = 90 … (iv)

Chapter 6

Lines and Angles

In ABP ,

BAP + ABP + APB = 180

[The sum of the interior angles in a triangle is180.]

90 + APB = 180

[From equation (iv)]

APB 180 90 = 90

Hence, APB 90

Question: 8

The angles of a triangle are in the ratio 2 : 3 : 4. Find the

angles of the triangle.

Solution:

Given,

The ratio of the angles of a triangle is 2 : 3 : 4.

Let the angles of a triangle be A, B and C .

A 2x B 3x , and C 4x

In ABC, A B C 180

Chapter 6

Lines and Angles

[The sum of the angles of a triangle is180 ]

2x 3x 4x = 180

9x 180

x 20

A 2x 2 20 40

B 3x 3 20 60

and C 4x 4 20 80

Question: 9

A triangle ABC is right angled at A. L is a point on BC

such thatAL BC . Prove that BAL ACB .

Solution:

Given:

In ABC, A 90 and AL BC

To prove:

BAL ACB

Chapter 6

Lines and Angles

Proof:

In ABC and LAC ,

BAC ALC … (i)

and ABC ABL ..… (ii) [Common angle]

Adding equations (i) and (ii), we get

BAC ABC ALC + ABL ..… (iii)

In ABC, BAC ACB ABC 180

[The sum of the angles of a triangle is180 .]

BAC ABC 180 ACB ..… (iv)

In ABL, ABL ALB BAL 180

[The sum of the angles of a triangle is180 ]

ABL ALC 180 BAL ..… (v)

Chapter 6

Lines and Angles

[ ALC ALB 90 ]

Substituting the value from equation (iv) and (v) in

equation (iii), we get

180 ACB 180 BAL

ACB BAL


Question: 10

Two lines are respectively perpendicular to two parallel

lines. Show that they are parallel to each other.

Solution:

Given:

Two lines m and n are parallel and another two lines p

and q are perpendicular to m and n respectively.

Or p m and p n, q m and q n

To prove:

Chapter 6

Lines and Angles

p q

Proof:

Since,m n and p is perpendicular to m and n.

So,

p is perpendicular to m …(i)

p is perpendicular to n …(ii)

Since, m ∥ n and q is perpendicular to m and n.

So,

q is perpendicular to m …(iii)

q is perpendicular to n …(iv)

From the equations (i) and (iii) [Or from (ii) and (iv)], we

get,

p q .

Chapter 6

Lines and Angles

[If two lines are perpendicular to the same line, the lines

are parallel to each other.]

Hence, it isp q .

Chapter 6

Lines and Angles

Exercise: 6.4

Question: 1

If two lines intersect, prove that the vertically opposite

angles are equal.

Solution:

Given:

Two lines AB and CD intersect at point O.

To prove:

(i) AOC BOD

(ii) AOD BOC

Proof:

(i) Since, ray OA stands on line CD,

Chapter 6

Lines and Angles

AOC AOD 180 … (i)

[Linear pair axiom]

Similarly, ray OD stands on line AB.

AOD BOD 180 … (ii)


AOC AOD AOD BOD

AOC BOD

(ii) Since, ray OD stands on line AB,

AOD BOD 180 … (iii)

[Linear pair axiom]

Similarly, ray OB stands on line CD.

DOB BOC 180 … (iv)

From equations (iii) and (iv), we get

AOD BOD DOB BOC

AOD BOC

Chapter 6

Lines and Angles

Question: 2

Bisectors of interior B and exterior ACD of

ABC intersect at the point T.

Prove that:

1BTC BAC

2 .

Solution:

Given:

In ABC , the bisectors of ABC and ACD meet at point

T.

Construction:

Produce BC to D.

To Prove:

1BTC BAC

2

Chapter 6

Lines and Angles

Proof:

In ABC, ACD is an exterior angle.

ACD = ABC + CAB

[The exterior angle of a triangle is equal to the sum of two

opposite interior angles.]

1 1 1ACD CAB + ABC

2 2 2


1 1TCD CAB + ABC

2 2 … (i)

[∵ CT is a bisector of1

ACD ACD TCD2

]

In BTC ,

TCD BTC + CBT

[The exterior angle of a triangle is equal to the sum of two

interior opposite angles.]

1TCD BTC + ABC

2 … (ii)

[∵ BT bisector of 1

ABC CBT ABC2

]

Chapter 6

Lines and Angles


1 1 1CAB + ABC BTC ABC

2 2 2

1CAB BTC

2

or 1

BAC BTC2


Question: 3

A transversal intersects two parallel lines. Prove that

the bisectors of any pair of corresponding angles so

formed are parallel.

Solution:

Given:

Two lines DE and QR are parallel. A transversal intersects

the lines DE and QR at A and B respectively. Also, BP and

AF are the bisectors of angles ABR and CAE

respectively.

To prove:

Chapter 6

Lines and Angles

BP AF

Proof:

Given, DE QR

CAE ABR

[Corresponding angles]

1 1CAE ABR

2 2


CAF ABP

[∵ BP and AF are the bisectors of angles ABR and

CAE respectively]

This implies that AF is parallel to BP as the corresponding

angles CAF and ABP are equal.

Hence, AF BP .

Chapter 6

Lines and Angles

Question: 4

Prove that through a given point, we can draw only one

perpendicular on a given line.

[Hint: Use proof by contradiction].

Solution:

Given:

Consider a line l and a point P.

To prove:

Only one perpendicular can be drawn from P to l.

Construction:

Suppose that two lines PA and PB are passing through

the point P and they are perpendicular to l.

Proof:

Chapter 6

Lines and Angles

In APB ,

PAB P PBA 180

[The angle sum property of a triangle]

90 P 90 180 (since PA is perpendicular to

l and PB is perpendicular to l)

P 180 180

P 0

(Which is possible only when the lines PA and PB

coincide)

Hence, only one perpendicular line can be drawn through

a given point.

Question: 5

Prove that two lines that are respectively perpendicular

to two intersecting lines intersect each other.

[Hint: Use proof by contradiction].

Solution:

Given:

Chapter 6

Lines and Angles

Let lines l and m be two intersecting lines. Again, let n

and p be another two lines which are perpendicular to the

intersecting lines.

To prove:

Two lines n and p intersect at a point.

Proof:

Let us consider lines n and p are not intersecting,

and then it means they are parallel to each other

i.e., n p … (i)

Since, lines n and p are perpendicular to l and m

respectively, but from equation (i) n p , it is implied that

l m , it is a contradiction.

Thus, our assumption is wrong.

Hence, lines n and p intersect at a point.

Question: 6

Prove that a triangle must have at least two acute

angles.

Chapter 6

Lines and Angles

Solution:

Given:

ABC is a triangle

To prove:

ABC must have two acute angles.

Proof:

Let us consider the following cases:

Case I:

When two angles are90 :

Suppose two angles are B 90 and C 90 .

The sum of all three angles of a triangle is180 .

A B C 180

A 90 90 180

A 180 180 0 , which is not possible.

Hence, this case is rejected.

Case II:

When two angle are obtuse:

Chapter 6

Lines and Angles

Suppose two angles B and C are more than90 ,

the sum of all three angles of a triangle is180 .

A B C 180

A 180 ( B C) 180 − (an angle greater than180 )

A = negative angle, which is not possible.

Hence, this case is also rejected.

Case III:

When one angle is 90 and the other is obtuse:

Suppose B 90 and C is obtuse.

The sum of all three angles of a triangle is180 .

A B C 180

A 180 (90 C)

90 C

= negative angle, which is not possible.

Hence, this case is also rejected.

Case IV:

Chapter 6

Lines and Angles

When two angles are acute, then the sum of two angles is

less than 180 so that the third angle is also acute.

Hence, a triangle must have at least two acute angles.

Question: 7

In Fig. 6.17, Q R, PA is the bisector of QPR and

PM QR . Prove that

1

APM Q R2

.

Solution:

Given:

Q R, PA is the bisector of QPR and PM QR .

To prove:

1

APM Q R2

.

Chapter 6

Lines and Angles

Proof:

Since, PA is the bisector of QPR

QPA APR … (i)

In PQM, Q PMQ QPM 180

[𝐴ngle sum property of triangles]

Q 90 QPM 180

[ PMQ 90 ]

Q 90 QPM … (ii)

In PMR, PMR R RPM 180

[Angle sum property of triangles]

90 R RPM 180

[ PMQ 90 ]

R 180 90 RPM

R 90 RPM … (iii)

Subtracting equations (iii) from (ii), we get

Q R (90 QPM) (90 RPM)

Q R RPM QPM

Chapter 6

Lines and Angles

Q R ( RPA APM) ( QPA APM) … (iv)

Q R QPA APM QPA APM

[𝑈sing equation (i)]

Q R 2 APM

1APM ( Q R)

2

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