# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 7 – Lines And Angles

## Chapter 7 – Lines And Angles Exercise Ex. 7A

Define the following terms:

(i) Angle (ii) Interior of an angle (iii) Obtuse angle (iv) Reflex angle (v) Complementary angles (vi) Supplementary angles

(i) Angle: Two rays having a common end point form an angle.

(ii) Interior of an angle: The interior of AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.

(iii) Obtuse angle: An angle whose measure is more than 90^{o} but less than 180^{o}, is called an obtuse angle.

(iv) Reflex angle: An angle whose measure is more than 180^{o} but less than 360^{o} is called a reflex angle.

(v) Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90^{o}.

(vi) Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180^{o}.

Find

the angle whose complement is one-third of its supplement.

Let

the required angle be x^{o}

Then,

its complement is 90^{o} – x^{o} and its supplement is 180^{o}

– x^{o}

_{}

_{} The required angle is 45^{o}.

Two complementary angles are in the ratio 4: 5. Find the angles.

Let the two required angles be x^{o} and 90^{o} – x^{o}.

Then _{}

_{} 5x = 4(90 – x)

_{} 5x = 360 – 4x

_{} 5x + 4x = 360

_{} 9x = 360

_{} _{}

Thus, the required angles are 40^{o} and 90^{o} – x^{o} = 90^{ o} – 40^{o} = 50^{o}.

Find the value of x for which the

angles (2x – 5)° and (x –

10)°

are the complementary angles.

(2x

– 5)°

and (x – 10)° are complementary angles.

∴ (2x

– 5)°

+ (x – 10)° = 90°

⇒

2x – 5°

+ x – 10° = 90°

⇒

3x – 15°

= 90°

⇒

3x = 105°

⇒

x = 35°

Find the complement of each of the

following angle:

55°

Complement of 55° = 90° – 55° = 35°

Find the complement of each of the following angles.

16^{o}

Complement of 16^{o} = 90 – 16^{o} = 74^{o}

Find the complement of each of the

following angle:

90°

Complement of 90° = 90° – 90° = 0°

Find the complement of each of the following angles.

46^{o} 30

Complement of 46^{o} 30′ = 90^{o} – 46^{o} 30′ = 43^{o} 30′

Find the supplement of each of the

following angle:

42°

Supplement of 42° = 180° – 42° =

138°

Find the supplement of each of the

following angle:

90°

Supplement of 90° = 180° – 90° =

90°

Find the supplement of each of the

following angle:

124°

Supplement of 124° = 180° – 124° =

56°

Find the supplement of each of the following angles.

75^{o} 36′

Supplement of 75^{o} 36′ = 180^{o} – 75^{o} 36′ = 104^{o} 24′

Find

the measure of an angle which is

(i)

equal to its complement,

(ii) equal to its supplement.

(i)

Let the required angle be x^{o}

Then,

its complement = 90^{o} – x^{o}

_{}

_{} The measure of an angle which is equal to

its complement is 45^{o}.

(ii)

Let the required angle be x^{o}

Then,

its supplement = 180^{o} – x^{o}

_{}

_{} The measure of an angle which is equal to

its supplement is 90^{o}.

Find

the measure of an angle which is 36^{o} more than its complement.

Let

the required angle be x^{o}

Then

its complement is 90^{o} – x^{o}

_{}

_{} The measure of an angle which is 36^{o}

more than its complement is 63^{o}.

Find the measure of an angle which

is 30°

less than its supplement.

Let

the measure of the required angle = x°

Then, measure of its supplement =

(180 – x)°

It is given that

x°

= (180 – x)° – 30°

⇒

x°

= 180°

– x°

– 30°

⇒

2x°

= 150°

⇒

x°

= 75°

Hence, the measure of the required

angle is 75°.

Find

the angle which is four times its complement.

Let

the required angle be x^{o}

Then,

its complement = 90^{o} – x^{o}

_{}

_{} The required angle is 72^{o}.

Find

the angle which is five times its supplement.

Let

the required angle be x^{o}

Then,

its supplement is 180^{o} – x^{o}

_{}

_{} The required angle is 150^{o}.

Find the angle whose supplement is four times its complement.

Let the required angle be x^{o}

Then, its complement is 90^{o} – x^{o} and its supplement is 180^{o} – x^{o}

_{That is we have,}

_{} The required angle is 60^{o}.

## Chapter 7 – Lines And Angles Exercise Ex. 7B

In the given figure, AOB is a straight line. Find the value of x.

Since _{}BOC and _{}COA form a linear pair of angles, we have

_{}BOC + _{}COA = 180^{o}

_{} x^{o }+ 62^{o} = 180^{o}

_{} x = 180 – 62

_{} x = 118^{o}

If

two straight lines intersect each other in such a way that one of the angles

formed measures 90^{o}, show that each of the remaining angles

measures 90^{o}.

Let

two straight lines AB and CD intersect at O and let _{}AOC = 90^{o}.

Now, _{}AOC = _{}BOD

[Vertically opposite angles]

_{} _{}BOD = 90^{o}

Also,

as _{}AOC and _{}AOD form a linear pair.

_{} 90^{o} + _{}AOD = 180^{o}

_{} _{}AOD = 180^{o} – 90^{o} = 90^{o}

Since, _{}BOC = _{}AOD

[Verticallty opposite angles]

_{} _{}BOC = 90^{o}

Thus,

each of the remaining angles is 90^{o}.

Two

lines AB and CD intersect at a point O such that _{}BOC +_{}AOD = 280^{o}, as shown in the figure. Find

all the four angles.

Since,

_{}AOD and _{}BOC are vertically opposite angles.

_{}AOD

= _{}BOC

Now,

_{}AOD + _{}BOC = 280^{o} [Given]

_{} _{}AOD + _{}AOD = 280^{o}

_{} 2_{}AOD = 280^{o}

_{} _{}AOD = _{}

_{} _{}BOC = _{}AOD = 140^{o}

As,

_{}AOC and _{}AOD form a linear pair.

So, _{}AOC + _{}AOD = 180^{o}

_{} _{}AOC + 140^{o} = 180^{o}

_{} _{}AOC = 180^{o} – 140^{o} = 40^{o}

Since,

_{}AOC and _{}BOD are vertically opposite angles.

_{}AOC

= _{}BOD

_{} _{}BOD = 40^{o}

_{} _{}BOC = 140^{o}, _{}AOC = 40^{o} , _{}AOD = 140^{o} and _{}BOD = 40^{o}.

Two lines AB and CD intersect each

other at a point O such that ∠AOC : ∠AOD = 5 : 7. Find all the angles.

Let

∠AOC = 5x and ∠AOD = 7x

Now,

∠AOC + ∠AOD = 180° (linear pair of angles)

⇒ 5x + 7x = 180°

⇒ 12x = 180°

⇒ x = 15°

⇒ ∠AOC = 5x = 5(15°) = 75° and ∠AOD = 7x = 7(15°) = 105°

Now,

∠AOC = ∠BOD (vertically opposite angles)

⇒ ∠BOD = 75°

Also,

∠AOD = ∠BOC (vertically opposite angles)

⇒ ∠BOC = 105°

In the given figure, three lines

AB, CD and EF intersect at a point O such that ∠AOE

= 35°

and ∠BOD

= 40°. Find the measure of ∠AOC, ∠BOF, ∠COF and ∠DOE.

∠BOD = 40°

⇒ AOC = ∠BOD = 40° (vertically opposite angles)

∠AOE = 35°

⇒ ∠BOF = ∠AOE = 35° (vertically opposite angles)

∠AOB is a

straight angle.

⇒ ∠AOB = 180°

⇒ ∠AOE + ∠EOD + ∠BOD = 180°

⇒ 35° + ∠EOD + 40° = 180°

⇒ ∠EOD + 75° = 180°

⇒ ∠EOD = 105°

Now,

∠COF = ∠EOD = 105° (vertically opposite angles)

In the given figure, the two lines

AB and CD intersect at a point O such that ∠BOC

= 125°.

Find the values of x, y and z.

∠AOC + ∠BOC = 180° (linear pair of angles)

⇒ x + 125 = 180°

⇒ x = 55°

Now,

∠AOD = ∠BOC (vertically opposite angles)

⇒ y = 125°

Also,

∠BOD = ∠AOC (vertically opposite angles)

⇒ z = 55°

If

two straight lines intersect each other then prove that the ray opposite to

the bisector of one of the angles so formed bisects the vertically opposite

angle.

Given : AB and CD are two lines which are

intersecting at O. OE is a ray bisecting the _{}BOD. OF is a ray opposite to ray OE.

To Prove: _{}AOF

= _{}COF

Proof : Since _{} are two

opposite rays, _{} is a straight

line passing through O.

_{} _{}AOF = _{}BOE

and _{}COF = _{}DOE

[Vertically opposite angles]

But _{}BOE = _{}DOE (Given)

_{} _{}AOF = _{}COF

Hence, proved.

Prove

that the bisectors of two adjacent supplementary angles include a right

angle.

Given: _{}is

the bisector of _{}BCD and _{}is the bisector of _{}ACD.

To Prove: _{}ECF

= 90^{o}

Proof: Since _{}ACD and _{}BCD forms a linear pair.

_{}ACD + _{}BCD = 180^{o}

^{}

_{}ACE + _{}ECD + _{}DCF + _{}FCB = 180^{o}

_{} _{}ECD + _{}ECD + _{}DCF + _{}DCF = 180^{o}

because

_{}ACE = _{}ECD

and _{}DCF = _{}FCB

_{} 2(_{}ECD) + 2 (_{}CDF) = 180^{o}

_{} 2(_{}ECD + _{}DCF) = 180^{o}

_{} _{}ECD + _{}DCF = _{}

_{} _{}ECF = 90^{o} (Proved)

In the adjoining figure, AOB is a

straight line. Find the value of x. Hence, find ∠AOC

and ∠BOD.

∠AOB is a

straight angle.

⇒ ∠AOB = 180°

⇒ ∠AOC + ∠COD + ∠BOD = 180°

⇒ (3x – 7)° + 55° + (x + 20)° = 180°

⇒ 4x + 68° = 180°

⇒ 4x = 112°

⇒ x = 28°

Thus,

∠AOC = (3x – 7)° = 3(28°) – 7° = 84° – 7° = 77°

And,

∠BOD = (x + 20)° = 28° + 20° = 48°

In the given figure, AOB is a straight line. Find the value of x. Hence, find _{}AOC, _{}COD and _{}BOD.

Since _{}BOD and _{}DOA from a linear pair of angles.

_{} _{}BOD + _{}DOA = 180^{o}

_{} _{}BOD + _{}DOC + _{}COA = 180^{o}

_{} x^{o} + (2x – 19)^{o} + (3x + 7)^{o} = 180^{o}

_{} 6x – 12 = 180

_{} 6x = 180 + 12 = 192

_{ }

_{} x = 32

_{} _{}AOC = (3x + 7)^{o} = (3 _{} 32 + 7)^{o} = 103^{o}

_{} _{}COD = (2x – 19)^{o} = (2 _{} 32 – 19)^{o} = 45^{o}

and _{}BOD = x^{o} = 32^{o}

In the given figure, x: y: z = 5: 4: 6. If XOY is a straight line, find the values of x, y and z.

x: y: z = 5: 4: 6

The sum of their ratios = 5 + 4 + 6 = 15

But x + y + z = 180^{o}

[Since, XOY is a straight line]

So, if the total sum of the measures is 15, then the measure of x is 5.

If the sum of angles is 180^{o}, then, measure of _{}

And, if the total sum of the measures is 15, then the measure of y is 4.

If the sum of the angles is 180^{o}, then, measure of _{}

And _{}z = 180^{o} – _{}x – _{}y

= 180^{o} – 60^{o} – 48^{o}

= 180^{o} – 108^{o} = 72^{o}

_{} x = 60, y = 48 and z = 72.

In

the given figure, what value of x will make AOB, a straight line?

AOB

will be a straight line, if two adjacent angles form a linear pair.

_{}BOC

+ _{}AOC = 180^{o}

_{} (4x – 36)^{o} + (3x + 20)^{o}

= 180^{o}

_{} 4x – 36 + 3x + 20 = 180

_{} 7x – 16 = 180^{o}

_{} 7x = 180 + 16 = 196

_{} _{}

_{} The value of x = 28.

Two

lines AB and CD intersect at O. If _{}AOC = 50^{o}, find _{}AOD, _{}BOD and _{}BOC.

Since

_{}AOC and _{}AOD form a linear pair.

_{}AOC

+ _{}AOD = 180^{o}

_{} 50^{o} + _{}AOD = 180^{o}

_{} _{}AOD = 180^{o} – 50^{o} = 130^{o}

_{}AOD

and _{}BOC are vertically opposite angles.

_{} _{}AOD = _{}BOC

_{} _{}BOC = 130^{o}

_{}BOD

and _{}AOC are vertically opposite angles.

_{}BOD

= _{}AOC

_{} _{}BOD = 50^{o}

In the given figure, three coplanar lines AB,CD and EF intersect at a point O, forming angles as shown. Find the values of x,y,z and t.

Since _{}COE and _{}DOF are vertically opposite angles, we have,

_{}COE = _{}DOF

_{} _{}z = 50^{o}

Also _{}BOD and _{}COA are vertically opposite angles.

So, _{}BOD = _{}COA

_{} _{}t = 90^{o}

As _{}COA and _{}AOD form a linear pair,

_{}COA + _{}AOD = 180^{o}

_{} _{}COA + _{}AOF + _{}FOD = 180^{o} [_{}t = 90^{o}]

_{} t + x + 50^{o} = 180^{o}

_{} 90^{o} + x^{o} + 50^{o} = 180^{o}

_{} x + 140 = 180

_{} x = 180 – 140 = 40

Since _{}EOB and _{}AOF are vertically opposite angles

So, _{}EOB = _{}AOF

_{} y = x = 40

Thus, x = 40 = y = 40, z = 50 and t = 90

In

the given figure, three coplanar lines AB,CD and EF intersect at a point O.

Find the value of x. Hence, find _{}AOD, _{}COE and _{}AOE.

Since

_{}COE and _{}EOD form a linear pair of angles.

_{} _{}COE + _{}EOD = 180^{o}

_{} _{}COE + _{}EOA + _{}AOD = 180^{o}

_{} 5x + _{}EOA + 2x = 180

_{} 5x + _{}BOF + 2x = 180

[_{}EOA and _{}BOF are vertically opposite angles so, _{}EOA = _{}BOF]

_{} 5x + 3x + 2x = 180

_{} 10x = 180

_{} x = 18

Now _{}AOD = 2x^{o} = 2 _{} 18^{o}

= 36^{o}

_{}COE = 5x^{o} = 5 _{} 18^{o}

= 90^{o}

and, _{}EOA = _{}BOF = 3x^{o} = 3 _{} 18^{o}

= 54^{o}

Two

adjacent angles on a straight line are in the ratio 5 : 4. Find the measure

of each one of these angles.

Let

the two adjacent angles be 5x and 4x.

Now,

since these angles form a linear pair.

So, 5x + 4x = 180^{o}

_{} 9x = 180^{o}

_{} _{}

_{} The required angles are 5x = 5x = 5 _{} 20^{o}

= 100^{o}

and 4x = 4 _{} 20^{o}

= 80^{o}

## Chapter 7 – Lines And Angles Exercise Ex. 7C

In the given figure, l ∥

m and a transversal t cuts them. If ∠1 = 120°, find the measure of each of the remaining marked angles.

Given, ∠1

= 120°

Now, ∠1

+ ∠2

= 180° (linear pair)

⇒

120° + ∠2 = 180°

⇒ ∠2

= 60°

∠1

= ∠3 (vertically opposite

angles)

⇒ ∠3

= 120°

Also, ∠2

= ∠4 (vertically opposite angles)

⇒ ∠4

= 60°

Line

l ∥ line m and line t is a transversal.

⇒ ∠5

= ∠1

= 120° (corresponding angles)

∠6

= ∠2

= 60° (corresponding angles)

∠7

= ∠3

= 120° (corresponding angles)

∠8

= ∠4

= 60° (corresponding angles)

In the given figure, AB || CD. Find the value of x.

Through C draw FG || AE

Now, since CG || BE and CE is a transversal.

So, _{}GCE = _{}CEA = 20^{o} [Alternate angles]

_{} _{}DCG = 130^{o} – _{}GCE

= 130^{o} – 20^{o} = 110^{o}

Also, we have AB || CD and FG is a transversal.

So, _{}BFC = _{}DCG = 110^{o} [Corresponding angles]

As, FG || AE, AF is a transversal.

_{}BFG = _{}FAE [Corresponding angles]

_{} x^{o} = _{}FAE = 110^{o}.

Hence, x = 110

In the given figure, AB || PQ. Find the values of x and y.

Since AB || PQ and EF is a transversal.

So, _{}CEB = _{}EFQ [Corresponding angles]

_{} _{}EFQ = 75^{o}

_{} _{}EFG + _{}GFQ = 75^{o}

_{} 25^{o} + y^{o} = 75^{o}

_{} y = 75 – 25 = 50

Also, _{}BEF + _{}EFQ = 180^{o} [sum of consecutive interior angles is 180^{o}]

_{} BEF = 180^{o} – _{}EFQ

= 180^{o} – 75^{o}

_{ }BEF = 105^{o}

_{} _{}FEG + _{}GEB = _{}BEF = 105^{o}

_{} _{}FEG = 105^{o} – _{}GEB = 105^{o} – 20^{o} = 85^{o}

In _{}EFG we have,

x^{o} + 25^{o} + _{}FEG = 180^{o}

Hence, x = 70.

In the given figure, AB || CD. Find the value of x.

Since AB || CD and AC is a transversal.

So, _{}BAC + _{}ACD = 180^{o} [sum of consecutive interior angles is 180^{o}]

_{} _{}ACD = 180^{o} – _{}BAC

= 180^{o} – 75^{o} = 105^{o}

_{}ECF = _{}ACD [Vertically opposite angles]

_{} _{}ECF = 105^{o}

Now in _{}CEF,

ECF + CEF + EFC =180^{o}

_{} 105^{o} + x^{o} + 30^{o} = 180^{o}

_{} x = 180 – 30 – 105 = 45

_{Hence, x = 45.}

In the given figure, AB || CD. Find the value of x.

Since AB || CD and PQ a transversal.

So, _{}PEF = _{}EGH [Corresponding angles]

_{} _{}EGH = 85^{o}

_{}EGH and _{}QGH form a linear pair.

So, _{}EGH + _{}QGH = 180^{o}

_{} _{}QGH = 180^{o} – 85^{o} = 95^{o}

Similarly, _{}GHQ + 115^{o} = 180^{o}

_{} _{}GHQ = 180^{o} – 115^{o} = 65^{o}

In _{}GHQ, we have,

x^{o} + 65^{o} + 95^{o} = 180^{o}

_{} x = 180 – 65 – 95 = 180 – 160

_{} x = 20

In the given figure, AB || CD. Find the values of x, y and z.

Since AB || CD and BC is a transversal.

So, _{}ABC = _{}BCD

_{} x = 35

Also, AB || CD and AD is a transversal.

So, _{}BAD = _{}ADC

_{} z = 75

In _{}ABO, we have,

_{ }x^{o} + 75^{o} + y^{o} = 180^{o}

_{} 35 + 75 + y = 180

_{} y = 180 – 110 = 70

_{} x = 35, y = 70 and z = 75.

In the given figure, AB || CD. Prove that p + q – r = 180.

Through F, draw KH || AB || CD

Now, KF || CD and FG is a transversal.

_{}_{}KFG = _{}FGD = r^{o} (i)

[alternate angles]

Again AE || KF, and EF is a transversal.

So,_{}AEF + _{}KFE = 180^{o}

_{}_{}KFE = 180^{o} – p^{o} (ii)

Adding (i) and (ii) we get,

_{}KFG + _{}KFE = 180 – p + r

_{}_{}EFG = 180 – p + r

_{}q = 180 – p + r

i.e.,p + q – r = 180

In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

_{}PRQ = x^{o} = 60^{o} [vertically opposite angles]

Since EF || GH, and RQ is a transversal.

So, _{}x = _{}y [Alternate angles]

_{} y = 60

AB || CD and PR is a transversal.

So, [Alternate angles]

_{} [since ]

_{}x + _{}QRD = 110^{o}

_{} _{}QRD = 110^{o} – 60^{o} = 50^{o}

In _{}QRS, we have,

_{}QRD + t^{o} + y^{o} = 180^{o}

_{} 50 + t + 60 = 180

_{} t = 180 – 110 = 70

Since, AB || CD and GH is a transversal

So, z^{o} = t^{o} = 70^{o} [Alternate angles]

_{} x = 60 , y = 60, z = 70 and t = 70

In the given figure, AB ∥

CD and a transversal t cuts them at E and F respectively. If EG and FG are

the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF

= 90°.

AB ∥

CD and a transversal t cuts them at E and F respectively.

⇒ ∠BEF

+ ∠DFE

= 180° (interior angles)

⇒ ∠GEF

+ ∠GFE

= 90° ….(i)

Now, in ΔGEF,

by angle sum property

∠GEF

+ ∠GFE

+ ∠EGF

= 180°

⇒

90° + ∠EGF = 180° ….[From (i)]

⇒ ∠EGF

= 90°

In the given figure, AB ∥

CD and a transversal t cuts them at E and F respectively. If EP and FQ are

the bisectors of ∠AEF and ∠EFD respectively, prove that EP ∥

FQ.

Since

AB ∥ CD and t is a transversal, we

have

∠AEF

= ∠EFD (alternate angles)

⇒ ∠PEF

= ∠EFQ

But, these are alternate interior

angles formed when the transversal EF cuts EP and FQ.

∴

EP ∥

FQ

In the given figure, l ∥

m and a transversal t cuts them. If ∠7 = 80°, find the measure of each of the remaining marked angles.

Given, ∠7

= 80°

Now, ∠7

+ ∠8

= 180° (linear pair)

⇒

80° + ∠8 = 180°

⇒ ∠8

= 100°

∠7

= ∠5 (vertically opposite

angles)

⇒ ∠5

= 80°

Also, ∠6

= ∠8 (vertically opposite angles)

⇒ ∠6

= 100°

Line

l ∥ line m and line t is a transversal.

⇒ ∠1

= ∠5

= 80° (corresponding angles)

∠2

= ∠6

= 100° (corresponding angles)

∠3

= ∠7

= 80° (corresponding angles)

∠4

= ∠8

= 100° (corresponding angles)

In the given figure, BA ∥

ED and BC ∥ EF. Show that ∠ABC = ∠DEF.

Construction:

Produce DE to meet BC at Z.

Now,

AB ∥ DZ and BC is the transversal.

⇒ ∠ABC

= ∠DZC (corresponding angles) ….(i)

Also,

EF ∥ BC and DZ is the transversal.

⇒ ∠DZC

= ∠DEF (corresponding angles) ….(ii)

From (i) and (ii), we have

∠ABC

= ∠DEF

In the given figure, BA ∥

ED and BC ∥ EF. Show that ∠ABC + ∠DEF = 180°.

Construction:

Produce ED to meet BC at Z.

Now,

AB ∥ EZ and BC is the transversal.

⇒ ∠ABZ

+ ∠EZB

= 180° (interior

angles)

⇒ ∠ABC

+ ∠EZB

= 180° ….(i)

Also,

EF ∥ BC and EZ is the transversal.

⇒ ∠BZE

= ∠ZEF (alternate angles)

⇒ ∠BZE

= ∠DEF ….(ii)

From (i) and (ii), we have

∠ABC

+ ∠DEF

= 180°

In the given figure, m and n are

two plane mirrors perpendicular to each other. Show that the incident ray CA

is parallel to the reflected ray BD.

Let

the normal to mirrors m and n intersect at P.

Now,

OB ⊥ m, OC ⊥ n and m ⊥ n.

⇒ OB ⊥ OC

⇒ ∠APB = 90°

⇒ ∠2 + ∠3 = 90° (sum of acute angles of a right triangle

is 90°)

By

the laws of reflection, we have

∠1 = ∠2 and ∠4 = ∠3 (angle of incidence = angle of

reflection)

⇒ ∠1 + ∠4 = ∠2 + ∠3 = 90°

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ ∠CAB + ∠ABD = 180°

But,

∠CAB and ∠ABD are consecutive

interior angles formed, when the transversal AB cuts CA and BD.

∴ CA ∥ BD

In the figure given below, state which lines are

parallel and why?

In

the given figure,

∠BAC = ∠ACD = 110°

But,

these are alternate angles when transversal AC cuts AB and CD.

Hence,

AB ∥ CD.

Two lines are respectively perpendicular to two

parallel lines. Show that they are parallel to each other.

Let

the two parallel lines be m and n.

Let

p ⊥ m.

⇒ ∠1 = 90°

Let

q ⊥ n.

⇒ ∠2 = 90°

Now,

m ∥ n and p is a transversal.

⇒ ∠1 = ∠3 (corresponding angles)

⇒ ∠3 = 90°

⇒ ∠3 = ∠2 (each 90°)

But,

these are corresponding angles, when transversal n cuts lines p and q.

∴ p ∥ q.

Hence,

two lines which are perpendicular to two parallel

lines, are parallel to each other.

In the given figure, l ∥

m and a transversal t cuts them. If ∠1 : ∠2 = 2 : 3, find the measure of each of the marked angles.

Given, ∠1

: ∠2

= 2 : 3

Now, ∠1

+ ∠2

= 180° (linear pair)

⇒

2x + 3x = 180°

⇒

5x = 180°

⇒

x = 36°

⇒ ∠1

= 2x = 72° and ∠2 = 3x = 108°

∠1

= ∠3 (vertically opposite

angles)

⇒ ∠3

= 72°

Also, ∠2

= ∠4 (vertically opposite angles)

⇒ ∠4

= 108°

Line

l ∥ line m and line t is a transversal.

⇒ ∠5

= ∠1

= 72° (corresponding angles)

∠6

= ∠2

= 108° (corresponding angles)

∠7

= ∠3

= 72° (corresponding angles)

∠8

= ∠4

= 108° (corresponding angles)

For what value of x will the lines l and m be parallel to each other?

_{}

Lines l and m will be parallel if 3x – 20 = 2x + 10

[Since, if corresponding angles are equal, lines are parallel]

_{}3x – 2x = 10 + 20

_{}x = 30

For what value of x will the lines

l and m be parallel to each other?

*Question modified, back answer

incorrect.

For

lines l and m to be parallel to each other, the corresponding angles (3x + 5)° and (4x)° should be

equal.

⇒

(3x + 5)° = 4x°

⇒

x = 5°

In the given figure, AB || CD and BC || ED. Find the value of x.

Since AB || CD and BC is a transversal.

So, _{}BCD = _{}ABC = x^{o} [Alternate angles]

As BC || ED and CD is a transversal.

_{}BCD + _{}EDC = 180^{o}

^{ }BCD + 75^{o} =180^{o}

_{} _{}BCD = 180^{o} – 75^{o} = 105^{o}

ABC = 105^{o} [since _{}BCD = _{}ABC]

_{} x^{o} = _{}ABC = 105^{o}

Hence, x = 105.

In the given figure, AB || CD || EF. Find the value of x.

Since AB || CD and BC is a transversal.

So, _{}ABC = _{}BCD [atternate interior angles]

_{}70^{o} = x^{o} + _{}ECD(i)

Now, CD || EF and CE is transversal.

So,_{}ECD + _{}CEF = 180^{o} [sum of consecutive interior angles is 180^{o}]

_{}_{}ECD + 130^{o} = 180^{o}

_{}_{}ECD = 180^{o} – 130^{o} = 50^{o}

Putting _{}ECD = 50^{o} in (i) we get,

70^{o} = x^{o} + 50^{o}

_{}x = 70 – 50 = 20

In the give figure, AB ∥

CD. Find the values of x, y and z.

AB

∥ CD and EF is transversal.

⇒ ∠AEF

= ∠EFG (alternate angles)

Given, ∠AEF

= 75°

⇒ ∠EFG

= y = 75°

Now, ∠EFC

+ ∠EFG

= 180° (linear pair)

⇒

x + y = 180°

⇒

x + 75° = 180°

⇒

x = 105°

∠EGD

= ∠EFG

+ ∠FEG (Exterior angle property)

⇒

125° = y + z

⇒

125° = 75° + z

⇒

z = 50°

Thus, x = 105°, y = 75° and z = 50°

Ineach of the figures given below, AB || CD. Find the value of x in each case.

_{}

Through E draw EG || CD. Now since EG||CD and ED is a transversal.

So,_{}GED = _{}EDC = 65^{o}[Alternate interior angles]

Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So,_{}BEG = _{}ABE = 35^{o}[Alternate interior angles]

So,_{}DEB = x^{o}

_{}_{}BEG + _{}GED = 35^{o} + 65^{o} = 100^{o}.

Hence, x = 100.

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Through O draw OF||CD.

Now since OF || CD and OD is transversal.

_{}CDO + _{}FOD = 180^{o}

[sum of consecutive interior angles is 180^{o}]

_{}25^{o} + _{}FOD = 180^{o}

_{}_{}FOD = 180^{o} – 25^{o} = 155^{o}

As OF || CD and AB || CD [Given]

Thus, OF || AB and OB is a transversal.

So,_{}ABO + _{}FOB = 180^{o }[sum of consecutive interior angles is 180^{o}]

_{}55^{o} + _{}FOB = 180^{o}

_{}_{}FOB = 180^{o} – 55^{o} = 125^{o}

Now, x^{o} = _{}FOB + _{}FOD = 125^{o} + 155^{o} = 280^{o}.

Hence, x = 280.

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

_{}FEC + _{}ECD = 180^{o}

[sum of consecutive interior angles is 180^{o}]

_{}_{}FEC + 124^{o} = 180^{o}

_{}_{}FEC = 180^{o} – 124^{o} = 56^{o}

Since EF || CD and AB ||CD

So, EF || AB and AE is a trasveral.

So,_{}BAE + _{}FEA = 180^{o}

[sum of consecutive interior angles is 180^{o}]

_{}116^{o} + _{}FEA = 180^{o}

_{}_{}FEA = 180^{o} – 116^{o} = 64^{o}

Thus,x^{o} = _{}FEA + _{}FEC

= 64^{o} + 56^{o} = 120^{o}.

Hence, x = 120.

## Chapter 7 – Lines And Angles Exercise MCQ

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

other two angles, then the triangle is

(a) an

isosceles triangle

(b) an

obtuse triangle

(c) an

equilateral triangle

(d) a

right triangle

Correct

option: (d)

In

a right triangle, one angle is 90° and the sum of

acute angles of a right triangle is 90°.

The

measure of an angle is five times its complement. The angle measures

- 25°
- 35°
- 65°
- 75°

Two

complementary angles are such that twice the measure of the one is equal to

three times the measure of the other. The larger of the two measures

- 72°
^{o} - 54°
- 63°
- 36°

In

the given figure, AOB is a straight line. If ∠AOC

= 4x°

and ∠BOC

= 5x°,

then ∠AOC

=?

In

the given figure, AOB is a straight line. If ∠AOC

= (3x + 10)

° and ∠BOC

= (4x – 26)

°, then ∠BOC =?

- 96°
- 86°
- 76°
- 106°

In

the given figure, AOB is a straight line. If ∠AOC

= (3x – 10)

°, ∠COD

= 50°

and ∠BOD

= (x +20)

°, then ∠AOC =?

- 40°
- 60°
- 80°
- 50°

Which

of the following statements is false?

- Through a given

point, only one straight line can be drawn - Through two given

points, it is possible to draw one and only one straight line. - Two straight lines

can intersect only at one point - A line segment can be

produced to any desired length.

Correct option: (a)

Option (a) is false, since

through a given point we can draw an infinite number of straight lines.

An

angle is one-fifth of its supplement. The measure of the angle is

- 15°
- 30°
- 75°
- 150°

In

the adjoining figure, AOB is straight line. If x:y:z = 4:5:6, then y = ?

- 60°
- 80°
- 48°
- 72°

In

the given figure, straight lines AB and CD intersect at O. If ∠AOC

=ϕ,

∠BOC

= θ

and θ

= 3

ϕ, then

ϕ =?

- 30°
- 40°
- 45°
- 60°

In

the given figure, straight lines AB and CD intersect at O. If ∠AOC

+ ∠BOD

= 130°,

then ∠AOD

=?

- 65°
- 115°
- 110°
- 125°

An exterior angle of a triangle is 110°

and its two interior opposite angles are equal. Each of these equal angles is

(a) 70°

(b) 55°

(c) 35°

(d)

Correct

option: (b)

Let

each interior opposite angle be x.

Then,

x + x = 110° (Exterior angle property of a triangle)

⇒ 2x = 110°

⇒ x = 55°

In

the given figure AB is a mirror, PQ is the incident ray and and QR is the

reflected ray. If ∠PQR = 108°,

then ∠

AQP =?

- 72°
- 18°
- 36°
- 54°

In the given figure, AB ∥

CD, If ∠BAO

= 60°

and ∠OCD

= 110°

then ∠AOC

= ?

(a) 70°

(b) 60°

(c) 50°

(d) 40°

Correct

option: (c)

Let

∠AOC = x°

Draw

YOZ ∥ CD ∥ AB.

Now,

YO ∥ AB and OA is the transversal.

⇒ ∠YOA = ∠OAB = 60° (alternate angles)

Again,

OZ ∥ CD and OC is the transversal.

⇒ ∠COZ + ∠OCD = 180° (interior angles)

⇒ ∠COZ + 110° = 180°

⇒ ∠COZ = 70°

Now, ∠YOZ = 180° (straight angle)

⇒ ∠YOA

+ ∠AOC + ∠COZ

= 180°

⇒ 60° + x + 70° = 180°

⇒ x = 50°

⇒ ∠AOC

= 50°

In

the given figure, AB ‖ CD. If ∠AOC

= 30°

and ∠OAB

= 100°,

then ∠OCD

=?

- 130°
- 150°
- 80°
- 100°

In

the given figure, AB ‖ CD. If ‖CAB

= 80^{o} and ∠EFC=

25°,

then ∠CEF

=?

- 65°
- 55°
- 45°
- 75°

In

the given figure, AB ‖ CD, CD ‖

EF and y:z = 3:7, then x = ?

- 108°
- 126°
- 162°
- 63°

In

the given figure, AB ‖ CD. If ∠APQ

= 70°

and ∠RPD

= 120°,

then ∠QPR

=?

- 50°
- 60°
- 40°
- 35°

In

the given figure AB ‖ CD. If ∠EAB

= 50°

and ∠ECD=60°,

then ∠AEB

=?

- 50°
- 60°
- 70°
- 50°

In

the given figure, ∠OAB = 75°,

∠OBA=55°

and ∠OCD

= 100°.

Then ∠ODC=?

- 20°
- 25°
- 30°
- 35°

In

the adjoining figure y =?

- 36°
- 54°
- 63°
- 72°

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

- Acute angled
- Obtuse angled
- Right angled
- an isosceles triangle

If one of the angles of triangle is 130°

then the angle between the bisector of the other two angles can be

(a) 50°

(b) 65°

(c) 90°

(d) 155

Correct

option: (d)

Let

∠A = 130°

In

ΔABC, by angle

sum property,

∠B + ∠C + ∠A = 180°

⇒ ∠B + ∠C + 130° = 180°

⇒ ∠B + ∠C = 50°

In the given figure, AOB is a straight line. The value of

x is

(a) 12

(b) 15

(c) 20

(d) 25

Correct

option: (b)

AOB

is a straight line.

⇒ ∠AOB = 180°

⇒ 60° + 5x° + 3x° = 180°

⇒ 60° + 8x° = 180°

⇒ 8x° = 120°

⇒ x = 15°

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is

(a) 120°

(b) 100°

(c) 80°

(d) 60°

Correct

option: (c)

By

angle sum property,

2x

+ 3x + 4x = 180°

⇒ 9x = 180°

⇒ x = 20°

Hence,

largest angle = 4x = 4(20°) = 80°

In the given figure, ∠OAB

= 110°

and ∠BCD

= 130°

then ∠ABC

is equal to

(a) 40°

(b) 50°

(c) 60°

(d) 70°

Correct

option: (c)

Through

B draw YBZ ∥ OA ∥ CD.

Now, OA ∥

YB and AB is the transversal.

⇒ ∠OAB

+ ∠YBA

= 180° (interior angles

are supplementary)

⇒

110° + ∠YBA = 180°

⇒ ∠YBA

= 70°

Also, CD ∥

BZ and BC is the transversal.

⇒ ∠DCB

+ ∠CBZ

= 180° (interior angles

are supplementary)

⇒

130° + ∠CBZ = 180°

⇒ ∠CBZ

= 50°

Now, ∠YBZ

= 180° (straight angle)

⇒ ∠YBA

+ ∠ABC

+ ∠CBZ

= 180°

⇒

70° + x + 50° = 180°

⇒

x = 60°

⇒ ∠ABC

= 60°

If

two angles are complements of each other, then each angle is

- An acute angle
- An obtuse angle
- A right angle
- A reflex angle

Correct option: (a)

Two angles are said to be complementary, if the sum of their measures is 90°.

Clearly, the measures of each of the angles have to be less than 90°.

Hence, each angle is an acute angle.

An

angle which measures more than 180° but less than 360°,

is called

- An acute angle
- An obtuse angle
- A straight angle
- A reflex angle

Correct option: (d)

An angle which measures more than 180^{o} but less than 360^{o }is called a reflex angle.