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

Chapter 7 – Lines And Angles Exercise Ex. 7A

Question 1

Define the following terms:

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

Solution 1

(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 90o but less than 180o, is called an obtuse angle.

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

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

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

Question 2

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

Solution 2

Let
the required angle be xo

Then,
its complement is 90o – xo and its supplement is 180o
– xo

The required angle is 45o.

Question 3

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

Solution 3

Let the two required angles be xo and 90o – xo.

Then

5x = 4(90 – x)

5x = 360 – 4x

5x + 4x = 360

9x = 360

Thus, the required angles are 40o and 90o – xo = 90 o – 40o = 50o.

Question 4

Find the value of x for which the
angles (2x – 5 and (x –
10)
°
are the complementary angles.

Solution 4

(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°

Question 5

Find the complement of each of the
following angle:

55°

Solution 5

Complement of 55° = 90° – 55° = 35°

Question 6

Find the complement of each of the following angles.

16o

Solution 6

Complement of 16o = 90 – 16o = 74o

Question 7

Find the complement of each of the
following angle:

90°

Solution 7

Complement of 90° = 90° – 90° = 0°

Question 8

Find the complement of each of the following angles.

46o 30

Solution 8

Complement of 46o 30′ = 90o – 46o 30′ = 43o 30′

Question 9

Find the supplement of each of the
following angle:

42°

Solution 9

Supplement of 42° = 180° – 42° =
138°

Question 10

Find the supplement of each of the
following angle:

90°

Solution 10

Supplement of 90° = 180° – 90° =
90°

Question 11

Find the supplement of each of the
following angle:

124°

Solution 11

Supplement of 124° = 180° – 124° =
56°

Question 12

Find the supplement of each of the following angles.

75o 36′

Solution 12

Supplement of 75o 36′ = 180o – 75o 36′ = 104o 24′

Question 13

Find
the measure of an angle which is

(i)
equal to its complement,
(ii) equal to its supplement.

Solution 13

(i)
Let the required angle be xo

Then,
its complement = 90o – xo

The measure of an angle which is equal to
its complement is 45o.

(ii)
Let the required angle be xo

Then,
its supplement = 180o – xo

The measure of an angle which is equal to
its supplement is 90o.

Question 14

Find
the measure of an angle which is 36o more than its complement.

Solution 14

Let
the required angle be xo

Then
its complement is 90o – xo

The measure of an angle which is 36o
more than its complement is 63o.

Question 15

Find the measure of an angle which
is 30
°
less than its supplement.

Solution 15

Let
the measure of the required angle = x
°

Then, measure of its supplement =
(180 – x)
°

It is given that

x°
= (180 – x)° – 30°

= 180°
– x°
– 30°

2x°
= 150°

= 75°

Hence, the measure of the required
angle is 75
°.

Question 16

Find
the angle which is four times its complement.

Solution 16

Let
the required angle be xo

Then,
its complement = 90o – xo

The required angle is 72o.

Question 17

Find
the angle which is five times its supplement.

Solution 17

Let
the required angle be xo

Then,
its supplement is 180o – xo

The required angle is 150o.

Question 18

Find the angle whose supplement is four times its complement.

Solution 18

Let the required angle be xo

Then, its complement is 90o – xo and its supplement is 180o – xo

That is we have,

The required angle is 60o.

Chapter 7 – Lines And Angles Exercise Ex. 7B

Question 1

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

Solution 1

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

BOC + COA = 180o

xo + 62o = 180o

x = 180 – 62

x = 118o

Question 2

If
two straight lines intersect each other in such a way that one of the angles
formed measures 90o, show that each of the remaining angles
measures 90o.

Solution 2

Let
two straight lines AB and CD intersect at O and let AOC = 90o.

Now, AOC = BOD
[Vertically opposite angles]

BOD = 90o

Also,
as AOC and AOD form a linear pair.

90o + AOD = 180o

AOD = 180o – 90o = 90o

Since, BOC = AOD
[Verticallty opposite angles]

BOC = 90o

Thus,
each of the remaining angles is 90o.

Question 3

Two
lines AB and CD intersect at a point O such that BOC +AOD = 280o, as shown in the figure. Find
all the four angles.

Solution 3

Since,
AOD and BOC are vertically opposite angles.

AOD
= BOC

Now,
AOD + BOC = 280o [Given]

AOD + AOD = 280o

2AOD = 280o

AOD =

BOC = AOD = 140o

As,
AOC and AOD form a linear pair.

So, AOC + AOD = 180o

AOC + 140o = 180o

AOC = 180o – 140o = 40o

Since,
AOC and BOD are vertically opposite angles.

AOC
= BOD

BOD = 40o

BOC = 140o, AOC = 40o , AOD = 140o and BOD = 40o.

Question 4

Two lines AB and CD intersect each
other at a point O such that
AOC : ∠AOD = 5 : 7. Find all the angles.

Solution 4

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°

Question 5

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.

Solution 5

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)

Question 6

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.

Solution 6

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°

Question 7

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.

Solution 7

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.

Question 8

Prove
that the bisectors of two adjacent supplementary angles include a right
angle.

Solution 8

Given: is
the bisector of BCD and is the bisector of ACD.

To Prove: ECF
= 90o

Proof: Since ACD and BCD forms a linear pair.

ACD + BCD = 180o

ACE + ECD + DCF + FCB = 180o

ECD + ECD + DCF + DCF = 180o

because
ACE = ECD

and DCF = FCB

2(ECD) + 2 (CDF) = 180o

2(ECD + DCF) = 180o

ECD + DCF =

ECF = 90o (Proved)

Question 9

In the adjoining figure, AOB is a
straight line. Find the value of x. Hence, find
∠AOC
and ∠BOD.

Solution 9

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°

Question 10

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

Solution 10

Since BOD and DOA from a linear pair of angles.

BOD + DOA = 180o

BOD + DOC + COA = 180o

xo + (2x – 19)o + (3x + 7)o = 180o

6x – 12 = 180

6x = 180 + 12 = 192

x = 32

AOC = (3x + 7)o = (3 32 + 7)o = 103o

COD = (2x – 19)o = (2 32 – 19)o = 45o

and BOD = xo = 32o

Question 11

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

Solution 11

x: y: z = 5: 4: 6

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

But x + y + z = 180o

[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 180o, 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 180o, then, measure of

And z = 180ox – y

= 180o – 60o – 48o

= 180o – 108o = 72o

x = 60, y = 48 and z = 72.

Question 12

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

Solution 12

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

BOC
+ AOC = 180o

(4x – 36)o + (3x + 20)o
= 180o

4x – 36 + 3x + 20 = 180

7x – 16 = 180o

7x = 180 + 16 = 196

The value of x = 28.

Question 13

Two
lines AB and CD intersect at O. If AOC = 50o, find AOD, BOD and BOC.

Solution 13

Since
AOC and AOD form a linear pair.

AOC
+ AOD = 180o

50o + AOD = 180o

AOD = 180o – 50o = 130o

AOD
and BOC are vertically opposite angles.

AOD = BOC

BOC = 130o

BOD
and AOC are vertically opposite angles.

BOD
= AOC

BOD = 50o

Question 14

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.

Solution 14

Since COE and DOF are vertically opposite angles, we have,

COE = DOF

z = 50o

Also BOD and COA are vertically opposite angles.

So, BOD = COA

t = 90o

As COA and AOD form a linear pair,

COA + AOD = 180o

COA + AOF + FOD = 180o [t = 90o]

t + x + 50o = 180o

90o + xo + 50o = 180o

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

Question 15

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.

Solution 15

Since
COE and EOD form a linear pair of angles.

COE + EOD = 180o

COE + EOA + AOD = 180o

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 = 2xo = 2 18o
= 36o

COE = 5xo = 5 18o
= 90o

and, EOA = BOF = 3xo = 3 18o
= 54o

Question 16

Two
adjacent angles on a straight line are in the ratio 5 : 4. Find the measure
of each one of these angles.

Solution 16

Let
the two adjacent angles be 5x and 4x.

Now,
since these angles form a linear pair.

So, 5x + 4x = 180o

9x = 180o

The required angles are 5x = 5x = 5 20o
= 100o

and 4x = 4 20o
= 80o

Chapter 7 – Lines And Angles Exercise Ex. 7C

Question 1

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.

Solution 1

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)

Question 2

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

Solution 2

Through C draw FG || AE

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

So, GCE = CEA = 20o            [Alternate angles]

DCG = 130oGCE

= 130o – 20o = 110o

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

So, BFC = DCG = 110o          [Corresponding angles]

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

BFG = FAE                           [Corresponding angles]

xo = FAE = 110o.

Hence, x = 110

Question 3

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

Solution 3

Since AB || PQ and EF is a transversal.

So, CEB = EFQ                 [Corresponding angles]

EFQ = 75o

EFG + GFQ = 75o

25o + yo = 75o

y = 75 – 25 = 50

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

BEF = 180oEFQ

= 180o – 75o

BEF = 105o

FEG + GEB = BEF = 105o

FEG = 105oGEB = 105o – 20o = 85o

In EFG we have,

xo + 25o + FEG = 180o

Hence, x = 70.

Question 4

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

Solution 4

Since AB || CD and AC is a transversal.

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

ACD = 180oBAC

= 180o – 75o = 105o

ECF = ACD                     [Vertically opposite angles]

ECF = 105o

Now in CEF,

ECF + CEF + EFC =180o

105o + xo + 30o = 180o

x = 180 – 30 – 105 = 45

Hence, x = 45.

Question 5

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

Solution 5

Since AB || CD and PQ a transversal.

So, PEF = EGH [Corresponding angles]

EGH = 85o

EGH and QGH form a linear pair.

So, EGH + QGH = 180o

QGH = 180o – 85o = 95o

Similarly, GHQ + 115o = 180o

GHQ = 180o – 115o = 65o

In GHQ, we have,

xo + 65o + 95o = 180o

x = 180 – 65 – 95 = 180 – 160

x = 20

Question 6

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

Solution 6

Since AB || CD and BC is a transversal.

So, ABC = BCD

x = 35

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

z = 75

In ABO, we have,

xo + 75o + yo = 180o

35 + 75 + y = 180

y = 180 – 110 = 70

x = 35, y = 70 and z = 75.

Question 7

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

Solution 7

Through F, draw KH || AB || CD

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

KFG = FGD = ro (i)

[alternate angles]

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

So,AEF + KFE = 180o

KFE = 180o – po (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

Question 8

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

Solution 8

PRQ = xo = 60o            [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 = 110o

QRD = 110o – 60o = 50o

In QRS, we have,

QRD + to + yo = 180o

50 + t + 60 = 180

t = 180 – 110 = 70

Since, AB || CD and GH is a transversal

So, zo = to = 70o [Alternate angles]

x = 60 , y = 60, z = 70 and t = 70

Question 9

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°.

Solution 9

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°

Question 10

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.

Solution 10

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

Question 11

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.

Solution 11

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)

Question 12

In the given figure, BA
ED and BC ∥ EF. Show that ∠ABC = ∠DEF.

Solution 12

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

Question 13

In the given figure, BA
ED and BC ∥ EF. Show that ∠ABC + ∠DEF = 180°.

Solution 13

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°

Question 14

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.

Solution 14

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

Question 15

In the figure given below, state which lines are
parallel and why?

Solution 15

In
the given figure,

BAC = ∠ACD = 110°

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

Hence,
AB
∥ CD.

Question 16

Two lines are respectively perpendicular to two
parallel lines. Show that they are parallel to each other.

Solution 16

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.

Question 17

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.

Solution 17

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)

Question 18

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

Solution 18

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

Question 19

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

incorrect.

Solution 19

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°

Question 20

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

Solution 20

Since AB || CD and BC is a transversal.

So, BCD = ABC = xo     [Alternate angles]

As BC || ED and CD is a transversal.

BCD + EDC = 180o

BCD + 75o =180o

BCD = 180o – 75o = 105o

ABC = 105o                 [since BCD = ABC]

xo = ABC = 105o

Hence, x = 105.

Question 21

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

Solution 21

Since AB || CD and BC is a transversal.

So, ABC = BCD                [atternate interior angles]

70o = xo + ECD(i)

Now, CD || EF and CE is transversal.

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

ECD + 130o = 180o

ECD = 180o – 130o = 50o

Putting ECD = 50o in (i) we get,

70o = xo + 50o

x = 70 – 50 = 20

Question 22

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

Solution 22

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°

Question 23

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

Solution 23

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

So,GED = EDC = 65o[Alternate interior angles]

Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So,BEG = ABE = 35o[Alternate interior angles]

So,DEB = xo

BEG + GED = 35o + 65o = 100o.

Hence, x = 100.

Question 24

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

Solution 24

Through O draw OF||CD.

Now since OF || CD and OD is transversal.

CDO + FOD = 180o

[sum of consecutive interior angles is 180o]

25o + FOD = 180o

FOD = 180o – 25o = 155o

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

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

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

55o + FOB = 180o

FOB = 180o – 55o = 125o

Now, xo = FOB + FOD = 125o + 155o = 280o.

Hence, x = 280.

Question 25

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

Solution 25

Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

FEC + ECD = 180o

[sum of consecutive interior angles is 180o]

FEC + 124o = 180o

FEC = 180o – 124o = 56o

Since EF || CD and AB ||CD

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

So,BAE + FEA = 180o

[sum of consecutive interior angles is 180o]

116o + FEA = 180o

FEA = 180o – 116o = 64o

Thus,xo = FEA + FEC

= 64o + 56o = 120o.

Hence, x = 120.

Chapter 7 – Lines And Angles Exercise MCQ

Question 1

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

Solution 1

Correct
option: (d)

In
a right triangle, one angle is 90
° and the sum of
acute angles of a right triangle is 90°.

Question 2

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

1. 25°
2. 35°
3. 65°
4. 75°
Solution 2

Question 3

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

1. 72°o
2. 54°
3. 63°
4. 36°
Solution 3

Question 4

In
the given figure, AOB is a straight line. If ∠AOC
= 4x°
and ∠BOC
= 5x°,
then ∠AOC
=?

Solution 4

Question 5

In
the given figure, AOB is a straight line. If ∠AOC
= (3x + 10)
° and ∠BOC
= (4x – 26)
°, then ∠BOC =?

1. 96°
2. 86°
3. 76°
4. 106°

Solution 5

Question 6

In
the given figure, AOB is a straight line. If ∠AOC
= (3x – 10)
°, ∠COD
= 50°
and ∠BOD
= (x +20)
°, then ∠AOC =?

1. 40°
2. 60°
3. 80°
4. 50°
Solution 6

Question 7

Which
of the following statements is false?

1. Through a given
point, only one straight line can be drawn
2. Through two given
points, it is possible to draw one and only one straight line.
3. Two straight lines
can intersect only at one point
4. A line segment can be
produced to any desired length.
Solution 7

Correct option: (a)

Option (a) is false, since
through a given point we can draw an infinite number of straight lines.

Question 8

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

1. 15°
2. 30°
3. 75°
4. 150°
Solution 8

Question 9

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

1. 60°
2. 80°
3. 48°
4. 72°

Solution 9

Question 10

In
the given figure, straight lines AB and CD intersect at O. If ∠AOC
=ϕ,
∠BOC
= θ
and θ
= 3
ϕ, then
ϕ =?

1. 30°
2. 40°
3. 45°
4. 60°

Solution 10

Question 11

In
the given figure, straight lines AB and CD intersect at O. If ∠AOC
+ ∠BOD
= 130°,
then ∠AOD
=?

1. 65°
2. 115°
3. 110°
4. 125°
Solution 11

Question 12

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)

Solution 12

Correct
option: (b)

Let
each interior opposite angle be x.

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

2x = 110°

x = 55°

Question 13

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 =?

1. 72°
2. 18°
3. 36°
4. 54°

Solution 13

Question 14

In the given figure, AB
CD, If ∠BAO
= 60°
and ∠OCD
= 110°
then ∠AOC
= ?

(a) 70°

(b) 60°

(c) 50°

(d) 40°

Solution 14

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°

Question 15

In
the given figure, AB ‖ CD. If ∠AOC
= 30°
and ∠OAB
= 100°,
then ∠OCD
=?

1. 130°
2. 150°
3. 80°
4. 100°

Solution 15

Question 16

In
the given figure, AB ‖ CD. If ‖CAB
= 80o and ∠EFC=
25°,
then ∠CEF
=?

1. 65°
2. 55°
3. 45°
4. 75°

Solution 16

Question 17

In
the given figure, AB ‖ CD, CD ‖
EF and y:z = 3:7, then x = ?

1. 108°
2. 126°
3. 162°
4. 63°

Solution 17

Question 18

In
the given figure, AB ‖ CD. If ∠APQ
= 70°
and ∠RPD
= 120°,
then ∠QPR
=?

1. 50°
2. 60°
3. 40°
4. 35°

Solution 18

Question 19

In
the given figure AB ‖ CD. If ∠EAB
= 50°
and ∠ECD=60°,
then ∠AEB
=?

1. 50°
2. 60°
3. 70°
4. 50°

Solution 19

Question 20

In
the given figure, ∠OAB = 75°,
∠OBA=55°
and ∠OCD
= 100°.
Then ∠ODC=?

1. 20°
2. 25°
3. 30°
4. 35°

Solution 20

Question 21

In

1. 36°
2. 54°
3. 63°
4. 72°

Solution 21

Question 22

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

1. Acute angled
2. Obtuse angled
3. Right angled
4. an isosceles triangle
Solution 22

Question 23

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

Solution 23

Correct
option: (d)

Let
∠A = 130°

In
ΔABC, by angle
sum property,

B + ∠C + ∠A = 180°

∠B + ∠C + 130° = 180°

∠B + ∠C = 50°

Question 24

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

(a) 12

(b) 15

(c) 20

(d) 25

Solution 24

Correct
option: (b)

AOB
is a straight line.

∠AOB = 180°

60° + 5x° + 3x° = 180°

60° + 8x° = 180°

8x° = 120°

x = 15°

Question 25

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°

Solution 25

Correct
option: (c)

By
angle sum property,

2x
+ 3x + 4x = 180
°

9x = 180°

x = 20°

Hence,
largest angle = 4x = 4(20
°) = 80°

Question 26

In the given figure, ∠OAB
= 110°
and ∠BCD
= 130°
then ∠ABC
is equal to

(a) 40°

(b) 50°

(c) 60°

(d) 70°

Solution 26

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°

Question 27

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

1. An acute angle
2. An obtuse angle
3. A right angle
4. A reflex angle
Solution 27

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.

Question 28

An
angle which measures more than 180° but less than 360°,
is called

1. An acute angle
2. An obtuse angle
3. A straight angle
4. A reflex angle
Solution 28

Correct option: (d)

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

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