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

Page / Exercise

## Chapter 7 - Lines And Angles MCQ

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°.

Solution 2

Correct option: (b)

Let each interior opposite angle be x.

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

2x = 110°

x = 55°

Solution 3

Solution 4

Correct option: (d)

Let A = 130°

In ΔABC, by angle sum property,

B + C + A = 180°

B + C + 130° = 180°

B + C = 50°

Solution 5

Correct option: (b)

AOB is a straight line.

AOB = 180°

60° + 5x° + 3x° = 180°

60° + 8x° = 180°

8x° = 120°

x = 15°

Solution 6

Correct option: (c)

By angle sum property,

2x + 3x + 4x = 180°

9x = 180°

x = 20°

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

Solution 7

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°

Solution 8

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.

Solution 9

Correct option: (d)

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

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Correct option: (a)

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

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

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°

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

## Chapter 7 - Lines And Angles Ex. 7A

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.

Solution 2(ii)

Complement of 16o = 90 - 16o = 74o

Solution 2(iv)

Complement of 46o 30' = 90o - 46o 30' = 43o 30'

Solution 2(i)

Complement of 55° = 90° - 55° = 35°

Solution 2(iii)

Complement of 90° = 90° - 90° = 0°

Solution 3(iv)

Supplement of 75o 36' = 180o - 75o 36' = 104o 24'

Solution 3(i)

Supplement of 42° = 180° - 42° = 138°

Solution 3(ii)

Supplement of 90° = 180° - 90° = 90°

Solution 3(iii)

Supplement of 124° = 180° - 124° = 56°

Solution 4

(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.

Solution 5

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.

Solution 6

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°.

Solution 7

Let the required angle be xo

Then, its complement = 90o - xo

The required angle is 72o.

Solution 8

Let the required angle be xo

Then, its supplement is 180o - xo

The required angle is 150o.

Solution 9

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.

Solution 10

Let the required angle be xo

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

The required angle is 45o.

Solution 11

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.

Solution 12

(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°

## Chapter 7 - Lines And Angles Ex. 7B

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

Solution 2

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°

Solution 3

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

Solution 4

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 = 180o - x - y

= 180o - 60o - 48o

= 180o - 108o = 72o

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

Solution 5

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.

Solution 6

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

Solution 7

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

Solution 8

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

Solution 9

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

Solution 10

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.

Solution 11

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.

Solution 12

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°

Solution 13

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)

Solution 14

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°

Solution 15

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.

Solution 16

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)

## Chapter 7 - Lines And Angles Ex. 7C

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)

Solution 2

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)

Solution 3

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)

Solution 4

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

Solution 5

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°

Solution 6

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.

Solution 7

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

Solution 8

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°

Solution 9(i)

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.

Solution 9(ii)

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.

Solution 9(iii)

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.

Solution 10

Through C draw FG || AE

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

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

DCG = 130o - GCE

= 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

Solution 11

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 = 180o - EFQ

= 180o - 75o

BEF = 105o

FEG + GEB = BEF = 105o

FEG = 105o - GEB = 105o - 20o = 85o

In EFG we have,

xo + 25o + FEG = 180o

Hence, x = 70.

Solution 12

Since AB || CD and AC is a transversal.

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

ACD = 180o - BAC

= 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.

Solution 13

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

Solution 14

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.

Solution 16

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

Solution 17

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

Solution 18

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°

Solution 19

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

Solution 20

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

Solution 21

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°

Solution 22

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

Solution 23

In the given figure,

BAC = ACD = 110°

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

Hence, AB CD.

Solution 24

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.

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