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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 7 - Lines And Angles

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Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 7 - Lines And Angles.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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Exercise/Page

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 7 - Lines And Angles Page/Excercise 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

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 7 - Lines And Angles Page/Excercise 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° 

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 7 - Lines And Angles Page/Excercise 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)

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 7 - Lines And Angles Page/Excercise 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.

So, BAD = ADC

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. 

TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 7 - Lines And Angles for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 7 - Lines And Angles now.

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