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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 8 - Triangles

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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 8 - Triangles.

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 8 - Triangles Page/Excercise MCQ

Solution 1

Solution 2

Correct option: (c)

A - B = 42° 

A = B + 42° 

B - C = 21° 

C = B - 21° 

In ΔABC,

A + B + C = 180° 

B + 42° + B + B - 21° = 180° 

3B = 159

B = 53° 

Solution 3

Correct option: (b)

ACD = B + A (Exterior angle property)

110° = 50° + A

A = 60° 

Solution 4

Correct option: (d)

Solution 5

Correct option: (a)

Solution 6

Solution 7

Correct option: (b)

EAF = CAD (vertically opposite angles)

CAD = 30° 

In ΔABD, by angle sum property

A + B + D = 180° 

(x + 30)° + (x + 10)° + 90° = 180° 

2x + 130° = 180° 

2x = 50° 

x = 25° 

Solution 8

Correct option: (a)

ABF + ABC = 180° (linear pair)

x + ABC = 180° 

ABC = 180° - x

 

ACG + ACB = 180° (linear pair)

y + ACB = 180° 

ACB = 180° - y

 

In ΔABC, by angle sum property

ABC + ACB + BAC = 180° 

(180° - x) + (180° - y) + BAC = 180° 

BAC - x - y + 180° = 0

BAC = x + y - 180° 

 

Now, EAD = BAC (vertically opposite angles)

z = x + y - 180°  

Solution 9

Correct option: (b)

In ΔOAC, by angle sum property

OCA + COA + CAO = 180° 

80° + 40° + CAO = 180° 

CAO = 60° 

 

CAO + OAE = 180° (linear pair)

60° + x = 180° 

x = 120° 

 

COA = BOD (vertically opposite angles)

BOD = 40° 

 

In ΔOBD, by angle sum property

OBD + BOD + ODB = 180° 

OBD + 40° + 70° = 180° 

OBD = 70° 

 

OBD + DBF = 180° (linear pair)

70° + y = 180° 

y = 110° 

 

x + y = 120° + 110° = 230° 

Solution 10

Solution 11

Solution 12

Correct option: (a)

ACB + ACD = 180° (linear pair)

5y + 7y = 180° 

12y = 180° 

y = 15° 

 

Now, ACD = ABC + BAC (Exterior angle property)

7y = x + 3y

7(15°) = x + 3(15°)

105° = x + 45° 

x = 60° 

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 8 - Triangles Page/Excercise 8

Solution 1

Since, sum of the angles of a triangle is 180o

A + B + C = 180o

A + 76o + 48o = 180o

A = 180o - 124o = 56o

A = 56o

Solution 2

Let the measures of the angles of a triangle are (2x)o, (3x)o and (4x)o.

Then, 2x + 3x + 4x = 180         [sum of the angles of a triangle is 180o ]     

9x = 180

The measures of the required angles are:

2x = (2 20)o = 40o

3x = (3 20)o = 60o

4x = (4 20)o = 80o

Solution 3

Let 3A = 4B = 6C = x (say)

Then, 3A = x

A =

4B = x

and 6C = x

C =

As A + B + C = 180o

A =

B =

C =

Solution 4

A + B = 108o [Given]

But as A, B and C are the angles of a triangle,

A + B + C = 180o

108o + C = 180o

C = 180o - 108o = 72o

Also, B + C = 130o [Given]

B + 72o = 130o

B = 130o - 72o = 58o

Now as, A + B = 108o

A + 58o = 108o

A = 108o - 58o = 50o

A = 50o, B = 58o and C = 72o.

Solution 5

Since. A , B and C are the angles of a triangle .

So, A + B + C = 180o

Now, A + B = 125o [Given]

125o + C = 180o

C = 180o - 125o = 55o

Also, A + C = 113o [Given]

A + 55o = 113o

A = 113o - 55o = 58o

Now as A + B = 125o

58o + B = 125o

B = 125o - 58o = 67o

A = 58o, B = 67o and C = 55o.

Solution 6

Since, P, Q and R are the angles of a triangle.

So,P + Q + R = 180o(i)

Now,P - Q = 42o[Given]

P = 42o + Q(ii)

andQ - R = 21o[Given]

R = Q - 21o(iii)

Substituting the value of P and R from (ii) and (iii) in (i), we get,

42o + Q + Q + Q - 21o = 180o

3Q + 21o = 180o

3Q = 180o - 21o = 159o

Q =

P = 42o + Q

= 42o + 53o = 95o

R = Q - 21o

= 53o - 21o = 32o

P = 95o, Q = 53o and R = 32o.

Solution 7

Given that the sum of the angles A and B of a ABC is 116o, i.e., A + B = 116o.

Since, A + B + C = 180o

So, 116o + C = 180o

C = 180o - 116o = 64o

Also, it is given that:

A - B = 24o

A = 24o + B

Putting, A = 24o + B in A + B = 116o, we get,

24o + B + B = 116o

2B + 24o = 116o

2B = 116o - 24o = 92o

B =

Therefore, A = 24o + 46o = 70o

A = 70o, B = 46o and C = 64o.

Solution 8

Let the two equal angles, A and B, of the triangle be xo each.

We know,

A + B + C = 180o

xo + xo + C = 180o

2xo + C = 180o(i)

Also, it is given that,

C = xo + 18o(ii)

Substituting C from (ii) in (i), we get,

2xo + xo + 18o = 180o

3xo = 180o - 18o = 162o

x =

Thus, the required angles of the triangle are 54o, 54o and xo + 18o = 54o + 18o = 72o.

Solution 9

Let C be the smallest angle of ABC.

Then, A = 2C and B = 3C

Also, A + B + C = 180o

2C + 3C + C = 180o

6C = 180o

C = 30o

So, A = 2C = 2 30o = 60o

B = 3C = 3 30o = 90o

The required angles of the triangle are 60o, 90o, 30o.

Solution 10

Let ABC be a right angled triangle and C = 90o

Since, A + B + C = 180o

A + B = 180o - C = 180o - 90o = 90o

Suppose A = 53o

Then, 53o + B = 90o

B = 90o - 53o = 37o

The required angles are 53o, 37o and 90o.

Solution 11

Let ABC be a right angled triangle and C = 90o

Since, A + B + C = 180o

A + B = 180o - C = 180o - 90o = 90o

Suppose A = 53o

Then, 53o + B = 90o

B = 90o - 53o = 37o

The required angles are 53o, 37o and 90o.

Solution 12

Let ABC be a triangle.

So, begin mathsize 12px style angle end styleA < begin mathsize 12px style angle end styleB + begin mathsize 12px style angle end styleC

Adding begin mathsize 12px style angle end styleA to both sides of the inequality,

begin mathsize 12px style rightwards double arrow end stylebegin mathsize 12px style angle end styleA

Solution 13

Let ABC be a triangle and B > A + C

Since, A + B + C = 180o

A + C = 180o - B

Therefore, we get,

B > 180o - B

Adding B on both sides of the inequality, we get,

B + B > 180o - B + B

2B > 180o

B >

i.e., B > 90o which means B is an obtuse angle.

ABC is an obtuse angled triangle.

Solution 14

Since ACB and ACD form a linear pair.

So, ACB + ACD = 180o

ACB + 128o = 180o

ACB = 180o - 128 = 52o

Also, ABC + ACB + BAC = 180o

43o + 52o + BAC = 180o

95o + BAC = 180o

BAC = 180o - 95o = 85o

ACB = 52o and BAC = 85o.

Solution 15

As DBA and ABC form a linear pair.

So,DBA + ABC = 180o

106o + ABC = 180o

ABC = 180o - 106o = 74o

Also, ACB and ACE form a linear pair.

So,ACB + ACE = 180o

ACB + 118o = 180o

ACB = 180o - 118o = 62o

In ABC, we have,

ABC + ACB + BAC = 180o

74o + 62o + BAC = 180o

136o + BAC = 180o

BAC = 180o - 136o = 44o

In triangle ABC, A = 44o, B = 74o and C = 62o

Solution 16

(i) EAB + BAC = 180o [Linear pair angles]

110o + BAC = 180o

BAC = 180o - 110o = 70o

Again, BCA + ACD = 180o [Linear pair angles]

BCA + 120o = 180o

BCA = 180o - 120o = 60o

Now, in ABC,

ABC + BAC + ACB = 180o

xo + 70o + 60o = 180o

x + 130o = 180o

x = 180o - 130o = 50o

x = 50

(ii)

In ABC,

A + B + C = 180o

30o + 40o + C = 180o

70o + C = 180o

C = 180o - 70o = 110o

Now BCA + ACD = 180o [Linear pair]

110o + ACD = 180o

ACD = 180o - 110o = 70o

In ECD,

ECD + CDE + CED = 180o

70o + 50o + CED = 180o

120o + CED = 180o

CED = 180o - 120o = 60o

Since AED and CED from a linear pair

So, AED + CED = 180o

xo + 60o = 180o

xo = 180o - 60o = 120o

x = 120

(iii)

EAF = BAC [Vertically opposite angles]

BAC = 60o

In ABC, exterior ACD is equal to the sum of two opposite interior angles.

So, ACD = BAC + ABC

115o = 60o + xo

xo = 115o - 60o = 55o

x = 55

(iv)

Since AB || CD and AD is a transversal.

So, BAD = ADC

ADC = 60o

In ECD, we have,

E + C + D = 180o

xo + 45o + 60o = 180o

xo + 105o = 180o

xo = 180o - 105o = 75o

x = 75

(v)

In AEF,

Exterior BED = EAF + EFA

100o = 40o + EFA

EFA = 100o - 40o = 60o

Also, CFD = EFA [Vertically Opposite angles]

CFD = 60o

Now in FCD,

Exterior BCF = CFD + CDF

90o = 60o + xo

xo = 90o - 60o = 30o

x = 30

(vi)

In ABE, we have,

A + B + E = 180o

75o + 65o + E = 180o

140o + E = 180o

E = 180o - 140o = 40o

Now, CED = AEB [Vertically opposite angles]

CED = 40o

Now, in CED, we have,

C + E + D = 180o

110o + 40o + xo = 180o

150o + xo = 180o

xo = 180o - 150o = 30o

x = 30

Solution 17

AB CD and AC is the transversal.

BAC = ACD = 60° (alternate angles)

i.e. BAC = GCH = 60° 

 

Now, DHF = CHG = 50° (vertically opposite angles)

 

In ΔGCH, by angle sum property,

GCH + CHG + CGH = 180° 

60° + 50° + CGH = 180° 

CGH = 70° 

 

Now, CGH + AGH = 180° (linear pair)

70° + AGH = 180° 

AGH = 110° 

Solution 18

Produce CD to cut AB at E.

Now, in BDE, we have,

Exterior CDB = CEB + DBE

xo = CEB + 45o     .....(i)

In AEC, we have,

Exterior CEB = CAB + ACE

= 55o + 30o = 85o

Putting CEB = 85o in (i), we get,

xo = 85o + 45o = 130o

x = 130

Solution 19

The angle BAC is divided by AD in the ratio 1 : 3.

Let BAD and DAC be y and 3y, respectively.

As BAE is a straight line,

BAC + CAE = 180o        [linear pair]

BAD + DAC +  CAE = 180o

y + 3y + 108o = 180o

4y = 180o - 108o = 72o

Now, in ABC,

ABC + BCA + BAC = 180o

y + x + 4y = 180o

[Since, ABC = BAD (given AD = DB) and BAC = y + 3y = 4y]

5y + x = 180

5 18 + x = 180

90 + x = 180

x = 180 - 90 = 90

Solution 20

Given : A ABC in which BC, CA and AB are produced to D, E and F respectively.

To prove : Exterior DCA + Exterior BAE + Exterior FBD = 360o

Proof : Exterior DCA = A + B(i)

Exterior FAE = B + C(ii)

Exterior FBD = A + C(iii)

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

Ext. DCA + Ext. FAE + Ext. FBD

= A + B + B + C + A + C

= 2A +2B + 2C

= 2 (A + B + C)

= 2 180o

[Since, in triangle the sum of all three angle is 180o]

= 360o

Hence, proved.

Solution 21

In ACE, we have,

A + C + E = 180o (i)

In BDF, we have,

B + D + F = 180o (ii)

Adding both sides of (i) and (ii), we get,

A + C+E + B + D + F = 180o + 180o

A + B + C + D + E + F = 360o.

Solution 22

In ΔABC, by angle sum property,

A + B + C = 180° 

A + 70° + 20° = 180° 

A = 90° 

 

In ΔABM, by angle sum property,

BAM + ABM + AMB = 180° 

BAM + 70° + 90° = 180° 

BAM = 20° 

 

Since AN is the bisector of A,

 

Now, MAN + BAM = BAN

MAN + 20° = 45° 

MAN = 25° 

Solution 23

BAD EF and EC is the transversal.

AEF = CAD (corresponding angles)

CAD = 55° 

 

Now, CAD + CAB = 180° (linear pair)

55° + CAB = 180° 

CAB = 125° 

 

In ΔABC, by angle sum property,

ABC + CAB + ACB = 180° 

ABC + 125° + 25° = 180° 

ABC = 30° 

Solution 24

In the given ABC, we have,

A : B : C = 3 : 2 : 1

Let A = 3x, B = 2x, C = x. Then,

A + B + C = 180o

3x + 2x + x = 180o

6x = 180o

x = 30o

A = 3x = 3 30o = 90o

B = 2x = 2 30o = 60o

and, C = x = 30o

Now, in ABC, we have,

Ext ACE = A + B = 90o + 60o = 150o

ACD + ECD = 150o

 ECD = 150o - ACD 

 ECD = 150o - 90o    [since ]

  ECD= 60o


Solution 25

FGH + FGE = 180° (linear pair)

120° + y = 180° 

y = 60° 

 

AB DF and BD is the transversal.

ABC = CDE (alternate angles)

CDE = 50° 

 

BD FG and DF is the transversal.

EFG = CDE (alternate angles)

EFG = 50° 

 

In ΔEFG, by angle sum property,

FEG + FGE + EFG = 180° 

x + y + 50° = 180° 

x + 60° + 50° = 180° 

x = 70° 

Solution 26

AB CD and EF is the transversal.

AEF = EFD (alternate angles)

AEF = EFG + DFG

65° = EFG + 30° 

EFG = 35° 

 

In ΔGEF, by angle sum property,

GEF + EGF + EFG = 180° 

x + 90° + 35° = 180° 

x = 55° 

Solution 27

AB CD and AE is the transversal.

BAE = DOE (corresponding angles)

DOE = 65° 

 

Now, DOE + COE = 180° (linear pair)

65° + COE = 180° 

COE = 115° 

 

In ΔOCE, by angle sum property,

OEC + ECO + COE = 180° 

20° + ECO + 115° = 180° 

ECO = 45° 

Solution 28

AB CD and EF is the transversal.

EGB = GHD (corresponding angles)

GHD = 35° 

 

Now, GHD = QHP (vertically opposite angles)

QHP = 35° 

 

In DQHP, by angle sum property,

PQH + QHP + QPH = 180° 

PQH + 35° + 90° = 180° 

PQH = 55° 

Solution 29

AB CD and GE is the transversal.

EGF + GED = 180° (interior angles are supplementary)

EGF + 130° = 180° 

EGF = 50° 

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 8 - Triangles 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 8 - Triangles now.

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