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

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.

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)

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, A < B + C

Adding A to both sides of the inequality,

A

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.

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.

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°

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