Chapter 8 : Triangles - R S Aggarwal And V Aggarwal Solutions for Class 9 Maths CBSE

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Chapter 8 - Triangles Excercise MCQ

Question 1

Solution 1

Question 2

In a ΔABC, if A - B = 42° and B - C = 21° then B = ?

(a) 32° 

(b) 63° 

(c) 53° 

(d) 95° 

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° 

Question 3

In a ΔABC, side BC is produced to D. If ABC = 50° and ACD = 110° then A = ?

 

  

(a) 160° 

(b) 60° 

(c) 80° 

(d) 30° 

Solution 3

Correct option: (b)

ACD = B + A (Exterior angle property)

110° = 50° + A

A = 60° 

Question 4

Side BC of ΔABC has been produced to D on left hand side and to E on right hand side such that ABD = 125° and ACE = 130°. Then A = ?

  1. 50°
  2. 55°
  3. 65°
  4. 75°

 

Solution 4

Correct option: (d)

Question 5

In the given figure, sides CB and BA of ΔABC have been produced to D and E respectively such that ABD = 110° and CAE = 135°. Than ACB =?

  1. 65°
  2. 45°
  3. 55°
  4. 35°

Solution 5

Correct option: (a)

Question 6

The sides BC, CA and AB of ΔABC have been produced to D,E and F respectively.  ∠BAE + ∠CBF + ACD =?

  1. 240°
  2. 300°
  3. 320°
  4. 360°

Solution 6

Question 7

In the given figure, EAD BCD. Ray FAC cuts ray EAD at a point A such that EAF = 30°. Also, in ΔBAC, BAC = x° and ABC = (x + 10)°. Then, the value of x is

 

  

(a) 20

(b) 25

(c) 30

(d) 35

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° 

Question 8

In the given figure, two rays BD and CE intersect at a point A. The side BC of ΔABC have been produced on both sides to points F and G respectively. If ABF = x°, ACG = y° and DAE = z° then z = ?

 

 

  

(a) x + y - 180

(b) x + y + 180

(c) 180 - (x + y)

(d) x + y + 360° 

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°  

Question 9

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively. such that OAE = x° and DBF = y°.

 

  

 

If OCA = 80°, COA = 40° and BDO = 70° then x° + y° = ?

(a) 190° 

(b) 230° 

(c) 210° 

(d) 270° 

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° 

Question 10

In a ΔABC it is given that A:B:C = 3:2:1 and ∠ACD = 90o. If is produced to E Then ECD =?

  1. 60°
  2. 50°
  3. 40°
  4. 25°

Solution 10

Question 11

In the given figure , BO and CO are the bisectors of B and C respectively. If A = 50°, then BOC= ?

  1. 130°
  2. 100°
  3. 115°
  4. 120°

Solution 11

Question 12

In the given figure, side BC of ΔABC has been produced to a point D. If A = 3y°, B = x°, C = 5y° and ACD = 7y°. Then, the value of x is

 

  

(a) 60

(b) 50

(c) 45

(d) 35

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° 

Chapter 8 - Triangles Excercise Ex. 8

Question 1

In ABC, if B = 76o and C = 48o, find A.

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

Question 2

The angles of a triangle are in the ratio 2:3:4. Find the angles.

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

Question 3

In ABC, if 3A = 4B = 6C, calculate A, B and C.

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 =

Question 4

In ABC, if A + B = 108o and B + C = 130o, find A, B and 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.

Question 5

In ABC, A + B = 125o and A + C = 113o. Find A, B and C.

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.

Question 6

In PQR, if P - Q = 42o and Q - R = 21o, find P, Q and R.

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.

Question 7

The sum of two angles of a triangle is 116o and their difference is 24o. Find the measure of each angle of the triangle.

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.

Question 8

Two angles of a triangle are equal and the third angle is greater than each one of them by 18o. Find the angles.

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.

Question 9

Of the three angles of triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

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.

Question 10

In a right-angled triangle, one of the acute angles measures 53o. Find the measure of each angle of the triangle.

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.

Question 11

In a right-angled triangle, one of the acute angles measures 53o. Find the measure of each angle of the triangle.

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.

Question 12

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

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

begin mathsize 12px style rightwards double arrow end stylebegin mathsize 12px style angle end styleA < 180o [Since begin mathsize 12px style angle end styleA + begin mathsize 12px style angle end styleB + begin mathsize 12px style angle end styleC = 180o]

begin mathsize 12px style rightwards double arrow angle straight A less than 180 to the power of straight o over 2 equals 90 to the power of straight o end style 

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

begin mathsize 12px style rightwards double arrow end stylebegin mathsize 12px style angle end styleB < 90o

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

begin mathsize 12px style rightwards double arrow end stylebegin mathsize 12px style angle end styleC < 90o

begin mathsize 12px style therefore increment end styleABC is an acute angled triangle.

Question 13

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse angled.

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.

Question 14

In the given figure, side BC of ABC is produced to D. If ACD = 128o and ABC = 43o, find BAC and ACB.

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.

Question 15

In the given figure, the side BC of ABC has been produced on the right-hand side from B to D and  on the right-hand side from C and E. If ABD = 106o and ACE= 118o, find the measure of each angle of the triangle.

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

Question 16

Calculate the value of x in each of the following figures.

(i)

(ii)

(iii)

Given: AB || CD

(vi)

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

Question 17

In the figure given alongside, AB CD, EF BC, BAC = 60° and DHF = 50°. Find GCH and AGH.

 

  

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° 

Question 18

Calculate the value of x in the given figure.

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

Question 19

In the given figure, AD divides BAC in the ratio 1: 3 and AD = DB. Determine the value of x.

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

Question 20

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so fomed is equal to four right angles.

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.

Question 21

In the given figure, show that

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

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.

Question 22

In the given figure, AM BC and AN is the bisector of A. If ABC = 70° and ACB = 20°, find MAN.

 

  

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° 

Question 23

In the given figure, BAD EF, AEF = 55° and ACB = 25°, find ABC.

 

  

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° 

Question 24

In the given figure, ABC is a triangle in which A : B : C = 3 : 2 : 1 and AC CD. Find the measure of

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


Question 25

In the given figure, AB DE and BD FG such that ABC = 50° and FGH = 120°. Find the values of x and y.

  

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° 

Question 26

In the given figure, AB CD and EF is a transversal. If AEF = 65°, DFG = 30°, EGF = 90° and GEF = x°, Find the value of x.

 

  

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° 

Question 27

In the given figure, AB CD, BAE = 65° and OEC = 20°. Find ECO.

  

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° 

Question 28

In the given figure, AB CD and EF is a transversal, cutting them at G and H respectively. If EGB = 35° and QP EF, find the measure of PQH.

  

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° 

Question 29

In the given figure, AB CD and EF AB. If EG is the transversal such that GED = 130°, find EGF.

 

  

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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