RS Agarwal Solution | Class 10th | Chapter-7 |Triangles | Edugrown

Exercise 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° 

⇒ 3∠B = 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) 35Solution 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 = 90^o. If it 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) 35Solution 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° 

Exercise Ex. 8

Question 1

In ABC, if B = 76^o and C = 48^o, find A.Solution 1

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

A + B + C = 180^o

 A + 76^o + 48^o = 180^o

 A = 180^o – 124^o = 56^o

 A = 56^oQuestion 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 180^o ]     

 9x = 180

 The measures of the required angles are:

2x = (2  20)^o = 40^o

3x = (3  20)^o = 60^o

4x = (4  20)^o = 80^oQuestion 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 = 180^o

 A = 

B = 

C = Question 4

In ABC, if A + B = 108^o and B + C = 130^o, find A, B and C.Solution 4

A + B = 108^o [Given]

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

A + B + C = 180^o

 108^o + C = 180^o

 C = 180^o – 108^o = 72^o

Also, B + C = 130^o [Given]

 B + 72^o = 130^o

 B = 130^o – 72^o = 58^o

Now as, A + B = 108^o

 A + 58^o = 108^o

 A = 108^o – 58^o = 50^o

 A = 50^o, B = 58^o and C = 72^o.Question 5

In ABC, A + B = 125^o and A + C = 113^o. Find A, B and C.Solution 5

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

So, A + B + C = 180^o

Now, A + B = 125^o [Given]

 125^o + C = 180^o

 C = 180^o – 125^o = 55^o

Also, A + C = 113^o [Given]

 A + 55^o = 113^o

 A = 113^o – 55^o = 58^o

Now as A + B = 125^o

 58^o + B = 125^o

 B = 125^o – 58^o = 67^o

 A = 58^o, B = 67^o and C = 55^o.Question 6

In PQR, if P – Q = 42^o and Q – R = 21^o, find P, Q and R.Solution 6

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

So,P + Q + R = 180^o(i)

Now,P – Q = 42^o[Given]

P = 42^o + Q(ii)

andQ – R = 21^o[Given]

R = Q – 21^o(iii)

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

42^o + Q + Q + Q – 21^o = 180^o

3Q + 21^o = 180^o

3Q = 180^o – 21^o = 159^o

Q = 

P = 42^o + Q

= 42^o + 53^o = 95^o

R = Q – 21^o

= 53^o – 21^o = 32^o

P = 95^o, Q = 53^o and R = 32^o.Question 7

The sum of two angles of a triangle is 116^o and their difference is 24^o. 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 116^o, i.e., A + B = 116^o.

Since, A + B + C = 180^o

So, 116^o + C = 180^o

 C = 180^o – 116^o = 64^o

Also, it is given that:

A – B = 24^o

 A = 24^o + B

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

24^o + B + B = 116^o

 2B + 24^o = 116^o

 2B = 116^o – 24^o = 92^o

 B = 

Therefore, A = 24^o + 46^o = 70^o

 A = 70^o, B = 46^o and C = 64^o.Question 8

Two angles of a triangle are equal and the third angle is greater than each one of them by 18^o. Find the angles.Solution 8

Let the two equal angles, A and B, of the triangle be x^o each.

We know,

A + B + C = 180^o

x^o + x^o + C = 180^o

2x^o + C = 180^o(i)

Also, it is given that,

C = x^o + 18^o(ii)

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

2x^o + x^o + 18^o = 180^o

3x^o = 180^o – 18^o = 162^o

x = 

Thus, the required angles of the triangle are 54^o, 54^o and x^o + 18^o = 54^o + 18^o = 72^o.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 = 180^o

 2C + 3C + C = 180^o

 6C = 180^o

 C = 30^o

So, A = 2C = 2  30^o = 60^o

B = 3C = 3  30^o = 90^o

 The required angles of the triangle are 60^o, 90^o, 30^o.Question 10

In a right-angled triangle, one of the acute angles measures 53^o. Find the measure of each angle of the triangle.Solution 10

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

Since, A + B + C = 180^o

 A + B = 180^o – C = 180^o – 90^o = 90^o

Suppose A = 53^o

Then, 53^o + B = 90^o

 B = 90^o – 53^o = 37^o

 The required angles are 53^o, 37^o and 90^o.Question 11

In a right-angled triangle, one of the acute angles measures 53^o. Find the measure of each angle of the triangle.Solution 11

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

Since, A + B + C = 180^o

 A + B = 180^o – C = 180^o – 90^o = 90^o

Suppose A = 53^o

Then, 53^o + B = 90^o

 B = 90^o – 53^o = 37^o

 The required angles are 53^o, 37^o and 90^o.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, A < B + C

Adding A to both sides of the inequality,

2 A < A + B + C

2 A < 180^o [Since A + B + C = 180^o]

Similarly, B <A + C

B < 90^o

and C < A + B

C < 90^o

ABC 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 = 180^o

 A + C = 180^o – B

_{Therefore, we get,}

B > 180^o – B

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

B + B > 180^o – B + B

 2B > 180^o

 B > 

i.e., B > 90^o 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 = 128^o and ABC = 43^o, find BAC and ACB.

Solution 14

Since ACB and ACD form a linear pair.

So, ACB + ACD = 180^o

 ACB + 128^o = 180^o

 ACB = 180^o – 128 = 52^o

Also, ABC + ACB + BAC = 180^o

 43^o + 52^o + BAC = 180^o

 95^o + BAC = 180^o

 BAC = 180^o – 95^o = 85^o

 ACB = 52^o and BAC = 85^o.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 = 106^o and ACE= 118^o, find the measure of each angle of the triangle.

Solution 15

As DBA and ABC form a linear pair.

So,DBA + ABC = 180^o

106^o + ABC = 180^o

ABC = 180^o – 106^o = 74^o

Also, ACB and ACE form a linear pair.

So,ACB + ACE = 180^o

ACB + 118^o = 180^o

ACB = 180^o – 118^o = 62^o

In ABC, we have,

ABC + ACB + BAC = 180^o

74^o + 62^o + BAC = 180^o

136^o + BAC = 180^o

BAC = 180^o – 136^o = 44^o

In triangle ABC, A = 44^o, B = 74^o and C = 62^oQuestion 16

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

(i) 

(ii) 

(iii) 

Given: AB || CD

(vi) Solution 16

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

110^o + BAC = 180^o

 BAC = 180^o – 110^o = 70^o

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

 BCA + 120^o = 180^o

 BCA = 180^o – 120^o = 60^o

Now, in ABC,

ABC + BAC + ACB = 180^o

x^o + 70^o + 60^o = 180^o

 x + 130^o = 180^o

 x = 180^o – 130^o = 50^o

 x = 50

(ii)

In ABC,

A + B + C = 180^o

 30^o + 40^o + C = 180^o

 70^o + C = 180^o

 C = 180^o – 70^o = 110^o

Now BCA + ACD = 180^o [Linear pair]

 110^o + ACD = 180^o

 ACD = 180^o – 110^o = 70^o

In ECD,

ECD + CDE + CED = 180^o

 70^o + 50^o + CED = 180^o

 120^o + CED = 180^o

 CED = 180^o – 120^o = 60^o

Since AED and CED from a linear pair

So, AED + CED = 180^o

 x^o + 60^o = 180^o

 x^o = 180^o – 60^o = 120^o

 x = 120

(iii)

EAF = BAC [Vertically opposite angles]

 BAC = 60^o

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

So, ACD = BAC + ABC

 115^o = 60^o + x^o

 x^o = 115^o – 60^o = 55^o

 x = 55

(iv)

Since AB || CD and AD is a transversal.

So, BAD = ADC

 ADC = 60^o

In ECD, we have,

E + C + D = 180^o

 x^o + 45^o + 60^o = 180^o

 x^o + 105^o = 180^o

 x^o = 180^o – 105^o = 75^o

 x = 75

(v)

In AEF,

Exterior BED = EAF + EFA

 100^o = 40^o + EFA

 EFA = 100^o – 40^o = 60^o

Also, CFD = EFA [Vertically Opposite angles]

 CFD = 60^o

Now in FCD,

Exterior BCF = CFD + CDF

 90^o = 60^o + x^o

 x^o = 90^o – 60^o = 30^o

 x = 30

(vi)

In ABE, we have,

A + B + E = 180^o

 75^o + 65^o + E = 180^o

 140^o + E = 180^o

 E = 180^o – 140^o = 40^o

Now, CED = AEB [Vertically opposite angles]

 CED = 40^o

Now, in CED, we have,

C + E + D = 180^o

 110^o + 40^o + x^o = 180^o

 150^o + x^o = 180^o

 x^o = 180^o – 150^o = 30^o

 x = 30Question 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

 x^o = CEB + 45^o     …..(i)

In  AEC, we have,

Exterior CEB = CAB + ACE

= 55^o + 30^o = 85^o

Putting CEB = 85^o in (i), we get,

x^o = 85^o + 45^o = 130^o

 x = 130Question 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 = 180^o        [linear pair]

 BAD + DAC +  CAE = 180^o

 y + 3y + 108^o = 180^o

 4y = 180^o – 108^o = 72^o

Now, in ABC,

ABC + BCA + BAC = 180^o

y + x + 4y = 180^o

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

 5y + x = 180

 5  18 + x = 180

 90 + x = 180

 x = 180 – 90 = 90Question 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 = 360^o

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)

= 2 180^o

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

= 360^o

Hence, proved.Question 21

In the given figure, show that

A + B + C + D + E + F = 360^o.

Solution 21

In ACE, we have,

A + C + E = 180^o (i)

In BDF, we have,

B + D + F = 180^o (ii)

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

A + C+E + B + D + F = 180^o + 180^o

A + B + C + D + E + F = 360^o.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 = 180^o

 3x + 2x + x = 180^o

 6x = 180^o

 x = 30^o

 A = 3x = 3  30^o = 90^o

B = 2x = 2  30^o = 60^o

and, C = x = 30^o

Now, in ABC, we have,

Ext ACE = A + B = 90^o + 60^o = 150^o

 ACD + ECD = 150^o

 ECD = 150^o – ACD 

 ECD = 150^o – 90^o    [since , ]

  ECD= 60^o

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°