RD Sharma solution class 7 chapter 14 Lines and Angles Objective Type Exercise

Objective Type Exercise 

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Question 1:

The sum of an angle and one third of its supplementary angle is 90°. The measure of the angle is
(a) 135°
(b) 120°
(c) 60°
(d) 45°

Answer 1:

Let the required angle be x.
Now, supplementary of the required angle = 180^∘ − x
Then,
$$\begin{gathered} x+\frac{1}{3}\left(180°-x\right)=90° \\ \Rightarrow 3x+180°-x=270° \\ \Rightarrow 2x=90° \\ \Rightarrow x=45° \end{gathered}$$
Hence, the correct answer is option (d).

Question 2:

If angles of a linear pair are equal, then the measure of each angle is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 2:

Let the required angle be x
Now, Sum of linear pair angles = 180^∘
⇒ x + x = 180^∘
⇒ 2x = 180^∘
⇒ x = 90^∘
Hence, the correct answer is option (d).

Question 3:

Two complemntary angles are in the ratio 2 : 3. The measure of the larger angle is
(a) 60°
(b) 54°
(c) 66°
(d) 48°

Answer 3:

Let the angles be 2x and 3x.
Now, 2x + 3x = 90^∘
⇒ 5x = 90^∘
⇒ x = 18^∘ 
∴ Larger angle = 3x = 3 × 18^∘ = 54^∘ 
Hence, the correct answer is option (b).

Question 4:

An angle is thrice its supplement. The measure of the angle is
(a) 120°
(b) 105°
(c) 135°
(d) 150°

Answer 4:

Let the required angle be x. 
Then,
$$\begin{gathered} x=3\left(180°-x\right) \\ \Rightarrow x=540°-3x \\ \Rightarrow 4x=540° \\ \Rightarrow x=135° \end{gathered}$$
Hence, the correct answer is option (c).

Question 5:

In Fig. 88 PR is a straight line and ∠PQS : ∠SQR = 7 : 5. The measure of ∠SQR is
(a) 60°
(b) ${\left(62\frac{1}{2}\right)}^{°}$
(c) ${\left(67\frac{1}{2}\right)}^{°}$
(d) 75°

Answer 5:

Let the measures of the angle ∠PQS and ∠SQR be 7x and 5x.
Now, ∠PQS + ∠SQR = 180^∘             [Linear pair angles]
⇒ 7x + 5x = 180^∘
⇒ 12x = 180^∘
⇒ x = 15^∘ 
∴ ∠SQR = 5x = 5 × 15^∘ = 75^∘ 
Hence, the correct answer is option (d).

Question 6:

The sum of an angle and half of its complementary angle is 75°. The measure of the angle is
(a) 40°
(b) 50°
(c) 60°
(d) 80°

Answer 6:

Let the required angle be x.
Now, complementnary of the required angle = 90^∘ − x
Then,
$$\begin{gathered} x+\frac{1}{2}\left(90°-x\right)=75° \\ \Rightarrow 2x+90°-x=150° \\ \Rightarrow x=150-90° \\ \Rightarrow x=60° \end{gathered}$$
Hence, the correct answer is option (c).

Question 7:

∠A is an obtuse angle. The measure of ∠A and twice its supplementary differ by 30°. Then ∠A can be
(a) 150°
(b) 110°
(c) 140°
(d) 120°

Answer 7:

Supplementary of ∠A = 180^∘ − ∠A
Now,
∠A + 30^∘ = 2(180^∘ − ∠A)
⇒ ∠A + 30^∘ = 360^∘ − 2∠A
⇒ 3∠A = 360^∘ − 30^∘
⇒ 3∠A = 330^∘
⇒ ∠A = 110^∘
Hence, the correct answer is option (b).

Question 8:

An angle is double of its supplement. The measure of the angle is
(a) 60°
(b) 120°
(c) 40°
(d) 80°

Answer 8:

Let the required angle be x.
Now, supplementary of the required angle = 180^∘ − x
Then,
$$\begin{gathered} x=2\left(180°-x\right) \\ \Rightarrow x=360°-2x \\ \Rightarrow 3x=360° \\ \Rightarrow x=120° \end{gathered}$$
Hence, the correct answer is option (b).

Question 9:

The measure of an angle which is its own complement is
(a) 30°
(b) 60°
(c) 90°
(d) 45°

Answer 9:

Let the required angle be x.
Now, complementary of the required angle = 90^∘ − x
Then,
$$\begin{gathered} x=90°-x \\ \Rightarrow x=90°-x \\ \Rightarrow 2x=90° \\ \Rightarrow x=45° \end{gathered}$$
Hence, the correct answer is option (d).

Question 10:

Two supplementary angles are in the ratio 3 : 2. The smaller angle measures
(a) 108°
(b) 81°
(c) 72°
(d) 68°

Answer 10:

Let the angles be 3x and 2x.
Now, 3x + 2x = 180^∘
⇒ 5x = 180^∘
⇒ x = 36^∘ 
∴ Smaller angle = 2x = 2 × 36^∘ = 72^∘ 
Hence, the correct answer is option (c).

Question 11:

In Fig. 89, the value of x is
(a) 75
(b) 65
(c) 45
(d) 55

Answer 11:

∠AOC and ∠BOC = 180^∘             [∵ Linear pair angles]
⇒ 44^∘+ (2x + 6)^∘ = 180^∘
⇒ (2x + 6)^∘ = 136^∘
⇒ 2x + 6 = 136
⇒ 2x = 130
⇒ x = 65
Hence, the correct answer is option (b).

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Question 12:

In Fig. 90, AOB is a straight line and the ray OCstands on it. The value of x is
(a) 16
(b) 26
(c) 36
(d) 46

Answer 12:

∠AOC + ∠BOC = 180^∘             [∵ Linear pair angles]
⇒ (2x + 15)^∘ + (3x + 35)^∘ = 180^∘
⇒ (5x + 50)^∘ = 180^∘
⇒ 5x + 50 = 180
⇒ 5x = 130
⇒ x = 26
Hence, the correct answer is option (b).

Question 13:

In Fig. 91, AOB is a straight line and 4x = 5y. The value of x is
(a) 100
(b) 105
(c) 110
(d) 115

Answer 13:

∠AOC + ∠BOC = 180^∘             [∵ Linear pair angles]
⇒ y^∘ + x^∘ = 180^∘
⇒ y + x = 180
$$\begin{gathered} \Rightarrow \frac{4x}{5}+x=180 \left[\because 4x=5y\Rightarrow y=\frac{4x}{5}\right] \\ \Rightarrow 4x+5x=180\times 5 \\ \Rightarrow 9x=180\times 5 \\ \Rightarrow x=100 \end{gathered}$$
Hence, the correct answer is option (a).

Question 14:

In Fig. 92, AOB is a straight line such that ∠AOC = (3x + 10)°, ∠COD = 50° and ∠BOD = (x − 8)°. The value of x is
(a) 32
(b) 36
(c) 42
(d) 52

Answer 14:

∠AOC + ∠COD + ∠BOD = 180^∘             [AOB is a straight line]
⇒ (3x + 10)^∘ + 50^∘ + (x − 8)^∘ = 180^∘
⇒ 3x + 10 + 50 + x − 8 = 180
⇒ 4x + 52 = 180
⇒ 4x = 128
⇒ x = 32
Hence, the correct answer is option (a).

Question 15:

In Fig. 93, if AOC is a straight line, then x =
(a) 42°
(b) 52°
(c) 142°
(d) 38°

Answer 15:

∠AOD + ∠DOB + ∠BOC = 180^∘             [∵ AOC is a straight line]
⇒ 38^∘ + x + 90^∘ = 180^∘
⇒ x + 128^∘ = 180^∘
⇒ x = 52^∘
Hence, the correct answer is option (b).

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Question 16:

In Fig. 94, if  ∠AOC is a straight line, then the value of x is
(a) 15
(b) 18
(c) 20
(d) 16

Answer 16:

∠AOD + ∠DOB + ∠BOC = 180^∘             [ AOC is a straight line]
⇒ 2x^∘ + 90^∘ + 3x^∘ = 180^∘
⇒ 5x^∘ + 90^∘ = 180^∘
⇒ 5x = 90
⇒ x = 18
Hence, the correct answer is option (b).

Question 17:

In Fig. 95, if AB, CD and EF are straight lines, then x =
(a) 5
(b) 10
(c) 20
(d) 30

Answer 17:

Let all the lines intersect at O.


∠COF = ∠DOE = 4x^∘                              [Vertically opposite angles]
∠AOC + ∠COF + ∠BOF = 180^∘             [AOB is a straight line]
⇒ 2x^∘ + 4x^∘ + 3x^∘ = 180^∘
⇒ 9x^∘ = 180^∘
⇒ 9x = 180
⇒ x = 20
Hence, the correct answer is option (c).

Question 18:

In Fig. 96, if AB, CD and EF are straight lines, then x + y + z =
(a) 180
(b) 203
(c) 213
(d) 134

Answer 18:

∠DAE + ∠BAD + ∠BAF = 180^∘             [EAF is a straight line]
⇒ 3x^∘ + 49^∘ + 62^∘ = 180^∘
⇒ 3x^∘ + 111^∘ = 180^∘
⇒ 3x^∘ = 69^∘
⇒ 3x = 69
⇒ x = 23
Now, ∠CAE + ∠CAF = 180^∘             [∵ EAF is a straight line]
⇒ z^∘ + y^∘ = 180^∘
⇒ z + y = 180
Now, x + y + z = 23 + 180 = 203
Hence, the correct answer is option (b).

Question 19:

In Fig. 97, if AB is parallel to CD, then the value of ∠BPE is
(a) 106°
(b) 76°
(c) 74°
(d) 84°

Answer 19:

Since, AB || CD
∴ ∠BPQ = ∠PQC           [Alternate interior angles]
⇒ (3x + 34)^∘ = (5x − 14)^∘
⇒ 3x + 34 = 5x − 14
⇒ 48 = 2x
⇒ x = 24
∴ ∠BPQ = (3 × 24 + 34)^∘ = 106^∘
∠BPQ + ∠BPE = 180^∘             [EF is a straight line]
⇒ 106^∘ + ∠BPE = 180^∘
⇒ ∠BPE = 74^∘
Hence, the correct answer is option (c).

Page-14.29

Question 20:

In Fig. 98, if AB is parallel to CO and EF is a transversal, then x =
(a) 19
(b) 29
(c) 39
(d) 49

Answer 20:

Let the line EF intersect AB and CD at P and Q respectively.


Since, AB || CD
∴ ∠BPQ + ∠PQD = 180^∘         (Angles on the same side of a transversal line are supplementary)
⇒ (7x − 12)^∘ + (4x + 17)^∘ = 180^∘
⇒ 7x − 12 + 4x + 17 = 180
⇒ 11x + 5 = 180
⇒ 11x = 175
⇒ x = 15.90

Disclaimer: No option is correct.

Question 21:

In Fig. 99, AB || CD and EF is a transversal intersecting ABand CO at Pand Q respectively. The measure of  ∠DPQ is
(a) 100^∘
(b) 80^∘
(c) 110^∘
(d) 70^∘

Answer 21:

∠BQF = ∠AQP = (4x)^∘             [Vertically opposite angles]
Since, AB || CD
∴ ∠AQP + ∠CPQ = 180^∘         [Angles on the same side of a transversal line are supplementary]
⇒ (4x)^∘ + (5x)^∘ = 180^∘
⇒ 9x  = 180
⇒ x = 20
∴ ∠BQF = (4 × 20)^∘ = 80^∘
 Now, ∠BQF = ∠DPQ = 80^∘          [Corresponding angles]
Hence, the correct answer is option (b).

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Question 22:

In Fig. 100, AB || CO and EF is a transversal intersecting AB and CD at P and Q respective. The measure of ∠OOP is
(a) 65
(b) 25
(c) 115
(d) 105

Answer 22:

∠BPE = ∠APQ = (5x − 10)^∘        [Vertically opposite angles]
Since, AB || CD
∴ ∠APQ + ∠CQP = 180^∘            [Angles on the same side of a transversal line are supplementary]
⇒ (5x − 10)^∘ + (3x − 10)^∘ = 180^∘
⇒ 8x − 20  = 180
⇒ 8x = 200
⇒ x = 25
∴ ∠BPE = (5 × 25 − 10)^∘ = 115^∘
 Now, ∠BPE = ∠DQP = 115^∘          [Corresponding angles]
Hence, the correct answer is option (c).

Question 23:

In Fig. 101, AB || CD and EF is a transversal. The value of y − x is
(a) 30
(b) 35
(c) 95
(d) 25

Answer 23:

Since, AB || CD
∴ ∠BPQ = ∠DQF         [Corresponding angles]
⇒ (5x − 20)^∘ = (3x + 40)^∘
⇒ 5x − 20 = 3x + 40
⇒ 2x = 60
⇒ x = 30
∴ ∠BPQ = (5 × 30 − 20 )^∘ = 130^∘
Now, ∠APE  = ∠BPQ           [Vertically opposite angles]
⇒ 2y^∘ = 130^∘
⇒ y = 65
∴ y − x = 65 − 30 = 35
Hence, the correct answer is option (b).

Question 24:

In Fig. 102, AB || CD || EF, ∠ABG = 110°, ∠GCO = 100° and ∠BGC = x°. The value of x is
(a) 35
(b) 50
(c) 30
(d) 40

Answer 24:

Since, AB || EG
∴ ∠ABG + ∠EGB = 180^∘         (Angles on the same side of a transversal line are supplementary)
⇒ 110^∘ + ∠EGB = 180^∘
⇒ ∠EGB = 70^∘
Again, CD || GF
∴ ∠DCG + ∠FGC = 180^∘         (Angles on the same side of a transversal line are supplementary)
⇒ 100^∘ + ∠FGC = 180^∘
⇒ ∠FGC = 80^∘
Now, ∠EGB + ∠BGC +∠FGC = 180^∘  
⇒ 70^∘ + x^∘ + 80^∘ = 180^∘
⇒ 150^∘+ x^∘ = 180^∘
⇒ x^∘ = 30^∘
⇒ x = 30
Hence, the correct answer is option (c).

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Question 25:

In Fig. 103, PO || RS and ∠PAB = 60° and ∠ACS = 100°. Then, ∠BAC =
(a) 40°
(b) 60°
(c) 80°
(d) 50°

Answer 25:

Since, PQ || RS
∴ ∠PAC = ∠ACS = 100^∘      [Corresponding angles]
Now, ∠PAC = 100^∘
⇒ ∠PAB + ∠BAC = 100^∘ 
⇒ 60^∘ + ∠BAC = 100^∘ 
⇒ ∠BAC = 40^∘ 
Hence, the correct answer is option (a).

Question 26:

In Fig. 104, AB || CO, ∠OAB = 150° and ∠OCO = 120°. Then, ∠AOC =
(a) 80°
(b) 90°
(c) 70°
(d) 100°

Answer 26:

Construction: Draw a line OE from the point O parallel to AB and CD


Since, AB || OE
∴ ∠BAO + ∠AOE = 180^∘         [Angles on the same side of a transversal line are supplementary]
⇒ 150^∘ + ∠AOE = 180^∘
⇒ ∠AOE = 30^∘
Again, CD || OE
∴ ∠DCO + ∠COE = 180^∘         [Angles on the same side of a transversal line are supplementary]
⇒ 120^∘ + ∠COE = 180^∘
⇒ ∠COE = 60^∘
Now, ∠AOC = ∠AOE + ∠COE
= 30^∘ + 60^∘
= 90^∘
Hence, the correct answer is option (b).

Question 27:

In Fig. 105, if AOB and COD are straight lines. Then, x + y =
(a) 120
(b) 140
(c) 100
(d) 160

Answer 27:

∠AOD + ∠BOD = 180^∘         [Linear pair angles]
⇒ (7x − 20)^∘ + 3x^∘ = 180^∘
⇒ 7x − 20 + 3x = 180
⇒ 10x = 200
⇒ x = 20
∴ ∠AOD = (7 × 20 − 20)^∘ = 120^∘
Now∠AOD = ∠BOC = 120^∘                [Vertically opposite angles]
∴ y = 120
Now, x + y = 20 + 120
= 140
Hence, the correct answer is option (b).

Question 28:

In Fig. 106, the value of x is
(a) 22
(b) 20
(c) 21
(d) 24

Answer 28:

(8x − 41)^∘ + (3x)^∘ + (3x + 10)^∘ + (4x − 5)^∘= 360^∘
⇒ 8x − 41 + 3x + 3x + 10 + 4x − 5 = 360
⇒ 18x − 36 = 360
⇒ 18x = 396
⇒ x = 22
Hence, the correct answer is option (a).

Page-14.32

Question 29:

In Fig. 107, if AOBand COD are straight lines, then
(a) x = 29, y = 100
(b) x = 110, y = 29
(c) x = 29, y = 110
(d) x = 39, y = 110

Answer 29:

∠AOD + ∠BOD = 180^∘         [Linear pair angles]
⇒ y^∘ + 70^∘ = 180^∘
⇒ y^∘ = 110^∘
⇒ y = 110
Now, ∠AOC = ∠BOD = 70^∘                [Vertically opposite angles]
Now, ∠AOC + ∠COE + ∠EOB + ∠BOD + ∠AOD = 360^∘            [Complete angle]
⇒ 70^∘ + 28^∘ + (3x − 5)^∘ + 70^∘ + 110^∘ = 360^∘
⇒ (3x)^∘ + 273^∘ = 360^∘
⇒ 3x = 87
⇒ x = 29
Hence, the correct answer is option (c).

Question 30:

In Fig. 108, if AB || CD then the value of x is
(a) 87
(b) 93
(c) 147
(d) 141

Answer 30:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠FCD + Reflex∠FCD = 360^∘         (Complete angle)
⇒ ∠FCD + 273^∘ =  360^∘
⇒ ∠FCD = 87^∘
Since, PQ || CD
∴∠QFC + ∠FCD = 180^∘                 (Angles on the same side of a transversal line are supplementary)
⇒ ∠QFC + 87^∘ = 180^∘
⇒ ∠QFC = 93^∘
Now, ∠ABF = ∠BFQ              (Corresponding angles)
= ∠BFC + ∠QFC
= 54^∘ + 93^∘
= 147^∘
∴ x^∘ = 147^∘
⇒ x = 147
Hence, the correct answer is option (c).

Question 31:

In Fig. 109, if AB || CD then the value of x is
(a) 34
(b) 124
(c) 24
(d) 158

Answer 31:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠QEC + ∠ECD = 180^∘                 [Angles on the same side of a transversal line are supplementary]
⇒ ∠QEC + 56^∘ = 180^∘
⇒ ∠QEC = 124^∘
Now, ∠BEQ + ∠QEC = ∠BEC      
⇒ ∠BEQ + 124^∘ = 158^∘
⇒ ∠BEQ = 34^∘
Now, ∠ABE = ∠BEQ = 34^∘              [Corresponding angles]
∴ x^∘ = 34^∘
⇒ x = 34
Hence, the correct answer is option (a).

Question 32:

In Fig. 110, if AB || CD. The value of x is
(a) 122
(b) 238
(c) 58
(d) 119

Answer 32:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
Since, PQ || CD
∴ ∠EFC = ∠FEQ = 37^∘                 [Alternate angles]
Now, ∠AEQ + ∠FEQ = ∠AEF    
⇒ ∠AEQ + 37^∘ = 95^∘
⇒ ∠AEQ = 58^∘
Since, PQ || AB
∴∠EAB + ∠AEQ = 180^∘                 [Angles on the same side of a transversal line are supplementary]
⇒ ∠EAB + 58^∘ = 180^∘
⇒ ∠EAB = 122^∘
∠EAB + Reflex∠EAB = 360^∘              [Complete angle]
∴ 122^∘ + (2x)^∘ = 360^∘
⇒ 2x = 238
⇒ x = 119
Hence, the correct answer is option (d).

Page-14.33

Question 33:

In Fig. 111, if AB || CO then x =
(a) 154
(b) 139
(c) 144
(d) 164

Answer 33:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
Since, PQ || AB
∴ ∠AME + ∠QEM = 180^∘                [Angles on the same side of a transversal line are supplementary]
⇒ 139^∘ + ∠QEM = 180^∘ 
⇒ ∠QEM = 41^∘
Now, ∠QEM + ∠DEQ = ∠MED   
⇒ 41^∘ + ∠DEQ = 67^∘
⇒ ∠DEQ = 26^∘
Now, ∠PED + ∠DEQ = 180^∘                 [Linear Pair angles]
⇒ ∠PED + 26^∘ = 180^∘
⇒ ∠PED = 154^∘
Since, PQ || AB
∴ x^∘ = ∠PED                                         [Corresponding angles]
⇒ x^∘ = 154^∘
⇒ x = 154
Hence, the correct answer is option (a).

Question 34:

In Fig. 112, if AB || CD, then x =
(a) 32
(b) 42
(c) 52
(d) 31

Answer 34:

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠CDP + Reflex∠CDP = 360^∘              [Complete angle]
∴∠CDP + 249^∘ = 360^∘
⇒ ∠CDP = 111^∘
Since, PQ || AB
∴ ∠BAP = ∠APQ                                         [Alternate angles]
⇒ ∠BAP = 28^∘
Now, ∠APQ + ∠QPD = ∠APD  
⇒ 28^∘ + ∠QPD = (2x + 13)^∘
⇒ ∠QPD = (2x + 13)^∘ − 28^∘
Since, PQ || CD
∴ ∠QPD + ∠CDP = 180^∘                [Angles on the same side of a transversal line are supplementary]
⇒ (2x + 13)^∘ − 28^∘ + 111^∘ = 180^∘ 
⇒ 2x + 13 − 28 + 111 = 180
⇒ 2x = 84
⇒ x = 42
Hence, the correct answer is option (b).

Question 35:

In Fig. 113 if AC || OF and AB || CE, then
(a) x = 145, y = 223
(c) x = 135, y = 233
(b) x = 223, y = 145
(d) x = 233, y = 135

Answer 35:

Construction: Produce FD towards D to the point M
∠DCA + Reflex∠DCA = 360^∘              [Complete angle]
∴∠DCA + (y + 15)^∘ = 360^∘
⇒ ∠DCA = 345^∘ − y^∘
Now,
∠MDC = ∠EDF = 58^∘                                    [Vertically Opposite angles]
Since, MF || AC
∴ ∠MDC + ∠QPD = 180^∘                                [Angles on the same side of a transversal line are supplementary]
⇒ 58^∘ + 345^∘ − y^∘ = 180^∘
⇒ y = 223
∴ ∠DCA = 345^∘ − 223^∘ = 122^∘
Again, ∠BAC + Reflex∠BAC = 360^∘              [Complete angle]
∴∠BAC + (2x + 12)^∘ = 360^∘
⇒ ∠DCA = 348^∘ − (2x)^∘
Since, AB || CD
∴ ∠DCA + ∠DCA = 180^∘                [Angles on the same side of a transversal line are supplementary]
⇒ 348^∘ − (2x)^∘ + 122^∘ = 180^∘ 
⇒ (2x)^∘ = 290^∘
⇒ x = 145
Hence, the correct answer is option (a).