RD Sharma 2019 2020 Solutions for Class 7 Maths Chapter 14 Lines And Angles are provided here with simple step-by-step explanations. These solutions for Lines And Angles are extremely popular among class 7 students for Maths Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2019 2020 Book of class 7 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2019 2020 Solutions. All RD Sharma 2019 2020 Solutions for class 7 Maths are prepared by experts and are 100% accurate.

Page No 14.10:

Question 31:

In Fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.

Answer:

BOD + DOF + FOA = 180°        (Linear pair)
FOA = u = 180°-90°-50°=40°
FOA=x=40°    (Vertically opposite angles)
BOD=z=90°    (Vertically opposite angles)
EOC=y=50°    (Vertically opposite angles)

Page No 14.10:

Question 32:

In Fig., find the values of x, y and z.

Answer:

y=25°       Vertically opposite anglesSince x+y=180°         Linear pairx=180°-25°=155°z=x=155°        Vertically opposite angles



Page No 14.20:

Question 1:

In Fig., line n is a transversal to lines l and m. Identify the following:

(i) Alternate and corresponding angles in Fig. (i).
(ii) Angles alternate to ∠d and ∠g and angles corresponding to angles ∠f and ∠h in Fig. (ii).
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Fig. (iii).
(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. (ii).

Answer:

(i) Figure (i)
Corresponding angles:
EGB and GHD
HGB and FHD
EGA and GHC
AGH and CHF
Alternate angles:
EGB and CHF
HGB and CHG
EGA and FHD
AGH and GHD

(ii) Figure (ii)
Alternate angle to d is e.
Alternate angle to g is b.
Also,
Corresponding angle to f is c.
Corresponding angle to h is a.

(iii) Figure (iii)
Angle alternate to PQR is QRA.
Angle corresponding to RQF is ARB.
Angle alternate to POE is ARB.

(iv) Figure (ii)
Pair of interior angles are
a and e
d and f
Pair of exterior angles are
b and h
c and g

Page No 14.20:

Question 2:

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠CMQ = 60°, find all other angles in the figure.

Answer:

ALM = CMQ = 60°        (Corresponding angles)
LMD = CMQ = 60°        (Vertically opposite angles)
ALM = PLB = 60°          (Vertically opposite angles)
Since
CMQ + QMD = 180°     (Linear pair)
 QMD = 180°-60°=120°
QMD = MLB = 120°        (Corresponding angles)
QMD = CML = 120°        (Vertically opposite angles)
MLB = ALP = 120°          (Vertically opposite angles)

Page No 14.20:

Question 3:

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.

Answer:

In the given Fig., AB || CD.
ALM=LMD=35°     Alternate interior anglesSince PLA+ALM=180°     Linear pairPLA=180°-35°=145° 

Page No 14.20:

Question 4:

The line n is transversal to line l and m in Fig. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.

Answer:

In this given Fig., line l || m.
Here,
Alternate angle to 13 is 7.
Corresponding angle to 15 is 7.
Alternate angle to 15 is 5.



Page No 14.21:

Question 5:

In Fig., line l || m and n is a transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.

Answer:

In the given figure, l || m.
Here,
1+2=180°     Linear pair 2=180°-1=180°-40°=140°5=1=40°        Corresponding angles3=1=40°        Vertically opposite angles7=3=40°        Corresponding angles7=5=40°        Vertically opposite angles
Also,
2=6=140°        Corresponding angles2=4=140°        Vertically opposite angles4=8=140°        Corresponding angles8=6=40°          Vertically opposite angles
Thus,
2=8, 3=5, 6=4, 1=7
Hence, alternate angles are equal.

Page No 14.21:

Question 6:

In Fig., line l || m and a transversal n cuts them at P and Q respectively. If ∠1 = 75°, find all other angles.

Answer:

In the given figure, l || m, n is a transversal line and ∠1 = 75°.
Thus, we have:
1+2=180°        Linear pair2=180°-1=180°-75°=105°1=5=75°         Corresponding angles1=3=75°             Vertically opposite angles5=7=75°             Vertically opposite anglesNow, 2=6=105°         Corresponding angles6=8=105°          Vertically opposite angles2=4=105°          Vertically opposite angles

Page No 14.21:

Question 7:

In Fig., AB || CD and a transversal PQ cuts them at L and M respectively. If ∠QMD = 100°, find all other angles.

Answer:

In the given figure, AB || CD, PQ is a transversal line and QMD = 100°.
Thus, we have:
DMQ + QMC = 180°    (Linear pair)
QMC=180°-DMQ=180°-100°=80°
Thus,
DMQ = BLM = 100°         (Corresponding angles)
DMQ = CML = 100°         (Vertically opposite angles)
BLM = PLA = 100°           (Vertically opposite angles)
Also,
CMQ = ALM = 80°         (Corresponding angles)
CMQ = DML = 80°         (Vertically opposite angles)
ALM = PLB = 80°           (Vertically opposite angles)

Page No 14.21:

Question 8:

In Fig., l || m and p || q. Find the values of x, y, z, t.

Answer:

In the given figure, l || m and p || q.
Thus, we have:
z=80°                (Vertically opposite angles)
z=t=80°       (Corresponding angles)
z=y=80°       (Corresponding angles)
x=y=80°       (Corresponding angles)
 

Page No 14.21:

Question 9:

In Fig., line l || m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.

Answer:


In the given figure, ∠1 = 120° and ∠2 =100°.
Since l || m, so
2=5=100°              Alternate interior angles5+3=180°                Linear pair3=180°-5=180°-100°=80°          
Also,
1+6=180°          Linear pair6=180°-1=180°-120°=60°
We know that the sum of all the angles of triangle is 180°.
6+3+4=180°60°+80°+4=180°140°+4=180°4=180°-140°=40°

Page No 14.21:

Question 10:

In Fig., line l || m. Find the values of a, b, c, d. Give reasons.

Answer:

In the given figure, line l || m.
Thus, we have:
a=110°        Vertically opposite anglesb=a=110°               Corresponding anglesd=85°           Vertically opposite anglesc=d=85°                  Corresponding angles



Page No 14.22:

Question 11:

In Fig., AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

Answer:

In the given figure, AB || CD and t is a transversal line.
Now, let:
1=3x2=2x
Thus, we have:
1+2=180°      Linear pair 3x+2x=180°5x=180°x=180°5=36°Thus,1=3×36°=108°2=2×36°=72°
Now,
1=5=108°      Corresponding angles1=3=108°      Vertically opposite angles5=7=108°      Vertically opposite angles2=6=72°        Corresponding angles4=2=72°        Vertically opposite angles8=6=72°        Vertically opposite angles

Page No 14.22:

Question 12:

In Fig., l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.

Answer:

In the given figure, l || m || n and p is a transversal line.
Thus, we have:
4+60°=180°       Linear pair4=180°-60°=120°4=1=120°       Corresponding angles1=2=120°        Corresponding angles 3=2=120°         Vertically opposite anglesThus,1=2=3=120° 

Page No 14.22:

Question 13:

In Fig., if l || m || n and ∠1 = 60°, find ∠2.

Answer:

In the given figure, l || m || n and ∠1 = 60°.
Thus, we have:
3=1=60°     Corresponding angleNow,3+4=180°    Linear pair4=180°-3=180°-60°=120°2=4=120°      Alternate interior angles

Page No 14.22:

Question 14:

In Fig., if AB || CD and CD || EF, find ∠ACE.

Answer:

In the given figure, AB || CD and CD || EF.
Extend line CE to E'.

Thus, we have:
BAC=ACD=70°              Alternate anglesNow,3+CEF=180°                   Linear pair3=180°-CEF=180°-130°=50°Since CD||EF, then2=3=50°                Corresponding anglesACE=ACD-2=70°-50°=20°

Page No 14.22:

Question 15:

In Fig., if l || m, n || p and ∠1 = 85°, find ∠2.

Answer:



In the given figure, l || m, n || p and ∠1 = 85°.
Now, let ∠4 be the adjacent angle of ∠2.
Thus, we have:
3=1=85°        Corresponding angles
3+2=180°       (Sum of interior angles on the same side of the transversal)
2=180°-3=180°-85°=95°

Page No 14.22:

Question 16:

In Fig., a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l || m?

Answer:

We know that if the alternate exterior angles of two lines are equal, then the lines are parallel.
In the given figure, 1 and 7 are alternate exterior angles, but they are not equal.
 1 770°80°

Therefore, lines l and m are not parallel.



Page No 14.23:

Question 17:

In Fig., a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?

Answer:

 2 = 3 = 65°        (Vertically opposite angles)   
 8 = 6 = 65°         (Vertically opposite angles) 
∴ 3 = 6
l || m                       (Two lines are parallel if the alternate angles formed with the transversal are equal) 

Page No 14.23:

Question 18:

In Fig., show that AB || EF.

Answer:

Extend line CE to E'.


BAC=57°=22°+35°=ACE+ECD AB||CDHere, E'EF+FEC=180°    Linear pairE'EF=180°-FEC=180°-145°=35°=ECD EF||CDThus, AB||CD ||EF 

Page No 14.23:

Question 19:

In Fig., AB || CD. Find the values of x, y, z.

Answer:

x+125°=180°             (Linear pair)
x=180°-125°=55°

z=125°            (Corresponding angles)
x+z=180°   (Sum of adjacent interior angles is 180°)
x+125°=180°x=180°-125°=55°

x+y=180°   (Sum of adjacent interior angles is 180°)
55°+y=180°y=180°-55°=125°

Page No 14.23:

Question 20:

In Fig., find out ∠PXR, if PQ || RS.

Answer:

Draw a line parallel to PQ passing through X.


Here,
PQX=PXF=70° and SRX=RXF=50°      (Alternate interior angles)
∵ PQ || RS || XF
∴ PXR=PXF+FXR=70°+50°=120°

Page No 14.23:

Question 21:

In Fig., we have

(i) ∠MLY = 2 ∠LMQ, find ∠LMQ.
(ii) ∠XLM = (2x − 10)° and ∠LMQ = x + 30°, find x.
(iii) ∠XLM = ∠PML, find ∠ALY
(iv) ∠ALY = (2x − 15)°, and ∠LMQ = (x + 40)°, find x

Answer:

(i)
LMQ=ALY          Corresponding anglesMLY+ ALY=180°             Linear pair   2ALY+ALY=180°3ALY=180°ALY=180°3=60° LMQ=60°

(ii)
XLM=LMQ                Alternate interior angles2x-10°=x+30°2x-x=30°+10°x=40°

(iii)
ALX=LMP      Corresponding anglesALX+XLM=180°         Linear pairXLM=LMP         GivenLMP+LMP=180°2LMP=180° LMP=180° 2=90° XLM=LMP=90°ALY=XLM       Vertically opposite anglesALY=90°   

(iv)
ALY=LMQ          Corresponding angles2x-15°=x+40°2x-x=40°+15°x=55°

Page No 14.23:

Question 22:

In Fig., DE || BC. Find the values of x and y.

Answer:

ABC = DAB       (Alternate interior angles)
 x=40°

ACB = EAC       (Alternate interior angles)
 y=55°



Page No 14.24:

Question 23:

In Fig., line AC || line DE and ∠ABD = 32°. Find out the angles x and y if ∠E = 122°.

 

Answer:

BDE=ABD=32°            Alternate interior anglesBDE+y=180°      Linear pair 32°+y=180°y=180°-32°=148°

ABE=E=122°         (Alternate interior angle)ABD+DBE=122°32°+x=122°x=122°-32°=90°

Page No 14.24:

Question 24:

In Fig., side BC of ∆ABC has been produced to D and CE || BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD and ∠ACD.

Answer:

ABC = ECD = 55°          (Corresponding angles)
BAC = ACE = 65°          (Alternate interior angles)
Now, ACD = ACE + ECD
⇒ ACD = 55° + 65° = 120° 

Page No 14.24:

Question 25:

In Fig., line CAAB || line CR and line PR || line BD. Find ∠x, ∠y and ∠z.

Answer:

Since CA ⊥ AB,
x=90°
We know that the sum of all the angles of triangle is 180°.
In APQ,QAP+APQ+PQA=180°90°+APQ+20°=180°110°+APQ=180°APQ=180°-110°=70°
PBC = APQ = 70°            (Corresponding angles)
Since PRC+z=180°           Linear pair
z=180°-70°=110°    APQ=PRC   Alternate interior angles 

Page No 14.24:

Question 26:

In Fig., PQ || RS. Find the value of x.
   

Answer:





RCD+RCB=180° Linear pairRCB=180°-130°=50°In ABC, BAC+ABC+BCA=180°       Angle sum propertyBAC=180°-55°-50°=75°



 

Page No 14.24:

Question 27:

In Fig., AB || CD and AE || CF; ∠FCG = 90° and ∠BAC = 120°. Find the values of x, y and z.

Answer:

BAC = ACG = 120°          (Alternate interior angle)
∴ ACF + FCG = 120°  
ACF = 120° − 90° = 30°

DCA + ACG = 180°            (Linear pair)
x = 180° − 120° = 60°

BAC + BAE + EAC = 360°
CAE = 360° − 120° − (60° + 30°) = 150°             (BAE =  DCF)



Page No 14.25:

Question 28:

In Fig., AB || CD and AC || BD. Find the values of x, y, z.

Answer:

(i) Since AC || BD and CD || AB, ABCD is a parallelogram.
CAB + ACD = 180°     (Sum of adjacent angles of a parallelogram)
∴ ACD = 180° − 65° = 115°
CAD = CDB = 65°         (Opposite angles of a parallelogram)
ACD = DBA = 115°       (Opposite angles of a parallelogram)

(ii) Here,
AC || BD and CD || AB
DAC = x = 40°            (Alternate interior angle)
DAB = y = 35°            (Alternate interior angle)

Page No 14.25:

Question 29:

In Fig., state which lines are parallel and why?

Answer:

Let F be the point of intersection of line CD and the line passing through point E.



Since ACD and CDE are alternate and equal angles, so
ACD = 100° = CDE
∴ AC || EF

Page No 14.25:

Question 30:

In Fig. 87, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.

Answer:

   


Construction:
 Let G be the point of intersection of lines BC and DE.

∵ AB || DE and BC || EF

ABC=DGC=DEF=75°  (Corresponding angles)​



Page No 14.26:

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:

Let the required angle be x.
Now, supplementary of the required angle = 180∘ − x
Then,
x+13180°-x=90°3x+180°-x=270°2x=90°x=45°
Hence, the correct answer is option (d).

Page No 14.26:

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:

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).

Page No 14.26:

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:

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).

Page No 14.26:

Question 4:

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

Answer:

Let the required angle be x
Then,
x=3180°-xx=540°-3x4x=540°x=135°
Hence, the correct answer is option (c).

Page No 14.26:

Question 5:

In Fig. 88 PR is a straight line and ∠PQS : ∠SQR = 7 : 5. The measure of ∠SQR is
(a) 60°
(b) 6212°
(c) 6712°
(d) 75°

Answer:

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).

Page No 14.26:

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:

Let the required angle be x.
Now, complementnary of the required angle = 90∘ − x
Then,
x+1290°-x=75°2x+90°-x=150°x=150-90°x=60°
Hence, the correct answer is option (c).

Page No 14.26:

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:

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).

Page No 14.26:

Question 8:

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

Answer:

Let the required angle be x.
Now, supplementary of the required angle = 180∘ − x
Then,
x=2180°-xx=360°-2x3x=360°x=120°
Hence, the correct answer is option (b).

Page No 14.26:

Question 9:

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

Answer:

Let the required angle be x.
Now, complementary of the required angle = 90∘ − x
Then,
x=90°-xx=90°-x2x=90°x=45°
Hence, the correct answer is option (d).

Page No 14.26:

Question 10:

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

Answer:

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).

Page No 14.26:

Question 11:

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

Answer:

∠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).



Page No 14.27:

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:

∠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).

Page No 14.27:

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:

∠AOC + ∠BOC = 180∘             [∵ Linear pair angles]
y∘ + x∘ = 180∘
y + x = 180
4x5+x=180               4x=5yy=4x54x+5x=180×59x=180×5x=100
Hence, the correct answer is option (a).

Page No 14.27:

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:

∠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).

Page No 14.27:

Question 15:

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

Answer:

∠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).



Page No 14.28:

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:

∠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).

Page No 14.28:

Question 17:

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

Answer:

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).

Page No 14.28:

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:

∠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).

Page No 14.28:

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:

Since, AB || CD
∴ ∠BPQ = ∠PQC           [Alternate interior angles]
⇒ (3x + 34)∘ = (5− 14)∘
⇒ 3x + 34 = 5− 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 No 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:

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.

Page No 14.29:

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:

∠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∘
⇒ 9 = 180
x = 20
∴ ∠BQF = (4 × 20)∘ = 80∘
 Now, ∠BQF = ∠DPQ = 80∘          [Corresponding angles]
Hence, the correct answer is option (b).



Page No 14.30:

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:

∠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).

Page No 14.30:

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:

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).

Page No 14.30:

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:

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).



Page No 14.31:

Question 25:

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

Answer:

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).

Page No 14.31:

Question 26:

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

Answer:

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).

Page No 14.31:

Question 27:

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

Answer:

∠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).

Page No 14.31:

Question 28:

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

Answer:

(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 No 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:

∠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).

Page No 14.32:

Question 30:

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

Answer:



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).

Page No 14.32:

Question 31:

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

Answer:



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).

Page No 14.32:

Question 32:

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

Answer:



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 No 14.33:

Question 33:

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

Answer:



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).

Page No 14.33:

Question 34:

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

Answer:



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).

Page No 14.33:

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:



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).



Page No 14.6:

Question 1:

Write down each pair of adjacent angles shown in Fig.

Answer:

Adjacent angles are the angles that have a common vertex and a common arm.
Following are the adjacent angles in the given figure:

DOC and BOCCOB and BOA

Page No 14.6:

Question 2:

In Fig., name all the pairs of adjacent angles.

Answer:

In figure (i), the adjacent angles are:

EBA andABCACB and BCFBAC and CAD

In figure (ii), the adjacent angles are:

BAD and DAC
BDA and CDA

Page No 14.6:

Question 3:

In figure, write down: (i) each linear pair (ii) each pair of vertically opposite angles.

Answer:

(i) Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
1 and 3
1 and 2
4 and 3
4 and 2
5 and 6
5 and 7
6 and 8
7 and 8

(ii) Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
1 and 4
2 and 3
5 and 8
6 and 7



Page No 14.7:

Question 4:

Are the angles 1 and 2 given in Fig. adjacent angles?

Answer:

No, because they have no common vertex.

Page No 14.7:

Question 5:

Find the complement of each of the following angles:
(i) 35°
(ii) 72°
(iii) 45°
(iv) 85°

Answer:

Two angles are called complementary angles if the sum of those angles is 90°.

Complementary angles of the following angles are:

i 90°-35°=55°ii 90°-72°=18°iii 90°-45°=45°iv 90°-85°=5°

Page No 14.7:

Question 6:

Find the supplement of each of the following angles:
(i) 70°
(ii) 120°
(iii) 135°
(iv) 90°

Answer:

Two angles are called supplementary angles if the sum of those angles is 180°.
Supplementary angles of the following angles are:

(i) 180° − 70° = 110°
(ii) 180° − 120° = 60°
(iii) 180° − 135° = 45°
(iv) 180° − 90° = 90°

Page No 14.7:

Question 7:

Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 25°, 65°
(ii) 120°, 60°
(iii) 63°, 27°
(iv) 100°, 80°

Answer:

Since
(i) 25°+65°=90° , therefore this is complementary pair of angle. (ii) 120°+ 60°= 180°, therefore this is supplementary pair of angle.(iii) 63°+27°= 90°, therefore this is complementary pair of angle.(iv) 100°+ 80°= 180° , therefore this is supplementary pair of angle.

Therefore, (i) and (iii) are the pairs of complementary angles and (ii) and (iv) are the pairs of supplementary angles.

Page No 14.7:

Question 8:

Can two angles be supplementary, if both of them be
(i) obtuse?
(ii) right?
(iii) acute?

Answer:

(i) No, two obtuse angles cannot be supplementary.
(ii) Yes, two right angles can be supplementary. (90°+90°=180°)
(iii) No, two acute angles cannot be supplementary.

Page No 14.7:

Question 9:

Name the four pairs of supplementary angles shown in Fig.

Answer:

Following are the supplementary angles:
AOC and COB
BOC and DOB
BOD and DOA
AOC and DOA

Page No 14.7:

Question 10:

In Fig., A, B, C are collinear points and ∠DBA = ∠EBA.

(i) Name two linear pairs
(ii) Name two pairs of supplementary angles.

Answer:

(i) Linear pairs:
ABD and DBC
ABE and EBC

Because every linear pair forms supplementary angles, these angles are:
ABD and DBC
ABE and EBC

Page No 14.7:

Question 11:

If two supplementary angles have equal measure, what is the measure of each angle?

Answer:

Let x and y be two supplementary angles that are equal.
x=y
According to the question,
x+y=180°x+x=180°2x=180°x=180°2=90°x=y=90°

Page No 14.7:

Question 12:

If the complement of an angle is 28°, then find the supplement of the angle.

Answer:

Let x be the complement of the given angle 28°.
 x+28°=90°x=90°-28°=62°
So, supplement of the angle = 180°-62°=118°

Page No 14.7:

Question 13:

In Fig. 19, name each linear pair and each pair of vertically opposite angles:

Answer:

Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.

1 and 2
2 and 3
3 and 4
1 and 4
5 and 6
6 and 7
7 and 8
8 and 5
9 and 10
10 and 11
11 and 12
12 and 9

Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
1 and 3
4 and 2
5 and 7
6 and 8
9 and 11
10 and 12

Page No 14.7:

Question 14:

In Fig., OE is the bisector of ∠BOD. If ∠1 = 70°, find the magnitudes of ∠2, ∠3 and ∠4.

Answer:

Since OE is the bisector of BOD,
DOE=EOB2+1+EOB=180°                 Linear Pair2+21=180°              1=EOB2=180°-21=180°-2×70°=180°-140°=40°
4=2=40°                Vertically opposite angles3=DOB=1+EOB=70°+70°=140°            3=DOB Vertically opposite angles

Page No 14.7:

Question 15:

One of the angles forming a linear pair is a right angle. What can you say about its other angle?

Answer:

One angle of a linear pair is the right angle, i.e., 90°.
∴ The other angle = 180°â€‹ 90° = 90​°

Page No 14.7:

Question 16:

One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?

Answer:

If one of the angles of a linear pair is obtuse, then the other angle should be acute; only then can their sum be 180°.

Page No 14.7:

Question 17:

One of the angles forming a linear pair is an acute angle. What kind of angle is the other?

Answer:

In a linear pair, if one angle is acute, then the other angle should be obtuse. Only then their sum can be 180°.



Page No 14.8:

Question 18:

Can two acute angles form a linear pair?

Answer:

No, two acute angles cannot form a linear pair because their sum is always less than 180°.

Page No 14.8:

Question 19:

If the supplement of an angle is 65°; then find its complement.

Answer:

Let be the required angle.
Then, we have: 
x + 65° = 180°
x = 180° - 65° = 115°

The complement of angle cannot be determined.

Page No 14.8:

Question 20:

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

Answer:

(i)
Since BOA+BOC=180°         (Linear pair)
 x=180°-BOA=180°-60°=120°

(ii)
Since QOP+QOR=180°         Linear pair2x+3x=180°5x=180°x=180°5=36°

(iii)
Since LOP+PON+NOM=180°         Linear pairPON=180°-LOP-NOMx=180°-35°-60°x=180°-95°=85°

(iv)
Since COD+DOE+EOA+AOB+BOC=360°         Sum of all angles at a point83°+92°+75°+47°+x=360°297°+x=360°x=360°-297°=63°

(v)
2x°+x°+2x°+3x°=180°8x=180x=1808=22.5°

(vi)
3x°=105°x=1053=35°

Page No 14.8:

Question 21:

In Fig. 22, it being given that ∠1 = 65°, find all other angles.

Answer:

1=3          (Vertically opposite angles)
3=65°
Since 1+2=180°       (Linear pair)
2=180°-65°=115°
2=4          (Vertically opposite angles)
4=2=115° and 3=65°



Page No 14.9:

Question 22:

In Fig., OA and OB are opposite rays:


(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?

Answer:

AOC + BOC = 180°                   (Linear pair)
2y+5+3x=180°3x+2y=175°
(i) If x = 25°, then
3×25°+2y=175°75°+2y=175°2y=175°-75°=100°y=100°2=50°
(ii) If y = 35°, then
3x+2×35°=175°3x+70°=175°3x=175°-70°=105°x=105°3=35°

Page No 14.9:

Question 23:

In Fig., write all pairs of adjacent angles and all the linear pairs.

Answer:

Adjacent angles:

DOA and DOCDOC and BOC

AOD and DOBBOC and AOC

Linear pairs of angles:

AOD and DOBBOC and AOC

Page No 14.9:

Question 24:

In Fig. 25, find ∠x. Further find ∠BOC, ∠COD and ∠AOD.

Answer:

AOD+DOC+COB=180°(Linear pair)(x+10)°+x°+(x+20)°=180°3x+30°=180°3x=180°-30°3x=150°x=150°3=50°
BOC=x+20°=50°+20°=70°COD=x=50°AOD=x+10°=50°+10°=60°

Page No 14.9:

Question 25:

How many pairs of adjacent angles are formed when two lines intersect in a point?

Answer:

If two lines intersect at a point, then four adjacent pairs are formed, and those pairs are linear as well.

Page No 14.9:

Question 26:

How many pairs of adjacent angles, in all, can you name in Fig.?

Answer:

There are 10 adjacent pairs in the given figure; they are:
EOD and DOCCOD and BOCCOB and BOA
AOB and BODBOC and COECOD and COADOE and DOB
EOD and DOAEOC and AOCAOB and BOE

Page No 14.9:

Question 27:

In Fig., determine the value of x.

Answer:

AOB+BOC=180°           Linear pair3x+3x=180°6x=180°x=180°6=30°

Page No 14.9:

Question 28:

In Fig., AOC is a line, find x.

Answer:

AOB+BOC=180°                Linear pair70°+2x=180°2x=180°-70°=110°x=110°2=55°

Page No 14.9:

Question 29:

In Fig., POS is a line, find x.

Answer:

QOP+QOR+ROS=180°       (Angles on a straight line)

60°+4x+40°=180°100°+4x=180°4x=180°-100°=80°x=80°4=20°

Page No 14.9:

Question 30:

In Fig., lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.

Answer:

z=x=45°       Vertically opposite anglesNow,x+y=180°      Linear pairy=180°-45°=135°u=y=135°       Vertically opposite angles

 



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