RD Sharma 2017 Solutions for Class 7 Math 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 Math Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2017 Book of class 7 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2017 Solutions. All RD Sharma 2017 Solutions for class 7 Math 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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