RD Sharma solution class 7 chapter 14 Lines and Angles Exercise 14.2

Exercise 14.2

Page-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 1:

(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

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 2:

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)

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 3:

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

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 4:

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-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 5:

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.

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 6:

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

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 7:

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)

Question 8:

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

Answer 8:

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)
 

Question 9:

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

Answer 9:


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°

Question 10:

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

Answer 10:

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-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 11:

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

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 12:

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° 

Question 13:

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

Answer 13:

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

Question 14:

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

Answer 14:

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°

Question 15:

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

Answer 15:



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°

Question 16:

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

Answer 16:

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-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 17:

 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) 

Question 18:

In Fig., show that AB || EF.

Answer 18:

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 

Question 19:

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

Answer 19:

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°

Question 20:

In Fig., find out ∠PXR, if PQ || RS.

Answer 20:

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°

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 21:

(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°

Question 22:

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

Answer 22:

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

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

Page-14.24

Question 23:

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

 

Answer 23:

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°

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 24:

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

Question 25:

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

Answer 25:

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 

Question 26:

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

Answer 26:





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



 

Question 27:

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

Answer 27:

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-14.25

Question 28:

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

Answer 28:

(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)

Question 29:

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

Answer 29:

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

Question 30:

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

Answer 30:

   


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

∵ AB || DE and BC || EF

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

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