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Page No 198:

Question 1:

Define the following terms:
(i) Angle
(ii) Interior of an angle
(iii) Obtuse angle
(iv) Reflex angle
(v) Complementary angles
(vi) Supplementary angles

Answer:

(i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented as AOB.


(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.


(iii) An angle greater than 90° but less than 180° is called an obtuse angle.



(iv) An angle greater than 180° but less than 360° is called a reflex angle.



(v) Two angles are said to be complementary if the sum of their measures is 90°.

(vi) Two angles are said to be supplementary if the sum of their measures is 180°.

Page No 198:

Question 2:

Find the complement of each of the following angles.
(i) 55°
(ii) 16°
(iii) 90°
(iv) 23 of a right angle

Answer:


Two angles whose sum is 90° are called complementary angles.

(i) Complement of 55° = 90° − 55° = 35°

(ii) Complement of 16°=90-16°=74°

(iii) Complement of 90° = 90° − 90° = 0°

(iv) 23 of a right angle90×23°=60°
Complement of 23 of a right angle=90-60°=30°

Page No 198:

Question 3:

Find the supplement of each of the following angles.
(i) 42°
(ii) 90°
(iii) 124°
(iv) 35 of a right angle

Answer:


Two angles whose sum is 180° are called supplementary angles.

(i) Supplement of 42° = 180° − 42° = 138°

(ii) Supplement of 90° = 180° − 90° = 90°

(iii) Supplement of 124° = 180° − 124° = 56°

(iv) 35 of a right angle35×90=54°
Supplement of 35 of a right angle=180-54°=126°

Page No 198:

Question 4:

Find the measure of an angle which is
(i) equal to its complement
(ii) equal to its supplement

Answer:

(i) Let the measure of the required angle be x°.
Then, in case of complementary angles:
x+x=90°2x=90°x=45°
Hence, measure of the angle that is equal to its complement is 45°.

(ii) Let the measure of the required angle be x°â€‹.
Then, in case of supplementary angles:
x+x=180°2x=180°x=90°
Hence, measure of the angle that is equal to its supplement is 90°.

Page No 198:

Question 5:

Find the measure of an angle which is 36° more than its complement.

Answer:

Let the measure of the required angle be x°.
Then, measure of its complement =90-x°.
Therefore,
x-90°-x=36°2x=126°x=63°
Hence, the measure of the required angle is 63°.

Page No 198:

Question 6:

Find the measure of an angle which is 30° less than its supplement.

Answer:


Let the measure of the angle be x°.

∴ Supplement of x° = 180° − x°

It is given that,

(180° − x°) − x° = 30°

⇒ 180° − 2x°= 30°

⇒ 2x° = 180° − 30° = 150°

x° = 75°

Thus, the measure of the angle is 75°.

Page No 198:

Question 7:

Find the angle which is four times its complement.

Answer:

Let the measure of the required angle be x.
Then, measure of its complement =90°-x.
Therefore,
x=90°-x4x=360°-4x5x=360°x=72°
Hence, the measure of the required angle is 72°.

Page No 198:

Question 8:

Find the angle which is five times its supplement.

Answer:

Let the measure of the required angle be x.
Then, measure of its supplement =180°-x.
Therefore,
x=180°-x5x=900°-5x6x=900°x=150°
Hence, the measure of the required angle is 150°.

Page No 198:

Question 9:

Find the angle whose supplement is four times its complement.

Answer:

Let the measure of the required angle be x°.
Then, measure of its complement =90-x°.
And, measure of its supplement=180-x°.
Therefore,
180-x=490-x180-x=360-4x3x=180x=60
Hence, the measure of the required angle is 60°.

Page No 198:

Question 10:

Find the angle whose complement is one-third of its supplement.

Answer:

Let the measure of the required angle be x°.
Then, the measure of its complement =90-x°.
And the measure of its supplement=180-x°.
Therefore,
90-x=13180-x390-x=180-x270-3x=180-x2x=90x=45
Hence, the measure of the required angle is 45°.

Page No 198:

Question 11:

Two complementary angles are in the ratio 4 : 5. Find the angles.

Answer:

Let the two angles be 4x and 5x, respectively.
Then,
4x+5x=909x=90x=10°
Hence, the two angles are 4x=4×10°=40° and 5x=5×10°=50°.

Page No 198:

Question 12:

Find the value of x for which the angles (2x – 5)° and (x – 10)° are the complementary angles.

Answer:


Two angles whose sum is 90° are called complementary angles.

It is given that the angles (2x – 5)° and (x – 10)° are the complementary angles.

∴ (2x – 5)° + (x – 10)° = 90°

⇒ 3x° – 15° = 90°

⇒ 3x° = 90° + 15° = 105°

⇒ x° = 105°3 = 35°

Thus, the value of x is 35.



Page No 206:

Question 1:

In the adjoining figure, AOB is a straight line. Find the value of x.

Answer:

We know that the sum of angles in a linear pair is 180°.
Therefore,
AOC+BOC=180°62°+x°=180°x°=180°-62°x=118°
Hence, the value of x is 118°.

Page No 206:

Question 2:

In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ∠AOC and ∠BOD.

Answer:

As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°.
Therefore,
 AOC+COD+BOD=180°3x-7°+55°+x+20°=1804x=112°x=28°
Hence,
AOC=3x-7
         =3×28-7=77°
and BOD=x+20
                =28+20=48°



Page No 207:

Question 3:

In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ∠AOC, ∠COD and ∠BOD.

Answer:

AOB is a straight line. Therefore,
AOC+COD+BOD=180°3x+7°+2x-19°+x°=180°6x=192°x=32°
Therefore,
AOC=3×32°+7=103°COD=2×32°-19=45° andBOD=32°

Page No 207:

Question 4:

In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.

Answer:

Let x=5a, y=4a and z=6a
XOY is a straight line. Therefore,

XOP+POQ+YOQ=180°5a+4a+6a=180°15a=180°a=12°
Therefore,

      x5×12°=60°      y4×12°=48°and z6×12°=72°

Page No 207:

Question 5:

In the adjoining figure, what value of x will make AOB a straight line?

Answer:

AOB will be a straight line if
3x+20+4x-36=180°7x=196°x=28°
Hence, x = 28 will make AOB a straight line.

Page No 207:

Question 6:

Two lines AB and CD intersect at O. If ∠AOC = 50°, find ∠AOD, ∠BOD and ∠BOC.

Answer:

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, AOC=BOD=50°
Let AOD=BOC=x°
Also, we know that the sum of all angles around a point is 360°.
Therefore, 
AOC+AOD+BOD+BOC=360°50+x+50+x=360°2x=260°x=130°
Hence, AOD=BOC=130°
Therefore, AOD=130°, BOD=50° and BOC=130°.

Page No 207:

Question 7:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.

Answer:

We know that if two lines intersect, then the vertically opposite angles are equal.
BOD=AOC=90°
Hence, t=90°
Also, 
DOF=COE=50°
Hence, z=50°
Since, AOB is a straight line, we have:
AOC+COE+BOE=180°90+50+y=180°140+y=180°y=40°
Also,
BOE=AOF=40°
Hence, x=40°
 x=40°, y=40°, z=50° and t=90°

Page No 207:

Question 8:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ∠AOD, ∠COE and ∠AOE.

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.
DOF=COE=5x°AOD=BOC=2x° andAOE=BOF=3x°

Since, AOB is a straight line, we have:
AOE+COE+BOC=180°3x+5x+2x=180°10x=180°x=18°

Therefore,
AOD=2×18°=36°COE=5×18°=90°AOE=3×18°=54°

Page No 207:

Question 9:

Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.

Answer:

Let the two adjacent angles be 5x and 4x, respectively.
Then,
5x+4x=180°9x=180°x=20°
Hence, the two angles are 5×20°=100° and 4×20°=80°.

Page No 207:

Question 10:

If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.

AOC=90°. Then, AOC=BOD=90°.
And let BOC=AOD=x
Also, we know that the sum of all angles around a point is 360°
AOC+BOD+AOD+BOC=360°90°+90°+x+x=360°2x=180°x=90°
Hence, BOC=AOD=90°
AOC=BOD=BOC=AOD=90°
Hence, the measure of each of the remaining angles is 90o.



Page No 208:

Question 11:

Two lines AB and CD intersect at a point O, such that ∠BOC + ∠AOD = 280°, as shown in the figure. Find all the four angles.

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let BOC=AOD=x°
Then,
x+x=2802x=280x=140°BOC=AOD=140°
Also, let AOC=BOD=y°
We know that the sum of all angles around a point is 360°.
AOC+BOC+BOD+AOD=360°y+140+y+140=360°2y=80°y=40°
Hence, AOC=BOD=40°
BOC=AOD=140° and AOC=BOD=40°

Page No 208:

Question 12:

Two lines AB and CD intersect each other at a point O such that ∠AOC : ∠AOD = 5 : 7. Find all the angles.

Answer:


Let ∠AOC = 5k and ∠AOD = 7k, where k is some constant.

Here, ∠AOC and ∠AOD form a linear pair.

∴ ∠AOC + ∠AOD = 180º

⇒ 5k + 7k = 180º

⇒ 12k = 180º

⇒ k = 15º

∴ ∠AOC = 5k = 5 × 15º = 75º

∠AOD = 7= 7 × 15º = 105º

Now, ∠BOD = ∠AOC = 75º       (Vertically opposite angles)

∠BOC = ∠AOD = 105º       (Vertically opposite angles)

Page No 208:

Question 13:

In the given figure, three lines AB, CD and EF intersect at a point O such that ∠AOE = 35° and ∠BOD = 40°. Find the measure of ∠AOC, ∠BOF, ∠COF and ∠DOE.

Answer:


In the given figure, 

∠AOC = ∠BOD = 40º      (Vertically opposite angles)

∠BOF = ∠AOE = 35º       (Vertically opposite angles)

Now, ∠EOC and ∠COF form a linear pair.

∴ ∠EOC + ∠COF = 180º

⇒ (∠AOE + ∠AOC) + ∠COF = 180º

⇒ 35º + 40º + ∠COF = 180º

⇒ 75º + ∠COF = 180º

⇒ ∠COF = 180º − 75º = 105º

Also,  ∠DOE = ∠COF = 105º       (Vertically opposite angles)

Page No 208:

Question 14:

In the given figure, the two lines AB and CD intersect at a point O such that ∠BOC = 125°. Find the values of x, y and z.

Answer:


Here, ∠AOC and ∠BOC form a linear pair.

∴ ∠AOC + ∠BOC = 180º

⇒ xº + 125º = 180º

⇒ xº = 180º − 125ºâ€‹ = 55ºâ€‹

Now, 

∠AOD = ∠BOC = 125º       (Vertically opposite angles)

∴ yº = 125ºâ€‹

∠BOD = ∠AOC = 55º         (Vertically opposite angles)

∴ zº = 55ºâ€‹

Thus, the respective values of xy and z are 55, 125 and 55.

Page No 208:

Question 15:

If two straight lines intersect each other, then prove that the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Answer:

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect AOC. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let COE=1, AOE=2, BOF=3 and DOF=4.
We know that vertically-opposite angles are equal.
1=4 and 2=3
But, 1=2    [Since OE bisects AOC ]
4=3
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Page No 208:

Question 16:

Prove that the bisectors of two adjacent supplementary angles include a right angle.

Answer:

Let AOB denote a straight line and let AOC and BOC be the supplementary angles.
Then, we have:
AOC=x° and BOC=180-x°
Let OE bisect AOC and OF bisect BOC.
Then, we have:
AOE=COE=12x° andBOF=FOC=12180-x°
Therefore,
COE+FOC=12x+12180°-x
                       =12x+180°-x=12180°=90°



Page No 223:

Question 1:

In the given figure, l || m and a transversal t cuts them. If ∠1 = 120°, find the measure of each of the remaining marked angles.

Answer:

We have, 1=120°. Then,
1=5   Corresponding angles5=120°1=3   Vertically-opposite angles3=120°

5=7   Vertically-opposite angles7=120°1+2=180°   Since AFB is a straight line120°+2=180°

2=60°2=4   Vertically-opposite angles4=60°2=6   Corresponding angles

6 =60°6=8   Vertically-opposite angles8=60°1=120°, 2=60°, 3=120°, 4=60°, 5=120°,6 =60°, 7=120° and 8=60°

Page No 223:

Question 2:

In the given figure, l || m and a transversal t cuts them. If ∠7 = 80°, find the measure of each of the remaining marked angles.

Answer:


In the given figure, ∠7 and ∠8 form a linear pair.

∴ ∠7 + ∠8 = 180º

⇒ 80º + ∠8 = 180º

⇒ ∠8 = 180º − 80º = 100º

Now, 

∠6 = ∠8 = 100º       (Vertically opposite angles)

∠5 = ∠7 = 80º         (Vertically opposite angles)

It is given that, || m and is a transversal.

∴ ∠1 = ∠5 = 80º      (Pair of corresponding angles)

∠2 = ∠6 = 100º         (Pair of corresponding angles)

∠3 = ∠7 = 80º           (Pair of corresponding angles)

∠4 = ∠8 = 100º         (Pair of corresponding angles)

Page No 223:

Question 3:

In the given figure, l || m and a transversal t cuts them. If ∠1 : ∠2 = 2 : 3, find the measure of each of the marked angles.

Answer:


Let ∠1 = 2k and ∠2 = 3k, where k is some constant.

Now, ∠1 and ∠2 form a linear pair.

∴ ∠1 + ∠2 = 180º

⇒ 2k + 3k = 180º

⇒ 5k = 180º

k = 36º

∴ ∠1 = 2k = 2 × 36º = 72º

∠2 = 3k = 3 × 36º = 108º

Now, 

∠3 = ∠1 = 72º         (Vertically opposite angles)

∠4 = ∠2 = 108º       (Vertically opposite angles)

It is given that, || m and is a transversal.

∴ ∠5 = ∠1 = 72º       (Pair of corresponding angles)

∠6 = ∠2 = 108º         (Pair of corresponding angles)

∠7 = ∠1 = 72º           (Pair of alternate exterior angles)

∠8 = ∠2 = 108º         (Pair of alternate exterior angles)

Page No 223:

Question 4:

For what value of x will the line l and m be parallel to each other?

Answer:

For the lines l and m to be parallel
3x-20=2x+10   Corresponding Anglesx=30
 



Page No 224:

Question 5:

For what value of x will the lines l and m be parallel to each other?

Answer:

3x+5+4x=180   Consecutive Interior Angles7x=175x=25

Page No 224:

Question 6:

In the given figure, AB || CD and BC || ED. Find the value of x.

Answer:

BCED and CD is the transversal.
Then,

BCD+CDE=180°   Angles on the same side of a transversal line are supplementaryBCD+75=180BCD=105°

ABCD and BC is the transversal.

ABC=BCD  (alternate angles) x°=105°x=105

Page No 224:

Question 7:

In the given figure, AB || CD || EF. Find the value of x.

Answer:

EFCD and CE is the transversal.
Then,
ECD+CEF=180°   Consecutive Interior AnglesECD+130°=180°ECD=50°
Again, ABCD and BC is the transversal.
Then,
ABC=BCD   Alternate Interior Angles70°=x+50°  BCD=BCE+ECDx=20°

Page No 224:

Question 8:

In the given figure, AB || CD. Find the values of x, y and z.

Answer:



ABCD and let EF and EG be the transversals.
Now, ABCD  and EF is the transversal.
Then,
AEF=EFG   Alternate Anglesy°=75°y=75
Also,
EFC+EFD=180°   Since CFD is a straight linex+y=180x+75=180x=105
And,
EGF+EGD=180°   Since CFGD is a straight lineEGF+125=180EGF=55°
We know that the sum of angles of a triangle is 180°
EFG+GEF+EGF=180°y+z+55=18075+z+55=180z=50x=105, y=75 and z=50

Page No 224:

Question 9:

In each of the figures given below, AB || CD. Find the value of x in each case.

Answer:

(i)

Draw EFABCD.
Now, ABEF and BE is the transversal.
Then,
ABE=BEF   Alternate Interior AnglesBEF=35°
Again, EFCD and DE is the transversal.
Then,
DEF=FEDFED=65°x°=BEF+FED     =35+65°     =100°or, x=100

(ii)

Draw EOABCD.
Then, EOB+EOD=x°
Now, EOAB and BO is the transversal.
EOB+ABO=180°   Consecutive Interior AnglesEOB+55°=180°EOB=125°
Again, EOCD and DO is the transversal.
EOD+CDO=180°   Consecutive Interior AnglesEOD+25°=180°EOD=155°
Therefore,
x°=EOB+EOD =125+155° =280°or, x=280

(iii)

Draw EFABCD.
Then, AEF+CEF=x°
Now, EFAB and AE is the transversal.
 AEF+BAE=180°   Consecutive Interior Angles AEF+116=180AEF=64°
Again, EFCD and CE is the transversal.
CEF+ECD=180°   Consecutive Interior AnglesCEF+124=180CEF=56°

Therefore,
x°=AEF+CEF  =64+56°  =120°or, x=120



Page No 225:

Question 10:

In the given figures, AB || CD. Find the value of x.

Answer:


Draw EFABCD.
EFCD and CE is the transversal.
Then,
ECD+CEF=180°   Angles on the same side of a transversal line are supplementary130°+CEF=180°CEF=50°
Again, EFAB and AE is the transversal.
Then,
BAE+AEF=180°  Angles on the same side of a transversal line are supplementaryx°+20°+50°=180°   AEF=AEC+CEFx°+70°=180°x°=110°x=110

Page No 225:

Question 11:

In the given figure, AB || PQ. Find the values of x and y.

Answer:



Given, ABPQ.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
CEB+BEG+GEF=180°   Since CD is a straight line75°+20°+GEF=180°GEF=85°
We know that the sum of angles of a triangle is 180°.
GEF+EGF+EFG=18085°+x+25°=180°110°+x=180°x=70°
And
FEG+BEG=DFQ   Corresponding Angles85°+20°=DFQDFQ=105°EFG+GFQ+DFQ=180°   Since CD is a straight line25°+y+105°=180°y=50°x=70° and y=50°

Page No 225:

Question 12:

In the given figure, AB || CD. Find the value of x.

Answer:

ABCD and AC is the transversal.
Then,
BAC+ACD=180°   Consecutive Interior Angles75+ACD=180ACD=105°
And,
ACD=ECF   Vertically-Opposite AnglesECF=105°
We know that the sum of the angles of a triangle is 180°.
ECF+CFE+CEF=180°105°+30°+x=180°135°+x=180°x=45°

Page No 225:

Question 13:

In the given figure, AB || CD. Find the value of x.

Answer:

ABCD and PQ is the transversal.
Then,
PEF=EGH   Corresponding AnglesEGH=85°
And,
EGH+QGH=180°   Since PQ is a straight line85°+QGH=180°QGH=95°
Also,
CHQ+GHQ=180°   Since CD is a straight line115°+GHQ=180°GHQ=65°
We know that the sum of angles of a triangle is 180°.
QGH+GHQ+GQH=180°95°+65°+x=180°x=20°x=20°



Page No 226:

Question 14:

In the given figure, AB || CD. Find the values of x, y and z.

Answer:

ADC=DAB   Alternate Interior Anglesz=75°ABC=BCD   Alternate Interior Anglesx=35°
We know that the sum of the angles of a triangle is 180°.
35°+y+75°=180°y=70°x=35°, y=70° and z=75°.

Page No 226:

Question 15:

In the given figure, AB || CD. Prove that ∠BAE − ∠DCE = ∠AEC.

Answer:


Draw EFABCD through E.
Now, EFAB and AE is the transversal.
Then, BAE+AEF=180°   Angles on the same side of a transversal line are supplementary
Again, EFCD and CE is the transversal.
Then,
DCE+CEF=180°   Angles on the same side of a transversal line are supplementaryDCE+AEC+AEF=180°DCE+AEC+180°-BAE=180°BAE-DCE=AEC

Page No 226:

Question 16:

In the given figure, AB || CD. Prove that p + qr = 180.

Answer:


Draw PFQABCD.
Now, PFQAB and EF is the transversal.
Then,
AEF+EFP=180°.....(1)                                                     Angles on the same side of a transversal line are supplementary
Also, PFQCD.

PFG=FGD=r°Alternate  Anglesand EFP=EFG-PFG=q°-r°putting the value of EFP in eqn. (i)we get,p°+q°-r°=180°p+q-r=180

Page No 226:

Question 17:

In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

Answer:

In the given figure,
x=60°   Vertically-Opposite AnglesPRQ=SQR   Alternate Anglesy=60°APR=PQS   Corresponding Angles110°=PQR+60°   PQS=PQR+RQSPQR=50°
PQR+RQS+BQS=180°   Since AB is a straight line50°+60°+z=180°110°+z=180°z=70°
DSH=z   Corresponding AnglesDSH=70°DSH=t   Vertically-Opposite Anglest=70° x=60°, y=60°, z=70° and t=70°.

Page No 226:

Question 18:

In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF = 90°.

Answer:


It is given that, AB || CD and is a transversal.

 ∠BEF + ∠EFD = 180°     .....(1)     (Sum of the interior angles on the same side of a transversal is supplementary)

EG is the bisector of ∠BEF.    (Given)


∴ ∠BEG = ∠GEF = 12∠BEF

⇒ ∠BEF = 2∠GEF                .....(2)

Also, FG is the bisector of ∠EFD.    (Given)


∴ ∠EFG = ∠GFD = 12∠EFD

⇒ ∠EFD = 2∠EFG                .....(3)

From (1), (2) and (3), we have

2∠GEF + 2∠EFG = 180°

⇒ 2(
∠GEF + ∠EFG) = 180°

⇒ 
∠GEF + ∠EFG = 90°            .....(4)

In âˆ†EFG,

∠GEF + ∠EFG + ∠EGF = 180°          (Angle sum property)

⇒ 90° + ∠EGF = 180°                         [Using (4)]

⇒ ∠EGF = 180° − 90° = 90°



Page No 227:

Question 19:

In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP || FQ .



 

Answer:


It is given that, AB || CD and t is a transversal.

∴ ∠AEF = ∠EFD           .....(1)         (Pair of alternate interior angles)

EP is the bisectors of ∠AEF.        (Given)

∴ ∠AEP = ∠FEP = 12∠AEF

⇒ ∠AEF = 2∠FEP          .....(2)

Also, FQ is the bisectors of ∠EFD.

∴ ∠EFQ = ∠QFD = 12∠EFD

⇒ ∠EFD = 2∠EFQ         .....(3)

From (1), (2) and (3), we have

2∠FEP = 2∠EFQ

⇒ ∠FEP = ∠EFQ

Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal. 

∴ EP || FQ        (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)

Page No 227:

Question 20:

In the given figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF.

Answer:


It is given that, BA || ED and BC || EF.

Construction: Extend DE such that it intersects BC at J. Also, extend FE such that it intersects AB at H.



Now, BA || JD and BC is a transversal.

∴ ∠ABC = ∠DJC      .....(1)       (Pair of corresponding angles)

Also, BC || HF and DJ is a transversal.

∴ ∠DJC = ∠DEF      .....(2)       (Pair of corresponding angles)

From (1) and (2), we have

∠ABC = ∠DEF

Page No 227:

Question 21:

In the given figure, BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°.

Answer:


It is given that, BA || ED and BC || EF.

Construction: Extend ED such that it intersects BC at G. 



Now, BA || GE and BC is a transversal.

∴ ∠ABC = ∠EGC      .....(1)       (Pair of corresponding angles)

Also, BC || EF and EG is a transversal.

∴ ∠EGC + ∠GEF = 180°      .....(2)       (Interior angles on the same side of the transversal are supplementary)

From (1) and (2), we have

​∠ABC + ∠GEF = 180°        

Or ∠ABC + ∠DEF = 180°          

Page No 227:

Question 22:

In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.

Answer:


AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.

It is given that the two plane mirrors are perpendicular to each other.

Therefore, BP || OA and AP || OB.

So, BP ⊥ AP       (OA ⊥ OB)

⇒ ∠APB = 90°    .....(1)

In âˆ†APB,

​∠2 + ∠3 + ∠APB = 180°      (Angle sum property)

∴ ∠2 + ∠3 + 90° = 180°       [Using (1)]

⇒ ∠2 + ∠3 = 180° − 90° = 90°

⇒ 2∠2 + 2∠3 = 2 × 90° = 180°        .....(2)

By law of reflection, we have

∠1 = ∠2  and ∠3 = ∠4                     .....(3)       (Angle of incidence = Angle of reflection)   

From (2) and (3), we have

∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ ∠BAC + ∠ABD = 180°            (∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)

Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of the transversal are supplementary.

∴ CA || BD    

Page No 227:

Question 23:

In the figure given below, state which lines are parallel and why?

Answer:


Here, ∠BAC = ∠ACD = 110°

Thus, lines AB aand CD are intersected by a transversal AC such that the pair of alternate angles are equal.

∴ AB || CD     (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)

Thus, line AB is parallel to line CD.

Also, ∠ACD + ∠CDE = 110° + 80° = 190° ≠ 180°

If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.

Therefore, line AC is not parallel to line DE.      



Page No 228:

Question 24:

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Answer:



Let the two lines m and n be respectively perpendicular to the two parallel lines p and q.
To prove: is parallel to n.
Proof: Since, m is perpendicular to p
1=90°
Also, ​n is perpendicular to 
3=90°
Since p and are parallel and is a transversal line 
 2=1=90°              [Corresponding angles]
Also, 2=3=90°
We know that if two corresponding angles are equal then the two lines containing them must be parallel.
Therefore, the lines m and n are parallel to each other.



Page No 231:

Question 1:

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle

Answer:


Let âˆ†ABC be such that ∠A = ∠B + ∠C.

In âˆ†ABC,

∠A + ∠B + ∠C = 180º    (Angle sum property)

⇒ ∠A + ∠A = 180º         (∠A = ∠B + ∠C)

⇒ 2∠A = 180º

⇒ ∠A = 90º

Therefore, âˆ†ABC is a right triangle.

Thus, if one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle.

Hence, the correct answer is option (d).



Page No 232:

Question 2:

An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Each of these equal angles is
(a) 70°
(b) 55°
(c) 35°
(d) 27 1°2

Answer:


Let the measure of each of the two equal interior opposite angles of the triangle be x.

In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.

∴ xx = 110°

⇒ 2x = 110°

⇒ x = 55°

Thus, the measure of each of these equal angles is 55°.

Hence, the correct answer is option (b).

Page No 232:

Question 3:

The angles of a triangle are in the ration 3 : 5 : 7. The triangle is
(a) acute-angled
(b) obtuse-angled
(c) right-angled
(d) an isosceles triangle

Answer:

(a) acute-angled

Let the angles measure 3x°, 5x° and 7x°.
Then,
3x+5x+7x=180°15x=180°x=12°
Therefore, the angles are 312°=36°, 512°=60° and 712°=84°.
Hence, the triangle is acute-angled.

Page No 232:

Question 4:

If one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles can be
(a) 50°
(b) 65°
(c) 90°
(d) 155°

Answer:


Let âˆ†ABC be such that ∠A = 130°.



Here, BP is the bisector of ∠B and CP is the bisector of ∠C.

∴ ∠ABP = ∠PBC = 12∠B                .....(1)

Also, ∠ACP = ∠PCB = 12∠C           .....(2)

In âˆ†ABC,

∠A + ∠B + ∠C = 180°         (Angle sum property)

⇒ 130° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° − 130° = 50°

12∠B + 12∠C = 12 × 50° = 25°

⇒ ∠PBC + ∠PCB = 25°     .....(3)     [Using (1) and (2)]

In âˆ†PBC,

∠PBC + ∠PCB + ∠BPC = 180°         (Angle sum property)

⇒ 25° + ∠BPC = 180°                       [Using (3)]

⇒ ∠BPC = 180° − 25° = 155°

Thus, if one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles is 155°.

Hence, the correct answer is option (d).

Page No 232:

Question 5:

In the given figure, AOB is a straight line. The value of x is

(a) 12
(b) 15
(c) 20
(d) 25

Answer:


It is given that, AOB is a straight line.

∴ 60º + (5xº + 3xº) = 180º           (Linear pair)

⇒ 8xº = 180º − 60º = 120º

⇒ xº = 15º

Thus, the value of x is 15.

Hence, the correct answer is option (b).

Page No 232:

Question 6:

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is
(a) 12°
(b) 100°
(c) 80°
(d) 60°
 

Answer:


Suppose âˆ†ABC be such that ∠A : ∠B : ∠C = 2 : 3 : 4.

Let ∠A = 2k, ∠B = 3and ∠C = 4k, where k is some constant.

In ∆ABC,

∠A + ∠B + ∠C = 180º      (Angle sum property)

⇒ 2k + 3k + 4k = 180º

⇒ 9k = 180º

⇒ k = 20º

∴ Measure of the largest angle = 4k = 4 × 20º = 80º

Hence, the correct answer is option (c).

Page No 232:

Question 7:

In the given figure, ∠OAB = 110° and ∠BCD = 130° then ∠ABC is equal to

(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer:


In the given figure, OA || CD.

Construction: Extend OA such that it intersects BC at E.



Now, OE || CD and BC is a transversal.

∴ ∠AEC = ∠BCD = 130°       (Pair of corresponding angles)

Also, ∠OAB + ∠BAE = 180°         (Linear pair)

∴ 110° + ∠BAE = 180°

⇒ ∠BAE = 180° − 110° = 70°

In âˆ†ABE,

∠AEC = ∠BAE + ∠ABE                (In a triangle, exterior angle is equal to the sum of two opposite interior angles)

∴ 130° = 70° + x°

⇒ x° = 130° − 70° = 60°

Thus, the measure of angle ∠ABC is 60°.

Hence, the correct answer is option (c).

Page No 232:

Question 8:

If two angles are complements of each other, then each angle is
(a) an acute angle
(b) an obtuse angle
(c) a right angle
(d) a reflex angle

Answer:

(a) an acute angle

If two angles are complements of each other, that is, the sum of their measures is 90°, then each angle is an acute angle.

Page No 232:

Question 9:

An angle which measures more than 180° but less than 360°, is called
(a) an acute angle
(b) an obtuse angle
(c) a straight angle
(d) a reflex angle

Answer:


An angle which measures more than 180° but less than 360° is called a reflex angle.

Hence, the correct answer is option (d).

Page No 232:

Question 10:

The measure of an angle is five times its complement. The angle measures
(a) 25°
(b) 35°
(c) 65°
(d) 75°

Answer:

(d) 75°

Let the measure of the required angle be x°.
Then, the measure of its complement will be 90-x°.
x=590-xx=450-5x6x=450x=75



Page No 233:

Question 11:

Two complementary angles are such that twice the measure of one is equal to three times the measure of the other. The measure of larger angle is
(a) 72°
(b) 54°
(c) 63°
(d) 36°

Answer:

(b) 54°

Let the measure of the required angle be x°.
Then, the measure of its complement will be90-x°.
2x=390-x2x=270-3x5x=270x=54

Page No 233:

Question 12:

In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 100°

Answer:

(c) 80°

We have :
AOC+BOC=180°   Since AOB is a straight line4x+5x=180°9x=180°x=20°AOC=4×20°=80°

Page No 233:

Question 13:

In the given figure, AOB is a straight line. If ∠AOC = (3x + 10)° and ∠BOC (4x − 26)°, then ∠BOC = ?
(a) 96°
(b) 86°
(c) 76°
(d) 106°

Answer:

(b) 86°

We have :
AOC+BOC=180°   Since AOB is a straight line 3x+10+4x-26=180°7x=196°x=28°

BOC=4×28-26°Hence, BOC=86°.

Page No 233:

Question 14:

In the given figure, AOB is a straight line. If ∠AOC = (3x − 10)°, ∠COD = 50° and ∠BOD = (x + 20)°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 50°

Answer:

(c) 80°

We have :
AOC+COD+BOD=180°   Since AOB is a straight line 3x-10+50+x+20=1804x=120x=30

AOC=3×30-10°AOC=80°

Page No 233:

Question 15:

Which of the following statements is false?
(a) Through a given point, only one straight line can be drawn.
(b) Through two given points, it is possible to draw one and only one straight line.
(c) Two straight lines can intersect at only one point.
(d) A line segment can be produced to any desired length.

Answer:

(a) Through a given point, only one straight line can be drawn.

Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.

Page No 233:

Question 16:

An angle is one-fifth of its supplement. The measure of the angle is
(a) 15°
(b) 30°
(c) 75°
(d) 150°

Answer:

(b) 30°

Let the measure of the required angle be x°
Then, the measure of its supplement will be 180-x°
x=15180°-x5x=180°-x6x=180°x=30°

Page No 233:

Question 17:

In the adjoining figure, AOB is a straight line. If x : y : z = 4 : 5 : 6, then y = ?
(a) 60°
(b) 80°
(c) 48°
(d) 72°

Answer:

(a) 60°

Let  
AOB=x°=4a°, COB=y°=5a° and BOD=z°=6a°
Then, we have:
AOB+COB+BOD=180°   Since AOB is a straight line4a+5a+6a=180° 15a=180°a=12°y=5×12°=60°

Page No 233:

Question 18:

In the given figure, straight lines AB and CD intersect at O. If AOC = ɸ, ∠BOC = θ and θ = 3ɸ, then ɸ = ?
(a) 30°
(b) 40°
(c) 45°
(d) 60°

Answer:

(c) 45°

We have :
θ+ϕ=180°   AOD is a straight line3ϕ+ϕ=180°   θ=3ϕ4ϕ=180°ϕ=45°



Page No 234:

Question 19:

In the given figure, straight lines AB and CD intersect at O. If AOC + ∠BOD = 130°, then ∠AOD = ?
(a) 65°
(b) 115°
(c) 110°
(d) 125°

Answer:

(b) 115°

We have :
AOC=BOD   Vertically-Opposite AnglesAOC+BOD=130°AOC+AOC=130°   AOC=BOD2AOC=130°AOC=65°
Now,
AOC+AOD=180°  COD is a straight line 65°+AOD=180°AOC=115°

Page No 234:

Question 20:

In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, find ∠AQP.
(a) 72°
(b) 18°
(c) 36°
(d) 54°

Answer:

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let AQP=BQR=x°
Now,
AQP+PQR+BQR=180°   AQB is a straight line x+108+x=180°2x=72°x=36°AQP=36°

Page No 234:

Question 21:

In the given figure, AB || CD. If BAO = 60° and ∠OCD = 110°, then ∠AOC = ?
(a) 70°
(b) 60°
(c) 50°
(d) 40°

Answer:

(c) 50°

Draw EOFABCD.
Now, EOAB and OA is the transversal.
EOA=OAB=60°  Alternate Interior Angles 
Also,
OFCD and OC is the transversal.
COF+OCD=180°   Angles on the same side of a transversal line are supplementaryCOF+110°=180°COF=70°
Now,
EOA+AOC+COF=180°   EOF is a straight line60°+AOC+70°=180°AOC=50°

Page No 234:

Question 22:

In the given figure, AB || CD. If AOC = 30° and ∠OAB = 100°, then ∠OCD = ?
(a) 130°
(b) 150°
(c) 80°
(d) 100°

Answer:

(a) 130°

Draw OEABCD
Now, OEAB and OA is the transversal.
OAB+AOE=180°   Angles on the same side of a transversal line are supplementaryOAB+AOC+COE=180°100°+30°+COE=180°COE=50°
Also, 
OECD and OC is the transversal.
OCD+COE=180°   Angles on the same side of a transversal line are supplementaryOCD+50°=180°OCD=130°

Page No 234:

Question 23:

In the given figure, AB || CD. If CAB = 80° and ∠EFC = 25°, then ∠CEF = ?
(a) 65°
(b) 55°
(c) 45°
(d) 75°

Answer:

(c) 45°

ABCD and AF is the transversal.
DCF=CAB=80°   Corresponding Angles
Side EC of triangle EFC is produced to D.
CEF+EFC=DCFCEF+25°=80°CEF=55°

Page No 234:

Question 24:

In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, x = ?
(a) 108°
(b) 126°
(c) 162°
(d) 63°

Answer:

(b) 126°
Let y=3a° and z=7a°
Let the transversal intersect AB at P, CD at O and EF at Q.

Then, we have:
COP=DOF=y   Vertically-Opposite AnglesOQF+DOQ=180°   Consecutive Interior Angles3a+7a=180°10a=180°a=18°y=3×18°=54°
Also,
APO+COP=180°x+54°=180°x=126°



Page No 235:

Question 25:

In the given figure, AB || CD. If APQ = 70° and ∠PRD = 120°, then ∠QPR = ?
(a) 50°
(b) 60°
(c) 40°
(d) 35°

Answer:

(a) 50°

ABCD and PQ is the transversal.
PQR=APQ=70°   Alternate Interior Angles
Side QR of traingle PQR is produced to D.
PQR+QPR=PRD70°+QPR=120°QPR=50°

Page No 235:

Question 26:

In the given figure, AB || CD. If EAB = 50° and ∠ECD = 60°, then ∠AEB = ?
(a) 50°
(b) 60°
(c) 70°
(d) 55°

Answer:

(c) 70°

ABCD and BC is the transversal.
ABE=BCD=60°   Alternate Internal Angles
In ABE, we have:
EAB+ABE+AEB=180°   Sum of the angles of a triangle50°+60°+AEB=180°AEB=70°

Page No 235:

Question 27:

In the given figure, ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?
(a) 20°
(b) 25°
(c) 30°
(d) 35°

Answer:

(c) 30°
In OAB, we have:
OAB+OBA+AOB=180°   Sum of the angles of a triangle75°+55°+AOB=180°AOB=50°
COD=AOB=50°   Vertically-Opposite Angles
In OCD, we have:
COD+OCD+ODC=180°   Sum of the angles of a triangle50°+100°+ODC=180°ODC=30°

Page No 235:

Question 28:

In the adjoining figure, y = ?
(a) 36°
(b) 54°
(c) 63°
(d) 72°

Answer:

(b) 54°

We have:
3x+72=180°   AOB is a straight line3x=108x=36
Also,
AOC+COD+BOD=180°   AOB is a straight line36°+90°+y=180°y=54°



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