RD Sharma solution class 7 chapter 14 Lines and Angles Exercise 14.1

Exercise 14.1

Page-14.6

Question 1:

Write down each pair of adjacent angles shown in Fig.

Answer 1:

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

Question 2:

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

Answer 2:

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

Question 3:

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

Answer 3:

(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-14.7

Question 4:

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

Answer 4:

No, because they have no common vertex.

Question 5:

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

Answer 5:

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°

Question 6:

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

Answer 6:

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°

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 7:

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.

Question 8:

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

Answer 8:

(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.

Question 9:

Name the four pairs of supplementary angles shown in Fig.

Answer 9:

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

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 10:

(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

Question 11:

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

Answer 11:

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°

Question 12:

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

Answer 12:

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°

Question 13:

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

Answer 13:

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

Question 14:

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

Answer 14:

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

Question 15:

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

Answer 15:

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

Question 16:

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

Answer 16:

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°.

Question 17:

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

Answer 17:

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

Page-14.8

Question 18:

Can two acute angles form a linear pair?

Answer 18:

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

Question 19:

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

Answer 19:

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

The complement of angle cannot be determined.

Question 20:

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

Answer 20:

(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°

Question 21:

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

Answer 21:

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-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 22:

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°

Question 23:

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

Answer 23:

Adjacent angles:

DOA and DOCDOC and BOC

AOD and DOBBOC and AOC

Linear pairs of angles:

AOD and DOBBOC and AOC

Question 24:

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

Answer 24:

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°

Question 25:

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

Answer 25:

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

Question 26:

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

Answer 26:

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

Question 27:

In Fig., determine the value of x.

Answer 27:

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

Question 28:

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

Answer 28:

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

Question 29:

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

Answer 29:

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

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

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 30:

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

 

Page-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 31:

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)

Question 32:

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

Answer 32:

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

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