RD Sharma Solutions For Class 7 Maths Chapter 14 Lines And Angles Exercise 14.1

RD Sharma Solutions for Class 7 Maths Exercise 14.1 of Chapter 14 Lines And Angles in PDF is available here. Students can refer and download it from the available links. This exercise has thirty-two questions. RD Sharma Solutions for Class 7 provide solutions for all topics covered in this chapter. Exercise 14.1 mainly deals with a pair of angles and their properties. Let us have a look at some of the important topics present in this exercise.

• Pairs of angles
• Definition and meaning of adjacent angles
• Definition and meaning of linear pair
• Vertically opposite angles
• Definition of angles at a point
• Concept of complementary angles
• Concept of supplementary angles

RD Sharma Solutions for Class 7 Maths Chapter 14 – Lines and Angles Exercise 14.1

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Exercise 14.1 Page No: 14.6

1. Write down each pair of adjacent angles shown in fig. 13.

Solution:

The angles that have common vertex and a common arm are known as adjacent angles

Therefore the adjacent angles in given figure are:

∠DOC and ∠BOC

∠COB and ∠BOA

2. In Fig. 14, name all the pairs of adjacent angles.

Solution:

The angles that have common vertex and a common arm are known as adjacent angles.

In fig (i), the adjacent angles are

∠EBA and ∠ABC

∠ACB and ∠BCF

∠BAC and ∠CAD

In fig (ii), the adjacent angles are

∠BAD and ∠DAC

∠BDA and ∠CDA

3. In fig. 15, write down

(i) Each linear pair

(ii) Each pair of vertically opposite angles.

Solution:

(i) The 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) The two angles formed by two intersecting lines and have no common arms are called vertically opposite angles.

∠1 and ∠4

∠2 and ∠3

∠5 and ∠8

∠6 and ∠7

4. Are the angles 1 and 2 given in Fig. 16 adjacent angles?

Solution:

No, because they don’t have common vertex.

5. Find the complement of each of the following angles:

(i) 35^o

(ii) 72^o

(iii) 45^o

(iv) 85^o

Solution:

(i) The two angles are said to be complementary angles if the sum of those angles is 90^o

Complementary angle for given angle is

90^o – 35^o = 55^o

(ii) The two angles are said to be complementary angles if the sum of those angles is 90^o

Complementary angle for given angle is

90⁰ – 72^o = 18^o

(iii) The two angles are said to be complementary angles if the sum of those angles is 90^o

Complementary angle for given angle is

90^o – 45^o = 45^o

(iv) The two angles are said to be complementary angles if the sum of those angles is 90^o

Complementary angle for given angle is

90^o – 85^o = 5^o

6. Find the supplement of each of the following angles:

(i) 70^o

(ii) 120^o

(iii) 135^o

(iv) 90^o

Solution:

(i) The two angles are said to be supplementary angles if the sum of those angles is 180^o

Therefore supplementary angle for the given angle is

180^o – 70^o = 110^o

(ii) The two angles are said to be supplementary angles if the sum of those angles is 180^o

Therefore supplementary angle for the given angle is

180^o – 120^o = 60^o

(iii) The two angles are said to be supplementary angles if the sum of those angles is 180^o

Therefore supplementary angle for the given angle is

180^o – 135^o = 45^o

(iv) The two angles are said to be supplementary angles if the sum of those angles is 180^o

Therefore supplementary angle for the given angle is

180^o – 90^o = 90^o

7. Identify the complementary and supplementary pairs of angles from the following pairs:

(i) 25^o, 65^o

(ii) 120^o, 60^o

(iii) 63^o, 27^o

(iv) 100^o, 80^o

Solution:

(i) 25^o + 65^o = 90^o so, this is a complementary pair of angle.

(ii) 120^o + 60^o = 180^o so, this is a supplementary pair of angle.

(iii) 63^o + 27^o = 90^o so, this is a complementary pair of angle.

(iv) 100^o + 80^o = 180^o so, this is a supplementary pair of angle.

8. Can two obtuse angles be supplementary, if both of them be

(i) Obtuse?

(ii) Right?

(iii) Acute?

Solution:

(i) No, two obtuse angles cannot be supplementary

Because, the sum of two angles is greater than 90^o so their sum will be greater than 180^o

(ii) Yes, two right angles can be supplementary

Because, 90^o + 90^o = 180^o

(iii) No, two acute angle cannot be supplementary

Because, the sum of two angles is less than 90^o so their sum will also be less than 90^o

9. Name the four pairs of supplementary angles shown in Fig.17.

Solution:

The two angles are said to be supplementary angles if the sum of those angles is 180^o

The supplementary angles are

∠AOC and ∠COB

∠BOC and ∠DOB

∠BOD and ∠DOA

∠AOC and ∠DOA

10. In Fig. 18, A, B, C are collinear points and ∠DBA = ∠EBA.

(i) Name two linear pairs.

(ii) Name two pairs of supplementary angles.

Solution:

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

Therefore linear pairs are

∠ABD and ∠DBC

∠ABE and ∠EBC

(ii) We know that every linear pair forms supplementary angles, these angles are

∠ABD and ∠DBC

∠ABE and ∠EBC

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

Solution:

Let p and q be the two supplementary angles that are equal

The two angles are said to be supplementary angles if the sum of those angles is 180^o

∠p = ∠q

So,

∠p + ∠q = 180^o

∠p + ∠p = 180^o

2∠p = 180^o

∠p = 180^o/2

∠p = 90^o

Therefore, ∠p = ∠q = 90^o

12. If the complement of an angle is 28^o, then find the supplement of the angle.

Solution:

Given complement of an angle is 28^o

Here, let x be the complement of the given angle 28^o

Therefore, ∠x + 28^o = 90^o

∠x = 90^o – 28^o

= 62^o

So, the supplement of the angle = 180^o – 62^o

= 118^o

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

Solution:

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

Therefore linear pairs are listed below:

∠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

The two angles are said to be vertically opposite angles if the two intersecting lines have no common arms.

Therefore supplement of the angle are listed below:

∠1 and ∠3

∠4 and ∠2

∠5 and ∠7

∠6 and ∠8

∠9 and ∠11

∠10 and ∠12

14. In Fig. 20, OE is the bisector of ∠BOD. If ∠1 = 70^o, find the magnitude of ∠2, ∠3 and ∠4.

Solution:

Given, ∠1 = 70^o

∠3 = 2(∠1)

= 2(70^o)

∠3 = 140^o

∠3 = ∠4

As, OE is the angle bisector,

∠DOB = 2(∠1)

= 2(70^o)

= 140^o

∠DOB + ∠AOC + ∠COB +∠AOD = 360^o [sum of the angle of circle = 360^o]

140^o + 140^o + 2(∠COB) = 360^o

Since, ∠COB = ∠AOD

2(∠COB) = 360^o – 280^o

2(∠COB) = 80^o

∠COB = 80^o/2

∠COB = 40^o

Therefore, ∠COB = ∠AOD = 40^o

The angles are, ∠1 = 70^o, ∠2 = 40^o, ∠3 = 140^o and ∠4 = 40^o

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

Solution:

Given one of the angle of a linear pair is the right angle that is 90^o

We know that linear pair angle is 180^o

Therefore, the other angle is

180^o – 90^o = 90^o

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

Solution:

Given one of the angles of a linear pair is obtuse, then the other angle should be acute, because only then their sum will be 180^o.

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

Solution:

Given one of the Angles of a linear pair is acute, then the other angle should be obtuse, only then their sum will be 180^o.

18. Can two acute angles form a linear pair?

Solution:

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

19. If the supplement of an angle is 65^o, then find its complement.

Solution:

Let x be the required angle

So, x + 65^o = 180^o

x = 180^o – 65^o

x = 115^o

The two angles are said to be complementary angles if the sum of those angles is 90^o here it is more than 90^o therefore the complement of the angle cannot be determined.

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

Solution:

(i)  We know that ∠BOA + ∠BOC = 180^o

60^o + x^o = 180^o

x^o = 180^o – 60^o

x^o = 120^o

(ii) We know that ∠POQ + ∠QOR = 180^o

3x^o + 2x^o = 180^o

5x^o = 180^o

x^o = 180^o/5

x^o = 36^o

(iii) We know that ∠LOP + ∠PON + ∠NOM = 180^o

Since, 35^o + x^o + 60^o = 180^o

x^o = 180^o – 35^o – 60^o

x^o = 180^o – 95^o

x^o = 85^o

(iv) We know that ∠DOC + ∠DOE + ∠EOA + ∠AOB+ ∠BOC = 360^o

83^o + 92^o + 47^o + 75^o + x^o = 360^o

x^o + 297^o = 360^o

x^o = 360^o – 297^o

x^o = 63^o

(v) We know that ∠ROS + ∠ROQ + ∠QOP + ∠POS = 360^o

3x^o + 2x^o + x^o + 2x^o = 360^o

8x^o = 360^o

x^o = 360^o/8

x^o = 45^o

(vi)  Linear pair: The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180^o

Therefore 3x^o = 105^o

x^o = 105^o/3

x^o = 35^o

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

Solution:

Given from the figure 22, ∠1 = ∠3 are the vertically opposite angles

Therefore, ∠3 = 65^o

Here, ∠1 + ∠2 = 180° are the linear pair [The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180^o]

Therefore, ∠2 = 180^o – 65^o

= 115^o

∠2 = ∠4 are the vertically opposite angles [from the figure]

Therefore, ∠2 = ∠4 = 115^o

And ∠3 = 65^o

22. In Fig. 23, OA and OB are opposite rays:

(i) If x = 25^o, what is the value of y?

(ii) If y = 35^o, what is the value of x?

Solution:

(i) ∠AOC + ∠BOC = 180^o [The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180^o]

2y + 5⁰ + 3x = 180^o

3x + 2y = 175^o

Given If x = 25^o, then

3(25^o) + 2y = 175^o

75^o + 2y = 175^o

2y = 175^o – 75^o

2y = 100^o

y = 100^o/2

y = 50^o

(ii) ∠AOC + ∠BOC = 180^o [The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180^o]

2y + 5 + 3x = 180^o

3x + 2y = 175^o

Given If y = 35^o, then

3x + 2(35^o) = 175^o

3x + 70^o = 175^o

3x = 175⁰ – 70^o

3x = 105^o

x = 105^o/3

x = 35^o

23. In Fig. 24, write all pairs of adjacent angles and all the liner pairs.

Solution:

Pairs of adjacent angles are:

∠DOA and ∠DOC

∠BOC and ∠COD

∠AOD and ∠BOD

∠AOC and ∠BOC

Linear pairs: [The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180^o]

∠AOD and ∠BOD

∠AOC and ∠BOC

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

Solution:

(x + 10)^o + x^o + (x + 20)^o = 180^o[linear pair]

On rearranging we get

3x^o + 30^o = 180^o

3x^o = 180^o – 30^o

3x^o = 150^o

x^o = 150^o/3

x^o = 50^o

Also given that

∠BOC = (x + 20)^o

= (50 + 20)^o

= 70^o

∠COD = 50^o

∠AOD = (x + 10)^o

= (50 + 10)^o

= 60^o

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

Solution:

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

26. How many pairs of adjacent angles, in all, can you name in Fig. 26?

Solution:

There are 10 adjacent pairs formed in the given figure, they are

∠EOD and ∠DOC

∠COD and ∠BOC

∠COB and ∠BOA

∠AOB and ∠BOD

∠BOC and ∠COE

∠COD and ∠COA

∠DOE and ∠DOB

∠EOD and ∠DOA

∠EOC and ∠AOC

∠AOB and ∠BOE

27. In Fig. 27, determine the value of x.

Solution:

From the figure we can write as ∠COB + ∠AOB = 180^o [linear pair]

3x^o + 3x^o = 180^o

6x^o = 180^o

x^o = 180^o/6

x^o = 30^o

28. In Fig.28, AOC is a line, find x.

Solution:

From the figure we can write as

∠AOB + ∠BOC = 180^o [linear pair]

Linear pair

2x + 70^o = 180^o

2x = 180^o – 70^o

2x = 110^o

x = 110^o/2

x = 55^o

29. In Fig. 29, POS is a line, find x.

Solution:

From the figure we can write as angles of a straight line,

∠QOP + ∠QOR + ∠ROS = 180^o

60^o + 4x + 40^o = 180^o

On rearranging we get, 100^o + 4x = 180^o

4x = 180^o – 100^o

4x = 80^o

x = 80^o/4

x = 20^o

30. In Fig. 30, lines l₁ and l₂ intersect at O, forming angles as shown in the figure. If x = 45^o, find the values of y, z and u.

Solution:

Given that, ∠x = 45^o

From the figure we can write as

∠x = ∠z = 45^o

Also from the figure, we have

∠y = ∠u

From the property of linear pair we can write as

∠x + ∠y + ∠z + ∠u = 360^o

45^o + 45^o + ∠y + ∠u = 360^o

90^o + ∠y + ∠u = 360^o

∠y + ∠u = 360^o – 90^o

∠y + ∠u = 270^o (vertically opposite angles ∠y = ∠u)

2∠y = 270^o

∠y = 135^o

Therefore, ∠y = ∠u = 135^o

So, ∠x = 45^o, ∠y = 135^o, ∠z = 45^o and ∠u = 135^o

31. In Fig. 31, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u

Solution:

Given that, ∠x + ∠y + ∠z+ ∠u + 50^o + 90^o = 360^o

Linear pair, ∠x + 50^o + 90^o = 180^o

∠x + 140^o = 180^o

On rearranging we get

∠x = 180^o – 140^o

∠x = 40^o

From the figure we can write as

∠x = ∠u = 40^o are vertically opposite angles

∠z = 90^o is a vertically opposite angle

∠y = 50^o is a vertically opposite angle

Therefore, ∠x = 40^o, ∠y = 50^o, ∠z = 90^o and ∠u = 40^o

32. In Fig. 32, find the values of x, y and z.

Solution:

∠y = 25^o vertically opposite angle

From the figure we can write as

∠x = ∠z are vertically opposite angles

∠x + ∠y + ∠z + 25^o = 360^o

∠x + ∠z + 25^o + 25^o = 360^o

On rearranging we get,

∠x + ∠z + 50^o = 360^o

∠x + ∠z = 360^o – 50^o [∠x = ∠z]

2∠x = 310^o

∠x = 155^o

And, ∠x = ∠z = 155^o

Therefore, ∠x = 155^o, ∠y = 25^o and ∠z = 155^o