RD Sharma Solutions for Class 7 Maths Exercise 14.1 of Chapter 14 Lines And Angles in PDF are available here. Students can refer and download it from the available links. This exercise has thirty-two questions. RD Sharma Solutions for Class 7 provides solutions for all topics covered in this exercise. This exercise 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

## Download the PDF of RD Sharma Solutions For Class 7 Chapter 14 – Lines And Angles Exercise 14.1

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 14 – Lines And Angles Exercise 14.1

**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:

âˆ DOC and âˆ BOC

âˆ COB and âˆ 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

âˆ EBA and âˆ ABC

âˆ ACB and âˆ BCF

âˆ BAC and âˆ CAD

In fig (ii), the adjacent angles are

âˆ BAD and âˆ DAC

âˆ BDA and âˆ 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 â€“ common arms are two opposite rays.

âˆ 1 and âˆ 3

âˆ 1 and âˆ 2

âˆ 4 and âˆ 3

âˆ 4 and âˆ 2

âˆ 5 and âˆ 6

âˆ 5 and âˆ 7

âˆ 6 and âˆ 8

âˆ 7 and âˆ 8

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

âˆ 1 and âˆ 4

âˆ 2 and âˆ 3

âˆ 5 and âˆ 8

âˆ 6 and âˆ 7

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

**Solution:**

No, because they donâ€™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} â€“ 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

90^{0} â€“ 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 }â€“ 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} â€“ 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 question is

180^{o} â€“ 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 question is

180^{o} â€“ 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 question is

180^{o} â€“ 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 question is

180^{o} â€“ 90^{o} = 90^{o}

Â

**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

âˆ AOC and âˆ COB

âˆ BOC and âˆ DOB

âˆ BOD and âˆ DOA

âˆ AOC and âˆ DOA

**10. In Fig. 18, A, B, C are collinear points and âˆ DBA = âˆ 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

âˆ ABD and âˆ DBC

âˆ ABE and âˆ EBC

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

âˆ ABD and âˆ DBC

âˆ ABE and âˆ 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}

âˆ p = âˆ q

So,

âˆ p + âˆ q = 180^{o}

âˆ p + âˆ p = 180^{o}

2âˆ p = 180^{o}

âˆ p = 180^{o}/2

âˆ p = 90^{o}

Therefore, âˆ p = âˆ 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, âˆ x + 28^{o} = 90^{o}

âˆ x = 90^{o} â€“ 28^{o}

= 62^{o}

So, the supplement of the angle = 180^{o} â€“ 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:

âˆ 1 and âˆ 2

âˆ 2 and âˆ 3

âˆ 3 and âˆ 4

âˆ 1 and âˆ 4

âˆ 5 and âˆ 6

âˆ 6 and âˆ 7

âˆ 7 and âˆ 8

âˆ 8 and âˆ 5

âˆ 9 and âˆ 10

âˆ 10 and âˆ 11

âˆ 11 and âˆ 12

âˆ 12 and âˆ 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:

âˆ 1 and âˆ 3

âˆ 4 and âˆ 2

âˆ 5 and âˆ 7

âˆ 6 and âˆ 8

âˆ 9 and âˆ 11

âˆ 10 and âˆ 12

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

** **

** **

**Solution:**

Given, âˆ 1 = 70^{o}

âˆ 3 = 2(âˆ 1)

= 2(70^{o})

âˆ 3 = 140^{o}

âˆ 3 = âˆ 4

As, OE is the angle bisector,

âˆ DOB = 2(âˆ 1)

= 2(70^{o})

= 140^{o}

âˆ DOB + âˆ AOC + âˆ COB +âˆ DOB = 360^{o} [sum of the angle of circle = 360^{o}]

140^{o} + 140^{o} + 2(âˆ COB) = 360^{o}

Since, âˆ COB = âˆ AOD

2(âˆ COB) = 360^{o} â€“ 280^{o}

2(âˆ COB) = 80^{o}

âˆ COB = 80^{o}/2

âˆ COB = 40^{o}

Therefore, âˆ COB = âˆ AOB = 40^{o}

The angles are, âˆ 1 = 70^{o}, âˆ 2 = 40^{o}, âˆ 3 = 140^{o} and âˆ 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} â€“ 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} â€“ 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) Â We know that âˆ BOA + âˆ BOC = 180^{o}

^{o}]

60^{o} + x^{o} = 180^{o}

x^{o} = 180^{o} â€“ 60^{o}

x^{o} = 120^{o}

(ii) We know that âˆ POQ + âˆ QOR = 180^{o}

^{o}]

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

5x^{o} = 180^{o}

x^{o} = 180^{o}/5

x^{o} = 36^{o}

(iii) We know that âˆ LOP + âˆ PON + âˆ NOM = 180^{o}

^{o}]

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

x^{o} = 180^{o} â€“ 35^{o} â€“ 60^{o}

x^{o} = 180^{o} â€“ 95^{o}

x^{o} = 85^{o}

(iv) We know that âˆ DOC + âˆ DOE + âˆ EOA + âˆ AOB+ âˆ 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} â€“ 297^{o}

x^{o} = 63^{o}

(v) We know that âˆ ROS + âˆ ROQ + âˆ QOP + âˆ 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) Â Linear pair: The two adjacent angles are said to form a linear pair of angles if their nonâ€“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} = 45^{o}

**21. In Fig. 22, it being given that âˆ 1 = 65 ^{o}, find all other angles.**

** **

**Solution:**

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

Therefore, âˆ 3 = 65^{o}

Here, âˆ 1 + âˆ 2 = 180Â° are the linear pair [The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180^{o}]

Therefore, âˆ 2 = 180^{o} â€“ 65^{o}

= 115^{o}

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

Therefore, âˆ 2 = âˆ 4 = 115^{o}

And âˆ 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) âˆ AOC + âˆ BOC = 180^{o} [The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180^{o}]

2y + 5^{0} + 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} â€“ 75^{o}

2y = 100^{o}

y = 100^{o}/2

y = 50^{o}

(ii) âˆ AOC + âˆ BOC = 180^{o} [The two adjacent angles are said to form a linear pair of angles if their nonâ€“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 = 175^{0} â€“ 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:

âˆ DOA and âˆ DOC

âˆ BOC and âˆ COD

âˆ AOD and âˆ BOD

âˆ AOC and âˆ BOC

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

âˆ AOD and âˆ BOD

âˆ AOC and âˆ BOC

**24. In Fig. 25, find âˆ x. Further find âˆ BOC, âˆ COD and âˆ 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} â€“ 30^{o}

3x^{o} = 150^{o}

x^{o} = 150^{o}/3

x^{o} = 50^{o}

Also given that

âˆ BOC = (x + 20)^{o}

= (50 + 20)^{o}

= 70^{o}

âˆ COD = 50^{o}

âˆ 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

âˆ EOD and âˆ DOC

âˆ COD and âˆ BOC

âˆ COB and âˆ BOA

âˆ AOB and âˆ BOD

âˆ BOC and âˆ COE

âˆ COD and âˆ COA

âˆ DOE and âˆ DOB

âˆ EOD and âˆ DOA

âˆ EOC and âˆ AOC

âˆ AOB and âˆ BOE

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

**Solution:**

From the figure we can write as âˆ COB + âˆ 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

âˆ AOB + âˆ BOC = 180^{o} [linear pair]

Linear pair

2x + 70^{o} = 180^{o}

2x = 180^{o} â€“ 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,

âˆ QOP + âˆ QOR + âˆ ROS = 180^{o}

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

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

4x = 180^{o} â€“ 100^{o}

4x = 80^{o}

x = 80^{o}/4

x = 20^{o}

**30. In Fig. 30, lines l _{1 }and l_{2} 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, âˆ x = 45^{o}

From the figure we can write as

âˆ x = âˆ z = 45^{o}

Also from the figure, we have

âˆ y = âˆ u

From the property of linear pair we can write as

âˆ x + âˆ y + âˆ z + âˆ u = 360^{o}

45^{o} + 45^{o} + âˆ y + âˆ u = 360^{o}

90^{o} + âˆ y + âˆ u = 360^{o}

âˆ y + âˆ u = 360^{o} â€“ 90^{o}

âˆ y + âˆ u = 270^{o}

âˆ y + âˆ z = 270^{o}

2âˆ z = 270^{o}

âˆ z = 135^{o}

Therefore, âˆ y = âˆ u = 135^{o}

So, âˆ x = 45^{o}, âˆ y = 135^{o}, âˆ z = 45^{o} and âˆ 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, âˆ x + âˆ y + âˆ z+ âˆ u + 50^{o} + 90^{o} = 360^{o}

Linear pair, âˆ x + 50^{o} + 90^{o} = 180^{o}

âˆ x + 140^{o} = 180^{o}

On rearranging we get

âˆ x = 180^{o} â€“ 140^{o}

âˆ x = 40^{o}

From the figure we can write as

âˆ x = âˆ u = 40^{o} are vertically opposite angles

âˆ z = 90^{o} is a vertically opposite angle

âˆ y = 50^{o} is a vertically opposite angle

Therefore, âˆ x = 40^{o}, âˆ y = 50^{o}, âˆ z = 90^{o} and âˆ u = 40^{o}

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

** **

**Solution:**

âˆ y = 25^{o} vertically opposite angle

From the figure we can write as

âˆ x = âˆ y are vertically opposite angles

âˆ x + âˆ y + âˆ z + 25^{o} = 360^{o}

âˆ x + âˆ z + 25^{o} + 25^{o} = 360^{o}

On rearranging we get,

âˆ x + âˆ z + 50^{o} = 360^{o}

âˆ x + âˆ z = 360^{o} â€“ 50^{o} [âˆ x = âˆ z]

2âˆ x = 310^{o}

âˆ x = 155^{o}

And, âˆ x = âˆ z = 155^{o}

Therefore, âˆ x = 155^{o}, âˆ y = 25^{o} and âˆ z = 155^{o}

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