NCERT Exemplar Solutions for Class 7 Maths Chapter 5 Lines and Angles

NCERT Exemplar Solutions for Class 7 Maths Chapter 5 Lines and Angles is an essential material as it offers a wide range of questions that test the students’ understanding of concepts. Our expert personnel have answered these questions in order to assist you with your exam preparation to attain good marks in the subject. Students who wish to score good marks in Maths practise NCERT Exemplar Solutions for Class 7 Maths.

Chapter 5 – Lines and Angles solutions are available for download in pdf format and provide solutions to all the questions provided in NCERT Exemplar Class 7 Maths Chapter 5. Now, let us have a look at the concepts explained in this Chapter.

  • Related angles
    • Complementary Angles
    • Supplementary Angles
    • Adjacent Angles
    • Linear Pair
    • Vertically Opposite Angles
  • Pairs of Lines
    • Intersecting Lines
    • Transversal
    • Angles made by a Transversal
    • Transversal of Parallel Lines

Download the PDF of NCERT Exemplar Solutions for Class 7 Maths Chapter 5 Lines and Angles

 

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Access answers to Maths NCERT Exemplar Solutions for Class 7 Maths Chapter 5 Lines and Angles

Exercise Page: 128

In questions 1 to 41, there are four options out of which one is correct. Write the correct one.

1. The angles between North and West and South and East are

(a) complementary (b) supplementary

(c) both are acute (d) both are obtuse

Solution:-

(b) supplementary

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 1

The angles between North and West is 90o, angle between South and East is 90o as shown in the figure above. So, 90o + 90o = 180o.

Then, the angles between North and West and South and East are supplementary.

When the sum of the measures of two angles is 180°, the angles are called supplementary angles.

2. Angles between South and West and South and East are

(a) vertically opposite angles (b) complementary angles

(c) making a linear pair (d) adjacent but not supplementary

Solution:-

(c) making a linear pair

A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.

3. In Fig. 5.9, PQ is a mirror, AB is the incident ray and BC is the reflected ray. If ∠ ABC = 46o, then ∠ ABP is equal to

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 2

(a) 44o (b) 67o (c) 13o (d) 62o

Solution:-

(b) 67o

As we know that, the angle formed by the incident ray and angle formed by the reflected ray is equal.

From the given figure,

PQ is a straight line,

So, ∠ABP + ∠ABC + ∠CBQ = 180o

Let us assume the ∠ABP = ∠CBQ = x

Then,

x + 46o + x = 180o

2x + 46o = 180o

2x = 180o – 46o

2x = 134o

x = 134o/2

x = 67o

Therefore, the ∠ABP = ∠CBQ = 67o

4. If the complement of an angle is 79o, then the angle will be of

(a) 1o (b) 11o (c) 79o (d) 101o

Solution:-

(b) 11o

When the sum of the measures of two angles is 90°, the angles are called complementary angles. Each of them is called complement of the other.

The given complement of an angle is 79o

Let the measure of its angle be xo.

Then,

x + 79o = 90o

x = 90o – 79o

x = 11o

Hence, the measure of its angle is 11o.

5. Angles which are both supplementary and vertically opposite are

(a) 95o, 85o (b) 90o, 90o (c) 100o, 80o (d) 45o, 45o

Solution:-

(b) 90o, 90o

When the sum of the measures of two angles is 180°, the angles are called supplementary angles.

6. The angle which makes a linear pair with an angle of 61o is of

(a) 29o (b) 61o (c) 122o (d) 119o

Solution:-

(d) 119o

A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.

We know that, measure of sum of adjacent angles is equal to 180o.

Let the measure of other angle be xo.

Then,

x + 61o = 180o

x = 180o – 61o

x = 119o

7. The angles x and 90o – x are

(a) supplementary (b) complementary

(c) vertically opposite (d) making a linear pair

Solution:-

(b) complementary

When the sum of the measures of two angles is 90o, the angles are called complementary angles.

x + 90o – x = 90o

90o = 90o

LHS = RHS

8. The angles x – 10o and 190o – x are

(a) interior angles on the same side of the transversal

(b) making a linear pair

(c) complementary

(d) supplementary

Solution:-

(d) supplementary

When the sum of the measures of two angles is 180o, the angles are called supplementary angles.

x – 10o + 190o – x = 180o

190o – 10 = 180o

180o = 180o

LHS = RHS

9. In Fig. 5.10, the value of x is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 3

(a) 110o (b) 46o (c) 64o (d) 150o

Solution:-

(d) 150o

Sum of all angle about a point given in the figure are equal to 360o.

Then, 100o + 46o + 64o + x = 360o

210o + x = 360o

x = 360o – 210o

x = 150o

10. In Fig. 5.11, if AB || CD, ∠ APQ = 50o and ∠PRD = 130o, then ∠ QPR is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 4

(a) 130o (b) 50o (c) 80o (d) 30o

Solution:-

(c) 80o

We know that, ∠APR = ∠PRD … [because interior alternate angles]

∠APQ + ∠QPR = 130o

50o + ∠QPR = 130o

∠QPR = 130o – 50o

∠QPR = 80o

11. In Fig. 5.12, lines l and m intersect each other at a point. Which of the following is false?

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 5

(a) ∠a = ∠b (b) ∠d = ∠c

(c) ∠a + ∠d = 180o (d) ∠a = ∠d

Solution:-

(d) ∠a = ∠d

∠a ≠ ∠d

∠a = ∠b [because vertically opposite angles]

∠d = ∠c [because vertically opposite angles]

∠a + ∠d = 180o

12. If angle P and angle Q are supplementary and the measure of angle P is 60o, then the measure of angle Q is

(a) 120o (b) 60o (c) 30o (d) 20o

Solution:-

(a) 120o

When the sum of the measures of two angles is 180o, the angles are called supplementary angles.

P + Q = 180o

60o + Q = 180o

Q = 180o – 60o

Q = 120o

13. In Fig. 5.13, POR is a line. The value of a is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 6

(a) 40o (b) 45o (c) 55o (d) 60o

Solution:-

(a) 40o

We know that, when the sum of the measures of two angles is 180o, the angles are called supplementary angles.

(3a + 5)o + (2a – 25)o = 180o

3a + 5 + 2a – 25 = 180o

5a – 20 = 180o

5a = 180o + 20

5a = 200

a = 200/5

a = 40o

14. In Fig. 5.14, POQ is a line. If x = 30°, then ∠ QOR is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 7

(a) 90o (b) 30o (c) 150o (d) 60o

Solution:-

(a) 90o

Sum of all angle about a straight line given in the figure are equal to 180o.

Then, 30o + 2y + 3y = 180o

30o + 5y = 180o

5y = 180o – 30o

5y = 150o

y = 150/5

y = 30o

So, 2y = 2 × 30 = 60o

3y = 3 × 30 = 90o

Therefore, ∠ QOR = 90o

15. The measure of an angle which is four times its supplement is

(a) 36o (b) 144o (c) 16o (d) 64o

Solution:-

(b) 144o

We know that, when the sum of the measures of two angles is 180o, the angles are called supplementary angles.

Let us assume the angle be x.

Then, its supplement angle = (180o – x)

As per the condition given in the question, x = 4 (180o – x)

x = 720o – 4x

x + 4x = 720o

5x = 720o

x = 720o/5

x =144o

16. In Fig. 5.15, the value of y is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 8

(a) 30o (b) 15o (c) 20o (d) 22.5o

Solution:-

(c) 20o

Sum of all angle about a straight line given in the figure are equal to 180o.

Then, 6y + y + 2y = 180o

9y = 180o

y = 180/9

y = 20o

So, value of y is 20o.

17. In Fig. 5.16, PA || BC || DT and AB || DC. Then, the values of a and b are respectively.

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 9

(a) 60o, 120o (b) 50o,130o (c) 70o,110o (d) 80o,100o

Solution:-

(b) 50o,130o

We know that, ∠PAB = ∠ABC = 50o … [because interior alternate angles]

Given, AB || DC so consider it as parallelogram,

In parallelogram adjacent angles of a parallelogram are supplementary.

So, ∠ABC + ∠BCD = 180o

50o + ∠BCD = 180o

∠BCD = 180o – 50o

∠BCD = 130o

∠BCD = ∠CDT = 130o  … [because interior alternate angles]

Therefore, a = 50o and b = 130o

18. The difference of two complementary angles is 30o. Then, the angles are

(a) 60o, 30o (b) 70o, 40o (c) 20o, 50o (d) 105o, 75o

Solution:-

(a) 60o, 30o

When the sum of the measures of two angles is 90o, the angles are called complementary angles.

So, 60o + 30o = 90o

As per the condition in the question, 60o – 30o = 30o

19. In Fig. 5.17, PQ || SR and SP || RQ. Then, angles a and b are respectively

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 10

(a) 20o, 50o (b) 50o, 20o (c) 30o, 50o (d) 45o, 35o

Solution:-

(a) 20o, 50o

∠QRP = ∠RPS = 50o  … [because interior alternate angles]

∠SRP = ∠RPQ = 20o … [because interior alternate angles]

Therefore, angle a = 50o and angle b = 20o

20. In Fig. 5.18, a and b are

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 11

(a) alternate exterior angles

(b) corresponding angles

(c) alternate interior angles

(d) vertically opposite angles

Solution:-

(c) alternate interior angles

21. If two supplementary angles are in the ratio 1: 2, then the bigger angle is

(a) 120o (b) 125o (c) 110o (d) 90o

Solution:-

(a) 120o

We know that, when the sum of the measures of two angles is 180o, the angles are called supplementary angles.

Let us assume two angle be 1x and 2x.

1x + 2x = 180o

3x = 180o

x = 180o/3

x = 60o

Then the bigger angle is 2x = 2 × 60o = 120o

22. In Fig. 5.19, ∠ROS is a right angle and ∠POR and ∠QOS are in the ratio 1: 5. Then, ∠QOS measures

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 12

(a) 150o (b) 75o (c) 45o (d) 60o

Solution:-

(b) 75o

Sum of all angle about a straight line given in the figure are equal to 180o.

Given, ∠ROS is a right angle = 90o

Let us assume ∠POR = x and ∠QOS = 5x.

Then, ∠POR + ∠ROS + ∠QOS = 180o

x + 90+ 5x = 180o

6x = 180o – 90o

6x = 90o

x = 90o/6

x = 15o

So, ∠QOS measures = 5x = 5 × 15o = 75o

23. Statements a and b are as given below:

a : If two lines intersect, then the vertically opposite angles are equal.

b : If a transversal intersects, two other lines, then the sum of two interior angles on the same side of the transversal is 180o.

Then

(a) Both a and b are true (b) a is true and b is false

(c) a is false and b is true (d) both a and b are false

Solution:-

(b) a is true and b is false

24. 24. For Fig. 5.20, statements p and q are given below:

p : a and b are forming a linear pair.

q : a and b are forming a pair of adjacent angles.

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 13

Then,

(a) both p and q are true

(b) p is true and q is false

(c) p is false and q is true

(d) both p and q are false

Solution:-

(a) both p and q are true

25. In Fig. 5.21, ∠AOC and ∠ BOC form a pair of

(a) vertically opposite angles

(b) complementary angles

(c) alternate interior angles

(d) supplementary angles

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 14

Solution:-

(d) supplementary angles

26. In Fig. 5.22, the value of a is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 15

(a) 20o (b) 15o

(c) 5o (d) 10o

Solution:-

(d) 10o

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 16

∠AOF = ∠COD = 90o [because vertically opposite angles]

Sum of all angle about a straight line given in the figure are equal to 180o.

Then, ∠BOC + ∠COD + ∠DOE = 180o

40o + 90o + 5a = 180o

130o + 5a = 180o

5a = 180o – 130o

5a = 50o

a = 50/5

a = 10o

27. In Fig. 5.23, if QP || SR, the value of a is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 17

(a) 40o (b) 30o (c) 90o (d) 80o

Solution:-

(c) 90o

To find out the value of ‘a’, draw a line XY, to cut at ‘a’.

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 18

So, XY || SR

∠XTS = ∠TSR = 30o … [because interior alternate angles]

∠PQT = ∠QTX = 60o … [because interior alternate angles]

Then, a = ∠XTS + ∠QTX

= 30o + 60o

= 90o

28. In which of the following figures, a and b are forming a pair of adjacent angles?

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 19

Solution:-

In figure (d) a and b are forming a pair of adjacent angles.

29. In a pair of adjacent angles, (i) vertex is always common, (ii) one arm is always common, and (iii) uncommon arms are always opposite rays

Then

(a) All (i), (ii) and (iii) are true

(b) (iii) is false

(c) (i) is false but (ii) and (iii) are true

(d) (ii) is false

Solution:-

(b) (iii) is false

Two angles are called adjacent angles, if they have a common vertex and a common arm but no common interior points.

30. In Fig. 5.25, lines PQ and ST intersect at O. If ∠POR = 90° and x : y = 3 : 2, then z is equal to

(a) 126o (b) 144o (c) 136o (d) 154o

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 20

Solution:-

(b) 144o

Sum of all angle about a straight line given in the figure are equal to 180o.

PQ is a straight line.

Then, ∠POR + ∠ROT + ∠TOQ = 180o

Given, x : y = 3 : 2

Let us assume x = 3a, y = 2a

90o + 3a + 2a = 180o

90o + 5a = 180o

5a = 180o – 90o

5a = 90o

a = 90/5

a = 18o

So, x = 3a = 3 × 18 = 54o

Y = 2a = 2 × 18 = 36o

From the figure SOT is a straight line,

Then, Z + Y = 180o

Z + 36o = 180o

Z = 180o – 36o

Z = 144o

31. In Fig. 5.26, POQ is a line, then a is equal to

(a) 35o (b) 100o (c) 80o (d) 135o

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 21

Solution:-

(c) 80o

From the figure POQ is a straight line,

Then, 100 + a = 180o

a = 180o – 100

a= 80o

32. Vertically opposite angles are always

(a) supplementary (b) complementary

(c) adjacent (d) equal

Solution:-

(d) equal

33. In Fig. 5.27, a = 40o. The value of b is

(a) 20o (b) 24o (c) 36o (d) 120o

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 22

Solution:-

(a) 20o

Given, a = 40o

Then, 2a = 2 × 40 = 80o

From the figure, angel formed on the straight line are equal to 180o,

Then, 5b + 2a = 180o

5b + 80o = 180o

5b= 180o – 80o

5b = 100o

b = 100/5

b = 20o

34. If an angle is 60o less than two times of its supplement, then the greater angle is

(a) 100o (b) 80o (c) 60o (d) 120o

Solution:-

(a) 100o

Let us assume the angle be P.

Then, its supplement is 180o – P

As per the condition in the question,

P = 2(180o – P) – 60o

P = 360o – 2P – 60o

P + 2P = 300o

3P = 300o

P = 300/3

P = 100o

So, its supplement is 180o – P = 180o – 100o = 80o

Therefore, the greater angle is 100o.

35. In Fig. 5.28, PQ || RS. If ∠1=(2a+b)o and ∠6=(3a–b)o, then the measure of ∠2 in terms of b is

(a) (2+b)o (b) (3–b)o (c) (108–b)o (d) (180–b)o

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 23

Solution: –

(c) (108–b)o

From the question it is given that, ∠1 = (2a + b)o and ∠6 = (3a – b)o

Since ∠5 and ∠6 these two forms a linear pair

Then,

∠5 = (180-3a + b)o … [equation 1]

∠5 = ∠1 = (180-3a + b)o [Because Corresponding angles] …equation (2)

On equating equation (1) and equation (2) we get,

2a + b = 180-3a + b

 5a = 180

 a = 360

Since ∠1 and ∠2 forms a linear pair so

∠2 = 1800– 2a-b

36. In Fig. 5.29, PQ||RS and a : b = 3 : 2. Then, f is equal to

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 24

(a) 36o (b) 108o (c) 72o (d) 144o

Solution: –

(b) 108o

From the figure, PQ||RS.

From the question it is given that, a: b = 3: 2

So, let us assume a = 3m and b = 2m

We know that, sum of angles on the straight line is equal to 180o

Then, ∠a + ∠b = 180o

3m + 2m = 180o

5m = 180o

m = 180o/5

m = 36o

So, a = 3m = 3 × 36o = 108o

b = 2m = 2 × 36o = 72

Therefore, ∠a = ∠f = 108o [because corresponding angles]

37. In Fig. 5.30, line l intersects two parallel lines PQ and RS. Then, which one of the following is not true?

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 25

(a) ∠1 = ∠3 (b) ∠2 = ∠4

(c) ∠6 = ∠7 (d) ∠4 = ∠8

Solution:-

(d) ∠4 = ∠8

Because, ∠4 ≠ ∠8

38. In Fig. 5.30, which one of the following is not true?

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 26

(a) ∠1 + ∠5 = 180o

(b) ∠2 + ∠5 = 180o

(c) ∠3 + ∠8 = 180o

(d) ∠2 + ∠3 = 180o

Solution:-

(d) ∠2 + ∠3 = 180o

We know that, interior opposite angles are equal

∠2 = ∠3

39. In Fig. 5.30, which of the following is true?

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 27

(a) ∠1 = ∠5 (b) ∠4 = ∠8 (c) ∠5 = ∠8 (d) ∠3 = ∠7

Solution:-

(c) ∠5 = ∠8

From the figure, PQ||RS

∠5 = ∠8 [interior alternate angles are equal]

40. In Fig. 5.31, PQ||ST. Then, the value of x + y is

(a) 125o (b) 135o (c) 145o (d) 120o

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 28

Solution: –

(b) 135o

From the figure, PQ is a straight line

We know that, sum of angle on the straight is equal to 180o.

Then,

y + ∠ PQR = 1800 

y + 1300 = 1800

y = 50o

Then,

∠ QOS = ∠ TSO [Co-interior angle]

x = 850

x + y = 135

41. In Fig. 5.32, if PQ||RS and QR||TS, then the value a is

NCERT Exemplar Class 7 Maths Solutions Chapter 5 Image 29

(a) 95o (b) 90o (c) 85o (d) 75o

Solution:-

(a) 95o

We know that, corresponding angles are equal

So,

∠ RQP = ∠ TSR = 850 (Corresponding angles)

a + ∠ TSR = 1800

∠a = 95

In questions 42 to 56, fill in the blanks to make the statements true.

42. If sum of measures of two angles is 90o, then the angles are _________.

Solution:-

If sum of measures of two angles is 90o, then the angles are complementary.

43. If the sum of measures of two angles is 180o, then they are _________.

Solution:-

If the sum of measures of two angles is 180o, then they are supplementary.

44. A transversal intersects two or more than two lines at _________ points.

Solution:-

A transversal intersects two or more than two lines at distinct points.

If a transversal intersects two parallel lines, then (Q. 45 to 48).

45. sum of interior angles on the same side of a transversal is .

Solution:-

Sum of interior angles on the same side of a transversal is 180o.

46. Alternate interior angles have one common .

Solution:-

Alternate interior angles have one common arm.

47. Corresponding angles are on the side of the transversal.

Solution:-

Corresponding angles are on the same side of the transversal.

48. Alternate interior angles are on the side of the transversal.

Solution:-

Alternate interior angles are on the opposite side of the transversal

49. Two lines in a plane which do not meet at a point anywhere are called lines.

Solution:-

Two lines in a plane which do not meet at a point anywhere are called parallel lines.

50. Two angles forming a __________ pair are supplementary.

Solution:-

Two angles forming a parallel pair are supplementary.

51. The supplement of an acute is always __________ angle.

Solution:-

The supplement of an acute is always obtuse angle.

52. The supplement of a right angle is always _________ angle.

Solution:-

The supplement of a right angle is always right angle.

53. The supplement of an obtuse angle is always _________ angle.

Solution:-

The supplement of an obtuse angle is always acute angle.

54. In a pair of complementary angles, each angle cannot be more than _________.

Solution:-

In a pair of complementary angles, each angle cannot be more than 90o.

55. An angle is 45o. Its complementary angle will be __________ .

Solution:-

An angle is 45o. Its complementary angle will be 45o.

56. An angle which is half of its supplement is of __________.

Solution:-

An angle which is half of its supplement is of 60o.

Let us assume the angle be p, and supplement be 2p

p + 2p = 1800

3p = 1800

p = 600

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