Selina Solutions Concise Maths Class 7 Chapter 14 Lines and Angles (Including Construction of Angles) are designed by highly knowledgeable faculty at BYJU’S, keeping in mind the understanding capacity of students. The PDF of solutions prepare students to solve problems without any difficulty. It also improves confidence among students to appear for the annual exam. Selina Solutions Concise Maths Class 7 Chapter 14 Lines and Angles (Including Construction of Angles), PDF links are given here, with a free download option.

Chapter 14 deals with basic terms, types of lines, types of angles and the construction of angles. In the solutions, students are provided with the steps to be followed in the construction of various structures.

## Selina Solutions Concise Maths Class 7 Chapter 14: Lines and Angles (Including Construction of Angles) Download PDF

### Exercises of Selina Solutions Concise Maths Class 7 Chapter 14 – Lines and Angles (Including Construction of Angles)

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#### Exercise 14A page: 162

**1. State, true or false:**

**(i) A line segment 4 cm long can have only 2000 points in it.
(ii) A ray has one end point and a line segment has two end-points.
(iii) A line segment is the shortest distance between any two given points.
(iv) An infinite number of straight lines can be drawn through a given point.
(v) Write the number of end points in
(a) a line segment AB (b) a ray AB (c) a line AB
(vi) Out of which one has a fixed length?
(vii) How many rays can be drawn through a fixed point O?
(viii) How many lines can be drawn through three
(a) collinear points?
(b) non-collinear points?
(ix) Is 40° the complement of 60°?
(x) Is 45° the supplement of 45°?**

**Solution:**

(i) False.

It contains infinite number of points.

(ii) True.

(iii) True.

(iv) True.

(v) (a) 2 (b) 1 (c) 0

(vi) has fixed length.

(vii) Infinite rays can be drawn through a fixed point O.

(viii) (a) 1 line can be drawn through three collinear points.

(b) 3 lines can be drawn through three non-collinear points.

(ix) False.

40^{o} is the complement of 50^{o} as 40^{o} + 50^{o} = 90^{o}

(x) False.

45^{o} is the supplement of 135^{o} not 45^{o}.

**2. In which of the following figures, are ∠AOB and ∠AOC adjacent angles? Give, in each case, reason for your answer. **

**Solution:**

If ∠AOB and ∠AOC are adjacent angle, they have OA as their common arm.

(i) From the figure

OB is the common arm

∠AOB and ∠AOC are not adjacent angles.

(ii) From the figure

OC is the common arm

∠AOB and ∠AOC are not adjacent angles.

(iii) From the figure

OA is the common arm

∠AOB and ∠AOC are adjacent angles.

(iv) From the figure

OB is the common arm

∠AOB and ∠AOC are not adjacent angles.

**3. In the given figure, AOC is a straight line.
Find: (i) x (ii) ∠AOB (iii) ∠BOC**

**Solution:**

We know that

∠AOB and ∠COB are linear pairs

It can be written as

∠AOB + ∠COB = 180^{o}

Substituting the values

x + 25^{o} + 3x + 15^{o} = 180^{o}

By further calculation

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

So we get

4x = 180 – 40 = 140^{o}

(i) x = 140/4 = 35^{o}

(ii) ∠AOB = x + 25

Substituting the value of x

∠AOB = 25 + 35 = 60^{o}

(iii) ∠BOC = 3x + 15^{o}

Substituting the value of x

∠BOC = (3 × 35) + 15

∠BOC = 120^{o}

**4. Find y in the given figure.**

**Solution:**

Here AOC is a straight line

We can write it as

∠AOB + ∠BOD + ∠DOC = 180^{o}

Substituting the values

y + 150 – x + x = 180

By further calculation

y + 150 = 180

So we get

y = 180 – 150 = 30^{o}

**5. In the given figure, find ∠PQR.**

**Solution:**

Here SQR is a straight line

We can write it as

∠SQT + ∠TQP + ∠PQR = 180^{o}

Substituting the values

x + 70 + 20 – x + ∠PQR = 180^{o}

By further calculation

90^{o} + ∠PQR = 180^{o}

So we get

∠PQR = 180^{o} – 90^{o} = 90^{o}

**6. In the given figure, p ^{o} = q^{o} = r^{o}, find each.**

**Solution:**

We know that

p^{o} + q^{o} + r^{o} = 180^{o} is a straight angle

It is given that

p^{o} = q^{o} = r^{o}

We can write it as

p^{o} + p^{o} + p^{o} = 180^{o}

3p = 180

p = 180/3 = 60^{o}

Therefore, p^{o} = q^{o} = r^{o} = 60^{o}

**7. In the given figure, if x = 2y, find x and y.**

**Solution:**

It is given that

x = 2y

For a straight angle

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

Substituting the values

2y + y = 180

By further calculation

3y = 180

y = 180/3 = 60^{o}

x = 2y = 2 × 60^{o} = 120^{o}

**8. In the adjoining figure, if b ^{o} = a^{o} + c^{o}, find b.**

**Solution:**

It is given that

b^{o} = a^{o} + c^{o}

For a straight angle

a^{o} + b^{o} + c^{o} = 180^{o}

Substituting the values

b^{o} + b^{o} = 180^{o}

2b^{o} = 180^{o}

b^{o} = 180/2 = 90^{o}

**9. In the given figure, AB is perpendicular to BC at B.
Find : (i) the value of x.
(ii) the complement of angle x.**

**Solution:**

(i) From the figure

AB || BC at B

Here ∠ABC = 90^{o}

Substituting the values

x + 20 + 2x + 1 + 7x – 11 = 90

By further calculation

10x + 10 = 90

10x = 90 – 10 = 80

x = 80/10 = 8^{o}

(ii) The complement of angle x = 90 – x

So we get

= 90 – 8 = 82^{o}

**10. Write the complement of:**

**(i) 25 ^{o}**

**(ii) 90 ^{o}**

**(iii) a ^{o}**

**(iv) (x + 5) ^{o}**

**(v) (30 – a) ^{o}**

**(vi) ½ of a right angle**

**(vii) 1/3 of 180 ^{o}**

**(viii) 21 ^{o} 17’**

**Solution:**

(i) The complement of 25^{o} = 90^{o} – 25^{o} = 65^{o}

(ii) The complement of 90^{o} = 90^{o} – 90^{o} = 0

(iii) The complement of a^{o} = 90^{o} – a^{o}

(iv) The complement of (x + 5)^{o} = 90^{o} – (x + 5)^{o}

By further calculation

= 90^{o} – x – 5^{o}

= 85^{o} – x

(v) The complement of (30 – a)^{o} = 90^{o} – (30 – a)^{o}

By further calculation

= 90^{o} – 30^{o} + a^{o}

= 60^{o} + a^{o}

(vi) The complement of ½ of a right angle = 90^{o} – ½ of a right angle

So we get

= 90^{o} – ½ × 90^{o}

= 90^{o} – 45^{o}

= 45^{o}

(vii) The complement of 1/3 of 180^{o} = 90^{o} – 1/3 of 180^{o}

By further calculation

= 90^{o} – 60^{o}

= 30^{o}

(viii) The complement of 21^{o} 17’ = 90^{o} – 21^{o} 17’

So we get

= 68^{o} 43’

**11. Write the supplement of:**

**(i) 100°
(ii) 0°
(iii) x°
(iv) (x + 35)°
(v) (90 +a + b)°
(vi) (110 – x – 2y)°
(vii) 1/5 of a right angle
(viii) 80° 49′ 25″**

**Solution:**

(i) The supplement of 100° = 180 – 100 = 80^{0}

(ii) The supplement of 0° = 180 – 0 = 180^{o}

(iii) The supplement of x° = 180^{0} – x^{0}

(iv) The supplement of (x + 35)° = 180^{0} – (x + 35)^{0}

We can write it as

= 180 – x – 35

= 145^{0} – x^{0}

(v) The supplement of (90 + a + b)° = 180^{0} – (90 + a + b)^{0}

We can write it as

= 180 – 90 – a – b

So we get

= 90^{0} – a^{0} – b^{0}

= (90 – a – b)^{0}

(vi) The supplement of (110 – x – 2y)° = 180^{0} – (110 – x – 2y)°

We can write it as

= 180 – 110 + x + 2y

= 70^{0} + x^{0} + 2y^{0}

(vii) The supplement of 1/5 of a right angle = 180^{0} – 1/5 of a right angle

We can write it as

= 180^{0} – 1/5 × 90^{0}

So we get

= 180^{0} – 18^{0}

= 162^{0}

(viii) The supplement of 80° 49′ 25″ = 180^{0} – 80° 49′ 25″

We know that 1^{0} = 60’ and 1’ = 60”

So we get

= 99^{0} 10’ 35”

**12. Are the following pairs of angles complementary?**

**(i) 10° and 80°
(ii) 37° 28′ and 52° 33′
(iii) (x+ 16)°and (74 – x)°
(iv) 54° and 2/5 of a right angle.**

**Solution:**

(i) 10° and 80°

Yes, they are complementary angles as their sum = 10^{0} + 80^{0} = 90^{0}

(ii) 37° 28′ and 52° 33′

No, they are not complementary angles as their sum is not equal to 90^{0}

37° 28′ + 52° 33′ = 90^{0}1’

(iii) (x+ 16)° and (74 – x)°

Yes, they are complementary angles as their sum = x + 16 + 74 – x = 90^{0}

(iv) 54° and 2/5 of a right angle

We can write it as

= 54^{0} and 2/5 × 90^{0}

= 54^{0} and 36^{0}

Yes, they are complementary angles as their sum = 54 + 36 = 90^{0}

**13. Are the following pairs of angles supplementary?**

**(i) 139 ^{o} and 39^{o}**

**(ii) 26 ^{o}59’ and 153^{o}1’**

**(iii) 3/10 of a right angle and 4/15 of two right angles**

**(iv) 2x ^{o} + 65^{o} and 115^{o} – 2x^{o}**

**Solution:**

(i) 139^{o} and 39^{o}

No, they are not supplementary angles as their sum is not equal to 180^{0}

139^{0} + 39^{0} = 178^{0}

(ii) 26^{o}59’ and 153^{o}1’

Yes, they are supplementary angles as their sum = 26^{o}59’ + 153^{o}1’ = 180^{0}

(iii) 3/10 of a right angle and 4/15 of two right angles

We can write it as

= 3/10 of 90^{0} and 4/15 of 180^{0}

= 27^{0} and 48^{0}

No, they are not supplementary angles as their sum is not equal to 180^{0}

27^{0} + 48^{0} = 75^{0}

(iv) 2x^{o} + 65^{o} and 115^{o} – 2x^{o}

Yes they are supplementary angles as their sum = 2x + 65 + 115 – 2x = 180^{0}

**14. If 3x + 18° and 2x + 25° are supplementary, find the value of x.**

**Solution:**

It is given that 3x + 18° and 2x + 25° are supplementary

We can write it as

3x + 18° + 2x + 25° = 180^{0}

By further calculation

5x + 43^{0} = 180^{0}

So we get

5x = 180 – 43 = 137^{0}

x = 137/5 = 27.4^{0} or 27^{0} 24’

**15. If two complementary angles are in the ratio 1:5, find them.**

**Solution:**

It is given that two complementary angles are in the ratio 1:5

Consider x and 5x as the angles

We can write it as

x + 5x = 90^{0}

6x = 90^{0}

So we get

x = 90/6 = 15^{0}

Here the angles will be 15^{0} and 15 × 5 = 75^{0}

#### Exercise 14B page: 166

**1. In questions 1 and 2, given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:**

**(i) ∠3 and ∠6
(ii) ∠2 and ∠4
(iii) ∠3 and ∠7
(iv) ∠2 and ∠7
(v) ∠4 and∠6
(vi) ∠1 and ∠8
(vii) ∠1 and ∠5
(viii) ∠1 and ∠4
(ix) ∠5 and ∠7**

**Solution:**

(i) ∠3 and ∠6 are interior alternate angles.

(ii) ∠2 and ∠4 are adjacent angles.

(iii) ∠3 and ∠7 are corresponding angles.

(iv) ∠2 and ∠7 are exterior alternate angles.

(v) ∠4 and∠6 are allied or co-interior angles.

(vi) ∠1 and ∠8 are exterior alternate angles.

(vii) ∠1 and ∠5 are corresponding angles.

(viii) ∠1 and ∠4 are vertically opposite angles.

(ix) ∠5 and ∠7 are adjacent angles.

**2. (i) ∠1 and ∠4
(ii) ∠4 and ∠7
(iii) ∠10 and ∠12
(iv) ∠7 and ∠13
(v) ∠6 and ∠8
(vi) ∠11 and ∠8
(vii) ∠7 and ∠9
(viii) ∠4 and ∠5
(ix) ∠4 and ∠6
(x) ∠6 and ∠7
(xi) ∠2 and ∠13**

**Solution:**

(i) ∠1 and ∠4 are vertically opposite angles.

(ii) ∠4 and ∠7 are interior alternate angles.

(iii) ∠10 and ∠12 are vertically opposite angles.

(iv) ∠7 and ∠13 are corresponding angles.

(v) ∠6 and ∠8 are vertically opposite angles.

(vi) ∠11 and ∠8 are allied or co-interior angles.

(vii) ∠7 and ∠9 are vertically opposite angles.

(viii) ∠4 and ∠5 are adjacent angles.

(ix) ∠4 and ∠6 are allied or co-interior angles.

(x) ∠6 and ∠7 are adjacent angles.

(xi) ∠2 and ∠13 are allied or co-interior angles.

**3. In the following figures, the arrows indicate parallel lines. State which angles are equal. Give reasons.**

**Solution:**

(i) From the figure (i)

a = b are corresponding angles

b = c are vertically opposite angles

a = c are alternate angles

So we get

a = b = c

(ii) From the figure (ii)

x = y are vertically opposite angles

y = l are alternate angles

x = l are corresponding angles

1 = n are vertically opposite angles

n = r are corresponding angles

So we get

x = y = l = n = r

Similarly

m = k are vertically opposite angles

k = q are corresponding angles

Hence, m = k = q.

**4. In the given figure, find the measure of the unknown angles:**

**Solution:**

From the figure

a = d are vertically opposite angles

d = f are corresponding angles

f = 110^{0} are vertically opposite angles

So we get

a = d = f = 110^{0}

We know that

e + 110^{0} = 180^{0} are linear pair of angles

e = 180 – 110 = 70^{0}

b = c are vertically opposite angles

b = e are corresponding angles

e = g are vertically opposite angles

So we get

b = c = e = g = 70^{0}

Therefore, a = 110^{0}, b = 70^{0}, c = 70^{0}, d = 110^{0}, e = 70^{0}, f = 110^{0} and g = 70^{0}.

**5. Which pair of the dotted line, segments, in the following figures, are parallel. Give reason:**

**Solution:**

(i) From the figure (i)

If the lines are parallel we get 120^{0} + 50^{0} = 180^{0}

There are co-interior angles where 170^{0} ≠ 180^{0}

Therefore, they are not parallel lines.

(ii) From the figure (ii)

∠1 = 45^{0} are vertically opposite angles

We know that the lines are parallel if

∠1 + 135^{0} = 180^{0} are co-interior angles

Substituting the values

45^{0} + 135^{0} = 180^{0}

180^{0} = 180^{0} which is true

Therefore, the lines are parallel.

(iii) From the figure (iii)

The lines are parallel if corresponding angles are equal

Here 120^{0} ≠ 130^{0}

Hence, lines are not parallel.

(iv) ∠1 = 110^{0} are vertically opposite angles

We know that if lines are parallel

∠1 + 70^{0} = 180^{0} are co-interior angles

Substituting the values

110^{0} + 70^{0} = 180^{0}

180^{0} = 180^{0} which is correct

Therefore, the lines are parallel.

(v) ∠1 + 100^{0} = 180^{0}

So we get

∠1 = 180^{0} – 100^{0} = 80^{0} which is a linear pair

Here the lines l_{1} and l_{2} are parallel if ∠1 = 70^{0}

But here ∠1 = 80^{0} which is not equal to 70^{0}

So the lines l_{1} and l_{2} are not parallel

Now, l_{3} and l_{5} will be parallel if corresponding angles are equal

But here, the corresponding angles 80^{0} ≠ 70^{0}

Hence, l_{3} and l_{5} are not parallel.

We know that, in l_{2} and l_{4}

∠1 = 80^{0} are alternate angles

80^{0} = 80^{0} which is true

Hence, l_{2} and l_{4} are parallel.

(vi) Two lines are parallel if alternate angles are equal

Here, 50^{0} ≠ 40^{0 }which is not true

Hence, the lines are not parallel.

**6. In the given figures, the directed lines are parallel to each other. Find the unknown angles.**

**Solution:**

(i) If the lines are parallel

a = b are corresponding angles

a = c are vertically opposite angles

a = b = c

Here b = 60^{0} are vertically opposite angles

Therefore, a = b = c = 60^{0}

(ii) If the lines are parallel

x = z are corresponding angles

z + y = 180^{0} is a linear pair

y = 55^{0} are vertically opposite angles

Substituting the values

z + 55^{0} = 180^{0}

z = 180 – 55 = 125^{0}

If x = z we get x = 125^{0}

Therefore, x = 125^{0}, y = 55^{0} and z = 125^{0}.

(iii) If the lines are parallel

c = 120^{0} (corresponding angles)

a + 120^{0} = 180^{0} are co-interior angles

a = 180 – 120 = 60^{0}

We know that a = b are vertically opposite angles

So b = 60^{0}

Therefore, a = b = 60^{0} and c = 120^{0}.

(iv) If the lines are parallel

x = 50^{0} are alternate angles

y + 120^{0} = 180^{0} are co-interior angles

y = 180 – 120 = 60^{0}

We know that

x + y + z = 360^{0} are angles at a point

Substituting the values

50 + 60 + z = 360

By further calculation

110 + z = 360

z = 360 – 110 = 250^{0}

Therefore, x = 50^{0}, y = 60^{0} and z = 250^{0}.

(v) If the lines are parallel

x + 90^{0} = 180^{0} are co-interior angles

x = 180^{0} – 90^{0} = 90^{0}

∠2 = x

∠2 = 90^{0}

We know that the sum of angles of a triangle

∠1 + ∠2 + 30^{0} = 180^{0}

Substituting the values

∠1 + 90^{0} + 30^{0} = 180^{0}

By further calculation

∠1 + 120^{0} = 180^{0}

∠1 = 180 – 120 = 60^{0}

Here ∠1 = k are vertically opposite angles

k = 60^{0}

Here ∠1 = z are corresponding angles

z = 60^{0}

Here k + y = 180^{0} are co-interior angles

Substituting the values

60^{0} + y = 180^{0}

y = 180 – 60 = 120^{0}

Therefore, x = 90^{0}, y = 120^{0}, z = 60^{0}, k = 60^{0}.

**7. Find x, y and p is the given figures:**

**Solution:**

(i) From the figure (i)

The lines are parallel

x = z are corresponding angles

y = 40^{0} are corresponding angles

We know that

x + 40^{0} + 270^{0} = 360^{0} are the angles at a point

So we get

x + 310^{0} = 360^{0}

x = 360 – 310 = 50^{0}

So z = x = 50^{0}

Here p + z = 180^{0} is a linear pair

By substituting the values

p + 50^{0} = 180^{0}

p = 180 – 50 = 130^{0}

Therefore, x = 50^{0}, y = 40^{0}, z = 50^{0} and p = 130^{0}.

(ii) From the figure (ii)

The lines are parallel

y = 110^{0} are corresponding angles

We know that

25^{0} + p + 110^{0} = 180^{0} are angles on a line

p + 135^{0} = 180^{0}

p = 180 – 135 = 45^{0}

We know that the sum of angles of a triangle

x + y + 25^{0} = 180^{0}

x + 110^{0} + 25^{0} = 180^{0}

By further calculation

x + 135^{0} = 180^{0}

x = 180 – 135 = 45^{0}

Therefore, x = 45^{0}, y = 110^{0} and p = 45^{0}.

**8. Find x in the following cases:**

**Solution:**

(i) From the figure (i)

The lines are parallel

2x + x = 180^{0} are co-interior angles

3x = 180^{0}

x = 180/3 = 60^{0}

(ii) From the figure (ii)

The lines are parallel

4x + ∠1 = 180^{0} are co-interior angles

∠1 = 5x are vertically opposite angles

Substituting the values

4x + 5x = 180^{0}

So we get

9x = 180^{0}

x = 180/9 = 20^{0}

(iii) From the figure (iii)

The lines are parallel

∠1 + 4x = 180^{0} are co-interior angles

∠1 = x are vertically opposite angles

Substituting the values

x + 4x = 180^{0}

5x = 180^{0}

So we get

x = 180/5 = 36^{0}

(iv) From the figure (iv)

The lines are parallel

2x + 5 + 3x + 55 = 180^{0} are co-interior angles

5x + 60^{0} = 180^{0}

By further calculation

5x = 180 – 60 = 120^{0}

So we get

x = 120/5 = 24^{0}

(v) From the figure (v)

The lines are parallel

∠1 = 2x + 20^{0} are alternate angles

∠1 + 3x + 25^{0} = 180^{0} is a linear pair

Substituting the values

2x + 20^{0} + 3x + 25^{0} = 180^{0}

5x + 45^{0} = 180^{0}

So we get

5x = 180 – 45 = 135^{0}

x = 135/5 = 27^{0}

(vi) From the figure (vi)

Construct a line parallel to the given parallel lines

∠1 = 4x and ∠2 = 6x are corresponding angles

∠1 + ∠2 = 130^{0}

Substituting the values

4x + 6x = 130^{0}

10x = 130^{0}

So we get

x = 130

e + 110^{0} = 180^{0} are linear pair of angles

e = 180 – 110 = 70^{0}

b = c are vertically opposite angles

b = e are corresponding angles

e = g are vertically opposite angles

So we get

b = c = e = g = 70^{0}

Therefore, a = 110^{0}, b = 70^{0}, c = 70^{0}, d = 110^{0}, e = 70^{0}, f = 110^{0} and g = 70^{0}.

**5. Which pair of the dotted line, segments, in the following figures, are parallel. Give reason:**

**Solution:**

(i) From the figure (i)

If the lines are parallel we get 120^{0} + 50^{0} = 180^{0}

There are co-interior angles where 170^{0} ≠ 180^{0}

Therefore, they are not parallel lines.

(ii) From the figure (ii)

∠1 = 45^{0} are vertically opposite angles

We know that the lines are parallel if

∠1 + 135^{0} = 180^{0} are co-interior angles

Substituting the values

45^{0} + 135^{0} = 180^{0}

180^{0} = 180^{0} which is true

Therefore, the lines are parallel.

(iii) From the figure (iii)

The lines are parallel if corresponding angles are equal

Here 120^{0} ≠ 130^{0}

Hence, lines are not parallel.

(iv) ∠1 = 110^{0} are vertically opposite angles

We know that if lines are parallel

∠1 + 70^{0} = 180^{0} are co-interior angles

Substituting the values

110^{0} + 70^{0} = 180^{0}

180^{0} = 180^{0} which is correct

Therefore, the lines are parallel.

(v) ∠1 + 100^{0} = 180^{0}

So we get

∠1 = 180^{0} – 100^{0} = 80^{0} which is a linear pair

Here the lines l_{1} and l_{2} are parallel if ∠1 = 70^{0}

But here ∠1 = 80^{0} which is not equal to 70^{0}

So the lines l_{1} and l_{2} are not parallel

Now, l_{3} and l_{5} will be parallel if corresponding angles are equal

But here, the corresponding angles 80^{0} ≠ 70^{0}

Hence, l_{3} and l_{5} are not parallel.

We know that, in l_{2} and l_{4}

∠1 = 80^{0} are alternate angles

80^{0} = 80^{0} which is true

Hence, l_{2} and l_{4} are parallel.

(vi) Two lines are parallel if alternate angles are equal

Here, 50^{0} ≠ 40^{0 }which is not true

Hence, the lines are not parallel.

**6. In the given figures, the directed lines are parallel to each other. Find the unknown angles.**

**Solution:**

(i) If the lines are parallel

a = b are corresponding angles

a = c are vertically opposite angles

a = b = c

Here b = 60^{0} are vertically opposite angles

Therefore, a = b = c = 60^{0}

(ii) If the lines are parallel

x = z are corresponding angles

z + y = 180^{0} is a linear pair

y = 55^{0} are vertically opposite angles

Substituting the values

z + 55^{0} = 180^{0}

z = 180 – 55 = 125^{0}

If x = z we get x = 125^{0}

Therefore, x = 125^{0}, y = 55^{0} and z = 125^{0}.

(iii) If the lines are parallel

c = 120^{0} (corresponding angles)

a + 120^{0} = 180^{0} are co-interior angles

a = 180 – 120 = 60^{0}

We know that a = b are vertically opposite angles

So b = 60^{0}

Therefore, a = b = 60^{0} and c = 120^{0}.

(iv) If the lines are parallel

x = 50^{0} are alternate angles

y + 120^{0} = 180^{0} are co-interior angles

y = 180 – 120 = 60^{0}

We know that

x + y + z = 360^{0} are angles at a point

Substituting the values

50 + 60 + z = 360

By further calculation

110 + z = 360

z = 360 – 110 = 250^{0}

Therefore, x = 50^{0}, y = 60^{0} and z = 250^{0}.

(v) If the lines are parallel

x + 90^{0} = 180^{0} are co-interior angles

x = 180^{0} – 90^{0} = 90^{0}

∠2 = x

∠2 = 90^{0}

We know that the sum of angles of a triangle

∠1 + ∠2 + 30^{0} = 180^{0}

Substituting the values

∠1 + 90^{0} + 30^{0} = 180^{0}

By further calculation

∠1 + 120^{0} = 180^{0}

∠1 = 180 – 120 = 60^{0}

Here ∠1 = k are vertically opposite angles

k = 60^{0}

Here ∠1 = z are corresponding angles

z = 60^{0}

Here k + y = 180^{0} are co-interior angles

Substituting the values

60^{0} + y = 180^{0}

y = 180 – 60 = 120^{0}

Therefore, x = 90^{0}, y = 120^{0}, z = 60^{0}, k = 60^{0}.

**7. Find x, y and p is the given figures:**

**Solution:**

(i) From the figure (i)

The lines are parallel

x = z are corresponding angles

y = 40^{0} are corresponding angles

We know that

x + 40^{0} + 270^{0} = 360^{0} are the angles at a point

So we get

x + 310^{0} = 360^{0}

x = 360 – 310 = 50^{0}

So z = x = 50^{0}

Here p + z = 180^{0} is a linear pair

By substituting the values

p + 50^{0} = 180^{0}

p = 180 – 50 = 130^{0}

Therefore, x = 50^{0}, y = 40^{0}, z = 50^{0} and p = 130^{0}.

(ii) From the figure (ii)

The lines are parallel

y = 110^{0} are corresponding angles

We know that

25^{0} + p + 110^{0} = 180^{0} are angles on a line

p + 135^{0} = 180^{0}

p = 180 – 135 = 45^{0}

We know that the sum of angles of a triangle

x + y + 25^{0} = 180^{0}

x + 110^{0} + 25^{0} = 180^{0}

By further calculation

x + 135^{0} = 180^{0}

x = 180 – 135 = 45^{0}

Therefore, x = 45^{0}, y = 110^{0} and p = 45^{0}.

**8. Find x in the following cases:**

**Solution:**

(i) From the figure (i)

The lines are parallel

2x + x = 180^{0} are co-interior angles

3x = 180^{0}

x = 180/3 = 60^{0}

(ii) From the figure (ii)

The lines are parallel

4x + ∠1 = 180^{0} are co-interior angles

∠1 = 5x are vertically opposite angles

Substituting the values

4x + 5x = 180^{0}

So we get

9x = 180^{0}

x = 180/9 = 20^{0}

(iii) From the figure (iii)

The lines are parallel

∠1 + 4x = 180^{0} are co-interior angles

∠1 = x are vertically opposite angles

Substituting the values

x + 4x = 180^{0}

5x = 180^{0}

So we get

x = 180/5 = 36^{0}

(iv) From the figure (iv)

The lines are parallel

2x + 5 + 3x + 55 = 180^{0} are co-interior angles

5x + 60^{0} = 180^{0}

By further calculation

5x = 180 – 60 = 120^{0}

So we get

x = 120/5 = 24^{0}

(v) From the figure (v)

The lines are parallel

∠1 = 2x + 20^{0} are alternate angles

∠1 + 3x + 25^{0} = 180^{0} is a linear pair

Substituting the values

2x + 20^{0} + 3x + 25^{0} = 180^{0}

5x + 45^{0} = 180^{0}

So we get

5x = 180 – 45 = 135^{0}

x = 135/5 = 27^{0}

(vi) From the figure (vi)

Construct a line parallel to the given parallel lines

∠1 = 4x and ∠2 = 6x are corresponding angles

∠1 + ∠2 = 130^{0}

Substituting the values

4x + 6x = 130^{0}

10x = 130^{0}

So we get

x = 130/10 = 13^{0}

Exercise 14C PAGE: 172

**1. Using ruler and compasses, construct the following angles:
(i)30°
(ii)15°
(iii) 75°
(iv) 180°
(v) 165°
Solution:**

(i) 30°

Steps of Construction:

1. Construct a line segment BC.

2. Taking B as centre and a suitable radius construct an arc which meets BC at the point P.

3. Taking P as centre and same radius cut off the arc at the point Q.

4. Consider P and Q as centre construct two arcs which intersect each other at the point R.

5. Now join BR and produce it to point A forming ∠ABC = 30^{0}

(ii) 15^{0}

Steps of Construction:

1. Construct a line segment BC.

2. Taking B as centre and a suitable radius construct an arc which meets BC at the point P.

3. Taking P as centre and same radius cut off the arc at the point Q.

4. Consider P and Q as centres, construct two arcs which intersect each other at the point D and join BD.

5. Taking P and R as centre construct two more arcs which intersect each other at the point S.

6. Now join BS and produce it to point A.

∠ABC = 15^{0}

(iii) 75°

Steps of Construction:

1. Construct a line segment BC.

2. Taking B as centre and suitable radius construct an arc and cut off PQ then QR of the same radius.

3. Taking Q and R as centre, construct two arcs which intersect each other at the point S.

4. Now join SB.

5. Taking Q and D as centre construct two arcs which intersect each other at the point T.

6. Now join BT and produce it to point A.

∠ABC = 75°

(iv) 180°

Steps of Construction:

1. Construct a line segment BC.

2. Taking B as centre and some suitable radius construct an arc which meets BC at the point P.

3. Taking P as centre and with same radius cut off the arcs PQ, QR and RS.

4. Now join BS and produce it to point A.

∠ABC = 180°

(v) 165°

Steps of Construction:

1. Construct a line segment BC.

2. Taking B as centre and some suitable radius construct an arc which meets BC at the point P.

3. Taking P as centre and same radius cut off arcs at PQ, QR and then RS.

4. Now join SB and produce it to point E.

5. Taking R and S as centres construct two arcs which intersect each other at the point M.

6. Taking T and S as centres construct two arcs which intersect each other at the point L.

7. Now join BL and produce it to point A.

∠ABC = 165°

**2. Draw ∠ABC = 120°. Bisect the angle using ruler and compasses only. Measure each angle so obtained and check whether the angles obtained on bisecting ∠ABC are equal or not.**

**Solution:**

Steps of Construction:

1. Construct a line segment BC.

2. Taking B as centre and some suitable radius construct an arc which meets BC at the point P.

3. Taking P as centre and with same radius cut off the arcs PQ and QR.

4. Now join BR and produce it to point A

∠ABC = 120°

5. Taking P and R as centres construct two arcs which intersect each other at the point S.

6. Now join BS and produce it to point D.

Here BD is the bisector of ∠ABC

By measuring each angle we get to know that is it 60^{0}

Yes, both the angles are of equal measure.

**3. Draw a line segment PQ = 6 cm. Mark a point A in PQ so that AP = 2 cm. At point A, construct angle QAR = 60°.**

**Solution:**

Steps of Construction:

1. Construct a line segment PQ = 6cm.

2. Now mark point A on PQ so that AP = 2 cm.

3. Taking A as centre and some suitable radius construct an arc which meets AQ at the point C.

4. Taking C as centre and with same radius cut the arc CB.

5. Now join AB and produce it to point R.

∠QAR = 60^{0}.

**4. Draw a line segment AB = 8 cm. Mark a point P in AB so that AP = 5 cm. At P, construct angle APQ = 30°.**

**Solution:**

Steps of Construction:

1. Construct a line segment AB = 8 cm.

2. Now mark the point P on AB such that AP = 5 cm.

3. Taking P as centre and some suitable radius construct an arc which meets AB at L.

4. Taking L as centre and same radius cut the arc LM.

5. Now bisect the arc LM at the point N.

6. Join PN and produce it to point Q.

∠APQ = 30^{0}

**5. Construct an angle of 75° and then bisect it.**

**Solution:**

Steps of Construction:

1. Construct a line segment BC.

2. At the point B construct an angle ABC which is equal to 75^{0}.

3. Taking P and T as centres construct arcs which intersect each other at the point L.

4. Now join BL and produce it to point D.

Here BD bisects ∠ABC.

**6. Draw a line segment of length 6 .4 cm. Draw its perpendicular bisector.**

**Solution:**

Steps of Construction:

1. Construct a line segment AB = 6.4 cm.

2. Taking A and B as centres and with some suitable radius construct arcs which intersect each other at the points S and R.

3. Now join SR which intersects AB at the point Q.

Here PQR is the perpendicular bisector of the line segment AB.

**7. Draw a line segment AB = 5.8 cm. Mark a point P in AB such that PB = 3.6 cm. At P, draw perpendicular to AB.**

**Solution:**

Steps of Construction:

1. Construct a line segment AB = 5.8 cm.

2. Now mark a point P on the line segment AB such that PB = 3.6 cm.

3. Taking P as centre and some suitable radius construct an arc which meets AB at the point L.

4. Taking L as centre and same radius cut off the arcs LM and MN.

5. Now bisect the arc MN at the point S.

6. Join PS and produce it to point Q.

Here PQ is perpendicular to the line segment AB at point P.

**8. In each case, given below, draw a line through point P and parallel to AB:**

**Solution:**

Steps of Construction:

1. From the point P construct a line segment meeting AB

2. Taking Q as centre and some suitable radius construct an arc CD.

3. Taking P as centre and same radius construct another arc which meets PQ at the point E.

4. Taking E as centre and radius equal to CD cut this arc at the point F.

5. Now join PF and produce it to both sides to L and M.

Here the line LM is parallel to the given line AB.