Selina Solutions Concise Maths Class 7 Chapter 14 Lines and Angles (Including Construction of Angles) Exercise 14A are created by subject matter experts after conducting research on each concept. Fundamental concepts, types of lines and angles and the method of determining them are the topics discussed in this exercise. Students can improve their logical and analytical thinking abilities by using the solutions PDF. To score well in the annual exam, Selina Solutions Concise Maths Class 7 Chapter 14 Lines and Angles (Including Construction of Angles) Exercise 14A, PDF links are provided here for free download.

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

### Access other exercises of Selina Solutions Concise Maths Class 7 Chapter 14: Lines and Angles (Including Construction of Angles)

### Access Selina Solutions Concise Maths Class 7 Chapter 14: Lines and Angles (Including Construction of Angles) Exercise 14A

#### 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}