Selina Solutions Concise Mathematics Class 6 Chapter 24 Angles

Selina Solutions Concise Mathematics Class 6 Chapter 24 Angles help students understand the fundamental concepts, which is vital in higher classes as well. Angle is formed, when two rays meet at a point. Important terms from the exam point of view are explained in brief in this chapter. The solutions are well structured by the subject matter experts in a descriptive manner, for better understanding of students. For further details, students can access Selina Solutions Concise Mathematics Class 6 Chapter 24 Angles PDF, from the links which are found below

Chapter 24 provides students basic concepts of Angles such as parts of Angles, comparison of Angles, kinds of Angles and pairs of Angles. Students can use solutions PDF both in exercise and chapter wise format, in both online and offline, whenever required.

Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles Download PDF

 

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Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles

Exercise 24(A) Solutions

Exercise 24(B) Solutions

Access Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles

Exercise 24(A)

1. For each angle given below, write the name of the vertex, the names of the arms and the name of the angle.

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 1

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 2

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 3

(iv) Name the angles marked by letters a, b, c, x and y.

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 4

Solution:

(i) In the given figure,

Vertex = O

Arms = OA and OB

Angle = ∠AOB or ∠BOA or ∠O

(ii) In the given figure,

Vertex = Q

Arms = QP and QR

Angle = ∠PQR or ∠RQP or ∠Q

(iii) In the given figure,

Vertex = M

Arms = MN and ML

Angle = ∠LMN or ∠NML or ∠M

(iv) a = ∠AOE

b = ∠AOB

c = ∠BOC

d = ∠COD

e = ∠DOE

2. Name the points:

(i) in the interior of the angle PQR,

(ii) in the exterior of the angle PQR.

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 5

Solution:

(i) The points in the interior of the angle = a, b and x

(ii) The points in the exterior of the angle = d, m, n, s and t

3. In the given figure, figure out the number of angles formed within the arms OA and OE.

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 6

Solution:

In the given figure, the angles within the arms OA and OE are as follows:

(i) ∠AOE

(ii) ∠AOD

(iii) ∠AOC

(iv) ∠AOB

(v) ∠BOC

(vi) ∠BOD

(vii) ∠BOE

(viii) ∠COD

(ix) ∠COE and

(x) ∠DOE

4. Add:

(i) 290 16’23” and 80 27’12”

(ii) 90 45’56” and 730 8’ 15”

(iii) 560 38’ and 270 42’ 30”

(iv) 470 and 610 17’4”

Solution:

(i) 290 16’23” + 80 27’12”

290 16’23”

80 27’12” +

_____________

370 43’ 35”

_____________

Hence, addition of 290 16’23” and 80 27’12” = 370 43’ 35”

(ii) 90 45’56” + 730 8’ 15”

90 45’56”

730 8’ 15” +

____________

820 53’ 71”

____________

Hence, addition of 90 45’56” and 730 8’ 15” = 820 53’ 71”

(iii) 560 38’ + 270 42’ 30”

560 38’

270 42’ 30” +

_____________

830 80’ 30”

_____________

Hence, addition of 560 38’ and 270 42’ 30” = 830 80’ 30”

(iv) 470 + 610 17’4”

470

610 17’4” +

____________

1080 17’ 4”

_____________

Hence, addition of 470 and 610 17’4” = 1080 17’ 4”

5. In the figure, given below name:

(i) three pairs of adjacent angles

(ii) two acute angles

(iii) two obtuse angles

(iv) two reflex angles

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 7

Solution:

(i) In the given figure, the three pairs of adjacent angles are as follows:

∠AOB and ∠BOC ,

∠BOC and ∠COD

∠COD and ∠DOA

(ii) In the given figure, the two acute angles are

∠AOB and ∠AOD

(iii) In the given figure, the two obtuse angles are

∠BOC and ∠COD

(iv) In the given figure, the two reflex angles are

∠AOB and ∠COB

6. In the given figure; PQR is a straight line. If:

(i) ∠SQR = 750; find ∠PQS.

(ii) ∠PQS = 1100; find ∠RQS

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 8

Solution:

(i) Given PQR is a straight line

In the given figure,

∠PQS + ∠SQR = 1800 {linear pair of angles}

∠PQ S + 75 = 180

On further calculation, we get,

∠PQS = 180 – 75

∠PQS = 1050

(ii) Given PQR is a straight line

∠PQS + ∠RQS = 1800

1100 + ∠RQS = 1800

On further calculation, we get,

∠RQS = 1800 – 1100

∠RQS = 700

7. In the given figure; AOC is a straight line. If angle AOB = 500, angle AOE = 900 and angle COD = 250; find the measure of:

(i) angle BOC

(ii) angle EOD

(iii) obtuse angle BOD

(iv) reflex angle BOD

(v) reflex angle COE

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 9

Solution:

(i) Given

AOC is a straight line

∠AOB = 500

∠AOE = 900

∠COD = 250

To find the measure of ∠BOC

∠AOB + ∠BOC = 1800 (Linear pairs of angle)

500 + ∠BOC = 1800

On further calculation, we get,

∠BOC = 1800 – 500

∠BOC = 1300

(ii) Given

AOC is a straight line

∠AOB = 500

∠AOE = 900

∠COD = 250

To find the measure ∠EOD

∠EOD + ∠COD = 900 (Since ∠AOE = 900)

∠EOD + 250 = 900

∠EOD = 900 – 250

We get,

∠EOD = 650

(iii) Given

AOC is a straight line

∠AOB = 500

∠AOE = 900

∠COD = 250

To find the measure of obtuse angle BOD,

∠BOD = ∠BOC + ∠COD

Substituting the value of ∠BOC and ∠COD, we get,

∠BOD = 1300 + 250

We get,

∠BOD = 1550

(iv) Given

AOC is a straight line

∠AOB = 500

∠AOE = 900

∠COD = 250

To find the measure of reflex angel BOD

∠BOD = 3600 – ∠BOD

= 3600 – 1550

We get,

∠BOD = 2050

(iv) Given

AOC is a straight line

∠AOB = 500

∠AOE = 900

∠COD = 250

To find the measure of reflex angle COE

∠COE = 3600 – ∠COE

= 3600 – (∠COD + ∠EOD)

= 3600 – (250 + 650)

We get,

= 3600 – 900

∠COE = 2700

8. In the given figure if:

(i) a = 1300; find b

(ii) b = 200; find a

(iii) a = 5 / 3 right angle, find b

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 10

Solution:

(i) From figure,

a + b = 3600

Substitute the value of a in above equation

1300 + b = 3600

b = 3600 – 1300

We get,

b = 2300

(ii) From figure,

a + b = 3600

a + 2000 = 3600

a = 3600 – 2000

We get,

a = 1600

(iii) From figure,

a = 5 / 3 right angle

= 5 / 3 × 900

We get,

a = 1500

Now,

a + b = 3600

Substitute the value of a in above equation

1500 + b = 3600

b = 3600 – 1500

b = 2100

9. In the given diagram, ABC is a straight line

(i) If x = 530, find y

(ii) If y =
Selina Solutions Concise Mathematics Class 6 Chapter 24 - 11 right angles; find x.

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 12

Solution:

(i) From the figure,

Given that ABC is a straight line

Hence,

∠ABD + ∠DBC = 1800 {Linear pair of angles}

x + y = 1800

530 + y = 1800

y = 1800 – 530

y = 1270

(ii) From given figure,

x + y = 1800

x +
Selina Solutions Concise Mathematics Class 6 Chapter 24 - 13right angles = 1800

x + 3 / 2 × 900 = 1800

On further calculation, we get

x + 1350 = 1800

x = 1800 – 1350

x = 450

10. In the given figure, AOB is a straight line. Find the value of x and also answer each of the following:

(i) ∠AOP = …….

(ii) ∠BOP = …….

(iii) which angle is obtuse?

(iv) which angle is acute?

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 14

Solution:

Given from figure,

∠AOP = x + 300

∠BOP = x – 300

Now,

∠AOP + ∠BOP = 1800 (Since ∠AOB is a straight line)

(x + 300)+ (x – 300) = 1800

2x = 1800

We get,

x = 900

(i) ∠AOP = x + 300

= 900 + 300

= 1200

(ii) ∠BOP = x – 300

= 900 – 300

We get,

= 600

(iii) The obtuse angle is ∠AOP

(iv) The acute angle is ∠BOP

11.In the given figure, PQR is a straight line. Find x. Then complete the following:

(i) ∠AQB = ……………

(ii) ∠BQP = ……………

(iii) ∠AQR = …………….

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 15

Solution:

Given

PQR is a straight line

∠AQP = x + 200

∠AQB = 2x + 100

∠BQR = x – 100

Since PQR is a straight line

∠AQP + ∠AQB + ∠BQR = 1800

(x + 200) + (2x + 100) + (x – 100) = 1800

We get,

4x + 200 = 1800

4x = 1800 – 200

4x = 1600

x = 1600 / 4

We get,

x = 400

(i) ∠AQB = 2x + 100

= 2 × 400 + 100

= 800 + 100

= 900

Similarly,

∠AQP = x + 200

∠AQP = 400 + 200

∠AQP = 600

∠BQR = x – 100

∠BQR = 400 – 100

∠BQR = 300

(ii) ∠BQP = ∠AQP + ∠AQB

= 600 + 900

= 1500

(iii) ∠AQR = ∠AQB + ∠BQR

= 900 + 300

= 1200

12. In the given figure, lines AB and CD intersect at point O.

(i) Find the value of ∠a.

(ii) Name all the pairs of vertically opposite angles.

(iii) Name all the pairs of adjacent angles

(iv) Name all the reflex angles formed and write the measure of each.

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 16

Solution:

Given

AB and CD intersect each other at point C

∠AOC = 680

(i) Here, AOB is a line

Hence,

∠AOC + ∠BOC = 1800 (Linear pairs of angles)

680 + a = 1800

a = 1800 – 680

a = 1120

(ii) The pairs of vertically opposite angles are,

∠AOC and ∠BOD and ∠BOC and ∠AOD

(iii) The pairs of adjacent angles are,

∠AOC and ∠BOC, ∠BOC and ∠BOD, ∠BOD and ∠DOA, ∠DOA and ∠AOC

(iv) The reflex angles in the given figure are,

∠BOC and ∠DOA,

Reflex angle BOC = 1800 + 680

= 2480

Reflex angle DOA = 1800 + 680

= 2480

13. In the given figure:

(i) If ∠AOB = 450, ∠BOC = 300 and ∠AOD = 1100;

Find: angles COD and BOD

(ii) If ∠BOC = ∠DOC = 340 and ∠AOD = 1200;

Find: angle AOB and angle AOC

(iii) If ∠AOB = ∠BOC = ∠COD = 380

Find: reflex angle AOC and reflex angle AOD

Selina Solutions Concise Mathematics Class 6 Chapter 24 - 17

Solution:

(i) ∠COD = ∠AOD – ∠AOC

= ∠AOD – (∠AOB + ∠BOC)

= 1100 – (450 + 300)

We get,

= 1100 – 750

= 350

Hence,

∠COD = 350

∠BOD = ∠BOC + ∠COD

= 300 + 350

= 650

Hence,

∠BOD = 650

(ii) ∠AOB = ∠AOD – ∠BOD

= ∠AOD – (∠BOC + ∠COD)

= 1200 – (340 + 340)

We get,

= 1200 – 680

= 520

Hence,

∠AOB = 520

∠AOC = ∠AOB + ∠BOC

= 520 + 340

= 860

Hence,

∠AOC = 860

(iii) Reflex angel AOC = 3600 – ∠AOC

= 3600 – (∠AOB + ∠BOC)

= 3600 – (380 + 380)

= 3600 – 760

We get,

= 2840

Hence,

∠AOC = 2840

Reflex angle AOD = 3600 – ∠AOD

= 3600 – (∠AOB + ∠BOC + ∠COD)

= 3600 – (380 + 380 + 380)

= 3600 – 1140

We get,

= 2460

Hence,

∠AOC = 2460


Exercise 24(B)

1. Write the complement angle of:

(i) 450

(ii) x0

(iii) (x – 10)0

(iv) 200 + y0

Solution:

(i) The complement angle of 450 is,

= 900 – 450

= 450

Therefore, the complement angle of 450 is 450

(ii) x0

The complement angle of x0 is,

= 900 – x0

= (90 – x)0

Therefore, the complement angle of x0 is (90 – x)0

(iii) The complement angle of (x – 10)0 is,

= 900 – (x – 10)0

= 900 – x0 + 100

= 1000 – x0

= (100 – x)0

Therefore, the complement of (x – 10)0 is (100 – x)0

(iv) The complement angle of 200 + y0 is,

= 900 – (200 + y0)

= 900 – 200 – y0

We get,

= 700 – y0

= (70 – y)0

2. Write the supplement angle of:

(i) 490

(ii) 1110

(iii) (x – 30)0

(iv) 200 + y0

Solution:

(i) The supplement angle of 490 is,

= 1800 – 490

= 1310

Hence, the supplement angle of 490 is 1310

(ii) The supplement angle of 1110 is,

= 1800 – 1110

= 690

Hence, the supplement angle of 1110 is 690

(iii) The supplement angle of (x – 30)0 is,

= 1800 – (x – 30)0

= 1800 – x0 + 300

= 2100 – x0

= (210 – x)0

Hence, the supplement angle of (x – 30)0 is (210 – x)0

(iv) The supplement angle of 200 + y0 is,

= 1800 – (200 + y0)

= 1800 – 200 – y0

= 1600 – y0

= (160 – y)0

Hence, the supplement angle of 200 + y0 is (160 – y)0

3. Write the complement angle of:

(i) 1 / 2 of 600

(ii) 1 / 5 of 1600

(iii) 2 / 5 of 700

(iv) 1 / 6 of 900

Solution:

(i) The complement angle of (1 / 2 of 600) is,

= 900 – (1 / 2 × 600)

We get,

= 900 – 300

= 600

Therefore, the complement angle of (1 / 2 of 600) is 600

(ii) The complement angle of (1 / 5 of 1600) is,

= 900 – (1 / 5 × 1600)

We get,

= 900 – 320

= 580

Therefore, the complement angle of (1 / 5 of 1600) is 580

(iii) The complement angle of (2 / 5 of 700) is,

= 900 – (2 / 5 × 700)

We get,

= 900 – 280

= 620

Therefore, the complement of (2 / 5 of 700) is 620

(iv) The complement angle of (1 / 6 of 900) is,

= 900 – (1 / 6 × 900)

We get,

= 900 – 150

= 750

Therefore, the complement of (1 / 6 of 900) is 750

4.

(i) 50% of 1200

(ii) 1 / 3 of 1500

(iii) 60% of 1000

(iv) 3 / 4 of 1600

Solution:

(i) Supplement angle of 50% of 1200 is,

= 1800 – (50% of 1200)

= 1800 – [(1200 × 50) / 100]

We get,

= 1800 – 600

= 1200

Hence, supplement angle of 50% of 1200 is 1200

(ii) Supplement angle of (1 / 3 of 1500) is,

= 1800 – (1 / 3 × 1500)

We get,

= 1800 – 500

= 1300

Hence, supplement angle of (1 / 3 of 1500) is 1300

(iii) Supplement angle of 60% of 1000 is,

= 1800 – (60% of 1000)

= 1800 – [(60 × 100) / 100]

We get,

= 1800 – 600

= 1200

Hence, the supplement angle of (60% of 1000) is 1200

(iv) Supplement angle of 3 / 4 of 1600

= 1800 – (3 / 4 of 1600)

We get,

= 1800 – 1200

= 600

Hence, the supplement angle of (3 / 4 of 1600) is 600

5. Find the angle:

(i) that is equal to its complement?

(ii) that is equal to its supplement?

Solution:

(i) The angle equal to its complement is 450

(ii) The angle equal to its supplement is 900

6. Two complementary angles are in the ratio 7: 8. Find the angles

Solution:

Given

Two complementary angles are in the ratio 7: 8

Let the two complementary angles be 7x and 8x

Hence,

7x + 8x = 900

15x = 900

We get,

x = 900 / 15

x = 60

So, two complementary angles are

7x = 7 × 60

= 420

8x = 8 × 60

= 480

Therefore, two complementary angles are 420 and 480

7. Two supplementary angles are in the ratio 7: 11. Find the angles

Solution:

Given

Two supplementary angles are in the ratio 7: 11

Let the two supplementary angles be 7x and 11x

Hence,

7x + 11x = 1800

18x = 1800

x = 1800 / 18

We get,

x = 100

So, two supplementary angles are

7x = 7 × 100

= 700

11x = 11 × 100

= 1100

Therefore, two supplementary angles are 700 and 1100

8. The measures of two complementary angles are (2x – 7)0 and (x + 4)0. Find x.

Solution:

Given

(2x – 7)0 and (x + 4)0 are two complementary angles

We know that,

Sum of two complementary angles = 900

Hence,

(2x – 7)0 + (x + 4)0 = 900

2x – 7 + x + 4 = 900

3x – 3 = 900

3x = 900 + 30

3x = 930

x = 930 / 3

We get,

x = 310

Therefore, the value of x = 310

9. The measures of two supplementary angles are (3x + 15)0 and (2x + 5)0. Find x.

Solution:

Given

(3x + 15)0 and (2x + 5)0 are two supplementary angles

We know that,

Sum of two supplementary angles = 1800

Hence,

(3x + 15)0 + (2x + 5)0 = 1800

3x + 15 + 2x + 5 = 1800

5x + 200 = 1800

5x = 1800 – 200

5x = 1600

x = 1600 / 5

We get,

x = 320

Therefore, the value of x is 320

10. For an angle x0, find:

(i) the complementary angle

(ii) the supplementary angle

(iii) the value of x0 if its supplementary angle is three times its complementary angle.

Solution:

For an angle x0

(i) Complementary angle of x0 is,

= (900 – x)

(ii) Supplementary angle of x0 is,

= (1800 – x)

(iii) As per the statement,

Supplementary angle = 3 (Complementary angle)

1800 – x = 3 (900 – x)

1800 – x = 2700 – 3x

– x + 3x = 2700 – 1800

2x = 900

x = 900 / 2

We get,

x = 450

Therefore, the value of x is 450

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