What is the angle subtended at the centre of a circle of radius 5 cm by an arc length 45 cm?

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Solution:

Given,

Radius = 4 cm

Angle subtended at the centre = 30°

Length of arc = θ/360 × 2πr 

Length of arc = 30/360 × 2π × 4 cm 

                      = 2π/3 

Therefore, the length of arc that subtends an angle of 30o degree is 2π/3 cm

Question 2. Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length 5π/3 cm. 

Solution:

Length of arc = 5π/3 cm

Length of arc = θ/360 × 2πr cm

5π/3 cm = θ/360 × 2πr cm

θ = 60°

Therefore, the angle subtended at the centre of circle is 60°

Question 3. An arc of length 20π cm subtends an angle of 144° at the centre of a circle. Find the radius of the circle.

Solution:

Length of arc = 20π cm

θ = Angle subtended at the centre of circle = 144°

Length of arc = θ/360 × 2πr cm

θ/360 × 2πr cm = 144/360 × 2πr cm = 4π/5 × r cm

20π cm = 4π/5 × r cm

r = 25 cm.

Therefore, the radius of the circle is 25 cm.

Question 4. An arc of length 15 cm subtends an angle of 45° at the centre of a circle. Find in terms of π, the radius of the circle. 

Solution:

Length of arc = 15 cm

θ = Angle subtended at the centre of circle = 45°

Length of arc = θ/360 × 2πr cm

                = 45/360 × 2πr cm

15 cm = 45/360 × 2π × r cm

15 = πr/4

Radius = 15×4/ π = 60/π

Therefore, the radius of the circle is 60/π cm.

Solution:

Radius = a cm

Length of arc = aπ/4 cm

θ = angle subtended at the centre of circle

Length of arc = θ/360 × 2πr cm

θ/360 × 2πa cm = aπ/4 cm

θ = 360/ (2 x 4)

θ = 45°

Therefore, the angle subtended at the centre of circle is 45°

Question 6. A sector of a circle of radius 4 cm subtends an angle of 30°. Find the area of the sector. 

Solution:

Radius = 4 cm

Angle subtended at the centre O = 30°

Area of the sector = θ/360 × πr2

                             = 30/360 × π42 

                             = 1/12 × π16 

                             = 4π/3 cm2 

                             = 4.19 cm2 

Therefore, the area of the sector of the circle = 4.19 cm2 

Question 7. A sector of a circle of radius 8 cm contains an angle of 135o. Find the area of sector. 

Solution:

Radius = 8 cm

Angle subtended at the centre O = 135°

Area of the sector = θ/360 × πr2

Area of the sector = 135/360 × π82

                             = 24π cm2

                                       = 75.42 cm2

Therefore, area of the sector calculated = 75.42 cm2

Question 8. The area of a sector of a circle of radius 2 cm is π cm2. Find the angle contained by the sector. 

Solution:

Radius = 2 cm

Area of sector of circle = π cm2

Area of the sector = θ/360 × πr2

                             = θ/360 × π22

                            = πθ/90

π  = π θ/90

θ = 90°

Therefore, the angle subtended at the centre of circle is 90°

Question 9. The area of a sector of a circle of radius 5 cm is 5π cm2. Find the angle contained by the sector. 

Solution:

Radius = 5 cm

Area of sector of circle = 5π cm2

Area of the sector = θ/360 × πr2

                             = θ/360 × π52

                             = 5πθ/72

5π  = 5πθ/72

θ = 72°

Therefore, the angle subtended at the centre of circle is 72°

Question 10. Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.

Solution:

Radius = 5 cm

Length of arc = 3.5 cm

Length of arc = θ/360 × 2πr cm

                     = θ/360 × 2π(5)

3.5 = θ/360 × 2π(5)

3.5 = 10π × θ/360

θ = 360 x 3.5/ (10π)

θ = 126/ π

Area of the sector = θ/360 × πr2

                             = (126/ π)/ 360 × π(5)2

                             = 126 x 25 / 360 

                             = 8.75

Therefore, the area of the sector = 8.75 cm2

Question 11. In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of the arc and area of the sector. 

Solution:

Radius = 35 cm

Angle subtended at the centre = 72°

Length of arc = θ/360 × 2πr cm

                      = 72/360 × 2π(35)

                      = 14π 

                      = 14(22/7) 

                      = 44 cm

Area of the sector = θ/360 × πr2

                              = 72/360 × π 352

                              = (0.2) x (22/7) x 35 × 35

                             = 0.2 × 22 × 5 × 35

Area of the sector = (35 × 22) = 770 cm2

Length of arc = 44cm

Question 12. The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector. 

Solution:

Perimeter of sector includes length of arc and two radii

Radius = 5.7 cm = OA = OB

Perimeter of the sector = 27.2 m

Length of arc = θ/360 × 2πr m

Perimeter = l + 2r

Perimeter of the sector = θ/360 × 2πr + OA + OB

27.2 = θ/360 × 2π x 5.7 cm + 5.7 + 5.7

27.2 – 11.4 = θ/360 × 2π x 5.7

15.8 = θ/360 × 2π x 5.7

θ = 158.8°

Area of the sector = θ/360 × πr2

Area of the sector = 158.8/360 × π 5.72

Area of the sector = 45.03 m2

Question 13. The perimeter of a certain sector of a circle of radius is 5.6 m and 27.2 m. Find the area of the sector.

Solution:

Radius of the circle = 5.6 m = OA = OB

Perimeter of the sector = Perimeter = l + 2r = 27.2

Length of arc = θ/360 × 2πr cm

θ/360 × 2πr cm + OA + OB = 27.2 m

θ/360 × 2πr cm + 5.6 + 5.6 = 27.2 m

θ = 163.64°

Area of the sector = θ/360 × πr2

Area of the sector = 163.64/360 × π 5.62 

                             = 44.8

Therefore, the area of the sector = 44.8 m2

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Solution: 

(i) 7 and 13

Let A be the Arithmetic Mean of 7 and 13.

Then,
7, A and 13 are in A.P.

Now,
A – 7 = 13 – A

2A = 13 + 7

A = 20/2 = 10

∴ A.M. = 10

(ii) 12 and -8

Let A be the Arithmetic Mean of 12 and -8.

Then, 12, A and -8 are in A.P.

Now,

A – 12 = – 8 – A

2A = 12 + 8

A = 2

∴ A.M. = 2

(iii) (x – y) and (x + y)

Let A be the Arithmetic Mean of (x – y) and (x + y).

Then, (x – y), A and (x + y) are in A.P.

Now,

A – (x – y) = (x + y) – A

2A = x + y + x – y

A = x

∴ A.M. = x

Question 2: Insert 4 A.M.s between 4 and 19

Solution:

Let A1, A2, A3, A4 be the 4 A.M.s Between 4 and 19.

Then, 4, A1, A2, A3, A4, 19 are in A.P. of 6 terms.

We know,

An = a + (n – 1).d

a = 4

Then,

a6 = 19 = 4 + (6 – 1).d

∴ d = 3

Now,

A1 = a + d = 4 + 3 = 7

A2 = A1 + d = 7 + 3 = 10

A3 = A2 + d = 10 + 3 = 13

A4 = A3 + d = 13 + 3 = 16

∴ The 4 A.M.s between 4 and 16 are 7, 10, 13 and 16.

Question 3: Insert 7 A.M.s between 2 and 17

Solution:

Let A1, A2, A3, A4, A5, A6, A7 be the 7 A.M.s between 2 and 17.

Then, 2, A1, A2, A3, A4, A5, A6, A7, 17 are in A.P. of 9 terms.

We know,

An = a + (n – 1).d

a = 2

Then,

a9 = 17 = 2 + (9 – 1).d

17 = 2 + 9d – d

17 = 2 + 8d

8d = 17 – 2

8d = 15

∴ d = 15/8

Now,

A1 = a + d = 2 + 15/8 = 31/8

A2 = A1 + d = 31/8 + 15/8 = 46/8

A3 = A2 + d = 46/8 + 15/8 = 61/8

A4 = A3 + d = 61/8 + 15/8 = 76/8

A5 = A4 + d = 76/8 + 15/8 = 91/8

A6 = A5 + d = 91/8 + 15/8 = 106/8

A7 = A6 + d = 106/8 + 15/8 = 121/8

∴ The 7 AMs between 2 and 7 are 31/8, 46/8, 61/8, 76/8, 91/8, 106/8 and 121/8.

Question 4: Insert six A.M.s between 15 and – 13

Solution:

Let A1, A2, A3, A4, A5, A6 be the 6 A.M.s between 15 and –13.

Then, 15, A1, A2, A3, A4, A5, A6, –13 are in A.P. series.

We know,

An = a + (n – 1)d

a = 15

Then,

a8 = -13 = 15 + (8 – 1)d

-13 = 15 + 7d

7d = -13 – 15

7d = -28

∴ d = -4

So,

A1 = a + d = 15 – 4 = 11

A2 = A1 + d = 11 – 4 = 7

A3 = A2 + d = 7 – 4 = 3

A4 = A3 + d = 3 – 4 = -1

A5 = A4 + d = -1 – 4 = -5

A6 = A5 + d = -5 – 4 = -9

∴ The 6 A.M.s between 15 and -13 are 11, 7, 3, -1, -5 and -9.

Question 5: There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3: 1. Find the value of n.

Solution:

Let the A.P. series be 3, A1, A2, A3, …….., An, 17.

Given,

An/A1 = 3/1

We know total terms in AP are n + 2.

So, 17 is the (n + 2)th term.

We know,

An = a + (n – 1)d

a = 3

Then,

An = 17, a = 3

An = 17 = 3 + (n + 2 – 1)d

17 = 3 + (n + 1)d

17 – 3 = (n + 1)d

14 = (n + 1)d

∴ d = 14/(n + 1).

Now,

An = 3 + 14/(n + 1) = (17n + 3)/(n + 1)

A1 = 3 + d = (3n + 17)/(n + 1)

Since,

An/A1 = 3/1

(17n + 3)/(3n + 17) = 3/1

17n + 3 = 3(3n + 17)

17n + 3 = 9n + 51

17n – 9n = 51 – 3

8n = 48

n = 48/8

= 6

∴ There are 6 terms in the A.P. series.

Question 6: Insert A.M.s between 7 and 71 in such a way that the 5th A.M. is 27. Find the number of A.M.s.

Solution:

Let the series be 7, A1, A2, A3, …….., An, 71

We know total terms in the A.P. series is n + 2.

So, 71 is the (n + 2)th term

We know,

An = a + (n – 1)d

a = 7

Then,

5th A.M. = A6 = a + (6 – 1)d

a + 5d = 27

∴ d = 4

So,

71 = (n + 2)th term

71 = a + (n + 2 – 1).d

71 = 7 + n.(4)

n = 15

∴ There are 15 terms in the A.P. series.

Question 7: If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.

Solution:

Let a and b be the first and last terms.

The series be a, A1, A2, A3, …….., An, b.

We know, Mean of a and b = (a + b)/2

Mean of A1 and An = (A1 + An)/2

∴ A1 = a + d

∴ An = a – d

So, AM = (a + d + b – d) / 2 = (a + b) / 2

A.M. between A2 and An-1 = (a + 2d + b – 2d) / 2 = (a + b) / 2

Similarly, (a + b) / 2 is constant for all such numbers.

∴ Hence, A.M. = (a + b) / 2

Question 8: If x, y, z are in A.P. and A1is the A.M. of x and y, and A2 is the A.M. of y and z, then prove that the A.M. of A1 and A2 is y.

Solution:

Given that,

A1 = A.M. of x and y

A2 = A.M. of y and z

So, 

A1 = (x + y)/2

A2 = (y + z)/2

AM of A1 and A2 = (A1 + A2)/2

= [(x + y)/2 + (y + z)/2]/2

= (x + y + y + z)/4

= (x + 2y + z)/4  ……(i)

Since x, y, z are in AP,

y = (x + z)/2  ……(ii)

From (i) and (ii),

A.M. = [{x + z/2} + {(x + 2y + z)/4}]/2

= (y + y)/2

= 2y/2 = y

A.M. = y [Hence proved]

Question 9: Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Solution:

Let A1, A2, A3, A4, A5 be the 5 numbers between 8 and 26

Then, 8, A1, A2, A3, A4, A5, 26 are in the A.P. series

We know,

An = a + (n – 1)d

a = 8

Then,

a7 = 26 = 8 + (7 – 1)d

26 = 8 + 6d

6d = 26 – 8

6d = 18

∴ d = 18/6 = 3

So,

A1 = a + d = 8 + 3 = 11

A2 = A1 + d = 11 + 3 = 14

A3 = A2 + d = 14 + 3 = 17

A4 = A3 + d = 17 + 3 = 20

A5 = A4 + d = 20 + 3 = 23

∴ So, the A.P. series is 8, 11, 14, 17, 20, 23 and 26.

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Solution:

To prove:

A’ – B’ = B – A

Firstly show that

A’ – B’ ⊆ B – A

Let, x ∈ A’ – B’

⇒ x ∈ A’ and x ∉ B’

⇒ x ∉ A and x ∈ B (since, A ∩ A’ = ϕ)

⇒ x ∈ B – A

It is true for all x ∈ A’ – B’

Therefore,

A’ – B’ = B – A

Hence, Proved.

Question 2: For any two sets A and B, prove the following:

(i) A ∩ (A‘ ∪ B) = A ∩ B

(ii) A – (A – B) = A ∩ B

(iii) A ∩ (A ∪ B’) = ϕ

(iv) A – B = A Δ (A ∩ B)

Solution:

(i) A ∩ (A’ ∪ B) = A ∩ B

Here, LHS A ∩ (A’ ∪ B)

Upon expanding

(A ∩ A’) ∪ (A ∩ B)

We know, (A ∩ A’) =ϕ

⇒ ϕ ∪ (A∩ B)

⇒ (A ∩ B)

Therefore,

LHS = RHS

Hence, proved.

(ii) A – (A – B) = A ∩ B

For any sets A and B we have De-Morgan’s law

(A ∪ B)’ = A’ ∩ B’, (A ∩ B) ‘ = A’ ∪ B’

Take LHS

= A – (A–B)

= A ∩ (A–B)’

= A ∩ (A∩B’)’

= A ∩ (A’ ∪ B’)’) (since, (B’)’ = B)

= A ∩ (A’ ∪ B)

= (A ∩ A’) ∪ (A ∩ B)

= ϕ ∪ (A ∩ B) (since, A ∩ A’ = ϕ)

= (A ∩ B) (since, ϕ ∪ x = x, for any set)

= RHS

Therefore,

LHS = RHS

Hence, proved.

(iii) A ∩ (A ∪ B’) = ϕ

Here, LHS A ∩ (A ∪ B’)

= A ∩ (A ∪ B’)

= A ∩ (A’∩ B’) {By De–Morgan’s law}

= (A ∩ A’) ∩ B’ (since, A ∩ A’ = ϕ)

= ϕ ∩ B’

= ϕ (since, ϕ ∩ B’ = ϕ)

= RHS

Therefore,

LHS = RHS

Hence, proved.

(iv) A – B = A Δ (A ∩ B)

Here, RHS A Δ (A ∩ B)

A Δ (A ∩ B) (since, E Δ F = (E–F) ∪ (F–E))

= (A – (A ∩ B)) ∪ (A ∩ B –A) (since, E – F = E ∩ F’)

= (A ∩ (A ∩ B)’) ∪ (A ∩ B ∩ A’)

= (A ∩ (A’ ∪ B’)) ∪ (A ∩ A’ ∩ B) {by using De-Morgan’s law and associative law}

= (A ∩ A’) ∪ (A ∩ B’) ∪ (ϕ ∩ B) (by using distributive law)

= ϕ ∪ (A ∩ B’) ∪ ϕ

= A ∩ B’ (since, A ∩ B’ = A – B)

= A – B

= LHS

Therefore,

LHS = RHS

Hence, Proved

Question 3: If A, B, C are three sets such that A ⊂ B, then prove that C – B ⊂ C – A.

Solution:

Given: A ⊂ B

To prove: C – B ⊂ C – A

Let’s assume, x ∈ C – B

⇒ x ∈ C and x ∉ B

⇒ x ∈ C and x ∉ A

⇒ x ∈ C – A

Thus, x ∈ C–B ⇒ x ∈ C – A

This is true for all x ∈ C – B

Therefore,

C – B ⊂ C – A

Hence, proved.

Question 4: For any two sets A and B, prove that

(i) (A ∪ B) – B = A – B

(ii) A – (A ∩ B) = A – B

(iii) A – (A – B) = A ∩ B

(iv) A ∪ (B – A) = A ∪ B

(v) (A – B) ∪ (A ∩ B) = A

Solution:

(i) (A ∪ B) – B = A – B

Let’s assume LHS (A ∪ B) – B

= (A–B) ∪ (B–B)

= (A–B) ∪ ϕ (since, B–B = ϕ)

= A–B (since, x ∪ ϕ = x for any set)

= RHS

Hence, proved.

(ii) A – (A ∩ B) = A – B

Let’s assume LHS A – (A ∩ B)

= (A–A) ∩ (A–B)

= ϕ ∩ (A – B) (since, A-A = ϕ)

= A – B

= RHS

Hence, proved.

(iii) A – (A – B) = A ∩ B

Let’s assume LHS A – (A – B)

Let, x ∈ A – (A–B) = x ∈ A and x ∉ (A–B)

x ∈ A and x ∉ (A ∩ B)

= x ∈ A ∩ (A ∩ B)

= x ∈ (A ∩ B)

= (A ∩ B)

= RHS

Hence, proved.

(iv) A ∪ (B – A) = A ∪ B

Let’s assume LHS A ∪ (B – A)

Let, x ∈ A ∪ (B –A) ⇒ x ∈ A or x ∈ (B – A)

⇒ x ∈ A or x ∈ B and x ∉ A

⇒ x ∈ B

⇒ x ∈ (A ∪ B) (since, B ⊂ (A ∪ B))

This is true for all x ∈ A ∪ (B–A)

∴ A ∪ (B–A) ⊂ (A ∪ B) —(equation 1)

Contrarily,

Let x ∈ (A ∪ B) ⇒ x ∈ A or x ∈ B

⇒ x ∈ A or x ∈ (B–A) (since, B ⊂ (A ∪ B))

⇒ x ∈ A ∪ (B–A)

∴ (A ∪ B) ⊂ A ∪ (B–A) —(equation 2)

From equation 1 and 2 we get,

A ∪ (B – A) = A ∪ B

Hence, proved.

(v) (A – B) ∪ (A ∩ B) = A

Let’s assume LHS (A – B) ∪ (A ∩ B)

Let, x ∈ A

Then either x ∈ (A–B) or x ∈ (A ∩ B)

⇒ x ∈ (A–B) ∪ (A ∩ B)

∴ A ⊂ (A – B) ∪ (A ∩ B) —(equation 1)

Contrarily,

Let x ∈ (A–B) ∪ (A ∩ B)

⇒ x ∈ (A–B) or x ∈ (A ∩ B)

⇒ x ∈ A and x ∉ B or x ∈ B

⇒ x ∈ A

(A–B) ∪ (A ∩ B) ⊂ A —(equation 2)

∴ From equation (1) and (2), We get

(A–B) ∪ (A ∩ B) = A

Hence, proved