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**Brilliant’s Composite Mathematics Class 8 Solutions Chapter 16 Circumference and Area of a Circle**

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 16, Circumference and Area of a Circle. Here students can easily find step by step solutions of all the problems for Circumference and Area of a Circle, Exercise 16.1, 16.2 and 16.3 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

**Circumference and Area of a Circle Exercise 16.1 Solution**

**Question no – (1)**

**Solution :**

**(a) Circumference,**

= 2 π r = 2 × 22/7 × 2.1

= 2 × 22 × 0.3

= 13.2 cm

**(b) Circumference,**

2 π r = 2 × 22/7 × 4.2

= 2 × 22 × 0.6

= 26.4 cm

**(c) Circumference,**

2 π r = 2 × 22/7 × 2.8

= 2 × 22 × 0.4

= 17.6 cm

**Question no – (2)**

**Solution :**

**(a)** Diameter = 2π = 7 cm

**∴** radius 7/2 = 3.5 cm

**∴** Circumference,

= 2πr = 2 × 22/7 × 3.5

= 12 cm

**Question no –(3)**

**Solution :**

**(a)** 2 π r = 132

or, 2 × 22/7 × π

= 132 × 7/2 × 22

= 21 cm

**∴** Radius = 21 cm,

**∴** Diameter = 21 × 2

= 42 cm

**(b)** 2 π r = 16.5

Or, 2 × 22/7 × π = 16.5

Or, π = 16.5 × 7/2 × 22

= 115 × 7/4

= 2.625 m

= 5. 25 cm

**Question no – (4)**

**Solution :**

Let, the diameters be 2x, 6x units

Then the radiuses are 5x/2, 6x/2 units

**∴** Ratio of the circumferences,

= 2 π (5x/2) : 2 π (6x/2)

= 5 : 6

**Question no – (5)**

**Solution :**

The circumference,

= 2 (12.2 + 6.5)

= 2 × 18.7

= 37.4 cm

**∴** Now , 2 π r = 37.4

or, π = 37.4 × 7/22 × 2

= 1.7 × 7/2

= 5.95 cm

**Question no – (6)**

**Solution :**

Diameter = 63 cm

**∴** Radius = 63/2 cm

**∴** Distance travelled = Circumference × number of revolution

= 2 × 22/7 × 63/3 × 600

= 22 × 9 × 600

= 118800 cm

= 118800/100

= 118.8 m

**Question no – (7)**

**Solution :**

Diameter = 56 cm .

**∴** Distance travelled = circumference × number of revolutions

= 2 × 22/7 × 28 × 1000

= 176000 m

= 176000/100

= 1760 m .

**Question no – (8)**

**Solution :**

Circumference,

= 2 π r = 2 × 3.14 × 384000

= 2411520 km .

**Question – (9)**

**Solution :**

Difference of circumference = 2 π R₁ – 2 π R₂

= 2 π (R₁ – R₂)

= 2 × 22/7 × (112 – 105)

= 2 × 22/7 × 7

= 44 m .

**Question no – (10)**

**Solution :**

Diameter = 2 π = 1.26 m .

**∴** Distance = circumference × number at revolutions

= 2 π r × 560

= 1.26 × 22/7 × 560

= 1.8 × 22 × 560

= 22176 m .

**Question no – (11)**

**Solution :**

Circumference,

= 2 × 22/7 × 42

= 264 cm .

**∴** Side of a square

= 264/4

= 66 cm .

**Question no – (12)**

**Solution :**

Distance = Circumference × Number of revolutions

= 2 × 22/7 × 49/2 × 4

= 616 m.

**Question no – (13)**

**Solution :**

Number of revolutions of the first wheel

= distance/circumference

= 960000/2 × 22/7 × 10

= 7 × 24000/11

Number of revolutions of the second wheel

= Distance/Circumference

= 960000/2 × 22/7 × 16

= 7 × 15000/11

**∴** Difference between number of revolutions

= 7 × 24000/11 – 7 × 15000/11

= 7 × 9000/11

= 5727.272

**Circumference and Area of a Circle Exercise 16.2 Solution**

**Question no – (1)**

**Solution :**

**(a)** Area = πr ² = 22/7 × 2.1

= 13.86 cm²

**(b)** πr² = 22/7 × 3.5 × 3.5

= 38.5 m²

**Question no – (2)**

**Solution :**

**(a)** Diameter = 2π = 14 cm

**∴** Radius = 7 cm .

**∴** πr ² = 22/7 × 7 × 7

= 154 cm²

**Question no – (3)**

**Solution :**

2πr = 264

or, π =264 × 7/2 × 22

= 6 × 7

= 42 m.

**∴** Area = πr² = 22/7 × 42 × 42

= 5544 cm² .

**Question no – (4)**

**Solution :**

2πr = 22 or , 2 × = 22 × 7/22 = 7 m .

Area = πr ² = 22/7 × (7/2)²

= 22/7 × 49/4

= 77/2

= 38.5 m²

**Question no – (5)**

**Solution :**

2πr² = 154

or , π² = 154 × 7/22

or , π² = 7 × 7

**∴** π = 7 cm .

**∴** Circumference

= 2πr = 2 × 22/7 × 7

= 44 cm .

**Question no – (6)**

**Solution :**

π² = 616

or, π² = 116 × 7/22

or, π² = 4 × 49 = π = 2 × 7

= 14 cm

**∴** Diameter,

= 14 × 2

= 28 cm .

**Question no – (7)**

**Solution :**

**(a)** πr² = 38.5

Or, π² = 38.5 × 22/7 = 121

Or, π = 11 cm

**∴** Circumference,

= 2πr = 2 × 22/7 × 11

= 22 × 3.14

= 69.08 cm

**Question no – (8)**

**Solution :**

**(a)** 2 π r = 132

Or , π = 132 × 7/22 = 42 .

**∴** πr² = 22/7 × 42 × 42

= 5544 cm²

**Question no – (9)**

**Solution :**

Diameter = 2π = 5 cm

**∴** radius = π = 2.5 cm

**∴** Area = πr² = 3.14 × (2.5)

= 19.625 cm²

**Question no – (10)**

**Solution :**

Let the radil be 2x. 3x units

**(i)** Ratio of their circumference

= 2π (2x) : 2π (3x)

= 2 : 3

**(ii)** Ratio of their areas,

= π (2x)² : π (3x)

= 4 πx² : 9 πx²

= 4 : 9

**Question on – ( 12)**

**Solution :**

Let, R and π be the radii

πR πR²/πr² = 100/1

Or, R²/r² = 100/1

Or, (R/π )² = (10/1)²

Or, R/r = 10/1

Or 2πR/2 πr = 10/1

**∴** Ratio of their circumference = 10 : 1

**Question no – (13)**

**Solution :**

Side of the square of perimeter 44 cm = 11 cm

**∴** Area = 11²

= 121 cm²

Now, radius of the circle of circumference 44 cm = 7 cm .

**∴** Area,

= 22/7 × 7²

= 154 cm²

**∴** Difference of areas,

= 154 – 121

= 33 cm²

**Question no – (14)**

**Solution :**

Length of the Wire,

= 2 × 22/7 × 21

= 132 m .

**∴** Side of the square

= 132/4

= 33 m

**∴** Area of the square

= 33²

= 1089 m²

**Question no – (15)**

**Solution :**

Radius of the circular park,

= 84/2

= 42 m.

**∴** Area of the road,

= π (42 + 3.5)² – π (42)²

= π [(42.5)² – (42)²]

= 22/7 × 87.5 × 3.5

= 22 × 12.5 × 3.5

= 962.5 m².

**∴** Cost of construction,

= 962.5 × 25

= 24062.5 rupees.

**Question no – (16)**

**Solution :**

Combined area = π (5)² + π (12)² = πR²

Or, π (52 + 122) = πR²

Or, R² = 25 + 144 = 169

**∴** R = √169

= 13 cm.

**Question no – (17)**

**Solution :**

Area of the garden,

= 120 × 80

= 9600 m².

Area of the circular tank,

= πr² = 22/7 × 14 × 14

= 616 m²

**∴**** Cost of turfing,**

= (9600 – 616) × 3

= 9484 × 3

= 28452 Rs

**Question no – (18)**

**Solution :**

Cost of levelling,

= π [(77)² – (70)²] × 2.50

= 22/7 × 147 × 7 × 2.50

= 8085 rupees.

**Question no – (19)**

**Solution :**

Let the radius be r,

**∴** We have 2 πr × 1500 = 4500

Or, 2πr = 3

Or, r = 3 × 7/2 × 22 = 478 m

= 47.8 cm.

**Question no – (20)**

**Solution :**

Diameter = 30.8 cm,

**∴** radius = 15.4 cm.

**∴** Area = 1/2 πr²

= 1/2 × 22/7 × 15.4 × 15.4

= 372.68 cm²

Perimeter, πr + 2r

= 22/7 × 15.4 + 2 × 15.4

= 48.4 + 30.8

= 79.2 cm.

**Circumference and Area of a Circle Exercise 16.3 Solution**

**Question no – (1) **

**Solution : **

**(a)** length of arc = 2πr × 30/360

= 2 × 22/7 × 14 × 1/12

= 22/3 cm.

**(b)** 2 × 22/7 × 90°/360° × 14

= 22 cm.

**(c)** 2 × 12 × 14 × 135/360

= 33 Cm.

**Question no – (3)**

**Solution :**

Angle = length of area/circumference × 360°

= 11/2 × 22/7 × 21 × 360°

= 30°

**Question no – (5)**

**Solution :**

**(i)** Area = πr^{2} × sector angle/360°

= 22/7 × (14)^{2} × 45/360

= 77 cm^{2}.

**(ii)** Area = 22/7 × (42)^{2} × 60/360

= 924 cm^{2}.

**(iii)** 22/7 × (3.5)^{2 }× 100/360

= 10.69 cm^{2}.

**Question no – (7)**

**Solution :**

**(i)** Area = πr^{2} × arc length/2πr

= r × arc length/2

= 35 × 14/2

= 245 cm^{2}.

**(ii)** Area = r × arc length/2 = 15 × 13.6/2

= 15 × 6.8

= 102 cm^{2}

**Question no – (8)**

**Solution :**

Area = πr^{2} × 360 – 45/360

= 22/7 × (2)^{2} × 315/360

= 11 cm^{2}.

**Question no – (9)**

**Solution :**

Area of the sector = sector angle/360 × Area of circle

∴ sector angle = Area of sector/Area of circle × 360

= (1/6 × 360)°

= 60°

**Question no – (10)**

**Solution :**

Let the circumference be C cm.

**∴** 22/c × 360 = 18

Or, c = 22/18 × 360

= 440 m.

**Question no – (11)**

**Solution : **

We have , πr^{2} × 36/360 = 3.85

Or, r^{2} = 3.85 × 10 × 7/22

**∴** r = 3.5 cm.

**∴** length of the arc = 2πr × 36/360

= 2 × 22/7 × 3.5 × 1/10

= 2.2 cm.

**Question no – (12)**

**Solution :**

In 60 minutes, the minute hand sweep 360 degrees

**∴** In 15 minutes, it sweeps 90 degrees

**∴** Area = 22/7 × 8.4 × 8.4 × 90/360

= 55.44 cm^{2}.

**Question no – (13)**

**Solution :**

In 5 minutes, the minute hand sweeps 30 degrees

**∴** Area = 22/7 × (√21)^{2} × 30/360

= 11/2

= 55 cm^{2}.

**Question no – (14)**

**Solution :**

Remaining area = (50)^{2} – π (7)^{2}

= 2500 – 154

= 2346 m^{2}.

**Question no – (15)**

**Solution : **

Area = πr^{2} × 60/360

= 22/7 × (10)2 × 1/6

= 3.14 × 100 × 1/6

= 52.38 m^{2}.

**Question no – (16)**

**Solution :**

Perimeter of sector = l + 2r = 27.2

Or, l = 27.2 – 11.2 = 16

**∴** Area of sector = 1/2 × (length of arc) × radius

= 1/2 × 16 × 5.6

= 44.8 cm^{2}.

**Question no – (17)**

**Solution :**

Radius = 84/2 = 42 cm.

**∴** Length of the arc = 2πr^{2} × 150/360

= 2 × 22/7 × 42 × 5/12 = 110 cm.

Area = πr^{2} × 150/360

= 22/7 × 42 × 5/12

= 2310 cm^{2}.

**Question no – (18)**

**Solution :**

Required area,

= √3/4 × (4)^{2} – 3 × 22/7 × (2)^{2} × 60/360

= 4 √3 – 6.28

= 0.64 cm^{2}.

**Question no – (20)**

**Solution :**

Area = πr^{2} × 120/360

= 22/7 × 62 × 1/3

= 12 × 3.14

= 37.68 cm^{2}.

**Question no – (21)**

**Solution : **

Area of minor segment = πr^{2} × 90/360

= 22/7 × 100 × 1/4

= (25 × 3.14) cm^{2}

**Previous Chapter Solution : **