# Brilliant’s Composite Mathematics Class 8 Solutions Chapter 16

## 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

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

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

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

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 :

= 84/2

= 42 m.

= 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 :

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,

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 = πr2 × sector angle/360°

= 22/7 × (14)2 × 45/360

= 77 cm2.

(ii) Area = 22/7 × (42)2 × 60/360

= 924 cm2.

(iii) 22/7 × (3.5)× 100/360

= 10.69 cm2.

Question no – (7)

Solution :

(i) Area = πr2 × arc length/2πr

= r × arc length/2

= 35 × 14/2

= 245 cm2.

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

= 15 × 6.8

= 102 cm2

Question no – (8)

Solution :

Area = πr2 × 360 – 45/360

= 22/7 × (2)2 × 315/360

= 11 cm2.

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 , πr2 × 36/360 = 3.85

Or, r2 = 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 cm2.

Question no – (13)

Solution :

In 5 minutes, the minute hand sweeps 30 degrees

Area = 22/7 × (√21)2 × 30/360

= 11/2

= 55 cm2.

Question no – (14)

Solution :

Remaining area = (50)2 – π (7)2

= 2500 – 154

= 2346 m2.

Question no – (15)

Solution :

Area = πr2 × 60/360

= 22/7 × (10)2 × 1/6

= 3.14 × 100 × 1/6

= 52.38 m2.

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 cm2.

Question no – (17)

Solution :

Radius = 84/2 = 42 cm.

Length of the arc = 2πr2 × 150/360

= 2 × 22/7 × 42 × 5/12 = 110 cm.

Area = πr2 × 150/360

= 22/7 × 42 × 5/12

= 2310 cm2.

Question no – (18)

Solution :

Required area,

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

= 4 √3 – 6.28

= 0.64 cm2.

Question no – (20)

Solution :

Area = πr2 × 120/360

= 22/7 × 62 × 1/3

= 12 × 3.14

= 37.68 cm2.

Question no – (21)

Solution :

Area of minor segment = πr2 × 90/360

= 22/7 × 100 × 1/4

= (25 × 3.14) cm2

Previous Chapter Solution :

Updated: May 29, 2023 — 9:52 am