RS Aggarwal Solutions: Area of Circle, Sector and Segment- 4 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

Let the radius if circle be r
Given, Area of circle = 38.5 cm²
Area of circle = πr²
⇒ πr² = 38.5
Since π = 22/7
⇒ 22/7 × r²= 38.5
⇒ r² = 38.5 × (7/22)
⇒ r² = 12.25
⇒ r = √12.25 = 3.5 cm
∴ Radius of circle = 3.5 cm
Circumference of circle = 2πr
= 2 × 22/7 × 3.5 cm
= 22 cm
∴ Circumference of the circle is 22 cm
Let the radius if circle be r
Given, Area of circle = 38.5 cm²
Area of circle = πr²
πr² = 38.5
Since = 22/7
∴ πr² = 38.5
⇒ 22/7 × r² = 38.5
⇒ r² = 12.25
⇒ r = √12.25 = 3.50cm
∴ Radius of circle = 3.5 cm
Circumference of circle = 2πr
= 2 × 22/7 × 3.5
= 22 cm
∴ Circumference of the circle is 22 cm

Let the radius if circle be r
Given, Area of circle = 49π cm²
Area of circle = πr²
πr² = 49π
Since = 22/7
∴ πr² = 49π
⇒ r² = 49
⇒ r = √49 = 7 cm
∴ Radius of circle = 7 cm
Circumference of circle = 2πr
= 2 × π × 7 cm
= 14π cm
∴ Circumference of the circle is 14π cm

Let the radius if circle be r
Circumference of circle = 2πr
Difference between the circumference and radius of a circle = 37 cm
⇒ 2πr – r = 37 cm
⇒ 2 × 22/7 × r – r = 37 cm
⇒ 44/7 × r – r = 37 cm
⇒ (44/7 – 1) × r = 37 cm
⇒ 37/7 × r = 37 cm
⇒ r = 37 × 7/37
⇒ r = 7 cm
Area of circle = πr²
= 22/7 × 7 × 7 cm²
= 22/7 × 49 cm² = 22 × 7 cm² = 154 cm²
∴ Area of the circle is 154 cm²

Let the radius if circular field be r
Perimeter of circular field = 2πr
Perimeter of circular field = 242 m
⇒ 2πr = 242 m
⇒ 2 × 22/7 × r = 242 m
⇒ r = 242 × 1/2 × 7/22 = 38.5 m
∴ Radius of circular field = 38.5 m
Area of the field = πr²
= 22/7 × 38.5² m²
= 22/7 × 1482.5 m² = 4658.5 m²
∴ Area of the field = 4658.5 m²

Let the radius if circle be r
Area of circle = A = πr²
Radius increases by 40%
So, New Radius r’ = r + 40/100 × r = 1.4r
New Area of circle = A’ = πr’² = π × (1.4r)²
= 1.96πr2
Percentage increase in area = 
=  = .96 × 100 = 96
∴ Increase in area = 96%

Let the radius if circle be r
Area of circle = A = πr²
Radius decreases by 30%
So, New Radius r’ = r - 30/100 × r = 0.7r
New Area of circle = A’ = πr’² = π × (0.7r)²
= .49πr²
Percentage decrease in area 

∴ Decrease in area = 51%

Let the length of the side of the square be a
Let the radius if circle be r
Area of a square = a²
Area of circle = πr²
Area of a square = Area of a circle
a² = πr²
a = √π × r
Perimeter of circle = 2πr
Perimeter of square = 4a
= 4√π r


Ratio of perimeter of circle and square = √π : 2

Let the bigger circle be C₁ and other circles be C₂ and C₃
Radius of circle C₁ = r₁
Diameter of circle C₂ = 36 cm
Radius of circle C₂ = r₂ = 36/2 cm = 18cm
Diameter of circle C₃ = 20 cm
Radius of circle C₃ = r₃ = 20/2 cm = 10 cm
Circumference of circle C₂ = 2πr₂
= 2 × π × 18 cm = 36π cm
Circumference of circle C₃ = 2πr₃
= 2 × π × 10 cm = 20π cm
Circumference of circle C₁ = Circumference of circle C₂ + Circumference of circle C₃
⇒ 2πr₁ = 2πr₂ + 2πr₃
⇒ 2πr₁ = 36π + 20π
⇒ 2πr₁ = 56π
⇒ r₁ = 28 cm
Radius of circle C₁ = r₁ = 28 cm

Let the bigger circle be C₁ and other circles be C₂ and C₃
Radius of circle C₁ = r₁
Radius of circle C₂ = r₂ = 24 cm
Radius of circle C₃ = r₃ = 7 cm
Area of circle C₂ = 
= π × 24² = 576π
Area of circle C₃ = 
= π × 7² = 49π
Area of circle C₁ = Area of circle C₂ + Area of circle C₃



r₁ = 25 cm
Diameter of new circle = 25 × 2 cm = 50cm

Let the length of the side of the square be a
Let the radius if circle be r
Perimeter of circle = 2πr
Perimeter of square = 4a
Perimeter of circle = Perimeter of square
⇒ 2πr = 4a
a = π × r/2
Area of a square = a²
Area of circle = πr²

Ratio of area of square to circle = π: 4

Let three circles be C₁, C₂ and C
Area of circle C = Area of circle C₁ + Area of circle C₂

Let three circles be C₁, C₂ and C
Circumference of circle C = Circumference of circle C₁ + Circumference of circle C₂
⇒ 2πR = 2πR₁ + 2πR₂
R = R₁ + R₂

Let the length of the side of the square be a
Let the radius if circle be r
Perimeter of circle = 2πr
Perimeter of square = 4a
Perimeter of circle = Perimeter of square
2πr = 4a
a = π × r/2
Area of a square = a²
= (π × r/2)² = π/4 × πr²
Area of circle = πr²
Seeing the co-efficient of πr²
1 > π/4 ∴ πr² > π/4 × πr²
So, (area of the circle) > (area of the square)

Radius of circle 1 = r₁ = 19 cm
Radius of circle 2 = r₂ = 16 cm

= π (19² – 16²) cm²
= 22/7 × 105
= 330 cm²

Let the radius of circle 1 & 2 be R₁ and R₂ respectively
Area of circle 1 = 1386 cm²

R₁ = 21 cm
Area of circle 2 = 962.5 cm²

R₂ = 17.5 cm
Width of the ring = R₁ – R₂ = 21 -17.5 = 3.5 cm

Circumference of circle C₁ = 2πr₁
Circumference of circle C₂ = 2πr₂



∴ Ratio of two circles = 9: 16

Ratio of their circumferences = 3: 2

Radius of wheel = r = 0.25 m
Distance the wheel travels = 11 km = 11000 m
In 1 revolution wheel travels 2πr distance
No. of revolutions a wheel makes = 

= 7000 revolutions

Diameter of wheel = 40 cm
Radius of wheel = r = 40/2 cm = 20 cm
Distance the wheel travels = 176 m = 17600 cm
In 1 revolution wheel travels 2πr distance
No. of revolutions a wheel makes = 

= 140 revolutions

Distance the wheel travels = 88 km = 88000 m
In 1 revolution wheel travels 2πr distance
No. of revolutions a wheel makes = 
No. of revolutions a wheel makes = 1000

r = 14 m
Radius of wheel = 14 m
Diameter of wheel = 2 × 14 m = 28 m

Area of a sector of angle θº of a circle with radius R = area of circle × (θ/360)

Length of an arc of a sector of angle θº of a circle with radius R
= Circumference of circle × θ/360

Length of the minute hand of a clock = 21 cm
∴ Radius = R = 21 cm
In 1 minute, minute hand sweeps 6°
So, in 10 minutes, minute hand will sweep 10 × 6° = 60°
Area swept by minute hand in 10 minutes = Area of a sector of angle θº of a circle with radius R =  = 22/7 × 21 × 21 × 60/360 = 231 cm²

Radius of Circle = R = 10 cm

Area of minor Segment = Area of sector subtending 90° – Area of triangle ABC
Area of sector subtending 90° =  = 3.14 × 10 × 10 × 90/360 cm²
= 78.5 cm²
Area of triangle ABC = 1/2 × AC × BC
= 1/2 × 10 × 10 cm² = 50 cm²
Area of Minor segment = 78.5 cm² – 50 cm²
= 28.5 cm²

Radius of Circle = R = 21 cm
Angle Subtended by the arc = 60°
Length of an arc of a sector of angle θº of a circle with radius R = 
Length of arc = 2 × 22/7 × 21 × 60/360 cm = 22 cm

Radius of Circle = R = 14 cm
Angle Subtended by the arc = θ = 120°
Area of sector subtending 120° =  = 22/7 × 14 × 14 × 120/360 cm²
= 205.33 cm²
In Triangle ABC
AC = BC = 14 cm = R
Area of triangle ABC = 1/2 × base × height
= 2 × 1/2 × R sin θ/2 × R × cos θ/2
= 2 × 1/2 × 14 × 14 × sin 60° × cos 60°
= 84.77 cm²
Area of Segment = Area of sector subtending 120° - Area of triangle ABC
= 205.33 – 84.77 cm² = 120.56 cm²