 # R S Aggarwal Solutions for Class 10 Maths Chapter 16 Area of Circle, Sector and Segment

R S Aggarwal solutions for class 10 Chapter 16 helps students to solve all the questions on the topics Area of Circle, Sector and Segment. Detailed solutions to the R S Aggarwal textbook questions are provided here so that students can compare their answers to the sample responses. This study material contains all the exercise questions provided in the textbook.

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RS Aggarwal Class 10 Maths Solutions for Chapter 16 and clear concepts.

### Access solution to Maths R S Aggarwal Chapter 16 – Area of Circle, Sector and Segment

Get detailed solutions for all the questions listed under the below exercises:

Exercise 16A Solutions : 30 Questions (Short Answers)

Exercise 16B Solutions : 27 Questions (Short Answers)

## Exercise 16A

Question 1:
Solution:

Given: Circumference of the circle = 39.6 cm.

We know, Circumference of circle = 2πr

Where, r = Radius of the circle

⇨ 2πr = 39.6

r = 39.6/2π

r = 6.3

(put value of π = 22/7)

Area of the circle = πr2

Where, r = radius of the circle

⇨ Area of the circle = π(6.3)2

= 124.74

So, area of circle is 124.74 cm2.

Question 2:
Solution:

Area of a circle = 98.56 cm2

We know, Area of the circle = πr2

Where, r = radius of the circle

So, πr2 = 98.56

Put π = 22/7

r2 = 98.56/(22/7) = 31.36

r = 5.6 cm

Circumference of circle = 2πr = 2 ×22/7×5.6 = 35.2 cm

Question 3:
Solution:

Given: circumference of a circle exceeds its diameter by 45 cm.

⇨ Circumference of circle = 45 + Diameter of circle

2π r = 45 + 2r

2r(π – 1) = 45

2r(22/7 – 1) = 45

15/7 r = 45/2

r = 10.5

Circumference of a circle = 2πr = 2 x 22/7 x 10.5 = 66 cm

Question 4:
Solution:

Let the square be of side ‘a’ cm and radius of the circle be ‘r’

Area enclosed by the square = 484 cm2

Also, we know that Area of square = Side × Side

Area of the square = a2

⇨ a2 = 484

⇨ a = √484

⇨ a = 22

Therefore, side of square is 22 cm.

Perimeter of square = 4 x side = 4 x 22 = 88

From statement, circumference of the circle = Perimeter of square

2πr = 88

r = (88 x 7)/(2×22) = 14

Radius of the circle = 14 cm

Area of circle = πr2 = 22/7 x 14 x 14 = 616 cm2.

Question 5:
Solution:

Area of an equilateral triangle = 121√3 cm2 (given)

We know that, Area of an equilateral triangle = √3/4 (a2), where “a” is the side of a triangle. The area enclosed by the circle is 346.5 cm2

Question 6:
Solution:

The length of a chain used as the boundary of a semicircular park = 108 m

i.e. circumference of a semicircular park = 108 m

Let the radius of semicircular park = r

Which implies, Now,

Area of a semicircular park = πr2 = 22/7 x 21 x 21 = 1386

Area of the park = 1/2 (area of semicircular park) = 1/2 x 1386 = 693

Therefore, Area of the park is 693 m2.

Question 7:
Solution:

The sum of the radii of two circles is 7 cm (given)

Difference of circumferences of two circles is 8 cm (given)

Let radii of circles be x and (7-x)

Then,

2πx – 2π(7 – x) = 8

2πx – 14π + 2πx = 8

2πx = 26 …..(1)

Which is circumference of one circle.

Circumference of the another circles = 2π(7 – x) = 14π – 2πx

= 14π – 26

= 14 x 22/7 – 26

= 18

Therefore, Circumference of the circles are 26 cm and 18 cm.

Question 8:
Solution: Radius of outer ring = R = 23 cm

Radius of inner ring = r = 12 cm

Area of outer circle = πR2

= 22/7 x 23 x 23

= 1662.5 cm2

Area of inner circle = π r2

= 22/7 x 12 x 12

= 452.2 cm2

Area of Ring = outer area – inner area

= 1662.5 – 452.2

= 1210

Area of the ring is 1210 cm2

Question 9:
Solution:

Width of the path = 8 m

Inner radius of the circular park = 17 m

Outer radius of the circular park = (17 + 8) = 25 m

Now,

Area of the path = π[(252 – (172)]

= 22/7 x 42 x 8

= 1056

Area of the path is 1056 m2.

(ii) A park is in the shape of a circle of diameter 7m. It is surrounded by a path of width 0.7m. Find the expenditure of cementing the path at the rate of ₹ 110 per sq.m.

Solution:

Diameter of park = 7 m

so, radius of park = r1 = 3.5 m

Width of path = 0.7 m

Bigger radius of park = r2 = 0.7 + 3.5 = 4.2 m

Now, Area of path = area of bigger circle -area of smaller circle

= Pi r22 – Pi r12

= Pi(r22 – r12)

= 22/7 (4.22– 3.5 2)

= 22/7(17.64 – 12.25)

=16.94 m2

Also,

Cost of expenditure = rate × area

= 110 × 16.94

= 1864.40

Cost of Expenditure of cementing the path is ₹ 1864.40.

Question 10:
Solution:

Let r and R be the radii of inner circle and outer boundaries respectively.

Then,

Circumference of inner circle = 2πr = 352 and

Circumference of outer circle = 2πR = 396

r = 352/2π and R = 396/2π

Width of the track = (R – r) = 396/2π – 352/2π

= 44/2π

= 44/2 x 7/22

= 7

Width of the track is 7 m.

Again, R + r = 396/2π + 352/2π = 748/2π

Area of the track = π(R2 – r2)

= π(R – r) (R+r)

= π x 7 x 748/2π

= 2618

Therefore, the area of the track is 2618 m2.

Question 11:
Solution: Question 12:
Solution: The radius of the circle = 10 cm.

Chord PQ subtends an angle = 60 degrees at the centre.

PQO is an equilateral triangle, O being the centre of the circle.

Area of minor segment = θ/360 x π r2 – 1/2 r2 sinθ

= 60/360 x 22/7 x 10 x 10 – 1/2 x 10 x 10 x √3/2

= 9.03

Area of circle = π r2 = 22/7 x 10 x 10 = 314 (approx.)

Area of minor segment = Area of circle – Area of minor segment

= 314 – 9.03

= 304.97 cm2 (approx.)

Question 13:
Solution: Question 14:
Solution:

Radius of a circle = 7 cm

Area of circle = πr2 = 22/7 x 7 x 7 = 154 cm2 Area of major segment = Area of circle – Area of minor segment

= 154 – 14

= 140

Therefore, Area of major segment is 140 cm2.

Question 15:
Solution:

The lengths of the arcs cut off from a circle of radius (r ) 12 cm by a chord 12 cm long.

Which shows it form a equilateral triangle POQ, and ∠POQ = 60 degrees Area of minor segment = Area of sector – Area of triangle …(i)

Area of triangle = √3/4 (side)2

= √3/4 (12)2

= 62.28 cm2

Area of sector = θ/360 x πr2

= 60/360 x 3.14 x 12 x 12

= 75.36 cm2

Equation (1) implies

Area of minor segment = 75.36 – 62.28 = 13.08 cm2

Therefore, length of major arc is 62.8 cm and of minor arc is 12.56 cm and area of minor segment is 13.08 cm2.

Question 16:
Solution:

Radius of circle = 5√2 cm

Chord = 10 cm

To find: Areas of both the segments

Now,

Area of minor segment = (area of sector OACBO) – (area of ∆OAB)

= 39.25 – 25 = 14.25 cm2 Question 17:
Solution: Question 18:
Solution:

Radius of circle = r = 30 cm

Area of minor segment = Area of sector – Area of triangle …(1)

Area of major segment = Area of circle – Area of minor segment …(2)

Area of sector = θ/360 x πr2

= 60/360 x 3.14 x 30 x 30

= 471 cm2

Area of triangle = √3/4 (side)2 (Since it form a equilateral triangle)

= √3/4 x 30 x 30

= 389.7 cm2

(1) ⇨

Area of minor segment = 471 – 389.7 = 81.3 cm2

(2) ⇨

Area of major segment = π(302) – 81.3 = 2744.7 cm2

Area of major segment is 2744.7 cm2 and of minor segment is 81.3 cm2.

Question 19:
Solution:

Radius of circle = 10.5 cm

Let x cm be the major arc, then x/5 cm be the length of minor arc.

Circumference of circle = x + x/5 = 6x/5 cm

We know, Circumference of circle = 2π r = 2 x 22/7 x 10.5

This implies,

6x/5 = 2 x 22/7 x 10.5

x = 55 cm

Area of major sector = 1/2 x 55 x 10.5 = 288.75

The area of major sector is 288.75 cm2.

Question 20:
Solution:

The short and long hands of a clock are 4 cm and 6 cm long respectively.

In an hour the hour hand completes one rotation in 12 hours therefore in 24 hours it will complete 2 rotations.

Similarly, the minute hand completes one rotation therefore in 24 hours the minute hand will complete 24 rotations.

Distance travelled by its tip in 2 days = 4(circumference of the circle with radius 4 cm)

= (4 × 2π × 4) cm = 32π cm

(As, in 2 days, the short hand will complete 4 rounds)

In 2 days, the long hand will complete 48 rounds length moved by its tip = 48(circumference of the circle with radius 6 cm)

= (48 x 2π x 6) cm

= 576 π cm

Now,

Sum of the lengths moved

= (32π + 576π)

= 608π cm

= (608 x 3.14) cm

= 1909.12 cm

Question 21:
Solution:

Circumference = 88 cm

2πr = 88

r = 14 cm

Area of quadrant of a circle = 1/4 π r2

= 1/4 x 22/7 x 14 x 14

= 154 cm2

Question 22:
Solution:

At radius = 16 m, cow can graze the area of plot is πr2

= 22/7 x 16 x 16

= 804.5 m2

At radius = 23 m, cow can graze the area of plot is πr2

= 22/7 x 23 x 23

= 1662.57 m2

Additional ground area = 1662.57 – 804.5 = 858 m2

Question 23:
Solution:

Length of rectangular field = L = 70 m

Breadth of rectangular field = B = 52 m

So, Area of the field = L x B

= 70×52 = 3640

Area of the field is 3640 m2

And, Area of quadrant of radius 21 m = 1/4 x 22/7 x 21 x 21

= 346.5 m2

Area available for grazing = (Area of the field) – (Area of quadrant of radius 21 m)

= 3640 – 346.5

= 3293.5 m2

Thus, 3293.5 m2 area is left ungrazed.

Question 24:
Solution:

Each angle of equilateral triangle = 60^0

Side of an equilateral triangle = 12 m

Length of the rope = 7 m Area of an equilateral triangle = √3/4 (side)2 = √3/4 x 12 x 12 = 62.35 m2

Area of sector with radius 7 m = 60/360 x 22/7 x 7 x 7 = 25.66 m2

Now,

Area which cannot be grazed by horse = Area of an equilateral triangle – Area of sector with radius 7 m

= 62.35 m2 – 25.66 m2

= 36.68 m2

Question 25:
Solution: Question 26:
Solution: Question 27:
Solution:

Side of a square = 10 cm

Diameter of the inscribed circle = 10 cm

Radius of the inscribed circle = 5 cm

Diameter of the circumscribed circle = Diagonal of the square

Radius circumscribed circle = 5√2 cm

(i) the area of the inscribed circle

= 22/7 x 5 x 5 = 78.57 cm2

(ii) the area of the circumscribed circle

= 22/7 x 5√2 x 5√2 = 157.14 cm2

Question 28:
Solution:

Diagonal of square = diameter of circle = 2r cm, where r is radius of circle

Area of circle = πr2 cm2

Area of Square = 1/2 x diagonal2

= 1/2 x 4r2 = 2r2 cm

Now, ratio of the areas of the circle and the square

= πr2/2r2

= π/2

Therefore, the required ratio is π:2.

Question 29:
Solution: Area of circle = π r2 (where r be the radius of circle)

= 22/7 x r2

Given: area of a circle = 154 cm2

22/7 x r2 = 154

r2 = 154 x 7/22

or r = 7

Radius of the circle is 7 cm

Let us consider each side of the equilateral triangle is “a” and height is h, then

r = h/3

or h = 3r = 3 x 7 = 21

Height of the triangle is 21 cm

From right triangle APC:

h2 = a2 – (a/2)2

h2 = 3a2/4

h = √3/2 (a)

21 = √3/2 x a (putting value of h)

a = 14√3 cm

Perimeter of the triangle = Sum of all the sides = 3a = 3 x 14√3

= 42 x 1.73

= 72.66 cm

Question 30:
Solution:

Radius of the wheel = r = 42 cm (given)

Circumference of wheel = Circumference of circle = 2πr = 2 x 22/7 x 42

= 264

Circumference of wheel is 264 cm.

Again, distance travelled by wheel = 19.8 km = 1980000 cm

(Convert km into cm)

Now, number of revolutions taken by a wheel = 1980000/264 = 7500

## Exercise 16B Page No: 722

Question 1:
Solution:

Circumference of a circle – Radius = 37cm (given)

We know that, Circumference of a circle = 2 π r (where r = radius)

2 π r – r = 37

2 x 22/7 x r – r = 37

(37/7)r = 37

r = 7 cm

Therefore, circumference of a circle = 2 π r = 2 x 22/7 x 7 = 44 cm

Question 2:
Solution:

Circumference of a circle = 22 cm

That is, 2 π r = 22

2 x 22/7 x r = 22

r = 7/2 cm

Area of quadrant of circle = ¼ π r2 = ¼ x 22/7 x 7/2 x 7/2 = 77/8

Area of quadrant of circle is 77/8 cm2

Question 3:
Solution:

Area of circle = Area of circle of diameter 10 cm + Area of circle of diameter 24 cm

or Area of circle = Area of circle of radius 5 cm + Area of circle of radius 12 cm

πr2 = π(5)2 + π(12)2

πr2 = 25 π + 144π = 169π

r = 13 So, required diameter is 26 cm.

Question 4:
Solution:

Area of circle = 2 x circumference of circle

πr2 = 2 x 2πr

r = 4

Diameter of circle = 2r = 8 cm

Question 5:
Solution:

Given, square circumscribes a circle of radius a cm.

Side of the square = 2 x radius of circle = 2a cm

Now, perimeter of the square = (4 x 2a) = 8a cm

Perimeter of the square is 8a cm.

Question 6:
Solution:

Diameter of circle = 42 cm

Radius = 42/2 = 21 cm

Central angle = 600

We know that,

Length of the arc = θ/360 (2πr)

= 60/360 x 2 x 22/7 x 21

= 22

Length of the arc is 22 cm

Question 7:
Solution:

Area of circle = Area of circle of radius 4 cm + Area of circle of radius 3 cm

Area of circle = π(4)2 + π(3)2

π r2 = 16π + 9π

π r2 = 25π

r = 5

Radius of circle = 5 cm

Diameter of circle = 2r = 10 cm

Question 8:
Solution:

Circumference of circle = 8π

2πr = 8π

r = 4

Area of circle =πr2 = π(4)2 = 16 π

Question 9:
Solution:

Diameter of the semicircular protractor = 14 cm

Radius = 14/2 cm = 7 cm

Perimeter of semicircle = πr + d

Perimeter of semicircular protractor = 22/7 x7 + 14 = 22 + 14 = 36 cm

The perimeter of the semicircular protractor is 36 cm.

Question 10:
Solution:

Perimeter of circle = Area of circle (given)

2πr = πr2

(where r = radius of circle)

r = 2

The radius of a circle is 2 cm

Question 11:
Solution:

Circumference of Circle = Circumference of circle with radius 19 cm + Circumference of circle with radius 9 cm

2πr = 2π(19) + 2π(9)

2πr = 38π + 18 π

r = 28

Radius of the circle is 28 cm.

Question 12:
Solution:

Area of Circle = Area of circle with radius 8 cm + Area of circle with radius 6 cm

Πr2 = π(8)2 + π(6)2

Πr2 = 64π + 36 π

r2 = 100

or r = 10

Radius of the circle is 10 cm.

Question 13:
Solution:

Radius = r = 6 cm and θ = 300

Area of sector = θ/360 (πr2)

= 30/360 x 3.14 x (6)2

= 9.42 cm2

Question 14:
Solution:

Radius = r = 21 cm and θ = 600

Length of the arc = θ/360 (2πr)

= 60/360 x 2 x 22/7 x 21

= 22 cm

Question 15:
Solution:

Ratio of circumferences of two circles = 2:3 (given)

Let the two circles be C1 and C2 with radii r1 and r2.

Circumference of circle = 2πr

Circumference of C1 = 2πr1 and

Circumference of C2 = 2πr2

(Circumference of C1) / (Circumference of C2) = (2πr1)/(2πr2)

2/3 = r1/r2

Now,

(area of C1) / (area of C2) = (πr12)/(πr22)

= {(r1)/(r2)}2

= (2/3)2

=4/9

Therefore, the required ratio is 4:9.

Question 16:
Solution:

Ratio of areas of two circles = 4:9 (given)

Let the two circles be C1 and C2 with radii r1 and r2.

Area of circle = πr2

Area of C1 = πr12 and

Area of C2 = πr22

(Area of C1) / (Area of C2) = (πr12)/(πr22)

4/9 = r12/r22

or (r1)/(r2) = 2/3

Now,

(Circumference of C1) / (Circumference of C2) = (2πr1)/(2πr2)

= (r1)/(r2)

= (2/3)

= 2/3

Therefore, the required ratio is 2:3.

Question 17:
Solution:

A square is inscribed in a circle (given)

Let r be the radius of circle and ‘x’ be the side of the square.

So, length of the diagonal = 2r

Length of side of square = x = diagonal/√2 = 2r/√2 = √2r

Area of square = (side)2 = ( x)2 = √2r × √2r = 2r2

Area of circle = πr2

Ratio of areas of circle and square = (area of the circle)/(area of square)

= πr2 / 2r2

= π/2

Hence, the ratio of areas of circle and square is π:2.

Question 18:
Solution:

Circumference of a circle = 8 cm

Central angle = 720

Now, Circumference of a circle = 2πr

8 = 2πr

r = 14/11 cm

Area of sector = θ/360 x (πr2)

= 72/360 x 22/7 x 14/11 x 14/11

= 1.02 cm2

Question 19:
Solution:

A pendulum swings through an angle of 30^0 and describes an arc 8.8 cm in length.

Length of the pendulum = Radius of sector of the circle

Arc length = 8.8

θ/360 (2πr) = 8.8

30/360 x 2 x 22/7 x r = 8.8

r = 16.8

Therefore, the length of the pendulum is 16.8 cm.

Question 20:
Solution:

The minute hand of a clock is 15 cm long

Angle described by the minute hand in 60 minutes = 360°

Angle described by minute hand in 20 minutes = 360/60 x 20 = 1200

So, area swept by it in 20 minutes = Area of the sector having central angle 120o and radius 15 cm

= θ/360 (πr2)

= 120/360 x 22/7 x 15 x 15

= 235.5

Therefore, the area swept by minute hand in 20 minutes is 235.5 cm2.

Question 21:
Solution:

Area of the sector = 17.6 cm2 (given)

We know, Area of the sector = θ/360 (πr2) square units

This implies,

θ/360 (πr2) = 17.6

56/360 x 2/7 x r2 = 17.6

r2 = 36

or r = 6

Radius of the circle is 6 cm.

Question 22:
Solution:

Area of the sector of a circle = 69.3 cm2

Area of the sector = θ/360 (πr2)

This implies,

θ/360 (πr2) = 69.3

θ/360 x 22/7 x 10.5 x 10.5 = 69.3

θ = 72

Therefore, central angle of the sector is 72 degrees.

Question 23:
Solution:

Perimeter of a sector of circle = 31 cm

Arc length = 31 – (6.5 + 6.5) = 18 cm

Now, Area of sector = 1/2 x Arc length x radius

= 1/2 x 18 x 6.5

= 58.5 cm2

Question 24:
Solution:

Radius of a circle = 17.5 cm

Length of arc of circle = 44 cm

Now,

Area of Sector = 1/2 x Arc length x radius

= 1/2 x 44 x 17.5

= 385

Area of Sector is 385 cm2

Question 25:
Solution:

Length of the rectangular cardboard = 14 cm and

Breadth of the rectangular cardboard = 7 cm

Area of cardboard = Area of rectangle = length × breadth = 14×7 = 98 cm2

Let the two circles with equal radii and maximum area have a radius r each.

2r = 7 or r = 7/2 cm

Again,

Area of two circular cut outs = 2 × πr2 = 2 x 22/7 x (7/2)2 = 77 cm2

Now,

The area of remaining cardboard = Area of cardboard – Area of two circular cut outs

= 98 – 77

= 21

Therefore, area of remaining cardboard is 21 cm2.

Question 26:
Solution:

Side of square = 4 cm

Radius of circle = 1 cm

Area of square = (side)2 = 4 x 4 = 16 cm2

Area of four quadrants of circle = 4 (1/4 x 3.14 x 1 x 1) = 3.14 cm2

Area of circle with diameter 2 cm = πr2 = 3.14 x 1 x 1 = 3.14 cm2

Now,

Area of the shaded region = Area of square – (Area of four quadrants of circle + Area of circle with diameter 2 cm)

= 16 – (3.14 + 3.14)

= 9.72

Therefore, the Area of the shaded region is 9.72 cm2.

Question 27:
Solution:

Length of rectangular sheet of paper = 40 cm

Breadth of rectangular sheet of paper = 28 cm

Radius of the semicircular cut out = 14 cm Area of rectangular sheet of paper = Area of rectangle = length × breadth = 40 × 28 = 1120 cm2

Area of semicircular cut out = 1/2 πr2

= 1/2 x 22/7 x 14 x 14

= 308 cm2

Now,

Area of remaining sheet of paper = Area of rectangular sheet of paper – Area of semicircular cut out

= 1120 – 308

= 812

Therefore, area of remaining sheet of paper is 812 cm2.

## R S Aggarwal Solutions for Class 10 Maths Chapter 16 Area of Circle, Sector and Segment Topics

In this chapter students will study important concepts on quadratic equations as listed below:

• Area of Circle, Sector and Segment introduction
• Circle and its parts
• Problems on the sector and segment parts
• Areas of combinations of plane figures
• Real-life Problems

### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 16 Area of Circle, Sector and Segment

1. R S Aggarwal Solutions assist students in their preparations, provide detailed answers to all questions.

2. Easy and simple language is used.

3. Step by step problem-solving approach is used to help students clear their concepts on Area of Circle, Sector and Segment.

4. Easy for quick revision.

5. This study material prepared is based on the latest CBSE syllabus by Maths subject experts.