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# NCERT Solution for Class 10th Maths Chapter 12 Areas Related To Circles

This chapter is an integral part of the mathematics syllabus and deals with calculating areas related to circles, such as the area of a sector, the area of a segment, and the length of an arc.
The chapter provides students with the fundamental concepts of circles, such as the formulae for calculating the circumference and area of a circle. The chapter also covers various practical applications of circles-related areas, making it an essential topic for students to learn.
The NCERT Solutions provided here have been designed to help students understand the concepts covered in this chapter clearly and concisely. These solutions have been prepared by experienced teachers and subject matter experts, considering the latest CBSE guidelines and syllabus.
The solutions provide step-by-step explanations to all the questions in the NCERT textbook, ensuring that students understand the concepts thoroughly. The solutions also provide tips and tricks to solve problems quickly and accurately.
We hope that these NCERT Solutions will help students prepare for the Class 10 board exams and also help them build a strong foundation for higher studies in mathematics.

## Answers of Math NCERT Solution for Class 10th Maths Chapter 12 Areas Related To Circles

Unit 12

Areas Related to Circles

Exercise 12.1

Q1:

The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Radius (r1) of 1st circle = 19 cm Radius (r2) or 2nd circle = 9 cm Let the radius of 3rd circle be r.

Circumference of 1st circle = 2πr1 = 2π (19) = 38π Circumference of 2nd circle = 2πr2 = 2π (9) = 18π Circumference of 3rd circle = 2πr

Given that,

Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle 2πr = 38π + 18π = 56π

Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.

Q2:

The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Radius (r1) of 1st circle = 8 cm Radius (r2) of 2nd circle = 6 cm Let the radius of 3rd circle be r.

Area of 1st circle

Area of 2nd circle

Given that,

Area of 3rd circle = Area of 1st circle + Area of 2nd circle

However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm.

Q3:

Given figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands are 10.5 cm wide. Find the area of each of the five scoring regions.

Radius (r1) of gold region (i.e., 1st circle)

Given that each circle is 10.5 cm wider than the previous circle. Therefore, radius (r2) of 2nd circle = 10.5 + 10.5

21 cm

Radius (r3) of 3rd circle = 21 + 10.5

= 31.5 cm

Radius (r4) of 4th circle = 31.5 + 10.5

= 42 cm

Radius (r5) of 5th circle = 42 + 10.5

= 52.5 cm

Area of gold region = Area of 1st circle

Area of red region = Area of 2nd circle – Area of 1st circle

Area of blue region = Area of 3rd circle – Area of 2nd circle

Area of black region = Area of 4th circle – Area of 3rd circle

Area of white region = Area of 5th circle – Area of 4th circle

Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm2, 1039.5 cm2, 1732.5 cm2, 2425.5 cm2, and 3118.5 cm2 respectively.

Q4:

The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km? per hour?

Diameter of the wheel of the car = 80 cm Radius (r) of the wheel of the car = 40 cm Circumference of wheel = 2πr

= 2π (40) = 80π cm Speed of car = 66 km/hour

Distance travelled by the car in 10 minutes

= 110000 × 10 = 1100000 cm

Let the number of revolutions of the wheel of the car be n. n × Distance travelled in 1 revolution (i.e., circumference)

= Distance travelled in 10 minutes

Therefore, each wheel of the car will make 4375 revolutions.

Q5:

Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

1. 2 units         (B) π units         (C)     4 units         (D)    7 units

Let the radius of the circle be r.

Circumference of circle = 2πr Area of circle = πr

Given that, the circumference of the circle and the area of the circle are equal. This implies 2πr = πr2

2 = r

Therefore, the radius of the circle is 2 units. Hence, the correct answer is A.

Exercise 12.2

Q1:

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Let OACB be a sector of the circle making 60° angle at centre O of the circle.

Area of sector of angle

Area of sector

Therefore, the area of the sector of the circle making 60° at the centre of the circle is

Q2:

Find the area of a quadrant of a circle whose circumference is 22 cm.

Let the radius of the circle be r.

Circumference = 22 cm

2πr = 22

Quadrant of circle will subtend 90° angle at the centre of the circle.

Area of such quadrant of the circle

*Q3:

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.

In 5 minutes, minute hand will rotate =

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.

Area of sector of angle θ =

Area of sector of 30°

Therefore, the area swept by the minute hand in 5 minutes is

Q4:

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

1.         Minor segment
2.         Major sector [Use π = 3.14]

Let AB be the chord of the circle subtending 90° angle at centre O of the circle.

Area of major sector OADB =

Area of minor sector OACB =

Area of ΔOAB =

= 50 cm2

Area of minor segment ACB = Area of minor sector OACB – Area of ΔOAB = 78.5 – 50 = 28.5 cm2

Q5:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

1.         The length of the arc
2.     Area of the sector formed by the arc
3. Area of the segment forced by the corresponding chord

Radius (r) of circle = 21 cm

Angle subtended by the given arc = 60°

Length of an arc of a sector of angle θ =

Length of arc ACB =

= 22 cm

Area of sector OACB =

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠OAB + ∠AOB + ∠OBA = 180° 2∠OAB + 60° = 180°

∠OAB = 60°

Therefore, ΔOAB is an equilateral triangle.

Area of ΔOAB =

Area of segment ACB = Area of sector OACB – Area of ΔOAB

Q6:

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.

[Use π = 3.14 and

Radius (r) of circle = 15 cm

Area of sector OPRQ =

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180° 2 ∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.

Area of ΔOPQ =

Area of segment PRQ = Area of sector OPRQ – Area of ΔOPQ

= 117.75 – 97.3125

= 20.4375 cm2

Area of major segment PSQ = Area of circle – Area of segment PRQ

*Q7:

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

[Use π = 3.14 and

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST. SV = VT

In ΔOVS,

Area of ΔOST =

Area of sector OSUT =

Area of segment SUT = Area of sector OSUT – Area of ΔOST

= 150.72 – 62.28

= 88.44 cm2

Q8:

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see the given figure). Find

1. The area of that part of the field in which the horse can graze.
2. The increase in the grazing area of the rope were 10 m long instead of 5 m. [Use Ãâ‚¬ = 3.14]

From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius. Area that can be grazed by horse = Area of sector OACB

Area that can be grazed by the horse when length of rope is 10 m long

Increase in grazing area = (78.5 – 19.625) m2

= 58.875 m2

*Q9:

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find.

1.         The total length of the silver wire required.
2.     The area of each sector of the brooch

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Circumference of brooch = 2πr

= 110 mm

Length of wire required = 110 + 5 × 35

= 110 + 175 = 285 mm

It can be observed from the figure that each of 10 sectors of the circle is subtending 36° at the centre of the circle.

Therefore, area of each sector =

Q10:

An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending    at the centre of the assumed flat circle.

Area between two consecutive ribs of circle

Q11:

A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

It can be observed from the figure that each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.

Area of such sector

*Q12:

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships warned. [Use π = 3.14]

It can be observed from the figure that the lighthouse spreads light across a sector of 80° in a circle of 16.5 km radius.

Area of sector OACB

Q13:

A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs.0.35 per cm2. [Use ]

It can be observed that these designs are segments of the circle.

Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute at the centre of the circle.

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠AOB = 60°

∠OAB + ∠OBA + ∠AOB = 180° 2∠OAB = 180° – 60° = 120°

∠OAB = 60°

Therefore, ΔOAB is an equilateral triangle.

Area of ΔOAB =

Area of sector OAPB

Area of segment APB = Area of sector OAPB – Area of ΔOAB

Cost of making 1 cm2 designs = Rs 0.35

Cost of making 464.76 cm2 designs = Rs 162.68 Therefore, the cost of making such designs is Rs 162.68.

Q14:

Tick the correct answer in the following:

Area of a sector of angle p (in degrees) of a circle with radius R is

(A)         (B)         (C)         (D)

We know that area of sector of angle

Area of sector of angle

Hence, (D) is the correct answer.

Q15:

Tick the correct answer in the following:

Area of a sector of angle p (in degrees) of a circle with radius R is

(A)         (B)         (C)         (D)

We know that area of sector of angle

Area of sector of angle

Hence, (D) is the correct answer.

Exercise 12.3

Q1:

Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Radius of inner circle = 7 cm Radius of outer circle = 14 cm

Area of shaded region = Area of sector OAFC – Area of sector OBED

=40°360°×π(14)2 – 40°

*Q2:

Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

It can be observed from the figure that the radius of each semicircle is 7 cm.

Area of each semi-circle

Area of square ABCD = (Side)2 = (14)2 = 196 cm2 Area of the shaded region

= Area of square ABCD – Area of semi-circle APD – Area of semi-circle BPC

= 196 – 77 – 77 = 196 – 154 = 42 cm

Q3:    Find the area of the shaded region in the given figure, where a circular arc of     radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12     cm as centre.

We know that each interior angle of an equilateral triangle is of measure 60°.

Area of sector OCDE

Area of

Area of circle

Area of shaded region = Area of ΔOAB + Area of circle – Area of sector OCDE

*Q4:

From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the given figure. Find the area of the remaining portion of the

square.

Each quadrant is a sector of 90° in a circle of 1 cm radius.

Area of square = (Side)2 = (4)2 = 16 cm2 Area of circle = πr2 = π (1)2

Area of the shaded region = Area of square – Area of circle – 4 × Area of quadrant

Q5:

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design (Shaded region).

Radius (r) of circle = 32 cm AD is the median of ABC.

In ΔABD,

Area of equilateral triangle,

Area of circle = πr2

Area of design = Area of circle – Area of ΔABC

*Q6:

In the given figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded

region.

Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius.

Area of each sector

Area of square ABCD = (Side)2 = (14)2 = 196 cm

Area of shaded portion = Area of square ABCD – 4 × Area of each sector

Therefore, the area of the shaded portion is 42 cm2.

Q7:

The given figure depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

1.         The distance around the track along its inner edge
2.     The area of the track

Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA

Area of the track = (Area of GHIJ – Area of ABCD) + (Area of semi-circle HKI – Area of semi-circle BEC) + (Area of semi-circle GLJ – Area of semi-circle

AFD)

Therefore, the area of the track is 4320 m2.

*Q8:

In the given figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Radius (r1) of larger circle = 7 cm

Area of smaller circle

Area of semicircle AECFB of larger circle

Area of

= Area of smaller circle + Area of semi-circle AECFB – Area of ΔABC

*Q9:

The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (See the given figure). Find the area of shaded region. [Use π = 3.14 and

Let the side of the equilateral triangle be a.

Area of equilateral triangle = 17320.5 cm2

Each sector is of measure 60°.

Area of shaded region = Area of equilateral triangle – 3 × Area of each sector

Q10:

On a square handkerchief, nine circular designs each of radius 7 cm are made (see the given figure). Find the area of the remaining portion of the handkerchief.

From the figure, it can be observed that the side of the square is 42 cm.

Area of square = (Side)2 = (42)2 = 1764 cm

Area of each circle = πr2

Area of 9 circles = 9 × 154 = 1386 cm2

Area of the remaining portion of the handkerchief = 1764 – 1386

= 378 cm2

Q11:

In the given figure, OACB is a quadrant of circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

1. Since OACB is a quadrant, it will subtend 90° angle at O.

1. Area of ΔOBD

Area of the shaded region = Area of quadrant OACB – Area of ΔOBD

Q12:

In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [Use 3.14]

In ΔOAB,

OB2 = OA2 + AB2

= (20)2 + (20)2

Area of OABC = (Side)2 = (20)2 = 400 cm

Area of shaded region = Area of quadrant OPBQ – Area of OABC

= (628 – 400) cm

= 228 cm2

Q13:

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the given figure). If ∠AOB = 30°, find the area of the shaded region.

Area of the shaded region = Area of sector OAEB – Area of sector OCFD

Q14:

In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

As ABC is a quadrant of the circle, ∠BAC will be of measure 90 º.

In ΔABC,

BC2 = AC2 + AB

= (14)2 + (14)2

Radius (r1) of semi-circle drawn on

Area of

Area of sector

= 154 – (154

= 98 cm2

*Q15:

Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each.

The designed area is the common region between two sectors BAEC and DAFC.

Area of sector

Area of ΔBAC

Area of the designed portion = 2 × (Area of segment AEC)

= 2 × (Area of sector BAEC – Area of ΔBAC)

Q16:

Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each.

The designed area is the common region between two sectors BAEC and DAFC.

Area of sector

Area of ΔBAC

Area of the designed portion = 2 × (Area of segment AEC)

= 2 × (Area of sector BAEC – Area of ΔBAC)

=2×3527-32cm2

=2567cm2

## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

The 12th Chapter of NCERT Solutions for Class 10 Maths covers the concepts of perimeter (circumference) and area of a circle and applies this knowledge in finding the areas of two special ‘parts’ of a circular region known as sector and segment.

Areas Related to Circles is a part of Mensuration, and this Unit holds a total weightage of 10 marks in the final exam. In the final examination, one question is sometimes asked from this chapter.

List of Exercises in class 10 Maths Chapter 12

Exercise 12.1 Solutions (5 Solved Questions)

Exercise 12.2 Solutions (14 Solved Questions)

Exercise 12.3 Solutions (16 Solved Questions)

NCERT solutions for Class 10 Maths Chapter 12 is about parts of circle, their measurements and areas of plane figures. SWC’s subject experts have prepared solutions for each question adhering to the CBSE syllabus.

Area related to circles, the chapter of Class 10 consists of important topics such as;

### Key Features of NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

• NCERT Solutions help students strengthen their concepts on circle related areas.
• Questions are explained using diagrams which make learning more interactive and comprehensive.
• Easy and understandable language used in NCERT solutions.
• Step by step solving approach helps students to clear their basics.
• Helps students to solve complex problems at their own pace.

Conclusion

The NCERT Solutions for Class 10 Maths Chapter 12, Areas Related to Circles, provides a comprehensive and detailed explanation of the fundamental concepts of circles and their applications in calculating the areas related to circles. The chapter covers various topics, such as the formulae for calculating the circumference and area of a circle, the area of a sector, the area of a segment, and the length of an arc.
Experts have carefully crafted these solutions to provide step-by-step explanations of all the questions in the NCERT textbook. The solutions also offer helpful tips and tricks to help students solve problems quickly and accurately.
Using these solutions, students can strengthen their understanding of the subject, improve their problem-solving skills, and prepare effectively for the Class 10 board exams. These solutions are an excellent resource for students who wish to excel in mathematics and build a strong foundation for higher studies.
Overall, the NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles are indispensable for students looking to score well in mathematics and enhance their analytical and mathematical abilities.

### What is the importance of studying NCERT Solutions for Class 10 Maths Chapter 12?

The solutions of NCERT Class 10 Maths Chapter 12 helps students in procuring a summary of the question paper pattern, including a variety of questions, such as mostly repeating questions, concise and long answer type questions, multiple-choice questions, marks etc. The numerous you solve, the more assurance you will get towards your success.

### Mention concepts that are important from an exam perspective in NCERT Solutions for Class 10 Maths Chapter 12?

The important from an exam perspective in NCERT Solutions for Class 10 Maths Chapter 12 Introduction of area of circle, perimeter and area of a circle, areas of sector and segment of a circle, areas of combinations of plane figures and lastly explains the summary of the whole chapter.

### How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 12?

There are 3 exercises in NCERT Solutions for Class 10 Maths Chapter 12. First exercise has 5 questions, second exercise has 14 exercises and last and final exercise has 16 questions based on perimeter and area of a circle, areas of sector and segment of a circle and areas of combinations of plane figure.