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NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 - Areas related to circles

NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 - Areas related to circles

Edited By Komal Miglani | Updated on Apr 29, 2025 11:04 PM IST

In a circle, the smooth curve lines that join two end points is called an arc, and a line segment that joins two points on the circle is called a chord. The portion of the arc enclosed by two radii is called the sector of the circle, and the portion enclosed between a portion of the chord is called the segment. There are two types of sectors in the circle: Minor sector and Major sector. The smaller portion of a sector or a small area of arc that is enclosed by two radii is called the minor sector, and the larger area of arc that is enclosed by two radii is called the major sector. The area of the sector of angle Θ= Θ360∘×πr2 and The length of the arc is given by Length of arc = Θ360∘×2πr

This Story also Contains
  1. Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1
  2. Assess NCERT Solutions for Class 10 Maths chapter 11 exercise 11.1
  3. Topics Covered in Chapter 11 Area Related to Circles: Exercise 11.1
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 - Areas related to circles
NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 - Areas related to circles

Questions based on the area of a sector out of some given conditions like radius and angle of the sector, and are like a base for the exam. 10th class Maths exercise 11.1 answers are designed as per the students' demand, covering comprehensive, step-by-step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in-depth understanding of concepts. Students can find all exercises together using the link provided below. These NCERT solutions are created by a subject matter expert at Careers360, considering the latest syllabus and pattern of CBSE 2025-26. The Class 10 Maths chapter 11 exercise 11.1 covers topics like the concept of finding major and minor segments/sectors subtended by the chord of the circle. Also, some properties and technical terms of circular curves are enclosed with some special conditions. Students can also go through the NCERT Books according to their standards and subjects.

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Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1

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Assess NCERT Solutions for Class 10 Maths chapter 11 exercise 11.1

Q1. Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.

Answer:

We know that the area of a sector having radius r and angle Θ is given by:-

Area=Θ360×πr2

Thus, the area of the given sector is:-

= 60360×π×62

= $\frac{132}{7}cm2

Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer:

We are given the circumference of the circle.

Thus,

2πr = 22

r=11πcm

Also, we know that the area of a sector is given by :

Area=Θ360×πr2

It is given that we need to find the area of a quadrant, thus Θ = 90∘

Hence, the area becomes:-

= 90360×π(11π)2

= 778cm2

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer:

The minute hand rotates 360 o in one hour.

We need to find a rotation in 5 minutes. :-

= 360∘60×5 = 30∘

The area of the sector is given by :

Area = Θ360∘×πr2Area= 30360∘×π×142Area= 1543 cm2

Hence, the area swept by the minute hand in 5 minutes is 1543 cm2.

Q4. A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding : (i) minor segment

Answer:

The angle in the minor sector is 90 o .

Thus, the area of the sector is given by:-

Area = Θ360∘×πr2Area= 90∘360∘×π×102Area= 110014 cm2 = 78.5 cm2

Now the area of a triangle is:-

Area = 12bh = 12×10×10 = 50 cm2

Thus, the area of the minor segment = Area of the sector - Area of a triangle

or = 78.5 − 50 = 28.5 cm2

Q4. A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding : (ii) major sector. (Use π = 3.14)

Answer:

The area of the major sector can be found directly by using the formula :

Area = Θ360∘×πr2

In this case, the angle is 360 o - 90 o = 270 o .

Thus, the area is: -

= 270∘360∘×π×10×10= 330014 = 235.7 cm2

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find: (i) the length of the arc

Answer:

The length of the arc is given by:-

Length of arc = Θ360∘×2πr

= 60∘360∘×2×π×21= 22 cm

Hence, the length of the arc is 22 cm.

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find: (ii) the area of the sector formed by the arc

Answer:

We know that the area of the sector is given by:-

Area = Θ360∘ ×πr2

= 60360∘ ×π×212= 231 cm2

Thus, the area of the sector is 231 cm 2.

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find: (iii) area of the segment formed by the corresponding chord

Answer:

For the area of the segment, we need to subtract the area of the triangle attached from the area of the arc.

Thus, consider the triangle:-

It is given that the angle of arc is 60 o , or we can say that all angles are 60 o (since two sides are equal). Hence, it is an equilateral triangle.

Area of the triangle is:-

= 34×a2= 34×212= 44134 cm2

Hence, the area of the segment is:-

= (231 − 44134) cm2

Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the areas of the corresponding minor and major segments of the circle.
(Use π=3.14 and 3=1.73 )

Answer:

The area of the sector is :

= Θ360∘×πr2= 60∘360∘×π×152= 117.85 cm2

Now consider the triangle, the angle of the sector is 60 0.

This implies it is an equilateral triangle. (As two sides are equal, so will have the same angle. This is possible only when all angles are equal, i.e., 60o .)

Thus, the area of the triangle is:-

= 34×152

or = 56.253 = 97.31 cm2

Hence area of the minor segment : =\117.85 − 97.31 = 20.53 cm2

And the area of the major segment is :

= πr2 − 20.53

or = π×152 − 20.53

or = 707.14 − 20.53

or = 686.6 cm2

Q7. A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding segment of the circle.
(Use π=3.14 and 3=1.73 )

Answer:

For the area of the segment, we need the area of the sector and the area of the associated triangle.

So, the area of the sector is :

= 120∘360∘×π×122

or = 150.72 cm2

Now, consider the triangle:-

Draw a perpendicular from the centre of the circle to the base of the triangle (let it be h).

Using geometry, we can write,

hr = cos⁡60∘

or h = 6 cm

Similarly, b2r = sin⁡60∘

or b = 123 cm

Thus, the area of the triangle is :

= 12×123×6

or = 62.28 cm2

Hence the area of segment is: = 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 Fig.). Find (i) the area of that part of the field in which the horse can graze.

1636091304687

Answer:

The part grazed by the horse is given by = Area of the sector

Area = Θ360∘ ×πr2

= 90∘360∘ ×π×52

= 19.62 m2

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 Fig.). Find (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π=3.14 )

1636091352998

Answer:

When the length of the rope is 10 m, the area grazed will be:-

Area = Θ360∘ ×πr2

= 90∘360∘ ×π×102

= 25π m2

Hence, the change in the grazing area is given by :

= 25π − 25π4 = 58.85 m2

Q9. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. Find: (i) the total length of the silver wire required.

1636091383511

Answer:

The total wire required will be for 5 diameters and the circumference of the brooch.

The circumference of the brooch:-

= 2πr= 2×π×352= 110 mm

Hence the total wire required will be:- = 110 mm + 5×35 mm = 285 mm .

Q9. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find : (ii) the area of each sector of the brooch.

1636091389686

Answer:

The total number of lines present in the brooch is 10 (lines starting from the centre).

Thus, the angle of each sector is 36o.

The area of the sector is given by:-

Area = Θ360∘ ×πr2Area= 36∘360∘ ×π×(352)2Area= 3854 mm2

Q10. An umbrella has 8 ribs that are equally spaced (see Fig. ). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

1636091407785

Answer:

It is given that the umbrella has 8 ribs, so the angle of each sector is 45o.

Thus, the area of the sector is given by:-

Area = Θ360∘ ×πr2Area= 45∘360∘ ×π×452Area= 2227528 cm2

Hence, the area between two consecutive ribs is 2227528 cm2.

Q11. A car has two wipers that do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Answer:

The area cleaned by one wiper is:-

Area = Θ360∘ ×πr2

or = 115∘360∘ ×π×252

or = 158125252 cm2

Hence, the required area (area cleaned by both blades) is given by:-

= 2×158125252 = 158125126 cm2

Q12. To warn ships of underwater rocks, a lighthouse spreads a red-colored light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Answer:

The area of the sector is given by:-

Area = Θ360∘ ×πr2

In this case, the angle is 80 o .

Thus, the area is:-

= 80∘360∘ ×π×16.52

or = 189.97 Km2

Q13. A round table cover has six equal designs as shown in Fig. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm 2. (Use 3=1.7 )

Answer:

The angle of each of the six sectors is 60 o at the centre. (∵360∘6 = 60∘)

Area of the sector is given by:-

Area = Θ360∘ ×πr2

or = 60∘360∘ ×π×782

or = 410.66 cm2

And the area of the equilateral triangle associated with the segment:-

= 34×a2

or = 34×282 = 333.2 cm2

Hence the area of segment is : = 410.66 − 333.2 = 77.46 cm2

Thus the total area of design is : =6×77.46 = 464.76 cm2

So, the total cost for the design is:- = 0.35×464.76 = Rs.162.66

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) p180×2πR

(B) p180×πR2

(C) p360×2πR

(D) p720×2πR2

Answer:

We know that the area of the sector is given by:-

Area = Θ360∘ ×πr2

= p360∘ ×πr2

Hence, option (d) is correct.

Topics Covered in Chapter 11 Area Related to Circles: Exercise 11.1

  • Area of a Circle: The area of the circle is the enclosed region covered by its boundaries. The formula for calculating the area of the circle is πr2.
  • Circumference of a Circle: The circumference of the circle is the total length of the boundary of the circle, and it is calculated using the formula of 2πr.
  • Area of a Sector: The part or region of the circle that is covered by two radii and the corresponding arc is called the area of the sector, and it is calculated using the formula Θ360∘ ×πr2
  • Length of an Arc: The length of the arc on the circumference of the circle is called the length of the arc, and it is calculated using the formula Θ360∘×2πr.

Also, see-

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NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

1. What are the important points to be covered in exercise 11.1?

The exercise deals with a question related to the area of circles. These are important topics which will be important and covered in higher classes  

2. What is the formula of the area of a segment of the circle?

Area of a segment of circle = area of the corresponding sector – the area of the corresponding triangle

3. From How many points we can draw a circle, and can we trace these points in straight line and form triangle with them?

From three non-collinear points we can draw a circle, talking about triangle then yes we can make depending upon coordinates of point we  draw a triangle out of it.

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