NCERT Solutions Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.2

NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.2 will assist the students in enhancing their knowledge about the concepts beyond the perimeter and area of basic circles. The exercise is largely based on explaining and applying the sector and segment concepts of a circle. Here, students will learn to compute the areas of sectors and segments, which is a core area of two-dimensional geometry. The skills acquired in this part of chapter 11 will not only assist students in answering questions during exams but also in real-world applications. Therefore, let's proceed to learn about this key topic. 

1.0Download NCERT Solutions Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.2 : Free PDF

2.0Introduction to Sectors and Segments of a Circle:

Before jumping to the formulas, let's discuss what sectors and segments of a circle are. A circle in geometry is cut into various regions. Among these regions, the two most significant ones are the sector and the segment. These can be defined as: 

Sector:

A sector of a circle is the region between two radii and the corresponding arc. It is like a slice of a circle. Sectors can be classified according to the angle at the centre into:

• Major Sector: A major sector is created when the central angle between two radii of a circle is more than 180°.
• Minor Sector: A minor sector is smaller in size, i.e., the central angle subtended between two radii is less than 180°.

Segment:

A segment of a circle refers to the area enclosed by a chord and the corresponding arc of the circle. Just like in sectors, the circle is divided into two segments by the chord:

1. Major Segment: It is the major segment of the circle created by a chord, having an angle subtending greater than 180°.
2. Minor Segment: The smaller segment of the circle is created by a chord, with the subtended angle less than 180°.

3.0Exercise 11.2 - Overview: Key Concepts

In exercise 11.2, students will be answering different questions regarding the area of a sector and segments. So, let's discuss the formulas and important concepts for this crucial topic of chapter 12:

The Area of a Sector:

The area of a sector is dependent on the central angle (θ) and the radius of the circle (r). The formula derived to calculate the area of a sector is determined by the fraction of the circle occupied by the sector. It is given by:

$\text{Area of a Sector}=\frac{\theta }{360}\times \pi r^2$

The formula mentioned above assists in finding any problem concerning the sector (major or minor) of a circle if the radius and central angle are known for the sector.

Length of an Arc:

The length of a sector's arc is the length of the curved boundary's circumference associated with it. This is also a valuable concept of practice for solving all kinds of questions. The length of an arc formula can be expressed as:

$\text{Length of an Arc}=\frac{\theta }{360}\times 2\pi r$

The Area of a Segment:

A circle segment is a region bordered by the chord and its related arc. In order to find the area of a segment, the area of the triangle traced by the radii and the chord must be deducted from the sector area. The formula for it is:

The area of a Segment= Area of the sector - Area of triangle

Here, the area of the triangle can be determined using various known formulas for its area. For instance, if the length of all the sides is known, then the area of a triangle can be determined using Heron's formula.

To improve your understanding and performance in this chapter, start solving the NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.2. It will provide you with a solid foundation for both your exams and advanced geometry studies.

NCERT Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.2 : Detailed Solutions

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

Solutions:

Radius, r = 6 cm; sector angle, θ = 60 degrees.

Area of the sector = (θ/360) × πr²

= (60/360) × (22/7) × (6)² cm²

= (1/6) × (22/7) × 36 cm²

= (132/7) cm²

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

Solutions:

Let the radius of the circle = r.

Therefore, 2πr = 22.

=> 2 × (22/7) × r = 22.

=> r = 22 × (7/22) × (1/2) = 7/2 cm.

Here, θ = 90°.

Therefore, Area of the (1/4)th part of the circle or area of quadrant of the circle = (θ/360) × πr²

= (90/360) × (22/7) × (7/2)² cm²

= (1/4) × (22/7) × (7/2) × (7/2) cm²

= (77/8) cm²

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

Solutions:

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

In 5 minutes, the minute hand will rotate at an angle:

= (360° / 60) × 5 = 30°

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 = (θ / 360°) × πr²

Area of sector of 30° = (30° / 360°) × (22 / 7) × 14 × 14

= (22 / 12) × 2 × 14 = (11 × 14) / 3 = 154 / 3 cm²

Therefore, the area swept by the minute hand in 5 minutes is 154 / 3 cm².

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)

Solutions:

Here, the radius of the circle is r = 10 cm. The sector angle of the minor sector made corresponding to the chord AB is 90°.

Now, the area of the minor sector is:

(90/360) × πr²

= (1/4) × π × (10)² cm²

= (1/4) × 3.14 × 100 cm²

= 314/4 cm²

= 78.5 cm²

Then, the area of the minor segment is:

The area of the minor sector - The area of triangle OAB

= 78.5 cm² - (1/2) × OA × OB (since angle AOB = 90°)

= 78.5 cm² - (1/2) × 10 × 10 cm²

= (78.5 - 50) cm²

= 28.5 cm²

The area of the major sector is:

((360 - 90)/360) × πr²

= (270/360) × 3.14 × (10)² cm²

= (3/4) × 314 cm²

= (3 × 157)/2 cm²

= 235.5 cm²

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

(i) The length of the arc

(ii) Area of the sector formed by the arc

(iii) Area of the segment formed by the corresponding chord

Solution:

Here, radius = 21 cm and θ = 60°

(i) Circumference of the circle = 2πr

= 2 × (22/7) × 21 cm

= 2 × 22 × 3 cm

= 132 cm

Length of arc APB = (θ/360°) × 2πr = (60°/360°) × 132 cm

= (1/6) × 132 cm = 22 cm

(ii) Area of the sector with sector angle 60° = (60°/360°) × πr²

= (60°/360°) × (22/7) × 21 × 21 cm²

= 11 × 21 cm² = 231 cm²

(iii) Area of the segment APB = [Area of the sector AOB] - [Area of ΔAOB]

In ΔAOB, OA = OB = 21 cm

Therefore, ∠A = ∠B = 60° [∵ ∠O = 60°]

This means AOB is an equilateral triangle.

Therefore, AB = 21 cm

Area of ΔAOB = (√3 / 4) × (side)²

= (√3 / 4) × 21 × 21 cm² = (441√3 / 4) cm²

From (1) and (2), we have:

Area of segment = [231 cm²] - [(441√3 / 4) cm²]

= (231 - (441√3 / 4)) cm²

= 40.047 cm²

6. 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 √3 = 1.73)

Solution:

Here, radius (r) = 15 cm and sector angle (θ) = 60°

Area of the sector = (θ/360°) × πr²

= (60°/360°) × (314/100) × 15 × 15 cm²

= (11775/100) cm² = 117.75 cm²

Since ∠O = 60° and OA = OB = 15 cm, AOB is an equilateral triangle.

=> AB = 15 cm and ∠A = 60°

Draw OM ⊥ AB, in ΔAMO

∴ OM/OA = sin 60° = √3/2

=> OM = OA × √3/2 = (15√3)/2 cm

Now, ar(ΔAOB) = (1/2) × AB × OM

= (1/2) × 15 × (15√3)/2 cm² = (225√3)/4 cm²

= (225 × 1.73)/4 cm² = 97.3125 cm²

Now area of the minor segment

= (Area of minor sector) - (ar ΔAOB)

= (117.75 - 97.3125) cm² = 20.4375 cm²

Area of the major segment

= (Area of the circle) - (Area of the minor segment)

= πr² - 20.4375 cm²

= [(314/100) × 15²] - 20.4375 cm²

= 706.5 - 20.4375 cm² = 686.0625 cm²

7. 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 √3 = 1.73)

Solution:

Here θ = 120° and r = 12 cm

∴ Area of the sector = (θ/360°) × πr²

= (120/360) × (314/100) × 12 × 12 cm²

= (314 × 4 × 12)/100 cm² = 15072/100 cm²

= 150.72 cm²

Now, area of triangle AOB = (1/2) × AB × OM 

[Since OM is perpendicular to AB]

In triangle OAB, angle O = 120 degrees.

Therefore, angle A + angle B = 180 degrees - 120 degrees = 60 degrees.

Because OB = OA = 12 cm,

angle A = angle B = 30 degrees.

So, OM/OA = sin 30 degrees = 1/2.

This implies OM = OA × (1/2).

Therefore, OM = 12 × (1/2) = 6 cm.

And AM/OA = cos 30 degrees = √3/2.

This implies AM = (√3/2) × OA = (√3/2) × 12 = 6√3 cm.

Therefore, AB = 2(AM) = 12√3 cm.

Now, from (ii),

Area of triangle AOB = (1/2) × AB × OM

= (1/2) × 12√3 × 6 cm² = 36√3 cm²

= 36 × 1.73 cm² = 62.28 cm².

From (i) and (iii),

Area of the minor segment = (Area of sector) - (Area of triangle AOB)

= (150.72 cm²) - (62.28 cm²) = 88.44 cm².

8. 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. Find

(i) The area of that part of the field in which the horse can graze.

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

Solution:

Here, r = 5 m, θ = 90°

(i) The required area (Grazing area for horse) = The area of the sector OAB

= (90/360) × πr² = (1/4) × 3.14 × (5)² m²

= (1/4) × 78.50 m² = 19.625 m²

(ii) Now, the radius for the sector OCD = 10 m and sector angle = 90°.

The area of the sector OCD

= (90/360) × π × (10)² m²

= (1/4) × 3.14 × 100 m² = 78.5 m²

Therefore, the increase of grazing area

= The area of sector OCD - The area of sector OAB = 78.5 m² - 19.625 m²

= 58.875 m²

9. A brooch is made with silver wire in the form of a circle with a diameter 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.

(ii) the area of each sector of the brooch.

Solution:

Diameter of the circle = 35 mm

Therefore, Radius (r) = 35/2 mm

(i) Circumference = 2πr

= 2 × (22/7) × (35/2) mm = 22 × 5 = 110 mm

Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors = 35 mm

Therefore, Length of 5 pieces = 5 × 35 = 175 mm

Therefore, Total length of the silver wire = 110 + 175 mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,

Sector angle θ = 360°/10 = 36°

Area of each sector = (θ/360°) × πr²

= (36°/360°) × (22/7) × (35/2) × (35/2) mm²

= (11 × 35) / 4 mm²

= 385/4 mm²

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

Solution:

Here, radius (r) = 45 cm.

Since the circle is divided into 8 equal parts,

Sector angle corresponding to each part θ = 360°/8 = 45°.

Therefore, Area of a sector (part) = (θ/360°) × πr² = (45°/360°) × (22/7) × 45 × 45 cm²

= (11 × 45 × 45) / (4 × 7) cm² = 22275/28 cm².

Therefore, the required area between the two ribs = 22275/28 cm².

11. A car has two wipers which 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.

Solution:

Here, one blade of a wiper sweeps a sector area of a circle of radius 25 cm.

Sector angle = 115°.

i.e., r = 25 cm and θ = 115°.

The area covered by one blade = (115/360) × π × (25)² cm².

Then, the area covered by two blades = 2 × (115/360) × (22/7) × 625 cm²

= (23/18) × (11/7) × 625 cm² = 158125/126 cm².

12. To warn ships of 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 are warned. (Use π = 3.14)

Solution:

Here, Radius (r) = 16.5 km and Sector angle (θ) = 80°.

Therefore, Area of the sea surface over which the ships are warned = (θ/360°) × πr²

= (80°/360°) × (314/100) × (165/10) × (165/10) km²

= (157 × 11 × 11) / 100 km² = 18997/100 km² = 189.97 km².

13. A round table cover has six equal designs as shown in the 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 √3 = 1.7)

Solution:

Here, r = 28 cm. θ = 360° / 6 = 60°

In the figure, triangle OAB is equilateral, having side 28 cm.

The area of one shaded designed portion = The area of the sector OAB - The area of the triangle OAB

= { (60 / 360) × π × (28)² - (√3 / 4) × (28)² } cm²

= { (1 / 6) × (22 / 7) × 28 × 28 - (1.7 / 4) × 28 × 28 } cm²

= { (11 / 3) × 112 - 1.7 × 196 } cm²

= { (1232 / 3) - 333.2 } cm²

The total area of six designed portions = 6 × { (1232 / 3) - 333.2 } cm²

= 2464 - 1999.2 cm² = 464.8 cm²

The total cost of making the designs at the rate of ₹ 0.35 per cm² = ₹ 0.35 × 464.8 = ₹ 162.68.

14. Tick the correct answer in the following: Area of a sector of angle p (in degree) of a circle with radius R is.

(A) (p / 180) × 2πR

(B) (p / 180) × πR²

(C) (p / 360) × 2πR

(D) (p / 720) × 2πR²

Solution:

(D) Here, radius (r) = R

Angle of sector (θ) = p°

Therefore, Area of the sector = (θ / 360) × πr² = (p / 360°) × πR²

= (2 / 2) × { (p / 360°) × πr² } = (p / 720°) × 2πR²

4.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.2

• Strengthens the use of formulas like area of a sector, arc length, and area of a segment.
• Covers frequently asked questions in board exams for better scoring.
• Important for NTSE, Olympiads, and other entrance exams with topics on mensuration.
• Summarized solutions with step-by-step explanations make last-minute revisions easier.