# NCERT Solutions for Class 10 Maths Exercise 12.2 Chapter 12 Areas Related To Circles

NCERT Solutions involve detailed solutions of all exercise questions. Get free NCERT Solutions for Maths Chapter 12, Exercise 12.2 crafted by subject expert as per NCERT guidelines. Class 10 Maths Chapter 12 Area Related to Circles Exercise 12.2 Questions and answers helps students to clear their concepts and get good marks in exams. Exercise-wise, thorough solutions to the questions of the NCERT textbooks are provided with the aim of helping students compare their answers. NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles are prepared by experts to provide accurate and easy solutions for all questions covered in Exercise 12.2.

### Access Other Exercise Solutions of Class 10 Maths Chapter 12 Areas Related to Circles

Exercise 12.1 Solutions : 5 Solved Questions
Exercise 12.3 Solutions : 16 Solved Questions

### Access Answers to NCERT Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.2

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

Solution:

It is given that the angle of the sector is 60Â°

We know that the area of sector = (Î¸/360Â°)Ã—Ï€r2

âˆ´ Area of the sector with angle 60Â° = (60Â°/360Â°)Ã—Ï€r2 cm2

= (36/6)Ï€ cm2

= 6Ã—22/7 cm2 = 132/7 cm2

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

Solution:

Circumference of the circle, C = 22 cm (given)

It should be noted that a quadrant of a circle is a sector which is making an angle of 90Â°.

Let the radius of the circle = r

As C = 2Ï€r = 22,

R = 22/2Ï€ cm = 7/2 cm

âˆ´ Area of the quadrant = (Î¸/360Â°) Ã— Ï€r2

Here, Î¸ = 90Â°

So, A = (90Â°/360Â°) Ã— Ï€ r2 cm2

= (49/16) Ï€ cm2

= 77/8 cm2 = 9.6 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.

Solution:

Length of minute hand = radius of the clock (circle)

âˆ´ Radius (r) of the circle = 14 cm (given)

Angle swept by minute hand in 60 minutes = 360Â°

So, the angle swept by the minute hand in 5 minutes = 360Â° Ã— 5/60 = 30Â°

We know,

Area of a sector = (Î¸/360Â°) Ã— Ï€r2

Now, area of the sector making an angle of 30Â° = (30Â°/360Â°) Ã— Ï€r2 cm2

= (1/12) Ã— Ï€142

= (49/3)Ã—(22/7) cm2

= 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 Ï€ = 3.14)

Solution:

Here AB be the chord which is subtending an angle 90Â° at the center O.

It is given that the radius (r) of the circle = 10 cm

(i) Area of minor sector = (90/360Â°)Ã—Ï€r2

= (Â¼)Ã—(22/7)Ã—102

Or, Area of minor sector = 78.5 cm2

Also, area of Î”AOB = Â½Ã—OBÃ—OA

Here, OB and OA are the radii of the circle i.e. = 10 cm

So, area of Î”AOB = Â½Ã—10Ã—10

= 50 cm2

Now, area of minor segment = area of minor sector â€“ area of Î”AOB

= 78.5 â€“ 50

= 28.5 cm2

(ii) Area of major sector = Area of circle â€“ Area of minor sector

= (3.14Ã—102)-78.5

= 235.5 cm2

5. In a circle of radius 21 cm, an arc subtends an angle of 60Â° 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:

Given,

Î¸ = 60Â°

(i) Length of an arc = Î¸/360Â°Ã—Circumference(2Ï€r)

âˆ´ Length of an arc AB = (60Â°/360Â°)Ã—2Ã—(22/7)Ã—21

= (1/6)Ã—2Ã—(22/7)Ã—21

Or Arc AB Length = 22cm

(ii) It is given that the angle subtend by the arc = 60Â°

So, area of the sector making an angle of 60Â° = (60Â°/360Â°)Ã—Ï€ r2 cm2

= 441/6Ã—22/7 cm2

Or, the area of the sector formed by the arc APB is 231 cm2

(iii) Area of segment APB = Area of sector OAPB â€“ Area of Î”OAB

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60Â°, Î”OAB is an equilateral triangle. So, its area will be âˆš3/4Ã—a2 sq. Units.

Area of segment APB = 231-(âˆš3/4)Ã—(OA)2

= 231-(âˆš3/4)Ã—212

Or, Area of segment APB = [231-(441Ã—âˆš3)/4] cm2

6. A chord of a circle of radius 15 cm subtends an angle of 60Â° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use Ï€ = 3.14 and âˆš3 = 1.73)

Solution:

Given,

Î¸ = 60Â°

So,

Area of sector OAPB = (60Â°/360Â°)Ã—Ï€r2 cm2

= 225/6 Ï€cm2

Now, Î”AOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60Â°

So, Area of Î”AOB = (âˆš3/4) Ã—a2

Or, (âˆš3/4) Ã—152

âˆ´ Area of Î”AOB = 97.31 cm2

Now, area of minor segment APB = Area of OAPB â€“ Area of Î”AOB

Or, area of minor segment APB = ((225/6)Ï€ â€“ 97.31) cm2 = 20.43 cm2

And,

Area of major segment = Area of circle â€“ Area of segment APB

Or, area of major segment = (Ï€Ã—152) â€“ 20.4 = 686.06 cm2

7. A chord of a circle of radius 12 cm subtends an angle of 120Â° at the centre. Find the area of the corresponding segment of the circle. (Use Ï€ = 3.14 and âˆš3 = 1.73)

Solution:

Now, draw a perpendicular OD on chord AB and it will bisect chord AB.

Now, the area of the minor sector = (Î¸/360Â°)Ã—Ï€r2

= (120/360)Ã—(22/7)Ã—122

= 150.72 cm2

Consider the Î”AOB,

âˆ  OAB = 180Â°-(90Â°+60Â°) = 30Â°

We know OD bisects AB. So,

AB = 2Ã—AD = 12âˆš3 cm

Now, sin 30Â° = OD/OA

Or, Â½ = OD/12

âˆ´ OD = 6 cm

So, the area of Î”AOB = Â½ Ã— base Ã— height

Here, base = AB = 12âˆš3 and

Height = OD = 6

So, area of Î”AOB = Â½Ã—12âˆš3Ã—6 = 36âˆš3 cm = 62.28 cm2

âˆ´ Area of the corresponding Minor segment = Area of the Minor sector â€“ Area of Î”AOB

= 150.72 cm2â€“ 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 (see Fig. 12.11). 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 Ï€ = 3.14)

Solution:

As the horse is tied at one end of a square field, it will graze only a quarter (i.e. sector with Î¸ = 90Â°) of the field with radius 5 m.

Here, the length of rope will be the radius of the circle i.e. r = 5 m

It is also known that the side of square field = 15 m

(i) Area of circle = Ï€r2 = 22/7 Ã— 52 = 78.5 m2

Now, the area of the part of the field where the horse can graze = Â¼ (the area of the circle) = 78.5/4 = 19.625 m2

(ii) If the rope is increased to 10 m,

Area of circle will be = Ï€r2 =22/7Ã—102 = 314 m2

Now, the area of the part of the field where the horse can graze = Â¼ (the area of the circle)

= 314/4 = 78.5 m2

âˆ´ Increase in the grazing area = 78.5 m2 â€“ 19.625 m2 = 58.875 m2

9. 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 Fig. 12.12. Find:

(i) the total length of the silver wire required.

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

Solution:

Diameter (D) = 35 mm

Total number of diameters to be considered= 5

Now, the total length of 5 diameters that would be required = 35Ã—5 = 175

Circumference of the circle = 2Ï€r

Or, C = Ï€D = 22/7Ã—35 = 110

Area of the circle = Ï€r2

Or, A = (22/7)Ã—(35/2)2 = 1925/2 mm2

(i) Total length of silver wire required = Circumference of the circle + Length of 5 diameter

= 110+175 = 285 mm

(ii) Total Number of sectors in the brooch = 10

So, the area of each sector = total area of the circle/number of sectors

âˆ´ Area of each sector = (1925/2)Ã—1/10 = 385/4 mm2

10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

The radius (r) of the umbrella when flat = 45 cm

So, the area of the circle (A) = Ï€r2 = (22/7)Ã—(45)2 =6364.29 cm2

Total number of ribs (n) = 8

âˆ´ The area between the two consecutive ribs of the umbrella = A/n

6364.29/8 cm2

Or, The area between the two consecutive ribs of the umbrella = 795.5 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 115Â°. Find the total area cleaned at each sweep of the blades.

Solution:

Given,

Sector angle (Î¸) = 115Â°

The total area of the sector made by wiper = 2Ã—(Î¸/360Â°)Ã—Ï€ r2

= 2Ã—(115/360)Ã—(22/7)Ã—252

= 2Ã—158125/252 cm2

= 158125/126 = 1254.96 cm2

12. To warn ships for underwater rocks, a lighthouse spreads a red colored light over a sector of angle 80Â° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

(Use Ï€ = 3.14)

Solution:

Let O bet the position of Lighthouse.

Given, radius (r) = 16.5 km

Sector angle (Î¸) = 80Â°

Now, the total area of the sea over which the ships are warned = Area made by the sector

Or, Area of sector = (Î¸/360Â°)Ã—Ï€r2

= (80Â°/360Â°)Ã—Ï€r2 km2

= 189.97 km2

13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of â‚¹ 0.35 per cm2 . (Use âˆš3 = 1.7)

Solution:

Total number of equal designs = 6

AOB= 360Â°/6 = 60Â°

Radius of the cover = 28 cm

Cost of making design = â‚¹ 0.35 per cm2

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60Â°, Î”AOB is an equilateral triangle. So, its area will be (âˆš3/4)Ã—a2 sq. units

Here, a = OA

âˆ´ Area of equilateral Î”AOB = (âˆš3/4)Ã—282 = 333.2 cm2

Area of sector ACB = (60Â°/360Â°)Ã—Ï€r2 cm2

= 410.66 cm2

So, area of a single design = area of sector ACB â€“ area of Î”AOB

= 410.66 cm2 â€“ 333.2 cm2 = 77.46 cm2

âˆ´ Area of 6 designs = 6Ã—77.46 cm2 = 464.76 cm2

So, total cost of making design = 464.76 cm2 Ã—Rs.0.35 per cm2

= Rs. 162.66

14. Tick the correct solution in the following:

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

(A) p/180 Ã— 2Ï€R

(B) p/180 Ã— Ï€ R2

(C) p/360 Ã— 2Ï€R

(D) p/720 Ã— 2Ï€R2

Solution:

The area of a sector = (Î¸/360Â°)Ã—Ï€r2

Given, Î¸ = p

So, area of sector = p/360Ã—Ï€R2

Multiplying and dividing by 2 simultaneously,

= (p/360)Ã—2/2Ã—Ï€R2

= (2p/720)Ã—2Ï€R2

So, option (D) is correct.

This exercise mainly deals with the Areas of Sector and Segment of a Circle. In this exercise, students shall deal with problems related to the major sector, minor sector and circle segments. Exercise answers have been structured in a logical and easy language for quick revisions during examination or tests.