NCERT Solutions Maths Class 10 Exercise 12.2

# NCERT Solutions Maths Class 10 Exercise 12.2

## Maths Class 10 Exercise 12.2

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

Solution: Here, radius (r) = 6 cm and  θ = 60°

= 132/7 cm2

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

Solution: It is given that, circumference (2πr) = 22 cm

⇒ 2 × 22/7 × r = 22

⇒ r = 7/2 cm

We know that for a quadrant of the circle, θ = 90°

= 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.

Solution: Here, r = 14 cm and θ = 90°/3 = 30°

Therefore, the area swept by the minute hand = = 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)

Solution: (i) Here, r = 10 cm and θ = 90°

= 78.5 cm2

Area of OAB = ½ × Base × Height

= ½ × 10 × 10

= 50 cm2

So, the area of minor segment = area of minor sector – area of OAB

= 78.5 – 50

= 28.5 cm2

(ii) For major sector, radius = 10 cm and θ = 360° 90° = 270°

= 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: Given, r = 21 cm and θ = 60°

= 22 cm

= 231 cm2

(iii) Area of segment formed by corresponding chord

Area of the sector – Area of OAB

Area of segment = 231 – Area of OAB        ………. (i)

In right-angled OMA and OMB,

OA = OB               [Radii of the same circle]

OM = OM            [Common]

OMA  OMB        [RHS congruence criterion]

AM = BM                    [By CPCT]

Therefore, M is the mid-point of AB and AOM = BOM

AOM = BOM

= ½ AOB = ½ × 60° = 30°

Therefore, in right-angled OMA,

cos 30° = OM/OA

√3/2 = OM/21

OM = 21√3/2 cm

Also, sin 30° = AM/OA

½ = AM/21

AM = 21/2 cm

Therefore, AB = 2 AM

= 2 × 21/2 cm

= 21 cm

Area of OAB = ½ × AB × OM

= ½ × 21 × 21√3/2

= 441√3/4 cm2

From equation (i), we have

Area of segment formed by corresponding chord = (231 – 441√3/4) cm2

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

Solution: Here, r = 15 cm and θ = 60°

= 117.75 cm2

To find the area of AOB, draw OM AB.

In right-angled OMA and OMB, we have

OA = OB                      [Radii of the same circle]

OM = OM                   [Common]

Therefore, ∆OMA OMB [RHS congruence criterion]

So, AM = BM             [By CPCT]

AM = BM = ½ AB and AOM = BOM = ½ AOB = ½ × 60° = 30°

In right-angled OMA, cos 30° = OM/OA

√3/2 = OM/15

OM = 15√3/2 cm

Also, sin 30° = AM/OA

⇒ ½ = AM/15

AM = 15/2 cm

2 AM = 2 × 15/2 = 15 cm

AB = 15 cm

Since, area of AOB = ½ × AB × OM

= ½ × 15 × 15√3/2 = 225√3/4

= (225 × 1.73)/4

= 97.3125 cm2

Therefore, the area of minor segment = Area of minor sector – Area of AOB

= 117.75 – 97.3125 = 20.4375 cm2

And, area of major segment = πr2 – Area of minor segment

= 3.14 × 15 ×15 – 20.4375

= 706.5 – 20.4375

= 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, r = 12 cm and θ = 120°

= 150.72 cm2

To find the area of AOB, draw OM AB.

In right-angled OMA and OMB, we have

OA = OB                      [Radii of the same circle]

OM = OM                    [Common]

OMA  OMB         [RHS congruence criterion]

Therefore, AM = BM  [By CPCT]

AM = BM = ½ AB and AOM = BOM = ½ AOB = ½ × 120° = 60°

In right-angled OMA, cos 60° = OM/OA

⇒ ½ = OM/12

OM = 6 cm

Also, sin 60° = AM/OA

√3/2 = AM/12

AM = 6√3 cm

2 AM = 2 × 6√3 = 12√3 cm

AB = 12√3 cm

Now, the area of AOB = ½ × AB × OM

= ½ × 12√3 × 6

36√3

= 36 × 1.73

= 62.28 cm2

Therefore, the area of corresponding segment = Area of corresponding sector – Area of AOB

= 150.72 – 62.28 = 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 figure). 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: (i) The horse can graze the area of a quadrant with radius 5 m.

= 19.625 m2

78.5 m2

Then, the increase in grazing area = 78.5 – 19.625

= 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 figure. Find:

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

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

Solution: (i) Diameter of the circular brooch = 35 mm

Radius of the circular brooch = 35/2 mm

Circumference of the circular brooch = 2πr = 2 × 22/7 × 35/2

= 110 mm

The length of 5 diameters = 35 × 5 = 175 mm

The total length of the silver wire required = Circumference + Length of 5 diameters

= 110 + 175 = 285 mm

(ii) To find the area of each sector, r = 35/2 mm and θ = 360°/10 = 36°

= 385/4 mm2

10. 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.

Solution: Here, r = 45 cm and θ = 360°/8 = 45°

Area between two consecutive ribs of the umbrella = Area of a sector of the circle

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: The area cleaned by the blade is in the shape of a sector.

Here, r = 25 cm and θ = 115°

The total area cleaned at each sweep of the blades

12. 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 are warned. (Use π = 3.14)

Solution: Here, r = 16.5 km and θ = 80°

The area of sea over which the ships are warned = = 189.97 km2

13. 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 √3 = 1.7)

Solution: Here, r = 28 cm and θ = 360°/6 = 60°

= 1232/3

= 410.67 cm2

To find the area of AOB, draw OM AB.

In right-angled OMA and OMB, we have

OA = OB                      [Radii of the same circle]

OM = OM                    [Common]

OMA  OMB         [RHS congruence criterion]

Therefore, AM = BM  [By CPCT]

AM = BM = ½ AB and AOM = BOM = ½ AOB = ½ × 60° = 30°

In right-angled OMA, cos 30° = OM/OA

√3/2 = OM/28

OM = 14√3 cm

Also, sin 30° = AM/OA

½ = AM/28

AM = 14 cm

2 AM = 2 × 14 = 28 cm

AB = 28 cm

Now, the area of AOB = ½ × AB × OM

= ½ × 28 × 14√3

196√3

= 196 × 1.7

= 333.2 cm2

Therefore, the area of minor segment = Area of minor sector – Area of AOB

= 410.67 – 333.2 = 77.47 cm2

So, the area of one design = 77.47 cm2

Area of six designs = 77.47 × 6 = 464.82 cm2

Cost of making the designs = 464.82 × 0.35 = Rs. 162.68

14. Tick the correct answer in the following:

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

Solution: (D) Given, r = R and θ = p