Class 10 - Chapter 11: Areas Related to Circles

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°.

Area of sector = (θ/360°) × πr²

= (60/360) × (22/7) × 6 × 6

= (1/6) × (22/7) × 36

= 132/7 cm²

Area of the sector = 132/7 cm²

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

Circumference = 2πr

22 = 2 × (22/7) × r

r = 7/2 cm

Area of circle = πr²

= (22/7) × (7/2)²

= (22/7) × 49/4

= 38.5 cm²

Area of quadrant

= (1/4) × 38.5

= 9.625 cm²

Area of quadrant = 9.625 cm²

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

Radius = 14 cm

In 60 minutes, the minute hand sweeps 360°.

In 5 minutes, angle swept

= (360 × 5)/60

= 30°

Area swept

= (30/360) × (22/7) × 14²

= (1/12) × (22/7) × 196

= 154/3 cm²

≈ 51.33 cm²

Area swept = 154/3 cm² ≈ 51.33 cm²

Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

(i) Minor Sector

θ = 90°

Area of minor sector

= (90/360) × 3.14 × 10²

= (1/4) × 314

= 78.5 cm²

(ii) Major Sector

Area of circle

= 3.14 × 10²

= 314 cm²

Area of major sector

= 314 − 78.5

= 235.5 cm²

Minor Sector = 78.5 cm²

Major Sector = 235.5 cm²

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.

(i) Length of the Arc

Length of arc

= (60/360) × 2πr

= (1/6) × 2 × (22/7) × 21

= 22 cm

(ii) Area of the Sector

Area of sector

= (60/360) × (22/7) × 21²

= 231 cm²

(iii) Area of the Segment

Area of triangle formed by two radii

= (√3/4) × 21²

= 441√3/4 cm²

Area of segment

= Area of sector − Area of triangle

= 231 − (441√3/4)

≈ 40.04 cm²

Length of arc = 22 cm

Area of sector = 231 cm²

Area of segment ≈ 40.04 cm²

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.

Area of sector

= (60/360) × 3.14 × 15²

= 117.75 cm²

Area of triangle

= (√3/4) × 15²

= (1.73/4) × 225

= 97.31 cm²

Area of minor segment

= 117.75 − 97.31

= 20.44 cm²

Area of circle

= 3.14 × 15²

= 706.50 cm²

Area of major segment

= 706.50 − 20.44

= 686.06 cm²

Minor Segment = 20.44 cm²

Major Segment = 686.06 cm²

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.

Area of sector

= (120/360) × 3.14 × 12²

= 150.72 cm²

Area of triangle

= (1/2) × r² × sin120°

= (1/2) × 12² × (√3/2)

= 72 × (1.73/2)

= 62.28 cm²

Area of segment

= 150.72 − 62.28

= 88.44 cm²

Area of the segment = 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.

The horse can graze only inside the square.

Since the horse is tied at a corner, the grazing region is a quadrant of radius 5 m.

Area grazed

= (1/4) × πr²

= (1/4) × (22/7) × 5²

= 275/14 m²

≈ 19.64 m²

Area grazed by the horse = 275/14 m² ≈ 19.64 m²

Q8 (ii). Find the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

When rope length = 5 m:

Area grazed

= (1/4) × π × 5²

= (1/4) × 3.14 × 25

= 19.625 m²

When rope length = 10 m:

Area grazed

= (1/4) × π × 10²

= (1/4) × 3.14 × 100

= 78.5 m²

Increase in grazing area

= 78.5 − 19.625

= 58.875 m²

Increase in grazing area = 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.

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

Diameter of circle = 35 mm

Circumference of circle

= πd

= (22/7) × 35

= 110 mm

Length of 5 diameters

= 5 × 35

= 175 mm

Total length of silver wire

= 110 + 175

= 285 mm

Total length of silver wire = 285 mm

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

Radius = 35/2

= 17.5 mm

Area of circle

= πr²

= (22/7) × 17.5 × 17.5

= 962.5 mm²

The circle is divided into 10 equal sectors.

Area of each sector

= 962.5/10

= 96.25 mm²

Area of each sector = 96.25 mm²

Q10. 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 two consecutive ribs of the umbrella.

Number of sectors = 8

Angle of each sector

= 360°/8

= 45°

Radius = 45 cm

Area between two consecutive ribs

= (45/360) × (22/7) × 45²

= (1/8) × (22/7) × 2025

= 6364.29/8

= 795.54 cm²

Area between two consecutive ribs = 795.54 cm²

Q11. 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.

Length of each blade (radius)

= 25 cm

Angle swept

= 115°

Area cleaned by one wiper

= (115/360) × (22/7) × 25²

= (115/360) × (22/7) × 625

≈ 627.48 cm²

Since there are two wipers and they do not overlap,

Total area cleaned

= 2 × 627.48

≈ 1254.96 cm²

Total area cleaned at each sweep = 1254.96 cm²

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 are warned. (Use π = 3.14)

Radius of sector (r) = 16.5 km

Angle of sector (θ) = 80°

Area of sector

= (θ/360) × πr²

= (80/360) × 3.14 × (16.5)²

= (2/9) × 3.14 × 272.25

= 189.97 km²

Area of the sea warned = 189.97 km²

Q13. A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm². (Use √3 = 1.7)

The shaded figure is a regular hexagon inscribed in a circle.

Radius of circle = 28 cm

Side of regular hexagon = Radius = 28 cm

Area of the Circle

= πr²

= (22/7) × 28 × 28

= 2464 cm²

Area of the Regular Hexagon

Area of regular hexagon

= 6 × (√3/4) × a²

= 6 × (1.7/4) × 28²

= 6 × 333.2

= 1999.2 cm²

Area of Six Designs

= Area of circle − Area of hexagon

= 2464 − 1999.2

= 464.8 cm²

Cost of Making Designs

= 464.8 × 0.35

= ₹162.68

Cost of making the designs = ₹162.68

Q14. Tick the correct answer.

Area of a sector of angle p° in a circle of radius R is:

Area of sector

= (p/360) × πR²

= (p/720) × 2πR²

Therefore, the correct option is:

(D) (p/720) × 2πR²