AREAS RELATED TO CIRCLES

EXERCISE 11.1

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

Solution:

Formula: Area of sector = (θ / 360°) × π × r²

Here, θ = 60° and r = 6 cm.

Area = (60 / 360) × π × (6)²
= (1 / 6) × π × 36
= 6π cm²

Take π = 22/7:
Area = 6 × 22/7 = 132/7 cm²

Answer: 132/7 cm².

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

Solution:

1. Circumference of circle = 2πr
22 = 2 × (22/7) × r
22 = (44/7) × r
r = (22 × 7) / 44
r = 7/2 = 3.5 cm

2. Area of circle = πr² = (22/7) × (3.5)²
= (22/7) × (12.25)
= 38.5 cm²

3. Area of a quadrant = (1/4) × Area of circle
= (1/4) × 38.5
= 9.625 cm²

Answer: Area of the quadrant = 77/8 cm² (or 9.625 cm²).

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Q3: Find the area swept by the minute hand of a clock in 5 minutes, if its length is 14 cm.

Solution:

1. The length of the minute hand = radius of the circle = r = 14 cm.

2. In 60 minutes, the minute hand completes a full circle (360°).
Therefore, in 5 minutes, it covers:
(5/60) × 360° = 30°.

3. Area of a sector = (θ / 360°) × π × r²
= (30 / 360) × π × (14)²
= (1 / 12) × π × 196
= (196 / 12) × π
= (49 / 3) × π

4. Taking π = 22/7:
Area = (49 / 3) × (22 / 7)
= 154 / 3 cm²

Answer: The area swept by the minute hand in 5 minutes = 154/3 cm² (or 51⅓ cm²).

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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 segment
(ii) major sector (Use π = 3.14)

Solution:

Step 1. Radius of circle = 10 cm, central angle θ = 90°.

(i) Area of minor segment

Minor segment = Area of sector – Area of triangle.

• Area of sector = (θ / 360) × π × r²
  = (90 / 360) × 3.14 × (10)²
  = (1 / 4) × 3.14 × 100
  = 78.5 cm²

• Area of triangle (formed by two radii and chord, with angle 90° at the centre):
  = 1/2 × r × r × sin 90°
  = 1/2 × 10 × 10 × 1
  = 50 cm²

So, Minor segment = 78.5 – 50 = 28.5 cm²

(ii) Area of major sector

Major sector = Area of circle – Area of minor sector.

• Area of circle = π × r² = 3.14 × 100 = 314 cm²

• Major sector = 314 – 78.5 = 235.5 cm²

Answer:
(i) Minor segment = 28.5 cm²
(ii) Major sector = 235.5 cm²

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Q5: 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 central angle θ = 60°.

Common fraction factor: θ/360 = 60/360 = 1/6.

(i) Length of the arc
Length = (θ / 360) × 2πr = (1/6) × 2π × 21 = (2π × 21) / 6 = 7π cm.
Using π = 22/7, length = 7 × 22/7 = 22 cm.

Answer (i): 7π cm, which equals 22 cm (using π = 22/7).

(ii) Area of the sector
Area of sector = (θ / 360) × πr² = (1/6) × π × 21² = (1/6) × π × 441 = (441/6)π = (147/2)π cm².
Using π = 22/7 gives area = (147/2) × 22/7 = 231 cm².

Answer (ii): (147/2)π cm², which equals 231 cm² (using π = 22/7).

(iii) Area of the segment (minor segment)
Segment area = Area of sector − Area of triangle formed by the two radii and the chord.

Area of triangle (isosceles with sides 21,21 and included angle 60°) = 1/2 × r² × sinθ
= 1/2 × 21² × sin 60° = 1/2 × 441 × (√3/2) = (441√3) / 4 cm².

So exact area of segment = (147/2)π − (441√3) / 4.

Numeric value (use π = 22/7 and √3 ≈ 1.732):
Sector area = 231 cm² (from part ii).
Triangle area ≈ (441 × 1.732) / 4 = 110.25 × 1.732 ≈ 190.95 cm².
Segment area ≈ 231 − 190.95 = 40.05 cm² (rounded to two decimal places).

Answer (iii): Exact form = (147/2)π − (441√3) / 4 cm²;
Numeric ≈ 40.05 cm² (using π = 22/7 and √3 ≈ 1.732).

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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 of the circle. (Use π = 3.14 and √3 = 1.73)

Solution:

Step 1: Area of the sector

Formula: Area of sector = (θ / 360) × π × r²

Here, θ = 60°, r = 15 cm.

Area of sector = (60 / 360) × 3.14 × (15)²
= (1 / 6) × 3.14 × 225
= 117.75 cm²

Step 2: Area of the triangle (formed by 2 radii and the chord)

Formula: Area = (1/2) × r² × sinθ

= (1/2) × (15)² × sin 60°
= (1/2) × 225 × (√3 / 2)
= (225 / 2) × (1.73 / 2)
= 112.5 × 0.865
= 97.3125 cm²

Step 3: Area of the minor segment

Minor segment = Area of sector − Area of triangle
= 117.75 − 97.3125
= 20.4375 cm²

Step 4: Area of the major segment

Area of circle = π × r² = 3.14 × 225 = 706.5 cm²

Major segment = Area of circle − Area of minor segment
= 706.5 − 20.4375
= 686.0625 cm²

Final Answers:
(i) Minor segment = 20.4375 cm²
(ii) Major segment = 686.0625 cm²

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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 of the circle.
(Use π = 3.14 and √3 = 1.73)

Solution:

Step 1: Area of the sector

Formula: Area of sector = (θ / 360) × π × r²

θ = 120°, r = 12 cm

Area of sector = (120 / 360) × 3.14 × (12)²
= (1 / 3) × 3.14 × 144
= 150.72 cm²

Step 2: Area of the triangle (formed by 2 radii and the chord)

Formula: Area = (1/2) × r² × sinθ

= (1/2) × 12² × sin 120°
= (1/2) × 144 × (√3 / 2)
= 72 × (1.73 / 2)
= 72 × 0.865
= 62.28 cm²

Step 3: Area of the segment

Segment area = Area of sector − Area of triangle
= 150.72 − 62.28
= 88.44 cm²

Final Answer:
The area of the segment = 88.44 cm²

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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 rope (see Fig. 11.8(refer book)). 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:

When the horse is tied at a corner and the rope length is less than the side of the square, the horse can graze only inside a quarter of a circle whose radius equals the rope length.

(i) Rope = 5 m.
Area of full circle of radius 5 = π × 5² = π × 25.
Area of quarter circle = (1/4) × 25π = 25π / 4.
Using π = 3.14: Area = (25 × 3.14) / 4 = 78.5 / 4 = 19.625 m².

So grazing area with 5 m rope = 25π/4 m² ≈ 19.625 m².

(ii) Rope = 10 m.
Area of full circle of radius 10 = π × 10² = 100π.
Area of quarter circle = (1/4) × 100π = 25π.
Using π = 3.14: Area = 25 × 3.14 = 78.5 m².

Increase in grazing area = area with 10 m rope − area with 5 m rope
= 78.5 − 19.625 = 58.875 m².

So increase in grazing area = 75π/4 m² ≈ 58.875 m².

Answers:
(i) Grazing area with 5 m rope = 25π/4 m² ≈ 19.625 m².
(ii) Increase when rope is 10 m = 75π/4 m² ≈ 58.875 m².

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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 as shown in Fig. 11.9 (refer book). Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Solution:

Given: diameter = 35 mm.
Radius r = 35 ÷ 2 = 35/2 mm.

Use π = 22/7.

(i) Total length of silver wire

1. Wire for the circumference = π × d
= (22/7) × 35
= 22 × 5
= 110 mm.

2. Wire for the 5 diameters = 5 × 35 = 175 mm.

3. Total length = circumference + diameters
= 110 + 175 = 285 mm.

Answer (i): 285 mm.

(ii) Area of each sector

1. Area of full circle = π × r²
r² = (35/2)² = 1225/4.
Area = (22/7) × (1225/4).

Simplify: 1225 ÷ 7 = 175, so area = 22 × 175 ÷ 4 = 3850/4 = 1925/2 mm².

2. Number of equal sectors = 10.
Area of each sector = (Area of circle) ÷ 10
= (1925/2) ÷ 10 = 1925/20 = 385/4 mm².

Answer (ii): 385/4 mm² (which is 96.25 mm²).

Final Answers:
(i) Total length of wire = 285 mm.
(ii) Area of each sector = 385/4 mm² (= 96.25 mm²).

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Q10: An umbrella has 8 ribs which are equally spaced (see Fig. 11.10 (refer book)). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

Radius of umbrella = 45 cm

Step 1: Area of the whole umbrella (circle)
Area of circle = π × r²
= 22/7 × 45 × 45
= 22/7 × 2025
= 44550/7 cm²

Step 2: Area of one sector
The umbrella is divided into 8 equal sectors.

Area of one sector = (Total area) ÷ 8
= (44550/7) ÷ 8
= 44550 / 56 cm²

Simplifying,
= 22275 / 28 cm²

Final Answer:
The area between two consecutive ribs of the umbrella = 22275/28 cm²

(or as a mixed fraction: 794 23/28 cm²)

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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.

Solution:

Given: radius (blade length) = 25 cm, central angle for each wiper = 115°.

1. Area of one sector (one wiper)
Area of sector = (angle / 360) × π × r²
= (115 / 360) × π × 25²
= (115 / 360) × π × 625
= (71875 / 360) × π
Reduce factor 5: = (14375 / 72) × π cm².

2. Total area for two wipers (they do not overlap)
Total area = 2 × area of one sector
= 2 × (14375 / 72) × π
= (14375 / 36) × π cm². (This is the exact answer in terms of π.)

3. Numeric value using π = 22/7
Total area = (14375 / 36) × (22 / 7)
= 316250 / 252
Reduce by 2: = 158125 / 126 cm².

Decimal value ≈ 1254.96 cm² (rounded to two decimal places).

Answers:
Exact (in terms of π): (14375/36) × π cm².
Numeric (using π = 22/7): 158125/126 cm² (≈ 1254.96 cm²).

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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)

Solution:

Here, radius of the sector = 16.5 km
Angle of the sector = 80°

Formula:
Area of sector = (θ / 360) × π × r²

1. Substitute values:
Area = (80 / 360) × 3.14 × (16.5)²

2. Square radius:
(16.5)² = 272.25

So, Area = (80 / 360) × 3.14 × 272.25

3. Simplify:
= (2 / 9) × 3.14 × 272.25
= 0.222… × 3.14 × 272.25

= 189.44 km² (approx).

Final Answer:
The area of the sea over which ships are warned = 189.44 km² (approx).

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Q13: A round table cover has six equal designs as shown in Fig. 11.11 (refer book). 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)

Solution:

Radius of the cover = 28 cm

Step 1: Area of the circle
Area of circle = πr²
= (22/7) × 28 × 28
= (22/7) × 784
= 2464 cm²

Step 2: Area of the regular hexagon inside the circle
For a regular hexagon,
Area = (3√3 / 2) × a²
Here, side a = 28 cm (same as radius).

So,
Area = (3 × 1.7 / 2) × 28²
= (5.1 / 2) × 784
= 2.55 × 784
= 1999.2 cm²

Step 3: Area of 6 designs
Area of designs = Area of circle – Area of hexagon
= 2464 – 1999.2
= 464.8 cm²

Step 4: Cost of making the designs
Cost = Area × Rate
= 464.8 × ₹0.35
= ₹162.68

Final Answer:
The cost of making the designs = ₹162.68

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Q14: Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is

Answer:
(p / 720) × 2πR²

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