11.1-Areas of Sector and Segment of a Circle
11.1-Areas of Sector and Segment of a Circle Important Formulae
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Grade 10 → Math → Areas Related to Circles → 11.1-Areas of Sector and Segment of a Circle
- Describe sector and segment of a circle in order to differentiate between the two.
- Describe minor and major segment of a circle in order to differentiate between the two.
- Apply the formula of area of sector and segment of a circle in order to compute the area of a specified region.
- Calculate the length of an arc of a circle in order to comment whether it is the major arc or minor arc.
The study of areas related to circles includes important concepts such as sectors and segments. Understanding these concepts is essential for solving various geometric problems involving circles.
1. Sector of a Circle
A sector is a portion of a circle enclosed by two radii and the arc between them. It resembles a "slice" of the circle. The area of a sector can be calculated using the formula:
$$\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2$$
where:
- $\theta$ is the angle in degrees of the sector.
- $r$ is the radius of the circle.
2. Example of Sector Area Calculation
For example, consider a sector with a radius of 7 cm and an angle of 60 degrees:
Using the formula:
$$\text{Area} = \frac{60}{360} \times \pi \times (7)^2$$
Calculating this gives:
$$\text{Area} = \frac{1}{6} \times \pi \times 49 \approx 25.67 \, \text{cm}^2$$
3. Segment of a Circle
A segment of a circle is the region enclosed by a chord and the arc that subtends the chord. The area of a segment can be calculated using the area of the sector minus the area of the triangle formed by the two radii and the chord.
4. Area of a Segment
The formula for the area of a segment is given by:
$$\text{Area of Segment} = \text{Area of Sector} - \text{Area of Triangle}$$
Where:
- Area of Sector is calculated using the previously mentioned formula.
- Area of Triangle can be calculated using the formula:
$$\text{Area} = \frac{1}{2} \times r^2 \times \sin \theta$$
5. Example of Segment Area Calculation
For example, consider a segment with a radius of 7 cm and an angle of 60 degrees:
- First, calculate the area of the sector:
- Next, calculate the area of the triangle:
- Now, calculate the area of the segment:
$$\text{Area of Sector} = \frac{60}{360} \times \pi \times (7)^2 \approx 25.67 \, \text{cm}^2$$
$$\text{Area of Triangle} = \frac{1}{2} \times 7^2 \times \sin(60^\circ) = \frac{1}{2} \times 49 \times \frac{\sqrt{3}}{2} \approx 21.22 \, \text{cm}^2$$
$$\text{Area of Segment} = 25.67 - 21.22 \approx 4.45 \, \text{cm}^2$$
6. Key Concepts
When working with sectors and segments, keep these key points in mind:
- The area of a sector is directly proportional to the angle $\theta$.
- The area of a segment decreases as the angle decreases.
- The area of the triangle can be calculated using trigonometric functions when dealing with angles.
7. Application of Areas of Sector and Segment
Understanding the areas of sectors and segments is crucial in various fields such as engineering, architecture, and any field that involves circular objects. It also aids in solving real-world problems, such as finding the area of land with circular plots or designing components that involve circular shapes.
Representation of the area of the swept sector between the indicated angles.
Olmar arranz, CC BY-SA 4.0, via Wikimedia Commons
Solved Example: 11-1-01
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Area of a Sector of a Circle
Given:
- Radius (r) = 6 cm
- Angle of the sector (θ) = 60°
Formula for the area of a sector:
Area = (θ/360) × π × r²
Substituting the given values:
Area = (60/360) × π × (6)²
Calculating:
- 60/360 = 1/6
- (6)² = 36
- Area = (1/6) × π × 36
- Area = 6π cm²
Approximate value (using π ≈ 3.14):
Area ≈ 6 × 3.14 = 18.84 cm²
Solved Example: 11-1-02
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Area of a Quadrant of a Circle
Given:
- Circumference = 22 cm
Formula for the circumference of a circle:
C = 2πr
Substituting the given circumference:
22 = 2πr
Solving for r:
- r = 22 / (2π)
- r = 11 / π cm
Formula for the area of a circle:
Area = πr²
Area of the whole circle:
Area = π × (11/π)²
Calculating:
- Area = π × (121/π²)
- Area = 121/π cm²
Area of the quadrant:
Area of Quadrant = (1/4) × Area of Circle
Area of Quadrant = (1/4) × (121/π)
Area of Quadrant = 121/(4π) cm²
Solved Example: 11-1-03
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Area Swept by the Minute Hand of a Clock
Given:
- Length of the minute hand (r) = 14 cm
- Time = 5 minutes
Formula for the area of a sector:
Area = (θ/360) × πr²
First, find the angle (θ) swept by the minute hand in 5 minutes:
Since the minute hand completes 360° in 60 minutes:
θ = (5/60) × 360
Calculating θ:
- θ = (1/12) × 360
- θ = 30°
Now substitute the values into the area formula:
Area = (30/360) × π × (14)²
Calculating:
- 30/360 = 1/12
- (14)² = 196
- Area = (1/12) × π × 196
- Area = 196/12 × π
- Area = 49/3 × π cm²
Solved Example: 11-1-04
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 $\pi$ = 3.14)
Solution:
Area of Minor Segment and Major Sector
Given:
- Radius (r) = 10 cm
- Angle (θ) = 90° (right angle)
1. Area of the Sector:
Formula for the area of a sector:
Area = (θ/360) × πr²
Substituting the values:
Area of Sector = (90/360) × 3.14 × (10)²
Calculating:
- 90/360 = 1/4
- (10)² = 100
- Area of Sector = (1/4) × 3.14 × 100
- Area of Sector = 78.5 cm²
2. Area of the Triangle:
Formula for the area of the triangle formed by the radius and chord:
Area = (1/2) × base × height
In this case, the triangle is an isosceles right triangle:
Base = Height = r = 10 cm
Calculating the area:
- Area of Triangle = (1/2) × 10 × 10
- Area of Triangle = 50 cm²
3. Area of the Minor Segment:
Formula:
Area of Minor Segment = Area of Sector - Area of Triangle
Substituting the values:
Area of Minor Segment = 78.5 - 50
Area of Minor Segment = 28.5 cm²
4. Area of the Major Sector:
Formula:
Area of Major Sector = Area of Circle - Area of Minor Sector
Area of Circle:
Area of Circle = πr²
Area of Circle = 3.14 × (10)²
Area of Circle = 314 cm²
Calculating the Area of the Major Sector:
Area of Major Sector = 314 - 78.5
Area of Major Sector = 235.5 cm²
Solved Example: 11-1-05
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:
Arc Length, Area of Sector, and Area of Segment
Given:
- Radius (r) = 21 cm
- Angle (θ) = 60°
1. Length of the Arc:
Formula for the length of the arc:
Length of Arc = (θ/360) × 2πr
Substituting the values:
Length of Arc = (60/360) × 2 × 3.14 × 21
Calculating:
- 60/360 = 1/6
- Length of Arc = (1/6) × 2 × 3.14 × 21
- Length of Arc = (1/6) × 131.88
- Length of Arc = 21.98 cm
2. Area of the Sector:
Formula for the area of a sector:
Area = (θ/360) × πr²
Substituting the values:
Area of Sector = (60/360) × 3.14 × (21)²
Calculating:
- Area of Sector = (1/6) × 3.14 × 441
- Area of Sector = 3.14 × 73.5
- Area of Sector = 230.79 cm²
3. Area of the Segment:
Formula for the area of the segment:
Area of Segment = Area of Sector - Area of Triangle
Area of Triangle:
In this case, the triangle is formed by the two radii and the chord:
Formula for the area of the triangle (using sine):
Area = (1/2) × r² × sin(θ)
Calculating:
Area of Triangle = (1/2) × (21)² × sin(60°)
Area of Triangle = (1/2) × 441 × (√3/2)
Area of Triangle = 220.5 × (√3/2)
Area of Triangle ≈ 220.5 × 0.866 = 191.13 cm²
Now substituting to find the Area of the Segment:
Area of Segment = 230.79 - 191.13
Area of Segment ≈ 39.66 cm²
Solved Example: 11-1-06
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 $\pi$ =3.14 and $\sqrt{3}$ =1.73)
Solution:
Areas of Minor and Major Segments
Given:
- Radius (r) = 15 cm
- Angle (θ) = 60°
1. Area of the Sector:
Formula for the area of a sector:
Area of Sector = (θ/360) × πr²
Substituting the values:
Area of Sector = (60/360) × 3.14 × (15)²
Calculating:
- 60/360 = 1/6
- (15)² = 225
- Area of Sector = (1/6) × 3.14 × 225
- Area of Sector = 117.5 cm²
2. Area of the Triangle:
Formula for the area of the triangle formed by the radius and chord:
Area = (1/2) × r² × sin(θ)
Calculating:
Area of Triangle = (1/2) × (15)² × sin(60°)
Area of Triangle = (1/2) × 225 × (√3/2)
Area of Triangle = (1/2) × 225 × (1.73/2)
Area of Triangle = (1/2) × 225 × 0.865
Area of Triangle ≈ 97.3125 cm²
3. Area of the Minor Segment:
Formula:
Area of Minor Segment = Area of Sector - Area of Triangle
Substituting the values:
Area of Minor Segment = 117.5 - 97.3125
Area of Minor Segment ≈ 20.1875 cm²
4. Area of the Major Segment:
Formula:
Area of Major Segment = Area of Circle - Area of Minor Segment
Area of Circle:
Area of Circle = πr²
Area of Circle = 3.14 × (15)²
Area of Circle = 3.14 × 225
Area of Circle = 706.5 cm²
Calculating the Area of the Major Segment:
Area of Major Segment = 706.5 - 20.1875
Area of Major Segment ≈ 686.3125 cm²
Solved Example: 11-1-07
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 $\pi$ =3.14 and $\sqrt{3}$ =1.73)
Solution:
Area of the Segment of a Circle
Given:
- Radius (r) = 12 cm
- Angle (θ) = 120°
1. Area of the Sector:
Formula for the area of a sector:
Area of Sector = (θ/360) × πr²
Substituting the values:
Area of Sector = (120/360) × 3.14 × (12)²
Calculating:
- 120/360 = 1/3
- (12)² = 144
- Area of Sector = (1/3) × 3.14 × 144
- Area of Sector = 150.72 cm²
2. Area of the Triangle:
Formula for the area of the triangle formed by the radius and chord:
Area = (1/2) × r² × sin(θ)
Calculating:
Area of Triangle = (1/2) × (12)² × sin(120°)
Area of Triangle = (1/2) × 144 × (√3/2)
Area of Triangle = (1/2) × 144 × (1.73/2)
Area of Triangle = (1/2) × 144 × 0.865
Area of Triangle ≈ 62.28 cm²
3. Area of the Segment:
Formula:
Area of Segment = Area of Sector - Area of Triangle
Substituting the values:
Area of Segment = 150.72 - 62.28
Area of Segment ≈ 88.44 cm²
Solved Example: 11-1-08
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. 11.8). 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 $\pi$ = 3.14)
Solution:
Grazing Area of a Horse Tied to a Peg
Given:
- Side of the square field = 15 m
- Length of the rope (r) = 5 m
1. Area where the horse can graze:
The horse can graze in a quarter circle because it is tied at a corner.
Formula for the area of a circle:
Area = πr²
Calculating the area of the quarter circle:
Area of Quarter Circle = (1/4) × πr²
Area of Quarter Circle = (1/4) × 3.14 × (5)²
Area of Quarter Circle = (1/4) × 3.14 × 25
Area of Quarter Circle = (1/4) × 78.5
Area of Quarter Circle = 19.625 m²
2. Increase in the grazing area if the rope were 10 m long:
New length of the rope (r) = 10 m
Calculating the area of the new quarter circle:
Area of New Quarter Circle = (1/4) × π(10)²
Area of New Quarter Circle = (1/4) × 3.14 × 100
Area of New Quarter Circle = (1/4) × 314
Area of New Quarter Circle = 78.5 m²
Calculating the increase in grazing area:
Increase in Grazing Area = Area of New Quarter Circle - Area of Original Quarter Circle
Increase in Grazing Area = 78.5 - 19.625
Increase in Grazing Area ≈ 58.875 m²
Solved Example: 11-1-09
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. Find :
(i) The total length of the silver wire required.
(ii) The area of each sector of the brooch.
Solution:
Total Length of Silver Wire and Area of Each Sector
Given:
- Diameter of the circle = 35 mm
- Radius (r) = Diameter/2 = 17.5 mm
1. Total Length of the Silver Wire Required:
Length of the circular wire (circumference):
Circumference = πd
Circumference = 3.14 × 35
Circumference = 109.9 mm
Length of the 5 diameters:
Length of 5 Diameters = 5 × Diameter
Length of 5 Diameters = 5 × 35
Length of 5 Diameters = 175 mm
Calculating the total length of the silver wire:
Total Length = Circumference + Length of 5 Diameters
Total Length = 109.9 + 175
Total Length = 284.9 mm
2. Area of Each Sector of the Brooch:
Area of the circle:
Area = πr²
Area = 3.14 × (17.5)²
Area = 3.14 × 306.25
Area = 960.625 mm²
Area of each sector:
Area of Each Sector = Total Area / Number of Sectors
Area of Each Sector = 960.625 / 10
Area of Each Sector = 96.0625 mm²
Solved Example: 11-1-10
An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Solution:
Area Between Two Consecutive Ribs of an Umbrella
Given:
- Number of ribs (n) = 8
- Radius of the umbrella (r) = 45 cm
The area of the entire umbrella (circle) is calculated using the formula:
Area = πr²
Substituting the values:
Area = π × (45)² = π × 2025 = 2025π cm²
Since there are 8 ribs, the area between two consecutive ribs will be:
Area between two ribs = Total Area / Number of ribs
Area between two ribs = (2025π) / 8 cm²
Calculating this gives:
Area between two ribs = 253.125π cm²
Solved Example: 11-1-11
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:
Total Area Cleaned by Car Wipers
Given:
- Length of each wiper blade (r) = 25 cm
- Angle of sweep (θ) = 115°
To find the area cleaned by one wiper, we use the formula for the area of a sector:
Area = (θ / 360°) × πr²
Substituting the values for one wiper:
Area = (115° / 360°) × π × (25)²
Area = (115 / 360) × π × 625
Area = (115 × 625π) / 360
Area = (71875π) / 360 cm²
Now, to find the total area cleaned by both wipers:
Total Area = 2 × Area of one wiper
Total Area = 2 × (71875π / 360) cm²
Total Area = (143750π) / 360 cm²
Solved Example: 11-1-12
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 $\pi$ = 3.14)
Solution:
Area of Sea Warned by Lighthouse
Given:
- Angle of the sector (θ) = 80°
- Distance of the light (r) = 16.5 km
To find the area of the sector, we use the formula:
Area = (θ / 360°) × πr²
Substituting the values:
Area = (80 / 360) × 3.14 × (16.5)²
Area = (80 / 360) × 3.14 × 272.25
Area = (80 × 3.14 × 272.25) / 360
Area = (6867.0) / 360
Area ≈ 19.1 km²
Solved Example: 11-1-13
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 Rs. 0.35 per cm$^2$. (Use $\sqrt{3}$ = 1.7)
Solution:
Cost of Making Designs on a Round Table Cover
Given:
- Radius of the cover (r) = 28 cm
- Number of designs = 6
- Cost per cm² = Rs. 0.35
First, we calculate the area of the round table cover:
Area = πr²
Using π = 3.14:
Area = 3.14 × (28)²
Area = 3.14 × 784
Area = 2467.76 cm²
Since there are 6 equal designs, the area for one design is:
Area of one design = Total Area / Number of designs
Area of one design = 2467.76 / 6
Area of one design ≈ 411.29 cm²
Now, to find the cost of making one design:
Cost of one design = Area of one design × Cost per cm²
Cost of one design = 411.29 × 0.35
Cost of one design ≈ Rs. 143.95
Thus, the total cost for six designs is:
Total Cost = 6 × Cost of one design
Total Cost = 6 × 143.95
Total Cost ≈ Rs. 863.70
Solved Example: 11-1-14
Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is:
Solution:
Area of a Sector of a Circle
Tick the correct answer:
- A) (πR²p) / 360
- B) (pR²) / 360
- C) (p/360) × πR²
- D) (360/p) × πR²