Quick Facts
- Area of a circle:
A = πr² - Area of a sector:
A = (θ/360°) × πr² - Area of a quadrant:
A = ¼πr² - Area of a semicircle:
A = ½πr² - Area of a segment:
Area of segment = Area of sector − Area of triangle - Major sector angle:
360° − minor sector angle - A sector is the region enclosed by two radii and an arc.
- A segment is the region bounded by an arc and the chord joining the endpoints of that arc.
Story
Imagine the minute hand of a clock moving around the clock face. It sweeps out a portion of a circle. If the hand moves for a few minutes, it does not cover the whole circular region—it covers a sector.
Now imagine drawing a chord across a circle. The smaller region between the chord and the corresponding arc is called a segment.
These ideas help us calculate the areas covered by clock hands, car wipers, circular gardens, curved windows and many other objects.
In Exercise 6.3, we move from the familiar area of a circle to sectors and segments and learn how angles determine the portion of the circular area involved. The textbook introduces the sector-area formula using the rotational symmetry of a circle.
Key Terms
| Term | Meaning |
|---|---|
| Circle | A set of points at a fixed distance from a centre. |
| Radius (r) | Distance from the centre to any point on the circle. |
| Sector | A part of a circular region enclosed by two radii and an arc. |
| Quadrant | A sector with a central angle of 90°. |
| Minor Sector | The smaller sector formed by an angle less than 180°. |
| Major Sector | The larger sector formed by the remaining part of the circle. |
| Chord | A line segment joining two points on a circle. |
| Segment | Region bounded by a chord and its corresponding arc. |
| Minor Segment | The smaller region between a chord and its minor arc. |
| Major Segment | The larger region between a chord and its major arc. |
| Central Angle | The angle made by two radii at the centre of a circle. |
Questions
Question 1
Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Solution
Given:
- Radius, r = 7 cm
- Angle, θ = 60°
- π = 22/7
Formula:
Area of sector = (θ/360°) × πr²
Substituting:
= (60/360) × (22/7) × 7²
= 1/6 × 22/7 × 49
= 77/3
= 25⅔ cm²
Answer
Area of the sector = 25⅔ cm²
Question 2
Find the area of a quadrant of a circle whose circumference is 44 cm.
Solution
Given:
Circumference = 44 cm
Formula:
Circumference = 2πr
Therefore,
44 = 2 × 22/7 × r
44 = 44r/7
So,
r = 7 cm
A quadrant is one-fourth of a circle.
Therefore,
Area of quadrant = ¼πr²
= ¼ × 22/7 × 7²
= ¼ × 154
= 38.5 cm²
Answer
Area of the quadrant = 38.5 cm²
Question 3
The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Solution
The minute hand completes:
360° in 60 minutes
Therefore, in 10 minutes it sweeps:
θ = (10/60) × 360°
= 60°
The radius of the swept sector is the length of the minute hand:
r = 7 cm
Area swept:
= (60/360) × 22/7 × 7²
= 1/6 × 154
= 77/3 cm²
= 25⅔ cm²
Answer
Area swept = 25⅔ cm²
Question 4
A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding:
(i) minor sector
(ii) major sector
Use π ≈ 3.14.
(i) Minor Sector
Given:
- r = 10 cm
- θ = 90°
- π = 3.14
Area of minor sector:
= (90/360) × 3.14 × 10²
= ¼ × 314
= 78.5 cm²
(ii) Major Sector
The major sector angle is:
360° − 90° = 270°
Area of major sector:
= (270/360) × 3.14 × 10²
= ¾ × 314
= 235.5 cm²
Answer
(i) Minor sector = 78.5 cm²
(ii) Major sector = 235.5 cm²
Check:
78.5 + 235.5 = 314 cm², which is the area of the complete circle.
Question 5
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.
Step 1: Area of the Minor Sector
r = 15 cm
θ = 60°
Area of minor sector:
= (60/360) × 3.14 × 15²
= 1/6 × 3.14 × 225
= 117.75 cm²
Step 2: Area of the Triangle
The two radii and the chord form an equilateral triangle because the two radii are 15 cm and the included angle is 60°.
Area of an equilateral triangle:
Area = (√3/4)a²
Therefore,
= 1.73/4 × 15²
= 1.73/4 × 225
= 97.3125 cm²
Step 3: Minor Segment
Minor segment = Minor sector − Triangle
= 117.75 − 97.3125
= 20.4375 cm²
Step 4: Major Segment
Area of complete circle:
= πr²
= 3.14 × 225
= 706.5 cm²
Major segment:
= Area of circle − Minor segment
= 706.5 − 20.4375
= 686.0625 cm²
Answer
Minor segment = 20.4375 cm²
Major segment = 686.0625 cm²
Important: Do not subtract the triangle from the whole circle when finding the minor segment. The minor segment is specifically the minor sector minus the triangle.
Question 6
A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.
Solution
For each wiper:
- r = 28 cm
- θ = 120°
- π = 22/7
Area swept by one wiper:
= (120/360) × 22/7 × 28²
= 1/3 × 22/7 × 784
= 616/3 cm²
Since there are two non-overlapping wipers:
Total area:
= 2 × 616/3
= 1232/3
= 410⅔ cm²
Answer
Total area cleaned = 410⅔ cm²
Question 7
A chord of a circle of radius r subtends an angle of 60° at the centre. Show that the area of the corresponding minor segment is
πr²/6 − (√3/4)r².
Proof
The minor segment is obtained by removing the triangle from the corresponding minor sector.
Area of the 60° sector
Area of sector:
= (60/360) × πr²
= πr²/6
Area of the triangle
The two radii are both r and the included angle is 60°. Therefore, the triangle is equilateral with side r.
Area of the equilateral triangle:
= √3/4 × r²
Therefore
Area of minor segment:
= Area of sector − Area of triangle
= πr²/6 − √3r²/4
Hence,
Area of minor segment = πr²/6 − (√3/4)r²
Answer
Proved.
Question 8
An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is 3√3/(4π) ≈ 0.413.
Solution
For an equilateral triangle inscribed in a circle, each side subtends 60° at the centre.
The side of the equilateral triangle is:
a = √3r
Area of triangle:
= √3/4 × a²
= √3/4 × (√3r)²
= √3/4 × 3r²
= 3√3r²/4
Area of circle:
= πr²
Therefore,
Ratio:
= (3√3r²/4) ÷ πr²
= 3√3/(4π)
Using π ≈ 22/7 and √3 ≈ 1.73:
3 × 1.73 / (4 × 22/7)
≈ 0.413
Answer
Ratio = 3√3/(4π) ≈ 0.413
Question 9
A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is 2/π ≈ 0.637.
Solution
For a square inscribed in a circle, the diagonal of the square is equal to the diameter of the circle.
Therefore:
Diagonal = 2r
For a square:
Diagonal = side × √2
So,
side × √2 = 2r
Therefore,
side = √2r
Area of square:
= (√2r)²
= 2r²
Area of circle:
= πr²
Therefore,
Ratio:
= 2r² / πr²
= 2/π
Using π ≈ 22/7:
2/(22/7)
= 14/22
= 7/11
≈ 0.636
Thus, to three decimal places:
≈ 0.637
Answer
Ratio = 2/π ≈ 0.637
Question 10
A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is 3√3/(2π) ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
Solution
Join the centre of the circle to all six vertices of the regular hexagon.
This divides the hexagon into six equilateral triangles, each having side r.
Area of one equilateral triangle:
= √3/4 × r²
Therefore, area of six triangles:
= 6 × √3/4 × r²
= 3√3r²/2
So,
Area of hexagon:
= 3√3r²/2
Area of circle:
= πr²
Therefore,
Ratio:
= (3√3r²/2) ÷ πr²
= 3√3/(2π)
Using π ≈ 22/7 and √3 ≈ 1.73:
= approximately 0.827
Why is it twice Question 8?
Question 8 gives:
3√3/(4π)
Question 10 gives:
3√3/(2π)
Since:
1/2 = 2 × 1/4
we have:
3√3/(2π) = 2 × [3√3/(4π)]
Hence, the answer to Question 10 is exactly twice the answer to Question 8.
Answer
Ratio = 3√3/(2π) ≈ 0.827
Memory Tricks
1. Sector Formula
Remember:
“Angle over 360, multiplied by πr².”
Area of sector = θ/360 × πr²
2. Segment Formula
Think:
SEGMENT = SECTOR − TRIANGLE
So:
Area of segment = Area of sector − Area of triangle
3. Common Circle Fractions
| Region | Angle | Fraction of Circle |
|---|---|---|
| Full circle | 360° | 1 |
| Semicircle | 180° | 1/2 |
| Quadrant | 90° | 1/4 |
| 60° sector | 60° | 1/6 |
| 120° sector | 120° | 1/3 |
| 270° sector | 270° | 3/4 |
4. Inscribed Shapes Trick
- Equilateral triangle: area ratio = 3√3/(4π)
- Square: area ratio = 2/π
- Regular hexagon: area ratio = 3√3/(2π)
The hexagon can be divided into 6 equilateral triangles, which makes its area easy to calculate.
Case Study
Case Study: Designing a Circular Wiper System
A car has two non-overlapping wipers. Each blade has a length of 28 cm and sweeps through 120°.
Questions
1. What type of circular region does one wiper sweep?
A. Circle
B. Sector
C. Segment
D. Semicircle
Answer: Sector
2. What fraction of the complete circle is swept by one wiper?
120°/360° = 1/3
3. What is the area swept by one wiper?
Area = 1/3 × 22/7 × 28²
= 616/3 cm²
4. What is the total area cleaned by both wipers?
= 2 × 616/3
= 1232/3 cm²
= 410⅔ cm²
Learning Outcome
This case shows how the sector-area formula can be applied to a practical situation. The same mathematical idea can be used for rotating arms, sprinkler systems, radar sweeps and other circular motions.
The wiper problem appears as Question 6 in the exercise set.
Quiz
A sector has radius 7 cm and central angle 60°. Using π = 22/7, what is its area?
1 of 5
A
154 cm²
B
25⅔ cm²
C
38.5 cm²
D
77 cm²Next
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