Area of Segment of a Circle
A segment of a circle is the region enclosed between a chord and the arc it cuts off. Every chord (except the diameter) divides a circle into two unequal segments — a minor segment and a major segment.
This topic is part of Chapter 12 — Areas Related to Circles in the CBSE Class 10 syllabus. The area of a segment is calculated using the area of the corresponding sector minus the area of the triangle formed by the chord and the two radii.
Segment problems frequently appear in board exams as 3-mark and 5-mark questions, often combined with sector area and arc length calculations.
What is Area of Segment of a Circle?
Definition: A segment of a circle is the region between a chord and the arc intercepted by that chord.
- Minor segment: The smaller region between a chord and its minor arc.
- Major segment: The larger region between a chord and its major arc.
- When the chord is a diameter, both segments are equal — each is a semicircle.
- Minor segment + Major segment = Area of the full circle.
Key distinction:
- A sector is bounded by two radii and an arc (like a slice of pizza).
- A segment is bounded by a chord and an arc (the sector minus the triangle).
Area of Segment of a Circle Formula
Area of Minor Segment:
Area of Segment = Area of Sector − Area of Triangle
= (θ/360) × πr² − (1/2) × r² × sin θ
Where:
- θ = central angle of the sector (in degrees)
- r = radius of the circle
- sin θ = sine of the central angle
Area of Major Segment:
Area of Major Segment = πr² − Area of Minor Segment
For common angles:
| Angle θ | Area of Triangle (using two radii) |
|---|---|
| 60° | (√3/4) × r² (equilateral triangle) |
| 90° | (1/2) × r² (right triangle with legs = r) |
| 120° | (√3/4) × r² |
Derivation and Proof
Deriving the area of a segment:
- Draw a circle with centre O and radius r.
- Let AB be a chord that subtends angle θ at the centre.
- The sector OAB is the region bounded by radii OA, OB, and arc AB.
- Area of sector OAB = (θ/360) × πr².
- The triangle OAB is formed by the two radii OA, OB, and chord AB.
- Area of triangle OAB = (1/2) × r × r × sin θ = (1/2)r² sin θ.
- The minor segment is the region inside the sector but outside the triangle.
- Area of minor segment = Area of sector − Area of triangle.
- = (θ/360) × πr² − (1/2)r² sin θ.
For θ = 90°:
- Area of sector = (90/360) × πr² = πr²/4
- Area of triangle = (1/2) × r × r = r²/2
- Area of minor segment = πr²/4 − r²/2
For θ = 60°:
- Triangle OAB is equilateral (OA = OB = r, angle = 60°, so AB = r).
- Area of sector = πr²/6
- Area of equilateral triangle = (√3/4)r²
- Area of minor segment = πr²/6 − (√3/4)r²
Types and Properties
Type 1: Finding Area of Minor Segment
- Given radius and central angle, compute sector area minus triangle area.
Type 2: Finding Area of Major Segment
- Major segment area = Total circle area − Minor segment area.
Type 3: Segment with θ = 90°
- Triangle is right-angled with legs equal to the radius.
- Area of triangle = r²/2.
Type 4: Segment with θ = 60°
- Triangle is equilateral with side = r.
- Area of triangle = (√3/4)r².
Type 5: Shaded Region Problems
- Combinations where segments must be identified within overlapping circles or within squares/rectangles.
Methods
Steps to find the area of a segment:
- Identify the radius (r) and the central angle (θ).
- Calculate the area of the sector: (θ/360) × πr².
- Calculate the area of the triangle formed by the two radii and the chord.
- For standard angles, use known formulas. For other angles, use (1/2)r² sin θ.
- Area of minor segment = Sector area − Triangle area.
- For major segment: Subtract the minor segment area from πr².
Triangle area for common angles:
- θ = 60°: Equilateral triangle → Area = (√3/4)r²
- θ = 90°: Right triangle with legs r → Area = r²/2
- θ = 120°: Use (1/2)r² sin 120° = (1/2)r²(√3/2) = (√3/4)r²
Common Mistakes:
- Forgetting to subtract the triangle area from the sector area.
- Using the wrong triangle formula for the given angle.
- Confusing segment with sector — a sector includes the triangle, a segment does not.
Solved Examples
Example 1: Minor Segment with θ = 90°
Problem: Find the area of the minor segment of a circle with radius 14 cm, where the central angle is 90°. (Use π = 22/7)
Solution:
Given:
- r = 14 cm, θ = 90°
Step 1: Area of sector
- = (90/360) × πr² = (1/4) × (22/7) × 196 = (1/4) × 616 = 154 cm²
Step 2: Area of triangle
- = (1/2) × r × r = (1/2) × 14 × 14 = 98 cm²
Step 3: Area of minor segment
- = 154 − 98 = 56 cm²
Answer: The area of the minor segment is 56 cm².
Example 2: Minor Segment with θ = 60°
Problem: Find the area of the minor segment of a circle with radius 21 cm, where the central angle is 60°. (Use π = 22/7, √3 = 1.73)
Solution:
Given:
- r = 21 cm, θ = 60°
Step 1: Area of sector
- = (60/360) × (22/7) × 21² = (1/6) × (22/7) × 441 = (1/6) × 1386 = 231 cm²
Step 2: Area of equilateral triangle
- = (√3/4) × r² = (1.73/4) × 441 = 0.4325 × 441 = 190.73 cm²
Step 3: Area of minor segment
- = 231 − 190.73 = 40.27 cm²
Answer: The area of the minor segment is 40.27 cm².
Example 3: Major Segment
Problem: Find the area of the major segment of a circle with radius 7 cm, where the minor segment has a central angle of 90°. (Use π = 22/7)
Solution:
Given:
- r = 7 cm, θ = 90°
Step 1: Area of full circle
- = πr² = (22/7) × 49 = 154 cm²
Step 2: Area of minor segment
- Sector area = (1/4) × 154 = 38.5 cm²
- Triangle area = (1/2) × 7 × 7 = 24.5 cm²
- Minor segment = 38.5 − 24.5 = 14 cm²
Step 3: Area of major segment
- = 154 − 14 = 140 cm²
Answer: The area of the major segment is 140 cm².
Example 4: Segment with θ = 120°
Problem: A chord subtends an angle of 120° at the centre of a circle of radius 7 cm. Find the area of the corresponding minor segment. (Use π = 22/7, √3 = 1.73)
Solution:
Given:
- r = 7 cm, θ = 120°
Step 1: Area of sector
- = (120/360) × (22/7) × 49 = (1/3) × 154 = 51.33 cm²
Step 2: Area of triangle
- = (1/2)r² sin 120° = (1/2)(49)(√3/2) = (49√3)/4 = (49 × 1.73)/4 = 21.19 cm²
Step 3: Area of minor segment
- = 51.33 − 21.19 = 30.14 cm²
Answer: The area of the minor segment is 30.14 cm².
Example 5: Shaded Region: Two Overlapping Segments
Problem: Two equal circles of radius 7 cm intersect each other such that each passes through the centre of the other. Find the area of the overlapping region. (Use π = 22/7, √3 = 1.73)
Solution:
Given:
- r = 7 cm for both circles
- Distance between centres = 7 cm (each passes through the other's centre)
- The angle subtended at each centre by the chord of intersection = 120°
Steps:
- The overlapping region = 2 × (minor segment with θ = 120°)
- Area of sector (120°) = (120/360) × (22/7) × 49 = 154/3 = 51.33 cm²
- Area of triangle = (√3/4) × 49 = 1.73 × 49/4 = 21.19 cm²
- Area of one minor segment = 51.33 − 21.19 = 30.14 cm²
- Overlapping area = 2 × 30.14 = 60.28 cm²
Answer: The area of the overlapping region is 60.28 cm².
Example 6: Segment Area: Chord Given
Problem: A chord of length 14 cm is drawn in a circle of radius 14 cm. Find the area of the minor segment. (Use π = 22/7, √3 = 1.73)
Solution:
Given:
- r = 14 cm, chord length = 14 cm
Finding θ:
- Since chord = radius, triangle OAB has OA = OB = AB = 14 cm → equilateral triangle.
- Therefore θ = 60°.
Steps:
- Area of sector = (60/360) × (22/7) × 196 = (1/6) × 616 = 102.67 cm²
- Area of equilateral triangle = (√3/4) × 196 = 1.73 × 49 = 84.77 cm²
- Area of minor segment = 102.67 − 84.77 = 17.9 cm²
Answer: The area of the minor segment is 17.9 cm².
Example 7: Segment in a Quadrant
Problem: From a square of side 14 cm, a quadrant (quarter circle) of radius 14 cm is cut out from one corner. Find the area of the remaining region. (Use π = 22/7)
Solution:
Given:
- Side of square = 14 cm
- Quadrant radius = 14 cm, θ = 90°
Steps:
- Area of square = 14² = 196 cm²
- Area of quadrant (sector with θ = 90°) = (1/4) × (22/7) × 196 = 154 cm²
- Remaining area = 196 − 154 = 42 cm²
Answer: The remaining area is 42 cm².
Example 8: Area Between Two Concentric Segments
Problem: Two concentric circles have radii 7 cm and 14 cm. A sector of angle 90° is cut from both. Find the area of the region between the two sectors (the segment-shaped ring). (Use π = 22/7)
Solution:
Given:
- R = 14 cm (outer), r = 7 cm (inner), θ = 90°
Steps:
- Area of outer sector = (90/360) × (22/7) × 196 = 154 cm²
- Area of inner sector = (90/360) × (22/7) × 49 = 38.5 cm²
- Area of ring-sector = 154 − 38.5 = 115.5 cm²
Answer: The area between the two sectors is 115.5 cm².
Example 9: Total Segment Area in Three Equal Parts
Problem: Three equal chords divide a circle of radius 7 cm into three equal arcs. Find the area of each minor segment. (Use π = 22/7, √3 = 1.73)
Solution:
Given:
- r = 7 cm
- Three equal arcs → each central angle = 360°/3 = 120°
Steps:
- Area of sector (120°) = (120/360) × (22/7) × 49 = (1/3) × 154 = 51.33 cm²
- Area of triangle = (1/2)(7)(7) sin 120° = (49/2)(√3/2) = 49√3/4 = 21.19 cm²
- Area of each minor segment = 51.33 − 21.19 = 30.14 cm²
Answer: Each minor segment has area 30.14 cm².
Example 10: Percentage of Segment Area
Problem: What percentage of the area of a circle of radius 7 cm is the minor segment corresponding to a central angle of 90°? (Use π = 22/7)
Solution:
Given:
- r = 7 cm, θ = 90°
Steps:
- Area of circle = (22/7) × 49 = 154 cm²
- Area of sector = 154/4 = 38.5 cm²
- Area of triangle = (1/2)(7)(7) = 24.5 cm²
- Area of segment = 38.5 − 24.5 = 14 cm²
- Percentage = (14/154) × 100 = 9.09%
Answer: The minor segment is 9.09% of the circle's area.
Real-World Applications
Engineering:
- Cross-sections of pipes, tunnels, and arches often involve segments of circles.
- Calculating the flow area of partially filled circular pipes uses the segment area formula.
Architecture:
- Stained glass windows, arched doorways, and dome structures involve circular segments.
Astronomy:
- The illuminated portion of the Moon during different phases can be approximated as circular segments.
Manufacturing:
- Cutting circular sheets into segments for specific shapes and designs.
Key Points to Remember
- A segment is the region between a chord and its arc.
- Area of minor segment = Area of sector − Area of triangle.
- Area of major segment = πr² − Area of minor segment.
- Minor segment + Major segment = Area of full circle.
- For θ = 60°: triangle is equilateral, area = (√3/4)r².
- For θ = 90°: triangle is right-angled, area = r²/2.
- For θ = 120°: area of triangle = (√3/4)r².
- When chord = radius, the central angle is 60°.
- Do NOT confuse sector with segment: sector includes the triangle, segment does not.
- Always identify whether the problem asks for the minor or major segment.
Practice Problems
- Find the area of the minor segment of a circle with radius 21 cm where the central angle is 60°. (Use π = 22/7, √3 = 1.73)
- Find the area of the major segment of a circle with radius 14 cm and central angle 90°. (Use π = 22/7)
- A chord subtends a right angle at the centre of a circle with radius 10 cm. Find the area of the minor segment. (Use π = 3.14)
- Two chords of a circle of radius 7 cm subtend angles of 60° and 120° at the centre. Find the difference in the areas of their minor segments.
- Find the area of the shaded region if two equal chords of a circle of radius 14 cm each subtend 90° at the centre and are on opposite sides.
- A circular table top of radius 35 cm has a segment cut off by a chord subtending 60° at the centre. Find the area of the remaining table top.
- Prove that the minor segment area for central angle θ can be written as (r²/2)(πθ/180 − sin θ).
- Find the area of the region enclosed between two concentric circles of radii 7 cm and 14 cm, bounded by two radii enclosing a 60° angle.
Frequently Asked Questions
Q1. What is a segment of a circle?
A segment is the region between a chord and the arc it intercepts. The minor segment is the smaller part, and the major segment is the larger part.
Q2. How do you find the area of a segment?
Area of segment = Area of corresponding sector − Area of the triangle formed by the chord and the two radii. Formula: (θ/360)πr² − (1/2)r² sin θ.
Q3. What is the difference between a sector and a segment?
A sector is bounded by two radii and an arc (pie-slice shape). A segment is bounded by a chord and an arc. Sector = Segment + Triangle.
Q4. How do you find the area of the major segment?
Area of major segment = Area of full circle − Area of minor segment = πr² − (sector area − triangle area).
Q5. What happens when the chord is a diameter?
When the chord is a diameter, it subtends 180° at the centre. Both segments become semicircles with equal areas of πr²/2.
Q6. Why do we subtract the triangle area from the sector area?
The sector contains both the segment and the triangle formed by the two radii and the chord. Subtracting the triangle leaves only the segment (the curved region between the chord and the arc).
Q7. What is the segment area when the central angle is 90°?
Area = (1/4)πr² − (1/2)r² = r²(π/4 − 1/2). For r = 7 cm with π = 22/7, this is 38.5 − 24.5 = 14 cm².
Q8. How do you find the central angle if the chord length equals the radius?
If chord = radius, then the triangle formed (OAB with OA = OB = AB = r) is equilateral. The central angle is 60°.
Q9. Can a segment be larger than a semicircle?
Yes. A major segment is always larger than a semicircle when the central angle of the minor arc is less than 180°.
Q10. Is the segment formula used in the CBSE Class 10 board exam?
Yes. Questions on segments typically carry 3-5 marks and involve finding the area of minor or major segments with standard angles like 60°, 90°, or 120°.
