Tenth Grade Mathematics: Areas of Sectors and Segments of a Circle
When a circular pie is sliced from the center or sectioned by a straight cut, the resulting pieces represent the two fundamental sub-regions of a circle: sectors and segments. A sector is bounded by two radii and an arc, while a segment is bounded by a chord and an arc. Calculating their areas requires understanding fractional proportionality relative to the complete 360-degree rotation of a circle and synthesizing circular geometry with triangle trigonometry.
The Area and Arc Length of a Sector
A sector of a circle is the portion of the circular plane enclosed by two radii and the intercepted arc between them.
Geometry of a Sector:
/ / / | O | Central Angle = theta (in degrees)
| / \ | Radius = r
\/ \/
A --- B
Because a full circle corresponds to a central angle of 360 degrees and encloses an area of pi * r^2, a sector with a central angle of theta degrees represents a fractional part of the circle equal to theta / 360:
Theorem (Sector Area and Arc Length):
1. Length of an Arc of a Sector (l):
theta
l = --------- * (2 * pi * r)
360°
2. Area of a Sector (A):
theta
A = --------- * (pi * r^2)
360°
Alternative Relation between Area and Arc Length:
Notice that:
A = (theta / 360) * pi * r^2 = (1/2) * r * [(theta / 360) * 2*pi*r] = (1/2) * l * r
Area = (1/2) * l * r
This remarkable formula shows that the area of a sector resembles the area of a triangle: (1/2) * Base * Height, where the curved arc length l acts as the base and the radius r acts as the height!
Classification of Sectors:
- Minor Sector: Sector with central angle theta < 180°.
- Major Sector: Sector with central angle theta > 180°.
Area(Major Sector) = Area(Circle) - Area(Minor Sector)
= pi * r^2 - (theta / 360) * pi * r^2
The Area of a Segment of a Circle
A segment of a circle is the region bounded by a chord and the intercepted arc.
Geometry of a Segment:
/ / O / / \ \ Sector OAB
| / \ |
|A-----+B Chord AB
\_____/ Segment (Shaded below chord)
Notice that the minor segment is obtained by subtracting the area of triangle OAB from the area of sector OAB:
Area(Minor Segment) = Area(Sector OAB) - Area(triangle OAB)
Similarly, the area of the major segment is the remainder of the circle:
Area(Major Segment) = Area(Circle) - Area(Minor Segment)
Trigonometric Area Formula for Triangle OAB
To evaluate Area(triangle OAB) where OA = OB = r and Angle AOB = theta:
Theorem: The area of an isosceles triangle with legs r and included angle theta is:
Area(triangle OAB) = (1/2) * r^2 * sin(theta)
Comprehensive Minor Segment Area Formula:
theta
Area(Segment) = [----- * pi * r^2] - [(1/2) * r^2 * sin(theta)]
360°
= r^2 * [ (pi * theta / 360°) - (1/2) * sin(theta) ]
Standard Values for (1/2)*r^2*sin(theta):
1. For theta = 60°: Area(triangle) = (1/2) * r^2 * sin(60°) = (sqrt(3)/4) * r^2
2. For theta = 90°: Area(triangle) = (1/2) * r^2 * sin(90°) = (1/2) * r^2
3. For theta = 120°: Area(triangle) = (1/2) * r^2 * sin(120°) = (sqrt(3)/4) * r^2
Detailed Worked Example: Segment with 120° Angle
A chord of a circle of radius 12 cm subtends an angle of 120 degrees at the center. Find the area of the corresponding minor segment of the circle. (Use pi = 3.14 and sqrt(3) = 1.732).
Step 1: Radius r = 12 cm, Central Angle theta = 120°.
Step 2: Area of Sector OAB:
Area(Sector) = (120 / 360) * pi * r^2 = (1 / 3) * 3.14 * 12^2
= (1 / 3) * 3.14 * 144 = 48 * 3.14 = 150.72 cm^2.
Step 3: Area of Triangle OAB:
Area(triangle) = (1/2) * r^2 * sin(120°)
Recall sin(120°) = sin(180° - 60°) = sin(60°) = sqrt(3) / 2.
Area(triangle) = (1/2) * 144 * (sqrt(3) / 2) = 36 * sqrt(3)
= 36 * 1.732 = 62.352 cm^2.
Step 4: Area of Minor Segment:
Area(Segment) = Area(Sector) - Area(triangle)
= 150.72 - 62.352 = 88.368 cm^2.
The area of the minor segment is 88.37 cm^2.
Chapter Exercises
Exercise 1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60 degrees.
Exercise 2: A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding: (a) minor segment, and (b) major sector. (Use pi = 3.14).
Exercise 3: In a circle of radius 21 cm, an arc subtends an angle of 60 degrees at the center. Find: (a) the length of the arc, (b) area of the sector formed by the arc, and (c) area of the segment formed by the corresponding chord.
Solutions and Explanations
Solution 1: Given radius r = 6 cm and central angle theta = 60°. The area of the sector is given by: Area = (theta / 360) * pi * r^2 = (60 / 360) * (22/7) * 6^2 = (1 / 6) * (22/7) * 36 = 6 * (22/7) = 132 / 7 = 18.86 cm^2.
Solution 2: Given r = 10 cm, theta = 90°. For part (a), Area of sector = (90 / 360) * pi * r^2 = (1/4) * 3.14 * 100 = 78.5 cm^2. Since theta = 90°, triangle OAB is a right-angled isosceles triangle: Area(triangle OAB) = (1/2) * Base * Height = (1/2) * 10 * 10 = 50 cm^2. The area of the minor segment is: Area(Segment) = Area(Sector) - Area(triangle) = 78.5 - 50 = 28.5 cm^2. For part (b), the major sector has central angle 360° - 90° = 270°: Area(Major Sector) = (270 / 360) * pi * r^2 = (3/4) * 3.14 * 100 = 3 * 78.5 = 235.5 cm^2.
Solution 3: Given r = 21 cm, theta = 60°. For part (a), length of arc l = (theta / 360) * (2 * pi * r) = (60 / 360) * 2 * (22/7) * 21 = (1/6) * 2 * 22 * 3 = (6 / 6) * 22 = 22 cm. For part (b), area of sector = (theta / 360) * pi * r^2 = (60 / 360) * (22/7) * 21^2 = (1/6) * (22/7) * 441 = (1/6) * 22 * 63 = 231 cm^2. For part (c), area of triangle OAB with theta = 60°: since OA = OB = 21 cm and Angle AOB = 60°, triangle OAB is an equilateral triangle: Area(triangle OAB) = (sqrt(3)/4) * side^2 = (sqrt(3)/4) * 21^2 = (441 * sqrt(3)) / 4 cm^2. The area of the segment is Area(Sector) - Area(triangle) = 231 - (441*sqrt(3))/4 cm^2 (approximately 231 - 190.95 = 40.05 cm^2).