A segment of a circle is the region (area) bounded by a chord of the circle and the arc corresponding to that chord. It can also be obtained by subtracting the area of the triangle formed by the chord from the area of the corresponding sector.
Types of Segments
There are two types of segments that are:
Major Segment | When a chord divides a circle into two parts, the larger part is called the major segment. |
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Minor Segment | The segment which is smaller in a circle is called the minor segment. |
Special Case: Semicircle
Other than these two types, there is one more special case of segments when both the segments become an equation i.e., semicircle. Semicircle is the largest segment in a circle as the diameter is the largest chord of the circle.
Area of Segment of a Circle
The area of the segment is the difference between the area of the sector and the area of the triangle. In the above figure area of the segment is determined by subtracting the area of the triangle from the area of the sector.
Area of the segment = Area of the sector - Area of the triangle
Formula for Area of Segment of a Circle
- Area of segment (when θ in radians) = (1/2) × r2(θ - sinθ)
- Area of segment (when θ in degrees) = (1/2) × r2 [(π/180) θ - sinθ]
Perimeter of Segment of a Circle
The perimeter of the segment is the sum of the length of the chord and the length of the arc.
Perimeter of Segment = Length of Chord + Length of the Arc
Formula for Perimeter of Segment of a Circle
- Perimeter of segment (when θ in radians) = rθ + 2rsin(θ/2)
- Perimeter of segment (when θ in degrees) = rθ(π/180) + 2rsin(θ/2)
Alternate Segment Theorem
The alternate segment theorem states that the angle formed by the point of contact of the chord and tangent is equal to the angle formed by the chord of the alternate segment. (angle a = angle b)
Angles in the Same Segment Theorem
Angles in the same segment theorem states that the angles formed by the same segment are equal. (angle a = angle b)
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Solved Examples
Problem 1: Find the area of the segment given that the area of the sector and the area of the triangle is 10 cm2 and 6 cm2.
Solution :
Area of the segment = Area of the sector - Area of the triangle
= 10 - 6
Area of the segment = 4 cm2
Problem 2: Find the perimeter of the segment if the length of the arc is 15 cm and the length of the chord is 10 cm.
Solution :
Perimeter of the segment = Length of the chord + Length of the arc
⇒ Perimeter of the segment = 15 + 10
⇒ Perimeter of the segment = 25 cm
Problem 3: Find the area of the segment if the radius of the circle is 15 cm and subtended angle is 30°.
Solution :
Area of segment (when θ in degrees) = (1/2) × r2 [(π/180) θ - sinθ]
⇒ Area of segment = (1/2) × 152 [(π/180) 30° - sin30°]
⇒ (1/2) × 225 (
⇒ Area of segment = 2.65 cm2
Problem 4: Find the area of the circle if the area of the major and minor segments is 10 cm2 and 2 cm2.
Solution:
Area of the circle = Area of the major segment + Area of the minor segment
⇒ Area of the circle = 10 + 2
⇒ Area of the circle = 12 cm2
Problem 5: Find the perimeter of segment given the radius of the circle 25 cm and angle subtended by segment is 60°.
Solution:
Perimeter of segment (when θ in degrees) = rθ(π/180) - 2rsin(θ/2)
⇒ Perimeter of segment (when θ in degrees) = 25× 60(π/180) - 2× 25 × sin(60/2)
⇒ Perimeter of segment (when θ in degrees) = 25× (π/3) - 50 × sin(30)
⇒ Perimeter of segment (when θ in degrees) = 1.18 cm
Problem 6: Find the area of the segment if the height of the triangle is 10 cm and the angle subtended is 60°.
Solution:
We have to find radius of the circle to find the area of segment.
In triangle AOB, OP is the height (perpendicular) which bisects angle AOB.
∠ AOP = ∠ BOP = 30°
So, in right-angled triangle BOP
sin 30° = OP / OB
⇒ OB = 10 / sin 30°
⇒ OB (radius of circle) = 20 cm
Thus, Area of segment (when θ in degrees) = (1/2) × r2 [(π/180) θ - sinθ]
⇒ Area of segment (when θ in degrees) = (1/2) × 202 [(π/180) 60 - sin60]
⇒ Area of segment (when θ in degrees) = 109.44 cm2